CHAPTER TWO 3 wk
7d
24 h 3600 s 1000 ms
= 18144 . × 10 9 ms 1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s (b) = 25.98 mi / h ⇒ 26.0 mi / h 3.2808 ft 1 h
2.1 (a)
(c)
2.2 (a)
554 m 4 1d 1h d ⋅ kg 24 h 60 min
1 kg 108 cm 4 = 3.85 × 10 4 cm 4 / min⋅ g 1000 g 1 m 4
1 m 1 h 760 mi = 340 m / s h 0.0006214 mi 3600 s 1 m3 = 57.5 lb m / ft 3 35.3145 ft 3
(b)
921 kg 2.20462 lb m m3 1 kg
(c)
5.37 × 10 3 kJ 1 min 1000 J min 60 s 1 kJ
1.34 × 10 -3 hp = 119.93 hp ⇒ 120 hp 1 J/s
2.3 Assume that a golf ball occupies the space equivalent to a 2 in × 2 in × 2 in cube. For a classroom with dimensions 40 ft × 40 ft × 15 ft : 40 × 40 × 15 ft 3 (12) 3 in 3 1 ball n balls = . × 10 6 ≈ 5 million balls = 518 ft 3 2 3 in 3 The estimate could vary by an order of magnitude or more, depending on the assumptions made. 2.4 4.3 light yr 365 d 24 h 1 yr
3600 s 1.86 × 10 5 mi
1d
1 h
1
s
3.2808 ft 0.0006214 mi
1 step = 7 × 1016 steps 2 ft
2.5 Distance from the earth to the moon = 238857 miles 238857 mi
1
m
0.0006214 mi
1 report 0.001 m
= 4 × 1011 reports
2.6
19 km 1000 m 0.0006214 mi 1000 L = 44.7 mi / gal 1 L 1 km 1 m 264.17 gal Calculate the total cost to travel x miles. Total Cost
Total Cost
American
European
= $14,500 +
= $21,700 +
$1.25 1 gal gal 28 mi
x (mi)
= 14,500 + 0.04464 x
$1.25 1 gal x (mi) = 21,700 + 0.02796 x gal 44.7 mi
Equate the two costs ⇒ x = 4.3 × 10 5 miles
2-1
2.7 5320 imp. gal
106 cm3
14 h 365 d
plane ⋅ h
1 d
1 yr
0.965 g
220.83 imp. gal
1
cm
3
1 kg
1 tonne
1000 g
1000 kg
tonne kerosene plane ⋅ yr
= 1.188 × 105
4.02 × 109 tonne crude oil 1 tonne kerosene
plane ⋅ yr
7 tonne crude oil 1.188 × 10 tonne kerosene 5
yr
= 4834 planes ⇒ 5000 planes
2.8 (a) (b) (c)
2.9
2.10 2.11
32.1714 ft / s 2
25.0 lb m
1
lb f
32.1714 lb m ⋅ ft / s 2 25 N
1 kg ⋅ m/s 2
1 9.8066 m/s 2
10 ton
1N
1 lb m 5 × 10
-4
50 × 15 × 2 m 3
500 lb m
= 2.5493 kg ⇒ 2.5 kg
980.66 cm / s 2
1000 g ton
= 25.0 lb f
1 g ⋅ cm / s 2
2.20462 lb m
35.3145 ft 3 1 m3
85.3 lb m 1 ft 3
1
1 m3
kg
2.20462 lb m
1 dyne
11.5 kg
= 9 × 10 9 dynes
32.174 ft 1 lb f = 4.5 × 10 6 lb f 2 2 1 s 32.174 lb m / ft ⋅ s
≈ 5 × 10 2
FG 1 IJ FG 1 IJ ≈ 25 m H 2 K H 10K
3
(a) mdisplaced fluid = mcylinder ⇒ ρ f V f = ρ cVc ⇒ ρ f hπr 2 = ρ c Hπr 2
ρc
ρfh
(30 cm − 14.1 cm)(100 . g / cm 3 ) ρc = = = 0.53 g / cm 3 H 30 cm ρ H (30 cm)(0.53 g / cm 3 ) (b) ρ f = c = = 171 . g / cm 3 (30 cm - 20.7 cm) h
2.12
Vs =
πR 2 H 3
⇒ Vf =
; Vf =
3
−
πr 2 h 3
;
πh Rh
2
2
f
H H−
h3 H2
3
2
= ρs
H3 = ρs H 3 − h3
h
3
r
2
2
ρf h
R r R = ⇒r = h H h H
FG IJ = πR FG H − h IJ − 3 3 H HK 3 H H K πR F πR H h I ⇒ρ =ρ H− G J 3 H 3 H K
πR H 2
ρ f V f = ρ sVs ⇒ ρ f = ρs
πR 2 H
H
H
2
ρs
s
R
1
1−
FG h IJ H HK
2-2
3
ρf
2.13
Say h( m) = depth of liquid
y y= 1 dA y=y=1––1+h h xx
⇒ A(m 2 )
1− y
dA = dy ⋅
∫
h
x = 1– y 2 y= –1
dA
2
−1+ h
( )=2 ∫
dx = 2 1 − y dy ⇒ A m 2
2
1 − y 2 dy
−1
− 1− y 2
⇓
1m
Table of integrals or trigonometric substitution
π 2 A m 2 = y 1 − y 2 + sin −1 y ⎤⎥ = ( h − 1) 1 − ( h − 1) + sin −1 ( h − 1) + ⎦ −1 2 h −1
( )
b g
W N =
4 m × A( m 2 ) 0.879 g 10 6 cm 2 cm
3
1m
E Substitute for A L W b N g = 3.45 × 10 Mbh − 1g 1 − bh − 1g N 4
2.14
3
1 kg
9.81 N
3
kg N
10 g
= 3.45 × 10 4 A
g g0
2
b g π2 OPQ
+ sin −1 h − 1 +
1 lb f = 1 slug ⋅ ft / s 2 = 32.174 lb m ⋅ ft / s 2 ⇒ 1 slug = 32.174 lb m 1 1 poundal = 1 lb m ⋅ ft / s 2 = lb f 32.174 (a) (i) On the earth: 175 lb m 1 slug M= = 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W= = 5.63 × 10 3 poundals 2 s 1 lb m ⋅ ft / s 2 (ii) On the moon 175 lb m 1 slug M= = 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W= = 938 poundals 2 6 s 1 lb m ⋅ ft / s 2 (b) F = ma ⇒ a = F / m =
355 poundals 25.0 slugs
1 lb m ⋅ ft / s 2 1 poundal
= 0.135 m / s 2
2-3
1 slug 32.174 lb m
1m 3.2808 ft
2.15 (a) F = ma ⇒ 1 fern = (1 bung)(32.174 ft / s 2 )
⇒
FG 1IJ = 5.3623 bung ⋅ ft / s H 6K
2
1 fern 5.3623 bung ⋅ ft / s 2
3 bung 32.174 ft 1 fern = 3 fern 2 6 s 5.3623 bung ⋅ ft / s 2 On the earth: W = (3)( 32.174) / 5.3623 = 18 fern
(b) On the moon: W =
2.16 (a) ≈ (3)(9) = 27
(b)
(2.7)(8.632) = 23 (c) ≈ 2 + 125 = 127
(d)
2.365 + 125.2 = 127.5 2.17 R ≈
4.0 × 10−4 ≈ 1× 10−5 40 (3.600 ×10−4 ) / 45 = 8.0 × 10−6 ≈
≈ 50 × 10 3 − 1 × 10 3 ≈ 49 × 10 3 ≈ 5 × 10 4 4.753 × 10 4 − 9 × 10 2 = 5 × 10 4
(7 ×10−1 )(3 × 105 )(6)(5 × 104 ) ≈ 42 × 102 ≈ 4 × 103 (Any digit in range 2-6 is acceptable) (3)(5 × 106 )
Rexact = 3812.5 ⇒ 3810 ⇒ 3.81× 103 2.18 (a) A: R = 731 . − 72.4 = 0.7 o C X=
. + 72.6 + 72.8 + 73.0 72.4 + 731 = 72.8 o C 5
s=
(72.4 − 72.8) 2 + (731 . − 72.8) 2 + (72.6 − 72.8) 2 + (72.8 − 72.8) 2 + (73.0 − 72.8) 2 5−1
= 0.3o C B: R = 1031 . − 97.3 = 58 . oC X=
97.3 + 1014 . + 98.7 + 1031 . + 100.4 = 100.2 o C 5
s=
(97.3 − 100.2) 2 + (1014 . − 100.2) 2 + (98.7 − 100.2) 2 + (1031 . − 100.2) 2 + (100.4 − 100.2) 2 5−1
= 2.3o C
(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate.
2-4
2.19 (a)
12
X=
∑X
12
i
i =1
C min=
= 73.5 s= 12 = X − 2 s = 73.5 − 2(1.2) = 711 .
∑ ( X − 735. )
2
i =1
= 12 .
12 − 1
C max= = X + 2 s = 735 . + 2(12 . ) = 75.9
(b) Joanne is more likely to be the statistician, because she wants to make the control limits stricter. (c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor temperature (failure of reactor control system), problems with the color measurement system, operator carelessness 2.20 (a), (b) 1 2 (a) Run 134 131 X Mean(X) 131.9 Stdev(X) 2.2 127.5 Min 136.4 Max (b) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14
X 128 131 133 130 133 129 133 135 137 133 136 138 135 139
Min 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5
3 129
Mean 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9
4 5 6 7 8 9 10 11 12 13 14 15 133 135 131 134 130 131 136 129 130 133 130 133
Max 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4
140 138 136 134 132 130 128 126 0
5
10
15
(c) Beginning with Run 11, the process has been near or well over the upper quality assurance limit. An overhaul would have been reasonable after Run 12.
2.21 (a) Q ' =
2.36 × 10−4 kg ⋅ m 2
(b) Q 'approximate ≈
h
2.20462 lb 3.28082 ft 2 m2
kg
1
h
3600 s
(2 × 10−4 )(2)(9) ≈ 12 × 10( −4−3) ≈ 1.2 × 10−6 lb ⋅ ft 2 / s 3 × 103
Q 'exact =1.56 × 10−6 lb ⋅ ft 2 / s = 0.00000156 lb ⋅ ft 2 / s
2-5
2.22 N Pr = N Pr ≈
Cpμ
=
k
0.583 J / g ⋅ o C
1936 lb m
0.286 W / m ⋅ C −1
1 h 3.2808 ft
ft ⋅ h
o
3600 s
1000 g
m 2.20462 lb m
(6 × 10 )(2 × 10 )(3 × 10 ) 3 × 10 ≈ ≈ 15 . × 10 3 . The calculator solution is 163 . × 10 3 −1 3 2 (3 × 10 )(4 × 10 )(2) 3
3
3
2.23 Re = Re ≈
2.24 (a)
Duρ
μ
=
0.48 ft
1
m
2.067 in
1 m
1 kg 10 6 cm 3
0.805 g
s 3.2808 ft 0.43 × 10 −3 kg / m ⋅ s 39.37 in
cm 3
1000 g
1 m3
(5 × 10 −1 )(2)(8 × 10 −1 )(10 6 ) 5 × 101− ( −3) ≈ ≈ 2 × 10 4 ⇒ the flow is turbulent 3 (3)(4 × 10)(10 3 )(4 × 10 −4 )
kg d p y D
1/ 3
⎛ μ ⎞ = 2.00 + 0.600 ⎜ ⎟ ⎝ ρD ⎠
⎛ d p uρ ⎞ ⎜ ⎟ ⎝ μ ⎠
1/ 2
1/ 3
⎡ ⎤ 1.00 × 10−5 N ⋅ s/m 2 = 2.00 + 0.600 ⎢ ⎥ −5 3 2 ⎣ (1.00 kg/m )(1.00 × 10 m / s) ⎦ k g (0.00500 m)(0.100) = 44.426 ⇒ = 44.426 ⇒ k g 1.00 × 10−5 m 2 / s
1/ 2
⎡ (0.00500 m)(10.0 m/s)(1.00 kg/m3 ) ⎤ ⎢ ⎥ (1.00 × 10−5 N ⋅ s/m 2 ) ⎣ ⎦ = 0.888 m / s
(b) The diameter of the particles is not uniform, the conditions of the system used to model the equation may differ significantly from the conditions in the reactor (out of the range of empirical data), all of the other variables are subject to measurement or estimation error. (c) dp (m)
y
0.005 0.010 0.005 0.005 0.005
0.1 0.1 0.1 0.1 0.1
D (m2/s) μ (N-s/m2) ρ (kg/m3) u (m/s) 1.00E-05 1.00E-05 1 10 1.00E-05 1.00E-05 1 10 2.00E-05 1.00E-05 1 10 1.00E-05 2.00E-05 1 10 1.00E-05 1.00E-05 1 20
kg 0.889 0.620 1.427 0.796 1.240
2.25 (a) 200 crystals / min ⋅ mm; 10 crystals / min ⋅ mm 2
200 crystals 0.050 in 25.4 mm 10 crystals 0.050 2 in 2 − min ⋅ mm in min ⋅ mm 2 238 crystals 1 min = 238 crystals / min ⇒ = 4.0 crystals / s 60 s min
(25.4) 2 mm 2 in 2
(b) r =
b g
(c) D mm =
b g
D ′ in
FG H
IJ K
crystals 60 s 25.4 mm crystals = 25.4 D ′ ; r = r′ = 60r ′ s 1 min min 1 in
b
g b
⇒ 60r ′ = 200 25.4 D ′ − 10 25.4 D ′
g
2
2-6
b g
⇒ r ′ = 84.7 D ′ − 108 D ′
2
2.26 (a) 70.5 lb m / ft 3 ; 8.27 × 10 -7 in 2 / lb f ⎡8.27 × 10−7 in 2 9 × 106 N 14.696 lbf / in 2 ⎤ (b) ρ = (70.5 lb m / ft 3 )exp ⎢ ⎥ lbf m 2 1.01325 × 105 N/m 2 ⎥⎦ ⎢⎣ 70.57 lb m 35.3145 ft 3 1 m3 1000 g = = 1.13 g/cm3 3 3 6 3 ft m 10 cm 2.20462 lbm
(c)
F lb IJ = ρ ′ g ρG H ft K cm F lb IJ = P' N PG H in K m m 3
3
f 2
1 lb m
28,317 cm 3
453.593 g
1 ft 3
0.2248 lb f
12
2
m2
39.37 2 in 2
1N
d
= 62.43ρ ′ = 145 . × 10 −4 P '
id
i
d
i
⇒ 62.43ρ ′ = 70.5 exp 8.27 × 10 −7 1.45 × 10 −4 P ' ⇒ ρ ′ = 113 . exp 120 . × 10 −10 P '
P ' = 9.00 × 10 6 N / m 2 ⇒ ρ ' = 113 . exp[(1.20 × 10 −10 )(9.00 × 10 6 )] = 113 . g / cm 3 cm = 16.39V ' ; t bsg = 3600t ′b hr g d i V ' din i 28,317 1728 in ⇒ 16.39V ' = expb3600t ′ g ⇒ V ' = 0.06102 expb3600t ′ g
2.27 (a) V cm 3 =
3
3
3
(b) The t in the exponent has a coefficient of s-1. 2.28 (a) 3.00 mol / L, 2.00 min -1 (b) t = 0 ⇒ C = 3.00 exp[(-2.00)(0)] = 3.00 mol / L
t = 1 ⇒ C = 3.00 exp[(-2.00)(1)] = 0.406 mol / L 0.406 − 3.00 For t=0.6 min: (0.6 − 0) + 3.00 = 14 . mol / L Cint = 1− 0 Cexact = 3.00 exp[(-2.00)(0.6)] = 0.9 mol / L For C=0.10 mol/L:
t int = t exact
1− 0 (010 . − 3.00) + 0 = 112 . min 0.406 − 3 1 C 1 0.10 =ln = - ln = 1.70 min 2.00 3.00 2 3.00
(c) 3.5 C exact vs. t
3 C (mol/L)
2.5 2
(t=0.6, C=1.4)
1.5 1
(t=1.12, C=0.10)
0.5 0 0
1
2
t (min)
2-7
p* =
2.29 (a)
(b)
60 − 20 (185 − 166.2) + 20 = 42 mm Hg 199.8 − 166.2
c
MAIN PROGRAM FOR PROBLEM 2.29 IMPLICIT REAL*4(A–H, 0–Z) DIMENSION TD(6), PD(6) DO 1 I = 1, 6 READ (5, *) TD(I), PD(I) 1 CONTINUE WRITE (5, 902) 902 FORMAT (‘0’, 5X, ‘TEMPERATURE VAPOR PRESSURE’/6X, * ‘ (C) (MM HG)’/) DO 2 I = 0, 115, 5 T = 100 + I CALL VAP (T, P, TD, PD) WRITE (6, 903) T, P 903 FORMAT (10X, F5.1, 10X, F5.1) 2 CONTINUE END SUBROUTINE VAP (T, P, TD, PD) DIMENSION TD(6), PD(6) I=1 1 IF (TD(I).LE.T.AND.T.LT.TD(I + 1)) GO TO 2 I=I+1 IF (I.EQ.6) STOP GO TO 1 2 P = PD(I) + (T – TD(I))/(TD(I + 1) – TD(I)) * (PD(I + 1) – PD(I)) RETURN END OUTPUT DATA 98.5 1.0 TEMPERATURE VAPOR PRESSURE 131.8 5.0 (C) (MM HG) 100.0 1.2 # # 215.5 100.0 105.0 1.8 # # 215.0 98.7
2.30 (b) ln y = ln a + bx ⇒ y = ae bx b = (ln y 2 − ln y1 ) / ( x 2 − x1 ) = (ln 2 − ln 1) / (1 − 2) = −0.693 ln a = ln y − bx = ln 2 + 0.63(1) ⇒ a = 4.00 ⇒ y = 4.00e −0.693 x
(c) ln y = ln a + b ln x ⇒ y = ax b b = (ln y 2 − ln y1 ) / (ln x 2 − ln x1 ) = (ln 2 − ln 1) / (ln 1 − ln 2) = −1
ln a = ln y − b ln x = ln 2 − ( −1) ln(1) ⇒ a = 2 ⇒ y = 2 / x (d) ln( xy ) = ln a + b( y / x) ⇒ xy = aeby / x ⇒ y = (a / x)eby / x [can't get y = f ( x)] b = [ln( xy ) 2 − ln( xy )1 ]/[( y / x) 2 − ( y / x)1 ] = (ln 807.0 − ln 40.2) /(2.0 − 1.0) = 3 ln a = ln( xy ) − b( y / x) = ln 807.0 − 3ln(2.0) ⇒ a = 2 ⇒ xy = 2e3 y / x [can't solve explicitly for y ( x)]
2-8
2.30 (cont’d) (e) ln( y 2 / x ) = ln a + b ln( x − 2) ⇒ y 2 / x = a ( x − 2) b ⇒ y = [ax ( x − 2) b ]1/ 2
b = [ln( y 2 / x ) 2 − ln( y 2 / x ) 1 ] / [ln( x − 2) 2 − ln( x − 2) 1 ] . ) = 4.33 = (ln 807.0 − ln 40.2) / (ln 2.0 − ln 10 ln a = ln( y 2 / x ) − b( x − 2) = ln 807.0 − 4.33 ln(2.0) ⇒ a = 40.2 ⇒ y 2 / x = 40.2( x − 2) 4.33 ⇒ y = 6.34 x 1/ 2 ( x − 2) 2.165 2.31 (b) Plot y 2 vs. x 3 on rectangular axes. Slope = m, Intcpt = − n (c)
1 1 a 1 = + x ⇒ Plot vs. ln( y − 3) b b ln( y − 3)
x [rect. axes], slope =
a 1 , intercept = b b
(d)
1 1 = a ( x − 3) 3 ⇒ Plot vs. ( x − 3) 3 [rect. axes], slope = a , intercept = 0 2 2 ( y + 1) ( y + 1) OR 2 ln( y + 1) = − ln a − 3 ln( x − 3) Plot ln( y + 1) vs. ln( x − 3) [rect.] or (y + 1) vs. (x - 3) [log] 3 ln a ⇒ slope = − , intercept = − 2 2 (e) ln y = a x + b
Plot ln y vs.
x [rect.] or y vs.
x [semilog ], slope = a, intercept = b
(f) log10 ( xy ) = a ( x 2 + y 2 ) + b Plot log10 ( xy ) vs. ( x 2 + y 2 ) [rect.] ⇒ slope = a, intercept = b
(g)
x b x 1 vs. x 2 [rect.], slope = a , intercept = b = ax + ⇒ = ax 2 + b ⇒ Plot y x y y OR
b 1 1 b 1 1 vs. 2 [rect.] , slope = b, intercept = a = ax + ⇒ = a + 2 ⇒ Plot y x xy xy x x
2-9
2.32 (a) A plot of y vs. R is a line through ( R = 5 , y = 0.011 ) and ( R = 80 , y = 0169 . ).
0.18 0.16 0.14 0.12 y
0.1 0.08 0.06 0.04 0.02 0 0
20
40
60
80
100
R
y=aR+b
U| V| W
. − 0.011 0169 = 2.11 × 10 −3 80 − 5 ⇒ y = 2.11 × 10 −3 R + 4.50 × 10 −4 −3 −4 b = 0.011 − 2.11 × 10 5 = 4.50 × 10
a=
ib g
d
ib g
d
(b) R = 43 ⇒ y = 2.11 × 10 −3 43 + 4.50 × 10 −4 = 0.092 kg H 2 O kg
b1200 kg hgb0.092 kg H O kgg = 110 kg H O h 2
2
2.33 (a) ln T = ln a + b ln φ ⇒ T = aφ b
b = (ln T2 − ln T1 ) / (ln φ 2 − ln φ 1 ) = (ln 120 − ln 210) / (ln 40 − ln 25) = −119 . ln a = ln T − b ln φ = ln 210 − ( −119 . ) ln(25) ⇒ a = 9677.6 ⇒ T = 9677.6φ −1.19
b
(b) T = 9677.6φ −1.19 ⇒ φ = 9677.6 / T
b
g
T = 85o C ⇒ φ = 9677.6 / 85
0.8403
b g T = 290 C ⇒ φ = b9677.6 / 290g T = 175o C ⇒ φ = 9677.6 / 175 o
g
0.8403
. L/s = 535
0.8403
= 29.1 L / s
0.8403
= 19.0 L / s
(c) The estimate for T=175°C is probably closest to the real value, because the value of temperature is in the range of the data originally taken to fit the line. The value of T=290°C is probably the least likely to be correct, because it is farthest away from the date range.
2-10
ln ((CA-CAe)/(CA0-CAe))
2.34 (a) Yes, because when ln[(C A − C Ae ) / (C A0 − C Ae )] is plotted vs. t in rectangular coordinates, the plot is a straight line. 0
50
100
150
200
0 -0.5 -1 -1.5 -2 t (m in)
Slope = -0.0093 ⇒ k = 9.3 × 10-3 min −1
(b) ln[(C A − C Ae ) /(C A0 − C Ae )] = − kt ⇒ C A = (C A0 − C Ae )e − kt + C Ae
C A = (0.1823 − 0.0495)e − (9.3×10 C =m /V ⇒ m =CV =
−3
)(120)
+ 0.0495 = 9.300 × 10-2 g/L
9.300 × 10-2 g 30.5 gal 28.317 L = 10.7 g L 7.4805 gal
2.35 (a) ft 3 and h -2 , respectively (b) ln(V) vs. t2 in rectangular coordinates, slope=2 and intercept= ln(353 . × 10 −2 ) ; or V(logarithmic axis) vs. t2 in semilog coordinates, slope=2, intercept= 353 . × 10−2 (c) V ( m3 ) = 100 . × 10 −3 exp(15 . × 10 −7 t 2 ) 2.36 PV k = C ⇒ P = C / V k ⇒ ln P = ln C − k lnV 8.5
lnP
8 7.5 7 6.5 6 2.5
3
lnP = -1.573(lnV ) + 12.736
3.5
4
lnV
k = − slope = − ( −1573 . ) = 1573 . (dimensionless) Intercept = ln C = 12.736 ⇒ C = e12.736 = 3.40 × 105 mm Hg ⋅ cm4.719 G − GL 1 G −G G −G = ln K L + m ln C = ⇒ 0 = K L C m ⇒ ln 0 m G0 − G K L C G − GL G − GL ln (G 0 -G )/(G -G L )= 2 .4 8 3 5 ln C - 1 0 .0 4 5
3 ln(G 0-G)/(G-G L )
2.37 (a)
2 1 0 -1 3 .5
4
4 .5
5
ln C
2-11
5 .5
2.37 (cont’d)
m = slope = 2.483 (dimensionless) Intercept = ln K L = −10.045 ⇒ K L = 4.340 × 10 −5 ppm-2.483
G − 180 . × 10 −3 = 4.340 × 10 −5 (475) 2.483 ⇒ G = 1806 × 10 −3 . 3.00 × 10 −3 − G C=475 ppm is well beyond the range of the data.
(b) C = 475 ⇒
2.38 (a) For runs 2, 3 and 4: Z = aV b p c ⇒ ln Z = ln a + b lnV + c ln p
ln( 35 . ) = ln a + b ln(102 . ) + c ln(9.1)
b = 0.68 ⇒ c = −1.46
ln(2.58) = ln a + b ln(102 . ) + c ln(112 . )
a = 86.7 volts ⋅ kPa 1.46 / (L / s) 0.678
ln(3.72) = ln a + b ln(175 . ) + c ln(112 . )
. Slope=b, Intercept= ln a + c ln p (b) When P is constant (runs 1 to 4), plot ln Z vs. lnV 2
lnZ
1.5 1 0.5 0 -1
-0.5
0
lnZ = 0.5199lnV + 1.0035
0.5
1
1.5
lnV
b = slope = 0.52
Intercept = lna + c ln P = 10035 . When V is constant (runs 5 to 7), plot lnZ vs. lnP. Slope=c, Intercept= ln a + c lnV 2
lnZ
1.5 1 0.5 0 1.5
1.7
lnZ = -0.9972lnP + 3.4551
1.9
2.1
2.3
lnP
c = slope = −0.997 ⇒ 10 . Intercept = lna + b lnV = 3.4551
Z
Plot Z vs V b P c . Slope=a (no intercept) 7 6 5 4 3 2 1 0.05
Z = 31.096VbPc
0.1
0.15
0.2
Vb Pc
a = slope = 311 . volt ⋅ kPa / (L / s) .52
The results in part (b) are more reliable, because more data were used to obtain them.
2-12
2.39 (a)
sxy = sxx =
a= b=
∑x y
1 n
∑x
1 n
sx =
n
1 n
i i
= [(0.4)(0.3) + (2.1)(19 . ) + (31 . )( 3.2)] / 3 = 4.677
i =1 n
= (0.32 + 19 . 2 + 3.2 2 ) / 3 = 4.647
2 i
i =1 n
∑
xi = (0.3 + 1.9 + 3.2) / 3 = 18 . ; sy =
i =1
sxy − sx s y
b g
sxx − sx
2
=
sxx s y − sxy sx
b g
sxx − sx
2
1 n
n
∑y
i
= (0.4 + 2.1 + 31 . ) / 3 = 1867 .
i =1
4.677 − (18 . )(1.867) = 0.936 4.647 − (18 . )2 ( 4.647)(1867 . ) − (4.677)(18 . ) = 0.182 2 4.647 − (18 . )
=
. y = 0.936 x + 0182 (b) a =
sxy sxx
=
4.677 = 1.0065 ⇒ y = 1.0065x 4.647
4
y
3
y = 0.936x + 0.182
2
y = 1.0065x
1 0 0
1
2
3
4
x
2.40 (a) 1/C vs. t. Slope= b, intercept=a
a = Intercept = 0.082 L / g
3 2.5 2 1.5 1 0.5 0
2 1.5 C
1/C
(b) b = slope = 0.477 L / g ⋅ h;
1 0.5 0
0
1
1/C = 0.4771t + 0.0823
2
3
4
5
6
1
t
C
2 C-fitted
3
4
5
t
(c) C = 1 / (a + bt ) ⇒ 1 / [0.082 + 0.477(0)] = 12.2 g / L
t = (1 / C − a ) / b = (1 / 0.01 − 0.082) / 0.477 = 209.5 h (d) t=0 and C=0.01 are out of the range of the experimental data. (e) The concentration of the hazardous substance could be enough to cause damage to the biotic resources in the river; the treatment requires an extremely large period of time; some of the hazardous substances might remain in the tank instead of being converted; the decomposition products might not be harmless.
2-13
2.41 (a) and (c)
y
10
1 0.1
1
10
100
x
(b) y = ax b ⇒ ln y = ln a + b ln x; Slope = b, Intercept = ln a ln y = 0.1684ln x + 1.1258 2
ln y
1.5 1 0.5
b = slope = 0.168
0 -1
0
1
2 ln x
3
4
5
Intercept = ln a = 11258 . ⇒ a = 3.08
2.42 (a) ln(1-Cp/CA0) vs. t in rectangular coordinates. Slope=-k, intercept=0 (b)
600
0
800
ln(1-Cp/Cao)
400
ln(1-Cp/Cao)
0 200 0 -1 -2 -3 -4 ln(1-Cp/Cao) = -0.0062t
100
400
500
t
Lab 1
600
400
600
-4
-6 ln(1-Cp/Cao) = -0.0111t
t
Lab 2
k = 0.0111 s-1
800
0
0 ln(1-Cp/Cao)
ln(1-Cp/Cao)
200
300
-2
k = 0.0062 s-1
0
200
0
-2 -4
200
400
600
800
0 -2 -4
-6 ln(1-Cp/Cao)= -0.0064t
-6 ln(1-Cp/Cao) = -0.0063t t
t
Lab 3
k = 0.0063 s-1
Lab 4
k = 0.0064 s-1
(c) Disregarding the value of k that is very different from the other three, k is estimated with the average of the calculated k’s. k = 0.0063 s-1 (d) Errors in measurement of concentration, poor temperature control, errors in time measurements, delays in taking the samples, impure reactants, impurities acting as catalysts, inadequate mixing, poor sample handling, clerical errors in the reports, dirty reactor.
2-14
2.43 yi = axi ⇒ φ (a ) =
n
∑
di2
=
i =1
⇒a=
i =1
2.44
i
i =1
− axi
g
2
dφ ⇒ =0= da
∑ 2b y n
i
i =1
g
− axi xi ⇒
n
∑y x
i i
i =1
−a
n
∑x
2 i
i =1
n
n
∑
∑by n
yi xi /
∑x
2 i
i =1
DIMENSION X(100), Y(100) READ (5, 1) N C N = NUMBER OF DATA POINTS 1FORMAT (I10) READ (5, 2) (X(J), Y(J), J = 1, N 2FORMAT (8F 10.2) SX = 0.0 SY = 0.0 SXX = 0.0 SXY = 0.0 DO 100J = 1, N SX = SX + X(J) SY = SY + Y(J) SXX = SXX + X(J) ** 2 100SXY = SXY + X(J) * Y(J) AN = N SX = SX/AN SY = SY/AN SXX = SXX/AN SXY = SXY/AN CALCULATE SLOPE AND INTERCEPT A = (SXY - SX * SY)/(SXX - SX ** 2) B = SY - A * SX WRITE (6, 3) 3FORMAT (1H1, 20X 'PROBLEM 2-39'/) WRITE (6, 4) A, B 4FORMAT (1H0, 'SLOPEb -- bAb =', F6.3, 3X 'INTERCEPTb -- b8b =', F7.3/) C CALCULATE FITTED VALUES OF Y, AND SUM OF SQUARES OF RESIDUALS SSQ = 0.0 DO 200J = 1, N YC = A * X(J) + B RES = Y(J) - YC WRITE (6, 5) X(J), Y(J), YC, RES 5FORMAT (3X 'Xb =', F5.2, 5X /Yb =', F7.2, 5X 'Y(FITTED)b =', F7.2, 5X * 'RESIDUALb =', F6.3) 200SSQ = SSQ + RES ** 2 WRITE (6, 6) SSQ 6FORMAT (IH0, 'SUM OF SQUARES OF RESIDUALSb =', E10.3) STOP END $DATA 5 1.0 2.35 1.5 5.53 2.0 8.92 2.5 12.15 3.0 15.38 SOLUTION: a = 6.536, b = −4.206
2-15
=0
2.45 (a) E(cal/mol), D0 (cm2/s) (b) ln D vs. 1/T, Slope=-E/R, intercept=ln D0. (c) Intercept = ln D0 = -3.0151 ⇒ D0 = 0.05 cm2 / s .
3.0E-03
2.9E-03
2.8E-03
2.7E-03
2.6E-03
2.5E-03
2.4E-03
2.3E-03
2.2E-03
2.1E-03
2.0E-03
Slope = − E / R = -3666 K ⇒ E = (3666 K)(1.987 cal / mol ⋅ K) = 7284 cal / mol
-10.0 ln D
-11.0 -12.0 -13.0 -14.0
ln D = -3666(1/T) - 3.0151
1/T
(d) Spreadsheet T 347 374.2 396.2 420.7 447.7 471.2
D 1.34E-06 2.50E-06 4.55E-06 8.52E-06 1.41E-05 2.00E-05
1/T 2.88E-03 2.67E-03 2.52E-03 2.38E-03 2.23E-03 2.12E-03 Sx Sy Syx Sxx -E/R ln D0
lnD (1/T)*(lnD) -13.5 -0.03897 -12.9 -0.03447 -12.3 -0.03105 -11.7 -0.02775 -11.2 -0.02495 -10.8 -0.02296 2.47E-03 -12.1 -3.00E-02 6.16E-06 -3666 -3.0151
D0
7284
E
0.05
2-16
(1/T)**2 8.31E-06 7.14E-06 6.37E-06 5.65E-06 4.99E-06 4.50E-06
CHAPTER THREE 3.1
(a) m =
(b) m =
16 × 6 × 2 m3 1000 kg ≈ 2 × 10 5 2 103 ≈ 2 × 105 kg 3 m
b
8 oz 2s
106 cm3
1 qt
gb gb gd i
1g
32 oz 1056.68 qt cm
3
≈
4 × 106
b3 × 10gd10 i 3
≈ 1 × 102 g / s
(c) Weight of a boxer ≈ 220 lb m 12 × 220 lb m 1 stone Wmax ≥ ≈ 220 stones 14 lb m
dictionary
(d)
V= ≈
πD 2 L 4
=
2
314 . 4.5 ft 4
d
2
800 miles 5880 ft 7.4805 gal 1 barrel 1 mile 1 ft 3 42 gal
i d
i
3 × 4 × 5 × 8 × 10 2 × 5 × 10 3 × 7 4 × 4 × 10
(e) (i) V ≈
≈ 1 × 10 7 barrels
6 ft × 1 ft × 0.5 ft 28,317 cm3 ≈ 3 × 3 × 104 ≈ 1 × 105 cm3 3 1 ft
(ii) V ≈
150 lb m
1 ft 3
28,317 cm3
62.4 lb m
1 ft 3
≈
150 × 3 × 104 ≈ 1 × 105 cm3 60
(f) SG ≈ 105 .
3.2
995 kg 1 lb m 0.028317 m3 = 62.12 lb m / ft 3 3 3 m 0.45359 kg 1 ft
(a) (i)
(ii)
995 kg / m3
62.43 lb m / ft 3 1000 kg / m3
= 62.12 lb m / ft 3
(b) ρ = ρ H2 O × SG = 62.43 lb m / ft 3 × 5.7 = 360 lb m / ft 3 3.3
(a)
(b)
(c)
50 L
0.70 × 103 kg
1 m3
m3 103 L 1150 kg min 10 gal
= 35 kg
m3 1000 L 1 min = 27 L s 0.7 × 1000 kg 1 m3 60 s 1 ft 3
2 min 7.481 gal
0.70 × 62.43 lb m 1 ft 3
≅ 29 lb m / min
3-1
3.3 (cont’d) (d) Assuming that 1 cm3 kerosene was mixed with Vg (cm3 ) gasoline
d
i d i 1dcm kerosenei ⇒ 0.82dg kerosenei d0.70V + 0.82idg blendi = 0.78 ⇒ V SG = V + 1dcm blend i
Vg cm3gasoline ⇒ 0.70Vg g gasoline 3
g
3
g
=
g
Volumetric ratio =
3.4
In France: In U.S.:
3.5
0.82 − 0.78 3 = 0.5 0 cm 0.78 − 0.70
Vgasoline 0.50 cm3 3 3 = 3 = 0.50 cm gasoline / cm kerosene Vkerosene 1 cm
50.0 kg
L 5 Fr $1 = $68.42 0.7 × 10 . kg 1L 5.22 Fr 50.0 kg L 1 gal $1.20 = $22.64 0.70 × 10 . kg 3.7854 L 1 gal
VB ( ft 3 / h ), m B ( lb m / h )
V ( ft 3 / h), SG = 0.850
VH ( ft 3 / h ), m H ( lb m / h )
700 lb m / h
700 lb m ft 3 = 1319 . ft 3 / h (a) V = h 0.850 × 62.43 lb m 3 V ft 0.879 × 62.43 lb m kg / h m B = B = 54.88V B h ft 3 m = V 0.659 × 62.43 = 4114 . V kg / h H
d i bg d hb
b
g
H
H
b
g
g
VB + VH = 1319 . ft 3 / h m B + m H = 54.88VB + 4114 . VH = 700 lb m ⇒ V = 114 . ft 3 / h ⇒ m = 628 lb / h benzene B
B
m
VH = 1.74 ft 3 / h ⇒ m H = 71.6 lb m / h hexane (b) – No buildup of mass in unit. – ρ B and ρ H at inlet stream conditions are equal to their tabulated values (which are o
strictly valid at 20 C and 1 atm.) – Volumes of benzene and hexane are additive. – Densitometer gives correct reading.
3-2
3.6
(a) V =
195.5 kg H 2SO 4
1 kg solution
L
0.35kg H 2SO 4 12563 . × 1000 . kg
= 445 L
(b) 195.5 kg H 2 SO 4
L 18255 . × 1.00 kg 195.5 kg H 2 SO 4 0.65 kg H 2 O L + = 470 L 0.35 kg H 2 SO 4 1.000 kg 470 − 445 % error = × 100% = 5.6% 445
Videal =
3.7
b gE
b
Buoyant force up = Weight of block down
g
Mass of oil displaced + Mass of water displaced = Mass of block
b
g
b
g
ρ oil 0.542 V + ρ H O 1 − 0.542 V = ρ c V 2
. g / cm3 ⇒ ρ oil = 3.325 g / cm3 From Table B.1: ρ c = 2.26 g / cm3 , ρ w = 100 moil = ρ oil × V = 3.325 g / cm3 × 35.3 cm3 = 117.4 g moil + flask = 117.4 g + 124.8 g = 242 g 3.8
b g
b
Buoyant force up = Weight of block down
g
⇒ Wdisplaced liquid = Wblock ⇒ ( ρVg ) disp. Liq = ( ρVg ) block
b g
b g
Expt. 1: ρ w 15 . A g = ρB 2A g ⇒ ρB = ρw × ρ w =1.00 g/cm3
bg
15 . 2
b g
ρ B = 0.75 g / cm3 ⇒ SG
B
= 0.75
b g
b g
Expt. 2: ρ soln A g = ρ B 2 A g ⇒ ρ soln = 2 ρ B = 15 . g / cm3 ⇒ SG 3.9
= 15 .
soln
Let ρ w = density of water. Note: ρ A > ρ w (object sinks)
d
Volume displaced: Vd 1 = Ab hsi = Ab hp1 − hb1
hs 1
WA + WB
Archimedes ⇒
hρ1
hb1
ρ wVd 1 g
weight of displaced water
d
Subst. (1) for Vd 1 , solve for h p1 − hb1
Before object is jettisoned
WA + WB pw gAb
h p1 − hb1 =
i
(2)
bi g
bg
for b p 1 − hb 1
Vw = Ap h p1 −
d
bg
subst. 3 for h p 1 in
b 2 g, solve for h
b1
hb1 =
WA + WB V W + WB ⇒ h p1 = w + A pw g Ap pw gAp
b
W + WB Vw + A Ap pw g
g LM 1 MN A
−
p
3-3
1 Ab
OP PQ
(1)
= WA + WB
Volume of pond water: Vw = Ap h p1 − Vd 1 ⇒Vw = Ap h p1 − Ab h p1 − hb1 subst. 2
i
(3) (4)
i
3.9 (cont’d) hs 2
WB WA
WA ρ Ag
(5)
Volume displaced by boat: Vd 2 = Ab h p 2 − hb 2
(6)
Let V A = volume of jettisoned object = hρ2
h b2
After object is jettisoned
d
Archimedes ⇒ ρ WVd 2 g = WB Subst. for Vd 2
h p 2 − hb 2 =
E , solve for dh
WB pw gAb
⇒ hp 2 = hp 2
bg
subst. 8
⇒
bg
for h p 2 in 7 , solve for hb 2
(a)
p2
− hb 2
i (7)
Volume of pond water: Vw = Ap h p 2 − Vd 2 − V A solve for
i
b5g, b6g & b7 g
Vw = Ap h p 2 −
Vw WB WA + + Ap pw gAp p A gAp hb 2 =
WB W − A pw g p A g
(8)
Vw WB WA WB + + − Ap pw gAp p A gAp pw gAb
(9)
Change in pond level ( 8 ) − ( 3) W ⎡ 1 1 ⎤ WA ( pW − p A ) ρW < ρ A A hp 2 − hp1 = − ⎯⎯⎯⎯ → 0 > 0
5g b WA V p A Ap 1 1 1 = A 1+ −1 > 0 − + Ap g p A Ap pW Ap pW Ab Ap pW Ab
2.93 kg CaCO 3 L CaCO 3
(b) Wbag = ρ bulkVg =
0.70 L CaCO 3 L total
= 2.05 kg / L
2.05 kg 50 L 9.807 m / s2
1N
= 100 . × 103 N
L 1 kg ⋅ m / s Neglected the weight of the bag itself and of the air in the filled bag. 2
(c) The limestone would fall short of filling three bags, because – the powder would pack tighter than the original particles. – you could never recover 100% of what you fed to the mill.
3-4
3.11 (a) Wb = mb g =
122.5 kg 9.807 m / s2
1N 1 kg ⋅ m / s2
= 1202 N
(1202 N - 44.0 N) 1 kg ⋅ m / s2 Wb − WI = = 119 L ρwg 0.996 kg / L × 9.807 m / s2 1N m 122.5 kg = 103 ρb = b = . kg / L Vb 119 L
Vb =
m f + mnf = mb
(b)
xf =
mf mb
(1)
⇒ m f = mb x f
(2)
d
(1),(2) ⇒ mnf = mb 1 − x f V f + Vnf = Vb ⇒
b2 g,b 3g
⇒ mb
(c) x f =
Fx GH ρ
+
f
1 / ρ f − 1 / ρ nf
ρf
b
ρ nf
1 / ρ b − 1 / ρ nf
+
I=m JK ρ
1− xf
f
mf
i
mnf
ρ nf
=
⇒ xf
b
=
(3) mb
ρb
F1 GH ρ
−
f
1
ρ nf
I= 1 − 1 JK ρ ρ b
⇒ xf =
nf
1 / ρ b − 1 / ρ nf 1 / ρ f − 1 / ρ nf
1 / 103 . − 1 / 1.1 = 0.31 1 / 0.9 − 1 / 1.1
(d) V f + Vnf + Vlungs + Vother = Vb
mf
ρf
+
mnf
mb
+ Vlungs + Vother =
ρ nf
m f = mb x f mnf = mb (1− x f )
mb
Fx GH ρ
f
−
ρb
1− xf
f
ρ nf
I + (V JK
lungs
+ Vother ) = mb
F1− 1I GH ρ ρ JK b
nf
F 1 − 1 I = 1 − 1 − V +V GH ρ ρ JK ρ ρ m F 1 − 1 I − F V + V I F 1 1 I F 12. + 01. I GH ρ ρ JK GH m JK GH 1.03 − 11. JK − GH 122.5 JK ⇒x = = = 0.25 F1− 1I FG 1 − 1 IJ GH ρ ρ JK H 0.9 11. K ⇒ xf
lungs
f
nf
b
nf
b
lungs
b
nf
other
other
b
f
f
nf
3-5
Conc. (g Ile/100 g H2O)
3.12 (a) 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0.987
y = 545.5x - 539.03 R2 = 0.9992
0.989
0.991
0.993
0.995
0.997
Density (g/cm3)
. ρ − 539.03 From the plot above, r = 5455 (b) For ρ = 0.9940 g / cm3 ,
m Ile =
150 L
0.994 g
h
cm
3
r = 3.197 g Ile / 100g H 2 O 1000 cm3
3.197 g Ile
L
1 kg
103.197 g sol 1000 g
= 4.6 kg Ile / h
(c) The measured solution density is 0.9940 g ILE/cm3 solution at 50oC. For the calculation of Part (b) to be correct, the density would have to be changed to its equivalent at 47oC. Presuming that the dependence of solution density on T is the same as that of pure water, the solution density at 47oC would be higher than 0.9940 g ILE/cm3. The ILE mass flow rate calculated in Part (b) is therefore too low. 3.13 (a)
Mass Flow Rate (kg/min)
1.20 1.00
y = 0.0743x + 0.1523 R 2 = 0.9989
0.80 0.60 0.40 0.20 0.00 0.0
2.0
4.0
6.0
8.0
Rotameter Reading
3-6
10.0
12.0
3.13 (cont’d)
b g
From the plot, R = 5.3 ⇒ m = 0.0743 5.3 + 01523 . = 0.55 kg / min (b) Rotameter Collection Collected Volume Reading Time (cm3) (min) 2 1 297 2 1 301 4 1 454 4 1 448 6 0.5 300 6 0.5 298 8 0.5 371 8 0.5 377 10 0.5 440 10 0.5 453
Mass Flow Rate (kg/min) 0.297 0.301 0.454 0.448 0.600 0.596 0.742 0.754 0.880 0.906
b
Difference Duplicate (Di)
Mean Di
0.004 0.006 0.004
0.0104
0.012 0.026
g
1 0.004 + 0.006 + 0.004 + 0.012 + 0.026 = 0.0104 kg / min 5 . Di ) kg / min = 0.610 ± 0.018 kg / min 95% confidence limits: (0.610 ± 174 Di =
There is roughly a 95% probability that the true flow rate is between 0.592 kg / min and 0.628 kg / min . 3.14 (a) (b)
(c)
(d)
(e)
(f)
(g)
(h)
15.0 kmol C 6 H 6 15.0 kmol C 6 H 6
15,000 mol C 6 H 6 15,000 mol C 6 H 6
78.114 kg C 6 H 6 . × 103 kg C 6 H 6 = 117 kmol C 6 H 6 1000 mol = 15 . × 104 mol C 6 H 6 kmol lb - mole = 33.07 lb - mole C 6 H 6 453.6 mol 6 mol C 1 mol C 6 H 6
15,000 mol C 6 H 6
= 90,000 mol C
6 mol H = 90,000 mol H 1 mol C 6 H 6
90,000 mol C 12.011 g C mol C
= 1.08 × 106 g C
90,000 mol H 1.008 g H = 9.07 × 104 g H mol H 15,000 mol C 6 H 6
6.022 × 1023 mol
= 9.03 × 1027 molecules of C 6 H 6
3-7
3.15 (a) m =
(b) n =
175 m3
1000 L
h
m
0.866 kg 3
L
1h 60 min
= 2526 kg / min
2526 kg 1000 mol 1 min = 457 mol / s min 92.13 kg 60 s
(c) Assumed density (SG) at T, P of stream is the same as the density at 20oC and 1 atm 3.16 (a)
200.0 kg mix 0150 . kg CH 3OH kmol CH 3OH 1000 mol = 936 mol CH 3OH kg mix 32.04 kg CH 3OH 1 kmol
(b) m mix =
3.17
M=
m N 2
100.0 lb - mole MA h
0.25 mol N 2
74.08 lb m MA
1 lb m mix
1 lb - mole MA 0.850 lb m MA 28.02 g N 2
+
0.75 mol H 2
mol N 2 3000 kg kmol 0.25 kmol N 2 = h 8.52 kg kmol feed
= 8715 lb m / h
2.02 g H 2
= 8.52 g mol mol H 2 28.02 kg N 2 = 2470 kg N 2 h kmol N 2
3.18 M suspension = 565 g − 65 g = 500 g , M CaCO 3 = 215 g − 65 g = 150 g (a) V = 455 mL min , m = 500 g min (b) ρ = m / V = 500 g / 455 mL = 110 . g mL (c) 150 g CaCO 3 / 500 g suspension = 0.300 g CaCO 3 g suspension 3.19
Assume 100 mol mix. mC2 H 5OH =
10.0 mol C 2 H 5OH
46.07 g C 2 H 5OH
= 461 g C 2 H 5OH mol C 2 H 5OH 75.0 mol C 4 H 8 O 2 88.1 g C 4 H 8 O 2 = 6608 g C 4 H 8 O 2 mC4 H 8O 2 = mol C 4 H 8O 2 15.0 mol CH 3COOH 60.05 g CH 3COOH mCH 3COOH = = 901 g CH 3COOH mol CH 3COOH 461 g = 0.0578 g C 2 H 5OH / g mix xC2 H 5OH = 461 g + 6608 g + 901 g 6608 g = 0.8291 g C 4 H 8 O 2 / g mix xC 4 H 8 O 2 = 461 g + 6608 g + 901 g 901 g = 0113 . g CH 3COOH / g mix xCH 3COOH = 461 g + 6608 g + 901 g 461 g + 6608 g + 901 g MW = = 79.7 g / mol 100 mol 25 kmol EA 100 kmol mix 79.7 kg mix m= = 2660 kg mix 75 kmol EA 1 kmol mix
3-8
3.20 (a) Unit Crystallizer Filter Dryer
Function Form solid gypsum particles from a solution Separate particles from solution Remove water from filter cake 0.35 kg C aSO 4 ⋅ 2 H 2 O = 0 .35 kg C aSO 4 ⋅ 2 H 2 O L slurry
(b) m gypsum = 1 L slurry
L CaSO4 ⋅ 2H2O = 0151 . L CaSO4 ⋅ 2H2O 2.32 kg CaSO4 ⋅ 2H2O 0.35 kg gypsum 136.15 kg CaSO 4 CaSO 4 in gypsum: m = = 0.277 kg CaSO 4 172.18 kg gypsum
Vgypsum =
0.35 kg CaSO4 ⋅ 2H2O
CaSO 4 in soln.: m =
. g L sol b1− 0151
0.35 kg gypsum
(c) m =
% recovery =
1.05 kg 0.209 kg CaSO 4 L
100.209 kg sol
0.209 g CaSO 4 0.05 kg sol = 3.84 × 10 -5 kg CaSO 4 0.95 kg gypsum 100.209 g sol
0.277 g + 3.84 × 10 -5 g × 100% = 99.3% 0.277 g + 0.00186 g
3.21 CSA: FB:
= 0.00186 kg CaSO 4
45.8 L 0.90 kg min L 55.2 L 0.75 kg min
kmol = 0.5496 75 kg kmol = 0.4600 90 kg
L
U| |V || W
kmol mol CSA 0.5496 min ⇒ = 1.2 kmol mol FB 0.4600 min
She was wrong. The mixer would come to a grinding halt and the motor would overheat. 3.22 (a)
150 mol EtOH 6910 g EtO H
V =
46.07 g EtOH = 6910 g EtOH mol EtOH 0.600 g H 2 O = 10365 g H 2 O 0.400 g EtOH
6910 g EtOH
789 g EtOH
SG =
(b) V ′ =
L
(6910 +10365) g 19.1 L
L 1000 g
( 6910 + 10365) g mix
% error =
+
10365 g H 2 O
L 1000 g H 2 O
= 0.903
L = 18.472 L ⇒ 18.5 L 935.18 g
(19.123 − 18.472 ) L × 100% = 3.5% 18.472 L
3-9
= 19.123 L ⇒ 19.1 L
3.23
0.09 mol CH 4
16.04 g 0.91 mol Air 29.0 g Air + = 27.83 g mol mol mol 700 kg kmol 0.090 kmol CH 4 = 2.264 kmol CH 4 h h 27.83 kg 1.00 kmol mix 2.264 kmol CH 4 0.91 kmol air = 22.89 kmol air h h 0.09 kmol CH 4 M =
5% CH 4 ⇒
2.264 kmol CH 4 h
Dilution air required:
0.95 kmol air = 43.01 kmol air h 0.05 kmol CH 4
b43.01 - 22.89g kmol air h
1000 mol 1 kmol
= 20200 mol air h
20.20 kmol Air 29 kg Air Product gas: 700 kg + = 1286 kg h h
3.24
h
kmol Air
43.01 kmol Air
0.21 kmol O2
32.00 kg O2
h
h
1.00 kmol Air
1 kmol O2
1286 kg total
mi2 ∑V ≠ρ i
Not helpful.
xi =
B: 1
ρ
xi
∑
mi mi 1 = M Vi M
mi Vi 1 1 V = ∑ Vi = M = ρ Correct. m M i i xi 0.60 0.25 0.15 = + + = 1.091 ⇒ ρ = 0.917 g / cm 3 ρ i 0.791 1.049 1.595
∑ρ =
kg O2 kg
mi m M , ρi = i , ρ = M Vi V
∑ xi ρi = ∑
A:
= 0.225
=
∑M
3.25 (a) Basis: 100 mol N 2 ⇒ 20 mol CH 4
R|20 × 80 = 64 mol CO 25 ⇒S |T 20 × 40 = 32 mol CO 25
2
N total = 100 + 20 + 64 + 32 = 216 mol
32 64 = 0.15 m ol C O / m ol , x C O 2 = = 0.30 m ol C O 2 / m ol 216 216 100 20 = = 0.09 mol CH 4 / mol , x N 2 = = 0.46 mol N 2 / mol 216 216
xCO =
x CH 4
(b) M = ∑ yi M i = 015 . × 28 + 0.30 × 44 + 0.09 × 16 + 0.46 × 28 = 32 g / mol
3-10
3.26 (a) Samples Species
MW
k
Peak Mole Mass moles Area Fraction Fraction 3.6 0.156 0.062 0.540 2.8 0.233 0.173 0.804 2.4 0.324 0.353 1.121 1.7 0.287 0.412 0.991
mass
1
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
2
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
7.8 2.4 5.6 0.4
0.249 0.146 0.556 0.050
0.111 0.123 0.685 0.081
1.170 0.689 2.615 0.233
18.767 20.712 115.304 13.554
3
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
3.4 4.5 2.6 0.8
0.146 0.371 0.349 0.134
0.064 0.304 0.419 0.212
0.510 1.292 1.214 0.466
8.180 38.835 53.534 27.107
4
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
4.8 2.5 1.3 0.2
0.333 0.332 0.281 0.054
0.173 0.324 0.401 0.102
0.720 0.718 0.607 0.117
11.549 21.575 26.767 6.777
5
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
6.4 7.9 4.8 2.3
0.141 0.333 0.329 0.197
0.059 0.262 0.380 0.299
0.960 2.267 2.242 1.341
15.398 68.178 98.832 77.933
(b) REAL A(10), MW(10), K(10), MOL(10), MASS(10), MOLT, MASST INTEGER N, ND, ID, J READ (5, *) N CN-NUMBER OF SPECIES READ (5, *) (MW(J), K(J), J = 1, N) READ (5, *) ND DO 20 ID = 1 , ND READ (5, *)(A(J), J = 1, N) MOLT = 0. 0 MASST = 0. 0 DO 10 J = 1, N MOL(J) = MASS(J) = MOL(J) * MW(J) MOLT = MOLT + MOL(J) MASST = MASST + MASS(J) 10 CONTINUE DO 15 J = 1, N MOL(J) = MOL(J)/MOLT MASS(J) = MASS(J)/MASST 15 CONTINUE WRITE (6, 1) ID, (J, MOL(J), MASS (J), J = 1, N) 20 CONTINUE 1 FORMAT (' SAMPLE: `, I3, /, ∗ ' SPECIES MOLE FR. MASS FR.', /,
3-11
8.662 24.164 49.416 57.603
3.26 (cont’d) ∗ 10(3X, I3, 2(5X, F5.3), /), /) END $DATA ∗ 4 16. 04 0. 150 30. 07 0. 287 44 . 09 0. 467 58. 12 0. 583 5 3. 6 2. 8 2. 4 1. 7 7 . 8 2. 4 5. 6 0. 4 3. 4 4 . 5 2. 6 0. 8 4 . 8 2. 5 1. 3 0. 2 6 . 4 7. 9 4 . 8 2. 3 [OUTPUT] SAMPLE: 1 SPECIES MOLE FR MASS FR 1 0.156 0.062 2 3 4 SAMPLE: 2 (ETC.)
3.27 (a)
0.233 0.324 0.287
(8.7 × 10 6 × 0.40) kg C
0.173 0.353 0.412
44 kg CO 2 = 1.28 × 10 7 kg CO 2 ⇒ 2.9 × 105 kmol CO 2 12 kg C
(11 . × 10 6 × 0.26) kg C 28 kg CO 12 kg C ( 3.8 × 10 5 × 0.10) kg C
m=
= 6.67 × 10 5 kg CO ⇒ 2.38 × 10 4 kmol CO
16 kg CH 4 = 5.07 × 10 4 kg CH 4 ⇒ 3.17 × 10 3 kmol CH 4 12 kg C
(1.28 × 10 7 + 6.67 × 10 5 + 5.07 × 10 4 ) kg 1 metric ton
M =
1000 kg
∑y
i
= 13,500
metric tons yr
M i = 0.915 × 44 + 0.075 × 28 + 0.01 × 16 = 42.5 g / mol
3.28 (a) Basis: 1 liter of solution 1000 mL
1.03 g 5 g H 2 SO 4 mL 100 g
mol H 2 SO 4 = 0.525 mol / L ⇒ 0.525 molar solution 98.08 g H 2 SO 4
3-12
3.28 (cont’d) (b) t = V = V 55 gal
55 gal
3.7854 L
min
60 s
gal
87 L
min
3.7854 L 10 3 mL 1.03 g gal
1L
= 144 s
0.0500 g H 2 SO 4
1 lbm
g
453.59 g
mL
= 23.6 lb m H 2 SO 4
m 3 1 min V 87 L = = 0.513 m / s A min 1000 L 60 s (π × 0.06 2 / 4 ) m 2 L 45 m t= = = 88 s u 0.513 m / s
(c) u =
3.29 (a) n1 (mol/min)
n2 (mol/min)
0.180 mol C6H14/mol 0.820 mol N2/mol
0.050 mol C6H14/mol 0.950 mol N2/mol
1.50 L C6H14(l)/min n 3 (mol C6H14(l)/min)
n3 =
. L 0.659 kg 1000 mol 150 min
L
86.17 kg
. mol / min = 1147
UV W
|RS |T
Hexane balance: 0.180n1 = 0050 . n2 + 1147 . (mol C6 H14 / min) solve n1 = 838 . mol / min ⇒ n2 = 72.3 mol / min Nitrogen balance: 0.820n1 = 0950 . n2 (mol N2 / min) (b) Hexane recovery =
30 mL 3.30
n3 1147 . × 100% = × 100% = 76% n1 0180 . 838 .
b g
1L
0.030 mol
172 g
103 mL
lL
1 mol
= 0155 . g Nauseum
3-13
3.31 (a) kt is dimensionless ⇒ k (min -1 ) (b) A semilog plot of CA vs. t is a straight line ⇒ ln CA = ln CAO − kt
ln(CA)
1 0 -1 -2 -3 -4 -5
y = -0.4137x + 0.2512 R2 = 0.9996
0.0
5.0 t (m in)
10.0
k = 0.414 min −1
ln CAO = 02512 . ⇒ CAO = 1286 . lb - moles ft 3
FG 1b - molesIJ = C′ mol 28.317 liter H ft K liter 1 ft t ′bsg 1 min t bming = = t ′ 60 60 s
(c) C A
3
A
2.26462 lb - moles
3
1000 mol
= 0.06243C A′
C A = C A 0 exp(− kt )
b
g
0.06243C A′ = 1334 . exp −0.419t ′ 60
drop primes
⇒
b
g
b
C A mol / L = 214 . exp −0.00693t
g
t = 200 s ⇒ C A = 5.30 mol / L 3.32 (a)
(b)
2600 mm Hg
14.696 psi = 50.3 psi 760 mm Hg
275 ft H 2 O 101.325 kPa = 822.0 kPa 33.9 ft H 2 O
3.00 atm 101325 . × 105 N m2 12 m2 (c) = 30.4 N cm2 2 2 1 atm 100 cm (d)
280 cm Hg 10 mm 101325 . × 106 dynes cm2 1002 cm2 dynes = 3.733 × 1010 2 2 m2 1 cm 760 mm Hg 1 m
(e) 1 atm −
20 cm Hg 10 mm 1 atm = 0.737 atm 1 cm 760 mm Hg
3-14
3.32 (cont’d) (f)
(g)
b
25.0 psig 760 mm Hg gauge 14.696 psig
b25.0 + 14.696gpsi
g = 1293 mm Hg bgaugeg b g
760 mm Hg = 2053 mm Hg abs 14.696 psi
b
(h) 325 mm Hg − 760 mm Hg = −435 mm Hg gauge
(i)
Eq. (3.4-2) ⇒ h =
g
2 ft 3 P 35.0 lbf 144 in = ρg in2 1 ft 2 1.595x62.43 lbm
s2
32.174 lbm ⋅ ft
100 cm
32.174 ft
s ⋅ lbf
3.2808 ft
2
= 1540 cm CCl4
3.33 (a) Pg = ρgh =
0.92 × 1000 kg 9.81 m / s2 m3
h (m)
1N 1 kPa 2 1 kg ⋅ m / s 103 N / m2
. Pg (kPa) ⇒ h (m) = 0111
h
Pg Pg = 68 kPa ⇒ h = 0111 . × 68 = 7.55 m
FG H
moil = ρV = 0.92 × 1000
IJ FG K H
IJ K
kg 16 2 3 × 7 . 55 × π × m = 14 . × 10 6 kg 4 m3
(b) Pg + Patm = Ptop + ρgh
b
g b g
68 + 101 = 115 + 0.92 × 1000 × 9.81 / 103 h ⇒ h = 5.98 m 3.34 (a) Weight of block = Sum of weights of displaced liquids ρ h + ρ 2 h2 (h1 + h2 ) Aρ b g = h1 Aρ 1 g + h2 Aρ 2 g ⇒ ρ b = 1 1 h1 + h2 (b)
Ptop = Patm + ρ1gh0 , Pbottom = Patm + ρ1g(h0 + h1) + ρ2 gh2 , Wb = ρb (h1 + h2 ) A
⇒Fdown = ( Patm + ρ1gh0 ) A + ρb (h1 + h2 ) A , Fup = [ Patm + ρ1g(h0 + h1) + ρ2 gh2 ]A Fdown = Fup ⇒ ρb (h1 + h2 ) A = ρ1gh1 A + ρ2 gh2 A ⇒ Wblock = Wliquid displaced
3-15
3.35
b
g
Δ P = Patm + ρgh − Pinside
= 1 atm − 1 atm +
F=
3.36
. g1000 kg b105 m3
9.8066 m 150 m 12 m2 1N s2 1002 cm2 1 kg ⋅ m / s2
FG H
IJ K
lb f 022481 . 154 N 65 cm2 4 N = 100 . × 10 × = 2250 lb f 1N cm2
14 . × 62.43 lb m 1 ft 3 2.3 × 106 gal m = ρV = = 2.69 × 107 lb m 3 ft 7.481 gal P = P0 + ρgh . × 62.43 lb m 32.174 ft 30 ft 1 lb f 12 ft 2 lb f 14 = 14.7 2 + in ft 3 s2 32.174 lb m ⋅ ft / s2 12 2 in 2 = 32.9 psi — Structural flaw in the tank. — Tank strength inadequate for that much force. — Molasses corroded tank wall
3.37 (a) mhead =
π × 24 2 × 3 in 3
W = mhead g =
4 392 lb m
1 ft 3 8.0 × 62.43 lb m = 392 lb m 3 3 12 in ft 3 1 lb f 32.174 ft / s 2 = 392 lb f 32.174 lb m ⋅ ft / s 2
⎡⎣( 30 + 14.7 ) ⎤⎦ lb f π × 202 in 2 Fnet = Fgas − Fatm − W = in 2 4 −
14.7 lbf π × 242 in 2 − 392 lb f = 7.00 × 103 lbf 2 in 4
The head would blow off.
7.000 × 10 lbf F Initial acceleration: a = net = 392 lb m mhead 3
32.174 lb m ⋅ ft/s 2 1 lb f
= 576 ft/s 2
(b) Vent the reactor through a valve to the outside or a hood before removing the head.
3-16
3.38 (a)
a
Pa = ρgh + Patm , Pb = Patm If the inside pressure on the door equaled Pa , the force on the door would be F = Adoor ( Pa − Pb ) = ρghAdoor Since the pressure at every point on the door is greater than Pa , Since the pressure at every point on the door is greater than Pa , F >ρghAdoor
b
2m 1m
(b) Assume an average bathtub 5 ft long, 2.5 ft wide, and 2 ft high takes about 10 min to fill. V 5 × 25 . × 2 ft 3 = 25 Vtub = ≈ . ft 3 / min ⇒ V = 5 × 25 . = 125 . ft 3 / min 10 min t
For a full room, h = 7 m
(i)
⇒F >
1000 kg 9.81 m 3
m
2
s
7 m 2 m2
1N 1 kg ⋅ m/s
2
⇒ F > 1.4 ×105 N
The door will break before the room fills (ii)
d i dP i
3.39 (a) Pg g
If the door holds, it will take 35.3145 ft 3 5 × 15 × 10 m3 V t fill = room = V 12.5 ft 3 / min 1 m3 He will not have enough time.
b
tap
=
25 m H 2 O
junction
g
1h = 31 h 60 min
101.3 kPa
= 245 kPa 10.33 m H 2 O 25 + 5 m H 2 O 101.3 kPa = = 294 kPa 10.33 m H 2 O
b
g
(b) Air in the line. (lowers average density of the water.) (c) The line could be clogged, or there could be a leak between the junction and the tap. 3.40
Pabs = 800 mm Hg Pgauge = 25 mm Hg Patm = 800 − 25 = 775 mm Hg
3-17
b
g
3.41 (a) P1 + ρ A g h1 + h2 = P2 + ρ B gh1 + ρ C gh2
b
g
b
g
⇒ P1 − P2 = ρ B − ρ A gh1 + ρ C − ρ A gh2
LMb10. − 0.792g g 981 cm 30.0 cm + b137 . − 0.792g g cm cm s N F 1 dyne I F I = 123.0 kPa 101325 . kPa ×G J G H 1 g ⋅ cm / s K H 1.01325 × 10 dynes / cm JK
(b) P1 = 121 kPa +
3
2
2
3.42
3
6
OP Q
981 cm 24.0 cm s2
2
(a) Say ρt (g/cm3) = density of toluene, ρm (g/cm3) = density of manometer fluid ρ t g (500 − h + R ) = ρ m gR ⇒ R =
500 − h
ρm −1 ρt
. cm (i) Hg: ρ t = 0.866, ρ m = 13.6, h = 150 cm ⇒ R = 238 . , h = 150 cm ⇒ R = 2260 cm (ii) H 2 O: ρ t = 0.866, ρ m = 100
Use mercury, because the water manometer would have to be too tall. (b) If the manometer were simply filled with toluene, the level in the glass tube would be at the level in the tank. Advantages of using mercury: smaller manometer; less evaporation. (c) The nitrogen blanket is used to avoid contact between toluene and atmospheric oxygen, minimizing the risk of combustion. 3.43
b
P g 7.23 g F P − ρ gIJ b26 cmg P − P = d ρ − ρ i gb26 cmg = G H 7.23 m K Patm = ρ f g 7.23 m ⇒ ρ f =
atm
atm
a
b
f
w
w
F756 mmHg 1 m −1000 kg 9.81 m/s GH 7.23 m 100 cm m
2
=
3
Ib g JK
1N 760 mmHg 1m 26 cm 1 kg⋅ m/s2 1.01325×105 N m2 100 cm
⇒ Pa − Pb = 81 . mm Hg
3.44 (a) Δh = 900 − hl =
75 . psi 760 mm Hg 14.696 psi
= 388 mm Hg ⇒ hl = 900 − 388=512 mm
(b) Δh = 388 − 25 × 2 = 338 mm ⇒ Pg =
338 mm Hg
14.696 psi 760 mm Hg
3-18
= 6.54 psig
3.45 (a) h = L sin θ (b) h = 8.7 cm sin 15° = 2.3 cm H 2 O = 23 mm H 2 O
b
g b g
3.46 (a) P = Patm − Poil − PHg
920 kg 9.81 m / s2 = 765 − 365 − m3 = 393 mm Hg
0.10 m
1N 760 mm Hg 2 1 kg ⋅ m / s 1.01325 × 105 N / m2
(b) — Nonreactive with the vapor in the apparatus. — Lighter than and immiscible with mercury. — Low rate of evaporation (low volatility).
c
h
3.47 (a) Let ρ f = manometer fluid density 110 . g cm 3 , ρ ac = acetone density
c0.791 g cm h 3
d
i
Differential manometer formula: ΔP = ρ f − ρ ac gh
. − 0791 . gg 981 cm h (mm) 1 cm 1 dyne g b110 cm s 10 mm 1 g⋅ cm/ s = 0.02274 hb mmg V b mL sg 62 87 107 123 138 151 hb mmg 5 10 15 20 25 30 ΔPb mm Hgg 0.114 0.227 0.341 0.455 0.568 0.682
b
ΔP mm Hg =
3
2
2
b g
(b) lnV = n ln ΔP + ln K 6
ln(V)
5.5
y = 0.4979x + 5.2068
5 4.5 4 -2.5
-2
-1.5 -1 ln( P)
-0.5
0
b g
. ln ΔP + 52068 . From the plot above, ln V = 04979 ml s ⇒ n = 04979 . ≈ 05 . , ln K = 5.2068 ⇒ K = 183 mm Hg
b
3-19
g
0.5
760 mm Hg 1.01325×106 dyne/ cm2
3.47 (cont’d)
b
b
gb g
g
(c) h = 23 ⇒ ΔP = 0.02274 23 = 0.523 mm Hg ⇒ V = 183 0.523 132 mL 0.791 g s
mL
= 104 g s
104 g s
1 mol 58.08 g
0.5
= 132 mL s
= 180 . mol s
. = 544° R / 18 . = 303 K − 273 = 30°C 3.48 (a) T = 85° F + 4597 . = 474° R − 460 = 14° F (b) T = −10° C + 273 = 263 K × 18 (c) ΔT =
(d)
85° C 10 . °K 85° C 18 . °F 85° C 1.8° R = 85° K; = 153° F; = 153° R . °C . °C 10 1° C 10
150° R 1° F 150° R 1.0° C 150° R 1.0D K = 150° F; = 83.3° K; = 83.3° C 1° R 1.8° R 1.8° R
3.49 (a) T = 0.0940 × 1000D FB + 4.00 = 98.0D C ⇒ T = 98.0 × 1.8 + 32 = 208D F (b) ΔT (D C) = 0.0940ΔT (D FB) = 0.94D C ⇒ ΔT (K) = 0.94 K
0.94D C 1.8D F ΔT (D F) = . D F ⇒ ΔT (D R) = 169 . DR = 169 D 1.0 C
D D (c) T1 = 15D C ⇒ 100D L ; T2 = 43 C ⇒1000 L T (D C) = aT (D L) + b a=
⇒
b43 − 15gD C = 0.0311FG D C IJ ; H D LK b1000 - 100gD L
. DC b = 15 − 0.0311 × 100 = 119
T (D C) = 0.0311T (D L) + 11.9 and T (D L) =
1 ⎡0.0940T (o FB)+4.00-11.9 ⎤⎦ = 3.023T (o FB)-254 0.0311 ⎣
(d) Tbp = −88.6D C ⇒ 184.6 K ⇒ 332.3D R ⇒ -127.4D F ⇒ −9851 . D FB ⇒ −3232 D L (e) ΔT = 50.0D L ⇒ 1.56D C ⇒ 16.6D FB ⇒ 156 . K ⇒ 2.8D F ⇒ 2.8D R
3-20
3.50
bT g = 100° C bT g (a) V b mVg = aT b° Cg + b b H 2O
m AgCl
5.27 = 100a + b
⇒
= 455° C
a = 0.05524 mV ° C
24.88 = 455a + b b = −0.2539 mV V mV = 0.05524T ° C − 0.2539
b g
b g
⇓ T ° C = 1810 . V mV + 4.596
b g
b g
. mV→136 . mV ⇒1856 . °C →2508 . °C ⇒ (b) 100 3.51 (a) ln T = ln K + n ln R n=
b
b
g
. − 1856 . °C dT 2508 = = 326 . °C / s dt 20 s
T = KR n
g = 1184 .
ln 250.0 110.0
b
g
ln 40.0 20.0
. ln K = ln 1100 . − 1184 . (ln200 . ) = 1154 . ⇒ K = 3169 . ⇒T = 3169 . R1184
F 320 IJ (b) R = G H 3169 . K
1/1.184
= 49.3
(c) Extrapolation error, thermocouple reading wrong. 3.52 (a) PV = 0.08206nT
b g Pbatmg = . 14696
. P′ psig + 14696
b g b
g
n mol = n ′ lb - moles ×
bg d i
. ft 3 28317 , V L = V ′ ft × L 3
T ′(D F) − 32 453.59 mol . , T(D K) = + 27315 lb − moles 1.8
b P′ + 14.696g × V ′ × 28317 453.59 L (T ′ − 32) O . . P = 0.08206 × n ′ × ×M + 27315 14.696 1 N 1.8 Q
⇒
b
g
b
g
⇒ P ′ + 14.696 × V ′ =
0.08206 × 14.696 × 453.59 × n ′ × T ′ + 459.7 28.317 × 18 .
b
b
g
⇒ P′ + 14.696 V ′ = 1073 . n′ T ′ + 459.7
3-21
g
3.52 (cont’d) (b) ntot ′ =
b500 + 14.696g × 35. = 0.308 lb - mole 10.73 × b85 + 459.7g
mCO =
(c) T ′ =
0308 . lb - mole 0.30 lb - mole CO 28 lb m CO = 2.6 lbm CO lb - mole lb - mole CO
b3000 + 14.696g × 35. − 459.7 = 2733D F 10.73 × 0.308
b g
b
g
3.53 (a) T ° C = a × r ohms + b
UV ⇒ 100 = 33028 . a + bW 0 = 23624 . a +b
a = 10634 . b = −25122 .
b g
b g
⇒ T ° C = 10634 . r ohms − 25122 .
FG kmol IJ = n ′ (kmol) 1 min = n ′ H s K min 60 s 60 P′bmm Hgg 1 atm P′ Pbatmg = = 760 mm Hg 760
(b) n
bg b g
, T K = T ′ ° C + 27316 .
F I GH JK
m3 1 min V ′ m3 V =V ′ = s min 60 s 60
b g d b g
i
. P′ mm Hg V ′ m3 min 0016034 . P′ V ′ n ′ 12186 = ⇒ n ′ = . 60 T ′ ° C + 27316 . 760 T ′ + 27316 60 (c) T = 10.634r − 25122 .
r1 = 26159 . ⇒ T1 = 26.95° C ⇒ r2 = 26157 . ⇒ T2 = 26.93° C r3 = 44.789 ⇒ T3 = 2251 . °C P (mm Hg) = h + Patm = h + (29.76 in Hg)
FG 760 mm Hg IJ = h + 755.9 H 29.92 in Hg K
h1 = 232 mm ⇒ P1 = 987.9 mm Hg . mm Hg ⇒ h2 = 156 mm ⇒ P2 = 9119 h3 = 74 mm ⇒ P3 = 829.9 mm Hg
3-22
3.53 (cont’d)
b0.016034gb987.9gb947 60g = 0.8331 kmol CH 26.95 + 27316 . b0.016034gb9119. gb195g = 9.501 kmol air min =
(d) n1 =
n2
4
min
26.93 + 27316 . n3 = n1 + n2 = 10.33 kmol min (e) V3 =
(f)
b
g b
gb
g
n3 T2 + 27316 . 10.33 2251 . + 27316 . = = 387 m3 min 0.016034 P3 0.016034 829.9
b
gb
0.8331 kmol CH 4 16.04 kg CH 4 min kmol 0.21× 9.501 kmol O2 32.0 kg O2
min xCH4 =
kmol O2
g
kg CH 4 min 0.79 × 9.501 kmol N2 28.0 kg N2
= 13.36
+
min
13.36 kg CH 4 min = 0.0465 kg CH 4 kg (13.36 + 274) kg / min
REAL, MW, T, SLOPE, INTCPT, KO, E REAL TIME (100), CA (100), TK (100), X (100), Y(100) INTEGER IT, N, NT, J READ 5, ∗ MW, NT DO 10 IT=1, NT READ 5, ∗ TC, N TK(IT) = TC + 273.15 READ 5, ∗ (TIME (J), CA (J), J = 1 , N) DO 1 J=1, N CA J = CA J / MW
3.54
b g b g b g
bg bg XbJg = TIMEbJg YbJg = 1./CAbJg
1
CONTINUE CALL LS (X, Y, N, SLOPE, INTCPT)
b g
K IT = SLOPE
WRITE (E, 2) TK (IT), (TIME (J), CA (J), J = 1 , N) WRITE (6, 3) K (IT) 10 CONTINUE DO 4 J=1, NT X J = 1./TK J
bg bg YbJg = LOGcKbJgh
3-23
kmol N2
= 274
kg air min
3.54 (cont’d)
4 CONTINUE CALL LS (X, Y, NT, SLOPE, INTCPT) KO = EXP INTCPT
b
2
3 5
10
g
E = −8.314 = SLOPE WRITE (6, 5) KO, E FORMAT (' TEMPERATURE (K): ', F6.2, / * ' TIME CA', /, * ' (MIN) (MOLES)', / * 100 (IX, F5.2, 3X, F7.4, /)) FORMAT (' K (L/MOL – MIN): ', F5.3, //) FORMAT (/, ' KO (L/MOL – MIN) : ', E 12.4, /, ' E (J/MOL): ', E 12.4) END SUBROUTINE LS (X, Y, N, SLOPE, INTCPT) REAL X(100), Y(100), SLOPE, INTCPT, SX, SY, SXX, SXY, AN INTEGER N, J SX=0 SY=0 SXX=0 SXY=0 DO 10 J=1,N SX = SX + X(J) SY = SY + Y(J) SXX = SXX + X(J)**2 SXY = SXY + X(J)*Y(J) CONTINUE AN = N SX = SX/AN SY = SY/AN SXX = SXX/AN SXY = SXY/AN SLOPE = (SXY – SX*SY)/(SXX – SX**2) INTCPT = SY – SLOPE*SX RETURN END $ DATA 65.0 94.0 10.0 20.0 30.0 40.0 50.0
4 6 8.1 4.3 3.0 2.2 1.8
[OUTPUT] TEMPERATURE (K): 367.15 TIME CA (MIN) (MOLS/L) 10.00 0.1246 20.00 0.0662 30.00 0.0462 40.00 0.0338
3-24
3.54 (cont’d)
60.0
1.5
50.00 0.0277 60.00 0.0231
b
g
K L / MOL ⋅ MIN : 0.707 110. 10.0 20.0 30.0 40.0 50.0 60.0
6 3.5 1.8 1.2 0.92 0.73 0.61
127.
6
# # ETC
bat 94°Cg
TEMPERATURE (K): 383.15 #
b
g
K L / MOL ⋅ MIN : 1.758 #
b g E bJ / MOLg: 0.6690E + 05
K0 L / MOL − MIN : 0.2329E + 10
3-25
CHAPTER FOUR 4.1
a.
Continuous, Transient
b.
Input – Output = Accumulation No reactions ⇒ Generation = 0, Consumption = 0
6.00 c.
4.2
t=
dn kg kg dn kg − 3.00 = ⇒ = 3.00 dt dt s s s
100 . m3 1000 kg 1 s = 333 s 1 m3 3.00 kg
a.
Continuous, Steady State
b.
k = 0 ⇒ C A = C A0
c.
Input – Output – Consumption = 0 Steady state ⇒ Accumulation = 0 A is a reactant ⇒ Generation = 0
k = ∞ ⇒ CA = 0
FG IJ FG IJ FG IJ FG IJ H K H K H K H K
FG IJ H K
3 m3 mol m C mol + kVC mol ⇒ C = C A 0 V CA0 V = A A A kV s m3 s m3 s 1+ V
4.3
b
v kg / h m
a. 100 kg / h 0.550 kg B / kg 0.450 kg T / kg
g
0.850 kg B / kg 0.150 kg T / kg
b
l kg / h m
g
Input – Output = 0 Steady state ⇒ Accumulation = 0 No reaction ⇒ Generation = 0, Consumption = 0
0.106 kg B / kg 0.894 kg T / kg
v + m l (1) Total Mass Balance: 100.0 kg / h = m v + 0106 l (2) Benzene Balance: 0.550 × 100.0 kg B / h = 0.850m . m
v = 59.7 kg h, m l = 40.3 kg h Solve (1) & (2) simultaneously ⇒ m b.
The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by masses (kg). The balance equations are also identical (initial input = final output).
c.
Possible explanations ⇒ a chemical reaction is taking place, the process is not at steady state, the feed composition is incorrect, the flow rates are not what they are supposed to be, other species are in the feed stream, measurement errors.
4- 1
4.4
b.
b
n (mol) 0.500 mol N 2 mol 0.500 mol CH 4 mol
c.
100.0 g / s
n E =
b g bg C H gg bg C H gg
x E g C2 H 6 g xP xB
d.
3
8
4
10
b
b
g
b
b g R|n blb - mole DA sg U| S| 0.21 lb - moleO lb - mole DA V| T 0.79 lb - mole N lb - mole DAW
g
100 x E g C 2 H 6 1 lb m lb - mole C2 H 6 3600 s s h 453593 . g 30 lb m C2 H 6
= 26.45x E lb - mole C 2 H 6 / h
g
nO2 = 0.21n2 ( lb-mole O 2 / s )
n1 lb - mole H 2 O s 2
xH2O =
2
2
e.
g
0.500 n mol N 2 28 g N 2 1 kg = 0.014 n kg N 2 mol N 2 1000 g
xO2 =
n ( mol )
n1 ⎛ lb-mole H 2 O ⎞ n1 + n2 ⎜⎝ lb-mole ⎟⎠
0.21n2 ⎛ lb-mole O 2 ⎞ n1 + n2 ⎜⎝ lb-mole ⎟⎠
nN2O4 = n ⎡⎣0.600 − yNO2 ⎤⎦ ( mol N 2 O4 )
0.400 mol NO mol yNO2 ( mol NO 2 mol ) 0.600 − yNO2 ( mol N 2 O 4 mol ) 4.5
a.
Basis: 1000 lbm C3H8 / h fresh feed (Could also take 1 h operation as basis flow chart would be as below except that all / h would be deleted.) 1000 lb m C3H 8 / h
b
b
n 6 lb m / h
n 7 lb m / h
g
0.02 lb m C3H8 / lb m 0.98 lb m C3H 6 / lb m
g
0.97 lb m C3H8 / lb m 0.03 lb m C3H 6 / lb m
Still
Compressor
b n blb n blb n blb
g C H / hg CH / h g H / hg
b b
n1 lb m C3H 8 / h
n1 lb m C3H 8 / h
Reactor
2
m
3
m
4
m
3
4
2
Note: the compressor and the off gas from the absorber are not mentioned explicitly in the process description, but their presence should be inferred.
b b
n3 lb m CH 4 / h n 4 lb m H 2 / h
g
g
b
n5 lb m / h
g
Stripper
Absorber
b b n blb
n1 lb m C3H 8 / h
g g
n 2 lb m C3H 6 / h 5
4- 2
g g
n 2 lb m C3H 6 / h
6
m
oil / h
g
4.5 (cont’d) b. Overall objective: To produce C3H6 from C3H8. Preheater function: Raise temperature of the reactants to raise the reaction rate. Reactor function: Convert C3H8 to C3H6. Absorption tower function: Separate the C3H8 and C3H6 in the reactor effluent from the other components. Stripping tower function: Recover the C3H8 and C3H6 from the solvent. Distillation column function: Separate the C3H5 from the C3H8. 4.6
a.
3 independent balances (one for each species)
b.
1 , m 3 , m 5 , x2 , y2 , y4 , z4 ) 7 unknowns ( m – 3 balances – 2 mole fraction summations 2 unknowns must be specified
c.
y2 = 1 − x2
FG kg A IJ = m + b1200gb0.70g FG kg A IJ H hK H hK F kg I F kg I + 5300 G J = m + 1200 + m G J Overall Balance: m H hK H hK F kg BIJ = 1200 y + 0.60m FG kg BIJ + 5300 x G B Balance: 0.03m H hK H hK A Balance: 5300 x2
3
1
3
1
5
2
4
5
z4 = 1 − 0.70 − y4 4.7
a.
3 independent balances (one for each species)
b.
Water Balance:
bg b g
R g 0.995 g H O 400 g 0.885 g H 2O m 2 R = 356 g min = ⇒m g min g min
Acetic Acid Balance:
F g CH OOH IJ = 0.005m . gG b400gb0115 H min K 3
R
E = 461g min ⇒m
FG g CH OOH IJ H min K 3
FG g IJ = m + m FG g IJ ⇒ m = 417 g min H min K H min K F g IJ = b0.096gb461g FG g IJ ⇒ 44 g min = 44 g min . gb400g − b0.005gb356g G b0115 H min K H min K
C + 400 Overall Balance: m c.
E + 0.096m
R
4- 3
E
C
4.7 (cont’d) d.
CH 3COOH
H 2O some CH3COOH CH3COOH H 2O C 4 H9OH
Extractor
C4 H9OH CH3COOH
Distillation Column
C4 H 9OH
4.8
a. 120 eggs/min 0.30 broken egg/egg 0.70 unbroken egg/egg
X-large: 25 broken eggs/min 35 unbroken eggs/min 45 Large: n 1 broken eggs/min n 2 unbroken eggs/min
b.
120 = 25 + 45 + n1 + n2 ( eggs min ) ⇒ n1 + n2 = 50 ⎫ n1 = 11 ⎪ ⇒ ⎬ ( 0.30 )(120 ) = 25 + n1 ⎪⎭ n2 = 39
c.
n1 + n2 = 50 large eggs min
b
g
n1 large eggs broken/50 large eggs = 11 50 = 0.22
4.9
b
g
d.
22% of the large eggs (right hand) and 25 70 ⇒ 36% of the extra-large eggs (left hand) are broken. Since it does not require much strength to break an egg, the left hand is probably poorly controlled (rather than strong) relative to the right. Therefore, Fred is right-handed.
a.
m1 lb m strawberries
b
015 . lb m S / lb m 0.85 lb m W / lb m
c
m2 lb m S sugar
g
b
m3 lb m W evaporated
g
1.00 lb m jam 0.667 lb m S / lb m
h
0.333 lb m W / lb m
b.
c.
3 unknowns ( m1 , m2 , m3 ) – 2 balances – 1 feed ratio 0 DF Feed ratio: m1 / m 2 = 45 / 55 (1) S balance: 0.15 m1 + m 2 = 0.667 (2) Solve simultaneously ⇒ m1 = 0.49 lb m strawberries, m 2 = 0.59 lb m sugar
4- 4
4.10
a.
300 gal
b g
m1 lb m
b g
0.750 lb m C 2 H 5OH / lb m
m3 lb m
0.250 lb m H 2O / lb m
0.600 lb m C 2 H 5OH / lb m 0.400 lb m H 2O / lb m
b g m b lb g
4 unknowns ( m1 , m2 ,V40 , m3 ) – 2 balances – 2 specific gravities 0 DF
V40 gal 2
m
0.400 lb m C 2 H 5OH / lb m 0.600 lb m H 2 O / lb m
b.
m1 =
300 gal
1ft 3 0.877 × 62.4 lb m = 2195 lb m 7.4805 gal ft 3
Overall balance: m1 + m2 = m3 C2H5OH balance: 0.750m1 + 0.400m2 = 0.600m3 Solve (1) & (2) simultaneously ⇒ m2 = 1646 lb m, , m3 = 3841 lb m V40 = 4.11
a.
b
1646 lb m
n1 mol / s
ft 3 7.4805 gal = 207 gal 0.952 × 62.4lb m 1ft 3
0.0403 mol C3H 8 / mol
b
n3 mol / s
0.9597 mol air / mol
b
3 unknowns ( n1 , n2 , n3 ) – 2 balances 1 DF
g
n 2 mol air / s
(1) (2)
g
0.0205 mol C3H 8 / mol
g
0.9795 mol air / mol
0.21 mol O 2 / mol 0.79 mol N 2 / mol
b.
b
g
Propane feed rate: 0.0403n1 = 150 ⇒ n1 = 3722 mol / s
b
g
Propane balance: 0.0403n1 = 0.0205n3 ⇒ n3 = 7317 mol / s
b
g
Overall balance: 3722 + n2 = 7317 ⇒ n2 = 3600 mol / s c.
> . The dilution rate should be greater than the value calculated to ensure that ignition is not possible even if the fuel feed rate increases slightly.
4- 5
4.12
a.
b
kg / h m
g
0.960 kg CH3OH / kg
1000 kg / h 0.500 kg CH 3OH / kg 0.500 kg H 2O / kg
,x ) 2 unknowns ( m – 2 balances 0 DF
0.040 kg H 2O / kg 673 kg / h
b
g 1 − x b kg H O / kgg x kg CH 3OH / kg 2
b.
+ 673 ⇒ m = 327 kg / h Overall balance: 1000 = m
b g
b g b g
Methanol balance: 0.500 1000 = 0.960 327 + x 673 ⇒ x = 0.276 kg CH 3OH / kg Molar flow rates of methanol and water: 673 kg 0.276 kg CH3OH 1000 g mol CH3OH = 5.80 × 103 mol CH3OH / h h kg kg 32.0 g CH3OH
673 kg 0.724 kg H 2O 1000 g mol H 2O = 2.71 × 104 mol H 2O / h h kg kg 18 g H 2O Mole fraction of Methanol: 5.80 × 103 = 0176 . mol CH 3OH / mol 5.80 × 103 + 2.71 × 104 c.
4.13
Analyzer is wrong, flow rates are wrong, impurities in the feed, a reaction is taking place, the system is not at steady state.
a.
Product Feed
Reactor
Reactor effluent
1239 kg
Purifier
2253 kg R = 388
2253 kg
R = 583
W aste
b g
m w kg
R = 140
Analyzer Calibration Data 1
xp
x p = 0.000145R
1.364546
0.1
0.01 100
1000
R
4- 6
4.13 (cont’d) b. Effluent: x = 0.000145 388 1.3645 = 0.494 kg P / kg p
b g Product: x = 0.000145b583g = 0.861 kg P / kg Waste: x = 0.000145b140g = 0123 . kg P / kg 0.861b1239g Efficiency = × 100% = 95.8% 0.494b2253g 1.3645
p
1.3645
p
c.
Mass balance on purifier: 2253 = 1239 + mw ⇒ mw = 1014 kg P balance on purifier: Input: 0.494 kg P / kg 2253 kg = 1113 kg P
b
gb g Output: b0.861 kg P / kggb1239 kgg + b0123 . kg P / kggb1014 kgg = 1192 kg P
The P balance does not close . Analyzer readings are wrong; impure feed; extrapolation beyond analyzer calibration data is risky -- recalibrate; get data for R > 583; not at steady state; additional reaction occurs in purifier; normal data scatter. 4.14
a.
b
g
n1 lb- mole/ h . 00100 lb- mole H2O/ lb- mole . 09900 lb- mole DA/ lb- mole
b d i
b
g
n3 lb- mole/ h . lb- mole H2O/ lb- mole 0100
g
. lb- mole DA/ lb- mole 0900
n2 lb- mole HO/ h 2 3 v2 ft / h
4 unknowns ( n1 , n2 , n3 , v ) – 2 balances – 1 density – 1 meter reading = 0 DF Assume linear relationship: v = aR + b Slope: a =
v2 − v1 96.9 − 40.0 = 1.626 = R2 − R1 50 − 15
b g
Intercept: b = va − aR1 = 40.0 − 1.626 15 = 15.61
b g
c
v2 = 1.626 95 + 15.61 = 170 ft / h n 2 =
3
h
b
170 ft 3 62 .4 lb m lb - mol = 589 lb - moles H 2 O / h h ft 3 18.0 lb m
g
(1) DA balance: 0.9900n1 = 0.900n3 (2) Overall balance: n1 + n2 = n3 Solve (1) & (2) simultaneously ⇒ n1 = 5890 lb - moles / h, n 3 = 6480 lb - moles / h b.
Bad calibration data, not at steady state, leaks, 7% value is wrong, v − R relationship is not linear, extrapolation of analyzer correlation leads to error.
4- 7
4.15
a.
b
kg / s m
g
100 kg / s
0.900 kg E / kg
0.600 kg E / kg
0100 . kg H 2 O / kg
0.050 kg S / kg 0.350 kg H 2 O / kg
b
kg / s m
g
, xE , xS ) 3 unknowns ( m – 3 balances 0 DF
b g x b kg S / kg g 1 − x − x b kg H O / kg g x E kg E / kg S
E
b.
S
2
b
g = 0100 . b kgS / kgg
⇒ m = 50.0 kg / s Overall balance: 100 = 2m
b g b g
S balance: 0.050 100 = xS 50 ⇒ xS
b g
b g b g kg E in bottom stream 0.300b50g kg E in bottom stream = = 0.25 kg E in feed 0.600b100g kg E in feed
E balance: 0.600 100 = 0.900 50 + x E 50 ⇒ x E = 0.300 kg E / kg
c.
bg bg bg lnb x / x g lnb0.400 / 0100 . g b= = = 1491 . lnb R / R g lnb38 / 15g lnbag = lnb x g − b lnb R g = lnb0100 . g − 1491 . lnb15g = −6.340 ⇒ a = 1764 . × 10 x = aRb ⇒ ln x = ln a + b ln R 2
1
2
1
1
−3
1
x = 1764 . × 10−3 R1.491
F x I F 0.900 IJ R=G J =G H a K H 1764 . × 10 K 1 b
−3
d.
1 1.491
. = 655
Device not calibrated – recalibrate. Calibration curve deviates from linearity at high mass fractions – measure against known standard. Impurities in the stream – analyze a sample. Mixture is not all liquid – check sample. Calibration data are temperature dependent – check calibration at various temperatures. System is not at steady state – take more measurements. Scatter in data – take more measurements.
4- 8
4.16
a.
b.
b
4.00 mol H 2SO 4 0.098 kg H 2SO 4 L of solution = 0.323 kg H 2SO 4 / kg solution L of solution mol H 2SO 4 1213 . kg solution
bg
5 unknowns ( v1 , v2 , v3 , m2 , m3 ) – 2 balances – 3 specific gravities 0 DF
v1 L
bg m b kg g
100 kg
v3 L
0.200 kg H 2SO 4 / kg
3
0.800 kg H 2 O / kg
g
0.323 kg H 2SO 4 / kg
SG = 1139 .
0.677 kg H 2 O / kg SG = 1.213
bg m b kg g v2 L 2
0.600 kg H 2SO 4 / kg 0.400 kg H 2 O / kg SG = 1.498
UV ⇒ m = 44.4 kg m = 144 kg Water balance: 0.800b100g + 0.400m = 0.677m W Overall mass balance: 100 + m2 = m3
2
2
v1 =
100 kg
v2 =
44.4 kg
3
3
L = 87.80 L20%solution 1139 . kg
L = 29.64 L 60% solution 1498 . kg
v1 87.80 L 20%solution = = 2.96 v2 29.64 L 60% solution c.
4.17
1250 kg P 44.4 kg 60%solution L = 257 L / h h 144 kg P 1498 . kgsolution
b g
m1 kg @$18 / kg 0.25 kg P / kg 0.75 kg H2O / kg
100 . kg 017 . kg P/ kg 0.83 kg H2O / kg
b g
m2 kg @$10 / kg 012 . kg P / kg 0.88 kg H 2O / kg Overall balance: m1 + m2 = 100 .
(1)
(2) b g Solve (1) and (2) simultaneously ⇒ m = 0.385 kg 25% paint, m = 0.615 kg12% paint Cost of blend: 0.385b$18.00g + 0.615b$10.00g = $13.08 per kg Selling price: 110 . b$13.08g = $14.39 per kg
. m2 = 017 . 100 . Pigment balance: 0.25m1 + 012 1
4- 9
2
4.18
b
a.
gb
m1 kg H 2O 85% of entering water
100 kg 0.800 kgS / kg 0.200 kg H 2O / kg
g
b g m b kg H Og m2 kgS 3
2
b
gb g Sugar balance: m = 0.800b100g = 80.0 kgS
85% drying: m1 = 0.850 0.200 100 = 17.0 kg H 2O 2
Overall balance: 100 = 17 + 80 + m3 ⇒ m3 = 3 kg H 2O 3 kg H 2O xw = = 0.0361 kg H 2O / kg 3 + 80 kg
b
g
m1 17 kg H 2O = = 0.205 kg H 2O / kg wet sugar m2 + m3 80 + 3 kg
b
b.
g
1000 tons wet sugar 3 tons H 2 O = 30 tons H 2 O / day day 100 tons wet sugar 1000 tons WS 0.800 tons DS 2000 lb m $0.15 365 days = $8.8 × 107 per year day ton WS ton lb m year
c.
b
g
1 x w1 + x w 2 +...+ x w10 = 0.0504 kg H 2 O / kg 10 1 2 2 SD = x w1 − x w +...+ x w10 − x w = 0.00181 kg H 2 O / kg 9 Endpoints = 0.0504 ± 3 0.00181 xw =
b
g
b
b
g
g
Lower limit = 0.0450, Upper limit = 0.0558
4.19
d.
The evaporator is probably not working according to design specifications since x w = 0.0361 < 0.0450 .
a.
v1 m 3
c h m b kg H O g 2
1
SG = 1.00
d i m b kg suspension g v3 m 3 3
d i
v2 m
3
SG = 1.48
5 unknowns ( v1 , v2 , v3 , m1 , m3 ) – 1 mass balance – 1 volume balance – 3 specific gravities 0 DF
400 kg galena SG = 7.44
Total mass balance: m1 + 400 = m3
(1)
4- 10
4.19 (cont’d)
Assume volume additivity:
b g
m1 kg
b g
m kg m3 400 kg m 3 m3 + = 3 (2) 1000 kg 7440 kg 1480 kg
Solve (1) and (2) simultaneously ⇒ m1 = 668 kg H 2O, m3 = 1068 kg suspension v1 =
4.20
668 kg
m3 = 0.668 m 3 water fed to tank 1000 kg
b.
Specific gravity of coal < 1.48 < Specific gravity of slate
c.
The suspension begins to settle. Stir the suspension. 1.00 < Specific gravity of coal < 1.48
a.
b
n1 mol / h
g
b
n 2 mol / h
b
g
g
0.040 mol H 2O / mol
x mol H 2O / mol
0.960 mol DA / mol
1 − x mol DA / mol
b
b
n3 mol H 2O adsorbed / h
g
g
97% of H 2O in feed
Adsorption rate: n3 =
. − 3.40g kg b354
b
5h
g
mol H 2O = 1556 . mol H 2O / h 0.0180 kg H 2O
97% adsorbed: 156 . = 0.97 0.04n1 ⇒ n1 = 401 . mol / h . − 1556 . = 38.54 mol / h Total mole balance: n1 = n 2 + n3 ⇒ n 2 = 401
Water balance: 0.040 ( 40.1) = 1.556 + x ( 38.54 ) ⇒ x = 1.2 × 10−3 ( mol H 2 O/mol )
4.21
b.
The calcium chloride pellets have reached their saturation limit. Eventually the mole fraction will reach that of the inlet stream, i.e. 4%.
a.
300 lb m / h 0.55 lb m H 2SO 4 / lb m
b
0.45 lb m H 2O / lb m
b
B lb m / h m
C lb m / h m
g
g
0.75 lb m H 2SO 4 / lb m
0.90 lb m H 2SO 4 / lb m
0.25 lb m H 2O / lb m
010 . lb m H 2O / lb m
B = m C Overall balance: 300 + m
(1)
B = 0.75m C H2SO4 balance: 0.55 300 + 0.90m B = 400 lb m / h, m C = 700 lb m / h Solve (1) and (2) simultaneously ⇒ m
(2)
b g
4- 11
4.21 (cont’d) b.
500 − 150 A = 7.78 RA − 44.4 RA − 25 ⇒ m 70 − 25 800 − 200 B − 200 = B = 15.0 RB − 100 m RB − 20 ⇒ m 60 − 20 ln 100 − ln 20 ln x − ln 20 = Rx − 4 ⇒ ln x = 0.2682 Rx + 1923 . ⇒ x = 6.841e 0.2682 Rx 10 − 4 300 + 44.4 400 + 100 mA = 300 ⇒ RA = = 44.3, mB = 400 ⇒ RB = = 333 ., 7.78 15.0 1 55 x = 55% ⇒ Rx = = 7.78 ln 0.268 6.841
b b
A − 150 = m
g g
b
g
FG H
c.
IJ K
A + m B = m C Overall balance: m
b
g
A + 0.90m B = 0.75m C = 0.75 m A + m B ⇒ m B = H2SO4 balance: 0.01xm
d
0.75 − 0.01 6.841e
⇒ 15.0 RB − 100 =
d
⇒ RB = 2.59 − 0.236e
0.2682 Rx
iR
0.2682 Rx
i b7.78 R
A
b
n A kmol / h
100 kg / h
0.90 kmol N 2 / kmol
n P kmol / h
b
g
b g
0.2682 7.78
. e j44.3 + 135
b g − 813 . = 333 .
0.2682 7.78
g
010 . kmol H 2 / kmol
n B kmol / h
015 .
+ 135 . e0.2682 Rx − 813 .
e
a.
− 44.4
b
g
0.20 kmol H 2 / kmol
g
0.80 kmol N 2 / kmol
0.50 kmol H 2 / kmol 0.50 kmol N 2 / kmol
b
g
b
A
015 .
Check: RA = 44.3, Rx = 7.78 ⇒ RB = 2.59 − 0.236e
4.22
A
b0.75 − 0.01xgm
g
MW = 0.20 2.016 + 0.80 28.012 = 22.813 kg / kmol 100 kg kmol = 4.38 kmol / h h 22.813 kg Overall balance: n A + n B = 4.38
(1)
. n A + 0.50 n B = 0.20 4.38 H2 balance: 010
(2)
⇒ n P =
b g
. kmol / h Solve (1) and (2) simultaneously ⇒ n A = 3.29 kmol / h, n B = 110
4- 12
4.22 (cont’d) b.
n P =
m P 22.813
m P 22.813 x m H2 balance: x A n A + x B n B = P P 22.813
Overall balance: n A + n B =
⇒
c.
d. 4.23
Trial 1 2 3 4 5 6 7 8 9 10 11 12
n A =
b b
m P x B − x P 22.813 x B − x A
XA 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10
g g
XB 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50
XP 0.10 0.20 0.30 0.40 0.50 0.60 0.10 0.20 0.30 0.40 0.50 0.60
mP 100 100 100 100 100 100 250 250 250 250 250 250
g g
nA 4.38 3.29 2.19 1.10 0.00 -1.10 10.96 8.22 5.48 2.74 0.00 -2.74
nB 0.00 1.10 2.19 3.29 4.38 5.48 0.00 2.74 5.48 8.22 10.96 13.70
The results of trials 6 and 12 are impossible since the flow rates are negative. You cannot blend a 10% H2 mixture with a 50% H2 mixture and obtain a 60% H2 mixture. Results are the same as in part c. Venous blood 195.0 ml / min 1.75 mg urea / ml
Arterial blood 200.0 ml / min 1.90 mg urea / ml
Dialysate
b b
Dialyzing fluid 1500 ml / min
a.
b b
m P x P − x A 22.813 x B − x A
n B =
g
v ml / min c mg urea / ml
g
Water removal rate: 200.0 − 195.0 = 5.0 ml / min
b
g
b
g
. 200.0 − 175 . 195.0 = 38.8 mg urea / min Urea removal rate: 190 b.
c.
v = 1500 + 5.0 = 1505ml / min 38.8 mg urea/min c= = 0.0258 mg urea/ml 1505 ml/min
b2.7 − 11. g mg removed
1 min
ml
38.8 mg removed
4- 13
10 3 ml 5.0 L 1L
= 206 min (3.4 h)
4.24
a.
b
n1 kmol / min
g b
20.0 kg CO 2 / min
b
n 2 kmol / min
n3 kmol / min
g
g
0.023 kmol CO 2 / kmol
0.015 kmol CO 2 / kmol
20.0 kg CO 2 kmol = 0.455 kmol CO 2 / min min 44.0 kg CO 2 Overall balance: 0.455 + n 2 = n3 CO2 balance: 0.455 + 0.015n 2 = 0.023n3 Solve (1) and (2) simultaneously ⇒ n 2 = 55.6 kmol / min, n3 = 561 . kmol / min n1 =
b.
u=
150 m = 8.33 m / s 18 s
1 561 . kmol m3 1 min s A = πD 2 = ⇒ D = 108 . m 4 min 0123 . kmol 60 s 8.33 m
b
g
Spectrophotometer calibration: C = kA ====> C μg / L = 3.333 A
4.25
A = 0.9 C =3
b
gb g
Dye concentration: A = 018 . ⇒ C = 3333 . 018 . = 0.600 μg / L Dye injected =
b
0.60 cm 3
1L 5.0 mg 103 μ g = 3.0 μ g 103 cm 3 1L 1 mg
g bg
⇒ 3.0 μ g V L = 0.600 μ g / L ⇒ V = 5.0 L 4.26
a.
1000 L B / min
b g V d m / min i n b kmol / min g y b kmol SO / kmolg 1 − y b kmol A / kmolg 2 kg B / min m 3
1 1
1
2
1
b g y b kmol SO / kmolg 1 − y b kmol A / kmolg b kg / min g m x b kg SO / kgg 1 − x b kg B / kgg n3 kmol / min 3
2
3
4
4
2
4
4- 14
(1) (2)
4.26 (cont’d)
2 , m 4 , x4 , y1 , y3 ) 8 unknowns ( n1 , n3 , v1 , m – 3 material balances – 2 analyzer readings – 1 meter reading – 1 gas density formula – 1 specific gravity 0 DF b.
Orifice meter calibration: A log plot of V vs. h is a line through the points h1 = 100, V1 = 142 and h2 = 400, V2 = 290 .
d
i
d
ln V = b ln h + ln a ⇒ V = ah ln V2 V1 ln 290 142 b= = = 0.515 ln h2 h1 ln 400 100 b
h b g b
d b
g g ln a = ln V − b ln h = lnb142g − 0.515 ln 100 = 2.58 ⇒ a = e 1
1
2 .58
= 13.2 ⇒ V = 13.2h 0.515
Analyzer calibration: ln y = bR + ln a ⇒ y = aebR b=
b
ln y 2 y1 R2 − R1
ln a = ln y1 − bR1
E
a = 5.00 × 10 −4
c.
U| 90 − 20 || = lnb0.00166g − 0.0600b20g = −7.60V ⇒ y = 5.00 × 10 || |W
. 0.00166g g = lnb01107 = 0.0600
n1 =
e 0.0600 R
b g = 207.3 m min b12.2g b150 + 14.7g 14.7 batmg = 0.460 mol / L = 0.460 kmol / m = b75 + 460g 18. bKg E
h1 = 210 mm ⇒ V1 = 13.2 210
ρ feed gas
−4
0.515
3
207.3 m 3 0.460 kmol = 95.34 kmol min min m3
b g expb0.0600 × 116 . g = 0.00100 kmol SO
R1 = 82.4 ⇒ y1 = 500 . × 10−4 exp 0.0600 × 82.4 = 0.0702 kmol SO2 kmol R3 = 116 . ⇒ y3 = 500 . × 10−4 2 = m
1000 L B 130 . kg = 1300 kg / min min LB
4- 15
2
kmol
3
i
4.26 (cont’d) A balance: 1 − 0.0702 95.34 = 1 − 0.00100 n3 ⇒ n3 = 88.7 kmol min
b
SO2
gb g b g x balance: b0.0702gb9534 . g( 64.0 kg / kmol) = b0.00100gb88.7g( 64) + m
4 4
(1)
4 (1 − x4 ) B balance: 1300 = m (2) 4 = 1723 kg / min, x4 = 0.245 kg SO2 absorbed / kg Solve (1) and (2) simultaneously ⇒ m 4 x4 = 422 kg SO 2 / min SO2 removed = m
4.27
d.
Decreasing the bubble size increases the bubble surface-to-volume ratio, which results in a higher rate of transfer of SO2 from the gas to the liquid phase.
a.
V2 m 3 / min
b g y b kmolSO / kmol g 1 − y b kmol A / kmol g
d i b kg B / min g m
n 3 kmol / min
3
d i n b kmol / min g y b kmolSO / kmol g 1 − y b kmol A / kmol g V1 m 3 / min
R3
b g x b kgSO kg g 1 − x b kg B / kg g 4 kg / min m
1
1
2
3
2
2
4
1
2
4
P1 , T1 , R1 , h1
b.
2 , n3 , y3 , R3 , m 4 , x4 ) 14 unknowns ( n1 ,V1 , y1 , P1 , T1 , R1 , h1 ,V2 , m – 3 material balances – 3 analyzer and orifice meter readings – 1 gas density formula (relates n1 and V1 ) 2 and V2 ) – 1 specific gravity (relates m 6 DF
b
g b
g
A balance: 1 − y1 n1 = 1 − y3 n3 SO2 balance: y1n1 = y3n3 +
b
g
(1)
4 x4 m 64 kgSO 2 / kmol
(2)
2 = 1 − x4 m 4 B balance: m
Calibration formulas:
(3)
y1 = 5.00 × 10−4 e0.060 R1
(4)
−4 0.060 R3
y3 = 5.00 × 10 e V = 13.2h 0.515 1
Gas density formula: n1 =
b
1
g
(6)
12.2 P1 + 14.7 / 14.7
bT + 460g / 18. 1
(5)
V1
(7)
b g
(8)
kg m m3 Liquid specific gravity: SG = 130 . ⇒ V2 = 2 h 1300 kg
4- 16
4.27 (cont’d) T1 c.
75 °F
y1
0.07 kmol SO2/kmol
150 psig
V1
207 m3/h
h1
210 torr
n1
95.26 kmol/h
R1
82.4
P1
x4 (kg SO2/kg)
Trial 1 2 3 4 5 6 7 8 9 10
y3 (kmol SO2/kmol) V2 (m3/h) n3 (kmol/h)
0.10 0.10 0.10 0.10 0.10 0.20 0.20 0.20 0.20 0.20
0.050 0.025 0.010 0.005 0.001 0.050 0.025 0.010 0.005 0.001
0.89 1.95 2.56 2.76 2.92 0.39 0.87 1.14 1.23 1.30
93.25 90.86 89.48 89.03 88.68 93.25 90.86 89.48 89.03 88.68
m4 (kg/h) 1283.45 2813.72 3694.78 3982.57 4210.72 641.73 1406.86 1847.39 1991.28 2105.36
m2 (kg/h) 1155.11 2532.35 3325.31 3584.31 3789.65 513.38 1125.49 1477.91 1593.03 1684.29
3
V 2 (m /h)
V2 vs. y3 3 .5 0 3 .0 0 2 .5 0 2 .0 0 1 .5 0 1 .0 0 0 .5 0 0 .0 0 0 .0 0 0
0.0 2 0
0 .0 4 0
0 .06 0
y 3 ( k m o l S O 2 /k m o l) x4 = 0 .1 0
x4 = 0 .20
For a given SO2 feed rate removing more SO2 (lower y3) requires a higher solvent feed rate ( V2 ). For a given SO2 removal rate (y3), a higher solvent feed rate ( V ) tends to a more dilute 2
SO2 solution at the outlet (lower x4). d. 4.28
Answers are the same as in part c. Maximum balances: Overall - 3, Unit 1 - 2; Unit 2 - 3; Mixing point - 3 3 Overall mass balance ⇒ m 1 Mass balance - Unit 1 ⇒ m A balance - Unit 1 ⇒ x1 2 Mass balance - mixing point ⇒ m A balance - mixing point ⇒ x2 C balance - mixing point ⇒ y2
4- 17
4.29
a.
100 mol / h 0.300 mol B / mol 0.250 mol T / mol 0.450 mol X / mol
b
n 2 mol / h
b
g
b g Column 1 x b mol T / molg 1 − x − x b mol X / molg n b mol / h g
n 4 mol / h
0.940 mol B / mol 0.060 mol T / mol
x B 2 mol B / mol T2
B2
Column 2
T2
3
0.020 mol T / mol 0.980 mol X / mol
b
n5 mol / h
b g x b mol T / molg 1 − x − x b mol X / molg T5
Column 1 96% X recovery: 0.96 0.450 100 = 0.98n3
gb g
(2)
B balance: 0.300 100 = x B 2 n 2
(3) + 0.020n3
Column 2 97% B recovery: 0.97 x B 2 n 2 = 0.940n 4
b.
(1)
Total mole balance: 100 = n 2 + n3 2 T 2n
(4)
(5)
Total mole balance: n 2 = n 4 + n5
(6)
B balance: x B 2 n 2 = 0.940 n 4 + x B5 n5
(7)
T balance: xT 2 n 2 = 0.060n 4 + xT 5 n5
(8)
(1) ⇒ n 3 = 44.1 mol / h
(2) ⇒ n 2 = 55.9 mol / h
(3) ⇒ x B 2 = 0.536 mol B / mol (5) ⇒ n 4 = 30.95 mol / h
(4) ⇒ x T 2 = 0.431 mol T / mol ( 6) ⇒ n5 = 24.96 mol / h
(7) ⇒ x B5 = 0.036 mol B / mol
(8) ⇒ x T 5 = 0.892 mol T / mol
b
g
0.940 30.95 × 100% = 97% 0.300 100
b g 0.892b24.96g × 100 = 89% Overall toluene recovery: 0.250b100g
Overall benzene recovery:
T5
Column 2: 4 unknowns ( n3 , n4 , n5 , y x ) – 3 balances – 1 recovery of B in top (97%) 0 DF
Column 1 4 unknowns ( n 2 , n3 , x B 2 , xT 2 ) –3 balances – 1 recovery of X in bot. (96%) 0 DF
b g T balance: 0.250b100g = x
g
x B5 mol B / mol
B5
b
4- 18
g
4.30
a.
0.035 kg S / kg 0.965 kg H 2O / kg
b
w kg H 2O / h m
b.
b b
3 kg / h m
100 kg / h
g
x3 kg S / kg
1
b
1 − x3 kg H 2O / kg
b
w kg H 2O / h 0100 . m
g
b g x b kg S / kgg 1 − x b kg H O / kgg
4
g
4
4
g
b
m 10 ( kg / h) 0.050 kg S/kg 0.950 kg H2O/kg
m w ( kg H 2 O / h)
b g
10 Salt balance: 0.035 100 = 0.050m w + m 10 Overall balance: 100 = m H2O yield: Yw =
b b
g
w kg H 2O recovered m 96.5 kg H 2O in fresh feed
g
First 4 evaporators
b b
g
m4 kg / h x 4 kg S/ kg 1 − x4 kg H2 O / kg
100 kg/ h 0.035 kg S/ kg 0.965 kg H2 O / kg
w
g
Overall balance: 100 = 4 0100 . m w + m 4
b g
4 Salt balance: 0.035 100 = x4 m c.
g
b g 4 × 0100 . m b kg H O / hg
b
2
w kg H 2O / h 0100 . m
Overall process 100 kg/h 0.035 kg S/kg 0.965 kg H2O/kg
b
4 kg / h m
g
Yw = 0.31 x 4 = 0.0398
4- 19
2
g
10 kg / h m
10
g
0.050 kg S / kg 0.950 kg H 2O / kg
b
w kg H 2O / h 0100 . m
g
4.31
b g
a.
2n1 mol
Condenser
0.97 mol B / mol 0.03 mol T / mol
b g
b g
100 mol 0.50 mol B / mol 0.50 mol T / mol
n1 mol
n1 mol (89.2% of Bin feed )
0.97 mol B / mol
0.97 mol B / mol 0.03 mol T / mol
0.03 mol T / mol
Still
b gb y b mol B / molg 1 − y b mol T / molg
n 4 mol 45% of feed to reboiler
g
B
B
b g z b mol B / molg 1 − z b mol T / molg n 2 mol B
b g x b mol B / molg 1 − x b mol T / molg n3 mol
Reboiler
B
B
B
Overall process:
Condenser:
3 unknowns ( n1 , n3 , x B ) Still: 5 unknowns ( n1 , n2 , n4 , y B , z B ) – 2 balances – 2 balances – 1 relationship (89.2% recovery) 3 DF 0 DF
1 unknown ( n1 ) – 0 balances 1 DF
Reboiler:
6 unknowns ( n2 , n3 , n4 , x B , y B , z B ) – 2 balances – 2 relationships (2.25 ratio & 45% vapor) 3 DF
Begin with overall process. b.
Overall process 89.2% recovery: 0.892 0.50 100 = 0.97 n1
b gb g
Overall balance: 100 = n1 + n3
b g
B balance: 0.50 100 = 0.97 n1 + x B n3 Reboiler
e j = 2.25 / b1 − x g
yB / 1 − yB Composition relationship:
xB
B
Percent vaporized: n 4 = 0.45n 2
(1)
Mole balance: n 2 = n3 + n 4
(2)
(Solve (1) and (2) simultaneously.) B balance: z B n 2 = x B n3 + y B n 4
4- 20
4.31 (cont’d) c. B fraction in bottoms: x B = 0100 . mol B / mol
Moles of overhead: n1 = 46.0 mol Recovery of toluene:
4.32
Moles of bottoms: n3 = 54.0 mol
. gb54.02g b1 − x gn × 100% = b1 − 010 × 100% = 97% 0.50b100g 0.50b100g B
3
a.
b
m3 kg H 2O
g
Bypass
Mixing point
b g
Basis: 100 kg 100 kg 0.12 kg S / kg
Evaporator
m1 kg
0.88 kg H2O / kg
012 . kg S / kg 0.88 kg H2O / kg
b g
b g
m4 kg
m5 kg
0.58 kg S / kg 0.42 kg H 2O / kg
0.42 kg S / kg 0.58 kg H2O / kg
b g
m2 kg
012 . kg S / kg 0.88 kg H2O / kg
Overall process:
Evaporator:
2 unknowns ( m3 , m5 ) – 2 balances 0 DF
3 unknowns ( m1 , m3 , m4 ) – 2 balances 1 DF
Bypass:
Mixing point:
2 unknowns ( m1 , m2 ) – 1 independent balance 1 DF
3 unknowns ( m2 , m4 , m5 ) – 2 balances 1 DF
b g
. 100 = 0.42m5 Overall S balance: 012 Overall mass balance: 100 = m3 + m5 Mixing point mass balance: m4 + m2 = m5
(1)
Mixing point S balance: 0.58m4 + 012 . m2 = 0.42m5
(2)
Solve (1) and (2) simultaneously Bypass mass balance: 100 = m1 + m2 b.
m1 = 90.05 kg, m2 = 9.95 kg, m3 = 714 . kg, m4 = 18.65 kg, m5 = 28.6 kg product Bypass fraction:
c.
m2 = 0.095 100
Over-evaporating could degrade the juice; additional evaporation could be uneconomical; a stream consisting of 90% solids could be hard to transport.
4- 21
4.33
a.
b
4 kg Cr / h m
b
1 kg / h m
g
0.0515 kg Cr / kg 0.9485 kg W / kg
b
2 kg / h m
g
g
b g x b kg Cr / kgg 1 − x b kg W / kgg 5 kg / h m
0.0515 kg Cr / kg 0.9485 kg W / kg
Treatment Unit
b
3 kg / h m
5
5
b g x b kg Cr / kgg 1 − x b kg W / kgg 6 kg / h m 6
6
g
0.0515 kg Cr / kg 0.9485 kg W / kg
b.
b
1 = 6000 kg / h ⇒ m 2 = 4500 kg / h maximum allowed value m 3 = 6000 − 4500 = 1500 kg / h Bypass point mass balance: m
b
gb
g
g
4 = 0.95 0.0515 4500 = 220.2 kg Cr / h 95% Cr removal: m 5 = 4500 − 220.2 = 4279.8 kg / h Mass balance on treatment unit: m 0.0515 4500 − 220.2 = 0.002707 kg Cr / kg Cr balance on treatment unit: x5 = 4779.8 6 = 1500 + 4279.8 = 5779.8 kg / h Mixing point mass balance: m
b
Mixing point Cr balance: x6 =
c.
b g
g
b
g
0.0515 1500 + 0.0002707 4279.8 = 0.0154 kg Cr / kg 5779.8
m 1 (kg/h) m 2 (kg/h) m 3 (kg/h) m 4 (kg/h) m 5 (kg/h) 1000 1000 0 48.9 951 2000 2000 0 97.9 1902 3000 3000 0 147 2853 4000 4000 0 196 3804 5000 4500 500 220 4280 6000 4500 1500 220 4280 7000 4500 2500 220 4280 8000 4500 3500 220 4280 9000 4500 4500 220 4280 10000 4500 5500 220 4280
4- 22
m 6 (kg/h) x5 0.00271 951 0.00271 1902 0.00271 2853 0.00271 3804 0.00271 4780 0.00271 5780 0.00271 6780 0.00271 7780 0.00271 8780 0.00271 9780
x6 0.00271 0.00271 0.00271 0.00271 0.00781 0.0154 0.0207 0.0247 0.0277 0.0301
4.33 (cont’d)
x 6 (kg Cr/kg)
m 1 vs. x 6 0.03500 0.03000 0.02500 0.02000 0.01500 0.01000 0.00500 0.00000 0
2000
4000
6000
8000 10000 12000
m 1 (kg/h)
d.
4.34
Cost of additional capacity – installation and maintenance, revenue from additional recovered Cr, anticipated wastewater production in coming years, capacity of waste lagoon, regulatory limits on Cr emissions.
a.
b
175 kg H 2O / s 45% of water fed to evaporator
b
1 kg / s m
g
b b
4 kg K 2SO 4 / s m
0196 kg K 2SO 4 / kg .
5 kg H 2O / s m
g
b b
g
6 kg K 2SO 4 / s m 7 kg H 2O / s m
Evaporator
g
g
0.804 kg H 2O / kg
g
Crystallizer Filter
Filter cake
b
2 kg K 2SO 4 / s 10 m
g
b g RS0.400 kg K SO / kg UV T0.600 kg H O / kg W 2 kgsoln / s m 2
Filtrate
b
3 kg / s m
4
2
g
0.400 kg K 2SO 4 / kg 0.600 kg H 2 O / kg
Let K = K2SO4, W = H2 Basis: 175 kg W evaporated/s
1, m 2) Overall process: 2 unknowns ( m - 2 balances 0 DF Evaporator:
Mixing point:
4, m 5, m 6, m 7 ) 4 unknowns ( m – 2 balances – 1 percent evaporation 1 DF
Crystallizer:
1, m 3, m 4, m 5 ) 4 unknowns ( m - 2 balances 2 DF
2, m 3, m 6, m 7 ) 4 unknowns ( m – 2 balances 2 DF
U| verify that each |V chosen subsystem involves ,m | no more than two Balances around mixing point ⇒ m ,m |W unknown variables Balances around evaporator ⇒ m
1, m 2 Strategy: Overall balances ⇒ m 5 % evaporation ⇒ m
3
6
4- 23
4
7
4.34 (cont’d)
Overall mass balance: m 1 = 175 + 10m 2 + m 2 Overall K balance: . m 1 = 10m 2 + 0.400m 2 0196 Production rate of crystals = 10m 2
U| V| W
45% evaporation: 175 kg evaporated min = 0.450m 5 W balance around mixing point: 0.804m 1 + 0.600m 3 = m 5 Mass balance around mixing point: m 1 + m 3 = m 4 + m 5 K balance around evaporator: m 6 = m 4 W balance around evaporator: m 5 = 175 + m 7 Mole fraction of K in stream entering evaporator =
b.
1 = 221 kg / s Fresh feed rate: m
m 4 m 4 + m 5
bg
2 = 416 . kg K s s Production rate of crystals = 10m Recycle ratio:
c.
b
g
3 kg recycle s m 352.3 kg recycle = = 160 . 1 kg fresh feed s 220.8 kg fresh feed m
b
g
Scale to 75% of capacity. Flow rate of stream entering evaporator = 0.75(398 kg / s) = 299 kg / s 46.3% K, 53.7% W
d.
Drying . Principal costs are likely to be the heating cost for the evaporator and the dryer and the cooling cost for the crystallizer.
4- 24
4.35
a.
Overall objective: Separate components of a CH4-CO2 mixture, recover CH4, and discharge CO2 to the atmosphere. Absorber function: Separates CO2 from CH4. Stripper function: Removes dissolved CO2 from CH3OH so that the latter can be reused.
b.
The top streams are liquids while the bottom streams are gases. The liquids are heavier than the gases so the liquids fall through the columns and the gases rise.
c.
b
n1 mol / h
g
b g n b mol CO / h g n 5 mol N 2 / h
0.010 mol CO 2 / mol
6
0.990 mol CH 4 / mol
100 mol / h 0.300 mol CO 2 / mol
b
n 2 mol / h
Absorber
g
Stripper
b
n 5 mol N 2 / h
0.005 mol CO 2 / mol
0.700 mol CH 4 / mol
2
g
0.995 mol CH 3OH / mol
b g n b mol CH OH / h g n 3 mol CO 2 / h 4
Overall:
Stripper:
3
3 unknowns ( n1 , n5 , n 6 ) – 2 balances 1 DF
Absorber:
4 unknowns ( n1 , n 2 , n3 , n 4 ) – 3 balances 1 DF
4 unknowns ( n2 , n3 , n4 , n5 ) – 2 balances – 1 percent removal (90%) 1 DF
b0.700gb100g bmol CH / hg = 0.990n Overall mole balance: 100b mol / h g = n + n
Overall CH4 balance:
4
1
1
6
Percent CO2 stripped: 0.90 n3 = n 6 Stripper CO2 balance: n3 = n 6 + 0.005n 2 Stripper CH3OH balance: n 4 = 0.995n 2 d.
n1 = 70.71 mol / h , n 2 = 6510 . mol / h, n3 = 32.55 mol CO 2 / h, n 4 = 647.7 mol CH3OH / h, n 6 = 29.29 mol CO 2 / h 30.0 − 0.010n1 = 0.976 mol CO 2 absorbed / mol fed Fractional CO2 absorption: f CO 2 = 30.0
4- 25
4.35 (cont’d)
Total molar flow rate of liquid feed to stripper and mole fraction of CO2: n3 n3 + n 4 = 680 mol / h, x3 = = 0.0478 mol CO 2 / mol n3 + n 4 e.
Scale up to 1000 kg/h (=106 g/h) of product gas:
b
g
b
g
MW1 = 0.01 44 g CO 2 / mol + 0.99 16 g CH 4 / mol = 16.28 g / mol
. × 10 mol / h bn g = d10. × 10 g / hib16.28 g / molg = 6142 . × 10 mol / h) / (70.71 mol / h) = 8.69 × 10 bn g = b100 mol / hg (6142 6
4
1 new
4
feed new
4.36
4
mol / h
f.
Ta < Ts The higher temperature in the stripper will help drive off the gas. Pa > Ps The higher pressure in the absorber will help dissolve the gas in the liquid.
g.
The methanol must have a high solubility for CO2, a low solubility for CH4, and a low volatility at the stripper temperature.
a.
Basis: 100 kg beans fed
e
m kg C H 5 6 14
e
m kg C H 1 6 14
j
300 kg C 6 H14
Ex
j
Condenser
b g b g y b kg oil / kgg 1 − x − y b kg C H m2 kg
2
2
2
6
b g b g 1 − y b kg C H m4 kg
F
x2 kg S / kg
14 / kg
13.0 kg oil 87.0 kg S
Ev
y4 kg oil / kg 6
4
g
14
/ kg
g
g
b g
m3 kg
0.75 kg S / kg
b
y3 kg oil / kg
b
g
0.25 − y3 kg C 6 H14 / kg
Overall:
b
m6 kg oil
4 unknowns ( m1 , m3 , m6 , y3 ) – 3 balances 1 DF
Extractor:
g
3 unknowns ( m2 , x2 , y2 ) – 3 balances 0 DF
2 unknowns ( m1 , m5 ) Evaporator: 4 unknowns ( m4 , m5 , m6 , y4 ) – 1 balance – 2 balances 1 DF 2 DF Filter: 7 unknowns ( m2 , m3 , m4 , x2 , y2 , y3 , y4 ) – 3 balances – 1 oil/hexane ratio 3 DF Mixing Pt:
Start with extractor (0 degrees of freedom) Extractor mass balance: 300 + 87.0 + 13.0 kg = m2
4- 26
4.36
(cont’d) Extractor S balance: 87.0 kg S = x2 m2
Extractor oil balance: 13.0 kg oil = y2 m2 Filter S balance: 87.0 kg S = 0.75m3
b g
Filter mass balance: m2 kg = m3 + m4 Oil / hexane ratio in filter cake:
y3 0.25 − y3
=
y2 1 − x2 − y2
Filter oil balance: 13.0 kg oil = y3m3 + y4 m4
b
g
Evaporator hexane balance: 1 − y4 m4 = m5 Mixing pt. Hexane balance: m1 + m5 = 300 kg C6 H14 Evaporator oil balance: y4 m4 = m6 b.
b
g
m6 118 . kg oil = = 0118 kg oil / kg beans fed . 100 100 kg beans fed m 28 kg C6 H14 Fresh hexane feed = 1 = = 0.28 kg C 6 H14 / kg beans fed 100 100 kg beans fed m 272 kg C 6 H14 recycled Recycle ratio = 5 = = 9.71 kg C 6 H14 recycled / kg C 6 H14 fed m1 28 kg C6 H14 fed
Yield =
b b
c.
g
g
Lower heating cost for the evaporator and lower cooling cost for the condenser.
b
4.37
g
m lb m dirt 1 98 lb m dry shirts 3 lb m Whizzo
100 lbm 2 lbm dirt 98 lbm dry shirts
b
m lb m Whizzo 2
g
Tub
b g
Filter
b g
m lb m 3 0.03 lb m dirt / lb m
m lb m 4 013 . lb m dirt / lb m
0.97 lb m Whizzo / lb m
0.87 lb m Whizzo / lb m
b g
m lb m 5 0.92 lb m dirt / lb m 0.08 lb m Whizzo / lb m
b g b g blb Whizzo/ lb g
m6 lbm 1 − x lb m dirt / lbm x
m
m
Strategy 95% dirt removal ⇒ m1 ( = 5% of the dirt entering) Overall balances: 2 allowed (we have implicitly used a clean shirt balance in labeling the chart) ⇒ m2 , m5 (solves Part (a))
4- 27
4.37
(cont’d)
b g around the filter b m , m , x g , but the tub only involves 2 b m , m g and 2 balances are Balances around the mixing point involve 3 unknowns m3 , m6 , x , as do balances 4
6
3
4
allowed for each subsystem. Balances around tub ⇒ m3 , m4 Balances around mixing point ⇒ m6 , x (solves Part (b)) a.
b gb g Overall dirt balance: 2.0 = 010 . + b0.92gm ⇒ m = 2.065 lb dirt . lb Overall Whizzo balance: m = 3 + b0.08gb2.065g blb Whizzog = 317
95% dirt removal: m1 = 0.05 2.0 = 010 . lb m dirt 5
5
2
b.
m
m
. + 013 . m4 Tub dirt balance: 2 + 0.03m3 = 010 Tub Whizzo balance: 0.97m3 = 3 + 0.87m4 Solve (1) & (2) simultaneously ⇒ m3 = 20.4 lb m , m4 = 19.3 lb m Mixing pt. mass balance: 317 . + m6 = 20.4 lb m ⇒ m6 = 17.3 lb m Mixing pt. Whizzo balance:
m
Whizzo (1) (2)
3.17 + x (17.3) = ( 0.97 )( 20.4 ) ⇒ x = 0.961 lb m Whizzo/lb m ⇒ 96% Whizzo, 4% dirt
4.38
a.
2720 kg S mixer 3 Discarded C 3L kg L C 3S kg S 3300 kg S
C 2L kg L C 2S kg S
Filter 3 F 3L kg L F 3S kg S
620 kg L mixer 1 Filter 1 F 1L F 1S
mixer 2 C 1L kg L C 1S kg S kg L kg S
Filter 2 F 2L kg L F 2S kg S To holding tank
mixer filter 1: balance: mixer filter 2: balance: mixer filter 3: balance:
b g 620 = 6.2 + C ⇒ 0.01b6138 . +F g= F U | 6138 . +F = F +C V⇒ |W 0.01C = F 0.01 620 = F1L ⇒ 2L
. kg L C1L = 6138 F2 L = 6.2 kg L
3L
C2 L = 613.7 kg L
1L
3L
3L
2L
F1L = 6.2 kg L
. kg L F3L = 61 613.7 = 6.1+ C3L ⇒ C3L = 607.6 kg L
2L
3L
4- 28
4.38 (cont’d)
Solvent m f 1:
b g 3300 = 495 + F ⇒ U| 015 . b495 + F g = C 495 + F = C + F | V⇒ 015 . b2720 + C g = C | 2720 + C = F + C |W 015 . 3300 = C1S ⇒
balance: m f 2:
C1S = 495 kg S
2S
F1S = 2805 kg S C2 S = 482.6 kg S
2S
F2 S = 2734.6 kg S
1S
3S
balance:
3S
m f 3:
2S
2S
balance:
3S
2S
C3S = 480.4 kg S
3S
F3S = 2722.2 kg S
3S
Holding Tank Contents 6.2 + 6.2 = 12.4 kg leaf 2805 + 2734.6 = 5540 kg solvent b.
b g
5540 kgS 0165 . kg E / kg 0.835 kg W / kg
b g
Q0 kg
QR kg
. kg E / kg Extraction 013 0.15kg F / kg Unit
b g Q b kg Fg
Steam Stripper
b g
b g Q b kg D g Q b kg Fg QE kg E
F
0.026 kg F / kg 0.774 kg W / kg QB kg
0.855kg W / kg
QD kg D
0.200 kg E / kg
0.013 kg E / kg
D
0.987 kg W / kg
F
b
Q3 kg steam Mass of D in Product:
1 kg D 1000 kg leaf
620 kg leaf
g
= 0.62 kg D = QD
b g
Water balance around extraction unit: 0.835 5540 = 0.855QR ⇒ QR = 5410 kg Ethanol balance around extraction unit: 0165 . 5540 = 013 . 5410 + QE ⇒ QE = 211 kg ethanol in extract
b
c.
g
b
g
b
g
F balance around stripper 0.015 5410 = 0.026Q0 ⇒ Q0 = 3121 kg mass of stripper overhead product
b
g
b
g
E balance around stripper 013 . 5410 = 0.200 3121 + 0.013QB ⇒ QB = 6085 kg mass of stripper bottom product
b
g
b g
W balance around stripper
b
g
b
b
g
b
g
g
0.855 5410 + QS = 0.774 3121 + 0.987 6085 ⇒ QS = 3796 kg steam fed to stripper 4.39
a.
C2 H 2 + 2 H 2 → C2 H 6 2 mol H 2 react / mol C 2 H 2 react 0.5 kmol C 2 H 6 formed / kmol H 2 react
4- 29
4.39 (cont’d) b. nH 2
nC2 H2
. < 2.0 ⇒ H 2 is limiting reactant = 15
. mol H 2 fed ⇒ 10 . mol C 2 H 2 fed ⇒ 0.75 mol C 2 H 2 required (theoretical) 15 . mol fed − 0.75 mol required 10 % excess C 2 H 2 = . × 100% = 333% 0.75 mol required c.
4 × 106 tonnes C 2 H 6 1 yr 1 day 1 h 1000 kg 1 kmol C 2 H 6 2 kmol H 2 2.00 kg H 2 yr 300 days 24 h 3600 s tonne 30.0 kg C 2 H 6 1 kmol C 2 H 6 1 kmol H 2 = 20.6 kg H 2 / s
4.40
d.
The extra cost will be involved in separating the product from the excess reactant.
a.
4 NH 3 + 5 O 2 → 4 NO + 6 H 2O 5 lb - mole O 2 react = 125 . lb - mole O 2 react / lb - mole NO formed 4 lb - mole NO formed
b.
dn i
O 2 theoretical
=
100 kmol NH3 5 kmol O 2 = 125 kmol O 2 4 kmol NH 3 h
d i
40% excess O 2 ⇒ nO 2 c.
fed
b
g
. 125 kmol O 2 = 175 kmol O 2 = 140
b50.0 kg NH gb1 kmol NH / 17 kg NH g = 2.94 kmol NH . kmol O b100.0 kg O gb1 kmol O / 32 kg O g = 3125 F n I = 3125 F n I = 5 = 125 . NH3 N2 + 3H22NÆ 2 2NH 3 (1-yp) (1-fsp) n1 (mol N2) (1-yp) (1-fsp) 3n1 (mol H2) (1-yp) n2 (mol I)
1 mol (1-XI0)/4 (mol N2 / mol) 3/4 (1-XI0) (mol H2 / mol) XI0 (mol I / mol)
nr (mol) n1 (mol N2) 3n1 (mol H2) n2 (mol I)
Reactor
4-52
(1-fsp) n1 (mol N2) (1-fsp) 3n1 (mol H2) n2 (mol I)
nr (mol) (1-fsp) n1 (mol N2) (1-fsp) 3n1 (mol H2) n2 (mol I) 2 fsp n1 (mol NH3)
yp (1-fsp) n1 (mol N2) yp (1-fsp) 3n1 (mol H2) yp n2 (mol I)
Condenser
np (mol) 2 fsp n1 (mol NH3)
4.61 (cont’d) At mixing point:
N2: (1-XI0)/4 + (1-yp)(1-fsp) n1 = n1 I: XI0 + (1-yp) n2 = n2 Total moles fed to reactor: nr = 4n1 + n2 Moles of NH3 produced: np = 2fspn1 Overall N2 conversion:
b.
(1 − X I0 ) / 4 − y p (1 − f sp )n 1 (1 − X I0 ) / 4
× 100%
XI0 = 0.01 fsp = 0.20 yp = 0.10 n1 = 0.884 mol N2
nr = 3.636 mol fed
n2 = 0.1 mol I
np = 0.3536 mol NH3 produced N2 conversion = 71.4%
c.
Recycle: recover and reuse unconsumed reactants. Purge: avoid accumulation of I in the system.
d.
Increasing XI0 results in increasing nr, decreasing np, and has no effect on fov. Increasing fsp results in decreasing nr, increasing np, and increasing fov. Increasing yp results in decreasing nr, decreasing np, and decreasing fov. Optimal values would result in a low value of nr and fsp, and a high value of np, this would give the highest profit. XI0 0.01 0.05 0.10 0.01 0.01 0.01 0.10 0.10 0.10
fsp 0.20 0.20 0.20 0.30 0.40 0.50 0.20 0.20 0.20
yp 0.10 0.10 0.10 0.10 0.10 0.10 0.20 0.30 0.40
nr 3.636 3.893 4.214 2.776 2.252 1.900 3.000 2.379 1.981
4-53
np 0.354 0.339 0.321 0.401 0.430 0.450 0.250 0.205 0.173
fov 71.4% 71.4% 71.4% 81.1% 87.0% 90.9% 55.6% 45.5% 38.5%
4.62
a.
i - C 4 H 10 + C 4 H 8 = C 8 H 18
D
Basis: 1-hour operation
n 2 (n-C 4 H10 ) n 3 (i-C 4 H 10) n 1 (C 8 H18 ) m4 (91% H 2 SO4 )
F decanter
E
Units of n: kmol Units of m: kg
still
n 1 (C 8 H18 ) n 2 (n-C 4 H10 ) n 3 (i-C 4 H 10)
n 5 (n-C 4 H10 ) n 6 (i-C 4 H 10) n 7 (C 8 H18 ) m8 (91% H 2 SO 4 )
reactor
C B
n 1 (C 8 H18 ) n 2 (n-C 4 H10 )
P
m4 (kg 91% H 2 SO4 )
40000 kg A n 0 kmol 0.25 i-C4 H10 0.50 n-C4 H10 0.25 C4 H 8
n 3 (i-C 4 H 10)
Calculate moles of feed
b gb
g b gb
g
M = 0.25 M L − C4 H10 + 0.50 M n − C4 H10 + 0.25 M C4 H 8 = 0.75 5812 . + 0.25 5610 .
= 57.6 kg kmol
b
gb
g
n0 = 40000 kg 1 kmol 57.6 kg = 694 kmol
b gb g
Overall n - C 4 H 10 balance: n2 = 0.50 694 = 347 kmol n - C 4 H 10 in product C 8 H 18 balance:
n1 =
b0.25gb694g kmol C H 4
8
react 1 mol C 8 H 18 = 1735 . kmol C 8 H 8 in product 1 mol C 4 H 8
b
At (A), 5 mol i - C 4 H 10 1 mole C 4 H 8 ⇒ n mol i - C 4 H 10
b
g = b5gb0.25gb694g = 867.5 kmol A
i -C 4 H 10 at
moles C 4 H 8 at A=173.5
g
Note: n mol C 4 H 8 = 173.5 at (A), (B) and (C) and in feed
b gb g
i - C 4 H 10 balance around first mixing point ⇒ 0.25 694 + n3 = 867.5 ⇒ n3 = 694 kmol i - C 4 H 10 recycled from still
At C, 200 mol i - C4 H10 mol C4 H 8
b
⇒ n mol i - C4 H10
g = b200gb1735. g = 34,700 kmol i - C H 4
C
4-54
10
b A g and b Bg
4.62 (cont’d)
i - C 4 H 10 balance around second mixing point ⇒ 867.5 + n6 = 34,700 ⇒ n6 = 33,800 kmol C 4 H 10 in recycle E Recycle E: Since Streams (D) and (E) have the same composition,
b g = n bmoles i - C H g ⇒ n bmoles n - C H g n bmoles i - C H g bmoles C H g = n ⇒ n = 8460 kmol C H bmoles C H g n
n5 moles n - C 4 H 10 n2 n7 n1
4
E
6
4
10 E
10 D
3
4
10 D
8
18 E
6
8
18 D
3
4
7
5
= 16,900 kmol n - C 4 H 10
18
Hydrocarbons entering reactor:
kg I b347 + 16900gbkmol n - C H g FGH 58.12 kmol JK kg I F . kg IJ + 1735. kmol C H FG 5610 + b867.5 + 33800gb kmol i - C H g G 5812 . H kmol K H kmol JK kg I F + 8460 kmol C H G 114.22 J = 4.00 × 10 kg . H kmol K H SO solution entering reactor 4.00 × 10 kg HC 2 kg H SO baq g = band leaving reactor g 1 kg HC = 8.00 × 10 kg H SO baq g m b H SO in recycleg n b n - C H in recycleg = 8.00 × 10 b H SO leaving reactor g n + n b n - C H leaving reactor g ⇒ m = 7.84 × 10 kg H SO baq g in recycle E 4
10
4
10
4
8
6
8
2
18
6
4
2
4
6
2
8
2
4
4
5
4
10
5
4
10
6
2
2
4
6
8
2
4
m4 = H 2 SO 4 entering reactor − H 2 SO 4 in E
b g
= 16 . × 10 5 kg H 2 SO 4 aq recycled from decanter
b g d ib g d16. × 10 ib0.09gkg H O b1 kmol 18.02 kgg = 799 kmol H O from decanter
⇒ 16 . × 10 5 0.91 kg H 2 SO 4 1 kmol 98.08 kg = 1480 kmol H 2 SO 4 in recycle 5
2
2
Summary: (Change amounts to flow rates) Product: 173.5 kmol C 8 H 18 h , 347 kmol n - C 4 H 10 h Recycle from still: 694 kmol i - C 4 H 10 h Acid recycle: 1480 kmol H 2 SO 4 h , 799 kmol H 2 O h Recycle E: 16,900 kmol n - C 4 H 10 h , 33,800 kmol L - C 4 H 10 h , 8460 kmol C 8 H 18 h, 7.84 × 10 6 kg h 91% H 2 SO 4 ⇒ 72,740 kmol H 2 SO 4 h , 39,150 kmol H 2 O h
4-55
4.63
a.
A balance on ith tank (input = output + consumption) v L min C A, i −1 mol L = vC Ai + kC Ai C Bi mol liter ⋅ min V L
b
g
b
g
b
gbg
E ÷ v, note V / v = τ
C A , i −1 = C Ai + kτ C Ai C Bi B balance. By analogy, C B , i −1 = C Bi + kτ C Ai C Bi Subtract equations ⇒ C Bi − C Ai = C B , i −1 − C A, i −1
=
A from balances on bi −1g tank
C B , i − 2 − C A, i − 2 =… = C B 0 − C A0
st
b.
C Bi − C Ai = C B 0 − C A 0 ⇒ C Bi = C Ai + C B 0 − C A 0 . Substitute in A balance from part (a).
b
g
2 0 C A, i −1 = C Ai + kτ C Ai C Ai + C B 0 − C A0 . Collect terms in C Ai , C 1Ai , C Ai . 2 C Ai
b
g
kτ + C AL 1 + kτ C B 0 − C A0 − C A , i −1 = 0 ⇒α
2 C AL
Solution: C Ai =
b
g
+ β C AL + γ = 0 where α = kτ , β = 1 + kτ C B 0 − C A 0 , γ = − C A, i −1
− β + β 2 − 4αγ
(Only + rather than ±: since αγ is negative and the 2α negative solution would yield a negative concentration.)
c. k= v= V= CA0 = CB0 = alpha = beta =
36.2 5000 2000 0.0567 0.1000 14.48 1.6270
N 1 2 3 4 5 6 7 8 9 10 11 12 13 14
gamma -5.670E-02 -2.791E-02 -1.512E-02 -8.631E-03 -5.076E-03 -3.038E-03 -1.837E-03 -1.118E-03 -6.830E-04 -4.182E-04 -2.565E-04 -1.574E-04 -9.667E-05 -5.939E-05
CA(N) 2.791E-02 1.512E-02 8.631E-03 5.076E-03 3.038E-03 1.837E-03 1.118E-03 6.830E-04 4.182E-04 2.565E-04 1.574E-04 9.667E-05 5.939E-05 3.649E-05
xA(N) 0.5077 0.7333 0.8478 0.9105 0.9464 0.9676 0.9803 0.9880 0.9926 0.9955 0.9972 0.9983 0.9990 0.9994
(xmin = 0.50, N = 1), (xmin = 0.80, N = 3), (xmin = 0.90, N = 4), (xmin = 0.95, N = 6), (xmin = 0.99, N = 9), (xmin = 0.999, N = 13). As xmin → 1, the required number of tanks and hence the process cost becomes infinite. d.
(i) k increases ⇒ N decreases (faster reaction ⇒ fewer tanks) (ii) v increases ⇒ N increases (faster throughput ⇒ less time spent in reactor ⇒ lower conversion per reactor)
(iii) V increases ⇒ N decreases (larger reactor ⇒ more time spent in reactor ⇒ higher conversion per reactor)
4-56
4.64
a.
Basis: 1000 g gas Species
m (g)
MW
n (mol)
mole % (wet)
mole % (dry)
C3H8
800
44.09
18.145
77.2%
87.5%
C4H10
150
58.12
2.581
11.0%
12.5%
H2O
50
18.02
2.775
11.8%
Total
1000
23.501
100%
100%
Total moles = 23.50 mol, Total moles (dry) = 20.74 mol Ratio: 2.775 / 20.726 = 0.134 mol H 2 O / mol dry gas
b.
C3H8 + 5 O2 → 3 CO2 + 4 H2O, C4H10 + 13/2 O2 → 4 CO2 + 5 H2O Theoretical O2: C3H8: C 4 H 10 :
5 kmol O 2 100 kg gas 80 kg C 3 H 8 1 kmol C 3 H 8 = 9.07 kmol O 2 / h h 100 kg gas 44.09 kg C 3 H 8 1 kmol C 3 H 8 6.5 kmol O 2 100 kg gas 15 kg C 4 H 10 1 kmol C 4 H 10 = 1.68 kmol O 2 / h h 100 kg gas 58.12 kg C 4 H 10 1 kmol C 4 H 10
Total: (9.07 + 1.68) kmol O2/h = 10.75 kmol O2/h Air feed rate:
10.75 kmol O 2 1 kmol Air 1.3 kmol air fed = 66.5 kmol air / h h .21 kmol O 2 1 kmol air required
The answer does not change for incomplete combustion 4.65
5 L C 6 H 14 0.659 kg C 6 H 14 1000 mol C 6 H 14 = 38.3 mol C 6 H 14 L C 6 H 14 86 kg C 6 H 14 4 L C 7 H 16 0.684 kg C 7 H 16 1000 mol C 7 H 16 = 27.36 mol C 7 H 16 L C 7 H 16 100 kg C 7 H 16
C6H14 +19/2 O2 → 6 CO2 + 7 H2O
C6H14 +13/2 O2 → 6 CO + 7 H2O
C7H16 + 11 O2 → 7 CO2 + 8 H2O
C7H16 + 15/2 O2 → 7 CO + 8 H2O
Theoretical oxygen:
38.3 mol C 6 H 14
9.5 mol O 2 27.36 mol C 7 H 16 + mol C 6 H 14
11 mol O 2 = 665 mol O 2 required mol C 7 H 16
O2 fed: (4000 mol air )(.21 mol O2 / mol air) = 840 mol O2 fed Percent excess air:
840 − 665 × 100% = 26.3% excess air 665
4-57
4.66
CO +
1 O 2 → CO 2 2
H2 +
1 O2 → H 2O 2
175 kmol/h 0.500 kmol N2/kmol x (kmol CO/mol) (0.500–x) (kmol H2/kmol) 20% excess air
Note: Since CO and H 2 each require 0 .5 mol O 2 / mol fuel for complete combustion, we can calculate the air feed rate without determining x CO . We include its calculation for illustrative purposes.
b
g
A plot of x vs. R on log paper is a straight line through the points R1 = 10.0, x1 = 0.05 and
bR
2
g
= 99.7, x 2 = 10 . . ln x = b ln R + ln a
@
x = a Rb R = 38.3 ⇒ x = 0.288
b
g b
g ln a = lnb10 . g − 1303 . lnb99.7g = −6.00 a = expb −6.00g = 2.49 × 10 −3
b = ln 10 . 0.05 ln 99.7 10.0 = 1.303 . ⇒ x = 2.49 × 10 −3 R1303
moles CO mol
Theoretical O : 175 kmol 0.288 kmol CO 0.5 kmol O 2 2 h kmol kmol CO kmol O 2 2 = 43.75 h kmol kmol H h 2 43.75 kmol O required 1 kmol air 1.2 kmol air fed kmol air 2 Air fed: = 250 h 0.21 kmol O 1 kmol air required h 2 +
4.67
a.
CH 4 + 2O 2 7 C2 H 6 + O2 2 C 3 H 8 + 5O 2 13 C 4 H 10 + O 2 2
Theoretical O 2 :
→ CO 2 + 2H 2 O
→ 3CO 2 + 4H 2 O
b g
b g
0.5 kmol O
17% excess air na (kmol air/h) 0.21 O2 0.79 N2
→ 4CO 2 + 5H 2 O
0.944 100 kmol CH 4 h
2
100 kmol/h 0.944 CH4 0.0340 C2H6 0.0060 C3H8 0.0050 C4H10
→ 2CO 2 + 3H 2 O
0.0060 100 kmol C 3 H 8 h = 207.0 kmol O 2 h +
0.212 kmol H
175 kmol
b g
2 kmol O 2 0.0340 100 kmol C 2 H 6 + 1 kmol CH 4 h
3.5 kmol O 2 1 kmol C 2 H 6
0.0050 100 kmol C 4 H 10 5 kmol O 2 + 1 kmol C 3 H 3 h
6.5 kmol O 2 1 kmol C 4 H 10
b g
4-58
4.67 (cont’d)
207.0 kmol O 2 h
Air feed rate: n f =
b
1 kmol air 0.21 kmol O 2
gb
gb
1.17 kmol air fed = 1153 kmol air h kmol air req.
g
b.
na = n f 2 x1 + 35 . x 2 + 5x 3 + 6.5x 4 1 + Pxs 100 1 0.21
c.
n f = aR f , (n f = 75.0 kmol / h, R f = 60) ⇒ n f = 125 . Rf n a = bRa , (n a = 550 kmol / h, Ra = 25) ⇒ n a = 22.0 Ra xi = kAi ⇒
∑x
i
=k
∑A
i
i
=1 ⇒ k =
i
1
∑A
i
i
⇒ xi =
Ai
∑A
, i = CH 4 , C 2 H 4 , C 3 H 8 , C 4 H 10
i
i
4.68
Run 1 2 3
Pxs 15% 15% 15%
Rf 62 83 108
A1 248.7 305.3 294.2
A2 19.74 14.57 16.61
A3 6.35 2.56 4.78
A4 1.48 0.70 2.11
Run 1 2 3
nf 77.5 103.8 135.0
x1 0.900 0.945 0.926
x2 0.0715 0.0451 0.0523
x3 0.0230 0.0079 0.0150
x4 0.0054 0.0022 0.0066
na 934 1194 1592
d.
Either of the flowmeters could be in error, the fuel gas analyzer could be in error, the flowmeter calibration formulas might not be linear, or the stack gas analysis could be incorrect.
a.
C4H10 + 13/2 O2 → 4 CO2 + 5 H2O Basis:
100 mol C4H10
nCO2 (mol CO2) nH2O (mol H2O) nC4H10 (mol C4H10) nO2 (mol O2) nN2 (mol N2)
Pxs (% excess air) nair (mol air) 0.21 O2 0.79 N2 D.F. analysis 6 unknowns (n, n1, n2, n3, n4, n5) -3 atomic balances (C, H, O) -1 N2 balance -1 % excess air -1 % conversion 0 D.F.
4-59
Ra 42.4 54.3 72.4
4.68 (cont’d) b.
i) Theoretical oxygen = (100 mol C4H10)(6.5 mol O2/mol C4H10) = 650 mol O2
n air = (650 mol O 2 )(1 mol air / 0.21 mol O 2 ) = 3095 mol air 100% conversion ⇒ n C4H10 = 0 , nO 2 = 0
U| 73.1% N b gb g = b100 mol C H react gb4 mol CO mol C H g = 400 mol CO V12.0% CO = b100 mol C H react gb5 mol H O mol C H g = 500 mol H O |W14.9% H O
n N2 = 0.79 3095 mol = 2445 mol nCO2 n H2O
4
4
2
10
2
10
2
4
4
10
10
2
2
2
2
ii) 100% conversion ⇒ nC4H10 = 0 20% excess ⇒ nair = 1.2(3095) = 3714 mol (780 mol O2, 2934 mol N2) Exit gas: 400 mol CO2
10.1% CO2
500 mol H2O
12.6% H2O
130 mol O2
3.3% O2
2934 mol N2
74.0% N 2
iii) 90% conversion ⇒ nC4H10 = 10 mol C4H10 (90 mol C4H10 react, 585 mol O2 consumed) 20% excess: nair = 1.2(3095) = 3714 mol (780 mol O2, 2483 mol N2) Exit gas: 0.3% C4H10
10 mol C4H10 360 mol CO2
9.1% CO2
450 mol H2O (v)
4.69
a.
11.4% H2O
195 mol O2
4.9% O2
2934 mol N2
74.3% N 2
C3H8 + 5 O2 → 3 CO2 + 4 H2O
H2 +1/2 O2 → H2O
C3H8 + 7/2 O2 → 3 CO + 4 H2O Basis: 100 mol feed gas 100 mol 0.75 mol C3H8 0.25 mol H2
n1 (mol C3H8) n2 (mol H2) n3 (mol CO2) n4 (mol CO) n5 (mol H2O) n6 (mol O2) n7 (mol N2)
n0 (mol air) 0.21 mol O2/mol 0.79 mol N2/mol Theoretical oxygen:
75 mol C 3 H 8
5 mol O 2 25 mol H 2 0.50 mol O 2 + = 387.5 mol O 2 mol C 3 H 8 mol H 2
4-60
4.69 (cont’d)
Air feed rate: n0 =
387.5 mol O 2 1 kmol air 1.25 kmol air fed = 2306.5 mol air h 0.21 kmol O 2 1 kmol air req' d.
. (75 mol C 3 H 8 ) = 7.5 mol C 3 H 8 90% propane conversion ⇒ n1 = 0100 (67.5 mol C 3 H 8 reacts) . (25 mol C 3 H 8 ) = 3.75 mol H 2 85% hydrogen conversion ⇒ n2 = 0150 95% CO 2 selectivity ⇒ n3 =
0.95(67.5 mol C 3 H 8 react) 3 mol CO 2 generated mol C 3 H 8 react
= 192.4 mol CO 2 5% CO selectivity ⇒ n3 =
0.05(67.5 mol C 3 H 8 react) 3 mol CO generated = 101 . mol CO mol C 3 H 8 react
FG H
H balance: (75 mol C 3 H 8 ) 8
IJ K
mol H + ( 25 mol H 2 )(2) mol C 3 H 8
= (7.5 mol C 3 H 8 )(8) + (3.75 mol H 2 )(2) + n5 ( mol H 2 O)(2) ⇒ n5 = 2912 . mol H 2 O mol O ) = (192.4 mol CO 2 )(2) mol O 2 + (101 . mol CO)(1) + (2912 . mol H 2 O)(1) + 2n6 ( mol O 2 ) ⇒ n6 = 1413 . mol O 2
O balance: (0.21 × 2306.5 mol O 2 )(2
N 2 balance: n7 = 0.79(2306.5) mol N 2 = 1822 mol N 2 Total moles of exit gas = (7.5 + 3.75 + 192.4 + 10.1 + 291.2 + 141.3 + 1822) mol = 2468 mol CO concentration in exit gas =
b.
101 . mol CO × 10 6 = 4090 ppm 2468 mol
If more air is fed to the furnace, (i)
more gas must be compressed (pumped), leading to a higher cost (possibly a larger pump, and greater utility costs)
(ii) The heat released by the combustion is absorbed by a greater quantity of gas, and so the product gas temperature decreases and less steam is produced.
4-61
4.70
a.
C5H12 + 8 O2 → 5 CO2 + 6 H2O Basis: 100 moles dry product gas n1 (mol C5H12)
100 mol dry product gas (DPG) 0.0027 mol C5H12/mol DPG 0.053 mol O2/mol DPG 0.091 mol CO2/mol DPG 0.853 mol N2/mol DPG n3 (mol H2O)
Excess air n2 (mol O2) 3.76n2 (mol N2)
3 unknowns (n1, n2, n3) -3 atomic balances (O, C, H) -1 N2 balance -1 D.F. ⇒ Problem is overspecified b.
N2 balance: 3.76 n2 = 0.8533 (100) ⇒ n2 = 22.69 mol O2 C balance: 5 n1 = 5(0.0027)(100) + (0.091)(100) ⇒ n1 = 2.09 mol C5H12 H balance: 12 n1 = 12(0.0027)(100) + 2n3 ⇒ n3 = 10.92 mol H2O O balance: 2n2 = 100[(0.053)(2) + (0.091)(2)] + n3 ⇒ 45.38 mol O = 39.72 mol O Since the 4th balance does not close, the given data cannot be correct.
c.
n1 (mol C5H12)
100 mol dry product gas (DPG) 0.00304 mol C5H12/mol DPG 0.059 mol O2/mol DPG 0.102 mol CO2/mol DPG 0.836 mol N2/mol DPG n3 (mol H2O)
Excess air n2 (mol O2) 3.76n2 (mol N2)
N2 balance: 3.76 n2 = 0.836 (100) ⇒ n2 = 22.2 mol O2 C balance: 5 n1 = 100 (5*0.00304 + 0.102) ⇒ n1 = 2.34 mol C5H12 H balance: 12 n1 = 12(0.00304)(100) + 2n3 ⇒ n3 = 12.2 mol H2O O balance: 2n2 = 100[(0.0590)(2) + (0.102)(2)] + n3 ⇒ 44.4 mol O = 44.4 mol O √ Fractional conversion of C5H12:
2.344 − 100 × 0.00304 = 0.870 mol react/mol fed 2.344
Theoretical O2 required: 2.344 mol C5H12 (8 mol O2/mol C5H12) = 18.75 mol O2 % excess air:
22.23 mol O 2 fed - 18.75 mol O 2 required × 100% = 18.6% excess air 18.75 mol O 2 required
4-62
4.71
a.
12 L CH 3 OH 1000 ml 0.792 g mol = 296.6 mol CH 3 OH / h h L ml 32.04 g CH3OH + 3/2 O2 → CO2 +2 H2O, CH3OH + O2 → CO +2 H2O
n 2 ( mol dry gas / h) 0.0045 mol CH3OH(v)/mol DG 0.0903 mol CO2/mol DG 0.0181 mol CO/mol DG x (mol N2/mol DG) (0.8871–x) (mol O2/mol DG) n 3 ( mol H 2 O(v) / h)
296.6 mol CH3OH(l)/h
n1 (mol O 2 / h) 3.76n1 (mol N 2 / h)
4 unknowns (n1 , n 2 , n 3 , x ) – 4 balances (C, H, O, N2) = 0 D.F. b.
Theoretical O2: 296.6 (1.5) = 444.9 mol O2 / h C balance: 296.6 = n 2 (0.0045 + 0.0903 + 0.0181) ⇒ n 2 = 2627 mol/h H balance: 4 (296.6) = n 2 (4*0.0045) + 2 n 3 ⇒ n 3 = 569.6 mol H2O / h
O balance : 296.6 + 2n1 = 2627[0.0045 + 2(0.0903) + 0.0181 + 2(0.8871 - x)] + 569.6 N2 balance: 3.76 n 1 = x ( 2627) Solving simultaneously ⇒ n1 = 574.3 mol O 2 / h, x = 0.822 mol N 2 / mol DG Fractional conversion: % excess air:
574.3 − 444.9 × 100% = 29.1% 444.9
Mole fraction of water:
4.72
296.6 − 2627(0.0045) = 0.960 mol CH 3 OH react/mol fed 296.6
569.6 mol H 2 O = 0.178 mol H 2 O/mol (2627 + 569.6) mol
c.
Fire, CO toxicity. Vent gas to outside, install CO or hydrocarbon detector in room, trigger alarm if concentrations are too high
a.
G.C. Say ns mols fuel gas constitute the sample injected into the G.C. If xCH 4 and xC2 H 6 are the mole fractions of methane and ethane in the fuel, then
b g b n b molg x b mol CH
gb g = 20 molgb1 mol C 1 mol CH g 85
ns mol xC2 H 6 mol C 2 H 2 mol 2 mol C 1 mol C 2 H 6
E
s
CH 4
b bmol CH
4
4
xC2 H 6 mol C 2 H 6 mol fuel xCH 4
4
mol fuel
g
g = 01176 mole C H . 2
4-63
6
mole CH 4 in fuel gas
4.72 (cont’d)
b1.134 g H Ogb1 mol 18.02 gg = 0126 .
mole H 2 O 0.50 mol product gas mole product gas Basis: 100 mol product gas. Since we have the most information about the product stream composition, we choose this basis now, and would subsequently scale to the given fuel and air flow rates if it were necessary (which it is not). 2
Condensation measurement:
CH 4 + 2O 2 → CO 2 + 2H 2 O 7 C 2 H 6 + O 2 → 2CO 2 + 3H 2 O 2 100 mol dry gas / h
n1 (mol CH4 ) 0.1176 n1 (mol C2H6) n2 (mol CO2)
0.126 mol H2O / mol 0..874 mol dry gas / mol 0.119 mol CO2 / mol D.G. x (mol N2 / mol) (0.881-x) (mol O2 / mol D.G.)
n3 (mol O2 / h) 376 n3 (mol N2 / h)
Strategy: H balance ⇒ n ; 1
N 2 balance
C balance ⇒ n 2 ;
b gb
g b gb
gb g
O balance
UV ⇒ n , x W 3
H balance: 4n1 + 6 01176 . n1 = 100 0126 . 2 ⇒ n1 = 5.356 mol CH 4 in fuel ⇒ 0.1176(5.356) = 0.630 mol C2H6 in fuel
b gb
g
b gb
gb
g
C balance: 5.356 + 2 0.630 + n2 = 100 0.874 0119 . ⇒ n2 = 3.784 mol CO 2 in fuel Composition of fuel: 5.356 mol CH 4 , 0.630 mol C 2 H 6 , 3.784 mols CO 2 ⇒ 0.548 CH 4 , 0.064 C 2 H 6 , 0.388 CO 2
b gb
g
N 2 balance: 3.76n3 = 100 0.874 x
b gb
g
b gb
g b gb
gb g
b
O balance: 2 3.784 + 2n3 = 100 0.126 + 100 0.874 2 0119 . + 0.881 − x
g
Solve simultaneously: n3 = 18.86 mols O 2 fed , x = 0.813 5.356 mol CH 4 2 mol O 2 0.630 mol C 2 H 6 3.5 mol O 2 Theoretical O 2 : + 1 mol CH 4 1 mol CH 4
= 12.92 mol O 2 required Desired O2 fed:
(5.356 + 0.630 + 3.784) mol fuel 7 mol air 0.21 mol O 2 = 14.36 mol O2 1 mol fuel mol air
Desired % excess air:
b.
Actual % excess air:
14.36 − 12.92 × 100% = 11% 12.92
18.86 − 12.92 × 100% = 46% 12.92
Actual molar feed ratio of air to fuel:
(18.86 / 0.21) mol air = 9 :1 9.77 mol feed
4-64
4.73
a.
C3H8 +5 O2 → 3 CO2 + 4 H2O, C4H10 + 13/2 O2 → 4 CO2 + 5 H2O Basis 100: mol product gas n1 (mol C3H8) n2 (mol C4H10)
100 mol 0.474 mol H2O/mol x (mol CO2/mol) (0.526–x) (mol O2/mol)
n3 (mol O2)
Dry product gas contains 69.4% CO2 ⇒
x 69.4 = ⇒ x = 0.365 mol CO 2 /mol 0.526 − x 30.6
3 unknowns (n1, n2, n3) – 3 balances (C, H, O) = 0 D.F. O balance: 2 n3 = 152.6 ⇒ n3 = 76.3 mol O2 n1 = 7.1 mol C 3 H 8 C balance : 3 n1 + 4 n 2 = 36.5 ⎫ ⇒ 65.1% C 3 H 8 , 34.9% C 4 H10 ⎬⇒ H balance : 8 n1 + 10 n 2 = 94.8⎭ n 2 = 3.8 mol C 4 H10 b.
nc=100 mol (0.365 mol CO2/mol)(1mol C/mol CO2) = 365 mol C nh = 100 mol (0.474 mol H2O/mol)(2mol H/mol H2O)=94.8 mol H
⇒ 27.8%C, 72.2% H From a: 7.10 mol C 3 H 8
3.80 mol C 4 H10 4 mol C 3 mol C + mol C 3 H 8 mol C 4 H10
7.10 mol C 3 H 8 11 mol (C + H) 3.80 mol C 4 H10 14 mol (C + H) + mol C 3 H 8 mol C 4 H10 4.74
Basis: 100 kg fuel oil Moles of C in fuel:
100 kg 0.85 kg C 1 kmol C = 7.08 kmol C kg 12.01 kg C
Moles of H in fuel:
100 kg 0.12 kg H 1 kmol H = 12.0 kmol H kg 1 kg H
Moles of S in fuel:
100 kg 0.017 kg S 1 kmol S = 0.053 kmol S kg 32.064 kg S
1.3 kg non-combustible materials (NC)
4-65
× 100% = 27.8% C
4.74 (cont’d)
100 kg fuel oil 7.08 kmol C 12.0 kmol H 0.053 kmol S 1.3 kg NC (s) 20% excess air n1 (kmol O2) 3.76 n1 (kmol N2)
C + O2 → CO2 C + 1/2 O2 → CO 2H + 1/2 O2 → H2O S + O2 → SO2
n2 (kmol N2) n3 (kmol O2) n4 (kmol CO2) (8/92) n4 (kmol CO) n5 (kmol SO2) n6 (kmol H2O)
Theoretical O2: 7.08 kmol C 1 kmol O 2 12 kmol H .5 kmol O 2 0.053 kmol S 1 kmol O 2 + + = 10.133 kmol O 2 1 kmol S 2 kmol H 1 kmol C
20 % excess air: n1 = 1.2(10.133) = 12.16 kmol O2 fed O balance: 2 (12.16) = 2 (6.5136) + 0.5664 + 2 (0.053) + 6 + 2 n3 ⇒ n3 = 2.3102 kmol O2 C balance: 7.08 = n4+8n4/92 ⇒ n4 = 6.514 mol CO2 ⇒ 8 (6.514)/92 = 0.566 mol CO S balance: n5 = 0.53 kmol SO2 H balance: 12 = 2n6 ⇒ n6 = 6.00 kmol H2O N2 balance: n2 = 3.76(12.16) = 45.72 kmol N2 Total moles of stack gas = (6.514 + 0.566 + 0.053 + 6.00 + 2.310 + 45.72) kmol = 61.16 kmol ⇒ 10.7% CO, 0.92% CO, 0.087% SO 2 , 9.8% H 2 O, 3.8% O 2 , 74.8% N 2
4.75
a. Basis: 5000 kg coal/h; 50 kmol air min = 3000 kmol air h 5000 kg coal / h 0.75 kg C / kg 0.17 kg H / kg 0.02 kg S / kg 0.06 kg ash / kg
C + 02 --> CO2 2H + 1/2 O2 -->H2O S + O2 --> SO2 C + 1/2 O2 --> CO
3000 kmol air / h 0.21 kmol O2 / kmol 0.79 kmol N2 / kmol
n1 (kmol O2 / h) n2 (kmol N2 / h) n3 (kmol CO2 / h) 0.1 n3 (kmol CO / h) n4 (kmol SO2 / h) n5 (kmol H2O / h)
mo kg slag / h
Theoretical O 2 : C:
0.75 5000 kg C
b g
1 kmol C
1 kmol O 2
h
12.01 kg C
1 kmol C
4-66
= 312.2 kmol O 2 h
4.75 (cont’d)
H:
S:
b g
0.17 5000 kg H 1 kmol H 1 kmol H 2 O h
101 . kg H
2 kmol H
0.02 5000 kg S
b g
1 kmol S
1 kmol O 2
h
32.06 kg S
1 kmol S
1 kmol O 2 2 kmol H 2 O
= 210.4 kmol O 2 h
= 3.1 kmol O2/h
Total = (312.2+210.4 + 3.1) kmol O2/h = 525.7 kmol O 2 h
b
g
O 2 fed = 0.21 3000 = 630 kmol O 2 h Excess air:
630 − 525.7 × 100% = 19.8% excess air 525.7
b. Balances: 0.94 0.75 5000 kg C react 1 kmol C = n 3 + 0.1n 3 C: h 12.01 kg C
b gb gb g
⇒ n 3 = 266.8 kmol CO 2 h , 01 . n 3 = 26.7 kmol CO h
H:
. gb5000g kg H b017
S:
(from part a)
N2 :
O:
h
1 kmol H 1 kmol H 2 O 101 . kg H
b
2 kmol H
3.1 kmol O 2 for SO 2
g
= n5 ⇒ n5 = 420.8 kmol H 2 O h
1 kmol SO 2 1 kmol O 2
h
. kmol SO 2 h = n 4 ⇒ n 4 = 31
b0.79gb3000g kmol N h = n ⇒ n = 2370 kmol N h b0.21g(3000)b2g = 2n + 2b266.8g + 1b26.68g + 2b31. g + b1gb420.8g 2
2
2
2
1
⇒ n1 = 136.4 kmol O 2 / h Stack gas total = 3223 kmol h Mole fractions: x CO = 26.7 3224 = 8.3 × 10 −3 mol CO mol
x SO 2 = 31 . 3224 = 9.6 × 10 −4 mol SO 2 mol
c.
1 SO 2 + O 2 → SO 3 2 SO 3 + H 2 O → H 2SO 4 3.1 kmol SO 2 1 kmol SO 3 1 kmol H 2SO 4 h 1 kmol SO 3 1 kmol SO 2
4-67
98.08 kg H 2SO 4 = 304 kg H 2SO 4 h kmol H 2SO 4
4.76 a.
Basis: 100 g coal as received (c.a.r.). Let a.d.c. denote air-dried coal; v.m. denote volatile matter 100 g c.a. r. 1.147 g a.d.c. 1.207 g c.a. r.
95.03 g a.d.c
= 95.03 g air - dried coal; 4.97 g H 2 O lost by air drying
. ggH O b1.234 − 1204 = 2.31 g H O lost in second drying step 2
2
1.234 g a.d.c.
Total H 2 O = 4.97 g + 2.31 g = 7.28 g moisture 95.03 g a.d.c
. − 0.811g g b v. m.+ H Og b1347 − 2.31 g H O = 3550 . g volatile matter 2
2
1.347 g a.d.c.
95.03 g a.d.c
0.111 g ash 1.175 g a.d.c.
= 8.98 g ash
b
g
Fixed carbon = 100 − 7.28 − 3550 . − 8.98 g = 48.24 g fixed carbon
7.28 g moisture 48.24 g fixed carbon
7.3% moisture 48.2% fixed carbon 35.50 g volatile matter ⇒ 35.5% volatile matter 8.98 g ash 9.0% ash 100 g coal as received b. Assume volatile matter is all carbon and hydrogen.
C + CO 2 → CO 2 : 2H +
1 mol O 2 1 mol C
1 mol C 10 3 g 1 mol air = 396.5 mol air kg C 12.01 g C 1 kg 0.21 mol O 2
0.5 mol O 2 1 O2 → H 2O : 2 2 mol H
Air required:
1 mol H 10 3 g 1 mol air = 1179 mol air kg H 1.01 g H 1 kg 0.21 mol O 2
1000 kg coal 0.482 kg C 396.5 mol air kg coal kg C 1000 kg 0.355 kg v. m. 6 kg C 396.5 mol air kg 7 kg v. m. kg C 1000 kg 0.355 kg v. m. 1 kg H 1179 mol air + = 3.72 × 10 5 mol air kg 7 kg v. m. kg H +
4-68
4.77
a.
Basis 100 mol dry fuel gas. Assume no solid or liquid products!
n1 (mol C) n2 (mol H) n3 (mol S)
100 mol dry gas C + 02 --> CO2 C + 1/2 O2 --> CO 2H + 1/2 O2 -->H2O S + O2 --> SO2
n4 (mol O2) (20% excess)
0.720 mol CO2 / mol 0.0257 mol CO / mol 0.000592 mol SO2 / mol 0.254 mol O2 / mol n5 (mol H2O (v))
⎫ ⎪ O balance : 2 n 4 = 100 [ 2(0.720) + 0.0257 + 2 (0.000592) + 2 (0.254)] + n 5 ⎬ ⎪ 20 % excess O 2 : (1.20) (74.57 + 0.0592 + 0.25 n 2 ] = n 4 ⎭
H balance : n 2 = 2 n 5
⇒ n2 = 183.6 mol H, n4 = 144.6 mol O2, n5 = 91.8 mol H2O Total moles in feed: 258.4 mol (C+H+S) ⇒ 28.9% C, 71.1% H, 0.023% S
4.78
Basis: 100 g oil Stack SO 2 , N 2 , O 2, CO 2, H 2O (612.5 ppm SO 2)
x n 3 mol SO 2 (N2 , O2 , CO2 , H 2 O)
0.10 (1 – x ) n 5 mol SO2 (N2 , O2 , CO2 , H 2 O)
100 g oil 0.87 g C/g 0.10 g H/g 0.03 g S/g n 1 mol O2 3.76 n 1 mol N2 (25% excess)
furnace Alkaline solution (1 – x ) n 5 mol SO 2 (N2 , O2 , CO2 , H 2 O) n 2 mol N 2 n 3 mol O 2 n 4 mol CO2 n 5 mol SO 2 n 6 mol H 2 O
CO 2 : H 2 O:
b g
0.87 100 g C
scrubber
0.90 (1 – x ) n 5 mol SO2
FG H
7.244 mol O 2 1 mol C 1 mol CO 2 ⇒ n4 = 7.244 mol CO 2 consumed 1 mol C 12.01 g C
b g
FG H
IJ K
2.475 mol O 2 0.10 100 g H 1 mol H 1 mol H 2 O ⇒ n6 = 4.95 mol H 2 O consumed 2 mol H 101 . gH
4-69
IJ K
4.78 (cont’d)
SO 2 :
b g
FG H
0.0956 mol O 2 1 mol S 1 mol SO 2 ⇒ n5 = 0.0936 mol SO 2 consumed 32.06 g S 1 mol S
0.03 100 g S
b
g
IJ K
25% excess O 2 : n1 = 125 . 7.244 + 2.475 + 0.0936 ⇒ 12.27 mol O 2
b
g
O 2 balance: n3 = 12.27 mol O 2 fed − 7.244 + 2.475 + 0.0936 mol O 2 consumed = 2.46 mol O 2
b
g
N 2 balance: n 2 = 3.76 12.27 mol = 4614 . mol N 2
b
g
SO 2 in stack SO 2 balance around mixing point :
F H
I K
b gb
g
b
x 0.0936 + 010 . 1 − x 0.0936 = 0.00936 + 0.0842 x mol SO 2 n5
g
Total dry gas in stack (Assume no CO2 , O2 , or N 2 is absorbed in the scrubber)
b
g
b
g
7.244 + 2.46+ 4614 . + 0.00936 + 0.0842 x = 5585 . + 0.0842 x mol dry gas
b CO g b O g 2
bN g
2
bSO g
2
b
2
g
612.5 ppm SO 2 dry basis in stack gas
0.00936 + 0.0842 x 612.5 = ⇒ x = 0.295 ⇒ 30% bypassed 5585 . + 0.0842 x 10 . × 10 6 Basis: 100 mol stack gas
4.79
n 1 (mol C) n 2 (mol H) n 3 (mol S) n 4 (mol O2 ) 3.76 n 4 (mol O2 )
a.
b gb b gb
C + O2 → CO2 1 2H + O2 → H 2 O 2 S + O 2 → SO 2
100 mol 0.7566 N 2 0.1024 CO2 0.0827 H 2 O 0.0575 O 2 0.000825 SO 2
g gb g
C balance: n1 = 100 01024 . = 10.24 mol C mol C 10.24 mol C ⇒ = 0.62 H balance: n2 = 100 0.0827 2 = 16.54 mol H mol H 16.54 mol H The C/H mole ratio of CH 4 is 0.25, and that of C2 H 6 is 0.333; no mixture of the two could have a C/H ratio of 0.62, so the fuel could not be the natural gas.
b.
b gb
g
S balance: n 3 = 100 0.000825 = 0.0825 mol S
b10.24 mol Cgb12.0 g 1 molg = 122.88 g CU| 122.88 = 7.35 g C g H b16.54 mol Hgb1.01 g 1 molg = 16.71 g HV ⇒ 216.65.71 b0.0825 mol Sgb32.07 g 1 molg = 2.65 g S |W 142.24 × 100% = 1.9% S
4-70
⇒ No. 4 fuel oil
4.80
a.
Basis: 1 mol CpHqOr 1 mol CpHqOr no (mol S) Xs (kg s/ kg fuel)
C + 02 --> CO2 2H + 1/2 O2 -->H2O S + O2 --> SO2
P (% excess air) n1 (mol O2) 3.76 n1 (mol N2)
n2 (mol CO2) n3 (mol SO2) n4 (mol O2) 3.76 n1 (mol N2) n5 (mol H2O (v))
Hydrocarbon mass: p (mol C) ( 12 g / mol) = 12 p (g C) q (mol H) (1 g / mol) =
q (g H)
⇒ (12 p + q + 16 r) g fuel
r (mol O) (16 g / mol) = 16 r (g O) S in feed: n o=
(12 p + q + 16r) g fuel
Theoretical O2:
X s (g S) X (12 p + q + 16 r) 1 mol S = s (mol S) (1) (1 - X s ) (g fuel) 32.07 g S 32.07(1 - X s )
p (mol C) 1 mol O 2 q (mol H) 0.5 mol O 2 ( r mol O) 1 mol O 2 + − 1 mol C 2 mol H 2 mol O
= (p + 1/4 q − 1/2 r) mol O 2 required % excess ⇒ n1 = (1 + P/100) (p +1/4 q – ½ r) mol O2 fed
(2)
C balance: n2 = p
(3)
H balance: n5 = q/2
(4)
S balance: n3 = n0
(5)
O balance: r + 2n1 = 2n2 + 2n3 + 2n4 + n5 ⇒ n4 = ½ (r+2n1-2n2-2n3-n5)
(6)
Given: p = 0.71, q= 1.1, r = 0.003, Xs = 0.02 P = 18% excess air (1) ⇒ n0 = 0.00616 mol S
(5) ⇒ n3 = 0.00616 mol SO2
(2) ⇒ n1 = 1.16 mol O2 fed
(6) ⇒ n4 = 0.170 mol O2
(3) ⇒ n2 = 0.71 mol CO2
(4) ⇒ n5 = 0.55 mol H2O
(3.76*1.16) mol N2 = 4.36 mol N2 Total moles of dry product gas = n2 + n3 + n4 + 3.76 n1=5.246 mol dry product gas Dry basis composition yCO2 = (0.710 mol CO2/ 5.246 mol dry gas) * 100% = 13.5% CO2 yO2 = (0.170 / 5.246) * 100% = 3.2% O2 yN2 = (4.36 / 5.246) * 100% = 83.1% N2 ySO2 = (0.00616 / 5.246) * 106 = 1174 ppm SO2
4-71
CHAPTER FIVE 5.1
Assume volume additivity 1
Av. density (Eq. 5.1-1):
m
a.
A
ρ
+ m0 ⇒ m = = mt
A
mass of tank at time t
mass of empty tank
=
0.400 0.600 + ⇒ ρ = 0.719 kg L 0.703 kg L 0.730 kg L
A
A
ρO
ρD
b250 − 150gkg = 14.28 kg min bm = mass flow rate of liquidg b10 − 3g min
1L / min) = 14.28 kg / min) = m(kg ⇒ V(L ⇒ V = 19.9 L min min 0.719 kg ρ ( kg / L)
bg
= 150 − 14.28 3 = 107 kg b. m0 = m(t) - mt
5.2
b
g
void volume of bed: 100 cm3 − 2335 . − 184 cm3 = 50.5 cm3 porosity: 50.5 cm3 void 184 cm3 total = 0.274 cm3 void cm3 total bulk density: 600 g 184 cm3 = 3.26 g cm3
b
g
absolute density: 600 g 184 − 50.5 cm3 = 4.49 g cm3 5.3 C 6 H 6 (l )
B (kg / min) m = 20.0 L / min V B
(kg / min) m (L / min) V
C 7 H 8 (l )
T (kg / min) m (L / min) V T 2 2 . m = ΔV = πD Δh = π (5.5 m) 015 V = 0.0594 m3 / min Δt 4 60 min 4 Δt Assume additive volumes =V -V = 59.4 − 20.0 L / min = 39.4 L / min V T B
b
g
0.879 kg 20.0 L 0.866 kg 39.4 L . kg / min + = 517 L min L min m (0.879 kg / L)(20.0 L / min) xB = B = = 0.34 kg B / kg m (517 . kg / min) + ρ ⋅V = = ρB ⋅ V m B T T
5-1
a.
b.
1
ρ sl
=
xc
+
ρc
F IF I e jge jhbmgGH 11 N JK GH 11 Pa JK = ρ gh kg
m s2
m3
kg⋅m s2
sl
N m2
b1 − x g ⇒ check units! c
ρl
1 kg crystals / kg slurry kg liquid / kg slurry = + kg slurry / L slurry kg crystals / L crystals kg liquid / L liquid L slurry L crystals L liquid L slurry = + = kg slurry kg slurry kg slurry kg slurry ΔP 2775 = = 1415 kg / m3 c. i.) ρ sl = 9.8066 0.200 gh ii.)
1
ρ sl
xc
=
ρc
b gb g b1- x g ⇒ x FG 1 − 1 IJ = FG 1 + ρ Hρ ρ K Hρ c
c
l
F 1 GG 1415 kg / m H
xc =
c
3
−
l
−
sl
I J 12 . d1000 kg / m i JK
IJ ρ K 1
l
1
3
= 0.316 kg crystals / kg slurry F I GG 2.3d10001kg / m i − 12. d10001kg / m iJJ H K 3
msl
iii.) Vsl =
=
ρ sl
3
175 kg 1000 L = 123.8 L 3 1415 kg / m m3
b
gb
g
iv.) mc = x c msl = 0.316 kg crystals / kg slurry 175 kg slurry = 55.3 kg crystals v.) mCuSO 4 =
55.3 kg CuSO 4 ⋅ 5H 2 O 1 kmol 1 kmol CuSO 4 159.6 kg = 35.4 kg CuSO 4 249 kg 1 kmol CuSO 4 ⋅ 5H 2 O 1 kmol
b
g
b
gb
g
vi.) ml = 1 − x c msl = 0.684 kg liquid / kg slurry 175 kg slurry = 120 kg liquid solution vii.)
Vl =
h(m) ρ l(kg/m^3) ρ c(kg/m^3) ΔP(Pa) xc ρ sl(kg/m^3)
0.2 1200 2300 2353.58 0 1200.00
ml
ρl
=
120 kg
1000 L = 100 L m3
b1.2gd1000 kg / m i 3
d. 2411.24 0.05 1229.40
2471.80 0.1 1260.27
2602.52 0.2 1326.92
2747.84 0.3 1401.02
2772.61 0.316 1413.64
2910.35 0.4 1483.87
3093.28 0.5 1577.14
Effect of Slurry Density on Pressure Measurement
0.6 Solids Fraction
5.4
U| V| W
P1 = P0 + ρ sl gh1 P2 = P0 + ρ sl gh 2 ⇒ ΔP = P1 − P2 = ρ sl h = h1 − h 2
0.5 0.4 0.3
ΔP = 2775, ρ = 0.316
0.2 0.1 0 2300.00
2500.00
2700.00
2900.00
Pressure Difference (Pa)
5-2
3100.00
5.4 (cont’d) e.
b
g d
i
Basis: 1 kg slurry ⇒ x c kg crystals , Vc m3 crystals =
b1- x gbkg liquid g, V dm c
l
3
b
g
x c kg crystals
d
ρ c kg / m
3
i
g i b1-ρx dgbkgkg/ mliquid i
liquid =
c
3
l
ρ sl =
1 kg
bV + V gdm i 3
c
5.5
=
l
xc
ρc
+
1 1 − xc
b
ρl
g
Assume Patm = 1 atm 3 = RT ⇒ V = 0.08206 m ⋅ atm 313.2 K 1 kmol = 0.0064 m3 mol PV kmol ⋅ K 4.0 atm 103 mol
ρ=
5.6
a.
1 mol 0.0064 m air
V=
mol 103 g
= 4.5 kg m3
b3.06L - 2.8Lg × 100% = 9.3% 2.8L
. bar Assume Patm = 1013 a.
PV = nRT ⇒ n =
b.
. gbar b10 + 1013
b
20.0 m3 kmol ⋅ K 28.02 kg N 2 = 249 kg N 2 kmol 25 + 273.2 K 0.08314 m3 ⋅ bar
g
T P n PV nRT = ⇒ n = V⋅ s ⋅ ⋅ s Ps Vs n s RTs T Ps Vs
n=
5.8
1 kg
nRT 1.00 mol 0.08206 L ⋅ atm 373.2 K = = 3.06 L P mol ⋅ K 10 atm
b. % error = 5.7
29.0 g
3
20.0 m3
273K 298.2K
. gbar b10 + 1013 1.013 bar
1 kmol 28.02 kg N 2 = 249 kg N 2 3 kmol 22.415 m STP
b g
a.
R=
Ps Vs 1 atm 22.415 m3 atm ⋅ m3 = = 8.21 × 10 −2 n sTs 1 kmol 273 K kmol ⋅ K
b.
R=
Ps Vs 1 atm 760 torr 359.05 ft 3 torr ⋅ ft 3 = = 555 n sTs 1 lb - mole 1 atm 492 D R lb - mole ⋅D R
5-3
5.9
P = 1 atm +
10 cm H 2 O
1m 1 atm = 101 . atm 2 10 cm 10.333 m H 2 O
3 = 2.0 m = 0.40 m3 min = 400 L min T = 25D C = 298.2 K , V 5 min = n mol / min ⋅ MW g / mol m
b
g
= m
a.
= m
b.
b
g
L 28.02 PV 1.01 atm 400 min ⋅ MW = L⋅atm RT 0.08206 mol⋅K 298.2 K
400
L min
28.02 273 K 1 mol 298.2 K 22.4 L STP
b g
g mol
g mol
= 458 g min
= 458 g min
P u F mI V dm si = nRT uG J = ⇒ H s K Ad m i π D 4 u 3
5.10 Assume ideal gas behavior:
2
2
2
=
1
T2 P1 D12 nR ⋅ ⋅ ⋅ 2 nR T1 P2 D 2
(1.80 + 1.013) bar ( 7.50 cm ) = 165 2 (1.53 + 1.013) bar ( 5.00 cm ) 2
T P D 2 60.0 m 333.2K u 2 = u1 2 1 12 = T1P2 D 2 sec 300.2K
b
m sec
g
. + 100 . atm 5 L PV 100 = = 0.406 mol L⋅atm 0.08206 mol⋅K 300 K RT MW = 13.0 g 0.406 mol = 32.0 g mol ⇒ Oxygen
5.11 Assume ideal gas behavior: n =
5.12 Assume ideal gas behavior: Say m t = mass of tank, n g = mol of gas in tank
b gU|V ⇒ n = 0.009391 mol CO : 37.440 g = m + n b44.1 g molg W| m = 37.0256 g b37.062 − 37.0256gg = 3.9 g mol ⇒ Helium unknown: MW = 37.289 g = m t + n g 28.02 g mol
N2:
2
t
g
g
t
0.009391 mol
5.13 a.
b.
3 3 cm3 STP min = Δ V liters 273K 763 mm Hg 10 cm = 925.3 Δ V V std Δt 1L Δ t min 296.2K 760 mm Hg
b g
b g
U| |V straight line plot ||φ = 0.031EV + 0.93 W 22.4 litersbSTPg 10 cm = 224 cm
φ
cm3 STP min V std
5.0 9.0 12.0
139 268 370
= 0.010 mol N 2 V std min
d
std
3
1 mole
1L
i
φ = 0.031 224 cm3 / min + 0.93 = 7.9 5-4
3
3
/ min
b Fρ I si ⋅ G J Hρ K
/ kmol) PM g nbkmolgM(kg ====> Vb Lg RT n P = V RT
5.14 Assume ideal gas behavior ρ kg L =
d
12
i d
V2 cm s = V1 cm 3
3
1
= V1 P1M1T2 P2 M 2 T1
12
2
LM N
cm3 758 mm Hg 28.02 g mol 323.2K s 1800 mm Hg 2.02 g mol 298.2K
OP Q
12
a.
VH 2 = 350
b.
M = 0.25M CH 4 + 0.75M C3H 8 = 0.25 16.05 + 0.75 44.11 = 37.10 g mol
cm3 Vg = 350 s
b gb
g b gb
LM b758gb28.02gb323.2g OP N b1800gb37.10gb298.2g Q
= 205 cm3 s
= 881 cm3 s
g
12
5.15 a. Reactor
Δh soap
b.
n CO 2
d
2 πR 2 Δh π 0.012 m PV = ⇒V= = 4 Δt RT
n CO2 =
i
2
. m 60 s 12 = 11 . × 10 −3 m3 / min 7.4 s min
755 mm Hg 1 atm 1.1× 10-3 m3 / min 1000 mol = 0.044 mol/min 3 m ⋅atm 300 K 1 kmol 0.08206 kmol ⋅K 760 mm Hg
5.16 air = 10.0 kg / h m
n air (kmol / h)
n (kmol / h) y CO (kmol CO 2 / kmol) 2
3 V CO 2 = 20.0 m / h
n CO (kmol / h) 2
150 o C, 1.5 bar
Assume ideal gas behavior 10.0 kg 1 kmol n air = = 0.345 kmol air / h h 29.0 kg air n CO 2 =
PV 15 . bar 100 kPa 20.0 m3 / h = = 0.853 kmol CO 2 / h 3 RT 8.314 mkmol⋅kPa 1 bar 423.2 K ⋅K
y CO 2 × 100% =
b
0.853 kmol CO 2 / h × 100% = 712% . 0.853 kmol CO 2 / h + 0.345 kmol air / h
g
5-5
5.17 Basis: Given flow rates of outlet gas. Assume ideal gas behavior 1 (kg / min) m 0.70 kg H 2 O / kg 0.30 kg S / kg
311 m 3 / min, 83o C, 1 atm n 3 (kmol / min) 0.12 kmol H 2 O / kmol 0.88 kmol dry air / kmol
n 2 (kmol air / min) (m 3 / min) V 2 167 o C, - 40 cm H 2 O gauge
4 (kg S / min) m
a.
n 3 =
1 atm
311 m3
356.2K
min
kmol ⋅ K 0.08206 m3 ⋅ atm
H 2 O balance : 0.70 m1 =
10.64 kmol 0.12 kmol H 2 O 18.02 kg kmol kmol min
1 = 32.9 kg min milk ⇒m
b g
= 10.64 kmol min
b
g
4 ⇒m 4 = 9.6 kg S min S olids balance: 0.30 32.2 kg min = m
Dry air balance : n 2 = 0.88 (10.64 kmol min ) ⇒ n 2 = 9.36 kmol min air
= V 2
9.36 kmol 0.08206 m3 ⋅ atm kmol ⋅ K
min
440K
(1033 − 40 ) cm H 2O
1033 cm H 2 O 1 atm
= 352 m3 air min (m3 / s) 352 m3 1 min V u air (m/min)= air = A (m 2 ) min 60 s
π
4
⋅ (6 m) 2
= 0.21 m/s
b. If the velocity of the air is too high, the powdered milk would be blown out of the reactor by the air instead of falling to the conveyor belt.
5.18 SG CO 2
5.19 a.
ρ CO2 = = ρ air
PM CO2 RT PM air RT
=
M CO 2 M air
=
44 kg / kmol = 152 . 29 kg / kmol
x CO 2 = 0.75 x air = 1 − 0.75 = 0.25 Since air is 21% O 2 , x O 2 = (0.25)(0.21) = 0.0525 = 5.25 mole% O 2
b.
mCO 2 = n ⋅ x CO 2 ⋅ M CO 2
b
g
2 × 1.5 × 3 m3 0.75 kmol CO 2 44.01 kg CO 2 1 atm = =12 kg m3 ⋅atm kmol kmol CO 2 298.2 K 0.08206 kmol ⋅K
More needs to escape from the cylinder since the room is not sealed.
5-6
5.19 (cont’d) c. With the room closed off all weekend and the valve to the liquid cylinder leaking, if a person entered the room and closed the door, over a period of time the person could die of asphyxiation. Measures that would reduce hazards are:
1. Change the lock so the door can always be opened from the inside without a key. 2. Provide ventilation that keeps air flowing through the room. 3. Install a gas monitor that sets off an alarm once the mole% reaches a certain amount. 4. Install safety valves on the cylinder in case of leaks.
5.20 n CO 2 =
15.7 kg
1 kmol = 0.357 kmol CO 2 44.01 kg
b
Assume ideal gas behavior, negligible temperature change T = 19° C = 292.2 K
a.
g
P1V n1RT n1 P 102kPa = ⇒ = 1 = P2 V n1 + 0.357 RT n1 + 0.357 P2 3.27 × 103 kPa
b
g
⇒ n1 = 0.0115 kmol air in tank b. Vtank =
ρf = c.
n1RT 0.0115 kmol 292.2 K 8.314 m3 ⋅ kPa 103 L = = 274 L P1 m3 102kPa kmol ⋅ K
15700 g CO 2 + 11.5 mol air ⋅ (29.0 g air / mol) = 58.5 g / L 274 L
CO 2 sublimates ⇒ large volume change due to phase change ⇒ rapid pressure rise. Sublimation causes temperature drop; afterwards, T gradually rises back to room temperature, increase in T at constant V ⇒ slow pressure rise.
b
gb
g
5.21 At point of entry, P1 = 10 ft H 2 O 29.9 in. Hg 33.9 ft H 2 O + 28.3 in. Hg = 37.1 in. Hg .
At surface, P2 = 28.3 in. Hg, V2 = bubble volume at entry 1 x x 0.20 0.80 Mean Slurry Density: = solid + solution = + 3 ρ sl ρ solid ρ solution (12 . )(100 . g / cm ) (100 . g / cm3 ) = 0.967
cm 3 1.03 g 2.20 lb 5 × 10 −4 ton 10 6 cm 3 ⇒ ρ sl = = 4.3 × 10 −3 ton / gal g 1 lb 264.17 gal cm 3 1000 g
a.
300 ton gal 40.0 ft 3 (STP) 534.7 o R 29.9 in Hg = 2440 ft 3 / hr hr 4.3 × 10 −3 ton 1000 gal 492 o R 37.1 in Hg
b.
3π P2 V2 nRT V P = ⇒ 2 = 1⇒ P1V1 nRT V1 P2 4 3π 4
% change =
e j e j
D2 3 2 D1 3 2
=
37.1 ⇒ D 32 = 1.31D13 28.3
b2.2 - 2.0g mm × 100 = 10% 2.0 mm
5-7
==> D D1 = 2 mm
2
= 2.2 mm
5.22 Let B = benzene n1 , n 2 , n 3 = moles in the container when the sample is collected, after the helium is added, and after the gas is fed to the GC. n inj = moles of gas injected
n B , n air , n He = moles of benzene and air in the container and moles of helium added n BGC , m BGC = moles, g of benzene in the GC y B = mole fraction of benzene in room air a.
P1V1 = n1RT1 (1 ≡ condition when sample was taken): P1 = 99 kPa, T1 = 306K n1 =
99 kPa 2 L mol ⋅ K = 0.078 mol = n air + n B kPa 101.3 atm 306 K .08206 L ⋅ atm
P2 V2 = n 2 RT2 (2 ≡ condition when charged with He): P2 = 500 kPa, T2 = 306K
n2 =
500 kPa 2 L mol ⋅ K = 0.393 mol = n air + n B + n He kPa 101.3 atm 306 K .08206 L ⋅ atm
P3 V3 = n 3 RT3 (3 ≡ final condition in lab): P3 = 400 kPa, T3 = 296K n3 =
mol ⋅ K 400 kPa 2 L = 0.325 mol = (n air + n B + n He ) − n inj kPa 101.3 atm 296 K .08206 L ⋅ atm
n inj = n 2 − n 3 = 0.068 mol n B = n BGC × y B (ppm) =
n2 0.393 mol m BGC (g B) 1 mol = = 0.0741 ⋅ m BGC n inj 0.068 mol 78.0 g
nB 0.0741 ⋅ m BGC × 106 = × 106 = 0.950 × 106 ⋅ m BGC n1 0.078
U| | ) = 0.749 ppm VThe avg. is below the PEL | ) = 0.864 ppm| W
9 am: y B = (0.950 × 106 )(0.656 × 10 −6 ) = 0.623 ppm 1 pm: y B = (0.950 × 106 )(0.788 × 10 −6 5 pm: y B = (0.950 × 106 )(0.910 × 10 −6
b. Helium is used as a carrier gas for the gas chromatograph, and to pressurize the container so gas will flow into the GC sample chamber. Waiting a day allows the gases to mix sufficiently and to reach thermal equilibrium. c. (i) It is very difficult to have a completely evacuated sample cylinder; the sample may be dilute to begin with. (ii) The sample was taken on Monday after 2 days of inactivity at the plant. A reading should be taken on Friday. (iii) Helium used for the carrier gas is less dense than the benzene and air; therefore, the sample injected in the GC may be Herich depending on where the sample was taken from the cylinder. (iv) The benzene may not be uniformly distributed in the laboratory. In some areas the benzene concentration could be well above the PEL.
5- 8
b g
4 3 5.23 Volume of balloon = π 10 m = 4189 m3 3 Moles of gas in balloon
b
g
n kmol =
a.
492° R 3 atm
4189 m3
1 kmol
b g = 515.9 kmol
535° R 1 atm 22.4 m3 STP
He in balloon:
b
gb
g
m = 515.9 kmol ⋅ 4.003 kg kmol = 2065 kg He mg =
b.
2065 kg 9.807 m 1N = 20,250 N 2 s 1 kg ⋅ m / s2
dP dP
iV = n RT ⇒ n iV = n RT
gas in balloon air displaced
Fbuoyant
gas
air
=
air
Pair 1 atm ⋅ n gas = ⋅ 515.9 kmol = 172.0 kmol Pgas 3 atm
Fbuoyant = Wair displaced =
172.0 kmol 29.0 kg 9.807 m
Since balloon is stationary, Wtotal
1N = 48,920 N s 1 kg2⋅m s 2
1 kmol
∑F = 0 1
Fcable
Fcable = Fbuoyant − Wtotal = 48920 N −
b2065 + 150gkg
9.807 m 1 N = 27,20 s2 1 kg2⋅m s
c. When cable is released, Fnet
dAi = 27200 N = M ⇒a=
b
tot a
27200 N 1 kg ⋅ m / s2 = 12.3 m s2 2065 + 150 kg N
g
d. When mass of displaced air equals mass of balloon + helium the balloon stops rising. Need to know how density of air varies with altitude. e. The balloon expands, displacing more air ⇒ buoyant force increases ⇒ balloon rises until decrease in air density at higher altitudes compensates for added volume. 5.24 Assume ideal gas behavior, Patm = 1 atm
a.
3 PN VN 5.7 atm 400 m / h PN VN = Pc Vc ⇒ Vc = = = 240 m3 h 9.5 atm Pc
b.
Mass flow rate before diversion: 400 m3 h
273 K 5.7 atm 303 K
1 atm
1 kmol
44.09 kg
22.4 m ( STP )
kmol
3
5- 9
= 4043
kg C3 H 6 h
5.24 (cont’d)
Monthly revenue:
( 4043 c.
kg h )( 24 h day )( 30 days month )( $0.60 kg ) = $1,747,000 month
Mass flow rate at Noxious plant after diversion: 400 m3 hr
273 K 2.8 atm
1 kmol
44.09 kg
303 K
22.4 m3
kmol
1 atm
= 1986 kg hr
Propane diverted = ( 4043 − 1986 ) kg h = 2057 kg h 5.25 a. PHe = y He ⋅ P = 0.35 ⋅ (2.00 atm) = 0.70 atm
PCH 4 = y CH 4 ⋅ P = 0.20 ⋅ (2.00 atm) = 0.40 atm PN 2 = y N 2 ⋅ P = 0.45 ⋅ (2.00 atm) = 0.90 atm b. Assume 1.00 mole gas 4.004 g 0.35 mol He = 140 . g He mol
U| || 16.05 g I F 0.20 mol CH G H mol JK = 3.21 g CH V|17.22 g ⇒ mass fraction CH F 28.02 g IJ = 12.61 g N || 0.45 mol N G H mol K W FG H
IJ K
4
2
4
4
=
3.21 g . = 0186 17.22 g
2
g of gas = 17.2 g / mol mol P MW 2.00 atm 17.2 kg / kmol m n MW = = = = = 115 . kg / m3 m3 ⋅atm V V RT 0.08206 kmol⋅K 363.2 K
c.
MW =
d.
ρ gas
d i d i b e
gb
jb
g g
5.26 a. It is safer to release a mixture that is too lean to ignite. If a mixture that is rich is released in the atmosphere, it can diffuse in the air and the C3H8 mole fraction can drop below the UFL, thereby producing a fire hazard. b.
fuel-air mixture n 1 ( mol / s) y C3H 8 = 0.0403 mol C 3 H 8 / mol n C3H 8 = 150 mol C 3 H 8 / s
n 3 ( mol / s) 0.0205 mol C 3 H 8 / mol
diluting air n 2 ( mol / s) n 1 =
150 mol C 3 H 8 mol = 3722 mol / s s 0.0403 mol C 3 H 8
Propane balance: 150 = 0.0205 ⋅ n 3 ⇒ n 3 = 7317 mol / s
5- 10
5.26 (cont’d)
Total mole balance: n 1 + n 2 = n 3 ⇒ n 2 = 7317 − 3722 = 3595 mol air / s c.
b g
n 2 = 13 . n 2
min
= 4674 mol / s
U| |V || W
3 = 4674 mol / s 8.314 m ⋅ Pa 398.2 K = 118 m3 / s V 2 mol ⋅ K 131,000 Pa V m3 diluting air 2 = 1.41 V m3 fuel gas 3722 mol 8.314 m3 ⋅ Pa 298.2 K 1 3 = 83.9 m / s V1 = s mol ⋅ K 110000 Pa
y2 =
150 mol / s 150 mol / s = × 100% = 18% . n 1 + n 2 3722 mol / s + 4674 mol / s
b
g
d. The incoming propane mixture could be higher than 4.03%. If n 2 = n 2 min , fluctuations in the air flow rate would lead to temporary explosive
b g
conditions. 5.27
Basis: (12 breaths min )( 500 mL air inhaled breath ) = 6000 mL inhaled min 24 o C, 1 atm 6000 mL / min
lungs
n in (mol / min) 0.206 O 2 0.774 N 2 0.020 H 2 O
a.
n in =
6000 mL
blood
1L
273K
37 o C, 1 atm n out (mol / min) 0.151 O 2 0.037 CO 2 0.750 N 2 0.062 H 2 O
1 mol
b g = 0.246 mol min
3
min
10 mL 297K 22.4 L STP
b
gb
g
N 2 balance: 0.774 0.246 = 0.750n out ⇒ n out = 0.254 mol exhaled min O 2 transferred to blood:
b0.246gb0.206g − b0.254gb0.151g bmol O
2
g
min 32.0 g mol
= 0.394 g O 2 min CO 2 transferred from blood:
b0.254gb0.037g bmol CO
2
g
min 44.01 g mol
= 0.414 g CO 2 min H 2 O transferred from blood:
b0.254gb0.062g − b0.246gb0.020g bmol H O ming 18.02 g mol 2
= 0.195 g H 2 O min
5- 11
5.27 (cont’d)
PVin n RT = in in PVout n out RTout
FG IJ FG T IJ = FG 0.254 mol min IJ FG 310K IJ = 1078 H K H T K H 0.246 mol min K H 297K K . mL exhaled ml inhaled . g H O lost ming − b0.394 g O gained ming = 0.215 g min b0.414 g CO lost ming + b0195 ⇒
b.
Vout n = out Vin n in
out in
2
2
2
STACK
5.28
Ta (K) Ts (K) M s (g/mol) M a (g/mol) Ps (Pa) Pc (Pa)
2L (M)
L(m)
PM RT
Ideal gas: ρ =
b g
b g
LM N
OP Q
D = ρgL
b.
M s = 0.18 44.1 + 0.02 32.0 + 0.80 28.0 = 310 . g mol , Ts = 655K ,
combust.
− ρgL
Pa M a PM P gL M a M a gL − a s gL = a − RTa RTs R Ta Ts
a.
stack
=
b gb g b gb g b gb g
Pa = 755 mm Hg M a = 29.0 g mol , Ta = 294 K , L = 53 m D=
755 mm Hg
1 atm
53.0 m 9.807 m
760 mm Hg
×
s
2
kmol - K 0.08206 m3 − atm
LM 29.0 kg kmol − 31.0 kg kmol OP × FG 1 N IJ = 323 N 1033 cm H O 655K N 294K Q H 1 kg ⋅ m / s K m 1.013 × 10 N m 2
2
2
5
2
= 3.3 cm H 2 O
b.
b g
P MW MWCCl2O = 98.91 g / mol ρ CCl2O 98.91 = = 3.41 29.0 RT ρ air Phosgene, which is 3.41 times more dense than air, will displace air near the ground. 2 π D in L π 2 Vtube = = 0.635 cm - 2 0.0559 cm 15.0 cm = 3.22 cm3 4 4
5.29 a. ρ =
=======>
b g
b
m CCl O = Vtube ⋅ ρ CCl O = 2
c.
n CCl 2 O(l) =
2
3.22 cm3
gb
g
1L 1 atm 98.91 g / mol = 0.0131 g 3 L⋅atm 296.2 K 10 cm 0.08206 mol⋅K 3
3.22 cm3 137 . × 1000 . g mol = 0.0446 mol CCl 2 O 3 cm 98.91 g
5- 12
5.29 (cont’d)
PV 1 atm 2200 ft 3 28.317 L mol ⋅ K = = 2563 mol air 3 RT 296.2K .08206 L ⋅ atm ft
n air = n CCl 2 O n air
=
0.0446 = 17.4 × 10 −6 = 17.4 ppm 2563
The level of phosgene in the room exceeded the safe level by a factor of more than 100. Even if the phosgene were below the safe level, there would be an unsafe level near the floor since phosgene is denser than air, and the concentration would be much higher in the vicinity of the leak. d. Pete’s biggest mistake was working with a highly toxic substance with no supervision or guidance from an experienced safety officer. He also should have been working under a hood and should have worn a gas mask. 5.30 CH 4 + 2O 2 → CO 2 + 2 H 2 O 7 C 2 H 6 + O 2 → 2CO 2 + 3H 2 O 2 C 3 H 8 + 5O 2 → 3CO 2 + 4 H 2 O 1450 m 3 / h @ 15o C, 150 kPa n 1 (kmol / h)
0.86 CH 4 , 0.08 C 2 H 6 , 0.06 C 3 H 8 n 2 (kmol air / h) 8% excess, 0.21 O 2 , 0.79 N 2
n 1 =
1450 m3
273.2K
b101.3 + 150gkPa
1 kmol
h
288.2K
101.3 kPa
22.4 m3 STP
b g = 152 kmol h
Theoretical O 2 :
LM F MN GH
IJ K
FG H
IJ K
. 349.6 kmol O 2 = 108 Air flow: V air h
1 kmol Air
FG H
152 kmol 2 kmol O 2 3.5 kmol O 2 5 kmol O 2 + 0.08 + 0.06 0.86 h kmol CH 4 kmol C 2 H 6 kmol C 3 H 8
b
g
0.21 kmol O 2
5- 13
IJ OP = 349.6 kmol h O K PQ
b g = 4.0 × 10
22.4 m3 STP kmol
4
b g
2
m3 STP h
5.31 Calibration formulas
bT = 25.0; R = 14g , bT = 35.0, R = 27g ⇒ Tb° Cg = 0.77R + 14.2 dP = 0; R = 0i , dP = 20.0, R = 6i ⇒ P bkPag = 3.33R dV = 0; R = 0i , dV = 2.0 × 10 , R = 10i ⇒ V dm hi = 200R dV = 0; R = 0h , dV = 1.0 × 10 , R = 25i ⇒ V dm hi = 4000R T
g
T
p
g
T
r
gauge
p
3
F
p
F
A
A
A
3
F
F
5
F
3
A
A
A
(m 3 / h), T, P V F g
CH 4 + 2O 2 → CO 2 + 2 H 2 O 7 C 2 H 6 + O 2 → 2CO 2 + 3H 2 O 2 C 3 H 8 + 5O 2 → 3CO 2 + 4 H 2 O 13 C 4 H 10 + O 2 → 4CO 2 + 5H 2 O 2
n F (kmol / h) x A (mol CH 4 / mol) x B (mol C 2 H 6 / mol) x C (mol C 3 H 8 / mol) x D (mol n - C 4 H 10 / mol) x E (mol i - C 4 H 10 / mol) ( m 3 / h) (STP) V A n A (kmol / h) 0.21 mol O 2 / mol 0.79 mol N 2 / mol
n F =
d
m3 h V F
i
273.2K
dP + 101.3ikPa
1 kmol
g
bT + 273.2gK 101.3 kPa d P + 101.3i kmol 0.12031V FG IJ = T + 273 b g H h K F
b g
22.4 m3 STP
g
Theoretical O 2 :
dn i
o 2 Th
b
c
gh
= n F 2x A + 3.5x B + 5x C + 6.5 x D + x E kmol O 2 req. h
Air feed: n A =
dn i
o 2 Th
kmol O 2 req.
1 kmol air
h
0.21 kmol O 2
FG P IJ dn i H 100K bkmol air hg d22.4 m bSTPg kmoli = 22.4n = 4.762 1 +
= n V A a
b1 + P
x
g
100 kmol feed
1 kmol req.
x
o 2 Th
3
A
b g
m3 STP h
RT T(C) Rp Pg(kPa) Rf xa xb xc xd xe PX(%) nF nO2, th nA Vf (m3/h) Va (m3/h) Ra 25.0 7.25 0.81 0.08 0.05 0.04 0.02 1450 22506.2 5.63 23.1 32.0 7.5 15 72.2 183.47 1004.74 64.3 5.8 0.58 0.31 0.06 0.05 0.00 1160 29697.8 7.42 7.5 20.0 19.3 23 78.9 226.4 1325.8 52.6 2.45 0.00 0.00 0.65 0.25 0.10 490 22022.3 5.51 46.5 50.0 15.8 33 28.1 155.2 983.1 10.0 1200 30439.2 7.6 21 30.4 3 6 0.02 0.4 0.35 0.1 0.13 15 53.0 248.1 1358.9 13.3 1400 29283.4 7.3 23 31.9 4 7 0.45 0.12 0.23 0.16 0.04 15 63.3 238.7 1307.3 16.7 1800 32721.2 8.2 25 33.5 5 9 0.5 0.3 0.1 0.04 0.06 15 83.4 266.7 1460.8 20.0 10 0.5 0.3 0.1 0.04 0.06 2000 37196.7 9.3 27 35.0 6 15 94.8 303.2 1660.6
5- 14
5.32 NO + 21 O 2 ⇔ NO 2 1 mol 0.20 mol NO / mol 0.80 mol air / mol 0.21 O 2 0.79 N 2 P0 = 380 kPa
R| S| T
a.
n1 (mol NO) n 2 (mol O 2 ) n 3 (mol N 2 ) n 4 (mol NO 2 ) Pf (kPa)
U| V| W
Basis: 1.0 mol feed 90% NO conversion: n1 = 0.10(0.20) = 0.020 mol NO ⇒ NO reacted = 0.18 mol 018 . mol NO 0.5 mol O 2 = 0.0780 mol O 2 mol NO N 2 balance: n 3 = 0.80(0.79) = 0.632 mol N 2
O 2 balance: n 2 = 0.80(0.21) −
n4 =
y NO
018 . mol NO 1 mol NO 2 = 018 . mol NO 2 ⇒ n f = n1 + n 2 + n 3 + n 4 = 0.91 mol 1 mol NO 0.020 mol NO mol NO = = 0.022 0.91 mol mol
y O 2 = 0.086
mol O 2 mol N 2 mol NO 2 y N 2 = 0.695 y NO 2 = 0198 . mol mol mol
FG H
IJ K
Pf V n f RT n 0.91 mol = ⇒ Pf = P0 f = 380 kPa = 346 kPa P0 V n 0 RT n0 1 mol b.
n f = n0
Pf 360 kPa = (1 mol) = 0.95 mol 380 kPa P0
n i = n i0 + υ iξ
E
n1 (mol NO) = 0.20 − ξ n 2 ( mol O 2 ) = (0.21)(0.80) − 0.5ξ n 3 (mol N 2 ) = (0.79)(0.80) n 4 ( mol NO 2 ) = ξ n f = 1 − 0.5ξ = 0.95 ⇒ ξ = 010 . mol O 2 , n 3 = 0.632 mol N 2 , ⇒ n1 = 010 . mol NO, n 2 = 0118 . n 4 = 010 . mol NO 2 ⇒ y NO = 0105 . , y O 2 = 0124 . , y N 2 = 0.665, y NO 2 = 0105 .
NO conversion = P (atm) =
Kp =
b0.20 - n g × 100% = 50% 1
0.20
360 kPa = 355 . atm 101.3 kPa atm (y NO 2 P)
( y NO P)( y O 2 P)
0.5
=
(y NO 2 ) 0.5
( y NO )( y O 2 ) P
0.5
5- 15
=
b
0.105
(0105 . ) 0.124
g b3.55g 0.5
1
0.5
= 151 . atm 2
5.33
Liquid composition: 49.2 kg M 1 kmol = 0.437 kmol M 112.6 kg
100 kg liquid ⇒
0.481 kmol M / kmol
29.6 kg D 1 kmol = 0.201 kmol D 147.0 kg
⇒ 0.221 kmol D / kmol
21.2 kg B 1 kmol = 0.271 kmol B 78.12 kg
0.298 kmol B / kmol
0.909 kmol a.
Basis: 1 kmol C 6 H 6 fed V1 (m 3 ) @ 40o C, 120 kPa n1 (kmol) 0.920 HCl 0.080 Cl 2 1 kmol C 6 H 6 ( 78.12 kg)
n 0 (kmol Cl 2 )
n 2 (kmol) 0.298 C 6 H 6 0.481 C 6 H 5Cl 0.221 C 6 H 4 Cl 2
C 6 H 6 + Cl 2 → C 6 H 5Cl + HCl
C balance:
1 kmol C 6 H 6
C 6 H 5Cl + Cl 2 → C 6 H 5Cl 2 + HCl
6 kmol C = n 2 0.298 × 6 + 0.481 × 6 + 0.221 × 6 1 kmol C 6 H 6
⇒ n 2 = 100 . kmol H balance:
1 kmol C 6 H 6
b
g
6 kmol H = n1 0.920 (1) 1 kmol C 6 H 6
+ n 2 0.298 × 6 + 0.481 × 5 + 0.221 × 4 ⇒ n1 = 100 . kmol V1 =
n1RT 100 . kmol 1013 . kPa 0.08206 m3 ⋅ atm 313.2 K = 217 = . m3 120 kPa 1 atm P kmol ⋅ K
⇒ b.
217 . m3 V1 = = 0.278 m3 / kg B . kg B m B 7812
2 4⋅V gas ( m3 / s) = u(m / s) ⋅ A(m2 ) = u ⋅ πd ⇒ d 2 = V gas 4 π ⋅u
d2 =
B0 ( kg B) 0.278 m3 4m s 1 min 104 cm2 B0 (cm2 ) = 5.90m kg B π (10) m 60 s min m2
b g
B0 ⇒ d(cm) = 2.43 ⋅ m
1 2
c. Decreased use of chlorinated products, especially solvents.
5- 16
5.34 Vb ( m 3 / min) @900D C, 604 mtorr 60% DCS conversion n 1 (mol DCS / min) n 2 (mol N 2 O / min) n b (mol / min) n 3 (mol N 2 / min) n 4 (mol HCl(g) / min)
3.74 SCMM
U| V| W
n a ( mol / min) 0.220 DCS 0.780 N 2 O
SiH 2 Cl 2(g) + 2 N 2 O (g) → SiO 2(s) + 2 N 2(g) + 2 HCl (g) a.
n a =
3.74 m3 (STP) 103 mol =167 mol / min min 22.4 m3 (STP)
mol DCS I gFGH 0.220 mol JK b167 mol / ming = 14.7 mol DCS / min mol DCS DCS reacted: b0.60gb0.220gb167g = 22.04 mol DCS reacted / min min
b
60% conversion: n 1 = 1- 0.60
b g molminN O
N 2 O balance: n 2 = 0.780 167 − N 2 balance: n 3 =
2
22.04 mol DCS 2 mol N 2 O = 8618 . mol N 2 O / min mol DCS min
22.04 mol DCS 2 mol N 2 = 44.08 mol N 2 / min min mol DCS
HCl balance: n 4 =
22.04 mol DCS 2 mol HCl = 44.08 mol HCl / min mol DCS min
n B = n 1 + n 2 + n 3 + n 4 = 189 mol / min = ⇒V B b.
n B RT 189 mol 62.36 L ⋅ torr 0.001 m3 1173 K = = 2.29 × 104 m3 / min P min mol ⋅ K L 0.604 torr
n 1 14.7 mol DCS/min P= ⋅ 604 mtorr=47.0 mtorr n B 189 mol/min n 86.2 mol N 2 O/min = x N 2O ⋅ P= 2 P= ⋅ 604 mtorr=275.5 mtorr n B 189 mol/min
p DCS = x DCS ⋅ P=
p N2O
r=3.16 × 10-8 ⋅ p DCS ⋅ p N 2O 0.65 = 3.16 × 10-8 ( 47.0
)( 275.5 )0.65 = 5.7 × 10−5
mol SiO 2 m2 ⋅ s
MW 5.7 × 10−5 mol SiO 2 60 s 120 min 60.09 g/mol 1010 A c. h(A)=r ⋅t⋅ = 2 6 3 ρSiO2 min m ⋅s 2.25 × 10 g/m 1 m (Table B.1)
=1.1 × 10 A 5
The films will be thicker closer to the entrance where the lower conversion yields higher pDCS and p N 2 O values, which in turn yields a higher deposition rate.
5-17
5.35
Basis: 100 kmol dry product gas n1 (kmol C x H y ) m1 (kg C x H y ) kmol dry gasU R|100 |V 0.105 CO S|0.053 O |W T0.842 N
(m 3 ) V 2
n 2 (kmol air) 0.21 O 2 0.79 N 2
2
2
2
30 o C, 98 kPa
a.
n 3 (kmol H 2 O)
N 2 balance: 0.79n 2 = 0.842(100) ⇒ n 2 = 106.6 kmol air
b
g
b
g b
g
O balance: 2 0.21n 2 = 100 2 0105 . + 2 0.053 + n 3 ⇒ n 3 = 1317 . kmol H 2 O
C balance:
d
n1 kmol C x H y
i x bkmol Cg = 100b0105 . g ⇒ n x = 10.5 dkmol C H i
b1g
1
x
y
b2 g
n 3 =13.17
H balance: n1y = 2n 3 ====> n1y = 26.34
b g b g yx = 2610.34.5 = 2.51 mol H / mol C fed: 0.21b106.6 kmol air g = 22.4 kmol in excess = 5.3 kmol ⇒ Theoretical O = b22.4 - 5.3g kmol = 17.1 kmol
Divide 2 by 1 ⇒
O2
O2
2
% excess = b.
V2 = m1 =
5.3 kmol O 2 × 100% = 31% excess air 17.1 kmol O 2
106.6 kmol N 2 22.4 m3 (STP) 1013 . kPa 303 K = 2740 m3 kmol 98 kPa 273 K
b
g
b
g
n1x kmol C 12.0 kg n1y kmol H 101 . kg n1x=10.5 + =====> m1 = 152.6 kg kmol n1y=26.34 kmol
V2 2740 m3 air m3 air = 18.0 = m1 152.6 kg fuel kg fuel 5.36 3N 2 H 4 → 6xH 2 + (1 + 2 x)N 2 + (4 − 4 x)NH 3 a. 0 ≤ x ≤ 1 b.
n N2H4 =
50 L 0.82 kg 1 kmol = 128 . kmol L 32.06 kg
LM 6x kmol H + b1 + 2xg kmol N + b4 − 4xg kmol NH OP 3 kmol N H N 3 kmol N H 3 kmol N H Q 128 . = . x + 2.13 kmol b6x + 1 + 2x + 4 − 4xg = 1707 3 n product = 128 . kmol N 2 H 4
2
2
2
4
5-18
2
4
3
2
4
5.36 (cont’d) nproduct 2.13 2.30 2.47 2.64 2.81 2.98 3.15 3.32 3.50 3.67 3.84
Vp (L) 15447.92 16685.93 17923.94 19161.95 20399.96 21637.97 22875.98 24113.99 25352.00 26590.01 27828.02
Volume of Product Gas 30000.00
25000.00
20000.00 V (L)
x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
15000.00
10000.00
5000.00
0.00 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
c.
Hydrazine is a good propellant because as it decomposes generates a large number of moles and hence a large volume of gas.
5.37 A (g A / h) m
c
h
C A (g A / m 3 )
m3 / h V air
a. (i) Cap left off container of liquid A and it evaporates into room, (ii) valve leak in cylinder with A in it, (iii) pill of liquid A which evaporates into room, (iv) waste containing A poured into sink, A used as cleaning solvent.
A b. m
c. d.
yA
FG kg A IJ H hK
FG kg A IJ H hK C e j⋅ V mol A = = mol air M e j ⋅ n A =m
in
d h V air
out
gA A m3 gA A mol
yA = 50 ×10−6
=V air
air
F m I C FG kg A IJ GH h JK H m K 3
A
3
===================> m PV CA = A ; nair = k⋅Vair RT
yA =
A RT m k ⋅ Vair M A P
A = 90 . g/ h m
m ⋅Pa 8.314 mol 9.0 g / h 293 K m RT ⋅K = A = = 83 m3 / h 3 −6 kyA MA P 0.5 50 × 10 101.3 × 10 Pa 104.14 g / mol 3
min
d
i
Concentration of styrene could be higher in some areas due to incomplete mixing (high concentrations of A near source); 9.0 g/h may be an underestimate; some individuals might be sensitive to concentrations < PEL. e. Increase in the room temperature could increase the volatility of A and hence the rate of : At higher T, need evaporation from the tank. T in the numerator of expression for V air a greater air volume throughput for y to be < PEL.
5-19
C3 H 6 + H 2 ⇔ C3 H 8
5.38 Basis: 2 mol feed gas
U| V| W
n p (mol C 3 H 8 ) (1 - n p )(mol C 3 H 6 ) n 2 = n p + 2(1 − n p ) = 2 − n p (1 - n p )(mol H 2 )
1 mol C 3 H 6 1 mol H 2 25D C, 32 atm
235D C, P2
a. At completion, n p = 1 mol , n 2 = 2 − 1 = 1 mol
1 mol 508K 32.0 atm P2 V n 2 RT2 n T = ⇒ P2 = 2 2 P1 = = 27.3 atm 2 mol 298K P1V n1RT1 n1 T1 b. P2 = 35.1 atm
n2 =
35.1 atm 298K 2 mol P2 T1 n1 = = 1.29 mol 32.0 atm 508K P1 T2
1.29 = 2 − n p ⇒ n p = 0.71 mol C 3 H 8 produced
b
g
⇒ 1- 0.71 = 0.29 mol C 3 H 6 unreacted ⇒ 71% conversion of propylene c. n2 1.009 1.028 1.083 1.101 1.156 1.174 1.211 1.229 1.248 1.266 1.285 1.358 1.431 1.468
C3H8 prod. 0.99075 0.9724 0.91735 0.899 0.84395 0.8256 0.7889 0.77055 0.7522 0.73385 0.7155 0.6421 0.5687 0.532
%conv. 99.075 97.24 91.735 89.9 84.395 82.56 78.89 77.055 75.22 73.385 71.55 64.21 56.87 53.2
Pressure vs Fraction Conversion 120
100
80 % conversion
P2 (atm) 27.5 28.0 29.5 30.0 31.5 32.0 33.0 33.5 34.0 34.5 35.0 37.0 39.0 40.0
%conv.
60
40
20
0 25.0
27.0
29.0
31.0
33.0 Pressure (atm)
5-20
35.0
37.0
39.0
41.0
Convert fuel composition to molar basis
5.39
Basis: 100 g ⇒
b b
UV W
g g
97.2 mol % CH 4 95 g CH 4 1 mol 16.04 g = 5.92 mol CH 4 ⇒ 2.8 mol % C 2 H 6 5 g C 2 H 6 1 mol 30.07 g = 017 . mol C 2 H 6
500 m3 / h
n 2 (kmol CO 2 / h) n 3 (kmol H 2 O / h)
n 1 (mol / h) 0.972 CH 4
n 4 (kmol O 2 / h) n 5 (kmol N 2 / h)
0.028 C 2 H 6 D 40 C, 1.1 bar
(SCMH) V air 25% excess air
n 1 =
11 . bar 500 m3 P1V 1 = RT1 313K h
kmol ⋅ K . kmol h = 211 0.08314 m3 ⋅ bar 7 C 2 H 6 + O 2 → 2CO 2 + 3H 2 O 2
CH 4 + 2O 2 → CO 2 + 2 H 2 O Theoretical O 2 =
LM N
21.1 kmol 0.972 kmol CH 4 kmol h +
Air Feed:
0.028 kmol C 2 H 6 kmol
b
125 . 431 . kmol O 2 h
g
2 kmol O 2 1 kmol CH 4
OP = 431. kmol O h Q 22.4 m bSTPg = 5700 SCMH
3.5 kmol O 2 1 kmol C 2 H 6
1 kmol Air 0.21 kmol O 2
2
3
1 kmol
5.40 Basis: 1 m3 gas fed @ 205° C, 1.1 bars Ac = acetone 1 m 3 @205D C, 1.1 bar
n 3 (kmol), 10D C, 40 bar
condenser
n1 (kmol) y1 (kmol Ac / kmol) (1 - y1 )(kmol air / kmol) p AC = 0.100 bar
y 3 (kmol Ac / kmol) (1- y 3 )(kmol air / kmol) p AC = 0.379 bar
n 2 (kmol Ac(l))
a.
n1 = y1 =
1.00 m3
273K
110 . bars
1 kmol
b g = 0.0277 kmol
478K 10132 . bars 22.4 m3 STP
0.100 bar 0.379 bar = 0.0909 kmol Ac kmol , y 3 = = 9.47 × 10 −3 kmol Ac kmol 1.1 bars 40.0 bars
b
gb
g
Air balance: 0.0277 0.910 = (1 − 9.47 × 10 −3 )n 3 ⇒ n 3 = 0.0254 kmol
Mole balance: 0.0277 = 0.0254 + n 2 ⇒ n 2 = 0.0023 kmol Ac condensed Acetone condensed =
0.0023 kmol Ac 58.08 kg Ac = 0133 . kg acetone condensed 1 kmol Ac
5-21
5.40 (cont’d)
Product gas volume =
b.
b g
0.0254 kmol 22.4 m3 STP
. 283K 10132 bars 273K
40.0 bars
= 0.0149 m3
20.0 m3 effluent 0.0277 kmol feed 0.0909 kmol Ac 58.08 kg Ac = 196 kg Ac h h 0.0149 m3 effluent kmol feed kmol Ac
5.41 Basis: 1.00 × 10 6 gal. wastewater day. Neglect evaporation of water. Effluent gas: 68D F, 21.3 psia(assume)
1.00 × 10 6 gal / day
n1 (lb-moles H2O/day) 0.03n1 (lb-moles NH3 /day)
n2 (lb-moles air/day) n3 (lb-moles NH3 /day)
300 × 10 6 ft 3 air / day
Effluent liquid
n1 (lb-moles H2O/day) n4 (lb-moles NH3 /day)
D
68 F, 21.3 psia n2 (lb-moles air/day)
a.
Density of wastewater: Assume ρ = 62.4 lb m ft 3 1 ft 3 ⎡ n1 lb-moles H 2 O 18.02 lb m 0.03n1 lb m NH3 17.03 lb m ⎤ + × ⎢ day 1 lb-mole day 1 lb-mole ⎥⎦⎥ 62.4 lb m ⎣⎢ = 1.00 × 106
7.4805 gal 1 ft 3
gal day
⇒ n1 = 4.50 × 105 lb-moles H 2 O fed day , 0.03n1 = 1.35 × 104 lb-moles NH 3 fed day
n2 =
300 × 106 ft 3 day
492D R
21.3 psi
1 lb-mole
527.7 R 14.7 psi 359 ft 3 ( STP ) D
= 1.13 × 106 lb-moles air day
93% stripping: n3 = 0.93 × 13500 lb-moles NH 3 fed day = 12555 lb-moles NH3 day Volumetric flow rate of effluent gas 6 3 PVout nout RT nout 300 × 10 ft = ⇒ Vout = Vin = nin RT nin PVin day
(1.13 × 10
6
)
+ 12555 lb-moles day
1.13 × 106 lb-moles day
= 303 × 106 ft 3 day Partial pressure of NH 3 = y NH 3 P =
12555 lb - moles NH 3 day
d1.129 × 10
= 0.234 psi
5-22
6
i
+ 12555 lb - moles day
× 213 . psi
5.42 Basis: 2 liters fed / min
Cl ads.=
2.0 L soln 1130 g 0.12 g NaOH 1 mol 0.23 NaOH ads. 1 mol Cl 2 mol = 0.013 60 min L g soln 40.0 g mol NaOH 2 mol NaOH min
2 L / min @ 23D C, 510 mm H 2 O n 1 (mol / min) y (mol Cl 2 / mol) (1 - y)(mol air / mol)
n 2 (mol air / mol)
0.013 mol Cl 2 / min
b g = b10.33 + 0.510g m H O = 10.84 m H O
Assume Patm = 10.33 m H 2 O ⇒ Pabs n 1 =
2L
273K 10.84 m H 2 O
2
1 mol
b g = 0.0864 mol min
min 296K 10.33 m H 2 O 22.4 L STP
Cl balance: 0.0864y = 0.013 ⇒ y = 0150 .
5.43
2
in
mol Cl 2 ,∴ specification is wrong mol
(L / min) @ 65o C, 1 atm V 3
125 L / min @ 25o C, 105 kPa
n 1 ( mol / min) y1 (mol H 2 O / mol) 1- y1 ( mol dry gas / mol) 0.235 mol C 2 H 6 / mol DG 0.765 mol C 2 H 4 / mol DG
R|b g S| T
n C2 H 6 (mol C 2 H 6 / min) n C2 H 4 (mol C 2 H 4 / min) n air (mol air / min) n H 2 O (mol H 2 O / min)
U| V| W
355 L / min air @ 75o C, 115 kPa n 2 (mol / min) y 2 ( mol H 2 O / mol) (1- y 2 )( mol dry air / mol)
a.
Hygrometer Calibration ln y = bR + ln a b=
b
ln y1 y 2 R 2 − R1
dy = ae i bR
g = lnd0.2 10 i = 0.08942 −4
90 − 5
bg
ln a = ln y1 − bR1 = ln 10 −4 − 0.08942 5 ⇒ a = 6.395 × 10 −5 ⇒ y = 6.395 × 10 −5 e 0.08942R
b. n 1 =
n 2 =
125 L 273K 105 kPa 1 mol = 5.315 mol min wet gas min 298K 101 kPa 22.4 L STP
b g
355 L 273K 115 kPa 1 mol = 14.156 mol min wet air min 348K 101 kPa 22.4 L STP
b g
R 1 = 86.0 → y1 = 0.140 , R 2 = 12.8 → y 2 = 2.00 × 10 −4 mol H 2 O mol
5-23
5.43 (cont’d)
b
C 2 H 6 balance: n C2 H 6 = 5.315 mol min
C H I gFGH b1 − 0.140g molmolDG IJK FGH 0.235 mol J mol DG K 2
6
= 1.07 mol C 2 H 6 min
b gb gb g Dry air balance: n = b14.156gd1 − 2.00 × 10 i = 14.15 mol DA min Water balance: n = b5.315gb0.140g + b14.156gd1.00 × 10 i = 0.746 mol H O min n = b1.07 + 3.50 + 14.15g mol min = 18.72 mol min , n = b18.72 + 0.746g = 19.47 mol min = 19.47 mol min 22.4 L bSTPg 338K = 540 liters min V C 2 H 4 balance: n C2 H 4 = 5.315 0.860 0.765 = 3.50 mol C 2 H 4 min −4
air
−4
H 2O
2
dry product gas total
3
mol
Dry basis composition: c.
p H 2O = y H 2Ol ⋅ P =
273K
FG 1.07 IJ × 100% = 5.7% C H , 18.7% C H , 75% dry air H 18.72 K 2
6
2
4
0.746 mol H 2 O × 1 atm = 0.03832 atm 19.47 mol
y H 2 O = 0.03832 ⇒ R =
FG H
IJ K
1 0.03832 ln = 71.5 0.08942 6.395 × 10−5
5.44 CaCO 3 → CaO + CO 2
n CO 2 =
1350 m3 273K 1 kmol = 12.92 kmol CO 2 h h 1273K 22.4 m3 STP
b g
12.92 kmol CO 2 1 kmol CaCO 3 100.09 kg CaCO 3 h 1 kmol CO 2 1 kmol CaCO 3 1362 kg limestone 0.17 kg clay = 279 kg clay h h 0.83 kg limestone Weight % Fe 2 O 3
b g
kg Fe 2 O 3 kg clay
279 0.07 × 100% = 18% Fe 2 O 3 . 1362 + 279 − 12.92 44.1
kg limestone kg clay
b g
kg CO 2 evolved
5-24
1 kg limestone = 1362 kg limestone h 0.95 kg CaCO 3
5.45
Basis:
R|864.7 g C b1 mol 12.01 gg = 72.0 mol C |116.5 g H b1 mol 1.01 gg = 115.3 mol H 1 kg Oil ⇒ S ||13.5 g S b1 mol 32.06 gg = 0.4211 mol S T5.3 g I 5.3 g I n 1 (mol CO 2 ) n 2 (mol CO) n 3 (mol H 2 O) n 4 (mol SO 2 ) n 5 (mol O 2 ) n 6 (mol N 2 )
72.0 mol C 115.3 mol H 0.4211 mol S 5.3 g I
C + O 2 → CO 2 1 C + O 2 → CO 2 S + O 2 → SO 2 1 2H + O 2 → H 2 O 2
n a (mol), 0.21 O 2 , 0.79 N 2 15% excess air 175D C, 180 mm Hg (gauge)
a.
Theoretical O 2 : 72.0 mol C 1 mol O
2+ 1 mol C
+
115.3 mol H
0.25 mol O
2
1 mol H
0.4211 mol S 1 mol O
2 = 101.2 mol O 2 1 mol S
(
1.15 101.2 mol O Air Fed:
2
)
1 mol Air 0.21 mol O
554 mol Air 1 kg oil
22.4 liter ( STP ) mol
1 m3 3
10 liter
= 554 mol Air = n
a
2
448K
760 mm Hg
273K
940 mm Hg
= 16.5 m 3 air kg oil
b. S balance: n 4 = 0.4211 mol SO 2 H balance: 115.3 = 2n 3 ⇒ n 3 = 57.6 mol H 2 O
b g
C balance: 0.95 72.0 = n1 ⇒ n1 = 68.4 mol CO 2 ⇒ 0.05(72.0) = n 2 = 3.6 mol CO N 2 balance: 0.79 ( 554 ) =n 6 ⇒ n 6 = 437.7 mol N 2
O balance: 0.21( 554 ) 2=57.6+3.6+2(68.4)+2 ( 0.4211) +2n 5 ⇒ n 5 = 16.9 mol O 2 Total moles ( excluding inerts ) dry basis:
wet basis:
3.6 mol CO 527 mol 3.6 mol CO 585 mol
wet: 585 mols
= 6.8 × 10 −3
mol CO mol
,
× 106 = 6150 ppm CO ,
5-25
dry: 527 mols
0.4211 mol SO 2 527 mol
= 7.2 × 10−4
0.4211 mol SO 2 585 mol
mol SO 2 mol
× 106 = 720 ppm SO 2
bg
5.46 Basis: 50.4 liters C 5 H 12 l min
50.4 L C5 H12 (l ) / min n 1 (kmol C5 H12 / min)
heater
n 1 , n 2
Combustion chamber
(L / min) 15% excess air, V air n 2 kmol air 0.21 O 2 0.79 N 2 336 K, 208.6 kPa (gauge)
n 3 (kmol C 5 H 12 / min) n 4 (kmol O 2 / min) n 5 (kmol N 2 / min) n 6 (kmol CO 2 / min) n 7 (kmol H 2 O / min)
Condenser
C5 H 12 + 802 → 5CO 2 + 6H 2 O a.
50.4 L 0.630 kg 1 kmol = 0.440 kmol min min L 72.15 kg
n 3 =
3175 . kg 1 kmol = 0.044 kmol / min min 72.15 kg
n 2 =
n 4 (kmol O 2 / min) n 5 (kmol N 2 / min) n 6 (kmol CO 2 / min)
(L/min) V liq m=3.175 kg C5 O12 / min n 3 (kmol C5 O12 / min) n 7 (kmol H 2 O(l ) / min)
n 1 =
frac. convert =
(L/min), 275 K, 1 atm V gas
0.440 - 0.044 kmol × 100 = 90% C5 H 12 converted 0.440
0.440 kmol C5 H12 1.15 ( 8 kmol O 2 ) 1 mol air = 19.28 kmol air min min kmol C5 H12 0.21 mol O 2
b g
= 19.28 kmol 22.4 L STP V air min mol
336K
101 kPa 1000 mol = 173000 L min 273K 309.6 kPa kmol
n 4 = [(0.21)(19.28) − (0.90)(0.440)(8)]
kmol O 2 = 0.882 kmol O 2 / min min
n 5 =
19.28 kmol air 0.79 kmol N 2 = 15.23 kmol N 2 / min min kmol air
n 6 =
0.90(0.440 kmol C5 H 12 ) 5 kmol CO 2 . kmol CO 2 / min = 198 min kmol C5 H 12
= 0.882+15.23+1.98 kmol 22.4 L(STP) 275 K 1000 mol = 4.08 × 105 L/min V gas min mol 273 K kmol
5-26
5.46 (cont’d)
n 7 =
0.9(0.440 kmol C5 H 12 ) 6 kmol H 2 O = 2.38 kmol H 2 O(l ) / min min kmol C5 H 12
Condensate: 0.044 kmol 72.15 kg L V = 5.04 L min C5 H 12 = min kmol 0.630 kg L 2.38 kmol 18.02 kg V = 42.89 L min H 2O = min kmol 1 kg
Assume volume additivity (liquids are immiscible) = 5.04 + 42.89 = 47.9 L min V liq
b. C5 H 12 (l )
bg
C 5 H 12 l
bg
H 2O l
bg
H 2O l
5.47 n air (kmol / min), 25D C, 1 atm 0.21 O 2 0.79 N 2 n 0 (kmol / min)
n 1 (kmol H 2S / min) Furnace
H 2 S + 23 O 2 → SO 2 + H 2 O
0.20 kmol H 2S / mol 0.80 kmol CO 2 / mol
Reactor
n 2 (kmol H 2 S / min)
2H 2 S + SO 2 → 3S(g) + 2 H 2 O
10.0 m 3 / min @ 380D C, 205 kPa n exit (kmol / min) n 3 (kmol N 2 / min) n 4 (kmol H 2 O / min) n 5 (kmol CO 2 / min) n 6 (kmol S / min)
n exit =
PV 205 kPa 10.0 m3 / min = = 0.377 kmol / min m3 ⋅kPa RT 8.314 kmol 653 K ⋅K
b g
n 1 = 0.20 n 0 / 3 = 0.0667n 0 ; n 2 = 2 n 1 = 0.133n 0
5-27
5.47 (cont’d)
Air feed to furnace: n air =
0.0667 n 0 (kmol H 2 S fed) 15 . kmol O 2 1 kmol air (min) 1 kmol H 2 S 0.21 kmol O 2
= 0.4764 n 0 kmol air / min Overall N 2 balance: n 3 =
Overall S balance: n 6 =
0.4764 n 0 (kmol air) 0.79 kmol N 2 = 0.3764 n 0 ( kmol N 2 / min) (min) min
0.200n 0 (kmol H 2S) 1 kmol S = 0.200n 0 (kmol S / min) (min) 1 kmol H 2S
Overall CO 2 balance: n 5 = 0.800n 0 (kmol CO 2 / min) Overall H balance:
0.200n 0 (kmol H 2 S) 2 kmol H n kmol H 2 O 2 kmol H = 4 (min) 1 kmol H 2 S min 1 kmol H 2 O
⇒ n 4 = 0.200n 0 (kmol H 2 O / min)
b
g
n exit = n 0 0.376 + 0.200 + 0.200 + 0.800 = 0.377 kmol / min ⇒ n 0 = 0.24 kmol / min n air = 0.4764(0.24 kmol air / min) = 0114 . kmol air / min
5.48 Basis: 100 kg ore fed ⇒ 82.0 kg FeS2 (s), 18.0 kg I.
b
gb
g
n FeS2 fed = 82.0 kg FeS2 1 kmol / 120.0 kg = 0.6833 kmol FeS2
100 kg ore 06833 . kmol FeS2 18 kg I
Vout m3 (STP) n 2 (kmol SO 2 ) n 3 (kmol SO 3 ) n 4 (kmol O 2 ) n5 (kmol N 2 )
40% excess air n 1 (kmol) 0.21 O 2 0.79 N 2 V1 m 3 (STP)
m6 (kg FeS2 ) m7 (kg Fe 2 O 3 ) 18 kg I
2 FeS2(s) + 112 O 2(g) → Fe 2 O 3(s) + 4SO 2(g) 2 FeS2(s) + 152 O 2(g) → Fe 2 O 3(s) + 4SO 3(g)
a.
n1 =
0.6833 kmol FeS2 7.5 kmol O 2 1 kmol air req'd 140 . kmol air fed = 17.08 kmol air 2 kmol FeS2 0.21 kmol O 2 kmol air req'd
b
gb
g
V1 = 17.08 kmol 22.4 SCM / kmol = 382 SCM / 100 kg ore n2 =
(0.85)(0.40)0.6833 kmol FeS2 4 kmol SO 2 = 0.4646 kmol SO 2 2 kmol FeS2
5-28
5.48 (cont’d)
n3 =
(0.85)( 0.60)0.6833 kmol FeS2 4 kmol SO 2 = 0.6970 kmol SO 3 2 kmol FeS2
n = ( 0.21 × 17.08 ) kmol O fed − 4 2
.4646 kmol SO
2
5.5 kmol O 4 kmol SO
2
2
.697 kmol SO 7.5 kmol O 3 2 = 1.641 kmol O 2 4 kmol SO 3 n 5 = 0.79 × 17.08 kmol N 2 = 13.49 kmol N 2 −
b
g
Vout = ⎡⎣( 0.4646+0.6970+1.641+13.49 ) kmol⎤⎦ [ 22.4 SCM (STP)/kmol] = 365 SCM/100 kg ore fed ySO2 =
0.4646 kmol SO 2 × 100% = 2.9%; ySO3 = 4.3%; y O2 = 10.1%; y N 2 = 82.8% 16.285 kmol
b.
e j
Product gas, T o C
Converter
0.4646 kmol SO 2 0.697 kmol SO 3 1633 . kmol O 2 13.49 kmol N 2
n SO 2 (kmol) n SO 3 (kmol) n O 2 (kmol) n N 2 (kmol)
Let ξ (kmol) = extent of reaction n SO2 = 0.4646 − ξ n SO3 = 0.697 + ξ n O2 = 1.641 − 12 ξ n N2 = 13.49 n=16.29- 12 ξ K p (T)=
⎫ 0.4646 − ξ 0.697 + ξ ⎪ ySO2 = , ySO3 = 1 16.29- 2 ξ 16.29- 12 ξ ⎪ ⎬⇒ 1.641 − 12 ξ 13.49 ⎪ y O2 = , y N2 = 1 ⎪ 16.29- 2 ξ 16.29- 12 ξ ⎭
P ⋅ ySO2 (P ⋅ yO2 )
(0.697 + ξ ) (16.29 − 12 ξ ) 2 1
P ⋅ ySO3 1 2
⇒
(0.4646 − ξ ) (1.641 − ξ ) 1 2
1 2
-1
⋅ P 2 = K p (T)
P=1 atm, T=600o C, K p = 9.53 atm 2 ⇒ ξ = 0.1707 kmol -1
⇒ n SO2 = 0.2939 kmol ⇒ fSO2 =
( 0.4646 − 0.2939 ) kmol SO 2 0.4646 kmol SO 2 fed
reacted
= 0.367
P=1 atm, T=400o C, K p = 397 atm 2 ⇒ ξ = 0.4548 kmol -1
⇒ n SO2 = 0.0098 kmol ⇒ fSO2 = 0.979 The gases are initially heated in order to get the reaction going at a reasonable rate. Once the reaction approaches equilibrium the gases are cooled to produce a higher equilibrium conversion of SO2.
5-29
5.48 (cont’d) c.
SO 3 leaving converter: (0.6970 + 0.4687) kmol = 1.156 kmol 1.156 kmol SO 3 1 kmol H 2 SO 4 98 kg H 2 SO 4 = 113.3 kg H 2 SO 4 min 1 kmol SO 3 kmol
⇒
Sulfur in ore:
0.683 kmol FeS 2 2 kmol S 32.1 kg S = 438 . kg S kmol FeS2 kmol
113.3 kg H 2SO 4 kg H 2SO 4 = 2.59 43.8 kg S kg S
100% conv.of S:
0.683 kmol FeS2 2 kmol S 1 kmol H 2SO 4 98 kg = 133.9 kg H 2SO 4 kmol FeS2 1 kmol S kmol
133.9 kg H 2SO 4 kg H 2SO 4 = 3.06 43.8 kg S kg S
⇒
The sulfur is not completely converted to H2SO4 because of (i) incomplete oxidation of FeS2 in the roasting furnace, (ii) incomplete conversion of SO2 to SO3 in the converter. 5.49 N 2 O 4 ⇔ 2 NO 2
dP
gauge
i
+ 1.00 V
n0 =
b.
n1 = mol NO 2 , n 2 = mol N 2 O 4
RT0
=
b2.00 atmgb2.00 Lg b473Kgb0.08206 L ⋅ atm mol - Kg = 0.103 mol NO
a.
FG n IJ P ⇒ K = n P n bn + n g Hn +n K Ideal gas equation of state ⇒ PV = b n + n gRT ⇒ n + n = PV / RT b1g p NO 2 = y NO 2 P =
FG n IJ P , p Hn +n K
2
1
1
N 2O4
=
2 1
2
p
2
1
1
2
2
2
1
1
2
2
Stoichiometric equation ⇒ each mole of N 2 O 4 present at equilibrium represents a loss of two moles of NO 2 from that initially present ⇒ n1 + 2n 2 = 0.103
b3g ,
Solve (1) and (2) ⇒ n1 = 2(PV / RT) − 0.103
b2 g
b4 g
n 2 = 0.103 − (PV / RT)
Substitute (3) and (4) in the expression for K p , and replace P with Pgauge + 1
b2n − 0.103g dP n b0.103 − n g 2
t
gauge
t
T(K) 350 335 315 300
Pgauge(atm) 0.272 0.111 -0.097 -0.224
i
+ 1 where n t =
dP
gauge
nt Kp(atm) 0.088568 5.46915 0.080821 2.131425 0.069861 0.525954 0.063037 0.164006
(1/T) ln(Kp) 0.002857 1.699123 0.002985 0.756791 0.003175 -0.64254 0.003333 -1.80785
i
+1 V
RT
t
⇒ nt =
V=2 L
2
i
T
y = -7367x + 22.747 R2 = 1
1 0 -1 -2 0.0028
0.003
0.0032 1/T
5-30
d
24.37 Pg + 1
Variation of Kp with Temperature
ln Kp
Kp =
0.0034
5.49 (cont’d) c.
A semilog plot of K p vs.
1 is a straight line. Fitting the line to the exponential law T
yields ln K p = −
FG H
IJ K
a = 7.567 × 109 atm 7367 −7367 + 22.747 ⇒ K p = 7.567 × 109 exp ⇒ T T b = 7367K 10.00 atm
5.50 n 1 (kmol A / h) n 2 (kmol H 2 / h)
5.00 kmol S / h n 4 (kmol A / h) n 5 (kmol H 2 / h)
3n 3 (kmol A / h) n 3 (kmol H 2 / h)
5.00 kmol S / h
n 4 (kmol A / h) n 5 (kmol H 2 / h) Vrcy (SCMH)
A + H2
S
Overall A balance: n1 =
5.00 kmol S 1 kmol A react = 5.00 kmol A / h h 1 kmol S form
Overall H 2 balance: n2 =
5.00 kmol S 1 kmol H 2 react = 5.00 kmol H 2 / h h 1 kmol S form
Extent of reaction equations: n i = n i0 + ν iξ A + H2 ↔ S n 4 = 3n 3 − ξ H 2 : n 5 = n 3 − ξ S: 5.00 = ξ =====> n 4 = 3n 3 − 5.00 n 5 = n 3 − 5.00 n S = 5.00 A:
n tot = 4 n 3
U| |V ⇒ p | − 5.00|W
A
= yA P =
n 4 P= n tot
p H2 = y H2 P = pS = yS P = Kp =
b
g
3n 3 - 5.00 10.0 4 n 3 − 5.00
n 5 n - 5.00 P= 3 10.0 n tot 4 n 3 − 5.00
5.00 10.0 4 n 3 − 5.00
5.00 4n 3 − 5.00 pS . = = 0100 ⇒ n 3 = 11.94 kmol H 2 / h p A p H 2 10.0 3n 3 − 5.00 n 3 − 5.00
b
gb
g
n 4 = 3(11.94) - 5.00 = 30.82 kmol A / h n 5 = 1194 . − 5.00 = 6.94 kmol H 2 / h
b
g
d
i
= 30.82 + 6.94 kmol / h 22.4 m 3 (STP) / kmol = 846 SCMH V rcy
5-31
5.51 n 4 (kmol CO / h) n 5 (kmol H 2 / h)
Reactor n 1 (kmol CO / h) n 2 (kmol H 2 / h)
Separator
100 kmol CO / h
n 4 (kmol CO/h) n 5 (kmol H 2 /h) n 6 (kmol CH 3 OH/h) T, P
n 3 (kmol H 2 / h) T (K), P (kPa)
H xs (% H 2 excess)
n 6 (kmol CH 3OH/h)
a. Balances on reactor ⇒ 4 equations in n3 , n4 , n5 , and n6 . 5.0% XS H2: n3 = C balance:
100 kmol CO fed 2 kmol H 2 reqd 1.05 kmol H 2 fed kmol H 2 = 210 h 1 kmol CO fed 1 kmol H 2 reqd h
100 kmol CO 1 kmol C = n (1) + n6 (1) ⇒ 100 = n4 + n6 h 1 kmol CO 4
(1)
H balance: 210(2) = n5 (2) + n6 (4) ⇒ 210 = n5 + 2n6
(2)
(O balance: 100 = n4 + n6 ⇒ identical to C balance ⇒ not independent) (1) ⇒ n4 = 100 − n6 , (2) ⇒ n5 = 210 − 2n6 ntot = n4 + n5 + n6 = (100 − n6 ) + (210 − 2n6 ) + n6 = 310 − 2n6
9143.6 ⎛ ⎞ ⎜ 21.225+ 500 K − 7.492ln ( 500K ) ⎟ K p ( T=500K ) = 1.390 × 10 exp ⎜ = 9.11 × 10−7 kPa -2 ⎟ ⎜ +4.076 × 10-3 ( 500K ) -1.161 × 10-8 ( 500K )2 ⎟ ⎝ ⎠ n 6 (1) − ( 3) 310 − 2 n 6 yMP yM 2 Kp = ⇒ K P = ====> p 2 2 2 100 − n 6 210 − 2 n 6 y CO P y H 2 P y CO y H 2 310 − 2 n 6 310 − 2 n 2 −4
d
i
d i
b
K p P 2 = 9.11 × 10 −7 kPa -2 5000 kPa
g
2
= 22.775 =
b
b b
b
gb gb
n 6 310 − 2 n 6
g
g
6
2
b100 − n gb210 − 2n g 6
Solving for n6 ⇒ n6 = 75.7 kmol CH 3OH/h , n4 = 100 − n6 = 24.3 kmol CO/h n5 = 210 − 2n6 = 58.6 kmol H 2 / h Overall C balance: n1 (1) = n6 (1) ⇒ n1 = 75.7 kmol CO/h Overall H balance: n2 (2) = n6 (4) ⇒ n 2 = 151 kmol H 2 /h 3
22.4 m (STP) Vrec = ( n4 + n5 ) = 1860 SCMH kmol
5-32
6
2
g g
5.51 (cont’d) b. P(kPa) 1000 5000 10000 5000 5000 5000 5000 5000 5000
`
T(K) Hxs(%) 500 5 500 5 500 5 400 5 500 5 600 5 500 0 500 5 500 10
n6(kmol ntot M/h) (kmol/h) KpcE8 25.55 258.90 9.1E-01 9.00 292.00 2.3E-01 86.72 136.56 9.1E+01 98.93 112.15 7.8E+03 75.68 158.64 2.3E+01 14.58 280.84 4.1E-01 73.35 153.30 2.3E+01 75.68 158.64 2.3E+01 77.77 164.45 2.3E+01
Kp(T)E8 9.1E+01 9.1E+01 9.1E+01 3.1E+04 9.1E+01 1.6E+00 9.1E+01 9.1E+01 9.1E+01
KpP^2 0.91 22.78 91.11 7849.77 22.78 0.41 22.78 22.78 22.78
KpP^2- n1(kmol CO/h) KpcP^2 1.3E-05 25.55 2.3E+01 9.00 4.9E-03 86.72 3.2E-08 98.93 3.4E-03 75.68 -2.9E-04 14.58 9.8E-03 73.35 3.4E-03 75.68 -3.1E-03 77.77
n3(kmol n4(kmol n5(kmol H2/h) H2/h) CO/h) 210 74.45 158.90 210 91.00 192.00 210 13.28 36.56 210 1.07 12.15 210 24.32 58.64 210 85.42 180.84 200 26.65 53.30 210 24.32 58.64 220 22.23 64.45 n2(kmol H2/h) 51.10 18.00 173.44 197.85 151.36 29.16 146.70 151.36 155.55
Vrec (SCMH) 5227 6339 1116 296 1858 5964 1791 1858 1942
c. Increase yield by raising pressure, lowering temperature, increasing Hxs. Increasing the pressure raises costs because more compression is needed. d. If the temperature is too low, a low reaction rate may keep the reaction from reaching equilibrium in a reasonable time period. e. Assumed that reaction reached equilibrium, ideal gas behavior, complete condensation of methanol, not steady-state measurement errors.
5.52
CO 2 ⇔ CO + 21 O 2
1.0 mol CO 2 1.0 mol O 2 1.0 mol N 2 T = 3000 K, P = 5.0 atm
K1 =
dp
1/ 2
i = 0.3272 atm
1/2
p CO 2 1 1 O 2 + 2 N 2 ⇔ NO 2 p NO K2 = = 0.1222 1/ 2 p N 2 p O2
d
1 A ⇔ B+ C 2 1 1 C+ D= E 2 2
CO p O 2
i
A − CO 2 , B − CO , C − O 2 , D − N 2 , E − NO
ξ 1 - extent of rxn 1
n A0 = n C0 = n D0 = 1 , n B0 = n E0 = 0
ξ 2 - extent of rxn 2
5-33
5.52 (cont’d) nA = 1 − ξ1 nB = ξ1 1 1 nC = 1 + ξ1 − ξ 2 2 2 1 nD = 1− ξ2 2 nE = ξ2 1 6 + ξ1 n tot = 3 + ξ 1 = 2 2
U| || y = n n = 2 b1 − ξ g b 6 + ξ g | yy == b22ξ + ξb 6 −+ ξξ gg b 6 + ξ g V| || yy == 2b 2ξ − ξb 6 g+ bξ6 g+ ξ g || W A B
A
1
tot
1
C
1
D
2
2
1
b
g b5g K = p b gb g ⇒ 0.3272b1 − ξ gb6 + ξ g = 2.236ξ b2 + ξ − ξ g y y1 2 1+ 1 −1 2ξ 2 + ξ 1 − ξ 2 = B C pb 2 g = 1 yA 2 1 − ξ1 6 + ξ1
p CO p1O22
1
CO 2
12
12
12
12
1
K2 =
d
p NO p O2 p N2
b
i
12
=
2 + ξ1 − ξ 2 ⇒ 01222 .
= 0.3272
12
1
yE 12 12 yC y D
pi = yiP
1
1
2
E
1
1
1
p1−1 2 −1 2 =
g b2 − ξ g 12
12
2
1
b2 + ξ
(1)
2
2ξ 2 1 − ξ2
g b2 − ξ g 12
12
= 01222 .
2
= 2ξ 2
(2)
. , Solve (1) and (2) simultaneously with E-Z Solve ⇒ ξ 1 = 0.20167, ξ 2 = 012081
b
yA = 2 1 − ξ1
g b6 + ξ g = 0.2574 mol CO 1
2
mol
y B = 0.0650 mol CO mol
y D = 0.3030 mol N 2 mol y E = 0.0390 mol NO mol
y C = 0.3355 mol O 2 mol n 4 (kmol / h)
5.53 a.
0.04 O 2 0.96 N 2
PX=C8 H10 , TPA=C8 H 6 O 4 , S=Solvent (m 3 / h) @105o C, 5.5 atm V 3 n 3O (kmol O 2 / h) n 3N (kmol N 2 / h) n 3W (kmol H 2 O(v) / h) (m3 / h) at 25o C, 6.0 atm V 2 n 2 (kmol / h) 0.21 O 2 0.79 N 2
condenser
n 3W (kmol H 2 O(v) / h) (m 3 / h) V 3W
reactor
n 1 (kmol PX / h) ( n 1 + n 3p ) kmol PX / h s (kg S / h) m 3 kg S / kg PX n 3p (kmol PX / h) s (kg S / h) m
5-34
n 3p ( kmolPX / h) 100 kmol TPA / h s (kg S / h) m
separator
100 mol TPA / s
5.53 (cont’d) b. Overall C balance: n 1
c.
FG kmol PX IJ 8 kmol C = 100 kmol TPA H h K kmol PX h
O 2 consumed =
8 kmol C ⇒ n 1 = 100 kmol PX / h kmol TPA
100 kmol TPA 3 kmol O 2 = 300 kmol O2 /h h 1 kmol TPA
kmol O 2 ⎫ +0.04n 4 ⎪ n 2 = 1694 kmol air/h h ⎬ ⇒ n 4 = 1394 kmol/h ⎪ Overall N 2 balance: 0.79n 2 = 0.96n 4 ⎭ 100 kmol TPA 2 kmol H 2 O Overall H 2 O balance: n 3W = = 200 kmol H 2 O / h h 1 kmol TPA
Overall O 2 balance: 0.21n 2 = 300
3 = n 2 RT = 1694 kmol 0.08206 m ⋅ atm 298 K = 6.90 × 103 m3 air/h V 2 P h kmol ⋅ K 6.0 atm 3 = ( n 3W + n 4 ) RT = ( 200+1394 ) kmol 0.08206 m ⋅ atm 378 K = 8990 m3 /h V 3 P h kmol ⋅ K 5.5 atm 3 = 200 kmol H 2 O (l) 18.0 kg 1 m = 3.60 m3 H O(l) / h leave condenser V 3W 2 h kmol 1000 kg
d
i
n 1 =100
d. 90% single pass conversion ⇒ n 3p = 0.10 n 1 + n 3p ====> n 3p = 111 . kmol PX / h
m recycle = m S + m 3 P =
(100 + 11.1) kg PX 106 kg PX 3 kg S 11.1 kmol PX 106 kg PX + h 1 kmol PX kg PX h 1 kmol PX
= 3.65 × 104 kg/h e. O2 is used to react with the PX. N2 does not react with anything but enters with O2 in the air. The catalyst is used to accelerate the reaction and the solvent is used to disperse the PX. f.
The stream can be allowed to settle and separate into water and PX layers, which may then be separated.
5.54
n 1 (kmol CO / h), n 3 (kmol H 2 / h), 0.10n 2 (kmol H 2 / h)
Separator n 6 (kmol CO / h) n 7 (kmol H 2 / h) n 8 (kmol CO 2 / h) 2 kmol N 2 / h
0.90 n 2 2 kmol N 2 / h
n 1 , n 2 , n 3 2 kmol N 2 / h 0.300 kmol CO / kmol 0.630 kmol H 2 / kmol 0.020 kmol N 2 / kmol 0.050 kmol CO 2 / kmol
Reactor
n 1 (kmol CO / h) n 2 (kmol H 2 / h) n 3 (kmol CO 2 / h) n 4 (kmol M / h) n 5 (kmol H 2 O / h) 2 kmol N 2 / h
5-35
Separator
n 4 (kmol M / h) n 5 (kmol H 2 O / h)
5.54 (cont’d)
CO + 2H 2 ⇔ CH 3OH(M) CO 2 + 3H 2 ⇔ CH 3OH + H 2 O a.
Let ξ 1 ( kmol / h) = extent of rxn 1, ξ 2 ( kmol / h) = extent of rxn 2 n 1 = 30 - ξ 1 n 2 = 63 - 2ξ 1 − 3ξ 2
CO: H2:
U| M: n = ξ + ξ || P⋅y P⋅y H O: n = ξ V| ⇒ dK i = P ⋅ y P ⋅Py ⋅ y , dK i = bP ⋅ y gd P ⋅ y i d id i d i N : n = 2 || n = 100 - 2ξ − 2ξ W n − 2ξ − 2ξ g = 84.65 (1) dK i ⋅ P = n nF n I = bbξ30+−ξξ gbgb100 63 − 2ξ − 3ξ g G J n H n K FG n IJ FG n IJ (2) . dK i ⋅ P = FH nn IKFH nn IK = ξ bbξ5 −+ξξ gbgb63100− 2−ξ2ξ− 3−ξ2ξg g = 1259 GH n JK GH n JK CO 2 : n 3 = 5 - ξ 2
2 2
4
1
5
2
2
M
p 1
CO
N2
tot
CO 2
H2
3
H2
4
2
1
tot
p 1
2
1
2
2
2
2
H 2O
2
1
2
M
p 2
2
1
2
tot
tot
1
4
5
tot
tot
3
2
tot
tot
p 2
1
2
2
2
2
1
1
2 2
3
2
1
2
Solve (1) and (2) for ξ 1 , ξ 2 ⇒ ξ 1 = 25.27 kmol / h ξ 2 = 0.0157 kmol / h n 1 = 30.0 − 25.27 = 4.73 kmol CO / h
⇒
9.98% CO
n 2 = 63.0 − 2(25.27) − 3(0.0157) = 12.4 kmol H 2 / h
26.2% H 2
n 3 = 5.0 − 0.0157 = 4.98 kmol CO 2 / h
10.5% CO 2
n 4 = 25.27 + 0.0157 = 25.3 kmol M / h
⇒
n 5 = 0.0157 = 0.0157 kmol H 2 O / h n total = 49.4 kmol / h
53.4% M 0.03% H 2 O
UV W
n 6 = 25.4 kmol CO / h C balance: n 4 = 25.3 kmol / h ⇒ n 8 = 0.02 kmol CO 2 / h O balance: n 6 + 2 n 8 = n 4 + n 5 = 25.44 mol / s H balance: 2n 7 = 2(0.9 n 2 ) + 4 n 4 + 2 n 5 = 123.7 ⇒ n 7 = 618 . mol H 2 / s b. (n 4 ) process = 237 kmol M / h ⇒ Scale Factor =
237 kmol M / h 25.3 kmol / h
5-36
5.54 (cont’d)
237 kmol / h I F 22.4 m (STP) I gFGH 25.3 J kmol JK = 18,700 SCMH kmol / h K GH F 237 kmol / h IJ = 444 kmol / h Reactor effluent flow rate: b 49.4 kmol / hgG H 25.3 kmol / sK F kmol IJ FG 22.4 m (STP) IJ = 9946 SCMH ⇒ V G 444 H h K H kmol K
b
3
Process feed: 25.4 + 618 . + 0.02 + 2.0
3
std
c.
. kPa 9950 m 3 (STP) 473.2 K 1013 ⇒ Vactual = = 354 m 3 / h h 273.2 K 4925 kPa 3 = V = 354 m / h 1000 L 1 kmol = 0.8 L / mol V n 444 kmol / h m 3 1000 mol (5.2-36)
< 20 L / mol ====> ideal gas approximation is poor V from n using the ideal gas equation of state is likely Most obviously, the calculation of V to lead to error. In addition, the reaction equilibrium expressions are probably strictly valid only for ideal gases, so that every calculated quantity is likely to be in error.
5.55 a.
RTc PV B Bo + ωB1 = 1+ ⇒ B = RT Pc V From Table B.1 for ethane: Tc = 305.4 K, Pc = 48.2 atm From Table 5.3 -1 ω = 0.098 0.422 0.422 = −0.333 Bo = 0.083 − 1.6 = 0.083 − 1.6 Tr 308.2 K 305.4 K . . 0172 0172 − 4.2 = 0139 − = −0.0270 B1 = 0139 . . 4 .2 Tr 308.2 K 305.4K RTc 0.08206 L ⋅ atm 305.4 K B(T) = −0.333 − 0.098 0.0270 B o + ωB 1 = Pc mol ⋅ K 48.2 atm L / mol = −01745 .
b
b
g
e
j
e
j
g
FG H
b
IJ K
g
2 PV mol ⋅ K 2 - B = 10.0 atm −V V − V + 0.1745 = 0 RT 308.2K 0.08206 L ⋅ atm = ⇒V
b
gb
1 ± 1 - 4 0.395 mol / L 01745 . L / mol
b
2 0.395 mol / L
g
g = 2.343 L / mol, 0.188 L / mol
Videal = RT / P = 0.08206 × 308.2 / 10.0 = 2.53, so the second solution is
b. c.
likely to be a mathematical artifact. 10.0 atm 2.343 L / mol PV = = 0.926 z= L⋅atm RT 0.08206 mol 308.2K ⋅K = m
V 1000 L mol 30.0 g 1 kg = 12.8 kg / h MW = h 2.343 L mol 1000 g V
5-37
5.56
RTc PV B = 1+ ⇒ B = Bo + ωB1 RT Pc V
b
g
b g T bC H g = 369.9 K, P = 42.0 atm From Table 5.3 -1 ω bCH OH g = 0.559, ω bC H g = 0.152 From Table B.1 Tc CH 3OH = 513.2 K, Pc = 78.50 atm c
3
8
c
3
3
8
0.422 0.422 = 0.083 − = −0.619 1.6 1.6 Tr 373.2K 513.2K 0.422 0.422 = −0.333 Bo (C 3 H 8 ) = 0.083 − 1.6 = 0.083 − 1.6 Tr 373.2K 369.9K 0172 0172 . . . − 4.2 = 0139 . − = −0.516 B1 ( CH 3OH) = 0139 4 .2 Tr 373.2K 513.2K 0172 . 0172 . . − 4.2 = 0139 . − = −0.0270 B1 ( C 3 H 8 ) = 0139 4 .2 Tr 373.2K 369.9K RTc Bo + ωB1 B(CH 3OH) = Pc Bo (CH 3OH) = 0.083 −
e
j
e
j
e
j
e
b
j
g
0.08206 L ⋅ atm 513.2K −0.619 − 0.559 0.516 = −0.4868 mol ⋅ K 78.5 atm RTc B(C 3 H 8 ) = Bo + ωB1 Pc
b
= B mix =
h
0.08206 L ⋅ atm 369.9 K −0.333 − 0152 . 0.0270 = −0.2436 mol ⋅ K 42.0 atm
b
c
i
b
g
L mol
g
∑∑y y B i
b
c
=
j
ij
d
⇒B ij = 0.5 B ii + B jj
j
g
h
L mol
i
B ij = 0.5 −0.4868 − 0.2436 L / mol = -0.3652 L / mol
b gb gb
g
g b gb gb
g b gb gb
g
B mix = 0.30 0.30 −0.4868 + 2 0.30 0.70 −0.3652 + 0.70 0.70 −0.2436 = −0.3166 L / mol
FG H
IJ K
2 PV mol ⋅ K 2 - B = 10.0 atm −V V − V + 0.3166 = 0 mix RT 373.2K 0.08206 L ⋅ atm = Solve for V:V
b
gb
1 ± 1- 4 0.326 mol / L 0.3166 L / mol
b
2 0.326 mol / L
g
g = 2.70 L / mol, 0.359 L / mol
RT 0.08206 L ⋅ atm 373.2 K V = = 3.06 L / mol ⇒ V ideal = virial = 2.70 L / mol P mol ⋅ K 10.0 atm = 2.70 L / mol = Vn V
15.0 kmol CH 3OH / h 1000 mol 1 m3 =135 m3 / h 0.30 kmol CH 3OH / kmol 1 kmol 1000 L
5-38
5.57 a.
van der Waals equation: P =
d
i
d
RT a2 − 2 -b V V
i
V - b ⇒ PV 3 − PV 2 b = RTV 2 − aV + ab Multiply both sides by V 2
b
g
3 + -Pb - RT V 2 + aV - ab = 0 PV
c 3 = P = 50.0 atm
b
g b
gb
g c
c 2 = -Pb - RT = −50.0 atm 0.0366 L / mol − 0.08206 . atm ⋅ L / mol c 1 = − a = 133 2
hb223 Kg = −201. L ⋅ atm / mol
2
ib
d
. atm ⋅ L2 / mol 2 0.0366 L / mol c 0 = − ab = - 133 = −0.0487
L⋅atm mol⋅K
g
atm ⋅ L3 mol 3
RT 0.08206 L ⋅ atm 223 K b. V = = 0.366 L / mol ideal = P mol ⋅ K 50.0 atm
c. T(K) 223 223 223 223 223
P(atm) 1.0 10.0 50.0 100.0 200.0
c3
c2
1.0 10.0 50.0 100.0 200.0
c1
-18.336 -18.6654 -20.1294 -21.9594 -25.6194
c0
1.33 1.33 1.33 1.33 1.33
-0.0487 -0.0487 -0.0487 -0.0487 -0.0487
V(ideal) V (L/mol) (L/mol) 18.2994 18.2633 1.8299 1.7939 0.3660 0.3313 0.1830 0.1532 0.0915 0.0835
f(V) 0.0000 0.0000 0.0008 -0.0007 0.0002
d. 1 eq. in 1 unknown - use Newton-Raphson.
=0 . gV-.0487 b1g ⇒ gdV i = 50.0V + b-20.1294gV + b133 3
2
∂g 2 − 40.259 V + 1.33 = 150V ∂V solve −g Eq. (A.2-14) ⇒ ad = − g ⇒ d = a
Eq. (A.2-13) ⇒ a =
(k+1) = V (k) + d Guess V (1) = V Then V ideal = 0.3660 L / mol .
1 2 3 4
(k) V 0.3660 0.33714 0.33137 0.33114
5-39
(k +1) V 0.33714 0.33137 0.33114 0.33114 converged
% error 0.2 2.0 10.5 19.4 9.6
b
d
5.58 C 3 H 8 : TC = 369.9 K
Specific Volume
PC = 42.0 atm 4.26 × 106 Pa
5.0 m3
44.09 kg
1 kmol
75 kg
1 kmol
103 mol
i
ω = 0152 .
= 2.93 × 10 −3 m3 mol
Calculate constants a=
0.42747
b=
0.08664
d8.314 m ⋅ Pa mol ⋅ Ki b369.9 Kg 2
3
4.26 × 10 Pa 6
2
= 0.949 m6 ⋅ Pa mol 2
d8.314 m ⋅ Pa mol ⋅ Ki b369.9 Kg = 6.25 × 10 3
4.26 × 106 Pa
b
g
b
0152 0152 m = 0.48508 + 155171 − 015613 . . . .
e
α = 1 + 0.717 1 − 298.2 369.9
j
2
g
2
−5
m3 mol
= 0.717
= 115 .
SRK Equation:
d8.314 m ⋅ Pa mol ⋅ Kib298.2 Kg − d2.93 × 10 − 6.25 × 10 i m mol 2.93 × 10
d
115 . 0.949 m6 ⋅ Pa mol 2
3
P=
−3
−5
3
−3
d
i
i
m3 mol 2.93 × 10 −3 + 6.25 × 10 −5 m3 mol
⇒ P = 7.40 × 106 Pa ⇒ 7.30 atm P=
Ideal:
ib
d
(8.35 − 7.30) atm × 100% = 14.4% 7.30 atm
Percent Error:
PC = 72.9 atm ω = 0.225
TC = 304.2 K
5.59 CO 2 :
g
3 RT 8.314 m ⋅ Pa mol ⋅ K 298.2 K = = 8.46 × 106 Pa ⇒ 8.35 atm 3 −3 2.93 × 10 m mol V
TC = 151.2 K PC = 48.0 atm ω = −0.004 = 35.0 L / 50.0 mol = 0.70 L mol P = 510 . atm , V
Ar:
Calculate constants (use R = 0.08206 L ⋅ atm mol ⋅ K ) L2 ⋅ atm L , m = 0.826 , b = 0.0297 , α = 1 + 0.826 1 − T 304.2 2 mol mol L2 ⋅ atm L a = 137 . , m = 0.479 , b = 0.0224 , α = 1 + 0.479 1 − T 151.2 2 mol mol
e e
CO 2 : a = 3.65 Ar:
bg
f T =
e
RT a − 1 + m 1 − T TC V−b V V+b
d
i
j
2
−P=0
Use E-Z Solve. Initial value (ideal gas): L L ⋅ atm Tideal = 510 . atm 0.70 0.08206 = 435.0 K mol mol ⋅ K
gFGH E - Z Solve ⇒ bT g b
max CO 2
IJ FG K H
= 455.4 K ,
bT g
max Ar
IJ K
= 431.2 K
5-40
j j
2
2
b
g
5.60 O 2 : TC = 154.4 K ; PC = 49.7 atm ; ω = 0.021 ; T = 208.2 K 65° C ; P = 8.3 atm ;
= 250 kg h ; R = 0.08206 L ⋅ atm mol ⋅ K m . L2 ⋅ atm mol 2 ; b = 0.0221 L mol ; m = 0.517 ; α = 0.840 SRK constants: a = 138
d i dVRT− bi − V dVaα+ bi − P = 0=====> V = 2.01 L / mol E-Z Solve
= f V
SRK equation:
kmol 103 mol 2.01 L = 250 kg ⇒V = 15,700 L h h 32.00 kg 1 kmol mol W
5.61
∑F
y
= PCO 2 ⋅ A - W = 0
e
where W = mg = 5500 kg 9.81
m s2
j = 53900 N
PCO 2 ⋅ A
a.
PCO 2 =
W A piston
=
53900 N π 4
. mg b015
2
b. SRK equation of state: P =
1 atm = 301 . atm 1.013 × 105 N / m2 RT αa − +b V-b V V
d i d
i
For CO 2 : Tc = 304.2, Pc = 72.9 atm , ω = 0.225 . a = 3.654 m6 ⋅ atm / kmol 2 , b = 0.02967 m3 / kmol, m = 0.8263, α ( 25o C) = 1016 . ge3.654 e0.08206 jb298.2 Kg − b1016 j 301 . atm = eV - 0.02967 j V dV + 0.02967i m3 ⋅atm kmol⋅K
m6 ⋅atm kmol 2
m6 kmol 2
m3 kmol
E-Z Solve
= 0.675 m 3 / kmol =====> V
b g Vbafter expansiong = 0.030 m
V before expansion = 0.030 m3
mCO 2 =
3
b
g b15. mg = 0.0565 m
+ π4 015 . m
2
3
44.01 kg V 0.0565 m3 = 3.68 kg MW = 3 0.675 m / kmol kmol V
mCO 2 (initially) =
PV 1 atm 0.030 m3 44.01 kg MW = = 0.0540 kg 3 m ⋅atm 298.2 K RT kmol 0.08206 kmol ⋅K
mCO 2 (added) = 3.68 - 0.0540 kg = 3.63 kg
5-41
5.61 (cont’d) c.
W = 53,900 N
V
h
add 3.63 kg CO 2
n o (kmol) Vo (m3 ) 1 atm, 25o C
ho
========> n (kmol) P (atm), 25o C
ho
d(m)
d(m)
Given T, Vo , h, find d V Initial: n o = o Po = 1 RT V πd 2 h 3.63 (kg) Final: V = Vo + , n = no + = o + 0.0825 4 44 (kg / kmol) RT
b
Vo +
g
πd 2 h
4 =V= V V n o + 0.0825 RT W RT RT 53,900 αa αa P= = − ⇒ 2 = − V +b V +b A piston V - b V πd / 4 V - b V
i d i in b1g ⇒ one equation in one unknown. Substitute expression for V
b1g
d
Solve for d .
5.62 a. Using ideal gas assumption:
Pg =
35.3 lb m O 2 1 lb - mole 10.73 ft 3 ⋅ psia 509.7 o R nRT − Patm = − 14.7 psia = 2400 psig V 32.0 lb m lb - mole ⋅ o R 2.5 ft 3
b. SRK Equation of state: P =
αa RT − -b V V +b V
d i d
i
3 ft 3 ˆ = 2.5 ft 32.0 lb m / lb-mole = 2.27 V ideal 35.3 lb m lb-mole
(Use as a first estimate when solving the SRK equation) For O 2 : Tc = 277.9 o R, Pc = 730.4 psi, ω = 0.021 a = 52038 .
ft 6 ⋅ psi ft 3 = 0 . 3537 , m = 0.518, α 50o F = 0.667 , b lb - mole lb - mole 2
d
i
e10.73 jd509.7 Ri − b0.667ge52038. b2400 + 14.7g psi = V - 0.3537 dV + 0.3537i V d i ft 3 ⋅psi
o
lb-mole⋅ R o
ft 6 ⋅psi
lb-mole 2
ft 3 lb-mole
= 2.139 ft 3 / lb - mole E - Z Solve ⇒ V
5-42
j
ft 6 lb-mole 2
5.62 (cont’d)
mO 2 =
32.0 lb m V 2.5 ft 3 MW = = 37.4 lb m 3 2.139 ft / lb - mole lb - mole V
Ideal gas gives a conservative estimate. It calls for charging less O2 than the tank can safely hold. c. 1. 2. 3. 4.
Pressure gauge is faulty The room temperature is higher than 50°F Crack or weakness in the tank Tank was not completely evacuated before charging and O2 reacted with something in the tank 5. Faulty scale used to measure O2 6. The tank was mislabeled and did not contain pure oxygen.
5.63 a.
SRK Equation of State: P =
αa RT − +b V-b V V
d i d
i
d id i i = PV dV - bid V + bi − RTV dV + bi + αad V - bi = 0 f dV i = PV − RTV + dαa - b P - bRTiV - αab = 0 f dV
V -b V +b : ⇒ multiply both sides of the equation by V
3
2
2
b. Problem 5.63-SRK Equation Spreadsheet Species Tc(K) Pc(atm) ω a b m
CO2 304.2 R=0.08206 m^3 atm/kmol K 72.9 0.225 3.653924 m^6 atm/kmol^2 0.029668 m^3/kmol 0.826312
f(V)=B14*E14^3-0.08206*A14*E14^2+($B$7*C14-$B$8^2*B14-$B$8*0.08206*A14)*E14-C14*$B$7*$B$8 T(K) 200 250 300 300 300
P(atm) 6.8 12.3 6.8 21.5 50.0
alpha 1.3370 1.1604 1.0115 1.0115 1.0115
V(ideal) 2.4135 1.6679 3.6203 1.1450 0.4924
V(SRK) 2.1125 1.4727 3.4972 1.0149 0.3392
f(V) 0.0003 0.0001 0.0001 0.0000 0.0001
c. E-Z Solve solves the equation f(V)=0 in one step. Answers identical to VSRK values in part b. d. REAL T, P, TC, PC, W, R, A, B, M, ALP, Y, VP, F, FP INTEGER I CHARACTER A20 GAS DATA R 10.08206/ READ (5, *) GAS WRITE (6, *) GAS 10 READ (5, *) TC, PC, W
5-43
5.63 (cont’d)
READ (5, *) T, P IF (T.LT.Q.) STOP R = 0.42747 *R*R/PC*TC*TC B = 0.08664 *R*TC/PC . − W∗015613 . M = 0.48508 + W = 155171
d
c b
b
g
g
hi
ALP = 1.+ M∗ 1 − T / TC ∗∗0.5 ∗∗2 . VP = R∗ T / P DO 20 I = 7, 15 V = VP F = R * T/(V – B) – ALP * A/V/(V + B) – P FP = ALP * A * (2. * V + B)/V/V/(V + B) ** 2 – R * T/(V – B) ** 2. VP = V – F/FP IF (ABS(VP – V)/VP.LT.0.0001) GOTO 30 20 CONTINUE WRITE (6, 2) 2 FORMAT ('DID NOT CONVERGE') STOP 30 WRITE (6, 3) T, P, VP 3 FORMAT (F6.1, 'K', 3X, F5.1, 'ATM', 3X, F5.2, 'LITER/MOL') GOTO 10 END $ DATA CARBON 304.2 200.0 250.0 300.0 –1
72.9 6.8 12.3 21.5 0.
DIOXIDE 0.225
RESULTS CARBON DIOXIDE 200.0 K 6.8 ATM 250.0 K 12.3 ATM 300.0 K 6.8 ATM 300.0 K 21.5 ATM 300.0 K 50.0 ATM 5.64 a.
b.
2.11 LITER/MOL 1.47 LITER/MOL 3.50 LITER/MOL 1.01 LITER/MOL 0.34 LITER/MOL
b
g
Tr = 40 + 273.2 126.2 = 2.48 N 2 : TC = 126.2 K ⇒ 10 atm 40 MPa PC = 33.5 atm Pr = . = 1178 33.5 atm 1.013 MPa
b
gb g
g
U| V| W
Tr = −200 + 273.2 5.26 + 8 = 5.52 He: TC = 5.26 K ⇒ PC = 2.26 atm Pr = 350 2.26 + 8 = 34.11
b
↑ Newton’s correction
5-44
Fig. 5.4-4
. ⇒ z = 12
U| V| W
Fig. 5.4-4
⇒ z = 1.6
5.65 a.
d
i
ρ kg / m3 = =
b.
m (kg) (MW)P = RT V (m3 ) 30 kg kmol 9.0 MPa 10 atm = 69.8 kg m3 3 m ⋅ atm 465 K 0.08206 kmol⋅K 1.013 MPa
UV P = 9.0 4.5 = 2.0 W
. Tr = 465 310 = 15
Fig. 5.4-3
⇒ z = 0.84
r
ρ=
(MW)P 69.8 kg m3 = = 831 . kg m3 zRT 0.84
5.66 Moles of CO 2 :
100 lb m CO 2 1 lb - mole CO 2 44.01 lb m CO 2
= 2.27 lb - moles
TC = 304.2 K ⎫ (1600 + 14.7 ) psi 1 atm = 1.507 ⎬ ⇒ Pr = P PC = 72.9 atm 14.7 psi PC = 72.9 atm ⎭ ˆ 10.0 ft 3 72.9 atm lb-mole ⋅ °R 1K ˆ = VPC = V = 0.80 r 3 RTC 2.27 lb-moles 304.2 K 0.7302 ft ⋅ atm 1.8 °R Fig. 5.4-3: Pr = 1507 . , Vr = 0.80 ⇒ z = 0.85
T=
10.0 ft 3 lb - mole⋅° R 1 atm PV 1614.7 psi = 779° R = 320 ° F = 3 znR 0.85 2.27 lb - moles 0.7302 ft ⋅ atm 14.7 psi
PC = 49.7 atm
Pr1 Tr2 Pr2
V2 = V1 V2 =
U|z = 1.00 (Fig. 5.4 - 2) V = 1 49.7 = 0.02 |W = 358 154.4 = 2.23 U| Vz = 1.61 bFig. 5.4 - 4g = 1000 49.7 = 20.12 |W
Tr1 = 298 154.4 = 1.93
5.67 O : T = 154.4 K 2 C
1
2
z 2 T2 P1 z1 T1 P2
127 m3 1.61 358 K 1 atm = 0.246 m3 h h 1.00 298 K 1000 atm
b
g
Tr = 27 + 273.2 154.4 = 1.94 Pr1 = 175 49.7 = 3.52 ⇒ z1 = 0.95
5.68 O 2 : TC = 154.4 K
PC = 49.7 atm
(Fig. 5.3-2)
Pr2 = 1.1 49.7 = 0.02 ⇒ z 2 = 1.00 n1 − n 2 =
FG H
IJ K
FG H
IJ K
10.0 L mol ⋅ K V P1 P2 175 atm 11 . atm − = − = 74.3 mol O 2 300.2 K 0.08206 L ⋅ atm 0.95 RT z1 z 2 1.00
5- 45
5.69 a.
= V = 50.0 mL 44.01 g = 4401 V . mL / mol n 5.00 g mol RT 82.06 mL ⋅ atm 1000 K P= = = 186 atm mol ⋅ K 440.1 mL / mol V
b. For CO 2 : Tc = 304.2 K, Pc = 72.9 atm T 1000 K Tr = = = 3.2873 Tc 304.2 K VP 4401 . mL 72.9 atm mol ⋅ K c = = 1.28 Vr ideal = RTc mol 304.2 K 82.06 mL ⋅ atm
Figure 5.4-3: Vr ideal = 1.28 and Tr = 3.29 ⇒ z=1.02 P= c.
zRT 1.02 82.06 mL ⋅ atm mol 1000 K = = 190 atm ˆ mol ⋅ K 440.1 mL V
a = 3.654 × 10 6 mL2 ⋅ atm / mol 2 , b = 29.67 mL / mol, m = 0.8263, α (1000 K ) = 0.1077
j = 198 atm c82.06 hb1000 Kg − b0.1077ge3.654 × 10 P= b440.1- 29.67g 440.1b440.1 + 29.67g 6 mL2 ⋅atm mol 2
mL⋅atm mol⋅K
mL2 mol 2
mL mol
5.70 a. The tank is being purged in case it is later filled with a gas that could ignite in the presence of O2. b. Enough N2 needs to be added to make x O 2 = 10 × 10 −6 . Since the O2 is so dilute at this
condition, the properties of the gas will be that of N2. Tc = 126.2 K, Pc = 335 . atm, Tr = 2.36
n initial = n1 =
1 atm 5000 L PV = = 204.3 mol L⋅atm RT 0.08206 mol⋅K 298.2 K
n O 2 = 204.3 mol air n O2 n2
FG 0.21 mol O IJ = 42.9 mol O H mol air K 2
2
= 10 × 10 −6 ⇒ n 2 = 4.29 × 10 −6 mol
5000 L . × 10 -3 L / mol = 116 4.29 × 10 6 mol . × 10 −3 L . atm mol ⋅ K 335 ideal = VPc = 116 . × 10 −3 = 38 V r RTc mol 0.08206 L ⋅ atm 126.2 K ⇒ not found on compressibility charts = V
Ideal gas: P =
RT 0.08206 L ⋅ atm 298.2 K = 2.1 × 10 4 atm = −3 ⋅ mol K . × 10 L / mol 116 V
The pressure required will be higher than 2.1 × 10 4 atm if z ≥ 1, which from Fig. 5.3 - 3 is very likely.
ib
d
g
n added = 4.29 × 106 − 204.3 ≅ 4.29 × 106 mol N 2 0.028 kg N 2 / mol = 120 . × 105 kg N 2
5- 46
5.70 (cont’d) c.
143 . kmol N 2
143 . kmol N 2
n initial = 0.204 kmol y O = 0.21 kmol O 2 / kmol 2
143 . kmol N 2 y1
143 . kmol N 2 y2
Fig 5.4-2
N 2 at 700 kPa gauge = 7.91 atm abs. ⇒ Pr = 0.236, Tr = 2.36 =======> z = 0.99
n2 =
P2 V 7.91 atm 5000 L . = = 1633 kmol L⋅atm zRT 0.99 0.08206 mol ⋅K 298.2 K
y1 =
0.21 0.204 y init n init = = 0.026 1634 . 1.634
y2 =
y 1 n init n init = y init 1634 . 1.634
b g FG H
IJ K
= 0.0033
FG y IJ FG n IJ ⇒ n = H y K = 4.8 ⇒ Need at least 5 stages H 1634 K . F n IJ lnG H 1.634 K . kmol N gb28.0 kg / kmolg = 200 kg N = 5b143 ln
n
y n = y init
2
n
init
init
init
Total N 2
2
2
d. Multiple cycles use less N2 and require lower operating pressures. The disadvantage is that it takes longer.
5.71
= MW a. m
b.
UV W
Tc = 369.9 K = 665.8o R ⇒ Tr = 0.85 Pc = 42.0 atm ⇒ Pr = 016 .
= 60.4 m
F GH
I JK
PV SPV SPV 44.09 lb m / lb - mol SPV = MW ⇒ Cost ($ / h) = mS = = 60.4 3 ft ⋅atm RT RT T T 0.7302 lb-mol ⋅o R Fig. 5.4-2
⇒ z = 0.91
m PV ideal = ideal = 110 . m zT z
⇒ Delivering 10% more than they are charging for (undercharging their customer)
5- 47
5.72 a.
For N 2 : Tc = 126.20 K = 227.16o R, Pc = 335 . atm
U| |V || W
609.7 o R = 2.68 227.16o R ⇒ z = 1.02 600 psia 1 atm = 1.2 Pr = . atm 14.7 psia 335
After heater: Tr =
n =
150 SCFM = 0.418 lb - mole / min 359 SCF / lb - mole
. 0.418 lb - mole 10.73 ft 3 ⋅ psia 609.7 o R = zRTn = 102 V = 4.65 ft 3 / min P min lb - mole ⋅o R 600 psia b. tank =
0.418 lb - mole 28 lb m / lb - mole 60 min 24 h 7 days 2 weeks min h day week 0.81 62.4 lb m / ft 3
b g
= 4668 ft 3 = 34,900 gal 5.73 a.
For CO: Tc = 133.0 K, Pc = 34.5 atm 300 K = 2.26 133.0 K 2514.7 psia 1 atm Pr1 = = 5.0 34.5 atm 14.7 psia
Initially: Tr1 =
U| |V || W
Fig. 5.4-3
⇒ z = 1.02
2514.7 psia 150 L 1 atm mol ⋅ K = 1022 mol 1.02 300 K 14.7 psia 0.08206 L ⋅ atm
n1 =
U| |V || W
300 K = 2.26 133.0 K 2258.7 psia 1 atm Pr1 = = 4.5 34.5 atm 14.7 psia
After 60h: Tr1 =
n2 = n leak b.
Fig. 5.4-3
⇒ z = 1.02
mol ⋅ K 2259.7 psia 150 L 1 atm = 918 mol 1.02 300 K 14.7 psia 0.08206 L ⋅ atm n − n2 = 173 = 1 . mol / h 60 h
n 2 = y 2 n air = y 2 t min =
1 atm 30.7 m3 1000 L PV 200 × 10 −6 mol CO = = 0.25 mol L⋅atm RT mol air 0.08206 mol m3 ⋅K 300 K
n2 0.25 mol = = 014 . h n leak 1.73 mol / h ⇒ t min would be greater because the room is not perfectly sealed
c. (i) CO may not be evenly dispersed in the room air; (ii) you could walk into a high concentration area; (iii) there may be residual CO left from another tank; (iv) the tank temperature could be higher than the room temperature, and the estimate of gas escaping could be low.
5- 48
5.74 CH 4 : Tc = 190.7 K , Pc = 458 . atm
C 2 H 6 : Tc = 305.4 K , Pc = 48.2 atm C 2 H 4 : Tc = 2831 . K , Pc = 50.5 atm
b gb g b gb g b gb g Pseudocritical pressure: P ′ = b0.20gb45.8g + b0.30gb48.2g + b0.50gb50.5g = 48.9 atm U| b90 + 273.2gK = 134 Reduced temperature: T = . |V ⇒ z = 0.71 2713 . K 200 bars 1 atm Reduced pressure: P = = 4.04 | |W 48.9 atm 1.01325 bars Mean molecular weight of mixture: M = b0.20gM + b0.30gM + b0.50gM = b0.20gb16.04g + b0.30gb30.07g + b0.50gb28.05g
. = 2713 . K Pseudocritical temperature: Tc′ = 0.20 190.7 + 0.30 305.4 + 0.50 2831 c
r
Figure 5.4-3
r
CH 4
C2 H 6
C2 H 4
= 26.25 kg kmol V=
znRT 0.71 10 kg 1 kmol 0.08314 m 3 ⋅ bar = P 26.25 kg kmol ⋅ K
b90 + 273g K = 0.041 m (41 L) 3
200 bars
UV T′ = 0.10b309.5g + 0.90b126.2g = 144.5 K N O: T = 309.5 K, P = 71.7 atmW P ′ = 0.10b71.7g + 0.90b33.5g = 37.3 atm M = 0.10b44.02g + 0.90b28.02g = 29.62 n = 5.0 kgb1 kmol 29.62 kgg = 0.169 kmol = 169 mol a. T = b24 + 273.2g 144.5 = 2.06 U| 37.3 atm mol ⋅ K V ⇒ z = 0.97bFig. 5.4 - 3g = 30 L V = 0.56| 169 mol 144.5 K 0.08206 L ⋅ atm W
5.75 N 2 : Tc = 126.2 K, PC = 33.5 atm 2
c
C
c c
r
r
P=
b.
0.97 169 mol 297.2 K 0.08206 L ⋅ atm mol ⋅ K
30 L
U|V g |W
= 133 atm ⇒ 132 atm gauge
Pr = 273 37.3 = 7.32 ⇒ z = 1.14 Fig. 5.4 - 3 = 0.56 from a. V r
b
T=
b
g
273 atm 30 L mol ⋅ K = 518 K ⇒ 245° C 1.14 169 mol 0.08206 L ⋅ atm
5- 49
b g b g
UV W
b
g g
5.76 CO: Tc = 133.0 K, Pc = 34.5 atm Tc′ = 0.60 133.0 + 0.40 33 + 8 = 96.2 K H 2 : Tc = 33 K, Pc = 12.8 atm Pc′ = 0.60 34.5 + 0.40 12.8 + 8 = 29.0 atm
b
b
U| 1 atm V ⎯ ⎯⎯⎯→ z ≈ 1.01 = 4.69 | 14.7 psi W g
Tr = 150 + 273.2 96.2 = 4.4
Turbine inlet:
Pr =
2000 psi 29.0 atm
Fig. 5.4-1
Turbine exit: Tr = 373.2 96.2 = 3.88
⇒ z=1.0
Pr = 1 29.0 = 0.03
Pin V z nRTin P z T ft 3 14.7 psia 1.01 423.2K in = in ⇒ Vin = Vout × out in in = 15,000 z out n RTout Pin z out Tout min 2000 psia 1.00 373.2 Pout V out = 126 ft 3 / min If the ideal gas equation of state were used, the factor 1.01 would instead be 1.00 ⇒ −1% error
UV T′ = 0.97b133.0g + 0.03b304.2g = 138.1 K CO : T = 304.2 K, P = 72.9 atmW P ′ = 0.97b34.5g + 0.03b72.9g = 35.7 atm = 524.8 psi Initial: T = 303.2 138.1 = 2.2 U V ⎯⎯⎯⎯→ z = 0.97 P = 2014.7 524.8 = 3.8W
5.77 CO: Tc = 133.0 K, Pc = 34.5 atm 2
c
c
c c
r
Fig. 5.4-3
1
r
Final: Pr = 1889.7 524.8 = 3.6 ⇒ z1 = 0.97 Total moles leaked:
n1 − n 2 =
FG P − P IJ V = b2000 − 1875gpsi 0.97 H z z K RT
1 atm
mol ⋅ K
1
2
30.0 L
1
2
303 K 14.7 psi 0.08206 L ⋅ atm
= 10.6 mol leaked
b g
Moles CO leaked: 0.97 10.6 = 10.3 mol CO Total moles in room:
Mole% CO in room =
24.2 m3 103 L 273 K 1m
3
1 mol
b g = 973.4 mol
303 K 22.4 L STP
10.3 mol CO × 100% = 10% . CO 973.4 mol
5- 50
CO + 2H 2 → CH 3OH
5.78 Basis: 54.5 kmol CH 3OH h
n 1 (kmol CO / h) 2n 1 (kmol H 2 / h) 644 K 34.5 MPa
Catalyst Bed
Condenser
CO, H 2
54.5 kmol CH 3OH (l ) / h
a.
n 1 =
54.5 kmol CH 3OH 1 kmol CO react 1 kmol CO fed = 218 kmol h CO h 1 kmol CH 3OH 0.25 kmol CO react
b g
b
g
2n 1 = 2 218 = 436 kmol H 2 h ⇒ 218 + 436 = 654 kmol h (total feed) CO: Tc = 133.0 K
Pc = 34.5 atm
H 2 : Tc = 33 K
Pc = 12.8 atm ⇓ Newton’s corrections
b
g b
g
b g b
g
Tc′ =
1 2 133.0 + 33 + 8 = 717 . K 3 3
Pc′ =
1 2 34.5 + 12.8 + 8 = 25.4 atm 3 3
U| V| W
Tr = 644 71.7 = 8.98 5.4-4 34.5 MPa 10 atm ⎯Fig. ⎯⎯ ⎯→ z 1 = 1.18 Pr = = 13.45 24.5 atm 1.013 MPa 1.18 654 kmol 644 K 0.08206 m3 ⋅ atm 1.013 MPa V = = 120 m3 h feed h 34.5 MPa kmol ⋅ K 10 atm Vcat =
120 m3 h
1 m3 cat 25,000 m3 / h
= 0.0048 m3 catalyst (4.8 L)
b. CO, H 2 n 4 kmol CO / h 2n 4 kmol H 2 / h
54.5 kmol CH 3OH (l ) / h
Overall C balance ⇒ n 4 = 54.5 mol CO h Fresh feed:
54.5 kmol CO h 109.0 kmol H 2 h 163.5 kmol feed gas h
1.18 163.5 kmol 644 K 0.08206 m3 ⋅ atm 1.013 MPa V = = 29.9 m3 h feed h 34.5 MPa kmol ⋅ K 10 atm
5- 51
5.79
H 2 : Tc = (33.3 + 8) K = 41.3 K
1 - butene: Tc = 419.6 K
Pc = (12.8 + 8) atm = 20.8 atm
Pc = 39.7 atm
Tc ' = 0.15(413 . K) + 0.85(419.6 K) = 362.8 K
UV P ' = 0.27 W
Tr ' = 0.89
Pc ' = 0.15(20.8 atm) + 0.85(39.7 atm) = 36.9 atm
Fig. 5.4-2
⇒ z = 0.86
r
0.86 35 kmol 0.08206 m3 ⋅ atm 323 K 1 h = znRT V = = 1.33 m3 / min P h kmol ⋅ K 10 atm 60 min
d
F I FG IJ d i GH JK H K
i FG 100 cmIJ = 10.6 cm gH m K
4 133 . m 3 / min m3 m 4V πd 2 2 V =u A m =u× ⇒d= = min min 4 πu π 150 m / min 5.80
CH 4 :
b
Tc = 190.7 K Pc = 45.8 atm
C2 H 4 : Tc = 283.1 K Pc = 50.5 atm C2 H 6 : Tc = 305.4 K Pc = 48.2 atm
U| V P ' = 015 . ( 458 . atm) + 0.60(50.5 atm) + 0.25(48.2 atm) = 49.2 atm =====> P ' = 35 . |W T=90o C
Tc ' = 015 . (190.7 K) + 0.60(283.1 K) + 0.25(305.4 K) = 274.8 K ====>
Tr ' = 132 .
P=175 bar
c
r
5.4-3 ⎯Fig. ⎯⎯ ⎯→ z = 0.67
F I FG IJ d i FG GH JK H K H
3 m = u m A m 2 = 10 m V s s s
n = 5.81
IJ FG 60 s IJ π b0.02 mg K H min K 4
2
= 0188 .
m3 min
175 bar 1 atm PV kmol ⋅ K 0188 . m3 / min = = 163 . kmol / min 363 K zRT 0.67 1.013 bar .08206 m3 ⋅ atm
N2:
Tc = 126.2 K = 227.16o R Pc = 335 . atm
acetonitrile: Tc = 548 K = 986.4 o R
Pc = 47.7 atm Fig. 5.4-3
Tank 1 (acetonitrile): T1 = 550 o F, P1 = 4500 psia ⇒ Tr1 = 1.02 Pr1 = 6.4 ⇒ z 1 = 0.80 ⇒ n 1 =
P1 V1 306 atm 0.200 ft 3 = z 1 RT1 0.80 1009.7 o R
lb - mole ⋅ o R = 0.104 lb - mole 0.7302 ft 3 ⋅ atm Fig. 5.4-3
Tank 2 (N 2 ): T2 = 550 o F, P2 = 10 atm ⇒ Tr2 = 4.4 , Pr2 = 6.4 ⇒ z 2 = 1.00 ⇒ n 2 =
P2 V2 10.0 atm 2.00 ft 3 = z 2 RT2 1.00 1009.7 o R
5- 52
lb - mole ⋅ o R = 0.027 lb - mole .7302 ft 3 ⋅ atm
5.81 (cont’d)
. I .027 I FG 0104 J 986.4 R + FGH 00131 J 227.16 R = 830 R H 0131 . K . K . I F 0104 F 0.027 IJ 33.5 atm = 44.8 atm 47.7 atm + G P '= G J H 0131 K H 0131 . . K
Final: Tc ' =
o
o
o
o
⎯T=550 ⎯⎯ ⎯F →
Tr ' = 122 .
c
dV i r
P=
=
ideal
' Fig. 5.4-2 VP 2.2 ft 3 44.8 atm lb - mole ⋅ o R c = = 1.24 ⇒ z = 0.85 RTc ' 0.131 lb - mole 830 o R 0.7302 ft 3 ⋅ atm
znRT 0.85 0131 . lb - mole .7302 ft 3 ⋅ atm 1009.7 o R = = 37.3 atm V lb - mole ⋅ o R 2.2 ft 3
5.82
3.48 g C a H b O c , 26.8o C, 499.9 kPa n c (mol C), n H (mol H), n O (mol O)
1 L @483.4 o C, 1950 kPa n p (mol) 0.387 mol CO 2 / mol 0.258 mol O 2 / mol 0.355 mol H 2 O / mol
n O 2 (mol O 2 ) 26.8o C, 499.9 kPa
a.
d
i
Volume of sample: 3.42 g 1 cm3 159 . g = 2.15 cm3 O 2 in Charge: n O2
d
1.000 L − 2.15 cm 3 10 −3 L km 3 = L ⋅ atm 0.08206 mol ⋅ K
i
499.9 kPa
1 atm
300 K
101.3 kPa
= 0.200 mol O 2
Product 1.000 L 1950 kPa 1 atm L ⋅ atm = 0.310 mol product 756.6 K 101.3 kPa 0.08206 mol ⋅ K Balances: np =
b
g
b
g b
g
O: 2 0.200 + n O = 0.310 2 0.387 + 2 0.258 + 0.355 ⇒ n O = 0.110 mol O in sample
b
g
C: n C = 0.387 0.310 = 0.120 mol C in sample
b
gb
g
H: n H = 2 0.355 0.310 = 0.220 mol H in sample Assume c = 1 ⇒ a = 0.120 0.110 = 1.1 b = 0.220 0.110 = 2 Since a, b, and c must be integers, possible solutions are (a,b,c) = (11,20,10), (22,40,20), etc. b.
b g
b g
MW = 12.01a + 1.01b + 16.0c = 12.01 1.1c + 1.01 2c + 16.0c = 31.23c 300 < MW < 350 ⇒ c = 10 ⇒ C 11 H 20 O 10
5- 53
bg
C5 H 10 +
5.83 Basis: 10 mL C5 H 10 l charged to reactor
15 O 2 → 5CO 2 + 5H 2 O 2
bg
10 mL C5 H 10 l n1 (mol C5 H 10 )
n 2 (mol air) 0.21 O 2 0.79 N 2 27 o C, 11.2 L, Po (bar)
a.
bg
10.0 mL C5 H 10 l
n1 =
Stoichiometric air: n 2 = Po =
n 3 (mol CO 2 ) n 4 mol H 2 O(v) n 5 (mol N 2 ) 75.3 bar (gauge), Tad
b
0.745 g
1 mol
mL
70.13 g
g
d Ci o
= 0.1062 mol C5 H 10
0.1062 mol C5 H 10
7.5 mol O 2 1 mol C 2 H 10
1 mol air = 3.79 mol air 0.21 mol O 2
nRT 3.79 mol 0.08314 L ⋅ bar 300K = 8.44 bars = 11.2 L mol ⋅ K V
(We neglect the C5 H 10 that may be present in the gas phase due to evaporation) Initial gauge pressure = 8.44 bar − 1 bar = 7.44 bar b.
U| || V| || W
5 mol CO 2 = 0.531 mol CO 2 1 mol C 5 H 10 0.531 mol CO 2 1 mol H 2 O ⇒ 4.052 mol product gas n4 = = 0.531 mol H 2 O 1 mol CO 2 n 5 = 0.79 3.79 = 2.99 mol N 2 n3 =
0.1062 mol C 5 H 10
b g
CO 2 : y 3 = 0.531 / 4.052 = 0.131 mol CO 2 / mol, Tc = 304.2 K Pc = 72.9 atm H 2 O: y 4 = 0.531 / 4.052 = 0.131 mol H 2 O / mol, N2:
y 5 = 2.99 / 4.052 = 0.738 mol N 2 / mol,
Tc = 647.4 K Pc = 218.3 atm Tc = 126.2 K Pc = 33.5 atm
. (304.2 K) + 0.131(647.4 K) + 0.738(126.2 K) = 217.8 K Tc ' = 0131 . (72.9 atm) + 0.131(218.3 atm) + 0.738(33.5 atm) = 62.9 atm ⇒ Pr ' = 121 . Pc ' = 0131 mol ⋅ K ideal = VPc ' = 11.2 L 62.9 atm V = 9.7 ⇒ z ≈ 1.04 (Fig. 5.4 - 3) r RTc ' 4.052 mol 217.8 K .08206 L ⋅ atm T=
b
g
75.3 + 1 bars 112 . L PV mol ⋅ K = = 2439 K - 273 = 2166o C znR 1.04 4.052 mol 0.08314 L ⋅ bar
5- 54
CHAPTER SIX 6.1
a.
AB: Heat liquid - -V ≈ constant BC: Evaporate liquid - -V increases, system remains at point on vapor - liquid equilibrium curve as long as some liquid is present. T = 100 o C. CD: Heat vapor - -T increases, V increases .
b. Point B: Neglect the variation of the density of liquid water with temperature, so ρ = 1.00 g/mL and VB = 10 mL
Point C: H2O (v, 100°C) n=
10 mL
1.00 g 1 mol = 0.555 mol mL 18.02 g
PCVC = nRTC ⇒ VC =
6.2
nRTC 0.555 mol 0.08206 L ⋅ atm 373 K = = 17 L 1 atm mol ⋅ K PC
a. Pfinal = 243 mm Hg . Since liquid is still present, the pressure and temperature must lie on the
vapor-liquid equilibrium curve, where by definition the pressure is the vapor pressure of the species at the system temperature. b. Assuming ideal gas behavior for the vapor,
m(vapor) = m(liquid) =
(3.000 - 0.010) L mol ⋅ K 243 mm Hg 1 atm 119.39 g = 4.59 g (30 + 273.2) K 0.08206 L ⋅ atm 760 mm Hg mol 10 mL
1.489 g mL
= 14.89 g
m total = m(vapor) + m(liquid) = 19.5 g
x vapor =
6.3
a.
4.59 = 0.235 g vapor / g total 19.48
log 10 p ∗ = 7.09808 −
1238.71 = 2.370 ⇒ p * = 10 2.370 = 234.5 mm Hg 45 + 217
d
i
* * ΔH v 1 ΔH v ln p2 / p1 b. ln p = − + B⇒− = = 1 R T R − T11 T2 ∗
B = ln( p1* ) +
b
g
ln 760 / 118.3
b
1 77 .0 + 273.2 K
g
−
ΔH v / R 4151 K = ln 118.3 + = 18.49 T1 29.5 + 273.2 K
b
g b
g
6-1
b
1 29.5+ 273.2 K
g
= −4151K
6.3 (cont’d)
ln p ∗ (45o C) = −
b
4151 + 18.49 ⇒ p ∗ = 2310 . mm Hg 45 + 273.2
g
. − 234.5 2310 error × 100% = −15% . 234.5 c.
p∗ =
FG 118.3 − 760IJ b45 − 29.5g + 118.3 = 327.7 mm Hg H 29.5 − 77 K
327.7 − 234.5 × 100% = 39.7% error 234.5
b
g
1 (rect. scale) on semilog paper T + 273.2 ⇒ straight line: slope = −7076K , intercept = 2167 .
Plot p ∗ log scale vs
b
g
ln p ∗ mm Hg =
b
7076 K Δ Hv = 7076K ⇒ Δ H v = R
g
OP Q
8.314 J 1 kJ = 58.8 kJ mol mol ⋅ K 10 3 J
ln p* = A/T(K) + B p*(mm Hg) 5 20 40 100 400 760
1/T(K) 0.002834 0.002639 0.002543 0.002410 0.002214 0.002125
ln(p*) p*(fitted) 1.609 5.03 2.996 20.01 3.689 39.26 4.605 101.05 5.991 403.81 6.633 755.13
7 6 5 4 3 2 1 0
1/T
6-2
0.003
0.0026
0.0024
0.0022
y = -7075.9x + 21.666
0.002
o
T( C) 79.7 105.8 120.0 141.8 178.5 197.3
ln(p*)
6.5
LM N
−7076 −7076 . ⇒ p ∗ mm Hg = exp . + 2167 + 2167 o T ( C) + 273.2 T ( C) + 273.2 o
0.0028
6.4
T(oC) p*(fitted) 50 0.80 80 5.12 110 24.55 198 760.00 230 2000.00 Least confidence (Extrapolated)
6.6
a.
p*(mm Hg) =758.9 + hright -hleft -3 42.7 3.17×10 34.9 58.9 3.01×10-3 78.9 -3 68.3 2.93×10 122.9 77.9 2.85×10-3 184.9 88.6 2.76×10-3 282.9 -3 98.3 2.69×10 404.9 105.8 2.64×10-3 524.9 ΔH v −51438 . K b.Plot is linear, ln p∗ = − + B ⇒ ln p∗ = + 19.855 RT T
T(°C)
1/T(K)
At the normal boiling point, p∗ = 760 mmHg ⇒ Tb = 116° C 8.314 J 5143.8 K 1 kJ Δ H v = = 42.8 kJ mol mol ⋅ K 10 3 J
c. Yes — linearity of the ln p∗ vs 1 / T plot over the full range implies validity. 6.7
a.
b
g
ln p∗ = a T + 273.2 + b ⇒ y = ax + b
b
y = ln p∗ ; x = 1 T + 273.2
g
Perry' s Handbook, Table 3 - 8:
T1 = 39.5° C , p1 ∗ = 400 mm Hg ⇒ x1 = 31980 . × 10 −3 , y1 = 5.99146 T2 = 56.5° C , p2 ∗ = 760 mm Hg ⇒ x 2 = 3.0331 × 10 −3 , y 2 = 6.63332
T = 50° C ⇒ x = 3.0941 × 10 −3 x − x1 y 2 − y1 = 6.39588 ⇒ p∗ 50° C = e 6.39588 = 599 mm Hg y = y1 + x 2 − x1
FG H
IJ b K
g
b
Cox chart b. 50° C = 122° F ⎯⎯⎯⎯ → p∗ =
c.
6.8
log p∗ = 7.02447 −
b
g
12 psi 760 mm Hg = 625 mm Hg 14.6 psi
11610 . = 2.7872 ⇒ p∗ = 10 2.7872 = 613 mm Hg 50 + 224
g
Estimate p∗ 35° C : Assume ln p∗ =
a + b , interpolate given data. T K
b g
U| ln p∗ b35° Cg = − 6577.1 + 25.97 = 4.630 35 + 273.2 |V ⇒ − − 6577.1 a b = ln p ∗− = lnb50g + = 25.97 | = 102.5 mm Hg p∗ b35° Cg = e |W 25 + 273.2 T
a=
b
g=
ln p2 ∗ p1 ∗ 1 T2
1 T1
b
g
ln 200 50
1 45+ 273.2
= −6577.1
1 25+ 273.2
4 .630
1
1
Moles in gas phase: n =
150 mL
= 8.0 × 10
b −4
273 K 102.5 mm Hg 1L 1 mol 3 35 + 273.2 K 760 mm Hg 10 mL 22.4 L STP
g
mol
6-3
b g
6.9
a.
m=2 π =2⇒ F =2+2−2=2.
Two intensive variable values (e.g., T & P) must be
specified to determine the state of the system. 1209.6 b. log p∗ MEK = 6.97421 − = 2.5107 ⇒ p∗ MEK = 10 2.5107 = 324 mm Hg 55 + 216. Since vapor & liquid are in equilibrium p MEK = p∗ MEK = 324 mm Hg ⇒ y MEK = p MEK / P = 324 1200 = 0.27 > 0115 . The vessel does not constitute an explosion hazard.
6.10 a. The solvent with the lower flash point is easier to ignite and more dangerous. The solvent with a flash point of 15°C should always be prevented from contacting air at room temperature. The other one should be kept from any heating sources when contacted with air. b. At the LFL, y M = 0.06 ⇒ p M = p *M = 0.06 × 760 mm Hg = 45.60 mm Hg 1473.11 Antoine ⇒ log 10 45.60 = 7.87863 ⇒ T = 6.85° C T + 230 c. The flame may heat up the methanol-air mixture, raising its temperature above the flash point. 6.11 a. At the dew point, p ∗ ( H 2 O) = p( H 2 O) = 500 × 0.1 = 50 mm Hg ⇒ T = 38.1° C from Table B.3.
b. VH2O =
30.0 L
273 K 500 mm Hg 1 mol 0.100 mol H2 O 18.02 g 1 cm3 =134 . cm3 (50 + 273) K 760 mm Hg 22.4 L (STP) mol mol g
c. (iv) (the gauge pressure)
6-4
6.12 a.
b g = 110° C , p ∗ = 755 mm Hg − b577 − 222gmm Hg = 400 mm Hg
T1 = 58.3° C , p1 ∗ = 755 mm Hg − 747 − 52 mm Hg = 60 mm Hg
T2
2
ln p∗ = a=
a +b T K
b g
b
g=
b
ln p2 ∗ p1 ∗ 1 T2
−
b = ln p1 ∗−
1 T1
g
ln 400 60 1 110 + 273.2
−
1 58.3+ 273.2
= −46614 .
b g
a 46614 . = ln 60 + = 18156 . T1 58.3 + 273.2
T=130oC=403.2 K −46614 . + 18156 . T ln p∗ 130° C = 6.595 ⇒ p∗ 130° C = e 6.595 = 7314 . mm Hg
ln p∗ =
b
b.
g
b
g
Basis: 100 mol feed gas CB denotes chlorobenzene. n1 mol @ 58.3°C, 1atm y1 (mol CB(v)/mol) (sat’d) (1-y1) (mol air/mol)
100 mol @ 130°C, 1atm y0 (mol CB(v)/mol) (sat’d) (1-y0) (mol air/mol)
n2 mol CB (l)
b
g
Saturation condition at inlet: y o P = pCB ∗ 130° C ⇒ y o =
731 mm Hg = 0.962 mol CB mol 760 mm Hg
b
60 mm Hg = 0.0789 mol CB mol g 760 mm Hg Air balance: 100b1 − y g = n b1 − y g ⇒ n = b100gb1 − 0.962g b1 − 0.0789g = 4.126 mol Total mole balance: 100 = n + n ⇒ n = 100 − 4.126 = 9587 . mol CBbl g
Saturation condition at outlet: y1 P = pCB ∗ 58.3° C ⇒ y1 = o
1
1
1
% condensation:
2
1
2
95.87 mol CB condensed × 100% = 99.7% 0.962 × 100 mol CB feed
b
g
c. Assumptions: (1) Raoult’s law holds at initial and final conditions; (2) CB is the only condensable species (no water condenses); (3) Clausius-Clapeyron estimate is accurate at 130°C. 6.13 T = 78° F = 25.56° C , Pbar = 29.9 in Hg = 759.5 mm Hg , hr = 87%
b
y H 2 O P = 0.87 p∗ 25.56° C
g
Table B.3
yH 2 O =
0.87 ( 24.559 mm Hg )
( )
759.5 mm Hg
Dew Point: p ∗ Tdp = yp = 0.0281( 759.5 ) = 21.34 mm Hg
6-5
Table B.3
= 0.0281 mol H 2 O mol air
Tdp = 23.2°C
6.13 (cont’d)
hm =
0.0281 = 0.0289 mol H 2 O mol dry air 1 − 0.0281
ha =
0.0289 mol H 2 O 18.02 g H 2 O mol dry air = 0.0180 g H 2 O g dry air mol dry air mol H 2 O 29.0 g dry air
hp =
hm 0.0289 × 100% = × 100% = 86.5% 24.559 [ 759.5 − 24.559] p ∗ ( 25.56°C ) ⎡⎣ P − p ∗ ( 25.56°C ) ⎤⎦
6.14 Basis I : 1 mol humid air @ 70° F (21.1° C), 1 atm, hr = 50%
b
hr = 50% ⇒ y H 2 O P = 0.50 p H 2O ∗ 21.1° C Table B.3
Mass of air:
y H 2O =
mol H 2 O 0.50 × 18.765 mm Hg = 0.012 760.0 mm Hg mol
0.012 mol H 2 O 18.02 g 1 mol
Volume of air: Density of air =
1 mol
g 0.988 mol dry air 29.0 g
+
1 mol
= 28.87 g
b g b273.2 + 21.1gK = 24.13 L
22.4 L STP 1 mol
273.2K
28.87 g = 1196 . g L 24.13 L
Basis II: 1 mol humid air @ 70° F (21.1° C), 1 atm, hr = 80%
b
hr = 80% ⇒ y H 2 O P = 0.80 p H 2O ∗ 21.1° C Table B.3
Mass of air:
y H 2O =
mol H 2 O 0.80 × 18.765 mm Hg = 0.020 760.0 mm Hg mol
0.020 mol H 2 O 18.02 g 1 mol
Volume of air: Density of air =
1 mol
g
+
0.980 mol dry air 29.0 g 1 mol
b g b273.2 + 21.1gK = 24.13 L
22.4 L STP 1 mol
273.2K
28.78 g = 1193 . g L 24.13 L
Basis III: 1 mol humid air @ 90° F (32.2° C), 1 atm, hr = 80%
b
hr = 80% ⇒ y H 2 O P = 0.80 p H 2 O ∗ 32.2° C Table B.3
y H 2O =
g
mol H 2 O 0.80 × 36.068 mm Hg = 0.038 760.0 mm Hg mol
6-6
= 28.78 g
6.14 (cont’d)
Mass of air:
0.038 mol H 2 O 18.02 g 0.962 mol dry air 29.0 g + = 28.58 g 1 mol 1 mol
Volume of air: Density =
b g b273.2 + 32.2gK = 25.04 L
1 mol
22.4 L STP 1 mol
273.2K
28.58 g = 1141 . g L 25.04 L
Increase in T ⇒ increase in V ⇒ decrease in density Increase in hr ⇒ more water (MW = 18), less dry air (MW = 29) ⇒ decrease in m ⇒ decrease in density Since the density in hot, humid air is lower than in cooler, dryer air, the buoyancy force on the ball must also be lower. Therefore, the statement is wrong.
b
6.15 a. hr = 50% ⇒ y H 2O P = 0.50 p H 2 O ∗ 90° C Table B.3
y H 2O =
g
0.50 × 525.76 mm Hg = 0.346 mol H 2 O / mol 760.0 mm Hg
b g
d i
Dew Point: y H 2 O p = p∗ Tdp = 0.346 760 = 262.9 mm Hg
Table B.3
Tdp = 72.7° C
Degrees of Superheat = 90 − 72.7 = 17.3° C of superheat
b. Basis:
1 m 3 feed gas 10 3 L 273K m
Saturation Condition: y1 =
3
mol
b g = 33.6 mol
363K 22.4 L STP
b
p H* 2 O 25° C
g = 23.756 = 0.0313 mol H O mol
760 P Dry air balance: 0.654 ( 33.6 ) = n1 (1 − 0.0313) ⇒ n1 = 22.7 mol
2
Total mol balance: 33.6=22.7+n2 ⇒ n2 = 10.9 mol H 2 O condense/m3
b
g
c. y H 2 O P = p∗ 90° C ⇒ P =
p * (90° C) 525.76 mmHg = = 1520 mm Hg = 2.00 atm y H 2O 0.346
6-7
6.16 T = 90° F = 32.2° C , p = 29.7 in Hg = 754.4 mm Hg , hr = 95%
Basis: 10 gal water condensed/min ncondensed =
1 ft 3
10 gal H 2 O min
62.43 lb m
7.4805 gal
ft
18.02 lb m
y2 (lb-mol H2O (v)/lb-mol) (sat’d) (1-y2) (lb-mol DA/lb-mol) 40oF (4.4oC), 754 mm Hg
y1 (lb-mol H2O (v)/lb-mol) (1-y1) (lb-mol DA/lb-mol) hr=95%, 90oF (32.2oC), 29.7 in Hg (754 mm Hg)
4.631 lb-moles H2O (l)/min
b
95% hr at inlet: y H 2 O P = 0.95 p∗ 32.2° C y H 2O =
= 4.631 lb-mole/min
n2 (lb - moles / min)
V1 (ft 3 / m in) n1 (lb - m oles / m in)
Table B.3
1 lb-mol
3
b
g
0.95 36.068 mm Hg 754.4 mm Hg
Raoult's law: y2 P = p* ( 4.4°C )
g = 0.045 lb - mol H O lb - mol 2
Table B.3
y2 =
6.274 = 0.00817 lb-mol H 2 O lb-mol 754.4
Mole balance : n1 = n2 + 4.631 ⎫ ⎧n1 = 124.7 lb-moles/min ⎬⇒ ⎨ Water balance : 0.045n1 = 0.00817 n2 + 4.631⎭ ⎩n2 = 120.1 lb-moles/min
Volume in: V =
124.7 lb-moles 359 ft 3 (STP) (460+90)o R 760 mm Hg min
lb-moles
492o R
754 mm Hg
= 5.04 × 104 ft 3 / min
6.17 a. Assume no water condenses and that the vapor at 15°C can be treated as an ideal gas.
p final =
760 mm Hg
(15 + 273) K (200 + 273) K
= 462.7 mm Hg ⇒ ( p H 2 O ) final = 0.20 × 462.7 = 92.6 mm Hg
p * (15° C) = 12.79 mm Hg < p H 2 O . Impossible ⇒ condensation occurs. Tfinal 288 K = (0.80 × 760) mm Hg × = 370.2 mm Hg 473 K Tinitial = 370.2 + 12.79 = 383 mm Hg
( pair ) final = ( pair ) initial
P = p H 2 O + pair b. Basis:
mol 1 L 273 K = 0.0258 mol 473 K 22.4 L (STP)
6-8
6.17 (cont’d) n1 mol @ 15°C, 383.1 mm Hg y1 (mol H2O (v)/mol) (sat’d) (1-y1) (mol dry air/mol)
0.0258 mols @ 200°C, 760 mm Hg 0.200 H2O mol /mol 0.800 mol air/mol
n2 mol H2O (l)
Saturation Condition: y1 =
b
p H* 2 O 15° C P
b
g = 12.79 mm Hg = 0.03339 mol H O mol 2
3831 . mm Hg
g b
g
Dry air balance: 0.800 0.0258 = n1 1 − 0.03339 ⇒ n1 = 0.02135 mol
c.
Total mole balance: 0.0258 = 0.02135 + n2 ⇒ n2 = 0.00445 mol Mass of water condensed =
0.00445 mol
18.02 g = 0.0802 g mol
6.18 Basis: 1 mol feed n2 (mol), 15.6°C, 3 atm y 2 (mol H2 O (v)/mol)(sat'd) (1 – y 2) (mol DA/mol)
3
V1 (m ) 1 mol, 90°C, 1 atm 0.10 mol H 2O (v)/mol 0.90 mol dry air/mol
heat
100°C, 3 atm n2 (mol) 3 V2 (m )
n3 (mol) H 2 O( l ), 15.6°C, 3 atm
Saturation: y 2 =
b
p H* 2O 15.6° C
g
Table B.3
P
y2 =
bg b g H O mol balance: 0.10b1g = 0.00583b0.9053g + n
13.29 mm Hg atm = 0.00583 3 atm 760 mm Hg
Dry air balance: 0.90 1 = n2 1 − 0.00583 ⇒ n2 = 0.9053 mol 2
Fraction H 2 O condensed:
3
⇒ n3 = 0.0947 mol
0.0947 mol condensed = 0.947 mol condense mol fed . mol fed 0100
b
g
hr =
y 2 P × 100% 0.00583 3 atm = × 100% = 175% . p∗ 100° C 1 atm
V2 =
0.9053 mol 22.4 L STP mol
b
g
b g
b g
1 mol 22.4 L STP V1 = mol
373K 1 atm 1 m 3 = 9.24 × 10 −3 m 3 outlet air @ 100° C 273K 3 atm 10 3 L
363K 1 m 3 = 2.98 × 10 −2 m 3 feed air @ 90° C 3 273K 10 L
V2 9.24 × 10 −3 m 3 outlet air = = 0.310 m 3 outlet air m 3 feed air V1 2.98 × 10 −2 m 3 feed air
6-9
6.19 Liquid H 2 O initially present:
Saturation at outlet: y H 2 O = ⇒
25 L 100 . kg
1 kmol 18.02 kg
L
b
p H* 2 O 25° C P
g=
bg
= 1387 . kmol H 2 O l
23.76 mm Hg = 0.0208 mol H 2 O mol air 15 . × 760 mm Hg
0.0208 = 0.0212 mol H 2 O mol dry air 1 − 0.0208
b g
15 L STP 1 mol = 0.670 mol dry air min min 22.4 L STP 0.670 mol dry air 0.0212 mol H 2 O = 0.0142 mol H 2 O min Evaporation Rate: min mol dry air Flow rate of dry air:
Complete Evaporation:
1.387 kmol 10 3 mol 1h min = 1628 h kmol 0.0142 mol 60 min
6.20 a. Daily rate of octane use =
( SG ) C8 H18
b. Δp =
b g
π 4
⋅ 30 2 ⋅ (18 − 8) =
7.069 × 10 3 ft 3 day
b67.8 daysg
7.481 gal = 5.288 × 10 4 gal / day 3 ft
5.288 × 10 4 gal 1 ft 3 0.703 × 62.43 lb m day 7.481 gal ft 3 = 0.703 ⇒ = 3.10 × 10 5 lb m C 8 H 18 / day
0.703 × 62.43 lb m ft
32.174 ft
3
s
2
1 lb f 32.174
(18-8) ft
lb m ⋅ft s
2
1 ft 2 144 in
29.921 in Hg 2
14.696 lb f / in 2
= 6.21 in Hg
14.696 psi = 0.40 lb f / in 2 = poctane = y octane P 760 mm Hg Octane lost to environment = octane vapor contained in the vapor space displaced by liquid during refilling.
c. Table B.4: pC* 8 H18 (90 o F) =
Volume:
5.288 × 10 4 gal
20.74 mm Hg
1 ft 3 = 7069 ft 3 7.481 gal
(16.0 + 14.7) psi 7069 ft 3 pV = = 36.77 lb - moles RT 10.73 ft 3 ⋅ psi / (lb - mole ⋅ o R) (90 + 460) o R pC H 0.40 psi Mole fraction of C 8 H 18 : y = 8 18 = = 0.0130 lb - mole C 8 H 18 / lb - mole P (16.0 + 14.7) psi
Total moles: n =
Octane lost = 0.0130(36.77) lb - mole = 0.479 lb - mole ( = 55 lb m = 25 kg) d. A mixture of octane and air could ignite.
6-10
* * 6.21 a. Antoine equation ⇒ ptol (85o F) = ptol (29.44 o C) = 35.63 mmHg = ptol
Mole fraction of toluene in gas: y =
ptol 35.63 mmHg = = 0.0469 lb - mole toluene / lb - mole 760 mmHg P
yPV RT 0.0469 lb - mole tol
Toluene displaced = yntotal = =
lb - mole
0.7302
1 ft 3
900 gal
1 atm ft 3 ⋅ atm lb - mole ⋅ o R
(85 + 460) o R 7.481 gal
92.13 lb m tol lb - mole
= 1.31 lb m toluene displaced b. Basis: 1mol 0.0469 mol C7H8(v)/mol 0.9531 mol G/mol
nV (mol) y (mol C7H8(v)/mol) (1-y) (mol G/mol) T(oF), 5 atm
Assume G is noncondensable
nL [mol C7H8 (l)] 90% of C7H8 in feed
90% condensation ⇒ n L = 0.90(0.0469)(1) mol C 7 H 8 = 0.0422 mol C 7 H 8 (l ) Mole balance: 1 = nV + 0.0422 ⇒ nV = 0.9578 mol Toluene balance: 0.0469(1) = y (0.9578) + 0.0422 ⇒ y = 0.004907 mol C 7 H 8 / mol Raoult’s law:
* ptol = yP = ( 0.004907)(5 × 760) = 18.65 mmHg = ptol (T )
Antoine equation:
T=
B − C ( A − log10 p* ) 1346.773 − 219.693(6.95805 − log10 18.65) = = 17.11o C=62.8o F * 6.95805 − log10 18.65 A − log10 p
6.22 a. Molar flow rate: n =
100 m 3 VP = RT h
kmol ⋅ K 2 atm = 6.53 kmol / h 3 82.06 × 10 m ⋅ atm (100 + 273) K -3
b. Antoine Equation:
1175.817 = 3.26601 100+224.867 ⇒ p* = 1845 mm Hg
* log10 pHex (100°C)=6.88555-
0.150(2.00) atm 760 mm Hg * ⇒ not saturated = 228 mm Hg < p Hex atm 1175.817 * pHex = 2.35793 ⇒ T = 34.8°C (T ) = 228 mm Hg ⇒ log10 228=6.88555T+224.867
p Hex = y Hex ⋅ P =
6-11
6.22 (cont’d) c. 6.53 kmol/h 0.15 C6H14 (v) 0.85 N2
nV (kmol/h) y (kmol C6H14 (v)/kmol), sat’d (1-y) (kmol N2/kmol) T (oC), 2 atm
nL (kmol C6H14 (l)/h) 80% of C6H14 in feed
80% condensation: Mole balance: Hexane balance: Raoult’s law: Antoine equation:
n L = 0.80(015 . )(6.53 kmol / h) = 0.7836 kmol C 6 H 14 (l ) / h 6.53 = nV + 0.7836 ⇒ nV = 5.746 kmol / h 015 . (6.53) = y (5.746) + 0.7836 ⇒ y = 0.03409 kmol C 6 H 14 / kmol * p Hex = yP = (0.03409)( 2 × 760 mmHg) = 51.82 mmHg = p Hex (T ) 1175.817 log10 51.82 = 6.88555 − ⇒ T = 2.52o C T + 224.867
6.23 Let H=n-hexane a. n0 ( kmol / min) y0 (kmol H(v)/kmol (1-y0) (kmol N2/kmol) 80oC, 1 atm, 50% rel. sat’n
Condenser
n1 ( kmol / min) 0.05 kmol H(v)/kmol, sat’d 0.95 kmol N2/kmol T (oC), 1 atm
1.50 kmol H(l)/min
50% relative saturation at inlet: y o P = 0.500 p H* (80 o C) Table B.4
yo =
( 0.500)(1068 mmHg) = 0.703 kmolH / kmol 760 mmHg
Saturation at outlet: 0.05 P = p *H (T1 ) ⇒ p H* (T1 ) = 0.05(760 mmHg) = 38 mmHg Antoine equation: log10 38 = 6.88555 −
1175.817 ⇒ T1 = −3.26o C T1 + 224.867
UV RS W T
Mole balance: n 0 = n1 + 150 . n 0 = 2.18 kmol / min ⇒ N 2 balance: (1 − 0.703) n0 = 0.95n1 n1 = 0.682 kmol / min (0.95)0.682 kmol 22.4 m 3 (STP) N2 volume: VN 2 = = 14.5 SCMM min kmol
6-12
6.23 (cont’d) b.
Assume no condensation occurs during the compression
2.18 kmol/min 0.703 H(v) 0.297 N2 80oC, 1 atm
Compressor
V1 (m3 / min) 0.682 kmol/min 0.05 H(v), sat’d 0.95 N2 T1 (oC), 10 atm
V0 ( m 3 / min) 2.18 kmol/min 0.703 H(v) 0.297 N2 T0 (oC), 10 atm, 50% R.S.
Condenser
1.5 kmol H(l)/min
50% relative saturation at condenser inlet: . × 10 4 mmHg 0.500 p H* (T0 ) = 0.703(7600 mmHg) ⇒ p H* (T0 ) = 1068 Saturation at outlet: 0.050(7600 mmHg) = 380 mmHg = p H* (T1 ) Volume ratio:
Antoine
Antoine
T0 = 187 o C T1 = 48.2° C
V1 n1 RT1 / P n1 (T1 + 273.2) 0.682 kmol/min 321 K m3 out = = = × = 0.22 3 V0 n0 RT0 / P n0 (T0 + 273.2) 2.18 kmol/min 460 K m in
c. The cost of cooling to −3.26o C (installed cost of condenser + utilities and other operating costs) vs. the cost of compressing to 10 atm and cooling at 10 atm. 6.24 a.
Maximum mole fraction of nonane achieved if all the liquid evaporates and none escapes. (SG)nonane n max =
15 L C 9 H 20 (l ) 0.718 × 1.00 kg L C 9 H 20
kmol 128.25 kg
= 0.084 kmol C 9 H 20
Assume T = 25o C, P = 1 atm n gas =
2 × 10 4 L 273 K 1 kmol = 0.818 kmol 3 298 K 22.4 × 10 L(STP)
y max =
n max 0.084 kmol C 9 H 20 = = 010 . kmol C 9 H 20 / kmol (10 mole%) 0.818 kmol n gas
As the nonane evaporates, the mole fraction will pass through the explosive range (0.8% to 2.9%). The answer is therefore yes . The nonane will not spread uniformly—it will be high near the sump as long as liquid is present (and low far from the sump). There will always be a region where the mixture is explosive at some time during the evaporation. b. ln p * = −
A +B T
T1 = 258 . o C = 299 K, p1* = 5.00 mmHg T2 = 66.0 o C = 339 K, p2* = 40.0 mmHg
6-13
6.24 (cont’d)
ln(40.0 / 5.00) 5269 5269 ⇒ A = 5269, B = ln(5.00) + = 19.23 ⇒ p * = exp(19.23 − ) 1 1 T ( K) 299 − 339 299 At lower explosion limit, y = 0.008 kmol C 9 H 20 / kmol ⇒ p * ( T ) = yP = (0.008)( 760 mm Hg) −A =
= 6.08 mm Hg
Formula for p*
T = 302 K = 29 o C
c. The purpose of purge is to evaporate and carry out the liquid nonane. Using steam rather than air is to make sure an explosive mixture of nonane and oxygen is never present in the tank. Before anyone goes into the tank, a sample of the contents should be drawn and analyzed for nonane.
6.25 Basis: 24 hours of breathing n0 (mol H2 O) 23°C, 1 atm n1 (mol) @ hr = 10% 0.79 mol N 2/mol y 1 (mol H 2 O/mol) + O2 , CO2
Lungs
O2
Air inhaled: n1 =
37°C, 1 atm n2 (mol), saturated 0.75 mol N 2/mol y 2 (mol H 2 O/mol) + O2 , CO2
CO2
12 breaths 500 ml 1 liter min
3
breath
273K
10 ml
1 mol
b23 + 273gK
60 min 24 hr
b g
22.4 liter STP
1 hr
1 day
= 356 mol inhaled day
Inhaled air - -10% r. h.: y1 = Inhaled air - -50% r. h.: y1 =
b
g = 010 . b2107 . mm Hgg = 2.77 × 10
010 . p∗ H 2O 23° C P
mol H 2 O mol
−2
mol H 2 O mol
760 mm Hg
b
g = 0.50b2107 . mm Hgg = 139 . × 10
0.50 p∗ H 2 O 23° C P
−3
760 mm Hg
H 2 O balance: n0 = n2 y 2 − n1 y1 ⇒ (n0 ) 10% rh − ( n0 ) 50% rh = (n1 y1 ) 50% − (n1 y1 ) 10%
FG H
= 356
mol day
IJ L(0.0139 − 0.00277) mol H O OFG 18.0 g IJ = 71 g / day mol PQH 1 mol K K MN 2
Although the problem does not call for it, we could also calculate that n2 = 375 mol exhaled/day, y2 = 0.0619, and the rate of weight loss by breathing at 23oC and 50% relative humidity is n0 (18) = (n2y2 - n1y1)18 = 329 g/day.
6-14
6.26 a. To increase profits and reduce pollution. b. Assume condensation occurs. A=acetone n 1 mol @ To C, 1 at m y1 mol A(v)/ mol (sat’d) (1-y1 ) mol N2 /mo l
1 mo l @ 90o C, 1 atm 0.20 mol A(v)/ mol 0.80 mol N2 /mo l
n 2 mol A(l)
For cooling water at 20oC
(
)
log10 p*A 20 o C = 7.11714 −
d
(
)
1210.595 = 2.26824 ⇒ p*A 20 o C = 184.6 mmHg 20 + 229.664
i
Saturation: y1 ⋅ P = p *A 20 o C ⇒ y1 =
184.6 = 0.243 > 0.2 , so no saturation occurs. 760
For refrigerant at –35oC
(
)
log10 p*A −35 o C = 7.11714 −
(
)
1210.595 = 0.89824 ⇒ p*A −35 o C = 7.61 mmHg −35 + 229.664
(Note: -35oC is outside the range of validity of the Antoine equation coefficients in Table B.4. An alternative is to look up the vapor pressure of acetone at that temperature in a handbook. The final result is almost identical.) 7.61 = 0.0100 d i 760 N mole balance: 1b0.8g = n b1 − 0.01g ⇒ n = 0.808 mol
Saturation: y1 ⋅ P = p *A −35o C ⇒ y1 = 2
1
1
. mol Total mole balance: 1 = 0.808 + n2 ⇒ n2 = 0192 Percentage acetone recovery: c. d.
0.192 × 100% = 96% 2
Costs of acetone, nitrogen, cooling tower, cooling water and refrigerant The condenser temperature could never be as low as the initial cooling fluid temperature because heat is transferred between the condenser and the surrounding environment. It will lower the percentage acetone recovery.
6-15
6.27
Basis:
12500 L 1 mol 273 K 103000 Pa = 528.5 mol / h h 22.4 L(STP) 293 K 101325 Pa
n o (mol/h) @ 35o C, 103 KPa y0 [mol H2O(v)/mol] H2 O(v)/ mol y1– o mol y0 (mol DA/mol) ) mol DA/mo l (1-y o hr=90% h r =90%
528.5 (mo l/h) @ 20o C, 103 KPa y1 [mol H2O(v)/mol] H2 O(v)/ mol (sat’d) y11–mol DA/mol) y1 (mol (1-y1 ) mol DA/mo l
H2O(l)/h] 2[mol H2O(l)/h n 2nmol
Inlet: yo =
(
hr ⋅ pH* 2O 35 o C P
Outlet: y1 =
p H* 2 O
) = 0.90 × 42.175 mmHg
101325 Pa = 0.04913 mol H 2 O/mol 760 mmHg
103000 Pa
d20 Ci = 17.535 mmHg 101325 Pa = 0.02270 mol H O / mol o
2
103000 Pa 760 mmHg P Dry air balance: 1 − 0.04913 no = 1 − 0.02270 528.5 ⇒ no = 543.2 mol / h
b
g b
gb
g
543.2 mol 22.4 L(STP) 308 K 101325 Pa = 13500 L / h h mol 273 K 103000 Pa Total balance: 543.2 = 528.5 + n2 ⇒ n2 = 14.7 mol / h
Inlet air:
14.7 mol 18.02 g H 2 O 1 kg = 0.265 kg / h h 1 mol H 2 O 1000 g
Condensation rate:
6.28
Basis:
10000 ft 3 1 lb - mol 492 o R 29.8 in Hg = 24.82 lb - mol / min min 359 ft 3 (STP) 550 o R 29.92 in Hg
n1 lb-mole/min 40oF, 29.8 in.Hg y1 [lb-mole H2O(v)/lb-mole] 1- y1 (lb-mole DA/mol)
24.82 lb-mole/min 90oF, 29.8 in.Hg y0 [lb-mole H2O(v)/mol 1- y0 (lb-mole DA/mol) hr = 88%
n1 lb-mole/min 65oF, 29.8 in.Hg y1 [lb-mole H2O(v)/lb-mole] 1- y1 (lb-mole DA/lb-mole)
n2 [lb-mole H2O(l)/min]
Inlet: y o =
d
i = 0.88b36.07 mmHgg
hr ⋅ p H* 2 O 90 o F
Outlet: y1 =
P
29.8 in Hg
d
i = 6.274 mmHg
p H* 2 O 40 o F P
29.8 in Hg
b
1 in Hg = 0.0419 lb - mol H 2 O / lb - mol 25.4 mmHg
1 in Hg = 0.00829 lb - mol H 2 O / lb - mol 25.4 mmHg
g b
g
Dry air balance: 24.82 1 − 0.0419 = n1 1 − 0.00829 ⇒ n1 = 23.98 lb - mol / min Total balance: 24.82 = 23.98 + n2 ⇒ n2 = 0.84 lb - mole / min
6-16
6.28 (cont’d)
0.84 lb - mol 18.02 lb m 1 ft 3 7.48 gal = 181 . gal / min min lb − mol 62.4 lb m 1 ft 3
Condensation rate:
Air delivered @ 65oF:
23.98 lb - mol 359 ft 3 (STP) 525o R 29.92 in Hg = 9223 ft 3 / min o min 1 lb − mol 492 R 29.8 in Hg
6.29 Basis: 100 mol product gas
no mol, 32oC, 1 atm yo mol H2O(v)/mol (1-yo) mol DA/mol hr=70%
100 mol, T1, 1 atm
100 mol, 25oC,1 atm
y1 mol H2O(v)/mol, (sat’d) (1-y1) mol DA/mol
y1 mol H2O(v)/mol, (1-y1) mol DA/mol hr=55%
(mol HH22O(l)) O(l)/min nn22lb-mol
Outlet: y1 =
d
hr ⋅ p H* 2 O 25o C
i = 0.55b23.756g = 0.0172 mol H O / mol 2
760
P
b g b g d32 Ci = 0.70b35.663g = 0.0328 mol H O / mol
Saturation at T1 : 0.0172 760 = 13.07 = p H* 2 O T1 ⇒ T1 = 15.3o C Inlet: y o =
hr ⋅
o
p H* 2 O
P
2
760
b
g
b
g
Dry air balance: no 1 − 0.0328 = 100 1 − 0.0172 ⇒ no = 1016 . mol Total balance: 1016 . + n2 = 100.0 ⇒ n2 = −1.6 mol (i.e. removed) kg H 2 O removed :
kg dry air:
Ratio:
b
16 . mol 18.02 g 1 kg = 0.0288 kg H 2 O 1 mol 1000 g
g
100 1 − 0.0172 mol 29.0 g 1 kg = 2.85 kg dry air 1 mol 1000 g
0.0288 = 0.0101 kg H 2 O removed / kg dry air 2.85
6-17
6.30 a.
Room air − T = 22° C , P = 1 atm , hr = 40% :
y1 P = 0.40 p ∗H2O ( 22°C ) ⇒ y1 =
( 0.40 )19.827 mm Hg = 0.01044 mol H O 2
760 mm Hg
mol
Second sample − T = 50° C , P = 839 mm Hg , saturated: y2 P = p ∗H2 O ( 50°C ) ⇒ y2 =
92.51 mm Hg = 0.1103 mol H 2 O mol 839 mm Hg
. , H 2 = 48 ln y = bH + ln a ⇔ y = ae bH , y1 = 0.01044, H1 = 5 , y 2 = 01103
b=
b
ln y 2 y1
0.01044g . g = lnb01103 = 0.054827
H 2 − H1
48 − 5
b
g b gb g expb0.054827 H g
b
g
ln a = ln y1 − bH1 = ln 0.01044 − 0.054827 5 = −4.8362 ⇒ a = exp −4.8362 = 7.937 × 10 −3 ⇒ y = 7.937 × 10 −3 b.
Basis:
1 m 3 delivered air
b
273K 1 k mol 22 + 273 K 22.4m 3 STP
g
b g
10 3 mol = 4131 . mol air delivered 1 kmol o 41.31 mol, 41.4 mol, 22o22 C,1C, at 1matm
41.31 mol, T, 1 at m n o mol, 35o C, 1 at m
moll H22O(v)/mo O(v)/mol, 0.0104 mo l, (sat’d) sat’d mo l DA/ mol 0.09896 0.9896 mol DA/mol
yo mol H2 O(v)/ mol (1-yo ) mol DA/mo l H=30
l 0.0104 l H2HO(v)/mo 0.0104momol 2O(v)/mol 0.09896 mol 0.9896 mo moll DA/ DA/mol
n 1 mol H2 O(l)
Saturation condition prior to reheat stage: yH2 O P = pH* 2 O (T ) ⇒ ( 0.01044 )( 760 mm Hg ) = 7.93 mm Hg ⇒ T = 7.8°C (from Table B.3)
bg
Part a
Humidity of outside air: H = 30 ⇒ y 0 = 0.0411 mol H 2 O mol Overall dry air balance: n0 (1 − y0 ) = 41.31( 0.9896 ) ⇒ n0 =
( 41.31)( 0.9896 ) = 42.63 mol (1 − 0.0411)
Overall water balance: n0 y0 = n1 + ( 41.31)( 0.0104 ) ⇒ n1 = ( 42.63)( 0.0411) − ( 41.31)( 0.0104 ) = 1.32 mol H 2 O condensed
Mass of condensed water =
1.32 mol H 2 O 18.02 g H 2 O 1 mol H 2 O
1 kg 10 3 g
= 0.024 kg H 2 O condensed m 3 air delivered
6-18
6.31 a.
Basis: n 0 mol feed gas . S = solvent , G = solvent - free gas n1 (mol) @ Tf (°C), P4 (mm Hg) y1 [mol S(v)/mol] (sat’d) (1–y1) (mol G/mol)
n0 (mol) @ T0 (°C), P0 (mm Hg) y0 (mol S/mol) (1-y0) (mol G/mol) Td0 (°C) (dew point)
n2 (mol S (l))
b g
Inlet dew point = T0 ⇒ y o Po = p∗ Tdo ⇒ y o =
b g
p∗ Tdo Po
d i
Saturation condition at outlet: y1 Pf = p∗ T f ⇒ y1 =
Fractional condensation of S = f ⇒ n2 = n0 y 0 f Total mole balance: n 0 = n1 + n2 ⇒ n1 = n 0 − n2
(1)
d i
p∗ T f
(2)
Pf
b gP n fp∗ bT g −
(1) ⎯⎯ → n2 = n0 fp∗ T0
bg
Eq. 3 for n1
⇒
n1 = n 0
b gb g
0
0
do
Po
S balance: n0 y 0 = n1 y1 + n2 (1) - (4)
b g = LMn P MN
n 0 p∗ Tdo
0
−
o
⇒
b g OPFG p∗ dT iIJ + n fp∗ bT g P PQGH P JK p
n 0 fp∗ Tdo
f
o
f
do
f
do
o
do
o
b1 − f g p∗ bT g = LM1 − fp∗ bT g OP p∗ dT i P MN P PQ P o
0
LM fp∗ eTdo j OP p∗ d T i 1 − MM Po PP Q N = p∗ T b1 − f g eP do j f
⇒ Pf
f
o
b.
Condensation of ethylbenzene fromnitrogen Antoine constants for ethylbenzene A= 6.9565 B= 1423.5 C= 213.09 Run T0 P0 Td0 f 1 2 3 4
50 50 50 50
765 765 765 765
40 40 40 40
0.95 0.95 0.95 0.95
Tf
p* (Td0) p*(Tf)
45 40 35 20
21.472 21.472 21.472 21.472
6-19
27.60 21.47 16.54 7.07
Pf 19139 14892 11471 4902
Crefr Ccomp Ctot 2675 4700 8075 26300
107027 109702 83329 88029 64239 72314 27582 53882
(3)
(4)
6.31 (cont’d) When Tf decreases, Pf decreases. Decreasing temperature and increasing pressure both to c. increase the fractional condensation. When you decrease Tf, less compression is required to achieve a specified fractional condensation. d.
6.32 a.
A lower Tf requires more refrigeration and therefore a greater refrigeration cost (Crefr). However, since less compression is required at the lower temperature, Ccomp is lower at the lower temperature. Similarly, running at a higher Tf lowers the refrigeration cost but raises the compression cost. The sum of the two costs is a minimum at an intermediate temperature. Basis : 120 m 3 min feed @ 1000 o C(1273K), 35 atm . Use Kay’s rule.
b g P batmg bT g
Cmpd. Tc K
c
c corr .
bP g
c corr
H2
33.2
12.8
41.3
20.8
CO CO 2
133.0 304.2
34.5 72.9
− −
− −
CH 4
190.7
. 458
−
−
∑y T P′ = ∑ y P
Tc′ =
i ci
c
i
ci
bApply Newton' s corrections for H g 2
b g b g b g b g = 0.40b20.8g + 0.35b34.5g + 0.20b72.9g + 0.05b458 . g = 37.3 atm
. + 0.35 133.0 + 0.20 304.2 + 0.05 190.7 = 133.4 K = 0.40 413
Feed gas to cooler Tr = 1273 K 133.4 K = 9.54
⎫ Generalized compressibility charts (Fig. 5.4-3) ⎬ Pr = 35.0 atm 37.3 atm = 0.94 ⎭ ⇒ z = 1.02
1.02 V = 35 atm
8.314 N ⋅ m mol ⋅ K
120 m 3
m ol 3.04 × 10
m in
−3
m
3
1273 K
1 km ol 10 3 m ol
1 atm = 3.04 × 10 − 3 m 3 mol 101325 N m 3
= 39 .5 km ol m in
Feed gas to absorber Tr = 283 K 133.4 K = 2.12
⎫ Generalized compressibility charts (Fig. 5.4-3) ⎬ Pr = 35.0 atm 37.3 atm = 0.94 ⎭ ⇒ z = 0.98
Vˆ =
0.98
8.314 N ⋅ m mol ⋅ K
35 atm
V=
39.5 kmol 103 mol min
1 kmol
283 K
1 atm 101325 N m
6.50 × 10-4 m3 mol
6-20
3
= 25.7
= 6.50 × 10−4 m 3 mol m3 min
6.32 (cont’d)
1.2(39.5) kmol/min
n1 (kmol/min), 261 K, 35 atm
39.5 kmol/min, 283K, 35 atm
yMeOH sat’d yH2 yCH4 (2% of feed) yCO
0.40 mol H2/mol 0.35 mol CO/mol 0.20 mol CO2/mol 0.05 mol CH4/mol
n2 (kmol/min), liquid xMeOH xCO2 xCH4 (98% of feed)
Saturation at Outlet: y McOH =
b
p∗ MeOH 261K P
= 4.97 × 10 y McOH =
n MeOH
n MeOH + n H 2 + nCH 4
A
= input
b.
A
= 0.02 of input
g = 10
b
7 .87863−1473.11 −12 + 2300
b
g
mm Hg
g
35 atm 760 mm Hg atm
−4
+ nCO
mol MeOH mol =
A = input
n MeOH
n MeOH + 39.5(0.40 + 0.02 ( 0.05) + 0.35)
E
n MeOH = 0.0148 kmol min MeOH in gas
The gas may be used as a fuel. CO2 has no fuel value, so that the cost of the added energy required to pump it would be wasted.
6-21
6.33 n0 (kmol/min wet air) @ 28°C, 760 mmHg
y1 (mol H2O/mol) (1-y1) (mol dry air/mol) 50% rel. sat.
n1 (kmol/min wet air) @ 80°C, 770
y2 (mol H2O/mol) (1-y2) (mol dry air/mol) Tdew point = 40.0oC
1500 kg/min wet pulp
m 1 (kg/min wet pulp)
0.75 /(1 + 0.75) kg H2O/kg 1/1.75 kg dry pulp/kg
0.0015 kg H2O/kg 0.9985 kg dry pulp/kg
Dry pulp balance: 1500 ×
1 = m 1 (1 − 0.0015) ⇒ m 1 = 858 kg / min 1 + 0.75
50% rel. sat’n at inlet: y1 P = 0.50 pH* 2O (28o C) ⇒ y1 = 0.50(28.349 mm Hg)/(760 mm Hg) = 0.0187 mol H 2 O/mol 40 C dew point at outlet: y 2 P = o
p H* 2 O (40 o C)
⇒ y 2 = (55.324 mm Hg) / (770 mm Hg) = 0.0718 mol H 2 O / mol
Mass balance on dry air: n 0 (1 − 0.0187) = n1 (1 − 0.0718)
(1)
Mass balance on water: n 0 ( 0.0187 )(18.0 kg / kmol ) + 1500 ( 0.75 / 1.75) = n 1 ( 0.0718 )(18 ) + 858 ( 0.0015) ( 2 ) Solve (1) and (2) ⇒ n 0 = 622.8 kmol / min, n1 = 658.4 kmol / min Mass of water removed from pulp: [1500(0.75/1.75)–858(.0015)]kg H2O = 642 kg / min 622.8 kmol 22.4 m 3 (STP) (273 + 28) K = 1.538 × 10 4 m 3 / min Air feed rate: V0 = min kmol 273 K
6-22
6.34
Basis: 500 lb m hr dried leather (L) n1 (lb - moles / h)@130o F, 1 atm n0 (lb - moles dry air / h)@140o F, 1 atm
y1 (lb - moles H2 O / lb - mole) (1- y1 )(lb - moles dry air / lb - mole)
0 (lb m / h) m 0.61 lb m H2 O(l) / lb m
500 lb m / h 0.06 lb m H2 O(l) / lbm 0.94 lb m L / lb
0.39 lb m L / lb m
b gb g
Dry leather balance: 0.39m0 = 0.94 500 ⇒ m0 = 1205 lb m wet leather hr
b
g
Humidity of outlet air: y1 P = 0.50 p∗ H 2 O 130° F ⇒ y1 =
b gb
g
a
H 2 O balance: 0.61 1205 lb m hr = ( 0.06 ) 500 lb
b
E
m
mol H 2 O 0.50(115 mm Hg) = 0.0756 760 mmHg mol
fb
hr +
g
0.0756n1 lb - moles H 2 O 18.02 lb m hr 1 lb - mole
n1 = 517.5 lb - moles hr
g 359 ft bSTPg b140 + 460g° R = 2.09 × 10
Dry air balance: n0 = 1 − 0.0756 (517.5) lb - moles hr = 478.4 lb - moles hr Vinlet =
6.35 a.
478.4 lb - moles hr
3
5
492° R
1 lb - mole
ft 3 hr
Basis: 1 kg dry solids n1 (kmol)N 2, 85°C
n2 (kmol) 80°C, 1 atm y 2 (mol Hex/mol) (1 – y 2) (mols N 2 /mol) 70% rel. sat.
dryer 1.00 kg solids 0.78 kg Hex
condenser
n3 (kmol) 28°C, 5.0 atm y 3 (mol Hex/mol) sat'd (1 – y 3) (mols N 2 /mol) n4 (kmol) Hex(l)
0.05 kg Hex 1.00 kg solids
Mol Hex in gas at 80°C:
b0.78 − 0.05gkg
kmol = 8.47 × 10 −3 kmol Hex 86.17 kg
Antoine eq.
↓
70% rel. sat.: y2 =
0.70 p ∗hex ( 80°C ) P
=
( 0.70 )106.88555 − 1175.817 (80+ 224.867 ) 760
6-23
= 0.984 mol Hex mol
6.35 (cont’d)
n2 =
8.47 × 10 −3 kmol Hex
1 kmol = 0.0086 kmol 0.984 kmol Hex
b
g
N 2 balance on dryer: n1 = 1 − 0.984 0.0086 = 1376 . × 10 −4 kmol Antoine Eq.
↓
Saturation at outlet: y3 =
p ∗hex ( 28°C ) P
=
10
6.88555 − 1175.817 ( 28 + 224.867 )
b
5 ( 760 )
= 0.0452 mol Hex mol
g
Overall N 2 balance: 1.376 × 10 -4 = n3 1 − 0.0452 ⇒ n3 = 144 . × 10 −4 kmol Mole balance on condenser: 0.0086 = 144 . × 10 −4 + n4 ⇒ n4 = 0.0085 kmol Fractional hexane recovery:
0.0085 kmol cond. 86.17 kg = 0.939 kg cond. kg feed 0.78 kg feed kmol
b. Basis: 1 kg dry solids
0.9n
3
heater
0.9n 3 (kmol) @ 28°C, 5.0 atm
y3 (1 – y3)
n 1 (kmol)N 2 85°C dryer 1.00 kg solids 0.78 kg Hex
y 3 (mol Hex/mol) sat'd (1 – y 3) (mol N 2/mol)
n 2 (kmol) 80°C, 1 atm y 2 (mol Hex/mol) (1 – y 2) (mols N2 /mol) 70% rel. sat.
condenser
n3 (kmol) y3 (1 – y 3)
0.1n3
n4 (kmol) Hex(l)
0.05 kg Hex 1.00 kg solids
Mol Hex in gas at 80°C: 8.47x10-3 + 0.9n3(0.0452) = n2(0.984)
(1)
N2 balance on dryer: n1 + 0.9n3 (1 − 0.0452) = n2 (1 − 0.984)
( 2)
Overall N2 balance: n1 = 0.1n3 (1 − 0.0452)
(3)
⎧n1 = 1.38 × 10 kmol ⎪ Equations (1) to (3) ⇒ ⎨n2 = 0.00861 kmol ⎪ −4 ⎩n3 = 1.44 × 10 kmol 1.376 × 10-4 − 1.38 × 10−5 Saved fraction of nitrogen= × 100% = 90% 1.376 × 10−4 −5
Introducing the recycle leads to added costs for pumping (compression) and heating.
6-24
6.36 b. m 1 (lbm/h)
300 lbm/h wet product 0.2 = 0167 . lb m T(l) / lb m 1 + 0.2 0.833 lb m D / lb m
0.02 / (102 . ) = 0.0196 lb m T(l) / 0.9804 lb m D / lb m
Dryer
@ 200OF, y3 (lb-mole T/lb-mole) (1-y3)( lb-mole N2/lb-mole) n3 (lb-mole/h)
n1 (lb-mole/h)
y1 (lb-mole T(v)/lb-mole) (1–y1) (lb-mole N2/lbmole) T=toluene 70% r.s.,150oF, 1.2 atm D=dry solids
Heater
n3 (lb-mole/h)
y3 (lb-mole T(v)/lb-mole) (1-y3) (lb-mole N2/lb-mole)
Condenser
Eq.@ 90OF, 1atm
n2 ( lb-mole T(l)/h
Strategy: Overall balance⇒ m 1 & n 2 ; Relative saturation⇒y1;, Gas and liquid equilibrium⇒y3 Balance over the condenser⇒ n1 & n 3
UV RS W T
Toluene Balance: 300 × 0167 . = m 1 × 0.0196 + n 2 × 92.13 m 1 = 255 lb m / h ⇒ Dry Solids Balance: 300 × 0.833 = m 1 × 0.9804 n 2 = 0.488 lb - mole / h 70% relative saturation of dryer outlet gas:
pC* 7 H8 (150O F=65.56O C)=10
(6.95805−
y1 P = 0.70 pC* 7 H8 (150 O F) ⇒ y1 =
1346.773 ) 65.56 + 219.693
0.70 pC* 7 H8 P
=
= 172.47 mmHg
(0.70)(172.47) = 01324 . lb - mole T(v) / lb - mole 12 . × 760
Saturation at condenser outlet: pC* 7 H8 (90O F=32.22O C)=10 y3 =
pC* 7 H8 P
=
(6.95805−
1346.773 ) 65.56 + 219.693
= 40.90 mmHg
40.90 = 0.0538 mol T(v)/mol 760
UV RS W T
Condenser Toluene Balance: n1 × 01324 . = 0.488 + n 3 × 0.0538 n1 = 5.875 lb - mole / h ⇒ . ) = n 3 × (1 − 0.0538) Condenser N 2 Balance: n1 × (1 − 01324 n 3 = 5.387 lb - mole / h
6-25
6.36 (cont’d) Circulation rate of dry nitrogen = 5.875 × (1 - 0.1324) =
5.097 lb - mole
lb - mole
h
28.02 lb m
= 0182 . lb m / h
Vinlet =
6.37
b g
5.387 lb - moles 359 ft 3 STP hr
(200 + 460)° R 492° R
1 lb - mole
C 6 H 14 +
Basis: 100 mol C 6 H 14
= 2590 ft 3 h
19 O 2 → 6CO 2 + 7H 2 O 2
100 mol C 6H 14
n1 (mol) dry gas, 1 atm 0.821 mol N 2/mol D.G. 0.069 mol CO2/mol D.G. 0.069 mol CO 2 /mol D.G. 0.021 mol CO/mol D.G. 0.021 mol CO/mol D.G. 0.00265 mol C6H14/mol 0.086 molOO/mol) 2/mol D.G. x (mol 2 0.00265 mol C 6HN14/mol D.G. (0.907–x) (mol 2/mol) n2n (mol H O) (mol H2 O)
n0 (mol) air 0.21 mol O 2/mol 0.79 mol N 2/mol
L O C balance: 6b100g = n M0.069 + 0.021+ 6b0.00265gP ⇒ n = 5666 mol dry gas MN b g b g b g PQ 100 − 0.00265b5666g mol reacted Conversion: × 100% = 85.0% 100 mol fed H balance: 14b100g = 2n + 5666b14gb0.00265g ⇒ n = 595 mol H O p∗ dT i 595 Dew point: y = = ⇒ p∗ dT i = 72.2 mm Hg ⇒ 595 + 5666 760 mm Hg 2
1
CO 2
CO
2
1
C 6 H 14
2
2
2
Table B.3
dp
dp
H 2O
N 2 balance: 0.79n0 = 5666(0.907 − x) O balance: 0.21(n0 )(2) = 5666[(0.069)(2) + 0.021 + 2 x) + 595
Solve simultaneously to obtain n0 = 5888 mol air, x = 0.086 mol O2/mol Theoretical air:
Excess air:
100 mol C 2 H 14
19 mol O 2 2 mol C 2 H 14
1 mol air = 4524 mol air 0.21 mol O 2
5888 − 4524 × 100% = 30.2% excess air 4524
6-26
. °C Tdp = 451
6.38 Basis: 1 mol outlet gas/min n 0 ( mol / min) y 0 ( mol CH 4 / mol) (1 − y 0 ( mol C 2 H 6 / mol)
1 mol / min @ 573K, 105 kPa y1 (mol CO 2 / mol)
n1 (mol O 2 / min)
y 2 (mol H 2 O / mol) (1 − y1 − y 2 ) mol N 2 / mol
3.76n1 (mol N 2 / min)
CH 4 + 2O 2 → CO 2 + 2H 2 O
pCO 2 = 80 mmHg ⇒ y1 =
C2 H 6 +
7 O 2 → 2CO 2 + 3H 2 O 2
80 mmHg 101325 Pa = 01016 mol CO 2 / mol . 105000 Pa 760 mmHg
b
g
100% O2 conversion : 2no yo + 7 no 1 − yo = n1
(1)
. C balance: no yo + 2no 1 − yo = 01016
(2)
. n1 = 1 − y1 − y2 N2 balance: 376
(3)
H balance: 4no yo + 6no 1 − yo = 2 y2
(4)
b
2
g
b
g
R|n = 0.0770 mol | y = 0.6924 mol CH / mol Solve equations 1 to 4 ⇒ S . mol O ||n = 01912 . mol H O / mol Ty = 01793 Dew point: 01793 . b105000g Pa 760 mmHg = 141.2 mmHg ⇒ T p dT i = o
4
o
* H2 O
1
2
2
2
dp
dp
101325 Pa
b
6.39 Basis: 100 mol dry stack gas n P (mol C 3 H 8) n B (mol C 4H10 ) n out (mol) 0.21 O2 0.79 N2
P = 780 mm Hg Stack gas: Tdp = 46.5°C 100 mol dry gas 0.000527 mol C 3 H 8/mol 0.000527 mol C 4H 10/mol 0.0148 mol CO/mol 0.0712 mol CO 2/mol + O 2, N 2
nw (mol H2O)
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
C 4 H 10 +
6-27
g
= 58.8 o C Table B.3
13 O 2 → 4CO 2 + 5H 2 O 2
6.39 (cont’d)
b
g
Dew point = 46.5° C ⇒ y w P = p∗ w 46.5° C ⇒ y w = But yw =
mol H 2 O 77.6 mm Hg = 0.0995 780 mm Hg mol
nw = 0.0995 ⇒ nw = 11.05 mol H 2 O (Rounding off strongly affects the result) 100 + nw
b gb
gb g b
gb g
C balance: 3n p + 4n B = 100 0.000527 3 + 0.000527 4 + 0.0148 + 0.0712 ⇒
b1g
3n p + 4n B = 8.969
H balance: 8n p + 10nB = (100 ) ⎡⎣( 0.000527 )( 8 ) + ( 0.000527 )(10 ) ⎤⎦ + (11.05 )( 2 )
( 2)
⇒ 8n p + 10nB = 23.047
⎧ 49% C3 H8 ⎪⎧n p = 1.25 mol C3 H8 ⎪⎫ ⎪ Solve (1) & ( 2 ) simultaneously: ⇒ ⎨ ⎬ ⇒⎨ ⎪⎩nB = 1.30 mol C4 H10 ⎪⎭ ⎪⎩ 51% C 4 H10 ( Answers may vary ± 8% due to loss of precision )
6.40 a.
L1 (lb - mole C 10 H 22 / h)
L 2 (lb - mole / h) x 2 (lb - mole C 3 H 8 / lb - mole) 1 − x 2 (lb - mole C 10 H 22 / lb - mole)
G 1 (lb - mole / h) y1 (lb - mole C 3 H 8 / lb - mole) 1 − y1 (lb - mole N 2 / lb - mole)
G 2 = 1 lb - mole / h 0.07 (lb - mole C 3 H 8 / lb - mole) 0.93 (lb - mole N 2 / lb - mole)
Basis: G 2 = 1 lb - mole h feed gas
b gb g
b
g
b
g = b1 − 0.985gb1gb0.07g ⇒ G y
b1g b2 g
N 2 balance: 1 0.93 = G 1 1 − y1 ⇒ G 1 1 − y1 = 0.93 98.5% propane absorption ⇒ G 1 y1
b1g & b2g ⇒ G
1
1 1
= 1.05 × 10 −3
= 0.93105 lb - mol h , y1 = 1128 × 10 −3 mol C 3 H 8 mol .
Assume G 2 − L 2 streams are in equilibrium
From Cox Chart (Figure 6.1-4), p *C3 H8 (80 o F ) = 160 lb / in 2 = 10.89 atm
b
g
Raoult' s law: x 2 p∗ C3H 8 80° F = 0.07 p ⇒ x 2 =
b gb g
b0.07gb10. atmg = 0.006428 mol H O 2
10.89 atm
Propane balance: 0.07 1 = G 1 y1 + L 2 x 2 ⇒ L 2 =
b
gd
mol
0.07 − 0.93105 1128 × 10 −3 .
i
0.006428
= 10.726 lb - mole h Decane balance: L1 = 1 − x 2 L 2 = 1 − 0.006428 10.726 = 10.66 lb - mole h
b
⇒
gd h b d L / G h 1
2 min
gb
g
= 10.7 mol liquid feed / mol gas feed
6-28
6.40 (cont’d) b. The flow rate of propane in the exiting liquid must be the same as in Part (a) [same feed rate and fractional absorption], or n C3H 8 =
10.726 lb - mole 0.006428 lb - mole C 3 H 3 h
lb - mole
= 0.06895 lb - mol C 3 H 8 h
The decane flow rate is 1.2 x 10.66 = 12.8 lb-moles C10H22/h ⇒ x2 =
b
0.06895 lb - mole C 3 H 8 h = 0.00536 lb - mole C 3 H 8 / lb - mole 0.06895 + 12.8 lb - moles h
g
c. Increasing the liquid/gas feed ratio from the minimum value decreases the size (and hence the cost) of the column, but increases the raw material (decane) and pumping costs. All three costs would have to be determined as a function of the feed ratio. 6.41 a. Basis: 100 mol/s liquid feed stream
Let B = n - butane , HC = other hydrocarbons n4 (mol/s) @ 30°C, 1 atm y4 (mol B/mol) (1-y4) (mol N2/mol)
100 mol/s @ 30oC, 1 atm xB =12.5 mol B/s 87.5 mol other hydrocarbon/s
n3 (mol N2/s)
88.125 mol/s 0.625 mol B/s (5% of B fed) 87.5 mol HC/s
p *B (30 o C) ≅ 41 lb / in 2 = 2120 mm Hg (from Figure 6.1-4)
x B p *B (30 o C) 0125 . × 2120 = = 0.3487 P 760 95% n-butane stripped: n 4 ⋅ 0.3487 = 12.5 0.95 ⇒ n 4 = 34.06 mol / s Total mole balance: 100 + n3 = 34.06 + 88.125 ⇒ n3 = 22.18 mol/s Raoult' s law: y 4 P = x B p B* ( 30 o C) ⇒ y 4 =
b
⇒
g b gb g
mol gas fed 22.18 mol/s = = 0.222 mol gas fed/mol liquid fed mol liquid fed 100 mol/s
b. If y 4 = 0.8 × 0.3487 = 0.2790 , following the same steps as in Part (a),
b
g b gb g
95% n-butane is stripped: n 4 ⋅ 0.2790 = 12.5 0.95 ⇒ n 4 = 42.56 mol / s Total mole balance: 100 + n 3 = 42.56 + 88125 . ⇒ n 3 = 30.68 mol / s mol gas fed 30.68 mol/s = = 0.307 mol gas fed/mol liquid fed ⇒ mol liquid fed 100 mol/s c. When the N2 feed rate is at the minimum value calculated in (a), the required column length is infinite and hence so is the column cost. As the N2 feed rate increases for a given liquid feed rate, the column size and cost decrease but the cost of purchasing and compressing (pumping) the N2 increases. To determine the optimum gas/liquid feed ratio, you would need to know how the column size and cost and the N2 purchase and compression costs depend on the N2 feed rate and find the rate at which the cost is a minimum.
6-29
6.42 Basis: 100 mol NH 3 Preheated air 100 mol NH 3 780 kPa sat'd
N2 O2
converter
n3 n4 n5 n6
n 1 (mol) O2 3.76 n 1 (mol) N 2 n 2 (mol) H2 O 1 atm, 30°C h r = 0.5
a. i)
absorber
(mol NO) (mol N 2) (mol O2 ) (mol H 2O)
55 wt% HNO 3 (aq ) n 8 (mol HNO 3 ) n 9 (mol H 2O)
n7 (mol H 2O)
b g
NH 3 feed: P = P∗ Tsat = 820 kPa = 6150 mm Hg = 8.09 atm Antoine: log 10 6150 = 7.55466 − 1002.711 Tsat + 247.885 ⇒ Tsat = 18.4° C = 291.6 K
b
g
b
Table B.1 ⇒ VNH 3 =
g
UV W
. atm ⇒ Pr = 8.09 / 1113 . = 0.073 Pc = 1113 ⇒ z = 0.92 . K ⇒ Tr = 2916 . / 4055 . = 0.72 Tc = 4055
b
0.92 100 mol
g
(Fig. 5.3-1)
8.314 Pa 2916 . K = 0.272 m 3 NH 3 3 mol - K 820 × 10 Pa
Air feed: NH 3 + 2O 2 → HNO 3 + H 2 O n1 =
100 mol NH 3
Water in Air: y H 2 O =
b
2 mol O 2 = 200 mol O 2 mol NH 3
hr ⋅ p * 30° C p
⇒ 0.02094 =
g = 0.500 × 31824 . = 0.02094 760 n2 4.76( 200 )
n2 +
⇒ n2 = 20.36 mol H 2 O
A
( 4 .76 mol air mol O 2 )
Vair =
b g
b g
4.76 200 + 20.36 mol 22.4 L STP 1 mol
303K
1 m3
273K 10 3 L
= 24.2 m 3 air
ii) Reactions: 4 NH 3 + 5O 2 → 4 NO + 6H 2 O , 4 NH 3 + 3O 2 → 2 N 2 + 6H 2 O
Balances on converter NO: n3 =
97 mol NH 3
4 mol NO = 97 mol NO 4 mol NH 3
6-30
6.42 (cont’d)
b g
N 2 : n4 = 3.76 2.00 mol + O 2 : n5 = 200 mol −
3 mol NH 3
2 mol N 2
4 mol NH 3 5 mol O 2
97 mol NH 3
= 753.5 mol N 2
4 mol NH 3 −
3 mol NH 3
3 mol O 2 4 mol NH 3
H 2 O: n6 = 20.36 mol +
= 76.5 mol O 2
6 mol H 2 O
100 mol NH 3
4 mol NH 3
= 170.4 mol H 2 O
⇒ n total = (97 + 7535 . + 76.5 + 170.4) mol = 1097 mol converter effluent 8.8% NO, 68.7% N 2 , 7.0% O 2 , 15.5% H 2 O iii) Reaction: 4 NO + 3O 2 + 2 H 2 O → 4 HNO 3 HNO 3 bal. in absorber: n8 =
H 2 O in product: n9 =
97 mol NO react 4 mol HNO 3 4 mol NO
97 mol HNO 3
63.02 g HNO 3 mol
= 277.56 mol H 2 O
b
gb g
b
= 97 mol HNO 3
1 mol H 2 O 45 g H 2 O 55 g HNO 3 18.02 g H 2 O
gb gb
H balance on absorber: 170.4 2 + 2n7 = 97 + 277.6 2 mol H
g
⇒ n7 = 155.7 mol H 2 O added VH 2 O = b.
bg
155.7 mol H 2 O 18.02 g H 2 O 1 cm 3 1 m3 = 2.81 × 10 −3 m 3 H 2 O l 6 3 1 mol 1 g 10 cm
M acid in old basis =
97 mol HNO3
63.02 g HNO3 277.6 mol H 2 O 18.02 g H 2 O + mol mol
= 11115 g = 11.115 kg Scale factor =
b1000 metric tonsgb1000 kg metric tong = 8.997 × 10 11.115 kg
d
id i = d8.997 × 10 id24.2 m air i = 2.18 × 10 m air = d8.997 × 10 id2.81 × 10 m H Oi = 253 m H Obl g
VNH 3 = 8.997 × 10 4 0.272 m 3 NH 3 = 2.45 × 10 4 m 3 NH 3 Vair
VH 2O
3
4
4
6
−3
3
3
3
2
6-31
2
4
6.43 a.
Basis: 100 mol feed gas 100 mol 0.10 mol NH3 /mol 0.90 mol G/mol
G = NH3 -free gas Absorber
n 1 (mol H2 O( l))
n 2 (mol) in equilibrium y A (mol NH 3 /mol) at 10°C(50°F) y W (mol H 2O/mol) and 1 atm y G (mol G/mol) n 3 (mol) x A (mol NH 3 /mol) (1 – x A) (mol H 2O/mol)
Composition of liquid effluent . Basis: 100 g solution Perry, Table 2.32, p. 2-99: T = 10oC (50oF), ρ = 0.9534 g/mL ⇒ 0.120 g NH3/g solution
⇒
12.0 g NH 3 88.0 g H 2 O = 4.89 mol NH 3 = 0.706 mol NH 3 , (18.0 g / 1 mol) (17.0 g / 1 mol)
⇒ 12.6 mole% NH 3 (aq), 87.4 mole% H 2 O(l) Composition of gas effluent
g U| . = 0155 psiabTable 2 - 21gV |W = 14.7 psia b
. psia Table 2 - 23 p NH 3 = 121 . ⎯ ⎯⎯→ T = 50 F, x A = 0126 Perry
o
pH 2O p total
. / 14.7 = 0.0823 mol NH 3 mol y A = 121 . / 14.7 = 0.0105 mol H 2 O mol ⇒ yW = 0155 y G = 1 − y A − yW = 0.907 mol G mol
b gb g b gb g b0.907g = 99.2 mol absorbed = b100gb010 . g − b99.2gb0.0823g = 184 . mol NH
G balance: 100 0.90 = n2 y G ⇒ n2 = 100 0.90
NH 3
in
% absorption =
3
out
1.84 mol absorbed × 100% = 18.4% . mol fed 100 010
b gb g
b. If the slip stream or densitometer temperature were higher than the temperature in the contactor, dissolved ammonia would come out of solution and the calculated solution composition would be in error. 6.44 a.
15% oleum: Basis - 100kg 15 kg SO 3 +
85 kg H 2 SO 4
1 kmol H 2 SO 4 98.08 kg H 2 SO 4
⇒ 84.4% SO 3
6-32
1 kmol SO 3 1 kmol H 2 SO 4
80.07 kg SO 3 = 84.4 kg 1 kmol SO 3
6.44 (cont’d) b.
Basis 1 kg liquid feed n o (mol), 40o C, 1.2 at m
n 1 (mol), 40o C, 1.2 at m
0.90 mol SO3 /mol 0.10 mol G/ mol
y1 mol SO3 /mo l (1-y1 ) mol G/ mol Equilib riu m @ 40o C
1 kg 98% H2 SO4
m1 (kg) 15% oleu m
0.98 kg SO3 0.02 kg H2 O
0.15 kg SO3 /kg 0.85 kg H2 SO4 /kg
b
pSO 3 40° C, 84.4%
g = 115 . . × 10 = 151
−3 mol SO 3 mol 760 P ii) 2.02 kg H 2.02 kg H 0.98 kg H 2 SO 4 0.02 kg H 2 O H balance: + 98.08 kg H 2 SO 4 18.02 kg H 2 O 0.85 m1 H 2 SO 4 2.02 kg H = ⇒ m1 = 128 . kg 98.08 kg H 2 SO 4 But since the feed solution has a mass of 1 kg, 0.28 kg SO 3 10 3 g 1 mol SO 3 absorbed = 128 . − 1.0 kg = = 350 . mol kg 80.07 g ⇒ 3.5 mol = n0 − n1 . × 10 −3 n1 G balance: 0.10n0 = 1 − 151
i)
y1 =
b
g
d
i
E
n0 = 3.89 mol n1 = 0.39 mol
V=
b g
3.89 mol 22.4 L STP 1 kg liquid feed mol
313K 1 atm 1 m 3 273K 1.2 atm 10 3 L
= 8.33 × 10 -2 m 3 kg liquid feed 6.45 a. Raoult’s law can be used for water and Henry’s law for nitrogen. b. Raoult’s law can be used for each component of the mixture, but Henry’s law is not valid here. c. Raoult’s law can be used for water, and Henry’s law can be used for CO2. 6.46 pB∗ (100°C ) = 10∗∗ ( 6.89272 − 1203.531 (100 + 219.888 ) ) = 1350.1 mm Hg pT∗ (100°C ) = 10∗∗ ( 6.95805 − 1346.773 (100 + 219.693) ) = 556.3 mm Hg
Raoult's Law: yB P = xB pB∗ ⇒
yB = yT =
0.40 (1350.1) 10 ( 760 )
0.60 ( 556.3) 10 ( 760 )
= 0.0711 mol Benzene mol
= 0.0439 mol Toluene mol
yN 2 = 1 − 0.0711 − 0.0439 = 0.885 mol N 2 mol
6-33
6.47 N 2 - Henry' s law: Perry' s Chemical Engineers' Handbook, Page. 2 - 127, Table 2 - 138
b
g
⇒ H N 2 80° C = 12.6 × 10 4 atm mole fraction
b gd i . mm Hg 3551 1 atm H O - Raoult' s law: p b80° Cg = = 0.467 atm 760 mm Hg = d x id p i = b0.997gb0.467g = 0.466 atm ⇒ p ⇒ p N 2 = x N 2 H N 2 = 0.003 12.6 × 10 4 = 378 atm ∗ H 2O
2
H 2O
H 2O
∗ H 2O
Total pressure: P = p N 2 + p H 2 O = 378 + 0.466 = 378.5 atm Mole fractions: yH 2O = pH2O P = 0.466 / 378.5 = 1.23 × 10−3 mol H 2 O mol gas yN 2 = 1 − yH 2O = 0.999 mol N 2 mol gas
b
g
6.48 H 2 O - Raoult' s law: p H∗ 2 O 70° C =
233.7 mm Hg
b
1 atm = 0.3075 atm 760 mm Hg
gb
g
⇒ p H 2 O = x H 2 O p H∗ 2O = 1 − x m 0.3075 Methane − Henry' s law: p m = x m ⋅ H m
Total pressure: P = p m + p H 2 O = x m ⋅ 6.66 × 10 4 + (1 − x m )(0.3075) = 10 ⇒ x m = 1.46 × 10 −4 mol CH 4 / mol 6.49 a.
Moles of water: n H 2O =
1000 cm 3
1g cm
mol
3
18.02 g
= 55.49 mol
Moles of nitrogen: nN 2 =
(1 - 0.334) × 14.1 cm 3 (STP)
1 mol
1L
22.4 L (STP) 1000 cm 3
= 4.192 × 10 −4 mol
Moles of oxygen: n O2 =
(0.334) ⋅ 14.1 cm 3 (STP)
mol
L
22.4 L (STP) 1000 cm
3
= 2.102 × 10 −4 mol
Mole fractions of dissolved gases: nN 2 mol N 2 4.192 × 10−4 xN 2 = = = 7.554 × 10−6 −4 −4 nH 2O + nN 2 + nO2 55.49 + 4.192 × 10 + 2.102 × 10 mol xO2 =
nO2 nH 2O + nN2 + nO2
=
2.102 × 10−4 = 3.788 × 10−6 mol O 2 / mol 55.49 + 4.192 × 10−4 + 2.102 × 10−4
6-34
6.49 (cont’d) Henry' s law
Nitrogen: H N 2 = Oxygen: H O 2 = b.
p N2
=
x N2 pO2 x O2
=
0.79 ⋅ 1 = 1.046 × 10 5 atm / mole fraction 7.554 × 10 −6
0.21 ⋅ 1 = 5544 . × 10 4 atm / mole fraction 3.788 × 10 −6
Mass of oxygen dissolved in 1 liter of blood: m O2 =
2.102 × 10 -4 mol 32.0 g
Mass flow rate of blood: m blood =
c.
6.50 a.
mol 0.4 g O 2
= 6.726 × 10 −3 g
min
1 L blood 6.72 × 10 -3 g O 2
= 59 L blood / min
Assumptions: (1) The solubility of oxygen in blood is the same as it is in pure water (in fact, it is much greater) (2) The temperature of blood is 36.9°C.
bg
Basis: 1 cm 3 H 2 O l
1 g H 2 O 1 mol
H2O =1.0 ⎯(SG) ⎯⎯⎯ ⎯→
18.0 g
= 0.0555 mol H 2 O
b g
0.0901 cm 3 STP CO 2
( SC) CO2 = 0.0901
⎯ ⎯⎯⎯⎯ ⎯→
1 mol 3
b g
22,400 cm STP
= 4.022 × 10 −6 mol CO 2
d4.022 × 10 i mol CO = 7.246 × 10 mol CO mol d0.0555 + 4.022 × 10 i mol 1 atm p =x H ⇒ H b20° Cg = = 13800 atm mole fraction 7.246 × 10 For simplicity, assume n ≈ n b molg x =p H = b35 . atmg b13800 atm mole fractiong = 2.536 × 10 mol CO mol −6
2
pCO 2 = 1 atm ⇒ x CO 2 =
CO 2
b.
CO 2
CO 2
−5
2
−6
−5
CO 2
total
H 2O
−4
CO 2
CO 2
nCO 2 =
2
12 oz
1L 33.8 oz
10 3 g H 2 O 1 mol H 2 O 1L
18.0 g H 2 O
2.536 × 10 −4 mol CO 2
44.0 g CO 2
1 mol H 2 O
1 mol CO 2
= 0.220 g CO 2 c.
V=
0.220 g CO 2
b g b273 + 37gK = 0127 . L = 127 cm
1 mol CO 2
22.4 L STP
44.0 g CO 2
1 mol
6-35
273K
3
6.51 a.
– SO2 is hazardous and should not be released directly into the atmosphere, especially if the analyzer is inside. – From Henry’s law, the partial pressure of SO2 increases with the mole fraction of SO2 in the liquid, which increases with time. If the water were never replaced, the gas leaving the bubbler would contain 1000 ppm SO2 (nothing would be absorbed), and the mole fraction of SO2 in the liquid would have the value corresponding to 1000 ppm SO2 in the gas phase.
b
g
b
b. Calculate x mol SO 2 mol in terms of r g SO 2 100 g H 2 O
b g r (g SO )b1 mol 64.07 gg = 0.01561r (mol SO ) 0.01561r F mol SO I ⇒x = J G 5.55 + 0.01561r H mol K
g
Basis: 100 g H 2 O 1 mol 18.02 g = 5.55 mol H 2 O 2
2
2
SO 2
From this relation and the given data, pSO 2 = 0 mmHg ⇔ xSO 2 = 0 mol SO 2 mol 1.4 x 10–3 2.8 x 10–3 4.2 x 10–3 5.6 x 10–3
42 85 129 176
A plot of pSO 2 vs. xSO 2 is a straight line. Fitting the line using the method of least squares (Appendix A.1) yields
dp
SO 2
= HSO 2 xSO 2
i
, H SO 2 = 3136 . × 104
mm Hg mole fraction
c. 100 ppm SO 2 ⇒ ySO = 100 mol SO 2 = 1.00 × 10−4 mol SO 2 2 mol 106 mols gas
(
)
⇒ pSO2 = ySO2 P = 1.0 × 10−4 ( 760 mm Hg ) = 0.0760 mm Hg Henry's law ⇒ xSO 2 =
Since xSO 2
H SO 2
=
0.0760 mm Hg 3.136 × 104 mm Hg mole fraction
= 2.40 × 10 −6 mol SO 2 mol is so low, we may assume for simplicity that Vfinal ≈ Vinitial = 140 L , and
nfinal ≈ ninitial = ⇒ nSO 2 =
pSO 2
bg
140 L 103 g H 2 O l 1L
1 mol 18 g
= 7.78 × 103 moles
7.78 × 10 mol solution 2.40 × 10 −6 mol SO 2 3
1 mol solution
= 0.0187 mol SO 2 dissolved
0.0187 mol SO 2 dissolved . × 10 −4 mol SO 2 L = 134 140 L
yH 2 O = Raoult’s law for water:
xH 2O pH* 2O (30o C) P
=
mol H 2 O(v) (1)(31.824 mm Hg) = 0.419 760 mm Hg mol
yair = 1 − ySO2 − yH 2O = 0.958
mol dry air mol
d. Agitate/recirculate the scrubbing solution, change it more frequently. Add a base to the solution to react with the absorbed SO2.
6-36
6.52
Raoult’s law + Antoine equation (S = styrene, T = toluene):
yS P = xS pS∗ ⇒ xS =
0.650(150 mm Hg) 10
yT P = xT pT∗ ⇒ xT =
7.06623 − 1507.434 (T + 214.985 )
0.350(150 mm Hg) 10
6.95334 − 1343.943 (T + 219.377 )
0.65(150)
1 = xS + xT =
7.06623 − 1507.434 (T + 214.985 )
0.35(150)
+
6.95334 − 1343.943 ( T + 219.377 )
10 10 ⇒ T = 86.0°C (Determine using E-Z Solve or a spreadsheeet)
xS =
0.65(150) 10
7.06623 − 1507.434 ( 86.0 + 214.985 )
= 0.853 mol styrene/mol
xT = 1 − xS = 1 − 0.853 = 0.147 mol toluene/mol 6.89272 − 1205.531 ( 85+ 219.888 )
= 881.6 mm Hg
6.95805 − 1346.773 ( 85+ 219.693)
= 345.1 mm Hg
6.53 pB∗ ( 85°C ) = 10
pT∗ ( 85°C ) = 10
Raoult's Law: yB P = xB pB∗ ⇒
yB = yT =
0.35 ( 881.6 ) 10 ( 760 )
0.65 ( 345.1) 10 ( 760 )
= 0.0406 mol Benzene mol = 0.0295 mol Toluene mol
yN 2 = 1 − 0.0406 − 0.0295 = 0.930 mol N 2 mol 6.54 a. From the Cox chart, at 77° F, p *P = 140 psig, p *nB = 35 psig, p *iB = 51 psig
Total pressure P=x p ⋅ p*p +x nB ⋅ p*nB +x iB ⋅ p*iB = 0.50(140) + 0.30(35) + 0.20(51) = 91 psia ⇒ 76 psig P < 200 psig, so the container is technically safe. * * b. From the Cox chart, at 140° F, pP* = 300 psig, pnB = 90 psig, piB = 120 psig
Total pressure P = 0.50(300) + 0.30( 90) + 0.20(120) ≅ 200 psig The temperature in a room will never reach 140oF unless a fire breaks out, so the container is adequate. ∗ 6.55 a. Antoine: pnp (120°C ) = 10
pip∗ (120°C ) = 10
6.84471 − 1060.793 (120 + 231.541)
6.73457 − 992.019 (120 + 231.541)
= 6717 mm Hg
= 7883 mm Hg
(Note: We are using the Antoine equation at 120oC, well above the validity ranges in Table B.4 for n-pentane and isopentane, so that all calculated vapor pressures must be considered rough estimates. To get more accuracy, we would need to find a vapor pressure correlation valid at higher temperatures.) When the first bubble of vapor forms,
6-37
6.55 (cont’d)
xnp = 0.500 mol n - C5 H 12 (l) / mol
xip = 0.500 mol i -C5 H12 (l)/mol
* Total pressure: P =xnp ⋅ pnp +xip ⋅ pip* = 0.50(6717) + 0.50(7883) = 7300 mm Hg * xnp ⋅ pnp
ynp =
P
=
0.500(6717) = 0.46 mol n-C5 H12 (v)/mol 7300
yip = 1 − ynp = 1 − 0.46 = 0.54 mol i -C5 H12 (v)/mol
When the last drop of liquid evaporates, ynp = 0.500 mol n - C5 H 12 (v) / mol xnp + xip = xnp =
ynp P * pnp (120o C)
+
yip P * pip (120o C)
yip = 0.500 mol i -C5 H12 (v)/mol =
0.500 P 0.500 P + = 1 ⇒ P = 7291 mm Hg 6725 7960
0.5 * 7250 mm Hg = 0.54 mol n-C5 H12 (l)/mol 6717 mm Hg
xip = 1 − xnp = 1 − 0.54 = 0.46 mol i -C5 H12 (l)/mol b. When the first drop of liquid forms,
ynp = 0.500 mol n - C5 H 12 (v) / mol
yip = 0.500 mol i - C12 H 12 (v) / mol
P = (1200 + 760) = 1960 mm Hg 0.500 P 0.500 P 980 980 xnp + xip = * + * = 6.84471−1060.793/(T + 231.541) + 6.73457 −992.019 /(T + 231.541) = 1 dp dp pnp (Tdp ) pip (Tdp ) 10 10 ⇒ Tdp = 63.1o C ∗ pnp = 10
pip∗ = 10 xnp =
6.84471−1060.793 ( 63.1+ 231.541)
6.73457 −992.019 ( 63.1+ 231.541)
= 1758 mm Hg
= 2215 mm Hg
0.5 *1960 mm Hg = 0.56 mol n-C5 H12 /mol * pnp (63.1o C)
xip = 1 − xnp = 1 − 0.56 = 0.44 mol i -C5 H12 /mol When the last bubble of vapor condenses, xnp = 0.500 mol n - C5 H 12 (l) / mol
xip = 0.500 mol i - C5 H 12 (l) / mol
* Total pressure: P =xnp ⋅ pnp +xip ⋅ pip*
⇒ 1960 = (0.5)10
xnp ⋅
6.84471−1060.793 (T + 231.541)
* pnp (62.6o C)
+ (0.5)10
6.73457 −992.019 /(Tbp + 231.541)
0.5(1734) = 0.44 mol n-C5 H12 (v)/mol P 1960 yip = 1 − ynp = 1 − 0.44 = 0.56 mol i -C5 H12 (v)/mol
ynp =
=
6-38
⇒ T = 62.6°C
B = benzene, T = toluene
6.56
nv ( mol / min) at 80o C, 3 atm 10 L(STP)/min n N 2 ( mol / min) n N 2 =
xB [mol B(l)/mol]
yN2 (mol N2/mol) yB [mol B(v)/mol]
xT [mol T(l)/mol]
yT [mol T(v)/mol]
10.0 L(STP) / min = 0.4464 mol N 2 / min 22.4 L(STP) / mol 6.89272 −1203.531 ( 80 + 219.888 )
= 757.6 mm Hg
6.95805 −1346.773 ( 80 + 219.693)
= 291.2 mm Hg
Antoine: pB∗ ( 80°C ) = 10 pT∗ ( 80°C ) = 10
a. Initially, xB = 0.500, xT = 0.500.
N 2 balance: 0.4464 mol N 2 / min = n v (1 − 0166 . − 0.0639) ⇒ n v = 0.5797 mol / min mol I F mol B I mol B(v) FG 0166 . = 0.0962 J G J K H K H min mol min mol I F BI F = G 0.5797 J = 0.0370 molminT(v) J G 0.0639 mol H min K H mol K
⇒ n B0 = 0.5797 nT 0
b. Since benzene is evaporating more rapidly than toluene, xB decreases with time and xT (= 1–xB) increases. c. Since xB decreases, yB (= xBpB*/P) also decreases. Since xT increases, yT (= xTpT*/P) also increases. 6.57 a. P =
∗ xhex phex
dT i + bp
∗ xhep phep
dT i bp
, yi =
d i
xi pi∗ Tbp P
, Antoine equation for pi∗
6.88555 −1175.817 /(Tbp + 224.867) ⎤ + 0.500 ⎡106.90253−1267.828 /(Tbp + 216.823) ⎤ 760 mm Hg = 0.500 ⎡10 ⎣ ⎦ ⎣ ⎦ E-Z Solve or Goal Seek ⇒ Tbp = 80.5° C ⇒ yhex = 0.713, yhep = 0.287
b.
xi =
yi P pi∗
dT i dp
⇒
∑x i
i
=P
∑
i
yi
pi∗
dT i
=1
dp
0.30 0.30 ⎡ ⎤ 760 mmHg ⎢ 6.88555−1175.817 /(T + 224.867) + 6.90253−1267.828/(T + 216.823) ⎥ = 1 dp dp 10 ⎣10 ⎦
. ° C ⇒ xhex = 0.279, xhep = 0.721 E-Z Solve or Goal Seek ⇒ Tdp = 711
6-39
6.58
a.
f (T ) = P −
N
∑
xi pi* (T )
= 0 ⇒ T , where
pi* (T )
FG A − B IJ T +C K = 10H i
i
i
i =1
yi (i = 1,2,", N ) =
xi pi* ( T ) P
b. Calculation of Bubble Points A B Benzene 6.89272 1203.531 Ethylbenzene 6.95650 1423.543 Toluene 6.95805 1346.773
C 219.888 213.091 219.693
P(mmHg)= 760 xB 0.226 0.443 0.226
xEB 0.443 0.226 0.226
Tbp(oC) 108.09 96.47 104.48
xT 0.331 0.331 0.548
b
g
d i
When x B = 1 pure benzene , Tbp = Tbp
b
g
= 801 . oC
C6 H 6
d i
When x EB = 1 pure ethylbenzene , Tbp = Tbp
b
g
d i
When xT = 1 pure toluene , Tbp = Tbp
C7 H8
pB pEB pT f(T) 378.0 148.2 233.9 -0.086 543.1 51.6 165.2 0.11 344.0 67.3 348.6 0.07
C8 H 10
= 136.2 o C ⇒ Tbp , EB > Tbp ,T > Tbp , B
= 110.6o C
Mixture 1 contains more ethylbenzene (higher boiling point) and less benzene (lower bp) than Mixture 2, and so (Tbp)1 > (Tbp)2 . Mixture 3 contains more toluene (lower bp) and less ethylbenzene (higher bp) than Mixture 1, and so (Tbp)3 < (Tbp)1. Mixture 3 contains more toluene (higher bp) and less benzene (lower bp) than Mixture 2, and so (Tbp)3 > (Tbp)2
6-40
a. Basis: 150.0 L/s vapor mixture
6.59
n 1 (mol/s) @ T(oC), 1100 mm Hg
0.600 mol B(v)/mol 0.400 mol H(v)/mol
n 0 (mol/s)@120°C, 1 atm
0.500 mol B(v)/mol 0.500 mol H(v)/mol
n 2 (mol/s) x2 [mol B(l)/mol] (1- x2) [mol H(l)/mol]
Gibbs phase rule : F=2+c-π =2+2-2=2
Since the composition of the vapor and the pressure are given, the information is enough. Equations needed: Mole balances on butane and hexane, Antoine equation and Raoult’s law for butane and hexane b. Molar flow rate of feed: n 0 =
150.0 L 273 K mol = 4.652 mol/s s 393 K 22.4 L (STP)
Raoult's law for butane: 0.600(1100)=x 2 ⋅ 106.82485−943.453/(T + 239.711) 6.88555 −1175.817 /(T + 224.867)
Raoult's law for hexane: 0.400(1100)=(1-x 2 ) ⋅ 10 Mole balance on butane: 4.652(0.5)=n 1 ⋅ 0.6 + n 2 ⋅ x 2 Mole balance on hexane: 4.652(0.5)=n 1 ⋅ 0.4 + n 2 ⋅ (1 − x 2 ) c. From (1) and (2), 1=
(1) (2) (3) (4)
1100(0.6) 1100(0.4) + 943.453 1175.817 10 **(6.82485 − ) 10 **(6.88555 − ) T + 239.711 T + 224.867
⇒ T = 57.0 ° C x2 =
1100(0.6) 10
6.82485 − 943.453/(57.0 + 239.711)
= 0.149 mol butane /mol
Solving (3) and (4) simultaneously ⇒ n1 = 3.62 mol C 4 H10 /s; n2 = 1.03 mol C6 H14 /s d.
6.60
Assumptions: (1) Antoine equation is accurate for the calculation of vapor pressure; (2) Raoult’s law is accurate; (3) Ideal gas law is valid.
P = n-pentane, H = n-hexane 170.0 kmol/h, T1a (oC), 1 atm
85.0 kmol/h, T1b (oC), 1 n 0 (kmol/h)
0.98 mol P(l)/mol 0.02 mol H(l)/mol
0.45 kmol P(l)/kmol 0.55 kmol H(l)/kmol
n 2 (kmol/h) (l),
x2 (kmol P(l)/kmol) (1- x2) (kmol H(l)/kmol)
6-41
6.60 (cont’d) a. Molar flow rate of feed: n 0 (0.45)(0.95) = 85(0.98) ⇒ n 0 = 195 kmol / h
Total mole balance : 195 = 85.0 + n 2 ⇒ n 2 = 110 kmol / h Pentane balance: 195(0.45) = 85.0(0.98) + 110 ⋅ x 2 ⇒ x 2 = 0.0405 mol P / mol b. Dew point of column overhead vapor effluent:
Eq. 6.4-7, Antoine equation 0.98(760) 0.02(760) ⇒ + 6.88555−1175.817 /(T + 224.687) = 1 ⇒ T1a = 37.3o C 6.84471−1060.793/(T1a + 231.541) 1a 10 10 Flow rate of column overhead vapor effluent. Assuming ideal gas behavior, 170 kmol 0.08206 m 3 ⋅ atm (273.2 + 37.3) K = 4330 m 3 / h Vvapor = h kmol ⋅ K 1 atm Flow rate of liquid distillate product. Table B.1 ⇒ ρ P = 0.621 g / mL, ρ H = 0.659 g / mL
0.98(85) kmol P 72.15 kg P L Vdistillate = h kmol P 0.621 kg P +
L 0.02(85) kmol H 86.17 kg H = 9.9 × 10 3 L / h h kmol H 0.659 kg H
c. Reboiler temperature. 0.04 ⋅ 106.84471−1060.793/(T2 + 231.541) + 0.96 ⋅ 106.88555−1175.817 /(T2 + 224.867) = 760 ⇒ T2 =66.6°C
Boilup composition. y2 =
x2 pP* (66.6o C) 0.04 ⋅ 106.84471−1060.793/(66.6+ 231.541) = = 0.102 mol P(v)/mol P 760 ⇒ (1 - y 2 ) = 0.898 mol H(v) / mol
d. Minimum pipe diameter
F I GH JK
m3 V s
= u max
⇒ Dmin =
FG mIJ × πD H sK 4
4Vvapor
π ⋅ u max
2 min
=
(m2 )
4 4330 m 3 / h 1 h = 0.39 m (39 cm) π 10 m / s 3600 s
Assumptions: Ideal gas behavior, validity of Raoult’s law and the Antoine equation, constant temperature and pressure in the pipe connecting the column and the condenser, column operates at steady state.
6-42
Condenser
6.61 a. F (mol) x 0 (mol butane/mol)
V (mol) 0.96 mol butane/mol R (mol) x 1 (mol butane/mol)
T P
Partial condenser: 40° C is the dew point of a 96% C 4 H 10 − 4% C 5 H 12 vapor mixture at P = Pmin Total condenser: 40° C is the bubble point of a 96% C 4 H 10 - 4% C 5 H 12 liquid mixture at P = Pmin
Dew Point: 1 = (Raoult's Law)
∑x =∑ p i
yi P
∗ i
Antoine Eq. for
b40° Cg pi∗
⇒ Pmin =
bC H 4
10
i
b
pi∗ 40° C
FG 6.82485− 943.453 IJ g = 10H 40+239.711K
Antoine Eq. for pi∗ ( C5 H12 ) = 10 ⇒ Pmin =
∑y
1
1060.793 ⎞ ⎛ ⎜ 6.84471− ⎟ 40 + 231.541 ⎠ ⎝
g
= 2830.70 mmHg = 867.22 mmHg
b
1 = 2595.63 mm Hg partial condenser 0.96 2830.70 + 0.04 867.22
g
b40° Cg P = 0.96b2830.70g + 0.04b867.22g = 2752.16 mm Hg b total condenser g
Bubble Point: P =
∑y P=∑x p i
i
∗ i
b. V = 75 kmol / h , R V = 15 . ⇒ R = 75 × 15 . kmol / h = 112.5 kmol / h
Feed and product stream compositions are identical: y = 0.96 kmol butane kmol
Total balance: F = 75 + 112.5 = 187.5 kmol / h c.
Total balance as in b. R = 112.5 kmol / h
UV P = 2596 mm Hg gb gW x = 0.8803 mol butane mol = 112.5b0.8803g + 0.96b75g ⇒ x = 0.9122 mol butane mol reflux
b b
g
Equilibrium: 0.96 P = x1 2830.70 Raoult' s law 0.04 P = 1 − x1 867.22
b
g
Butane balance: 187.5x 0
6.62 a.
b.
Raoult' s law:
(
)
yi pi∗ y x p ∗ P p ∗A = ⇒ α AB = A A = ∗A = = α AB xi P yB xB p B P p B∗ 1507.434 ⎞ ⎛ ⎜ 7.06623− ⎟ 85 + 214.985 ⎠ 1423.543 ⎞ ⎛ ⎜ 6.95650 − ⎟ 85+ 213.091 ⎠
p*EB (85o C) = 10⎝ pB (85 C) = 10 o
1
0
p*S 85o C = 10⎝
*
F = 187.5 kmol / h
1203.531 ⎞ ⎛ ⎜ 6.89272 − ⎟ 85 + 219.888 ⎠ ⎝
= 109.95 mm Hg = 151.69 mm Hg
= 881.59 mm Hg
6-43
6.62 (cont’d)
p*S 109.95 p*B 881.59 0.725 , = = α = = = 5.812 B,EB p*EB 151.69 p*EB 151.69
α S,EB =
Styrene − ethylbenzene is the more difficult pair to separate by distillation because α S,EB is closer to 1 than is α B,EB .
c.
α ij =
yi xi yj xj
y j =1− yi x j =1− xi
⇒α
ij
d. α B , EB = 5810 . ⇒ yB =
α ij xi yi xi ⇒ yi = (1 − y i ) 1 − xi 1 + α ij − 1 x i
b
g
x B α B , EB 1 + (α B , EB − 1) x B
d
=
i
581 . xB , P = x B p *B + (1 − x B ) p *EB 1 + 4.81x B
bg bg
xB yB P 6.63 a.
=
0.0 0.2 0.4 0.6 0.8 10 . mol B l mol 0.0 0.592 0.795 0.897 0.959 10 . mol B v mol 152 298 444 5900 736 882 mmHg
Since benzene is more volatile, the fraction of benzene will increase moving up the column. For ideal stages, the temperature of each stage corresponds to the bubble point temperature of the liquid. Since the fraction of benzene (the more volatile species) increases moving up the column, the temperature will decrease moving up the column.
b. Stage 1: n l = 150 mol / h, n v = 200 mol / h ; x1 = 0.55 mol B mol ⇒ 0.45 mol S mol ;
y 0 = 0.65 mol B mol ⇒ 0.35 mol S mol Bubble point T: P =
∑x p i
∗ i
bT g
P1 = (0.400 × 760) mmHg = ( 0.55 )106.89272−1203.531/(T + 219.888) + ( 0.45 )107.06623−1507.434 /(T + 214.985) E-Z Solve ⎯⎯⎯⎯ → T1 = 67.6o C
⇒ y1 =
b g=
x1 p ∗B T
b g
0.55 508
= 0.920 mol B mol ⇒ 0.080 mol S mol P 0.400 × 760 B balance: y 0 n v + x 2 n l = y1n v + x1n l ⇒ x 2 = 0.910 mol B mol ⇒ 0.090 mol S mol
Stage 2: Solve ( 0.400 × 760) mmHg = 0.910 p *B (T2 ) + 0.090 p S* ( T2 ) ⎯E-Z ⎯⎯ ⎯→ T2 = 55.3o C
y2 =
b
g = 0.991 mol B mol ⇒ 0.009 mol S mol
0.910 3310 .
760 × 0.400 B balance: y1n v + x 3 n l = y 2 n v + x 2 n l ⇒ x 3 ≈ 1 mol B mol ⇒ ≈ 0 mol S mol c. In this process, the styrene content is less than 5% in two stages. In general, the calculation of part b would be repeated until (1–yn) is less than the specified fraction. 6-44
6.64 Basis: 100 mol/s gas feed. H=hexane. nGN (mol/s)
200 mol oil/s
nL (mol/s) xi–1 (mol H/mol)
yN (mol H/mol) (1–yN) (mol N2/mol) 100 mol/s 0.05 mol H/mol 0.95 mol N2/mol
a.
nG (mol/s) yi (mol H/mol) Stage i
nL1 (mol/s) x1 (mol H/mol) (99.5% of H in feed) (1–x1) (mol oil/mol)
nL (mol/s) xi (mol H/mol)
nG (mol/s) yi+1 (mol H/mol)
⎫⎪ nGN = 95.025 mol s ⎬⇒ 99.5% absorption: 0.05 (100 )( 0.005 ) = y N nGN ⎪⎭ y N = 2.63 × 10−4 mol H(v) mol Mole Balance: 100 + 200 = 95.025 + n L1 ⇒ nL1 = 205 mol s N 2 balance: 0.95 (100 ) = (1 − y N ) nGN
Hexane Balance: 0.05 (100 ) = 2.63 × 10−4 ( 95.025 ) + x1 ( 204.99 ) ⇒ x1 = 0.0243 mol H(l) mol n L =
1 1 ( 200 + 205) ⇒ nL = 202.48 mol s , nG = (100 + 95.025 ) ⇒ nG = 97.52 mol s 2 2
B
Antoine
b.
b
g
b
g
y1 = x1 p H∗ 50° C / P = 0.0243 403.73 / 760 = 0.0129 mol H(v) mol H balance on 1st Stage: y 0 n v + x 2 n l = y1n v + x1n l ⇒ x 2 = 0.00643 mol H(l) mol
c. The given formulas follow from Raoult’s law and a hexane balance on Stage i. d. Hexane Absorption P= y0= nGN= A=
760 0.05 95.025 6.88555
T 30
p*(T) 187.1
i 0 1 2 3
x(i) 2.43E-02 3.10E-03 5.86E-04
PR= x1= nL1= B=
y(i) 5.00E-02 5.98E-03 7.63E-04 1.44E-04
1 0.0243 204.98 1175.817
yN= 2.63E-04 nG= 97.52 nL= C= 224.867
T 50
p*(T) 405.3059
i 0 1 2 3 4 5
x(i) 2.43E-02 6.46E-03 1.88E-03 7.01E-04 3.99E-04
6-45
y(i) 5.00E-02 1.30E-02 3.45E-03 1.00E-03 3.74E-04 2.13E-04
202.48
T 70
p*(T) 790.5546
i 0 1 2 3 4 5 ... 21
x(i)
y(i) 5.00E-02 2.43E-02 2.53E-02 1.24E-02 1.29E-02 6.43E-03 6.69E-03 3.44E-03 3.58E-03 1.94E-03 2.02E-03 ... ... 4.38E-04 4.56E-04
6.64 (cont’d) e. If the column is long enough, the liquid flowing down eventually approaches equilibrium with the entering gas. At 70oC, the mole fraction of hexane in the exiting liquid in equilibrium with the mole fraction in the entering gas is 4.56x10–4 mol H/mol, which is insufficient to bring the total hexane absorption to the desired level. To reach that level at 70oC, either the liquid feed rate must be increased or the pressure must be raised to a value for which the final mole fraction of hexane in the vapor is 2.63x10–4 or less. The solution is Pmin = 1037 mm Hg. 6.65 a. Intersection of vapor curve with y B = 0.30 at T = 104° C ⇒ 13% B(l), 87%T(l) b. T = 100°C ⇒ xB = 0.24 mol B mol ( liquid ) , yB = 0.46 mol B mol ( vapor ) n V (mol vapor) 0.46 mol B(v)/mol n L (mol liquid) 0.24 mol B(l)/mol
Basis: 1 mol 0.30 mol B(v)/mol
Balances
UV ⇒ n W n
Total moles: 1 = nV + n L B:
0.30 = 0.46nV + 0.24n L
L
= 0.727 mol
V
= 0.273 mol
⇒
nV mol vapor = 0.375 mol liquid nL
c. Intersection of liquid curve with x B = 0.3 at T = 98° C ⇒ 50% B(v), 50%T(v) 6.66 a.
P = 798 mm Hg, y B = 0.50 mol B(v) mol
b.
P = 690 mm Hg, xB = 0.15 mol B(l) mol
c.
P = 750 mm Hg, yB = 0.43 mol B(v) mol, xB = 0.24 mol B(l) mol nV (mol) 0.43 mol B/mol nL (mol) 0.24 mol B/mol
3 mol B 7 mol T
Mole bal.: 10 = nV + n L B bal.:
3 = 0.43nV + 0.24n L
UV ⇒ n W n
V
= 316 . mol
L
= 6.84 mol
⇒
nv mol vapor = 0.46 mol liquid nl
Answers may vary due to difficulty of reading chart. d. i)
P = 1000 mm Hg ⇒ all liquid . Assume volume additivity of mixture components. V=
3 mol B 78.11 g B mol B
10 −3 L 0.879 g B
+
7 mol T 92.13 g T
ii) 750 mmHg. Assume liquid volume negligible
6-46
mol T
10 −3 L 0.866 g T
= 10 . L
6.66 (cont’d)
V=
3.16 mol vapor 0.08206 L ⋅ atm 373 K 760 mm Hg − 0.6 L = 97.4 L mol ⋅ K 750 mm Hg 1 atm (Liquid volume is about 0.6 L)
iii) 600 mm Hg
v=
10 mol vapor 0.08206 L ⋅ atm 373K 760 mm Hg = 388 L mol ⋅ K 600 mm Hg 1 atm
6.67 a. M = methanol n V (mol) y (mol M (v)mol) n L (mol) x (mol M (l)/mol)
n f (mol) x F (mol M (l)/mol)
Mole balance:
n f = nV + n L
MeOH balance: x F n f = ynV + xn L x F = 0.4, x = 0.23, y = 0.62 ⇒ f =
UV ⇒ x W
F nV
+ x F n L = ynV + xn L ⇒ f =
0.4 − 0.23 = 0.436 0.62 − 0.23
b. Tmin = 75o C, f = 0 , Tmax = 87 o C, f = 1 6.68 a.
Txy diagram (P=1 atm) 80 75 T(oC)
70
Vapor
65 liquid
60 55 50 0
0.2
0.4
0.6
Mole fraction of Acetone
b.
x A = 0.47; y A = 0.66
6-47
0.8
1
nV x F − x = nL y−x
6.68 (cont’d) c. (i) x A = 0.34; y A = 0.55
(ii) Mole bal.: 1 = nV + nL A bal.:
⎫⎪ ⎬ ⇒ nV = 0.762 mol vapor, nL = 0.238 mol liquid ∂ 0.50 = 0.55nV + 0.34nL ⎪⎭ ⇒ 76.2 mole% vapor
(iii) ρ A( l ) = 0.791 g/cm3 , ρ E(l) = 0.789 g/cm3 ⇒ ρl ≈ 0.790 g/cm3 (To be more precise, we could convert the given mole fractions to mass fractions and calculate the weighted average density of the mixture, but since the pure component densities are almost identical there is little point in doing all that.) M A = 58.08 g/mol, M E = 46.07 g/mol
⇒ M l = ( 0.34 )( 58.08 ) + (1 − 0.34 )( 46.07 ) = 50.15 g/mol
Basis: 1 mol liquid ⇒ (0.762 mol vapor / 0.238 mol liquid) = 3.2 mol vapor Liquid volume:
Vl =
(1 mol)(5015 . g / mol) = 63.48 cm 3 3 (0.790 g / cm )
Vapor volume: Vv =
3.2 mol 22400 cm 3 (STP) (65 + 273)K mol
Volume percent of vapor =
273K
= 88,747 cm 3
88,747 × 100% = 99.9 volume% vapor 88747 + 63.48
d. For a basis of 1 mol fed, guess T, calculate nV as above; if nV ≠ 0.20, pick new T. T 65 °C 64.5 °C e.
xA 0.34 0.36
yA 0.55 0.56
fV 0.333 0.200
Raoult' s law: y i P = x i pi* ⇒ P = x A p *A + x E p E*
760 = 0.5 × 10
7.11714 −1210.595/(Tbp + 229.664)
8.11220−1592.864/(Tbp + 226.184)
+ 0.5 × 10
⇒ Tbp = 66.16o C
xp*A 0.5 × 107.11714−1210.595/(66.25+ 229.664) y= = = 0.696 mol acetone/mol P 760 ΔTbp 66.25 − 61.8 The actual Tbp = 61.8o C ⇒ = × 100% = 7.20% error in Tbp 61.8 Tbp (real) yA = 0.674 ⇒
ΔyA 0.696 − 0.674 = × 100% = 3.3% error in yA yA (real) 0.674
Acetone and ethanol are not structurally similar compounds (as are, for example, pentane and hexane or benzene and toluene). There is consequently no reason to expect Raoult’s law to be valid for acetone mole fractions that are not very close to 1.
6-48
6.69 a. B = benzene, C = chloroform. At 1 atm, (Tbp)B = 80.1oC, (Tbp)C = 61.0oC
The Txy diagram should look like Fig. 6.4-1, with the curves converging at 80.1oC when xC = 0 and at 61.0oC when xC = 1. (See solution to part c.) b. Txy Diagram for an Ideal Binary Solution A B C 6.90328 1163.03 227.4 Chloroform 6.89272 1203.531 219.888 Benzene 760 P(mmHg)= x 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
T 80.10 78.92 77.77 76.66 75.58 74.53 73.51 72.52 71.56 70.62 69.71 68.82 67.95 67.11 66.28 65.48 64.69 63.93 63.18 62.45 61.73
y 0 0.084 0.163 0.236 0.305 0.370 0.431 0.488 0.542 0.593 0.641 0.686 0.729 0.770 0.808 0.844 0.879 0.911 0.942 0.972 1
p1 0 63.90 123.65 179.63 232.10 281.34 327.61 371.15 412.18 450.78 487.27 521.68 554.15 585.00 614.02 641.70 667.76 692.72 716.27 738.72 760
p2 760 696.13 636.28 580.34 527.86 478.59 432.30 388.79 347.85 309.20 272.79 238.38 205.83 175.10 145.94 118.36 92.17 67.35 43.75 21.33 0
p1+p2 760 760.03 759.93 759.97 759.96 759.93 759.91 759.94 760.03 759.99 760.07 760.06 759.98 760.10 759.96 760.06 759.93 760.07 760.03 760.05 760
T xy diagram (P =1 atm ) 85
75
V apor
o
T( C)
80
70 Liquid 65 60 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
M ole fraction of chloroform
6-49
0.8
0.9
1
6.69 (cont’d) d. Txy diagram (P=1 atm) 85
T(oC)
80
yc
xc
75 70 x
y
65 60 0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
1
Mole fraction of choloroform
Raoult’s law: Tbp = 71o C, y = 0.58 ⇒
71 − 75.3 ΔT = × 100% = −5.7% error in Tbp 75.3 Tactual Δy y actual
=
0.58 − 0.60 × 100% = −3.33% error in y 0.60
Benzene and chloroform are not structurally similar compounds (as are, for example, pentane and hexane or benzene and toluene). There is consequently no reason to expect Raoult’s law to be valid for chloroform mole fractions that are not very close to 1.
d i b
g d i
6.70 P ≈ 1 atm = 760 mm Hg = x m pm* Tbp + 1 − x m p *P Tbp
760 = 0.40 × 10
7.87863−1473.11/(Tbp + 230)
+ 0.60 × 10
7.74416 −1437.686 /(Tbp +198.463)
E-Z Solve ⎯⎯⎯⎯ → T = 79.9o C
We assume (1) the validity of Antoine’s equation and Raoult’s law, (ii) that pressure head and surface tension effects on the boiling point are negligible. The liquid temperature will rise until it reaches 79.9 °C, where boiling will commence. The escaping vapor will be richer in methanol and thus the liquid composition will become richer in propanol. The increasing fraction of the less volatile component in the residual liquid will cause the boiling temperature to rise.
6-50
6.71 Basis: 1000 kg/h product nH4 (mol H 2 /h) E = C2 H5 OH ( M = 46.05) A = CH 3 CHO ( M = 44.05)
scrubber n3 (mol/h) y A3 (mol A/mol), sat'd y E3 (mol E/mol), sat'd y H3 (mol H 2 /h) vapor, –40°C
P = 760 mm Hg Fresh feed n0 (mol E/h) nA1 (mol A/h) nE1 (mol E/h) 280°C
reactor
nA2 (mol A/h) nE2 (mol E/h) nH2 (mol H 2/h)
condenser
nC (mol/h) 0.550 A 0.450 E liquid, –40°C
Scrubbed Hydrocarbons nA4 (mol A/h) nE4 (mol E/h)
still
Product 1000 kg/h np (mol/h) 0.97 A 0.03 E
nr (mol/h) 0.05 A 0.95 E
Strategy
a.
d i
•
Calculate molar flow rate of product n p from mass flow rate and composition
•
Calculate y A3 and y E3 from Raoult’s law: y H3 = 1 − y A3 − y E3 . Balances about the still involve fewest unknowns (n c and n r )
•
Total mole balance about still ⇒ n c , n r A balance about still
•
A, E and H 2 balances about scrubber ⇒ n A4 , n E4 , and n H4 in terms of n 3 . Overall atomic balances on C, H, and O now involve only 2 unknowns ( n 0 , n 3 )
•
Overall C balance ⇒ n 0 , n 3 Overall H balance
• • • •
A balance about fresh feed-recycle mixing point ⇒ n A1 E balance about fresh feed-recycle mixing point ⇒ n E1 A, E, H 2 balances about condenser n A2 , n E2 , n H2 All desired quantities may now be calculated from known molar flow rates.
UV W
UV W
Molar flow rate of product
b gb
g b gb
g
M = 0.97 M A + 0.03 M E = 0.97 44.05 + 0.03 46.05 = 44.11 g mol n p =
1000 kg 1 kmol = 22.67 kmol h h 44.11 kg
Table B.4 (Antoine) ⇒
pA* ( −40°C ) = 44.8 mm Hg pE* ( −40°C ) = 0.360 mm Hg
Note: The calculations that follow can at best be considered rough estimates, since we are using the Antoine correlations of Table B.4 far outside their temperature ranges of validity. Raoult’s law ⇒ yA3 =
0.550 pA* ( −40 °C ) P
=
0.550(44.8) = 0.03242 kmol A/kmol 760
6-51
6.71 (cont’d)
yE3 =
0.450 pE* ( −40 °C )
yH3
0.450(0.360) = 2.13 × 10−4 kmol E kmol P 760 = 1 − yA3 − yE3 = 0.9674 kmol H 2 kmol =
UV W
n r = 29.5 kmol / h recycle Mole balance about still: n c = n p + n r ⇒ n c = 22.67 + n r ⇒ A balance about still: 0.550n c = 0.97(22.67) + 0.05n r n c = 52.1 kmol / h A balance about scrubber: nA4 = n3 yA3 = 0.03242n3
(1)
E balance about scrubber: nE4 = n3 yE3 = 2.13 × 10−4 n3
(2)
H 2 balance about scrubber: nH4 = n3 yH3 = 0.9764n3
(3)
Overall C balance: n 0 (mol E) 2 mol C = n A4 2 + n E4 2 + 0.97n p 2 + 0.03n p 2 h 1 mol E
b gb g b gb g d
ib g d
ib g
⇒ n 0 = n A 4 + n E 4 + 22.67
(4)
Overall H balance:
b gb g b gb g
6n 0 = 2n H4 + 4n A4 + 6n E4 + n p 0.97 4 + 0.03 6
(5)
Solve (1)–(5) simultaneously (E-Z Solve): n0 = 23.4 kmol E/h (fresh feed), nH 4 = 22.7 kmol H2 /h (in off-gas)
n3 = 23.3 kmol/h, nA 4 = 0.755 kmol A/h, nE 4 = 0.00496 kmol E/h A balance about feed mixing point: nA1 = 0.05nr = 1.475 kmol A h E balance about feed mixing point: nE1 = n0 + 0.95nr = 51.5 kmol E h E balance about condenser: nE2 = n3 yE3 + 0.450nc = 23.5 kmol E h Ideal gas equation of state : Vreactor feed = b.
(1.47 + 51.5 ) kmol
Overall conversion =
22.4 m3 ( STP ) ( 273+280 ) K
h
1 kmol
n0 − 0.03n p
× 100% =
273K
23.4 − ( 0.03)( 22.67 )
= 2.40 × 103 m3 h
× 100% = 97% n0 23.4 n − n 51.5 − 23.5 × 100% = 54% Single-pass conversion = E1 E2 × 100% = nE1 51.5 Feed rate of A to scrubber:
nA4 =0.76 kmol A/h
Feed rate of E to scrubber:
nE4 = 0.0050 kmol E h
6-52
6.72 a. G = dry natural gas, W = water n 3 (lb - mole G / d) n 4 (lb - mole W / d) 10 lb m W / 10 6 SCF gas 90 o F, 500 psia
Absorber n 7 (lb - mole W / d)
FG lb - mole TEG IJ H d K F lb - mole W IJ n G H d K
4.0 × 10 6 SCF / d 4 × 80 = 320 lb m W / d n1 (lb - mole G / d) n 2 [lb - mole W(v) / d]
n 5
Distillation Column
6
FG lb - mole TEGIJ H d K F lb - mole WIJ n G H d K n5 8
Overall system D. F. analysis:
Water feed rate: n 2 =
5 unknowns (n1 , n 2 , n 3 , n 4 , n 7 ) − 2 feed specifications (total flow rate, flow rate of water) − 1 water content of dried gas −2 balances (W, G) 0 D. F.
320 lb m W 1 lb - mole = 17.78 lb - moles W / d d 18.0 lb m
Dry gas feed rate: 4.0 × 106 SCF 1 lb - mole lb - moles W − 17.78 = 1112 n1 = . × 104 lb - moles G / d d d 359 SCF
. × 10 4 lb - moles G / d Overall G balance: n1 = n 3 ⇒ n 3 = 1112 Flow rate of water in dried gas: n 4 =
(n 3 + n 4 ) lb - moles d
359 SCF gas 10 lb m W 1 lb - mole W lb - mole 10 6 SCF 18.0 lb m
3 =1.112 ×10 ⎯n⎯⎯⎯ ⎯→ n 4 = 2.218 lb - mole W(l) / d
4
Overall W balance: n 7 =
(17.78 − 2.218) lb - moles W
18.0 lb m
d
1 lb - mole
6-53
= 280
lb m W × d
F 1 ft I = 4.5 ft W GH 62.4 lb JK d 3
3
m
6.72 (cont’d) b. Mole fraction of water in dried gas =
yw =
n 4 2.218 lb - moles W / d lb - moles W(v) = = 1.99 × 10 −4 4 lb - mole n 3 + n 4 (2.218 + 1.112 × 10 ) lb - moles / d
Henry’s law: ywP = Hwxw ⇒
( x w ) max =
(199 . × 10 −4 )(500 psia)(1 atm / 14.7 psia) lb - mole dissolved W = 0.0170 0.398 atm / mole fraction lb - mole solution
c. Solvent/solute mole ratio
37 lb m TEG 1 lb - mole TEG 18.0 lb m W n5 lb - mole TEG = = 4.434 lb m W 150.2 lb m TEG 1 lb m W n2 − n4 lb - mole W absorbed ⇒ n5 = 4.434(17.78 − 2.22) = 69.0 lb - moles TEG / d (xw)in = 0.80(0.0170) = 0.0136
n8 lb-mole W n5 = 69.0 = ⎯⎯⎯⎯ → n8 = 0.951 lb-mole W/d n5 + n8 lb-mole
Solvent stream entering absorber
= m
0.951 lb - moles W 18.0 lb m 69.0 lb - moles TEG + d lb - mole d
150.2 lb m lb - mole
= 1.04 × 104 lb m / d W balance on absorber n6 = (17.78 + 0.95 − 2.22) lb-moles W/d = 16.51 lb-moles W/d 16.51 lb-moles W/d ⇒ ( xw )out = = 0.19 lb-mole W/lb-mole (16.51 + 69.9) lb-moles/d
d. The distillation column recovers the solvent for subsequent re-use in the absorber. 6.73 Basis: Given feed rates G1 G3 G2 100 mol/h 200 mol air/h n1 (mol/h) 0.999 H 2 0.96 H2 0.04 H2 S, sat'd 0.001 H 2S 1.8 atm absorber stripper L2 0°C L1 40°C n3 (mol/h) n4 (mol/h) 0.002 H 2S x 3 (mol H 2 S/mol) (1 – x 3) (mol solvent/mol) 0.998 solvent 0°C heater
6-54
G4 200 mol air/h n2 mol H 2S/mol 0.40°C, 1 at m
n3 (mol/h) x 3 (mol H 2S/mol) (1 – x 3) (mol solvent/mol) 40°C
6.73 (cont’d)
b gb
g
Equilibrium condition: At G1, p H 2S = 0.04 18 . atm = 0.072 atm ⇒ x3 =
p H 2S H H 2S
=
0.072 atm = 2.67 × 10 −3 mole H 2 S mole 27 atm mol fraction
Strategy: Overall H 2 and H 2 S balances ⇒ n1 , n 2 n 2 + air flow rate ⇒ volumetric flow rate at G4 H 2 S and solvent balances around absorber ⇒ n 3 , n 4 0.998n 4 = solvent flow rate
b100gb0.96g = 0.999n ⇒ n Overall H S balance: b100gb0.04g = 0.001n + n Overall H 2 balance:
1
2
1
1
= 961 . mol h n1 = 96.1
⇒ n 2 = 3.90 mol H 2 S h
2
Volumetric flow rate at stripper outlet
b
g
b g b273 + 40gK = 5240 L hr
200 + 3.90 mol 22.4 liters STP VG4 = h 1 mol
273 K
H 2 S and solvent balances around absorber:
b100gb0.04g + 0.002n = 0.001n 0.998n = n d1 − 2.67 × 10 i 4
1
+ n 3 x 3 ⇒ n 4 = 1.335n 3 − 1952
−3
4
3
Solvent flow rate = 0.998n 4 = 5820 mol solvent h 6.74 Basis: 100 g H 2 O Sat'd solution @ 60°C 100 g H 2 O 16.4 g NaHCO 3
Sat'd solution @ 30°C 100 g H 2 O 11.1 g NaHCO3 ms (g NaHCO3 ( s))
bg
NaHCO 3 balance ⇒ 16.4 = 111 . + ms ⇒ ms = 5.3 g NaHCO 3 s % crystallization =
5.3 g crystallized × 100% = 32.3% 16.4 g fed
6.75 Basis: 875 kg/h feed solution m1 (kg H 2 O(v )/h)
875 kg/h x 0 (kg KOH/kg) (1 – x0) (kg H 2O/kg)
Sat'd solution 10°C m2 (kg H2 O(1)/h) 1.03 m2 (kg KOH/h)
m3 (kg KOH-2H 2O( s )/h) 60% of KOH in feed
6-55
|UV ⇒ n W|
3
≈ n 4 = 5830 mol h
6.75 (cont’d)
Analysis of feed: 2KOH + H 2 SO 4 → K 2 SO 4 + 2H 2 O x0 =
bg
22.4 mL H 2 SO 4 l 5 g feed soln
1L
0.85 mol H 2 SO 4
2 mol KOH
56.11 g KOH
L
1 mol H 2 SO 4
1 mol KOH
3
10 mL
= 0.427 g KOH g feed 60% recovery: 875 ( 0.427 )( 0.60 ) = 224.2 kg KOH h m3 =
224.2 kg KOH 92.15 kg KOH ⋅ 2H 2 O = 368.2 kg KOH ⋅ 2H 2 O h (143.8 kg H 2 O h ) h 56.11 kg KOH
KOH balance: 0.427 ( 875 ) = 224.2 + 1.03m2 ⇒ m2 = 145.1 kg h Total mass balance: 875 = 368.2 + 2.03 (145.1) + m1 ⇒ m1 = 212 kg H 2 O h evaporated 6.76 a.
R 0 30 45 g A dissolved CA 0 0.200 0.300 mL solution Plot CA vs. R ⇒ CA = R / 150
CA =
500 mol 1.10 g = 550 g (160 g A, 390 g S) ml The initial solution is saturated at 10.2 °C. 160 g A Solubility @ 10.2 °C = = 0.410 g A g S = 410 . g A 100 g S @ 10.2° C 390 g S 17.5 150 g A 1 mL soln At 0°C, R = 17.5 ⇒ CA = = 0.106 g A g soln mL soln 110 . g soln Thus 1 g of solution saturated at 0°C contains 0.106 g A & 0.894 g S. 0106 . gA Solubility @ 0°C = 0118 g A g S = 118 . . g A 100 g S @ 0° C 0.894 g S 390 g S 11.8 g A Mass of solid A: 160 g A − = 114 g A s 100 g S
b. Mass of solution:
bg
c.
g A remaining in soln
g A initial
0.5 × 390 g S 11.8 g A 160 − 114 g A − = 23.0 g A s 100 g S
b
g
bg
6.77 a. Table 6.5-1 shows that at 50oF (10.0oF), the salt that crystallizes is MgSO 4 ⋅ 7 H 2 O , which
contains 48.8 wt% MgSO4. b. Basis: 1000 kg crystals/h.
m 0 (g/h) sat’d solution @ 130oF
m 1 (g/h) sat’d solution @ 50oF 0.23 g MgSO4/g 0.77 g H2O/g
0.35 g MgSO4/g 0.65 g H2O/g
1000 kg MgSO4·7H2O(s)/h
6-56
6.77 (cont’d)
m 0 = 2150 kg feed / h Mass balance: m 0 = m 1 + 1000 kg / h MgSO 4 balance: 0.35m0 = 0.23m 1 + 0.488(1000) kg MgSO 4 / h ⇒ m = 1150 kg soln / h 1 The crystals would yield 0.488 × 1000 kg / h = 488 6.78
kg anhydrous MgSO 4 h
Basis: 1 lbm feed solution. Figure 6.5-1 ⇒ a saturated KNO3 solution at 25oC contains 40 g KNO3/100 g H2O ⇒ x KNO 3 =
40 g KNO 3 = 0.286 g KNO 3 / g = 0.286 lb m KNO 3 / lb m x (40 + 100) g solution 1 lbm solution @ 80oC 0.50 lbm KNO3/lbm 0.50 lbm H2O/lbm
m1(lbm) sat’d solution @ 25oC 0.286 lbm KNO3/lbm soln 0.714 lbm H2O/lbm soln m2 [lbm KNO3(s)]
Mass balance: 1 lb m = m1 + m2 KNO3 balance: 0.50 lb m KNO3 = 0.286m1 + m2
⇒
m1 = 0.700 lb m solution / lb m feed m2 = 0.300 lb m crystals / lb m feed
0.300 lb m crystals / lb m feed = 0.429 lb m crystals / lb m solution 0.700 lb m solution / lb m feed
Solid / liquid mass ratio =
6.79 a. Basis: 1000 kg NaCl(s)/h.
Figure 6.5-1 ⇒ a saturated NaCl solution at 80oC contains 39 g NaCl/100 g H2O ⇒ x NaCl =
39 g NaCl = 0.281 g NaCl / g = 0.281 kg NaCl / kg (39 + 100) g solution m 2 [kg H 2 O(v) / h]
m 0 (kg/h) solution
m 1 (kg/h) sat’d solution @ 80oC 0.281 kg NaCl/kg soln 0.719 kg H2O/kg soln 1000 kg NaCl(s)/h
0.100 kg NaCl/kg 0.900 kg H2O/kg
0 = m 1 + m 2 Mass balance: m 1 + m 2 NaCl balance: 0.100 kg NaCl = 0281 . m Solid / liquid mass ratio =
⇒
1 = 0.700 lb m solution / lb m feed m 2 = 0.300 lb m crystals / lb m feed m
0.300 lb m crystals / lb m feed = 0.429 lb m crystals / lb m solution 0.700 lb m solution / lb m feed
The minimum feed rate would be that for which all of the water in the feed evaporates to produce solid NaCl at the specified rate. In this case
6-57
6.79 (cont’d)
0100 . (m 0 ) min = 1000 kg NaCl / h ⇒ (m 0 ) min = 10,000 kg / min Evaporation rate: m 2 = 9000 kg H 2 O / h Exit solution flow rate: m 1 = 0
m 2 [kg H 2 O(v) / h]
b.
m 0 (kg/h) solution
m 1 (kg/h) sat’d solution @ 80oC 0.281 kg NaCl/kg soln 0.719 kg H2O/kg soln 1000 kg NaCl(s)/h
0.100 kg NaCl/kg 0.900 kg H2O/kg
40% solids content in slurry ⇒ 1000
kg NaCl kg = 0.400(m 1 ) max ⇒ (m 1 ) max = 2500 h h
0 = 0281 0 = 7025 kg / h NaCl balance: 0.100m . (2500) ⇒ m 0 = 2500 + m 2 ⇒ m 2 = 4525 kg H 2 O evaporate / h Mass balance: m 6.80 Basis: 1000 kg K 2 Cr2 O 7 ( s) h . Let K = K 2 Cr2 O 7 , A = dry air, S = solution, W = water.
Composition of saturated solution:
0.20 kg K 0.20 kg K ⇒ = 01667 . kg K kg soln kg W 1+ 0.20 kg soln
b
g
n2 (mol / h) y2 (mol W(v) / mol)
m e [kg W(v) / h)
(1− y2 )(mol A/ mol) . oC 90o C, 1 atm, Tdp = 392
1 (kg / h) m f (kg / h) m
f +m r (kg / h) m
CRYSTALLIZERCENTRIFUGE
0.210 kg K/ kg
0.90 kg K(s) / kg
DRYER
1000 kg K(s) / h
0.10 kg soln / kg 0.1667 kg K / kg
0.790 kg W(l) / kg
0.8333 kg W/ kg na (mol A / h)
m r (kg recycle / h) 0.1667 kg K / kg 0.8333 kg W / kg
b
g
* 39.2° C ⇒ y 2 = Dryer outlet gas: y 2 P = pW
53.01 mm Hg = 0.0698 mol W mol 760 mm Hg
f = 1000 kg K h ⇒ m f = 4760 kg h feed solution Overall K balance: 0.210m
6-58
6.80 (cont’d)
b
gb
g
1 + 01667 1 = 1090 kg h . 010 . m 1 = 1000 kg h ⇒ m K balance on dryer: 0.90m Mass balance around crystallizer-centrifuge
m f + m r = m e + m 1 + m r ⇒ me = 4760 − 1090 = 3670 kg h water evaporated 95% solution recycled ⇒ m r =
b0.10 × 1090g kg
h not recycled
95 kg recycled 5 kg not recycled
= 2070 kg h recycled Water balance on dryer
. gb1090g kg W h b0.8333gb010 = 0.0698n 18.01 × 10
−3
kg mol
2
⇒ n 2 = 7.225 × 10 4 mol h
Dry air balance on dryer
na =
b1 − 0.0698g7.225 × 10
4
b g
b g
mol 22.4 L STP . × 10 6 L STP h = 151 h 1 mol
6-59
6.81. Basis : 100 kg liquid feed. Assume Patm=1 atm n 2w (kmol H 2 O )(sat' d)
100 kg Feed 0.07 kg Na 2 CO 3 / kg
Reactor Reactor
0.93 kg H 2 O / kg
e
n 2c (kmol CO 2 ) n 2a (kmol Air) 70 o C, 3 atm(absolute)
n1 (kmol) 0.70 kmol CO 2 / kmol m 3 ( kg NaHCO 3 (s))
0.30 kmol Air / kmol
R|m (kg solution) U| S|0.024 kg NaHCO / kgV| T0.976 kg H O / kg W 4
3
2
Filtrate m5 (kg)
Filter
0.024 kg NaHCO 3 / kg 0.976 kg H 2 O / kg Filter cake m 6 (kg) 0.86 kg NaHCO 3 (s) / kg
R|0.14 kg solution U| S|0.024 kg NaHCO / kgV| T0.976 kg H O / kg W 3
2
Degree of freedom analysis: Reactor 6 unknowns (n1, n2, y2w, y2c, m3, m4) –4 atomic species balances (Na, C, O, H) –1 air balance –1 (Raoult's law for water) 0 DF
Filter 2 unknowns –2 balances 0 DF
Na balance on reactor
100 kg 0.07 kg Na 2 CO3 kg
46 kg Na 106 kg Na 2 CO 3
=
⇒ 3.038 = 0.2738(m3 + 0.024m4 )
(1)
Air balance: 0.300 n1 = n2 a
( 2)
(m3 + 0.024m4 ) kg NaHCO 3
23 kg Na 84 kg NaHCO 3
C balance on reactor :
n1 (kmol) 0.700 kmol CO 2 12 kg C 100 kg 0.07 kg Na 2 CO 3 12 kg C + kmol 1 kmol CO2 kg 106 kg Na 2 CO 3 = (n2c )(12) + (m3 + 0.024m2 )(
12 ) ⇒ 8.40n1 + 0.7924 = 12n2c + 01429 . (m3 + 0.024m4 ) 84
H balance : 2 1 2 ) = (n2 w )(2) + (m3 + 0.024m4 )( ) + 0.976m4 ( ) 18 84 18 ⇒ 10.33 = 2n2 w + 0.01190(m3 + 0.024m4 ) + 01084 . m4 ( 4) (100)(0.93)(
6-60
(3)
6.81(cont'd)
O balance (not counting O in the air): 48 16 n1 (0.700)(932) + 100 (0.07)( ) + 100 (0.93)( ) 106 18 48 16 = (n2 w )(16) + n2 c (32) + (m3 + 0.024m4 )( ) + 0.976m4 ( ) 84 18 ⇒ 22.4n1 + 8584 . = 16n2 w + 32n2 c + 0.5714(m3 + 0.024m4 ) + 0.8676m4
(5)
Raoult's Law :
y w P = p w* (70 o C ) ⇒ ⇒ n2 w = 01025 . (n2 w
n2 w 233.7 mm Hg = n2 w + n2 c + n2 a (3 * 760) mm Hg + n2 c + n2 a )
(6)
Solve (1)-(6) simultaneously with E-Z solve (need a good set of starting values to converge). n1 = 0.8086 kmol, n2a = 0.2426 kmol air, n2c = 0.500 kmol CO 2 , n2w = 0.0848 kmol H 2 O(v), m3 = 8.874 kg NaHCO 3 (s), m4 = 92.50 kg solution
NaHCO3 balance on filter: m3 + 0.024m4 = 0.024m5 + m6 [0.86 + ( 014 . )(0.024)] m3 = 8.874
11.09 = 0.024m 5 + 0.8634m 6
(7)
m4 = 92 .50
. = m5 + m6 Mass Balance on filter: 8.874 + 92.50 = 1014 Solve (7) & (8) ⇒ Scale factor =
m5 = 91.09 kg filtrate m6 = 10.31 kg filter cake
(8)
⇒ (0.86)(10.31) = 8.867 kg NaHCO 3 (s)
500 kg / h = 56.39 h −1 8.867 kg
(a) Gas stream leaving reactor
U| V| W
R| |S || T
46.7kmol / h n 2w = (0.0848)(56.39) = 4.78 kmol H 2 O(v) / h 0.102 kmol H 2 O(v) / kmol ⇒ n 2c = (0.500)(56.39) = 28.2 kmol O 2 / h 0.604 kmol CO 2 / kmol n 2a = (0.2426)(56.39) = 13.7 kmol air / h 0.293 kmol Air / kmol
n RT V2 = 2 = P
(46.7 kmol / h)(0.08206 3 atm
m 3 atm )(343 K) kmol ⋅ K = 438 m 3 / h
56.39 × 0.8086 kmol 22.4 m 3 (STP) 1h (b) Gas feed rate: V1 = = 17.0 SCMM h kmol 60 min
6-61
6.81(cont'd) (c) Liquid feed: (100)(56.39) = 5640 kg / h
To calculate V , we would need to know the density of a 7 wt% aqueous Na2CO3 solution. (d) If T dropped in the filter, more solid NaHCO3 would be recovered and the residual solution would contain less than 2.4% NaHCO3. (e) Henry's law
Benefit: Higher pressure ⇒ greater pCO2
higher concentration of CO 2 in solution
⇒ higher rate of reaction ⇒ smaller reactor needed to get the same conversion ⇒ lower cost Penalty: Higher pressure ⇒ greater cost of compressing the gas (purchase cost of compressor, power consumption) 6.82 600 lb m / h Dissolution Dissolution 0.90 MgSO4 ⋅ 7H 2 O Dissolution Tank Tank Tank 010 . I 1 (lb m H 2 O / h) m
Filter I
R|m (lb soln / h) |U S|0.32 kg MgSO / kgV| T0.68 kg H O / kg W 2
4
4
6000 lb m I / h
6 (lb m / h) m 0.23 lb m MgSO 4 / lb m 0.77 lb m H 2 O / lb m
110o F
R|300 lb soln / hU| S|0.32 MgSO V| T0.68 H O W m
m
2
6000 lb m I / h
2
3 (lb m so ln/ h) m 0.32 MgSO 4
0.68 H 2 O
4 (lb m MgSO 4 ⋅ 7 H 2 O / h) m
Filter II
R|m (lb |S0.23 lb ||0.77 lb T 5
m soln) m MgSO 4 / lb m m H 2 O / lb m
U| |V || W
Crystallizer
4 (lb m MgSO 4 ⋅ 7 H 2 O) m
R|0.05m S|0.23 lb T0.77 lb
4
(lb m soln) m MgSO 4 / lb m m H 2 O / lb m
U| V| W
a. Heating the solution dissolves all MgSO4; filtering removes I, and cooling recrystallizes MgSO4 enabling subsequent recovery. (b) Strategy: Do D.F analysis.
6-62
6.82(cont'd)
UV ⇒ m , m balance W
Overall mass balance Overall MgSO 4
1
4
Diss. tank overall mass balance Diss. tank MgSO 4 balance
UV ⇒ m , m W 2
6
( MW) MgSO4 = (24.31 + 32.06 + 64.00) = 120.37, ( MW) MgSO4 ⋅7H2O = (120.37 + 7 * 18.01) = 246.44 Overall MgSO4 balance:
0.90 lb m MgSO 4 ⋅ 7H 2 O 120.37 lb m MgSO 4 lb m 246.44 lb m MgSO 4 ⋅ 7H 2 O = (300 lb m / h)(0.32 lb m MgSO 4 / lb m ) + m 4 (120.37 / 246.44) + 0.05m 4 (0.23)
60,000 lb m h
⇒ m 4 = 5.257 x10 4 lb m crystals / h m 4 = 5.257 x104 lb m / h
Overall mass balance: 60,000 + m 1 = 6300 + 105 . m 4
m 1 = 1494 lb m H 2 O / h
c.
Diss. tank overall mass balance: Diss. tank MgSO 4 balance: ⇒
60,000 + m 1 + m 6 = m 2 + 6000 54,000(120.37 / 246.44) + 0.23m 6 = 0.32m 2
UV W
. m 2 = 1512 x10 5 lb m / h m 6 = 9.575x10 4 lb m / h recycle
Recycle/fresh feed ratio =
9.575x10 4 lb m / h = 64 lb m recycle / lb m fresh feed 1494 lb m / h
6.83 a. n 1 (kmol CO 2 / h)
Cryst Filter
1000 kg H 2 SO 4 / h (10 wt%) 1000 kg HNO 3 / h w (kg H 2 O / h) m
2 (kg CaSO4 / h) m 3 (kg Ca(NO3 )2 / h) m
Filter cake 5 (kg / h) m 0.96 kg CaSO 4 (s) / kg
4 (kg H2O/ h) m
0.04 kg soln / kg
0 (kg CaCO 3 / h) m 0 (kg solution / h) 2m
0 (kg CaCO 3 / h) m
8 (kg soln / h) m
0 (kg solution / h) 2m
R| X (kg CaSO / kg) U| Solution composition: S 500 X (kg H O / kg) |T(1 − 501X )(kg Ca(NO ) / kg)V|W a
4
a
a
6-63
2
3 2
6.83 (cont’d) b. Acid is corrosive to pipes and other equipment in waste water treatment plant. c. Acid feed:
1000 kg H 2 SO 4 / h = 0.10 ⇒ m w = 8000 kg H 2 O / h (2000 + m w ) kg / h
Overall S balance: 1000 kg H 2 SO 4
32 kg S
h
98 kg H 2 SO 4 +
=
5 (kg / h) (0.96 + 0.04 X a ) (kg CaSO 4 ) m
32 kg S
kg
136 kg CaSO 4
32 kg S m 8 (kg / h) X a (kg CaSO 4 ) kg 136 kg CaSO 4
5 (0.96 + 0.04 X a ) + 0.2353m 8 X a ⇒ 3265 . = 0.2353m
(1)
Overall N balance: 1000 kg HNO 3
14 kg N
h
63 kg HNO 3 +
=
0.04m 5 (kg / h) (1 − 501X a ) (kg Ca(NO 3 ) 2 ) kg
m 8 (kg / h) (1 − 501X a ) (kg Ca(NO 3 ) 2 ) kg
28 kg N 164 kg Ca(NO 3 ) 2
28 kg N 164 kg Ca(NO 3 ) 2
⇒ 222.2 = 0.00683m 5 (1 − 501X a ) + 0171 . m 8 (1 − 501X a )
(2)
Overall Ca balance:
0 (kg / h) m
40 kg Ca 100 kg CaCO 3
+ + +
=
5 (kg / h) (0.96 + 0.04X a ) (kg CaSO 4 ) m kg
5 (kg / h) (1 − 501X a ) (kg Ca(NO 3 ) 2 ) 0.04m kg
40 kg Ca 136 kg CaSO 4
40 kg Ca 164 kg Ca(NO 3 ) 2
8 (kg / h) X a (kg CaSO 4 ) m kg
40 kg Ca 136 kg CaSO 4
8 (kg / h) (1 − 501X a ) (kg Ca(NO 3 ) 2 ) m kg
40 kg Ca 164 kg Ca(NO 3 ) 2
⇒ 0.40m 0 = 0.294m 5 (0.96 + 0.04 X a ) + 0.00976m 5 (1 − 501 X a ) + 0.294m 8 X a + 0.244m 8 (1 − 501 X a )
(3)
Overall C balance :
0 (kg / h) m
12 kg C n 1 (kmol CO 2 / h) 1 kmol C = 100 kg CaCO 3 1 kmol CO 2
⇒ 0.01m 0 = n1
( 4)
6-64
12 kg C 1 kmol C
6.83 (cont’d)
Overall H balance : 1000 (kg H 2SO4 )
2 kg H
+
1000 kg HNO3
h
98 kg H 2SO4
5 X a + 5556 8 Xa ⇒ 92517 . = 2.22m . m
(5)
1kg H
h
+
w (kg / h) m
2 kg H
63 kg HNO3 18 kg H 2 O 5 (kg / h) 500 X a (kg H 2 O) 2 kg H 8 (kg / h) 500 X a (kg H 2 O) 2 kg H m 0.04m = + kg 18 kg H 2 O kg 18 kg H 2 O
Solve eqns. (1)-(5) simultaneously, using E-Z Solve. m 0 = 1812.5 kg CaCO 3 (s) / h,
m 5 = 1428.1 kg / h,
m 8 = 9584.9 kg soln / h,
n1 = 18.1 kmol CO 2 / h(v),
X a = 0.00173 kg CaSO 4 / kg
Recycle stream = 2 * m 0 = 3625 kg soln / h
CaSO U . R| 0.00173(kg CaSO / kg) U| R||0173% || S| 500 * 0.00173(kg H O / kg) V| ⇒ S|86.5% H O V| T(1 − 501* 0.00173)(kg Ca(NO ) / kg)W |T13.3% Ca(NO ) |W 4
4 2
2
3 2
d.
3 2
From Table B.1, for CO2: Tc = 304.2 K , Pc = 72.9 atm T (40 + 273.2) K ⇒ Tr = = = 103 . , Tc 304.2
Pr =
30 atm = 0.411 72.9 atm
From generalized compressibility chart (Fig. 5.4-2): zRT 0.86 0.08206 L ⋅ atm 313.2 K L z = 0.86 ⇒ V = = = 0.737 mol ⋅ K 30 atm P mol CO 2 Volumetric flow rate of CO2: 18.1 kmol CO 2 V = n1 * V = h
e.
0.737 L
1000 mol
mol CO 2
1 kmol
= 1.33x10 4 L / h
Solution saturated with Ca(NO3)2: ⇒
1 − 501X a (kg Ca(NO 3 ) 2 / kg) = 1.526 ⇒ X a = 0.00079 kg CaSO 4 / kg 500Xa (kg H 2 O / kg)
Let m 1 (kg HNO3/h) = feed rate of nitric acid corresponding to saturation without crystallization.
6-65
6.83 (cont’d)
Overall S balance: 1000 kg H2SO4
32 kg S
h
98 kg H2SO4
=
5 (kg / h) (0.96 + (0.04)(0.00079)) (kg CaSO4 ) m kg
32 kg S 136 kg CaSO4
8 (kg / h) 0.00079 (kg CaSO4 ) m 32 kg S kg 136 kg CaSO4 5 + 0.000186m 8 ⇒ 326.5 = 0.226m +
(1')
Overall N balance: 1 (kg HNO3 ) 5 (kg / h) (1 − (501)(0.00079)) (kg Ca(NO3 ) 2 ) m 14 kg N 0.04m 28 kg N = h 63kg HNO3 kg 164 kg Ca(NO3 ) 2 8 (kg / h) (1 − (501)(0.00079)) (kg Ca(NO3 ) 2 ) m 28 kg N + kg 164 kg Ca(NO3 ) 2 1 = 000413 5 + 0103 8 m . m . . m (2') ⇒ 0222
Overall H balance: 1000 (kg H 2 SO 4 )
2 kg H
h
98 kg H 2 SO 4
+
8000 (kg / h)
2 kg H 18 kg H 2 O
= +
+
m 1 kg HNO 3
1 kg H
h
63 kg HNO 3
0.04m 5 (kg / h) 500(0.00079) (kg H 2 O) kg
18 kg H 2 O
m 8 (kg / h) 500(0.00079) (kg H 2 O) kg
2 kg H
2 kg H 18 kg H 2 O
⇒ 909.30 + 0.0159m 1 = 0.00175m 5 + 0.0439m 8
(3' )
Solve eqns (1')-(3') simultaneously using E-Z solve: m 1 = 1155 . x10 4 kg / h;
5 = 1424 . x10 3 kg / h; m
Maximum ratio of nitric acid to sulfuric acid in the feed =
1155 . x10 4 kg / h = 115 . kg HNO 3 / kg H 2 SO 4 1000 kg / h
6-66
8 = 2.484 x10 4 kg / h m
6.84
550.0 ml 0.879 g
Moles of benzene (B): ⇒ x DP =
bg
U| |V 1 mol = 6.19 mol | |W 78.11 g
56.0 g = 0.363 mol 154.2 g mol
Moles of diphenyl (DP):
ml
0.363 = 0.0544 mol DP mol 6.19 + 0.363
bg
b
g
p B* T = (1 − x DP ) p B* T = 0.945 120.67 mm Hg = 114.0 mm Hg
b
g b0.0554g = 3.6 K = 3.6 C ⇒ T 8.314b273.2 + 801 .g = . K = 1.85 C b0.0554g = 185
8.314 273.2 + 55 . RT 2 ΔTm = m0 x DP = 9837 ΔH m RTb02 ΔTbp = x DP ΔH
2
o
m
. − 3.6 = 19 . °C = 55
2
o
30,765
v
⇒ Tb = 801 . + 185 . = 82.0 ° C
6.85
Tm 0 = 0.0o C, ΔTm = 4.6o C=4.6K Eq. 6.5-5 ⎯⎯⎯ → xu = Table B.1
ΔTm ΔHˆ m
Eq. (6.5-4) ⇒ ΔTb =
R (Tm 0 )
2
=
(4.6K)(600.95 J/mol) (8.314 J/mol ⋅ K)(273.2K) 2
= 0.0445 mol urea/mol
RTb 0 2 (8.314)(373.2) 2 xu = 0.0445 = 1.3K = 1.3o C ΔHˆ v 40, 656
1000 grams of this solution contains mu (g urea) and (1000 – mu) (g water) nu1 (mol urea) =
mu1 (g)
nw1 (mol water) =
60.06 g/mol
(1000 − mu1 )(g) 18.02 g/mol
mu1
xu1 = 0.0445 =
(mol urea) 60.06 ⇒ mu1 = 134 g urea, mw1 = 866 g water (1000 − mu1 ) ⎤ ⎡ mu1 ⎢⎣ 60.06 + 18.02 ⎥⎦ (mol solution)
ΔTb = 3.0o C = 3.0K ⇒ xu 2 =
ΔTb ΔHˆ v R (Tb 0 )
2
=
(3.0K)(40,656 J/mol) (8.314 J/mol ⋅ K)(373.2K) 2
mu 2
= 0.105 mol urea/mol
(mol urea) 60.06 xu 2 = 0.105 = ⇒ mu 2 = 339 g urea 866 ⎤ ⎡ mu 2 ⎢⎣ 60.06 + 18.02 ⎥⎦ (mol solution) ⇒ Add (339-134) g urea = 205 g urea
6-67
6.86 x aI =
b0.5150 gg b1101. g molg b0.5150 gg b110.1 g molg+ b100.0 gg b94.10 g molg = 0.00438 mol solute mol
II RTm20 x sI ΔTmI mol solute 0.49° C II I ΔTm xs ⇒ ΔTm = = II ⇒ x s = x s = 0.00438 = 0.00523 II I 0.41° C mol solution xs ΔTm ΔTm ΔH m
⇒
b1 − 0.00523g mol solvent 0.00523 mol solute
ΔH m = 6.87 a.
94.10 g solvent 0.4460 g solute 1 mol solvent
b
b g
2
ΔH vI ΔH vII + B , ln p s* Tbs = − +B RTb 0 RTbs
b g
Assume ΔH vI ≅ ΔH vII ; T0 Ts ≅ T02
IJ K ΔH Δ T p bT g = b1 − x g p bT g ⇒ lnb1 − x g ≈ − x = − RT
b g
b g
⇒ ln Ps* Tb 0 − ln P0* Tbs = −
b. Raoult’s Law:
= 83.50 g solute mol
g b0.00523g = 6380 J mol = 6.38 kJ / mol
8.314 273.2 − 5.00 RTm20 xs = 0.49 ΔTm
ln p s* Tb 0 = −
95.60 g solvent
* s
* 0
b0
FG H
ΔH v 1 ΔH v Tbs − Tb 0 1 − ≅ R Tb 0 Tbs R Tb20 v
bs
2 b0
6.88 m 1 (g styrene) 90 g ethylbenezene 100 g EG 90 g ethylbenzene 30 g styrene
m 2 (g styrene) 100 g EG
Styrene balance: m1 + m2 = 30 g styrene Equilibrium relation:
b
FG H
m2 m1 = 0.19 100 + m2 90 + m1
IJ K
solve simultaneously
m1 = 25.6 g styrene in ethylbenzene phase m2 = 4.4 g styrene in ethylene glycol phase
6-68
RTb20 ⇒ ΔTb = x ΔH v
6.89 Basis: 100 kg/h.
A=oleic acid; C=condensed oil; P=propane
100 kg / h 0.05 kg A / kg
95.0 kg C / h 2 kg A / h m
0.95 kg C / kg
3 kg A / h m 1 kg P / h m
1 kg P / h m
a. 90% extraction: m 3 = (0.09)(0.05)(100 kg / h) = 4.5 kg A / h
Balance on oleic acid: (0.05)(100) = m 2 + 4.5 kg A / h ⇒ m 2 = 0.5 kg A / h Equilibrium condition:
015 . =
0.5 / (n1 + 0.5) ⇒ n1 = 73.2 kg P / h 4.5 / (4.5 + 95)
b. Operating pressure must be above the vapor pressure of propane at T=85oC=185oF * Figure 6.1-4 ⇒ p propane = 500 psi = 34 atm c. Other less volatile hydrocarbons cost more and/or impose greater health or environmental hazards. 6.90 a. Benzene is the solvent of choice. It holds a greater amount of acetic acid for a given mass fraction of acetic acid in water.
Basis: 100 kg feed.
A=Acetic acid, W=H2O, H=Hexane, B=Benzene
100 (kg) 0.30 kg A / kg
m1 (kg) 0.10 kg A / kg
0.70 kg W / kg
0.90 kg W / kg
m 2 (kg A)
m H (kg H)
m H (kg H) or m B (kg B)
or m B (kg B)
Balance on W:
100 * 0.70 = m1 * 0.90 ⇒ m1 = 77.8 kg
Balance on A:
100 * 0.30 = m2 + 77.8 * 010 . ⇒ m2 = 22.2 kg
Equilibrium for H: KH =
m2 / (m2 + mH ) 22.2 / ( 22.2 + mH ) = = 0.017 ⇒ mH = 130 . x10 4 kg H xA 010 .
Equilibrium for B: KB =
m2 / ( m2 + mB ) 22.2 / (22.2 + m B ) = = 0.098 ⇒ m B = 2.20 x10 3 kg B xA 010 .
(b) Other factors in picking solvent include cost, solvent volatility, and health, safety, and environmental considerations.
6-69
6.91
a. Basis: 100 g feed ⇒ 40 g acetone, 60 g H 2 O. A = acetone, H = n - C 6 H 14 , W = water 40 g A 60 g W
e 1 (g A) 60 g W
25°C
100 g H
75 g H
100 g H r 1 (g A)
b
25°C
xA in H phase / xA in W phase = 0.343 x = mass fraction Balance on A − stage 1: Equilibrium condition − stage 1:
75 g H r 2 (g A)
g
U| e = 27.8 g acetone b g = 0.343V ⇒ r = 12.2 g acetone |W e b60 + e g 27.8 = e + r U| r = 7.2 g acetone r b75 + r g ⇒ = 0.343V| e = 20.6 g acetone e b60 + e g W
40 = e1 + r1 r1 100 + r1
1
1
1
Balance on A − stage 2:
e 2 (g A) 60 g W
1
2
2
2
Equilibrium condition − stage 2: % acetone not extracted =
2
2
2
2
2
20.6 g A remaining × 100% = 515% . 40 g A fed
b. 40 g A 60 g W
e1 g A 60 g W
r1 g A 175 g H
175 g H
Balance on A − stage 1: Equilibrium condition − stage 1: % acetone not extracted =
c.
40 g A 60 g W
U| r = 17.8 g acetone b g = 0.343V ⇒ e = 22.2g acetone |W b60 + e g
40.0 = e1 + r1 r1 175 + r1 e1
1
1
22.2 g A remaining × 100% = 555% . 40 g A fed 19.4 g A 60 g W
20.6 g A m (g H)
m (g H)
Equilibrium condition:
1
20.6 / (m + 20.6) = 0.343 ⇒ m = 225 g hexane 19.4 / (60 + 19.4)
d. Define a function F=(value of recovered acetone over process lifetime)-(cost of hexane over process lifetime) – (cost of an equilibrium stage x number of stages). The most costeffective process is the one for which F is the highest.
6-70
6.92
a. P--penicillin; Ac--acid solution; BA--butyl acetate; Alk--alkaline solution
Broth Mixing tank
100 kg 0.015 P 0.985 Ac
m1 (kg BA) Extraction Unit I
Acid
D.F. analysis: Extraction Unit I 3 unknown (m1, m2p, m3p) –1 balance (P) –1 distribution coefficient –1 (90% transfer) 0 DF
m3P (kg P) 98.5 (kg Ac) pH=2.1 m4 (kg Alk)
Extraction II m6P (kg P) m1 (kg BA)
m5P (kg P) m4 (kg Alk) pH=5.8
Extraction Unit II (consider m1, m3p) 3 unknowns –1 balance (P) –1 distribution coefficient –1 (90% transfer) 0 DF
b. In Unit I, 90% transfer ⇒ m3 P = 0.90(15 . ) = 135 . kg P P balance: 15 . = m2 P + 135 . ⇒ m2 P = 015 . kg P . / (135 . + m1 ) 135 ⇒ m1 = 34.16 kg BA pH=2.1 ⇒ K = 25.0 = . / (015 . + 98.5) 015
. kg P In Unit II, 90% transfer: m5 P = 0.90(m3 P ) = 1215 P balance: m3 P = 1215 . + m6 P ⇒ m6 P = 0135 . kg P m / (m6 P + 34.16) ⇒ m4 = 29.65 kg Alk . = 6P pH=5.8 ⇒ K = 010 1215 . / (1.215 + m4 ) m1 34.16 kg BA = = 0.3416 kg butyl acetate / kg acidified broth 100 100 kg broth m4 29.65 kg Alk = = 0.2965 kg alkaline solution / kg acidified broth 100 100 kg broth Mass fraction of P in the product solution: m5 P 1.215 P xP = = = 0.394 kg P / kg m4 + m5 P (29.65 + 1.215) kg c. (i). The first transfer (low pH) separates most of the P from the other broth constituents, which are not soluble in butyl acetate. The second transfer (high pH) moves the penicillin back into an aqueous phase without the broth impurities. (ii). Low pH favors transfer to the organic phase, and high pH favors transfer back to the aqueous phase. (iii).The penicillin always moves from the raffinate solvent to the extract solvent.
6-71
6.93
W = water, A = acetone, M = methyl isobutyl ketone x W = 0.20 x A = 0.33 x M = 0.47
U| V| ⇒ W
Figure 6.6-1
Phase 1: x W = 0.07, x A = 0.35, x M = 0.58 Phase 2: x W = 0.71, x A = 0.25, x M = 0.04
Basis: 1.2 kg of original mixture, m1=total mass in phase 1, m2=total mass in phase 2.
H 2 O Balance: Acetone balance:
R| S| T
m1 = 0.95 kg in MIBK - rich phase 12 . * 0.20 = 0.07m1 + 0.71m2 ⇒ m2 = 0.24 kg in water - rich phase 12 . * 0.33 = 0.35m1 + 0.25m2
6.94 Basis: Given feeds: A = acetone, W = H2O, M=MIBK
Overall system composition:
U| b g V 3500 g b20 wt% A, 80 wt% M g ⇒ 700 g A, 2800 g M |W 2200 g A U | ⇒ 3500 g WV ⇒ 25.9% A, 41.2% W, 32.9% M 2800 g M |W
5000 g 30 wt% A, 70 wt% W ⇒ 1500 g A, 3500 g W
Fig. 6.6-1
Phase 1: 31% A, 63% M, 6% W Phase 2: 21% A, 3% M, 76% W
Let m1=total mass in phase 1, m2=total mass in phase 2. H 2 O Balance: Acetone balance: 6.95
R| S| T
m1 = 4200 g in MIBK - rich phase 3500 = 0.06m1 + 0.76m2 ⇒ m2 = 4270 g in water - rich phase 2200 = 0.31m1 + 0.21m2
A=acetone, W = H2O, M=MIBK 41.0 lb m / h x A,1 , x W,1 , 0.70
32 lb m / h x AF (lb m A / lb m ) x WF (lb m W / lb m )
2 lb m / h m x A ,2 , x W ,2 , x M ,2
1 (lb m M / h) m
Figure 6.6-1⇒ Phase 1: x M = 0.700 ⇒ x w ,1 = 0.05; x A,1 = 0.25 ; Phase 2: x w ,2 = 0.81; x A,2 = 0.81; x M ,2 = 0.03 Overall mass balance: MIBK balance:
UV W
. lb m MIBK / h m 1 = 281 32.0 lb m / h + m 1 = 410 . lb m h + m 2 ⇒ . * 0.7 + m 2 * 0.03 m 1 = 410 m 2 = 19.1 lb m h
6-72
6.96 a. Basis: 100 kg; A=acetone, W=water, M=MIBK
System 1: x a,org = 0.375 mol A, x m,org = 0.550 mol M, x w,org = 0.075 mol W x a,aq = 0.275 mol A, x m,aq = 0.050 mol M, x w,aq = 0.675 mol W
UV W
maq,1 = 417 . kg Mass balance: maq ,1 + morg ,1 = 100 ⇒ Acetone balance: maq ,1 * 0.275 + morg ,1 * 0.375 = 33.33 morg ,1 = 58.3 kg System 2: x a,org = 0.100 mol A, x m,org = 0.870 mol M, x w,org = 0.030 mol W x a,aq = 0.055 mol A, x m,aq = 0.020 mol M, x w,aq maq ,2 + morg ,2 = 100
Mass balance:
Acetone balance: maq ,2 * 0.055 + morg ,2 b. K a ,1 =
x a ,org ,1 x a ,aq ,1
=
0.375 = 136 . ; 0.275
K a ,2 =
x a ,org ,2 x a ,aq ,2
=
= 0.925 mol W
UV ⇒ m * 0100 . = 9W m
aq,2
= 22.2 kg
org,2
= 77.8 kg
0.100 = 182 . 0.055
High Ka to extract acetone from water into MIBK; low Ka to extract acetone from MIBK into water. c.
β aw,1 =
x a ,org / x w ,org x a ,aq / x w ,aq
=
0.375 / 0.075 0100 . / 0.040 = 12.3; β = = 418 . aw,2 0.055 / 0.920 0.275 / 0.675
If water and MIBK were immiscible, x w ,org = 0 ⇒ β d.
aw
→∞
Organic phase= extract phase; aqueous phase= raffinate phase
β a ,w =
( x a / x w ) org
=
( x a / x w ) aq
( x a ) org / ( x a ) aq ( x w ) org / ( x w ) aq
=
Ka Kw
When it is critically important for the raffinate to be as pure (acetone-free) as possible. 6.97 Basis: Given feed rates: A = acetone, W = water, M=MIBK
e 1
(kg / h)
x1A (kg A / kg)
x 2A (kg A / kg)
x1W (kg W / kg)
x 2W (kg W / kg)
x1M (kg M / kg)
x 2M (kg M / kg)
200 kg / h 0.30 kg A / kg
e 2
(kg / h)
Stage I
0.70 kg M / kg
r1 (kg / h) y1A (kg A / kg) y1W (kg W / kg) y1M (kg M / kg)
300 kg W / h
6-73
Stage II IIS 300 kg W / h
r2 (kg / h) y 2A (kg A / kg) y 2W (kg W / kg) y 2M (kg M / kg)
6.97(cont'd)
Overall composition of feed to Stage 1:
b200gb0.30g = 60 kg A h U| 500 kg h 200 − 60 = 140 kg M h V ⇒ 12% A, 28% M, 60% W 300 kg W h |W Figure 6.6-1 ⇒
Extract: x1A = 0.095, x1W = 0.880, x1M = 0.025 Raffinate: y1A = 015 . , y1W = 0.035, y1M = 0.815
Mass balance Acetone balance:
R| S| T
e1 = 273 kg / h 500 = e1 + r1 ⇒ r1 = 227 kg / h 60 = 0.095e1 + 015 . r1
Overall composition of feed to Stage 2:
U| 527 kg h . g = 34 kg A h b227gb015 b227gb0.815g = 185 kg M h V ⇒ 6.5% A, 35.1% MIBK, 58.4% W b227gb0.035g + 300 = 308 kg W h|W Figure 6.6-1 ⇒
Extract: x 2 A = 0.04, x 2 W = 0.94, x 2 M = 0.02 Raffinate: y 2 A = 0.085, y 2 W = 0.025, y 2 M = 0.89
Mass balance:
527 = e2 + r2
Acetone balance:
34 = 0.04e2 + 0.085r2
⇒
R|e = 240 kg / h S|r = 287 kg / h T 2
2
Acetone removed: [60 − ( 0.085)(287)] kg A removed / h = 0.59 kg acetone removed / kg fed 60 kg A / h in feed Combined extract: Overall flow rate = e1 + e2 = 273 + 240 = 513 kg / h Acetone:
( x1 A e1 + x 2 A e2 ) kg A
=
0.095 * 273 + 0.04 * 240 = 0.069 kg A / kg 513
Water:
( x1w e1 + x 2 w e2 ) kg W 0.88 * 273 + 0.94 * 240 = = 0.908 kg W / kg e1 + e2 513
MIBK:
( x1 M e1 + x 2 M e2 ) kg M 0.025 * 273 + 0.02 * 240 = = 0.023 kg M / kg (e1 + e2 ) kg 513
6-74
6.98. a.
1.50 L / min 25o C, 1atm, rh = 25% n0 (mol / min)
M (g gel) Ma (g H2O)
y0 (mol H2 O / mol) (1- y0 ) (mol dry air / mol) n 0 =
(1 atm)(1.50 L / min) PV = = 0.06134 mol / min RT (0.08206 L ⋅ atm / mol ⋅ K)(298 K)
r.h.=25%⇒
pH 2 O pH* 2O (25o C)
= 0.25
Silica gel saturation condition:
Water feed rate: ⇒ m H
2O
=
y0 =
X * = 12.5
0.25 p H* 2 O (25o C ) p
pH 2O p H* 2 O
=
= 12.5 * 0.25 = 3125 .
g H 2 O ads 100 g silica gel
0.25( 23.756 mm Hg) mol H 2 O = 0.00781 760 mm Hg mol
0.06134 mol 0.00781 mol H 2 O 18.01 g H 2 O = 0.00863 g H 2 O / min min mol mol H 2 O
Adsorption in 2 hours = (0.00863 g H 2 O / min)(120min) = 1.035 g H 2 O Saturation condition:
1.035 g H 2 O 3.125 g H 2 O = ⇒ M = 33.1 g silica gel M (g silica gel) 100 g silica gel
Assume that all entering water vapor is adsorbed throughout the 2 hours and that P and T are constant. b. Humid air is dehumidified by being passed through a column of silica gel, which absorbs a significant fraction of the water in the entering air and relatively little oxygen and nitrogen. The capacity of the gel to absorb water, while large, is not infinite, and eventually the gel reaches its capacity. If air were still fed to the column past this point, no further dehumidification would take place. To keep this situation from occurring, the gel is replaced at or (preferably) before the time when it becomes saturated. 6.99 a.
Let c = CCl4 Relative saturation = 0.30 ⇒
pc * pc (34 o C)
⇒ pc = 0.30 * (169 mm Hg) = 50.7 mm Hg
b. Initial moles of gas in tank:
n0 =
1 atm 50.0 L P0V0 . mol = = 1985 RT0 0.08206 L ⋅ atm / mol ⋅ K 307 K
Initial moles of CCl4 in tank: nc 0 = y c 0 n0 =
pc 0 50.7 mm Hg × 1985 . mol = 0.1324 mol CCl 4 n0 = 760 mm Hg P0
6-75
6.99 (cont’d)
50% CCl4 adsorbed ⇒ nc = 0.500nc 0 = 0662 mol CCl 4 (= nads)
. − 0.0662) mol = 1.919 mol Total moles in tank: n tot = n0 − nads = (1985 Pressure in tank. Assume T = T0 and V = V0. P=
FG H
IJ FG 760 mm Hg IJ = 735 mm Hg K H atm K
n tot RT0 (1919 . )(0.08206)(307) = atm V0 50.0
yC =
nc 0.0662 mol CCl 4 mol CCl 4 = = 0.0345 n tot 1.919 mol mol
⇒ pc = 0.0345(760 mm Hg) = 26.2 mm Hg
c. Moles of air in tank: na = n0 − nc 0 = (1.985 − 01324 . ) mol air = 1.853 mol air yc =
nc mol CCl 4 . = 0.001 ⇒ nc = 1854 × 10 −3 mol CCl 4 . nc + 1853 mol
. ⇒ n tot = nc + nair = 1854 mol
LM n RT OP = 1854 × 10 mol . 50.0 L N V Q −3
pc = y c P = 0.001
0
tot
0
0.08206 L ⋅ atm 307 K 760 mm mol ⋅ K
1 atm
= 0.710 mm Hg X*
FG g CCl IJ = 0.0762 p H g carbon K 1 + 0.096 p 4
⇒ X* =
c
c
Mass of CCl4 adsorbed
mads = (nc 0 − nc )( MW ) c =
g CCl 4 adsorbed 0.0762(0.710) = 0.0506 g carbon 1 + 0.096(0.710)
− 0.001854) mol CCl 4 (01324 .
153.85 g 1 mol CCl 4
= 20.3 mol CCl 4 adsorbed 20.3 g CCl 4 ads = 400 g carbon Mass of carbon required: mc = g CCl 4 ads 0.0506 g carbon
a.
β β X * = K F p NO ⇒ ln X * = ln K F + β ln p NO 2 2
ln(PNO2)
6.100
y = 1.406x - 1.965
2 1.5 1 0.5 0 -0.5 -1 -1.5 0
1
2 ln(X*)
6-76
3
6.100 (cont’d) .406 .406 ln X * = 1406 . ln p NO2 − 1.965 ⇒ X * = e −1.965 p 1NO . = 0140 p 1NO 2 2
K F = 0.140 (kg NO 2 / 100 kg gel)(mm Hg) −1.406 ; β = 1406 .
b. Mass of silica gel : mg =
π * (0.05 m) 2 (1m) 10 3 L 0.75 kg gel 1m 3
L
= 5.89 kg gel
Maximum NO2 adsorbed : p NO 2 in feed = 0.010(760 mm Hg) = 7.60 mm Hg mads =
0.140(7.60) 1.406 kg NO 2 100 kg gel
5.89 kg gel
= 0.143 kg NO 2
Average molecular weight of feed : MW = 0.01( MW ) NO2 + 0.99( MW ) air = (0.01)(46.01) + (0.99)(29.0) = 29.17
kg kmol
Mass feed rate of NO2: m =
8.00 kg 1 kmol 0.01 kmol NO 2 h 29.17 kg kmol tb =
Breakthrough time:
46.01 kg NO 2 kg NO 2 = 0.126 kmol NO 2 h
0.143 kg NO 2 = 1.13 h = 68 min 0.126 kg NO 2 / h
c. The first column would start at time 0 and finish at 1.13 h, and would not be available for another run until (1.13+1.50) = 2.63 h. The second column could start at 1.13 h and finish at 2.26 h. Since the first column would still be in the regeneration stage, a third column would be needed to start at 2.26 h. It would run until 3.39 h, at which time the first column would be available for another run. The first few cycles are shown below on a Gantt chart. Run
Regenerate
Column 1 0
1.13
2.63
3.39
4.52
6.02
Column 2 1.13
2.26
3.76
4.52
5.65
Column 3 2.26
6-77
3.39
4.89
5.65
6.78
Let S=sucrose, I=trace impurities, A=activated carbon
Add m A (kg A)
mS (kg S)
mS (kg S) m I (kg I)
m I0 (kg I) R0 (color units / kg S)
R (color units / kg S)
Come to equilibrium
V (L)
V (L) m A (kg A)
m IA (kg I adsorbed)
Assume
no sucrose is adsorbed • solution volume (V) is not affected by addition of the carbon m a. R(color units/kg S) = kCi (kg I / L) = k I (1) V •
mIA = mI 0 − mI
k ⇒ ΔR = k (Ci 0 − Ci ) = (m I 0 − mI ) V
ΔR =
kmIA V
km IA / V m ΔR x100% = x100 = 100 IA mI 0 R0 kmI 0 / V m Equilibrium adsorption ratio: X i* = IA mA Normalized percentage color removal:
% removal of color =
υ=
m m % removal ( 3) 100 m IA / m I 0 = 100 IA S = m A / mS m A / mS m A mI 0 m mI 0 ⇒ υ = 100X *i S ⇒ X i* = υ 100mS mI 0 (1),( 5)
Freundlich isotherm X i* = K F Ci β
⇒ υ=
100mS K F mI 0 k
β
9.500 9.000
(2) (3) (4)
(5)
mI 0 R υ = KF ( )β 100mS k
R β = K F' R β
A plot of ln υ vs. ln R should be linear: slope = β ;
ln v
6.101
y = 0.4504x + 8.0718
8.500 8.000 0.000 1.000 2.000 3.000 ln R
6-78
intercept = lnK 'F
6.101 (cont’d)
ln υ = 0.4504 ln p NO2 + 8.0718 ⇒ υ = e 8.0718 R 0.4504 = 3203R 0.4504
⇒ K F' = 3203, β = 0.4504 b. 100 kg 48% sucrose solution ⇒ m S = 480 kg
95% reduction in color
⇒ R = 0.025(20.0) = 0.50 color units / kg sucrose
υ = K F' R β = 3203(0.50) 0.4504 = 2344 ⇒ 2344 =
97.5 % color reduction = ⇒ m A = 20.0 kg carbon m A / mS m A / 480
6-79
CHAPTER SEVEN
7.1
0.80 L 35 . × 10 4 kJ 0.30 kJ work 1h 1 kW = 2.33 kW ⇒ 2.3 kW h L 1 kJ heat 3600 s 1 k J s
2.33 kW 10 3 W 1.341 × 10 −3 hp 1 kW 7.2
1W
= 312 . hp ⇒ 3.1 hp
All kinetic energy dissipated by friction mu 2 2 5500 lbm 552 miles 2 = 2 h2 = 715 Btu
(a) E k =
52802 ft 2 12 mile 2
12 h 2 36002 s 2
1 lbf 32.174 lbm ⋅ ft / s2
9.486 × 10 −4 Btu 0.7376 ft ⋅ lb f
(b)
3 × 108 brakings 715 Btu 1 day 1h 1W 1 MW = 2617 MW −4 day braking 24 h 3600 s 9.486 × 10 Btu/s 106 W ⇒ 3000 MW
7.3
(a) Emissions: 1000 sacks Paper ⇒ Plastic ⇒
2000 sacks
1000 sacks
sack 16 oz ( 0.0045 + 0.0146) oz
(724 + 905) Btu sack
Plastic ⇒
1 lb m
= 6.41 lb m
1 lb m
sack 16 oz
Energy: Paper ⇒
(0.0510 + 0.0516) oz
2000 sacks
= 2.39 lb m
= 163 . × 10 6 Btu
(185 + 464) Btu = 1.30 × 10 6 Btu sack
(b) For paper (double for plastic)
Materials for 400 sacks
Raw Materials Acquisition and Production
Sack Production and Use
7- 1
1000 sacks
400 sacks
Disposal
7.3 (cont’d)
Emissions: Paper ⇒
400 sacks
0.0510 oz
1 lb m
sack 16 oz
+
1000 sacks
0.0516 oz
1 lb m
sack 16 oz
= 4.5 lb m
⇒ 30% reduction Plastic ⇒
800 sacks
0.0045 oz
1 lb m
sack 16 oz
2000 sacks
+
0.0146 oz
1 lb m
sack 16 oz
= 2.05 lb m
⇒ 14% reduction Energy: Paper ⇒
400 sacks
Plastic ⇒
(c) .
724 Btu 1000 sacks + sack
800 sacks
905 Btu = 119 . × 10 6 Btu; 27% reduction sack
185 Btu 2000 sacks + sack
3 × 10 8 persons
1 sack person - day
1 day
464 Btu = 1.08 × 10 6 Btu; 17% reduction sack
1h
24 h
649 Btu
1J
1 MW
1 sack 9.486 × 10
3600 s
-4
Btu 10 6 J / s
= 2,375 MW . (2,375 MW) = 404 MW Savings for recycling: 017 (d) Cost, toxicity, biodegradability, depletion of nonrenewable resources.
7.4
1 ft 3
(a) Mass flow rate: m =
3.00 gal min
7.4805 gal
Stream velocity: u =
3.00 gal
1728 in 3
Kinetic energy: E k =
min
1 ft
3
2
2
1
b g
7.4805 gal Π 0.5
mu 2 0.330 lb m = 2 s
d
(0.792)(62.43) lb m
. g ft b1225 2
2
s
2
in
1 min 60 s
1 ft
1 min
12 in
60 s
−3
f
(b) Heat losses in electrical circuits, friction in pump bearings.
7- 2
. = 1225 ft s
ft ⋅ lb f 1 1 lb f = 7.70 × 10−3 2 s 2 32.174 lb m ⋅ ft / s
. × 10 hp I = 140 . × 10 iFGH 01341 .7376 ft ⋅ lb / sJK
= 7.70 × 10−3 ft ⋅ lb f / s
= 0.330 lb m s
−5
hp
7.5
(a) Mass flow rate: 42.0 m π 0.07 m m = s 4
b
g
2
10 3 L 273 K 1m
3
2 127.9 g 1 kg mu E k = = 2 2 s 1000 g (b)
130 kPa
29 g
273 K
b g
573 K 101.3 kPa 22.4 L STP 42.0 2 m2 s
1N
130 kPa
1J
1 kg ⋅ m /
2
127.9 g 1 mol 673 K 101.3 kPa 22.4 L ( STP ) s
1 mol
s2
1 mol
mol
= 127.9 g s
= 113 J s
N⋅m
1 m3 103
29 g
4
L π (0.07)2 m2
= 49.32 m s
2 127.9 g 1 kg 49.32 2 m2 1N 1J mu . J/s E k = = = 1558 2 2 2 2 s 1000 g s 1 kg ⋅ m / s N ⋅ m ΔE k = E k (400 D C) - E k (300 D C) = (155.8 - 113) J / s = 42.8 J / s ⇒ 43 J / s (c) Some of the heat added goes to raise T (and hence U) of the air 7.6
(a)
ΔE p = mgΔz =
1 gal
(b) E k = − ΔE p ⇒
. ft −10 ft 1 lbf 1 ft 3 62.43 lb m 32174 . ft ⋅ lb f = −834 3 7.4805 gal 1 ft s2 32.174 lbm ⋅ ft / s2
b g
b g
mu 2 = mg − Δz ⇒ u = 2 g − Δz 2
12
LM FG NH
= 2 32.174
IJ b gOP K Q
ft 10 ft s2
12
= 25.4
ft s
(c) False 7.7 (a)
ΔE k ⇒ positive When the pressure decreases, the volumetric flow rate increases, and hence the velocity increases. ΔE ⇒ negative The gas exits at a level below the entrance level. p
(b) m =
b g
5 m π 1.5 cm 2 s 2
1 m3 10 4 cm 2
273 K 10 bars 1 kmol 303 K 1.01325 bars 22.4 m 3 STP
b g
16.0 kg CH 4 1 kmol
= 0.0225 kg s PoutVout nRT V u (m/s) ⋅ A(m2) P P = ⇒ out = in ⇒ out = in 2 PinVin nRT Vin Pout uin (m/s) ⋅ A(m ) Pout ⇒ uout = uin
Pin 10 bar = 5( m s) = 5.555 m s Pout 9 bar
1
2 ΔE k = m (uout − uin2 ) =
2
0.5(0.0225) kg (5.5552 − 5.0002 )m 2 1N 1W 2 2 s s 1 kg ⋅ m/s 1 N ⋅ m/s
= 0.0659 W ( zout − zin ) = ΔE p = mg
0.0225 kg 9.8066 m -200 m s s
= −44.1 W
7- 3
1N 1W 2 kg ⋅ m/s 1 N ⋅ m/s
7.8
Δz = ΔE p = mg
105 m3 103 L 1 kg H2O 981 1N 1 J 2.778 × 10−7 kW⋅ h . m −75 m h 1 m3 1L s2 1 kg ⋅ m/ s2 1 N ⋅ m 1J
. × 104 kW⋅ h h = −204
The maximum energy to be gained equals the potential energy lost by the water, or 2.04 × 10 4 kW ⋅ h 24 h 7 days = 3.43 × 10 6 kW ⋅ h week (more than sufficient) h 1 day 1 week 7.9
(b) Q − W = ΔU + ΔE k + ΔE p
b g = 0 b no height changeg
ΔE k = 0 system is stationary ΔE p
Q − W = ΔU , Q < 0, W > 0 (c) Q − W = ΔU + ΔE k + ΔE p
b
g
b
g
Q = 0 adiabatic , W = 0 no moving parts or generated currents ΔE k = 0 system is stationary ΔE p = 0 no height change
b b
g
g
ΔU = 0 (d). Q − W = ΔU + ΔE k + ΔE p
b
g
W = 0 no moving parts or generated currents ΔE k = 0 system is stationary ΔE p = 0 no height change Q = ΔU , Q < 0
b b
g
g
Even though the system is isothermal, the occurrence of a chemical reaction assures that ΔU ≠ 0 in a non-adiabatic reactor. If the temperature went up in the adiabatic reactor, heat must be transferred from the system to keep T constant, hence Q < 0 .
7.10 4.00 L, 30 °C, 5.00 bar ⇒ V (L), T (°C), 8.00 bar (a). Closed system:
ΔU + ΔE k + ΔE p = Q − W
RSΔE |TΔE
k p
b b
= 0 initial / final states stationary = 0 by assumption
g
g
ΔU = Q − W (b)
Constant T ⇒ ΔU = 0 ⇒ Q = W =
−7.65 L ⋅ bar
8.314 J = −765 J 0.08314 L ⋅ bar
(c) Adiabatic ⇒ Q = 0 ⇒ ΔU = −W = 7.65 L ⋅ bar > 0, Tfinal > 30° C
7- 4
transferred from gas to surroundings
7.11
bg
π 3
2
1 m2 = 2.83 × 10 −3 m 2 10 4 cm 2 (a) Downward force on piston: A=
cm 2
Fd = Patm A + mpiston+weight g =
1 atm 1.01325 × 105 N / m2 2.83 × 10 −3 m2 atm
+
24.50 kg 9.81 m
d
s
i d
Upward force on piston: Fu = APgas = 2.83 × 10 −3 m 2 Pg N m 2
1N 1 kg ⋅ m / s2
2
= 527 N
i
Equilibrium condition: Fu = Fd ⇒ 2.83 × 10 −3 m2 ⋅ P0 = 527 ⇒ P0 = 186 . × 10 5 N m 2 = 186 . × 10 5 Pa
V0 =
303 K 1.01325 × 105 Pa 0.08206 L ⋅ atm nRT 1.40 g N 2 1 mol N 2 = 0.677 L = P0 28.02 g 1.86 × 105 Pa 1 atm mol ⋅ K
(b) For any step, ΔU + ΔE k + ΔE p = Q − W ⇒ ΔU = Q − W ΔE k = 0 ΔE p = 0
Step 1: Q ≈ 0 ⇒ ΔU = −W Step 2: ΔU = Q − W As the gas temperature changes, the pressure remains constant, so that V = nRT Pg must vary. This implies that the piston moves, so that W is not zero. Overall: Tinitial = Tfinal ⇒ ΔU = 0 ⇒ Q − W = 0 In step 1, the gas expands ⇒ W > 0 ⇒ ΔU < 0 ⇒ T decreases
b gd
i b gb gb g
id
. 101325 . × 10 5 2.83 × 10 −3 + 4.50 9.81 1 = 331 N (units (c) Downward force Fd = 100 as in Part (a)) F 331 N = = 116 . × 10 5 N m 2 −3 2 A 2.83 × 10 m P . × 10 5 Pa 186 = 1.08 L Since T0 = T f = 30° C , Pf V f = P0V0 ⇒ V f = V0 0 = 0.677 L Pf . × 10 5 Pa 116
Final gas pressure Pf =
b
Distance traversed by piston =
b
gb
b
g
ΔV 1.08 − 0.677 L = A
g
g
1 m3 103 L
2.83 × 10 −3 m2
m . = 0142
. ⇒ W = Fd = 331 N 0142 m = 47 N ⋅ m = 47 J Since work is done by the gas on its surroundings, W = +47 J ⇒ Q = +47 J Q −W = 0
(heat transferred to gas) 32.00 g 4.684 cm3 103 L 7.12 V = = 01499 . L mol mol g 106 cm3 41.64 atm 0.1499 L 8.314 J / (mol ⋅ K) H = U + PV = 1706 J mol + = 2338 J mol mol 0.08206 L ⋅ atm / (mol ⋅ K)
7- 5
7.13
d
i
Ref state U = 0 ⇒ liquid Bromine @ 300 K, 0.310 bar
(a)
(b) ΔU = U final − U initial = 0.000 − 28.24 = −28.24 kJ mol
d i
Δ H = ΔU + Δ PV = ΔU + PΔV (Pressure Constant)
ΔHˆ = −28.24 kJ mol +
b
0.310 bar
( 0.0516 − 79.94) L
8.314 J
1 kJ
mol 0.08314 L ⋅ bar 103 J
gb
= −30.7 kJ mol
g
Δ H = nΔ H = 5.00 mol −30.7 kJ / mol = −15358 . kJ ⇒ −154 kJ
b
g b
g
(c) U independent of P ⇒ U 300 K, 0.205 bar = U 300 K, 0.310 bar = 28.24 kJ mol U 340 K, Pf = U 340 K, 1.33 bar = 29.62 kJ mol
d
i b
g
ΔU = U final − U initial
E
kJ mol ΔU = 29.62 − 28.24 = 1380 . = P' V' ⇒ V' = PV / P' V changes with pressure. At constant temperature ⇒ PV (T = 300K, P = 0.205 bar) = (0.310 bar)(79.94 L / mol) = 120.88 L / mol V' 0.205 bar 5.00 L 1 mol n= = 0.0414 mol 120.88 L ΔU = nΔU = 0.0414 mol 138 . kJ / mol = 0.0571 kJ
b
gb
g
ΔU + ΔE k + ΔE p = Q − W ⇒ Q = 0.0571 kJ 0
0
0
(d) Some heat is lost to the surroundings; the energy needed to heat the wall of the container is being neglected; internal energy is not completely independent of pressure. 7.14 (a) By definition H = U + PV ; ideal gas PV = RT ⇒ H = U + RT
b g bg
b g bg
bg
U T , P = U T ⇒ H T , P = U T + RT = H T independent of P . cal 50 K cal 1987 + = 3599 cal mol (b) Δ H = ΔU + RΔT = 3500 mol ⋅ K mol Δ H = nΔ H = 2.5 mol 3599 cal / mol = 8998 cal ⇒ 9.0 × 10 3 cal
b
gb
7.15 ΔU + ΔE k + ΔE p = Q − Ws
b b
g
g
Δ E k = 0 no change in m and u Δ E p = 0 no elevation change Ws = PΔV since energy is transferred from the system to the surroundings
g
b
ΔU = Q − W ⇒ ΔU = Q − PΔV ⇒ Q = ΔU + PΔV = Δ (U + PV ) = ΔH
7- 6
g
b b
g
7.16. (a) Δ E k = 0 u1 = u2 = 0 Δ E p = 0 no elevation change
g
ΔP = 0 (the pressure is constant since restraining force is constant, and area is constrant) Ws = PΔV the only work done is expansion work H = 34980 + 355 . T (J / mol), V1 = 785 cm3, T1 = 400 K, P = 125 kPa, Q = 83.8 J
b
.
g
125 × 103 Pa 785 cm3 1 m3 PV = = 0.0295 mol RT 8.314 m3 ⋅ Pa / mol ⋅ K 400 K 106 cm3 -H ) = 0.0295 mol 34980 + 35.5T - 34980 - 35.5(400K) (J / mol) Q = ΔH = n(H 2 1 2 n=
83.8 J = 0.0295 35.5T2 - 35.5(400) ⇒ T2 = 480 K nRT 0.0295 mol 8.314 m3 ⋅ Pa 106 cm3 480 K = 941 cm3 = 125 × 105 Pa 1 m3 mol ⋅ K P 125 × 105 N (941 - 785)cm3 1 m3 = 19.5 J ii ) W = PΔV = 106 cm3 m2 iii ) Q = ΔU + PΔV ⇒ ΔU = Q − ΔPV = 838 . J − 19.5 J = 64.3 J i) V =
(b) ΔEp = 0 7.17 (a) "The gas temperature remains constant while the circuit is open." (If heat losses could occur, the temperature would drop during these periods.) (b) ΔU + Δ E p + Δ E R = Q Δt − W Δt Δ E p = 0, Δ E k = 0, W = 0, U ( t = 0) = 0 Q =
0.90 × 1.4 W
1 J s 1W
= 1.26 J s
U (J ) = 126 . t Moles in tank: n = PV RT =
1 atm
b
2.10 L 25 + 273 K
g
1 mol ⋅ K = 0.0859 mol 0.08206 L ⋅ atm
U 126 . t (J) U = = = 14.67t n 0.0859 mol
Thermocouple calibration: T = aE + b
b g
b g
T ° C = 181 . E mV + 4.51 T = 0 , E =−0.249 T =100 , E = 5.27
U = 14.67t 0 440 880 1320 T = 181 . E + 4.51 25 45 65 85 (c) To keep the temperature uniform throughout the chamber. (d) Power losses in electrical lines, heat absorbed by chamber walls. (e) In a closed container, the pressure will increase with increasing temperature. However, at the low pressures of the experiment, the gas is probably close to ideal ⇒ U = f T only.
bg
Ideality could be tested by repeating experiment at several initial pressures ⇒ same results.
7- 7
7.18 (b) ΔH + ΔE k + ΔE p = Q − W s (The system is the liquid stream.)
c c
h
Δ E k = 0 no change in m and u Δ E p = 0 no elevation change Ws = 0 no moving parts or generated currents
c
h
h
Δ H = Q , Q > 0 (c) ΔH + ΔE k + ΔE p = Q − W s (The system is the water)
c
h
Δ H = 0 T and P ~ constant Δ E k = 0 no change in m and u Q = 0 no Δ T between system and surroundings
c
c
h
b
g
h
ΔE p = −W s , W s > 0 for water system
(d) ΔH + ΔE k + ΔE p = Q − W s (The system is the oil)
c
Δ E k =0 no velocity change
h
ΔH + Δ E p = Q − W s Q < 0 (friction loss); W s < 0 (pump work). (e) ΔH + ΔE k + ΔE p = Q − W s (The system is the reaction mixture)
c
h
Δ E k = Δ E p = 0 given ΔWs = 0 no moving parts or generated current
h
1.25 m3 273 K
1 mol
c
Δ H = Q , Q pos. or neg. depends on reaction
7.19 (a) molar flow:
min
122 kPa
b g
423 K 101.3 kPa 22.4 L STP
103 L 1 m3
= 43.4 mol min
Δ H + Δ E k + Δ E p = Q − W s
c
h
Δ E k = Δ E p = 0 given Ws = 0 no moving parts
c
h
43.37 mol 1 min Q = Δ H = nΔ H = min 60s
3640 J kW = 2.63 kW mol 10 3 J / s
(b) More information would be needed. The change in kinetic energy would depend on the cross-sectional area of the inlet and outlet pipes, hence the internal diameter of the inlet and outlet pipes would be needed to answer this question.
7- 8
b g
7.20 (a) H = 1.04 T ° C − 25
H in kJ kg
H out = 104 . 34.0 − 25 = 9.36 kJ kg H = 1.04 30.0 − 25 = 5.20 kJ kg
n (m ol/ s) N2 30 o C
in
Δ H = 9.36 − 5.20 = 4.16 kJ kg Δ H + Δ E + Δ E = Q − W k
p
s
c
h h
Δ E k = Δ E p = 0 assumed Ws = 0 no moving parts
c
P= 11 0 kP a Q =1 .25 k W
Q = Δ H = nΔ H ⇒ n =
34 o C
1.25 kW kg 1 kJ / s 103 g 1 mol Q = = 10.7 mol s Δ H 4.16 kJ kW 1 kg 28.02 g
b g
10.7 mol 22.4 L STP ⇒ V = s mol
303 K 1013 . kPa 273 K
110 kPa
= 2455 . L / s ⇒ 246 L s
(b) Some heat is lost to the surroundings, some heat is needed to heat the coil, enthalpy is assumed to depend linearly on temperature and to be independent of pressure, errors in measured temperature and in wattmeter reading.
U| V| W
. 7.21 (a) H = aT + b a = H 2 − H1 = 129.8 − 258 = 5.2 ⇒ H kJ kg = 5.2T ° C − 130.2 T2 − T1 50 − 30 b = H 1 − aT1 = 258 . − 5.2 30 = −130.2
b gb g
b
g
b g
130.2 H = 0 ⇒ Tref = = 25° C 5.2
b g
Table B.1 ⇒ S . G.
b
g
bg
C 6 H 14 l
b
1 m3 = 0.659 ⇒ V = = 152 . × 10 −3 m 3 kg 659 kg
g
U kJ kg = H − PV = 5.2T − 130.2 kJ / kg −
b
1 atm 1.0132 × 105 N / m2 1.52 × 10 −3 m3 1
atm
1
kg
1J
1 kJ
1 N ⋅ m 103 J
g
⇒ U kJ kg = 5.2T − 130.4
(b) Energy balance: Q = ΔU = ΔE k , Δ E p , W = 0
20 kg [(5.2 × 20 - 130.4) - (5.2 × 80 - 130.4)] kJ
Average rate of heat removal =
1 6240 kJ 1 min = 20.8 kW 5 min 60 s
7- 9
kg
= −6240 kJ
7.22
m (kg/s) 260°C, 7 bars H = 2974 kJ/kg u0 = 0
m (kg/s) 200°C, 4 bars H = 2860 kJ/kg u (m/s)
ΔH + ΔE k + ΔE p = Q − Ws ΔE p = Q = W s = 0
2 mu H out − H in ΔE k =− ΔH ⇒ =− m 2
d
b
i
g
(2) 2974 − 2860 kJ 103 N ⋅ m 1 kg ⋅ m / s2 m2 u 2 = 2 H in − H out = = 2.28 × 105 2 ⇒ u = 477 m / s kg 1 kJ 1N s
d
i
7.23 (a) 5 L/min
5 L/min 100 mm Hg (gauge)
0 mm Hg (gauge)
Qin
Qout
Since there is only one inlet stream and one outlet stream, and m in = m out ≡ m , Eq. (7.4-12) may be written m Δz = Q − W s m ΔU + m Δ PV + Δ u 2 + mg 2 ΔU = 0 (given )
d i
d i
a
f
Pout − Pin = VΔP m ΔPV = mV Δu = 0 (assume for incompressible fluid ) 2
Δz = 0 W s = 0 (all energy other than flow work included in heat terms) Q = Q in − Q out
VΔP = Q in − Q out
b
g
5 L 100 − 0 mm Hg 1 atm 8.314 J = 66.7 J min (b) Flow work: VΔP = min 760 mm Hg 0.08206 liter ⋅ atm 5 ml O 2 20.2 J Heat input: Q in = = 101 J min min 1 ml O 2 Efficiency:
V Δ P 66.7 J min = × 100% = 66% 101 J min Q in
7- 10
7.24 (a) ΔH + ΔE k + ΔE p = Q − W s ; ΔE k , ΔE p , W s = 0 ⇒ ΔH = Q
b b
g
H 400° C, 1 atm = 3278 kJ kg (Table B.7) H 100° C, sat' d ⇒ 1 atm = 2676 kJ kg (Table B.5)
g
100 kg H 2 O(v)/s 100o C, saturated
100 kg H 2 O(v)/s 400o C, 1 atm Q (kW)
100 kg Q = s
b3278 − 2676.0gkJ
10 3 J
kg
1 kJ
= 6.02 × 10 7 J s
(b) ΔU + ΔE k + ΔE p = Q − W ; ΔE k , ΔE p , W = 0 ⇒ ΔU = Q kJ m3 Table B.5 ⇒ Uˆ (100 ° C, 1 atm ) = 2507 , Vˆ (100 ° C, 1 atm ) = 1.673 = Vˆ ( 400 ° C, Pfinal ) kg kg
Interpolate in Table B.7 to find P at which Vˆ =1.673 at 400oC, and then interpolate again to find Uˆ at 400oC and that pressure: ⎛ 3.11 − 1.673 ⎞ o Vˆ = 1.673 m 3 /g ⇒ Pfinal = 1.0 + 4.0 ⎜ ⎟ = 3.3 bar , Uˆ (400 C, 3.3 bar) = 2966 kJ/kg ⎝ 3.11 − 0.617 ⎠
(
)
⇒ Q = ΔU = mΔ Uˆ = 100 kg [( 2966 − 2507 ) kJ kg ] 10 3 J kJ = 4.59 × 10 7 J
The difference is the net energy needed to move the fluid through the system (flow work). (The energy change associated with the pressure change in Part (b) is insignificant.)
bg
c
h
7.25 H H 2 O l , 20° C = 83.9 kJ kg (Table B.5)
b
g
H steam , 2 0 bars, sat' d = 2 7 97 .2 k J kg
(Table B.6)
m [kg H 2 O(l)/h]
m [kg H 2 O(v)/h]
20o C
20 bar (sat'd) Q =0.65(813 kW) = 528 kW
(a) ΔH + ΔE k + ΔE p = Q − W s ; ΔE k , ΔE p , W s = 0 ⇒ ΔH = Q ΔH = m ΔH
m =
528 kW Q = Δ H
b
gd
kg
b2797.2 − 83.9 gkJ
1 kJ / s 1 kW
3600 s 1
h
= 701 kg h
i
(b) V = 701 kg h 0.0995 m 3 kg = 69.7 m 3 h sat' d steam @ 20 bar
A
Table B.6
701 kg / h 103 g / kg 485.4 K 0.08314 L ⋅ bar 1 m3 nRT = (c) V = = 78.5 m3 / h 18.02 g / mol 20 bar mol ⋅ K 103 L P
The calculation in (b) is more accurate because the steam tables account for the effect of pressure on specific enthalpy (nonideal gas behavior). (d) Most energy released goes to raise the temperature of the combustion products, some is transferred to the boiler tubes and walls, and some is lost to the surroundings.
7- 11
c
bg
h
7.26 H H 2 O l , 24° C, 10 bar = 100.6 kJ kg (Table B.5 for saturated liquid at 24oC; assume H
independent of P).
b
g
H 10 bar, sat' d steam = 2776.2 kJ kg (Table B.6) ⇒ Δ H = 2776.2 − 100.6 = 2675.6 kJ kg
[kg H2O(l)/h] m
[kg H2O(v)/h] m 15,000 m3/h @10 bar (sat'd)
24oC, 10 bar Q (kW) m =
15000 m 3
kg
4 01943 m 3 = 7.72 × 10 kg h .
h
A
b Table 8.6 g
d
i
Energy balance ΔE p , W s = 0 : ΔH + ΔE k = Q
ΔE k = E kfinal − E kinitial Δ E k =
f mu
E kinitial ≈0
7.72 × 10 4 kg
2
=
2
h
ΔE k = E kfinal
d15,000 m h i 0.15 π 4 2
3
2
2
2
A
m
1
h3
1 3
2
3600 s
1J 5
1 kg ⋅ m 2 / s 2 = 5.96x10 J/s
3
A =π D 2 4 = 5.96 × 10 5 J / s
1h 7.72 × 10 4 kg 2675.6 kJ 5.96 × 10 5 J 1 kJ Q = m ΔH + ΔE k = + h kg 3600 s s 10 3 J = 57973 kJ s = 5.80 × 10 4 kW 228 g/min 25oC
7.27 (a)
228 g/min T(oC) Q ( kW)
=0
228 g 1 min ( H out − H in ) J Energy balance: Q = ΔH ⇒ Q W = min 60 s g ΔE x , ΔE p , Ws =0
b g
b g
b g
⇒ H out J g = 0.263Q W
b g
T °C H J g = 0.263Q W
b g
b
g
(b) H = b T − 25
b g
b g
25 26.4 0 4.47
27.8 29.0 32.4 9.28 13.4 24.8
Fit to data by least squares (App. A.1)
b=
∑ H i
i
bT − 25g ∑ bT − 25g i
i
i
2
= 3.34
b g
⇒ H J g = 3.34 T ° C − 25
b
g
350 kg 10 3 g 1 min 3.34 40 − 20 J (c) Q = Δ H = min kg 60 s g
kW ⋅ s 10 3 J
= 390 kW heat input to liquid
(d) Heat is absorbed by the pipe, lost through the insulation, lost in the electrical leads.
7- 12
7.28 m w [ kg H 2 O(v) / min] 3 bar, sat' d
m w [ kg H 2 O(l) / min] 27 o C
Q ( kW)
m e [ kg C 2 H 6 / min] 16 o C, 2.5 bar
m e [ kg C 2 H 6 / min] 93 o C, 2.5 bar
3 3 (a) C H mass flow: m = 795 m 10 L 2.50 bar 2 6 e min m 3 289 K
1
K - mol 30.01 g
0.08314 L - bar
1 kg
mol 1000 g
= 2.487 × 103 kg min H ei = 941 kJ kg , H ef = 1073 kJ kg Energy Balance on C 2 H 6 : ΔE p , W s = 0, ΔE k ≅ 0 ⇒ Q = ΔH
LMb N
g OPQ
2.487 × 103 kJ 1 min kJ kg Q = 2.487 × 103 1073 − 941 = = 5.47 × 103 kW min min 60 s kg
b g bliquid, 27° Cg = 1131. kJ kg (Table B.5)
(b) H s1 3.00 bar, sat' d vapor = 2724.7 kJ kg (Table B.6)
H s2
Assume that heat losses to the surroundings are negligible, so that the heat given up by the
d
condensing steam equals the heat transferred to the ethane 5.47 × 10 3 kW
d
Energy balance on H 2 O: Q = ΔH = m H s2 − H s1 ⇒ m =
Q H s2 − H s1
b
=
−5.47 × 10 3 kJ
gd
s
i
i kg
b1131. − 2724.7gkJ
i
= 2.09 kg s steam
⇒ Vs = 2.09 kg / s 0.606 m 3 kg = 1.27 m 3 s
A
Table B.6
Too low. Extra flow would make up for the heat losses to surroundings. (c) Countercurrent flow Cocurrent (as depicted on the flowchart) would not work, since it
would require heat flow from the ethane to the steam over some portion of the exchanger. (Observe the two outlet temperatures)
7- 13
7.29
250 kg H2 O(v )/min 40 bar, 500°C H 1 (kJ/kg)
Heat exchanger
250 kg/min 5 bar, T 2 (°C), H 2 (kJ/kg)
Turbine
W s =1500 kW
b b
250 kg/min 5 bar, 500°C H3 (kJ/kg)
Q(kW)
g
H 2 O v , 40 bar, 500° C : H 1 = 3445 kJ kg (Table B.7) H O v , 5 bar, 500° C : H = 3484 kJ kg (Table B.7) 2
g
3
(a) Energy balance on turbine: ΔE p = 0, Q = 0, ΔE k ≅ 0
d
i
ΔH = −W s ⇒ m H 2 − H 1 = −W s ⇒ H 2 = H 1 − W s m =
3445 kJ 1500 kJ − s kg
min 60 s = 3085 kJ kg 250 kg 1 min
H = 3085 kJ kg and P = 5 bars ⇒ T = 310° C (Table B.7)
(b) Energy balance on heat exchanger: ΔE p = 0, W s = 0, ΔE k ≅ 0
d
i
250 kg H 3 − H 2 = Q = ΔH = m min
b3484 − 3085gkJ
1 min 1 kW = 1663 kW kg 60 s 1 kJ / s
(c) Overall energy balance: ΔE p = 0, ΔE k ≅ 0
d
i
ΔH = Q − W s ⇒ m s H 3 − H 1 = Q − W s
b3484 − 3445gkJ
250 kg Q = ΔH + ΔW s = min
kg
1 min 1 kW 1500 kJ 1 kW + s 1 kJ / s 60 s 1 kJ / s
= 1663 kW √
b g H Obv , 5 bar, 310° Cg: V
(d) H 2 O v , 40 bar, 500° C : V1 = 0.0864 m 3 kg (Table B.7) 2
u1 = u2 =
2
= 0.5318 m 3 kg (Table B.7)
250 kg 1 min 0.0864 m 3 min
60 s
kg
250 kg min 0.5318 m 3 min 60 s kg
1 0.5 π 4 m 2 2
1 0.5 π 4 m 2 2
250 kg 1 1 min m 2 ΔE k = u2 − u12 = 2 min 2 60 s
. ms = 183
= 113 . ms
. g b11.3g − b183
= 0.26 kW 0. yv would be less (less water evaporates) because some of the energy that would have
vaporized water instead is converted to kinetic energy. (c) Pf = 39.8 bar (pressure at which the water is still liquid, but has the same enthalpy as the feed) (d) Since enthalpy does not change, then when Pf ≥ 39.8 bar the temperature cannot increase, because a higher temperature would increase the enthalpy. Also, when Pf ≥ 39.8 bar , the product is only liquid ⇒ no evaporation occurs.
7-17
7.36 (cont’d) 0.4 Tf (C)
y
0.3 0.2 0.1 0 0
20
40
60
300 250 200 150 100 50 0
80
1
5
10
Pf (bar)
15
20
25
30
36 39.8 60
Pf (bar)
7.37 10 m3, n moles of steam(v), 275°C, 15 bar ⇒ 10 m3, n moles of water (v+l), 1.2 bar 10.0 m3 H2O (v)
10.0 m3
min (kg) 275oC, 1.5 , 15bar bar
mv [kg H2O (v)] ml [kg H2O (l)] Q
(a) P=1.2 bar, saturated,
1.2 bar, saturated
Table B.6
(b) Total mass of water: min =
10 m 3
T2 = 104.8 D C
1 kg = 55 kg 0.1818 m 3
Mass Balance: mv + ml = 55.0 Volume additivity: Vv + Vl = 10.0 m 3 = mv (1428 . m 3 / kg) + ml (0.001048 m 3 / kg) ⇒ mv = 7.0 kg, ml = 48.0 kg condensed (c) Table B.7 ⇒ U in = 2739.2 kJ / kg; Vin = 0.1818 m 3 / kg
R| S| T
3 Table B.6 ⇒ U l = 439.2 kJ / kg; Vl = 0.001048 m / kg U v = 2512.1 kJ / kg; Vv = 1.428 m 3 / kg
Energy balance: Q = ΔU = mv U v + ml U l − minU in ΔE p , ΔE k , W = 0
= [(7.0)(2512.1 kJ / kg) + (48.0)( 439.2) - 55 kg (2739.2)] kJ = −1.12 × 10 5 kJ 7.38 (a) Assume both liquid and vapor are present in the valve effluent. 1 kg H 2 O(v ) / s 15 bar, Tsat + 150o C
m l [ kg H 2 O(l ) / s] m v [ kg H 2 O( v ) / s] 1.0 bar, saturated
(b) Table B.6 ⇒ Tsat'n (15 bar) = 198.3o C ⇒ Tin = 348.3o C Table B.7 ⇒ H in = H (348.3D C, 15 bar) ≈ 3149 kJ / kg Table B.6 ⇒ H l (1.0 bar, sat' d) = 417.5 kJ / kg; H v (1.0 bar, sat' d) = 2675.4 kJ / kg
7-18
7.38 (cont’d)
Energy balance: ΔH = 0 ⇒ m l H l + m v H v − m in H in = 0 ΔE p , ΔE k ,Q , Ws = 0
⇒ m in H in = m l H l + m v H v
m v + m l
3149 kJ / kg = m l ( 417.5) + (1 − m l )( 2675.4)
There is no value of m l between 0 and 1 that would satisfy this equation. (For any value in this range, the right-hand side would be between 417.5 and 2675.4). The two-phase assumption is therefore incorrect; the effluent must be pure vapor. m in = m out = 1 3149 kJ / kg = H (1 bar, Tout ) (c) Energy balance ⇒ m out H out = m in H in Table B.7
Tout ≈ 337 D C
(This answer is only approximate, since ΔE k is not zero in this process). 7.39 Basis: 40 lb m min circulation (a) Expansion valve R = Refrigerant 12 40 lbm R(l)/min
40 lb m / min x v lb m R ( v ) / lb m
93.3 psig, 86°F H = 27.8 Btu/lbm
(1 − x v ) lb m R( l ) / lb m H v = 77.8 Btu / lb m , H l = 9.6 Btu / lb m
Energy balance: ΔE p , W s , Q = 0, neglect ΔE k ⇒ ΔH =
bg
40 X v lb m R v min
b
g
bg
77.8 Btu 40 1 − X v lb m R l + lb m min
E
∑ n H − ∑ n H i
i
out
9.6 Btu 40 lb m − min lb m
b
i
i
=0
in
27.8 Btu =0 lb m
g
X v = 0.267 26.7% evaporates
(b) Evaporator coil 40 lbm /min 0.267 R(v ) 0.733 R( l ) 11.8 psig, 5°F H v = 77.8 Btu/lbm , H l = 9.6 Btu/lbm
40 lb m R( v )/min 11.8 psig, 5°F H = 77.8 Btu/lbm
Energy balance: ΔE p , W s = 0, neglect ΔE k ⇒ Q = ΔH
40 lb m Q = min
b gb
g
bg
77.8 Btu 40 0.267 lb m R v − lb m min = 2000 Btu min
7-19
b gb
g
bg
40 0.733 lb m R l 77.8 Btu − lb m min
9.6 Btu lb m
7.39 (cont’d) (c) We may analyze the overall process in several ways, each of which leads to the same
result. Let us first note that the net rate of heat input to the system is
Q = Q evaporator − Q condenser = 2000 − 2500 = −500 Btu min and the compressor work Wc represents the total work done on the system. The system is
b g
closed (no mass flow in or out). Consider a time interval Δt min . Since the system is at steady state, the changes ΔU , ΔE k and ΔE p over this time interval all equal zero. The total heat input is Q Δt , the work input is W Δt , and (Eq. 8.3-4) yields c
−500 Btu 1 min 1.341 × 10 −3 hp . hp Q Δt − W c Δt = 0 ⇒ W c = Q = = 118 min 60 s 9.486 × 10 −4 Btu s
7.40 Basis: Given feed rates n1 (mol / h)
nC 3H 8 (mol C 3 H 8 / h) nC 4 H10 (mol C 4 H 10 / h) 227 o C
0.2 C 3 H 8 0.8 C 4 H 10 0o C, 1.1 atm
n2 (mol / h) 0.40 C 3 H 8 0.60 C 4 H 10 25o C, 1.1 atm
Q (kJ / h)
Molar flow rates of feed streams: 300 L 1.1 atm 1 mol = 14.7 mol h n1 = hr 1 atm 22.4 L STP
b g
n 2 =
200 L 273 K 1.1 atm 1 mol = 9.00 mol h hr 298 K 1 atm 22.4 L STP
b g
14.7 mol 0.20 mol C 3 H 8 9.00 mol 0.40 mol C 3 H 8 + h mol h mol = 6.54 mol C 3 H 8 h Total mole balance: n C4 H10 = (14.7 + 9.00 − 6.54) mol C 4 H 20 h = 17.16 mol C 4 H 20 h Propane balance ⇒ n C3H 8 =
Energy balance: ΔE p , W s = 0, neglect ΔE k ⇒ Q = ΔH Q = ΔH =
∑ N H − ∑ N H i
i
i
out
−
in
b0.40 × 9.00g mol C H 3
h
8
i
=
6.54 mol C 3 H 8 h
b
20.685 kJ 17.16 mol C 4 H 10 + mol h
g
1.772 kJ 0.60 × 9.00 mol C 4 H 10 − mol h
( H i = 0 for components of 1st feed stream)
7-20
27.442 kJ mol
2.394 kJ = 587 kJ h mol
510 m 3 273 K 10 3 L 1 mol min 291 K m 3 22.4 L STP
b g
7.41 Basis: (a)
.
n0 (kmol/min) 38°C, h r = 97% y 0 (mol H 2 O/mol) (1 –yx0) (mol dry air/mol)
1 kmol = 214 . kmol min 10 3 mol
21.4 kmol/min 18°C, sat'd y 1 (mol H 2 O/mol) (1 – y 1) (mol dry air)
.
n2 (kmol H 2O(l )/mol) 18°C
Q (kJ/min)
Inlet condition: yo =
hr PH∗2O ( 38°C ) P PH∗2O (18°C )
Outlet condition: y1 =
P
b
=
=
g b
0.97 ( 49.692 mm Hg ) 760 mm Hg
= 0.0634 mol H 2 O mol
15.477 mm Hg = 0.0204 mol H 2 O mol 760 mm Hg
g
Dry air balance: 1 − 0.0634 n o = 1 − 0.0204 214 . ⇒ n o = 22.4 kmol min
b
g
b
g
Water balance: 0.0634 22.4 = n 2 + 0.0204 21.4 ⇒ n 2 = 0.98 kmol min 0.98 kmol 18.02 kg = 18 kg / min H 2 O condenses min kmol
b
g
b
g
(b). Enthaphies: H air 38° C = 0.0291 38 − 25 = 0.3783 kJ mol
b g b g bv, 38° Cg = 2570.8 kJkg 101 kgg 18.02molg = 46.33 kJ molU| | bv, 18° Cg = 2534.5 kgkJ 101 kgg 18.02molg = 45.67 kJ mol |VTable B.5 || 75.5 kJ 1 kg 18.02 g bl, 18° Cg = kg 10 g mol = 1.36 kJ mol |W H air 18° C = 0.0291 18 − 25 = −0.204 kJ mol
H H 2 O H H 2 O H H 2 O
3
3
3
Energy balance: ΔE , W = 0, ΔE ≅ 0 p
s
gd ib g +b0.0204gd214 . × 10 ib45.67g + d0.98 × 10 ib136 . g − b1 − 0.0634gd22.4 × 10 ib0.3783g −b0.0634gd22.4 × 10 ib46.33g = −5.67 × 10 kJ min k
Q = ΔH =
∑ n H − ∑ n H i
i
i
out
i
b
⇒ Q = 1 − 0.0204 214 . × 10 3 −0.204
in
3
3
3
3
4
4 ⇒ 5.67 × 10 kJ 60 min 0.9486 Btu 1 ton cooling = 270 tons of cooling min h kJ 12000 Btu
7-21
7.42 Basis: 100 mol feed n2 (mol), 63.0°C 0.98 A(v ) 0.02 B(v )
A - Acetone B - Acetic Acid
Qc (cal)
0.5n2 (mol) 0.98 A(l ) 0.02 B(l )
100 mol, 67.5°C 0.65 A(l ) 0.35 B(l )
56.8°C
n5 (mol), 98.7°C 0.544 A(v ) 0.456 B(v )
0.5n2 (mol) 0.98 A(l ) 0.02 B(l )
n5 (mol), 98.7°C 0.155 A(l ) 0.845 B(l )
Qr (cal)
(a) Overall balances: Total moles: 100 = 0.5n2 + n5 n2 = 120 mol A: 0.65 100 = 0.98 0.5n2 + 0155 . n5 n5 = 40 mol
b g
b
g
UV W
b g b g 0.155b40g = 6.2 mol A 0.845b40g = 338 . mol B
Product flow rates: Overhead 0.5 120 0.98 = 58.8 mol A 0.5 120 0.02 = 1.2 mol B
Bottoms
Overall energy balance: Q = ΔH = ΔE , W = 0, ΔE ≅ 0 p
2
∑ n H − ∑ n H i
i
out
x
i
i
in
interpolate in table
↓
interpolate in table
↓
⇒ Q = 58.8 ( 0 ) + 1.2 ( 0 ) + 6.2 (1385 ) + 33.8 (1312 ) − 65 ( 354 ) − 35 ( 335 ) = 1.82 × 10 4 cal
b g 2b12 . g = 2.4 mols B
(b) Flow through condenser: 2 58.8 = 117.6 mols A
Energy balance on condenser: Qc = ΔH ΔE , W = 0, ΔE ≅ 0 p
b
3
k
g
b
g
Qc = 117.6 0 − 7322 + 2.4 0 − 6807 = −8.77 × 10 5 cal heat removed from condenser Assume negligible heat transfer between system & surroundings other than Qc & Qr
(
)
Qr = Q − Qc = 1.82 × 104 − −8.77 × 105 = 8.95 × 105 cal heat added to reboiler
7.43
1.96 kg, P1= 10.0 bar, T1
2.96 kg, P3= 7.0 bar, T3=250oC
1.00 kg, P2= 7.0 bar, T2 Q= 0
7-22
7.43 (cont’d) (a) T2 = T ( P = 7.0 bar, sat' d steam) = 165.0 o C H 3 ( H 2 O(v ), P = 7.0 bar, T = 250 o C) = 2954 kJ kg (Table B.7) H 2 ( H 2 O(v ), P = 7.0 bar, sat' d) = 2760 kJ kg (Table B.6) Energy balance ΔE , Q , W , ΔE ≅ 0 p
s
k
ΔH = 0 = 2.96 H 3 − 196 . H 1 − 10 . H 2 ⇒ 196 . H 1 = 2.96 kg(2954 kJ / kg) - 1.0 kg(2760 kJ / kg) ⇒ H (10.0 bar, T ) = 3053 kJ / kg ⇒ T ≅ 300 D C 1
1
1
(b) The estimate is too low. If heat is being lost the entering steam temperature would have to be higher for the exiting steam to be at the given temperature. 7.44
T1 = T ( P = 3.0 bar, sat' d.) = 133.5D C
(a)
Vapor
Vl ( P = 3.0 bar, sat' d.) = 0.001074 m / kg V ( P = 3.0 bar, sat' d.) = 0.606 m 3 / kg 3
P=3 bar
v
Liquid
0.001074 m 3 1000 L 165 kg Vl = = 177.2 L kg m3 Vspace = 200.0 L - 177.2 L = 22.8 L mv =
22.8 L
m=165.0 kg V=200.0 L Pmax=20 bar
3
1 m 1 kg = 0.0376 kg 1000 L 0.606 m 3
(b) P = Pmax = 20.0 bar;
mtotal = 165.0 + 0.0376 = 165.04 kg
T1 = T ( P = 20.0 bar, sat' d.) = 212.4 D C Vl ( P = 20.0 bar, sat' d.) = 0.001177 m 3 / kg; Vv ( P = 20.0 bar, sat' d.) = 0.0995 m 3 / kg V = m V + m V ⇒ m V + (m − m )V total
l l
v v
⇒ 200.0 L
l l
total
l
v
3
1 m = m kg(0.001177 m 3 / kg) + (165.04 - m ) kg(0.0995 m 3 / kg) l l 1000 L
⇒ ml = 164.98 kg; mv = 0.06 kg Vl =
0.001177 m 3 kg
mevaporated =
1000 L 164.98 kg = 194.2 L; m3
Vspace = 200.0 L - 194.2 L = 5.8 L
(0.06 - 0.04) kg 1000 g = 20 g kg
(c) Energy balance Q = ΔU = U ( P = 20.0 bar, sat' d) − U ( P = 3.0 bar, sat' d) ΔE , W , ΔE ≅ 0 p
s
k
U l ( P = 20.0 bar, sat' d.) = 906.2 kJ / kg; U v ( P = 20.0 bar, sat' d.) = 2598.2 kJ / kg U ( P = 3.0 bar, sat' d.) = 561.1 kJ / kg; U ( P = 3.0 bar, sat' d.) = 2543 kJ / kg l
v
Q = 0.06 kg(2598.2 kJ / kg) + 164.98 kg(906.2 kJ / kg) - 0.04 kg(2543 kJ / kg)
− 165.0 kg (561.1 kJ / kg) = 5.70 × 10 4 kJ Heat lost to the surroundings, energy needed to heat the walls of the tank
7-23
7.44 (cont’d) (d) (i) The specific volume of liquid increases with the temperature, hence the same mass of liquid water will occupy more space; (ii) some liquid water vaporizes, and the lower density of vapor leads to a pressure increase; (iii) the head space is smaller as a result of the changes mentioned above. (e) – Using an automatic control system that interrupts the heating at a set value of pressure – A safety valve for pressure overload. – Never leaving a tank under pressure unattended during operations that involve temperature and pressure changes. 7.45 Basis: 1 kg wet steam (a) 1 kg H2 O 20 bars 0.97 kg H2 O(v) 0.03 kg H2 O(l) H1 (kJ/kg)
1 kg H 2O,(v) 1 atm
1 kg H2 O Tamb, 1 atm
H2 (kJ/kg)
Q=0
b b
Q
U|Vb |W
g g
Enthalpies: H v , 20 bars, sat' d = 2797.2 kJ kg Table B.7 H l , 20 bars, sat' d = 908.6 kJ kg
g
b
g
b
g
Energy balance on condenser: ΔH = 0 ⇒ H 2 = H 1 = 0.97 2797.2 + 0.03 908.6 ΔE , ΔE , Q , W =0 p
K
3
⇒ H 2 = 2740 kJ / kg
Table B.7
T ≈ 132 o C
(b) As the steam (which is transparent) moves away from the trap, it cools. When it reaches its saturation temperature at 1 atm, it begins to condense, so that T = 100° C . The white
plume is a mist formed by liquid droplets.
bg
bg
1 quart 1 m3 1000 kg = 0.2365 kg H 2 O l 32 oz 1057 quarts m3 (For simplicity, we assume the beverage is water)
7.46 Basis:
8 oz H 2 O l
0.2365 kg H2O (l) 18°C m (kg H2O (s)) 32°F (0°C)
(m + 0.2365) (kg H2O (l)) 4°C
Assume P = 1 atm
Enthalpies (from Table B.5): Hˆ ( H 2 O(l ), 18°C ) = 75.5 kJ/kg; Hˆ ( H 2 O(l ), 4°C ) = 16.8 kJ/kg; Hˆ ( H 2 O(s), 0°C ) =-348 kJ/kg Energy balance ( closed isobaric system ) : ⇒ ΔH =
∑ n Hˆ − ∑ n Hˆ i
out
ΔE p , ΔEk , Q, W = 0
i
i
i
=0
in
⇒ (m + 0.2365) kg(16.8 kJ / kg) = 0.2365 kg(75.5 kJ / kg) + m kg (-348 kJ / kg) ⇒ m = 0.038 kg = 38 g ice
7-24
7.47 (a) When T = 0 o C, H = 0, ⇒ Tref = 0 o C (b) Energy Balance-Closed System: ΔU = 0 ΔE , ΔE , Q , W = 0 k
p
25 g Fe, 175°C 25 g Fe 1000 g H2O Tf (°C)
1000 g H2O(l) 20°C
d i
b g d i b g 4.13dT − 175ical 4.184 J = 432 T − 175 J
U Fe T f + U H 2O T f − U Fe 175° C − U H 2 O 20° C, 1 atm = 0 or ΔU Fe + ΔU H 2 O = 0 ΔU Fe =
25.0 g
f
g
f
cal
d i
e
j
1.0 L 10 3 g U H 2 O T f − 83.9 J = 1000 U H 2 O T f − 83.9 J 1 L g . × 10 5 = f T f = 0 ⇒ 432T f + 1000U H 2 O T f − 160
Table B.5 ⇒ ΔU H 2 O =
d i
⇒
Tf ° C
30
f Tf
−2.1 × 10
d i
d i
j
d i
40 4
e
+2.5 × 10
7-25
35 4
34
1670 −2612
Interpolate
T f = 34.6° C
7.48
I
II
H 2 O(v ) 760 mm Hg 100°C
H 2 O(v ) (760 + 50.1) mm Hg Tf ⇒ 1.08 bar sat'd ⇒ Tf = 101.8°C (Table 8.5) H 2 O( l ), Tf
⇒
H 2 O( l ), 100 °C T0
Tf
Energy balance - closed system: ΔE p , ΔE K , W , Q = 0
ΔU = 0 = mvIIU vII + mlIIU lII + mbIIU bII − mvI U vI − mlI U lI − mbI U bI
Vl Vv U l U
b
I 101 . bar, 100° C 1044 . 1673 419.0 2506.5
bL kgg bL kgg bL kgg bL kgg
v
v -vapor l -liquid b -block
. bar, 101.8° Cg g II b108 1046 . 1576 426.6 2508.6
Initial vapor volume: VvI = 20.0 L − 5.0 L −
b
50 kg
g
1L 8.92 kg
bg
= 14.4 L H 2 O v
bg
Initial vapor mass: mvI = 14.4 L 1673 L kg = 8.61 × 10 −3 kg H 2 O v
b g = 0.36b101.8g = 36.6 kJ kg
bg
Initial liquid mass: mlI = 5.0 L 1.044 L kg = 4.79 kg H 2 O l Final energy of bar: U bII
Assume negligible change in volume & liquid ⇒ VvII = 14.4 L
b
g
bg
g b g
b
Final vapor mass: mvII = 14.4 L 1576 L kg = 9.14 × 10 −3 kg H 2 O v Initial energy of the bar:
d
b
g
b
g
b
gi
1 9.14 × 10 −3 2508.6 + 4.79 426.6 + 5.0 36.6 − 8.61 × 10 −3 2506.5 − 4.79 419.0 5.0 kg = 441 . kJ kg 44.1 kJ / kg (a) Oven Temperature: To = = 122.5° C 0.36 kJ / kg ⋅ o C U bI =
H 2 O evaporated = mvII − mvI = 9.14 × 10 −3 kg - 8.61 × 10 −3 kg = 5.30 × 10 −4 kg = 0.53 g (b) U bI = 44.1 + 8.3 5.0 = 458 . kJ kg To = 458 . 0.36 = 127.2° C (c) Meshuggeneh forgot to turn the oven on ( To < 100° C )
7-26
7.49 (a) Pressure in cylinder = P=
30.0 kg
weight of piston + atmospheric pressure area of piston
b100 cmg 1 bmg
9.807 N
400.0 cm2
2
2
kg
2
10 . bar 105 N m2
+
1 atm 1.013 bar atm
. bar = 108
. °C ⇒ Tsat = 1018
Heat required to bring the water and block to the boiling point
d b
g
b
gi
d b g
b
Q = ΔU = mw U wl 108 . bar, sat' d − U wl l, 20° C + m Al U Al Tsat − U Al 20° C =
7.0 kg
b426.6 − 83.9gkJ + 3.0 kg
gi
[0.94(1018 . − 20)]kJ = 2630 kJ kg
kg 2630 kJ < 3310 kJ ⇒ Sufficient heat for vaporization
V = 1046 L kg , U l = 426.6 kJ kg . (b) T f = Tsat = 1018 . ° C . Table B.5 ⇒ l Vv = 1576 L kg , U v = 2508.6 kJ kg 7.0 kg H 2 O(l ) H = 426.6 kJ / kg V = 1.046 L / kg
mv (kg H 2 O(v )) 1576 L/kg, 2508.6 kJ/kg
T ≡ 101.8°C P ≡ 1.08 bars
Q (kJ)
1.046 L/kg, 426.6 kJ/kg ml (kg H 2 O(l )) W (kJ)
(Since the Al block stays at the same temperature in this stage of the process, we can ignore it -i.e., U in = U out ) Water balance: 7.0 = ml + mv (1) Work done by the piston: W = F Δz = w piston + Patm A Δ z =
OPb AΔ zg = PΔV ⇒ W = b1.08 bar g 1576m + 1.046m − b1.046gb7.0g L Q 8.314 J / mol ⋅ K 1 kJ × = b170.2 m + 0.113m − 0.7908g kJ 0.08314 liter - bar / mol ⋅ K 10 J
LM w + P NA
atm
v
3
l
v
l
Energy balance: ΔU = Q − W Q ΔU W
⇒ 2508.6mv + 426.6m L − 426.6 7 = (3310 − 2630) − (170.2mv + 0.113m L − 0.7908) ⇒ 2679mv + 426.7m L − 3667 = 0 (2) Solving (1) and (2) simultaneously yields mv = 0.302 kg , ml = 6.698 kg
bg
b gb g Vapor volume = b0.302 kggb1576 L kgg = 476 L vapor . gL ΔV 7.01 + 476 − b7.0gb1046 Piston displacement: Δz = = Liquid volume = 6.698 kg 1046 . L kg = 7.01 L liquid
10 3 cm3
1
= 1190 cm 1 L 400 cm2 ⇒ All 3310 kJ go into the block before a measurable amount is transferred to the A
(c) Tupper
b
g b
g
water. Then ΔU AL = Q ⇒ 3.0 kg 0.94 Tu − 20 kJ kg = 3310 ⇒ Tu = 1194° C if melting is o
neglected. In fact, the bar would melt at 660 C.
7-27
7.50
100 . L H 2 O( v ), 25o C m v1 (kg)
not all the liquid UV Assume Eq. at W Tis vaporized. , P . m = kg H O vaporized.
m v2 [kg H 2 O(v)] = m v1 + m e
f
o
f
e
2
m L2 [kg H 2 O(l)] = m L1 + m e
4.00 L H 2 O(l ), 25 C m L1 (kg)
Q=2915 kJ
Initial conditions: Table B.5 ⇒ U L1 = 104.8 kJ kg , VL1 = 1.003 L kg P = 0.0317 bar T = 25° C, sat' d ⇒ U = 2409.9 kJ kg , V = 43,400 L kg
b
gb
v1 −5
g
mv1 = 1.00 l 43400 l kg = 2.304 × 10 Energy balance:
d
L1
b
kg , m LI = 4.00 l
i d i b
g b1.003 l kgg = 3.988 kg
g d i d
ib
ΔU = Q ⇒ 2.304 × 10 −5 + me U v T f + 3.988 − me U L T f − 2.304 × 10 −5 2409.9
b
g
− 3.988 (104.8) = 2915 kJ
d
g d i i dEi b 3333 − d2.304 × 10 iU − 3.988U =
⇒ 2.304 × 10 −5 + me U v T f + 3.988 − me U v T f = 3333 −5
⇒ me
F GG H
v
L
U v − U L
I J d i b A JK A 5.00 − d2.304 × 10 iV − 3.988V =
(1)
g d i
V L + Vv = Vtan k ⇒ 2.304 × 10 −5 + me VL T f + 3.988 − me VL T f = 5.00 L kg
liters kg
−5
⇒ me
b1g − b2g ⇒ f dT i f
v
b2g
L
Vv − VL 3333 − 2.304 × 10 −5 U v T f − 3.988U L T f = U − U
d
i d i
d
v
L
d i
i
5.00 − 2.304 × 10 −5 Vv − 3.988VL − =0 V − V v
L
d i Find T
Table 8.5
Procedure: Assume T f Tf U v U L 2014 . 2593.8 856.7 198.3 2592.4 842.9 195.0 2590.8 828.5 196.4 25915 . 834.6
bg
Eq 1
bg
⇒ U v , U L , Vv , VL ⇒ f T f
Vv 123.7 131.7 140.7 136.9
f
d i
such that f T f = 0
VL f 1159 . . × 10 −2 −512 1154 . −1.93 × 10 −2 1149 134 . . × 10 −2 1151 . −4.03 × 10 −4 ⇒ T f ≅ 196.4° C, Pf = 14.4 bars
me = 2.6 × 10 −3 kg ⇒ 2.6 g evaporated
or Eq 2
7-28
g
7.51.
Basis: 1 mol feed B = benzene T = toluene
nV (mol vapor) y B(mol B(v)/mol) (1 – y B ) (mol T(v)/mol)
1 mol @ 130°C z B (mol B(l)/mol) (1 – z B )(mol T(l)/mol)
in equilibrium at T(°C), P(mm Hg)
nL (mol liquid) x B(mol B(l)/mol) (1 – x B ) (mol T(l)/mol)
(a) 7 variables: (nV , y B , n L , x B , Q, T , P) –2 equilibrium equations –2 material balances –1 energy balance 2 degrees of freedom. If T and P are fixed, we can calculate nV , y B , n L , x B , and Q. (b) Mass balance: nV + n L = 1 ⇒ nV = 1 − n2 Benzene balance: z B = nV y B + n L x B
(1) (2)
bg d i d i C H bv g: dT = 80, H = 4161 . i , dT = 120, H = 45.79i ⇒ H = 01045 . T + 33.25 C H bl g: dT = 0, H = 0i , dT = 111, H = 18.58i ⇒ H = 01674 . T C H bv g: dT = 89, H = 49.18i , dT = 111, H = 52.05i ⇒ H = 01304 . T + 37.57 C 6 H 6 l : T = 0, H = 0 , T = 80, H = 10.85 ⇒ H BL = 01356 . T 6
6
7
8
7
8
BV
TL
TV
Energy balance: ΔE p , Ws = 0, neglect ΔE k
b
g g b g
b
g
bg
b g
Q = ΔH = nV y B H BV + nV 1 − y B H TV + n L x B H BL + n L 1 − x B H TL − 1 z B H BL TF − 1 1 − z H T
b gb
Raoult' s Law:
B
TL
(3) (4) (5) (6)
(7)
F
y B P = x B p B*
(8)
(1 - y B ) P = (1 − x B ) pT* Antoine Equation. For T= 90°C and P=652 mmHg: pB* (90o C) = 10[6.89272−1203.531/(90+ 219.888)] = 1021 mmHg pT* (90o C) = 10[6.95805−1346.773/(90 + 219.693)] = 406.7 mmHg Adding equations (8) and (9) ⇒ P = x B pB* + (1 − x B ) pT* ⇒ x B =
P − pT* pB*
−
pT*
=
P − pT* pB*
x B p*B
−
pT*
=
652 − 406.7 = 0.399 mol B(l) / mol 1021- 406.7
0.399(1021 mmHg) = = 0.625 mol B(v) / mol P 652 mmHg z − xB 0.5 − 0.399 Solving (1) and (2) ⇒ nV = B = = 0.446 mol vapor y B − x B 0.625 − 0.399 n L = 1 − nV = 1 − 0.446 = 0.554 mol liquid yB =
7-29
(9)
7.51 (cont’d)
Substituting (3), (4), (5), and (6) in (7) ⇒ Q = 0.446(0.625)[01045 . (90) + 33.25] + 0.446(1 − 0.625)[01304 . (90) + 37.57] + 0.554(0.399)[01356 . (90)] + 0.554(1 − 0.399)[01674 . (90)] − 0.5[ 01356 . (130)] . (130)] ⇒ Q = 814 . kJ / mol − 0.5[01674 (c). If PPmax, all the output is liquid. (d) At P=652 mmHg it is necessary to add heat to achieve the equilibrium and at P=714 mmHg, it is necessary to release heat to achieve the equilibrium. The higher the pressure, there is more liquid than vapor, and the liquid has a lower enthalpy than the equilibrium vapor: enthalpy out < enthalpy in. zB 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
T 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90
P 652 714 582 590 600 610 620 630 640 650 660 670 680 690 700 710
pB 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021 1021
pT 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7 406.7
xB 0.399 0.500 0.285 0.298 0.315 0.331 0.347 0.364 0.380 0.396 0.412 0.429 0.445 0.461 0.477 0.494
(e). Pmax = 714 mmHg, Pmin = 582 mmHg nV vs. P 1
nV
0.8 0.6 0.4 0.2 0 582
632
682
732
P (mm Hg)
nV = 0.5 @ P ≅ 640 mmHg
7-30
yB 0.625 0.715 0.500 0.516 0.535 0.554 0.572 0.589 0.606 0.622 0.638 0.653 0.668 0.682 0.696 0.710
nV 0.446 -0.001 0.998 0.925 0.840 0.758 0.680 0.605 0.532 0.460 0.389 0.318 0.247 0.176 0.103 0.029
nL 0.554 1.001 0.002 0.075 0.160 0.242 0.320 0.395 0.468 0.540 0.611 0.682 0.753 0.824 0.897 0.971
Q 8.14 -6.09 26.20 23.8 21.0 18.3 15.8 13.3 10.9 8.60 6.31 4.04 1.78 -0.50 -2.80 -5.14
ΔP
Δu 2 + + gΔz = 0 ρ 2
7.52 (a). Bernoulli equation:
ΔP
ρ
=
d0.977 × 10
−5
i
− 15 . × 10 5 Pa 1 N / m 2
1.12 × 10 3 kg
Pa
bg
m3
= −46.7
m2 s2
gΔz = (9.8066 m / s 2 ) 6 m = 58.8 m 2 / s 2 Δu 2 = 46.7 − 58.8 m 2 / s 2 ⇒ u22 = u12 + 2 −12.1 m 2 / s 2 2 2 = 5.00 m 2 / s 2 − (2)(12.1) m 2 / s 2 = 0.800 m 2 / s 2 ⇒ u2 = 0.894 m / s
b
Bernoulli ⇒
g
d
i
b g
d
i
(b). Since the fluid is incompressible, V m 3 s = π d 12 u1 4 = π d 22 u2 4
b g
u2 = 6 cm u1
0.894 m s = 2.54 cm 5.00 m s
d i b g
d i b g
⇒ d1 = d 2
d
i
A 7.53 (a). V m 3 s = A1 m 2 u1 m s = A2 m 2 u2 m s ⇒ u2 = u1 1 A2 (b). Bernoulli equation (Δz = 0)
d
ρ u22 − u12 Δu 2 + = 0 ⇒ Δ P = P2 − P1 = − ρ 2 2
ΔP
A1 = 4 A2
u2 = 4u1
i
Multiply both sides by − 1 Substitute u 2 = 16u1 2
2
2
Multiply top and bottom of right - hand side by A1 2 2 2 note V = A1 u1
P1 − P2 =
d
i
(c) P1 − P2 = ρ Hg − ρ H 2 O gh = 2
V =
b g
2 π 7.5
2 2
15
cm4 1
15ρV 2 2 A12
15ρ H 2 OV 2 2 A12
⇒ V 2 =
m4
9.8066 m
108 cm4
s2
⇒ V = 0.044 m 3 s = 44 L s
7-31
F GH
I JK
2 A12 gh ρ Hg −1 ρ H 2O 15
38 cm
1m 102 cm
b13.6 − 1g = 1955 × 10 .
−3
m6 s2
b g
7.54 (a). Point 1 - surface of fluid . P1 = 31 . bar , z1 = +7 m , u1 = 0 m s
b g
Point 2 - discharge pipe outlet . P2 = 1 atm , z 2 = 0 m , u2 = ? Δρ
ρ
=
gΔz =
b =1.013 bar g
b1.013 − 31. gbar
10 5 N 1 m3 = −2635 . m 2 s2 m 2 ⋅ bar 0.792 × 10 3 kg
−7 m
9.8066 m s2
Bernoulli equation ⇒
= −68.6 m 2 s 2
ΔP Δu 2 =− − gΔz = 263.5 + 68.6 m 2 s 2 = 332.1 m 2 s 2 2 ρ
b
Δu = u 2 − 0 2
2
g
2
u22 = 2(332.1 m 2 s 2 ) = 664.2 m 2 s 2 ⇒ u2 = 258 . m/s
π (100 . 2 ) cm 2 V = 4
2580 cm 1 L 60 s = 122 L / min 1 s 10 3 cm 3 1 min
(b) The friction loss term of Eq. (7.7-2), which was dropped to derive the Bernoulli equation, becomes increasingly significant as the valve is closed. 7.55 Point 1 - surface of lake . P1 = 1 atm , z1 = 0 , u1 = 0
bg
Point 2 - pipe outlet . P2 = 1 atm , z 2 = z ft
u2 =
V 95 gal = min A
1 ft 3 7.4805 gal
FG L = Z = 2 zIJ H sin 30° K
b
bP = P g F = 0.041b2 z g ft ⋅ lb lb
Pressure drop: Δ P ρ = 0 Friction loss:
1 2 in 2 . π 0.5 × 1049
1
g
144 in 2 1 ft 2
1 min = 35.3 ft s 60 s
2
f
m
= 0.0822 z (ft ⋅ lb f lb m )
-8 hp 0.7376 ft ⋅ lb f / s W Shaft work: s = m 1.341 × 10 −3 hp
1 min 7.4805 gal 95 gal
1 ft
3
1 ft 3
60 s
62.4 lb m
1 min
= −333 ft ⋅ lb f lb m Kinetic energy: Δ u 2 2 =
b35.3g
2
2 Potential energy: gΔz =
b
g
Eq. 7.7 - 2 ⇒
ΔP
ρ
+
32.174 ft s2
− 0 2 ft 2 s
2
bg
z ft
1
lb f
32.174 lb m ⋅ ft / s 2
= 19.4 ft ⋅ lb f lb m
b
1 lb f = z ft ⋅ lb f lb m 32.174 lb m ⋅ ft / s 2
g
−W s Δu 2 + gΔz + F = ⇒ 19.4 + z + 0.082 z = 333 ⇒ z = 290 ft m 2
7-32
7.56 Point 1 - surface of reservoir . P1 = 1 atm (assume), u1 = 0 , z1 = 60 m Point 2 - discharge pipe outlet . P2 = 1 atm (assume), u2 = ? , z 2 = 0
ΔP ρ = 0
d h
V A Δu 2 u22 = = 2 2 2
2
=
V 2 (m 6 / s 2 )
= 3.376V 2 −65 m
9.8066 m s2
gΔz =
b g bN ⋅ m kgg π 35
(2)
1 2 2
cm 4
ρ
1 m4
1 kg ⋅ m / s 2
1 m3 = 800 V N ⋅ m kg 1000 kg
b
d i Mechanical energy balance: neglect F b Eq. 7.7 - 2g +
1 N
1N = −637 N ⋅ m kg 1 kg ⋅ m / s 2
6 s W s 0.80 × 10 W 1 N ⋅ m / s = V m3 W m
ΔP
10 8 cm 4
g
−W s 800 T + E 127 Δu 2 . m 3 60 s + gΔz = ⇒ 3.376V 2 − 637 = − ⇒ = 76.2 m 3 min V= s 1 min 2 m V
Include friction (add F > 0 to left side of equation) ⇒ V increases.
b g
7.57 (a). Point 1: Surface at fluid in storage tank, P1 = 1 atm , u1 = 0 , z1 = H m
Point 2 (just within pipe): Entrance to washing machine. P2 = 1 atm , z 2 = 0
u2 = ΔP
ρ
600 L min π 4.0 cm
b
= 0;
gΔz =
Δu 2 u22 = 2 2
9.807 m s
10 3 cm 3 1 min 1m = 7.96 m s 4 1 L 60 s 100 cm
g 7.96 m sg =b 2
2
c0 − Hbmgh
2
Bernoulli Equation:
2
ΔP
ρ
+
1 J = 31.7 J / kg 1 kg ⋅ m 2 / s 2 1 J
1 kg ⋅ m 2 / s 2
= −9.807 H (J / kg)
Δu 2 + gΔz = 0 ⇒ H = 3.23 m 2
(b). Point 1: Fluid in washing machine. P1 = 1 atm , u1 ≈ 0 , z1 = 0 Point 2: Entrance to storage tank (within pipe). P2 = 1 atm , u2 = 7.96 m s , z 2 = 3.23 m ΔP
ρ
= 0;
J J J Δu 2 ; gΔz = 9.807 3.23 − 0 = 317 = 317 . . ; F = 72 kg 2 kg kg
b
LM N
g
ΔP Δu 2 Mechanical energy balance: W s = − m + + gΔz + F ρ 2 600 L 0.96 kg 1 min ⇒ W s = − min L 60 s
b31.7 + 31.7 + 72g J
OP Q
1 kW = −1.30 kW kg 10 3 J s (work applied to the system)
Rated Power = 130 . kW 0.75 = 1.7 kW
7-33
7.58 Basis: 1000 liters of 95% solution . Assume volume additivity. xi 0.95 0.05 1 l Density of 95% solution: = = + = 0.804 ⇒ ρ = 124 . kg liter ρ ρ 1 . 26 100 . kg b Eq. 6.1-1g i
∑
Density of 35% solution:
Mass of 95% solution:
1
ρ
=
0.35 0.65 l + = 0.9278 ⇒ ρ = 108 . kg liter 126 . 100 . kg
1000 liters 1.24 kg = 1240 kg liter
G = glycerol W = water 1240 kg (1000 L) 0.95 G 0.05 W
m2 (kg) 0.60 G 0.40 W 23 m
m1 (kg) 0.35 G 0.65 W
5 cm I.D.
UV ⇒ m = 1740 kg 35% solution b gb g b gb g b gb gW m = 2980 kg 60% solution
Mass balance: 1240 + m1 = m2 Glycerol balance: 0.95 1240 + 0.35 m1 = 0.60 m2 Volume of 35% solution added =
1740 kg
b
g
1
2
1L = 1610 L 1.08 kg
⇒ Final solution volume = 1000 + 1610 L = 2610 L Point 1. Surface of fluid in 35% solution storage tank. P1 = 1 atm , u1 = 0 , z1 = 0 Point 2. Exit from discharge pipe. P2 = 1 atm , z 2 = 23 m u2 =
1610 L 13 min
Δ P ρ = 0, gΔz =
1 m 3 1 min 1 2 3 10 L 60 s π 2.5 cm 2
b g
b g
2 Δu 2 Δu22 1.051 m 2 / s 2 = = 2 2 (2)
9.8066 m s2
Mass flow rate: m =
23 m
1 N = 0.552 N ⋅ m kg 1 kg ⋅ m / s 2
1N = 225.6 N ⋅ m kg , F = 50 J kg = 50 N ⋅ m kg 1 kg ⋅ m / s 2
1740 kg 1 min = 2.23 kg s 13 min 60 s
b
Mechanical energy balance Eq. 7.7 - 2
LM N
10 4 cm 2 = 1.051 m s 1 m2
g
OP Q
2.23 kg ΔP Δu 2 W s = − m + + gΔz + F = − ρ 2 s
b0.552 + 225.6 + 50gN ⋅ m
= −0.62 kW ⇒ 0.62 kW delivered to fluid by pump.
7-34
kg
1J
1 kW
1 N ⋅ m 10 3 J s
CHAPTER EIGHT 8.1
a.
U (T ) = 25.96T + 0.02134T 2 J / mol U (0 o C) = 0 J / mol U (100 o C) = 2809 J / mol
o C) = 0) Tref = 0 o C (since U(0
b. We can never know the true internal energy. U (100 o C) is just the change from U (0 o C) to U (100 o C) . c.
Q − W = ΔU + ΔE k + ΔE p ΔE k = 0, ΔE p = 0, W = 0
Q = ΔU = (3.0 mol)[(2809 − 0) J / mol] = 8428 J ⇒ 8400 J
d.
Cv =
F ∂U I GH ∂T JK
ΔU =
z
= V
dU = [25.96 + 0.04268T ] J / (mol⋅ o C) dT
T2
z
F GG H
100
Cv (T )dT =
(25.96 + 0.04268T )dT = 25.96T + 0.04268
0
T1
T2 2
OP PQ
100
0
I JJ J / mol K
ΔU = (3.0 mol) ⋅ ΔU (J / mol) = (3.0 mol) ⋅ [25.96(100 − 0) + 0.02134(100 2 − 0)] (J / mol) = 8428 J ⇒ 8400 J
8.2
a.
b
g
b
gb
Cv = C p − R ⇒ Cv = 35.3 + 0.0291T [J / (mol⋅° C)] − 8.314 [J / (mol ⋅ K)] 1 K 1° C
g
⇒ Cv = 27.0 + 0.0291T [J / (mol⋅° C)] 100
b.
ΔHˆ =
∫ C p dT = 35.3T ]25 + 0.0291 100
25
c.
ΔU =
z
100
Cv dT =
25
8.3
z
100
z
100
T2 ⎤ ⎥ = 2784 J mol 2 ⎦ 25
100
C p dT −
25
b
g
25
d.
H is a state property
a.
Cv [ kJ / (mol⋅ o C)] = 0.0252 + 1547 . × 10 −5 T − 3.012 × 10 −9 T 2 n=
gb
RdT = ΔH − RΔT = 2784 − 8.314 100 − 25 = 2160 J mol
PV (2.00 atm)(3.00 L) = = 0.245 mol RT (0.08206[atm ⋅ L / (mol ⋅ K)](298 K)
Q1 = nΔU 1 = (0.245 mol) ⋅ Q2 = nΔU 2 = (0.245) ⋅ Q3 = nΔU 3 = (0.245) ⋅
z z
z
1000
0.0252 dT ( kJ / mol) = 6.02 kJ
25
1000
× 10 −5 T ] dT = 7.91 kJ [0.0252 + 1547 .
25 1000
× 10 −5 T − 3.012 × 10 −9 T 2 ] dT = 7.67 kJ [0.0252 + 1547 .
25
6.02 - 7.67 × 100% = −215% . 7.67 7.91- 7.67 % error in Q2 = × 100% = 313% . 7.67 % error in Q1 =
8-1
8.3 (cont’d) b.
C p = Cv + R
C p [ kJ / (mol⋅ o C)] = (0.0252 + 1547 . × 10 −5 T − 3.012 × 10 −9 T 2 ) + 0.008314 = 0.0335 + 1547 . × 10 −5 T − 3.012 × 10 −9 T 2
z
T2
Q = ΔH = n C P dT T1
z
1000
= (0.245 mol) ⋅
[0.0335 + 1547 . × 10 −5 T − 3.012 × 10 −9 T 2 ] dT [kJ / (mol⋅ o C)] = 9.65 × 10 3 J
25
Piston moves upward (gas expands).
8.4
c.
The difference is the work done on the piston by the gas in the constant pressure process.
a.
(C )
b.
dC i
( 40 C ) = 0.1265 + 23.4 × 10 ( 40 ) = 0.1360 [kJ/(mol ⋅ K)] −5
o
p C H (l ) 6 6
b g b40° Cg = 0.07406 + 32.95 × 10
−5
p C H v 6 6
b40g − 25.20 × 10 b40g −8
2
b g
+ 77.57 × 10 −12 40
3
= 0.08684 [kJ / (mol⋅ o C)]
c.
. × 10 b313g − 4.891 × 10 b313g dC i b g b313 Kg = 0.01118 + 1095
d.
32.95 × 10 ΔH C6 H6 bv g = 0.07406T + 2
e.
8.5
−5
2
p C s
1095 . × 10 ΔH C b sg = 0.01118T + 2
−5
−5
−2
= 0.009615 [ kJ / (mol ⋅ K)]
2520 . × 10 −8 3 77.57 × 10 −12 4 T − T + T 3 4 2
T + 4.891 × 10 T 2
2
−1
OP PQ
OP PQ
300
= 3171 . kJ mol 40
573
= 3.459 kJ / mol 313
H 2 O (v, 100 o C, 1 atm) → H 2 O (v, 350 o C, 100 bar) a. H = 2926 kJ kg − 2676 kJ kg = 250 kJ kg
b.
H =
z
350
0.03346 + 0.6886 × 10 −5 T + 0.7604 × 10 −8 T 2 − 3593 . × 10 −12 T 3 dT
100
= 8.845 kJ mol ⇒ 491.4 kJ kg
Difference results from assumption in (b) that H is independent of P. The numerical difference is ΔH for H 2 O v, 350° C, 1 atm → H 2 O v, 350° C, 100 bar
b
8.6
b.
g
b
z
g
80
dC i
p n − C H (l) 6 14
= 0.2163 kJ / (mol⋅ o C) ⇒ ΔH = [0.2163] dT = 1190 . kJ / mol 25
The specific enthalpy of liquid n-hexane at 80oC relative to liquid n-hexane at 25oC is 11.90 kJ/mol c.
dC i
p n − C H (v) [ kJ 6 14
ΔH =
z
/ (mol⋅ o C)] = 013744 . + 40.85 × 10 −5 T − 23.92 × 10 −8 T 2 + 57.66 × 10 −12 T 3
0
[013744 . + 40.85 × 10 −5 T − 23.92 × 10 −8 T 2 + 57.66 × 10 −12 T 3 ] dT = −110.7 kJ / mol
500
The specific enthalpy of hexane vapor at 500oC relative to hexane vapor at 0oC is 110.7 kJ/mol. The specific enthalpy of hexane vapor at 0oC relative to hexane vapor at 500oC is –110.7 kJ/mol.
8-2
8.7
b g 181. T ′b° Fg − 32 = 0.5556T ′b° Fg − 17.78 C bcal mol⋅° Cg = 6.890 + 0.001436 0.5556T ′b° Fg − 17.78 = 6.864 + 0.0007978T ′b° Fg cal 453.6 mol 1 Btu 1° C = b100 C ′ b Btu lb - mole⋅° Fg C . gC = mol⋅° C 1 lb - mole 252 cal 1.8° F E T °C = p
p
p
b
g
p
drop primes
b g
C p Btu lb - mole⋅° F = 6.864 + 0.0007978T ° F
8.8
. − 01031 . b01588 g T = 01031 . + . + 0.000557T [kJ / (mol⋅ bT g = 01031 100 55.0 L 789 g 1 mol F 0.000557 O . Q = ΔH = T+ T P 01031 G s 1 L 46.07 g H 2 Q
dC i
p CH CH OH(l) 3 2
o
C)]
78.5
2
20
kJ mol
= 941.9 × 7.636 kJ / s = 7193 kW
8.9
a.
kJ mol
b
g
Q = ΔH = 5,000 mol s ⋅
z
200
0.03360 + 1367 . × 10 −5 T − 1607 . × 10 −8 T 2 + 6.473 × 10 −12 T 3 dT
100
= 17,650 kW
b.
b
gb
gb
Q = ΔU = ΔH − ΔPV = ΔH − nRΔT = 17,650 kJ − 5.0 kmol ⋅ 8.314 [kJ / (kmol ⋅ K)] ⋅ 100 K = 13,490 kJ
c.
8.10 a. b.
The difference is the flow work done on the gas in the continuous system. Qadditional = heat needed to raise temperature of vessel wall + heat that escapes from wall to surroundings. C p is a constant, i. e. C p is independent of T. Q mΔT (16.73 - 6.14) kJ
Q = mC p ΔT ⇒ C p = Cp =
Q 1 L 86.17 g 10 3 J = = 0.223 kJ / (mol ⋅ K) mΔT (2.00 L)(3.10 K) 659 g 1 mol 1 kJ
Table B.2 ⇒ C p = 0.216 kJ / (mol⋅ o C) = 0.216 kJ / (mol ⋅ K)
FG ∂H IJ = FG ∂U IJ H ∂T K H ∂T K F ∂U I = dU = F ∂U I depends only on T, G H ∂T JK dT GH ∂T JK
PV = RT
8.11
a∂ ∂T f
H = U + PV =====> H = U + RT =====> P
p
But since U
p
8-3
+ R ⇒ Cp = p
FG ∂U IJ H ∂T K
+R p
≡ Cv ⇒ C p = Cv + R V
g
8.12 a.
dC i
= 75.4 kJ / (kmol⋅ o C) =75.4 kJ/(kmol.oC) V = 1230 L ,
p H O(l) 2
n=
Vρ 1230 L 1 kg 1 kmol = = 68.3 kmol M 1 L 18 kg
zd
T2
n⋅ Q Q = = t
b.
Cp
i
H 2 O(l)
dT
T`
=
t
68.3 kmol 75.4 kJ (40 − 29) o C 1 h = 1967 . kW 8h 3600 s kmol⋅ o C
Q total = Q to the surroundings + Q to water , Q to the surroundings = 1967 . kW
z
40
n ⋅ C P ( H2 O ) dT
Q Q to water = to water = t
=
29
t
68.3 kmol 75.4 kJ / (kmol⋅ o C) 11 o C = 5.245 kW 3h 3600 s / h
Q total = 7.212 kW ⇒ E total = 7.212 kW × 3 h = 21.64 kW ⋅ h
c.
Cost heating up from 29 o C to 40 o C = 21.64 kW ⋅ h × $0.10 / (kW ⋅ h) = $2.16
Costkeeping temperature constant for 13 h = 1.967 kW × 13 h × $0.10/(kW ⋅ h)=$2.56 Costtotal = $2.16 + $2.56 = $4.72 d. If the lid is removed, more heat will be transferred into the surroundings and lost, resulting in higher cost. 8.13 a.
ΔH N
o o 2 (25 C) → N 2 (700 C)
b.
ΔH H
2 (800
c.
ΔH CO
d.
ΔH O
8.14 a.
o
F) → H 2 (77 o F)
2 (300
2 (970
o
o
= H N = H H
C) → CO 2 (1250o C)
m = 300 kg / min n =
Q = n ⋅ ΔH = n ⋅
z
T2
2 (77
o
o
2 (0
o
F)
C)
F)
= H CO
= H O
F) → O 2 (0o F)
2 (700
− H N
− H H
2 (1250
− H O
o
b g = b0 − 5021g = −5021 Btu / lb - mol − H = b63.06 − 1158 . g = 5148 . kJ mol = b−539 − 6774g = −7313 Btu / lb - mol
o 2 (25 C)
2 (800
C)
2 (970
o
= 20.59 − 0 = 20.59 kJ mol
F)
CO 2 (300o C)
o
F)
300 kg 1 min 1000 g 1 mol = 178.5 mol / s min 60 s 1 kg 28.01 g
C p dT
T1
z
50
= (178.5 mol / s) ⋅
b
[0.02895 + 0.411 × 10 −5 T + 0.3548 × 10 −8 T 2 − 2.22 × 10 −12 T 3 ] dT [kJ / mol]
450
g
= (178.5 mol / s) −12.076 [kJ / mol] = −2,156 kW
b. 8.15 a.
Q = n ⋅ ΔH = n ⋅ H (50o C) − H ( 450o C)
= (178.5 mol / s)(0.73-12.815[kJ / mol]) = −2,157 kW
n = 250 mol / h
250 mol (2676 − 3697) kJ 1 kg 1 h 18.02 g Q = nΔH = = −1.278 kW h 1 kg 1000 g 3600 s 1 mol
i)
Q = nΔH = n ⋅
z
T2
T1
ii) =
C p dT
250 mol 1 h h 3600 s
z
100
[003346 . . . . × 10−12 T 3 ] = −1.274 kW + 06880 × 10−5 T + 07604 × 10−8 T 2 − 3593
600
8-4
8.15 (cont’d)
b
g
250 mol Q = ⋅ 2.54 − 20.91 [kJ / mol] = −1276 . kW 3600 s b. Method (i) is most accurate since it takes into account the dependence of enthalpy on pressure and (ii) and (iii) do not. c. The enthalpy change for steam going from 10 bar to 1 atm at 600oC.
iii)
8.16 Assume ideal gas behavior, so that pressure changes do not affect ΔH . n =
200 ft 3 492 o R 12 . atm 1 lb - mol = 0.6125 lb - mole / h o h 1 537 R atm 359 ft 3 (STP)
b
g
lb - mole Q = nΔH = (0.6125 ) ⋅ (2993 − 0) [Btu / lb - mole] = 1833 Btu / h h
8.17 a.
50 kg 1.14 kJ kg⋅° C
b50 − 10g° C = 2280 kJ
b.
(C )
p Na CO 2 3
≈ 2 (C p )
Na
+ ( C p ) + 3 ( C p ) = 2 ( 0.026 ) + 0.0075 + 3 ( 0.017 ) = 0.1105 kJ mol ⋅ °C C
50,000 g 0.1105 kJ mol⋅° C % error =
8.18
dC i dC i
p C H O(l) 6 14
O
1 mol 105.99 g
b50 − 10g° C = 2085 kJ
2085 − 2280 × 100% = −8.6% error 2280
b
g b
g b
g
= 6 0.012 + 14 0.018 + 1 0.025 = 0.349 kJ / (mol⋅ o C) (Kopp’s Rule)
p CH COCH (l) 3 3
= 01230 . + 18.6 × 10 −5 T kJ (mol⋅° C)
Assume ΔH mix ≅ 0 ↓ C 6 H 14 O
↓ CH 3 COCH 3
0.30 ( 0.1230+18.6 × 10 T ) kJ 1 mol 0.70 ( 0.349 ) kJ 1 mol + mol ⋅ °C 58.08 g mol ⋅ °C 102.17 g −5
C pm =
= [0.003026 + 9.607 × 10−7 T] kJ (g ⋅ °C) 20 ΔHˆ = ∫ [0.003026 + 9.607 × 10−7 T] dT = −0.07643 kJ g 45
8.19 Assume ideal gas behavior, ΔH mix ≅ 0
g b g ΔH = dC i dT = 10.08 kJ / mol, ΔH = dC i dT = 14.49 kJ / mol 2 L1 OF 1000 g IJ FG 1 mol IJ = 433 kJ kg H = M b14.49 kJ / molg + b10.08 kJ / molgPG 3 N3 QH 1 kg K H 26.68 g K Mw = O2
b
g 1 2 16.04 + 32.00 = 26.68 mol 3 3
z
350
25
p O 2
CH 4
8-5
z
350
25
p CH 4
8.20 n =
1000 m 3 1 min 273 K min
1 kmol
b g = 0.6704 kmol s = 670.4 mol / s
303 K 22.4 m 3 STP
60 s
Energy balance on air: Table B.8 for ΔH
Q = ΔH = nΔH
Solar energy required = Area required = 8.21
Q=
670.4 mol 0.73 kJ s
mol
1 kW 1 kJ s
489.4 kW heating 1 kW solar energy 0.3 kW heating
= 489.4 kW
= 1631 kW
1631 kW 1000 W 1 m 2 = 1813 m 2 1 kW 900 W
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
n fuel =
. × 10 5 SCFH 1 lb - mol 135 lb - mol = 376 3 h h 359 ft
n air =
376 lb − mol 5 lb - mol O 2 1 lb - mol air 115 lb − mol . = 103 . × 10 4 h 1b - mol C 3 H 8 0.211b - mol O 2 h T2
Q =ΔH =n ⋅ ∫ C p dT T1
lb − mol ⎞ ⎛ = ⎜1.03 × 104 ⎟⋅ h ⎝ ⎠
302
∫ [0.02894 + 0.4147 × 10
−5
T + 0.3191 × 10−8 T 2 − 1.965 × 10−12 T 3 ] dT
0
1.03 × 10 lb-mol 8.954 kJ 453.593 mol 9.486 × 10-1 Btu =3.97 × 107 Btu/h h mol lb-mol kJ 4
= 8.22
Basis: 100 mol feed (95 mol CH4 and 5 mol C2H6) 7 CH 4 + 2O 2 → CO 2 + 2H 2 O C 2 H 6 + O 2 → 2CO 2 + 3H 2 O 2
a.
n O 2 = 125 . ⋅
LM 95 mol CH MN
4
OP PQ
2 mol O 2 5 mol C 2 H 6 35 . mol O 2 = 259.4 mol O 2 + 1 mol CH 4 1 mol C 2 H 6
Product Gas:
CO 2 : 95(1)+5(2)=105 mol CO 2
H 2 O : 95(2)+5(3)=205 mol H 2 O
O 2 : 259.4-95(2)-5(3.5)=51.9 mol O 2
N 2 : 3.76(259.4)=975 mol N 2
Energy balance (enthalpies from Table B.8) ΔH CO 2 = H (CO , 450o C) − H (CO , 900o C) = 18.845 − 42.94 = −24.09 kJ / mol 2
ΔH H 2 O = H (H ΔH O 2 = H (O
2 O,
2,
ΔH N 2 = H (N
2,
2
450o C)
−H (H
450o C)
−H (O
450o C)
−H (N
2,
2 O,
900o C)
. − 33.32 = −18.20 kJ / mol = 1512
900o C)
= 13.375 − 28.89 = −15.51 kJ / mol
900o C)
= 12.695 − 27.19 = −14.49 kJ / mol
2,
Q = ΔH = 105(-24.09) + 205(-18.20) + 51.9(-15.51) + 975(-14.49) Q = 21,200 kJ / 100 mol feed
(40 o C) = 167.5 kJ / kg; H b. From Table B.5: H liq vap (50 bars) = 2794.2 kJ / kg; = n(2794.2 -167.5) = 21200 ⇒ n = 8.07 kg / 100 mol feed Q = n ⋅ ΔH
8-6
8.22 (cont’d) c.
From part (b), 8.07 kg steam is produced per 100 mol feed 1250 kg steam 01 . kmol feed 1 h n feed = = 4.30 × 10 −3 kmol / s h 8.07 kg steam 3600 s 723 K 4.30 mol feed 1336.9 mol product gas 8.314 Pa ⋅ m 3 Vproduct gas = = 3.41 m 3 / s 5 s 100 mol feed mol ⋅ K 1.01325 × 10 Pa
d. Steam produced from the waste heat boiler is used for heating, power generation, or process application. Without the waste heat boiler, the steam required will have to be produced with additional cost to the plant.
Assume ΔH mix ≅ 0 ⇒ ΔH = ΔH C10 H12 O2 + ΔH C6 H6
8.23
d i
Kopp’s rule: C p ΔH C10 H12 O2 = ΔH C6 H6 =
C10 H12 O2
e
j
e
= 10(12) + 12(18) + 2(25) = 386 J mol⋅ o C = 2.35 J g⋅ o C
20.0 L 1021 g 1 kJ 2.35 J (71 − 25) o C = 2207 kJ L 10 3 J g⋅ o C
15.0 L 879 g 1 mol ⋅ L 78.11 g
LM N
z
348
j
OP Q
[0.06255 + 23.4 × 10 −5 T] dT = 1166 kJ
298
ΔH = 2207 + 1166 = 3373 kJ
8.24 a.
100 mol C3 H8 @ 40 o C, 250 kPa
100 mol C3 H8 @ 240 o C, 250 kPa
VP 1 (m3 )
VP 2 (m3 ) mw kg H2 O(v) @ 300 o C, 5.0 bar
mw kg H2 O(l, sat‘d) @ 5.0 bar Vw2 (m3 )
Vw1 (m3 )
b. References: H2O (l, 0.01 oC), C3H8 (gas, 40 oC)
z
240
C3H8: H in = 0 kJ / mol; H out = CpC H dT =1936 . kJ mol (Cp from Table B.2) 40
3 8
H 2 O : Hˆ in = 3065 kJ/kg (Table B.7); Hˆ out = 640.1 kJ/kg (Table B.6) c.
ΔHˆ C3 H8 = 19.36 kJ/mol, ΔHˆ w = (640.1 − 3065) kJ/kg = −2425 kJ/kg Q = ΔH = 100ΔH C3 H 8 + mw ΔH w = 0 ⇒ mw = 0.798 kg
b
g
From Table B.7: Vsteam 5.0 bar, 300° C = 0.522 m 3 kg 0.008314 m ⋅ kPa (mol ⋅ K) 313 K VC3 H8 40° C, 250 kPa = = 0.0104 m 3 mol C 3 H 8 250 kPa
b
g
0.798 kg steam 0.522 m3 steam 100 mol C3 H8 d.
e.
Q = mw ΔHˆ w =
1 kg steam 0.798 kg steam 100 mol C3 H8
3
1 mol C3 H8 0.0104 m3 C3 H8 2425 kJ
= 0.400 m 3 steam m 3 C3 H8
1 mol C3 H8
kg steam 0.0104 m 3 C3 H 8
= 1860
kJ m 3 C3 H8 fed
A lower outlet temperature for propane and a higher outlet temperature for steam.
8-7
8.25 a. 5500 L(ST P)/ min CH3 OH (v) 65o C
n 2 mol/min CH3 O H (v) 260o C
n 2 (mol/ min) mw kg/ min H2 O(l, s at‘d) @ 90o C
mw kg/ min H2 O(v, s at‘d) @ 300o C
Vw 2 (m3 /min )
n2 =
Vw 1 (m3 /min )
5500 L(STP) 1 mol = 245.5 mol CH 3OH(v)/min min 22.4 L(STP)
An energy balance on the unit is then written, using Tables B.5 and B.6 for the specific enthalpies of the outlet and inlet water, respectively, and Table B.2 for the heat capacity of methanol vapor. The only unknown is the flow rate of water, which is calculated to be 1.13 kg H 2 O/min. b.
8.26 a.
kg ⎞ ⎛ kJ ⎞ ⎛ 1 min ⎞ ⎛ 1 kW ⎞ ⎛ Q = ⎜ 1.13 ⎟⎜ ⎟ ⎜ 2373.9 ⎟⎜ ⎟ = 44.7 kW min ⎠ ⎝ kg ⎠ ⎝ 60 sec ⎠ ⎝ 1 kJ/s ⎠ ⎝ 100 mol/s (30o C) 0.100 mol H2 O(v)/ mo l 0.100 mol CO/ mol 0.800 mol CO2 /mol
n 2 mol/s (30o C) 0.020 mol H2 O(v)/ mo l y 2 mol CO/s (mol CO/mol) (0.980-y 2 ) mol (molCO CO2 /s 2/mol)
m3 kg humid air/s (50o C)
m4 kg humid air/s (30 (48ooC) C)
H 2O(v) only
(0.002 /1.002 ) kg H2 O(v)/kg humid air (1.000 /1.002 ) kg dry air/ kg humid air
y 4 kg H2 O(v)/kg humid air (1-y 4 ) kg dry air/kg humid air
Basis: 100 mol gas mixture/s 5 unknowns: n2, m3, m4, y2, y4 – 4 independent material balances, H2O(v), CO, CO2 , dry air – 1 energy balance equation 0 degrees of freedom ( all unknowns may be determined) b.
|UV W|
(1) CO balance: (100)(0.100) = n 2 y 2 mol CO / mol ⇒ n 2 = 9184 . mol / s, x 2 = 01089 . (2) CO 2 balance: (100)(0.800) = n 2 (1 − y 2 ) 1000 . = m4 (1 − y 4 ) 1002 . (100)(0.100)(18) (0.020)(18) 0.002 (4) H 2 O balance: + m 3 = 9184 + m 4 y 4 . 1000 . 1002 1000 References: CO, CO2, H2O(v), air at 25oC ( H values from Table B.8 ) substance n out ( mol / s) n in ( mol / s) H in (kJ / mol) (3) Dry air balance: m3
H2O(v) CO CO2
10 10 80
0.169 0.146 0.193
91.84(0.020) 10 80
H out (kJ / mol) 0.169 0.146 0.193
H2O(v) dry air
m3(0.002/1.002)(1000/18) m3(1.000/1.002) (1000/29)
0.847 0.727
m4y4(1000/18) m4(1-y4) (1000/29)
0.779 0.672
8-8
8.26 (cont’d) (5) Energy balance:
. FG 0.002 IJ FG 1000IJ (0.847) + m FG 1000 IJ FG 1000 IJ (0.727) H 1002 H 1002 K H 18 K K H 29 K . . F 1000IJ + m (1 − y )(0.672)FG 1000 IJ . ) + m y (0.779)G = 91.84(0.020)(0169 H 18 K H 29 K
. ) + m3 10(0169
3
4 4
4
4
Solve Eqs. (3)–(5) simultaneously ⇒ m3 = 2.55 kg/s, m4 = 2.70 kg/s, y4 = 0.0564 kg H2O/kg 2.55 kg humid air / s kg humid air = 0.0255 100 mol gas / s mol gas
Mole fraction of water :
kg H 2 O 00564 .
=.0963
(1-.0564) kg dry air kmol DA 18 kg H 2 O ⇒
Relative humidity:
29 kg DA 1 kmol H 2 O
0.0963 kmol H 2 O (1 + 00963 . ) kmol humid air
p H 2O p H* 2 O
e48 Cj o
= 00878 .
kmol H 2 O kmol DA
kmol H 2 O kmol humid air
(0.0878)(760 mm Hg) × 100% = 79.7% 83.71 mm Hg
=
c. The membrane must be permeable to water, impermeable to CO, CO2, O2, and N2, and both durable and leakproof at temperatures up to 50oC. 8.27 a.
y H 2O =
b
p * 57° C P
b g
28.5 m 3 STP h
g = 129.82 mm Hg = 0171 . mol H O mol 2
760 mm Hg ↓ 1 mol = 1270 mol h ⇒ 217.2 mol H 2 O h 0.0224 m 3 STP
b g
b3.91 kg H O hg
R| 89.5 mol CO h mol dry gas |110.5 mol CO h 1270 − 217.2 = 1053 =======> S h || 5.3 mol O h T847.6 mol N h given
2
percentages
2
2
1270 mol/h, 620°C 425°C m (kg H2 O( l )/h), 20°C
References for enthalpy calculations:
e
j
CO, CO 2 , O 2 , N 2 at 25°C (Table B.8); H 2 O l , 0.01o C (steam tables) substance
n in
CO CO 2 O2 N2
89.5 110.6 5.3 847.6
H 2O v
3.91 m
bg H Obl g
H in 18.22 27.60 19.10 18.03
3749 83.9
n out
89.5 110.6 5.3 847.6
H out 12.03 17.60 12.54 11.92
3.91 + m 3330 ---
2
8-9
U| n in mol h V| H in kJ mol W UV n in kg h W H in kJ kg
2
8.27 (cont’d) ΔH =
∑ n H − ∑ n H i
i
i
out
i
= 0 ⇒ −8504 + 3246m = 0 ⇒ m = 2.62 kg h
in
b. When cold water contacts hot gas, heat is transferred from the hot gas to the cold water lowering the temperature of the gas (the object of the process) and raising the temperature of the water.
b gb
g
. 5.294 mm Hg = 0.7941 mm Hg 8.28 2°C, 15% rel. humidity ⇒ p H 2 O = 015
dy i H 2O
inhaled
n inhaled =
. × 10 b0.7941g b760g = 1045
5500 ml 273 K
−3
mol H 2 O mol inhaled air
1 liter
1 mol
b g = 0.2438 mol air inhaled min
3
min
275 K 10 ml 22.4 liters STP
Saturation at 37 °C ⇒ y H 2 O =
b
p * 37° C
g
760 mm Hg
=
47.067 = 0.0619 mol H 2 O mol exhaled dry gas 760
0.2438 mol/min 2oC
n2 kmol/min 37oC
1.045 x 10-3 H2O 0.999 dry gas
0.0619 H2O 0.9381 dry gas n1 mol H2O(l)/min 22o C
Mass of dry gas inhaled (and exhaled) =
b0.2438gb0.999gmol dry gas
29.0 g
min
mol
b0.999gb0.2438g = 0.9381 n
Dry gas balance:
= 7.063 g min
⇒ n 2 = 0.2596 mols exhaled min
2
. × 10 j + n = b0.2596gb0.0619g ⇒ n = 0.0158 mol H O min b0.2438ge1045 References for enthalpy calculations: H Obl g at triple point, dry gas at 2 °C −3
H 2 O balance:
1
1
2
2
substance
m in
Dry gas H 2O v
7.063 0.00459 0.285
bg H Obl g 2
Q = ΔH =
H in 0 2505 92.2
m out 7.063 0.290 —
H out 36.75 2569 —
m in g min H in J g
m H2 O = 18.02nH2 O Hˆ H2 O from Table 8.4 Hˆ dry gas = 1.05 (T − 2 )
∑ m H − ∑ m H i
out
i
i
in
i
=
966.8 J 60 min 24 hr min
1 hr
8-10
1 day
= 1.39 × 10 6 J day
8.29 a.
bg
75 liters C 2 H 5 OH l
bg
789 g 1 mol = 1284 mol C 2 H 3 OH l liter 46.07 g
e
j 1 mol = 3054 mol H Obl g 18.01 g
(C p ) CH 3OH = 01031 . + 0.557 × 10 −3 T kJ / (mol⋅ o C) (fitting the two values in Table B.2)
bg
55 L H 2 O l
1000 g liter
2
b
(C p ) H 2 O = 0.0754 kJ mol⋅° C
g
1284 mol C2H5 OH(l) (70.0oC) 1284 mol C2H5 OH (l) (To C) 3054 mol H2O(l) (20.0oC)
3054 mol H2O(l) (To C) T
T
0 = 1284 ∫ ( 0.1031 + 0.557 × 10−3 T ) dT + 3054 ∫ ( 0.0754 ) dT
Q = ΔU ≅ ΔH ( liquids ) ⎫⎪ ⎬ ⇒ ⇓ Integrate, solve quadratic equation Q = 0 ( adiabatic ) ⎪⎭ T=44.3 o C 70
25
b. 1. 2. 3. 4. 5.
Heat of mixing could affect the final temperature. Heat loss to the outside (not adiabatic) Heat absorbed by the flask wall & thermometer Evaporation of the liquids will affect the final temperature. Heat capacity of ethanol may not be linear; heat capacity of water may not be constant 6. Mistakes in measured volumes & initial temperatures of feed liquids 7. Thermometer is wrong
8.30 a. 1515 L/s air 500oC, 835 tor, Tdp=30o C
1515 L/s air , 1 atm 110 g/s H2O(v)
110 g/s H2O, T=25oC
Let n1 (mol / s) be the molar flow rate of dry air in the air stream, and n 2 (mol / s) be the molar flow rate of H2O in the air stream. n1 + n 2 =
mol ⋅ K 1515 L 835 mm Hg = 26.2 mol / s s 773 K 62.36 L ⋅ mm Hg
n 2 mmHg p * (30 o C) 31824 . =y= = = 0.0381 mol H 2 O / mol air 835 mmHg n1 + n 2 Ptotal ⇒ n1 = 25.2 mol dry air / s; n 2 = 10 . mol H 2 O / s
8-11
8.30 (cont’d) References: H2O (l, 25oC), Air (v, 25oC) substances n in (mol / s) H in (kJ / mol) dry air 25.2 14.37 H2O(v)
z z
1.0
dC i dC i
100
p H O ( l ) dT 2
25 500
100
H2O(l)
25.2
z
zd zd
--
z
25
100
25 T
0
zd T
100
ΔH = 0 = n out ⋅ H out − n in ⋅ H in
z
7.1
+ H vap
p H O ( v ) dT 2
6.1
H out (kJ / mol)
n out (mol / s)
i i
Cp
Cp
Cp
i
air
H2 O ( l )
H2 O ( v )
dT
dT + H vap dT
--
b25.2gFGH dC i dT IJK + b7.1gFGH dC i dT + H + dC i dT IJK F dC i dT + H + dC i dT IJ = 0 −b25.2gb14.37g − b100 . gG H K T
25
p air
z
100
25
p H O(l ) 2
vap
100
25
p H O(l ) 2
vap
z
T
100
p H O(v ) 2
500
100
p H O(v ) 2
Integrate, solve : T = 139 o C
b.
139 139 Q = − ( 25.2 ) ∫ ( C p ) dT − (1.00 ) ∫ ( C p ) 500
air
500
H 2O ( v )
dT = −290 kW
This heat goes to vaporize the entering liquid water and bring it to the final temperature of 139oC. c. When cold water contacts hot air, heat is transferred from the air to the cold water mist, lowering the temperature of the gas and raising the temperature of the cooling water.
8-12
8.31
520 kg NH 3 103 g 1 mol 1h Basis: = 8.48 mol NH 3 s h 1 kg 17.03 g 3600 s 8.48 mol NH 3 /s 25°C n 1 (mol air/s) T °C
n 2 (mol/s) 0.100 NH 3 0.900 air 600°C
Q = –7 kW
NH 3 balance: 8.48 = 0100 . n2 ⇒ n2 = 84.8 mol s
b
gb g
Air balance: n1 = 0.900 84.8 = 76.3 mol air s
bg
References for enthalphy calculations: NH 3 g , air at 25° C NH 3
H in = 0.0 H out =
z
600
25
C p from
dC i
⇒ Hout p NH dT Table B.2 3
= 25.62 kJ mol
Air: C p ( J mol ⋅ °C ) = 0.02894 + 0.4147 × 10−5 T + 0.3191 × 10−8 T 2 − 1.965 × 10−12 T 3 T Hˆ in = ∫ C p dT 25
= ( −0.4913 × 10−12 T 4 + 0.1064 × 10−8 T 3 + 0.20735 × 10−5 T 2 + 0.02894T − 0.7248 ) ( kJ mol )
600 Hˆ out = ∫ C p dT = 17.55 kJ mol 25
Energy balance: Q = ΔH =
∑ n Hˆ − ∑ n Hˆ i
i
i
out
i
in
⇓
−7 kJ s = ( 8.48 mols NH 3 s )( 25.62 kJ mol ) + ( 76.3 mols air s )(17.55 kJ mol ) − ( 8.48 )( 0.0 ) − ( 76.3) ( −0.4913 × 10−12 T 4 + 0.1064 × 10−8 T 3 + 0.20735 × 10−5 T 2 + 0.02894T − 0.7248 )
Solve for T by trial-and-error, E-Z Solve, or Excel/Goal Seek ⇒ T = 691o C 8.32 a.
Basis: 100 mol/s of natural gas. Let M represent methane, and E for ethane Stack gas (900oC)
100 mol/s 0.95 mol M/mol 0.05 mol E/mol
Furnace
n3 n4 n5 n6
mol mol mol mol
CO2/s H2O/s O2/s N2/s
Stack gas (ToC)
Heat Exchanger
n3 n4 n5 n6
mol mol mol mol
CO2/s H2O/s O2/s N2/s
air (245oC)
20 % excess air (20oC)
n1 mol O2/s n2 mol N2/s
n1 mol O2/s n2 mol N2/s
CH 4 + 2O 2 → CO 2 + 2H 2 O
b g
C 2 H 6 + 7 / 2 O 2 → 2CO 2 + 3H 2 O
8-13
8.32 (cont’d)
⎡ 95 mol M 2 mol O 2 4.76 mol air 5 mol E 3.5 mol O 2 4.76 mol air ⎤ nair = 1.2 ⎢ + ⎥ s 1 mol M mol O 2 s 1 mol E mol O 2 ⎦ ⎣ nair = 1185 mol air/s n1 = 0.21 × 1185 = 249 mol O 2 /s, n2 = 0.79 × 1185 = 936 mol N 2 /s n3 =
95 mol M 1 mol CO 2 5 mol E 2 mol CO 2 + = 105 mol CO 2 /s s 1 mol E s 1 mol M
n4 =
95 mol M 2 mol H 2 O 5 mol E 3 mol H 2 O + = 205 mol H 2 O/s s 1 mol M s 1 mol E
95 mol M 2 mol O 2 5 mol E 3.5 mol O 2 + = 41.5 mol O 2 /s s 1 mol M s 1 mol E n6 = n2 = 936 mol N 2 /s n5 = 249 −
Energy balance on air: 245 mol air ⎞⎛ kJ ⎞ kJ ⎛ Q = nair ∫ (C p ) air dT = ⎜1185 ⎟⎜ 6.649 ⎟ = 7879 ( = 7879 kW) 20 s ⎠⎝ mol air ⎠ s ⎝
Energy balance on stack gas: 6
(
)
T Q = −ΔH = −∑ ni ∫ ( C p ) dT i =3
−7879 = n3 ∫
T
900
900
(C )
p CO 2
i
dT + n4 ∫
T
900
(C )
p H O (v ) 2
dT + n5 ∫
T
900
(C )
p O 2
dT + n6 ∫
T
900
(C )
p N 2
dT
Substitute for the heat capacities (Table B.2), integrate, solve for T using E-Z Solve⇒ T = 732 o C b.
350 m 3 (STP) mol 1000 L 1 h = 4.34 mol / s h 22.4 L(STP) m 3 3600 s 4.34 mol / s = 0.0434 100 mol / s Q ′ = 0.0434 7851 = 341 kW
Scale factor =
b g
8.33 a.
z
b
g b
g
100 . + 4 351 . + 38.4 + 42.0 + 2 36.7 + 40.2 43.9 = 23100 J mol 335 3 150 mol 23100 J 1 kW Q = ΔH = nΔH = = 3465 kW s mol 1000 J / s ΔH =
600
0
C p dT =
b. The method of least squares (Equations A1-4 and A1-5) yields (for X = T , y = C p )
z
b g
Cp = 0.0334 + 1732 . × 10−5 T ° C kJ (mol ⋅° C) ⇒ Q = 150
600
0
0.0334 + 1732 . × 10−5 T dT = 3474 kW
The estimates are exactly identical; in general, (a) would be more reliable, since a linear fit is forced in (b). 8.34 a.
e
j
ln C p = bT 1 2 + ln a ⇒ C p = a exp bT 1 2 , b=
ln C p 2 C p1 T2 − T1
T1 = 7.1 , C p1 = 0.329 ,
= 0.0473
⇒a=e ln a = ln C p1 − b T1 = −14475 .
−1.4475
8-14
U| |V ⇒ C = 0.235| |W
p
T2 = 17.3 , C p 2 = 0.533
e
= 0.235 exp 0.0473T 1 2
j
8.34 (cont’d) b.
z
150
e
j
0.235 exp 0.0473T 1 2 dT =
1800
1
20
30
40 2 200
b0.235gb2g RSexpe0.473T jLT MN 0.0473 T 12
12
−
1 .0473
OPUV QW
150
= −1730 cal g
1800
DIMENSIONS CP(101), NPTS(2) WRITE (6, 1) FORMAT (1H1, 20X'SOLUTION TO PROBLEM 8.37'/) NPTS(1) = 51 NPTS(2) = 101 DO 200K = 1, 2 N = NPTS (K) NM1 = N – 1 NM2 = N – 2 DT = (150.0 – 1800.0)/FLOAT (NM1) T = 1800.0 DO 20 J = 1, N CP (J) = 0.235*EXP(0.0473*SQRT(T)) T = T + DT SUMI = 0.0 DO 30 J = 2, NM1, 2 SUMI = SUMI + CP(J) SUM2 = 0.0 DO 40 J = 3, NM2, 2 SUM2 = SUM2 + CP (J) DH = DT*(CP(1) + 4.0 = SUM1 + 2.0 = SUM2 + CP(N))/3.0 WRITE (6, 2) N, DH FORMAT (1H0, 5XI3, 'bPOINT INTEGRATIONbbbDELTA(H)b= ', E11.4,'bCAL/G') CONTINUE STOP END
Solution: N = 11 ⇒ ΔH = −1731 cal g N = 101 ⇒ ΔH = −1731 cal g Simpson's rule with N = 11 thus provides an excellent approximation 8.35 a.
U| 175 kg 1000 g 1 mol 56.9 kJ 1 min M .W . = 62.07 g / mol V ⇒ Q = ΔH = = 2670 kW min kg 62.07 g mol 60 s | ΔH = 56.9 kJ / mol W
m = 175 kg / min
v
b. The product stream will be a mixture of vapor and liquid. c. The product stream will be a supercooled liquid. The stream goes from state A to state B as shown in the following phase diagram.
P B
A
T
8-15
(T ) = 28.85 kJ / mol Table B.1 ⇒ Tb = 68.74 o C, ΔH v b is not a function of pressure Assume: n - hexane vapor is an ideal gas, i.e. ΔH
8.36 a.
bC H g B ΔH bC H g 6
6
1
14 l, 68.74 o C
ΔH 1 = ΔH 2 =
z z
68.74
20 200 68.74
bC H g A ΔH b g ⎯ ⎯⎯⎯→ bC H g ΔH
Total ⎯ ⎯⎯ ⎯→
14 l, 20o C
6
T ΔH v b
14 v, 200o C
6
2
14 v, 68.74 o C
0.2163 dT = 10.54 kJ / mol . + 40.85 × 10 −5 T − 23.92 × 10 −8 T 2 + 57.66 × 10 −9 T 3 dT 013744
ΔH 2 = 24.66 kJ / mol ΔH Total = ΔH 1 + ΔH 2 + ΔH v Tb = 10.54 + 24.66 + 28.85 = 64.05 kJ / mol
b g
b.
ΔH = −64.05 kJ / mol
c.
U 200 o C , 2 atm = H − PV
e
j
Assume ideal gas behavior ⇒ PV = RT = 3.93 kJ / mol U = 64.05 − 3.93 = 6012 . kJ / mol
8.37
Tb = 100.00° C
e j B ΔH H O el, 100 Cj H 2 O l, 50o C
b g
ΔH v tb = 40.656 kJ mol
e
50o C ΔH v
j
⎯ ⎯⎯⎯ ⎯→
1
o
2
ΔH 1 =
z z
e j A ΔH H O e v, 100 Cj
H 2 O v, 50o C 2
e
100o C ΔH v
j
⎯ ⎯⎯⎯⎯→
o
2
100
C pH 2 Ob l g dT = 3.77 kJ mol
25
ΔH 2 =
25
C pH 2 Ob v g dT = −169 . kJ mol
100
b
Table B.1
B
g
ΔH v 50° C = 3.77 + 40.656− 169 . = 42.7 kJ mol
Steam table:
( 2547.3 − 104.8) kJ
18.01 g
1 kg
kg
1 mol
1000 g
= 44.0 kJ mol
The first value uses physical properties of water at 1 atm (Tables B.1, B.2, and B.8), while the heat of vaporization at 50oC in Table B.5 is for a pressure of 0.1234 bar (0.12 atm). The difference is ΔH for liquid water going from 50oC and 0.1234 bar to 50oC and 1 atm plus ΔH for water vapor going from 50oC and 1 atm to 50oC and 0.1234 bar. 8.38
1.75 m3 2.0 min
879 kg m3
1 kmol 1000 mol 1 min = 164.1 mol/s 78.11 kg 1 kmol 60 s
b g
Tb = 801 . ° C , ΔH v Tb = 30.765 kJ mol
8-16
8.38 (cont’d)
e j ⎯⎯→ B ΔH l −Δ H C H e v, 80.1 Cj ⎯ ⎯→ C 6 H 6 v, 580 o C 1
6
6
ΔH 1 =
z z
2
v
o
e j A ΔH el, 80.1 Cj
C 6 H 6 l, 25o C
C6H 6
o
80.1
C pC 6 H 6 b v g dT = −77.23 kJ mol
580
ΔH 2 =
298
C pC 6 H 6 b l g dT = −7.699 kJ mol
353.1
d i . x10 Q = ΔH = nΔH = b164.1 mol / sgb −115.7 kJ / molg = −190 ΔH = ΔH 1 − ΔH v 801 . o C + ΔH 2 = −115.7 kJ / mol
−4
kW
Antoine
8.39
UV W
b
B
g
35° C PV∗ 25° C 176.0 mm Hg ⇒ yCCl 4 = 015 . = 015 . = 0.0347 mol CCl 4 mol 15% relative saturation 1 atm 760 mm Hg ( ΔH v ) CCl 4
Table B.1
=
30.0
10 mol 0.0347 mol CCl 4 kJ ⇒ Q = ΔH = min mol mol
30.0 kJ = 10.4 kJ min mol CCl 4
Time to Saturation 6 kg carbon 0.40 g CCl 4 g carbon
8.40 a.
b
g
1 mol CCl 4 153.84 g CCl 4
b
1 mol gas 0.0347 mol CCl 4
g
CO 2 g, 20° C → CO 2 s, − 78.4° C : ΔH =
z
−78.4
20
1 min = 45.0 min 10 mol gas
b gdT − ΔH sub b−78.4° Cg
dC i
p CO g 2
In the absence of better heat capacity data; we use the formula given in Table B.2 (which is strictly applicable only above 0° C ). −78.4 kJ ΔH ≈ .03611 + 4.233 × 10 −5 T − 2.887 × 10 −8 T 2 + 7.464 × 10 −12 T 3 dT 20 mol
z
FG IJ H K
− 6030
cal
4.184 × 10 −3 kJ
mol
1 cal
Q = ΔH = nΔH =
= −28.66 kJ mol
300 kg CO 2
10 3 g
1 mol
28.66 kJ removed
h
1 kg
44.01 g
mol CO 2
(or 6.23 × 10 7 cal hr or 72.4 kW ) b. According to Figure 6.1-1b, Tfusion=-56oC Q = ΔH = nΔH where, ΔH = Q = n
z
−56
20
LMz dC i N
p CO (v) dT 2
−56
20
e j z dC i dT +ΔH e −56 Cj + z dC i dT OPQ
dC i
p CO (v) 2
+ΔH v −56o C + o
v
−78.4
−56
p CO (l) dT 2
−78.4
−56
8-17
p CO (l) 2
= 195 . × 10 5 k J h
C p = a + bT
8.41 a.
U| . + 0.01765T b Kg V| ⇒ C bJ mol ⋅ Kg = 4512 a = 53.94 − b0.01765gb500g = 4512 . |W NaCl b s, 300 Kg → NaClb s, 1073 Kg → NaClbl , 1073 Kg L b4512 O J + 30.21 kJ ΔH = C dT + ΔH b1073 Kg = M . + 0.01765T gdT P N Q mol mol b=
53.94 − 50.41 = 0.01765 500 − 300
p
z
ps
300
= 7.44 × 10
Q = ΔU = n
b.
z
1073
z
m
4
300
10 3 J 1 kJ
J mol
b
1073
300
1073
Cv dT + ΔU m 1073 K
g
Cv ≈ C p ΔU m ≅ ΔHm
Q ≈ ΔH = nΔH =
200 kg 10 3 g 1 kg
t=
c. 8.42
2.55 × 10 8 J
s
1 mol
74450 J
58.44 g
mol
1 kJ
0.85 × 3000 kJ 10 3 J
= 2.55 × 10 8 J
= 100 s
ΔH v = 35.98 kJ mol , Tb = 136.2° C = 409.4 K , Pc = 37.0 atm , Tc = 619.7 K (from Table B.1)
b
gb
g
b
Trouton's rule: ΔH v ≈ 0.088Tb = 0.088 409.4 K = 36.0 kJ mol 01% error . Chen's rule:
LM MN
g
FG T IJ − 0.0327 + 0.0297 log P OP HT K PQ = 35.7 kJ mol (–0.7% error) ΔH ≈ FT I 107 . −G J HT K F 619.7 − 373.2 IJ = 38.2 kJ mol Watson’s correlation : ΔH b100° Cg ≈ 35.98G H 619.7 − 409.4 K Tb 0.0331
v
b
c
10
c
b c
0.38
v
8.43
b g b g Trouton's Rule ⇒ ΔH b200° Cg = 0.088b200 + 273.2g = 41.6 kJ mol C H N bl , 25° Cg → C H Nbl , 200° Cg → C H Nbv , 200° Cg
C 7 H 2 N : Kopp's Rule ⇒ C p ≈ 7 0.012 + 12 0.018 + 0.033 = 0.333 k J (mol ⋅° C) v
7
12
7
12
7
12
200
ΔHˆ =
kJ ∫ C dT + ΔHˆ ( 200°C ) ≈ 0.333(200 − 25) mol p
v
25
8-18
+ 41.6
kJ = 100 kJ mol mol
8.44 a.
b g
Antoine equation: Tb ° C = Watson Correction:
1211033 . − 220.790 = 261 . °C 6.90565 − log 100
b g F 562.6 − 299.3IJ ΔH b261 . ° Cg = 30.765G H 562.6 − 3531. K
0.38
= 33.6 kJ mol
v
b
g
b
g
b. Antoine equation: Tb 50 mm Hg = 118 . ° C ; Tb 150 mm Hg = 35.2° C ln p 2 p1 ΔH v Clausius-Clapeyron: ln p = − + C ⇒ ΔH v = − R 1 T2 − 1 T1 RT ΔH v = −0.008314
c.
R| S| T
b
Δ H v (80.1°C)
C 6 H 6 ( l , 80.1°C)
zd zd
i
Δ H 2
C6 H 6 (v , 80.1°C)
C p dT = 7.50 kJ mol
26.1
ΔH 2 =
U| V| W
g
C6 H 6 (v , 26.1°C)
Δ H 1
80.1
g
ln 150 50 kJ = 34.3 kJ mol mol ⋅ K 1 308.4 K − 1 285.0 K
C 6 H 6 ( l , 26.1°C)
ΔH 1 =
b
26.1
l
i
C p dT = −4.90 kJ mol
80.1
b
v
g
ΔH v 261 . ° C = 7.50 + 30.765 − 4.90 = 33.4 kJ mol
8.45 a. Tout = 49.3oC. The only temperature at which a pure species can exist as both vapor and liquid at 1 atm is the normal boiling point, which from Table B.1 is 49.3oC for cyclopentane. b. Let n f , n v , and n l denote the molar flow rates of the feed, vapor product, and liquid product streams, respectively.
Ideal gas equation of state n f =
1550 L 273 K s
1 mol
423 K 22.4 L(STP)
= 44.66 mol C 5 H 10 (v) / s
55% condensation: n l = 0.550(44.66 mol / s) = 24.56 mol C 5 H 10 (l) / s Cyclopentane balance ⇒ n v = (44.66 − 24.56) mol C 5 H 10 / s = 20.10 mol C 5 H 10 (v) / s Reference: C5H10(l) at 49.3oC n in (mol/s)
H in (kJ/mol)
n out (mol/s)
H out (kJ/mol)
C5H10 (l)
—
—
24.56
0
C5H10 (v)
44.66
H f
20.10
H v
Substance
H i = ΔH v +
z
Ti
49.3o C
8-19
C p dT
8.45 (cont’d)
Substituting for ΔH v from Table B.1 and for C p from Table B.2 ⇒ H = 38.36 kJ / mol, H = 27.30 kJ / mol f
v
Energy balance: Q =
8.46 a.
∑n
out H out
−
∑n
in H in
= −116 . × 10 3 kJ / s = −116 . × 10 3 kW
Basis: 100 mol humid air fed n 2 (mol), 20o C, 1 atm
Q(kJ)
y 2 (mol H2 O/ mol), sat’d 1-y 2 (mol d ry air/ mo l)
100 mol y 1 (mol H2 O/ mol) 1-y 1 (mol d ry air/ mo l) 50o C, 1 at m, 2o superheat
n 3 (mol H2 O(l))
There are five unknowns (n2, n3, y1, y2, Q) and five equations (two independent material balances, 2oC superheat, saturation at outlet, energy balance). The problem can be solved. b.
2° C superheat ⇒ y1 =
b
p∗ 48° C
g
p
saturation at outlet ⇒ y2 =
b
p∗ 20° C
g
p
b gb g b g H O balance: b100gb y g = bn gb y g + n
dry air balance: 100 1 − y1 = n2 1 − y2 2
1
2
2
b
c.
3
g
b
References: Air 25° C , H 2 O l, 20° C
H 2 = =
25
Cp
100
20
100
20 50
H 2 =
20
25
−
−
H 2O v
air
Cp
dT =
H 2 O(l)
50
25
b
g
n in mol
n2 ⋅ y 2
H 4
H in kJ mol
n3
0
b
1
g
3
0.02894 + 0.4147 × 10 −5 T + 0.3191 × 10 −8 T 2 − 1965 × 10 −12 T 3 dT .
e
j
dT + ΔH v 100o C +
z
50
100
dC i
p H O(v) dT 2
0.0754 dT + 40.656 +
Cp
100
20
H 2
100 ⋅ 1 − y1
H out H
n out
0.03346 + 0.688 × 10 −5 T + 0.7604 × 10 −8 T 2 − 3593 × 10 −12 T 3 dT .
100
H 3 =
100 ⋅ y1
Air
zd i z z d i z z zd i z d i 50
n2 ⋅ 1 − y 2
n in
2
H 1 =
H in H
Substance
bg H Obl g
g
air
Cp
dT
H 2 O(l)
e
j
dT + ΔH v 100o C +
z
20
100
dC i
8-20
p H O(v) dT 2
8.46 (cont’d) c.
Q = ΔH = ∑ ni H i − ∑ ni H i out
Vair =
in
100 mol 8.314 Pa ⋅ m 3 323 K mol ⋅ K 101325 . × 105 Pa
∑ n H − ∑ n H i
⇒
d.
i
i
i
Q out in = Vair 100 mol 8.314 Pa ⋅ m 3 323 K mol ⋅ K × 10 5 Pa 101325 .
2° C superheat ⇒ y1 =
b
p∗ 48° C p
saturation at outlet ⇒ y2 =
g = 83.71 mm Hg = 0110 . mol H O mol 2
760 mm Hg
b
p∗ 20° C p
g = 17.535 mm Hg = 0.023 mol H O mol 2
760 mm Hg
b gb g b g H O balance: b100gb0110 . g = b9110 . gb0.023g + n ⇒ n
dry air balance: 100 1 − 0110 . = n2 1 − 0.023 ⇒ n2 = 9110 . mol 2
3
3
=
8.90 mol H 2 O 0.018 kg 1 mol
= 0160 . kg H 2 O condensed
Q = ΔH = ∑ ni H i − ∑ ni H i = −480.5 kJ out
Vair =
in
100 mol 8.314 Pa ⋅ m 3 323 K = 2.65 m 3 mol ⋅ K 101325 × 10 5 Pa .
⇒ ⇒
0160 . kg H 2 O condensed 3
2.65 m air fed −480.5 kJ 3
= 0.0604 kg H 2 O condensed / m 3 air fed
= −181 kJ / m 3 air fed
2.65 m air fed
e. f.
Solve equations with E-Z Solve. Q=
−181 kJ 250 m 3 air fed 1 h 1 kW = −12.6 kW h 3600 s 1 kJ / s m 3 air fed
8-21
8.47 Basis:
226 m 3
10 3 mol
273 K
b g
309 K 22.415 m 3 STP
min
= 8908 mol humid air min . DA = Dry air
Q ( kJ / min)
8908 mol / min y 0 [ mol H 2 O(v) / mol] (1- y 0 )(mol DA / mol)
n1 ( mol / min) y1 [ mol H 2 O(v) / mol] (1- y1 )(mol DA / mol)
36 o C, 1 atm, 98% rel. hum.
10 o C, 1 atm, saturated
n 2 [ mol H 2 O(l) / min], 10 o C
a. Degree of freedom analysis: 5 unknowns – (1 relative humidity + 2 material balances + 1 saturation
condition at outlet + 1 energy balance) = 0 degrees of freedom.
B
Table B.3
44.563 mm Hg) = 0.0575 mol H O(v) mol b36° Cg ⇒ y = 0.98(760 mm Hg Outlet air: y = p (10 C) / P = b9.209 mm Hgg b760 mm Hgg = 0.0121 mol H O(v) mol Air balance: b1 − 0.0575g(8908 mol / min) = b1 − 0.0121gn ⇒ n = 8499 mol / min F mol IJ = 0.0121(8499 mol ) + n ⇒ n = 409 mol H O(l) min H O balance: 0.0575G 8908 H min K min References: H Obl, triple point g, air b77° Fg
b. Inlet air: y 0 P =
0.98 p w* ∗
0
2
o
2
1
1
2
1
2
2
2
2
Substance nin Hˆ in nout Hˆ out Air 8396 0.3198 8396 −0.4352 n in mol min . H O (v) 512 46.2 103 45.3 Hˆ in kJ/mol 2
H2O (l )
−
−
409
0.741
Air: H from Table B.8 H 2 O: H ( kJ / kg) from Table B.5 × (0.018 kg / mol) Energy balance:
Q = ΔH = ∑ni Hˆ i − ∑ni Hˆ i = out
in
−2.50 ×104 kJ 60 min 9.486 ×10−4 Btu 1 ton = 119 tons min 1 h 0.001 kJ −12000 Btu h
8-22
8.48 746.7 m 3 outlet gas / h 3 atm
Basis:
1 kmol
b g
1 atm 22.4 m 3 STP
= 100.0 kmol / h
100 kmol/h @ 0o C, 3 atm yout (kmol C6 H14 (v)/kmol), sat'd (1 − yout )(kmol N 2 /kmol)
n1 (kmol/h) @ 75 C, 3 atm o
yin (kmol C6 H14 (v)/kmol), 90% sat'd (1 − yin )(kmol N 2 /kmol)
n2 (kmol C6 H14 (l)/h), 0o C
Antoine:
log pv∗ = 6.88555 −
1175.817 224.867 + T
pv∗ ( 0°C ) = 45.24 mm Hg, pv∗ ( 75°C ) = 920.44 mm Hg
b g = 45.24 = 0.0198 kmol C H kmol , P 3b760g 0.90 p b75° Cg b0.90gb920.44g kmol C H y = = = 0.363 P 3b760g kmol N balance: n b1 − 0.363g = 100b1 − 0.0198g ⇒ n = 153.9 kmol h C H balance: b153.9gb0.363g = b100gb0.0198g + n ⇒ n = 5389 . kmol C H bl g h . kmol h condenseg b0.363 × 153.9gb kmol h in feed g × 100% = 96.5% Percent Condensation: b5389 y out =
p v∗ 0° C
6
14
∗ v
6
14
in
2
6
1
1
14
2
References: N2(25oC), n-C6H14(l, 0oC) Substance n H n in
N2
bg bl g
n - C 6 H 14 r n - C 6 H 14
b
98000
in
. 146
out
g
−
2000
33.33
53800
0.0
N 2 : H = C p T − 25 , n − C 6 H 14 (v): H =
z
68.7
6
n in mol h H in kJ mol
b g
C pA dT + ΔH v 68.7 +
0
z T
C pv dT
68.7
Energy balance: Q = Δ H = ( −2.64 × 10 6 kJ h)(1 h / 3600 s) ⇒ −733 kW
∑ n H − ∑ n H i
out
i
i
i
in
8-23
14
H out
98000 −0.726
55800 44.75 −
2
8.49 Let A denote acetone.
Q ( kW)
W s = −25.2 kW
n1 (mol / s) @ − 18 o C, 5 atm y1 [mol A(v) / mol], sat' d (1 − y1 )( mol air / mol)
142 L / s @ 150 o C, 1.3 atm n 0 ( mol / s) y 0 [mol A(v) / mol], sat' d (1 − y 0 )( mol air / mol)
n 2 [ mol A(l) / s]@−18 o C, 5 atm
6 unknowns ( n 0 , n1 , n 2 , y 0 , y1 , Q ) –2 material balances –1 equation of state for feed gas –1 sampling result for feed gas –1 saturation condition at outlet –1 energy balance 0 degrees of freedom
a. Degree of freedom analysis:
b.
Ideal gas equation of state
Raoult’s law
P0V0 RT0
(1) n 0 =
(2) y1 =
p *A ( −18 o C) 5 atm
(Antoine equation for p *A )
Feed stream analysis (3)
y0
Air balance: n1 =
FG mol A IJ = H mol K
[(4.973 − 4.017) g A][1 mol A / 58.05 g] [(3.00 L) P0 / RT0 ] mol feed gas
n 0 (1 − y 0 ) (1 − y1 )
(4)
Acetone balance: n 2 = n 0 y 0 − n1 y1 (5) o
o
Reference states: A(l, –18 C), air(25 C)
Hˆ in
(mol/s)
(kJ/mol)
(mol/s)
(kJ/mol)
A(l)
−
n2
A(v)
n0 y0
− Hˆ
n1 y1
0 Hˆ
air
n0 (1 − y0 )
Hˆ a 0
n1 (1 − y1 )
Hˆ a1
(6) H A(v) (T ) =
z
56o C
−18 C o
A0
(C p ) A(l) dT + ( ΔH v ) A + Table B.2
(7)
nout
nin
Substance
Tab le B.1
Hˆ out
A1
z
T
56o C
(C p ) A(v) dT Ta ble B.2
H air (T ) from Table B.8
(8) Q = W s +
∑ n
out H out
−
∑ n
in H in
(W s = −25.2 kJ / s)
8-24
8.49 (cont’d) c.
(2) ⇒ y1 = 6.58 × 10−3 mol A(v)/mol outlet gas
(1) ⇒ n0 = 5.32 mol feed gas/s
(3) ⇒ y0 = 0.147 mol A(v)/mol feed gas (4) ⇒ n1 = 4.57 mol outlet gas/s
(5) ⇒ n2 = 0.75 mol A(l)/s
(6) ⇒ Hˆ A0 = 48.1 kJ/mol, Hˆ A1 = 34.0 kJ/mol (7) ⇒ Hˆ a 0 = 3.666 kJ/mol, Hˆ a1 = −1.245 kJ/mol (8) ⇒ Q = −84.1 kW
8.50
a. Feed:
3 m π (35)2 cm2
1 m2 4
s
273 K 2
10 cm
850 torr
1 kmol
103 mol
3
(273+40)K 760 torr 22.4 m (STP) 1 kmol
= 50.3
mol s
Assume outlet gas is at 850 mm Hg.
n2 (mol C6 H14 (v)/s), sat'd at T (o C) & 850 torr n3 (mol air/s)
Q (kW) 50.3 mol/s @ 40o C, 850 mm Hg yo (mol C6 H14 (v)/mol) (1 − yo )(mol air/mol)
n1 (mol C6 H14 (l)/s), T (o C)
Tdp = 25o C
60% of hexane in feed
Degree-of-freedom analysis 6 unknowns ( y0 , n1 , n2 , n3 , T , Q ) – 2 independent material balances – 2 Raoult’s law (for feed and outlet gases) – 1 60% recovery equation – 1 energy balance 0 degrees of freedom ⇒ All unknowns can be calculated. b. Let H = C6H14
(T )
dp feed
Antoine equation, Table B.4
= 25 °C ⇒ y0 =
60% recovery ⇒ n1 =
pH* ( 25°C )
0.600
P
=
151 mm Hg = 0.178 mol H mol 850 mm Hg
( 50.3)( 0.178 ) mols H feed s
= 5.37 mol H ( l ) s
Hexane balance: (0.178)(50.3) = 5.37 + n2 ⇒ n2 = 3.58 mol H ( v ) s
8-25
8.50 (cont’d) Air balance: n3 = ( 50.3)(1 − 0.178 ) = 41.3 mol air s Mole fraction of hexane in outlet gas:
pH ( T ) n2 3.58 = = ⇒ pH (T ) = 67.8 mm Hg n2 + n3 ( 3.58 + 41.3) 850 mm Hg Table B.4 → T = 7.8°C Saturation at outlet: pH* (T ) = pH (T ) = 67.8 mm Hg ⎯⎯⎯⎯
b
g
Reference states: C 6 H 14 l, 7.8° C , air (25°C)
8.95
H in 37.5
3.58
H out 32.7
—
—
5.37
0
41.3
0.435
41.3
–0.499
Substance
n in
C6 H14 ( v ) C6 H14 ( l ) Air
C6 H14 ( v ) : H =
z
68 .74
n out
b
g
T
C pl dT + Δ H v 68.74 ° C +
7 .8
z
n in mol/s H in kJ/mol
C pv dT ,
68 .74
C p from Table B.2 Δ H from Table B.1 v
Air: H from Table B.8 Energy balance: Q = ΔH =
∑ n H − ∑ n H i
i
i
out
c.
u ⋅ A = u'⋅ A' ; A =
π ⋅ D2 4
; D' =
i
=
−257 kJ s 1 kW cooling
in
U|V |W
1 D ⇒ u' = 4 ⋅ u = 12.0 m / s 2
8-26
−1 kJ s
= 257 kW
8.51 n v ( mol / min) @ 65 o C, P0 (atm) y[ mol P(v) / mol], sat' d (1- y )(mol H(v) / mol) 100 mol / s @80 o C, 5.0 atm 0.500 mol P(l) / mol 0.500 mol H(l) / mol
Q ( kJ / s) n l ( mol / min) @ 65o C, P0 (atm) 0.41 mol P(l) / mol 0.59 mol H(l) / mol
a. Degree of freedom analysis
5 unknowns – 2 material balances – 2 equilibrium relations (Raoult’s law) at outlet – 1 energy balance = 0 degrees of freedom Antoine equation (Table B.4) ⇒ p *P (65 o C) = 1851 mm Hg, p *H (65 o C) = 675 mm Hg Raoult' s law for pentane and hexane 0.410 p *P (65 o C) = yP0
y = 0.656 mol P(v) / mol
⇒
0.590 p *H (65 o C) = (1 − y ) P0
P0 = 1157 mm Hg (1.52 atm)
Total mole balance: 100 mol = n v + n l Pentane balance: 50 mole P = 0.656n v + 0.410n l Ideal gas equation of state: Vv =
n v RT 36.6 mol = P0 s
Fractional vaporization of propane: f =
⇒
n v = 36.6 mol vapor / s n l = 63.4 mol liquid / s
0.08206 L ⋅ atm mol ⋅ K
H in
P(v)
−
P(l)
50
H(v)
−
H(l)
50
n out
−
H out
24.0 24.33
2.806 26.0 −
z z
Tb o
65 C
Liquid: H (T) =
0
n in mol s H in kJ / mol
12.6 29.05
3.245 37.4
Vapor: H (T ) =
0
C pl dT + ΔH v (Tb ) +
z
T
Tb
C pv dT
T
65o C
C pl dT
Tb and ΔH v from Table B.1, C p from Table B.2
Energy balance:
. atm 152
= 667 L / s
(0.656 × 36.6) mol P(v)/s mol P vaporized = 0.480 50.0 mol P(l) fed/s mol fed
References: P(l), H(l) at 65 o C Substance n in
(65 + 273)K
Q = ∑ nout Hˆ out − ∑ nin Hˆ in = 647 kW
8-27
8.52 a.
B=benzene; T=toluene n 2 mol/s 95o C 1320 mo l/s 25o C
0.735 mol B/ mo l 0.265 mol T/ mol
0.500 mol B/ mo l 0.500 mol T/ mol
n 3 mol/s 95o C 0.425 mol B/ mo l 0.575 mol T/ mol
Q
UV RS W T
Total mole balance: 1320 = n 2 + n 3 n = 319 mol / s ⇒ 2 Benzene balance: 1320(0.500) = n 2 (0.735) + n 3 (0.425) n 3 = 1001 mol / s
References: B(l, 25oC), T(l, 25oC) n in (mol / s) H in ( kJ / mol) n out (mol / s) H out ( kJ / mol)
Substance B(l) B(v) T(l) T(v)
660 -660 --
0 -0 --
425 234 576 85
Q = ∑ ni Hˆ i − ∑ ni Hˆ i = 2.42 × 104 kW out
b.
9.838 39.91 11.78 46.06
in
e
j
e
j
Antoine equation (Table B.4) ⇒ p *B 95 o C = 1176 torr , p T* 95 o C = 476.9 torr Raoult' s law Benzene: Toluene:
b0.425gb1176g = b0.735g P ⇒ P = 680 torr U|V ⇒ P ≠ P' b0.575gb476.9g = b0.265g P' ⇒ P' = 1035 torr W|
⇒ Analyses are inconsistent.
Possible reasons: The analyses are wrong; the evaporator had not reached steady state when the samples were taken; the vapor and liquid product streams are not in equilibrium; Raoult’s law is invalid at the system conditions (not likely).
bg b gb g b gb g C H Obl g — C = b5gb12g + b12gb18g + 25 = 301 J mol Trouton’s rule — Eq. (8.4-3): ΔH = b0109 . gb113 + 273g = 42.1 kJ mol Eq. (8.4-5) ⇒ ΔH = b0.050gb52 + 273g = 16.25 k J mol
8.53 Kopp’s rule (Table B.10):
C 5 H 12 O s — C p = 5 7.5 + 12 9.6 + 17 = 170 J mol 5
12
p
v
m
Basis:
235 m 3
273 K
1 kmol 3
h
b g
389 K 22.4 m STP
10 3 mol
1h
1 kmol
3600 s
= 2.05 mol s
Neglect enthalpy change for the vapor transition from 116°C to 113°C.
b
g
b
g Obs, 25° Cg
b
C 5 H 12 O v , 113° C → C 5 H 12 O l , 113° C → C 5 H 12 O v , 52° C
b
g
→ C 5 H 12 O s, 52° C → C 5 H 12
8-28
g
8.53 (cont’d)
b
g
b
g 1 kJ kJ J − 16.2 − b301gb61g + b170gb27g × = −813 . kJ mol mol mol 10 J
ΔH = − ΔH v + C pl 52 − 113 − ΔH m + C ps 25 − 52 = −42.1
kJ mol
3
Required heat transfer: Q = ΔH = nΔH =
2.05 mol −813 . kJ s
1 kW
mol
1 kJ s
= −167 kW
8.54
Basis: 100 kg wet film ⇒ a.
95 kg dry film
0.5 kg acetone remain in film
90% A evaporation
5 kg acetone
4.5 kg acetone exit in gas phase
95 kg DF 0.5 kg C3 H 6 O( l ) Tf 2 n 1 mol air 4.5 kg C3 H 6 O( v) (40% sat'd) Ta2 = 49°C, 1.0 atm
95 kg DF 5 kg C3 H 6 O( l ) Tf 1 = 35°C n 1 mol air Ta1 , 1.01 atm
Antoine equation (Table B.4) ⇒ p C* 3H 6O = 59118 . mm Hg 4.5 kg C 3 H 6 O
⇒y=
1 kmol
10 3 mol
58.08 kg
kmol
b
bg
= 77.5 mol C 3 H 6 O v in exit gas
b g
g
171.6 mol 22.4 L STP 0.40 59118 . mm Hg 77.5 = ⇒ n1 = 77.5 + n1 760 mm Hg mol
b
g
b
g b
References: Air 25° C , C 3 H 6 O l , 35° C , DF 35° C
b.
n in
H in
n out
DF
95
0
95
86.1
0
8.6
—
—
77.5
32.3
dC i
171.6
0.70
bg Obv g
C 6 H 14 O l
C 6 H 14
z
Ta1
dC i dT + ΔH
+
171.6
Air
25
H A(v) =
z
86
p l
35
p air dT
zd
49 v
Cp
i dT , v
H out
95 kg DF
i − 35i
1.33 T f 2 − 35
b
f2
i
i
i
p air dT
=
out
⇒
z
25
c.
dC i
Ta1 = 120° C ⇒
z
Ta 1
25
n in mol H in kJ/mol
g
d
dC i
z
i
. (T f 2 − 35) + 2623.4 − 1716 . = 126.4 T f 2 − 35 + 111
in Ta1
n in kg H in kJ/kg
86
∑ n H − ∑ n H i
kg DF
H DF = C p T − 35
Energy balance ΔH =
b g
L STP
g
Substance
d 0.129 dT
= 405 .
d
i
127.5 T f 2 − 35 + 2623.4
p air dT
1716 .
d
i
= 2.78 kJ mol ⇒ T f 2 − 35 ° C = −16.8° C
8-29
Ta1
25
dC i
p air dT
=0
8.54 (cont’d) T&E
T&E
d. T f 2 = 34° C ⇒ Ta1 = 506° C , T f 2 = 36° C ⇒ Ta1 = 552° C e.
8.55
In an adiabatic system, when a liquid evaporates, the temperature of the remaining condensed phase drops. In this problem, the heat transferred from the air goes to (1) vaporize 90% of the acetone in the feed; (2) raise the temperature of the remaining wet film above what it would be if the process were adiabatic. If the feed air temperature is above about 530 °C, enough heat is transferred to keep the film above its inlet temperature of 35 °C; otherwise, the film temperature drops.
b
g
Tset p = 200 psia ≈ 100° F (Cox chart – Fig. 6.1-4)
a. Basis:
3.00 × 10 3 SCF 1 lb - mole = 8.357 lb ⋅ mole h C 3 H 8 h 359 SCF
8.357 lb-mole C3H8(v)/h 200 psia, 100oF
8.357 lb-mole C3H8(l)/h 200 psia, 100oF
Q
- mole H 2 O(l) / h m(lb 70oF
- mole H 2 O(l) / h m(lb 85oF
The outlet water temperature is 85oF. It must be less than the outlet propane temperature; otherwise, heat would be transferred from the water to the propane near the outlet, causing vaporization rather than condensation of the propane. b. Energy balance on propane: Table B.1
B Btu Q = ΔH = − nΔH v = 8.357 lb − moles −18.77 kJ 0.9486 Btu 453.593 mol = −6.75 × 10 4 h h mol kJ 1 lb ⋅ mole Energy balance on cooling water: Assume no heat loss to surroundings. p ΔT ⇒ m = Q = ΔH = mC
6.75 × 104 Btu lb m ⋅ °F lb cooling water = 4500 m h 1.0 Btu 15 °F h
8.56
m 2 [kg H 2 O(v)/h]@100o C, 1 atm 1000 kg/h, 30oC 0.200 kg solids/kg 0.800 kg H2O(l)/kg
m 3 (kg/h) @ 100o C 0.350 kg solids/kg 0.650 kg H 2 O(l)/kg m 1 [ kg H 2 O(v) / h], 1.6 bar, sat' d
a.
Solids balance: 200 = 0.35m3
b
. H 2 O balance: 800 = m2 + 0.65 5714
m 1 [ kg H 2 O(l) / h], 1.6 bar, sat' d
g
⇒ m3 = 5714 . kg h slurry
bg
⇒ m2 = 428.6 kg h H 2 O v
8-30
8.56 (cont’d)
References: Solids (0.01°C), H 2 O (l, 0.01oC) Substance m in m out H in
H out
bg H Obv g
200 800 —
62.85 125.7 —
200 571.4 428.6
209.6 419.1 2676
H 2 O , 1.6 bar
m 1
2696.2
m 1
475.4
Solids H 2O l 2
E.B. Q = ΔH =
∑ m Hˆ − ∑ m Hˆ i
i
i
out
i
m ( kg h ) H H 2 O from steam tables
b
H kJ kg
g
= 0 ⇒ 1.315 × 106 − 2221m 1 = 0 ⇒ m 1 = 592 kg steam h
in
b.
( 592.0 − 428.6 ) = 163
c.
The cost of compressing and reheating the steam vs. the cost of obtaining it externally.
8.57 Basis: 15,000 kg feed/h.
kg h additional steam
A = acetone, B = acetic acid, C = acetic anhydride Q c (kJ/h) 2 n 1 (kg A(v )/h) 329 K
15000 kg/h 0.46 A 0.27 B 0.27 C 348 K, 1 atm
condenser
n 1 (kg A(l )/h) 303 K
n 1 (kg A(l )/h) 303 K still
1% of A in feed n 2 (kg A(l )/h) n 3 (kg B( l )/h) Q r (kJ/h) n 4 (kg C( l )/h) 398 K reboiler
a.
b gb gb
g
n 2 = 0.01 0.46 15,000 kg h = 69 kg A h
b gb g Acetic anhydride balance: n = b0.27gb15,000g = 4050 kg h Acetone balance: b0.46gb15,000g = n + 69 ⇒ n = 6831 kg h ` Acetic acid balance: n 3 = 0.27 15,000 = 4050 kg B h 4
1
1
⇓ Distillate product: 6831 kg acetone h 8169 kg h 0.8% acetone Bottoms product: 69 + 4050 + 4050 kg h = 49.6% acetic acid 49.6% acetic anhydride
b
g
b. Energy balance on condenser
8-31
8.57 (cont’d)
b
g
b
g
b
C 3 H 6 O v , 329 K → C 3 H 6 O l , 329 K → C 3 H 6 O l , 303 K
b
z
g
303
g C dT = −520.6 + b2.3gb−26g = −580.4 kJ kg b2 × 6831gkg −580.4 kJ = −7.93 × 10 kJ h = ΔH = nΔH =
ΔH = − ΔH v 329 K +
pl
329
Q c
c.
6
h
kg
Overall process energy balance Reference states: A(l), B(l), C(l) at 348 K (All H m = 0 ) Substance n in H n out H in
out
b g — 0 6831 –103.5 115.0 69 0 — A bl, 398 Kg 4050 109.0 0 — B bl, 398 Kg 4050 113 0 — C bl , 398 Kg J C ≈ b4 × 12g + b6 × 18g + b3 × 25g Acetic anhydride (l): mol⋅° C A l , 303 K
p
n in kg/h H in kJ/kg
1 mol 10 3 g 1 kJ 102.1 g 1 kg 10 3 J
= 2.3 kJ kg⋅° C H T = C p T − 348 (all substances)
bg
b
g
Q = ΔH ⇒ Q c + Q r =
∑ n H − ∑ n H i
out
i
i
in
i
⇒ Q r = −Q c +
A=0
∑ n H = e7.93 × 10 i
i
6
out
. × 10 6 kJ h = 813
(We have neglected heat losses from the still.) d.
H 2 O (saturated at ≈ 11 bars): ΔH v = 1999 kJ kg (Table 8.6) 813 . × 10 6 kJ h Q r = n H 2 O ΔH v ⇒ n H 2 O = = 4070 kg steam h 1999 kJ kg
8.58 Basis: 5000 kg seawater/h a.
S = Salt n 3 (kg H 2 O(l )/h @ 4 bars) 2738 kJ/kg
n 4 kg H 2 O(v )/h @ 0.2 bars 2610 kJ/kg
n 2 (kg H 2 O(v )/h @ 0.6 bars) 2654 kJ/kg n 1 (kg/h @ 0.6 bars) 0.055 S 0.945 H 2 O(l ) 360 kJ/kg
5000 kg/h @ 300 K 0.035 S 0.965 H 2O( l) 113.1 kJ/kg
n 5 (kg H 2 O(l )/h @ 4 bars) 605 kJ/kg
b
n 3 (kg/h @ 0.2 bars) x (kg S/kg) (1 – x) (kg H2 O(l )/hr) 252 kJ/kg n 2 (kg H 2 O(l )/h @ 0.6 bars) 360 kJ/kg
gb g
b. S balance on 1st effect: 0.035 5000 = 0.055n1 ⇒ n1 = 3182 kg h
Mass balance on 1st effect: 5000 = 3182 + n 2 ⇒ n 2 = 1818 kg h
8-32
j
+ 2.00 × 105 kJ h
8.58 (cont’d)
Energy balance on 1st effect:
b gb
g b gb g b gb = 2534 kg H Obv g h
g b gb g
ΔH = 0 ⇒ n 2 2654 + n1 360 + n 5 605 − 2738 − 5000 1131 . =0 n 5
n1 = 3182 n2 =1818
c.
2
Mass balance on 2nd effect: 3182 = n 3 + n 4 (1)
bΔH = 0g bn gb2610g + bn gb252g + bn gb360 − 2654g − bn gb360g = 0 E n = 3182, n = 1818
Energy balance on 2nd effect: 4
3
2
1
1
2
5.316 × 10 = 252n 3 + 2610n 4 6
(2)
Solve (1) and (2) simultaneously: n 3 = 1267 kg h brine solution
bg
n 4 = 1915 kg h H 2 O v
b
g
Production rate of fresh water = n 2 + n 4 = 1818 + 1915 = 3733 kg h fresh water
b
gb g
Overall S balance: 0.035 5000 = 1267 x ⇒ x = 0138 . kg salt kg
d. The entering steam must be at a higher temperature (and hence a higher saturation pressure) than that of the liquid to be vaporized for the required heat transfer to take place.
e.
n 5 (kg H 2 O(v )/h) 2738 kJ/kg
3733 kg/h H 2 O(v ) @ 0.2 bar 2610 kJ/kg n 1 (kg brine/h @ 0.2 bar 252 kJ/kg
5000 kg/h 0.035 S 0.965 H 2 O(l ) 113.1 kJ/kg Q3
n 5 (kg H 2 O(l )/h) 605 kJ/kg
Mass balance: 5000 = 3733 + n1 ⇒ n1 = 1267 kg h
dΔH = 0i b3733gb2610g + b1267gb252g + n b605 − 2738g − b5000gb1131. g = 0 ⇒ n = 4452 kg H Obv g h
Energy balance:
5
5
2
Which costs more: the additional 1918 kg/hr fresh steam required for the single-stage process, or the construction and maintenance of the second effect?
8-33
8.59 a.
b0.035gb5000g = 583 kg h
Salt balance: x L 7 n L 7 = x L1 n L1 ⇒ n L1 =
0.30 Fresh water produced: n L 7 − n L1 = 5000 − 583 = 4417 kg fresh water h
b. Final result given in Part (d). c.
Salt balance on i th effect:
b g bx g
n Li x Li = n L
i +1
L i +1
bn g b x g L i +1
⇒ x Li =
L i +1
(1)
nθ Li
Energy balance on i th effect:
b g e H j
ΔH = 0 ⇒ n vi H vi + n v ⇒
b g n v
L −1
=
b g e H j + n H − bn g e H j e H j − e H j
L −1
n vi H vi
v
Li v
b g = bn g + bn g ⇒ bn g
Mass balance on i − 1 n Li
v i −1
L i −1
th
L −1 Li
+ n Li H Li − n L L i +1
L
i −1
L
L +1
L
L +1
b g e H j
− n v
L −1
v
L −1
=0
(2)
i +1
L −1
effect: L i −1
b g
= n Li − n v
(3)
i −1
d.
Fresh steam Effect 1 Effect 2 Effect 3 Effect 4 Effect 5 Effect 6 Effect (7)
P (bar) 2.0 0.9 0.7 0.5 0.3 0.2 0.1 1.0
T (K) 393.4 369.9 363.2 354.5 342.3 333.3 319.0 300.0
nL (kg/h) --584 1518 2407 3216 3950 4562 5000
8-34
xL --0.2997 0.1153 0.0727 0.0544 0.0443 0.0384 0.0350
nV (kg/h) 981 934 889 809 734 612 438 ---
HL (kJ/kg) 504.7 405.2 376.8 340.6 289.3 251.5 191.8 113.0
HV (kJ/kg) 2706.3 2670.9 2660.1 2646.0 2625.4 2609.9 2584.8 ---
8.60 a.
dC i = dC i p v
p l
b g ≈ dC i
= 20 cal (mol⋅° C) ; Cv
p v
v
b
− R ≈ 10 − 2
g molcal⋅° C = 8 cal (mol⋅° C)
b.
n0 (mol N2)
n0 (mol N2)
o
3.00 L@ 93 C, 1 atm
n2 [mol A(v)] 85oC, P(atm)
n1 (mol A(l)
n3 [mol A(l)] o
85oC, P(atm)
0.70 mL, 93 C n0 =
3.00 L
n1 =
70.0 mL
b
273 K 1 mol . = 0100 mol N 2 273 + 93 K 22.4 L STP
g
b g
bg
0.90 g 1 mol . mol A l = 15 mL 42 g
Energy balance ⇒ ΔU = 0 ⇒
∑ n U − ∑ n U i
i
i
out
c.
i
=0
in
b g b gb
g
References: N 2 g , A l 85° C, 1 atm
Substance n in U in n out U out 010 39.8 010 0 N2 . . n in mol 15 160 n 3 0 Al . U in cal mol Av n 2 20050 − −
bg bg
b g b g b g Abv , 85° Cg: U = 20b90 − 85g + 20,000 + 10b85 − 90g = 20050 cal mol ΔU = 0 ⇒ n b20050g − b010 . gb39.8g − b15 . gb160g = 0 ⇒ n = 0.012 mol A evaporate A l , 93° C and N 2 g , 93° C : U = Cv 93 − 85 A( v )
v1
v1
0.012 mol A 42 g A
⇒
mol A
= 0.51 g evaporate
d. Ideal gas equation of state
P=
bn
0
g
+ n 2 RT V
=
0.112 mol 3.00 liters
b273 + 85gK
0.08206 L ⋅ atm = 1097 . atm mol ⋅ K
Raoult’s law
b
g
p ∗A 85° C = y A P =
0.012 mol 1097 atm . n2 P= = 0.117 atm 0.112 mol n0 + n2
8-35
b= 89.3 mmHgg
8.61 (a) i)
FG mIJ HV K
Expt 1 ⇒
=
b4.4553 − 3.2551gkg = 0.600 kg ⇒ bSGg 2.000 L
liquid
L
b
liquid
= 0.600
g
ii) Expt 2 ⇒ Mass of gas = 3.2571 − 3.2551 kg = 0.0020 kg = 2.0 g
Moles of gas =
b763 − 500gmm Hg
2.000 L 273 K 363 K
1 mol = 0.0232 mol 22.4 liters STP
b g
760 mm Hg
b g b0.0232 molg = 86 g mol
Molecular weight = 2.0 g
iii) Expt. 1 ⇒ n =
b liquid g
2.000 liters 10 3 cm 3 1 liter
0.600 g 1 mol = 14 mol cm 3 86 g
Energy balance: The data show that Cv is independent of temperature Q = ΔU = nCv ΔT
b g
⇒ Cv
liquid
b g
⇒ Cv
liquid
=
Q 800 J = = 24 J mol ⋅
[email protected] K nΔT 14 mols 2.4 K
b gb g 800 J = b14 molsgb2.4 Kg = 24 J mol ⋅
[email protected] K
≡ 24 J mol ⋅ K
bg
Expt. 2 ⇒ n = 0.0232 mol from ii
b vapor g
Cv = a + bT ⇒ Q = 0.0232
z
T2
T1
LM N LM N
LM N
b (a + bT ) dT = 0.0232 a (T2 − T1 ) + (T22 − T12 ) 2
b 130 . J = 0.0232 a(366.9 - 363.0) + (366.9 2 − 363.02 ) 2 b 1.30 J = 0.0232 a(492.7 - 490.0) + (492.7 2 − 490.02 ) 2
b g
⇒ Cv
vapor
bg
OP U| Q V ⇒ a = −4.069 OP| b = 0.05052 Q|W
(J / mol ⋅ K) = −4.069 + 0.05052T K
iv) Liquid: C p ≈ Cv ≡ 24 J mol ⋅ K Vapor: Assuming ideal gas behavior, C p = Cv + R = Cv + 8.314 J mol ⋅ K
b
g
bg
⇒ C p J mol ⋅ K = 4.245 + 0.05052T K
b
g
v) Expt. 3 ⇒ T = 315K , p ∗ = 763 − 564 mm Hg = 199 mm Hg ∗
T = 334 K , p = 401 mm Hg T = 354 K , p ∗ = 761 mm Hg T = 379 K , p ∗ = 1521 mm Hg
8-36
OP Q
8.61 (cont’d)
Plot p ∗ (log scale) vs. 1 T (linear scale); straight line fit yields −3770 ln p ∗ = + 17.28 or p∗ = 3196 . × 107 exp − 3770 T T K
b
bg
g
b g
1 17.28 − ln 760 = = 2.824 × 10 −3 K −1 ⇒ Tb = 354 K A T 3770 Part v b
vi) p ∗ = 760 mm Hg ⇒
vii)
ΔH v = 3770 K ⇒ ΔH v = 3770 K 8.314 J mol ⋅ K ⇒ ΔH v = 31,300 J mol R PartA v
bg
b
gb
g
3.5 L feed 273 K 1 mol = 0.0836 mol s feed gas s 510 K 22.4l STP Let A denote the drug
b g
(b) Basis:
.
0.0836 mol/s @ 510 K 0.20 A 0.80 N 2
n 1 [mol A(v)/s] . n 2 [mol N 2 /s] T(K), saturated with A
.
Q(kW)
n 3 (mols A(l )/s), 90% of A in feed T(K)
b
gb g 90% condensation: n = b0.900gb0.200 × 0.0836g = 0.01505 mol Abl g s n = b0100 . gb0.200 × 0.0836g = 167 . × 10 mol Abv g s N 2 balance: n2 = 0.800 0.0836 mol s = 0.0669 mol N 2 s 3
−3
1
Partial pressure of A in outlet gas: pA =
b
n1 . × 10 −3 mol 167 (760 mm Hg) = 18.5 mm Hg = p∗A T P= 0.0686 mol n1 + n2
bg
g
E Part (a) - (v) 1 17.28 − lnb18.5g = = 3.81 × 10 3770
T
−3
K −1
⇓ T = 262 K
bg
(c) Reference states: N 2 , A l at 262 K
substance
nin
H in
nout
N2
0.0669
7286
0.0669
bg Abl g
Av
H out
0.0167 37575 1.67 × 10
−
−
n in mol s 31686 H in J mol
0
−3
0.01505
8-37
0
8.61 (cont’d)
b
g
N 2 510 K : H N 2 (510K) - H N 2 ( 262 K) = H N 2 (237 o C) - H N 2 ( −11o C) Table B.8
B
= [6.24 - (-1.05)] kJ / mol = 7.286 kJ / mol = 7286 J / mol
b
g
b
g
A(v, 262K): H = C pl Tb − 262 + ΔH v 359 K +
z
262
C pv dT
Tb
Part (a) results for Tb , C pl , C pv , ΔH v
b
LM N
g
T H = 24 354 − 262 + 31300 + 4.245 + 0.05052 2
b
g
b
g
A(v, 510K): H = C pl Tb − 262 + ΔH v 354 K +
z
510
Tb
OP Q
2 262
= 31686 J mol
354
C pv dT = 37575 J mol
−1060 J s 1 kW cooling Energy balance: Q = ΔH = ∑ ni H i − ∑ ni H i = = 106 . kW −103 kJ s out in
8.62 a. Basis: 50 kg wet steaks/min D.M. = dry meat m1 (kg H 2 O(v )/min) (96% of H 2 O in feed) 60°C
50 kg/min @ –26°C 0.72 H 2O( s) 0.28 D.M. Q(kW)
m2 (kg D.M./min) m3 (kg H 2 O(l )/min) 50°C
96% vaporization: m 1 = 0.96 0.72 × 50 kg min = 34.56 kg H 2 O v min
b g bg m = 0.04b0.72 × 50 kg ming = 144 . kg H O bl g min Dry meat balance: m = b0.28gb50g = 14.0 kg D. M. min Reference states: Dry meat at −26° C , H Obl, 0° Cg 3
2
2
2
substance dry meat H 2 O s, − 26° C H 2 O l , 50° C H 2 O v , 60° C
b
b b
g g
m in H in m out H out 14.0 0 14.0 105 m in kg min H in kJ kg 36.0 −390 − − . 144 209 − − 34.56 2599 − −
g
1.38 kJ 76° C = 105 kJ kg Dry meat: H 50° C = C p 50 − −26 = kg ⋅ C°
b
b g H Obs, − 26° Cg: H Obl , 0° Cg → H Obs, 0° Cg → H Obs, 2
g
2
2
8-38
2
− 26° C
g
8.62 (cont’d)
z
b g
−6.01 kJ
−26
ΔH = − ΔH m 0° C + C p dT =
1 mol 10 3 g 2.17 kJ 18.02 g 1 kg + kg⋅° C
mol
A
0
−26° C
= −390 kJ kg
Table B.1
b
b
g
g b b50 − 0g° C
H 2 O l, 50° C : H 2 O l , 0° C → H 2 O l , 50° C
z
0.0754 kJ 50 ΔH = C p dT = mol ° C
1 mol 1000 g 18.02 g 1 kg = 209 kJ kg
A
0
g
Table B.2
b
g
b
g
b
g
b
g
b
H 2 O v , 60° C : H 2 O l , 0° C → H 2 O l , 100° C → H 2 O v , 100° C → H 2 O v , 60° C
0.0754 kJ ΔH = mol⋅° C
b100 − 0g° C
A
+ 40.656
Ad
Table B.2
=
kJ + mol
Table B.1 ΔH v
46.830 kJ
1 mol
1000 g
mol
18.02 g
1 kg
zd 60
100
i
Cp
i
H 2 O(v)
A
g
dT
Table B.2
= 2599 kJ kg
Energy balance:
Q = ΔH =
∑
mi H i −
out
∑
mi H i =
1.06 × 10 5 kJ 1 min min
in
60 s
1 kW 1 kJ s
= 1760 kW
8.63 Basis: 20,000 kg/h ice crystallized. S = solids in juice. W = water . . m1 (kg/h) juice 0.12 solids(S) 0.88 H 2O( l )(W) 20°C
Qf preconcentrate .m (kg/h) . m5 (kg/h) product Slurry(10% ice), –7°C 2 freezer filter x 2 (kg S/kg) 20,000 kg W( s )/h 0.45 kg S/kg . m4 kg residue/h 0.55 kg W/kg (1 – x 2) (kg W/kg) 0.45 kg S/kg 20,000 kg W( s )/h . kg W( l )/kg m4 (kg/h), 0.45 S, 0.55 W .m (kg/h),0.55 0°C 3 separator 0.45 kg S/kg 20,000 kg W( s )/h 0.45 kg W( l )/kg
(a) 10% ice in slurry ⇒
20000 10 = ⇒ m 4 = 180000 kg h concentrate leaving freezer 90 m 4
UV W
m 1 = 27273 kg h feed Overall S balance: 012 . m 1 = 0.45m 5 ⇒ m 5 = 7273 kg h concentrate product Overall mass balance: m 1 = m 5 + 20000 Mass balance on filter: 20000 + m 4 + m 5 + 20000 + m 6
⇒
m 4 =180000 m 5 = 7273
m 6 = 172730 kg h recycle
Mass balance on mixing point: 27273 + 172730 = m 2 ⇒ m 2 = 2.000 × 105 kg h preconcentrate
8-39
8.63 (Cont’d)
S balance on mixing point: 012 . 27273 + 0.45 172730 = 2.000 × 105 X 2 ⇒ X 2 ⋅ 100% = 40.5% S
b gb
g b gb
g
(b) Draw system boundary for every balance to enclose freezer and mixing point (Inputs: fresh feed and recycle streams; output; slurry leaving freezer)
bg
Refs: S, H 2 O l at −7° C substance
m in
H in
m out
H out
12% soln
27273
108
−
−
45% soln 172730
28
180000
0
−
20000
−337
bg
−
H 2O s
b b
g g
m kg h H kJ kg
bg b g kJ kg H = − ΔH b − T ° Cg ≈ − ΔH b0° Cg
Solutions: H T = 4.00 T − −7 Ice:
m
m
= −6.0095 kJ mol ⇒ −337 kJ kg
D Table B.1
−1452 × 107 kJ . 1h 1 kW E.B. Q c = ΔH = ∑ m i H i − ∑ m i H i = = −4030 kW h 3600 s 1 kJ s out in
d
8.64 a. B=n-butane, I=iso-butane, hf=heating fluid. (C ) = 2.62 kJ / kg⋅ o C p hf 24.5 kmol/h @ 10oC, P (bar) 0.35 kmol B(l)/h
i
24.5 kmol/h @ 180oC 0.35 kmol B(l)/h
Q ( kW) m (kg HF / h), T( o C)
m (kg HF / h), 215 o C
From the Cox chart (Figure 6.1-4)
d
i
d
i
p B* 10 o C = 22 psi, pI* 10 o C = 32 psi p min = p B + pI = x B p B* + x I pI* = 28.5 psi b.
d
i
d
i
FG 1.01325 bar IJ = 1.96 bar H 14.696 psi K d
Hv H1 B l, 10 o C ⎯Δ⎯ ⎯ → B v, 10 o C ⎯Δ⎯ ⎯ → B v, 180 o C
d
i
d
i
d
Hv H2 I l, 10 o C ⎯Δ⎯ ⎯ → I v, 10 o C ⎯Δ⎯ ⎯ → I v, 180 o C
i
i
Assume temperature remains constant during vaporization. Assume mixture vaporizes at 10oC i.e. won’t vaporize at respective boiling points as a pure component.
8-40
8.64 (cont’d)
References: B(l, 10oC), I(l, 10oC) substance nin mol / h H in kJ / mol
b
B (l) B (v) I (l) I (v)
g
b
8575 -15925 --
z z
b
g
n out mol / h
0 -0 --
g
-8575 -15925
b
H out kJ / mol
g
-42.21 -41.01
d H i = dΔH i + dC i = 42.21 kJ / mol . kJ / mol d H i = dΔH i + dC i = 4101 ΔH = ∑ n H − ∑ n H = 8575b42.21g − 15825b41.01g out
v
B
B
180
p B
10
180
out
v
I
i
I
i
out
p I
10
i
i
in
ΔH = 1015 . × 10 6 kJ / h c.
d
Q = 1.015 × 10 6 kJ / h = m hf 2.62 kJ / kg⋅ o C
i b215 − 45g C o
m hf = 2280 kg / h d.
. × 10 b2540 kg / hg 2.62 kJ / dkg⋅ Ci b215 − 45g C = 1131 o
o
6
kJ / h
Heat transfer rate = 1131 . × 10 6 − 1015 . × 10 6 = 116 . × 10 5 kJ / h e. The heat loss leads to a pumping cost for the additional heating fluid and a greater heating cost to raise the additional fluid back to 215oC. f. Adding the insulation reduces the costs given in part (e). The insulation is probably preferable since it is a one-time cost and the other costs continue as long as the process runs. The final decision would depend on how long it would take for the savings to make up for the cost of buying and installing the insulation. 8.65 (a) Basis: 100 g of mixture, SGBenzene=0.879: SGToluene=0.866
50 g 50 g + = (0.640 + 0.542) mol = 1183 . mol 78.11 g / mol 92.13 g / mol 50 g 50 g = + = 114.6 cm3 3 0.879 g / cm 0.866 g / cm3
ntotal =
Vtotal
dx i f
C6 H 6
=
0.640 mol C 6 H 6 = 0.541 mol C 6 H 6 mol 1.183 mol
Actual feed:
32.5 m3 106 cm3 h
1 m
3
1183 . mol mixture 3
1h
114.6 cm mixture 3600 s
= 9319 . mol / s
T = 90° C ⇒ pC∗ 6 H 6 = 1021 mm Hg , pC∗ 7 H 8 = 407 mm Hg (from Table 6.1-1)
b
gb g b
gb g
Raoult' s law: ptot = x C6 H 6 pC∗ 6 H 6 + x C7 H 8 pC∗ 7 H 8 = 0.541 1021 + 0.459 407 =
739.2 mmHg
1 atm = 0.973 atm ⇒ P0 > 0.973 atm 760 mmHg
8-41
8.65 (cont’d) (b) T = 75° C ⇒ pC∗ 6 H 6 = 648 mm Hg , pC∗ 7 H 8 = 244 mm Hg (from Table 6.1-1)
b
gb g b
gb g
Raoult's law ⇒ ptank = xC6 H 6 pC∗ 6 H 6 + xC7 H 8 pC∗ 7 H 8 = 0.439 648 + 0.561 244
b
g
= 284 + 137 mm Hg = 421 mmHg ⇒ Ptank = 0.554 atm yC 6 H 6 =
bg
284 mm Hg = 0.675 mol C 6 H 6 v mol 421 mm Hg n v (mol/s), 75°C 0.675 C 6H 6 (v ) 0.554 atm 0.325 C 7H 8 (v ) n L (mol/s), 75°C 0.439 C6 H6 (l ) 0.541 C7H8 (l )
93.19 mol/s 0.541 C 6H 6( l ) 0.459 C 7 H 8 (l ) 90°C, P0 atm
UV ⇒ n + 0.439n W n
Mole balance: 9319 . = nv + n L
b
gb
g
C 6 H 6 balance: 0.541 9319 . = 0.675nv
bg
L
v
= 40.27 mol vapor s
L
= 52.92 mol liquid s
bg
(c) Reference states: C 6 H 6 l , C 6 H 6 l at 75° C
Substance
n in
H in
H out
n out
b g − − 27.18 310. n in mol s 0 C H bl g 50.41 2.16 23.23 H in kJ mol − − 13.09 35.3 C H bv g 0 C H bl g 42.78 2.64 29.69 C H bl , 90° Cg: H = b0.144gb90 − 75g = 2.16 kJ mol C H bl , 90° Cg: H = b0.176gb90 − 75g = 2.64 kJ mol C H bv , 75° Cg: H = b0144 . gb801 . − 75g + 30.77 + z 0.074 + 0.330 × 10 A C6H 6 v 6
6
7
8
7
8
6
6
7
8
6
6
75
b
ΔH v 80 .1°C
80.1
g
−3
T dT
= 310 . kJ mol
b
g
b
gb
g
C 7 H 8 v , 75° C : H = 0176 . 110.6 − 75 + 33.47 +
z
75
110.6
0.0942 + 0.380 × 10 −3 T dT
= 35.3 kJ mol Energy balance: Q = ΔH =
∑ n H − ∑ n H i
out
i
i
in
i
=
1082 kJ 1 kW = 1082 kW s 1 kJ s
(d) The feed composition changed; the chromatographic analysis is wrong; the heating rate changed; the system is not at steady state; Raoult’s law and/or the Antoine equation are only approximations; the vapor and liquid streams are not in equilibrium. (e) Heat is required to vaporize a liquid and heat is lost from any vessel for which T>Tambient. If insufficient heat is provided to the vessel, the temperature drops. To run the experiment isothermally, a greater heating rate is required.
8-42
8.66 a. Basis: 1 mol feed/s n V mo l vapor/s @ T, P 1 mo l/s @ TFo C
y mol A/mol (1-y) mol B/mo l
xF mol A/mol (1-xF) mol B/mo l
n L mol vapor/s @ T, P
vapor and liquid streams in equilibrium
x mol A/mol (1-x) mol B/mo l
bg b g bg b T g ⇒ y = x ⋅ p bT g
Raoult's law ⇒ x ⋅ p ∗A T + 1 − x ⋅ p ∗B T = P ⇒ x = pA = y ⋅ P = x ⋅
p ∗A
bg bT g − p bT g
P − p B∗ T p ∗A
∗ B
∗ A
(2)
P
Mole balance: 1 = n L + nV ⇒ nV = 1 − n L A balance:
b x gb1g = y ⋅ n F
V
Energy balance: ΔH =
(4)
for nv from (4) + x ⋅ n L ⎯Substitute ⎯⎯⎯⎯⎯⎯ ⎯→ n L =
∑ n H − ∑ n H i
out
i
i
(1)
i
y − xF y−x
=0
(3) (5)
in
b. Tref(deg.C) = 25 Compound n-pentane n-hexane
A 6.84471 6.88555
B 1060.793 1175.817
C 231.541 224.867
xF Tf(deg.C) P(mm Hg) HAF(kJ/mol) HBF(kJ/mol)
0.5 110 760 16.6 18.4
0.5 110 1000 16.6 18.4
0.5 150 1000 24.4 27.0
T(deg.C) pA*(mm Hg) pB*(mm Hg) x y nL(mol/s) nV(mol/s) HAL(kJ/mol) HBL(kJ/mol) HAV(kJ/mol) HBV(kJ/mol) DH(kJ/s)
51.8 1262 432 0.395 0.656 0.598 0.402 5.2 5.8 31.4 42.4 0.00
60.0 1609 573 0.412 0.663 0.648 0.352 6.8 7.6 32.5 43.7 0.00
62.3 1714 617 0.349 0.598 0.394 0.606 7.3 8.0 32.8 44.1 0.00
al 0.195 0.216
8-43
av 0.115 0.137
bv 3.41E-04 4.09E-04
Tbp 36.07 68.74
DHv 25.77 28.85
8.66 (cont’d) c. C*
C* 1 C*
C*
C*
C*
2
20 25 3 30
PROGRAM FOR PROBLEM 8.66 IMPLICIT REAL (N) READ (5, 1) A1, B1, C1, A2, B2, C2 ANTOINE EQUATION COEFFICIENTS FOR A AND B FORMAT (8F10.4) READ (5, 1) TRA, TRB ARBITRARY REFERENCE TEMPERATURES (DEG.C.) FOR A AND B READ (5, 1) CAL, TBPA, DHVA, CAV1, CAV2 READ (5, 1) CBL, TBPB, DHVB, CBV1, CBV2 CP(LIQ, KS/MBL-DEG.C.), NORMAL BOILING POINT (DEG.C), HEAT OF VAPORIZATION (KJ/MOL), COEFFICIENTS OF CP(VAP., KJ/MOL-DEG.C) = CV1 + CV2*T(DEG.C) READ (5, 1) XF, TF, P MOLE FRACTION OF A IN FEED, FEED TEMP.(DEG.C), EVAPORATOR PRESSURE (MMHG) WRITE (6, 2) TF, XF, P FORMAT (1H0, 'FEEDbATb', F6.1, 'bDEG.CbCONTAINSb', F6.3,' bMOLESbA/MOLEbT *OTAL'//1X'EVAPORATORbPRESSUREb=', E11.4, 'bMMbHG'/) ITER = 0 DT = 0.5 HAF = CAL*(TF – TRA) HBF = CBL*(TF – TRB) F1 = XF*HAF + (1.0 – XF)*HBF F2 = CAL*(TBPA – TRA) + DHVA – CAV1*TBPA – 0.5*CAV2*TBPA**2 F3 = CBL*(TBPB – TRB) + DHVB – CBV1*TBPB – 0.5*CBV2*TBPB**2 T = TF INTER = ITER + 1 IF(ITER – 200) 30, 30, 25 WRITE (6, 3) FORMAT (1H0, 'NO CONVERGENCE') STOP PAV = 10.0** (A1 – B1/(T + C1)) PAV = 10.0** (A2 – B2/(T + C2)) XL = (P – PBV)/(PAV – PBV) XV = XL*PAV/P NL = (XV – XF)/(XV – XL) NV = 1.0 – NL IF (XL.LE.00.OR.XL.GE.1.0.OR.NL.LE.0.0.OR.NL.GE.1.0) GO TO 45 HAL = CAL*(T – TRA) HBL = CBL*(T – TRB) HAV = F2 + CAV1*T + 0.5*CAV2*T**2 HBV = F3 + CBV1*T + 0.5*CBV2*T**2
8-44
8.66(cont’d)
DELH = NL *(XL*HAL + (1.0 – XL)*HBL) + NV*(XV*HAV + (1.0 – XV)*HBV) – F1 WRITE (6, 4) T, NL, NV, DELH 4 FORMAT (1Hb, 5X' Tb=', F6.1, 3X' NLb=', F7.4, 3X' NVb=', F7.4, 3X'DELHb =',* E11.4) WRITE (6, 5) PAV, PBV, XL, HAL, HBL, XV, HAV, HBV 5 FORMAT (1Hb, 5X' PAV, PBVb=', 2F8.1, 3X' XL, HAL, HBLb=', F7.4, 2E13.4,3X' XV, HAV, HBVb=', F7.4, 2E13.4/) IF (DELH) 50, 50, 40 40 DHOLD = DELH TOLD = T 45 T = T – DT GO TO 20 50 T = (T*DHOLD – TOLD*DELH)/(DHOLD – DELH) PAV = 10.0**(A1 – B1/(T + C1)) PBV = 10.0**(A2 – B2/(T + C2)) XL = (P – PBV)/(PAV – PBV) XV = XL * PAV/P NL = (XV – XF)/(XV – XL) NV = 1.0 – NL WRITE (6, 6) T, NL, XL, NV, XV 6 FORMAT (1H0, 'PROCEDUREbCONVERGED'//3X'EVAPORATORb TEMPERATUREb=', F6. *1//3X' LIQUIDbPRODUCTb--', F6.3, 'bMOLEbCONTAININGb', F6.3, 'bMOLEbA/ *MOLEbTOTAL'//3X' VAPORbPRODUCTb--', F6.3, MOLEbCONTAININGb,' F6.3, *'bMOLEbA/MOLEb TOTAL') STOP END $DATA (Fields of 10 Columns) Solution: Tevaportor = 52.2° C
d
i
d
i
n L = 0.552 mol, x C5H12 nv = 0.448 mol, x C5H12
liquid
= 0.383 mol C 5 H 12 mol liquid
vapor
= 0.644 mol C 5 H 12 mol liquid
8-45
8.67 Basis:
2500 kmol product 1 kmol condensate h
.25 kmol product
= 10,000 kmol h fed to condenser
.
m1 , (kg/h) at T1 1090 kmol/h C 3 H 8 (v ) 7520 kmol/h i -C 4H10 (v ) 1390 kmol/h n -C 4H10 (v ) saturated vapor at Tf, P
1090 kmol/h C 3 H 8 ( l ) 7520 kmol/h i -C 4H10 ( l ) P (mm Hg) 1390 kmol/h -C H ( ) n 4 10 l T out
.
m1 (kg/h) at 2 T2
(a) Refrigerant: Tout = 0 o C , T1 = T2 = −6 o C .
Antoine constants C3H 8 i − C 4 H 10 n − C 4 H 10
A 7.58163 6.78866 6.82485
B 1133.65 899.617 943.453
C 283.26 241.942 239.711
Calculate P for Tout = Tbubble pt. P = ∑ xi pi* ( 0°C ) = 0.109 ( 3797 mm Hg ) + 0.752 (1176 mm Hg ) + 0.139 ( 775 mm Hg ) i
⇒ P = 1406 mm Hg Dew pt. T f = Tdp ⇒ f (T f ) = 1 − P ∑ i
bg
yi
p (T f * i
)
= 0 trial & error to find T f ⇒ T f = 5.00o C
bg
Refs: C 3 H 8 l , C 4 H 10 l at 0 °C, Refrigerant @ –6°C
b g
Assume: ΔH v Tb , Table B.1 substance C3H 8 i − C 4 H 10
n − C 4 H 10 Refrigerant
nin
H in
1090 19110 7520 21740 1390 22760
m 1
0
nout
H out
1090 7520 1390
0 0 0
n (kmol/h) H (kJ/kmol)
m 1
151
m (kg/h) H (kJ/kmol)
↓
U| H bvapor g = ΔH b0° Cg + V| C dT bTable B.2g W 2
z
v
4 .95
p
0
UVH = ΔH W
v
E.B.: ΔH = ∑ ni H i − ∑ ni H i = 0 ⇒ 151m 1 − 2.16 × 106 = 0 ⇒ m 1 = 143 . × 106 kg h refrigerant out
in
8-46
8.67 (cont’d) (b) Cooling water: Tout = 40° C , T2 = 34°C , T1 = 25°C
P = ∑ xi pi* ( 40°C ) = 0.109 (11,877 ) + 0.752 ( 3961) + 0.139 ( 2831) = 4667 mm Hg i
f (T f ) = 1 − P ∑
bg
i
T +E yi = 0 ⇒ T f = 45.7°C p (T f ) * i
bg
bg
Refs: C 3 H 8 l , C 4 H 10 l @ 40°C, H 2 O l @ 25°C. ΔH = 0 ⇒ 37.7m 1 − 2.17 × 10 8 = 0 ⇒ m 1 = 5.74 × 10 6 kg H 2 O / h (c) Cost of refrigerant pumping and recompression, cost of cooling water pumping, cost of maintaining system at the higher pressure of part (b). 8.68 Basis: 100 mol leaving conversion reactor H 2 O(v) 3.1 bars, sat'd
n 3 (mol O 2 ) 3.76 n 3 (mol N 2 )
H 2 O( l ) 45°C
conversion 100 mol, 600°C, 1 atm 145°C 100°C reactor 0.199 mol HCHO/mol n 4 (mol H 2 O( v)) 0.0834 mol CH 3OH/mol 0.303 mol N 2/mol n 1 (mol CH3 OH(l )) n 2 (mol CH3 OH(l )) 0.0083 mol O 2/mol 0.050 mol H 2/mol m w1 (kg H 2 O(l )) m w2 (kg H 2 O(l )) n 8 (mol CH3 OH(l )) 0.356 mol H 2O( v)/mol 3.1 bars, sat'd 30°C Q (kJ) CH 3 OH( l ), 1 atm, sat'd
2.5n 8 (mol CH3 OH) n (mol HCHO) distillation 6a absorption (l ) n 6b (mol CH3 OH( l )) n 6c (mol H 2 O( l )) sat'd, 1 atm 88°C, 1 atm Product solution n 7 (mol) 0.37 g HCHO/g (x 1 mol/min) Absorber off-gas m w3 (kg H 2 O(l )) n 5a (mol N 2 ) 30°C 0.01 g CH 3OH/g (x 2 mol/min) 20oC 0.82 g H 3 O/g (x 3 mol/min) n 5b (mol O 2 ) n 5 c (mol H 2 ) n 5d (mol H 2 O(v )), sat'd n 5 e (mol HCHO(v )), 200 ppm 27°C, 1 atm
a. Strategy C balance on conversion reactor ⇒ n2 , N 2 balance on conversion reactor ⇒ n3 H balance on conversion reactor ⇒ n4 , (O balance on conversion reactor to check consistency) N 2 balance on absorber ⇒ n5a , O 2 balance on absorber ⇒ n5b H 2 balance on absorber ⇒ n5e
UV ⇒ n 200 ppm HCHO in absorber off - gasW H 2 O saturation of absorber off - gas
5d
8-47
, n5b
8.68 (cont’d)
HCHO balance on absorber ⇒ n6a , CH 3 OH balance on absorber ⇒ n6b Wt. fractions of product solution ⇒ x1 , x 2 , x 3 HCHO balance on distillation column ⇒ n7 CH 3 OH balance on distillation column ⇒ n8 CH 3 OH balance on recycle mixing point ⇒ n1 Energy balance on waste heat boiler ⇒ mw1 , E.B. on cooler ⇒ mw2 Energy balance on reboiler ⇒ Q C balance on conversion reactor: n2 = 19.9 mol HCHO + 8.34 mol CH 3 OH = 28.24 mol CH 3 OH N 2 balance on conversion reactor: 3.76n3 = 30.3 ⇒ n3 = 8.06 mol O 2 , 3.76 × 8.06 = 30.3 mol N 2 feed H balance on conversion reactor:
bg
bg
bg
bg bg
bg
n4 2 + 28.24 4 − 19.9 2 + 8.34 4 + 5 2 + 35.6 2 ⇒ n4 = 20.7 mol H 2 O fed O balance: 65.1 mol O in, 65.5 mol O out. Accept (precision error) N 2 balance on absorber: 30.3 = n5a ⇒ n5a = 30.3 mol N 2 O 2 balance on absorber: 0.83 = n5b ⇒ n5b = 0.83 mol O 2 H 2 balance on absorber: 5.00 = n5c ⇒ n5c = 5.00 mol H 2 H 2 O saturation of off - gas:
yw =
b
p w* 27° C P
g = LM 26.739 mm Hg = n N 760 mm Hg 30.3 + 0.83 + 5.00 + n 5d
g U| | 200 ppm HCHO in off gas: V| n 200 ⇒ = 2| |W 3613 . +n +n 10
5d
+ n5e
b
OP Q
⇒ n5d = 0.03518 3613 . + n5d + n5e 1
solve
⇒
mol H 2 O n5d = 1318 . n5e = 7.49 × 10 −3 mol HCHO
5e
6
5d
5e
Moles of absorber off-gas = n5a + n5b + n5c + n5e = 37.46 mol off - gas HCHO balance on absorber: 19.9 = n6a + 7.49 × 10 −3 ⇒ n6a − 19.89 mol HCHO CH 3 OH balance on absorber: 8.34 = n6b ⇒ n6b = 8.34 mol CH 3 OH Product solution
U| |V || W
%MW
. x1 = 0.262 mol HCHO mol Basis - 100 g ⇒ 37.0 g HCHO ⇒ 1232 mol HCHO 1.0 g CH 3 OH ⇒ 0.031 mol CH 3 OH ⇒ x 2 = 0.006 mol CH 3 OH mol x 3 = 0.732 mol H 2 O mol 62.0 g H 2 O ⇒ 3.441 mol H 2 O
8-48
8.68 (cont’d)
HCHO balance on distillation column (include the condenser + reflux stream within the system for this and the next balance):
19.89 = 0.262n7 ⇒ n7 = 75.9 mol product CH 3 OH balance on distillation column:
b g
8.34 = 0.006 75.9 + n8 ⇒ n8 = 7.88 mol CH 3 OH CH 3 OH balance on recycle mixing point:
n1 + n8 = n2 ⇒ n1 = 28.24 − 7.83 = 20.36 mol CH 3 OH fresh feed Summary of requested material balance results:
bg
n1 = 20.4 mol CH 3 OH l fresh feed n2 = 75.9 mol product solution
bg
n3 = 7.88 mol CH 3 OH l recycle n4 = 37.5 mol absorber off - gas Waste heat boiler:
b
g
b
g
bg
Refs: HCHO v, 145° C , CH 3 OH v, 145° C ; N 2 , O 2 , H 2 , H 2 O v at 25°C for product
b
g
gas, H 2 O l, triple point for boiler water substance
nin
H in
nout
H out
HCHO CH 3 OH
19.9 8.34 30.3 0.83 5.0 35.6
22.55 32.02 17.39 18.41 16.81 20.91
19.9 8.34 30.3 0.83 5.0 35.6
0 0 3.51 3.60 3.47 4.09
mw1
566.2
mw1 2726.32 m (kg) H (kJ/kg)
N2 O2 H2 H 2O H 2O (boiler) E.B. ΔH =
∑ n H − ∑ n H i
out
i
i
i
UVH = C dT W U| |VH = C bT g T − 25 || W UVH from steam tables W
z T
n (mol) H (kJ/mol)
p
145
p
= 0 ⇒ −1814 + 2160mw1 = 0 ⇒ mw1 = 0.84 kg 3.1 bar steam
in
8-49
8.68 (cont’d)
b
g
Gas cooler: Same refs. as above for product gas, H 2 O l, 30° C for cooling water substance
nin
H in
nout
H out
HCHO CH 3 OH
19.9 8.34 30.3 0.83 5.0 35.6
0 0 3.51 3.60 3.47 4.09
19.9 8.34 30.3 0.83 5.0 35.6
–1.78 –2.38 2.19 2.24 2.16 2.54
mw 2
0
mw 2
62.76
N2 O2 H2 H 2O H 2O (coolant) E.B. ΔH =
∑ n H − ∑ n H i
i
i
out
i
n (mol) H (kJ/mol)
m (kg) H (kJ/kg)
H = 4.184
b
g
kJ T − 30 ° C kg⋅° C
= 0 ⇒ −1581 . + 62.6mw 2 = 0 ⇒ mw 2 = 2.52 kg cooling water
in
b gb g Q = − nΔH b1 atmg = −b27.58 molgb35.27 kJ molg
Condenser: CH 3 OH condensed = n8 + 2.5n8 = 3.5 7.88 = 27.58 mol CH 3 OH condensed E.B.:
b.
v
= −973 kJ (transferred from condenser)
3.6 × 10 4 tonne / y
10 6 g
1 yr
1d
1 metric ton 350 d 24 h . × 10 b0.37gd4.286 × 10 i = 1586 ⇒ b0.01gd4.286 × 10 i = 4.286 × 10 b0.62gd4.286 × 10 i = 2.657 × 10
= 4.286 × 10 6 g h product soln
6
6
g HCHO h ⇒ 5.281 × 10 4 mol HCHO h
6
6
g CH 3 OH h ⇒ 1338 mol CH 3 OH h
6
6
× 10 5 mol H 2 O h g H 2 O h ⇒ 1475 .
⇒ 2.016 × 10 5 mol h ⇒ Scale factor =
U| V| |W
2.016 × 105 mol h = 2657 h −1 75.9 mol
8.69 (a) For 24°C and 50% relative humidity, from Figure 8.4-1, Absolute humidity = 0.0093 kg water / kg DA, Humid volume ≈ 0.856 m 3 / kg DA
Specific enthalpy = (48 - 0.2) kJ / kg DA = 47.8 kJ / kg DA , Dew point = 13o C, Twb = 17 o C (b) 24 o C (Tdb ) (c) 13o C (Dew point) (d) Water evaporates, causing your skin temperature to drop. Tskin ≈ 13o C (Twb ). At 98%
R.H. the rate of evaporation would be lower, Tskin would be closer to Tambient , and you would not feel as cold.
8-50
Vroom = 141 ft 3 . DA = dry air.
8.70
mDA = ha =
140 ft 3
lb - mol⋅ o R 29 lb m DA 1 atm . lb m DA = 101 lb - mol 550 o R 0.7302 ft 3 ⋅ atm
0.205 lb m H 2 O = 0.0203 lb m H 2 O / lb m DA 101 . lb m DA
From the psychrometric chart, Tdb = 90 o F, ha = 0.0903 h r = 67%
Twb = 80.5o F
Tdew point = 77.3o F
8.71
Tdb = 35° C Tab = 27° C
8.72 a.
= 44.0 − 011 H . ≅ 43.9 Btu / lb m
⇒ hr = 55% He wins Fig. 8.4-1
Tdb = 40° C, Tdew point = 20° C
b. Mass of dry air: mda =
Mass of water: c.
= 14.3 ft 3 / lb DA V m
⇒
hr = 33%, ha = 0.0148 kg H 2 O kg dry air Twb = 255 . °C
1 m 3 1 kg dry air = 2.2 × 10 −3 kg dry air 10 3 L 0.92 m 3 ↑ from Fig. 8.4-1
2.00 L
2.2 × 10 −3 kg dry air 0.0148 kg H 2 O 10 3 g = 0.033 g H 2 O 1 kg dry air 1 kg
b g b g H b20° C, saturated g ≈ 57.5 kJ kg dry air (both values from Fig. 8.4-1) 2.2 × 10 kg dry air b57.5 − 77.4g kJ 10 J ΔH = = −44 J
H 40° C, 33% relative humidity ≈ 78.0 − 0.65 kJ kg dry air = 77.4 kJ kg dry air
−3
3
40→ 20
kg dry air
1 kJ
d. Energy balance: closed system
2.2 × 10 −3 kg dry air 10 3 g 1 mol 0.033 g H 2 O 1 mol + = 0.078 mol 18 g 1 kg 29 g Q = ΔU = nΔU = n ΔH − RΔT = ΔH − nRΔT n=
d
= −44 J −
i
0.078 mol 8.314 J
b20 − 40g° C
mol ⋅ K
1K 1° C
8-51
= −31 J(23 J transferred from the air)
400 kg 2.44 kg water
= 10.0 kg water evaporates / min 97.56 kg air 10 kg H 2 O min (b) ha = = 0.025 kg H 2 O kg dry air , Tdb = 50° C 400 kg dry air min
8.73 (a)
min
Fig. 8.4-1
b
g
H = 116 − 11 . = 115 kJ kg dry air , Twb = 33° C, hr = 32%, Tdew point = 28.5° C
(c) Tdb = 10° C , saturated ⇒ ha = 0.0077 kg H 2 O kg dry air , H = 29.5 kJ kg dry air
(d)
b0.0250 − 0.0077g kg H O = 6.92 kg H O min condense
400 kg dry air
2
min
2
kg dry air
bg
References: Dry air at 0° C, H 2 O l at 0° C substance
m in
H in
m out
H out
Air
400
115
400
29.5
—
—
6.92
42
bg
H 2O l
b
g
b
g
m air in kg dry air/min, m H 2O in kg/min H air in kJ/kg dry air, H H 2 O in kJ/kg
H 2 O l , 0° C → H 2 O l , 20° C :
H =
75.4
J
1 mol
mol⋅° C
18 g
b10 − 0g° C
1 kJ
103 g
= 42 kJ kg 103 J 1 kg −34027.8 kJ 1 min 1 kW Q = ΔH = ∑ m i Hˆ i − ∑ m i Hˆ i = = −565 kW min 60 s 1 kJ/s out in (e)
T>50°C, because the heat required to evaporate the water would be transferred from the air, causing its temperature to drop. To calculate (Tair)in, you would need to know the flow rate, heat capacity and temperature change of the solids.
8.74 a. Outside air: Tdb = 87° F , hr = 80% ⇒ ha = 0.0226 lb m H 2 O lb m D.A. , H = 455 . − 0.01 = 455 . Btu lb m D.A.
Room air: Tdb = 75° F , hr = 40% ⇒ ha = 0.0075 lb m H 2 O lb m D.A. , H = 26.2 − 0.02 = 26.2 Btu lb D.A. m
Delivered air: Tdb = 55° F , ha = 0.0075 lb m H 2 O lb m D.A. ⇒ H = 214 . − 0.02 = 214 . Btu lb m D.A. , V = 13.07 ft 3 lb m D.A. Dry air delivered:
1,000 ft 3 1 lb m D.A. = 76.5 lb m D.A. min min 13.07 ft 2
H 2 O condensed: 76.5 lb m D.A. min
b0.0226 − 0.0075g lb
m
H 2O
lb m D.A.
8-52
= 12 . lb m H 2 O min condensed
8.74 (cont’d)
The outside air is first cooled to a temperature at which the required amount of water is condensed, and the cold air is then reheated to 55°F. Since ha remains constant in the second step, the condition of the air following the cooling step must lie at the intersection of the ha = 0.0075 line and the saturation curve ⇒ T = 49° F
b
g
References: Same as Fig. 8.4-2 [including H 2 O l, 32° F ] H in
m out H out
substance
m in
Air
76.5 45.5 76.5 21.4 m air in lb m D.A./min
b
g
H 2 O l, 49° F
—
—
1.2 17.0 H in Btu/ lb D.A. air m m in lb /min, H H 2O
Q = ΔH =
b76.5g 214. − 455. + 1.2(17.0) (Btu) min
m
H 2O
in Btu/ lb m
60 min 1 ton cooling −12,000 Btu h 1h
= 9.1 tons cooling b. 6 7
(76.5 lb m DA/min)
hr = 40%, ha = 0.0075 lb m H 2 O/lb m DA 75o F, 26.2 Btu/lb m DA
1 7
(76.5 lb m DA/min)
hr = 80%, ha = 0.0226 lb m H 2 O/lb m DA
Coolerreheater
87o F, 45.5 Btu/lb m DA
76.5 lb m DA/min ha = 0.0075 lb m H 2 O/lb m DA
Lab
55o F, 21.4 Btu/lb m DA
Q lab m H2 O (kg H 2 O(l)/min)
Q (tons)
Water balance on cooler-reheater (system shown as dashed box in flow chart) 1 7
lb m H2O ⎞ 6 lb m DA ⎞ ⎛ ⎛ ⎟ + ( 76.5 ⎜ 76.5 ⎟ ⎜ 0.0226 min ⎠ ⎝ lb m DA ⎠ 7 ⎝
)( 0.0075) = (76.5)(0.0075) + m H O
⇒ m H2 O = 0.165 kg H 2 O condensed/min
8-53
2
8.74 (cont’d)
Energy balance on cooler-reheater References: Same as Fig. 8.4-2 [including H2O(l, 32oF)] m in Hˆ in 10.93 45.5 65.57 26.2
Substance Fresh air feed Recirculated air feed Delivered air o
Condensed water (49 F)
Hˆ out — — 21.4
—
—
76.5
—
—
0.165 17.0
Q = ΔH = ∑ m i Hˆ i − ∑ m i Hˆ i = out
m out — —
in
Percent saved by recirculating =
m DA in lb m dry air/min Hˆ air in Btu/lb m dry air m H2 O(l) in lb m /min Hˆ H2 O(l) in Btu/lb m
−575.3 Btu 60 min 1 ton cooling = 2.9 tons min 1h -12,000 Btu/h (9.1 tons − 2.9 tons) × 100% = 68% 9.1 tons
Once the system reaches steady state, most of the air passing through the conditioner is cooler than the outside air, and (more importantly) much less water must be condensed (only the water in the fresh feed). c. Total recirculation could eventually lead to an unhealthy depletion of oxygen and buildup of carbon dioxide in the laboratory. 8.75 Basis: 1 kg wet chips. DA = dry air, DC = dry chips Outlet air: Tdb=38oC, Twb=29oC m2a (kg DA) m2w [kg H2O(v)]
Inlet air: 11.6 m3(STP), Tdb=100oC m1a (kg DA)
1 kg wet chips, 19oC 0.40 kg H2O(l)/kg 0.60 kg DC/kg
m3c (kg dry chips) m3w [kg H2O(l)] T (oC)
(a) Dry air: m1a =
b g
11.6 m 3 STP DA
1 kmol 3
b g
22.4 m STP
29.0 kg 1 kmol
= 15.02 kg DA = m 2a
Outlet air: Fig. 8.4-1 → Hˆ 2 = (95.3 − 0.48) = 94.8 (Tdb = 38°C, Twb = 29°C) ⎯⎯⎯⎯
b
kJ kg DA
g
Water in outlet air: m2 w = ha2 m2 a = 0.0223 15.02 = 0.335 kg H 2 O (b) H 2 O balance: 0.400 kg = 0.335 kg + m3w ⇒ m3w = 0.065 kg H 2 O
8-54
ha2 = 0.0223
kg H2 O kg DA
8.75 (cont’d)
Moisture content of exiting chips: 0.065 kg water × 100% = 9.8% < 15% ∴ meets design specification 0.600 kg dry chips + 0.065 kg water
bg
(c) References: Dry air, H 2 O l , dry chips @ 0°C.
substance
H in
min
Air H 2O l dry chips
15.02 100.2 0.400 79.5 0.600 39.9
bg
H out
mout
15.02 94.8 mair in kg DA, H air in kJ/kg DA 0.065 4.184T m in kg DC, H in in kJ/kg DC 0.6 2.10T
Energy Balance: ΔH = ∑ mout Hˆ out − ∑ min Hˆ in =0 ⇒ −136.8 + 1.532T = 0 ⇒ T = 89.3°C
Tdb = 45° C hr = 10% b. Twb = 210 . °C hr = 60%
8.76 a.
H 2 O added:
8.77
Tas = Twb = 210 . °C Tdb = 26.8° C 15 kg air 1 kg D.A. min 1.0059 kg air
11.3 m3 1 kg D.A. 0.92 m3
Fig. 8.4-1
b0.0142 − 0.0059g kg H O = 012 . kg H O min 2
2
1 kg D.A.
V = 0.92 m 3 kg D.A. , Twb = 22° C ha = 0.0050 kg H 2 O kg D.A.
= 12.3 kg D.A. min
Outlet air: Twb = Tas = 22° C saturated
Evaporation:
ha = 0.0142 kg H 2 O kg DA
Fig. 8.4-1
Inlet air: Tdb = 50° C Tdew pt. = 4° C
min
ha = 0.0059 kg H 2 O kg DA
Fig. 8.4-1
12.3 kg D.A. min
⇒ T = 22° C ha = 0.0165 kg H 2 O kg D.A.
b0.0165 − 0.0050g kg H O = 014 . kg H O min 2
kg D.A.
8-55
2
8.78 a.
UV = 4° C W
bh g
Tdb = 45° C Tdew point
a in
= 0.0050 kg H 2 O kg D.A.
Twb = 20.4° C, V = 0.908 m 3 kg D.A.
Fig. 8.4-1
b g
Twb = Tas = 20.4° C, saturated ⇒ ha
out
= 0.0151 kg H 2 O kg D.A.
b. Basis: 1 kg entering sugar (S) solution m1 (kg D.A.) 0.0050 kg H2O/kg DA
m1 (kg D.A.) 0.0151 kg H2O(v)/kg
1 kg 0.05 kg S/kg 0.95 kg H2O/kg
m2 (kg) 0.20 kg S/kg 0.80 kg H2O/kg
b gb g b g Water balance: bm gb0.0050g + b1gb0.95g = bm gb0.0151g + b0.25gb0.80g Sugar balance: 0.05 1 = 0.20 m2 ⇒ m2 = 0.25 kg 1
1
m1 = 74 kg dry air ⇒
1 lb m D.A. ha1 (lb m H 2O) T d = 20°F h r = 70%
Inlet air (A):
Coil bank
C
D
Spray 1 lb m D.A. 1 lb m D.A. ha2(lb m H 2O) chamber ha3(lb m H 2O) T d = 75°F H2 O
Tdb = 20° F hr = 70%
Outlet air (D):
0.908 m 3 = 67 m 3 1 kg D.A.
74 kg dry air
B
A
8.79
V=
UV W
Tdb = 70° F hr = 35%
Fig. 8.4-2
a. Inlet of spray chamber (B):
1 lb m D.A. ha3(lb m H 2O) T d = 70°F h r = 35%
ha1 ≈ 0.0017 lb m H 2 O lb m D.A. V ≈ 12.2 ft 3 lb m D.A.
Fig. 8.4-2
UV W
Coil bank
ha 3 = 0.0054 lb m H 2 O lb m D.A.
UV W
ha = 0.0017 lb m H 2 O lb m D.A. ⇒ Twb = 49.5° F Tdb = 75° F
The state of the air at (C) must lie on the same adiabatic saturation curve as does the state at (B), or Twb = 49.5° F . Thus, Outlet of spray chamber (C):
UV W
ha = 0.0054 lb m H 2 O lb m D.A. ⇒ hr = 52% Twb = 49.5° F
At point C, Tdb = 58.5° F
b.
bh
a3
g
− ha1 lb m H 2 O evaporate lb m DA
b
g
lb m DA 0.0054 − 0.0017 lb H O = = 3.0 × 10 −4 m3 2 3 V A ft inlet air 12.2 ft air
d
i
8-56
8.79 (cont’d)
( 20 - 6.4) Btu / lb m dry air c. QBA = ΔH = H B − H A ≅ = 1.1 Btu / ft 3 12.2 ft 3 / lb m dry air QDC = ΔH = H D − H C ≅
(23 - 20) Btu / lb m dry air 3
12.2 ft / lb m dry air
= 0.25 Btu / ft 3
d. 70% 52% 35%
C
D
A
B
58.5
20
70
75
8.80 Basis: 1 kg D.A. a.
1 kg D.A. ha1(kg H 2 O/kg D.A.) Tdb = 40°C, Tab = 18°C
1 kg D.A. ha2(kg H 2 O/kg D.A.) 20°C,
m w kg H2 O
Tdb = 40° C
⇒ ha1 = 0.0039 kg H 2 O kg D.A. Twb = 18° C Tdb = 20° C ⇒ ha 2 = 0.0122 kg H 2 O kg D.A. Outlet air: Twb = 18° C adiabatic humidification
Inlet air:
b
b gb g b gb g
g
b
g
Overall H 2 O balance: mw + 1 ha1 = 1 ha 2 ⇒ mn = 0.0122 − 0.0039 kg H 2 O kg D.A. = 0.0083 kg H 2 O kg D.A. b.
ma (lb m H2 O/h) T=15o C, sat’d
1250 kg/h T=37o C, h r=50%
mc (lb m H2 O/h) liquid, 12°C Qc (Btu/h)
8-57
8.80 (cont’d)
Inlet air:
Tdb = 37° C hr = 50%
Moles dry air: m a =
RSh = 0.0198 kg H O kg DA TH = b88.5 - 0.5g kJ kg DA = 88.0 kJ kg DA
Fig. 8.4-1
⇒
2
a1 1
1250 kg 1 kg DA = 1226 kg DA h h 1.0198 kg
Outlet air: Tdb = 15° C, sat' d
RSh = 0.0106 kg H O kg DA TH = 42.1 kJ kg DA 1226 kg DA b0.0198 − 0.0106g kg H O = Fig. 8.4-1
2
a
⇒
2
Overall water balance ⇒ m c
2
h
kg DA
= 113 . kg H 2 O h withdrawn
bg
Reference states for enthalpy calculations: H 2 O l , dry air at 0oC. (Cp)H2O(l) = 4.184
b
z
g
12
H 2 O l , 12° C : H = C p dT = 50.3 kJ / kg 0
Overall system energy balance: Q c = ΔH =
∑ m H − ∑ m H i
i
out
=
i
i
in
LM113. kg H O 50.3 kJ + 1226 kg DA b42.1 − 88g kJ OPFG 1 h IJ FG 1 kW IJ h kg DA QH 3600 s K H 1 kJ / s K N h kg H O 2
2
= −155 . kW ΔH =
8.81 8.82 a.
b.
400 mol NH 3
−78.2 kJ mol NH 3
b
g
b
= −31,280 kJ
g
b
g
HCl g , 25° C , H 2 O l , 25° C → HCl 25° C, r = 5 . B.11 ΔH = ΔH s 25° C, r = 5 ⎯Table ⎯⎯ ⎯→ ΔH = −64.05 kJ mol HCl
b
g HClbaq, r = ∞g → HClbr = 5g, H Obl g ΔH = ΔH b25° C, n = 5g − ΔH b25° C, n = ∞ g = b −64.05 + 7514 . g kJ mol HCl = 11.09 kJ / mol HCl 2
s
s
8-58
kJ kg ⋅ o C
8.83 Basis: 100 mol solution ⇒ 20 mol NaOH, 80 mol H2O
80 mol H 2 O = 4.00 mol H 2 O mol NaOH 20 mol NaOH
⇒ r=
bg
Refs: NaOH(s), H 2 O l @25° C nout
H out
20.0 0.0
−
−
80.0 0.0
−
−
−
20.0 −34.43 ← n in mol NaOH
substance
nin
NaOH s
bg H Obl g NaOHbr = 4.00g 2
b
H in
−
n in mol H in kJ mol
g
H NaOH, r = 4.00 = −34.43 kJ mol NaOH (Table B.11) ΔH = ∑ ni H i − ∑ ni H i = (20)( −34.43) = out
Q=
−688.6 kJ 9.486 × 10 −4 Btu 10 −3 kJ
in
b
−653.2 Btu
g
b
103 g
g
20.0 40.00 + 80.0 18.01 g 2.20462 lb m
= −653.2 Btu
= −132.3 Btu lb m product solution
8.84 Basis: 1 liter solution
n H 2SO 4 = mtotal = nH 2O =
FG H
IJ K
1 L 8 g - eq 1 mol 0.09808 kg = 0.392 kg H 2 SO 4 = 4 mol H 2 SO 4 × L 2 g - eq 1 mol
1 L 1.230 kg = 1.230 kg solution L . − 0.392gkg H O b1230 2
1000 mol H 2 O
= 46.5 mol H 2 O
18.02 kg H 2 O
⇒r =
n H 2O n H 2SO 4
=
mol H 2 O 46.49 mol H 2 O = 11.6 mol H 2 SO 4 4 mol H 2 SO 4
d
i
d
i
d
H 2 SO 4 aq, r = ∞,25o C → H 2 SO 4 aq, r = 11.6, 25o C + H 2 O l , 25o C ΔH 1 = ΔH s ( r = 11.6) − ΔH s (r = ∞)
LMn N H b H SO , r = 116 . , 60° Cg = 2
=
4
RS T
Table B.11
=
H 2SO 4 ΔH1
+m
( −67.6 + 96.19) = 28.6
z
60
25
OP Q
kJ mol H 2 SO 4
C p dT kJ
n H 2SO ( mol H 2 SO 4 )
4 mol H 2 SO 4 1 4 mol H 2 SO 4
4
28.6 kJ 1.230 kg + mol H 2 SO 4
= 60.9 kJ mol H 2 SO 4
8-59
i
3.00 kJ kg⋅° C
b60 − 25g° CUV W
d
i
8.85 2 mol H 2SO 4 = 0.30 2.00 + nH 2 O ⇒ nH 2 O = 4.67 mol H 2 O ⇒ r = a.
For this closed constant pressure system,
b
g
2 mol H 2SO 4 Q = ΔH = nH 2SO 4 ΔH s 25° C, r = 2.33 =
b. msolution =
2 mol H 2SO 4
98.08 g H 2SO 4 mol
b
g b280.6 + 150gg −88.6 kJ +
+
ΔH = 0 ⇒ nH 2SO 4 ΔH s 25° C, r = 2.33 + m
8.86 a.
mol H 2 O 4.67 = 2.33 mol H 2SO 4 2
Basis:
z
−44.28 kJ mol H 2SO 4
= −88.6 kJ
4.67 mol H 2 O 18.0 g H 2 O mol T
25
= 280.2 g
C p dT = 0
3.3 J g⋅° C
bT − 25g° C
1 kJ = 0 ⇒ T = 87° C 1000 J
e j
1 L product solution 1.12 10 3 g = 1120 g solution L
1 L 8 mol HCl 36.47 g HCl = 292 g HCl L mol HCl 46.0 mol H2 O(l, 25°C)
8.0 mo l HCl(g , 20°C, 790 mm Hg)
1 L HCl (aq)
1120 g − 292 g = 828 g H 2 O 828 g H 2 O
n=
mol = 46.0 mol H 2 O 18.0 g
46.0 mol H 2 O = 5.75 mol H 2 O mol HCl 8.0 mol HCl
Assume all HCl is absorbed Volume of gas:
b g
b g
8 mol 293 K 760 mm Hg 22.4 L STP = 185 liter STP gas feed L HCl solution 273 K 790 mm Hg mol
b. Ref: 25° C nin
substance
bg bg
H 2O l
HCl g
b
g
HCl n = 5.75
H in
nout − 46.0 0.0 − 8.0 −015 . − − 8.0
H out − − −59.07
n in mol H in kJ mol
8-60
8.86 (cont’d)
b
g
b
g
1 H HCl, n = 5.75 = ΔH s 25° C, n = 5.75 + nHCl = −64.87 kJ mol +
e
j
H HCl, 20o C =
z
20
25
z
40
mC p dT
25
1120 g 0.66 cal g⋅° C
8 mols
b40 − 25g° C
4.184 J kJ cal 103 J
. 0.02913 − 01341 × 10 −5 T + 0.9715 × 10 −8 T 2 − 4.335 × 10 −12 T 3 dT
= -0.15 kJ / mol Q = ΔH = −471 kJ L product
c.
e j b
g
Q = 0 = ΔH = 8 H − 8 −015 .
b
g
o 1120 g 0.66 cal T − 25 C 4.184 J 1 kJ . = H = −64.87 + −015 8 mol g⋅o C cal 1000 J
T = 192 o C
8.87 Basis: Given solution feed rate
.
.
n a (mol air/min) 200°C, 1.1 bars
n a (mol air/min) . n 1 (mol H 2O( v)/min) saturated @ 50°C, 1 atm
150 mol/min solution 0.001 NaOH 0.999 H 2O 25°C
.
n 2 (mol/min) @ 50°C 0.05 NaOH 0.95 H 2O
b
gb g H O balance: b0.999gb150g = n + 0.95b3.0g ⇒ n = 147 mol H O min n Raoult’s law: y P = P = p b50° Cg = 92.51 mm Hg ⇒ n + n NaOH balance: 0.001 150 = 0.05n2 ⇒ n2 = 3.0 mol min 2
1
1
Table B.4
∗ H 2O
1
H 2O
1
2
n1 =147 P = 760
a
b g
1061 mol 22.4 L STP Vinlet air = min 1 mol
bg
n a = 1061
mol air min
473 K 1013 . bars = 37,900 L min 273 K 1.1 bars
bg ΔH b25° Cg = −42.47 kJ mol NaOH
References for enthalpy calculations: H 2 O l , NaOH s , air @ 25° C
999 mol H 2 O Table B.11 ⇒ 1 mol NaOH
0.1% solution @ 25°C: r =
5% solution @ 50°C: r = Solution mass: m =
b
g
b
b
g
95 mol H 2 O 19 mol H 2 O kJ = ⇒ ΔH s 25° C = −42.81 mol NaOH 5 mol NaOH mol NaOH
1 mol NaOH 40.0 g 19 mol H 2 O 18.0 g g solution + = 382 1 mol 1 mol mol NaOH
g
H 50° C = ΔH s 25° C + m = −42.81
s
z
50
25
C p dT
382 g 4.184 J kJ + mol NaOH mol NaOH 1 g⋅° C
8-61
b50 − 25g° C
1 kJ = −2.85 kJ 103 J
8.87 (cont’d)
Air @ 200°C: Table B.8 ⇒ H = 515 . kJ mol Air (dry) @ 50°C: Table B.8 ⇒ H = 0.73 kJ mol
b
g
2592 − 104.8 kJ 1 kg 18.0 g H 2 O v , 50° C : Table B.5 ⇒ H = = 44.81 kJ mol kg 103 g 1 mol
b
g
substance
b g H Obv g
nin
H in
nout
H out
NaOH aq
0.15
−42.47
0.15
−
−
147
−2.85 n in mol min 44.81 H in kJ mol
Dry air
1061
. 515
1061
0.73
2
Energy balance: Q = ΔH = ∑ ni H i − ∑ ni H i = 1900 kJ min transferred to unit
b neglect ΔE g
out
n
in
8.88 a. Basis: 1 L 4.00 molar H2SO4 solution (S.G. = 1.231)
4.00 mol H 2 SO 4 1231 − 392.3 = 838.7 g H 2 O 1 L 1231 g = 1231 g ⇒ ⇒ = 392.3 g H 2 SO 4 L = 46.57 mol H 2 O B.11 ⇒ r = 1164 ⎯⎯ ⎯→ ΔH s = −67.6 kJ / mol H 2 SO 4 . mol H 2 O / mol H 2 SO 4 ⎯Table
b
g
b
Ref: H 2 O l , 25° C , H 2 SO 4 25° C substance H 2O l H 2 SO 4 l H 2 SO 4 25° C, n = 1164 .
bg bg
b
b
g
nin H in 46.57 0.0754 T − 25 4.00 0 − −
b
g
g
b
g
gb
nout H out n in mol − − H in kJ mol − − 4.00 −67.6
g
Q = ΔH = 0 = 4.00 −67.6 − 46.57 0.0754 T − 25 ⇒ T = −52° C (The water would not be liquid at this temperature ⇒ impossible alternative!)
b
g
b
b. Ref: H 2 O l , 25° C , H 2 SO 4 25° C
substance H 2O l H 2O s H 2 SO 4 (l ) . H 2 SO 4 25° C, n = 1164
bg bg
b
b
g
g
nin H in 0.0754 0 − 25 nl n s −6.01 + 0.0754 0 − 25 4.00 0
b
g
b
g
g
nout H out n in mols − − H in kJ mol − − − − 4.00 −67.61
ΔH m H 2 O, 0° C = 6.01 kJ mol
A
Table B.1
UV ⇒ n ΔH = 0 = 4.00b −67.61g − n b −1885 . g − b46.57 − n gb −7.895gW n ⇒ 2914 . g H ObAg + 547.3 g H Ob sg@0° C nA + n s = 46.57
l
2
l
2
8-62
l s
= 1618 . mol liquid H 2 O = 30.39 mol ice
8.89 P2 O5 + 3H 2 O → 2H 3 PO 4
B P 2n
mol H 3 PO 4
a.
wt% P2 O 5 =
b
g × 100% ,
n 14196 . mt
wt% H 3 PO 4 =
B
g H 3 PO 4 mol
b98.00g × 100%
mc
A
g total
where n = mol P2 O5 and mt = total mass . wt% H 3 PO 4 =
b
g wt% P O
2 98.00 14196 .
2
5
= 1381 . wt% P2 O 5
b. Basis: 1 lb m feed solution 28 wt% P2 O5 ⇒ 38.67 wt% H 3 PO 4 m1 (lbm H2 O(v )), T , 3.7 psia 1 lb msolution, 125°F 0.3867 lb mH 3PO 4 0.6133 lb mH 2O
m2 (lbm solution), T 0.5800 lb mH 3PO 4/lb 0.4200 lb mH 2O/lb m
m
H 3 PO 4 balance: 0.3867 = 0.5800m2 ⇒ m2 0.667 lb m solution
bg
Total balance: 1 = m1 + m2 ⇒ m1 = 0.3333 lb m H 2 O r
bg
Evaporation ratio: 0.3333 lb m H 2 O v lb m feed solution c. Condensate:
b
P = 37 . psia 0.255 bar
g
Table B.6
⇒ Tsat = 654 . o C =149 o F, Vliq =
m =
0.00102 m3 353145 . ft 3 / m3 ft 3 = 0.0163 lb m H 2 O(l) kg 2.205 lb m / kg
100 tons feed 2000 lb m 1 lb m H 2 O 1 day = 46.3 lb m / min 1 ton day 3 lb m (24 × 60) min
46.3 lb m V = min
0.0163 ft 3
7.4805 gal
lb m
ft 3
= 5.65 gal condensate / min
Heat of condensation process:
46.3lbm H2O(v)/min
46.3lbm H2O(l)/min
(149+37)°F, 3.7 psia
149°F, 3.7 psia
.
Q (Btu/min)
8-63
8.89 (cont’d)
R| ||H Table B.6 ⇒ S ||H |T
F GG H
H2 O ( l ) (149
o
I JJ = 1141 Btu / lb kJ kg K
Btu
o o H2 O ( v ) (186 F = 85.6 C) = (2652 kJ / kg) 0.4303
b
lb m
m
g
F = 65.4 o C) = (274 kJ / kg) 0.4303 = 118 Btu / lb m
LM N
lb Btu Q = m ΔH = (46.3 m ) (118 − 1141) min lb m
OP = −47,360 Btu / min Q
⇒ 4.74 × 10 4 Btu min available at 149 o F
bg
bg
d. Refs: H 3 PO 4 l , H 2 O l @77° F
min H in mout H out 100 . 13.95 m in lb m − − 0.667 34.13 H in Btu lb m − − 0.3333 1099 − −
substance H 3 PO 4 28% H 3 PO 4 42% H 2O v
b g b g bg
b
g
H H 3 PO 4 , 28% = +
g
H H 3 PO 4 , 42% = +
0.705 Btu lb m ⋅° F
1 lb - mole H 3 PO 3 98.00 lb m H 3 PO 4
b125 − 77g° F = 13.95 Btu lb
0.705 Btu lb m ⋅° F
b
−5040 Btu lb - mole H 3 PO 4
−5040 Btu lb - mole H 3 PO 4
m
g b
soln
1 lb - mole H 3 PO 4 98.00 lb m H 3 PO 4
b186.7 − 77g° F = 34.13 Btu lb
b g b
0.3867 lb m H 3 PO 4 1.00 lb m soln
m
0.5800 lb m H 3 PO 4 1.00 lb m sol.
soln
g b
g
H H 2 O = H 3.7 psia, 186° F − H l , 77° F = 2652 − 104.7 kJ kg ⇒ 1096 Btu lb m At 27.6 psia (=1.90 bar), Table B.6 ⇒ ΔH v = 2206 kJ / kg = 949 Btu / lb m ΔH = ∑ ni H i − ∑ ni H i = 375 Btu = msteam ΔH v ⇒ msteam = out
⇒ ⇒
in
375 Btu = 0.395 lb m steam 949 Btu / lb m
0.395 lb m steam 100 × 2000 lb m H 3 PO 4
1 day
lb m 28% H 3 PO 4
24 h
day
= 3292 lb m steam / h
3292 lb m steam lb m steam . = 118 (46.3 × 60) lb m H 2 O evaporated / h lb m H 2 O evaporated
8-64
8.90 Basis: 200 kg/h feed solution. A = NaC 2 H 3 O 2
.
n 1 (kmol H2 O(v )/h) 50°C, 16.9% of H 2O in feed 200 kg/h @ 60°C . n 0 (kmol/h) 0.20 A 0.80 H 2O
Product slurry @ 50°C . n 2 (kmol A-3H 2 O(v )/h) . n 3 (kmol solution/h) 0.154 A 0.896 H 2 O
Q (kJ/hr)
a. Average molecular weight of feed solution: M = 0.200 M A + 0.800 M H 2 O
b
gb g b
gb g
= 0.200 82.0 + 0.800 18.0 = 30.8 kg k Molar flow rate of feed: n0 =
200 kg 1 kmol = 6.49 kmol h h 30.8 kg
b
gb gb g b0.20gb6.49 kmol hg = n bkmol A ⋅ 3 H Og
bg
b. 16.9% evaporation ⇒ n1 = 0169 . 0.80 6.49 kmol h = 0.877 kmol H 2 O v h
A balance:
2
E
h
⇒ n2 + 0154 . n3 = 130 .
b gb
1 mole A
2
g
H 2 O balance: 0.80 6.49 kmol h = 0.877 +
1 mole A ⋅ 3 H 2 O
+ 0154 . n3 (1)
b
n2 kmol A ⋅ 3 H 2 O h
+ 0.846n3 ⇒ 3n2 + 0.846n3 = 4.315
bg b g
g
3 moles H 2 O 1 mole A ⋅ 3 H 2 O
bg
b2g
Solve 1 and 2 simultaneously ⇒ n2 = 113 . kmol A ⋅ 3H 2 O s h n3 = 1095 . kmol solution h Mass flow rate of crystals
bg
1.13 kmol A ⋅ 3H 2 O 136 kg A ⋅ 3H 2 O 154 kg NaC 2 H 3 O 2 ⋅ 3H 2 O s = h 1 kmol h Mass flow rate of product solution
b
gb g
bg
200 kg feed 154 kg crystals 0.877 18.0 kg H 2 O v − − = 30 kg solution h h h h c.
bg
bg
References for enthalpy calculations: NaC 2 H 3 O 2 s , H 2 O l @25° C
b
g
Feed solution: nH = n A ΔH s 25° C + m
b g
z
60
25
C p dT (form solution at 25° C , heat to 60° C )
. × 104 kJ 200 kg 3.5 kJ 0.20 6.49 kmol A −171 nH = + hr kg⋅° C h kmol A
8-65
b60 − 25g° C = 2300 kJ h
8.90 (cont’d)
b
g
Product solution: nH = n A ΔH s 25° C + m
b0.154g1.095 kmol A =
z
50
25
z
50
25
C p dT (hydrate at 25° C , heat to 50° C )
bg
1.13 kmol A ⋅ 3H 2 O s h = −36700 kJ h
=
b
z
LM N
g
H 2 O v , 50° C : nΔH = n ΔH v + =
OP Q
50
25
b neglect ΔE g
C p dT (vaporize at 25° C , heat to 50° C )
b gb
i
g
4.39 × 10 4 + 32.4 50 − 25 kJ
∑ n H − ∑ n H i
out
R
b50 − 25g° C
−3.66 × 10 4 kJ 154 kg 1.2 kJ + h kg⋅° C kmol
0.877 kmol H 2 O h
Energy balance: Q = ΔH =
b50 − 25g° C
−171 . × 10 4 kJ 30 kg 3.5 kJ + h kg⋅° C kmol A
h = −259 kJ h
Crystals: nH = n A ΔH hydration + m
C p dT
i
i
b
= 39200 kJ h
g b
g
= −259 − 36700 + 39200 − 2300
kJ h
in
= −60 kJ h (Transfer heat from unit)
U| mL |V ⇒ r = 500 . mol H O mol H SO 84.2 mL H Oblg 100 . g | = 84.2 g H Oblg ⇒ 4.678 mol H Oblg |W mL
8.91 50 mL H 2SO4
1834 g .
mol H 2SO4 = 917 . g H 2SO4 ⇒ 0935 .
2
2
2
2
2
Ref: H 2 O , H 2SO 4 @ 25 °C H ( H 2 O(l ), 15o C) = [0.0754 kJ / (mol ⋅ o C)](15 − 25) o C = − 0.754 kJ / mol
b
g
(91.7 + 84.2) g kJ + H H 2 SO 4 , r = 5.00 = −58.03 mol 0.935 mol H 2 SO 4
2.43 J
bT − 25g° C
1 kJ
g⋅° C
10 3 J
= ( −69.46 + 0.457T )( kJ / mol H 2 SO 4 ) substance
nin
bg
H in
H out
nout
H 2O l
4.678 –0.754 — — n in mol 0.935 0.0 — — H 2SO 4 H in kJ/mol — 0.935 −69.46 + 0.457T n mol H 3SO 4 H 2SO 4 r = 4.00 —
b
g b
b
g
b
g
b
g
g
Energy Balance: ΔH = 0 = 0.935 −69.46 + 0.457T − 4.678 −0.754 ⇒ T = 144 ° C Conditions: Adiabatic, negligible heat absorbed by the solution container.
8-66
4
8.92 a. mA (g A) @ TA0 (oC) nA (mol A)
nS (mol solution) @ Tmax (oC)
mB (g B) @ TB0 (oC) nB (mol B) Refs: A(l), B(l) @ 25 °C substance nin H in nout A nA H —
H out
A
—
n in mol H in J / mol
B
nB H B
—
—
S
—
nA
H S (J mol A)
—
mA (g A) m , nB = B M A (g A / mol A) MB
Moles of feed materials: n A (mol A) = Enthalpies of feeds and product
H A = m A C pA ( T A 0 − 25 o C), H B = m B C pB ( TB 0 − 25 o C) r (mol B mol A) = n B n A =
H S
FG J IJ = H mol A K n
A
mB / M B mA / M A
LMn 1 M ( mol A) M MM+ N
A ( mol
FG J IJ H mol A K F J I × (T )( g soln) × C G H g soln ⋅ C JK
A) × Δ H m ( r )
(m A + m B
ps
max
o
− 25)( o
OP PP C) P PQ
1 ⇒ H S = n A Δ H m ( r ) + ( m A + m B ) C ps ( Tmax − 25) nA
Energy balance ΔH = n A H S − n A H A − n B H B = 0
bg
b
g
b
g
mA ΔHm r + (m A + mB )C ps (Tmax − 25) − m A C pA TA0 − 25 − mB C pB TB 0 − 25 = 0 MA m m A C pA TA0 − 25 + mB C pB TB 0 − 25 − A ΔH m r MA ⇒ Tmax = 25 + (m A + mB )C ps ⇒
b
g
b
g
bg
Conditions for validity: Adiabatic mixing; negligible heat absorbed by the solution container, negligible dependence of heat capacities on temperature between 25oC and TA0 for A, 25oC and TB0 for B, and 25oC and Tmax for the solution.
b.
m A = 100.0 g mB = 225.0 g
b
g UV |W
M A = 40.00 TA 0 = 25° C C pA = ? irrelevant mol H 2 O ⇒ r = 5.00 M B = 18.01 TB 0 = 40° C C pB = 4.18 J (g⋅° C) mol NaOH
C ps = 3.35 J (g⋅° C)
b
g
ΔH m n = 5.00 = −37,740 J mol A ⇒ Tmax = 125° C
8-67
8.93 Refs: Sulfuric acid and water @ 25 °C b.
substance
nin
H in
H2SO4 H2O H 2 SO 4 aq
1 r
M A C pA T0 − 25 M w C pw T0 − 25 —
b g
b b
—
g g
H out
nout
— — 1
— n in mol H in J/mol — ΔH m r + M A + rM w C ps Ts − 25
bg b
g b
g
(J/mol H2SO4)
bg b g b g b g = ΔH br g + b98 + 18r gC bT − 25g − (98C + 18rC )bT 1 ( 98C + 18rC )bT − 25g − ΔH br g ⇒ T = 25 + (98 + 18r )C
b
g
ΔH = 0 = ΔH m r + M A + rM w C ps Ts − 25 − M A C pa T0 − 25 − rM w C pw T0 − 25 m
ps
s
s
pa
pa
pw
pw
0
0
g
− 25
m
ps
c.
H2O(l) H2SO4 r 0.5 1 1.5 2 3 4 5 10 25 50 100
Cp (J/mol-K) 75.4 185.6
Cp (J/g-K) 4.2 1.9
Cps 1.58 1.85 1.89 1.94 2.1 2.27 2.43 3.03 3.56 3.84 4
ΔH m (r ) -15,730 -28,070 -36,900 -41,920 -48,990 -54,060 -58,030 -67,030 -72,300 -73,340 -73,970
Ts 137.9 174.0 200.2 205.7 197.8 184.0 170.5 121.3 78.0 59.6 50.0
250
Ts
200 150 100 50 0 0.1
1
10
100
r
d. Some heat would be lost to the surroundings, leading to a lower final temperature.
8-68
8.94 a.
Ideal gas equation of state n A 0 = P0V g / RT0 Total moles of B: n B 0 ( mol B) =
(1)
b
gd
Vl ( L) × SG B × 1 kg / L 10 3 g / kg
i
(2)
M B (g / mol B)
Total moles of A: n Ao = n Av + n Al Henry’s Law: r
FG mol A(l) IJ = k p H mol B K s
(3) A
⇒
b
g
n Al n Av RT = c0 + c1T nB0 Vg
(4)
Solve (3) and (4) for nAl and nAv.
b
nB 0 RT c0 + c1T Vg
n Al =
g
LM1 + n RT bc + c T gOP MN V PQ n = LM1 + n RT bc + c T gOP PQ MN V
(5)
B0
0
1
g
n Av
Ao
(6)
B0
0
1
g
Ideal gas equation of state P=
n Av RT ( 6 ) n A 0 RT = Vg Vg + nB 0 RT c0 + c1T
b
g
(7)
b g bg
Refs: A g , B l @ 298 K substance
bg Bb l g
Ag
Solution
nin
U in
n Ao nB0
neq
U eq
M A CvA T0 − 298
n Av
M A CvA T − 298
M B CvB
—
—
n Al
U 1 (kJ/mol A)
—
b bT
0
g − 298g
—
b
g b
b
g
n in mol U in kJ/mol
g
1 U 1 = ΔU s + n Al M A + n B 0 M B Cvs T − 298 n Al
E.B.: ΔU = 0 =
∑ n U − ∑ n U i
out
c
i
i
i
in
b
g hb g b d−ΔU i + bn C + n C gbT − 298g n C + bn M + n M gC
gb
g
0 = n Av CvA + n Al M A + nB M B Cvs T − 298 + n Al ΔU s − n Ao CvA + n B CvB T0 − 298 ⇒ T = 298 +
n Al
s
Av
Ao
vA
Al
vA
B
A
8-69
vB
B
0
B
vs
8.94 (cont’d) b. Vt 20.0
MA 47.0
CvA 0.831
MB 26.0
CvB 3.85
SGB 1.76
Vl 3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0
T0 300 300 300 300 330 330 330 330
P0 1.0 5.0 10.0 20.0 1.0 5.0 10.0 20.0
Vg 17.0 17.0 17.0 17.0 17.0 17.0 17.0 17.0
nB0 203.1 203.1 203.1 203.1 203.1 203.1 203.1 203.1
nA0 0.691 3.453 6.906 13.811 0.628 3.139 6.278 12.555
c0 c1 0.00154 -1.60E-06 T 301.4 307.0 313.9 327.6 331.3 336.4 342.8 355.3
nA(v) 0.526 2.624 5.234 10.414 0.473 2.359 4.709 9.381
Dus -174000
Cvs 3.80
nA(l) 0.164 0.828 1.671 3.397 0.155 0.779 1.569 3.174
P 0.8 3.9 7.9 16.5 0.8 3.8 7.8 16.1
c.
C*
REAL R, NB, T0, P0, VG, C, D, DUS, MA, MB, CVA, CVB, CVS REAL NA0, T, DEN, P, NAL, NAV, NUM, TN INTEGER K R = 0.08206 1 READ (5, *) NB IF (NB.LT.0) STOP READ (1, *) T0, P0, VG, C, D, DUS, MA, MB, CVA, CVB, CVS WRITE (6, 900) NA0 = P0 * VG/R/T0 T = 1.1 * T0 K=1 10 DEN = VG/R/T/NB + C + D * T P = NA0/NB/DEN NAL = (C + D * T) * NA0/DEN NAV = VG/R/T/NB * NA0/DEN NUM = NAL * (–DUS) + (NA0 * CVA + NB * CVB) * (TO – 298) DEN = NAV * CVA + (NAL * MA + NB * MB) * CVS TN = 298 + NUM/DEN WRITE (6, 901) T, P, NAV, NAL, TN IF (ABS(T – TN).LT.0.01) GOTO 20 K=K+1 T = TN IF (K.LT.15) GOTO 10 WRITE (6, 902) STOP 20 WRITE (6, 903) GOTO 1 900 FORMAT ('T(assumed) P Nav Nal T(calc.)'/ * ' (K) (atm) (mols) (mols) (K)') 901 FORMAT (F9.2, 2X, F6.3, 2X, F7.3, 2X, F7.3, 2X, F7.3, 2) 902 FORMAT (' *** DID NOT CONVERGE ***') 903 FORMAT ('CONVERGENCE'/) END $ DATA 300 291 10.0 15.0 1.54E–3 –2.6E–6 –74 35.0 18.0 0.0291 0.0754 4.2E–03
8-70
Tcalc 301.4 307.0 313.9 327.6 331.3 336.4 342.8 355.3
8.94 (cont’d)
300 291 35.0 –1
50.0 18.0
Program Output T (assumed) (K) 321.10 296.54 296.57
15.0 0.0291
1.54E–3 0.0754
–2.6E–6 4.2E–03
–74
P (atm) 8.019 7.415 7.416
Nav (mols) 4.579 4.571 4.571
Nal (mols) 1.703 1.711 1.711
T(calc.) (K) 296.542 296.568 296.568
P (atm) 40.093 39.676 39.680
Nav (mols) 22.895 22.885 22.885
Nal (mols) 8.573 8.523 8.523
T(calc.) (K) 316.912 316.942 316.942
Convergence T (assumed) (K) 320.10 316.91 316.94
8.95 Q=0
350 mL 85% H2SO4 ma(g), 60 oF, ρ=1.78
30% H2SO4 ms(g), T(oF)
H2O, Vw(mL), mw(g), 60 oF
a.
Vw =
350 mL feed
. 178 g 1 mL feed
. g H 2 O added 1 mL water 0.85(70 / 30) − 015 g feed 1 g water
= 1140 mL H 2 O b.
Fig. 8.5-1 ⇒ Hˆ a ≈ −103 Btu/lb m ;
Water: Hˆ water ≈ 27 Btu/lb m
Mass Balance: mp=mf+mw=(350 mL)(1.78 g/mL)+(1142 mL)(1 g/mL)=623+1142=1765 g Energy Balance:
c.
m Hˆ + mw Hˆ w ΔH = 0 = m p Hˆ product − ma Hˆ a − mw Hˆ w ⇒ Hˆ s = f f mp − + (623)( 103) (1140)(27) ⇒ Hˆ product = = −18.9 Btu/lb m 1765
T ( Hˆ = −18.9 Btu/lb m ,30%) ≈ 130o F
d. When acid is added slowly to water, the rate of temperature change is slow: few isotherms are crossed on Fig. 8.5-1 when xacid increases by, say, 0.10. On the other hand, a change from xacid=1 to xacid=0.9 can lead to a temperature increase of 200°F or more.
8-71
8.96 a. 2.30 lb m 15.0 wt% H 2 SO 4 = −10 Btu / lb @ 77 o F ⇒ H 1
m
U| || V| ⎯⎯⎯⎯⎯⎯→ m ( lb || W adiabatic mixing
3
m2 (lb m ) 80.0 wt% H 2 SO 4 = −120 Btu / lb @ 60o F ⇒ H 2
60.0 wt% H 2 SO 4 @ T o F, H 3
|UV ⇒ |RSm mass balance: 2.30b0.150g + m b0.800g = m (0.600) W| |m T
Total mass balance: H 2 SO 4
m
m)
2.30 + m2 = m3
2
3
2
= 517 . lb m (80%)
3
= 7.47 lb m (60%)
b. Adiabatic mixing ⇒ Q = ΔH = 0
. gb −120g = 0 ⇒ H b7.47gH − b2.30gb−10g − b517 3
3
= −861 . Btu / lb m
E Figure 8.5 - 1
T = 140 o F c.
d
i
H 60 wt%, 77 o F = −130 Btu / lb m
d
i
b
gb
g
Q = m3 H 60 wt%, 77 o F − H 3 = 7.475 −130 + 861 . = −328 Btu d. Add the concentrated solution to the dilute solution . The rate of temperature rise is
much lower (isotherms are crossed at a lower rate) when moving from left to right on Figure 8.5-1. 8.97 a.
b.
x NH 3 = 0.30
Fig. 8.5-2
y NH 3 = 0.96 lb m NH 3 lb m vapor , T = 80° F
Basis: 1 lb m system mass
⇒ 0.90 lb m liquid
⇒ 010 . lb m vapor Mass fractions: zNH 3 =
b0.27 + 0.096glb
m
NH 3
1 lb m
x NH 3 = 0.30
0.27 lb m NH 3 0.63 lb m H 2 O
x NH 3 = 0.96
0.096 lb m NH 3 0.004 lb m H 2 O
= 0.37 lb m NH 3 lb m
1 − 0.37 = 0.63 lb m H 2 O lb m 0.90 lb m liquid 0.10 lb m vapor 670 Btu −25 Btu Enthalpy: H = + = 44 Btu lb m 1 lb m 1 lb m 1 lb m liquid 1 lb m vapor
8-72
8.98
T = 140° F
Fig. 8.5-2
Vapor: 80% NH 3 , 20% H 2 O
C
Liquid: 14% NH 3 , 86% H 2 O
A B
Basis: 250 g system mass ⇒ mv (g vapor), mL (g liquid)
.14
.60
.80 x NH3
Mass Balance: mv + mL = 250 . mL = ( 0.60)(250) ⇒ mv = 175 g, mL = 75g NH3 Balance: 0.80m g + 014
b gb g = b014 . gb75 gg = 10.5 g NH , 64.5 g H O Liquid
Vapor: mNH 3 = 0.80 175 g = 140 g NH 3 , 35 g H 2 O Liquid: mNH 3
3
2
8.99 Basis: 200 lb m feed h
m v (lb m h) xv(lbm NH3(g)/lbm) H v ( Btu lb m )
200 lbm/h 0.70 lbm NH3(aq)/lbm 0.30 lbm H2O(l)/lbm
m l (lb m h)
H f = −50 Btu lb m
xl[lbm NH3(aq)/lbm]
in equilibrium at 80oF
H l ( Btu lb m )
Q ( Btu h)
Figure 8.5-2 ⇒ Mass fraction of NH 3 in vapor: xv = 0.96 lb m NH 3 lb m Mass fraction of NH 3 in liquid: xl = 0.30 lb m NH 3 lb m Specific enthalpies: H v = 650 Btu lb m , H l = −30 Btu lb m
UV ⇒ m = 120 lb h vapor b0.70gb200g = 0.96m + 0.30m W m = 80 lb h liquid 200 = m v + m l
Mass balance: Ammonia balance:
m
v
v
l
l
m
Energy balance: Neglect ΔE k . Q = ΔH =
∑ m H i
out
i
120 lb m − m f H f = h = 86,000
650 Btu 80 lb m + lb m h
Btu h
8-73
−30 Btu
lb m
−
200 lb m h
−50 Btu
lb m
CHAPTER NINE 4 NH 3 (g) + 5O 2 (g) → 4NO(g) + 6H 2 O(g) ΔH o = −904.7 kJ / mol
9.1
r
a.
When 4 g-moles of NH3(g) and 5 g-moles of O2(g) at 25°C and 1 atm react to form 4 g-moles of NO(g) and 6 g-moles of water vapor at 25°C and 1 atm, the change in enthalpy is -904.7 kJ.
b.
Exothermic at 25°C. The reactor must be cooled to keep the temperature constant. The temperature would increase under adiabatic conditions. The energy required to break the molecular bonds of the reactants is less than the energy released when the product bonds are formed.
c.
5 O 2 (g) → 2NO(g) + 3H 2 O(g) 2 Reducing the stoichiometric coefficients of a reaction by half reduces the heat of reaction by half. 904.7 ΔH ro = − = −452.4 kJ / mol 2 5 3 NO(g) + H 2 O(g) → NH 3 (g) + O 2 (g) 4 2 Reversing the reaction reverses the sign of the heat of reaction. Also reducing the stoichiometric coefficients to one-fourth reduces the heat of reaction to one-fourth. ( −904.7) ΔH ro = − = +226.2 kJ / mol 4
d.
e.
2 NH 3 (g) +
NH3 = 340 g/s m n NH3 =
340 g
1 mol
s
17.03 g
= ΔH= Q
ˆo n NH3 ΔH r
ν NH
=
= 20.0 mol/s
20.0 mol NH 3 s
3
−904.7 kJ 4 mol NH 3
= −4.52 × 103 kJ/s
The reactor pressure is low enough to have a negligible effect on enthalpy. f.
Yes. Pure water can only exist as vapor at 1 atm above 100°C, but in a mixture of gases, it can exist as vapor at lower temperatures.
9.2 C 9 H 20 (l) + 14O 2 (g) → 9CO 2 (g) +10H 2 O(l) ΔH o = −6124 kJ / mol r
a.
When 1 g-mole of C9H20(l) and 14 g-moles of O2(g) at 25°C and 1 atm react to form 9 g-moles of CO2(g) and 10 g-moles of water vapor at 25°C and 1 atm, the change in enthalpy is -6124 kJ.
b.
Exothermic at 25°C. The reactor must be cooled to keep the temperature constant. The temperature would increase under adiabatic conditions. The energy required to break the molecular bonds of the reactants is less than the energy released when the product bonds are formed.
c.
= ΔH = Q
0 n C 9 H 20 ΔH r
ν C9 H 20
=
25.0 mol C 9 H 20 s
−6124 kJ
1 kW
1 mol C 9 H 20 1 kJ / s
9-1
= −153 . × 105 kW
9.2 (cont'd) Heat Output = 1.53×105 kW. The reactor pressure is low enough to have a negligible effect on enthalpy. d.
C 9 H 20 (g) + 14O 2 (g) → 9CO 2 (g) +10H 2 O(l) ΔH o = −6171 kJ / mol
(1)
C 9 H 20 (l) + 14O 2 (g) → 9CO 2 (g) +10H 2 O(l) ΔH o = −6124 kJ / mol
(2)
r
r
(2) − (1) ⇒ C 9 H 20 (l) → C 9 H 20 (g) ΔH o (C H ,25D C) = − 6124 kJ / mol − ( −6171 kJ / mol) = 47 kJ / mol v
e.
9.3
a.
9
20
Yes. Pure n-nonane can only exist as vapor at 1 atm above 150.6°C, but in a mixture of gases, it can exist as a vapor at lower temperatures.
Exothermic. The reactor will have to be cooled to keep the temperature constant. The temperature would increase under adiabatic conditions. The energy required to break the reactant bonds is less than the energy released when the product bonds are formed.
b.
b g 192 O bgg → 6CO bgg + 7H Obgg b1g ΔH = ? 19 C H blg + O bgg → 6CO bgg + 7 H Oblg b2g ΔH = ΔH = −1791 × 10 Btu lb - mole . 2 C H bgg → C H blg b3g ΔH = −e ΔH j = −13,550 Btu lb - mole H Oblg → H Obgg b4g ΔH = e ΔH j = 18,934 Btu lb - mole . × 10 Btu lb - mole b1g = b2g + b3g + 7 × b4g ⇒ ΔH = ΔH + ΔH + 7ΔH = −1672 C 6 H 14 g + 6
14
6
14
2
2
2
6
2
2
14
o r
2
2
2
3
2
v
4
v
H 2O
6
1
M O2 =32.0
m = 120 lb m / s
Q = ΔH =
⇒
nO2 ΔHˆ ro vO2
bg
=
6
C 2 H 14
Hess's law
c.
o r
2
3
4
n = 3.75 lb - mole / s.
3.75 lb-mole/s −1.672 × 106 Btu = −6.60 × 105 Btu/s ( from reactor ) 9.5 1 lb-mole O 2
bg
bg
bg
bg
CaC 2 s + 5H 2 O l → CaO s + 2CO 2 g + 5H 2 g , ΔH ro = 69.36 kJ kmol
9.4 a.
Endothermic. The reactor will have to be heated to keep the temperature constant. The temperature would decrease under adiabatic conditions. The energy required to break the reactant bonds is more than the energy released when the product bonds are formed. b. ΔU ro = ΔH ro
LM OP kJ 8.314 J − RT M ∑ ν − ∑ ν P = 69.36 − mol mol ⋅ K MM PP N Q i
gaseous products
i
gaseous reactants
= 52.0 kJ mol
9-2
1 kJ 298 K 103 J
b7 − 0g
9.4 (cont’d) ΔU ro is the change in internal energy when 1 g - mole of CaC2 (s) and 5 g - moles of H2 O(l) at 25D C and 1 atm react to form 1 g - mole of CaO(s), 2 g - moles of CO2 (g) and 5 g - moles of H2 (g) at 25D C and 1 atm.
c.
Q = ΔU =
nCaC 2 ΔU ro vCaC 2
=
150 g CaC 2
1 mol 52.0 kJ = 121.7 kJ 64.10 g 1 mol CaC 2
Heat must be transferred to the reactor. 9.5 a.
Given reaction = (1) – (2)
Hess's law
⇒
b
g
ΔH ro = ΔH ro1 − ΔH ro2 = 1226 − 18,935 Btu lb - mole = −17,709 Btu lb - mole
b.
Given reaction = (1) – (2)
Hess's law
⇒
b
g
ΔH ro = ΔH ro1 − ΔH ro2 = −121,740 + 104,040 Btu lb - mole = −17,700 Btu lb - mole
⎛ ⎝
Hess's law
9.6
a. b.
9.7
Reaction (3) = 0.5 × (1) − ( 2 ) ⇒ ΔHˆ ro = 0.5 ⎜ −326.2
Reactions (1) and (2) are easy to carry out experimentally, but it would be very hard to decompose methanol with only reaction (3) occurring.
bg
bg
bg
a.
N 2 g + O 2 g → 2NO g ,
b.
n − C5 H 12 g +
bg
F B I = 2G 90.37 kJ molJ = 180.74 kJ mol JK GH Table B.1
ΔH ro
e j
=2
ΔH fo
NO(g)
bg bg bg ΔH = 5e ΔH j e j bg e j b g . g − b−146.4g kJ mol = −21212 . kJ mol = b5gb −110.52g + b6gb −28584 19 C H blg + O bgg → 6CO bgg + 7H Obgg 2 ˆ ΔH = 6 ( ΔHˆ ) + 7 ( ΔHˆ ) − ( ΔHˆ ) ( ) () o r
c.
kJ ⎞ ⎛ kJ ⎞ kJ ⎟ − ⎜ −285.8 ⎟ = 122.7 mol ⎠ ⎝ mol ⎠ mol
6
o f
11 O 2 g → 5CO g + 6H 2 O l 2 + 6 ΔH fo − ΔH fo
CO(g)
14
2
o r
o f
n − C5 H 12 g
H 2O l
2
2
o f
CO 2
o f
H2 O g
C6 H14 l
= ⎡⎣( 6 )( −393.5 ) + 7 ( −241.83) − ( −198.8 ) ⎤⎦ kJ mol = −3855 kJ mol d.
Na 2SO 4 (l) + 4CO(g) → Na 2S(l) + 4CO 2 (g) + 4 ΔH o − ΔH o ΔH o = ΔH o r
e j
− 4e ΔH j e j e j . g − b −1384.5 + 24.3g − 4( −110.52 = ( −373.2 + 6.7) + b4gb −3935 f
Na 2S( l )
f
CO 2 ( g )
f
9-3
Na 2SO 4 ( l )
o f
CO(g )
kJ mol = −138.2 kJ mol
9.8
a.
⇒ e ΔH j = −385.76 + 52.28 = −333.48 kJ mol e j ΔH = −276.2 − 92.31 + 333.48 = −35.03 kJ mol b g + e ΔH j b g − e ΔH j Given reaction = b1g + b2g ⇒ −385.76 − 35.03 = −420.79 kJ mol o r2
b.
c.
e j = e ΔH j
ΔH ro1 = ΔH fo
C 2 H 2 Cl 4 ( l )
o f
− ΔH fo o f
C 2 HCl 3 l
300 mol C 2 HCl 3 Q = ΔH = h Heat is evolved.
o f
C2 H 4 ( g)
o f
HCl g
−420.79 kJ mol
C 2 H 2 Cl 4 ( l )
C 2 H 2 Cl 4 (l )
b
= −126 . × 105 kJ h = −35 kW
g
9.9 a.
5 O 2 (g) → 2CO 2 (g) + H 2 O(l) 2
C 2 H 2 ( g) +
ΔH co = −1299.6 kJ mol
The enthalpy change when 1 g-mole of C2H2(g) and 2.5 g-moles of O2(g) at 25°C and 1 atm react to form 2 g-moles of CO2(g) and 1 g-mole of H2O(l) at 25°C and 1 atm is -1299.6 kJ. b.
e j
ΔH co = 2 ΔH fo
B
Table B.1
b
=
c.
CO 2 ( g )
g b
b g − e ΔH f j C H b g g o
H 2O l
2
g b
2
kJ kJ = −1299.6 g mol mol
2 −3935 . + −28584 . − 226.75
d
(i) ΔH ro = ΔH fo
B
Table B.1
=
d
i
B
Table B.1
=
C 2 H 2 ( g) +
C2 H 6 ( g)
d
− ΔH fo
i
bg
C2 H 2 g
kJ kJ . = −3114 b−84.67g − b226.75g mol mol
(ii) ΔH ro = ΔH co
d.
e j
+ ΔH fo
i
C2 H 2 ( g )
d
+ 2 ΔH co
i
H2 (g)
d
− ΔH co
i
bg
C2 H 6 g
kJ kJ . = −3114 b−1299.6g + 2(−285.84) − b−1559.9g mol mol
5 O 2 (g) → 2CO 2 (g) + H 2 O(l) 2
(1)
1 O 2 (g) → H 2 O(l) 2 7 C 2 H 6 (g) + O 2 (g) → 2CO 2 (g) + 3H 2 O(l) 2
ΔH co1 = −1299.6 kJ mol
(2) ΔH co2 = −28584 . kJ mol
H 2 (g) +
(3) ΔH co3 = −1559.9 kJ mol
The acetylene dehydrogenation reaction is (1) + 2 × (2) − (3) Hess's law
⇒
ΔH ro = ΔH co1 + 2 × ΔH co2 − ΔH co3
b
g
. ) − ( −1559.9) kJ mol = −3114 . kJ / mol = −1299.6 + 2( −28584
9-4
9.10
a.
bg
C8 H 18 l +
bg
bg
25 O 2 (g) → 8CO 2 g + 9H 2 O g 2
ΔH ro = −4850 kJ / mol
When 1 g-mole of C8H18(l) and 12.5 g-moles of O2(g) at 25°C and 1 atm react to form 8 g-moles of CO2(g) and 9 g-moles of H2O(g), the change in enthalpy equals -4850 kJ. b.
Energy balance on reaction system (not including heated water):
b
g
b
ΔE k , ΔE p , W = 0 ⇒ Q = ΔU = n mol C 8 H 18 consumed ΔU co kJ mol
g
(Cp ) H 2 O(l) from Table B.2 = 75.4 × 10 −3 kJ / mol.D C − Q = mH 2 O (Cp ) H 2 O(l) ΔT =
Q = ΔU ⇒ −89.4 kJ = ⇒
ΔU co
1.00 kg
75.4 × 10 −3 kJ 21.34° C
1 mol 18.0 × 10 −3 kg
mol.D C
2.01 g C 8 H 18 consumed 1 mol C8 H 18 114.2 g
= 89.4 kJ
ΔU co (kJ) 1 mol C8 H 18
= −5079 kJ mol
ΔH co = ΔU co
LM OP + RT M ∑ ν − ∑ ν P MM PP N Q i
gaseous products
=− 5079 kJ mol +
i
gaseous reactants
8.314 J 1 kJ 298 K mol ⋅ K 103 J
b8 + 9 − 12.5g
⇒ ΔHˆ co = −5068 kJ mol % difference = c.
(−5068) − ( −4850) × 100 = − 4.3 % −5068
e j b g + 9eΔH j b g − eΔH j b g ⇒ ( ΔHˆ ) = ⎡8 ( −393.5 ) + 9 ( −241.83) + 5068⎤⎦ kJ/mol = −256.5 kJ/mol () ⎣
ΔH co = 8 ΔH fo
CO 2 g
o f
o ff
H 2O g
o f
C8 H 18 l
C8 H18 l
There is no practical way to react carbon and hydrogen such that 2,3,3-trimethylpentane is the only product.
9-5
9.11
a.
bg
bg
n − C 4 H10 g → i − C 4 H10 g
Basis: 1 mol feed gas
0.930 mol n-C4H10
(nn- C4H10)out
0.050 mol i-C4H10
( ni-C4H10)out
0.020 mol HCl
0.020 mol HCl
149°C
Q(kJ/mol)
149°C
(n n-CH 4 H10 ) out = 0.930(1 − 0.400) = 0.560 mol (n i-CH 4 H10 ) out = 0.050 + 0.930 × 0.400 = 0.420 mol
ξ =
(n n-C4H10 ) out − (n n-C4H10 ) in
0.560 − 0.930 = 0.370 mol 1
=
ν n-C H10 4
e j
b.
ΔH ro = ΔH fo
c.
References: n − C 4 H10
i − C 4 H 10
substance
e j ⇒ ΔH = −134.5 − b−124.7g bgg, i − C H bgg at 25° C
− ΔH fo
Table B.1
4
H in
n in (mol)
b kJ molg
n − C 4 H 10
1
H 1
i − C 4 H 10
−
−
H 1 =
LM MN
z
B
Table B.2
149
25
Cp
o r
n − C 4 H 10
kJ mol = −9.8 kJ mol
10
n out (mol)
H out
b kJ molg
0.600
H 1
0.400
H 2
O kJ dT P PQ mol = 14.29 kJ mol
H 2 =
LM MN
z
B
Table B.2
149
25
C p dT
OP kJ PQ mol = 14.14 kJ mol
Q = ΔH = ξ [ΔHˆ ro + ∑ ni Hˆ i − ∑ ni Hˆ i ] = 0.370 ⎡− ⎣ 9.8 + (1)(14.142 ) − (1)(14.287 ) ⎤⎦ kJ out
in
= −3.68 kJ For 325 mol/h fed, Q = d.
−9.8 kJ 325 mol feed 1h 1 kW = −0.90 kW 1 mol feed h 3600 s 1 kJ/s
−3.68 kJ ΔHˆ r (149°C ) = = −9.95 kJ/mol 0.370 mol
9-6
9.12
a. 1 m3 at 298K, 3.00 torr
Products at 1375K, 3.00 torr
n0 (mol) 0.111 mol SiH4/mol 0.8889 mol O2/mol
n1 (mol O2) n2 (mol SiO2) n3 (mol H2)
SiH 4 ( g) + O 2 (g) → SiO 2 (s) + 2H 2 (g)
Ideal Gas Equation of state : no =
1 m3
273 K 3.00 torr 1 mol = 0.1614 mol 298 K 760 torr 22.4 × 10-3 m3
ni = nio + ν iξ SiH 4 : 0=0.1111(0.1614 mol) − ξ ⇒ ξ = 0.0179 mol O 2 : n1 = 0.8889(0.1614 mol) − ξ = 0.1256 mol O 2 SiO 2 : n2 = ξ = 0.0179 mol SiO 2 H 2 : n3 =2ξ =0.0358 mol H 2 b.
ΔH ro = ( ΔH fo ) SiO 2 (s) − ( ΔH fo ) SiH 4 ( g) = [ −851 − ( −61.9)] kJ mol = −789.1 kJ / mol
References : SiH 4 (g), O 2 (g),SiO 2 (g), H 2 (g) at 298 K
Substance
Hˆ in
nin
nout
Hˆ out
(mol h) ( kJ mol) (mol h) ( kJ mol) SiH 4
0.0179
0
−
O2
0.1435
0
0.1256
SiO 2
−
−
H2
−
−
0.0179 0.0358
− Hˆ
1
Hˆ 2 Hˆ
3
B
Table B.8
O 2 (g,1375K): H 1 = H O 2 (1102 o C) = 3614 . kJ / mol SiO 2 (s,1375K): H 2 =
z
1375
(C p ) SiO 2 ( s) dT = 79.18 kJ / mol
298
B
Table B.8
H 2 (g,1375K): H 3 = H H 2 (1102 o C) = 32.35 kJ / mol c.
Q = ΔH = ξ ΔHˆ ro + ∑ ni Hˆ i −∑ ni Hˆ i = −7.01 kJ/m3 feed out
Q=
−7.01 kJ 27.5 m m3 h
3
in
1h 1 kW = −0.0536 kW (transferred from reactor) 3600 s 1 kJ/s
9-7
9.13
a.
bg bg
bg
bg
Fe 2 O 3 s + 3C s → 2 Fe s + 3CO g , ΔH r (77 D F) = 2.111 × 105 Btu lb - mole
Basis:
2000 lb m Fe 1 lb - mole
= 3581 . lb - moles Fe produced 55.85 lb m 53.72 lb - moles CO produced
17.9 lb - moles Fe 2 O 3 fed 53.72 lb - moles C fed
17.9 lb-moles Fe2O3 (s) 77° F
35.81 lb-moles Fe (l) 2800° F
53.72 lb-moles C 77° F
53.72 lb-moles CO(g) 570° F Q (Btu/ton Fe)
b.
bg bg bg
Substance
b
b g Febl,2800° Fg CObg,570° Fg
H in
nin
g
Fe 2 O 3 s,77° F
Fe(l,2800D F): H 1 =
z
53.72
0
−
−
3581 .
− H
−
−
53.72
H 2
2794
77
dC i
bg
p Fe s
CO(g,570D F): H 2 = H CO (570D F)
Q = ΔH =
nFe ΔH ro
ν Fe
H out
nout
(lb - moles) (Btu lb - mole) (lb - moles) (Btu lb - mole) 17.91 0 − −
C s,77° F
c.
bg
References: Fe 2 O 3 s , C s , Fe s , CO g at 77° F
+
b
i
g
dT + ΔH m 2794° F +
z
1
dC i
2800
2794
b gdT = 28400
p Fe l
Btu lb - mole
= 3486 Btu lb - mole A I FH interpolating from Table B.9 K
∑ n H − ∑ n H out
−
i
i
i
in
. ge2.111 × 10 j b3581 = + b3581 . gb28400g + b53.72gb3486g − 0 = 4.98 × 10 5
2
d.
6
Btu / ton Fe produced
Effect of any pressure changes on enthalpy are neglected. Specific heat of Fe(s) is assumed to vary linearly with temperature from 77°F to 570°F. Specific heat of Fe(l) is assumed to remain constant with temperature. Reaction is complete. No vaporization occurs.
9-8
9.14
a.
bg
C 7 H 16 g → C 6 H 5CH 3 (g) + 4 H 2 (g)
Basis: 1 mol C7H16 1 mol C7H16
1 mol C6H5CH3
400°C
4 mol H2 400° C
Q (kJ/mol)
bg b g
References: C s , H 2 g at 25° C
b.
H in
substance nin
H out
nout
bmolg bkJ molg bmolg bkJ molg
C 7 H 16 C7 H 8 H2
H 1 − −
1 − −
− 1 4
− H2 H
3
⎡ 400 ↓ ⎤ C7 H16 ( g,400°C ) : Hˆ 1 = (ΔHˆ fD )C7 H16 (g) + ⎢ ∫ C p dT ⎥ ⎢⎣ 25 ⎥⎦ = ( − 187.8 +91.0) kJ/mol= −96.8 kJ/mol 0.2427
b
g
C 6 H 5 CH 3 g,400° C : H 2 = ( ΔH fD ) C6 H 5CH 3 ( g) +
LM MN
z
B
Table B.2
400
25
C p dT
OP PQ
= (+50 + 60.2) kJ / mol = 110.2 kJ / mol
b
B
Table B.8
g
H 2 g,400° C : H 3 = H H 2 (400D C) = 10.89 kJ mol
c.
Q = ΔH = ∑ ni Hˆ i − ∑ ni Hˆ i out
in
= ⎡⎣(1)(110.2 ) + ( 4 )(10.89 ) − (1)( −96.8 ) ⎤⎦ kJ = 251 kJ (transferred to reactor) d.
ΔHˆ r (400D C)=
251 kJ = 251 kJ/mol 1 mol C7 H16 react
9-9
9.15
a.
bCH g Obgg → CH bgg + H bgg + CObgg 3 2
4
2
Moles charged: (Assume ideal gas)
b g
2.00 liters 273 K 350 mm Hg 1 mol = 0.01286 mol CH 3 2 O 873 K 760 mm Hg 22.4 liters STP
b g
b g
Let x = fraction CH 3 2 O decomposed (Clearly x CS1). Then CS2 decreases with increasing t as well as CS1. Finally dCS2/dt approaches zero as t→∞. Therefore, CS2 increases until it reaches a maximum value, then it decreases. The plot of CS3 vs. t begins at (t=0, CS3=0). When t=0, the slope (=dCS3/dt) is 0.04(0 − 0) = 0 . As t increases, CS2 increases (CS3 < CS2)⇒ dCS3/dt =0.04(CS2-CS3) becomes positive ⇒ CS2 increases with increasing t until dCS3/dt changes to negative (CS3 > CS1). Finally dCS3/dt approaches zero as t→∞. Therefore, CS3 increases until it reaches a maximum value then it decreases. c. 3
CS1, CS2, CS3 (g/L)
2.5 2 CS1 1.5 CS2
1
CS3
0.5 0 0
20
40
60
80
100
120
140
160
t (s)
11.29 a. (i) Rate of generation of B in the 1st reaction: rB1 = 2r1 = 0.2C A
(ii) Rate of consumption of B in the 2nd reaction: − rB 2 = r2 = 0.2CB2 b. Mole Balance on A: Accumulation=-Consumption
E
d ( C AV ) dC A . C AV ⇒ . CA = −01 = −01 dt dt . mol / L t = 0, C A0 = 100
Mole Balance on B: Accumulation= Generation-Consumption
E
d ( CBV ) dCB = 0.2C AV − 0.2CB2V ⇒ = 0.2C A − 0.2CB2 dt dt t = 0, CB 0 = 0 mol / L
11-23
11.29 (cont’d) c. 2
CA, CB, CC
CC
1 CB CA 0 t
The plot of CA vs. t begins at (t=0, CA=1). When t=0, the slope (=dCA/dt) is −01 . × 1 = −01 . . As t increases, CA decreases ⇒ dCA/dt=-0.1CA becomes less negative, approaches zero as t→∞. CA→0 as t→∞. The curve is therefore concave up. The plot of CB vs. t begins at (t=0, CB=0). When t=0, the slope (=dCB/dt) is 0.2(1 − 0) = 0.2 . As t increases, CB increases, CA decreases ( C B2 < CA)⇒ dCB/dt =0.2(CA- C B2 ) becomes less positive until dCB/dt changes to negative ( CB2 > CA). Then CB decreases with increasing t as well as CA. Finally dCB/dt approaches zero as t→∞. Therefore, CB increases first until it reaches a maximum value, then it decreases. CB→0 as t→∞. The plot of CC vs. t begins at (t=0, CC=0). When t=0, the slope (=dCC/dt) is 0.2(0) = 0 . As t increases, CB increases ⇒ dCc/dt =0.2 C B2 becomes positive also increases with increasing t ⇒ CC increases faster until CB decreases with increasing t ⇒ dCc/dt =0.2 CB2 becomes less positive, approaches zero as t→∞ so CC increases more slowly. Finally CC→2 as t→∞. The curve is therefore S-shaped.
CA, CB, CC (mol/L)
d. 2.2 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0
CC
CB CA
0
10
20
30 t (s)
11-24
40
50
11.30 a. When x = 1, y = 1 . y=
a ax x =1, y =1 ⇒ 1= ⇒ a = 1+ b 1+ b x+b pC5 H12 = yP = xp *C5 H12 ( 46o C ) ⇒ y =
b. Raoult’s Law:
p *C5 H12 (46o C) = 10
Antoine Equation:
(6.84471−
1060.793 ) 46 + 231.541
o
xp *C5 H12 ( 46 C )
⇒y=
P
=
xp *C5 H12 ( 46o C ) P
= 1053 mm Hg
0.7 × 1053 = 0.970 760
0.70a R| y = ax . 0.970 = ""(1) |Ra = 1078 ⇒S 0.70 + b S| x + b TFrom part (a), a = 1+ b"""""""""""(2) |Tb = 0.078 x=0.70, y=0.970
c. Mole Balance on Residual Liquid: Accumulation=-Output
E
dN L = − nV dt t = 0, N L = 100 mol Balance on Pentane: Accumulation=-Output
E
ax dx dN L d(NLx) = − nV y ⇒ x + NL = − nV x+b dt dt dt dN L / dt = − nV
E
FG H
n ax dx =− V −x NL x + b dt t = 0, x = 0.70
IJ K
d. Energy Balance: Consumption=Input
E
nV ΔH vap = Q
ΔH vap =27.0 kJ/ mol
t = 0, N L = 100 mol
From part (c),
dN L = − nV dt nV Q 27.0 = NL Qt 100 27.0
nV =
b
Q 27.0 kJ / mol
N L = 100 − nV t = 100 −
Substitute this expression into the equation for dx/dt from part (c):
11-25
g
Qt 27.0
11.30 (cont’d)
IJ K
FG H
FG H
dx n ax Q 27.0 ax =− V −x =− −x dt NL x + b x+b Qt 100 27.0 x(0) = 0.70
IJ K
e. 1 0.9 0.8 y (Q=1.5 kJ/s)
0.7 x, y
0.6 x (Q=1.5 kJ/s)
0.5 0.4
x (Q=3 kJ/s)
0.3
y (Q=3 kJ/s)
0.2 0.1 0 0
200
400
600
800
1000 1200 1400 1600 1800
t(s)
f. The mole fractions of pentane in the vapor product and residual liquid continuously decrease over a run. The initial and final mole fraction of pentane in the vapor are 0.970 and 0, respectively. The higher the heating rate, the faster x and y decrease.
11-26
