SOLUTION - Fundamentals of Heat and Mass Transfer 5th Edition

PROBLEM 1.1 KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal conductivity k and inner temperature, T1. FIND: The outer temperature of the wall, T2. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: The rate equation for conduction through the wall is given by Fourier’s law,

q cond = q x = q ′′x ⋅ A = -k

T −T dT ⋅ A = kA 1 2 . dx L

Solving for T2 gives

T2 = T1 −

q cond L . kA

Substituting numerical values, find

T2 = 415$ C -

3000W × 0.025m 0.2W / m ⋅ K × 10m2

T2 = 415$ C - 37.5$ C T2 = 378$ C. COMMENTS: Note direction of heat flow and fact that T2 must be less than T1.


To,h), while the other surface (x = 0) is maintained at To. Also, wall experiences uniform volumetric heating q such that the maximum steady-state temperature will exceed T∞. FIND: (a) Sketch temperature distribution (T vs. X) for following conditions: initial (t ≤ 0), steadystate (t → ∞), and two intermediate times; also show distribution when there is no heat flow at the x = L boundary, (b) Sketch the heat flux q ′′x vs. t at the boundaries x = 0 and L.

1

6

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Uniform volumetric generation, (4) To < T∞ and q large enough that T(x,∞) > T∞. ANALYSIS: (a) The initial and boundary conditions for the wall can be written as

Initial (t ≤ 0):

T(x,0) = To

Uniform temperature

Boundary:

x = 0 T(0,t) = To

Constant temperature

x=L

−k

 

1 6

∂T = h T L, t − T∞ ∂ x x=L

Convection process.

The temperature distributions are shown on the T-x coordinates below. Note the special condition when the heat flux at (x = L) is zero.

1 6

1 6

(b) The heat flux as a function of time at the boundaries, q ′′x 0, t and q ′′x L, t , can be inferred from the temperature distributions using Fourier’s law.

COMMENTS: Since T ( x,∞ ) > T∞ and T∞ > To , heat transfer at both boundaries must be out of the

 = 0. wall. Hence, it follows from an overall energy balance on the wall that + q′′x ( 0, ∞ ) − q′′x ( L,∞ ) + qL

PROBLEM 2.49 KNOWN: Plane wall, initially at a uniform temperature To, has one surface (x = L) suddenly exposed to a convection process (T∞ < To, h), while the other surface (x = 0) is maintained at To. Also, wall experiences uniform volumetric heating q such that the maximum steady-state temperature will exceed T∞. FIND: (a) Sketch temperature distribution (T vs. x) for following conditions: initial (t ≤ 0), steadystate (t → ∞), and two intermediate times; identify key features of the distributions, (b) Sketch the heat flux q ′′x vs. t at the boundaries x = 0 and L; identify key features of the distributions.

1

6

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) Uniform volumetric generation, (4) T∞ < To and q large enough that T(x,∞) > To. ANALYSIS: (a) The initial and boundary conditions for the wall can be written as

Initial (t ≤ 0):

T(x,0) = To

Boundary: x=L

Uniform temperature Constant temperature

x = 0 T(0,t) = To ∂T −k = h T L, t − T∞ ∂ x x=L

 

1 6

Convection process.

The temperature distributions are shown on the T-x coordinates below. Note that the maximum temperature occurs under steady-state conditions not at the midplane, but to the right toward the surface experiencing convection. The temperature gradients at x = L increase for t > 0 since the convection heat rate from the surface increases as the surface temperature increases.

1 6

1 6

(b) The heat flux as a function of time at the boundaries, q ′′x 0, t and q ′′x L, t , can be inferred from the temperature distributions using Fourier’s law. At the surface x = L, the convection heat flux at t = 0 is q ′′x ( L, 0 ) = h ( To − T∞ ). Because the surface temperature dips slightly at early times, the convection heat flux decreases slightly, and then increases until the steady-state condition is reached. For the steady-state condition, heat transfer at both boundaries must be out of the wall. It follows from  = 0. an overall energy balance on the wall that + q′′x ( 0, ∞ ) − q′′x ( L, ∞ ) + qL

PROBLEM 2.50 KNOWN: Interfacial heat flux and outer surface temperature of adjoining, equivalent plane walls. FIND: (a) Form of temperature distribution at representative times during the heating process, (b) Variation of heat flux with time at the interface and outer surface. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties. ANALYSIS: (a) With symmetry about the interface, consideration of the temperature distribution may be restricted to 0 ≤ x ≤ L. During early stages of the process, heat transfer is into the material from the outer surface, as well as from the interface. During later stages and the eventual steady state, heat is transferred from the material at the outer surface. At steady-state, dT/dx = − ( q′′o 2 ) k = const . and T(0,t) = To + ( q′′o 2 ) L k .

(b) At the outer surface, the heat flux is initially negative, but increases with time, approaching q′′o /2. It is zero when dT dx x = L = 0 .

PROBLEM 2.51 KNOWN: Temperature distribution in a plane wall of thickness L experiencing uniform volumetric heating q having one surface (x = 0) insulated and the other exposed to a convection process characterized by T∞ and h. Suddenly the volumetric heat generation is deactivated while convection continues to occur. FIND: (a) Determine the magnitude of the volumetric energy generation rate associated with the initial condition, (b) On T-x coordinates, sketch the temperature distributions for the initial condition (T ≤ 0), the steady-state condition (t → ∞), and two intermediate times; (c) On q′′x - t coordinates, sketch the variation with time of the heat flux at the boundary exposed to the convection process, q ′′x ( L, t ) ; calculate the corresponding value of the heat flux at t = 0; and (d) Determine the amount of 2

energy removed from the wall per unit area (J/m ) by the fluid stream as the wall cools from its initial to steady-state condition. SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, and (3) Uniform internal volumetric heat generation for t < 0. ANALYSIS: (a) The volumetric heating rate can be determined by substituting the temperature distribution for the initial condition into the appropriate form of the heat diffusion equation.

d  dT  q  + =0 dx  dx  k

where

d q q (0 + 2bx ) + = 0 + 2b + = 0 dx k k

(

T ( x, 0 ) = a + bx 2

)

q = −2kb = −2 × 90 W / m ⋅ K −1.0 ×104°C / m 2 = 1.8 × 106 W / m3


> w representing the distance normal to the page,

)

(

hP h ⋅ 2 h ⋅2 2 500 W m 2 ⋅ K × 2 4 × 10−3 mm = 0.0533 ⋅ ∆x 2 ≈ ∆x 2 = ∆x = 3 − kA c k ⋅⋅w kw 50 W m ⋅ K × 6 ×10 m Node 1: Node n: Node 12:

100 + T2 + 0.0533 × 30 − ( 2 + 0.0533) T1 = 0 or

-2.053T1 + T2 = -101.6

Tn +1 + Tn −1 + 1.60 − 2.0533Tn = 0 or T11 + ( 0.0533 2 ) 30 − ( 0.0533 2 + 1) T12 = 0 or

Tn −1 − 2.053Tn + Tn −1 = −1.60 T11 − 1.0267T12 = −0.800

Using matrix notation, Eq. 4.52, where [A] [T] = [C], the A-matrix is tridiagonal and only the non-zero terms are shown below. A matrix inversion routine was used to obtain [T]. Tridiagonal Matrix A Nonzero Terms a1,1 a1,2 a2,1 a2,2 a2,3 a3,2 a3,3 a3,4 a4,3 a4,4 a4,5 a5,4 a5,5 a5,6 a6,5 a6,6 a6,7 a7,6 a7,7 a7,8 a8,7 a8,8 a8,9 a9,8 a9,9 a9,10 a10,9 a10,10 a10,11 a11,10 a11,11 a11,12 a12,11 a12,12 a12,13

1 1 1 1 1 1 1 1 1 1 1

Column Matrices Values -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -2.053 -1.027

1 1 1 1 1 1 1 1 1 1 1 1

Node 1 2 3 4 5 6 7 8 9 10 11 12

C -101.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -1.6 -0.8

T 85.8 74.5 65.6 58.6 53.1 48.8 45.5 43.0 41.2 39.9 39.2 38.9

The assumption of one-dimensional heat conduction is justified when Bi ≡ h(w/2)/k < 0.1. Hence, with Bi = 500 W/m2⋅K(3 × 10-3 m)/50 W/m⋅K = 0.03, the assumption is reasonable. Continued...

PROBLEM 4.72 (Cont.) (b) The fin heat rate can be most easily found from an energy balance on the control volume about Node 0, T −T  ∆x  q′f = q1′ + q′conv = k ⋅ w 0 1 + h  2  ( T0 − T∞ ) ∆x  2 

(

q′f = 50 W m ⋅ K 6 × 10−3 m

)

(100 − 85.8 )$ C 4 × 10

−3



+ 500 W m2 ⋅ K  2 ⋅

 

m

4 × 10−3 m  2

$  (100 − 30 ) C  

q′f = (1065 + 140 ) W m = 1205 W m . From Eq. 3.76, the fin heat rate is q = ( hPkA c )

1/ 2


Tm) is appropriate, k 0.634 W/m ⋅ K 0.4 h i = 0.023Re4/5 = 0.023 (12,611 )4 / 5 ( 4.16 )0.4 = 1230 W/m 2 ⋅K. D Pr D 0.040 m For the external flow, ReD = VD/ν = 100 m/s × 0.040 m/38.79 × 10 -6 m 2 / s = 1.031 ×105 and from Table 7.4, C = 0.26 and m = 0.6; Pr ≤ 10, n = 0.37, and Pr ≈Prs , n h o = ( k/D ) CRem D Pr ( Pr/Prs )

1/4

ho =

40.7 × 10−3 W/m ⋅ K

(

× 0.26 1.031 × 105

)

0.6

( 0.684) 0.37 ( 1)1 / 4 = 234 W/m2 ⋅ K.

0.040 m 2 Substituting numerical values into Eq. 8.46 with P = πD and U = 197 W/m ⋅K, T∞ − Tm,o & p = exp − PLU/mc T∞ − Tm,i

(

)

(1)

 π × 0.040 m × 4 m  o Tm,o = 225o C − ( 225 − 30 ) Cexp  − ×197 W/m 2 ⋅ K  = 47.6o C  0.25 kg/s × 4179 J/kg ⋅ K  COMMENTS: Note the assumed value of Tm,o to evaluate water properties was reasonable. Using Eq. (1), replacing T∞ and U with Ts and hi, respectively, find Ts = 63.2°C; hence, Prs (Ts ) ≈ 0.687. The assumption that Pr ≈ Prs in the Zhukauskas relation is reasonable.


350°C.

PROBLEM 10.29 KNOWN: Horizontal, stainless steel bar submerged in water at 25°C. FIND: Heat rate per unit length of the bar. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Film pool boiling, (3) Water at 1 atm. PROPERTIES: Table A-6, Water, liquid (1 atm, Tsat = 100°C): ρ l = 957.9 kg/m3, hfg = 2257 3 kJ/kg; Table A-6, Water, vapor, (Tf = (Ts + Tsat )/2 ≈ 450K): ρ v = 4.81 kg/m , cp,v = 2560 J/kg⋅K, µv -6 2 = 14.85 × 10 N⋅s/m , kv = 0.0331 W/m⋅K. ANALYSIS: The heat rate per unit length is

q′s = q s / l = q′′π D = hπ D ( Ts − Tsat ) = h π D ∆Te

where ∆Te = (250-100)°C = 150°C. Note from the boiling curve of Figure 10.4, that film boiling will occur. From Eq. 10.10, 3 4 / 3 + h h1/3 h 4 / 3 = hconv or h = hconv + h rad if hconv > h rad . rad

4

(

)

To estimate the convection coefficient, use Eq. 10.9, 1/4

 g ( ρ − ρ ) h′ D3  h conv D l v fg  Nu D = = C kv  ν v k v ∆Te   

where C = 0.62 for the horizontal cylinder and h ′fg = h fg + 0.8cp,v ( Ts − Tsat ) . Find 1/4

h conv =

0.0331W/m ⋅ K 0.050m

 9.8m/s 2 ( 957.9 − 4.81 ) k g / m 3 2257 × 10 3 + 0.8 × 2560J/kg ⋅ K × 150 K  ( 0.050m)3     0.62  −6 2   14.85 × 10 /4.81 m / s × 0.0331 W / m ⋅ K × 150K  

(

)

hconv = 273 W / m 2 ⋅ K. To estimate the radiation coefficient, use Eq. 10.11, 4 ε σ Ts4 − Tsat 0.50 × 5.67 ×10 −8 W / m 2 ⋅ K 4 5234 − 3734 K 4 hrad = = = 1 1 W / m 2 ⋅ K. Ts − Tsat 150K Since hconv > hrad, the simpler form of Eq. 10.10 is appropriate. Find, h =  273 + 3 / 4 ×11 W / m2 ⋅ K = 2 8 1 W / m2 ⋅ K.

(



)

(

(

)

)



Using the rate equation, find q′s = 281W/m 2 ⋅ K × π × ( 0.050m ) ×150K = 6.62kW/m. COMMENTS: The effect of the water being subcooled (T = 25°C < Tsat ) is considered to be negligible.


120°C, the element is operating in the film-boiling (FB) regime. The electrical power dissipation per unit length is q′s = h (π D )( Ts − Tsat ) (1) where the total heat transfer coefficient is 4/3 + h 1/ 3 h 4 / 3 = h conv rad h The convection coefficient is given by the correlation, Eq. 10.9, with C = 0.62,

(2)

1/ 4

 g ( ρ − ρ ) h′ D3  h conv D " v fg  = C kv  ηv k v ( Ts − Tsat )   

(3) 1/ 4

 9.8 m / s 2 (833.9 − 12.54 ) kg / m3 × 2.905 × 106 J / kg ⋅ K (0.005 m )3   h conv = 0.62  6 2 −   1.31× 10 m / s × 0.04186 W / m ⋅ K (350 − 100 ) K   h conv = 626 W / m 2 ⋅ K −8

The radiation coefficient, Eq. (10.11), with σ = 5.67 × 10

2

4

W/m ⋅K , is Continued …..

h rad =

h rad =

(

4 εσ Ts4 − Tsat

(Ts − Tsat )

PROBLEM 10.30 (Cont.)

)

(

)

0.25σ 6234 − 3734 K 4

(350 − 100 ) K

= 4.5 W / m 2 ⋅ K

Substituting numerical values into Eq. (2) for h, and into Eq. (1) for q′s , find

h = 630 W / m 2 ⋅ K q′s = 630 W / m 2 ⋅ K (π × 0.005 m )(350 − 100 ) K = 2473 W / m


1800, so indeed the flow is turbulent, and using Eq. (4) or (3), find h L = 5645 W m 2 ⋅ K . From the rate equations (1) and (2), the heat transfer and condensation rates are q′ = 5645 W m 2 ⋅ K × 2.5m 100 − 54 K = 649k W m

(

)

<
Reδ , the flow is not turbulent, but wavy-laminar. Now the procedure follows that of Example 10.3. For L = 1.25 m with wavy-laminar flow, Eq. 10.38 is the appropriate correlation. The calculations yield these results: Reδ = 1372 h L = 5199 W m 2 ⋅ K

q′ = 299 kW m


2300, the flow is turbulent and since flow is assumed to be fully developed, use the DittusBoelter correlation with n = 0.4 for heating, 0.4 Nu D = 0.023 Re0.8 = 0.023 ( 75, 638 ) D Pr

0.8

h = Nu D

k D

(3.58 )0.4 = 306.4

= 306.4 × 0.648 W m ⋅ K ( 0.1m ) = 1985 W m 2 ⋅ K .


10). The DittusBoelter correlation with n = 0.4 is appropriate, 0.4 Nu D = hi Di k f = 0.023 Re0.8 D Prf = 0.023 × ( 21, 673 )

0.8

(5.2 )0.4 = 130.9 Continued...

PROBLEM 11.35 (Cont.) k 0.620W m ⋅ K × 130.9 = 6057 W m 2 ⋅ K . h i = f Nu D = − 3 Di 13.4 × 10 m Substituting numerical values into Eq. (1), the overall heat transfer coefficient is

(

)

  15.9 × 10−3 m 2 15.9 15.9 1 1   "n + + × Uo = 13, 500 W m 2 ⋅ K 2 115 W m ⋅ K 13.4 13.4 6057 W m ⋅ K    U o = 7.407 × 10−5 + 1.183 × 10−5 + 19.590 × 10−5 



−1



−1