1-1
Chapter 1 INTRODUCTION AND BASIC CONCEPTS Thermodynamics 1-1C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the average behavior of large groups of particles. 1-2C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle. 1-3C There is no truth to his claim. It violates the second law of thermodynamics.
Mass, Force, and Units 1-4C Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit. One pound-force is the force required to accelerate a mass of 32.174 lbm by 1 ft/s2. In other words, the weight of a 1-lbm mass at sea level is 1 lbf. 1-5C Kg-mass is the mass unit in the SI system whereas kg-force is a force unit. 1-kg-force is the force required to accelerate a 1-kg mass by 9.807 m/s2. In other words, the weight of 1-kg mass at sea level is 1 kg-force. 1-6C There is no acceleration, thus the net force is zero in both cases.
1-7 A plastic tank is filled with water. The weight of the combined system is to be determined. Assumptions The density of water is constant throughout. Properties The density of water is given to be ρ = 1000 kg/m3. Analysis The mass of the water in the tank and the total mass are
mtank = 3 kg 3
V =0.2 m
mw =ρV =(1000 kg/m )(0.2 m ) = 200 kg 3
3
mtotal = mw + mtank = 200 + 3 = 203 kg Thus, 1N = 1991 N W = mg = (203 kg)(9.81 m/s 2 ) 2 1 kg ⋅ m/s
H2O
1-2
1-8 The interior dimensions of a room are given. The mass and weight of the air in the room are to be determined. Assumptions The density of air is constant throughout the room. Properties The density of air is given to be ρ = 1.16 kg/m3. Analysis The mass of the air in the room is 3
ROOM AIR
3
m = ρV = (1.16 kg/m )(6 × 6 × 8 m ) = 334.1 kg
6X6X8 m3
Thus, 1N W = mg = (334.1 kg)(9.81 m/s 2 ) 1 kg ⋅ m/s 2
= 3277 N
1-9 The variation of gravitational acceleration above the sea level is given as a function of altitude. The height at which the weight of a body will decrease by 1% is to be determined. z Analysis The weight of a body at the elevation z can be expressed as W = mg = m(9.807 − 3.32 × 10−6 z )
In our case, W = 0.99Ws = 0.99mgs = 0.99(m)(9.807)
Substituting, 0.99(9.81) = (9.81 − 3.32 × 10 −6 z) → z = 29,539 m
0 Sea level
1-10E An astronaut took his scales with him to space. It is to be determined how much he will weigh on the spring and beam scales in space. Analysis (a) A spring scale measures weight, which is the local gravitational force applied on a body: 1 lbf W = mg = (150 lbm)(5.48 ft/s 2 ) 32.2 lbm ⋅ ft/s 2
= 25.5 lbf
(b) A beam scale compares masses and thus is not affected by the variations in gravitational acceleration. The beam scale will read what it reads on earth, W = 150 lbf
1-11 The acceleration of an aircraft is given in g’s. The net upward force acting on a man in the aircraft is to be determined. Analysis From the Newton's second law, the force applied is 1N F = ma = m(6 g) = (90 kg)(6 × 9.81 m/s 2 ) 1 kg ⋅ m/s 2
= 5297 N
1-3
1-12 [Also solved by EES on enclosed CD] A rock is thrown upward with a specified force. The acceleration of the rock is to be determined. Analysis The weight of the rock is 1N W = mg = (5 kg)(9.79 m/s 2 ) 1 kg ⋅ m/s 2
= 48.95 N
Then the net force that acts on the rock is Fnet = Fup − Fdown = 150 − 48.95 = 101.05 N
From the Newton's second law, the acceleration of the rock becomes a=
F 101.05 N 1 kg ⋅ m/s = m 5 kg 1N
2
Stone
= 20.2 m/s 2
1-13 EES Problem 1-12 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below. W=m*g"[N]" m=5"[kg]" g=9.79"[m/s^2]" "The force balance on the rock yields the net force acting on the rock as" F_net = F_up - F_down"[N]" F_up=150"[N]" F_down=W"[N]" "The acceleration of the rock is determined from Newton's second law." F_net=a*m "To Run the program, press F2 or click on the calculator icon from the Calculate menu" SOLUTION a=20.21 [m/s^2] F_down=48.95 [N] F_net=101.1 [N] F_up=150 [N] g=9.79 [m/s^2] m=5 [kg] W=48.95 [N]
1-4
1-14 Gravitational acceleration g and thus the weight of bodies decreases with increasing elevation. The percent reduction in the weight of an airplane cruising at 13,000 m is to be determined. Properties The gravitational acceleration g is given to be 9.807 m/s2 at sea level and 9.767 m/s2 at an altitude of 13,000 m. Analysis Weight is proportional to the gravitational acceleration g, and thus the percent reduction in weight is equivalent to the percent reduction in the gravitational acceleration, which is determined from ∆g 9.807 − 9.767 × 100 = × 100 = 0.41% %Reduction in weight = %Reduction in g = 9.807 g Therefore, the airplane and the people in it will weight 0.41% less at 13,000 m altitude. Discussion Note that the weight loss at cruising altitudes is negligible.
Systems, Properties, State, and Processes 1-15C The radiator should be analyzed as an open system since mass is crossing the boundaries of the system. 1-16C A can of soft drink should be analyzed as a closed system since no mass is crossing the boundaries of the system. 1-17C Intensive properties do not depend on the size (extent) of the system but extensive properties do. 1-18C For a system to be in thermodynamic equilibrium, the temperature has to be the same throughout but the pressure does not. However, there should be no unbalanced pressure forces present. The increasing pressure with depth in a fluid, for example, should be balanced by increasing weight. 1-19C A process during which a system remains almost in equilibrium at all times is called a quasiequilibrium process. Many engineering processes can be approximated as being quasi-equilibrium. The work output of a device is maximum and the work input to a device is minimum when quasi-equilibrium processes are used instead of nonquasi-equilibrium processes. 1-20C A process during which the temperature remains constant is called isothermal; a process during which the pressure remains constant is called isobaric; and a process during which the volume remains constant is called isochoric. 1-21C The state of a simple compressible system is completely specified by two independent, intensive properties. 1-22C Yes, because temperature and pressure are two independent properties and the air in an isolated room is a simple compressible system. 1-23C A process is said to be steady-flow if it involves no changes with time anywhere within the system or at the system boundaries. 1-24C The specific gravity, or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C, for which ρH2O = 1000 kg/m3). That is, SG = ρ / ρ H2O . When specific gravity is known, density is determined from ρ = SG × ρ H2O .
1-5
1-25 EES The variation of density of atmospheric air with elevation is given in tabular form. A relation for the variation of density with elevation is to be obtained, the density at 7 km elevation is to be calculated, and the mass of the atmosphere using the correlation is to be estimated. Assumptions 1 Atmospheric air behaves as an ideal gas. 2 The earth is perfectly sphere with a radius of 6377 km, and the thickness of the atmosphere is 25 km. Properties The density data are given in tabular form as
ρ, kg/m3 1.225 1.112 1.007 0.9093 0.8194 0.7364 0.6601 0.5258 0.4135 0.1948 0.08891 0.04008
1.4 1.2 1 3
z, km 0 1 2 3 4 5 6 8 10 15 20 25
ρ, kg/m
r, km 6377 6378 6379 6380 6381 6382 6383 6385 6387 6392 6397 6402
0.8 0.6 0.4 0.2 0 0
5
10
15
20
z, km
25
Analysis Using EES, (1) Define a trivial function rho= a+z in equation window, (2) select new parametric table from Tables, and type the data in a two-column table, (3) select Plot and plot the data, and (4) select plot and click on “curve fit” to get curve fit window. Then specify 2nd order polynomial and enter/edit equation. The results are: ρ(z) = a + bz + cz2 = 1.20252 – 0.101674z + 0.0022375z2
for the unit of kg/m3,
(or, ρ(z) = (1.20252 – 0.101674z + 0.0022375z2)×109 for the unit of kg/km3) where z is the vertical distance from the earth surface at sea level. At z = 7 km, the equation would give ρ = 0.60 kg/m3. (b) The mass of atmosphere can be evaluated by integration to be m=
∫
V
ρdV =
∫
h
z =0
(a + bz + cz 2 )4π (r0 + z ) 2 dz = 4π
[
∫
h
z =0
(a + bz + cz 2 )(r02 + 2r0 z + z 2 )dz
= 4π ar02 h + r0 (2a + br0 )h 2 / 2 + (a + 2br0 + cr02 )h 3 / 3 + (b + 2cr0 )h 4 / 4 + ch 5 / 5
]
where r0 = 6377 km is the radius of the earth, h = 25 km is the thickness of the atmosphere, and a = 1.20252, b = -0.101674, and c = 0.0022375 are the constants in the density function. Substituting and multiplying by the factor 109 for the density unity kg/km3, the mass of the atmosphere is determined to be m = 5.092×1018 kg Discussion Performing the analysis with excel would yield exactly the same results. EES Solution for final result: a=1.2025166 b=-0.10167 c=0.0022375 r=6377 h=25 m=4*pi*(a*r^2*h+r*(2*a+b*r)*h^2/2+(a+2*b*r+c*r^2)*h^3/3+(b+2*c*r)*h^4/4+c*h^5/5)*1E+9
1-6
Temperature 1-26C The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the same temperature reading, even if they are not in contact. 1-27C They are celsius(°C) and kelvin (K) in the SI, and fahrenheit (°F) and rankine (R) in the English system. 1-28C Probably, but not necessarily. The operation of these two thermometers is based on the thermal expansion of a fluid. If the thermal expansion coefficients of both fluids vary linearly with temperature, then both fluids will expand at the same rate with temperature, and both thermometers will always give identical readings. Otherwise, the two readings may deviate.
1-29 A temperature is given in °C. It is to be expressed in K. Analysis The Kelvin scale is related to Celsius scale by T(K] = T(°C) + 273 Thus, T(K] = 37°C + 273 = 310 K
1-30E A temperature is given in °C. It is to be expressed in °F, K, and R. Analysis Using the conversion relations between the various temperature scales, T(K] = T(°C) + 273 = 18°C + 273 = 291 K T(°F] = 1.8T(°C) + 32 = (1.8)(18) + 32 = 64.4°F T(R] = T(°F) + 460 = 64.4 + 460 = 524.4 R
1-31 A temperature change is given in °C. It is to be expressed in K. Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales. Thus, ∆T(K] = ∆T(°C) = 15 K
1-32E A temperature change is given in °F. It is to be expressed in °C, K, and R. Analysis This problem deals with temperature changes, which are identical in Rankine and Fahrenheit scales. Thus, ∆T(R) = ∆T(°F) = 45 R The temperature changes in Celsius and Kelvin scales are also identical, and are related to the changes in Fahrenheit and Rankine scales by ∆T(K) = ∆T(R)/1.8 = 45/1.8 = 25 K and ∆T(°C) = ∆T(K) = 25°C
1-33 Two systems having different temperatures and energy contents are brought in contact. The direction of heat transfer is to be determined. Analysis Heat transfer occurs from warmer to cooler objects. Therefore, heat will be transferred from system B to system A until both systems reach the same temperature.
1-7
Pressure, Manometer, and Barometer 1-34C The pressure relative to the atmospheric pressure is called the gage pressure, and the pressure relative to an absolute vacuum is called absolute pressure. 1-35C The atmospheric pressure, which is the external pressure exerted on the skin, decreases with increasing elevation. Therefore, the pressure is lower at higher elevations. As a result, the difference between the blood pressure in the veins and the air pressure outside increases. This pressure imbalance may cause some thin-walled veins such as the ones in the nose to burst, causing bleeding. The shortness of breath is caused by the lower air density at higher elevations, and thus lower amount of oxygen per unit volume. 1-36C No, the absolute pressure in a liquid of constant density does not double when the depth is doubled. It is the gage pressure that doubles when the depth is doubled. 1-37C If the lengths of the sides of the tiny cube suspended in water by a string are very small, the magnitudes of the pressures on all sides of the cube will be the same. 1-38C Pascal’s principle states that the pressure applied to a confined fluid increases the pressure throughout by the same amount. This is a consequence of the pressure in a fluid remaining constant in the horizontal direction. An example of Pascal’s principle is the operation of the hydraulic car jack. 1-39C The density of air at sea level is higher than the density of air on top of a high mountain. Therefore, the volume flow rates of the two fans running at identical speeds will be the same, but the mass flow rate of the fan at sea level will be higher.
1-40 The pressure in a vacuum chamber is measured by a vacuum gage. The absolute pressure in the chamber is to be determined. Analysis The absolute pressure in the chamber is determined from Pabs = Patm − Pvac = 92 − 35 = 57 kPa
Pabs
Patm = 92 kPa
35 kPa
1-8
1-41E The pressure in a tank is measured with a manometer by measuring the differential height of the manometer fluid. The absolute pressure in the tank is to be determined for the cases of the manometer arm with the higher and lower fluid level being attached to the tank . Assumptions incompressible.
The
fluid
in
the
manometer
is
Properties The specific gravity of the fluid is given to be SG = 1.25. The density of water at 32°F is 62.4 lbm/ft3 (Table A-3E)
Air 28 in
Analysis The density of the fluid is obtained by multiplying its specific gravity by the density of water,
ρ = SG × ρ H 2 O = (1.25)(62.4 lbm/ft 3 ) = 78.0 lbm/ft 3
SG = 1.25
Patm = 12.7 psia
The pressure difference corresponding to a differential height of 28 in between the two arms of the manometer is 1ft 2 1 lbf = 1.26 psia ∆P = ρgh = (78 lbm/ft3 )(32.174 ft/s 2 )(28/12 ft) 2 2 32.174 lbm ⋅ ft/s 144 in
Then the absolute pressures in the tank for the two cases become: (a) The fluid level in the arm attached to the tank is higher (vacuum): Pabs = Patm − Pvac = 12.7 − 1.26 = 11.44 psia
(b) The fluid level in the arm attached to the tank is lower: Pabs = Pgage + Patm = 12.7 + 1.26 = 13.96 psia
Discussion Note that we can determine whether the pressure in a tank is above or below atmospheric pressure by simply observing the side of the manometer arm with the higher fluid level.
1-9
1-42 The pressure in a pressurized water tank is measured by a multi-fluid manometer. The gage pressure of air in the tank is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus we can determine the pressure at the air-water interface. Properties The densities of mercury, water, and oil are given to be 13,600, 1000, and 850 kg/m3, respectively. Analysis Starting with the pressure at point 1 at the air-water interface, and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach point 2, and setting the result equal to Patm since the tube is open to the atmosphere gives P1 + ρ water gh1 + ρ oil gh2 − ρ mercury gh3 = Patm
Air 1
Solving for P1, P1 = Patm − ρ water gh1 − ρ oil gh2 + ρ mercury gh3
h1
or, h3
P1 − Patm = g ( ρ mercury h3 − ρ water h1 − ρ oil h2 )
Water
Noting that P1,gage = P1 - Patm and substituting,
h2
P1,gage = (9.81 m/s 2 )[(13,600 kg/m 3 )(0.46 m) − (1000 kg/m 3 )(0.2 m) 1N − (850 kg/m 3 )(0.3 m)] 1 kg ⋅ m/s 2 = 56.9 kPa
1 kPa 1000 N/m 2
Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly.
1-43 The barometric reading at a location is given in height of mercury column. The atmospheric pressure is to be determined. Properties The density of mercury is given to be 13,600 kg/m3. Analysis The atmospheric pressure is determined directly from Patm = ρgh 1N = (13,600 kg/m 3 )(9.81 m/s 2 )(0.750 m) 1 kg ⋅ m/s 2 = 100.1 kPa
1 kPa 1000 N/m 2
1-10
1-44 The gage pressure in a liquid at a certain depth is given. The gage pressure in the same liquid at a different depth is to be determined. Assumptions The variation of the density of the liquid with depth is negligible. Analysis The gage pressure at two different depths of a liquid can be expressed as P1 = ρgh1
and
P2 = ρgh2
Taking their ratio, h1
P2 ρgh2 h2 = = P1 ρgh1 h1
1
h2
Solving for P2 and substituting gives P2 =
2
h2 9m P1 = (28 kPa) = 84 kPa h1 3m
Discussion Note that the gage pressure in a given fluid is proportional to depth.
1-45 The absolute pressure in water at a specified depth is given. The local atmospheric pressure and the absolute pressure at the same depth in a different liquid are to be determined. Assumptions The liquid and water are incompressible. Properties The specific gravity of the fluid is given to be SG = 0.85. We take the density of water to be 1000 kg/m3. Then density of the liquid is obtained by multiplying its specific gravity by the density of water,
ρ = SG × ρ H 2O = (0.85)(1000 kg/m 3 ) = 850 kg/m 3 Analysis (a) Knowing the absolute pressure, the atmospheric pressure can be determined from Patm = P − ρgh
1 kPa = (145 kPa) − (1000 kg/m 3 )(9.81 m/s 2 )(5 m) 1000 N/m 2 = 96.0 kPa
Patm h P
(b) The absolute pressure at a depth of 5 m in the other liquid is P = Patm + ρgh
1 kPa = (96.0 kPa) + (850 kg/m 3 )(9.81 m/s 2 )(5 m) 1000 N/m 2 = 137.7 kPa
Discussion Note that at a given depth, the pressure in the lighter fluid is lower, as expected.
1-11
1-46E It is to be shown that 1 kgf/cm2 = 14.223 psi . Analysis Noting that 1 kgf = 9.80665 N, 1 N = 0.22481 lbf, and 1 in = 2.54 cm, we have 0.22481 lbf 1 kgf = 9.80665 N = (9.80665 N ) 1N
= 2.20463 lbf
and 2
2.54 cm = 14.223 lbf/in 2 = 14.223 psi 1 kgf/cm 2 = 2.20463 lbf/cm 2 = (2.20463 lbf/cm 2 ) 1 in
1-47E The weight and the foot imprint area of a person are given. The pressures this man exerts on the ground when he stands on one and on both feet are to be determined. Assumptions The weight of the person is distributed uniformly on foot imprint area. Analysis The weight of the man is given to be 200 lbf. Noting that pressure is force per unit area, the pressure this man exerts on the ground is (a) On both feet:
P=
200 lbf W = = 2.78 lbf/in 2 = 2.78 psi 2 A 2 × 36 in 2
(b) On one foot:
P=
W 200 lbf = = 5.56 lbf/in 2 = 5.56 psi A 36 in 2
Discussion Note that the pressure exerted on the ground (and on the feet) is reduced by half when the person stands on both feet.
1-48 The mass of a woman is given. The minimum imprint area per shoe needed to enable her to walk on the snow without sinking is to be determined. Assumptions 1 The weight of the person is distributed uniformly on the imprint area of the shoes. 2 One foot carries the entire weight of a person during walking, and the shoe is sized for walking conditions (rather than standing). 3 The weight of the shoes is negligible. Analysis The mass of the woman is given to be 70 kg. For a pressure of 0.5 kPa on the snow, the imprint area of one shoe must be W mg (70 kg)(9.81 m/s 2 ) 1N = = 2 P P 0.5 kPa 1 kg ⋅ m/s
1 kPa = 1.37 m 2 1000 N/m 2 Discussion This is a very large area for a shoe, and such shoes would be impractical to use. Therefore, some sinking of the snow should be allowed to have shoes of reasonable size. A=
1-12
1-49 The vacuum pressure reading of a tank is given. The absolute pressure in the tank is to be determined. Properties The density of mercury is given to be ρ = 13,590 kg/m3. Analysis The atmospheric (or barometric) pressure can be expressed as Patm = ρ g h
1N = (13,590 kg/m 3 )(9.807 m/s 2 )(0.750 m) 1 kg ⋅ m/s 2 = 100.0 kPa
1 kPa 1000 N/m 2
Pabs
15 kPa
Patm = 750 mmHg
Then the absolute pressure in the tank becomes Pabs = Patm − Pvac = 100.0 − 15 = 85.0 kPa
1-50E A pressure gage connected to a tank reads 50 psi. The absolute pressure in the tank is to be determined. Properties The density of mercury is given to be ρ = 848.4 lbm/ft3. Analysis The atmospheric (or barometric) pressure can be expressed as Patm = ρ g h
1 lbf = (848.4 lbm/ft 3 )(32.2 ft/s 2 )(29.1/12 ft) 32.2 lbm ⋅ ft/s 2 = 14.29 psia
1 ft 2 144 in 2
Pabs
50 psi
Then the absolute pressure in the tank is Pabs = Pgage + Patm = 50 + 14.29 = 64.3 psia
1-51 A pressure gage connected to a tank reads 500 kPa. The absolute pressure in the tank is to be determined. Analysis The absolute pressure in the tank is determined from
Pabs
Pabs = Pgage + Patm = 500 + 94 = 594 kPa Patm = 94 kPa
500 kPa
1-13
1-52 A mountain hiker records the barometric reading before and after a hiking trip. The vertical distance climbed is to be determined. 780 mbar
Assumptions The variation of air density and the gravitational acceleration with altitude is negligible. Properties The density of air is given to be ρ = 1.20 kg/m3.
h=?
Analysis Taking an air column between the top and the bottom of the mountain and writing a force balance per unit base area, we obtain Wair / A = Pbottom − Ptop
930 mbar
( ρgh) air = Pbottom − Ptop 1N (1.20 kg/m 3 )(9.81 m/s 2 )(h) 1 kg ⋅ m/s 2
It yields
1 bar 100,000 N/m 2
= (0.930 − 0.780) bar
h = 1274 m
which is also the distance climbed.
1-53 A barometer is used to measure the height of a building by recording reading at the bottom and at the top of the building. The height of the building is to be determined. Assumptions The variation of air density with altitude is negligible. Properties The density of air is given to be ρ = 1.18 kg/m3. The density of mercury is 13,600 kg/m3.
730 mmHg
Analysis Atmospheric pressures at the top and at the bottom of the building are h
Ptop = ( ρ g h) top
1N = (13,600 kg/m 3 )(9.807 m/s 2 )(0.730 m) 1 kg ⋅ m/s 2 = 97.36 kPa
1 kPa 1000 N/m 2
755 mmHg
Pbottom = ( ρ g h) bottom
1N = (13,600 kg/m 3 )(9.807 m/s 2 )(0.755 m) 1kg ⋅ m/s 2 = 100.70 kPa
1 kPa 1000 N/m 2
Taking an air column between the top and the bottom of the building and writing a force balance per unit base area, we obtain Wair / A = Pbottom − Ptop ( ρgh) air = Pbottom − Ptop 1N (1.18 kg/m 3 )(9.807 m/s 2 )(h) 1 kg ⋅ m/s 2
It yields
h = 288.6 m
which is also the height of the building.
1 kPa 1000 N/m 2
= (100.70 − 97.36) kPa
1-14
1-54 EES Problem 1-53 is reconsidered. The entire EES solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below. P_bottom=755"[mmHg]" P_top=730"[mmHg]" g=9.807 "[m/s^2]" "local acceleration of gravity at sea level" rho=1.18"[kg/m^3]" DELTAP_abs=(P_bottom-P_top)*CONVERT('mmHg','kPa')"[kPa]" "Delta P reading from the barometers, converted from mmHg to kPa." DELTAP_h =rho*g*h/1000 "[kPa]" "Equ. 1-16. Delta P due to the air fluid column height, h, between the top and bottom of the building." "Instead of dividing by 1000 Pa/kPa we could have multiplied rho*g*h by the EES function, CONVERT('Pa','kPa')" DELTAP_abs=DELTAP_h SOLUTION Variables in Main DELTAP_abs=3.333 [kPa] DELTAP_h=3.333 [kPa] g=9.807 [m/s^2] h=288 [m] P_bottom=755 [mmHg] P_top=730 [mmHg] rho=1.18 [kg/m^3]
1-55 A diver is moving at a specified depth from the water surface. The pressure exerted on the surface of the diver by water is to be determined. Assumptions The variation of the density of water with depth is negligible. Properties The specific gravity of seawater is given to be SG = 1.03. We take the density of water to be 1000 kg/m3. Patm Analysis The density of the seawater is obtained by multiplying its specific gravity by the density of water which is taken to be 1000 Sea kg/m3: h ρ = SG × ρ H 2 O = (1.03)(1000 kg/m 3 ) = 1030 kg/m3 P The pressure exerted on a diver at 30 m below the free surface of the sea is the absolute pressure at that location: P = Patm + ρgh
1 kPa = (101 kPa) + (1030 kg/m 3 )(9.807 m/s 2 )(30 m) 1000 N/m 2 = 404.0 kPa
1-15
1-56E A submarine is cruising at a specified depth from the water surface. The pressure exerted on the surface of the submarine by water is to be determined. Assumptions The variation of the density of water with depth is negligible. Properties The specific gravity of seawater is given to be SG = 1.03. The density of water at 32°F is 62.4 lbm/ft3 (Table A-3E). Analysis The density of the seawater is obtained by multiplying its specific gravity by the density of water, 3
ρ = SG × ρ H 2O = (1.03)(62.4 lbm/ft ) = 64.27 lbm/ft
Patm Sea h P
3
The pressure exerted on the surface of the submarine cruising 300 ft below the free surface of the sea is the absolute pressure at that location: P = Patm + ρgh
1 lbf = (14.7 psia) + (64.27 lbm/ft 3 )(32.2 ft/s 2 )(175 ft) 32.2 lbm ⋅ ft/s 2 = 92.8 psia
1 ft 2 144 in 2
1-57 A gas contained in a vertical piston-cylinder device is pressurized by a spring and by the weight of the piston. The pressure of the gas is to be determined. Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield PA = Patm A + W + Fspring
Fspring
Thus, P = Patm +
mg + Fspring
A (4 kg)(9.81 m/s 2 ) + 60 N 1 kPa = (95 kPa) + 1000 N/m 2 35 × 10 − 4 m 2 = 123.4 kPa
Patm
P W = mg
1-16
1-58 EES Problem 1-57 is reconsidered. The effect of the spring force in the range of 0 to 500 N on the pressure inside the cylinder is to be investigated. The pressure against the spring force is to be plotted, and results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. g=9.807"[m/s^2]" P_atm= 95"[kPa]" m_piston=4"[kg]" {F_spring=60"[N]"} A=35*CONVERT('cm^2','m^2')"[m^2]" W_piston=m_piston*g"[N]" F_atm=P_atm*A*CONVERT('kPa','N/m^2')"[N]" "From the free body diagram of the piston, the balancing vertical forces yield:" F_gas= F_atm+F_spring+W_piston"[N]" P_gas=F_gas/A*CONVERT('N/m^2','kPa')"[kPa]" Fspring [N] 0 55.56 111.1 166.7 222.2 277.8 333.3 388.9 444.4 500
Pgas [kPa] 106.2 122.1 138 153.8 169.7 185.6 201.4 217.3 233.2 249.1
260 240
P gas [kPa]
220 200 180 160 140 120 100 0
100
200
300
F
[N]
spring
400
500
1-17
1-59 [Also solved by EES on enclosed CD] Both a gage and a manometer are attached to a gas to measure its pressure. For a specified reading of gage pressure, the difference between the fluid levels of the two arms of the manometer is to be determined for mercury and water. Properties The densities of water and mercury are given to be ρwater = 1000 kg/m3 and be ρHg = 13,600 kg/m3.
80 kPa
Analysis The gage pressure is related to the vertical distance h between the two fluid levels by Pgage = ρ g h → h =
AIR
h
Pgage
ρg
(a) For mercury, h=
Pgage
ρ Hg g
=
1 kN/m 2 (13,600 kg/m 3 )(9.81 m/s 2 ) 1 kPa
1000 kg/m ⋅ s 2 1 kN
= 0.60 m
1 kN/m 2 (1000 kg/m 3 )(9.81 m/s 2 ) 1 kPa
1000 kg/m ⋅ s 2 1 kN
= 8.16 m
80 kPa
(b) For water, h=
Pgage
ρ H 2O g
=
80 kPa
1-18
1-60 EES Problem 1-59 is reconsidered. The effect of the manometer fluid density in the range of 800 to 13,000 kg/m3 on the differential fluid height of the manometer is to be investigated. Differential fluid height against the density is to be plotted, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. Function fluid_density(Fluid$) If fluid$='Mercury' then fluid_density=13600 else fluid_density=1000 end {Input from the diagram window. If the diagram window is hidden, then all of the input must come from the equations window. Also note that brackets can also denote comments - but these comments do not appear in the formatted equations window.} {Fluid$='Mercury' P_atm = 101.325 "kpa" DELTAP=80 "kPa Note how DELTAP is displayed on the Formatted Equations Window."} g=9.807 "m/s2, local acceleration of gravity at sea level" rho=Fluid_density(Fluid$) "Get the fluid density, either Hg or H2O, from the function" "To plot fluid height against density place {} around the above equation. Then set up the parametric table and solve." DELTAP = RHO*g*h/1000 "Instead of dividing by 1000 Pa/kPa we could have multiplied by the EES function, CONVERT('Pa','kPa')" h_mm=h*convert('m','mm') "The fluid height in mm is found using the built-in CONVERT function." P_abs= P_atm + DELTAP
hmm [mm] 10197 3784 2323 1676 1311 1076 913.1 792.8 700.5 627.5
ρ [kg/m3] 800 2156 3511 4867 6222 7578 8933 10289 11644 13000
Manometer Fluid Height vs Manometer Fluid Density 11000 8800
] m m [
6600 4400
m m
h
2200 0 0
2000
4000
6000
8000
ρ [kg/m^3]
10000 12000 14000
1-19
1-61 The air pressure in a tank is measured by an oil manometer. For a given oil-level difference between the two columns, the absolute pressure in the tank is to be determined. Properties The density of oil is given to be ρ = 850 kg/m3. Analysis The absolute pressure in the tank is determined from P = Patm + ρgh
1kPa = (98 kPa) + (850 kg/m 3 )(9.81m/s 2 )(0.60 m) 1000 N/m 2 = 103 kPa
0.60 m
AIR
Patm = 98 kPa
1-62 The air pressure in a duct is measured by a mercury manometer. For a given mercury-level difference between the two columns, the absolute pressure in the duct is to be determined. Properties The density of mercury is given to be ρ = 13,600 kg/m3. Analysis (a) The pressure in the duct is above atmospheric pressure since the fluid column on the duct side is at a lower level. (b) The absolute pressure in the duct is determined from P = Patm + ρgh
1N = (100 kPa) + (13,600 kg/m 3 )(9.81 m/s 2 )(0.015 m) 1 kg ⋅ m/s 2 = 102 kPa
AIR
1 kPa 1000 N/m 2
15 mm
P
1-63 The air pressure in a duct is measured by a mercury manometer. For a given mercury-level difference between the two columns, the absolute pressure in the duct is to be determined. Properties The density of mercury is given to be ρ = 13,600 kg/m3. Analysis (a) The pressure in the duct is above atmospheric pressure since the fluid column on the duct side is at a lower level.
AIR
(b) The absolute pressure in the duct is determined from P = Patm + ρgh
1N = (100 kPa) + (13,600 kg/m 3 )(9.81 m/s 2 )(0.045 m) 1 kg ⋅ m/s 2 = 106 kPa
P
1 kPa 1000 N/m 2
45 mm
1-20
1-64 The systolic and diastolic pressures of a healthy person are given in mmHg. These pressures are to be expressed in kPa, psi, and meter water column. Assumptions Both mercury and water are incompressible substances. Properties We take the densities of water and mercury to be 1000 kg/m3 and 13,600 kg/m3, respectively. Analysis Using the relation P = ρgh for gage pressure, the high and low pressures are expressed as 1 kPa 1N = 16.0 kPa Phigh = ρghhigh = (13,600 kg/m 3 )(9.81 m/s 2 )(0.12 m) 2 2 1 kg ⋅ m/s 1000 N/m 1 kPa 1N = 10.7 kPa Plow = ρghlow = (13,600 kg/m 3 )(9.81 m/s 2 )(0.08 m) 2 2 ⋅ 1 kg m/s 1000 N/m
Noting that 1 psi = 6.895 kPa, 1 psi = 2.32 psi Phigh = (16.0 Pa) 6.895 kPa
and
1 psi = 1.55 psi Plow = (10.7 Pa) 6.895 kPa
For a given pressure, the relation P = ρgh can be expressed for mercury and water as P = ρ water gh water and P = ρ mercury ghmercury . Setting these two relations equal to each other and solving for water height gives
P = ρ water ghwater = ρ mercury ghmercury
→
hwater =
ρ mercury ρ water
hmercury
h
Therefore, hwater, high = hwater, low =
ρ mercury ρ water ρ mercury ρ water
hmercury, high = hmercury, low =
13,600 kg/m 3 1000 kg/m 3
13,600 kg/m 3 1000 kg/m 3
(0.12 m) = 1.63 m
(0.08 m) = 1.09 m
Discussion Note that measuring blood pressure with a “water” monometer would involve differential fluid heights higher than the person, and thus it is impractical. This problem shows why mercury is a suitable fluid for blood pressure measurement devices.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
1-21
1-65 A vertical tube open to the atmosphere is connected to the vein in the arm of a person. The height that the blood will rise in the tube is to be determined. Assumptions 1 The density of blood is constant. 2 The gage pressure of blood is 120 mmHg. Properties The density of blood is given to be ρ = 1050 kg/m3. Analysis For a given gage pressure, the relation P = ρgh can be expressed for mercury and blood as P = ρ blood ghblood and P = ρ mercury ghmercury .
Blood
h
Setting these two relations equal to each other we get P = ρ blood ghblood = ρ mercury ghmercury
Solving for blood height and substituting gives hblood =
ρ mercury ρ blood
hmercury =
13,600 kg/m 3 1050 kg/m 3
(0.12 m) = 1.55 m
Discussion Note that the blood can rise about one and a half meters in a tube connected to the vein. This explains why IV tubes must be placed high to force a fluid into the vein of a patient.
1-66 A man is standing in water vertically while being completely submerged. The difference between the pressures acting on the head and on the toes is to be determined. Assumptions Water is an incompressible substance, and thus the density does not change with depth.
hhead
Properties We take the density of water to be ρ =1000 kg/m3. Analysis The pressures at the head and toes of the person can be expressed as Phead = Patm + ρghhead
and
Ptoe = Patm + ρgh toe
where h is the vertical distance of the location in water from the free surface. The pressure difference between the toes and the head is determined by subtracting the first relation above from the second,
htoe
Ptoe − Phead = ρgh toe − ρghhead = ρg (h toe − hhead )
Substituting, 1N Ptoe − Phead = (1000 kg/m 3 )(9.81 m/s 2 )(1.80 m - 0) 1kg ⋅ m/s 2
1kPa 1000 N/m 2
= 17.7 kPa
Discussion This problem can also be solved by noting that the atmospheric pressure (1 atm = 101.325 kPa) is equivalent to 10.3-m of water height, and finding the pressure that corresponds to a water height of 1.8 m.
1-22
1-67 Water is poured into the U-tube from one arm and oil from the other arm. The water column height in one arm and the ratio of the heights of the two fluids in the other arm are given. The height of each fluid in that arm is to be determined. Assumptions Both water and oil are incompressible substances.
oil
Water
Properties The density of oil is given to be ρ = 790 kg/m . We take the density of water to be ρ =1000 kg/m3. 3
ha
Analysis The height of water column in the left arm of the monometer is given to be hw1 = 0.70 m. We let the height of water and oil in the right arm to be hw2 and ha, respectively. Then, ha = 4hw2. Noting that both arms are open to the atmosphere, the pressure at the bottom of the U-tube can be expressed as Pbottom = Patm + ρ w gh w1
hw1
hw2
Pbottom = Patm + ρ w gh w2 + ρ a gha
and
Setting them equal to each other and simplifying,
ρ w gh w1 = ρ w gh w2 + ρ a gha
→
ρ w h w1 = ρ w h w2 + ρ a ha
→
h w1 = h w2 + ( ρ a / ρ w )ha
Noting that ha = 4hw2, the water and oil column heights in the second arm are determined to be 0.7 m = h w2 + (790/1000) 4 hw 2 →
h w2 = 0.168 m
0.7 m = 0.168 m + (790/1000)ha →
ha = 0.673 m
Discussion Note that the fluid height in the arm that contains oil is higher. This is expected since oil is lighter than water.
1-68 The hydraulic lift in a car repair shop is to lift cars. The fluid gage pressure that must be maintained in the reservoir is to be determined. Assumptions The weight of the piston of the lift is negligible. Analysis Pressure is force per unit area, and thus the gage pressure required is simply the ratio of the weight of the car to the area of the lift, Pgage
W mg = = A πD 2 / 4 =
(2000 kg)(9.81 m/s 2 ) 1 kN = 278 kN/m 2 = 278 kPa 2 2 π (0.30 m) / 4 1000 kg ⋅ m/s
W = mg Patm
P
Discussion Note that the pressure level in the reservoir can be reduced by using a piston with a larger area.
1-23
1-69 Fresh and seawater flowing in parallel horizontal pipelines are connected to each other by a double Utube manometer. The pressure difference between the two pipelines is to be determined. Assumptions 1 All the liquids are incompressible. 2 The effect of air column on pressure is negligible.
Air
Properties The densities of seawater and mercury are given to be ρsea = 1035 kg/m3 and ρHg = 13,600 kg/m3. We take the density of water to be ρ w =1000 kg/m3. Analysis Starting with the pressure in the fresh water pipe (point 1) and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the sea water pipe (point 2), and setting the result equal to P2 gives
hsea
Fresh Water
Sea Water
hair hw hHg
P1 + ρ w gh w − ρ Hg ghHg − ρ air ghair + ρ sea ghsea = P2
Mercury
Rearranging and neglecting the effect of air column on pressure, P1 − P2 = − ρ w gh w + ρ Hg ghHg − ρ sea ghsea = g ( ρ Hg hHg − ρ w hw − ρ sea hsea )
Substituting, P1 − P2 = (9.81 m/s 2 )[(13600 kg/m 3 )(0.1 m) 1 kN − (1000 kg/m 3 )(0.6 m) − (1035 kg/m 3 )(0.4 m)] 1000 kg ⋅ m/s 2
= 3.39 kN/m 2 = 3.39 kPa
Therefore, the pressure in the fresh water pipe is 3.39 kPa higher than the pressure in the sea water pipe. Discussion A 0.70-m high air column with a density of 1.2 kg/m3 corresponds to a pressure difference of 0.008 kPa. Therefore, its effect on the pressure difference between the two pipes is negligible.
1-24
1-70 Fresh and seawater flowing in parallel horizontal pipelines are connected to each other by a double Utube manometer. The pressure difference between the two pipelines is to be determined. Assumptions All the liquids are incompressible. Properties The densities of seawater and mercury are given to be ρsea = 1035 kg/m3 and ρHg = 13,600 kg/m3. We take the density of water to be ρ w =1000 kg/m3. The specific gravity of oil is given to be 0.72, and thus its density is 720 kg/m3. Analysis Starting with the pressure in the fresh water pipe (point 1) and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the sea water pipe (point 2), and setting the result equal to P2 gives
Oil
hsea
Fresh Water
Sea Water
hoil hw hHg
P1 + ρ w gh w − ρ Hg ghHg − ρ oil ghoil + ρ sea ghsea = P2
Mercury
Rearranging, P1 − P2 = − ρ w gh w + ρ Hg ghHg + ρ oil ghoil − ρ sea ghsea = g ( ρ Hg hHg + ρ oil hoil − ρ w hw − ρ sea hsea )
Substituting, P1 − P2 = (9.81 m/s 2 )[(13600 kg/m 3 )(0.1 m) + (720 kg/m 3 )(0.7 m) − (1000 kg/m 3 )(0.6 m) 1 kN − (1035 kg/m 3 )(0.4 m)] 1000 kg ⋅ m/s 2
= 8.34 kN/m 2 = 8.34 kPa
Therefore, the pressure in the fresh water pipe is 8.34 kPa higher than the pressure in the sea water pipe.
1-25
1-71E The pressure in a natural gas pipeline is measured by a double U-tube manometer with one of the arms open to the atmosphere. The absolute pressure in the pipeline is to be determined. Assumptions 1 All the liquids are incompressible. 2 The effect of air column on pressure is negligible. 3 The pressure throughout the natural gas (including the tube) is uniform since its density is low.
Air
Properties We take the density of water to be ρw = 62.4 lbm/ft3. The specific gravity of mercury is given to be 13.6, and thus its density is ρHg = 13.6×62.4 = 848.6 lbm/ft3. Analysis Starting with the pressure at point 1 in the natural gas pipeline, and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives
10in
Water
hw hHg Natural gas
P1 − ρ Hg ghHg − ρ water ghwater = Patm
Solving for P1, P1 = Patm + ρ Hg ghHg + ρ water gh1
Mercury
Substituting, 1 lbf P = 14.2 psia + (32.2 ft/s 2 )[(848.6 lbm/ft 3 )(6/12 ft) + (62.4 lbm/ft 3 )(27/12 ft)] 32.2 lbm ⋅ ft/s 2 = 18.1psia
1 ft 2 144 in 2
Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly. Also, it can be shown that the 15-in high air column with a density of 0.075 lbm/ft3 corresponds to a pressure difference of 0.00065 psi. Therefore, its effect on the pressure difference between the two pipes is negligible.
1-26
1-72E The pressure in a natural gas pipeline is measured by a double U-tube manometer with one of the arms open to the atmosphere. The absolute pressure in the pipeline is to be determined. Assumptions 1 All the liquids are incompressible. 2 The pressure throughout the natural gas (including the tube) is uniform since its density is low.
Oil
Properties We take the density of water to be ρ w = 62.4 lbm/ft3. The specific gravity of mercury is given to be 13.6, and thus its density is ρHg = 13.6×62.4 = 848.6 lbm/ft3. The specific gravity of oil is given to be 0.69, and thus its density is ρoil = 0.69×62.4 = 43.1 lbm/ft3.
Water
hw
Analysis Starting with the pressure at point 1 in the natural gas pipeline, and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives
hHg Natural gas
hoil
P1 − ρ Hg ghHg + ρ oil ghoil − ρ water gh water = Patm
Solving for P1,
Mercury
P1 = Patm + ρ Hg ghHg + ρ water gh1 − ρ oil ghoil
Substituting, P1 = 14.2 psia + (32.2 ft/s 2 )[(848.6 lbm/ft 3 )(6/12 ft) + (62.4 lbm/ft 3 )(27/12 ft) 1 lbf − (43.1 lbm/ft 3 )(15/12 ft)] 32.2 lbm ⋅ ft/s 2 = 17.7 psia
1 ft 2 144 in 2
Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly.
1-27
1-73 The gage pressure of air in a pressurized water tank is measured simultaneously by both a pressure gage and a manometer. The differential height h of the mercury column is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air-water interface is the same as the indicated gage pressure. Properties We take the density of water to be ρw =1000 kg/m3. The specific gravities of oil and mercury are given to be 0.72 and 13.6, respectively. Analysis Starting with the pressure of air in the tank (point 1), and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives P1 + ρ w gh w − ρ Hg ghHg − ρ oil ghoil = Patm 80 kPa
Rearranging, P1 − Patm = ρ oil ghoil + ρ Hg ghHg − ρ w gh w
AIR
hoil
or, P1,gage
ρw g
= SG oil hoil + SG Hg hHg − h w
Water
hw
Substituting, 1000 kg ⋅ m/s 2 80 kPa (1000 kg/m 3 )(9.81 m/s 2 ) 1 kPa. ⋅ m 2
Solving for hHg gives 58.2 cm.
hHg
= 0.72 × (0.75 m) + 13.6 × hHg − 0.3 m
hHg = 0.582 m. Therefore, the differential height of the mercury column must be
Discussion Double instrumentation like this allows one to verify the measurement of one of the instruments by the measurement of another instrument.
1-28
1-74 The gage pressure of air in a pressurized water tank is measured simultaneously by both a pressure gage and a manometer. The differential height h of the mercury column is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air-water interface is the same as the indicated gage pressure. Properties We take the density of water to be ρ are given to be 0.72 and 13.6, respectively.
w
=1000 kg/m3. The specific gravities of oil and mercury
Analysis Starting with the pressure of air in the tank (point 1), and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives P1 + ρ w gh w − ρ Hg ghHg − ρ oil ghoil = Patm 40 kPa
Rearranging, P1 − Patm = ρ oil ghoil + ρ Hg ghHg − ρ w gh w
AIR
or,
hoil P1,gage
ρw g
= SG oil hoil + SG
Hg h Hg
Water
− hw
hw
Substituting, 1000 kg ⋅ m/s 2 40 kPa 3 2 2 (1000 kg/m )(9.81 m/s ) 1 kPa. ⋅ m
Solving for hHg gives 28.2 cm.
hHg
= 0.72 × (0.75 m) + 13.6 × hHg − 0.3 m
hHg = 0.282 m. Therefore, the differential height of the mercury column must be
Discussion Double instrumentation like this allows one to verify the measurement of one of the instruments by the measurement of another instrument.
1-75 The top part of a water tank is divided into two compartments, and a fluid with an unknown density is poured into one side. The levels of the water and the liquid are measured. The density of the fluid is to be determined. Assumptions 1 Both water and the added liquid are incompressible substances. 2 The added liquid does not mix with water. Properties We take the density of water to be ρ =1000 kg/m3. Analysis Both fluids are open to the atmosphere. Noting that the pressure of both water and the added fluid is the same at the contact surface, the pressure at this surface can be expressed as
Fluid Water hf
hw
Pcontact = Patm + ρ f ghf = Patm + ρ w gh w
Simplifying and solving for ρf gives
ρ f ghf = ρ w ghw →
ρf =
hw 45 cm ρw = (1000 kg/m 3 ) = 562.5 kg/m 3 hf 80 cm
Discussion Note that the added fluid is lighter than water as expected (a heavier fluid would sink in water).
1-29
1-76 A double-fluid manometer attached to an air pipe is considered. The specific gravity of one fluid is known, and the specific gravity of the other fluid is to be determined. Assumptions 1 Densities of liquids are constant. 2 The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air-water interface is the same as the indicated gage pressure.
Air P = 76 kPa
Properties The specific gravity of one fluid is given to be 13.55. We take the standard density of water to be 1000 kg/m3. Analysis Starting with the pressure of air in the tank, and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface where the oil tube is exposed to the atmosphere, and setting the result equal to Patm give Pair + ρ 1 gh1 − ρ 2 gh2 = Patm
→
40 cm
Fluid 1 SG1
22 cm
Fluid 2 SG2
Pair − Patm = SG 2 ρ w gh2 − SG1 ρ w gh1
Rearranging and solving for SG2, SG 2 = SG 1
1000 kg ⋅ m/s 2 h1 Pair − Patm 0.22 m 76 − 100 kPa + = 13.55 + ρ w gh2 0.40 m (1000 kg/m 3 )(9.81 m/s 2 )(0.40 m) 1 kPa. ⋅ m 2 h2
= 5.0
Discussion Note that the right fluid column is higher than the left, and this would imply above atmospheric pressure in the pipe for a single-fluid manometer.
1-30
1-77 The fluid levels in a multi-fluid U-tube manometer change as a result of a pressure drop in the trapped air space. For a given pressure drop and brine level change, the area ratio is to be determined. Assumptions 1 All the liquids are incompressible. 2 Pressure in the brine pipe remains constant. 3 The variation of pressure in the trapped air space is negligible. Properties The specific gravities are given to be 13.56 for mercury and 1.1 for brine. We take the standard density of water to be ρw =1000 kg/m3.
A Air
Water Area, A1
B Brine pipe SG=1.1
Analysis It is clear from the problem statement and the figure that the brine pressure is much higher than the air pressure, and when the air pressure drops by 0.7 kPa, the pressure difference between the brine and the air space increases also by the same amount.
Mercury SG=13.56
∆hb = 5 mm Area, A2
Starting with the air pressure (point A) and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the brine pipe (point B), and setting the result equal to PB before and after the pressure change of air give Before:
PA1 + ρ w gh w + ρ Hg ghHg, 1 − ρ br gh br,1 = PB
After:
PA2 + ρ w gh w + ρ Hg ghHg, 2 − ρ br gh br,2 = PB
Subtracting, PA2 − PA1 + ρ Hg g∆hHg − ρ br g∆h br = 0 →
PA1 − PA2 = SG Hg ∆hHg − SG br ∆h br = 0 ρwg
(1)
where ∆hHg and ∆h br are the changes in the differential mercury and brine column heights, respectively, due to the drop in air pressure. Both of these are positive quantities since as the mercury-brine interface drops, the differential fluid heights for both mercury and brine increase. Noting also that the volume of mercury is constant, we have A1 ∆hHg,left = A2 ∆hHg, right and PA2 − PA1 = −0.7 kPa = −700 N/m 2 = −700 kg/m ⋅ s 2 ∆h br = 0.005 m ∆hHg = ∆hHg,right + ∆hHg,left = ∆hbr + ∆hbr A2 /A 1 = ∆hbr (1 + A2 /A 1 )
Substituting, 700 kg/m ⋅ s 2 (1000 kg/m 3 )(9.81 m/s 2 )
It gives A2/A1 = 0.134
= [13.56 × 0.005(1 + A2 /A1 ) − (1.1× 0.005)] m
1-31
1-78 A multi-fluid container is connected to a U-tube. For the given specific gravities and fluid column heights, the gage pressure at A and the height of a mercury column that would create the same pressure at A are to be determined. Assumptions 1 All the liquids are incompressible. 2 The multi-fluid container is open to the atmosphere. Properties The specific gravities are given to be 1.26 for glycerin and 0.90 for oil. We take the standard density of water to be ρw =1000 kg/m3, and the specific gravity of mercury to be 13.6. Analysis Starting with the atmospheric pressure on the top surface of the container and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach point A, and setting the result equal to PA give
70 cm
Oil SG=0.90
30 cm
Water
20 cm
Glycerin SG=1.26
A
90 cm
Patm + ρ oil ghoil + ρ w gh w − ρ gly ghgly = PA
15 cm
Rearranging and using the definition of specific gravity, PA − Patm = SG oil ρ w ghoil + SG w ρ w gh w − SG gly ρ w ghgly
or PA,gage = gρ w (SG oil hoil + SG w hw − SG gly hgly )
Substituting, 1 kN PA,gage = (9.81 m/s 2 )(1000 kg/m 3 )[0.90(0.70 m) + 1(0.3 m) − 1.26(0.70 m)] 1000 kg ⋅ m/s 2
= 0.471 kN/m 2 = 0.471 kPa
The equivalent mercury column height is hHg =
PA,gage
ρ Hg g
=
1000 kg ⋅ m/s 2 1 kN (13.6)(1000 kg/m 3 )(9.81 m/s 2 ) 0.471 kN/m 2
= 0.00353 m = 0.353 cm
Discussion Note that the high density of mercury makes it a very suitable fluid for measuring high pressures in manometers.
1-32
Solving Engineering Problems and EES 1-79C Despite the convenience and capability the engineering software packages offer, they are still just tools, and they will not replace the traditional engineering courses. They will simply cause a shift in emphasis in the course material from mathematics to physics. They are of great value in engineering practice, however, as engineers today rely on software packages for solving large and complex problems in a short time, and perform optimization studies efficiently.
1-80 EES Determine a positive real root of the following equation using EES: 2x3 – 10x0.5 – 3x = -3 Solution by EES Software (Copy the following line and paste on a blank EES screen to verify solution): 2*x^3-10*x^0.5-3*x = -3 Answer: x = 2.063 (using an initial guess of x=2)
1-81 EES Solve the following system of 2 equations with 2 unknowns using EES: x3 – y2 = 7.75 3xy + y = 3.5 Solution by EES Software (Copy the following lines and paste on a blank EES screen to verify solution): x^3-y^2=7.75 3*x*y+y=3.5 Answer x=2 y=0.5
1-82 EES Solve the following system of 3 equations with 3 unknowns using EES: 2x – y + z = 5 3x2 + 2y = z + 2 xy + 2z = 8 Solution by EES Software (Copy the following lines and paste on a blank EES screen to verify solution): 2*x-y+z=5 3*x^2+2*y=z+2 x*y+2*z=8 Answer x=1.141, y=0.8159, z=3.535
1-83 EES Solve the following system of 3 equations with 3 unknowns using EES: x2y – z = 1 x – 3y0.5 + xz = - 2 x+y–z=2 Solution by EES Software (Copy the following lines and paste on a blank EES screen to verify solution): x^2*y-z=1 x-3*y^0.5+x*z=-2 x+y-z=2 Answer x=1, y=1, z=0
1-33
1-84E EES Specific heat of water is to be expressed at various units using unit conversion capability of EES. Analysis The problem is solved using EES, and the solution is given below. EQUATION WINDOW "GIVEN" C_p=4.18 [kJ/kg-C] "ANALYSIS" C_p_1=C_p*Convert(kJ/kg-C, kJ/kg-K) C_p_2=C_p*Convert(kJ/kg-C, Btu/lbm-F) C_p_3=C_p*Convert(kJ/kg-C, Btu/lbm-R) C_p_4=C_p*Convert(kJ/kg-C, kCal/kg-C)
FORMATTED EQUATIONS WINDOW GIVEN C p = 4.18
[kJ/kg-C]
ANALYSIS kJ/kg–K
C p,1 =
Cp ·
1 ·
C p,2 =
Cp ·
0.238846 ·
C p,3 =
Cp ·
0.238846 ·
C p,4 =
Cp ·
0.238846 ·
kJ/kg–C
SOLUTION WINDOW C_p=4.18 [kJ/kg-C] C_p_1=4.18 [kJ/kg-K] C_p_2=0.9984 [Btu/lbm-F] C_p_3=0.9984 [Btu/lbm-R] C_p_4=0.9984 [kCal/kg-C]
Btu/lbm–F kJ/kg–C Btu/lbm–R kJ/kg–C kCal/kg–C kJ/kg–C
1-34
Review Problems 1-85 A hydraulic lift is used to lift a weight. The diameter of the piston on which the weight to be placed is to be determined. Assumptions 1 The cylinders of the lift are vertical. 2 There are no leaks. 3 Atmospheric pressure act on both sides, and thus it can be disregarded. Analysis Noting that pressure is force per unit area, the pressure on the smaller piston is determined from F m1g P1 = 1 = A1 πD12 / 4 =
F1
Weight 2500 kg
25 kg
F2
10 cm
D2
(25 kg)(9.81 m/s 2 ) 1 kN 2 2 π (0.10 m) / 4 1000 kg ⋅ m/s
= 31.23 kN/m 2 = 31.23 kPa From Pascal’s principle, the pressure on the greater piston is equal to that in the smaller piston. Then, the needed diameter is determined from → D 2 = 1.0 m Discussion Note that large weights can be raised by little effort in hydraulic lift by making use of Pascal’s principle. P1 = P2 =
F2 m2 g (2500 kg)(9.81 m/s 2 ) 1 kN 2 = → = 31 . 23 kN/m 2 2 A2 πD 2 2 / 4 πD 2 / 4 1000 kg ⋅ m/s
1-86 A vertical piston-cylinder device contains a gas. Some weights are to be placed on the piston to increase the gas pressure. The local atmospheric pressure and the mass of the weights that will double the pressure of the gas are to be determined. Assumptions Friction between the piston and the cylinder is negligible. Analysis The gas pressure in the piston-cylinder device initially depends on the local atmospheric pressure and the weight of the piston. Balancing the vertical forces yield
WEIGTHS
GAS
(5 kg)(9.81 m/s 2 ) 1 kN = 95.66 kN/m 2 = 95.7 kPa 2 2 A π (0.12 m )/4 1000 kg ⋅ m/s The force balance when the weights are placed is used to determine the mass of the weights (m piston + m weights ) g P = Patm + A (5 kg + m weights )(9.81 m/s 2 ) 1 kN 200 kPa = 95.66 kPa + → m weights = 115.3 kg 2 2 π (0.12 m )/4 1000 kg ⋅ m/s A large mass is needed to double the pressure. Patm = P −
m piston g
= 100 kPa −
1-35
1-87 An airplane is flying over a city. The local atmospheric pressure in that city is to be determined. Assumptions The gravitational acceleration does not change with altitude. Properties The densities of air and mercury are given to be 1.15 kg/m3 and 13,600 kg/m3. Analysis The local atmospheric pressure is determined from Patm = Pplane + ρgh 1 kN = 58 kPa + (1.15 kg/m 3 )(9.81 m/s 2 )(3000 m) 1000 kg ⋅ m/s 2 The atmospheric pressure may be expressed in mmHg as hHg =
= 91.84 kN/m 2 = 91.8 kPa
Patm 91.8 kPa 1000 Pa 1000 mm = = 688 mmHg ρg (13,600 kg/m 3 )(9.81 m/s 2 ) 1 kPa 1 m
1-88 The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a body at different locations are to be determined. Analysis The weight of an 80-kg man at various locations is obtained by substituting the altitude z (values in m) into the relation
1N W = mg = (80kg)(9.807 − 3.32 × 10 −6 z m/s 2 ) 1kg ⋅ m/s 2 Sea level: Denver: Mt. Ev.:
(z = 0 m): W = 80×(9.807-3.32x10-6×0) = 80×9.807 = 784.6 N (z = 1610 m): W = 80×(9.807-3.32x10-6×1610) = 80×9.802 = 784.2 N (z = 8848 m): W = 80×(9.807-3.32x10-6×8848) = 80×9.778 = 782.2 N
1-89 A man is considering buying a 12-oz steak for $3.15, or a 320-g steak for $2.80. The steak that is a better buy is to be determined. Assumptions The steaks are of identical quality. Analysis To make a comparison possible, we need to express the cost of each steak on a common basis. Let us choose 1 kg as the basis for comparison. Using proper conversion factors, the unit cost of each steak is determined to be
12 ounce steak:
$3.15 16 oz 1 lbm = $9.26/kg Unit Cost = 12 oz 1 lbm 0.45359 kg
320 gram steak: $2.80 1000 g = $8.75/kg Unit Cost = 320 g 1 kg Therefore, the steak at the international market is a better buy.
1-36
1-90 The thrust developed by the jet engine of a Boeing 777 is given to be 85,000 pounds. This thrust is to be expressed in N and kgf. Analysis Noting that 1 lbf = 4.448 N and 1 kgf = 9.81 N, the thrust developed can be expressed in two other units as
Thrust in N:
4.448 N 5 Thrust = (85,000 lbf ) = 3.78 × 10 N 1 lbf
Thrust in kgf:
1 kgf 4 Thrust = (37.8 × 10 5 N ) = 3.85 × 10 kgf 9.81 N
1-91E The efficiency of a refrigerator increases by 3% per °C rise in the minimum temperature. This increase is to be expressed per °F, K, and R rise in the minimum temperature. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the increase in efficiency is (a) 3% for each K rise in temperature, and (b), (c) 3/1.8 = 1.67% for each R or °F rise in temperature.
1-92E The boiling temperature of water decreases by 3°C for each 1000 m rise in altitude. This decrease in temperature is to be expressed in °F, K, and R. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the decrease in the boiling temperature is (a) 3 K for each 1000 m rise in altitude, and (b), (c) 3×1.8 = 5.4°F = 5.4 R for each 1000 m rise in altitude.
1-93E The average body temperature of a person rises by about 2°C during strenuous exercise. This increase in temperature is to be expressed in °F, K, and R. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the rise in the body temperature during strenuous exercise is (a) 2 K (b) 2×1.8 = 3.6°F (c) 2×1.8 = 3.6 R
1-37 1-94E Hyperthermia of 5°C is considered fatal. This fatal level temperature change of body temperature is to be expressed in °F, K, and R. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the fatal level of hypothermia is (a) 5 K (b) 5×1.8 = 9°F (c) 5×1.8 = 9 R
1-95E A house is losing heat at a rate of 4500 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. The rate of heat loss is to be expressed per °F, K, and R of temperature difference between the indoor and the outdoor temperatures. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the rate of heat loss from the house is (a) 4500 kJ/h per K difference in temperature, and (b), (c) 4500/1.8 = 2500 kJ/h per R or °F rise in temperature.
1-96 The average temperature of the atmosphere is expressed as Tatm = 288.15 – 6.5z where z is altitude in km. The temperature outside an airplane cruising at 12,000 m is to be determined. Analysis Using the relation given, the average temperature of the atmosphere at an altitude of 12,000 m is determined to be Tatm = 288.15 - 6.5z = 288.15 - 6.5×12 = 210.15 K = - 63°C Discussion This is the “average” temperature. The actual temperature at different times can be different.
1-97 A new “Smith” absolute temperature scale is proposed, and a value of 1000 S is assigned to the boiling point of water. The ice point on this scale, and its relation to the Kelvin scale are to be determined. Analysis All linear absolute temperature scales read zero at absolute zero pressure, and are constant multiples of each other. For example, T(R) = 1.8 T(K). That is, S K multiplying a temperature value in K by 1.8 will give the same temperature in R. 1000 373.15 The proposed temperature scale is an acceptable absolute temperature scale since it differs from the other absolute temperature scales by a constant only. The boiling temperature of water in the Kelvin and the Smith scales are 315.15 K and 1000 K, respectively. Therefore, these two temperature scales are related to each other by 1000 T (S ) = T ( K ) = 2.6799 T(K ) 373.15 The ice point of water on the Smith scale is T(S)ice = 2.6799 T(K)ice = 2.6799×273.15 = 732.0 S 0
1-38
1-98E An expression for the equivalent wind chill temperature is given in English units. It is to be converted to SI units. Analysis The required conversion relations are 1 mph = 1.609 km/h and T(°F) = 1.8T(°C) + 32. The first thought that comes to mind is to replace T(°F) in the equation by its equivalent 1.8T(°C) + 32, and V in mph by 1.609 km/h, which is the “regular” way of converting units. However, the equation we have is not a regular dimensionally homogeneous equation, and thus the regular rules do not apply. The V in the equation is a constant whose value is equal to the numerical value of the velocity in mph. Therefore, if V is given in km/h, we should divide it by 1.609 to convert it to the desired unit of mph. That is, Tequiv (° F) = 914 . − [914 . − Tambient (° F)][0.475 − 0.0203(V / 1609 . ) + 0.304 V / 1609 . ]
or Tequiv (° F) = 914 . − [914 . − Tambient (° F)][0.475 − 0.0126V + 0.240 V ]
where V is in km/h. Now the problem reduces to converting a temperature in °F to a temperature in °C, using the proper convection relation: 1.8Tequiv (° C ) + 32 = 914 . − [914 . − (18 . Tambient (° C ) + 32)][0.475 − 0.0126V + 0.240 V ]
which simplifies to Tequiv (° C) = 33.0 − (33.0 − Tambient )(0.475 − 0.0126V + 0.240 V )
where the ambient air temperature is in °C.
1-39
1-99E EES Problem 1-98E is reconsidered. The equivalent wind-chill temperatures in °F as a function of wind velocity in the range of 4 mph to 100 mph for the ambient temperatures of 20, 40, and 60°F are to be plotted, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Obtain V and T_ambient from the Diagram Window" {T_ambient=10 V=20} V_use=max(V,4) T_equiv=91.4-(91.4-T_ambient)*(0.475 - 0.0203*V_use + 0.304*sqrt(V_use)) "The parametric table was used to generate the plot, Fill in values for T_ambient and V (use Alter Values under Tables menu) then use F3 to solve table. Plot the first 10 rows and then overlay the second ten, and so on. Place the text on the plot using Add Text under the Plot menu." Tambient [F] -25 -20 -15 -10 -5 0 5 10 15 20 -25 -20 -15 -10 -5 0 5 10 15 20 -25 -20 -15 -10 -5 0 5 10 15 20 -25 -20 -15 -10 -5 0 5 10 15 20
V [mph] 10 10 10 10 10 10 10 10 10 10 20 20 20 20 20 20 20 20 20 20 30 30 30 30 30 30 30 30 30 30 40 40 40 40 40 40 40 40 40 40
20
W ind Chill Tem perature
10 0 -10
W ind speed =10 m ph
-20
T W indChill
Tequiv [F] -52 -46 -40 -34 -27 -21 -15 -9 -3 3 -75 -68 -61 -53 -46 -39 -32 -25 -18 -11 -87 -79 -72 -64 -56 -49 -41 -33 -26 -18 -93 -85 -77 -69 -61 -54 -46 -38 -30 -22
-30
20 m ph
-40 -50
30 m ph -60 -70
40 m ph
-80 -30
-20
-10
0
10
20
80
100
T am bient
60 50
Tamb = 60F
40 30
] F[
vi u q e
20
Tamb = 40F 10
T
0 -10
Tamb = 20F -20 0
20
40
60
V [mph]
1-40
1-100 One section of the duct of an air-conditioning system is laid underwater. The upward force the water will exert on the duct is to be determined. Assumptions 1 The diameter given is the outer diameter of the duct (or, the thickness of the duct material is negligible). 2 The weight of the duct and the air in is negligible. Properties The density of air is given to be ρ = 1.30 kg/m3. We take the density of water to be 1000 kg/m3. D =15 cm Analysis Noting that the weight of the duct and the air in it is L = 20 m negligible, the net upward force acting on the duct is the buoyancy FB force exerted by water. The volume of the underground section of the duct is V = AL = (πD 2 / 4) L = [π (0.15 m) 2 /4](20 m) = 0.353 m 3
Then the buoyancy force becomes 1 kN = 3.46 kN FB = ρgV = (1000 kg/m 3 )(9.81 m/s 2 )(0.353 m 3 ) 1000 kg ⋅ m/s 2 Discussion The upward force exerted by water on the duct is 3.46 kN, which is equivalent to the weight of a mass of 353 kg. Therefore, this force must be treated seriously.
1-101 A helium balloon tied to the ground carries 2 people. The acceleration of the balloon when it is first released is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. Properties The density of air is given to be ρ = 1.16 kg/m3. The density of helium gas is 1/7th of this. Analysis The buoyancy force acting on the balloon is
V balloon = 4π r 3 /3 = 4 π(5 m) 3 /3 = 523.6 m 3 FB = ρ air gV balloon
1N = (1.16 kg/m 3 )(9.81m/s 2 )(523.6 m 3 ) 1 kg ⋅ m/s 2 The total mass is
= 5958 N
Helium balloon
1.16 m He = ρ HeV = kg/m 3 (523.6 m 3 ) = 86.8 kg 7 m total = m He + m people = 86.8 + 2 × 70 = 226.8 kg
The total weight is 1N W = m total g = (226.8 kg)(9.81 m/s 2 ) 1 kg ⋅ m/s 2 Thus the net force acting on the balloon is Fnet = FB − W = 5958 − 2225 = 3733 N
= 2225 N
Then the acceleration becomes a=
Fnet 3733 N 1kg ⋅ m/s 2 = 1N m total 226.8 kg
= 16.5 m/s 2
m = 140 kg
1-41
1-102 EES Problem 1-101 is reconsidered. The effect of the number of people carried in the balloon on acceleration is to be investigated. Acceleration is to be plotted against the number of people, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given Data:" rho_air=1.16"[kg/m^3]" "density of air" g=9.807"[m/s^2]" d_balloon=10"[m]" m_1person=70"[kg]" {NoPeople = 2} "Data suppied in Parametric Table" "Calculated values:" rho_He=rho_air/7"[kg/m^3]" "density of helium" r_balloon=d_balloon/2"[m]" V_balloon=4*pi*r_balloon^3/3"[m^3]" m_people=NoPeople*m_1person"[kg]" m_He=rho_He*V_balloon"[kg]" m_total=m_He+m_people"[kg]" "The total weight of balloon and people is:" W_total=m_total*g"[N]" "The buoyancy force acting on the balloon, F_b, is equal to the weight of the air displaced by the balloon." F_b=rho_air*V_balloon*g"[N]" "From the free body diagram of the balloon, the balancing vertical forces must equal the product of the total mass and the vertical acceleration:" F_b- W_total=m_total*a_up NoPeople 1 2 3 4 5 6 7 8 9 10
30 25 20
a up [m /s^2]
Aup [m/s2] 28.19 16.46 10.26 6.434 3.831 1.947 0.5204 -0.5973 -1.497 -2.236
15 10 5 0 -5 1
2
3
4
5
6
NoPeople
7
8
9
10
1-42
1-103 A balloon is filled with helium gas. The maximum amount of load the balloon can carry is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. Properties The density of air is given to be ρ = 1.16 kg/m3. The density of helium gas is 1/7th of this. Analysis In the limiting case, the net force acting on the balloon will be zero. That is, the buoyancy force and the weight will balance each other: W = mg = FB m total
Helium balloon
F 5958 N = B = = 607.3 kg g 9.81 m/s 2
Thus, m people = m total − m He = 607.3 − 86.8 = 520.5 kg
m
1-104E The pressure in a steam boiler is given in kgf/cm2. It is to be expressed in psi, kPa, atm, and bars. Analysis We note that 1 atm = 1.03323 kgf/cm2, 1 atm = 14.696 psi, 1 atm = 101.325 kPa, and 1 atm = 1.01325 bar (inner cover page of text). Then the desired conversions become:
In atm:
1 atm P = (92 kgf/cm 2 ) 1.03323 kgf/cm 2
= 89.04 atm
In psi:
1 atm P = (92 kgf/cm 2 ) 1.03323 kgf/cm 2
14.696 psi = 1309 psi 1 atm
In kPa:
1 atm P = (92 kgf/cm 2 ) 1.03323 kgf/cm 2
101.325 kPa = 9022 kPa 1 atm
In bars:
1 atm P = (92 kgf/cm 2 ) 1.03323 kgf/cm 2
1.01325 bar = 90.22 bar 1 atm
Discussion Note that the units atm, kgf/cm2, and bar are almost identical to each other.
1-43
1-105 A barometer is used to measure the altitude of a plane relative to the ground. The barometric readings at the ground and in the plane are given. The altitude of the plane is to be determined. Assumptions The variation of air density with altitude is negligible. Properties The densities of air and mercury are given to be ρ = 1.20 kg/m3 and ρ = 13,600 kg/m3. Analysis Atmospheric pressures at the location of the plane and the ground level are Pplane = ( ρ g h) plane 1 kPa 1N = (13,600 kg/m 3 )(9.81 m/s 2 )(0.690 m) 1 kg ⋅ m/s 2 1000 N/m 2 = 92.06 kPa Pground = ( ρ g h) ground
1N = (13,600 kg/m 3 )(9.81 m/s 2 )(0.753 m) 1 kg ⋅ m/s 2 = 100.46 kPa
1 kPa 1000 N/m 2
Taking an air column between the airplane and the ground and writing a force balance per unit base area, we obtain Wair / A = Pground − Pplane
h
0 Sea level
( ρ g h ) air = Pground − Pplane 1N (1.20 kg/m 3 )(9.81 m/s 2 )(h) 1 kg ⋅ m/s 2 It yields h = 714 m which is also the altitude of the airplane.
1 kPa 1000 N/m 2
= (100.46 − 92.06) kPa
1-106 A 10-m high cylindrical container is filled with equal volumes of water and oil. The pressure difference between the top and the bottom of the container is to be determined. Properties The density of water is given to be ρ = 1000 kg/m3. The specific gravity of oil is given to be 0.85. Analysis The density of the oil is obtained by multiplying its specific gravity by the density of water, 3
ρ = SG × ρ H 2 O = (0.85)(1000 kg/m ) = 850 kg/m
3
The pressure difference between the top and the bottom of the cylinder is the sum of the pressure differences across the two fluids, ∆Ptotal = ∆Poil + ∆Pwater = ( ρgh) oil + ( ρgh) water
[
Oil SG = 0.85 h = 10 m Water
]
1 kPa = (850 kg/m 3 )(9.81 m/s 2 )(5 m) + (1000 kg/m 3 )(9.81 m/s 2 )(5 m) 1000 N/m 2 = 90.7 kPa
1-44
1-107 The pressure of a gas contained in a vertical piston-cylinder device is measured to be 250 kPa. The mass of the piston is to be determined. Assumptions There is no friction between the piston and the cylinder. Patm Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield W = PA − Patm A mg = ( P − Patm ) A
1000 kg/m ⋅ s 2 (m)(9.81 m/s 2 ) = (250 − 100 kPa)(30 × 10 − 4 m 2 ) 1kPa It yields m = 45.9 kg
P
W = mg
1-108 The gage pressure in a pressure cooker is maintained constant at 100 kPa by a petcock. The mass of the petcock is to be determined. Assumptions There is no blockage of the pressure release valve. Patm Analysis Atmospheric pressure is acting on all surfaces of the petcock, which balances itself out. Therefore, it can be disregarded in calculations if we use the gage pressure as the cooker pressure. A force balance on the petcock (ΣFy = 0) yields W = Pgage A P W = mg − 6 2 2 Pgage A (100 kPa)(4 × 10 m ) 1000 kg/m ⋅ s = m= g 1 kPa 9.81 m/s 2 = 0.0408 kg
1-109 A glass tube open to the atmosphere is attached to a water pipe, and the pressure at the bottom of the tube is measured. It is to be determined how high the water will rise in the tube. Properties The density of water is given to be ρ = 1000 kg/m3. Analysis The pressure at the bottom of the tube can be expressed as P = Patm + ( ρ g h) tube
Solving for h, P − Patm h= ρg 1 kg ⋅ m/s 2 1N (1000 kg/m 3 )(9.81 m/s 2 ) = 2.34 m
=
(115 − 92) kPa
Patm= 92 kPa 1000 N/m 2 1 kPa
Water
h
1-45
1-110 The average atmospheric pressure is given as Patm = 101.325(1 − 0.02256 z )5.256 where z is the altitude in km. The atmospheric pressures at various locations are to be determined. Analysis The atmospheric pressures at various locations are obtained by substituting the altitude z values in km into the relation Patm = 101325 . (1 − 0.02256z )5.256
Atlanta: Denver: M. City: Mt. Ev.:
(z = 0.306 km): Patm = 101.325(1 - 0.02256×0.306)5.256 = 97.7 kPa (z = 1.610 km): Patm = 101.325(1 - 0.02256×1.610)5.256 = 83.4 kPa (z = 2.309 km): Patm = 101.325(1 - 0.02256×2.309)5.256 = 76.5 kPa (z = 8.848 km): Patm = 101.325(1 - 0.02256×8.848)5.256 = 31.4 kPa
1-111 The air pressure in a duct is measured by an inclined manometer. For a given vertical level difference, the gage pressure in the duct and the length of the differential fluid column are to be determined. Assumptions The manometer fluid is an incompressible substance. Properties The density of the liquid is given to be ρ = 0.81 kg/L = 810 kg/m3. Fresh Water Analysis The gage pressure in the duct is determined from L Pgage = Pabs − Patm = ρgh 8 cm 1 Pa 1N 3 2 = (810 kg/m )(9.81 m/s )(0.08 m) 35° 1 kg ⋅ m/s 2 1 N/m 2 = 636 Pa
The length of the differential fluid column is L = h / sin θ = (8 cm ) / sin 35° = 13.9 cm Discussion Note that the length of the differential fluid column is extended considerably by inclining the manometer arm for better readability.
1-112E Equal volumes of water and oil are poured into a U-tube from different arms, and the oil side is pressurized until the contact surface of the two fluids moves to the bottom and the liquid levels in both arms become the same. The excess pressure applied on the oil side is to be determined. Assumptions 1 Both water and oil are incompressible substances. 2 Oil does not mix with water. 3 The cross-sectional area of the U-tube Blown is constant. Water air 3 Properties The density of oil is given to be ρoil = 49.3 lbm/ft . We take 3 the density of water to be ρw = 62.4 lbm/ft . Oil Analysis Noting that the pressure of both the water and the oil is the same at the contact surface, the pressure at this surface can be 30 in expressed as Pcontact = Pblow + ρ a gha = Patm + ρ w gh w
Noting that ha = hw and rearranging, Pgage,blow = Pblow − Patm = ( ρ w − ρ oil ) gh
1 lbf = (62.4 - 49.3 lbm/ft 3 )(32.2 ft/s 2 )(30/12 ft) 32.2 lbm ⋅ ft/s 2 = 0.227 psi
1 ft 2 144 in 2
Discussion When the person stops blowing, the oil will rise and some water will flow into the right arm. It can be shown that when the curvature effects of the tube are disregarded, the differential height of water will be 23.7 in to balance 30-in of oil.
1-46
1-113 It is given that an IV fluid and the blood pressures balance each other when the bottle is at a certain height, and a certain gage pressure at the arm level is needed for sufficient flow rate. The gage pressure of the blood and elevation of the bottle required to maintain flow at the desired rate are to be determined. Assumptions 1 The IV fluid is incompressible. 2 The IV bottle is open to the atmosphere. Properties The density of the IV fluid is given to be ρ = 1020 kg/m3. Patm Analysis (a) Noting that the IV fluid and the blood pressures balance each other when the bottle is 1.2 m above the arm level, the gage pressure of IV Bottle the blood in the arm is simply equal to the gage pressure of the IV fluid at a depth of 1.2 m, Pgage, arm = Pabs − Patm = ρgharm - bottle 1 kPa 1 kN = (1020 kg/m 3 )(9.81 m/s 2 )(1.20 m) 1000 kg ⋅ m/s 2 1 kN/m 2 1.2 m = 12.0 k Pa
(b) To provide a gage pressure of 20 kPa at the arm level, the height of the bottle from the arm level is again determined from Pgage, arm = ρgharm- bottle to be harm - bottle =
Pgage, arm
ρg
1000 kg ⋅ m/s 2 1 kN/m 2 = 2.0 m 1 kPa 1 kN (1020 kg/m 3 )(9.81 m/s 2 ) Discussion Note that the height of the reservoir can be used to control flow rates in gravity driven flows. When there is flow, the pressure drop in the tube due to friction should also be considered. This will result in raising the bottle a little higher to overcome pressure drop. =
20 kPa
1-47
1-114 A gasoline line is connected to a pressure gage through a double-U manometer. For a given reading of the pressure gage, the gage pressure of the gasoline line is to be determined. Assumptions 1 All the liquids are incompressible. 2 The effect of air column on pressure is negligible. Properties The specific gravities of oil, mercury, and gasoline are given to be 0.79, 13.6, and 0.70, respectively. We take the density of water to be ρw = 1000 kg/m3. Analysis Starting with the pressure indicated by the pressure gage and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the gasoline pipe, and setting the result equal to Pgasoline gives
Pgage = 370 kPa Oil
45 cm
Gasoline
Air 22 cm 50 cm 10 cm Water Mercury
Pgage − ρ w gh w + ρ oil ghoil − ρ Hg ghHg − ρ gasoline ghgasoline = Pgasoline
Rearranging, Pgasoline = Pgage − ρ w g (hw − SG oil hoil + SG Hg hHg + SG gasoline hgasoline )
Substituting, Pgasoline = 370 kPa - (1000 kg/m 3 )(9.81 m/s 2 )[(0.45 m) − 0.79(0.5 m) + 13.6(0.1 m) + 0.70(0.22 m)] 1 kPa 1 kN × 2 2 1000 kg m/s ⋅ 1 kN/m = 354.6 kPa Therefore, the pressure in the gasoline pipe is 15.4 kPa lower than the pressure reading of the pressure gage. Discussion Note that sometimes the use of specific gravity offers great convenience in the solution of problems that involve several fluids.
1-48
1-115 A gasoline line is connected to a pressure gage through a double-U manometer. For a given reading of the pressure gage, the gage pressure of the gasoline line is to be determined. Assumptions 1 All the liquids are incompressible. 2 The effect of air column on pressure is negligible. Properties The specific gravities of oil, mercury, and gasoline are given to be 0.79, 13.6, and 0.70, respectively. We take the density of water to be ρw = 1000 kg/m3. Analysis Starting with the pressure indicated by the pressure gage and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the gasoline pipe, and setting the result equal to Pgasoline gives
Pgage = 180 kPa Oil
45 cm
Gasoline
Air 22 cm 50 cm 10 cm Water Mercury
Pgage − ρ w ghw + ρ oil ghoil − ρ Hg ghHg − ρ gasoline ghgasoline = Pgasoline
Rearranging, Pgasoline = Pgage − ρ w g (hw − SG oil hoil + SG Hg hHg + SG gasoline hgasoline )
Substituting, Pgasoline = 180 kPa - (1000 kg/m 3 )(9.807 m/s 2 )[(0.45 m) − 0.79(0.5 m) + 13.6(0.1 m) + 0.70(0.22 m)] 1 kPa 1 kN × 1000 kg ⋅ m/s 2 1 kN/m 2 = 164.6 kPa Therefore, the pressure in the gasoline pipe is 15.4 kPa lower than the pressure reading of the pressure gage. Discussion Note that sometimes the use of specific gravity offers great convenience in the solution of problems that involve several fluids.
1-49
1-116E A water pipe is connected to a double-U manometer whose free arm is open to the atmosphere. The absolute pressure at the center of the pipe is to be determined. Assumptions 1 All the liquids are incompressible. 2 The solubility of the liquids in each other is negligible. Properties The specific gravities of mercury and oil are given to be 13.6 and 0.80, respectively. We take the density of water to be ρw = 62.4 lbm/ft3. Analysis Starting with the pressure at the center of the water pipe, and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρgh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives
Oil
Oil
35 in
40 in 60 in 15 in
Water
Pwater pipe − ρ water gh water + ρ oil ghoil − ρ Hg ghHg − ρ oil ghoil = Patm
Mercury
Solving for Pwater pipe, Pwater pipe = Patm + ρ water g (hwater − SG oil hoil + SG Hg hHg + SG oil hoil ) Substituting, Pwater pipe = 14.2 psia + (62.4 lbm/ft 3 )(32.2 ft/s 2 )[(35/12 ft) − 0.8(60/12 ft) + 13.6(15/12 ft) 1 lbf + 0.8(40/12 ft)] × 32.2 lbm ⋅ ft/s 2 = 22.3 psia
1 ft 2 144 in 2
Therefore, the absolute pressure in the water pipe is 22.3 psia. Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly.
1-50 1-117 The temperature of the atmosphere varies with altitude z as T = T0 − βz , while the gravitational
acceleration varies by g ( z ) = g 0 /(1 + z / 6,370,320) 2 . Relations for the variation of pressure in atmosphere are to be obtained (a) by ignoring and (b) by considering the variation of g with altitude. Assumptions The air in the troposphere behaves as an ideal gas. Analysis (a) Pressure change across a differential fluid layer of thickness dz in the vertical z direction is dP = − ρgdz From the ideal gas relation, the air density can be expressed as ρ = dP = −
P P . Then, = RT R (T0 − β z )
P gdz R (T0 − βz )
Separating variables and integrating from z = 0 where P = P0 to z = z where P = P,
∫
P
P0
dP =− P
∫
z
0
gdz R (T0 − βz )
Performing the integrations. T − βz g P ln ln 0 = P0 Rβ T0 Rearranging, the desired relation for atmospheric pressure for the case of constant g becomes g
β z βR P = P0 1 − T0 (b) When the variation of g with altitude is considered, the procedure remains the same but the expressions become more complicated, g0 P dz dP = − R(T0 − βz ) (1 + z / 6,370,320) 2
Separating variables and integrating from z = 0 where P = P0 to z = z where P = P,
∫
P
P0
dP =− P
z
g 0 dz
0
R (T0 − β z )(1 + z / 6,370,320) 2
∫
Performing the integrations, ln P
P P0
g 1 1 1 + kz = 0 − ln 2 Rβ (1 + kT0 / β )(1 + kz ) (1 + kT0 / β ) T0 − β z 2
z
0
2
where R = 287 J/kg⋅K = 287 m /s ⋅K is the gas constant of air. After some manipulations, we obtain 1 g0 1 1 + kz + P = P0 exp − ln R ( β + kT0 ) 1 + 1 / kz 1 + kT0 / β 1 − βz / T0
where T0 = 288.15 K, β = 0.0065 K/m, g0 = 9.807 m/s2, k = 1/6,370,320 m-1, and z is the elevation in m.. Discussion When performing the integration in part (b), the following expression from integral tables is used, together with a transformation of variable x = T0 − βz , dx
∫ x(a + bx)
2
=
a + bx 1 1 − ln a (a + bx ) a 2 x
Also, for z = 11,000 m, for example, the relations in (a) and (b) give 22.62 and 22.69 kPa, respectively.
1-51
1-118 The variation of pressure with density in a thick gas layer is given. A relation is to be obtained for pressure as a function of elevation z. Assumptions The property relation P = Cρ n is valid over the entire region considered. Analysis The pressure change across a differential fluid layer of thickness dz in the vertical z direction is given as, dP = − ρgdz
Also, the relation P = Cρ n can be expressed as C = P / ρ n = P0 / ρ 0n , and thus
ρ = ρ 0 ( P / P0 ) 1 / n Substituting, dP = − gρ 0 ( P / P0 ) 1 / n dz
Separating variables and integrating from z = 0 where P = P0 = Cρ 0n to z = z where P = P,
∫
P
P0
∫
z
( P / P0 ) −1 / n dP = − ρ 0 g dz 0
Performing the integrations. ( P / P0 ) −1 / n +1 P0 − 1/ n + 1
P
= − ρ 0 gz P0
→
P P 0
( n −1) / n
−1 = −
n − 1 ρ 0 gz n P0
Solving for P, n /( n −1)
n − 1 ρ 0 gz P = P0 1 − n P0 which is the desired relation. Discussion The final result could be expressed in various forms. The form given is very convenient for calculations as it facilitates unit cancellations and reduces the chance of error.
1-52
1-119 A pressure transducers is used to measure pressure by generating analogue signals, and it is to be calibrated by measuring both the pressure and the electric current simultaneously for various settings, and the results are tabulated. A calibration curve in the form of P = aI + b is to be obtained, and the pressure corresponding to a signal of 10 mA is to be calculated. Assumptions Mercury is an incompressible liquid. Properties The specific gravity of mercury is given to be 13.56, and thus its density is 13,560 kg/m3. Analysis For a given differential height, the pressure can be calculated from P = ρg∆h
For ∆h = 28.0 mm = 0.0280 m, for example, 1 kN P = 13.56(1000 kg/m 3 )(9.81 m/s 2 )(0.0280 m) 2 1000 kg ⋅ m/s Repeating the calculations and tabulating, we have
1 kPa 1 kN/m 2 = 3.75 kPa
∆h(mm)
28.0
181.5
297.8
413.1
765.9
1027
1149
1362
1458
1536
P(kPa)
3.73
24.14
39.61
54.95
101.9
136.6
152.8
181.2
193.9
204.3
I (mA)
4.21
5.78
6.97
8.15
11.76
14.43
15.68
17.86
18.84
19.64
A plot of P versus I is given below. It is clear that the pressure varies linearly with the current, and using EES, the best curve fit is obtained to be P = 13.00I - 51.00
Multimeter
(kPa) for 4.21 ≤ I ≤ 19.64 .
For I = 10 mA, for example, we would get P = 79.0 kPa. Pressure transducer
225
Valve
180
P, kPa
135
Pressurized Air, P
90
Rigid container
45
0 4
6
8
10
12
I, mA
14
16
18
∆h
20
Discussion Note that the calibration relation is valid in the specified range of currents or pressures.
Manometer Mercury SG=13.5 6
1-53
Fundamentals of Engineering (FE) Exam Problems
1-120 Consider a fish swimming 5 m below the free surface of water. The increase in the pressure exerted on the fish when it dives to a depth of 45 m below the free surface is (a) 392 Pa (b) 9800 Pa (c) 50,000 Pa (d) 392,000 Pa (e) 441,000 Pa
Answer (d) 392,000 Pa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1000 "kg/m3" g=9.81 "m/s2" z1=5 "m" z2=45 "m" DELTAP=rho*g*(z2-z1) "Pa" "Some Wrong Solutions with Common Mistakes:" W1_P=rho*g*(z2-z1)/1000 "dividing by 1000" W2_P=rho*g*(z1+z2) "adding depts instead of subtracting" W3_P=rho*(z1+z2) "not using g" W4_P=rho*g*(0+z2) "ignoring z1"
1-121 The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 96.0 and 98.0 kPa. If the density of air is 1.0 kg/m3, the height of the building is (a) 17 m (b) 20 m (c) 170 m (d) 204 m (e) 252 m
Answer (d) 204 m Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1.0 "kg/m3" g=9.81 "m/s2" P1=96 "kPa" P2=98 "kPa" DELTAP=P2-P1 "kPa" DELTAP=rho*g*h/1000 "kPa" "Some Wrong Solutions with Common Mistakes:" DELTAP=rho*W1_h/1000 "not using g" DELTAP=g*W2_h/1000 "not using rho" P2=rho*g*W3_h/1000 "ignoring P1" P1=rho*g*W4_h/1000 "ignoring P2"
1-54 1-122 An apple loses 4.5 kJ of heat as it cools per °C drop in its temperature. The amount of heat loss from the apple per °F drop in its temperature is (a) 1.25 kJ (b) 2.50 kJ (c) 5.0 kJ (d) 8.1 kJ (e) 4.1 kJ
Answer (b) 2.50 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Q_perC=4.5 "kJ" Q_perF=Q_perC/1.8 "kJ" "Some Wrong Solutions with Common Mistakes:" W1_Q=Q_perC*1.8 "multiplying instead of dividing" W2_Q=Q_perC "setting them equal to each other"
1-123 Consider a 2-m deep swimming pool. The pressure difference between the top and bottom of the pool is (a) 12.0 kPa (b) 19.6 kPa (c) 38.1 kPa (d) 50.8 kPa (e) 200 kPa
Answer (b) 19.6 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1000 "kg/m^3" g=9.81 "m/s2" z1=0 "m" z2=2 "m" DELTAP=rho*g*(z2-z1)/1000 "kPa" "Some Wrong Solutions with Common Mistakes:" W1_P=rho*(z1+z2)/1000 "not using g" W2_P=rho*g*(z2-z1)/2000 "taking half of z" W3_P=rho*g*(z2-z1) "not dividing by 1000"
1-55
1-124 At sea level, the weight of 1 kg mass in SI units is 9.81 N. The weight of 1 lbm mass in English units is (a) 1 lbf (b) 9.81 lbf (c) 32.2 lbf (d) 0.1 lbf (e) 0.031 lbf
Answer (a) 1 lbf Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=1 "lbm" g=32.2 "ft/s2" W=m*g/32.2 "lbf" "Some Wrong Solutions with Common Mistakes:" gSI=9.81 "m/s2" W1_W= m*gSI "Using wrong conversion" W2_W= m*g "Using wrong conversion" W3_W= m/gSI "Using wrong conversion" W4_W= m/g "Using wrong conversion"
1-125 During a heating process, the temperature of an object rises by 20°C. This temperature rise is equivalent to a temperature rise of (b) 52°F (c) 36 K (d) 36 R (e) 293 K (a) 20°F
Answer (d) 36 R Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T_inC=20 "C" T_inR=T_inC*1.8 "R" "Some Wrong Solutions with Common Mistakes:" W1_TinF=T_inC "F, setting C and F equal to each other" W2_TinF=T_inC*1.8+32 "F, converting to F " W3_TinK=1.8*T_inC "K, wrong conversion from C to K" W4_TinK=T_inC+273 "K, converting to K"
1-126 … 1-129 Design, Essay, and Experiment Problems
KJ
2-1
Chapter 2 ENERGY, ENERGY TRANSFER, AND GENERAL ENERGY ANALYSIS Forms of Energy 2-1C In electric heaters, electrical energy is converted to sensible internal energy. 2-2C The forms of energy involved are electrical energy and sensible internal energy. Electrical energy is converted to sensible internal energy, which is transferred to the water as heat. 2-3C The macroscopic forms of energy are those a system possesses as a whole with respect to some outside reference frame. The microscopic forms of energy, on the other hand, are those related to the molecular structure of a system and the degree of the molecular activity, and are independent of outside reference frames. 2-4C The sum of all forms of the energy a system possesses is called total energy. In the absence of magnetic, electrical and surface tension effects, the total energy of a system consists of the kinetic, potential, and internal energies. 2-5C The internal energy of a system is made up of sensible, latent, chemical and nuclear energies. The sensible internal energy is due to translational, rotational, and vibrational effects. 2-6C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in daily life. 2-7C The mechanical energy is the form of energy that can be converted to mechanical work completely and directly by a mechanical device such as a propeller. It differs from thermal energy in that thermal energy cannot be converted to work directly and completely. The forms of mechanical energy of a fluid stream are kinetic, potential, and flow energies.
2-2
2-8 A river is flowing at a specified velocity, flow rate, and elevation. The total mechanical energy of the river water per unit mass, and the power generation potential of the entire river are to be determined. Assumptions 1 The elevation given is the elevation of the free surface of the river. 2 The velocity given is the average velocity. 3 The mechanical energy of water at the turbine exit is negligible. Properties We take the density of water to be ρ = 3 m/s 1000 kg/m3. River Analysis Noting that the sum of the flow energy and the potential energy is constant for a given fluid body, we can take the elevation of the entire river 90 m water to be the elevation of the free surface, and ignore the flow energy. Then the total mechanical energy of the river water per unit mass becomes (3 m/s) 2 1 kJ/kg V 2 = 0.887 kJ/kg = (9.81 m/s 2 )(90 m) + 1000 m 2 /s 2 2 2 The power generation potential of the river water is obtained by multiplying the total mechanical energy by the mass flow rate, m& = ρV& = (1000 kg/m 3 )(500 m 3 /s) = 500,000 kg/s emech = pe + ke = gh +
W&max = E& mech = m& emech = (500,000 kg/s)(0.887 kJ/kg) = 444,000 kW = 444 MW
Therefore, 444 MW of power can be generated from this river as it discharges into the lake if its power potential can be recovered completely. Discussion Note that the kinetic energy of water is negligible compared to the potential energy, and it can be ignored in the analysis. Also, the power output of an actual turbine will be less than 444 MW because of losses and inefficiencies.
2-9 A hydraulic turbine-generator is to generate electricity from the water of a large reservoir. The power generation potential is to be determined. Assumptions 1 The elevation of the reservoir remains constant. 2 The mechanical energy of water at the turbine exit is negligible. Analysis The total mechanical energy water in a reservoir possesses is equivalent to the potential 120 m energy of water at the free surface, and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow Turbine Generator rate. 1 kJ/kg emech = pe = gz = (9.81 m/s 2 )(120 m) = 1.177 kJ/kg 2 2 1000 m /s
Then the power generation potential becomes 1 kW W& max = E& mech = m& e mech = (1500 kg/s)(1.177 kJ/kg) = 1766 kW 1 kJ/s Therefore, the reservoir has the potential to generate 1766 kW of power. Discussion This problem can also be solved by considering a point at the turbine inlet, and using flow energy instead of potential energy. It would give the same result since the flow energy at the turbine inlet is equal to the potential energy at the free surface of the reservoir.
2-3
2-10 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass and the power generation potential are to be determined. Assumptions The wind is blowing steadily at a constant uniform velocity. Properties The density of air is given to be ρ = 1.25 kg/m3.
Wind 10 m/s
Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate: e mech = ke =
Wind turbine 60 m
V 2 (10 m/s ) 2 1 kJ/kg = = 0.050 kJ/kg 2 2 1000 m 2 /s 2
m& = ρVA = ρV
πD 2 4
= (1.25 kg/m 3 )(10 m/s)
π (60 m) 2 4
= 35,340 kg/s
W& max = E& mech = m& e mech = (35,340 kg/s)(0.050 kJ/kg) = 1770 kW
Therefore, 1770 kW of actual power can be generated by this wind turbine at the stated conditions. Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions.
2-11 A water jet strikes the buckets located on the perimeter of a wheel at a specified velocity and flow rate. The power generation potential of this system is to be determined. Assumptions Water jet flows steadily at the specified speed and flow rate. Analysis Kinetic energy is the only form of harvestable mechanical energy the water jet possesses, and it can be converted to work entirely. Therefore, the power potential of the water jet is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate: Shaf V 2 (60 m/s) 2 1 kJ/kg emech = ke = = = 1.8 kJ/kg 2 2 2 2 1000 m /s Nozzl W&max = E& mech = m& emech 1 kW = (120 kg/s)(1.8 kJ/kg) = 216 kW V 1 kJ/s Therefore, 216 kW of power can be generated by this water jet at the stated conditions. Discussion An actual hydroelectric turbine (such as the Pelton wheel) can convert over 90% of this potential to actual electric power.
2-4
2-12 Two sites with specified wind data are being considered for wind power generation. The site better suited for wind power generation is to be determined. Assumptions 1The wind is blowing steadily at specified velocity during specified times. 2 The wind power generation is negligible during other times. Properties We take the density of air to be ρ = 1.25 kg/m3 (it does not affect the final answer). Wind Analysis Kinetic energy is the only form of mechanical turbine energy the wind possesses, and it can be converted to work Wind entirely. Therefore, the power potential of the wind is its V, m/s kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate. Considering a unit flow area (A = 1 m2), the maximum wind power and power generation becomes e mech, 1 = ke1 =
V12 (7 m/s ) 2 1 kJ/kg = 2 2 1000 m 2 /s 2
e mech, 2 = ke 2 =
= 0.0245 kJ/kg
V 22 (10 m/s ) 2 1 kJ/kg = = 0.050 kJ/kg 2 2 1000 m 2 /s 2
W& max, 1 = E& mech, 1 = m& 1e mech, 1 = ρV1 Ake1 = (1.25 kg/m 3 )(7 m/s)(1 m 2 )(0.0245 kJ/kg) = 0.2144 kW W& max, 2 = E& mech, 2 = m& 2 e mech, 2 = ρV 2 Ake 2 = (1.25 kg/m 3 )(10 m/s)(1 m 2 )(0.050 kJ/kg) = 0.625 kW
since 1 kW = 1 kJ/s. Then the maximum electric power generations per year become E = W& ∆t = (0.2144 kW)(3000 h/yr) = 643 kWh/yr (per m 2 flow area) max, 1
max, 1
1
E max, 2 = W& max, 2 ∆t 2 = (0.625 kW)(2000 h/yr) = 1250 kWh/yr (per m 2 flow area)
Therefore, second site is a better one for wind generation. Discussion Note the power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the average wind velocity is the primary consideration in wind power generation decisions.
2-5
2-13 A river flowing steadily at a specified flow rate is considered for hydroelectric power generation by collecting the water in a dam. For a specified water height, the power generation potential is to be determined. Assumptions 1 The elevation given is the elevation of the free surface of the river. 2 The mechanical energy of water at the turbine exit is negligible. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate.
River 50 m
1 kJ/kg e mech = pe = gz = (9.81 m/s 2 )(50 m) = 0.4905 kJ/kg 1000 m 2 /s 2
The mass flow rate is m& = ρV& = (1000 kg/m 3 )(240 m 3 /s) = 240,000 kg/s
Then the power generation potential becomes 1 MW W& max = E& mech = m& e mech = (240,000 kg/s)(0.4905 kJ/kg) = 118 MW 1000 kJ/s
Therefore, 118 MW of power can be generated from this river if its power potential can be recovered completely. Discussion Note that the power output of an actual turbine will be less than 118 MW because of losses and inefficiencies.
2-14 A person with his suitcase goes up to the 10th floor in an elevator. The part of the energy of the elevator stored in the suitcase is to be determined. Assumptions 1 The vibrational effects in the elevator are negligible. Analysis The energy stored in the suitcase is stored in the form of potential energy, which is mgz. Therefore, 1 kJ/kg ∆E suitcase = ∆PE = mg∆z = (30 kg )(9.81 m/s 2 )(35 m) = 10.3 kJ 1000 m 2 /s 2
Therefore, the suitcase on 10th floor has 10.3 kJ more energy compared to an identical suitcase on the lobby level. Discussion Noting that 1 kWh = 3600 kJ, the energy transferred to the suitcase is 10.3/3600 = 0.0029 kWh, which is very small.
2-6
Energy Transfer by Heat and Work 2-15C Energy can cross the boundaries of a closed system in two forms: heat and work. 2-16C The form of energy that crosses the boundary of a closed system because of a temperature difference is heat; all other forms are work. 2-17C An adiabatic process is a process during which there is no heat transfer. A system that does not exchange any heat with its surroundings is an adiabatic system. 2-18C It is a work interaction. 2-19C It is a work interaction since the electrons are crossing the system boundary, thus doing electrical work. 2-20C It is a heat interaction since it is due to the temperature difference between the sun and the room. 2-21C This is neither a heat nor a work interaction since no energy is crossing the system boundary. This is simply the conversion of one form of internal energy (chemical energy) to another form (sensible energy). 2-22C Point functions depend on the state only whereas the path functions depend on the path followed during a process. Properties of substances are point functions, heat and work are path functions. 2-23C The caloric theory is based on the assumption that heat is a fluid-like substance called the "caloric" which is a massless, colorless, odorless substance. It was abandoned in the middle of the nineteenth century after it was shown that there is no such thing as the caloric.
2-7
Mechanical Forms of Work 2-24C The work done is the same, but the power is different. 2-25C The work done is the same, but the power is different.
2-26 A car is accelerated from rest to 100 km/h. The work needed to achieve this is to be determined. Analysis The work needed to accelerate a body the change in kinetic energy of the body, 100,000 m 2 1 1 1 kJ 2 2 = 309 kJ − 0 Wa = m(V2 − V1 ) = (800 kg) 2 2 2 2 3600 s 1000 kg ⋅ m /s
2-27 A car is accelerated from 10 to 60 km/h on an uphill road. The work needed to achieve this is to be determined. Analysis The total work required is the sum of the changes in potential and kinetic energies, Wa =
and
60,000 m 2 10,000 m 2 1 1 1 kJ = 175.5 kJ − m V22 − V12 = (1300 kg) 2 2 2 2 3600 s 3600 s 1000 kg ⋅ m /s
(
)
1 kJ = 510.0 kJ Wg = mg (z2 − z1 ) = (1300 kg)(9.81 m/s 2 )(40 m) 2 2 1000 kg ⋅ m /s
Thus, Wtotal = Wa + Wg = 175.5 + 510.0 = 686 kJ
2-28E The engine of a car develops 450 hp at 3000 rpm. The torque transmitted through the shaft is to be determined. Analysis The torque is determined from T=
550 lbf ⋅ ft/s 450 hp W&sh = 788 lbf ⋅ ft = 2πn& 2π (3000/60 )/s 1 hp
2-8
2-29 A linear spring is elongated by 20 cm from its rest position. The work done is to be determined. Analysis The spring work can be determined from Wspring =
1 1 k ( x22 − x12 ) = (70 kN/m)(0.22 − 0) m 2 = 1.4 kN ⋅ m = 1.4 kJ 2 2
2-30 The engine of a car develops 75 kW of power. The acceleration time of this car from rest to 100 km/h on a level road is to be determined. Analysis The work needed to accelerate a body is the change in its kinetic energy, 100,000 m 2 1 kJ 1 1 2 2 − 0 Wa = m V2 − V1 = (1500 kg) 2 2 2 3600 s 2 1000 kg ⋅ m /s
(
)
Thus the time required is
∆t =
Wa 578.7 kJ = = 7.72 s 75 kJ/s W& a
This answer is not realistic because part of the power will be used against the air drag, friction, and rolling resistance.
= 578.7 kJ
2-9
2-31 A ski lift is operating steadily at 10 km/h. The power required to operate and also to accelerate this ski lift from rest to the operating speed are to be determined. Assumptions 1 Air drag and friction are negligible. 2 The average mass of each loaded chair is 250 kg. 3 The mass of chairs is small relative to the mass of people, and thus the contribution of returning empty chairs to the motion is disregarded (this provides a safety factor). Analysis The lift is 1000 m long and the chairs are spaced 20 m apart. Thus at any given time there are 1000/20 = 50 chairs being lifted. Considering that the mass of each chair is 250 kg, the load of the lift at any given time is Load = (50 chairs)(250 kg/chair) = 12,500 kg Neglecting the work done on the system by the returning empty chairs, the work needed to raise this mass by 200 m is 1 kJ W g = mg (z 2 − z1 ) = (12,500 kg)(9.81 m/s 2 )(200 m) 1000 kg ⋅ m 2 /s 2
= 24,525 kJ
At 10 km/h, it will take ∆t =
distance 1 km = = 0.1 h = 360 s velocity 10 km / h
to do this work. Thus the power needed is W& g =
Wg ∆t
=
24,525 kJ 360 s
= 68.1 kW
The velocity of the lift during steady operation, and the acceleration during start up are 1 m/s V = (10 km/h) = 2.778 m/s 3.6 km/h a=
∆V 2.778 m/s - 0 = = 0.556 m/s 2 ∆t 5s
During acceleration, the power needed is 1 kJ/kg 1 1 W& a = m(V 22 − V12 ) / ∆t = (12,500 kg) (2.778 m/s) 2 − 0 1000 m 2 /s 2 2 2
(
)
/(5 s) = 9.6 kW
Assuming the power applied is constant, the acceleration will also be constant and the vertical distance traveled during acceleration will be h=
1 2 1 200 m 1 at sin α = at 2 = (0.556 m/s2 )(5 s) 2 (0.2) = 1.39 m 2 2 1000 m 2
and 1 kJ/kg W& g = mg (z 2 − z1 ) / ∆t = (12,500 kg)(9.81 m/s 2 )(1.39 m) 1000 kg ⋅ m 2 /s 2
Thus, W& total = W& a + W& g = 9.6 + 34.1 = 43.7 kW
/(5 s) = 34.1 kW
2-10
2-32 A car is to climb a hill in 10 s. The power needed is to be determined for three different cases. Assumptions Air drag, friction, and rolling resistance are negligible. Analysis The total power required for each case is the sum of the rates of changes in potential and kinetic energies. That is, W& total = W& a + W& g
(a) W& a = 0 since the velocity is constant. Also, the vertical rise is h = (100 m)(sin 30°) = 50 m. Thus, 1 kJ W& g = mg ( z 2 − z1 ) / ∆t = (2000 kg)(9.81 m/s 2 )(50 m) 1000 kg ⋅ m 2 /s 2
and
/(10 s) = 98.1 kW
W& total = W& a + W& g = 0 + 98.1 = 98.1 kW
(b) The power needed to accelerate is
[
1 1 W&a = m(V22 − V12 ) / ∆t = (2000 kg) (30 m/s )2 − 0 2 2
and
] 1000 kg1 kJ⋅ m /s 2
2
/(10 s) = 90 kW
W& total = W& a + W& g = 90 + 98.1 = 188.1 kW
(c) The power needed to decelerate is
[
1 1 W&a = m(V22 − V12 ) / ∆t = (2000 kg) (5 m/s)2 − (35 m/s )2 2 2
and
] 1000 kg1 kJ⋅ m /s
2
2
/(10 s) = −120 kW
W& total = W& a + W& g = −120 + 98.1 = −21.9 kW (breaking power)
2-33 A damaged car is being towed by a truck. The extra power needed is to be determined for three different cases. Assumptions Air drag, friction, and rolling resistance are negligible. Analysis The total power required for each case is the sum of the rates of changes in potential and kinetic energies. That is, W& total = W& a + W& g
(a) Zero. (b) W& a = 0 . Thus, ∆z = mgV z = mgV sin 30 o W& total = W& g = mg ( z 2 − z1 ) / ∆t = mg ∆t 50,000 m 1 kJ/kg = (1200 kg)(9.81m/s 2 ) 1000 m 2 /s 2 (0.5) = 81.7 kW 3600 s
(c) W& g = 0 . Thus, 90,000 m 2 1 kJ/kg 1 1 2 2 & & /(12 s) = 31.3 kW − 0 Wtotal = Wa = m(V2 − V1 ) / ∆t = (1200 kg) 3600 s 1000 m 2 /s 2 2 2
2-11
The First Law of Thermodynamics 2-34C No. This is the case for adiabatic systems only. 2-35C Warmer. Because energy is added to the room air in the form of electrical work. 2-36C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass transport.
2-37 Water is heated in a pan on top of a range while being stirred. The energy of the water at the end of the process is to be determined. Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible. Analysis We take the water in the pan as our system. This is a closed system since no mass enters or leaves. Applying the energy balance on this system gives E −E 1in424out 3
=
Net energy transfer by heat, work, and mass
∆E system 1 424 3
Change in internal, kinetic, potential, etc. energies
Qin + Wsh,in − Qout = ∆U = U 2 − U 1 30 kJ + 0.5 kJ − 5 kJ = U 2 − 10 kJ U 2 = 35.5 kJ
Therefore, the final internal energy of the system is 35.5 kJ.
2-38E Water is heated in a cylinder on top of a range. The change in the energy of the water during this process is to be determined. Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible. Analysis We take the water in the cylinder as the system. This is a closed system since no mass enters or leaves. Applying the energy balance on this system gives E −E 1in424out 3
Net energy transfer by heat, work, and mass
=
∆E system 1 424 3
Change in internal, kinetic, potential, etc. energies
Qin − Wout − Qout = ∆U = U 2 − U 1 65 Btu − 5 Btu − 8 Btu = ∆U ∆U = U 2 − U 1 = 52 Btu
Therefore, the energy content of the system increases by 52 Btu during this process.
2-12
2-39 A classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be determined. Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Analysis The total cooling load of the room is determined from = Q& + Q& + Q& Q& cooling
lights
people
heat gain
where Q& lights = 10 × 100 W = 1 kW Q& people = 40 × 360 kJ / h = 4 kW
Room
Q& heat gain = 15,000 kJ / h = 4.17 kW
Substituting, Q&
cooling
15,000 kJ/h = 1 + 4 + 4.17 = 9.17 kW
40 people 10 bulbs
·
Qcool
Thus the number of air-conditioning units required is 9.17 kW = 1.83 → 2 units 5 kW/unit
2-40 An industrial facility is to replace its 40-W standard fluorescent lamps by their 35-W high efficiency counterparts. The amount of energy and money that will be saved a year as well as the simple payback period are to be determined. Analysis The reduction in the total electric power consumed by the lighting as a result of switching to the high efficiency fluorescent is Wattage reduction = (Wattage reduction per lamp)(Number of lamps) = (40 - 34 W/lamp)(700 lamps) = 4200 W Then using the relations given earlier, the energy and cost savings associated with the replacement of the high efficiency fluorescent lamps are determined to be Energy Savings = (Total wattage reduction)(Ballast factor)(Operating hours) = (4.2 kW)(1.1)(2800 h/year) = 12,936 kWh/year Cost Savings = (Energy savings)(Unit electricity cost) = (12,936 kWh/year)($0.08/kWh) = $1035/year The implementation cost of this measure is simply the extra cost of the energy efficient fluorescent bulbs relative to standard ones, and is determined to be Implementation Cost = (Cost difference of lamps)(Number of lamps) = [($2.26-$1.77)/lamp](700 lamps) = $343 This gives a simple payback period of Implementation cost $343 = = 0.33 year (4.0 months) Simple payback period = Annual cost savings $1035 / year Discussion Note that if all the lamps were burned out today and are replaced by high-efficiency lamps instead of the conventional ones, the savings from electricity cost would pay for the cost differential in about 4 months. The electricity saved will also help the environment by reducing the amount of CO2, CO, NOx, etc. associated with the generation of electricity in a power plant.
2-13
2-41 The lighting energy consumption of a storage room is to be reduced by installing motion sensors. The amount of energy and money that will be saved as well as the simple payback period are to be determined. Assumptions The electrical energy consumed by the ballasts is negligible. Analysis The plant operates 12 hours a day, and thus currently the lights are on for the entire 12 hour period. The motion sensors installed will keep the lights on for 3 hours, and off for the remaining 9 hours every day. This corresponds to a total of 9×365 = 3285 off hours per year. Disregarding the ballast factor, the annual energy and cost savings become Energy Savings = (Number of lamps)(Lamp wattage)(Reduction of annual operating hours) = (24 lamps)(60 W/lamp )(3285 hours/year) = 4730 kWh/year Cost Savings = (Energy Savings)(Unit cost of energy) = (4730 kWh/year)($0.08/kWh) = $378/year The implementation cost of this measure is the sum of the purchase price of the sensor plus the labor, Implementation Cost = Material + Labor = $32 + $40 = $72 This gives a simple payback period of Simple payback period =
Implementation cost $72 = = 0.19 year (2.3 months) Annual cost savings $378 / year
Therefore, the motion sensor will pay for itself in about 2 months.
2-42 The classrooms and faculty offices of a university campus are not occupied an average of 4 hours a day, but the lights are kept on. The amounts of electricity and money the campus will save per year if the lights are turned off during unoccupied periods are to be determined. Analysis The total electric power consumed by the lights in the classrooms and faculty offices is E& lighting, classroom = (Power consumed per lamp) × (No. of lamps) = (200 × 12 × 110 W) = 264,000 = 264 kW E& lighting, offices = (Power consumed per lamp) × (No. of lamps) = (400 × 6 × 110 W) = 264,000 = 264 kW E& lighting, total = E& lighting, classroom + E& lighting, offices = 264 + 264 = 528 kW
Noting that the campus is open 240 days a year, the total number of unoccupied work hours per year is Unoccupied hours = (4 hours/day)(240 days/year) = 960 h/yr Then the amount of electrical energy consumed per year during unoccupied work period and its cost are Energy savings = ( E& )( Unoccupied hours) = (528 kW)(960 h/yr) = 506,880 kWh lighting, total
Cost savings = (Energy savings)(Unit cost of energy) = (506,880 kWh/yr)($0.082/kWh) = $41,564/yr
Discussion Note that simple conservation measures can result in significant energy and cost savings.
2-14
2-43 A room contains a light bulb, a TV set, a refrigerator, and an iron. The rate of increase of the energy content of the room when all of these electric devices are on is to be determined. Assumptions 1 The room is well sealed, and heat loss from the room is negligible. 2 All the appliances are kept on. Analysis Taking the room as the system, the rate form of the energy balance can be written as E& − E& 1in424out 3
Rate of net energy transfer by heat, work, and mass
=
→
dE system / dt 14243
dE room / dt = E& in
Rate of change in internal, kinetic, potential, etc. energies
since no energy is leaving the room in any form, and thus E& out = 0 . Also,
ROOM
E& in = E& lights + E& TV + E& refrig + E& iron = 100 + 110 + 200 + 1000 W = 1410 W
Electricity
Substituting, the rate of increase in the energy content of the room becomes
- Lights - TV - Refrig - Iron
dE room / dt = E& in = 1410 W
Discussion Note that some appliances such as refrigerators and irons operate intermittently, switching on and off as controlled by a thermostat. Therefore, the rate of energy transfer to the room, in general, will be less.
2-44 A fan is to accelerate quiescent air to a specified velocity at a specified flow rate. The minimum power that must be supplied to the fan is to be determined. Assumptions The fan operates steadily. Properties The density of air is given to be ρ = 1.18 kg/m3. Analysis A fan transmits the mechanical energy of the shaft (shaft power) to mechanical energy of air (kinetic energy). For a control volume that encloses the fan, the energy balance can be written as E& − E& 1in424out 3
Rate of net energy transfer by heat, work, and mass
= dE system / dt ©0 (steady) = 0 144424443
→
E& in = E& out
Rate of change in internal, kinetic, potential, etc. energies
V2 W& sh, in = m& air ke out = m& air out 2
where m& air = ρV& = (1.18 kg/m 3 )(4 m 3 /s) = 4.72 kg/s
Substituting, the minimum power input required is determined to be V2 (10 m/s) 2 1 J/kg W& sh, in = m& air out = (4.72 kg/s) = 236 J/s = 236 W 2 2 1 m 2 /s 2
Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be considerably higher because of the losses associated with the conversion of mechanical shaft energy to kinetic energy of air.
2-15
2-45E A fan accelerates air to a specified velocity in a square duct. The minimum electric power that must be supplied to the fan motor is to be determined. Assumptions 1 The fan operates steadily. 2 There are no conversion losses. Properties The density of air is given to be ρ = 0.075 lbm/ft3. Analysis A fan motor converts electrical energy to mechanical shaft energy, and the fan transmits the mechanical energy of the shaft (shaft power) to mechanical energy of air (kinetic energy). For a control volume that encloses the fan-motor unit, the energy balance can be written as E& − E& 1in424out 3
Rate of net energy transfer by heat, work, and mass
= dE system / dt ©0 (steady) = 0 144424443
→
E& in = E& out
Rate of change in internal, kinetic, potential, etc. energies
V2 W& elect, in = m& air ke out = m& air out 2
where m& air = ρVA = (0.075 lbm/ft3 )(3 × 3 ft 2 )(22 ft/s) = 14.85 lbm/s
Substituting, the minimum power input required is determined to be V2 (22 ft/s) 2 W& in = m& air out = (14.85 lbm/s) 2 2
1 Btu/lbm 2 2 25,037 ft /s
= 0.1435 Btu/s = 151 W
since 1 Btu = 1.055 kJ and 1 kJ/s = 1000 W. Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be considerably higher because of the losses associated with the conversion of electrical-to-mechanical shaft and mechanical shaft-to-kinetic energy of air.
2-16
2-46 A water pump is claimed to raise water to a specified elevation at a specified rate while consuming electric power at a specified rate. The validity of this claim is to be investigated. Assumptions 1 The water pump operates steadily. 2 Both 2 the lake and the pool are open to the atmosphere, and the flow velocities in them are negligible. Properties We take the density of water to be ρ = 1000 kg/m3 = 1 kg/L. Pool Analysis For a control volume that encloses the pump30 m motor unit, the energy balance can be written as Pump E& in − E& out = dE system / dt ©0 (steady) = 0 14243 144424443 1 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass
potential, etc. energies
Lake
E& in = E& out W& in + m& pe 1 = m& pe 2
→
W& in = m& ∆pe = m& g ( z 2 − z1 )
since the changes in kinetic and flow energies of water are negligible. Also, m& = ρV& = (1 kg/L)(50 L/s) = 50 kg/s Substituting, the minimum power input required is determined to be 1 kJ/kg W&in = m& g ( z2 − z1 ) = (50 kg/s)(9.81 m/s 2 )(30 m) = 14.7 kJ/s = 14.7 kW 2 2 1000 m /s
which is much greater than 2 kW. Therefore, the claim is false. Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be considerably higher than 14.7 kW because of the losses associated with the conversion of electrical-to-mechanical shaft and mechanical shaft-to-potential energy of water.
2-47 A gasoline pump raises the pressure to a specified value while consuming electric power at a specified rate. The maximum volume flow rate of gasoline is to be determined. Assumptions 1 The gasoline pump operates steadily. 2 The changes in kinetic and potential energies across the pump are negligible. Analysis For a control volume that encloses the pump-motor unit, the energy balance can be written as E& − E& out = dEsystem / dt ©0 (steady) = 0 → E& in = E& out 1in 424 3 144424443 Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc. energies
5.2 kW
W&in + m& ( Pv )1 = m& ( Pv ) 2 → W&in = m& ( P2 − P1 )v = V& ∆P
since m& = V&/v and the changes in kinetic and potential energies of PUMP Motor gasoline are negligible, Solving for volume flow rate and substituting, the maximum flow rate is determined to be Pump W& 5.2 kJ/s 1 kPa ⋅ m 3 inlet = 1.04 m 3 /s V&max = in = ∆P 5 kPa 1 kJ Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the volume flow rate will be less because of the losses associated with the conversion of electricalto-mechanical shaft and mechanical shaft-to-flow energy.
2-17
2-48 The fan of a central heating system circulates air through the ducts. For a specified pressure rise, the highest possible average flow velocity is to be determined. Assumptions 1 The fan operates steadily. 2 The changes in kinetic and potential energies across the fan are negligible. Analysis For a control volume that encloses the fan unit, the energy balance can be written as E& − E& 1in424out 3
Rate of net energy transfer by heat, work, and mass
= dE system / dt ©0 (steady) = 0 144424443
→
E& in = E& out
Rate of change in internal, kinetic, potential, etc. energies
W& in + m& ( Pv ) 1 = m& ( Pv ) 2 → W& in = m& ( P2 − P1 )v = V& ∆P
since m& = V&/v and the changes in kinetic and potential energies of gasoline are negligible, Solving for volume flow rate and substituting, the maximum flow rate and velocity are determined to be
V&max = V max =
W& in 60 J/s 1 Pa ⋅ m 3 = ∆P 50 Pa 1 J
V&max Ac
=
∆P = 50 Pa Air V m/s
1
= 1.2 m 3 /s
2 D = 30 cm 60 W
V&max 1.2 m 3 /s = = 17.0 m/s πD 2 / 4 π (0.30 m) 2 /4
Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the velocity will be less because of the losses associated with the conversion of electrical-tomechanical shaft and mechanical shaft-to-flow energy.
2-49E The heat loss from a house is to be made up by heat gain from people, lights, appliances, and resistance heaters. For a specified rate of heat loss, the required rated power of resistance heaters is to be determined. Assumptions 1 The house is well-sealed, so no air enters or heaves the house. 2 All the lights and appliances are kept on. 3 The house temperature remains constant. Analysis Taking the house as the system, the energy balance can be written as E& − E& out 1in 424 3
Rate of net energy transfer by heat, work, and mass
=
dEsystem / dt ©0 (steady) 144424443
=0
→
E& in = E& out
HOUSE
Rate of change in internal, kinetic, potential, etc. energies
where E& out = Q& out = 60,000 Btu/h and E& in = E& people + E& lights + E& appliance + E& heater = 6000 Btu/h + E& heater
Energy
- Lights - People - Appliance - Heaters
Qout
Substituting, the required power rating of the heaters becomes 1 kW E& heater = 60,000 − 6000 = 54,000 Btu/h = 15.8 kW 3412 Btu/h
Discussion When the energy gain of the house equals the energy loss, the temperature of the house remains constant. But when the energy supplied drops below the heat loss, the house temperature starts dropping.
2-18
2-50 An inclined escalator is to move a certain number of people upstairs at a constant velocity. The minimum power required to drive this escalator is to be determined. Assumptions 1 Air drag and friction are negligible. 2 The average mass of each person is 75 kg. 3 The escalator operates steadily, with no acceleration or breaking. 4 The mass of escalator itself is negligible. Analysis At design conditions, the total mass moved by the escalator at any given time is Mass = (30 persons)(75 kg/person) = 2250 kg The vertical component of escalator velocity is Vvert = V sin 45° = (0.8 m/s)sin45°
Under stated assumptions, the power supplied is used to increase the potential energy of people. Taking the people on elevator as the closed system, the energy balance in the rate form can be written as E& − E& out 1in 424 3
Rate of net energy transfer by heat, work, and mass
=
dEsystem / dt 14243
Rate of change in internal, kinetic, potential, etc. energies
=0
→
∆Esys E& in = dEsys / dt ≅ ∆t
∆PE mg∆z = W&in = = mgVvert ∆t ∆t
That is, under stated assumptions, the power input to the escalator must be equal to the rate of increase of the potential energy of people. Substituting, the required power input becomes 1 kJ/kg = 12.5 kJ/s = 12.5 kW W&in = mgVvert = (2250 kg)(9.81 m/s 2 )(0.8 m/s)sin45° 2 2 1000 m /s
When the escalator velocity is doubled to V = 1.6 m/s, the power needed to drive the escalator becomes 1 kJ/kg = 25.0 kJ/s = 25.0 kW W&in = mgVvert = (2250 kg)(9.81 m/s 2 )(1.6 m/s)sin45° 2 2 1000 m /s
Discussion Note that the power needed to drive an escalator is proportional to the escalator velocity.
2-19
2-51 A car cruising at a constant speed to accelerate to a specified speed within a specified time. The additional power needed to achieve this acceleration is to be determined. Assumptions 1 The additional air drag, friction, and rolling resistance are not considered. 2 The road is a level road. Analysis We consider the entire car as the system, except that let’s assume the power is supplied to the engine externally for simplicity (rather that internally by the combustion of a fuel and the associated energy conversion processes). The energy balance for the entire mass of the car can be written in the rate form as E& − E& out 1in 424 3
Rate of net energy transfer by heat, work, and mass
=
dEsystem / dt 14243
=0
→
Rate of change in internal, kinetic, potential, etc. energies
∆Esys E& in = dEsys / dt ≅ ∆t
∆KE m(V 22 − V12 ) / 2 = W& in = ∆t ∆t
since we are considering the change in the energy content of the car due to a change in its kinetic energy (acceleration). Substituting, the required additional power input to achieve the indicated acceleration becomes V 2 − V12 (110/3.6 m/s) 2 - (70/3.6 m/s) 2 W& in = m 2 = (1400 kg) 2 ∆t 2(5 s)
1 kJ/kg 1000 m 2 /s 2
= 77.8 kJ/s = 77.8 kW
since 1 m/s = 3.6 km/h. If the total mass of the car were 700 kg only, the power needed would be V 2 − V12 (110/3.6 m/s) 2 - (70/3.6 m/s) 2 W& in = m 2 = (700 kg) 2 ∆t 2(5 s)
1 kJ/kg 1000 m 2 /s 2
= 38.9 kW
Discussion Note that the power needed to accelerate a car is inversely proportional to the acceleration time. Therefore, the short acceleration times are indicative of powerful engines.
2-20
Energy Conversion Efficiencies 2-52C Mechanical efficiency is defined as the ratio of the mechanical energy output to the mechanical energy input. A mechanical efficiency of 100% for a hydraulic turbine means that the entire mechanical energy of the fluid is converted to mechanical (shaft) work. 2-53C The combined pump-motor efficiency of a pump/motor system is defined as the ratio of the increase in the mechanical energy of the fluid to the electrical power consumption of the motor, W& pump E& mech,out − E& mech,in ∆E& mech,fluid η pump-motor = η pumpη motor = = = W& W& W& elect,in
elect,in
elect,in
The combined pump-motor efficiency cannot be greater than either of the pump or motor efficiency since both pump and motor efficiencies are less than 1, and the product of two numbers that are less than one is less than either of the numbers. 2-54C The turbine efficiency, generator efficiency, and combined turbine-generator efficiency are defined as follows:
η turbine =
W& shaft,out Mechanical energy output = Mechanical energy extracted from the fluid | ∆E& mech,fluid |
η generator =
Electrical power output W& elect,out = Mechanical power input W& shaft,in
η turbine-gen = η turbineηgenerator = & E
W&elect,out − E&
mech,in
mech,out
=
W&elect,out | ∆E& mech,fluid |
2-55C No, the combined pump-motor efficiency cannot be greater that either of the pump efficiency of the motor efficiency. This is because η pump- motor = η pumpη motor , and both η pump and η motor are less than one, and a number gets smaller when multiplied by a number smaller than one.
2-21
2-56 A hooded electric open burner and a gas burner are considered. The amount of the electrical energy used directly for cooking and the cost of energy per “utilized” kWh are to be determined. Analysis The efficiency of the electric heater is given to be 73 percent. Therefore, a burner that consumes 3-kW of electrical energy will supply
η gas = 38% η electric = 73% Q& utilized = (Energy input) × (Efficiency) = (3 kW)(0.73) = 2.19 kW
of useful energy. The unit cost of utilized energy is inversely proportional to the efficiency, and is determined from Cost of utilized energy =
Cost of energy input $0.07 / kWh = = $0.096/kWh Efficiency 0.73
Noting that the efficiency of a gas burner is 38 percent, the energy input to a gas burner that supplies utilized energy at the same rate (2.19 kW) is Q& utilized 2.19 kW Q&input, gas = = = 5.76 kW (= 19,660 Btu/h) Efficiency 0.38 since 1 kW = 3412 Btu/h. Therefore, a gas burner should have a rating of at least 19,660 Btu/h to perform as well as the electric unit. Noting that 1 therm = 29.3 kWh, the unit cost of utilized energy in the case of gas burner is determined the same way to be Cost of utilized energy =
Cost of energy input $1.20 /( 29.3 kWh) = = $0.108/kWh Efficiency 0.38
2-57 A worn out standard motor is replaced by a high efficiency one. The reduction in the internal heat gain due to the higher efficiency under full load conditions is to be determined. Assumptions 1 The motor and the equipment driven by the motor are in the same room. 2 The motor operates at full load so that fload = 1. Analysis The heat generated by a motor is due to its inefficiency, and the difference between the heat generated by two motors that deliver the same shaft power is simply the difference between the electric power drawn by the motors, = W& = (75 × 746 W)/0.91 = 61,484 W W& /η in, electric, standard
shaft
motor
W& in, electric, efficient = W& shaft / η motor = (75 × 746 W)/0.954 = 58,648 W
Then the reduction in heat generation becomes Q& reduction = W& in, electric, standard − W& in, electric, efficient = 61,484 − 58,648 = 2836 W
2-22
2-58 An electric car is powered by an electric motor mounted in the engine compartment. The rate of heat supply by the motor to the engine compartment at full load conditions is to be determined. Assumptions The motor operates at full load so that the load factor is 1. Analysis The heat generated by a motor is due to its inefficiency, and is equal to the difference between the electrical energy it consumes and the shaft power it delivers, W& in, electric = W& shaft / η motor = (90 hp)/0.91 = 98.90 hp Q& generation = W& in, electric − W& shaft out = 98.90 − 90 = 8.90 hp = 6.64 kW
since 1 hp = 0.746 kW. Discussion Note that the electrical energy not converted to mechanical power is converted to heat.
2-59 A worn out standard motor is to be replaced by a high efficiency one. The amount of electrical energy and money savings as a result of installing the high efficiency motor instead of the standard one as well as the simple payback period are to be determined. Assumptions The load factor of the motor remains constant at 0.75. Analysis The electric power drawn by each motor and their difference can be expressed as W& electric in, standard = W& shaft / η standard = (Power rating)(Load factor) / η standard W& electric in, efficient = W& shaft / η efficient = (Power rating)(Load factor) / η efficient Power savings = W& electric in, standard − W& electric in, efficient = (Power rating)(Load factor)[1 / η standard − 1 / η efficient ]
where ηstandard is the efficiency of the standard motor, and ηefficient is the efficiency of the comparable high efficiency motor. Then the annual energy and cost savings associated with the installation of the high efficiency motor are determined to be Energy Savings = (Power savings)(Operating Hours) = (Power Rating)(Operating Hours)(Load Factor)(1/ηstandard- 1/ηefficient) = (75 hp)(0.746 kW/hp)(4,368 hours/year)(0.75)(1/0.91 - 1/0.954) = 9,290 kWh/year
η old = 91.0% η new = 95.4%
Cost Savings = (Energy savings)(Unit cost of energy) = (9,290 kWh/year)($0.08/kWh) = $743/year The implementation cost of this measure consists of the excess cost the high efficiency motor over the standard one. That is, Implementation Cost = Cost differential = $5,520 - $5,449 = $71 This gives a simple payback period of Simple payback period =
Implementation cost $71 = = 0.096 year (or 1.1 months) Annual cost savings $743 / year
Therefore, the high-efficiency motor will pay for its cost differential in about one month.
2-23
2-60E The combustion efficiency of a furnace is raised from 0.7 to 0.8 by tuning it up. The annual energy and cost savings as a result of tuning up the boiler are to be determined. Assumptions The boiler operates at full load while operating. Analysis The heat output of boiler is related to the fuel energy input to the boiler by Boiler output = (Boiler input)(Combustion efficiency)
or
Q& out = Q& inη furnace
The current rate of heat input to the boiler is given to be Q& in, current = 3.6 × 10 6 Btu/h . Then the rate of useful heat output of the boiler becomes Q& out = (Q& in η furnace ) current = (3.6 × 10 6 Btu/h)(0.7) = 2.52 × 10 6 Btu/h
The boiler must supply useful heat at the same rate after the tune up. Therefore, the rate of heat input to the boiler after the tune up and the rate of energy savings become
Boiler 70% 3.6×106 Btu/h
Q& in, new = Q& out / η furnace, new = (2.52 × 10 6 Btu/h)/0.8 = 3.15 × 10 6 Btu/h Q& in, saved = Q& in, current − Q& in, new = 3.6 × 10 6 − 3.15 × 10 6 = 0.45 × 10 6 Btu/h
Then the annual energy and cost savings associated with tuning up the boiler become Energy Savings = Q& in, saved (Operation hours) = (0.45×106 Btu/h)(1500 h/year) = 675×106 Btu/yr Cost Savings = (Energy Savings)(Unit cost of energy) = (675×106 Btu/yr)($4.35 per 106 Btu) = $2936/year Discussion Notice that tuning up the boiler will save $2936 a year, which is a significant amount. The implementation cost of this measure is negligible if the adjustment can be made by in-house personnel. Otherwise it is worthwhile to have an authorized representative of the boiler manufacturer to service the boiler twice a year.
2-24
2-61E EES Problem 2-60E is reconsidered. The effects of the unit cost of energy and combustion efficiency on the annual energy used and the cost savings as the efficiency varies from 0.6 to 0.9 and the unit cost varies from $4 to $6 per million Btu are the investigated. The annual energy saved and the cost savings are to be plotted against the efficiency for unit costs of $4, $5, and $6 per million Btu. Analysis The problem is solved using EES, and the solution is given below. "Knowns:" eta_boiler_current = 0.7 eta_boiler_new = 0.8 Q_dot_in_current = 3.6E+6 "[Btu/h]" DELTAt = 1500 "[h/year]" UnitCost_energy = 5E-6 "[dollars/Btu]" "Analysis: The heat output of boiler is related to the fuel energy input to the boiler by Boiler output = (Boiler input)(Combustion efficiency) Then the rate of useful heat output of the boiler becomes" Q_dot_out=Q_dot_in_current*eta_boiler_current "[Btu/h]" "The boiler must supply useful heat at the same rate after the tune up. Therefore, the rate of heat input to the boiler after the tune up and the rate of energy savings become " Q_dot_in_new=Q_dot_out/eta_boiler_new "[Btu/h]" Q_dot_in_saved=Q_dot_in_current - Q_dot_in_new "[Btu/h]" "Then the annual energy and cost savings associated with tuning up the boiler become" EnergySavings =Q_dot_in_saved*DELTAt "[Btu/year]" CostSavings = EnergySavings*UnitCost_energy "[dollars/year]" "Discussion Notice that tuning up the boiler will save $2936 a year, which is a significant amount. The implementation cost of this measure is negligible if the adjustment can be made by in-house personnel. Otherwise it is worthwhile to have an authorized representative of the boiler manufacturer to service the boiler twice a year. " CostSavings [dollars/year] -4500 0 3375 6000
ηboiler,new
EnergySavings [Btu/year] -9.000E+08 0 6.750E+08 1.200E+09
0.6 0.7 0.8 0.9
CostSavings [dollars/year]
8000
6000
4000
2000
1,250x109
Unit Cost of Energy 4E-6 $/Btu 5E-6 $/Btu 6E-6 $/Btu
0
-2000
-4000
-6000 0.6
0.65
0.7
0.75
η
boiler,new
0.8
0.85
0.9
] r a e y/ ut B[ s g ni v a S y g r e n E
8,000x108
Unit Cost = $5/106 Btu
3,500x108 -1,000x108 -5,500x108 -1,000x109 0,6
0,65
0,7
0,75
ηboiler;new
0,8
0,85
0,9
2-25
2-62 Several people are working out in an exercise room. The rate of heat gain from people and the equipment is to be determined. Assumptions The average rate of heat dissipated by people in an exercise room is 525 W. Analysis The 8 weight lifting machines do not have any motors, and thus they do not contribute to the internal heat gain directly. The usage factors of the motors of the treadmills are taken to be unity since they are used constantly during peak periods. Noting that 1 hp = 746 W, the total heat generated by the motors is Q& = ( No. of motors) × W& ×f ×f /η motors
motor
load
usage
motor
= 4 × (2.5 × 746 W) × 0.70 × 1.0/0.77 = 6782 W
The heat gain from 14 people is = ( No. of people) × Q& Q& people
person
= 14 × (525 W) = 7350 W
Then the total rate of heat gain of the exercise room during peak period becomes Q& = Q& + Q& = 6782 + 7350 = 14,132 W total
motors
people
2-63 A classroom has a specified number of students, instructors, and fluorescent light bulbs. The rate of internal heat generation in this classroom is to be determined. Assumptions 1 There is a mix of men, women, and children in the classroom. 2 The amount of light (and thus energy) leaving the room through the windows is negligible. Properties The average rate of heat generation from people seated in a room/office is given to be 100 W. Analysis The amount of heat dissipated by the lamps is equal to the amount of electrical energy consumed by the lamps, including the 10% additional electricity consumed by the ballasts. Therefore, Q& = (Energy consumed per lamp) × (No. of lamps) lighting
Q& people
= (40 W)(1.1)(18) = 792 W = ( No. of people) × Q& = 56 × (100 W) = 5600 W person
Then the total rate of heat gain (or the internal heat load) of the classroom from the lights and people become Q& total = Q& lighting + Q& people = 792 + 5600 = 6392 W
2-64 A room is cooled by circulating chilled water through a heat exchanger, and the air is circulated through the heat exchanger by a fan. The contribution of the fan-motor assembly to the cooling load of the room is to be determined. Assumptions The fan motor operates at full load so that fload = 1. Analysis The entire electrical energy consumed by the motor, including the shaft power delivered to the fan, is eventually dissipated as heat. Therefore, the contribution of the fan-motor assembly to the cooling load of the room is equal to the electrical energy it consumes, Q& internal generation = W& in, electric = W& shaft / η motor = (0.25 hp)/0.54 = 0.463 hp = 345 W
since 1 hp = 746 W.
2-26
2-65 A hydraulic turbine-generator is generating electricity from the water of a large reservoir. The combined turbine-generator efficiency and the turbine efficiency are to be determined. Assumptions 1 The elevation of the reservoir remains constant. 2 The mechanical energy of water at the turbine exit is negligible. Analysis We take the free surface of the reservoir to be point 1 and the turbine exit to be point 2. We also take the turbine exit as the reference level (z2 = 0), and thus the potential energy at points 1 and 2 are pe1 = gz1 and pe2 = 0. The flow energy P/ρ at both points is zero since both 1 and 2 are open to the atmosphere (P1 = P2 = Patm). Further, the kinetic energy at both points is zero (ke1 = ke2 = 0) since the water at point 1 is essentially motionless, and the kinetic energy of water at turbine exit is assumed to be negligible. The potential energy of water at point 1 is 1 kJ/kg pe1 = gz1 = (9.81 m/s 2 )(70 m) = 0.687 kJ/kg 1000 m 2 /s 2
Then the rate at which the mechanical energy of the fluid is supplied to the turbine become
1
∆E& mech,fluid = m& (emech,in − emech,out ) = m& ( pe1 − 0) = m& pe1 = (1500 kg/s)(0.687 kJ/kg) = 1031 kW
The combined turbine-generator and the turbine efficiency are determined from their definitions,
η turbine-gen = η turbine =
750 kW
70 m
W& elect,out 750 kW = = 0.727 or 72.7% & | ∆E mech,fluid | 1031 kW
Generator
Turbine
2
W& shaft,out 800 kW = = 0.776 or 77.6% & 1031 kW | ∆E mech,fluid |
Therefore, the reservoir supplies 1031 kW of mechanical energy to the turbine, which converts 800 kW of it to shaft work that drives the generator, which generates 750 kW of electric power. Discussion This problem can also be solved by taking point 1 to be at the turbine inlet, and using flow energy instead of potential energy. It would give the same result since the flow energy at the turbine inlet is equal to the potential energy at the free surface of the reservoir.
2-27
2-66 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass, the power generation potential, and the actual electric power generation are to be determined. Assumptions 1 The wind is blowing steadily at a constant uniform velocity. 2 The efficiency of the wind turbine is independent of the wind speed. Properties The density of air is given to be ρ = 1.25 kg/m3. Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate: emech = ke =
Wind
Wind turbine
12 m/s
50 m
V 2 (12 m/s) 2 1 kJ/kg = = 0.072 kJ/kg 2 2 2 2 1000 m /s
m& = ρVA = ρV
πD 2 4
= (1.25 kg/m 3 )(12 m/s)
π (50 m)2 4
= 29,450 kg/s
W&max = E& mech = m& emech = (29,450 kg/s)(0.072 kJ/kg) = 2121 kW
The actual electric power generation is determined by multiplying the power generation potential by the efficiency, W&elect = η wind turbineW&max = (0.30)(2121 kW) = 636 kW
Therefore, 636 kW of actual power can be generated by this wind turbine at the stated conditions. Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions.
2-28
2-67 EES Problem 2-66 is reconsidered. The effect of wind velocity and the blade span diameter on wind power generation as the velocity varies from 5 m/s to 20 m/s in increments of 5 m/s, and the diameter varies from 20 m to 80 m in increments of 20 m is to be investigated. Analysis The problem is solved using EES, and the solution is given below. D1=20 "m" D2=40 "m" D3=60 "m" D4=80 "m" Eta=0.30 rho=1.25 "kg/m3" m1_dot=rho*V*(pi*D1^2/4); W1_Elect=Eta*m1_dot*(V^2/2)/1000 "kW" m2_dot=rho*V*(pi*D2^2/4); W2_Elect=Eta*m2_dot*(V^2/2)/1000 "kW" m3_dot=rho*V*(pi*D3^2/4); W3_Elect=Eta*m3_dot*(V^2/2)/1000 "kW" m4_dot=rho*V*(pi*D4^2/4); W4_Elect=Eta*m4_dot*(V^2/2)/1000 "kW" D, m 20
V, m/s 5 10 15 20 5 10 15 20 5 10 15 20 5 10 15 20
40
60
80
Welect, kW 7 59 199 471 29 236 795 1885 66 530 1789 4241 118 942 3181 7540
m, kg/s 1,963 3,927 5,890 7,854 7,854 15,708 23,562 31,416 17,671 35,343 53,014 70,686 31,416 62,832 94,248 125,664
8000 D = 80 m
7000
6000
WElect
5000 D = 60 m
4000
3000 D = 40 m
2000
1000 0 4
D = 20 m
6
8
10
12
14
V, m/s
16
18
20
2-29
2-68 A wind turbine produces 180 kW of power. The average velocity of the air and the conversion efficiency of the turbine are to be determined. Assumptions The wind turbine operates steadily. Properties The density of air is given to be 1.31 kg/m3. Analysis (a) The blade diameter and the blade span area are Vtip = D= πn&
A=
πD 2 4
=
1 m/s (250 km/h) 3.6 km/h = 88.42 m 1 min π (15 L/min) 60 s
π (88.42 m) 2 4
= 6140 m 2
Then the average velocity of air through the wind turbine becomes V=
42,000 kg/s m& = = 5.23 m/s ρA (1.31 kg/m 3 )(6140 m 2 )
(b) The kinetic energy of the air flowing through the turbine is KE& =
1 1 m& V 2 = (42,000 kg/s)(5.23 m/s) 2 = 574.3 kW 2 2
Then the conversion efficiency of the turbine becomes 180 kW W& η= = = 0.313 = 31.3% KE& 574.3 kW Discussion Note that about one-third of the kinetic energy of the wind is converted to power by the wind turbine, which is typical of actual turbines.
2-30
2-69 Water is pumped from a lake to a storage tank at a specified rate. The overall efficiency of the pumpmotor unit and the pressure difference between the inlet and the exit of the pump are to be determined. √ Assumptions 1 The elevations of the tank and the lake remain constant. 2 Frictional losses in the pipes are negligible. 3 The changes in kinetic energy are negligible. 4 The elevation difference across the pump is negligible. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis (a) We take the free surface of the lake to be point 1 and the free surfaces of the storage tank to be point 2. We also take the lake surface as the reference level (z1 = 0), and thus the potential energy at points 1 and 2 are pe1 = 0 and pe2 = gz2. The flow energy at both points is zero since both 1 and 2 are open to the atmosphere (P1 = P2 = Patm). Further, the kinetic energy at both points is zero (ke1 = ke2 = 0) since the water at both locations is essentially stationary. The mass flow rate of water and its potential energy at point 2 are
2 Storage tank
20 m
Pump
1
m& = ρV& = (1000 kg/m 3 )(0.070 m 3/s) = 70 kg/s
1 kJ/kg pe 2 = gz 2 = (9.81 m/s 2 )(20 m) = 0.196 kJ/kg 1000 m 2 /s 2
Then the rate of increase of the mechanical energy of water becomes ∆E& mech,fluid = m& (e mech,out − e mech,in ) = m& ( pe 2 − 0) = m& pe 2 = (70 kg/s)(0.196 kJ/kg) = 13.7 kW
The overall efficiency of the combined pump-motor unit is determined from its definition,
η pump -motor =
∆E& mech,fluid 13.7 kW = = 0.672 or 67.2% 20.4 kW W& elect,in
(b) Now we consider the pump. The change in the mechanical energy of water as it flows through the pump consists of the change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy are negligible. Also, this change must be equal to the useful mechanical energy supplied by the pump, which is 13.7 kW: ∆E& mech,fluid = m& (e mech,out − e mech,in ) = m&
P2 − P1
ρ
= V&∆P
Solving for ∆P and substituting, ∆P =
∆E& mech,fluid 13.7 kJ/s 1 kPa ⋅ m 3 = 0.070 m 3 /s 1 kJ V&
= 196 kPa
Therefore, the pump must boost the pressure of water by 196 kPa in order to raise its elevation by 20 m. Discussion Note that only two-thirds of the electric energy consumed by the pump-motor is converted to the mechanical energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor.
2-31
2-70 Geothermal water is raised from a given depth by a pump at a specified rate. For a given pump efficiency, the required power input to the pump is to be determined. Assumptions 1 The pump operates steadily. 2 Frictional losses in the pipes are negligible. 3 The changes in kinetic energy are negligible. 4 The geothermal water is exposed to the atmosphere and thus its free surface is at atmospheric pressure.
2
Properties The density of geothermal water is given to be ρ = 1050 kg/m3. Analysis The elevation of geothermal water and thus its potential energy changes, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of geothermal water is equal to the change in its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. That is, ∆E& = m& ∆e = m& ∆pe = m& g∆z = ρV&g∆z mech
200 m
1 Pump
mech
1N = (1050 kg/m 3 )(0.3 m 3 /s)(9.81 m/s 2 )(200 m) 1 kg ⋅ m/s 2
1 kW 1000 N ⋅ m/s = 618.0 kW
Then the required power input to the pump becomes
W& pump, elect =
∆E& mech
η pump-motor
=
618 kW = 835 kW 0.74
Discussion The frictional losses in piping systems are usually significant, and thus a larger pump will be needed to overcome these frictional losses.
2-71 An electric motor with a specified efficiency operates in a room. The rate at which the motor dissipates heat to the room it is in when operating at full load and if this heat dissipation is adequate to heat the room in winter are to be determined. Assumptions The motor operates at full load. Analysis The motor efficiency represents the fraction of electrical energy consumed by the motor that is converted to mechanical work. The remaining part of electrical energy is converted to thermal energy and is dissipated as heat. Q& dissipated = (1 − η motor )W& in, electric = (1 − 0.88)(20 kW) = 2.4 kW
which is larger than the rating of the heater. Therefore, the heat dissipated by the motor alone is sufficient to heat the room in winter, and there is no need to turn the heater on. Discussion Note that the heat generated by electric motors is significant, and it should be considered in the determination of heating and cooling loads.
2-32
2-72 A large wind turbine is installed at a location where the wind is blowing steadily at a certain velocity. The electric power generation, the daily electricity production, and the monetary value of this electricity are to be determined. Assumptions 1 The wind is blowing steadily at a constant uniform velocity. 2 The efficiency of the Wind wind turbine is independent of the wind speed. turbine Wind Properties The density of air is given to be ρ = 1.25 8 m/s kg/m3. 100 m Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate: e mech = ke =
V 2 (8 m/s ) 2 1 kJ/kg = = 0.032 kJ/kg 2 2 1000 m 2 /s 2
m& = ρVA = ρV
πD 2 4
= (1.25 kg/m 3 )(8 m/s)
π (100 m) 2 4
= 78,540 kg/s
W& max = E& mech = m& e mech = (78,540 kg/s)(0.032 kJ/kg) = 2513 kW
The actual electric power generation is determined from W& =η W& = (0.32)(2513 kW) = 804.2 kW elect
wind turbine
max
Then the amount of electricity generated per day and its monetary value become Amount of electricity = (Wind power)(Operating hours)=(804.2 kW)(24 h) =19,300 kWh Revenues = (Amount of electricity)(Unit price) = (19,300 kWh)($0.06/kWh) = $1158 (per day) Discussion Note that a single wind turbine can generate several thousand dollars worth of electricity every day at a reasonable cost, which explains the overwhelming popularity of wind turbines in recent years.
2-73E A water pump raises the pressure of water by a specified amount at a specified flow rate while consuming a known amount of electric power. The mechanical efficiency of the pump is to be determined. Assumptions 1 The pump operates steadily. 2 The changes in velocity and elevation across the pump are negligible. 3 Water is incompressible. Analysis To determine the mechanical efficiency of the pump, we need to know the increase in the mechanical energy of the fluid as it flows through the pump, which is ∆E& mech,fluid = m& (emech,out − emech,in ) = m& [( Pv ) 2 − ( Pv )1 ] = m& ( P2 − P1 )v
∆P = 1.2 psi 3 hp PUMP
Pump inlet
1 Btu = 1.776 Btu/s = 2.51 hp = V& ( P2 − P1 ) = (8 ft 3 /s)(1.2 psi) 3 5.404 psi ⋅ ft since 1 hp = 0.7068 Btu/s, m& = ρV& = V& / v , and there is no change in kinetic and potential energies of the
fluid. Then the mechanical efficiency of the pump becomes ∆E& mech,fluid 2.51 hp η pump = = = 0.838 or 83.8% 3 hp W& pump, shaft Discussion The overall efficiency of this pump will be lower than 83.8% because of the inefficiency of the electric motor that drives the pump.
2-33
2-74 Water is pumped from a lower reservoir to a higher reservoir at a specified rate. For a specified shaft power input, the power that is converted to thermal energy is to be determined. Assumptions 1 The pump operates steadily. 2 The elevations of the reservoirs remain constant. 3 The changes in kinetic energy are negligible. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. That is,
2 Reservoir 45 m
Pump
1 Reservoir
∆E& mech = m& ∆e mech = m& ∆pe = m& g∆z = ρV&g∆z 1N = (1000 kg/m 3 )(0.03 m 3 /s)(9.81 m/s 2 )(45 m) 2 1 kg ⋅ m/s
1 kW 1000 N ⋅ m/s = 13.2 kW
Then the mechanical power lost because of frictional effects becomes
W& frict = W& pump, in − ∆E& mech = 20 − 13.2 kW = 6.8 kW Discussion The 6.8 kW of power is used to overcome the friction in the piping system. The effect of frictional losses in a pump is always to convert mechanical energy to an equivalent amount of thermal energy, which results in a slight rise in fluid temperature. Note that this pumping process could be accomplished by a 13.2 kW pump (rather than 20 kW) if there were no frictional losses in the system. In this ideal case, the pump would function as a turbine when the water is allowed to flow from the upper reservoir to the lower reservoir and extract 13.2 kW of power from the water.
2-34
2-75 A pump with a specified shaft power and efficiency is used to raise water to a higher elevation. The maximum flow rate of water is to be determined.
2
Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the reservoirs is constant. 3 We assume the flow in the pipes to be frictionless since the maximum flow rate is to be determined, Properties We take the density of water to be ρ = 1000 kg/m3.
PUMP
Analysis The useful pumping power (the part converted to mechanical energy of water) is
15 m
W& pump, u = η pump W& pump, shaft = (0.82 )(7 hp) = 5.74 hp
1
The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. That is,
Water
∆E& mech = m& ∆e mech = m& ∆pe = m& g∆z = ρV&g∆z
Noting that ∆E& mech = W& pump,u , the volume flow rate of water is determined to be
V& =
W& pump, u
ρgz 2
=
745.7 W 1 N ⋅ m/s 1 kg ⋅ m/s 2 = 0.0291 m3 /s 1 hp 1 W 1 N (1000 kg/m )(9.81 m/s )(15 m) 5.74 hp
3
2
Discussion This is the maximum flow rate since the frictional effects are ignored. In an actual system, the flow rate of water will be less because of friction in pipes.
2-76 The available head of a hydraulic turbine and its overall efficiency are given. The electric power output of this turbine is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation of the reservoir remains constant. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis The total mechanical energy the water in a reservoir possesses is equivalent to the potential energy of water at the free surface, and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. Therefore, the actual power produced by the turbine can be expressed as
85 m Eff.=91% Turbine
Generator
W& turbine = η turbine m& ghturbine = η turbine ρV&ghturbine
Substituting, 1N 1 kW W& turbine = (0.91)(1000 kg/m3 )(0.25 m3 /s)(9.81 m/s 2 )(85 m) = 190 kW 2 1 kg ⋅ m/s 1000 N ⋅ m/s Discussion Note that the power output of a hydraulic turbine is proportional to the available elevation difference (turbine head) and the flow rate.
2-35
2-77 A pump is pumping oil at a specified rate. The pressure rise of oil in the pump is measured, and the motor efficiency is specified. The mechanical efficiency of the pump is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference across the pump is negligible. Properties The density of oil is given to be ρ = 860 kg/m3. Analysis Then the total mechanical energy of a fluid is the sum of the potential, flow, and kinetic energies, and is expressed per unit mass as emech = gh + Pv + V 2 / 2 . To determine the mechanical efficiency of the pump, we need to know the increase in the mechanical energy of the fluid as it flows through the pump, which is V2 V2 ∆E& mech,fluid = m& (e mech,out − e mech,in ) = m& ( Pv ) 2 + 2 − ( Pv ) 1 − 1 2 2
& V 2 − V12 = V ( P2 − P1 ) + ρ 2 2
since m& = ρV& = V& / v , and there is no change in the potential energy of the fluid. Also, V1 = V2 =
V& A1
V& A2
=
V&
πD12 / 4
=
=
V& πD 22 / 4
0.1 m3 /s
π (0.08 m)2 / 4
=
2
35 kW
= 19.9 m/s
0.1 m 3 /s
π (0.12 m) 2 / 4
PUMP
= 8.84 m/s
Substituting, the useful pumping power is determined to be
Motor
Pump inlet 1
W& pump,u = ∆E& mech,fluid (8.84 m/s) 2 − (19.9 m/s) 2 = (0.1 m 3 /s) 400 kN/m 2 + (860 kg/m 3 ) 2 = 26.3 kW
1 kN 1000 kg ⋅ m/s 2
1 kW 1 kN ⋅ m/s
Then the shaft power and the mechanical efficiency of the pump become W& pump,shaft = η motor W& electric = (0.90)(35 kW) = 31.5 kW
η pump =
W& pump, u W&
pump, shaft
=
26.3 kW = 0.836 = 83.6% 31.5 kW
Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor efficiencies, which is 0.9×0.836 = 0.75.
2-36
2-78E Water is pumped from a lake to a nearby pool by a pump with specified power and efficiency. The mechanical power used to overcome frictional effects is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the lake and the free surface of the pool is constant. 3 The average flow velocity is constant since pipe diameter is constant. Properties We take the density of water to be ρ = 62.4 lbm/ft3. Analysis The useful mechanical pumping power delivered to water is W& pump, u = η pump W& pump = (0.73)(12 hp) = 8.76 hp
The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. That is,
Pool
2
Pump
35 ft Lake
1
∆E& mech = m& ∆emech = m& ∆pe = m& g∆z = ρV&g∆z
Substituting, the rate of change of mechanical energy of water becomes 1 lbf 1 hp ∆E& mech = (62.4 lbm/ft 3 )(1.2 ft 3 /s)(32.2 ft/s 2 )(35 ft ) = 4.76 hp 2 ⋅ 550 lbf ft/s 32.2 lbm ft/s ⋅
Then the mechanical power lost in piping because of frictional effects becomes W&frict = W& pump, u − ∆E& mech = 8.76 − 4.76 hp = 4.0 hp Discussion Note that the pump must supply to the water an additional useful mechanical power of 4.0 hp to overcome the frictional losses in pipes.
2-37
Energy and Environment 2-79C Energy conversion pollutes the soil, the water, and the air, and the environmental pollution is a serious threat to vegetation, wild life, and human health. The emissions emitted during the combustion of fossil fuels are responsible for smog, acid rain, and global warming and climate change. The primary chemicals that pollute the air are hydrocarbons (HC, also referred to as volatile organic compounds, VOC), nitrogen oxides (NOx), and carbon monoxide (CO). The primary source of these pollutants is the motor vehicles. 2-80C Smog is the brown haze that builds up in a large stagnant air mass, and hangs over populated areas on calm hot summer days. Smog is made up mostly of ground-level ozone (O3), but it also contains numerous other chemicals, including carbon monoxide (CO), particulate matter such as soot and dust, volatile organic compounds (VOC) such as benzene, butane, and other hydrocarbons. Ground-level ozone is formed when hydrocarbons and nitrogen oxides react in the presence of sunlight in hot calm days. Ozone irritates eyes and damage the air sacs in the lungs where oxygen and carbon dioxide are exchanged, causing eventual hardening of this soft and spongy tissue. It also causes shortness of breath, wheezing, fatigue, headaches, nausea, and aggravate respiratory problems such as asthma. 2-81C Fossil fuels include small amounts of sulfur. The sulfur in the fuel reacts with oxygen to form sulfur dioxide (SO2), which is an air pollutant. The sulfur oxides and nitric oxides react with water vapor and other chemicals high in the atmosphere in the presence of sunlight to form sulfuric and nitric acids. The acids formed usually dissolve in the suspended water droplets in clouds or fog. These acid-laden droplets are washed from the air on to the soil by rain or snow. This is known as acid rain. It is called “rain” since it comes down with rain droplets. As a result of acid rain, many lakes and rivers in industrial areas have become too acidic for fish to grow. Forests in those areas also experience a slow death due to absorbing the acids through their leaves, needles, and roots. Even marble structures deteriorate due to acid rain. 2-82C Carbon dioxide (CO2), water vapor, and trace amounts of some other gases such as methane and nitrogen oxides act like a blanket and keep the earth warm at night by blocking the heat radiated from the earth. This is known as the greenhouse effect. The greenhouse effect makes life on earth possible by keeping the earth warm. But excessive amounts of these gases disturb the delicate balance by trapping too much energy, which causes the average temperature of the earth to rise and the climate at some localities to change. These undesirable consequences of the greenhouse effect are referred to as global warming or global climate change. The greenhouse effect can be reduced by reducing the net production of CO2 by consuming less energy (for example, by buying energy efficient cars and appliances) and planting trees. 2-83C Carbon monoxide, which is a colorless, odorless, poisonous gas that deprives the body's organs from getting enough oxygen by binding with the red blood cells that would otherwise carry oxygen. At low levels, carbon monoxide decreases the amount of oxygen supplied to the brain and other organs and muscles, slows body reactions and reflexes, and impairs judgment. It poses a serious threat to people with heart disease because of the fragile condition of the circulatory system and to fetuses because of the oxygen needs of the developing brain. At high levels, it can be fatal, as evidenced by numerous deaths caused by cars that are warmed up in closed garages or by exhaust gases leaking into the cars.
2-38
2-84E A person trades in his Ford Taurus for a Ford Explorer. The extra amount of CO2 emitted by the Explorer within 5 years is to be determined. Assumptions The Explorer is assumed to use 940 gallons of gasoline a year compared to 715 gallons for Taurus. Analysis The extra amount of gasoline the Explorer will use within 5 years is Extra Gasoline
= (Extra per year)(No. of years) = (940 – 715 gal/yr)(5 yr) = 1125 gal
Extra CO2 produced
= (Extra gallons of gasoline used)(CO2 emission per gallon) = (1125 gal)(19.7 lbm/gal) = 22,163 lbm CO2
Discussion Note that the car we choose to drive has a significant effect on the amount of greenhouse gases produced.
2-85 A power plant that burns natural gas produces 0.59 kg of carbon dioxide (CO2) per kWh. The amount of CO2 production that is due to the refrigerators in a city is to be determined. Assumptions The city uses electricity produced by a natural gas power plant. Properties 0.59 kg of CO2 is produced per kWh of electricity generated (given). Analysis Noting that there are 200,000 households in the city and each household consumes 700 kWh of electricity for refrigeration, the total amount of CO2 produced is Amount of CO 2 produced = (Amount of electricity consumed)(Amount of CO 2 per kWh) = (200,000 household)(700 kWh/year household)(0.59 kg/kWh) = 8.26 × 107 CO 2 kg/year = 82,600 CO2 ton/year
Therefore, the refrigerators in this city are responsible for the production of 82,600 tons of CO2.
2-86 A power plant that burns coal, produces 1.1 kg of carbon dioxide (CO2) per kWh. The amount of CO2 production that is due to the refrigerators in a city is to be determined. Assumptions power plant.
The city uses electricity produced by a coal
Properties 1.1 kg of CO2 is produced per kWh of electricity generated (given). Analysis Noting that there are 200,000 households in the city and each household consumes 700 kWh of electricity for refrigeration, the total amount of CO2 produced is Amount of CO 2 produced = ( Amount of electricity consumed)(Amount of CO 2 per kWh) = (200,000 household)(700 kWh/household)(1.1 kg/kWh) = 15.4 × 10 7 CO 2 kg/year = 154,000 CO 2 ton/year
Therefore, the refrigerators in this city are responsible for the production of 154,000 tons of CO2.
2-39
2-87E A household uses fuel oil for heating, and electricity for other energy needs. Now the household reduces its energy use by 20%. The reduction in the CO2 production this household is responsible for is to be determined. Properties The amount of CO2 produced is 1.54 lbm per kWh and 26.4 lbm per gallon of fuel oil (given). Analysis Noting that this household consumes 11,000 kWh of electricity and 1500 gallons of fuel oil per year, the amount of CO2 production this household is responsible for is Amount of CO 2 produced = (Amount of electricity consumed)(Amount of CO 2 per kWh) + (Amount of fuel oil consumed)(Amount of CO 2 per gallon) = (11,000 kWh/yr)(1.54 lbm/kWh) + (1500 gal/yr)(26.4 lbm/gal) = 56,540 CO 2 lbm/year
Then reducing the electricity and fuel oil usage by 15% will reduce the annual amount of CO2 production by this household by Reduction in CO 2 produced = (0.15)(Current amount of CO 2 production) = (0.15)(56,540 CO 2 kg/year) = 8481 CO 2 lbm/year
Therefore, any measure that saves energy also reduces the amount of pollution emitted to the environment.
2-88 A household has 2 cars, a natural gas furnace for heating, and uses electricity for other energy needs. The annual amount of NOx emission to the atmosphere this household is responsible for is to be determined. Properties The amount of NOx produced is 7.1 g per kWh, 4.3 g per therm of natural gas, and 11 kg per car (given). Analysis Noting that this household has 2 cars, consumes 1200 therms of natural gas, and 9,000 kWh of electricity per year, the amount of NOx production this household is responsible for is Amount of NO x produced = ( No. of cars)(Amount of NO x produced per car) + ( Amount of electricity consumed)(Amount of NO x per kWh) + ( Amount of gas consumed)(Amount of NO x per gallon) = (2 cars)(11 kg/car) + (9000 kWh/yr)(0.0071 kg/kWh) + (1200 therms/yr)(0.0043 kg/therm) = 91.06 NOx kg/year
Discussion Any measure that saves energy will also reduce the amount of pollution emitted to the atmosphere.
2-40
Special Topic: Mechanisms of Heat Transfer 2-89C The three mechanisms of heat transfer are conduction, convection, and radiation. 2-90C No. It is purely by radiation. 2-91C Diamond has a higher thermal conductivity than silver, and thus diamond is a better conductor of heat. 2-92C In forced convection, the fluid is forced to move by external means such as a fan, pump, or the wind. The fluid motion in natural convection is due to buoyancy effects only. 2-93C Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a blackbody at the same temperature. Absorptivity is the fraction of radiation incident on a surface that is absorbed by the surface. The Kirchhoff's law of radiation states that the emissivity and the absorptivity of a surface are equal at the same temperature and wavelength. 2-94C A blackbody is an idealized body that emits the maximum amount of radiation at a given temperature, and that absorbs all the radiation incident on it. Real bodies emit and absorb less radiation than a blackbody at the same temperature.
2-95 The inner and outer surfaces of a brick wall are maintained at specified temperatures. The rate of heat transfer through the wall is to be determined. Assumptions 1 Steady operating conditions exist since the surface Brick temperatures of the wall remain constant at the specified values. 2 Thermal properties of the wall are constant. Q=? Properties The thermal conductivity of the wall is given to be k = 0.69 W/m⋅°C. 30 cm Analysis Under steady conditions, the rate of heat transfer through the wall is 5°C 20°C T ∆ (20 − 5) ° C = (0.69 W/m ⋅ °C)(5 × 6 m 2 ) = 1035 W Q& cond = kA L 0.3 m
2-96 The inner and outer surfaces of a window glass are maintained at specified temperatures. The amount of heat transferred through the glass in 5 h is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Thermal properties of the glass are constant. Properties The thermal conductivity of the glass is given to be k = 0.78 W/m⋅°C. Glass Analysis Under steady conditions, the rate of heat transfer through the glass by conduction is ∆T (10 − 3)°C = (0.78 W/m ⋅ °C)(2 × 2 m 2 ) = 4368 W Q& cond = kA L 0.005 m Then the amount of heat transferred over a period of 5 h becomes Q = Q& ∆t = (4.368 kJ/s)(5 × 3600s) = 78,600 kJ
10°C
3°C
cond
If the thickness of the glass is doubled to 1 cm, then the amount of heat transferred will go down by half to 39,300 kJ.
0.5 cm
2-41
2-97 EES Reconsider Prob. 2-96. Using EES (or other) software, investigate the effect of glass thickness on heat loss for the specified glass surface temperatures. Let the glass thickness vary from 0.2 cm to 2 cm. Plot the heat loss versus the glass thickness, and discuss the results. Analysis The problem is solved using EES, and the solution is given below. FUNCTION klookup(material$) If material$='Glass' then klookup:=0.78 If material$='Brick' then klookup:=0.72 If material$='Fiber Glass' then klookup:=0.043 If material$='Air' then klookup:=0.026 If material$='Wood(oak)' then klookup:=0.17 END L=2"[m]" W=2"[m]" {material$='Glass' T_in=10"[C]" T_out=3"[C]" k=0.78"[W/m-C]" t=5"[hr]" thickness=0.5"[cm]"} k=klookup(material$)"[W/m-K]" A=L*W"[m^2]" Q_dot_loss=A*k*(T_in-T_out)/(thickness*convert(cm,m))"[W]" Q_loss_total=Q_dot_loss*t*convert(hr,s)*convert(J,kJ)"[kJ]" Qloss,total [kJ] 196560 98280 65520 49140 39312 32760 28080 24570 21840 19656
Thickness [cm] 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
200000
Heat loss through glass "w all" in 5 hours
Q loss,total [kJ]
160000
120000
80000
40000
0 0.2
0.4
0.6
0.8
1
1.2
1.4
thickness [cm ]
1.6
1.8
2
2-42
2-98 Heat is transferred steadily to boiling water in the pan through its bottom. The inner surface temperature of the bottom of the pan is given. The temperature of the outer surface is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the pan remain constant at the specified values. 2 Thermal properties of the aluminum pan are constant. Properties The thermal conductivity of the aluminum is given to be k = 237 W/m⋅°C. Analysis The heat transfer surface area is A = π r² = π(0.1 m)² = 0.0314 m² Under steady conditions, the rate of heat transfer through the bottom of the pan by conduction is ∆T T −T 105°C Q& = kA = kA 2 1 L L T − 105o C Substituting, 500 W = (237 W / m⋅o C)(0.0314 m2 ) 2 500 W 0.4 cm 0.004 m which gives
T2 = 105.3°C
2-99 A person is standing in a room at a specified temperature. The rate of heat transfer between a person and the surrounding Q& air by convection is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 The environment is at Ts =34°C a uniform temperature. Analysis The heat transfer surface area of the person is A = πDL = π(0.3 m)(1.70 m) = 1.60 m² Under steady conditions, the rate of heat transfer by convection is Q& conv = hA∆T = (15 W/m 2 ⋅ °C)(1.60 m 2 )(34 − 20)°C = 336 W
2-100 A spherical ball whose surface is maintained at a temperature of 70°C is suspended in the middle of a room at 20°C. The total rate of heat transfer from the ball is to be determined. Assumptions 1 Steady operating conditions exist since the ball surface and the surrounding air and surfaces remain at constant Air temperatures. 2 The thermal properties of the ball and the 20°C convection heat transfer coefficient are constant and uniform. 70°C Properties The emissivity of the ball surface is given to be ε = 0.8. Analysis The heat transfer surface area is . A = πD² = 3.14x(0.05 m)² = 0.007854 m² Q D = 5 cm Under steady conditions, the rates of convection and radiation heat transfer are Q& = hA∆T = (15 W/m 2 ⋅o C)(0.007854 m 2 )(70 − 20)o C = 5.89 W conv
Q& rad = εσA(Ts4 − To4 ) = 0.8(0.007854 m 2 )(5.67 × 10−8 W/m 2 ⋅ K 4 )[(343 K) 4 − (293 K) 4 ] = 2.31 W
Therefore, Q&
total
= Q& conv + Q& rad = 5.89 + 2.31 = 8.20 W
2-43
2-101 EES Reconsider Prob. 2-100. Using EES (or other) software, investigate the effect of the convection heat transfer coefficient and surface emissivity on the heat transfer rate from the ball. Let the heat transfer coefficient vary from 5 W/m2.°C to 30 W/m2.°C. Plot the rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 0.1, 0.5, 0.8, and 1, and discuss the results. Analysis The problem is solved using EES, and the solution is given below. sigma=5.67e-8"[W/m^2-K^4]" {T_sphere=70"[C]" T_room=20"[C]" D_sphere=5"[cm]" epsilon=0.1 h_c=15"[W/m^2-K]"} A=4*pi*(D_sphere/2)^2*convert(cm^2,m^2)"[m^2]" Q_dot_conv=A*h_c*(T_sphere-T_room)"[W]" Q_dot_rad=A*epsilon*sigma*((T_sphere+273)^4-(T_room+273)^4)"[W]" Q_dot_total=Q_dot_conv+Q_dot_rad"[W]" Qtotal [W] 2.252 4.215 6.179 8.142 10.11 12.07
hc [W/m2-K] 5 10 15 20 25 30
15 13
] W [
l at ot
Q
11
ε = 1.0
9 7
ε = 0.1
5 3 1 5
10
15
20
hc [W/m^2-K]
25
30
2-44
2-102 Hot air is blown over a flat surface at a specified temperature. The rate of heat transfer from the air to the plate is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by 80°C radiation is not considered. 3 The convection heat transfer coefficient Air is constant and uniform over the surface. Analysis Under steady conditions, the rate of heat transfer by convection is 30°C = hA∆T = (55 W/m 2 ⋅ °C)(2 × 4 m 2 )(80 − 30)o C = 22,000 W = 22 kW Q& conv
2-103 A 1000-W iron is left on the iron board with its base exposed to the air at 20°C. The temperature of the base of the iron is to be determined in steady operation. Assumptions 1 Steady operating conditions exist. 2 The Iron thermal properties of the iron base and the convection heat 1000 W transfer coefficient are constant and uniform. 3 The temperature of the surrounding surfaces is the same as the temperature of the surrounding air. Properties The emissivity of the base surface is given to be ε = 0.6. Analysis At steady conditions, the 1000 W of energy supplied to the iron will be dissipated to the surroundings by convection and radiation heat transfer. Therefore, = Q& + Q& = 1000 W Q& total
where
conv
rad
Q& conv = hA∆T = (35 W/m 2 ⋅ K)(0.02 m 2 )(Ts − 293 K) = 0.7(Ts − 293 K) W
and Q& rad = εσA(Ts4 − To4 ) = 0.6(0.02 m 2 )(5.67 × 10 −8 W / m 2 ⋅ K 4 )[Ts4 − (293 K) 4 ] = 0.06804 × 10 −8 [Ts4 − (293 K) 4 ] W
Substituting,
1000 W = 0.7(Ts − 293 K ) + 0.06804 × 10 −8 [Ts4 − (293 K) 4 ]
Solving by trial and error gives
Ts = 947 K = 674°C
Discussion We note that the iron will dissipate all the energy it receives by convection and radiation when its surface temperature reaches 947 K. 2-104 The backside of the thin metal plate is insulated and the front side is exposed to solar radiation. The surface temperature of the plate is to be determined when it stabilizes. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the insulated side of the plate is negligible. 3 The heat transfer coefficient is constant and uniform over the plate. 4 Heat loss by radiation is negligible. Properties The solar absorptivity of the plate is given to be α = 0.6. Analysis When the heat loss from the plate by convection equals the solar radiation absorbed, the surface temperature of the plate can be determined 700 W/m2 from Q& = Q& solar absorbed
conv
αQ& solar = hA(Ts − To ) 2
2 o
0.6 × A × 700W/m = (50W/m ⋅ C) A(Ts − 25)
α = 0.6 25°C
2-45
2-105 EES Reconsider Prob. 2-104. Using EES (or other) software, investigate the effect of the convection heat transfer coefficient on the surface temperature of the plate. Let the heat transfer coefficient vary from 10 W/m2.°C to 90 W/m2.°C. Plot the surface temperature against the convection heat transfer coefficient,
2-46
2-106 A hot water pipe at 80°C is losing heat to the surrounding air at 5°C by natural convection with a heat transfer coefficient of 25 W/ m2.°C. The rate of heat loss from the pipe by convection is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 The convection heat transfer coefficient is constant and uniform over the surface.
80°C
Analysis The heat transfer surface area is A = (πD)L = 3.14x(0.05 m)(10 m) = 1.571 m² Under steady conditions, the rate of heat transfer by convection is
D = 5 cm L = 10 m
Q Air, 5°C
Q& conv = hA∆T = (25 W/m 2 ⋅ °C)(1.571 m 2 )(80 − 5)°C = 2945 W = 2.95 kW
2-107 A spacecraft in space absorbs solar radiation while losing heat to deep space by thermal radiation. The surface temperature of the spacecraft is to be determined when steady conditions are reached.. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal properties of the spacecraft are constant. Properties The outer surface of a spacecraft has an emissivity of 0.8 and an absorptivity of 0.3. Analysis When the heat loss from the outer surface of the spacecraft by radiation equals the solar radiation absorbed, the surface temperature can be determined from Q& solar absorbed = Q& rad 4 αQ& solar = εσA(Ts4 − Tspace )
0.3 × A × (1000 W/m 2 ) = 0.8 × A × (5.67 × 10−8 W/m 2 ⋅ K 4 )[Ts4 − (0 K) 4 ]
Canceling the surface area A and solving for Ts gives Ts = 285 K
1000 W/m2 α = 0.3 ε = 0.8
2-47
2-108 EES Reconsider Prob. 2-107. Using EES (or other) software, investigate the effect of the surface emissivity and absorptivity of the spacecraft on the equilibrium surface temperature. Plot the surface temperature against emissivity for solar absorptivities of 0.1, 0.5, 0.8, and 1, and discuss the results. Analysis The problem is solved using EES, and the solution is given below. "Knowns" sigma=5.67e-8"[W/m^2-K^4]" "The following variables are obtained from the Diagram Window." {T_space=10"[C]" S=1000"[W/m^2]" alpha_solar=0.3 epsilon=0.8} "Solution" "An energy balance on the spacecraft gives:" Q_dot_solar=Q_dot_out "The absorbed solar" Q_dot_solar =S*alpha_solar "The net leaving radiation leaving the spacecraft:" Q_dot_out=epsilon*sigma*((T_spacecraft+273)^4-(T_space+273)^4) ε
140
Tspacecraft [C] 218.7 150 117.2 97.2 83.41 73.25 65.4 59.13 54 49.71
T spacecraft [C]
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Surface em issivity = 0.8 120 100 80 60 40 20 0.1
0.2
0.3
0.4
0.5
α
solar
225
solar absorptivity = 0.3
T spacecraft [C]
185
145
105
65
25 0.1
0.2
0.3
0.4
0.5
ε
0.6
0.7
0.8
0.9
0.6
1
0.7
0.8
0.9
1
2-48
2-109 A hollow spherical iron container is filled with iced water at 0°C. The rate of heat loss from the sphere and the rate at which ice melts in the container are to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Heat transfer through the shell is one-dimensional. 3 Thermal properties of the iron shell are constant. 4 The inner surface of the shell is at the same temperature as the iced water, 0°C. Properties The thermal conductivity of iron is k = 80.2 W/m⋅°C (Table 2-3). The heat of fusion of water is at 1 atm is 333.7 kJ/kg. 5°C Analysis This spherical shell can be approximated as a plate of thickness 0.4 cm and surface area A = πD² = 3.14×(0.2 m)² = 0.126 m² Then the rate of heat transfer through the shell by conduction is ∆T (5 − 0)°C Q& cond = kA = (80.2 W/m⋅o C)(0.126 m 2 ) = 12,632 W L 0.004 m
0.4 cm
Iced water 0°C
Considering that it takes 333.7 kJ of energy to melt 1 kg of ice at 0°C, the rate at which ice melts in the container can be determined from Q& 12.632 kJ/s m& ice = = = 0.038 kg/s hif 333.7 kJ/kg Discussion We should point out that this result is slightly in error for approximating a curved wall as a plain wall. The error in this case is very small because of the large diameter to thickness ratio. For better accuracy, we could use the inner surface area (D = 19.2 cm) or the mean surface area (D = 19.6 cm) in the calculations.
2-110 The inner and outer glasses of a double pane window with a 1-cm air space are at specified temperatures. The rate of heat transfer through the window is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Heat transfer through the window is one-dimensional. 3 Thermal properties of the air are constant. 4 The air trapped between the two glasses is still, and thus heat transfer is by conduction only. Properties The thermal conductivity of air at room temperature is k = 0.026 W/m.°C (Table 2-3). Analysis Under steady conditions, the rate of heat transfer through the window by conduction is Q& cond
∆T (18 − 6)o C = kA = (0.026 W/m⋅o C)(2 × 2 m 2 ) 0.01 m L = 125 W = 0.125 kW
18°C
Air ·
Q
1cm
6°C
2-49
2-111 Two surfaces of a flat plate are maintained at specified temperatures, and the rate of heat transfer through the plate is measured. The thermal conductivity of the plate material is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the plate remain constant at the specified values. 2 Heat transfer through the plate is one-dimensional. 3 Thermal properties of the plate are constant.
Plate 2 cm 0°C
100°C Analysis The thermal conductivity is determined directly from the steady one-dimensional heat conduction relation to be T −T (Q& / A) L (500 W/m 2 )(0.02 m) Q& = kA 1 2 → k = = = 0.1 W/m.°C L T1 − T2 (100 - 0)°C
500 W/m2
Review Problems
2-112 The weight of the cabin of an elevator is balanced by a counterweight. The power needed when the fully loaded cabin is rising, and when the empty cabin is descending at a constant speed are to be determined. Assumptions 1 The weight of the cables is negligible. 2 The guide rails and pulleys are frictionless. 3 Air drag is negligible. Analysis (a) When the cabin is fully loaded, half of the weight is balanced by the counterweight. The power required to raise the cabin at a constant speed of 1.2 m/s is 1N 1 kW mgz = 4.71 kW = mgV = (400 kg )(9.81 m/s 2 )(1.2 m/s) W& = 2 ∆t 1 kg ⋅ m/s 1000 N ⋅ m/s
If no counterweight is used, the mass would double to 800 kg and the power would be 2×4.71 = 9.42 kW. (b) When the empty cabin is descending (and the counterweight is ascending) there is mass imbalance of 400150 = 250 kg. The power required to raise this mass at a constant speed of 1.2 m/s is
Counter weight
1N mgz 1 kW = 2.94 kW W& = = mgV = (250 kg )(9.81 m/s 2 )(1.2 m/s) 2 ∆t 1 kg ⋅ m/s 1000 N ⋅ m/s
If a friction force of 800 N develops between the cabin and the guide rails, we will need F z 1 kW = 0.96 kW W&friction = friction = FfrictionV = (800 N )(1.2 m/s ) ∆t 1000 N ⋅ m/s
of additional power to combat friction which always acts in the opposite direction to motion. Therefore, the total power needed in this case is W& total = W& + W& friction = 2.94 + 0.96 = 3.90 kW
Cabin
2-50
2-113 A decision is to be made between a cheaper but inefficient natural gas heater and an expensive but efficient natural gas heater for a house. Assumptions The two heaters are comparable in all aspects other than the initial cost and efficiency. Analysis Other things being equal, the logical choice is the heater that will cost less during its lifetime. The total cost of a system during its lifetime (the initial, operation, maintenance, etc.) can be determined by performing a life cycle cost analysis. A simpler alternative is to determine the simple payback period. The annual heating cost is given to be $1200. Noting that the existing heater is 55% efficient, only 55% of that energy (and thus money) is delivered to the house, and the rest is wasted due to the inefficiency of the heater. Therefore, the monetary value of the heating load of the house is Cost of useful heat = (55%)(Current annual heating cost) = 0.55×($1200/yr)=$660/yr
Gas Heater η1 = 82% η2 = 95%
This is how much it would cost to heat this house with a heater that is 100% efficient. For heaters that are less efficient, the annual heating cost is determined by dividing $660 by the efficiency: 82% heater:
Annual cost of heating = (Cost of useful heat)/Efficiency = ($660/yr)/0.82 = $805/yr
95% heater:
Annual cost of heating = (Cost of useful heat)/Efficiency = ($660/yr)/0.95 = $695/yr
Annual cost savings with the efficient heater = 805 - 695 = $110 Excess initial cost of the efficient heater = 2700 - 1600 = $1100 The simple payback period becomes Simple payback period =
Excess initial cost $1100 = = 10 years Annaul cost savings $110 / yr
Therefore, the more efficient heater will pay for the $1100 cost differential in this case in 10 years, which is more than the 8-year limit. Therefore, the purchase of the cheaper and less efficient heater is a better buy in this case.
2-51
2-114 A wind turbine is rotating at 20 rpm under steady winds of 30 km/h. The power produced, the tip speed of the blade, and the revenue generated by the wind turbine per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 The wind turbine operates continuously during the entire year at the specified conditions. Properties The density of air is given to be ρ = 1.20 kg/m3. Analysis (a) The blade span area and the mass flow rate of air through the turbine are A = πD 2 / 4 = π (80 m) 2 / 4 = 5027 m 2 1000 m 1 h V = (30 km/h) = 8.333 m/s 1 km 3600 s m& = ρAV = (1.2 kg/m 3 )(5027 m 2 )(8.333 m/s) = 50,270 kg/s
Noting that the kinetic energy of a unit mass is V2/2 and the wind turbine captures 35% of this energy, the power generated by this wind turbine becomes 1 1 kJ/kg 1 W& = η m& V 2 = (0.35) (50,270 kg/s )(8.333 m/s ) 2 = 610.9 kW 2 2 1000 m 2 /s 2
(b) Noting that the tip of blade travels a distance of πD per revolution, the tip velocity of the turbine blade for an rpm of n& becomes V tip = πDn& = π (80 m)(20 / min) = 5027 m/min = 83.8 m/s = 302 km/h
(c) The amount of electricity produced and the revenue generated per year are Electricity produced = W& ∆t = (610.9 kW)(365 × 24 h/year) = 5.351× 10 6 kWh/year Revenue generated = (Electricity produced)(Unit price) = (5.351× 10 6 kWh/year)($0.06/kWh) = $321,100/year
2-52
2-115 A wind turbine is rotating at 20 rpm under steady winds of 25 km/h. The power produced, the tip speed of the blade, and the revenue generated by the wind turbine per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 The wind turbine operates continuously during the entire year at the specified conditions. Properties The density of air is given to be ρ = 1.20 kg/m3. Analysis (a) The blade span area and the mass flow rate of air through the turbine are A = πD 2 / 4 = π (80 m)2 / 4 = 5027 m 2 1000 m 1 h V = (25 km/h) = 6.944 m/s 1 km 3600 s m& = ρAV = (1.2 kg/m3 )(5027 m 2 )(6.944 m/s) = 41,891 kg/s
Noting that the kinetic energy of a unit mass is V2/2 and the wind turbine captures 35% of this energy, the power generated by this wind turbine becomes 1 1 1 kJ/kg W& = η m& V 2 = (0.35) (41,891 kg/s)(6.944 m/s) 2 = 353.5 kW 2 2 2 2 1000 m /s
(b) Noting that the tip of blade travels a distance of πD per revolution, the tip velocity of the turbine blade for an rpm of n& becomes Vtip = πDn& = π (80 m)(20 / min) = 5027 m/min = 83.8 m/s = 302 km/h
(c) The amount of electricity produced and the revenue generated per year are Electricity produced = W& ∆t = (353.5 kW)(365 × 24 h/year) = 3,096,660 kWh/year Revenue generated = (Electricity produced)(Unit price) = (3,096,660 kWh/year)($0.06/kWh) = $185,800/year
2-53
2-116E The energy contents, unit costs, and typical conversion efficiencies of various energy sources for use in water heaters are given. The lowest cost energy source is to be determined. Assumptions The differences in installation costs of different water heaters are not considered. Properties The energy contents, unit costs, and typical conversion efficiencies of different systems are given in the problem statement. Analysis The unit cost of each Btu of useful energy supplied to the water heater by each system can be determined from Unit cost of useful energy =
Unit cost of energy supplied Conversion efficiency
Substituting, 1 ft 3 −6 1025 Btu = $21.3 × 10 / Btu
Natural gas heater:
Unit cost of useful energy =
$0.012/ft 3 0.55
Heating by oil heater:
Unit cost of useful energy =
$1.15/gal 1 gal = $15.1× 10 − 6 / Btu 0.55 138,700 Btu
Electric heater:
Unit cost of useful energy =
$0.084/kWh) 1 kWh −6 = $27.4 × 10 / Btu 0.90 3412 Btu
Therefore, the lowest cost energy source for hot water heaters in this case is oil.
2-117 A home owner is considering three different heating systems for heating his house. The system with the lowest energy cost is to be determined. Assumptions The differences in installation costs of different heating systems are not considered. Properties The energy contents, unit costs, and typical conversion efficiencies of different systems are given in the problem statement. Analysis The unit cost of each Btu of useful energy supplied to the house by each system can be determined from Unit cost of useful energy =
Unit cost of energy supplied Conversion efficiency
Substituting, Natural gas heater:
Unit cost of useful energy =
$1.24/therm 1 therm = $13.5 × 10 −6 / kJ 0.87 105,500 kJ
Heating oil heater:
Unit cost of useful energy =
$1.25/gal 1 gal = $10.4 × 10 − 6 / kJ 0.87 138,500 kJ
Electric heater:
Unit cost of useful energy =
$0.09/kWh) 1 kWh −6 = $25.0 × 10 / kJ 1.0 3600 kJ
Therefore, the system with the lowest energy cost for heating the house is the heating oil heater.
2-54
2-118 The heating and cooling costs of a poorly insulated house can be reduced by up to 30 percent by adding adequate insulation. The time it will take for the added insulation to pay for itself from the energy it saves is to be determined. Assumptions It is given that the annual energy usage of a house is $1200 a year, and 46% of it is used for heating and cooling. The cost of added insulation is given to be $200.
Heat loss
Analysis The amount of money that would be saved per year is determined directly from Money saved = ($1200 / year)(0.46)(0.30) = $166 / yr
Then the simple payback period becomes Payback period =
House
Cost $200 = = 1.2 yr Money saved $166/yr
Therefore, the proposed measure will pay for itself in less than one and a half year.
2-119 Caulking and weather-stripping doors and windows to reduce air leaks can reduce the energy use of a house by up to 10 percent. The time it will take for the caulking and weather-stripping to pay for itself from the energy it saves is to be determined. Assumptions It is given that the annual energy usage of a house is $1100 a year, and the cost of caulking and weather-stripping a house is $50. Analysis The amount of money that would be saved per year is determined directly from Money saved = ($1100 / year)(0.10) = $110 / yr
Then the simple payback period becomes Payback period =
Cost $50 = = 0.45 yr Money saved $110/yr
Therefore, the proposed measure will pay for itself in less than half a year.
2-120 It is estimated that 570,000 barrels of oil would be saved per day if the thermostat setting in residences in winter were lowered by 6°F (3.3°C). The amount of money that would be saved per year is to be determined. Assumptions The average heating season is given to be 180 days, and the cost of oil to be $40/barrel. Analysis The amount of money that would be saved per year is determined directly from (570,000 barrel/day)(180 days/year)($40/barrel) = $4,104,000 ,000
Therefore, the proposed measure will save more than 4-billion dollars a year in energy costs.
2-55
2-121 A TV set is kept on a specified number of hours per day. The cost of electricity this TV set consumes per month is to be determined. Assumptions 1 The month is 30 days. 2 The TV set consumes its rated power when on. Analysis The total number of hours the TV is on per month is Operating hours = (6 h/day)(30 days) = 180 h Then the amount of electricity consumed per month and its cost become Amount of electricity = (Power consumed)(Operating hours)=(0.120 kW)(180 h) =21.6 kWh Cost of electricity = (Amount of electricity)(Unit cost) = (21.6 kWh)($0.08/kWh) = $1.73 (per month) Properties Note that an ordinary TV consumes more electricity that a large light bulb, and there should be a conscious effort to turn it off when not in use to save energy.
2-122 The pump of a water distribution system is pumping water at a specified flow rate. The pressure rise of water in the pump is measured, and the motor efficiency is specified. The mechanical efficiency of the pump is to be determined. Assumptions 1 The flow is steady. 2 The elevation difference across the pump is negligible. 3 Water is incompressible. Analysis From the definition of motor efficiency, the mechanical (shaft) power delivered by the he motor is
15 kW
PUMP
W& pump,shaft = η motor W& electric = (0.90 )(15 kW) = 13.5 kW
Motor
Pump To determine the mechanical efficiency of the pump, we need to inlet know the increase in the mechanical energy of the fluid as it flows through the pump, which is ∆E& mech,fluid = m& (e mech,out − e mech,in ) = m& [( Pv ) 2 − ( Pv ) 1 ] = m& ( P2 − P1 )v = V& ( P2 − P1 ) 1 kJ = (0.050 m 3 /s)(300 - 100 kPa) = 10 kJ/s = 10 kW 1 kPa ⋅ m 3
since m& = ρV& = V& / v and there is no change in kinetic and potential energies of the fluid. Then the pump efficiency becomes
η pump =
∆E& mech,fluid 10 kW = = 0.741 or 74.1% & W pump, shaft 13.5 kW
Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor efficiencies, which is 0.9×0.741 = 0.667.
2-56
2-123 The available head, flow rate, and efficiency of a hydroelectric turbine are given. The electric power output is to be determined. Assumptions 1 The flow is steady. 2 Water levels at the reservoir and the discharge site remain constant. 3 Frictional losses in piping are negligible. Properties We take the density of water to be ρ = 1000 kg/m3 = 1 kg/L. Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate.
1
120 m
1 kJ/kg emech = pe = gz = (9.81 m/s 2 )(120 m) = 1.177 kJ/kg 1000 m 2 /s 2
ηoverall = 80%
Generator
Turbine
2
The mass flow rate is m& = ρV& = (1000 kg/m 3 )(100 m 3 /s) = 200,000 kg/s
Then the maximum and actual electric power generation become 1 MW W&max = E& mech = m& emech = (100,000 kg/s)(1.177 kJ/kg) = 117.7 MW 1000 kJ/s W& =η W& = 0.80(117.7 MW) = 94.2 MW electric
overall
max
Discussion Note that the power generation would increase by more than 1 MW for each percentage point improvement in the efficiency of the turbine–generator unit.
2-57
2-124 An entrepreneur is to build a large reservoir above the lake level, and pump water from the lake to the reservoir at night using cheap power, and let the water flow from the reservoir back to the lake during the day, producing power. The potential revenue this system can generate per year is to be determined. Assumptions 1 The flow in each direction is steady and incompressible. 2 The elevation difference between the lake and the reservoir can be taken to be constant, and the elevation change of reservoir during charging and discharging is disregarded. 3 Frictional losses in piping are negligible. 4 The system operates every day of the year for 10 hours in each mode. Properties We take the density of water to be ρ = 1000 kg/m3. Analysis The total mechanical energy of water in an upper reservoir relative to water in a lower reservoir is equivalent to the potential energy of water at the free surface of this reservoir relative to free surface of the lower reservoir. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and m& gz for a given mass flow rate. This also represents the minimum power required to pump water from the lower reservoir to the higher reservoir.
2 Reservoir
Pumpturbine
40 m Lake
1
W& max, turbine = W& min, pump = W& ideal = ∆E& mech = m& ∆e mech = m& ∆pe = m& g∆z = ρV&g∆z
1N = (1000 kg/m 3 )(2 m 3 /s)(9.81 m/s 2 )(40 m) ⋅ m/s 2 1 kg
1 kW 1000 N ⋅ m/s = 784.8 kW
The actual pump and turbine electric powers are
W& pump, elect =
W& ideal
η pump-motor
=
784.8 kW = 1046 kW 0.75
W& turbine = η turbine-genW& ideal = 0.75(784.8 kW) = 588.6 kW Then the power consumption cost of the pump, the revenue generated by the turbine, and the net income (revenue minus cost) per year become Cost = W& pump, elect ∆t × Unit price = (1046 kW)(365 × 10 h/year)($0.03/kWh) = $114,500/year Reveue = W& turbine ∆t × Unit price = (588.6 kW)(365 × 10 h/year)($0.08/kWh) = $171,900/year Net income = Revenue – Cost = 171,900 –114,500 = $57,400/year Discussion It appears that this pump-turbine system has a potential to generate net revenues of about $57,000 per year. A decision on such a system will depend on the initial cost of the system, its life, the operating and maintenance costs, the interest rate, and the length of the contract period, among other things.
2-58
2-125 A diesel engine burning light diesel fuel that contains sulfur is considered. The rate of sulfur that ends up in the exhaust and the rate of sulfurous acid given off to the environment are to be determined. Assumptions 1 All of the sulfur in the fuel ends up in the exhaust. 2 For one kmol of sulfur in the exhaust, one kmol of sulfurous acid is added to the environment. Properties The molar mass of sulfur is 32 kg/kmol. Analysis The mass flow rates of fuel and the sulfur in the exhaust are m& fuel =
m& air (336 kg air/h) = = 18.67 kg fuel/h AF (18 kg air/kg fuel)
m& Sulfur = (750 × 10 -6 )m& fuel = (750 × 10 -6 )(18.67 kg/h) = 0.014 kg/h
The rate of sulfurous acid given off to the environment is m& H2SO3 =
M H2SO3 2 × 1 + 32 + 3 × 16 m& Sulfur = (0.014 kg/h) = 0.036 kg/h M Sulfur 32
Discussion This problem shows why the sulfur percentage in diesel fuel must be below certain value to satisfy regulations.
2-126 Lead is a very toxic engine emission. Leaded gasoline contains lead that ends up in the exhaust. The amount of lead put out to the atmosphere per year for a given city is to be determined. Assumptions 35% of lead is exhausted to the environment. Analysis The gasoline consumption and the lead emission are Gasoline Consumption = (10,000 cars)(15,000 km/car - year)(10 L/100 km) = 1.5 × 107 L/year Lead Emission = (GaolineConsumption )mlead f lead = (1.5 × 107 L/year)(0.15 × 10-3 kg/L)(0.35) = 788 kg/year
Discussion Note that a huge amount of lead emission is avoided by the use of unleaded gasoline.
2-59
Fundamentals of Engineering (FE) Exam Problems
2-127 A 2-kW electric resistance heater in a room is turned on and kept on for 30 min. The amount of energy transferred to the room by the heater is (a) 1 kJ (b) 60 kJ (c) 1800 kJ (d) 3600 kJ (e) 7200 kJ
Answer (d) 3600 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). We= 2 "kJ/s" time=30*60 "s" We_total=We*time "kJ" "Some Wrong Solutions with Common Mistakes:" W1_Etotal=We*time/60 "using minutes instead of s" W2_Etotal=We "ignoring time"
2-128 In a hot summer day, the air in a well-sealed room is circulated by a 0.50-hp (shaft) fan driven by a 65% efficient motor. (Note that the motor delivers 0.50 hp of net shaft power to the fan). The rate of energy supply from the fan-motor assembly to the room is (a) 0.769 kJ/s (b) 0.325 kJ/s (c) 0.574 kJ/s (d) 0.373 kJ/s (e) 0.242 kJ/s
Answer (c) 0.574 kJ/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Eff=0.65 W_fan=0.50*0.7457 "kW" E=W_fan/Eff "kJ/s" "Some Wrong Solutions with Common Mistakes:" W1_E=W_fan*Eff "Multiplying by efficiency" W2_E=W_fan "Ignoring efficiency" W3_E=W_fan/Eff/0.7457 "Using hp instead of kW"
2-60 2-129 A fan is to accelerate quiescent air to a velocity to 12 m/s at a rate of 3 m3/min. If the density of air is 1.15 kg/m3, the minimum power that must be supplied to the fan is (a) 248 W (b) 72 W (c) 497 W (d) 216 W (e) 162 W
Answer (a) 248 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1.15 V=12 Vdot=3 "m3/s" mdot=rho*Vdot "kg/s" We=mdot*V^2/2 "Some Wrong Solutions with Common Mistakes:" W1_We=Vdot*V^2/2 "Using volume flow rate" W2_We=mdot*V^2 "forgetting the 2" W3_We=V^2/2 "not using mass flow rate"
2-130 A 900-kg car cruising at a constant speed of 60 km/h is to accelerate to 100 km/h in 6 s. The additional power needed to achieve this acceleration is (a) 41 kW (b) 222 kW (c) 1.7 kW (d) 26 kW (e) 37 kW
Answer (e) 37 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=900 "kg" V1=60 "km/h" V2=100 "km/h" Dt=6 "s" Wa=m*((V2/3.6)^2-(V1/3.6)^2)/2000/Dt "kW" "Some Wrong Solutions with Common Mistakes:" W1_Wa=((V2/3.6)^2-(V1/3.6)^2)/2/Dt "Not using mass" W2_Wa=m*((V2)^2-(V1)^2)/2000/Dt "Not using conversion factor" W3_Wa=m*((V2/3.6)^2-(V1/3.6)^2)/2000 "Not using time interval" W4_Wa=m*((V2/3.6)-(V1/3.6))/1000/Dt "Using velocities"
2-61
2-131 The elevator of a large building is to raise a net mass of 400 kg at a constant speed of 12 m/s using an electric motor. Minimum power rating of the motor should be (a) 0 kW (b) 4.8 kW (c) 47 kW (d) 12 kW (e) 36 kW
Answer (c) 47 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=400 "kg" V=12 "m/s" g=9.81 "m/s2" Wg=m*g*V/1000 "kW" "Some Wrong Solutions with Common Mistakes:" W1_Wg=m*V "Not using g" W2_Wg=m*g*V^2/2000 "Using kinetic energy" W3_Wg=m*g/V "Using wrong relation"
2-132 Electric power is to be generated in a hydroelectric power plant that receives water at a rate of 70 m3/s from an elevation of 65 m using a turbine–generator with an efficiency of 85 percent. When frictional losses in piping are disregarded, the electric power output of this plant is (a) 3.9 MW (b) 38 MW (c) 45 MW (d) 53 MW (e) 65 MW
Answer (b) 38 MW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Vdot=70 "m3/s" z=65 "m" g=9.81 "m/s2" Eff=0.85 rho=1000 "kg/m3" We=rho*Vdot*g*z*Eff/10^6 "MW" "Some Wrong Solutions with Common Mistakes:" W1_We=rho*Vdot*z*Eff/10^6 "Not using g" W2_We=rho*Vdot*g*z/Eff/10^6 "Dividing by efficiency" W3_We=rho*Vdot*g*z/10^6 "Not using efficiency"
2-62
2-133 A 75 hp (shaft) compressor in a facility that operates at full load for 2500 hours a year is powered by an electric motor that has an efficiency of 88 percent. If the unit cost of electricity is $0.06/kWh, the annual electricity cost of this compressor is (a) $7382 (b) $9900 (c) $12,780 (d) $9533 (e) $8389
Answer (d) $9533 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Wcomp=75 "hp" Hours=2500 “h/year” Eff=0.88 price=0.06 “$/kWh” We=Wcomp*0.7457*Hours/Eff Cost=We*price "Some Wrong Solutions with Common Mistakes:" W1_cost= Wcomp*0.7457*Hours*price*Eff “multiplying by efficiency” W2_cost= Wcomp*Hours*price/Eff “not using conversion” W3_cost= Wcomp*Hours*price*Eff “multiplying by efficiency and not using conversion” W4_cost= Wcomp*0.7457*Hours*price “Not using efficiency”
2-134 Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarter of the time and the unit cost of electricity is $0.09/kWh, the electricity cost of this refrigerator per month (30 days) is (a) $3.56 (b) $5.18 (c) $8.54 (d) $9.28 (e) $20.74
Answer (b) $5.18 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). We=0.320 "kW" Hours=0.25*(24*30) "h/year" price=0.09 "$/kWh" Cost=We*hours*price "Some Wrong Solutions with Common Mistakes:" W1_cost= We*24*30*price "running continuously"
2-63 2-135 A 2-kW pump is used to pump kerosene (ρ = 0.820 kg/L) from a tank on the ground to a tank at a higher elevation. Both tanks are open to the atmosphere, and the elevation difference between the free surfaces of the tanks is 30 m. The maximum volume flow rate of kerosene is (a) 8.3 L/s (b) 7.2 L/s (c) 6.8 L/s (d) 12.1 L/s (e) 17.8 L/s
Answer (a) 8.3 L/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). W=2 "kW" rho=0.820 "kg/L" z=30 "m" g=9.81 "m/s2" W=rho*Vdot*g*z/1000 "Some Wrong Solutions with Common Mistakes:" W=W1_Vdot*g*z/1000 "Not using density"
2-136 A glycerin pump is powered by a 5-kW electric motor. The pressure differential between the outlet and the inlet of the pump at full load is measured to be 211 kPa. If the flow rate through the pump is 18 L/s and the changes in elevation and the flow velocity across the pump are negligible, the overall efficiency of the pump is (a) 69% (b) 72% (c) 76% (d) 79% (e) 82%
Answer (c) 76% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). We=5 "kW" Vdot= 0.018 "m3/s" DP=211 "kPa" Emech=Vdot*DP Emech=Eff*We
2-64
The following problems are based on the optional special topic of heat transfer 2-137 A 10-cm high and 20-cm wide circuit board houses on its surface 100 closely spaced chips, each generating heat at a rate of 0.08 W and transferring it by convection to the surrounding air at 40°C. Heat transfer from the back surface of the board is negligible. If the convection heat transfer coefficient on the surface of the board is 10 W/m2.°C and radiation heat transfer is negligible, the average surface temperature of the chips is (a) 80°C (b) 54°C (c) 41°C (d) 72°C (e) 60°C
Answer (a) 80°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=0.10*0.20 "m^2" Q= 100*0.08 "W" Tair=40 "C" h=10 "W/m^2.C" Q= h*A*(Ts-Tair) "W" "Some Wrong Solutions with Common Mistakes:" Q= h*(W1_Ts-Tair) "Not using area" Q= h*2*A*(W2_Ts-Tair) "Using both sides of surfaces" Q= h*A*(W3_Ts+Tair) "Adding temperatures instead of subtracting" Q/100= h*A*(W4_Ts-Tair) "Considering 1 chip only"
2-138 A 50-cm-long, 0.2-cm-diameter electric resistance wire submerged in water is used to determine the boiling heat transfer coefficient in water at 1 atm experimentally. The surface temperature of the wire is measured to be 130°C when a wattmeter indicates the electric power consumption to be 4.1 kW. Then the heat transfer coefficient is (b) 137 W/m2.°C (c) 68,330 W/m2.°C (d) 10,038 W/m2.°C (a) 43,500 W/m2.°C 2 (e) 37,540 W/m .°C
Answer (a) 43,500 W/m2.°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). L=0.5 "m" D=0.002 "m" A=pi*D*L "m^2" We=4.1 "kW" Ts=130 "C" Tf=100 "C (Boiling temperature of water at 1 atm)" We= h*A*(Ts-Tf) "W" "Some Wrong Solutions with Common Mistakes:" We= W1_h*(Ts-Tf) "Not using area" We= W2_h*(L*pi*D^2/4)*(Ts-Tf) "Using volume instead of area" We= W3_h*A*Ts "Using Ts instead of temp difference"
2-65
2-139 A 3-m2 hot black surface at 80°C is losing heat to the surrounding air at 25°C by convection with a convection heat transfer coefficient of 12 W/m2.°C, and by radiation to the surrounding surfaces at 15°C. The total rate of heat loss from the surface is (a) 1987 W (b) 2239 W (c) 2348 W (d) 3451 W (e) 3811 W
Answer (d) 3451 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). sigma=5.67E-8 "W/m^2.K^4" eps=1 A=3 "m^2" h_conv=12 "W/m^2.C" Ts=80 "C" Tf=25 "C" Tsurr=15 "C" Q_conv=h_conv*A*(Ts-Tf) "W" Q_rad=eps*sigma*A*((Ts+273)^4-(Tsurr+273)^4) "W" Q_total=Q_conv+Q_rad "W" "Some Wrong Solutions with Common Mistakes:" W1_Ql=Q_conv "Ignoring radiation" W2_Q=Q_rad "ignoring convection" W3_Q=Q_conv+eps*sigma*A*(Ts^4-Tsurr^4) "Using C in radiation calculations" W4_Q=Q_total/A "not using area"
2-140 Heat is transferred steadily through a 0.2-m thick 8 m by 4 m wall at a rate of 1.6 kW. The inner and outer surface temperatures of the wall are measured to be 15°C to 5°C. The average thermal conductivity of the wall is (a) 0.001 W/m.°C (b) 0.5 W/m.°C (c) 1.0 W/m.°C (d) 2.0 W/m.°C (e) 5.0 W/m.°C
Answer (c) 1.0 W/m.°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=8*4 "m^2" L=0.2 "m" T1=15 "C" T2=5 "C" Q=1600 "W" Q=k*A*(T1-T2)/L "W" "Some Wrong Solutions with Common Mistakes:" Q=W1_k*(T1-T2)/L "Not using area" Q=W2_k*2*A*(T1-T2)/L "Using areas of both surfaces" Q=W3_k*A*(T1+T2)/L "Adding temperatures instead of subtracting" Q=W4_k*A*L*(T1-T2) "Multiplying by thickness instead of dividing by it"
2-66
2-141 The roof of an electrically heated house is 7 m long, 10 m wide, and 0.25 m thick. It is made of a flat layer of concrete whose thermal conductivity is 0.92 W/m.°C. During a certain winter night, the temperatures of the inner and outer surfaces of the roof are measured to be 15°C and 4°C, respectively. The average rate of heat loss through the roof that night was (a) 41 W (b) 177 W (c) 4894 W (d) 5567 W (e) 2834 W
Answer (e) 2834 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=7*10 "m^2" L=0.25 "m" k=0.92 "W/m.C" T1=15 "C" T2=4 "C" Q_cond=k*A*(T1-T2)/L "W" "Some Wrong Solutions with Common Mistakes:" W1_Q=k*(T1-T2)/L "Not using area" W2_Q=k*2*A*(T1-T2)/L "Using areas of both surfaces" W3_Q=k*A*(T1+T2)/L "Adding temperatures instead of subtracting" W4_Q=k*A*L*(T1-T2) "Multiplying by thickness instead of dividing by it"
2-142 … 2-148 Design and Essay Problems
KJ
3-1
Chapter 3 PROPERTIES OF PURE SUBSTANCES Pure Substances, Phase Change Processes, Property Diagrams 3-1C Yes. Because it has the same chemical composition throughout. 3-2C A liquid that is about to vaporize is saturated liquid; otherwise it is compressed liquid. 3-3C A vapor that is about to condense is saturated vapor; otherwise it is superheated vapor. 3-4C No. 3-5C No. 3-6C Yes. The saturation temperature of a pure substance depends on pressure. The higher the pressure, the higher the saturation or boiling temperature. 3-7C The temperature will also increase since the boiling or saturation temperature of a pure substance depends on pressure. 3-8C Because one cannot be varied while holding the other constant. In other words, when one changes, so does the other one. 3-9C At critical point the saturated liquid and the saturated vapor states are identical. At triple point the three phases of a pure substance coexist in equilibrium. 3-10C Yes. 3-11C Case (c) when the pan is covered with a heavy lid. Because the heavier the lid, the greater the pressure in the pan, and thus the greater the cooking temperature. 3-12C At supercritical pressures, there is no distinct phase change process. The liquid uniformly and gradually expands into a vapor. At subcritical pressures, there is always a distinct surface between the phases.
Property Tables 3-13C A given volume of water will boil at a higher temperature in a tall and narrow pot since the pressure at the bottom (and thus the corresponding saturation pressure) will be higher in that case. 3-14C A perfectly fitting pot and its lid often stick after cooking as a result of the vacuum created inside as the temperature and thus the corresponding saturation pressure inside the pan drops. An easy way of removing the lid is to reheat the food. When the temperature rises to boiling level, the pressure rises to atmospheric value and thus the lid will come right off.
3-2
3-15C The molar mass of gasoline (C8H18) is 114 kg/kmol, which is much larger than the molar mass of air that is 29 kg/kmol. Therefore, the gasoline vapor will settle down instead of rising even if it is at a much higher temperature than the surrounding air. As a result, the warm mixture of air and gasoline on top of an open gasoline will most likely settle down instead of rising in a cooler environment 3-16C Ice can be made by evacuating the air in a water tank. During evacuation, vapor is also thrown out, and thus the vapor pressure in the tank drops, causing a difference between the vapor pressures at the water surface and in the tank. This pressure difference is the driving force of vaporization, and forces the liquid to evaporate. But the liquid must absorb the heat of vaporization before it can vaporize, and it absorbs it from the liquid and the air in the neighborhood, causing the temperature in the tank to drop. The process continues until water starts freezing. The process can be made more efficient by insulating the tank well so that the entire heat of vaporization comes essentially from the water. 3-17C Yes. Otherwise we can create energy by alternately vaporizing and condensing a substance. 3-18C No. Because in the thermodynamic analysis we deal with the changes in properties; and the changes are independent of the selected reference state. 3-19C The term hfg represents the amount of energy needed to vaporize a unit mass of saturated liquid at a specified temperature or pressure. It can be determined from hfg = hg - hf . 3-20C Yes; the higher the temperature the lower the hfg value. 3-21C Quality is the fraction of vapor in a saturated liquid-vapor mixture. It has no meaning in the superheated vapor region. 3-22C Completely vaporizing 1 kg of saturated liquid at 1 atm pressure since the higher the pressure, the lower the hfg . 3-23C Yes. It decreases with increasing pressure and becomes zero at the critical pressure. 3-24C No. Quality is a mass ratio, and it is not identical to the volume ratio. 3-25C The compressed liquid can be approximated as a saturated liquid at the given temperature. Thus vT ,P ≅ v f @ T .
3-26 [Also solved by EES on enclosed CD] Complete the following table for H2 O: T, °C 50 120.21 250 110
P, kPa 12.352 200 400 600
v, m3 / kg 4.16 0.8858 0.5952 0.001051
Phase description Saturated mixture Saturated vapor Superheated vapor Compressed liquid
3-3
3-27 EES Problem 3-26 is reconsidered. The missing properties of water are to be determined using EES, and the solution is to be repeated for refrigerant-134a, refrigerant-22, and ammonia. Analysis The problem is solved using EES, and the solution is given below. $Warning off {$Arrays off} Procedure Find(Fluid$,Prop1$,Prop2$,Value1,Value2:T,p,h,s,v,u,x,State$) "Due to the very general nature of this problem, a large number of 'if-then-else' statements are necessary." If Prop1$='Temperature, C' Then T=Value1 If Prop2$='Temperature, C' then Call Error('Both properties cannot be Temperature, T=xxxF2',T) if Prop2$='Pressure, kPa' then p=value2 h=enthalpy(Fluid$,T=T,P=p) s=entropy(Fluid$,T=T,P=p) v=volume(Fluid$,T=T,P=p) u=intenergy(Fluid$,T=T,P=p) x=quality(Fluid$,T=T,P=p) endif if Prop2$='Enthalpy, kJ/kg' then h=value2 p=Pressure(Fluid$,T=T,h=h) s=entropy(Fluid$,T=T,h=h) v=volume(Fluid$,T=T,h=h) u=intenergy(Fluid$,T=T,h=h) x=quality(Fluid$,T=T,h=h) endif if Prop2$='Entropy, kJ/kg-K' then s=value2 p=Pressure(Fluid$,T=T,s=s) h=enthalpy(Fluid$,T=T,s=s) v=volume(Fluid$,T=T,s=s) u=intenergy(Fluid$,T=T,s=s) x=quality(Fluid$,T=T,s=s) endif if Prop2$='Volume, m^3/kg' then v=value2 p=Pressure(Fluid$,T=T,v=v) h=enthalpy(Fluid$,T=T,v=v) s=entropy(Fluid$,T=T,v=v) u=intenergy(Fluid$,T=T,v=v) x=quality(Fluid$,T=T,v=v) endif if Prop2$='Internal Energy, kJ/kg' then u=value2 p=Pressure(Fluid$,T=T,u=u) h=enthalpy(Fluid$,T=T,u=u) s=entropy(Fluid$,T=T,u=u) v=volume(Fluid$,T=T,s=s) x=quality(Fluid$,T=T,u=u) endif if Prop2$='Quality' then x=value2
3-4
p=Pressure(Fluid$,T=T,x=x) h=enthalpy(Fluid$,T=T,x=x) s=entropy(Fluid$,T=T,x=x) v=volume(Fluid$,T=T,x=x) u=IntEnergy(Fluid$,T=T,x=x) endif Endif If Prop1$='Pressure, kPa' Then p=Value1 If Prop2$='Pressure, kPa' then Call Error('Both properties cannot be Pressure, p=xxxF2',p) if Prop2$='Temperature, C' then T=value2 h=enthalpy(Fluid$,T=T,P=p) s=entropy(Fluid$,T=T,P=p) v=volume(Fluid$,T=T,P=p) u=intenergy(Fluid$,T=T,P=p) x=quality(Fluid$,T=T,P=p) endif if Prop2$='Enthalpy, kJ/kg' then h=value2 T=Temperature(Fluid$,p=p,h=h) s=entropy(Fluid$,p=p,h=h) v=volume(Fluid$,p=p,h=h) u=intenergy(Fluid$,p=p,h=h) x=quality(Fluid$,p=p,h=h) endif if Prop2$='Entropy, kJ/kg-K' then s=value2 T=Temperature(Fluid$,p=p,s=s) h=enthalpy(Fluid$,p=p,s=s) v=volume(Fluid$,p=p,s=s) u=intenergy(Fluid$,p=p,s=s) x=quality(Fluid$,p=p,s=s) endif if Prop2$='Volume, m^3/kg' then v=value2 T=Temperature(Fluid$,p=p,v=v) h=enthalpy(Fluid$,p=p,v=v) s=entropy(Fluid$,p=p,v=v) u=intenergy(Fluid$,p=p,v=v) x=quality(Fluid$,p=p,v=v) endif if Prop2$='Internal Energy, kJ/kg' then u=value2 T=Temperature(Fluid$,p=p,u=u) h=enthalpy(Fluid$,p=p,u=u) s=entropy(Fluid$,p=p,u=u) v=volume(Fluid$,p=p,s=s) x=quality(Fluid$,p=p,u=u) endif if Prop2$='Quality' then x=value2 T=Temperature(Fluid$,p=p,x=x) h=enthalpy(Fluid$,p=p,x=x) s=entropy(Fluid$,p=p,x=x) v=volume(Fluid$,p=p,x=x)
3-5
u=IntEnergy(Fluid$,p=p,x=x) endif Endif If Prop1$='Enthalpy, kJ/kg' Then h=Value1 If Prop2$='Enthalpy, kJ/kg' then Call Error('Both properties cannot be Enthalpy, h=xxxF2',h) if Prop2$='Pressure, kPa' then p=value2 T=Temperature(Fluid$,h=h,P=p) s=entropy(Fluid$,h=h,P=p) v=volume(Fluid$,h=h,P=p) u=intenergy(Fluid$,h=h,P=p) x=quality(Fluid$,h=h,P=p) endif if Prop2$='Temperature, C' then T=value2 p=Pressure(Fluid$,T=T,h=h) s=entropy(Fluid$,T=T,h=h) v=volume(Fluid$,T=T,h=h) u=intenergy(Fluid$,T=T,h=h) x=quality(Fluid$,T=T,h=h) endif if Prop2$='Entropy, kJ/kg-K' then s=value2 p=Pressure(Fluid$,h=h,s=s) T=Temperature(Fluid$,h=h,s=s) v=volume(Fluid$,h=h,s=s) u=intenergy(Fluid$,h=h,s=s) x=quality(Fluid$,h=h,s=s) endif if Prop2$='Volume, m^3/kg' then v=value2 p=Pressure(Fluid$,h=h,v=v) T=Temperature(Fluid$,h=h,v=v) s=entropy(Fluid$,h=h,v=v) u=intenergy(Fluid$,h=h,v=v) x=quality(Fluid$,h=h,v=v) endif if Prop2$='Internal Energy, kJ/kg' then u=value2 p=Pressure(Fluid$,h=h,u=u) T=Temperature(Fluid$,h=h,u=u) s=entropy(Fluid$,h=h,u=u) v=volume(Fluid$,h=h,s=s) x=quality(Fluid$,h=h,u=u) endif if Prop2$='Quality' then x=value2 p=Pressure(Fluid$,h=h,x=x) T=Temperature(Fluid$,h=h,x=x) s=entropy(Fluid$,h=h,x=x) v=volume(Fluid$,h=h,x=x) u=IntEnergy(Fluid$,h=h,x=x) endif endif If Prop1$='Entropy, kJ/kg-K' Then
3-6
s=Value1 If Prop2$='Entropy, kJ/kg-K' then Call Error('Both properties cannot be Entrolpy, h=xxxF2',s) if Prop2$='Pressure, kPa' then p=value2 T=Temperature(Fluid$,s=s,P=p) h=enthalpy(Fluid$,s=s,P=p) v=volume(Fluid$,s=s,P=p) u=intenergy(Fluid$,s=s,P=p) x=quality(Fluid$,s=s,P=p) endif if Prop2$='Temperature, C' then T=value2 p=Pressure(Fluid$,T=T,s=s) h=enthalpy(Fluid$,T=T,s=s) v=volume(Fluid$,T=T,s=s) u=intenergy(Fluid$,T=T,s=s) x=quality(Fluid$,T=T,s=s) endif if Prop2$='Enthalpy, kJ/kg' then h=value2 p=Pressure(Fluid$,h=h,s=s) T=Temperature(Fluid$,h=h,s=s) v=volume(Fluid$,h=h,s=s) u=intenergy(Fluid$,h=h,s=s) x=quality(Fluid$,h=h,s=s) endif if Prop2$='Volume, m^3/kg' then v=value2 p=Pressure(Fluid$,s=s,v=v) T=Temperature(Fluid$,s=s,v=v) h=enthalpy(Fluid$,s=s,v=v) u=intenergy(Fluid$,s=s,v=v) x=quality(Fluid$,s=s,v=v) endif if Prop2$='Internal Energy, kJ/kg' then u=value2 p=Pressure(Fluid$,s=s,u=u) T=Temperature(Fluid$,s=s,u=u) h=enthalpy(Fluid$,s=s,u=u) v=volume(Fluid$,s=s,s=s) x=quality(Fluid$,s=s,u=u) endif if Prop2$='Quality' then x=value2 p=Pressure(Fluid$,s=s,x=x) T=Temperature(Fluid$,s=s,x=x) h=enthalpy(Fluid$,s=s,x=x) v=volume(Fluid$,s=s,x=x) u=IntEnergy(Fluid$,s=s,x=x) endif Endif if x1 then State$='in the superheated region.' If (x0) then State$='in the two-phase region.' If (x=1) then State$='a saturated vapor.' if (x=0) then State$='a saturated liquid.'
3-7
end "Input from the diagram window" {Fluid$='Steam' Prop1$='Temperature' Prop2$='Pressure' Value1=50 value2=101.3} Call Find(Fluid$,Prop1$,Prop2$,Value1,Value2:T,p,h,s,v,u,x,State$) T[1]=T ; p[1]=p ; h[1]=h ; s[1]=s ; v[1]=v ; u[1]=u ; x[1]=x "Array variables were used so the states can be plotted on property plots." ARRAYS TABLE h KJ/kg 2964.5
P kPa 400
s kJ/kgK 7.3804
T C 250
u KJ/kg 2726.4
x
v m3/kg 0.5952
100
Steam
700 600 500 400
] C [ T
300
8600 kPa 2600 kPa
200
500 kPa
100 0 0,0
45 kPa
1,0
2,0
3,0
4,0
5,0
6,0
7,0
8,0
9,0
10,0
s [kJ/kg-K]
Steam
700 600 500
] C [ T
400 300
8600 kPa 2600 kPa
200
500 kPa
100 0 10-4
45 kPa
10-3
10-2
10-1
100
v [m3/kg]
101
102
103
3-8
Steam
105
104 250 C
103
] a P k[ P
170 C
110 C
102
75 C
101
100 10-3
10-2
10-1
100
101
102
v [m3/kg]
Steam
105
104 250 C
103
] a P k[ P
170 C
110 C
102
75 C
101
100 0
500
1000
1500
2000
2500
3000
h [kJ/kg]
Steam
4000
8600 kPa
] g k/ J k[ h
3500
2600 kPa
3000
500 kPa
2500
45 kPa
2000 1500 1000 500 0 0,0
1,0
2,0
3,0
4,0
5,0
6,0
s [kJ/kg-K]
7,0
8,0
9,0
10,0
3-9
3-28E Complete the following table for H2 O: P, psia T, °F 300 67.03 40 267.22 500 120 400 400
u, Btu / lbm 782 236.02 1174.4 373.84
Phase description Saturated mixture Saturated liquid Superheated vapor Compressed liquid
3-29E EES Problem 3-28E is reconsidered. The missing properties of water are to be determined using EES, and the solution is to be repeated for refrigerant-134a, refrigerant-22, and ammonia. Analysis The problem is solved using EES, and the solution is given below. "Given" T[1]=300 [F] u[1]=782 [Btu/lbm] P[2]=40 [psia] x[2]=0 T[3]=500 [F] P[3]=120 [psia] T[4]=400 [F] P[4]=420 [psia] "Analysis" Fluid$='steam_iapws' P[1]=pressure(Fluid$, T=T[1], u=u[1]) x[1]=quality(Fluid$, T=T[1], u=u[1]) T[2]=temperature(Fluid$, P=P[2], x=x[2]) u[2]=intenergy(Fluid$, P=P[2], x=x[2]) u[3]=intenergy(Fluid$, P=P[3], T=T[3]) x[3]=quality(Fluid$, P=P[3], T=T[3]) u[4]=intenergy(Fluid$, P=P[4], T=T[4]) x[4]=quality(Fluid$, P=P[4], T=T[4]) "x = 100 for superheated vapor and x = -100 for compressed liquid" Solution for steam T, ºF P, psia 300 67.028 267.2 40 500 120 400 400
x 0.6173 0 100 -100
u, Btu/lbm 782 236 1174 373.8
3-30 Complete the following table for H2 O: T, °C 120.21 140 177.66 80 350.0
P, kPa 200 361.53 950 500 800
h, kJ / kg 2045.8 1800 752.74 335.37 3162.2
x 0.7 0.565 0.0 -----
Phase description Saturated mixture Saturated mixture Saturated liquid Compressed liquid Superheated vapor
3-10
3-31 Complete the following table for Refrigerant-134a: T, °C -8 30 -12.73 80
v, m3 / kg
P, kPa 320 770.64 180 600
0.0007569 0.015 0.11041 0.044710
Phase description Compressed liquid Saturated mixture Saturated vapor Superheated vapor
3-32 Complete the following table for Refrigerant-134a: T, °C 20 -12 86.24 8
P, kPa 572.07 185.37 400 600
u, kJ / kg 95 35.78 300 62.26
Phase description Saturated mixture Saturated liquid Superheated vapor Compressed liquid
3-33E Complete the following table for Refrigerant-134a: T, °F 65.89 15 10 160 110
P, psia 80 29.759 70 180 161.16
h, Btu / lbm 78 69.92 15.35 129.46 117.23
x 0.566 0.6 ----1.0
Phase description Saturated mixture Saturated mixture Compressed liquid Superheated vapor Saturated vapor
3-34 Complete the following table for H2 O: T, °C 140 155.46 125 500
P, kPa 361.53 550 750 2500
v, m3 / kg 0.05 0.001097 0.001065 0.140
Phase description Saturated mixture Saturated liquid Compressed liquid Superheated vapor
u, kJ / kg 1450 2601.3 805.15 3040
Phase description Saturated mixture Saturated vapor Compressed liquid Superheated vapor
3-35 Complete the following table for H2 O: T, °C 143.61 220 190 466.21
P, kPa 400 2319.6 2500 4000
3-11
3-36 A rigid tank contains steam at a specified state. The pressure, quality, and density of steam are to be determined. Properties At 220°C vf = 0.001190 m3/kg and vg = 0.08609 m3/kg (Table A-4). Analysis (a) Two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. The pressure of the steam is the saturation pressure at the given temperature. Then the pressure in the tank must be the saturation pressure at the specified temperature, P = Tsat @220°C = 2320 kPa
(b) The total mass and the quality are determined as
Vf 1/3 × (1.8 m3 ) mf = = = 504.2 kg v f 0.001190 m3/kg mg =
Steam 1.8 m3 220°C
Vg 2/3 × (1.8 m3 ) = = 13.94 kg v g 0.08609 m3/kg
mt = m f + mg = 504.2 + 13.94 = 518.1 kg x=
mg mt
=
13.94 = 0.0269 518.1
(c) The density is determined from
v = v f + x(v g − v f ) = 0.001190 + (0.0269)(0.08609) = 0.003474 m 3 /kg ρ=
1
v
=
1 = 287.8 kg/m 3 0.003474
3-37 A piston-cylinder device contains R-134a at a specified state. Heat is transferred to R-134a. The final pressure, the volume change of the cylinder, and the enthalpy change are to be determined. Analysis (a) The final pressure is equal to the initial pressure, which is determined from P2 = P1 = Patm +
mp g
πD 2 /4
= 88 kPa +
(12 kg)(9.81 m/s 2 ) 1 kN = 90.4 kPa 2 2 π (0.25 m) /4 1000 kg.m/s
(b) The specific volume and enthalpy of R-134a at the initial state of 90.4 kPa and -10°C and at the final state of 90.4 kPa and 15°C are (from EES)
v1 = 0.2302 m3/kg v 2 = 0.2544 m3/kg
h1 = 247.76 kJ/kg h2 = 268.16 kJ/kg
The initial and the final volumes and the volume change are
V1 = mv 1 = (0.85 kg)(0.2302 m 3 /kg) = 0.1957 m 3 V 2 = mv 2 = (0.85 kg)(0.2544 m 3 /kg) = 0.2162 m 3 ∆V = 0.2162 − 0.1957 = 0.0205 m 3
(c) The total enthalpy change is determined from ∆H = m(h2 − h1 ) = (0.85 kg)(268.16 − 247.76) kJ/kg = 17.4 kJ/kg
R-134a 0.85 kg -10°C
Q
3-12
3-38E The temperature in a pressure cooker during cooking at sea level is measured to be 250°F. The absolute pressure inside the cooker and the effect of elevation on the answer are to be determined. Assumptions Properties of pure water can be used to approximate the properties of juicy water in the cooker. Properties The saturation pressure of water at 250°F is 29.84 psia (Table A-4E). The standard atmospheric pressure at sea level is 1 atm = 14.7 psia. Analysis The absolute pressure in the cooker is simply the saturation pressure at the cooking temperature, Pabs = Psat@250°F = 29.84 psia
It is equivalent to
H2O 250°F
1 atm = 2.03 atm Pabs = 29.84 psia 14.7 psia
The elevation has no effect on the absolute pressure inside when the temperature is maintained constant at 250°F.
3-39E The local atmospheric pressure, and thus the boiling temperature, changes with the weather conditions. The change in the boiling temperature corresponding to a change of 0.3 in of mercury in atmospheric pressure is to be determined. Properties The saturation pressures of water at 200 and 212°F are 11.538 and 14.709 psia, respectively (Table A-4E). One in. of mercury is equivalent to 1 inHg = 3.387 kPa = 0.491 psia (inner cover page). Analysis A change of 0.3 in of mercury in atmospheric pressure corresponds to 0.491 psia = 0.147 psia ∆P = (0.3 inHg) 1 inHg
P±0.3 inHg
At about boiling temperature, the change in boiling temperature per 1 psia change in pressure is determined using data at 200 and 212°F to be (212 − 200)°F ∆T = = 3.783 °F/psia ∆P (14.709 − 11.538) psia
Then the change in saturation (boiling) temperature corresponding to a change of 0.147 psia becomes ∆Tboiling = (3.783 °F/psia)∆P = (3.783 °F/psia)(0.147 psia) = 0.56°F
which is very small. Therefore, the effect of variation of atmospheric pressure on the boiling temperature is negligible.
3-13
3-40 A person cooks a meal in a pot that is covered with a well-fitting lid, and leaves the food to cool to the room temperature. It is to be determined if the lid will open or the pan will move up together with the lid when the person attempts to open the pan by lifting the lid up. Assumptions 1 The local atmospheric pressure is 1 atm = 101.325 kPa. 2 The weight of the lid is small and thus its effect on the boiling pressure and temperature is negligible. 3 No air has leaked into the pan during cooling. Properties The saturation pressure of water at 20°C is 2.3392 kPa (Table A-4). Analysis Noting that the weight of the lid is negligible, the reaction force F on the lid after cooling at the pan-lid interface can be determined from a force balance on the lid in the vertical direction to be PA +F = PatmA or, F = A( Patm − P) = (πD 2 / 4)( Patm − P ) =
π (0.3 m) 2
P
(101,325 − 2339.2) Pa
4 = 6997 m 2 Pa = 6997 N (since 1 Pa = 1 N/m 2 )
2.3392 kPa
Patm = 1 atm
The weight of the pan and its contents is W = mg = (8 kg)(9.81 m/s 2 ) = 78.5 N
which is much less than the reaction force of 6997 N at the pan-lid interface. Therefore, the pan will move up together with the lid when the person attempts to open the pan by lifting the lid up. In fact, it looks like the lid will not open even if the mass of the pan and its contents is several hundred kg.
3-41 Water is boiled at sea level (1 atm pressure) in a pan placed on top of a 3-kW electric burner that transfers 60% of the heat generated to the water. The rate of evaporation of water is to be determined. Properties The properties of water at 1 atm and thus at the saturation temperature of 100°C are hfg = 2256.4 kJ/kg (Table A-4). Analysis The net rate of heat transfer to the water is Q& = 0.60 × 3 kW = 1.8 kW
Noting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, the rate of evaporation of water is determined to be Q& 1.8 kJ/s m& evaporation = = = 0.80 × 10 −3 kg/s = 2.872 kg/h hfg 2256.4 kJ/kg
H2O 100°C
3-14
3-42 Water is boiled at 1500 m (84.5 kPa pressure) in a pan placed on top of a 3-kW electric burner that transfers 60% of the heat generated to the water. The rate of evaporation of water is to be determined. Properties The properties of water at 84.5 kPa and thus at the saturation temperature of 95°C are hfg = 2269.6 kJ/kg (Table A-4). Analysis The net rate of heat transfer to the water is Q& = 0.60 × 3 kW = 1.8 kW
H2O 95°C
Noting that it takes 2269.6 kJ of energy to vaporize 1 kg of saturated liquid water, the rate of evaporation of water is determined to be 1.8 kJ/s Q& = = 0.793 × 10− 3 kg/s = 2.855 kg/h m& evaporation = h fg 2269.6 kJ/kg
3-43 Water is boiled at 1 atm pressure in a pan placed on an electric burner. The water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined. Properties The properties of water at 1 atm and thus at a saturation temperature of Tsat = 100°C are hfg = 2256.5 kJ/kg and vf = 0.001043 m3/kg (Table A-4). Analysis The rate of evaporation of water is mevap = m& evap =
Vevap (πD 2 / 4) L [π (0.25 m)2 / 4](0.10 m) = = = 4.704 kg 0.001043 vf vf mevap ∆t
=
H2O 1 atm
4.704 kg = 0.001742 kg/s 45 × 60 s
Then the rate of heat transfer to water becomes Q& = m& evap h fg = (0.001742 kg/s)(2256.5 kJ/kg) = 3.93 kW
3-44 Water is boiled at a location where the atmospheric pressure is 79.5 kPa in a pan placed on an electric burner. The water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined. Properties The properties of water at 79.5 kPa are Tsat = 93.3°C, hfg = 2273.9 kJ/kg and vf = 0.001038 m3/kg (Table A-5). Analysis The rate of evaporation of water is m evap = m& evap =
V evap vf m evap ∆t
= =
(πD 2 / 4) L
vf
[π (0.25 m) 2 / 4](0.10 m) = = 4.727 kg 0.001038
4.727 kg = 0.001751 kg/s 45 × 60 s
Then the rate of heat transfer to water becomes Q& = m& evap h fg = (0.001751 kg/s)(2273.9 kJ/kg) = 3.98 kW
H2O 79.5 kPa
3-15
3-45 Saturated steam at Tsat = 30°C condenses on the outer surface of a cooling tube at a rate of 45 kg/h. The rate of heat transfer from the steam to the cooling water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The condensate leaves the condenser as a saturated liquid at 30°C. Properties The properties of water at the saturation temperature of 30°C are hfg = 2429.8 kJ/kg (Table A4). Analysis Noting that 2429.8 kJ of heat is released as 1 kg of saturated 30°C vapor at 30°C condenses, the rate of heat transfer from the steam to the cooling water in the tube is determined directly from D = 3 cm L = 35 m Q& = m& evap h fg = (45 kg/h)(2429.8 kJ/kg) = 109,341 kJ/h = 30.4 kW
3-46 The average atmospheric pressure in Denver is 83.4 kPa. The boiling temperature of water in Denver is to be determined. Analysis The boiling temperature of water in Denver is the saturation temperature corresponding to the atmospheric pressure in Denver, which is 83.4 kPa: (Table A-5) T =
[email protected] kPa = 94.6°C
3-47 The boiling temperature of water in a 5-cm deep pan is given. The boiling temperature in a 40-cm deep pan is to be determined. Assumptions Both pans are full of water. Properties The density of liquid water is approximately ρ = 1000 kg/m3. Analysis The pressure at the bottom of the 5-cm pan is the saturation 40 cm pressure corresponding to the boiling temperature of 98°C: 5 cm P = Psat@98o C = 94.39 kPa (Table A-4) The pressure difference between the bottoms of two pans is 1 kPa ∆P = ρ g h = (1000 kg/m 3 )(9.807 m/s 2 )(0.35 m) 1000 kg/m ⋅ s 2 Then the pressure at the bottom of the 40-cm deep pan is P = 94.39 + 3.43 = 97.82 kPa Then the boiling temperature becomes Tboiling =
[email protected] kPa = 99.0°C (Table A-5)
= 3.43 kPa
3-48 A cooking pan is filled with water and covered with a 4-kg lid. The boiling temperature of water is to be determined. Analysis The pressure in the pan is determined from a force balance on the lid, PA = PatmA + W Patm or, mg P = Patm + A (4 kg)(9.81 m/s 2 ) 1 kPa = (101 kPa) + 2 2 π (0.1 m) 1000 kg/m ⋅ s P = 102.25 kPa W = mg The boiling temperature is the saturation temperature corresponding to this pressure, (Table A-5) T = Tsat @102.25 kPa = 100.2°C
3-16
3-49 EES Problem 3-48 is reconsidered. Using EES (or other) software, the effect of the mass of the lid on the boiling temperature of water in the pan is to be investigated. The mass is to vary from 1 kg to 10 kg, and the boiling temperature is to be plotted against the mass of the lid. Analysis The problem is solved using EES, and the solution is given below. "Given data" {P_atm=101[kPa]} D_lid=20 [cm] {m_lid=4 [kg]} "Solution" "The atmospheric pressure in kPa varies with altitude in km by the approximate function:" P_atm=101.325*(1-0.02256*z)^5.256 "The local acceleration of gravity at 45 degrees latitude as a function of altitude in m is given by:" g=9.807+3.32*10^(-6)*z*convert(km,m) "At sea level:" z=0 "[km]" A_lid=pi*D_lid^2/4*convert(cm^2,m^2) W_lid=m_lid*g*convert(kg*m/s^2,N) P_lid=W_lid/A_lid*convert(N/m^2,kPa) P_water=P_lid+P_atm T_water=temperature(steam_iapws,P=P_water,x=0) 100.9
Twater [C] 100.1 100.1 100.2 100.3 100.4 100.5 100.6 100.7 100.7 100.8
100.8 100.7 100.6
T w ater [C]
mlid [kg] 1 2 3 4 5 6 7 8 9 10
100.5 100.4 100.3 100.2 100.1 100 1
2
3
4
5
m
lid
6
7
8
9
10
[kg]
110
105
90
85
80
75
70 0
80 70 60 50 40 30 0
1
2
3
4
5
z [km]
6
7
8
P w ater [kPa]
90
T w ater [C]
T w ater [C]
95
P w ate r [kPa]
100
mass of lid = 4 kg
100
mass of lid = 4 kg
1
2
3
4
5
6
7
8
9
z [km]
9
Effect of altitude on boiling pressure of w ater in pan w ith lid
Effect of altitude on boiling temperature of water in pan with lid
3-17
3-50 A vertical piston-cylinder device is filled with water and covered with a 20-kg piston that serves as the lid. The boiling temperature of water is to be determined. Analysis The pressure in the cylinder is determined from a force balance on the piston, PA = PatmA + W Patm
or, P = Patm +
mg A
= (100 kPa) + = 119.61 kPa
(20 kg)(9.81 m/s 2 ) 1 kPa 2 2 0.01 m 1000 kg/m ⋅ s
P
W = mg
The boiling temperature is the saturation temperature corresponding to this pressure, T = Tsat @119.61 kPa = 104.7°C
(Table A-5)
3-51 A rigid tank that is filled with saturated liquid-vapor mixture is heated. The temperature at which the liquid in the tank is completely vaporized is to be determined, and the T-v diagram is to be drawn. Analysis This is a constant volume process (v = V /m = constant), H2O 75°C
and the specific volume is determined to be
v=
V m
=
2.5 m 3 = 0.1667 m 3 /kg 15 kg
When the liquid is completely vaporized the tank will contain saturated vapor only. Thus,
v 2 = v g = 0.1667 m 3 /kg The temperature at this point is the temperature that corresponds to this vg value, T = Tsat @v
g = 0.1667
m 3 /kg
T 2
1
= 187.0°C (Table A-4)
v
3-52 A rigid vessel is filled with refrigerant-134a. The total volume and the total internal energy are to be determined. Properties The properties of R-134a at the given state are (Table A-13). P = 800 kPa u = 327.87 kJ/kg T = 120 o C v = 0.037625 m 3 /kg
Analysis The total volume and internal energy are determined from
V = mv = (2 kg)(0.037625 m 3 /kg) = 0.0753 m 3 U = mu = (2 kg)(327.87 kJ/kg) = 655.7 kJ
R-134a 2 kg 800 kPa 120°C
3-18
3-53E A rigid tank contains water at a specified pressure. The temperature, total enthalpy, and the mass of each phase are to be determined. Analysis (a) The specific volume of the water is
v=
V m
=
5 ft 3 = 1.0 ft 3/lbm 5 lbm
At 20 psia, vf = 0.01683 ft3/lbm and vg = 20.093 ft3/lbm (Table A-12E). Thus the tank contains saturated liquid-vapor mixture since vf < v < vg , and the temperature must be the saturation temperature at the specified pressure, T = Tsat @ 20 psia = 227.92 °F
(b) The quality of the water and its total enthalpy are determined from x=
v −v f v fg
=
1.0 − 0.01683 = 0.04897 20.093 − 0.01683
h = h f + xh fg = 196.27 + 0.04897 × 959.93 = 243.28 Btu/lbm
H = mh = (5 lbm)(243.28 Btu/lbm) = 1216.4 Btu (c) The mass of each phase is determined from
H2O 5 lbm 20 psia
m g = xmt = 0.04897 × 5 = 0.245 lbm m f = mt + m g = 5 − 0.245 = 4.755 lbm
3-54 A rigid vessel contains R-134a at specified temperature. The pressure, total internal energy, and the volume of the liquid phase are to be determined. Analysis (a) The specific volume of the refrigerant is
v=
V m
=
0.5 m 3 = 0.05 m 3 /kg 10 kg
At -20°C, vf = 0.0007362 m3/kg and vg = 0.14729 m3/kg (Table A-11). Thus the tank contains saturated liquid-vapor mixture since vf < v < vg , and the pressure must be the saturation pressure at the specified temperature, P = Psat @ − 20o C = 132.82 kPa
(b) The quality of the refrigerant-134a and its total internal energy are determined from x=
v −v f v fg
=
0.05 − 0.0007362 = 0.3361 0.14729 − 0.0007362
u = u f + xu fg = 25.39 + 0.3361× 193.45 = 90.42 kJ/kg U = mu = (10 kg)(90.42 kJ/kg) = 904.2 kJ
(c) The mass of the liquid phase and its volume are determined from m f = (1 − x)mt = (1 − 0.3361) × 10 = 6.639 kg
V f = m f v f = (6.639 kg)(0.0007362 m3/kg) = 0.00489 m 3
R-134a 10 kg -20°C
3-19
3-55 [Also solved by EES on enclosed CD] A piston-cylinder device contains a saturated liquid-vapor mixture of water at 800 kPa pressure. The mixture is heated at constant pressure until the temperature rises to 350°C. The initial temperature, the total mass of water, the final volume are to be determined, and the Pv diagram is to be drawn. Analysis (a) Initially two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. Then the temperature in the tank must be the saturation temperature at the specified pressure, T = Tsat @800 kPa = 170.41°C
(b) The total mass in this case can easily be determined by adding the mass of each phase, mf = mg =
Vf vf Vg vg
= =
0.1 m 3 0.001115 m 3 /kg 0.9 m 3 0.24035 m 3 /kg
= 89.704 kg P
= 3.745 kg
mt = m f + m g = 89.704 + 3.745 = 93.45 kg
1
2
(c) At the final state water is superheated vapor, and its specific volume is P2 = 800 kPa 3 v = 0.35442 m /kg T2 = 350 o C 2
(Table A-6)
Then,
V 2 = mt v 2 = (93.45 kg)(0.35442 m 3 /kg) = 33.12 m 3
v
3-20
3-56 EES Problem 3-55 is reconsidered. The effect of pressure on the total mass of water in the tank as the pressure varies from 0.1 MPa to 1 MPa is to be investigated. The total mass of water is to be plotted against pressure, and results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. P[1]=800 [kPa] P[2]=P[1] T[2]=350 [C] V_f1 = 0.1 [m^3] V_g1=0.9 [m^3] spvsat_f1=volume(Steam_iapws, P=P[1],x=0) "sat. liq. specific volume, m^3/kg" spvsat_g1=volume(Steam_iapws,P=P[1],x=1) "sat. vap. specific volume, m^3/kg" m_f1=V_f1/spvsat_f1 "sat. liq. mass, kg" m_g1=V_g1/spvsat_g1 "sat. vap. mass, kg" m_tot=m_f1+m_g1 V[1]=V_f1+V_g1 spvol[1]=V[1]/m_tot "specific volume1, m^3" T[1]=temperature(Steam_iapws, P=P[1],v=spvol[1])"C" "The final volume is calculated from the specific volume at the final T and P" spvol[2]=volume(Steam_iapws, P=P[2], T=T[2]) "specific volume2, m^3/kg" V[2]=m_tot*spvol[2] Steam
10 5
P1 [kPa] 100 200 300 400 500 600 700 800 900 1000
10 4
350 C
10 3
P [kPa]
mtot [kg] 96.39 95.31 94.67 94.24 93.93 93.71 93.56 93.45 93.38 93.34
1
2 P=800 kPa
10 2
10 1
10 0 10 -3
10 -2
10 -1
10 0 3
v [m /kg]
96.5 96
m tot [kg]
95.5 95 94.5 94 93.5 93 100
200
300
400
500
600
P[1] [kPa]
700
800
900
1000
10 1
10 2
3-21
3-57E Superheated water vapor cools at constant volume until the temperature drops to 250°F. At the final state, the pressure, the quality, and the enthalpy are to be determined. Analysis This is a constant volume process (v = V/m = constant), and the initial specific volume is determined to be P1 = 180 psia 3 v = 3.0433 ft /lbm T1 = 500 o F 1 3
(Table A-6E) H2O 180 psia 500°F
3
At 250°F, vf = 0.01700 ft /lbm and vg = 13.816 ft /lbm. Thus at the final state, the tank will contain saturated liquid-vapor mixture since vf < v < vg , and the final pressure must be the saturation pressure at the final temperature, P = Psat @ 250o F = 29.84 psia
T
1
(b) The quality at the final state is determined from x2 =
v 2 −v f v fg
=
3.0433 − 0.01700 = 0.219 13.816 − 0.01700
(c) The enthalpy at the final state is determined from h = h f + xh fg = 218.63 + 0.219 × 945.41 = 426.0 Btu/lbm
2
v
3-22
3-58E EES Problem 3-57E is reconsidered. The effect of initial pressure on the quality of water at the final state as the pressure varies from 100 psi to 300 psi is to be investigated. The quality is to be plotted against initial pressure, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below.
T[1]=500 [F] P[1]=180 [psia] T[2]=250 [F] v[ 1]=volume(steam_iapws,T=T[1],P=P[1]) v[2]=v[1] P[2]=pressure(steam_iapws,T=T[2],v=v[2]) h[2]=enthalpy(steam_iapws,T=T[2],v=v[2]) x[2]=quality(steam_iapws,T=T[2],v=v[2]) Steam
1400
x2 0.4037 0.3283 0.2761 0.2378 0.2084 0.1853 0.1665 0.1510 0.1379 0.1268
1200
1.21.31.4 1.5 Btu/lbm-R
T [°F]
1000 800 600
1600 psia 780 psia
400
1
180 psia
2
29.82 psia
200 0 10 -2
0.050.1 0.2 0.5
10 -1
10 0
10 1
10 2
3
v [ft /lb ] m
0.45 0.4 0.35
x[2]
P1 [psia] 100 122.2 144.4 166.7 188.9 211.1 233.3 255.6 277.8 300
0.3 0.25 0.2 0.15 0.1 100
140
180
220
P[1] [psia]
260
300
10 3
10 4
3-23
3-59 A piston-cylinder device that is initially filled with water is heated at constant pressure until all the liquid has vaporized. The mass of water, the final temperature, and the total enthalpy change are to be determined, and the T-v diagram is to be drawn. Analysis Initially the cylinder contains compressed liquid (since P > Psat@40°C) that can be approximated as a saturated liquid at the specified temperature (Table A-4), T
v 1 ≅ v f@40°C = 0.001008 m 3 /kg h1 ≅ hf@40°C = 167.53 kJ/kg
2
(a) The mass is determined from
H2O 40°C 200 kPa
1
V 0.050 m 3 m= 1 = = 49.61 kg v 1 0.001008 m 3 /kg
v
(b) At the final state, the cylinder contains saturated vapor and thus the final temperature must be the saturation temperature at the final pressure, T = Tsat @ 200 kPa = 120.21°C
(c) The final enthalpy is h2 = hg @ 200 kPa = 2706.3 kJ/kg. Thus,
∆H = m(h2 − h1 ) = (49.61 kg)(2706.3 − 167.53)kJ/kg = 125,943 kJ
3-60 A rigid vessel that contains a saturated liquid-vapor mixture is heated until it reaches the critical state. The mass of the liquid water and the volume occupied by the liquid at the initial state are to be determined. Analysis This is a constant volume process (v = V /m = constant) to the critical state, and thus the initial specific volume will be equal to the final specific volume, which is equal to the critical specific volume of water,
v 1 = v 2 = v cr = 0.003106 m 3 /kg
(last row of Table A-4)
The total mass is m=
3
0.3 m V = = 96.60 kg v 0.003106 m 3 /kg
T
CP H2O 150°C
At 150°C, vf = 0.001091 m3/kg and vg = 0.39248 m3/kg (Table A-4). Then the quality of water at the initial state is x1 =
v1 −v f v fg
=
0.003106 − 0.001091 = 0.005149 0.39248 − 0.001091
vcr
v
Then the mass of the liquid phase and its volume at the initial state are determined from m f = (1 − x1 )mt = (1 − 0.005149)(96.60) = 96.10 kg
V f = m f v f = (96.10 kg)(0.001091 m3/kg) = 0.105 m 3
3-24
3-61 The properties of compressed liquid water at a specified state are to be determined using the compressed liquid tables, and also by using the saturated liquid approximation, and the results are to be compared. Analysis Compressed liquid can be approximated as saturated liquid at the given temperature. Then from Table A-4, T = 100°C ⇒
v ≅ v f @ 100°C = 0.001043 m 3 /kg (0.72% error) u ≅ u f @ 100°C = 419.06 kJ/kg h ≅ h f @ 100°C = 419.17 kJ/kg
(1.02% error) (2.61% error)
From compressed liquid table (Table A-7),
v = 0.001036 m 3 /kg P = 15 MPa u = 414.85 kJ/kg T = 100°C h = 430.39 kJ/kg The percent errors involved in the saturated liquid approximation are listed above in parentheses.
3-62 EES Problem 3-61 is reconsidered. Using EES, the indicated properties of compressed liquid are to be determined, and they are to be compared to those obtained using the saturated liquid approximation. Analysis The problem is solved using EES, and the solution is given below.
Fluid$='Steam_IAPWS' T = 100 [C] P = 15000 [kPa] v = VOLUME(Fluid$,T=T,P=P) u = INTENERGY(Fluid$,T=T,P=P) h = ENTHALPY(Fluid$,T=T,P=P) v_app = VOLUME(Fluid$,T=T,x=0) u_app = INTENERGY(Fluid$,T=T,x=0) h_app_1 = ENTHALPY(Fluid$,T=T,x=0) h_app_2 = ENTHALPY(Fluid$,T=T,x=0)+v_app*(P-pressure(Fluid$,T=T,x=0)) SOLUTION Fluid$='Steam_IAPWS' h=430.4 [kJ/kg] h_app_1=419.2 [kJ/kg] h_app_2=434.7 [kJ/kg] P=15000 [kPa] T=100 [C] u=414.9 [kJ/kg] u_app=419.1 [kJ/kg] v=0.001036 [m^3/kg] v_app=0.001043 [m^3/kg]
3-25
3-63E A rigid tank contains saturated liquid-vapor mixture of R-134a. The quality and total mass of the refrigerant are to be determined. Analysis At 50 psia, vf = 0.01252 ft3/lbm and vg = 0.94791 ft3/lbm (Table A-12E). The volume occupied by the liquid and the vapor phases are
V f = 3 ft 3 and V g = 12 ft 3
R-134a 15 ft3 50 psia
Thus the mass of each phase is mf = mg =
Vf vf Vg vg
= =
3 ft 3 0.01252 ft 3 /lbm 12 ft 3 0.94791 ft 3 /lbm
= 239.63 lbm = 12.66 lbm
Then the total mass and the quality of the refrigerant are mt = mf + mg = 239.63 + 12.66 = 252.29 lbm x=
mg mt
=
12.66 lbm = 0.05018 252.29 lbm
3-64 Superheated steam in a piston-cylinder device is cooled at constant pressure until half of the mass condenses. The final temperature and the volume change are to be determined, and the process should be shown on a T-v diagram. Analysis (b) At the final state the cylinder contains saturated liquid-vapor mixture, and thus the final temperature must be the saturation temperature at the final pressure, T = Tsat@1 MPa = 179.88°C
(Table A-5)
(c) The quality at the final state is specified to be x2 = 0.5. The specific volumes at the initial and the final states are P1 = 1.0 MPa 3 v = 0.25799 m /kg T1 = 300 o C 1
P2 = 1.0 MPa x2 = 0.5
H2O 300°C 1 MPa
(Table A-6)
v 2 = v f + x2v fg = 0.001127 + 0.5 × (0.19436 − 0.001127) = 0.09775 m3/kg
Thus, ∆V = m(v 2 − v 1 ) = (0.8 kg)(0.09775 − 0.25799)m 3 /kg = −0.1282 m 3
T 1
2
v
3-26
3-65 The water in a rigid tank is cooled until the vapor starts condensing. The initial pressure in the tank is to be determined. Analysis This is a constant volume process (v = V /m = constant), and the initial specific volume is equal to the final specific volume that is
v 1 = v 2 = v g @150°C = 0.39248 m 3 /kg
(Table A-4)
since the vapor starts condensing at 150°C. Then from Table A-6, T1 = 250°C P = 0.60 MPa 3 v1 = 0.39248 m /kg 1
T °C H2O T1= 250°C P1 = ?
250
1
150 2
v
3-66 Water is boiled in a pan by supplying electrical heat. The local atmospheric pressure is to be estimated. Assumptions 75 percent of electricity consumed by the heater is transferred to the water. Analysis The amount of heat transfer to the water during this period is Q = fEelect time = (0.75)(2 kJ/s)(30 × 60 s) = 2700 kJ
The enthalpy of vaporization is determined from h fg =
Q 2700 kJ = = 2269 kJ/kg m boil 1.19 kg
Using the data by a trial-error approach in saturation table of water (Table A-5) or using EES as we did, the saturation pressure that corresponds to an enthalpy of vaporization value of 2269 kJ/kg is Psat = 85.4 kPa which is the local atmospheric pressure.
3-27
3-67 Heat is supplied to a rigid tank that contains water at a specified state. The volume of the tank, the final temperature and pressure, and the internal energy change of water are to be determined. Properties The saturated liquid properties of water at 200°C are: vf = 0.001157 m3/kg and uf = 850.46 kJ/kg (Table A-4). Analysis (a) The tank initially contains saturated liquid water and air. The volume occupied by water is
V1 = mv 1 = (1.4 kg)(0.001157 m 3 /kg) = 0.001619 m 3 which is the 25 percent of total volume. Then, the total volume is determined from
V =
1 (0.001619) = 0.006476 m 3 0.25
(b) Properties after the heat addition process are
v2 =
V m
=
0.006476 m3 = 0.004626 m3 / kg 1.4 kg
v 2 = 0.004626 m3 / kg x2 = 1
T2 = 371.3 °C
P2 = 21,367 kPa u2 = 2201.5 kJ/kg
(Table A-4 or A-5 or EES)
(c) The total internal energy change is determined from ∆U = m(u 2 − u1 ) = (1.4 kg)(2201.5 - 850.46) kJ/kg = 1892 kJ
3-68 Heat is lost from a piston-cylinder device that contains steam at a specified state. The initial temperature, the enthalpy change, and the final pressure and quality are to be determined. Analysis (a) The saturation temperature of steam at 3.5 MPa is
[email protected] MPa = 242.6°C (Table A-5) Then, the initial temperature becomes T1 = 242.6+5 = 247.6°C Also,
P1 = 3.5 MPa h1 = 2821.1 kJ/kg T1 = 247.6°C
(Table A-6)
Steam 3.5 MPa
(b) The properties of steam when the piston first hits the stops are P2 = P1 = 3.5 MPa h2 = 1049.7 kJ/kg 3 x2 = 0 v 2 = 0.001235 m /kg
(Table A-5)
Then, the enthalpy change of steam becomes ∆h = h2 − h1 = 1049.7 − 2821.1 = -1771 kJ/kg
(c) At the final state
v 3 = v 2 = 0.001235 m3/kg P3 = 1555 kPa T3 = 200°C
x3 = 0.0006
(Table A-4 or EES)
The cylinder contains saturated liquid-vapor mixture with a small mass of vapor at the final state.
Q
3-28
Ideal Gas 3-69C Propane (molar mass = 44.1 kg/kmol) poses a greater fire danger than methane (molar mass = 16 kg/kmol) since propane is heavier than air (molar mass = 29 kg/kmol), and it will settle near the floor. Methane, on the other hand, is lighter than air and thus it will rise and leak out. 3-70C A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to its critical temperature and pressure. 3-71C Ru is the universal gas constant that is the same for all gases whereas R is the specific gas constant that is different for different gases. These two are related to each other by R = Ru / M, where M is the molar mass of the gas. 3-72C Mass m is simply the amount of matter; molar mass M is the mass of one mole in grams or the mass of one kmol in kilograms. These two are related to each other by m = NM, where N is the number of moles.
3-73 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. Assumptions At specified conditions, helium behaves as an ideal gas. Properties The universal gas constant is Ru = 8.314 kPa.m3/kmol.K. The molar mass of helium is 4.0 kg/kmol (Table A-1). Analysis The volume of the sphere is 4 3
4 3
V = π r 3 = π (3 m) 3 = 113.1 m 3 Assuming ideal gas behavior, the mole numbers of He is determined from N=
(200 kPa)(113.1 m3 ) PV = = 9.28 kmol RuT (8.314 kPa ⋅ m3/kmol ⋅ K)(293 K)
Then the mass of He can be determined from m = NM = (9.28 kmol)(4.0 kg/kmol) = 37.15 kg
He D=6m 20°C 200 kPa
3-29
3-74 EES Problem 3-73 is to be reconsidered. The effect of the balloon diameter on the mass of helium contained in the balloon is to be determined for the pressures of (a) 100 kPa and (b) 200 kPa as the diameter varies from 5 m to 15 m. The mass of helium is to be plotted against the diameter for both cases. Analysis The problem is solved using EES, and the solution is given below. "Given Data" {D=6 [m]} {P=200 [kPa]} T=20 [C] P=200 [kPa] R_u=8.314 [kJ/kmol-K] "Solution" P*V=N*R_u*(T+273) V=4*pi*(D/2)^3/3 m=N*MOLARMASS(Helium) D [m] 5 6.111 7.222 8.333 9.444 10.56 11.67 12.78 13.89 15
m [kg] 21.51 39.27 64.82 99.57 145 202.4 273.2 359 461 580.7
600
500
400
] g k[
300
P=200 kPa
m 200
P=100 kPa
100
0 5
7
9
11
D [m]
13
15
3-30
3-75 An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must be bled off to reduce the temperature to the original value are to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis Initially, the absolute pressure in the tire is P1 = Pg + Patm = 210 + 100 = 310kPa
Tire 25°C
Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire can be determined from P1V1 P2V 2 T 323 K = → P2 = 2 P1 = (310 kPa) = 336 kPa T1 T2 T1 298 K
Thus the pressure rise is ∆P = P2 − P1 = 336 − 310 = 26 kPa
The amount of air that needs to be bled off to restore pressure to its original value is m1 =
P1V (310 kPa)(0.025 m3 ) = = 0.0906 kg RT1 (0.287 kPa ⋅ m3/kg ⋅ K)(298 K)
m2 =
P1V (310 kPa)(0.025 m3 ) = = 0.0836 kg RT2 (0.287 kPa ⋅ m3/kg ⋅ K)(323 K) ∆m = m1 − m2 = 0.0906 − 0.0836 = 0.0070 kg
3-76E An automobile tire is under inflated with air. The amount of air that needs to be added to the tire to raise its pressure to the recommended value is to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant. Properties The gas constant of air is R = 0.3704 psia.ft3/lbm.R (Table A-1E). Analysis The initial and final absolute pressures in the tire are P1 = Pg1 + Patm = 20 + 14.6 = 34.6 psia P2 = Pg2 + Patm = 30 + 14.6 = 44.6 psia Treating air as an ideal gas, the initial mass in the tire is m1 =
Tire 0.53 ft3 90°F 20 psig
(34.6 psia)(0.53 ft 3 ) P1V = = 0.0900 lbm RT1 (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(550 R)
Noting that the temperature and the volume of the tire remain constant, the final mass in the tire becomes m2 =
(44.6 psia)(0.53 ft 3 ) P2V = = 0.1160 lbm RT2 (0.3704 psia ⋅ ft 3/lbm ⋅ R)(550 R)
Thus the amount of air that needs to be added is ∆m = m2 − m1 = 0.1160 − 0.0900 = 0.0260 lbm
3-31
3-77 The pressure and temperature of oxygen gas in a storage tank are given. The mass of oxygen in the tank is to be determined. Assumptions At specified conditions, oxygen behaves as an ideal gas Properties The gas constant of oxygen is R = 0.2598 kPa.m3/kg.K (Table A-1).
Pg = 500 kPa
Analysis The absolute pressure of O2 is P = Pg + Patm = 500 + 97 = 597 kPa Treating O2 as an ideal gas, the mass of O2 in tank is determined to be m=
(597 kPa)(2.5 m 3 ) PV = = 19.08 kg RT (0.2598 kPa ⋅ m 3 /kg ⋅ K)(28 + 273)K
O2
V = 2.5 m3 T = 28°C
3-78E A rigid tank contains slightly pressurized air. The amount of air that needs to be added to the tank to raise its pressure and temperature to the recommended values is to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tank remains constant. Properties The gas constant of air is R = 0.3704 psia.ft3/lbm.R (Table A-1E). Analysis Treating air as an ideal gas, the initial volume and the final mass in the tank are determined to be
V = m2 =
m1 RT1 (20 lbm)(0.3704 psia ⋅ ft 3 /lbm ⋅ R)(530 R) = = 196.3 ft 3 P1 20 psia P2V (35 psia)(196.3 ft 3 ) = = 33.73 lbm RT2 (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(550 R)
Air, 20 lbm 20 psia 70°F
Thus the amount of air added is ∆m = m2 − m1 = 33.73 − 20.0 = 13.73 lbm
3-79 A rigid tank contains air at a specified state. The gage pressure of the gas in the tank is to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis Treating air as an ideal gas, the absolute pressure in the tank is determined from P=
mRT
V
=
(5 kg)(0.287 kPa ⋅ m 3 /kg ⋅ K)(298 K) 0.4 m 3
Thus the gage pressure is Pg = P − Patm = 1069.1 − 97 = 972.1 kPa
= 1069.1 kPa
Pg Air 400 L 25°C
3-32
3-80 Two rigid tanks connected by a valve to each other contain air at specified conditions. The volume of the second tank and the final equilibrium pressure when the valve is opened are to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis Let's call the first and the second tanks A and B. Treating air as an ideal gas, the volume of the second tank and the mass of air in the first tank are determined to be m RT
V B = 1 1 = P1 B
(5 kg)(0.287 kPa ⋅ m3/kg ⋅ K)(308 K) = 2.21 m 3 200 kPa
PV (500 kPa)(1.0 m 3 ) m A = 1 = = 5.846 kg 3 RT1 A (0.287 kPa ⋅ m /kg ⋅ K)(298 K)
Thus,
V = V A + V B = 1.0 + 2.21 = 3.21 m3 m = m A + mB = 5.846 + 5.0 = 10.846 kg
A Air V=1 m3 T=25°C P=500 kPa
Then the final equilibrium pressure becomes P2 =
mRT2
V
=
(10.846 kg)(0.287 kPa ⋅ m3 /kg ⋅ K)(293 K) 3.21 m3
= 284.1 kPa
B
×
Air m=5 kg T=35°C P=200 kPa
3-33
Compressibility Factor 3-81C It represent the deviation from ideal gas behavior. The further away it is from 1, the more the gas deviates from ideal gas behavior. 3-82C All gases have the same compressibility factor Z at the same reduced temperature and pressure. 3-83C Reduced pressure is the pressure normalized with respect to the critical pressure; and reduced temperature is the temperature normalized with respect to the critical temperature.
3-84 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1, R = 0.4615 kPa·m3/kg·K,
Tcr = 647.1 K,
Pcr = 22.06 MPa
Analysis (a) From the ideal gas equation of state,
v=
RT (0.4615 kPa ⋅ m 3 /kg ⋅ K)(673 K) = = 0.03106 m 3 /kg (17.6% error) P (10,000 kPa)
(b) From the compressibility chart (Fig. A-15), 10 MPa P = = 0.453 Pcr 22.06 MPa Z = 0.84 673 K T TR = = = 1.04 Tcr 647.1 K PR =
Thus,
v = Zv ideal = (0.84)(0.03106 m 3 /kg) = 0.02609 m 3 /kg (1.2% error) (c) From the superheated steam table (Table A-6), P = 10 MPa T = 400°C
} v = 0.02644 m /kg 3
H2O 10 MPa 400°C
3-34
3-85 EES Problem 3-84 is reconsidered. The problem is to be solved using the general compressibility factor feature of EES (or other) software. The specific volume of water for the three cases at 10 MPa over the temperature range of 325°C to 600°C in 25°C intervals is to be compared, and the %error involved in the ideal gas approximation is to be plotted against temperature. Analysis The problem is solved using EES, and the solution is given below. P=10 [MPa]*Convert(MPa,kPa) {T_Celsius= 400 [C]} T=T_Celsius+273 "[K]" T_critical=T_CRIT(Steam_iapws) P_critical=P_CRIT(Steam_iapws) {v=Vol/m} P_table=P; P_comp=P;P_idealgas=P T_table=T; T_comp=T;T_idealgas=T v_table=volume(Steam_iapws,P=P_table,T=T_table) "EES data for steam as a real gas" {P_table=pressure(Steam_iapws, T=T_table,v=v)} {T_sat=temperature(Steam_iapws,P=P_table,v=v)} MM=MOLARMASS(water) R_u=8.314 [kJ/kmol-K] "Universal gas constant" R=R_u/MM "[kJ/kg-K], Particular gas constant" P_idealgas*v_idealgas=R*T_idealgas "Ideal gas equation" z = COMPRESS(T_comp/T_critical,P_comp/P_critical) P_comp*v_comp=z*R*T_comp "generalized Compressibility factor" Error_idealgas=Abs(v_table-v_idealgas)/v_table*Convert(, %) Error_comp=Abs(v_table-v_comp)/v_table*Convert(, %) Errorcomp [%] 6.088 2.422 0.7425 0.129 0.6015 0.8559 0.9832 1.034 1.037 1.01 0.9652 0.9093
Errorideal gas [%] 38.96 28.2 21.83 17.53 14.42 12.07 10.23 8.755 7.55 6.55 5.712 5
TCelcius [C] 325 350 375 400 425 450 475 500 525 550 575 600
40
Percent Error [%]
Specific Volum e
35
Steam at 10 MPa
30
Ideal Gas
25
Compressibility Factor
20 15 10 5 0 300
350
400
T
450
Celsius
500
[C]
550
600
3-35
3-86 The specific volume of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R-134a tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1, R = 0.08149 kPa·m3/kg·K,
Tcr = 374.2 K,
Pcr = 4.059 MPa
Analysis (a) From the ideal gas equation of state,
v=
RT (0.08149 kPa ⋅ m3/kg ⋅ K)(343 K) = = 0.03105 m3 /kg P 900 kPa
(13.3% error )
(b) From the compressibility chart (Fig. A-15), 0.9 MPa P = = 0.222 Pcr 4.059 MPa Z = 0.894 343 K T = = 0.917 TR = Tcr 374.2 K PR =
R-134a 0.9 MPa 70°C
Thus,
v = Zv ideal = (0.894)(0.03105 m 3 /kg) = 0.02776 m 3 /kg
(1.3%error)
(c) From the superheated refrigerant table (Table A-13),
}
P = 0.9 MPa v = 0.027413 m3 /kg T = 70°C
3-87 The specific volume of nitrogen gas is to be determined using the ideal gas relation and the compressibility chart. The errors involved in these two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of nitrogen are, from Table A-1, R = 0.2968 kPa·m3/kg·K,
Tcr = 126.2 K,
Pcr = 3.39 MPa
Analysis (a) From the ideal gas equation of state,
v=
RT (0.2968 kPa ⋅ m 3 /kg ⋅ K)(150 K) = = 0.004452 m 3 /kg P 10,000 kPa
(86.4% error)
(b) From the compressibility chart (Fig. A-15), 10 MPa P = = 2.95 Pcr 3.39 MPa Z = 0.54 150 K T = = 1.19 TR = Tcr 126.2 K
N2 10 MPa 150 K
PR =
Thus,
v = Zv ideal = (0.54)(0.004452 m 3 /kg) = 0.002404 m 3 /kg
(0.7% error)
3-36
3-88 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1, R = 0.4615 kPa·m3/kg·K,
Tcr = 647.1 K,
Pcr = 22.06 MPa
Analysis (a) From the ideal gas equation of state,
v=
RT (0.4615 kPa ⋅ m3/kg ⋅ K)(723 K) = = 0.09533 m3 /kg P 3500 kPa
(3.7% error)
(b) From the compressibility chart (Fig. A-15), H2O 3.5 MPa
P 3.5 MPa = = 0.159 Pcr 22.06 MPa Z = 0.961 T 723 K TR = = = 1.12 Tcr 647.1 K PR =
450°C
Thus,
v = Zv ideal = (0.961)(0.09533 m 3 /kg) = 0.09161 m 3 /kg
(0.4% error)
(c) From the superheated steam table (Table A-6), P = 3.5 MPa T = 450°C
} v = 0.09196 m /kg 3
3-89E The temperature of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R-134a tables. Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1E, R = 0.10517 psia·ft3/lbm·R,
Tcr = 673.6 R,
Pcr = 588.7 psia
Analysis (a) From the ideal gas equation of state, T=
Pv (400 psia)(0.1386 ft 3 /lbm) = = 527.2 R R (0.10517 psia ⋅ ft 3 /lbm ⋅ R)
(b) From the compressibility chart (Fig. A-15a), TR = 1.03 3 (0.1386 ft /lbm)(588.7 psia) v actual 1.15 vR = = = RTcr / Pcr (0.10517 psia ⋅ ft 3/lbm ⋅ R)(673.65 R) PR =
P 400 psia = = 0.678 Pcr 588.7 psia
Thus, T = TR Tcr = 1.03 × 673.6 = 693.8 R
(c) From the superheated refrigerant table (Table A-13E), P = 400 psia
v = 0.13853 ft 3 /lbm
T = 240°F (700 R)
3-37
3-90 The pressure of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R-134a tables. Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1, R = 0.08149 kPa·m3/kg·K,
Tcr = 374.2 K,
Pcr = 4.059 MPa
Analysis The specific volume of the refrigerant is
v=
V m
=
0.016773 m 3 = 0.016773 m 3 /kg 1 kg R-134a 3 0.016773 m /kg
(a) From the ideal gas equation of state, P=
RT
v
=
(0.08149 kPa ⋅ m3 /kg ⋅ K)(383 K) 0.016773 m3 /kg
= 1861 kPa
110°C
(b) From the compressibility chart (Fig. A-15), PR = 0.39 v actual 0.016773 m3/kg = = 2.24 vR = RTcr /Pcr (0.08149 kPa ⋅ m3/kg ⋅ K)(374.2 K)/(4059 kPa) TR =
T 383 K = = 1.023 Tcr 374.2 K
Thus, P = PR Pcr = (0.39)(4059 kPa) = 1583 kPa
(c) From the superheated refrigerant table (Table A-13), T = 110 o C P = 1600 kPa v = 0.016773 m 3 /kg
3-91 Somebody claims that oxygen gas at a specified state can be treated as an ideal gas with an error less than 10%. The validity of this claim is to be determined. Properties The critical pressure, and the critical temperature of oxygen are, from Table A-1, Tcr = 154.8 K
and
Pcr = 5.08 MPa
Analysis From the compressibility chart (Fig. A-15), 3 MPa P = = 0.591 Pcr 5.08 MPa Z = 0.79 160 K T TR = = = 1.034 Tcr 154.8 K PR =
Then the error involved can be determined from Error =
v − v ideal 1 1 = 1− = 1− = −26.6% Z 0.79 v
Thus the claim is false.
O2 3 MPa 160 K
3-38
3-92 The percent error involved in treating CO2 at a specified state as an ideal gas is to be determined. Properties The critical pressure, and the critical temperature of CO2 are, from Table A-1, Tcr = 304.2K and Pcr = 7.39MPa
Analysis From the compressibility chart (Fig. A-15), P 3 MPa = = 0.406 Pcr 7.39 MPa Z = 0.80 T 283 K TR = = = 0.93 Tcr 304.2 K PR =
CO2 3 MPa 10°C
Then the error involved in treating CO2 as an ideal gas is Error =
v − v ideal 1 1 = 1− = 1− = −0.25 or 25.0% Z 0.80 v
3-93 The % error involved in treating CO2 at a specified state as an ideal gas is to be determined. Properties The critical pressure, and the critical temperature of CO2 are, from Table A-1, Tcr = 304.2 K and Pcr = 7.39 MPa
Analysis From the compressibility chart (Fig. A-15), P 7 MPa = = 0.947 Pcr 7.39 MPa Z = 0.84 T 380 K TR = = = 1.25 Tcr 304.2 K PR =
Then the error involved in treating CO2 as an ideal gas is Error =
v − v ideal 1 1 = 1− = 1− = −0.190 or 19.0% Z 0.84 v
CO2 7 MPa 380 K
3-39
3-94 CO2 gas flows through a pipe. The volume flow rate and the density at the inlet and the volume flow rate at the exit of the pipe are to be determined. 3 MPa 500 K 2 kg/s
CO2
450 K
Properties The gas constant, the critical pressure, and the critical temperature of CO2 are (Table A-1) R = 0.1889 kPa·m3/kg·K,
Tcr = 304.2 K,
Pcr = 7.39 MPa
Analysis (a) From the ideal gas equation of state,
V&1 =
m& RT1 (2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K) = = 0.06297 m 3 /kg (2.1% error) (3000 kPa) P1
ρ1 =
P1 (3000 kPa) = = 31.76 kg/m 3 (2.1% error) RT1 (0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K)
V&2 =
m& RT2 (2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(450 K) = = 0.05667 m 3 /kg (3.6% error) (3000 kPa) P2
(b) From the compressibility chart (EES function for compressibility factor is used) P1 3 MPa = = 0.407 Pcr 7.39 MPa Z1 = 0.9791 T1 500 K = = = 1.64 Tcr 304.2 K
PR = TR ,1
P2 3 MPa = = 0.407 Pcr 7.39 MPa Z 2 = 0.9656 T2 450 K = = = 1.48 Tcr 304.2 K
PR = TR , 2
Thus,
V&1 =
Z 1 m& RT1 (0.9791)(2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(500 K) = = 0.06165 m 3 /kg (3000 kPa) P1
ρ1 =
P1 (3000 kPa) = = 32.44 kg/m 3 3 Z 1 RT1 (0.9791)(0.1889 kPa ⋅ m /kg ⋅ K)(500 K)
V&2 =
Z 2 m& RT2 (0.9656)(2 kg/s)(0.1889 kPa ⋅ m 3 /kg ⋅ K)(450 K) = = 0.05472 m 3 /kg (3000 kPa) P2
3-40
Other Equations of State 3-95C The constant a represents the increase in pressure as a result of intermolecular forces; the constant b represents the volume occupied by the molecules. They are determined from the requirement that the critical isotherm has an inflection point at the critical point.
3-96 The pressure of nitrogen in a tank at a specified state is to be determined using the ideal gas, van der Waals, and Beattie-Bridgeman equations. The error involved in each case is to be determined. Properties The gas constant, molar mass, critical pressure, and critical temperature of nitrogen are (Table A-1) R = 0.2968 kPa·m3/kg·K,
M = 28.013 kg/kmol,
Tcr = 126.2 K,
Pcr = 3.39 MPa
Analysis The specific volume of nitrogen is
v=
V m
=
3.27 m3 = 0.0327 m3/kg 100 kg
N2 3 0.0327 m /kg 175 K
(a) From the ideal gas equation of state, P=
RT
v
=
(0.2968 kPa ⋅ m 3 /kg ⋅ K)(175 K) 0.0327 m 3 /kg
= 1588 kPa (5.5% error)
(b) The van der Waals constants for nitrogen are determined from a=
27 R 2 Tcr2 (27)(0.2968 kPa ⋅ m 3 / kg ⋅ K) 2 (126.2 K) 2 = = 0.175 m 6 ⋅ kPa / kg 2 (64)(3390 kPa) 64 Pcr
b=
RTcr (0.2968 kPa ⋅ m 3 / kg ⋅ K)(126.2 K) = = 0.00138 m 3 / kg 8 Pcr 8 × 3390 kPa
P=
0.2968 × 175 0.175 RT a − 2 = − = 1495 kPa (0.7% error) v −b v 0.0327 − 0.00138 (0.0327) 2
Then,
(c) The constants in the Beattie-Bridgeman equation are a 0.02617 A = Ao 1 − = 136.23151 − = 132.339 0.9160 v b − 0.00691 B = Bo 1 − = 0.050461 − = 0.05084 0.9160 v c = 4.2 × 10 4 m 3 ⋅ K 3 /kmol
since v = Mv = (28.013 kg/kmol)(0.0327 m3/kg) = 0.9160 m3/kmol . Substituting, P=
RuT c A 1 − 3 (v + B ) − 2 2 v vT v
8.314 × 175 4.2 × 104 1 − (0.9160 + 0.05084) − 132.3392 2 3 (0.9160) 0.9160 × 175 (0.9160) = 1504 kPa (0.07% error)
=
3-41
3-97 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation, van der Waals equation, and the steam tables. Properties The gas constant, critical pressure, and critical temperature of steam are (Table A-1) R = 0.4615 kPa·m3/kg·K,
Tcr = 647.1 K,
Pcr = 22.06 MPa
Analysis The specific volume of steam is
v=
V m
=
1 m3 = 0.3520 m 3 /kg 2.841 kg
H2O 3 1m 2.841 kg 0.6 MPa
(a) From the ideal gas equation of state, T =
Pv (600 kPa)(0.352 m 3/kg) = = 457.6 K R 0.4615 kPa ⋅ m3 /kg ⋅ K
(b) The van der Waals constants for steam are determined from a=
27 R 2Tcr2 (27)(0.4615 kPa ⋅ m 3 /kg ⋅ K) 2 (647.1 K) 2 = = 1.705 m 6 ⋅ kPa/kg 2 64 Pcr (64)(22,060 kPa)
b=
RTcr (0.4615 kPa ⋅ m 3 /kg ⋅ K)(647.1 K) = = 0.00169 m 3 /kg 8 Pcr 8 × 22,060 kPa
T=
1 1 1.705 a 600 + P + 2 (v − b ) = 0.4615 R (0.3520) 2 v
Then,
(c) From the superheated steam table (Tables A-6), P = 0.6 MPa
v = 0.3520 m 3 /kg
T = 200°C
(= 473 K)
(0.352 − 0.00169 ) = 465.9 K
3-42
3-98 EES Problem 3-97 is reconsidered. The problem is to be solved using EES (or other) software. The temperature of water is to be compared for the three cases at constant specific volume over the pressure range of 0.1 MPa to 1 MPa in 0.1 MPa increments. The %error involved in the ideal gas approximation is to be plotted against pressure. Analysis The problem is solved using EES, and the solution is given below. Function vanderWaals(T,v,M,R_u,T_cr,P_cr) v_bar=v*M "Conversion from m^3/kg to m^3/kmol" "The constants for the van der Waals equation of state are given by equation 3-24" a=27*R_u^2*T_cr^2/(64*P_cr) b=R_u*T_cr/(8*P_cr) "The van der Waals equation of state gives the pressure as" vanderWaals:=R_u*T/(v_bar-b)-a/v_bar**2 End m=2.841[kg] Vol=1 [m^3] {P=6*convert(MPa,kPa)} T_cr=T_CRIT(Steam_iapws) P_cr=P_CRIT(Steam_iapws) v=Vol/m P_table=P; P_vdW=P;P_idealgas=P T_table=temperature(Steam_iapws,P=P_table,v=v) "EES data for steam as a real gas" {P_table=pressure(Steam_iapws, T=T_table,v=v)} {T_sat=temperature(Steam_iapws,P=P_table,v=v)} MM=MOLARMASS(water) R_u=8.314 [kJ/kmol-K] "Universal gas constant" R=R_u/MM "Particular gas constant" P_idealgas=R*T_idealgas/v "Ideal gas equation" "The value of P_vdW is found from van der Waals equation of state Function" P_vdW=vanderWaals(T_vdW,v,MM,R_u,T_cr,P_cr) Error_idealgas=Abs(T_table-T_idealgas)/T_table*Convert(, %) Error_vdW=Abs(T_table-T_vdW)/T_table*Convert(, %) P [kPa] 100 200 300 400 500 600 700 800 900 1000
Tideal gas [K] 76.27 152.5 228.8 305.1 381.4 457.6 533.9 610.2 686.4 762.7
Ttable [K] 372.8 393.4 406.7 416.8 425 473 545.3 619.1 693.7 768.6
TvdW [K] 86.35 162.3 238.2 314.1 390 465.9 541.8 617.7 693.6 769.5
Errorideal gas [K] 79.54 61.22 43.74 26.8 10.27 3.249 2.087 1.442 1.041 0.7725
3-43
100 90
van der W aals Ideal gas
80
Err or vdW [ % ]
70 60 50 40 30 20 10 0 100
200
300
400
500
600
700
800
900
1000
P [kPa] T vs. v for Steam at 600 kPa 1000 900 800
Steam Table
T [K]
700
Ideal Gas
600
van der Waals
500 600 kPa
400
T vs v for Steam at 6000 kPa
300 200 10 -3
0.05 0.1 0.2
10 -2
10 -1
10 0
10 1
0.5
1000 10 2
10 3
3
900
Steam Table
800
v [m /kg]
Ideal Gas
T [K]
700
van der W aals
600 500
6000 kPa
T table [K]
400 800 750 700 650 600 550 500 450 400 350 300 250 200 150 100 50 100
300 200 10 -3
Steam table van der W aals Ideal gas
200
300
400
500
10 -2
10 -1
10 0
3
v [m /kg]
600
P [kPa]
700
800
900
1000
10 1
10 2
10 3
3-44
3-99E The temperature of R-134a in a tank at a specified state is to be determined using the ideal gas relation, the van der Waals equation, and the refrigerant tables. Properties The gas constant, critical pressure, and critical temperature of R-134a are (Table A-1E) R = 0.1052 psia·ft3/lbm·R,
Tcr = 673.6 R,
Pcr = 588. 7 psia
Analysis (a) From the ideal gas equation of state, T=
Pv (100 psia)(0.54022 ft 3/lbm) = = 513.5 R R 0.1052 psia ⋅ ft 3/lbm ⋅ R
(b) The van der Waals constants for the refrigerant are determined from
Then,
a=
27 R 2Tcr2 (27)(0.1052 psia ⋅ ft 3 /lbm ⋅ R) 2 (673.6 R) 2 = = 3.591 ft 6 ⋅ psia/lbm 2 64 Pcr (64)(588.7 psia)
b=
RTcr (0.1052 psia ⋅ ft 3 /lbm ⋅ R)(673.6 R) = = 0.0150 ft 3 /lbm 8 Pcr 8 × 588.7 psia
T=
1 1 3.591 a 100 + (0.54022 − 0.0150) = 560.7 R P + 2 (v − b ) = 0.1052 R (0.54022) 2 v
(c) From the superheated refrigerant table (Table A-13E), P = 100 psia
v = 0.54022 ft 3/lbm
T = 120°F (580R)
3-100 [Also solved by EES on enclosed CD] The pressure of nitrogen in a tank at a specified state is to be determined using the ideal gas relation and the Beattie-Bridgeman equation. The error involved in each case is to be determined. Properties The gas constant and molar mass of nitrogen are (Table A-1) R = 0.2968 kPa·m3/kg·K and M = 28.013 kg/kmol Analysis (a) From the ideal gas equation of state, P=
RT
v
=
(0.2968 kPa ⋅ m3 /kg ⋅ K)(150 K) 0.041884 m3 /kg
= 1063 kPa (6.3% error)
N2 3 0.041884 m /kg 150 K
(b) The constants in the Beattie-Bridgeman equation are a 0.02617 A = Ao 1 − = 136.23151 − = 133.193 1.1733 v b − 0.00691 B = Bo 1 − = 0.050461 − = 0.05076 1.1733 v c = 4.2 × 10 4 m 3 ⋅ K 3 /kmol
since
v = Mv = (28.013 kg/kmol)(0.041884 m 3 /kg) = 1.1733 m 3 /kmol .
Substituting, P=
4.2 × 10 4 RuT c A 8.314 × 150 1 1 B ( ) (1.1733 + 0.05076) − 133.1932 − v + − = − 2 3 2 2 3 (1.1733) 1.1733 × 150 v vT v (1.1733)
= 1000.4 kPa (negligible error)
3-45
3-101 EES Problem 3-100 is reconsidered. Using EES (or other) software, the pressure results of the ideal gas and Beattie-Bridgeman equations with nitrogen data supplied by EES are to be compared. The temperature is to be plotted versus specific volume for a pressure of 1000 kPa with respect to the saturated liquid and saturated vapor lines of nitrogen over the range of 110 K < T < 150 K. Analysis The problem is solved using EES, and the solution is given below. Function BeattBridg(T,v,M,R_u) v_bar=v*M "Conversion from m^3/kg to m^3/kmol" "The constants for the Beattie-Bridgeman equation of state are found in text" Ao=136.2315; aa=0.02617; Bo=0.05046; bb=-0.00691; cc=4.20*1E4 B=Bo*(1-bb/v_bar) A=Ao*(1-aa/v_bar) "The Beattie-Bridgeman equation of state is" BeattBridg:=R_u*T/(v_bar**2)*(1-cc/(v_bar*T**3))*(v_bar+B)-A/v_bar**2 End T=150 [K] v=0.041884 [m^3/kg] P_exper=1000 [kPa] T_table=T; T_BB=T;T_idealgas=T P_table=PRESSURE(Nitrogen,T=T_table,v=v) "EES data for nitrogen as a real gas" {T_table=temperature(Nitrogen, P=P_table,v=v)} M=MOLARMASS(Nitrogen) R_u=8.314 [kJ/kmol-K] "Universal gas constant" R=R_u/M "Particular gas constant" P_idealgas=R*T_idealgas/v "Ideal gas equation" P_BB=BeattBridg(T_BB,v,M,R_u) "Beattie-Bridgeman equation of state Function" Ptable [kPa] 1000 1000 1000 1000 1000 1000 1000 160 150 140
Pidealgas [kPa] 1000 1000 1000 1000 1000 1000 1000
v [m3/kg] 0.01 0.02 0.025 0.03 0.035 0.04 0.05
TBB [K] 91.23 95.52 105 116.8 130.1 144.4 174.6
Tideal gas [K] 33.69 67.39 84.23 101.1 117.9 134.8 168.5
Nitrogen, T vs v for P=1000 kPa Ideal Gas Beattie-Bridgem an EES Table Value
130
T [K]
PBB [kPa] 1000 1000 1000 1000 1000 1000 1000
120 110 1000 kPa
100 90 80 70 10 -3
10 -2 3
v [m /kg]
10 -1
Ttable [K] 103.8 103.8 106.1 117.2 130.1 144.3 174.5
3-46
Special Topic: Vapor Pressure and Phase Equilibrium 3-102 A glass of water is left in a room. The vapor pressures at the free surface of the water and in the room far from the glass are to be determined. Assumptions The water in the glass is at a uniform temperature. Properties The saturation pressure of water is 2.339 kPa at 20°C, and 1.706 kPa at 15°C (Table A-4). Analysis The vapor pressure at the water surface is the saturation pressure of water at the water temperature, Pv , water surface = Psat @ Twater = Psat@15°C = 1.706 kPa H2O 15°C
Noting that the air in the room is not saturated, the vapor pressure in the room far from the glass is Pv , air = φPsat @ Tair = φPsat@20°C = (0.6)(2.339 kPa) = 1.404 kPa
3-103 The vapor pressure in the air at the beach when the air temperature is 30°C is claimed to be 5.2 kPa. The validity of this claim is to be evaluated. Properties The saturation pressure of water at 30°C is 4.247 kPa (Table A-4).
30°C
Analysis The maximum vapor pressure in the air is the saturation pressure of water at the given temperature, which is
WATER
Pv , max = Psat @ Tair = Psat@30°C = 4.247 kPa
which is less than the claimed value of 5.2 kPa. Therefore, the claim is false.
3-104 The temperature and relative humidity of air over a swimming pool are given. The water temperature of the swimming pool when phase equilibrium conditions are established is to be determined. Assumptions The temperature and relative humidity of air over the pool remain constant. Properties The saturation pressure of water at 20°C is 2.339 kPa (Table A-4). Analysis The vapor pressure of air over the swimming pool is Pv , air = φPsat @ Tair = φPsat@20°C = (0.4)(2.339 kPa) = 0.9357 kPa
Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface. Therefore,
Patm, 20°C
POOL
Pv , water surface = Pv, air = 0.9357 kPa
and
Twater = Tsat @ Pv = Tsat @ 0.9357
kPa
= 6.0°C
Discussion Note that the water temperature drops to 6.0°C in an environment at 20°C when phase equilibrium is established.
3-47
3-105 Two rooms are identical except that they are maintained at different temperatures and relative humidities. The room that contains more moisture is to be determined. Properties The saturation pressure of water is 2.339 kPa at 20°C, and 4.247 kPa at 30°C (Table A-4). Analysis The vapor pressures in the two rooms are Room 1:
Pv1 = φ1 Psat @ T1 = φ1 Psat@30°C = (0.4)(4.247 kPa) = 1.699 kPa
Room 2:
Pv 2 = φ 2 Psat @ T2 = φ 2 Psat@20°C = (0.7)(2.339 kPa) = 1.637 kPa
Therefore, room 1 at 30°C and 40% relative humidity contains more moisture.
3-106E A thermos bottle half-filled with water is left open to air in a room at a specified temperature and pressure. The temperature of water when phase equilibrium is established is to be determined. Assumptions The temperature and relative humidity of air over the bottle remain constant. Properties The saturation pressure of water at 70°F is 0.3633 psia (Table A-4E). Analysis The vapor pressure of air in the room is Pv , air = φPsat @ Tair = φPsat@70°F = (0.35)(0.3633 psia) = 0.1272 psia
Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface. Therefore, Pv , water surface = Pv , air = 0.1272 psia
Thermos bottle 70°F 35%
and Twater = Tsat @ Pv = Tsat @ 0.1272
psia
= 41.1°F
Discussion Note that the water temperature drops to 41°F in an environment at 70°F when phase equilibrium is established.
3-107 A person buys a supposedly cold drink in a hot and humid summer day, yet no condensation occurs on the drink. The claim that the temperature of the drink is below 10°C is to be evaluated. Properties The saturation pressure of water at 35°C is 5.629 kPa (Table A-4). Analysis The vapor pressure of air is Pv , air = φPsat @ Tair = φPsat@35°C = (0.7)(5.629 kPa) = 3.940 kPa
The saturation temperature corresponding to this pressure (called the dew-point temperature) is
35°C 70%
Tsat = Tsat @ Pv =
[email protected] kPa = 28.7°C
That is, the vapor in the air will condense at temperatures below 28.7°C. Noting that no condensation is observed on the can, the claim that the drink is at 10°C is false.
3-48
Review Problems 3-108 The cylinder conditions before the heat addition process is specified. The pressure after the heat addition process is to be determined. Assumptions 1 The contents of cylinder are approximated by the air properties. 2 Air is an ideal gas.
Combustion chamber 1.8 MPa 450°C
Analysis The final pressure may be determined from the ideal gas relation P2 =
T2 1300 + 273 K P1 = (1800 kPa) = 3916 kPa T1 450 + 273 K
3-109 A rigid tank contains an ideal gas at a specified state. The final temperature is to be determined for two different processes. Analysis (a) The first case is a constant volume process. When half of the gas is withdrawn from the tank, the final temperature may be determined from the ideal gas relation as T2 =
m1 P2 100 kPa T1 = (2) (600 K) = 400 K m 2 P1 300 kPa
(b) The second case is a constant volume and constant mass process. The ideal gas relation for this case yields P2 =
Ideal gas 300 kPa 600 K
T2 400 K P1 = (300 kPa) = 200 kPa T1 600 K
3-110 Carbon dioxide flows through a pipe at a given state. The volume and mass flow rates and the density of CO2 at the given state and the volume flow rate at the exit of the pipe are to be determined. Analysis (a) The volume and mass flow rates may be determined from ideal gas relation as
3 MPa 500 K 0.4 kmol/s
CO2
V&1 =
N& Ru T1 (0.4 kmol/s)(8.314 kPa.m 3 /kmol.K)(500 K) = = 0.5543 m 3 /s 3000 kPa P
m& 1 =
(3000 kPa)(0.5543 m 3 / s ) P1V&1 = = 17.60 kg/s RT1 (0.1889 kPa.m3 /kg.K)(500 K)
The density is
ρ1 =
m& 1 (17.60 kg/s) = = 31.76 kg/m 3 3 & V1 (0.5543 m /s)
(b) The volume flow rate at the exit is
V&2 =
N& Ru T2 (0.4 kmol/s)(8.314 kPa.m 3 /kmol.K)(450 K) = = 0.4988 m 3 /s P 3000 kPa
450 K
3-49
3-111 A piston-cylinder device contains steam at a specified state. Steam is cooled at constant pressure. The volume change is to be determined using compressibility factor. Properties The gas constant, the critical pressure, and the critical temperature of steam are R = 0.4615 kPa·m3/kg·K, Tcr = 647.1 K,
Pcr = 22.06 MPa
Analysis The exact solution is given by the following: P = 200 kPa 3 v1 = 1.31623 m /kg T1 = 300°C P = 200 kPa 3 v 2 = 0.95986 m /kg T2 = 150°C
Steam 0.2 kg 200 kPa 300°C
(Table A-6)
Q
∆Vexact = m(v1 − v 2 ) = (0.2 kg)(1.31623 − 0.95986)m3 /kg = 0.07128 m3
Using compressibility chart (EES function for compressibility factor is used) P1 0.2 MPa = = 0.0091 Pcr 22.06 MPa Z 1 = 0.9956 T1 300 + 273 K = = = 0.886 Tcr 647.1 K
PR = TR ,1
0.2 MPa P2 = = 0.0091 Pcr 22.06 MPa Z 2 = 0.9897 T2 150 + 273 K = = = 0.65 647.1 K Tcr
PR = TR , 2
V1 =
Z1mRT1 (0.9956)(0.2 kg)(0.4615 kPa ⋅ m 3 /kg ⋅ K)(300 + 273 K) = = 0.2633 m3 (200 kPa) P1
V2 =
Z 2 mRT2 (0.9897)(0.2 kg)(0.4615 kPa ⋅ m3 /kg ⋅ K)(150 + 273 K) = = 0.1932 m3 P2 (200 kPa)
∆V chart = V1 − V 2 = 0.2633 − 0.1932 = 0.07006 m3 ,
Error : 1.7%
3-112 The cylinder conditions before the heat addition process is specified. The temperature after the heat addition process is to be determined. Assumptions 1 The contents of cylinder is approximated by the air properties. 2 Air is an ideal gas. Analysis The ratio of the initial to the final mass is m1 AF 22 22 = = = m2 AF + 1 22 + 1 23
The final temperature may be determined from ideal gas relation T2 =
3 m1 V 2 22 150 cm T1 = 3 m 2 V1 23 75 cm
(950 K) = 1817 K
Combustion chamber 950 K 75 cm3
3-50
3-113 (a) On the P-v diagram, the constant temperature process through the state P= 300 kPa, v = 0.525 m3/kg as pressure changes from P1 = 200 kPa to P2 = 400 kPa is to be sketched. The value of the temperature on the process curve on the P-v diagram is to be placed.
SteamIAPWS
106
105
104
] a P k[ P
103
102
400 300 200
2 133.5°C
1
101
0.525 100 10-4
10-3
10-2
10-1
100
101
102
3
v [m /kg]
(b) On the T-v diagram the constant specific vol-ume process through the state T = 120°C, v = 0.7163 m3/kg from P1= 100 kPa to P2 = 300 kPa is to be sketched.. For this data set, the temperature values at states 1 and 2 on its axis is to be placed. The value of the specific volume on its axis is also to be placed.
SteamIAPWS
700 600
300 kPa
500
] C °[ T
198.7 kPa
400
100 kPa 300 200
2
100 0 10-3
1 0.7163 10-2
10-1
100 3
v [m /kg]
101
102
3-51
3-114 The pressure in an automobile tire increases during a trip while its volume remains constant. The percent increase in the absolute temperature of the air in the tire is to be determined. Assumptions 1 The volume of the tire remains constant. 2 Air is an ideal gas. Properties The local atmospheric pressure is 90 kPa.
TIRE 200 kPa 3 0.035 m
Analysis The absolute pressures in the tire before and after the trip are P1 = Pgage,1 + Patm = 200 + 90 = 290 kPa P2 = Pgage,2 + Patm = 220 + 90 = 310 kPa
Noting that air is an ideal gas and the volume is constant, the ratio of absolute temperatures after and before the trip are P1V 1 P2V 2 T P 310 kPa = → 2 = 2 = = 1.069 T1 T2 T1 P1 290 kPa
Therefore, the absolute temperature of air in the tire will increase by 6.9% during this trip.
3-115 A hot air balloon with 3 people in its cage is hanging still in the air. The average temperature of the air in the balloon for two environment temperatures is to be determined. Assumptions Air is an ideal gas. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis The buoyancy force acting on the balloon is
V balloon = 4π r 3 / 3 = 4π (10m)3 /3 = 4189m3 ρcool air =
90 kPa P = = 1.089 kg/m3 RT (0.287 kPa ⋅ m3/kg ⋅ K)(288 K)
FB = ρcool air gV balloon 1N = 44,700 N = (1.089 kg/m3 )(9.8 m/s 2 )(4189 m3 ) 2 1kg ⋅ m/s Hot air balloon D = 20 m
The vertical force balance on the balloon gives FB = W hot air + Wcage + W people = (m hot air + m cage + m people ) g
Patm = 90 kPa T = 15°C
Substituting, 1N 44,700 N = (mhot air + 80 kg + 195 kg)(9.8 m/s 2 ) 2 1 kg ⋅ m/s
which gives m hot air = 4287 kg
mcage = 80 kg
Therefore, the average temperature of the air in the balloon is T=
PV (90 kPa)(4189 m3 ) = = 306.5 K mR (4287 kg)(0.287 kPa ⋅ m3/kg ⋅ K)
Repeating the solution above for an atmospheric air temperature of 30°C gives 323.6 K for the average air temperature in the balloon.
3-52
3-116 EES Problem 3-115 is to be reconsidered. The effect of the environment temperature on the average air temperature in the balloon when the balloon is suspended in the air is to be investigated as the environment temperature varies from -10°C to 30°C. The average air temperature in the balloon is to be plotted versus the environment temperature. Analysis The problem is solved using EES, and the solution is given below. "Given Data:" "atm---atmosphere about balloon" "gas---heated air inside balloon" g=9.807 [m/s^2] d_balloon=20 [m] m_cage = 80 [kg] m_1person=65 [kg] NoPeople = 6 {T_atm_Celsius = 15 [C]} T_atm =T_atm_Celsius+273 "[K]" P_atm = 90 [kPa] R=0.287 [kJ/kg-K] P_gas = P_atm T_gas_Celsius=T_gas - 273 "[C]" "Calculated values:" P_atm= rho_atm*R*T_atm "rho_atm = density of air outside balloon" P_gas= rho_gas*R*T_gas "rho_gas = density of gas inside balloon" r_balloon=d_balloon/2 V_balloon=4*pi*r_balloon^3/3 m_people=NoPeople*m_1person m_gas=rho_gas*V_balloon m_total=m_gas+m_people+m_cage "The total weight of balloon, people, and cage is:" W_total=m_total*g "The buoyancy force acting on the balloon, F_b, is equal to the weight of the air displaced by the balloon." F_b=rho_atm*V_balloon*g "From the free body diagram of the balloon, the balancing vertical forces must equal the product of the total mass and the vertical acceleration:" F_b- W_total=m_total*a_up a_up = 0 "The balloon is hanging still in the air" 100
Tatm,Celcius [C] -10 -5 0 5 10 15 20 25 30
Tgas,Celcius [C] 17.32 23.42 29.55 35.71 41.89 48.09 54.31 60.57 66.84
90 80
] C [ s ui sl e C, s a g
T
70
9 people
60
6 people
50
3 people
40 30 20 10 0 -10
-5
0
5
10
15
Tatm,Celsius [C]
20
25
30
3-53
3-117 A hot air balloon with 2 people in its cage is about to take off. The average temperature of the air in the balloon for two environment temperatures is to be determined. Assumptions Air is an ideal gas. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K.
Hot air balloon D = 18 m
Analysis The buoyancy force acting on the balloon is
V balloon = 4π r 3 / 3 = 4π (9 m) 3 /3 = 3054 m 3 ρ coolair =
93 kPa P = = 1.137 kg/m 3 RT (0.287 kPa ⋅ m 3 /kg ⋅ K)(285 K)
FB = ρ coolair gV balloon
1N = (1.137 kg/m 3 )(9.8 m/s 2 )(3054 m 3 ) 1 kg ⋅ m/s 2
Patm = 93 kPa T = 12°C
= 34,029 N
The vertical force balance on the balloon gives FB = W hot air + Wcage + W people = (m hot air + m cage + m people ) g
mcage = 120 kg
Substituting, 1N 34,029 N = (mhot air + 120 kg + 140 kg)(9.81 m/s 2 ) 2 1 kg ⋅ m/s
which gives m hot air = 3212 kg
Therefore, the average temperature of the air in the balloon is T=
PV (93 kPa)(3054 m3 ) = = 308 K mR (3212 kg)(0.287 kPa ⋅ m3/kg ⋅ K)
Repeating the solution above for an atmospheric air temperature of 25°C gives 323 K for the average air temperature in the balloon.
3-118E Water in a pressure cooker boils at 260°F. The absolute pressure in the pressure cooker is to be determined. Analysis The absolute pressure in the pressure cooker is the saturation pressure that corresponds to the boiling temperature, P = Psat @ 260o F = 35.45 psia
H2O 260°F
3-54
3-119 The refrigerant in a rigid tank is allowed to cool. The pressure at which the refrigerant starts condensing is to be determined, and the process is to be shown on a P-v diagram. Analysis This is a constant volume process (v = V /m = constant), and the specific volume is determined to be
v=
V m
=
0.117 m 3 = 0.117 m 3 /kg 1 kg
R-134a 240 kPa P
When the refrigerant starts condensing, the tank will contain saturated vapor only. Thus,
1
v 2 = v g = 0.117 m 3 /kg
2
The pressure at this point is the pressure that corresponds to this vg value, P2 = Psat @v
g = 0.117
m 3 /kg
v
= 169 kPa
3-120 The rigid tank contains saturated liquid-vapor mixture of water. The mixture is heated until it exists in a single phase. For a given tank volume, it is to be determined if the final phase is a liquid or a vapor. Analysis This is a constant volume process (v = V /m = constant), and thus the final specific volume will be equal to the initial specific volume,
v 2 = v1
H2O
V=4L
3
The critical specific volume of water is 0.003106 m /kg. Thus if the final specific volume is smaller than this value, the water will exist as a liquid, otherwise as a vapor.
V = 4 L →v =
V m
V = 400 L →v =
=
V m
m = 2 kg T = 50°C
0.004 m3 = 0.002 m3/kg < v cr Thus, liquid. 2 kg =
0.4 m3 = 0.2 m3/kg > v cr . Thus, vapor. 2 kg
3-121 Superheated refrigerant-134a is cooled at constant pressure until it exists as a compressed liquid. The changes in total volume and internal energy are to be determined, and the process is to be shown on a T-v diagram. Analysis The refrigerant is a superheated vapor at the initial state and a compressed liquid at the final state. From Tables A-13 and A-11, P1 = 1.2 MPa T1 = 70°C
kJ/kg } vu == 277.21 0.019502 m /kg
P2 = 1.2 MPa T2 = 20°C
} vu
1
1
2 2
T 1
3
≅ u f @ 20o C = 78.86 kJ/kg ≅ v f @ 20o C = 0.0008161 m3/kg
2
v
Thus, (b)
∆V = m(v 2 − v 1 ) = (10 kg)(0.0008161 − 0.019502) m 3 /kg = −0.187 m 3
(c)
∆U = m(u 2 − u1 ) = (10 kg)(78.86 − 277.21) kJ/kg = −1984 kJ
R-134a 70°C 1.2 MPa
3-55
3-122 Two rigid tanks that contain hydrogen at two different states are connected to each other. Now a valve is opened, and the two gases are allowed to mix while achieving thermal equilibrium with the surroundings. The final pressure in the tanks is to be determined. Properties The gas constant for hydrogen is 4.124 kPa·m3/kg·K (Table A-1). Analysis Let's call the first and the second tanks A and B. Treating H2 as an ideal gas, the total volume and the total mass of H2 are A B V = V A + V B = 0.5 + 0.5 = 1.0 m 3 H2
PV (600 kPa)(0.5 m 3 ) m A = 1 = = 0.248 kg 3 RT1 A (4.124 kPa ⋅ m /kg ⋅ K)(293 K)
V = 0.5 m3 T=20°C P=600 kPa
3
PV (150 kPa)(0.5 m ) m B = 1 = = 0.060 kg 3 RT1 B (4.124 kPa ⋅ m /kg ⋅ K)(303 K) m = m A + m B = 0.248 + 0.060 = 0.308kg
×
H2
V = 0.5 m3 T=30°C P=150 kPa
Then the final pressure can be determined from P=
mRT2
V
=
(0.308 kg)(4.124 kPa ⋅ m 3 /kg ⋅ K)(288 K) 1.0 m 3
= 365.8 kPa
3-123 EES Problem 3-122 is reconsidered. The effect of the surroundings temperature on the final equilibrium pressure in the tanks is to be investigated. The final pressure in the tanks is to be plotted versus the surroundings temperature, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given Data" V_A=0.5 [m^3] T_A=20 [C] P_A=600 [kPa] V_B=0.5 [m^3] T_B=30 [C] P_B=150 [kPa] {T_2=15 [C]} "Solution" R=R_u/MOLARMASS(H2) R_u=8.314 [kJ/kmol-K] V_total=V_A+V_B m_total=m_A+m_B P_A*V_A=m_A*R*(T_A+273) P_B*V_B=m_B*R*(T_B+273) P_2*V_total=m_total*R*(T_2+273) T2 [C] -10 -5 0 5 10 15 20 25 30
380 370
P 2 [kPa]
P2 [kPa] 334.4 340.7 347.1 353.5 359.8 366.2 372.5 378.9 385.2
390
360 350 340 330 -10
-5
0
5
10
T
2
[C]
15
20
25
30
3-56
3-124 A large tank contains nitrogen at a specified temperature and pressure. Now some nitrogen is allowed to escape, and the temperature and pressure of nitrogen drop to new values. The amount of nitrogen that has escaped is to be determined. Properties The gas constant for nitrogen is 0.2968 kPa·m3/kg·K (Table A-1). Analysis Treating N2 as an ideal gas, the initial and the final masses in the tank are determined to be m1 =
P1V (600 kPa)(20 m 3 ) = = 136.6 kg RT1 (0.2968kPa ⋅ m 3 /kg ⋅ K)(296 K)
m2 =
P2V (400 kPa)(20 m 3 ) = = 92.0 kg RT2 (0.2968 kPa ⋅ m 3 /kg ⋅ K)(293 K)
Thus the amount of N2 that escaped is ∆m = m1 − m 2 = 136.6 − 92.0 = 44.6 kg
N2 600 kPa 23°C 20 m3
3-125 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation, the generalized chart, and the steam tables. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1, R = 0.4615 kPa ⋅ m3/kg ⋅ K,
Tcr = 647.1 K,
Pcr = 22.06 MPa
Analysis (a) From the ideal gas equation of state, P=
RT
v
=
(0.4615 kPa ⋅ m3/kg ⋅ K)(673 K) = 15,529 kPa 0.02 m3/kg
(b) From the compressibility chart (Fig. A-15a), PR = 0.57 3 (0.02 m /kg)(22,060 kPa) v actual vR = = = 1.48 RTcr / Pcr (0.4615 kPa ⋅ m3/kg ⋅ K)(647.1 K) P = PR Pcr = 0.57 × 22,060 = 12,574 kPa 673 K T TR = = = 1.040 Tcr 647.1 K
Thus,
H2O 0.02 m3/kg 400°C
(c) From the superheated steam table, T = 400°C
P = 12,576 kPa
v = 0.02 m 3 /kg
(from EES)
3-126 One section of a tank is filled with saturated liquid R-134a while the other side is evacuated. The partition is removed, and the temperature and pressure in the tank are measured. The volume of the tank is to be determined. Analysis The mass of the refrigerant contained in the tank is m=
since
V1 0.01 m 3 = = 11.82 kg v 1 0.0008458 m 3 /kg
v 1 = v f @ 0.8MPa = 0.0008458 m 3 /kg
At the final state (Table A-13), P2 = 400 kPa v = 0.05421 m3/kg 2 T2 = 20°C
}
Thus,
R-134a P=0.8 MPa V =0.01 m3
V tank = V 2 = mv 2 = (11.82 kg)(0.05421 m 3 /kg) = 0.641 m 3
Evacuated
3-57
3-127 EES Problem 3-126 is reconsidered. The effect of the initial pressure of refrigerant-134 on the volume of the tank is to be investigated as the initial pressure varies from 0.5 MPa to 1.5 MPa. The volume of the tank is to be plotted versus the initial pressure, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given Data" x_1=0.0 Vol_1=0.01[m^3] P_1=800 [kPa] T_2=20 [C] P_2=400 [kPa] "Solution" v_1=volume(R134a,P=P_1,x=x_1) Vol_1=m*v_1 v_2=volume(R134a,P=P_2,T=T_2) Vol_2=m*v_2
P1 [kPa] 500 600 700 800 900 1000 1100 1200 1300 1400 1500
Vol2 [m3] 0.6727 0.6612 0.6507 0.641 0.6318 0.6231 0.6148 0.6068 0.599 0.5914 0.584
m [kg] 12.41 12.2 12 11.82 11.65 11.49 11.34 11.19 11.05 10.91 10.77
0.68
0.66
3
]
0.64
m [ l2 o V
0.62
0.6
0.58 500
700
900
1100
P1 [kPa]
1300
1500
3-58
3-128 A propane tank contains 5 L of liquid propane at the ambient temperature. Now a leak develops at the top of the tank and propane starts to leak out. The temperature of propane when the pressure drops to 1 atm and the amount of heat transferred to the tank by the time the entire propane in the tank is vaporized are to be determined. Properties The properties of propane at 1 atm are Tsat = -42.1°C, ρ = 581 kg / m 3 , and hfg = 427.8 kJ/kg (Table A-3). Analysis The temperature of propane when the pressure drops to 1 atm is simply the saturation pressure at that temperature, T = Tsat @1 atm = −42.1° C
Propane 5L
The initial mass of liquid propane is m = ρV = (581 kg/m3 )(0.005 m3 ) = 2.905 kg
20°C
The amount of heat absorbed is simply the total heat of vaporization, Qabsorbed = mh fg = (2.905 kg)(427.8 kJ / kg) = 1243 kJ
Leak
3-129 An isobutane tank contains 5 L of liquid isobutane at the ambient temperature. Now a leak develops at the top of the tank and isobutane starts to leak out. The temperature of isobutane when the pressure drops to 1 atm and the amount of heat transferred to the tank by the time the entire isobutane in the tank is vaporized are to be determined. Properties The properties of isobutane at 1 atm are Tsat = -11.7°C, ρ = 593.8 kg / m 3 , and hfg = 367.1 kJ/kg (Table A-3). Analysis The temperature of isobutane when the pressure drops to 1 atm is simply the saturation pressure at that temperature, T = Tsat @1 atm = −11.7° C Isobutane 5L
The initial mass of liquid isobutane is m = ρV = (593.8 kg/m 3 )(0.005 m 3 ) = 2.969kg
20°C
The amount of heat absorbed is simply the total heat of vaporization, Qabsorbed = mh fg = (2.969 kg)(367.1 kJ / kg) = 1090 kJ
Leak
3-130 A tank contains helium at a specified state. Heat is transferred to helium until it reaches a specified temperature. The final gage pressure of the helium is to be determined. Assumptions 1 Helium is an ideal gas. Properties The local atmospheric pressure is given to be 100 kPa. Analysis Noting that the specific volume of helium in the tank remains constant, from ideal gas relation, we have P2 = P1
T2 (300 + 273)K = 169.0 kPa = (10 + 100 kPa) (100 + 273)K T1
Then the gage pressure becomes Pgage,2 = P2 − Patm = 169.0 − 100 = 69.0 kPa
Helium 100ºC 10 kPa gage
Q
3-59
3-131 A tank contains argon at a specified state. Heat is transferred from argon until it reaches a specified temperature. The final gage pressure of the argon is to be determined. Assumptions 1 Argon is an ideal gas. Properties The local atmospheric pressure is given to be 100 kPa. Analysis Noting that the specific volume of argon in the tank remains constant, from ideal gas relation, we have P2 = P1
Argon 600ºC 200 kPa gage
T2 (300 + 273)K = 196.9 kPa = (200 + 100 kPa) (600 + 273)K T1
Q
Then the gage pressure becomes Pgage,2 = P2 − Patm = 196.9 − 100 = 96.9 kPa
3-132 Complete the following table for H2 O: P, kPa 200 270.3
T, °C 30 130
v, m3 / kg 0.001004 -
u, kJ/kg 125.71 -
200 300
400 133.52
1.5493 0.500
2967.2 2196.4
500
473.1
0.6858
3084
Phase description Compressed liquid Insufficient information Superheated steam Saturated mixture, x=0.825 Superheated steam
3-133 Complete the following table for R-134a: P, kPa 320 1000
T, °C -12 39.37
v, m3 / kg
u, kJ/kg
0.0007497 -
35.72 -
140 180
40 -12.73
0.17794 0.0700
263.79 153.66
200
22.13
0.1152
249
Phase description Compressed liquid Insufficient information Superheated vapor Saturated mixture, x=0.6315 Superheated vapor
3-60
3-134 (a) On the P-v diagram the constant temperature process through the state P = 280 kPa, v = 0.06 m3/kg as pressure changes from P1 = 400 kPa to P2 = 200 kPa is to be sketched. The value of the temperature on the process curve on the P-v diagram is to be placed. R134a
105
104
] a P k[ P
103
102
400 280 200
1 -1.25°C
2
0.06
101 10-4
10-3
10-2
10-1
100
101
3
v [m /kg]
(b) On the T-v diagram the constant specific volume process through the state T = 20°C, v = 0.02 m3/kg from P1 = 1200 kPa to P2 = 300 kPa is to be sketched. For this data set the temperature values at states 1 and 2 on its axis is to be placed. The value of the specific volume on its axis is also to be placed. R134a
250 200 150
] C °[ T
100
1
75
50
1200 kPa
20 0 0.6
572 kPa 300 kPa
2
-50
0.02 -100 10-4
10-3
10-2 3
v [m /kg]
10-1
100
3-61
Fundamentals of Engineering (FE) Exam Problems 3-135 A rigid tank contains 6 kg of an ideal gas at 3 atm and 40°C. Now a valve is opened, and half of mass of the gas is allowed to escape. If the final pressure in the tank is 2.2 atm, the final temperature in the tank is (a) 186°C (b) 59°C (c) -43°C (d) 20°C (e) 230°C Answer (a) 186°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). "When R=constant and V= constant, P1/P2=m1*T1/m2*T2" m1=6 "kg" P1=3 "atm" P2=2.2 "atm" T1=40+273 "K" m2=0.5*m1 "kg" P1/P2=m1*T1/(m2*T2) T2_C=T2-273 "C" "Some Wrong Solutions with Common Mistakes:" P1/P2=m1*(T1-273)/(m2*W1_T2) "Using C instead of K" P1/P2=m1*T1/(m1*(W2_T2+273)) "Disregarding the decrease in mass" P1/P2=m1*T1/(m1*W3_T2) "Disregarding the decrease in mass, and not converting to deg. C" W4_T2=(T1-273)/2 "Taking T2 to be half of T1 since half of the mass is discharged" 3-136 The pressure of an automobile tire is measured to be 190 kPa (gage) before a trip and 215 kPa (gage) after the trip at a location where the atmospheric pressure is 95 kPa. If the temperature of air in the tire before the trip is 25°C, the air temperature after the trip is (a) 51.1°C (b) 64.2°C (c) 27.2°C (d) 28.3°C (e) 25.0°C Answer (a) 51.1°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). "When R, V, and m are constant, P1/P2=T1/T2" Patm=95 P1=190+Patm "kPa" P2=215+Patm "kPa" T1=25+273 "K" P1/P2=T1/T2 T2_C=T2-273 "C" "Some Wrong Solutions with Common Mistakes:" P1/P2=(T1-273)/W1_T2 "Using C instead of K" (P1-Patm)/(P2-Patm)=T1/(W2_T2+273) "Using gage pressure instead of absolute pressure" (P1-Patm)/(P2-Patm)=(T1-273)/W3_T2 "Making both of the mistakes above" W4_T2=T1-273 "Assuming the temperature to remain constant"
3-62 3-137 A 300-m3 rigid tank is filled with saturated liquid-vapor mixture of water at 200 kPa. If 25% of the mass is liquid and the 75% of the mass is vapor, the total mass in the tank is (a) 451 kg (b) 556 kg (c) 300 kg (d) 331 kg (e) 195 kg Answer (a) 451 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V_tank=300 "m3" P1=200 "kPa" x=0.75 v_f=VOLUME(Steam_IAPWS, x=0,P=P1) v_g=VOLUME(Steam_IAPWS, x=1,P=P1) v=v_f+x*(v_g-v_f) m=V_tank/v "kg" "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" T=TEMPERATURE(Steam_IAPWS,x=0,P=P1) P1*V_tank=W1_m*R*(T+273) "Treating steam as ideal gas" P1*V_tank=W2_m*R*T "Treating steam as ideal gas and using deg.C" W3_m=V_tank "Taking the density to be 1 kg/m^3"
3-138 Water is boiled at 1 atm pressure in a coffee maker equipped with an immersion-type electric heating element. The coffee maker initially contains 1 kg of water. Once boiling started, it is observed that half of the water in the coffee maker evaporated in 18 minutes. If the heat loss from the coffee maker is negligible, the power rating of the heating element is (a) 0.90 kW (b) 1.52 kW (c) 2.09 kW (d) 1.05 kW (e) 1.24 kW Answer (d) 1.05 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_1=1 "kg" P=101.325 "kPa" time=18*60 "s" m_evap=0.5*m_1 Power*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Power*time=m_evap*h_g "Using h_g" W2_Power*time/60=m_evap*h_g "Using minutes instead of seconds for time" W3_Power=2*Power "Assuming all the water evaporates"
3-63 3-139 A 1-m3 rigid tank contains 10 kg of water (in any phase or phases) at 160°C. The pressure in the tank is (a) 738 kPa (b) 618 kPa (c) 370 kPa (d) 2000 kPa (e) 1618 kPa Answer (b) 618 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V_tank=1 "m^3" m=10 "kg" v=V_tank/m T=160 "C" P=PRESSURE(Steam_IAPWS,v=v,T=T) "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" W1_P*V_tank=m*R*(T+273) "Treating steam as ideal gas" W2_P*V_tank=m*R*T "Treating steam as ideal gas and using deg.C"
3-140 Water is boiling at 1 atm pressure in a stainless steel pan on an electric range. It is observed that 2 kg of liquid water evaporates in 30 minutes. The rate of heat transfer to the water is (a) 2.51 kW (b) 2.32 kW (c) 2.97 kW (d) 0.47 kW (e) 3.12 kW Answer (a) 2.51 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_evap=2 "kg" P=101.325 "kPa" time=30*60 "s" Q*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Q*time=m_evap*h_g "Using h_g" W2_Q*time/60=m_evap*h_g "Using minutes instead of seconds for time" W3_Q*time=m_evap*h_f "Using h_f"
3-64
3-141 Water is boiled in a pan on a stove at sea level. During 10 min of boiling, its is observed that 200 g of water has evaporated. Then the rate of heat transfer to the water is (a) 0.84 kJ/min (b) 45.1 kJ/min (c) 41.8 kJ/min (d) 53.5 kJ/min (e) 225.7 kJ/min Answer (b) 45.1 kJ/min Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_evap=0.2 "kg" P=101.325 "kPa" time=10 "min" Q*time=m_evap*h_fg "kJ" h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Q*time=m_evap*h_g "Using h_g" W2_Q*time*60=m_evap*h_g "Using seconds instead of minutes for time" W3_Q*time=m_evap*h_f "Using h_f"
3-142 A rigid 3-m3 rigid vessel contains steam at 10 MPa and 500°C. The mass of the steam is (a) 3.0 kg (b) 19 kg (c) 84 kg (d) 91 kg (e) 130 kg Answer (d) 91 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=3 "m^3" m=V/v1 "m^3/kg" P1=10000 "kPa" T1=500 "C" v1=VOLUME(Steam_IAPWS,T=T1,P=P1) "Some Wrong Solutions with Common Mistakes:" R=0.4615 "kJ/kg.K" P1*V=W1_m*R*(T1+273) "Treating steam as ideal gas" P1*V=W2_m*R*T1 "Treating steam as ideal gas and using deg.C"
3-65
3-143 Consider a sealed can that is filled with refrigerant-134a. The contents of the can are at the room temperature of 25°C. Now a leak developes, and the pressure in the can drops to the local atmospheric pressure of 90 kPa. The temperature of the refrigerant in the can is expected to drop to (rounded to the nearest integer) (a) 0°C (b) -29°C (c) -16°C (d) 5°C (e) 25°C Answer (b) -29°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 "C" P2=90 "kPa" T2=TEMPERATURE(R134a,x=0,P=P2) "Some Wrong Solutions with Common Mistakes:" W1_T2=T1 "Assuming temperature remains constant"
3-144 … 3-146 Design, Essay and Experiment Problems 3-144 It is helium.
KJ
4-1
Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS Moving Boundary Work 4-1C It represents the boundary work for quasi-equilibrium processes. 4-2C Yes. 4-3C The area under the process curve, and thus the boundary work done, is greater in the constant pressure case. 4-4C 1 kPa ⋅ m 3 = 1 k(N / m 2 ) ⋅ m 3 = 1 kN ⋅ m = 1 kJ
4-5 A piston-cylinder device contains nitrogen gas at a specified state. The boundary work is to be determined for the polytropic expansion of nitrogen. Properties The gas constant for nitrogen is 0.2968 kJ/kg.K (Table A-2). Analysis The mass and volume of nitrogen at the initial state are m=
V2 =
P1V1 (130 kPa)(0.07 m 3 ) = = 0.07802 kg RT1 (0.2968 kJ/kg.K)(120 + 273 K)
N2 130 kPa 120°C
mRT2 (0.07802 kg)(0.2968 kPa.m 3 /kg.K)(100 + 273 K) = = 0.08637 m 3 P2 100 kPa
The polytropic index is determined from P1V1n = P2V 2n →(130 kPa)(0.07 m 3 ) n = (100 kPa)(0.08637 m 3 ) n → n = 1.249
The boundary work is determined from Wb =
P2V 2 − P1V 1 (100 kPa)(0.08637 m 3 ) − (130 kPa)(0.07 m 3 ) = = 1.86 kJ 1− n 1 − 1.249
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-2
4-6 A piston-cylinder device with a set of stops contains steam at a specified state. Now, the steam is cooled. The compression work for two cases and the final temperature are to be determined. Analysis (a) The specific volumes for the initial and final states are (Table A-6) P1 = 1 MPa 3 v1 = 0.30661 m /kg T1 = 400°C
P2 = 1 MPa 3 v 2 = 0.23275 m /kg T2 = 250°C
Noting that pressure is constant during the process, the boundary work is determined from Wb = mP (v1 − v 2 ) = (0.3 kg)(1000 kPa)(0.30661 − 0.23275)m3/kg = 22.16 kJ
(b) The volume of the cylinder at the final state is 60% of initial volume. Then, the boundary work becomes
Steam 0.3 kg 1 MPa 400°C
Wb = mP (v1 − 0.60v1 ) = (0.3 kg)(1000 kPa)(0.30661 − 0.60 × 0.30661)m3/kg = 36.79 kJ
The temperature at the final state is P2 = 0.5 MPa
T2 = 151.8°C (Table A-5) v 2 = (0.60 × 0.30661) m /kg 3
4-7 A piston-cylinder device contains nitrogen gas at a specified state. The final temperature and the boundary work are to be determined for the isentropic expansion of nitrogen. Properties The properties of nitrogen are R = 0.2968 kJ/kg.K , k = 1.4 (Table A-2a) Analysis The mass and the final volume of nitrogen are m=
P1V1 (130 kPa)(0.07 m 3 ) = = 0.07802 kg RT1 (0.2968 kJ/kg.K)(120 + 273 K)
N2 130 kPa 120°C
P1V1k = P2V 2k →(130 kPa)(0.07 m 3 )1.4 = (100 kPa)V 21.4 →V 2 = 0.08443 m 3
The final temperature and the boundary work are determined as T2 =
P2V 2 (100 kPa)(0.08443 m 3 ) = = 364.6 K mR (0.07802 kg)(0.2968 kPa.m 3 /kg.K)
Wb =
P2V 2 − P1V 1 (100 kPa)(0.08443 m 3 ) − (130 kPa)(0.07 m 3 ) = 1.64 kJ = 1− k 1 − 1.4
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Q
4-3
4-8 Saturated water vapor in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-4 through A-6) P1 = 300 kPa 3 v 1 = v g @ 300 kPa = 0.60582 m /kg Sat. vapor P2 = 300 kPa 3 v 2 = 0.71643 m /kg T2 = 200°C
P (kPa 1
300
2
Analysis The boundary work is determined from its definition to be Wb,out =
∫
2
1
V
P dV = P (V 2 − V1 ) = mP(v 2 − v1 )
1 kJ = (5 kg)(300 kPa)(0.71643 − 0.60582) m3/kg 3 1 kPa ⋅ m = 165.9 kJ
Discussion The positive sign indicates that work is done by the system (work output).
4-9 Refrigerant-134a in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-11 through A-13) P1 = 900 kPa 3 v 1 = v f @ 900 kPa = 0.0008580 m /kg P2 = 900 kPa 3 v 2 = 0.027413 m /kg T2 = 70°C Sat. liquid
P (kPa) 900
1
2
Analysis The boundary work is determined from its definition to be m=
V1 0.2 m 3 = = 233.1 kg v 1 0.0008580 m 3 /kg
and Wb,out =
∫
2
1
P dV = P (V 2 − V1 ) = mP(v 2 − v1 )
1 kJ = (233.1 kg)(900 kPa)(0.027413 − 0.0008580)m3/kg 3 1 kPa ⋅ m = 5571 kJ
Discussion The positive sign indicates that work is done by the system (work output).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
v
4-4
4-10 EES Problem 4-9 is reconsidered. The effect of pressure on the work done as the pressure varies from 400 kPa to 1200 kPa is to be investigated. The work done is to be plotted versus the pressure. Analysis The problem is solved using EES, and the solution is given below. "Knowns" Vol_1L=200 [L] x_1=0 "saturated liquid state" P=900 [kPa] T_2=70 [C] "Solution" Vol_1=Vol_1L*convert(L,m^3) "The work is the boundary work done by the R-134a during the constant pressure process." W_boundary=P*(Vol_2-Vol_1)
125
Vol_1=m*v_1 v_1=volume(R134a,P=P,x=x_1) Vol_2=m*v_2 v_2=volume(R134a,P=P,T=T_2)
100 75
] C °[ T
"Plot information:" v[1]=v_1 v[2]=v_2 P[1]=P P[2]=P T[1]=temperature(R134a,P=P,x=x_1) T[2]=T_2 P [kPa] 400 500 600 700 800 900 1000 1100 1200
R134a
150
"The mass is:"
Wboundary [kJ] 6643 6405 6183 5972 5769 5571 5377 5187 4999
2
50 25
1
900 kPa
0 -25 -50 10-4
10-3
10-2
10-1
3
v [m /kg] R134a
105
104
] a P k[ P
103
2 1
102
101 10-4
10-3
10-2
10-1
3
v [m /kg]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-5
7250 6800
] J k[
y r a d n u o b
W
P = 800 kPa
6350 5900 5450 5000 50
60
70
80
90
100
110
120
130
T[2] [C] 7500
T2 = 100 C
7150
] J k[
y r a d n u o b
W
6800 6450 6100 5750 400
500
600
700
800
900
1000 1100 1200
P [kPa] 7000
6500
] J k[
y r a d n u o b
W
T2 = 70 C
6000
5500
5000
4500 400
500
600
700
800
900
1000 1100 1200
P [kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-6
4-11E Superheated water vapor in a cylinder is cooled at constant pressure until 70% of it condenses. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-4E through A-6E) P1 = 40 psia 3 v 1 = 15.686 ft /lbm T1 = 600°F P2 = 40 psia v 2 = v f + x 2v fg x 2 = 0.3 = 0.01715 + 0.3(10.501 − 0.01715)
P (psia) 2
40
1
= 3.1623 ft 3 /lbm
v
Analysis The boundary work is determined from its definition to be Wb,out =
∫
2
1
P dV = P (V 2 − V1 ) = mP(v 2 − v1 )
1 Btu = (16 lbm)(40 psia)(3.1623 − 15.686)ft 3/lbm 3 5.4039 psia ⋅ ft = −1483 Btu
Discussion The negative sign indicates that work is done on the system (work input).
4-12 Air in a cylinder is compressed at constant temperature until its pressure rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1 The process is quasi-equilibrium. 2 Air is an ideal gas. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).
P
Analysis The boundary work is determined from its definition to be 2 Wb,out =
∫
2
1
P dV = P1V1 ln
V2 P = mRT ln 1 V1 P2
= (2.4 kg)(0.287 kJ/kg ⋅ K)(285 K)ln
T = 12°C 150 kPa 600 kPa
1
= −272 kJ
Discussion The negative sign indicates that work is done on the system (work input).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
V
4-7
4-13 Nitrogen gas in a cylinder is compressed at constant temperature until its pressure rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1 The process is quasi-equilibrium. 2 Nitrogen is an ideal gas. Analysis The boundary work is determined from its definition to be Wb,out =
2
∫
1
P
P V P dV = P1V1 ln 2 = P1V1 ln 1 P2 V1
2
150 kPa 1 kJ = (150 kPa)(0.2 m3 ) ln 3 800 kPa 1 kPa ⋅ m = −50.2 kJ
T = 300 K 1
V
Discussion The negative sign indicates that work is done on the system (work input).
4-14 A gas in a cylinder is compressed to a specified volume in a process during which the pressure changes linearly with volume. The boundary work done during this process is to be determined by plotting the process on a P-V diagram and also by integration. Assumptions The process is quasi-equilibrium. Analysis (a) The pressure of the gas changes linearly with volume, and thus the process curve on a P-V diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus, P1 = aV 1 + b = (−1200 kPa/m 3 )(0.42 m 3 ) + (600 kPa) = 96 kPa P2 = aV 2 + b = (−1200 kPa/m 3 )(0.12 m 3 ) + (600 kPa) = 456 kPa
P (kPa) P2
and P1 + P2 (V 2 −V1 ) 2 1 kJ (96 + 456)kPa (0.12 − 0.42)m 3 = 1 kPa ⋅ m 3 2 = −82.8 kJ
2
1
P1
Wb,out = Area =
P = aV + b
0.12
0.42
(b) The boundary work can also be determined by integration to be Wb,out =
∫
2
∫
2
1
(aV + b)dV = a
V 22 −V12
+ b(V 2 −V1 ) 2 (0.12 2 − 0.42 2 )m 6 = (−1200 kPa/m 3 ) + (600 kPa)(0.12 − 0.42)m 3 2 = −82.8 kJ 1
P dV =
GAS P = aV + b
Discussion The negative sign indicates that work is done on the system (work input).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
V
(m3)
4-8
4-15E A gas in a cylinder is heated and is allowed to expand to a specified pressure in a process during which the pressure changes linearly with volume. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis (a) The pressure of the gas changes linearly with volume, and thus the process curve on a P-V diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus, At state 1:
P (psia)
P1 = aV1 + b 15 psia = (5 psia/ft 3 )(7 ft 3 ) + b
P = aV + b 2
100
b = −20 psia
At state 2:
1
15
P2 = aV 2 + b 100 psia = (5 psia/ft 3 )V 2 + (−20 psia)
V 3 (ft )
7
V 2 = 24 ft 3 and, Wb,out = Area =
P1 + P2 1 Btu (100 + 15)psia (24 − 7)ft 3 (V 2 −V1 ) = 5.4039 psia ⋅ ft 3 2 2
= 181 Btu
Discussion The positive sign indicates that work is done by the system (work output).
4-16 [Also solved by EES on enclosed CD] A gas in a cylinder expands polytropically to a specified volume. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis The boundary work for this polytropic process can be determined directly from V P2 = P1 1 V 2
n
0.03 m 3 = (150 kPa) 0.2 m 3
1.3
P (kPa ) 150
= 12.74 kPa
and, Wb,out =
∫
2
1
P dV =
P2V 2 − P1V1 1− n
(12.74 × 0.2 − 150 × 0.03) kPa ⋅ m 3 = 1 − 1.3 = 6.51 kJ
1
PV
1 kJ 1 kPa ⋅ m 3
2
0.03
0.2
Discussion The positive sign indicates that work is done by the system (work output).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
V (m3)
4-9
4-17 EES Problem 4-16 is reconsidered. The process described in the problem is to be plotted on a P-V diagram, and the effect of the polytropic exponent n on the boundary work as the polytropic exponent varies from 1.1 to 1.6 is to be plotted. Analysis The problem is solved using EES, and the solution is given below. Function BoundWork(P[1],V[1],P[2],V[2],n) "This function returns the Boundary Work for the polytropic process. This function is required since the expression for boundary work depens on whether n=1 or n1" If n1 then BoundWork:=(P[2]*V[2]-P[1]*V[1])/(1-n)"Use Equation 3-22 when n=1" else BoundWork:= P[1]*V[1]*ln(V[2]/V[1]) "Use Equation 3-20 when n=1" endif end "Inputs from the diagram window" {n=1.3 P[1] = 150 [kPa] V[1] = 0.03 [m^3] V[2] = 0.2 [m^3] Gas$='AIR'} "System: The gas enclosed in the piston-cylinder device." "Process: Polytropic expansion or compression, P*V^n = C" P[2]*V[2]^n=P[1]*V[1]^n "n = 1.3" "Polytropic exponent" "Input Data" W_b = BoundWork(P[1],V[1],P[2],V[2],n)"[kJ]" "If we modify this problem and specify the mass, then we can calculate the final temperature of the fluid for compression or expansion" m[1] = m[2] "Conservation of mass for the closed system" "Let's solve the problem for m[1] = 0.05 kg" m[1] = 0.05 [kg] "Find the temperatures from the pressure and specific volume." T[1]=temperature(gas$,P=P[1],v=V[1]/m[1]) T[2]=temperature(gas$,P=P[2],v=V[2]/m[2])
8 n 1.1 1.156 1.211 1.267 1.322 1.378 1.433 1.489 1.544 1.6
Wb [kJ] 7.776 7.393 7.035 6.7 6.387 6.094 5.82 5.564 5.323 5.097
7.5 7
] J k[ b
W
6.5 6 5.5 5 1.1
1.2
1.3
n
1.4
1.5
1.6
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-10
4-18 Nitrogen gas in a cylinder is compressed polytropically until the temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1 The process is quasi-equilibrium. 2 Nitrogen is an ideal gas. Properties The gas constant for nitrogen is R = 0.2968 kJ/kg.K (Table A-2a) Analysis The boundary work for this polytropic process can be determined from
P 2
P V − PV mR(T2 − T1 ) = P dV = 2 2 1 1 = 1 1− n 1− n (2 kg)(0.2968 kJ/kg ⋅ K)(360 − 300)K = 1 − 1.4 = −89.0 kJ
∫
Wb,out
2
PVn = C 1
V
Discussion The negative sign indicates that work is done on the system (work input).
4-19 [Also solved by EES on enclosed CD] A gas whose equation of state is v ( P + 10 / v 2 ) = Ru T expands in a cylinder isothermally to a specified volume. The unit of the quantity 10 and the boundary work done during this process are to be determined. Assumptions The process is quasi-equilibrium. Analysis (a) The term 10 / v is added to P.
2
P
must have pressure units since it
Thus the quantity 10 must have the unit kPa·m6/kmol2. T = 300 K
(b) The boundary work for this process can be determined from P=
Ru T
v
−
10
v
2
=
Ru T NRu T 10 N 2 10 − = − V / N (V / N ) 2 V V2
2
4
and 2 NRuT − 10 N dV = NRuT ln V 2 + 10 N 2 1 − 1 2 1 1 V1 V V 2 V1 V 4 m3 = (0.5 kmol)(8.314 kJ/kmol ⋅ K)(300 K)ln 2 m3 1 1 1 kJ + (10 kPa ⋅ m 6 /kmol2 )(0.5kmol)2 − 3 3 3 4 m 2 m 1 kPa m ⋅ = 864 kJ
Wb,out =
∫
2
P dV =
∫
2
Discussion The positive sign indicates that work is done by the system (work output).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
V
4-11
4-20 EES Problem 4-19 is reconsidered. Using the integration feature, the work done is to be calculated and compared, and the process is to be plotted on a P-V diagram. Analysis The problem is solved using EES, and the solution is given below. "Input Data" N=0.5 [kmol] v1_bar=2/N "[m^3/kmol]" v2_bar=4/N "[m^3/kmol]" T=300 [K] R_u=8.314 [kJ/kmol-K] "The quation of state is:" v_bar*(P+10/v_bar^2)=R_u*T "P is in kPa" "using the EES integral function, the boundary work, W_bEES, is" W_b_EES=N*integral(P,v_bar, v1_bar, v2_bar,0.01) "We can show that W_bhand= integeral of Pdv_bar is (one should solve for P=F(v_bar) and do the integral 'by hand' for practice)." W_b_hand = N*(R_u*T*ln(v2_bar/v1_bar) +10*(1/v2_bar-1/v1_bar)) "To plot P vs v_bar, define P_plot =f(v_bar_plot, T) as" {v_bar_plot*(P_plot+10/v_bar_plot^2)=R_u*T} " P=P_plot and v_bar=v_bar_plot just to generate the parametric table for plotting purposes. To plot P vs v_bar for a new temperature or v_bar_plot range, remove the '{' and '}' from the above equation, and reset the v_bar_plot values in the Parametric Table. Then press F3 or select Solve Table from the Calculate menu. Next select New Plot Window under the Plot menu to plot the new data." vplot 4 4.444 4.889 5.333 5.778 6.222 6.667 7.111 7.556 8
P vs v bar
650 600
1
550 500
T = 300 K
450
P plot [kPa]
Pplot 622.9 560.7 509.8 467.3 431.4 400.6 373.9 350.5 329.9 311.6
400 350
2
300 250
Area = W boundary
200 150 100 50 0 3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
v plot [m ^3/km ol]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
8.5
4-12
4-21 CO2 gas in a cylinder is compressed until the volume drops to a specified value. The pressure changes during the process with volume as P = aV −2 . The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium.
P
Analysis The boundary work done during this process is determined from Wb,out =
2
∫ PdV = ∫ 1
2
1 a 1 − 2 dV = − a V V 2 V1
2
1
1 1 = −(8 kPa ⋅ m 6 ) − 0.1 m 3 0.3 m 3 = −53.3 kJ
P = aV--2
1 kJ 1 kPa ⋅ m 3
1 V 3 (m )
0.3
0.1
Discussion The negative sign indicates that work is done on the system (work input).
4-22E Hydrogen gas in a cylinder equipped with a spring is heated. The gas expands and compresses the spring until its volume doubles. The final pressure, the boundary work done by the gas, and the work done against the spring are to be determined, and a P-V diagram is to be drawn. Assumptions 1 The process is quasi-equilibrium. 2 Hydrogen is an ideal gas. Analysis (a) When the volume doubles, the spring force and the final pressure of H2 becomes Fs = kx 2 = k P2 = P1 +
15 ft 3 ∆V = (15,000 lbf/ft) = 75,000 lbf A 3 ft 2
Fs 75,000 lbf = (14.7 psia) + A 3 ft 2
1 ft 2 144 in 2
P
= 188.3 psia
(b) The pressure of H2 changes linearly with volume during this process, and thus the process curve on a P-V diagram will be a straight line. Then the boundary work during this process is simply the area under the process curve, which is a trapezoid. Thus, Wb,out = Area = =
2 1
15
30
V (ft3)
P1 + P2 (V 2 −V 1 ) 2
1 Btu (188.3 + 14.7)psia (30 − 15)ft 3 5.40395 psia ⋅ ft 3 2
= 281.7 Btu
(c) If there were no spring, we would have a constant pressure process at P = 14.7 psia. The work done during this process would be Wb,out, no spring =
2
∫ PdV = P(V 1
2
−V 1 )
1 Btu = (14.7 psia)(30 − 15) ft 3 5.40395 psia ⋅ ft 3
= 40.8 Btu
Thus, Wspring = Wb − Wb,no spring = 281.7 − 40.8 = 240.9 Btu
Discussion The positive sign for boundary work indicates that work is done by the system (work output).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-13
4-23 Water in a cylinder equipped with a spring is heated and evaporated. The vapor expands until it compresses the spring 20 cm. The final pressure and temperature, and the boundary work done are to be determined, and the process is to be shown on a P-V diagram. Assumptions The process is quasi-equilibrium. Analysis (a) The final pressure is determined from P3 = P2 +
Fs (100 kN/m)(0.2 m) 1 kPa kx = P2 + = (250 kPa) + 1 kN/m 2 A A 0.1 m 2
= 450 kPa
The specific and total volumes at the three states are T1 = 25°C
3 v1 ≅ v f @ 25o C = 0.001003 m /kg P1 = 250 kPa
V1 = mv1 = (50 kg)(0.001003 m3/kg) = 0.05 m3
P 3
1
V 2 = 0.2 m3
2
V3 = V 2 + x23 Ap = (0.2 m3 ) + (0.2 m)(0.1 m 2 ) = 0.22 m3 v3 =
V3 m
=
v
0.22 m3 = 0.0044 m3/kg 50 kg
At 450 kPa, vf = 0.001088 m3/kg and vg = 0.41392 m3/kg. Noting that vf < v3 < vg , the final state is a saturated mixture and thus the final temperature is T3 = Tsat@450 kPa = 147.9°C
(b) The pressure remains constant during process 1-2 and changes linearly (a straight line) during process 2-3. Then the boundary work during this process is simply the total area under the process curve, Wb,out = Area = P1 (V 2 −V1 ) +
P2 + P3 (V 3 −V 2 ) 2
(250 + 450) kPa 1 kJ (0.22 − 0.2)m 3 = (250 kPa)(0.2 − 0.05)m 3 + 2 1 kPa ⋅ m 3
= 44.5 kJ
Discussion The positive sign indicates that work is done by the system (work output).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-14
4-24 EES Problem 4-23 is reconsidered. The effect of the spring constant on the final pressure in the cylinder and the boundary work done as the spring constant varies from 50 kN/m to 500 kN/m is to be investigated. The final pressure and the boundary work are to be plotted against the spring constant. Analysis The problem is solved using EES, and the solution is given below.
P[3]=P[2]+(Spring_const)*(V[3] - V[2]) "P[3] is a linear function of V[3]" "where Spring_const = k/A^2, the actual spring constant divided by the piston face area squared" "Input Data" P[1]=150 [kPa] m=50 [kg] T[1]=25 [C] P[2]=P[1] V[2]=0.2 [m^3] A=0.1[m^2] k=100 [kN/m] DELTAx=20 [cm] Spring_const=k/A^2 "[kN/m^5]" V[1]=m*spvol[1] spvol[1]=volume(Steam_iapws,P=P[1],T=T[1]) V[2]=m*spvol[2] V[3]=V[2]+A*DELTAx*convert(cm,m) V[3]=m*spvol[3] "The temperature at state 2 is:" T[2]=temperature(Steam_iapws,P=P[2],v=spvol[2]) "The temperature at state 3 is:" T[3]=temperature(Steam_iapws,P=P[3],v=spvol[3]) Wnet_other = 0 W_out=Wnet_other + W_b12+W_b23 W_b12=P[1]*(V[2]-V[1]) "W_b23 = integral of P[3]*dV[3] for Deltax = 20 cm and is given by:" W_b23=P[2]*(V[3]-V[2])+Spring_const/2*(V[3]-V[2])^2 k [kN/m] 50 100 150 200 250 300 350 400 450 500
P3 [kPa] 350 450 550 650 750 850 950 1050 1150 1250
Wout [kJ] 43.46 44.46 45.46 46.46 47.46 48.46 49.46 50.46 51.46 52.46
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-15
Steam
105 104
] a P k[ P
103
3 138.9°C
102
1
111.4°C
2
101 25°C
100 10-4
10-3
10-2
10-1
100
101
102
3
v [m /kg] 1300
1100
] a P k[ ] 3[ P
900
700
500
300 50
100
150
200
250
300
350
400
450
500
350
400
450
500
k [kN/m] 54 52 50
] J k[
48
W
46
t u o
44 42 50
100
150
200
250
300
k [kN/m]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-16
4-25 Several sets of pressure and volume data are taken as a gas expands. The boundary work done during this process is to be determined using the experimental data. Assumptions The process is quasi-equilibrium. Analysis Plotting the given data on a P-V diagram on a graph paper and evaluating the area under the process curve, the work done is determined to be 0.25 kJ.
4-26 A piston-cylinder device contains nitrogen gas at a specified state. The boundary work is to be determined for the isothermal expansion of nitrogen. Properties The properties of nitrogen are R = 0.2968 kJ/kg.K , k = 1.4 (Table A-2a). Analysis We first determine initial and final volumes from ideal gas relation, and find the boundary work using the relation for isothermal expansion of an ideal gas
V1 =
mRT (0.25 kg)(0.2968 kJ/kg.K)(120 + 273 K) = = 0.2243 m 3 P1 (130 kPa)
V2 =
mRT (0.25 kg)(0.2968 kJ/kg.K)(120 + 273 K) = = 0.2916 m 3 P2 (100 kPa)
V Wb = P1V1 ln 2 V1
0.2916 m 3 = (130 kPa)(0.2243 m 3 ) ln 0.2243 m 3
N2 130 kPa 120°C
= 7.65 kJ
4-27 A piston-cylinder device contains air gas at a specified state. The air undergoes a cycle with three processes. The boundary work for each process and the net work of the cycle are to be determined. Properties The properties of air are R = 0.287 kJ/kg.K , k = 1.4 (Table A-2a). Analysis For the isothermal expansion process:
V1 = V2 =
mRT (0.15 kg)(0.287 kJ/kg.K)(350 + 273 K) = = 0.01341 m 3 P1 (2000 kPa) mRT (0.15 kg)(0.287 kJ/kg.K)(350 + 273 K) = = 0.05364 m 3 P2 (500 kPa)
Air 2 MPa 350°C
0.05364 m3 V = 37.18 kJ Wb,1− 2 = P1V1 ln 2 = (2000 kPa)(0.01341 m3 ) ln 0.01341 m3 V1
For the polytropic compression process: P2V 2n = P3V 3n →(500 kPa)(0.05364 m 3 )1.2 = (2000 kPa)V 31.2 →V 3 = 0.01690 m 3 Wb , 2 − 3 =
P3V 3 − P2V 2 (2000 kPa)(0.01690 m 3 ) − (500 kPa)(0.05364 m 3 ) = = -34.86 kJ 1− n 1 − 1.2
For the constant pressure compression process: Wb,3−1 = P3 (V 1 −V 3 ) = (2000 kPa)(0.01341 − 0.01690)m 3 = -6.97 kJ
The net work for the cycle is the sum of the works for each process Wnet = Wb,1− 2 + Wb,2−3 + Wb,3−1 = 37.18 + (−34.86) + (−6.97) = -4.65 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-17
Closed System Energy Analysis
4-28 A rigid tank is initially filled with superheated R-134a. Heat is transferred to the tank until the pressure inside rises to a specified value. The mass of the refrigerant and the amount of heat transfer are to be determined, and the process is to be shown on a P-v diagram. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions. Analysis (a) We take the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E −E 1in424out 3
=
Net energy transfer by heat, work, and mass
∆E system 1 424 3
R-134a 160 kPa
Change in internal, kinetic, potential, etc. energies
Qin = ∆U = m(u 2 − u1 )
(since W = KE = PE = 0)
Using data from the refrigerant tables (Tables A-11 through A13), the properties of R-134a are determined to be P1 = 160 kPa v f = 0.0007437, v g = 0.12348 m 3 /kg x1 = 0.4 u fg = 190.27kJ/kg u f = 31.09,
v 1 = v f + x1v fg = 0.0007437 + 0.4(0.12348 − 0.0007437) = 0.04984 m 3 /kg u1 = u f + x1u fg = 31.09 + 0.4(190.27) = 107.19 kJ/kg P2 = 700 kPa u 2 = 376.99 kJ/kg (Superheated vapor) (v 2 = v 1 )
P
2
Then the mass of the refrigerant is determined to be m=
V1 0.5 m 3 = = 10.03 kg v 1 0.04984 m 3 /kg
(b) Then the heat transfer to the tank becomes
1
v
Qin = m(u 2 − u1 ) = (10.03 kg)(376.99 − 107.19) kJ/kg = 2707 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-18
4-29E A rigid tank is initially filled with saturated R-134a vapor. Heat is transferred from the refrigerant until the pressure inside drops to a specified value. The final temperature, the mass of the refrigerant that has condensed, and the amount of heat transfer are to be determined. Also, the process is to be shown on a P-v diagram. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions. Analysis (a) We take the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E −E 1in424out 3
=
Net energy transfer by heat, work, and mass
∆E system 1 424 3
Change in internal, kinetic, potential, etc. energies
− Qout = ∆U = m(u 2 − u1 )
(since W = KE = PE = 0)
Qout = m(u1 − u 2 )
Using data from the refrigerant tables (Tables A-11E through A-13E), the properties of R-134a are determined to be P1 = 160 psia v 1 = v g @160 psia = 0.29316 ft 3 /lbm sat. vapor u1 = u g @160 psia = 108.50 Btu/lbm P2 = 50 psia v f = 0.01252, v g = 0.94791 ft 3 /lbm (v 2 = v 1 ) u f = 24.832, u fg = 75.209 Btu/lbm
R-134a 160 psia Sat. vapor
The final state is saturated mixture. Thus, T2 = Tsat @ 50 psia = 40.23°F (b) The total mass and the amount of refrigerant that has condensed are m=
V1 20 ft 3 = = 68.22 lbm v 1 0.29316 ft 3 /lbm v 2 − v f 0.29316 − 0.01252
x2 =
v fg
=
0.94791 − 0.01252
= 0.300
P 1
m f = (1 − x 2 )m = (1 − 0.300)(68.22 lbm) = 47.75 lbm 2
Also, u 2 = u f + x 2 u fg = 24.832 + 0.300(75.209) = 47.40 Btu/lbm
(c) Substituting, Qout = m(u1 − u 2 ) = (68.22 lbm)(108.50 − 47.40) Btu/lbm = 4169 Btu
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
v
4-19
4-30 An insulated rigid tank is initially filled with a saturated liquid-vapor mixture of water. An electric heater in the tank is turned on, and the entire liquid in the tank is vaporized. The length of time the heater was kept on is to be determined, and the process is to be shown on a P-v diagram. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The device is well-insulated and thus heat transfer is negligible. 3 The energy stored in the resistance wires, and the heat transferred to the tank itself is negligible. Analysis We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E − Eout 1in424 3
Net energy transfer by heat, work, and mass
=
∆Esystem 1 424 3
H 2O V = const.
Change in internal, kinetic, potential, etc. energies
We,in = ∆U = m(u2 − u1 )
(since Q = KE = PE = 0)
VI∆t = m(u2 − u1 )
We
The properties of water are (Tables A-4 through A-6) P1 = 100kPa v f = 0.001043, v g = 1.6941 m3/kg x1 = 0.25 u f = 417.40, u fg = 2088.2 kJ/kg
v1 = v f + x1v fg = 0.001043 + [0.25 × (1.6941 − 0.001043)] = 0.42431 m3/kg
T 2
u1 = u f + x1u fg = 417.40 + (0.25 × 2088.2) = 939.4 kJ/kg
v 2 = v1 = 0.42431 m3/kg sat.vapor
1
u2 = u g @ 0.42431m 3 /kg = 2556.2 kJ/kg
Substituting, 1000 VA (110 V)(8 A)∆t = (5 kg)(2556.2 − 939.4)kJ/kg 1 kJ/s ∆t = 9186 s ≅ 153.1 min
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
v
4-20
4-31 EES Problem 4-30 is reconsidered. The effect of the initial mass of water on the length of time required to completely vaporize the liquid as the initial mass varies from 1 kg to 10 kg is to be investigated. The vaporization time is to be plotted against the initial mass. Analysis The problem is solved using EES, and the solution is given below. PROCEDURE P2X2(v[1]:P[2],x[2]) Fluid$='Steam_IAPWS' If v[1] > V_CRIT(Fluid$) then P[2]=pressure(Fluid$,v=v[1],x=1) x[2]=1 else P[2]=pressure(Fluid$,v=v[1],x=0) x[2]=0 EndIf End
350 300
∆ t m in [m in]
250
"Knowns" {m=5 [kg]} P[1]=100 [kPa] y=0.75 "moisture" Volts=110 [V] I=8 [amp]
200 150 100 50 0 1
2
3
4
5
6
7
8
9
10
"Solution" m [kg] "Conservation of Energy for the closed tank:" E_dot_in-E_dot_out=DELTAE_dot E_dot_in=W_dot_ele "[kW]" W_dot_ele=Volts*I*CONVERT(J/s,kW) "[kW]" E_dot_out=0 "[kW]" DELTAE_dot=m*(u[2]-u[1])/DELTAt_s "[kW]" DELTAt_min=DELTAt_s*convert(s,min) "[min]" "The quality at state 1 is:" Fluid$='Steam_IAPWS' x[1]=1-y u[1]=INTENERGY(Fluid$,P=P[1], x=x[1]) "[kJ/kg]" v[1]=volume(Fluid$,P=P[1], x=x[1]) "[m^3/kg]" T[1]=temperature(Fluid$,P=P[1], x=x[1]) "[C]" "Check to see if state 2 is on the saturated liquid line or saturated vapor line:" Call P2X2(v[1]:P[2],x[2]) u[2]=INTENERGY(Fluid$,P=P[2], x=x[2]) "[kJ/kg]" v[2]=volume(Fluid$,P=P[2], x=x[2]) "[m^3/kg]" T[2]=temperature(Fluid$,P=P[2], x=x[2]) "[C]" S te a m
700
m [kg] 1 2 3 4 5 6 7 8 9 10
600
500 T [°C]
∆tmin [min] 30.63 61.26 91.89 122.5 153.2 183.8 214.4 245 275.7 306.3
400
300
200
2 4 37 .9 kP a
100
0 1 0 -3
1 00 kP a
1 0 .05
1 0 -2
1 0 -1
100 3
0 .1
0.2
101
0 .5
102
10 3
v [m /k g ]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-21
4-32 One part of an insulated tank contains compressed liquid while the other side is evacuated. The partition is then removed, and water is allowed to expand into the entire tank. The final temperature and the volume of the tank are to be determined. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The tank is insulated and thus heat transfer is negligible. 3 There are no work interactions. Analysis We take the entire contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E − Eout 1in 424 3
=
Net energy transfer by heat, work, and mass
∆Esystem 1 424 3
Change in internal, kinetic, potential, etc. energies
0 = ∆U = m(u2 − u1 )
Evacuate (since W = Q = KE = PE = 0)
Partition
u1 = u2
The properties of water are (Tables A-4 through A-6)
H2O
3
P1 = 600 kPa v 1 ≅ v f @60°C = 0.001017 m /kg T1 = 60°C u1 ≅ u f @ 60°C = 251.16 kJ/kg
We now assume the final state in the tank is saturated liquid-vapor mixture and determine quality. This assumption will be verified if we get a quality between 0 and 1. P2 = 10 kPa v f = 0.001010, v g = 14.670 m3/kg (u2 = u1 ) u f = 191.79, u fg = 2245.4 kJ/kg x2 =
u2 − u f u fg
=
251.16 − 191.79 = 0.02644 2245.4
Thus, T2 = =Tsat @ 10 kPa = 45.81 °C
v 2 = v f + x 2v fg = 0.001010 + [0.02644 × (14.670 − 0.001010 )] = 0.38886 m 3 /kg and,
V = mv2 =(2.5 kg)(0.38886 m3/kg) = 0.972 m3
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-22
4-33 EES Problem 4-32 is reconsidered. The effect of the initial pressure of water on the final temperature in the tank as the initial pressure varies from 100 kPa to 600 kPa is to be investigated. The final temperature is to be plotted against the initial pressure. Analysis The problem is solved using EES, and the solution is given below. "Knowns" m=2.5 [kg] {P[1]=600 [kPa]} T[1]=60 [C] P[2]=10 [kPa] "Solution" Fluid$='Steam_IAPWS' "Conservation of Energy for the closed tank:" E_in-E_out=DELTAE E_in=0 E_out=0 DELTAE=m*(u[2]-u[1]) u[1]=INTENERGY(Fluid$,P=P[1], T=T[1]) v[1]=volume(Fluid$,P=P[1], T=T[1]) T[2]=temperature(Fluid$,P=P[2], u=u[2]) T_2=T[2] v[2]=volume(Fluid$,P=P[2], u=u[2]) 700 V_total=m*v[2]
Steam
600 500
T2 [C] 45.79 45.79 45.79 45.79 45.79 45.79
T [°C]
P1 [kPa] 100 200 300 400 500 600
400 300 200 600 kPa
100 0 10 -4
2
1
10 -3
10 kPa
10 -2
0.05
10 -1
10 0
0.1
0.2
10 1
0.5
10 2
3
v [m /kg]
50 40
T 2 [C]
30 20 10 0 100
200
300
400
500
600
P[1] [kPa] PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10 3
4-23
4-34 A cylinder is initially filled with R-134a at a specified state. The refrigerant is cooled at constant pressure. The amount of heat loss is to be determined, and the process is to be shown on a T-v diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E − Eout = ∆Esystem 1in 424 3 1 424 3 Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc. energies
− Qout − Wb,out = ∆U = m(u2 − u1 )
(since KE = PE = 0)
− Qout = m(h2 − h1 )
since ∆U + Wb = ∆H during a constant pressure quasiequilibrium process. The properties of R-134a are (Tables A-11 through A-13) P1 = 800 kPa h1 = 306.88 kJ/kg T1 = 70°C P2 = 800 kPa h2 = h f @15°C = 72.34 kJ/kg T2 = 15°C Substituting,
Q
R-134a 800 kPa T
1
2
v
Qout = - (5 kg)(72.34 - 306.88) kJ/kg = 1173 kJ
4-35E A cylinder contains water initially at a specified state. The water is heated at constant pressure. The final temperature of the water is to be determined, and the process is to be shown on a T-v diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The thermal energy stored in the cylinder itself is negligible. 3 The compression or expansion process is quasiequilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E −E = ∆E system 1in424out 3 1 424 3 Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc. energies
Qin − Wb,out = ∆U = m(u 2 − u1 )
(since KE = PE = 0)
Qin = m(h2 − h1 )
since ∆U + Wb = ∆H during a constant pressure quasi-equilibrium process. The properties of water are (Tables A-6E)
v1 =
V1 m
=
2 ft 3 = 4 ft 3 /lbm 0.5 lbm
P1 = 120 psia h1 = 1217.0 Btu/lbm v 1 = 4 ft 3 /lbm Substituting, 200 Btu = (0.5 lbm)(h2 − 1217.0)Btu/lbm
Q
H2O 120 psia
T 2 1
h2 = 1617.0 Btu/lbm
Then, P2 = 120 psia
T2 = 1161.4°F h2 = 1617.0 Btu/lbm
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
v
4-24
4-36 A cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically as it is stirred by a paddle-wheel at constant pressure. The voltage of the current source is to be determined, and the process is to be shown on a P-v diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The cylinder is well-insulated and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E −E 1in424out 3
=
Net energy transfer by heat, work, and mass
∆E system 1 424 3
Change in internal, kinetic, potential, etc. energies
We,in + W pw,in − W b,out = ∆U
H2O
(since Q = KE = PE = 0)
P = const.
We,in + W pw,in = m(h2 − h1 ) ( VI∆t ) + W pw,in = m(h2 − h1 )
Wpw
We
since ∆U + Wb = ∆H during a constant pressure quasi-equilibrium process. The properties of water are (Tables A-4 through A-6) P1 = 175 kPa h1 = h f @175 kPa = 487.01 kJ/kg 3 sat.liquid v1 = v f @175 kPa = 0.001057 m /kg P2 = 175 kPa h2 = h f + x2 h fg = 487.01 + (0.5 × 2213.1) = 1593.6 kJ/kg x2 = 0.5 m=
V1 0.005 m3 = = 4.731 kg v1 0.001057 m3/kg
P
Substituting, VI∆t + (400kJ) = (4.731 kg)(1593.6 − 487.01)kJ/kg VI∆t = 4835 kJ V=
1000 VA 4835 kJ = 223.9 V (8 A)(45 × 60 s) 1 kJ/s
1
2
v
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-25
4-37 A cylinder is initially filled with steam at a specified state. The steam is cooled at constant pressure. The mass of the steam, the final temperature, and the amount of heat transfer are to be determined, and the process is to be shown on a T-v diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E −E 1in424out 3
Net energy transfer by heat, work, and mass
=
∆E system 1 424 3
Change in internal, kinetic, potential, etc. energies
− Qout − Wb,out = ∆U = m(u 2 − u1 )
(since KE = PE = 0)
H2O 1 MPa 450°C
− Qout = m(h2 − h1 )
since ∆U + Wb = ∆H during a constant pressure quasiequilibrium process. The properties of water are (Tables A-4 through A-6) P1 = 1 MPa v1 = 0.33045 m3/kg T2 = 450°C h1 = 3371.3 kJ/kg m=
Q
T 1
3
V1 2.5 m = = 7.565 kg v1 0.33045 m3/kg
(b) The final temperature is determined from P2 = 1 MPa T2 = Tsat @1 MPa = 179.9 °C sat. vapor h2 = hg@1 MPa = 2777.1 kJ/kg
2
v
(c) Substituting, the energy balance gives Qout = - (7.565 kg)(2777.1 – 3371.3) kJ/kg = 4495 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-26
4-38 [Also solved by EES on enclosed CD] A cylinder equipped with an external spring is initially filled with steam at a specified state. Heat is transferred to the steam, and both the temperature and pressure rise. The final temperature, the boundary work done by the steam, and the amount of heat transfer are to be determined, and the process is to be shown on a P-v diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The thermal energy stored in the cylinder itself is negligible. 3 The compression or expansion process is quasiequilibrium. 4 The spring is a linear spring. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. Noting that the spring is not part of the system (it is external), the energy balance for this stationary closed system can be expressed as E −E 1in424out 3
=
Net energy transfer by heat, work, and mass
∆E system 1 424 3
Change in internal, kinetic, potential, etc. energies
Qin − Wb,out = ∆U = m(u 2 − u1 )
Q
H2O 200 kPa 200°C
(since KE = PE = 0)
Qin = m(u 2 − u1 ) + Wb,out
The properties of steam are (Tables A-4 through A-6) P1 = 200 kPa v 1 = 1.08049 m 3 /kg T1 = 200°C u1 = 2654.6 kJ/kg
P
V 0.5 m 3 = 0.4628 kg m= 1 = v 1 1.08049 m 3 /kg v2 =
V2 m
=
2 1
0.6 m 3 = 1.2966 m 3 /kg 0.4628 kg
v
P2 = 500 kPa
T2 = 1132°C 3 v 2 = 1.2966 m /kg u 2 = 4325.2 kJ/kg
(b) The pressure of the gas changes linearly with volume, and thus the process curve on a P-V diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus, Wb = Area =
P1 + P2 (V 2 −V1 ) = (200 + 500)kPa (0.6 − 0.5)m 3 1 kJ 3 2 2 1 kPa ⋅ m
= 35 kJ
(c) From the energy balance we have Qin = (0.4628 kg)(4325.2 - 2654.6)kJ/kg + 35 kJ = 808 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4-27
4-39 EES Problem 4-38 is reconsidered. The effect of the initial temperature of steam on the final temperature, the work done, and the total heat transfer as the initial temperature varies from 150°C to 250°C is to be investigated. The final results are to be plotted against the initial temperature. Analysis The problem is solved using EES, and the solution is given below. "The process is given by:" "P[2]=P[1]+k*x*A/A, and as the spring moves 'x' amount, the volume changes by V[2]-V[1]." P[2]=P[1]+(Spring_const)*(V[2] - V[1]) "P[2] is a linear function of V[2]" "where Spring_const = k/A, the actual spring constant divided by the piston face area" "Conservation of mass for the closed system is:" m[2]=m[1] "The conservation of energy for the closed system is" "E_in - E_out = DeltaE, neglect DeltaKE and DeltaPE for the system" Q_in - W_out = m[1]*(u[2]-u[1]) DELTAU=m[1]*(u[2]-u[1]) "Input Data" P[1]=200 [kPa] V[1]=0.5 [m^3] "T[1]=200 [C]" P[2]=500 [kPa] V[2]=0.6 [m^3]
50
Fluid$='Steam_IAPWS' m[1]=V[1]/spvol[1] spvol[1]=volume(Fluid$,T=T[1], P=P[1]) u[1]=intenergy(Fluid$, T=T[1], P=P[1]) spvol[2]=V[2]/m[2]
W out [kJ]
40
30
20
10
0 150
170
190
210
230
T[1] [C]
"The final temperature is:" T[2]=temperature(Fluid$,P=P[2],v=spvol[2]) u[2]=intenergy(Fluid$, P=P[2], T=T[2]) Wnet_other = 0 W_out=Wnet_other + W_b "W_b = integral of P[2]*dV[2] for 0.5 W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4])
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-45
η 0.1 0.1644 0.1814 0.1806 0.1702 0.1533 0.131 0.1041 0.07272 0.03675
Bwr 0.5229 0.6305 0.7038 0.7611 0.8088 0.85 0.8864 0.9192 0.9491 0.9767
Pratio 2 4 6 8 10 12 14 16 18 20
Wc [kW] 1818 4033 5543 6723 7705 8553 9304 9980 10596 11165
Wnet [kW] 1659 2364 2333 2110 1822 1510 1192 877.2 567.9 266.1
Wt [kW] 3477 6396 7876 8833 9527 10063 10496 10857 11164 11431
Qin [kW] 16587 14373 12862 11682 10700 9852 9102 8426 7809 7241
1500 Air Standard Brayton Cycle Pressure ratio = 8 and Tmax = 1160K 3
1000
] K [ T
4
2 2s
4s
500 800 kPa 100 kPa
0 5.0
1
5.5
6.0
6.5
7.0
7.5
s [kJ/kg-K] 0.25
2500
0.20
η
η , y c n ei ci ff e el c y C
0.15
0.00 2
1500 ] W k[
η = 0.82 t η = 0.75 c
0.10
0.05
2000
Wnet
1000
Tmax=1160 K
6
8
10 12 Pratio
14
W
500
Note Pratio for maximum work and η
4
t e n
16
18
0 20
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-46
9-75 A simple Brayton cycle with air as the working fluid has a pressure ratio of 8. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) Using the compressor and turbine efficiency relations, P T2 s = T1 2 P1
(k −1) / k
= (310 K )(8)0.4/1.4 = 561.5 K
(k −1) / k
0.4/1.4
P 1 = 640.4 K T4 s = T3 4 = (1160 K ) 8 P3 c p (T2 s − T1 ) T − T1 h −h η C = 2s 1 = → T2 = T1 + 2 s ηC h2 − h1 c p (T2 − T1 )
T 3
1160 K
qin 2s
310 K
1
2
qout
4s
561.5 − 310 = 310 + = 645.3 K 0.75
ηT =
(b)
c p (T3 − T4 ) h3 − h4 → T4 = T3 − η T (T3 − T4 s ) = h3 − h4 s c p (T3 − T4 s ) = 1160 − (0.82)(1160 − 640.4 ) = 733.9 K
q in = h3 − h2 = c p (T3 − T2 ) = (1.005 kJ/kg ⋅ K )(1160 − 645.3)K = 517.3 kJ/kg q out = h4 − h1 = c p (T4 − T1 ) = (1.005 kJ/kg ⋅ K )(733.9 − 310)K = 426.0 kJ/kg wnet,out = q in − q out = 517.3 − 426.0 = 91.3 kJ/kg
(c)
η th =
wnet,out q in
=
91.3 kJ/kg = 17.6% 517.3 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4
s
9-47
9-76 A gas turbine power plant that operates on the simple Brayton cycle with air as the working fluid has a specified pressure ratio. The required mass flow rate of air is to be determined for two cases. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) Using the isentropic relations, P T2 s = T1 2 P1
P = T3 4 P3
T4 s
(k −1) / k
(k −1) / k
T
= (300 K )(12 )0.4/1.4 = 610.2 K 1 = (1000 K ) 12
3
1000 K 2s
300 K
0.4/1.4
= 491.7 K
1
2 4s
ws,C,in = h2 s − h1 = c p (T2 s − T1 ) = (1.005 kJ/kg ⋅ K )(610.2 − 300)K = 311.75 kJ/kg ws,T, out = h3 − h4 s = c p (T3 − T4 s ) = (1.005 kJ/kg ⋅ K )(1000 − 491.7 )K = 510.84 kJ/kg ws, net,out = ws,T,out − ws,C,in = 510.84 − 311.75 = 199.1 kJ/kg m& s =
W& net,out ws, net,out
=
70,000 kJ/s = 352 kg/s 199.1 kJ/kg
(b) The net work output is determined to be wa,net,out = wa,T,out − wa,C,in = η T ws,T, out − ws,C,in / η C = (0.85)(510.84 ) − 311.75 0.85 = 67.5 kJ/kg
m& a =
W& net,out wa,net,out
=
70,000 kJ/s = 1037 kg/s 67.5 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4
s
9-48
9-77 A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The power delivered by this plant is to be determined assuming constant and variable specific heats. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas. T
Analysis (a) Assuming constant specific heats, P T2 s = T1 2 P1 P T4 s = T3 4 P3
η th = 1 −
(k −1) / k
(k −1) / k
= (290 K )(8)0.4/1.4 = 525.3 K
3
1100 K
qin 2
1 = (1100 K ) 8
0.4/1.4
= 607.2 K
290 K
4 1
qout
c p (T4 − T1 )
q out T −T 607.2 − 290 = 1− = 1− 4 1 = 1− = 0.448 q in c p (T3 − T2 ) T3 − T2 1100 − 525.3
W& net,out = η th Q& in = (0.448)(35,000 kW ) = 15,680 kW
(b) Assuming variable specific heats (Table A-17), T1 = 290 K → Pr 2 =
h1 = 290.16 kJ/kg Pr1 = 1.2311
P2 Pr = (8)(1.2311) = 9.8488 → h2 = 526.12 kJ/kg P1 1
T3 = 1100 K →
h3 = 1161.07 kJ/kg Pr3 = 167.1
P4 1 Pr = (167.1) = 20.89 → h4 = 651.37 kJ/kg P3 3 8 q h −h 651.37 − 290.16 = 1 − out == 1 − 4 1 = 1 − = 0.431 q in h3 − h2 1161.07 − 526.11
Pr 4 =
η th
W& net,out = η T Q& in = (0.431)(35,000 kW ) = 15,085 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-49
9-78 An actual gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (a) Using the isentropic relations, T1 = 300 K
→
h1 = 300.19 kJ / kg
T2 = 580 K
→
h2 = 586.04 kJ / kg
P 700 rp = 2 = =7 P1 100
T 950 kJ/kg
580 K
q in = h3 − h2 → h3 = 950 + 586.04 = 1536.04kJ/kg
300 K
2s 1
3
2 4s
4
→ Pr3 = 474.11 Pr 4 =
P4 1 Pr = (474.11) = 67.73 → h4 s = 905.83 kJ/kg P3 3 7
wC,in = h2 − h1 = 586.04 − 300.19 = 285.85 kJ/kg wT,out = η T (h3 − h4 s ) = (0.86 )(1536.04 − 905.83) = 542.0 kJ/kg wC,in
rbw =
(b)
wnet.out = wT,out − wC,in = 542.0 − 285.85 = 256.15 kJ/kg
η th =
wT,out
wnet,out q in
=
285.85 kJ/kg = 52.7% 542.0 kJ/kg
Thus,
=
256.15 kJ/kg = 27.0% 950 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-50
9-79 A gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes T are negligible. 4 Air is an ideal gas with constant specific heats. 3 Properties The properties of air at room temperature are cp = 1.005 950 kJ/kg kJ/kg·K and k = 1.4 (Table A-2). 580 K 2s 2 Analysis (a) Using constant specific heats, P 700 rp = 2 = =7 4s 4 300 K P1 100 1
s
q in = h3 − h2 = c p (T3 − T2 ) → T3 = T2 + q in /c p
= 580 K + (950 kJ/kg )/ (1.005 kJ/kg ⋅ K )
= 1525.3 K T4s
P = T3 4 P3
(k −1)/k
1 = (1525.3 K ) 7
0.4/1.4
= 874.8 K
wC,in = h2 − h1 = c p (T2 − T1 ) = (1.005kJ/kg ⋅ K )(580 − 300 )K = 281.4 kJ/kg wT,out = η T (h3 − h4 s ) = η T c p (T3 − T4 s ) = (0.86 )(1.005 kJ/kg ⋅ K )(1525.3 − 874.8)K = 562.2 kJ/kg wC,in
rbw =
(b)
wnet,out = wT,out − wC,in = 562.2 − 281.4 = 280.8 kJ/kg
η th =
=
281.4 kJ/kg = 50.1% 562.2 kJ/kg
Thus,
wT,out wnet,out q in
=
280.8 kJ/kg = 29.6% 950 kJ/kg
9-80E A gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The net power output of the plant is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis Using variable specific heats for air, →
h3 = 504.71 Btu / lbm
T
T4 = 1200 R → P 120 rp = 2 = =8 P1 15
h4 = 291.30 Btu / lbm
2000 R
T3 = 2000 R
1200 R
Q& out = m& (h4 − h1 ) → h1 = 291.30 − 6400/40 = 131.30 Btu/lbm → Pr1 = 1.474 Pr 2 =
3
2s 1
2 4s 6400 Btu/s
P2 Pr = (8)(1.474 ) = 11.79 → h2 s = 238.07 Btu/lbm P1 1
W& C,in = m& (h2 s − h1 ) / η C = (40 lbm/s)(238.07 − 131.30 )/ (0.80 ) = 5339 Btu/s W& T,out = m& (h3 − h4 ) = (40 lbm/s)(504.71 − 291.30)Btu/lbm = 8536 Btu/s W& net,out = W& T,out − W& C,in = 8536 − 5339 = 3197 Btu/s = 3373 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
4
s
9-51
9-81E A gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The compressor efficiency for which the power plant produces zero net work is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis Using variable specific heats, T3 = 2000 R
→
T4 = 1200 R
→
h3 = 504.71 Btu / lbm h4 = 291.30 Btu / lbm
T 3
2000 R
1200 R
· Wnet = 0
4
4s 1
P2 120 = =8 P1 15
rp =
2
2s
s
Q& out = m& (h4 − h1 ) → h1 = 291.30 − 6400/40 = 131.30 Btu/lbm → Pr1 = 1.474 Pr 2 =
Then,
P2 Pr = (8)(1.474) = 11.79 → h2 s = 238.07 Btu/lbm P1 1
W& C,in = W& T,out → m& (h2 s − h1 ) / η C = m& (h3 − h4 )
ηC =
h2 s − h1 238.07 − 131.30 = = 50.0% h3 − h4 504.71 − 291.30
9-82 A 32-MW gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The mass flow rate of air through the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis Using variable specific heats, T1 = 310 K → Pr 2 =
T
h1 = 310.24 kJ/kg Pr1 = 1.5546
P2 Pr = (8)(1.5546 ) = 12.44 → h2 s = 562.26 kJ/kg P1 1
T3 = 900 K →
h3 = 932.93 kJ/kg
3
900 K
2s
310 K
Pr3 = 75.29
2
· Wnet =
32 MW 4s
4
1
P4 1 Pr3 = (75.29 ) = 9.411 → h4 s = 519.32 kJ/kg P3 8 = wT,out − wC,in = η T (h3 − h4 s ) − (h2 s − h1 ) / η C
Pr 4 = wnet,out
and
= (0.86 )(932.93 − 519.32 ) − (562.26 − 310.24 ) / (0.80 ) = 40.68 kJ/kg
m& =
W& net,out wnet,out
=
32,000 kJ/s = 786.6 kg/s 40.68 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-52
9-83 A 32-MW gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The mass flow rate of air through the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis Using constant specific heats, P T2 s = T1 2 P1
P T4 s = T3 4 P3
(k −1) / k
(k −1) / k
= (310 K )(8)0.4/1.4 = 561.5 K 1 = (900 K ) 8
T 3
900 K
0.4/1.4
= 496.8 K
wnet,out = wT,out − wC,in = η T c p (T3 − T4 s ) − c p (T2 s − T1 ) / η C
2s
310 K
2
· Wnet =
32 MW 4s
4
1
= (1.005 kJ/kg ⋅ K )[(0.86 )(900 − 496.8) − (561.5 − 310)/ (0.80 )]K = 32.5 kJ/kg
and m& =
W& net,out wnet,out
=
32,000 kJ/s = 984.6 kg/s 32.5 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-53
9-84 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio, and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis (a) For this problem, we use the properties from EES software. Remember that Combustion for an ideal gas, enthalpy is a function of chamber temperature only whereas entropy is functions of 3 both temperature and pressure. 2 1.2 MPa Process 1-2: Compression T1 = 30°C → h1 = 303.60 kJ/kg
Compress.
T1 = 30°C s1 = 5.7159 kJ/kg ⋅ K P1 = 100 kPa P2 = 1200 kPa h2 s = 617.37 kJ/kg s 2 = s1 = 5.7159 kJ/kg.K
ηC =
1
100 kPa 30°C
Turbine 500°C
4
h2 s − h1 617.37 − 303.60 → 0.82 = → h2 = 686.24 kJ/kg h2 − h1 h2 − 303.60
Process 3-4: Expansion T4 = 500°C → h4 = 792.62 kJ/kg
ηT =
h3 − h4 h − 792.62 → 0.88 = 3 h3 − h4 s h3 − h4 s
We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with the isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The solution by hand would require a trial-error approach. h_3=enthalpy(Air, T=T_3) s_3=entropy(Air, T=T_3, P=P_2) h_4s=enthalpy(Air, P=P_1, s=s_3)
The mass flow rate is determined from P V& (100 kPa)(150/60 m 3 / s) m& = 1 1 = = 2.875 kg/s RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (30 + 273 K )
(
)
The net power output is W& C,in = m& (h2 − h1 ) = (2.875 kg/s)(686.24 − 303.60)kJ/kg = 1100 kW W& T,out = m& (h3 − h4 ) = (2.875 kg/s)(1404.7 − 792.62)kJ/kg = 1759 kW W& net = W& T,out − W& C,in = 1759 − 1100 = 659 kW
(b) The back work ratio is W& C,in 1100 kW rbw = = = 0.625 W& T,out 1759 kW (c) The rate of heat input and the thermal efficiency are Q& = m& (h − h ) = (2.875 kg/s)(1404.7 − 686.24)kJ/kg = 2065 kW in
3
2
W& 659 kW η th = net = = 0.319 & 2065 kW Qin
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-54
Brayton Cycle with Regeneration 9-85C Regeneration increases the thermal efficiency of a Brayton cycle by capturing some of the waste heat from the exhaust gases and preheating the air before it enters the combustion chamber. 9-86C Yes. At very high compression ratios, the gas temperature at the turbine exit may be lower than the temperature at the compressor exit. Therefore, if these two streams are brought into thermal contact in a regenerator, heat will flow to the exhaust gases instead of from the exhaust gases. As a result, the thermal efficiency will decrease. 9-87C The extent to which a regenerator approaches an ideal regenerator is called the effectiveness ε, and is defined as ε = qregen, act /qregen, max. 9-88C (b) turbine exit. 9-89C The steam injected increases the mass flow rate through the turbine and thus the power output. This, in turn, increases the thermal efficiency since η = W / Qin and W increases while Qin remains constant. Steam can be obtained by utilizing the hot exhaust gases.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-55
9-90E A car is powered by a gas turbine with a pressure ratio of 4. The thermal efficiency of the car and the mass flow rate of air for a net power output of 95 hp are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 The ambient air is 540 R and 14.5 psia. 4 The effectiveness of the regenerator is 0.9, and the isentropic efficiencies for both the compressor and the turbine are 80%. 5 The combustion gases can be treated as air. Properties The properties of air at the compressor and turbine inlet temperatures can be obtained from Table A-17E. Analysis The gas turbine cycle with regeneration can be analyzed as follows: T1 = 540 R →
h1 = 129.06 Btu/lbm Pr1 = 1.386
P Pr 2 = 2 Pr1 = (4 )(1.386 ) = 5.544 → h2 s = 192.0 Btu/lbm P1 T3 = 2160 R → Pr 4 =
T qin
2160 R 5
h3 = 549.35 Btu/lbm 2s
Pr3 = 230.12
P4 1 Pr3 = (230.12 ) = 57.53 → h4 s = 372.2 Btu/lbm P3 4
3
540 R
2
4s
1
4
s
and
η comp =
h2 s − h1 192.0 − 129.06 → 0.80 = → h2 = 207.74 Btu/lbm h2 − h1 h2 − 129.06
η turb =
h3 − h4 549.35 − h4 → 0.80 = → h4 = 407.63 Btu/lbm h3 − h4 s 549.35 − 372.2
Then the thermal efficiency of the gas turbine cycle becomes q regen = ε (h4 − h2 ) = 0.9(407.63 − 207.74) = 179.9 Btu/lbm q in = (h3 − h2 ) − q regen = (549.35 − 207.74) − 179.9 = 161.7 Btu/lbm wnet,out = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 ) = (549.35 − 407.63) − (207.74 − 129.06) = 63.0 Btu/lbm
η th =
wnet,out q in
=
63.0 Btu/lbm = 0.39 = 39% 161.7 Btu/lbm
Finally, the mass flow rate of air through the turbine becomes m& air =
W& net 0.7068 Btu/s 95 hp = 1.07 lbm/s = wnet 63.0 Btu/lbm 1 hp
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-56
9-91 [Also solved by EES on enclosed CD] The thermal efficiency and power output of an actual gas turbine are given. The isentropic efficiency of the turbine and of the compressor, and the thermal efficiency of the gas turbine modified with a regenerator are to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. 3 The mass flow rates of air and of the combustion gases are the same, and the properties of combustion gases are the same as those of air. Properties The properties of air are given in Table A-17. Analysis The properties at various states are T1 = 20°C = 293 K →
h1 = 293.2 kJ/kg Pr1 = 1.2765
T
Pr 4 =
h3 = 1710.0 kJ/kg
3
5
P Pr 2 = 2 Pr1 = (14.7 )(1.2765) = 18.765 → h2 s = 643.3 kJ/kg P1 T3 = 1288°C = 1561 K →
qin
1561 K
2s
293 K
2
4s
4
1
Pr3 = 712.5
s
P4 1 Pr3 = → h4 s = 825.23 kJ/kg (712.5) = 48.47 P3 14.7
The net work output and the heat input per unit mass are W& 159,000 kW 3600 s wnet = net = = 372.66 kJ/kg m& 1,536,000 kg/h 1 h w 372.66 kJ/kg q in = net = = 1038.0 kJ/kg 0.359 η th q in = h3 − h2 → h2 = h3 − q in = 1710 − 1038 = 672.0 kJ/kg q out = q in − wnet = 1038.0 − 372.66 = 665.34 kJ/kg q out = h4 − h1 → h4 = q out + h1 = 665.34 + 293.2 = 958.54 kJ/kg → T4 = 650°C
Then the compressor and turbine efficiencies become
ηT =
h3 − h4 1710 − 958.54 = = 0.849 h3 − h4 s 1710 − 825.23
ηC =
h2 s − h1 643.3 − 293.2 = = 0.924 h2 − h1 672 − 293.2
When a regenerator is added, the new heat input and the thermal efficiency become q regen = ε (h4 − h2 ) = (0.80)(958.54 - 672.0) = 286.54 kJ/kg q in, new = q in − q regen = 1038 − 286.54 = 751.46 kJ/kg
η th,new =
wnet 372.66 kJ/kg = = 0.496 q in, new 751.46 kJ/kg
Discussion Note an 80% efficient regenerator would increase the thermal efficiency of this gas turbine from 35.9% to 49.6%.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-57
9-92 EES Problem 9-91 is reconsidered. A solution that allows different isentropic efficiencies for the compressor and turbine is to be developed and the effect of the isentropic efficiencies on net work done and the heat supplied to the cycle is to be studied. Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows: "Input data" T[3] = 1288 [C] Pratio = 14.7 T[1] = 20 [C] P[1]= 100 [kPa] {T[4]=589 [C]} {W_dot_net=159 [MW] }"We omit the information about the cycle net work" m_dot = 1536000 [kg/h]*Convert(kg/h,kg/s) {Eta_th_noreg=0.359} "We omit the information about the cycle efficiency." Eta_reg = 0.80 Eta_c = 0.892 "Compressor isentorpic efficiency" Eta_t = 0.926 "Turbien isentropic efficiency" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = W_dot_compisen/W_dot_comp "compressor adiabatic efficiency, W_dot_comp > W_dot_compisen" "Conservation of energy for the compressor for the isentropic case: E_dot_in - E_dot_out = DELTAE_dot=0 for steady-flow" m_dot*h[1] + W_dot_compisen = m_dot*h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" m_dot*h[1] + W_dot_comp = m_dot*h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 E_dot_in - E_dot_out =DELTAE_dot_cv =0 for steady flow" m_dot*h[2] + Q_dot_in_noreg = m_dot*h[3] q_in_noreg=Q_dot_in_noreg/m_dot h[3]=ENTHALPY(Air,T=T[3]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[3] /Pratio s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4])"T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = W_dot_turb /W_dot_turbisen "turbine adiabatic efficiency, W_dot_turbisen > W_dot_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 E_dot_in -E_dot_out = DELTAE_dot_cv = 0 for steady-flow"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-58 m_dot*h[3] = W_dot_turbisen + m_dot*h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4]) "Actual Turbine analysis:" m_dot*h[3] = W_dot_turb + m_dot*h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Cycle analysis" "Using the definition of the net cycle work and 1 MW = 1000 kW:" W_dot_net*1000=W_dot_turb-W_dot_comp "kJ/s" Eta_th_noreg=W_dot_net*1000/Q_dot_in_noreg"Cycle thermal efficiency" Bwr=W_dot_comp/W_dot_turb"Back work ratio" "With the regenerator the heat added in the external heat exchanger is" m_dot*h[5] + Q_dot_in_withreg = m_dot*h[3] q_in_withreg=Q_dot_in_withreg/m_dot h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[2])/(h[4]-h[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" m_dot*h[2] + m_dot*h[4]=m_dot*h[5] + m_dot*h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6]) P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=W_dot_net*1000/Q_dot_in_withreg "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6] ηt
ηc
ηth,noreg
ηth,withreg
0.7 0.75 0.8 0.85 0.9 0.95 1
0.892 0.892 0.892 0.892 0.892 0.892 0.892
0.2309 0.2736 0.3163 0.359 0.4016 0.4443 0.487
0.3405 0.3841 0.4237 0.4599 0.493 0.5234 0.5515
Qinnoreg [kW] 442063 442063 442063 442063 442063 442063 442063
Qinwithreg [kW] 299766 314863 329960 345056 360153 375250 390346
Wnet [kW] 102.1 120.9 139.8 158.7 177.6 196.4 215.3
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-59
T-s Diagram for Gas Turbine with Regeneration 1700 1500
1470 kPa
3
1300 1100 900
] C [ T
100 kPa
700
5 2
500
2s
300 100 -100 4.5
6
4 4s
1 5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
0.95
1
s [kJ/kg-K]
220 200
] W k[
t e n
W
180 160 140 120 100 0.7
0.75
0.8
0.85
0.9
ηt 0.6
450000
415000
0.55
no regeneration
0.5 0.45
380000
n,i t o d
345000
ht
with regeneration
at E
Q
310000
275000 0.7
with regeneration
0.4 0.35 0.3
no regeneration
0.25
0.75
0.8
0.85
ηt
0.9
0.95
1
0.2 0.7
0.75
0.8
0.85
0.9
0.95
ηt
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
1
9-60
9-93 An ideal Brayton cycle with regeneration is considered. The effectiveness of the regenerator is 100%. The net work output and the thermal efficiency of the cycle are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis Noting that this is an ideal cycle and thus the compression and expansion processes are isentropic, we have → T1 = 300 K Pr 2 =
h1 = 300.19 kJ/kg Pr1 = 1.386
P2 → h2 = 579.87 kJ/kg Pr = (10 )(1.386 ) = 13.86 P1 1
→ T3 = 1200 K
1200 K
qin 5
h3 = 1277.79 kJ/kg
P4 1 → h4 = 675.85 kJ/kg Pr = (238) = 23.8 P3 3 10 = h2 − h1 = 579.87 − 300.19 = 279.68 kJ/kg
3 4
2
Pr3 = 238
Pr 4 = wC,in
T
300 K
1
wT,out = h3 − h4 = 1277.79 − 675.85 = 601.94 kJ/kg
Thus, wnet = wT,out − wC,in = 601.94 − 279.68 = 322.26 kJ/kg
Also,
ε = 100%
→
h5 = h4 = 675.85 kJ / kg
q in = h3 − h5 = 1277.79 − 675.85 = 601.94 kJ/kg
and
η th =
wnet 322.26 kJ/kg = = 53.5% q in 601.94 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-61
9-94 EES Problem 9-93 is reconsidered. The effects of the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle are to be studied. Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows: "Input data" T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8 "Compressor isentorpic efficiency" Eta_t =0.9 "Turbien isentropic efficiency" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" h[1] + w_compisen = h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" h[1] + w_comp = h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" h[2] + q_in_noreg = h[3] h[3]=ENTHALPY(Air,T=T[3]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[3] /Pratio s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4])"T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" h[3] = w_turbisen + h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4]) "Actual Turbine analysis:" h[3] = w_turb + h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Cycle analysis" w_net=w_turb-w_comp Eta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency" Bwr=w_comp/w_turb"Back work ratio"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-62
"With the regenerator the heat added in the external heat exchanger is" h[5] + q_in_withreg = h[3] h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[2])/(h[4]-h[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" h[2] + h[4]=h[5] + h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6]) P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "[%]" "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6] ηc
ηt
ηth,noreg
ηth,withreg
0.6 0.65 0.7 0.75 0.8 0.85 0.9
0.9 0.9 0.9 0.9 0.9 0.9 0.9
14.76 20.35 24.59 27.91 30.59 32.79 34.64
13.92 20.54 26.22 31.14 35.44 39.24 42.61
wnet [kJ/kg] 75.4 111.3 142 168.6 192 212.5 230.8
Air
1600
a kP 0 0 10
1400 3
1200
] K [ T
qinwithreg [kJ/kg] 541.6 541.6 541.6 541.6 541.6 541.6 541.6
qinnoreg [kJ/kg] 510.9 546.8 577.5 604.2 627.5 648 666.3
a kP 0 0 1
1000 800
2
5
2s
600
6
4s
4
400 200 4.5
1 5.0
5.5
6.0
6.5
7.0
7.5
s [kJ/kg-K]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-63 45 40 35
η c = 0.8 With regeneration
30
ht
η
25
No regeneration
20 15 10 0.7
0.75
0.8
0.85
0.9
0.95
1
0.95
1
0.95
1
ηt 275 230
] g k/ J k[
t e n
w
η c = 0.8
185 140 95 50 0.7
0.75
0.8
0.85
0.9
ηt 650 600 550
ni
q
No regeneration With regeneration
500
η c = 0.8
450 400 0.7
0.75
0.8
0.85
0.9
ηt PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-64
45 40
η t = 0.9
35
With regeneration
30
ht
25
η
No regeneration
20 15 10 0.6
0.65
0.7
0.75
0.8
0.85
0.9
ηc 250 215
] g k/ J k[
t e n
w
η t = 0.9
180 145 110 75 0.6
0.65
0.7
0.75
0.8
0.85
0.9
ηc 680 660
η t = 0.9
640 620
No regeneration
600
ni
q
580
With regeneration
560 540 520 500 0.6
0.65
0.7
0.75
0.8
0.85
0.9
ηc
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-65
9-95 An ideal Brayton cycle with regeneration is considered. The effectiveness of the regenerator is 100%. The net work output and the thermal efficiency of the cycle are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis Noting that this is an ideal cycle and thus the compression and expansion processes are isentropic, we have P T2 = T1 2 P1 P T4 = T3 4 P3
(k −1) / k
(k −1) / k
= (300 K )(10 )0.4/1.4 = 579.2 K 1 = (1200 K ) 10
T 1200 K
qin 5
3 4
2
0.4/1.4
= 621.5 K
300 K
1
ε = 100% → T5 = T4 = 621.5 K and T6 = T2 = 579.2 K η th = 1 −
c p (T6 − T1 ) q out T −T 579.2 − 300 = 1− 6 1 = 1− = 0.517 = 1− q in c p (T3 − T5 ) T3 − T5 1200 − 621.5
T or η th = 1 − 1 T3
( k −1) / k 300 (1.4 −1)/1.4 r p = 1− = 0.517 (10) 1200
Then, wnet = w turb, out − wcomp, in = (h3 − h4 ) − (h2 − h1 ) = c p [(T3 − T4 ) − (T2 − T1 )] = (1.005 kJ/kg.K)[(1200 - 621.5) - (579.2 - 300)]K = 300.8 kJ/kg
or, wnet = η th q in = η th (h3 − h5 ) = η th c p (T3 − T5 ) = (0.517)(1.005 kJ/kg.K)(1200 - 621.5) = 300.6 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-66
9-96 A Brayton cycle with regeneration using air as the working fluid is considered. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17.
T qin
1150 K
Analysis (a) The properties of air at various states are T1 = 310 K →
5
h1 = 310.24 kJ/kg Pr1 = 1.5546
3
2s
310 K
4s
2 6
1
P Pr2 = 2 Pr1 = (7 )(1.5546) = 10.88 → h2 s = 541.26 kJ/kg P1
ηC =
h3 = 1219.25 kJ/kg Pr3 = 200.15
Pr4 =
P4 1 Pr3 = (200.15) = 28.59 → h4 s = 711.80 kJ/kg P3 7
ηT =
h3 − h4 → h4 = h3 − η T (h3 − h4 s ) = 1219.25 − (0.82 )(1219.25 − 711.80) = 803.14 kJ/kg h3 − h4 s
(b)
s
h2 s − h1 → h2 = h1 + (h2 s − h1 ) / η C = 310.24 + (541.26 − 310.24 )/ (0.75) = 618.26 kJ/kg h2 − h1
T3 = 1150 K →
Thus,
4
T4 = 782.8 K
wnet = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 )
= (1219.25 − 803.14 ) − (618.26 − 310.24 ) = 108.09 kJ/kg
(c)
ε=
h5 − h 2 → h5 = h2 + ε (h4 − h2 ) h4 − h2 = 618.26 + (0.65)(803.14 − 618.26) = 738.43 kJ/kg
Then, q in = h3 − h5 = 1219.25 − 738.43 = 480.82 kJ/kg
η th =
wnet 108.09 kJ/kg = = 22.5% 480.82 kJ/kg q in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-67
9-97 A stationary gas-turbine power plant operating on an ideal regenerative Brayton cycle with air as the working fluid is considered. The power delivered by this plant is to be determined for two cases. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas. 3 Kinetic and potential energy changes are negligible. Properties When assuming constant specific heats, the properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). When assuming variable specific heats, the properties of air are obtained from Table A-17. Analysis (a) Assuming constant specific heats, P T2 = T1 2 P1
P T4 = T3 4 P3
(k −1) / k
(k −1) / k
= (290 K )(8)0.4/1.4 = 525.3 K 1 = (1100 K ) 8
T 1100 K
0.4/1.4
2
= 607.2 K 290 K
ε = 100% → T5 = T4 = 607.2 K and T6 = T2 = 525.3 K η th = 1 −
1
75,000 kW
3
5
4 6
qout
c p (T6 − T1 ) q out T −T 525.3 − 290 = 0.5225 = 1− = 1− 6 1 = 1− q in c p (T3 − T5 ) T3 − T5 1100 − 607.2
W& net = η T Q& in = (0.5225)(75,000 kW ) = 39,188 kW
(b) Assuming variable specific heats, T1 = 290K → Pr 2 =
h1 = 290.16 kJ/kg Pr1 = 1.2311
P2 Pr = (8)(1.2311) = 9.8488 → h2 = 526.12 kJ/kg P1 1
T3 = 1100K → Pr 4 =
h3 = 1161.07 kJ/kg Pr3 = 167.1
P4 1 Pr = (167.1) = 20.89 → h4 = 651.37 kJ/kg P3 3 8
ε = 100% → h5 = h4 = 651.37 kJ/kg and h6 = h2 = 526.12 kJ/kg q out h −h 526.12 − 290.16 = 1− 6 1 = 1− = 0.5371 q in h3 − h5 1161.07 − 651.37 = η T Q& in = (0.5371)(75,000 kW ) = 40,283 kW
η th = 1 − W& net
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-68
9-98 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17.
T
Analysis (a) The properties at various states are r p = P2 / P1 = 800 / 100 = 8 → h1 = 300.19 kJ/kg T1 = 300 K → h2 = 586.04 kJ/kg T2 = 580 K → h3 = 1277.79 kJ/kg T3 = 1200 K Pr3 = 238.0
qin
1200 K
3
5
580 K
2s
300 K
1
2
4s
4
6
P4 1 → h4 s = 719.75 kJ/kg Pr = (238.0) = 29.75 P3 3 8 h −h η T = 3 4 → h4 = h3 − η T (h3 − h4 s ) h3 − h4 s = 1277.79 − (0.86 )(1277.79 − 719.75) = 797.88 kJ/kg q regen = ε (h4 − h2 ) = (0.72 )(797.88 − 586.04 ) = 152.5 kJ/kg Pr 4 =
(b)
wnet = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 ) = (1277.79 − 797.88) − (586.04 − 300.19 ) = 194.06 kJ/kg q in = (h3 − h2 ) − q regen = (1277.79 − 586.04) − 152.52 = 539.23 kJ/kg
η th =
wnet 194.06 kJ/kg = = 36.0% q in 539.23 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-69
9-99 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) Using the isentropic relations and turbine efficiency, T r p = P2 / P1 = 800 / 100 = 8 (k −1) / k
0.4 / 1.4
P 1 = (1200 K ) = 662.5 K T4 s = T3 4 P 8 3 c p (T3 − T4 ) h −h → T4 = T3 − η T (T3 − T4 s ) ηT = 3 4 = h3 − h4 s c p (T3 − T4 s ) = 1200 − (0.86)(1200 − 662.5)
3
qin
1200 K 5
580 K
2s
300 K
1
2
4s
4
6
s
= 737.8 K
q regen = ε (h4 − h2 ) = ε c p (T4 − T2 ) = (0.72 )(1.005 kJ/kg ⋅ K )(737.8 − 580 )K = 114.2 kJ/kg
(b)
wnet = wT,out − wC,in = c p (T3 − T4 ) − c p (T2 − T1 )
= (1.005 kJ/kg ⋅ K )[(1200 − 737.8) − (580 − 300 )]K = 183.1 kJ/kg
q in = (h3 − h2 ) − q regen = c p (T3 − T2 ) − q regen = (1.005 kJ/kg ⋅ K )(1200 − 580 )K − 114.2 = 508.9 kJ/kg
η th =
wnet 183.1 kJ/kg = = 36.0% 508.9 kJ/kg q in
9-100 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. T Analysis (a) The properties of air at various states are 3 r p = P2 / P1 = 800 / 100 = 8 qin 1200 K 5
T1 = 300K → h1 = 300.19kJ/kg T2 = 580K → h2 = 586.04kJ/kg
580 K
2s
T3 = 1200K → h3 = 1277.79kJ/kg Pr3 = 238.0
300 K
1
2
4s
4
6
s
P4 1 Pr3 = (238.0) = 29.75 → h4 s = 719.75 kJ/kg P3 8 h −h η T = 3 4 → h4 = h3 − η T (h3 − h4 s ) = 1277.79 − (0.86 )(1277.79 − 719.75) = 797.88 kJ/kg h3 − h4 s Pr 4 =
q regen = ε (h3 − h2 ) = (0.70)(797.88 − 586.04) = 148.3 kJ/kg
(b)
wnet = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 ) = (1277.79 − 797.88) − (586.04 − 300.19) = 194.06 kJ/kg q in = (h3 − h2 ) − q regen = (1277.79 − 586.04 ) − 148.3 = 543.5 kJ/kg
η th =
wnet 194.06 kJ/kg = = 35.7% 543.5 kJ/kg q in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-70
Brayton Cycle with Intercooling, Reheating, and Regeneration 9-101C As the number of compression and expansion stages are increased and regeneration is employed, the ideal Brayton cycle will approach the Ericsson cycle. 9-102C (a) decrease, (b) decrease, and (c) decrease. 9-103C (a) increase, (b) decrease, and (c) decrease. 9-104C (a) increase, (b) decrease, (c) decrease, and (d) increase. 9-105C (a) increase, (b) decrease, (c) increase, and (d) decrease. 9-106C Because the steady-flow work is proportional to the specific volume of the gas. Intercooling decreases the average specific volume of the gas during compression, and thus the compressor work. Reheating increases the average specific volume of the gas, and thus the turbine work output. 9-107C (c) The Carnot (or Ericsson) cycle efficiency.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-71
9-108 An ideal gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine since this is an ideal cycle. Then, → T1 = 300 K Pr 2 =
h1 = 300.19 kJ/kg Pr1 = 1.386
P2 → h2 = h4 = 411.26 kJ/kg Pr = (3)(1.386) = 4.158 P1 1
→ T5 = 1200 K
T 1200 K
qin 9
h5 = h7 = 1277.79 kJ/kg Pr5 = 238
P 1 → h6 = h8 = 946.36 kJ/kg Pr6 = 6 Pr5 = (238) = 79.33 P5 3 wC,in = 2(h2 − h1 ) = 2(411.26 − 300.19 ) = 222.14 kJ/kg
300 K
4
2
3
1
5
7
6
8
10
wT,out = 2(h5 − h6 ) = 2(1277.79 − 946.36) = 662.86 kJ/kg
Thus,
rbw =
wC,in wT,out
=
222.14 kJ/kg = 33.5% 662.86 kJ/kg
q in = (h5 − h4 ) + (h7 − h6 ) = (1277.79 − 411.26 ) + (1277.79 − 946.36 ) = 1197.96 kJ/kg wnet = wT,out − wC,in = 662.86 − 222.14 = 440.72 kJ/kg
η th =
wnet 440.72 kJ/kg = = 36.8% q in 1197.96 kJ/kg
(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes q regen = ε (h8 − h4 ) = (0.75)(946.36 − 411.26 ) = 401.33 kJ/kg q in = q in,old − q regen = 1197.96 − 401.33 = 796.63 kJ/kg
η th =
wnet 440.72 kJ/kg = = 55.3% q in 796.63 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-72
9-109 A gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. Then, → h1 = 300.19 kJ/kg T1 = 300 K Pr1 = 1.386 Pr 2 =
T
P2 → h2 s = h4 s = 411.26 kJ/kg Pr = (3)(1.386) = 4.158 P1 1
h −h η C = 2 s 1 → h2 = h4 = h1 + (h2 s − h1 ) / η C h2 − h1 = 300.19 + (411.26 − 300.19 ) / (0.80 ) = 439.03 kJ/kg
5
qin 9
4 3
4
2s
2
10
1
→ h5 = h7 = 1277.79 kJ/kg T5 = 1200 K Pr5 = 238 P6 1 → h6 = h8 = 946.36 kJ/kg Pr5 = (238) = 79.33 P5 3 h −h η T = 5 6 → h6 = h8 = h5 − η T (h5 − h6 s ) h5 − h6 s = 1277.79 − (0.85)(1277.79 − 946.36 )
Pr6 =
= 996.07 kJ/kg wC,in = 2(h2 − h1 ) = 2(439.03 − 300.19 ) = 277.68 kJ/kg
wT,out = 2(h5 − h6 ) = 2(1277.79 − 996.07 ) = 563.44 kJ/kg
Thus,
rbw =
wC,in wT,out
=
277.68 kJ/kg = 49.3% 563.44 kJ/kg
q in = (h5 − h4 ) + (h7 − h6 ) = (1277.79 − 439.03) + (1277.79 − 996.07 ) = 1120.48 kJ/kg wnet = wT,out − wC,in = 563.44 − 277.68 = 285.76 kJ/kg
η th =
wnet 285.76 kJ/kg = = 25.5% q in 1120.48 kJ/kg
(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes q regen = ε (h8 − h4 ) = (0.75)(996.07 − 439.03) = 417.78 kJ/kg q in = q in,old − q regen = 1120.48 − 417.78 = 702.70 kJ/kg
η th =
7
6 8 6s 8
wnet 285.76 kJ/kg = = 40.7% q in 702.70 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-73
9-110 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The minimum mass flow rate of air needed to develop a specified net power output is to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis The mass flow rate will be a minimum when the cycle is ideal. That is, the turbine and the compressors are isentropic, the regenerator has an effectiveness of 100%, and the compression ratios across each compression or expansion stage are identical. In our case it is rp = 9 = 3. Then the work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. → h1 = 300.19 kJ/kg, Pr1 = 1.386 T1 = 300 K Pr 2 =
P2 Pr = (3)(1.386) = 4.158 → h2 = h4 = 411.26 kJ/kg P1 1
T5 = 1200 K → h5 = h7 = 1277.79 kJ/kg, Pr5 = 238
T 1200 K
P6 1 → h6 = h8 = 946.36 kJ/kg Pr = (238) = 79.33 P5 5 3 = 2(h2 − h1 ) = 2(411.26 − 300.19 ) = 222.14 kJ/kg
Pr6 = wC,in
wT,out = 2(h5 − h6 ) = 2(1277.79 − 946.36) = 662.86 kJ/kg
300 K
4
2
3
1
5
7
6
8
s
wnet = wT, out − wC,in = 662.86 − 222.14 = 440.72 kJ/kg m& =
W& net 110,000 kJ/s = = 249.6 kg/s wnet 440.72 kJ/kg
9-111 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The minimum mass flow rate of air needed to develop a specified net power output is to be determined. Assumptions 1 Argon is an ideal gas with constant specific heats. 2 Kinetic and potential energy changes are negligible. Properties The properties of argon at room temperature are cp = 0.5203 kJ/kg.K and k = 1.667 (Table A2a). Analysis The mass flow rate will be a minimum when the cycle is ideal. That is, the turbine and the compressors are isentropic, the regenerator has an effectiveness of 100%, and the compression ratios across each compression or expansion stage are identical. In our case it is rp = 9 = 3. Then the work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. P T2 = T1 2 P1
P T6 = T5 6 P5
(k −1) / k
(k −1) / k
= (300 K )(3)0.667/1.667 = 465.6 K T
1 = (1200 K ) 3
0.667/1.667
= 773.2 K
1200 K
wC,in = 2(h2 − h1 ) = 2c p (T2 − T1 ) = 2(0.5203 kJ/kg ⋅ K )(465.6 − 300 )K = 172.3 kJ/kg wT,out = 2(h5 − h6 ) = 2c p (T5 − T6 ) = 2(0.5203 kJ/kg ⋅ K )(1200 − 773.2)K = 444.1 kJ/kg
300 K
4
2
3
1
5
7
6
8
wnet = wT,out − wC,in = 444.1 − 172.3 = 271.8 kJ/kg m& =
W& net 110,000 kJ/s = = 404.7 kg/s wnet 271.8 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-74
Jet-Propulsion Cycles 9-112C The power developed from the thrust of the engine is called the propulsive power. It is equal to thrust times the aircraft velocity. 9-113C The ratio of the propulsive power developed and the rate of heat input is called the propulsive efficiency. It is determined by calculating these two quantities separately, and taking their ratio. 9-114C It reduces the exit velocity, and thus the thrust.
9-115E A turbojet engine operating on an ideal cycle is flying at an altitude of 20,000 ft. The pressure at the turbine exit, the velocity of the exhaust gases, and the propulsive efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are cp = 0.24 Btu/lbm.R and k = 1.4 (Table A-2Ea). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V1 = 900 ft/s. Ideally, the air will leave the diffuser with a negligible velocity (V2 ≅ 0). Diffuser: T
E& in − E& out = ∆E& system ©0 (steady)
5
h1 + V12 / 2 = h2 + V 22 / 2 V 22
3 ©0
− V12 2 0 = c p (T2 − T1 ) − V12 / 2 0 = h2 − h1 +
T2 = T1 +
2 1
1 Btu/lbm V12 (900 ft/s )2 = 470 + (2)(0.24 Btu/lbm ⋅ R ) 25,037 ft 2 /s 2 2c p
T P2 = P1 2 T1
Compressor:
4
qin
E& in = E& out
k / (k −1)
537.3 R = (7 psia ) 470 R
6
qout
s
= 537.4 R
1.4/0.4
= 11.19 psia
( )
P3 = P4 = r p (P2 ) = (13)(11.19 psia ) = 145.5 psia P T3 = T2 3 P2
(k −1) / k
= (537.4 R )(13)0.4/1.4 = 1118.3 R
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-75
Turbine: wcomp,in = w turb,out → h3 − h2 = h4 − h5 → c p (T3 − T2 ) = c p (T4 − T5 )
or, T5 = T4 − T3 + T2 = 2400 − 1118.3 + 537.4 = 1819.1 R T P5 = P4 5 T4
k / (k −1)
1819.1 R = (145.5 psia ) 2400 R
1.4/0.4
= 55.2 psia
(b) Nozzle: P T6 = T5 6 P5
(k −1) / k
7 psia = (1819.1 R ) 55.2 psia
0.4/1.4
= 1008.6 R
E& in − E& out = ∆E& system ©0 (steady) E& in = E& out h5 + V52 / 2 = h6 + V 62 / 2 ©0
V62 − V52 2 0 = c p (T6 − T5 ) + V62 / 2 0 = h6 − h5 +
or, V6 =
/s 2 1 Btu/lbm
(2)(0.240 Btu/lbm ⋅ R )(1819.1 − 1008.6)R 25,037 ft
2
= 3121 ft/s
(c) The propulsive efficiency is the ratio of the propulsive work to the heat input, w p = (V exit − Vinlet )V aircraft
q in
1 Btu/lbm = 79.8 Btu/lbm = [(3121 − 900 ) ft/s ](900 ft/s ) 25,037 ft 2 /s 2 = h4 − h3 = c p (T4 − T3 ) = (0.24 Btu/lbm ⋅ R )(2400 − 1118.3)R = 307.6 Btu/lbm
ηp =
wp q in
=
79.8 Btu/lbm = 25.9% 307.6 Btu/lbm
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-76
9-116E A turbojet engine operating on an ideal cycle is flying at an altitude of 20,000 ft. The pressure at the turbine exit, the velocity of the exhaust gases, and the propulsive efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air are given in Table A-17E. Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V1= 900 ft/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0). Diffuser: T1 = 470 R → h1 = 112.20 Btu/lbm Pr1 = 0.8548
T
E& in − E& out = ∆E& system ©0 (steady)
5 3
E& in = E& out
2
h1 + V12 / 2 = h2 + V 22 / 2 0 = h2 − h1 + h2 = h1 +
V 22
1
©0
− V12 2
V12 (900 ft/s )2 = 112.20 + 2 2
Pr 2 P2 = P1 Pr 1
Compressor:
4
qin
1 Btu/lbm 25,037 ft 2 /s 2
6
qout
s
= 128.48 Btu/lbm → Pr 2 = 1.3698
= (7 psia ) 1.3698 = 11.22 psia 0.8548
( )
P3 = P4 = r p (P2 ) = (13)(11.22 psia ) = 145.8 psia P 145.8 → h3 = 267.56 Btu/lbm Pr3 = 3 Pr2 = (1.368) = 17.80 P 11.22 2 → Turbine: T4 = 2400 R
h4 = 617.22 Btu/lbm Pr 4 = 367.6
wcomp,in = w turb,out h3 − h2 = h4 − h5
or, → Pr5 = 142.7 h5 = h4 − h3 + h2 = 617.22 − 267.56 + 128.48 = 478.14 Btu/lbm Pr5 P5 = P4 Pr 4
= (145.8 psia ) 142.7 = 56.6 psia 367.6
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-77
(b) Nozzle: P 7 psia = 17.66 Pr6 = Pr5 6 = (142.7) → h6 = 266.93 Btu/lbm 56.6 psia P5 E& in − E& out = ∆E& system ©0 (steady) E& in = E& out h5 + V52 / 2 = h6 + V 62 / 2 0 = h6 − h5 +
V62 − V52 2
©0
or, V6 = 2(h5 − h6 ) =
/s 2 1 Btu/lbm
(2)(478.14 − 266.93)Btu/lbm 25,037 ft
2
= 3252 ft/s
(c) The propulsive efficiency is the ratio of the propulsive work to the heat input, w p = (V exit − Vinlet )Vaircraft
q in
1 Btu/lbm = 84.55 Btu/lbm = [(3252 − 900) ft/s)](900 ft/s) 25,037 ft 2 /s 2 = h4 − h3 = 617.22 − 267.56 = 349.66 Btu/lbm
ηp =
wp q in
=
84.55 Btu/lbm = 24.2% 349.66 Btu/lbm
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-78
9-117 A turbojet aircraft flying at an altitude of 9150 m is operating on the ideal jet propulsion cycle. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0). Diffuser: ©0 (steady) E& − E& = ∆E& → E& = E& in
out
system
in
out
h1 + V12 / 2 = h2 + V 22 / 2 → 0 = h2 − h1 +
V 22
©0
0 = c p (T2 − T1 ) − V12 / 2 T2 = T1 +
· Qi
k / (k −1)
4 5
3
V12 (320 m/s)2 1 kJ/kg = 241 K + (2)(1.005 kJ/kg ⋅ K ) 1000 m 2 /s 2 2c p
T P2 = P1 2 T1 Compressor: P3 = P4 =
291.9 K = (32 kPa ) 241 K
= 291.9 K
2
6
1
1.4/0.4
= 62.6 kPa
(r p )(P2 ) = (12)(62.6 kPa ) = 751.2 kPa
P T3 = T2 3 P2
Turbine:
T
− V12 2
(k −1) / k
= (291.9 K )(12)0.4/1.4 = 593.7 K
wcomp,in = w turb,out → h3 − h2 = h4 − h5 → c p (T3 − T2 ) = c p (T4 − T5 )
or, T5 = T4 − T3 + T2 = 1400 − 593.7 + 291.9 = 1098.2K
Nozzle:
(k −1) / k
P 32 kPa = (1400 K ) T6 = T4 6 751.2 kPa P4 E& in − E& out = ∆E& system ©0 (steady) → E& in = E& out
0.4/1.4
= 568.2 K
h5 + V52 / 2 = h6 + V 62 / 2 0 = h6 − h5 +
or, (b) (c)
V6 =
V62 − V52 2
©0
→ 0 = c p (T6 − T5 ) + V 62 / 2
/s 2 1 kJ/kg
(2)(1.005 kJ/kg ⋅ K )(1098.2 − 568.2)K 1000 m
2
= 1032 m/s
1 kJ/kg = 13,670 kW W& p = m& (V exit − Vinlet )V aircraft = (60 kg/s )(1032 − 320 )m/s(320 m/s ) 1000 m 2 /s 2 Q& in = m& (h4 − h3 ) = m& c p (T4 − T3 ) = (60 kg/s )(1.005 kJ/kg ⋅ K )(1400 − 593.7 )K = 48,620 kJ/s
m& fuel =
Q& in 48,620 kJ/s = = 1.14 kg/s HV 42,700 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-79
9-118 A turbojet aircraft is flying at an altitude of 9150 m. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0). Diffuser: E& in − E& out = ∆E& system ©0 (steady)
T
h1 + V12 / 2 = h2 + V 22 / 2
T2 = T1 +
©0
2
k / (k −1)
5 5s 6
1
V12 (320 m/s)2 1 kJ/kg = 241 K + (2)(1.005 kJ/kg ⋅ K ) 1000 m 2 /s 2 2c p
T P2 = P1 2 T1
Compressor:
V 22
3
− V12 2 0 = c p (T2 − T1 ) − V12 / 2 0 = h2 − h1 +
4
· Qi
E& in = E& out
291.9 K = (32 kPa ) 241 K
= 291.9 K
1.4/0.4
= 62.6 kPa
( )
P3 = P4 = r p (P2 ) = (12 )(62.6 kPa ) = 751.2 kPa P T3s = T2 3 P2
(k −1) / k
= (291.9 K )(12 )0.4/1.4 = 593.7 K
h3s − h2 c p (T3s − T2 ) = h3 − h 2 c p (T3 − T2 )
ηC =
T3 = T2 + (T3s − T2 ) / η C = 291.9 + (593.7 − 291.9)/ (0.80 ) = 669.2 K
Turbine: wcomp,in = w turb,out → h3 − h2 = h4 − h5 → c p (T3 − T2 ) = c p (T4 − T5 )
or, T5 = T4 − T3 + T2 = 1400 − 669.2 + 291.9 = 1022.7 K
ηT =
c p (T4 − T5 ) h4 − h5 = h4 − h5 s c p (T4 − T5 s )
T5 s = T4 − (T4 − T5 ) / η T = 1400 − (1400 − 1022.7 ) / 0.85 = 956.1 K T P5 = P4 5 s T4
k / (k −1)
956.1 K = (751.2 kPa ) 1400 K
1.4/0.4
= 197.7 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-80
Nozzle: (k −1) / k
P 32 kPa T6 = T5 6 = (1022.7 K ) P 197.7 kPa 5 E& in − E& out = ∆E& system ©0 (steady)
0.4/1.4
= 607.8 K
E& in = E& out h5 + V52 / 2 = h6 + V 62 / 2 ©0
V 2 − V52 0 = h6 − h5 + 6 2 0 = c p (T6 − T5 ) + V62 / 2
or, V6 =
/s 2 1 kJ/kg
(2)(1.005 kJ/kg ⋅ K )(1022.7 − 607.8)K 1000 m
2
= 913.2 m/s
(b)
W& p = m& (Vexit − Vinlet )Vaircraft
(c)
Q& in = m& (h4 − h3 ) = m& c p (T4 − T3 ) = (60 kg/s )(1.005 kJ/kg ⋅ K )(1400 − 669.2 )K = 44,067 kJ/s
1 kJ/kg = (60 kg/s )(913.2 − 320 )m/s(320 m/s ) 1000 m 2 /s 2 = 11,390 kW
m& fuel =
Q& in 44,067 kJ/s = = 1.03 kg/s HV 42,700 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-81
9-119 A turbojet aircraft that has a pressure rate of 12 is stationary on the ground. The force that must be applied on the brakes to hold the plane stationary is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the nozzle exit. Properties The properties of air are given in Table A17. T
Analysis (a) Using variable specific heats for air,
qin
→ h1 = 300.19 kJ/kg Compressor: T1 = 300 K
4
Pr1 = 1.386
Pr 2 =
3
2
P2 Pr = (12 )(1.386 ) = 16.63 → h2 = 610.65 kJ/kg P1 1
5 1
Q& in = m& fuel × HV = (0.2 kg/s )(42,700 kJ/kg ) = 8540 kJ/s q in =
Q& in 8540 kJ/s = = 854 kJ/kg m& 10 kg/s
q in = h3 − h2 → h3 = h2 + q in = 610.65 + 854 = 1464.65 kJ/kg → Pr3 = 396.27
Turbine: wcomp,in = w turb,out → h2 − h1 = h3 − h4
or, h4 = h3 − h2 + h1 = 1464.65 − 610.65 + 300.19 = 741.17 kJ/kg
Nozzle: P 1 Pr5 = Pr3 5 = (396.27 ) = 33.02 → h5 = 741.79 kJ/kg 12 P3 E& in − E& out = ∆E& system ©0 (steady) E& in = E& out h4 + V 42 / 2 = h5 + V52 / 2 V 2 − V 42 0 = h5 − h4 + 5 2
©0
or, V5 = 2(h4 − h5 ) =
/s 2 1 kJ/kg
(2)(1154.19 − 741.17 )kJ/kg 1000 m
2
= 908.9 m/s
1N Brake force = Thrust = m& (Vexit − Vinlet ) = (10 kg/s )(908.9 − 0 )m/s 1 kg ⋅ m/s 2
= 9089 N
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-82
9-120 EES Problem 9-119 is reconsidered. The effect of compressor inlet temperature on the force that must be applied to the brakes to hold the plane stationary is to be investigated. Analysis Using EES, the problem is solved as follows: P_ratio = 12 T_1 = 27 [C] T[1] = T_1+273 "[K]" P[1]= 95 [kPa] P[5]=P[1] Vel[1]=0 [m/s] V_dot[1] = 9.063 [m^3/s] HV_fuel = 42700 [kJ/kg] m_dot_fuel = 0.2 [kg/s] Eta_c = 1.0 Eta_t = 1.0 Eta_N = 1.0 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) v[1]=volume(Air,T=T[1],P=P[1]) m_dot = V_dot[1]/v[1] "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) Q_dot_in = m_dot_fuel*HV_fuel m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" {P_ratio= P[3] /P[4]} T_s[4]=TEMPERATURE(Air,h=h_s[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" {h_s[4]=ENTHALPY(Air,T=T_s[4])} "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" T[4]=TEMPERATURE(Air,h=h[4]) P[4]=pressure(Air,s=s_s[4],h=h_s[4]) "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" W_dot_net = 0 [kW]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-83
"Exit nozzle analysis:" s[4]=entropy('air',T=T[4],P=P[4]) s_s[5]=s[4] "For the ideal case the entropies are constant across the nozzle" T_s[5]=TEMPERATURE(Air,s=s_s[5], P=P[5]) "T_s[5] is the isentropic value of T[5] at nozzle exit" h_s[5]=ENTHALPY(Air,T=T_s[5]) Eta_N=(h[4]-h[5])/(h[4]-h_s[5]) m_dot*h[4] = m_dot*(h_s[5] + Vel_s[5]^2/2*convert(m^2/s^2,kJ/kg)) m_dot*h[4] = m_dot*(h[5] + Vel[5]^2/2*convert(m^2/s^2,kJ/kg)) T[5]=TEMPERATURE(Air,h=h[5]) s[5]=entropy('air',T=T[5],P=P[5]) "Brake Force to hold the aircraft:" Thrust = m_dot*(Vel[5] - Vel[1]) "[N]" BrakeForce = Thrust "[N]" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) s[2]=entropy('air',T=T[2],P=P[2]) Air 1600
m [kg/s]
T3 [K]
T1 [C]
11.86 11.41 10.99 10.6 10.24 9.9
1164 1206 1247 1289 1330 1371
-20 -10 0 10 20 30
1400
3
4s 95
kP
11
1000
a
40
kP
a
1200
T [K]
Brake Force [N] 9971 9764 9568 9383 9207 9040
800 5s
2s
600 400 1
200 4.5
5.0
5.5
6.0
6.5
7.0
7.5
s [kJ/kg-K] 10000
BrakeForce [N]
9800
9600
9400
9200
9000 -20
-10
0
10
20
30
T 1 [C]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-84
9-121 Air enters a turbojet engine. The thrust produced by this turbojet engine is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air are given in Table A-17. Analysis We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 300 m/s. Taking the entire engine as our control volume and writing the steady-flow energy balance yield T1 = 280 K
→
h1 = 28013 . kJ / kg
T2 = 700 K
→
h2 = 713.27 kJ / kg
E& in − E& out = ∆E& system ©0 (steady) E& in = E& out Q& in + m& (h1 + V12 / 2) = m& (h2 + V 22 / 2)
15,000 kJ/s 7°C 300 m/s 16 kg/s
427°C 1
V 2 − V12 Q& in = m& h2 − h1 + 2 2 V 2 − (300 m/s )2 15,000 kJ/s = (16 kg/s )713.27 − 280.13 + 2 2
2
1 kJ/kg 1000 m 2 /s 2
It gives V 2 = 1048 m/s Thus, Fp = m& (V2 − V1 ) = (16 kg/s )(1048 − 300 )m/s = 11,968 N
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-85
Second-Law Analysis of Gas Power Cycles 9-122 The total exergy destruction associated with the Otto cycle described in Prob. 9-34 and the exergy at the end of the power stroke are to be determined. Analysis From Prob. 9-34, qin = 750, qout = 357.62 kJ/kg, T1 = 300 K, and T4 = 774.5 K. The total exergy destruction associated with this Otto cycle is determined from q 357.62 kJ/kg 750 kJ/kg q = 245.12 kJ/kg x destroyed = T0 out − in = (300 K ) − 300 K 2000 K TL TH
Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke (state 4) is determined from
φ 4 = (u 4 − u 0 ) − T0 (s 4 − s 0 ) + P0 (v 4 − v 0 ) where u 4 − u 0 = u 4 − u1 = q out = 357.62kJ/kg
v 4 −v 0 = v 4 −v1 = 0 s 4 − s 0 = s 4 − s1 = s 4o − s1o − R ln
T Tv P4 = s 4o − s1o − Rln 4 1 = s 4o − s1o − R ln 4 T1 T1v 4 P1
= 2.6823 − 1.70203 − (0.287 kJ/kg ⋅ K ) ln
774.5 K = 0.7081 kJ/kg ⋅ K 300 K
Thus,
φ 4 = (357.62 kJ/kg ) − (300 K )(0.7081 kJ/kg ⋅ K ) + 0 = 145.2 kJ/kg
9-123 The total exergy destruction associated with the Diesel cycle described in Prob. 9-47 and the exergy at the end of the compression stroke are to be determined. Analysis From Prob. 9-47, qin = 1019.7, qout = 445.63 kJ/kg, T1 = 300 K, v1 = 0.906 m3/kg, and v 2 = v 1 / r = 0.906 / 12 = 0.0566 m3/kg. The total exergy destruction associated with this Otto cycle is determined from q 445.63 kJ/kg 1019.7 kJ/kg q = 292.7 kJ/kg − x destroyed = T0 out − in = (300 K ) 300 K 2000 K TL TH
Noting that state 1 is identical to the state of the surroundings, the exergy at the end of the compression stroke (state 2) is determined from
φ 2 = (u 2 − u 0 ) − T0 (s 2 − s 0 ) + P0 (v 2 − v 0 ) = (u 2 − u1 ) − T0 (s 2 − s1 ) + P0 (v 2 − v 1 )
1 kJ = (643.3 − 214.07 ) − 0 + (95 kPa )(0.0566 − 0.906 )m 3 /kg 1 kPa ⋅ m 3 = 348.6 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-86
9-124E The exergy destruction associated with the heat rejection process of the Diesel cycle described in Prob. 9-49E and the exergy at the end of the expansion stroke are to be determined. Analysis From Prob. 9-49E, qout = 158.9 Btu/lbm, T1 = 540 R, T4 = 1420.6 R, and v 4 = v 1. At Tavg = (T4 + T1)/2 = (1420.6 + 540)/2 = 980.3 R, we have cv,avg = 0.180 Btu/lbm·R. The entropy change during process 4-1 is s1 − s4 = cv ln
v ©0 540 R T1 + R ln 1 = (0.180 Btu/lbm ⋅ R )ln = −0.1741 Btu/lbm ⋅ R v4 1420.6 R T4
Thus, qR, 41 158.9 Btu/lbm = (540R ) − 0.1741 Btu/lbm ⋅ R + = 64.9 Btu/lbm xdestroyed, 41 = T0 s1 − s4 + T 540 R R Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke (state 4) is determined from
φ 4 = (u 4 − u 0 ) − T0 (s 4 − s 0 ) + P0 (v 4 − v 0 ) where u 4 − u 0 = u 4 − u1 = q out = 158.9 Btu/lbm ⋅ R
v 4 −v 0 = v 4 −v1 = 0 s 4 − s 0 = s 4 − s1 = 0.1741 Btu/lbm ⋅ R
Thus,
φ 4 = (158.9 Btu/lbm ) − (540R )(0.1741 Btu/lbm ⋅ R ) + 0 = 64.9 Btu/lbm Discussion Note that the exergy at state 4 is identical to the exergy destruction for the process 4-1 since state 1 is identical to the dead state, and the entire exergy at state 4 is wasted during process 4-1.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-87
9-125 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-73 is to be determined. Analysis From Prob. 9-73, qin = 584.62 kJ/kg, qout = 478.92 kJ/kg, and → s1o = 1.73498kJ/kg ⋅ K
T1 = 310K
h2 = 646.3kJ/kg → s 2o = 2.47256kJ/kg ⋅ K T3 = 1160K
→ s 3o = 3.13916kJ/kg ⋅ K
h4 = 789.16kJ/kg → s 4o = 2.67602kJ/kg ⋅ K
Thus, P x destroyed,12 = T0 s gen,12 = T0 (s 2 − s1 ) = T0 s 2o − s1o − Rln 2 = P1 = (290 K )(2.47256 − 1.73498 − (0.287 kJ/kg ⋅ K )ln(8)) = 40.83 kJ/kg q R ,23 x destroyed, 23 = T0 s gen,23 = T0 s 3 − s 2 + TR
P = T0 s 3o − s 2o − Rln 3 P2
©0
− q in TH
+
584.62 kJ/kg = 87.35 kJ/kg = (290 K ) 3.13916 − 2.47256 − 1600 K P x destroyed, 34 = T0 s gen,34 = T0 (s 4 − s 3 ) = T0 s 4o − s 3o − Rln 4 = P3 = (290 K )(2.67602 − 3.13916 − (0.287 kJ/kg ⋅ K )ln(1/8)) = 38.76 kJ/kg q R ,41 P = T0 s1o − s 4o − Rln 1 x destroyed, 41 = T0 s gen,41 = T0 s1 − s 4 + TR P4
©0
+
q out TL
478.92 kJ/kg = 206.0 kJ/kg = (290 K )1.73498 − 2.67602 + 310 K
9-126 The total exergy destruction associated with the Brayton cycle described in Prob. 9-93 and the exergy at the exhaust gases at the turbine exit are to be determined. Analysis From Prob. 9-93, qin = 601.94, qout = 279.68 kJ/kg, and h6 = 579.87 kJ/kg. The total exergy destruction associated with this Otto cycle is determined from q 279.68 kJ/kg 601.94 kJ/kg q = 179.4 kJ/kg − x destroyed = T0 out − in = (300 K ) 300 K 1800 K TL TH
Noting that h0 = h@ determined from
300 K
= 300.19 kJ/kg, the stream exergy at the exit of the regenerator (state 6) is
V2 φ 6 = (h6 − h0 ) − T0 (s 6 − s 0 ) + 6 2
©0
+ gz 6 ©0
P6 Ê0 = 2.36275 − 1.70203 = 0.66072 kJ / kg ⋅ K P1
where
s6 − s0 = s6 − s1 = s6o − s1o − R ln
Thus,
φ6 = 579.87 − 300.19 − (300 K )(0.66072 kJ/kg ⋅ K ) = 81.5 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-88
9-127 EES Problem 9-126 is reconsidered. The effect of the cycle pressure on the total irreversibility for the cycle and the exergy of the exhaust gas leaving the regenerator is to be investigated. Analysis Using EES, the problem is solved as follows: "Input data" T_o = 300 [K] T_L = 300 [K] T_H = 1400 [K] T[3] = 1200 [K] {Pratio = 10} T[1] = 300 [K] C_P=1.005 [kJ/kg-K] P[1]= 100 [kPa] P_o=P[1] Eta_reg = 1.0 Eta_c =1.0"Compressor isentorpic efficiency" Eta_t =1.0"Turbien isentropic efficiency" MM=MOLARMASS(Air) R=R_u/MM R_u=8.314 [kJ/kmol-K] C_V=C_P - R k=C_P/C_V "Isentropic Compressor anaysis" "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] "T_s[2] is the isentropic value of T[2] at compressor exit" T_s[2]=T[1]*(Pratio)^((k-1)/k) Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_s[2]-T[1]) "Actual compressor analysis:" w_comp = C_P*(T[2]-T[1]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" q_in_noreg = C_P*(T[3]-T[2]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "Turbine analysis" "For the ideal case the entropies are constant across the turbine" P[4] = P[3] /Pratio T_s[4]=T[3]*(1/Pratio)^((k-1)/k) "T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen=C_P*(T[3] - T_s[4]) "Actual Turbine analysis:" w_turb= C_P*(T[3]-T[4]) "Cycle analysis" w_net=w_turb-w_comp "[kJ/kg]" Eta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency" Bwr=w_comp/w_turb "Back work ratio"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-89 "With the regenerator the heat added in the external heat exchanger is" q_in_withreg = C_P*(T[3]-T[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (T[5]-T[2])/(T[4]-T[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" "h[2] + h[4]=h[5] + h[6]" T[2] + T[4]=T[5] + T[6] P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "[%]" "Irreversibility associated with the Brayton cycle is determined from:" q_out_withreg = q_in_withreg - w_net i_withreg = T_o*(q_out_withreg/T_L - q_in_withreg/T_H) q_out_noreg = q_in_noreg - w_net i_noreg = T_o*(q_out_noreg/T_L - q_in_noreg/T_H) "Neglecting the ke and pe of the exhaust gases, the exergy of the exhaust gases at the exit of the regenerator is:" "Psi_6 = (h[6] - h_o) - T_o(s[6] - s_o)" Psi_exit_withreg = C_P*(T[6] - T_o) - T_o*(C_P*ln(T[6]/T_o)-R*ln(P[6]/P_o)) Psi_exit_noreg = C_P*(T[4] - T_o) - T_o*(C_P*ln(T[4]/T_o)-R*ln(P[4]/P_o)) inoreg
iwithreg
Pratio
270.8 244.5 223 205 189.6 176.3 164.6 154.2 144.9
97.94 113.9 128.8 142.7 155.9 168.4 180.2 191.5 202.3
6 7 8 9 10 11 12 13 14
Ψexit,noreg [kJ/kg] 157.8 139.9 125.5 113.6 103.6 95.05 87.62 81.11 75.35
Ψexit,withreg [kJ/kg] 47.16 56.53 65.46 74 82.17 90.02 97.58 104.9 111.9
ηth,noreg [%] 40.05 42.63 44.78 46.61 48.19 49.58 50.82 51.93 52.94
ηth,withreg [%] 58.3 56.42 54.73 53.18 51.75 50.41 49.17 47.99 46.88
Air
1600 1400
3
T [K]
1200
1000 kPa
1000 800
2
100 kPa
5
2s
600
4s
4
6 400 1 200 4.5
5.0
5.5
6.0
6.5
7.0
7.5
s [kJ/kg-K]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-90 275
No regeneration
i [kJ/kg]
235
195
155
115
75 6
W ith regeneration
7
8
9
10
11
12
13
14
Pratio 160
No regeneration
Psi exit [kJ/kg]
140 120 100 80 60
W ith regeneration
40 6
7
8
9
10
11
12
13
14
13
14
Pratio 60
W ith regeneration 56
η th [%]
52
48
44
40 6
No regeneration
7
8
9
10
11
12
Pratio
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-91
9-128 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-98 and the exergy at the end of the exhaust gases at the exit of the regenerator are to be determined. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis From Prob. 9-98, qin = 539.23 kJ/kg, qout = 345.17 kJ/kg, and T1 = 300 K
→ s1o = 1.70203 kJ/kg ⋅ K
h2 = 586.04 kJ/kg → s 2o = 2.37348 kJ/kg ⋅ K T3 = 1200 K
→ s 3o = 3.17888 kJ/kg ⋅ K
h4 = 797.88 kJ/kg → s 4o = 2.68737 kJ/kg ⋅ K h5 = 738.56 kJ/kg → s 5o = 2.60833 kJ/kg ⋅ K
and, from an energy balance on the heat exchanger, h5 − h2 = h4 − h6 → h6 = 797.88 − 738.56 + 586.04 = 645.36 kJ/kg → s 6o = 2.47108 kJ/kg ⋅ K
Thus, P x destroyed,12 = T0 s gen,12 = T0 (s 2 − s1 ) = T0 s 2o − s1o − Rln 2 P1 = (300 K )(2.37348 − 1.70203 − (0.287 kJ/kg ⋅ K )ln(8)) = 22.40 kJ/kg P x destroyed, 34 = T0 s gen,34 = T0 (s 4 − s 3 ) = T0 s 4o − s 3o − Rln 4 P3 = (300 K )(2.68737 − 3.17888 − (0.287kJ/kg ⋅ K )ln (1/8)) = 31.59 kJ/kg
[(
) (
x destroyed, regen = T0 s gen,regen = T0 [(s 5 − s 2 ) + (s 6 − s 4 )] = T0 s 5o − s 2o + s 6o − s 4o
)]
= (300 K )(2.60833 − 2.37348 + 2.47108 − 2.68737 ) = 5.57 kJ/kg
q R,53 x destroyed, 53 = T0 s gen,53 = T0 s 3 − s 5 − TR
P ©0 q in = T0 s 3o − s 5o − Rln 3 − P5 T H
539.23 kJ/kg = 42.78 kJ/kg = (300 K ) 3.17888 − 2.60833 − 1260 K q R,61 P = T0 s1o − s 6o − Rln 1 x destroyed, 61 = T0 s gen,61 = T0 s1 − s 6 + T P R 6
©0
+
q out TL
345.17 kJ/kg = 114.5 kJ/kg = (300 K )1.70203 − 2.47108 + 300 K
Noting that h0 = h@ determined from
300 K
= 300.19 kJ/kg, the stream exergy at the exit of the regenerator (state 6) is
φ6 = (h6 − h0 ) − T0 (s6 − s0 ) +
V62 2
©0
+ gz6©0 P6 P1
©0
where
s 6 − s 0 = s 6 − s1 = s 6o − s1o − Rln
Thus,
φ 6 = 645.36 − 300.19 − (300 K )(0.76905 kJ/kg ⋅ K ) = 114.5 kJ/kg
= 2.47108 − 1.70203 = 0.76905 kJ/kg ⋅ K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-92
9-129 A gas-turbine plant uses diesel fuel and operates on simple Brayton cycle. The isentropic efficiency of the compressor, the net power output, the back work ratio, the thermal efficiency, and the second-law efficiency are to be determined. Assumptions 1 The air-standard assumptions are Diesel fuel Combustion applicable. 2 Kinetic and potential energy chamber changes are negligible. 3 Air is an ideal gas with 3 constant specific heats. 700 2 Properties The properties of air at 500ºC = 773 kPa K are cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.357 (Table ATurbine Compres 2b). Analysis (a) The isentropic efficiency of the compressor may be determined if we first 100 4 1 calculate the exit temperature for the isentropic kPa case ( k −1) / k
(1.357-1)/1.357
P 700 kPa = 505.6 K T2 s = T1 2 = (303 K ) 100 kPa P1 T −T (505.6 − 303)K η C = 2s 1 = = 0.881 T2 − T1 (533 − 303)K (b) The total mass flowing through the turbine and the rate of heat input are m& 12.6 kg/s = 12.6 kg/s + 0.21 kg/s = 12.81 kg/s m& t = m& a + m& f = m& a + a = 12.6 kg/s + 60 AF Q& in = m& f q HVη c = (0.21 kg/s)(42,000 kJ/kg)(0.97) = 8555 kW
The temperature at the exit of combustion chamber is Q& in = m& c p (T3 − T2 ) → 8555 kJ/s = (12.81 kg/s)(1.093 kJ/kg.K)(T3 − 533)K → T3 = 1144 K The temperature at the turbine exit is determined using isentropic efficiency relation ( k −1) / k
(1.357-1)/1.357 P 100 kPa = 685.7 K T4 s = T3 4 = (1144 K ) 700 kPa P3 T − T4 (1144 − T4 )K ηT = 3 → 0.85 = → T4 = 754.4 K T3 − T4 s (1144 − 685.7)K The net power and the back work ratio are W&C,in = m& a c p (T2 − T1 ) = (12.6 kg/s)(1.093 kJ/kg.K)(533 − 303)K = 3168 kW
W&T, out = m& c p (T3 − T4 ) = (12.81 kg/s)(1.093 kJ/kg.K)(1144 − 754.4)K = 5455 kW W& net = W&T, out − W&C,in = 5455 − 3168 = 2287 kW W& 3168 kW rbw = & C,in = = 0.581 WT,out 5455 kW
(c) The thermal efficiency is
η th =
W& net 2287 kW = = 0.267 8555 kW Q& in
The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is, T 303 K η max = 1 − 1 = 1 − = 0.735 T3 1144 K and
η II =
η th 0.267 = = 0.364 η max 0.735
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-93
9-130 A modern compression ignition engine operates on the ideal dual cycle. The maximum temperature in the cycle, the net work output, the thermal efficiency, the mean effective pressure, the net power output, the second-law efficiency of the cycle, and the rate of exergy of the exhaust gases are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are cp = 1.110 kJ/kg·K, cv = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.349 (Table A-2b). Analysis (a) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are r=
Vc + V d V + 0.0028 m3 →14 = c →V c = 0.0002154 m3 = V 2 = V x Vc Vc
V1 = V c +V d = 0.0002154 + 0.0028 = 0.003015 m 3 = V 4
P
Process 1-2: Isentropic compression v T2 = T1 1 v 2 v P2 = P1 1 v 2
k −1
3
x
= (328 K )(14 )1.349-1 = 823.9 K
2
k
= (95 kPa )(14 )1.349 = 3341 kPa
Qin
4
Process 2-x and x-3: Constant-volume and constant pressure heat addition processes: T x = T2
Qout
Px 9000 kPa = (823.9 K) = 2220 K P2 3341 kPa
1
q 2- x = cv (T x − T2 ) = (0.823 kJ/kg.K)(2220 − 823.9)K = 1149 kJ/kg
V
q 2− x = q x -3 = c p (T3 − T x ) → 1149 kJ/kg = (0.823 kJ/kg.K)(T3 − 2220)K → T3 = 3254 K
(b)
q in = q 2− x + q x -3 = 1149 + 1149 = 2298 kJ/kg
V3 =V x
T3 3254 K = (0.0002154 m 3 ) = 0.0003158 m 3 Tx 2220 K
Process 3-4: isentropic expansion. k −1
1.349-1
V T4 = T3 3 V 4
0.0003158 m 3 = (3254 K ) 0.003015 m 3
V P4 = P3 3 V 4
0.0003158 m 3 = (9000 kPa ) 0.003015 m 3
k
= 1481 K 1.349
= 428.9 kPa
Process 4-1: constant voume heat rejection. q out = cv (T4 − T1 ) = (0.823 kJ/kg ⋅ K )(1481 − 328)K = 948.7 kJ/kg
The net work output and the thermal efficiency are wnet,out = q in − q out = 2298 − 948.7 = 1349 kJ/kg
η th =
wnet,out q in
=
1349 kJ/kg = 0.587 2298 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-94
(c) The mean effective pressure is determined to be m=
P1V1 (95 kPa)(0.003015 m 3 ) = = 0.003043 kg RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (328 K )
(
MEP =
mwnet,out
V 1 −V 2
)
=
(0.003043 kg)(1349 kJ/kg) kPa ⋅ m 3 (0.003015 − 0.0002154)m 3 kJ
= 1466 kPa
(d) The power for engine speed of 3500 rpm is 3500 (rev/min) 1 min n& W& net = mwnet = (0.003043 kg)(1349 kJ/kg) = 120 kW 2 (2 rev/cycle) 60 s
Note that there are two revolutions in one cycle in four-stroke engines. (e) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). We take the dead state temperature and pressure to be 25ºC and 100 kPa.
η max = 1 − and
η II =
T0 (25 + 273) K =1− = 0.908 T3 3254 K
η th 0.587 = = 0.646 η max 0.908
The rate of exergy of the exhaust gases is determined as follows T P x 4 = u 4 − u 0 − T0 ( s 4 − s 0 ) = cv (T4 − T0 ) − T0 c p ln 4 − R ln 4 T0 P0 1481 428.9 = (0.823)(1481 − 298)− (298) (1.110 kJ/kg.Kln − 0.287 ln = 567.6 kJ/kg 298 100 3500 (rev/min) 1 min n& X& 4 = mx 4 = (0.003043 kg)(567.6 kJ/kg) = 50.4 kW 2 (2 rev/cycle) 60 s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-95
9-131 A gas-turbine plant operates on the regenerative Brayton cycle. The isentropic efficiency of the compressor, the effectiveness of the regenerator, the air-fuel ratio in the combustion chamber, the net power output, the back work ratio, the thermal efficiency, the second law efficiency, the exergy efficiencies of the compressor, the turbine, and the regenerator, and the rate of the exergy of the combustion gases at the regenerator exit are to be determined. Regenerator Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with 6 constant specific heats. 5 Combustion Properties The properties of air at 500ºC = 773 chamber K are cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K, 100 kPa R = 0.287 kJ/kg·K, and k = 1.357 (Table A30°C 4 700 kPa 400°C 1 2b). 871°C 260°C 2 3 Analysis (a) For the compressor and the turbine: k −1
1.357 -1
P k 700 kPa 1.357 = 505.6 K T2 s = T1 2 = (303 K ) P 100 kPa 1 T −T (505.6 − 303)K η C = 2s 1 = = 0.881 (533 − 303)K T2 − T1
Compress.
Turbine
( k −1) / k
(1.357-1)/1.357 P 100 kPa = (1144 K ) = 685.6 K T4 s = T3 4 700 kPa P3 T − T4 (1144 − T4 )K ηT = 3 → 0.85 = → T4 = 754.4 K T3 − T4 s (1144 − 685.6)K (b) The effectiveness of the regenerator is T − T2 (673 − 533)K ε regen = 5 = = 0.632 T4 − T2 (754.4 − 533)K (c) The fuel rate and air-fuel ratio are Q& in = m& f q HVη c = (m& f + m& a )c p (T3 − T5 )
m& f (42,000 kJ/kg)(0.97) = (m& f + 12.6)(1.093 kJ/kg.K)(1144 − 673)K → m& f = 0.1613 kg/s AF =
Also,
m& a 12.6 = = 78.14 m& f 0.1613
m& = m& a + m& f = 12.6 + 0.1613 = 12.76 kg/s Q& = m& q η = (0.1613 kg/s)(42,000 kJ/kg)(0.97) = 6570 kW in
f
HV c
(d) The net power and the back work ratio are W& C,in = m& a c p (T2 − T1 ) = (12.6 kg/s)(1.093 kJ/kg.K)(533 − 303)K = 3168 kW W&T, out = m& c p (T3 − T4 ) = (12.76 kg/s)(1.093 kJ/kg.K)(1144 − 754.4)K = 5434 kW W& net = W&T, out − W&C,in = 5434 − 3168 = 2267 kW W& C,in 3168 kW = = 0.583 rbw = & 5434 kW WT,out
(e) The thermal efficiency is W& 2267 kW η th = net = = 0.345 & 6570 kW Qin (f) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-96
η max = 1 − and
η II =
T1 303 K =1− = 0.735 T3 1144 K
η th 0.345 = = 0.469 η max 0.735
(g) The exergy efficiency for the compressor is defined as the ratio of stream exergy difference between the inlet and exit of the compressor to the actual power input: T P ∆X& C = m& a [h2 − h1 − T0 ( s 2 − s1 )] = m& a c p (T2 − T1 ) − T0 c p ln 2 − R ln 2 T1 P1
η II,C
533 700 = (12.6)(1.093)(533 − 303)− (303) (1.093)ln − 0.287 ln = 2943 kW 303 100 ∆X& C 2943 kW = = = 0.929 W& C,in 3168 kW
The exergy efficiency for the turbine is defined as the ratio of actual turbine power to the stream exergy difference between the inlet and exit of the turbine: T P ∆X& T = m& c p (T3 − T4 ) − T0 c p ln 3 − R ln 3 T4 P4 700 1144 = (12.76)(1.093)(1144 − 754.4)− (303) (1.093)ln = 5834 kW − 0.287 ln 100 754.4 W& T,in 5434 kW = = 0.932 ∆X& T 5834 kW An energy balance on the regenerator gives m& a c p (T5 − T2 ) = m& c p (T4 − T6 )
η II ,T =
(12.6)(1.093)(673 − 533) = (12.76)(1.093)(754.4 − T6 ) → T6 = 616.2 K The exergy efficiency for the regenerator is defined as the ratio of the exergy increase of the cold fluid to the exergy decrease of the hot fluid: T ∆X& regen,hot = m& c p (T4 − T6 ) − T0 c p ln 4 − 0 T6 754.4 = (12.76)(1.093)(754.4 − 616.2)− (303) (1.093)ln − 0 = 1073 kW 616.2 T ∆X& regen,cold = m& c p (T5 − T2 ) − T0 c p ln 5 − 0 T2 673 = (12.76)(1.093)(673 − 533)− (303) (1.093)ln − 0 = 954.8 kW 533 ∆X& regen,cold 954.8 kW η II ,T = = = 0.890 1073 kW ∆X& regen,hot
The exergy of the combustion gases at the regenerator exit: T X& 6 = m& c p (T6 − T0 ) − T0 c p ln 6 − 0 T0 616.2 = (12.76)(1.093)(616.2 − 303)− (303) (1.093)ln − 0 = 1351 kW 303
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-97
Review Problems
9-132 A turbocharged four-stroke V-16 diesel engine produces 3500 hp at 1200 rpm. The amount of power produced per cylinder per mechanical and per thermodynamic cycle is to be determined. Analysis Noting that there are 16 cylinders and each thermodynamic cycle corresponds to 2 mechanical cycles, we have (a) wmechanical = =
Total power produced (No. of cylinders)(No. of mechanical cycles) 42.41 Btu/min 3500 hp (16 cylinders)(1200 rev/min) 1 hp
= 7.73 Btu/cyl ⋅ mech cycle (= 8.16 kJ/cyl ⋅ mech cycle)
(b) wthermodynamic =
Total power produced (No. of cylinders)(No. of thermodynamic cycles)
=
42.41 Btu/min 3500 hp (16 cylinders)(1200/2 rev/min) 1 hp
= 15.46 Btu/cyl ⋅ therm cycle (= 16.31 kJ/cyl ⋅ therm cycle)
9-133 A simple ideal Brayton cycle operating between the specified temperature limits is considered. The pressure ratio for which the compressor and the turbine exit temperature of air are equal is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air is k =1.4 (Table A-2). Analysis We treat air as an ideal gas with constant specific heats. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as P T2 = T1 2 P1
(k −1) / k
P T4 = T3 4 P3
( )
= T1 rp (k −1) / k
(k −1) / k
1 = T3 rp
T
k / 2 (k −1)
1500 K = 300 K
qin 2
Setting T2 = T4 and solving for rp gives T r p = 3 T1
3
T3
(k −1) / k
T1 1.4/0.8
4 1
qout
= 16.7
s
Therefore, the compressor and turbine exit temperatures will be equal when the compression ratio is 16.7.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-98
9-134 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work output and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17.
P
Analysis (b) We treat air as an ideal gas with variable specific heats,
q23
→ u1 = 214.07 kJ/kg T1 = 300 K
q12
h1 = 300.19 kJ/kg
1
300 kPa P P2v 2 P1v 1 (300 K ) = → T2 = 2 T1 = P1 T2 T1 100 kPa = 900 K → u 2 = 674.58 kJ/kg h2 = 932.93 kJ/kg T3 = 1300 K → u 3 = 1022.82 kJ/kg h3 = 1395.97 kJ/kg, Pr 3 = 330.9 Pr 4 =
3
2
100 kPa P4 (330.9) = 110.3 Pr3 = P3 300 kPa
4
qout
v
T 3
q23 q12 1
2 4
qout
→ h4 = 1036.46 kJ/kg
q in = q12,in + q 23,in = (u 2 − u1 ) + (h3 − h2 )
= (674.58 − 214.07 ) + (1395.97 − 932.93) = 923.55 kJ/kg
q out = h4 − h1 = 1036.46 − 300.19 = 736.27 kJ/kg wnet = q in − q out = 923.55 − 736.27 = 187.28 kJ/kg
(c)
η th =
wnet 187.28kJ/kg = = 20.3% 923.55kJ/kg q in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-99
9-135 All four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work output and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). P
Analysis (b) Process 3-4 is isentropic: P T4 = T3 4 P3
(k −1) / k
1 = (1300 K ) 3
q23
0.4/1.4
= 949.8 K
3
2
q12
300 kPa P2v 2 P1v 1 P (300 K ) = 900 K = → T2 = 2 T1 = T2 T1 P1 100 kPa
1
4
qout
v
q in = q12,in + q 23,in = (u 2 − u1 ) + (h3 − h2 ) = c v (T2 − T1 ) + c p (T3 − T2 ) = (0.718 kJ/kg ⋅ K )(900 − 300)K + (1.005 kJ/kg ⋅ K )(1300 − 900 )K = 832.8 kJ/kg
T
q out = h4 − h1 = c p (T4 − T1 ) = (1.005 kJ/kg ⋅ K )(949.8 − 300 )K = 653 kJ/kg
q12
4
wnet = q in − q out = 832.8 − 653 = 179.8 kJ/kg
(c)
η th
3
q23 2
qout
1
w 179.8 kJ/kg = net = = 21.6% q in 832.8 kJ/kg
s
9-136 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17.
P
Analysis (b) We treat air as an ideal gas with variable specific heats,
qin
2
T1 = 300 K → u1 = 214.07 kJ/kg
3
qout
Pr1 = 1.386 1
700 kPa P (1.386) = 9.702 Pr2 = 2 Pr1 = → h2 = 523.90 kJ/kg P1 100 kPa 700 kPa P P3v 3 P1v 1 (300 K ) = 2100 K = → Tmax = T3 = 3 T1 = P1 T3 T1 100 kPa T3 = 2100 K → u 3 = 1775.3 kJ/kg
v T qin
3
2
h3 = 2377.7 kJ/kg
(c)
q in = h3 − h2 = 2377.7 − 523.9 = 1853.8 kJ/kg
1
qout
q out = u 3 − u1 = 1775.3 − 214.07 = 1561.23kJ/kg
η th = 1 −
q out 1561.23 kJ/kg = 1− = 15.8% q in 1853.8 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-100
9-137 All three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). P
Analysis (b) We treat air as an ideal gas with constant specific heats. Process 1-2 is isentropic: P T2 = T1 2 P1
(k −1) / k
700 kPa = (300 K ) 100 kPa
3
0.4/1.4
qout
= 523.1 K 1
v
700 kPa P3v 3 P1v1 P (300 K ) = 2100 K = → Tmax = T3 = 3 T1 = T3 T1 P1 100 kPa
T qin
(c) q in = h3 − h2 = c p (T3 − T2 ) = (1.005 kJ/kg ⋅ K )(2100 − 523.1)K = 1584.8 kJ/kg
3
2
q out = u 3 − u1 = c v (T3 − T1 ) = (0.718 kJ/kg ⋅ K )(2100 − 300)K = 1292.4 kJ/kg
η th = 1 −
qin
2
q out 1292.4 kJ/kg = 1− = 18.5% q in 1584.8 kJ/kg
1
qout
s
9-138 A Carnot cycle executed in a closed system uses air as the working fluid. The net work output per cycle is to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis (a) The maximum temperature is determined from
η th = 1 −
T
TL 300 K → 0.60 = 1 − → T H = 750 K TH TH ©0
P2 700 kPa = −(0.287 kJ/kg ⋅ K )ln P1 1000 kPa = 0.1204 kJ/kg ⋅ K
s 2 − s1 = s 2o − s1o − R ln
1
1 MPa
2 700 kPa
ηth = 60% 300 K
4
3
W net = m(s 2 − s1 )(T H − T L ) = (0.0025 kg )(0.1024 kJ/kg ⋅ K )(750 − 300)K = 0.115 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-101
9-139 [Also solved by EES on enclosed CD] A four-cylinder spark-ignition engine with a compression ratio of 8 is considered. The amount of heat supplied per cylinder, the thermal efficiency, and the rpm for a net power output of 60 kW are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17. Analysis (a) Process 1-2: isentropic compression. P
→ u1 = 206.91 kJ/kg T1 = 290 K
v r1 = 676.1 v r2 =
3
v2 1 1 v r = v r = (676.1) = 84.51 v1 1 r 1 8
1800 K
Qin
→ u 2 = 475.11 kJ/kg
2
Process 2-3: v = constant heat addition.
4 1
Qout
v
T3 = 1800 K → u 3 = 1487.2 kJ/kg
v r3 = 3.994 m=
(
)
P1V1 (98 kPa ) 0.0006 m 3 = = 7.065 × 10 − 4 kg 3 RT1 0.287 kPa ⋅ m /kg ⋅ K (290 K )
(
(
)
)
Qin = m(u 3 − u 2 ) = 7.065 × 10 − 4 kg (1487.2 − 475.11)kJ/kg = 0.715 kJ
(b) Process 3-4: isentropic expansion.
v r4 =
v4 v r = rv r3 = (8)(3.994 ) = 31.95 → u 4 = 693.23 kJ/kg v3 3
Process 4-1: v = constant heat rejection.
(
)
Qout = m(u 4 − u1 ) = 7.065 × 10 -4 kg (693.23 − 206.91)kJ/kg = 0.344 kJ W net = Qin − Qout = 0.715 − 0.344 = 0.371 kJ
η th = (c)
n& = 2
W net 0.371 kJ = = 51.9% 0.715 kJ Qin
60 s W& net 60 kJ/s = 4852 rpm = (2 rev/cycle) 4 × (0.371 kJ/cycle) 1 min n cyl W net,cyl
Note that for four-stroke cycles, there are two revolutions per cycle.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-102
9-140 EES Problem 9-139 is reconsidered. The effect of the compression ratio net work done and the efficiency of the cycle is to be investigated. Also, the T-s and P-v diagrams for the cycle are to be plotted. Analysis Using EES, the problem is solved as follows: "Input Data" T[1]=(17+273) [K] P[1]=98 [kPa] T[3]=1800 [K] V_cyl=0.6 [L]*Convert(L, m^3) r_v=8 "Compression ratio" W_dot_net = 60 [kW] N_cyl=4 "number of cyclinders" v[1]/v[2]=r_v "The first part of the solution is done per unit mass." "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] s[2]=entropy(air, T=T[2], v=v[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] "Conservation of energy for process 1 to 2: no heat transfer (s=const.) with work input" w_in = DELTAu_12 DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" s[3]=entropy(air, T=T[3], P=P[3]) {P[3]*v[3]/T[3]=P[2]*v[2]/T[2]} P[3]*v[3]=R*T[3] v[3]=v[2] "Conservation of energy for process 2 to 3: the work is zero for v=const, heat is added" q_in = DELTAu_23 DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=entropy(air,T=T[4],P=P[4]) s[4]=s[3] P[4]*v[4]/T[4]=P[3]*v[3]/T[3] {P[4]*v[4]=R*T[4]} "Conservation of energy for process 3 to 4: no heat transfer (s=const) with work output" - w_out = DELTAu_34 DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" v[4]=v[1] "Conservation of energy for process 2 to 3: the work is zero for v=const; heat is rejected" - q_out = DELTAu_41 DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) w_net = w_out - w_in Eta_th=w_net/q_in*Convert(, %) "Thermal efficiency, in percent" "The mass contained in each cylinder is found from the volume of the cylinder:" V_cyl=m*v[1] "The net work done per cycle is:" W_dot_net=m*w_net"kJ/cyl"*N_cyl*N_dot"mechanical cycles/min"*1"min"/60"s"*1"thermal cycle"/2"mechanical cycles"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-103
ηth [%] 42.81 46.39 49.26 51.63 53.63 55.35 56.85
wnet [kJ/kg] 467.1 492.5 509.8 521.7 529.8 535.2 538.5
rv 5 6 7 8 9 10 11
Air Otto Cycle P-v Diagram 8000 3 s = const
2 1000 ] a P k[ P
4
100
1
50 10-2
10-1
1800 K
290 K
100
101
102
v [m3/kg]
Air Otto Cycle T-s Diagram 2500 4866 kPa
2000
3 1500
] K [ T
98 kPa 1000
4 2
500
v = const
1 0 4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
s [kJ/kg-K]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-104 540 530
w net [kJ/kg]
520 510 500 490 480 470 460 5
6
7
8
r
9
10
11
v
58 56 54
η th [%]
52 50 48 46 44 42 5
6
7
8
r
9
10
11
v
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-105
9-141 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17. Analysis (a) Process 1-2: isentropic compression.
P
→ u1 = 214.07 kJ/kg T1 = 300 K
v r1 = 621.2 v r2
3
qin
v 1 1 = 2 v r1 = v r1 = (621.2) = 67.52 → T2 = 708.3 K r v1 9.2
4 2
u 2 = 518.9 kJ/kg
1
qout
708.3 K P2v 2 P1v 1 v T (98 kPa ) = 2129 kPa = → P2 = 1 2 P1 = (9.2 ) T2 T1 v 2 T1 300 K
Process 2-3: v = constant heat addition. P3v 3 P2v 2 P = → T3 = 3 T2 = 2T2 = (2 )(708.3) = 1416.6 K → u 3 = 1128.7 kJ/kg T3 T2 P2 v r3 = 8.593 q in = u 3 − u 2 = 1128.7 − 518.9 = 609.8 kJ/kg
(b) Process 3-4: isentropic expansion.
v r4 =
v4 v r = rv r3 = (9.2)(8.593) = 79.06 → u 4 = 487.75 kJ/kg v3 3
Process 4-1: v = constant heat rejection. q out = u 4 − u1 = 487.75 − 214.07 = 273.7 kJ/kg wnet = q in − q out = 609.8 − 273.7 = 336.1 kJ/kg wnet 336.1 kJ/kg = = 55.1% q in 609.8 kJ/kg
(c)
η th =
(d)
v max = v 1 = v min = v 2 = MEP =
(
)
RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (300 K ) = = 0.879 m 3 /kg P1 98 kPa
v max r
1 kPa ⋅ m 3 wnet wnet 336.1 kJ/kg = = v 1 − v 2 v 1 (1 − 1 / r ) 0.879 m 3 /kg (1 − 1/9.2 ) 1 kJ
(
)
= 429 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
v
9-106
9-142 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a) Process 1-2 is isentropic compression: v T2 = T1 1 v 2
k −1
P
= (300 K )(9.2 )0.4 = 728.8 K
3
728.8 K v T P2v 2 P1v 1 (98 kPa ) = 2190 kPa = → P2 = 1 2 P1 = (9.2 ) v 2 T1 T2 T1 300 K
qin
4 2
Process 2-3: v = constant heat addition.
1
qout
v
P3v 3 P2v 2 P = → T3 = 3 T2 = 2T2 = (2 )(728.8) = 1457.6 K T3 T2 P2 q in = u 3 − u 2 = cv (T3 − T2 ) = (0.718 kJ/kg ⋅ K )(1457.6 − 728.8)K = 523.3 kJ/kg
(b) Process 3-4: isentropic expansion. v T4 = T3 3 v 4
k −1
1 = (1457.6 K ) 9.2
0.4
= 600.0 K
Process 4-1: v = constant heat rejection. q out = u 4 − u1 = cv (T4 − T1 ) = (0.718 kJ/kg ⋅ K )(600 − 300 )K = 215.4 kJ/kg wnet = q in − q out = 523.3 − 215.4 = 307.9 kJ/kg wnet 307.9 kJ/kg = = 58.8% q in 523.3 kJ/kg
(c)
η th =
(d)
v max = v 1 = v min = v 2 = MEP =
(
)
RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (300 K ) = = 0.879 m 3 /kg 98 kPa P1
v max r
1 kPa ⋅ m 3 wnet wnet 307.9 kJ/kg = = v 1 − v 2 v 1 (1 − 1 / r ) 0.879 m 3 /kg (1 − 1/9.2 ) 1 kJ
(
)
= 393 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-107
9-143 An engine operating on the ideal diesel cycle with air as the working fluid is considered. The pressure at the beginning of the heat-rejection process, the net work per cycle, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17. Analysis (a) The compression and the cutoff ratios are r=
V1 1200 cm3 = = 16 V2 75 cm3
P
V3 150 cm3 = =2 V 2 75 cm3
rc =
2
qin
Process 1-2: isentropic compression.
3
4
T1 = 290 K → u1 = 206.91 kJ/kg
1
qout
v r1 = 676.1 v r2 =
v
v2 1 1 v r1 = v r1 = (676.1) = 42.256 → T2 = 837.3 K 16 v1 r
h2 = 863.03 kJ/kg
Process 2-3: P = constant heat addition. P3v 3 P2v 2 v = → T3 = 3 T2 = 2T2 = (2 )(837.3) = 1674.6 K T3 T2 v2 → h3 = 1848.9 kJ/kg
v r3 = 5.002 Process 3-4: isentropic expansion.
v r4 =
v4 v r 16 v r 3 = 4 v r 3 = v r 3 = (5.002) = 40.016 → T4 = 853.4 K v3 2v 2 2 2
u4 = 636.00 kJ/kg
Process 4-1: v = constant heat rejection. 853.4 K P4v 4 P1v1 T (100 kPa ) = 294.3 kPa = → P4 = 4 P1 = T4 T1 T1 290 K
(b)
m=
(
)
P1V 1 (100 kPa ) 0.0012 m 3 = = 1.442 × 10 −3 kg RT1 0.287 kPa ⋅ m 3 /kg ⋅ K (290 K )
(
( − u ) = (1.442 × 10
)
) kg )(636.00 − 206.91)kJ/kg = 0.619 kJ
Qin = m(h3 − h2 ) = 1.442 × 10 -3 kg (1848.9 − 863.08) = 1.422 kJ Qout = m(u 4
1
-3
W net = Qin − Qout = 1.422 − 0.619 = 0.803 kJ
(c)
MEP =
1 kPa ⋅ m 3 W net W net 0.803 kJ = = V1 −V 2 V1 (1 − 1 / r ) 0.0012m 3 (1 − 1/16 ) 1 kJ
(
)
= 714 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-108
9-144 An engine operating on the ideal diesel cycle with argon as the working fluid is considered. The pressure at the beginning of the heat-rejection process, the net work per cycle, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Argon is an ideal gas with constant specific heats. Properties The properties of argon at room temperature are cp = 0.5203 kJ/kg.K, cv = 0.3122 kJ/kg·K, R = 0.2081 kJ/kg·K and k = 1.667 (Table A-2). Analysis (a) The compression and the cutoff ratios are r=
V1 1200 cm3 = = 16 V2 75 cm3
rc =
V3 150 cm3 = =2 V 2 75 cm3
P
2
Qin
3
Process 1-2: isentropic compression. V T2 = T1 2 V1
k −1
4
= (290 K )(16)
0.667
= 1843 K
1
Qout
v
Process 2-3: P = constant heat addition. P3v 3 P2v 2 v = → T3 = 3 T2 = 2T2 = (2)(1843) = 3686 K T3 T2 v2
Process 3-4: isentropic expansion. V T4 = T3 3 V4
k −1
2V = T3 2 V4
k −1
2 = T3 r
k −1
2 = (3686 K ) 16
0.667
= 920.9 K
Process 4-1: v = constant heat rejection. 920.9 K P4v 4 P1v1 T (100 kPa ) = 317.6 kPa = → P4 = 4 P1 = T4 T1 T1 290 K
(b)
m=
(
)
P1V1 (100 kPa ) 0.0012 m3 = = 1.988 × 10− 3 kg RT1 0.2081 kPa ⋅ m3 /kg ⋅ K (290 K )
(
)
( − T ) = (1.988 × 10
) kg )(0.3122 kJ/kg ⋅ K )(920.9 − 290 )K = 0.392 kJ
Qin = m(h3 − h2 ) = mc p (T3 − T2 ) = 1.988 × 10 -3 kg (0.5203 kJ/kg ⋅ K )(3686 − 1843)K = 1.906 kJ Qout = m(u 4 − u1 ) = mcv (T4
1
-3
W net = Qin − Qout = 1.906 − 0.392 = 1.514 kJ
(c)
MEP =
1 kPa ⋅ m 3 W net W net 1.514 kJ = = V1 −V 2 V1 (1 − 1 / r ) 0.0012 m 3 (1 − 1/16 ) 1 kJ
(
)
= 1346 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-109
9-145E An ideal dual cycle with air as the working fluid with a compression ratio of 12 is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm.R, cv = 0.171 Btu/lbm.R, and k = 1.4 (Table A-2E). Analysis The mass of air is m=
(
)
P1V1 (14.7 psia ) 75/1728 ft 3 = = 3.132 × 10− 3 lbm 3 RT1 ( ) 0.3704 psia ⋅ ft /lbm ⋅ R 550 R
(
)
P
Process 1-2: isentropic compression. V T2 = T1 1 V2
k −1
x 2
1.1 Btu
3
0.3 Btu
= (550 R )(12)0.4 = 1486 R
4 1
Qout
v
Process 2-x: v = constant heat addition, Q 2 − x ,in = m(u x − u 2 ) = mcv (T x − T2 )
(
)
0.3 Btu = 3.132 × 10 −3 lbm (0.171 Btu/lbm ⋅ R )(Tx − 1486 )R → T x = 2046 R
Process x-3: P = constant heat addition. Q x −3,in = m(h3 − h x ) = mc p (T3 − T x )
(
)
→ T3 = 3509 R 1.1 Btu = 3.132 × 10 −3 lbm (0.240 Btu/lbm ⋅ R )(T3 − 2046 )R P3V 3 PxV x V T 3509 R = → rc = 3 = 3 = = 1.715 T3 Tx V x T x 2046 R
Process 3-4: isentropic expansion. V T4 = T3 3 V 4
k −1
1.715V 1 = T3 V4
k −1
1.715 = T3 r
k −1
1.715 = (3509 R ) 12
0.4
= 1611 R
Process 4-1: v = constant heat rejection. Qout = m(u 4 − u1 ) = mcv (T4 − T1 )
(
)
= 3.132 × 10 −3 lbm (0.171 Btu/lbm ⋅ R )(1611 − 550 )R = 0.568 Btu
η th = 1 −
Qout 0.568 Btu = 1− = 59.4% Qin 1.4 Btu
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-110
9-146 An ideal Stirling cycle with air as the working fluid is considered. The maximum pressure in the cycle and the net work output are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a) The entropy change during process 1-2 is s 2 − s1 =
T
qin = 900 kJ/kg
350 K
q12 900 kJ/kg = = 0.5 kJ/kg ⋅ K TH 1800 K
2
1
1800 K
4
qout
3
s
and s 2 − s1 = cv ln
T2 T1
©0
+ Rln
v2 v v → 0.5 kJ/kg ⋅ K = (0.287 kJ/kg ⋅ K ) ln 2 → 2 = 5.710 v1 v1 v1
1800 K v T P3v 3 P1v 1 v T = 5873 kPa = → P1 = P3 3 1 = P3 2 1 = (200 kPa )(5.710 ) T3 T1 v 1 T3 v 1 T3 350 K
(b)
T wnet = η th q in = 1 − L TH
350 K q in = 1 − (900 kJ/kg ) = 725 kJ/kg 1800 K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-111
9-147 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis The properties at various states are T1 = 300 K
→
T
h1 = 300.19 kJ / kg Pr 1 = 1.386
T3 = 1300 K →
h3 = 1395.97 kJ / kg Pr 3 = 330.9
3′ 2′
3
qin
2 4 1
For rp = 6,
qout
P Pr2 = 2 Pr1 = (6 )(1.386 ) = 8.316 → h2 = 501.40 kJ/kg P1 Pr4 =
P4 1 Pr3 = (330.9) = 55.15 → h4 = 855.3 kJ/kg P3 6
q in = h3 − h2 = 1395.97 − 501.40 = 894.57 kJ/kg q out = h4 − h1 = 855.3 − 300.19 = 555.11 kJ/kg wnet = q in − q out = 894.57 − 555.11 = 339.46 kJ/kg
η th =
wnet 339.46 kJ/kg = = 37.9% q in 894.57 kJ/kg
For rp = 12, Pr2 =
P2 Pr = (12 )(1.386 ) = 16.63 → h2 = 610.6 kJ/kg P1 1
Pr4 =
P4 1 Pr3 = (330.9 ) = 27.58 → h4 = 704.6 kJ/kg P3 12
q in = h3 − h2 = 1395.97 − 610.60 = 785.37 kJ/kg q out = h4 − h1 = 704.6 − 300.19 = 404.41 kJ/kg wnet = q in − q out = 785.37 − 404.41 = 380.96 kJ/kg
η th =
wnet 380.96 kJ/kg = = 48.5% q in 785.37 kJ/kg
Thus, (a) (b)
∆wnet = 380.96 − 339.46 = 41.5 kJ/kg (increase) ∆η th = 48.5% − 37.9% = 10.6%
(increase)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-112
9-148 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis Processes 1-2 and 3-4 are isentropic. Therefore, For rp = 6, P T2 = T1 2 P1
P T4 = T3 4 P3
(k −1) / k
= (300 K )(6)0.4/1.4 = 500.6 K
T 3′
(k −1) / k
1 = (1300 K ) 6
2′
0.4/1.4
= 779.1 K
3
qin
2
q in = h3 − h2 = c p (T3 − T2 )
= (1.005 kJ/kg ⋅ K )(1300 − 500.6 )K = 803.4 kJ/kg
4 1
qout
q out = h4 − h1 = c p (T4 − T1 )
= (1.005 kJ/kg ⋅ K )(779.1 − 300)K = 481.5 kJ/kg
wnet = q in − q out = 803.4 − 481.5 = 321.9 kJ/kg
η th =
wnet 321.9 kJ/kg = = 40.1% q in 803.4 kJ/kg
For rp = 12, P T2 = T1 2 P1
P T4 = T3 4 P3
(k −1) / k
(k −1) / k
= (300 K )(12 )0.4/1.4 = 610.2 K 1 = (1300 K ) 12
0.4/1.4
= 639.2 K
q in = h3 − h2 = c p (T3 − T2 )
= (1.005 kJ/kg ⋅ K )(1300 − 610.2)K = 693.2 kJ/kg
q out = h4 − h1 = c p (T4 − T1 )
= (1.005 kJ/kg ⋅ K )(639.2 − 300)K = 340.9 kJ/kg
wnet = q in − q out = 693.2 − 340.9 = 352.3 kJ/kg
η th =
wnet 352.3 kJ/kg = = 50.8% q in 693.2 kJ/kg
Thus, (a)
∆wnet = 352.3 − 321.9 = 30.4 kJ/kg (increase)
(b)
∆η th = 50.8% − 40.1% = 10.7%
(increase)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-113
9-149 A regenerative Brayton cycle with helium as the working fluid is considered. The thermal efficiency and the required mass flow rate of helium are to be determined for 100 percent and 80 percent isentropic efficiencies for both the compressor and the turbine. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with constant specific heats. Properties The properties of helium are cp = 5.1926 kJ/kg.K T and k = 1.667 (Table A-2). 3 Analysis (a) Assuming ηT = ηC = 100%, qin 1800 K P T2 s = T1 2 P1 P T4 s = T3 4 P3
ε=
(k −1) / k
(k −1) / k
5
= (300 K )(8)0.667/1.667 = 689.4 K 2s
1 = (1800 K ) 8
0.667/1.667
= 783.3 K
300 K
2
4s
4
6
1
h5 − h2 c p (T5 − T2 ) = → T5 = T2 + ε (T4 − T2 ) = 689.4 + (0.75)(783.3 − 689.4 ) = 759.8 K h4 − h2 c p (T4 − T2 )
wnet = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 ) = c p [(T3 − T4 ) − (T2 − T1 )]
= (5.1926 kJ/kg ⋅ K )[(1800 − 783.3) − (689.4 − 300)]K = 3257.3 kJ/kg W& 60,000 kJ/s = 18.42 kg/s m& = net = wnet 3257.3 kJ/kg
q in = h3 − h5 = c p (T3 − T5 ) = (5.1926 kJ/kg ⋅ K )(1800 − 759.8)K = 5401.3 kJ/kg
η th =
wnet 3257.3 kJ/kg = = 60.3% q in 5401.3 kJ/kg
(b) Assuming ηT = ηC = 80%, P T2 s = T1 2 P1
ηC =
ε=
(k −1) / k
= (300 K )(8)0.667/1.667 = 689.4 K
h2 s − h1 c p (T2 s − T1 ) = → T2 = T1 + (T2 s − T1 ) / η C = 300 + (689.4 − 300) / (0.80) = 786.8 K h2 − h1 c p (T2 − T1 )
P T4 s = T3 4 P3
ηT =
(k −1) / k
1 = (1800 K ) 8
0.667/1.667
= 783.3 K
c p (T3 − T4 ) h3 − h4 = → T4 = T3 − η T (T3 − T4 s ) = 1800 − (0.80)(1800 − 783.3) = 986.6 K h3 − h4 s c p (T3 − T4 s ) h5 − h2 c p (T5 − T2 ) → T5 = T2 + ε (T4 − T2 ) = 786.8 + (0.75)(986.6 − 786.8) = 936.7 K = h4 − h2 c p (T4 − T2 )
wnet = wT,out − wC,in = (h3 − h4 ) − (h2 − h1 ) = c p [(T3 − T4 ) − (T2 − T1 )]
= (5.1926 kJ/kg ⋅ K )[(1800 − 986.6 ) − (786.8 − 300 )]K = 1695.9 kJ/kg
m& =
W&net 60,000 kJ/s = = 35.4 kg/s wnet 1695.9 kJ/kg
qin = h3 − h5 = c p (T3 − T5 ) = (5.1926 kJ/kg ⋅ K )(1800 − 936.7 )K = 4482.8 kJ/kg
η th =
wnet 1695.9 kJ/kg = = 37.8% qin 4482.8 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-114
9-150 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. P T4 s = T2 s = T1 2 P1
(k −1) / k
= (300 K )(3.5)0.4/1.4 = 429.1 K
T
c p (T2 s − T1 ) h −h → T4 = T2 = T1 + (T2 s − T1 ) / η C η C = 2s 1 = h2 − h1 c p (T2 − T1 ) = 300 + (429.1 − 300) / (0.78) = 465.5 K T9 s = T7 s
ηT =
P = T6 7 P6
(k −1) / k
1 = (1200 K ) 3.5
6 5
4
0.4/1.4
= 838.9 K
3
4
2s
8
7 9 7s 9
2 1
c p (T6 − T7 ) h6 − h7 = → T9 = T7 = T6 − η T (T6 − T7 s ) h6 − h7 s c p (T6 − T7 s ) = 1200 − (0.86 )(1200 − 838.9) = 889.5 K
ε=
h5 − h4 c p (T5 − T4 ) = → T5 = T4 + ε (T9 − T4 ) h9 − h4 c p (T9 − T4 ) = 465.5 + (0.72 )(889.5 − 465.5) = 770.8 K
wC,in = 2(h2 − h1 ) = 2c p (T2 − T1 ) = 2(1.005 kJ/kg ⋅ K )(465.5 − 300 )K = 332.7 kJ/kg wT,out = 2(h6 − h7 ) = 2c p (T6 − T7 ) = 2(1.005 kJ/kg ⋅ K )(1200 − 889.5)K = 624.1 kJ/kg
Thus,
rbw =
wC,in wT,out
=
332.7 kJ/kg = 53.3% 624.1 kJ/kg
q in = (h6 − h5 ) + (h8 − h7 ) = c p [(T6 − T5 ) + (T8 − T7 )]
= (1.005 kJ/kg ⋅ K )[(1200 − 770.8) + (1200 − 889.5)]K = 743.4 kJ/kg
wnet = wT,out − wC,in = 624.1 − 332.7 = 291.4 kJ/kg
η th =
wnet 291.4 kJ/kg = = 39.2% 743.4 kJ/kg q in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-115
9-151 EES Problem 9-150 is reconsidered. The effect of the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle is to be investigated. Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows: "Input data" T[6] = 1200 [K] T[8] = T[6] Pratio = 3.5 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = T[1] Eta_reg = 0.72 "Regenerator effectiveness" Eta_c =0.78 "Compressor isentorpic efficiency" Eta_t =0.86 "Turbien isentropic efficiency" "LP Compressor:" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = w_compisen_LP/w_comp_LP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the LP compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" h[1] + w_compisen_LP = h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" h[1] + w_comp_LP = h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "HP Compressor:" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the HP compressor" P[4] = Pratio*P[3] P[3] = P[2] s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4]) "T_s[4] is the isentropic value of T[4] at compressor exit" Eta_c = w_compisen_HP/w_comp_HP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" h[3] + w_compisen_HP = h_s[4] h[3]=ENTHALPY(Air,T=T[3]) h_s[4]=ENTHALPY(Air,T=T_s[4]) "Actual compressor analysis:" h[3] + w_comp_HP = h[4]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-116 h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Intercooling heat loss:" h[2] = q_out_intercool + h[3] "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" h[4] + q_in_noreg = h[6] h[6]=ENTHALPY(Air,T=T[6]) P[6]=P[4]"process 4-6 is SSSF constant pressure" "HP Turbine analysis" s[6]=ENTROPY(Air,T=T[6],P=P[6]) s_s[7]=s[6] "For the ideal case the entropies are constant across the turbine" P[7] = P[6] /Pratio s_s[7]=ENTROPY(Air,T=T_s[7],P=P[7])"T_s[7] is the isentropic value of T[7] at HP turbine exit" Eta_t = w_turb_HP /w_turbisen_HP "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" h[6] = w_turbisen_HP + h_s[7] h_s[7]=ENTHALPY(Air,T=T_s[7]) "Actual Turbine analysis:" h[6] = w_turb_HP + h[7] h[7]=ENTHALPY(Air,T=T[7]) s[7]=ENTROPY(Air,T=T[7], P=P[7]) "Reheat Q_in:" h[7] + q_in_reheat = h[8] h[8]=ENTHALPY(Air,T=T[8]) "HL Turbine analysis" P[8]=P[7] s[8]=ENTROPY(Air,T=T[8],P=P[8]) s_s[9]=s[8] "For the ideal case the entropies are constant across the turbine" P[9] = P[8] /Pratio s_s[9]=ENTROPY(Air,T=T_s[9],P=P[9])"T_s[9] is the isentropic value of T[9] at LP turbine exit" Eta_t = w_turb_LP /w_turbisen_LP "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" h[8] = w_turbisen_LP + h_s[9] h_s[9]=ENTHALPY(Air,T=T_s[9]) "Actual Turbine analysis:" h[8] = w_turb_LP + h[9] h[9]=ENTHALPY(Air,T=T[9]) s[9]=ENTROPY(Air,T=T[9], P=P[9]) "Cycle analysis" w_net=w_turb_HP+w_turb_LP - w_comp_HP - w_comp_LP q_in_total_noreg=q_in_noreg+q_in_reheat Eta_th_noreg=w_net/(q_in_total_noreg)*Convert(, %) "[%]" "Cycle thermal efficiency" PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-117 Bwr=(w_comp_HP + w_comp_LP)/(w_turb_HP+w_turb_LP)"Back work ratio" "With the regenerator, the heat added in the external heat exchanger is" h[5] + q_in_withreg = h[6] h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[4] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[4])/(h[9]-h[4]) "Energy balance on regenerator gives h[10] and thus T[10] as:" h[4] + h[9]=h[5] + h[10] h[10]=ENTHALPY(Air, T=T[10]) s[10]=ENTROPY(Air,T=T[10], P=P[10]) P[10]=P[9] "Cycle thermal efficiency with regenerator" q_in_total_withreg=q_in_withreg+q_in_reheat Eta_th_withreg=w_net/(q_in_total_withreg)*Convert(, %) "[%]" "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6] s_s[8]=s[8] T_s[8]=T[8] s_s[10]=s[10] T_s[10]=T[10] ηC
ηreg
ηt
0.78 0.78 0.78 0.78 0.78 0.78 0.78 0.78 0.78
0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
0.86 0.86 0.86 0.86 0.86 0.86 0.86 0.86 0.86
ηth,noreg [%] 27.03 27.03 27.03 27.03 27.03 27.03 27.03 27.03 27.03
ηth,withreg [%] 36.59 37.7 38.88 40.14 41.48 42.92 44.45 46.11 47.88
qin,total,noreg [kJ/kg] 1130 1130 1130 1130 1130 1130 1130 1130 1130
qin,total,withreg [kJ/kg] 834.6 810 785.4 760.8 736.2 711.6 687 662.4 637.8
wnet [kJ/kg] 305.4 305.4 305.4 305.4 305.4 305.4 305.4 305.4 305.4
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-118
Air
1600
a kP 5 2 12
1400 6
1200
] K [ T
a kP 0 35
8
a kP 0 10
1000 5
800 600
4
2
10
400
1
3
200 4.5
5.0
9
7
5.5
6.0
6.5
7.0
7.5
s [kJ/kg-K] 50
η
45
η
40 35
= 0.72
reg c
= 0.78
W ith regeneration
η th
30 25 20
No regeneration
15 10 0.6
0.65
0.7
0.75
0.8
η
0.85
0.9
0.95
1
t
1200 1100
q in,total
1000 900
η
reg
η
No regeneration
c
= 0.72 = 0.78
W ith regeneration
800 700 600 0.6
0.65
0.7
0.75
0.8
η
0.85
0.9
0.95
1
t
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-119 450
η
400
reg
w net [kJ/kg]
350
η
c
= 0.72 = 0.78
300 250 200 150 100 0.6
0.65
0.7
0.75
0.8
η
0.85
0.9
0.95
1
t
50
η 45
c
η
= 0.78 = 0.86
t
η th
40
W ith regenertion
35
No regeneration 30
25 0.6
0.65
0.7
0.75
0.8
η
0.85
0.9
0.95
1
reg
50 45 40
η η
reg t
= 0.72
= 0.86 W ith regeneration
η th
35 30 25
No regeneration
20 15 0.6
0.65
0.7
0.75
0.8
η
0.85
0.9
0.95
1
c
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-120
9-152 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with constant specific heats. Properties The properties of helium at room temperature are cp = 5.1926 kJ/kg.K and k = 1.667 (Table A-2). Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. P T4 s = T2 s = T1 2 P1
(k −1) / k
= (300 K )(3.5)0.667/1.667 = 495.2 K
T
c p (T2 s − T1 ) h −h → T4 = T2 = T1 + (T2 s − T1 ) / η C η C = 2s 1 = h2 − h1 c p (T2 − T1 ) = 300 + (495.2 − 300) / (0.78) = 550.3 K T9 s = T7 s
ηT =
ε=
P = T6 7 P6
(k −1) / k
1 = (1200 K ) 3.5
qin 5
4
0.667/1.667
= 726.9 K
3
4
2s
6
7 9 7s 9
2 1
c p (T6 − T7 ) h6 − h7 = → T9 = T7 = T6 − η T (T6 − T7 s ) h6 − h7 s c p (T6 − T7 s ) = 1200 − (0.86 )(1200 − 726.9) = 793.1 K h5 − h4 c p (T5 − T4 ) = → T5 = T4 + ε (T9 − T4 ) h9 − h4 c p (T9 − T4 ) = 550.3 + (0.72 )(793.1 − 550.3) = 725.1 K
wC,in = 2(h2 − h1 ) = 2c p (T2 − T1 ) = 2(5.1926 kJ/kg ⋅ K )(550.3 − 300 )K = 2599.4 kJ/kg wT,out = 2(h6 − h7 ) = 2c p (T6 − T7 ) = 2(5.1926 kJ/kg ⋅ K )(1200 − 793.1)K = 4225.7 kJ/kg
Thus,
rbw =
wC,in wT,out
=
2599.4 kJ/kg = 61.5% 4225.7 kJ/kg
q in = (h6 − h5 ) + (h8 − h7 ) = c p [(T6 − T5 ) + (T8 − T7 )]
= (5.1926 kJ/kg ⋅ K )[(1200 − 725.1) + (1200 − 793.1)]K = 4578.8 kJ/kg
wnet = wT,out − wC,in = 4225.7 − 2599.4 = 1626.3 kJ/kg
η th =
8
wnet 1626.3 kJ/kg = = 35.5% 4578.8 kJ/kg q in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-121
9-153 An ideal regenerative Brayton cycle is considered. The pressure ratio that maximizes the thermal efficiency of the cycle is to be determined, and to be compared with the pressure ratio that maximizes the cycle net work. Analysis Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as (k −1) / k
P T2 = T1 2 P1
P T4 = T3 4 P3
(k −1) / k
( )
= T1 r p (k −1) / k 1 = T3 rp
(k −1) / k
T
= T3 r p (1− k ) / k 5
Then, q in
qin
( T (r (
) − 1)
= h3 − h5 = c p (T3 − T5 ) = c p (T3 − T4 ) = c p T3 1 − r p(1− k ) / k
q out = h6 − h1 = c p (T6 − T1 ) = c p (T2 − T1 ) = c p
(
1
k −1) / k p
wnet = q in − q out = c p T3 − T3 r p(1− k ) / k − T1 r p(k −1) / k + T1
)
2
3 4
6
1
s
To maximize the net work, we must have
∂ wnet k −1 1− k = c p − T3 r p (1− k ) / k − 1 − T1 r p (k −1) / k − 1 = 0 ∂ rp k k Solving for rp gives T rp = 1 T3
Similarly,
η th
k / 2 (1− k )
( (
) )
c p T1 r p (k −1) / k − 1 q out = 1− = 1− q in c p T3 1 − r p (1− k ) / k
which simplifies to
η th = 1 −
T1 (k −1) / k rp T3
When rp = 1, the thermal efficiency becomes ηth = 1 - T1/T3, which is the Carnot efficiency. Therefore, the efficiency is a maximum when rp = 1, and must decrease as rp increases for the fixed values of T1 and T3. Note that the compression ratio cannot be less than 1, and the factor rp (k −1) / k
is always greater than 1 for rp > 1. Also note that the net work wnet = 0 for rp = 1. This being the case, the pressure ratio for maximum thermal efficiency, which is rp = 1, is always less than the pressure ratio for maximum work.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-122
9-154 An ideal gas-turbine cycle with one stage of compression and two stages of expansion and regeneration is considered. The thermal efficiency of the cycle as a function of the compressor pressure ratio and the high-pressure turbine to compressor inlet temperature ratio is to be determined, and to be compared with the efficiency of the standard regenerative cycle. Analysis The T-s diagram of the cycle is as shown in the figure. If the overall pressure ratio of the cycle is rp, which is the pressure ratio across the compressor, then the pressure ratio across each turbine stage in the ideal case becomes √ rp. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as P T5 = T2 = T1 2 P1
(k −1) / k
P T7 = T4 = T3 4 P3 P T6 = T5 6 P5
( )
(k −1) / k
= T1 rp (k −1) / k
(k −1) / k
1 = T3 r p
1 = T5 r p
Then,
T qin
3
7 4 2
5
1
qout
6
(k −1) / k
= T3rp
(1− k ) / 2 k
s
(k −1) / k
= T2rp (1− k ) / 2 k = T1rp (k −1) / k rp (1− k ) / 2 k = T1rp (k −1) / 2 k
( − T ) = c T (r (
) − 1)
q in = h3 − h7 = c p (T3 − T7 ) = c p T3 1 − r p (1− k ) / 2 k q out = h6 − h1 = c p (T6
and thus
η th
1
( (
p 1
p
k −1) / 2 k
) )
c p T1 r p (k −1) / 2 k − 1 q out = 1− = 1− q in c p T3 1 − r p (1− k ) / 2 k
which simplifies to
η th = 1 −
T1 (k −1) / 2 k rp T3
The thermal efficiency of the single stage ideal regenerative cycle is given as
η th = 1 −
T1 (k −1) / k rp T3
Therefore, the regenerative cycle with two stages of expansion has a higher thermal efficiency than the standard regenerative cycle with a single stage of expansion for any given value of the pressure ratio rp.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-123
9-155 A spark-ignition engine operates on an ideal Otto cycle with a compression ratio of 11. The maximum temperature and pressure in the cycle, the net work per cycle and per cylinder, the thermal efficiency, the mean effective pressure, and the power output for a specified engine speed are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis (a) First the mass in one cylinder is determined as follows r=
V c +V d V + (0.0018) / 4 → 11 = c →V c = 0.000045 m 3 for one cylinder Vc Vc
V1 = V c +V d = 0.000045 + 0.00045 = 0.000495 m 3 m=
P1V1 (90 kPa)(0.000495 m3 ) = = 0.0004805 kg RT1 0.287 kPa ⋅ m3 /kg ⋅ K (323 K )
(
)
For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure. Process 1-2: Isentropic compression → u1 = 230.88 kJ/kg T1 = 50°C = 323 K T1 = 50°C = 323 K s1 = 5.8100 kJ/kg ⋅ K P1 = 90 kPa
v1 =
V1
=
m
Qin
3
0.000495 m = 1.0302 m 3 /kg 0.0004805 kg
V 2 = V c = 0.000045 m 3 v2 =
V2 m
3 P
=
2
0.000045 m 3 = 0.09364 m 3 /kg 0.0004805 kg
4 Qout 1
V
s 2 = s1 = 5.8100 kJ/kg.K T2 = 807.3 K v 2 = 0.09364 m 3 / kg T2 = 807.3 K → u 2 = 598.33 kJ/kg P2 = P1
0.000495 m 3 V 1 T2 = (90 kPa) 0.000045 m 3 V 2 T1
807.3 K = 2474 kPa 323 K
Process 2-3: constant volume heat addition Qin = m(u 3 − u 2 ) → 1.5 kJ = (0.0004805 kg)(u 3 − 598.33)kJ/kg → u 3 = 3719.8 kJ/kg u 2 = 598.33 kJ/kg → T3 = 4037 K T P3 = P2 3 T2
4037 K = (2474 kPa) = 12,375 kPa 807.3 K
T3 = 4037 K s 3 = 7.3218 kJ/kg ⋅ K P3 = 12,375 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-124
(b) Process 3-4: isentropic expansion. s 4 = s 3 = 7.3218 kJ/kg.K T = 2028 K v 4 = v 1 = 1.0302 m 3 / kg 4 T4 = 2028 K → u 4 = 1703.6 kJ/kg P4 = P3
V 3 T4 1 2028 K = (12,375 kPa) = 565 kPa V 4 T3 11 4037 K
Process 4-1: constant voume heat rejection Qout = m(u 4 − u1 ) = (0.0004805 kg)(1703.6 − 230.88)kJ/kg = 0.7077 kJ
The net work output and the thermal efficiency are Wnet,out = Qin − Qout = 1.5 − 0.7077 = 0.792 kJ (per cycle per cylinder)
η th =
Wnet,out Qin
=
0.792 kJ = 0.528 1.5 kJ
(c) The mean effective pressure is determined to be MEP =
Wnet,out
V 1 −V 2
=
kPa ⋅ m 3 (0.000495 − 0.000045)m 3 kJ 0.7923 kJ
= 1761 kPa
(d) The power for engine speed of 3000 rpm is (3000 rev/min) 1 min n& W& net = n cylWnet = (4 cylinder)(0.792 kJ/cylinder - cycle) = 79.2 kW 2 (2 rev/cycle) 60 s
Note that there are two revolutions in one cycle in four-stroke engines.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-125
9-156 A gas-turbine plant operates on the regenerative Brayton cycle with reheating and intercooling. The back work ratio, the net work output, the thermal efficiency, the second-law efficiency, and the exergies at the exits of the combustion chamber and the regenerator are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K. Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure. Optimum intercooling and reheating pressure: P2 = P1 P4 = (100)(1200) = 346.4 kPa Process 1-2, 3-4: Compression T1 = 300 K → h1 = 300.43 kJ/kg
T
T1 = 300 K s1 = 5.7054 kJ/kg ⋅ K P1 = 100 kPa P2 = 346.4 kPa h2 s = 428.79 kJ/kg s 2 = s1 = 5.7054 kJ/kg.K
ηC =
h2 s − h1 428.79 − 300.43 → 0.80 = → h2 = 460.88 kJ/kg h2 − h1 h2 − 300.43
5
qin
6 8 6s 8
9
4 3
4
2s
2
10
1
→ h3 = 350.78 kJ/kg T3 = 350 K T3 = 350 K s 3 = 5.5040 kJ/kg ⋅ K P3 = 346.4 kPa P4 = 1200 kPa h4 s = 500.42 kJ/kg s 4 = s 3 = 5.5040 kJ/kg.K
ηC =
h 4 s − h3 500.42 − 350.78 → 0.80 = → h4 = 537.83 kJ/kg h4 − h3 h4 − 350.78
Process 6-7, 8-9: Expansion → h6 = 1514.9 kJ/kg T6 = 1400 K T6 = 1400 K s 6 = 6.6514 kJ/kg ⋅ K P6 = 1200 kPa P7 = 346.4 kPa h7 s = 1083.9 kJ/kg s 7 = s 6 = 6.6514 kJ/kg.K
ηT =
7
1514.9 − h7 h6 − h7 → h7 = 1170.1 kJ/kg → 0.80 = 1514.9 − 1083.9 h6 − h7 s
→ h8 = 1395.6 kJ/kg T8 = 1300 K T8 = 1300 K s 8 = 6.9196 kJ/kg ⋅ K P8 = 346.4 kPa P9 = 100 kPa h9 s = 996.00 kJ/kg s 9 = s8 = 6.9196 kJ/kg.K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-126
ηT =
h8 − h9 1395.6 − h9 → 0.80 = → h9 = 1075.9 kJ/kg h8 − h9 s 1395.6 − 996.00
Cycle analysis: wC,in = h2 − h1 + h4 − h3 = 460.88 − 300.43 + 537.83 − 350.78 = 347.50 kJ/kg wT,out = h6 − h7 + h8 − h9 = 1514.9 − 1170.1 + 1395.6 − 1075.9 = 664.50 kJ/kg rbw =
wC,in wT,out
=
347.50 = 0.523 664.50
wnet = wT,out − wC,in = 664.50 − 347.50 = 317.0 kJ/kg
Regenerator analysis:
ε regen =
h9 − h10 1075.9 − h10 → 0.75 = → h10 = 672.36 kJ/kg h9 − h4 1075.9 − 537.83
h10 = 672.36 K s10 = 6.5157 kJ/kg ⋅ K P10 = 100 kPa q regen = h9 − h10 = h5 − h4 → 1075.9 − 672.36 = h5 − 537.83 → h5 = 941.40 kJ/kg
(b)
q in = h6 − h5 = 1514.9 − 941.40 = 573.54 kJ/kg
η th =
wnet 317.0 = = 0.553 q in 573.54
(c) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,
η max = 1 − and
η II =
T1 300 K =1− = 0.786 T6 1400 K
η th 0.553 = = 0.704 η max 0.786
(d) The exergies at the combustion chamber exit and the regenerator exit are x 6 = h6 − h0 − T0 ( s 6 − s 0 ) = (1514.9 − 300.43)kJ/kg − (300 K )(6.6514 − 5.7054)kJ/kg.K = 930.7 kJ/kg x10 = h10 − h0 − T0 ( s10 − s 0 ) = (672.36 − 300.43)kJ/kg − (300 K )(6.5157 − 5.7054)kJ/kg.K = 128.8 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-127
9-157 The electricity and the process heat requirements of a manufacturing facility are to be met by a cogeneration plant consisting of a gas-turbine and a heat exchanger for steam production. The mass flow rate of the air in the cycle, the back work ratio, the thermal efficiency, the rate at which steam is produced in the heat exchanger, and the utilization efficiency of the cogeneration plant are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) For this problem, we use the properties of air from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure. Process 1-2: Compression
350°C
P2 = 1200 kPa h2 s = 617.37 kJ/kg s 2 = s1 = 5.7159 kJ/kg.K
1.2 MPa
ηC =
Compress.
1
Heat exchanger
3
T1 = 30°C → h1 = 303.60 kJ/kg T1 = 30°C s1 = 5.7159 kJ/kg ⋅ K P1 = 100 kPa
5
Combustion chamber
2
25°C
4 Turbine
500°C Sat. vap. 200°C
100 kPa 30°C
h2 s − h1 617.37 − 303.60 → 0.82 = → h2 = 686.24 kJ/kg h2 − h1 h2 − 303.60
Process 3-4: Expansion T4 = 500°C → h4 = 792.62 kJ/kg
ηT =
h3 − h4 h − 792.62 → 0.82 = 3 h3 − h4 s h3 − h4 s
We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together with the isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The solution by hand would require a trial-error approach. h_3=enthalpy(Air, T=T_3) s_3=entropy(Air, T=T_3, P=P_2) h_4s=enthalpy(Air, P=P_1, s=s_3)
Also,
T5 = 350°C → h5 = 631.44 kJ/kg
The inlet water is compressed liquid at 25ºC and at the saturation pressure of steam at 200ºC (1555 kPa). This is not available in the tables but we can obtain it in EES. The alternative is to use saturated liquid enthalpy at the given temperature. Tw1 = 25°C hw1 = 106.27 kJ/kg P1 = 1555 kPa Tw2 = 200°C hw2 = 2792.0 kJ/kg x2 = 1
The net work output is wC,in = h2 − h1 = 686.24 − 303.60 = 382.64 kJ/kg wT,out = h3 − h4 = 1404.7 − 792.62 = 612.03 kJ/kg wnet = wT,out − wC,in = 612.03 − 382.64 = 229.39 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-128
The mass flow rate of air is W& 800 kJ/s m& a = net = = 3.487 kg/s wnet 229.39 kJ/kg (b) The back work ratio is rbw =
wC,in wT,out
=
382.64 = 0.625 612.03
The rate of heat input and the thermal efficiency are Q& in = m& a (h3 − h2 ) = (3.487 kg/s)(1404.7 − 686.24)kJ/kg = 2505 kW
η th =
W& net 800 kW = = 0.319 & 2505 kW Qin
(c) An energy balance on the heat exchanger gives m& a (h4 − h5 ) = m& w (hw2 − hw1 )
(3.487 kg/s)(792.62 − 631.44)kJ/kg = m& w (2792.0 − 106.27)kJ/kg → m& w = 0.2093 kg/s
(d) The heat supplied to the water in the heat exchanger (process heat) and the utilization efficiency are Q& p = m& w (hw 2 − hw1 ) = (0.2093 kg/s)(2792.0 − 106.27)kJ/kg = 562.1 kW
εu =
W& net + Q& p 800 + 562.1 = = 0.544 2505 kW Q& in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-129
9-158 A turbojet aircraft flying is considered. The pressure of the gases at the turbine exit, the mass flow rate of the air through the compressor, the velocity of the gases at the nozzle exit, the propulsive power, and the propulsive efficiency of the cycle are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). T 4 Analysis (a) For this problem, we use the properties from EES software. qin Remember that for an ideal gas, enthalpy is a function of temperature 5 only whereas entropy is functions of both temperature and pressure. 3 Diffuser, Process 1-2: T1 = −35°C → h1 = 238.23 kJ/kg
2 1
6
qout
V2 V2 h1 + 1 = h2 + 2 2 2 2 (900/3.6 m/s) 1 kJ/kg (15 m/s) 2 1 kJ/kg (238.23 kJ/kg) + → h2 = 269.37 kJ/kg = h2 + 2 2 2 2 2 2 1000 m /s 1000 m /s h2 = 269.37 kJ/kg s 2 = 5.7951 kJ/kg ⋅ K P2 = 50 kPa
Compressor, Process 2-3: P3 = 450 kPa
h3s = 505.19 kJ/kg s 3 = s 2 = 5.7951 kJ/kg.K
ηC =
h3s − h2 505.19 − 269.37 → 0.83 = → h3 = 553.50 kJ/kg h3 − h2 h3 − 269.37
Turbine, Process 3-4: T4 = 950°C → h4 = 1304.8 kJ/kg h3 − h2 = h4 − h5 → 553.50 − 269.37 = 1304.8 − h5 → h5 = 1020.6 kJ/kg
where the mass flow rates through the compressor and the turbine are assumed equal.
ηT =
h4 − h5 1304.8 − 1020.6 → 0.83 = → h5 s = 962.45 kJ/kg h4 − h5 s 1304.8 − h5 s
T4 = 950°C s 4 = 6.7725 kJ/kg ⋅ K P4 = 450 kPa h5 s = 962.45 kJ/kg
P5 = 147.4 kPa s 5 = s 4 = 6.7725 kJ/kg ⋅ K
(b) The mass flow rate of the air through the compressor is m& =
W& C 500 kJ/s = = 1.760 kg/s h3 − h2 (553.50 − 269.37) kJ/kg
(c) Nozzle, Process 5-6: h5 = 1020.6 kJ/kg s 5 = 6.8336 kJ/kg ⋅ K P5 = 147.4 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
9-130
P6 = 40 kPa
h6 s = 709.66 kJ/kg s 6 = s 5 = 6.8336 kJ/kg.K
ηN =
h5 − h6 1020.6 − h6 → 0.83 = → h6 = 762.52 kJ/kg h5 − h6 s 1020.6 − 709.66 h5 +
V52 V2 = h6 + 6 2 2
(1020.6 kJ/kg) + 0 = 762.52 kJ/kg +
V62 1 kJ/kg →V6 = 718.5 m/s 2 1000 m 2 /s 2
where the velocity at nozzle inlet is assumed zero. (d) The propulsive power and the propulsive efficiency are 1 kJ/kg W& p = m& (V6 − V1 )V1 = (1.76 kg/s)(718.5 m/s − 250 m/s)(250 m/s) = 206.1 kW 2 2 1000 m /s Q& in = m& (h4 − h3 ) = (1.76 kg/s)(1304.8 − 553.50)kJ/kg = 1322 kW
ηp =
W& p 206.1 kW = = 0.156 Q& in 1322 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-131
9-159 EES The effect of variable specific heats on the thermal efficiency of the ideal Otto cycle using air as the working fluid is to be investigated. The percentage of error involved in using constant specific heat values at room temperature for different combinations of compression ratio and maximum cycle temperature is to be determined. Analysis Using EES, the problem is solved as follows: Procedure ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) "For Air:" C_V = 0.718 [kJ/kg-K] k = 1.4 T2 = T[1]*r_comp^(k-1) P2 = P[1]*r_comp^k q_in_23 = C_V*(T[3]-T2) T4 = T[3]*(1/r_comp)^(k-1) q_out_41 = C_V*(T4-T[1]) Eta_th_ConstProp = (1-q_out_41/q_in_23)*Convert(, %) "[%]" "The Easy Way to calculate the constant property Otto cycle efficiency is:" Eta_th_easy = (1 - 1/r_comp^(k-1))*Convert(, %) "[%]" END "Input Data" T[1]=300 [K] P[1]=100 [kPa] "T[3] = 1000 [K]" r_comp = 12 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for process 1 to 2" q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" v[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-132 V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent" Call ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) PerCentError = ABS(Eta_th - Eta_th_ConstProp)/Eta_th*Convert(, %) "[%]" PerCentErro r [%] 3.604 6.681 9.421 11.64
rcomp
ηth [%]
ηth,ConstProp [%]
ηth,easy [%]
T3 [K
12 12 12 12
60.8 59.04 57.57 56.42
62.99 62.99 62.99 62.99
62.99 62.99 62.99 62.99
1000 1500 2000 2500
Percent Error = |η
th
-η |/η th,ConstProp th
4.3
PerCentError [%]
4.2
T
max
= 1000 K
4.1 4 3.9 3.8 3.7 3.6 6
7
8
9
r
10
11
12
comp
15
PerCentError [%]
r
comp
12.8
=6 =12
10.6
8.4
6.2
4 1000
1200
1400
1600
1800
2000
2200
2400
2600
T[3] [K] PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-133
9-160 EES The effects of compression ratio on the net work output and the thermal efficiency of the Otto cycle for given operating conditions is to be investigated. Analysis Using EES, the problem is solved as follows: "Input Data" T[1]=300 [K] P[1]=100 [kPa] T[3] = 2000 [K] r_comp = 12 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for process 1 to 2" q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" v[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-134
ηth [%] 45.83 48.67 51.03 53.02 54.74 56.24 57.57 58.75 59.83 60.8
wnet [kJ/kg] 567.4 589.3 604.9 616.2 624.3 630 633.8 636.3 637.5 637.9
rcomp 6 7 8 9 10 11 12 13 14 15
640 630
w net [kJ/kg]
620 610 600 590 580 570 560 6
7
8
9
10
11
r
12
13
14
15
com p
62.5
59
η th [%]
55.5
52
48.5
45 6
7
8
9
10
11
12
13
14
15
r comp
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-135
9-161 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycle is to be investigated. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be determined. Analysis Using EES, the problem is solved as follows: P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s] Eta_c = 100/100 Eta_t = 100/100 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4])
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-136
Bwr
η
Pratio
0.254 0.2665 0.2776 0.2876 0.2968 0.3052 0.313 0.3203 0.3272 0.3337 0.3398 0.3457 0.3513 0.3567 0.3618 0.3668 0.3716 0.3762 0.3806 0.385 0.3892 0.3932 0.3972 0.401 0.4048 0.4084 0.412 0.4155 0.4189 0.4222
0.3383 0.3689 0.3938 0.4146 0.4324 0.4478 0.4615 0.4736 0.4846 0.4945 0.5036 0.512 0.5197 0.5269 0.5336 0.5399 0.5458 0.5513 0.5566 0.5615 0.5663 0.5707 0.575 0.5791 0.583 0.5867 0.5903 0.5937 0.597 0.6002
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34
Wc [kW] 175.8 201.2 223.7 244.1 262.6 279.7 295.7 310.6 324.6 337.8 350.4 362.4 373.9 384.8 395.4 405.5 415.3 424.7 433.8 442.7 451.2 459.6 467.7 475.5 483.2 490.7 498 505.1 512.1 518.9
Wnet [kW] 516.3 553.7 582.2 604.5 622.4 637 649 659.1 667.5 674.7 680.8 685.9 690.3 694.1 697.3 700 702.3 704.3 705.9 707.2 708.3 709.2 709.8 710.3 710.6 710.7 710.8 710.7 710.4 710.1
725
Wt [kW] 692.1 754.9 805.9 848.5 885 916.7 944.7 969.6 992.1 1013 1031 1048 1064 1079 1093 1106 1118 1129 1140 1150 1160 1169 1177 1186 1194 1201 1209 1216 1223 1229
Qin [kW] 1526 1501 1478 1458 1439 1422 1406 1392 1378 1364 1352 1340 1328 1317 1307 1297 1287 1277 1268 1259 1251 1243 1234 1227 1219 1211 1204 1197 1190 1183
0.65
700
0.6
675 0.55
650
] W k[
625
0.5
600
0.45 ht
W
575
t e n
η
0.4
550 0.35
525 500 5
10
15
20
25
30
0.3 35
Pratio
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-137
9-162 EES The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycle is to be investigated assuming adiabatic efficiencies of 85 percent for both the turbine and the compressor. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be determined. Analysis Using EES, the problem is solved as follows: P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s] Eta_c = 85/100 Eta_t = 85/100 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2])
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-138
s[4]=entropy('air',T=T[4],P=P[4])
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-139
η
Pratio
0.2551 0.2764 0.2931 0.3068 0.3182 0.3278 0.3361 0.3432 0.3495 0.355 0.3599 0.3643 0.3682 0.3717 0.3748 0.3777 0.3802 0.3825 0.3846 0.3865
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Wnet [kW] 381.5 405 421.8 434.1 443.3 450.1 455.1 458.8 461.4 463.2 464.2 464.7 464.7 464.4 463.6 462.6 461.4 460 458.4 456.6
Wt [kW] 588.3 641.7 685 721.3 752.2 779.2 803 824.2 843.3 860.6 876.5 891.1 904.6 917.1 928.8 939.7 950 959.6 968.8 977.4
470
0.4
460 450
0.38
W net
η th
W net [kW ]
440
0.36 0.34
430
0.32
420
0.3
410 400 390 380 5
Qin [kW] 1495 1465 1439 1415 1393 1373 1354 1337 1320 1305 1290 1276 1262 1249 1237 1225 1214 1202 1192 1181
9
13
17
P
P ratio for
0.28
W
0.26
net,m ax 21
th
0.3515 0.3689 0.3843 0.3981 0.4107 0.4224 0.4332 0.4433 0.4528 0.4618 0.4704 0.4785 0.4862 0.4937 0.5008 0.5077 0.5143 0.5207 0.5268 0.5328
Wc [kW] 206.8 236.7 263.2 287.1 309 329.1 347.8 365.4 381.9 397.5 412.3 426.4 439.8 452.7 465.1 477.1 488.6 499.7 510.4 520.8
η
Bwr
0.24 25
ratio
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-140
9-163 EES The effects of pressure ratio, maximum cycle temperature, and compressor and turbine inefficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated. Constant specific heats at room temperature are to be used. Analysis Using EES, the problem is solved as follows: Procedure ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) "For Air:" C_V = 0.718 [kJ/kg-K] k = 1.4 T2 = T[1]*r_comp^(k-1) P2 = P[1]*r_comp^k q_in_23 = C_V*(T[3]-T2) T4 = T[3]*(1/r_comp)^(k-1) q_out_41 = C_V*(T4-T[1]) Eta_th_ConstProp = (1-q_out_41/q_in_23)*Convert(, %) "[%]" "The Easy Way to calculate the constant property Otto cycle efficiency is:" Eta_th_easy = (1 - 1/r_comp^(k-1))*Convert(, %) "[%]" END "Input Data" T[1]=300 [K] P[1]=100 [kPa] {T[3] = 1000 [K]} r_comp = 12 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for process 1 to 2" q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" v[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] "Conservation of energy for process 2 to 3" q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3])
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-141
"Process 4-1 is constant volume heat rejection" V[4] = V[1] "Conservation of energy for process 4 to 1" q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent" Call ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) PerCentError = ABS(Eta_th - Eta_th_ConstProp)/Eta_th*Convert(, %) "[%]" PerCentError [%] 3.604 6.681 9.421 11.64
ηth [%] 60.8 59.04 57.57 56.42
rcomp 12 12 12 12
P e rc e n t E rro r = | η
ηth,ConstProp [%] 62.99 62.99 62.99 62.99
th
ηth,easy [%] 62.99 62.99 62.99 62.99
T3 [K] 1000 1500 2000 2500
- η |/η th ,C o n s tP ro p th
4 .3
PerCentError [%]
4 .2
T
m ax
= 1000 K
4 .1
4
3 .9
3 .8
3 .7
3 .6 6
7
8
9
r
10
11
12
com p
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-142 15
PerCentError [%]
r
comp
12.8
=6 =12
10.6
8.4
6.2
4 1000
1200
1400
1600
1800
2000
2200
2400
2600
T[3] [K]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-143
9-164 EES The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated. Variable specific heats are to be used. Analysis Using EES, the problem is solved as follows: "Input data - from diagram window" {P_ratio = 8} {T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 800 [K] m_dot = 1 [kg/s] Eta_c = 75/100 Eta_t = 82/100} "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2])
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-144
s[4]=entropy('air',T=T[4],P=P[4])
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-145
Bwr 0.5229 0.6305 0.7038 0.7611 0.8088 0.85 0.8864 0.9192 0.9491 0.9767
η
Pratio
0.1 0.1644 0.1814 0.1806 0.1702 0.1533 0.131 0.1041 0.07272 0.03675
2 4 6 8 10 12 14 16 18 20
Wc [kW] 1818 4033 5543 6723 7705 8553 9304 9980 10596 11165
Wnet [kW] 1659 2364 2333 2110 1822 1510 1192 877.2 567.9 266.1
Wt [kW] 3477 6396 7876 8833 9527 10063 10496 10857 11164 11431
Qin [kW] 16587 14373 12862 11682 10700 9852 9102 8426 7809 7241
1500 Air Standard Brayton Cycle Pressure ratio = 8 and T max = 1160K
3
T [K]
1000
4
2 2s
4s
500 800 kPa 100 kPa
0 5.0
1
5.5
6.0
6.5
7.0
7.5
s [kJ/kg-K]
0.25
2500
η
Cycle efficiency,
η
W
0.15
η η
0.10
0.05
0.00 2
T Note P
4
ratio
6
2000 net
1500 t c
m ax
= 0.82 = 0.75
1000
=1160 K
500
for m axim um w ork and η
8
10
12 P
14
W net [kW ]
0.20
16
18
0 20
ratio
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-146
9-165 EES The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with helium as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: Function hFunc(WorkFluid$,T,P) "The EES functions teat helium as a real gas; thus, T and P are needed for helium's enthalpy." IF WorkFluid$ = 'Air' then hFunc:=enthalpy(Air,T=T) ELSE hFunc: = enthalpy(Helium,T=T,P=P) endif END Procedure EtaCheck(Eta_th:EtaError$) If Eta_th < 0 then EtaError$ = 'Why are the net work done and efficiency < 0?' Else EtaError$ = '' END "Input data - from diagram window" {P_ratio = 8} {T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 800 [K] m_dot = 1 [kg/s] Eta_c = 0.8 Eta_t = 0.8 WorkFluid$ = 'Helium'} "Inlet conditions" h[1]=hFunc(WorkFluid$,T[1],P[1]) s[1]=ENTROPY(WorkFluid$,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(WorkFluid$,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=hFunc(WorkFluid$,T_s[2],P[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=hFunc(WorkFluid$,T[3],P[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(WorkFluid$,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(WorkFluid$,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=hFunc(WorkFluid$,T_s[4],P[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta_th=W_dot_net/Q_dot_in"Cycle thermal efficiency"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-147
Call EtaCheck(Eta_th:EtaError$) Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4]) Bwr 0.5229 0.6305 0.7038 0.7611 0.8088 0.85 0.8864 0.9192 0.9491 0.9767
η
Pratio
0.1 0.1644 0.1814 0.1806 0.1702 0.1533 0.131 0.1041 0.07272 0.03675
2 4 6 8 10 12 14 16 18 20
Wc [kW] 1818 4033 5543 6723 7705 8553 9304 9980 10596 11165
Wnet [kW] 1659 2364 2333 2110 1822 1510 1192 877.2 567.9 266.1
Wt [kW] 3477 6396 7876 8833 9527 10063 10496 10857 11164 11431
Qin [kW] 16587 14373 12862 11682 10700 9852 9102 8426 7809 7241
1500 B rayto n C ycle P re ssure ratio = 8 an d T m ax = 116 0K
3
T [K]
1000
4
2 2s
4s
500 800 kP a 100 kP a
0 5.0
1
5.5
6.0
6.5
7.0
7.5
s [kJ/kg-K ]
0.25
Brayton Cycle using Air m air = 20 kg/s
0.20
η
2000
Wnet
η , y c n ei ci ff e el c y C
2500
0.15
0.10
0.05
0.00 2
1500
] W k[ 1000 et
η = 0.82 t η = 0.75 c Tmax=1160 K
n
W 500
Note Pratio for maximum work and η
4
6
8
10 12 Pratio
14
16
18
0 20
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-148
9-166 EES The effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a regenerative Brayton cycle with air as the working fluid is to be investigated. Constant specific heats at room temperature are to be used. Analysis Using EES, the problem is solved as follows: "Input data for air" C_P = 1.005 [kJ/kg-K] k = 1.4 "Other Input data from the diagram window" {T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8"Compressor isentorpic efficiency" Eta_t =0.9"Turbien isentropic efficiency"} "Isentropic Compressor anaysis" T_s[2] = T[1]*Pratio^((k-1)/k) P[2] = Pratio*P[1] "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_s[2]-T[1]) "Actual compressor analysis:" w_comp = C_P*(T[2]-T[1]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" q_in_noreg = C_P*(T[3] - T[2]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "Turbine analysis" P[4] = P[3] /Pratio T_s[4] = T[3]*(1/Pratio)^((k-1)/k) "T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen = C_P*(T[3] - T_s[4]) "Actual Turbine analysis:" w_turb = C_P*(T[3] - T[4]) "Cycle analysis" w_net=w_turb-w_comp Eta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" Bwr=w_comp/w_turb "Back work ratio"
"Cycle thermal efficiency"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-149
"With the regenerator the heat added in the external heat exchanger is" q_in_withreg = C_P*(T[3] - T[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (T[5]-T[2])/(T[4]-T[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" T[2] + T[4]=T[5] + T[6] P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "[%]" ηc
ηt
0.6 0.65 0.7 0.75 0.8 0.85 0.9
0.9 0.9 0.9 0.9 0.9 0.9 0.9
ηth,noreg [%] 14.76 20.35 24.59 27.91 30.59 32.79 34.64
ηth,withreg [%] 13.92 20.54 26.22 31.14 35.44 39.24 42.61
qin,noreg [kJ/kg] 510.9 546.8 577.5 604.2 627.5 648 666.3
qin,withreg [kJ/kg] 541.6 541.6 541.6 541.6 541.6 541.6 541.6
wnet [kJ/kg] 75.4 111.3 142 168.6 192 212.5 230.8
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-150
9-167 EES The effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a regenerative Brayton cycle with air as the working fluid is to be investigated. Variable specific heats are to be used. Analysis Using EES, the problem is solved as follows: "Input data" "Input data from the diagram window" {T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8 "Compressor isentorpic efficiency" Eta_t =0.9 "Turbien isentropic efficiency"} "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" h[1] + w_compisen = h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" h[1] + w_comp = h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" h[2] + q_in_noreg = h[3] h[3]=ENTHALPY(Air,T=T[3]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[3] /Pratio s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4])"T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-151
h[3] = w_turbisen + h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4]) "Actual Turbine analysis:" h[3] = w_turb + h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Cycle analysis" w_net=w_turb-w_comp Eta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" Bwr=w_comp/w_turb"Back work ratio"
"Cycle thermal efficiency"
"With the regenerator the heat added in the external heat exchanger is" h[5] + q_in_withreg = h[3] h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[2])/(h[4]-h[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" h[2] + h[4]=h[5] + h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6]) P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "[%]" Air
1600 1400
3
1200
T [K]
"The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6]
1000 kPa
1000 800
2
100 kPa
5
2s
600
4s
4
6 400 1 200 4.5
5.0
5.5
6.0
6.5
7.0
s [kJ/kg-K]
ηc
ηt
0.6 0.65 0.7 0.75 0.8 0.85 0.9
0.9 0.9 0.9 0.9 0.9 0.9 0.9
ηth,noreg [%] 14.76 20.35 24.59 27.91 30.59 32.79 34.64
ηth,withreg [%] 13.92 20.54 26.22 31.14 35.44 39.24 42.61
qin,noreg [kJ/kg] 510.9 546.8 577.5 604.2 627.5 648 666.3
qin,withreg [kJ/kg] 541.6 541.6 541.6 541.6 541.6 541.6 541.6
wnet [kJ/kg] 75.4 111.3 142 168.6 192 212.5 230.8
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
7.5
9-152
45
η
40
c
= 0.8
W ith regeneration
35 30
η th
25
No regeneration
20 15 10 0.7
0.75
0.8
0.85
η
0.9
0.95
1
0.9
0.95
1
t
275
η
w net [kJ/kg]
230
c
= 0.8
185
140
95
50 0.7
0.75
0.8
0.85
η
t
650
600
q in
550
No regeneration W ith regeneration
500
η
450
400 0.7
0.75
0.8
0.85
η
c
= 0.8
0.9
0.95
1
t
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-153 45
η
40
t
= 0.9
W ith regeneration
35 30
η th
25
No regeneration
20 15 10 0.6
0.65
0.7
0.75
η
0.8
0.85
0.9
c
250
w net [kJ/kg]
215
η
t
= 0.9
180
145
110
75 0.6
0.65
0.7
0.75
η
0.8
0.85
0.9
c
680 660
η
t
= 0.9
640 620
No regeneration
q in
600 580
W ith regeneration
560 540 520 500 0.6
0.65
0.7
0.75
η
0.8
0.85
0.9
c
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-154
9-168 EES The effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a regenerative Brayton cycle with helium as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: "Input data for helium" C_P = 5.1926 [kJ/kg-K] k = 1.667 "Other Input data from the diagram window" {T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8 "Compressor isentorpic efficiency" Eta_t =0.9 "Turbien isentropic efficiency"} "Isentropic Compressor anaysis" T_s[2] = T[1]*Pratio^((k-1)/k) P[2] = Pratio*P[1] "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_s[2]-T[1]) "Actual compressor analysis:" w_comp = C_P*(T[2]-T[1]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" q_in_noreg = C_P*(T[3] - T[2]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "Turbine analysis" P[4] = P[3] /Pratio T_s[4] = T[3]*(1/Pratio)^((k-1)/k) "T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen = C_P*(T[3] - T_s[4]) "Actual Turbine analysis:" w_turb = C_P*(T[3] - T[4]) "Cycle analysis" w_net=w_turb-w_comp
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-155
Eta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" Bwr=w_comp/w_turb "Back work ratio"
"Cycle thermal efficiency"
"With the regenerator the heat added in the external heat exchanger is" q_in_withreg = C_P*(T[3] - T[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (T[5]-T[2])/(T[4]-T[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" T[2] + T[4]=T[5] + T[6] P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "[%]" ηc
ηt
0.6 0.65 0.7 0.75 0.8 0.85 0.9
0.9 0.9 0.9 0.9 0.9 0.9 0.9
ηth,noreg [%] 14.76 20.35 24.59 27.91 30.59 32.79 34.64
ηth,withreg [%] 13.92 20.54 26.22 31.14 35.44 39.24 42.61
qin,noreg [kJ/kg] 510.9 546.8 577.5 604.2 627.5 648 666.3
qin,withreg [kJ/kg] 541.6 541.6 541.6 541.6 541.6 541.6 541.6
wnet [kJ/kg] 75.4 111.3 142 168.6 192 212.5 230.8
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-156
9-169 EES The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and air as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: "Input data for air" C_P = 1.005 [kJ/kg-K] k = 1.4 "Nstages is the number of compression and expansion stages" Nstages = 1 T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness" Eta_c =1.0 "Compressor isentorpic efficiency" Eta_t =1.0 "Turbine isentropic efficiency" R_p = Pratio^(1/Nstages) "Isentropic Compressor anaysis" T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 "T_2s is the isentropic value of T_2 at compressor exit" Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_2s-T_1) "Actual compressor analysis:" w_comp = C_P*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:" w_comp_total = Nstages*w_comp "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" "The heat added in the external heat exchanger + the reheat between turbines is" q_in_total = C_P*(T_6 - T_5) +(Nstages - 1)*C_P*(T_8 - T_7) "Reheat is assumed to occur until:" T_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit" T_7s = T_6*(1/R_p)^((k-1)/k) "Turbine adiabatic efficiency, w_turbisen > w_turb" Eta_t = w_turb /w_turbisen "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen = C_P*(T_6 - T_7s) "Actual Turbine analysis:" w_turb = C_P*(T_6 - T_7) w_turb_total = Nstages*w_turb "Cycle analysis"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-157
w_net=w_turb_total-w_comp_total "[kJ/kg]" Bwr=w_comp/w_turb "Back work ratio" P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 "The regenerator effectiveness gives T_5 as:" Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10 as:" T_4 + T_9=T_5 + T_10 "Cycle thermal efficiency with regenerator" Eta_th_regenerative=w_net/q_in_total*Convert(, %) "[%]" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T_6 and T_1 for this problem." Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]" ηth,Regenerative [%] 49.15 64.35 68.32 70.14 72.33 73.79 74.05 74.18
ηth,Ericksson [%] 75 75 75 75 75 75 75 75
Nstages 1 2 3 4 7 15 19 22
80
70
Ericsson η th [%]
60
Ideal Regenerative Brayton
50
40 0
2
4
6
8
10
12
14
16
18
20
22
24
Nstages
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-158
9-170 EES The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and helium as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: "Input data for Helium" C_P = 5.1926 [kJ/kg-K] k = 1.667 "Nstages is the number of compression and expansion stages" {Nstages = 1} T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness" Eta_c =1.0 "Compressor isentorpic efficiency" Eta_t =1.0 "Turbine isentropic efficiency" R_p = Pratio^(1/Nstages) "Isentropic Compressor anaysis" T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 "T_2s is the isentropic value of T_2 at compressor exit" Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_2s-T_1) "Actual compressor analysis:" w_comp = C_P*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:" w_comp_total = Nstages*w_comp "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" "The heat added in the external heat exchanger + the reheat between turbines is" q_in_total = C_P*(T_6 - T_5) +(Nstages - 1)*C_P*(T_8 - T_7) "Reheat is assumed to occur until:" T_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit" T_7s = T_6*(1/R_p)^((k-1)/k) "Turbine adiabatic efficiency, w_turbisen > w_turb" Eta_t = w_turb /w_turbisen "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen = C_P*(T_6 - T_7s) "Actual Turbine analysis:" w_turb = C_P*(T_6 - T_7) w_turb_total = Nstages*w_turb
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-159
"Cycle analysis" w_net=w_turb_total-w_comp_total Bwr=w_comp/w_turb "Back work ratio" P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 "The regenerator effectiveness gives T_5 as:" Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10 as:" T_4 + T_9=T_5 + T_10 "Cycle thermal efficiency with regenerator" Eta_th_regenerative=w_net/q_in_total*Convert(, %) "[%]" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T_6 and T_1 for this problem." Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]" ηth,Ericksson [%] 75 75 75 75 75 75 75 75
ηth,Regenerative [%] 32.43 58.9 65.18 67.95 71.18 73.29 73.66 73.84
Nstages 1 2 3 4 7 15 19 22
80
70
Ericsson
η th [%]
60
Ideal Regenerative Brayton
50
40
30 0
2
4
6
8
10
12
14
16
18
20
22
24
Nstages
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-160
Fundamentals of Engineering (FE) Exam Problems 9-171 An Otto cycle with air as the working fluid has a compression ratio of 8.2. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 24% (b) 43% (c) 52% (d) 57% (e) 75% Answer (d) 57% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). r=8.2 k=1.4 Eta_Otto=1-1/r^(k-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/r "Taking efficiency to be 1/r" W2_Eta = 1/r^(k-1) "Using incorrect relation" W3_Eta = 1-1/r^(k1-1); k1=1.667 "Using wrong k value"
9-172 For specified limits for the maximum and minimum temperatures, the ideal cycle with the lowest thermal efficiency is (a) Carnot (b) Stirling (c) Ericsson (d) Otto (e) All are the same Answer (d) Otto
9-173 A Carnot cycle operates between the temperatures limits of 300 K and 2000 K, and produces 600 kW of net power. The rate of entropy change of the working fluid during the heat addition process is (a) 0 (b) 0.300 kW/K (c) 0.353 kW/K (d) 0.261 kW/K (e) 2.0 kW/K Answer (c) 0.353 kW/K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=300 "K" TH=2000 "K" Wnet=600 "kJ/s" Wnet= (TH-TL)*DS "Some Wrong Solutions with Common Mistakes:" W1_DS = Wnet/TH "Using TH instead of TH-TL" W2_DS = Wnet/TL "Using TL instead of TH-TL" W3_DS = Wnet/(TH+TL) "Using TH+TL instead of TH-TL"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-161
9-174 Air in an ideal Diesel cycle is compressed from 3 L to 0.15 L, and then it expands during the constant pressure heat addition process to 0.30 L. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 35% (b) 44% (c) 65% (d) 70% (e) 82% Answer (c) 65% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V1=3 "L" V2= 0.15 "L" V3= 0.30 "L" r=V1/V2 rc=V3/V2 k=1.4 Eta_Diesel=1-(1/r^(k-1))*(rc^k-1)/k/(rc-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-(1/r1^(k-1))*(rc^k-1)/k/(rc-1); r1=V1/V3 "Wrong r value" W2_Eta = 1-Eta_Diesel "Using incorrect relation" W3_Eta = 1-(1/r^(k1-1))*(rc^k1-1)/k1/(rc-1); k1=1.667 "Using wrong k value" W4_Eta = 1-1/r^(k-1) "Using Otto cycle efficiency"
9-175 Helium gas in an ideal Otto cycle is compressed from 20°C and 2.5 L to 0.25 L, and its temperature increases by an additional 700°C during the heat addition process. The temperature of helium before the expansion process is (a) 1790°C (b) 2060°C (c) 1240°C (d) 620°C (e) 820°C Answer (a) 1790°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 V1=2.5 V2=0.25 r=V1/V2 T1=20+273 "K" T2=T1*r^(k-1) T3=T2+700-273 "C" "Some Wrong Solutions with Common Mistakes:" W1_T3 =T22+700-273; T22=T1*r^(k1-1); k1=1.4 "Using wrong k value" W2_T3 = T3+273 "Using K instead of C" W3_T3 = T1+700-273 "Disregarding temp rise during compression" W4_T3 = T222+700-273; T222=(T1-273)*r^(k-1) "Using C for T1 instead of K"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-162 9-176 In an ideal Otto cycle, air is compressed from 1.20 kg/m3 and 2.2 L to 0.26 L, and the net work output of the cycle is 440 kJ/kg. The mean effective pressure (MEP) for this cycle is (a) 612 kPa (b) 599 kPa (c) 528 kPa (d) 416 kPa (e) 367 kPa Answer (b) 599 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho1=1.20 "kg/m^3" k=1.4 V1=2.2 V2=0.26 m=rho1*V1/1000 "kg" w_net=440 "kJ/kg" Wtotal=m*w_net MEP=Wtotal/((V1-V2)/1000) "Some Wrong Solutions with Common Mistakes:" W1_MEP = w_net/((V1-V2)/1000) "Disregarding mass" W2_MEP = Wtotal/(V1/1000) "Using V1 instead of V1-V2" W3_MEP = (rho1*V2/1000)*w_net/((V1-V2)/1000); "Finding mass using V2 instead of V1" W4_MEP = Wtotal/((V1+V2)/1000) "Adding V1 and V2 instead of subtracting"
9-177 In an ideal Brayton cycle, air is compressed from 95 kPa and 25°C to 800 kPa. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 46% (b) 54% (c) 57% (d) 39% (e) 61% Answer (a) 46% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=95 "kPa" P2=800 "kPa" T1=25+273 "K" rp=P2/P1 k=1.4 Eta_Brayton=1-1/rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/rp "Taking efficiency to be 1/rp" W2_Eta = 1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-1/rp^((k1-1)/k1); k1=1.667 "Using wrong k value"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-163
9-178 Consider an ideal Brayton cycle executed between the pressure limits of 1200 kPa and 100 kPa and temperature limits of 20°C and 1000°C with argon as the working fluid. The net work output of the cycle is (a) 68 kJ/kg (b) 93 kJ/kg (c) 158 kJ/kg (d) 186 kJ/kg (e) 310 kJ/kg Answer (c) 158 kJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=100 "kPa" P2=1200 "kPa" T1=20+273 "K" T3=1000+273 "K" rp=P2/P1 k=1.667 Cp=0.5203 "kJ/kg.K" Cv=0.3122 "kJ/kg.K" T2=T1*rp^((k-1)/k) q_in=Cp*(T3-T2) Eta_Brayton=1-1/rp^((k-1)/k) w_net=Eta_Brayton*q_in "Some Wrong Solutions with Common Mistakes:" W1_wnet = (1-1/rp^((k-1)/k))*qin1; qin1=Cv*(T3-T2) "Using Cv instead of Cp" W2_wnet = (1-1/rp^((k-1)/k))*qin2; qin2=1.005*(T3-T2) "Using Cp of air instead of argon" W3_wnet = (1-1/rp^((k1-1)/k1))*Cp*(T3-T22); T22=T1*rp^((k1-1)/k1); k1=1.4 "Using k of air instead of argon" W4_wnet = (1-1/rp^((k-1)/k))*Cp*(T3-T222); T222=(T1-273)*rp^((k-1)/k) "Using C for T1 instead of K"
9-179 An ideal Brayton cycle has a net work output of 150 kJ/kg and a backwork ratio of 0.4. If both the turbine and the compressor had an isentropic efficiency of 85%, the net work output of the cycle would be (a) 74 kJ/kg (b) 95 kJ/kg (c) 109 kJ/kg (d) 128 kJ/kg (e) 177 kJ/kg Answer (b) 95 kJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). wcomp/wturb=0.4 wturb-wcomp=150 "kJ/kg" Eff=0.85 w_net=Eff*wturb-wcomp/Eff "Some Wrong Solutions with Common Mistakes:" W1_wnet = Eff*wturb-wcomp*Eff "Making a mistake in Wnet relation" W2_wnet = (wturb-wcomp)/Eff "Using a wrong relation" W3_wnet = wturb/eff-wcomp*Eff "Using a wrong relation"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-164
9-180 In an ideal Brayton cycle, air is compressed from 100 kPa and 25°C to 1 MPa, and then heated to 1200°C before entering the turbine. Under cold air standard conditions, the air temperature at the turbine exit is (a) 490°C (b) 515°C (c) 622°C (d) 763°C (e) 895°C Answer (a) 490°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=100 "kPa" P2=1000 "kPa" T1=25+273 "K" T3=1200+273 "K" rp=P2/P1 k=1.4 T4=T3*(1/rp)^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp "Using wrong relation" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T1+800-273 "Disregarding temp rise during compression"
9-181 In an ideal Brayton cycle with regeneration, argon gas is compressed from 100 kPa and 25°C to 400 kPa, and then heated to 1200°C before entering the turbine. The highest temperature that argon can be heated in the regenerator is (a) 246°C (b) 846°C (c) 689°C (d) 368°C (e) 573°C Answer (e) 573°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 Cp=0.5203 "kJ/kg.K" P1=100 "kPa" P2=400 "kPa" T1=25+273 "K" T3=1200+273 "K" "The highest temperature that argon can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2=T1*rp^((k-1)/k) T4=T3/rp^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T2-273 "Taking compressor exit temp as the answer" PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-165
9-182 In an ideal Brayton cycle with regeneration, air is compressed from 80 kPa and 10°C to 400 kPa and 175°C, is heated to 450°C in the regenerator, and then further heated to 1000°C before entering the turbine. Under cold air standard conditions, the effectiveness of the regenerator is (a) 33% (b) 44% (c) 62% (d) 77% (e) 89% Answer (d) 77% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" P1=80 "kPa" P2=400 "kPa" T1=10+273 "K" T2=175+273 "K" T3=1000+273 "K" T5=450+273 "K" "The highest temperature that the gas can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2check=T1*rp^((k-1)/k) "Checking the given value of T2. It checks." T4=T3/rp^((k-1)/k) Effective=(T5-T2)/(T4-T2) "Some Wrong Solutions with Common Mistakes:" W1_eff = (T5-T2)/(T3-T2) "Using wrong relation" W2_eff = (T5-T2)/(T44-T2); T44=(T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_eff = (T5-T2)/(T444-T2); T444=T3/rp "Using wrong relation for T4"
9-183 Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle with regeneration between the temperature limits of 20°C and 900°C. If the specific heat ratio of the working fluid is 1.3, the highest thermal efficiency this gas turbine can have is (a) 38% (b) 46% (c) 62% (d) 58% (e) 97% Answer (c) 62% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.3 rp=6 T1=20+273 "K" T3=900+273 "K" Eta_regen=1-(T1/T3)*rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-((T1-273)/(T3-273))*rp^((k-1)/k) "Using C for temperatures instead of K" W2_Eta = (T1/T3)*rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k1-1)/k1); k1=1.4 "Using wrong k value (the one for air)"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
9-166
9-184 An ideal gas turbine cycle with many stages of compression and expansion and a regenerator of 100 percent effectiveness has an overall pressure ratio of 10. Air enters every stage of compressor at 290 K, and every stage of turbine at 1200 K. The thermal efficiency of this gas-turbine cycle is (a) 36% (b) 40% (c) 52% (d) 64% (e) 76% Answer (e) 76% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 rp=10 T1=290 "K" T3=1200 "K" Eff=1-T1/T3 "Some Wrong Solutions with Common Mistakes:" W1_Eta = 100 W2_Eta = 1-1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k-1)/k) "Using wrong relation" W4_Eta = T1/T3 "Using wrong relation"
9-185 Air enters a turbojet engine at 260 m/s at a rate of 30 kg/s, and exits at 800 m/s relative to the aircraft. The thrust developed by the engine is (a) 8 kN (b) 16 kN (c) 24 kN (d) 20 kN (e) 32 kN Answer (b) 16 kN Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Vel1=260 "m/s" Vel2=800 "m/s" Thrust=m*(Vel2-Vel1)/1000 "kN" m= 30 "kg/s" "Some Wrong Solutions with Common Mistakes:" W1_thrust = (Vel2-Vel1)/1000 "Disregarding mass flow rate" W2_thrust = m*Vel2/1000 "Using incorrect relation"
9-186 ··· 9-191 Design and Essay Problems.
KJ PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-1
Chapter 10 VAPOR AND COMBINED POWER CYCLES Carnot Vapor Cycle 10-1C Because excessive moisture in steam causes erosion on the turbine blades. The highest moisture content allowed is about 10%. 10-2C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heat transfer processes to two-phase systems to maintain isothermal conditions severely limits the maximum temperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisture content which causes erosion, and (3) it is not practical to design a compressor that will handle two phases.
10-3E A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the quality at the end of the heat rejection process, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We note that TH = Tsat @180 psia = 373.1°F = 833.1 R TL = Tsat @14.7 psia = 212.0°F = 672.0 R
T
and ηth,C = 1 −
TL 672.0 R = 1− = 19.3% TH 833.1 R
1 180 psia 2 qin
(b) Noting that s4 = s1 = sf @ 180 psia = 0.53274 Btu/lbm·R, x4 =
s4 − s f s fg
=
0.53274 − 0.31215 = 0.153 1.44441
14.7 psia 4
3
(c) The enthalpies before and after the heat addition process are h1 = h f @ 180 psia = 346.14 Btu/lbm
h2 = h f + x 2 h fg = 346.14 + (0.90 )(851.16 ) = 1112.2 Btu/lbm
Thus, q in = h2 − h1 = 1112.2 − 346.14 = 766.0 Btu/lbm
and,
wnet = η th q in = (0.1934)(766.0 Btu/lbm) = 148.1 Btu/lbm
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-2
10-4 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the amount of heat rejected, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = 250°C = 523 K and TL = Tsat @ 20 kPa = 60.06°C = 333.1 K, the thermal efficiency becomes
η th,C = 1 −
TL 333.1 K = 1− = 0.3632 = 36.3% 523 K TH
(b) The heat supplied during this cycle is simply the enthalpy of vaporization , q in = h fg @ 250oC = 1715.3 kJ/kg
T
2
1
250°C
qin
Thus, q out
20 kPa
333.1 K T (1715.3 kJ/kg ) = 1092.3 kJ/kg = q L = L q in = TH 523 K
4
qout
3 s
(c) The net work output of this cycle is wnet = η th q in = (0.3632 )(1715.3 kJ/kg ) = 623.0 kJ/kg
10-5 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the amount of heat rejected, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = 250°C = 523 K and TL = Tsat @ 10 kPa = 45.81°C = 318.8 K, the thermal efficiency becomes
η th, C = 1 −
TL 318.8 K =1− = 39.04% TH 523 K
(b) The heat supplied during this cycle is simply the enthalpy of vaporization , q in = h fg @ 250°C = 1715.3 kJ/kg
T
250°C
qin
Thus, q out
318.8 K T (1715.3 kJ/kg ) = 1045.6 kJ/kg = q L = L q in = TH 523 K
2
1
10 kPa 4
qout
3
(c) The net work output of this cycle is wnet = η th q in = (0.3904)(1715.3 kJ/kg ) = 669.7 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-3
10-6 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the pressure at the turbine inlet, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The thermal efficiency is determined from
η th, C = 1 −
T
TL 60 + 273 K = 1− = 46.5% TH 350 + 273 K
(b) Note that s2 = s3 = sf + x3sfg
350°C
1
2
4
3
= 0.8313 + 0.891 × 7.0769 = 7.1368 kJ/kg·K Thus ,
60°C
T2 = 350°C
P2 ≅ 1.40 MPa (Table A-6) s 2 = 7.1368 kJ/kg ⋅ K
s
(c) The net work can be determined by calculating the enclosed area on the T-s diagram, s 4 = s f + x 4 s fg = 0.8313 + (0.1)(7.0769) = 1.5390 kJ/kg ⋅ K
Thus,
wnet = Area = (TH − TL )(s 3 − s 4 ) = (350 − 60 )(7.1368 − 1.5390) = 1623 kJ/kg
The Simple Rankine Cycle 10-7C The four processes that make up the simple ideal cycle are (1) Isentropic compression in a pump, (2) P = constant heat addition in a boiler, (3) Isentropic expansion in a turbine, and (4) P = constant heat rejection in a condenser. 10-8C Heat rejected decreases; everything else increases. 10-9C Heat rejected decreases; everything else increases. 10-10C The pump work remains the same, the moisture content decreases, everything else increases. 10-11C The actual vapor power cycles differ from the idealized ones in that the actual cycles involve friction and pressure drops in various components and the piping, and heat loss to the surrounding medium from these components and piping. 10-12C The boiler exit pressure will be (a) lower than the boiler inlet pressure in actual cycles, and (b) the same as the boiler inlet pressure in ideal cycles. 10-13C We would reject this proposal because wturb = h1 - h2 - qout, and any heat loss from the steam will adversely affect the turbine work output. 10-14C Yes, because the saturation temperature of steam at 10 kPa is 45.81°C, which is much higher than the temperature of the cooling water.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-4
10-15 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle and the net power output of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 50 kPa = 340.54 kJ/kg
T
v 1 = v f @ 50 kPa = 0.001030 m 3 /kg w p ,in = v 1 (P2 − P1 )
(
3
)
1 kJ = 0.001030 m 3 /kg (3000 − 50 ) kPa 1 kPa ⋅ m 3 = 3.04 kJ/kg
h2 = h1 + w p ,in = 340.54 + 3.04 = 343.58 kJ/kg
2
3 MPa qin 50 kPa
1
qout
4
P3 = 3 MPa h3 = 2994.3 kJ/kg T3 = 300 °C s 3 = 6.5412 kJ/kg ⋅ K s4 − s f P4 = 50 kPa 6.5412 − 1.0912 = = 0.8382 x4 = s 4 = s3 s fg 6.5019
h4 = h f + x 4 h fg = 340.54 + (0.8382)(2304.7 ) = 2272.3 kJ/kg
Thus, q in = h3 − h2 = 2994.3 − 343.58 = 2650.7 kJ/kg q out = h4 − h1 = 2272.3 − 340.54 = 1931.8 kJ/kg wnet = q in − q out = 2650.7 − 1931.8 = 718.9 kJ/kg
and
η th = 1 − (b)
q out 1931.8 = 1− = 27.1% q in 2650.7
W& net = m& wnet = (35 kg/s )(718.9 kJ/kg ) = 25.2 MW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-5
10-16 A steam power plant that operates on a simple ideal Rankine cycle is considered. The quality of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg
v1 = v f @ 10 kPa = 0.00101 m 3 /kg w p ,in = v1 (P2 − P1 )
(
3
)
1 kJ = 0.00101 m /kg (10,000 − 10 kPa ) 1 kPa ⋅ m 3 = 10.09 kJ/kg 3
T
2
h2 = h1 + w p ,in = 191.81 + 10.09 = 201.90 kJ/kg
P3 = 10 MPa h3 = 3375.1 kJ/kg T3 = 500 °C s 3 = 6.5995 kJ/kg ⋅ K
10 MPa qin 10 kPa
1
qout
4
s 4 − s f 6.5995 − 0.6492 P4 = 10 kPa = = 0.7934 x4 = s 4 = s3 s fg 7.4996
h4 = h f + x 4 h fg = 191.81 + (0.7934)(2392.1) = 2089.7 kJ/kg
(b)
q in = h3 − h2 = 3375.1 − 201.90 = 3173.2 kJ/kg q out = h4 − h1 = 2089.7 − 191.81 = 1897.9 kJ/kg wnet = q in − q out = 3173.2 − 1897.9 = 1275.4 kJ/kg
and
η th = (c)
m& =
wnet 1275.4 kJ/kg = = 40.2% q in 3173.2 kJ/kg
W& net 210,000 kJ/s = = 164.7 kg/s wnet 1275.4 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-6
10-17 A steam power plant that operates on a simple nonideal Rankine cycle is considered. The quality of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg
T
v 1 = v f @ 10 kPa = 0.00101 m 3 /kg w p ,in = v 1 (P2 − P1 ) / η p
(
)
1 kJ = 0.00101 m /kg (10,000 − 10 kPa ) 1 kPa ⋅ m 3 = 11.87 kJ/kg h2 = h1 + w p ,in = 191.81 + 11.87 = 203.68 kJ/kg 3
/ (0.85)
2
2
10 MPa qin
3
10 kPa 1
P3 = 10 MPa h3 = 3375.1 kJ/kg T3 = 500 °C s 3 = 6.5995 kJ/kg ⋅ K
qout
4 4
s 4s − s f P4 s = 10 kPa 6.5995 − 0.6492 = = 0.7934 x 4s = s 4s = s3 s 7.4996 fg
h4 s = h f + x 4 h fg = 191.81 + (0.7934)(2392.1) = 2089.7 kJ/kg
ηT =
h3 − h4 → h4 = h3 − η T (h3 − h4 s ) h3 − h4 s = 3375.1 − (0.85)(3375.1 − 2089.7 ) = 2282.5 kJ/kg
h4 − h f 2282.5 − 191.81 = = 0.874 x4 = h4 = 2282.5 kJ/kg h fg 2392.1 P4 = 10 kPa
(b)
qin = h3 − h2 = 3375.1 − 203.68 = 3171.4 kJ/kg qout = h4 − h1 = 2282.5 − 191.81 = 2090.7 kJ/kg wnet = qin − qout = 3171.4 − 2090.7 = 1080.7 kJ/kg
and ηth =
(c)
m& =
wnet 1080.7 kJ/kg = = 34.1% qin 3171.5 kJ/kg
W&net 210,000 kJ/s = = 194.3 kg/s wnet 1080.7 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-7
10-18E A steam power plant that operates on a simple ideal Rankine cycle between the specified pressure limits is considered. The minimum turbine inlet temperature, the rate of heat input in the boiler, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
T
Analysis (a) From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = h f @ 2 psia = 94.02 Btu/lbm
2
3
v1 = v f @ 2 psia = 0.01623 ft /lbm w p ,in = v1 (P2 − P1 )
(
1
)
1 Btu = 0.01623 ft 3/lbm (1250 − 2 psia ) 3 5.4039 psia ⋅ ft = 3.75 Btu/lbm
1250 psia · Qin 2 psia · Qout
3
4 x4 = 0.9
h2 = h1 + w p ,in = 94.02 + 3.75 = 97.77 Btu/lbm h4 = h f + x 4 h fg = 94.02 + (0.9)(1021.7 ) = 1013.6 Btu/lbm
s 4 = s f + x 4 s fg = 0.17499 + (0.9)(1.74444 ) = 1.7450 Btu/lbm ⋅ R P3 = 1250 psia h3 = 1693.4 Btu/lbm s3 = s 4 T3 = 1337°F
(b)
Q& in = m& (h3 − h2 ) = (75 lbm/s)(1693.4 − 97.77 ) = 119,672 Btu/s
(c)
Q& out = m& (h4 − h1 ) = (75 lbm/s)(1013.6 − 94.02) = 68,967 Btu/s Q& 68,967 Btu/s η th = 1 − out = 1 − = 42.4% & 119,672 Btu/s Qin
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-8
10-19E A steam power plant operates on a simple nonideal Rankine cycle between the specified pressure limits. The minimum turbine inlet temperature, the rate of heat input in the boiler, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
T
Analysis (a) From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = h f @ 2 psia = 94.02 Btu/lbm
v 1 = v f @ 2 psia = 0.01623 ft 3 /lbm w p ,in = v 1 (P2 − P1 ) / η P
(
2s
)
1 Btu = 0.01623 ft 3 /lbm (1250 − 2 psia ) 5.4039 psia ⋅ ft 3 = 4.41 Btu/lbm
/ 0.85
2
1
1250 psia · Qin 2 psia · Qout
3
4s 4 x4 = 0.9
s
h2 = h1 + w p ,in = 94.02 + 4.41 = 98.43 Btu/lbm h4 = h f + x 4 h fg = 94.02 + (0.9)(1021.7 ) = 1013.6 Btu/lbm
s 4 = s f + x 4 s fg = 0.17499 + (0.9)(1.74444 ) = 1.7450 Btu/lbm ⋅ R
The turbine inlet temperature is determined by trial and error , Try 1:
P3 = 1250 psia h3 = 1439.0 Btu/lbm T3 = 900°F s 3 = 1.5826 Btu/lbm.R x 4s =
s 4s − s f s fg
=
s3 − s f s fg
=
1.5826 − 0.17499 = 0.8069 1.74444
h4 s = h f + x 4 s h fg = 94.02 + (0.8069 )(1021.7 ) = 918.4 Btu/lbm
ηT = Try 2:
h3 − h4 1439.0 − 1013.6 = = 0.8171 h3 − h4 s 1439.0 − 918.4
P3 = 1250 psia h3 = 1498.6 Btu/lbm T3 = 1000°F s 3 = 1.6249 Btu/lbm.R x4s =
s4s − s f s fg
=
s3 − s f s fg
=
1.6249 − 0.17499 = 0.8312 1.74444
h4 s = h f + x 4 s h fg = 94.02 + (0.8312)(1021.7 ) = 943.3 Btu/lbm
ηT =
h3 − h4 1498.6 − 1013.6 = = 0.8734 h3 − h4 s 1498.6 − 943.3
By linear interpolation, at ηT = 0.85 we obtain T3 = 958.4°F. This is approximate. We can determine state 3 exactly using EES software with these results: T3 = 955.7°F, h3 = 1472.5 Btu/lbm. (b)
Q& in = m& (h3 − h2 ) = (75 lbm/s)(1472.5 − 98.43) = 103,055 Btu/s
(c)
Q& out = m& (h4 − h1 ) = (75 lbm/s)(1013.6 − 94.02 ) = 68,969 Btu/s Q& 68,969 Btu/s η th = 1 − out = 1 − = 33.1% & 103,055 Btu/s Qin
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-9
10-20 A 300-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The overall plant efficiency and the required rate of the coal supply are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 25 kPa = 271.96 kJ/kg
T
v 1 = v f @ 25 kPa = 0.001020 m 3 /kg w p ,in = v 1 (P2 − P1 )
(
)
1 kJ = 0.00102 m 3 /kg (5000 − 25 kPa ) 1 kPa ⋅ m 3 = 5.07 kJ/kg
3
h2 = h1 + w p ,in = 271.96 + 5.07 = 277.03 kJ/kg
2
5 MPa · Qin 25 kPa
1
· Qout
4
P3 = 5 MPa h3 = 3317.2 kJ/kg T3 = 450°C s 3 = 6.8210 kJ/kg ⋅ K s 4 − s f 6.8210 − 0.8932 P4 = 25 kPa = = 0.8545 x4 = s fg 6.9370 h4 = h f + x 4 h fg = 271.96 + (0.8545)(2345.5) = 2276.2 kJ/kg
s 4 = s3
The thermal efficiency is determined from qin = h3 − h2 = 3317.2 − 277.03 = 3040.2 kJ/kg qout = h4 − h1 = 2276.2 − 271.96 = 2004.2 kJ/kg
and
η th = 1 − Thus,
q out 2004.2 = 1− = 0.3407 3040.2 q in
η overall = η th ×η comb ×η gen = (0.3407 )(0.75)(0.96 ) = 24.5%
(b) Then the required rate of coal supply becomes W& net 300,000 kJ/s Q& in = = = 1,222,992 kJ/s η overall 0.2453 and Q& 1,222,992 kJ/s 1 ton = 0.04174 tons/s = 150.3 tons/h m& coal = in = C coal 29,300 kJ/kg 1000 kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-10
10-21 A solar-pond power plant that operates on a simple ideal Rankine cycle with refrigerant-134a as the working fluid is considered. The thermal efficiency of the cycle and the power output of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant tables (Tables A-11, A-12, and A-13), h1 = h f @ 0.7 MPa = 88.82 kJ/kg
v 1 = v f @ 0.7 MPa = 0.0008331 m 3 /kg
T
w p ,in = v 1 (P2 − P1 )
(
)
1 kJ = 0.0008331 m 3 /kg (1400 − 700 kPa ) 1 kPa ⋅ m 3 = 0.58 kJ/kg h2 = h1 + w p ,in = 88.82 + 0.58 = 89.40 kJ/kg
1.4 MPa 2
3
qin R-134a 0.7 MPa
1
P3 = 1.4 MPa h3 = h g @ 1.4 MPa = 276.12 kJ/kg sat.vapor s 3 = s g @ 1.4 MPa = 0.9105 kJ/kg ⋅ K
qout
4
s 4 − s f 0.9105 − 0.33230 P4 = 0.7 MPa = = 0.9839 x4 = s 4 = s3 s fg 0.58763 h4 = h f + x 4 h fg = 88.82 + (0.9839)(176.21) = 262.20 kJ/kg
Thus , q in = h3 − h2 = 276.12 − 89.40 = 186.72 kJ/kg q out = h4 − h1 = 262.20 − 88.82 = 173.38 kJ/kg wnet = q in − q out = 186.72 − 173.38 = 13.34 kJ/kg
and
η th = (b)
wnet 13.34 kJ/kg = = 7.1% q in 186.72 kJ/kg
W& net = m& wnet = (3 kg/s )(13.34 kJ/kg ) = 40.02 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-11
10-22 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg T
v 1 = v f @ 10 kPa = 0.00101 m 3 /kg
w p ,in = v 1 (P2 − P1 )
(
)
1 kJ = 0.00101 m 3 /kg (7,000 − 10 kPa ) 1 kPa ⋅ m 3 = 7.06 kJ/kg
h2 = h1 + w p ,in = 191.81 + 7.06 = 198.87 kJ/kg
3 2
7 MPa qin 10 kPa
1
qout
4
P3 = 7 MPa h3 = 3411.4 kJ/kg T3 = 500°C s 3 = 6.8000 kJ/kg ⋅ K s 4 − s f 6.8000 − 0.6492 P4 = 10 kPa = = 0.8201 x4 = s fg 7.4996 h4 = h f + x 4 h fg = 191.81 + (0.8201)(2392.1) = 2153.6 kJ/kg
s 4 = s3
Thus,
q in = h3 − h2 = 3411.4 − 198.87 = 3212.5 kJ/kg q out = h4 − h1 = 2153.6 − 191.81 = 1961.8 kJ/kg wnet = q in − q out = 3212.5 − 1961.8 = 1250.7 kJ/kg
and
η th =
(b)
m& =
wnet 1250.7 kJ/kg = = 38.9% q in 3212.5 kJ/kg
W&net 45,000 kJ/s = = 36.0 kg/s wnet 1250.7 kJ/kg
(c) The rate of heat rejection to the cooling water and its temperature rise are Q& out = m& q out = (35.98 kg/s )(1961.8 kJ/kg ) = 70,586 kJ/s Q& out 70,586 kJ/s ∆Tcooling water = = = 8.4°C (m& c) cooling water (2000 kg/s )(4.18 kJ/kg ⋅ °C )
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-12
10-23 A steam power plant operates on a simple nonideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg
T
v 1 = v f @ 10 kPa = 0.00101 m 3 /kg w p ,in = v 1 (P2 − P1 ) / η p
(
)
1 kJ = 0.00101 m 3 /kg (7,000 − 10 kPa ) 1 kPa ⋅ m 3 = 8.11 kJ/kg h2 = h1 + w p ,in = 191.81 + 8.11 = 199.92 kJ/kg
/ (0.87 )
2
7 MPa qin
2
3
10 kPa 1
qout
4 4
P3 = 7 MPa h3 = 3411.4 kJ/kg T3 = 500°C s 3 = 6.8000 kJ/kg ⋅ K s 4 − s f 6.8000 − 0.6492 P4 = 10 kPa = = 0.8201 x4 = s 4 = s3 s fg 7.4996
h4 s = h f + x 4 h fg = 191.81 + (0.820 )(2392.1) = 2153.6 kJ/kg
ηT = Thus,
h3 − h4 → h4 = h3 − ηT (h3 − h4 s ) h3 − h4 s = 3411.4 − (0.87 )(3411.4 − 2153.6) = 2317.1 kJ/kg
qin = h3 − h2 = 3411.4 − 199.92 = 3211.5 kJ/kg qout = h4 − h1 = 2317.1 − 191.81 = 2125.3 kJ/kg wnet = qin − qout = 3211.5 − 2125.3 = 1086.2 kJ/kg
and
η th =
(b)
m& =
wnet 1086.2 kJ/kg = = 33.8% q in 3211.5 kJ/kg
W&net 45,000 kJ/s = = 41.43 kg/s wnet 1086.2 kJ/kg
(c) The rate of heat rejection to the cooling water and its temperature rise are Q& out = m& q out = (41.43 kg/s )(2125.3 kJ/kg ) = 88,051 kJ/s ∆Tcooling water =
Q& out (m& c) cooling water
=
88,051 kJ/s = 10.5°C (2000 kg/s )(4.18 kJ/kg ⋅ °C)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-13
10-24 The net work outputs and the thermal efficiencies for a Carnot cycle and a simple ideal Rankine cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Rankine cycle analysis: From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg
v1 = v f @ 20 kPa = 0.001017 m 3 /kg w p ,in = v1 (P2 − P1 )
(
T
Rankine cycle
)
1 kJ = 0.001017 m3 /kg (10,000 − 20 ) kPa 3 1 kPa m ⋅ = 10.15 kJ/kg
h2 = h1 + w p ,in = 251.42 + 10.15 = 261.57 kJ/kg
3 2
P3 = 10 MPa h3 = 2725.5 kJ/kg x3 = 1 s 3 = 5.6159 kJ/kg ⋅ K
4
1
s
s4 − s f P4 = 20 kPa 5.6159 − 0.8320 = = 0.6761 x4 = s 4 = s3 7.0752 s fg
h4 = h f + x 4 h fg = 251.42 + (0.6761)(2357.5) = 1845.3 kJ/kg
q in = h3 − h2 = 2725.5 − 261.57 = 2463.9 kJ/kg q out = h4 − h1 = 1845.3 − 251.42 = 1594.0 kJ/kg wnet = q in − q out = 2463.9 − 1594.0 = 869.9 kJ/kg
η th = 1 −
q out 1594.0 = 1− = 0.353 2463.9 q in
(b) Carnot Cycle analysis: P3 = 10 MPa h3 = 2725.5 kJ/kg x3 = 1 T3 = 311.0 °C T2 = T3 = 311.0 °C h2 = 1407.8 kJ/kg x2 = 0 s 2 = 3.3603 kJ/kg ⋅ K x1 =
s1 − s f
= P1 = 20 kPa s fg h = h +x h s1 = s 2 f 1 fg 1
3.3603 − 0.8320 = 0.3574 7.0752
T
Carnot cycle 2
3
1
4 s
= 251.42 + (0.3574)(2357.5) = 1093.9 kJ/kg q in = h3 − h2 = 2725.5 − 1407.8 = 1317.7 kJ/kg q out = h4 − h1 = 1845.3 − 1093.9 = 751.4 kJ/kg wnet = q in − q out = 1317.7 − 752.3 = 565.4 kJ/kg
η th = 1 −
q out 751.4 = 1− = 0.430 1317.7 q in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-14
10-25 A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Properties The specific heat of geothermal water is taken to be 4.18 kJ/kg.ºC. Analysis (a) We need properties of isobutane, which are not available in the book. However, we can obtain the properties from EES. Turbine:
4
P3 = 3250 kPa h3 = 761.54 kJ/kg T3 = 147°C s3 = 2.5457 kJ/kg ⋅ K
h3 − h4 761.54 − 689.74 = = 0.788 h3 − h4 s 761.54 − 670.40
1
turbine pump
P4 = 410 kPa h4 s = 670.40 kJ/kg s 4 = s3 P4 = 410 kPa h4 = 689.74 kJ/kg T4 = 179.5°C
ηT =
air-cooled condenser
3
2 heat exchanger
Geothermal water in
Geothermal water out T
(b) Pump: h1 = h f @ 410 kPa = 273.01 kJ/kg 3.25 MPa
v1 = v f @ 410 kPa = 0.001842 m3 /kg w p ,in = v1 (P2 − P1 ) / η P
(
2s
)
1 kJ / 0.90 = 0.001842 m3 /kg (3250 − 410 ) kPa 1 kPa ⋅ m3 = 5.81 kJ/kg
3
qin
2
410 kPa 1
qout
4s 4
h2 = h1 + w p ,in = 273.01 + 5.81 = 278.82 kJ/kg W& T,out = m& (h3 − h4 ) = (305.6 kJ/kg)(761.54 − 689.74)kJ/kg = 21,941 kW
W& P,in = m& (h2 − h1 ) = m& wp,in = (305.6 kJ/kg)(5.81 kJ/kg) = 1777 kW W& net = W& T,out − W& P,in = 21,941 − 1777 = 20,165 kW
Heat Exchanger: Q& in = m& geocgeo (Tin − Tout ) = (555.9 kJ/kg)(4.18 kJ/kg.°C)(160 − 90)°C = 162,656 kW
(c)
η th =
W& net 20,165 = = 0.124 = 12.4% & 162,656 Qin
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-15
10-26 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of steam through the turbine, the isentropic efficiency of the turbine, the power output from the turbine, and the thermal efficiency of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6) T1 = 230°C h1 = 990.14 kJ/kg x1 = 0 h2 − h f P2 = 500 kPa 990.14 − 640.09 = = 0.1661 x2 = h2 = h1 = 990.14 kJ/kg h fg 2108
3 steam turbine
The mass flow rate of steam through the turbine is m& 3 = x 2 m& 1 = (0.1661)(230 kg/s) = 38.20 kg/s
2
condenser
(b) Turbine: 6
P3 = 500 kPa h3 = 2748.1 kJ/kg x3 = 1 s 3 = 6.8207 kJ/kg ⋅ K P4 = 10 kPa h4 s = 2160.3 kJ/kg
s 4 = s3
Flash chamber production well
1
P4 = 10 kPa h4 = h f + x 4 h fg = 191.81 + (0.90)(2392.1) = 2344.7 kJ/kg x 4 = 0.90
ηT =
4
separator
5
reinjection well
h3 − h4 2748.1 − 2344.7 = = 0.686 h3 − h4 s 2748.1 − 2160.3
(c) The power output from the turbine is W& T,out = m& 3 (h3 − h4 ) = (38.20 kJ/kg)(2748.1 − 2344.7)kJ/kg = 15,410 kW
(d) We use saturated liquid state at the standard temperature for dead state enthalpy T0 = 25°C h0 = 104.83 kJ/kg x0 = 0 E& in = m& 1 (h1 − h0 ) = (230 kJ/kg)(990.14 − 104.83)kJ/kg = 203,622 kW
η th =
W& T,out 15,410 = = 0.0757 = 7.6% & 203,622 E in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-16
10-27 A double-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The temperature of the steam at the exit of the second flash chamber, the power produced from the second turbine, and the thermal efficiency of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6) T1 = 230°C h1 = 990.14 kJ/kg x1 = 0 P2 = 500 kPa x 2 = 0.1661 h2 = h1 = 990.14 kJ/kg
3
m& 3 = x2 m& 1 = (0.1661)(230 kg/s) = 38.20 kg/s m& 6 = m& 1 − m& 3 = 230 − 0.1661 = 191.80 kg/s P3 = 500 kPa h3 = 2748.1 kJ/kg x3 = 1 P4 = 10 kPa h4 = 2344.7 kJ/kg x 4 = 0.90
8 4
separator
2 6
P6 = 500 kPa h6 = 640.09 kJ/kg x6 = 0 P7 = 150 kPa T7 = 111.35 °C h7 = h6 x 7 = 0.0777
steam turbine
Flash chamber
production well
7
condenser separator
Flash chamber
1
5
9
reinjection well
P8 = 150 kPa h8 = 2693.1 kJ/kg x8 = 1
(b) The mass flow rate at the lower stage of the turbine is m& 8 = x7 m& 6 = (0.0777)(191.80 kg/s) = 14.90 kg/s
The power outputs from the high and low pressure stages of the turbine are W&T1, out = m& 3 (h3 − h4 ) = (38.20 kJ/kg)(2748.1 − 2344.7)kJ/kg = 15,410 kW W&T2, out = m& 8 (h8 − h4 ) = (14.90 kJ/kg)(2693.1 − 2344.7)kJ/kg = 5191 kW
(c) We use saturated liquid state at the standard temperature for the dead state enthalpy T0 = 25°C h0 = 104.83 kJ/kg x0 = 0 E& in = m& 1 (h1 − h0 ) = (230 kg/s)(990.14 − 104.83)kJ/kg = 203,621 kW
η th =
W& T, out 15,410 + 5193 = = 0.101 = 10.1% 203,621 E& in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-17
10-28 A combined flash-binary geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of isobutane in the binary cycle, the net power outputs from the steam turbine and the binary cycle, and the thermal efficiencies for the binary cycle and the combined plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6) T1 = 230°C h1 = 990.14 kJ/kg x1 = 0 P2 = 500 kPa x 2 = 0.1661 h2 = h1 = 990.14 kJ/kg m& 3 = x2 m& 1 = (0.1661)(230 kg/s) = 38.20 kg/s m& 6 = m& 1 − m& 3 = 230 − 38.20 = 191.80 kg/s P3 = 500 kPa h3 = 2748.1 kJ/kg x3 = 1
3 separator
P4 = 10 kPa h4 = 2344.7 kJ/kg x 4 = 0.90 P6 = 500 kPa h6 = 640.09 kJ/kg x6 = 0 T7 = 90°C h7 = 377.04 kJ/kg x7 = 0
condenser
4
9
1
isobutane turbine
2
BINARY CYCLE
8
pump
heat exchanger flash chamber
1
5
air-cooled condenser
6
The isobutene properties are obtained from EES: P8 = 3250 kPa h8 = 755.05 kJ/kg T8 = 145°C P9 = 400 kPa h9 = 691.01 kJ/kg T9 = 80°C
steam turbine
1
7
production well
reinjection well
P10 = 400 kPa h10 = 270.83 kJ/kg 3 x10 = 0 v 10 = 0.001839 m /kg w p ,in = v10 (P11 − P10 ) / η p
(
)
1 kJ / 0.90 = 0.001819 m3 /kg (3250 − 400 ) kPa 1 kPa ⋅ m3 = 5.82 kJ/kg.
h11 = h10 + w p ,in = 270.83 + 5.82 = 276.65 kJ/kg
An energy balance on the heat exchanger gives m& 6 (h6 − h7 ) = m& iso (h8 − h11 ) (191.81 kg/s)(640.09 - 377.04)kJ/kg = m& iso (755.05 - 276.65)kJ/kg → m& iso = 105.46 kg/s
(b) The power outputs from the steam turbine and the binary cycle are W&T,steam = m& 3 (h3 − h4 ) = (38.19 kJ/kg)(2748.1 − 2344.7)kJ/kg = 15,410 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-18
W& T,iso = m& iso (h8 − h9 ) = (105.46 kJ/kg)(755.05 − 691.01)kJ/kg = 6753 kW W& net,binary = W& T,iso − m& iso w p ,in = 6753 − (105.46 kg/s)(5.82 kJ/kg ) = 6139 kW
(c) The thermal efficiencies of the binary cycle and the combined plant are Q& in, binary = m& iso (h8 − h11 ) = (105.46 kJ/kg)(755.05 − 276.65)kJ/kg = 50,454 kW
η th,binary =
W& net, binary 6139 = = 0.122 = 12.2% & 50 ,454 Qin, binary
T0 = 25°C h0 = 104.83 kJ/kg x0 = 0 E& in = m& 1 (h1 − h0 ) = (230 kJ/kg)(990.14 − 104.83)kJ/kg = 203,622 kW
η th,plant =
W& T,steam + W& net, binary 15,410 + 6139 = = 0.106 = 10.6% 203,622 E& in
The Reheat Rankine Cycle 10-29C The pump work remains the same, the moisture content decreases, everything else increases. 10-30C The T-s diagram of the ideal Rankine cycle with 3 stages of reheat is shown on the side. The cycle efficiency will increase as the number of reheating stages increases.
T 3
5
7 9 II III
6
8
I 4 2 1
10
s
10-31C The thermal efficiency of the simple ideal Rankine cycle will probably be higher since the average temperature at which heat is added will be higher in this case.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-19
10-32 [Also solved by EES on enclosed CD] A steam power plant that operates on the ideal reheat Rankine cycle is considered. The turbine work output and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg
T 3
v1 = v f @ 20 kPa = 0.001017 m3 /kg w p ,in = v1 (P2 − P1 )
(
)
1 kJ = 0.001017 m3 /kg (8000 − 20 kPa ) 1 kPa ⋅ m3 = 8.12 kJ/kg
h2 = h1 + w p ,in = 251.42 + 8.12 = 259.54 kJ/kg
5
8 MPa 4 2 20 kPa 1
6
P3 = 8 MPa h3 = 3399.5 kJ/kg T3 = 500°C s3 = 6.7266 kJ/kg ⋅ K P4 = 3 MPa h4 = 3105.1 kJ/kg s4 = s3 P5 = 3 MPa h5 = 3457.2 kJ/kg T5 = 500°C s5 = 7.2359 kJ/kg ⋅ K s6 − s f 7.2359 − 0.8320 = = 0.9051 P6 = 20 kPa x6 = s fg 7.0752 s6 = s5 h6 = h f + x6 h fg = 251.42 + (0.9051)(2357.5) = 2385.2 kJ/kg
The turbine work output and the thermal efficiency are determined from wT,out = (h3 − h4 ) + (h5 − h6 ) = 3399.5 − 3105.1 + 3457.2 − 2385.2 = 1366.4 kJ/kg
and
q in = (h3 − h2 ) + (h5 − h4 ) = 3399.5 − 259.54 + 3457.2 − 3105.1 = 3492.0 kJ/kg
wnet = wT ,out − w p ,in = 1366.4 − 8.12 = 1358.3 kJ/kg
Thus,
η th =
wnet 1358.3 kJ/kg = = 38.9% 3492.5 kJ/kg q in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-20
10-33 EES Problem 10-32 is reconsidered. The problem is to be solved by the diagram window data entry feature of EES by including the effects of the turbine and pump efficiencies and reheat on the steam quality at the low-pressure turbine exit Also, the T-s diagram is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "Input Data - from diagram window" {P[6] = 20 [kPa] P[3] = 8000 [kPa] T[3] = 500 [C] P[4] = 3000 [kPa] T[5] = 500 [C] Eta_t = 100/100 "Turbine isentropic efficiency" Eta_p = 100/100 "Pump isentropic efficiency"} "Pump analysis" function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$='' if (x6>1) then x6$='(superheated)' if (x6 h g (superheated vapor )
From steam tables at 10 kPa we read T6 = 88.1°C. (b)
wT, out = (h3 − h4 ) + (h5 − h6 ) = 3375.1 − 2902.0 + 3479.1 − 2664.8 = 1287.4 kJ/kg qin = (h3 − h2 ) + (h5 − h4 ) = 3375.1 − 202.43 + 3479.1 − 2902.0 = 3749.8 kJ/kg wnet = wT, out − wp,in = 1287.4 − 10.62 = 1276.8 kJ/kg
Thus the thermal efficiency is
η th =
wnet 1276.8 kJ/kg = = 34.1% 3749.8 kJ/kg q in
(c) The mass flow rate of the steam is W& 80,000 kJ/s = 62.7 kg/s m& = net = wnet 1276.9 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-23
10-35 A steam power plant that operates on the ideal reheat Rankine cycle is considered. The quality (or temperature, if superheated) of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg
T
v 1 = v f @ 10 kPa = 0.00101 m 3 /kg w p ,in = v 1 (P2 − P1 )
(
)
1 kJ = 0.00101 m 3 /kg (10,000 − 10 kPa ) 1 kPa ⋅ m 3 = 10.09 kJ/kg
3
10 MPa 4 2
h2 = h1 + w p ,in = 191.81 + 10.09 = 201.90 kJ/kg P3 = 10 MPa h3 = 3375.1 kJ/kg T3 = 500°C s3 = 6.5995 kJ/kg ⋅ K
5
10 kPa 1
6
P4 = 1 MPa h4 = 2783.8 kJ/kg s4 = s3 P5 = 1 MPa h5 = 3479.1 kJ/kg T5 = 500°C s5 = 7.7642 kJ/kg ⋅ K s6 − s f 7.7642 − 0.6492 = = 0.9487 (at turbine exit ) P6 = 10 kPa x6 = 7.4996 s fg s6 = s5 h6 = h f + x6 h fg = 191.81 + (0.9487 )(2392.1) = 2461.2 kJ/kg
(b)
wT, out = (h3 − h4 ) + (h5 − h6 ) = 3375.1 − 2783.7 + 3479.1 − 2461.2 = 1609.3 kJ/kg qin = (h3 − h2 ) + (h5 − h4 ) = 3375.1 − 201.90 + 3479.1 − 2783.7 = 3868.5 kJ/kg wnet = wT, out − w p ,in = 1609.4 − 10.09 = 1599.3 kJ/kg
Thus the thermal efficiency is
η th =
wnet 1599.3 kJ/kg = = 41.3% 3868.5 kJ/kg q in
(c) The mass flow rate of the steam is W& 80,000 kJ/s m& = net = = 50.0 kg/s wnet 1599.3 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-24
10-36E A steam power plant that operates on the ideal reheat Rankine cycle is considered. The pressure at which reheating takes place, the net power output, the thermal efficiency, and the minimum mass flow rate of the cooling water required are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = hsat @ 1 psia = 69.72 Btu/lbm
T
v 1 = v sat @ 1 psia = 0.01614 ft 3 /lbm
3
T1 = Tsat @ 1 psia = 101.69°F w p ,in = v 1 (P2 − P1 )
(
5
800 psia 4
)
1 Btu = 0.01614 ft 3 /lbm (800 − 1 psia ) 5.4039 psia ⋅ ft 3 = 2.39 Btu/lbm
2 1 psia 1
h2 = h1 + w p ,in = 69.72 + 2.39 = 72.11 Btu/lbm
6
s
P3 = 800 psia h 3 = 1456 . 0 Btu/lbm T 3 = 900 ° F s 3 = 1 . 6413 Btu/lbm ⋅ R h 4 = h g @ s g = s 4 = 1178 . 5 Btu/lbm (sat. vapor ) P4 = Psat @ s g = s 4 = 62.23 psia (the reheat pressure) s4 = s3
P5 = 62 . 23 psia h 5 = 1431 . 4 Btu/lbm T 5 = 800 ° F s 5 = 1 . 8985 Btu/lbm ⋅ R s6 − s f 1 . 8985 − 0 . 13262 = = 0 . 9572 P6 = 1 psia x 6 = s 1 . 84495 fg s6 = s5 h 6 = h f + x 6 h fg = 69 . 72 + (0 . 9572 )(1035 . 7 ) = 1061 . 0 Btu/lbm
(b)
q in = (h3 − h2 ) + (h5 − h4 ) = 1456.0 − 72.11 + 1431.4 − 1178.5 = 1636.8 Btu/lbm q out = h6 − h1 = 1061.0 − 69.72 = 991.3 Btu/lbm
Thus,
η th = 1 −
q out 991.3 Btu/lbm = 1− = 39.4% 1636.8 Btu/lbm q in
(c) The mass flow rate of the cooling water will be minimum when it is heated to the temperature of the steam in the condenser, which is 101.7°F,
(
)
Q& out = Q& in − W& net = (1 − η th )Q& in = (1 − 0.3943) 6 × 10 4 Btu/s = 3.634 × 10 4 Btu/s Q& 3.634 × 10 4 Btu/s = 641.0 lbm/s m& cool = out = c∆T (1.0 Btu/lbm ⋅ °F)(101.69 − 45)°F
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-25
10-37 A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressure limits is considered. The pressure at which reheating takes place, the total rate of heat input in the boiler, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = hsat @ 10 kPa = 191.81 kJ/kg
T
v 1 = v sat @ 10 kPa = 0.00101 m 3 /kg w p ,in = v 1 (P2 − P1 )
(
)
1 kJ = 0.00101 m /kg (15,000 − 10 kPa ) 1 kPa ⋅ m 3 = 15.14 kJ/kg 3
3
h2 = h1 + w p ,in = 191.81 + 15.14 = 206.95 kJ/kg
5
15 MPa 4 2 10 kPa 1
6
P3 = 15 MPa h3 = 3310.8 kJ/kg T3 = 500°C s3 = 6.3480 kJ/kg ⋅ K P6 = 10 kPa h6 = h f + x6 h fg = 191.81 + (0.90 )(2392.1) = 2344.7 kJ/kg s6 = s5 s6 = s f + x6 s fg = 0.6492 + (0.90 )(7.4996) = 7.3988 kJ/kg ⋅ K T5 = 500°C P5 = 2161 kPa (the reheat pressure) s5 = s6 h5 = 3466.53 kJ/kg P4 = 2.161 MPa h4 = 2817.2 kJ/kg s4 = s3
(b) The rate of heat supply is Q& in = m& [(h3 − h2 ) + (h5 − h4 )]
= (12 kg/s )(3310.8 − 206.95 + 3466.53 − 2817.2 )kJ/kg = 45,038 kW
(c) The thermal efficiency is determined from Thus,
Q& out = m& (h6 − h1 ) = (12 kJ/s )(2344.7 − 191.81)kJ/kg = 25,835 kJ/s
η th = 1 −
Q& out 25,834 kJ/s = 1− = 42.6% & 45,039 kJ/s Q in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-26
10-38 A steam power plant that operates on a reheat Rankine cycle is considered. The condenser pressure, the net power output, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), P3 = 12.5 MPa h3 = 3476.5 kJ/kg T3 = 550°C s3 = 6.6317 kJ/kg ⋅ K P4 = 2 MPa h4 s = 2948.1 kJ/kg s4 s = s3 ηT =
3 Turbine
Boiler 4
6
h3 − h4 h3 − h4 s
5 2
→h4 = h3 − ηT (h3 − h4 s )
= 3476.5 − (0.85)(3476.5 − 2948.1) = 3027.3 kJ/kg
Condenser
Pump 1
T
P5 = 2 MPa h5 = 3358.2 kJ/kg T5 = 450°C s5 = 7.2815 kJ/kg ⋅ K 12.5 MPa
P6 = ?
h6 = x6 = 0.95
4s 2s
P6 = ? h6 s = s6 = s5
5
3 4
2
P=?
1
h −h ηT = 5 6 → h6 = h5 − ηT (h5 − h6 s ) h5 − h6 s = 3358.2 − (0.85)(3358.2 − 2948.1) = 3027.3 kJ/kg
6s 6
s
The pressure at state 6 may be determined by a trial-error approach from the steam tables or by using EES from the above equations: P6 = 9.73 kPa, h6 = 2463.3 kJ/kg, (b) Then, h1 = h f @ 9.73 kPa = 189.57 kJ/kg
v1 = v f @ 10 kPa = 0.001010 m3 /kg w p ,in = v1 (P2 − P1 ) / η p
(
)
1 kJ / (0.90 ) = 0.00101 m3 /kg (12,500 − 9.73 kPa ) 3 1 kPa ⋅ m = 14.02 kJ/kg
h2 = h1 + w p ,in = 189.57 + 14.02 = 203.59 kJ/kg
Cycle analysis: q in = (h3 − h2 ) + (h5 − h4 ) = 3476.5 − 3027.3 + 3358.2 − 2463.3 = 3603.8 kJ/kg q out = h6 − h1 = 3027.3 − 189.57 = 2273.7 kJ/kg W& net = m& (q in − q out ) = (7.7 kg/s)(3603.8 - 2273.7)kJ/kg = 10,242 kW
(c) The thermal efficiency is q 2273.7 kJ/kg η th = 1 − out = 1 − = 0.369 = 36.9% 3603.8 kJ/kg q in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-27
Regenerative Rankine Cycle 10-39C Moisture content remains the same, everything else decreases. 10-40C This is a smart idea because we waste little work potential but we save a lot from the heat input. The extracted steam has little work potential left, and most of its energy would be part of the heat rejected anyway. Therefore, by regeneration, we utilize a considerable amount of heat by sacrificing little work output. 10-41C In open feedwater heaters, the two fluids actually mix, but in closed feedwater heaters there is no mixing. 10-42C Both cycles would have the same efficiency. 10-43C To have the same thermal efficiency as the Carnot cycle, the cycle must receive and reject heat isothermally. Thus the liquid should be brought to the saturated liquid state at the boiler pressure isothermally, and the steam must be a saturated vapor at the turbine inlet. This will require an infinite number of heat exchangers (feedwater heaters), as shown on the T-s diagram.
T Boiler inlet qin
Boiler exit
qout
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-28
10-44 A steam power plant that operates on an ideal regenerative Rankine cycle with an open feedwater heater is considered. The net work output per kg of steam and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg T
v 1 = v f @ 20 kPa = 0.001017 m 3 /kg w pI ,in = v 1 (P2 − P1 )
(
)
1 kJ = 0.001017 m 3 /kg (400 − 20 kPa ) 1 kPa ⋅ m 3 = 0.39 kJ/kg
5
h2 = h1 + w pI ,in = 251.42 + 0.39 = 251.81 kJ/kg P3 = 0.4 MPa h3 = h f @ 0.4 MPa = 604.66 kJ/kg 3 sat.liquid v 3 = v f @ 0.4 MPa = 0.001084 m /kg
(
qin
)
1 kJ w pII ,in = v 3 (P4 − P3 ) = 0.001084 m 3 /kg (6000 − 400 kPa ) 1 kPa ⋅ m 3 h4 = h3 + w pII ,in = 604.66 + 6.07 = 610.73 kJ/kg
4
6 MPa y
3 0.4 MPa
6
2
1-y
20 kPa 1
qout
7
= 6.07 kJ/kg
P5 = 6 MPa h5 = 3302.9 kJ/kg T5 = 450°C s 5 = 6.7219 kJ/kg ⋅ K s6 − s f 6.7219 − 1.7765 = = 0.9661 P6 = 0.4 MPa x 6 = 5.1191 s fg s 6 = s5 h6 = h f + x 6 h fg = 604.66 + (0.9661)(2133.4) = 2665.7 kJ/kg s7 − s f 6.7219 − 0.8320 = = 0.8325 P7 = 20 kPa x 7 = 7.0752 s fg s7 = s5 h7 = h f + x 7 h fg = 251.42 + (0.8325)(2357.5) = 2214.0 kJ/kg
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heater. Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 , E& in − E& out = ∆E& system ©0 (steady) = 0 E& in = E& out
∑ m& h = ∑ m& h i i
e e
→ m& 6 h6 + m& 2 h2 = m& 3h3 → yh6 + (1 − y )h2 = 1(h3 )
where y is the fraction of steam extracted from the turbine ( = m& 6 / m& 3 ). Solving for y, y=
h3 − h2 604.66 − 251.81 = 0.1462 = h6 − h2 2665.7 − 251.81
Then, q in = h5 − h4 = 3302.9 − 610.73 = 2692.2 kJ/kg
q out = (1 − y )(h7 − h1 ) = (1 − 0.1462)(2214.0 − 251.42 ) = 1675.7 kJ/kg
And wnet = q in − q out = 2692.2 − 1675.7 = 1016.5 kJ/kg (b) The thermal efficiency is determined from q 1675.7 kJ/kg η th = 1 − out = 1 − = 37.8% 2692.2 kJ/kg q in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-29
10-45 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwater heater is considered. The net work output per kg of steam and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg T
v 1 = v f @ 20 kPa = 0.001017 m 3 /kg w pI ,in = v 1 (P2 − P1 )
(
5
)
1 kJ = 0.001017 m /kg (6000 − 20 kPa ) 1 kPa ⋅ m 3 = 6.08 kJ/kg 3
qin 9 84
2
(
0.4 MPa
y 6 1-y
3
h2 = h1 + w pI ,in = 251.42 + 6.08 = 257.50 kJ/kg P3 = 0.4 MPa h3 = h f @ 0.4 MPa = 604.66 kJ/kg 3 sat. liquid v 3 = v f @ 0.4 MPa = 0.001084 m /kg
6 MPa
20 kPa 1
qou
7 s
)
1 kJ = 6.07 kJ/kg w pII ,in = v 3 (P9 − P3 ) = 0.001084 m3/kg (6000 − 400 kPa ) 3 ⋅ 1 kPa m h9 = h3 + w pII ,in = 604.66 + 6.07 = 610.73 kJ/kg
h8 = h3 + v 3 (P8 − P3 ) = h9 = 610.73 kJ/kg Also, h4 = h9 = h8¸ = 610.73 kJ/kg since the two fluid streams which are being mixed have the same enthalpy. P5 = 6 MPa h5 = 3302.9 kJ/kg T5 = 450°C s 5 = 6.7219 kJ/kg ⋅ K s 6 − s f 6.7219 − 1.7765 = = 0.9661 P6 = 0.4 MPa x 6 = s fg 5.1191 s 6 = s5 h6 = h f + x 6 h fg = 604.66 + (0.9661)(2133.4) = 2665.7 kJ/kg s 7 − s f 6.7219 − 0.8320 = = 0.8325 P7 = 20 kPa x 7 = s fg 7.0752 s 7 = s5 h7 = h f + x 7 h fg = 251.42 + (0.8325)(2357.5) = 2214.0 kJ/kg
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heater. Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 , ©0 (steady) E& − E& = ∆E& =0 in
out
system
E& in = E& out
∑ m& h = ∑ m& h i i
e e
→ m& 2 (h8 − h2 ) = m& 6 (h6 − h3 ) → (1 − y )(h8 − h2 ) = y (h6 − h3 )
where y is the fraction of steam extracted from the turbine ( = m& 6 / m& 5 ). Solving for y, y=
Then,
h8 − h2
(h6 − h3 ) + (h8 − h2 )
=
610.73 − 257.50 = 0.1463 2665.7 − 604.66 + 610.73 − 257.50
q in = h5 − h4 = 3302.9 − 610.73 = 2692.2 kJ/kg
q out = (1 − y )(h7 − h1 ) = (1 − 0.1463)(2214.0 − 251.42) = 1675.4 kJ/kg
And wnet = q in − q out = 2692.2 − 1675.4 = 1016.8 kJ/kg (b) The thermal efficiency is determined from q 1675.4 kJ/kg η th = 1 − out = 1 − = 37.8% 2692.2 kJ/kg qin PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-30
10-46 A steam power plant operates on an ideal regenerative Rankine cycle with two open feedwater heaters. The net power output of the power plant and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis 7 Turbine
Boiler
T 7
8
6 10 MPa
10
6
9 fwh
Condenser
fwh I 4
5
P III
P II
3
0.2 MPa
y
5 kPa
2 3
2
8
5 0.6 MPa
4
1
1
1-y 9 1-y-z 10 s
PI
(a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 5 kPa = 137.75 kJ/kg
v 1 = v f @ 5 kPa = 0.001005 m 3 /kg
(
)
1 kJ w pI ,in = v 1 (P2 − P1 ) = 0.001005 m 3 /kg (200 − 5 kPa ) 1 kPa ⋅ m 3 h =h +w = 137.75 + 0.20 = 137.95 kJ/kg 2
1
pI ,in
= 0.20 kJ/kg
P3 = 0.2 MPa h3 = h f @ 0.2 MPa = 504.71 kJ/kg 3 sat.liquid v 3 = v f @ 0.2 MPa = 0.001061 m /kg
(
)
1 kJ w pII ,in = v 3 (P4 − P3 ) = 0.001061 m 3 /kg (600 − 200 kPa ) 1 kPa ⋅ m 3 = 0.42 kJ/kg
h4 = h3 + w pII ,in = 504.71 + 0.42 = 505.13 kJ/kg P5 = 0.6 MPa h5 = h f @ 0.6 MPa = 670.38 kJ/kg 3 sat.liquid v 5 = v f @ 0.6 MPa = 0.001101 m /kg
(
)
1 kJ w pIII ,in = v 5 (P6 − P5 ) = 0.001101 m 3 /kg (10,000 − 600 kPa ) 1 kPa ⋅ m 3 = 10.35 kJ/kg
h6 = h5 + w pIII ,in = 670.38 + 10.35 = 680.73 kJ/kg P7 = 10 MPa h7 = 3625.8 kJ/kg T7 = 600°C s 7 = 6.9045 kJ/kg ⋅ K P8 = 0.6 MPa h8 = 2821.8 kJ/kg s8 = s 7 x9 =
s9 − s f
=
6.9045 − 1.5302 = 0.9602 5.5968
s fg P9 = 0.2 MPa h9 = h f + x 9 h fg = 504.71 + (0.9602)(2201.6) s9 = s7 = 2618.7 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-31 s10 − s f 6.9045 − 0.4762 = = 0.8119 P10 = 5 kPa x10 = s fg 7.9176 s10 = s 7 h10 = h f + x10 h fg = 137.75 + (0.8119)(2423.0) = 2105.0 kJ/kg
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 , E& in − E& out = ∆E& system ©0 (steady) = 0 E& in = E& out
FWH-2:
∑ m& h = ∑ m& h i i
e e
→ m& 8 h8 + m& 4 h4 = m& 5 h5 → yh8 + (1 − y )h4 = 1(h5 )
&8 / m & 5 ). Solving for y, where y is the fraction of steam extracted from the turbine ( = m y=
FWH-1:
h5 − h4 670.38 − 505.13 = = 0.07133 h8 − h4 2821.8 − 505.13
∑ m& h = ∑ m& h i i
e e
→ m& 9 h9 + m& 2 h2 = m& 3 h3 → zh9 + (1 − y − z )h2 = (1 − y )h3
&9 / m & 5 ) at the second stage. Solving for z, where z is the fraction of steam extracted from the turbine ( = m z=
h3 − h2 (1 − y ) = 504.71 − 137.95 (1 − 0.07136) = 0.1373 h9 − h2 2618.7 − 137.95
Then, qin = h7 − h6 = 3625.8 − 680.73 = 2945.0 kJ/kg
qout = (1 − y − z )(h10 − h1 ) = (1 − 0.07133 − 0.1373)(2105.0 − 137.75) = 1556.8 kJ/kg wnet = qin − qout = 2945.0 − 1556.8 = 1388.2 kJ/kg
and W& net = m& wnet = (22 kg/s )(1388.2 kJ/kg ) = 30,540 kW ≅ 30.5 MW
(b)
η th = 1 −
q out 1556.8 kJ/kg = 1− = 47.1% q in 2945.0 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-32
10-47 [Also solved by EES on enclosed CD] A steam power plant operates on an ideal regenerative Rankine cycle with two feedwater heaters, one closed and one open. The mass flow rate of steam through the boiler for a net power output of 250 MW and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg
8
v 1 = v f @ 10 kPa = 0.00101 m 3 /kg w pI ,in = v 1 (P2 − P1 )
(
Turbine
Boiler
)
1 kJ = 0.00101 m 3 /kg (300 − 10 kPa ) 1 kPa ⋅ m 3 = 0.29 kJ/kg
h2 = h1 + w pI ,in = 191.81 + 0.29 = 192.10 kJ/kg
9
11
5
4
Closed fwh
P3 = 0.3 MPa h3 = h f @ 0.3 MPa = 561.43 kJ/kg 3 sat. liquid v 3 = v f @ 0.3 MPa = 0.001073 m /kg
10
3 P II
Condenser
2 6
(
)
7
T6 = T5 , P5 = 12.5 MPa → h5 = 727.83 kJ/kg P8 = 12.5 MPa h8 = 3476.5 kJ/kg T8 = 550°C s 8 = 6.6317 kJ/kg ⋅ K
1 PI
T
h4 = h3 + w pII ,in = 561.43 + 13.09 = 574.52 kJ/kg h6 = h7 = h f @ 0.8 MPa = 720.87 kJ/kg P6 = 0.8 MPa 3 v 6 = v f @ 0.8 MPa = 0.001115 m /kg sat. liquid T6 = Tsat @ 0.8 MPa = 170.4°C
1-y-z
z
Open fwh
w pII ,in = v 3 (P4 − P3 )
1 kJ = 0.001073 m 3 /kg (12,500 − 300 kPa ) 1 kPa ⋅ m 3 = 13.09 kJ/kg
y
8 12.5 5 MPa 6 0.8 MPa 9 4 y 2
3 7 1
0.3 MPa 10 z 1-y-z 10 kPa 11 s
s9 − s f 6.6317 − 2.0457 = = 0.9935 P9 = 0.8 MPa x 9 = s 4.6160 fg s 9 = s8 h9 = h f + x 9 h fg = 720.87 + (0.9935)(2047.5) = 2755.0 kJ/kg s10 − s f 6.6317 − 1.6717 = = 0.9323 P10 = 0.3 MPa x10 = s 5.3200 fg s10 = s8 h10 = h f + x10 h fg = 561.43 + (0.9323)(2163.5) = 2578.5 kJ/kg s11 − s f 6.6317 − 0.6492 = = 0.7977 P11 = 10 kPa x11 = s fg 7.4996 s11 = s8 h11 = h f + x11h fg = 191.81 + (0.7977 )(2392.1) = 2100.0 kJ/kg
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 , E& in − E& out = ∆E& system ©0 (steady) = 0 E& in = E& out
∑ m& h = ∑ m& h i i
e e
→ m& 9 (h9 − h6 ) = m& 5 (h5 − h4 ) → y (h9 − h6 ) = (h5 − h4 )
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-33 where y is the fraction of steam extracted from the turbine ( = m& 10 / m& 5 ). Solving for y, y=
h5 − h4 727.83 − 574.52 = = 0.0753 h9 − h6 2755.0 − 720.87
For the open FWH, E& in − E& out = ∆E& system ©0 (steady) = 0 E& in = E& out
∑ m& h = ∑ m& h i i
e e
→ m& 7 h7 + m& 2 h2 + m& 10 h10 = m& 3 h3 → yh7 + (1 − y − z )h2 + zh10 = (1)h3
&9 / m & 5 ) at the second stage. Solving for z, where z is the fraction of steam extracted from the turbine ( = m
z=
(h3 − h2 ) − y(h7 − h2 ) = 561.43 − 192.10 − (0.0753)(720.87 − 192.10) = 0.1381 h10 − h2
2578.5 − 192.10
Then, qin = h8 − h5 = 3476.5 − 727.36 = 2749.1 kJ/kg
qout = (1 − y − z )(h11 − h1 ) = (1 − 0.0753 − 0.1381)(2100.0 − 191.81) = 1500.1 kJ/kg wnet = qin − qout = 2749.1 − 1500.1 = 1249 kJ/kg
and m& =
(b)
W&net 250,000 kJ/s = = 200.2 kg/s wnet 1249 kJ/kg
η th = 1 −
q out 1500.1 kJ/kg = 1− = 45.4% q in 2749.1 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-34
10-48 EES Problem 10-47 is reconsidered. The effects of turbine and pump efficiencies on the mass flow rate and thermal efficiency are to be investigated. Also, the T-s diagram is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P[8] = 12500 [kPa] T[8] = 550 [C] P[9] = 800 [kPa] "P_cfwh=300 [kPa]" P[10] = P_cfwh P_cond=10 [kPa] P[11] = P_cond W_dot_net=250 [MW]*Convert(MW, kW) Eta_turb= 100/100 "Turbine isentropic efficiency" Eta_turb_hp = Eta_turb "Turbine isentropic efficiency for high pressure stages" Eta_turb_ip = Eta_turb "Turbine isentropic efficiency for intermediate pressure stages" Eta_turb_lp = Eta_turb "Turbine isentropic efficiency for low pressure stages" Eta_pump = 100/100 "Pump isentropic efficiency" "Condenser exit pump or Pump 1 analysis" Fluid$='Steam_IAPWS' P[1] = P[11] P[2]=P[10] h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid} v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" z*h[10] + y*h[7] + (1-y-z)*h[2] = 1*h[3] "Steady-flow conservation of energy" h[3]=enthalpy(Fluid$,P=P[3],x=0) T[3]=temperature(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" s[3]=entropy(Fluid$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[5]=P[8] P[4] = P[5] P[3]=P[10] v3=volume(Fluid$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[3]+w_pump2= h[4] "Steady-flow conservation of energy" s[4]=entropy(Fluid$,P=P[4],h=h[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) "Closed Feedwater Heater analysis" P[6]=P[9] y*h[9] + 1*h[4] = 1*h[5] + y*h[6] "Steady-flow conservation of energy" h[5]=enthalpy(Fluid$,P=P[6],x=0) "h[5] = h(T[5], P[5]) where T[5]=Tsat at P[9]" PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-35
T[5]=temperature(Fluid$,P=P[5],h=h[5]) "Condensate leaves heater as sat. liquid at P[6]" s[5]=entropy(Fluid$,P=P[6],h=h[5]) h[6]=enthalpy(Fluid$,P=P[6],x=0) T[6]=temperature(Fluid$,P=P[6],x=0) "Condensate leaves heater as sat. liquid at P[6]" s[6]=entropy(Fluid$,P=P[6],x=0) "Trap analysis" P[7] = P[10] y*h[6] = y*h[7] "Steady-flow conservation of energy for the trap operating as a throttle" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) "Boiler analysis" q_in + h[5]=h[8]"SSSF conservation of energy for the Boiler" h[8]=enthalpy(Fluid$, T=T[8], P=P[8]) s[8]=entropy(Fluid$, T=T[8], P=P[8]) "Turbine analysis" ss[9]=s[8] hs[9]=enthalpy(Fluid$,s=ss[9],P=P[9]) Ts[9]=temperature(Fluid$,s=ss[9],P=P[9]) h[9]=h[8]-Eta_turb_hp*(h[8]-hs[9])"Definition of turbine efficiency for high pressure stages" T[9]=temperature(Fluid$,P=P[9],h=h[9]) s[9]=entropy(Fluid$,P=P[9],h=h[9]) ss[10]=s[8] hs[10]=enthalpy(Fluid$,s=ss[10],P=P[10]) Ts[10]=temperature(Fluid$,s=ss[10],P=P[10]) h[10]=h[9]-Eta_turb_ip*(h[9]-hs[10])"Definition of turbine efficiency for Intermediate pressure stages" T[10]=temperature(Fluid$,P=P[10],h=h[10]) s[10]=entropy(Fluid$,P=P[10],h=h[10]) ss[11]=s[8] hs[11]=enthalpy(Fluid$,s=ss[11],P=P[11]) Ts[11]=temperature(Fluid$,s=ss[11],P=P[11]) h[11]=h[10]-Eta_turb_lp*(h[10]-hs[11])"Definition of turbine efficiency for low pressure stages" T[11]=temperature(Fluid$,P=P[11],h=h[11]) s[11]=entropy(Fluid$,P=P[11],h=h[11]) h[8] =y*h[9] + z*h[10] + (1-y-z)*h[11] + w_turb "SSSF conservation of energy for turbine" "Condenser analysis" (1-y-z)*h[11]=q_out+(1-y-z)*h[1]"SSSF First Law for the Condenser" "Cycle Statistics" w_net=w_turb - ((1-y-z)*w_pump1+ w_pump2) Eta_th=w_net/q_in W_dot_net = m_dot * w_net
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-36
ηturb 0.7 0.75 0.8 0.85 0.9 0.95 1
ηturb 0.7 0.75 0.8 0.85 0.9 0.95 1
ηth 0.3916 0.4045 0.4161 0.4267 0.4363 0.4452 0.4535
m [kg/s] 231.6 224.3 218 212.6 207.9 203.8 200.1
0.46
235 ηturb =ηpump
0.45
230
0.44
225
0.43
220 s] g/ k[ 215 m
ht
η
0.42 0.41
210
0.4
205
0.39 0.7
0.75
0.8
0.85
0.9
ηturb
0.95
200 1
Steam
600
8
500 400 ] C [ T
300 200 100 0 0
12500 kPa 800 kPa
5,6
9 10
3,4
7
300 kPa 11 10 kPa
1,2
2
4
6 8 s [kJ/kg-K]
10
12
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-37
10-49 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with an open feedwater heater. The mass flow rate of steam through the boiler and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), T h1 = h f @ 10 kPa = 191.81 kJ/kg
v 1 = v f @ 10 kPa = 0.00101 m 3 /kg
(
5
)
1 kJ w pI ,in = v 1 (P2 − P1 ) = 0.00101 m 3 /kg (800 − 10 kPa ) 1 kPa ⋅ m 3 = 0.80 kJ/kg
4
P3 = 0.8 MPa h3 = h f @ 0.8 MPa = 720.87 kJ/kg 3 sat.liquid v 3 = v f @ 0.8 MPa = 0.001115 m /kg
(
y
3
1-y
10 kPa
8
1
)
1 kJ = v 3 (P4 − P3 ) = 0.001115 m 3 /kg (10,000 − 800 kPa ) 1 kPa ⋅ m 3 = 10.26 kJ/kg
s
h4 = h3 + w pII ,in = 720.87 + 10.26 = 731.12 kJ/kg P5 = 10 MPa h5 = 3502.0 kJ/kg T5 = 550°C s 5 = 6.7585 kJ/kg ⋅ K
7
6
0.8 MPa
2
h2 = h1 + w pI ,in = 191.81 + 0.80 = 192.61 kJ/kg
w pII ,in
10 MPa
5 Turbine
Boiler
P6 = 0.8 MPa h6 = 2812.1 kJ/kg s6 = s5 P7 = 0.8 MPa h7 = 3481.3 kJ/kg T7 = 500°C s 7 = 7.8692 kJ/kg ⋅ K
6 1-y 6 Open fwh
4
P II
s8 − s f 7.8692 − 0.6492 = = 0.9627 P8 = 10 kPa x8 = s fg 7.4996 s8 = s 7 h8 = h f + x8 h fg = 191.81 + (0.9627 )(2392.1) = 2494.7 kJ/kg
7 y
8
Condenser 2
3
1 PI
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 , E& in − E& out = ∆E& system©0 (steady) = 0 → E& in = E& out
∑ m& h = ∑ m& h i i
e e
→ m& 6 h6 + m& 2 h2 = m& 3 h3 → yh6 + (1 − y )h2 = 1(h3 )
&6 / m & 3 ). Solving for y, where y is the fraction of steam extracted from the turbine ( = m y=
Then,
h3 − h2 720.87 − 192.61 = 0.2017 = h6 − h2 2812.1 − 192.61
q in = (h5 − h4 ) + (1 − y )(h7 − h6 ) = (3502.0 − 731.12 ) + (1 − 0.2017 )(3481.3 − 2812.1) = 3305.1 kJ/kg
q out = (1 − y )(h8 − h1 ) = (1 − 0.2017 )(2494.7 − 191.81) = 1838.5 kJ/kg wnet = q in − q out = 3305.1 − 1838.5 = 1466.6 kJ/kg W& net 80,000 kJ/s = = 54.5 kg/s wnet 1466.1 kJ/kg
and
m& =
(b)
η th =
wnet 1466.1 kJ/kg = = 44.4% q in 3305.1 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-38
10-50 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with a closed feedwater heater. The mass flow rate of steam through the boiler and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis
T
5 Turbine
Boiler
4
1-y
Mixing chamber
10
9
5
6 7
y Closed fwh
2
10 MPa 0.8 MPa
3
6 y
7 1-y
Condenser 10 kPa
2
3
P II
8
10 4 9
1
1
8
PI
s
(a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg
v1 = v f @ 10 kPa = 0.00101 m3/kg
(
)
1 kJ w pI ,in = v1 (P2 − P1 ) = 0.00101 m3/kg (10,000 − 10 kPa ) 1 kPa ⋅ m3 = 10.09 kJ/kg h2 = h1 + w pI ,in = 191.81 + 10.09 = 201.90 kJ/kg P3 = 0.8 MPa h3 = h f @ 0.8 MPa = 720.87 kJ/kg 3 sat.liquid v 3 = v f @ 0.8 MPa = 0.001115 m /kg
(
)
1 kJ w pII ,in = v 3 (P4 − P3 ) = 0.001115 m3/kg (10,000 − 800 kPa ) 1 kPa ⋅ m3 = 10.26 kJ/kg h4 = h3 + w pII ,in = 720.87 + 10.26 = 731.13 kJ/kg
Also, h4 = h9 = h10 = 731.12 kJ/kg since the two fluid streams that are being mixed have the same enthalpy. P5 = 10 MPa h5 = 3502.0 kJ/kg T5 = 550°C s5 = 6.7585 kJ/kg ⋅ K P6 = 0.8 MPa h6 = 2812.7 kJ/kg s6 = s5 P7 = 0.8 MPa h7 = 3481.3 kJ/kg T7 = 500°C s7 = 7.8692 kJ/kg ⋅ K s8 − s f 7.8692 − 0.6492 = = 0.9627 P8 = 10 kPa x8 = s fg 7.4996 s8 = s7 h8 = h f + x8h fg = 191.81 + (0.9627 )(2392.1) = 2494.7 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-39
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 , E& in − E& out = ∆E& system©0 (steady) = 0 E& in = E& out
∑ m& h = ∑ m& h i i
e e
→ m& 2 (h9 − h2 ) = m& 3 (h6 − h3 ) → (1 − y )(h9 − h2 ) = y (h6 − h3 )
&3 / m & 4 ). Solving for y, where y is the fraction of steam extracted from the turbine ( = m
y=
h9 − h2
(h6 − h3 ) + (h9 − h2 )
=
731.13 − 201.90 = 0.2019 2812.7 − 720.87 + 731.13 − 201.90
Then, q in = (h5 − h4 ) + (1 − y )(h7 − h6 ) = (3502.0 − 731.13) + (1 − 0.2019 )(3481.3 − 2812.7 ) = 3304.5 kJ/kg
q out = (1 − y )(h8 − h1 ) = (1 − 0.2019 )(2494.7 − 191.81) = 1837.9 kJ/kg wnet = q in − q out = 3304.5 − 1837.8 = 1466.6 kJ/kg
and m& =
(b)
W& net 80,000 kJ/s = = 54.5 kg/s wnet 1467.1 kJ/kg
η th = 1 −
q out 1837.8 kJ/kg = 1− = 44.4% q in 3304.5 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-40
10-51E A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater and two open feedwater heaters. The mass flow rate of steam through the boiler, the net power output of the plant, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis
7
High-P Turbine
Boiler
T
Low-P Turbine
8
6
Open fwh II
P III
10 Open fwh I
P II
11
z 2
5 4
6
9 1-y
y
3
7
12
1-y-z
4 2
Condenser
3 1
1
5
1500
y 8 1 - y 10
250 psia z 11 140 psia 9 40 psia 1-y-z 1 psia 12
PI
(a) From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = h f @ 1 psia = 69.72 Btu/lbm
v 1 = v f @ 1 psia = 0.01614 ft 3 /lbm
w pI ,in = v 1 (P2 − P1 )
(
)
1 Btu = 0.01614 ft 3 /lbm (40 − 1 psia ) 5.4039 psia ⋅ ft 3 = 0.12 Btu/lbm
h2 = h1 + w pI ,in = 69.72 + 0.12 = 69.84 Btu/lbm P3 = 40 psia h3 = h f @ 40 psia = 236.14 Btu/lbm 3 sat. liquid v 3 = v f @ 40 psia = 0.01715 ft /lbm w pII ,in = v 3 (P4 − P3 )
(
)
1 Btu = 0.01715 ft 3 /lbm (250 − 40 psia ) 5.4039 psia ⋅ ft 3 = 0.67 Btu/lbm
h4 = h3 + w pII ,in = 236.14 + 0.67 = 236.81 Btu/lbm P5 = 250 psia h5 = h f @ 250 psia = 376.09 Btu/lbm 3 sat. liquid v 5 = v f @ 250 psia = 0.01865 ft /lbm w pIII ,in = v 5 (P6 − P5 )
(
)
1 Btu = 0.01865 ft 3 /lbm (1500 − 250 psia ) 5.4039 psia ⋅ ft 3 = 4.31 Btu/lbm
h6 = h5 + w pIII ,in = 376.09 + 4.31 = 380.41 Btu/lbm
P7 = 1500 psia h7 = 1550.5 Btu/lbm T7 = 1100°F s 7 = 1.6402 Btu/lbm ⋅ R P8 = 250 psia h8 = 1308.5 Btu/lbm s8 = s 7 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
10-41
P9 = 140 psia h9 = 1248.8 Btu/lbm s9 = s7 P10 = 140 psia h10 = 1531.3 Btu/lbm T10 = 1000°F s10 = 1.8832 Btu/lbm ⋅ R P11 = 40 psia h11 = 1356.0 Btu/lbm s11 = s10 x12 =
s12 − s f
=
1.8832 − 0.13262 = 0.9488 1.84495
s fg P12 = 1 psia h12 = h f + x12 h fg = 69.72 + (0.9488)(1035.7 ) s12 = s10 = 1052.4 Btu/lbm
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 , E& in − E& out = ∆E& system©0 (steady) = 0 E& in = E& out
FWH-2:
∑ m& h = ∑ m& h i i
e e
→ m& 8 h8 + m& 4 h4 = m& 5 h5 → yh8 + (1 − y )h4 = 1(h5 )
&8 / m & 5 ). Solving for y, where y is the fraction of steam extracted from the turbine ( = m
y=
h5 − h4 376.09 − 236.81 = = 0.1300 h8 − h4 1308.5 − 236.81
E& in − E& out = ∆E& system©0 (steady) = 0 E& in = E& out
FWH-1
∑ m& h = ∑ m& h i i
e e
→ m& 11h11 + m& 2 h2 = m& 3 h3 → zh11 + (1 − y − z )h2 = (1 − y )h3
&9 / m & 5 ) at the second stage. Solving for z, where z is the fraction of steam extracted from the turbine ( = m
z=
h3 − h2 (1 − y ) = 236.14 − 69.84 (1 − 0.1300) = 0.1125 h11 − h2 1356.0 − 69.84
Then, q in = h7 − h6 + (1 − y )(h10 − h9 ) = 1550.5 − 380.41 + (1 − 0.1300)(1531.3 − 1248.8) = 1415.8 Btu/lbm
q out = (1 − y − z )(h12 − h1 ) = (1 − 0.1300 − 0.1125)(1052.4 − 69.72) = 744.4 Btu/lbm wnet = q in − q out = 1415.8 − 744.4 = 671.4 Btu/lbm
and m& =
Q& in 4 × 10 5 Btu/s = = 282.5 lbm/s q in 1415.8 Btu/lbm
(b)
1.055 kJ = 200.1 MW W& net = m& wnet = (282.5 lbm/s)(671.4 Btu/lbm) 1 Btu
(c)
η th = 1 −
q out 744.4 Btu/lbm = 1− = 47.4% q in 1415.8 Btu/lbm
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-42
10-52 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwater heater is considered. The temperature of the steam at the inlet of the closed feedwater heater, the mass flow rate of the steam extracted from the turbine for the closed feedwater heater, the net power output, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg
High-P turbine
5
v 1 = v f @ 20 kPa = 0.001017 m 3 /kg w pI ,in = v 1 (P2 − P1 ) / η p
Low-P turbine
Boiler
= (0.001017 m 3 /kg )(12,500 − 20 kPa ) / 0.88 = 14.43 kJ/kg
6
P3 = 1 MPa h3 = h f @ 1 MPa = 762.51 kJ/kg sat. liquid v 3 = v f @ 1 MPa = 0.001127 m 3 /kg
w pII ,in = v 3 (P11 − P3 ) / η p
1-y 9
7
h2 = h1 + w pI ,in = 251.42 + 14.43 = 265.85 kJ/kg
4
Mixing Cham.
10
8 y
Closed fwh
Cond.
2 11
PII
3
1 PI
= (0.001127 m 3 /kg )(12,500 − 1000 kPa ) / 0.88 = 14.73 kJ/kg h11 = h3 + w pII ,in = 762.51 + 14.73 = 777.25 kJ/kg
Also, h4 = h10 = h11 = 777.25 kJ/kg since the two fluid streams which are being mixed have the same enthalpy. P5 = 12.5 MPa h5 = 3476.5 kJ/kg s 5 = 6.6317 kJ/kg ⋅ K
T5 = 550°C
P6 = 5 MPa h6 s = 3185.6 kJ/kg s6 = s5
ηT =
h5 − h6 → h6 = h5 − η T (h5 − h6 s ) h5 − h6 s = 3476.5 − (0.88)(3476.5 − 3185.6 ) = 3220.5 kJ/kg
P7 = 5 MPa h7 = 3550.9 kJ/kg T7 = 550°C s 7 = 7.1238 kJ/kg ⋅ K P8 = 1 MPa h8 s = 3051.1 kJ/kg s8 = s 7
ηT =
h7 − h8 → h8 = h7 − η T (h7 − h8 s ) h7 − h8 s = 3550.9 − (0.88)(3550.9 − 3051.1) = 3111.1 kJ/kg
P8 = 1 MPa
T8 = 328°C h8 = 3111.1 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-43
P9 = 20 kPa h9 s = 2347.9 kJ/kg s9 = s7
ηT =
h7 − h9 → h9 = h7 − η T (h7 − h9 s ) h7 − h9 s = 3550.9 − (0.88)(3550.9 − 2347.9 ) = 2492.2 kJ/kg
The fraction of steam extracted from the low pressure turbine for closed feedwater heater is determined from the steady-flow energy balance equation applied to the feedwater heater. Noting that Q& ≅ W& ≅ ∆ke ≅ ∆pe ≅ 0 ,
(1 − y )(h10 − h2 ) = y(h8 − h3 ) (1 − y )(777.25 − 265.85) = y (3111.1 − 762.51) → y = 0.1788
The corresponding mass flow rate is m& 8 = ym& 5 = (0.1788)(24 kg/s) = 4.29 kg/s
(c) Then, q in = h5 − h4 + h7 − h6 = 3476.5 − 777.25 + 3550.9 − 3220.5 = 3029.7 kJ/kg
q out = (1 − y )(h9 − h1 ) = (1 − 0.1788)(2492.2 − 251.42 ) = 1840.1 kJ/kg
and W& net = m& (q in − q out ) = (24 kg/s)(3029.7 − 1840.1)kJ/kg = 28,550 kW
(b) The thermal efficiency is determined from
η th = 1 −
q out 1840.1 kJ/kg = 1− = 0.393 = 39.3% q in 3029.7 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-44
Second-Law Analysis of Vapor Power Cycles 10-53C In the simple ideal Rankine cycle, irreversibilities occur during heat addition and heat rejection processes in the boiler and the condenser, respectively, and both are due to temperature difference. Therefore, the irreversibilities can be decreased and thus the 2nd law efficiency can be increased by minimizing the temperature differences during heat transfer in the boiler and the condenser. One way of doing that is regeneration.
10-54 The exergy destructions associated with each of the processes of the Rankine cycle described in Prob. 10-15 are to be determined for the specified source and sink temperatures. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From Problem 10-15, s1 = s2 = s f @ 50 kPa = 1.0912 kJ/kg ⋅ K s3 = s4 = 6.5412 kJ/kg ⋅ K qin = 2650.72 kJ/kg qout = 1931.8 kJ/kg
Processes 1-2 and 3-4 are isentropic. Thus, i12 = 0 and i34 = 0. Also, qR , 23 − 2650.8 kJ/kg = (290 K ) 6.5412 − 1.0912 + = 1068 kJ/kg xdestroyed,23 = T0 s3 − s2 + 1500 K TR qR , 41 1931.8 kJ/kg = (290 K )1.0912 − 6.5412 + = 351.3 kJ/kg xdestroyed,41 = T0 s1 − s4 + 290 K TR
10-55 The exergy destructions associated with each of the processes of the Rankine cycle described in Prob. 10-16 are to be determined for the specified source and sink temperatures. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From Problem 10-16, s1 = s 2 = s f @10kPa = 0.6492 kJ/kg ⋅ K s 3 = s 4 = 6.5995 kJ/kg ⋅ K q in = 3173.2 kJ/kg q out = 1897.9 kJ/kg
Processes 1-2 and 3-4 are isentropic. Thus, i12 = 0 and i34 = 0. Also, q R ,23 − 3173.2 kJ/kg = (290 K ) 6.5995 − 0.6492 + = 1112.1 kJ/kg x destroyed,23 = T0 s 3 − s 2 + T 1500 K R q R ,41 1897.9 kJ/kg = (290 K ) 0.6492 − 6.5995 + = 172.3 kJ/kg x destroyed,41 = T0 s1 − s 4 + TR 290 K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-45
10-56 The exergy destruction associated with the heat rejection process in Prob. 10-22 is to be determined for the specified source and sink temperatures. The exergy of the steam at the boiler exit is also to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From Problem 10-22, s1 = s 2 = s f @10 kPa = 0.6492 kJ/kg ⋅ K s 3 = s 4 = 6.8000 kJ/kg ⋅ K h3 = 3411.4 kJ/kg q out = 1961.8 kJ/kg
The exergy destruction associated with the heat rejection process is q R ,41 1961.8 kJ/kg = (290 K ) 0.6492 − 6.8000 + = 178.0 kJ/kg x destroyed,41 = T0 s1 − s 4 + TR 290 K
The exergy of the steam at the boiler exit is simply the flow exergy,
ψ 3 = (h3 − h0 ) − T0 (s3 − s0 ) + = (h3 − h0 ) − T0 (s3 − s0 )
V32 2
©0
+ qz3©0
where
h0 = h@ (290 K , 100 kPa ) ≅ h f @ 290 K = 71.95 kJ/kg s0 = s@ (290 K , 100 kPa ) ≅ s f @ 290 K = 0.2533 kJ/kg ⋅ K
Thus,
ψ 3 = (3411.4 − 71.95) kJ/kg − (290 K )(6.800 − 0.2532 ) kJ/kg ⋅ K = 1440.9 kJ/kg
10-57 The exergy destructions associated with each of the processes of the reheat Rankine cycle described in Prob. 10-32 are to be determined for the specified source and sink temperatures. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From Problem 10-32, s1 = s2 = s f @ 20 kPa = 0.8320 kJ/kg ⋅ K s3 = s4 = 6.7266 kJ/kg ⋅ K s5 = s6 = 7.2359 kJ/kg ⋅ K q23,in = 3399.5 − 259.54 = 3140.0 kJ/kg q45,in = 3457.2 − 3105.1 = 352.1 kJ/kg qout = h6 − h1 = 2385.2 − 251.42 = 2133.8 kJ/kg
Processes 1-2, 3-4, and 5-6 are isentropic. Thus, i12 = i34 = i56 = 0. Also, qR , 23 − 3140.0 kJ/kg = (300 K ) 6.7266 − 0.8320 + xdestroyed,23 = T0 s3 − s2 + = 1245.0 kJ/kg TR 1800 K qR , 45 − 352.5 kJ/kg = (300 K ) 7.2359 − 6.7266 + = 94.1 kJ/kg xdestroyed,45 = T0 s5 − s4 + TR 1800 K qR ,61 2133.8 kJ/kg = (300 K ) 0.8320 − 7.2359 + xdestroyed,61 = T0 s1 − s6 + = 212.6 kJ/kg T 300 K R
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10-46
10-58 EES Problem 10-57 is reconsidered. The problem is to be solved by the diagram window data entry feature of EES by including the effects of the turbine and pump efficiencies. Also, the T-s diagram is to be plotted. Analysis The problem is solved using EES, and the solution is given below. function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$='' if (x6>1) then x6$='(superheated)' if (x6 W_dot_comp_isen" m_dot*h[1] + W_dot_comp_isen = m_dot*hs2"SSSF First Law for the isentropic compressor, assuming: adiabatic, ke=pe=0, m_dot is the mass flow rate in kg/s" h[1]=ENTHALPY(Air,T=T[1]) hs2=ENTHALPY(Air,T=Ts2) m_dot*h[1] + W_dot_comp = m_dot*h[2]"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,h=h[2],P=P[2]) "Heat Rejection Process 2-3, assumed SSSF constant pressure process" m_dot*h[2] + Q_dot_out = m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" h[3]=ENTHALPY(Air,T=T[3]) "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s4s=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[1] s4s=ENTROPY(Air,T=Ts4,P=P[4])"Ts4 is the isentropic value of T[4] at turbine exit" Eta_turb = W_dot_turb /W_dot_turb_isen "turbine adiabatic efficiency, W_dot_turb_isen > W_dot_turb" m_dot*h[3] = W_dot_turb_isen + m_dot*hs4"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0" hs4=ENTHALPY(Air,T=Ts4) m_dot*h[3] = W_dot_turb + m_dot*h[4]"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,h=h[4],P=P[4]) "Refrigeration effect:" m_dot*h[4] + Q_dot_Refrig = m_dot*h[1] "Cycle analysis" W_dot_in_net=W_dot_comp-W_dot_turb"External work supplied to compressor" COP= Q_dot_Refrig/W_dot_in_net "The following is for plotting data only:" Ts[1]=Ts2 ss[1]=s2s Ts[2]=Ts4 ss[2]=s4s PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-40
COP
ηcomp
ηturb
0.6937 0.9229 1.242 1.717
0.7 0.8 0.9 1
1 1 1 1
Winnet [kW] 9.612 7.224 5.368 3.882
QRefrig [kW] 6.667 6.667 6.667 6.667
7 6 5
] W k[
4
gi rf e R
Q
3
η turb
2
0.7
1
0.85 1.0
0 0,7
0,75
0,8
0,85
0,9
0,95
1
ηcomp 12 10
] W k[
t e n; ni
W
8 6 4 2 0 0,7
η turb 0.7 0.85 1.0 0,75
0,8
0,85
0,9
0,95
1
0,9
0,95
1
ηcomp 2
η turb 1,5
P O C
1
0.7 0.85 1.0
0,5 0 0,7
0,75
0,8
0,85
ηcomp
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-41
11-58E An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17E), T1 = 500 R
→
h1 = 119.48 Btu / lbm Pr 1 = 1.0590
T1 = 580 R
→
h3 = 138.66 Btu / lbm Pr 3 = 1.7800
T · QH 120°F 40°F
Thus, P 30 → T2 = 683.9 R Pr2 = 2 Pr1 = (1.0590) = 3.177 P1 10 h2 = 163.68 Btu/lbm Pr4 =
2
3
4
1 · QRefrig
P4 10 Pr3 = (1.7800) = 0.5933 → T4 = 423.4 R P3 30 h4 = 101.14 Btu/lbm
Then the rate of refrigeration is Q& refrig = m& (qL ) = m& (h1 − h4 ) = (0.5 lbm/s)(119.48 − 101.14 ) Btu/lbm = 9.17 Btu/s
(b) The net power input is determined from W& net, in = W& comp, in − W& turb, out
where W&comp,in = m& (h2 − h1 ) = (0.5 lbm/s)(163.68 − 119.48) Btu/lbm = 22.10 Btu/s W& turb,out = m& (h3 − h4 ) = (0.5 lbm/s)(138.66 − 101.14 ) Btu/lbm = 18.79 Btu/s
Thus,
W& net, in = 2210 . − 18.76 = 3.34 Btu / s = 4.73 hp
(c) The COP of this ideal gas refrigeration cycle is determined from Q& L 9.17 Btu / s COPR = = = 2.75 3.34 Btu / s W& net, in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
11-42
11-59 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A17), T1 = 285 K T1 = 320 K
→ →
. kJ / kg h1 = 28514 . Pr 1 = 11584 h3 = 320.29 kJ / kg Pr 3 = 1.7375
2 T
47°C 12°C
Thus, P 250 Pr2 = 2 Pr1 = → T2 s = 450.4 K (1.1584 ) = 5.792 P1 50 h2 s = 452.17 kJ/kg Pr4 =
P4 50 Pr3 = → T4 s = 201.8 K (1.7375) = 0.3475 P3 250 h4 s = 201.76 kJ/kg
ηT =
h3 − h 4 → h4 = h3 − ηT (h3 − h4 s ) h3 − h4 s = 320.29 − (0.85)(320.29 − 201.76)
· QH
2
3 1 · 4 QRefrig 4s
Also,
= 219.54 kJ/kg
Then the rate of refrigeration is Q& refrig = m& (qL ) = m& (h1 − h4 ) = (0.08 kg/s)(285.14 − 219.54 ) kJ/kg = 5.25 kW
(b) The net power input is determined from W& net, in = W& comp, in − W& turb, out
where W&comp,in = m& (h2 − h1 ) = m& (h2 s − h1 ) / ηC = (0.08 kg/s)[(452.17 − 285.14 ) kJ/kg]/ (0.80 ) = 16.70 kW W& turb,out = m& (h3 − h4 ) = (0.08 kg/s)(320.29 − 219.54 ) kJ/kg = 8.06 kW
Thus,
W& net, in = 16.70 − 8.06 = 8.64 kW
(c) The COP of this ideal gas refrigeration cycle is determined from Q& L 5.25 kW COPR = = = 0.61 8.64 kW W& net, in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
11-43
11-60 A gas refrigeration cycle with helium as the working fluid is considered. The minimum temperature in the cycle, the COP, and the mass flow rate of the helium are to be determined. Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2). Analysis (a) From the isentropic relations, P T2 s = T1 2 P1 T4 s
P = T3 4 P3
(k −1) / k
(k −1) / k
= (263K )(3)0.667 / 1.667 = 408.2K
2 T
50°C -10°C
1 = (323K ) 3
0.667 / 1.667
= 208.1K
· QH
2s
3 1 4 Q· Refrig 4s s
and
ηT =
h3 − h4 T − T4 = 3 → T4 = T3 − ηT (T3 − T4 s ) = 323 − (0.80 )(323 − 208.1) h3 − h4 s T3 − T4 s = 231.1 K = T min
h − h1 T2 s − T1 ηC = 2s → T2 = T1 + (T2 s − T1 ) / η C = 263 + (408.2 − 263) / (0.80 ) = h2 − h1 T2 − T1 = 444.5 K
(b) The COP of this ideal gas refrigeration cycle is determined from COPR = = = =
qL qL = wnet,in wcomp,in − wturb,out h1 − h4
(h2 − h1 ) − (h3 − h4 ) T1 − T4 (T2 − T1 ) − (T3 − T4 ) 263 − 231.1
(444.5 − 263) − (323 − 231.1)
= 0.356
(c) The mass flow rate of helium is determined from Q& refrig Q& refrig Q& refrig 18 kJ/s m& = = = = = 0.109 kg/s qL h1 − h4 c p (T1 − T4 ) (5.1926 kJ/kg ⋅ K )(263 − 231.1) K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-44
11-61 An ideal-gas refrigeration cycle with air as the working fluid is considered. The lowest temperature that can be obtained by this cycle, the COP, and the mass flow rate of air are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) The lowest temperature in the cycle occurs at the turbine exit. From the isentropic relations, P T2 = T1 2 P1
(k −1) / k
P T5 = T4 5 P4
= (266 K )(4 )0.4 / 1.4 = 395.3 K = 122.3°C
(k −1) / k
1 = (258 K ) 4
0.4 / 1.4
= 173.6 K = −99.4°C = Tmin
(b) From an energy balance on the regenerator, E& in − E& out = ∆E& system ©0 (steady) = 0 E& in = E& out
∑ m& h = ∑ m& h e e
i i
→ m& (h3 − h4 ) = m& (h1 − h6 )
or,
T
· QH 3
27°C -7°C -15°C
Qrege
5
6 · QRefrig
T6 = T1 − T3 + T4 = (− 7°C ) − 27°C + (− 15°C ) = −49°C
Then the COP of this ideal gas refrigeration cycle is determined from COPR = = = =
1
4
m& c p (T3 − T4 ) = m& c p (T1 − T6 ) → T3 − T4 = T1 − T6
or,
2
qL qL = wnet,in wcomp,in − w turb,out h6 − h5
(h2 − h1 ) − (h4 − h5 ) T6 − T5 (T2 − T1 ) − (T4 − T5 )
− 49°C − (− 99.4°C ) = 1.12 [122.3 − (− 7 )]°C − [− 15 − (− 99.4)]°C
(c) The mass flow rate is determined from Q& refrig Q& refrig Q& refrig 12 kJ/s m& = = = = = 0.237 kg/s qL h6 − h5 c p (T6 − T5 ) (1.005 kJ/kg ⋅ °C )[− 49 − (− 99.4)]°C
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
11-45
11-62 An ideal-gas refrigeration cycle with air as the working fluid is considered. The lowest temperature that can be obtained by this cycle, the COP, and the mass flow rate of air are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).
T
Analysis (a) The lowest temperature in the cycle occurs at the turbine exit. From the isentropic relations, P T2 s = T1 2 P1
(k −1) / k
P T5s = T4 5 P4
(k −1) / k
-7°C -15°C
Qrege
1
4
0.4 / 1.4
= 173.6 K = −99.4°C
2 2s
3
27°C
= (266 K )(4 )0.4 / 1.4 = 395.3 K = 122.3°C 1 = (258 K ) 4
· QH
5
·6 5 QRefrig
and ηT =
h4 − h5 T − T5 = 4 → T5 = T4 − ηT (T4 − T5s ) = −15 − (0.80 )(− 15 − (− 99.4 )) h4 − h5s T4 − T5s = −82.5°C = T min
h − h1 T2 s − T1 ηC = 2 s = → T2 = T1 + (T2 s − T1 ) / ηC = −7 + (122.3 − (− 7 )) / (0.75) h2 − h1 T2 − T1 = 165.4°C
(b) From an energy balance on the regenerator, E& in − E& out = ∆E& system©0 (steady) = 0 → E& in = E& out
∑ m& h = ∑ m& h e e
i i
→ m& (h3 − h4 ) = m& (h1 − h6 )
or, m& c p (T3 − T4 ) = m& c p (T1 − T6 ) → T3 − T4 = T1 − T6
or,
T6 = T1 − T3 + T4 = (− 7°C ) − 27°C + (− 15°C ) = −49°C
Then the COP of this ideal gas refrigeration cycle is determined from COPR = = =
qL qL = wnet,in wcomp,in − w turb,out h6 − h5 (h2 − h1 ) − (h4 − h5 ) T6 − T5
(T2 − T1 ) − (T4 − T5 ) − 49°C − (− 82.5°C ) = 0.32 = [165.4 − (− 7 )]°C − [− 15 − (− 82.5)]°C (c) The mass flow rate is determined from Q& refrig Q& refrig Q& refrig 12 kJ/s m& = = = = = 0.356 kg/s qL h6 − h5 c p (T6 − T5 ) (1.005 kJ/kg ⋅ °C )[− 49 − (− 82.5)]°C
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
11-46
11-63 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) From the isentropic relations, T2 s
P = T1 2 P1
ηC = 0.80 =
(k −1) / k
= (273.2 K )(5)0.4 / 1.4 = 432.4 K
h2 s − h1 T2 s − T1 = h2 − h1 T2 − T1
The temperature at state 4 can be determined by solving the following two equations simultaneously:
ηT =
Heat Exch.
432.4 − 273.2 → T2 = 472.5 K T2 − 273.2
P T5 s = T4 5 P4
(k −1) / k
1 = T4 5
.
QL
6 Regenerator
Heat Exch.
3 5
0.4 / 1.4
h4 − h5 T − 193.2 → 0.85 = 4 h4 − h5 s T4 − T5 s
2
QH
Turbine Compressor
Using EES, we obtain T4 = 281.3 K. An energy balance on the regenerator may be written as
T
m& c p (T3 − T4 ) = m& c p (T1 − T6 ) → T3 − T4 = T1 − T6
or, T6 = T1 − T3 + T4 = 273.2 − 308.2 + 281.3 = 246.3 K h3 − h4 T3 − T4 308.2 − 281.3 = = = 0.434 h3 − h6 T3 − T6 308.2 − 246.3
· QH
2
2s
3
35°C 0°C
The effectiveness of the regenerator is
ε regen =
1
.
4
Qregen
1
4 -80°C
5s
·6 5 QRefrig
(b) The refrigeration load is
s
Q& L = m& c p (T6 − T5 ) = (0.4 kg/s)(1.005 kJ/kg.K)(246.3 − 193.2)K = 21.36 kW
(c) The turbine and compressor powers and the COP of the cycle are W& C,in = m& c p (T2 − T1 ) = (0.4 kg/s)(1.005 kJ/kg.K)(472.5 − 273.2)K = 80.13 kW W& T,out = m& c p (T4 − T5 ) = (0.4 kg/s)(1.005 kJ/kg.K)(281.3 − 193.2)kJ/kg = 35.43 kW COP =
Q& L W&
net,in
=
Q& L W& C,in − W& T,out
=
21.36 = 0.478 80.13 − 35.43
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-47
(d) The simple gas refrigeration cycle analysis is as follows: T4 s
1 = T3 r
ηT =
(k −1) / k
1 = (308.2 K ) 5
0.4 / 1.4
= 194.6 K
T3 − T4 308.2 − T4 → 0.85 = → T4 = 211.6 K T3 − T4 s 308.2 − 194.6
2 T
35°C 0°C
Q& L = m& c p (T1 − T4 ) = (0.4 kg/s)(1.005 kJ/kg.K)(273.2 − 211.6)kJ/kg = 24.74 kW
· QH
2
3 1 · 4 QRefrig 4s
W&net,in = m& c p (T2 − T1 ) − m& c p (T3 − T4 )
= (0.4 kg/s)(1.005 kJ/kg.K)[(472.5 − 273.2) − (308.2 − 211.6)kJ/kg ] = 41.32 kW
COP =
Q& L W&
net,in
=
24.74 = 0.599 41.32
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
11-48
Absorption Refrigeration Systems 11-64C Absorption refrigeration is the kind of refrigeration that involves the absorption of the refrigerant during part of the cycle. In absorption refrigeration cycles, the refrigerant is compressed in the liquid phase instead of in the vapor form. 11-65C The main advantage of absorption refrigeration is its being economical in the presence of an inexpensive heat source. Its disadvantages include being expensive, complex, and requiring an external heat source. 11-66C In absorption refrigeration, water can be used as the refrigerant in air conditioning applications since the temperature of water never needs to fall below the freezing point. 11-67C The fluid in the absorber is cooled to maximize the refrigerant content of the liquid; the fluid in the generator is heated to maximize the refrigerant content of the vapor. 11-68C The coefficient of performance of absorption refrigeration systems is defined as
COPR =
QL Q desiredoutput = ≅ L requiredinput Qgen + Wpump,in Qgen
11-69C The rectifier separates the water from NH3 and returns it to the generator. The regenerator transfers some heat from the water-rich solution leaving the generator to the NH3-rich solution leaving the pump.
11-70 The COP of an absorption refrigeration system that operates at specified conditions is given. It is to be determined whether the given COP value is possible. Analysis The maximum COP that this refrigeration system can have is T TL 300 K 268 = 1 − COPR,max = 1 − 0 = 2.14 Ts T0 − TL 403 K 300 − 268
which is slightly greater than 2. Thus the claim is possible, but not probable.
11-71 The conditions at which an absorption refrigeration system operates are specified. The maximum COP this absorption refrigeration system can have is to be determined. Analysis The maximum COP that this refrigeration system can have is T COPR,max = 1 − 0 Ts
TL T − T L 0
298 K 273 = 1 − = 2.64 393 K 298 − 273
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-49
11-72 The conditions at which an absorption refrigeration system operates are specified. The maximum rate at which this system can remove heat from the refrigerated space is to be determined. Analysis The maximum COP that this refrigeration system can have is
Thus,
T TL 298 K 243 = 1 − COPR,max = 1 − 0 = 1.15 Ts T0 − TL 403 K 298 − 243 Q& = COP Q& = (1.15) 5 × 105 kJ/h = 5.75 × 105 kJ/h L, max
(
R, max gen
)
11-73E The conditions at which an absorption refrigeration system operates are specified. The COP is also given. The maximum rate at which this system can remove heat from the refrigerated space is to be determined. Analysis For a COP = 0.55, the rate at which this system can remove heat from the refrigerated space is Q& = COP Q& = (0.55) 10 5 Btu/h = 0.55 × 10 5 Btu/h L
R
(
gen
)
11-74 A reversible absorption refrigerator consists of a reversible heat engine and a reversible refrigerator. The rate at which the steam condenses, the power input to the reversible refrigerator, and the second law efficiency of an actual chiller are to be determined. Properties The enthalpy of vaporization of water at 200C is hfg = 1939.8 kJ/kg (Table A-4). Analysis (a) The thermal efficiency of the reversible heat engine is T (25 + 273.15) K η th,rev = 1 − 0 = 1 − = 0.370 Ts (200 + 273.15) K Ts T 0
The COP of the reversible refrigerator is TL (−10 + 273.15) K COPR, rev = = = 7.52 T0 − T L (25 + 273.15) − (−10 + 273.15) K The COP of the reversible absorption refrigerator is COPabs, rev = η th,rev COPR, rev = (0.370)(7.52) = 2.78
Rev. HE
Rev. Ref.
T0
TL
The heat input to the reversible heat engine is Q& L 22 kW Q& in = = = 7.911 kW COPabs,rev 2.78 Then, the rate at which the steam condenses becomes Q& 7.911 kJ/s m& s = in = = 0.00408 kg/s h fg 1939.8 kJ/kg
(b) The power input to the refrigerator is equal to the power output from the heat engine W& = W& =η Q& = (0.370)(7.911 kW ) = 2.93 kW in, R
out, HE
th, rev
in
(c) The second-law efficiency of an actual absorption chiller with a COP of 0.7 is COPactual 0.7 η II = = = 0.252 COPabs,rev 2.78
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-50
Special Topic: Thermoelectric Power Generation and Refrigeration Systems 11-75C The circuit that incorporates both thermal and electrical effects is called a thermoelectric circuit. 11-76C When two wires made from different metals joined at both ends (junctions) forming a closed circuit and one of the joints is heated, a current flows continuously in the circuit. This is called the Seebeck effect. When a small current is passed through the junction of two dissimilar wires, the junction is cooled. This is called the Peltier effect. 11-77C No. 11-78C No. 11-79C Yes. 11-80C When a thermoelectric circuit is broken, the current will cease to flow, and we can measure the voltage generated in the circuit by a voltmeter. The voltage generated is a function of the temperature difference, and the temperature can be measured by simply measuring voltages. 11-81C The performance of thermoelectric refrigerators improves considerably when semiconductors are used instead of metals. 11-82C The efficiency of a thermoelectric generator is limited by the Carnot efficiency because a thermoelectric generator fits into the definition of a heat engine with electrons serving as the working fluid.
11-83E A thermoelectric generator that operates at specified conditions is considered. The maximum thermal efficiency this thermoelectric generator can have is to be determined. Analysis The maximum thermal efficiency of this thermoelectric generator is the Carnot efficiency,
η th,max = η th,Carnot = 1 −
TL 550R = 1− = 31.3% 800R TH
11-84 A thermoelectric refrigerator that operates at specified conditions is considered. The maximum COP this thermoelectric refrigerator can have and the minimum required power input are to be determined. Analysis The maximum COP of this thermoelectric refrigerator is the COP of a Carnot refrigerator operating between the same temperature limits, COPmax = COPR,Carnot =
Thus, W& in, min =
1 1 = = 10.72 (TH / TL ) − 1 (293 K ) / (268 K ) − 1
Q& L 130 W = = 12.1 W COPmax 10.72
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-51
11-85 A thermoelectric cooler that operates at specified conditions with a given COP is considered. The required power input to the thermoelectric cooler is to be determined. Analysis The required power input is determined from the definition of COPR, Q& Q& L 180 W COPR = & L → W&in = = = 1200 W COP 0.15 W R
in
11-86E A thermoelectric cooler that operates at specified conditions with a given COP is considered. The required power input to the thermoelectric cooler is to be determined. Analysis The required power input is determined from the definition of COPR, COPR =
Q& L Q& L 20 Btu/min → W& in = = = 133.3 Btu/min = 3.14 hp & COPR 0.15 Win
11-87 A thermoelectric refrigerator powered by a car battery cools 9 canned drinks in 12 h. The average COP of this refrigerator is to be determined. Assumptions Heat transfer through the walls of the refrigerator is negligible. Properties The properties of canned drinks are the same as those of water at room temperature, ρ = 1 kg/L and cp = 4.18 kJ/kg·°C (Table A-3). Analysis The cooling rate of the refrigerator is simply the rate of decrease of the energy of the canned drinks, m = ρV = 9 × (1 kg/L)(0.350 L) = 3.15 kg Qcooling = mc∆T = (3.15 kg)(4.18 kJ/kg ⋅ °C)(25 - 3)°C = 290 kJ Q& cooling =
Qcooling ∆t
=
290 kJ = 0.00671 kW = 6.71 W 12 × 3600 s
The electric power consumed by the refrigerator is W& in = VI = (12 V)(3 A) = 36 W
Then the COP of the refrigerator becomes COP =
Q& cooling 6.71 W = = 0.186 ≈ 0.20 36 W W& in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-52
11-88E A thermoelectric cooler is said to cool a 12-oz drink or to heat a cup of coffee in about 15 min. The average rate of heat removal from the drink, the average rate of heat supply to the coffee, and the electric power drawn from the battery of the car are to be determined. Assumptions Heat transfer through the walls of the refrigerator is negligible. Properties The properties of canned drinks are the same as those of water at room temperature, cp = 1.0 Btu/lbm.°F (Table A-3E). Analysis (a) The average cooling rate of the refrigerator is simply the rate of decrease of the energy content of the canned drinks, Qcooling = mc p ∆T = (0.771 lbm)(1.0 Btu/lbm ⋅ °F)(78 - 38)°F = 30.84 Btu Q& cooling =
Qcooling ∆t
=
30.84 Btu 1055 J = 36.2 W 15 × 60 s 1 Btu
(b) The average heating rate of the refrigerator is simply the rate of increase of the energy content of the canned drinks, Q heating = mc p ∆T = (0.771 lbm)(1.0 Btu/lbm ⋅ °F)(130 - 75)°F = 42.4 Btu Q& heating =
Q heating ∆t
=
42.4 Btu 1055 J = 49.7 W 15 × 60 s 1 Btu
(c) The electric power drawn from the car battery during cooling and heating is W& in,cooling =
Q& cooling COPcooling
=
36.2 W = 181 W 0.2
COPheating = COPcooling + 1 = 0.2 + 1 = 1.2 W& in, heating =
Q& heating COPheating
=
49.7 W = 41.4 W 1.2
11-89 The maximum power a thermoelectric generator can produce is to be determined. Analysis The maximum thermal efficiency this thermoelectric generator can have is ηth,max = 1 −
TL 303 K = 1− = 0.142 TH 353 K
Thus, W&out, max = η th, maxQ&in = (0.142)(106 kJ/h) = 142,000 kJ/h = 39.4 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-53
Review Problems 11-90 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP, the condenser and evaporator pressures, and the net work input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The COP of this refrigeration cycle is determined from 1 1 COPR,C = = = 5.06 T (TH / TL ) − 1 (303 K )/ (253 K ) − 1 (b) The condenser and evaporative pressures are (Table A-11) Pevap = Psat @ − 20°C = 132.82 kPa 4 3 30°C Pcond = Psat @30°C = 770.64 kPa (c) The net work input is determined from h1 = h f + x1 h fg @ − 20°C = 25.49 + (0.15)(212.91) = 57.43 kJ/kg
( ) h2 = (h f + x 2 h fg )@ − 20°C = 25.49 + (0.80)(212.91) = 195.82 kJ/kg
-20°C
1
qL
2
s
q L = h2 − h1 = 195.82 − 57.43 = 138.4kJ/kg qL 138.4 kJ/kg = = 27.35 kJ/kg wnet,in = COPR 5.06
11-91 A large refrigeration plant that operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant, the power input to the compressor, and the mass flow rate of the cooling water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa h1 = h g @ 120 kPa = 236.97 kJ/kg s = s sat. vapor g @ 120 kPa = 0.94779 kJ/kg ⋅ K 1 T
P2 = 0.7 MPa h2 = 273.50 kJ/kg s 2 = s1
(T2 = 34.95°C)
P3 = 0.7 MPa h3 = h f @ 0.7 MPa = 88.82 kJ/kg sat. liquid
· QH 3 0.7 MPa
h4 ≅ h3 = 88.82 kJ/kg (throttling ) 0.12 MPa The mass flow rate of the refrigerant is determined from · 4 QL Q& L 100 kJ/s m& = = = 0.675 kg/s h1 − h4 (236.97 − 88.82) kJ/kg (b) The power input to the compressor is W& in = m& (h2 − h1 ) = (0.675 kg/s )(273.50 − 236.97 ) kJ/kg = 24.7 kW (c) The mass flow rate of the cooling water is determined from Q& H = m& (h2 − h3 ) = (0.675 kg/s )(273.50 − 88.82 ) kJ/kg = 124.7 kW Q& H 124.7 kJ/s = = 3.73 kg/s m& cooling = (c p ∆T ) water (4.18 kJ/kg ⋅ °C )(8°C )
2 · Win
1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
11-54
11-92 EES Problem 11-91 is reconsidered. The effect of evaporator pressure on the COP and the power input is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" "P[1]=120 [kPa]" P[2] = 700 [kPa] Q_dot_in= 100 [kW] DELTAT_cw = 8 [C] C_P_cw = 4.18 [kJ/kg-K] Fluid$='R134a' Eta_c=1.0 "Compressor isentropic efficiency" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" Wc=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+Wc=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" {h[2]=enthalpy(Fluid$,P=P[2],T=T[2]) } T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_c=m_dot*Wc "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] "energy balance on condenser" Q_dot_out=m_dot*Qout "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" Q_dot_in=m_dot*Q_in COP=Q_dot_in/W_dot_c "definition of COP" COP_plot = COP W_dot_in = W_dot_c m_dot_cw*C_P_cw*DELTAT_cw = Q_dot_out
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-55
P1 [kPa] 120 150 180 210 240 270 300 330 360 390
COP 4,056 4,743 5,475 6,27 7,146 8,126 9,235 10,51 11,99 13,75
WC [kW] 24,66 21,09 18,27 15,95 13,99 12,31 10,83 9,517 8,34 7,274
16
25 21
12 17 ]
P O C
W k[
8
13 c
W
9
4 100
150
200
250
300
350
5 400
P[1] [kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-56
11-93 A large refrigeration plant operates on the vapor-compression cycle with refrigerant-134a as the working fluid. The mass flow rate of the refrigerant, the power input to the compressor, the mass flow rate of the cooling water, and the rate of exergy destruction associated with the compression process are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa h1 = h g @ 120 kPa = 236.97 kJ/kg s = s sat. vapor g @ 120 kPa = 0.94779 kJ/kg ⋅ K 1 P2 = 0.7 MPa s 2 s = s1
h2 s = 273.50 kJ/kg
P3 = 0.7 MPa h3 = h f sat. liquid
@ 0.7 MPa
(T2 s
T · QH
= 34.95°C )
2
3 0.7 MPa
· Win
= 88.82 kJ/kg
h4 ≅ h3 = 88.82 kJ/kg (throttling )
The mass flow rate of the refrigerant is determined from Q& L 100 kJ/s m& = = = 0.675 kg/s h1 − h4 (236.97 − 88.82 ) kJ/kg
0.12 MPa 4
· QL
1
(b) The actual enthalpy at the compressor exit is
ηC = Thus,
h2 s − h1 → h2 = h1 + (h2 s − h1 ) / η C = 236.97 + (273.50 − 236.97 ) / (0.75) h2 − h1 = 285.67 kJ/kg
W& in = m& (h2 − h1 ) = (0.675 kg/s )(285.67 − 236.97 ) kJ/kg = 32.9 kW
(c) The mass flow rate of the cooling water is determined from and
Q& H = m& (h2 − h3 ) = (0.675 kg/s )(285.67 − 88.82) kJ/kg = 132.9 kW m& cooling =
Q& H 132.9 kJ/s = = 3.97 kg/s (c p ∆T ) water (4.18 kJ/kg ⋅ °C )(8°C )
The exergy destruction associated with this adiabatic compression process is determined from X& destroyed = T0 S&gen = T0 m& (s2 − s1 )
where P2 = 0.7 MPa s 2 = 0.98655 kJ/kg ⋅ K h2 = 285.67 kJ/kg
Thus, X& destroyed = (298 K )(0.675 kg/s )(0.98655 − 0.94779 ) kJ/kg ⋅ K = 7.80 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
11-57
11-94 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid is used to heat a house. The rate of heat supply to the house, the volume flow rate of the refrigerant at the compressor inlet, and the COP of this heat pump are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), h1 = h g @ 200 kPa = 244.46 kJ/kg P1 = 200 kPa s1 = s g @ 200 kPa = 0.93773 kJ/kg ⋅ K sat. vapor v 1 = v g @ 200 kPa = 0.099867 m 3 /kg P2 = 0.9 MPa s 2 = s1
T
h2 = 275.75 kJ/kg
P3 = 0.9 MPa h3 = h f sat. liquid
@ 0.9 MPa
House 3 0.9 MPa
= 101.61 kJ/kg
2 · Win
200 kPa 4
h4 ≅ h3 = 101.61 kJ/kg (throttling )
· QH
· QL
1
The rate of heat supply to the house is determined from
s
Q& H = m& (h2 − h3 ) = (0.32 kg/s )(275.75 − 101.61) kJ/kg = 55.73 kW
(b) The volume flow rate of the refrigerant at the compressor inlet is
(
)
V&1 = m& v 1 = (0.32 kg/s ) 0.099867 m 3 /kg = 0.0320 m 3 /s (c) The COP of t his heat pump is determined from COPR =
q L h2 − h3 275.75 − 101.61 = = = 5.57 win h2 − h1 275.75 − 244.46
11-95 A relation for the COP of the two-stage refrigeration system with a flash chamber shown in Fig. 1112 is to be derived. Analysis The coefficient of performance is determined from COPR =
qL win
where qL = (1 − x6 )(h1 − h8 ) with x6 =
h6 − h f h fg
win = wcompI,in + wcompII,in = (1 − x6 )(h2 − h1 ) + (1)(h4 − h9 )
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-58
11-96 A two-stage compression refrigeration system using refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the amount of heat removed from the refrigerated space, the compressor work, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The flashing chamber is adiabatic. Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables to be (Tables A-11, A-12, and A-13) h1 = 239.16 kJ/kg, h2 = 260.58 kJ/kg h3 = 255.55 kJ/kg,
T
h5 =95.47 kJ/kg,
h6 = 95.47 kJ/kg
h7 = 63.94 kJ/kg,
h8 = 63.94 kJ/kg
0.8 MPa 4
The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6, h6 − h f 95.47 − 63.94 = = 0.1646 x6 = 191.62 h fg
0.4 MPa
5
2
A 7
6 8
B
qL
(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber: ©0 (steady) E& − E& = ∆E& = 0 → E& = E& in
out
system
in
0.14 MPa
3 9 1
s
out
∑ m& h = ∑ m& h e e
i i
(1)h9 = x6 h3 + (1 − x 6 )h2 h9 = (0.1646)(255.55) + (1 − 0.1646)(260.58) = 259.75 kJ/kg
P9 = 0.4 MPa s 9 = 0.94168 kJ/kg ⋅ K h9 = 259.75 kJ/kg P4 = 0.8 MPa
h4 = 274.47 kJ/kg s 4 = s 9 = 0.94168 kJ/kg ⋅ K Then the amount of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are q L = (1 − x 6 )(h1 − h8 ) = (1 − 0.1646)(239.16 − 63.94 ) kJ/kg = 146.4 kJ/kg
Also,
win = wcompI,in + wcompII,in = (1 − x 6 )(h2 − h1 ) + (1)(h4 − h9 ) = (1 − 0.1646)(260.58 − 239.16 ) kJ/kg + (274.47 − 259.75) kJ/kg = 32.6 kJ/kg
(c) The coefficient of performance is determined from q 146.4 kJ/kg COPR = L = = 4.49 32.6 kJ/kg win
11-97 An aircraft on the ground is to be cooled by a gas refrigeration cycle operating with air on an open cycle. The temperature of the air leaving the turbine is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The specific heat ratio of air at room temperature is k = 1.4 (Table A-2). Analysis Assuming the turbine to be isentropic, the air temperature at the turbine exit is determined from
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-59
P T4 = T3 4 P3
(k −1) / k
100 kPa = (343 K ) 250 kPa
0.4 / 1.4
= 264 K = −9.0°C
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-60
11-98 A regenerative gas refrigeration cycle with helium as the working fluid is considered. The temperature of the helium at the turbine inlet, the COP of the cycle, and the net power input required are to be determined. Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2). Analysis (a) The temperature of the helium at the turbine inlet is determined from an energy balance on the regenerator, T
E& in − E& out = ∆E& system©0 (steady) = 0 E& in = E& out
∑ m& h = ∑ m& h e e
i i
→ m& (h3 − h4 ) = m& (h1 − h6 )
-10°C
1 4
m& c p (T3 − T4 ) = m& c p (T1 − T6 ) → T3 − T4 = T1 − T6 -25°C
T4 = T3 − T1 + T6 = 20°C − (− 10°C ) + (− 25°C ) = 5°C = 278 K
2
3
20°C
or, Thus,
· QH
5
· Qregen ·6 QRefrig
s
(b) From the isentropic relations, P T2 = T1 2 P1
(k −1) / k
P T5 = T4 5 P4
(k −1) / k
= (263 K )(3)0.667 / 1.667 = 408.2 K = 135.2°C 1 = (278 K ) 3
0.667 / 1.667
= 179.1 K = −93.9°C
Then the COP of this ideal gas refrigeration cycle is determined from COPR = =
h6 − h5 qL qL = = wnet,in wcomp,in − w turb,out (h2 − h1 ) − (h4 − h5 )
T6 − T5 − 25°C − (− 93.9°C ) = = 1.49 (T2 − T1 ) − (T4 − T5 ) [135.2 − (− 10)]°C − [5 − (− 93.9)]°C
(c) The net power input is determined from W& net,in = W& comp,in − W& turb,out = m& [(h2 − h1 ) − (h4 − h5 )] = m& c p [(T2 − T1 ) − (T4 − T5 )]
= (0.45 kg/s )(5.1926 kJ/kg ⋅ °C )([135.2 − (− 10 )] − [5 − (− 93.9 )]) = 108.2 kW
11-99 An absorption refrigeration system operating at specified conditions is considered. The minimum rate of heat supply required is to be determined. Analysis The maximum COP that this refrigeration system can have is T COPR, max = 1 − 0 Ts
Thus,
Q& gen,min =
T L T − T L 0
298K 263 = 1 − = 1.259 358K 298 − 263
Q& L 12 kW = = 9.53 kW COPR, max 1.259
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-61
11-100 EES Problem 11-99 is reconsidered. The effect of the source temperature on the minimum rate of heat supply is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" T_L = -10 [C] T_0 = 25 [C] T_s = 85 [C] Q_dot_L = 8 [kW] "The maximum COP that this refrigeration system can have is:" COP_R_max = (1-(T_0+273)/(T_s+273))*((T_L+273)/(T_0 - T_L)) "The minimum rate of heat supply is:" Q_dot_gen_min = Q_dot_L/COP_R_max Qgenmin [kW] 13.76 8.996 6.833 5.295 4.237 3.603 2.878 2.475
Ts [C] 50 65 80 100 125 150 200 250
14 12
] W k[ ni m ; n e g
Q
10 8 6 4 2 50
90
130
170
210
250
Ts [C]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-62
11-101 A house is cooled adequately by a 3.5 ton air-conditioning unit. The rate of heat gain of the house when the air-conditioner is running continuously is to be determined. Assumptions 1 The heat gain includes heat transfer through the walls and the roof, infiltration heat gain, solar heat gain, internal heat gain, etc. 2 Steady operating conditions exist. Analysis Noting that 1 ton of refrigeration is equivalent to a cooling rate of 211 kJ/min, the rate of heat gain of the house in steady operation is simply equal to the cooling rate of the air-conditioning system, Q& heat gain = Q& cooling = (3.5 ton)(211 kJ / min) = 738.5 kJ / min = 44,310 kJ / h
11-102 A room is cooled adequately by a 5000 Btu/h window air-conditioning unit. The rate of heat gain of the room when the air-conditioner is running continuously is to be determined. Assumptions 1 The heat gain includes heat transfer through the walls and the roof, infiltration heat gain, solar heat gain, internal heat gain, etc. 2 Steady operating conditions exist. Analysis The rate of heat gain of the room in steady operation is simply equal to the cooling rate of the airconditioning system, Q& heat gain = Q& cooling = 5,000 Btu / h
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-63
11-103 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) For this problem, we use the properties of air from EES: → h1 = 273.40 kJ/kg T1 = 0°C
.
QL
P1 = 100 kPa s1 = 5.6110 kJ/kg.K P2 = 500 kPa h2 s = 433.50 kJ/kg s 2 = s1
T1 = 0°C
h −h η C = 2s 1 h2 − h1
Heat Exch.
6
Regenerator
3
433.50 − 273.40 0.80 = h2 − 273.40
5
Heat Exch. 1
.
4
2
QH
h2 = 473.52 kJ/kg T3 = 35°C → h3 = 308.63 kJ/kg
Turbine
For the turbine inlet and exit we have
Compressor
T5 = −80°C → h5 = 193.45 kJ/kg → h4 = T4 = ?
ηT =
T
h4 − h5 h4 − h5 s
P5 = 100 kPa s1 = 5.6110 kJ/kg.K T1 = 0°C P4 = 500 kPa s4 = T4 = ?
· QH
2 2s
3
35°C 0°C
Qregen
1
4 -80°C
P2 = 500 kPa h5 s = s5 = s4
5s
·6 5 QRefrig
s
We can determine the temperature at the turbine inlet from EES using the above relations. A hand solution would require a trial-error approach. T4 = 281.8 K, h4 = 282.08 kJ/kg An energy balance on the regenerator gives h6 = h1 − h3 + h4 = 273.40 − 308.63 + 282.08 = 246.85 kJ/kg
The effectiveness of the regenerator is determined from
ε regen =
h3 − h4 308.63 − 282.08 = = 0.430 h3 − h6 308.63 − 246.85
(b) The refrigeration load is Q& L = m& (h6 − h5 ) = (0.4 kg/s)(246.85 − 193.45)kJ/kg = 21.36 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-64
(c) The turbine and compressor powers and the COP of the cycle are W& C,in = m& (h2 − h1 ) = (0.4 kg/s)(473.52 − 273.40)kJ/kg = 80.05 kW W& T,out = m& (h4 − h5 ) = (0.4 kg/s)(282.08 − 193.45)kJ/kg = 35.45 kW COP =
Q& L W&
net,in
=
Q& L W& C,in − W& T, out
=
21.36 = 0.479 80.05 − 35.45
(d) The simple gas refrigeration cycle analysis is as follows: h1 = 273.40 kJ/kg h2 = 473.52 kJ/kg
2 T
h3 = 308.63 kJ/kg P3 = 500 kPa s 3 = 5.2704 kJ/kg T3 = 35°C
35°C 0°C
P1 = 100 kPa h4 s = 194.52 kJ/kg.K
s 4 = s3
ηT =
· QH
2
3 1 4 Q· Refrig 4s s
h3 − h4 308.63 − h4 → 0.85 = → h4 = 211.64 kJ/kg h3 − h4 s 308.63 − 194.52
Q& L = m& (h1 − h4 ) = (0.4 kg/s)(273.40 − 211.64)kJ/kg = 24.70 kW W& net,in = m& (h2 − h1 ) − m& (h3 − h4 ) = (0.4 kg/s)[(473.52 − 273.40) − (308.63 − 211.64)kJ/kg ] = 41.25 kW COP =
Q& L 24.70 = = 0.599 & W net,in 41.25
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-65
11-104 An air-conditioner with refrigerant-134a as the refrigerant is considered. The temperature of the refrigerant at the compressor exit, the rate of heat generated by the people in the room, the COP of the airconditioner, and the minimum volume flow rate of the refrigerant at the compressor inlet are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant-134a tables (Tables A-11 through A-13) h = 259.30 kJ/kg P1 = 500 kPa 1 34°C v . 1 = 0.04112 kJ/kg x1 = 1 s = 0.9240 kJ/kg QH 1200 kPa
1
P2 = 1200 kPa h2 s = 277.39 h3 = h f@1200 kPa = 117.77 kJ/kg s 2 = s1
h4 = h3 = 117.77 kJ/kg h −h η C = 2s 1 h2 − h1 0.75 =
277.39 − 259.30 → h2 = 283.42 kJ/kg h2 − 259.30
Condenser 3
2
Expansion valve
Compressor 1
4
Evaporator . QL
P2 = 1200 kPa T2 = 54.5°C h2 = 283.42 kJ/kg
(b) The mass flow rate of the refrigerant is 1 m 3 1 min (100 L/min) 1000 L 60 s V& 1 m& = = 0.04053 kg/s = v1 0.04112 m 3 /kg The refrigeration load is Q& L = m& (h1 − h4 )
. Win
500 kPa
26°C T
= (0.04053 kg/s)(259.30 − 117.77)kJ/kg = 5.737 kW which is the total heat removed from the room. Then, the rate of heat generated by the people in the room is determined from Q& people = Q& L − Q& heat − Q& equip = (5.737 − 250 / 60 − 0.9) kW = 0.67 kW
2 · QH
2s · Win
3
4
· QL
1 s
(c) The power input and the COP are W& in = m& (h2 − h1 ) = (0.04053 kg/s)(283.42 − 259.30)kJ/kg = 0.9774 kW Q& 5.737 COP = L = = 5.87 & 0 .9774 Win (d) The reversible COP of the cycle is 1 1 COPrev = = = 37.38 T H / T L − 1 (34 + 273) /( 26 + 273) − 1 The corresponding minimum power input is Q& L 5.737 kW W& in, min = = = 0.1535 kW COPrev 37.38 The minimum mass and volume flow rates are W& in, min 0.1535 kW = = 0.006366 kg/s m& min = h2 − h1 (283.42 − 259.30)kJ/kg V& = m& v = (0.006366 kg/s)(0.04112 m 3 /kg) = (0.0002617 m 3 /s) = 15.7 L/min 1, min
min 1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-66
11-105 A heat pump water heater has a COP of 2.2 and consumes 2 kW when running. It is to be determined if this heat pump can be used to meet the cooling needs of a room by absorbing heat from it. Assumptions The COP of the heat pump remains constant whether heat is absorbed from the outdoor air or room air. Analysis The COP of the heat pump is given to be 2.2. Then the COP of the air-conditioning system becomes COPair-cond = COPheat pump − 1 = 2.2 − 1 = 1.2
Then the rate of cooling (heat absorption from the air) becomes Q& cooling = COPair-cond W& in = (1.2)(2 kW) = 2.4 kW = 8640 kJ / h
since 1 kW = 3600 kJ/h. We conclude that this heat pump can meet the cooling needs of the room since its cooling rate is greater than the rate of heat gain of the room.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-67
11-106 A vortex tube receives compressed air at 500 kPa and 300 K, and supplies 25 percent of it as cold air and the rest as hot air. The COP of the vortex tube is to be compared to that of a reversed Brayton cycle for the same pressure ratio; the exit temperature of the hot fluid stream and the COP are to be determined; and it is to be shown if this process violates the second law. Assumptions 1 The vortex tube is adiabatic. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Steady operating conditions exist. Properties The gas constant of air is 0.287 kJ/kg.K (Table A-1). The specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The enthalpy of air at absolute temperature T can be expressed in terms of specific heats as h = cpT. Analysis (a) The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of the same pressure ratio since the vortex tube involves vortices, which are highly irreversible. Owing to this irreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the one that can be obtained by the revered Brayton cycle. (b) We take the vortex tube as the system. This is a steady flow system with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equation for this system E& in = E& out for a unit mass flow rate at the inlet (m& 1 = 1 kg / s) can be expressed as m& 1 h1 = m& 2 h2 + m& 3 h3 m& 1 c p T1 = m& 2 c p T2 + m& 3 c p T3
Compressed air
1c p T1 = 0.25c p T2 + 0.75c p T3
Canceling cp and solving for T3 gives T − 0.25T2 T3 = 1 0.75 300 − 0.25 × 278 = = 307.3 K 0.75
1
Cold air 2
Warm air
3
Therefore, the hot air stream will leave the vortex tube at an average temperature of 307.3 K. (c) The entropy balance for this steady flow system S&in − S&out + S&gen = 0 can be expressed as with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equation for this system for a unit mass flow rate at the inlet (m& 1 = 1 kg / s) can be expressed S& gen = S& out − S& in = m& 2 s 2 + m& 3 s 3 − m& 1 s1 = m& 2 s 2 + m& 3 s 3 − (m& 2 + m& 3 ) s1 = m& 2 ( s 2 − s1 ) + m& 3 ( s 3 − s1 ) = 0.25( s 2 − s1 ) + 0.75( s 3 − s1 ) T P P T = 0.25 c p ln 2 − R ln 2 + 0.75 c p ln 3 − R ln 3 P1 T1 T1 P1
Substituting the known quantities, the rate of entropy generation is determined to be 100 kPa 278 K S& gen = 0.25 (1.005 kJ/kg.K) ln − (0.287 kJ/kg.K) ln 500 kPa 300 K 307.3 K 100 kPa + 0.75 (1.005 kJ/kg.K) ln − (0.287 kJ/kg.K) ln 300 K 500 kPa = 0.461 kW/K > 0
which is a positive quantity. Therefore, this process satisfies the 2nd law of thermodynamics.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-68
(d) For a unit mass flow rate at the inlet (m& 1 = 1 kg / s) , the cooling rate and the power input to the compressor are determined to Q& cooling = m& c (h1 − hc ) = m& c c p (T1 − Tc ) = (0.25 kg/s)(1.005 kJ/kg.K)(300 - 278)K = 5.53 kW W& comp,in = =
m& 0 RT0 (k − 1)η comp
P 1 P0
( k −1) / k
− 1
(1 kg/s)(0.287 kJ/kg.K)(300 K) 500 kPa (1.4 − 1)0.80 100 kPa
(1.4 −1) / 1.4
− 1 = 157.1 kW
Then the COP of the vortex refrigerator becomes Q& cooling 5.53 kW COP = = = 0.035 & Wcomp, in 157.1 kW The COP of a Carnot refrigerator operating between the same temperature limits of 300 K and 278 K is COPCarnot =
TL 278 K = = 12.6 TH − TL (300 − 278) K
Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnot refrigerator operating between the same temperature limits.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-69
11-107 A vortex tube receives compressed air at 600 kPa and 300 K, and supplies 25 percent of it as cold air and the rest as hot air. The COP of the vortex tube is to be compared to that of a reversed Brayton cycle for the same pressure ratio; the exit temperature of the hot fluid stream and the COP are to be determined; and it is to be shown if this process violates the second law. Assumptions 1 The vortex tube is adiabatic. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Steady operating conditions exist. Properties The gas constant of air is 0.287 kJ/kg.K (Table A-1). The specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The enthalpy of air at absolute temperature T can be expressed in terms of specific heats as h = cp T. Analysis (a) The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of the same pressure ratio since the vortex tube involves vortices, which are highly irreversible. Owing to this irreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the one that can be obtained by the revered Brayton cycle. (b) We take the vortex tube as the system. This is a steady flow system with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equation for this system E& in = E& out for a unit mass flow rate at the inlet (m& 1 = 1 kg / s) can be expressed as m& 1 h1 = m& 2 h2 + m& 3 h3 m& 1 c p T1 = m& 2 c p T2 + m& 3 c p T3
Compressed air
1c p T1 = 0.25c p T2 + 0.75c p T3
Canceling cp and solving for T3 gives T − 0.25T2 T3 = 1 0.75 300 − 0.25 × 278 = = 307.3 K 0.75
1
Cold air 2
Warm air
3
Therefore, the hot air stream will leave the vortex tube at an average temperature of 307.3 K. (c) The entropy balance for this steady flow system S&in − S&out + S&gen = 0 can be expressed as with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equation for this system for a unit mass flow rate at the inlet (m& 1 = 1 kg / s) can be expressed S& gen = S& out − S& in = m& 2 s 2 + m& 3 s 3 − m& 1 s1 = m& 2 s 2 + m& 3 s 3 − (m& 2 + m& 3 ) s1 = m& 2 ( s 2 − s1 ) + m& 3 ( s 3 − s1 ) = 0.25( s 2 − s1 ) + 0.75( s 3 − s1 ) T P P T = 0.25 c p ln 2 − R ln 2 + 0.75 c p ln 3 − R ln 3 P1 T1 P1 T1
Substituting the known quantities, the rate of entropy generation is determined to be 278 K 100 kPa S&gen = 0.25 (1.005 kJ/kg.K)ln − (0.287 kJ/kg.K)ln 300 K 600 kPa 307.3 K 100 kPa + 0.75 (1.005 kJ/kg.K)ln − (0.287 kJ/kg.K)ln 300 K 600 kPa = 0.513 kW/K > 0
which is a positive quantity. Therefore, this process satisfies the 2nd law of thermodynamics.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-70
(d) For a unit mass flow rate at the inlet (m& 1 = 1 kg / s) , the cooling rate and the power input to the compressor are determined to Q& cooling = m& c (h1 − hc ) = m& c c p (T1 − Tc ) = (0.25 kg/s)(1.005 kJ/kg.K)(300 - 278)K = 5.53 kW W& comp,in = =
m& 0 RT0 (k − 1)η comp
P 1 P0
( k −1) / k
− 1
(1 kg/s)(0.287 kJ/kg.K)(300 K) 600 kPa (1.4 − 1)0.80 100 kPa
(1.4 −1) / 1.4
− 1 = 179.9 kW
Then the COP of the vortex refrigerator becomes Q& cooling 5.53 kW COP = = = 0.031 & Wcomp, in 179.9 kW The COP of a Carnot refrigerator operating between the same temperature limits of 300 K and 278 K is COPCarnot =
TL 278 K = = 12.6 TH − TL (300 − 278) K
Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnot refrigerator operating between the same temperature limits.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-71
11-108 EES The effect of the evaporator pressure on the COP of an ideal vapor-compression refrigeration cycle with R-134a as the working fluid is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P[1]=100 [kPa] P[2] = 1000 [kPa] Fluid$='R134a' Eta_c=0.7 "Compressor isentropic efficiency" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" W_c=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+W_c=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] "energy balance on condenser" "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" "Coefficient of Performance:" COP=Q_in/W_c "definition of COP" COP 1.851 2.863 4.014 5.462 7.424
ηc 0.7 0.7 0.7 0.7 0.7
P1 [kPa] 100 200 300 400 500
10
η comp
8
1.0 0.7
6
P O C
4 2 0 100
150
200
250
300
350
400
450
P[1] [kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
500
11-72
11-109 EES The effect of the condenser pressure on the COP of an ideal vapor-compression refrigeration cycle with R-134a as the working fluid is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P[1]=120 [kPa] P[2] = 400 [kPa] Fluid$='R134a' Eta_c=0.7 "Compressor isentropic efficiency" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" W_c=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+W_c=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] "energy balance on condenser" "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" "Coefficient of Performance:" COP=Q_in/W_c "definition of COP"
COP 4.935 3.04 2.258 1.803 1.492
ηc 0.7 0.7 0.7 0.7 0.7
8
P2 [kPa] 400 650 900 1150 1400
7
η comp
6
1.0 0.7
5
P O C
4 3 2 1 0 400
600
800
1000
1200
1400
P[2] [kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-73
Fundamentals of Engineering (FE) Exam Problems
11-110 Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation dome between the pressure limits of 140 kPa and 800 kPa. R-134a changes from saturated vapor to saturated liquid during the heat rejection process. The net work input for this cycle is (a) 28 kJ/kg (b) 34 kJ/kg (c) 49 kJ/kg (d) 144 kJ/kg (e) 275 kJ/kg Answer (a) 28 kJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=800 "kPa" P2=140 "kPa" h_fg=ENTHALPY(R134a,x=1,P=P1)-ENTHALPY(R134a,x=0,P=P1) TH=TEMPERATURE(R134a,x=0,P=P1)+273 TL=TEMPERATURE(R134a,x=0,P=P2)+273 q_H=h_fg COP=TH/(TH-TL) w_net=q_H/COP "Some Wrong Solutions with Common Mistakes:" W1_work = q_H/COP1; COP1=TL/(TH-TL) "Using COP of regrigerator" W2_work = q_H/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K" W3_work = h_fg3/COP; h_fg3= ENTHALPY(R134a,x=1,P=P2)-ENTHALPY(R134a,x=0,P=P2) "Using h_fg at P2" W4_work = q_H*TL/TH "Using the wrong relation"
11-111 A refrigerator removes heat from a refrigerated space at –5°C at a rate of 0.35 kJ/s and rejects it to an environment at 20°C. The minimum required power input is (a) 30 W (b) 33 W (c) 56 W (d) 124 W (e) 350 W Answer (b) 33 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TH=20+273 TL=-5+273 Q_L=0.35 "kJ/s" COP_max=TL/(TH-TL) w_min=Q_L/COP_max "Some Wrong Solutions with Common Mistakes:" W1_work = Q_L/COP1; COP1=TH/(TH-TL) "Using COP of heat pump" W2_work = Q_L/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K" W3_work = Q_L*TL/TH "Using the wrong relation" W4_work = Q_L "Taking the rate of refrigeration as power input"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-74
11-112 A refrigerator operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 120 kPa and 800 kPa. If the rate of heat removal from the refrigerated space is 32 kJ/s, the mass flow rate of the refrigerant is (a) 0.19 kg/s (b) 0.15 kg/s (c) 0.23 kg/s (d) 0.28 kg/s (e) 0.81 kg/s Answer (c) 0.23 kg/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=120 "kPa" P2=800 "kPa" P3=P2 P4=P1 s2=s1 Q_refrig=32 "kJ/s" m=Q_refrig/(h1-h4) h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 "Some Wrong Solutions with Common Mistakes:" W1_mass = Q_refrig/(h2-h1) "Using wrong enthalpies, for W_in" W2_mass = Q_refrig/(h2-h3) "Using wrong enthalpies, for Q_H" W3_mass = Q_refrig/(h1-h44); h44=ENTHALPY(R134a,x=0,P=P4) "Using wrong enthalpy h4 (at P4)" W4_mass = Q_refrig/h_fg; h_fg=ENTHALPY(R134a,x=1,P=P2) - ENTHALPY(R134a,x=0,P=P2) "Using h_fg at P2"
11-113 A heat pump operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. If the mass flow rate of the refrigerant is 0.193 kg/s, the rate of heat supply by the heat pump to the heated space is (a) 3.3 kW (b) 23 kW (c) 26 kW (d) 31 kW (e) 45 kW Answer (d) 31 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=320 "kPa" P2=1200 "kPa" P3=P2 P4=P1 s2=s1 m=0.193 "kg/s" Q_supply=m*(h2-h3) "kJ/s" h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-75
h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 "Some Wrong Solutions with Common Mistakes:" W1_Qh = m*(h2-h1) "Using wrong enthalpies, for W_in" W2_Qh = m*(h1-h4) "Using wrong enthalpies, for Q_L" W3_Qh = m*(h22-h4); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)" W4_Qh = m*h_fg; h_fg=ENTHALPY(R134a,x=1,P=P1) - ENTHALPY(R134a,x=0,P=P1) "Using h_fg at P1"
11-114 An ideal vapor compression refrigeration cycle with R-134a as the working fluid operates between the pressure limits of 120 kPa and 1000 kPa. The mass fraction of the refrigerant that is in the liquid phase at the inlet of the evaporator is (a) 0.65 (b) 0.60 (c) 0.40 (d) 0.55 (e) 0.35 Answer (b) 0.60 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=120 "kPa" P2=1000 "kPa" P3=P2 P4=P1 h1=ENTHALPY(R134a,x=1,P=P1) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 x4=QUALITY(R134a,h=h4,P=P4) liquid=1-x4 "Some Wrong Solutions with Common Mistakes:" W1_liquid = x4 "Taking quality as liquid content" W2_liquid = 0 "Assuming superheated vapor" W3_liquid = 1-x4s; x4s=QUALITY(R134a,s=s3,P=P4) "Assuming isentropic expansion" s3=ENTROPY(R134a,x=0,P=P3)
11-115 Consider a heat pump that operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. The coefficient of performance of this heat pump is (a) 0.17 (b) 1.2 (c) 3.1 (d) 4.9 (e) 5.9 Answer (e) 5.9 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=320 "kPa" P2=1200 "kPa" P3=P2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-76
P4=P1 s2=s1 h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 COP_HP=qH/Win Win=h2-h1 qH=h2-h3 "Some Wrong Solutions with Common Mistakes:" W1_COP = (h1-h4)/(h2-h1) "COP of refrigerator" W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH" W3_COP = (h22-h3)/(h22-h1); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)"
11-116 An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limits of 80 kPa and 280 kPa. Air is cooled to 35°C before entering the turbine. The lowest temperature of this cycle is (a) –58°C (b) -26°C (c) 0°C (d) 11°C (e) 24°C Answer (a) –58°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 P1= 80 "kPa" P2=280 "kPa" T3=35+273 "K" "Mimimum temperature is the turbine exit temperature" T4=T3*(P1/P2)^((k-1)/k) - 273 "Some Wrong Solutions with Common Mistakes:" W1_Tmin = (T3-273)*(P1/P2)^((k-1)/k) "Using C instead of K" W2_Tmin = T3*(P1/P2)^((k-1)) - 273 "Using wrong exponent" W3_Tmin = T3*(P1/P2)^k - 273 "Using wrong exponent"
11-117 Consider an ideal gas refrigeration cycle using helium as the working fluid. Helium enters the compressor at 100 kPa and –10°C and is compressed to 250 kPa. Helium is then cooled to 20°C before it enters the turbine. For a mass flow rate of 0.2 kg/s, the net power input required is (a) 9.3 kW (b) 27.6 kW (c) 48.8 kW (d) 93.5 kW (e) 119 kW Answer (b) 27.6 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-77
k=1.667 Cp=5.1926 "kJ/kg.K" P1= 100 "kPa" T1=-10+273 "K" P2=250 "kPa" T3=20+273 "K" m=0.2 "kg/s" "Mimimum temperature is the turbine exit temperature" T2=T1*(P2/P1)^((k-1)/k) T4=T3*(P1/P2)^((k-1)/k) W_netin=m*Cp*((T2-T1)-(T3-T4)) "Some Wrong Solutions with Common Mistakes:" W1_Win = m*Cp*((T22-T1)-(T3-T44)); T22=T1*P2/P1; T44=T3*P1/P2 "Using wrong relations for temps" W2_Win = m*Cp*(T2-T1) "Ignoring turbine work" W3_Win=m*1.005*((T2B-T1)-(T3-T4B)); T2B=T1*(P2/P1)^((kB-1)/kB); T4B=T3*(P1/P2)^((kB1)/kB); kB=1.4 "Using air properties" W4_Win=m*Cp*((T2A-(T1-273))-(T3-273-T4A)); T2A=(T1-273)*(P2/P1)^((k-1)/k); T4A=(T3273)*(P1/P2)^((k-1)/k) "Using C instead of K"
11-118 An absorption air-conditioning system is to remove heat from the conditioned space at 20°C at a rate of 150 kJ/s while operating in an environment at 35°C. Heat is to be supplied from a geothermal source at 140°C. The minimum rate of heat supply required is (a) 86 kJ/s (b) 21 kJ/s (c) 30 kJ/s (d) 61 kJ/s (e) 150 kJ/s Answer (c) 30 kJ/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=20+273 "K" Q_refrig=150 "kJ/s" To=35+273 "K" Ts=140+273 "K" COP_max=(1-To/Ts)*(TL/(To-TL)) Q_in=Q_refrig/COP_max "Some Wrong Solutions with Common Mistakes:" W1_Qin = Q_refrig "Taking COP = 1" W2_Qin = Q_refrig/COP2; COP2=TL/(Ts-TL) "Wrong COP expression" W3_Qin = Q_refrig/COP3; COP3=(1-To/Ts)*(Ts/(To-TL)) "Wrong COP expression, COP_HP" W4_Qin = Q_refrig*COP_max "Multiplying by COP instead of dividing"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
11-78
11-119 Consider a refrigerator that operates on the vapor compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 160 kPa, and exits at 800 kPa and 50°C, and leaves the condenser as saturated liquid at 800 kPa. The coefficient of performance of this refrigerator is (a) 2.6 (b) 1.0 (c) 4.2 (d) 3.2 (e) 4.4 Answer (d) 3.2 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=160 "kPa" P2=800 "kPa" T2=50 "C" P3=P2 P4=P1 h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,T=T2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 COP_R=qL/Win Win=h2-h1 qL=h1-h4 "Some Wrong Solutions with Common Mistakes:" W1_COP = (h2-h3)/(h2-h1) "COP of heat pump" W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH" W3_COP = (h1-h4)/(h2s-h1); h2s=ENTHALPY(R134a,s=s1,P=P2) "Assuming isentropic compression"
11-120 ··· 11-129 Design and Essay Problems
KJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-1
Chapter 12 THERMODYNAMIC PROPERTY RELATIONS Partial Derivatives and Associated Relations 12-1C
z dz
∂x ≡ dx (∂z)y
∂y ≡ dy dz = (∂z ) x + (∂z ) y
(∂z)x
y
y + dy dy
x
y
dx x +dx
x 12-2C For functions that depend on one variable, they are identical. For functions that depend on two or more variable, the partial differential represents the change in the function with one of the variables as the other variables are held constant. The ordinary differential for such functions represents the total change as a result of differential changes in all variables. 12-3C (a) (∂x)y = dx ; (b) (∂z) y ≤ dz; and (c) dz = (∂z)x + (∂z) y 12-4C Only when (∂z/∂x) y = 0. That is, when z does not depend on y and thus z = z(x). 12-5C It indicates that z does not depend on y. That is, z = z(x). 12-6C Yes. 12-7C Yes.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-2
12-8 Air at a specified temperature and specific volume is considered. The changes in pressure corresponding to a certain increase of different properties are to be determined. Assumptions Air is an ideal gas Properties The gas constant of air is R = 0.287 kPa·m3/kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = RT/v. Noting that R is a constant and P = P(T,v), R dT RT dv ∂P ∂P dP = − dT + dv = v v2 ∂T v ∂v T
(a) The change in T can be expressed as dT ≅ ∆T = 400 × 0.01 = 4.0 K. At v = constant,
(dP )v
=
R dT
v
=
(0.287 kPa ⋅ m 3 /kg ⋅ K)(4.0 K) 0.90 m 3 /kg
= 1.276 kPa
(b) The change in v can be expressed as dv ≅ ∆v = 0.90 × 0.01 = 0.009 m3/kg. At T = constant,
(dP )T
=−
RT dv
v2
=−
(0.287 kPa ⋅ m 3 /kg ⋅ K)(400K)(0.009 m 3 /kg) (0.90 m 3 /kg) 2
= −1.276 kPa
(c) When both v and T increases by 1%, the change in P becomes dP = (dP)v + (dP )T = 1.276 + (−1.276) = 0
Thus the changes in T and v balance each other.
12-9 Helium at a specified temperature and specific volume is considered. The changes in pressure corresponding to a certain increase of different properties are to be determined. Assumptions Helium is an ideal gas Properties The gas constant of helium is R = 2.0769 kPa·m3/kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = RT/v. Noting that R is a constant and P = P(T, v ), R dT RT dv ∂P ∂P dP = − dT + dv = v v2 ∂T v ∂v T
(a) The change in T can be expressed as dT ≅ ∆T = 400 × 0.01 = 4.0 K. At v = constant,
(dP )v
=
R dT
v
=
(2.0769 kPa ⋅ m 3 /kg ⋅ K)(4.0 K) 0.90 m 3 /kg
= 9.231 kPa
(b) The change in v can be expressed as d v ≅ ∆ v = 0.90 × 0.01 = 0.009 m3/kg. At T = constant,
(dP )T
=−
RT dv
v2
=
(2.0769 kPa ⋅ m 3 /kg ⋅ K)(400 K)(0.009 m 3 ) (0.90 m 3 /kg) 2
= −9.231 kPa
(c) When both v and T increases by 1%, the change in P becomes dP = (dP) v + (dP) T = 9.231 + (−9.231) = 0
Thus the changes in T and v balance each other.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-3
12-10 It is to be proven for an ideal gas that the P = constant lines on a T- v diagram are straight lines and that the high pressure lines are steeper than the low-pressure lines. Analysis (a) For an ideal gas Pv = RT or T = Pv/R. Taking the partial derivative of T with respect to v holding P constant yields P ∂T = ∂ v P R
which remains constant at P = constant. Thus the derivative (∂T/∂v)P, which represents the slope of the P = const. lines on a T-v diagram, remains constant. That is, the P = const. lines are straight lines on a T-v diagram. (b) The slope of the P = const. lines on a T-v diagram is equal to P/R, which is proportional to P. Therefore, the high pressure lines are steeper than low pressure lines on the T-v diagram.
T P = const
v
12-11 A relation is to be derived for the slope of the v = constant lines on a T-P diagram for a gas that obeys the van der Waals equation of state. Analysis The van der Waals equation of state can be expressed as T=
a 1 P + 2 (v − b ) R v
Taking the derivative of T with respect to P holding v constant, 1 v −b ∂T = (1 + 0 )(v − b ) = R ∂P v R
which is the slope of the v = constant lines on a T-P diagram.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-4
12-12 Nitrogen gas at a specified state is considered. The cp and cv of the nitrogen are to be determined using Table A-18, and to be compared to the values listed in Table A-2b. Analysis The cp and cv of ideal gases depends on temperature only, and are expressed as cp(T) = dh(T)/dT and cv(T) = du(T)/dT. Approximating the differentials as differences about 400 K, the cp and cv values are determined to be dh(T ) ∆h(T ) ≅ c p (400 K ) = dT T = 400 K ∆T T ≅ 400 K =
h(410 K ) − h(390 K ) (410 − 390)K
=
(11,932 − 11,347)/28.0 kJ/kg (410 − 390)K
h
cp
= 1.045 kJ/kg ⋅ K
(Compare: Table A-2b at 400 K → cp = 1.044 kJ/kg·K)
T
∆u (T ) du (T ) cv (400K ) = ≅ dT T = 400 K ∆T T ≅ 400 K =
u (410 K ) − u (390 K ) (410 − 390)K
=
(8,523 − 8,104)/28.0 kJ/kg = 0.748 kJ/kg ⋅ K (410 − 390)K
(Compare: Table A-2b at 400 K → cv = 0.747 kJ/kg·K)
12-13E Nitrogen gas at a specified state is considered. The cp and cv of the nitrogen are to be determined using Table A-18E, and to be compared to the values listed in Table A-2Eb. Analysis The cp and cv of ideal gases depends on temperature only, and are expressed as cp(T) = dh(T)/dT and cv(T) = du(T)/dT. Approximating the differentials as differences about 600 R, the cp and cv values are determined to be ∆h(T ) dh(T ) c p (600 R ) = ≅ dT T = 600 R ∆T T ≅ 600 R =
h(620 R ) − h(580 R ) (620 − 580)R
=
(4,307.1 − 4,028.7)/28.0 Btu/lbm = 0.249 Btu/lbm ⋅ R (620 − 580) R
(Compare: Table A-2Eb at 600 R → cp = 0.248 Btu/lbm·R ) ∆u (T ) du (T ) cv (600 R ) = ≅ dT T = 600 R ∆T T ≅ 600 R =
u (620 R ) − u (580 R ) (620 − 580)R
=
(3,075.9 − 2,876.9)/28.0 Btu/lbm = 0.178 Btu/lbm ⋅ R (620 − 580) R
(Compare: Table A-2Eb at 600 R → cv = 0.178 Btu/lbm·R)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-5
12-14 The state of an ideal gas is altered slightly. The change in the specific volume of the gas is to be determined using differential relations and the ideal-gas relation at each state. Assumptions The gas is air and air is an ideal gas. Properties The gas constant of air is R = 0.287 kPa·m3/kg·K (Table A-1). Analysis (a) The changes in T and P can be expressed as dT ≅ ∆T = (404 − 400)K = 4 K dP ≅ ∆P = (96 − 100)kPa = −4 kPa
The ideal gas relation Pv = RT can be expressed as v = RT/P. Note that R is a constant and v = v (T, P). Applying the total differential relation and using average values for T and P, R dT RT dP ∂v ∂v − dv = dT + dP = P ∂T P ∂P T P2 4K (402 K)(−4 kPa) = (0.287 kPa ⋅ m 3 /kg ⋅ K) − 98 kPa (98 kPa) 2 = (0.0117 m 3 /kg) + (0.04805 m 3 /kg) = 0.0598 m 3 /kg
(b) Using the ideal gas relation at each state,
v1 =
RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(400 K) = = 1.1480 m 3 /kg P1 100 kPa
v2 =
RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(404 K) = = 1.2078 m 3 /kg P2 96 kPa
Thus, ∆v = v 2 − v1 = 1.2078 − 1.1480 = 0.0598 m3 /kg
The two results are identical.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-6
12-15 Using the equation of state P(v-a) = RT, the cyclic relation, and the reciprocity relation at constant v are to be verified. Analysis (a) This equation of state involves three variables P, v, and T. Any two of these can be taken as the independent variables, with the remaining one being the dependent variable. Replacing x, y, and z by P, v, and T, the cyclic relation can be expressed as ∂P ∂v ∂T = −1 ∂v T ∂T P ∂P v
where RT P − RT ∂P → =− = 2 v −a v −a ∂v T (v − a ) RT R v ∂ v= +a → = P ∂T P P P (v − a) v −a ∂T T= → = R R ∂P v P=
Substituting, P R v − a ∂P ∂v ∂T = − = −1 ∂ T P ∂ − a P R ∂ v v T P v
which is the desired result. (b) The reciprocity rule for this gas at v = constant can be expressed as 1 ∂P = ∂T v (∂T / ∂P)v P(v − a) v −a ∂T T= → = R R ∂P v RT R ∂P → P= = ∂ v −a T v −a v
We observe that the first differential is the inverse of the second one. Thus the proof is complete.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-7
The Maxwell Relations
12-16 The validity of the last Maxwell relation for refrigerant-134a at a specified state is to be verified. Analysis We do not have exact analytical property relations for refrigerant-134a, and thus we need to replace the differential quantities in the last Maxwell relation with the corresponding finite quantities. Using property values from the tables about the specified state, ∂s ? ∂v =− ∂P T ∂T P ? ∆s ∆v ≅− ∆ P T =80°C ∆T P =1200 kPa
s1400 kPa − s1000 kPa (1400 − 1000 )kPa
? v 100°C − v 60°C ≅ − T =80°C (100 − 60 )°C
P =1200kPa
(1.0056 − 1.0458)kJ/kg ⋅ K ? (0.022442 − 0.018404)m 3 /kg ≅− (1400 − 1000)kPa (100 − 60)°C − 1.005 × 10 − 4 m 3 /kg ⋅ K ≅ −1.0095 × 10 − 4 m 3 /kg ⋅ K
since kJ ≡ kPa·m³, and K ≡ °C for temperature differences. Thus the last Maxwell relation is satisfied.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-8
12-17 EES Problem 12-16 is reconsidered. The validity of the last Maxwell relation for refrigerant 134a at the specified state is to be verified. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" T=80 [C] P=1200 [kPa] P_increment = 200 [kPa] T_increment = 20 [C] P[2]=P+P_increment P[1]=P-P_increment T[2]=T+T_increment T[1]=T-T_increment DELTAP = P[2]-P[1] DELTAT = T[2]-T[1] v[1]=volume(R134a,T=T[1],P=P) v[2]=volume(R134a,T=T[2],P=P) s[1]=entropy(R134a,T=T,P=P[1]) s[2]=entropy(R134a,T=T,P=P[2]) DELTAs=s[2] - s[1] DELTAv=v[2] - v[1] "The partial derivatives in the last Maxwell relation (Eq. 11-19) is associated with the Gibbs function and are approximated by the ratio of ordinary differentials:" LeftSide =DELTAs/DELTAP*Convert(kJ,m^3-kPa) "[m^3/kg-K]" "at T = Const." RightSide=-DELTAv/DELTAT "[m^3/kg-K]" "at P = Const." SOLUTION DELTAP=400 [kPa] DELTAs=-0.04026 [kJ/kg-K] DELTAT=40 [C] DELTAv=0.004038 [m^3/kg] LeftSide=-0.0001007 [m^3/kg-K] P=1200 [kPa] P[1]=1000 [kPa] P[2]=1400 [kPa] P_increment=200 [kPa]
RightSide=-0.000101 [m^3/kg-K] s[1]=1.046 [kJ/kg-K] s[2]=1.006 [kJ/kg-K] T=80 [C] T[1]=60 [C] T[2]=100 [C] T_increment=20 [C] v[1]=0.0184 [m^3/kg] v[2]=0.02244 [m^3/kg]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-9
12-18E The validity of the last Maxwell relation for steam at a specified state is to be verified. Analysis We do not have exact analytical property relations for steam, and thus we need to replace the differential quantities in the last Maxwell relation with the corresponding finite quantities. Using property values from the tables about the specified state, ∂s ? ∂v =− ∂P T ∂T P ? ∆s ∆v ≅− ∆ P T =800°F ∆T P = 400psia
s 450 psia − s 350 psia (450 − 350)psia
? v 900° F − v 700° F ≅ − T =800°F (900 − 700 )°F
P = 400psia
(1.6706 − 1.7009)Btu/lbm ⋅ R ? (1.9777 − 1.6507)ft 3 /lbm ≅− (450 − 350)psia (900 − 700)°F − 1.639 × 10 −3 ft 3 /lbm ⋅ R ≅ −1.635 × 10 −3 ft 3 /lbm ⋅ R
since 1 Btu ≡ 5.4039 psia·ft3, and R ≡ °F for temperature differences. Thus the fourth Maxwell relation is satisfied.
12-19 Using the Maxwell relations, a relation for (∂s/∂P)T for a gas whose equation of state is P(v-b) = RT is to be obtained. Analysis This equation of state can be expressed as v =
RT + b . Then, P
R ∂v = ∂ T P P
From the fourth Maxwell relation, R ∂s ∂v = − =− P ∂P T ∂T P
12-20 Using the Maxwell relations, a relation for (∂s/∂v)T for a gas whose equation of state is (P-a/v2)(v-b) = RT is to be obtained. Analysis This equation of state can be expressed as P =
RT
+
a
v −b v 2
. Then,
R ∂P = ∂T v v − b
From the third Maxwell relation, R ∂s ∂P = = ∂v T ∂T v v − b
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-10
12-21 Using the Maxwell relations and the ideal-gas equation of state, a relation for (∂s/∂v)T for an ideal gas is to be obtained. Analysis The ideal gas equation of state can be expressed as P =
RT
v
. Then,
R ∂P = ∂ v T v
From the third Maxwell relation, R ∂s ∂P = = ∂v T ∂T v v
The Clapeyron Equation 12-22C It enables us to determine the enthalpy of vaporization from hfg at a given temperature from the P, v, T data alone. 12-23C It is exact. 12-24C It is assumed that vfg ≅ vg ≅ RT/P, and hfg ≅ constant for small temperature intervals.
12-25 Using the Clapeyron equation, the enthalpy of vaporization of refrigerant-134a at a specified temperature is to be estimated and to be compared to the tabulated data. Analysis From the Clapeyron equation, dP h fg = Tv fg dT sat ∆P ≅ T (v g − v f ) @ 40°C ∆T sat,40°C Psat @ 42°C − Psat @38°C = T (v g − v f ) @ 40°C 42°C − 38°C
(1072.8 − 963.68)kPa = (40 + 273.15 K)(0.019952 − 0.0008720 m 3 /kg) 4K = 163.00 kJ/kg
The tabulated value of hfg at 40°C is 163.00 kJ/kg.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-11
12-26 EES Problem 12-25 is reconsidered. The enthalpy of vaporization of refrigerant 134-a as a function of temperature over the temperature range -20 to 80°C by using the Clapeyron equation and the refrigerant 134-a data in EES is to be plotted. Analysis The problem is solved using EES, and the results are tabulated and plotted below. "Input Data:" T=30 [C] T_increment = 5 [C]
0.35
T[2]=T+T_increment T[1]=T-T_increment P[1] = pressure(R134a,T=T[1],x=0) P[2] = pressure(R134a,T=T[2],x=0) DELTAP = P[2]-P[1] DELTAT = T[2]-T[1] v_f=volume(R134a,T=T,x=0) v_g=volume(R134a,T=T,x=1) h_f=enthalpy(R134a,T=T,x=0) h_g=enthalpy(R134a,T=T,x=1) h_fg=h_g - h_f v_fg=v_g - v_f
0.3
] % [ r o rr E t n e cr e P
0.25 0.2 0.15 0.1 0.05 -20
0
20
40
60
80
T [C]
"The Clapeyron equation (Eq. 12-22) provides a means to calculate the enthalpy of vaporization, h_fg at a given temperature by determining the slope of the saturation curve on a P-T diagram and the specific volume of the saturated liquid and satruated vapor at the temperature." h_fg_Clapeyron=(T+273.2)*v_fg*DELTAP/DELTAT*Convert(m^3-kPa,kJ) PercentError=ABS(h_fg_Clapeyron-h_fg)/h_fg*Convert(, %) "[%]" hfg [kJ/kg] 212.91 205.96 198.60 190.73 182.27 173.08 163.00 151.79 139.10 124.32 106.35
hfg,Clapeyron [kJ/kg] 213.68 206.56 199.05 191.07 182.52 173.28 163.18 151.96 139.26 124.47 106.45
PercentError [%] 0.3593 0.2895 0.2283 0.1776 0.1394 0.1154 0.1057 0.1081 0.1166 0.1195 0.09821
T [C] -20 -10 0 10 20 30 40 50 60 70 80
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-12
12-27 Using the Clapeyron equation, the enthalpy of vaporization of steam at a specified pressure is to be estimated and to be compared to the tabulated data. Analysis From the Clapeyron equation, dP h fg = Tv fg dT sat ∆P ≅ T (v g − v f ) @300 kPa ∆T sat, 300 kPa (325 − 275)kPa = Tsat @300 kPa (v g − v f ) @300 kPa Tsat @325 kPa − Tsat @275 kPa 50 kPa 3 = (133.52 + 273.15 K)(0.60582 − 0.001073 m /kg) − ° (136.27 130.58) C = 2161.1 kJ/kg
The tabulated value of hfg at 300 kPa is 2163.5 kJ/kg.
12-28 The hfg and sfg of steam at a specified temperature are to be calculated using the Clapeyron equation and to be compared to the tabulated data. Analysis From the Clapeyron equation, dP h fg = Tv fg dT sat ∆P ≅ T (v g − v f ) @120°C ∆T sat,120°C Psat @125°C − Psat @115°C = T (v g − v f ) @120°C 125°C − 115°C
(232.23 − 169.18)kPa = (120 + 273.15 K)(0.89133 − 0.001060 m 3 /kg) 10 K = 2206.8 kJ/kg
Also, s fg =
h fg T
=
2206.8 kJ/kg = 5.6131 kJ/kg ⋅ K (120 + 273.15)K
The tabulated values at 120°C are hfg = 2202.1 kJ/kg and sfg = 5.6013 kJ/kg·K.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-13
12-29E [Also solved by EES on enclosed CD] The hfg of refrigerant-134a at a specified temperature is to be calculated using the Clapeyron equation and Clapeyron-Clausius equation and to be compared to the tabulated data. Analysis (a) From the Clapeyron equation, dP h fg = Tv fg dT sat ∆P ≅ T (v g − v f ) @ 50°F ∆T sat, 50° F Psat @ 60 ° F − Psat @ 40 °F = T (v g − v f ) @ 50°F 60°F − 40°F
(72.152 − 49.776) psia = (50 + 459.67 R)(0.79136 − 0.01270 ft 3 /lbm) 20 R = 444.0 psia ⋅ ft 3 /lbm = 82.16 Btu/lbm
(0.2% error)
since 1 Btu = 5.4039 psia·ft3. (b) From the Clapeyron-Clausius equation, P ln 2 P1
h fg 1 1 ≅ − R T1 T2 sat sat
h fg 72.152 psia 1 1 ≅ ln − 49 . 776 psia 0.01946 Btu/lbm R 40 459 . 67 R 60 459 . 67 R ⋅ + + h fg = 93.80 Btu/lbm (14.4% error)
The tabulated value of hfg at 50°F is 82.00 Btu/lbm.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-14
12-30 EES The enthalpy of vaporization of steam as a function of temperature using Clapeyron equation and steam data in EES is to be plotted. Analysis The enthalpy of vaporization is determined using Clapeyron equation from h fg ,Clapeyron = Tv fg
∆P ∆T
At 100ºC, for an increment of 5ºC, we obtain T1 = T − Tincrement = 100 − 5 = 95°C T2 = T + Tincrement = 100 + 5 = 105°C P1 = Psat @ 95°C = 84.61 kPa P2 = Psat @ 105°C = 120.90 kPa ∆T = T2 − T1 = 105 − 95 = 10°C ∆P = P2 − P1 = 120.90 − 84.61 = 36.29 kPa
v f @ 100°C = 0.001043 m 3 /kg v g @ 100°C = 1.6720 m 3 /kg v fg = v g − v f = 1.6720 − 0.001043 = 1.6710 m 3 /kg Substituting, h fg ,Clapeyron = Tv fg
36.29 kPa ∆P = (100 + 273.15 K)(1.6710 m 3 /kg) = 2262.8 kJ/kg 10 K ∆T
The enthalpy of vaporization from steam table is h fg @ 100°C = 2256.4 m 3 /kg
The percent error in using Clapeyron equation is PercentError =
2262.8 − 2256.4 × 100 = 0.28% 2256.4
We repeat the analysis over the temperature range 10 to 200ºC using EES. Below, the copy of EES solution is provided: "Input Data:" "T=100" "[C]" T_increment = 5"[C]" T[2]=T+T_increment"[C]" T[1]=T-T_increment"[C]" P[1] = pressure(Steam_iapws,T=T[1],x=0)"[kPa]" P[2] = pressure(Steam_iapws,T=T[2],x=0)"[kPa]" DELTAP = P[2]-P[1]"[kPa]" DELTAT = T[2]-T[1]"[C]" v_f=volume(Steam_iapws,T=T,x=0)"[m^3/kg]" v_g=volume(Steam_iapws,T=T,x=1)"[m^3/kg]" h_f=enthalpy(Steam_iapws,T=T,x=0)"[kJ/kg]" h_g=enthalpy(Steam_iapws,T=T,x=1)"[kJ/kg]" h_fg=h_g - h_f"[kJ/kg-K]" v_fg=v_g - v_f"[m^3/kg]"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-15 "The Clapeyron equation (Eq. 11-22) provides a means to calculate the enthalpy of vaporization, h_fg at a given temperature by determining the slope of the saturation curve on a P-T diagram and the specific volume of the saturated liquid and satruated vapor at the temperature." h_fg_Clapeyron=(T+273.15)*v_fg*DELTAP/DELTAT*Convert(m^3-kPa,kJ)"[kJ/kg]" PercentError=ABS(h_fg_Clapeyron-h_fg)/h_fg*100"[%]" hfg [kJ/kg]
hfg,Clapeyron [kJ/kg]
2477.20 2429.82 2381.95 2333.04 2282.51 2229.68 2173.73 2113.77 2014.17 1899.67 1765.50
2508.09 2451.09 2396.69 2343.47 2290.07 2235.25 2177.86 2116.84 2016.15 1900.98 1766.38
PercentErro r [%] 1.247 0.8756 0.6188 0.4469 0.3311 0.25 0.1903 0.1454 0.09829 0.06915 0.05015
T [C] 10 30 50 70 90 110 130 150 180 210 240
2600 2500
hfg calculated by Clapeyron equation
2400
] g k/ J k[
g f
h
2300 2200
hfg calculated by EES
2100 2000 1900 1800 1700 0
50
100
150
200
250
T [C]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-16
12-31 The sublimation pressure of water at -30ºC is to be determined using Clapeyron-Clasius equation and the triple point data of water. Analysis The sublimation pressure may be determined using Clapeyron-Clasius equation from Psub,CC ln P1
hig 1 1 = R T − T 2 1
where the triple point properties of water are P1 = 0.6117 kPa and T1 = 0.01ºC = 273.16 K (first line in Table A-4). Also, the enthalpy of sublimation of water at -30ºC is determined from Table A-8 to be 2838.4 kJ/kg. Substituting, Psub,CC hig 1 1 = − ln P1 R T1 T2 Psub,CC 2838.4 kJ/kg 1 1 = ln − 0 . 6117 kPa 0.4615 kJ/kg.K 273 . 16 K 30 273 . 15 K − + Psub,CC = 0.03799 kPa
The sublimation pressure of water at -30ºC is given in Table A-8 to be 0.03802 kPa. Then, the error involved in using Clapeyron-Clasius equation becomes PercentError =
0.03802 − 0.03799 × 100 = 0.08% 0.03802
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-17
General Relations for du, dh, ds, cv, and cp 12-32C Yes, through the relation
∂c p ∂P
2 = −T ∂ v ∂T 2 T
P
12-33 It is to be shown that the enthalpy of an ideal gas is a function of temperature only and that for an incompressible substance it also depends on pressure. Analysis The change in enthalpy is expressed as ∂v dh = c P dT + v − T dP ∂T P
For an ideal gas v = RT/P. Then, R ∂v = v − T = v −v = 0 P ∂T P
v −T Thus,
dh = c p dT
To complete the proof we need to show that cp is not a function of P either. This is done with the help of the relation
∂c p ∂P
2 = −T ∂ v ∂T 2 T
P
For an ideal gas, R ∂v = ∂ T P P
and
∂ 2v ∂T 2
= ∂ ( R / P ) = 0 P ∂T P
Thus, ∂c P ∂P
= 0 T
Therefore we conclude that the enthalpy of an ideal gas is a function of temperature only. For an incompressible substance v = constant and thus ∂v/∂T = 0. Then,
dh = c p dT + v dP Therefore we conclude that the enthalpy of an incompressible substance is a function of temperature and pressure.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-18
12-34 General expressions for ∆u, ∆h, and ∆s for a gas that obeys the van der Waals equation of state for an isothermal process are to be derived. Analysis (a) For an isothermal process dT = 0 and the general relation for ∆u reduces to ∆u = u 2 − u1 =
∫
T2
T1
cv dT +
v2
∫v
1
∂P T ∂T − P dv = v
v2
∫v
1
∂P T ∂T − P dv v
The van der Waals equation of state can be expressed as P=
RT a R ∂P − 2 → = v −b v ∂T v v − b
Thus, RT RT a a ∂P − + 2 = 2 T −P= v −b v −b v v ∂T v
Substituting, ∆u =
v2
∫v
1
1 1 dv = a − v v1 v 2 a
2
(b) The enthalpy change ∆h is related to ∆u through the relation ∆h = ∆u + P2v 2 − P1v 1
where Pv =
RTv a − v −b v
Thus, v v 1 1 P2v 2 − P1v 1 = RT 2 − 1 + a − v 2 − b v1 − b v1 v 2
Substituting, 1 v v 1 + RT 2 − 1 ∆h = 2a − v1 v 2 v 2 − b v1 − b
(c) For an isothermal process dT = 0 and the general relation for ∆s reduces to ∆s = s 2 − s1 =
∫
T2
T1
cv dT + T
v2
∫v
1
∂P dv = ∂T v
v2
∫v
1
∂P dv ∂T v
Substituting (∂P/∂T)v = R/(v - b), ∆s =
v2
∫v
1
v −b R dv = Rln 2 v −b v1 − b
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-19
12-35 General expressions for ∆u, ∆h, and ∆s for a gas whose equation of state is P(v-a) = RT for an isothermal process are to be derived. Analysis (a) A relation for ∆u is obtained from the general relation ∆u = u 2 − u1 =
∫
T2
T1
∂P T ∂T − P dv v
v2
∫v
cv dT +
1
The equation of state for the specified gas can be expressed as RT R ∂P → = v −a ∂T v v − a
P=
Thus, RT ∂P −P = P−P = 0 T −P = v −a ∂T v
∫
∆u =
Substituting,
T2
T1
cv dT
(b) A relation for ∆h is obtained from the general relation
∆h = h2 − h1 =
∫
T2
T1
c P dT +
∫
P2
P1
∂v v − T dP ∂T P
The equation of state for the specified gas can be expressed as
v=
RT R ∂v +a → = ∂ P T P P
Thus, R ∂v = v − T = v − (v − a ) = a P ∂T P
v −T Substituting,
∆h =
∫
T2
T1
c p dT +
∫
P2
P1
a dP =
∫
T2
T1
c p dT + a(P2 − P1 )
(c) A relation for ∆s is obtained from the general relation ∆s = s 2 − s1 =
T2
cp
T1
T
∫
dT −
∫
P2
P1
∂v dP ∂T P
Substituting (∂v/∂T)P = R/T, ∆s =
T2
cp
T1
T
∫
dT −
∫
P2
P1
R dP = P P
T2
cp
T1
T
∫
dT − R ln
P2 P1
For an isothermal process dT = 0 and these relations reduce to ∆u = 0,
∆h = a(P2 − P1 ),
and
∆s = − Rln
P2 P1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-20
12-36 General expressions for (∂u/∂P)T and (∂h/∂v)T in terms of P, v, and T only are to be derived. Analysis The general relation for du is ∂P du = cv dT + T − P dv ∂T v
Differentiating each term in this equation with respect to P at T = constant yields ∂P ∂v ∂u ∂P ∂v ∂v = 0 + T − P = T − P ∂P T ∂T v ∂P T ∂P T ∂T v ∂P T
Using the properties P, T, v, the cyclic relation can be expressed as ∂P ∂T ∂v ∂P ∂v ∂v → = −1 = − ∂T v ∂v P ∂P T ∂T v ∂P T ∂T P
Substituting, we get ∂u ∂v ∂v = −T − P ∂ P ∂ T T P ∂P T
The general relation for dh is
∂v dh = c p dT + v − T dP ∂T P Differentiating each term in this equation with respect to v at T = constant yields
∂h ∂v ∂P ∂P ∂v ∂P = 0 + v − T =v −T ∂v T ∂T P ∂v T ∂v T ∂T P ∂v T Using the properties v, T, P, the cyclic relation can be expressed as ∂v ∂T ∂P ∂v ∂P ∂T → = −1 = − ∂T P ∂P v ∂v T ∂T P ∂v T ∂P v
Substituting, we get ∂h ∂P ∂T =v +T v v ∂ ∂ T T ∂P v
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-21
12-37 Expressions for the specific heat difference cp-cv for three substances are to be derived. Analysis The general relation for the specific heat difference cp - cv is 2
∂v ∂P c p − cv = −T ∂T P ∂v T (a) For an ideal gas Pv = RT. Then,
v= P=
RT R ∂v → = ∂ P T P P RT
v
RT P ∂P → =− 2 =− v ∂v T v
Substituting, 2
P R TR c p − cv = −T − = R=R v P Pv a (b) For a van der Waals gas P + 2 (v − b ) = RT . Then, v
T=
1 1 2a 1 a a ∂T → P + 2 (v − b ) = − 3 (v − b ) + P + 2 R R ∂v P R v v v 2a (b − v ) T = + v −b Rv 3 1 ∂v = ( )+ T 2 a b − v ∂T P 3 v −b Rv
Inverting,
Also,
P=
RT
−
a
v −b v 2
RT 2a ∂P → + 3 =− 2 ∂ v T (v − b ) v
Substituting, 1 c p − cv = T 2a (b − v ) T + v −b Rv 3
2
RT 2a − 3 2 (v − b ) v
(c) For an incompressible substance v = constant and thus (∂v /∂T)P = 0. Therefore, c p − cv = 0
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-22
12-38 The specific heat difference cp-cv for liquid water at 15 MPa and 80°C is to be estimated. Analysis The specific heat difference cp - cv is given as 2
∂v ∂P c p − cv = −T ∂T P ∂v T Approximating differentials by differences about the specified state, 2
∆P ∆v c p − cv ≅ −T ∆ T P =15 MPa ∆v T =80°C v 100°C − v 60°C = −(80 + 273.15 K ) (100 − 60)°C
2
(20 − 10)MPa P =15 MPa v 20MPa − v 10 MPa
(0.0010361 − 0.0010105)m 3 /kg = −(353.15 K ) 40 K
2
T =80°C
10,000 kPa (0.0010199 − 0.0010244)m 3 /kg
= 0.3114 kPa ⋅ m 3 /kg ⋅ K = 0.3214 kJ/kg ⋅ K
12-39E The specific heat difference cp-cv for liquid water at 1000 psia and 150°F is to be estimated. Analysis The specific heat difference cp - cv is given as 2
∂v ∂P c p − cv = −T ∂T P ∂v T Approximating differentials by differences about the specified state, 2
∆v ∆P c p − cv ≅ −T ∆T P =1000psia ∆v T =150° F v 175° F − v 125° F = −(150 + 459.67 R ) (175 − 125)°F
2
(1500 − 500)psia P =1000 psia v 1500 psia − v 500 psia
(0.016427 − 0.016177)ft 3 /lbm = −(609.67 R) 50 R
2
T =150° F
1000 psia (0.016267 − 0.016317)ft 3 /lbm
= 0.3081 psia ⋅ ft 3 /lbm ⋅ R = 0.0570 Btu/lbm ⋅ R (1 Btu = 5.4039 psia ⋅ ft 3 )
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-23
12-40 Relations for the volume expansivity β and the isothermal compressibility α for an ideal gas and for a gas whose equation of state is P(v-a) = RT are to be obtained. Analysis The volume expansivity and isothermal compressibility are expressed as β=
1 ∂v 1 ∂v and α = − v ∂T P v ∂P T
(a) For an ideal gas v = RT/P. Thus, R ∂v = ∂T P P
→ β =
1 R 1 = v P T
v 1 v 1 RT ∂v → α = − − = =− 2 =− v ∂ P P T P P P (b) For a gas whose equation of state is v = RT/P + a, R ∂v = ∂T P P
→ β =
1 R R = v P RT + aP
1 v −a v −a v −a RT ∂v → α = − − = =− 2 =− ∂ P P v P Pv T P
12-41 The volume expansivity β and the isothermal compressibility α of refrigerant-134a at 200 kPa and 30°C are to be estimated. Analysis The volume expansivity and isothermal compressibility are expressed as β=
1 ∂v 1 ∂v and α = − v ∂T P v ∂P T
Approximating differentials by differences about the specified state, β≅ =
1 ∆v 1 v 40°C − v 20°C = v ∆T P = 200 kPa v (40 − 20)°C
P = 200 kPa
(0.12322 − 0.11418)m 3 /kg = 0.00381 K −1 20 K 0.11874 m 3 /kg 1
and α≅− =−
1 ∆v 1 v 240 kPa − v 180 kPa = − v ∆P T =30°C v (240 − 180)kPa
T =30°C
(0.09812 − 0.13248)m 3 /kg = 0.00482 kPa −1 60 kPa 0.11874 m 3 /kg 1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-24
The Joule-Thomson Coefficient 12-42C It represents the variation of temperature with pressure during a throttling process. 12-43C The line that passes through the peak points of the constant enthalpy lines on a T-P diagram is called the inversion line. The maximum inversion temperature is the highest temperature a fluid can be cooled by throttling. 12-44C No. The temperature may even increase as a result of throttling. 12-45C Yes. 12-46C No. Helium is an ideal gas and h = h(T) for ideal gases. Therefore, the temperature of an ideal gas remains constant during a throttling (h = constant) process.
12-47 The equation of state of a gas is given to be P(v-a) = RT. It is to be determined if it is possible to cool this gas by throttling. Analysis The equation of state of this gas can be expressed as
v=
RT R ∂v +a → = P ∂T P P
Substituting into the Joule-Thomson coefficient relation, R 1 1 ∂v v − T (v − v + a ) = − a < 0 v − T = − = − cp P cp cp ∂T P Therefore, this gas cannot be cooled by throttling since µ is always a negative quantity. µ=−
1 cp
12-48 Relations for the Joule-Thompson coefficient and the inversion temperature for a gas whose equation of state is (P+a/v2) v = RT are to be obtained. Analysis The equation of state of this gas can be expressed as a a v 2a 1 2av T RTv − 2a ∂T + = → P + 2 = − 3 + P + 2 = − R R R v ∂ v v Rv 2 v Rv 2 P v Substituting into the Joule-Thomson coefficient relation, T=
v
µ=−
1 cp
∂v 1 v − T = − ∂ T c P p
2 2 av v − RTv =− − RT v 2 a c ( 2 a − RTv ) p
The temperature at µ = 0 is the inversion temperature, 2 av µ=− =0 → v = 0 c p (2a − RTv ) Thus the line of v = 0 is the inversion line. Since it is not physically possible to have v = 0, this gas does not have an inversion line.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-25
12-49 The Joule-Thompson coefficient of steam at two states is to be estimated. Analysis (a) The enthalpy of steam at 3 MPa and 300°C is h = 2994.3 kJ/kg. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as ∆T ∂T µ= ≅ P ∂ h ∆P h = 2994.3 kJ/kg
Considering a throttling process from 3.5 MPa to 2.5 MPa at h = 2994.3 kJ/kg, the Joule-Thomson coefficient is determined to be
T3.5 MPa − T2.5 MPa µ = (3.5 − 2.5) MPa
(306.3 − 294)°C = = 12.3 °C/MPa h = 2994.3 kJ/kg (3.5 − 2.5) MPa
(b) The enthalpy of steam at 6 MPa and 500°C is h = 3423.1 kJ/kg. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as ∂T ∆T µ= ≅ ∂P h ∆P h =3423.1 kJ/kg
Considering a throttling process from 7.0 MPa to 5.0 MPa at h = 3423.1 kJ/kg, the Joule-Thomson coefficient is determined to be T7.0 MPa − T5.0 MPa (504.8 − 495.1)°C = = 4.9 °C/MPa (7.0 5.0) MPa (7.0 − 5.0) MPa − h =3423.1 kJ/kg
µ =
12-50E [Also solved by EES on enclosed CD] The Joule-Thompson coefficient of nitrogen at two states is to be estimated. Analysis (a) The enthalpy of nitrogen at 200 psia and 500 R is, from EES, h = -10.564 Btu/lbm. Note that in EES, by default, the reference state for specific enthalpy and entropy is 0 at 25ºC (77ºF) and 1 atm. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as ∆T ∂T µ= ≅ ∂P h ∆P h = −10.564 Btu/lbm
Considering a throttling process from 210 psia to 190 psia at h = -10.564 Btu/lbm, the Joule-Thomson coefficient is determined to be T190 psia − T210 psia µ = (190 − 210) psia
(499.703 − 500.296) R = = 0.0297 R/psia (190 − 210) psia h = −10.564 Btu/lbm
(b) The enthalpy of nitrogen at 2000 psia and 400 R is, from EES, h = -55.321 Btu/lbm. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as ∆T ∂T µ= ≅ ∂P h ∆P h = −55.321 Btu/lbm
Considering a throttling process from 2010 psia to 1990 psia at h = -55.321 Btu/lbm, the Joule-Thomson coefficient is determined to be T1999 psia − T2001 psia µ = (1990 − 2010) psia
(399.786 - 400.213) R = = 0.0213 R/psia (1990 − 2010) psia h = −55.321 Btu/lbm
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-26
12-51E EES Problem 12-50E is reconsidered. The Joule-Thompson coefficient for nitrogen over the pressure range 100 to 1500 psia at the enthalpy values 100, 175, and 225 Btu/lbm is to be plotted. Analysis The problem is solved using EES, and the results are tabulated and plotted below. Gas$ = 'Nitrogen' {P_ref=200 [psia] T_ref=500 [R] P= P_ref} h=50 [Btu/lbm] {h=enthalpy(Gas$, T=T_ref, P=P_ref)} dP = 10 [psia] T = temperature(Gas$, P=P, h=h) P[1] = P + dP P[2] = P - dP T[1] = temperature(Gas$, P=P[1], h=h) T[2] = temperature(Gas$, P=P[2], h=h) Mu = DELTAT/DELTAP "Approximate the differential by differences about the state at h=const." DELTAT=T[2]-T[1] DELTAP=P[2]-P[1]
h = 100 Btu/lbm P [psia] µ [R/psia] 100 0.003675 275 0.003277 450 0.002899 625 0.00254 800 0.002198 975 0.001871 1150 0.001558 1325 0.001258 1500 0.0009699 0.004 0.003 0.002 h = 100 Btu/lbm
0.001
] ai s p/ R [ µ
0 -0.001 -0.002
h = 175 Btu/lbm
-0.003 -0.004
h = 225 Btu/lbm
-0.005 -0.006 0
200
400
600
800
1000
1200
1400
1600
P [psia]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-27
12-52 The Joule-Thompson coefficient of refrigerant-134a at a specified state is to be estimated. Analysis The enthalpy of refrigerant-134a at 0.7 MPa and T = 50°C is h = 288.53 kJ/kg. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as ∂T ∆T µ= ≅ ∂ P h ∆P h = 288.53 kJ/kg
Considering a throttling process from 0.8 MPa to 0.6 MPa at h = 288.53 kJ/kg, the Joule-Thomson coefficient is determined to be T0.8 MPa − T0.6 MPa µ = (0.8 − 0.6)MPa
(51.81 − 48.19)°C = 18.1 °C/MPa = (0.8 − 0.6)MPa h = 288.53 kJ/kg
12-53 Steam is throttled slightly from 1 MPa and 300°C. It is to be determined if the temperature of the steam will increase, decrease, or remain the same during this process. Analysis The enthalpy of steam at 1 MPa and T = 300°C is h = 3051.6 kJ/kg. Now consider a throttling process from this state to 0.8 MPa, which is the next lowest pressure listed in the tables. The temperature of the steam at the end of this throttling process will be
P = 0.8 MPa T2 = 297.52°C h = 3051.6 kJ/kg Therefore, the temperature will decrease.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-28
The ∆h, ∆u, and ∆s of Real Gases
12-54C It is the variation of enthalpy with pressure at a fixed temperature. 12-55C As PR approaches zero, the gas approaches ideal gas behavior. As a result, the deviation from ideal gas behavior diminishes. 12-56C So that a single chart can be used for all gases instead of a single particular gas.
12-57 The enthalpy of nitrogen at 175 K and 8 MPa is to be determined using data from the ideal-gas nitrogen table and the generalized enthalpy departure chart. Analysis (a) From the ideal gas table of nitrogen (Table A-18) we read
h = 5083.8 kJ/kmol = 181.48 kJ/kg ( M N 2 = 28.013 kg/kmol) at the specified temperature. This value involves 44.4% error. (b) The enthalpy departure of nitrogen at the specified state is determined from the generalized chart to be
N2 175 K 8 MPa
T 175 = = 1.387 (hideal − h ) T , P Tcr 126.2 → Z h = = 1 .6 P 8 Ru Tcr PR = = = 2.360 Pcr 3.39
TR =
and
Thus, h = hideal − Z h Ru Tcr = 5083.8 − [(1.6)(8.314)(126.2)] = 3405.0 kJ/kmol
or, h=
3405.0 kJ/kmol h = = 121.6 kJ/kg (3.1%error) M 28.013 kg/kmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-29
12-58E The enthalpy of nitrogen at 400 R and 2000 psia is to be determined using data from the ideal-gas nitrogen table and the generalized enthalpy departure chart. Analysis (a) From the ideal gas table of nitrogen (Table A-18E) we read h = 2777.0 Btu/lbmol = 99.18 Btu/lbm ( M N 2 = 28 lbm/lbmol)
at the specified temperature. This value involves 44.2% error. (b) The enthalpy departure of nitrogen at the specified state is determined from the generalized chart to be (Fig. A-29)
N2 400 R 2000 psia
400 T = = 1.761 (hideal − h ) T , P Tcr 227.1 → Z h = = 1.18 2000 P Ru Tcr PR = = = 4.065 492 Pcr
TR =
and
Thus, h = hideal − Z h Ru Tcr = 2777.0 − [(1.18)(1.986 )(227.1)] = 2244.8 Btu/lbmol
or, h=
2244.8 Btu/lbmol h = = 80.17 Btu/lbm M 28 lbm/lbmol
(54.9% error)
12-59 The errors involved in the enthalpy and internal energy of CO2 at 350 K and 10 MPa if it is assumed to be an ideal gas are to be determined. Analysis (a) The enthalpy departure of CO2 at the specified state is determined from the generalized chart to be (Fig. A-29) 350 T = = 1.151 (hideal − h ) T , P Tcr 304.2 → Z h = = 1.5 10 P Ru Tcr = = 1.353 PR = Pcr 7.39
TR =
and
Thus,
CO2 350 K 10 MPa
h = hideal − Z h Ru Tcr = 11,351 − [(1.5)(8.314)(304.2)] = 7,557kJ/kmol
and, Error =
(hideal − h ) T , P h
=
11,351 − 7,557 = 50.2% 7,557
(b) At the calculated TR and PR the compressibility factor is determined from the compressibility chart to be Z = 0.65. Then using the definition of enthalpy, the internal energy is determined to be u = h − Pv = h − ZRu T = 7557 − [(0.65)(8.314)(350)] = 5,666kJ/kmol
and, Error =
u ideal − u 8,439 − 5,666 = = 48.9% u 5,666
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-30
12-60 The enthalpy and entropy changes of nitrogen during a process are to be determined assuming ideal gas behavior and using generalized charts. Analysis (a) Using data from the ideal gas property table of nitrogen (Table A-18), (h2 − h1 ) ideal = h2,ideal − h1,ideal = 9306 − 6,537 = 2769 kJ/kmol
and ( s 2 − s1 ) ideal = s 2o − s1o − Ru ln
P2 12 = 193.562 − 183.289 − 8.314 × ln = 4.510 kJ/kmol ⋅ K P1 6
(b) The enthalpy and entropy departures of nitrogen at the specified states are determined from the generalized charts to be (Figs. A-29, A-30) T1 225 = = 1.783 Tcr 126.2 → Z h1 = 0.6 and Z s1 = 0.25 P1 6 = = = 1.770 Pcr 3.39
T R1 = PR1
and T2 320 = = 2.536 Tcr 126.2 → Z h 2 = 0.4 and Z s 2 = 0.15 P2 12 = = = 2.540 Pcr 3.39
TR 2 = PR 2
Substituting, h2 − h1 = Ru Tcr ( Z h1 − Z h 2 ) + (h2 − h1 ) ideal
= (8.314)(126.2)(0.6 − 0.4 ) + 2769 = 2979 kJ/kmol
s 2 − s1 = Ru ( Z s1 − Z s 2 ) + ( s 2 − s1 ) ideal
= (8.314)(0.25 − 0.15) + 4.510 = 5.341 kJ/kmol ⋅ K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-31
12-61 The enthalpy and entropy changes of CO2 during a process are to be determined assuming ideal gas behavior and using generalized charts. Analysis (a) Using data from the ideal gas property table of CO2 (Table A-20),
250 K
7 MPa
CO2
280 K
12 MPa
(h2 − h1 ) ideal = h2,ideal − h1,ideal = 8,697 − 7,627 = 1,070 kJ/kmol ( s 2 − s1 ) ideal = s 2o − s1o − Ru ln
Thus,
P2 12 = 211.376 − 207.337 − 8.314 × ln = −0.442 kJ/kmol ⋅ K P1 7
(h2 − h1 ) ideal 1,070 kJ/kmol = = 24.32 kJ/kg M 44 kg/kmol (s − s ) − 0.442 kJ/kmol = 2 1 ideal = = −0.0100 kJ/kg ⋅ K M 44 kg/kmol
(h2 − h1 ) ideal = ( s 2 − s1 ) ideal
(b) The enthalpy and entropy departures of CO2 at the specified states are determined from the generalized charts to be (Figs. A-29, A-30) T1 250 = = 0.822 Tcr 304.2 → Z h1 = 5.5 and Z s1 = 5.3 P1 7 = = = 0.947 Pcr 7.39
T R1 = PR1
and
T2 280 = = 0.920 Tcr 304.2 → Z h 2 = 5.0 and Z s 2 = 4.2 P2 12 = = = 1.624 Pcr 7.39
TR 2 = PR 2
Thus, h2 − h1 = RTcr ( Z h1 − Z h 2 ) + (h2 − h1 ) ideal = (0.1889)(304.2 )(5.5 − 5.0 ) + 24.32 = 53.05 kJ/kg s 2 − s1 = R( Z s1 − Z s 2 ) + ( s 2 − s1 ) ideal = (0.1889)(5.3 − 4.2) − 0.010 = 0.198 kJ/kg ⋅ K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-32
12-62 Methane is compressed adiabatically by a steady-flow compressor. The required power input to the compressor is to be determined using the generalized charts. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis The steady-flow energy balance equation for this compressor can be expressed as E& in − E& out = ∆E& system ©0 (steady) = 0 E& in = E& out
10 MPa 110°C
W& C,in + m& h1 = m& h2 W& C,in = m& (h2 − h1 )
The enthalpy departures of CH4 at the specified states are determined from the generalized charts to be (Fig. A-29) T1 263 = = 1.376 Tcr 191.1 → Z h1 = 0.21 P 2 = 1 = = 0.431 Pcr 4.64
CH4
· = 0.55 kg/s m
T R1 = PR1
and
2 MPa -10 °C
T2 383 = = 2.00 Tcr 191.1 → Z h 2 = 0.50 P2 10 = = = 2.155 Pcr 4.64
TR 2 = PR 2
Thus, h2 − h1 = RTcr ( Z h1 − Z h 2 ) + (h2 − h1 ) ideal
= (0.5182)(191.1)(0.21 − 0.50 ) + 2.2537(110 − (− 10)) = 241.7 kJ/kg
Substituting, W& C,in = (0.55 kg/s )(241.7 kJ/kg ) = 133 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-33
12-63 [Also solved by EES on enclosed CD] Propane is compressed isothermally by a piston-cylinder device. The work done and the heat transfer are to be determined using the generalized charts. Assumptions 1 The compression process is quasi-equilibrium. 2 Kinetic and potential energy changes are negligible. Analysis (a) The enthalpy departure and the compressibility factors of propane at the initial and the final states are determined from the generalized charts to be (Figs. A-29, A-15) T1 373 = = 1.008 Tcr 370 → Z h1 = 0.28 and Z 1 = 0.92 P1 1 = = = 0.235 Pcr 4.26
T R1 = PR1
and
T2 373 = = 1.008 Tcr 370 → Z h 2 = 1.8 and Z 2 = 0.50 P2 4 = = = 0.939 Pcr 4.26
TR 2 = PR 2
Propane 1 MPa 100 °C
Treating propane as a real gas with Zavg = (Z1+Z2)/2 = (0.92 + 0.50)/2 = 0.71, Pv = ZRT ≅ Z avg RT = C = constant
Then the boundary work becomes
∫
2
wb,in = − P dv = − 1
∫
2
1
C
v
dv = −C ln
v2 Z RT / P2 Z P = Z avg RT ln 2 = − Z ave RT ln 2 1 v1 Z 1 RT / P1 Z 1 P2
= −(0.71)(0.1885 kJ/kg ⋅ K )(373 K )ln
(0.50)(1) = 99.6 kJ/kg (0.92)(4)
Also, h2 − h1 = RTcr ( Z h1 − Z h 2 ) + (h2 − h1 ) ideal = (0.1885)(370 )(0.28 − 1.8) + 0 = −106 kJ/kg u 2 − u1 = (h2 − h1 ) − R( Z 2 T2 − Z 1T1 ) = −106 − (0.1885)[(0.5)(373) − (0.92 )(373)] = −76.5 kJ/kg
Then the heat transfer for this process is determined from the closed system energy balance to be E in − E out = ∆E system q in + wb,in = ∆u = u 2 − u1
q in = (u 2 − u1 ) − wb,in = −76.5 − 99.6 = −176.1 kJ/kg → q out = 176.1 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Q
12-34
12-64 EES Problem 12-63 is reconsidered. This problem is to be extended to compare the solutions based on the ideal gas assumption, generalized chart data and real fluid (EES) data. Also, the solution is to be extended to carbon dioxide, nitrogen and methane. Analysis The problem is solved using EES, and the solution is given below. Procedure INFO(Name$, T[1] : Fluid$, T_critical, p_critical) If Name$='Propane' then T_critical=370 ; p_critical=4620 ; Fluid$='C3H8'; goto 10 endif If Name$='Methane' then T_critical=191.1 ; p_critical=4640 ; Fluid$='CH4'; goto 10 endif If Name$='Nitrogen' then T_critical=126.2 ; p_critical=3390 ; Fluid$='N2'; goto 10 endif If Name$='Oxygen' then T_critical=154.8 ; p_critical=5080 ; Fluid$='O2'; goto 10 endif If Name$='CarbonDioxide' then T_critical=304.2 ; p_critical=7390 ; Fluid$='CO2' ; goto 10 endif If Name$='n-Butane' then T_critical=425.2 ; p_critical=3800 ; Fluid$='C4H10' ; goto 10 endif 10: If T[1] 0. Using the Tds relation:
dh = T ds + v dP →
(1) P = constant:
∂ h = T ∂ s P
(2) T = constant:
∂h ∂P = T +v ∂s T ∂s T
But the 4th Maxwell relation: Substituting:
dh dP = T +v ds ds
∂P ∂T = − ∂s T ∂v P
1 ∂T ∂h =T − = T −v β ∂v P ∂s T
Therefore, the slope of P = constant lines is greater than the slope of T = constant lines. (3) v = constant:
∂P ∂h = T + v (a) ∂s v ∂s v
From the ds relation: Divide by dP holding v constant:
ds =
cv ∂P dT + dv T ∂T v c ∂s = v T ∂P v
∂T ∂P v
or
T ∂P ∂P = (b) ∂s v cv ∂T v
Using the properties P, T, v, the cyclic relation can be expressed as 1 β ∂v ∂P ∂P ∂P ∂T ∂v → = = (− β v ) = − = −1 − αv α ∂T P ∂v T ∂T v ∂T v ∂v P ∂P T
(c )
where we used the definitions of α and β. Substituting (b) and (c) into (a), Tβ v ∂h ∂P >T = T +v = T + cv α ∂s v ∂s v
Here α is positive for all phases of all substances. T is the absolute temperature that is also positive, so is cv. Therefore, the second term on the right is always a positive quantity since β is given to be positive. Then we conclude that the slope of P = constant lines is less than the slope of v = constant lines.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-44
12-72 Using the cyclic relation and the first Maxwell relation, the other three Maxwell relations are to be obtained. Analysis (1) Using the properties P, s, v, the cyclic relation can be expressed as ∂P ∂s ∂v = −1 ∂s v ∂v P ∂P s
Substituting the first Maxwell relation,
∂T ∂P = − , ∂v s ∂s v
∂T ∂s ∂v ∂T ∂s ∂T ∂v − → → = −1 =1 = ∂v s ∂v P ∂P s ∂P s ∂v P ∂P s ∂s P
(2) Using the properties T, v, s, the cyclic relation can be expressed as ∂T ∂v ∂s = −1 ∂v s ∂s T ∂T v
Substituting the first Maxwell relation,
∂P ∂T = − , ∂ v ∂s v s
∂P ∂v ∂s ∂P ∂v ∂s ∂P − → → = −1 =1 = ∂s v ∂s T ∂T v ∂T v ∂s T ∂v T ∂T v
(3) Using the properties P, T, v, the cyclic relation can be expressed as ∂P ∂T ∂v = −1 ∂T v ∂v P ∂P T
Substituting the third Maxwell relation,
∂P ∂s , = ∂ v T ∂T v
∂s ∂T ∂v ∂s ∂T ∂s ∂v → → = − = −1 = −1 ∂v T ∂v P ∂P T ∂P T ∂v P ∂P T ∂T P
12-73 It is to be shown that the slope of a constant-pressure line on an h-s diagram is constant in the saturation region and increases with temperature in the superheated region. Analysis For P = constant, dP = 0 and the given relation reduces to dh = Tds, which can also be expressed as
∂ h = T ∂ s P
h P = const.
Thus the slope of the P = constant lines on an h-s diagram is equal to the temperature. (a) In the saturation region, T = constant for P = constant lines, and the slope remains constant. (b) In the superheat region, the slope increases with increasing temperature since the slope is equal temperature.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
s
12-45
12-74 The relations for ∆u, ∆h, and ∆s of a gas that obeys the equation of state (P+a/v2)v = RT for an isothermal process are to be derived. Analysis (a) For an isothermal process dT = 0 and the general relation for ∆u reduces to ∆u = u 2 − u1 =
∫
T2
cv dT +
T1
v2
∫v
1
∂P T ∂T − P dv = v
v2
∫v
1
∂P T ∂T − P dv v
For this gas the equation of state can be expressed as P=
RT
v
−
a
v
2
R ∂P → = ∂T v v
Thus, RT RT a a ∂P T − + 2 = 2 −P = T ∂ v v v v v
Substituting,
∆u =
v2
∫v
1
a
v
2
1 1 dv = a − v1 v 2
(b) The enthalpy change ∆h is related to ∆u through the relation ∆h = ∆u + P2v 2 − P1v 1
where
Pv = RT −
a
v
Thus, a P2v 2 − P1v 1 = RT − v2
Substituting,
1 1 a − RT − = a − v1 v1 v 2
1 1 ∆h = 2a − v1 v 2
(c) For an isothermal process dT = 0 and the general relation for ∆s reduces to ∆s = s 2 − s1 =
∫
T2
T1
cv dT + T
v2
∫v
1
∂P dv = ∂T v
v2
∫v
1
∂P dv ∂T v
Substituting (∂P/∂T)v = R/v, ∆s =
v2
R
1
v
∫v
dv = Rln
v2 v1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-46
12-75 It is to be shown that ∂v ∂P ∂P ∂v cv = −T and c p = T ∂ T ∂ T s v ∂T s ∂T P
Analysis Using the definition of cv , ∂s ∂s ∂P cv = T = T ∂T v ∂P v ∂T v ∂s ∂v Substituting the first Maxwell relation = − , ∂P v ∂T s ∂v ∂P cv = −T ∂T s ∂T v
Using the definition of cp, ∂s ∂v ∂s cp = T = T T ∂ ∂v P ∂T P P ∂P ∂s Substituting the second Maxwell relation , = ∂v P ∂T s ∂P ∂v cp = T ∂T s ∂T P
12-76 The Cp of nitrogen at 300 kPa and 400 K is to be estimated using the relation given and its definition, and the results are to be compared to the value listed in Table A-2b. Analysis (a) We treat nitrogen as an ideal gas with R = 0.297 kJ/kg·K and k = 1.397. Note that PT-k/(k-1) = C = constant for the isentropic processes of ideal gases. The cp relation is given as ∂P ∂v cp = T ∂T s ∂T P
v=
RT R ∂v → = ∂ P T P P
(
)
k k kP ∂P → P = CT k /( k −1) CT k /( k −1) −1 = PT − k /( k −1) T k /( k −1) −1 = = ∂ T k − 1 k − 1 T ( k − 1) s
Substituting, kP R kR 1.397(0.297 kJ/kg ⋅ K) = c p = T = 1.045 kJ/kg ⋅ K = 1.397 − 1 T (k − 1) P k − 1 ∂h (b) The cp is defined as cp = . Replacing the differentials by differences, ∂T P h(410 K ) − h(390 K ) (11,932 − 11,347 )/28.0 kJ/kg ∆h cp ≅ = 1.045 kJ/kg ⋅ K = = (410 − 390)K (410 − 390)K ∆T P =300 kPa
(Compare: Table A-2b at 400 K → cp = 1.044 kJ/kg·K)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-47
12-77 The temperature change of steam and the average Joule-Thompson coefficient during a throttling process are to be estimated. Analysis The enthalpy of steam at 4.5 MPa and T = 300°C is h = 2944.2 kJ/kg. Now consider a throttling process from this state to 2.5 MPa. The temperature of the steam at the end of this throttling process is P = 2.5 MPa T2 = 273.72°C h = 2944.2 kJ/kg Thus the temperature drop during this throttling process is ∆T = T2 − T1 = 273.72 − 300 = −26.28°C The average Joule-Thomson coefficient for this process is determined from
(273.72 − 300)°C = 13.14°C/MPa ∂T ∆T µ= = ≅ (2.5 − 4.5)MPa ∂ P ∆ P h h =3204.7 kJ/kg
12-78 The initial state and the final temperature of argon contained in a rigid tank are given. The mass of the argon in the tank, the final pressure, and the heat transfer are to be determined using the generalized charts. Analysis (a) The compressibility factor of argon at the initial state is determined from the generalized chart to be Ar T1 173 -100 °C = = 1.146 T R1 = 1 MPa Tcr 151.0 Z 1 = 0.95 and Z h1 = 0.18 P 1 PR1 = 1 = = 0.206 Pcr 4.86 Q Then, Pv = ZRT → v = m=
ZRT (0.95)(0.2081 kPa ⋅ m 3 /kg ⋅ K)(173 K) = = 0.0342 m 3 /kg 1000 kPa P
1.2 m 3 V = = 35.1 kg v 0.0342 m 3 /kg
(b) The specific volume of argon remains constant during this process, v2 = v 1. Thus, PR2 = 0.315 Z 2 = 0.99 3 v2 0.0342 m /kg Z ≅ 0 = = = 5 . 29 h2 RTcr / Pcr (0.2081 kPa ⋅ m 3 /kg ⋅ K)(151 K)(4860 kPa)
T R2 =
v R2
T2 273 = = 1.808 Tcr 151.0
P2 = PR2 Pcr = (0.315)(4860) = 1531 kPa
(c) The energy balance relation for this closed system can be expressed as Ein − Eout = ∆Esystem Qin = ∆U = m(u2 − u1 )
Qin = m[h2 − h1 − ( P2v 2 − P1v1 )] = m[h2 − h1 − R ( Z 2T2 − Z1T1 )]
where h2 − h1 = RTcr Z h1 − Z h2 + (h2 − h1 )ideal = (0.2081)(151)(0.18 − 0) + 0.5203(0 − (− 100 )) = 57.69 kJ/kg
(
Thus,
)
Qin = (35.1 kg )[57.69 − (0.2081 kJ/kg ⋅ K)[(0.99)(273) − (0.95)(173)]K ] = 1251 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-48
12-79 Argon enters a turbine at a specified state and leaves at another specified state. Power output of the turbine and exergy destruction during this process are to be determined using the generalized charts. Properties The gas constant and critical properties of Argon are R = 0.2081 kJ/kg.K, Tcr = 151 K, and Pcr = 4.86 MPa (Table A-1). Analysis (a) The enthalpy and entropy departures of argon at the specified states are determined from the generalized charts to be T 600 P1 = 7 MPa = 3.97 T R1 = 1 = Tcr 151 T1 = 600 K Z h1 ≅ 0 and Z s1 ≅ 0 V1 = 100 m/s 60 kW P1 7 PR1 = = = 1.44 Pcr 4.86 Thus argon behaves as an ideal gas at turbine inlet. Also, Ar · = 5 kg/s T2 280 m = = 1.85 T R2 = · Tcr 151 W Z h2 = 0.04 and Z s2 = 0.02 P 1 T0 = 25°C PR2 = 2 = = 0.206 Pcr 4.86 P2 = 1 MPa h2 − h1 = RTcr Z h1 − Z h2 + (h2 − h1 )ideal T2 = 280 K Thus, V2 = 150 m/s = (0.2081)(151)(0 − 0.04 ) + 0.5203(280 − 600 ) = −167.8 kJ/kg The power output of the turbine is to be determined from the energy balance equation, = 0 (steady) → E& = E& E& − E& = ∆E&
(
in
)
out
m& (h1 + V12
system
/ 2) =
m& (h2 + V 22
in
out
/ 2) + Q& out + W& out
V 2 − V12 & W& out = −m& (h2 − h1 ) + 2 − Qout 2
Substituting, (150 m/s) 2 − (100 m/s) 2 1 kJ/kg − 60 kJ/s = 747.8 kW W& out = −(5 kg/s) − 167.8 + 1000 m 2 /s 2 2 (b) Under steady conditions, the rate form of the entropy balance for the turbine simplifies to Ê0 =0 S& − S& + S& = ∆S& in
out
gen
Q& & 1 − ms & 2 − out + S&gen = 0 ms Tb,out
system
→
Q& & ( s2 − s2 ) + out S&gen = m T0
The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,
where
Q& X& destroyed = T0 S& gen = T0 m& ( s 2 − s 2 ) + out T0 s 2 − s1 = R Z s1 − Z s2 + (s 2 − s1 )ideal
and
(s 2 − s1 )ideal = c p ln
Thus,
s 2 − s1 = R Z s1 − Z s2 + (s 2 − s1 )ideal = (0.2081)[0 − (0.02)] + 0.0084 = 0.0042 kJ/kg ⋅ K
(
Substituting,
(
)
T2 P 280 1 − R ln 2 = 0.5203 ln − 0.2081 ln = 0.0084 kJ/kg ⋅ K T1 600 P1 7
)
60 kW = 66.3 kW X& destroyed = (298 K ) (5 kg/s )(0.0042 kJ/kg ⋅ K ) + 298 K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-49
12-80 EES Problem 12-79 is reconsidered. The problem is to be solved assuming steam is the working fluid by using the generalized chart method and EES data for steam. The power output and the exergy destruction rate for these two calculation methods against the turbine exit pressure are to be plotted. Analysis The problem is solved using EES, and the results are tabulated and plotted below.
" Input Data " T[1]=600 [K] P[1]=7000 [kPa] Vel[1]=100 [m/s] T[2]=455 [K] P[2]=1000 [kPa] Vel[2]=150 [m/s] Q_dot_out=60 [kW] T_o=25+273 "[K]" m_dot=5 [kg/s] Name$='Steam_iapws' T_critical=647.3 [K] P_critical=22090 [kPa] Fluid$='H2O' R_u=8.314 M=molarmass(Fluid$) R=R_u/M "****** IDEAL GAS SOLUTION ******" "State 1" h_ideal[1]=enthalpy(Fluid$,T=T[1]) "Enthalpy of ideal gas" s_ideal[1]=entropy(Fluid$, T=T[1], P=P[1]) "Entropy of ideal gas" "State 2" h_ideal[2]=enthalpy(Fluid$,T=T[2]) "Enthalpy of ideal gas" s_ideal[2]=entropy(Fluid$, T=T[2], P=P[2]) "Entropy of ideal gas" "Conservation of Energy, Steady-flow: " "E_dot_in=E_dot_out" m_dot*(h_ideal[1]+Vel[1]^2/2*convert(m^2/s^2,kJ/kg))=m_dot*(h_ideal[2]+Vel[2]^2/2*convert(m^2 /s^2,kJ/kg))+Q_dot_out+W_dot_out_ideal "Second Law analysis:" "S_dot_in-S_dot_out+S_dot_gen = 0" m_dot*s_ideal[1] - m_dot*s_ideal[2] - Q_dot_out/T_o + S_dot_gen_ideal = 0 "Exergy Destroyed:" X_dot_destroyed_ideal = T_o*S_dot_gen_ideal "***** COMPRESSABILITY CHART SOLUTION ******" "State 1" Tr[1]=T[1]/T_critical Pr[1]=P[1]/P_critical Z[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure" h_chart[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts" DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure" s_chart[1]=s_ideal[1]-DELTAs[1] "Entropy of real gas using charts" "State 2" PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-50 Tr[2]=T[2]/T_critical Pr[2]=P[2]/P_critical Z[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure" DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure" h_chart[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts" s_chart[2]=s_ideal[2]-DELTAs[2] "Entropy of real gas using charts" "Conservation of Energy, Steady-flow: " "E_dot_in=E_dot_out" m_dot*(h_chart[1]+Vel[1]^2/2*convert(m^2/s^2,kJ/kg))=m_dot*(h_chart[2]+Vel[2]^2/2*convert(m^ 2/s^2,kJ/kg))+Q_dot_out+W_dot_out_chart "Second Law analysis:" "S_dot_in-S_dot_out+S_dot_gen = 0" m_dot*s_chart[1] - m_dot*s_chart[2] - Q_dot_out/T_o + S_dot_gen_chart = 0 "Exergy Destroyed:" X_dot_destroyed_chart = T_o*S_dot_gen_chart"[kW]" "***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****" "At state 1" h_ees[1]=enthalpy(Name$,T=T[1],P=P[1]) s_ees[1]=entropy(Name$,T=T[1],P=P[1]) "At state 2" h_ees[2]=enthalpy(Name$,T=T[2],P=P[2]) s_ees[2]=entropy(Name$,T=T[2],P=P[2]) "Conservation of Energy, Steady-flow: " "E_dot_in=E_dot_out" m_dot*(h_ees[1]+Vel[1]^2/2*convert(m^2/s^2,kJ/kg))=m_dot*(h_ees[2]+Vel[2]^2/2*convert(m^2/s ^2,kJ/kg))+Q_dot_out+W_dot_out_ees "Second Law analysis:" "S_dot_in-S_dot_out+S_dot_gen = 0" m_dot*s_ees[1] - m_dot*s_ees[2] - Q_dot_out/T_o + S_dot_gen_ees= 0 "Exergy Destroyed:" X_dot_destroyed_ees = T_o*S_dot_gen_ees
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-51
P2 [kPa] 100 200 300 400 500 600 700 800 900 1000
T2 [K] 455 455 455 455 455 455 455 455 455 455
Woutchart [kW] 713.3 725.2 737.3 749.5 761.7 774.1 786.5 799.1 811.8 824.5
Woutees [kW] 420.6 448.1 476.5 505.8 536.1 567.5 600 633.9 669.3 706.6
Woutideal [kW] 1336 1336 1336 1336 1336 1336 1336 1336 1336 1336
Xdestroyedchart [kW] 2383 1901 1617 1415 1256 1126 1014 917.3 831 753.1
Xdestroyedees [kW] 2519 2029 1736 1523 1354 1212 1090 980.1 880.6 788.4
Xdestroyedideal [kW] 2171 1694 1416 1218 1064 939 833 741.2 660.2 587.7
1600
] W k[ s e e; t u o
Solution Method EES Chart Ideal gas
1200
800
W
400 100
200
300
400
500
600
700
800
900 1000
P[2] [kPa] 2800 2400
] W k[ d e y o rt s e d
X
Solution Method EES Charts Ideal Gas
2000 1600 1200 800 400 100
200
300
400
500
600
700
800
900
1000
P[2] [kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-52
12-81E Argon gas enters a turbine at a specified state and leaves at another specified state. The power output of the turbine and the exergy destruction associated with the process are to be determined using the generalized charts. Properties The gas constant and critical properties of argon are R = 0.04971 Btu/lbm.R, Tcr = 272 R, and Pcr = 705 psia (Table A-1E). Analysis (a) The enthalpy and entropy departures of argon at the specified states are determined from the generalized charts to be T 1000 P1 = 1000 psia = 3.68 T R1 = 1 = T1 = 1000 R 272 Tcr Z h1 ≅ 0 and Z s1 ≅ 0 80 Btu/s V1 = 300 ft/s P1 1000 = = 1.418 PR1 = Pcr 705 Thus argon behaves as an ideal gas at turbine inlet. Also, Ar ·m = 12 lbm/s T2 500 = = 1.838 T R2 = · W Tcr 272 Z = 0 . 04 and Z = 0 . 02 h2 s2 P 150 = 0.213 PR2 = 2 = Pcr 705 P2 = 150 psia T2 = 500 R Thus , V2 = 450 ft/s h2 − h1 = RTcr Z h1 − Z h2 + (h2 − h1 )ideal
(
)
= (0.04971)(272 )(0 − 0.04 ) + 0.1253(500 − 1000) = −63.2 Btu/lbm
The power output of the turbine is to be determined from the energy balance equation, = 0 (steady) → E& = E& E& − E& = ∆E& in
out
system
in
out
V 2 − V12 & → W& out = −m& (h2 − h1 ) + 2 m& (h1 + V12 / 2) = m& (h2 + V 22 / 2) + Q& out + W& out − Qout 2 (450 ft/s) 2 − (300 ft/s) 2 W& out = −(12 lbm/s) − 63.2 + 2 = 651.4 Btu/s = 922 hp
1 Btu/lbm 25,037 ft 2 /s 2
− 80 Btu/s
(b) Under steady conditions, the rate form of the entropy balance for the turbine simplifies to Ê0 S& − S& + S& = ∆S& =0 in
out
gen
system
Q& Q& m& s1 − m& s 2 − out + S& gen = 0 → S& gen = m& ( s 2 − s 2 ) + out Tb,out T0
The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , Q& X& destroyed = T0 S& gen = T0 m& ( s 2 − s 2 ) + out T0
(
and
)
s 2 − s1 = R Z s1 − Z s2 + (s 2 − s1 )ideal
where
(s 2 − s1 )ideal = c p ln
(
)
T2 P 500 150 − R ln 2 = 0.1253ln − 0.04971 ln = 0.00745 Btu/lbm ⋅ R 1000 T1 P1 1000
Thus s 2 − s1 = R Z s1 − Z s2 + (s 2 − s1 )ideal = (0.04971)[0 − (0.02)] + 0.00745 = 0.00646 Btu/lbm ⋅ R Substituting,
80 Btu/s = 121.5 Btu/s X& destroyed = (535 R ) (12 lbm/s)(0.00646 Btu/lbm ⋅ R ) + 535 R
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-53
12-82 An adiabatic storage tank that is initially evacuated is connected to a supply line that carries nitrogen. A valve is opened, and nitrogen flows into the tank. The final temperature in the tank is to be determined by treating nitrogen as an ideal gas and using the generalized charts, and the results are to be compared to the given actual value. Assumptions 1 Uniform flow conditions exist. 2 Kinetic and potential energies are negligible. Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min − mout = ∆msystem Energy balance:
→
Ein − Eout = ∆Esystem
Combining the two balances:
mi = m2
(since mout = minitial = 0)
→ 0 + mi hi = m2 u2
u2 = hi
(a) From the ideal gas property table of nitrogen, at 225 K we read
N2
10 MPa 225 K
u2 = hi = h@ 225 K = 6,537 kJ / kmol V1 = 0.2 m3
The temperature that corresponds to this u2 value is T2 = 314.8 K
Initially evacuated
(7.4% error)
(b) Using the generalized enthalpy departure chart, hi is determined to be Ti 225 = = 1.78 hi ,ideal − hi Tcr 126.2 = 0.9 Z h ,i = Pi Ru Tcr 10 = = = 2.95 Pcr 3.39
T R ,i = PR ,i
(Fig. A-29)
Thus, hi = hi ,ideal − 0.9 Ru Tcr = 6,537 − (0.9 )(8.314 )(126.2) = 5,593 kJ/kmol
and u 2 = hi = 5,593 kJ/kmol
Try T2 = 280 K. Then at PR2 = 2.95 and TR2 = 2.22 we read Z2 = 0.98 and (h2,ideal − h2 ) / Ru Tcr = 0.55 Thus, h2 = h2,ideal − 0.55Ru Tcr = 8,141 − (0.55)(8.314)(126.2) = 7,564 kJ/kmol u 2 = h2 − ZRu T2 = 7,564 − (0.98)(8.314)(280) = 5,283 kJ/kmol
Try T2 = 300 K. Then at PR2 = 2.95 and TR2 = 2.38 we read Z2 = 1.0 and (h2,ideal − h2 ) / Ru Tcr = 0.50 Thus, h2 = h2,ideal − 0.50 Ru Tcr = 8,723 − (0.50 )(8.314)(126.2 ) = 8,198 kJ/kmol u 2 = h2 − ZRu T2 = 8,198 − (1.0)(8.314 )(300) = 5,704 kJ/kmol
By linear interpolation, T2 = 294.7 K
(0.6% error)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-54
12-83 It is to be shown that
dv
v
= β dT − α dP . Also, a relation is to be obtained for the ratio of specific
volumes v 2/ v 1 as a homogeneous system undergoes a process from state 1 to state 2. Analysis We take v = v (P, T). Its total differential is ∂v ∂v dv = dP dT + ∂P T ∂T P
Dividing by v, dv
v
=
1 ∂v 1 ∂v dP dT + v ∂T P v ∂P T
Using the definitions of α and β, dv
v
= β dT − α dP
Taking α and β to be constants, integration from 1 to 2 yields ln
v2 = β (T2 − T1 ) − α (P2 − P1 ) v1
which is the desired relation.
12-84 It is to be shown that
dv
v
= β dT − α dP . Also, a relation is to be obtained for the ratio of specific
volumes v 2/ v 1 as a homogeneous system undergoes an isobaric process from state 1 to state 2. Analysis We take v = v (P, T). Its total differential is ∂v ∂v dv = dP dT + T ∂ ∂P T P
which, for a constant pressure process, reduces to ∂v dv = dT ∂T P
Dividing by v, dv
v
=
1 ∂v dT v ∂T P
Using the definition of β, dv
v
= β dT
Taking β to be a constant, integration from 1 to 2 yields ln
or
v2 = β (T2 − T1 ) = v1 v2 = exp[β (T2 − T1 )] v1
which is the desired relation.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-55
12-85 The volume expansivity of water is given. The change in volume of water when it is heated at constant pressure is to be determined. Properties The volume expansivity of water is given to be 0.207×10-6 K-1 at 20°C. Analysis We take v = v (P, T). Its total differential is ∂v ∂v dv = dT + dP ∂ T ∂P T P
which, for a constant pressure process, reduces to ∂v dv = dT ∂T P
Dividing by v and using the definition of β, dv
v
=
1 ∂v dT = β dT v ∂T P
Taking β to be a constant, integration from 1 to 2 yields ln
v2 = β (T2 − T1 ) v1
or
v2 = exp[β (T2 − T1 )] v1 Substituting the given values and noting that for a fixed mass V2/V1 = v2/v1,
(
) [(
)
V 2 = V 1 exp[β (T2 − T1 )] = 1 m 3 exp 0.207 × 10 −6 K −1 (30 − 10)°C = 1.00000414 m
]
3
Therefore, ∆V = V 2 −V1 = 1.00000414 − 1 = 0.00000414 m 3 = 4.14 cm 3
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-56
12-86 The volume expansivity of copper is given at two temperatures. The percent change in the volume of copper when it is heated at atmospheric pressure is to be determined. Properties The volume expansivity of copper is given to be 49.2×10-6 K-1 at 300 K, and be 54.2×10-6 K-1 at 500 K Analysis We take v = v (P, T). Its total differential is ∂v ∂v dv = dP dT + T ∂ ∂P T P
which, for a constant pressure process, reduces to ∂v dv = dT ∂T P
Dividing by v and using the definition of β, dv
v
=
1 ∂v dT = β dT v ∂T P
Taking β to be a constant, integration from 1 to 2 yields ln
v2 = β (T2 − T1 ) v1
or
v2 = exp[β (T2 − T1 )] v1 The average value of β is
(
)
β ave = (β1 + β 2 ) / 2 = 49.2 × 10 −6 + 54.2 × 10 −6 / 2 = 51.7 × 10 −6 K −1 Substituting the given values,
[(
)
]
v2 = exp[β (T2 − T1 )] = exp 51.7 × 10 − 6 K −1 (500 − 300)K = 1.0104 v1 Therefore, the volume of copper block will increase by 1.04 percent.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-57
12-87 It is to be shown that the position of the Joule-Thompson coefficient inversion curve on the T-P plane is given by (∂Z/∂T)P = 0. Analysis The inversion curve is the locus of the points at which the Joule-Thompson coefficient µ is zero, µ=
1 cp
∂v T ∂T − v = 0 P
which can also be written as ZRT ∂v =0 T − P ∂T P
(a)
since it is given that
v=
ZRT P
(b)
Taking the derivative of (b) with respect to T holding P constant gives R ∂Z ∂ (ZRT / P ) ∂v = T = + Z ∂T ∂T P P P ∂T P
Substituting in (a), ZRT TR ∂Z T =0 + Z − P ∂T P P ∂Z T +Z −Z =0 ∂T P ∂Z =0 ∂T P
which is the desired relation.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-58 12-88 It is to be shown that for an isentropic expansion or compression process Pv k = constant. It is also to be shown that the isentropic expansion exponent k reduces to the specific heat ratio cp/cv for an ideal gas. Analysis We note that ds = 0 for an isentropic process. Taking s = s(P, v), the total differential ds can be expressed as ∂s ∂s ds = dP + (a) dv = 0 ∂v P ∂P v We now substitute the Maxwell relations below into (a) ∂s ∂P ∂s ∂v = = − and v P T ∂ ∂ ∂ P ∂T s v s to get ∂P ∂v − dv = 0 dP + ∂T s ∂T s Rearranging, ∂P ∂T ∂P dP − → dP − dv = 0 dv = 0 v T ∂ ∂ ∂v s s s dP 1 ∂P − dv = 0 P P ∂v s We now define isentropic expansion exponent k as v ∂P k=− P ∂v s
Dividing by P,
(b)
dP dv +k =0 P v Taking k to be a constant and integrating, ln P + k ln v = constant → ln Pv k = constant Thus, Pv k = constant To show that k = cp/cv for an ideal gas, we write the cyclic relations for the following two groups of variables: (s, T , v ) → ∂s ∂v ∂T = −1 → cv ∂v ∂T = −1 (c) T ∂s T ∂v s ∂T v ∂s T ∂v s
Substituting in (b),
c p ∂P ∂T ∂P ∂T → = −1 (d ) = −1 T ∂s T ∂P s ∂T P ∂s T ∂P s
(s, T , P ) → ∂s
∂s ∂s cv = T and c p = T T ∂ ∂T P v Setting Eqs. (c) and (d) equal to each other, c p ∂P ∂T c ∂v ∂T = v T ∂s T ∂P s T ∂s T ∂v s or, c p ∂s ∂P ∂v ∂T ∂s ∂v ∂P ∂T ∂v ∂P = = = cv ∂P T ∂T s ∂s T ∂v s ∂P ∂s T ∂T ∂v s ∂P T ∂v s
where we used the relations
v ∂v ∂ (RT / P ) = =− ∂P P ∂P T T cp v ∂P Substituting, =− =k cv P ∂v s which is the desired relation. but
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-59
12-89 EES The work done by the refrigerant 134a as it undergoes an isothermal process in a closed system is to be determined using the tabular (EES) data and the generalized charts. Analysis The solution using EES built-in property data is as follows: T1 = 60°C u1 = 135.65 kJ/kg P1 = 3 MPa s1 = 0.4828 kJ/kg.K T2 = 60°C
u 2 = 280.35 kJ/kg P2 = 0.1 MPa s 2 = 1.2035 kJ/kg.K
∆s EES = s 2 − s1 = 1.2035 − 0.4828 = 0.7207 kJ/kg.K q EES = T1 ∆s EES = (60 + 273.15 K )(0.7207 kJ/kg.K ) = 240.11 kJ/kg wEES = q EES − (u 2 − u1 ) = 240.1 − (280.35 − 135.65) = 95.40 kJ/kg
For the generalized chart solution we first determine the following factors using EES as T1 333.15 = = 0.8903 Tcr 374.2 → Z1 = 0.1292, Z h1 = 4.475 and Z s1 = 4.383 P 3 PR1 = 1 = = 0.7391 Pcr 4.059
TR1 =
T2 333.15 = = 0.8903 374.2 Tcr → Z 2 = 0.988, Z h 2 = 0.03091 and Z s 2 = 0.02281 P 0.1 = 2 = = 0.02464 Pcr 4.059
TR 2 = PR 2
Then, ∆h1 = Z h1 RTcr = (4.475)(0.08148 kJ/kg.K)(374.2 K) = 136.43 kJ/kg ∆s1 = Z s1 R = (4.383)(0.08148 kJ/kg.K) = 0.3572 kJ/kg.K ∆h2 = Z h 2 RTcr = (0.03091)(0.08148 kJ/kg.K)(374.2 K) = 0.94 kJ/kg ∆s 2 = Z s 2 R = (0.02281)(0.08148 kJ/kg.K) = 0.001858 kJ/kg.K ∆s ideal = R ln
P2 0.1 = (0.08148 kJ/kg.K)ln = 0.2771 kJ/kg ⋅ K P1 3
∆s chart = ∆s ideal − (∆s 2 − ∆s1 ) = 0.2771 − (0.001858 − 0.3572) = 0.6324 kJ/kg ⋅ K q chart = T1 ∆s chart = (60 + 273.15 K )(0.6324 kJ/kg.K ) = 210.70 kJ/kg ∆u chart = ∆hideal − (∆h2 − ∆h1 ) − ( Z 2 RT2 − Z 1 RT1 )
= 0 − (0.94 − 136.43) − [(0.988)(0.08148)(333) − (0.1292)(0.08148)(333)] = 112.17 kJ/kg
wchart = q chart − ∆u chart = 210.70 − 112.17 = 98.53 kJ/kg
The copy of the EES solution of this problem is given next.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-60
"Input data" T_critical=T_CRIT(R134a) "[K]" P_critical=P_CRIT(R134a) "[kpa]" T[1]=60+273.15"[K]" T[2]=T[1]"[K]" P[1]=3000"[kPa]" P[2]=100"[kPa]" R_u=8.314"[kJ/kmol-K]" M=molarmass(R134a) R=R_u/M"[kJ/kg-K]" "***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****" "For the isothermal process, the heat transfer is T*(s[2] - s[1]):" DELTAs_EES=(entropy(R134a,T=T[2],P=P[2])-entropy(R134a,T=T[1],P=P[1])) q_EES=T[1]*DELTAs_EES s_2=entropy(R134a,T=T[2],P=P[2]) s_1=entropy(R134a,T=T[1],P=P[1]) "Conservation of energy for the closed system:" DELTAu_EES=intEnergy(R134a,T=T[2],p=P[2])-intEnergy(R134a,T=T[1],P=P[1]) q_EES-w_EES=DELTAu_EES u_1=intEnergy(R134a,T=T[1],P=P[1]) u_2=intEnergy(R134a,T=T[2],p=P[2]) "***** COMPRESSABILITY CHART SOLUTION ******" "State 1" Tr[1]=T[1]/T_critical pr[1]=p[1]/p_critical Z[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical"Enthalpy departure" Z_h1=ENTHDEP(Tr[1], Pr[1]) DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure" Z_s1=ENTRDEP(Tr[1], Pr[1]) "State 2" Tr[2]=T[2]/T_critical Pr[2]=P[2]/P_critical Z[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical"Enthalpy departure" Z_h2=ENTHDEP(Tr[2], Pr[2]) DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure" Z_s2=ENTRDEP(Tr[2], Pr[2]) "Entropy Change" DELTAs_ideal= -R*ln(P[2]/P[1]) DELTAs_chart=DELTAs_ideal-(DELTAs[2]-DELTAs[1]) "For the isothermal process, the heat transfer is T*(s[2] - s[1]):" q_chart=T[1]*DELTAs_chart "Conservation of energy for the closed system:" DELTAh_ideal=0 DELTAu_chart=DELTAh_ideal-(DELTAh[2]-DELTAh[1])-(Z[2]*R*T[2]-Z[1]*R*T[1]) q_chart-w_chart=DELTAu_chart PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-61
SOLUTION DELTAh[1]=136.43 DELTAh[2]=0.94 DELTAh_ideal=0 DELTAs[1]=0.3572 DELTAs[2]=0.001858 DELTAs_chart=0.6324 [kJ/kg-K] DELTAs_EES=0.7207 [kJ/kg-K] DELTAs_ideal=0.2771 [kJ/kg-K] DELTAu_chart=112.17 DELTAu_EES=144.7 M=102 [kg/kmol] P[1]=3000 [kPa] P[2]=100 [kPa] pr[1]=0.7391 Pr[2]=0.02464 P_critical=4059 [kpa] q_chart=210.70 [kJ/kg] q_EES=240.11 [kJ/kg] R=0.08148 [kJ/kg-K
R_u=8.314 [kJ/kmol-K] s_1=0.4828 [kJ/kg-K] s_2=1.2035 [kJ/kg-K] T[1]=333.2 [K] T[2]=333.2 [K] Tr[1]=0.8903 Tr[2]=0.8903 T_critical=374.2 [K] u_1=135.65 [kJ/kg] u_2=280.35 [kJ/kg] w_chart=98.53 [kJ/kg] w_EES=95.42 [kJ/kg] Z[1]=0.1292 Z[2]=0.988 Z_h1=4.475 Z_h2=0.03091 Z_s1=4.383 Z_s2=0.02281
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-62
12-90 The heat transfer, work, and entropy changes of methane during a process in a piston-cylinder device are to be determined assuming ideal gas behavior, using generalized charts, and real fluid (EES) data. Analysis The ideal gas solution: (Properties are obtained from EES) T1 = 100°C → h1 = −4492 kJ/kg T1 = 100°C, P1 = 4 MPa → s1 = 10.22 kJ/kg.K State 1: u = h − RT = (−4492) − (0.5182)(100 + 273.15) = −4685 kJ/kg 1 1 1
v1 = R
T1 100 + 273.15 K 3 = (0.5182 kJ/kg.K ) = 0.04834 m /kg P1 4000 kPa
T2 = 350°C → h2 = −3770 kJ/kg T2 = 350°C, P2 = 4 MPa → s 2 = 11.68 kJ/kg.K State 2: u = h − RT = (−3770) − (0.5182)(350 + 273.15) = −4093 kJ/kg 2 2 2
v2 = R
T2 350 + 273.15 K 3 = (0.5182 kJ/kg.K ) = 0.08073 m /kg P2 4000 kPa
wideal = P(v 2 − v 1 ) = (4000 kPa)(0.08073 - 0.04834)m 3 /kg = 129.56 kJ/kg q ideal = wideal + (u 2 − u1 ) = 129.56 + [(−4093) − (−4685)] = 721.70 kJ/kg ∆s ideal = s 2 − s1 = 11.68 − 10.22 = 1.46 kJ/kg
For the generalized chart solution we first determine the following factors using EES as T1 373 = = 1.227 Tcr 304.2 → Z 1 = 0.9023, Z h1 = 0.4318 and Z s1 = 0.2555 P1 4 = = = 0.5413 Pcr 7.39
T R1 = PR1
T2 623 = = 2.048 Tcr 304.2 → Z 2 = 0.995, Z h 2 = 0.1435 and Z s 2 = 0.06446 P2 4 = = = 0.5413 Pcr 7.39
TR 2 = PR 2
State 1: ∆h1 = Z h1 RTcr = (0.4318)(0.5182 kJ/kg.K)(304.2 K) = 68.07 kJ/kg h1 = h1,ideal − ∆h1 = (−4492) − 68.07 = −4560 kJ/kg u1 = h1 − Z 1 RT1 = (−4560) − (0.9023)(0.5182)(373.15) = −4734 kJ/kg
v 1 = Z1 R
T1 373.15 = (0.9023)(0.5182) = 0.04362 m 3 /kg P1 4000
∆s1 = Z s1 R = (0.2555)(0.5182 kJ/kg.K) = 0.1324 kJ/kg.K s1 = s1,ideal − ∆s1 = 10.22 − 0.1324 = 10.09 kJ/kg.K
State 2: ∆h2 = Z h 2 RTcr = (0.1435)(0.5182 kJ/kg.K)(304.2 K) = 22.62 kJ/kg h2 = h2,ideal − ∆h2 = (−3770) − 22.62 = −3793 kJ/kg u 2 = h2 − Z 2 RT2 = (−3793) − (0.995)(0.5182)(623.15) = −4114 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-63
v2 = Z2R
T2 623.15 = (0.995)(0.5182) = 0.08033 m 3 /kg P2 4000
∆s 2 = Z s 2 R = (0.06446)(0.5182 kJ/kg.K) = 0.03341 kJ/kg.K s 2 = s 2,ideal − ∆s 2 = 11.68 − 0.03341 = 11.65 kJ/kg.K
Then, wchart = P(v 2 − v 1 ) = (4000 kPa)(0.08033 - 0.04362)m 3 /kg = 146.84 kJ/kg q chart = wchart + (u 2 − u1 ) = 146.84 + [(−4114) − (−4734)] = 766.84 kJ/kg ∆s chart = s 2 − s1 = 11.65 − 10.09 = 1.56 kJ/kg
The solution using EES built-in property data is as follows:
v = 0.04717 m 3 /kg T1 = 100°C 1 u1 = −39.82 kJ/kg P1 = 4 MPa s1 = −1.439 kJ/kg.K v = 0.08141 m 3 /kg T2 = 350°C 2 u 2 = 564.52 kJ/kg P2 = 4 MPa s 2 = 0.06329 kJ/kg.K wEES = P(v 2 − v 1 ) = (4000 kPa)(0.08141 - 0.04717)m 3 /kg = 136.96 kJ/kg q EES = wEES + (u 2 − u1 ) = 136.97 + [564.52 − (−39.82)] = 741.31 kJ/kg ∆s EES = s 2 − s1 = 0.06329 − (−1.439) = 1.50 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-64
Fundamentals of Engineering (FE) Exam Problems
12-91 A substance whose Joule-Thomson coefficient is negative is throttled to a lower pressure. During this process, (select the correct statement) (a) the temperature of the substance will increase. (b) the temperature of the substance will decrease. (c) the entropy of the substance will remain constant. (d) the entropy of the substance will decrease. (e) the enthalpy of the substance will decrease. Answer (a) the temperature of the substance will increase.
12-92 Consider the liquid-vapor saturation curve of a pure substance on the P-T diagram. The magnitude of the slope of the tangent line to this curve at a temperature T (in Kelvin) is (a) proportional to the enthalpy of vaporization hfg at that temperature, (b) proportional to the temperature T, (c) proportional to the square of the temperature T, (d) proportional to the volume change vfg at that temperature, (e) inversely proportional to the entropy change sfg at that temperature, Answer (a) proportional to the enthalpy of vaporization hfg at that temperature,
12-93 Based on the generalized charts, the error involved in the enthalpy of CO2 at 350 K and 8 MPa if it is assumed to be an ideal gas is (a) 0 (b) 20% (c) 35% (d) 26% (e) 65% Answer (d) 26% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T=350 "K" P=8000 "kPa" Pcr=P_CRIT(CarbonDioxide) Tcr=T_CRIT(CarbonDioxide) Tr=T/Tcr Pr=P/Pcr Z=COMPRESS(Tr, Pr) hR=ENTHDEP(Tr, Pr)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
12-65
12-94 Based on data from the refrigerant-134a tables, the Joule-Thompson coefficient of refrigerant-134a at 0.8 MPa and 100°C is approximately (a) 0 (b) -5°C/MPa (c) 11°C/MPa (d) 8°C/MPa (e) 26°C/MPa Answer (c) 11°C/MPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=100 "C" P1=800 "kPa" h1=ENTHALPY(R134a,T=T1,P=P1) Tlow=TEMPERATURE(R134a,h=h1,P=P1+100) Thigh=TEMPERATURE(R134a,h=h1,P=P1-100) JT=(Tlow-Thigh)/200
12-95 For a gas whose equation of state is P(v - b) = RT, the specific heat difference cp – cv is equal to (a) R (b) R – b (c) R + b (d) 0 (e) R(1 + v/b) Answer (a) R Solution The general relation for the specific heat difference cp - cv is 2
∂v ∂P c p − cv = −T ∂T P ∂v T For the given gas, P(v - b) = RT. Then,
v=
RT R ∂v +b → = P ∂T P P
P=
RT RT P ∂P → =− =− 2 v −b v −b ∂v T (v − b)
Substituting, 2
2
P TR R c p − cv = −T − =R = ( − v v − b) P b P
12-96 ··· 12-98 Design and Essay Problems
KJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-1
Chapter 13 GAS MIXTURES Composition of Gas Mixtures 13-1C It is the average or the equivalent gas constant of the gas mixture. No. 13-2C No. We can do this only when each gas has the same mole fraction. 13-3C It is the average or the equivalent molar mass of the gas mixture. No. 13-4C The mass fractions will be identical, but the mole fractions will not. 13-5C Yes. 13-6C The ratio of the mass of a component to the mass of the mixture is called the mass fraction (mf), and the ratio of the mole number of a component to the mole number of the mixture is called the mole fraction (y). 13-7C From the definition of mass fraction, M mi N M = i i = y i i mm N m M m Mm
mf i =
13-8C Yes, because both CO2 and N2O has the same molar mass, M = 44 kg/kmol.
13-9 A mixture consists of two gases. Relations for mole fractions when mass fractions are known are to be obtained . Analysis The mass fractions of A and B are expressed as mf A =
MA mA N M = A A = yA Mm mm N m M m
and
mf B = y B
MB Mm
Where m is mass, M is the molar mass, N is the number of moles, and y is the mole fraction. The apparent molar mass of the mixture is
Mm =
mm N A M A + N B M B = = y AM A + yB M B Nm Nm
Combining the two equation above and noting that y A + y B = 1 gives the following convenient relations for converting mass fractions to mole fractions, yA =
MB M A (1 / mf A − 1) + M B
and
yB = 1 − y A
which are the desired relations.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-2
13-10 The molar fractions of the constituents of moist air are given. The mass fractions of the constituents are to be determined. Assumptions The small amounts of gases in air are ignored, and dry air is assumed to consist of N2 and O2 only. Properties The molar masses of N2, O2, and H2O are 28.0, 32.0, and 18.0 kg/kmol, respectively (Table A1). Analysis The molar mass of moist air is M=
∑y M i
i
= 0.78 × 28.0 + 0.20 × 32.0 + 0.02 × 18 = 28.6 kg / kmol
Then the mass fractions of constituent gases are determined to be N2 :
mf N 2 = y N 2
O2 :
mfO 2 = yO 2
M N2 M M O2 M
mf H 2 O = y H 2 O
H 2O :
= (0.78)
28.0 = 0.764 28.6
= (0.20)
32.0 = 0.224 28.6
M H 2O M
= (0.02)
Moist air 78% N2 20% O2 2% H2 O (Mole fractions)
18.0 = 0.013 28.6
Therefore, the mass fractions of N2, O2, and H2O in the air are 76.4%, 22.4%, and 1.3%, respectively.
13-11 The molar fractions of the constituents of a gas mixture are given. The gravimetric analysis of the mixture, its molar mass, and gas constant are to be determined. Properties The molar masses of N2, and CO2 are 28.0 and 44.0 kg/kmol, respectively (Table A-1) Analysis Consider 100 kmol of mixture. Then the mass of each component and the total mass are N N 2 = 60 kmol → m N 2 = N N 2 M N 2 = (60 kmol)(28 kg/kmol) = 1680 kg N CO 2 = 40 kmol → m CO 2 = N CO 2 M CO 2 = (40 kmol)(44 kg/kmol) = 1760 kg m m = m N 2 + m CO 2 = 1680 kg + 1760 kg = 3440 kg
Then the mass fraction of each component (gravimetric analysis) becomes mf N 2 = mf CO 2 =
m N2 mm m CO 2 mm
=
1680 kg = 0.488 or 48.8% 3440 kg
=
mole 60% N2 40% CO2
1760 kg = 0.512 or 51.2% 3440 kg
The molar mass and the gas constant of the mixture are determined from their definitions, Mm =
mm 3,440 kg = = 34.40 kg / kmol N m 100 kmol
Rm =
Ru 8.314 kJ / kmol ⋅ K = = 0.242 kJ / kg ⋅ K Mm 34.4 kg / kmol
and
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-3
13-12 The molar fractions of the constituents of a gas mixture are given. The gravimetric analysis of the mixture, its molar mass, and gas constant are to be determined. Properties The molar masses of O2 and CO2 are 32.0 and 44.0 kg/kmol, respectively (Table A-1) Analysis Consider 100 kmol of mixture. Then the mass of each component and the total mass are N O 2 = 60 kmol → m O 2 = N O 2 M O 2 = (60 kmol)(32 kg/kmol) = 1920 kg N CO 2 = 40 kmol → m CO 2 = N CO 2 M CO 2 = (40 kmol)(44 kg/kmol) = 1760 kg m m = m O 2 + m CO 2 = 1920 kg + 1760 kg = 3680 kg
Then the mass fraction of each component (gravimetric analysis) becomes mf O 2 = mf CO 2 =
mO2 mm m CO 2 mm
=
1920 kg = 0.522 or 52.2% 3680 kg
=
mole 60% O2 40% CO2
1760 kg = 0.478 or 47.8% 3680 kg
The molar mass and the gas constant of the mixture are determined from their definitions, Mm =
mm 3680 kg = = 36.80 kg/kmol N m 100 kmol
and Rm =
Ru 8.314 kJ/kmol ⋅ K = = 0.226 kJ/kg ⋅ K Mm 36.8 kg/kmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-4
13-13 The masses of the constituents of a gas mixture are given. The mass fractions, the mole fractions, the average molar mass, and gas constant are to be determined. Properties The molar masses of O2, N2, and CO2 are 32.0, 28.0 and 44.0 kg/kmol, respectively (Table A-1) Analysis (a) The total mass of the mixture is m m = m O 2 + m N 2 + m CO 2 = 5 kg + 8 kg + 10 kg = 23 kg
Then the mass fraction of each component becomes mf O 2 = mf N 2 = mf CO 2 =
mO2 mm
=
5 kg = 0.217 23 kg
=
8 kg = 0.348 23 kg
m N2 mm m CO 2
=
mm
5 kg O2 8 kg N2 10 kg CO2
10 kg = 0.435 23 kg
(b) To find the mole fractions, we need to determine the mole numbers of each component first, N O2 = N N2 = N CO 2 =
mO2 M O2 m N2 M N2
=
5 kg = 0.156 kmol 32 kg/kmol
=
8 kg = 0.286 kmol 28 kg/kmol
m CO 2
=
M CO 2
10 kg = 0.227 kmol 44 kg/kmol
Thus, N m = N O 2 + N N 2 + N CO 2 = 0.156 kmol + 0.286 kmol + 0.227 kmol = 0.669 kmol
and y O2 = y N2 = y CO 2 =
N O2 Nm N N2 Nm N CO 2 Nm
=
0.156 kmol = 0.233 0.699 kmol
=
0.286 kmol = 0.428 0.669 kmol
=
0.227 kmol = 0.339 0.669 kmol
(c) The average molar mass and gas constant of the mixture are determined from their definitions: Mm =
mm 23 kg = = 34.4 kg/kmol N m 0.669 kmol
and Rm =
Ru 8.314 kJ/kmol ⋅ K = = 0.242 kJ/kg ⋅ K 34.4 kg/kmol Mm
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-5
13-14 The mass fractions of the constituents of a gas mixture are given. The mole fractions of the gas and gas constant are to be determined. Properties The molar masses of CH4, and CO2 are 16.0 and 44.0 kg/kmol, respectively (Table A-1) Analysis For convenience, consider 100 kg of the mixture. Then the number of moles of each component and the total number of moles are → N CH 4 = m CH 4 = 75 kg → N CO 2 = m CO 2 = 25 kg
m CH 4 M CH 4 m CO 2 M CO 2
=
75 kg = 4.688 kmol 16 kg/kmol
mass
=
25 kg = 0.568 kmol 44 kg/kmol
75% CH4 25% CO2
N m = N CH 4 + N CO 2 = 4.688 kmol + 0.568 kmol = 5.256 kmol
Then the mole fraction of each component becomes y CH 4 = y CO 2 =
N CH 4 Nm N CO 2 Nm
=
4.688 kmol = 0.892 or 89.2% 5.256 kmol
=
0.568 kmol = 0.108 or 10.8% 5.256 kmol
The molar mass and the gas constant of the mixture are determined from their definitions, Mm =
mm 100 kg = = 19.03 kg/kmol N m 5.256 kmol
and Rm =
Ru 8.314 kJ/kmol ⋅ K = = 0.437 kJ/kg ⋅ K Mm 19.03 kg/kmol
13-15 The mole numbers of the constituents of a gas mixture are given. The mass of each gas and the apparent gas constant are to be determined. Properties The molar masses of H2, and N2 are 2.0 and 28.0 kg/kmol, respectively (Table A-1) Analysis The mass of each component is determined from N H 2 = 8 kmol → m H 2 = N H 2 M H 2 = (8 kmol)(2.0 kg/kmol) = 16 kg N N 2 = 2 kmol → m N 2 = N N 2 M N 2 = (2 kmol)(28 kg/kmol) = 56 kg
8 kmol H2 2 kmol N2
The total mass and the total number of moles are m m = m H 2 + m N 2 = 16 kg + 56 kg = 72 kg N m = N H 2 + N N 2 = 8 kmol + 2 kmol = 10 kmol
The molar mass and the gas constant of the mixture are determined from their definitions, Mm =
mm 72 kg = = 7.2 kg/kmol N m 10 kmol
and Rm =
Ru 8.314 kJ/kmol ⋅ K = = 1.155 kJ/kg ⋅ K Mm 7.2 kg/kmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-6
13-16E The mole numbers of the constituents of a gas mixture are given. The mass of each gas and the apparent gas constant are to be determined. Properties The molar masses of H2, and N2 are 2.0 and 28.0 lbm/lbmol, respectively (Table A-1E). Analysis The mass of each component is determined from N H 2 = 5 lbmol → m H 2 = N H 2 M H 2 = (5 lbmol)(2.0 lbm/lbmol) = 10 lbm N N 2 = 4 lbmol → m N 2 = N N 2 M N 2 = (4 lbmol)(28 lbm/lbmol) = 112 lbm
The total mass and the total number of moles are m m = m H 2 + m N 2 = 10 lbm + 112 lbm = 122 lbm
5 lbmol H2 4 lbmol N2
N m = N H 2 + N N 2 = 5 lbmol + 4 lbmol = 9 lbmol
The molar mass and the gas constant of the mixture are determined from their definitions, m 122 lbm Mm = m = = 13.56 lbm/lbmol N m 9 lbmol and
Rm =
Ru 1.986 Btu/lbmol ⋅ R = = 0.1465 Btu/lbm ⋅ R Mm 13.56 lbm/lbmol
13-17 The mass fractions of the constituents of a gas mixture are given. The volumetric analysis of the mixture and the apparent gas constant are to be determined. Properties The molar masses of O2, N2 and CO2 are 32.0, 28, and 44.0 kg/kmol, respectively (Table A-1) Analysis For convenience, consider 100 kg of the mixture. Then the number of moles of each component and the total number of moles are mO2 20 kg m O 2 = 20 kg → N O 2 = = = 0.625 kmol M O 2 32 kg/kmol m N 2 = 20 kg → N N 2 =
mN2 M N2
→ N CO 2 = m CO 2 = 50 kg
30 kg = = 1.071 kmol 28 kg/kmol
m CO 2 M CO 2
=
50 kg = 1.136 kmol 44 kg/kmol
mass
20% O2 30% N2 50% CO2
N m = N O 2 + N N 2 + N CO 2 = 0.625 + 1.071 + 1.136 = 2.832 kmol
Noting that the volume fractions are same as the mole fractions, the volume fraction of each component becomes N O 2 0.625 kmol = = 0.221 or 22.1% y O2 = 2.832 kmol Nm y N2 =
N N2 Nm
y CO 2 =
=
N CO 2 Nm
1.071 kmol = 0.378 or 37.8% 2.832 kmol =
1.136 kmol = 0.401 or 40.1% 2.832 kmol
The molar mass and the gas constant of the mixture are determined from their definitions, m 100 kg Mm = m = = 35.31 kg/kmol N m 2.832 kmol and
Rm =
Ru 8.314 kJ/kmol ⋅ K = = 0.235 kJ/kg ⋅ K Mm 35.31 kg/kmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-7
P-v-T Behavior of Gas Mixtures 13-18C Normally yes. Air, for example, behaves as an ideal gas in the range of temperatures and pressures at which oxygen and nitrogen behave as ideal gases. 13-19C The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if existed alone at the mixture temperature and volume. This law holds exactly for ideal gas mixtures, but only approximately for real gas mixtures. 13-20C The volume of a gas mixture is equal to the sum of the volumes each gas would occupy if existed alone at the mixture temperature and pressure. This law holds exactly for ideal gas mixtures, but only approximately for real gas mixtures. 13-21C The P-v-T behavior of a component in an ideal gas mixture is expressed by the ideal gas equation of state using the properties of the individual component instead of the mixture, Pivi = RiTi. The P-v-T behavior of a component in a real gas mixture is expressed by more complex equations of state, or by Pivi = ZiRiTi, where Zi is the compressibility factor. 13-22C Component pressure is the pressure a component would exert if existed alone at the mixture temperature and volume. Partial pressure is the quantity yiPm, where yi is the mole fraction of component i. These two are identical for ideal gases. 13-23C Component volume is the volume a component would occupy if existed alone at the mixture temperature and pressure. Partial volume is the quantity yiVm, where yi is the mole fraction of component i. These two are identical for ideal gases. 13-24C The one with the highest mole number. 13-25C The partial pressures will decrease but the pressure fractions will remain the same. 13-26C The partial pressures will increase but the pressure fractions will remain the same. 13-27C No. The correct expression is “the volume of a gas mixture is equal to the sum of the volumes each gas would occupy if existed alone at the mixture temperature and pressure.” 13-28C No. The correct expression is “the temperature of a gas mixture is equal to the temperature of the individual gas components.” 13-29C Yes, it is correct. 13-30C With Kay's rule, a real-gas mixture is treated as a pure substance whose critical pressure and temperature are defined in terms of the critical pressures and temperatures of the mixture components as Pcr′ , m =
∑y P
i cr ,i
and Tcr′ , m =
∑yT
i cr ,i
The compressibility factor of the mixture (Zm) is then easily determined using these pseudo-critical point values.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-8
13-31 A tank contains a mixture of two gases of known masses at a specified pressure and temperature. The volume of the tank is to be determined. Assumptions Under specified conditions both O2 and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Analysis The total number of moles is
8 kmol O2 10 kmol CO2
N m = N O 2 + N CO 2 = 8 kmol + 10 kmol = 18 kmol
Then
290 K 150 kPa
N RT (18 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(290 K) Vm = m u m = = 289.3 m 3 Pm 150 kPa
13-32 A tank contains a mixture of two gases of known masses at a specified pressure and temperature. The volume of the tank is to be determined. Assumptions Under specified conditions both O2 and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Analysis The total number of moles is
8 kmol O2 10 kmol CO2
N m = N O 2 + N CO 2 = 8 kmol + 10 kmol = 18 kmol
Then
400 K 150 kPa
N RT (18 kmol)(8.314 kPa ⋅ m 3/kmol ⋅ K)(400 K) Vm = m u m = = 399.1 m 3 Pm 150 kPa
13-33 A tank contains a mixture of two gases of known masses at a specified pressure and temperature. The mixture is now heated to a specified temperature. The volume of the tank and the final pressure of the mixture are to be determined. Assumptions Under specified conditions both Ar and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Analysis The total number of moles is
0.5 kmol Ar 2 kmol N2
N m = N Ar + N N 2 = 0.5 kmol + 2 kmol = 2.5 kmol
And
Vm
N R T (2.5 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(280 K) = m u m = = 23.3 m 3 Pm 250 kPa
Q
280 K 250 kPa
Also, T P2V 2 P1V1 400 K = → P2 = 2 P1 = (250 kPa ) = 357.1 kPa T1 280 K T2 T1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-9
13-34 The masses of the constituents of a gas mixture at a specified pressure and temperature are given. The partial pressure of each gas and the apparent molar mass of the gas mixture are to be determined. Assumptions Under specified conditions both CO2 and CH4 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 kg/kmol, respectively (Table A-1) Analysis The mole numbers of the constituents are mCO 2 1 kg mCO 2 = 1 kg → N CO 2 = = = 0.0227 kmol MCO 2 44 kg / kmol 1 kg CO2 mCH 4 = 3 kg
→
N CH 4 =
mCH 4 MCH 4
=
3 kg = 0.1875 kmol 16 kg / kmol
N m = NCO 2 + NCH 4 = 0.0227 kmol + 0.1875 kmol = 0.2102 kmol yCO 2 =
N CO 2 Nm N CH 4
=
3 kg CH4 300 K 200 kPa
0.0227 kmol = 0108 . 0.2102 kmol
0.1875 kmol = 0.892 Nm 0.2102 kmol Then the partial pressures become PCO 2 = y CO 2 Pm = (0.108)(200 kPa ) = 21.6 kPa yCH 4 =
=
PCH 4 = y CH 4 Pm = (0.892)(200 kPa ) = 178.4 kPa
The apparent molar mass of the mixture is m 4 kg Mm = m = = 19.03 kg / kmol N m 0.2102 kmol
13-35E The masses of the constituents of a gas mixture at a specified pressure and temperature are given. The partial pressure of each gas and the apparent molar mass of the gas mixture are to be determined. Assumptions Under specified conditions both CO2 and CH4 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 lbm/lbmol, respectively (Table A-1E) Analysis The mole numbers of gases are m CO 2 1 lbm m CO 2 = 1 lbm → N CO 2 = = = 0.0227 lbmol M CO 2 44 lbm/lbmol 1 lbm CO2 m CH 4 = 3 lbm → N CH 4 =
m CH 4 M CH 4
=
3 lbm = 0.1875 lbmol 16 lbm/lbmol
N m = N CO 2 + N CH 4 = 0.0227 lbmol + 0.1875 lbmol = 0.2102 lbmol y CO 2 =
N CO 2 Nm N CH 4
=
3 lbm CH4 600 R 20 psia
0.0227 lbmol = 0.108 0.2102 lbmol
0.1875 lbmol = 0.892 Nm 0.2102 lbmol Then the partial pressures become PCO 2 = y CO 2 Pm = (0.108)(20 psia ) = 2.16 psia y CH 4 =
=
PCH 4 = y CH 4 Pm = (0.892)(20 psia ) = 17.84 psia
The apparent molar mass of the mixture is m 4 lbm Mm = m = = 19.03 lbm/lbmol N m 0.2102 lbmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-10
13-36 The masses of the constituents of a gas mixture at a specified temperature are given. The partial pressure of each gas and the total pressure of the mixture are to be determined. Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Analysis The partial pressures of constituent gases are PN 2
(0.6 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K)(300 K) mRT = 178.1 kPa = = 0.3 m 3 V N2
(0.4 kg)(0.2598 kPa ⋅ m 3 /kg ⋅ K)(300 K) mRT PO 2 = = 103.9 kPa = 0.3 m 3 V O2
0.3 m3 0.6 kg N2 0.4 kg O2 300 K
and Pm = PN 2 + PO 2 = 178.1 kPa + 103.9 kPa = 282.0 kPa
13-37 The volumetric fractions of the constituents of a gas mixture at a specified pressure and temperature are given. The mass fraction and partial pressure of each gas are to be determined. Assumptions Under specified conditions all N2, O2 and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of N2, O2 and CO2 are 28.0, 32.0, and 44.0 kg/kmol, respectively (Table A-1) Analysis For convenience, consider 100 kmol of mixture. Then the mass of each component and the total mass are N N 2 = 65 kmol → m N 2 = N N 2 M N 2 = (65 kmol)(28 kg/kmol) = 1820 kg N O 2 = 20 kmol → m O 2 = N O 2 M O 2 = (20 kmol)(32 kg/kmol) = 640 kg
N CO 2 = 15 kmol → m CO 2 = N CO 2 M CO 2 = (15 kmol)(44 kg/kmol) = 660 kg m m = m N 2 + m O 2 + m CO 2 = 1820 kg + 640 kg + 660 kg = 3120 kg
65% N2 20% O2 15% CO2 350 K 300 kPa
Then the mass fraction of each component (gravimetric analysis) becomes mf N 2 = mfO 2 = mfCO 2 =
mN 2 mm mO 2 mm mCO 2 mm
=
1820 kg = 0.583 or 58.3% 3120 kg
=
640 kg = 0.205 or 20.5% 3120 kg
=
660 kg = 0.212 or 21.2% 3120 kg
For ideal gases, the partial pressure is proportional to the mole fraction, and is determined from PN 2 = y N 2 Pm = (0.65)(300 kPa ) = 195 kPa PO 2 = y O 2 Pm = (0.20)(300 kPa ) = 60 kPa
PCO 2 = y CO 2 Pm = (0.15)(300 kPa ) = 45 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-11
13-38 The masses, temperatures, and pressures of two gases contained in two tanks connected to each other are given. The valve connecting the tanks is opened and the final temperature is measured. The volume of each tank and the final pressure are to be determined. Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases, and the mixture as an ideal gas mixture Properties The molar masses of N2 and O2 are 28.0 and 32.0 kg/kmol, respectively (Table A-1) Analysis The volumes of the tanks are (1 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K)(298 K) mRT = 0.295 m 3 = 300 kPa P N2
V N2 =
(3 kg)(0.2598 kPa ⋅ m 3 /kg ⋅ K)(298 K) mRT = 0.465 m 3 = 500 kPa P O2
V O2 =
1 kg N2
3 kg O2
25°C 300 kPa
25°C 500 kPa
V total = V N 2 + V O 2 = 0.295 m 3 + 0.465 m 3 = 0.76 m 3 Also, m N 2 = 1 kg → N N 2 = m O 2 = 3 kg → N O 2 =
m N2 M N2 mO2 M O2
=
1 kg = 0.03571 kmol 28 kg/kmol
=
3 kg = 0.09375 kmol 32 kg/kmol
N m = N N 2 + N O 2 = 0.03571 kmol + 0.09375 kmol = 0.1295 kmol
Thus, NRu T Pm = V
(0.1295 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(298 K) = = 422.2 kPa 0.76 m 3 m
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-12
13-39 The volumes, temperatures, and pressures of two gases forming a mixture are given. The volume of the mixture is to be determined using three methods. Analysis (a) Under specified conditions both O2 and N2 will considerably deviate from the ideal gas behavior. Treating the mixture as an ideal gas, PV N O 2 = RuT
(8000 kPa)(0.3 m 3 ) = = 1.443 kmol 3 O 2 (8.314 kPa ⋅ m /kmol ⋅ K)(200 K)
PV N N 2 = RuT
(8000 kPa)(0.5 m 3 ) = = 2.406 kmol 3 N 2 (8.314 kPa ⋅ m /kmol ⋅ K)(200 K)
N m = N O 2 + N N 2 = 1.443 kmol + 2.406 kmol = 3.849 kmol
Vm =
0.3 m3 O2 200 K 8 MPa 0.5 m3 N2 200 K 8 MPa
N2 + O2 200 K 8 MPa
N m RuTm (3.849 kmol)(8.314 kPa ⋅ m3 /kmol ⋅ K)(200 K) = = 0.8 m 3 8000 kPa Pm
(b) To use Kay's rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of O2 and N2 from Table A-1. But we first need to determine the Z and the mole numbers of each component at the mixture temperature and pressure (Fig. A-15), Z O 2 = 0.77 8 MPa = = 1.575 5.08 MPa
T R ,O 2 =
Tm 200 K = = 1.292 Tcr,O 2 154.8 K
PR ,O 2 =
Pm Pcr,O 2
O 2:
TR, N 2 =
N 2: PR , N 2 =
Z N 2 = 0.863 8 MPa = = 2.360 3.39 MPa
Tm
=
Tcr, N 2 Pm Pcr, N 2
200K = 1.585 126.2K
PV N O 2 = ZRu T
(8000 kPa)(0.3 m 3 ) = = 1.874 kmol 3 O 2 (0.77)(8.314 kPa ⋅ m /kmol ⋅ K)(200 K)
PV N N 2 = ZRu T
(8000 kPa)(0.5 m 3 ) = = 2.787 kmol 3 N 2 (0.863)(8.314 kPa ⋅ m /kmol ⋅ K)(200 K)
N m = N O 2 + N N 2 = 1.874 kmol + 2.787 kmol = 4.661 kmol
The mole fractions are y O2 = y N2 = Tcr′ , m =
N O2 Nm N N2 Nm
=
1.874kmol = 0.402 4.661kmol
=
2.787kmol = 0.598 4.661kmol
∑yT
i cr ,i
= y O 2 Tcr ,O 2 + y N 2 Tcr , N 2
= (0.402)(154.8K) + (0.598)(126.2K) = 137.7K Pcr′ , m =
∑y P
i cr ,i
= y O 2 Pcr ,O 2 + y N 2 Pcr , N 2
= (0.402)(5.08MPa) + (0.598)(3.39MPa) = 4.07 MPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-13
Then, TR = PR =
Thus,
Vm =
Tm
Z m = 0.82 8 MPa = = 1.966 4.07 MPa
=
' Tcr, O2
Pm ' Pcr, O2
200 K = 1.452 137.7 K
(Fig. A-15)
Z m N m Ru Tm (0.82)(4.661 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(200 K) = = 0.79 m 3 Pm 8000 kPa
(c) To use the Amagat’s law for this real gas mixture, we first need the Z of each component at the mixture temperature and pressure, which are determined in part (b). Then,
Thus,
Zm =
∑y Z
Vm =
Z m N m Ru Tm (0.83)(4.661 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(200 K) = = 0.80 m 3 Pm 8000 kPa
i
i
= y O 2 Z O 2 + y N 2 Z N 2 = (0.402 )(0.77 ) + (0.598)(0.863) = 0.83
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-14
13-40 [Also solved by EES on enclosed CD] The mole numbers, temperatures, and pressures of two gases forming a mixture are given. The final temperature is also given. The pressure of the mixture is to be determined using two methods. Analysis (a) Under specified conditions both Ar and N2 will considerably deviate from the ideal gas behavior. Treating the mixture as an ideal gas, Initial state : P1V1 = N1RuT1 N 2T2 (4)(200 K) P1 = (5 MPa ) = 18.2 MPa P2 = Final state : P2V 2 = N 2 RuT2 N1T1 (1)(220 K)
(b) Initially, TR = PR =
Z Ar = 0.90 (Fig. A-15) 5 MPa = = 1.0278 4.86 MPa
T1 220 K = = 1.457 Tcr,Ar 151.0 K P1 Pcr,Ar
1 kmol Ar 220 K 5 MPa
3 kmol N2 190 K 8 MPa
Then the volume of the tank is
V =
ZN Ar Ru T (0.90)(1 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(220 K) = = 0.33 m 3 P 5000 kPa
After mixing, V m / N Ar v Ar = = PR = 0.90 Ru Tcr,Ar / Pcr,Ar Ru Tcr,Ar / Pcr,Ar 3 (0.33 m )/(1 kmol) 1 . 278 = = (8.314 kPa ⋅ m 3 /kmol ⋅ K)(151.0 K)/(4860 kPa)
(Fig. A-15)
Tcr, N 2 v N2 V m / N N2 = = PR = 3.75 Ru Tcr, N 2 / Pcr, N 2 Ru Tcr, N 2 / Pcr, N 2 3 (0.33 m )/(3 kmol) = = 0.355 (8.314 kPa ⋅ m 3 /kmol ⋅ K)(126.2 K)/(3390 kPa)
(Fig. A-15)
T R ,Ar =
Ar:
V R , Ar
TR, N 2 =
N 2:
V R, N 2
Tm 200K = = 1.325 Tcr,Ar 151.0K
Tm
=
200K = 1.585 126.2K
Thus, PAr = ( PR Pcr ) Ar = (0.90)(4.86 MPa) = 4.37 MPa PN 2 = ( PR Pcr ) N 2 = (3.75)(3.39 MPa) = 12.7 MPa
and Pm = PAr + PN 2 = 4.37 MPa + 12.7 MPa = 17.1 MPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-15
13-41 EES Problem 13-40 is reconsidered. The effect of the moles of nitrogen supplied to the tank on the final pressure of the mixture is to be studied using the ideal-gas equation of state and the compressibility chart with Dalton's law. Analysis The problem is solved using EES, and the solution is given below. "Input Data" R_u = 8.314 [kJ/kmol-K] "universal Gas Constant" T_Ar = 220 [K] P_Ar = 5000 [kPa] "Pressure for only Argon in the tank initially." N_Ar = 1 [kmol] {N_N2 = 3 [kmol]} T_mix = 200 [K] T_cr_Ar=151.0 [K] "Critical Constants are found in Table A.1 of the text" P_cr_Ar=4860 [kPa] T_cr_N2=126.2 [K] P_cr_N2=3390 [kPa] "Ideal-gas Solution:" P_Ar*V_Tank_IG = N_Ar*R_u*T_Ar "Apply the ideal gas law the gas in the tank." P_mix_IG*V_Tank_IG = N_mix*R_u*T_mix "Ideal-gas mixture pressure" N_mix=N_Ar + N_N2 "Moles of mixture" "Real Gas Solution:" P_Ar*V_Tank_RG = Z_Ar_1*N_Ar*R_u*T_Ar "Real gas volume of tank" T_R=T_Ar/T_cr_Ar "Initial reduced Temp. of Ar" P_R=P_Ar/P_cr_Ar "Initial reduced Press. of Ar" Z_Ar_1=COMPRESS(T_R, P_R ) "Initial compressibility factor for Ar" P_Ar_mix*V_Tank_RG = Z_Ar_mix*N_Ar*R_u*T_mix "Real gas Ar Pressure in mixture" T_R_Ar_mix=T_mix/T_cr_Ar "Reduced Temp. of Ar in mixture" P_R_Ar_mix=P_Ar_mix/P_cr_Ar "Reduced Press. of Ar in mixture" Z_Ar_mix=COMPRESS(T_R_Ar_mix, P_R_Ar_mix ) "Compressibility factor for Ar in mixture" P_N2_mix*V_Tank_RG = Z_N2_mix*N_N2*R_u*T_mix "Real gas N2 Pressure in mixture" T_R_N2_mix=T_mix/T_cr_N2 "Reduced Temp. of N2 in mixture" P_R_N2_mix=P_N2_mix/P_cr_N2 "Reduced Press. of N2 in mixture" Z_N2_mix=COMPRESS(T_R_N2_mix, P_R_N2_mix ) "Compressibility factor for N2 in mixture" P_mix=P_R_Ar_mix*P_cr_Ar +P_R_N2_mix*P_cr_N2 "Mixture pressure by Dalton's law. 23800" 225000
Pmix [kPa] 9009 13276 17793 23254 30565 41067 56970 82372 126040 211047
Pmix,IG [kPa] 9091 13636 18182 22727 27273 31818 36364 40909 45455 50000
180000
P m ix [kPa]
NN2 [kmol] 1 2 3 4 5 6 7 8 9 10
Solution Method
135000
Chart
Ideal Gas 90000
45000
0 1
2
3
4
5
6
7
8
9
N N2 [kmol]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
10
13-16
13-42E The mole numbers, temperatures, and pressures of two gases forming a mixture are given. For a specified final temperature, the pressure of the mixture is to be determined using two methods. Properties The critical properties of Ar are Tcr = 272 R and Pcr = 705 psia. The critical properties of N2 are Tcr = 227.1 R and Pcr = 492 psia (Table A-1E). Analysis (a) Under specified conditions both Ar and N2 will considerably deviate from the ideal gas behavior. Treating the mixture as an ideal gas, P1V1 = N1RuT1 N 2T2 (4)(360 R) P1 = (750 psia ) = 2700 psia P2 = P2V 2 = N 2 RuT2 N1T1 (1)(400 R)
Initial state : Final state :
(b) Initially, TR = PR =
Z Ar = 0.90 (Fig. A-15) 750 psia = = 1.07 705 psia
T1 400 R = = 1.47 Tcr,Ar 272 R P1 Pcr,Ar
1 lbmol Ar 400 R 750 psia
3 lbmol N2 340 R 1200 psia
Then the volume of the tank is
V =
ZN Ar Ru T (0.90)(1 lbmol)(10.73 psia ⋅ ft 3 /lbmol ⋅ R)(400 R) = = 5.15 ft 3 750 psia P
After mixing, V m / N Ar v Ar = = PR = 0.82 Ru Tcr,Ar / Pcr,Ar Ru Tcr,Ar / Pcr,Ar 3 (5.15 ft )/(1 lbmol) = = 1.244 (10.73 psia ⋅ ft 3 /lbmol ⋅ R)(272 R)/(705 psia)
(Fig. A-15)
Tcr, N 2 v N2 V m / N N2 = = PR = 3.85 Ru Tcr, N 2 / Pcr, N 2 Ru Tcr, N 2 / Pcr, N 2 3 (5.15 ft )/(3 lbmol) 0 . 347 = = (10.73 psia ⋅ ft 3 /lbmol ⋅ R)(227.1 R)/(492 psia)
(Fig. A-15)
T R , Ar =
Ar:
v R , Ar
TR, N 2 =
N 2:
v R, N 2
Tm 360 R = = 1.324 Tcr,Ar 272 R
Tm
=
360 R = 1.585 227.1 R
Thus, PAr = ( PR Pcr ) Ar = (0.82)(705 psia) = 578 psia PN 2 = ( PR Pcr ) N 2 = (3.85)(492 psia) = 1894 psia
and Pm = PAr + PN 2 = 578 psia + 1894 psia = 2472 psia
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-17
Properties of Gas Mixtures 13-43C Yes. Yes (extensive property). 13-44C No (intensive property). 13-45C The answers are the same for entropy. 13-46C Yes. Yes (conservation of energy). 13-47C We have to use the partial pressure. 13-48C No, this is an approximate approach. It assumes a component behaves as if it existed alone at the mixture temperature and pressure (i.e., it disregards the influence of dissimilar molecules on each other.)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-18
13-49 Oxygen, nitrogen, and argon gases are supplied from separate tanks at different temperatures to form a mixture. The total entropy change for the mixing process is to be determined. Assumptions Under specified conditions all N2, O2, and argon can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of O2, N2, and Ar are 32.0, 28.0, and 40.0 kg/kmol, respectively (TableA-1). The properties of Argon are R = 0.2081 kJ/kg.K and cp = 0.5203 kJ/kg.K (Table A-2). Analysis Note that volume fractions are equal to mole fractions in ideal gas mixtures. The partial pressures in the mixture are O2 PO 2 , 2 = y O 2 Pm = (0.21)(200 kPa) = 42 kPa 10°C PN 2 , 2 = y N 2 Pm = (0.78)(200 kPa) = 156 kPa
N2 60°C
PAr, 2 = y Ar Pm = (0.01)(200 kPa) = 2 kPa
The molar mass of the mixture is determined to be M m = yO 2 M O 2 + y N 2 M N 2 + y Ar M Ar = (0.21)(32 kg/kmol) + (0.78)(28) + (0.01)(40) = 28.96 kg/kmol
21% O2 78% N2 21% Ar
200 kPa
Ar 200°C
The mass fractions are M O2 32 kg/kmol mf O 2 = y O 2 = (0.21) = 0.2320 28.96 kg/kmol Mm M N2
mf N 2 = y N 2 mf Ar = y Ar
Mm
= (0.78)
28 kg/kmol = 0.7541 28.96 kg/kmol
M Ar 40 kg/kmol = (0.01) = 0.0138 28.96 kg/kmol Mm
The final temperature of the mixture is needed. The conservation of energy on a unit mass basis for steady flow mixing with no heat transfer or work allows calculation of mixture temperature. All components of the exit mixture have the same common temperature, Tm. We obtain the properties of O2 and N2 from EES: ein = eout mf O 2 h@ 10°C + mf N 2 h@ 60°C + mf Ar c p , ArTAr,1 = mf O 2 h@ Tm + mf N 2 h@ Tm + mf Ar c p , ArTm (0.2320)(−13.85) + (0.7541)(36.47) + (0.0138)(0.5203)(200) = (0.2320)h@ Tm + (0.7541)h@ Tm + (0.0138)(0.5203)Tm
Solving this equation with EES gives Tm = 50.4ºC. The entropies of O2 and N2 are obtained from EES to be T = 10°C, P = 200 kPa → s O 2 ,1 = 6.1797 kJ/kg.K T = 50.4°C, P = 42 kPa → s O 2 ,2 = 6.7082 kJ/kg.K T = 60°C, P = 200 kPa → s N 2 ,1 = 6.7461 kJ/kg.K T = 50.4°C, P = 156 kPa → s N 2 , 2 = 6.7893 kJ/kg.K
The entropy changes are ∆s O 2 = s O 2 , 2 − s O 2 ,1 = 6.7082 − 6.1797 = 0.5284 kg/kg.K ∆s N 2 = s N 2 , 2 − s N 2 ,1 = 6.7893 − 6.7461 = 0.04321 kg/kg.K ∆s Ar = c p ln
T2 P 50.4 + 273 2 − R ln 2 = (0.5203) ln − (0.2081) ln = 0.7605 kJ/kg.K T1 P1 200 + 273 200
The total entropy change is ∆s total = mf O 2 ∆s O 2 + mf O 2 ∆s O 2 + mf Ar ∆s Ar = (0.2320)(0.5284) + (0.7541)(0.04321) + (0.0138)(0.7605) = 0.1656 kJ/kg.K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-19
13-50 Volumetric fractions of the constituents of a mixture are given. The mixture undergoes an adiabatic compression process. The makeup of the mixture on a mass basis and the internal energy change per unit mass of mixture are to be determined. Assumptions Under specified conditions all CO2, CO, O2, and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties 1 The molar masses of CO2, CO, O2, and N2 are 44.0, 28.0, 32.0, and 28.0 kg/kmol, respectively (Table A-1). 2 The process is reversible. Analysis Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the molar mass of the mixture is determined to be M m = y CO 2 M CO 2 + y CO M CO + y O 2 M O 2 + y N 2 M N 2 = (0.15)(44) + (0.05)(28) + (0.10)(32) + (0.70)(28) = 30.80 kg/kmol The mass fractions are M CO 2 44 kg/kmol mf CO 2 = y CO 2 = (0.15) = 0.2143 30.80 kg/kmol Mm
mf CO = y CO mf O 2 = y O 2
M CO 28 kg/kmol = (0.05) = 0.0454 30.80 kg/kmol Mm M O2 Mm M N2
= (0.10)
32 kg/kmol = 0.1039 30.80 kg/kmol
15% CO2 5% CO 10% O2 70% N2 300 K, 1 bar
28 kg/kmol = 0.6364 30.80 kg/kmol Mm The final pressure of mixture is expressed from ideal gas relation to be T T P2 = P1 r 2 = (100 kPa )(8) 2 = 2.667T2 (Eq. 1) 300 K T1 since the final temperature is not known. We assume that the process is reversible as well being adiabatic (i.e. isentropic). Using Dalton’s law to find partial pressures, the entropies at the initial state are determined from EES to be: → s CO 2 ,1 = 5.2190 kJ/kg.K T = 300 K, P = (0.2143 × 100) = 21.43 kPa mf N 2 = y N 2
= (0.70)
→ s CO,1 = 79483 kJ/kg.K T = 300 K, P = (0.04545 × 100) = 4.55 kPa → s N 2 ,1 = 6.9485 kJ/kg.K T = 300 K, P = (0.1039 × 100) = 10.39 kPa → s O 2 ,1 = 7.0115 kJ/kg.K T = 300 K, P = (0.6364 × 100) = 63.64 kPa
The final state entropies cannot be determined at this point since the final pressure and temperature are not known. However, for an isentropic process, the entropy change is zero and the final temperature and the final pressure may be determined from ∆s total = mf CO 2 ∆s CO 2 + mf CO ∆s CO + mf O 2 ∆s O 2 + mf N 2 ∆s N 2 = 0 and using Eq. (1). The solution may be obtained using EES to be T2 = 631.4 K, P2 = 1684 kPa The initial and final internal energies are (from EES) u CO 2 , 2 = −8734 kJ/kg u CO 2 ,1 = −8997 kJ/kg u CO, 2 = −3780 kJ/kg u CC,1 = −4033 kJ/kg → → T2 = 631.4 K T1 = 300 K u O 2 , 2 = 156.8 kJ/kg u O 2 ,1 = −76.24 kJ/kg u N 2 , 2 = 163.9 kJ/kg u N 2 ,1 = −87.11 kJ/kg, The internal energy change per unit mass of mixture is determined from ∆u mixture = mf CO 2 (u CO 2 , 2 − u CO 2 ,1 ) + mf CO (u CO, 2 − u CO,1 ) + mf O 2 (u O 2 , 2 − u O 2 ,1 ) + mf N 2 (u N 2 , 2 − u N 2 ,1 ) = 0.2143[(−8734) − (−8997)] + 0.0454[(−3780) − (−4033)] + 0.1039[156.8 − (−76.24)]6 + 0.6364[163.9 − (−87.11)]
= 251.8 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-20
13-51 Propane and air mixture is compressed isentropically in an internal combustion engine. The work input is to be determined. Assumptions Under specified conditions propane and air can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of C3H8 and air are 44.0 and 28.97 kg/kmol, respectively (TableA-1). Analysis Given the air-fuel ratio, the mass fractions are determined to be mf air = mf C3H8
AF 16 = = 0.9412 AF + 1 17 1 1 = = = 0.05882 AF + 1 17
The molar mass of the mixture is determined to be Mm =
mf air M air
1 1 = = 29.56 kg/kmol mf C3H8 0.9412 0.05882 + + 28.97 kg/kmol 44.0 kg/kmol M C3H8
Propane Air 95 kPa 30ºC
The mole fractions are y air = mf air
Mm 29.56 kg/kmol = (0.9412) = 0.9606 28.97 kg/kmol M air
y C3H8 = mf C3H8
Mm 29.56 kg/kmol = (0.05882) = 0.03944 44.0 kg/kmol M C3H8
The final pressure is expressed from ideal gas relation to be P2 = P1 r
T2 T2 = (95 kPa )(9.5) = 2.977T2 T1 (30 + 273.15) K
(1)
since the final temperature is not known. Using Dalton’s law to find partial pressures, the entropies at the initial state are determined from EES to be: T = 30°C, P = (0.9606 × 95) = 91.26 kPa → s air ,1 = 5.7417 kJ/kg.K T = 30°C, P = (0.03944 × 95) = 3.75 kPa → s C3H8 ,1 = 6.7697 kJ/kg.K
The final state entropies cannot be determined at this point since the final pressure and temperature are not known. However, for an isentropic process, the entropy change is zero and the final temperature and the final pressure may be determined from ∆s total = mf air ∆s air + mf C3H8 ∆s C3H 8 = 0
and using Eq. (1). The solution may be obtained using EES to be T2 = 654.9 K, P2 = 1951 kPa The initial and final internal energies are (from EES) T1 = 30°C →
u air ,1 = 216.5 kJ/kg u C3H8 ,1 = −2404 kJ/kg
T2 = 654.9 K →
u air , 2 = 477.1 kJ/kg u C3H8 , 2 = −1607 kJ/kg
Noting that the heat transfer is zero, an energy balance on the system gives q in + win = ∆u m → win = ∆u m
where
∆u m = mf air (u air,2 − u air,1 ) + mf C3H8 (u C3H8 ,2 − u C3H 8 ,1 )
Substituting,
win = ∆u m = (0.9412)(477.1 − 216.5) + (0.05882)[(−1607) − (−2404)] = 292.2 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-21
13-52 The moles, temperatures, and pressures of two gases forming a mixture are given. The mixture temperature and pressure are to be determined. Assumptions 1 Under specified conditions both CO2 and H2 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 The tank is insulated and thus there is no heat transfer. 3 There are no other forms of work involved. Properties The molar masses and specific heats of CO2 and H2 are 44.0 kg/kmol, 2.0 kg/kmol, 0.657 kJ/kg.°C, and 10.183 kJ/kg.°C, respectively. (Tables A-1 and A-2b). Analysis (a) We take both gases as our system. No heat, work, or mass crosses the system boundary, therefore this is a closed system with Q = 0 and W = 0. Then the energy balance for this closed system reduces to E in − E out = ∆E system 0 = ∆U = ∆U CO 2 + ∆U H 2
0 =[mcv (Tm − T1 )]CO + [mcv (Tm − T1 )]H 2
2
Using cv values at room temperature and noting that m = NM, the final temperature of the mixture is determined to be
CO2
H2
2.5 kmol
7.5 kmol
200 kPa 27°C
400 kPa 40°C
(2.5 × 44 kg )(0.657 kJ/kg ⋅ °C)(Tm − 27°C) + (7.5 × 2 kg )(10.183 kJ/kg ⋅ °C)(Tm − 40°C ) = 0 Tm = 35.8°C (308.8 K ) (b) The volume of each tank is determined from NRu T1 (2.5 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(300 K) = = 31.18 m 3 200 kPa P1 CO 2
V CO 2 =
NRu T1 (7.5 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(313 K) = = 48.79 m 3 400 kPa P 1 H2
V H 2 = Thus,
V m = V CO2 + V H 2 = 31.18 m 3 + 48.79 m 3 = 79.97 m 3 N m = N CO 2 + N H 2 = 2.5 kmol + 7.5 kmol = 10.0 kmol
and
Pm =
N m Ru Tm (10.0 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(308.8 K) = = 321 kPa Vm 79.97 m 3
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-22
13-53 The temperatures and pressures of two gases forming a mixture are given. The final mixture temperature and pressure are to be determined. Assumptions 1 Under specified conditions both Ne and Ar can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 There are no other forms of work involved. Properties The molar masses and specific heats of Ne and Ar are 20.18 kg/kmol, 39.95 kg/kmol, 0.6179 kJ/kg.°C, and 0.3122 kJ/kg.°C, respectively. (Tables A-1 and A-2). Analysis The mole number of each gas is PV (100 kPa)(0.45 m3 ) = 0.0185 kmol N Ne = 1 1 = 3 RuT1 Ne (8.314 kPa ⋅ m /kmol ⋅ K)(293 K) PV (200 kPa)(0.45 m3 ) = 0.0335 kmol N Ar = 1 1 = 3 RuT1 Ar (8.314 kPa ⋅ m /kmol ⋅ K)(323 K)
Ne
Ar
100 kPa 20°C
200 kPa 50°C
Thus,
15 kJ
N m = N Ne + N Ar = 0.0185 kmol + 0.0335 kmol = 0.0520 kmol
(a) We take both gases as the system. No work or mass crosses the system boundary, therefore this is a closed system with W = 0. Then the conservation of energy equation for this closed system reduces to E in − E out = ∆E system − Qout = ∆U = ∆U Ne + ∆U Ar → − Qout = [mcv (Tm − T1 )]Ne + [mcv (Tm − T1 )]Ar
Using cv values at room temperature and noting that m = NM, the final temperature of the mixture is determined to be −15 kJ = (0.0185 × 20.18 kg )(0.6179 kJ/kg ⋅ °C )(Tm − 20°C )
+ (0.0335 × 39.95 kg )(0.3122 kJ/kg ⋅ °C )(Tm − 50°C )
Tm = 16.2°C (289.2 K )
(b) The final pressure in the tank is determined from Pm =
N m Ru Tm (0.052 kmol)(8.314 kPa ⋅ m 3 / kmol ⋅ K)(289.2 K) = = 138.9 kPa Vm 0.9 m 3
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-23
13-54 The temperatures and pressures of two gases forming a mixture are given. The final mixture temperature and pressure are to be determined. Assumptions 1 Under specified conditions both Ne and Ar can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 There are no other forms of work involved. Properties The molar masses and specific heats of Ne and Ar are 20.18 kg/kmol, 39.95 kg/kmol, 0.6179 kJ/kg.°C, and 0.3122 kJ/kg.°C, respectively. (Tables A-1 and A-2b). Analysis The mole number of each gas is PV (100 kPa)(0.45 m3 ) = 0.0185 kmol N Ne = 1 1 = 3 RuT1 Ne (8.314 kPa ⋅ m /kmol ⋅ K)(293 K) PV (200 kPa)(0.45 m3 ) = 0.0335 kmol N Ar = 1 1 = 3 RuT1 Ar (8.314 kPa ⋅ m /kmol ⋅ K)(323 K)
Ne
Ar
100 kPa 20°C
200 kPa 50°C
Thus,
8 kJ
N m = N Ne + N Ar = 0.0185 kmol + 0.0335 kmol = 0.0520 kmol
(a) We take both gases as the system. No work or mass crosses the system boundary, therefore this is a closed system with W = 0. Then the conservation of energy equation for this closed system reduces to E in − E out = ∆E system − Qout = ∆U = ∆U Ne + ∆U Ar
− Qout = [mcv (Tm − T1 )]Ne + [mcv (Tm − T1 )]Ar
Using cv values at room temperature and noting that m = NM, the final temperature of the mixture is determined to be −8 kJ = (0.0185 × 20.18 kg )(0.6179 kJ/kg ⋅ °C )(Tm − 20°C )
+ (0.0335 × 39.95 kg )(0.3122 kJ/kg ⋅ °C )(Tm − 50°C )
Tm = 27.0°C (300.0 K )
(b) The final pressure in the tank is determined from Pm =
N m RuTm
Vm
=
(0.052 kmol)(8.314 kPa ⋅ m3 /kmol ⋅ K)(300.0 K) 0.9 m3
= 144.1 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-24
13-55 [Also solved by EES on enclosed CD] The temperatures and pressures of two gases forming a mixture in a mixing chamber are given. The mixture temperature and the rate of entropy generation are to be determined. Assumptions 1 Under specified conditions both C2H6 and CH4 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 The mixing chamber is insulated and thus there is no heat transfer. 3 There are no other forms of work involved. 3 This is a steady-flow process. 4 The kinetic and potential energy changes are negligible. Properties The specific heats of C2H6 and CH4 are 1.7662 kJ/kg.°C and 2.2537 kJ/kg.°C, respectively. (Table A-2b). 20°C Analysis (a) The enthalpy of ideal gases is independent of 9 kg/s C2H6 pressure, and thus the two gases can be treated independently even after mixing. Noting that W& = Q& = 0 , the steady-flow 200 kPa energy balance equation reduces to E& in − E& out = ∆E& system ©0 (steady) = 0 45°C 4.5 kg/s
E& in = E& out
∑ m& h = ∑ m& h 0 = ∑ m& h − ∑ m& h = m& 0 = [m& c (T − T )] + [m& c i i
CH4
e e e e
p
e
i i
i
(he − hi )C H p (Te − Ti )]CH
C2H6
C2H6
2
6
+ m& CH 4 (he − hi )CH
4
4
Using cp values at room temperature and substituting, the exit temperature of the mixture becomes 0 = (9 kg/s )(1.7662 kJ/kg ⋅ °C )(Tm − 20°C ) + (4.5 kg/s )(2.2537 kJ/kg ⋅ °C )(Tm − 45°C )
Tm = 29.7°C (302.7 K ) (b) The rate of entropy change associated with this process is determined from an entropy balance on the mixing chamber, S&in − S&out + S&gen = ∆S&system Ê0 = 0
& ( s1 − s2 )]C2 H6 + [m & ( s1 − s2 )]CH4 + S&gen = 0 → S&gen = [m & ( s2 − s1 )]C2 H 6 + [m & ( s2 − s1 )]CH 4 [m
The molar flow rate of the two gases in the mixture is 4.5 kg/s m& = = 0.2813 kmol/s N& CH 4 = M CH 4 16 kg/kmol
9 kg/s m& = = 0.3 kmol/s N& C 2 H 6 = M C 2 H 6 30 kg/kmol
Then the mole fraction of each gas becomes 0.3 0.2813 yC 2 H 6 = = 0.516 yCH 4 = = 0.484 0.3 + 0.2813 0.3 + 0.2813 Thus, y Pm,2 T T ( s 2 − s1 ) C 2 H 6 = c p ln 2 − R ln = c p ln 2 − R ln y T P T 1 1 C H 1 C2H6 2 6 = (1.7662 kJ/kg ⋅ K) ln
302.7 K − (0.2765 kJ/kg ⋅ K) ln(0.516) = 0.240 kJ/kg ⋅ K 293 K
y Pm,2 T ( s 2 − s1 ) CH 4 = c p ln 2 − R ln T1 P1 = (2.2537 kJ/kg ⋅ K) ln
T = c p ln 2 − R ln y T1 CH 4 CH 4
302.7 K − (0.5182 kJ/kg ⋅ K) ln(0.484) = 0.265 kJ/kg ⋅ K 318 K
Noting that Pm, 2 = Pi, 1 = 200 kPa and substituting, S&gen = (9 kg/s )(0.240 kJ/kg ⋅ K ) + (4.5 kg/s )(0.265 kJ/kg ⋅ K ) = 3.353 kW/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-25
13-56 EES Problem 13-55 is reconsidered. The effect of the mass fraction of methane in the mixture on the mixture temperature and the rate of exergy destruction is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input from the Diagram Window" {Fluid1$='C2H6' Fluid2$='CH4' m_dot_F1=9 [kg/s] m_dot_F2=m_dot_F1/2 T1=20 [C] T2=45 [C] P=200 [kPa]} {mf_F2=0.1} {m_dot_total =13.5 [kg/s] m_dot_F2 =mf_F2*m_dot_total} m_dot_total = m_dot_F1 + m_dot_F2 T_o = 25 [C] "Conservation of energy for this steady-state, steady-flow control volume is" E_dot_in=E_dot_out E_dot_in=m_dot_F1*enthalpy(Fluid1$,T=T1) +m_dot_F2 *enthalpy(Fluid2$,T=T2) E_dot_out=m_dot_F1*enthalpy(Fluid1$,T=T3) +m_dot_F2 *enthalpy(Fluid2$,T=T3) "For entropy calculations the partial pressures are used." Mwt_F1=MOLARMASS(Fluid1$) N_dot_F1=m_dot_F1/Mwt_F1 Mwt_F2=MOLARMASS(Fluid2$) N_dot_F2=m_dot_F2 /Mwt_F2 N_dot_tot=N_dot_F1+N_dot_F2 y_F1=IF(fluid1$,Fluid2$,N_dot_F1/N_dot_tot,1,N_dot_F1/N_dot_tot) y_F2=IF(fluid1$,Fluid2$,N_dot_F2/N_dot_tot,1,N_dot_F2/N_dot_tot) "If the two fluids are the same, the mole fractions are both 1 ." "The entropy change of each fluid is:" DELTAs_F1=entropy(Fluid1$, T=T3, P=y_F1*P)-entropy(Fluid1$, T=T1, P=P) DELTAs_F2=entropy(Fluid2$, T=T3, P=y_F2*P)-entropy(Fluid2$, T=T2, P=P) "And the entropy generation is:" S_dot_gen=m_dot_F1*DELTAs_F1+m_dot_F2*DELTAs_F2 "Then the exergy destroyed is:" X_dot_destroyed = (T_o+273)*S_dot_gen mfF2 0.01 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.99
T3 [C] 95.93 502.5 761.4 948.5 1096 1219 1324 1415 1497 1570 1631
Xdestroyed [kW] 20.48 24.08 27 29.2 30.92 32.3 33.43 34.38 35.18 35.87 36.41
38
2000
36
] C [ 3 T
34
1600
32
] W 1200k[
30
28
F1 = C2H6
26
F2 = CH4
X
mtotal = 13.5 kg/s
24
800
400
22
20 0
0,2
0,4
mfF2
0,6
d e y or t s e d
0,8
0 1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-26
13-57 An equimolar mixture of helium and argon gases expands in a turbine. The isentropic work output of the turbine is to be determined. Assumptions 1 Under specified conditions both He and Ar can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 The turbine is insulated and thus there is no heat transfer. 3 This is a steady-flow process. 4 The kinetic and potential energy changes are negligible. Properties The molar masses and specific heats of He and Ar are 4.0 kg/kmol, 40.0 kg/kmol, 5.1926 kJ/kg.°C, and 0.5203 kJ/kg.°C, respectively. (Table A-1 and Table A-2). Analysis The Cp and k values of this equimolar mixture are determined from M m = ∑ y i M i = y He M He + y Ar M Ar = 0.5 × 4 + 0.5 × 40 = 22 kg/kmol mf i =
mi mm
=
2.5 MPa 1300 K
Ni M i
y M = i i NmM m Mm
He-Ar turbine
y M y M c p,m = ∑ mf i c p,i = He He c p,He + Ar Ar c p,Ar Mm Mm 0.5 × 4 kg/kmol (5.1926 kJ/kg ⋅ K ) + 0.5 × 40 kg/kmol (0.5203 kJ/kg ⋅ K ) = 22 kg/kmol 22 kg/kmol = 0.945 kJ/kg ⋅ K
200 kPa
and km = 1.667
since k = 1.667 for both gases.
Therefore, the He-Ar mixture can be treated as a single ideal gas with the properties above. For isentropic processes, P T2 = T1 2 P1
(k −1) / k
200 kPa = (1300 K ) 2500 kPa
0.667/1.667
= 473.2 K
From an energy balance on the turbine, E& in − E& out = ∆E& system ©0 (steady) = 0 E& in = E& out h1 = h2 + wout wout = h1 − h2
wout = c p (T1 − T2 ) = (0.945 kJ/kg ⋅ K )(1300 − 473.2 )K = 781.3 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-27
13-58E [Also solved by EES on enclosed CD] A gas mixture with known mass fractions is accelerated through a nozzle from a specified state to a specified pressure. For a specified isentropic efficiency, the exit temperature and the exit velocity of the mixture are to be determined. Assumptions 1 Under specified conditions both N2 and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 The nozzle is adiabatic and thus heat transfer is negligible. 3 This is a steadyflow process. 4 Potential energy changes are negligible. Properties The specific heats of N2 and CO2 are cp,N2 = 0.248 Btu/lbm.R, cv,N2 = 0.177 Btu/lbm.R, cp,CO2 = 0.203 Btu/lbm.R, and cv,CO2 = 0.158 Btu/lbm.R. (Table A-2E). Analysis (a) Under specified conditions both N2 and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. The cp, cv, and k values of this mixture are determined from c p,m =
∑ mf c
cv ,m =
∑ mf cv
i p ,i
= mf N 2 c p , N 2 + mf CO 2 c p ,CO 2
= (0.8)(0.248) + (0.2)(0.203) = 0.239 Btu/lbm ⋅ R i
,i
= mf N 2 cv , N 2 + mf CO 2 cv ,CO 2
= (0.8)(0.177 ) + (0.2)(0.158) = 0.173 Btu/lbm ⋅ R
km =
c p,m cv , m
=
90 psia 1800 R
80% N2 20% CO2
12 psia
0.239 Btu/lbm ⋅ R = 1.382 0.173 Btu/lbm ⋅ R
Therefore, the N2-CO2 mixture can be treated as a single ideal gas with above properties. Then the isentropic exit temperature can be determined from P T2 s = T1 2 P1
(k −1) / k
12 psia = (1800 R ) 90 psia
0.382/1.382
= 1031.3 R
From the definition of adiabatic efficiency,
ηN =
c p (T1 − T2 ) h1 − h2 1,800 − T2 = → 0.92 = → T2 = 1092.8 R h1 − h2 s c p (T1 − T2 s ) 1,800 − 1031.3
(b) Noting that, q = w = 0, from the steady-flow energy balance relation, E& in − E& out = ∆E& system ©0 (steady) = 0 E& in = E& out h1 + V12 / 2 = h2 + V 22 / 2 0 = c p (T2 − T1 ) +
V 22 − V12 2
©0
25,037 ft 2 /s 2 V 2 = 2c p (T1 − T2 ) = 2(0.239 Btu/lbm ⋅ R )(1800 − 1092.8) R 1 Btu/lbm
= 2,909 ft/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-28
13-59E EES Problem 13-58E is reconsidered. The problem is first to be solved and then, for all other conditions being the same, the problem is to be resolved to determine the composition of the nitrogen and carbon dioxide that is required to have an exit velocity of 2000 ft/s at the nozzle exit. Analysis The problem is solved using EES, and the solution is given below. "Input Data" mf_N2 = 0.8 "Mass fraction for the nitrogen, lbm_N2/lbm_mix" mf_CO2 = 0.2 "Mass fraction for the carbon dioxide, lbm_CO2/lbm_mix" T[1] = 1800 [R] P[1] = 90 [psia] Vel[1] = 0 [ft/s] P[2] = 12 [psia] Eta_N =0.92 "Nozzle adiabatic efficiency" "Enthalpy property data per unit mass of mixture:" " Note: EES calculates the enthalpy of ideal gases referenced to the enthalpy of formation as h = h_f + (h_T - h_537) where h_f is the enthalpy of formation such that the enthalpy of the elements or their stable compounds is zero at 77 F or 537 R, see Chapter 14. The enthalpy of formation is often negative; thus, the enthalpy of ideal gases can be negative at a given temperature. This is true for CO2 in this problem." h[1]= mf_N2* enthalpy(N2, T=T[1]) + mf_CO2* enthalpy(CO2, T=T[1]) h[2]= mf_N2* enthalpy(N2, T=T[2]) + mf_CO2* enthalpy(CO2, T=T[2]) "Conservation of Energy for a unit mass flow of mixture:" "E_in - E_out = DELTAE_cv Where DELTAE_cv = 0 for SSSF" h[1]+Vel[1]^2/2*convert(ft^2/s^2,Btu/lbm) - h[2] - Vel[2]^2/2*convert(ft^2/s^2,Btu/lbm) =0 "SSSF energy balance" "Nozzle Efficiency Calculation:" Eta_N=(h[1]-h[2])/(h[1]-h_s2) h_s2= mf_N2* enthalpy(N2, T=T_s2) + mf_CO2* enthalpy(CO2, T=T_s2) "The mixture isentropic exit temperature, T_s2, is calculated from setting the entropy change per unit mass of mixture equal to zero." DELTAs_mix=mf_N2 * DELTAs_N2 + mf_CO2 * DELTAs_CO2 DELTAs_N2 = entropy(N2, T=T_s2, P=P_2_N2) - entropy(N2, T=T[1], P=P_1_N2) DELTAs_CO2 = entropy(CO2, T=T_s2, P=P_2_CO2) - entropy(CO2, T=T[1], P=P_1_CO2) DELTAs_mix=0 "By Dalton's Law the partial pressures are:" P_1_N2 = y_N2 * P[1]; P_1_CO2 = y_CO2 * P[1] P_2_N2 = y_N2 * P[2]; P_2_CO2 = y_CO2 * P[2] "mass fractions, mf, and mole fractions, y, are related by:" M_N2 = molarmass(N2) M_CO2=molarmass(CO2) y_N2=mf_N2/M_N2/(mf_N2/M_N2 + mf_CO2/M_CO2) y_CO2=mf_CO2/M_CO2/(mf_N2/M_N2 + mf_CO2/M_CO2)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-29
SOLUTION of the stated problem DELTAs_CO2=-0.04486 [Btu/lbm-R] DELTAs_N2=0.01122 [Btu/lbm-R] h[1]=-439.7 [Btu/lbm] h_s2=-628.8 [Btu/lbm] mf_N2=0.8 [lbm_N2/lbm_mix] M_N2=28.01 [lbm/lbmol] P[2]=12 [psia] P_1_N2=77.64 [psia] P_2_N2=10.35 [psia] T[2]=1160 [R] Vel[1]=0 [ft/s] y_CO2=0.1373 [ft/s]
DELTAs_mix=0 [Btu/lbm-R] Eta_N=0.92 h[2]=-613.7 [Btu/lbm] mf_CO2=0.2 [lbm_CO2/lbm_mix] M_CO2=44.01 [lbm/lbmol] P[1]=90 [psia] P_1_CO2=12.36 [psia] P_2_CO2=1.647 [psia] T[1]=1800 [R] T_s2=1102 [R] Vel[2]=2952 [ft/s] y_N2=0.8627 [lbmol_N2/lbmol_mix]
SOLUTION of the problem with exit velocity of 2600 ft/s DELTAs_CO2=-0.005444 [Btu/lbm-R] DELTAs_N2=0.05015 [Btu/lbm-R] h[1]=-3142 [Btu/lbm] h_s2=-3288 [Btu/lbm] mf_N2=0.09793 [lbm_N2/lbm_mix] M_N2=28.01 [lbm/lbmol] P[2]=12 [psia] P_1_N2=13.11 [psia] P_2_N2=1.748 [psia] T[2]=1323 [R] Vel[1]=0 [ft/s] y_CO2=0.8543 [ft/s]
DELTAs_mix=0 [Btu/lbm-R] Eta_N=0.92 h[2]=-3277 [Btu/lbm] mf_CO2=0.9021 [lbm_CO2/lbm_mix] M_CO2=44.01 [lbm/lbmol] P[1]=90 [psia] P_1_CO2=76.89 [psia] P_2_CO2=10.25 [psia] T[1]=1800 [R] T_s2=1279 [R] Vel[2]=2600 [ft/s] y_N2=0.1457 [lbmol_N2/lbmol_mix]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-30
13-60 A piston-cylinder device contains a gas mixture at a given state. Heat is transferred to the mixture. The amount of heat transfer and the entropy change of the mixture are to be determined. Assumptions 1 Under specified conditions both H2 and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 Kinetic and potential energy changes are negligible. Properties The constant pressure specific heats of H2 and N2 at 450 K are 14.501 kJ/kg.K and 1.049 kJ/kg.K, respectively. (Table A-2b). Analysis (a) Noting that P2 = P1 and V2 = 2V1, P2V 2 P1V1 2V = → T2 = 1 T1 = 2T1 = (2 )(300 K ) = 600 K V1 T2 T1
0.5 kg H2 1.6 kg N2 100 kPa 300 K
Also P = constant. Then from the closed system energy balance relation, Ein − Eout = ∆Esystem Qin − Wb,out = ∆U →
Qin = ∆H
Q
since Wb and ∆U combine into ∆H for quasi-equilibrium constant pressure processes.
[
]
[
]
Qin = ∆H = ∆H H 2 + ∆H N 2 = mc p ,avg (T2 − T1 ) H + mc p ,avg (T2 − T1 ) N 2
2
= (0.5 kg )(14.501 kJ/kg ⋅ K )(600 − 300)K + (1.6 kg )(1.049 kJ/kg ⋅ K )(600 − 300)K = 2679 kJ
(b) Noting that the total mixture pressure, and thus the partial pressure of each gas, remains constant, the entropy change of the mixture during this process is T P ©0 T ∆S H 2 = [m(s 2 − s1 )]H 2 = m H 2 c p ln 2 − R ln 2 = m H 2 c p ln 2 T1 P1 T1 H H2 2 = (0.5 kg )(14.501 kJ/kg ⋅ K )ln = 5.026 kJ/K
600 K 300 K
T P ©0 T ∆S N 2 = [m(s 2 − s1 )]N 2 = m N 2 c p ln 2 − R ln 2 = m N 2 c p ln 2 T1 P T 1 1 N2 = (1.6 kg )(1.049 kJ/kg ⋅ K )ln = 1.163 kJ/K
N2
600 K 300 K
∆S total = ∆S H 2 + ∆S N 2 = 5.026 kJ/K + 1.163 kJ/K = 6.19 kJ/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-31
13-61 The states of two gases contained in two tanks are given. The gases are allowed to mix to form a homogeneous mixture. The final pressure, the heat transfer, and the entropy generated are to be determined. Assumptions 1 Under specified conditions both O2 and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 The tank containing oxygen is insulated. 3 There are no other forms of work involved. Properties The constant volume specific heats of O2 and N2 are 0.658 kJ/kg.°C and 0.743 kJ/kg.°C, respectively. (Table A-2). Analysis (a) The volume of the O2 tank and mass of the nitrogen are mRT
1 = V1,O 2 = P 1 O 2
m N2
(1 kg)(0.2598 kPa ⋅ m 3 /kg ⋅ K)(288 K) = 0.25 m 3 300 kPa
PV (500 kPa)(2 m 3 ) = 1 1 = = 10.43 kg 3 RT1 N 2 (0.2968 kPa ⋅ m /kg ⋅ K)(323 K)
V total = V1,O 2 + V 1, N 2 = 0.25 m 3 + 2.0 m 3 = 2.25 m 3
O2
N2
1 kg 15°C 300 kPa
2 m3 50°C 500 kPa
Also, m O 2 = 1 kg → N O 2 =
mO 2 M O2
m N 2 = 10.43 kg → N N 2 =
=
Q
1 kg = 0.03125 kmol 32 kg/kmol
m N2 M N2
=
10.43 kg = 0.3725 kmol 28 kg/kmol
N m = N N 2 + N O 2 = 0.3725 kmol + 0.03125 kmol = 0.40375 kmol
Thus,
NRu T Pm = V
(0.40375 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(298 K) = = 444.6 kPa 2.25 m 3 m
(b) We take both gases as the system. No work or mass crosses the system boundary, and thus this is a closed system with W = 0. Taking the direction of heat transfer to be from the system (will be verified), the energy balance for this closed system reduces to E in − E out = ∆E system − Qout = ∆U = ∆U O 2 + ∆U N 2 → Qout = [mcv (T1 − Tm )]O + [mcv (T1 − Tm )]N 2
2
Using cv values at room temperature (Table A-2), the heat transfer is determined to be Qout = (1 kg )(0.658 kJ/kg ⋅ °C )(15 − 25)°C + (10.43 kg )(0.743 kJ/kg ⋅ °C )(50 − 25)°C = 187.2 kJ
(from the system)
(c) For and extended system that involves the tanks and their immediate surroundings such that the boundary temperature is the surroundings temperature, the entropy balance can be expressed as S in − S out + S gen = ∆S system −
Qout + S gen = m( s 2 − s1 ) Tb,surr S gen = m( s 2 − s1 ) +
Qout Tsurr
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-32
The mole fraction of each gas is yO 2 = yN 2 =
NO2 Nm NN2 Nm
=
0.03125 = 0.077 0.40375
=
0.3725 = 0.923 0.40375
Thus, y Pm,2 T ( s 2 − s1 ) O 2 = c p ln 2 − Rln T1 P1 = (0.918 kJ/kg ⋅ K) ln
O2
298 K (0.077)(444.6 kPa) − (0.2598 kJ/kg ⋅ K) ln 288 K 300 kPa
= 0.5952 kJ/kg ⋅ K y Pm,2 T ( s 2 − s1 ) N 2 = c p ln 2 − Rln T1 P1 = (1.039 kJ/kg ⋅ K) ln
N2
298 K (0.923)(444.6 kPa) − (0.2968 kJ/kg ⋅ K) ln 323 K 500 kPa
= −0.0251 kJ/kg ⋅ K
Substituting, S gen = (1 kg )(0.5952 kJ/kg ⋅ K ) + (10.43 kg )(− 0.0251 kJ/kg ⋅ K ) +
187.2 kJ = 0.962 kJ/K 298 K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-33
13-62 EES Problem 13-61 is reconsidered. The results obtained assuming ideal gas behavior with constant specific heats at the average temperature, and using real gas data obtained from EES by assuming variable specific heats over the temperature range are to be compared. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" T_O2[1] =15 [C]; T_N2[1] =50 [C] T[2] =25 [C]; T_o = 25 [C] m_O2 = 1 [kg]; P_O2[1]=300 [kPa] V_N2[1]=2 [m^3]; P_N2[1]=500 [kPa] R_u=8.314 [kJ/kmol-K]; MM_O2=molarmass(O2) MM_N2=molarmass(N2); P_O2[1]*V_O2[1]=m_O2*R_u/MM_O2*(T_O2[1]+273) P_N2[1]*V_N2[1]=m_N2*R_u/MM_N2*(T_N2[1]+273) V_total=V_O2[1]+V_N2[1]; N_O2=m_O2/MM_O2 N_N2=m_N2/MM_N2; N_total=N_O2+N_N2 P[2]*V_total=N_total*R_u*(T[2]+273); P_Final =P[2] "Conservation of energy for the combined system:" E_in - E_out = DELTAE_sys E_in = 0 [kJ] E_out = Q DELTAE_sys=m_O2*(intenergy(O2,T=T[2]) - intenergy(O2,T=T_O2[1])) + m_N2*(intenergy(N2,T=T[2]) - intenergy(N2,T=T_N2[1])) P_O2[2]=P[2]*N_O2/N_total P_N2[2]=P[2]*N_N2/N_total "Entropy generation:" - Q/(T_o+273) + S_gen = DELTAS_O2 + DELTAS_N2 DELTAS_O2 = m_O2*(entropy(O2,T=T[2],P=P_O2[2]) - entropy(O2,T=T_O2[1],P=P_O2[1])) DELTAS_N2 = m_N2*(entropy(N2,T=T[2],P=P_N2[2]) - entropy(N2,T=T_N2[1],P=P_N2[1])) "Constant Property (ConstP) Solution:" -Q_ConstP=m_O2*Cv_O2*(T[2]-T_O2[1])+m_N2*Cv_N2*(T[2]-T_N2[1]) Tav_O2 =(T[2]+T_O2[1])/2 Cv_O2 = SPECHEAT(O2,T=Tav_O2) - R_u/MM_O2 Tav_N2 =(T[2]+T_N2[1])/2 Cv_N2 = SPECHEAT(N2,T=Tav_N2) - R_u/MM_N2 - Q_ConstP/(T_o+273) + S_gen_ConstP = DELTAS_O2_ConstP + DELTAS_N2_ConstP DELTAS_O2_ConstP = m_O2*( SPECHEAT(O2,T=Tav_O2)*LN((T[2]+273)/(T_O2[1]+273))R_u/MM_O2*LN(P_O2[2]/P_O2[1])) DELTAS_N2_ConstP = m_N2*( SPECHEAT(N2,T=Tav_N2)*LN((T[2]+273)/(T_N2[1]+273))R_u/MM_N2*LN(P_N2[2]/P_N2[1])) SOLUTION Cv_N2=0.7454 [kJ/kg-K] [kJ]DELTAS_N2=-0.262 [kJ/K] DELTAS_O2=0.594 [kJ/K] E_in=0 [kJ] MM_N2=28.01 [kg/kmol] m_O2=1 [kg] [kmol]N_total=0.4036 [kmol] P_N2[1]=500 [kPa] [kPa]P_O2[2]=34.42 [kPa] R_u=8.314 [kJ/kmol-K] [kJ]Tav_N2=37.5 [C] T_N2[1]=50 [C] V_N2[1]=2 [m^3]
Cv_O2=0.6627 [kJ/kg-K] DELTAE_sys=-187.7 DELTAS_N2_ConstP=-0.2625 [kJ/K] DELTAS_O2_ConstP=0.594 [kJ/K] E_out=187.7 [kJ] MM_O2=32 [kg/kmol] m_N2=10.43 [kg] N_N2=0.3724 [kmol] N_O2=0.03125 P[2]=444.6 [kPa] P_Final=444.6 [kPa] P_N2[2]=410.1 [kPa] P_O2[1]=300 Q=187.7 [kJ] Q_ConstP=187.8 [kJ] S_gen=0.962 [kJ] S_gen_ConstP=0.9616 Tav_O2=20 [C] T[2]=25 [C] T_o=25 [C] T_O2[1]=15 [C] V_O2[1]=0.2494 [m^3] V_total=2.249 [m^3/kg]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-34
13-63 Heat is transferred to a gas mixture contained in a piston cylinder device. The initial state and the final temperature are given. The heat transfer is to be determined for the ideal gas and non-ideal gas cases. Properties The molar masses of H2 and N2 are 2.0, and 28.0 kg/kmol. (Table A-1). Analysis From the energy balance relation, E in − E out = ∆E Qin − Wb,out = ∆U
(
Qin = ∆H = ∆H H 2 + ∆H N 2 = N H 2 h2 − h1
)H
2
(
+ N N 2 h2 − h1
)N
2
since Wb and ∆U combine into ∆H for quasi-equilibrium constant pressure processes mH 2 6 kg NH2 = = = 3 kmol M H 2 2 kg / kmol NN2 =
mN 2 MN 2
=
6 kg H2 21 kg N2 5 MPa 160 K
Q
21 kg = 0.75 kmol 28 kg / kmol
(a) Assuming ideal gas behavior, the inlet and exit enthalpies of H2 and N2 are determined from the ideal gas tables to be H2 : h1 = h@160 K = 4,535.4 kJ / kmol, h2 = h@ 200 K = 5,669.2 kJ / kmol N2 :
h1 = h@160 K = 4,648 kJ / kmol,
h2 = h@ 200 K = 5,810 kJ / kmol
Thus, Qideal = 3 × (5,669.2 − 4,535.4 ) + 0.75 × (5,810 − 4,648) = 4273 kJ (b) Using Amagat's law and the generalized enthalpy departure chart, the enthalpy change of each gas is determined to be Tm,1 160 = = 4.805 T R1 ,H 2 = Tcr,H 2 33.3 Z h1 ≅ 0 Pm 5 = = 3.846 PR1 ,H 2 = PR2 ,H 2 = (Fig. A-29) H2: Pcr,H 2 1.30 Zh ≅ 0 2 Tm , 2 200 = = 6.006 T R2 , H 2 = Tcr,H 2 33.3 Thus H2 can be treated as an ideal gas during this process. Tm,1 160 = = 1.27 T R1 , N 2 = Tcr, N 2 126.2 Z h1 = 1.3 Pm 5 = = 1.47 PR1 , N 2 = PR2 , N 2 = N2: Pcr, N 2 3.39 Z h = 0.7 2 Tm, 2 200 = = 1.58 T R2 , N 2 = Tcr, N 2 126.2 Therefore, h2 − h1
(
(Fig. A-29)
)H = (h2 − h1 )H ,ideal = 5,669.2 − 4,535.4 = 1,133.8kJ/kmol
(h2 − h1 )N
2
2
2
(
) (
)
= RuTcr Z h1 − Z h2 + h2 − h1 ideal = (8.314kPa ⋅ m3/kmol ⋅ K)(126.2K)(1.3 − 0.7) + (5,810 − 4,648)kJ/kmol = 1,791.5kJ/kmol
Substituting, Qin = (3 kmol)(1,133.8 kJ/kmol) + (0.75 kmol)(1,791.5 kJ/kmol) = 4745 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-35
13-64 Heat is transferred to a gas mixture contained in a piston cylinder device discussed in previous problem. The total entropy change and the exergy destruction are to be determined for two cases. Analysis The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the piston-cylinder device and its immediate surroundings so that the boundary temperature of the extended system is the environment temperature at all times. It gives S in − S out + S gen = ∆Ssystem Qin Tboundary
+ S gen = ∆S water
→
S gen = m( s2 − s1 ) −
Qin Tsurr
Then the exergy destroyed during a process can be determined from its definition X destroyed = T0 S gen . (a) Noting that the total mixture pressure, and thus the partial pressure of each gas, remains constant, the entropy change of a component in the mixture during this process is T P ©0 T ∆S i = mi c p ln 2 − R ln 2 = mi c p ,i ln 2 T P T1 1 1 i
Assuming ideal gas behavior and using cp values at the average temperature, the ∆S of H2 and N2 are determined from ∆S H 2 ,ideal = (6 kg )(13.60 kJ/kg ⋅ K ) ln
200 K = 18.21 kJ/K 160 K
∆S N 2 ,ideal = (21 kg )(1.039 kJ/kg ⋅ K ) ln
200 K = 4.87 kJ/K 160 K
and 4273 kJ = 8.98 kJ/K 303 K = (303 K )(8.98 kJ/K ) = 2721 kJ
S gen = 18.21 kJ/K + 4.87 kJ/K − X destroyed = T0 S gen
(b) Using Amagat's law and the generalized entropy departure chart, the entropy change of each gas is determined to be T R1 ,H 2 =
H2:
Tm,1 Tcr,H 2
PR1 ,H 2 = PR2 , H 2 T R2 , H 2 =
Tm , 2 Tcr,H 2
Z s1 ≅ 1 Pm 5 = = = 3.846 Pcr,H 2 1.30 Zs ≅1 2 200 = = 6.006 33.3
=
160 = 4.805 33.3
(Table A-30)
Thus H2 can be treated as an ideal gas during this process. T R1 , N 2 =
N2:
Tm,1 Tcr, N 2
PR1 , N 2 = PR2 , N 2 T R2 , N 2 =
Tm, 2 Tcr, N 2
Z s1 = 0.8 Pm 5 = = = 1.475 Pcr, N 2 3.39 Z s = 0.4 2 200 = = 1.585 126.2
=
160 = 1.268 126.2
(Table A-30)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-36
Therefore, ∆S H 2 = ∆S H 2 ,ideal = 18.21 kJ/K
(
)
∆S N 2 = N N 2 Ru Z s1 − Z s2 + ∆S N 2 ,ideal = (0.75 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(0.8 − 0.4) + (4.87 kJ/K ) = 7.37 kJ/K ∆S surr =
Qsurr − 4745 kJ = = −15.66 kJ/K T0 303 K
and 4745 kJ = 9.92 kJ/K 303 K = (303 K )(9.92 kJ/K ) = 3006 kJ
S gen = 18.21 kJ/K + 7.37 kJ/K − X destroyed = T0 S gen
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-37
13-65 Air is compressed isothermally in a steady-flow device. The power input to the compressor and the rate of heat rejection are to be determined for ideal and non-ideal gas cases. Assumptions 1 This is a steady-flow process. 2 The kinetic and potential energy changes are negligible. Properties The molar mass of air is 29.0 kg/kmol. (Table A-1). Analysis The mass flow rate of air can be expressed in terms of the mole numbers as 200 K 8 MPa
m& 2.90 kg / s N& = = = 0.10 kmol / s M 29.0 kg / kmol
(a) Assuming ideal gas behavior, the ∆h and ∆s of air during this process is ∆h = 0 (isothermal process ) ∆s = c p ln
T2 T1
©0
− Ru ln
79% N2 21% O2
P2 P = − Ru ln 2 P1 P1
= −(8.314 kJ/kg ⋅ K ) ln
8 MPa = −5.763 kJ/kmol ⋅ K 4 MPa
200 K 4 MPa
Disregarding any changes in kinetic and potential energies, the steady-flow energy balance equation for the isothermal process of an ideal gas reduces to Ê0 (steady) E& in − E& out = ∆E& system =0
E& in = E& out & = Q& + Nh & W& in + Nh 1 2 out W& in − Q& out = N& ∆h Ê0 = 0
→
W& in = Q& out
Also for an isothermal, internally reversible process the heat transfer is related to the entropy change by Q = T∆S = NT∆s , Q& = N& T∆s = (0.10 kmol/s )(200 K )(− 5.763 kJ/kmol ⋅ K ) = −115.3 kW → Q& out = 115.3 kW
Therefore, W& in = Q& out = 115.3 kW
(b) Using Amagat's law and the generalized charts, the enthalpy and entropy changes of each gas are determined from h2 − h1 = Ru Tcr ( Z h1 − Z h2 ) + (h2 − h1 ) ideal
Ê0
s2 − s1 = Ru ( Z s1 − Z s2 ) + (s2 − s1 ) ideal
where Pcr, N 2 Z h1 = 0.4, Z s1 = 0.2 T 220 T R1 = T R2 = m = = 1.74 Tcr, N 2 126.2 Z h = 0.8, Z s = 0.35 2 2 Pm,2 8 PR2 = = = 2.36 Pcr, N 2 3.39 PR1 =
N2:
Pm,1
=
4 = 1.18 3.39
(Tables A-29 and A-30)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-38
Pcr,O 2 Z h1 = 0.4, Z s1 = 0.25 Tm 220 T R1 = T R2 = = = 1.421 Tcr,O 2 154.8 Z h = 1.0, Z s = 0.5 2 2 Pm,2 8 PR2 = = = 1.575 Pcr,O 2 5.08 PR1 =
O2:
Pm,1
=
4 = 0.787 5.08
(Tables A-29 and A-30)
Then, h2 − h1 = y i ∆hi = y N 2 (h2 − h1 ) N 2 + y O 2 (h2 − h1 ) O 2 = (0.79)(8.314)(126.2)(0.4 − 0.8) + (0.21)(8.314)(154.8)(0.4 − 1.0) + 0 = −494kJ/kmol s 2 − s1 = y i ∆s i = y N 2 ( s 2 − s1 ) N 2 + y O 2 ( s 2 − s1 ) O 2 = (0.79)(8.314)(0.2 − 0.35) + (0.21)(8.314)(0.25 − 0.5) + (−5.763) = −7.18kJ/kmol ⋅ K
Thus, Q& out = − N& T∆s = −(0.10 kmol/s)(200 K )(− 7.18 kJ/kmol ⋅ K ) = 143.6 kW E& in − E& out = ∆E& system ©0 (steady) = 0 E& in = E& out W& in + N& h1 = Q& out + N& h2 W& in = Q& out + N& (h2 − h1 ) → W& in = 143.6 kW + (0.10kmol/s)(−494kJ/kmol) = 94.2 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-39
13-66 EES Problem 13-65 is reconsidered. The results obtained by assuming ideal behavior, real gas behavior with Amagat's law, and real gas behavior with EES data are to be compared. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" y_N2 = 0.79 y_O2 = 0.21 T[1]=200 [K] "Inlet temperature" T[2]=200 [K] "Exit temmperature" P[1]=4000 [kPa] P[2]=8000 [kPa] m_dot = 2.9 [kg/s] R_u = 8.314 [kJ/kmol-K] DELTAe_bar_sys = 0 "Steady-flow analysis for all cases" m_dot = N_dot * (y_N2*molarmass(N2)+y_O2*molarmass(O2)) "Ideal gas:" e_bar_in_IG - e_bar_out_IG = DELTAe_bar_sys e_bar_in_IG =w_bar_in_IG + h_bar_IG[1] e_bar_out_IG = q_bar_out_IG +h_bar_IG[2] h_bar_IG[1] = y_N2*enthalpy(N2,T=T[1]) + y_O2*enthalpy(O2,T=T[1]) h_bar_IG[2] = y_N2*enthalpy(N2,T=T[2]) + y_O2*enthalpy(O2,T=T[2]) "The pocess is isothermal so h_bar_IG's are equal. q_bar_IG is found from the entropy change:" q_bar_out_IG = -T[1]*DELTAs_IG s_IG[2]= y_N2*entropy(N2,T=T[2],P=y_N2*P[2]) + y_O2*entropy(O2,T=T[2],P=y_O2*P[2]) s_IG[1] =y_N2*entropy(N2,T=T[1],P=y_N2*P[1]) + y_O2*entropy(O2,T=T[1],P=y_O2*P[1]) DELTAs_IG =s_IG[2]-s_IG[1] Q_dot_out_IG=N_dot*q_bar_out_IG W_dot_in_IG=N_dot*w_bar_in_IG "EES:" PN2[1]=y_N2*P[1] PO2[1]=y_O2*P[1] PN2[2]=y_N2*P[2] PO2[2]=y_O2*P[2] e_bar_in_EES - e_bar_out_EES = DELTAe_bar_sys e_bar_in_EES =w_bar_in_EES + h_bar_EES[1] e_bar_out_EES = q_bar_out_EES+h_bar_EES[2] h_bar_EES[1] = y_N2*enthalpy(Nitrogen,T=T[1], P=PN2[1]) + y_O2*enthalpy(Oxygen,T=T[1],P=PO2[1]) h_bar_EES[2] = y_N2*enthalpy(Nitrogen,T=T[2],P=PN2[2]) + y_O2*enthalpy(Oxygen,T=T[2],P=PO2[2]) q_bar_out_EES = -T[1]*DELTAs_EES DELTAs_EES =y_N2*entropy(Nitrogen,T=T[2],P=PN2[2]) + y_O2*entropy(Oxygen,T=T[2],P=PO2[2]) y_N2*entropy(Nitrogen,T=T[1],P=PN2[1]) - y_O2*entropy(Oxygen,T=T[1],P=PO2[1]) Q_dot_out_EES=N_dot*q_bar_out_EES W_dot_in_EES=N_dot*w_bar_in_EES "Amagat's Rule:" Tcr_N2=126.2 [K] "Table A.1" Tcr_O2=154.8 [K] Pcr_N2=3390 [kPa] "Table A.1" Pcr_O2=5080 [kPa] e_bar_in_Zchart - e_bar_out_Zchart = DELTAe_bar_sys e_bar_in_Zchart=w_bar_in_Zchart + h_bar_Zchart[1] e_bar_out_Zchart =q_bar_out_Zchart + h_bar_Zchart[2] q_bar_out_Zchart = -T[1]*DELTAs_Zchart Q_dot_out_Zchart=N_dot*q_bar_out_Zchart W_dot_in_Zchart=N_dot*w_bar_in_Zchart "State 1by compressability chart" Tr_N2[1]=T[1]/Tcr_N2 Pr_N2[1]=y_N2*P[1]/Pcr_N2 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-40 Tr_O2[1]=T[1]/Tcr_O2 Pr_O2[1]=y_O2*P[1]/Pcr_O2 DELTAh_bar_1_N2=ENTHDEP(Tr_N2[1], Pr_N2[1])*R_u*Tcr_N2 "Enthalpy departure, N2" DELTAh_bar_1_O2=ENTHDEP(Tr_O2[1], Pr_O2[1])*R_u*Tcr_O2 "Enthalpy departure, O2" h_bar_Zchart[1]=h_bar_IG[1]-(y_N2*DELTAh_bar_1_N2+y_O2*DELTAh_bar_1_O2) "Enthalpy of real gas using charts" DELTAs_N2[1]=ENTRDEP(Tr_N2[1], Pr_N2[1])*R_u "Entropy departure, N2" DELTAs_O2[1]=ENTRDEP(Tr_O2[1], Pr_O2[1])*R_u "Entropy departure, O2" s[1]=s_IG[1]-(y_N2*DELTAs_N2[1]+y_O2*DELTAs_O2[1]) "Entropy of real gas using charts" "State 2 by compressability chart" Tr_N2[2]=T[2]/Tcr_N2 Pr_N2[2]=y_N2*P[2]/Pcr_N2 Tr_O2[2]=T[2]/Tcr_O2 Pr_O2[2]=y_O2*P[2]/Pcr_O2 DELTAh_bar_2_N2=ENTHDEP(Tr_N2[2], Pr_N2[2])*R_u*Tcr_N2 "Enthalpy departure, N2" DELTAh_bar_2_O2=ENTHDEP(Tr_O2[2], Pr_O2[2])*R_u*Tcr_O2 "Enthalpy departure, O2" h_bar_Zchart[2]=h_bar_IG[2]-(y_N2*DELTAh_bar_2_N2+y_O2*DELTAh_bar_2_O2) "Enthalpy of real gas using charts" DELTAs_N2[2]=ENTRDEP(Tr_N2[2], Pr_N2[2])*R_u "Entropy departure, N2" DELTAs_O2[2]=ENTRDEP(Tr_O2[2], Pr_O2[2])*R_u "Entropy departure, O2" s[2]=s_IG[2]-(y_N2*DELTAs_N2[2]+y_O2*DELTAs_O2[2]) "Entropy of real gas using charts" DELTAs_Zchart = s[2]-s[1] "[kJ/kmol-K]" SOLUTION DELTAe_bar_sys=0 [kJ/kmol] DELTAh_bar_1_O2=147.6 DELTAh_bar_2_O2=299.5 DELTAs_IG=-5.763 [kJ/kmol-K] DELTAs_N2[2]=3.644 DELTAs_O2[2]=1.094 e_bar_in_EES=-2173 [kJ/kmol] e_bar_in_Zchart=-2103 e_bar_out_IG=-1633 [kJ/kmol] h_bar_EES[1]=-3235 h_bar_IG[1]=-2785 h_bar_Zchart[1]=-3181 m_dot=2.9 [kg/s] Pcr_N2=3390 [kPa] P[1]=4000 [kPa] PN2[1]=3160 PO2[1]=840 Pr_N2[1]=0.9322 Pr_O2[1]=0.1654 q_bar_out_EES=1446 [kJ/kmol] q_bar_out_Zchart=1462 Q_dot_out_IG=115.9 [kW] R_u=8.314 [kJ/kmol-K] s[2]=147.8 s_IG[2]=150.9 Tcr_O2=154.8 [K] T[2]=200 [K] Tr_N2[2]=1.585 Tr_O2[2]=1.292 w_bar_in_IG=1153 [kJ/kmol] W_dot_in_EES=106.8 [kW] W_dot_in_Zchart=108.3 [kW] y_O2=0.21
DELTAh_bar_1_N2=461.2 DELTAh_bar_2_N2=907.8 DELTAs_EES=-7.23 [kJ/kmol-K] DELTAs_N2[1]=1.831 DELTAs_O2[1]=0.5361 DELTAs_Zchart=-7.312 [kJ/kmol-K] e_bar_in_IG=-1633 [kJ/kmol] e_bar_out_EES=-2173 [kJ/kmol] e_bar_out_Zchart=-2103 h_bar_EES[2]=-3619 h_bar_IG[2]=-2785 h_bar_Zchart[2]=-3565 N_dot=0.1005 [kmol/s] Pcr_O2=5080 [kPa] P[2]=8000 [kPa] PN2[2]=6320 PO2[2]=1680 Pr_N2[2]=1.864 Pr_O2[2]=0.3307 q_bar_out_IG=1153 [kJ/kmol] Q_dot_out_EES=145.3 [kW] Q_dot_out_Zchart=147 [kW] s[1]=155.1 s_IG[1]=156.7 Tcr_N2=126.2 [K] T[1]=200 [K] Tr_N2[1]=1.585 Tr_O2[1]=1.292 w_bar_in_EES=1062 [kJ/kmol] w_bar_in_Zchart=1078 [kJ/kmmol] W_dot_in_IG=115.9 [kW] y_N2=0.79
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-41
13-67 The volumetric fractions of the constituents of a mixture of products of combustion are given. The average molar mass of the mixture, the average specific heat, and the partial pressure of the water vapor in the mixture are to be determined. Assumptions Under specified conditions all N2, O2, H2O, and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2, H2O, O2, and N2 are 44.0, 18.0, 32.0, and 28.0 kg/kmol, respectively (Table A-1). The specific heats of CO2, H2O, O2, and N2 at 600 K are 1.075, 2.015, 1.003, and 1.075 kJ/kg.K, respectively (Table A-2b). The specific heat of water vapor at 600 K is obtained from EES. Analysis For convenience, consider 100 kmol of mixture. Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the average molar mass of the mixture is determined to be Mm =
N CO 2 M CO 2 + N H 2O M H 2O + N O 2 M O 2 + N N 2 M N 2 N CO 2 + N H 2O + N O 2 + N N 2
(4.89 kmol)(44 kg/kmol) + (6.50)(18) + (12.20)(32) + (76.41)(28) = (4.89 + 6.50 + 12.20 + 76.41) kmol = 28.62 kg/kmol
The average specific heat is determined from
c p,m =
76.41% N2 12.20% O2 6.50% H2O 4.89% CO2 600 K 200 kPa
N CO 2 c p ,CO 2 M CO 2 + N H 2O c p ,H 2 O M H 2 O + N O 2 c p ,O 2 M O 2 + N N 2 c p , N 2 M N 2 N CO 2 + N H 2O + N O 2 + N N 2
(4.89 kmol)(1.075 kJ/kg.K)(44 kg/kmol) + (6.50)(2.015)(18) + (12.20)(1.003)(32) + (76.41)(1.075)(28) (4.89 + 6.50 + 12.20 + 76.41) kmol = 31.59 kJ/kmol.K
=
The partial pressure of the water in the mixture is yv =
N H 2O N CO 2 + N H 2O + N O 2 + N N 2
=
6.50 kmol = 0.0650 (4.89 + 6.50 + 12.20 + 76.41) kmol
Pv = y v Pm = (0.0650)(200 kPa) = 13.0 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-42
13-68 The volumetric fractions of the constituents of a mixture are given. The makeup of the mixture on a mass basis and the enthalpy change per unit mass of mixture during a process are to be determined. Assumptions Under specified conditions all N2, O2, and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2, O2, and N2 are 44.0, 32.0, and 28.0 kg/kmol, respectively (Table A1). Analysis Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the molar mass of the mixture is determined to be M m = y CO 2 M CO 2 + y O 2 M O 2 + y N 2 M N 2 = (0.20)(44 kg/kmol) + (0.10)(32) + (0.70)(28) = 31.60 kg/kmol
The mass fractions are mf CO 2 = y CO 2 mf O 2 = y O 2 mf N 2 = y N 2
M CO 2 Mm
M O2 Mm
= (0.20)
44 kg/kmol = 0.2785 31.60 kg/kmol
= (0.10)
32 kg/kmol = 0.1013 31.60 kg/kmol
= (0.70)
28 kg/kmol = 0.6203 31.60 kg/kmol
M N2 Mm
70% N2 10% O2 20% CO2 T1 = 300 K T2 = 500 K
The enthalpy change of each gas and the enthalpy change of the mixture are (from Tables A-18-20) ∆hCO 2 = ∆hO 2 = ∆hN 2 =
h@ 500 K − h@ 300 K M CO 2
h@ 500 K − h@ 300 K M O2 h@ 500 K − h@ 300 K M N2
=
(17,678 − 9431) kJ/kmol = 187.43 kJ/kg 44 kg/kmol
=
(14,770 − 8736) kJ/kmol = 188.56 kJ/kg 32 kg/kmol
=
(14,581 − 8723) kJ/kmol = 209.21 kJ/kg 28 kg/kmol
∆hm = mf CO 2 ∆hCO 2 + mf O 2 ∆hO 2 + mf N 2 ∆h N 2 = (0.2785)(187.43) + (0.1013)(188.56) + (0.6203)(209.21) = 201.1 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-43
Special Topic: Chemical Potential and the Separation Work of Mixtures 13-69C No, a process that separates a mixture into its components without requiring any work (exergy) input is impossible since such a process would violate the 2nd law of thermodynamics. 13-70C Yes, the volume of the mixture can be more or less than the sum of the initial volumes of the mixing liquids because of the attractive or repulsive forces acting between dissimilar molecules. 13-71C The person who claims that the temperature of the mixture can be higher than the temperatures of the components is right since the total enthalpy of the mixture of two components at the same pressure and temperature, in general, is not equal to the sum of the total enthalpies of the individual components before mixing, the difference being the enthalpy (or heat) of mixing, which is the heat released or absorbed as two or more components are mixed isothermally. 13-72C Mixtures or solutions in which the effects of molecules of different components on each other are negligible are called ideal solutions (or ideal mixtures). The ideal-gas mixture is just one category of ideal solutions. For ideal solutions, the enthalpy change and the volume change due to mixing are zero, but the entropy change is not. The chemical potential of a component of an ideal mixture is independent of the identity of the other constituents of the mixture. The chemical potential of a component in an ideal mixture is equal to the Gibbs function of the pure component.
13-73 Brackish water is used to produce fresh water. The minimum power input and the minimum height the brackish water must be raised by a pump for reverse osmosis are to be determined. Assumptions 1 The brackish water is an ideal solution since it is dilute. 2 The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is also 12°C. Properties The molar masses of water and salt are Mw = 18.0 kg/kmol and Ms = 58.44 kg/kmol. The gas constant of pure water is Rw = 0.4615 kJ/kg⋅K (Table A-1). The density of fresh water is 1000 kg/m3. Analysis First we determine the mole fraction of pure water in brackish water using Eqs. 13-4 and 13-5. Noting that mfs = 0.00078 and mfw = 1- mfs = 0.99922, Mm =
1 1 1 = = = 18.01 kg/kmol mfi mf s mf w 0.00078 0.99922 + + Mi Ms Mw 58.44 18.0
∑
yi = mfi
Mm Mi
→
yw = mf w
Mm 18.01 kg/kmol = (0.99922) = 0.99976 Mw 18.0 kg/kmol
The minimum work input required to produce 1 kg of freshwater from brackish water is wmin, in = RwT0 ln(1 / yw ) = (0.4615 kJ/kg ⋅ K)(285.15 K) ln(1/0.99976) = 0.03159 kJ/kg fresh water
Therefore, 0.03159 kJ of work is needed to produce 1 kg of fresh water is mixed with seawater reversibly. Therefore, the required power input to produce fresh water at the specified rate is 1 kW W& min, in = ρV&wmin, in = (1000 kg/m 3 )(0.280 m 3 /s)(0.03159 kJ/kg) = 8.85 kW 1 kJ/s
The minimum height to which the brackish water must be pumped is ∆z min =
wmin,in g
0.03159 kJ/kg 1 kg.m/s 2 = 9.81 m/s 2 1 N
1000 N.m = 3.22 m 1 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-44
13-74 A river is discharging into the ocean at a specified rate. The amount of power that can be generated is to be determined. Assumptions 1 The seawater is an ideal solution since it is dilute. 2 The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is also 15°C. Properties The molar masses of water and salt are Mw = 18.0 kg/kmol and Ms = 58.44 kg/kmol. The gas constant of pure water is Rw = 0.4615 kJ/kg⋅K (Table A-1). The density of river water is 1000 kg/m3. Analysis First we determine the mole fraction of pure water in ocean water using Eqs. 13-4 and 13-5. Noting that mfs = 0.035 and mfw = 1- mfs = 0.965, Mm =
1 1 1 = = = 18.45 kg/kmol mfi mf s mf w 0.035 0.965 + + Mi Ms Mw 58.44 18.0
∑
yi = mfi
Mm Mi
→
yw = mf w
Mm 18.45 kg/kmol = (0.965) = 0.9891 Mw 18.0 kg/kmol
The maximum work output associated with mixing 1 kg of seawater (or the minimum work input required to produce 1 kg of freshwater from seawater) is wmax, out = RwT0 ln(1 / yw ) = (0.4615 kJ/kg ⋅ K)(288.15 K)ln(1/0.9891) = 1.46 kJ/kg fresh water
Therefore, 1.46 kJ of work can be produced as 1 kg of fresh water is mixed with seawater reversibly. Therefore, the power that can be generated as a river with a flow rate of 400,000 m3/s mixes reversibly with seawater is 1 kW 6 W& max out = ρV&wmax out = (1000 kg/m 3 )(4 × 10 5 m 3 /s)(1.46 kJ/kg) = 582 × 10 kW 1 kJ/s
Discussion This is more power than produced by all nuclear power plants (112 of them) in the U.S., which shows the tremendous amount of power potential wasted as the rivers discharge into the seas.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-45
13-75 EES Problem 13-74 is reconsidered. The effect of the salinity of the ocean on the maximum power generated is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Properties:" M_w = 18.0 [kg/kmol] "Molar masses of water" M_s = 58.44 [kg/kmol] "Molar masses of salt" R_w = 0.4615 [kJ/kg-K] "Gas constant of pure water" roh_w = 1000 [kg/m^3] "density of river water" V_dot = 4E5 [m^3/s] T_0 = 15 [C] "Analysis: First we determine the mole fraction of pure water in ocean water using Eqs. 13-4 and 13-5. " mf_s = 0.035 "mass fraction of the salt in seawater = salinity" mf_w = 1- mf_s "mass fraction of the water in seawater" "Molar mass of the seawater is:" M_m=1/(mf_s/m_s+mf_w/M_w) "Mole fraction of the water is:" y_w=mf_w*M_m/M_w "The maximum work output associated with mixing 1 kg of seawater (or the minimum work input required to produce 1 kg of freshwater from seawater) is:" w_maxout =R_w*(T_0+273.15)*ln(1/y_w) "[kJ/kg fresh water]" "The power that can be generated as a river with a flow rate of 400,000 m^3/s mixes reversibly with seawater is" W_dot_max=roh_w*V_dot*w_maxout "Discussion This is more power than produced by all nuclear power plants (112 of them) in the US., which shows the tremendous amount of power potential wasted as the rivers discharge into the seas." 9.000 x 10 8
0 0.01 0.02 0.03 0.04 0.05
Wmax [kW] 0 1.652E+08 3.333E+08 5.043E+08 6.783E+08 8.554E+08
8.000 x 10 8 7.000 x 10 8
W max [kw ]
mfs
6.000 x 10 8 5.000 x 10 8 4.000 x 10 8 3.000 x 10 8 2.000 x 10 8 1.000 x 10 8 0 0
0.01
0.02
0.03
0.04
0.05
m fs
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-46
13-76E Brackish water is used to produce fresh water. The mole fractions, the minimum work inputs required to separate 1 lbm of brackish water and to obtain 1 lbm of fresh water are to be determined. Assumptions 1 The brackish water is an ideal solution since it is dilute. 2 The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is equal to the water temperature. Properties The molar masses of water and salt are Mw = 18.0 lbm/lbmol and Ms = 58.44 lbm/lbmol. The gas constant of pure water is Rw = 0.1102 Btu/lbm⋅R (Table A-1E). Analysis (a) First we determine the mole fraction of pure water in brackish water using Eqs. 13-4 and 135. Noting that mfs = 0.0012 and mfw = 1- mfs = 0.9988, Mm =
1 1 1 = = = 18.015 lbm/lbmol mfi mf s mf w 0.0012 0.9988 + + Mi Ms Mw 58.44 18.0
∑
yi = mfi
Mm Mi
→
yw = mf w
Mm 18.015 lbm/lbmol = (0.9988) = 0.99963 Mw 18.0 lbm/lbmol
y s = 1 − y w = 1 − 0.99963 = 0.00037
(b) The minimum work input required to separate 1 lbmol of brackish water is wmin,in = − R wT0 ( y w ln y w + y s ln y s ) = −(0.1102 Btu/lbmol.R)(525 R)[0.99963 ln(0.99963) + 0.00037 ln(0.00037)] = −0.191 Btu/lbm brackish water
(c) The minimum work input required to produce 1 lbm of freshwater from brackish water is wmin, in = RwT0 ln(1 / yw ) = (0.1102 Btu/lbm ⋅ R)(525 R)ln(1/0.99963) = 0.0214 Btu/lbm fresh water
Discussion Note that it takes about 9 times work to separate 1 lbm of brackish water into pure water and salt compared to producing 1 lbm of fresh water from a large body of brackish water.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-47
13-77 A desalination plant produces fresh water from seawater. The second law efficiency of the plant is to be determined. Assumptions 1 The seawater is an ideal solution since it is dilute. 2 The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is equal to the seawater temperature. Properties The molar masses of water and salt are Mw = 18.0 kg/kmol and Ms = 58.44 kg/kmol. The gas constant of pure water is Rw = 0.4615 kJ/kg⋅K (Table A-1). The density of river water is 1000 kg/m3. Analysis First we determine the mole fraction of pure water in seawater using Eqs. 13-4 and 13-5. Noting that mfs = 0.032 and mfw = 1- mfs = 0.968, Mm =
1 1 1 = = = 18.41 kg/kmol mfi mf s mf w 0.032 0.968 + + Mi Ms Mw 58.44 18.0
∑
yi = mfi
Mm Mi
→
yw = mf w
Mm 18.41 kg/kmol = (0.968) = 0.9900 Mw 18.0 kg/kmol
The maximum work output associated with mixing 1 kg of seawater (or the minimum work input required to produce 1 kg of freshwater from seawater) is wmax, out = R w T0 ln(1 / y w ) = (0.4615 kJ/kg ⋅ K)(283.15 K)ln(1/0.990) = 1.313 kJ/kg fresh water
The power that can be generated as 1.4 m3/s fresh water mixes reversibly with seawater is 1 kW W& max out = ρV&wmax out = (1000 kg/m 3 )(1.4 m 3 /s)(1.313 kJ/kg) = 1.84 kW 1 kJ/s
Then the second law efficiency of the plant becomes
η II =
W& min,in 1.83 MW = = 0.216 = 21.6% 8.5 MW W& in
13-78 The power consumption and the second law efficiency of a desalination plant are given. The power that can be produced if the fresh water produced is mixed with the seawater reversibly is to be determined. Assumptions 1 This is a steady-flow process. 2 The kinetic and potential energy changes are negligible. Analysis From the definition of the second law efficiency W& W&rev ηII = & rev → 0.18 = → W&rev = 0.594 MW 3.3 MW W actual
which is the maximum power that can be generated.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-48
Review Problems 13-79 The molar fractions of constituents of air are given. The gravimetric analysis of air and its molar mass are to be determined. Assumptions All the constituent gases and their mixture are ideal gases. Properties The molar masses of O2, N2, and Ar are 32.0, 28.0, and 40.0 kg/kmol. (Table A-1). Analysis For convenience, consider 100 kmol of air. Then the mass of each component and the total mass are N O 2 = 21 kmol → m O 2 = N O 2 M O 2 = (21 kmol)(32 kg/kmol) = 672 kg
N N 2 = 78 kmol → m N 2 = N N 2 M N 2 = (78 kmol)(28 kg/kmol) = 2184 kg N Ar = 1 kmol → m Ar = N Ar M Ar = (1 kmol)(40 kg/kmol) = 40 kg m m = m O 2 + m N 2 + m Ar = 672 kg + 2184 kg + 40 kg = 2896 kg
AIR 21% O2 78% N2 1% Ar
Then the mass fraction of each component (gravimetric analysis) becomes mf O 2 = mf N 2 = mf Ar =
mO 2 mm
=
672 kg = 0.232 or 23.2% 2896 kg
=
2184 kg = 0.754 or 75.4% 2896 kg
mN2 mm
m Ar 40 kg = = 0.014 or 1.4% 2896 kg mm
The molar mass of the mixture is determined from its definitions, Mm =
mm 2,896 kg = = 28.96 kg / kmol N m 100 kmol
k
13-80 Using Amagat’s law, it is to be shown that Z m =
∑y Z
i i
for a real-gas mixture.
i =1
Analysis Using the compressibility factor, the volume of a component of a real-gas mixture and of the volume of the gas mixture can be expressed as
Vi =
Z i N i Ru Tm Z N R T and V m = m m u m Pm Pm
Amagat's law can be expressed asV m = Z m N m Ru Tm = Pm
Simplifying,
∑
Zm Nm =
Dividing by Nm, Z m =
∑V (T i
m , Pm
).
Substituting,
Zi Ni Ru Tm Pm
∑Z N i
i
∑yZ
i i
where Zi is determined at the mixture temperature and pressure.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-49
k
13-81 Using Dalton’s law, it is to be shown that Z m =
∑y Z
i i
for a real-gas mixture.
i =1
Analysis Using the compressibility factor, the pressure of a component of a real-gas mixture and of the pressure of the gas mixture can be expressed as ZNRT Z N RT Pi = i i u m and Pm = m m u m Vm Vm Dalton's law can be expressed as Pm = Z m N m Ru Tm = Vm
Simplifying,
∑
Zm Nm =
Dividing by Nm, Z m =
∑ P (T i
m , Vm
) . Substituting,
Zi Ni Ru Tm Vm
∑Z N i
i
∑yZ
i i
where Zi is determined at the mixture temperature and volume.
13-82 A mixture of carbon dioxide and nitrogen flows through a converging nozzle. The required make up of the mixture on a mass basis is to be determined. Assumptions Under specified conditions CO2 and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2 and N2 are 44.0 and 28.0 kg/kmol, respectively (Table A-1). The specific heat ratios of CO2 and N2 at 500 K are kCO2 = 1.229 and kN2 = 1.391 (Table A-2). Analysis The molar mass of the mixture is determined from M m = y CO 2 M CO 2 + y N 2 M N 2 The molar fractions are related to each other by y CO 2 + y N 2 = 1 The gas constant of the mixture is given by R Rm = u Mm
CO2 N2
500 K 360 m/s
The specific heat ratio of the mixture is expressed as k = mf CO 2 k CO 2 + mf N 2 k N 2 The mass fractions are mf CO 2 = y CO 2 mf N 2 = y N 2
M CO 2 Mm
M N2 Mm
The exit velocity equals the speed of sound at 500 K 1000 m 2 /s 2 Vexit = kR m T 1 kJ/kg
Substituting the given values and known properties and solving the above equations simultaneously using EES, we find mf CO 2 = 0.838, mf N 2 = 0.162
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-50
13-83 The volumetric fractions of the constituents of combustion gases are given. The mixture undergoes a reversible adiabatic expansion process in a piston-cylinder device. The work done is to be determined. Assumptions Under specified conditions all CO2, H2O, O2, and N2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2, H2O, O2, and N2 are 44.0, 18.0, 32.0, and 28.0 kg/kmol, respectively (Table A-1). Analysis Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the molar mass of the mixture is determined to be M m = y CO 2 M CO 2 + y H 2O M H 2O + y O 2 M O 2 + y N 2 M N 2 = (0.0489)(44) + (0.0650)(18) + (0.1220)(32) + (0.7641)(28) = 28.63 kg/kmol The mass fractions are M CO 2 44 kg/kmol mf CO 2 = y CO 2 = (0.0489) = 0.07516 Mm 28.63 kg/kmol
mf H 2O = y H 2O mf O 2 = y O 2 mf N 2 = y N 2
M H 2O Mm
M O2 Mm M N2 Mm
= (0.0650)
18 kg/kmol = 0.0409 28.63 kg/kmol
= (0.1220)
32 kg/kmol = 0.1363 28.63 kg/kmol
= (0.7641)
28 kg/kmol = 0.7476 28.63 kg/kmol
4.89% CO2 6.5% H2O 12.2% O2 76.41% N2 1800 K, 1 MPa
Using Dalton’s law to find partial pressures, the entropies at the initial state are determined from EES as T = 1800 K, P = (0.0489 × 1000) = 48.9 kPa → s CO 2 ,1 = 7.0148 kJ/kg.K T = 1800 K, P = (0.0650 × 1000) = 65 kPa → s H 2O,1 = 14.590 kJ/kg.K T = 1800 K, P = (0.1220 × 1000) = 122 kPa → s N 2 ,1 = 8.2570 kJ/kg.K T = 1800 K, P = (0.7641× 1000) = 764.1 kPa → s O 2 ,1 = 8.2199 kJ/kg.K
The final state entropies cannot be determined at this point since the final temperature is not known. However, for an isentropic process, the entropy change is zero and the final temperature may be determined from ∆s total = mf CO 2 ∆s CO 2 + mf H 2 O ∆s H 2C + mf O 2 ∆s O 2 + mf N 2 ∆s N 2 = 0 The solution may be obtained using EES to be T2 = 1253 K The initial and final internal energies are (from EES) u CO 2 ,1 = −7478 kJ/kg u H C,1 = −10,779 kJ/kg → 2 T1 = 1800 K u O 2 ,1 = 1147 kJ/kg u N 2 ,1 = 1214 kJ/kg,
u CO 2 , 2 = −8102 kJ/kg → T2 = 1253 K
u H 2C, 2 = −11,955 kJ/kg u O 2 , 2 = 662.8 kJ/kg u N 2 , 2 = 696.5 kJ/kg
Noting that the heat transfer is zero, an energy balance on the system gives q in − wout = ∆u m → wout = −∆u m
where ∆u m = mf CO 2 (u CO 2 , 2 − u CO 2 ,1 ) + mf CO (u H 2 O, 2 − u H 2 O,1 ) + mf O 2 (u O 2 , 2 − u O 2 ,1 ) + mf N 2 (u N 2 ,2 − u N 2 ,1 ) Substituting, wout = −∆u m = −0.07516[(−8102) − (−7478)] − 0.0409[(−11,955) − (−10,779)] − 0.1363[662.8 − 1147] − 0.7476[696.5 − 1214]
= 547.8 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-51
13-84 The mole numbers, pressure, and temperature of the constituents of a gas mixture are given. The volume of the tank containing this gas mixture is to be determined using three methods. Analysis (a) Under specified conditions both N2 and CH4 will considerably deviate from the ideal gas behavior. Treating the mixture as an ideal gas gives N m = N N 2 + N CH 4 = 2 kmol + 6 kmol = 8 kmol
and
Vm =
N m Ru Tm (8 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(200 K) = = 1.11 m 3 12,000 kPa Pm
(b) To use Kay's rule, we first need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of N2 and CH4 from Table A-1, y N2 = Tcr′ , m =
N N2
2 kmol = 0.25 8 kmol
=
Nm
and
y CH 4 =
∑yT
= y N 2 Tcr , N 2 + y CH 4 Tcr ,CH 4
∑y P
= y N 2 Pcr , N 2 + y CH 4 Pcr ,CH 4
i cr ,i
N CH 4 Nm
=
2 kmol N2 6 kmol CH4 200 K 12 MPa
6 kmol = 0.75 8 kmol
= (0.25)(126.2 K) + (0.75)(191.1K) = 174.9K Pcr′ , m =
i cr ,i
= (0.25)(3.39 MPa) + (0.75)(4.64 MPa) = 4.33 MPa
Then, TR = PR =
200 = 1.144 174.9 Z m = 0.47 12 = = 2.77 4.33
Tm
=
Tcr' , m Pm Pcr' ,m
(Fig. A-15)
Thus,
Vm =
Z m N m Ru Tm = Z mV ideal = (0.47)(1.11 m 3 ) = 0.52 m 3 Pm
(c) To use the Amagat's law for this real gas mixture, we first need to determine the Z of each component at the mixture temperature and pressure, Tm 200 = = 1.585 Tcr, N 2 126.2 Z N 2 = 0.85 Pm 12 = = = 3.54 Pcr, N 2 3.39
TR, N 2 =
N2: PR , N 2
T R ,CH 4 =
CH4: PR ,CH 4
Tm
Tcr,CH 4 Pm = Pcr,CH 4
200 = 1.047 191.1 Z CH 4 = 0.37 12 = = 2.586 4.64
(Fig. A-15)
=
(Fig. A-15)
Mixture: Zm =
∑y Z
Vm =
Z m N m Ru Tm = Z mV ideal = (0.49)(1.11 m 3 ) = 0.544 m 3 Pm
i
i
= y N 2 Z N 2 + y CH 4 Z CH 4 = (0.25)(0.85) + (0.75)(0.37 ) = 0.49
Thus,
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-52
13-85 A stream of gas mixture at a given pressure and temperature is to be separated into its constituents steadily. The minimum work required is to be determined. Assumptions 1 Both the N2 and CO2 gases and their mixture are ideal gases. 2 This is a steady-flow process. 3 The kinetic and potential energy changes are negligible. Properties The molar masses of N2 and CO2 are 28.0 and 44.0 kg/kmol. (Table A-1). Analysis The minimum work required to separate a gas mixture into its components is equal to the reversible work associated with the mixing process, which is equal to the exergy destruction (or irreversibility) associated with the mixing process since N2 18°C
X destroyed = W rev,out − Wact ,u Ê0 = W rev,out = T0 S gen
where Sgen is the entropy generation associated with the steady-flow mixing process. The entropy change associated with a constant pressure and temperature adiabatic mixing process is determined from s gen =
∑ ∆s
i
= − Ru
∑y
i
50% N2 50% CO2 18°C
100 kPa
ln y i = −(8.314 kJ/kmol ⋅ K )[0.5 ln(0.5) + 0.5 ln(0.5)]
CO2 18°C
= 5.763 kJ/kmol ⋅ K Mm = s gen =
∑y M i
s gen Mm
=
i
= (0.5)(28 kg/kmol) + (0.5)(44 kg/kmol) = 36 kg/kmol
5.763 kJ/kmol ⋅ K = 0.160 kJ/kg ⋅ K 36 kg/kmol
x destroyed = T0 s gen = (291 K )(0.160 kJ/kg ⋅ K ) = 46.6 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-53
13-86 A gas mixture is heated during a steady-flow process. The heat transfer is to be determined using two approaches. Assumptions 1 Steady flow conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis Noting that there is no work involved, the energy balance for this gas mixture can be written, on a unit mole basis, as Ê0 (steady) E& in − E& out = ∆E& system =0
E& in = E& out
180 K
8 MPa
qin + h1 = h2 qin = ∆h
1O2+3N2
210 K
Q
Also, y O2 = 0.25 and y N 2 = 0.75 .
(a) Assuming ideal gas behavior, the inlet and exit enthalpies of O2 and N2 are determined from the ideal gas tables to be O 2 : h1 = h@180 K = 5239.6 kJ/kmol, h2 = h@ 210 K = 6112.9 kJ/kmol N 2 : h1 = h@180 K = 5229 kJ/kmol, h2 = h@ 210 K = 6,100.5 kJ/kmol
Thus, q in, ideal =
∑ y ∆h i
i
= y O 2 (h2 − h1 ) O 2 + y N 2 (h2 − h1 ) N 2
= (0.25)(6,112.9 − 5,239.6) + (0.75)(6,100.5 − 5,229) = 872.0 kJ/kmol
(b) Using the Kay's rule, the gas mixture can be treated as a pseudo-pure substance whose critical temperature and pressure are Tcr′ , m =
∑yT
= y O 2 Tcr,O 2 + y N 2 Tcr , N 2
∑y P
= y O 2 Pcr ,O 2 + y N 2 Pcr , N 2
i cr ,i
= (0.25)(154.8 K) + (0.75)(126.2 K) = 133.4 K Pcr′ , m =
i cr ,i
= (0.25)(5.08 MPa) + (0.75)(3.39 MPa) = 3.81 MPa
Then, 180 = 1.349 Tcr ,m 133.4 Z h1 = 1.4 P 8 = PR , 2 = m = = 2.100 Pcr , m 3.81 Z h = 1.1 2 Tm , 2 210 = = = 1.574 Tcr ,m 133.4
T R ,1 = PR ,1 T R,2
Tm,1
=
(Fig. A-29)
The heat transfer in this case is determined from q in = h2 − h1 = Ru Tcr ( Z h1 − Z h2 ) + (h2 − h1 ) ideal = Ru Tcr ( Z h1 − Z h2 ) + q ideal = (8.314 kJ/kmol ⋅ K)(133.4 K)(1.4 − 1.1) + (872 kJ/kmol) = 1205 kJ/kmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-54
13-87 EES Problem 13-86 is reconsidered. The effect of the mole fraction of oxygen in the mixture on heat transfer using real gas behavior with EES data is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" y_N2/y_O2 =3 T[1]=180 [K] "Inlet temperature" T[2]=210 [K] "Exit temmperature" P=8000 [kPa] R_u = 8.34 [kJ/kmol-K] "Solution is done on a unit mole of mixture basis:" y_N2 + y_O2 =1 DELTAe_bar_sys = 0 "Steady-flow analysis for all cases" "Ideal gas:" e_bar_in_IG - e_bar_out_IG = DELTAe_bar_sys e_bar_in_IG =q_bar_in_IG + h_bar_1_IG e_bar_out_IG = h_bar_2_IG h_bar_1_IG = y_N2*enthalpy(N2,T=T[1]) + y_O2*enthalpy(O2,T=T[1]) h_bar_2_IG = y_N2*enthalpy(N2,T=T[2]) + y_O2*enthalpy(O2,T=T[2]) "EES:" P_N2 = y_N2*P P_O2 = y_O2*P e_bar_in_EES - e_bar_out_EES = DELTAe_bar_sys e_bar_in_EES =q_bar_in_EES + h_bar_1_EES e_bar_out_EES = h_bar_2_EES h_bar_1_EES = y_N2*enthalpy(Nitrogen,T=T[1], P=P_N2) + y_O2*enthalpy(Oxygen,T=T[1],P=P_O2) h_bar_2_EES = y_N2*enthalpy(Nitrogen,T=T[2],P=P_N2) + y_O2*enthalpy(Oxygen,T=T[2],P=P_O2) "Kay's Rule:" Tcr_N2=126.2 [K] "Table A.1" Tcr_O2=154.8 [K] Pcr_N2=3390 [kPa] "Table A.1" Pcr_O2=5080 [kPa] Tcr_mix=y_N2*Tcr_N2+y_O2*Tcr_O2 Pcr_mix=y_N2*Pcr_N2+y_O2*Pcr_O2 e_bar_in_Zchart - e_bar_out_Zchart = DELTAe_bar_sys e_bar_in_Zchart=q_bar_in_Zchart + h_bar_1_Zchart e_bar_out_Zchart = h_bar_2_Zchart "State 1by compressability chart" Tr[1]=T[1]/Tcr_mix Pr[1]=P/Pcr_mix DELTAh_bar_1=ENTHDEP(Tr[1], Pr[1])*R_u*Tcr_mix "Enthalpy departure" h_bar_1_Zchart=h_bar_1_IG-DELTAh_bar_1 "Enthalpy of real gas using charts" "State 2 by compressability chart" Tr[2]=T[2]/Tcr_mix Pr[2]=Pr[1] DELTAh_bar_2=ENTHDEP(Tr[2], Pr[2])*R_u*Tcr_mix "Enthalpy departure" h_bar_2_Zchart=h_bar_2_IG-DELTAh_bar_2 "Enthalpy of real gas using charts"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-55
qinEES [kJ/kmol] 1320 1241 1171 1144 1123 1099 1105 1144 1221 1343 1518 1722
qinIG [kJ/kmol] 659.6 693.3 730.7 749.4 768.1 805.5 842.9 880.3 917.7 955.2 992.6 1026
qinZchart [kJ/kmol] 1139 1193 1255 1287 1320 1387 1459 1534 1615 1702 1797 1892
yO2 0.01 0.1 0.2 0.25 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.99
2000
q in [kJ/km ol]
1800 1600
Ideal Gas EES Com p. Chart
1400 1200 1000 800 600 0
0.2
0.4
0.6
0.8
1
yO2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-56
13-88 A gas mixture is heated during a steady-flow process, as discussed in the previous problem. The total entropy change and the exergy destruction are to be determined using two methods. Analysis The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the piston-cylinder device and its immediate surroundings so that the boundary temperature of the extended system is the environment temperature at all times. It gives
180 K
8 MPa
1O2+3N2
210 K
Q
S in − S out + S gen = ∆Ssystem Qin + S gen = ∆Ssystem Tboundary
→
S gen = m( s2 − s1 ) −
Qin Tsurr
Then the exergy destroyed during a process can be determined from its definition X destroyed = T0 S gen . (a) Noting that the total mixture pressure, and thus the partial pressure of each gas, remains constant, the entropy change of a component in the mixture during this process is T P ©0 T ∆s i = c p ln 2 − Ru ln 2 = Mc p ln 2 T P T1 1 1 i
Assuming ideal gas behavior and cp values at room temperature (Table A-2), the ∆s of O2 and N2 are determined from ∆s O 2 ,ideal = (32 kg/kmol)(0.918 kJ/kg ⋅ K ) ln
210 K = 4.52 kJ/kmol ⋅ K 180 K
∆s N 2 ,ideal = (28 kg/kmol)(1.039 kJ/kg ⋅ K ) ln
210 K = 4.48 kJ/kmol ⋅ K 180 K
∆s sys,ideal =
∑ y ∆s i
i
= y O2 ∆s O2 + y N 2 ∆s N 2
= (0.25)(4.52 kJ/kmol ⋅ K ) + (0.75)(4.48 kJ/kmol ⋅ K ) = 4.49 kJ/kmol ⋅ K
and 872 kJ/kmol = 1.61 kJ/kmol ⋅ K 303 K = (303 K )(1.61 kJ/kmol ⋅ K ) = 488 kJ/kmol
s gen = 4.49 kJ/kmol ⋅ K − x destroyed = T0 s gen
(b) Using the Kay's rule, the gas mixture can be treated as a pseudo-pure substance whose critical temperature and pressure are Tcr′ , m =
∑yT
= y O 2 Tcr ,O 2 + y N 2 Tcr , N 2
∑y P
= y O 2 Pcr ,O 2 + y N 2 Pcr , N 2
i cr ,i
= (0.25)(154.8 K) + (0.75)(126.2 K) = 133.4 K Pcr′ , m =
i cr ,i
= (0.25)(5.08 MPa) + (0.75)(3.39 MPa) = 3.81 MPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-57
Then, 180 = 1.349 Tcr ,m 133.4 Z s1 = 0.8 Pm 8 = PR , 2 = = = 2.100 Pcr , m 3.81 Z s = 0.45 2 Tm , 2 210 = = = 1.574 Tcr ,m 133.4
T R ,1 = PR ,1 T R,2
Tm,1
=
(Fig. A-30)
Thus, ∆s sys = Ru ( Z s1 − Z s2 ) + ∆s sys,ideal = (8.314 kJ/kmol ⋅ K)(0.8 − 0.45) + (4.49 kJ/kmol ⋅ K) = 7.40 kJ/kmol ⋅ K
and 1204.7 kJ/kmol = 3.41 kJ/kmol ⋅ K 303 K = (303 K )(3.41 kJ/kmol ⋅ K ) = 1034 kJ/kmol
s gen = 7.40 kJ/kmol ⋅ K − x destroyed = T0 s gen
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-58
13-89 The masses, pressures, and temperatures of the constituents of a gas mixture in a tank are given. Heat is transferred to the tank. The final pressure of the mixture and the heat transfer are to be determined. Assumptions He is an ideal gas and O2 is a nonideal gas. Properties The molar masses of He and O2 are 4.0 and 32.0 kg/kmol. (Table A-1) Analysis (a) The number of moles of each gas is m He 4 kg = = 1 kmol M He 4.0 kg/kmol
N He =
mO2
N O2 =
M O2
4 kg He 8 kg O2
8 kg = = 0.25 kmol 32 kg/kmol
170 K 7 MPa
N m = N He + N O 2 = 1 kmol + 0.25 kmol = 1.25 kmol
Q
Then the partial volume of each gas and the volume of the tank are He:
V He =
N He Ru T1 (1 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(170 K) = = 0.202 m 3 Pm,1 7000 kPa Pm,1
PR1 =
O2: T R1
Pcr ,O 2 T1 = Tcr ,O 2
V O2 =
7 = 1.38 5.08 Z 1 = 0.53 170 = = 1.10 154.8 =
ZN O 2 Ru T1 Pm,1
=
(Fig. A-15)
(0.53)(0.25 kmol)(8.314 kPa ⋅ m 3 /kg ⋅ K)(170 K) = 0.027 m 3 7000 kPa
V tank = V He + V O 2 = 0.202 m 3 + 0.027 m 3 = 0.229 m 3 The partial pressure of each gas and the total final pressure is He:
PHe,2 =
v R ,O 2
V tank
=
(1 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(220 K) 0.229 m 3
= 7987 kPa
Tcr,O 2 v O2 V m / N O2 = = PR = 0.39 Ru Tcr,O 2 / Pcr,O 2 Ru Tcr,O 2 / Pcr,O 2 (0.229 m 3 )/(0.25 kmol) = = 3 . 616 (8.314 kPa ⋅ m 3 /kmol ⋅ K)(154.8 K)/(5080 kPa)
T R2 =
O2:
N He Ru T2
T2
=
220 = 1.42 154.8
(Fig. A-15)
PO 2 = (PR Pcr )O = (0.39 )(5080 kPa ) = 1981 kPa = 1.981 MPa 2
Pm,2 = PHe + PO 2 = 7.987 MPa + 1.981 MPa = 9.97 MPa
(b) We take both gases as the system. No work or mass crosses the system boundary, therefore this is a closed system with no work interactions. Then the energy balance for this closed system reduces to Ein − Eout = ∆Esystem Qin = ∆U = ∆U He + ∆U O 2
He:
∆U He = mcv (Tm − T1 ) = (4 kg )(3.1156 kJ/kg ⋅ K )(220 − 170 )K = 623.1 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-59
O2: T R1 = 1.10 Z = 2.2 PR1 = 1.38 h1 T R2 = 1.42 9.97 Z h = 1.2 = 1.963 2 PR2 = 5.08
(Fig. A-29)
h2 − h1 = Ru Tcr ( Z h1 − Z h2 ) + (h2 − h1 ) ideal = (8.314 kJ/kmol ⋅ K)(154.8 K)(2.2 − 1.2) + (6404 − 4949)kJ/kmol = 2742 kJ/kmol
Also, PHe,1 =
N He Ru T1
V tank
=
PO 2 ,1 = Pm,1 − PHe,1
(1 kmol)(8.314 kPa ⋅ m 3 /kg ⋅ K)(170 K)
0.229 m 3 = 7000 kPa − 6172 kPa = 828 kPa
= 6,172 kPa
Thus, ∆U O 2 = N O 2 (h2 − h1 ) − ( P2V 2 − P1V 1 ) = N O 2 (h2 − h1 ) − ( PO 2 ,2 − PO 2 ,1 )V tank = (0.25 kmol)(2742 kJ/kmol) − (1981 − 828)(0.229)kPa ⋅ m 3 = 421.5 kJ
Substituting, Qin = 623.1 kJ + 421.5 kJ = 1045 kJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-60
13-90 A mixture of carbon dioxide and methane expands through a turbine. The power produced by the mixture is to be determined using ideal gas approximation and Kay’s rule. Assumptions The expansion process is reversible and adiabatic (isentropic). Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 kg/kmol and respectively. The critical properties are 304.2 K, 7390 kPa for CO2 and 191.1 K and 4640 kPa for CH4 (Table A-1). Analysis The molar mass of the mixture is determined to be M m = y CO 2 M CO 2 + y CH 4 M CH 4 = (0.60)(44) + (0.40)(16) = 32.80 kg/kmol
The gas constant is
1600 K 800 kPa 10 L/s
R 8.314 kJ/kmol.K R= u = = 0.2533 kJ/kg.K Mm 32.8 kg/kmol
The mass fractions are mf CO 2 = y CO 2 mf CH 4 = y CH 4
M CO 2 Mm M CH 4 Mm
60% CO2 40% CH4
44 kg/kmol = (0.60) = 0.8049 32.8 kg/kmol = (0.40)
16 kg/kmol = 0.1951 32.8 kg/kmol
100 kPa
Ideal gas solution: Using Dalton’s law to find partial pressures, the entropies at the initial state are determined from EES to be: T = 1600 K, P = (0.60 × 800) = 480 kPa → s CO 2 ,1 = 6.424 kJ/kg.K T = 1600 K, P = (0.40 × 800) = 320 kPa → s CH 4 ,1 = 17.188 kJ/kg.K
The final state entropies cannot be determined at this point since the final temperature is not known. However, for an isentropic process, the entropy change is zero and the final temperature may be determined from ∆s total = mf CO 2 ∆s CO 2 + mf CH 4 ∆s CH 4 0 = mf CO 2 ( s CO 2 , 2 − s CO 2 ,1 ) + mf CH 4 ( s CH 4 , 2 − s CH 4 ,1 )
The solution is obtained using EES to be T2 = 1243 K The initial and final enthalpies and the changes in enthalpy are (from EES) T1 = 1600 K →
hCO 2 ,1 = −7408 kJ/kg
u CH 4 ,1 = 747.4 kJ/kg
T2 = 1243 K →
hCO 2 , 2 = −7877 kJ/kg
u CH 4 , 2 = −1136 kJ/kg
Noting that the heat transfer is zero, an energy balance on the system gives Q& in − W& out = m& ∆hm → W& out = − m& ∆hm
where ∆hm = mf CO 2 (hCO 2 ,2 − hCO 2 ,1 ) + mf CH 4 (hCH 4 , 2 − hCH 4 ,1 )
= (0.8049)[(−7877) − (−7408)] + (0.1951)[(−1136) − (747.4)] = −745.9 kJ/kg
The mass flow rate is m& =
Substituting,
P1V&1 (800 kPa)(0.010 m 3 /s) = = 0.01974 kg/s RT1 (0.2533 kJ/kg.K)(1600 K) W& out = m& ∆hm = −(0.01974)(−745.9 kJ/kg) = 14.72 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-61
Kay’s rule solution: The critical temperature and pressure of the mixture is Tcr = y CO 2 Tcr,CO 2 + y CH 4 Tcr, CH = (0.60)(304.2 K) + (0.40)(191.1 K) = 259.0 K 4
Pcr = y CO 2 Pcr,CO 2 + y CH 4 Pcr, CH = (0.60)(7390 kPa) + (0.40)(4640 kPa) = 6290 kPa 4
State 1 properties: T1 1600 K = = 6.178 Z 1 = 1.002 Tcr 259.0 K Z h1 = −0.01025 (from EES) P1 800 kPa = = = 0.127 Z s1 = 0.0001277 Pcr 6290 kPa
T R1 = PR1
∆h1 = Z h1 RTcr = (−0.01025)(0.2533 kJ/kg.K)(259.0 K) = −0.6714 kJ/kg h1 = mf CO 2 hCO 2 ,1 + mf CH 4 hCH 4 ,1 − ∆h1 = (0.8049)(−7408) + (0.1951)(747.1) − (−0.6714) = −5813 kJ/kg ∆s1 = Z s1 R = (0.0001277)(0.2533 kJ/kg.K) = 0.00003234 kJ/kg.K s1 = mf CO 2 s CO 2 ,1 + mf CH 4 s CH 4 ,1 − ∆s1 = (0.8049)(6.424) + (0.1951)(17.188) − (0.00003234) = 8.529 kJ/kg.K
The final state entropies cannot be determined at this point since the final temperature is not known. However, for an isentropic process, the entropy change is zero and the final temperature may be determined from ∆s total = mf CO 2 ∆s CO 2 + mf CH 4 ∆s CH 4 0 = mf CO 2 ( s CO 2 , 2 − s CO 2 ,1 ) + mf CH 4 ( s CH 4 , 2 − s CH 4 ,1 )
The solution is obtained using EES to be T2 = 1243 K The initial and final enthalpies and the changes in enthalpy are Z = −0.00007368 h2 (from EES) Z s 2 = 0.0001171 P2 100 kPa = = = 0.016 Pcr 6290 kPa
TR 2 = PR 2
T2 1243 K = = 4.80 Tcr 259.0 K
∆h2 = Z h 2 RTcr = (−0.000007368)(0.2533 kJ/kg.K)(259.0 K) = −0.04828 kJ/kg h2 = mf CO 2 hCO 2 ,2 + mf CH 4 hCH 4 , 2 − ∆h2 = (0.8049)(−7877) + (0.1951)(−1136) − (−0.4828) = −6559 kJ/kg
Noting that the heat transfer is zero, an energy balance on the system gives Q& in − W& out = m& ∆hm → W& out = −m& (h2 − h1 )
where the mass flow rate is m& =
P1V&1 (800 kPa)(0.010 m 3 /s) = = 0.01970 kg/s Z1 RT1 (1.002)(0.2533 kJ/kg.K)(1600 K)
Substituting, W& out = −(0.01970 kg/s)[(−6559) − (−5813) kJ/kg ] = 14.71 kW
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-62
13-91 Carbon dioxide and oxygen contained in one tank and nitrogen contained in another tank are allowed to mix during which heat is supplied to the gases. The final pressure and temperature of the mixture and the total volume of the mixture are to be determined. Assumptions Under specified conditions CO2, N2, and O2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2, N2, and O2 are 44.0, 28.0, and 32.0 kg/kmol, respectively (TableA1). The gas constants of CO2, N2, and O2 are 0.1889, 0.2968, 2598 kJ/kg.K, respectively (Table A-2). Analysis The molar mass of the mixture in tank 1 are M m = y CO 2 M CO 2 + y O 2 M O 2 = (0.625)(44) + (0.375)(32) = 39.5 kg/kmol The gas constant in tank 1 is R 8.314 kJ/kmol.K R1 = u = = 0.2104 kJ/kg.K Mm 39.5 kg/kmol
Qin = 100 kJ
The volumes of the tanks and the total volume are m RT (5 kg)(0.2104 kJ/kg.K)(30 + 273 K) V1 = 1 1 1 = = 2.551 m 3 P1 125 kPa
V2 =
m2 R2T2 (10 kg)(0.2968 kJ/kg.K)(15 + 273 K) = = 4.276 m 3 P2 200 kPa
V total = V1 + V 2 = 2.551 + 4.276 = 6.828 m3
N2 10 kg 15°C 200 kPa
62.5% CO2 37.5% O2 5 kg 30°C 125 kPa
The mass fractions in tank 1 are M CO 2 M O2 44 kg/kmol 32 kg/kmol mf CO 2 ,1 = y CO 2 = (0.625) = 0.6963 mf O 2 ,1 = y O 2 = (0.375) = 0.3037 39.5 kg/kmol 39.5 kg/kmol Mm Mm The masses in tank 1 and the total mass after mixing are m CO 2 ,1 = mf CO 2 ,1 m1 = (0.6963)(5 kg) = 3.481 kg m total = m1 + m 2 = 5 + 10 = 15 kg m O 2 ,1 = mf O 2 ,1 m1 = (0.3037)(5 kg) = 1.519 kg The mass fractions of the combined mixture are m CO 2 ,1 3.481 m O 2 ,1 1.519 mf CO 2 , 2 = = = 0.2321 mf O 2 , 2 = = = 0.1012 15 15 m total m total The initial internal energies are uCO 2 ,1 = 2 − 8995 kJ/kg T1 = 30°C → uO 2 ,1 = −74.16 kJ/kg
mf N 2 , 2 =
m2 10 = = 0.6667 m total 15
T1 = 15°C → u N 2 ,1 = −84.77 kJ/kg
Noting that there is no work interaction, an energy balance gives Qin = m total ∆u m Qin / m total = mf CO 2 ,2 (u CO 2 ,2 − u CO 2 ,1 ) + mf O 2 ,2 (u O 2 ,2 − u O 2 ,1 ) + mf N 2 ,2 (u N 2 ,2 − u N 2 ,1 )
[
]
[
]
[
(100 kJ)/(15 kg) = (0.2321) u CO 2 ,2 − (−8995) + (0.1012) u O 2 ,2 − (−74.16) + (0.6667) u N 2 ,2 − (−84.77)
]
The internal energies after the mixing are a function of mixture temperature only. Using EES, the final temperature of the mixture is determined to be Tmix = 312.4 K The gas constant of the final mixture is R mix = mf CO 2 , 2 RCO 2 + mf O 2 , 2 RO 2 + mf N 2 , 2 R N 2 = (0.2321)(0.1889) + (0.1012)(0.2598) + (0.6667)(0.2968) = 0.2680 kg/kmol
The final pressure is determined from ideal gas relation to be m R T (15 kg)(0.2680 kJ/kg.K)(312.4 K) Pmix = total mix mix = = 184 kPa V total 6.828 m 3 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-63
13-92 EES A program is to be written to determine the mole fractions of the components of a mixture of three gases with known molar masses when the mass fractions are given, and to determine the mass fractions of the components when the mole fractions are given. Also, the program is to be run for a sample case. Analysis The problem is solved using EES, and the solution is given below. Procedure Fractions(Type$,A$,B$,C$,A,B,C:mf_A,mf_B,mf_C,y_A,y_B,y_C) {If Type$ ('mass fraction' OR 'mole fraction' ) then Call ERROR('Type$ must be set equal to "mass fraction" or "mole fraction".') GOTO 10 endif} Sum = A+B+C If ABS(Sum - 1) > 0 then goto 20 MM_A = molarmass(A$) MM_B = molarmass(B$) MM_C = molarmass(C$) If Type$ = 'mass fraction' then mf_A = A mf_B = B mf_C = C sumM_mix = mf_A/MM_A+ mf_B/MM_B+ mf_C/MM_C y_A = mf_A/MM_A/sumM_mix y_B = mf_B/MM_B/sumM_mix y_C = mf_C/MM_C/sumM_mix GOTO 10 endif if Type$ = 'mole fraction' then y_A = A y_B = B y_C = C MM_mix = y_A*MM_A+ y_B*MM_B+ y_C*MM_C mf_A = y_A*MM_A/MM_mix mf_B = y_B*MM_B/MM_mix mf_C = y_C*MM_C/MM_mix GOTO 10 Endif Call ERROR('Type$ must be either mass fraction or mole fraction.') GOTO 10 20: Call ERROR('The sum of the mass or mole fractions must be 1') 10: END "Either the mole fraction y_i or the mass fraction mf_i may be given by setting the parameter Type$='mole fraction' when the mole fractions are given or Type$='mass fraction' is given" {Input Data in the Diagram Window} {Type$='mole fraction' A$ = 'N2' B$ = 'O2' C$ = 'Argon' A = 0.71 "When Type$='mole fraction' A, B, C are the mole fractions" B = 0.28 "When Type$='mass fraction' A, B, C are the mass fractions" C = 0.01} Call Fractions(Type$,A$,B$,C$,A,B,C:mf_A,mf_B,mf_C,y_A,y_B,y_C) SOLUTION A=0.71 C=0.01 mf_C=0.014 y_B=0.280
A$='N2' B=0.28 C$='Argon' mf_A=0.680 Type$='mole fraction' y_C=0.010
B$='O2' mf_B=0.306 y_A=0.710
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-64
13-93 EES A program is to be written to determine the apparent gas constant, constant volume specific heat, and internal energy of a mixture of 3 ideal gases when the mass fractions and other properties of the constituent gases are given. Also, the program is to be run for a sample case. Analysis The problem is solved using EES, and the solution is given below. T=300 [K] A$ = 'N2' B$ = 'O2' C$ = 'CO2' mf_A = 0.71 mf_B = 0.28 mf_C = 0.01 R_u = 8.314 [kJ/kmol-K] MM_A = molarmass(A$) MM_B = molarmass(B$) MM_C = molarmass(C$) SumM_mix = mf_A/MM_A+ mf_B/MM_B+ mf_C/MM_C y_A = mf_A/MM_A/SumM_mix y_B = mf_B/MM_B/SumM_mix y_C = mf_C/MM_C/SumM_mix MM_mix = y_A*MM_A+ y_B*MM_B+ y_C*MM_C R_mix = R_u/MM_mix C_P_mix=mf_A*specheat(A$,T=T)+mf_B*specheat(B$,T=T)+mf_C*specheat(C$,T=T) C_V_mix=C_P_mix - R_mix u_mix=C_V_mix*T h_mix=C_P_mix*T SOLUTION A$='N2' B$='O2' C$='CO2' C_P_mix=1.006 [kJ/kg-K] C_V_mix=0.7206 [kJ/kg-K] h_mix=301.8 [kJ/kg] mf_A=0.71 mf_B=0.28 mf_C=0.01 MM_A=28.01 [kg/kmol] MM_B=32 [kg/kmol] MM_C=44.01 [kg/kmol] MM_mix=29.14 [kg/kmol] R_mix=0.2854 [kJ/kg-K] R_u=8.314 [kJ/kmol-K] SumM_mix=0.03432 T=300 [K] u_mix=216.2 [kJ/kg] y_A=0.7384 y_B=0.2549 y_C=0.00662
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-65
13-94 EES A program is to be written to determine the entropy change of a mixture of 3 ideal gases when the mole fractions and other properties of the constituent gases are given. Also, the program is to be run for a sample case. Analysis The problem is solved using EES, and the solution is given below. T1=300 [K] T2=600 [K] P1=100 [kPa] P2=500 [kPa] A$ = 'N2' B$ = 'O2' C$ = 'Argon' y_A = 0.71 y_B = 0.28 y_C = 0.01 MM_A = molarmass(A$) MM_B = molarmass(B$) MM_C = molarmass(C$) MM_mix = y_A*MM_A+ y_B*MM_B+ y_C*MM_C mf_A = y_A*MM_A/MM_mix mf_B = y_B*MM_B/MM_mix mf_C = y_C*MM_C/MM_mix DELTAs_mix=mf_A*(entropy(A$,T=T2,P=y_B*P2)entropy(A$,T=T1,P=y_A*P1))+mf_B*(entropy(B$,T=T2,P=y_B*P2)entropy(B$,T=T1,P=y_B*P1))+mf_C*(entropy(C$,T=T2,P=y_C*P2)entropy(C$,T=T1,P=y_C*P1)) SOLUTION A$='N2' B$='O2' C$='Argon' DELTAs_mix=12.41 [kJ/kg-K] mf_A=0.68 mf_B=0.3063 mf_C=0.01366 MM_A=28.01 [kg/kmol] MM_B=32 [kg/kmol] MM_C=39.95 [kg/kmol] MM_mix=29.25 [kJ/kmol] P1=100 [kPa] P2=500 [kPa] T1=300 [K] T2=600 [K] y_A=0.71 y_B=0.28 y_C=0.01
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-66
Fundamentals of Engineering (FE) Exam Problems 13-95 An ideal gas mixture whose apparent molar mass is 36 kg/kmol consists of nitrogen N2 and three other gases. If the mole fraction of nitrogen is 0.30, its mass fraction is (a) 0.15 (b) 0.23 (c) 0.30 (d) 0.39 (e) 0.70 Answer (b) 0.23 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). M_mix=36 "kg/kmol" M_N2=28 "kg/kmol" y_N2=0.3 mf_N2=(M_N2/M_mix)*y_N2 "Some Wrong Solutions with Common Mistakes:" W1_mf = y_N2 "Taking mass fraction to be equal to mole fraction" W2_mf= y_N2*(M_mix/M_N2) "Using the molar mass ratio backwords" W3_mf= 1-mf_N2 "Taking the complement of the mass fraction"
13-96 An ideal gas mixture consists of 2 kmol of N2 and 6 kmol of CO2. The mass fraction of CO2 in the mixture is (a) 0.175 (b) 0.250 (c) 0.500 (d) 0.750 (e) 0.825 Answer (e) 0.825 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). N1=2 "kmol" N2=6 "kmol" N_mix=N1+N2 MM1=28 "kg/kmol" MM2=44 "kg/kmol" m_mix=N1*MM1+N2*MM2 mf2=N2*MM2/m_mix "Some Wrong Solutions with Common Mistakes:" W1_mf = N2/N_mix "Using mole fraction" W2_mf = 1-mf2 "The wrong mass fraction"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-67
13-97 An ideal gas mixture consists of 2 kmol of N2 and 4 kmol of CO2. The apparent gas constant of the mixture is (a) 0.215 kJ/kg⋅K (b) 0.225 kJ/kg⋅K (c) 0.243 kJ/kg⋅K (d) 0.875 kJ/kg⋅K (e) 1.24 kJ/kg⋅K Answer (a) 0.215 kJ/kg⋅K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Ru=8.314 "kJ/kmol.K" N1=2 "kmol" N2=4 "kmol" MM1=28 "kg/kmol" MM2=44 "kg/kmol" R1=Ru/MM1 R2=Ru/MM2 N_mix=N1+N2 y1=N1/N_mix y2=N2/N_mix MM_mix=y1*MM1+y2*MM2 R_mix=Ru/MM_mix "Some Wrong Solutions with Common Mistakes:" W1_Rmix =(R1+R2)/2 "Taking the arithmetic average of gas constants" W2_Rmix= y1*R1+y2*R2 "Using wrong relation for Rmixture" 13-98 A rigid tank is divided into two compartments by a partition. One compartment contains 3 kmol of N2 at 600 kPa pressure and the other compartment contains 7 kmol of CO2 at 200 kPa. Now the partition is removed, and the two gases form a homogeneous mixture at 300 kPa. The partial pressure of N2 in the mixture is (a) 75 kPa (b) 90 kPa (c) 150 kPa (d) 175 kPa (e) 225 kPa Answer (b) 90 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1 = 600 "kPa" P2 = 200 "kPa" P_mix=300 "kPa" N1=3 "kmol" N2=7 "kmol" MM1=28 "kg/kmol" MM2=44 "kg/kmol" N_mix=N1+N2 y1=N1/N_mix y2=N2/N_mix P_N2=y1*P_mix "Some Wrong Solutions with Common Mistakes:" W1_P1= P_mix/2 "Assuming equal partial pressures" W2_P1= mf1*P_mix; mf1=N1*MM1/(N1*MM1+N2*MM2) "Using mass fractions" W3_P1 = P_mix*N1*P1/(N1*P1+N2*P2) "Using some kind of weighed averaging"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-68
13-99 An 80-L rigid tank contains an ideal gas mixture of 5 g of N2 and 5 g of CO2 at a specified pressure and temperature. If N2 were separated from the mixture and stored at mixture temperature and pressure, its volume would be (a) 32 L (b) 36 L (c) 40 L (d) 49 L (e) 80 L Answer (d) 49 L Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V_mix=80 "L" m1=5 "g" m2=5 "g" MM1=28 "kg/kmol" MM2=44 "kg/kmol" N1=m1/MM1 N2=m2/MM2 N_mix=N1+N2 y1=N1/N_mix V1=y1*V_mix "L" "Some Wrong Solutions with Common Mistakes:" W1_V1=V_mix*m1/(m1+m2) "Using mass fractions" W2_V1= V_mix "Assuming the volume to be the mixture volume" 13-100 An ideal gas mixture consists of 3 kg of Ar and 6 kg of CO2 gases. The mixture is now heated at constant volume from 250 K to 350 K. The amount of heat transfer is (a) 374 kJ (b) 436 kJ (c) 488 kJ (d) 525 kJ (e) 664 kJ Answer (c) 488 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=250 "K" T2=350 "K" Cv1=0.3122; Cp1=0.5203 "kJ/kg.K" Cv2=0.657; Cp2=0.846 "kJ/kg.K" m1=3 "kg" m2=6 "kg" MM1=39.95 "kg/kmol" MM2=44 "kg/kmol" "Applying Energy balance gives Q=DeltaU=DeltaU_Ar+DeltaU_CO2" Q=(m1*Cv1+m2*Cv2)*(T2-T1) "Some Wrong Solutions with Common Mistakes:" W1_Q = (m1+m2)*(Cv1+Cv2)/2*(T2-T1) "Using arithmetic average of properties" W2_Q = (m1*Cp1+m2*Cp2)*(T2-T1)"Using Cp instead of Cv" W3_Q = (m1*Cv1+m2*Cv2)*T2 "Using T2 instead of T2-T1"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-69
13-101 An ideal gas mixture consists of 30% helium and 70% argon gases by mass. The mixture is now expanded isentropically in a turbine from 400°C and 1.2 MPa to a pressure of 200 kPa. The mixture temperature at turbine exit is (a) 195°C (b) 56°C (c) 112°C (d) 130°C (e) 400°C Answer (b) 56°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=400+273"K" P1=1200 "kPa" P2=200 "kPa" mf_He=0.3 mf_Ar=0.7 k1=1.667 k2=1.667 "The specific heat ratio k of the mixture is also 1.667 since k=1.667 for all componet gases" k_mix=1.667 T2=T1*(P2/P1)^((k_mix-1)/k_mix)-273 "Some Wrong Solutions with Common Mistakes:" W1_T2 = (T1-273)*(P2/P1)^((k_mix-1)/k_mix) "Using C for T1 instead of K" W2_T2 = T1*(P2/P1)^((k_air-1)/k_air)-273; k_air=1.4 "Using k value for air" W3_T2 = T1*P2/P1 "Assuming T to be proportional to P" 13-102 One compartment of an insulated rigid tank contains 2 kmol of CO2 at 20°C and 150 kPa while the other compartment contains 5 kmol of H2 gas at 35°C and 300 kPa. Now the partition between the two gases is removed, and the two gases form a homogeneous ideal gas mixture. The temperature of the mixture is (a) 25°C (b) 29°C (c) 22°C (d) 32°C (e) 34°C Answer (b) 29°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). N_H2=5 "kmol" T1_H2=35 "C" P1_H2=300 "kPa" N_CO2=2 "kmol" T1_CO2=20 "C" P1_CO2=150 "kPa" Cv_H2=10.183; Cp_H2=14.307 "kJ/kg.K" Cv_CO2=0.657; Cp_CO2=0.846 "kJ/kg.K" MM_H2=2 "kg/kmol" MM_CO2=44 "kg/kmol" m_H2=N_H2*MM_H2 m_CO2=N_CO2*MM_CO2 "Applying Energy balance gives 0=DeltaU=DeltaU_H2+DeltaU_CO2" 0=m_H2*Cv_H2*(T2-T1_H2)+m_CO2*Cv_CO2*(T2-T1_CO2)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-70
"Some Wrong Solutions with Common Mistakes:" 0=m_H2*Cp_H2*(W1_T2-T1_H2)+m_CO2*Cp_CO2*(W1_T2-T1_CO2) "Using Cp instead of Cv" 0=N_H2*Cv_H2*(W2_T2-T1_H2)+N_CO2*Cv_CO2*(W2_T2-T1_CO2) "Using N instead of mass" W3_T2 = (T1_H2+T1_CO2)/2 "Assuming averate temperature"
13-103 A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas at 50°C and 400 kPa. Now the gas expands at constant pressure until its volume doubles. The amount of heat transfer to the gas mixture is (a) 6.2 MJ (b) 42 MJ (c) 27 MJ (d) 10 MJ (e) 67 MJ Answer (e) 67 MJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). N_He=3 "kmol" N_Ar=7 "kmol" T1=50+273 "C" P1=400 "kPa" P2=P1 "T2=2T1 since PV/T=const for ideal gases and it is given that P=constant" T2=2*T1 "K" MM_He=4 "kg/kmol" MM_Ar=39.95 "kg/kmol" m_He=N_He*MM_He m_Ar=N_Ar*MM_Ar Cp_Ar=0.5203; Cv_Ar = 3122 "kJ/kg.C" Cp_He=5.1926; Cv_He = 3.1156 "kJ/kg.K" "For a P=const process, Q=DeltaH since DeltaU+Wb is DeltaH" Q=m_Ar*Cp_Ar*(T2-T1)+m_He*Cp_He*(T2-T1) "Some Wrong Solutions with Common Mistakes:" W1_Q =m_Ar*Cv_Ar*(T2-T1)+m_He*Cv_He*(T2-T1) "Using Cv instead of Cp" W2_Q=N_Ar*Cp_Ar*(T2-T1)+N_He*Cp_He*(T2-T1) "Using N instead of mass" W3_Q=m_Ar*Cp_Ar*(T22-T1)+m_He*Cp_He*(T22-T1); T22=2*(T1-273)+273 "Using C for T1" W4_Q=(m_Ar+m_He)*0.5*(Cp_Ar+Cp_He)*(T2-T1) "Using arithmetic averate of Cp"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-71
13-104 An ideal gas mixture of helium and argon gases with identical mass fractions enters a turbine at 1200 K and 1 MPa at a rate of 0.3 kg/s, and expands isentropically to 100 kPa. The power output of the turbine is (a) 478 kW (b) 619 kW (c) 926 kW (d) 729 kW (e) 564 kW Answer (b) 619 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=0.3 "kg/s" T1=1200 "K" P1=1000 "kPa" P2=100 "kPa" mf_He=0.5 mf_Ar=0.5 k_He=1.667 k_Ar=1.667 Cp_Ar=0.5203 Cp_He=5.1926 Cp_mix=mf_He*Cp_He+mf_Ar*Cp_Ar "The specific heat ratio k of the mixture is also 1.667 since k=1.667 for all componet gases" k_mix=1.667 T2=T1*(P2/P1)^((k_mix-1)/k_mix) -W_out=m*Cp_mix*(T2-T1) "Some Wrong Solutions with Common Mistakes:" W1_Wout= - m*Cp_mix*(T22-T1); T22 = (T1-273)*(P2/P1)^((k_mix-1)/k_mix)+273 "Using C for T1 instead of K" W2_Wout= - m*Cp_mix*(T222-T1); T222 = T1*(P2/P1)^((k_air-1)/k_air)-273; k_air=1.4 "Using k value for air" W3_Wout= - m*Cp_mix*(T2222-T1); T2222 = T1*P2/P1 "Assuming T to be proportional to P" W4_Wout= - m*0.5*(Cp_Ar+Cp_He)*(T2-T1) "Using arithmetic average for Cp"
13-105 Design and Essay Problem
KJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-1
Chapter 14 GAS-VAPOR MIXTURES AND AIR CONDITIONING Dry and Atmospheric Air, Specific and Relative Humidity 14-1C Yes; by cooling the air at constant pressure. 14-2C Yes. 14-3C Specific humidity will decrease but relative humidity will increase. 14-4C Dry air does not contain any water vapor, but atmospheric air does. 14-5C Yes, the water vapor in the air can be treated as an ideal gas because of its very low partial pressure. 14-6C The partial pressure of the water vapor in atmospheric air is called vapor pressure. 14-7C The same. This is because water vapor behaves as an ideal gas at low pressures, and the enthalpy of an ideal gas depends on temperature only. 14-8C Specific humidity is the amount of water vapor present in a unit mass of dry air. Relative humidity is the ratio of the actual amount of vapor in the air at a given temperature to the maximum amount of vapor air can hold at that temperature. 14-9C The specific humidity will remain constant, but the relative humidity will decrease as the temperature rises in a well-sealed room. 14-10C The specific humidity will remain constant, but the relative humidity will decrease as the temperature drops in a well-sealed room. 14-11C A tank that contains moist air at 3 atm is located in moist air that is at 1 atm. The driving force for moisture transfer is the vapor pressure difference, and thus it is possible for the water vapor to flow into the tank from surroundings if the vapor pressure in the surroundings is greater than the vapor pressure in the tank. 14-12C Insulations on chilled water lines are always wrapped with vapor barrier jackets to eliminate the possibility of vapor entering the insulation. This is because moisture that migrates through the insulation to the cold surface will condense and remain there indefinitely with no possibility of vaporizing and moving back to the outside. 14-13C When the temperature, total pressure, and the relative humidity are given, the vapor pressure can be determined from the psychrometric chart or the relation Pv = φPsat where Psat is the saturation (or boiling) pressure of water at the specified temperature and φ is the relative humidity.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-2
14-14 A tank contains saturated air at a specified temperature and pressure. The mass of dry air, the specific humidity, and the enthalpy of the air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The mass of dry air can be determined from the ideal gas relation for the dry air, ma =
PaV [(105 − 4.2469) kPa ](8 m 3 ) = 9.264 kg = R a T (0.287 kJ/kg.K)(30 + 273.15 K)
(b) The relative humidity of air is 100 percent since the air saturated. The vapor pressure is equal to the saturation pressure of water at 30ºC Pv = Pg = Psat @ 30°C = 4.2469 kPa
The specific humidity can be determined from
AIR 30°C 105 kPa 8 m3
0.622 Pv (0.622)(4.2469 kPa) ω= = = 0.0262 kg H 2 O/kg dry air (105 − 4.2469) kPa P − Pv
(c) The enthalpy of air per unit mass of dry air is determined from h = ha + ωhv ≅ c p T + ωh g @ 30°C = (1.005 kJ/kg ⋅ °C)(30°C) + (0.0262)(2555.6 kJ/kg) = 97.1 kJ/kg dry air
14-15 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative humidity, and the volume of the tank are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The specific humidity can be determined form its definition,
ω=
mv 0.3 kg = = 0.0143 kg H 2 O/kg dry air ma 21 kg
(b) The saturation pressure of water at 30°C is Pg = Psat @ 30°C = 4.2469 kPa
Then the relative humidity can be determined from
φ=
21 kg dry air 0.3 kg H2O vapor 30°C 100 kPa
(0.0143)(100 kPa) ωP = = 52.9% (0.622 + ω ) Pg (0.622 + 0.0143)(4.2469 kPa)
(c) The volume of the tank can be determined from the ideal gas relation for the dry air, Pv = φPg = (0.529)(4.2469 kPa) = 2.245 kPa Pa = P − Pv = 100 − 2.245 = 97.755 kPa
V =
m a R a T (21 kg)(0.287 kJ/kg ⋅ K)(303 K) = = 18.7 m 3 97.755 kPa Pa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-3
14-16 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative humidity, and the volume of the tank are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The specific humidity can be determined form its definition,
ω=
mv 0.3 kg = = 0.0143 kg H 2 O/kg dry air ma 21 kg
(b) The saturation pressure of water at 24°C is Pg = Psat @24°C = 2.986 kPa
Then the relative humidity can be determined from
φ=
21 kg dry air 0.3 kg H2O vapor 24°C 100 kPa
(0.0143)(100 kPa) ωP = = 75.2% (0.622 + ω ) Pg (0.622 + 0.0143)2.986 kPa
(c) The volume of the tank can be determined from the ideal gas relation for the dry air, Pv = φPg = (0.752)(2.986 kPa) = 2.245 kPa Pa = P − Pv = 100 − 2.245 = 97.755 kPa
V =
m a R a T (21 kg)(0.287 kJ/kg ⋅ K)(297 K) = = 18.3 m 3 97.755 kPa Pa
14-17 A room contains air at specified conditions and relative humidity. The partial pressure of air, the specific humidity, and the enthalpy per unit mass of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The partial pressure of dry air can be determined from Pv = φPg = φPsat @ 20°C = (0.85)(2.3392 kPa) = 1.988 kPa Pa = P − Pv = 98 − 1.988 = 96.01 kPa
(b) The specific humidity of air is determined from
ω=
0.622 Pv (0.622)(1.988 kPa) = = 0.0129 kg H 2 O/kg dry air P − Pv (98 − 1.988) kPa
AIR 20°C 98 kPa 85% RH
(c) The enthalpy of air per unit mass of dry air is determined from h = ha + ωhv ≅ c p T + ωh g = (1.005 kJ/kg ⋅ °C)(20°C) + (0.0129)(2537.4 kJ/kg) = 52.78 kJ/kg dry air
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-4
14-18 A room contains air at specified conditions and relative humidity. The partial pressure of air, the specific humidity, and the enthalpy per unit mass of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The partial pressure of dry air can be determined from Pv = φPg = φPsat @ 20°C = (0.85)(2.3392 kPa) = 1.988 kPa Pa = P − Pv = 85 − 1.988 = 83.01 kPa
(b) The specific humidity of air is determined from 0.622 Pv (0.622)(1.988 kPa) ω= = = 0.0149 kg H 2 O/kg dry air P − Pv (85 − 1.988) kPa
AIR 20°C 85 kPa 85% RH
(c) The enthalpy of air per unit mass of dry air is determined from h = ha + ωhv ≅ c p T + ωh g = (1.005 kJ/kg ⋅ °C)(20°C) + (0.0149)(2537.4 kJ/kg) = 57.90 kJ/kg dry air
14-19E A room contains air at specified conditions and relative humidity. The partial pressure of air, the specific humidity, and the enthalpy per unit mass of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The partial pressure of dry air can be determined from Pv = φPg = φPsat @ 70° F = (0.85)(0.36334 psia) = 0.309 psia AIR Pa = P − Pv = 14.6 − 0.309 = 14.291 psia 70°F 14.6 psia (b) The specific humidity of air is determined from 85% RH 0.622 Pv (0.622)(0.309 psia) ω= = = 0.0134 lbm H 2 O/lbm dry air P − Pv (14.6 − 0.309) psia (c) The enthalpy of air per unit mass of dry air is determined from h = ha + ωhv ≅ c p T + ωh g = (0.24 Btu/lbm ⋅ °F)(70°F) + (0.0134)(1091.8 Btu/lbm) = 31.43 Btu/lbm dry air
14-20 The masses of dry air and the water vapor contained in a room at specified conditions and relative humidity are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis The partial pressure of water vapor and dry air are determined to be Pv = φPg = φPsat @ 23°C = (0.50)(2.811 kPa) = 1.41 kPa Pa = P − Pv = 98 − 1.41 = 96.59 kPa
The masses are determined to be ma =
PaV (96.59 kPa)(240 m 3 ) = = 272.9 kg R a T (0.287 kPa ⋅ m 3 /kg ⋅ K)(296 K)
mv =
PvV (1.41 kPa)(240 m 3 ) = = 2.47 kg Rv T (0.4615 kPa ⋅ m 3 /kg ⋅ K)(296 K)
ROOM 240 m3 23°C 98 kPa 50% RH
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-5
Dew-point, Adiabatic Saturation, and Wet-bulb Temperatures 14-21C Dew-point temperature is the temperature at which condensation begins when air is cooled at constant pressure. 14-22C Andy’s. The temperature of his glasses may be below the dew-point temperature of the room, causing condensation on the surface of the glasses. 14-23C The outer surface temperature of the glass may drop below the dew-point temperature of the surrounding air, causing the moisture in the vicinity of the glass to condense. After a while, the condensate may start dripping down because of gravity. 14-24C When the temperature falls below the dew-point temperature, dew forms on the outer surfaces of the car. If the temperature is below 0°C, the dew will freeze. At very low temperatures, the moisture in the air will freeze directly on the car windows. 14-25C When the air is saturated (100% relative humidity). 14-26C These two are approximately equal at atmospheric temperatures and pressure.
14-27 A house contains air at a specified temperature and relative humidity. It is to be determined whether any moisture will condense on the inner surfaces of the windows when the temperature of the window drops to a specified value. Assumptions The air and the water vapor are ideal gases. Analysis The vapor pressure Pv is uniform throughout the house, and its value can be determined from Pv = φPg @ 25°C = (0.65)(3.1698 kPa) = 2.06 kPa
The dew-point temperature of the air in the house is
25°C
φ = 65% 10°C
Tdp = Tsat @ Pv = Tsat @ 2.06 kPa = 18.0°C
That is, the moisture in the house air will start condensing when the temperature drops below 18.0°C. Since the windows are at a lower temperature than the dew-point temperature, some moisture will condense on the window surfaces.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-6
14-28 A person wearing glasses enters a warm room at a specified temperature and relative humidity from the cold outdoors. It is to be determined whether the glasses will get fogged. Assumptions The air and the water vapor are ideal gases. Analysis The vapor pressure Pv of the air in the house is uniform throughout, and its value can be determined from Pv = φPg @ 25°C = (0.40)(3.1698 kPa) = 1.268 kPa
The dew-point temperature of the air in the house is
25°C φ = 40%
8°C
Tdp = Tsat @ Pv = Tsat @ 1.268 kPa = 10.5°C (from EES)
That is, the moisture in the house air will start condensing when the air temperature drops below 10.5°C. Since the glasses are at a lower temperature than the dew-point temperature, some moisture will condense on the glasses, and thus they will get fogged.
14-29 A person wearing glasses enters a warm room at a specified temperature and relative humidity from the cold outdoors. It is to be determined whether the glasses will get fogged. Assumptions The air and the water vapor are ideal gases. Analysis The vapor pressure Pv of the air in the house is uniform throughout, and its value can be determined from Pv = φPg @ 25°C = (0.30)(3.1698 kPa) = 0.95 kPa
The dew-point temperature of the air in the house is
25°C φ = 30%
8°C
Tdp = Tsat @ Pv = Tsat @ 0.95 kPa = 6.2°C (from EES)
That is, the moisture in the house air will start condensing when the air temperature drops below 6.2°C. Since the glasses are at a higher temperature than the dew-point temperature, moisture will not condense on the glasses, and thus they will not get fogged.
14-30E A woman drinks a cool canned soda in a room at a specified temperature and relative humidity. It is to be determined whether the can will sweat. Assumptions The air and the water vapor are ideal gases. Analysis The vapor pressure Pv of the air in the house is uniform throughout, and its value can be determined from Pv = φPg @ 80° F = (0.50)(0.50745 psia) = 0.254 psia
The dew-point temperature of the air in the house is
80°F 50% RH Cola 40°F
Tdp = Tsat @ Pv = Tsat @ 0.254 psia = 59.7°F (from EES)
That is, the moisture in the house air will start condensing when the air temperature drops below 59.7°C. Since the canned drink is at a lower temperature than the dew-point temperature, some moisture will condense on the can, and thus it will sweat.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-7
14-31 The dry- and wet-bulb temperatures of atmospheric air at a specified pressure are given. The specific humidity, the relative humidity, and the enthalpy of air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) We obtain the properties of water vapor from EES. The specific humidity ω1 is determined from c p (T2 − T1 ) + ω 2 h fg 2
ω1 =
h g1 − h f 2
where T2 is the wet-bulb temperature, and ω2 is determined from
ω2 = Thus,
0.622 Pg 2 P2 − Pg 2
=
95 kPa 25°C Twb = 17°C
(0.622)(1.938 kPa) = 0.01295 kg H 2 O/kg dry air (95 − 1.938) kPa
(1.005 kJ/kg ⋅ °C)(17 − 25)°C + (0.01295)(2460.6 kJ/kg) = 0.00963 kg H 2 O/kg dry air (2546.5 − 71.36) kJ/kg
ω1 =
(b) The relative humidity φ1 is determined from
φ1 =
ω 1 P1 (0.00963)(95 kPa) = = 0.457 or 45.7% (0.622 + ω 1 ) Pg1 (0.622 + 0.00963)(3.1698 kPa)
(c) The enthalpy of air per unit mass of dry air is determined from h1 = ha1 + ω1 hv1 ≅ c p T1 + ω1 h g1 = (1.005 kJ/kg ⋅ °C)(25°C) + (0.00963)(2546.5 kJ/kg) = 49.65 kJ/kg dry air
14-32 The dry- and wet-bulb temperatures of air in room at a specified pressure are given. The specific humidity, the relative humidity, and the dew-point temperature are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) We obtain the properties of water vapor from EES. The specific humidity ω1 is determined from c p (T2 − T1 ) + ω 2 h fg 2
ω1 =
h g1 − h f 2
where T2 is the wet-bulb temperature, and ω2 is determined from
ω2 = Thus,
ω1 =
0.622 Pg 2 P2 − Pg 2
=
(0.622)(1.819 kPa) = 0.01152 kg H 2 O/kg dry air (100 − 1.819) kPa
100 kPa 22°C Twb = 16°C
(1.005 kJ/kg ⋅ °C)(16 − 22)°C + (0.01152)(2463.0 kJ/kg) = 0.00903 kg H 2O/kg dry air (2541.1 − 67.17) kJ/kg
(b) The relative humidity φ1 is determined from
φ1 =
ω1 P1 (0.00903)(100 kPa) = = 0.541 or 54.1% (0.622 + ω 1 ) Pg1 (0.622 + 0.0091)(2.6452 kPa)
(c) The vapor pressure at the inlet conditions is Pv1 = φ1 Pg1 = φ1 Psat @ 22°C = (0.541)(2.6452 kPa) = 1.432 kPa
Thus the dew-point temperature of the air is Tdp = Tsat @ Pv = Tsat @ 1.432 kPa = 12.3°C
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-8
14-33 EES Problem 14-32 is reconsidered. The required properties are to be determined using EES at 100 and 300 kPa pressures. Analysis The problem is solved using EES, and the solution is given below. Tdb=22 [C] Twb=16 [C] P1=100 [kPa] P2=300 [kPa] h1=enthalpy(AirH2O;T=Tdb;P=P1;B=Twb) v1=volume(AirH2O;T=Tdb;P=P1;B=Twb) Tdp1=dewpoint(AirH2O;T=Tdb;P=P1;B=Twb) w1=humrat(AirH2O;T=Tdb;P=P1;B=Twb) Rh1=relhum(AirH2O;T=Tdb;P=P1;B=Twb) h2=enthalpy(AirH2O;T=Tdb;P=P2;B=Twb) v2=volume(AirH2O;T=Tdb;P=P2;B=Twb) Tdp2=dewpoint(AirH2O;T=Tdb;P=P2;B=Twb) w2=humrat(AirH2O;T=Tdb;P=P2;B=Twb) Rh2=relhum(AirH2O;T=Tdb;P=P2;B=Twb) SOLUTION h1=45.09 [kJ/kga] h2=25.54 [kJ/kga] P1=100 [kPa] P2=300 [kPa] Rh1=0.541 Rh2=0.243 Tdb=22 [C] Tdp1=12.3 [C] Tdp2=0.6964 [C] Twb=16 [C] v1=0.8595 [m^3/kga] v2=0.283 [m^3/kga] w1=0.009029 [kgv/kga] w2=0.001336 [kgv/kga]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-9
14-34E The dry- and wet-bulb temperatures of air in room at a specified pressure are given. The specific humidity, the relative humidity, and the dew-point temperature are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The specific humidity ω1 is determined from c p (T2 − T1 ) + ω 2 h fg 2 14.7 psia ω1 = 80°F h g1 − h f 2 Twb = 65°F where T2 is the wet-bulb temperature, and ω2 is determined from 0.622 Pg 2 (0.622)(0.30578 psia) ω2 = = = 0.01321 lbm H 2 O/lbm dry air P2 − Pg 2 (14.7 − 0.30578) psia Thus,
ω1 =
(0.24 Btu/lbm ⋅ °F)(65 − 80)°F + (0.01321)(1056.5 Btu/lbm) = 0.00974 lbm H 2O/lbm dry air (1096.1 − 33.08) Btu/lbm
(b) The relative humidity φ1 is determined from ω1 P1 (0.00974)(14.7 psia) φ1 = = = 0.447 or 44.7% (0.622 + ω1 ) Pg1 (0.622 + 0.00974)(0.50745 psia) (c) The vapor pressure at the inlet conditions is Pv1 = φ1 Pg1 = φ1 Psat @ 70°F = (0.447)(0.50745 psia) = 0.2268 psia Thus the dew-point temperature of the air is Tdp = Tsat @ Pv = Tsat @ 0.2268 psia = 56.6°F
(from EES)
14-35 Atmospheric air flows steadily into an adiabatic saturation device and leaves as a saturated vapor. The relative humidity and specific humidity of air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The exit state of the air is completely specified, and the total pressure is 98 kPa. The properties of the moist air at the exit state may be determined from EES to be h2 = 78.11 kJ/kg dry air
ω 2 = 0.02079 kg H 2 O/kg dry air The enthalpy of makeup water is hw 2 = h f@ 25°C = 104.83 kJ/kg
Water 25°C
(Table A - 4)
An energy balance on the control volume gives h1 + (ω 2 − ω1 )h w = h2 h1 + (0.02079 − ω1 )(104.83 kJ/kg) = 78.11 kJ/kg
Humidifier 35°C 98 kPa
AIR
25°C 98 kPa 100%
Pressure and temperature are known for inlet air. Other properties may be determined from this equation using EES. A hand solution would require a trial-error approach. The results are h1 = 77.66 kJ/kg dry air
ω1 = 0.01654 kg H 2 O/kg dry air φ1 = 0.4511
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-10
Psychrometric Chart 14-36C They are very nearly parallel to each other. 14-37C The saturation states (located on the saturation curve). 14-38C By drawing a horizontal line until it intersects with the saturation curve. The corresponding temperature is the dew-point temperature. 14-39C No, they cannot. The enthalpy of moist air depends on ω, which depends on the total pressure.
14-40 [Also solved by EES on enclosed CD] The pressure, temperature, and relative humidity of air in a room are specified. Using the psychrometric chart, the specific humidity, the enthalpy, the wet-bulb temperature, the dew-point temperature, and the specific volume of the air are to be determined. Analysis From the psychrometric chart (Fig. A-31) we read (a) ω = 0.0181 kg H 2 O / kg dry air (b) h = 78.4 kJ / kg dry air (c) Twb = 25.5°C (d) Tdp = 23.3°C (e) v = 0.890 m 3 / kg dry air
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-11
14-41 EES Problem 14-40 is reconsidered. The required properties are to be determined using EES. Also, the properties are to be obtained at an altitude of 1500 m. Analysis The problem is solved using EES, and the solution is given below. Tdb=32 [C] Rh=0.60 P1=101.325 [kPa] Z = 1500 [m] P2=101.325*(1-0.02256*Z*convert(m,km))^5.256 "Relation giving P as a function of altitude" h1=enthalpy(AirH2O,T=Tdb,P=P1,R=Rh) v1=volume(AirH2O,T=Tdb,P=P1,R=Rh) Tdp1=dewpoint(AirH2O,T=Tdb,P=P1,R=Rh) w1=humrat(AirH2O,T=Tdb,P=P1,R=Rh) Twb1=wetbulb(AirH2O,T=Tdb,P=P1,R=Rh) h2=enthalpy(AirH2O,T=Tdb,P=P2,R=Rh) v2=volume(AirH2O,T=Tdb,P=P2,R=Rh) Tdp2=dewpoint(AirH2O,T=Tdb,P=P2,R=Rh) w2=humrat(AirH2O,T=Tdb,P=P2,R=Rh) Twb2=wetbulb(AirH2O,T=Tdb,P=P2,R=Rh) SOLUTION
Psychrom etric Diagram for M oist Air (SI Units) 0.050 0.045
Pressure = 101.0 [kPa]
0.040 0.035
Hum idity Ratio
h1=78.37 [kJ/kg] h2=87.85 [kJ/kg] P1=101.3 [kPa] P2=84.55 [kPa] Rh=0.6 Tdb=32 [C] Tdp1=23.26 [C] Tdp2=23.26 [C] Twb1=25.55 [C] Twb2=25.27 [C] v1=0.8895 [m^3/kg] v2=1.072 [m^3/kg] w1=0.01804 [kg/kg] w2=0.02174 [kg/kg] Z=1500 [m]
0.8
0.030
30 C
0.025
0.6
0.020 20 C
0.015 0.010
10 C
0.2
0C
0.005 0.000 -10
0.4
-5
0
5
10
15
20
25
30
35
T [C]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
40
14-12
14-42 The pressure, temperature, and relative humidity of air in a room are specified. Using the psychrometric chart, the specific humidity, the enthalpy, the wet-bulb temperature, the dew-point temperature, and the specific volume of the air are to be determined. Analysis From the psychrometric chart (Fig. A-31) we read (a) ω = 0.0148 kg H 2 O / kg dry air (b) h = 63.9 kJ / kg dry air (c) Twb = 21.9°C (d) Tdp = 20.1°C (e) v = 0.868 m3 / kg dry air
14-43 EES Problem 14-42 is reconsidered. The required properties are to be determined using EES. Also, the properties are to be obtained at an altitude of 2000 m. Analysis The problem is solved using EES, and the solution is given below. Tdb=26 [C] Rh=0.70 P1=101.325 [kPa] Z = 2000 [m] P2=101.325*(1-0.02256*Z*convert(m,km))^5.256 "Relation giving P as a function of altitude" h1=enthalpy(AirH2O,T=Tdb,P=P1,R=Rh) v1=volume(AirH2O,T=Tdb,P=P1,R=Rh) Tdp1=dewpoint(AirH2O,T=Tdb,P=P1,R=Rh) w1=humrat(AirH2O,T=Tdb,P=P1,R=Rh) Twb1=wetbulb(AirH2O,T=Tdb,P=P1,R=Rh) h2=enthalpy(AirH2O,T=Tdb,P=P2,R=Rh) v2=volume(AirH2O,T=Tdb,P=P2,R=Rh) Tdp2=dewpoint(AirH2O,T=Tdb,P=P2,R=Rh) w2=humrat(AirH2O,T=Tdb,P=P2,R=Rh) Twb2=wetbulb(AirH2O,T=Tdb,P=P2,R=Rh) SOLUTION h1=63.88 [kJ/kg] P1=101.3 [kPa] Rh=0.7 Tdp1=20.11 [C] Twb1=21.87 [C] v1=0.8676 [m^3/kg] w1=0.0148 [kg/kg] Z=2000 [m]
h2=74.55 [kJ/kg] P2=79.49 [kPa] Tdb=26 [C] Tdp2=20.11 [C] Twb2=21.59 [C] v2=1.113 [m^3/kg] w2=0.01899 [kg/kg]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-13
14-44E The pressure, temperature, and relative humidity of air in a room are specified. Using the psychrometric chart, the specific humidity, the enthalpy, the wet-bulb temperature, the dew-point temperature, and the specific volume of the air are to be determined. Analysis From the psychrometric chart (Fig. A-31) we read (a) ω = 0.0165 lbm H 2 O / lbm dry air (b) h = 37.8 Btu / lbm dry air (c) Twb = 74.3°F (d) Tdp = 71.3°F (e) v = 14.0 ft 3 / lbm dry air
14-45E EES Problem 14-44E is reconsidered. The required properties are to be determined using EES. Also, the properties are to be obtained at an altitude of 5000 ft. Analysis The problem is solved using EES, and the solution is given below. Tdb=82 [F] Rh=0.70 P1=14.696 [psia] Z = 5000 [ft] Zeqv=Z*convert(ft,m) P2=101.325*(1-0.02256*Zeqv/1000)^5.256*convert(kPa,psia) "Relation giving P as a function of altitude" h1=enthalpy(AirH2O,T=Tdb,P=P1,R=Rh) v1=volume(AirH2O,T=Tdb,P=P1,R=Rh) Tdp1=dewpoint(AirH2O,T=Tdb,P=P1,R=Rh) w1=humrat(AirH2O,T=Tdb,P=P1,R=Rh) Twb1=wetbulb(AirH2O,T=Tdb,P=P1,R=Rh) h2=enthalpy(AirH2O,T=Tdb,P=P2,R=Rh) v2=volume(AirH2O,T=Tdb,P=P2,R=Rh) Tdp2=dewpoint(AirH2O,T=Tdb,P=P2,R=Rh) w2=humrat(AirH2O,T=Tdb,P=P2,R=Rh) Twb2=wetbulb(AirH2O,T=Tdb,P=P2,R=Rh) SOLUTION h1=37.78 [Btu/lbm] P1=14.7 [psia] Rh=0.7 Tdp1=71.25 [F] Twb1=74.27 [F] v1=14.02 [ft^3/lbm] w1=0.01647 [lbm/lbm] Z=5000 [ft]
h2=41.54 [Btu/lbm] P2=12.23 [psia] Tdb=82 [F] Tdp2=71.25 [F] Twb2=73.89 [F] v2=16.94 [ft^3/lbm] w2=0.0199 [lbm/lbm] Zeqv=1524 [m]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-14
14-46 The pressure and the dry- and wet-bulb temperatures of air in a room are specified. Using the psychrometric chart, the specific humidity, the enthalpy, the relative humidity, the dew-point temperature, and the specific volume of the air are to be determined. Analysis From the psychrometric chart (Fig. A-31) we read (a) ω = 0.0092 kg H 2 O / kg dry air (b) h = 47.6 kJ / kg dry air (c) φ = 49.6% (d) Tdp = 12.8°C (e) v = 0.855 m3 / kg dry air
14-47 EES Problem 14-46 is reconsidered. The required properties are to be determined using EES. Also, the properties are to be obtained at an altitude of 3000 m. Analysis The problem is solved using EES, and the solution is given below. Tdb=24 [C] Twb=17 [C] P1=101.325 [kPa] Z = 3000 [m] P2=101.325*(1-0.02256*Z*convert(m,km))^5.256 "Relation giving P as function of altitude" h1=enthalpy(AirH2O,T=Tdb,P=P1,B=Twb) v1=volume(AirH2O,T=Tdb,P=P1,B=Twb) Tdp1=dewpoint(AirH2O,T=Tdb,P=P1,B=Twb) w1=humrat(AirH2O,T=Tdb,P=P1,B=Twb) Rh1=relhum(AirH2O,T=Tdb,P=P1,B=Twb) h2=enthalpy(AirH2O,T=Tdb,P=P2,B=Twb) v2=volume(AirH2O,T=Tdb,P=P2,B=Twb) Tdp2=dewpoint(AirH2O,T=Tdb,P=P2,B=Twb) w2=humrat(AirH2O,T=Tdb,P=P2,B=Twb) Rh2=relhum(AirH2O,T=Tdb,P=P2,B=Twb) SOLUTION h1=47.61 [kJ/kg] P1=101.3 [kPa] Rh1=0.4956 Tdb=24 [C] Tdp2=14.24 [C] v1=0.8542 [m^3/kg] w1=0.009219 [kg/kg] Z=3000 [m]
h2=61.68 [kJ/kg] P2=70.11 [kPa] Rh2=0.5438 Tdp1=12.81 [C] Twb=17 [C] v2=1.245 [m^3/kg] w2=0.01475 [kg/kg]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-15
Human Comfort and Air-Conditioning 14-48C It humidifies, dehumidifies, cleans and even deodorizes the air. 14-49C (a) Perspires more, (b) cuts the blood circulation near the skin, and (c) sweats excessively. 14-50C It is the direct heat exchange between the body and the surrounding surfaces. It can make a person feel chilly in winter, and hot in summer. 14-51C It affects by removing the warm, moist air that builds up around the body and replacing it with fresh air. 14-52C The spectators. Because they have a lower level of activity, and thus a lower level of heat generation within their bodies. 14-53C Because they have a large skin area to volume ratio. That is, they have a smaller volume to generate heat but a larger area to lose it from. 14-54C It affects a body’s ability to perspire, and thus the amount of heat a body can dissipate through evaporation. 14-55C Humidification is to add moisture into an environment, dehumidification is to remove it. 14-56C The metabolism refers to the burning of foods such as carbohydrates, fat, and protein in order to perform the necessary bodily functions. The metabolic rate for an average man ranges from 108 W while reading, writing, typing, or listening to a lecture in a classroom in a seated position to 1250 W at age 20 (730 at age 70) during strenuous exercise. The corresponding rates for women are about 30 percent lower. Maximum metabolic rates of trained athletes can exceed 2000 W. We are interested in metabolic rate of the occupants of a building when we deal with heating and air conditioning because the metabolic rate represents the rate at which a body generates heat and dissipates it to the room. This body heat contributes to the heating in winter, but it adds to the cooling load of the building in summer. 14-57C The metabolic rate is proportional to the size of the body, and the metabolic rate of women, in general, is lower than that of men because of their smaller size. Clothing serves as insulation, and the thicker the clothing, the lower the environmental temperature that feels comfortable. 14-58C Sensible heat is the energy associated with a temperature change. The sensible heat loss from a human body increases as (a) the skin temperature increases, (b) the environment temperature decreases, and (c) the air motion (and thus the convection heat transfer coefficient) increases. 14-59C Latent heat is the energy released as water vapor condenses on cold surfaces, or the energy absorbed from a warm surface as liquid water evaporates. The latent heat loss from a human body increases as (a) the skin wetness increases and (b) the relative humidity of the environment decreases. The rate of evaporation from the body is related to the rate of latent heat loss by Q& latent = m& vapor h fg where hfg is the latent heat of vaporization of water at the skin temperature.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-16
14-60 An average person produces 0.25 kg of moisture while taking a shower. The contribution of showers of a family of four to the latent heat load of the air-conditioner per day is to be determined. Assumptions All the water vapor from the shower is condensed by the air-conditioning system. Properties The latent heat of vaporization of water is given to be 2450 kJ/kg. Analysis The amount of moisture produced per day is m& vapor = ( Moisture produced per person)(No. of persons) = (0.25 kg / person)(4 persons / day) = 1 kg / day
Then the latent heat load due to showers becomes Q& = m& h = (1 kg / day)(2450 kJ / kg) = 2450 kJ / day latent
vapor fg
14-61 There are 100 chickens in a breeding room. The rate of total heat generation and the rate of moisture production in the room are to be determined. Assumptions All the moisture from the chickens is condensed by the air-conditioning system. Properties The latent heat of vaporization of water is given to be 2430 kJ/kg. The average metabolic rate of chicken during normal activity is 10.2 W (3.78 W sensible and 6.42 W latent). Analysis The total rate of heat generation of the chickens in the breeding room is Q& = q& (No. of chickens) = (10.2 W / chicken)(100 chickens) = 1020 W gen, total
gen, total
The latent heat generated by the chicken and the rate of moisture production are (No. of chickens) Q& = q& gen, latent
gen, latent
= (6.42 W/chicken)(100 chickens) = 642 W = 0.642 kW
m& moisture =
Q& gen, latent h fg
=
0.642 kJ / s = 0.000264 kg / s = 0.264 g / s 2430 kJ / kg
14-62 A department store expects to have a specified number of people at peak times in summer. The contribution of people to the sensible, latent, and total cooling load of the store is to be determined. Assumptions There is a mix of men, women, and children in the classroom. Properties The average rate of heat generation from people doing light work is 115 W, and 70% of is in sensible form (see Sec. 14-6). Analysis The contribution of people to the sensible, latent, and total cooling load of the store are Q& people, total = (No. of people) × Q& person, total = 135 × (115 W) = 15,525 W Q& people, sensible = (No. of people) × Q& person, sensible = 135 × (0.7 × 115 W) = 10,868 W Q& people, latent = (No. of people) × Q& person, latent = 135 × (0.3 × 115 W) = 4658 W
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-17
14-63E There are a specified number of people in a movie theater in winter. It is to be determined if the theater needs to be heated or cooled. Assumptions There is a mix of men, women, and children in the classroom. Properties The average rate of heat generation from people in a movie theater is 105 W, and 70 W of it is in sensible form and 35 W in latent form. Analysis Noting that only the sensible heat from a person contributes to the heating load of a building, the contribution of people to the heating of the building is Q& people, sensible = (No. of people) × Q& person, sensible = 500 × (70 W) = 35,000 W = 119,420 Btu/h
since 1 W = 3.412 Btu/h. The building needs to be heated since the heat gain from people is less than the rate of heat loss of 130,000 Btu/h from the building.
14-64 The infiltration rate of a building is estimated to be 1.2 ACH. The sensible, latent, and total infiltration heat loads of the building at sea level are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air infiltrates at the outdoor conditions, and exfiltrates at the indoor conditions. 3 Excess moisture condenses at room temperature of 24°C. 4 The effect of water vapor on air density is negligible. Properties The gas constant and the specific heat of air are R = 0.287 kPa.m3/kg.K and cp = 1.005 kJ/kg⋅°C (Table A-2). The heat of vaporization of water at 24°C is h fg = h fg @ 24°C = 2444.1 kJ/kg (Table A-4). The properties of the ambient and room air are determined from the psychrometric chart (Fig. A-31) to be Tambient = 32º C⎫ = 0.0150 kg/kg dryair ⎬w φ ambient = 50% ⎭ ambient Troom = 24º C⎫ = 0.0093 kg/kg dryair ⎬w φ room = 50% ⎭ room
Analysis Noting that the infiltration of ambient air will cause the air in the cold storage room to be changed 1.2 times every hour, the air will enter the room at a mass flow rate of
ρ ambient =
P0 101.325 kPa = = 1.158 kg/m 3 RT0 (0.287 kPa.m 3 /kg.K)(32 + 273 K)
m& air = ρ ambientV room ACH = (1.158 kg/m 3 )(20 × 13 × 3 m 3 )(1.2 h -1 ) = 1084 kg/h = 0.301 kg/s
Then the sensible, latent, and total infiltration heat loads of the room are determined to be Q& = m& c (T −T ) = (0.301 kg/s)(1.005 kJ/kg.°C)(32 − 24)°C = 2.42 kW infiltration, sensible
air
p
ambient
room
Q& infiltration, latent = m& air ( wambient − wroom )h fg = (0.301 kg/s)(0.0150 − 0.0093)(2444.1 kJ/kg) = 4.16 kW Q& infiltration, total = Q& infiltration, sensible + Q& infiltration, latent = 2.42 + 4.16 = 6.58 kW
Discussion The specific volume of the dry air at the ambient conditions could also be determined from the psychrometric chart at ambient conditions.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-18
14-65 The infiltration rate of a building is estimated to be 1.8 ACH. The sensible, latent, and total infiltration heat loads of the building at sea level are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air infiltrates at the outdoor conditions, and exfiltrates at the indoor conditions. 3 Excess moisture condenses at room temperature of 24°C. 4 The effect of water vapor on air density is negligible. Properties The gas constant and the specific heat of air are R = 0.287 kPa.m3/kg.K and cp = 1.005 kJ/kg⋅°C (Table A-2). The heat of vaporization of water at 24°C is h fg = h fg @ 24°C = 2444.1 kJ/kg (Table A-4). The properties of the ambient and room air are determined from the psychrometric chart (Fig. A-31) to be Tambient = 32º C⎫ = 0.0150 kg/kg dryair ⎬w φ ambient = 50% ⎭ ambient Troom = 24º C⎫ = 0.0093 kg/kg dryair ⎬w φ room = 50% ⎭ room
Analysis Noting that the infiltration of ambient air will cause the air in the cold storage room to be changed 1.8 times every hour, the air will enter the room at a mass flow rate of
ρ ambient =
P0 101.325 kPa = = 1.158 kg/m 3 RT0 (0.287 kPa.m 3 /kg.K)(32 + 273 K)
m& air = ρ ambientV room ACH = (1.158 kg/m 3 )(20 × 13 × 3 m 3 )(1.8 h -1 ) = 1084 kg/h = 0.4514 kg/s
Then the sensible, latent, and total infiltration heat loads of the room are determined to be Q& = m& c (T −T ) = (0.4514 kg/s)(1.005 kJ/kg.°C)(32 − 24)°C = 3.63 kW infiltration, sensible
air
p
ambient
room
Q& infiltration, latent = m& air ( wambient − wroom )h fg = (0.4514 kg/s)(0.0150 − 0.0093)(2444.1 kJ/kg) = 6.24 kW Q& infiltration, total = Q& infiltration, sensible + Q& infiltration, latent = 3.63 + 6.24 = 9.87 kW
Discussion The specific volume of the dry air at the ambient conditions could also be determined from the psychrometric chart at ambient conditions.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-19
Simple Heating and cooling 14-66C Relative humidity decreases during a simple heating process and increases during a simple cooling process. Specific humidity, on the other hand, remains constant in both cases. 14-67C Because a horizontal line on the psychrometric chart represents a ω = constant process, and the moisture content ω of air remains constant during these processes.
14-68 Air enters a heating section at a specified state and relative humidity. The rate of heat transfer in the heating section and the relative humidity of the air at the exit are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air remains constant (ω1 = ω2) as it flows through the heating section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 95 kPa. The properties of the air are determined to be Pv1 = φ1 Pg1 = φ1 Psat @12°C = (0.3)(1.403 kPa) = 0.421 kPa Pa1 = P1 − Pv1 = 95 − 0.421 = 94.58 kPa RT (0.287 kPa ⋅ m 3 / kg ⋅ K)(285 K) v1 = a 1 = 94.58 kPa Pa1
Heating il 1
= 0.8648 m / kg dry air 3
ω1 =
95 kPa 12°C 30% RH
25°C 2 AIR
Heat
0.622 Pv1 0.622(0.421 kPa) = = 0.002768 kg H 2 O/kg dry air (= ω 2 ) (95 − 0.421) kPa P1 − Pv1
h1 = c p T1 + ω1hg1 = (1.005 kJ/kg ⋅ °C)(12°C) + (0.002768)(2522.9 kJ/kg) = 19.04 kJ/kg dry air
and
h2 = c p T2 + ω 2 hg 2 = (1.005 kJ/kg ⋅ °C)(25°C) + (0.002768)(2546.5 kJ/kg) = 32.17 kJ/kg dry air
Also, m& a1 =
V&1 6 m 3 / min = = 6.938 kg/min v 1 0.8648 m 3 / kg dry air
Then the rate of heat transfer to the air in the heating section is determined from an energy balance on air in the heating section to be Q& in = m& a (h2 − h1 ) = (6.938 kg/min)(32.17 − 19.04) kJ/kg = 91.1 kJ/min
(b) Noting that the vapor pressure of air remains constant (Pv2 = Pv1) during a simple heating process, the relative humidity of the air at leaving the heating section becomes
φ2 =
Pv 2 Pv 2 0.421 kPa = = = 0.133 or 13.3% Pg 2 Psat@25°C 3.1698 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-20
14-69E Air enters a heating section at a specified pressure, temperature, velocity, and relative humidity. The exit temperature of air, the exit relative humidity, and the exit velocity are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air remains constant (ω 1 = ω 2) as it flows through the heating section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31E) to be h1 = 15.3 Btu/lbm dry air
ω1 = 0.0030 lbm H 2O/lbm dry air (= ω2 ) v 1 = 12.9 ft 3 / lbm dry air
1
The mass flow rate of dry air through the heating section is m& a =
1
v1
1 atm 50°F 40% RH 25 ft/s 4 kW
V1 A1 1
=
D = 15 in 2
(25 ft/s)(π × (15/12) 2 /4 ft 2 ) = 2.38 lbm/s
3
(12.9 ft / lbm)
From the energy balance on air in the heating section, Q& = m& (h − h ) in
a
2
1
⎛ 0.9478 Btu/s ⎞ 4 kW⎜ ⎟ = (2.38 lbm/s)(h2 − 15.3)Btu/lbm 1 kW ⎝ ⎠ h2 = 16.9 Btu/lbm dry air
The exit state of the air is fixed now since we know both h2 and ω2. From the psychrometric chart at this state we read T2 = 56.6°F
(b)
φ 2 = 31.4% v 2 = 13.1 ft 3 / lbm dry air
(c) The exit velocity is determined from the conservation of mass of dry air, m& a1 = m& a 2 ⎯ ⎯→
V A V A V&1 V&2 = ⎯ ⎯→ 1 = 2 v1 v 2 v1 v2
Thus, V2 =
v2 13.1 V1 = (25 ft/s) = 25.4 ft/s v1 12.9
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-21
14-70 Air enters a cooling section at a specified pressure, temperature, velocity, and relative humidity. The exit temperature, the exit relative humidity of the air, and the exit velocity are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air remains constant (ω 1 = ω 2) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31) to be h1 = 55.0 kJ/kg dry air
1200 kJ/min
ω1 = 0.0089 kg H 2 O/kg dry air (= ω 2 ) v 1 = 0.877 m 3 / kg dry air The mass flow rate of dry air through the cooling section is m& a =
1
v1
1
V1 A1 1
=
32°C 30% 18 m/s
2 1 atm
AIR
(18 m/s)(π × 0.4 2 /4 m 2 ) = 2.58 kg/s
3
(0.877 m / kg)
From the energy balance on air in the cooling section, & (h − h ) − Q& = m out
a
2
1
−1200 / 60 kJ / s = (2.58 kg / s)( h2 − 55.0) kJ / kg h2 = 47.2 kJ / kg dry air
The exit state of the air is fixed now since we know both h2 and ω2. From the psychrometric chart at this state we read T2 = 24.4°C
(b)
φ 2 = 46.6% v 2 = 0.856 m 3 / kg dry air
(c) The exit velocity is determined from the conservation of mass of dry air, V A V A V& V& ⎯→ 1 = 2 ⎯ ⎯→ 1 = 2 m& a1 = m& a 2 ⎯
v1 v 2 v1 v2 v2 0.856 V2 = V1 = (18 m/s) = 17.6 m/s 0.877 v1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-22
14-71 Air enters a cooling section at a specified pressure, temperature, velocity, and relative humidity. The exit temperature, the exit relative humidity of the air, and the exit velocity are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air remains constant (ω 1 = ω 2) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31) to be h1 = 55.0 kJ/kg dry air
800 kJ/min
ω1 = 0.0089 kg H 2 O/kg dry air (= ω 2 ) v 1 = 0.877 m 3 / kg dry air The mass flow rate of dry air through the cooling section is m& a =
1
v1
1
V1 A1 1
=
32°C 30% 18 m/s
2 1 atm
AIR
(18 m/s)(π × 0.4 2 /4 m 2 ) = 2.58 kg/s
3
(0.877 m / kg)
From the energy balance on air in the cooling section, & (h − h ) − Q& = m out
a
2
1
−800 / 60 kJ / s = (2.58 kg / s)( h2 − 55.0) kJ / kg h2 = 49.8 kJ / kg dry air
The exit state of the air is fixed now since we know both h2 and ω2. From the psychrometric chart at this state we read T2 = 26.9°C
(b)
φ 2 = 40.0% v 2 = 0.862 m 3 / kg dry air
(c) The exit velocity is determined from the conservation of mass of dry air, V A V A V& V& ⎯→ 1 = 2 ⎯ ⎯→ 1 = 2 m& a1 = m& a 2 ⎯
v1 v 2 v1 v2 v2 0.862 V2 = V1 = (18 m/s) = 17.7 m/s 0.877 v1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-23
Heating with Humidification 14-72C To achieve a higher level of comfort. Very dry air can cause dry skin, respiratory difficulties, and increased static electricity.
14-73 Air is first heated and then humidified by water vapor. The amount of steam added to the air and the amount of heat transfer to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 311 . kJ / kg dry air ω 1 = 0.0064 kg H 2 O / kg dry air ( = ω 2 ) Heating h2 = 36.2 kJ / kg dry air coils h3 = 581 . kJ / kg dry air 1 atm ω 3 = 0.0129 kg H 2 O / kg dry air T3 = 25°C T1 = 15°C AIR φ 3 = 65% φ = 60% 1 Analysis (a) The amount of moisture in the air T2 = 20°C remains constant it flows through the heating 1 2 3 section (ω 1 = ω 2), but increases in the humidifying section (ω 3 > ω 2). The amount of steam added to the air in the heating section is ∆ω = ω 3 − ω 2 = 0.0129 − 0.0064 = 0.0065 kg H 2 O / kg dry air (b) The heat transfer to the air in the heating section per unit mass of air is qin = h2 − h1 = 36.2 − 311 . = 5.1 kJ / kg dry air
14-74E Air is first heated and then humidified by water vapor. The amount of steam added to the air and the amount of heat transfer to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31E) to be h1 = 17.0 Btu/lbm dry air Heating
ω1 = 0.0046 lbm H 2 O/lbm dry air h2 = 22.3 Btu/lbm dry air
ω 2 = ω1 = 0.0046 lbm H 2 O/lbm dry air h3 = 29.2 Btu/lbm dry air
ω 3 = 0.0102 lbm H 2 O/lbm dry air
coils 14.7 psia
T1 = 50°F φ 1 = 60%
AIR 1
T3 = 75°F φ 3 = 55%
T2 = 72°F 2
3
Analysis (a) The amount of moisture in the air remains constant it flows through the heating section (ω1 = ω2), but increases in the humidifying section (ω 3 > ω 2). The amount of steam added to the air in the heating section is ∆ω = ω 3 − ω 2 = 0.0102 − 0.0046 = 0.0056 lbm H 2 O/lbm dry air (b) The heat transfer to the air in the heating section per unit mass of air is q in = h2 − h1 = 22.3 − 17.0 = 5.3 Btu/lbm dry air
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-24
14-75 Air is first heated and then humidified by wet steam. The temperature and relative humidity of air at the exit of heating section, the rate of heat transfer, and the rate at which water is added to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 23.5 kJ/kg dry air
ω1 = 0.0053 kg H 2 O/kg dry air (= ω 2 ) v 1 = 0.809 m 3 /kg dry air h3 = 42.3 kJ/kg dry air
ω 3 = 0.0087 kg H 2 O/kg dry air Analysis (a) The amount of moisture in the air remains constant it flows through the heating section (ω 1 = ω 2), but increases in the humidifying section (ω 3 > ω 2). The mass flow rate of dry air is m& a =
Sat. vapor 100°C
Heating coils
Humidifie AIR
10°C 70% 35 m3/min
20°C 60%
1 atm 1
2
3
V&1 35 m3 / min = = 43.3 kg/min v1 0.809 m3 / kg
Noting that Q = W =0, the energy balance on the humidifying section can be expressed as E& in − E& out = ∆E& systemÊ0 (steady) = 0 E& in = E& out ∑ m& i hi = ∑ m& e he
⎯ ⎯→
m& w hw + m& a 2 h2 = m& a h3 (ω 3 − ω 2 )hw + h2 = h3
Solving for h2, h2 = h3 − (ω 3 − ω 2 )h g @ 100°C = 42.3 − (0.0087 − 0.0053)(2675.6) = 33.2 kJ/kg dry air
Thus at the exit of the heating section we have ω2 = 0.0053 kg H2O dry air and h2 = 33.2 kJ/kg dry air, which completely fixes the state. Then from the psychrometric chart we read T2 = 19.5°C
φ 2 = 37.8% (b) The rate of heat transfer to the air in the heating section is Q& in = m& a (h2 − h1 ) = (43.3 kg/min)(33.2 − 23.5) kJ/kg = 420 kJ/min
(c) The amount of water added to the air in the humidifying section is determined from the conservation of mass equation of water in the humidifying section, m& w = m& a (ω 3 − ω 2 ) = (43.3 kg/min)(0.0087 − 0.0053) = 0.15 kg/min
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-25
14-76 Air is first heated and then humidified by wet steam. The temperature and relative humidity of air at the exit of heating section, the rate of heat transfer, and the rate at which water is added to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Sat. vapor Analysis (a) The amount of moisture in the air also 100°C Heating remains constant it flows through the heating section (ω 1 coils Humidifier = ω 2), but increases in the humidifying section (ω 3 > ω 2). 10°C AIR The inlet and the exit states of the air are completely 20°C 70% specified, and the total pressure is 95 kPa. The properties 60% 3 95 kPa 35 m /min of the air at various states are determined to be 1 2 3 Pv1 = φ1 Pg1 = φ1 Psat @ 10°C = (0.70)(1.2281 kPa) = 0.860 kPa (= Pv 2 ) Pa1 = P1 − Pv1 = 95 − 0.860 = 94.14 kPa
v1 =
R a T1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(283 K) = = 0.863 m 3 / kg dry air 94.14 kPa Pa1
ω1 =
0.622 Pv1 0.622(0.86 kPa) = = 0.00568 kg H 2 O/kg dry air (= ω 2 ) (95 − 0.86) kPa P1 − Pv1
h1 = c p T1 + ω1 h g1 = (1.005 kJ/kg ⋅ °C)(10°C) + (0.00568)(2519.2 kJ/kg) = 24.36 kJ/kg dry air Pv 3 = φ3 Pg 3 = φ3 Psat @ 20°C = (0.60)(2.3392 kPa) = 1.40 kPa
ω3 =
0.622 Pv3 0.622(1.40 kPa) = = 0.00930 kg H 2O/kg dry air (95 − 1.40) kPa P3 − Pv 3
h3 = c pT3 + ω3hg 3 = (1.005 kJ/kg ⋅ °C)(20°C) + (0.0093)(2537.4 kJ/kg) = 43.70 kJ/kg dry air
Also,
m& a =
V&1 35 m 3 / min = = 40.6 kg/min v 1 0.863 m 3 / kg
Noting that Q = W = 0, the energy balance on the humidifying section gives E& in − E& out = ∆E& system Ê0 (steady) = 0 ⎯ ⎯→ E& in = E& out ∑ m& e he = ∑ m& i hi ⎯ ⎯→ m& w hw + m& a 2 h2 = m& a h3 ⎯ ⎯→(ω 3 − ω 2 )h w + h2 = h3 h2 = h3 − (ω 3 − ω 2 )h g @ 100°C = 43.7 − (0.0093 − 0.00568) × 2675.6 = 34.0 kJ/kg dry air
Thus at the exit of the heating section we have ω = 0.00568 kg H2O dry air and h2 = 34.0 kJ/kg dry air, which completely fixes the state. The temperature of air at the exit of the heating section is determined from the definition of enthalpy, h2 = c p T2 + ω 2 h g 2 ≅ c p T2 + ω 2 (2500.9 + 1.82T2 ) 34.0 = (1.005)T2 + (0.00568)(2500.9 + 1.82T2 ) Solving for h2, yields T2 = 19.5° C The relative humidity at this state is 0.859 kPa P Pv 2 φ2 = v 2 = = = 0.377 or 37.7% Pg 2 Psat @ 19.5°C 2.2759 kPa
(b) The rate of heat transfer to the air in the heating section becomes Q& in = m& a (h2 − h1 ) = (40.6 kg/min)(34.0 − 24.36) kJ/kg = 391 kJ/min (c) The amount of water added to the air in the humidifying section is determined from the conservation of mass equation of water in the humidifying section, &w = m & a (ω 3 − ω 2 ) = ( 40.6 kg / min)( 0.0093 − 0.00568) = 0.147 kg / min m
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-26
Cooling with Dehumidification 14-77C To drop its relative humidity to more desirable levels.
14-78 Air is cooled and dehumidified by a window air conditioner. The rates of heat and moisture removal from the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31) to be h1 = 86.3 kJ/kg dry air
ω1 = 0.0211 kg H 2 O/kg dry air
Cooling coils
v 1 = 0.894 m 3 /kg dry air T2 = 15°C φ 2 = 100%
and h2 = 42.0 kJ/kg dry air
ω 2 = 0.0107 kg H 2 O/kg dry air
2
1
Also, hw ≅ h f @ 15°C = 62.982 kJ/kg
(Table A-4)
T1 = 32°C φ 1 = 70% V1 = 2 m3/min
1 atm Condensate
15°C
Condensate removal
Analysis (a) The amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1). The mass flow rate of air is m& a1 =
V&1 2 m 3 / min = = 2.238 kg/min v 1 0.894 m 3 / kg dry air
Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: ∑ m& w,i = ∑ m& w ,e ⎯ ⎯→ m& a1ω 1 = m& a 2ω 2 + m& w m& w = m& a (ω 1 − ω 2 ) = (2.238 kg/min)(0.0211 − 0.0107) = 0.0233 kg/min
Energy Balance: E& in − E& out = ∆E& system ©0 (steady) = 0 E& in = E& out ∑ m& i hi = Q& out + ∑ m& e he Q& out = m& a1 h1 − (m& a 2 h2 + m& w h w ) = m& a (h1 − h2 ) − m& w h w Q& out = (2.238 kg/min)(86.3 − 42.0)kJ/kg − (0.0233 kg/min)(62.982 kJ/kg) = 97.7 kJ/min
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-27
14-79 Air is first cooled, then dehumidified, and finally heated. The temperature of air before it enters the heating section, the amount of heat removed in the cooling section, and the amount of heat supplied in the heating section are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air decreases due to dehumidification (ω 3 < ω 1), and remains constant during heating (ω 3 = ω 2). The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The intermediate state (state 2) is also known since φ2 = 100% and ω 2 = ω 3. Therefore, we can determined the properties of the air at all three states from the psychrometric chart (Fig. A-31) to be h1 = 95.2 kJ / kg dry air
ω 1 = 0.0238 kg H 2 O / kg dry air and h3 = 431 . kJ / kg dry air
Heating section
Cooling section T1 = 34°C φ 1 = 70%
ω 3 = 0.0082 kg H 2 O / kg dry air ( = ω 2 )
T2 1
Also, hw ≅ h f @ 10°C = 42.02 kJ/kg (Table A - 4)
2 w
T3 = 22°C φ 3 = 50%
1 atm AIR 3
10°C
h2 = 31.8 kJ/kg dry air T2 = 11.1°C
(b) The amount of heat removed in the cooling section is determined from the energy balance equation applied to the cooling section, E& in − E& out = ∆E& systemÊ0 (steady) = 0 E& in = E& out ∑ m& i hi = ∑ m& e he + Q& out,cooling Q& out,cooling = m& a1h1 − (m& a 2 h2 + m& w hw ) = m& a (h1 − h2 ) − m& w hw
or, per unit mass of dry air, q out,cooling = (h1 − h2 ) − (ω 1 − ω 2 )hw = (95.2 − 31.8) − (0.0238 − 0.0082)42.02 = 62.7 kJ/kg dry air
(c) The amount of heat supplied in the heating section per unit mass of dry air is qin,heating = h3 − h2 = 431 . − 31.8 = 11.3 kJ / kg dry air
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-28
14-80 [Also solved by EES on enclosed CD] Air is cooled by passing it over a cooling coil through which chilled water flows. The rate of heat transfer, the mass flow rate of water, and the exit velocity of airstream are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The saturation pressure of water at 35ºC is 5. 6291 kPa (Table A-4). Then the dew point temperature of the incoming air stream at 35°C becomes Tdp = Tsat @ Pv = Tsat @ 0.6×5.6291 kPa = 26°C (Table A-5) since air is cooled to 20°C, which is below its dew point temperature, some of the moisture in the air will condense. The amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. Then the properties of the air at both states are determined from the psychrometric chart (Fig. A-31) to be h1 = 90.3 kJ/kg dry air
ω1 = 0.0215 kg H 2 O/kg dry air
Water
v 1 = 0.904 m 3 /kg dry air and
T + 8°C Cooling coils
h2 = 57.5 kJ/kg dry air
ω 2 = 0.0147 kg H 2 O/kg dry air v 2 = 0.851 m 3 /kg dry air Also,
1
hw ≅ h f @ 20°C = 83.93 kJ/kg (Table A-4)
35°C 60% 120 m/min
AIR
20°C 2 Saturated
Then,
V&1 = V1 A1 = V1 m& a1 =
π D2 4
⎛ π (0.3 m) 2 = (120 m/min)⎜ ⎜ 4 ⎝
⎞ ⎟ = 8.48 m 3 / min ⎟ ⎠
V&1 8.48 m 3 / min = = 9.38 kg/min v 1 0.904 m 3 / kg dry air
Applying the water mass balance and the energy balance equations to the combined cooling and dehumidification section (excluding the water), Water Mass Balance: ∑ m& w,i = ∑ m& w ,e ⎯ ⎯→ m& a1ω 1 = m& a 2ω 2 + m& w m& w = m& a ( ω1 − ω2 ) = ( 9.38 kg/min)(0.0215 − 0.0147 ) = 0.064 kg/min Energy Balance: E& in − E& out = ∆E& systemÊ0 (steady) = 0 ⎯ ⎯→ E& in = E& out ∑ m& i hi = ∑ m& e he + Q& out ⎯ ⎯→ Qout = m& a1h1 − (m& a 2 h2 + m& w hw ) = m& a (h1 − h2 ) − m& w hw Q& out = (9.38 kg/min)(90.3 − 57.5)kJ/kg − (0.064 kg/min)(83.93 kJ/kg) = 302.3 kJ/min (b) Noting that the heat lost by the air is gained by the cooling water, the mass flow rate of the cooling water is determined from Q& cooling water = m& cooling water ∆h = m& cooling water c p ∆T m& cooling water =
Q& w 302.3 kJ/min = = 9.04 kg/min c p ∆T (4.18 kJ/kg ⋅ °C)(8°C)
(c) The exit velocity is determined from the conservation of mass of dry air, V A V A V& V& ⎯→ 1 = 2 ⎯ ⎯→ 1 = 2 m& a1 = m& a 2 ⎯
v1
V2 =
v2
v1
v2
v2 0.851 V1 = (120 m/min) = 113 m/min v1 0.904
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-29
14-81 EES Problem 14-80 is reconsidered. A general solution of the problem in which the input variables may be supplied and parametric studies performed is to be developed and the process is to be shown in the psychrometric chart for each set of input variables. Analysis The problem is solved using EES, and the solution is given below. "Input Data from the Diagram Window" {D=0.3 P[1] =101.32 [kPa] T[1] = 35 [C] RH[1] = 60/100 "%, relative humidity" Vel[1] = 120/60 "[m/s]" DELTAT_cw =8 [C] P[2] = 101.32 [kPa] T[2] = 20 [C]} RH[2] = 100/100 "%"
0.050 0.045
"Dry air flow rate, m_dot_a, is constant" Vol_dot[1]= (pi * D^2)/4*Vel[1] v[1]=VOLUME(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_a = Vol_dot[1]/v[1] oi t a "Exit vleocity" R y Vol_dot[2]= (pi * D^2)/4*Vel[2] t v[2]=VOLUME(AirH2O,T=T[2],P=P[2],R=RH[2])dii m_dot_a = Vol_dot[2]/v[2] m u H "Mass flow rate of the condensed water" m_dot_v[1]=m_dot_v[2]+m_dot_w w[1]=HUMRAT(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_v[1] = m_dot_a*w[1] w[2]=HUMRAT(AirH2O,T=T[2],P=P[2],R=RH[2]) m_dot_v[2] = m_dot_a*w[2]
Pressure = 101.0 [kPa]
0.040 0.035
0.8
0.030
30 C
0.025
0.6
0.020 20 C
0.015 0.010
10 C
0.005 0.000 -10
0.4
0.2
0C
-5
-0
5
10
15
20
25
30
35
40
T [C]
"SSSF conservation of energy for the air" m_dot_a *(h[1] + (1+w[1])*Vel[1]^2/2*Convert(m^2/s^2, kJ/kg)) + Q_dot = m_dot_a*(h[2] +(1+w[2])*Vel[2]^2/2*Convert(m^2/s^2, kJ/kg)) +m_dot_w*h_liq_2 h[1]=ENTHALPY(AirH2O,T=T[1],P=P[1],w=w[1]) h[2]=ENTHALPY(AirH2O,T=T[2],P=P[2],w=w[2]) h_liq_2=ENTHALPY(Water,T=T[2],P=P[2]) "SSSF conservation of energy for the cooling water" -Q_dot =m_dot_cw*Cp_cw*DELTAT_cw "Note: Q_netwater=-Q_netair" Cp_cw = SpecHeat(water,T=10,P=P[2])"kJ/kg-K"
RH1
ma
mw
mcw
Q [kW]
0.5 0.6 0.7 0.8 0.9
0.1574 0.1565 0.1556 0.1547 0.1538
0.0004834 0.001056 0.001629 0.002201 0.002774
0.1085 0.1505 0.1926 0.2346 0.2766
-3.632 -5.039 -6.445 -7.852 -9.258
Vel1 [m/s] 2 2 2 2 2
Vel2 [m/s] 1.894 1.883 1.872 1.861 1.85
T1 [C] 35 35 35 35 35
T2 [C] 20 20 20 20 20
w1
w2
0.01777 0.02144 0.02516 0.02892 0.03273
0.0147 0.0147 0.0147 0.0147 0.0147
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-30
14-82 Air is cooled by passing it over a cooling coil. The rate of heat transfer, the mass flow rate of water, and the exit velocity of airstream are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The dew point temperature of the incoming air stream at 35°C is
Water
Pv1 = φ1 Pg1 = φ1 Psat @ 35°C = (0.6)(5.6291 kPa) = 3.38 kPa
T + 8°C
Cooling coils
Tdp = Tsat @ Pv = Tsat @ 3.38 kPa = 25.9°C
Since air is cooled to 20°C, which is below its dew point temperature, some of the moisture in the air will condense.
1
35°C 60% 120 m/min
AIR
20°C 2 Saturated
The amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 95 kPa. Then the properties of the air at both states are determined to be Pa1 = P1 − Pv1 = 95 − 3.38 = 91.62 kPa
v1 =
Ra T1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(308 K) = = 0.965 m 3 / kg dry air 91.62 kPa Pa1
ω1 =
0.622 Pv1 0.622(3.38 kPa) = = 0.0229 kg H 2 O/kg dry air (95 − 3.38) kPa P1 − Pv1
h1 = c p T1 + ω1hg1 = (1.005 kJ/kg ⋅ °C)(35°C) + (0.0229)(2564.6 kJ/kg) = 93.90 kJ/kg dry air
and Pv 2 = φ 2 Pg 2 = (1.00) Psat @ 20°C = 2.3392 kPa
v2 =
Ra T2 (0.287 kPa ⋅ m 3 / kg ⋅ K)(293 K) = = 0.908 m 3 / kg dry air (95 − 2.339) kPa Pa 2
ω2 =
0.622 Pv 2 0.622(2.339 kPa) = = 0.0157 kg H 2 O/kg dry air (95 − 2.339) kPa P2 − Pv 2
h2 = c p T2 + ω 2 hg 2 = (1.005 kJ/kg ⋅ °C)(20°C) + (0.0157)(2537.4 kJ/kg) = 59.95 kJ/kg dry air
Also, hw ≅ h f @ 20°C = 83.915 kJ/kg
(Table A-4)
Then,
V&1 = V1 A1 = V1 m& a1 =
π D2 4
⎛ π (0.3 m) 2 = (120 m/min)⎜ ⎜ 4 ⎝
⎞ ⎟ = 8.48 m 3 / min ⎟ ⎠
V&1 8.48 m 3 / min = = 8.79 kg/min v 1 0.965 m 3 / kg dry air
Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section (excluding the water),
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-31
Water Mass Balance: ∑ m& w,i = ∑ m& w ,e
⎯ ⎯→
m& a1ω 1 = m& a 2ω 2 + m& w
m& w = m& a (ω 1 − ω 2 ) = (8.79 kg / min)(0.0229 − 0.0157) = 0.0633 kg / min
Energy Balance: E& in − E& out = ∆E& systemÊ0 (steady) = 0 E& in = E& out ∑ m& i hi = ∑ m& e he + Q& out → Q& out = m& a1h1 − (m& a 2 h2 + m& w hw ) = m& a (h1 − h2 ) − m& w hw Q& out = (8.79 kg/min)(93.90 − 59.94)kJ/kg − (0.0633 kg/min)(83.915 kJ/kg) = 293.2 kJ/min
(b) Noting that the heat lost by the air is gained by the cooling water, the mass flow rate of the cooling water is determined from Q& c ∆T = m& ∆h = m& cooling water
m& cooling water
cooling water
cooling water p
Q& w 293.2 kJ/min = = = 8.77 kg/min c p ∆T (4.18 kJ/kg ⋅ °C)(8°C)
(c) The exit velocity is determined from the conservation of mass of dry air, V& V& V A V A m& a1 = m& a 2 ⎯ ⎯→ 1 = 2 ⎯ ⎯→ 1 = 2
v1
V2 =
v2
v1
v2
v2 0.908 V1 = (120 m/min) = 113 m/min 0.965 v1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-45
Adiabatic Mixing of Airstreams 14-100C This will occur when the straight line connecting the states of the two streams on the psychrometric chart crosses the saturation line. 14-101C Yes.
14-102 Two airstreams are mixed steadily. The specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture are to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Properties Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31) to be h1 = 62.7 kJ/kg dry air
h2 = 31.9 kJ/kg dry air
ω1 = 0.0119 kg H 2 O/kg dry air and ω 2 = 0.0079 kg H 2 O/kg dry air v 1 = 0.882 m 3 /kg dry air
v 2 = 0.819 m 3 /kg dry air
1
Analysis The mass flow rate of dry air in each stream is m& a1 =
V&1 20 m 3 / min = = 22.7 kg/min v 1 0.882 m 3 / kg dry air
m& a 2 =
V&2 25 m 3 / min = = 30.5 kg/min v 2 0.819 m 3 / kg dry air
From the conservation of mass, & a3 = m & a1 + m & a 2 = ( 22.7 + 30.5) kg / min = 53.2 kg / min m
32°C 40% 20 m3/min P = 1 atm AIR
2
ω3 φ3 3 T3
25 m3/min 12°C 90%
The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: & a1 ω 2 − ω 3 h2 − h3 m = = & ma 2 ω 3 − ω 1 h3 − h1 . − h3 22.7 0.0079 − ω 3 319 = = 30.5 ω 3 − 0.0119 h3 − 62.7
which yields,
ω 3 = 0.0096 kg H 2O / kg dry air h3 = 45.0 kJ / kg dry air
These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: T3 = 20.6°C
φ3 = 63.4% v 3 = 0.845 m 3 /kg dry air Finally, the volume flow rate of the mixture is determined from
V&3 = m& a 3v 3 = (53.2 kg/min)(0.845 m 3 / kg) = 45.0 m 3 /min
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-46
14-103 Two airstreams are mixed steadily. The specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture are to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Analysis The properties of each inlet stream are determined to be Pv1 = φ1Pg1 = φ1Psat @ 32°C = (0.40)(4.760 kPa) = 1.90 kPa Pa1 = P1 − Pv1 = 90 − 1.90 = 88.10 kPa
1
(0.287 kPa ⋅ m / kg ⋅ K)(305 K) RT v1 = a 1 = = 0.994 m3 / kg dry air 88.10 kPa Pa1 3
ω1 =
32°C 40% 20 m3/min P = 90 kPa AIR
0.622 Pv1 0.622(1.90 kPa) = = 0.0134 kg H 2O/kg dry air (90 − 1.90) kPa P1 − Pv1
and
T3
3
h1 = c pT1 + ω1hg1 = (1.005 kJ/kg ⋅ °C)(32°C) + (0.0134)(2559.2 kJ/kg) = 66.45 kJ/kg dry air
ω3 φ3 3
2
25 m /min 12°C 90%
Pv 2 = φ 2 Pg 2 = φ 2 Psat@12°C = (0.90)(1.403 kPa) = 1.26 kPa Pa 2 = P2 − Pv 2 = 90 − 1.26 = 88.74 kPa
v2 =
R a T2 (0.287 kPa ⋅ m 3 / kg ⋅ K)(285 K) = = 0.922 m 3 / kg dry air Pa 2 88.74 kPa
ω2 =
0.622 Pv 2 0.622(1.26 kPa) = = 0.00883 kg H 2 O/kg dry air P2 − Pv 2 (90 − 1.26) kPa
h2 = c p T2 + ω 2 h g 2 = (1.005 kJ/kg ⋅ °C)(12°C) + (0.00883)(2522.9 kJ/kg) = 34.34 kJ/kg dry air
Then the mass flow rate of dry air in each stream is V& V&2 25 m 3 / min 20 m 3 / min & m& a1 = 1 = = m = = = 27.11 kg/min 20 . 12 kg/min 2 a v1 0.994 m 3 / kg dry air v 2 0.922 m3 / kg dry air From the conservation of mass, m& a 3 = m& a1 + m& a 2 = (20.12 + 27.11) kg/min = 47.23 kg/min The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: m& a1 ω2 − ω3 h2 − h3 20.12 0.00883 − ω3 34.34 − h3 = = ⎯ ⎯→ = = m& a 2 ω3 − ω1 h3 − h1 h3 − 66.45 27.11 ω3 − 0.0134 which yields ω3 = 0.0108 kg H 2O/kg dry air h3 = 48.02 kJ/kg dry air These two properties fix the state of the mixture. Other properties are determined from h3 = c pT3 + ω3hg 3 ≅ c pT3 + ω3 (2501.3 + 1.82T3 ) 48.02 kJ/kg = (1.005 kJ/kg ⋅ °C)T3 + (0.0108)(2500.9 + 1.82T3 ) kJ/kg ⎯⎯→ T3 = 20.5°C
ω3 =
0.622 Pv 3 0.622 Pv 3 ⎯ ⎯→ 0.0108 = ⎯ ⎯→ Pv3 = 1.54 kPa 90 − Pv3 P3 − Pv3
φ3 =
Pv 3 Pv 3 1.54 kPa = = = 0.639 or 63.9% Pg 3 Psat @ T3 2.41 kPa
Finally, Pa 3 = P3 − Pv 3 = 90 − 1.54 = 88.46 kPa
v3 =
Ra T3 (0.287 kPa ⋅ m 3 / kg ⋅ K)(293.5 K) = = 0.952 m 3 /kg dry air Pa 3 88.46 kPa
V&3 = m& a 3v 3 = (47.23 kg/min)(0.952 m 3 / kg) = 45.0 m 3 /min
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-47
14-104E Two airstreams are mixed steadily. The temperature, the specific humidity, and the relative humidity of the mixture are to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Properties The properties of each inlet stream are determined from the psychrometric chart (Fig. A-31E) to be h1 = 19.9 Btu/lbm dry air
ω1 = 0.0039 lbm H 2 O/lbm dry air
1
v 1 = 13.30 ft /lbm dry air 3
and
65°F 30% 900 ft3/min P = 1 atm AIR
h2 = 41.1 Btu/lbm dry air
ω 2 = 0.0200 lbm H 2 O/lbm dry air
v 2 = 14.04 ft 3 /lbm dry air Analysis The mass flow rate of dry air in each stream is V& 900 ft 3 / min m& a1 = 1 = = 67.7 lbm/min v 1 13.30 ft 3 / lbm dry air V& 300 ft 3 / min m& a 2 = 2 = = 21.4 lbm/min v 2 14.04 ft 3 / lbm dry air
2
ω3 φ3 3 T3
300 ft3/min 80°C 90%
The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: & a1 ω 2 − ω 3 h2 − h3 m = = & a 2 ω 3 − ω 1 h3 − h1 m 67.7 0.0200 − ω 3 411 . − h3 = = ω 3 − 0.0039 h3 − 19.9 214 .
which yields, (a)
ω 3 = 0.0078 lbm H 2O / lbm dry air h3 = 25.0 Btu / lbm dry air
These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: (b)
T3 = 68.7°F
(c)
φ 3 = 52.1%
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-48
14-105E EES Problem 14-104E is reconsidered. A general solution of the problem in which the input variables may be supplied and parametric studies performed is to be developed and the process is to be shown in the psychrometric chart for each set of input variables. Analysis The problem is solved using EES, and the solution is given below. "Input Data by Diagram Window:" {P=14.696 [psia] Tdb[1] =65 [F] Rh[1] = 0.30 V_dot[1] = 900 [ft^3/min] Tdb[2] =80 [F] Rh[2] = 0.90 V_dot[2] = 300 [ft^3/min]} P[1]=P P[2]=P[1] P[3]=P[1] "Energy balance for the steady-flow mixing process:" "We neglect the PE of the flow. Since we don't know the cross sectional area of the flow streams, we also neglect theKE of the flow." E_dot_in - E_dot_out = DELTAE_dot_sys DELTAE_dot_sys = 0 [kW] E_dot_in = m_dot[1]*h[1]+m_dot[2]*h[2] E_dot_out = m_dot[3]*h[3] "Conservation of mass of dry air during mixing:" m_dot[1]+m_dot[2] = m_dot[3] "Conservation of mass of water vapor during mixing:" m_dot[1]*w[1]+m_dot[2]*w[2] = m_dot[3]*w[3] m_dot[1]=V_dot[1]/v[1]*convert(1/min,1/s) m_dot[2]=V_dot[2]/v[2]*convert(1/min,1/s) h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) v[2]=VOLUME(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) w[2]=HUMRAT(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) Tdb[3]=TEMPERATURE(AirH2O,h=h[3],P=P[3],w=w[3]) Rh[3]=RELHUM(AirH2O,T=Tdb[3],P=P[3],w=w[3]) v[3]=VOLUME(AirH2O,T=Tdb[3],P=P[3],w=w[3]) m_dot[3]=V_dot[3]/v[3]*convert(1/min,1/s)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-49
AirH2O
0.045 Pressure = 14.7 [psia]
0.040 0.035 90°F
0.030
oi t a R y ti di m u H
0.8
0.025
80°F 0.6
2
0.020 70°F
0.015
0.4 60°F
0.010
50°F
3
40°F
0.005 0.000 30
0.2
1 44
58
72
86
100
T [°F]
SOLUTION DELTAE_dot_sys=0 E_dot_out=37.04 [kW] h[2]=41.09 [Btu/lb_m] m_dot[1]=1.127 [kga/s] m_dot[3]=1.483 [kga/s] P[1]=14.7 [psia] P[3]=14.7 [psia] Rh[2]=0.9 Tdb[1]=65 [F] Tdb[3]=68.68 [F] v[2]=14.04 [ft^3/lb_ma] V_dot[1]=900 [ft^3/min] V_dot[3]=1200 [ft^3/min] w[2]=0.01995 [lb_mv/lb_ma]
E_dot_in=37.04 [kW] h[1]=19.88 [Btu/lb_m] h[3]=24.97 [Btu/lb_m] m_dot[2]=0.3561 [kga/s] P=14.7 [psia] P[2]=14.7 [psia] Rh[1]=0.3 Rh[3]=0.5214 Tdb[2]=80 [F] v[1]=13.31 [ft^3/lb_ma] v[3]=13.49 [ft^3/lb_ma] V_dot[2]=300 [ft^3/min] w[1]=0.003907 [lb_mv/lb_ma] w[3]=0.007759 [lb_mv/lb_ma]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-50
14-106 A stream of warm air is mixed with a stream of saturated cool air. The temperature, the specific humidity, and the relative humidity of the mixture are to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Properties The properties of each inlet stream are determined from the psychrometric chart (Fig. A-31) to be h1 = 110.2 kJ/kg dry air
ω1 = 0.0272 kg H 2 O/kg dry air and h2 = 50.9 kJ/kg dry air
1
ω 2 = 0.0129 kg H 2 O/kg dry air Analysis The specific humidity and the enthalpy of the mixture can be determined from Eqs. 14-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: m& a1 ω 2 − ω 3 h2 − h3 = = m& a 2 ω 3 − ω1 h3 − h1
40°C 8 kg/s Twb1 = 32°C P = 1 atm AIR
2
ω3 φ3 3 T3
6 kg/s 18°C 100%
50.9 − h3 8.0 0.0129 − ω 3 = = 6.0 ω 3 − 0.0272 h3 − 110.2
which yields, (b)
ω 3 = 0.0211 kg H 2O / kg dry air h3 = 84.8 kJ / kg dry air
These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: (a)
T3 = 30.7° C
(c)
φ 3 = 75.1%
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-51
14-107 EES Problem 14-106 is reconsidered. The effect of the mass flow rate of saturated cool air stream on the mixture temperature, specific humidity, and relative humidity is to be investigated. Analysis The problem is solved using EES, and the solution is given below. P=101.325 [kPa] Tdb[1] =40 [C] Twb[1] =32 [C] m_dot[1] = 8 [kg/s] Tdb[2] =18 [C] Rh[2] = 1.0 m_dot[2] = 6 [kg/s] P[1]=P P[2]=P[1] P[3]=P[1] "Energy balance for the steady-flow mixing process:" "We neglect the PE of the flow. Since we don't know the cross sectional area of the flow streams, we also neglect theKE of the flow." E_dot_in - E_dot_out = DELTAE_dot_sys DELTAE_dot_sys = 0 [kW] E_dot_in = m_dot[1]*h[1]+m_dot[2]*h[2] E_dot_out = m_dot[3]*h[3] "Conservation of mass of dry air during mixing:" m_dot[1]+m_dot[2] = m_dot[3] "Conservation of mass of water vapor during mixing:" m_dot[1]*w[1]+m_dot[2]*w[2] = m_dot[3]*w[3] m_dot[1]=V_dot[1]/v[1]*convert(1/min,1/s) m_dot[2]=V_dot[2]/v[2]*convert(1/min,1/s) h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],B=Twb[1]) Rh[1]=RELHUM(AirH2O,T=Tdb[1],P=P[1],B=Twb[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) v[2]=VOLUME(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) w[2]=HUMRAT(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) Tdb[3]=TEMPERATURE(AirH2O,h=h[3],P=P[3],w=w[3]) Rh[3]=RELHUM(AirH2O,T=Tdb[3],P=P[3],w=w[3]) v[3]=VOLUME(AirH2O,T=Tdb[3],P=P[3],w=w[3]) Twb[2]=WETBULB(AirH2O,T=Tdb[2],P=P[2],R=RH[2]) Twb[3]=WETBULB(AirH2O,T=Tdb[3],P=P[3],R=RH[3]) m_dot[3]=V_dot[3]/v[3]*convert(1/min,1/s) m2 [kga/s] 0 2 4 6 8 10 12 14 16
Tdb3 [C] 40 35.69 32.79 30.7 29.13 27.91 26.93 26.13 25.45
Rh3 0.5743 0.6524 0.7088 0.751 0.7834 0.8089 0.8294 0.8462 0.8601
w3 [kgw/kga] 0.02717 0.02433 0.02243 0.02107 0.02005 0.01926 0.01863 0.01811 0.01768
40 38 36
] C [ ] 3[ b d T
34 32 30 28 26 24 0
2
4
6
8
10
12
14
m[2] [kga/s]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16
14-52
0.9 0.85
Rh[3]
0.8 0.75 0.7 0.65 0.6 0.55 0
2
4
6
8
10
12
14
16
m[2] [kga/s] 0.028
w [3] [kgw /kga]
0.026 0.024 0.022 0.02 0.018 0.016 0
2
4
6
8
10
12
14
16
m [2] [kga/s]
Wet Cooling Towers 14-108C The working principle of a natural draft cooling tower is based on buoyancy. The air in the tower has a high moisture content, and thus is lighter than the outside air. This light moist air rises under the influence of buoyancy, inducing flow through the tower. 14-109C A spray pond cools the warm water by spraying it into the open atmosphere. They require 25 to 50 times the area of a wet cooling tower for the same cooling load.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-53
14-110 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis (a) The mass flow rate of dry air through the tower remains constant (m& a1 = m& a 2 = m& a ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields Dry Air Mass Balance: ∑ m& a ,i = ∑ m& a ,e
⎯ ⎯→
m& a1 = m& a 2 = m& a
Water Mass Balance: & w ,i = ∑ m & w,e ∑m
⎯ ⎯→
2
&3 + m & a1ω 1 = m &4 + m & a 2ω 2 m
32°C 100%
&3 − m &4 = m & a (ω 2 − ω 1 ) = m & makeup m
Energy Balance: E& − E& = ∆E& in
out
Ê0 (steady) system
=0
E& in = E& out
3
WATER 40°C 90 kg/s
∑ m& i hi = ∑ m& e he since Q& = W& = 0 0 = ∑ m& e he − ∑ m& i hi 0 = m& a 2 h2 + m& 4 h4 − m& a1 h1 − m& 3 h3
System boundary 1
0 = m& a (h2 − h1 ) + (m& 3 − m& makeup )h4 − m& 3 h3 4
Solving for m& a , m& a =
m& 3 (h3 − h4 ) (h2 − h1 ) − (ω 2 − ω 1 )h4
25°C
From the psychrometric chart (Fig. A-31), h1 = 49.9 kJ/kg dry air
AIR
1 atm 23°C 60%
Makeup
ω1 = 0.0105 kg H 2 O/kg dry air v 1 = 0.853 m 3 /kg dry air and h2 = 110.7 kJ/kg dry air
ω 2 = 0.0307 kg H 2 O/kg dry air From Table A-4, h3 ≅ h f @ 40°C = 167.53 kJ/kg H 2 O h4 ≅ h f @ 25°C = 104.83 kJ/kg H 2 O
Substituting, m& a =
(90 kg/s)(167.53 − 104.83)kJ/kg = 96.2 kg/s (110.7 − 49.9) kJ/kg − (0.0307 − 0.0105)(104.83) kJ/kg
Then the volume flow rate of air into the cooling tower becomes V& = m& v = (96.2 kg/s)(0.854 m 3 / kg ) = 82.2 m 3 /s 1
a 1
(b) The mass flow rate of the required makeup water is determined from m& makeup = m& a (ω 2 − ω1 ) = (96.2 kg/s)(0.0307 − 0.0105) = 1.94 kg/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-54
14-111E Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis (a) The mass flow rate of dry air through the tower remains constant (m& a1 = m& a 2 = m& a ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass balance and the energy balance equations yields Dry Air Mass Balance: ∑ m& a ,i = ∑ m& a ,e
⎯ ⎯→
m& a1 = m& a 2 = m& a
Water Mass Balance: & w ,i = ∑ m & w,e ∑m
2
⎯ ⎯→
&3 + m & a1ω 1 = m &4 + m & a 2ω 2 m
95°F 100%
&3 − m &4 = m & a (ω 2 − ω 1 ) = m & makeup m
Energy Balance: E& − E& = ∆E& in
out
system
Ê0 (steady)
3
WATER
=0
E& in = E& out & i hi = ∑ m & e he (since Q& = W& = 0) ∑m & e he − ∑ m & i hi 0= ∑m
110°F 100 lbm/s System boundar
& a 2 h2 + m & 4 h4 − m & a1h1 − m & 3h3 0=m & a ( h2 − h1 ) + ( m &3 − m & makeup )h4 − m & 3h3 0=m
1 AIR 4
Solving for m& a , m& a =
m& 3 (h3 − h4 ) (h2 − h1 ) − (ω 2 − ω 1 )h4
80°F
1 atm 76°F 60%
Makeup
From the psychrometric chart (Fig. A-31), h1 = 30.9 Btu/lbm dry air
ω1 = 0.0115 lbm H 2 O/lbm dry air
v 1 = 13.76 ft 3 /lbm dry air and h2 = 63.2 Btu / lbm dry air
ω 2 = 0.0366 lbm H 2 O / lbm dry air From Table A-4E, h3 ≅ h f @ 110° F = 78.02 Btu/lbm H 2 O h4 ≅ h f @ 80° F = 48.07 Btu/lbm H 2 O
Substituting, m& a =
(100 lbm/s)(78.02 − 48.07)Btu/lbm = 96.3 lbm/s (63.2 − 30.9) Btu/lbm − (0.0366 − 0.0115)(48.07) Btu/lbm
Then the volume flow rate of air into the cooling tower becomes V& = m& v = (96.3 lbm/s)(13.76 ft 3 /lbm) = 1325 ft 3 /s 1
a 1
(b) The mass flow rate of the required makeup water is determined from & makeup = m & a (ω 2 − ω 1 ) = (96.3 lbm / s)(0.0366 − 0.0115) = 2.42 lbm / s m
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-55
14-112 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis (a) The mass flow rate of dry air through the tower remains constant (m& a1 = m& a 2 = m& a ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields Dry Air Mass Balance: ∑ m& a ,i = ∑ m& a ,e
⎯ ⎯→
m& a1 = m& a 2 = m& a
AIR 34°C 2 EXIT 90%
Water Mass Balance: ∑ m& w,i = ∑ m& w,e → m& 3 + m& a1ω1 = m& 4 + m& a 2 ω 2 m& 3 − m& 4 = m& a (ω 2 − ω1 ) = m& makeup
Energy Balance: E& − E& = ∆E& in
out
system
Ê0 (steady)
=0
E& in = E& out
WARM WATER
3
40°C 60 kg/s
& i hi = ∑ m & e he (since Q& = W& = 0) ∑m & e he − ∑ m & i hi 0= ∑m & a 2 h2 + m & 4 h4 − m & a1h1 − m & 3h3 0=m & a ( h2 − h1 ) + ( m &3 − m & makeup )h4 − m & 3h3 0=m
4
Solving for m& a , m& a =
m& 3 (h3 − h4 ) (h2 − h1 ) − (ω 2 − ω 1 )h4
COOL WATER
From the psychrometric chart (Fig. A-31), h1 = 44.7 kJ/kg dry air
1 AIR INLET 1 atm Tdb = 22°C Twb = 16°C
26°C Makeup water
ω1 = 0.0089 kg H 2 O/kg dry air
v 1 = 0.849 m 3 /kg dry air and h2 = 1135 . kJ / kg dry air
ω 2 = 0.0309 kg H 2 O / kg dry air From Table A-4, h3 ≅ h f @ 40°C = 167.53 kJ/kg H 2 O h4 ≅ h f @ 26°C = 109.01 kJ/kg H 2 O
Substituting, m& a =
(60 kg/s)(167.53 − 109.01)kJ/kg = 52.9 kg/s (113.5 − 44.7) kJ/kg − (0.0309 − 0.0089)(109.01) kJ/kg
Then the volume flow rate of air into the cooling tower becomes V& = m& v = (52.9 kg/s)(0.849 m 3 / kg ) = 44.9 m 3 /s 1
a 1
(b) The mass flow rate of the required makeup water is determined from & makeup = m & a (ω 2 − ω 1 ) = (52.9 kg / s)(0.0309 − 0.0089) = 1.16 kg / s m
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-56
14-113 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis (a) The mass flow rate of dry air through the tower remains constant (m& a1 = m& a 2 = m& a ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields Dry Air Mass Balance: 35°C 2 ∑ m& a ,i = ∑ m& a ,e ⎯ ⎯→ m& a1 = m& a 2 = m& a 100%
Water Mass Balance: ∑ m& w,i = ∑ m& w,e ⎯ ⎯→ m& 3 + m& a1ω1 = m& 4 + m& a 2ω2 m& 3 − m& 4 = m& a (ω2 − ω1 ) = m& makeup
Energy Balance: E& − E& = ∆E& in
out
Ê0 (steady) system
= 0 ⎯⎯→ E& in = E& out
∑ m& i hi = ∑ m& e he (since Q& = W& = 0) 0 = ∑ m& e he − ∑ m& i hi 0 = m& a 2 h2 + m& 4 h4 − m& a1h1 − m& 3h3
3
WATER 40°C 25 kg/s System boundary
0 = m& a (h2 − h1 ) + (m& 3 − m& makeup )h4 − m& 3h3 m& 3 (h3 − h4 ) (h2 − h1 ) − (ω 2 − ω 1 )h4 The properties of air at the inlet and the exit are Pv1 = φ1Pg1 = φ1Psat @ 20°C = (0.70)(2.3392 kPa) = 1.637 kPa
1 4
m& a =
Pa1 = P1 − Pv1 = 96 − 1.637 = 94.363 kPa
30°C
96 kPa 20°C 70%
Makeup
v1 =
RaT1 (0.287 kPa ⋅ m3 / kg ⋅ K)(293 K) = = 0.891 m 3 / kg dry air Pa1 94.363 kPa
ω1 =
0.622 Pv1 0.622(1.637 kPa) = = 0.0108 kg H 2O/kg dry air P1 − Pv1 (96 − 1.637) kPa
h1 = c pT1 + ω1hg1 = (1.005 kJ/kg ⋅ °C)(20°C) + (0.0108)(2537.4 kJ/kg) = 47.5 kJ/kg dry air
and
Pv 2 = φ2 Pg 2 = φ2 Psat @ 35°C = (1.00)(5.6291 kPa) = 5.6291 kPa
ω2 =
0.622 Pv 2 0.622(5.6291 kPa) = = 0.0387 kg H 2O/kg dry air P2 − Pv 2 (96 − 5.6291) kPa
h2 = c pT2 + ω2 hg 2 = (1.005 kJ/kg ⋅ °C)(35°C) + (0.0387)(2564.6 kJ/kg) = 134.4 kJ/kg dry air
From Table A-4, h3 ≅ h f @ 40°C = 167.53 kJ/kg H 2 O h4 ≅ h f @ 30°C = 125.74 kJ/kg H 2 O (25 kg/s)(167.53 − 125.74)kJ/kg = 12.53 kg/s (134.4 − 47.5) kJ/kg − (0.0387 − 0.0108)(125.74) kJ/kg Then the volume flow rate of air into the cooling tower becomes V&1 = m& av 1 = (12.53 kg/s)(0.891 m 3 / kg) = 11.2 m 3 /s (b) The mass flow rate of the required makeup water is determined from m& makeup = m& a (ω 2 − ω1 ) = (12.53 kg/s)(0.0387 − 0.0108) = 0.35 kg/s
Substituting,
AIR
m& a =
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-57
14-114 A natural-draft cooling tower is used to remove waste heat from the cooling water flowing through the condenser of a steam power plant. The mass flow rate of the cooling water, the volume flow rate of air into the cooling tower, and the mass flow rate of the required makeup water are to be determined. Assumptions 1 All processes are steady-flow and the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet and exit states of the moist air for the tower are completely specified. The properties may be determined from the psychrometric chart (Fig. A-31) or using EES psychrometric functions to be (we used EES) h1 = 50.74 kJ/kg dry air
ω1 = 0.01085 kg H 2 O/kg dry air v 1 = 0.8536 m 3 /kg dry air h2 = 142.83 kJ/kg dry air
ω 2 = 0.04112 kg H 2 O/kg dry air
T2 = 37°C φ 2 =100%
The enthalpies of cooling water at the inlet and exit of the condenser are (Table A-4) hw3 = h f@ 40°C = 167.53 kJ/kg
T1 = 23°C Twb1 = 18°C
AIR 2
1 Makeup water
hw 4 = h f@ 26°C = 109.01 kJ/kg
The steam properties for the condenser are (Steam tables) Ps1 = 200 kPa ⎫ ⎬h s1 = 504.71 kJ/kg x s1 = 0 ⎭ Ps 2 = 10 kPa
⎫ ⎬hs 2 = 2524.3 kJ/kg s s 2 = 7.962 kJ/kg.K ⎭ Ps 3 = 10 kPa ⎫ ⎬h s 3 = 191.81 kJ/kg x s1 = 0 ⎭ The mass flow rate of dry air is given by V& V&1 m& a = 1 = v 1 0.8536 m 3 /kg The mass flow rates of vapor at the inlet and exit of the cooling tower are V&1 m& v1 = ω1m& a = (0.01085) = 0.01271V&1 0.8536 V&1 m& v 2 = ω 2 m& a = (0.04112) = 0.04817V&1 0.8536 Mass and energy balances on the cooling tower give m& v1 + m& cw3 = m& v 2 + m& cw4 m& a h1 + m& cw3 hw3 = m& a h2 + m& cw4 h w4 The mass flow rate of the makeup water is determined from m& makeup = m& v 2 − m& v1 = m& cw3 − m& cw4
An energy balance on the condenser gives 0.18m& s h s1 + 0.82m& s hs 2 + m& cw4 h w4 + m& makeup hw 4 = m& s h s 3 + m& cw3 h w3 Solving all the above equations simultaneously with known and determined values using EES, we obtain m& cw3 = 1413 kg/s V& = 47,700 m 3 /min 1
m& makeup = 28.19 kg/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-58
Review Problems
14-115 Air is compressed by a compressor and then cooled to the ambient temperature at high pressure. It is to be determined if there will be any condensation in the compressed air lines. Assumptions The air and the water vapor are ideal gases. Properties The saturation pressure of water at 20°C is 2.3392 kPa (Table A-4).. Analysis The vapor pressure of air before compression is Pv1 = φ1 Pg = φ1 Psat @ 25°C = (0.50)(2.3392 kPa) = 1.17 kPa
The pressure ratio during the compression process is (800 kPa)/(92 kPa) = 8.70. That is, the pressure of air and any of its components increases by 8.70 times. Then the vapor pressure of air after compression becomes Pv 2 = Pv1 × (Pressure ratio) = (1.17 kPa)(8.70) = 10.2 kPa
The dew-point temperature of the air at this vapor pressure is Tdp = Tsat @ Pv 2 = Tsat @ 10.2 kPa = 46.1°C
which is greater than 20°C. Therefore, part of the moisture in the compressed air will condense when air is cooled to 20°C.
14-116E The error involved in assuming the density of air to remain constant during a humidification process is to be determined. Properties The density of moist air before and after the humidification process is determined from the psychrometric chart (Fig. A-31E) to be T1 = 80°F⎫ = 0.0729 lbm/ft 3 ⎬ ρ φ1 = 25% ⎭ air ,1 T1 = 80°F⎫ = 0.0716 lbm/ft 3 ⎬ρ φ1 = 75% ⎭ air , 2
Analysis The error involved as a result of assuming constant air density is then determined to be %Error =
∆ρ air
ρ air,1
× 100 =
(0.0729 − 0.0716) lbm/ft 3 0.0729 lbm/ft 3
× 100 = 1.7%
which is acceptable for most engineering purposes.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-59
14-117 Dry air flows over a water body at constant pressure and temperature until it is saturated. The molar analysis of the saturated air and the density of air before and after the process are to be determined. Assumptions The air and the water vapor are ideal gases. Properties The molar masses of N2, O2, Ar, and H2O are 28.0, 32.0, 39.9 and 18 kg / kmol, respectively (Table A-1). The molar analysis of dry air is given to be 78.1 percent N2, 20.9 percent O2, and 1 percent Ar. The saturation pressure of water at 25°C is 3.1698 kPa (Table A-4). Also, 1 atm = 101.325 kPa. Analysis (a) Noting that the total pressure remains constant at 101.32 kPa during this process, the partial pressure of air becomes P = Pair + Pvapor → Pair = P − Pvapor = 101.325 − 3.1698 = 98.155 kPa
Then the molar analysis of the saturated air becomes PH 2O
3.1698 = 0.0313 101.325 PN y N 2 ,dry Pdry air 0.781(98.155 kPa) y N2 = 2 = = = 0.7566 P P 101.325 PO y O ,dry Pdry air 0.209(98.155 kPa) y O2 = 2 = 2 = = 0.2025 P P 101.325 y Ar,dry Pdry air 0.01(98.155 kPa) P y Ar = Ar = = = 0.0097 P P 101.325 y H 2O =
P
=
Air 1 atm 25°C
Lake
(b) The molar masses of dry and saturated air are
∑ y M = 0.781 × 28.0 + 0.209 × 32.0 + 0.01 × 39.9 = 29.0 kg / kmol = ∑ y M = 0.7566 × 28.0 + 0.2025 × 32.0 + 0.0097 × 39.9 + 0.0313 × 18 = 28.62 kg / kmol
M dry air =
i
i
M sat. air
i
i
Then the densities of dry and saturated air are determined from the ideal gas relation to be
ρ dry air =
101.325 kPa P = = 1.186 kg/m 3 ( Ru / M dry air )T [(8.314 kPa ⋅ m³/kmol ⋅ K ) / 29.0 kg/kmol](25 + 273)K
ρ sat. air =
101.325 kPa P = = 1.170 kg/m 3 ( Ru / M sat air )T [(8.314 kPa ⋅ m³/kmol ⋅ K ) / 28.62 kg/kmol](25 + 273)K
Discussion We conclude that the density of saturated air is less than that of the dry air, as expected. This is due to the molar mass of water being less than that of dry air.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-60
14-118E The mole fraction of the water vapor at the surface of a lake and the mole fraction of water in the lake are to be determined and compared. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 60°F is 0.2564 psia (Table A-4E). Henry’s constant for air dissolved in water at 60ºF (289 K) is given in Table 16-2 to be H = 62,000 bar. Analysis The air at the water surface will be saturated. Therefore, the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at 60°F, Pvapor = Psat @60° F = 0.2564 psia
Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the air at the surface of the lake is determined to be y vapor =
Pvapor P
=
Air 13.8 psi
0.2564 psia = 0.0186 (or 1.86 percent) 13.8 psia
60°F
The partial pressure of dry air just above the lake surface is Lake
Pdry air = P − Pvapor = 13.8 − 0.2564 = 13.54 psia
Then the mole fraction of air in the water becomes y dry air, liquid side =
Pdry air, gasside H
=
1354 . psia(1 atm / 14.696 psia ) = 151 . × 10 −5 62,000 bar (1 atm / 1.01325 bar)
which is very small, as expected. Therefore, the mole fraction of water in the lake near the surface is y water,liquid side = 1 − y dry air, liquid side = 1 − 1.51×10 −5 ≅ 1.0
Discussion The concentration of air in water just below the air-water interface is 1.51 moles per 100,000 moles. The amount of air dissolved in water will decrease with increasing depth.
14-119 The mole fraction of the water vapor at the surface of a lake at a specified temperature is to be determined. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air at the lake surface is saturated. Properties The saturation pressure of water at 18°C is 2.065 kPa (Table A-4). Analysis The air at the water surface will be saturated. Therefore, the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at 18°C, Pvapor = Psat @18°C = 2.065 kPa
Assuming both the air and vapor to be ideal gases, the partial pressure and mole fraction of dry air in the air at the surface of the lake are determined to be
Air 100 kPa
Pdry air = P − Pvapor = 100 − 2.065 = 97.94 kPa y dry air =
Pdry air P
=
18°C
Lake
97.94 kPa = 0.979 (or 97.9%) 100 kPa
Therefore, the mole fraction of dry air is 97.9 percent just above the air-water interface.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-61
14-120E A room is cooled adequately by a 7500 Btu/h air-conditioning unit. If the room is to be cooled by an evaporative cooler, the amount of water that needs to be supplied to the cooler is to be determined. Assumptions 1 The evaporative cooler removes heat at the same rate as the air conditioning unit. 2 Water evaporates at an average temperature of 70°F. Properties The enthalpy of vaporization of water at 70°F is 1053.7 Btu/lbm (Table A-4E). Analysis Noting that 1 lbm of water removes 1053.7 Btu of heat as it evaporates, the amount of water that needs to evaporate to remove heat at a rate of 7500 Btu/h is determined from Q& = m& water h fg to be m& water =
Q& 7500 Btu/h = = 7.12 lbm/h h fg 1053.7 Btu/lbm
14-121E The required size of an evaporative cooler in cfm (ft3/min) for an 8-ft high house is determined by multiplying the floor area of the house by 4. An equivalent rule is to be obtained in SI units. Analysis Noting that 1 ft = 0.3048 m and thus 1 ft2 = 0.0929 m2 and 1 ft3 = 0.0283 m3, and noting that a flow rate of 4 ft3/min is required per ft2 of floor area, the required flow rate in SI units per m2 of floor area is determined to 1 ft 2 ↔ 4 ft 3 / min 0.0929 m 2 ↔ 4 × 0.0283 m 3 / min 1 m 2 ↔ 1.22 m 3 / min
Therefore, a flow rate of 1.22 m3/min is required per m2 of floor area.
14-122 A cooling tower with a cooling capacity of 440 kW is claimed to evaporate 15,800 kg of water per day. It is to be determined if this is a reasonable claim. Assumptions 1 Water evaporates at an average temperature of 30°C. 2 The coefficient of performance of the air-conditioning unit is COP = 3. Properties The enthalpy of vaporization of water at 30°C is 2429.8 kJ/kg (Table A-4). Analysis Using the definition of COP, the electric power consumed by the air conditioning unit when running is W& in =
Q& cooling COP
=
440 kW = 146.7 kW 3
Then the rate of heat rejected at the cooling tower becomes Q& rejected = Q& cooling + W& in = 440 + 146.7 = 586.7 kW
Noting that 1 kg of water removes 2429.8 kJ of heat as it evaporates, the amount of water that needs to evaporate to remove heat at a rate of 586.7 kW is determined from Q& rejected = m& water h fg to be m& water =
Q& rejected h fg
=
586.7 kJ/s = 0.2415 kg/s = 869.3 kg/h = 20,860 kg/day 2429.8 kJ/kg
In practice, the air-conditioner will run intermittently rather than continuously at the rated power, and thus the water use will be less. Therefore, the claim amount of 15,800 kg per day is reasonable.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-62
14-123E It is estimated that 190,000 barrels of oil would be saved per day if the thermostat setting in residences in summer were raised by 6°F (3.3°C). The amount of money that would be saved per year is to be determined. Assumptions The average cooling season is given to be 120 days, and the cost of oil to be $20/barrel. Analysis The amount of money that would be saved per year is determined directly from (190,000 barrel / day)(120 days / year)($20 / barrel) = $456,000,000 Therefore, the proposed measure will save about half-a-billion dollars a year.
14-124E Wearing heavy long-sleeved sweaters and reducing the thermostat setting 1°F reduces the heating cost of a house by 4 percent at a particular location. The amount of money saved per year by lowering the thermostat setting by 4°F is to be determined. Assumptions The household is willing to wear heavy long-sleeved sweaters in the house, and the annual heating cost is given to be $600 a year. Analysis The amount of money that would be saved per year is determined directly from ($600 / year)(0.04/° F)(4° F) = $96 / year Therefore, the proposed measure will save the homeowner about $100 during a heating season..
14-125 Shading the condenser can reduce the air-conditioning costs by up to 10 percent. The amount of money shading can save a homeowner per year during its lifetime is to be determined. Assumptions It is given that the annual air-conditioning cost is $500 a year, and the life of the airconditioning system is 20 years. Analysis The amount of money that would be saved per year is determined directly from ($500 / year)(20 years)(0.10) = $1000 Therefore, the proposed measure will save about $1000 during the lifetime of the system.
14-126 A tank contains saturated air at a specified state. The mass of the dry air, the specific humidity, and the enthalpy of the air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The air is saturated, thus the partial pressure of water vapor is equal to the saturation pressure at the given temperature, Pv = Pg = Psat @ 25°C = 3.1698 kPa Pa = P − Pv = 97 − 3.1698 = 93.83 kPa
Treating air as an ideal gas, PaV (93.83 kPa)(3 m3 ) = = 3.29 kg RaT (0.287 kPa ⋅ m3 / kg ⋅ K)(298 K) (b) The specific humidity of air is determined from 0.622 Pv (0.622)(3.1698 kPa) ω= = = 0.0210 kg H 2 O/kg dry air (97 − 3.1698) kPa P − Pv ma =
3 m3 25°C 97 kPa
(c) The enthalpy of air per unit mass of dry air is determined from h = ha + ωhv ≅ c p T + ωh g = (1.005 kJ/kg ⋅ °C)(25°C) + (0.0210)(2546.5 kJ/kg) = 78.6 kJ/kg dry air
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-63
14-127 EES Problem 14-126 is reconsidered. The properties of the air at the initial state are to be determined and the effects of heating the air at constant volume until the pressure is 110 kPa is to be studied. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" Tdb[1] = 25 [C] P[1]=97 [kPa] Rh[1]=1.0 P[2]=110 [kPa] Vol = 3 [m^3] w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) m_a=Vol/v[1] h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],w=w[1]) "Energy Balance for the constant volume tank:" E_in - E_out = DELTAE_tank DELTAE_tank=m_a*(u[2] -u[1]) E_in = Q_in E_out = 0 [kJ] u[1]=INTENERGY(AirH2O,T=Tdb[1],P=P[1],w=w[1]) u[2]=INTENERGY(AirH2O,T=Tdb[2],P=P[2],w=w[2]) "The ideal gas mixture assumption applied to the constant volume process yields:" P[1]/(Tdb[1]+273)=P[2]/(Tdb[2]+273) "The mass of the water vapor and dry air are constant, thus:" w[2]=w[1] Rh[2]=RELHUM(AirH2O,T=Tdb[2],P=P[2],w=w[2]) h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],w=w[2]) v[2]=VOLUME(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) PROPERTIES AT THE INITIAL STATE h[1]=78.67 [kJ/kga] m_a=3.289 [kga] v[1]=0.9121 [m^3/kga] w[1]=0.02101 [kgw/kga] 100 P2 [kPa] 97 99 101 103 105 107 109 110
Qin [kJ] 0 15.12 30.23 45.34 60.45 75.55 90.65 98.2
75
] J k[
50
ni
Q
25
0 96
98
100
102
104
106
108
110
P[2] [kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-64
14-128E Air at a specified state and relative humidity flows through a circular duct. The dew-point temperature, the volume flow rate of air, and the mass flow rate of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. Analysis (a) The vapor pressure of air is AIR Pv = φPg = φPsat @ 60°F = (0.50)(0.2564 psia) = 0.128 psia 15 psia Thus the dew-point temperature of the air is 50 f/s Tdp = Tsat @ Pv = Tsat @ 0.128 psia = 41.3°F (from EES) 60°F, 50% (b) The volume flow rate is determined from ⎛ π × (8 / 12 ft ) 2 ⎞ πD 2 ⎟ = 17.45 ft 3 /s V& = VA = V = (50 ft/s)⎜ ⎟ ⎜ 4 4 ⎠ ⎝ (c) To determine the mass flow rate of dry air, we first need to calculate its specific volume, Pa = P − Pv = 15 − 0.128 = 14.872 psia RaT1 (0.3704 psia ⋅ ft 3 / lbm ⋅ R)(520 R) = = 12.95 ft 3 / lbm dry air 14.872 psia Pa1 17.45 ft 3 / s V& m& a1 = 1 = = 1.35 lbm/s v1 12.95 ft 3 / lbm dry air
v1 = Thus,
14-129 Air enters a cooling section at a specified pressure, temperature, and relative humidity. The temperature of the air at the exit and the rate of heat transfer are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air also remains constant (ω 1 = ω 2 ) as it flows through the cooling section since the process involves no humidification or dehumidification. The total pressure is 97 kPa. The properties of the air at the inlet state are Pv1 = φ1Pg1 = φ1Psat @ 35°C = (0.3)(5.629 kPa) = 1.69 kPa Cooling coils
Pa1 = P1 − Pv1 = 97 − 1.69 = 95.31 kPa
v1 =
RaT1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(308 K) = Pa1 95.31 kPa
= 0.927 m3 / kg dry air
ω1 =
1
35°C 30% 6 m3/min
2 97 kPa
AIR
0.622 Pv1 0.622(1.69 kPa) = = 0.0110 kg H 2O/kg dry air (= ω2 ) P1 − Pv1 (97 − 1.69) kPa
h1 = c pT1 + ω1hg1 = (1.005 kJ/kg°C)(35°C) + (0.0110)(2564.6 kJ/kg) = 63.44 kJ/kg dry air
The air at the final state is saturated and the vapor pressure during this process remains constant. Therefore, the exit temperature of the air must be the dew-point temperature, Tdp = Tsat @ Pv = Tsat @ 1.69 kPa = 14.8°C (b) The enthalpy of the air at the exit is h2 = c pT2 + ω2 hg 2 = (1.005 kJ/kg ⋅ °C)(14.8°C) + (0.0110)(2528.1 kJ/kg) = 42.78 kJ/kg dry air Also
m& a =
V&1 6 m3 / s = = 6.47 kg/min v 1 0.927 m 3 / kg dry air
Then the rate of heat transfer from the air in the cooling section becomes Q& out = m& a (h1 − h2 ) = (6.47 kg/min)(63.44 − 42.78)kJ/kg = 134 kJ/min
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-65
14-130 The outdoor air is first heated and then humidified by hot steam in an air-conditioning system. The rate of heat supply in the heating section and the mass flow rate of the steam required in the humidifying section are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The amount of moisture in the air also remains constants it flows through the heating section (ω 1 = ω 2 ) , but increases in the humidifying section (ω 3 > ω 2 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Fig. A-31) to be h1 = 17.7 kJ/kg dry air
ω1 = 0.0030 kg H 2 O/kg dry air (= ω 2 )
Heating coils
v 1 = 0.807 m /kg dry air 3
h2 = 29.8 kJ / kg dry air ω 2 = ω 1 = 0.0030 kg H 2 O / kg dry air
1 atm
10°C 40% 22 m3/min
h3 = 52.9 kJ / kg dry air ω 3 = 0.0109 kg H 2 O / kg dry air
AIR 1
25°C 55%
22°C 2
3
Analysis (a) The mass flow rate of dry air is m& a =
22 m3 / min V&1 = = 27.3 kg/min v1 0.807 m3 / kg
Then the rate of heat transfer to the air in the heating section becomes & ( h − h ) = ( 27.3 kg / min)(29.8 − 17.7)kJ / kg = 330.3 kJ / min Q& = m in
a
2
1
(b) The conservation of mass equation for water in the humidifying section can be expressed as m& a 2ω 2 + m& w = m& a 3ω 3
or m& w = m& a (ω 3 − ω 2 )
Thus, & w = ( 27.3 kg / min)(0.0109 − 0.0030) = 0.216 kg / min m
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-66
14-131 Air is cooled and dehumidified in an air-conditioning system with refrigerant-134a as the working fluid. The rate of dehumidification, the rate of heat transfer, and the mass flow rate of the refrigerant are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The saturation pressure of water at 30ºC is 4.2469 kPa. Then the dew point temperature of the incoming air stream at 30°C becomes Tdp = Tsat @ Pv = Tsat @ 0.7×4.2469 kPa = 24°C
Since air is cooled to 20°C, which is below its dew point temperature, some moisture in air will condense. The mass flow rate of dry air remains constant during the entire process, but the amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. Then the properties of the air at both states are determined from the psychrometric chart (Fig. A-31) to be h1 = 78.3 kJ/kg dry air
ω1 = 0.0188 kg H 2 O/kg dry air
4
3
v 1 = 0.885 m 3 /kg dry air
R-134a
700 kPa x3 = 20%
700 kPa sat. vapor
and h2 = 57.5 kJ / kg dry air 1
ω 2 = 0.0147 kg H 2 O / kg dry air Also,
hw ≅ h f @ 20°C = 83.915 kJ/kg (Table A-4)
Then,
m& a1 =
30°C 70% 4 m3/min
AIR 2 1 atm
20°C
V&1 4 m 3 / min = = 4.52 kg/min v 1 0.885 m 3 / kg dry air
Applying the water mass balance and the energy balance equations to the combined cooling and dehumidification section (excluding the refrigerant), Water Mass Balance:
∑ m& w,i = ∑ m& w,e
⎯ ⎯→
m& a1ω 1 = m& a 2ω 2 + m& w
m& w = m& a (ω 1 − ω 2 ) = (4.52 kg / min)(0.0188 − 0.0147) = 0.0185 kg / min
(b) Energy Balance: E& in − E& out = ∆E& system Ê0 (steady) = 0 E& in = E& out & i hi = Q& out + ∑ m & e he ∑m
⎯ ⎯→
& a 2 h2 + m & w hw ) = m & a ( h1 − h2 ) − m & w hw & a1h1 − ( m Q& out = m
Q& out = (4.52 kg/min)(78.3 − 57.5)kJ/kg − (0.0185 kg/min)(83.915 kJ/kg) = 92.5 kJ/min
(c) The inlet and exit enthalpies of the refrigerant are h3 = h g + x 3 h fg = 88.82 + 0.2 × 176.21 = 124.06 kJ/kg h4 = h g @ 700 kPa = 265.03 kJ/kg
Noting that the heat lost by the air is gained by the refrigerant, the mass flow rate of the refrigerant becomes Q& R = m& R (h4 − h3 ) → m& R =
Q& R 92.5 kJ/min = = 0.66 kg/min h4 − h3 (265.03 − 124.06) kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-67
14-132 Air is cooled and dehumidified in an air-conditioning system with refrigerant-134a as the working fluid. The rate of dehumidification, the rate of heat transfer, and the mass flow rate of the refrigerant are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The dew point temperature of the incoming air stream at 30°C is Pv1 = φ1Pg1 = φ1Psat @ 30°C = (0.7)(4.247 kPa) = 2.973 kPa Tdp = Tsat @ Pv = Tsat @ 2.973 kPa = 24°C
Since air is cooled to 20°C, which is below its dew point temperature, some of the moisture in the air will condense.
4
3 R-134a
The amount of moisture in the air decreases due to dehumidification (ω 2 < ω 1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 95 kPa. The properties of the air at both states are determined to be
1
700 kPa x3 = 20%
30°C 70% 4 m3/min
700 kPa sat. vapor AIR
95 kPa
2
20°C
Pa1 = P1 − Pv1 = 95 − 2.97 = 92.03 kPa
v1 =
Ra T1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(303 K) = = 0.945 m 3 / kg dry air 92.03 kPa Pa1
ω1 =
0.622 Pv1 0.622(2.97 kPa) = = 0.0201 kg H 2 O/kg dry air (95 − 2.97) kPa P1 − Pv1
h1 = c p T1 + ω1hg1 = (1.005 kJ/kg ⋅ °C)(30°C) + (0.0201)(2555.6 kJ/kg) = 81.50 kJ/kg dry air
and Pv 2 = φ 2 Pg 2 = (1.00) Psat @ 20°C = 2.3392 kPa
ω2 =
0.622 Pv 2 0.622(2.3392 kPa) = = 0.0157 kg H 2 O/kg dry air P2 − Pv 2 (95 − 2.3392) kPa
h2 = c p T2 + ω 2 hg 2 = (1.005 kJ/kg ⋅ °C)(20°C) + (0.0157)(2537.4 kJ/kg) = 59.94 kJ/kg dry air
Also, hw ≅ h f @ 20°C = 83.915 kJ/kg
(Table A-4)
Then, m& a1 =
V&1 4 m3 / min = = 4.23 kg/min v1 0.945 m 3 / kg dry air
Applying the water mass balance and the energy balance equations to the combined cooling and dehumidification section (excluding the refrigerant), Water Mass Balance:
∑ m& w,i = ∑ m& w,e
⎯ ⎯→
m& a1ω 1 = m& a 2ω 2 + m& w
m& w = m& a (ω 1 − ω 2 ) = (4.23 kg / min)(0.0201 − 0.0157) = 0.0186 kg / min
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-68
(b) Energy Balance: E& − E& = ∆E& in
out
Ê0 (steady)
system
E& in = E& out & i hi = Q& out + ∑ m & e he ∑m
=0 ⎯ ⎯→
& a 2 h2 + m & w hw ) = m & a ( h1 − h2 ) − m & w hw & a1h1 − ( m Q& out = m
Q& out = (4.23 kg/min)(81.50 − 59.94)kJ/kg − (0.0186 kg/min)(83.915 kJ/kg) = 89.7 kJ/min
(c) The inlet and exit enthalpies of the refrigerant are h3 = h g + x 3 h fg = 88.82 + 0.2 × 176.21 = 124.06 kJ/kg h4 = h g @ 700 kPa = 265.03 kJ/kg
Noting that the heat lost by the air is gained by the refrigerant, the mass flow rate of the refrigerant is determined from Q& = m& (h − h ) R
R
4
3
Q& R 89.7 kJ/min m& R = = = 0.636 kg/min h4 − h3 (265.03 − 124.06) kJ/kg
14-133 Air is heated and dehumidified in an air-conditioning system consisting of a heating section and an evaporative cooler. The temperature and relative humidity of the air when it leaves the heating section, the rate of heat transfer in the heating section, and the rate of water added to the air in the evaporative cooler are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) Assuming the wet-bulb temperature of the air remains constant during the evaporative cooling process, the properties of air at various states are determined from the psychrometric chart (Fig. A-31) to be h1 = 23.5 kJ/kg dry air T1 = 10°C ⎫ ⎬ ω = 0.00532 kg/ H 2 O/kg dry air φ1 = 70% ⎭ 1 Water v 1 = 0.810 m 3 / kg Heating
ω 2 = ω1 Twb2 = Twb3
T2 = 28.3°C ⎫ ⎬ φ 2 = 22.3% ⎭ h ≅ h = 42.3 kJ / kg dry air 2 3
h = 42.3 kJ/kg dry air T3 = 20°C ⎫ 3 ⎬ ω = 0.00875 kg/ H 2O/kg dry air φ3 = 60% ⎭ 3 Twb3 = 15.1°C
coils 1 atm
10°C 70% 30 m3/min
AIR 1
20°C 60%
T2 2
3
(b) The mass flow rate of dry air is V& 30 m 3 / min m& a = 1 = = 37.0 kg/min v 1 0.810 m 3 / kg dry air Then the rate of heat transfer to air in the heating section becomes Q& = m& (h − h ) = (37.0 kg/min)(42.3 − 23.5)kJ/kg = 696 kJ/min in
a
2
1
(c) The rate of water added to the air in evaporative cooler is m& w, added = m& w3 − m& w2 = m& a (ω3 − ω2 ) = (37.0 kg/min)(0.00875 − 0.00532) = 0.127 kg/min
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-69
14-134 EES Problem 14-133 is reconsidered. The effect of total pressure in the range 94 to 104 kPa on the results required in the problem is to be studied. Analysis The problem is solved using EES, and the solution is given below. P=101.325 [kPa] Tdb[1] =10 [C] Rh[1] = 0.70 Vol_dot[1]= 50 [m^3/min] Tdb[3] = 20 [C] Rh[3] = 0.60 P[1]=P P[2]=P[1] P[3]=P[1] "Energy balance for the steady-flow heating process 1 to 2:" "We neglect the PE of the flow. Since we don't know the cross sectional area of the flow streams, we also neglect theKE of the flow." E_dot_in - E_dot_out = DELTAE_dot_sys DELTAE_dot_sys = 0 [kJ/min] E_dot_in = m_dot_a*h[1]+Q_dot_in E_dot_out = m_dot_a*h[2] "Conservation of mass of dry air during mixing: m_dot_a = constant" m_dot_a = Vol_dot[1]/v[1] "Conservation of mass of water vapor during the heating process:" m_dot_a*w[1] = m_dot_a*w[2] "Conservation of mass of water vapor during the evaporative cooler process:" m_dot_a*w[2]+m_dot_w = m_dot_a*w[3] "During the evaporative cooler process:" Twb[2] = Twb[3] Twb[3] =WETBULB(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) {h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],B=Twb[2])} h[2]=h[3] Tdb[2]=TEMPERATURE(AirH2O,h=h[2],P=P[2],w=w[2]) w[2]=HUMRAT(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) h[3]=ENTHALPY(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) w[3]=HUMRAT(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) mw [kg/min] 0.2112 0.2112 0.2111 0.2111 0.211 0.2109
Qin [kJ/min] 1119 1131 1143 1155 1168 1180
Rh2 0.212 0.2144 0.2167 0.219 0.2212 0.2233
Tdb2 [C] 29.2 29 28.82 28.64 28.47 28.3
P [kPa] 94 96 98 100 102 104
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-70
0.224 0.222
Rh[2]
0.22 0.218 0.216 0.214 0.212 94
96
98
100
102
104
102
104
P [kPa] 29.2 29.1
Tdb[2] [C]
29 28.9 28.8 28.7 28.6 28.5 28.4 28.3 94
96
98
100
P [kPa] 1180
0.2113
1170 0.2112
1160
] ni m / J k[ ni
] ni 0.2111m g/ k[
1150 1140 1130
0.211
Q
w
m
1120 1110 94
96
98
100
102
0.2109 104
P [kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-71
14-135 Air is heated and dehumidified in an air-conditioning system consisting of a heating section and an evaporative cooler. The temperature and relative humidity of the air when it leaves the heating section, the rate of heat transfer in the heating section, and the rate at which water is added to the air in the evaporative cooler are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) Assuming the wet-bulb temperature of the air remains constant during the evaporative cooling process, the properties of air at various states are determined to be Pv1 = φ1Pg1 = φ1Psat @ 10°C = (0.70)(1.2281 kPa) = 0.86 kPa Water Heating coils
Pa1 = P1 − Pv1 = 96 − 0.86 = 95.14 kPa
v1 =
RaT1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(283 K) = Pa1 95.14 kPa
= 0.854 m3 / kg dry air
10°C 70% 30 m3/min
96 kPa AIR
1
20°C 60%
T2 2
3
0.622 Pv1 0.622(0.86 kPa) ω1 = = = 0.00562 kg H 2O/kg dry air P1 − Pv1 (96 − 0.86) kPa h1 = c pT1 + ω1hg1 = (1.005 kJ/kg ⋅ °C)(10°C) + (0.00562)(2519.2 kJ/kg) = 24.21 kJ/kg dry air
and Pv 3 = φ3 Pg 3 = φ3 Psat @ 20°C = (0.60)(2.3392 kPa) = 1.40 kPa Pa 3 = P3 − Pv3 = 96 − 1.40 = 94.60 kPa
ω3 =
0.622 Pv3 0.622(1.40 kPa) = = 0.00923 kg H 2 O/kg dry air P3 − Pv 3 (96 − 1.40) kPa
h3 = c p T3 + ω3 hg 3 = (1.005 kJ/kg ⋅ °C)(20°C) + (0.00921)(2537.4 kJ/kg) = 43.52 kJ/kg dry air
Also, h2 ≅ h3 = 43.52 kJ/kg
ω 2 = ω1 = 0.00562 kg H 2 O/kg dry air Thus, h2 = c pT2 + ω2 hg 2 ≅ c pT2 + ω2 (2500.9 + 1.82T2 ) = (1.005 kJ/kg ⋅ °C)T2 + (0.00562)(2500.9 + 1.82T2 )
Solving for T2, T2 = 29.0°C ⎯ ⎯→ Pg 2 = Psat@29°C = 4.013 kPa
Thus,
φ2 =
ω 2 P2 (0.00562)(96) = = 0.214 or 21.4% (0.622 + ω 2 ) Pg 2 (0.622 + 0.00562)(4.013)
(b) The mass flow rate of dry air is V& 30 m 3 / min m& a = 1 = = 35.1 kg/min v 1 0.854 m 3 / kg dry air Then the rate of heat transfer to air in the heating section becomes Q& = m& (h − h ) = (35.1 kg/min)(43.52 − 24.21)kJ/kg = 679 kJ/min in
a
2
1
(c) The rate of water addition to the air in evaporative cooler is m& w, added = m& w3 − m& w 2 = m& a (ω3 − ω2 ) = (35.1 kg/min)(0.00923 − 0.00562) = 0.127 kg/min
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-72
14-136 Conditioned air is to be mixed with outside air. The ratio of the dry air mass flow rates of the conditioned- to-outside air, and the temperature of the mixture are to be determined. Assumptions 1 Steady operating conditions exist. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing chamber is adiabatic. Properties The properties of each inlet stream are determined from the psychrometric chart (Fig. A-31) to be h1 = 34.3 kJ / kg dry air
ω 1 = 0.0084 kg H 2 O / kg dry air
1 13°C 90%
and h2 = 68.5 kJ / kg dry air
ω 2 = 0.0134 kg H 2 O / kg dry air
P = 1 atm
Analysis The ratio of the dry air mass flow rates of the Conditioned air to the outside air can be determined from m& a1 ω 2 − ω 3 h2 − h3 = = m& a 2 ω 3 − ω1 h3 − h1
T3 3
34°C 40% 2
But state 3 is not completely specified. However, we know that state 3 is on the straight line connecting states 1 and 2 on the psychrometric chart. At the intersection point of this line and φ = 60% line we read (b)
T3 = 23.5° C
ω 3 = 0.0109 kg H 2 O / kg dry air h3 = 513 . kJ / kg dry air
Therefore, the mixture will leave at 23.5°C. The m& a1 / m& a 2 ratio is determined by substituting the specific humidity (or enthalpy) values into the above relation, (a)
m& a1 0.0134 − 0.0109 = = 1.00 m& a 2 0.0109 − 0.0084
Therefore, the mass flow rate of each stream must be the same.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-73
14-137 EES Problem 14-136 is reconsidered. The desired quantities are to be determined using EES at 1 atm and 80 kPa pressures. Analysis The problem is solved using EES, and the solution is given below. "Without loss of generality assume the mass flow rate of the outside air is m_dot[2] = 1 kg/s." P=101.325 [kPa] Tdb[1] =13 [C] "State 1 is the conditioned air" Rh[1] = 0.90 Tdb[2] =34 [C] "State 2 is the outside air" Rh[2] = 0.40 Rh[3] = 0.60 P[1]=P P[2]=P[1] P[3]=P[1] m_dot[2] = 1 [kg/s] MassRatio = m_dot[1]/m_dot[2] "Energy balance for the steady-flow mixing process:" "We neglect the PE of the flow. Since we don't know the cross sectional area of the flow streams, we also neglect theKE of the flow." E_dot_in - E_dot_out = DELTAE_dot_sys DELTAE_dot_sys = 0 [kW] E_dot_in = m_dot[1]*h[1]+m_dot[2]*h[2] E_dot_out = m_dot[3]*h[3] "Conservation of mass of dry air during mixing:" m_dot[1]+m_dot[2] = m_dot[3] "Conservation of mass of water vapor during mixing:" m_dot[1]*w[1]+m_dot[2]*w[2] = m_dot[3]*w[3] h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) v[2]=VOLUME(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) w[2]=HUMRAT(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) Tdb[3]=TEMPERATURE(AirH2O,h=h[3],P=P[3],R=Rh[3]) w[3]=HUMRAT(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) v[3]=VOLUME(AirH2O,T=Tdb[3],P=P[3],w=w[3])
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-74
AirH2O
0.050 Pressure = 101.3 [kPa]
0.045 0.040
35°C
0.035
oi t a R y ti di m u H
0.8
0.030
30°C 0.6
0.025 25°C
0.020
0.4
20°C
0.015 15°C
0.010
10°C
0.2
1
0.005 0.000 0
2
3
5
10
15
20
25
30
35
40
T [°C]
SOLUTION for P=1 atm (101.325 kPa) DELTAE_dot_sys=0 [kW] E_dot_out=102.9 [kW] h[2]=68.45 [kJ/kg] MassRatio=1.007 m_dot[2]=1 [kg/s] P=101.3 [kPa] P[2]=101.3 [kPa] Rh[1]=0.9 Rh[3]=0.6 Tdb[2]=34 [C] v[1]=0.8215 [m^3/kg] v[3]=0.855 [m^3/kg] w[2]=0.01336
E_dot_in=102.9 [kW] h[1]=34.26 [kJ/kg] h[3]=51.3 [kJ/kg] m_dot[1]=1.007 [kg/s] m_dot[3]=2.007 [kg/s] P[1]=101.3 [kPa] P[3]=101.3 [kPa] Rh[2]=0.4 Tdb[1]=13 [C] Tdb[3]=23.51 [C] v[2]=0.8888 [m^3/kg] w[1]=0.008387 w[3]=0.01086
SOLUTION for P=80 kPa DELTAE_dot_sys=0 E_dot_out=118.2 [kW] h[2]=77.82 [kJ/kga] MassRatio=1.009 m_dot[2]=1 [kga/s] P=80 [kPa] P[2]=80 [kPa] Rh[1]=0.9 Rh[3]=0.6 Tdb[2]=34 [C] v[1]=1.044 [m^3/kga] v[3]=1.088 [m^3/kga] w[2]=0.01701 [kgw/kga]
E_dot_in=118.2 [kW] h[1]=40 [kJ/kga] h[3]=58.82 [kJ/kga] m_dot[1]=1.009 [kga/s] m_dot[3]=2.009 [kga/s] P[1]=80 [kPa] P[3]=80 [kPa] Rh[2]=0.4 Tdb[1]=13 [C] Tdb[3]=23.51 [C] v[2]=1.132 [m^3/kga] w[1]=0.01066 [kgw/kga] w[3]=0.01382 [kgw/kga]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-75
14-138 [Also solved by EES on enclosed CD] Waste heat from the cooling water is rejected to air in a natural-draft cooling tower. The mass flow rate of the cooling water, the volume flow rate of air, and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis (a) The mass flow rate of dry air through the tower remains constant (m& a1 = m& a 2 = m& a ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation is made up later in the cycle using water at 27°C. Applying the mass balance and the energy balance equations yields Dry Air Mass Balance: ∑ m& a ,i = ∑ m& a ,e
⎯ ⎯→
m& a1 = m& a 2 = m& a
AIR EXIT
Water Mass Balance: & w ,i = ∑ m & w,e ∑m
⎯ ⎯→
2 37°C saturated
&3 + m & a1ω 1 = m &4 + m & a 2ω 2 m
&3 − m &4 = m & a (ω 2 − ω 1 ) = m & makeup m
Energy Balance: E& − E& = ∆E& in
Ê0 (steady) system
out
= 0 ⎯⎯→ E& in = E& out
∑ m& i hi = ∑ m& e he (since Q& = W& = 0) 0 = ∑ m& e he − ∑ m& i hi WARM 3 0 = m& a 2 h2 + m& 4 h4 − m& a1h1 − m& 3h3 WATER 0 = m& a (h2 − h1 ) + (m& 3 − m& makeup )h4 − m& 3h3
1 AIR INLET
27°C
Solving for m& a ,
4
m& 3 (h3 − h4 ) m& a = (h2 − h1 ) − (ω 2 − ω 1 )h4
COOL
ω1 = 0.0109 kg H 2O/kg dry air
and
v1 = 0.854 m3/kg dry air From Table A-4, h3 ≅ h f @ 42°C = 175.90 kJ/kg H 2 O m& a =
Tdb = 23°C Twb = 18°C
WATER
From the psychrometric chart (Fig. A-31), h1 = 50.8 kJ/kg dry air
Substituting
42°C
h2 = 143.0 kJ / kg dry air
ω 2 = 0.0412 kg H 2 O / kg dry air h4 ≅ h f @ 27°C = 113.19 kJ/kg H 2 O
m& 3 (175.90 − 113.19)kJ/kg &3 = 0.706 m (143.0 − 50.8) kJ/kg − (0.0412 − 0.0109)(113.25) kJ/kg
The mass flow rate of the cooling water is determined by applying the steady flow energy balance equation on the cooling water, & −m & & h − [m & −m & (ω − ω )]h & h − (m Q& =m )h = m waste
3 3
3
makeup
4
3 3
3
a
2
1
4
& 3h3 − m & 3[1 − 0.706( 0.0412 − 0.0109)]h4 = m & 3 ( h3 − 0.9786h4 ) =m 50,000 kJ/s = m& 3 (175.90 − 0.9786 × 113.19) kJ/kg ⎯ ⎯→ m& 3 = 768.1 kg/s
and
m& a = 0.706m& 3 = (0.706)(7681 . kg / s) = 542.3 kg / s
(b) Then the volume flow rate of air into the cooling tower becomes V& = m& v = (542.3 kg/s)(0.854 m 3 / kg ) = 463.1 m 3 /s 1
a 1
(c) The mass flow rate of the required makeup water is determined from & makeup = m & a (ω 2 − ω 1 ) = (542.3 kg / s)(0.0412 − 0.0109) = 16.4 kg / s m
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-76
14-139 EES Problem 14-138 is reconsidered. The effect of air inlet wet-bulb temperature on the required air volume flow rate and the makeup water flow rate is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P_atm =101.325 [kPa] T_db_1 = 23 [C] T_wb_1 = 18 [C] T_db_2 = 37 [C] RH_2 = 100/100 "%. relative humidity at state 2, saturated condition" Q_dot_waste = 50 [MW]*Convert(MW, kW) T_cw_3 = 42 [C] "Cooling water temperature at state 3" T_cw_4 = 27 [C] "Cooling water temperature at state 4" "Dry air mass flow rates:" "RH_1 is the relative humidity at state 1 on a decimal basis" v_1=VOLUME(AirH2O,T=T_db_1,P=P_atm,R=RH_1) T_wb_1 = WETBULB(AirH2O,T=T_db_1,P=P_atm,R=RH_1) m_dot_a_1 = Vol_dot_1/v_1 "Conservaton of mass for the dry air (ma) in the SSSF mixing device:" m_dot_a_in - m_dot_a_out = DELTAm_dot_a_cv m_dot_a_in = m_dot_a_1 m_dot_a_out = m_dot_a_2 DELTAm_dot_a_cv = 0 "Steady flow requirement" "Conservation of mass for the water vapor (mv) and cooling water for the SSSF process:" m_dot_w_in - m_dot_w_out = DELTAm_dot_w_cv m_dot_w_in = m_dot_v_1 + m_dot_cw_3 m_dot_w_out = m_dot_v_2+m_dot_cw_4 DELTAm_dot_w_cv = 0 "Steady flow requirement" w_1=HUMRAT(AirH2O,T=T_db_1,P=P_atm,R=RH_1) m_dot_v_1 = m_dot_a_1*w_1 w_2=HUMRAT(AirH2O,T=T_db_2,P=P_atm,R=RH_2) m_dot_v_2 = m_dot_a_2*w_2 "Conservation of energy for the SSSF cooling tower process:" "The process is adiabatic and has no work done, ngelect ke and pe" E_dot_in_tower - E_dot_out_tower = DELTAE_dot_tower_cv E_dot_in_tower= m_dot_a_1 *h[1] + m_dot_cw_3*h_w[3] E_dot_out_tower = m_dot_a_2*h[2] + m_dot_cw_4*h_w[4] DELTAE_dot_tower_cv = 0 "Steady flow requirement" h[1]=ENTHALPY(AirH2O,T=T_db_1,P=P_atm,w=w_1) h[2]=ENTHALPY(AirH2O,T=T_db_2,P=P_atm,w=w_2) h_w[3]=ENTHALPY(steam,T=T_cw_3,x=0) h_w[4]=ENTHALPY(steam,T=T_cw_4,x=0) "Energy balance on the external heater determines the cooling water flow rate:" E_dot_in_heater - E_dot_out_heater = DELTAE_dot_heater_cv E_dot_in_heater = Q_dot_waste + m_dot_cw_4*h_w[4] E_dot_out_heater = m_dot_cw_3 * h_w[3] DELTAE_dot_heater_cv = 0 "Steady flow requirement"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-77 "Conservation of mass on the external heater gives the makeup water flow rate." "Note: The makeup water flow rate equals the amount of water vaporized in the cooling tower." m_dot_cw_in - m_dot_cw_out = DELTAm_dot_cw_cv m_dot_cw_in = m_dot_cw_4 + m_dot_makeup m_dot_cw_out = m_dot_cw_3 DELTAm_dot_cw_cv = 0 "Steady flow requirement"
Vol1 [m3/s] 408.3 420.1 433.2 447.5 463.4 481.2 501.1 523.7 549.3 578.7
mmakeup [kgw/s] 16.8 16.72 16.63 16.54 16.43 16.31 16.18 16.03 15.87 15.67
mcw3 [kgw/s] 766.6 766.7 766.8 767 767.2 767.4 767.7 767.9 768.2 768.6
ma1 [kga/s] 481.9 495 509.4 525.3 542.9 562.6 584.7 609.7 638.1 670.7
Twb1 [C] 14 15 16 17 18 19 20 21 22 23
16.8 560
] s/ 3 ^ m [
16.6
] s 16.4 / w g 16.2 k[
520 480
p u 16.0 e k a m
l1
o V
440 400 14.0
15.8 m 15.6 16.0
18.0
20.0
22.0
Twb,1 [C]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-78
14-140 Atmospheric air enters an air-conditioning system at a specified pressure, temperature, and relative humidity. The heat transfer, the rate of condensation of water, and the mass flow rate of the refrigerant are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet and exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at the inlet and exit states may be determined from the psychrometric chart (Figure A31) or using EES psychrometric functions to be (we used EES) h1 = 78.24 kJ/kg dry air R-134a 350 kPa x = 1.0
ω1 = 0.01880 kg H 2O/kg dry air
350 kPa x = 0.20
v1 = 0.8847 m / kg dry air 3
Cooling coils
h2 = 27.45 kJ/kg dry air
ω2 = 0.002885 kg H 2O/kg dry air
T2 =20°C φ 2 = 20%
The mass flow rate of dry air is V& 4 m3/min m& a = 1 = = 4.521 kg/min 2 v1 0.8847 m3 The mass flow rates of vapor at the inlet and exit are m& v1 = ω1 m& a = (0.01880)(4.521 kg/min) = 0.0850 kg/min m& v 2 = ω 2 m& a = (0.002885)(4.521 kg/min) = 0.01304 kg/min An energy balance on the control volume gives m& a h1 = Q& out + m& a h2 + m& w h w2 where the the enthalpy of condensate water is hw 2 = h f@ 20°C = 83.91 kJ/kg (Table A - 4)
T1 =30°C φ 1 = 70% 4 m3/min
1 atm Condensate 1 20°C
Condensate removal
and the rate of condensation of water vapor is m& w = m& v1 − m& v 2 = 0.0850 − 0.01304 = 0.07196 kg/min Substituting, m& a h1 = Q& out + m& a h2 + m& w hw 2 (4.521 kg/min)(78.24 kJ/kg) = Q& + (4.521 kg/min)(27.45 kJ/kg) + (0.07196 kg/min)(83.91 kJ/kg) out
Q& out = 223.6 kJ/min = 3.727 kW The properties of the R-134a at the inlet and exit of the cooling section are PR1 = 350 kPa ⎫ ⎬h R1 = 97.56 kJ/kg 0,050 x R1 = 0.20 ⎭
PR 2 = 350 kPa ⎫ ⎬h R 2 = 253.34 kJ/kg x R 2 = 1.0 ⎭ Noting that the rate of heat lost from the air is received by the refrigerant, the mass flow rate of the refrigerant is determined from m& R h R1 + Q& in = m& R h R 2 Q& in m& R = h R 2 − h R1 223.6 kJ/min (253.34 − 97.56) kJ/kg = 1.435 kg/min =
AirH2O
Pressure = 101.3 [kPa]
0,045 0,040 35°C 0.8
0,035
oi t a R y ti di m u H
0,030 30°C 0.6
0,025 25°C
0,020
1
0.4
20°C
0,015 15°C
0,010
0.2
10°C
0,005 0,000 0
2 5
10
15
20
25
30
35
T [°C]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
40
14-79
14-141 An uninsulated tank contains moist air at a specified state. Water is sprayed into the tank until the relative humidity in the tank reaches a certain value. The amount of water supplied to the tank, the final pressure in the tank, and the heat transfer during the process are to be determined. Assumptions 1 Dry air and water vapor are ideal gases. 2 The kinetic and potential energy changes are negligible. Analysis The initial state of the moist air is completely specified. The properties of the air at the inlet state may be determined from the psychrometric chart (Figure A-31) or using EES psychrometric functions to be (we used EES) h1 = 49.16 kJ/kg dry air
ω1 = 0.005433 kg H 2 O/kg dry air v 1 = 0.6863 m 3 / kg dry air The initial mass in the tank is V 0.5 m 3 ma = 1 = = 0.7285 kg v 1 0.6863 m 3 The partial pressure of dry air in the tank is m RT (0.7285 kg)(0.287 kJ/kg.K)(35 + 273 K) Pa 2 = a a 2 = = 128.8 kPa V (0.5 m 3 ) Then, the pressure of moist air in the tank is determined from ω2 ⎞ ω ⎞ ⎛ ⎛ P2 = Pa 2 ⎜⎜1 + ⎟⎟ = (128.8 kPa)⎜⎜1 + 2 ⎟⎟ ⎝ 0.622 ⎠ ⎝ 0.622 ⎠ We cannot fix the final state explicitly by a hand-solution. However, using EES which has built-in functions for moist air properties, the final state properties are determined to be ω2 = 0.02446 kg H 2O/kg dry air P2 = 133.87 kPa h2 = 97.97 kJ/kg dry air
v 2 = 0.6867 m3 / kg dry air
The partial pressures at the initial and final states are Pv1 = φ1 Psat@35°C = 0.20(5.6291 kPa) = 1.126 kPa Pa1 = P 1 − Pv1 = 130 − 1.126 = 128.87 kPa Pv 2 = P 2 − Pa 2 = 133.87 − 128.81 = 5.07 kPa The specific volume of water at 35ºC is v w1 = v w2 = v g @35°C = 25.205 m 3 /kg
The internal energies per unit mass of dry air in the tank are u1 = h1 − Pa1v 1 − w1 Pv1v w1 = 49.16 − 128.87 × 0.6863 − 0.005433 × 1.126 × 25.205 = −39.44 kJ/kg u 2 = h2 − Pa 2v 2 − w 2 Pv 2v w 2 = 97.97 − 128.81 × 0.6867 − 0.02446 × 5.07 × 25.205 = 6.396 kJ/kg The enthalpy of water entering the tank from the supply line is hw1 = hf @50°C = 209.34 kJ/kg The internal energy of water vapor at the final state is u w 2 = u g @35°C = 2422.7 kJ/kg The amount of water supplied to the tank is m w = m a (ω 2 − ω1 ) = (0.7285 kg)(0.02446 - 0.005433) = 0.01386 kg An energy balance on the system gives E in = ∆E tank Qin + m w h w1 = m a (u 2 − u1 ) + m w u w2
Qin + (0.01386 kg )(209.34 kJ/kg) = (0.7285 kg)[6.396 - (-39.44)kJ/kg ] + (0.01386 kg)(2422.7 kJ/kg) Q& = 64.1 kJ in
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-80
14-142 Air flows steadily through an isentropic nozzle. The pressure, temperature, and velocity of the air at the nozzle exit are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (m& a1 = m& a 2 = m& a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis The inlet state of the air is completely specified, and the total pressure is 200 kPa. The properties of the air at the inlet state may be determined from the psychrometric chart (Figure A-31) or using EES psychrometric functions to be (we used EES) h1 = 57.65 kJ/kg dry air
Air 35ºC 200 kPa 50% RH
T2 P2 V2
ω1 = ω 2 = 0.008803 kg H 2 O/kg dry air (no condensation) s1 = s 2 = 5.613 kJ/kg.K dry air
(isentropic process)
We assume that the relative humidity at the nozzle exit is 100 percent since there is no condensation in the nozzle. Other exit state properties can be determined using EES built-in functions for moist air. The results are h2 = 42.53 kJ/kg dry air P2 = 168.2 kPa T2 = 20°C
An energy balance on the control volume gives the velocity at the exit h1 = h2 + (1 + ω 2 )
V 22 2
57.65 kJ/kg = 42.53 kJ/kg + (1 + 0.008803)
V 22 ⎛ 1 kJ/kg ⎞ ⎟ ⎜ 2 ⎝ 1000 m 2 /s 2 ⎠
V 2 = 173.2 m/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-81
Fundamentals of Engineering (FE) Exam Problems
14-143 A room is filled with saturated moist air at 25°C and a total pressure of 100 kPa. If the mass of dry air in the room is 100 kg, the mass of water vapor is (a) 0.52 kg (b) 1.97 kg (c) 2.96 kg (d) 2.04 kg (e) 3.17 kg Answer (d) 2.04 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 "C" P=100 "kPa" m_air=100 "kg" RH=1 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g P_air=P-P_v w=0.622*P_v/(P-P_v) w=m_v/m_air "Some Wrong Solutions with Common Mistakes:" W1_vmass=m_air*w1; w1=0.622*P_v/P "Using P instead of P-Pv in w relation" W2_vmass=m_air "Taking m_vapor = m_air" W3_vmass=P_v/P*m_air "Using wrong relation"
14-144 A room contains 50 kg of dry air and 0.6 kg of water vapor at 25°C and 95 kPa total pressure. The relative humidity of air in the room is (a) 1.2% (b) 18.4% (c) 56.7% (d) 65.2% (e) 78.0% Answer (c) 56.7% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 "C" P=95 "kPa" m_air=50 "kg" m_v=0.6 "kg" w=0.622*P_v/(P-P_v) w=m_v/m_air P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g "Some Wrong Solutions with Common Mistakes:" W1_RH=m_v/(m_air+m_v) "Using wrong relation" W2_RH=P_g/P "Using wrong relation"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-82
14-145 A 40-m3 room contains air at 30°C and a total pressure of 90 kPa with a relative humidity of 75 percent. The mass of dry air in the room is (a) 24.7 kg (b) 29.9 kg (c) 39.9 kg (d) 41.4 kg (e) 52.3 kg Answer (c) 39.9 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=40 "m^3" T1=30 "C" P=90 "kPa" RH=0.75 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g P_air=P-P_v R_air=0.287 "kJ/kg.K" m_air=P_air*V/(R_air*(T1+273)) "Some Wrong Solutions with Common Mistakes:" W1_mass=P_air*V/(R_air*T1) "Using C instead of K" W2_mass=P*V/(R_air*(T1+273)) "Using P instead of P_air" W3_mass=m_air*RH "Using wrong relation"
14-146 A room contains air at 30°C and a total pressure of 96.0 kPa with a relative humidity of 75 percent. The partial pressure of dry air is (a) 82.0 kPa (b) 85.8 kPa (c) 92.8 kPa (d) 90.6 kPa (e) 72.0 kPa Answer (c) 92.8 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=30 "C" P=96 "kPa" RH=0.75 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g P_air=P-P_v "Some Wrong Solutions with Common Mistakes:" W1_Pair=P_v "Using Pv as P_air" W2_Pair=P-P_g "Using wrong relation" W3_Pair=RH*P "Using wrong relation"
14-147 The air in a house is at 20°C and 50 percent relative humidity. Now the air is cooled at constant pressure. The temperature at which the moisture in the air will start condensing is (a) 8.7°C (b) 11.3°C (c) 13.8°C (d) 9.3°C (e) 10.0°C Answer (d) 9.3°C
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-83
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 "C" RH1=0.50 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH1=P_v/P_g T_dp=TEMPERATURE(Steam_IAPWS,x=0,P=P_v) "Some Wrong Solutions with Common Mistakes:" W1_Tdp=T1*RH1 "Using wrong relation" W2_Tdp=(T1+273)*RH1-273 "Using wrong relation" W3_Tdp=WETBULB(AirH2O,T=T1,P=P1,R=RH1); P1=100 "Using wet-bulb temperature"
14-148 On the psychrometric chart, a cooling and dehumidification process appears as a line that is (a) horizontal to the left, (b) vertical downward, (c) diagonal upwards to the right (NE direction) (d) diagonal upwards to the left (NW direction) (e) diagonal downwards to the left (SW direction) Answer (e) diagonal downwards to the left (SW direction)
14-149 On the psychrometric chart, a heating and humidification process appears as a line that is (a) horizontal to the right, (b) vertical upward, (c) diagonal upwards to the right (NE direction) (d) diagonal upwards to the left (NW direction) (e) diagonal downwards to the right (SE direction) Answer (c) diagonal upwards to the right (NE direction)
14-150 An air stream at a specified temperature and relative humidity undergoes evaporative cooling by spraying water into it at about the same temperature. The lowest temperature the air stream can be cooled to is (a) the dry bulb temperature at the given state (b) the wet bulb temperature at the given state (c) the dew point temperature at the given state (d) the saturation temperature corresponding to the humidity ratio at the given state (e) the triple point temperature of water Answer (a) the dry bulb temperature at the given state
14-151 Air is cooled and dehumidified as it flows over the coils of a refrigeration system at 85 kPa from 30°C and a humidity ratio of 0.023 kg/kg dry air to 15°C and a humidity ratio of 0.015 kg/kg dry air. If the mass flow rate of dry air is 0.7 kg/s, the rate of heat removal from the air is (a) 5 kJ/s (b) 10 kJ/s (c) 15 kJ/s (d) 20 kJ/s (e) 25 kJ/s Answer (e) 25 kJ/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
14-84
Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P=85 "kPa" T1=30 "C" w1=0.023 T2=15 "C" w2=0.015 m_air=0.7 "kg/s" m_water=m_air*(w1-w2) h1=ENTHALPY(AirH2O,T=T1,P=P,w=w1) h2=ENTHALPY(AirH2O,T=T2,P=P,w=w2) h_w=ENTHALPY(Steam_IAPWS,T=T2,x=0) Q=m_air*(h1-h2)-m_water*h_w "Some Wrong Solutions with Common Mistakes:" W1_Q=m_air*(h1-h2) "Ignoring condensed water" W2_Q=m_air*Cp_air*(T1-T2)-m_water*h_w; Cp_air = 1.005 "Using dry air enthalpies" W3_Q=m_air*(h1-h2)+m_water*h_w "Using wrong sign"
14-152 Air at a total pressure of 90 kPa, 15°C, and 75 percent relative humidity is heated and humidified to 25°C and 75 percent relative humidity by introducing water vapor. If the mass flow rate of dry air is 4 kg/s, the rate at which steam is added to the air is (a) 0.032 kg/s (b) 0.013 kg/s (c) 0.019 kg/s (d) 0.0079 kg/s (e) 0 kg/s Answer (a) 0.032 kg/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P=90 "kPa" T1=15 "C" RH1=0.75 T2=25 "C" RH2=0.75 m_air=4 "kg/s" w1=HUMRAT(AirH2O,T=T1,P=P,R=RH1) w2=HUMRAT(AirH2O,T=T2,P=P,R=RH2) m_water=m_air*(w2-w1) "Some Wrong Solutions with Common Mistakes:" W1_mv=0 "sine RH = constant" W2_mv=w2-w1 "Ignoring mass flow rate of air" W3_mv=RH1*m_air "Using wrong relation"
14-153 ··· 14-157 Design and Essay Problems
KJ PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-1
Chapter 15 CHEMICAL REACTIONS Fuels and Combustion 15-1C Gasoline is C8H18, diesel fuel is C12H26, and natural gas is CH4. 15-2C Nitrogen, in general, does not react with other chemical species during a combustion process but its presence affects the outcome of the process because nitrogen absorbs a large proportion of the heat released during the chemical process. 15-3C Moisture, in general, does not react chemically with any of the species present in the combustion chamber, but it absorbs some of the energy released during combustion, and it raises the dew point temperature of the combustion gases. 15-4C The dew-point temperature of the product gases is the temperature at which the water vapor in the product gases starts to condense as the gases are cooled at constant pressure. It is the saturation temperature corresponding to the vapor pressure of the product gases. 15-5C The number of atoms are preserved during a chemical reaction, but the total mole numbers are not. 15-6C Air-fuel ratio is the ratio of the mass of air to the mass of fuel during a combustion process. Fuelair ratio is the inverse of the air-fuel ratio. 15-7C No. Because the molar mass of the fuel and the molar mass of the air, in general, are different. Theoretical and Actual Combustion Processes 15-8C The causes of incomplete combustion are insufficient time, insufficient oxygen, insufficient mixing, and dissociation. 15-9C CO. Because oxygen is more strongly attracted to hydrogen than it is to carbon, and hydrogen is usually burned to completion even when there is a deficiency of oxygen. 15-10C It represent the amount of air that contains the exact amount of oxygen needed for complete combustion. 15-11C No. The theoretical combustion is also complete, but the products of theoretical combustion does not contain any uncombined oxygen. 15-12C Case (b).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-2
15-13 Methane is burned with the stoichiometric amount of air during a combustion process. The AF and FA ratios are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis This is a theoretical combustion process since methane is burned completely with stoichiometric amount of air. The stoichiometric combustion equation of CH4 is CH 4 + a th [O 2 + 3.76N 2 ] ⎯ ⎯→ CO 2 + 2H 2 O + 3.76a th N 2
O2 balance: Substituting,
a th = 1 + 1
⎯ ⎯→
a th = 2
CH4 Products Air stoichiometric
CH 4 + 2[O 2 + 3.76N 2 ] ⎯ ⎯→ CO 2 + 2H 2O + 7.52N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =
m air (2 × 4.76 kmol)(29 kg/kmol) = = 17.3 kg air/kg fuel m fuel (1 kmol)(12 kg/kmol) + (2 kmol)(2 kg/kmol)
The fuel-air ratio is the inverse of the air-fuel ratio, FA =
1 1 = = 0.0578 kg fuel/kg air AF 17.3 kg air/kg fuel
15-14 Propane is burned with 75 percent excess air during a combustion process. The AF ratio is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The combustion equation in this case can be written as C 3 H 8 + 1.75a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 3CO 2 + 4H 2 O + 0.75a th O 2 + (1.75 × 3.76)a th N 2
where ath is the stoichiometric coefficient for air. We have automatically accounted for the 75% excess air by using the C3H8 factor 1.75ath instead of ath for air. The stoichiometric amount of oxygen (athO2) will be used to oxidize the fuel, and the remaining Air excess amount (0.75athO2) will appear in the products as free oxygen. The coefficient ath is determined from the O2 balance, 75% excess O2 balance:
17 . 5a th = 3 + 2 + 0.75a th
Substituting,
C3H 8 + 8.75 O 2 + 3.76 N 2
⎯ ⎯→
Products
a th = 5
⎯ ⎯→ 3CO 2 + 4H 2O + 3.75O 2 + 32.9 N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =
m air (8.75 × 4.76 kmol)(29 kg/kmol) = = 27.5 kg air/kg fuel m fuel (3 kmol)(12 kg/kmol) + (4 kmol)(2 kg/kmol)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-3
15-15 Acetylene is burned with the stoichiometric amount of air during a combustion process. The AF ratio is to be determined on a mass and on a mole basis. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis This is a theoretical combustion process since C2H2 is burned completely with stoichiometric amount of air. The stoichiometric combustion equation of C2H2 is C 2 H 2 + a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 2CO 2 + H 2 O + 3.76a th N 2
O2 balance:
a th = 2 + 0.5
⎯ ⎯→
C2H2 Products
a th = 2.5
Substituting, C 2 H 2 + 2.5[O 2 + 3.76N 2 ] ⎯ ⎯→ 2CO 2 + H 2 O + 9.4N 2
100% theoretical air
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =
m air (2.5 × 4.76 kmol)(29 kg/kmol) = = 13.3 kg air/kg fuel m fuel (2 kmol)(12 kg/kmol) + (1 kmol)(2 kg/kmol)
On a mole basis, the air-fuel ratio is expressed as the ratio of the mole numbers of the air to the mole numbers of the fuel, AFmole basis =
N air (2.5 × 4.76) kmol = = 11.9 kmol air/kmol fuel N fuel 1 kmol fuel
15-16 Ethane is burned with an unknown amount of air during a combustion process. The AF ratio and the percentage of theoretical air used are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). C2H6 Analysis (a) The combustion equation in this case can be written as C 2 H 6 + a[O 2 + 3.76N 2 ] ⎯ ⎯→ 2CO 2 + 3H 2 O + 3O 2 + 3.76aN 2
O2 balance: Substituting,
a = 2 + 1.5 + 3 ⎯ ⎯→ a = 6.5
Products air
C 2 H 6 + 6.5[O 2 + 3.76N 2 ] ⎯ ⎯→ 2CO 2 + 3H 2 O + 3O 2 + 24.44N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =
m air (6.5 × 4.76 kmol)(29 kg/kmol) = = 29.9 kg air/kg fuel m fuel (2 kmol)(12 kg/kmol) + (3 kmol)(2 kg/kmol)
(b) To find the percent theoretical air used, we need to know the theoretical amount of air, which is determined from the theoretical combustion equation of C2H6, C 2 H 6 + a th O 2 + 3.76N 2
O2 balance: Then,
a th = 2 + 15 .
⎯ ⎯→ 2CO 2 + 3H 2O + 3.76a th N 2
⎯ ⎯→
Percent theoretical air =
m air,act m air, th
a th = 35 . =
N air,act N air, th
=
a 6.5 = = 186% a th 3.5
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-4
15-17E Ethylene is burned with 200 percent theoretical air during a combustion process. The AF ratio and the dew-point temperature of the products are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, and air are 12 lbm/lbmol, 2 lbm/lbmol, and 29 lbm/lbmol, respectively (Table A-1E). Analysis (a) The combustion equation in this case can be written as C2H4 C 2 H 4 + 2a th O 2 + 3.76N 2 ⎯ ⎯→ 2CO 2 + 2H 2 O + a th O 2 + (2 × 3.76)a th N 2 Products where ath is the stoichiometric coefficient for air. 200% It is determined from theoretical air O2 balance: 2a th = 2 + 1 + a th ⎯ ⎯→ a th = 3 Substituting,
C 2 H 4 + 6 O 2 + 3.76N 2
⎯⎯→ 2CO 2 + 2 H 2 O + 3O 2 + 22.56N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, m (6 × 4.76 lbmol)(29 lbm/lbmol) AF = air = = 29.6 lbm air/lbm fuel m fuel (2 lbmol)(12 lbm/lbmol) + (2 lbmol)(2 lbm/lbmol) (b) The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is,
Thus,
⎛ Nv ⎞ ⎛ 2 lbmol ⎞ ⎟P ⎟(14.5 psia ) = 0.981 psia =⎜ Pv = ⎜ ⎜ N prod ⎟ prod ⎜⎝ 29.56 lbmol ⎟⎠ ⎠ ⎝ Tdp = Tsat @ 0.981 psia = 101°F
15-18 Propylene is burned with 50 percent excess air during a combustion process. The AF ratio and the temperature at which the water vapor in the products will start condensing are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, and air are 12 kg/kmol, Products C3H6 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis (a) The combustion equation in this case can be 50% excess air written as C 3 H 6 + 1.5a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 3CO 2 + 3H 2 O + 0.5a th O 2 + (1.5 × 3.76)a th N 2
where ath is the stoichiometric coefficient for air. It is determined from O2 balance:
15 . a th = 3 + 15 . + 0.5a th
Substituting, C 3 H 6 + 6.75 O 2 + 3.76N 2
⎯ ⎯→
a th = 4.5
⎯⎯→ 3CO 2 + 3H 2 O + 2.25O 2 + 25.38N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, m (6.75 × 4.76 kmol)(29 kg/kmol) AF = air = = 22.2 kg air/kg fuel m fuel (3 kmol)(12 kg/kmol) + (3 kmol)(2 kg/kmol) (b) The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is,
Thus,
⎛ Nv ⎞ ⎛ 3 kmol ⎞ ⎟P ⎟(105 kPa ) = 9.367 kPa Pv = ⎜ =⎜ ⎜ N prod ⎟ prod ⎜⎝ 33.63 kmol ⎟⎠ ⎝ ⎠ Tdp = Tsat @9.367 kPa = 44.5°C
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-5
15-19 Propal alcohol C3H7OH is burned with 50 percent excess air. The balanced reaction equation for complete combustion is to be written and the air-to-fuel ratio is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. Properties The molar masses of C, H2, O2 and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The combustion equation in this case can be written as C 3 H 7 OH + 1.5a th [O 2 + 3.76N 2 ] ⎯ ⎯→ B CO 2 + D H 2 O + E O 2 + F N 2
where ath is the stoichiometric coefficient for air. We have automatically accounted for the 50% excess air by using the factor 1.5ath instead of ath for air. The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance:
B=3
Hydrogen balance:
2D = 8 ⎯ ⎯→ D = 4
Oxygen balance:
1 + 2 × 1.5a th = 2 B + D + 2 E 0.5a th = E
Nitrogen balance:
C3H7OH Products Air 50% eccess
1.5a th × 3.76 = F
Solving the above equations, we find the coefficients (E = 2.25, F = 25.38, and ath = 4.5) and write the balanced reaction equation as C 3 H 7 OH + 6.75[O 2 + 3.76N 2 ] ⎯ ⎯→3 CO 2 + 4 H 2 O + 2.25 O 2 + 25.38 N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =
m air (6.75 × 4.75 kmol)(29 kg/kmol) = = 15.51 kg air/kg fuel (3 × 12 + 8 × 1 + 1× 16) kg m fuel
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-6
15-20 Butane C4H10 is burned with 200 percent theoretical air. The kmol of water that needs to be sprayed into the combustion chamber per kmol of fuel is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. Properties The molar masses of C, H2, O2 and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The reaction equation for 200% theoretical air without the additional water is C 4 H 10 + 2a th [O 2 + 3.76N 2 ] ⎯ ⎯→ B CO 2 + D H 2 O + E O 2 + F N 2
where ath is the stoichiometric coefficient for air. We have automatically accounted for the 100% excess air by using the factor 2ath instead of ath for air. The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance:
B=4
Hydrogen balance:
2 D = 10 ⎯ ⎯→ D = 5
Oxygen balance:
2 × 2a th = 2 B + D + 2 E a th = E 2a th × 3.76 = F
Nitrogen balance:
C4H10 Products Air 200% theoretical
Solving the above equations, we find the coefficients (E = 6.5, F = 48.88, and ath = 6.5) and write the balanced reaction equation as C 4 H 10 + 13[O 2 + 3.76N 2 ] ⎯ ⎯→ 4 CO 2 + 5 H 2 O + 6.5 O 2 + 48.88 N 2
With the additional water sprayed into the combustion chamber, the balanced reaction equation is C 4 H 10 + 13[O 2 + 3.76N 2 ] + N v H 2 O ⎯ ⎯→ 4 CO 2 + (5 + N v ) H 2 O + 6.5 O 2 + 48.88 N 2
The partial pressure of water in the saturated product mixture at the dew point is Pv ,prod = Psat@60°C = 19.95 kPa
The vapor mole fraction is yv =
Pv ,prod Pprod
=
19.95 kPa = 0.1995 100 kPa
The amount of water that needs to be sprayed into the combustion chamber can be determined from yv =
N water 5 + Nv ⎯ ⎯→ 0.1995 = ⎯ ⎯→ N v = 9.796 kmol N total,product 4 + 5 + N v + 6.5 + 48.88
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-7
15-21 A fuel mixture of 20% by mass methane, CH4, and 80% by mass ethanol, C2H6O, is burned completely with theoretical air. The required flow rate of air is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, O2 and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The combustion equation in this case can be written as x CH 4 + y C 2 H 6 O + a th [O 2 + 3.76N 2 ] ⎯ ⎯→ B CO 2 + D H 2 O + F N 2
where ath is the stoichiometric coefficient for air. The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance:
x + 2y = B
Hydrogen balance:
4 x + 6 y = 2D
Oxygen balance:
2ath + y = 2 B + D
Nitrogen balance:
3.76a th = F
20% CH4 80% C2H6O Air
Products
100% theoretical
Solving the above equations, we find the coefficients as x = 0.4182
B = 1.582
y = 0.5818
D = 2.582
ath = 2.582
F = 9.708
Then, we write the balanced reaction equation as 0.4182 CH 4 + 0.5818 C 2 H 6O + 2.582 [O 2 + 3.76N 2 ] ⎯ ⎯→ 1.582 CO 2 + 2.582 H 2O + 9.708 N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF = =
m air m fuel (2.582 × 4.76 kmol)(29 kg/kmol) = 10.64 kg air/kg fuel (0.4182 kmol)(12 + 4 × 1)kg/kmol + (0.5818 kmol)(2 × 12 + 6 × 1 + 16)kg/kmol
Then, the required flow rate of air becomes m& air = AFm& fuel = (10.64)(31 kg/s) = 330 kg/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-8
15-22 Octane is burned with 250 percent theoretical air during a combustion process. The AF ratio and the dew-pint temperature of the products are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis (a) The combustion equation in this case can be written as C8 H18 + 2.5a th O 2 + 3.76N 2
⎯ ⎯→ 8CO 2 + 9H 2 O + 1.5a th O 2 + (2.5 × 3.76)a th N 2
where ath is the stoichiometric coefficient for air. It is determined from O2 balance:
2.5a th = 8 + 4.5 + 1.5a th ⎯ ⎯→ a th = 12.5
Substituting,
C 8 H 18 + 31.25[O 2 + 3.76N 2 ] → 8CO 2 + 9H 2 O + 18.75O 2 + 117.5N 2
Thus,
AF =
C8H18 Air 25°C
Combustion Products chamber P = 1 atm
m air (31.25 × 4.76 kmol)(29 kg/kmol) = = 37.8 kg air/kg fuel m fuel (8 kmol)(12 kg/kmol) + (9 kmol)(2 kg/kmol)
(b) The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is, ⎛ N ⎞ ⎛ 9 kmol ⎞ ⎟⎟(101.325 kPa ) = 5.951 kPa Pv = ⎜ v ⎟ Pprod = ⎜⎜ ⎜ N prod ⎟ ⎝ 153.25 kmol ⎠ ⎝ ⎠
Thus,
Tdp = Tsat @5.951 kPa = 36.0°C
15-23 Gasoline is burned steadily with air in a jet engine. The AF ratio is given. The percentage of excess air used is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The theoretical combustion equation in this case can be written as C8 H18 + a th O 2 + 3.76N 2
⎯ ⎯→ 8CO 2 + 9H 2 O + 3.76a th N 2
where ath is the stoichiometric coefficient for air. It is determined from O2 balance:
a th = 8 + 4.5
⎯ ⎯→
a th = 12.5
Gasoline (C8H18)
Jet engine
Products
Air
The air-fuel ratio for the theoretical reaction is determined by taking the ratio of the mass of the air to the mass of the fuel for, AFth =
m air,th m fuel
=
(12.5 × 4.76 kmol)(29 kg/kmol) = 15.14 kg air/kg fuel (8 kmol)(12 kg/kmol) + (9 kmol)(2 kg/kmol)
Then the percent theoretical air used can be determined from Percent theoretical air =
AFact AFth
=
18 kg air/kg fuel = 119% 15.14 kg air/kg fuel
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-9
15-24 Ethane is burned with air steadily. The mass flow rates of ethane and air are given. The percentage of excess air used is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The theoretical combustion equation in this case can be written as C 2 H 6 + a th O 2 + 3.76N 2
⎯ ⎯→ 2CO 2 + 3H 2O + 3.76a th N 2
C2H6 Combustion Products chamber
where ath is the stoichiometric coefficient for air. It is determined from O2 balance:
a th = 2 + 15 .
⎯ ⎯→
a th = 35 .
The air-fuel ratio for the theoretical reaction is determined by taking the ratio of the mass of the air to the mass of the fuel for, AFth =
m air,th m fuel
=
Air
(3.5 × 4.76 kmol)(29 kg/kmol) = 16.1 kg air/kg fuel (2 kmol)(12 kg/kmol) + (3 kmol)(2 kg/kmol)
The actual air-fuel ratio used is AFact =
m& air 176 kg/h = = 22 kgair/kgfuel m& fuel 8 kg/h
Then the percent theoretical air used can be determined from Percent theoretical air =
AFact AFth
=
22 kg air/kg fuel = 137% 16.1 kg air/kg fuel
Thus the excess air used during this process is 37%.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-10
15-25 Butane is burned with air. The masses of butane and air are given. The percentage of theoretical air used and the dew-point temperature of the products are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis (a) The theoretical combustion equation in this case can be written as C 4 H10 + a th O 2 + 3.76N 2
⎯ ⎯→ 4CO 2 + 5H 2O + 3.76a th N 2
where ath is the stoichiometric coefficient for air. It is determined from a th = 4 + 2.5
O2 balance:
⎯ ⎯→
m air,th m fuel
=
Combustion Products chamber
a th = 6.5
The air-fuel ratio for the theoretical reaction is determined by taking the ratio of the mass of the air to the mass of the fuel for, AFth =
C4H10
Air
(6.5 × 4.76 kmol)(29 kg/kmol) = 15.5 kg air/kg fuel (4 kmol)(12 kg/kmol) + (5 kmol)(2 kg/kmol)
The actual air-fuel ratio used is AFact =
mair 25 kg = = 25 kg air / kg fuel mfuel 1 kg
Then the percent theoretical air used can be determined from Percent theoretical air =
AFact AFth
=
25 kg air/kg fuel = 161% 15.5 kg air/kg fuel
(b) The combustion is complete, and thus products will contain only CO2, H2O, O2 and N2. The air-fuel ratio for this combustion process on a mole basis is AF =
N air m / M air (25 kg )/ (29 kg/kmol) = 50 kmol air/kmol fuel = air = (1 kg )/ (58 kg/kmol) N fuel m fuel / M fuel
Thus the combustion equation in this case can be written as C 4 H10 + (50/4.76 )[O 2 + 3.76N 2 ] ⎯ ⎯→ 4CO 2 + 5H 2 O + 4.0O 2 + 39.5N 2
The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is, ⎛ Nv Pv = ⎜ ⎜ N prod ⎝
⎞ ⎟ Pprod = ⎛⎜ 5 kmol ⎞⎟(90 kPa ) = 8.571 kPa ⎜ 52.5 kmol ⎟ ⎟ ⎝ ⎠ ⎠
Thus, Tdp = Tsat @8.571
kPa
= 42.8°C
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-11
15-26E Butane is burned with air. The masses of butane and air are given. The percentage of theoretical air used and the dew-point temperature of the products are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. 3 Combustion gases are ideal gases. Properties The molar masses of C, H2, and air are 12 lbm/lbmol, 2 lbm/lbmol, and 29 lbm/lbmol, respectively (Table A-1). Analysis (a) The theoretical combustion equation in this case can be written as C 4 H 10 + ath O 2 + 3.76N 2
⎯⎯→ 4CO 2 + 5H 2 O + 3.76ath N 2
where ath is the stoichiometric coefficient for air. It is determined from a th = 4 + 2.5
O2 balance:
⎯ ⎯→
m air,th m fuel
=
Combustion Products chamber
a th = 6.5
The air-fuel ratio for the theoretical reaction is determined by taking the ratio of the mass of the air to the mass of the fuel for, AFth =
C4H10
Air
(6.5 × 4.76 lbmol)(29 lbm/lbmol) = 15.5 lbm air/lbm fuel (4 lbmol)(12 lbm/lbmol) + (5 lbmol)(2 lbm/lbmol)
The actual air-fuel ratio used is AFact =
m air 25 lbm = = 25 lbm air/lbm fuel m fuel 1 lbm
Then the percent theoretical air used can be determined from Percent theoretical air =
AFact AFth
=
25 lbm air/lbm fuel = 161% 15.5 lbm air/lbm fuel
(b) The combustion is complete, and thus products will contain only CO2, H2O, O2 and N2. The air-fuel ratio for this combustion process on a mole basis is AF =
N air m / M air (25 lbm)/ (29 lbm/lbmol) = 50 lbmol air/lbmol fuel = air = (1 lbm)/ (58 lbm/lbmol) N fuel m fuel / M fuel
Thus the combustion equation in this case can be written as C 4 H 10 + (50/4.76 )[O 2 + 3.76N 2 ] ⎯ ⎯→ 4CO 2 + 5H 2 O + 4O 2 + 39.5N 2
The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is, ⎛ Nv Pv = ⎜ ⎜ N prod ⎝
⎞ ⎟ Pprod = ⎛⎜ 5 lbmol ⎞⎟(14.7 psia ) = 1.4 psia ⎜ 52.5 lbmol ⎟ ⎟ ⎝ ⎠ ⎠
Thus, Tdp = Tsat @1.4 psia = 113.2°F
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-12
15-27 The volumetric fractions of the constituents of a certain natural gas are given. The AF ratio is to be determined if this gas is burned with the stoichiometric amount of dry air. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, N2, O2, and air are 12 kg/kmol, 2 kg/kmol, 28 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis Considering 1 kmol of fuel, the combustion equation can be written as (0.65CH 4 + 0.08H 2 + 0.18N 2 + 0.03O 2 + 0.06CO 2 ) + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ xCO 2 + yH 2 O + zN 2
The unknown coefficients in the above equation are determined from mass balances, C : 0.65 + 0.06 = x
⎯ ⎯→ x = 0.71
H : 0.65 × 4 + 0.08 × 2 = 2 y
O 2 : 0.03 + 0.06 + a th = x + y / 2 N 2 : 0.18 + 3.76a th = z
Natural gas
⎯ ⎯→ y = 1.38
Combustion Products chamber
⎯ ⎯→ a th = 1.31
⎯ ⎯→ z = 5.106
Dry air
Thus, (0.65CH 4 + 0.08H 2 + 018 . N 2 + 0.03O 2 + 0.06CO 2 ) + 131 . (O 2 + 3.76N 2 ) ⎯⎯→ 0.71CO 2 + 138 . H 2 O + 5106 . N2
The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel, m air = (1.31× 4.76 kmol)(29 kg/kmol) = 180.8 kg
m fuel = (0.65 × 16 + 0.08 × 2 + 0.18 × 28 + 0.03 × 32 + 0.06 × 44 )kg = 19.2 kg
and AFth =
m air,th m fuel
=
180.8 kg = 9.42 kg air/kg fuel 19.2 kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-13
15-28 The composition of a certain natural gas is given. The gas is burned with stoichiometric amount of moist air. The AF ratio is to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, N2, O2, and air are 12 kg/kmol, 2 kg/kmol, 28 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis The fuel is burned completely with the stoichiometric amount of air, and thus the products will contain only H2O, CO2 and N2, but no free O2. The moisture in the air does not react with anything; it simply shows up as additional H2O in the products. Therefore, we can simply balance the combustion equation using dry air, and then add the moisture to both sides of the equation. Considering 1 kmol of fuel, the combustion equation can be written as (0.65CH 4 + 0.08H 2 + 0.18N 2 + 0.03O 2 + 0.06CO 2 ) + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ xCO 2 + yH 2O + zN 2
The unknown coefficients in the above equation are determined from mass balances,
C : 0.65 + 0.06 = x
⎯ ⎯→ x = 0.71
H : 0.65 × 4 + 0.08 × 2 = 2 y
O 2 : 0.03 + 0.06 + a th = x + y / 2 N 2 : 0.18 + 3.76a th = z
Natural gas
⎯ ⎯→ y = 1.38
Combustion Products chamber
⎯ ⎯→ a th = 1.31
⎯ ⎯→ z = 5.106
Moist air
Thus, (0.65CH 4 + 0.08H 2 + 018 . N 2 + 0.03O 2 + 0.06CO 2 ) + 1.31(O 2 + 3.76N 2 ) ⎯⎯→ 0.71CO 2 + 138 . H 2 O + 5106 . N2
Next we determine the amount of moisture that accompanies 4.76ath = (4.76)(1.31) = 6.24 kmol of dry air. The partial pressure of the moisture in the air is
Pv,in = φ air Psat@ 25°C = (0.85)(3.1698 kPa) = 2.694 kPa Assuming ideal gas behavior, the number of moles of the moisture in the air (Nv, in) is determined to be ⎛ Pv ,in N v ,in = ⎜⎜ ⎝ Ptotal
⎞ ⎛ 2.694 kPa ⎞ ⎟ N total = ⎜ ⎯→ N v,air = 0.17 kmol ⎜ 101.325 kPa ⎟⎟ 6.24 + N v ,in ⎯ ⎟ ⎝ ⎠ ⎠
(
)
The balanced combustion equation is obtained by substituting the coefficients determined earlier and adding 0.17 kmol of H2O to both sides of the equation, (0.65CH 4 + 0.08H 2 + 018 . N 2 + 0.03O 2 + 0.06CO 2 ) + 131 . (O 2 + 3.76N 2 ) + 017 . H 2O ⎯⎯→ 0.71CO 2 + 155 . H 2 O + 5106 . N2
The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel, m air = (1.31 × 4.76 kmol )(29 kg/kmol ) + (0.17 kmol × 18 kg/kmol ) = 183.9 kg
m fuel = (0.65 × 16 + 0.08 × 2 + 0.18 × 28 + 0.03 × 32 + 0.06 × 44 )kg = 19.2 kg
and AFth =
m air, th m fuel
=
183.9 kg = 9.58 kg air/kg fuel 19.2 kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-14
15-29 The composition of a gaseous fuel is given. It is burned with 130 percent theoretical air. The AF ratio and the fraction of water vapor that would condense if the product gases were cooled are to be determined. Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. Properties The molar masses of C, H2, N2, and air are 12 kg/kmol, 2 kg/kmol, 28 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis (a) The fuel is burned completely with excess air, and thus the products will contain H2O, CO2, N2, and some free O2. Considering 1 kmol of fuel, the combustion equation can be written as (0.60CH 4 + 0.30H 2 + 0.10N 2 ) + 1.3a th (O 2 + 3.76N 2 ) ⎯ ⎯→ xCO 2 + yH 2O + 0.3a th O 2 + zN 2
The unknown coefficients in the above equation are determined from mass balances, C : 0.60 = x
⎯ ⎯→ x = 0.60
H : 0.60 × 4 + 0.30 × 2 = 2 y O 2 : 1.3a th = x + y / 2 + 0.3a th N 2 : 0.10 + 3.76 × 1.3a th = z
⎯ ⎯→ y = 1.50 ⎯ ⎯→ a th = 1.35 ⎯ ⎯→ z = 6.70
Gaseous fuel Air
Combustion Products chamber
30% excess
Thus, (0.60CH 4 + 0.30H 2 + 0.10N 2 ) + 1.755(O 2 + 3.76N 2 ) ⎯ ⎯→ 0.6CO 2 + 1.5H 2O + 0.405O 2 + 6.7N 2
The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel, m air = (1.755 × 4.76 kmol )(29 kg/kmol ) = 242.3 kg
m fuel = (0.6 × 16 + 0.3 × 2 + 0.1× 28)kg = 13.0 kg
and AF =
mair 242.3 kg = = 18.6 kg air/kg fuel mfuel 13.0 kg
(b) For each kmol of fuel burned, 0.6 + 1.5 + 0.405 + 6.7 = 9.205 kmol of products are formed, including 1.5 kmol of H2O. Assuming that the dew-point temperature of the products is above 20°C, some of the water vapor will condense as the products are cooled to 20°C. If Nw kmol of H2O condenses, there will be 1.5 - Nw kmol of water vapor left in the products. The mole number of the products in the gas phase will also decrease to 9.205 - Nw as a result. Treating the product gases (including the remaining water vapor) as ideal gases, Nw is determined by equating the mole fraction of the water vapor to its pressure fraction, Nv P 1.5 − N w 2.3392 kPa = v ⎯ ⎯→ = ⎯ ⎯→ N w = 1.32 kmol N prod,gas Pprod 9.205 − N w 101.325 kPa
since Pv = Psat @ 20°C = 2.3392 kPa. Thus the fraction of water vapor that condenses is 1.32/1.5 = 0.88 or 88%.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-15
15-30 EES Problem 15-29 is reconsidered. The effects of varying the percentages of CH4, H2 and N2 making up the fuel and the product gas temperature are to be studied. Analysis The problem is solved using EES, and the solution is given below. Let's modify this problem to include the fuels butane, ethane, methane, and propane in pull down menu. Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: aCxHy+bH2+cN2 + (a*y/4 + a*x+b/2) (Theo_air/100) (O2 + 3.76 N2) a*xCO2 + ((a*y/2)+b) H2O + (c+3.76 (a*y/4 + a*x+b/2) (Theo_air/100)) N2 + (a*y/4 + a*x+b/2) (Theo_air/100 - 1) O2 T_prod is the product gas temperature. Theo_air is the % theoretical air. " Procedure H20Cond(P_prod,T_prod,Moles_H2O,M_other:T_DewPoint,Moles_H2O_vap,Moles_H2O_liq,Re sult$) P_v = Moles_H2O/(M_other+Moles_H2O)*P_prod T_DewPoint = temperature(steam,P=P_v,x=0) IF T_DewPoint = 100%, the solution assumes complete combustion.' {MolCO = 0 MolCO2 = x}
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-92 w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" T_air = 298 [K] Theo_air = 200 "%" Fuel$='CH4(g)'} T_fuel = 298 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=(xw)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x +y/4-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2 2500
Theoair [%] 75 90 100 120 150 200 300 500 800
Tprod [K] 2077 2287 2396 2122 1827 1506 1153 840.1 648.4
Adiabatic Flame Temp.
2100
for C8 H18 (liquid) ] K [ d or p
T
1700 1300 900 500 0
100
200
300
400
500
600
700
Theoair [%]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
800
15-93
15-109 EES A general program is to be written to determine the heat transfer during the complete combustion of a hydrocarbon fuel CnHm at 25°C in a steady-flow combustion chamber when the percent of excess air and the temperatures of air and the products are specified. Analysis The problem is solved using EES, and the solution is given below. Steady-flow combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> xCO2 + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + (x+y/4-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Steady-flow, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> (x-w)CO2 +wCO + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + ((x+y/4-z/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the product gas temperature, assuming no dissociation. Theo_air is the % theoretical air. " Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$,MM) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='Acetylene' h_fuel = 226730"Table A.26" MM=2*12+2*1 else If fuel$='C3H8(l)' then x=3; y=8; z=0 Name$='Propane(liq)' h_fuel = -103850-15060"Tables A.26 and A.27" MM=molarmass(C3H8) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='Octane(liq)' h_fuel = -249950"Table A.26" MM=8*12+18*1 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='Methane' h_fuel = enthalpy(CH4,T=T_fuel) MM=molarmass(CH4) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='Methyl alcohol' h_fuel = -200670"Table A.26" MM=1*12+4*1+1*16 endif; endif; endif; endif; endif end
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-94
Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" T_air = 298 [K] Theo_air = 200 [%] Fuel$='CH4(g)'} T_fuel = 298 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$,MM) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=(xw)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x +y/4-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) Q_out=(HR-HP)/MM "kJ/kg_fuel" Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2 SOLUTION for the sample calculation A_th=5 fuel$='C3H8(l)' HP=-149174 [kJ/kg] HR=-119067 [kJ/kg] h_fuel=-118910 MM=44.1 [kg/kmol] Moles_CO=0.000 Moles_CO2=3.000 Moles_H2O=4 Moles_N2=28.200 Moles_O2=2.500 MolO2=2.5 Name$='Propane(liq)' Q_out=682.8 [kJ/kg_fuel] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=150 [%] Th_air=1.500 T_air=298 [K] T_fuel=298 [K] T_prod=1800 [K] w=0 x=3 y=8
z=0
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-95
15-110 EES A general program is to be written to determine the adiabatic flame temperature during the complete combustion of a hydrocarbon fuel CnHm at 25°C in a steady-flow combustion chamber when the percent of excess air and its temperature are specified. Analysis The problem is solved using EES, and the solution is given below. Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Adiabatic, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) (x-w)CO2 +wCO + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + ((y/4 + xz/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 else If fuel$='C3H8(l)' then x=3; y=8; z=0 Name$='propane(liq)' h_fuel = -103850-15060 else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane(liq)' h_fuel = -249950 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-96
IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' {MolCO = 0 MolCO2 = x} w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" T_air = 298 [K] Theo_air = 200 [%] Fuel$='CH4(g)'} T_fuel = 298 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=(xw)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x +y/4-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2 SOLUTION for the sample calculation A_th=5 fuel$='C3H8(l)' HR=-119067 [kJ/kg] h_fuel=-118910 Moles_CO2=3.000 Moles_H2O=4 Moles_O2=2.500 MolO2=2.5 SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=150 [%] Th_air=1.500 T_fuel=298 [K] T_prod=1820 [K] x=3 y=8
HP=-119067 [kJ/kg] Moles_CO=0.000 Moles_N2=28.200 Name$='propane(liq)' T_air=298 [K] w=0 z=0
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-97
15-111 EES The adiabatic flame temperature of the fuels CH4(g), C2H2(g), CH3OH(g), C3H8(g), and C8H18(l) is to be determined. Analysis The problem is solved using EES, and the solution is given below.
Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" {"For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " 1+ 2*A_th=1*2+2*1"theoretical O balance"} "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730"Table A.26" else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950"Table A.26" else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670"Table A.26" endif; endif; endif; endif; endif end {"Input data from the diagram window" T_air = 298 [K] Theo_air = 200 [%] Fuel$='CH4(g)'} T_fuel = 298 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th = y/4 + x-z/2 Th_air = Theo_air/100 HR=h_fuel+ (y/4 + x-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(y/4 + x-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-98 HP=x*enthalpy(CO2,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(y/4 + x-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+(y/4 + x-z/2) *(Theo_air/100 - 1)*enthalpy(O2,T=T_prod) Moles_O2=(y/4 + x-z/2) *(Theo_air/100 - 1) Moles_N2=3.76*(y/4 + x-z/2)* (Theo_air/100) Moles_CO2=x Moles_H2O=y/2 T[1]=T_prod; xa[1]=Theo_air "array variable are plotted in Plot Window 1" SOLUTION for a sample calculation A_th=1.5 HR=-200733 [kJ/kg] Moles_H2O=2 Name$='methyl alcohol' T[1]=1540 T_prod=1540 [K] y=4
fuel$='CH3OH(g)' h_fuel=-200670 Moles_N2=11.280 Theo_air=200 [%] T_air=298 [K] x=1 z=1
HP=-200733 [kJ/kg] Moles_CO2=1 Moles_O2=1.500 Th_air=2 T_fuel=298 [K] xa[1]=200 [%]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-99
15-112 EES The minimum percent of excess air that needs to be used for the fuels CH4(g), C2H2(g), CH3OH(g), C3H8(g), and C8H18(l) if the adiabatic flame temperature is not to exceed 1500 K is to be determined. Analysis The problem is solved using EES, and the solution is given below.
Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" {"For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " 1+ 2*A_th=1*2+2*1"theoretical O balance"} "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 endif; endif; endif; endif; endif end {"Input data from the diagram window" T_air = 298 [K] Fuel$='CH4(g)'} T_fuel = 298 [K] Excess_air=Theo_air - 100 "[%]" Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th = y/4 + x-z/2 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-100 Th_air = Theo_air/100 HR=h_fuel+ (y/4 + x-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(y/4 + x-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=x*enthalpy(CO2,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(y/4 + x-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+(y/4 + x-z/2) *(Theo_air/100 - 1)*enthalpy(O2,T=T_prod) Moles_O2=(y/4 + x-z/2) *(Theo_air/100 - 1) Moles_N2=3.76*(y/4 + x-z/2)* (Theo_air/100) Moles_CO2=x Moles_H2O=y/2 T[1]=T_prod; xa[1]=Theo_air SOLUTION for a sample calculation A_th=2.5 fuel$='C2H2(g)' HR=226596 [kJ/kg] Moles_CO2=2 Moles_N2=24.09 Name$='acetylene' Th_air=2.563 T_air=298 [K] T_prod=1500 [K] xa[1]=256.3 z=0
Excess_air=156.251 [%] HP=226596 [kJ/kg] h_fuel=226730 Moles_H2O=1 Moles_O2=3.906 Theo_air=256.3 [%] T[1]=1500 [K] T_fuel=298 [K] x=2 y=2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-101
15-113 EES The minimum percentages of excess air that need to be used for the fuels CH4(g), C2H2(g), CH3OH(g), C3H8(g), and C8H18(l) AFOR adiabatic flame temperatures of 1200 K, 1750 K, and 2000 K are to be determined. Analysis The problem is solved using EES, and the solution is given below.
Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" {"For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " 1+ 2*A_th=1*2+2*1"theoretical O balance"} "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 endif; endif; endif; endif; endif end {"Input data from the diagram window" T_air = 298 [K] Fuel$='CH4(g)'} T_fuel = 298 [K] Excess_air=Theo_air - 100 "[%]" Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th = y/4 + x-z/2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-102 Th_air = Theo_air/100 HR=h_fuel+ (y/4 + x-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(y/4 + x-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=x*enthalpy(CO2,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(y/4 + x-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+(y/4 + x-z/2) *(Theo_air/100 - 1)*enthalpy(O2,T=T_prod) Moles_O2=(y/4 + x-z/2) *(Theo_air/100 - 1) Moles_N2=3.76*(y/4 + x-z/2)* (Theo_air/100) Moles_CO2=x Moles_H2O=y/2 T[1]=T_prod; xa[1]=Theo_air SOLUTION for a sample calculation A_th=5 fuel$='C3H8(g)' HR=-103995 [kJ/kg] Moles_CO2=3 Moles_N2=24.7 Name$='propane' Th_air=1.314 T_air=298 [K] T_prod=2000 [K] xa[1]=131.4 z=0
Excess_air=31.395 [%] HP=-103995 [kJ/kg] h_fuel=-103858 Moles_H2O=4 Moles_O2=1.570 Theo_air=131.4 [%] T[1]=2000 [K] T_fuel=298 [K] x=3 y=8
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-103
15-114 EES The adiabatic flame temperature of CH4(g) is to be determined when both the fuel and the air enter the combustion chamber at 25°C for the cases of 0, 20, 40, 60, 80, 100, 200, 500, and 1000 percent excess air. Analysis The problem is solved using EES, and the solution is given below.
Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Adiabatic, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) (x-w)CO2 +wCO + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + ((y/4 + xz/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-104 IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' {MolCO = 0 MolCO2 = x} w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" T_air = 298 [K] Theo_air = 200 [%] Fuel$='CH4(g)'} T_fuel = 298 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=(xw)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x +y/4-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Product temperature vs % excess air for CH4 Moles_H2O=y/2 3000 Theoair [%] 100 120 140 160 180 200 300 600 1100
Tprod [K] 2329 2071 1872 1715 1587 1480 1137 749.5 553
2500
] K [ d o r p
T
2000 1500 1000 500 0 100 200 300 400 500 600 700 800 900 1000 1100 Theoair [%]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-105
15-115 EES The rate of heat transfer is to be determined for the fuels CH4(g), C2H2(g), CH3OH(g), C3H8(g), and C8H18(l) when they are burned completely in a steady-flow combustion chamber with the theoretical amount of air. Analysis The problem is solved using EES, and the solution is given below. Steady-floe combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> xCO2 + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + (x+y/4-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Steady-flow, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> (x-w)CO2 +wCO + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + ((x+y/4-z/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the product gas temperature, assuming no dissociation. Theo_air is the % theoretical air. " Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$,MM) "This procedure takes the fuel name and returns the moles of C ,H and O and molar mass" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 MM=2*12+2*1 else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) MM=molarmass(C3H8) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950 MM=8*12+18*1 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) MM=molarmass(CH4) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 MM=1*12+4*1+1*16 endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-106 ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" m_dot_fuel = 0.1 [kg/s] T_air = 298 [K] Theo_air = 200 [%] Fuel$='CH4(g)'} T_fuel = 298 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$,MM) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=(xw)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x +y/4-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) HR =Q_out+HP "The heat transfer rate is:" Q_dot_out=Q_out/MM*m_dot_fuel "[kW]" Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2 SOLUTION for a sample calculation A_th=1.5 fuel$='CH3OH(g)' HP=-604942 [kJ/kg] HR=-200701 [kJ/kg] h_fuel=-200670 MM=32 Moles_CO=0.000 Moles_CO2=1.000 Moles_H2O=2 Moles_N2=5.640 Moles_O2=0.000 MolO2=0 m_dot_fuel=1 [kg/s] Name$='methyl alcohol' Q_dot_out=12633 [kW] Q_out=404241.1 [kJ/kmol_fuel] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_air=298 [K] T_fuel=298 [K] T_prod=1200 [K] w=0 x=1 y=4 z=1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-107
15-116 EES The rates of heat transfer are to be determined for the fuels CH4(g), C2H2(g), CH3OH(g), C3H8(g), and C8H18(l) when they are burned in a steady-flow combustion chamber with for 50, 100, and 200 percent excess air. Analysis The problem is solved using EES, and the solution is given below.
Steady-flow combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> xCO2 + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + (x+y/4-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Steady-flow, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> (x-w)CO2 +wCO + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + ((x+y/4-z/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the product gas temperature, assuming no dissociation. Theo_air is the % theoretical air. " Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$,MM) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 MM=2*12+2*1 else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) MM=molarmass(C3H8) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950 MM=8*12+18*1 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) MM=molarmass(CH4) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 MM=1*12+4*1+1*16 endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-108
ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" T_air = 298 [K] m_dot_fuel=1 [kg/s] Theo_air = 200 [%] Fuel$='CH4(g)'} T_fuel = 298 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$,MM) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=(xw)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x +y/4-z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) HR =Q_out+HP "The heat transfer rate is:" Q_dot_out=Q_out/MM*m_dot_fuel Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2
SOLUTION for a sample calculation A_th=12.5 fuel$='C8H18(l)' HP=-1.641E+06 [kJ/kg] HR=-250472 [kJ/kg] h_fuel=-249950 MM=114 [kg/kmol] Moles_CO=0.000 Moles_CO2=8.000 Moles_H2O=9 Moles_N2=94.000 Moles_O2=12.500 MolO2=12.5 m_dot_fuel=1 [kg/s] Name$='octane' Q_dot_out=12197 [kW] Q_out=1390433.6 [kJ/kmol_fuel] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=200 [%] Th_air=2.000 T_air=298 [K] T_fuel=298 [K] T_prod=1200 [K] w=0 x=8 y=18 z=0 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-109
15-117 EES The fuel among CH4(g), C2H2(g), C2H6(g), C3H8(g), and C8H18(l) that gives the highest temperature when burned completely in an adiabatic constant-volume chamber with the theoretical amount of air is to be determined. Analysis The problem is solved using EES, and the solution is given below.
Adiabatic Combustion of fuel CnHm with Stoichiometric Air at T_fuel =T_air=T_reac in a constant volume, closed system: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> xCO2 + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + (x+y/4-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Adiabatic, Incomplete Combustion of fuel CnHm with Stoichiometric Air at T_fuel =T_air=T_reac in a constant volume, closed system: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> (x-w)CO2 +wCO + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + ((x+y/4-z/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730"Table A.26" else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950"Table A.26" else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670"Table A.26" endif; endif; endif; endif; endif end
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-110
Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END {"Input data from the diagram window" Theo_air = 200 [%] Fuel$='CH4(g)'} T_reac = 298 [K] T_air = T_reac T_fuel = T_reac R_u = 8.314 [kJ/kmol-K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) UR=(h_fuel-R_u*T_fuel)+ (x+y/4-z/2) *(Theo_air/100) *(enthalpy(O2,T=T_air)R_u*T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *(enthalpy(N2,T=T_air)-R_u*T_air) UP=(x-w)*(enthalpy(CO2,T=T_prod)-R_u*T_prod)+w*(enthalpy(CO,T=T_prod)R_u*T_prod)+(y/2)*(enthalpy(H2O,T=T_prod)-R_u*T_prod)+3.76*(x+y/4-z/2)* (Theo_air/100)*(enthalpy(N2,T=T_prod)-R_u*T_prod)+MolO2*(enthalpy(O2,T=T_prod)R_u*T_prod) UR =UP "Adiabatic, constant volume conservation of energy" Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-111 SOLUTION for CH4 A_th=2 fuel$='CH4(g)' h_fuel=-74875 Moles_CO=0.000 Moles_CO2=1.000 Moles_H2O=2 Moles_N2=7.520 Moles_O2=0.000 MolO2=0 Name$='methane' R_u=8.314 [kJ/kmol-K] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_air=298 [K] T_fuel=298 [K] T_prod=2824 [K] T_reac=298 [K] UP=-100981 UR=-100981 w=0 x=1 y=4 z=0 SOLUTION for C2H2 A_th=2.5 fuel$='C2H2(g)' h_fuel=226730 Moles_CO=0.000 Moles_CO2=2.000 Moles_H2O=1 Moles_N2=9.400 Moles_O2=0.000 MolO2=0 Name$='acetylene' R_u=8.314 [kJ/kmol-K] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_air=298 [K] T_fuel=298 [K] T_prod=3535 [K] T_reac=298 [K] UP=194717 UR=194717 w=0 x=2 y=2 z=0 SOLUTION for CH3OH A_th=1.5 fuel$='CH3OH(g)' h_fuel=-200670 Moles_CO=0.000 Moles_CO2=1.000 Moles_H2O=2 Moles_N2=5.640 Moles_O2=0.000 MolO2=0 Name$='methyl alcohol' R_u=8.314 [kJ/kmol-K] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_air=298 [K] T_fuel=298 [K] T_prod=2817 [K] T_reac=298 [K] UP=-220869 UR=-220869 w=0 x=1 y=4 z=1 SOLUTION for C3H8 A_th=5 fuel$='C3H8(g)' h_fuel=-103858 Moles_CO=0.000 Moles_CO2=3.000 Moles_H2O=4 Moles_N2=18.800 Moles_O2=0.000 MolO2=0 Name$='propane' R_u=8.314 [kJ/kmol-K] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_air=298 [K] T_fuel=298 [K] T_prod=2909 [K] T_reac=298 [K] UP=-165406 UR=-165406 w=0 x=3 y=8 z=0 SOLUTION for C8H18 A_th=12.5 fuel$='C8H18(l)' h_fuel=-249950 Moles_CO=0.000 Moles_CO2=8.000 Moles_H2O=9 Moles_N2=47.000 Moles_O2=0.000 MolO2=0 Name$='octane' R_u=8.314 [kJ/kmol-K] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] Th_air=1.000 T_air=298 [K] T_fuel=298 [K] T_prod=2911 [K] T_reac=298 [K] UP=-400104 UR=-400104 w=0 x=8 y=18 z=0
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-112
Fundamentals of Engineering (FE) Exam Problems
15-118 A fuel is burned with 90 percent theoretical air. This is equivalent to (a) 10% excess air (b) 90% excess air (c) 10% deficiency of air (d) 90% deficiency of air (e) stoichiometric amount of air Answer (c) 10% deficiency of air Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). air_th=0.9 "air_th=air_access+1" air_th=1-air_deficiency
15-119 Propane C3H8 is burned with 150 percent theoretical air. The air-fuel mass ratio for this combustion process is (a) 5.3 (b) 10.5 (c) 15.7 (d) 23.4 (e) 39.3 Answer (d) 23.4 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). n_C=3 n_H=8 m_fuel=n_H*1+n_C*12 a_th=n_C+n_H/4 coeff=1.5 "coeff=1 for theoretical combustion, 1.5 for 50% excess air" n_O2=coeff*a_th n_N2=3.76*n_O2 m_air=n_O2*32+n_N2*28 AF=m_air/m_fuel
15-120 One kmol of methane (CH4) is burned with an unknown amount of air during a combustion process. If the combustion is complete and there are 2 kmol of free O2 in the products, the air-fuel mass ratio is (a) 34.3 (b) 17.2 (c) 19.0 (d) 14.9 (e) 12.1 Answer (a) 34.3 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). n_C=1 n_H=4
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-113
m_fuel=n_H*1+n_C*12 a_th=n_C+n_H/4 (coeff-1)*a_th=2 "O2 balance: Coeff=1 for theoretical combustion, 1.5 for 50% excess air" n_O2=coeff*a_th n_N2=3.76*n_O2 m_air=n_O2*32+n_N2*28 AF=m_air/m_fuel "Some Wrong Solutions with Common Mistakes:" W1_AF=1/AF "Taking the inverse of AF" W2_AF=n_O2+n_N2 "Finding air-fuel mole ratio" W3_AF=AF/coeff "Ignoring excess air"
15-121 A fuel is burned steadily in a combustion chamber. The combustion temperature will be the highest except when (a) the fuel is preheated. (b) the fuel is burned with a deficiency of air. (c) the air is dry. (d) the combustion chamber is well insulated. (e) the combustion is complete. Answer (b) the fuel is burned with a deficiency of air.
15-122 An equimolar mixture of carbon dioxide and water vapor at 1 atm and 60°C enter a dehumidifying section where the entire water vapor is condensed and removed from the mixture, and the carbon dioxide leaves at 1 atm and 60°C. The entropy change of carbon dioxide in the dehumidifying section is (a) –2.8 kJ/kg⋅K (b) –0.13 kJ/kg⋅K (c) 0 (d) 0.13 kJ/kg⋅K (e) 2.8 kJ/kg⋅K Answer (b) –0.13 kJ/kg⋅K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Cp_CO2=0.846 R_CO2=0.1889 T1=60+273 "K" T2=T1 P1= 1 "atm" P2=1 "atm" y1_CO2=0.5; P1_CO2=y1_CO2*P1 y2_CO2=1; P2_CO2=y2_CO2*P2 Ds_CO2=Cp_CO2*ln(T2/T1)-R_CO2*ln(P2_CO2/P1_CO2) "Some Wrong Solutions with Common Mistakes:" W1_Ds=0 "Assuming no entropy change" W2_Ds=Cp_CO2*ln(T2/T1)-R_CO2*ln(P1_CO2/P2_CO2) "Using pressure fractions backwards"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-114
15-123 Methane (CH4) is burned completely with 80% excess air during a steady-flow combustion process. If both the reactants and the products are maintained at 25°C and 1 atm and the water in the products exists in the liquid form, the heat transfer from the combustion chamber per unit mass of methane is (a) 890 MJ/kg (b) 802 MJ/kg (c) 75 MJ/kg (d) 56 MJ/kg (e) 50 MJ/kg Answer (d) 56 MJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T= 25 "C" P=1 "atm" EXCESS=0.8 "Heat transfer in this case is the HHV at room temperature," HHV_CH4 =55.53 "MJ/kg" LHV_CH4 =50.05 "MJ/kg" "Some Wrong Solutions with Common Mistakes:" W1_Q=LHV_CH4 "Assuming lower heating value" W2_Q=EXCESS*hHV_CH4 "Assuming Q to be proportional to excess air"
15-124 The higher heating value of a hydrocarbon fuel CnHm with m = 8 is given to be 1560 MJ/kmol of fuel. Then its lower heating value is (a) 1384 MJ/kmol (b) 1208 MJ/kmol (c) 1402 MJ/kmol (d) 1540 MJ/kmol (e) 1550 MJ/kmol Answer (a) 1384 MJ/kmol Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). HHV=1560 "MJ/kmol fuel" h_fg=2.4423 "MJ/kg, Enthalpy of vaporization of water at 25C" n_H=8 n_water=n_H/2 m_water=n_water*18 LHV=HHV-h_fg*m_water "Some Wrong Solutions with Common Mistakes:" W1_LHV=HHV - h_fg*n_water "Using mole numbers instead of mass" W2_LHV= HHV - h_fg*m_water*2 "Taking mole numbers of H2O to be m instead of m/2" W3_LHV= HHV - h_fg*n_water*2 "Taking mole numbers of H2O to be m instead of m/2, and using mole numbers"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-115
15-125 Acetylene gas (C2H2) is burned completely during a steady-flow combustion process. The fuel and the air enter the combustion chamber at 25°C, and the products leave at 1500 K. If the enthalpy of the products relative to the standard reference state is –404 MJ/kmol of fuel, the heat transfer from the combustion chamber is (a) 177 MJ/kmol (b) 227 MJ/kmol (c) 404 MJ/kmol (d) 631 MJ/kmol (e) 751 MJ/kmol Answer (d) 631 MJ/kmol Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). hf_fuel=226730/1000 "MJ/kmol fuel" H_prod=-404 "MJ/kmol fuel" H_react=hf_fuel Q_out=H_react-H_prod "Some Wrong Solutions with Common Mistakes:" W1_Qout= -H_prod "Taking Qout to be H_prod" W2_Qout= H_react+H_prod "Adding enthalpies instead of subtracting them"
15-126 Benzene gas (C6H6) is burned with 90 percent theoretical air during a steady-flow combustion process. The mole fraction of the CO in the products is (a) 1.6% (b) 4.4% (c) 2.5% (d) 10.0% (e) 16.7% Answer (b) 4.4% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). n_C=6 n_H=6 a_th=n_C+n_H/4 coeff=0.90 "coeff=1 for theoretical combustion, 1.5 for 50% excess air" "Assuming all the H burns to H2O, the combustion equation is C6H6+coeff*a_th(O2+3.76N2)----- (n_CO2) CO2+(n_CO)CO+(n_H2O) H2O+(n_N2) N2" n_O2=coeff*a_th n_N2=3.76*n_O2 n_H2O=n_H/2 n_CO2+n_CO=n_C 2*n_CO2+n_CO+n_H2O=2*n_O2 "Oxygen balance" n_prod=n_CO2+n_CO+n_H2O+n_N2 "Total mole numbers of product gases" y_CO=n_CO/n_prod "mole fraction of CO in product gases" "Some Wrong Solutions with Common Mistakes:" W1_yCO=n_CO/n1_prod; n1_prod=n_CO2+n_CO+n_H2O "Not including N2 in n_prod" W2_yCO=(n_CO2+n_CO)/n_prod "Using both CO and CO2 in calculations"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-116
15-127 A fuel is burned during a steady-flow combustion process. Heat is lost to the surroundings at 300 K at a rate of 1120 kW. The entropy of the reactants entering per unit time is 17 kW/K and that of the products is 15 kW/K. The total rate of exergy destruction during this combustion process is (a) 520 kW (b) 600 kW (c) 1120 kW (d) 340 kW (e) 739 kW Answer (a) 520 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). To=300 "K" Q_out=1120 "kW" S_react=17 "kW'K" S_prod= 15 "kW/K" S_react-S_prod-Q_out/To+S_gen=0 "Entropy balance for steady state operation, SinSout+Sgen=0" X_dest=To*S_gen "Some Wrong Solutions with Common Mistakes:" W1_Xdest=S_gen "Taking Sgen as exergy destruction" W2_Xdest=To*S_gen1; S_react-S_prod-S_gen1=0 "Ignoring Q_out/To"
15-128 ··· 15-133 Design and Essay Problems 15-129a Constant-volume vessels that store flammable gases are to be designed to withstand the rising pressures in case of an explosion. The safe design pressures for (a) acetylene, (b) propane, and (c) n-octane are to be determined for storage pressures slightly above the atmospheric pressure. Analysis (a) The final temperature (and pressure) in the tank will be highest when the combustion is complete, adiabatic, and stoichiometric. In addition, we assume the atmospheric pressure to be 100 kPa and the initial temperature in the tank to be 25°C. Then the initial pressure of the air-fuel mixture in the tank becomes 125 kPa. The combustion equation of C2H2(g) with stoichiometric amount of air is C 2 H 2 + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ 2CO 2 + H 2 O + 3.76a th N 2
where ath is the stoichiometric coefficient and is determined from the O2 balance, a th = 2 + 0.5 ⎯ ⎯→ a th = 2.5
Thus, C 2 H 2 + 2.5(O 2 + 3.76N 2 ) ⎯ ⎯→ 2CO 2 + H 2 O + 9.40N 2
The final temperature in the tank is determined from the energy balance relation E in − E out = ∆E system for reacting closed systems under adiabatic conditions (Q = 0) with no work interactions (W = 0), 0=
∑ N (h P
o f
+ h − h o − Pv
) − ∑ N (h P
R
o f
+ h − h o − Pv
)
R
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-117
Assuming both the reactants and the products to behave as ideal gases, all the internal energy and enthalpies depend on temperature only, and the Pv terms in this equation can be replaced by RuT. It yields
∑ N (h P
o f
+ hTP − h298K − Ru T
) = ∑ N (h P
R
o f
− Ru T
)
R
since the reactants are at the standard reference temperature of 25°C. From the tables,
Substance C2H2 O2 N2 H2O (g) CO2 Thus,
h fo kJ/kmol 226,730 0 0 -241,820 -393,520
h 298 K
kJ/kmol --8682 8669 9904 9364
(2)(− 393,520 + hCO − 9364 − 8.314 × TP ) + (1)(− 241,820 + hH O − 9904 − 8.314 × TP ) + (9.40 )(0 + h N − 8669 − 8.314 × T P ) = (1)(226,730 − 8.314 × 298) + (2.5)(0 − 8.314 × 298) + (9.40 )(0 − 8.314 × 298) 2
2
2
It yields 2hCO 2 + hH 2O + 9.40h N 2 − 103.094 × T P = 1,333,750 kJ
The temperature of the product gases is obtained from a trial and error solution, At 3200 K: 2hCO 2 + hH 2 O + 9.40h N 2 − 103.094 × T P = (2 )(174,695) + (1)(147,457 ) + (9.40)(108,830 ) − (103.094 )(3200) = 1,189,948 kJ (Lower than 1,333,750 kJ )
At 3250 K: 2hCO 2 + hH 2 O + 9.40h N 2 − 103.094 × T P = (2 )(177,822 ) + (1)(150,272 ) + (9.40 )(110,690 ) − (103.094 )(3250 ) = 1,211,347 kJ (Lower than 1,333,750 kJ )
By extrapolation,
TP = 3536 K
Treating both the reactants and the products as ideal gases, the final (maximum) pressure that can occur in the combustion chamber is determined to be N R T P1V N T (12.40 kmol)(3536 K ) (125 kPa ) = 1426 kPa = 1 u 1 ⎯ ⎯→ P2 = 2 2 P1 = (12.90 kmol)(298 K ) P2V N 2 R u T2 N 1T1
Then the pressure the tank must be designed for in order to meet the requirements of the code is P = (4 )(1426 kPa ) = 5704 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-118
15-129b The final temperature (and pressure) in the tank will be highest when the combustion is complete, adiabatic, and stoichiometric. In addition, we assume the atmospheric pressure to be 100 kPa and the initial temperature in the tank to be 25°C. Then the initial pressure of the air-fuel mixture in the tank becomes 125 kPa. The combustion equation of C3H8(g) with stoichiometric amount of air is C 3 H 8 + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ 3CO 2 + 4H 2 O + 3.76a th N 2
where ath is the stoichiometric coefficient and is determined from the O2 balance, a th = 3 + 2 ⎯ ⎯→ a th = 5
Thus,
C3H8 + 5(O 2 + 3.76N 2 ) ⎯ ⎯→ 3CO 2 + 4H 2O + 18.80N 2
The final temperature in the tank is determined from the energy balance relation E in − E out = ∆E system for reacting closed systems under adiabatic conditions (Q = 0) with no work interactions (W = 0), 0=
∑ N (h P
o f
+ h − h o − Pv
) − ∑ N (h R
P
o f
+ h − h o − Pv
)
R
Assuming both the reactants and the products to behave as ideal gases, all the internal energy and enthalpies depend on temperature only, and the Pv terms in this equation can be replaced by RuT. It yields
∑ N (h P
o f
+ hTP − h298 K − Ru T
) = ∑ N (h R
P
o f
− Ru T
)
R
since the reactants are at the standard reference temperature of 25°C. From the tables,
Substance C3H8 O2 N2 H2O (g) CO2 Thus,
h fo
h 298 K
kJ/kmol -103,850 0 0 -241,820 -393,520
kJ/kmol --8682 8669 9904 9364
(3)(− 393,520 + hCO − 9364 − 8.314 × TP ) + (4)(− 241,820 + hH O − 9904 − 8.314 × TP ) + (18.80 )(0 + h N − 8669 − 8.314 × T P ) = (1)(− 103,850 − 8.314 × 298) + (5)(0 − 8.314 × 298) + (18.80 )(0 − 8.314 × 298) 2
2
2
It yields 3hCO2 + 4hH 2O + 18.80hN 2 − 214.50 × TP = 2,213,231 kJ
The temperature of the product gases is obtained from a trial and error solution, At 2950 K:
3hCO 2 + 4hH 2O + 18.80h N 2 − 214.50 × T P = (3)(159,117 ) + (4)(133,486 ) + (18.80 )(99,556 ) − (214.50 )(2950 ) = 2,250,173 kJ (Higher than 2,213,231 kJ )
At 2900 K:
3hCO 2 + 4hH 2O + 18.80h N 2 − 214.50 × T P = (3)(156,009 ) + (4)(130,717 ) + (18.80 )(97,705) − (214.50 )(2900 ) = 2,205,699 kJ (Lower than 2,213,231 kJ )
By interpolation, TP = 2908 K Treating both the reactants and the products as ideal gases, the final (maximum) pressure that can occur in the combustion chamber is determined to be N R T P1V N T (25.80 kmol)(2908 K ) (125 kPa ) = 1269 kPa = 1 u 1 ⎯ ⎯→ P2 = 2 2 P1 = (24.80 kmol)(298 K ) P2V N 2 R u T2 N 1T1 Then the pressure the tank must be designed for in order to meet the requirements of the code is P = (4 )(1269 kPa ) = 5076 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-119
15-129c The final temperature (and pressure) in the tank will be highest when the combustion is complete, adiabatic, and stoichiometric. In addition, we assume the atmospheric pressure to be 100 kPa and the initial temperature in the tank to be 25°C. Then the initial pressure of the air-fuel mixture in the tank becomes 125 kPa. The combustion equation of C8H18(g) with stoichiometric amount of air is C 8 H 18 + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ 8CO 2 + 9H 2 O + 3.76a th N 2
where ath is the stoichiometric coefficient and is determined from the O2 balance, ath = 8 + 4.5 ⎯ ⎯→ ath = 12.5
Thus,
C8H18 + 12.5(O 2 + 3.76N 2 ) ⎯ ⎯→ 8CO 2 + 9H 2O + 47.0N 2
The final temperature in the tank is determined from the energy balance relation E in − E out = ∆E system for reacting closed systems under adiabatic conditions (Q = 0) with no work interactions (W = 0), 0=
∑ N (h P
o f
+ h − h o − Pv
) − ∑ N (h R
P
o f
+ h − h o − Pv
)
R
Assuming both the reactants and the products to behave as ideal gases, all the internal energy and enthalpies depend on temperature only, and the Pv terms in this equation can be replaced by RuT. It yields
∑ N (h P
o f
+ hTP − h298 K − Ru T
) = ∑ N (h R
P
o f
− Ru T
)
R
since the reactants are at the standard reference temperature of 25°C. From the tables,
Substance C8H18 O2 N2 H2O (g) CO2 Thus,
h fo
h 298 K
kJ/kmol -208,450 0 0 -241,820 -393,520
kJ/kmol --8682 8669 9904 9364
(8)(− 393,520 + hCO − 9364 − 8.314 × TP ) + (9)(− 241,820 + hH O − 9904 − 8.314 × TP ) + (47.0)(0 + h N − 8669 − 8.314 × T P ) = (1)(− 208,450 − 8.314 × 298) + (12.5)(0 − 8.314 × 298) + (47.0)(0 − 8.314 × 298) 2
2
2
It yields 8hCO 2 + 9hH 2 O + 47.0h N 2 − 532.10 × T P = 5,537,688 kJ
The temperature of the product gases is obtained from a trial and error solution, At 2950 K:
8hCO 2 + 9hH 2 O + 47.0h N 2 − 532.10 × T P = (8)(159,117 ) + (9)(133,486 ) + (47.0 )(99,556 ) − (532.10 )(2950) = 5,583,747 kJ (Higher than 5,534,220 kJ )
At 2900 K:
8hCO 2 + 9hH 2 O + 47.0h N 2 − 532.10 × T P = (8)(156,009 ) + (9 )(130,717 ) + (47.0 )(97,705) − (532.10 )(2900 ) = 5,473,570 kJ (Lower than 5,534,220 kJ )
By interpolation, TP = 2929 K Treating both the reactants and the products as ideal gases, the final (maximum) pressure that can occur in the combustion chamber is determined to be N R T P1V N T (64.0 kmol)(2929 K ) (125 kPa ) = 1300 kPa = 1 u 1 ⎯ ⎯→ P2 = 2 2 P1 = (60.5 kmol)(298 K ) P2V N 2 R u T2 N 1T1 Then the pressure the tank must be designed for in order to meet the requirements of the code is P = (4)(1300 kPa ) = 5200 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-120
15-130 A certain industrial process generates a liquid solution of ethanol and water as the waste product. The solution is to be burned using methane. A combustion process is to be developed to accomplish this incineration process with minimum amount of methane. Analysis The mass flow rate of the liquid ethanol-water solution is given to be 10 kg/s. Considering that the mass fraction of ethanol in the solution is 0.2, m& ethanol = (0.2 )(10 kg/s ) = 2 kg/s m& water = (0.8)(10 kg/s ) = 8 kg/s
Noting that the molar masses Methanol = 46 and Mwater = 18 kg/kmol and that mole numbers N = m/M, the mole flow rates become m& 2 kg/s N& ethanol = ethanol = = 0.04348 kmol/s M ethanol 46 kg/kmol m& 8 kg/s N& water = water = = 0.44444 kmol/s M water 18 kg/kmol
Note that N& water 0.44444 = = 10.222 kmol H 2 O/kmol C 2 H 5 OH N& ethanol 0.04348
That is, 10.222 moles of liquid water is present in the solution for each mole of ethanol. Assuming complete combustion, the combustion equation of C2H5OH (l) with stoichiometric amount of air is C 2 H 5 OH(l ) + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ 2CO 2 + 3H 2 O + 3.76a th N 2
where ath is the stoichiometric coefficient and is determined from the O2 balance, 1 + 2a th = 4 + 3 ⎯ ⎯→ a th = 3
Thus, C 2 H 5 OH(l ) + 3(O 2 + 3.76N 2 ) ⎯ ⎯→ 2CO 2 + 3H 2 O + 11.28N 2
Noting that 10.222 kmol of liquid water accompanies each kmol of ethanol, the actual combustion equation can be written as C 2 H 5 OH(l ) + 3(O 2 + 3.76N 2 ) + 10.222H 2 O(l ) ⎯ ⎯→ 2CO 2 + 3H 2 O(g ) + 11.28N 2 + 10.222H 2 O(l )
The heat transfer for this combustion process is determined from the steady-flow energy balance equation with W = 0, Q=
∑ N (h P
o f
+h −ho
) − ∑ N (h P
R
o f
+h −ho
)
R
Assuming the air and the combustion products to be ideal gases, we have h = h(T). We assume all the reactants to enter the combustion chamber at the standard reference temperature of 25°C. Furthermore, we assume the products to leave the combustion chamber at 1400 K which is a little over the required temperature of 1100°C. From the tables,
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
15-121
Substance C2H5OH (l) CH4 O2 N2 H2O (g) H2O (l) CO2
h fo kJ/kmol -277,690 -74,850 0 0 -241,820 -285,830 -393,520
h 298 K
h1400 K
kJ/kmol ----8682 8669 9904 --9364
kJ/kmol ----45,648 43,605 53,351 --65,271
Thus, Q = (2 )(−393,520 + 65,271 − 9364 ) + (3)(−241,820 + 53,351 − 9904) + (11.28)(0 + 43,605 − 8669) − (1)(− 277,690 ) − 0 − 0 + (10.222 )(− 241,820 + 53,351 − 9904) − (10.222 )(− 285,830 ) = 295,409 kJ/kmol of C2H5OH
The positive sign indicates that 295,409 kJ of heat must be supplied to the combustion chamber from another source (such as burning methane) to ensure that the combustion products will leave at the desired temperature of 1400 K. Then the rate of heat transfer required for a mole flow rate of 0.04348 kmol C2H5OH/s CO becomes Q& = N& Q = (0.04348 kmol/s )(295,409 kJ/kmol) = 12,844 kJ/s
Assuming complete combustion, the combustion equation of CH4(g) with stoichiometric amount of air is CH 4 + a th (O 2 + 3.76N 2 ) ⎯ ⎯→ CO 2 + 2H 2 O + 3.76a th N 2
where ath is the stoichiometric coefficient and is determined from the O2 balance, Thus,
a th = 1 + 1 ⎯ ⎯→ a th = 2 CH 4 + 2(O 2 + 3.76N 2 ) ⎯ ⎯→ CO 2 + 2H 2 O + 7.52N 2
The heat transfer for this combustion process is determined from the steady-flow energy balance E in − E out = ∆E system equation as shown above under the same assumptions and using the same mini table: Q = (1)(−393,520 + 65,271 − 9364 ) + (2 )(−241,820 + 53,351 − 9904 ) + (7.52 )(0 + 43,605 − 8669) − (1)(− 74,850 ) − 0 − 0 = −396,790 kJ/kmol of CH 4
That is, 396,790 kJ of heat is supplied to the combustion chamber for each kmol of methane burned. To supply heat at the required rate of 12,844 kJ/s, we must burn methane at a rate of 12,844 kJ/s Q& N& CH 4 = = = 0.03237 kmolCH 4 /s Q 396,790 kJ/kmol or, m& CH 4 = M CH 4 N& CH 4 = (16 kg/kmol)(0.03237 kmolCH 4 /s ) = 0.5179 kg/s Therefore, we must supply methane to the combustion chamber at a minimum rate 0.5179 kg/s in order to maintain the temperature of the combustion chamber above 1400 K.
KJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-1
Chapter 16 CHEMICAL AND PHASE EQUILIBRIUM The Kp and Equilibrium Composition of Ideal Gases 16-1C Because when a reacting system involves heat transfer, the increase-in-entropy principle relation requires a knowledge of heat transfer between the system and its surroundings, which is impractical. The equilibrium criteria can be expressed in terms of the properties alone when the Gibbs function is used. 16-2C No, the wooden table is NOT in chemical equilibrium with the air. With proper catalyst, it will reach with the oxygen in the air and burn. 16-3C They are PC C PDν D v
Kp =
PAν A PBvB
, K p = e − ∆G*(T ) / RuT
ν
and K p =
N CC N νDD ⎛ P ⎜ ⎜ N νAA N νBB ⎝ N total
⎞ ⎟ ⎟ ⎠
∆ν
where ∆ν = ν C + ν D −ν A −ν B . The first relation is useful in partial pressure calculations, the second in determining the Kp from gibbs functions, and the last one in equilibrium composition calculations. 16-4C (a) Kp1, (b) 1/Kp1, (c) Kp1, (d) Kp1, (e) Kp12. 16-5C (a) Kp1, (b) 1/Kp1, (c) Kp12, (d) Kp1, (e) 1/Kp13. 16-6C (a) No, because Kp depends on temperature only. (b) Yes, because the total mixture pressure affects the mixture composition. The equilibrium constant for the reaction CO + 21 O 2 ⇔ CO 2 can be expressed as ν
Kp =
CO 2 N CO 2
ν
ν
CO N CO N OO 2 2
⎛ P ⎜ ⎜N ⎝ total
⎞ ⎟ ⎟ ⎠
( vCO 2 −ν CO −ν O 2 )
The value of the exponent in this case is 1-1-0.5=-0.5, which is negative. Thus as the pressure increases, the term in the brackets will decrease. The value of Kp depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products (CO2) must increase, and the number of moles of the reactants (CO, O2) must decrease. 16-7C (a) No, because Kp depends on temperature only. (b) In general, the total mixture pressure affects the mixture composition. The equilibrium constant for the reaction N 2 + O 2 ⇔ 2NO can be expressed as ν
Kp =
NO N NO
ν
ν
N NN 2 N OO 2 2
2
⎛ P ⎜ ⎜N ⎝ total
⎞ ⎟ ⎟ ⎠
(ν NO −ν N 2 −ν O 2 )
The value of the exponent in this case is 2-1-1 = 0. Therefore, changing the total mixture pressure will have no effect on the number of moles of N2, O2 and NO.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-2
16-8C (a) The equilibrium constant for the reaction CO + 12 O 2 ⇔ CO 2 can be expressed as ν
Kp =
CO 2 N CO 2
ν
ν
CO N CO N OO 2 2
⎛ P ⎜ ⎜N ⎝ total
⎞ ⎟ ⎟ ⎠
(ν CO 2 −ν CO −ν O 2 )
Judging from the values in Table A-28, the Kp value for this reaction decreases as temperature increases. That is, the indicated reaction will be less complete at higher temperatures. Therefore, the number of moles of CO2 will decrease and the number moles of CO and O2 will increase as the temperature increases. (b) The value of the exponent in this case is 1-1-0.5=-0.5, which is negative. Thus as the pressure increases, the term in the brackets will decrease. The value of Kp depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products (CO2) must increase, and the number of moles of the reactants (CO, O2) must decrease. 16-9C (a) The equilibrium constant for the reaction N 2 ⇔ 2N can be expressed as ν
Kp =
N NN ⎛ P ⎜ ⎜ ν N NN 2 ⎝ N total 2
⎞ ⎟ ⎟ ⎠
(ν N −ν N 2 )
Judging from the values in Table A-28, the Kp value for this reaction increases as the temperature increases. That is, the indicated reaction will be more complete at higher temperatures. Therefore, the number of moles of N will increase and the number moles of N2 will decrease as the temperature increases. (b) The value of the exponent in this case is 2-1 = 1, which is positive. Thus as the pressure increases, the term in the brackets also increases. The value of Kp depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products (N) must decrease, and the number of moles of the reactants (N2) must increase. 16-10C The equilibrium constant for the reaction CO + 12 O 2 ⇔ CO 2 can be expressed as ν
Kp =
CO 2 N CO 2
ν
ν
CO N CO N OO 2 2
⎛ P ⎜ ⎜N ⎝ total
⎞ ⎟ ⎟ ⎠
(ν CO 2 −ν CO −ν O 2 )
Adding more N2 (an inert gas) at constant temperature and pressure will increase Ntotal but will have no direct effect on other terms. Then to keep the equation balanced, the number of moles of the products (CO2) must increase, and the number of moles of the reactants (CO, O2) must decrease. 16-11C The values of the equilibrium constants for each dissociation reaction at 3000 K are, from Table A-28, N 2 ⇔ 2N ⇔ ln K p = −22.359 H 2 ⇔ 2H ⇔ ln K p = −3.685
(greater than - 22.359)
Thus H2 is more likely to dissociate than N2.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-3
16-12 The equilibrium constant of the reaction H2 + 1/2O2 ↔ H2O is listed in Table A-28 at different temperatures. The data are to be verified at two temperatures using Gibbs function data. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G*( T )/ Ru T or ln K p = − ∆G *(T ) / Ru T
H2 + ½O2 ↔ H2O
where ∆G * (T )
= ν H 2O g H∗ 2 O (T ) −ν H 2
g H∗ 2
(T ) −ν O 2 g O∗ 2
(T )
25ºC
At 25°C, ∆G *(T ) = 1( −228,590) − 1(0) − 0.5(0) = −228,590 kJ / kmol
Substituting, ln K p = −(−228,590 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(298 K)] = 92.26
or K p = 1.12 × 10 40 (Table A - 28: ln K p = 92.21)
(b) At 2000 K, ∆G * (T ) = ν H 2 O g H∗ 2 O (T ) −ν H 2 g H∗ 2 (T ) −ν O 2 g O∗ 2 (T ) = ν H 2 O ( h − T s ) H 2 O − ν H 2 ( h − Ts ) H 2 − ν O 2 ( h − Ts ) O 2 = ν H 2 O [(h f + h2000 − h298 ) − Ts ] H 2O −ν H 2 [(h f + h2000 − h298 ) − Ts ] H 2 −ν O 2 [(h f + h2000 − h298 ) − Ts ] O 2 = 1× (−241,820 + 82,593 − 9904 − 2000 × 264.571) − 1× (0 + 61,400 − 8468 − 2000 × 188.297) − 0.5 × (0 + 67,881 − 8682) − 2000 × 268.655) = −135,556 kJ/kmol
Substituting, ln K p = −(−135,556 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(2000 K)] = 8.152
or K p = 3471 (Table A - 28 : ln K p = 8.145)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-4
16-13E The equilibrium constant of the reaction H2 + 1/2O2 ↔ H2O is listed in Table A-28 at different temperatures. The data are to be verified at two temperatures using Gibbs function data. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G*( T )/ Ru T or ln K p = − ∆G * (T ) / Ru T
where
H2 + ½O2 ↔ H2O ∆G * (T )
= ν H 2O g H∗ 2 O (T ) −ν H 2
g H∗ 2
(T ) −ν O 2 g O∗ 2
(T )
537 R
At 537 R, ∆G * (T ) = 1( −98,350) − 1(0) − 0.5(0) = −98,350 Btu / lbmol
Substituting, ln K p = − ( −98,350 Btu / lbmol) / [(1.986 Btu / lbmol ⋅ R)(537 R)] = 92.22
or K p = 1.12 × 10 40 (Table A - 28: ln K p = 92.21)
(b) At 3240 R, ∆G * (T ) = ν H 2 O g H∗ 2 O (T ) −ν H 2 g H∗ 2 (T ) −ν O 2 g O∗ 2 (T ) = ν H 2 O ( h − T s ) H 2 O − ν H 2 ( h − Ts ) H 2 − ν O 2 ( h − Ts ) O 2 = ν H 2 O [(h f + h3240 − h537 ) − Ts ] H 2O −ν H 2 [(h f + h3240 − h298 ) − Ts ] H 2 −ν O 2 [(h f + h3240 − h298 ) − Ts ] O 2 = 1× (−104,040 + 31,204.5 − 4258 − 3240 × 61.948) − 1× (0 + 23,484.7 − 3640.3 − 3240 × 44.125) − 0.5 × (0 + 25,972 − 3725.1 − 3240 × 63.224) = −63,385 Btu/lbmol
Substituting, ln K p = −(−63,385 Btu/lbmol)/[(1.986 Btu/lbmol.R)(3240 R)] = 9.85
or K p = 1.90 × 104 (Table A - 28: ln K p = 9.83)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-5
16-14 The equilibrium constant of the reaction CO + 1/2O2 ↔ CO2 at 298 K and 2000 K are to be determined, and compared with the values listed in Table A-28. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G*( T )/ Ru T or ln K p = − ∆G * (T ) / Ru T
where ∗ ∗ ∗ ∆G * (T ) = ν CO2 g CO2 (T ) −ν CO g CO (T ) −ν O2 g O2 (T )
CO + 1 O 2 ⇔ CO 2 2
298 K
At 298 K, ∆G * (T ) = 1(−394,360) − 1(−137,150) − 0.5(0) = −257,210 kJ/kmol
where the Gibbs functions are obtained from Table A-26. Substituting, ln K p = −
(−257,210 kJ/kmol) = 103.81 (8.314 kJ/kmol ⋅ K)(298 K) ln K p = 103.76
From Table A-28: (b) At 2000 K,
∗ ∗ ∗ ∆G * (T ) = ν CO2 g CO2 (T ) −ν CO g CO (T ) −ν O2 g O2 (T )
= ν CO2 (h − Ts ) CO2 −ν CO (h − Ts ) CO −ν O2 (h − Ts ) O2
= 1[(−302,128) − (2000)(309.00)] − 1[(−53,826) − (2000)(258.48)] − 0.5[(59,193) − (2000)(268.53)] = −110,409 kJ/kmol
The enthalpies at 2000 K and entropies at 2000 K and 101.3 kPa (1 atm) are obtained from EES. Substituting, ln K p = −
(−110,409 kJ/kmol) = 6.64 (8.314 kJ/kmol ⋅ K)(2000 K)
From Table A-28: ln K p = 6.635
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-6
16-15 EES The effect of varying the percent excess air during the steady-flow combustion of hydrogen is to be studied. Analysis The combustion equation of hydrogen with stoichiometric amount of air is H 2 + 0.5[O 2 + 3.76N 2 ] ⎯ ⎯→ H 2 O + 0.5(3.76) N 2
For the incomplete combustion with 100% excess air, the combustion equation is H 2 + (1 + Ex)(0.5)[O 2 + 3.76N 2 ] ⎯ ⎯→ 0.97 H 2 O + a H 2 + b O 2 + c N 2
The coefficients are to be determined from the mass balances Hydrogen balance:
2 = 0.97 × 2 + a × 2 ⎯ ⎯→ a = 0.03
Oxygen balance:
(1 + Ex) × 0.5 × 2 = 0.97 + b × 2
Nitrogen balance: (1 + Ex) × 0.5 × 3.76 × 2 = c × 2 Solving the above equations, we find the coefficients (Ex = 1, a = 0.03 b = 0.515, c = 3.76) and write the balanced reaction equation as H 2 + [O 2 + 3.76N 2 ] ⎯ ⎯→ 0.97 H 2 O + 0.03 H 2 + 0.515 O 2 + 3.76 N 2
Total moles of products at equilibrium are N tot = 0.97 + 0.03 + 0.515 + 3.76 = 5.275
The assumed equilibrium reaction is H 2 O ←⎯→ H 2 + 0.5O 2
The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G*( T )/ Ru T or ln K p = − ∆G * (T ) / Ru T
where ∗ ∗ ∗ ∆G * (T ) = ν H2 g H2 (Tprod ) + ν O2 g O2 (Tprod ) −ν H2O g H2O (Tprod )
and the Gibbs functions are defined as ∗ g H2 (Tprod ) = (h − Tprod s ) H2 ∗ g O2 (Tprod ) = (h − Tprod s ) O2 ∗ g H2O (Tprod ) = (h − Tprod s ) H2O
The equilibrium constant is also given by ⎛ P K p = ⎜⎜ ⎝ N tot
and
⎞ ⎟ ⎟ ⎠
1+ 0.5 −1
ab 0.5
⎛ 1 ⎞ =⎜ ⎟ 1 0.97 ⎝ 5.275 ⎠
0.5
(0.03)(0.515) 0.5 = 0.009664 0.97
ln K p = ln(0.009664) = −4.647
The corresponding temperature is obtained solving the above equations using EES to be Tprod = 2600 K
This is the temperature at which 97 percent of H2 will burn into H2O. The copy of EES solution is given next.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-7
"Input Data from parametric table:" {PercentEx = 10} Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The combustion equation of H2 with stoichiometric amount of air is H2 + 0.5(O2 + 3.76N2)=H2O +0.5(3.76)N2" "For the incomplete combustion with 100% excess air, the combustion equation is H2 + (1+EX)(0.5)(O2 + 3.76N2)=0.97 H2O +aH2 + bO2+cN2" "Specie balance equations give the values of a, b, and c." "H, hydrogen" 2 = 0.97*2 + a*2 "O, oxygen" (1+Ex)*0.5*2=0.97 + b*2 "N, nitrogen" (1+Ex)*0.5*3.76 *2 = c*2 N_tot =0.97+a +b +c "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is H2O=H2+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each H2mponent in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_H2O=Enthalpy(H2O,T=T_prod )-T_prod *Entropy(H2O,T=T_prod ,P=101.3) g_H2=Enthalpy(H2,T=T_prod )-T_prod *Entropy(H2,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_H2+0.5*g_O2-1*g_H2O "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(a^1*b^0.5)/(0.97^1)" sqrt(P/N_tot )*a *sqrt(b )=K_P *0.97 lnK_p = ln(k_P) 2625
ln Kp -5.414 -5.165 -5.019 -4.918 -4.844 -4.786 -4.739 -4.7 -4.667 -4.639
PercentEx [%] 10 20 30 40 50 60 70 80 90 100
Tprod [K] 2440 2490 2520 2542 2557 2570 2580 2589 2596 2602
2585
2545 d or p
T
2505
2465
2425 10
20
30
40
50 60 PercentEx
70
80
90
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
100
16-8
16-16 The equilibrium constant of the reaction CH4 + 2O2 ↔ CO2 + 2H2O at 25°C is to be determined. Analysis The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G*( T )/ Ru T or ln K p = − ∆G * (T ) / Ru T
where
CH4 + 2O2 ↔ CO2 + 2H2O 25°C
∗ ∗ ∆G * (T ) = ν CO 2 g CO (T ) + ν H 2 O g H∗ 2O (T ) −ν CH 4 g CH (T ) −ν O 2 g O∗ 2 (T ) 2 4
At 25°C, ∆G * (T ) = 1( −394,360) + 2( −228,590) − 1( −50,790) − 2(0) = −800,750 kJ / kmol
Substituting, ln K p = −(−800,750 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(298 K)] = 323.04
or
K p = 1.96 × 10 140
16-17 The equilibrium constant of the reaction CO2 ↔ CO + 1/2O2 is listed in Table A-28 at different temperatures. It is to be verified using Gibbs function data. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e − ∆G*(T ) / RuT or ln K p = − ∆G * (T ) / Ru T
where
∗ ∗ ∆G * (T ) = ν CO g CO (T ) + ν O 2 g O∗ 2 (T ) −ν CO 2 g CO (T ) 2
At 298 K,
CO2 ↔ CO + ½O2 298 K
∆G * (T ) = 1(−137,150) + 0.5(0) − 1(−394,360) = 257,210 kJ/kmol
Substituting, ln K p = −(257,210 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(298 K)] = -103.81
or
K p = 8.24 × 10 -46 (Table A - 28 : ln K p = −103.76)
(b) At 1800 K, ∗ ∗ ∆G * (T ) = ν CO g CO (T ) + ν O 2 g O∗ 2 (T ) −ν CO 2 g CO (T ) 2
= ν CO (h − Ts ) CO + ν O 2 (h − Ts ) O 2 −ν CO 2 (h − Ts ) CO 2 = ν CO [(h f + h1800 − h298 ) − Ts ] CO + ν O 2 [(h f + h1800 − h298 ) − Ts ] O 2 −ν CO 2 [(h f + h1800 − h298 ) − Ts ] CO 2 = 1× (−110,530 + 58,191 − 8669 − 1800 × 254.797) + 0.5 × (0 + 60,371 − 8682 − 1800 × 264.701) − 1× (−393,520 + 88,806 − 9364 − 1800 × 302.884) = 127,240.2 kJ/kmol
Substituting, or
ln K p = −(127,240.2 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(1800 K)] = −8.502
K p = 2.03 × 10 -4 (Table A - 28 : ln K p = −8.497)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-9
16-18 The equilibrium constant of the reaction H2O ↔ 1/2H2 + OH is listed in Table A-28 at different temperatures. It is to be verified at a given temperature using Gibbs function data. Analysis The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G*( T )/ Ru T or ln K p = − ∆G * (T ) / Ru T
where ∆G * (T )
= ν H 2 g H∗ 2
∗ (T ) + ν OH g OH (T ) −ν H 2O g H∗ 2O (T )
H2O ↔ ½H2 + OH 25°C
At 298 K, ∆G * (T ) = 0.5(0) + 1(34,280) − 1( −228,590) = 262,870 kJ / kmol
Substituting, ln K p = −(262,870 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(298 K)] = -106.10
or K p = 8.34 × 10 -47 (Table A - 28: ln K p = −106.21)
16-19 The temperature at which 5 percent of diatomic oxygen dissociates into monatomic oxygen at a specified pressure is to be determined. Assumptions 1 The equilibrium composition consists of O2 and O. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions can be written as Stoichiometric:
O 2 ⇔ 2O (thus ν O 2 = 1 and ν O = 2)
Actual:
O 2 ⇔ 0.95O 2 + 0{ .1O 1 424 3 react.
O2 ↔ 2O 5% 3 atm
prod.
The equilibrium constant Kp can be determined from ν O −ν O 2
Nν O ⎛ P ⎞ ⎟ K p = νO ⎜⎜ ⎟ N OO 2 ⎝ N total ⎠ 2
=
0.12 ⎛ 3 ⎞ ⎟ ⎜ 0.95 ⎝ 0.95 + 0.1 ⎠
2 −1
= 0.03008
From Table A-28, the temperature corresponding to this Kp value is T = 3133 K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-10
16-20 The temperature at which 5 percent of diatomic oxygen dissociates into monatomic oxygen at a specified pressure is to be determined. Assumptions 1 The equilibrium composition consists of O2 and O. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions can be written as Stoichiometric:
O 2 ⇔ 2O (thus ν O 2 = 1 and ν O = 2)
Actual:
O 2 ⇔ 0.95O 2 + 0{ .1O 1 424 3 react.
O2 ↔ 2O 5% 6 atm
prod.
The equilibrium constant Kp can be determined from ν
N O ⎛ P K p = νO ⎜⎜ N OO 2 ⎝ N total 2
ν O −ν O 2
⎞ ⎟ ⎟ ⎠
=
0.12 ⎛ 6 ⎞ ⎟ ⎜ 0.95 ⎝ 0.95 + 0.1 ⎠
2 −1
= 0.06015
From Table A-28, the temperature corresponding to this Kp value is T = 3152 K
16-21 [Also solved by EES on enclosed CD] Carbon monoxide is burned with 100 percent excess air. The temperature at which 97 percent of CO burn to CO2 is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written as Stoichiometric:
CO + 12 O 2 ⇔ CO 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )
Actual:
CO + 1(O 2 + 3.76 N 2 )
⎯ ⎯→
0.97 CO 2 + 0.03 CO + 0.515O 2 + 3.76 N 2 1424 3 14442444 3 12 4 4 3 product
reactants
inert
The equilibrium constant Kp can be determined from ν
Kp =
N COCO2 2
ν
νCO N CO N OO 2 2
⎛ P ⎞ ⎟⎟ ⎜⎜ ⎝ N total ⎠
( νCO 2 − νCO − νO 2 )
1−1.5
CO + ½O2 ↔ CO2 97 % 1 atm
0.97 1 ⎛ ⎞ ⎜ ⎟ 0.03 × 0.5150.5 ⎝ 0.97 + 0.03 + 0.515 + 3.76 ⎠ = 103.48 =
From Table A-28, the temperature corresponding to this Kp value is T = 2276 K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-11
16-22 EES Problem 16-21 is reconsidered. The effect of varying the percent excess air during the steadyflow process from 0 to 200 percent on the temperature at which 97 percent of CO burn into CO2 is to be studied. Analysis The problem is solved using EES, and the solution is given below. "To solve this problem, we need to give EES a guess value for T_prop other than the default value of 1. Set the guess value of T_prod to 1000 K by selecting Variable Infromation in the Options menu. Then press F2 or click the Calculator icon." "Input Data from the diagram window:" {PercentEx = 100} Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3 [kPa] R_u=8.314 [kJ/kmol-K] "The combustion equation of CO with stoichiometric amount of air is CO + 0.5(O2 + 3.76N2)=CO2 +0.5(3.76)N2" "For the incomplete combustion with 100% excess air, the combustion equation is CO + (!+EX)(0.5)(O2 + 3.76N2)=0.97 CO2 +aCO + bO2+cN2" "Specie balance equations give the values of a, b, and c." "C, Carbon" 1 = 0.97 + a "O, oxygen" 1 +(1+Ex)*0.5*2=0.97*2 + a *1 + b*2 "N, nitrogen" (1+Ex)*0.5*3.76 *2 = c*2 N_tot =0.97+a +b +c "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(a^1*b^0.5)/(0.97^1)" sqrt(P/N_tot )*a *sqrt(b )=K_P *0.97 lnK_p = ln(k_P) "Compare the value of lnK_p calculated by EES with the value of lnK_p from table A-28 in the text." PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-12
PercentEx [%] 0 20 40 60 80 100 120 140 160 180 200
Tprod [K] 2066 2194 2230 2250 2263 2273 2280 2285 2290 2294 2297
2350 2300 2250
] K [ d o r p
2200 2150
T
2100 2050 0
40
80
120
160
200
PercentEx [%]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-13
16-23E Carbon monoxide is burned with 100 percent excess air. The temperature at which 97 percent of CO burn to CO2 is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written as Stoichiometric:
CO + 12 O 2 ⇔ CO 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )
Actual:
CO + 1(O 2 + 3.76 N 2 )
⎯ ⎯→
0.97 CO 2 + 0.03 CO + 0.515O 2 + 3.76 N 2 1424 3 14442444 3 12 4 4 3 product
reactants
inert
The equilibrium constant Kp can be determined from ν
Kp =
N COCO2 2
ν
νCO N CO N OO2 2
⎛ P ⎜⎜ ⎝ N total
0.97
=
0.03 × 0.5150.5 = 103.48
⎞ ⎟⎟ ⎠
( νCO 2 − νCO − νO 2 )
1 ⎛ ⎞ ⎜ ⎟ ⎝ 0.97 + 0.03 + 0.515 + 3.76 ⎠
CO + ½O2 ↔ CO2 97 % 1 atm
1−1.5
From Table A-28, the temperature corresponding to this Kp value is T = 2276 K = 4097 R
16-24 Hydrogen is burned with 150 percent theoretical air. The temperature at which 98 percent of H2 will burn to H2O is to be determined. Assumptions 1 The equilibrium composition consists of H2O, H2, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written as Stoichiometric:
H 2 + 21 O 2 ⇔ H 2 O (thus ν H 2 O = 1, ν H 2 = 1, and ν O 2 = 21 )
Actual:
H 2 + 0.75(O 2 + 3.76 N 2 )
⎯ ⎯→
0.98 H 2 O + 0.02 H 2 + 0.26O 2 + 2.82 N 2 1424 3 144 42444 3 12 4 4 3 product
reactants
inert
The equilibrium constant Kp can be determined from ν
Kp = =
N HHO2O 2
ν H2
ν O2
2
2
NH NO
⎛ P ⎜ ⎜N ⎝ total
0.98
0.02 × 0.26 0.5 = 194.11
⎞ ⎟ ⎟ ⎠
(ν H 2 O −ν H 2 −ν O 2 )
1 ⎛ ⎞ ⎜ ⎟ 0 . 98 0 . 02 0 . 26 2 . 82 + + + ⎝ ⎠
H2 Combustion chamber
1−1.5
H2O, H2 O 2, N 2
Air
From Table A-28, the temperature corresponding to this Kp value is T = 2472 K.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-14
16-25 Air is heated to a high temperature. The equilibrium composition at that temperature is to be determined. Assumptions 1 The equilibrium composition consists of N2, O2, and NO. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are N 2 + 12 O 2 ⇔ NO (thus ν NO = 1, ν N 2 = 12 , and ν O 2 = 12 )
Stoichiometric:
1 2
Actual:
3.76 N 2 + O 2
⎯ ⎯→
x NO + y N + z O 123 1422432 prod.
reactants
N balance:
7.52 = x + 2y or y = 3.76 - 0.5x
O balance:
2 = x + 2z or z = 1 - 0.5x
Total number of moles:
Ntotal = x + y + z = x + 4.76- x = 4.76
AIR 2000 K 2 atm
The equilibrium constant relation can be expressed as
Kp =
NO N νNO
ν
ν
N NN2 2 N OO2 2
⎛ P ⎜⎜ ⎝ N total
⎞ ⎟⎟ ⎠
(ν NO −ν N 2 −ν O 2 )
From Table A-28, ln Kp = -3.931 at 2000 K. Thus Kp = 0.01962. Substituting, 0.01962 =
x (3.76 − 0.5 x) 0.5 (1 − 0.5 x) 0.5
⎛ 2 ⎞ ⎜ ⎟ ⎝ 4.76 ⎠
1−1
Solving for x, x = 0.0376 Then, y = 3.76-0.5x = 3.7412 z = 1-0.5x = 0.9812 Therefore, the equilibrium composition of the mixture at 2000 K and 2 atm is 0.0376NO + 3.7412N 2 + 0.9812O 2
The equilibrium constant for the reactions O 2 ⇔ 2O (ln Kp = -14.622) and N 2 ⇔ 2 N (ln Kp = -41.645) are much smaller than that of the specified reaction (ln Kp = -3.931). Therefore, it is realistic to assume that no monatomic oxygen or nitrogen will be present in the equilibrium mixture. Also the equilibrium composition is in this case is independent of pressure since ∆ν = 1 − 0.5 − 0.5 = 0 .
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-15
16-26 Hydrogen is heated to a high temperature at a constant pressure. The percentage of H2 that will dissociate into H is to be determined. Assumptions 1 The equilibrium composition consists of H2 and H. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions can be written as Stoichiometric: H 2 ⇔ 2H (thus ν H 2 = 1 and ν H = 2) H2 ⎯ ⎯→ { xH 2 + { yH
Actual:
prod.
react.
H balance: 2 = 2x + y or y = 2 - 2x Total number of moles: Ntotal = x + y = x + 2 - 2x = 2 - x The equilibrium constant relation can be expressed as
H2 3200 K 8 atm
ν H −ν H 2
Nν H ⎛ P ⎞ ⎟ K p = νH ⎜⎜ ⎟ N HH2 2 ⎝ N total ⎠
From Table A-28, ln Kp = -2.534 at 3200 K. Thus Kp = 0.07934. Substituting, 0.07934 =
(2 − 2 x) 2 x
⎛ 8 ⎞ ⎟ ⎜ ⎝2− x⎠
2 −1
Solving for x, x = 0.95 Thus the percentage of H2 which dissociates to H at 3200 K and 8 atm is 1 - 0.95 = 0.05 or 5.0%
16-27 Carbon dioxide is heated to a high temperature at a constant pressure. The percentage of CO2 that will dissociate into CO and O2 is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, and O2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are Stoichiometric: CO 2 ⇔ CO + 12 O 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 ) CO 2 ⎯ ⎯→ xCO 2 + y CO + z O 2 123 14243
Actual:
react.
CO2
products
2400 K 3 atm
C balance:
1= x+ y ⎯ ⎯→ y = 1 − x
O balance:
2 = 2 x + y + 2z ⎯ ⎯→ z = 0.5 − 0.5x
Total number of moles:
N total = x + y + z = 1.5 − 0.5 x
The equilibrium constant relation can be expressed as ν
CO NνCO N OO 2 ⎛ P ⎞ 2 ⎜ ⎟ Kp = ν CO 2 ⎜N ⎟ N CO ⎝ total ⎠ 2
(ν CO +ν O 2 −ν CO 2 )
From Table A-28, ln K p = −3.860 at 2400 K. Thus K p = 0.02107 Substituting, 1.5 −1
0.02107 =
(1 − x)(0.5 − 0.5 x)1 / 2 ⎛ 3 ⎞ ⎟ ⎜ x ⎝ 1.5 − 0.5 x ⎠
Solving for x, x = 0.936 Thus the percentage of CO2 which dissociates into CO and O2 is 1-0.936 = 0.064 or 6.4%
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-16
16-28 A mixture of CO and O2 is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, and O2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are Stoichiometric:
CO 2 ⇔ CO + 12 O 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )
Actual:
CO + 3O 2 ⎯ ⎯→ xCO 2 + y CO + z O 2 123 14243 react.
⎯ ⎯→
products
C balance:
1= x + y
y = 1− x
O balance:
7 = 2 x + y + 2 z or z = 3 − 0.5x
Total number of moles:
N total = x + y + z = 4 − 0.5x
1 CO 3 O2 2200 K 2 atm
The equilibrium constant relation can be expressed as ν
Kp =
CO 2 N CO 2
ν
ν
CO N CO N OO 2 2
⎛ P ⎜ ⎜N ⎝ total
⎞ ⎟ ⎟ ⎠
(ν CO 2 −ν CO −ν O 2 )
From Table A-28, ln K p = 5.120 at 2200 K. Thus K p = 167.34. Substituting, 167.34 =
x (1 − x)(3 − 0.5 x) 0.5
2 ⎛ ⎞ ⎜ ⎟ − x 4 0 . 5 ⎝ ⎠
1−1.5
Solving for x, x = 0.995 Then, y = 1 - x = 0.005 z = 3 - 0.5x = 2.5025 Therefore, the equilibrium composition of the mixture at 2200 K and 2 atm is 0.995CO 2 + 0.005CO + 2.5025O 2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-17
16-29E A mixture of CO, O2, and N2 is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are Stoichiometric:
CO + 12 O 2 ⇔ CO 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )
Actual:
2 CO + 2 O 2 + 6 N 2
⎯ ⎯→
x CO + y CO + z O 2 + 6 N 2 1232 14244 3 products
C balance:
2= x+ y
⎯ ⎯→
O balance:
6 = 2x + y + 2z
Total number of moles:
N total = x + y + z + 6 = 10 − 0.5x
reactants
:
inert
2 CO 2 O2 6 N2 4320 R 3 atm
y = 2− x
⎯ ⎯→
z = 2 − 0.5x
The equilibrium constant relation can be expressed as ν
Kp =
CO 2 N CO 2
ν
ν
CO N CO N OO 2 2
⎛ P ⎜ ⎜N ⎝ total
⎞ ⎟ ⎟ ⎠
(ν CO 2 −ν CO −ν O 2 )
From Table A-28, ln K p = 3.860 at T = 4320 R = 2400 K. Thus K p = 47.465. Substituting, 47.465 =
x (2 − x)(2 − 0.5 x) 0.5
3 ⎛ ⎞ ⎜ ⎟ 10 0 . 5 − x ⎝ ⎠
1−1.5
Solving for x, x = 1.930 Then, y = 2 - x = 0.070 z = 2 - 0.5x = 1.035 Therefore, the equilibrium composition of the mixture at 2400 K and 3 atm is 1.930CO 2 + 0.070CO + 1.035O 2 + 6N 2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-18
16-30 A mixture of N2, O2, and Ar is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of N2, O2, Ar, and NO. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are 3 N2 Stoichiometric: 12 N 2 + 12 O 2 ⇔ NO (thus ν NO = 1, ν N 2 = 12 , and ν O 2 = 12 ) 1 O2 3 N 2 + O 2 + 01 . Ar ⎯ ⎯→ x NO + y N 2 + z O 2 + 0.1 Ar Actual: 0.1 Ar 123 14243 123 2400 K prod. reactants inert 10 atm N balance: 6 = x + 2y ⎯ ⎯→ y = 3 − 0.5x O balance:
2 = x + 2z
⎯ ⎯→
z = 1 − 0.5x
Total number of moles:
N total = x + y + z + 01 . = 41 .
The equilibrium constant relation becomes, Kp =
ν NO N NO ν
ν
N NN 2 N OO2 2
2
⎛ P ⎜⎜ ⎝ N total
⎞ ⎟⎟ ⎠
(ν NO − ν N 2 − νO2 )
=
x y 0.5 z 0.5
⎛ P ⎜⎜ ⎝ N total
⎞ ⎟⎟ ⎠
1−0 .5−0 .5
From Table A-28, ln K p = −3.019 at 2400 K. Thus K p = 0.04885. Substituting, x ×1 (3 − 0.5 x) (1 − 0.5 x)0.5 x = 0.0823
0.04885 =
0.5
Solving for x, Then, y = 3 - 0.5x = 2.9589 z = 1 - 0.5x = 0.9589 Therefore, the equilibrium composition of the mixture at 2400 K and 10 atm is 0.0823NO + 2.9589N 2 + 0.9589O 2 + 0.1Ar
16-31 The mole fraction of sodium that ionizes according to the reaction Na ⇔ Na+ + e- at 2000 K and 0.8 atm is to be determined. Assumptions All components behave as ideal gases. Analysis The stoichiometric and actual reactions can be written as Stoichiometric:
Na ⇔ Na + + e - (thus ν Na = 1, ν Na + = 1 and ν e - = 1)
Actual:
Na ⎯ ⎯→ { x Na + y Na + + y e − 142 4 43 4 react.
Na balance:
products
Na ⇔ Na+ + e2000 K 0.8 atm
1 = x + y or y = 1 − x N total = x + 2 y = 2 − x
Total number of moles:
The equilibrium constant relation becomes, ν
Kp =
N NaNa N
ν
e-
e-
N νNaNa
⎛ P ⎜ ⎜N ⎝ total
⎞ ⎟ ⎟ ⎠
(ν
Na +
+ν - −ν Na ) e
=
y2 x
⎛ P ⎜ ⎜N ⎝ total
⎞ ⎟ ⎟ ⎠
1+1−1
(1 − x) 2 ⎛ 0.8 ⎞ ⎟ ⎜ x ⎝2− x⎠ Solving for x, x = 0.325 Thus the fraction of Na which dissociates into Na+ and e- is 1 - 0.325 = 0.675 or 67.5%
Substituting,
0.668 =
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-19
16-32 Liquid propane enters a combustion chamber. The equilibrium composition of product gases and the rate of heat transfer from the combustion chamber are to be determined. Assumptions 1 The equilibrium composition consists of CO2, H2O, CO, N2, and O2. 2 The constituents of the mixture are ideal gases. Analysis (a) Considering 1 kmol of C3H8, the stoichiometric combustion equation can be written as C 3 H 8 (l) + a th (O 2 + 3.76 N 2 ) ⎯ ⎯→ 3CO 2 + 4H 2 O + 3.76a th N 2
C3H8
where ath is the stoichiometric coefficient and is determined from the O2 balance,
25°C
2.5a th = 3 + 2 + 1.5a th
⎯ ⎯→
a th = 5
Air
Then the actual combustion equation with 150% excess air and some CO in the products can be written as C3H 8 ( l ) + 12.5( O 2 + 3.76 N 2 )
⎯ ⎯→
Combustion chamber 2 atm
CO 1200 K CO2 H2O O2 N2
12°C
xCO 2 + (3 − x )CO + (9 − 0.5x )O 2 + 4H 2 O + 47N 2
After combustion, there will be no C3 H8 present in the combustion chamber, and H2O will act like an inert gas. The equilibrium equation among CO2, CO, and O2 can be expressed as CO 2 ⇔ CO + 12 O 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )
and ν
Kp =
ν
CO N CO N OO 2 ⎛ P 2 ⎜ ⎜N ν CO 2 ⎝ total N CO 2
⎞ ⎟ ⎟ ⎠
(ν CO +ν O 2 −ν CO 2 )
where N total = x + (3 − x ) + (9 − 0.5x ) + 4 + 47 = 63 − 0.5x
From Table A-28, ln K p = −17.871 at 1200 K. Thus K p = 1.73 × 10 −8 . Substituting, 1.73 × 10 −8 =
(3 − x )(9 − 0.5 x ) 0.5 ⎛ 2 ⎞ ⎜ ⎟ x ⎝ 63 − 0.5 x ⎠
1.5−1
Solving for x, x = 2.9999999 ≅ 3.0
Therefore, the amount CO in the product gases is negligible, and it can be disregarded with no loss in accuracy. Then the combustion equation and the equilibrium composition can be expressed as C 3 H 8 ( l) + 12.5(O 2 + 3.76N 2 ) ⎯ ⎯→ 3CO 2 + 7.5O 2 + 4 H 2 O + 47N 2
and 3CO 2 + 7.5O 2 + 4H 2 O + 47N 2
(b) The heat transfer for this combustion process is determined from the steady-flow energy balance E in − E out = ∆E system on the combustion chamber with W = 0, − Qout =
∑ N (h P
o f
+h −ho
) − ∑ N (h P
R
o f
+ h −ho
)
R
Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, (The h fo of liquid propane is obtained by adding the hfg at 25°C to h fo of gaseous propane).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-20
Substance C3H8 (l) O2 N2 H2O (g) CO2
h fo kJ/kmol -118,910 0 0 -241,820 -393,520
h 285 K
h 298 K
h1200 K
kJ/kmol --8696.5 8286.5 -----
kJ/kmol --8682 8669 9904 9364
kJ/kmol --38,447 36,777 44,380 53,848
Substituting, − Qout = 3( −393,520 + 53,848 − 9364) + 4( −241,820 + 44,380 − 9904) + 7.5( 0 + 38,447 − 8682) + 47( 0 + 36,777 − 8669) − 1( −118,910 + h298 − h298 ) − 12.5( 0 + 8296.5 − 8682) − 47( 0 + 8186.5 − 8669) = −185,764 kJ / kmol of C3H 8
or
Qout = 185,764 kJ / kmol of C3H 8
The mass flow rate of C3H8 can be expressed in terms of the mole numbers as & m 12 . kg / min N& = = = 0.02727 kmol / min M 44 kg / kmol
Thus the rate of heat transfer is Q& out = N& × Qout = (0.02727 kmol/min)(185,746 kJ/kmol) = 5066 kJ/min
The equilibrium constant for the reaction
1 2
N 2 + 12 O 2 ⇔ NO is ln Kp = -7.569, which is very small. This
indicates that the amount of NO formed during this process will be very small, and can be disregarded.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-21
16-33 EES Problem 16-32 is reconsidered. It is to be investigated if it is realistic to disregard the presence of NO in the product gases. Analysis The problem is solved using EES, and the solution is given below. "To solve this problem, the Gibbs function of the product gases is minimized. Click on the Min/Max icon." For this problem at 1200 K the moles of CO are 0.000 and moles of NO are 0.000, thus we can disregard both the CO and NO. However, try some product temperatures above 1286 K and observe the sign change on the Q_out and the amout of CO and NO present as the product temperature increases." "The reaction of C3H8(liq) with excess air can be written: C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO The coefficients A_th and EX are the theoretical oxygen and the percent excess air on a decimal basis. Coefficients a, b, c, d, e, and f are found by minimiming the Gibbs Free Energy at a total pressure of the product gases P_Prod and the product temperature T_Prod. The equilibrium solution can be found by applying the Law of Mass Action or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, click on the Min/Max icon. There are six compounds present in the products subject to four specie balances, so there are two degrees of freedom. Minimize the Gibbs function of the product gases with respect to two molar quantities such as coefficients b and f. The equilibrium mole numbers a, b, c, d, e, and f will be determined and displayed in the Solution window." PercentEx = 150 [%] Ex = PercentEx/100 "EX = % Excess air/100" P_prod =2*P_atm T_Prod=1200 [K] m_dot_fuel = 0.5 [kg/s] Fuel$='C3H8' T_air = 12+273 "[K]" T_fuel = 25+273 "[K]" P_atm = 101.325 [kPa] R_u=8.314 [kJ/kmol-K] "Theoretical combustion of C3H8 with oxygen: C3H8 + A_th O2 = 3 C02 + 4 H2O " 2*A_th = 3*2 + 4*1 "Balance the reaction for 1 kmol of C3H8" "C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO" b_max = 3 f_max = (1+Ex)*A_th*3.76*2 e_guess=Ex*A_th 1*3 = a*1+b*1 "Carbon balance" 1*8=c*2 "Hydrogen balance" (1+Ex)*A_th*2=a*2+b*1+c*1+e*2+f*1 "Oxygen balance" (1+Ex)*A_th*3.76*2=d*2+f*1 "Nitrogen balance"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-22
"Total moles and mole fractions" N_Total=a+b+c+d+e+f y_CO2=a/N_Total; y_CO=b/N_Total; y_H2O=c/N_Total; y_N2=d/N_Total; y_O2=e/N_Total; y_NO=f/N_Total "The following equations provide the specific Gibbs function for each component as a function of its molar amount" g_CO2=Enthalpy(CO2,T=T_Prod)-T_Prod*Entropy(CO2,T=T_Prod,P=P_Prod*y_CO2) g_CO=Enthalpy(CO,T=T_Prod)-T_Prod*Entropy(CO,T=T_Prod,P=P_Prod*y_CO) g_H2O=Enthalpy(H2O,T=T_Prod)-T_Prod*Entropy(H2O,T=T_Prod,P=P_Prod*y_H2O) g_N2=Enthalpy(N2,T=T_Prod)-T_Prod*Entropy(N2,T=T_Prod,P=P_Prod*y_N2) g_O2=Enthalpy(O2,T=T_Prod)-T_Prod*Entropy(O2,T=T_Prod,P=P_Prod*y_O2) g_NO=Enthalpy(NO,T=T_Prod)-T_Prod*Entropy(NO,T=T_Prod,P=P_Prod*y_NO) "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance" Gibbs=a*g_CO2+b*g_CO+c*g_H2O+d*g_N2+e*g_O2+f*g_NO "For the energy balance, we adjust the value of the enthalpy of gaseous propane given by EES:" h_fg_fuel = 15060"[kJ/kmol]" "Table A.27" h_fuel = enthalpy(Fuel$,T=T_fuel)-h_fg_fuel "Energy balance for the combustion process:" "C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO" HR =Q_out+HP HR=h_fuel+ (1+Ex)*A_th*(enthalpy(O2,T=T_air)+3.76*enthalpy(N2,T=T_air)) HP=a*enthalpy(CO2,T=T_prod)+b*enthalpy(CO,T=T_prod)+c*enthalpy(H2O,T=T_prod)+d*enthal py(N2,T=T_prod)+e*enthalpy(O2,T=T_prod)+f*enthalpy(NO,T=T_prod) "The heat transfer rate is:" Q_dot_out=Q_out/molarmass(Fuel$)*m_dot_fuel "[kW]" SOLUTION a=3.000 [kmol] A_th=5 b=0.000 [kmol] b_max=3 c=4.000 [kmol] d=47.000 [kmol] e=7.500 [kmol] Ex=1.5 e_guess=7.5 f=0.000 [kmol] Fuel$='C3H8' f_max=94 Gibbs=-17994897 [kJ] g_CO=-703496 [kJ/kmol]
g_CO2=-707231 [kJ/kmol] g_H2O=-515974 [kJ/kmol] g_N2=-248486 [kJ/kmol] g_NO=-342270 [kJ/kmol] g_O2=-284065 [kJ/kmol] HP=-330516.747 [kJ/kmol] HR=-141784.529 [kJ/kmol] h_fg_fuel=15060 [kJ/kmol] h_fuel=-118918 [kJ/kmol] m_dot_fuel=0.5 [kg/s] N_Total=61.5 [kmol/kmol_fuel] PercentEx=150 [%] P_atm=101.3 [kPa] P_prod=202.7 [kPa]
Q_dot_out=2140 [kW] Q_out=188732 [kJ/kmol_fuel] R_u=8.314 [kJ/kmol-K] T_air=285 [K] T_fuel=298 [K] T_Prod=1200.00 [K] y_CO=1.626E-15 y_CO2=0.04878 y_H2O=0.06504 y_N2=0.7642 y_NO=7.857E-08 y_O2=0.122
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-23
16-34E A steady-flow combustion chamber is supplied with CO and O2. The equilibrium composition of product gases and the rate of heat transfer from the combustion chamber are to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, and O2. 2 The constituents of the mixture are ideal gases. Analysis (a) We first need to calculate the amount of oxygen used per lbmol of CO before we can write the combustion equation, RT (0.3831 psia ⋅ ft 3 / lbm ⋅ R)(560 R) = = 13.41 ft 3 / lbm v CO = 16 psia P V& 12.5 ft 3 /min m& CO = CO = = 0.932 lbm/min v CO 13.41 ft 3 /lbm
CO 100°F O2
Combustion chamber
3600 R
16 psia
77°F
Then the molar air-fuel ratio becomes (it is actually O2-fuel ratio) AF =
N O2 N fuel
m& O 2 / M O 2
=
m& fuel / M fuel
=
(0.7 lbm/min)/(32 lbm/lbmol) = 0.657 lbmol O 2 /lbmol fuel (0.932 lbm/min)/(28 lbm/lbmol)
Then the combustion equation can be written as CO + 0.657O 2
⎯ ⎯→
xCO 2 + (1 − x )CO + (0.657 − 0.5 x )O 2
The equilibrium equation among CO2, CO, and O2 can be expressed as CO 2 ⇔ CO + 12 O 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )
and ν
ν
CO N CO N OO 2 ⎛ P 2 ⎜ Kp = ⎜N ν CO 2 ⎝ total N CO 2
⎞ ⎟ ⎟ ⎠
(ν CO +ν O 2 −ν CO 2 )
where N total = x + (1 − x ) + ( 0.657 − 0.5x ) = 1657 − 0.5x . P = 16 / 14.7 = 1088 . atm
From Table A-28, ln K p = −6.635 at T = 3600 R = 2000 K. Thus K p = 1.314 × 10 −3 . Substituting, 1.314 × 10 −3 =
(1 − x)(0.657 − 0.5 x) 0.5 x
⎛ 1.088 ⎞ ⎜ ⎟ x 1 . 657 0 . 5 − ⎝ ⎠
1.5−1
Solving for x, x = 0.9966 Then the combustion equation and the equilibrium composition can be expressed as CO + 0.657O 2 ⎯ ⎯→ 0.9966CO 2 + 0.0034CO + 01587 . O2
and 0.9966CO 2 + 0.0034CO + 0.1587O 2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
CO CO2 O2
16-24
(b) The heat transfer for this combustion process is determined from the steady-flow energy balance E in − E out = ∆E system on the combustion chamber with W = 0, − Qout =
∑ N (h P
o f
+h −ho
) − ∑ N (h P
R
o f
+ h −ho
)
R
Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,
Substance CO O2 CO2
h fo Btu/lbmol -47,540 0 -169,300
h 537 R
h 560 R
h 3600 R
Btu/lbmol 3725.1 3725.1 4027.5
Btu/lbmol 3889.5 -----
Btu/lbmol 28,127 29,174 43,411
Substituting, − Qout = 0.9966( −169,300 + 43,411 − 4027.5) .) + 0.0034( −47,540 + 28,127 − 37251 . ( 0 + 29,174 − 37251 .) + 01587 . )− 0 − 1( −47,540 + 3889.5 − 37251 = −78,139 Btu / lbmol of CO
or
Qout = 78,139 Btu / lbm of CO
The mass flow rate of CO can be expressed in terms of the mole numbers as m& 0.932 lbm / min N& = = = 0.0333 lbmol / min 28 lbm / lbmol M
Thus the rate of heat transfer is Q& out = N& × Qout = (0.0333 lbmol/min)(78,139 Btu/lbmol) = 2602 Btu/min
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-25
16-35 Oxygen is heated during a steady-flow process. The rate of heat supply needed during this process is to be determined for two cases. Assumptions 1 The equilibrium composition consists of O2 and O. 2 All components behave as ideal gases. Analysis (a) Assuming some O2 dissociates into O, the dissociation equation can be written as ⎯ ⎯→
O2
x O 2 + 2(1 − x)O
The equilibrium equation among O2 and O can be expressed as O 2 ⇔ 2O (thus ν O 2 = 1 and ν O = 2)
Q&
Assuming ideal gas behavior for all components, the equilibrium constant relation can be expressed as Kp =
νO
NO ⎛ P ⎜ ⎜ ν N OO 2 ⎝ N total 2
where
ν O −ν O 2
⎞ ⎟ ⎟ ⎠
O2
O2, O
298 K
3000 K
N total = x + 2(1 − x ) = 2 − x
From Table A-28, ln K p = −4.357 at 3000 K. Thus K p = 0.01282. Substituting, 0.01282 =
(2 − 2 x) 2 ⎛ 1 ⎞ ⎜ ⎟ x ⎝2− x⎠
Solving for x gives
2 −1
x = 0.943
Then the dissociation equation becomes ⎯ ⎯→
O2
0.943 O 2 + 0114 . O
The heat transfer for this combustion process is determined from the steady-flow energy balance E in − E out = ∆E system on the combustion chamber with W = 0, Qin =
∑ N (h P
o f
+ h −ho
) − ∑ N (h P
R
o f
+ h −ho
)
R
Assuming the O2 and O to be ideal gases, we have h = h(T). From the tables,
Substance O O2
h fo kJ/kmol 249,190 0
h 298 K
h 3000 K
kJ/kmol 6852 8682
kJ/kmol 63,425 106,780
Substituting, Qin = 0.943( 0 + 106,780 − 8682) + 0114 . ( 249,190 + 63,425 − 6852) − 0 = 127,363 kJ / kmol O 2
The mass flow rate of O2 can be expressed in terms of the mole numbers as 0.5 kg/min m& N& = = = 0.01563 kmol/min M 32 kg/kmol
Thus the rate of heat transfer is Q& in = N& × Qin = (0.01563 kmol/min)(127,363 kJ/kmol) = 1990 kJ/min
(b) If no O2 dissociates into O, then the process involves no chemical reactions and the heat transfer can be determined from the steady-flow energy balance for nonreacting systems to be Q& in = m& (h2 − h1 ) = N& (h2 − h1 ) = (0.01563 kmol/min)(106,780 - 8682) kJ/kmol = 1533 kJ/min
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-26
16-36 The equilibrium constant, Kp is to be estimated at 2500 K for the reaction CO + H2O = CO2 + H2. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e − ∆G*(T ) / RuT or ln K p = −∆G * (T ) / Ru T
where ∗ ∗ ∗ ∗ ∆G * (T ) = ν CO2 g CO2 (T ) + ν H2 g H2 (T ) −ν CO g CO (T ) −ν H2O g H2O (T )
At 2500 K, ∗ ∗ ∗ ∗ ∆G * (T ) = ν CO2 g CO2 (T ) + ν H2 g H2 (T ) −ν CO g CO (T ) −ν H2O g H2O (T )
= ν CO2 (h − Ts ) CO2 + ν H2 (h − Ts ) H2 −ν CO (h − Ts ) CO −ν H2O (h − Ts ) H2O = 1[(−271,641) − (2500)(322.60)] + 1[(70,452) − (2500)(196.10)]
− 1[(−35,510) − (2500)(266.65)] − 1[(−142,891) − (2500)(276.18)]
= 37,525 kJ/kmol
The enthalpies at 2500 K and entropies at 2500 K and 101.3 kPa (1 atm) are obtained from EES. Substituting, ln K p = −
37,525 kJ/kmol = −1.8054 ⎯ ⎯→ K p = 0.1644 (8.314 kJ/kmol ⋅ K)(2500 K)
The equilibrium constant may be estimated using the integrated van't Hoff equation: ⎛ K p ,est ⎞ ⎟= ln⎜ ⎜ K p1 ⎟ ⎝ ⎠ ⎛ K p ,est ⎞ ⎟= ln⎜⎜ ⎟ 0 . 2209 ⎠ ⎝
hR Ru
⎛ 1 1⎞ ⎜⎜ − ⎟⎟ ⎝ TR T ⎠
1 ⎞ − 26,176 kJ/kmol ⎛ 1 − ⎯→ K p ,est = 0.1612 ⎜ ⎟⎯ 8.314 kJ/kmol.K ⎝ 2000 K 2500 K ⎠
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-27
16-37 A constant volume tank contains a mixture of H2 and O2. The contents are ignited. The final temperature and pressure in the tank are to be determined. Analysis The reaction equation with products in equilibrium is H2 + O2 ⎯ ⎯→ a H 2 + b H 2 O + c O 2
The coefficients are determined from the mass balances Hydrogen balance:
2 = 2a + 2b
Oxygen balance:
2 = b + 2c
The assumed equilibrium reaction is H 2 O ←⎯→ H 2 + 0.5O 2
The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G*( T )/ Ru T or ln K p = − ∆G * (T ) / Ru T
where ∗ ∗ ∗ ∆G * (T ) = ν H2 g H2 (Tprod ) + ν O2 g O2 (Tprod ) −ν H2O g H2O (Tprod )
and the Gibbs functions are given by ∗ g H2 (Tprod ) = (h − Tprod s ) H2 ∗ g O2 (Tprod ) = (h − Tprod s ) O2 ∗ g H2O (Tprod ) = (h − Tprod s ) H2O
The equilibrium constant is also given by Kp =
a 1c 0.5 ⎛ P ⎜ b 1 ⎜⎝ N tot
⎞ ⎟ ⎟ ⎠
1+ 0.5 −1
=
ac 0.5 b
⎛ P2 / 101.3 ⎞ ⎟⎟ ⎜⎜ ⎝ a+b+c ⎠
0.5
An energy balance on the tank under adiabatic conditions gives UR =UP
where U R = 1(hH2@25°C − Ru Treac ) + 1(hO2@25°C − Ru Treac ) = 0 − (8.314 kJ/kmol.K)(298.15 K) + 0 − (8.314 kJ/kmol.K)(298.15 K) = −4958 kJ/kmol U P = a (hH2@ Tprod − Ru Tprod ) + b(hH2O@ Tprod − Ru Tprod ) + c(hO2@ Tprod − Ru Tprod )
The relation for the final pressure is P2 =
N tot Tprod ⎛ a + b + c ⎞⎛⎜ Tprod ⎞⎟ P1 = ⎜ (101.3 kPa) ⎟⎜ N 1 Treac 2 ⎝ ⎠⎝ 298.15 K ⎟⎠
Solving all the equations simultaneously using EES, we obtain the final temperature and pressure in the tank to be Tprod = 3857 K P2 = 1043 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-28
Simultaneous Reactions 16-38C It can be expresses as “(dG)T,P = 0 for each reaction.” Or as “the Kp relation for each reaction must be satisfied.” 16-39C The number of Kp relations needed to determine the equilibrium composition of a reacting mixture is equal to the difference between the number of species present in the equilibrium mixture and the number of elements.
16-40 Two chemical reactions are occurring in a mixture. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of H2O, OH, O2, and H2. 2 The constituents of the mixture are ideal gases. H 2O,OH Analysis The reaction equation during this process can be expressed as H O⇒ H 2O
⎯ ⎯→
2
x H 2O + y H 2 + z O 2 + w OH
O 2 ,H 2
3400 K 1 atm
Mass balances for hydrogen and oxygen yield 2 = 2x + 2 y + w (1) H balance:
1 = x + 2z + w (2) O balance: The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the Kp relations) to determine the equilibrium composition of the mixture. They are H 2 O ⇔ H 2 + 12 O 2
(reaction 1)
H 2 O ⇔ H 2 + OH
(reaction 2)
1 2
The equilibrium constant for these two reactions at 3400 K are determined from Table A-28 to be
ln K P1 = −1891 .
⎯ ⎯→
K P1 = 015092 .
ln K P 2 = −1576 .
⎯ ⎯→
K P 2 = 0.20680
The Kp relations for these two simultaneous reactions are ν
K P1 =
ν
N HH 2 N OO 2 ⎛ P 2 2 ⎜ ⎜N ν H 2O ⎝ total NH O 2
where
⎞ ⎟ ⎟ ⎠
ν
(ν H 2 +ν O 2 −ν H 2 O )
and
K P2 =
ν
OH N HH 2 N OH ⎛ P 2 ⎜ ⎜N ν H 2O ⎝ total NH O 2
⎞ ⎟ ⎟ ⎠
(ν H 2 +ν OH −ν H 2 O )
N total = N H2O + N H2 + N O2 + N OH = x + y + z + w
Substituting, 1/ 2
0.15092 =
( y )( z )1 / 2 x
⎛ ⎞ 1 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠
0.20680 =
( w)( y )1/ 2 x
⎞ ⎛ 1 ⎟⎟ ⎜⎜ x + y + z + w ⎠ ⎝
(3) 1/ 2
(4)
Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 0.574 y = 0.308 z = 0.095 w = 0.236 Therefore, the equilibrium composition becomes 0.574H 2 O + 0.308H 2 + 0.095O 2 + 0.236OH
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-29
16-41 Two chemical reactions are occurring in a mixture. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and O. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→
2 CO 2 + O 2
x CO 2 + y CO + z O 2 + w O
CO2, CO, O2, O 3200 K 2 atm
Mass balances for carbon and oxygen yield C balance:
2= x+ y
(1)
O balance:
6 = 2x + y + 2z + w
(2)
The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the KP relations) to determine the equilibrium composition of the mixture. They are
CO 2 ⇔ CO + 12 O 2
(reaction 1)
O 2 ⇔ 2O
(reaction 2)
The equilibrium constant for these two reactions at 3200 K are determined from Table A-28 to be ln K P1 = −0.429
⎯ ⎯→
K P1 = 0.65116
ln K P 2 = −3.072
⎯ ⎯→
K P 2 = 0.04633
The KP relations for these two simultaneous reactions are ν
K P1
CO N νCO N OO 2 ⎛ P 2 ⎜ = ⎜N ν CO 2 ⎝ total N CO 2
K P2 =
νO
NO ⎛ P ⎜ ⎜ ν N OO 2 ⎝ N total 2
⎞ ⎟ ⎟ ⎠
(ν CO +ν O 2 −ν CO 2 )
ν O −ν O 2
⎞ ⎟ ⎟ ⎠
where
N total = N CO2 + N O2 + N CO + N O = x + y + z + w Substituting, 0.65116 =
( y )( z )1 / 2 x
0.04633 =
w2 z
⎛ ⎞ 2 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠
⎞ ⎛ 2 ⎟⎟ ⎜⎜ ⎝ x + y + z + w⎠
1/ 2
(3)
2 −1
(4)
Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 1.127
y = 0.873
z = 1.273
w = 0.326
Thus the equilibrium composition is 1.127CO 2 + 0.873CO + 1.273O 2 + 0.326O
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-30
16-42 Two chemical reactions are occurring at high-temperature air. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of O2, N2, O, and NO. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→
O 2 + 3.76 N 2
Heat
x N 2 + y NO + z O 2 + w O
AIR
Mass balances for nitrogen and oxygen yield N balance:
7.52 = 2 x + y
(1)
O balance:
2 = y + 2z + w
(2)
Reaction chamber, 2 atm
O2, N2, O, NO 3000 K
The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the Kp relations) to determine the equilibrium composition of the mixture. They are 1 2
N 2 + 21 O 2 ⇔ NO
(reaction 1)
O 2 ⇔ 2O
(reaction 2)
The equilibrium constant for these two reactions at 3000 K are determined from Table A-28 to be ln K P1 = −2.114
⎯ ⎯→
K P1 = 012075 .
ln K P 2 = −4.357
⎯ ⎯→
K P 2 = 0.01282
The KP relations for these two simultaneous reactions are ν
K P1 =
⎛ P ⎜ ⎜N ⎝ total
NO N NO
ν
ν
N NN 2 N OO 2 2
2
νO
K P2
N ⎛ P = νO ⎜⎜ O2 N O ⎝ N total 2
where
⎞ ⎟ ⎟ ⎠
(ν NO −ν N 2 −ν O 2 )
ν O −ν O 2
⎞ ⎟ ⎟ ⎠
N total = N N 2 + N NO + N O 2 + N O = x + y + z + w
Substituting, 0.12075 =
y x 0 .5 z 0 .5
w2 0.01282 = z
⎞ ⎛ 2 ⎟⎟ ⎜⎜ ⎝ x+ y+ z+w⎠
⎞ ⎛ 2 ⎟⎟ ⎜⎜ ⎝ x+ y+ z+w⎠
1− 0.5 − 0.5
(3)
2 −1
(4)
Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 3.656
y = 0.2086
z = 0.8162
w = 0.1591
Thus the equilibrium composition is 3.656N 2 + 0.2086NO + 0.8162O 2 + 0.1591O
The equilibrium constant of the reaction N 2 ⇔ 2N at 3000 K is lnKP = -22.359, which is much smaller than the KP values of the reactions considered. Therefore, it is reasonable to assume that no N will be present in the equilibrium mixture.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-31
16-43E [Also solved by EES on enclosed CD] Two chemical reactions are occurring in air. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of O2, N2, O, and NO. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→
O 2 + 3.76 N 2
Heat
x N 2 + y NO + z O 2 + w O
AIR
Mass balances for nitrogen and oxygen yield N balance:
7.52 = 2 x + y
(1)
O balance:
2 = y + 2z + w
(2)
Reaction chamber, 1 atm
O2, N2, O, NO 5400 R
The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the Kp relations) to determine the equilibrium composition of the mixture. They are 1 2
N 2 + 12 O 2 ⇔ NO
(reaction 1)
O 2 ⇔ 2O
(reaction 2)
The equilibrium constant for these two reactions at T = 5400 R = 3000 K are determined from Table A-28 to be ln K P1 = −2.114
⎯ ⎯→
. K P1 = 012075
ln K P 2 = −4.357
⎯ ⎯→
K P 2 = 0.01282
The KP relations for these two simultaneous reactions are ν
K P1 =
⎛ P ⎜ ⎜N ⎝ total
NO N NO
ν
ν
N NN 2 N OO 2 2
2
νO
K P2
N ⎛ P = νO ⎜⎜ O2 N O ⎝ N total 2
where
⎞ ⎟ ⎟ ⎠
(ν NO −ν N 2 −ν O 2 )
ν O −ν O 2
⎞ ⎟ ⎟ ⎠
N total = N N 2 + N NO + N O2 + N O = x + y + z + w
Substituting, 0.12075 =
0.01282 =
y x 0 . 5 z 0 .5 w2 z
⎛ ⎞ 1 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠
⎞ ⎛ 1 ⎟⎟ ⎜⎜ x + y + z + w ⎠ ⎝
1− 0.5− 0.5
(3)
2 −1
(4)
Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 3.658
y = 0.2048
z = 0.7868
w = 0.2216
Thus the equilibrium composition is 3.658N 2 + 0.2048NO + 0.7868O 2 + 0.2216O
The equilibrium constant of the reaction N 2 ⇔ 2N at 5400 R is lnKP = -22.359, which is much smaller than the KP values of the reactions considered. Therefore, it is reasonable to assume that no N will be present in the equilibrium mixture.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-32
14-44E EES Problem 16-43E is reconsidered. Using EES (or other) software, the equilibrium solution is to be obtained by minimizing the Gibbs function by using the optimization capabilities built into EES. This solution technique is to be compared with that used in the previous problem. Analysis The problem is solved using EES, and the solution is given below. "This example illustrates how EES can be used to solve multi-reaction chemical equilibria problems by directly minimizing the Gibbs function. 0.21 O2+0.79 N2 = a O2+b O + c N2 + d NO Two of the four coefficients, a, b, c, and d, are found by minimiming the Gibbs function at a total pressure of 1 atm and a temperature of 5400 R. The other two are found from mass balances. The equilibrium solution can be found by applying the Law of Mass Action to two simultaneous equilibrium reactions or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, select MinMax from the Calculate menu. There are four compounds present in the products subject to two elemental balances, so there are two degrees of freedom. Minimize Gibbs with respect to two molar quantities such as coefficients b and d. The equilibrium mole numbers of each specie will be determined and displayed in the Solution window. Minimizing the Gibbs function to find the equilibrium composition requires good initial guesses." "Data from Data Input Window" {T=5400 "R" P=1 "atm" } AO2=0.21; BN2=0.79 "Composition of air" AO2*2=a*2+b+d "Oxygen balance" BN2*2=c*2+d "Nitrogen balance" "The total moles at equilibrium are" N_tot=a+b+c+d y_O2=a/N_tot; y_O=b/N_tot; y_N2=c/N_tot; y_NO=d/N_tot "The following equations provide the specific Gibbs function for three of the components." g_O2=Enthalpy(O2,T=T)-T*Entropy(O2,T=T,P=P*y_O2) g_N2=Enthalpy(N2,T=T)-T*Entropy(N2,T=T,P=P*y_N2) g_NO=Enthalpy(NO,T=T)-T*Entropy(NO,T=T,P=P*y_NO) "EES does not have a built-in property function for monatomic oxygen so we will use the JANAF procedure, found under Options/Function Info/External Procedures. The units for the JANAF procedure are kgmole, K, and kJ so we must convert h and s to English units." T_K=T*Convert(R,K) "Convert R to K" Call JANAF('O',T_K:Cp`,h`,S`) "Units from JANAF are SI" S_O=S`*Convert(kJ/kgmole-K, Btu/lbmole-R) h_O=h`*Convert(kJ/kgmole, Btu/lbmole) "The entropy from JANAF is for one atmosphere so it must be corrected for partial pressure." g_O=h_O-T*(S_O-R_u*ln(Y_O)) R_u=1.9858 "The universal gas constant in Btu/mole-R " "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance." Gibbs=a*g_O2+b*g_O+c*g_N2+d*g_NO
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-33
d [lbmol] 0.002698 0.004616 0.007239 0.01063 0.01481 0.01972 0.02527 0.03132 0.03751 0.04361
b [lbmol] 0.00001424 0.00006354 0.0002268 0.000677 0.001748 0.004009 0.008321 0.01596 0.02807 0.04641
Gibbs [Btu/lbmol] -162121 -178354 -194782 -211395 -228188 -245157 -262306 -279641 -297179 -314941
yO2
yO
yNO
yN2
0.2086 0.2077 0.2062 0.2043 0.2015 0.1977 0.1924 0.1849 0.1748 0.1613
0.0000 0.0001 0.0002 0.0007 0.0017 0.0040 0.0083 0.0158 0.0277 0.0454
0.0027 0.0046 0.0072 0.0106 0.0148 0.0197 0.0252 0.0311 0.0370 0.0426
0.7886 0.7877 0.7863 0.7844 0.7819 0.7786 0.7741 0.7682 0.7606 0.7508
T [R] 3000 3267 3533 3800 4067 4333 4600 4867 5133 5400
0.050
O d n a O N f o n oi t c ar f el o M
0.040
0.030
NO
0.020
O 0.010
0.000 3000
3500
4000
4500
5000
5500
T [R] Discussion The equilibrium composition in the above table are based on the reaction in which the reactants are 0.21 kmol O2 and 0.79 kmol N2. If you multiply the equilibrium composition mole numbers above with 4.76, you will obtain equilibrium composition for the reaction in which the reactants are 1 kmol O2 and 3.76 kmol N2.This is the case in problem 16-43E.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-34
16-45 Water vapor is heated during a steady-flow process. The rate of heat supply for a specified exit temperature is to be determined for two cases. Assumptions 1 The equilibrium composition consists of H2O, OH, O2, and H2. 2 The constituents of the mixture are ideal gases. Q Analysis (a) Assuming some H2O dissociates into H2, O2, and O, the dissociation equation can be written as ⎯ ⎯→
H 2O
x H 2O + y H 2 + z O 2 + w OH
H2O
Mass balances for hydrogen and oxygen yield
298 K
H balance:
2 = 2x + 2 y + w
(1)
O balance:
1 = x + 2z + w
(2)
H2O, H2, O2, OH 3000 K
The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the KP relations) to determine the equilibrium composition of the mixture. They are
H 2 O ⇔ H 2 + 12 O 2
(reaction 1)
H 2 O ⇔ 12 H 2 + OH
(reaction 2)
The equilibrium constant for these two reactions at 3000 K are determined from Table A-28 to be ln K P1 = −3.086
⎯ ⎯→
K P1 = 0.04568
ln K P 2 = −2.937
⎯ ⎯→
K P 2 = 0.05302
The KP relations for these three simultaneous reactions are ν
K P1
ν
N HH 2 N OO 2 ⎛ P 2 2 ⎜ = ⎜N ν H 2O ⎝ total NH O 2
ν H2
K P2 =
ν
OH N H N OH ⎛ P 2 ⎜ ⎜N ν H 2O ⎝ total NH O 2
⎞ ⎟ ⎟ ⎠
(ν H 2 +ν O 2 −ν H 2 O )
⎞ ⎟ ⎟ ⎠
(ν H 2 +ν O 2 −ν H 2 O )
where
N total = N H 2O + N H 2 + N O 2 + N OH = x + y + z + w Substituting, 1/ 2
0.04568 =
( y )( z )1/ 2 x
⎞ ⎛ 1 ⎟⎟ ⎜⎜ x + y + z + w ⎠ ⎝
0.05302 =
( w)( y )1 / 2 x
⎞ ⎛ 1 ⎟⎟ ⎜⎜ ⎝ x + y + z + w⎠
(3) 1/ 2
(4)
Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 0.784
y = 0.162
z = 0.054
w = 0.108
Thus the balanced equation for the dissociation reaction is H 2O
⎯ ⎯→
0.784H 2 O + 0.162H 2 + 0.054O 2 + 0.108OH
The heat transfer for this dissociation process is determined from the steady-flow energy balance E in − E out = ∆E system with W = 0, Qin =
∑ N (h P
o f
+h −ho
) − ∑ N (h P
R
o f
+h −ho
)
R
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-35
Assuming the O2 and O to be ideal gases, we have h = h(T). From the tables,
Substance
H2O H2 O2 OH
hfo kJ/kmol -241,820 0 0 39,460
h 298 K
h 3000 K
kJ/kmol 9904 8468 8682 9188
kJ/kmol 136,264 97,211 106,780 98,763
Substituting, Qin = 0.784( −241,820 + 136,264 − 9904) + 0162 . ( 0 + 97,211 − 8468) + 0.054( 0 + 106,780 − 8682) + 0108 . (39,460 + 98,763 − 9188) − ( −241,820) = 184,909 kJ / kmol H 2 O
The mass flow rate of H2O can be expressed in terms of the mole numbers as 0.2 kg / min m& = = 0.01111 kmol / min N& = M 18 kg / kmol
Thus,
Q& in = N& × Qin = (0.01111 kmol/min)(184,909 kJ/kmol) = 2055 kJ/min (b) If no dissociates takes place, then the process involves no chemical reactions and the heat transfer can be determined from the steady-flow energy balance for nonreacting systems to be
& ( h2 − h1 ) = N& ( h2 − h1 ) Q& in = m = ( 0.01111 kmol / min)(136,264 − 9904) kJ / kmol = 1404 kJ / min
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-36
16-46 EES Problem 16-45 is reconsidered. The effect of the final temperature on the rate of heat supplied for the two cases is to be studied. Analysis The problem is solved using EES, and the solution is given below. "This example illustrates how EES can be used to solve multi-reaction chemical equilibria problems by directly minimizing the Gibbs function. H2O = x H2O+y H2+z O2 + w OH Two of the four coefficients, x, y, z, and w are found by minimiming the Gibbs function at a total pressure of 1 atm and a temperature of 3000 K. The other two are found from mass balances. The equilibrium solution can be found by applying the Law of Mass Action (Eq. 15-15) to two simultaneous equilibrium reactions or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, click on the Min/Max icon. There are four compounds present in the products subject to two elemental balances, so there are two degrees of freedom. Minimize Gibbs with respect to two molar quantities such as coefficient z and w. The equilibrium mole numbers of each specie will be determined and displayed in the Solution window. Minimizing the Gibbs function to find the equilibrium composition requires good initial guesses." "T_Prod=3000 [K]" P=101.325 [kPa] m_dot_H2O = 0.2 [kg/min] T_reac = 298 [K] T = T_prod P_atm=101.325 [kPa] "H2O = x H2O+y H2+z O2 + w OH" AH2O=1 "Solution for 1 mole of water" AH2O=x+z*2+w "Oxygen balance" AH2O*2=x*2+y*2+w "Hydrogen balance" "The total moles at equilibrium are" N_tot=x+y+z+w y_H2O=x/N_tot; y_H2=y/N_tot; y_O2=z/N_tot; y_OH=w/N_tot "EES does not have a built-in property function for monatomic oxygen so we will use the JANAF procedure, found under Options/Function Info/External Procedures. The units for the JANAF procedure are kgmole, K, and kJ." Call JANAF('OH',T_prod:Cp`,h`,S`) "Units from JANAF are SI" S_OH=S` h_OH=h` "The entropy from JANAF is for one atmosphere so it must be corrected for partial pressure." g_OH=h_OH-T_prod*(S_OH-R_u*ln(y_OH*P/P_atm)) R_u=8.314 "The universal gas constant in kJ/kmol-K " "The following equations provide the specific Gibbs function for three of the components." g_O2=Enthalpy(O2,T=T_prod)-T_prod*Entropy(O2,T=T_prod,P=P*y_O2) g_H2=Enthalpy(H2,T=T_prod)-T_prod*Entropy(H2,T=T_prod,P=P*y_H2) g_H2O=Enthalpy(H2O,T=T_prod)-T_prod*Entropy(H2O,T=T_prod,P=P*y_H2O) PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-37 "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance." Gibbs=x*g_H2O+y*g_H2+z*g_O2+w*g_OH "H2O = x H2O+y H2+z O2 + w OH" 1*Enthalpy(H2O,T=T_reac)+Q_in=x*Enthalpy(H2O,T=T_prod)+y*Enthalpy(H2,T=T_prod)+z*Enth alpy(O2,T=T_prod)+w*h_OH N_dot_H2O = m_dot_H2O/molarmass(H2O) Q_dot_in_Dissoc = N_dot_H2O*Q_in Q_dot_in_NoDissoc = N_dot_H2O*(Enthalpy(H2O,T=T_prod) - Enthalpy(H2O,T=T_reac))
Tprod [K] 2500 2600 2700 2800 2900 3000 3100 3200 3300 3400 3500
Qin,Dissoc [kJ/min] 1266 1326 1529 1687 1862 2053 2260 2480 2710 2944 3178
Qin,NoDissoc [kJ/min] 1098 1158 1219 1280 1341 1403 1465 1528 1590 1653 1716
3000
2500
] ni m / J k[
Qin,Dissoc 2000
ni
Qin,NoDissoc
Q
1500
1000 2500
2700
2900
3100
3300
3500
TProd [K]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-38
16-47 EES Ethyl alcohol C2H5OH (gas) is burned in a steady-flow adiabatic combustion chamber with 40 percent excess air. The adiabatic flame temperature of the products is to be determined and the adiabatic flame temperature as a function of the percent excess air is to be plotted. Analysis The complete combustion reaction in this case can be written as C 2 H 5 OH (gas) + (1 + Ex)a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 2 CO 2 + 3 H 2 O + ( Ex)(a th ) O 2 + f N 2
where ath is the stoichiometric coefficient for air. The oxygen balance gives 1 + (1 + Ex)a th × 2 = 2 × 2 + 3 × 1 + ( Ex)(a th ) × 2 The reaction equation with products in equilibrium is C 2 H 5 OH (gas) + (1 + Ex)a th [O 2 + 3.76N 2 ] ⎯ ⎯→ a CO 2 + b CO + d H 2 O + e O 2 + f N 2 + g NO
The coefficients are determined from the mass balances Carbon balance: 2 = a+b Hydrogen balance: Oxygen balance:
6 = 2d ⎯ ⎯→ d = 3 1 + (1 + Ex)a th × 2 = a × 2 + b + d + e × 2 + g
Nitrogen balance: (1 + Ex)a th × 3.76 × 2 = f × 2 + g Solving the above equations, we find the coefficients to be Ex = 0.4, ath = 3, a = 1.995, b = 0.004712, d = 3, e = 1.17, f = 15.76, g = 0.06428 Then, we write the balanced reaction equation as C 2 H 5 OH (gas) + 4.2[O 2 + 3.76N 2 ] ⎯ ⎯→ 1.995 CO 2 + 0.004712 CO + 3 H 2 O + 1.17 O 2 + 15.76 N 2 + 0.06428 NO
Total moles of products at equilibrium are N tot = 1.995 + 0.004712 + 3 + 1.17 + 15.76 = 21.99 The first assumed equilibrium reaction is CO 2 ←⎯→ CO + 0.5O 2
The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using ⎛ − ∆G1 * (Tprod ) ⎞ ⎟ K p1 = exp⎜ ⎜ ⎟ Ru Tprod ⎝ ⎠
Where
∗ ∗ ∗ ∆G1 * (Tprod ) = ν CO g CO (Tprod ) + ν O2 g O2 (Tprod ) −ν CO2 g CO2 (Tprod )
and the Gibbs functions are defined as ∗ (Tprod ) = (h − Tprod s ) CO g CO ∗ (Tprod ) = (h − Tprod s ) O2 g O2 ∗ (Tprod ) = (h − Tprod s ) CO2 g CO2
The equilibrium constant is also given by 1+ 0.5 −1
⎛ P ⎞ (0.004712)(1.17) 0.5 ⎛ 1 ⎞ ⎟ ⎜ = K p1 ⎜ ⎟ ⎜N ⎟ 1.995 ⎝ 21.99 ⎠ ⎝ tot ⎠ The second assumed equilibrium reaction is be 0.5 = a
0.5
= 0.0005447
0.5N 2 + 0.5O 2 ←⎯→ NO
Also, for this reaction, we have ∗ g NO (Tprod ) = (h − Tprod s ) NO ∗ g N2 (Tprod ) = (h − Tprod s ) N2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-39
∗ g O2 (Tprod ) = (h − Tprod s ) O2 ∗ ∗ ∗ ∆G 2 * (Tprod ) = ν NO g NO (Tprod ) −ν N2 g N2 (Tprod ) −ν O2 g O2 (Tprod )
⎛ − ∆G 2 * (Tprod ) ⎞ ⎟ K p 2 = exp⎜ ⎜ ⎟ R T u prod ⎝ ⎠ 1− 0.5 − 0.5
⎛ P ⎞ g ⎟ K p 2 = ⎜⎜ ⎟ o . 5 e f ⎝ N tot ⎠ A steady flow energy balance gives HR = HP
0
0.5
0.06428 ⎛ 1 ⎞ =⎜ = 0.01497 ⎟ ⎝ 21.99 ⎠ (1.17) 0.5 (15.76) 0.5
where H R = h fo fuel@25°C + 4.2hO2@25°C + 15.79h N2@25°C = (−235,310 kJ/kmol) + 4.2(0) + 15.79(0) = −235,310 kJ/kmol H P = 1.995hCO2@Tprod + 0.004712hCO@Tprod + 3hH2O@Tprod + 1.17 hO2@Tprod + 15.76h N2@Tprod + 0.06428h NO@Tprod
Solving the energy balance equation using EES, we obtain the adiabatic flame temperature Tprod = 1901 K The copy of entire EES solution including parametric studies is given next: "The reactant temperature is:" T_reac= 25+273 "[K]" "For adiabatic combustion of 1 kmol of fuel: " Q_out = 0 "[kJ]" PercentEx = 40 "Percent excess air" Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The complete combustion reaction equation for excess air is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=2 CO2 + 3 H2O + Ex*A_th O2 + f N2 " "Oxygen Balance for complete combustion:" 1 + (1+Ex)*A_th*2=2*2+3*1 + Ex*A_th*2 "The reaction equation for excess air and products in equilibrium is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=a CO2 + b CO+ d H2O + e O2 + f N2 + g NO" "Carbon Balance:" 2=a + b "Hydrogen Balance:" 6=2*d "Oxygen Balance:" 1 + (1+Ex)*A_th*2=a*2+b + d + e*2 +g "Nitrogen Balance:" (1+Ex)*A_th*3.76 *2= f*2 + g N_tot =a +b + d + e + f +g "Total kilomoles of products at equilibrium" "The first assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG_1 =1*g_CO+0.5*g_O2-1*g_CO2 PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-40
"The equilibrium constant is given by Eq. 15-14." K_P_1 = exp(-DELTAG_1 /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P_1 = (P/N_tot)^(1+0.5-1)*(b^1*e^0.5)/(a^1)" sqrt(P/N_tot) *b *sqrt(e) =K_P_1*a "The econd assumed equilibrium reaction is 0.5N2+0.5O2=NO" g_NO=Enthalpy(NO,T=T_prod )-T_prod *Entropy(NO,T=T_prod ,P=101.3) g_N2=Enthalpy(N2,T=T_prod )-T_prod *Entropy(N2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG_2 =1*g_NO-0.5*g_O2-0.5*g_N2 "The equilibrium constant is given by Eq. 15-14." K_P_2 = exp(-DELTAG_2 /(R_u*T_prod )) "The equilibrium constant is also given by Eq. 15-15." "K_ P_2 = (P/N_tot)^(1-0.5-0.5)*(g^1)/(e^0.5*f^0.5)" g=K_P_2 *sqrt(e*f) "The steady-flow energy balance is:" H_R = Q_out+H_P h_bar_f_C2H5OHgas=-235310 "[kJ/kmol]" H_R=1*(h_bar_f_C2H5OHgas ) +(1+Ex)*A_th*ENTHALPY(O2,T=T_reac)+(1+Ex)*A_th*3.76*ENTHALPY(N2,T=T_reac) "[kJ/kmol]" H_P=a*ENTHALPY(CO2,T=T_prod)+b*ENTHALPY(CO,T=T_prod)+d*ENTHALPY(H2O,T=T_prod)+e* ENTHALPY(O2,T=T_prod)+f*ENTHALPY(N2,T=T_prod)+g*ENTHALPY(NO,T=T_prod) "[kJ/kmol]"
ath
a
b
d
e
f
g
3 3 3 3 3 3 3 3 3 3
1.922 1.971 1.988 1.995 1.998 1.999 2 2 2 2
0.07779 0.0293 0.01151 0.004708 0.001993 0.0008688 0.0003884 0.0001774 0.00008262 0.00003914
3 3 3 3 3 3 3 3 3 3
0.3081 0.5798 0.8713 1.17 1.472 1.775 2.078 2.381 2.683 2.986
12.38 13.5 14.63 15.76 16.89 18.02 19.15 20.28 21.42 22.55
0.0616 0.06965 0.06899 0.06426 0.05791 0.05118 0.04467 0.03867 0.0333 0.02856
PercentEx [%] 10 20 30 40 50 60 70 80 90 100
Tprod [K] 2184 2085 1989 1901 1820 1747 1682 1621 1566 1516
2200
2100
2000
) K ( d or p
1900
1800
T
1700
1600
1500 10
20
30
40
50
60
70
80
90
100
PercentEx
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-41
Variations of Kp with Temperature
16-48C It enables us to determine the enthalpy of reaction hR from a knowledge of equilibrium constant KP. 16-49C At 2000 K since combustion processes are exothermic, and exothermic reactions are more complete at lower temperatures.
16-50 The hR at a specified temperature is to be determined using the enthalpy and KP data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The complete combustion equation of CO can be expressed as
CO + 12 O 2 ⇔ CO 2 The hR of the combustion process of CO at 2200 K is the amount of energy released as one kmol of CO is burned in a steady-flow combustion chamber at a temperature of 2200 K, and can be determined from hR =
∑ N (h P
o f
+h −ho
) − ∑ N (h P
R
o f
+h −ho
)
R
Assuming the CO, O2 and CO2 to be ideal gases, we have h = h(T). From the tables,
Substance CO2 CO O2
hfo kJ/kmol -393,520 -110,530 0
h 298 K
h 2200 K
kJ/kmol 9364 8669 8682
kJ/kmol 112,939 72,688 75,484
Substituting, hR = 1( −393,520 + 112,939 − 9364) − 1( −110,530 + 72,688 − 8669) − 0.5(0 + 75,484 − 8682) = −276,835 kJ / kmol
(b) The hR value at 2200 K can be estimated by using KP values at 2000 K and 2400 K (the closest two temperatures to 2200 K for which KP data are available) from Table A-28, ln
K P 2 hR ⎛ 1 h ⎛1 1 ⎞ 1 ⎞ ⎜⎜ − ⎟⎟ or ln K P 2 − ln K P1 ≅ R ⎜⎜ − ⎟ ≅ K P1 Ru ⎝ T1 T2 ⎠ Ru ⎝ T1 T2 ⎟⎠
3.860 − 6.635 ≅
hR 1 ⎞ ⎛ 1 − ⎜ ⎟ 8.314 kJ/kmol ⋅ K ⎝ 2000 K 2400 K ⎠
h R ≅ −276,856 kJ/kmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-42
16-51E The hR at a specified temperature is to be determined using the enthalpy and KP data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The complete combustion equation of CO can be expressed as
CO + 12 O 2 ⇔ CO 2 The hR of the combustion process of CO at 3960 R is the amount of energy released as one kmol of H2 is burned in a steady-flow combustion chamber at a temperature of 3960 R, and can be determined from hR =
∑ N (h P
o f
+h −ho
) − ∑ N (h P
R
o f
+h −ho
)
R
Assuming the CO, O2 and CO2 to be ideal gases, we have h = h (T). From the tables,
Substance CO2 CO O2
hfo Btu/lbmol -169,300 -47,540 0
h 537 R
h 3960 R
Btu/lbmol 4027.5 3725.1 3725.1
Btu/lbmol 48,647 31,256.5 32,440.5
Substituting, hR = 1( −169,300 + 48,647 − 4027.5) .) − 1( −47,540 + 31,256.5 − 37251 .) − 0.5(0 + 32,440.5 − 37251 = −119,030 Btu / lbmol
(b) The hR value at 3960 R can be estimated by using KP values at 3600 R and 4320 R (the closest two temperatures to 3960 R for which KP data are available) from Table A-28, ln
K P 2 hR ⎛ 1 h ⎛1 1 ⎞ 1 ⎞ ⎜⎜ − ⎟⎟ or ln K P 2 − ln K P1 ≅ R ⎜⎜ − ⎟ ≅ K P1 Ru ⎝ T1 T2 ⎠ Ru ⎝ T1 T2 ⎟⎠
3.860 − 6.635 ≅
hR 1 ⎞ ⎛ 1 − ⎜ ⎟ 1.986 Btu/lbmol ⋅ R ⎝ 3600 R 4320 R ⎠
h R ≅ −119,041 Btu/lbmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-43
16-52 The KP value of the combustion process H2 + 1/2O2 ⇔ H2O is to be determined at a specified temperature using hR data and KP value . Assumptions Both the reactants and products are ideal gases. Analysis The hR and KP data are related to each other by ln
K P 2 hR ⎛ 1 h ⎛1 1 ⎞ 1 ⎞ ⎜⎜ − ⎟⎟ or ln K P 2 − ln K P1 ≅ R ⎜⎜ − ⎟⎟ ≅ K P1 Ru ⎝ T1 T2 ⎠ Ru ⎝ T1 T2 ⎠
The hR of the specified reaction at 2400 K is the amount of energy released as one kmol of H2 is burned in a steady-flow combustion chamber at a temperature of 2400 K, and can be determined from hR =
∑ N (h P
o f
+h −ho
) − ∑ N (h P
R
o f
+h −ho
)
R
Assuming the H2O, H2 and O2 to be ideal gases, we have h = h (T). From the tables,
Substance H2O H2 O2
h fo kJ/kmol -241,820 0 0
h 298 K
h 2400 K
kJ/kmol 9904 8468 8682
kJ/kmol 103,508 75,383 83,174
Substituting, hR = 1( −241,820 + 103,508 − 9904) − 1(0 + 75,383 − 8468) − 0.5(0 + 83,174 − 8682) = −252,377 kJ / kmol
The KP value at 2600 K can be estimated from the equation above by using this hR value and the KP value at 2200 K which is ln KP1 = 6.768, ln K P 2 − 6.768 ≅
1 ⎞ − 252,377 kJ/kmol ⎛ 1 − ⎜ ⎟ 8.314 kJ/kmol ⋅ K ⎝ 2200 K 2600 K ⎠
ln K P 2 = 4.645
(Table A - 28: lnK P 2 = 4.648)
or K P2 = 104.1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-44
16-53 The hR value for the dissociation process CO2 ⇔ CO + 1/2O2 at a specified temperature is to be determined using enthalpy and Kp data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The dissociation equation of CO2 can be expressed as CO 2 ⇔ CO + 12 O 2
The hR of the dissociation process of CO2 at 2200 K is the amount of energy absorbed or released as one kmol of CO2 dissociates in a steady-flow combustion chamber at a temperature of 2200 K, and can be determined from hR =
∑ N (h P
o f
+h −ho
) − ∑ N (h P
R
o f
+h −ho
)
R
Assuming the CO, O2 and CO2 to be ideal gases, we have h = h (T). From the tables, h fo kJ/kmol -393,520 -110,530 0
Substance CO2 CO O2
h 298 K
h 2200 K
kJ/kmol 9364 8669 8682
kJ/kmol 112,939 72,688 75,484
Substituting, hR = 1( −110,530 + 72,688 − 8669) + 0.5( 0 + 75,484 − 8682) − 1( −393,520 + 112,939 − 9364) = 276,835 kJ / kmol
(b) The hR value at 2200 K can be estimated by using KP values at 2000 K and 2400 K (the closest two temperatures to 2200 K for which KP data are available) from Table A-28, ln
K P 2 hR ≅ K P1 Ru
⎛1 1 ⎜⎜ − ⎝ T1 T2
− 3.860 − (−6.635) ≅
⎞ h ⎟⎟ or ln K P 2 − ln K P1 ≅ R Ru ⎠
⎛1 1 ⎜⎜ − ⎝ T1 T2
⎞ ⎟⎟ ⎠
hR 1 ⎞ ⎛ 1 − ⎜ ⎟ 8.314 kJ/kmol ⋅ K ⎝ 2000 K 2400 K ⎠
h R ≅ 276,856 kJ/kmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-45
16-54 The hR value for the dissociation process O2 ⇔ 2O at a specified temperature is to be determined using enthalpy and KP data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The dissociation equation of O2 can be expressed as O 2 ⇔ 2O
The hR of the dissociation process of O2 at 3100 K is the amount of energy absorbed or released as one kmol of O2 dissociates in a steady-flow combustion chamber at a temperature of 3100 K, and can be determined from hR =
∑ N (h P
o f
+h −ho
) − ∑ N (h P
R
o f
+h −ho
)
R
Assuming the O2 and O to be ideal gases, we have h = h (T). From the tables, h fo kJ/kmol 249,190 0
Substance O O2
h 298 K
h 2900 K
kJ/kmol 6852 8682
kJ/kmol 65,520 110,784
Substituting, hR = 2(249,190 + 65,520 − 6852) − 1(0 + 110,784 − 8682) = 513,614 kJ/kmol
(b) The hR value at 3100 K can be estimated by using KP values at 3000 K and 3200 K (the closest two temperatures to 3100 K for which KP data are available) from Table A-28, ln
K P 2 hR ≅ K P1 Ru
⎛1 1 ⎜⎜ − ⎝ T1 T2
− 3.072 − (−4.357) ≅
⎞ h ⎟⎟ or ln K P 2 − ln K P1 ≅ R Ru ⎠
⎛1 1 ⎜⎜ − ⎝ T1 T2
⎞ ⎟⎟ ⎠
hR 1 ⎞ ⎛ 1 − ⎟ ⎜ 8.314 kJ/kmol ⋅ K ⎝ 3000 K 3200 K ⎠
h R ≅ 512,808 kJ/kmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-46
16-55 The enthalpy of reaction for the equilibrium reaction CH4 + 2O2 = CO2 + 2H2O at 2500 K is to be estimated using enthalpy data and equilibrium constant, Kp data. Analysis The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e − ∆G*(T ) / RuT or ln K p = −∆G * (T ) / Ru T
where ∗ ∗ ∗ ∗ ∆G * (T ) = ν CO2 g CO2 (T ) + ν H2O g H2O (T ) −ν CH4 g CH4 (T ) −ν O2 g O2 (T )
At T1 = 2500 - 10 = 2490 K: ∗ ∗ ∗ ∗ ∆G1 * (T ) = ν CO2 g CO2 (T1 ) + ν H2O g H2O (T1 ) −ν CH4 g CH4 (T1 ) −ν O2 g O2 (T1 )
= 1(−1.075 × 10 6 ) + 2(−830,577) − 1(−717,973) − 2(−611,582) = −794,929 kJ/kmol
At T2 = 2500 + 10 = 2510 K: ∗ ∗ ∗ ∗ ∆G 2 * (T ) = ν CO2 g CO2 (T2 ) + ν H2O g H2O (T2 ) −ν CH4 g CH4 (T2 ) −ν O2 g O2 (T2 )
= 1(−1.081× 10 6 ) + 2(−836,100) − 1(−724,516) − 2(−617,124) = −794,801 kJ/kmol
The Gibbs functions are obtained from enthalpy and entropy properties using EES. Substituting, ⎞ ⎛ − 794,929 kJ/kmol ⎟⎟ = 4.747 × 1016 K p1 = exp⎜⎜ − (8.314 kJ/kmol K)(2490 K) ⋅ ⎠ ⎝ ⎞ ⎛ − 794,801 kJ/kmol ⎟⎟ = 3.475 × 1016 K p 2 = exp⎜⎜ − ⎝ (8.314 kJ/kmol ⋅ K)(2510 K) ⎠
The enthalpy of reaction is determined by using the integrated van't Hoff equation: ⎛ K p2 ln⎜ ⎜ K p1 ⎝ ⎛ 3.475 × 1016 ln⎜⎜ 16 ⎝ 4.747 × 10
⎞ hR ⎛ 1 1 ⎞ ⎟= ⎜ − ⎟ ⎟ Ru ⎜⎝ T1 T2 ⎟⎠ ⎠ ⎞ hR 1 ⎞ ⎛ 1 ⎟= ⎯→ h R = −810,845 kJ/kmol ⎟ 8.314 kJ/kmol.K ⎜⎝ 2490 K − 2510 K ⎟⎠ ⎯ ⎠
The enthalpy of reaction can also be determined from an energy balance to be hR = H P − H R
where H R = 1hCH4 @ 2500 K + 2hO2 @ 2500 K = 96,668 + 2(78,377) = 253,422 kJ/kmol H P = 1hCO2 @ 2500 K + 2hH2O @ 2500 K = (−271,641) + 2(−142,891) = −557,423 kJ/kmol
The enthalpies are obtained from EES. Substituting, h R = H P − H R = (−557,423) − (253,422) = −810,845 kJ/kmol
which is identical to the value obtained using Kp data.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-47
Phase Equilibrium 16-56C No. Because the specific gibbs function of each phase will not be affected by this process; i.e., we will still have gf = gg. 16-57C Yes. Because the number of independent variables for a two-phase (PH=2), two-component (C=2) mixture is, from the phase rule, IV = C - PH + 2 = 2 - 2 + 2 = 2 Therefore, two properties can be changed independently for this mixture. In other words, we can hold the temperature constant and vary the pressure and still be in the two-phase region. Notice that if we had a single component (C=1) two phase system, we would have IV=1, which means that fixing one independent property automatically fixes all the other properties. 11-58C Using solubility data of a solid in a specified liquid, the mass fraction w of the solid A in the liquid at the interface at a specified temperature can be determined from mf A =
msolid msolid + m liquid
where msolid is the maximum amount of solid dissolved in the liquid of mass mliquid at the specified temperature. 11-59C The molar concentration Ci of the gas species i in the solid at the interface Ci, solid side (0) is proportional to the partial pressure of the species i in the gas Pi, gas side(0) on the gas side of the interface, and is determined from C i, solid side (0) = S × Pi, gas side (0)
(kmol/m3)
where S is the solubility of the gas in that solid at the specified temperature. 11-60C Using Henry’s constant data for a gas dissolved in a liquid, the mole fraction of the gas dissolved in the liquid at the interface at a specified temperature can be determined from Henry’s law expressed as yi, liquid side (0) =
Pi, gas side (0) H
where H is Henry’s constant and Pi,gas side(0) is the partial pressure of the gas i at the gas side of the interface. This relation is applicable for dilute solutions (gases that are weakly soluble in liquids).
16-61 It is to be shown that a mixture of saturated liquid water and saturated water vapor at 100°C satisfies the criterion for phase equilibrium. Analysis Using the definition of Gibbs function and enthalpy and entropy data from Table A-4, g f = h f − Ts f = (419.17 kJ/kg) − (373.15 K)(1.3072 kJ/kg ⋅ K) = −68.61 kJ/kg g g = h g − Ts g = (2675.6 kJ/kg) − (373.15 K)(7.3542 kJ/kg ⋅ K) = −68.62 kJ/kg
which are practically same. Therefore, the criterion for phase equilibrium is satisfied.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-48
16-62 It is to be shown that a mixture of saturated liquid water and saturated water vapor at 300 kPa satisfies the criterion for phase equilibrium. Analysis The saturation temperature at 300 kPa is 406.7 K. Using the definition of Gibbs function and enthalpy and entropy data from Table A-5, g f = h f − Ts f = (561.43 kJ/kg) − (406.7 K)(1.6717 kJ/kg ⋅ K) = −118.5 kJ/kg g g = h g − Ts g = (2724.9 kJ/kg) − (406.7 K)(6.9917 kJ/kg ⋅ K) = −118.6 kJ/kg
which are practically same. Therefore, the criterion for phase equilibrium is satisfied.
16-63 It is to be shown that a saturated liquid-vapor mixture of refrigerant-134a at -10°C satisfies the criterion for phase equilibrium. Analysis Using the definition of Gibbs function and enthalpy and entropy data from Table A-11, g f = h f − Ts f = (38.55 kJ/kg) − (263.15 K)(0.15504 kJ/kg ⋅ K) = −2.249 kJ/kg g g = h g − Ts g = (244.51 kJ/kg) − (263.15 K)(0.93766 kJ/kg ⋅ K) = −2.235 kJ/kg
which are sufficiently close. Therefore, the criterion for phase equilibrium is satisfied.
16-64 The number of independent properties needed to fix the state of a mixture of oxygen and nitrogen in the gas phase is to be determined. Analysis In this case the number of components is C = 2 and the number of phases is PH = 1. Then the number of independent variables is determined from the phase rule to be IV = C - PH + 2 = 2 - 1 + 2 = 3 Therefore, three independent properties need to be specified to fix the state. They can be temperature, the pressure, and the mole fraction of one of the gases.
16-65 A liquid-vapor mixture of ammonia and water in equilibrium at a specified temperature is considered. The composition of the liquid phase is given. The composition of the vapor phase is to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 30°C, Psat, H 2O = 4.247 kPa and Psat, NH 3 = 1167.4 kPa . Analysis The vapor pressures are PH 2 O = y f ,H 2O Psat, H 2O (T ) = 0.40(4.247 kPa) = 1.70 kPa
H2O + NH3 30°C
PNH 3 = y f , NH 3 Psat, NH 3 (T ) = 0.60(1167.4 kPa) = 700.44 kPa
Thus the total pressure of the mixture is Ptotal = PH 2O + PNH 3 = (1.70 + 700.44) kPa = 702.1 kPa Then the mole fractions in the vapor phase become PH O 1.70 kPa = 0.0024 or 0.24% y g , H 2O = 2 = Ptotal 702.1 kPa y g , NH 3 =
PNH 3 Ptotal
=
700.44 kPa = 0.9976 or 99.76% 702.1 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-49
16-66 A liquid-vapor mixture of ammonia and water in equilibrium at a specified temperature is considered. The composition of the liquid phase is given. The composition of the vapor phase is to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 25°C, Psat, H 2 O = 3.170 kPa and Psat, NH 3 = 1003.5 kPa . Analysis The vapor pressures are PH 2O = y f ,H 2O Psat,H 2O (T ) = 0.50(3.170 kPa) = 1.585 kPa
H2O + NH3 25°C
PNH3 = y f , NH 3 Psat, NH3 (T ) = 0.50(1003.5 kPa) = 501.74 kPa
Thus the total pressure of the mixture is Ptotal = PH 2O + PNH 3 = (1.585 + 501.74) kPa = 503.33 kPa Then the mole fractions in the vapor phase become PH O 1.585 kPa = 0.0031 or 0.31% y g , H 2O = 2 = Ptotal 503.33 kPa y g , NH3 =
PNH 3 Ptotal
=
501.74 kPa = 0.9969 or 99.69% 503.33 kPa
16-67 A liquid-vapor mixture of ammonia and water in equilibrium at a specified temperature is considered. The composition of the vapor phase is given. The composition of the liquid phase is to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 50°C, Psat, H 2O = 12.352 kPa and Psat, NH 3 = 2033.5 kPa. Analysis We have y g ,H 2O = 1% and y g , NH3 = 99% . For an ideal two-phase mixture we have y g , H 2 O Pm = y f , H 2 O Psat, H 2 O (T ) y g , NH 3 Pm = y f , NH 3 Psat, NH 3 (T )
H2O + NH3
y f , H 2 O + y f , NH 3 = 1
50°C
Solving for y f ,H 2O, y f ,H 2O =
y g ,H 2O Psat, NH3 y g , NH3 Psat,H 2O
(1 − y f ,H 2O ) =
(0.01)(2033.5 kPa) (1 − y f ,H 2O ) (0.99)(12.352 kPa)
It yields y f ,H 2O = 0.624 and y f , NH3 = 0.376
16-68 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture, the composition of each phase at a specified temperature and pressure is to be determined. Analysis From the equilibrium diagram (Fig. 16-21) we read Liquid: 37% O 2 and 63% N 2 Vapor: 10% O 2 and 90% N 2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-50
16-69 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture, the composition of each phase at a specified temperature and pressure is to be determined. Analysis From the equilibrium diagram (Fig. 16-21) we read Liquid: 30% N 2 and 70% O 2 Vapor: 66% N 2 and 34% O 2
16-70 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture at a specified pressure, the temperature is to be determined for a specified composition of the vapor phase. Analysis From the equilibrium diagram (Fig. 16-21) we read T = 82 K.
16-71 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture at a specified pressure, the temperature is to be determined for a specified composition of the liquid phase. Analysis From the equilibrium diagram (Fig. 16-21) we read T = 84 K.
16-72 A rubber plate is exposed to nitrogen. The molar and mass density of nitrogen in the iron at the interface is to be determined. Assumptions Rubber and nitrogen are in thermodynamic equilibrium at the interface. Properties The molar mass of nitrogen is M = 28.0 kg/kmol (Table A-1). The solubility of nitrogen in rubber at 298 K is 0.00156 kmol/m3⋅bar (Table 16-3). Analysis Noting that 250 kPa = 2.5 bar, the molar density of nitrogen in the rubber at the interface is determined to be C N 2 , solid side (0) = S × PN 2 , gas side = (0.00156 kmol/m 3 .bar )(2.5 bar) = 0.0039 kmol/m 3
It corresponds to a mass density of
ρ N 2 , solid side (0) = C N 2 , solid side (0) M N 2 = (0.0039 kmol/m 3 )(28 kg/kmol) = 0.1092 kg/m 3
That is, there will be 0.0039 kmol (or 0.1092 kg) of N2 gas in each m3 volume of iron adjacent to the interface.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-51
16-73 A rubber wall separates O2 and N2 gases. The molar concentrations of O2 and N2 in the wall are to be determined. Assumptions The O2 and N2 gases are in phase equilibrium with the rubber wall. Properties The molar mass of oxygen and nitrogen are 32.0 and 28.0 kg/kmol, respectively (Table A-1). The solubility of oxygen and nitrogen in rubber at 298 K are 0.00312 and 0.00156 kmol/m3⋅bar, respectively (Table 16-3). Analysis Noting that 500 kPa = 5 bar, the molar densities of oxygen and nitrogen in the rubber wall are determined to be
Rubber plate
C O 2 , solid side (0) = S × PO 2 , gas side = (0.00312 kmol/m 3 .bar )(5 bar) = 0.0156 kmol/m
3
CN 2 , solid side (0) = × PN 2 , gas side
O2 25°C 500 kPa
CO2 CN2
N2 25°C 500 kPa
= (0.00156 kmol / m3 . bar )(5 bar) = 0.0078 kmol / m 3
That is, there will be 0.0156 kmol of O2 and 0.0078 kmol of N2 gas in each m3 volume of the rubber wall.
16-74 A glass of water is left in a room. The mole fraction of the water vapor in the air and the mole fraction of air in the water are to be determined when the water and the air are in thermal and phase equilibrium. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is saturated since the humidity is 100 percent. 3 Air is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 27°C is 3.568 kPa (Table A-4). Henry’s constant for air dissolved in water at 27ºC (300 K) is given in Table 16-2 to be H = 74,000 bar. Molar masses of dry air and water are 29 and 18 kg/kmol, respectively (Table A-1). Analysis (a) Noting that air is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at 27°C,
Pvapor = Psat @ 27°C = 3.600 kPa (Table A-4) Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the air is determined to be y vapor =
Pvapor P
=
3.600 kPa = 0.0371 97 kPa
Air 27ºC 97 kPa φ = 100%
(b) Noting that the total pressure is 97 kPa, the partial pressure of dry air is
Pdry air = P − Pvapor = 97 − 3.600 = 93.4 kPa = 0.934 bar
Water 27ºC
From Henry’s law, the mole fraction of air in the water is determined to be y dry air,liquid side =
Pdry air,gas side H
=
0.934 bar = 1.26 ×10 − 5 74,000 bar
Discussion The amount of air dissolved in water is very small, as expected.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-52
16-75E Water is sprayed into air, and the falling water droplets are collected in a container. The mass and mole fractions of air dissolved in the water are to be determined. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is saturated since water is constantly sprayed into it. 3 Air is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 80°F is 0.5075 psia (Table A-4E). Henry’s constant for air dissolved in water at 80ºF (300 K) is given in Table 16-2 to be H = 74,000 bar. Molar masses of dry air and water are 29 and 18 lbm / lbmol, respectively (Table A-1). Analysis Noting that air is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at 80°F,
Pvapor = Psat @ 80°F = 0.5075 psia Then the partial pressure of dry air becomes
Pdry air = P − Pvapor = 14.3 − 0.5075 = 13.793 psia From Henry’s law, the mole fraction of air in the water is determined to be y dry air,liquid side =
Pdry air,gasside H
=
13.793 psia (1 atm / 14.696 psia ) = 1.29 ×10 −5 74,000 bar (1 atm/1.01325 bar)
which is very small, as expected. The mass and mole fractions of a mixture are related to each other by mf i =
mi N M Mi = i i = yi mm N m M m Mm
where the apparent molar mass of the liquid water - air mixture is
Mm =
∑y M i
i
= y liquid water M water + y dry air M dry air
≅ 1 × 18.0 + 0 × 29.0 ≅ 18.0 kg/kmol Then the mass fraction of dissolved air in liquid water becomes mf dry air, liquid side = y dry air, liquid side (0)
M dry air Mm
= (1.29 ×10 −5 )
29 = 2.08 ×10 −5 18
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-53
16-76 A carbonated drink in a bottle is considered. Assuming the gas space above the liquid consists of a saturated mixture of CO2 and water vapor and treating the drink as a water, determine the mole fraction of the water vapor in the CO2 gas and the mass of dissolved CO2 in a 300 ml drink are to be determined when the water and the CO2 gas are in thermal and phase equilibrium. Assumptions 1 The liquid drink can be treated as water. 2 Both the CO2 and the water vapor are ideal gases. 3 The CO2 gas and water vapor in the bottle from a saturated mixture. 4 The CO2 is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 27°C is 3.568 kPa (Table A-4). Henry’s constant for CO2 dissolved in water at 27ºC (300 K) is given in Table 16-2 to be H = 1710 bar. Molar masses of CO2 and water are 44 and 18 kg/kmol, respectively (Table A-1). Analysis (a) Noting that the CO2 gas in the bottle is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at 27°C,
Pvapor = Psat @ 27°C = 3.568 kPa
(more accurate EES value compared to interpolation value from Table A-4)
Assuming both CO2 and vapor to be ideal gases, the mole fraction of water vapor in the CO2 gas becomes y vapor =
Pvapor P
=
3.568 kPa = 0.0274 130 kPa
(b) Noting that the total pressure is 130 kPa, the partial pressure of CO2 is
PCO 2 gas = P − Pvapor = 130 − 3.568 = 126.4 kPa = 1.264 bar From Henry’s law, the mole fraction of CO2 in the drink is determined to be yCO 2 ,liquid side =
PCO 2 ,gas side H
=
1264 . bar = 7.39 × 10 −4 1710 bar
Then the mole fraction of water in the drink becomes y water, liquid side = 1 − y CO 2 , liquid side = 1 − 7.39 × 10 −4 = 0.9993
The mass and mole fractions of a mixture are related to each other by mf i =
mi N M Mi = i i = yi mm N m M m Mm
where the apparent molar mass of the drink (liquid water - CO2 mixture) is Mm =
∑y M i
i
= yliquid water M water + yCO 2 M CO 2 = 0.9993 × 18.0 + ( 7.39 × 10 −4 ) × 44 = 18.02 kg / kmol
Then the mass fraction of dissolved CO2 gas in liquid water becomes mf CO 2 , liquid side = y CO 2 , liquid side (0)
M CO 2 Mm
= 7.39 ×10 − 4
44 = 0.00180 18.02
Therefore, the mass of dissolved CO2 in a 300 ml ≈ 300 g drink is
mCO 2 = mf CO 2 m m = (0.00180)(300 g) = 0.54 g
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-54
Review Problems
16-77 The equilibrium constant of the dissociation process O2 ↔ 2O is given in Table A-28 at different temperatures. The value at a given temperature is to be verified using Gibbs function data. Analysis The KP value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G *( T )/ Ru T or ln K p = − ∆G * (T ) / Ru T
O2 ↔ 2O
where ∆G * (T ) = ν O g O* (T ) −ν O 2 g O* 2 (T )
2000 K
= ν O ( h − Ts ) O − ν O 2 ( h − Ts ) O 2 = ν O [(h f + h2000 − h298 ) − Ts ] O −ν O 2 [(h f + h2000 − h298 ) − Ts ] O 2 = 2 × (249,190 + 42,564 − 6852 − 2000 × 201.135) − 1× (0 + 67,881 − 8682 − 2000 × 268.655) = 243,375 kJ/kmol
Substituting,
ln K p = −(243,375 kJ/kmol)/[(8.314 kJ/kmol ⋅ K)(2000 K)] = −14.636 or K p = 4.4 × 10 −7
(Table A-28: ln KP = -14.622)
16-78 A mixture of H2 and Ar is heated is heated until 15% of H2 is dissociated. The final temperature of mixture is to be determined. Assumptions 1 The constituents of the mixture are ideal gases. 2 Ar in the mixture remains an inert gas. Analysis The stoichiometric and actual reactions can be written as
Stoichiometric:
H 2 ⇔ 2H (thus ν H 2 = 1 and ν H = 2)
Actual:
H 2 + Ar ⎯ ⎯→ 0{ .3H + 0.85H 2 + Ar { 1 424 3 inert prod
H 2 ⇔ 2H
Ar 1 atm
react.
The equilibrium constant KP can be determined from ν N H ⎛ P K p = νH ⎜⎜ N HH 2 ⎝ N total 2
ν H −ν H 2
⎞ ⎟ ⎟ ⎠
=
0.3 2 ⎛ 1 ⎞ ⎜ ⎟ 0.85 ⎝ 0.85 + 0.3 + 1 ⎠
2 −1
= 0.0492
From Table A-28, the temperature corresponding to this KP value is T = 3117 K.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-55
16-79 A mixture of H2O, O2, and N2 is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of H2O, O2, N2 and H2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are
Stoichiometric:
H 2 O ⇔ H 2 + 12 O 2 (thus ν H 2 O = 1, ν H 2 = 1, and ν O 2 = 12 )
Actual:
H 2 O + 2O 2 + 5 N 2 ⎯ ⎯→ x H 2 O + y H 2 + z O 2 + 5 N 2 123 14243 { react.
H balance:
2 = 2x + 2 y
O balance:
5 = x + 2z ⎯ ⎯→ z = 2.5 − 0.5 x
Total number of moles:
⎯ ⎯→
products
inert
1 H2O 2 O2 5 N2 2200 K 5 atm
y = 1− x
N total = x + y + z + 5 = 8.5 − 0.5 x
The equilibrium constant relation can be expressed as ν
ν
N HH 2 N OO 2 ⎛ P ⎞ 2 2 ⎜ ⎟ Kp = ν ⎜ ⎟ N HH2 O2 O ⎝ N total ⎠
(ν H 2 −ν O 2 −ν H 2 O )
=
1+ 0.5 −1
z 0 .5 ⎛ P ⎞ ⎜ ⎟ ⎜N ⎟ x ⎝ total ⎠
y
From Table A-28, lnKP = -6.768 at 2200 K. Thus KP = 0.00115. Substituting,
0.00115 =
(1 − x)(1.5 − 0.5 x) 0.5 ⎛ 5 ⎞ ⎜ ⎟ x ⎝ 8.5 − 0.5 x ⎠
0.5
Solving for x, x = 0.9981 Then, y = 1 - x = 0.0019 z = 2.5 - 0.5x = 2.00095 Therefore, the equilibrium composition of the mixture at 2200 K and 5 atm is 0.9981H 2 O + 0.0019H 2 + 2.00095O 2 + 5N 2
The equilibrium constant for the reaction H 2 O ⇔ OH + 21 H 2 is lnKP = -7.148, which is very close to the KP value of the reaction considered. Therefore, it is not realistic to assume that no OH will be present in equilibrium mixture.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-56
16-80 The mole fraction of argon that ionizes at a specified temperature and pressure is to be determined. Assumptions All components behave as ideal gases. Analysis The stoichiometric and actual reactions can be written as
Stoichiometric:
Ar ⇔ Ar + + e - (thus ν Ar = 1, ν Ar + = 1 and ν e - = 1)
Actual:
Ar ⎯ ⎯→ { xAr + y Ar + + ye − 14243 react.
Ar balance:
Ar ⇔ Ar + + e −
products
10,000 K 0.35 atm
1 = x + y or y = 1 − x N total = x + 2 y = 2 − x
Total number of moles:
The equilibrium constant relation becomes Kp =
N νArAr N N νArAr
ν
e-
e-
⎛ P ⎜ ⎜N ⎝ total
⎞ ⎟ ⎟ ⎠
(ν
Ar +
+ν - −ν Ar ) e
=
y2 x
⎛ P ⎜ ⎜N ⎝ total
⎞ ⎟ ⎟ ⎠
1+1−1
Substituting, 0.00042 =
(1 − x) 2 ⎛ 0.35 ⎞ ⎜ ⎟ x ⎝2− x⎠
Solving for x, x = 0.965 Thus the fraction of Ar which dissociates into Ar+ and e- is 1 - 0.965 = 0.035 or 3.5%
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-57
16-81 [Also solved by EES on enclosed CD] Methane gas is burned with stoichiometric amount of air during a combustion process. The equilibrium composition and the exit temperature are to be determined. Assumptions 1 The product gases consist of CO2, H2O, CO, N2, and O2. 2 The constituents of the mixture are ideal gases. 3 This is an adiabatic and steady-flow combustion process. Analysis (a) The combustion equation of CH4 with stoichiometric amount of O2 can be written as CH 4 + 2( O 2 + 3.76 N 2 ) ⎯ ⎯→ xCO 2 + (1 − x )CO + (0.5 − 0.5x )O 2 + 2H 2O + 7.52N 2 After combustion, there will be no CH4 present in the combustion chamber, and H2O will act like an inert gas. The equilibrium equation among CO2, CO, and O2 can be expressed as CO 2 ⇔ CO + 12 O 2 (thus ν CO 2 = 1, ν CO = 1, and ν O 2 = 12 )
and ν
CO N νCO N OO 2 ⎛ P ⎞ 2 ⎜ ⎟ Kp = ν CO 2 ⎜N ⎟ N CO total ⎠ ⎝ 2
where
CH4
(ν CO +ν O 2 −ν CO 2 )
25°C
N total = x + (1 − x ) + (15 . − 0.5 x ) + 2 + 7.52 = 12.02 − 0.5x
CO CO2 H2O O2 N2
Combustion chamber
Air
1 atm
1.5−1
(1 − x)(0.5 − 0.5 x) ⎛ 1 25°C ⎞ Kp = Substituting, ⎜ ⎟ x ⎝ 12.02 − 0.5 x ⎠ The value of KP depends on temperature of the products, which is yet to be determined. A second relation to determine KP and x is obtained from the steady-flow energy balance expressed as 0.5
0=
∑ N (h P
o f
+h −ho
) − ∑ N (h P
R
o f
+h −ho
)
R
⎯ ⎯→ 0 =
∑ N (h P
o f
+h −ho
) −∑ N P
o Rhf R
since the combustion is adiabatic and the reactants enter the combustion chamber at 25°C. Assuming the air and the combustion products to be ideal gases, we have h = h (T). From the tables,
hfo kJ/kmol -74,850 0 0 -241,820 -110,530 -393,520
Substance
CH4(g) N2 O2 H2O(g) CO CO2
h 298 K
kJ/kmol -8669 8682 9904 8669 9364
Substituting, 0 = x ( −393,520 + hCO 2 − 9364) + (1 − x )( −110,530 + hCO − 8669) + 2( −241,820 + hH 2 O − 9904) + (0.5 − 0.5x )(0 + hO 2 − 8682) + 7.52(0 + hN 2 − 8669) − 1( −74,850 + h298 − h298 ) − 0 − 0
which yields
x hCO 2 + (1 − x )hCO + 2 hH 2O + ( 0.5 − 0.5 x )hO 2 + 7.52 hN 2 − 279,344 x = 617,329
Now we have two equations with two unknowns, TP and x. The solution is obtained by trial and error by assuming a temperature TP, calculating the equilibrium composition from the first equation, and then checking to see if the second equation is satisfied. A first guess is obtained by assuming there is no CO in the products, i.e., x = 1. It yields TP = 2328 K. The adiabatic combustion temperature with incomplete combustion will be less. Take Tp = 2300 K
⎯ ⎯→
ln K p = −4.49
Take Tp = 2250 K
⎯ ⎯→
ln K p = −4.805
By interpolation,
⎯ ⎯→ ⎯ ⎯→
x = 0.870
⎯ ⎯→
RHS = 641,093
x = 0.893
⎯ ⎯→
RHS = 612,755
Tp = 2258 K and x = 0.889
Thus the composition of the equilibrium mixture is 0.889CO 2 + 0.111CO + 0.0555O 2 + 2H 2 O + 7.52N 2
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-58
16-82 EES Problem 16-81 is reconsidered. The effect of excess air on the equilibrium composition and the exit temperature by varying the percent excess air from 0 to 200 percent is to be studied. Analysis The problem is solved using EES, and the solution is given below. "Often, for nonlinear problems such as this one, good gusses are required to start the solution. First, run the program with zero percent excess air to determine the net heat transfer as a function of T_prod. Just press F3 or click on the Solve Table icon. From Plot Window 1, where Q_net is plotted vs T_prod, determnine the value of T_prod for Q_net=0 by holding down the Shift key and move the cross hairs by moving the mouse. Q_net is approximately zero at T_prod = 2269 K. From Plot Window 2 at T_prod = 2269 K, a, b, and c are approximately 0.89, 0.10, and 0.056, respectively." "For EES to calculate a, b, c, and T_prod directly for the adiabatic case, remove the '{ }' in the last line of this window to set Q_net = 0.0. Then from the Options menu select Variable Info and set the Guess Values of a, b, c, and T_prod to the guess values selected from the Plot Windows. Then press F2 or click on the Calculator icon." "Input Data" {PercentEx = 0} Ex = PercentEX/100 P_prod =101.3 [kPa] R_u=8.314 [kJ/kmol-K] T_fuel=298 [K] T_air=298 [K] "The combustion equation of CH4 with stoichiometric amount of air is CH4 + (1+Ex)(2)(O2 + 3.76N2)=CO2 +2H2O+(1+Ex)(2)(3.76)N2" "For the incomplete combustion process in this problem, the combustion equation is CH4 + (1+Ex)(2)(O2 + 3.76N2)=aCO2 +bCO + cO2+2H2O+(1+Ex)(2)(3.76)N2" "Specie balance equations" "O" 4=a *2+b +c *2+2 "C" 1=a +b N_tot =a +b +c +2+(1+Ex)*(2)*3.76 "Total kilomoles of products at equilibrium" "We assume the equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component as a function of its temperature at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 16-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 16-15." "K_ P = (P/N_tot)^(1+0.5-1)*(b^1*c^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(c )=K_P *a "Conservation of energy for the reaction, assuming SSSF, neglecting work , ke, and pe:" E_in - E_out = DELTAE_cv E_in = Q_net + HR "The enthalpy of the reactant gases is" HR=enthalpy(CH4,T=T_fuel)+ (1+Ex)*(2) *enthalpy(O2,T=T_air)+(1+Ex)*(2)*3.76 *enthalpy(N2,T=T_air) E_out = HP "The enthalpy of the product gases is" HP=a *enthalpy(CO2,T=T_prod )+b *enthalpy(CO,T=T_prod ) +2*enthalpy(H2O,T=T_prod )+(1+Ex)*(2)*3.76*enthalpy(N2,T=T_prod ) + c *enthalpy(O2,T=T_prod ) DELTAE_cv = 0 "Steady-flow requirement" Q_net=0 "For an adiabatic reaction the net heat added is zero."
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-59
PercentEx 0 20 40 60 80 100 120 140 160 180 200
Tprod [K] 2260 2091 1940 1809 1695 1597 1511 1437 1370 1312 1259
2400 2200 2000
] K [
1800
d or p
1600
T
1400 1200 0
40
80
120
160
200
Percent Excess Air [%] Coefficients for CO2, CO, and O2 vs Tprod 1.10 0.90
c, b, a : st n ei ci ff e o C
a CO2 b CO c O2
0.70 0.50 0.30 0.10 -0.10 1200
1400
1600
1800
2000
2200
2400
2600
Tprod, K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-60
16-83 A mixture of H2 and O2 in a tank is ignited. The equilibrium composition of the product gases and the amount of heat transfer from the combustion chamber are to be determined. Assumptions 1 The equilibrium composition consists of H2O, H2, and O2. 2 The constituents of the mixture are ideal gases. Analysis (a) The combustion equation can be written as H 2 + 0.5 O 2
⎯ ⎯→
x H 2 O + (1 − x )H 2 + (0.5 − 0.5x )O 2
H2O, H2, O2 2800 K 5 atm
The equilibrium equation among H2O, H2, and O2 can be expressed as H 2 O ⇔ H 2 + 12 O 2 (thus ν H 2 O = 1, ν H 2 = 1, and ν O 2 = 12 ) N total = x + (1 − x ) + ( 0.5 − 0.5x ) = 15 . − 0.5x
Total number of moles:
The equilibrium constant relation can be expressed as ν
ν
N HH 2 N OO 2 ⎛ P 2 2 ⎜ Kp = ⎜N ν H 2O ⎝ total NH O 2
⎞ ⎟ ⎟ ⎠
(ν H 2 +ν O 2 −ν H 2 O )
From Table A-28, lnKP = -3.812 at 2800 K. Thus KP = 0.02210. Substituting, 1+ 0.5−1
(1 − x )(0.5 − 0.5 x) 0.5 ⎛ 5 ⎞ ⎜ ⎟ x ⎝ 1.5 − 0.5 x ⎠ Solving for x, x = 0.944 Then the combustion equation and the equilibrium composition can be expressed as 0.0221 =
H 2 + 0.5O 2
and
⎯ ⎯→
0.944 H 2 O + 0.056H 2 + 0.028O 2
0.944H 2 O + 0.056H 2 + 0.028O 2
(b) The heat transfer can be determined from − Qout =
∑ N (h P
o f
+ h − h o − Pv
) − ∑ N (h R
P
o f
+ h − h o − Pv
)
R
Since W = 0 and both the reactants and the products are assumed to be ideal gases, all the internal energy and enthalpies depend on temperature only, and the Pv terms in this equation can be replaced by RuT. It yields − Qout =
∑ N (h P
o f
+ h2800 K − h298 K − Ru T
) − ∑ N (h P
R
o f
− Ru T
)
R
since reactants are at the standard reference temperature of 25°C. From the tables,
Substance H2 O2 H2O
h fo kJ/kmol 0 0 -241,820
h 298 K
h 2800 K
kJ/kmol 8468 8682 9904
kJ/kmol 89,838 98,826 125,198
Substituting, −Qout = 0.944(−241,820 + 125,198 − 9904 − 8.314 × 2800) + 0.056(0 + 89,838 − 8468 − 8.314 × 2800) + 0.028(0 + 98,826 − 8682 − 8.314 × 2800) − 1(0 − 8.314 × 298) − 0.5(0 − 8.314 × 298) = −132,574 kJ/kmol H 2
or
Qout = 132,574 kJ/mol H 2
The equilibrium constant for the reaction H 2 O ⇔ OH + 12 H 2 is ln KP = -3.763, which is very close to the KP value of the reaction considered. Therefore, it is not realistic to assume that no OH will be present in equilibrium mixture.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-61
16-84 A mixture of H2O and O2 is heated to a high temperature. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of H2O, OH, O2, and H2. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→
2H 2 O + 3O 2
H2O, OH, H2, O2 3600 K 8 atm
x H 2 O + y H 2 + z O 2 + w OH
Mass balances for hydrogen and oxygen yield H balance:
4 = 2x + 2 y + w
(1)
O balance:
8 = x + 2z + w
(2)
The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the KP relations) to determine the equilibrium composition of the mixture. They are H 2 O ⇔ H 2 + 21 O 2
(reaction 1)
H 2 O ⇔ 21 H 2 + OH
(reaction 2)
The equilibrium constant for these two reactions at 3600 K are determined from Table A-28 to be ln K P1 = −1392 .
⎯ ⎯→
K P1 = 0.24858
ln K P 2 = −1088 .
⎯ ⎯→
K P 2 = 0.33689
The KP relations for these two simultaneous reactions are ν
K P1
ν
N HH 2 N OO 2 ⎛ P 2 2 ⎜ = ⎜N ν H 2O ⎝ total NH O 2
ν H2
K P2
ν
OH N H N OH ⎛ P 2 ⎜ = ⎜N ν H 2O ⎝ total NH O 2
⎞ ⎟ ⎟ ⎠
(ν H 2 +ν O 2 −ν H 2 O )
⎞ ⎟ ⎟ ⎠
(ν H 2 +ν OH −ν H 2 O )
where N total = N H 2 O + N H 2 + N O 2 + N OH = x + y + z + w
Substituting, 1/ 2
( y )( z )1 / 2 0.24858 = x
⎛ ⎞ 8 ⎜⎜ ⎟⎟ x + y + z + w ⎝ ⎠
( w)( y )1 / 2 x
⎛ ⎞ 8 ⎜⎜ ⎟⎟ ⎝ x + y + z + w⎠
0.33689 =
(3) 1/ 2
(4)
Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 1.371
y = 0.1646
z = 2.85
w = 0.928
Therefore, the equilibrium composition becomes 1.371H 2O + 0.165H 2 + 2.85O 2 + 0.928OH
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-62
16-85 A mixture of CO2 and O2 is heated to a high temperature. The equilibrium composition is to be determined. Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and O. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as ⎯ ⎯→
3C 2O + 3O 2
x CO 2 + y CO + z O 2 + w O
CO2, CO, O2, O 3400 K 2 atm
Mass balances for carbon and oxygen yield C balance:
3= x+ y
(1)
O balance:
12 = 2 x + y + 2 z + w
(2)
The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the KP relations) to determine the equilibrium composition of the mixture. They are CO 2 ⇔ CO + 12 O 2
(reaction 1)
O 2 ⇔ 2O
(reaction 2)
The equilibrium constant for these two reactions at 3400 K are determined from Table A-28 to be ln K P1 = 0169 .
⎯ ⎯→
K P1 = 11841 .
ln K P 2 = −1935 .
⎯ ⎯→
K P 2 = 01444 .
The KP relations for these two simultaneous reactions are ν
ν
K P1
CO N CO N OO 2 ⎛ P 2 ⎜ = ⎜N ν CO 2 ⎝ total N CO 2
K P2 =
νO
NO ⎛ P ⎜ ⎜ ν N OO 2 ⎝ N total 2
⎞ ⎟ ⎟ ⎠
(ν CO +ν O 2 −ν CO 2 )
ν O −ν O 2
⎞ ⎟ ⎟ ⎠
where N total = N CO 2 + N O 2 + N CO + N O = x + y + z + w
Substituting, 1.1841 =
( y )( z )1 / 2 x
w2 0.1444 = z
⎞ ⎛ 2 ⎟⎟ ⎜⎜ ⎝ x+ y+ z +w⎠
⎞ ⎛ 2 ⎟⎟ ⎜⎜ ⎝ x + y + z + w⎠
1/ 2
(3)
2 −1
(4)
Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 1.313
y = 1.687
z = 3.187
w = 1.314
Thus the equilibrium composition is 1.313CO 2 + 1.687CO + 3.187O 2 + 1.314O
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-63
16-86 EES Problem 16-85 is reconsidered. The effect of pressure on the equilibrium composition by varying pressure from 1 atm to 10 atm is to be studied. Analysis The problem is solved using EES, and the solution is given below. "For EES to calculate a, b, c, and d at T_prod and P_prod press F2 or click on the Calculator icon. The EES results using the built in function data is not the same as the anwers provided with the problem. However, if we supply the K_P's from Table A-28 to ESS, the results are equal to the answer provided. The plot of moles CO vs. P_atm was done with the EES property data." "Input Data" P_atm = 2 [atm] P_prod =P_atm*101.3 R_u=8.314 [kJ/kmol-K] T_prod=3400 [K] P=P_atm "For the incomplete combustion process in this problem, the combustion equation is 3 CO2 + 3 O2=aCO2 +bCO + cO2+dO" "Specie balance equations" "O" 3*2+3*2=a *2+b +c *2+d*1 "C" 3*1=a*1 +b*1 N_tot =a +b +c +d "Total kilomoles of products at equilibrium" "We assume the equilibrium reactions are CO2=CO+0.5O2 O2=2O" "The following equations provide the specific Gibbs function (g=h-Ts) for each component as a function of its temperature at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "EES does not have a built-in property function for monatomic oxygen so we will use the JANAF procedure, found under Options/Function Info/External Procedures. The units for the JANAF procedure are kmol, K, and kJ. The values are calculated for 1 atm. The entropy must be corrected for other pressrues." Call JANAF('O',T_prod:Cp,h_O,s_O) "Units from JANAF are SI" "The entropy from JANAF is for one atmosphere and that's what we need for this approach." g_O=h_O-T_prod*s_O "The standard-state (at 1 atm) Gibbs functions are" DELTAG_1 =1*g_CO+0.5*g_O2-1*g_CO2 DELTAG_2 =2*g_O-1*g_O2 "The equilibrium constants are given by Eq. 15-14." {K_P_2=0.1444 "From Table A-28" K_P_1 = 0.8445}"From Table A-28" K_p_1 = exp(-DELTAG_1/(R_u*T_prod)) "From EES data" K_P_2 = exp(-DELTAG_2/(R_u*T_prod)) "From EES data" "The equilibrium constant is also given by Eq. 15-15." PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-64
"Write the equilibrium constant for the following system of equations: 3 CO2 + 3 O2=aCO2 +bCO + cO2+dO CO2=CO+0.5O2 O2=2O" "K_ P_1 = (P/N_tot)^(1+0.5-1)*(b^1*c^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(c )/a=K_P_1 "K_ P_2 = (P/N_tot)^(2-1)*(d^2)/(c^1)" P/N_tot *d^2/c =K_P_2 b [kmolCO] 1.968 1.687 1.52 1.404 1.315 1.244 1.186 1.136 1.093 1.055
Patm [atm] 1 2 3 4 5 6 7 8 9 10
2
1.8
] O lC
o m k[ b
1.6
1.4
1.2
1 1
2
3
4
5
6
7
8
9
10
Patm [atm]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-65
16-87 The hR at a specified temperature is to be determined using enthalpy and Kp data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The complete combustion equation of H2 can be expressed as H 2 + 12 O 2 ⇔ H 2 O
The hR of the combustion process of H 2 at 2400 K is the amount of energy released as one kmol of H2 is burned in a steady-flow combustion chamber at a temperature of 2400 K, and can be determined from hR =
∑ N (h P
o f
+h −ho
) − ∑ N (h P
R
o f
+h −ho
)
R
Assuming the H2O, H2, and O2 to be ideal gases, we have h = h (T). From the tables,
Substance H2O H2 O2
h fo kJ/kmol -241,820 0 0
h 298 K
h 2400 K
kJ/kmol 9904 8468 8682
kJ/kmol 103,508 75,383 83,174
Substituting, hR = 1( −241,820 + 103,508 − 9904) − 1(0 + 75,383 − 8468) − 0.5(0 + 83,174 − 8682) = −252,377 kJ / kmol
(b) The hR value at 2400 K can be estimated by using KP values at 2200 K and 2600 K (the closest two temperatures to 2400 K for which KP data are available) from Table A-28, ln
K P 2 hR ⎛ 1 h ⎛1 1 ⎞ 1 ⎞ ⎟ ⎟⎟ or ln K P 2 − ln K P1 ≅ R ⎜⎜ − ⎜⎜ − ≅ K P1 Ru ⎝ T1 T2 ⎠ Ru ⎝ T1 T2 ⎟⎠
4.648 − 6.768 ≅
hR 1 ⎞ ⎛ 1 − ⎜ ⎟ 8.314 kJ/kmol ⋅ K ⎝ 2200 K 2600 K ⎠
h R ≅ -252,047 kJ/kmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-66
16-88 EES Problem 16-87 is reconsidered. The effect of temperature on the enthalpy of reaction using both methods by varying the temperature from 2000 to 3000 K is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" T_prod=2400 [K] DELTAT_prod =25 [K] R_u=8.314 [kJ/kmol-K] T_prod_1 = T_prod - DELTAT_prod T_prod_2 = T_prod + DELTAT_prod "The combustion equation is 1 H2 + 0.5 O2 =>1 H2O" "The enthalpy of reaction H_bar_R using enthalpy data is:" h_bar_R_Enthalpy = HP - HR HP = 1*Enthalpy(H2O,T=T_prod ) HR = 1*Enthalpy(H2,T=T_prod ) + 0.5*Enthalpy(O2,T=T_prod ) "The enthalpy of reaction H_bar_R using enthalpy data is found using the following equilibruim data:" "The following equations provide the specific Gibbs function (g=h-Ts) for each component as a function of its temperature at 1 atm pressure, 101.3 kPa" g_H2O_1=Enthalpy(H2O,T=T_prod_1 )-T_prod_1 *Entropy(H2O,T=T_prod_1 ,P=101.3) g_H2_1=Enthalpy(H2,T=T_prod_1 )-T_prod_1 *Entropy(H2,T=T_prod_1 ,P=101.3) g_O2_1=Enthalpy(O2,T=T_prod_1 )-T_prod_1 *Entropy(O2,T=T_prod_1 ,P=101.3) g_H2O_2=Enthalpy(H2O,T=T_prod_2 )-T_prod_2 *Entropy(H2O,T=T_prod_2 ,P=101.3) g_H2_2=Enthalpy(H2,T=T_prod_2 )-T_prod_2 *Entropy(H2,T=T_prod_2 ,P=101.3) g_O2_2=Enthalpy(O2,T=T_prod_2 )-T_prod_2 *Entropy(O2,T=T_prod_2 ,P=101.3) "The standard-state (at 1 atm) Gibbs functions are" DELTAG_1 =1*g_H2O_1-0.5*g_O2_1-1*g_H2_1 DELTAG_2 =1*g_H2O_2-0.5*g_O2_2-1*g_H2_2 "The equilibrium constants are given by Eq. 15-14." K_p_1 = exp(-DELTAG_1/(R_u*T_prod_1)) "From EES data" K_P_2 = exp(-DELTAG_2/(R_u*T_prod_2)) "From EES data" "the entahlpy of reaction is estimated from the equilibrium constant K_p by using EQ 15-18 as:" ln(K_P_2/K_P_1)=h_bar_R_Kp/R_u*(1/T_prod_1 - 1/T_prod_2) PercentError = ABS((h_bar_R_enthalpy - h_bar_R_Kp)/h_bar_R_enthalpy)*Convert(, %)
Percent Error [%] 0.0002739 0.0002333 0.000198 0.0001673 0.0001405 0.0001173 0.00009706 0.00007957 0.00006448 0.00005154 0.0000405
Tprod [K] 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000
hREnthalpy [kJ/kmol] -251723 -251920 -252096 -252254 -252398 -252532 -252657 -252778 -252897 -253017 -253142
hRKp [kJ/kmol] -251722 -251919 -252095 -252254 -252398 -252531 -252657 -252777 -252896 -253017 -253142
-251500
DELTATprod = 25 K -251850
Enthalpy Data Kp Data
]l o -252200 m k/ J -252550 k[ R
h
-252900 -253250 2000
2200
2400
2600
2800
Tprod [k]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
3000
16-67
16-89 The KP value of the dissociation process O2 ⇔ 2O at a specified temperature is to be determined using the hR data and KP value at a specified temperature. Assumptions Both the reactants and products are ideal gases. Analysis The hR and KP data are related to each other by ln
h ⎛1 K P 2 hR ⎛ 1 1 ⎞ 1 ⎞ ⎜⎜ − ⎟⎟ or ln K P 2 − ln K P1 ≅ R ⎜⎜ − ⎟⎟ ≅ Ru ⎝ T1 T2 ⎠ K P1 Ru ⎝ T1 T2 ⎠
The hR of the specified reaction at 2800 K is the amount of energy released as one kmol of O2 dissociates in a steady-flow combustion chamber at a temperature of 2800 K, and can be determined from hR =
∑ N (h P
o f
+h −ho
) − ∑ N (h P
R
o f
+h −ho
)
R
Assuming the O2 and O to be ideal gases, we have h = h (T). From the tables, h fo kJ/kmol 249,190 0
Substance O O2
h 298 K
h 2800 K
kJ/kmol 6852 8682
kJ/kmol 59,241 98,826
Substituting, hR = 2(249,190 + 59,241 − 6852) − 1(0 + 98,826 − 8682) = 513,014 kJ / kmol
The KP value at 3000 K can be estimated from the equation above by using this hR value and the KP value at 2600 K which is ln KP1 = -7.521, ln K P 2 − (−7.521) = ln K P 2 = −4.357
or
513,014 kJ/kmol ⎛ 1 1 ⎞ − ⎜ ⎟ 8.314 kJ/kmol ⋅ K ⎝ 2600 K 3000 K ⎠
(Table A - 28 : ln K P 2 = −4.357)
K P2 = 0.0128
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-68
16-90 It is to be shown that when the three phases of a pure substance are in equilibrium, the specific Gibbs function of each phase is the same. Analysis The total Gibbs function of the three phase mixture of a pure substance can be expressed as G = m s g s + ml g l + m g g g
where the subscripts s, l, and g indicate solid, liquid and gaseous phases. Differentiating by holding the temperature and pressure (thus the Gibbs functions, g) constant yields dG = g s dm s + g l dm l + g g dm g
From conservation of mass, dms + dml + dmg = 0
mg ⎯ ⎯→
dms = −dml − dmg
ml
Substituting, dG = − g s (dml + dmg ) + g l dml + g g dmg
ms
Rearranging, dG = ( g l − g s )dml + ( g g − g s )dmg
For equilibrium, dG = 0. Also dml and dmg can be varied independently. Thus each term on the right hand side must be zero to satisfy the equilibrium criteria. It yields g l = g s and g g = g s
Combining these two conditions gives the desired result, gl = g s = g s
16-91 It is to be shown that when the two phases of a two-component system are in equilibrium, the specific Gibbs function of each phase of each component is the same. Analysis The total Gibbs function of the two phase mixture can be expressed as G = (ml1 g l1 + mg1 g g1 ) + (ml2 g l 2 + mg 2 g g 2 )
where the subscripts l and g indicate liquid and gaseous phases. Differentiating by holding the temperature and pressure (thus the Gibbs functions) constant yields dG = g l1dml1 + g g1dmg1 + g l2 dml2 + g g 2 dmg 2
From conservation of mass, dmg1 = − dml1 and dmg 2 = − dml 2
Substituting, dG = ( g l1 − g g1 )dml1 + ( g l2 − g g 2 )dml 2
mg1 mg2 ml 1 ml 2
For equilibrium, dG = 0. Also dml1 and dml2 can be varied independently. Thus each term on the right hand side must be zero to satisfy the equilibrium criteria. Then we have g l1 = g g1 and g l2 = g g 2
which is the desired result.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-69
16-92 A mixture of CO and O2 contained in a tank is ignited. The final pressure in the tank and the amount of heat transfer are to be determined. Assumptions 1 The equilibrium composition consists of CO2 and O2. 2 Both the reactants and the products are ideal gases. Analysis The combustion equation can be written as CO + 3 O 2
⎯ ⎯→
The heat transfer can be determined from − Qout =
∑ N (h P
o f
CO2, CO, O2
CO 2 + 2.5 O 2 + h − h o − Pv
) − ∑ N (h R
P
o f
+ h − h o − Pv
25°C 2 atm
)
R
Both the reactants and the products are assumed to be ideal gases, and thus all the internal energy and enthalpies depend on temperature only, and the Pv terms in this equation can be replaced by RuT. It yields − Qout =
∑ N (h P
o f
+ h500 K − h298 K − Ru T
) − ∑ N (h P
R
o f
− Ru T
)
R
since reactants are at the standard reference temperature of 25°C. From the tables, h fo kJ/kmol -110,530 0 -393,520
Substance CO O2 CO2
h 298 K
h 500 K
kJ/kmol 8669 8682 9364
kJ/kmol 14,600 14,770 17,678
Substituting, −Qout = 1(−393,520 + 17,678 − 9364 − 8.314 × 500) + 2.5(0 + 14,770 − 8682 − 8.314 × 500) − 3(0 − 8.314 × 298) − 1(−110,530 − 8.314 × 298) = −264,095 kJ/kmol CO
or
Qout = 264,095 kJ/kmol CO
The final pressure in the tank is determined from N R T N T P1V 3.5 500 K = 1 u 1 ⎯ ⎯→ P2 = 2 2 P1 = × (2 atm) = 2.94 atm N 1T1 4 298 K P2V N 2 Ru T2
The equilibrium constant for the reaction CO + 21 O 2 ⇔ CO 2 is ln KP = 57.62, which is much greater than 7. Therefore, it is not realistic to assume that no CO will be present in equilibrium mixture.
16-93 Using Henry’s law, it is to be shown that the dissolved gases in a liquid can be driven off by heating the liquid. Analysis Henry’s law is expressed as yi, liquid side (0) =
Pi, gas side (0) H
Henry’s constant H increases with temperature, and thus the fraction of gas i in the liquid yi,liquid decreases. Therefore, heating a liquid will drive off the dissolved gases in a liquid.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
side
16-70
16-94 A glass of water is left in a room. The mole fraction of the water vapor in the air at the water surface and far from the surface as well as the mole fraction of air in the water near the surface are to be determined when the water and the air are at the same temperature. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 25°C is 3.170 kPa (Table A-4). Henry’s constant for air dissolved in water at 25ºC (298 K) is given in Table 16-2 to be H = 71,600 bar. Molar masses of dry air and water are 29 and 18 kg/kmol, respectively (Table A-1). Analysis (a) Noting that the relative humidity of air is 70%, the partial pressure of water vapor in the air far from the water surface will be Pv , room air = φ Psat @ 25°C = (0.7)(3.170 kPa ) = 2.219 kPa
Air-water interface
Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the room air is y vapor =
Pvapor
=
P
2.219 kPa = 0.0222 100 kPa
(or 2.22%)
Water 25ºC
(b) Noting that air at the water surface is saturated, the partial pressure of water vapor in the air near the surface will simply be the saturation pressure of water at 25°C, Pv ,interface = Psat @ 25°C = 3.170 kPa . Then the mole fraction of water vapor in the air at the interface becomes y v , surface =
Pv , surface P
=
3.170 kPa = 0.0317 100 kPa
(or 3.17%)
(c) Noting that the total pressure is 100 kPa, the partial pressure of dry air at the water surface is Pair, surface = P − Pv , surface = 100 − 3.170 = 96.83 kPa
From Henry’s law, the mole fraction of air in the water is determined to be y dry air,liquid side =
Pdry air,gas side H
=
(96.83 / 100) bar = 1.35 ×10 − 5 71,600 bar
Discussion The water cannot remain at the room temperature when the air is not saturated. Therefore, some water will evaporate and the water temperature will drop until a balance is reached between the rate of heat transfer to the water and the rate of evaporation.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-71
16-95 A glass of water is left in a room. The mole fraction of the water vapor in the air at the water surface and far from the surface as well as the mole fraction of air in the water near the surface are to be determined when the water and the air are at the same temperature. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 25°C is 3.170 kPa (Table A-4). Henry’s constant for air dissolved in water at 25ºC (298 K) is given in Table 16-2 to be H = 71,600 bar. Molar masses of dry air and water are 29 and 18 kg/kmol, respectively (Table A-1). Analysis (a) Noting that the relative humidity of air is 40%, the partial pressure of water vapor in the air far from the water surface will be Pv , room air = φ Psat @ 25°C = (0.25)(3.170 kPa ) = 0.7925 kPa
Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the room air is y vapor =
Pvapor
=
P
0.7925 kPa = 0.0079 100 kPa
(or 0.79%)
(b) Noting that air at the water surface is saturated, the partial pressure of water vapor in the air near the surface will simply be the saturation pressure of water at 25°C, Pv ,interface = Psat @ 25°C = 3.170 kPa . Then the mole fraction of water vapor in the air at the interface becomes y v , surface =
Pv , surface P
=
3.170 kPa = 0.0317 100 kPa
(or 3.17%)
(c) Noting that the total pressure is 100 kPa, the partial pressure of dry air at the water surface is Pair, surface = P − Pv , surface = 100 − 3.170 = 96.83 kPa
From Henry’s law, the mole fraction of air in the water is determined to be y dry air,liquid side =
Pdry air,gas side H
=
(96.83 / 100) bar = 1.35 ×10 − 5 71,600 bar
Discussion The water cannot remain at the room temperature when the air is not saturated. Therefore, some water will evaporate and the water temperature will drop until a balance is reached between the rate of heat transfer to the water and the rate of evaporation.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-72
16-96 A 2-L bottle is filled with carbonated drink that is fully charged (saturated) with CO2 gas. The volume that the CO2 gas would occupy if it is released and stored in a container at room conditions is to be determined. Assumptions 1 The liquid drink can be treated as water. 2 Both the CO2 gas and the water vapor are ideal gases. 3 The CO2 gas is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 17°C is 1.938 kPa (Table A-4). Henry’s constant for CO2 dissolved in water at 17ºC (290 K) is H = 1280 bar (Table 16-2). Molar masses of CO2 and water are 44.01 and 18.015 kg/kmol, respectively (Table A-1). The gas constant of CO2 is 0.1889 kPa.m3/kg.K. Also, 1 bar = 100 kPa. Analysis In the charging station, the CO2 gas and water vapor mixture above the liquid will form a saturated mixture. Noting that the saturation pressure of water at 17°C is 1.938 kPa, the partial pressure of the CO2 gas is PCO 2 , gas side = P − Pvapor = P − Psat @ 17°C = 600 − 1.938 = 598.06 kPa = 5.9806 bar
From Henry’s law, the mole fraction of CO2 in the liquid drink is determined to be y CO 2 ,liquid side =
PCO 2 ,gas side H
=
5.9806 bar = 0.00467 1280 bar
Then the mole fraction of water in the drink becomes y water, liquid side = 1 − yCO 2 , liquid side = 1 − 0.00467 = 0.99533
The mass and mole fractions of a mixture are related to each other by wi =
mi N M Mi = i i = yi mm N m M m Mm
where the apparent molar mass of the drink (liquid water - CO2 mixture) is Mm =
∑y M i
i
= yliquid water Mwater + yCO 2 MCO 2
. kg / kmol = 0.99533 × 18.015 + 0.00467 × 44.01 = 1814
Then the mass fraction of dissolved CO2 in liquid drink becomes wCO 2 , liquid side = yCO 2 , liquid side (0)
M CO 2 Mm
= 0.00467
44.01 = 0.0113 1814 .
Therefore, the mass of dissolved CO2 in a 2 L ≈ 2 kg drink is mCO 2 = wCO 2 mm = 0.0113(2 kg) = 0.0226 kg
Then the volume occupied by this CO2 at the room conditions of 20°C and 100 kPa becomes
V =
mRT (0.0226 kg)(0.1889 kPa ⋅ m 3 / kg ⋅ K )(293 K) = = 0.0125 m 3 = 12.5 L 100 kPa P
Discussion Note that the amount of dissolved CO2 in a 2-L pressurized drink is large enough to fill 6 such bottles at room temperature and pressure. Also, we could simplify the calculations by assuming the molar mass of carbonated drink to be the same as that of water, and take it to be 18 kg/kmol because of the very low mole fraction of CO2 in the drink.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-73
16-97 EES Ethyl alcohol C2H5OH (gas) is burned in a steady-flow adiabatic combustion chamber with 40 percent excess air. The adiabatic flame temperature of the products is to be determined and the adiabatic flame temperature as a function of the percent excess air is to be plotted. Analysis The complete combustion reaction in this case can be written as C 2 H 5 OH (gas) + (1 + Ex)a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 2 CO 2 + 3 H 2 O + ( Ex)(a th ) O 2 + f N 2
where ath is the stoichiometric coefficient for air. The oxygen balance gives 1 + (1 + Ex)a th × 2 = 2 × 2 + 3 × 1 + ( Ex)(a th ) × 2
The reaction equation with products in equilibrium is C 2 H 5 OH (gas) + (1 + Ex)a th [O 2 + 3.76N 2 ] ⎯ ⎯→ a CO 2 + b CO + d H 2 O + e O 2 + f N 2
The coefficients are determined from the mass balances Carbon balance:
2 = a+b
Hydrogen balance:
6 = 2d ⎯ ⎯→ d = 3
Oxygen balance:
1 + (1 + Ex)a th × 2 = a × 2 + b + d + e × 2
Nitrogen balance: (1 + Ex)a th × 3.76 = f Solving the above equations, we find the coefficients to be Ex = 0.4, ath = 3, a = 1.995, b = 0.004938, d = 3, e = 1.202, f = 15.79 Then, we write the balanced reaction equation as C 2 H 5 OH (gas) + 4.2[O 2 + 3.76N 2 ] ⎯ ⎯→ 1.995 CO 2 + 0.004938 CO + 3 H 2 O + 1.202 O 2 + 15.79 N 2
Total moles of products at equilibrium are N tot = 1.995 + 0.004938 + 3 + 1.202 + 15.79 = 21.99
The assumed equilibrium reaction is CO 2 ←⎯→ CO + 0.5O 2
The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G*( T )/ Ru T or ln K p = − ∆G *(T ) / Ru T
where ∗ ∗ ∗ ∆G * (T ) = ν CO g CO (Tprod ) + ν O2 g O2 (Tprod ) −ν CO2 g CO2 (Tprod )
and the Gibbs functions are defined as ∗ g CO (Tprod ) = (h − Tprod s ) CO ∗ g O2 (Tprod ) = (h − Tprod s ) O2 ∗ g CO2 (Tprod ) = (h − Tprod s ) CO2
The equilibrium constant is also given by Kp =
be 0.5 a
⎛ P ⎜ ⎜N ⎝ tot
⎞ ⎟ ⎟ ⎠
1+ 0.5 −1
=
(0.004938)(1.202) 0.5 1.995
⎛ 1 ⎞ ⎜ ⎟ ⎝ 21.99 ⎠
0.5
= 0.0005787
A steady flow energy balance gives HR = HP
where
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-74
H R = h fo fuel@25°C + 4.2hO2@25°C + 15.79h N2@25°C = (−235,310 kJ/kmol) + 4.2(0) + 15.79(0) = −235,310 kJ/kmol H P = 1.995hCO2@Tprod + 0.004938hCO@Tprod + 3hH2O@Tprod + 1.202hO2@Tprod + 15.79h N2@Tprod
Solving the energy balance equation using EES, we obtain the adiabatic flame temperature to be Tprod = 1907 K
The copy of entire EES solution including parametric studies is given next: "The product temperature isT_prod" "The reactant temperature is:" T_reac= 25+273.15 "[K]" "For adiabatic combustion of 1 kmol of fuel: " Q_out = 0 "[kJ]" PercentEx = 40 "Percent excess air" Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The complete combustion reaction equation for excess air is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=2 CO2 + 3 H2O +Ex*A_th O2 + f N2" "Oxygen Balance for complete combustion:" 1 + (1+Ex)*A_th*2=2*2+3*1 + Ex*A_th*2 "The reaction equation for excess air and products in equilibrium is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=a CO2 + b CO+ d H2O + e O2 + f N2" "Carbon Balance:" 2=a + b "Hydrogen Balance:" 6=2*d "Oxygen Balance:" 1 + (1+Ex)*A_th*2=a*2+b + d + e*2 "Nitrogen Balance:" (1+Ex)*A_th*3.76 = f N_tot =a +b + d + e + f "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(b^1*e^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(e )=K_P *a "The steady-flow energy balance is:" H_R = Q_out+H_P h_bar_f_C2H5OHgas=-235310 "[kJ/kmol]"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-75
H_R=1*(h_bar_f_C2H5OHgas ) +(1+Ex)*A_th*ENTHALPY(O2,T=T_reac)+(1+Ex)*A_th*3.76*ENTHALPY(N2,T=T_reac) "[kJ/kmol]" H_P=a*ENTHALPY(CO2,T=T_prod)+b*ENTHALPY(CO,T=T_prod)+d*ENTHALPY(H2O,T=T_prod) +e*ENTHALPY(O2,T=T_prod)+f*ENTHALPY(N2,T=T_prod) "[kJ/kmol]"
a
ath
b
d
e
f
1.922 1.97 1.988 1.995 1.998 1.999 2 2 2 2
3 3 3 3 3 3 3 3 3 3
0.07809 0.03017 0.01201 0.004933 0.002089 0.0009089 0.000405 0.0001843 0.0000855 0.00004036
3 3 3 3 3 3 3 3 3 3
0.339 0.6151 0.906 1.202 1.501 1.8 2.1 2.4 2.7 3
12.41 13.54 14.66 15.79 16.92 18.05 19.18 20.3 21.43 22.56
PercentEx [%] 10 20 30 40 50 60 70 80 90 100
Tprod [K] 2191 2093 1996 1907 1826 1752 1685 1625 1569 1518
2200 2100 2000
) K ( d or p
1900 1800
T 1700 1600 1500 10
20
30
40
50
60
70
80
90
100
PercentEx
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-76
16-98 EES The natural log of the equilibrium constant as a function of temperature between 298 to 3000 K for the equilibrium reaction CO + H2O = CO2 + H2 is to be tabulated and compared to those given in Table A-228 Analysis The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using K p = e −∆G*( T )/ Ru T or ln K p = − ∆G *(T ) / Ru T
where ∗ ∗ ∗ ∗ ∆G * (T ) = ν CO2 g CO2 (T ) + ν H2 g H2 (T ) −ν CO g CO (T ) −ν H2O g H2O (T )
and the Gibbs functions are defined as ∗ g CO (Tprod ) = (h − Tprod s ) CO ∗ g H2O (Tprod ) = (h − Tprod s ) H2O ∗ g CO2 (Tprod ) = (h − Tprod s ) CO2 ∗ g H2 (Tprod ) = (h − Tprod s ) H2
The copy of entire EES solution with resulting parametric table is given next: {T_prod = 298 "[K]"} R_u=8.314"[kJ/kmol-K]" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, Tprod, at 1 atm pressure, 101.3 kPa" "For T_prod:" g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_H2=Enthalpy(H2,T=T_prod )-T_prod *Entropy(H2,T=T_prod ,P=101.3) g_H2O=Enthalpy(H2O,T=T_prod )-T_prod *Entropy(H2O,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO2+1*g_H2-1*g_CO-1*g_H2O "The equilibrium constant is given by:" K_p = exp(-DELTAG /(R_u*T_prod )) lnK_p=ln(k_p) Tprod [K] 298 500 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000
ln Kp 11,58 4,939 0,3725 -0,3084 -0,767 -1,092 -1,33 -1,51 -1,649 -1,759 -1,847 -1,918 -1,976
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-77
16-99 EES The percent theoretical air required for the combustion of octane such that the volume fraction of CO in the products is less than 0.1% and the heat transfer are to be determined. Also, the percent theoretical air required for 0.1% CO in the products as a function of product pressure is to be plotted. Analysis The complete combustion reaction equation for excess air is C 8 H18 + Pth a th [O 2 + 3.76N 2 ] ⎯ ⎯→ 8 CO 2 + 9 H 2 O + ( Pth − 1)a th O 2 + f N 2
The oxygen balance is Pth a th × 2 = 8 × 2 + 9 × 1 + ( Pth − 1)a th × 2
The reaction equation for excess air and products in equilibrium is C 8 H18 + Pth a th [O 2 + 3.76N 2 ] ⎯ ⎯→ a CO 2 + b CO + d H 2 O + e O 2 + f N 2
The coefficients are to be determined from the mass balances Carbon balance:
8=a+b
Hydrogen balance:
18 = 2d ⎯ ⎯→ d = 9
Oxygen balance:
Pth a th × 2 = a × 2 + b + d + e × 2
Nitrogen balance:
Pth a th × 3.76 = f
Volume fraction of CO must be less than 0.1%. That is, y CO =
b b = = 0.001 N tot a + b + d + e + f
The assumed equilibrium reaction is CO 2 ←⎯→ CO + 0.5O 2
The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data: ∗ g CO (Tprod ) = (h − Tprod s ) CO = (−53,826) − (2000)(258.48) = −570,781 kJ/kmol ∗ g O2 (Tprod ) = (h − Tprod s ) O2 = (59,193) − (2000)(268.53) = −477,876 kJ/kmol ∗ g CO2 (Tprod ) = (h − Tprod s ) CO2 = (−302,128) − (2000)(309.00) = −920,121 kJ/kmol
The enthalpies at 2000 K and entropies at 2000 K and 101.3 kPa are obtained from EES. Substituting, ∗ ∗ ∗ ∆G * (Tprod ) = ν CO g CO (Tprod ) + ν O2 g O2 (Tprod ) −ν CO2 g CO2 (Tprod )
= 1(−570,781) + 0.5(−477,876) − (−920,121) = 110,402 kJ/kmol ⎛ − ∆G * (Tprod ) ⎞ ⎟ = exp⎛⎜ − 110,402 ⎞⎟ = 0.001308 K p = exp⎜ ⎜ (8.314)(2000) ⎟ ⎜ Ru Tprod ⎟ ⎝ ⎠ ⎝ ⎠
The equilibrium constant is also given by Kp =
be 0.5 a
⎛ P ⎜ ⎜N ⎝ tot
⎞ ⎟ ⎟ ⎠
1+ 0.5 −1
=
be 0.5 a
⎛ Pprod / 101.3 ⎜ ⎜ a+b+d +e+ f ⎝
⎞ ⎟ ⎟ ⎠
1+ 0.5 −1
The steady flow energy balance gives H R = Qout + H P
where
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-78
H R = 1hC8H18 @ 298 K + Pth a th hO2 @ 298 K + ( Pth a th × 3.76)h N2 @ 298 K = (−208,459) + Pth a th (0) + ( Pth a th × 3.76)(0) = −208,459 kJ/kmol H P = ahCO2 @ 2000 K + bhCO @ 2000 K + dhH2O @ 2000 K + ehO2 @ 2000 K + f h N2 @ 2000 K = a(−302,128) + b(−53,826) + d (−169,171) + e(59,193) + f (56,115)
The enthalpies are obtained from EES. Solving all the equations simultaneously using EES, we obtain Pth = 1.024, a th = 12.5, a = 7.935, b = 0.06544, d = 9, e = 0.3289, f = 48.11 PercentTh = Pth × 100 = 1.024 × 100 = 102.4% Qout = 995,500 kJ/kmol C 8 H18
The copy of entire EES solution including parametric studies is given next: "The product temperature is:" T_prod = 2000 "[K]" "The reactant temperature is:" T_reac= 25+273 "[K]" "PercentTH is Percent theoretical air" Pth= PercentTh/100 "Pth = % theoretical air/100" P_prod = 5 "[atm]" *convert(atm,kPa)"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The complete combustion reaction equation for excess air is:" "C8H18+ Pth*A_th (O2 +3.76N2)=8 CO2 + 9 H2O +(Pth-1)*A_th O2 + f N2" "Oxygen Balance for complete combustion:" Pth*A_th*2=8*2+9*1 + (Pth-1)*A_th*2 "The reaction equation for excess air and products in equilibrium is:" "C8H18+ Pth*A_th (O2 +3.76N2)=a CO2 + b CO+ d H2O + e O2 + f N2" "Carbon Balance:" 8=a + b "Hydrogen Balance:" 18=2*d "Oxygen Balance:" Pth*A_th*2=a*2+b + d + e*2 "Nitrogen Balance:" Pth*A_th*3.76 = f N_tot =a +b + d + e + f "Total kilomoles of products at equilibrium" "The volume faction of CO in the products is to be less than 0.1%. For ideal gas mixtures volume fractions equal mole fractions." "The mole fraction of CO in the product gases is:" y_CO = 0.001 y_CO = b/N_tot "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-79
"K_ P = (P/N_tot)^(1+0.5-1)*(b^1*e^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(e )=K_P *a "The steady-flow energy balance is:" H_R = Q_out+H_P H_R=1*ENTHALPY(C8H18,T=T_reac)+Pth*A_th*ENTHALPY(O2,T=T_reac)+Pth*A_th*3.76*EN THALPY(N2,T=T_reac) "[kJ/kmol]" H_P=a*ENTHALPY(CO2,T=T_prod)+b*ENTHALPY(CO,T=T_prod)+d*ENTHALPY(H2O,T=T_pro d) +e*ENTHALPY(O2,T=T_prod)+f*ENTHALPY(N2,T=T_prod) "[kJ/kmol]"
Pprod [kPa] 100 300 500 700 900 1100 1300 1500 1700 1900 2100 2300
PercentTh [%] 112 104.1 102.4 101.7 101.2 101 100.8 100.6 100.5 100.5 100.4 100.3
112 110 108
% h Tt n e cr e P
106 104 102 100 0
500
1000
1500
2000
2500
Pprod [kPa]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-80
Fundamentals of Engineering (FE) Exam Problems
16-100 If the equilibrium constant for the reaction H2 + ½O2 → H2O is K, the equilibrium constant for the reaction 2H2O → 2H2 + O2 at the same temperature is (a) 1/K (b) 1/(2K) (c) 2K (d) K2 (e) 1/K2 Answer (e) 1/K2
16-101 If the equilibrium constant for the reaction CO + ½O2 → CO2 is K, the equilibrium constant for the reaction CO2 + 3N2 → CO + ½O2 + 3N2 at the same temperature is (a) 1/K (b) 1/(K + 3) (c) 4K (d) K (e) 1/K2 Answer (a) 1/K
16-102 The equilibrium constant for the reaction H2 + ½O2 → H2O at 1 atm and 1500°C is given to be K. Of the reactions given below, all at 1500°C, the reaction that has a different equilibrium constant is (a) H2 + ½O2 → H2O at 5 atm, (b) 2H2 + O2 → 2H2O at 1 atm, (c) H2 + O2 → H2O+ ½O2 at 2 atm, (d) H2 + ½O2 + 3N2 → H2O+ 3N2 at 5 atm, (e) H2 + ½O2 + 3N2 → H2O+ 3N2 at 1 atm, Answer (b) 2H2 + O2 → 2H2O at 1 atm,
16-103 Of the reactions given below, the reaction whose equilibrium composition at a specified temperature is not affected by pressure is (a) H2 + ½O2 → H2O (b) CO + ½O2 → CO2 (c) N2 + O2 → 2NO (d) N2 → 2N (e) all of the above. Answer (c) N2 + O2 → 2NO
16-104 Of the reactions given below, the reaction whose number of moles of products increases by the addition of inert gases into the reaction chamber at constant pressure and temperature is (a) H2 + ½O2 → H2O (b) CO + ½O2 → CO2 (c) N2 + O2 → 2NO (d) N2 → 2N (e) none of the above. Answer (d) N2 → 2N
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-81
16-105 Moist air is heated to a very high temperature. If the equilibrium composition consists of H2O, O2, N2, OH, H2, and NO, the number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is (e) 5 (a) 1 (b) 2 (c) 3 (d) 4 Answer (c) 3
16-106 Propane C3H8 is burned with air, and the combustion products consist of CO2, CO, H2O, O2, N2, OH, H2, and NO. The number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is (e) 5 (a) 1 (b) 2 (c) 3 (d) 4 Answer (d) 4
16-107 Consider a gas mixture that consists of three components. The number of independent variables that need to be specified to fix the state of the mixture is (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Answer (d) 4
16-108 The value of Henry’s constant for CO2 gas dissolved in water at 290 K is 12.8 MPa. Consider water exposed to air at 100 kPa that contains 3 percent CO2 by volume. Under phase equilibrium conditions, the mole fraction of CO2 gas dissolved in water at 290 K is (a) 2.3×10-4 (b) 3.0×10-4 (c) 0.80×10-4 (d) 2.2×10-4 (e) 5.6×10-4 Answer (a) 2.3×10-4 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). H=12.8 "MPa" P=0.1 "MPa" y_CO2_air=0.03 P_CO2_air=y_CO2_air*P y_CO2_liquid=P_CO2_air/H "Some Wrong Solutions with Common Mistakes:" W1_yCO2=P_CO2_air*H "Multiplying by H instead of dividing by it" W2_yCO2=P_CO2_air "Taking partial pressure in air"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
16-82
16-109 The solubility of nitrogen gas in rubber at 25°C is 0.00156 kmol/m3⋅bar. When phase equilibrium is established, the density of nitrogen in a rubber piece placed in a nitrogen gas chamber at 800 kPa is (b) 0.35 kg/m3 (c) 0.42 kg/m3 (d) 0.56 kg/m3 (e) 0.078 kg/m3 (a) 0.012 kg/m3 Answer (b) 0.35 kg/m3 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T=25 "C" S=0.00156 "kmol/bar.m^3" MM_N2=28 "kg/kmol" S_mass=S*MM_N2 "kg/bar.m^3" P_N2=8 "bar" rho_solid=S_mass*P_N2 "Some Wrong Solutions with Common Mistakes:" W1_density=S*P_N2 "Using solubility per kmol"
16-110 and 16-111 Design and Essay Problems
KJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-1
Chapter 17 COMPRESSIBLE FLOW Stagnation Properties 17-1C The temperature of the air will rise as it approaches the nozzle because of the stagnation process. 17-2C Stagnation enthalpy combines the ordinary enthalpy and the kinetic energy of a fluid, and offers convenience when analyzing high-speed flows. It differs from the ordinary enthalpy by the kinetic energy term. 17-3C Dynamic temperature is the temperature rise of a fluid during a stagnation process. 17-4C No. Because the velocities encountered in air-conditioning applications are very low, and thus the static and the stagnation temperatures are practically identical.
17-5 The state of air and its velocity are specified. The stagnation temperature and stagnation pressure of air are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Air is an ideal gas. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis The stagnation temperature of air is determined from T0 = T +
(470 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 245.9 K + ⎜ ⎟ = 355.8 K 2c p 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 /s 2 ⎠
Other stagnation properties at the specified state are determined by considering an isentropic process between the specified state and the stagnation state, ⎛T ⎞ P0 = P⎜ 0 ⎟ ⎝T ⎠
k /( k −1)
355.8 K ⎞ = (44 kPa) ⎛⎜ ⎟ ⎝ 245.9 K ⎠
1.4 /(1.4 −1)
= 160.3 kPa
Discussion Note that the stagnation properties can be significantly different than thermodynamic properties.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-2
17-6 Air at 300 K is flowing in a duct. The temperature that a stationary probe inserted into the duct will read is to be determined for different air velocities. Assumptions The stagnation process is isentropic. Properties The specific heat of air at room temperature is cp = 1.005 kJ/kg⋅K (Table A-2a). Analysis The air which strikes the probe will be brought to a complete stop, and thus it will undergo a stagnation process. The thermometer will sense the temperature of this stagnated air, which is the stagnation temperature, T0. It is determined from V2 T0 = T + 2c p (a)
T0 = 300 K +
⎛ 1 kJ/kg (1 m/s) 2 ⎜ 2 × 1.005 kJ/kg ⋅ K ⎜⎝ 1000 m 2 / s 2
(b)
T0 = 300 K +
(10 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎜ ⎟ = 300.1 K 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠
(c)
T0 = 300 K +
(100 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎜ ⎟ = 305.0 K 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠
⎞ ⎟ = 300.0 K ⎟ ⎠
AIR 300 K V
(1000 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎜ ⎟ = 797.5 K 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠ Discussion Note that the stagnation temperature is nearly identical to the thermodynamic temperature at low velocities, but the difference between the two is very significant at high velocities,
(d)
T0 = 300 K +
17-7 The states of different substances and their velocities are specified. The stagnation temperature and stagnation pressures are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Helium and nitrogen are ideal gases. Analysis (a) Helium can be treated as an ideal gas with cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2a). Then the stagnation temperature and pressure of helium are determined from (240 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 50°C + T0 = T + ⎜ ⎟ = 55.5°C 2c p 2 × 5.1926 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠ k / ( k −1)
1.667 / (1.667 −1)
k /( k −1)
1.4 /(1.4 −1)
⎛T ⎞ ⎛ 328.7 K ⎞ P0 = P⎜ 0 ⎟ = (0.25 MPa)⎜ = 0.261 MPa ⎟ T ⎝ 323.2 K ⎠ ⎝ ⎠ (b) Nitrogen can be treated as an ideal gas with cp = 1.039 kJ/kg·K and k =1.400. Then the stagnation temperature and pressure of nitrogen are determined from (300 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 50°C + T0 = T + ⎜ ⎟ = 93.3°C 2c p 2 × 1.039 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠ 366.5 K ⎞ ⎛T ⎞ P0 = P⎜ 0 ⎟ = (0.15 MPa)⎛⎜ = 0.233 MPa ⎟ T 323.2 K⎠ ⎝ ⎝ ⎠ (c) Steam can be treated as an ideal gas with cp = 1.865 kJ/kg·K and k =1.329. Then the stagnation temperature and pressure of steam are determined from T0 = T +
(480 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 350°C + ⎜ ⎟ = 411.8 °C = 685 K 2c p 2 × 1.865 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠ k /( k −1)
1.329 /(1.329 −1)
⎛T ⎞ ⎛ 685 K ⎞ P0 = P⎜⎜ 0 ⎟⎟ = (0.1 MPa)⎜ = 0.147 MPa ⎟ T ⎝ 623.2 K ⎠ ⎝ ⎠ Discussion Note that the stagnation properties can be significantly different than thermodynamic properties.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-3
17-8 The inlet stagnation temperature and pressure and the exit stagnation pressure of air flowing through a compressor are specified. The power input to the compressor is to be determined. Assumptions 1 The compressor is isentropic. 2 Air is an ideal gas. Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis The exit stagnation temperature of air T02 is determined from T02
⎛P = T01 ⎜⎜ 02 ⎝ P01
⎞ ⎟ ⎟ ⎠
( k −1) / k
900 ⎞ = (300.2 K)⎛⎜ ⎟ ⎝ 100 ⎠
(1.4 −1) / 1.4
900 kPa
= 562.4 K AIR 0.02 kg/s
From the energy balance on the compressor,
& W
W& in = m& (h20 − h01 )
or, W& in = m& c p (T02 − T01 ) = (0.02 kg/s)(1.005 kJ/kg ⋅ K)(562.4 − 300.2)K = 5.27 kW
100 kPa 27°C
Discussion Note that the stagnation properties can be used conveniently in the energy equation.
17-9E Steam flows through a device. The stagnation temperature and pressure of steam and its velocity are specified. The static pressure and temperature of the steam are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Steam is an ideal gas. Properties Steam can be treated as an ideal gas with cp = 0.445 Btu/lbm·R and k =1.329 (Table A-2Ea). Analysis The static temperature and pressure of steam are determined from T = T0 −
⎛ 1 Btu/lbm (900 ft/s) 2 V2 ⎜ = 700°F − 2c p 2 × 0.445 Btu/lbm ⋅ °F ⎜⎝ 25,037 ft 2 / s 2
⎛T P = P0 ⎜⎜ ⎝ T0
⎞ ⎟ ⎟ ⎠
k /( k −1)
⎛ 1123.6 R ⎞ = (120 psia)⎜ ⎟ ⎝ 1160 R ⎠
⎞ ⎟ = 663.6°F ⎟ ⎠
1.329 /(1.329 −1)
= 105.5 psia
Discussion Note that the stagnation properties can be significantly different than thermodynamic properties.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-4
17-10 The inlet stagnation temperature and pressure and the exit stagnation pressure of products of combustion flowing through a gas turbine are specified. The power output of the turbine is to be determined. Assumptions 1 The expansion process is isentropic. 2 Products of combustion are ideal gases. Properties The properties of products of combustion are given to be cp = 1.157 kJ/kg⋅K, R = 0.287 kJ/kg⋅K, and k = 1.33. Analysis The exit stagnation temperature T02 is determined to be ⎛P T02 = T01 ⎜⎜ 02 ⎝ P01
⎞ ⎟ ⎟ ⎠
( k −1) / k
0.1 ⎞ = (1023.2 K)⎛⎜ ⎟ ⎝ 1 ⎠
(1.33−1) / 1.33
= 577.9 K
kR k −1 1.33(0.287 kJ/kg ⋅ K) = 1.33 − 1 = 1.157 kJ/kg ⋅ K
c p = kc v = k (c p − R ) ⎯ ⎯→ c p =
Also,
1 MPa 750°C
W STEAM
100 kPa
From the energy balance on the turbine, − wout = (h20 − h01 )
or, wout = c p (T01 − T02 ) = (1.157 kJ/kg ⋅ K)(1023.2 − 577.9) K = 515.2 kJ/kg
Discussion Note that the stagnation properties can be used conveniently in the energy equation.
17-11 Air flows through a device. The stagnation temperature and pressure of air and its velocity are specified. The static pressure and temperature of air are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Air is an ideal gas. Properties The properties of air at an anticipated average temperature of 600 K are cp = 1.051 kJ/kg⋅K and k = 1.376 (Table A-2b). Analysis The static temperature and pressure of air are determined from T = T0 −
(570 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 673.2 − ⎜ ⎟ = 518.6 K 2c p 2 × 1.051 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠
and ⎛T P2 = P02 ⎜⎜ 2 ⎝ T02
⎞ ⎟ ⎟ ⎠
k /( k −1)
518.6 K ⎞ = (0.6 MPa)⎛⎜ ⎟ ⎝ 673.2 K ⎠
1.376 /(1.376 −1)
= 0.23 MPa
Discussion Note that the stagnation properties can be significantly different than thermodynamic properties.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-5
Speed of sound and Mach Number 17-12C Sound is an infinitesimally small pressure wave. It is generated by a small disturbance in a medium. It travels by wave propagation. Sound waves cannot travel in a vacuum. 17-13C Yes, it is. Because the amplitude of an ordinary sound wave is very small, and it does not cause any significant change in temperature and pressure. 17-14C The sonic speed in a medium depends on the properties of the medium, and it changes as the properties of the medium change. 17-15C In warm (higher temperature) air since c = kRT 17-16C Helium, since c = kRT and helium has the highest kR value. It is about 0.40 for air, 0.35 for argon and 3.46 for helium. 17-17C Air at specified conditions will behave like an ideal gas, and the speed of sound in an ideal gas depends on temperature only. Therefore, the speed of sound will be the same in both mediums. 17-18C In general, no. Because the Mach number also depends on the speed of sound in gas, which depends on the temperature of the gas. The Mach number will remain constant if the temperature is maintained constant.
17-19 The Mach number of an aircraft and the velocity of sound in air are to be determined at two specified temperatures. Assumptions Air is an ideal gas with constant specific heats at room temperature. Analysis (a) At 300 K air can be treated as an ideal gas with R = 0.287 kJ/kg·K and k = 1.4 (Table A-2a). Thus ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 347.2 m/s c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(300 K)⎜ ⎜ 1 kJ/kg ⎟ ⎝ ⎠
and
Ma =
V 280 m/s = = 0.81 c 347.2 m/s
(b) At 1000 K, ⎛ 1000 m 2 / s 2 c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(1000 K)⎜⎜ ⎝ 1 kJ/kg
and
Ma =
⎞ ⎟ = 634 m/s ⎟ ⎠
V 280 m/s = = 0.442 c 634 m/s
Discussion Note that a constant Mach number does not necessarily indicate constant speed. The Mach number of a rocket, for example, will be increasing even when it ascends at constant speed. Also, the specific heat ratio k changes with temperature, and the accuracy of the result at 1000 K can be improved by using the k value at that temperature (it would give k = 1.336, c = 619 m/s, and Ma = 0.452).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-6
17-20 Carbon dioxide flows through a nozzle. The inlet temperature and velocity and the exit temperature of CO2 are specified. The Mach number is to be determined at the inlet and exit of the nozzle. Assumptions 1 CO2 is an ideal gas with constant specific heats at room temperature. 2 This is a steady-flow process. Properties The gas constant of carbon dioxide is R = 0.1889 kJ/kg·K. Its constant pressure specific heat and specific heat ratio at room temperature are cp = 0.8439 kJ/kg⋅K and k = 1.288 (Table A-2a).
1200 K 50 m/s
Carbon dioxide
400 K
Analysis (a) At the inlet ⎛ 1000 m 2 / s 2 c1 = k1 RT1 = (1.288)(0.1889 kJ/kg ⋅ K)(1200 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 540.4 m/s ⎟ ⎠
Thus, Ma 1 =
V1 50 m/s = = 0.0925 c1 540.4 m/s
(b) At the exit, ⎛ 1000 m 2 / s 2 c 2 = k 2 RT2 = (1.288)(0.1889 kJ/kg ⋅ K)(400 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 312 m/s ⎟ ⎠
The nozzle exit velocity is determined from the steady-flow energy balance relation, 0 = h2 − h1 +
V 2 2 − V1 2 2
→
0 = c p (T2 − T1 ) +
0 = (0.8439 kJ/kg ⋅ K)(1200 − 400 K) +
V 2 2 − V1 2 2
V 2 2 − (50 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎯→ V 2 = 1163 m/s ⎜ ⎟⎯ 2 ⎝ 1000 m 2 / s 2 ⎠
Thus, Ma 2 =
V 2 1163 m/s = = 3.73 312 m/s c2
Discussion The specific heats and their ratio k change with temperature, and the accuracy of the results can be improved by accounting for this variation. Using EES (or another property database): →
c1 = 516 m/s,
V1 = 50 m/s,
At 400 K: cp = 0.9383 kJ/kg⋅K, k = 1.252 →
c2 = 308 m/s,
V2 = 1356 m/s,
At 1200 K: cp = 1.278 kJ/kg⋅K, k = 1.173
Ma1 = 0.0969 Ma2 = 4.41
Therefore, the constant specific heat assumption results in an error of 4.5% at the inlet and 15.5% at the exit in the Mach number, which are significant.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-7
17-21 Nitrogen flows through a heat exchanger. The inlet temperature, pressure, and velocity and the exit pressure and velocity are specified. The Mach number is to be determined at the inlet and exit of the heat exchanger. Assumptions 1 N2 is an ideal gas. 2 This is a steady-flow process. 3 The potential energy change is negligible. Properties The gas constant of N2 is R = 0.2968 kJ/kg·K. Its constant pressure specific heat and specific heat ratio at room temperature are cp = 1.040 kJ/kg⋅K and k = 1.4 (Table A-2a). ⎛ 1000 m 2 / s 2 c1 = k 1 RT1 = (1.400)(0.2968 kJ/kg ⋅ K)(283 K)⎜⎜ ⎝ 1 kJ/kg
Analysis Thus, Ma 1 =
120 kJ/kg
V1 100 m/s = = 0.292 c1 342.9 m/s
150 kPa 10°C 100 m/s
From the energy balance on the heat exchanger, qin = c p (T2 − T1 ) +
⎞ ⎟ = 342.9 m/s ⎟ ⎠
100 kPa 200 m/s
Nitrogen
V2 2 − V12 2
120 kJ/kg = (1.040 kJ/kg.°C)(T2 − 10°C) +
(200 m/s) 2 − (100 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎜ ⎟ 2 ⎝ 1000 m 2 / s 2 ⎠
It yields T2 = 111°C = 384 K ⎛ 1000 m 2 / s 2 c 2 = k 2 RT2 = (1.4 )(0.2968 kJ/kg ⋅ K)(384 K)⎜⎜ ⎝ 1 kJ/kg
Thus,
Ma 2 =
⎞ ⎟ = 399 m/s ⎟ ⎠
V 2 200 m/s = = 0.501 c 2 399 m/s
Discussion The specific heats and their ratio k change with temperature, and the accuracy of the results can be improved by accounting for this variation. Using EES (or another property database): →
c1 = 343 m/s,
V1 = 100 m/s,
Ma1 = 0.292
At 111°C cp = 1.041 kJ/kg⋅K, k = 1.399 →
c2 = 399 m/s,
V2 = 200 m/s,
Ma2 = 0.501
At 10°C : cp = 1.038 kJ/kg⋅K, k = 1.400
Therefore, the constant specific heat assumption results in no error at the inlet and at the exit in the Mach number.
17-22 The speed of sound in refrigerant-134a at a specified state is to be determined. Assumptions R-134a is an ideal gas with constant specific heats at room temperature. Properties The gas constant of R-134a is R = 0.08149 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.108. Analysis From the ideal-gas speed of sound relation, ⎛ 1000 m 2 / s 2 c = kRT = (1.108)(0.08149 kJ/kg ⋅ K)(60 + 273 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 173 m/s ⎟ ⎠
Discussion Note that the speed of sound is independent of pressure for ideal gases.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-8
17-23 The Mach number of a passenger plane for specified limiting operating conditions is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4 (Table A-2a). Analysis From the speed of sound relation ⎛ 1000 m 2 / s 2 c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(-60 + 273 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 293 m/s ⎟ ⎠
Thus, the Mach number corresponding to the maximum cruising speed of the plane is Ma =
V max (945 / 3.6) m/s = = 0.897 293 m/s c
Discussion Note that this is a subsonic flight since Ma < 1. Also, using a k value at -60°C would give practically the same result.
17-24E Steam flows through a device at a specified state and velocity. The Mach number of steam is to be determined assuming ideal gas behavior. Assumptions Steam is an ideal gas with constant specific heats. Properties The gas constant of steam is R = 0.1102 Btu/lbm·R. Its specific heat ratio is given to be k = 1.3. Analysis From the ideal-gas speed of sound relation, ⎛ 25,037 ft 2 / s 2 c = kRT = (1.3)(0.1102 Btu/lbm ⋅ R)(1160 R)⎜⎜ ⎝ 1 Btu/lbm
⎞ ⎟ = 2040.8 ft/s ⎟ ⎠
Thus, Ma =
V 900 ft/s = = 0.441 c 2040 ft/s
Discussion Using property data from steam tables and not assuming ideal gas behavior, it can be shown that the Mach number in steam at the specified state is 0.446, which is sufficiently close to the ideal-gas value of 0.441. Therefore, the ideal gas approximation is a reasonable one in this case.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-9
17-25E EES Problem 17-24e is reconsidered. The variation of Mach number with temperature as the temperature changes between 350 and 700°F is to be investigated, and the results are to be plotted. Analysis Using EES, this problem can be solved as follows: T=Temperature+460 R=0.1102 V=900 k=1.3 c=SQRT(k*R*T*25037) Ma=V/c
Mach number Ma 0.528 0.520 0.512 0.505 0.498 0.491 0.485 0.479 0.473 0.467 0.462 0.456 0.451 0.446 0.441
0.53 0.52 0.51 0.5 0.49
Ma
Temperature, T, °F 350 375 400 425 450 475 500 525 550 575 600 625 650 675 700
0.48 0.47 0.46 0.45 0.44 350
400
450
500
550
600
650
700
Temperature, °F
Discussion Note that for a specified flow speed, the Mach number decreases with increasing temperature, as expected.
17-26 The expression for the speed of sound for an ideal gas is to be obtained using the isentropic process equation and the definition of the speed of sound. Analysis The isentropic relation Pvk = A where A is a constant can also be expressed as k
⎛1⎞ P = A⎜ ⎟ = Aρ k ⎝v ⎠
Substituting it into the relation for the speed of sound, ⎛ ∂ ( Aρ ) k ⎛ ∂P ⎞ c 2 = ⎜⎜ ⎟⎟ = ⎜ ⎝ ∂ρ ⎠ s ⎜⎝ ∂ρ
⎞ ⎟ = kAρ k −1 = k ( Aρ k ) / ρ = k ( P / ρ ) = kRT ⎟ ⎠s
since for an ideal gas P = ρRT or RT = P/ρ. Therefore, c = kRT
which is the desired relation.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-10
17-27 The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air are R = 0.287 kJ/kg·K and k = 1.4 (Table A-2a). The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded. Analysis The final temperature of air is determined from the isentropic relation of ideal gases, ( k −1) / k
(1.4 −1) / 1.4 ⎛P ⎞ ⎛ 0.4 MPa ⎞ = 228.4 K T2 = T1 ⎜⎜ 2 ⎟⎟ = (333.2 K)⎜ ⎟ ⎝ 1.5 MPa ⎠ ⎝ P1 ⎠ Treating k as a constant, the ratio of the initial to the final speed of sound can be expressed as
Ratio =
c2 = c1
k1RT1 k2 RT2
=
T1 T2
=
333.2 = 1.21 228.4
Discussion Note that the speed of sound is proportional to the square root of thermodynamic temperature.
17-28 The inlet state and the exit pressure of helium are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions Helium is an ideal gas with constant specific heats at room temperature. Properties The properties of helium are R = 2.0769 kJ/kg·K and k = 1.667 (Table A-2a). Analysis The final temperature of helium is determined from the isentropic relation of ideal gases, ( k −1) / k
(1.667 −1) / 1.667 ⎛P ⎞ ⎛ 0.4 ⎞ = (333.2 K)⎜ = 196.3 K T2 = T1 ⎜⎜ 2 ⎟⎟ ⎟ ⎝ 1.5 ⎠ ⎝ P1 ⎠ The ratio of the initial to the final speed of sound can be expressed as
Ratio =
c2 = c1
k1RT1 k2 RT2
=
T1 T2
=
333.2 = 1.30 196.3
Discussion Note that the speed of sound is proportional to the square root of thermodynamic temperature.
17-29E The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air are R = 0.06855 Btu/lbm·R and k = 1.4 (Table A-2Ea). The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded. Analysis The final temperature of air is determined from the isentropic relation of ideal gases, ( k −1) / k
(1.4 −1) / 1.4 ⎛P ⎞ ⎛ 60 ⎞ = (659.7 R)⎜ = 489.9 R T2 = T1 ⎜⎜ 2 ⎟⎟ ⎟ ⎝ 170 ⎠ ⎝ P1 ⎠ Treating k as a constant, the ratio of the initial to the final speed of sound can be expressed as
Ratio =
c2 = c1
k1RT1 k2 RT2
=
T1 T2
=
659.7 = 1.16 489.9
Discussion Note that the speed of sound is proportional to the square root of thermodynamic temperature.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-11
One Dimensional Isentropic Flow 17-30C (a) The exit velocity remain constant at sonic speed, (b) the mass flow rate through the nozzle decreases because of the reduced flow area. 17-31C (a) The velocity will decrease, (b), (c), (d) the temperature, the pressure, and the density of the fluid will increase. 17-32C (a) The velocity will increase, (b), (c), (d) the temperature, the pressure, and the density of the fluid will decrease. 17-33C (a) The velocity will increase, (b), (c), (d) the temperature, the pressure, and the density of the fluid will decrease. 17-34C (a) The velocity will decrease, (b), (c), (d) the temperature, the pressure and the density of the fluid will increase. 17-35C They will be identical. 17-36C No, it is not possible. 17-37 Air enters a converging-diverging nozzle at specified conditions. The lowest pressure that can be obtained at the throat of the nozzle is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio of air at room temperature is k = 1.4 (Table A-2a). Analysis The lowest pressure that can be obtained at the throat is the critical pressure P*, which is determined from k /( k −1) 1.4 /(1.4 −1) 2 ⎞ 2 ⎞ = 0.634 MPa = (1.2 MPa)⎛⎜ P* = P0 ⎛⎜ ⎟ ⎟ ⎝ 1.4 + 1 ⎠ ⎝ k + 1⎠ Discussion This is the pressure that occurs at the throat when the flow past the throat is supersonic.
17-38 Helium enters a converging-diverging nozzle at specified conditions. The lowest temperature and pressure that can be obtained at the throat of the nozzle are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of helium are k = 1.667 and cp = 5.1926 kJ/kg·K (Table A-2a). Analysis The lowest temperature and pressure that can be obtained at the throat are the critical temperature T* and critical pressure P*. First we determine the stagnation temperature T0 and stagnation pressure P0, (100 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ T0 = T + = 800 K + ⎜ ⎟ = 801 K 2c p 2 × 5.1926 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠ ⎛T ⎞ P0 = P ⎜ 0 ⎟ ⎝T ⎠
Thus,
k /( k −1)
1.667 /(1.667 −1)
801 K ⎞ = 0.702 MPa = (0.7 MPa)⎛⎜ ⎟ ⎝ 800 K ⎠ 2 ⎛ 2 ⎞ ⎛ ⎞ T * = T0 ⎜ ⎟ = (801 K)⎜ ⎟ = 601 K ⎝ k +1⎠ ⎝ 1.667 + 1 ⎠ k /( k −1)
Helium
1.667 /(1.667 −1)
2 ⎞ 2 ⎞ = 0.342 MPa = (0.702 MPa)⎛⎜ P* = P0 ⎛⎜ ⎟ ⎟ ⎝ 1.667 + 1 ⎠ ⎝ k + 1⎠ Discussion These are the temperature and pressure that will occur at the throat when the flow past the throat is supersonic.
and
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-12
17-39 The critical temperature, pressure, and density of air and helium are to be determined at specified conditions. Assumptions Air and Helium are ideal gases with constant specific heats at room temperature. Properties The properties of air at room temperature are R = 0.287 kJ/kg·K, k = 1.4, and cp = 1.005 kJ/kg·K. The properties of helium at room temperature are R = 2.0769 kJ/kg·K, k = 1.667, and cp = 5.1926 kJ/kg·K (Table A-2a). Analysis (a) Before we calculate the critical temperature T*, pressure P*, and density ρ*, we need to determine the stagnation temperature T0, pressure P0, and density ρ0. T0 = 100°C + ⎛T ⎞ P0 = P ⎜ 0 ⎟ ⎝T ⎠
ρ0 =
(250 m/s) 2 V2 ⎛ 1 kJ/kg ⎞ = 100 + ⎟ = 131.1°C ⎜ 2c p 2 × 1.005 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠ k /( k −1)
404.3 K ⎞ = (200 kPa)⎛⎜ ⎟ ⎝ 373.2 K ⎠
1.4 /(1.4 −1)
= 264.7 kPa
P0 264.7 kPa = = 2.281 kg/m 3 RT0 (0.287 kPa ⋅ m 3 /kg ⋅ K)(404.3 K)
Thus, ⎛ 2 ⎞ ⎛ 2 ⎞ T * = T0 ⎜ ⎟ = (404.3 K)⎜ ⎟ = 337 K 1 k + ⎝ ⎠ ⎝ 1.4 + 1 ⎠ ⎛ 2 ⎞ P* = P0 ⎜ ⎟ ⎝ k +1⎠
k /( k −1)
⎛ 2 ⎞ = (264.7 kPa)⎜ ⎟ ⎝ 1.4 + 1 ⎠
ρ * = ρ 0 ⎛⎜
1 /( k −1)
⎛ 2 ⎞ = (2.281 kg/m 3 )⎜ ⎟ ⎝ 1.4 + 1 ⎠
2 ⎞ ⎟ ⎝ k +1⎠
(b) For helium,
T0 = T +
⎛T ⎞ P0 = P ⎜ 0 ⎟ ⎝T ⎠
ρ0 =
1.4 /(1.4 −1)
= 140 kPa
1 /(1.4 −1)
= 1.45 kg/m 3
(300 m/s) 2 V2 ⎛ 1 kJ/kg = 40 + ⎜ 2c p 2 × 5.1926 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2
k /( k −1)
321.9 K ⎞ = (200 kPa)⎛⎜ ⎟ ⎝ 313.2 K ⎠
1.667 /(1.667 −1)
⎞ ⎟ = 48.7°C ⎠
= 214.2 kPa
P0 214.2 kPa = = 0.320 kg/m 3 RT0 (2.0769 kPa ⋅ m 3 /kg ⋅ K)(321.9 K)
Thus, 2 ⎛ 2 ⎞ ⎛ ⎞ T * = T0 ⎜ ⎟ = (321.9 K)⎜ ⎟ = 241 K ⎝ k +1⎠ ⎝ 1.667 + 1 ⎠ ⎛ 2 ⎞ P* = P0 ⎜ ⎟ ⎝ k +1⎠
k /( k −1)
2 ⎛ ⎞ = (200 kPa)⎜ ⎟ ⎝ 1.667 + 1 ⎠
ρ * = ρ 0 ⎛⎜
1 /( k −1)
2 ⎛ ⎞ = (0.320 kg/m 3 )⎜ ⎟ ⎝ 1.667 + 1 ⎠
2 ⎞ ⎟ ⎝ k +1⎠
1.667 /(1.667 −1)
= 97.4 kPa
1 /(1.667 −1)
= 0.208 kg/m 3
Discussion These are the temperature, pressure, and density values that will occur at the throat when the flow past the throat is supersonic.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-13
17-40 Stationary carbon dioxide at a given state is accelerated isentropically to a specified Mach number. The temperature and pressure of the carbon dioxide after acceleration are to be determined. Assumptions Carbon dioxide is an ideal gas with constant specific heats. Properties The specific heat ratio of the carbon dioxide at 400 K is k = 1.252 (Table A-2b). Analysis The inlet temperature and pressure in this case is equivalent to the stagnation temperature and pressure since the inlet velocity of the carbon dioxide said to be negligible. That is, T0 = Ti = 400 K and P0 = Pi = 600 kPa. Then, ⎛ 2 T = T0 ⎜ ⎜ 2 + (k − 1)M 2 ⎝
⎞ ⎛ 2 ⎟ = (400 K)⎜ ⎟ ⎜ 2 + (1.252 - 1)(0.5) 2 ⎠ ⎝
⎞ ⎟ = 387.8 K ⎟ ⎠
and ⎛T P = P0 ⎜⎜ ⎝ T0
⎞ ⎟ ⎟ ⎠
k /( k −1)
⎛ 387.8 K ⎞ = (600 kPa)⎜ ⎟ ⎝ 400 K ⎠
1.252 /(1.252 −1)
= 514.3 kPa
Discussion Note that both the pressure and temperature drop as the gas is accelerated as part of the internal energy of the gas is converted to kinetic energy.
17-41 Air flows through a duct. The state of the air and its Mach number are specified. The velocity and the stagnation pressure, temperature, and density of the air are to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air at room temperature are R = 0.287 kPa.m3/kg.K and k = 1.4 (Table A-2a). Analysis The speed of sound in air at the specified conditions is ⎛ 1000 m 2 / s 2 c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(373.2 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 387.2 m/s ⎟ ⎠
Thus, V = Ma × c = (0.8)(387.2 m/s) = 310 m/s
AIR
Also,
ρ=
P 200 kPa = = 1.867 kg/m 3 RT (0.287 kPa ⋅ m 3 /kg ⋅ K)(373.2 K)
Then the stagnation properties are determined from ⎛ (k − 1)Ma 2 T0 = T ⎜ 1 + ⎜ 2 ⎝ ⎛T ⎞ P0 = P⎜ 0 ⎟ ⎝T ⎠
k /( k −1)
⎛ T0 ⎞ ⎟ ⎝T ⎠
ρ0 = ρ⎜
1 /( k −1)
2 ⎞ ⎛ ⎟ = (373.2 K)⎜1 + (1.4 - 1)(0.8) ⎟ ⎜ 2 ⎠ ⎝
⎛ 421.0 K ⎞ = (200 kPa)⎜ ⎟ ⎝ 373.2 K ⎠
1.4 /(1.4 −1)
⎛ 421.0 K ⎞ = (1.867 kg/m 3 )⎜ ⎟ ⎝ 373.2 K ⎠
⎞ ⎟ = 421 K ⎟ ⎠
= 305 kPa
1 /(1.4 −1)
= 2.52 kg/m 3
Discussion Note that both the pressure and temperature drop as the gas is accelerated as part of the internal energy of the gas is converted to kinetic energy.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-14
17-42 EES Problem 17-41 is reconsidered. The effect of Mach number on the velocity and stagnation properties as the Ma is varied from 0.1 to 2 are to be investigated, and the results are to be plotted. Analysis Using EES, the problem is solved as follows:
"Stagnation properties" T0=T*(1+(k-1)*Ma^2/2) P0=P*(T0/T)^(k/(k-1)) rho0=rho*(T0/T)^(1/(k-1))
1600 1400
V, T0, P0, and 100•ρ0
P=200 T=100+273.15 R=0.287 k=1.4 c=SQRT(k*R*T*1000) Ma=V/c rho=P/(R*T)
P0 1200 1000 800 600
T0
400
ρ0
200
V 0 0
0.4
0.8
1.2
1.6
Ma
Mach num. Ma 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0
Velocity, V, m/s 38.7 77.4 116.2 154.9 193.6 232.3 271.0 309.8 348.5 387.2 425.9 464.7 503.4 542.1 580.8 619.5 658.3 697.0 735.7 774.4
Stag. Temp, T0, K 373.9 376.1 379.9 385.1 391.8 400.0 409.7 420.9 433.6 447.8 463.5 480.6 499.3 519.4 541.1 564.2 588.8 615.0 642.6 671.7
Stag. Press, P0, kPa 201.4 205.7 212.9 223.3 237.2 255.1 277.4 304.9 338.3 378.6 427.0 485.0 554.1 636.5 734.2 850.1 987.2 1149.2 1340.1 1564.9
Stag. Density, ρ0, kg/m3 1.877 1.905 1.953 2.021 2.110 2.222 2.359 2.524 2.718 2.946 3.210 3.516 3.867 4.269 4.728 5.250 5.842 6.511 7.267 8.118
Discussion Note that as Mach number increases, so does the flow velocity and stagnation temperature, pressure, and density.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
2
17-15
17-43E Air flows through a duct at a specified state and Mach number. The velocity and the stagnation pressure, temperature, and density of the air are to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air are R = 0.06855 Btu/lbm.R = 0.3704 psia⋅ft3/lbm.R and k = 1.4 (Table A2Ea). Analysis The speed of sound in air at the specified conditions is ⎛ 25,037 ft 2 / s 2 c = kRT = (1.4)(0.06855 Btu/1bm ⋅ R)(671.7 R)⎜⎜ ⎝ 1 Btu/1bm
Thus,
⎞ ⎟ = 1270.4 ft/s ⎟ ⎠
V = Ma × c = (0.8)(1270.4 ft/s) = 1016 ft/s
Also,
ρ=
30 psia P = = 0.1206 1bm/ft 3 RT (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(671.7 R)
Then the stagnation properties are determined from ⎛ (k − 1)Ma 2 T0 = T ⎜⎜1 + 2 ⎝ ⎛T ⎞ P0 = P⎜ 0 ⎟ ⎝T ⎠
k /( k −1)
⎛ T0 ⎞ ⎟ ⎝T ⎠
ρ0 = ρ⎜
1 /( k −1)
⎞ ⎛ (1.4 - 1)(0.8) 2 ⎟ = (671.7 R)⎜1 + ⎟ ⎜ 2 ⎠ ⎝
⎛ 757.7 R ⎞ = (30 psia)⎜ ⎟ ⎝ 671.7 R ⎠
1.4 /(1.4 −1)
⎛ 757.7 R ⎞ = (0.1206 1bm/ft 3 )⎜ ⎟ ⎝ 671.7 R ⎠
⎞ ⎟ = 758 R ⎟ ⎠
= 45.7 psia 1 /(1.4 −1)
= 0.163 1bm/ft 3
Discussion Note that the temperature, pressure, and density of a gas increases during a stagnation process.
17-44 An aircraft is designed to cruise at a given Mach number, elevation, and the atmospheric temperature. The stagnation temperature on the leading edge of the wing is to be determined. Assumptions Air is an ideal gas. Properties The properties of air are R = 0.287 kPa.m3/kg.K, cp = 1.005 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The speed of sound in air at the specified conditions is ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 308.0 m/s c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(236.15 K)⎜ ⎜ 1 kJ/kg ⎟ ⎝ ⎠
Thus, V = Ma × c = (1.2)(308.0 m/s) = 369.6 m/s
Then, T0 = T +
(369.6 m/s) 2 ⎛ 1 kJ/kg ⎞ V2 = 236.15 + ⎜ ⎟ = 304.1 K 2c p 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠
Discussion Note that the temperature of a gas increases during a stagnation process as the kinetic energy is converted to enthalpy.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-16
Isentropic Flow Through Nozzles 17-45C (a) The exit velocity will reach the sonic speed, (b) the exit pressure will equal the critical pressure, and (c) the mass flow rate will reach the maximum value. 17-46C (a) None, (b) None, and (c) None. 17-47C They will be the same. 17-48C Maximum flow rate through a nozzle is achieved when Ma = 1 at the exit of a subsonic nozzle. For all other Ma values the mass flow rate decreases. Therefore, the mass flow rate would decrease if hypersonic velocities were achieved at the throat of a converging nozzle. 17-49C Ma* is the local velocity non-dimensionalized with respect to the sonic speed at the throat, whereas Ma is the local velocity non-dimensionalized with respect to the local sonic speed. 17-50C The fluid would accelerate even further instead of decelerating. 17-51C The fluid would decelerate instead of accelerating. 17-52C (a) The velocity will decrease, (b) the pressure will increase, and (c) the mass flow rate will remain the same. 17-53C No. If the velocity at the throat is subsonic, the diverging section will act like a diffuser and decelerate the flow.
17-54 It is to be explained why the maximum flow rate per unit area for a given ideal gas depends only on & max / A * = a P0 / T0 . P0 / T0 . Also for an ideal gas, a relation is to be obtained for the constant a in m
(
)
3
Properties The properties of the ideal gas considered are R = 0.287 kPa.m /kg⋅K and k = 1.4 (Table A-2a). Analysis The maximum flow rate is given by 2 ⎞ m& max = A * P0 k / RT0 ⎛⎜ ⎟ ⎝ k + 1⎠
or
(
m& max / A* = P0 / T0
)
( k +1) / 2 ( k −1)
2 ⎞ k / R ⎛⎜ ⎟ ⎝ k + 1⎠
( k +1) / 2 ( k −1)
For a given gas, k and R are fixed, and thus the mass flow rate depends on the parameter P0 / T0 .
(
)
m& max / A * can be expressed as a m& max / A* = a P0 / T0 where
⎛ 2 ⎞ a = k / R⎜ ⎟ ⎝ k + 1⎠
( k +1) / 2 ( k −1)
=
1.4 ⎛ 1000 m 2 / s 2 (0.287 kJ/kg.K)⎜⎜ ⎝ 1 kJ/kg
⎛ 2 ⎞ ⎜ ⎟ ⎞ ⎝ 1.4 + 1 ⎠ ⎟ ⎟ ⎠
2.4 / 0.8
= 0.0404 (m/s) K
Discussion Note that when sonic conditions exist at a throat of known cross-sectional area, the mass flow rate is fixed by the stagnation conditions.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-17
17-55 For an ideal gas, an expression is to be obtained for the ratio of the speed of sound where Ma = 1 to the speed of sound based on the stagnation temperature, c*/c0. Analysis For an ideal gas the speed of sound is expressed as c = kRT . Thus, c* = c0
kRT * kRT0
⎛T *⎞ ⎟ = ⎜⎜ ⎟ ⎝ T0 ⎠
1/ 2
⎛ 2 ⎞ =⎜ ⎟ ⎝ k +1⎠
1/ 2
Discussion Note that a speed of sound changes the flow as the temperature changes.
17-56 For subsonic flow at the inlet, the variation of pressure, velocity, and Mach number along the length of the nozzle are to be sketched for an ideal gas under specified conditions. Analysis Using EES and CO2 as the gas, we calculate and plot flow area A, velocity V, and Mach number Ma as the pressure drops from a stagnation value of 1400 kPa to 200 kPa. Note that the curve for A represents the shape of the nozzle, with horizontal axis serving as the centerline.
Mai < 1
T0=473 "K" m=3 "kg/s" rho_0=P0/(R*T0) rho=P/(R*T) rho_norm=rho/rho_0 "Normalized density" T=T0*(P/P0)^((k-1)/k) Tnorm=T/T0 "Normalized temperature" V=SQRT(2*Cp*(T0-T)*1000) V_norm=V/500 A=m/(rho*V)*500 C=SQRT(k*R*T*1000) Ma=V/C
Normalized A, Ma, P, V
k=1.289 Cp=0.846 "kJ/kg.K" R=0.1889 "kJ/kg.K" P0=1400 "kPa"
P A
Ma V
200
400
600
800
P, kPa
1000
1200
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
1400
17-18
17-57 For supersonic flow at the inlet, the variation of pressure, velocity, and Mach number along the length of the nozzle are to be sketched for an ideal gas under specified conditions. Analysis Using EES and CO2 as the gas, we calculate and plot flow area A, velocity V, and Mach number Ma as the pressure rises from 200 kPa at a very high velocity to the stagnation value of 1400 kPa. Note that the curve for A represents the shape of the nozzle, with horizontal axis serving as the centerline.
Mai > 1
T0=473 "K" m=3 "kg/s" rho_0=P0/(R*T0) rho=P/(R*T) rho_norm=rho/rho_0 "Normalized density" T=T0*(P/P0)^((k-1)/k) Tnorm=T/T0 "Normalized temperature" V=SQRT(2*Cp*(T0-T)*1000) V_norm=V/500 A=m/(rho*V)*500 C=SQRT(k*R*T*1000) Ma=V/C
Normalized A, Ma, P, V
k=1.289 Cp=0.846 "kJ/kg.K" R=0.1889 "kJ/kg.K" P0=1400 "kPa" P A
Ma V
200
400
600
800
P, kPa
1000
1200
Discussion Note that this problem is identical to the proceeding one, except the flow direction is reversed.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
1400
17-19
17-58 Air enters a nozzle at specified temperature, pressure, and velocity. The exit pressure, exit temperature, and exit-to-inlet area ratio are to be determined for a Mach number of Ma = 1 at the exit. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air are k = 1.4 and cp = 1.005 kJ/kg·K (Table A-2a). Analysis The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. We first determine the stagnation temperature and pressure, which remain constant throughout the nozzle since the flow is isentropic. V2 (150 m/s) 2 ⎛ 1 kJ/kg ⎞ T0 = Ti + i = 350 K + ⎜ ⎟ = 361.2 K 2c p 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠ i AIR
k /( k −1)
1.4 /(1.4 −1)
* Ma = 1
⎛T ⎞ 361.2 K ⎞ And P0 = Pi ⎜⎜ 0 ⎟⎟ = 0.223 MPa = (0.2 MPa)⎛⎜ ⎟ ⎝ 350 K ⎠ ⎝ Ti ⎠ From Table A-32 (or from Eqs. 17-18 and 17-19) at Ma = 1, we read T/T0 = 0.8333, P/P0 = 0.5283. Thus, T = 0.8333T0 = 0.8333(361.2 K) = 301 K and P = 0.5283P0 = 0.5283(0.223 MPa) = 0.118 MPa
Also, and
⎛ 1000 m 2 / s 2 c i = kRT i = (1.4 )(0.287 kJ/kg ⋅ K)(350 K)⎜⎜ ⎝ 1 kJ/kg V 150 m/s Ma i = i = = 0.40 c i 375 m/s
150 m/s
⎞ ⎟ = 375 m/s ⎟ ⎠
From Table A-32 at this Mach number we read Ai /A* = 1.5901. Thus the ratio of the throat area to the nozzle inlet area is A* 1 = = 0.629 Ai 15901 . Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.
17-59 Air enters a nozzle at specified temperature and pressure with low velocity. The exit pressure, exit temperature, and exit-to-inlet area ratio are to be determined for a Mach number of Ma = 1 at the exit. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio of air is k = 1.4 (Table A-2a). Analysis The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. The stagnation temperature and pressure in this case are identical to the inlet temperature and pressure since the inlet velocity is negligible. They remain constant throughout the nozzle since the flow is isentropic. T0 = Ti = 350 K P0 = Pi = 0.2 MPa From Table A-32 (or from Eqs. 17-18 and 17-19) at Ma i AIR * =1, we read T/T0 =0.8333, P/P0 = 0.5283. Thus, Ma = 1 Vi ≈ 0 T = 0.8333T0 = 0.83333(350 K) = 292 K and P = 0.5283P0 = 0.5283(0.2 MPa) = 0.106 MPa The Mach number at the nozzle inlet is Ma = 0 since Vi ≅ 0. From Table A-32 at this Mach number we read Ai/A* = ∞. Thus the ratio of the throat area to the nozzle inlet area is A* 1 = =0 Ai ∞ Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-20
17-60E Air enters a nozzle at specified temperature, pressure, and velocity. The exit pressure, exit temperature, and exit-to-inlet area ratio are to be determined for a Mach number of Ma = 1 at the exit. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air are k = 1.4 and cp = 0.240 Btu/lbm·R (Table A-2Ea). Analysis The properties of the fluid at the location where Ma =1 are the critical properties, denoted by superscript *. We first determine the stagnation temperature and pressure, which remain constant throughout the nozzle since the flow is isentropic. T0 = T +
⎛ 1 Btu/1bm ⎞ (450 ft/s) 2 Vi 2 ⎟ = 646.9 R ⎜ = 630 R + 2c p 2 × 0.240 Btu/lbm ⋅ R ⎜⎝ 25,037 ft 2 / s 2 ⎟⎠ i 450 ft/s
and
⎛T P0 = Pi ⎜⎜ 0 ⎝ Ti
⎞ ⎟⎟ ⎠
k /( k −1)
⎛ 646.9 K ⎞ = (30 psia)⎜ ⎟ ⎝ 630 K ⎠
1.4 /(1.4 −1)
AIR
* Ma = 1
= 32.9 psia
From Table A-32 (or from Eqs. 17-18 and 17-19) at Ma =1, we read T/T0 =0.8333, P/P0 = 0.5283. Thus, T = 0.8333T0 = 0.8333(646.9 R) = 539 R and
P = 0.5283P0 = 0.5283(32.9 psia) = 17.4 psia
Also,
⎛ 25,037 ft 2 / s 2 c i = kRT i = (1.4 )(0.06855 Btu/1bm ⋅ R)(630 R)⎜⎜ ⎝ 1 Btu/1bm
and
Ma i =
⎞ ⎟ = 1230 ft/s ⎟ ⎠
Vi 450 ft/s = = 0.3657 c i 1230 ft/s
From Table A-32 at this Mach number we read Ai/A* = 1.7426. Thus the ratio of the throat area to the nozzle inlet area is 1 A* = = 0.574 Ai 1.7426
Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.
17-61 Air enters a converging-diverging nozzle at a specified pressure. The back pressure that will result in a specified exit Mach number is to be determined. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio of air is k = 1.4 (Table A-2a). Analysis The stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. It remains constant throughout the nozzle since the flow is isentropic, P0 = Pi = 0.5 MPa From Table A-32 at Mae =1.8, we read Pe /P0 = 0.1740. Thus,
P = 0.1740P0 = 0.1740(0.5 MPa) = 0.087 MPa
Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.
i Vi ≈ 0
AIR
e Mae = 1.8
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-21
17-62 Nitrogen enters a converging-diverging nozzle at a given pressure. The critical velocity, pressure, temperature, and density in the nozzle are to be determined. Assumptions 1 Nitrogen is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of nitrogen are k = 1.4 and R = 0.2968 kJ/kg·K (Table A-2a). Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. They remain constant throughout the nozzle, P0 = Pi = 700 kPa
i Vi ≈ 0
T0 = Ti = 450 K
ρ0 =
N2
P0 700 kPa = = 5.241 kg/m 3 RT0 (0.2968 kPa ⋅ m 3 /kg ⋅ K)(450 K)
*
Critical properties are those at a location where the Mach number is Ma= 1. From Table A-32 at Ma =1, we read T/T0 =0.8333, P/P0 = 0.5283, and ρ/ρ0 = 0.6339. Then the critical properties become T* = 0.8333T0 = 0.8333(450 K) = 375 K P* = 0.52828P0 = 0.5283(700 kPa) = 370 MPa
ρ* = 0.63394ρ0 = 0.6339(5.241 kg/m3) = 3.32 kg/m3 Also, ⎛ 1000 m 2 /s 2 V * = c* = kRT * = (1.4)(0.2968 kJ/kg ⋅ K)(375.0 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 395 m/s ⎟ ⎠
Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.
17-63 An ideal gas is flowing through a nozzle. The flow area at a location where Ma = 2.4 is specified. The flow area where Ma = 1.2 is to be determined. Assumptions Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio is given to be k = 1.4. Analysis The flow is assumed to be isentropic, and thus the stagnation and critical properties remain constant throughout the nozzle. The flow area at a location where Ma2 = 1.2 is determined using A /A* data from Table A-32 to be Ma 1 = 2.4 :
A1 A1 25 cm 2 = 2.4031 ⎯ ⎯→ A* = = = 10.40 cm 2 A* 2.4031 2.4031
Ma 2 = 1.2 :
A2 = 1.0304 ⎯ ⎯→ A2 = (1.0304) A* = (1.0304)(10.40 cm 2 ) = 10.7 cm 2 A*
Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-22
17-64 An ideal gas is flowing through a nozzle. The flow area at a location where Ma = 2.4 is specified. The flow area where Ma = 1.2 is to be determined. Assumptions Flow through the nozzle is steady, one-dimensional, and isentropic. Analysis The flow is assumed to be isentropic, and thus the stagnation and critical properties remain constant throughout the nozzle. The flow area at a location where Ma2 = 1.2 is determined using the A /A* relation, A 1 ⎧⎛ 2 ⎞⎛ k − 1 ⎞⎫ Ma 2 ⎟⎬ = ⎟⎜1 + ⎨⎜ A * Ma ⎩⎝ k + 1 ⎠⎝ 2 ⎠⎭
( k +1) / 2 ( k −1)
For k = 1.33 and Ma1 = 2.4: A1 1 ⎧⎛ 2 ⎞⎛ 1.33 − 1 ⎫ = 2.4 2 ⎞⎟⎬ ⎟⎜1 + ⎨⎜ 2 A * 2.4 ⎩⎝ 1.33 + 1 ⎠⎝ ⎠⎭
and,
A* =
2.33 / 2×0.33
= 2.570
A1 25 cm 2 = = 9.729 cm 2 2.570 2.570
For k = 1.33 and Ma2 = 1.2: A2 1 ⎧⎛ 2 ⎞⎛ 1.33 − 1 2 ⎞⎫ = 1.2 ⎟⎬ ⎟⎜1 + ⎨⎜ 2 A * 1.2 ⎩⎝ 1.33 + 1 ⎠⎝ ⎠⎭
and
2.33 / 2×0.33
= 1.0316
A2 = (1.0316) A* = (1.0316)(9.729 cm 2 ) = 10.0 cm 2
Discussion Note that the compressible flow functions in Table A-32 are prepared for k = 1.4, and thus they cannot be used to solve this problem.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-23
17-65 [Also solved by EES on enclosed CD] Air enters a converging nozzle at a specified temperature and pressure with low velocity. The exit pressure, the exit velocity, and the mass flow rate versus the back pressure are to be calculated and plotted. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air are k = 1.4, R = 0.287 kJ/kg·K, and cp = 1.005 kJ/kg·K (Table A-2a). Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. They remain constant throughout the nozzle since the flow is isentropic., P0 = Pi = 900 kPa T0 = Ti = 400 K The critical pressure is determined to be 2 ⎞ P* = P0 ⎛⎜ ⎟ ⎝ k + 1⎠
k /( k −1)
2 ⎞ = (900 kPa)⎛⎜ ⎟ ⎝ 1.4 + 1 ⎠
1.4 / 0.4
i Vi ≈ 0
= 475.5 kPa
e
AIR
Then the pressure at the exit plane (throat) will be Pe = Pb
for
Pb ≥ 475.5 kPa
Pe = P* = 475.5 kPa
for
Pb < 475.5 kPa (choked flow)
Thus the back pressure will not affect the flow when 100 < Pb < 475.5 kPa. For a specified exit pressure Pe, the temperature, the velocity and the mass flow rate can be determined from ⎛P Te = T0 ⎜⎜ e ⎝ P0
Temperature
⎞ ⎟ ⎟ ⎠
( k −1) / k
⎛ P ⎞ = (400 K)⎜⎜ e ⎟⎟ ⎝ 900 ⎠
0.4 / 1.4
Pe
⎛ 1000 m 2 /s 2 Velocity V = 2c p (T0 − Te ) = 2(1.005 kJ/kg ⋅ K)(400 - Te )⎜⎜ ⎝ 1 kJ/kg
ρe =
Density Mass flow rate
Pe Pe = RTe (0.287 kPa ⋅ m3/kg ⋅ K )Te
⎞ ⎟ ⎟ ⎠ Pb Ve
m& = ρeVe Ae = ρeVe (0.001 m 2 )
c
The results of the calculations can be tabulated as Pb, kPa
Pe, kPa
Te, K
Ve, m/s
ρe, kg/m3
900 800 700 600 500 475.5 400 300 200 100
900 800 700 600 500 475.5 475.5 475.5 475.5 475.5
400 386.8 372.3 356.2 338.2 333.3 333.3 333.3 333.3 333.3
0 162.9 236.0 296.7 352.4 366.2 366.2 366.2 366.2 366.2
7.840 7.206 6.551 5.869 5.151 4.971 4.971 4.971 4.971 4.971
& , kg/s m 0 1.174 1.546 1.741 1.815 1.820 1.820 1.820 1.820 1.820
Pb
& m & max m
100
475.5
900
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Pb kPa
17-24
17-66 EES Problem 17-65 is reconsidered. Using EES (or other) software, The problem is to be solved for the inlet conditions of 1 MPa and 1000 K. Analysis Using EES, the problem is solved as follows: Procedure ExitPress(P_back,P_crit : P_exit, Condition$) If (P_back>=P_crit) then P_exit:=P_back "Unchoked Flow Condition" Condition$:='unchoked' else P_exit:=P_crit "Choked Flow Condition" Condition$:='choked' Endif End "Input data from Diagram Window" {Gas$='Air' A_cm2=10 P_inlet = 900"kPa" T_inlet= 400"K"} {P_back =475.5 "kPa"}
"Throat area, cm2"
A_exit = A_cm2*Convert(cm^2,m^2) C_p=specheat(Gas$,T=T_inlet) C_p-C_v=R k=C_p/C_v M=MOLARMASS(Gas$) "Molar mass of Gas$" R= 8.314/M "Gas constant for Gas$" "Since the inlet velocity is negligible, the stagnation temperature = T_inlet; and, since the nozzle is isentropic, the stagnation pressure = P_inlet." P_o=P_inlet "Stagnation pressure" T_o=T_inlet "Stagnation temperature" P_crit /P_o=(2/(k+1))^(k/(k-1)) "Critical pressure from Eq. 16-22" Call ExitPress(P_back,P_crit : P_exit, Condition$) T_exit /T_o=(P_exit/P_o)^((k-1)/k)
"Exit temperature for isentopic flow, K"
V_exit ^2/2=C_p*(T_o-T_exit)*1000 "Exit velocity, m/s" Rho_exit=P_exit/(R*T_exit)
"Exit density, kg/m3"
m_dot=Rho_exit*V_exit*A_exit
"Nozzle mass flow rate, kg/s"
"If you wish to redo the plots, hide the diagram window and remove the { } from the first 4 variables just under the procedure. Next set the desired range of back pressure in the parametric table. Finally, solve the table (F3). "
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-25
SOLUTION A_cm2=10 [cm^2] A_exit=0.001 [m^2] Condition$='choked' C_p=1.14 [kJ/kg-K] C_v=0.8532 [kJ/kg-K] Gas$='Air' k=1.336 M=28.97 [kg/kmol] m_dot=1.258 [kg/s] P_back=300 [kPA]
m [kg/s] 1.819 1.819 1.819 1.819 1.819 1.74 1.546 1.176 0
P_crit=539.2 [kPA] P_exit=539.2 [kPA] P_inlet=1000 [kPA] P_o=1000 [kPA] R=0.287 [kJ/kg-K] Rho_exit=2.195 [m^3/kg] T_exit=856 [K] T_inlet=1000 [K] T_o=1000 [K] V_exit=573 [m/s]
Pexit [kPa] 475.5 475.5 475.5 475.5 475.5 600 700 800 900
Texit [K] 333.3 333.3 333.3 333.3 333.3 356.2 372.3 386.8 400
Vexit [m/s] 366.1 366.1 366.1 366.1 366 296.6 236 163.1 0
ρexit [kg/m3] Pback [kPa] 4.97 100 4.97 200 4.97 300 4.97 400 4.97 475.5 5.868 600 6.551 700 7.207 800 7.839 900
2.0
1.6
Air, A=10 cm2
1.2
s g/ k , m
0.8
0.4
0.0 100
200
300
400
500
600
Pback, kPa
700
800
900
.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-26
17-67E Air enters a converging-diverging nozzle at a specified temperature and pressure with low velocity. The pressure, temperature, velocity, and mass flow rate are to be calculated in the specified test section. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air are k = 1.4 and R = 0.06855 Btu/lbm·R = 0.3704 psia·ft3/lbm·R (Table A2Ea). Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. They remain constant throughout the nozzle since the flow is isentropic. P0 = Pi = 150 psia T0 = Ti = 100°F = 560 R Then, ⎛ 2 T e = T0 ⎜ ⎜ 2 + (k − 1)Ma 2 ⎝
⎞ ⎛ 2 ⎟ = (560 R)⎜ ⎟ ⎜ 2 + (1.4 - 1)2 2 ⎠ ⎝
⎞ ⎟ = 311 R ⎟ ⎠
i Vi ≈ 0
AIR
e
and ⎛T Pe = P0 ⎜⎜ ⎝ T0
⎞ ⎟ ⎟ ⎠
k /( k −1)
⎛ 311 ⎞ = (150 psia)⎜ ⎟ ⎝ 560 ⎠
1.4 / 0.4
= 19.1 psia
Also,
ρe =
Pe 19.1 psia = = 0.166 1bm/ft 3 RTe (0.3704 psia.ft 3 /1bm.R)(311 R)
The nozzle exit velocity can be determined from Ve = Maece , where ce is the speed of sound at the exit conditions, ⎛ 25,037 ft 2 / s 2 Ve = Maece = Ma e kRTe = (2) (1.4)(0.06855 Btu/1bm.R)(311 R)⎜⎜ ⎝ 1 Btu/1bm
⎞ ⎟ = 1729 ft/s ⎟ ⎠
Finally, m& = ρe AeVe = (0. 166 1bm/ft3)(5 ft2 )(1729 ft/s) = 1435 1bm/s
Discussion Air must be very dry in this application because the exit temperature of air is extremely low, and any moisture in the air will turn to ice particles.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-27
Shock Waves and Expansion Waves 17-68C No, because the flow must be supersonic before a shock wave can occur. The flow in the converging section of a nozzle is always subsonic. 17-69C The Fanno line represents the states which satisfy the conservation of mass and energy equations. The Rayleigh line represents the states which satisfy the conservation of mass and momentum equations. The intersections points of these lines represents the states which satisfy the conservation of mass, energy, and momentum equations. 17-70C No, the second law of thermodynamics requires the flow after the shock to be subsonic.. 17-71C (a) decreases, (b) increases, (c) remains the same, (d) increases, and (e) decreases. 17-72C Oblique shocks occur when a gas flowing at supersonic speeds strikes a flat or inclined surface. Normal shock waves are perpendicular to flow whereas inclined shock waves, as the name implies, are typically inclined relative to the flow direction. Also, normal shocks form a straight line whereas oblique shocks can be straight or curved, depending on the surface geometry. 17-73C Yes, the upstream flow have to be supersonic for an oblique shock to occur. No, the flow downstream of an oblique shock can be subsonic, sonic, and even supersonic. 17-74C Yes. Conversely, normal shocks can be thought of as special oblique shocks in which the shock angle is β = π/2, or 90o. 17-75C When the wedge half-angle δ is greater than the maximum deflection angle θmax, the shock becomes curved and detaches from the nose of the wedge, forming what is called a detached oblique shock or a bow wave. The numerical value of the shock angle at the nose is be β = 90o. 17-76C When supersonic flow impinges on a blunt body like the rounded nose of an aircraft, the wedge half-angle δ at the nose is 90o, and an attached oblique shock cannot exist, regardless of Mach number. Therefore, a detached oblique shock must occur in front of all such blunt-nosed bodies, whether twodimensional, axisymmetric, or fully three-dimensional. 17-77C Isentropic relations of ideal gases are not applicable for flows across (a) normal shock waves and (b) oblique shock waves, but they are applicable for flows across (c) Prandtl-Meyer expansion waves.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-28
17-78 For an ideal gas flowing through a normal shock, a relation for V2/V1 in terms of k, Ma1, and Ma2 is to be developed. Analysis The conservation of mass relation across the shock is ρ1V1 = ρ 2V2 and it can be expressed as V2 ρ1 P / RT1 ⎛ P1 ⎞⎛ T2 = = 1 = ⎜ ⎟⎜ V1 ρ 2 P2 / RT2 ⎜⎝ P2 ⎟⎠⎜⎝ T1
⎞ ⎟⎟ ⎠
From Eqs. 17-35 and 17-38, V 2 ⎛ 1 + kMa 22 =⎜ V1 ⎜⎝ 1 + kMa 12
⎞⎛ 1 + Ma 12 (k − 1) / 2 ⎞ ⎟⎜ ⎟ ⎟⎜ 1 + Ma 2 (k − 1) / 2 ⎟ 2 ⎠⎝ ⎠
Discussion This is an important relation as it enables us to determine the velocity ratio across a normal shock when the Mach numbers before and after the shock are known.
17-79 Air flowing through a converging-diverging nozzle experiences a normal shock at the exit. The effect of the shock wave on various properties is to be determined. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane. Properties The properties of air are k = 1.4 and R = 0.287 kJ/kg·K (Table A-2a). Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Then, P01 = Pi = 1.5 MPa Shock T01 = Ti = 350 K wave Then, ⎛ 2 T1 = T01 ⎜ ⎜ 2 + (k − 1)Ma 2 1 ⎝
and
⎛T P1 = P01 ⎜⎜ 1 ⎝ T0
⎞ ⎟ ⎟ ⎠
⎞ ⎛ 2 ⎟ = (350 K)⎜ ⎜ 2 + (1.4 - 1)2 2 ⎟ ⎝ ⎠
k /( k −1)
⎛ 194.4 ⎞ = (1.5 MPa)⎜ ⎟ ⎝ 300 ⎠
⎞ ⎟ = 194.4 K ⎟ ⎠
i
AIR
1
2
Vi ≈ 0
1.4 / 0.4
= 0.1917 MPa
The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A-33. For Ma1 = 2.0 we read Ma 2 = 0.5774 ,
P02 P T = 0.7209, 2 = 4.5000, and 2 = 1.6875 P01 P1 T1
Then the stagnation pressure P02, static pressure P2, and static temperature T2, are determined to be P02 = 0.7209P01 = (0.7209)(1.5 MPa) = 1.081 MPa P2= 4.5000P1 = (4.5000)(0.1917 MPa) = 0.863 MPa T2 = 1.6875T1 = (1.6875)(194.4 K) = 328.1 K The air velocity after the shock can be determined from V2 = Ma2c2, where c2 is the velocity of sound at the exit conditions after the shock, ⎛ 1000 m 2 / s 2 V2 = Ma2c2 = Ma 2 kRT2 = (0.5774) (1.4)(0.287 kJ/kg ⋅ K)(328.1 K)⎜ ⎜ 1 kJ/kg ⎝
⎞ ⎟ = 209.6 m/s ⎟ ⎠
Discussion We can also solve this problem using the relations for normal shock functions. The results would be identical.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-29
17-80 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane. Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic, P01 = Pi = 2 MPa It is specified that A/A* =3.5. From Table A-32, Mach number and the pressure ratio which corresponds to this area ratio are the Ma1 =2.80 and P1/P01 = 0.0368. The pressure ratio across the shock for this Ma1 value is, from Table A-33, P2/P1 = 8.98. Thus the back pressure, which is equal to the static pressure at the nozzle exit, must be
shock wave i
AIR
1
2
Vi ≈ 0 Pb
P2 =8.98P1 = 8.98×0.0368P01 = 8.98×0.0368×(2 MPa) = 0.661 MPa Discussion We can also solve this problem using the relations for compressible flow and normal shock functions. The results would be identical.
17-81 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic, P01= Pi = 2 MPa It is specified that A/A* = 2. From Table A-32, the Mach number and the pressure ratio which corresponds to this area ratio are the Ma1 =2.20 and P1/P01 = 0.0935. The pressure ratio across the shock for this M1 value is, from Table A-33, P2/P1 = 5.48. Thus the back pressure, which is equal to the static pressure at the nozzle exit, must be
shock wave i
AIR
1
2
Vi ≈ 0 Pb
P2 =5.48P1 = 5.48×0.0935P01 = 5.48×0.0935×(2 MPa) = 1.02 MPa Discussion We can also solve this problem using the relations for compressible flow and normal shock functions. The results would be identical.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-30
17-82 Air flowing through a nozzle experiences a normal shock. The effect of the shock wave on various properties is to be determined. Analysis is to be repeated for helium under the same conditions. Assumptions 1 Air and helium are ideal gases with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Properties The properties of air are k = 1.4 and R = 0.287 kJ/kg·K, and the properties of helium are k = 1.667 and R = 2.0769 kJ/kg·K (Table A-2a). Analysis The air properties upstream the shock are
shock wave
Ma1 = 2.5, P1 = 61.64 kPa, and T1 = 262.15 K Fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions in Table A-33. For Ma1 = 2.5, Ma 2 = 0.513,
i
AIR
2
1
Ma1 = 2.5
P02 P T = 8.5262, 2 = 7.125, and 2 = 2.1375 P1 P1 T1
Then the stagnation pressure P02, static pressure P2, and static temperature T2, are determined to be P02 = 8.5261P1 = (8.5261)(61.64 kPa) = 526 kPa P2 = 7.125P1 = (7.125)(61.64 kPa) = 439 kPa T2 = 2.1375T1 = (2.1375)(262.15 K) = 560 K The air velocity after the shock can be determined from V2 = Ma2c2, where c2 is the speed of sound at the exit conditions after the shock, ⎛ 1000 m 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (0.513) (1.4)(0.287 kJ/kg ⋅ K)(560.3 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 243 m/s ⎟ ⎠
We now repeat the analysis for helium. This time we cannot use the tabulated values in Table A-33 since k is not 1.4. Therefore, we have to calculate the desired quantities using the analytical relations, ⎛ Ma 12 + 2 /( k − 1) ⎞ ⎟ Ma 2 = ⎜⎜ 2 ⎟ ⎝ 2Ma 1 k /( k − 1) − 1 ⎠
1/ 2
⎛ ⎞ 2.5 2 + 2 /(1.667 − 1) ⎟ = ⎜⎜ 2 ⎟ ⎝ 2 × 2.5 × 1.667 /(1.667 − 1) − 1 ⎠
1/ 2
= 0.553
P2 1 + kMa 12 1 + 1.667 × 2.5 2 = = = 7.5632 P1 1 + kMa 22 1 + 1.667 × 0.553 2 T2 1 + Ma 12 (k − 1) / 2 1 + 2.5 2 (1.667 − 1) / 2 = = = 2.7989 2 T1 1 + Ma 2 (k − 1) / 2 1 + 0.553 2 (1.667 − 1) / 2 P02 ⎛ 1 + kMa 12 =⎜ P1 ⎜⎝ 1 + kMa 22
⎞ ⎟ 1 + (k − 1)Ma 22 / 2 ⎟ ⎠
(
⎛ 1 + 1.667 × 2.5 2 = ⎜⎜ 2 ⎝ 1 + 1.667 × 0.553
Thus,
)k /(k −1)
(
)
⎞ 1.667 / 0.667 ⎟ 1 + (1.667 − 1) × 0.553 2 / 2 = 9.641 ⎟ ⎠
P02 = 11.546P1 = (9.641)(61.64 kPa) = 594 kPa P2 = 7.5632P1 = (7.5632)(61.64 kPa) = 466 kPa T2 = 2.7989T1 = (2.7989)(262.15 K) = 734 K
⎛ 1000 m 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT y = (0.553) (1.667)(2.0769 kJ/kg ⋅ K)(733.7 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 881 m/s ⎟ ⎠
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-31
17-83 Air flowing through a nozzle experiences a normal shock. The entropy change of air across the normal shock wave is to be determined. Assumptions 1 Air and helium are ideal gases with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Properties The properties of air are R = 0.287 kJ/kg·K and cp = 1.005 kJ/kg·K, and the properties of helium are R = 2.0769 kJ/kg·K and cp = 5.1926 kJ/kg·K (Table A-2a). Analysis The entropy change across the shock is determined to be s 2 − s1 = c p ln
T2 P − R ln 2 = (1.005 kJ/kg ⋅ K)ln(2.1375) - (0.287 kJ/kg ⋅ K)ln(7.125) = 0.200 kJ/kg ⋅ K T1 P1
For helium, the entropy change across the shock is determined to be s 2 − s1 = c p ln
T2 P − R ln 2 = (5.1926 kJ/kg ⋅ K)ln(2.7989) - (2.0769 kJ/kg ⋅ K)ln(7.5632) = 1.14 kJ/kg ⋅ K T1 P1
Discussion Note that shock wave is a highly dissipative process, and the entropy generation is large during shock waves.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-32
17-84E [Also solved by EES on enclosed CD] Air flowing through a nozzle experiences a normal shock. Effect of the shock wave on various properties is to be determined. Analysis is to be repeated for helium. Assumptions 1 Air and helium are ideal gases with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Properties The properties of air are k = 1.4 and R = 0.06855 Btu/lbm·R, and the properties of helium are k = 1.667 and R = 0.4961 Btu/lbm·R. Analysis The air properties upstream the shock are shock Ma1 = 2.5, P1 = 10 psia, and T1 = 440.5 R wave Fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in i AIR 2 1 Table A-33. For Ma1 = 2.5, Ma1 = 2.5 P P T Ma 2 = 0.513, 02 = 8.5262, 2 = 7.125, and 2 = 2.1375 P1 P1 T1 Then the stagnation pressure P02, static pressure P2, and static temperature T2, are determined to be P02 = 8.5262P1 = (8.5262)(10 psia) = 85.3 psia P2 = 7.125P1 = (7.125)(10 psia) = 71.3 psia T2 = 2.1375T1 = (2.1375)(440.5 R) = 942 R The air velocity after the shock can be determined from V2 = Ma2c2, where c2 is the speed of sound at the exit conditions after the shock, ⎛ 25,037 ft 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (0.513) (1.4)(0.06855 Btu/1bm ⋅ R)(941.6 R)⎜⎜ ⎝ 1 Btu/1bm
⎞ ⎟ = 772 ft/s ⎟ ⎠
We now repeat the analysis for helium. This time we cannot use the tabulated values in Table A-33 since k is not 1.4. Therefore, we have to calculate the desired quantities using the analytical relations, ⎛ Ma 12 + 2 /( k − 1) ⎞ ⎟ Ma 2 = ⎜⎜ 2 ⎟ 2 Ma k /( k 1 ) 1 − − 1 ⎝ ⎠
1/ 2
⎛ ⎞ 2.5 2 + 2 /(1.667 − 1) ⎟ = ⎜⎜ 2 ⎟ ⎝ 2 × 2.5 × 1.667 /(1.667 − 1) − 1 ⎠
1/ 2
= 0.553
P2 1 + kMa 12 1 + 1.667 × 2.5 2 = = = 7.5632 P1 1 + kMa 22 1 + 1.667 × 0.553 2 T2 1 + Ma 12 (k − 1) / 2 1 + 2.5 2 (1.667 − 1) / 2 = = = 2.7989 T1 1 + Ma 22 (k − 1) / 2 1 + 0.553 2 (1.667 − 1) / 2 P02 ⎛ 1 + kMa 12 =⎜ P1 ⎜⎝ 1 + kMa 22
⎞ ⎟ 1 + (k − 1)Ma 22 / 2 ⎟ ⎠
(
⎛ 1 + 1.667 × 2.5 2 = ⎜⎜ 2 ⎝ 1 + 1.667 × 0.553 Thus,
)k /(k −1)
⎞ 1.667 / 0.667 ⎟ 1 + (1.667 − 1) × 0.553 2 / 2 = 9.641 ⎟ ⎠
(
)
P02 = 11.546P1 = (9.641)(10 psia) = 594 psia P2 = 7.5632P1 = (7.5632)(10 psia) = 75.6 psia T2 = 2.7989T1 = (2.7989)(440.5 R) = 1233 R
⎛ 25,037 ft 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (0.553) (1.667)(0.4961 Btu/1bm.R)(1232.9 R)⎜⎜ ⎝ 1 Btu/1bm
⎞ ⎟ = 2794 ft/s ⎟ ⎠
Discussion This problem could also be solved using the relations for compressible flow and normal shock functions. The results would be identical.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-33
17-85E EES Problem 17-84E is reconsidered. The effects of both air and helium flowing steadily in a nozzle when there is a normal shock at a Mach number in the range 2 < Ma1 < 3.5 are to be studied. Also, the entropy change of the air and helium across the normal shock is to be calculated and the results are to be tabulated. Analysis Using EES, the problem is solved as follows: Procedure NormalShock(M_x,k:M_y,PyOPx, TyOTx,RhoyORhox, PoyOPox, PoyOPx) If M_x < 1 Then M_y = -1000;PyOPx=-1000;TyOTx=-1000;RhoyORhox=1000 PoyOPox=-1000;PoyOPx=-1000 else M_y=sqrt( (M_x^2+2/(k-1)) / (2*M_x^2*k/(k-1)-1) ) PyOPx=(1+k*M_x^2)/(1+k*M_y^2) TyOTx=( 1+M_x^2*(k-1)/2 )/(1+M_y^2*(k-1)/2 ) RhoyORhox=PyOPx/TyOTx PoyOPox=M_x/M_y*( (1+M_y^2*(k-1)/2)/ (1+M_x^2*(k-1)/2) )^((k+1)/(2*(k-1))) PoyOPx=(1+k*M_x^2)*(1+M_y^2*(k-1)/2)^(k/(k-1))/(1+k*M_y^2) Endif End Function ExitPress(P_back,P_crit) If P_back>=P_crit then ExitPress:=P_back If P_back 94 kPa
Therefore, the flow is choked, and the velocity at the exit of the hole is the sonic speed. Then the flow properties at the exit becomes
ρ0 =
P0 314 kPa = = 3.671 kg/m 3 RT0 (0.287 kPa ⋅ m 3 / kg ⋅ K )(298 K) 2 ⎞ ⎟ ⎝ k + 1⎠
ρ * = ρ 0 ⎛⎜ T* =
1 /( k −1)
2 ⎞ = (3.671 kg/m 3 )⎛⎜ ⎟ ⎝ 1.4 + 1 ⎠
1 /(1.4 −1)
= 2.327 kg/m 3
2 2 T0 = (298 K) = 248.3 K k +1 1.4 + 1
⎛ 1000 m 2 / s 2 V = c = kRT * = (1.4 )(0.287 kJ/kg ⋅ K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟(248.3 K) = 315.9 m/s ⎟ ⎠
Then the initial mass flow rate through the hole becomes m& = ρAV = (2.327 kg/m 3 )[π (0.004 m) 2 /4](315.9 m/s) = 0.00924 kg/s = 0.554 kg/min
Discussion The mass flow rate will decrease with time as the pressure inside the tire drops.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-65
17-119 The thrust developed by the engine of a Boeing 777 is about 380 kN. The mass flow rate of air through the nozzle is to be determined. Assumptions 1 Air is an ideal gas with constant specific properties. 2 Flow of combustion gases through the nozzle is isentropic. 3 Choked flow conditions exist at the nozzle exit. 4 The velocity of gases at the nozzle inlet is negligible. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1), and it can also be used for combustion gases. The specific heat ratio of combustion gases is k = 1.33 (Table 17-2). Analysis The velocity at the nozzle exit is the sonic velocity, which is determined to be ⎛ 1000 m 2 / s 2 V = c = kRT = (1.33)(0.287 kJ/kg ⋅ K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟(265 K) = 318.0 m/s ⎟ ⎠
Noting that thrust F is related to velocity by F = m& V , the mass flow rate of combustion gases is determined to be m& =
F 380,000 N ⎛⎜ 1 kg.m/s 2 = 318.0 m/s ⎜⎝ 1 N V
⎞ ⎟ = 1194.8 kg/s ⎟ ⎠
Discussion The combustion gases are mostly nitrogen (due to the 78% of N2 in air), and thus they can be treated as air with a good degree of approximation.
17-120 A stationary temperature probe is inserted into an air duct reads 85°C. The actual temperature of air is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The stagnation process is isentropic. Properties The specific heat of air at room temperature is cp = 1.005 kJ/kg⋅K (Table A-2a). Analysis The air that strikes the probe will be brought to a complete stop, and thus it will undergo a stagnation process. The thermometer will sense the temperature of this stagnated air, which is the stagnation temperature. The actual air temperature is determined from T = T0 −
(250 m/s) 2 V2 ⎛ 1 kJ/kg = 85°C − ⎜ 2c p 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2
⎞ ⎟ = 53.9°C ⎠
Discussion Temperature rise due to stagnation is very significant in high-speed flows, and should always be considered when compressibility effects are not negligible.
T 250 m/s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-66
17-121 Nitrogen flows through a heat exchanger. The stagnation pressure and temperature of the nitrogen at the inlet and the exit states are to be determined. Assumptions 1 Nitrogen is an ideal gas with constant specific properties. 2 Flow of nitrogen through the heat exchanger is isentropic.
Qin 150 kPa 10°C 100 m/s
Properties The properties of nitrogen are cp = 1.039 kJ/kg.K and k = 1.4 (Table A-2a).
Nitrogen
100 kPa 180 m/s
Analysis The stagnation temperature and pressure of nitrogen at the inlet and the exit states are determined from T01 = T1 +
V1 2 (100 m/s) 2 ⎛ 1 kJ/kg = 10°C + ⎜ 2c p 2 × 1.039 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2
⎛T ⎞ P01 = P1 ⎜⎜ 01 ⎟⎟ ⎝ T1 ⎠
k /( k −1)
⎛ 288.0 K ⎞ = (150 kPa)⎜ ⎟ ⎝ 283.2 K ⎠
⎞ ⎟ = 14.8°C ⎠
1.4 /(1.4 −1)
= 159.1 kPa
From the energy balance relation E in − E out = ∆E system with w = 0 V 22 − V1 2 + ∆pe ©0 2 (180 m/s) 2 − (100 m/s) 2 125 kJ/kg = (1.039 kJ/kg ⋅ °C)(T2 − 10°C) + 2 q in = c p (T2 − T1 ) +
⎛ 1 kJ/kg ⎞ ⎟ ⎜ ⎝ 1000 m 2 / s 2 ⎠
T2 = 119.5°C
and, T02 = T2 +
V2 2 (180 m/s) 2 ⎛ 1 kJ/kg ⎞ = 119.5°C + ⎟ = 135.1°C ⎜ 2c p 2 × 1.039 kJ/kg ⋅ °C ⎝ 1000 m 2 / s 2 ⎠
⎛T P02 = P2 ⎜⎜ 02 ⎝ T2
⎞ ⎟⎟ ⎠
k /( k −1)
⎛ 408.3 K ⎞ = (100 kPa)⎜ ⎟ ⎝ 392.7 K ⎠
1.4 /(1.4 −1)
= 114.6 kPa
Discussion Note that the stagnation temperature and pressure can be very different than their thermodynamic counterparts when dealing with compressible flow.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-67
17-122 An expression for the speed of sound based on van der Waals equation of state is to be derived. Using this relation, the speed of sound in carbon dioxide is to be determined and compared to that obtained by ideal gas behavior. Properties The properties of CO2 are R = 0.1889 kJ/kg·K and k = 1.279 at T = 50°C = 323.2 K (Table A2b). Analysis Van der Waals equation of state can be expressed as P = Differentiating,
RT a − 2 . v −b v
RT 2a ⎛ ∂P ⎞ + 3 ⎜ ⎟ = 2 v ⎝ ∂v ⎠ T (v − b)
Noting that ρ = 1 / v ⎯ ⎯→ dρ = −dv / v 2 , the speed of sound relation becomes
Substituting,
⎛ ∂P ⎞ ⎛ ∂P ⎞ c 2 = k⎜ ⎟ = v 2 k⎜ ⎟ ⎝ ∂r ⎠ T ⎝ ∂v ⎠ T c2 =
v 2 kRT 2ak − v (v − b) 2
Using the molar mass of CO2 (M = 44 kg/kmol), the constant a and b can be expressed per unit mass as a = 0.1882 kPa ⋅ m 6 /kg 2 and
b = 9.70 × 10 −4 m 3 / kg
The specific volume of CO2 is determined to be 200 kPa =
(0.1889 kPa ⋅ m 3 / kg ⋅ K)(323.2 K)
v − 0.000970 m 3 /kg
−
2 × 0.1882 kPa ⋅ m 6 / kg 2
v2
→ v = 0.300 m 3 / kg
Substituting, ⎛ (0.300 m 3 / kg) 2 (1.279)(0.1889 kJ/kg ⋅ K)(323.2 K) 1000 m 2 / s 2 ⎜ ⎜ 1 kJ/kg (0.300 − 0.000970 m 3 / kg ) 2 c=⎜ ⎜ 2(0.1882 kPa.m 6 / kg 3 )(1.279) 1000 m 2 / s 2 ⎜− (0.300 m 3 / kg) 2 1 kPa ⋅ m 3 /kg ⎝
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
1/ 2
= 271 m/s
If we treat CO2 as an ideal gas, the speed of sound becomes ⎛ 1000 m 2 / s 2 c = kRT = (1.279)(0.1889 kJ/kg ⋅ K)(323.2 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 279 m/s ⎟ ⎠
Discussion Note that the ideal gas relation is the simplest equation of state, and it is very accurate for most gases encountered in practice. At high pressures and/or low temperatures, however, the gases deviate from ideal gas behavior, and it becomes necessary to use more complicated equations of state.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-68
17-123 The equivalent relation for the speed of sound is to be verified using thermodynamic relations. ⎛ ∂P ⎞ ⎛ ∂P ⎞ Analysis The two relations are c 2 = ⎜⎜ ⎟⎟ and c 2 = k ⎜⎜ ⎟⎟ ∂ρ ⎝ ⎠s ⎝ ∂ρ ⎠ T From r = 1 / v ⎯ ⎯→ dr = −dv / v 2 . Thus, ⎛ ∂P ⎞ ⎛ ∂P ⎞ 2 ⎛ ∂P ∂T ⎞ 2 ⎛ ∂P ⎞ ⎛ ∂T ⎞ c 2 = ⎜ ⎟ = −v 2 ⎜ ⎟ = −v ⎜ ⎟ = −v ⎜ ⎟ ⎜ ⎟ ∂ r ∂ ∂ T ∂ v v ⎝ ⎠s ⎝ ⎠s ⎝ ⎠s ⎝ ∂T ⎠ s ⎝ ∂v ⎠ s From the cyclic rule, ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂s ⎞ ⎛ ∂P ⎞ ⎛ ∂s ⎞ ⎛ ∂P ⎞ ( P, T , s ) : ⎜ ⎯→ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎯ ⎟ = −⎜ ⎟ ⎜ ⎟ ∂ T ∂ s ∂ P ∂ T ⎝ ⎠s ⎝ ⎠P ⎝ ⎠T ⎝ ⎠s ⎝ ∂T ⎠ P ⎝ ∂s ⎠ T ⎛ ∂T ⎞ ⎛ ∂v ⎞ ⎛ ∂s ⎞ ⎛ ∂T ⎞ ⎛ ∂s ⎞ ⎛ ∂T ⎞ ⎯→ ⎜ (T , v , s ) : ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎯ ⎟ = −⎜ ⎟ ⎜ ⎟ ∂ ∂ s ∂ T ∂ v v ⎝ ⎠s ⎝ ⎠T ⎝ ⎠v ⎝ ⎠s ⎝ ∂v ⎠ T ⎝ ∂s ⎠ v Substituting, ⎛ ∂s ⎞ ⎛ ∂P ⎞ ⎛ ∂s ⎞ ⎛ ∂T ⎞ 2 ⎛ ∂s ⎞ ⎛ ∂T ⎞ ⎛ ∂P ⎞ c 2 = −v 2 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −v ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂s ⎠T ⎝ ∂v ⎠T ⎝ ∂s ⎠v ⎝ ∂T ⎠ P ⎝ ∂s ⎠v ⎝ ∂s ⎠T
Recall that cp
cv ⎛ ∂s ⎞ ⎛ ∂s ⎞ =⎜ and =⎜ ⎟ ⎟ T ⎝ ∂T ⎠ v T ⎝ ∂T ⎠ P Substituting, ⎛ c p ⎞⎛ T ⎞⎛ ∂P ⎞ 2 ⎛ ∂P ⎞ c 2 = −v 2 ⎜⎜ ⎟⎟⎜⎜ ⎟⎟⎜ ⎟ = −v k ⎜ ⎟ v T c ∂ ⎠T ⎝ ∂v ⎠T ⎝ ⎠⎝ v ⎠⎝
Replacing − dv / v 2 by dρ, ⎛ ∂P ⎞ ⎟⎟ c 2 = k ⎜⎜ ⎝ ∂ρ ⎠ T Discussion Note that the differential thermodynamic property relations are very useful in the derivation of other property relations in differential form.
17-124 For ideal gases undergoing isentropic flows, expressions for P/P*, T/T*, and ρ/ρ* as functions of k and Ma are to be obtained. T0 2 + (k − 1)Ma 2 T* 2 = Analysis Equations 17-18 and 17-21 are given to be and = T 2 T0 k + 1 ⎛ T0 T * ⎞ ⎛ 2 + (k − 1)Ma 2 ⎟ ⎜ ⎜ ⎜ T T ⎟=⎜ 2 0 ⎠ ⎝ ⎝ T k +1 Simplifying and inverting, = T * 2 + (k − 1)Ma 2
Multiplying the two,
From
P ⎛ T ⎞ =⎜ ⎟ P* ⎝T *⎠
k /( k −1)
⎯ ⎯→
k /( k −1)
⎞⎛ 2 ⎞ ⎟⎜ ⎟⎝ k + 1 ⎟⎠ ⎠
P ⎛ k +1 =⎜ P * ⎜⎝ 2 + (k − 1)Ma 2 k +1 ρ ⎛⎜ = ρ * ⎜⎝ 2 + (k − 1)Ma 2
(1) ⎞ ⎟ ⎟ ⎠
k /( k −1)
(2) k /( k −1)
⎞ ⎟ (3) ⎟ ⎠ Discussion Note that some very useful relations can be obtained by very simple manipulations.
From
ρ ⎛ ρ ⎞ ⎟ =⎜ ρ * ⎜⎝ ρ * ⎟⎠
⎯ ⎯→
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-69 17-125 It is to be verified that for the steady flow of ideal gases dT0/T = dA/A + (1-Ma2) dV/V. The effect of heating and area changes on the velocity of an ideal gas in steady flow for subsonic flow and supersonic flow are to be explained. Analysis We start with the relation Differentiating, We also have and
V2 = c p (T0 − T ) , 2
(1)
V dV = c p (dT0 − dT ) dρ
ρ dP
ρ
+
(2)
dA dV + =0 A V
(3)
+ V dV = 0
(4) dP dρ dT = + =0 ρ P T
Differentiating the ideal gas relation P = ρRT,
(5)
From the speed of sound relation,
c 2 = kRT = (k − 1)c p T = kP / ρ
(6)
Combining Eqs. (3) and (5),
dP dT dA dV − + + =0 P T A V
(7)
Combining Eqs. (4) and (6),
dP
ρ
=
dP kP / c 2
= −VdV
or,
dP k V 2 dV dV = − 2 V dV = −k 2 = − kMa 2 P V c c V
Combining Eqs. (2) and (6),
dT = dT0 − V
or,
(8)
dV cp
dT dT0 V 2 dV dT dT0 V2 dV dT0 dV = − = = − 2 = − (k − 1)Ma 2 T T c pT V T T T V c /(k − 1) V
(9)
dV dT0 dV dA dV − + (k − 1)Ma 2 + + =0 V T V A V
Combining Eqs. (7), (8), and (9),
− (k − 1)Ma 2
or,
dT0 dA dV = + − kMa 2 + (k − 1)Ma 2 + 1 T A V
Thus,
dT0 dA dV = + (1 − Ma 2 ) T A V
[
]
(10)
Differentiating the steady-flow energy equation q = h02 − h01 = c p (T02 − T01 )
δq = c p dT0
(11)
Eq. (11) relates the stagnation temperature change dT0 to the net heat transferred to the fluid. Eq. (10) relates the velocity changes to area changes dA, and the stagnation temperature change dT0 or the heat transferred. (a) When Ma < 1 (subsonic flow), the fluid will accelerate if the duck converges (dA < 0) or the fluid is heated (dT0 > 0 or δq > 0). The fluid will decelerate if the duck converges (dA < 0) or the fluid is cooled (dT0 < 0 or δq < 0). (b) When Ma > 1 (supersonic flow), the fluid will accelerate if the duck diverges (dA > 0) or the fluid is cooled (dT0 < 0 or δq < 0). The fluid will decelerate if the duck converges (dA < 0) or the fluid is heated (dT0 > 0 or δq > 0).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-70
17-126 A pitot tube measures the difference between the static and stagnation pressures for a subsonic airplane. The speed of the airplane and the flight Mach number are to be determined. Assumptions 1 Air is an ideal gas with constant specific heat ratio. 2 The stagnation process is isentropic. Properties The properties of air are R = 0.287 kJ/kg.K and k = 1.4 (Table A-2a). Analysis The stagnation pressure of air at the specified conditions is P0 = P + ∆P = 70.109 + 35 = 105.109 kPa
Then, P0 ⎛ (k − 1)Ma 2 = ⎜1 + 2 P ⎜⎝
It yields
⎞ ⎟ ⎟ ⎠
k / k −1
105.109 ⎛⎜ (1.4 − 1)Ma 2 ⎯ ⎯→ = 1+ 70.109 ⎜⎝ 2
⎞ ⎟ ⎟ ⎠
1.4 / 0.4
Ma = 0.783
The speed of sound in air at the specified conditions is ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 328.5 m/s c = kRT = (1.4)(0.287 kJ/kg ⋅ K)(268.65 K)⎜ ⎜ 1 kJ/kg ⎟ ⎝ ⎠
Thus, V = Ma × c = (0.783)(328.5 m/s) = 257.3 m/s
Discussion Note that the flow velocity can be measured in a simple and accurate way by simply measuring pressure.
17-127 The mass flow parameter m& RT0 / ( AP0 ) versus the Mach number for k = 1.2, 1.4, and 1.6 in the range of 0 ≤ Ma ≤ 1 is to be plotted. Analysis The mass flow rate parameter (m& RT0 ) / P0 A can be expressed as m& RT0 P0 A
Thus, Ma 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
⎛ 2 = Ma k ⎜⎜ 2 ⎝ 2 + (k − 1) M
k = 1.2 0 0.1089 0.2143 0.3128 0.4015 0.4782 0.5411 0.5894 0.6230 0.6424 0.6485
k = 1.4 0 0.1176 0.2311 0.3365 0.4306 0.5111 0.5763 0.6257 0.6595 0.6787 0.6847
⎞ ⎟ ⎟ ⎠
( k +1) / 2 ( k −1)
k = 1.6 0 0.1257 0.2465 0.3582 0.4571 0.5407 0.6077 0.6578 0.6916 0.7106 0.7164
0.75 k = 1.6 1.4 1.2
0.60
0.45
0.30
0.15
Ma 0
0.2
0.4
0.6
0.8
1.0
Discussion Note that the mass flow rate increases with increasing Mach number and specific heat ratio. It levels off at Ma = 1, and remains constant (choked flow).
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-71
17-128 Helium gas is accelerated in a nozzle. The pressure and temperature of helium at the location where Ma = 1 and the ratio of the flow area at this location to the inlet flow area are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of helium are R = 2.0769 kJ/kg⋅K, cp = 5.1926 kJ/kg⋅K, and k = 1.667 (Table A2a). Analysis The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. We first determine the stagnation temperature and pressure, which remain constant throughout the nozzle since the flow is isentropic. T0 = Ti +
Vi 2 (120 m/s) 2 ⎛ 1 kJ/kg ⎞ = 500 K + ⎜ ⎟ = 501.4 K 2c p 2 × 5.1926 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠ i 120 m/s
and ⎛T P0 = Pi ⎜⎜ 0 ⎝ Ti
⎞ ⎟ ⎟ ⎠
k /( k −1)
501.4 K ⎞ = (0.8 MPa)⎛⎜ ⎟ ⎝ 500 K ⎠
1.667 /(1.667 −1)
He
* Ma = 1
= 0.806 MPa
The Mach number at the nozzle exit is given to be Ma = 1. Therefore, the properties at the nozzle exit are the critical properties determined from 2 ⎛ ⎞ ⎛ 2 ⎞ T * = T0 ⎜ ⎟ = 376 K ⎟ = (501.4 K)⎜ ⎝ 1.667 + 1 ⎠ ⎝ k +1⎠ 2 ⎞ P* = P0 ⎛⎜ ⎟ k ⎝ + 1⎠
k /( k −1)
2 ⎞ = (0.806 MPa)⎛⎜ ⎟ 1.667 + 1 ⎝ ⎠
1.667 /(1.667 −1)
= 0.393 MPa
The speed of sound and the Mach number at the nozzle inlet are ⎛ 1000 m 2 / s 2 c i = kRT i = (1.667)(2.0769 kJ/kg ⋅ K)(500 K)⎜⎜ ⎝ 1 kJ/kg Ma i =
⎞ ⎟ = 1316 m/s ⎟ ⎠
Vi 120 m/s = = 0.0912 c i 1316 m/s
The ratio of the entrance-to-throat area is Ai A*
⎡⎛ 2 ⎞⎛ k − 1 ⎞⎤ Ma i2 ⎟⎥ ⎟⎜1 + ⎢⎜ 2 ⎠⎦ ⎣⎝ k + 1 ⎠⎝
( k +1) /[ 2 ( k −1)]
=
1 Ma i
=
1 ⎡⎛ 2 ⎞⎛ 1.667 − 1 ⎞⎤ (0.0912) 2 ⎟⎥ ⎟⎜1 + ⎢⎜ 0.0912 ⎣⎝ 1.667 + 1 ⎠⎝ 2 ⎠⎦
2.667 /( 2×0.667 )
= 6.20
Then the ratio of the throat area to the entrance area becomes A* 1 = = 0.161 Ai 6.20
Discussion The compressible flow functions are essential tools when determining the proper shape of the compressible flow duct.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-72
17-129 Helium gas enters a nozzle with negligible velocity, and is accelerated in a nozzle. The pressure and temperature of helium at the location where Ma = 1 and the ratio of the flow area at this location to the inlet flow area are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The entrance velocity is negligible. Properties The properties of helium are R = 2.0769 kJ/kg⋅K, cp = 5.1926 kJ/kg⋅K, and k = 1.667 (Table A2a). Analysis We treat helium as an ideal gas with k = 1.667. The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. The stagnation temperature and pressure in this case are identical to the inlet temperature and pressure since the inlet velocity is negligible. They remain constant throughout the nozzle since the flow is isentropic. T0 = Ti = 500 K P0 = Pi = 0.8 MPa The Mach number at the nozzle exit is given to be Ma = 1. Therefore, the properties at the nozzle exit are the critical properties determined from
i Vi ≈ 0
He
* Ma = 1
2 ⎛ 2 ⎞ ⎛ ⎞ T * = T0 ⎜ ⎟ = (500 K)⎜ ⎟ = 375 K ⎝ k +1⎠ ⎝ 1.667 + 1 ⎠ 2 ⎞ P* = P0 ⎛⎜ ⎟ ⎝ k + 1⎠
k /( k −1)
2 ⎞ = (0.8 MPa)⎛⎜ ⎟ ⎝ 1.667 + 1 ⎠
1.667 /(1.667 −1)
= 0.390 MPa
The ratio of the nozzle inlet area to the throat area is determined from Ai A*
=
1 Ma i
⎡⎛ 2 ⎞⎛ k − 1 2 ⎞⎤ ⎢⎜ k + 1 ⎟⎜1 + 2 Ma i ⎟⎥ ⎠⎝ ⎠⎦ ⎣⎝
( k +1) /[ 2 ( k −1)]
But the Mach number at the nozzle inlet is Ma = 0 since Vi ≅ 0. Thus the ratio of the throat area to the nozzle inlet area is A* 1 = =0 ∞ Ai
Discussion The compressible flow functions are essential tools when determining the proper shape of the compressible flow duct.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-73
17-130 EES Air enters a converging nozzle. The mass flow rate, the exit velocity, the exit Mach number, and the exit pressure-stagnation pressure ratio versus the back pressure-stagnation pressure ratio for a specified back pressure range are to be calculated and plotted. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air at room temperature are R = 0.287 kJ/kg⋅K, cp = 1.005 kJ/kg⋅K, and k = 1.4 (Table A-2a). Analysis The stagnation properties remain constant throughout the nozzle since the flow is isentropic. They are determined from V2 (180 m/s) 2 ⎛ 1 kJ/kg ⎞ T0 = Ti + i = 400 K + ⎜ ⎟ = 416.1 K 2c p 2 × 1.005 kJ/kg ⋅ K ⎝ 1000 m 2 / s 2 ⎠ and
⎛P Te = T0 ⎜⎜ e ⎝ P0
Temperature
⎞ ⎟ ⎟ ⎠
( k −1) / k
⎛ Pe ⎞ = (416.1 K)⎜ ⎟ ⎝ 1033.3 ⎠
c e = kRT
e
Pe
Pb
0.4 / 1.4
Ve
⎛ 1000 m 2 / s 2 = (1.4)(0.287 kJ/kg ⋅ K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ ⎟ ⎠
⎞ ⎟ ⎟ ⎠
Ma e = Ve / c e P Pe ρe = e = RTe (0.287 kPa ⋅ m 3 / kg ⋅ K )Te
Mach number Density Mass flow rate
e
180 m/s
⎛ 1000 m 2 /s 2 Velocity V = 2c p (T0 − Te ) = 2(1.005 kJ/kg ⋅ K)(416.1 − Te )⎜ ⎜ 1 kJ/kg ⎝
Speed of sound
Air
i
k /( k −1)
1.4 /(1.4 −1) ⎛T ⎞ 416.1 K ⎞ = (900 kPa)⎛⎜ = 1033.3 kPa P0 = Pi ⎜⎜ 0 ⎟⎟ ⎟ ⎝ 400 K ⎠ ⎝ Ti ⎠ The critical pressure is determined to be k /( k −1) 1.4 / 0.4 2 ⎞ 2 ⎞ = (1033.3 kPa)⎛⎜ = 545.9 kPa P* = P0 ⎛⎜ ⎟ ⎟ ⎝ k + 1⎠ ⎝ 1.4 + 1 ⎠ Then the pressure at the exit plane (throat) will be Pe = Pb for Pb ≥ 545.9 kPa Pe = P* = 545.9 kPa for Pb < 545.9 kPa (choked flow) Thus the back pressure will not affect the flow when 100 < Pb < 545.9 kPa. For a specified exit pressure Pe, the temperature, the velocity and the mass flow rate can be determined from
c
Pb
& m & max m
m& = ρ eV e Ae = ρ eVe (0.001 m 2 ) 100
Pb 900 kPa
475.5
The results of the calculations can be tabulated as Pb, kPa
Pb, P0
Pe, kPa
Pb, P0
Te, K
Ve, m/s
900 800 700 600 545.9 500 400 300 200 100
0.871 0.774 0.677 0.581 0.528 0.484 0.387 0.290 0.194 0.097
900 800 700 600 545.9 545.9 545.9 545.9 545.9 545.9
0.871 0.774 0.677 0.581 0.528 0.528 0.528 0.528 0.528 0.528
400.0 386.8 372.3 356.2 333.3 333.2 333.3 333.3 333.3 333.3
180.0 162.9 236.0 296.7 366.2 366.2 366.2 366.2 366.2 366.2
Ma 0.45 0.41 0.61 0.78 1.00 1.00 1.00 1.00 1.00 1.00
ρe, kg/m3
& kg / s m,
7.840 7.206 6.551 5.869 4.971 4.971 4.971 4.971 4.971 4.971
0 1.174 1.546 1.741 1.820 1.820 1.820 1.820 1.820 1.820
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-74
17-131 EES Steam enters a converging nozzle. The exit pressure, the exit velocity, and the mass flow rate versus the back pressure for a specified back pressure range are to be plotted. Assumptions 1 Steam is to be treated as an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The nozzle is adiabatic. Properties The ideal gas properties of steam are given to be R = 0.462 kJ/kg.K, cp = 1.872 kJ/kg.K, and k = 1.3. Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Since the flow is isentropic, they remain constant throughout the nozzle, P0 = Pi = 6 MPa T0 = Ti = 700 K The critical pressure is determined from to be 2 ⎞ P* = P0 ⎛⎜ ⎟ ⎝ k + 1⎠
k /( k −1)
2 ⎞ = (6 MPa)⎛⎜ ⎟ ⎝ 1.3 + 1 ⎠
1.3 / 0.3
= 3.274 MPa
for
Pe = P* = 3.274 MPa
Pb ≥ 3.274 MPa
Pb < 3.274 MPa (choked flow)
for
Thus the back pressure will not affect the flow when 3 < Pb < 3.274 MPa. For a specified exit pressure Pe, the temperature, the velocity and the mass flow rate can be determined from Temperature
⎛P Te = T0 ⎜⎜ e ⎝ P0
⎞ ⎟ ⎟ ⎠
( k −1) / k
⎛P ⎞ = (700 K)⎜ e ⎟ ⎝ 6 ⎠
Pb Ve
0.3 / 1.3
C
⎛ 1000 m 2 /s 2 Velocity V = 2c p (T0 − Te ) = 2(1.872 kJ/kg ⋅ K)(700 − Te )⎜⎜ ⎝ 1 kJ/kg
ρe =
Density Mass flow rate
e
Pe
Then the pressure at the exit plane (throat) will be Pe = Pb
STEAM
i Vi ≈ 0
⎞ ⎟ ⎟ ⎠
Pb
& m
Pe Pe = RTe (0.462 kPa ⋅ m 3 / kg ⋅ K )Te
& max m
2
m& = ρ eV e Ae = ρ eVe (0.0008 m )
The results of the calculations can be tabulated as follows: 3
Pb, MPa 6.0 5.5 5.0 4.5 4.0 3.5 3.274 3.0
Pe, MPa 6.0 5.5 5.0 4.5 4.0 3.5 3.274 3.274
Te, K 700 686.1 671.2 655.0 637.5 618.1 608.7 608.7
Ve, m/s 0 228.1 328.4 410.5 483.7 553.7 584.7 584.7
ρe, kg/m3
& kg / s m,
18.55 17.35 16.12 14.87 13.58 12.26 11.64 11.64
0 3.166 4.235 4.883 5.255 5.431 5.445 5.445
3.274
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
6
Pb MPa
17-75
17-132 An expression for the ratio of the stagnation pressure after a shock wave to the static pressure before the shock wave as a function of k and the Mach number upstream of the shock wave is to be found. Analysis The relation between P1 and P2 is ⎛ 1 + kMa 12 P2 1 + kMa 22 ⎜ P P = ⎯ ⎯→ = 2 1 ⎜ 1 + kMa 2 P1 1 + kMa 12 2 ⎝
⎞ ⎟ ⎟ ⎠
Substituting this into the isentropic relation
(
P02 = 1 + (k − 1)Ma 22 / 2 P1
)k /(k −1)
Then, P02 ⎛ 1 + kMa 12 =⎜ P1 ⎜⎝ 1 + kMa 22
⎞ ⎟ 1 + (k − 1)Ma 22 / 2 ⎟ ⎠
(
)k /(k −1)
where Ma 22 =
Ma 12 + 2 /( k − 1) 2kMa 22 /( k − 1) − 1
Substituting, P02 ⎛ (1 + kMa 12 )(2kMa 12 − k + 1) ⎞⎛ (k − 1)Ma 12 / 2 + 1 ⎞ ⎟ ⎟⎜1 + = ⎜⎜ ⎟⎜ 2kMa 2 /( k − 1) − 1 ⎟ P1 ⎝ kMa 12 (k + 1) − k + 3 1 ⎠ ⎠⎝
k /( k −1)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-76
17-133 Nitrogen entering a converging-diverging nozzle experiences a normal shock. The pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock are to be determined. The results are to be compared to those of air under the same conditions. Assumptions 1 Nitrogen is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The nozzle is adiabatic. Properties The properties of nitrogen are R = 0.2968 kJ/kg⋅K and k = 1.4 (Table A-2a). Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Assuming the flow before the shock to be isentropic, P01 = Pi = 700 kPa T01 = Ti = 300 K
shock wave
Then, ⎛ 2 T1 = T01 ⎜ ⎜ 2 + (k − 1)Ma 2 1 ⎝
⎞ ⎛ 2 ⎟ = (300 K)⎜ ⎜ 2 + (1.4 - 1)3 2 ⎟ ⎝ ⎠
⎞ ⎟ = 107.1 K ⎟ ⎠
i
N2
1
2
Vi ≈ 0 Ma1 = 3.0
and ⎛T ⎞ P1 = P01 ⎜⎜ 1 ⎟⎟ ⎝ T01 ⎠
k /( k −1)
⎛ 107.1 ⎞ = (700 kPa)⎜ ⎟ ⎝ 300 ⎠
1.4 / 0.4
= 19.06 kPa
The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A-14. For Ma1 = 3.0 we read Ma 2 = 0.4752,
P02 P T = 0.32834, 2 = 10.333, and 2 = 2.679 P01 P1 T1
Then the stagnation pressure P02, static pressure P2, and static temperature T2, are determined to be P02 = 0.32834 P01 = (0.32834 )(700 kPa) = 230 kPa P2 = 10.333P1 = (10.333)(19.06 kPa) = 197 kPa T2 = 2.679T1 = (2.679)(107.1 K) = 287 K
The velocity after the shock can be determined from V2 = Ma2c2, where c2 is the speed of sound at the exit conditions after the shock, ⎛ 1000 m 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (0.4752) (1.4)(0.2968 kJ/kg ⋅ K)(287 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 164 m/s ⎟ ⎠
Discussion For air at specified conditions k = 1.4 (same as nitrogen) and R = 0.287 kJ/kg·K. Thus the only quantity which will be different in the case of air is the velocity after the normal shock, which happens to be 161.3 m/s.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-77
17-134 The diffuser of an aircraft is considered. The static pressure rise across the diffuser and the exit area are to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the diffuser is steady, one-dimensional, and isentropic. 3 The diffuser is adiabatic. Properties Air properties at room temperature are R = 0.287 kJ/kg⋅K, cp = 1.005 kJ/kg·K, and k = 1.4 (Table A-2a). Analysis The inlet velocity is ⎛ 1000 m 2 / s 2 V1 = Ma 1c1 = M 1 kRT1 = (0.8) (1.4)(0.287 kJ/kg ⋅ K)(242.7 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 249.8 m/s ⎟ ⎠
Then the stagnation temperature and pressure at the diffuser inlet become T01 = T1 +
V1 2 (249.8 m/s) 2 ⎛ 1 kJ/kg = 242.7 + ⎜ 2c p 2(1.005 kJ/kg ⋅ K) ⎝ 1000 m 2 / s 2
⎛T ⎞ P01 = P1 ⎜⎜ 01 ⎟⎟ ⎝ T1 ⎠
k /( k −1)
273.7 K ⎞ = (41.1 kPa)⎛⎜ ⎟ ⎝ 242.7 K ⎠
1.4 /(1.4 −1)
⎞ ⎟ = 273.7 K ⎠
1 Ma1 = 0.8
AIR Diffuser
2 Ma2 = 0.3
= 62.6 kPa
For an adiabatic diffuser, the energy equation reduces to h01 = h02. Noting that h = cpT and the specific heats are assumed to be constant, we have T01 = T02 = T0 = 273.7 K
The isentropic relation between states 1 and 02 gives ⎛T P02 = P02 = P1 ⎜⎜ 02 ⎝ T1
⎞ ⎟⎟ ⎠
k /( k −1)
⎛ 273.72 K ⎞ = (41.1 kPa)⎜ ⎟ ⎝ 242.7 K ⎠
1.4 /(1.4 −1)
= 62.61 kPa
The exit velocity can be expressed as ⎛ 1000 m 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (0.3) (1.4)(0.287 kJ/kg ⋅ K) T2 ⎜⎜ ⎝ 1 kJ/kg
Thus
T2 = T02 −
⎞ ⎟ = 6.01 T2 ⎟ ⎠
V2 2 6.012 T2 m 2 /s 2 ⎛ 1 kJ/kg ⎞ = (273.7) − ⎜ ⎟ = 268.9 K 2(1.005 kJ/kg ⋅ K) ⎝ 1000 m 2 / s 2 ⎠ 2c p
Then the static exit pressure becomes ⎛T P2 = P02 ⎜⎜ 2 ⎝ T02
⎞ ⎟ ⎟ ⎠
k /( k −1)
⎛ 268.9 K ⎞ = (62.61 kPa)⎜ ⎟ ⎝ 273.7 K ⎠
1.4 /(1.4 −1)
= 58.85 kPa
Thus the static pressure rise across the diffuser is ∆P = P2 − P1 = 58.85 − 41.1 = 17.8 kPa
Also,
ρ2 =
P2 58.85 kPa = = 0.7626 kg/m 3 3 RT2 (0.287 kPa ⋅ m /kg ⋅ K)(268.9 K)
V 2 = 6.01 T2 = 6.01 268.9 = 98.6 m/s
Thus
A2 =
m&
ρ 2V 2
=
65 kg/s (0.7626 kg/m 3 )(98.6 m/s)
= 0.864 m 2
Discussion The pressure rise in actual diffusers will be lower because of the irreversibilities. However, flow through well-designed diffusers is very nearly isentropic.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-78
17-135 Helium gas is accelerated in a nozzle isentropically. For a specified mass flow rate, the throat and exit areas of the nozzle are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The nozzle is adiabatic. Properties The properties of helium are R = 2.0769 kJ/kg.K, cp = 5.1926 kJ/kg.K, k = 1.667 (Table A-2a). Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible, T01 = T1 = 500 K P01 = P1 = 1.0 MPa
The flow is assumed to be isentropic, thus the stagnation temperature and pressure remain constant throughout the nozzle, T02 = T01 = 500 K
He
1
*
2
Vi ≈ 0
P02 = P01 = 1.0 MPa
The critical pressure and temperature are determined from 2 ⎞ 2 ⎛ ⎞ = 375.0 K T * = T0 ⎛⎜ ⎟ = (500 K)⎜ ⎟ ⎝ k + 1⎠ ⎝ 1.667 + 1 ⎠ 2 ⎞ P* = P0 ⎛⎜ ⎟ k ⎝ + 1⎠
ρ* =
k /( k −1)
2 ⎞ = (1.0 MPa)⎛⎜ ⎟ 1.667 + 1 ⎝ ⎠
1.667 /(1.667 −1)
= 0.487 MPa
487 kPa P* = = 0.625 kg/m 3 RT * (2.0769 kPa ⋅ m 3 /kg ⋅ K)(375 K)
⎛ 1000 m 2 / s 2 V * = c* = kRT * = (1.667)(2.0769 kJ/kg ⋅ K)(375 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 1139.4 m/s ⎟ ⎠
Thus the throat area is 0.25 kg/s m& A* = = = 3.51× 10 − 4 m 2 = 3.51 cm 2 ρ * V * (0.625 kg/m 3 )(1139.4 m/s) At the nozzle exit the pressure is P2 = 0.1 MPa. Then the other properties at the nozzle exit are determined to be P0 ⎛ k − 1 ⎞ Ma 22 ⎟ = ⎜1 + P2 ⎝ 2 ⎠
k /( k −1)
⎯ ⎯→
1.0 MPa ⎛ 1.667 − 1 ⎞ Ma 22 ⎟ = ⎜1 + 2 0.1 MPa ⎝ ⎠
1.667 / 0.667
It yields Ma2 = 2.130, which is greater than 1. Therefore, the nozzle must be converging-diverging. ⎛ 2 T2 = T0 ⎜⎜ + k − 2 ( 1)Ma 22 ⎝
ρ2 =
⎞ ⎛ 2 ⎟ = (500 K )⎜ ⎜ 2 + (1.667 − 1) × 2.13 2 ⎟ ⎝ ⎠
⎞ ⎟ = 199.0 K ⎟ ⎠
P2 100 kPa = = 0.242 kg/m 3 RT2 (2.0769 kPa ⋅ m 3 /kg ⋅ K)(199 K)
⎛ 1000 m 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (2.13) (1.667)(2.0769 kJ/kg ⋅ K)(199 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 1768.0 m/s ⎟ ⎠
Thus the exit area is 0.25 kg/s m& A2 = = = 5.84 × 10 − 4 m 2 = 5.84 cm 2 ρ 2V 2 (0.242 kg/m 3 )(1768 m/s) Discussion Flow areas in actual nozzles would be somewhat larger to accommodate the irreversibilities.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-79
17-136E Helium gas is accelerated in a nozzle. For a specified mass flow rate, the throat and exit areas of the nozzle are to be determined for the cases of isentropic and 97% efficient nozzles. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The nozzle is adiabatic. Properties The properties of helium are R = 0.4961 Btu/lbm·R = 2.6809 psia·ft3/lbm·R, cp = 1.25 Btu/lbm·R, and k = 1.667 (Table A-2Ea). Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible, T01 = T1 = 900 R P01 = P1 = 150 psia
The flow is assumed to be isentropic, thus the stagnation temperature and pressure remain constant throughout the nozzle,
1
He
*
2
Vi ≈ 0
T02 = T01 = 900 R P02 = P01 = 150 psia
The critical pressure and temperature are determined from 2 2 ⎞ ⎛ ⎞ = 674.9 R T * = T0 ⎛⎜ ⎟ = (900 R)⎜ ⎟ ⎝ k + 1⎠ ⎝ 1.667 + 1 ⎠ 2 ⎞ P* = P0 ⎛⎜ ⎟ ⎝ k + 1⎠
ρ* =
k /( k −1)
2 ⎞ = (150 psia)⎛⎜ ⎟ ⎝ 1.667 + 1 ⎠
1.667 /(1.667 −1)
= 73.1 psia
73.1 psia P* = = 0.0404 1bm/ft 3 RT * (2.6809 psia ⋅ ft 3 /lbm ⋅ R)(674.9 R)
⎛ 25,037 ft 2 / s 2 V * = c* = kRT * = (1.667)(0.4961 Btu/lbm ⋅ R)(674.9 R)⎜⎜ ⎝ 1 Btu/1bm
and
A* =
⎞ ⎟ = 3738 ft/s ⎟ ⎠
m& 0.2 1bm/s = = 0.00132 ft 2 ρ * V * (0.0404 1bm/ft 3 )(3738 ft/s)
At the nozzle exit the pressure is P2 = 15 psia. Then the other properties at the nozzle exit are determined to be k /( k −1) 1.667 / 0.667 p0 ⎛ k − 1 150 psia ⎛ 1.667 − 1 ⎞ ⎞ Ma 22 ⎟ = ⎜1 + Ma 22 ⎟ ⎯ ⎯→ = ⎜1 + 2 p2 ⎝ 2 15 psia ⎝ ⎠ ⎠
It yields Ma2 = 2.130, which is greater than 1. Therefore, the nozzle must be converging-diverging. ⎛ 2 T2 = T0 ⎜⎜ + k − 2 ( 1)Ma 22 ⎝
ρ2 =
⎞ ⎛ 2 ⎟ = (900 R )⎜ ⎜ 2 + (1.667 − 1) × 2.13 2 ⎟ ⎝ ⎠
⎞ ⎟ = 358.1 R ⎟ ⎠
P2 15 psia = = 0.0156 1bm/ft 3 3 RT2 (2.6809 psia ⋅ ft /lbm ⋅ R)(358.1 R)
⎛ 25,037 ft 2 / s 2 V 2 = Ma 2 c 2 = Ma 2 kRT2 = (2.13) (1.667)(0.4961 Btu/lbm ⋅ R)(358.1 R)⎜⎜ ⎝ 1 Btu/1bm
⎞ ⎟ = 5800 ft/s ⎟ ⎠
Thus the exit area is A2 =
m& 0.2 lbm/s = = 0.00221 ft 2 3 ρ 2V 2 (0.0156 lbm/ft )(5800 ft/s)
Discussion Flow areas in actual nozzles would be somewhat larger to accommodate the irreversibilities.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-80
17-137 [Also solved by EES on enclosed CD] Using the compressible flow relations, the one-dimensional compressible flow functions are to be evaluated and tabulated as in Table A-32 for an ideal gas with k = 1.667. Properties The specific heat ratio of the ideal gas is given to be k = 1.667. Analysis The compressible flow functions listed below are expressed in EES and the results are tabulated. Ma * = Ma
k +1 2 + (k − 1)Ma 2
1 ⎡⎛ 2 ⎞⎛ k − 1 ⎞⎤ = Ma 2 ⎟⎥ ⎜ ⎟⎜1 + ⎢ * 2 Ma ⎣⎝ k + 1 ⎠⎝ ⎠⎦ A A
P ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ P0 ⎝ 2 ⎠
0.5( k +1) /( k −1)
− k /( k −1)
−1 /( k −1) ρ ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ ρ0 ⎝ 2 ⎠
T ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ T0 ⎝ 2 ⎠
−1
k=1.667 PP0=(1+(k-1)*M^2/2)^(-k/(k-1)) TT0=1/(1+(k-1)*M^2/2) DD0=(1+(k-1)*M^2/2)^(-1/(k-1)) Mcr=M*SQRT((k+1)/(2+(k-1)*M^2)) AAcr=((2/(k+1))*(1+0.5*(k-1)*M^2))^(0.5*(k+1)/(k-1))/M Ma
Ma*
A/A*
P/P0
T/T0
0 0.1153 0.2294 0.3413 0.4501 0.5547 0.6547 0.7494 0.8386 0.9222 1.0000 1.1390 1.2572 1.3570 1.4411 1.5117 1.5713 1.6216 1.6643 1.7007 1.7318 1.8895 1.9996
∞ 5.6624 2.8879 1.9891 1.5602 1.3203 1.1760 1.0875 1.0351 1.0081 1.0000 1.0267 1.0983 1.2075 1.3519 1.5311 1.7459 1.9980 2.2893 2.6222 2.9990 9.7920 ∞
ρ/ρ0
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 5.0 ∝
1.0000 0.9917 0.9674 0.9288 0.8782 0.8186 0.7532 0.6850 0.6166 0.5501 0.4871 0.3752 0.2845 0.2138 0.1603 0.1202 0.0906 0.0686 0.0524 0.0403 0.0313 0.0038 0
1.0000 0.9950 0.9803 0.9566 0.9250 0.8869 0.8437 0.7970 0.7482 0.6987 0.6495 0.5554 0.4704 0.3964 0.3334 0.2806 0.2368 0.2005 0.1705 0.1457 0.1251 0.0351 0
1.0000 0.9967 0.9868 0.9709 0.9493 0.9230 0.8928 0.8595 0.8241 0.7873 0.7499 0.6756 0.6047 0.5394 0.4806 0.4284 0.3825 0.3424 0.3073 0.2767 0.2499 0.1071 0
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-81
17-138 [Also solved by EES on enclosed CD] Using the normal shock relations, the normal shock functions are to be evaluated and tabulated as in Table A-33 for an ideal gas with k = 1.667. Properties The specific heat ratio of the ideal gas is given to be k = 1.667. Analysis The normal shock relations listed below are expressed in EES and the results are tabulated. Ma 2 =
P2 1 + kMa 12 2kMa 12 − k + 1 = = P1 1 + kMa 22 k +1
(k − 1)Ma 12 + 2 2kMa 12 − k + 1
T2 2 + Ma 12 (k − 1) = T1 2 + Ma 22 (k − 1)
ρ 2 P2 / P1 (k + 1)Ma 12 V = = = 1 , ρ 1 T2 / T1 2 + (k − 1)Ma 12 V 2 k +1
P02 Ma 1 ⎡1 + Ma 22 (k − 1) / 2 ⎤ 2( k −1) = ⎢ ⎥ P01 Ma 2 ⎢⎣1 + Ma 12 (k − 1) / 2 ⎦⎥
P02 (1 + kMa 12 )[1 + Ma 22 (k − 1) / 2] k /( k −1) = P1 1 + kMa 22
k=1.667 My=SQRT((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) PyPx=(1+k*Mx^2)/(1+k*My^2) TyTx=(1+Mx^2*(k-1)/2)/(1+My^2*(k-1)/2) RyRx=PyPx/TyTx P0yP0x=(Mx/My)*((1+My^2*(k-1)/2)/(1+Mx^2*(k-1)/2))^(0.5*(k+1)/(k-1)) P0yPx=(1+k*Mx^2)*(1+My^2*(k-1)/2)^(k/(k-1))/(1+k*My^2) Ma1 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 4.0 5.0 ∞
Ma2 1.0000 0.9131 0.8462 0.7934 0.7508 0.7157 0.6864 0.6618 0.6407 0.6227 0.6070 0.5933 0.5814 0.5708 0.5614 0.5530 0.5455 0.5388 0.5327 0.5273 0.5223 0.4905 0.4753 0.4473
P2/P1 1.0000 1.2625 1.5500 1.8626 2.2001 2.5626 2.9501 3.3627 3.8002 4.2627 4.7503 5.2628 5.8004 6.3629 6.9504 7.5630 8.2005 8.8631 9.5506 10.2632 11.0007 19.7514 31.0022 ∞
ρ2/ρ1 1.0000 1.1496 1.2972 1.4413 1.5805 1.7141 1.8415 1.9624 2.0766 2.1842 2.2853 2.3802 2.4689 2.5520 2.6296 2.7021 2.7699 2.8332 2.8923 2.9476 2.9993 3.3674 3.5703 3.9985
T2/T1 1.0000 1.0982 1.1949 1.2923 1.3920 1.4950 1.6020 1.7135 1.8300 1.9516 2.0786 2.2111 2.3493 2.4933 2.6432 2.7989 2.9606 3.1283 3.3021 3.4819 3.6678 5.8654 8.6834 ∞
P02/P01 1 0.999 0.9933 0.9813 0.9626 0.938 0.9085 0.8752 0.8392 0.8016 0.763 0.7243 0.6861 0.6486 0.6124 0.5775 0.5442 0.5125 0.4824 0.4541 0.4274 0.2374 0.1398 0
P02/P1 2.0530 2.3308 2.6473 2.9990 3.3838 3.8007 4.2488 4.7278 5.2371 5.7767 6.3462 6.9457 7.5749 8.2339 8.9225 9.6407 10.3885 11.1659 11.9728 12.8091 13.6750 23.9530 37.1723 ∞
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-82
17-139 The critical temperature, pressure, and density of an equimolar mixture of oxygen and nitrogen for specified stagnation properties are to be determined. Assumptions Both oxygen and nitrogen are ideal gases with constant specific heats at room temperature. Properties The specific heat ratio and molar mass are k = 1.395 and M = 32 kg/kmol for oxygen, and k = 1.4 and M = 28 kg/kmol for nitrogen (Tables A-1 and A-2). Analysis The gas constant of the mixture is M m = y O 2 M O 2 + y N 2 M N 2 = 0.5 × 32 + 0.5 × 28 = 30 kg/kmol Rm =
8.314 kJ/kmol ⋅ K Ru = = 0.2771 kJ/kg ⋅ K 30 kg/kmol Mm
The specific heat ratio is 1.4 for nitrogen, and nearly 1.4 for oxygen. Therefore, the specific heat ratio of the mixture is also 1.4. Then the critical temperature, pressure, and density of the mixture become 2 ⎞ ⎛ 2 ⎞ = 667 K T * = T0 ⎛⎜ ⎟ ⎟ = (800 K)⎜ k + 1 ⎝ 1.4 + 1 ⎠ ⎝ ⎠ 2 ⎞ P* = P0 ⎛⎜ ⎟ ⎝ k + 1⎠
ρ* =
k /( k −1)
2 ⎞ = (500 kPa)⎛⎜ ⎟ ⎝ 1.4 + 1 ⎠
1.4 /(1.4 −1)
= 264 kPa
264 kPa P* = = 1.43 kg/m 3 RT * (0.2771 kPa ⋅ m 3 /kg ⋅ K)(667 K)
Discussion If the specific heat ratios k of the two gases were different, then we would need to determine the k of the mixture from k = cp,m/cv,m where the specific heats of the mixture are determined from c p ,m = mf O 2 c p ,O 2 + mf N 2 c p , N 2 = ( y O 2 M O 2 / M m )c p ,O 2 + ( y N 2 M N 2 / M m )c p , N 2 cv , m = mf O 2 cv ,O 2 + mf N 2 cv , N 2 = ( y O 2 M O 2 / M m )cv ,O 2 + ( y N 2 M N 2 / M m )cv , N 2
where mf is the mass fraction and y is the mole fraction. In this case it would give c p ,m = (0.5 × 32 / 30) × 0.918 + (0.5 × 28 / 30) × 1.039 = 0.974 kJ/kg.K c p ,m = (0.5 × 32 / 30) × 0.658 + (0.5 × 28 / 30) × 0.743 = 0.698 kJ/kg.K
and
k = 0.974/0.698 = 1.40
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-83
17-140 EES Using EES (or other) software, the shape of a converging-diverging nozzle is to be determined for specified flow rate and stagnation conditions. The nozzle and the Mach number are to be plotted. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, onedimensional, and isentropic. 3 The nozzle is adiabatic. Properties The specific heat ratio of air at room temperature is 1.4 (Table A-2a). Analysis The problems is solved using EES, and the results are tabulated and plotted below. k=1.4 Cp=1.005 "kJ/kg.K" R=0.287 "kJ/kg.K" P0=1400 "kPa" T0=200+273 "K" m=3 "kg/s" rho_0=P0/(R*T0) rho=P/(R*T) T=T0*(P/P0)^((k-1)/k) V=SQRT(2*Cp*(T0-T)*1000) A=m/(rho*V)*10000 "cm2" C=SQRT(k*R*T*1000) Ma=V/C
2.5
2
Ma
1.5
1
0.5
Mach number Ma
1400 1350 1300 1250 1200 1150 1100 1050 1000 950 900 850 800 750 700 650 600 550 500 450 400 350 300 250 200 150 100
∞ 30.1 21.7 18.1 16.0 14.7 13.7 13.0 12.5 12.2 11.9 11.7 11.6 11.5 11.5 11.6 11.8 12.0 12.3 12.8 13.3 14.0 15.0 16.4 18.3 21.4 27.0
0 0.229 0.327 0.406 0.475 0.538 0.597 0.655 0.710 0.766 0.820 0.876 0.931 0.988 1.047 1.107 1.171 1.237 1.308 1.384 1.467 1.559 1.663 1.784 1.929 2.114 2.373
200
400
600
800
1000
1200
1400
P, kPa
50 45 40 35 2
Flow area A, cm2
Flow area A, cm
Pressure P, kPa
0 0
30 25 20 15 10 0
200
400
600
800
1000
1200
P, kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
1400
17-84
17-141 EES Using the compressible flow relations, the one-dimensional compressible flow functions are to be evaluated and tabulated as in Table A-32 for air. Properties The specific heat ratio is given to be k = 1.4 for air Analysis The compressible flow functions listed below are expressed in EES and the results are tabulated. Ma * = Ma A A*
=
1 Ma
k +1 2 + (k − 1)Ma 2
⎡⎛ 2 ⎞⎛ k − 1 2 ⎞⎤ ⎢⎜ k + 1 ⎟⎜1 + 2 Ma ⎟⎥ ⎠⎝ ⎠⎦ ⎣⎝
P ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ P0 ⎝ 2 ⎠
0.5( k +1) /( k −1)
− k /( k −1)
−1 /( k −1) ρ ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ ρ0 ⎝ 2 ⎠
T ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ T0 ⎝ 2 ⎠
−1
Air: k=1.4 PP0=(1+(k-1)*M^2/2)^(-k/(k-1)) TT0=1/(1+(k-1)*M^2/2) DD0=(1+(k-1)*M^2/2)^(-1/(k-1)) Mcr=M*SQRT((k+1)/(2+(k-1)*M^2)) AAcr=((2/(k+1))*(1+0.5*(k-1)*M^2))^(0.5*(k+1)/(k-1))/M Ma 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0
Ma* 1.0000 1.3646 1.6330 1.8257 1.9640 2.0642 2.1381 2.1936 2.2361 2.2691 2.2953 2.3163 2.3333 2.3474 2.3591 2.3689 2.3772 2.3843 2.3905
A/A* 1.0000 1.1762 1.6875 2.6367 4.2346 6.7896 10.7188 16.5622 25.0000 36.8690 53.1798 75.1343 104.1429 141.8415 190.1094 251.0862 327.1893 421.1314 535.9375
P/P0 0.5283 0.2724 0.1278 0.0585 0.0272 0.0131 0.0066 0.0035 0.0019 0.0011 0.0006 0.0004 0.0002 0.0002 0.0001 0.0001 0.0000 0.0000 0.0000
ρ/ρ0 0.6339 0.3950 0.2300 0.1317 0.0762 0.0452 0.0277 0.0174 0.0113 0.0076 0.0052 0.0036 0.0026 0.0019 0.0014 0.0011 0.0008 0.0006 0.0005
T/T0 0.8333 0.6897 0.5556 0.4444 0.3571 0.2899 0.2381 0.1980 0.1667 0.1418 0.1220 0.1058 0.0926 0.0816 0.0725 0.0647 0.0581 0.0525 0.0476
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-85
17-142 EES Using the compressible flow relations, the one-dimensional compressible flow functions are to be evaluated and tabulated as in Table A-32 for methane. Properties The specific heat ratio is given to be k = 1.3 for methane. Analysis The compressible flow functions listed below are expressed in EES and the results are tabulated. Ma * = Ma A A*
=
1 Ma
k +1 2 + (k − 1)Ma 2
⎡⎛ 2 ⎞⎛ k − 1 2 ⎞⎤ ⎢⎜ k + 1 ⎟⎜1 + 2 Ma ⎟⎥ ⎠⎝ ⎠⎦ ⎣⎝
P ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ P0 ⎝ 2 ⎠
0.5( k +1) /( k −1)
− k /( k −1)
−1 /( k −1) ρ ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ ρ0 ⎝ 2 ⎠
T ⎛ k −1 ⎞ = ⎜1 + Ma 2 ⎟ T0 ⎝ 2 ⎠
−1
Methane: k=1.3 PP0=(1+(k-1)*M^2/2)^(-k/(k-1)) TT0=1/(1+(k-1)*M^2/2) DD0=(1+(k-1)*M^2/2)^(-1/(k-1)) Mcr=M*SQRT((k+1)/(2+(k-1)*M^2)) AAcr=((2/(k+1))*(1+0.5*(k-1)*M^2))^(0.5*(k+1)/(k-1))/M Ma 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0
Ma* 1.0000 1.3909 1.6956 1.9261 2.0986 2.2282 2.3263 2.4016 2.4602 2.5064 2.5434 2.5733 2.5978 2.6181 2.6350 2.6493 2.6615 2.6719 2.6810
A/A* 1.0000 1.1895 1.7732 2.9545 5.1598 9.1098 15.9441 27.3870 45.9565 75.2197 120.0965 187.2173 285.3372 425.8095 623.1235 895.5077 1265.6040 1761.2133 2416.1184
P/P0 0.5457 0.2836 0.1305 0.0569 0.0247 0.0109 0.0050 0.0024 0.0012 0.0006 0.0003 0.0002 0.0001 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000
ρ/ρ0 0.6276 0.3793 0.2087 0.1103 0.0580 0.0309 0.0169 0.0095 0.0056 0.0033 0.0021 0.0013 0.0008 0.0006 0.0004 0.0003 0.0002 0.0001 0.0001
T/T0 0.8696 0.7477 0.6250 0.5161 0.4255 0.3524 0.2941 0.2477 0.2105 0.1806 0.1563 0.1363 0.1198 0.1060 0.0943 0.0845 0.0760 0.0688 0.0625
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-86
17-143 EES Using the normal shock relations, the normal shock functions are to be evaluated and tabulated as in Table A-33 for air. Properties The specific heat ratio is given to be k = 1.4 for air. Analysis The normal shock relations listed below are expressed in EES and the results are tabulated. Ma 2 =
(k − 1)Ma 12 + 2
P2 1 + kMa 12 2kMa 12 − k + 1 = = P1 1 + kMa 22 k +1
2kMa 12 − k + 1
ρ 2 P2 / P1 (k + 1)Ma 12 V = = = 1 , ρ 1 T2 / T1 2 + (k − 1)Ma 12 V 2
T2 2 + Ma 12 (k − 1) = T1 2 + Ma 22 (k − 1) k +1
P02 Ma 1 ⎡1 + Ma 22 (k − 1) / 2 ⎤ 2( k −1) = ⎢ ⎥ P01 Ma 2 ⎢⎣1 + Ma 12 (k − 1) / 2 ⎦⎥
P02 (1 + kMa 12 )[1 + Ma 22 (k − 1) / 2] k /( k −1) = P1 1 + kMa 22
Air: k=1.4 My=SQRT((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) PyPx=(1+k*Mx^2)/(1+k*My^2) TyTx=(1+Mx^2*(k-1)/2)/(1+My^2*(k-1)/2) RyRx=PyPx/TyTx P0yP0x=(Mx/My)*((1+My^2*(k-1)/2)/(1+Mx^2*(k-1)/2))^(0.5*(k+1)/(k-1)) P0yPx=(1+k*Mx^2)*(1+My^2*(k-1)/2)^(k/(k-1))/(1+k*My^2) Ma1 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0
Ma2 1.0000 0.7011 0.5774 0.5130 0.4752 0.4512 0.4350 0.4236 0.4152 0.4090 0.4042 0.4004 0.3974 0.3949 0.3929 0.3912 0.3898 0.3886 0.3876
P2/P1 1.0000 2.4583 4.5000 7.1250 10.3333 14.1250 18.5000 23.4583 29.0000 35.1250 41.8333 49.1250 57.0000 65.4583 74.5000 84.1250 94.3333 105.1250 116.5000
ρ2/ρ1 1.0000 1.8621 2.6667 3.3333 3.8571 4.2609 4.5714 4.8119 5.0000 5.1489 5.2683 5.3651 5.4444 5.5102 5.5652 5.6117 5.6512 5.6850 5.7143
T2/T1 1.0000 1.3202 1.6875 2.1375 2.6790 3.3151 4.0469 4.8751 5.8000 6.8218 7.9406 9.1564 10.4694 11.8795 13.3867 14.9911 16.6927 18.4915 20.3875
P02/P01 1 0.9298 0.7209 0.499 0.3283 0.2129 0.1388 0.0917 0.06172 0.04236 0.02965 0.02115 0.01535 0.01133 0.008488 0.006449 0.004964 0.003866 0.003045
P02/P1 1.8929 3.4133 5.6404 8.5261 12.0610 16.2420 21.0681 26.5387 32.6535 39.4124 46.8152 54.8620 63.5526 72.8871 82.8655 93.4876 104.7536 116.6634 129.2170
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-87
17-144 EES Using the normal shock relations, the normal shock functions are to be evaluated and tabulated as in Table A-14 for methane. Properties The specific heat ratio is given to be k = 1.3 for methane. Analysis The normal shock relations listed below are expressed in EES and the results are tabulated. Ma 2 =
(k − 1)Ma 12 + 2
P2 1 + kMa 12 2kMa 12 − k + 1 = = P1 1 + kMa 22 k +1
2kMa 12 − k + 1
ρ 2 P2 / P1 (k + 1)Ma 12 V = = = 1 , ρ 1 T2 / T1 2 + (k − 1)Ma 12 V 2
T2 2 + Ma 12 (k − 1) = T1 2 + Ma 22 (k − 1) k +1
P02 Ma 1 ⎡1 + Ma 22 (k − 1) / 2 ⎤ 2( k −1) = ⎢ ⎥ P01 Ma 2 ⎢⎣1 + Ma 12 (k − 1) / 2 ⎦⎥
P02 (1 + kMa 12 )[1 + Ma 22 (k − 1) / 2] k /( k −1) = P1 1 + kMa 22
Methane: k=1.3 My=SQRT((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) PyPx=(1+k*Mx^2)/(1+k*My^2) TyTx=(1+Mx^2*(k-1)/2)/(1+My^2*(k-1)/2) RyRx=PyPx/TyTx P0yP0x=(Mx/My)*((1+My^2*(k-1)/2)/(1+Mx^2*(k-1)/2))^(0.5*(k+1)/(k-1)) P0yPx=(1+k*Mx^2)*(1+My^2*(k-1)/2)^(k/(k-1))/(1+k*My^2) Ma1 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0
Ma2 1.0000 0.6942 0.5629 0.4929 0.4511 0.4241 0.4058 0.3927 0.3832 0.3760 0.3704 0.3660 0.3625 0.3596 0.3573 0.3553 0.3536 0.3522 0.3510
P2/P1 1.0000 2.4130 4.3913 6.9348 10.0435 13.7174 17.9565 22.7609 28.1304 34.0652 40.5652 47.6304 55.2609 63.4565 72.2174 81.5435 91.4348 101.8913 112.9130
ρ2/ρ1 1.0000 1.9346 2.8750 3.7097 4.4043 4.9648 5.4118 5.7678 6.0526 6.2822 6.4688 6.6218 6.7485 6.8543 6.9434 7.0190 7.0837 7.1393 7.1875
T2/T1 1.0000 1.2473 1.5274 1.8694 2.2804 2.7630 3.3181 3.9462 4.6476 5.4225 6.2710 7.1930 8.1886 9.2579 10.4009 11.6175 12.9079 14.2719 15.7096
P02/P01 1 0.9261 0.7006 0.461 0.2822 0.1677 0.09933 0.05939 0.03613 0.02243 0.01422 0.009218 0.006098 0.004114 0.002827 0.001977 0.001404 0.001012 0.000740
P02/P1 1.8324 3.2654 5.3700 8.0983 11.4409 15.3948 19.9589 25.1325 30.9155 37.3076 44.3087 51.9188 60.1379 68.9658 78.4027 88.4485 99.1032 110.367 122.239
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-88
17-145 Air flowing at a supersonic velocity in a duct is accelerated by cooling. For a specified exit Mach number, the rate of heat transfer is to be determined. Assumptions The assumptions associated with Rayleigh Q& flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant crossP01 = 240 kPa sectional area duct with negligible frictional effects) are Ma2 = 2 T01 = 350 K valid. Ma1 = 1.2 Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg⋅K, and R = 0.287 kJ/kg⋅K (Table A-2a). Analysis Knowing stagnation properties, the static properties are determined to be ⎛ k −1 ⎞ T1 = T01 ⎜1 + Ma 12 ⎟ 2 ⎝ ⎠
−1
⎛ k −1 ⎞ P1 = P01 ⎜1 + Ma 12 ⎟ 2 ⎝ ⎠
− k /( k −1)
ρ1 =
⎛ 1.4 - 1 2 ⎞ 1.2 ⎟ = (350 K)⎜1 + 2 ⎠ ⎝
−1
= 271.7 K
⎛ 1.4 - 1 2 ⎞ 1.2 ⎟ = (240 kPa)⎜1 + 2 ⎠ ⎝
−1.4 / 0.4
= 98.97 kPa
P1 98.97 kPa = = 1.269 kg/m 3 RT1 (0.287 kJ/kgK)(271.7 K)
Then the inlet velocity and the mass flow rate become ⎛ 1000 m 2 / s 2 c1 = kRT1 = (1.4)(0.287 kJ/kg ⋅ K)(271.7 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 330.4 m/s ⎟ ⎠
V1 = Ma 1c1 = 1.2(330.4 m/s ) = 396.5 m/s m& air = ρ1 Ac1V1 = (1.269 kg/m 3 )[π (0.20 m) 2 / 4](330.4 m/s) = 15.81 kg/s
The Rayleigh flow functions T0/T0* corresponding to the inlet and exit Mach numbers are (Table A-34): Ma1 = 1.8:
T01/T0* = 0.9787
Ma2 = 2:
T02/T0* = 0.7934
Then the exit stagnation temperature is determined to be T0 2 T02 / T0* 0.7934 = = = 0.8107 T0 1 T01 / T0* 0.9787
→ T02 = 0.8107T01 = 0.8107(350 K ) = 283.7 K
Finally, the rate of heat transfer is Q& = m& air c p (T02 − T01 ) = (15.81 kg/s )(1.005 kJ/kg ⋅ K )(283.7 − 350) K = -1053 kW
Discussion The negative sign confirms that the gas needs to be cooled in order to be accelerated. Also, it can be shown that the thermodynamic temperature drops to 158 K at the exit, which is extremely low. Therefore, the duct may need to be heavily insulated to maintain indicated flow conditions.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-89
17-146 Air flowing at a subsonic velocity in a duct is accelerated by heating. The highest rate of heat transfer without affecting the inlet conditions is to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 Inlet conditions (and thus the mass flow rate) remain constant. Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg⋅K, and R = 0.287 kJ/kg⋅K (Table A-2a).
Q&
Analysis Heat transfer will stop when the flow is choked, and thus Ma2 = V2/c2 = 1. The inlet density and stagnation temperature are
P1 = 400 kPa T1 = 360 K
ρ1 =
P1 400 kPa = = 3.872 kg/m 3 RT1 (0.287 kJ/kgK)(360 K)
Ma2 = 1
Ma1 = 0.4
⎛ k −1 ⎞ ⎛ 1.4 - 1 ⎞ T01 = T1 ⎜1 + Ma 12 ⎟ = (360 K)⎜1 + 0.4 2 ⎟ = 371.5 K 2 2 ⎝ ⎠ ⎝ ⎠
Then the inlet velocity and the mass flow rate become ⎛ 1000 m 2 / s 2 c1 = kRT1 = (1.4 )(0.287 kJ/kg ⋅ K)(360 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 380.3 m/s ⎟ ⎠
V1 = Ma 1c1 = 0.4(380.3 m/s ) = 152.1 m/s m& air = ρ 1 Ac1V1 = (3.872 kg/m 3 )(0.1× 0.1 m 2 )(152.1 m/s) = 5.890 kg/s
The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are T02/T0* = 1 (since Ma2 = 1) T01 T0*
=
(k + 1)Ma 12 [2 + (k − 1)Ma 12 ] (1 + kMa 12 ) 2
=
(1.4 + 1)0.4 2 [2 + (1.4 − 1)0.4 2 ] (1 + 1.4 × 0.4 2 ) 2
= 0.5290
Therefore, T0 2 T02 / T0* 1 = = * 0.5290 T0 1 T01 / T0
→ T02 = T01 / 0.5290 = (371.5 K ) / 0.5290 = 702.3 K
Then the rate of heat transfer becomes Q& = m& air c p (T02 − T01 ) = (5.890 kg/s )(1.005 kJ/kg ⋅ K )(702.3 − 371.5) K = 1958 kW
Discussion It can also be shown that T2 = 585 K, which is the highest thermodynamic temperature that can be attained under stated conditions. If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease. We can also solve this problem using the Rayleigh function values listed in Table A-34.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-90
17-147 Helium flowing at a subsonic velocity in a duct is accelerated by heating. The highest rate of heat transfer without affecting the inlet conditions is to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 Inlet conditions (and thus the mass flow rate) remain constant. Properties We take the properties of helium to be k = 1.667, cp = 5.193 kJ/kg⋅K, and R = 2.077 kJ/kg⋅K (Table A-2a). Analysis Heat transfer will stop when the flow is choked, and thus Ma2 = V2/c2 = 1. The inlet density and stagnation temperature are P 400 kPa ρ1 = 1 = = 0.5350 kg/m 3 RT1 (2.077 kJ/kgK)(360 K) ⎛ k −1 ⎞ ⎛ 1.667 - 1 ⎞ T01 = T1 ⎜1 + Ma 12 ⎟ = (360 K)⎜1 + 0.4 2 ⎟ = 379.2 K 2 2 ⎝ ⎠ ⎝ ⎠
Q& P1 = 400 kPa T1 = 360 K
Ma2 = 1
Ma1 = 0.4
Then the inlet velocity and the mass flow rate become ⎛ 1000 m 2 / s 2 c1 = kRT1 = (1.667)(2.077 kJ/kg ⋅ K)(360 K)⎜⎜ ⎝ 1 kJ/kg
⎞ ⎟ = 1116 m/s ⎟ ⎠
V1 = Ma 1c1 = 0.4(1116 m/s ) = 446.6 m/s m& air = ρ1 Ac1V1 = (0.5350 kg/m 3 )(0.1× 0.1 m 2 )(446.6 m/s) = 2.389 kg/s
The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are T02/T0* = 1 (since Ma2 = 1) T01 T0*
=
(k + 1)Ma 12 [2 + (k − 1)Ma 12 ] (1 + kMa 12 ) 2
=
(1.667 + 1)0.4 2 [2 + (1.667 − 1)0.4 2 ] (1 + 1.667 × 0.4 2 ) 2
= 0.5603
Therefore, T0 2 T02 / T0* 1 = = * 0.5603 T0 1 T01 / T0
→ T02 = T01 / 0.5603 = (379.2 K ) / 0.5603 = 676.8 K
Then the rate of heat transfer becomes Q& = m& air c p (T02 − T01 ) = (2.389 kg/s)(5.193 kJ/kg ⋅ K )(676.8 − 379.2) K = 3693 kW
Discussion It can also be shown that T2 = 508 K, which is the highest thermodynamic temperature that can be attained under stated conditions. If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease. Also, in the solution of this problem, we cannot use the values of Table A34 since they are based on k = 1.4.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-91
17-148 Air flowing at a subsonic velocity in a duct is accelerated by heating. For a specified exit Mach number, the heat transfer for a specified exit Mach number as well as the maximum heat transfer are to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 Inlet conditions (and thus the mass flow rate) remain constant. Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg⋅K, and R = 0.287 kJ/kg⋅K (Table A2a). Analysis The inlet Mach number and stagnation temperature are ⎛ 1000 m 2 / s 2 c1 = kRT1 = (1.4)(0.287 kJ/kg ⋅ K)(400 K)⎜⎜ ⎝ 1 kJ/kg Ma 1 =
V1 100 m/s = = 0.2494 c1 400.9 m/s
⎛ k −1 ⎞ T01 = T1⎜1 + Ma12 ⎟ 2 ⎝ ⎠ 1.4 1 ⎛ ⎞ 0.2494 2 ⎟ = (400 K)⎜1 + 2 ⎝ ⎠ = 405.0 K
⎞ ⎟ = 400.9 m/s ⎟ ⎠ q
P1 = 35 kPa T1 = 400 K
Ma2 = 0.8
V1 = 100 m/s
The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-34): Ma1 = 0.2494:
T01/T* = 0.2559
Ma2 = 0.8:
T02/T* = 0.9639
Then the exit stagnation temperature and the heat transfer are determined to be T0 2 T02 / T * 0.9639 = = = 3.7667 → T0 2 = 3.7667T01 = 3.7667(405.0 K ) = 1526 K T0 1 T01 / T * 0.2559 q = c p (T02 − T01 ) = (1.005 kJ/kg ⋅ K )(1526 − 405) K = 1126 kJ/kg
Maximum heat transfer will occur when the flow is choked, and thus Ma2 = 1 and thus T02/T* = 1. Then, T0 2 T02 / T * 1 = = → T0 2 = T01 / 0.2559 = 405.0 K ) / 0.2559 = 1583 K * 0.2559 T0 1 T01 / T q max = c p (T02 − T01 ) = (1.005 kJ/kg ⋅ K )(1583 − 405) K = 1184 kJ/kg
Discussion This is the maximum heat that can be transferred to the gas without affecting the mass flow rate. If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-92
17-149 Air flowing at sonic conditions in a duct is accelerated by cooling. For a specified exit Mach number, the amount of heat transfer per unit mass is to be determined. Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. Properties We take the properties of air to be k = 1.4, cp = 1.005 kJ/kg⋅K, and R = 0.287 kJ/kg⋅K (Table A2a). Analysis Noting that Ma1 = 1, the inlet stagnation temperature is T01
⎛ k −1 ⎞ ⎛ 1.4 - 1 2 ⎞ = T1 ⎜1 + Ma 12 ⎟ = (500 K)⎜1 + 1 ⎟ = 600 K 2 2 ⎝ ⎠ ⎝ ⎠
The Rayleigh flow functions T0/T0* corresponding to the inlet and exit Mach numbers are (Table A-34): Ma1 = 1:
T01/T0* = 1
Ma2 = 1.6:
T02/T0* = 0.8842
P01 = 420 kPa T01 = 500 K
q
Ma2 = 1.6
Ma1 = 1
Then the exit stagnation temperature and heat transfer are determined to be T0 2 T02 / T0* 0.8842 = = = 0.8842 T0 1 T01 / T0* 1
→ T02 = 0.8842T01 = 0.8842(600 K ) = 530.5 K
q = c p (T02 − T01 ) = (1.005 kJ/kg ⋅ K )(530.5 − 600) K = - 69.8 kJ/kg
Discussion The negative sign confirms that the gas needs to be cooled in order to be accelerated. Also, it can be shown that the thermodynamic temperature drops to 351 K at the exit
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-93
17-150 Saturated steam enters a converging-diverging nozzle with a low velocity. The throat area, exit velocity, mass flow rate, and exit Mach number are to be determined for isentropic and 90 percent efficient nozzle cases. Assumptions 1 Flow through the nozzle is steady and one-dimensional. 2 The nozzle is adiabatic. Analysis (a) The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Thus h10 = h1. At the inlet, h1 = (h f + x1 h fg ) @ 3 MPa = 1008.3 + 0.95 × 1794.9 = 2713.4 kJ/kg s1 = ( s f + x1 s fg ) @ 3 MPa = 2.6454 + 0.95 × 3.5402 = 6.0086 kJ/kg ⋅ K
At the exit, P2 = 1.2 MPa and s2 = s2s = s1 = 6.0086 kJ/kg·K. Thus, s2 = s f + x2 s fg → 6.0086 = 2.2159 + x2 (4.3058) → x2 = 0.8808
1 Vi ≈ 0
Steam
h2 = h f + x2 h fg = 798.33 + 0.8808 × 1985.4 = 2547.2 kJ/kg
t
2
a) ηN = 100% b) ηN = 90%
v 2 = v f + x2v fg = 0.001138 + 0.8808 × (0.16326 − 0.001138) = 0.1439 m3 / kg Then the exit velocity is determined from the steady-flow energy balance to be h1 +
V12 V2 V 2 − V12 = h2 + 2 → 0 = h2 − h1 + 2 2 2 2
Solving for V2, ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 576.7 m/s V2 = 2(h1 − h2 ) = 2(2713.4 - 2547.2)kJ/kg⎜ ⎜ 1 kJ/kg ⎟ ⎝ ⎠
The mass flow rate is determined from m& =
1
v2
A2V2 =
1 (16 × 10− 4 m 2 )(576.7 m/s) = 6.41 kg/s 3 0.1439 m / kg
The velocity of sound at the exit of the nozzle is determined from 1/ 2
⎛ ∂P ⎞ c=⎜ ⎟ ⎝ ∂r ⎠ s
1/ 2
⎛ ∆P ⎞ ⎟⎟ ≅ ⎜⎜ ⎝ ∆ (1 / v ) ⎠ s
The specific volume of steam at s2 = 6.0086 kJ/kg·K and at pressures just below and just above the specified pressure (1.1 and 1.3 MPa) are determined to be 0.1555 and 0.1340 m3/kg. Substituting, c2 =
(1300 − 1100) kPa
⎛ 1000 m 2 / s 2 ⎜ 3 ⎜ 1 ⎞ ⎛ 1 3 ⎝ 1 kPa ⋅ m − ⎟ kg/m ⎜ ⎝ 0.1340 0.1555 ⎠
⎞ ⎟ = 440.3 m/s ⎟ ⎠
Then the exit Mach number becomes Ma 2 =
V2 576.7 m/s = = 1.310 c2 440.3 m/s
The steam is saturated, and thus the critical pressure which occurs at the throat is taken to be Pt = P* = 0.576 × P01 = 0.576 × 3 = 1.728 MPa
Then at the throat, Pt = 1.728 MPa and s t = s1 = 6.0086 kJ/kg ⋅ K
Thus,
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-94
ht = 2611.4 kJ/kg v t = 0.1040 m3 / kg
Then the throat velocity is determined from the steady-flow energy balance, Ê0
h1 +
V12 2
= ht +
Vt 2 V2 → 0 = ht − h1 + t 2 2
Solving for Vt, ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 451.7 m/s Vt = 2(h1 − ht ) = 2(2713.4 − 2611.4)kJ/kg⎜ ⎜ 1 kJ/kg ⎟ ⎝ ⎠
Thus the throat area is m& v t (6.41 kg/s)(0.1040 m3 / kg) = = 14.75 × 10− 4 m 2 = 14.75 cm 2 (451.7 m/s) Vt
At =
(b) The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Thus h10 = h1. At the inlet, h1 = (h f + x1 h fg ) @ 3 MPa = 1008.3 + 0.95 × 1794.9 = 2713.4 kJ/kg s1 = ( s f + x1 s fg ) @ 3 MPa = 2.6454 + 0.95 × 3.5402 = 6.0086 kJ/kg ⋅ K
At state 2s, P2 = 1.2 MPa and s2 = s2s = s1 = 6.0086 kJ/kg·K. Thus,
1
Steam
Vi ≈ 0
s 2 s = s f + x 2 s s fg ⎯ ⎯→ 6.0086 = 2.2159 + x 2 s (4.3058) ⎯ ⎯→ x 2 s = 0.8808 h2 s = h f + x 2 s h fg = 798.33 + 0.8808 × 1985.4 = 2547.2 kJ/kg
t
a) ηN = 100% b) ηN = 90%
The enthalpy of steam at the actual exit state is determined from h01 − h2 2713.4 − h2 ⎯ ⎯→ 0.90 = ⎯ ⎯→ h2 = 2563.8 kJ/kg h01 − h2 s 2713.4 − 2547.2
ηN =
Therefore at the exit, P2 = 1.2 MPa and h2 = 2563.8 kJ/kg·K. Thus, h2 = h f + x2 h fg ⎯ ⎯→ 2563.8 = 798.33 + x2 (1985.4) ⎯ ⎯→ x2 = 0.8892 s2 = s f + x2 s fg = 2.2159 + 0.8892 × 4.3058 = 6.0447 v 2 = v f + x2v fg = 0.001138 + 0.8892 × (0.16326 − 0.001138) = 0.1453 kJ / kg
Then the exit velocity is determined from the steady-flow energy balance to be h1 +
V12 V2 V 2 − V12 = h2 + 2 → 0 = h2 − h1 + 2 2 2 2
Solving for V2, ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 547.1 m/s V2 = 2(h1 − h2 ) = 2(2713.4 − 2563.8)kJ/kg⎜ ⎜ 1 kJ/kg ⎟ ⎝ ⎠
The mass flow rate is determined from m& =
1
v2
A2V2 =
1 3
0.1453 m / kg
(16 × 10− 4 m 2 )(547.1 m/s) = 6.02 kg/s
The velocity of sound at the exit of the nozzle is determined from 1/ 2
⎛ ∂P ⎞ c = ⎜⎜ ⎟⎟ ⎝ ∂ρ ⎠ s
1/ 2
⎛ ∆P ⎞ ⎟⎟ ≅ ⎜⎜ ⎝ ∆ (1 / v ) ⎠ s
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
2
17-95
The specific volume of steam at s2 = 6.0447 kJ/kg·K and at pressures just below and just above the specified pressure (1.1 and 1.3 MPa) are determined to be 0.1570 and 0.1353 m3/kg. Substituting,
(1300 − 1100) kPa
⎛ 1000 m 2 / s 2 ⎞ ⎟ = 442.6 m/s ⎜ 3 ⎟ ⎜ 1 ⎞ ⎛ 1 3 ⎝ 1 kPa ⋅ m ⎠ − ⎜ ⎟ kg/m ⎝ 0.1353 0.1570 ⎠
c2 =
Then the exit Mach number becomes Ma 2 =
V2 547.1 m/s = = 1.236 c2 442.6 m/s
The steam is saturated, and thus the critical pressure which occurs at the throat is taken to be Pt = P* = 0.576 × P01 = 0.576 × 3 = 1.728 MPa
At state 2ts, Pts = 1.728 MPa and sts = s1 = 6.0086 kJ/kg·K. Thus, hts = 2611.4 kJ/kg. The actual enthalpy of steam at the throat is
ηN =
h01 − ht 2713.4 − ht ⎯ ⎯→ ht = 2621.6 kJ/kg ⎯ ⎯→ 0.90 = h01 − hts 2713.4 − 2611.4
Therefore at the throat, P2 = 1.728 MPa and ht = 2621.6 kJ/kg. Thus, vt = 0.1046 m3/kg. Then the throat velocity is determined from the steady-flow energy balance, Ê0
V2 h1 + 1 2
= ht +
Vt 2 V2 → 0 = ht − h1 + t 2 2
Solving for Vt, ⎛ 1000 m 2 / s 2 ⎞ ⎟ = 428.5 m/s Vt = 2(h1 − ht ) = 2(2713.4 − 2621.6)kJ/kg⎜ ⎜ 1 kJ/kg ⎟ ⎠ ⎝
Thus the throat area is At =
m& v t (6.02 kg/s)(0.1046 m3 / kg) = = 14.70 × 10− 4 m 2 = 14.70 cm 2 Vt (428.5 m/s)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-96
Fundamentals of Engineering (FE) Exam Problems
17-151 An aircraft is cruising in still air at 5°C at a velocity of 400 m/s. The air temperature at the nose of the aircraft where stagnation occurs is (a) 5°C (b) 25°C (c) 55°C (d) 80°C (e) 85°C Answer (e) 85°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" T1=5 "C" Vel1= 400 "m/s" T1_stag=T1+Vel1^2/(2*Cp*1000) "Some Wrong Solutions with Common Mistakes:" W1_Tstag=T1 "Assuming temperature rise" W2_Tstag=Vel1^2/(2*Cp*1000) "Using just the dynamic temperature" W3_Tstag=T1+Vel1^2/(Cp*1000) "Not using the factor 2"
17-152 Air is flowing in a wind tunnel at 15°C, 80 kPa, and 200 m/s. The stagnation pressure at a probe inserted into the flow stream is (a) 82 kPa (b) 91 kPa (c) 96 kPa (d) 101 kPa (e) 114 kPa Answer (d) 101 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" T1=15 "K" P1=80 "kPa" Vel1= 200 "m/s" T1_stag=(T1+273)+Vel1^2/(2*Cp*1000) "C" T1_stag/(T1+273)=(P1_stag/P1)^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" T11_stag/T1=(W1_P1stag/P1)^((k-1)/k); T11_stag=T1+Vel1^2/(2*Cp*1000) "Using deg. C for temperatures" T12_stag/(T1+273)=(W2_P1stag/P1)^((k-1)/k); T12_stag=(T1+273)+Vel1^2/(Cp*1000) "Not using the factor 2" T13_stag/(T1+273)=(W3_P1stag/P1)^(k-1); T13_stag=(T1+273)+Vel1^2/(2*Cp*1000) "Using wrong isentropic relation"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-97
17-153 An aircraft is reported to be cruising in still air at -20°C and 40 kPa at a Mach number of 0.86. The velocity of the aircraft is (a) 91 m/s (b) 220 m/s (c) 186 m/s (d) 280 m/s (e) 378 m/s Answer (d) 280 m/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" R=0.287 "kJ/kg.K" T1=-10+273 "K" P1=40 "kPa" Mach=0.86 VS1=SQRT(k*R*T1*1000) Mach=Vel1/VS1 "Some Wrong Solutions with Common Mistakes:" W1_vel=Mach*VS2; VS2=SQRT(k*R*T1) "Not using the factor 1000" W2_vel=VS1/Mach "Using Mach number relation backwards" W3_vel=Mach*VS3; VS3=k*R*T1 "Using wrong relation"
17-154 Air is flowing in a wind tunnel at 12°C and 66 kPa at a velocity of 230 m/s. The Mach number of the flow is (Problem changed, 2/2001) (a) 0.54 (b) 0.87 (c) 3.3 (d) 0.36 (e) 0.68 Answer (e) 0.68 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" R=0.287 "kJ/kg.K" T1=12+273 "K" P1=66 "kPa" Vel1=230 "m/s" VS1=SQRT(k*R*T1*1000) Mach=Vel1/VS1 "Some Wrong Solutions with Common Mistakes:" W1_Mach=Vel1/VS2; VS2=SQRT(k*R*(T1-273)*1000) "Using C for temperature" W2_Mach=VS1/Vel1 "Using Mach number relation backwards" W3_Mach=Vel1/VS3; VS3=k*R*T1 "Using wrong relation"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-98
17-155 Consider a converging nozzle with a low velocity at the inlet and sonic velocity at the exit plane. Now the nozzle exit diameter is reduced by half while the nozzle inlet temperature and pressure are maintained the same. The nozzle exit velocity will (a) remain the same. (b) double. (c) quadruple. (d) go down by half. (e) go down to one-fourth. Answer (a) remain the same.
17-156 Air is approaching a converging-diverging nozzle with a low velocity at 20°C and 300 kPa, and it leaves the nozzle at a supersonic velocity. The velocity of air at the throat of the nozzle is (a) 290 m/s (b) 98 m/s (c) 313 m/s (d) 343 m/s (e) 412 m/s Answer (c) 313 m/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 Cp=1.005 "kJ/kg.K" R=0.287 "kJ/kg.K" "Properties at the inlet" T1=20+273 "K" P1=300 "kPa" Vel1=0 "m/s" To=T1 "since velocity is zero" Po=P1 "Throat properties" T_throat=2*To/(k+1) P_throat=Po*(2/(k+1))^(k/(k-1)) "The velocity at the throat is the velocity of sound," V_throat=SQRT(k*R*T_throat*1000) "Some Wrong Solutions with Common Mistakes:" W1_Vthroat=SQRT(k*R*T1*1000) "Using T1 for temperature" W2_Vthroat=SQRT(k*R*T2_throat*1000); T2_throat=2*(To-273)/(k+1) "Using C for temperature" W3_Vthroat=k*R*T_throat "Using wrong relation"
17-157 Argon gas is approaching a converging-diverging nozzle with a low velocity at 20°C and 120 kPa, and it leaves the nozzle at a supersonic velocity. If the cross-sectional area of the throat is 0.015 m2, the mass flow rate of argon through the nozzle is (a) 0.41 kg/s (b) 3.4 kg/s (c) 5.3 kg/s (d) 17 kg/s (e) 22 kg/s Answer (c) 5.3 kg/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 Cp=0.5203 "kJ/kg.K"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-99
R=0.2081 "kJ/kg.K" A=0.015 "m^2" "Properties at the inlet" T1=20+273 "K" P1=120 "kPa" Vel1=0 "m/s" To=T1 "since velocity is zero" Po=P1 "Throat properties" T_throat=2*To/(k+1) P_throat=Po*(2/(k+1))^(k/(k-1)) rho_throat=P_throat/(R*T_throat) "The velocity at the throat is the velocity of sound," V_throat=SQRT(k*R*T_throat*1000) m=rho_throat*A*V_throat "Some Wrong Solutions with Common Mistakes:" W1_mass=rho_throat*A*V1_throat; V1_throat=SQRT(k*R*T1_throat*1000); T1_throat=2*(To273)/(k+1) "Using C for temp" W2_mass=rho2_throat*A*V_throat; rho2_throat=P1/(R*T1) "Using density at inlet"
17-158 Carbon dioxide enters a converging-diverging nozzle at 60 m/s, 310°C, and 300 kPa, and it leaves the nozzle at a supersonic velocity. The velocity of carbon dioxide at the throat of the nozzle is (a) 125 m/s (b) 225 m/s (c) 312 m/s (d) 353 m/s (e) 377 m/s Answer (d) 353 m/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.289 Cp=0.846 "kJ/kg.K" R=0.1889 "kJ/kg.K" "Properties at the inlet" T1=310+273 "K" P1=300 "kPa" Vel1=60 "m/s" To=T1+Vel1^2/(2*Cp*1000) To/T1=(Po/P1)^((k-1)/k) "Throat properties" T_throat=2*To/(k+1) P_throat=Po*(2/(k+1))^(k/(k-1)) "The velocity at the throat is the velocity of sound," V_throat=SQRT(k*R*T_throat*1000) "Some Wrong Solutions with Common Mistakes:" W1_Vthroat=SQRT(k*R*T1*1000) "Using T1 for temperature" W2_Vthroat=SQRT(k*R*T2_throat*1000); T2_throat=2*(T_throat-273)/(k+1) "Using C for temperature" W3_Vthroat=k*R*T_throat "Using wrong relation"
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
17-100
17-159 Consider gas flow through a converging-diverging nozzle. Of the five statements below, select the one that is incorrect: (a) The fluid velocity at the throat can never exceed the speed of sound. (b) If the fluid velocity at the throat is below the speed of sound, the diversion section will act like a diffuser. (c) If the fluid enters the diverging section with a Mach number greater than one, the flow at the nozzle exit will be supersonic. (d) There will be no flow through the nozzle if the back pressure equals the stagnation pressure. (e) The fluid velocity decreases, the entropy increases, and stagnation enthalpy remains constant during flow through a normal shock. Answer (c) If the fluid enters the diverging section with a Mach number greater than one, the flow at the nozzle exit will be supersonic.
17-160 Combustion gases with k = 1.33 enter a converging nozzle at stagnation temperature and pressure of 400°C and 800 kPa, and are discharged into the atmospheric air at 20°C and 100 kPa. The lowest pressure that will occur within the nozzle is (a) 26 kPa (b) 100 kPa (c) 321 kPa (d) 432 kPa (e) 272 kPa Answer (d) 432 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.33 Po=800 "kPa" "The critical pressure is" P_throat=Po*(2/(k+1))^(k/(k-1)) "The lowest pressure that will occur in the nozzle is the higher of the critical or atmospheric pressure." "Some Wrong Solutions with Common Mistakes:" W2_Pthroat=Po*(1/(k+1))^(k/(k-1)) "Using wrong relation" W3_Pthroat=100 "Assuming atmospheric pressure"
17-161 ··· 17-163 Design and Essay Problems
KJ
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
