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
⎢
⎥=
Ap g ⎣ p A pW ⎦
p A pW gAp
⇒ the pond level falls
(b)
Change in boat level
b 9 g−b 4 g
h p 2 − h p1 =
L
OP F I MM F F
PQ GH JK MM GH GH
N
LM
MN
⇒ the boat rises
3.10 (a) ρ bulk =
O
II PP
JK JK P
PQ
> 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
. <G
GH n JK 2..94 = 106
H n JK 4 .
3
3
2
2
2
O2
NH 3
3
3
2
O2
NH3
fed
stoich
⇒ O 2 is the limiting reactant
Required NH3:
3125
. kmol O 2 4 kmol NH 3
= 2.50 kmol NH 3
5 kmol O 2
% excess NH 3 =
2.94 − 2.50
× 100% = 17.6% excess NH 3
2.50
d i −v
Extent of reaction: nO 2 = nO2
Mass of NO:
4.41
a.
0
O2
b g
ξ ⇒ 0 = 3125
. − −5 ξ ⇒ ξ = 0.625 kmol = 625 mol
3125
.
kmol O 2 4 kmol NO 30.0 kg NO
= 75.0 kg NO
5 kmol O 2 1 kmol NO
By adding the feeds in stoichometric proportion, all of the H2S and SO2 would be consumed.
Automation provides for faster and more accurate response to fluctuations in the feed stream,
reducing the risk of release of H2S and SO2. It also may reduce labor costs.
4- 30
4.41 (cont’d)
b.
n c =
3.00 × 10 2 kmol 0.85 kmol H 2 S 1 kmol SO 2
= 127.5 kmol SO 2 / h
h
kmol
2 kmol H 2 S
c.
C a lib r a t io n C u r v e
1 .2 0
X (mol H 2 S/mol)
1 .0 0
0 .8 0
0 .6 0
0 .4 0
0 .2 0
0 .0 0
0 .0
2 0 .0
4 0 .0
6 0 .0
8 0 .0
1 0 0 .0
R a (m V )
X = 0.0199 Ra − 0.0605
b
d.
n c kmol SO 2 / h
b
n f kmol / h
b
g
x kmol H 2S / kmol
g
g
Blender
Flowmeter calibration:
n f = aR f
n f = 100 kmol / h , R f
Control valve calibration:
UVn
= 15 mV W
f
=
UV
W
20
Rf
3
n c = 25.0 kmol / h, R c = 10.0 mV
7
5
n c = Rc +
n c = 60.0 kmol / h, Rc = 25.0 mV
3
3
FG
H
IJ b
K
1
7
5 1 20
n f x ⇒ Rc + =
R f 0.0119 Ra − 0.0605
2
3
3 2 3
5
10
⇒ Rc =
R f 0.0119 Ra − 0.0605 −
7
7
Stoichiometric feed: n c =
b
n f = 3.00 × 10 2 kmol / h ⇒ R f =
g
3
n f = 45 mV
20
4- 31
g
4.41 (cont’d)
b gb
b g
e.
gb g
10
5
45 0.0119 76.5 − 0.0605 − = 53.9 mV
7
7
7
5
⇒ n c = 53.9 + = 127.4 kmol / h
3
3
Faulty sensors, computer problems, analyzer calibration not linear, extrapolation beyond
range of calibration data, system had not reached steady state yet.
Rc =
4.42
165 mol / s
b
x mol C 2 H 4 / mol
b
b
n mol / s
g
1 − x mol HBr / mol
g
0.310 mol C 2 H 4 / mol
mol HBr / mol
0173
.
0.517 mol C 2 H 5 Br / mol
g
C 2 H 4 + HBr → C 2 H 5 Br
C balance:
b
g
b
gb g b
gb g
165 mol x mol C 2 H 4 2 mol C
= n 0.310 2 + n 0.517 2
s
mol
mol C2 H 4
Br balance: 165 (1 − x )(1) = n ( 0.173)(1) + n ( 0.517 )(1)
(1)
(2)
(Note: An atomic H balance can be obtained as 2*(Eq. 2) + (Eq. 1) and so is not
independent)
Solve (1) and (2) simultaneously ⇒ n = 108.77 mol / s, x = 0.545 mol C 2 H 4 / mol
b g
⇒ 1 − x = 0.455 mol HBr / mol
Since the C2H4/HBr feed ratio (0.545/0.455) is greater than the stoichiometric ration (=1),
HBr is the limiting reactant .
bn g = b165 mol / sgb0.455 mol HBr / molg = 75.08 mol HBr
HBr fed
Fractional conversion of HBr =
( n )
( n )
C2 H 4
stoich
C2 H 4
fed
75.08 − ( 0.173)(108.8 )
75.08
= 0.749 mol HBr react/molfed
= 75.08molC 2 H 4
= (165mol/s )( 0.545molC2 H 4 /mol ) = 89.93molC2 H 4
89.93 − 75.08
= 19.8%
75.08
Extent of reaction: nC2 H5 Br = nC2 H5Br + vC2 H5 Brξ ⇒ (108.8 )( 0.517 ) = 0 + (1) ξ ⇒ ξ = 56.2 mol/s
% excess of C2 H 4 =
(
)
0
4- 32
4.43
a.
2HCl +
1
O 2 → Cl 2 + H 2 O
2
Basis: 100 mol HCl fed to reactor
b
g
n b mol O g
n b mol N g
n b mol Cl g
n b mol H Og
100 mol HCl
b
n1 mol air
n2 mol HCl
g
0.21 mol O 2 / mol
3
2
4
2
2
5
0.79 mol N 2 / mol
35% excess
2
6
mol O
= 25 mol O
bO gstoic = 100 mol HCl 0.5
2 mol HCl
35% excess air: 0.21n b mol O fed g = 135
. × 25 ⇒ n = 160.7 mol air fed
2
2
2
1
2
1
85% conversion ⇒ 85 mol HCl react ⇒ n2 = 15 mol HCl
n5 =
85 mol HCl react
b gb g
1 mol Cl 2
= 42.5 mol Cl 2
2 mol HCl
n6 = 85 1 2 = 42.5 mol H 2O
N 2 balance:
b160.7gb0.79g = n
4
⇒ n4 = 127 mol N 2
O balance:
b160.7gb0.21g mol O
2
42.5 mol H 2O 1 mol O
2 mol O
= 2n3 +
⇒ n3 = 12.5 mol O 2
1 mol H 2O
1 mol O 2
Total moles:
5
∑nj
= 239.5 mol ⇒
j= 2
mol O 2
mol N 2
15 mol HCl
mol HCl
= 0.063
, 0.052
, 0.530
,
239.5 mol
mol
mol
mol
0177
.
b.
mol Cl 2
mol H 2 O
, 0177
.
mol
mol
As before, n1 = 160.7 mol air fed , n2 = 15 mol HCl
1
2HCl + O 2 → Cl 2 + H 2O
2
b g
ni = ni
E
0
+ vi ξ
HCl: 15 = 100 − 2ξ ⇒ ξ = 42.5 mol
4- 33
4.43 (cont’d)
b g 21 ξ = 12.5 mol O
= 0.79b160.7g = 127 mol N
O 2 : n3 = 0.21 160.7 −
N 2 : n4
2
2
Cl 2 : n5 = ξ = 42.5 mol Cl 2
H 2 O: n6 = ξ = 42.5 mol H 2 O
c.
4.44
These molar quantities are the same as in part (a), so the mole fractions would also be the
same.
Use of pure O2 would eliminate the need for an extra process to remove the N2 from the
product gas, but O2 costs much more than air. The cheaper process will be the process of
choice.
b g
Fe O + 3H SO → Fe bSO g + 3H O
bTiOgSO + 2H O → H TiO bsg + H SO
H TiO bsg → TiO bsg + H O
FeTiO3 + 2H 2SO 4 → TiO SO 4 + FeSO 4 + 2H 2O
2
3
2
4
2
4
2
2
2
3
2
4 3
2
3
2
4
2
Basis: 1000 kg TiO2 produced
1000 kg TiO 2
kmol TiO 2
79.90 kg TiO 2
12.52 kmol FeTiO 3 dec.
1 kmol FeTiO 3
= 12.52 kmol FeTiO 3 decomposes
1 kmol TiO 2
1 kmol FeTiO 3 feed
0.89 kmol FeTiO 3 dec.
14.06 kmol FeTiO3
b
= 14.06 kmol FeTiO 3 fed
1 kmol Ti
47.90 kg Ti
1 kmol FeTiO 3
kmol Ti
g
= 6735
. kg Ti fed
673.5 kg Ti / M kg ore = 0.243 ⇒ M = 2772 kg ore fed
b
g
Ore is made up entirely of 14.06 kmol FeTiO3 + n kmol Fe 2O 3 (Assumption!)
n = 2772 kg ore −
638.1 kg Fe 2O 3
14.06 kmol FeTiO3 151.74 kg FeTiO3
= 6381
. kg Fe 2O 3
kmol FeTiO3
kmol Fe 2O 3
= 4.00 kmol Fe 2O 3
159.69 kg Fe 2O 3
14.06 kmol FeTiO3 2 kmol H2SO4 4.00 kmol FeTiO3 3 kmol H2SO4
+
= 4012
. kmol H2SO4
1 kmol FeTiO3
1 kmol Fe2O3
b
g
. 4012
. kmol H 2SO 4 = 6018
. kmol H 2SO 4 fed
50% excess: 15
Mass of 80% solution:
5902.4 kg H 2 SO 4 / M
60.18 kmol H 2SO 4
a
98.08 kg H 2SO 4
= 5902.4 kg H 2SO 4
1 kmol H 2SO 4
bkg solng = 0.80 ⇒ M a = 7380 kg 80% H SO
2
4- 34
4
feed
4.45
a.
Plot C (log scale) vs. R (linear scale) on semilog paper, get straight line through
d R = 10, C
1
1
i
= 0.30 g m 3 and
FH R
2
= 48, C2 = 2.67 g m 3
IK
ln C = bR + ln a ⇔ C = ae br
b=
b
g = 0.0575 , ln a = lnb2.67g − 0.0575b48g = −1.78 ⇒ a = e
ln 2.67 0.30
−1.78
48 − 10
⇒ C = 0169
. e 0.0575 R
i C ′(ftlb
E
d
C g m3 =
m)
453.6 g 35.31 ft 3
3
1 m3
1 lb m
d
= 0169
.
= 16,020C ′
i
16,020C ' = 0169
. e 0.0575 R ⇒ C ′ lb m SO 2 ft 3 = 1055
.
× 10 −5 e 0.0575 R
b.
d2867 ft sib60 s ming = 138 ft
3
1250 lb m min
d
3
lb m coal
i
R = 37 ⇒ C ′ lb m SO 2 ft 3 = 1055
.
× 10−5 eb
gb g = 8.86 × 10
0.0575 37
−5
lb m SO 2 ft 3
8.86 × 10−5 lb m SO 2
138 ft 3
lb SO 2
compliance achieved
= 0.012 < 0.018 m
3
lb m coal
ft
1 lb m coal
c.
S + O 2 → SO 2
1250 lb m coal 0.05 lb m S 64.06 lb m SO 2
min
1 lb m coal
32.06 lb m S
= 124.9 lb m SO 2 generated min
2867 ft 3 60 s 886
. × 10−5 lbm SO2
= 152
. lbm SO2 min in scrubbed gas
s
1 min
ft3
air
1250 lb m coal/min
62.5 lb m S/min
% removal =
d.
furnace
ash
b124.9 − 15.2g lb
scrubbing fluid
stack gas
124.9 lbm SO2 /min
scrubber scrubbed gas
15.2 lb m SO2 /min
liquid effluent
(124.9 – 15.2) lbm SO2 (absorbed)/min
SO 2 scrubbed min
× 100% = 88%
124.9 lb m SO 2 fed to scrubber min
m
The regulation was avoided by diluting the stack gas with fresh air before it exited from the
stack. The new regulation prevents this since the mass of SO2 emitted per mass of coal
burned is independent of the flow rate of air in the stack.
4- 35
4.46
a.
A + B ===== C + D
nA = nA − ξ
0
e
= en
= en
= en
j
− ξj n
ξj n
+ ξj n
nB = nB − ξ
y A = n A − ξ nT
nC = n C + ξ
yB
nD = nD + ξ
yC
0
0
0
nI = nI
Total nT = ∑ ni
At equilibrium:
yD
0
b
b
0
B0
T
C0 +
D0
gb
gb
T
T
g
g
nC0 + ξ c nD0 + ξ c
yC y D
=
= 4.87 ( nT ’s cancel)
y A yB
n A0 − ξ c nB0 − ξ c
b
c
gh b
g
387
. ξ 2c − nC0 + nD0 + 487
. nA0 + nB0 ξ c − nC0nD0 − 487
. nA0nB0 = 0
[aξ 2c + bξ c + c = 0]
a = 387
.
1
∴ξ c =
−b ± b2 − 4ac where b = − nC0 + nD0 + 487
. nA0 + nB0
2a
c = − nC0nD0 − 487
. nA0nB0
e
b.
Constants: a = 3.87
g
nA0 = 1 nB0 = 1 nC0 = nD0 = nI 0 = 0
Basis: 1 mol A feed
ξe =
b
j
b = −9.74 c = 4.87
(
1
9.74 ±
2 ( 3.87 )
( 9.74 )
2
)
− 4 ( 3.87 )( 4.87 ) ⇒ ξ e1 = 0.688
(ξ e 2 = 1.83 is also a solution but leads to a negative conversion )
Fractional conversion: X A ( = X B ) =
c.
nA0 − nA ξ e1
=
= 0.688
nA0
nA0
nB0 = 80, nC0 = nD0 = nJ 0 = 0
nC 0 = 0
nC = 70 = nC 0 + ξ c =======> ξ c = 70 mol
n A = n A0 − ξ c = n A0 − 70 mol
n B = n B 0 − ξ c = 80 − 70 = 10 mol
nC = nC 0 + ξ c = 70 mol
n D = n D0 + ξ c = 70 mol
4.87 =
b gb g
b
gb g
70 70
y C y D nC n D
=
⇒
= 4.87 ⇒ n A0 = 170.6 mol methanol fed
y A y B n A nB
n A0 − 70 10
4- 36
4.46 (cont’d)
Product gas n A = 170.6 − 70 = 100.6 mol
nB = 10 mol
nC = 70 mol
nD = 70 mol
U| y
|V ⇒ y
|| yy
W
A
= 0.401 mol CH3OH mol
B
= 0.040 mol CH3COOH mol
C
= 0.279 mol CH3COOCH 3 mol
D
= 0.279 mol H 2O mol
ntotal = 250.6 mol
4.47
d.
Cost of reactants, selling price for product, market for product, rate of reaction, need for
heating or cooling, and many other items.
a.
CO + H 2O ←
⎯→ CO 2 + H 2
(A)
(B)
(C)
(D)
b
g
n b mol H O g
n b mol CO g
n b mol H g
n b mol Ig
. mol
100
n A mol CO
0.20 mol CO / mol
. mol CO 2 / mol
010
B
2
2
C
0.40 mol H 2O / mol
D
0.30 mol I / mol
2
I
6 unknowns ( n A , nB , nC , nD , n I , ξ )
Degree of freedom analysis:
bg
– 4 expressions for ni ξ
– 1 balance on I
– 1 equilibrium relationship
0 DF
b.
Since two moles are prodcued for every two moles that react,
ntotal out = ntotal in = 100
. mol
b g b g
b g
n A = 0.20 − ξ
nB = 0.40 − ξ
nC = 010
. +ξ
nD = ξ
n I = 0.30
(1)
(2)
(3)
(4)
(5)
ntot = 100
. mol
At equilibrium:
b
y D = nD = ξ = 0110
.
mol H 2
c.
b
gb g
FG
H
IJ
K
010
. +ξ ξ
4020
yC y D nC nD
.
=
=
= 0.0247 exp
⇒ ξ = 0110
mol
0.20 − ξ 0.40 − ξ
1123
y A y B n A nB
b
/ molg
gb
The reaction has not reached equilibrium yet.
4- 37
g
4.47
(cont’d)
d.
T (K)
1223
1123
1023
923
823
723
623
673
698
688
x (CO)
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
x (H2O)
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
1123
1123
1123
1123
0.2
0.4
0.3
0.5
0.4
0.2
0.3
0.4
x (CO2)
Keq
Keq (Goal Seek) Extent of Reaction
0.6610
0.6610
0.2242
0.8858
0.8856
0.2424
1.2569
1.2569
0.2643
1.9240
1.9242
0.2905
3.2662
3.2661
0.3219
6.4187
6.4188
0.3585
15.6692
15.6692
0.3992
9.7017
9.7011
0.3785
7.8331
7.8331
0.3684
8.5171
8.5177
0.3724
0
0
0
0
0
0
0
0
0
0
0.1
0.1
0
0
0.8858
0.8858
0.8858
0.8858
0.8863
0.8857
0.8856
0.8867
0.1101
0.1100
0.1454
0.2156
y (H2)
0.224
0.242
0.264
0.291
0.322
0.358
0.399
0.378
0.368
0.372
0.110
0.110
0.145
0.216
The lower the temperature, the higher the extent of reaction. An equimolar feed ratio of
carbon monoxide and water also maximizes the extent of reaction.
4.48
a.
A + 2B → C
b g
g = lnd10.5 / 2.316 × 10 i = 11458
ln K e = ln A0 + E T K
b
ln K e1 / K e 2
E=
1 T1 − 1 T2
−4
1 373 − 1 573
ln A0 = ln K e1 − 11458 T1 = ln 10.5 − 11458 373 = −28.37 ⇒ A0 = 4.79 × 10−13
b gh
c
Ke = 4.79 × 10 −13 exp 11458 T K atm−2 ⇒ Ke (450K ) = 0.0548 atm−1
b.
n A = n A0 − ξ
nB = nB 0 − 2ξ
nC = nC 0 + ξ
nT = nT 0 − 2ξ
U|
|V
||
W
b
b
b
gb
gb
gb
g
g
g
y A = n A0 − ξ nT 0 − 2ξ
y B = nB 0 − 2ξ nT 0 − 2ξ
⇒
yC = nC 0 + ξ nT 0 − 2ξ
nT 0 = n A0 + nB 0 + nC 0
b
g
At equilibrium,
b
b
gb
gb
n + ξ e nT 0 − 2ξ e
yC 1
= C0
2
2
y A yB P
n A0 − ξ e nB 0 − 2ξ e
c.
g
g
2
2
bg
bg
1
= Ke T (substitute for K T from Part a.)
e
P2
Basis: 1 mol A (CO)
n A0 = 1 nB 0 = 1 nC 0 = 0 ⇒ nT 0 = 2 , P = 2 atm , T = 423K
b
ξ e 2 − 2ξ e
g
2
b1 − ξ gb1 − 2ξ g
e
e
2
b g
1
.
=0
= K e 423 = 0.278 atm -2 ⇒ ξ 2e − ξ e + 01317
4 atm 2
4- 38
4.48 (cont’d)
(For this particular set of initial conditions, we get a quadratic equation. In general, the
equation will be cubic.)
ξ e = 0156
.
, 0.844
Reject the second solution, since it leads to a negative nB .
. gh ⇒ y = 0.500
b
g c2 − 2b0156
y = c1 − 2b0156
. gh c2 − 2b0156
. gh ⇒ y = 0.408
y = b0 + 0156
. g c2 − 2b0156
. gh ⇒ y = 0.092
n −n
ξ
=
Fractional Conversion of CO b Ag =
n
n
y A = 1 − 0156
.
A
B
B
C
C
A0
A
A0
= 0156
.
mol A reacted / mol A feed
A0
Use the equations from part b.
d.
i)
ii)
iii)
iv)
*
1
2
Fractional conversion decreases with increasing fraction of CO.
Fractional conversion decreases with increasing fraction of CH3OH.
Fractional conversion decreases with increasing temperature.
Fractional conversion increases with increasing pressure.
REAL TRU, A, E, YA0, YC0, T, P, KE, P2KE, C0, C1, C2, C3, EK, EKPI,
FN, FDN, NT, CON, YA, YB, YC
INTEGER NIT, INMAX
TAU = 0.0001
INMAX = 10
A = 4.79E–13
E = 11458.
READ (5, *) YA0, YB0, YC0, T, P
KE = A * EXP(E/T)
P2KE = P*P*KE
C0 = YC0 – P2KE * YA0 * YB0 * YB0
C1 = 1. – 4. * YC0 + P2KE * YB0 * (YB0 + 4. * YA0)
C2 = 4. * (YC0 –1. – P2KE * (YA0 + YB0))
C3 = 4. * (1. + P2KE)
EK = 0.0
(Assume an initial value ξe = 0. 0 )
NIT = 0
FN = C0 + EK * (C1 + EK * (C2 + EK * C3)) FDN = C1 + EK * (2. * C2 +
EK * 3. * C3) EKPI = EK - FN/FDN NIT = NIT + 1 IF (NIT.EQ.INMAX)
GOTO 4 IF (ABS((EKPI – EK)/EKPI).LT.TAU) GOTO 2 EK =
EKPI GOTO 1
NT = 1. – 2. * EKPI
YA = (YA0 – EKPI)/NT
YB = (YB0 – 2. + EKPI)/NT
YC = (YC0 + EKPI)/NT
4- 39
4.48 (cont’d)
CON = EKPI/YA0 WRITE (6, 3) YA, YB, YC, CON STOP
WRITE (6, 5) INMAX, EKPI
FORMAT (' YA YB YC CON', 1, 4(F6.3, 1X)) FORMAT ('DID NOT
CONVERGE IN', I3, 'ITERATIONS',/,
*
'CURRENT VALUE = ', F6.3) END
$ DATA 0.5
0.5
0.0
423.
2.
RESULTS: YA = 0.500, YB = 0.408, YC = 0.092, CON = 0.156
4
3
Note: This will only find one root — there are two others that can only be found by
choosing different initial values of ξa
4.49
a.
CH 4 + O 2 ⎯⎯→ HCHO + H 2O
(1)
CH 4 + 2O 2 ⎯⎯→ CO 2 + 2H 2O
(2)
100 mol / s
b
g
n b mol O / s g
n b mol HCHO / sg
n b mol H O / s g
molCO
CO / gs)
nn b(mol
n1 mol CH 4 / s
0.50 mol CH 4 / mol
0.50 mol O 2 / mol
2
2
3
7 unknowns ( n1 , n 2 , n3 , n 4 , n5 , ξ 1 , ξ 2 )
– 5 equations for n ξ , ξ
i
e
1
2
4
2
55
2
2
j
2 DF
b.
n1 = 50 − ξ 1 − ξ 2
n = 50 − ξ − 2ξ
2
1
n3 = ξ 1
n = ξ + 2ξ
4
1
n5 = ξ 2
c.
(1)
(2)
2
(3)
(4)
2
(5)
Fractional conversion:
b50 − n g = 0.900 ⇒ n
1
1
50
= 5.00 mol CH / s
4
Fractional yield: n 3 = 0.855 ⇒ n 3 = 42.75 mol HCHO / s
50
U|
||
V|
||
W
y CH = 0.0500 mol CH 4 / mol
Equation 3 ⇒ ξ 1 = 42.75
4
y
O 2 = 0.0275 mol O 2 / mol
Equation 1 ⇒ ξ 2 = 2.25
Equation 2 ⇒ n 2 = 2.75 ⇒ y HCHO = 0.4275 mol HCHO / mol
y H O = 0.4725 mol H 2 O / mol
Equation 4 ⇒ n 4 = 47.25
2
y CO = 0.0225 mol CO 2 / mol
Equation 5 ⇒ n5 = 2.25
2
Selectivity: [(42.75 mol HCHO/s)/(2.25 mol CO 2 /s) = 19.0 mol HCHO/mol CO 2
4- 40
4.50
a.
Design for low conversion and feed ethane in excess. Low conversion and excess ethane
make the second reaction unlikely.
b.
C2H6 + Cl2 → C2H5Cl + HCl, C2H5Cl + Cl2 → C2H4Cl2 + HCl
Basis: 100 mol C2H5Cl produced
c.
n1 (mol C2H6)
100 mol C2H5Cl
n2 (mol Cl2)
n3 (mol C2H6)
n4 (mol HCl)
n5 (mol C2H5Cl2)
5 unknowns
–3 atomic balances
2 D.F.
Selectivity: 100 mol C 2 H 5 Cl = 14n5 (mol C 2 H 4 Cl 2 ) ⇒ n5 = 7.143 mol C 2 H 4 Cl 2
U| ⇒ n = 714.3 mol C H in
g
V
2n = 2b100g + 2n + 2b7.143g|W n = 114.3 mol C H out
6b714.3g = 5b100g + 6b114.3g + n + 4b7.143g ⇒ n = 607.1 mol HCl
2n = 100 + 607.1 + 2b7.143g ⇒ n = 114.3 mol Cl
b
15% conversion: 1 − 0.15 n1 = n3
C balance:
H balance:
Cl balance:
1
3
1
2
6
3
2
6
4
4
2
2
2
. mol Cl 2 / mol C2 H 6
Feed Ratio: 114.3 mol Cl 2 / 714.3 mol C2 H 6 = 016
Maximum possible amount of C2H5Cl:
114.3 mol Cl 2 1 mol C 2 H 5 Cl
n max =
= 114.3 mol C 2 H 5 Cl
1 mol Cl 2
Fractional yield of C2H5Cl:
4.51
nC2 H5Cl
n max
=
100 mol
= 0.875
114.3 mol
d.
Some of the C2H4Cl2 is further chlorinated in an undesired side reaction:
C2H5Cl2 + Cl2 → C2H4Cl3 + HCl
a.
C2H4 + H2O → C2H5OH, 2 C2H5OH → (C2H5)2O + H2O
Basis: 100 mol effluent gas
100 mol
0.433 mol C 2 H 4 / mol
n1 (mol C 2 H 4 )
n2 [mol H 2 O (v)]
3 unknowns
0.025 mol C 2 H 5OH / mol
0.0014 mol (C H ) O / mol
2 5 2
0.093 mol I / mol
n 3 (mol I)
-2 independent atomic balances
-1 I balance
0 D. F.
0.4476 mol H 2 O (v) / mol
b
(1) C balance: 2n1 = 100 2∗0.433 + 2∗0.025 + 4∗0.0014
b
g
g
(2) H balance: 4n1 + 2n2 = 100 4∗0.433 + 6∗0.025 + 10∗0.0014 + 2∗0.4476
b
g
(3) O balance: n2 = 100 0.025 + 0.0014 + 0.4476
Note; Eq. (1)∗2 + Eq. ( 3)∗2 = Eq. (2) ⇒2 independent atomic balances
(4) I balance: n3 = 9.3
4-41
4.51 (cont'd)
b.
(1) ⇒ n1 = 46.08 mol C 2 H 6
(3) ⇒ n2 = 47.4 mol H 2 O
(4) ⇒ n3 = 9.3 mol I
% conversion of C2H4:
U|
V| ⇒ Reactor feed contains 44.8% C H , 46.1% H O, 9.1% I
W
2
46.08 − 43.3
× 100% = 6.0%
46.08
d
6
If all C2H4 were converted and the second reaction did not occur, nC2 H5OH
d
⇒ Fractional Yield of C2H5OH: nC2 H5OH / nC2 H5OH
i
max
b
g
2
i
max
= 46.08 mol
= 2.5 / 46.08 = 0.054
Selectivity of C2H5OH to (C2H5)2O:
2.5 mol C 2 H 5 OH
= 17.9 mol C 2 H 5 OH / mol (C 2 H 5 ) 2 O
0.14 mol (C 2 H 5 ) 2 O
c.
4.52
Keep conversion low to prevent C2H5OH from being in reactor long enough to form
significant amounts of (C2H5)2O. Separate and recycle unreacted C2H4.
bg
bg
bg
bg
CaF2 s + H 2 SO 4 l → CaSO 4 s + 2HF g
1 metric ton acid
1000 kg acid
0.60 kg HF
= 600 kg HF
1 metric ton acid 1 kg acid
Basis: 100 kg Ore dissolved (not fed)
100 kg Ore d issolved
0.96 kg CaF 2 /kg
0.04 kg SiO 2/ kg
nA (kg 93% H2 SO4 )
0.93 H2 SO4 kg/ kg
0.07 H2 O kg/ kg
(kgCaSO
CaSO44))
nn1 1 (kg
n
(kg
HF)
n2 2 (kg HF)
(kgHH2SiF
nn3 3 (kg
66) )
4SiF
(kgHH
2SO
4) 4)
nn4 4 (kg
2SO
2O)
(kgHH
nn5 5 (kg
2 O)
Atomic balance - Si:
0.04 (100 ) kg SiO 2
28.1 kg Si
60.1 kg SiO 2
=
n3 (kg H 2SiF6 )
28.1 kg Si
⇒ n3 = 9.59 kg H 2SiF6
144.1 kg H 2SiF6
Atomic balance - F:
0.96 (100 ) kg CaF2
38.0 kg F
78.1 kg CaF2
+
9.59 kg H 2SiF6
=
n2 (kg HF)
114.0 kg F
144.1 kg H 2SiF6
19.0 kg F
20.0 kg HF
⇒ n2 = 41.2 kg HF
600 kg HF 100 kg ore diss. 1 kg ore feed
= 1533 kg ore
41.2 kg HF
0.95 kg ore diss.
4-42
4.53
a.
C 6 H 6 + Cl 2 → C 6 H 5 Cl + HCl
C 6 H 5 Cl + Cl 2 → C 6 H 4 Cl 2 + HCl
C 6 H 4 Cl 2 + Cl 2 → C 6 H 3 Cl 3 + HCl
Convert output wt% to mol%: Basis 100 g output
species
C6 H 6
C 6 H 5 Cl
C 6 H 4 Cl 2
C 6 H 3 Cl 3
g
65.0
32.0
2.5
0.5
Mol. Wt.
78.11
112.56
147.01
181.46
mol
0.832
0.284
0.017
0.003
mol %
73.2
25.0
1.5
0.3
total 1.136
Basis: 100 mol output
n1 (mol C6 H6 )
n2 (mol Cl 2)
n3 (mol I)
n 4 (mol HCl(g ))
n 3 (mol I)
65.0 mo l C6 H6
73.2 mol C H6
32.0 mo l C6 6 H
5 Cl
25.0 mol C6H5Cl
2.5 mo l C 6 H 4 Cl2
1.5 mol C6H4Cl2
0.5
mo l C
C6 H
H 3Cl
Cl 3
0.3 mol
6
b.
3
4 unknowns
-3 atomic balances
-1 wt% Cl 2 in feed
0 D.F.
3
C balance: 6n1 = 6 ( 73.2 + 25.0 + 1.5 + 0.3) ⇒ n1 = 100 mol C 6 H 6
H balance: 6 (100 ) = 6 ( 73.2 ) + 5 ( 25.0 ) + 4 (1.5 ) + 3 ( 0.3) + n4 ⇒ n4 = 28.9 mol HCl
Cl balance: 2n2 = 28.9 + 25.0 + 2 (1.5 ) + 3 ( 0.3) ⇒ n2 = 28.9 mol Cl2
Theoretical C 6 H 6 = 28.9 mol Cl 2 (1 mol C6 H 6 1 mol Cl2 ) = 28.9 mol C6 H 6
(100 − 28.9 ) 28.9 ×100% = 246% excess C6 H 6
Fractional Conversion: (100 − 73.2 ) 100 = 0.268 mol C 6 H 6 react/mol fed
Excess C 6 H 6 :
Yield: (25.0 mol C6 H 5 Cl) (28.9 mol C6 H 5 Cl maximum)=0.865
⎫
⎪
g gas
⎪
⎬ ⇒ 0.268
g liquid
⎛ 78.11 g C6 H 6 ⎞
⎪
Liquid feed: (100 mol C6 H 6 ) ⎜
⎟ = 7811 g liquid ⎪
⎝ mol C6 H 6 ⎠
⎭
Gas feed:
28.9 mol Cl2 70.91 g Cl2 1 g gas
= 2091 g gas
mole Cl2 0.98 g Cl2
c.
Low conversion ⇒ low residence time in reactor ⇒ lower chance of 2nd and 3rd reactions
occurring. Large excess of C 6 H 6 ⇒ Cl 2 much more likely to encounter C 6 H 6
than substituted C 6 H 6 ⇒ higher selectivity.
d.
Dissolve in water to produce hydrochloric acid.
e.
Reagent grade costs much more. Use only if impurities in technical grade mixture affect the
reaction rate or desired product yield.
4-43
4.54
a.
2CO 2 ⇔ 2CO + O 2
O 2 + N 2 ⇔ 2NO
2A ⇔ 2B + C
C + D ⇔ 2E
bn
bn
bn
bn
bn
n A = n A 0 − 2ξ e1
nB
nC
nD
nE
yA =
= n B 0 + 2ξ e 2
yB =
= nC 0 + ξ e1 − ξ e 2 ⇒ y C =
= n D0 − ξ e2
yD =
= n E 0 + 2ξ e 2
yE =
bn
ntotal = nT 0 + ξ e1
T0
− 2ξ e1
+ 2ξ e1
A0
B0
C0
D0
E0
g bn
g bn
T0
T0
+ ξ e1
+ ξ e1
gb
g
g
+ ξ e1 − ξ e 2 nT 0 + ξ e1
− 1ξ e 2 nT 0 + ξ e1
+ 2ξ e 2 nT 0 + ξ e1
gb
gb
g
g
g
= n A0 + n B 0 + nC 0 + n D 0 + n E 0
g
Equilibrium at 3000K and 1 atm
y B2 y C
y 2A
=
bn
g bn + ξ − ξ g = 01071
.
bn − 2ξ g bn + ξ g
B0
+ 2ξ e1
2
A0
e1
C0
2
e1
T0
e2
e1
( nE 0 + 2ξ e 2 )
yE2
=
= 0.01493
yC yD ( nA0 + ξ e1 − ξ e 2 )( nD 0 − ξ e 2 )
2
E
b
f 1 = 01071
.
n A 0 − 2ξ e1
b
g bn
2
T0
g b
gbn
g bn + ξ − ξ g = 0U| Defines functions
g − bn + 2ξ g = 0 V|W f bfξ bξ, ξ, ξg and
g
+ ξ e1 − n B 0 + 2ξ e1
f 2 = 0.01493 nC 0 + ξ e1 − ξ e 2
D0
− ξ e2
2
C0
e1
e2
2
E0
1
1
2
e2
2
1
2
b.
Given all nio’s, solve above equations for ξe1 and ξe2 ⇒ nA, nB, nC, nD, nE ⇒ yA, yB, yC, yD, yE
c.
nA0 = nC0 = nD0 = 0.333, nB0 = nE0 = 0 ⇒ ξe1 =0.0593, ξe2 = 0.0208
⇒ yA = 0.2027, yB = 0.1120, yC = 0.3510, yD = 0.2950, yE = 0.0393
d.
a11d 1 + a12 d 2 = − f 1
d1 =
bξ g
a12 f 2 − a 22 f 1
a11a 22 − a12 a 21
e1 new
= ξ e1 + d 1
a 21d 1 + a 22 d 2 = − f 2
d2 =
bξ g
a 21 f 1 − a11 f 2
a11a 22 − a12 a 21
e 2 new
= ξ e1 + d 2
(Solution given following program listing.)
.
1
30
IMPLICIT REAL * 4(N)
WRITE (6, 1)
FORMAT('1', 30X, 'SOLUTION TO PROBLEM 4.57'///)
READ (5, *) NA0, NB0, NC0, ND0, NE0
IF (NA0.LT.0.0)STOP
WRITE (6, 2) NA0, NB0, NC0, ND0, NE0
4-44
4.54 (cont’d)
2
3
100
4
120
FORMAT('0', 15X, 'NA0, NB0, NC0, ND0, NE0 *', 5F6.2/)
NTO = NA0 + NB0 + NC0 + ND0 + NE0
NMAX = 10
X1 = 0.1
X2 = 0.1
DO 100 J = 1, NMAX
NA = NA0 – X1 – X1
NB = NB0 + X1 + X1
NC = NC0 + X1 – X2
ND = ND0 – X2
NE = NE0 + X2 + X2
NAS = NA ** 2
NBS = NB ** 2
NES = NE ** 2
NT = NT0 + X1
F1 = 0.1071 * NAS * NT – NBS * NC
F2 = 0.01493 * NC * ND – NES
A11 = –0.4284 * NA * NT * 0.1071 * NAS – 4.0 * NB * NC – NBS
A12 = NBS
A21 = 0.01493 * ND
A22 = –0.01493 * (NC + ND) – 4.0 * NE
DEN = A11 * A22 – A12 * A21
D1 = (A12 * F2 – A22 * F1)/DEN
D2 = (A21 * F1 – A11 * F2)/DEN
X1C = X1 + D1
X2C = X2 + D2
WRITE (6, 3) J, X1, X2, X1C, X2C
FORMAT(20X, 'ITER *', I3, 3X, 'X1A, X2A =', 2F10.5, 6X, 'X1C, X2C =', * 2F10.5)
IF (ABS(D1/X1C).LT.1.0E–5.AND.ABS(D2/X2C).LT.1.0E–5) GOTO 120
X1 = X1C
X2 = X2C
CONTINUE
WRITE (6, 4) NMAX
FORMAT('0', 10X, 'PROGRAM DID NOT CONVERGE IN', I2, 'ITERATIONS'/)
STOP
YA = NA/NT
YB = NB/NT
YC = NC/NT
YD = ND/NT
YE = NE/NT
WRITE (6, 5) YA, YB, YC, YD, YE
5 FORMAT ('0', 15X, 'YA, YB, YC, YD, YE =', 1P5E14.4///)
GOTO 30
END
$DATA
0.3333 0.00 0.3333 0.3333 0.0
0.50
0.0 0.0
0.50
0.0
0.20
0.20 0.20
0.20
0.20
SOLUTION TO PROBLEM 4.54
NA0, NB0, NC0, ND0, NE0 = 0.33 0.00 0.33
ITER = 1 X1A, X2A = 0.10000 0.10000
ITER = 2 X1A, X2A = 0.06418 0.05181
ITER = 3 X1A, X2A = 0.05969 0.02486
4-45
0.33 0.00
X1C, X2C = 0.06418
X1C, X2C = 0.05969
X1C, X2C = 0.05937
0.05181
0.02986
0.02213
4.54 (cont’d)
ITER = 4 X1A, X2A = 0.05437
ITER = 5 X1A, X2A = 0.05931
ITER = 6 X1A, X2A = 0.05930
0.02213
0.02086
0.02083
YA, YB, YC, YD, YE =
2.0270E − 01
2.9501E − 01
NA0, NB0, NC0, ND0, NE0 = 0.20
ITER = 1 X1A, X2A = 0.10000
↓
ITER = 7 X1A, X2A = –0.02244
YA, YB, YC, YD, YE=
4.55
X1C, X2C = 0.05931
X1C, X2C = 0.05930
X1C, X2C = 0.05930
0.20
0.20
1.1197 E − 01
3.9319 E − 02
0.20
0.10000
0.02086
0.02083
0.02083
3.5100E − 01
0.20
X1C, X2C = 0.00012
–0.08339 X1C, X2C = –0.02244
0.00037
–0.08339
2.5051E − 01 1.5868E − 01 2.6693E − 01
2.8989E − 01 3.3991E − 02
(B)
a.
(1 − f ) m0 (kg/h)
xR = 0
Reactor: 99%
conv. of R
(A)
(P)
m0 (kg/h)
(1 − f ) m0 (kg/h)
m1 (kg/h)
mP (kg/h)
xRA (kg R/kg)
xRA (kg R/kg)
xR1 (kg R/kg)
0.0075 kg R/kg
fm0 (kg/h)
xRA (kg R/kg)
2(1 − f )m0 = m1
(1)
m1 xR1 = 0.01(1 − f )m0 xRA
(2)
Mass balance on reactor:
99% conversion of R:
m1 + fm0 = mP
Mass balance on mixing point:
R balance on mixing point:
m1 xR1 + fm0 xRA = 0.0075mP
(3)
(4)
The system has 6 unknowns (m0 , xRA , f , m1 , xR1 , mP ) and four independent equations relating
them, so there must be two degrees of freedom.
b.
2(1 − f )m0 = m1
m1 xR1 = 0.01(1 − f )m0 xRA
m1 + fm0 = mP
m1 xR1 + fm0 xRA = 0.0075mP
mP = 4850
E-Z Solve
⎯⎯⎯⎯
→
xRA = 0.0500
4-46
m0 = 2780 kg/h
f = 0.254 kg bypassed/kg fresh feed
4.55 (cont’d)
mP
4850
4850
4850
4850
4850
4850
4850
4850
4850
xRA
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
mA0
3327
3022
2870
2778
2717
2674
2641
2616
2596
mB0
1523
1828
1980
2072
2133
2176
2209
2234
2254
f
0.54
0.40
0.31
0.25
0.21
0.19
0.16
0.15
0.13
mP
2450
2450
2450
2450
2450
2450
2450
2450
2450
xRA
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
mA0
1663
1511
1435
1389
1359
1337
1321
1308
1298
mB0
762
914
990
1036
1066
1088
1104
1117
1127
f
0.54
0.40
0.31
0.25
0.22
0.19
0.16
0.15
0.13
f v s . x RA
f (kg bypass/kg fresh feed)
c.
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0.00
0.02
0.04
0.06
x R A (k g R / k g A )
4-47
0.08
0.10
0.12
4.56
a.
900 kg HCHO 1 kmol HCHO
= 30.0 kmol HCHO / h
h
30.03 kg HCHO
n (kmol CH OH / h)
1
3
30.0 kmol HCHO / h
n2 (kmol H 2 / h)
n3 (kmol CH 3OH / h)
% conversion:
30.0
= 0.60 ⇒ n1 = 50.0 kmol CH 3 OH / h
n1
b.
n (kmol CH OH / h)
1
3
30.0 kmol HCHO / h
30.0 kmol HCHO / h
n2 (kmol H 2 / h)
n2 (kmol H 2 / h)
n3 (kmol CH 3OH / h)
n (kmol CH OH / h)
3
3
Overall C balance: n1 (1) = 30.0 (1) ⇒ n1 = 30.0 kmol CH3OH/h (fresh feed)
Single pass conversion:
30.0
= 0.60 ⇒ n3 = 20.0 kmol CH 3OH / h
n1 + n3
n1 + n3 = 50.0 kmol CH3OH fed to reactor/h
4.57
c.
Increased xsp will (1) require a larger reactor and so will increase the cost of the reactor and
(2) lower the quantities of unreacted methanol and so will decrease the cost of the
separation. The plot would resemble a concave upward parabola with a minimum
around xsp = 60%.
a.
Convert effluent composition to molar basis. Basis: 100 g effluent:
10.6 g H 2
1 mol H 2
2.01 g H 2
64.0 g CO
= 5.25 mol H 2
1 mol CO
28.01 g CO
25.4 g CH 3OH
= 2.28 mol CO
1 mol CH 3 OH
32.04 g CH 3OH
= 0.793 mol CH 3OH
4-48
H : 0.631 mol H / mol
2
2
⇒
CO: 0.274 mol CO / mol
CH OH: 0.0953 mol CH OH / mol
3
3
4.57 (cont’d)
n4 (mol/min)
0.004 mol CH3OH(v)/mol
x (mol CO/mol)
(0.896 - x ) (mol H 2 / mol)
350 mol/ min
Reactor
n3 (mol CH 3OH(l)/min)
0.631 mol CH 3OH(v)/ mol
n1 (mol CO/min)
n2 (mol H2 / min)
Cond.
0.274 mol CO/ mol
CO + H 2 → CH 3OH 0.0953 mol H / mol
2
Condenser
Overall process
3 unknowns (n3 , n4 , x)
2 unknowns (n1 , n2 )
–3 balances
–2 independent atomic balances
0 degrees of freedom
0 degrees of freedom
Balances around condenser
⎫ n3 = 32.1 mol CH 3OH(l)/min
CO: 350 ∗ 0.274 = n4 ∗ x
⎪⎪
H : 350 ∗ 0.631 = n4 ∗ (0.996 − x) ⎬ ⇒ n4 = 318.7 mol recycle/min
2
⎪
x = .301 molCO/mol
CH OH : 350 ∗ 0.0953 = n3 + 0.004 ∗ n4 ⎪
3
⎭
Overall balances
C: n1 =n3 ⎫ n1 = 32.08 mol/min CO in feed
⎬⇒
H : 2n2 =4n3 ⎭ n2 = 64.16 mol/min H 2 in feed
Single pass conversion of CO:
Overall conversion of CO:
b.
(32.08 + 318.72 ∗ 0.3009) − 350 ∗ 0.274 × 100% = 25.07%
(32.08 + 318.72 ∗ 0.3009)
32.08 − 0
× 100% = 100%
32.08
–
Reactor conditions or feed rates drifting. (Recalibrate measurement instruments.)
–
Impurities in feed. (Re-analyze feed.)
–
Leak in methanol outlet pipe before flowmeter. (Check for it.)
4-49
4.58
a.
Basis: 100 kmol reactor feed/hr
n3 (kmol CH 4 /h)
100 kmol /h
Reactor
n1 (kmol CH 4 /h) 80 kmol CH4 /h
n2 (kmol Cl 2 /h) 20 kmol Cl2 /h
n3 (kmol CH 4 /h)
n4 (kmol HCl /h)
5n5 (kmol CH 3Cl /h)
n5 (kmol CH 2Cl 2 /h)
Cond.
Solvent
Absorb
n3 (kmol CH 4 /h)
n4 (kmol HCl /h)
n4 (kmol HCl /h)
5n5 (kmol CH 3Cl /h)
Still
5n5 (kmol CH 3Cl /h)
n5 (kmol CH 2Cl 2 /h)
n5 (kmol CH 2Cl 2 /h)
Overall process: 4 unknowns (n1, n2, n4, n5) -3 balances = 1 D.F.
Mixing Point: 3 unknowns (n1, n2, n3) -2 balances = 1 D.F.
Reactor: 3 unknowns (n3, n4, n5) -3 balances = 0 D.F.
Condenser: 3 unknowns (n3, n4, n5) -0 balances = 3 D.F.
Absorption column: 2 unknowns (n3, n4) -0 balances = 2 D.F.
Distillation Column: 2 unknowns (n4, n5) -0 balances = 2 D.F.
Atomic balances around reactor:
⎫
1) C balance : 80 = n 3 + 5n 5 + n 5
⎪
2) H balance : 320 = 4n 3 + n 4 + 15n 5 + 2n 5 ⎬ ⇒ Solve for n 3 , n 4 , n 5
⎪
3) Cl balance : 40 = n 4 + 5n 5 + 2n 5
⎭
CH4 balance around mixing point: n1 = (80 – n3)
Solve for n1
Cl2 balance: n2 = 20
b.
For a basis of 100 kmol/h into reactor
n1 = 17.1 kmol CH4/h
n4 = 20.0 kmol HCl/h
n2 = 20.0 kmol Cl2/h
5n5 = 14.5 kmol CH3Cl/h
n3 = 62.9 kmol CH4/h
c.
(1000 kg CH3Cl/h)(1 kmol/50.49 kg) = 19.81 kmol CH3Cl/h
Scale factor =
19.81 kmol CH 3Cl/h
14.5 kmol CH 3Cl/h
= 1.366
n tot = 50.6 kmol/h
n = (17.1)(1.366) = 23.3 kmol CH 4 /h ⎫
⇒
Fresh feed: 1
⎬
n 2 = ( 20.0)(1.366) = 27.3 kmol Cl 2 /h ⎭ 46.0 mol% CH 4 , 54.0 mole% Cl 2
Recycle: n3 = (62.9)(1.366) = 85.9 kmol CH4 recycled/h
4-50
4.59
a.
Basis: 100 mol fed to reactor/h ⇒ 25 mol O2/h, 75 mol C2H4/h
n1 (mol C 2H 4 //h)
n2 (mol O 2 /h)
Seperator
separator
reactor
nC2H4 ( mol C 2H 4 /h)
nO2 (mol O 2 /h)
75 mol C 2H 4 //h
25 mol O 2 /h
n1 (mol C 2H 4 //h)
n2 (mol O 2 /h)
n3 (mol C 2H 4O /h)
n4 (mol CO 2 /h)
n5 (mol H 2O /h)
n3 (mol C 2H 4O /h)
n4 (mol CO 2 /h)
n5 (mol H 2O /h)
Reactor
5 unknowns (n1 - n5)
-3 atomic balances
-1 - % yield
-1 - % conversion
0 D.F.
Strategy: 1. Solve balances around reactor to find n1- n5
2. Solve balances around mixing point to find nO2, nC2H4
(1) % Conversion ⇒ n1 = .800 * 75
(2) % yield: (.200)(75) mol C 2 H 4 ×
90 mol C 2 H 4 O
= n 3 (production rate of C 2 H 4 O)
100 mol C 2 H 4
(3) C balance (reactor): 150 = 2 n1 + 2 n3 + n4
(4) H balance (reactor): 300 = 4 n1 + 4 n3 + 2 n5
(5) O balance (reactor): 50 = 2 n2 + n3 + 2 n4 + n5
(6) O2 balance (mix pt): nO2 = 25 – n2
(7) C2H4 balance (mix pt): nC2H4 = 75 – n1
Overall conversion of C2H4: 100%
b.
c.
n1 = 60.0 mol C2H4/h
n5 = 3.00 mol H2O/h
n2 = 13.75 mol O2 /h
nO2 = 11.25 mol O2/h
n3 = 13.5 mol C2H4O/h
nC2H4 = 15.0 mol C2H4/h
n4 = 3.00 mol CO2/h
100% conversion of C2H4
Scale factor =
2000 lbm C 2 H 4 O 1 lb - mole C 2 H 4 O
h
lb − mol / h
= 3.363
h
44.05 lbm C 2 H 4 O 13.5 mol C 2 H 4 O
mol / h
nC2H4 = (3.363)(15.0) = 50.4 lb-mol C2H4/h
nO2 = (3.363)(11.25) = 37.8 lb-mol O2/h
4-51
4.60
a.
Basis: 100 mol feed/h. Put dots above all n’s in flow chart.
100 mol/h
n1 (mol /h)
32 mol CO/h
64 mol H 2 / h
4 mol N 2 / h
.13 mol N 2 /mol
reactor
n3n(mol
CHCH
OH / h)
2 (mol
3 3OH/h)
cond.
500 mol / h
x1 (mol N 2 /mol)
x2 (mol CO / mol)
1-x1-x2 (mol H 2 / h)
n3 (mol / h)
x1 (mol N 2 /mol)
x2 (mol CO / mol)
1-x1-x2 (mol H 2 / h)
Purge
Mixing point balances:
total: (100) + 500 = n1 ⇒ n1 = 600 mol/h
N2: 4 + x1 * 500 = .13 * 600 ⇒ x1 = 0.148 mol N2/mol
Overall system balances:
N2: 4 = .148 * n3 ⇒ n3 = 27 mol/h
Atomic C: 32 = n2 + x2*27
⇒
Atomic H: 2 * 64 = 4*24.3 + 2*(1-0.148-x2)*27
n2 = 24.3 mol CH3OH/h
x2 = 0.284 mol CO/mol
Overall CO conversion: 100*[32-0.284(27)]/32 = 76%
Single pass CO conversion: 24.3/ (32+.284*500) = 14%
b.
Recycle: To recover unconsumed CO and H2 and get a better overall conversion.
Purge: to prevent buildup of N2.
4.61
a.
+ 3H2 ->
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 =
V̂ideal =
αa
RT
−
-b V
V
+b
V
d i d
i
2.5 ft 3 32.0 lb m / lb-mole
ft 3
= 2.27
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 ⎭
V̂r =
ˆ
10.0 ft 3
72.9 atm
lb-mole ⋅ °R
1K
VP
C
=
= 0.80
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
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
m3 =
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
0.5 × 390 g S 11.8 g A
160 − 114 g A −
= 23.0 g A s
100 g S
b
g A initial
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 Û 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 << 1500 kW
7- 14
2
s
2
2
m2
1 kW ⋅ s
1N
1 kg ⋅ m / s
2
10 3 N ⋅ m
b
g
b
g
kJ
. h Ts − To = 300
7.30 (a) ΔE p , ΔE k , W s = 0 ⇒ Q = ΔH ⇒ − hA Ts − To = −300 kJ h ⇒ 18
h
(b) Clothed:
h = 8 ⇒ To = 13.4° C
Ts =34.2
Nude, immersed: h = 64 ⇒ To = 316
. ° C (Assuming Ts remains 34.2°C)
Ts =34.2
(c) The wind raises the effective heat transfer coefficient. (Stagnant air acts as a thermal
insulator —i.e., in the absence of wind, h is low.) For a given To , the skin temperature
must drop to satisfy the energy balance equation: when Ts drops, you feel cold.
7.31 Basis: 1 kg of 30°C stream
1 kg H2O(l)@30oC
3 kg H2O(l)@Tf(oC)
2 kg H2O(l)@90oC
b
g b
g
1
2
30 o C + 90 D C = 70 D C
3
3
(b) Internal Energy of feeds: U 30° C, liq. = 125.7 kJ kg
U 90° C, liq. = 376.9 kJ kg
(a) T f =
b
b
g
g
(Table B.5 - neglecting effect of P on H )
Energy Balance: Q - W = ΔU + ΔE p + ΔE k
b
g
U|V
|W
Q =W = ΔE p = ΔE k = 0
b
ΔU = 0
g
⇒ 3U f − (1 kg) 125.7 kJ / kg − (2 kg) 376.9 kJ / kg = 0
⇒ U f = 293.2 kJ kg ⇒ T f = 70.05° C (Table B.5)
Diff. =
7.32
70.05 − 70.00
× 100% = 0.07% (Any answer of this magnitude is acceptable).
70.05
.
m(kg/h)
. kg H2 O( v)/kg
0.85
0.15 kg H2 O( l)/kg
5 bar, saturated, T(oC)
P = 5 bars
(a) Table B.6
52.5 m3 H2O(v)/h
.
m(kg/h)
5 bar, T(oC)
.
Q (kW)
T = 1518
. ° C , H L = 6401
. kJ kg , H V = 2747.5 kJ kg
52.5 m3
bar, sat' d) = 0.375 m 3 / kg ⇒ m
=
V(5
h
b gb
g
1 kg
= 140 kg h
0.375 m3
(b) H 2 O evaporated = 015
. 140 kg h = 21 kg h
Energy balance: Q = ΔH =
21 kg
b2747.5 − 6401. gkJ
h
1h
1 kW
kg 3600 s 1 kJ s
7- 15
= 12 kW
7.33 (a) P = 5 bar
Tsaturation = 1518
. o C . At 75°C the discharge is all liquid
Table B.6
(b) Inlet: T=350°C, P=40 bar
Outlet: T=75°C, P=5 bar
Table B.7
Table B.7
H in = 3095 kJ / kg , Vin = 0.0665 m 3 / kg
H out = 314.3 kJ / kg , Vout = 1.03 × 10 -3 m 3 / kg
3
Vin 200 kg 1 min 0.0665 m / kg
uin =
=
= 5018
. m/s
π (0.075) 2 / 4 m 2
min 60 s
Ain
uout =
Vout 200 kg 1 min
=
min 60 s
Aout
0.00103 m 3 / kg
= 175
. m/s
π ( 0.05) 2 / 4 m 2
m
Energy balance: Q − W s ≈ ΔH + ΔE k = m ( H 2 − H 1 ) + (u22 − u12 )
2
200 kg 1 min (314-3095) kJ
200 kg 1 min (1.752 -50.182 ) m 2
Q − Ws =
+
min 60 s
kg
2 min 60 s
s2
= −13, 460 kW ( ⇒ 13,460 kW transferred from the turbine)
7.34 (a) Assume all heat from stream transferred to oil
4
Q = 1.00 × 10 kJ 1 min = 167 kJ s
min 60 s
100 kg oil/min
135°C
m (kg H2O(v)/s)
25 bars, sat'd
d
Energy balance on H 2 O: Q = ΔH = m H out − H in
i
100 kg oil/min
185°C
m (kg H2O(l)/s)
25 bars, sat'd
ΔE p , ΔE k , W s = 0
H (l , 25 bar, sat' d ) = 962.0 kJ kg , H (v , 25 bar, sat' d ) = 2800.9 kJ kg (Table B.6)
m =
Q
H out − H in
=
−167 kJ
s
Time between discharges:
(b) Unit Cost of Steam:
kg
b962.0 − 2800.9gkJ = 0.091 kg s
1200 g
1s
1 kg
= 13 s discharge
discharge 0.091 kg 10 3 g
$1
10 Btu
b2800.9 − 83.9g kJ
0.9486 Btu
= $2.6 × 10 −3 / kg
kg
kJ
6
Yearly cost:
1000 traps 0.091 kg stream 0.10 kg last 2.6 × 10 −3$ 3600 s 24 h 360 day
trap ⋅ s
kg stream
kg lost
h
day
year
= $7.4 × 10 5 / year
7-16
7.35 Basis: Given feed rate
200 kg H2O(v)/h
10 bar, sat’d, H = 2776.2 kJ / kg
n 3 [ kg H 2 O(v) / h]
10 bar, 250oC, H = 2943 kJ / kg
n 2 [ kg H 2 O(v) / h]
10 bar, 300 o C, H = 3052 kJ / kg
Q(kJ
/ h)
H from Table B.6 (saturated steam) or Table B.7 (superheated steam)
Mass balance: 200 + n 2 = n 3
(1)
b g
b
g b
g
Energy balance: Q = ΔH = n 3 2943 − 200 2776.2 − n 2 3052 , Q in kJ h
ΔE K , ΔE p , W = 0
(a) n 3 = 300 kg h
(b) Q = 0
(1), (2)
(1)
( 2)
n 2 = 100 kg h
(2)
Q = 2.25 × 10 4 kJ h
n 2 = 306 kg h , n 3 = 506 kg h
7.36 (a) Tsaturation @ 1.0 bar = 99.6 °C ⇒ T f = 99.6 D C
H 2 O (1.0 bar, sat' d) ⇒ H l = 417.5 kJ / kg, H v = 2675.4 kJ / kg
H 2 O (60 bar, 250 D C) = 1085.8 kJ / kg
Mass balance: mv + ml = 100 kg
Energy balance: ΔH = 0
(1)
ΔE K , Q , ΔE p , W = 0
⇒ mv H v + ml H l − m1 H 1 = mv H v + ml H l − (100 kg)(1085.8 kJ / kg) = 0
(1,2)
ml = 70.4 kg, mv = 29.6 kg ⇒ y v =
(2)
29.6 kg vapor
kg vapor
= 0.296
100 kg
kg
(b) T is unchanged. The temperature will still be the saturation temperature at the given final
pressure. The system undergoes expansion, so assuming the same pipe diameter, ΔE k > 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 P<Pmin, all the output is vapor. If P>Pmax, 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.
b
g
Q = ΔH = 5,000 mol s ⋅
z
kJ mol
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 ⋅ K@284.2 K
nΔT
14 mols 2.4 K
b
gb g
800 J
=
b14 molsgb2.4 Kg = 24 J mol ⋅ K@331.2 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<1 since Pf < 3 P0 )
0.01286(1 – x ) mol (CH3 )2O
0.01286 x mol CH4
0.01286 x mol H2
0.01286 x mol CO
0.01286 mol
(CH 3)2 O
600°C, 350 mm Hg
b g
600°C
875 mm Hg
b
g
Total moles in tank at t = 2h = 0.01286 1 − x + 3x = 0.01286 1 + 2 x mol
Pf V
P0V
b.
n f RT
=
n0 RT
⇒
nf
n0
=
Pf
P0
⇒
b
0.01286 1 + 2 x
0.01286
g = 875 ⇒ x = 0.75 ⇒ 75% decomposed
350
References: C ( s ) , H 2 ( g ) , O 2 ( g ) at 25D C
substance
bCH g Obgg
CH bgg
H bg g
CObgg
3 2
4
2
nin
nout
H in
H out
(mol) ( kJ / mol)
( mol)
(kJ / mol)
0.01286
H1
H 1
0.25 × 0.01286
−
−
H 2
0.75 × 0.01286
−
−
H 3
0.75 × 0.01286
−
−
H
0.75 × 0.01286
bCH g O(g,600 C): H
D
1
3 2
= ( ΔH fo ) b CH
g
3 2O
+
= −118 kJ mol
LM
MN
4
z
B
given
873
298
C p dT
OP J 1 kJ
. + 62.40) kJ / mol
PQ mol × 10 J = (−18016
3
⎡ 600 ↓
⎤
CH 4 (g,600 C):Hˆ 2 = (ΔHˆ fo )CH4 + ⎢ ∫ C p dT ⎥ = −74.85 + 29.46 = −45.39 kJ mol
⎢⎣ 25
⎥⎦
Table B.2
D
B
Table B.8
H 2 (g,600 C): H 3 = H H 2 (600 C) = 16.81 kJ mol
D
D
Table B.1
Table B.8
CO(g,600D C): H 4 = ( ΔH fo ) CO
c.
Table B.8
B
B
+ H CO (600D C) = − 110.52 + 17.57 kJ mol = −92.95 kJ mol
For the reaction of parts (a) and (b), the enthalpy change and extent of reaction are :
ΔH = ∑ nout Hˆ out − ∑ nin Hˆ in = [ −1.5515 − (−1.5175) ] kJ = −0.0340 kJ
9-10
9.15 (cont’d)
ξ=
(n CH4 )out − (n CH4 )in
ν CH
=
4
0.75 × 0.01286
mol = 0.009645 mol
1
−0.0340 kJ
ΔH = ξ ΔHˆ r ( 600°C ) ⇒ ΔHˆ r ( 600°C ) =
= −3.53 kJ/mol
0.009645
b
g
b
g
ΔU r 600° C = ΔH r 600° C − RT [
∑ν
i
d.
= −3.53 kJ mol −
8.314 J
∑ν
−
gaseous
products
i
]
gaseous
reactants
1 kJ
(1 + 1 + 1 − 1)
873 K
mol ⋅ K 10 J
3
= −18.0 kJ mol
Q = ξ ΔUˆ r ( 600°C ) = (0.009645 mol)( − 18.0 kJ/mol) = −0.174 kJ (transferred from reactor)
9.16
a.
SO 2 (g) +
Basis :
1
O 2 (g) → SO 3 (g)
2
l00 kg SO 3
10 3 mol SO 3
min
80.07 kg SO 3
= 1249 mol SO 3 min
o
mol SO /min
0 2 /min),2450 C
n0 (molnSO
450°C
100% excess
air, 450o C
100% excess
i
n1 (mol nO1 2 mol
/min)O2 /min
3.76 n 1 mol O2 /min
3.76n1 (mol N 2 /min)
450°C
1249mol
molSO
SO3/min
/min
1249
3s
nn02 (mol
mol SO
SO22/min
/min)
O 2/min
nn3 mol
3 (mol O 2 /min)
3.76 n 1 mol O2
3.76n1 (mol
N / min)
550°C / i 2
o
550 C
m.w (kg H 2O( l) /h)
25°C
m.w (kg H 2O(l) /h)
40°C
Assume low enough pressure for H to be independent of P.
SO 3 balance :
n0 (mol SO 2 fed) 0.65 mol SO 2 react 1 mol SO 3 produced
bGeneration = output g
min
1 mol SO 2 fed
1 mol SO 2 react
= 1249
mol SO 3
min
⇒ n0 = 1922 mol SO 2 / min fed
100% excess air: n1 =
1922 mol SO 2
0.5 mol O 2 reqd
min
1 mol SO 2
b g
b
N 2 balance : 3.76 1922 = 7227 mol / min in & out
g
b1 + 1g mol O
2
fed
1 mol O 2 reqd
g
b2gb1922g + b2gb1922g = b3gb1249g + b2gb673g + 2n
= 1922 mol O 2 min fed
b
65% conversion : n2 = 1922 1 − 0.65 mol s = 673 mol SO 2 min out
O balance:
9-11
3
⇒ n3 = 1298 mol / min out
9.16 (cont’d)
.
b.
.
(n SO 2 ) out − (n SO 2 ) in
.
Extent of reaction : ξ =
ν SO 2
=
673 − 1922
= 1249 mol / min
1
B
Table B.1
ΔH ro
=
( ΔH fo ) SO3 ( g)
−
( ΔH fo ) SO 2 ( g)
= − 39518
. − ( −296.9) = −99.28 kJ / mol
bg bg bg
bg
References : SO 2 g , O 2 g , N 2 g , SO 3 g at 25D C
O2
N2
SO 3
SO 2 (g,450 C) : H 1 =
nout
H out
( mol / min) ( kJ / mol) ( mol / min) ( kJ / mol)
1922
H 1
H 4
673
H 2
H 5
1922
1298
H3
H 6
7227
7227
H 7
−
−
1249
SO 2
D
H in
nin
Substance
z
B
Table B.2
450
C p dT = 19.62 kJ / mol
25
B
Table B.8
O 2 (g,450 C) = H 2 = H O 2 (450 C) = 13.36 kJ / mol
D
D
B
Table B.8
N 2 (g,450 C) = H 3 = H N 2 (450D C) = 12.69 kJ / mol
D
Out :
Table B.2
SO 2 (g,550 C) : Hˆ 4 = ∫
D
550
25
↓
C p dT = 24.79 kJ/mol
B
Table B.8
O 2 (g,550 C) = H 5 = H O 2 (550 C) = 16.71 kJ / mol
D
D
B
Table B.8
N 2 (g,550 C) = H 6 = H N 2 (550D C) = 1581
. kJ / mol
D
SO 3 (g,550 C) : H 7 =
D
z
B
Table B.2
550
25
C p dT = 35.34 kJ / mol
Q = Δ H = ξΔHˆ ro + ∑ ni Hˆ i − ∑ ni Hˆ i
out
in
= (1249 )( −98.28 ) + ( 673)( 24.796 ) + (179.8 )(16.711) + ( 7227 )(15.808 ) + (1249 )( 35.336 ) − (192
− 1922 (13.362 ) − ( 7227 )(12.691)
=
c.
−8.111 × 104 kJ 1 min 1 kW
= −1350 kW
min 60 s 1 kJ/s
Assume system is adiabatic, so that Q lost from reactor = Q gained by cooling water
9-12
9.16 (cont’d)
LM
MM A e
N
j
e
Q = ΔH = m w H w l, 40D C − H w l, 25D C
A
Table B.5
d.
Table B.5
FG IJ
H K
OP
jP
PQ
kJ
kg
kJ
= m w
167.5 − 104.8
⇒ m w = 1290 kg min cooling water
min
min
kg
⇒ 8111
. × 104
If elemental species were taken as references, the heats of formation of each molecular species would
have to be taken into account in the enthalpy calculations and the heat of reaction term would not have
been included in the calculation of ΔH .
bg
CO(g) + H 2 O v → H 2 (g) + CO 2 (g) ,
9.17
B
Table B.1
e j
ΔH ro = ΔH fo
a.
CO 2 ( g)
e j
− ΔH fo
CO(g)
e j
− ΔH fo
.
= − 4115
bg
H 2O v
b g
kJ
mol
b g
. mol h
Basis : 2.5 m 3 STP product gas h 1000 mol 22.4 m 3 STP = 1116
nn01 (mol CO/h)
CO/h)
25oC
25°C
n 2 (mol H 2 O(v )/h)
150°C
n 3 (mol CO2 /h)
n 4 (mol H 2 /h)
n 5 (mol H 2 O(v)/h),sat'd
15°C, 1 atm
n 6 (mol H 2 O( l )/h)
15°C, 1 atm
111.6 mol/h
condenser
0.40 mol H 2/mol
0.40 mol CO 2/mol
0.20 mol H 2O( v)/h
500°C
reactor
Q r (kW)
Q c (kW)
b gb
g
= 1116
. b2gb0.40g + b2gb0.20g mol h ⇒ n
. mol h = 44.64 mol CO h
C balance on reactor : n1 = 0.40 1116
H balance on reactor : 2n2
2
Steam theoretically required =
% excess steam =
44.64 mol CO 1 mol H 2 O
h
1 mol CO
bg
= 66.96 mol H 2 O v h
= 44.64 mol H 2 O
b66.96 − 44.64g mol h × 100% = 50% excess steam
44.64 mol h
b gb
g
= b0.40gb1116
. mol hg = 44.64 mol H
CO 2 balance on condenser : n3 = 0.40 1116
. mol h = 44.64 mol CO 2 h
H 2 balance on condenser: n4
2
h
Saturation of condenser outlet gas:
n5 mol H 2 O h
p ∗ 15° C
12.788 mm Hg
yH 2O = w
⇒
=
⇒ n5 = 153
. mol H 2 O v h
p
44.64 + 44.64 + n5 mol h
760 mm Hg
b
g
b
g
b
gb g
H O balance on condenser: b111.6gb0.20gmol H O h = 153
. + n
2
2
6
⇒ n6 = 20.8 mol H 2 O h condensed = 0.374 kg / h
9-13
bg
9.17 (cont’d)
b. Energy balance on condenser
D
References : H 2 (g), CO 2 (g) at 25 C, H 2 O at reference point of steam tables
n in
n out
H in
H out
mol / h kJ / mol mol / h kJ / mol
H 1
H 4
44.64
44.64
H 2
H 5
44.64
44.64
.
H 3
H 6
22.32
153
−
−
H 7
20.80
Substance
CO 2 (g)
H 2 ( g)
H 2O v
H 2O l
bg
bg
Enthalpies for CO2 and H2 from Table B.8
CO 2 (g,500 D C) : H 1 = H CO 2 (500 D C) = 2134
. kJ / mol
H 2 (g,500 D C) : H 2 = H H 2 (500 D C) = 1383
. kJ / mol
FG
H
IJ
K
18 kg
kJ
×
H 2 O(v,500 D C) : H 3 = 3488
= 62.86 kJ mol
kg
10 3 mol
CO 2 (g,15D C) : Hˆ 4 = Hˆ CO2 (15D C) = −0.552 kJ/mol
H 2 (g,15D C) : H 5 = H H 2 (15D C) = −0.432 kJ / mol
FG
H
IJ
K
18.0 kg
kJ
H 2 O(v,15D C) : H 6 = 2529
= 4552
×
. kJ mol
kg
10 3 mol
FG
H
IJ
K
18.0 kg
kJ
H 2 O(l,15D C) : H 7 = 62.9 ×
= 113
. kJ mol
kg 10 3 mol
Q = ΔH =
∑ n H − ∑ n H
i
out
i
i
i
=
b49.22 − 29718. g kJ
h
in
bheat transferred from condenserg
c.
1h
1 kW
3600 s 1 kJ s
= −0.812 kW
Energy balance on reactor :
References : H 2 (g), C(s), O 2 (g) at 25° C
Substance
CO(g)
H 2 O( v )
H2 g
CO 2 g
bg
bg
nin
nout
H in
H out
( mol / h) ( kJ / mol) ( mol / h) ( kJ / mol)
44.64
H 1
−
−
66.96
H2
22.32
H 3
−
−
44.64
H 4
−
−
44.64
H 5
CO(g,25D C) : H 1 = ( ΔH fD ) CO
Table B.1
=
−110.52 kJ / mol
H 2 O(v,150 D C) : H 2 = ( ΔH fD ) H 2 O(v) + H H 2 O (150 D C)
9-14
Tables B.1, B.8
=
−237.56 kJ mol
9.17 (cont’d)
H 2 O(v,500 D C) : H 3 = ( ΔH fD ) H 2 O(v) + H H 2O (500 D C)
H 2 (g,500 D C) : H 4 = H H 2 (500 D C)
∑ n H − ∑ n H
i
out
i
i
in
i
=
bheat transferred from reactor g
d.
=
−224.82 kJ mol
Table B.8
=
1383
. kJ / mol
CO 2 (g,500 D C) : H 5 = ( ΔH Df ) CO 2 + H CO 2 (500 D C)
Q = ΔH =
Tables B.1, B.8
Tables B.1, B.8
=
−372.16 kJ / mol
1h
1 kW
−2101383
. − ( −20839.96) kJ
= −0.0483 kW
h
3600 s 1 kJ s
Benefits
Preheating CO ⇒ more heat transferred from reactor (possibly generate additional steam for plant)
Cooling CO ⇒ lower cooling cost in condenser.
9-15
9.18
b.
References : FeO(s), CO(g), Fe(s), CO 2 (g) at 25o C
Substance
FeO
CO
Fe
CO 2
Q = ξ ΔH ro +
∑n
nin
nout
H in
H out
( mol) ( kJ / mol) ( mol) ( kJ / mol)
1.00
0
n1
H 1
n0
H0
n2
H 2
n3
H 3
−
−
n
H
−
−
4
out H out
−
∑n
4
in H in
⇒ Q = ξ ΔH ro + n1 H 1 + n2 H 2 + n3 H 3 + n4 H 4 − n0 H 0
Fractional Conversion : X =
(100
. − n1 )
⇒ n1 = 1 − X
.
100
1 mol CO
(1 − n1 ) mol FeO consumed
= (1 − n1 ) mol CO
1 mol FeO consumed
⇒ n2 = n0 − (1 − n1 ) = n0 − X
CO consumed :
1 mol Fe
(1 − n1 ) mol FeO consumed
= (1 − n1 ) mol Fe = X
1 mol FeO consumed
1 mol CO 2
(1 − n1 ) mol FeO consumed
= (1 − n1 ) mol CO 2 = X
CO 2 produced : n4 =
1 mol FeO consumed
Fe produced : n3 =
Extent of reaction : ξ =
H i =
z
(nCO ) out − (nCO ) in
ν CO
=
n2 − n0
1
=X
T
C pi dT
for i = 0,1,2,3,4
25
H 0 = 0.02761 ( T0 − 298) + 2.51 × 10 −6 (T0 2 − 298 2 )
⇒ H = ( −8.451 + 0.02761 T + 2.51 × 10 −6 T 2 ) kJ / mol
0
0
0
× 10 −6 (T 2 − 298 2 ) + 3188
× 10 2 (1 / T − 1 / 298)
H 1 = 0.0528 (T − 298) + 31215
.
.
⇒ H = ( −17.0814 + 0.0528 T + 31215
× 10 −6 T 2 + 3188
× 10 2 / T ) kJ / mol
.
.
1
H 2 = (0.02761 (T − 298) + 2.51 × 10 −6 (T 2 − 298 2 )
⇒ H = −8.451 + 0.02761 T + 2.51 × 10 −6 T 2 ) kJ / mol
2
H 3 = 0.01728 (T − 298) + 1.335 × 10 −5 (T 2 − 298 2 )
.
⇒ H = ( −6.335 + 0.01728 T + 1335
× 10 −5 T 2 ) kJ / mol
3
H 4 = 0.04326(T − 298) + 0.573 × 10 −5 (T 2 − 298 2 ) + 818
. × 10 2 (1 / T − 1 / 298)
⇒ H = ( −16145
.
+ 0.04326 T + 0.573 × 10 −5 T 2 + 8.18 × 10 2 / T ) kJ / mol
4
9- 16
9.18 (cont'd)
c.
n0 = 2.0 mol CO, T0 = 350 K, T = 550 K, and X = 0.700 mol FeO reacted/mol FeO fed
⇒ n1 = 1 − 0.7 = 0.3, n2 = 2 − 0.7 = 1.3, n3 = 0.7, n4 = 0.7, ξ = 0.7
Summary : Hˆ 0 = 1.520 kJ/mol, Hˆ 1 = 13.48 kJ/mol, Hˆ 2 = 7.494 kJ/mol,
Hˆ 3 = 7.207 kJ/mol, Hˆ 4 = 10.87 kJ/mol
ΔHˆ o = −16.48 kJ/mol
r
Q = (0.7)(−16.48) + (0.3)(13.48) + (1.3)(7.494) + (0.7)(7.207) + (0.7)(10.87) − (2)(1.520)
⇒ Q = 11.86 kJ
d.
no
To
400
400
400
400
400
400
400
400
X T
1
298
1
400
1
500
1
600
1
700
1
800
1
900
1 1000
Xi
1
1
1
1
1
1
1
1
no
1
1
1
1
1
1
1
1
To
298
400
500
600
700
800
900
1000
X
1
1
1
1
1
1
1
1
T
700
700
700
700
700
700
700
700
Xi
no
To
400
400
400
400
400
400
400
400
400
400
400
X
T
0 500
0.1 500
0.2 500
0.3 500
0.4 500
0.5 500
0.6 500
0.7 500
0.8 500
0.9 500
1 500
Xi
X
0.5
0.5
0.5
0.5
1
1
1
1
1
1
1
1
1
1
1
no To
0.5 400
0.6 400
0.8 400
1.0 400
T
400
400
400
400
n1
0
0
0
0
0
0
0
0
n2 n3 n4
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
H0
2.995
2.995
2.995
2.995
2.995
2.995
2.995
2.995
H1
H2
0
5.335
10.737
16.254
21.864
27.555
33.321
39.159
0
2.995
5.982
9.019
12.11
15.24
18.43
21.67
H3
0
2.713
5.643
8.839
12.303
16.033
20.031
24.295
0
4.121
8.553
13.237
18.113
23.152
28.339
33.663
Q
-19.48
-12.64
-5.279
2.601
10.941
19.71
28.895
38.483
H0
0
0
0
0
0
0
0
0
n2 n3 n4
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
2.995
5.982
9.019
12.11
15.24
18.43
21.67
H1
21.864
21.864
21.864
21.864
21.864
21.864
21.864
21.864
H2
12.11
12.11
12.11
12.11
12.11
12.11
12.11
12.11
H3
12.303
12.303
12.303
12.303
12.303
12.303
12.303
12.303
H4
18.113
18.113
18.113
18.113
18.113
18.113
18.113
18.113
Q
13.936
10.941
7.954
4.917
1.83
-1.308
-4.495
-7.733
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
n2 n3 n4
1
0
0
0.9 0.1 0.1
0.8 0.2 0.2
0.7 0.3 0.3
0.6 0.4 0.4
0.5 0.5 0.5
0.4 0.6 0.6
0.3 0.7 0.7
0.2 0.8 0.8
0.1 0.9 0.9
0
1
1
H0
2.995
2.995
2.995
2.995
2.995
2.995
2.995
2.995
2.995
2.995
2.995
H1
10.737
10.737
10.737
10.737
10.737
10.737
10.737
10.737
10.737
10.737
10.737
H2
5.55
5.55
5.55
5.55
5.55
5.55
5.55
5.55
5.55
5.55
5.55
H3
5.643
5.643
5.643
5.643
5.643
5.643
5.643
5.643
5.643
5.643
5.643
H4
8.533
8.533
8.533
8.533
8.533
8.533
8.533
8.533
8.533
8.533
8.533
Q
13.72
11.82
9.92
8.02
6.12
4.22
2.32
0.42
–1.48
–3.38
–5.28
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Xi
0.5
0.5
0.5
0.5
n1
0.5
0.5
0.5
0.5
n2 n3 n4
0.0 0.5 0.5
0.1 0.5 0.5
0.3 0.5 0.5
0.5 0.5 0.5
H0
2.995
2.995
2.995
2.995
H1
5.335
5.335
5.335
5.335
H2
2.995
2.995
2.995
2.995
H3
2.713
2.713
2.713
2.713
H4
4.121
4.121
4.121
4.121
Q
-3.653
-3.653
-3.653
-3.653
1
1
1
1
1
1
1
1
n1
1
1
1
1
1
1
1
1
n1
9- 17
H4
9.18 (cont'd)
400
400
400
400
400
0.5
0.5
0.5
0.5
0.5
400
400
400
400
400
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.7
0.9
1.1
1.3
1.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
2.995
2.995
2.995
2.995
2.995
50
40
30
20
10
0
-10
-20
-30
5.335
5.335
5.335
5.335
5.335
2.995
2.995
2.995
2.995
2.995
2.713
2.713
2.713
2.713
2.713
10
5
0
-5
-10
500
1000
0
1500
500
0
-1
-3
Q
Q
-2
-4
-5
-6
0
0.2
1000
1500
T o (K )
T (K )
0.4
0.6
0.8
0
-0.5
-1
-1.5
-2
-2.5
-3
-3.5
-4
1
0
0.5
1
X
a.
-3.653
-3.653
-3.653
-3.653
-3.653
15
0
9.19
4.121
4.121
4.121
4.121
4.121
20
Q
Q
1.2
1.4
1.6
1.8
2.0
1.5
2
2.5
no
Fermentor capacity : 550,000 gal
Solution volume : (0.9 × 550,000) = 495,000 gal
R|0.071 lb C H OH / lb solution
Final reaction mixture : S0.069 lb (yeast, other species) / lb
|T0.86 lb H O / lb solution
m
2
5
m
m
2
Mass of tank contents :
495,000 gal
Mass of ethanol produced :
m
solution
m
1 ft 3
65.52 lb m
7.4805 gal
1 ft 3
= 4335593 lb m
4.336 × 106 lb m solution 0.071 lb m C 2 H 5OH
⇒
3.078 × 105 lb m C 2 H 5OH
⇒
307827 lb m C 2 H 5OH
lb m solution
1 lb - mole C 2 H 5OH
46.1 lb m C 2 H 5OH
= 6677 lb - mole C 2 H 5OH
1 ft 3 C 2 H 5OH
7.4805 gal
49.67 lb m C 2 H 5OH
1 ft 3
9- 18
= 3.078 × 105 lb m C 2 H 5
= 46,360 gal C 2 H 5OH
9.19 (cont’d)
Makeup water required : 495,000 gal −
b.
46,360 gal C 2 H 5OH
25 gal mash
2.6 gal C 2 H 5OH
= 4.9 × 104 gal
46,360 gal C2 H5OH
1 bu
1 acre 1 batch 24 h 330 days
acres
= 175
. × 105
Acres reqd. :
1 batch
2.6 gal C2 H5OH 101 bu 8 h 1 day 1 year
year
ΔH co = −56491
C12 H 22 O11 (s) + 12O 2 (g) → 12CO 2 (g) + 11H 2 O(l)
. kJ / mol
o
o
o
o
ΔH c = 12 ΔH f (CO 2 ) + 11ΔH f ( H 2 O) − ΔH f (C12 H 22 O11 )
⇒ ΔH fo (C12 H 22 O11 ) = −221714
. kJ / mol
C12 H 22 O11 (s) + H 2 O(l) → 4C 2 H 5OH(l) + 4CO 2 (g)
c.
ΔH ro = 4 ΔH fo (C 2 H 5OH) + 4 ΔH fo (CO 2 ) − ΔH fo (C12 H 22 O11 ) − ΔH fo ( H 2 O) =− 184.5kJ / mol
. kJ 453.6 mol 0.9486 Btu
−1815
⇒ ΔH ro =
= −7.811 × 104 Btu / lb - mole
1 mol
1 lb - mole
1 kJ
Moles of maltose :
4.336 × 106 lb m solution 0.071 lb C 2 H 5OH 1 lb - mole C 2 H 5OH 1 lb - mole C12 H 22 O11
1 lb m solution
46.1 lb C 2 H 5OH
4 lb - mole C 2 H 5OH
= 1669 lb - moles C12 H 22 O11 ⇒ ξ = nC10H22O11 = 1669 lb - moles
Q = ξΔH r + mC p (95D F - 85D F)
Btu
Btu
) + (4.336 × 106 lb m )(0.95 D )(10D F)
lb - mole
lb - F
7
= −8.9 × 10 Btu ( heat transferred from reactor)
= (1669 lb - moles)( −7.811 × 104
d.
Brazil has a shortage of natural reserves of petroleum, unlike Venezuela.
9.20
a.
4NH 3 + 5O 2 → 4NO + 6H 2 O,
3
O 2 → N 2 + 3H 2 O
2
References: N 2 g , H 2 g , O 2 (g), at 25° C
2NH 3 +
bg bg
nin
Substance
NH 3
Air
NO
H 2O
N2
O2
H i = ΔH foi +
b
H in
nout
H out
(mol min) (kJ mol) (mol min) (kJ mol)
100
H 1
−
−
H2
−
−
900
H 3
−
−
90
H
−
−
150
4
z
−
−
716
−
−
69
H 5
H
T
25
g
C pi dT
NH 3 g, 25° C : H 1 = ( ΔH fo ) NH 3
Table B.1
B
=
−46.19 kJ mol
9- 19
6
9.20 (cont’d)
b
g
b
g
Table B.8
B
Air g, 150° C : H 2
=
3.67 kJ mol
NO g, 700° C : H 3 = 90.37 +
b
Table B.1,
Table B.8
g
B
H 2 O g, 700° C : H 4
b
g
g
Q = ΔH =
25
111.97 kJ mol
−216.91 kJ mol
B
=
20.59 kJ mol
Table B.8
O 2 g, 700° C : H 6
b.
B
=
C p dT
Table B.8
N 2 g, 700° C : H 5
b
=
z
Table B.1,Table B.2
700
B
=
21.86 kJ mol
∑ n H − ∑ n H
i
i
out
i
i
= −4890 kJ min × (1 min / 60s) = −815
. kW
in
(heat transferred from the reactor)
9.21
c.
If molecular species had been chosen as references for enthalpy calculations, the extents of each
reaction would have to be calculated and Equation 9.5-1b used to determine ΔH . The value of Q
would remain unchanged.
a.
Basis: 1 mol feed
1 mol at 310°C
0.537 C2H4 (v)
0.367 H20 (v)
0.096 N2(g)
Products at 310°C
n1 (mol C2H4 (v))
n2 (mol H2O(v))
0.096 mol N2 (g)
n3 (mol C2H5OH (v))
n4 (mol (C2H5)2O) (v))
C 2 H 4 ( v) + H 2 O(v) ⇔ C 2 H 5OH(v)
b
g
2C 2 H 5OH(v) ⇔ C 2 H 5 2 O(v) + H 2 O(v)
b
gb g
5% ethylene conversion: 0.537 0.05 = 0.02685 mol C 2 H 4 consumed
b gb
g
⇒ n1 = 0.95 0.537 = 0.510 mol C 2 H 4
90% ethanol yield:
n3 =
0.02685 mol C 2 H 4 consumed 0.9 mol C 2 H 5OH
1 mol C 2 H 4
b gb
g b gb
g b gb
= 0.02417 mol C 2 H 5OH
g
b
g
C balance : 2 0.537 = 2 0.510 + 2 0.02417 + 4n4 ⇒ n4 = 1415
.
× 10 −3 mol C 2 H 5 2 O
.
× 10
O balance : 0.367 = n2 + 0.02417 + 1415
9- 20
−3
⇒ n2 = 0.3414 mol H 2 O
9.21 (cont'd)
References: C ( s ) , H 2 ( g ) , O2 ( g ) at 25D C, N 2 ( g ) at 310D C
Hˆ out
(mol)
(kJ/mol)
Hˆ
C2 H 4
(mol) (kJ/mol)
0.537
Hˆ
H2O
N2
0.367
0.096
Hˆ 2
0
0.3414
0.096
C2 H 5 OH
−
−
0.02417
0.510
1
( C2 H 5 )2 O
−
−
C2 H 4 ( g, 310°C ) : Hˆ 1 = (ΔHˆ fo )C2 H4 + ∫
C p dT
b.
g
bC H g Obg, 310° Cg: H = eΔH j
2
5 2
o
f
4
(C 2 H 5 )O(l)
z
310
25
Energy balance: Q = ΔH =
C p dT
b
⇒
Table B.1
Table B.8
⇒
Table B.1
Table B.2
g
+ ΔH v 25° C +
= −204.2 kJ mol
3
−3
⇒
g
C 2 H 5OH g, 310° C : H 3 = ( ΔH fo ) C2 H 5OH(g) +
Hˆ 2
0
Hˆ
Hˆ 4
Table B.1 for ΔHˆ fo
Table B.2 for Cp
H 2 O g, 310° C : H 2 = ( ΔH fo ) H 2 O(v) + H H 2 O(v) (310D C)
b
1
1.415 × 10
310
25
b
nout
Hˆ in
nin
substance
z
( 52.28 + 16.41) = 68.69
kJ mol
. + 9.93g = −23190
. kJ mol
b−24183
. kJ mol
b−235.31 + 24.16g = −21115
310
25
b
C p dT = −272.8 + 26.05 + 42.52
g
∑ n H − ∑ n H =− 1.3 kJ ⇒ 1.3 kJ transferred from reactor mol feed
i
i
out
i
i
in
To suppress the undesired side reaction. Separation of unconsumed reactants from products and
recycle of ethylene.
9.22
C 6 H 5CH 3 + O 2 → C 6 H 5CHO + H 2 O
C 6 H 5CH 3 + 9O 2 → 7CO 2 + 4H 2 O
Basis: 100 lb-mole of C 6 H 5CH 3 fed to reactor.
100 lb-moles C6 H5 CH 3
n0 (lb-moles O 2 )
3.76n 0 (lb-moles N 2 )
350°F, 1 atm
V0 (ft3 )
reactor
Q(Btu)
jacket
mw(lbm H2 O( l )), 80°F
Vp (ft3 ) at 379°F, 1 atm
n1 (lb-moles C6 H5 CH3 )
n2 (lb-moles O 2 )
3.76n0 (lb-moles N 2 )
n3 (lb-moles C 6 H5 CHO)
n4 (lb-moles CO2 )
n5 (lb-moles H2 O)
mw(lbm H2 O( l )), 105°F
Strategy:
All material and energy balances will be performed for the assumed basis of 100 lb-mole
C 6 H 5CH 3 . The calculated quantities will then be scaled to the known flow rate of water in
b
g
the product gas 29.3 lb m 4 h .
9- 21
9.22 (cont'd)
Plan of attack: % excess air ⇒ n0
Ideal gas equation of state ⇒ V0
13% C6 H 5CHO formation ⇒ n3 Ideal gas equation of state ⇒ V p
0.5% CO 2 formation ⇒ n4
E.B. on reactor ⇒ Q
C balance ⇒ n1
E.B. on jacket ⇒ mw
H balance ⇒ n5
O balance ⇒ n2
Scale V0 , V p , Q, mw by n5 actual / n5 basis
100% excess air:
n0 =
100 lb - moles C 6 H 5CH 3
b
g
b g
b1 + 1gmole O
1 mol O 2 reqd
1 mole C 6 H 5CH 3
2
fed
b g
b g
= 200 lb - moles O 2
1 mol O 2 reqd
N 2 feed & output = 3.76 200 lb - moles N 2 = 752 lb - moles N 2
13% → C6H5CHO ⇒ n3 =
100 lb-moles C6H5CH3 0.13 mole C6H5CH3 react 1 mole C6H5CHO formed
1 mole C6H5CH3 fed
1 mole C6H5CH3 react
= 13 lb-moles C6H5CHO
0.5% → CO2 ⇒ n4 =
b100gb0.005glb - moles C H CH
6
5
3
react
7 moles CO2
= 35
. lb - moles CO2
1 mole C6 H5CH 3
a100faB7f lb - moles C = 7n + a13fa7f + a3.5fa1f ⇒ n = 86.5 lb - moles C H CH
b100gb8glb - moles H = b86.5gb8g + b13gb6g + 2n ⇒ n = 15.0 lb - moles H Obvg
b200gb2glb - moles O = 2n + b13gb1g + b35. gb2g + b15gb1g ⇒ n = 182.5 lb - moles O
mol C mole C7H8
C balance:
H balance:
O balance:
1
5
b100 + 200 + 752glb - moles
C7 H 8
Vp
O2
C 7 H 8O
CO 2
H 2O
N2
IJ
K
13 + 35
. + 15 + 752 lb - moles
D
492 D R
359 ft 3
1 lb - mole
9- 22
3
2
b g b350 + 460g R = 6.218 × 10
359 ft 3 STP
Ideal gas law – outlet:
5
2
2
1 lb - moles
FG 86.5+ 182.5+
=H
6
5
2
Ideal gas law − inlet:
V0 =
1
5
ft 3
b379 + 460g R = 6.443 × 10
D
492 D R
5
ft 3
9.22 (cont'd)
Energy balance on reactor (excluding cooling jacket)
bg b g b g b g
e
j
References : C s , H 2 g , O 2 g , N 2 g at 25D C 77 D F
nin
H in
nout
H out
C 6 H 5CH 3
O2
N2
100
200
752
H 1
H 2
H
86.5
182.5
752
H 4
H 5
H
C 6 H 5CHO
CO 2
H 2O
−
−
−
−
−
−
13
.
35
15
H 7
H 8
H
substance
blb - molesg bBtu lb - moleg blb - molesg bBtu lb - moleg
3
6
9
Enthalpies:
LB
O
430.28 Btu lb - mole
Btu
C H CH (g,T): H bT g = M ΔH b kJ molg ×
+ 31
T − 77g FP
b
MM
PP
1 kJ mol
1b - mole⋅° F
N
Q
Table B.1
6
5
D
o
f
3
C 6 H 5CH 3 (g,350D F): H 1 = 2.998 × 10 4 Btu lb - mole
C 6 H 5CH 3 (g,379 D F): H 4 = 3.088 × 104 Btu lb - mole
bg
b
g
D
C 6 H 5CHO(g,T): H T = −17200 + 31 T − 77 F Btu lb - mole
⇒ H 7 = −7.83 × 103 Btu lb - mole
B
Table B.9
e
D
e
D
e
D
e
D
j
O 2 g,350 F : H 2 = H O 2 (350 F) = 1972
.
× 103 Btu / lb − mole
D
B
Table B.9
j
.
N 2 g,350 F : H 3 = H N 2 (350D F) = 1911
× 103 Btu / lb − mole
B
Table B.9
j
O 2 g,379 F : H 5 = H O 2 (379 D F) = 2.186 × 103 Btu / lb − mole
B
Table B.9
j
N 2 g,379 F : H 6 = H N 2 (379 F) = 2.116 × 103 Btu / lb − mole
D
B
Table B.1 and B.9
e
j
H Obg,379° Fg: H
CO 2 g,379 F : H 8 =
D
2
( ΔH fD ) CO 2 ( g)
+ H CO 2 (379 F)
D
=
B
− 1664
× 105 Btu / lb − mole
.
Table B.1 and B.9
9
= ( ΔH fD ) H 2 O( g) + H H 2 O (379 D F)
=
− 1016
× 105 Btu / lb − mole
.
Energy Balance :
Q = ΔH =
∑ n H − ∑ n H
i
i
out
i
i
= −2.376 × 106 Btu
in
Energy balance on cooling jacket:
Q = ΔH = mw
⇓
z
105
80
dC i
b g dT
p H O l
2
Q = + 2.376 × 10 4 Btu , C p = 1.0 Btu (lb m ⋅ D F)
b g
2.376 × 10 6 Btu = mw lb m × 1.0
b
g
bg
Btu
D
× 105 − 80 F ⇒ mw = 9.504 × 10 4 lb m H 2 O l
lb m ⋅° F
9- 23
9.22(cont’d)
bn g
bn g
5 actual
Scale factor:
=
29.3 lb m H 2 O
5 basis
a.
d
id
i
= d6.443 × 10 ft id0.02711 h i = 175
. × 10
5
d
3
−1
d
a.
= 0.02711 h −1
b g
h b product g
id
i
id
i
4
ft 3
Q = −2.376 × 10 6 Btu 0.02711 h −1 = −6.44 × 10 4 Btu / h
w = 9504
m
. × 104 Btu 002711
.
h −1 =
9.23
1
18.016 lb m H 2 O 15.0 lb - moles H 2 O
V0 = 6.218 × 10 5 ft 3 0.02711 h −1 = 169
. × 10 4 ft 3 h feed
Vp
b.
1b - mole H 2 O
4h
1 ft 3
2577 lb m
h
7.4805 gal
62.4 lb m
1 ft
3
1h
. gal H 2 O min
= 515
60 min
CaCO 3 (s) → CaO(s) +CO 2 (g)
CaO(s)
900°C
CaCO3(s)
25°C
CO2(g)
900°C
Q (kJ)
10.0 kmol CaO(s) produced
1 mol
Basis : 1000 kg CaCO 3 =
= 10.0 kmol CaCO 3 ⇒ 10.0 kmol CO 2 (g) produced
0.100 kg
10.0 kmol CaCO 3 (s) fed
References: Ca(s), C(s), O2(g) at 25°C
1000 kg
Hˆ in
nin
Hˆ out
nout
Substance (mol) (kJ/mol) (mol) (kJ/mol)
CaCO3
104
Hˆ 1
−
−
4
CaO
10
−
−
Hˆ
2
−
CO 2
−
10
Hˆ 3
4
Table B.1
B
CaCO 3 (s, 25 C) : H 1 = ( ΔH fo ) CaCO 3 ( s)
=
o
z
− 1206.9 kJ / mol
Table B.1,
Table B.2
1173
CaO(s, 900o C) : H 2 = ( ΔH fo ) CaO( s) +
B
=
C p dT
( −635.6 + 48.54) kJ / mol = −587.06 kJ / mol
298
Table B.1,
Table B.8
CO 2 (g, 900o C) : H 3 = ( ΔH fo ) CO 2 ( g) + H CO 2 (900o C)
Energy balance: Q = ΔH =
B
=
F n H − n H I = 2.7 × 10
GH ∑ ∑ JK
i
out
i
i
in
9- 24
i
( −3935
. + 42.94) kJ / mol = −350.56 kJ / mol
6
kJ
9.23 (cont'd)
b. Basis : 1000 kg CaCO3 fed ⇒ 10.0 kmol CaCO3
CaCO 3 (s) → CaO(s) + CO 2 (g)
2CO + O 2 → 2CO 2
10 kmol CaCO3
25 oC
200 kmol at 900oC
0.75 N2
0.020 O2
0.090 CO
0.14 CO2
Product gas at 900oC
n2 (kmol CO2 )
n3 (kmol N2 )
n4 (kmol CO)
n1 [kmol CaO(s)]
10 kmol CaCO 3 react ⇒ n1 = 10.0 kmol CaO
n2 = (014
. )(200) +
10.0 kmol CaCO 3 react 1 kmol CO 2 4 kmol O 2 react 2 kmol CO 2
+
= 46 kmol CO 2
1 kmol O 2
1 kmol O 2
n3 = (0.75)(200) = 150 kmol N 2
C balance: (10.0)(1) + (200)(0.09)(1) + (200)(0.14)(1) = 46(1) + n4 (1) ⇒ n4 = 10.0 kmol CO
References : Ca(s), C(s), O 2 (g), N 2 (g) at 25D C
Hˆ in
nin
Hˆ out
nout
Substance (mol) (kJ/mol) (mol) (kJ/mol)
CaCO3
10.0
Hˆ 1
−
−
CaO
10
−
−
−587.06
CO 2
28
46
−350.56
−350.56
CO
18
10
Hˆ 2
Hˆ 2
O
4.0
Hˆ
−
−
2
3
N2
Hˆ 4
150
CaCO3 (s, 25 C) : Hˆ 1 = (ΔHˆ fo ) CaCO3 (s)
o
Hˆ 4
150
Table B.1
↓
= − 1206.9 kJ/mol
CO(g, 900o C) : Hˆ 1 = (ΔHˆ fo )CO(g) + Hˆ CO (900o C)
Table B.1,
Table B.8
↓
=
(−110.52 + 27.49) kJ/mol = −83.03 kJ/mol
Table B.8
↓
O 2 (g, 900 C) : Hˆ 2 = Hˆ O2 (900o C) = 28.89 kJ/mol
o
Table B.8
↓
N 2 (g, 900 C) : Hˆ 3 = Hˆ N2 (900 C) = 27.19 kJ/mol
o
Q = ΔH =
o
F n H − n H I = 0.44 × 10
GH ∑ ∑ JK
i
out
i
i
i
kJ
in
% reduction in heat requirement =
c.
6
2.7 × 106 − 0.44 × 106
2.7 × 106
× 100 = 838%
.
The hot combustion gases raise the temperature of the limestone, so that less heat from the outside
is needed to do so. Additional thermal energy is provided by the combustion of CO.
9- 25
9.24
a.
A+B→ C
2C → D + B
(1)
(2)
Basis: 1 mol
x AO (mol A / mol)
n A (mol A)
x BO (mol B / mol)
n B (mol B)
x IO (mol I / mol)
nC (mol C)
n D (mol D)
n I (mol I)
T ( D C)
Fractional conversion:
C generated: n0 =
fA =
mol A consumed x AO − n A
=
⇒ n A = x AO (1 − f A )
mol A feed
x AO
x A0 (mol A fed)
f A (mol A consumed) YC (mol C generated)
mol A fed
mol A consumed
⇒ nC = x AO f A YC
D generated: nD = 0.5 × mol C consumed = (1 2) × (mol A consumed − mol C out)
⇒ nD = (1 2)( x AO f A − nC )
Balance on B: mol B out = mol B in − mol B consumed in (1) + mol B generated in (2)
= mol B in − mol A consumed in (1) + mol D generated in (2)
⇒ n B = x BO − x AO f A + n D
Balance on I: mol I out = mol I in ⇒ n I = x IO
b.
c.
Species Formula
A
C2H4(v)
B
H2O(v)
C2H5OH(v)
C
D
C4H10)O(v
I
N2(g)
Tf
310
Tp
310
Species
A
B
C
D
I
n(in)
(mol)
0.537
0.367
0
0
0.096
Q(kJ) =
-1.31
DHf
52.28
-241.83
-235.31
-246.75
0
xA0
0.537
H(in)
(kJ/mol)
68.7
-231.9
-211.2
-204.2
9.4
a
0.04075
0.03346
0.06134
0.08945
0.02900
b
1.15E-04
6.88E-06
1.57E-04
4.03E-04
2.20E-05
xB0
0.367
xI0
0.096
n(out)
(mol)
0.510
0.341
0.024
0.001
0.096
c
-6.89E-08
7.60E-09
-8.75E-08
-2.24E-07
5.72E-09
fA
0.05
d
1.77E-11
-3.59E-12
1.98E-11
0
-2.87E-12
YC
0.90
H(out)
(kJ/mol)
68.7
-231.9
-211.2
-204.2
9.4
For T f = 125o C, Q = 7.90 kJ . Raising Tp, lowering fA, and raising YC all increase Q.
9- 26
9.25
a.
CH 4 ( g) + O 2 ( g) → HCHO(g) + H 2 O(g)
n3 (mol HCHO)
n4 (mol H2O)
10 L, 200 kPa
n0 (mol feed gas) at 25°C
0.851 mol CH4/mol
0.15 mol O2 /mol
n5 (mol CH4)
T (°C), P(kPa), 10L
Q (kJ)
Basis : n0 =
200 kPa 1000 Pa 10 L 10 −3 m 3
1 kPa
1L
1 mol K
8.314 m 3 Pa
298 K
= 0.8072 mol feed gas mixture
0.8072 mol feed gas mixture ⇒ (0.85)(0.8072) = 0.6861 mol CH 4 ,
⇒ (0.15)(0.8072) = 0.1211 mol O 2
CH 4 consumed :
1 mol CH 4
0.1211 mol O 2 fed
= 01211
.
mol CH 4
1 mol O 2 fed
⇒ n5 = (0.6861 − 01211
.
) mol CH 4 = 0.5650 mol CH 4
HCHO produced : n3 =
H 2 O produced : n4 =
1 mol HCHO
01211
.
mol CH 4 consumed
1 mol CH 4 consumed
1 mol H 2 O
01211
.
mol CH 4 consumed
1 mol CH 4 consumed
Extent of reaction : ξ =
( nO 2 ) out − ( nO 2 ) in
ν O2
=
.
0 − 01211
1
= 01211
.
mol HCHO
.
= 01211
mol H 2 O
= 01211
.
mol
References : CH 4 (g), O 2 (g), HCHO(g), H 2 O(g), at 25o C
Substance
CH 4
z
T
U i =
25
U in
nin
nout
mol kJ mol mol kJ mol
0.6861
0
0.5650
U 1
O2
01211
.
0
−
HCHO
H 2O
−
−
−
−
01211
.
01211
.
z
U out
−
U 2
U 3
T
(Cv ) i dT =
(C p − R) i dT
i = 1,2,3
25
Using (C p ) i from Table B.2 and R = 8.314 × 10 −3 kJ / mol ⋅ K:
× 10 −8 T 3 − 2.75 × 10 −12 T 4 − 0.6670) kJ / mol
.
U 1 = (0.02599 T + 2.7345 × 10 −5 T 2 + 01220
U 2 = (0.02597 T + 2.1340 × 10 −5 T 2 − 2.1735 × 10 −12 T 4 − 0.6623) kJ / mol
U = (0.02515 T + 0.3440 × 10 −5 T 2 + 0.2535 × 10 −8 T 3 − 0.8983 × 10 −12 T 4 − 0.6309) kJ / mol
3
9- 27
9.25 (cont’d)
Q=
100 J 85 s
s
1 kJ
1000 J
= 8.5 kJ
Table B.1
ΔH ro = (ΔH fo ) HCHO + (ΔH fo ) H2O − (ΔH fo ) CH4
B
=
. ) + (−24183
. ) − (−74.85)g kJ / mol
b(−11590
= −282.88 kJ / mol
ΔU ro = ΔH ro − RT (
∑
νi −
gaseous
products
= −282.88 kJ / mol −
∑ν
i
)
gaseous
reactants
298 K (1 + 1 − 1 − 1)
8.314 J
1 kJ
10 3 J
mol K
= −282.88 kJ / mol
Energy Balance :
Q = ξΔU ro +
∑ (n
i ) out (U i ) out
−
∑ (n
i ) in (U i ) in
= (0.1211)(−28288
. kJ / mol) +0.5650 U 1 + 01211
.
U 2 + 01211
.
U 3
Substitute for U 1 through U 3 and Q
0 = 0.02088 T + 1845
.
× 10 −5 T 2 + 0.09963 × 10 −8 T 3 − 1926
.
× 10 −12 T 4 − 43.29 kJ / mol
Solve for T using E - Z Solve ⇒ T = 1091o C = 1364 K
⇒ P = nRT / V =
0.8072 mol 8.314 m 3 ⋅ Pa 1364 K
1L
= 915 × 10 3 Pa = 915 kPa
mol ⋅ K
10 L 10 −3 m3
b.
Add heat to raise the reactants to a temperature at which the reaction rate is significant.
c.
Side reaction : CH 4 + 2O2 → CO2 + 2H 2 O. T would have been higher (more negative heat of
reaction for combustion of methane), volume and total moles would be the same, therefore
P = nRT / V would be greater.
9- 28
9.26
bg
a.
bg
1
O 2 (g) → C 2 H 4 O g
2
C 2 H 4 + 3O 2 → 2CO 2 + 2H 2 O
C2 H 4 g +
Basis: 2 mol C 2 H 4 fed to reactor
n 6 (mol CO2 )
n 7 (mol H 2 O(l ))
25°C
Qr (kJ)
heat
n1 (mol C 2H 4)
n2 (mol O 2 )
25°C
reactor
2 mol C2 H4
1 mol O2
450°C
n 3 (mol C 2H 4)
n 4 (mol O 2 )
separation
n 3 (mol C 2H 4)
process
n 4 (mol O 2 )
n 5 (mol C 2H 4O)
n 6 (mol CO2 )
n 7 (mol H 2 O)
450°C
n 5 (mol C 2H 4O(g))
25°C
. mol C 2 H 4
25% conversion ⇒ 0.500 mol C 2 H 4 consumed ⇒ n 3 = 150
70% yield ⇒ n5 =
0.500 mol C 2 H 4 consumed 0.700 mol C 2 H 4 O
b gb g b gb g b gb
g
1 mol C 2 H 4
= 0.350 mol C 2 H 4 O
. + 2 0.350 + n 6 ⇒ n 6 = 0.300 mol CO 2
C balance on reactor: 2 2 = 2 150
Water formed: n 7 =
0.300 mol CO 2
1 mol H 2 O
1 mol CO 2
b gb g
= 0.300 mol H 2 O
b gb g
= n + 2n = 0.300 + b2gb0.350g ⇒ n = 0.500 mol C H
= 2n + n + n = b2gb0.300g + b0.300g + b0.350g ⇒ n = 0.625 mol O
O balance on reactor: 2 1 = 2n 4 + 0.350 + 2 0.300 + 0.300 ⇒ n 4 = 0.375 mol O 2
Overall C balance: 2n1
Overall O balance: 2n 2
6
5
6
1
7
2
5
Feed stream: 44.4% C 2 H 4 , 55.6% O 2
4
2
Reactor inlet: 66.7% C 2 H 4 , 33.3% O 2
Recycle stream: 80.0% C 2 H 4 , 20.0% O 2
Reactor outlet: 53.1% C 2 H 4 , 13.3% O 2 , 12.4% C 2 H 4 O, 10.6% CO 2 , 10.6% H 2 O
Mass of ethylene oxide =
b.
0.350 mol C 2 H 4 O 44.05 g
1 mol
bg
1 kg
10 3 g
bg bg
= 0.0154 kg
References for enthalpy calculations : C s , H 2 g , O 2 g at 25° C
bg
H i T = ΔH ofi +
z
z
= ΔH 0f +
T
25
C p dT for C 2 H 4
T + 273
298
C p dT for C 2 H 4 O
= ΔH ofi + H i ( table B.8)
bg
= ΔH of for H 2 O l
bg
for O 2 , CO 2 , H 2 O g
9- 29
2
9.26 (cont’d)
Overall Process
H in
nin
Substance
Reactor
H out
nout
(mol) (kJ / mol) (mol) (kJ / mol)
−
−
0.500
52.28
C2 H4
−
O2
1
−
− 0350
.
−5100
.
C2 H 4 O
CO2
−
− 0300
.
−3935
.
H2 O l
−
− 0300
.
−28584
.
.
0625
C2 H4 O
0
Energy balance on process: Q = ΔH =
.
1337
−
− 0.350
−19.99
CO2
−
− 0.300
−374.66
H2O g
−
− 0.300
−226.72
∑ n H − ∑ n H
i
i
out
Energy balance on reactor: Q = ΔH =
H out
0.375
bg
i
nout
(mol) (kJ / mol) (mol) ( kJ / mol)
.
2
79.26
150
79.26
C2 H 4
−
O2
bg
.
1337
= −248 kJ
i
in
∑ n H − ∑ n H
i
out
c.
H in
nin
substance
i
i
i
= −236 kJ
in
Scale to 1500 kg C 2 H 4 O day :
C 2 H 4 O production for initial basis = (0.350 mol)(
⇒ Scale factor =
44.05 kg
10 3 mol
) = 0.01542 kg C 2 H 4 O
1500 kg day
= 9.73 × 10 4 day −1
0.01542 kg
U|
V| M = b0.500gb28.05 g C H molg + b0.625gb32.0 g O molg
W = 34.025 × 10 kg
kgje9.73 × 10 day j = 3310 kg day (44.4% C H , 55.6% O )
day j 1 day 1 hr
1 kW
= −279 kW
In initial basis, fresh feed contains
0.500 mol C 2 H 4
0.625 mol O 2
2
4
2
−3
e
Fresh feed rate = 34.025 × 10 −3
Qprocess =
b−248 kJ ge9.73 × 10
Qreactor =
b−236 kJ ge9.73 × 10
4
4
−1
2
−1
24 hr 3600 s 1 kJ s
4
day −1
j
1 day
1 hr
1 kW
24 hr 3600 s 1 kJ s
9- 30
= −265 kW
4
2
9.27
a.
1200 lb m C 9 H 12 1 lb - mole
= 10.0 lb - moles cumene produced h
h
120 lb m
Overall process :
Basis:
n 1 (lb-moles/h)
0.75 C 3H 6
0.25 C 4H 10
n 3 (lb-moles C 3 H 6 /h)
n 4 (lb-moles C 4 H10 /h)
n 2 (lb-moles C 6 H 6 /h)
10.0 lb-moles C9 H12 /h
bg
bg
bg
b
Benzene balance: n 2 =
10.0 lb - moles C 9 H12 produced 1 mole C 6 H 6 consumed
b input = consumption g
=
h
1 mole C 9 H12 produced
10.0 lb - moles C 6 H 6
78.1 lb m C 6 H 6
h
1 lb - mole
Propylene balance: 0.75n1 = n3 +
b input = output + consumption g
Mass flow rate of C 3 H 6 / C 4 H 10
1 mole C 3 H 6
h
1 mole C 9 H12
UV ⇒ n = 16.67 lb - moles h
n = 2.50 lb - moles C H h
b gW
b0.75gb16.67glb - moles C H 42.08 lb C H
feed =
1
3
4
10
Reactor :
6
m
3
6
1 lb - mole
58.12 lb m C 4 H 10
= 768 lb m h
1 lb - mole
a3 + 1fmoles fed to reactor = 40 lb - moles C H
10.0 lb - moles fresh feed
h
6
1 mole fresh feed
16.67 lb-moles/h @ 77oF
0.75 C3H6
0.25 C4H10
10.0 lb-moles C9H12/h
2.50 lb-moles C3H6/h
4.17 lb-moles C4H10/h
30.0 lb-moles C6H6/h
400oF
40.0 lb-moles C6H6/h
b.
6
h
h
Overhead from T1 ⇒
3
3
b0.25gb16.67glb - moles C H
Benzene feed rate =
= 781 lb m C 6 H 6 h
10.0 lb - moles C 9 H12
⇒ 0.75n1 = n3 + 10
20% C 3 H 6 unreacted⇒ n3 = 0.20 0.75n1
+
g
ΔH r 77° F = −39520 Btu lb - mole
C 3 H 6 l + C 6 H 6 l → C 9 H 12 l ,
6.67 lb - moles h
UV ⇒ 37.5%
C H
hW
62.5% C H
2.50 lb - moles C 3 H 6 h
4.17 lb - moles C 4 H 10
Heat exchanger :
Reactor effluent at 400°F
10.0 lb-moles C9H12 /h
2.50 lb-moles C 3H6 /h
4.17 lb-moles C 4H10 /h
30.0 lb-moles C6H6 /h
200°F
40.0 lb-moles C6H6 /h
77°F
T (°F)
9- 31
3
6
4
10
46.7 lb-moles/h
21.4% C9H12
5.4% C3H6
8.9% C4H10
64.3% C6H6
6
h
9.27 (cont'd)
Energy balance: ΔH = 0 ⇒
∑ n e H
i
i , out
j ∑ n C bT
− H i , in =
i
pi
out
− Tin
g
=0
i
(Assume adiabatic)
LM10 lb - moles C H
h
N
9
C 4 H10
B
b gb
C 3H 6
OPe
Q
B
120 lb m
0.40 Btu
200D F − 400D F + 2.50 42.08 0.57 200D F − 400
D
1 lb - mole 1b m ⋅ F
12
gb ge
j b A gb
j b gb
gb ge
gb ge
j
+ 4.17 5812
.
. 0.45 200D F − 400D F
0.55 200D F − 400D F + 30.0 7811
b A gb
C6H6 in
effluent
gb ge
j
+ 40.0 7811
. 0.45 T − 77 D F = 0 ⇒ T = 323° F
C6H6 fed
to reactor
(Refer to flow chart of Part b: T = 323° F )
References : C 3 H 6 l , C 4 H 10 l , C 6 H 6 l , C 9 H 12 l at 77° F
H i Btu lb - mole = C pi Btu lb m ⋅° F M i lb m lb - mole T − 77 ° F
b
g
bg
bg
b
H in
n in
Substance
bg
g b
bg
n out
gb
gb g
H out
(lb - mole / h) (Btu / lb - mole) (lb - mole / h) (Btu / lb - mole)
12.0
0
2.50
7750
4.17
0
4.17
10330
40.0
8650
30.0
11350
−
−
10.0
15530
C3H 6
C 4 H10
C6H6
C 9 H12
Energy balance on reactor :
n C H ΔH ro
Q = ΔH = 9 12
+
n i H i −
vC 9 H12
out
∑
∑ n H
i
i
in
b10.0gb−39520g + b2.50gb7750g + b4.17gb10330g + b30.0gb11350g + b10.0gb15530g
b1g
−b40.0gb8650g = −183000 Btu h b heat removalg
=
9.28
Basis :
a.
100 kg C8 H 8 10 3 g
h
1 kg
1 mol
104.15 g
= 960 mol h styrene produced
C8 H 10 (g) → C8 H 8 (g) + H 2 (g)
Overall system
n 2 (mol H2 /h)
Fresh feed
n 1 (mol C8H10/h)
960 mol C8H8 /h
Fresh feed rate: n1 =
bC H
8
10
balance
g
H 2 balance : n 2 =
960 mol C 8 H 8 1 mol C 8 H10
h
960 mol C 8 H10
h
= 960 mol C 8 H10 h fresh feed
1 mol C 8 H 8
1 mol H 2
= 960 mol H 2 h
1 mol C 8 H10
9- 32
9.28 (cont'd)
Reactor :
.
n 3 (mol C8H10 /h)
n 4 (mol H2O( v )/h)
600°C
n5 (mol C8H10 /h)
n 4 (mol H2O(v)/s)
v
960 (mol C8H8 /s)
960 (mol H2 /s)
560°C
Qc (kJ/h)
b
0.35n3 mol C8 H 10 react
h
⇒ n3 = 2740 mol C 8 H10 h fed to reactor
35% 1-pass conversion ⇒
a
g
1 mol C8 H 8
= 960 mol C8 H 8 h
1 mol C8 H 10
f
⇒ Recycle rate = 2740 − 960 = 1780 mol C 8 H10 h recycled
Reactor feed mixing point
2740 mol C8H10(v)/h
500oC
2740 mol C8H10(v)/h
n4 [mol H2O(v)/h]
600oC
n4 [mol H2O(v)/h]
700oC
b g
Energy balance: ΔH = 2740 ΔH C H + n 4 ΔH H O = 0 kJ h
8 10
2
b Neglect Q, ΔE g
k
ΔH C8 H10 =
ΔH H 2 O
LM
OP J 1 kJ
MM b118 + 0.30T gdT PP mol ⋅ C × 10 J = 28.3 kJ mol
N
Q
z
600
Table B.8
⇒
D
500
P =1 bar
= −3.9 kJ mol
a2740fa28.3f + n a−3.9f = 0 ⇒ n
Ethylbenzene preheater bA g :
4
b.
3
Cp
4
= 1.99 × 10 4 mol H 2 O / h
bg
960 mol fresh feed 1780 mol recycled 2740 mol EB l
at 25° C
+
=
h
h
h
2740 mol EB v
at 500° C
⇒
h
136
500
ΔH = C pi dT + ΔH v 136° C +
C pv dT = 20.2 + 36.0 + 77.7 kJ mol = 133.9 kJ mol
bg
z
a
25
2740 mol C 8 H10
Q A = ΔH =
h
Steam generator F :
f
z
a
136
133.9 kJ
mol C 8 H10
f
b
= 3.67 × 10 5 kJ h preheater
bg
19400 mol h H Obl, 25° Cg → 19400 mol h H Ob v, 700° C, 1 atmg
Table B.5 ⇒ H bl, 25° Cg = 104.8 kJ kg ;
Table B.7 ⇒ H b v, 700° C, 1 atm ≈ 1 bar g = 3928 kJ kg
2
2
9- 33
g
9.28 (cont'd)
19400 mol H 2 O 18.0 g 1 kg
Q F = ΔH =
h
1 mol 10 3 g
= 1.34 × 10
b
6
kJ h steam generator
a3928 − 104.8fkJ
kg
g
bg
Reactor C :
bg
bg bg
bg
References: C8 H 8 v , C8 H 10 v , H 2 g , H 2 O v at 600° C
e
j
H i 560D C =
zd
560
600
i
C pv i dT for C8 H 10 , C8 H 8
≈ H (T) for H 2 , H 2 O (interpolating from Table B.8)
Substance
C 8 H10
H 2O
C8H8
H2
H in
n in
n out
H out
(mol h ) (kJ mol) (mol h ) (kJ mol)
2740
0
1780
−11.68
19900
0
19900
−1.56
960
−
−
−10.86
.
960
−
−
−119
Energy balance :
960 mol C 8 H 8 produced
124.5 kJ
Q c = ΔH =
+
h
1 mol C 8 H 8
a
= 5.61 × 10 4 kJ h reactor
c.
i
out
i
i
in
H2 +
1
O2 → H 2O
2
CH 3 OH
O2 , N 2
H2
product gas
145°C
separation
units
a.
n f (mol/h) at 145°C, 1 atm
0.42 mol CH 3OH/mol
0.58 mol air/mol
0.21 mol O2 /mol air
0.79 mol N2 /mol air
n s mol H2 O(v )/h
saturated at 145°C
b.
i
This is a poorly designed process as shown. The reactor effluents are cooled to 25D C , and
then all but the hydrogen are reheated after separation. Probably less cooling is needed, and
in any case provisions for heat exchange should be included in the design.
CH 3OH → HCHO + H 2 ,
9.29
f
∑ n H − ∑ n H
reactor
reactor
product gas, 600°C
n 1 (mol CH3 OH/h)
n 2 (mol O 2 /h)
waste
n 3 (mol N 2 /h)
heat
boiler
n 4 (mol HCHO/h)
0.37 kg HCHO/h
n 5 (mol H 2 /h)
0.63 kg H 2O/h
n 6 (mol H 2 O/h)
m
(kg H
H22O(l)/h)
O(v )/h)
mb (kg H2 O(v )/h)
mbb(kg
30°C
sat'd at 3.1 bars
30oC
In the absence of data to the contrary, we assume that the separation of methanol from
formaldehyde is complete.
Methanol vaporizer:
bg
e j
The product stream, which contains 42 mole % CH 3OH v , is saturated at Tm D C and 1 atm.
9- 34
9.29 (cont'd)
b g b gb
g
b g
ym P = pm∗ Tm ⇒ 0.42 760 mmHg = 319.2 mmHg = pm∗ Tm
equation
⎯Antoine
⎯⎯⎯⎯
⎯→ p m∗ = 319.2 mmHg ⇒ Tm = 44.1D C
c.
Moles HCHO formed :
=
36 × 106 kg solution 0.37 kg HCHO
350 days
1 kg solution
1 kmol
1 day
30.03 kg HCHO
24 h
= 52.80
kmol HCHO
h
but if all the HCHO is recovered, then this equals n4 , or n4 = 52.80 kmol HCHO h
70% conversion :
52.80 kmol HCHO
h
1 kmol CH 3OH react
1 kmol CH 3OH fed
1 kmol feed gas
1 kmol HCHO formed 0.70 kmol CH 3OH react 0.42 kmol CH 3OH
= n f
⇒ n f = 179.59 kmol h
Methanol unreacted:
n1 =
b0.42gb179.59gkmol CH OH fed b1 − 0.70g kmol CH OH fed = 22.63 kmol CH OH
3
3
h
b
1 kmol CH 3OH fed
3
h
gb gb g
N 2 balance: n3 = 179.6 kmol h 0.58 0.79 = 82.29 kmol N 2 h
Four reactor stream variables remain unknown — n s , n2 , n5 , and n6 — and four relations are
available — H and O balances, the given H 2 content of the product gas (5%), and the energy
balance. The solution is tedious but straightforward.
b
gb gb g
b
gb g b gb g
H balance: 179.6 0.42 4 + 2ns = 22.63 4 + 52.8 2 + 2n5 + 2n6
⇒ n s = n 5 + n 6 − 52.80
b
gb gb g b
(1)
g
b gb g
O balance: 179.6 0.42 1 + 179.6 (0.58) 0.21 2 + n s = (22.63)(1) + 2n 2 + (52.80)(1) + n 6
⇒ n s = 2n 2 + n 6 − 43.75
(2)
n 5
= 0.05 ⇒ 19n 5 − n 2 − n 6 = 157.72
22.63 + n 2 + 82.29 + 52.89 + n 5 + n 6
H 2 content:
bg b g b g b g
References : C s , H 2 g , O 2 g , N 2 g at 25° C
H = ΔH fo +
z
Table B.2
T
25
B
C p dT
or Table B.8 for O 2 , N 2 and H 2
9- 35
(3)
9.29 (cont'd)
substance
H in
n in
H out
n out
kmol / h
kJ / kmol
kmol / h
kJ / kmol
CH 3 OH
75.43
−195220
22.63
−163200
O2
N2
.
2188
82.29
3620
3510
n2
82.29
18410
17390
H 2O
ns
−237740
n6
−220920
HCHO
−
−
52.80
−88800
H2
−
−
n5
16810
Energy Balance :
ΔH =
∑n H − ∑n H = 0 ⇒ 18410n
i
i
out
i
2
i
+ 16810n5 − 220920n6 + 237704ns = −7.406 × 106
(4)
in
We now have four equations in four unknowns. Solve using E-Z Solve.
n s =
bg
58.8 kmol H 2 O v
18.02 kg
h
1 kmol
= 1060 kg steam fed h
n2 = 2.26 kmol O 2 h , n5 = 1358
. kmol H 2 h , n6 = 98.00 kmol H 2 O h
Summarizing, the product gas component flow rates are 22.63 kmol CH3OH/h, 2.26 kmol O2/h,
82.29 kmol N2/h, 52.80 kmol HCHO/h, 13.58 kmol H2/h, and 98.02 kmol H2O/h
⇒
d.
272 kmol h product gas
8% CH 3 OH, 0.8% O 2 , 30% N 2 , 19% HCHO, 5% H 2 , 37% H 2 O
Energy balance on waste heat boiler. Since we have already calculated specific enthalpies of all
components of the product gas at the boiler inlet (at 600°C), and for all but two of them at the
boiler outlet (at 145°C), we will use the same reference states for the boiler calculation
bg b g b g b g
H Oblg at triple point for boiler water
Reference States: C s , H 2 g , O 2 g , N 2 g at 25° C for reactor gas
2
Substance
CH 3 OH
O2
N2
H 2O
HCHO
H2
H 2O
nin
Hˆ in
nout
Hˆ out
kmol/h
kJ/kmol
− 163200
18410
17390
− 220920
− 88800
16810
125.7
(kJ/kg)
mol
kJ/mol
− 195220
3620
3510
− 237730
− 111350
3550
2726.1
(kJ/kg)
22.63
2.26
82.29
98.02
52.80
13.58
mb
(kg/h)
9- 36
22.63
2.26
82.29
98.02
52.80
13.58
mb
(kg/h)
9.29 (cont'd)
Energy Balance :
ΔH =
∑ n H − ∑ n H
i
out
i
i
in
b
i
=0
g
⇒ mb 27261
. − 125.7 − 4.92 × 10 6 = 0
⇒ mb = 1892 kg steam h
9.30
a.
C 2 H 4 + HCl → C 2 H 5Cl
Basis:
bg
1600 kg C 2 H 5Cl l
h
n 3 (mol HCl(g)/h)
n 4 (mol C 2H 4 ( g)/h)
condenser
n 5 (mol C 2H 6 ( g)/h)
n 6 (mol C 2H 5 Cl( g)/h)
50°C
A
n 1 (mol HCl(g)/h)
0°C
C
n 3 (mol HCl(g)/h)
n 4 (mol C 2H 4 ( g)/h)
n 5 (mol C 2H 6 ( g)/h)
0°C
Cl(l)/h
6 (mol
2H5Cl(
g)/h)
n 6n(mol
CC2H
5
reactor
B
103 g 1 mol
= 24800 mol h C 2 H 5Cl
1 kg 64.52 g
n 2 (mol/h) at 0°C
0.93 C 2H 4
0.07 C 2H 6
D
( n 6 – 24,800) (mol C 2H 5 Cl( l)/h)
0°C
24,800 mol C2 H5 Cl(l )/h
Product composition data:
n3 = 0.015n1
b
b1g
b2 g
b3g
g
n4 = 0.015 0.93n2 = 0.01395n2
n5 = 0.07n2
Overall Cl balance :
b
n1 mol HCl h
g
b gb g b
gb g
b4g
1 mol Cl
= n3 1 + 24800 1
1 mol HCl
Solve (4) simultaneously with (1) ⇒ n1 = 25180 mol h = 2518
. kmol HCl fed / h
bg
n 3 = 378 mol HCl g h
Overall C balance :
b gb g b gb g
b gb
g
n2 0.93 2 + n2 0.07 2 = 2n4 + 2n5 + 2 24800
LM
N
OP b gb
Q
g
From Eqs. (2) and (3) ⇒ 2n2 0.93 + 0.07 − 0.0139 − 0.07 = 2 24800
n2 = 27070 mol fed h = 27.07 kmol h of Feed B
b.
U| 2.65 kmol / h of Product C
= 0.01395b27070g = 378 mol C H hV
14.3% HCl, 14.3% C H , 71.4% C H
= 0.07b27070g = 1895 mol C H h |W
n3 = 378 mol HCl h
n4
n5
2
2
4
6
9- 37
2
4
2
6
9.30 (cont'd)
c.
bg
bg
bg
bg
References : C 2 H 4 g , C 2 H 6 g , C 2 H 5 Cl g , HCl g at 0 D C
e
z
j
50
C 2 H 4 g, 50D C : H = C p dT
Table B.2
⇒
2.181 kJ mol
0
50
C2 H 6 ( g, 50D C ) : Hˆ = ∫ C p dT ⇒ 2.512 kJ mol
Table B.2
e
z
j
0
50
HCl g, 50D C : H = C p dT
Table B.2
⇒ 1.456 kJ mol
0
e j
C H Cleg, 50 Cj: H =
e j
C 2 H 5Cl l, 0D C : H = − Δ H v 0D C = −24.7 kJ mol
D
2
5
substance
HCl
C2 H 4
C2 H 6
C 2 H 5Cl
z
50
C pv dT = 2.709 kJ mol
0
nin
nout
H in
H out
mol
kJ / mol mol kJ / mol
.
0
378
1456
25180
25175
0
378
2.181
1895
0
1895 2.512
n6 − 24800 −24.7
n6
2.709
Energy balance:
ΔH = 0 ⇒
⇒
e j+
nA ΔH r 0D C
νA
∑ n H − ∑ n H
i
( 25180 − 378) mol HCl react
h
i
i
out
i
=0
in
−64.5 kJ
+ ( 378)(1.456) + ( 378)( 2.181) + (1895)( 2.512)
1 mol HCl
+ 2.709n6 − ( n6 − 24800)( −24.7 ) = 0 ⇒ n6 = 80490 mol C2 H5Cl h in reactor effluent
80490 mol condensed 24800 mol product
mol
−
= 55690
h
h
h
kmol recycled
= 55.7
h
C2 H 5 Cl recycled =
d.
C p is a linear function of temperature.
ΔH v is independent of temperature.
100% condensation of ethylbenzene in the heat exchanger is assumed.
Heat of mixing and influence of pressure on enthalpy is neglected.
Reactor is adiabatic.
No C2H4 or C2H6 is absorbed in the ethyl chloride product.
9.31
a.
4NH3(g) + 5O2(g) Æ 4NO(g) +6H2O(g)
Basis : 10 mol/s Feed gas
9- 38
ΔH ro = −904.7 kJ / mol
9.31 (cont'd)
4 mol / s NH 3
6 mol / s O 2
n3 (mol O 2 )
n4 (mol NO)
Tin = 200o C
n5 (mol H 2 O)
Tout
O 2 consumed :
5 mol O 2
4 mol NH 3 fed
4 mol NH 3
s
= 5 mol / s ⇒ n 3 = (6 − 1) mol O 2 / s = 1 mol O 2 /
NO produced : n 4 =
4 mol NO produced 4 mol NH 3 fed
H 2 O produced : n5 =
6 mol H 2 O produced 4 mol NH 3 fed
4 mol NH 3
s
4 mol NH 3
Extent of reaction : ξ =
s
(n NH 3 ) out − (n NH 3 ) in
ν NH 3
=
0−4
4
= 4 mol NO / s
= 6 mol H 2 O / s
= 1 mol / s
b.
Well-insulated reactor, so no heat loss
No absorption of heat by container wall
Neglect kinetic and potential energy changes;
No shaft work
No side reactions.
c.
References : NH 3 ( g), O 2 ( g), NO(g), H 2 O(g) at 25o C, 1atm
Substance
NH 3 (g)
O 2 ( g)
NO(g)
H 2 O(g)
H 1 =
z
nin
nout
H in
H out
( mol / s) ( kJ / mol) ( mol / s) ( kJ / mol)
−
−
4.00
H 1
6.00
100
H2
H 3
.
−
−
4.00
H 4
−
−
6.00
H
5
Table B.2
200
(C p ) NH 3 dT
B
=
Table B.8
6.74 kJ / mol,
H 2 = H O 2 (200o C)
B
=
5.31 kJ / mol
25
Using (Cp )i from Table B.2 :
Hˆ 3 = (0.0291 Tout + 0.5790 ×10−5 Tout 2 − 0.2025 ×10−8 Tout 3 + 0.3278 ×10−12 Tout 4 − 0.7311) kJ/mol
Hˆ 4 = (0.0295 Tout + 0.4094 ×10−5 Tout 2 − 0.0975 ×10−8 Tout 3 + 0.0913 ×10−12 Tout 4 − 0.7400) kJ/mol
Hˆ 5 = (0.03346 Tout + 0.3440 ×10−5 Tout 2 + 0.2535 ×10−8 Tout 3 − 0.8983 ×10−12 Tout 4 − 0.8387) kJ/mol
Energy Balance: ΔH = 0
ΔH = ξ ΔH ro +
5
∑
i =3
(ni ) out ( H i ) out −
2
∑ (n
i ) in ( Hi ) in
i =1
9- 39
9-31 (cont’d)
⇒ ΔH = ξΔHˆ ro + (1.00) Hˆ 3 + (4.00) Hˆ 4 + (6.00) Hˆ 5 − (4.00) Hˆ 1 − (6.00) Hˆ 2
⇓
o
Substitute for ξ , ΔHˆ r , and Hˆ 1 through Hˆ 6
ΔH = (0.3479 Tout + 4.28 × 10−5 Tout 2 + 0.9285 × 10−8 Tout 3 − 4.697 × 10−12 Tout 4 )
− 972.24 kJ/mol = 0
E-Z Solve ⇒ Tout = 2223 o C
d.
If only the first term from Table B.2 is used, H i =
z
T
(C pi )dT = C pi (T − 25)
25
H 1 = 0.03515(200 − 25) = 615
. kJ / mol, H 2 = 5.31 kJ / mol, H 3 = 0.0291(Tout − 25),
H 4 = 0.0295(Tout − 25), H 5 = 0.03346(Tout − 25)
E.B. ΔH = ξΔHˆ ro + (1.00) Hˆ 3 + (4.00) Hˆ 4 + (6.00) Hˆ 5 − (4.00) Hˆ 1 − (6.00) Hˆ 2 = 0
⇓
o
Substitute for ξ (=1 mol/s), ΔHˆ r ( = −904.7 kJ/mol) and Hˆ 1 through Hˆ 6
0=0.3479 Tout − 969.86 ⇒ Tout = 2788 o C ⇒ % error=
e.
9.32
2788o C − 2223o C
× 100 = 25%
2223 o C
If the higher temperature were used as the basis, the reactor design would be safer (but more
expensive).
Basis : 100 lb m coke fed
⇒ 84 lb m C ⇒ 7.00 lb - moles C fed ⇒ 7.00 lb - moles CO 2 fed
7.00 lb-moles CO2
400°F
7.00 lb-moles(84 lbm)C/hr
16 lb mash/hr
77°F
585,900 Btu
a.
bg
bg
bg
e77 Fj = eΔH j b g − 2eΔH j b g
. − b2gb −282.99g kJ
−39350
=
n 1 (lb-moles CO)
n 2 (lb-moles CO2 )
1830°F
n 3 lb-moles C( s )/hr
16 lb mash/hr
1830°F
C s + CO 2 g → 2CO g ,
ΔH ro
D
o
c
= 25°C
o
c
CO 2 g
CO g
mol
0.9486 Btu 453.6 mols
= 74,210 Btu lb - mole
1 kJ
1 lb - mole
Let x = fractional conversion of C and CO 2 :
E
n1 =
b
7.00 x lb - moles C reacted
b g
= 7.00b1 − x g lb - moles Cbsg
g
2 lb - moles CO formed
= 14.0 x lb - moles CO
1 lb - mole C reacted
n2 = 7.00 1 − x lb - moles CO 2
n3
bg
bg bg
D
References for enthalpy calculations: C s , CO 2 g , CO g , ash at 77 F
9- 40
9.32 (cont'd)
⇒ 3130 Btu lb - mole
b
g
CO bg,1830° Fg: H = H
(1830 F) ⇒ 20,880 Btu lb - mole
CObg,1830° Fg: H = H (1830 F) ⇒ 13,280 Btu lb - mole
0.24 Btu b1830 − 77g° F
Solid b1830° Fg: H =
= 420 Btu lb
lb ⋅ F
CO 2 g,400° F : H = H CO 2 (400D F)
Table B.9
Table B.9
D
2
CO 2
Table B.9
D
CO
D
m
m
Mass of solids (emerging)
=
b g
7.00 1 − x lb - moles C
12.0 lb m
1 lb - mole
substance
b
g
+ 16 lb m = 100 − 84 x lb m
nin
nout
H in
H out
(lb − moles) (Btu lb - mole) (lb − moles) (Btu lb - mole)
7.00
3130
7.00 1 − x
20,890
14.0 x
13,280
−
−
(lb m )
(Btu lb m )
(lb m )
(Btu lb m )
b g
CO 2
CO
solid
100
0
100 − 84 x
420
Extent of reaction: n CO = ( n CO ) o + ν COξ ⇒ 14.0 x = 2ξ ⇒ ξ (lb - moles) = 7.0 x
Energy balance:
Q = ΔH = ξ ΔH ro +
∑ n H − ∑ n H
i
out
i
i
i
in
a fa f
+a14.0 x fa13,280f + a100 − 84 x fa 420f − a7.00fa3130f
E
585,900 Btu =
7.0 x (lb - moles)
74,210 Btu
lb - mole
+ 7.00 1 − x 20,880
x = 0.801 ⇒ 80.1% conversion
b.
Advantages of CO. Gases are easier to store and transport than solids, and the product of the
combustion is CO2, which is a much lower environmental hazard than are the products of
coke combustion.
Disadvantages of CO. It is highly toxic and dangerous if it leaks or is not completely burned,
and it has a lower heating value than coke. Also, it costs something to produce it from coke.
9- 41
9.33
Basis :
17.1 m 3 10 3 L 273 K 5.00 atm
1 mol
a f = 3497 mol h feed
3
h
1m
298 K 1.00 atm 22.4 L STP
CO g + 2 H 2 g → CH 3OH g ,
bg
ΔH ro
bg
=e
j
ΔH fo
bg
b g − e ΔH j
o
f
CH 3OH g
= −90.68 kJ mol
CO(g)
3497 mol/h
n 1 (mol CH 3 OH /h)
0.333 mol CO/mol
n 2 (mol CO/h)
0.667 mol H 2/mol
n 3 (mol H 2 /h)
25°C, 5 atm
127°C, 5 atm
Q = –17.05 kW
Let f = fractional conversion of CO (which also equals the fractional conversion of H 2 , since
CO and H 2 are fed in stoichiometric proportion).
( 3497 )( 0. 333) mol CO feed
f ( mol react )
= 1166 f ( mol CO react)
mol feed
1166 f mol CO react 1 mol CH 3OH
= 1166 f mol CH 3OH h
CH 3OH produced : n1 =
1 mol CO
CO remaining : n 2 = 1166 1 − f mol CO h
1166 f mol CO react 2 mol H 2 react
H 2 remaining : n3 = 3497 0.667 mol H 2 fed −
1 mol CO react
CO reacted : =
a f
b gb g
= 2332b1 − f g mol H
h
2
bg
bg
Reference states : CO(g), H 2 g , CH 3OH g at 25°C
Substance
CO
H2
CH 3 OH
H in
n in
H out
n out
bmol hg bkJ molg bmol hg bkJ molg
1166
0
H
1166a1 − f f
H
2332
0
2332a1 − f f
1
2
−
−
H 3
1166 f
B
Table B.8
e
D
e
D
j
CO g,127 C : H 1 = H CO (127 D C) = 2.99 kJ mol
B
Table B.8
j
H 2 g,127 C : H 2 = H H 2 (127 D C) = 2.943 kJ mol
z
122
CH 3OH(g,127 C): H 3 =
D
B
Table B.2
C p dT = 5.009 kJ / mol
25
Energy balance : Q = ΔH = ξ ΔH ro +
∑ n H − ∑ n H
i
i
out
⇒
−17.05 kJ 3600 s
s
b
+ 2332 1 − f
1h
= (1166 f )( −90.68)
i
i
in
b
kJ
+ 1166 1 − f
h
g b2.993g + 1166 f b5.009g bkJ hg
b
g b2.99g
g
.
⇒ 1102
× 10 5 f = 7.173 × 10 4 ⇒ f = 0.651 mol CO or H 2 converted mol fed
9- 42
9.33 (cont’d)
b g
b
g
= 2332b1 − 0.651g = 813.9 mol h
E
n1 = 1166 0.651 = 759.1 mol h
n2 = 1166 1 − 0.651 = 406.9 mol h
n3
n tot = 1980
9.34
a.
b g
mol
1980 mol 22.4 L STP
⇒ Vout =
h
h
1 mol
bg bg
bg
1 m3
400 K 1.00 atm
273 K 5.00 atm 103 L
= 13.0 m 3 h
bg
CH 4 g + 4S g → CS 2 g + 2 H 2S g , ΔH r ( 700° C ) = −274 kJ mol
Basis : 1 mol of feed
1 mol at 700°C
0.20 mol CH 4/mol
0.80 mol S/mol
Product gas at 800°C
n1 (mol CS2)
n2 (mol H 2S)
n3 (mol CH 4)
n4 (mol S (v))
Reactor
Q = –41 kJ
Let f = fractional conversion of CH 4 (which also equals fractional conversion of S, since the
species are fed in stoichiometric proportion)
Moles CH 4 reacted = 0.20 f , Extent of reaction = ξ (mol) = 0.20 f
b
g
n 3 = 0.20 1 − f mol CH 4
n 4 = 0.80 mol S fed −
b
0.20 f mol CH 4 react
g
4 mol S react
1 mol CH 4 react
n1 =
0.20 f mol CH 4 react
n2 =
0.20 f mol CH 4 react
1 mol CS 2
1 mol CH 4
2 mol H 2 S
1 mol CH 4
b
g
= 0.80 1 − f mol S
= 0.20 f mol CS 2
= 0.40 f mol H 2 S
bg
References: CH 4 (g), S g , CS 2 (g), H 2S(g) at 700°C (temperature at which ΔH r is known)
substance nin
CH 4
S
H in
H out
nout
bmolg bkJ molg bmolg bkJ molg
0.20
0
H
0.20b1 − f g
H
0.80
0
0.80b1 − f g
1
2
CS 2
−
−
0.20 f
H 2S
−
−
0.40 f
H out = C pi (800 − 700 ) ⇒
b
H 3
H
4
g
CH 4 g, 800° C : H 1 = 7.14 kJ / mol
S g, 800° C : H = 3.64 kJ / mol
b
b
b
g
g
g
2
CS 2 g, 800° C : H 3 = 3.18 kJ / mol
H 2S g, 800° C : H 4 = 4.48 kJ / mol
9- 43
9.34 (cont’d)
Energy balance on reactor:
Q = ΔH = ξ ΔH r +
n i H i −
∑
=
b
out
gb
b1g
∑ n H
i
= 41
i
in
kJ
s
g + 0.20b1 − f gb7.140g + 0.80b1 − f gb3.640g + 0.20 f b3180
. g + 0.40 f b4.480g
0.20 f −274.0
⇒ f = 0.800
b.
0.04 mol CH4
0.16 mol S(l )
0.16 mol CS2
0.32 mol H2 S
200°C
0.20 mol CH4
0.80 mol S(l )
150°C
Q (kJ)
preheater
0.20 mol CH4
0.80 mol S( g)
T (°C)
0.20 mol CH4
0.80 mol S(l )
700°C
0.04 mol CH4
0.16 mol S(g )
0.16 mol CS2
0.32 mol H2S
800°C
System: Heat exchanger-preheater combination. Assume the heat exchanger is adiabatic, so
that the only heat transferred to the system from its surroundings is Q for the preheater.
bg
References : CH 4 (g), S l , CS 2 (g), H 2S(g) at 200°C
Substance
bCH g
bCH g
Sblg
Sbgg
4 150°,700°
4 800°,200°
H in
n in
n out
H out
bmolg bkJ molg bmolg bkJ molg
0.20
0.04
0.80
.
016
CS 2
016
.
H 2S
0.32
H 1
H
0.20
H 7
2
0.04
0
H 3
H
.
016
0.80
0
H 8
016
.
0
0.32
0
4
H 5
H
6
a
f
= dC i aT − 200f for Salf
af
F
I
= dC i G 444.6 − 200J + ΔH
bT g + dC i b g aT − 444.6f for Sbgg
a fH
K
H i = C pi T − 200 for all substances but S
p Sl
p Sl
Tb
v
= 83.7 kJ mol
b
9- 44
p Sg
9.34 (cont’d)
b
b
g
g
b
b
CH 4 g, 150° C : H 1 = − 3.57 kJ / mol
CH 4 g, 800° C : H 2 = 42.84 kJ / mol
S l, 150° C : H = − 1.47 kJ / mol
b
g
Sbg, 800° Cg: H
3
4
b
b g ∑ n H − ∑ n H
i
out
c.
4
b
g
g
7
S g, 700° C : H 8 = 100.19 kJ / mol
= 103.83 kJ / mol
Energy balance: Q kJ =
g
g
CS2 g, 800° C : H 5 = 19.08 kJ / mol
H 2 S g, 800° C : H 6 = 26.88 kJ / mol
CH g, 700° C : H = 35.7 kJ / mol
i
i
i
⇒ Q = 59.2 kJ ⇒ 59.2 kJ mol feed
in
The energy economy might be improved by insulating the reactor better. The reactor effluent will
emerge at a higher temperature and transfer more heat to the fresh feed in the first preheater,
lowering (and possibly eliminating) the heat requirement in the second preheater.
9- 45
9.35
Basis : 1 mol C 2 H 6 fed to reactor
n (mols) @ T (K), P atm
n C 2H6 (mol C 2H 6)
n C 2H4 (mol C 2H 4)
n H 2 (mol H 2)
1 mol C H
2 6
1273 K, P atm
a.
C2 H 6 ⇔ C2 H 4 + H 2 , K p =
x C2 H 4 x H 2
P = 7.28 × 10 6 exp[ −17,000 / T ( K )]
x C2 H 6
b
Fractional conversion = f mols C 2 H 6 react mol fed
U| x
n
= b1 − f gb mol C H g|
|V ⇒ x
n
= f b mol C H g
||
n = f b mol H g
|W x
n = 1 + f b molsg
=
C2 H 6
2
C2 H 4
2
H2
6
4
2
f
Kp =
x C2 H 4 x H 2
e1 − f jK
2
b.
P⇒ Kp =
x C2 H 4
p
= f
2
g
1 − f mol C 2 H 6
1+ f
mol
f mol C 2 H 4
C2 H 4 =
1+ f
mol
f mol H 2
H2 =
1 + f mol
ξ (mol) = f
C2 H 6
(1)
2
b1+ f g P
b1− f g
b1+ f g
2
F K I
P⇒ f =G
H P + K JK
=
f2P
f2
P
=
1− f 1+ f
1− f 2
b
gb
12
g
b2 g
p
p
bg
bg
bg
References : C 2 H 6 g , C 2 H 4 g , H 2 g at 1273 K
Energy balance:
b
g ∑ n H − ∑ n H
ΔH = 0 ⇒ ξ ΔH r 1273 K +
e H j
e H j
i
i
i
i
i
out
b
i
in
= 0 inlet temperature = reference temperature
in
out
=
z
g
T
1273
C pi dT
⇓ energy balance
b
g b
f ΔH r 1273 K kJ + 1 − f
z
g dC i
T
1273
p
dT + f
z
T
1273
C2 H 6
dC i
p C H dT
2 4
rearrange, reverse limits and change signs of integrals
1− f
=
f
b
g
ΔH r 1273K −
z
1273
z
dC i
dC i
p C H dT
2 4
T
−
p C H
2 6
dC i
dT
bg
φ T
1− f
1
= φ T ⇒ 1 − f = fφ T ⇒ f =
f
1+ φ T
bg
1273
T
1273
T
z
bg
b g b4 g
9-46
p H dT
2
b3g
+f
z
T
1273
dC i
p H dT
2
=0
9.35 (cont'd)
145600 −
bg
φT =
zb
g
1273
z
T
bg
⇒φ T =
c.
1273
−3
T
1273
T
z
e26.90 + 4.167 × 10 T jdT
T gdT
. + 01392
.
b1135
T dT −
9.419 + 01147
.
3052 + 36.2T + 0.05943T 2
127240 − 113
. T − 0.0696T 2
F K I = 1 ⇒ F K I − 1 = ψ bT g = 0
GH 1 + K JK 1 + φbT g GH 1 + K JK 1 + φbT g
φ bT g given by expression of Part b. K bT g given by Eq. (1)
12
12
p
p
p
p
p
d.
P
(atm)
0.01
0.05
0.1
0.5
1
5
10
T
(K)
794
847.4
872.3
932.8
960.3
1026
1055
f
0.518
0.47
0.446
0.388
0.36
0.292
0.261
Kp
(atm)
0.0037
0.0141
0.025
0.0886
0.1492
0.4646
0.7283
Phi
Psi
0.93152 -0.0001115
1.12964 -0.0002618
1.24028 0.00097743
1.57826
3.41E-05
1.77566
4.69E-05
2.42913 -2.57E-05
2.83692 -7.54E-05
Plot of T vs ln P
Plot of f vs. ln P
1100
0.6
0.5
1000
f
T(K)
0.4
900
0.3
0.2
800
0.1
0
700
-3
-2
-1
0
1
2
-3
-2
ln P(atm)
e.
-1
0
ln P(atm)
C **PROGRAM FOR PROBLEM 9-35
WRITE (5, 1)
1
FORMAT ('1', 20X, 'SOLUTION TO PROBLEM 9-35'//)
T = 1200.0
TLAST = 0.0
PSIL = 0.0
9-47
1
2
9.35 (cont'd)
C **DECREMENT BY 50 DEG. AND LOOK FOR A SIGN IN PSI
DO 10I =1, 20
CALL PSICAL (T, PHI, PSI)
IF ((PSIL*PSI).LT.0.0) GO TO 40
TLAST = T
PSIL = PSI
T = T – 50.
10
CONTINUE
40
IF (T.GE.0.0) GO TO 45
WRITE (3, 2)
2
FORMAT (1X, 'T LESS THAN ZERO -- ERROR')
STOP
C **APPLY REGULA-FALSI
45
DO 50 I = 1, 20
IF (I.NE.1) T2L = T2
T2 = (T*PSIL-TLAST*PSI)/(PSIL-PSI)
IF (ABS(T2-T2L).LT.0.01) GO TO 99
CALL PSICAL (T2, PHIT, PSIT)
IF (PSIT.EQ.0) GO TO 99
IF ((PBIT*PBIL).GT.0.0) PSIL = PSIT
IF ((PSIT*PSIL).GT.0.0) TLAST = T2
IF ((PSIT*PSI).GT.0.0) PSI = PSIT
IF ((PSIT*PSI).GT.0.0) T = T2
50
CONTINUE
IF (I.EQ.20) WRITE (3, 3)
3
FORMAT ('0', 'REGULA-FALSI DID NOT CONVERGE IN 20 ITERATIONS')
93
STOP
END
SUBROUTINE PSICAL (T, PHI, PSI)
REAL KF
PHI = (3052 + 36.2*T + 36.2*T + 0.05943*T**2)/(127240. – 11.35*T
* – 0.0636*T**2)
KP = 7.28E6*EXP(-17000./T)
FBI = SQRT((KP/(1. + KP)) – 1./12. + PHI)
WRITE (3, 1) T, PSI
1
FORMAT (6X, 'T =', F6.2, 4X, 'PSI =', E11,4)
RETURN
END
OUTPUT: SOLUTION TO PROBLEM 9-35
T = 1200.00 PSI = 0.8226E + 00
T = 1150.00 PSI = 0.7048E + 00
T = 1100.00 PSI = 0.5551E + 00
T = 1050.00 PSI = 0.3696E + 00
.
T = 1000.00 PSI = 01619
E + 00
T = 950.00 PSI = −0.3950E − 01
.
T = 959.80 PSI = −01824
E − 02
T = 960.25 PSI = −0.7671E − 04
T = 960.27 PSI = −0.3278E − 05
Solution: T = 960.3 K, f = 0.360 mol C 2 H 6 reacted mol fed
9-48
2CH 4 → C 2 H 2 + 3H 2
9.36
C 2 H 2 → 2C(s) + H 2
n1 (mol CH 4 / s)
Basis: 10.0 mol CH 4 (g)/s
n2 (mol C 2 H 2 / s)
n3 (mol H 2 (s)/s)
o
1500 C
n4 (mol C(s)/s)
1500o C
975 kW
a.
b
g
60% conversion ⇒ n1 = 10 1 − 0.600 = 4.00 mol CH 4 s
bg bg
bg bg
C balance: 10 1 = 4 1 + 2n 2 + n 4 ⇒ 2n 2 + n 4 = 6
H balance: 10 4 = 4 4 + 2n 2 + 2n 3 ⇒ 2n 2 + 2n 3 = 24
(1)
(2)
bg
References for enthalpy calculations : C(s), H 2 g at 25°C
Hi =
e
ΔH fo
Substance
bg
bg
bg
bg
CH 4 g
C2 H 2 g
H2 g
Cs
j
i
b
g
+ C pi 1500 − 25 , i = CH 4 , C 2 H 2 , C, H 2
n in
n out
H in
H out
( mol s) ( kJ mol) ( mol s) (kJ mol)
4
4168
.
10
41.68
n2
−
−
303.45
n 3
−
−
45.72
n 4
−
−
32.45
Energy Balance: Q = ΔH ⇒ 975 kJ / s =
∑ n H − ∑ n H
i
out
i
i
i
(3)
in
n 2 = 2.50 mol C 2 H 2 / s
Solve (1) - (3) simultaneously ⇒ n 3 = 9.50 mol H 2 / s
. mol C / s
n 4 = 100
Yield of acetylene =
b.
2.50 mol C 2 H 2 s
= 0.417 mol C 2 H 2 mol CH 4 consumed
6.00 mol CH 4 consumed s
If no side reaction,
n1 = 10.0(1 − 0.600) = 4.00 mol CH 4 / s
n 3 = 0 ⇒ n 2 = 3.00 mol C 2 H 2 / s, n 4 = 9.00 mol H 2 / s
Yield of acetylene =
3.00 mol C 2 H 2 s
= 0.500 mol C 2 H 2 mol CH 4 consumed
6.00 mol CH 4 consumed s
Reactor Efficiency =
0.417
= 0.834
0.500
9-49
9.37
bg
bg
bg bg
CObgg + H Ob vg → CO bgg + H bgg
C 3 H 8 g + 3H 2 O v → 3CO g + 7H 2 g
2
2
2
Basis : 1 mol C 3 H 8 fed
Heating gas
4.94 m3 at 1400°C, 1 atm
n g (mol)
n g (mol), 900°C
ng =
4.94 m 3 10 3 L
1m
3
Product gas, 800°C
n 1 (mol C 3H 8) = 0
n 2 (mol H 2O)
n 3 (mol CO)
n 4 (mol CO2 )
n 5 (mol H 2)
a
1 mol C 3H 8(g )
6 mol H 2 O( g )
125°C
273 K
1 mol
1673 K 22.4 L
= 35.99 mol heating gas
Let ξ 1 and ξ 2 be the extents of the two reactions.
n1 = 0
n4 = ξ 2
n1 = 1 − ξ 1 ⇒ ξ 1 = 1 mol
ξ 1 =1
ξ 1 =1
n 2 = 6 − 3ξ 1 − ξ 2 ⇒ n 2 = 3 − ξ 2
n5 = 7ξ 1 + ξ 2 ⇒ n5 = 7 + ξ 2
ξ 1 =1
n 3 = 3ξ 1 − ξ 2 ⇒ n 3 = 3 − ξ 2
bg bg
References : C(s), H 2 g , O 2 g at 25°C, heating gas at 900°C
z
T
H i = ΔH fio + C pi dT
for C 3 H 8
25
= Table B.8 for CO 2 , H 2 , H 2 O, CO
z
T
=
b
g
C p dT = C p T − 900 for heating gas
900
n in
H in
n out
H out
C3H8
mol
1
kJ / mol
−95.39
mol
0
kJ / mol
−
H 2O
6
−238.43 3 − ξ 2
−212.78
CO
−
−
3−ξ2
−86.39
CO 2
−
−
ξ2
−35615
.
H2
−
−
7+ξ2
22.85
200.00
35.99
0
Substance
heating gas 35.99
Energy Balance :
∑ n H − ∑ n H
i
out
i
i
i
= 0 ⇒ ξ 2 = 2.00 mol ⇒ n 2 = 1 mol H 2 O, n 3 = 1 mol CO,
in
n 4 = 1 mol CO 2 , n5 = 9 mol H 2 ⇒ 7.7 mol % H 2 O, 7.7% CO, 15.4% CO 2 , 69.2% H 2
9-50
9.38
a.
Any C consumed in reaction (2) is lost to reaction (1). Without the energy released by reaction
(2) to compensate for the energy consumed by reaction (1), the temperature in the adiabatic
reactor and hence the reaction rate would drop.
b.
Basis : 1.00 kg coal fed (+0.500 kg H20)
0.500 kg H20
⇒
1.0 kg coal
0.105 kg H2O/kg coal
0.226 kg ash/kg coal
0.669 kg combustible / kg coal
0.812 kg C / kg combustible
0.134 kg O / kg combustible
0.054 kg H / kg combustible
R|
|S
||
T
nf1 (mol C)
nf2 (mol O)
nf3 (mol H)
nf4 (mol H2O)
0.226 kg ash
U|
|V
||
W
n f 1 = [ (1.00)(0.669)(0.812) kg C][1 mol C / 12.01 × 10 −3 kg] = 45.23 mol C
n f 2 = (1.00)(0.669)(0.134) / 16.0 × 10 −3 = 5.6 mol O
n f 3 = (1.00)(0.669)(0.054) / 1.01 × 10 −3 = 35.77 mol H
n f 4 = [ (0.500 + 0.105) kg][1 mol H 2 O / 18.016 × 10 −3 kg] = 33.58 mol H 2 O
n0 (mol O2) 25°C
Product gas at 2500°C
n1 (mol CO2)
n2 (mol CO)
n3 (mol H2)
n4 (mol H2O)
1 kg coal + H2O, 25°C
45.23 mol C
5.60 mol O
35.77 mol H
33.58 mol H2O
0.226 mol kg ash
0.226 kg slag
2500°C
Reactive oxygen (O) available = (2n0 + 5.60) mol O
Oxygen consumed by H ( 2H+O → H2O) :
35.77 mol H 1 mol O
= 17.88 mol O
2 mol H
⇒ Reactive O remaining =(2n0 + 5.60) − 17.88 = (2n0 − 12.28) mol O
CO2 formed ( C+2O → CO2 ) : n1 =
C balance : 45.23=n1 + n2
(2n0 − 12.28) mol O 1 mol CO2
2 mol O
= (n0 − 6.14) mol CO2
n1 = n0 − 6.14
⇒
n2 = (51.37 − n0 ) mol CO
O balance : 2n0 + 5.60 + 33.58 = 2n1 + n2 + n4
H balance : 35.77+ 2(33.58)=2n3 + 2n4
9-51
n1 = n0 − 6.14
⇒
n2 = 51.37 − n0
n4 = n0 + 0.06
⇒
n4 = (n0 + 0.06) mol H 2 O
n3 = (51.37 − n0 ) mol H 2
9.38 (cont’d)
c.
1 kg coal contains 45.23 mol C and 35.77 mol H
⇒ 1 kg coal + nO 2 → 45.23 CO 2 + (35.77 / 2) mol H 2 O (l)
ΔH r = −21,400 kJ = 45.23( ΔH fo ) CO 2 + (35.77 / 2)( ΔH fo ) H 2 O(l) − ( ΔH fo ) coal
⇒ ( ΔH fo ) coal = −1510 kJ / kg
Re ferences : C(s), O 2 (g), H 2 (g), ash(s) at 25o C
nin
Hˆ in
nout
Hˆ out
(mol)
(kJ/mol)
(mol)
CO 2
−
−
n0 − 6.14
(kJ/mol)
Hˆ
CO
−
−
51.37 − n0
H2
−
51.37 − n0
Hˆ 2
Hˆ
n0 + 0.06
Hˆ 4
Substance
1
H2O
33.58
−
Hˆ
Coal
1 kg
−1510 kJ/kg
−
−
Ash(slag)
(in coal)
0
0.266 kg
Hˆ 5 (kJ/kg)
0
3
Hˆ i = ΔHˆ ofi + C pi (2500 − 25), i = 1,3
Hˆ 1 = −393.5 + 0.0508(2475) = −267.8 kJ/mol CO 2
Hˆ 2 = −110.52 + 0.0332(2475) = −28.35 kJ/mol CO
Hˆ 3 = 0.0300(2475) = 74.25 kJ/mol H 2
Hˆ 4 = −241.83 + 0.0395(2475)= − 144.07 kJ/mol H 2 O
Hˆ 5 = (ΔHˆ m )ash + 1.4(2475) = 710 + 1.4(2475) = 4175 kJ/kg ash
Energy Balance
ΔH = ∑ nout Hˆ out − ∑ nin Hˆ in = 0 ⇒ n0 = 35.4 mol O 2
9-52
9.39
Mass of H 2 SO 4 =
Mass of solution =
3 m3 10 3 L 1 mol H 2 SO 4
1 m3
L
3 m3 10 3 L 10 3 mL
1 m3
= 3000 mol H 2 SO 4
1.064 g
L
1 mL
FG 98.02 g IJ = 2.941× 10
H 1 mol K
5
g H 2 SO 4
= 3192
. × 10 6 g solution
⇒ Moles of H 2 O = (3192
.
× 10 6 − 2.941 × 10 5 )g H 2 O(
1 mol
) = 161
. × 10 5 mol H 2 O
18.02 g
FG mol H O IJ = 161
. × 10 mol H O
= 53.6 mol H O mol H SO
mol
H
SO
3000
mol H SO
H
K
kJ
= e ΔH j
+ e ΔH j
= b−81132
. − 73.39g
= −884.7 kJ mol
eΔH j b
g
b
g
bg
mol
5
2
n
2
2
2
f
4
2
o
f
H 2SO 4 aq., r =53.6
2
4
4
s
H 2SO 4 l
H 2SO 4 aq ., r =53.6
A
A
Table B.1
Table B.11
H = (3000 mol H 2 SO 4 )(-884.7 kJ / mol H 2 SO 4 ) = -2.65 × 10 6 kJ
9.40
e
HCl (aq): ΔH fo = ΔH fo
j
bg
HCl g
e
+ ΔH so
j
Tables B.1, B11
=
∞
− 92.31 − 7514
. = −167.45 kJ mol
B
Tables B1, B.11
e
j
NaOH (aq): ΔH fo = ΔH fo
bg
NaOH s
e
+ ΔH so
j
=
∞
− 426.6 − 42.89 = −469.49 kJ mol
B
Table B.1
NaCl (aq):
ΔH fo
b g
=
e
ΔH fo
b g
j
bg
NaCl s
+
e
ΔH so
b g
j
B
Given
= −4110
. + 4.87 = −4061
. kJ mol
∞
bg
HCl aq + NaOH aq → NaCl aq + H 2 O l
ΔH ro = −4061
. − 28584
. − −167.45 − −469.49 = −55.0 kJ mol
b
g b
HClbgg + NaOHbsg → NaClbsg + H Oblg
g
2
ΔH ro =
∑
v i ΔH fo −
∑
v i ΔH fo
reactants
products
b
g b
g
= −4110
. − 28584
. − −92.31 − −426.6
kJ mol = −177.9 kJ mol
The difference between the two calculated values equals
.
ΔH
− ΔH
− ΔH
{e j
s
9.41
a.
e j
s
NaCl
b g
HCl
e j
s
NaOH
b g
}
b g
bg
H 2 SO 4 aq + 2NaOH aq → Na 2 SO 4 aq + 2H 2 O l
Basis: 1 mol H 2 SO 4 soln ⇒
⇒
⇒
b
g
0.10 mol H 2 SO 4 × 98.08 g mol = 9.808 g H 2 SO 4
b
g
0.90 mol H 2 O × 18.02 g mol = 16.22 g H 2 O
U|
V|
W
26.03 g soln 1 cm 3
= 20.49 cm 3
127
. g
0.10 mol H 2 SO 4
2 mol NaOH
1 mol H 2 SO 4
b g
1 liter caustic soln 10 3 cm 3
= 66.67 cm 3 NaOH aq
3 mol NaOH
1L
9-53
9.41 (cont'd)
Volume ratio =
b.
66.67 cm 3 NaOH(aq)
= 3.25 cm 3 caustic solution / cm 3 acid solution
3
20.49 cm H 2 SO 4 (aq)
b g
H 2 SO 4 aq : r = 9 mol H 2 O / 1 mol H 2 SO 4
eΔH j
o
f
soln
e
= ΔH fo
j
. − 65.23g
= −877
b g + eΔH f j H SO baq., r = 9g = b−81132
mol
kJ
o
H 2SO 4 l
2
kJ mol H 2 SO 4
4
e
je
j
. g cm 3 = 75.34 g , and
NaOH(aq) : The solution fed contains 66.67 cm 3 113
b
g
⇒ b75.34 − 8.00g g H O ⇒ b67.39 g H Ogb1 mol 18.02 gg = 3.74 mol H O
(0.2 mol NaOH) 40.00 g mol = 8.00 g NaOH
2
2
2
⇒ r = 3.74 mol H 2 O 0.20 mol NaOH = 18.7 mol H 2 O / mol NaOH
eΔH j
soln
e j
Na SO baq g:
eΔH j = eΔH j
2
4
o
f
o
f
= ΔH fo
o
f
soln
b g + e ΔH s j NaOHbsgbaq., r =18.7g = b−426.6 − 42.8g mol = −469.4
kJ
o
NaOH s
bg
Na 2SO 4 s
e
+ ΔH fo
j
b g
Na 2SO 4 aq
b
= −1384.5 − 117
.
kJ mol NaOH
kJ
= −1385.7 kJ mol Na SO
g mol
2
Extent of reaction: (n H 2SO ) final = (n H 2SO 4 ) fed + ν H 2SO 4 ξ ⇒ 0 = 010
. mol − (1)ξ ⇒ ξ = 010
. mol
4
Energy Balance:
Q = ΔH = ξΔH ro = ξ ( ΔH fo ) Na 2SO 4 ( aq) + 2( ΔH fo ) H 2 O( l) − ( ΔH fo ) H 2SO 4 ( aq) − 2( ΔH fo ) NaOH ( aq)
= (0.10 mol) −1385.7 + 2( −28584
. ) − ( −876.55) −(2)( −469.4)
9.42
kJ
= −14.2 kJ
mol
Table B.1,
given
a.
e
NaCl(aq): ΔH fo = ΔH fo
j
bg
NaCl s
e
+ ΔH so
j
B
∞
=
b−4110. + 4.87gkJ / mol = −4061. kJ mol
NaOH(aq):
Table B.1
B
e j b g + eΔH j = b−426.6 − 42.89gkJ / mol = −469.5 kJ mol
1
1
NaClbaq g + H Oblg → H bgg + Cl bgg + NaOHbaq g
2
2
ΔH = −469.5 − b−4061
. g − b−28584
. g kJ mol = 222.44 kJ mol
ΔH fo = ΔH fo
2
o
s
NaOH s
2
∞
2
o
r
b.
8500 ktonne Cl 2
yr
10 3 tonne 10 3 kg 10 3 g 1 mol Cl 2
1 ktonne 1 tonne 1 kg 70.91 g Cl 2
10 3 J 2.778 × 10 −7 kW ⋅ h 1 MW ⋅ h
1 kJ
1J
10 3 kW ⋅ h
9-54
222.44 kJ
0.5 mol Cl 2
= 148
. × 10 7 MW ⋅ h / yr
4
9.43
a.
bg
bg
b
g
CaCl 2 s + 10H 2 O l → CaCl 2 aq , r = 10
bg
bg
b
b1g
b2g
g
⋅ 6H Obsg
CaCl 2 ⋅ 6H 2 O s + 4 H 2 O l → CaCl 2 aq , r = 10
(3)
b1g − b2g ⇒ CaCl bsg + 6H Oblg → CaCl
⇒ ΔH = ΔH − ΔH b Hess' s law g = −97.26 kJ mol
From (1), ΔH = e ΔH j
b
g − eΔH j b g
⇒ e ΔH j
b
g = b−64.85 − 794.96g kJ mol = −859.81 kJ mol
2
o
r3
b.
2
o
r1
2
2
o
r2
o
r1
o
f
o
f
b
9.44
ΔH ro1 = −64.85 kJ mol
ΔH ro2 = +32.41 kJ mol
Basis: 1 mol NH 4
o
f
CaCl 2 aq , r =10
CaCl 2 aq , r =10
g SO produced
4
2
2 mol NH3 (g)
75°C
1mol H2SO4 (aq)
25°C
bg
CaCl 2 s
1 mol (NH4)2SO4 (aq)
25°C
b g b
2NH 3 g + H 2 SO 4 aq → NH 4
g SO baqg
4
2
a.
References : Elements at 25°C
b
z
FG
H
75
g
IJ
K
. + 183
. kJ / mol = −44.36 kJ mol (Table B.1, B.2)
NH 3 g, 75° C : H = ΔH fo + C p dT = −4619
b
g
25
e
H 2 SO 4 aq , 25° C : H = ΔH fo
j
bNH g SO baq, 25° Cg: H = eΔH jb
4 2
o
f
4
b g = −907.51
H 2SO 4 aq
NH 4
kJ mol H 2 SO 4 (Ta.ble B.1)
g SO baq g = −11731.
2
b
kJ mol NH 4
g SO
4
2
4
(Table B.1)
Energy balance:
Q = ΔH =
∑ n H − ∑ n H = b1gb−11731. g − b2gb−44.36g − b1gb−907.51g kJ
i
i
out
= −177 kJ ⇒ 177
b.
i
i
in
kJ withdrawn
mol NH 4 2 SO 4 produced
b
g
1 mole % (NH 4 ) 2 SO 4 solution ⇒
1 mol (NH 4 ) 2 SO 4
99 mol H 2 O 18 g
132 g
= 132 g (NH 4 ) 2 SO 4
mol
1782 g H 2 O
mol
1914 g solution
The heat transferred from the reactor in part (a) now goes to heat the product solution from
25D C to Tfinal ⇒ 177 kJ =
c.
. g
1914
1 kg
3
10 g
=
4.184 kJ (T − 25) D C
D
kg C
⇒ Tfinal = 47.1D C
In a real reactor, the final solution temperature will be less than the value calculated in part b, due
to heat loss to the surroundings. The final temperature will therefore be less than 47.1oC.
9-55
9.45
a.
b g
b g
b g
b g
H 2 SO 4 aq + 2 NaOH aq → Na 2 SO 4 aq + 2 H 2 O aq
1 mol H 2SO 4
49 mol H 2O
25°C
Basis : 1 mol H 2 SO 4 fed
1 mol Na 2SO 4
89 mol H 2O
40°C
2 mol NaOH
38 mol H 2O
25°C
bg b g bg b g
H SO eaq , r = 49, 25 Cj:
L
O
nH = b1 mol H SO gMe ΔH j
+ ΔH b r = 49gPb kJ molg + 49e ΔH j
b
g
N
Q
= b1g −8113
. − 73.3 = −884.6 kJ + 49e ΔH j
bg
NaOHeaq , r = 19, 25 Cj:
L
O
nH = b2 mol NaOH gMe ΔH j
+ ΔH b r = 19gPb kJ molg + 38e ΔH j
bg
N
Q
= b2g −426.6 − 42.8 = −938.8 kJ + 38e ΔH j
bg
Na SO eaq , r = 89, 40 Cj:
Reference states : Na s , H 2 g , S s , O 2 g at 25°C
D
2
4
2
o
f
4
o
s
H 2SO 4 l
o
f
o
f
bg
H 2O l
H 2O l
D
o
f
o
s
NaOH s
o
f
o
f
bg
H 2O l
H 2O l
D
2
4
1 kmol Na 2 SO 4
142.0 kg
1 kmol
b
nH = 1 mol Na 2 SO 4
gLMNeΔH j
o
f
b
89 kmol H 2 O 18.02 kg
kg,
= 0142
.
Na 2 SO 4
1 kmol
e
+ ΔH so
j
Na 2SO 4
g
OP + 89eΔH j
Q
o
f
kg ⇒ 1746
kg
= 1604
.
.
b g + mC p b40 − 25g
H 2O l
ΔH fo =−1384.5 kJ mol Table B.1
ΔH so =−1.2 kJ mol
= 4.814 kJ (kg⋅D C)
m=1.746 kg, C p ≈ C p
H 2O l
FH IK
e
bg
nH = −1276 kJ + 89 ΔH fo
j
bg
H 2O l
Energy balance: Q = ΔH =
∑
ni H i −
out
Mass of acid fed
1 mol H 2 SO 4
98.08 g H 2 SO 4
1 mol
⇒
b.
∑
−285.84 kJ mol
e
ni H i = 547.4 + 2 ΔH fo
j
in
+
49 mol H 2 O 18.02 g H 2 O
1 mol
b g = −24.3 kJ
H 2O l
= 981 g = 0.981 kg
Q
−24.3 kJ
=
⇒ 24.8 kJ / kg acid transferred from reactor contents
M acid 0.981 kg acid
If the reactor is adiabatic, the heat transferred from the reactor of Part(a) instead goes to heat
the product solution from 40°C to T f
⇒ 24.3 × 10 J =
3
1746
.
kg 4.184 kJ
D
kg ⋅ C
9-56
dT
f
i
D
− 40 C
⇒ T f = 43D C
9.46
a.
b g
b g
b g
bg
H 2SO 4 aq + 2NaOH aq → Na 2SO 4 aq + 2H 2 O l
H 2SO4 solution: :
75 ml of 4M H 2SO 4 solution ⇒
4 mol H 2SO 4
1 L acid soln
1L
75 mL
= 0.30 mol H 2 SO 4
3
10 mL
. g mLg = 92.25 g, (0.3 mol H SO )b98.08 g molg = 29.42 g H SO
b75 mLgb123
⇒ b92.25 − 29.42g g H O ⇒ b62.83 g H Ogb1 mol 18.02 gg = 3.49 mol H O
2
2
4
2
2
4
2
⇒ r = 3.49 mol H 2 O 0.30 mol H 2 SO 4 = 1163
. mol H 2 O / mol H 2 SO 4
eΔH j
o
f
soln
e j
= ΔH fo
e j
+ ΔH fo
bg
H 2SO 4 l
Table B.1,
Table B.11
b
B
=
g
H 2SO 4 aq ., r =11.63
kJ
. − 67.42g
b−81132
mol
= −878.74 kJ mol H 2 SO 4
NaOH solution required:
0.30 m ol H 2 SO 4 2 m ol N aOH
1 m ol H 2 SO 4
b g
10 3 m L
= 50.00 m L NaOH aq
1L
1 L NaOH(aq)
12 m ol N aOH
. g mLg = 68.5 g
b50.00 mLgb137
12 mol NaOH
1L
50 mL
1 L NaOH(aq) 103 mL
b
g
40 g/ mol NaOH
= 0.60 mol NaOH
b
gb
g
⇒
24.00 g NaOH
⇒ 68.5 − 24.00 g H 2 O ⇒ 44.5 g H 2 O 1 mol 18.02 g = 2.47 mol H 2 O
⇒ r = 2.47 mol H 2 O 0.6 mol NaOH =
eΔH j
o
f
soln
e j
= ΔH fo
4.12 mol H 2 O
mol NaOH
. g
b g + eΔHs j NaOHbsgbaq., r =4.12g = b−426.6 − 3510
mol
kJ
o
NaOH s
= −46170
. kJ mol NaOH
b g
eΔH j = eΔH j
Na 2SO 4 aq :
o
f
soln
o
f
. g
= −1385.7
b g + eΔH f j Na SO baq g = b−1384.5 − 117
mol
kJ
o
Na 2SO 4 s
2
kJ mol Na 2SO 4
4
mtotal = total mass of reactants or products = (92.25g H 2SO4 soln + 68.5g NaOH) = 160.75g = 0.161 kg
Extent of reaction: (nH 2SO ) final = (nH 2SO 4 ) fed + ν H 2SO 4 ξ ⇒ 0 = 0.30 mol − (1)ξ ⇒ ξ = 0.30 mol
4
Standard heat of reaction
e j
ΔH ro = ΔH fo
b g + 2e ΔH f j H Oblg − eΔH f j H SO baq g − 2eΔH f j NaOHbaq g
o
Na 2SO 4 aq
o
2
o
2
4
Energy Balance : Q = ΔH = ξΔHro + mtotal C p (T − 25) C
F
GH
= (0.30 mol)(1552
. kJ / mol) + (0161
. kg) 4.184
b.
Volumes are additive.
Heat transferred to and through the container wall is negligible.
9-57
I (T − 25) C = 0 ⇒ T = 94 C
J
kg C K
kJ
9.47
Basis : 50,000 mol flue gas/h
50,000 mol/h
0.00300 SO2
0.997 N 2
50°C
n4 (mol SO2 /h)
n5 (mol N 2 /h)
35°C
n1 (mol solution/h)
0.100 (NH 4 ) 2 SO 3
0.900 H 2O( l )
25°C
n2 (mol NH4 HSO3 /h)
1.5n2 (mol (NH 4)2 SO3 /h)
n3 (mol H 2 O(l )/h)
35°C
a
fb
g
n = a0.997fb50,000 mol hg = 49,850 mol N h
N balance:
NH balance: b2gb0100
. gbn g = n + b15
. gb2gn ⇒ n = 20n U| n = 5400 mol h
V⇒
S balance:
0100
. n + b0.00300gb50,000g = 150
. + n + 15
. n |W n = 270 mol NH HSO
270 mol NH HSO produced
1 mol H O consumed
H O balance: n = b0900
. gb5400g −
h
2 mol NH HSO produced
= 4725 mol H Oblg h
90% SO2 removal: n4 = 0.100 0.00300 50,000 mol h = 15.0 mol SO2 h
2
5
2
+
4
1
2
2
1
1
2
4
2
1
2
4
2
2
3
3
h
2
3
4
3
2
Heat of reaction:
− e ΔH j
e j
b g − e ΔH j b g b g − e ΔH j
. g kJ mol = −47.3 kJ mol
= 2b −760g − b −890g − b −296.90g − b −28584
References : N bgg, SO bgg, b NH g SO baq g, NH HSO baq g, H Oblg at 25°C
. kJ mol ( C from Table B.2)
SO eg, 50 Cj: H = z dC i dT = 101
ΔHro = 2 ΔH fo
o
f
NH 4 HSO 4 aq
2
2
NH 4
4 2
o
f
SO 3 aq
2
3
4
o
f
SO 2 ( g )
3
H 2 O(l)
2
50
e
j
SO 2 g, 35 C : H =
e j
N eg, 35 Cj:
p
p SO
2
2
25
35
zd
Cp
25
i
SO 2
dT = 0.40 kJ mol
N 2 g, 50 C : H = 0.73 kJ mol (Table B.8)
2
H = 0.292 kJ mol
Entering solution: H = 0
Effluent solution at 35°C
b g
mg h =
+
270 mol NH 4 HSO 3
h
b
g
99 g
mol
1.5 × 270 mol NH 4 2 SO 3 116 g 4725 mol H 2 O 18 g
g
+
= 159,000
h
1 mol
h
h
mol
nH = mC p ΔT =
159,000 g 4 J
h
g⋅° C
a35 − 25f° C
1 kJ
= 6360 kJ / h
10 3 J
Extent of reaction:
(nNH 4 HSO3 ) out = (nNH 4 HSO 3 ) in + ν NH 4 HSO 3 ξ ⇒ 270 mol / h = 0 + 2ξ ⇒ ξ = 135 mol / h
9-58
9.47 (cont'd)
Energy balance: Q = ΔH = ξΔH ro +
Q=
+
9.48
a.
−47.3 kJ
135 mol
h
effluent solution
6360
bg
∑n H − ∑n H
i
i
out
i
i
in
N 2 out
a fa f a fa f
mol
−22,000 kJ
− a50,000fa0.003fa1.01f − a 49,850fa0.73f =
h
SO 2 out
+ 15 0.40 + 49,850 0.292
bg
1h
1 kW
= −6.11 kW
3600 s 1 kJ s
bg
CH 4 g + 2O 2 g → CO 2 (g) + 2H 2 O v
at 25° C
Table B.1 HHV
B
B
e j
− ΔHco
HHV = 890.36 kJ / mol, LHV =
− 2 ΔH v
H 2O
b
g
= 890.36 − 2 44.01 kJ mol
= 802.34 kJ mol CH 4
bg
7
C 2 H 4 g + O 2 (g) → 2CO 2 (g) + 3H 2 O(v)
2
a
f
a
f
HHV = 1559.9 kJ / mol, LHV = 1559.9 − 3 44.01 kJ mol = 1427.87 kJ mol C 2 H 6
bg
C 3 H 8 g + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(v)
HHV = 2220.0 kJ / mol, LHV = 2220.0 − 4 44.01 kJ mol = 2043.96 kJ mol C 3 H 8
b HHV g
natural gas
b
gb
g b
gb
g b
gb
g b
gb
g b
gb
= 0.875 890.36 kJ mol + 0.070 1559.9 kJ mol + 0.020 2200.00 kJ mol
g
= 933 kJ mol
b LHV g
natural gas
b
gb
= 0.875 802.34 kJ mol + 0.070 1427.87 kJ mol + 0.020 2043.96 kJ mol
g
= 843 kJ mol
b.
g I
g I
gFGH mol
JK + b0.070 mol C H gFGH 30.07 mol
JK
g I
g I
1 kg
F
F
+ b0.020 mol C H gG 44.09
H mol JK + b0.035 mol N gGH 28.02 mol JK ] × 10 g = 0.01800 kg
b
1 mol natural gas ⇒ [ 0.875 mol CH 4 16.04
3
⇒
c.
9.49
2
8
6
2
843 kJ
1 mol
mol
0.01800 kg
3
= 46800 kJ kg
The enthalpy change when 1 kg of the natural gas at 25oC is burned completely with oxygen at
25oC and the products CO2(g) and H2O(v) are brought back to 25oC.
B
Table B.1
bg
e j
C s + O 2 (g) → CO 2 (g), ΔHco = ΔH fo
CO 2 ( g)
B
=
. kJ
−3935
mol
1 mol 103 g
= −32,764 kJ kg C
12.01 g 1 kg
Table B.1
bg
S s + O 2 (g) → SO 2 (g),
ΔHco
=
e j
ΔH fo
MSO =32 .064
2
SO 2
= −296.90 kJ mol
B
B
⇒
− 9261 kJ / kg S
Table B.1
bg
bg
e j
1
H 2 g + O 2 (g) → H 2 O l , ΔHco = ΔH fo
2
9-59
M H =1.008
2
bg
H 2O l
= −28584
. kJ mol H 2
B
⇒ − 141,790 kJ kg H
9.49 (cont'd)
a.
x0 (kg O) 2 kg H
H available for combustion = total H – H in H 2 O ; latter is kg coal 16 kg O
FG
H
Eq. (9.6-3) ⇒ HHV = 32,764C + 141,790 H −
A
in water
IJ
K
O
+ 9261S
8
This formula does not take into account the heats of formation of the chemical constituents of
coal.
b.
b
C = 0. 758 , H = 0. 051, O = 0. 082 , S = 0. 016 ⇒ HHV
1 kg coal ⇒
g
Dulong
= 31,646 kJ kg coal
0.016 kg S 64.07 kg SO 2 formed
= 0. 0320 kg SO 2 kg coal
32.06 kg S burned
0.0320 kg SO 2 kg coal
. × 10 −6 kg SO 2 kJ
φ=
= 101
31,646 kJ kg coal
c.
Diluting the stack gas lowers the mole fraction of SO2, but does not reduce SO2 emission rates. The
dilution does not affect the kg SO2/kJ ratio, so there is nothing to be gained by it.
b
bg
g
CH 4 + 2O 2 → CO 2 + 2H 2 O l , HHV = − ΔHco = 890.36 kJ mol Table B.1
9.50
bg
7
C 2 H 6 + O 2 → 2CO 2 + 3H 2 O l , HHV = 1559. 9 kJ mol
2
1
CO + O 2 → CO 2 , HHV = 282. 99 kJ mol
2
2.000 L
273.2K
2323 mm Hg
1 mol
Initial moles charged:
= 0.25 mol
25 + 273.2 K 760 mm Hg 22.4 L STP
(Assume ideal gas)
a
f
a f
Average mol. wt.: (4.929 g) (0.25 mol) = 19.72 g / mol
c b
g
b
gh
MW = 19.72 ⇒ x b16.04 g mol CH g + x b30.07g + b1 − x − x gb28.01g = 19.72 b1g
HHV = 963.7 kJ mol ⇒ x b890.36g + x b1559.9g + b1 − x − x gb282.99g = 963.7 b2g
Let x1 = mol CH 4 mol gas, x2 = mol C 2 H 6 mol gas ⇒ 1 − x1 − x2 mol CO mol gas
1
4
1
2
2
1
1
2
2
Solving (1) & (2) simultaneously yields
x1 = 0.725 mol CH 4 mol, x 2 = 0188
.
mol C 2 H 6 mol, 1 − x1 − x 2 = 0.087 mol CO mol
9.51
a.
Basis : 1mol/s fuel gas
CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(v), ΔHco = −890.36 kJ / mol
7
C 2 H 6 (g) + O 2 (g) → 2CO 2 (g) + 3H 2 O(v), ΔHco = −1559.9 kJ / mol
2
Excess O2, 25°C
n2 , mol CO 2
n3 , mol H 2 O
1 mol/s fuel gas, 25°C
85% CH4
15% C2H6
n4 , mol O 2
25°C
9-60
9.51 (cont’d)
1 mol / s fuel gas ⇒ 0.85 mol CH 4 / s , 0.15 mol C 2 H 6 / s
Theoretical oxygen =
2 mol O 2
0.85 mol CH 4
1 mol CH 4
s
+
3.5 mol O 2
0.15 mol C 2 H 6
1 mol C 2 H 6
s
= 2.225 mol O 2 / s
Assume 10% excess O 2 ⇒ O 2 fed = 1.1 × 2.225 = 2.448 mol O 2 / s
b gb g b gb g
H balance : 2n = b0.85gb4g + b015
. gb6g ⇒ n = 2.15 mol H O / s
10% excess O ⇒ n = b01
. gb2.225g mol O / s = 0.223 mol O / s
C balance : n 2 = 0.85 1 + 015
2 ⇒ n 2 = 115
.
. mol CO 2 / s
3
3
2
4
2
2
2
Extents of reaction: ξ 1 = n CH 4 = 0.85 mol / s, ξ 2 = n C 2 H 6 = 015
. mol / s
bg
bg bg bg
bg
Reference states: CH 4 g , C 2 H 6 g , N 2 g , O 2 g , H 2 O l , CO 2 (g) at 25o C
(We will use the values of
ΔHco
bg
given in Table B.1, which are based on H 2 O l as a
combustion product, and so must choose the liquid as a reference state for water)
Substance
nin
nout
Hout
CH 4
C2 H 6
mol
0.85
015
.
kJ mol
0
0
mol
−
−
kJ mol
−
−
O2
2.225
0
0.223
0
CO 2
−
−
115
.
0
H 2O v
−
−
2.15
H1
bg
e
Hin
j
H1 = ΔH v 25o C = 44.01 kJ / mol
Energy Balance :
e j
Q = nCH 4 ΔHco
b
CH 4
e j
+ nC 2 H 6 ΔHco
C2 H 6
+
gb
∑n H − ∑n H
i
out
i
i
i
in
g b
+ b2.15 mol / s H Ogb44.01 kJ / molg = −896 kW
gb
= 0.85 mol / s CH 4 −890.36 kJ mol + 015
. mol / s C 2 H 6 −1559.9 kJ mol
g
2
⇒ − Q = 896 kW (transferred from reactor)
b.
Constant Volume Process. The flowchart and stoichiometry and material balance calculations are
the same as in part (a), except that amounts replace flow rates (mol instead of mol/s, etc.)
1 mol fuel gas ⇒ 0.85 mol CH 4 , 0.15 mol C 2 H 6
Theoretical oxygen = 2.225 mol O 2
Assume 10% excess O 2 ⇒ O 2 fed = 1.1 × 2.225 = 2.448 mol O 2
b gb g b gb g
H balance : 2n = b0.85gb4g + b015
. gb6g ⇒ n
10% excess O ⇒ n = b01
. gb2.225g mol O
C balance : n2 = 0.85 1 + 015
. 2 ⇒ n2 = 115
. mol CO 2
3
2
4
9-61
3
= 2.15 mol H 2 O
2
= 0.223 mol O 2
9.51 (cont'd)
bg
bg bg bg
bg
Reference states: CH 4 g , C 2 H 6 g , N 2 g , O 2 g , H 2 O l , CO 2 (g) at 25o C
For a constant volume process the heat released or absorbed is determined by the internal
energy of reaction.
nin
nout
U in
U out
Substance
mol kJ mol mol kJ mol
0.85
0
CH 4
−
−
015
0
C2 H 6
.
−
−
2.225
0
0.223
0
O2
115
0
CO 2
.
−
−
2.15
H 2O v
U1
−
−
bg
e
j
e
8.314 J
j
U 1 = ΔU v 25 o C = ΔH v 25 o C − RT = 44.01 kJ / mol −
Eq. (9.1-5) ⇒
ΔU co
=
ΔH co
∑ν
− RT (
i
e
⇒ ΔU co
j
CH 4
eΔU j
o
c
C2 H 6
b
g
= −890.36 kJ mol −
∑ν
−
gaseous
products
i
1 kJ
298 K
mol K 1000 J
= 4153
.
kJ
mol
)
gaseous
reactants
8.314 J 298 K (1+ 2 − 1 − 2)
1 kJ
3
= −890.36
kJ
mol
mol K
10 J
8.314 J 298 K (3 + 2 − 35
. − 1) 1 kJ
kJ
= −1559.9 kJ mol −
= −156114
.
3
mol
mol K
10 J
b
g
Energy balance:
e
j + n eΔU j + ∑ n U − ∑ n U
. mol / s C H gb −156114
. kJ molg
= b0.85 mol / s CH gb −890.36 kJ molg + b015
. kJ / molg = −902 kJ
+ b2.15 mol / s H Ogb4153
Q = ΔU = n CH 4 ΔU co
CH 4
o
c
C2 H 6
C2 H 6
i
i
i
out
4
i
in
2
6
2
⇒ − Q = 902 kJ (transferred from reactor)
9.52
c.
Since the O2 (and N2 if air were used) are at 25°C at both the inlet and outlet of this process, their
specific enthalpies or internal energies are zero and their amounts therefore have no effect on the
calculated values of ΔH and ΔU .
a.
n fuel ( − ΔHco ) = Ws − Ql (Rate of heat release due to combustion = shaft work + rate of heat loss)
V (gal)
28.317 L
h
7.4805 gal
100 hp
=
0.700 kg 10 3 g 49 kJ
L
1J/s
1.341 × 10 −3 hp
1 kg
1 kJ
10 3 J
g
3600 s
h
+
15 × 10 6 kJ
⇒ V = 2.5 gal / h
298 h
b.
The work delivered would be less since more of the energy released by combustion would go into
heating the exhaust gas.
c.
Heat loss increases as Ta decreases.
Lubricating oil becomes thicker, so more energy goes to overcoming friction.
9-62
9.53
a.
Energy balance: ΔU = 0 ⇒
a
f
b
n lb m fuel burned
b
g
ΔU co (Btu)
gb
ga
lb m
b
g
+ mCv Tout − 77° F = 0
f
⇒ 0.00215 ΔU co + 4.62 lb m 0.900 Btu lb m ⋅° F 87.06° F − 77.00° F = 0
⇒ ΔU co = −19500 Btu lb m
b.
The reaction for which we determined ΔU co is
1 lb m oil + aO 2 (g) → bCO 2 (g) + cH 2 O(v)
(1)
The higher heating value is ΔH r for the reaction
1 lb m oil + aO 2 ( g) → bCO 2 (g) + cH 2 O(l)
(2)
o
o
= ΔU c1
+ RT (b + c − a)
Eq. (9.1-5) on p. 441 ⇒ ΔHc1
o
o
= − ΔHc1
+ cΔH v (H 2 O, 77 F)
Eq. (9.6-1) on p. 462 ⇒ − ΔHc2
( HHV )
( LHV )
To calculate the higher heating value, we therefore need
a = lb - moles of O 2 that react with 1 lb m fuel oil
b = lb - moles of CO 2 formed when 1 lb m fuel oil is burned
c = lb - moles of H 2 O formed when 1 lb m fuel oil is burned
9.54
a.
bg
bg
3
e j
ΔH ro = ΔHco
CH 3 OH v + O 2 (g) → CO 2 (g) + 2H 2 O l
2
bg
CH 3OH v
= −764.0
kJ
mol
Basis : 1 mol CH 3 OH fed and burned
Q2(kJ)
1 mol CH 3OH( l )
25°C, 1.1 atm
vaporizer
1 mol CH 3OH( v)
100°C
1 atm
Q 1 (kJ)
reactor
n 0 (mol O2 )
3.76 n 0 (mol N2 )
100°C
Overall C balance:
1 mol CH 3OH
b
gb
1 mol C
g
1 mol CH 3OH
b
Effluent at 300°C, 1 atm
n p (mol dry gas)
0.048 mol CO 2/mol D.G.
0.143 mol O 2/mol D.G.
0.809 mol N 2/mol D.G.
n w (mol H 2O)
gb g
= n p 0.048 1 ⇒ n p = 20.83 mol dry gas
N 2 balance: 3.76n0 = 20.83 0.809 ⇒ n0 = 4.482 mol O 2
Theoretical O 2 :
% excess air =
b
b1 mol CH OHgb15. mol O
3
2
g
mol CH 3OH = 15
. mol O 2
(4.482 − 15
. ) mol O 2
× 100% = 200% excess air
15
. mol O 2
gb
g
bg
H balance: 1 mol CH 3OH 4 mol H 1 mol CH 3OH = nw 2 ⇒ nw = 2 mol H 2 O
(An atomic O balance ⇒ 9.96 mol O = 9.96 mol O , so that the results are consistent.)
pw∗ =
d i
nw
2 mol H 2 O
×P=
× 760 mm Hg = 66.58 mm Hg = pw∗ Tdp
2 + 20.83 mol
nw + n p
a
f
9-63
Table B.3
⇒
Tdp = 44.1° C
9.54 (cont'd)
b.
Energy balance on vaporizer:
LM
Q = ΔH = nΔH = 1 mol M
MN
OP kJ
C dT + ΔH +
C dT
PP mol = 40.33 kJ
A
A
A
Q
References : CH OHa vf, N (g), O (g), CO (g), H Oalf at 25° C
1
z
pl
25
v
Table B.2
CH 3 OH
2
16.85
100
pv
64.7
Table B.2
Table B.1
3
n in
(mol)
1.00
Substance
z
64.7
2
2
2
n out
(mol)
−
H in
(kJ / mol)
3.603
H out
(kJ / mol)
−
2.187
16.85
8118
.
4.482
2.235
2.98
8.470
CO 2
−
−
100
.
11578
.
H 2O
−
−
2.00
5358
.
N2
O2
af
H T = Hi for N 2 , O 2 , CO 2 (Table B.8)
d
af
i
= ΔHv 25 C + Hi for H 2 O v (Eq. 9.6 - 2a on p. 462, Table B.8)
=
z
af
T
25
C p dT for CH 3 OH v (Table B.2)
bg
(Note: H 2 O l was chosen as the reference state since the given value of ΔHco presumes liquid
water as the product.)
Extent of reaction: (nCH 3OH ) out = (nCH 3OH ) in + ν CH 3OH ξ ⇒ 0 = 1 mol − ξ ⇒ ξ = 1 mol
Energy balance on reactor: Q2 = ξΔHco +
b gb
g b
gb
g
b
∑n H − ∑n H
gb
i
out
i
g
i
i
in
= 1 −764.0 + 16.85 8118
.
+…− 4.482 2.235 kJ
bTable B.1g
= −534 kJ ⇒ 534 kJ transferred from reactor
9.55
a.
CH 4 + 2O 2 → CO 2 + 2H 2 O
3
CH 4 + O 2 → CO + 2H 2 O
2
Basis : 1000 mol CH 4 h fed
Q(kJ/s)
1000 mol CH4 /s
25°C
Stack gas, 400°C
n1 (mol CH4 /s)
n2 (mol O2 /s)
3.76n0 (mol N2 /s)
n3 (mol CO/s)
10 n3 (mol CO/s)
n4 (mol H2 O/s)
n0 (mol O2 /s)
3.76n0 (mol N2 /s)
100°C
90% combustion ⇒ n1 = 0.10 (1000 ) = 100 mol CH 4 s
Theoretical O2 required = 2000 mol/s
9-64
9.55 (cont’d)
10% excess O2 ⇒ O 2 fed=1.1(2000 mol/s)=2200 mol/s
C balance:
(1000 mol CH 4 s )(1 mol C
mol CH 4 ) = (100 )(1) + n3 (1) + 10n3 (1) ⇒ n3 = 81.8 mol CO s
⇒ 10n3 = 818 mol CO 2 s
H balance: (1000 )( 4 ) = (100 )( 4 ) + 2n4 ⇒ n4 = 1800 mol H 2 O s
O balance:
( 2200 )( 2 ) = 2n2 + (81.8)(1) + (818)( 2 ) + (1800 )(1) ⇒ n2 = 441 mol O2
s
af b g b g b g
References :C s , H 2 g , O 2 g , N 2 g at 25 C
Hˆ in
nin
Substance
( mol s ) ( kJ
CH 4
O2
N2
CO
CO 2
H2O
Bo
H = ΔH f +
( mol s ) ( kJ
−74.85
2.24
2.19
−
−
−
1000
2200
8272
−
−
−
Table B.1
mol )
z
Hˆ out
nout
100
441
8272
81.8
818
1800
mol )
−57.62
11.72
11.15
−99.27
−377.2
−228.63
B
Table B.2
T
25
C p dT for CH 4
B
Table B.8
= ΔHfo + Hi ( T) for others
Energy balance: Q = ΔH =
∑ n Hˆ − ∑ n Hˆ
i
out
b.
i
i
i
= −5.85 × 105 kJ s (kW)
in
(iii)
A (increases) ⇒ −Q A
%XS A ⇒ − Q B (more energy required to heat additional O and N to 400 C, therefore
less energy transferred.)
S
A ⇒ −Q A (reaction to form CO2 has a greater heat of combustion and so releases
(iv)
more thermal energy)
Tstack ⇒ − Q (more energy required to heat combustion products)
(i)
(ii)
Tair
o
2
2
CO 2 CO
A
B
9-65
CH 4 + 2O 2 → CO 2 + 2H 2 O, C 2 H 6 +
9.56
7
O 2 → 2 CO 2 + 3H 2 O
2
Basis : 100 mol stack gas. Assume ideal gas behavior.
n1 (mol CH4 )
n2 (mol C2 H6 )
Vf (m3 at 25°C, 1 atm)
100 mol at 800°C, 1 atm
0.0532 mol CO 2 /mol
0.0160 mol CO/mol
0.0732 mol O2 /mol
0.1224 mol H 2 O/mol
0.7352 mol N2 /mol
n3 (mol O2 )
3.76n3 (mol N2 )
200°C, 1 atm
a.
b gb g
C balance: n b1g + n b2g = b100gb0.0532gb1g + b100gb0.0160gb1gU| n = 3.72 mol CH
V| ⇒ n = 160
. mol C H
.
H balance: n b4g + n b6g = b100gb01224
gb2g
W
. gmol fuel gas 22.4 LbSTPg 298.2 K 1 m
b3.72 + 160
m
.
V =
= 0130
N 2 balance: 3.76n3 = 100 0.7352 mol N 2 ⇒ n3 = 19.55 mol O 2 fed
1
1
4
1
2
2
2
2
6
3
3
f
1 mol
Theoretical O 2 =
3.72 mol CH 4
3.72 mol CH 4
2
b.
2 mol O 2 1.60 mol C 2 H 6 3.5 mol O 2
+
= 13.04 mol O 2
1 mol CH 4
1 mol C 2 H 6
UV ⇒ 69.9 mole% CH
. mole% C H
1.60 mol C H W 301
b19.55 − 13.04gmol O in excess × 100% = 50% excess air
Fuel composition:
% Excess air:
273.2 K 103 L
4
2
6
6
2
13.04 mol O 2 required
bg b g b g b g
References : C s , H 2 g , O 2 g , N 2 g at 25° C
CH 4
n in
mol
3.72
H in
kJ / mol
−74.85
n out
mol
−
H out
kJ / mol
−
C2 H 6
.
160
−84.67
−
−
O2
19.55
5.31
7.32
25.35
N2
.
7352
.
513
.
7352
.
2386
CO
−
−
.
160
−86.39
CO 2
−
−
5.32
.
−3561
H2O
−
−
12.24 −212.78
Substance
9-66
9.56 (cont’d)
B
Table B.1
H = ΔH fo +
z
Table B.2, for
CH 4 , C2 H 6
B
T
C p dT
25
B
= ΔH fo + Hi (T) for O 2 , N 2 , CO, CO 2 , H 2 O v
Table B.8
bg
Energy balance:
Q = ΔH =
∑n H − ∑n H
i
out
i
i
in
i
=
−2764 kJ
0.130 m 3 fuel
9-67
= −2.13 × 104 kJ m3 fuel
9.57
Basis : 50000 lb m coal fed h ⇒
b0.730gb50000glb
mC
1b - mole C
h
12.01 lb m
= 3039 1b - mole C h
. = 2327 lb - moles H h (does not include H in water)
b0.047gb50000g 101
b0.037gb50000g 32.07 = 57.7 lb - moles S h
b0.068gb50000g 18.02 = 189 lb - moles H O h
. gb50000g = 5900 lb ash h
b0118
2
m
a.
50,000 lb m coal/h
3039 lb-moles C/h
2327 lb-moles H/h
57.7 lb-moles S/h
189 lb-moles H 2O/h
5900 lb m ash/h
77°F, 1 atm (assume)
Stack gas at 600°F, 1 atm (assume)
n 2 (lb-moles CO2 /h)
n 3 (lb-moles H2 O/h)
n 4 (lb-moles SO 2 /h)
n 5 (lb-moles O2 /h)
n 6 (lb-moles N2 /h)
m 7 (lbm fly ash/h)
n 1 (lb-moles air/h)
0.210 O 2
0.790 N 2
77°F, 1 atm (assume)
m 8 (lbm slag/h) at 600°F
0.287 lb mC/lb m
0.016 lb mS/lb m
0.697 lb mash/lb m
Feed rate of air :
b
g
O 2 required to oxidize carbon C + O 2 → CO 2 =
3039 lb - moles C 1 lb - mole O 2
h
1 lb - mole C
= 3039 lb - moles O 2 h
Air fed: n1 =
1.5 × 3039 lb - moles O 2 fed
1 mole air
h
0.210 mole O 2
= 21710 lb - moles air h
b
g
8 = 0.30 5900 lb m h ⇒ m 8 = 2540 lb m slag / h
30% ash in coal emerges in slag ⇒ 0.697m
b g
⇒ m 7 = 0.700 5900 = 4130 lb m fly ash h
b
g
b
gb g
C balance: 3039 lb - moles C h = n 2 + 0.287 2540 12.01
M CO 2 = 44.01
⇒ n 2 = 2978 lb - moles CO 2 h
b
131
. × 10 5 lb m CO 2 h
g b gb g
H balance: 2327 lb - moles H h + 189 2 = 2n 3
⇒ n 3 = 1352.5 lb - moles H 2 O h
b
M H 2 O =18.02
2.44 × 10 4 lb m H 2 O h
g
N 2 balance: n 6 = 0.790 21710 lb - moles h = 17150 lb - moles N 2 h
b
g bg
b
M N 2 = 28.02
4.81 × 105 lb m N 2 h
g
S balance: 57.7 lb - moles S h = 1 n 4 + 0.016 2540 32.06
⇒ n 4 = 56.4 lb - moles SO 2 h
b gb g b gb
M SO 2 = 64.2
3620 lb m SO 2 h
gb g b
gb g b
gb g b gb g
O balance: 189 1 + 0.21 21710 2 = 2978 2 + 1352.5 1 + 56.4 2 + 2n 5
bcoalg
bair g
bCO g
2
⇒ n 5 = 943 lb - moles O 2 h ⇒ 30200 lb m O 2 h
9- 68
eH O j
2
bSO g
2
bO g
2
9.57 (cont'd)
Summary of component mass flow rates
Stack gas at 600° F, 1 atm
2978 lb - moles CO 2 h ⇒
131000 lb m CO 2 h
1352.5 lb - moles H 2 O h ⇒ 24400 lb m H 2 O h
56.4 lb - moles SO 2 h ⇒
3620 lb m SO 2 h
943 lb - moles O 2 h ⇒
30200 lb m O 2 h
17150 lb - moles N 2 h ⇒
48100 lb m N 2 h
674,350 lbm stack gas/h
4130 lb m fly ash h
b
gb g
Check: 50000 + 21710 29
⇒
b679600g
in
b
⇔ 674350 + 2540
in
g
⇔ 676900
out
out
(0.4% roundoff error)
Total molar flow rate = 22480 lb - moles h at 600° F , 1 atm (excluding fly ash)
⇒V=
b.
a f
1060° R
= 1.74 × 10 7 ft 3 h
492° R
22480 lb - moles 359 ft 3 STP
h
1 lb - mole
References: Coal components, air at 77°F ⇒ ∑ ni H i = 0
in
Stack gas: nH =
674350 lb m
7.063 Btu
lb - mole⋅° F 28.02 lb m
h
2540 lb m
Slag: nH =
h
b600 − 77g° F = 8.90 × 10
1 lb - mole
0.22 Btu
lb m ⋅° F
b600 − 77g° F = 2.92 × 10
b
5
i
i
i
out
=
5 × 104 lb m
h
Btu h
Btu h
g ∑ n H − ∑ n H
Energy balance: Q = ΔH = n coal burned ΔH co 77° F +
7
i
in
. × 104 Btu
−18
+ 8.90 × 107 + 2.92 × 105 Btu h
lb m
e
j
. × 108 Btu h
= −811
Power generated =
. × 10 jBtu
b0.35ge811
8
e
Q = −811
. × 108 Btu h
⇒
j b5000 lb
1W
3600 s 9.486 × 10
h
c.
1 hr
m
−4
1 MW
Btu s 10 6 W
. MW
= 831
g
coal h = −162
. × 104 Btu lb m coal
1. 62 × 104 Btu lb m
− Q
= 0. 901
=
HHV 1. 80 × 104 Btu lb m
Some of the heat of combustion goes to vaporize water and heat the stack gas.
d.
− Q HHV would be closer to 1. Use heat exchange between the entering air and the stack gas.
9- 69
9.58
b.
Basis : 1 mol fuel gas/s
n 0 (mol O 2 s)
3.76n 0 (mol N 2 s)
Stack gas, Ts ( o C)
n O2 (mol O 2 / s)
3.76n 0 (mol N 2 / s)
o
Ta ( C)
n CO (mol CO / s)
rn CO (mol CO 2 / s)
1 mol / s @ 25o C
x m (mol CH 4 / mol)
n H 2 O (mol H 2 O / s)
n Ar (mol Ar / s)
x a (mol Ar / mol)
(1 − x m − x a ) (mol C 2 H 6 / mol)
CH 4 + 2O 2 → CO 2 + 2 H 2 O
C2 H 6 +
7
2
O 2 → 2CO 2 + 3H 2 O
Pxs
) 2 x m + 35
. (1 − x m − x a )
100
x + 2(1 − x m − x a )
C balance: x m + 2(1 − x m − x a ) = (1 + r )n CO ⇒ n CO = m
(1 + r )
Percent excess air: n 0 = (1 +
H balance: 4 x m + 6(1 − x m − x a ) = 2n H 2 O ⇒ n H 2 O = 2 x m + 3(1 − x m − x a )
O balance: 2n 0 = 2n O 2 + n CO + 2 r n CO + n H 2 O ⇒ n O 2 = n 0 − n CO (1 + 2r ) / 2 − n H 2 O / 2
References : C(s), H2(g), O2(g), N2(g) at 25°C
Substance
nin
H in
nout
CH 4
xm
0
−
C2 H 6
A
O2
(1 − xm − x A )
xA
no
−
xA
nO2
N2
CO
CO 2
H 2O
3.76no
−
−
−
0
0
H 1
H
H i = ( ΔH f ) i +
c.
3.76no
nCO
r nCO
nH 2 O
2
−
−
−
z
H out
−
−
H 3
H 4
H 5
H 6
H
7
H 8
Ta or Ts Table B.2
B
C p ,i dT
25
Given : x m = 0.85, x a = 0.05, Px s = 5%, r = 10.0, Ta = 150 o C, Ts = 700 o C
⇒ n o = 2.153, n CO = 0.0955, n H 2 O = 2.00, n O 2 = 01500
.
H 1 (kJ / mol) = 8.091, H 2 = 29.588, H 3 = 0.702, H 4 = 3.279,
H = 166.72, H = −8.567, H = −345.35, H = − 433.82
5
Energy balance: Q =
6
∑ n
7
out H out
−
8
∑ n
in H in
9- 70
= −655 kW
Xa
Pxs
r
Ta
Ts
Q
0
150
150
150
150
150
150
700
700
700
700
700
700
10
10
10
10
10
150
150
150
150
150
700
700
700
700
700
-
-200 0
996
905
813
722
631
-905
-869
-799
0.1
0.2
0.3
0.4
0.5
5
5
5
5
5
10
10
10
10
10
150
150
150
150
150
700
700
700
700
700
-893
-871
860
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
5
5
5
5
5
5
5
5
5
1
2
3
4
5
10
20
50
100
150
150
150
150
150
150
150
150
150
700
700
700
700
700
700
700
700
700
-722
-796
-834
-856
-871
-905
-924
-936
-941
0.1
0.1
0.1
0.1
0.1
5
5
5
5
5
10
10
10
10
10
25
100
150
200
250
700
700
700
700
700
-852
-883
-905
-926
-948
0.1
0.1
0.1
0.1
0.1
0.1
5
5
5
5
5
5
10
10
10
10
10
10
150 500
150 600
150 700
150 800
150 900
150 1000
-1014
-960
-905
-848
-790
-731
9- 71
0.2
0.4
0.6
0.8
1
1.2
80
100
120
0.4
0.5
0.6
80
100
120
200
250
300
800
1000
1200
-400
Q
-600
-800
-1000
-1200
Xm
0
0
-
20
40
60
Q 400
600
-800
1000
Pxs
-850
-860 0
0.1
0.2
0.3
-870
Q -880
-890
-900
-910
x
0
0
20
40
60
-200
-400
Q
10
10
10
10
10
10
-600
-800
-1000
r
-800
-850
Q
0.0
5
0.0
5
1
0.0
5
1
0.0
5
1
0.0
5
1
0.0
5
1
0.1
5
0.1 10
0.1 20
0.1 50
0.1 100
0
50
100
150
-900
-950
-1000
Ta
0
0
200
400
600
-500
Q
9.58 (cont'd)
d.
-1000
-1500
Ts
9.59
a.
Basis:
207.4 liters 273.2 K 1.1 atm
1 mol
= 10.0 mols s fuel gas to furnace
s
278.2 K 1.0 atm 22.4 liters STP
b g
H = C 6 H 14 ; M = CH 4
mw (kg H 2O( l)/s)
25°C
Qc (kW)
n0 (mol/s)
y 0 (mol C 6H 14/mol)
(1 – y 0) (mol CH 4/mol)
60°C, 1.2 atm
Tdp = 55°C
10.0 mol/s at 5°C, 1.1 atm
y2 (mol C 6H 14/mol)
(1 – y2) (mol CH 4/mol)
sat'd with C 6H 14
condenser
b
g
na (mol air/s) @ 200°C
0.21 mol O /mol
2
0.79 mol N /mol
2
100% excess
B
Antoine Eq.
Tdp = 55° C ⇒ y 0 P = p H 55° C
⇒ y0 =
Stack gas at 400°C, 1 atm
n3 (mol O 2/s)
n4 (mol N 2/s)
n5 (mol CO 2/s)
n6 (mol H 2O( v)/s)
reactor
nb (mol C 6H 14( l)/s)
α
mw (kg H 2O( v)/s)
10 bars, sat'd
=
483.3 mm Hg
483.3 mm Hg
= 0.530 mol C 6 H 14 mol ⇒ 0.470 mol CH 4 mol
1.2 × 760 mm Hg
Saturation at condenser outlet:
p ∗H 5° C
58.89 mm Hg
y2 =
=
= 0.070 mol C 6 H 14 mol = 0.93% mol CH 4 mol
P
1.1 × 760 mm Hg
b g
b
g
b
g
Methane balance on condenser: n0 1 − y0 = 10.0 1 − y2
Hexane balance on condenser: n0 y0 = n b + 10.0 y2
Volume of condensate =
bg
9.78 mol C 6 H 14 l
⇒
y2 = 0.070
⇒
n0 = 19.78 mol s
n b = 9.78 mol C 6 H 14 s condensed
n0 =19.78
y0 = 0.530
y2 = 0.070
cm 3
86.17 g
s
y0 = 0.530
1L
3
0.659 g 10 cm
mol
A
A
Table B.1
3600 s
3
1h
Table B.1
= 4600 L C 6 H 14 (l ) h
b.
e
j
e
References : CH 4 g, 5D C , C 6 H 14 l, 5D C
Substance
CH 4
bg
bl g
C 6 H 14 v
bg
H out
nout
(mol / s) (kJ / mol) (mol / s) (kJ / mol)
1985
9.30
0
9.30
.
10.48
41212
.
0.70
32.940
−
−
9.78
0
C 6 H 14
CH 4 g : H =
H in
nin
j
z
B
Table B.2
T
5
bg
C 6 H 14 v : H =
C p dT
Condenser energy balance: Q c = ΔH =
∑ n H − ∑ n H
i
out
9- 72
i
i
in
i
z
B
Table B.1
Tb
5
B
Table B.1
C pR dT + ΔH v +
= −427 kW
z
T
Tb
C pv dT
9.59 (cont'd)
CH 4 + 2O 2 → CO 2 + 2H 2 O , C 6 H 14 +
Theoretical O 2 :
2
O 2 → 6CO 2 + 7H 2 O
9.30 mol CH 4
2 mol O 2
s
1 mol CH 4
b g = 2 × bO g
balance: 0.79b240.95g = n ⇒ n
100% excess ⇒ O 2
N2
19
2 theor.
fed
4
4
+
0.70 mol C 6 H 14
9.5 mol O 2
s
1 mol C 6 H 14
= 25.3 mol O 2 s
⇒ 0.21na = 2 × 25.3 ⇒ na = 240.95 mol air s
= 190.35 mol N 2 s
C balance:
9.30 mol CH 4
1 mol C
s
1 mol CH 4
+
0.70 mol C 6 H 14
6 mol C
1 mol C 6 H 14
b
n5 mol CO 2
=
g
1 mol C
1 mol CO 2
⇒ n5 = 135
. mol CO 2 s
H balance:
b9.30 mol CH sgb4 mol H mol CH g + b0.70gb14g = n b2g ⇒ n = 235. mol H O
1
Since combustion is complete, bO g
= bO g
= bO g ⇒ n = 25.3 mol O s
2
References : Cbsg, H bgg, O bgg, N bgg at 25° C for reactor side, H Oblg at triple point for
4
4
6
2 remaining
2
2
6
2 excess
2
2 fed
2
3
2
2
steam side (reference state for steam tables)
H in
n in
Substance
n out
H out
mol / s
9.30
kJ / mol mol / s
−75553
.
−
kJ / mol
−
C 6 H 14 v
0.70
−170.07 −
−
O2
50.6
5.31
25.3
1172
.
N2
190.35
.
513
190.35
.
1115
CO 2
−
−
135
.
−377.15
H 2O v
−
−
235
.
−228.60
CH 4
bg
bg
H Ob boiler water g
2
m w (kg / s) 104.8
Table B.1 and B.2
B
=
bg
H T
z
m w ( kg / s) 2776.2
T
ΔH fo + C p dT
for CH 4 , C 6 H 14
25
Table B.1 and B.8
B
=
bg
bg
ΔH fo + H i T for O 2 , N 2 , CO 2 , H 2 O v
Energy balance on reactor (assume adiabatic):
ΔH =
∑ n H − ∑ n H
i
out
i
i
i
b
g
= 0 ⇒ −8468 + m w 2776.2 − 104.8 = 0 ⇒ m w = 3.2 kg steam s
in
9- 73
9.60
a.
CH 4 + 2O2 → CO2 + 2H 2O
Basis: 450 kmol CH 4 fed h
n a ( kmol air / h)@25 o C
o
0.21 kmol O 2 / kmol
0.79 kmol N 2 / kmol
450 kmol CH 4 / h @ 25 o C
Stack gas@300 C
n 1 (kmol CO 2 / h)
n 2 (kmol H 2 O / h)
Q ( kJ / h)
n 3 (kmol O 2 / h)
n 4 (kmol N 2 / h)
m w [kg H 2 O(l) / h]
m w [kg H 2 O(v) / h]
o
o
25 C
17 bar, 250 C
450 kmol CH 4
h
Air fed: n a =
2 kmol O 2 req' d 1.2 kmol O 2 fed
1 kmol air
1 kmol CH 4
1 kmol O 2 req' d 0.21 kmol O 2
= 5143 kmol air h
450 kmol / h CH 4 react ⇒ n1 = 450 kmol CO 2 h , n 2 = 900 kmol H 2 O h
b ge
j
N 2 balance: n 4 = 0.79 5143
.
× 10 6 mol h = 4060 kmol N 2 h
Molecular O 2 balance:
b gb g mol Oh fed − 450 kmol hCH
= 0.0805
y
450 kmol CO h U
|
.
900 kmol H O h |
|V ⇒ y = 0161
y = 0.726
4060 kmol N h |
| y = 0.0322
180 kmol O h |
W
n 3 = 0.21 5143
2
4
react
2 mol O 2
= 180 kmol O 2 h
1 mol CH 4
CO 2
2
2
H 2O
2
N2
2
O2
5590 kmol / h
Mean heat capacity of stack gas
Cp =
∑y C
i
pi
b
gb
g b
gb
g b
gb
g b
gb
= 0.0805 0.0423 + 0161
.
0.0343 + 0.726 0.0297 + 0.0322 0.0312
= 0.0315 kJ mol ⋅ D C
Energy balance on furnace (combustion side only)
bg
bg bg
bg
bg
References: CH 4 g , CO 2 g , O 2 g , N 2 g , H 2 O l at 25D C
n in
H in
n out H out
Substance (kmol / h) (kJ / kmol) (kJ / h)
CH 4
450
0
−
Air
5143
0
−
Stack gas
−
−
H p
Extent of reaction:
ξ = n CH 4 = 450 kmol / h
9- 74
g
9.60 (cont’d)
H p = n 2 ( ΔH v ) H
o
2 O(25 C)
+ n stack gas (C p ) stack gas (Tstack gas − 25o C)
3
5590 kmol 10 3 mol
= 180 kmol H 2 O 10 mol 44.01 kJ +
h
1 kmol
mol
h
1 kmol
7
= 5.63 × 10 kJ / h
Q = ΔH = ξ ( ΔH co ) CH 4 +
∑ n H − ∑ n H
i
i
=
i
i
FG 450 kmol IJ FG1000 mol IJ FG −890.36 kJ IJ + 5.63 × 10
H h K H kmol K H
mol K
out
0.0315 kJ (300 - 25) o C
mol ⋅ o C
in
7
kJ
kJ
= −3.44 × 10 8
h
h
Energy balance on steam boiler
LM FG IJ OP LMb2914 − 105g kJ OP
kg Q
N H KQ N
Table B.7 Table B.6
kg
kJ
= m w
Q = m w ΔH w ⇒ + 3.44 × 10 8
h
h
⇒ m w = 123
. × 10 5 kg steam / h
b.
n a (mol air/h) at
Ta (°C)
Stack gas
n 1 (mol CO 2/h)
n 1 (mol CO 2/h)
air
45 kmol CH 4 /h
n 2 (mol H O/h)
n 2 (mol H 2O/h)
2
furnace
25°C
n 3 (mol O 2/h)
n 3 (mol O 2/h)
preheater
n 4 (mol N 2/h)
n 4 (mol N 2/h)
300°C
150°C
mw (kg H 2O/h)
mw (kg H 2O/h)
Liquid, 25°C
vapor, 17 bars
n a (mol air/h) at 25°C
250°C
0.21 O 2
0.79 N 2
E.B. on overall process: The material balances and the energy balance are identical to those of part
(a), except that the stack gas exits at 150oC instead of 300oC.
b g b g b g b g bg
b
g
H Oalf at triple point (steam table reference) (steam tube side)
References: CH 4 g , CO 2 g , O 2 g , N 2 g , H 2 O l at 25D C furnace side
2
Substance
n in
(kmol / h)
H in
(kJ / kmol)
n out H out
(kJ / h)
CH 4
Air
Stack gas
450
5143
−
0
0
−
−
−
H p
H 2O
H p = n 2 ( ΔH v ) H
o
2 O(25 C)
m w ( kg / h) 105 kJ / kg
m w ( kg / h) 2914 kJ / kg
+ n stack gas (C p ) stack gas (Tstack gas − 25o C)
3
5590 kmol 10 3 mol
= 180 kmol H 2 O 10 mol 44.01 kJ +
h
1 kmol
mol
h
1 kmol
7
= 2.99 × 10 kJ / h
9- 75
0.0315 kJ (150 - 25) o C
mol ⋅ o C
9.60 (cont’d)
ΔH = ξ ( ΔH co ) CH 4 +
∑ n H − ∑ n H
i
i
i
i
=0
FG 450 kmol IJ FG1000 mol IJ FG −890.36 kJ IJ + 2.99 × 10 kJ
H h K H kmol K H
mol K
h
L F kg I OL
kJ O
+ Mm G J P Mb2914 − 105g P = 0 ⇒ m = 1.32 × 10 kg steam / h
H
K
h
kg
N
QN
Q
+ d ΔH i = 0
Energy balance on preheater: ΔH = d ΔH i
kJ
mol 0.0315 kJ b150 − 300g C
= −2.64 × 10
bΔH g = nC ΔT = 5590hkmol 101 kmol
h
mol ⋅ C
out
in
⇒
7
5
w
w
stack gas
air
D
3
stack gas
b − ΔH g
stack gas
b g
= ΔH
D
7
kJ
= 5.133
= n a H air (Ta ) ⇒ H air (Ta ) = 2.64 × 10 kJ / h 1 kmol
5143 kmol / h 10 3 mol
mol
air
Table B.8
.
kJ / mol
H air = 5133
c.
7
p
Ta = 199 o C
The energy balance on the furnace includes the term −
∑n
in H in .
If the air is preheated and the
stack gas temperature remains the same, this term and hence Q become more negative, meaning
that more heat is transferred to the boiler water and more steam is produced. The stack gas is a
logical heating medium since it is available at a high temperature and costs nothing.
9.61
Basis: 40000 kg coal h ⇒
b0.76 × 40000gkg C
h
Assume coal enters at 25°C
103 g 1 mol C
= 2.531 × 106 mol C h
1 kg 12.01 g
. j = 198
. × 10
b0.05 × 4000gkg H h e10 101
b0.08 × 4000gkg O h e10 16.0j = 2.00 × 10
. × 40000g = 4400 kg ash h
b011
3
6
3
5
mol H h
mol O h
4400 kg ash/h, 450°C
Q to steam
40,000 kg coal/h
2.531 ×10 6 mol C/h
1.98 ×10 6 mol H/h
2.00 ×10 5 mol O/h
4400 kg ash/h
25°C
Flue gas at 260°C
n3 (mol dry gas/h)
0.078 mol CO 2 /mol D.G.
0.012 mol CO/mol D.G.
0.114 mol O 2/mol D.G.
0.796 mol N 2/mol D.G.
n4 (mol H2 O/h)
furnace
Preheated air at Ta (°C)
a.
air at 30°C, 1 atm, hr = 30%
n1 (mol O2 /h)
3.76 n1 (mol N2 /h)
n2 (mol H2 O/h)
preheater
Cooled flue gas at 150°C
n3 (mol dry gas/h)
0.078 CO 2
0.012 CO
0.114 O 2
0.796 N 2
n4 (mol H2 O/h)
Overall system balances
C balance: 2.531 × 10 6 = 0.078n 3 + 0.012n 3 ⇒ n 3 = 2.812 × 10 7 mol h dry flue gas
b
ge
j
b ge
N 2 balance: 3.76n1 = 0.796 2.812 × 10 7 ⇒ n1 = 5.95 × 10 6 mol O 2 h 3.76 5.95 × 10 6
= 224 × 10 mol N 2 h
7
9- 76
j
9.61 (cont'd)
30% relative humidity (inlet air):
B
Table B.3
y H 2O P =
0.30 p H∗ 2 O
b30° Cg ⇒ 5.95 × 10
n 2
6
.
mm Hgg
b760 mm Hgg = 0.300 b31824
+ 2.24 × 10 7 + n 2
⇒ n 2 = 3.61 × 10 5 mol H 2 O h
Volumetric flow rate of inlet air:
e
b g
j
5.95 × 10 6 + 224 × 10 7 + 3.61 × 10 5 mol 22.4 liters STP
V =
h
1 mol
Air/fuel ratio:
1 m3
3
= 6.43 × 10 5 SCMH
10 liters
6.43 × 10 5 m 3 air h
= 161
. SCM air kg coal
40000 kg coal h
e
j
. × 10 6 mol H h + 2 3.61 × 10 5 mol H h = 2n 4 ⇒ n 4 = 1351
.
× 10 6 mol H 2 O h
H balance: 198
H in coal
H 2 O content of stack gas =
b.
H in water vapor
1357
× 10 6 mol H 2 O h
.
× 10
.
e1357
j
+ 2.812 × 10 7 mol h
6
× 100% = 4.6% H 2 O
Energy balance on stack gas in preheater
bg
References : CO 2 , CO, O 2 , N 2 , H 2 O v at 25D C
Substance
CO 2
CO
O2
N2
H2O
n in
mol h
2.193 × 10 6
0.337 × 10 6
3.706 × 10 6
22.38 × 10 6
1.357 × 10 6
H in
kJ mol
4.942
3669
3758
3655
4266
H i (T ) from Table B.8 for inlet
Q=
∑ n H − ∑ n H
i
i
out
i
i
n out
mol h
2.193 × 10 6
0.337 × 10 6
3.206 × 10 6
72.38 × 10 6
1.351 × 10 6
H i (T ) =
b
z
H out
kJ mol
9.738
6.961
7.193
6.918
8135
Table B.2
B
Cp
dT for outlet
g
= −101
. × 10 8 kJ h Heat transferred from stack gas
in
Air preheating
1.01 ×10 8 kJ/hr
2.83 ×10 7 mol dry air/h
3.61 ×10 5 mol H 2O/h
30°C
2.83 ×10 7 mol dry air/h
3.61 ×10 5 mol H 2O/h
T a (°C)
(We assume preheater is adiabatic, so that Qstack gas = − Qair )
Energy balance on air:
Q = ΔH ⇒ 101
. × 10
8
kJ hr =
∑
z
Ta
z
Ta
ni (C p ) i dT =
30
30
9- 77
Table B.2
n dry air
z
T
Table B.2
a
B
B
(C p ) dry air dT + n H 2 O (C p ) H 2 O dT
30
9.61 (cont'd)
⇒ 101
. × 10 8 = 8.31 × 10 5 (Ta − 30) + 59.92(Ta2 − 30 2 ) + 0.031(Ta3 − 30 3 ) − 142
. × 10 −5 (Ta4 − 30 4 )
D
⇒ Ta = 150 C
c.
4400 kg ash/h at 450°C
40,000 kg coal/h
25°C
Flue gas at 260°C
2193× 106 mol CO 2 /h
0.337 ×10 6 mo l CO/h
3.206 × 106 mo l O2 /h
22.38 × 106 mo l N2 /h
1.351 × 106 mo l H2 O( v)/h
5.95 ×10 6 mo l O2 /h
2.24 ×10 7 mo l N2 /h
3.61 ×10 5 mo l H2 O( v)/h
150°C(= 149.8°C)
m (kg H 2 O(l )/h)
50°C
m (kg H 2 O(v )/h)
30 bars, sat'd
bg
References for energy balance on furnace: CO 2 , CO, O 2 , N 2 , H 2 O l , coal at 25° C
bg
(Must choose H 2 O l since we are given the higher heating value of the coal.)
H in
0
−
3.758
3.655
−
−
48.28
substance
n in
40000
Coal
−
Ash
5.95 × 10 6
O2
N2
2.24 × 10 6
−
CO 2
−
CO
3.61 × 10 5
H2O
n out
−
4400
3.206 × 10 6
2.24 × 10 7
2.193 × 10 6
0.337 × 10 6
1351
.
× 10 6
H out
−
n kg h
412.25 H kJ kg
7.193
6.918
n mol h
9.738 H kJ mol
6.961
52.14
b
b
b
b
g
g
g
g
(Furnace only — exclude boiler water)
Heat transferred from furnace
Q = n coal ΔH io +
FG
H
= 4 × 10 4
∑ n H − ∑ n H
i
kg
h
= −8.76× 10
8
i
i
i
IJ FG −2.5 × 10 kJ IJ + FG 2.74 × 10
KH
kg K G
H
out
in
4
3
− 122
. × 10 8
A
H of preheated air
I kJ
JJ kg
K
kJ h
8
8
Heat transferred to boiler water: 0.60(8.76x10 kJ/h) = 5.25x10 kJ/h
b g FGH IJK c b g
h e bg j
L
O kJ kJ h = m M2802.3 − 209.3P
MN A A PQ kg ⇒ m = 2.02 × 10 kg steam h
kg Energy balance on boiler: Q kJ h = m
H H 2 O l , 30b, sat' d − H H 2 O l , 50 D C
h
⇒ 5.25 × 10 8
5
Table B.6
9- 78
Table B.5
9.62
CO + O 2 → CO 2 , ΔH co = −282.99 kJ mol
1
Basis : 1 mol CO burned.
2
1 mol CO
n 0 mol O2
3.76n 0 mol N2
25°C
a.
1 mol CO2
(n 0 – 0.5) mol O2
3.76n 0 mol N2
1400°C
b
g
1 mol CO react 0.5 mol O 2
= n0 − 0.5
1 mol CO
References: CO, CO 2 , O 2 , N 2 at 25D C
n in
n out
H in
H out
Substance mol
mol
kJ mol
kJ mol
Oxygen in product gas: n1 = n0 mol O 2 fed −
b g b
1
n0
3.76n 0
−
CO
O2
N2
CO 2
g b g b
−
n 0 − 0.5
3.76n 0
1
0
0
0
−
g
−
H 1
H 2
H 3
B
b
g
B
N bg,1400° Cg: H = H (1400 C) = 44.51 kJ mol
B
CO bg,1400° Cg: H = H
(1400 C) = 7189
. kJ mol
Table B.8
O 2 g,1400° C : H 1 = H O 2 (1400D C) = 47.07 kJ mol
Table B.8
D
2
2
N2
Table B.8
D
2
3
CO 2
E.B.:
ΔH = nCO ΔH co +
∑ n H − ∑ n H
i
i
i
out
i
b
g
b
g
= −282.99 + 47.07 n0 − 0.5 + 44.51 3.76n0 + 71.89 = 0
in
⇒ n0 = 1094
.
mol O 2
b
gb
g
Theoretical O 2 = 1 mol CO 0.5 mol O 2 mol CO = 0.500 mol O 2
1094
.
mol fed − 0.500 mol reqd.
× 100% = 119% excess oxygen
0.500 mol
Increase %XS air ⇒ Tad would decrease, since the heat liberated by combustion would go into
heating a larger quantity of gas (i.e., the additional N 2 and unconsumed O 2 ).
Excess oxygen:
b.
9.63
a.
Basis : 100 mol natural gas ⇒ 82 mol CH 4 , 18 mol C 2 H 6
CH (g) + 2O (g) → CO (g) + 2H O(v), ΔH o = −890.36 kJ / mol
4
2
2
2
c
7
C 2 H 6 (g) + O 2 (g) → 2CO 2 (g) + 3H 2 O(v), ΔH co = −1559.9 kJ / mol
2
82 mol CH4
18 mol C2H6
298 K
Stack gas at T(°C)
n2 (mol CO2)
n3 (mol H2O (v))
n4 (mol O2)
n5 (mol N2)
n0 (mol air) at 423 K
0.21 O2 (20% XS)
0.79 N2
9-79
9.63 (cont’d)
Theoretical oxygen =
Air fed : n1 =
2 mol O 2
82 mol CH 4
1 mol CH 4
1.2 × 227 mol O 2
1 mol air
0.21 mol O 2
gb g b gb g
= b82.00gb4g + b18.00gb6g ⇒ n
+
3.5 mol O 2
18 mol C 2 H 6
1 mol C 2 H 6
= 227 mol O 2
= 1297.14 mol air
b
C balance : n2 = 82.00 1 + 18.00 2 ⇒ n2 = 118.00 mol CO 2
H balance : 2n3
3
= 218.00 mol H 2 O
b gb g
20% excess air, complete combustion ⇒ n4 = 0.2 227 mol O 2 = 45.40 mol O 2
b gb
g
N 2 balance : n5 = 0.79 1297.14 = 1024.63 mol N 2
Extents of reaction: ξ 1 = nCH 4 = 82 mol, ξ 1 = nC 2 H 6 = 18 mol
bg
bg bg bg
bg
Reference states: CH 4 g , C 2 H 6 g , N 2 g , O 2 g , H 2 O l at 298 K
bg
(We will use the values of ΔH co given in Table B.1, which are based on H 2 O l as a combustion
product, and so must choose the liquid as a reference state for water.)
bg
b
g
H i T = C pi T − 298 K for all species but water
b
g
b
g
= ΔH v,H 2 O 298 K + C p , H 2 O T − 298 K for water
Substance
n in
mol
H in
kJ mol
CH 4
C2 H 6
O2
N2
CO 2
H2O v
82.00
18.00
272.40
1024.63
−
−
0
4.14
3.91
−
−
bg
H out
kJ mol
n out
mol
−
−
−
−
45.40
0.0331 T − 298
1024.63
0.0313 T − 298
118.00
0.0500 T − 298
218.00 44.013 + 0.0385 T − 298
b
b
b
b
g
g
g
g
Energy balance : ΔH = 0
e
ξ 1 ΔH co
b
j
CH 4
e
+ ξ 2 ΔH co
j
C2 H 6
+
∑ n H − ∑ n H
i
i
out
i
i
=0
in
gb
g b
gb
g
+ b45.40gb0.0331g + b1024.63gb0.0313g + b118.00gb0.0500g + b218.00gb0.0385g bT − 298g
+ b218.00gb44.01g − (272.40)(4.14) − (1024.63)(3.91) = 0
⇒ 82.00 mol CH 4 −890.36 kJ mol + 18.00 mol C 2 H 6 −1559.90 kJ mol
b.
Solving for T using E - Z Solve ⇒ T = 2317 K
Increase % excess air ⇒ Tout decreases. (Heat of combustion has more gas to heat)
% methane increases ⇒ Tout might decrease. (lower heat of combustion, but heat released goes
into heating fewer moles of gas.)
9-80
bg
bg
af
bg
C 3 H 6 O g + 4O 2 g → 3CO 2 g + 3H 2 O l , ΔH io = −1821.4 kJ mol
9.64
Basis :
b g
1410 m3 STP feed gas
103 mol
3
min
b g
1 min
22.4 m STP
60 s
= 1049 mol s feed gas
Stochiometric proportion:
b
g
1 mol C 3 H 6 O ⇒ 4 mol O 2 ⇒ 4 × 3.76 = 15.04 mol N 2 ⇒ 1 + 4 + 15.04 = 20.04 mol
yC 3 H 6O =
1 mol C 3 H 6 O
mol C 3 H 6 O
4
, yO 2 =
= 0.1996 mol O 2 mol
= 0. 0499
20.04
20.04 mol
mol
Preheat
Feed gas
1049 mol/s
0.0499 C 3 H 6 O
0.1996 O22
0.1496
0.7505 N 2
T f (°C), 150 mm Hg
Rel. satn = 12.2%
a.
Cooling
Feed gas
562°C
Product gas
n 1 (mol CO 2 /s)
n 2 (mol H 2O/s)
n 3 (mol N 2/s)
T a (°C)
Q1 (kW)
Product gas
350°C
Q2 (kW)
d i
p C∗ 3 H 6O T f
.
Relative saturation = 12.2% ⇒ y C 3H 6O P = 0122
⇒ p∗ =
b.
Reaction
b
gb
0.0499 1500 mm Hg
0.122
g = 61352
. mm Hg
Table B.4
T f = 50.0 o C
b
gb
g
b1049gb0.1996g = 209.4 mol O s
b1049gb0.7505g = 787.3 mol N s
n = b52.34gb3g = 157.0 mol CO s U 14.25 mole% CO
|
⇒ Product contains n = b52.34gb3g = 157.0 mol H O sV ⇒ 14.25% H O
|W 71.5% N
n = 787.3 mol N s
References : C H Obgg, O , N , H Oblg, CO at 25 C
Feed contains 1049 mol s 0.0499 C 3 H 6 O mol = 52.34 mol C 3 H 6 O s
2
2
1
2
2
2
3
2
2
2
2
D
3
6
2
Hˆ in
nin
Substance
2
(mols) (kJ/mol)
2
2
nout
Hˆ out
(mols)
(kJ/mol)
D
Ta
(562 C)
C3 H 6 O
O2
52.34
209.4
67.66
17.72
−
−
−
−
N2
CO 2
H2O
787.3
−
−
17.18
−
−
787.3
157.0
157.0
0.032 (Ta − 25 )
0.052 (Ta − 25 )
44.013 + 0.040 (Ta − 25 )
Energy balance on reactor:
ΔH = nC3 H 6O ΔH co +
∑ n H − ∑ n H
i
out
i
i
i
b g
= 0 kJ s
in
kJ ⎞
⎛
4
⇒ ( 5234 mol s) ⎜ −1821.1
⎟ + 39.638(Ta − 25) + 157.0( 44.013) − 2.078 ×10 = 0 ⇒ Ta = 2780°C
mol ⎠
⎝
9-81
9.64 (cont'd)
c.
bg
Preheating step: References: C 3 H 6 g , O 2 , N 2 at 25° C
n in
H in
n out
H out
( mol / s) (kJ / mol) (mol / s) ( kJ / mol)
(562 D C)
(50 D C)
52.34
315
52.34
67.66
.
209.4
0.826
209.4
17.72
787.3
0.775
787.3
16.65
Substance
C3H 6O
O2
N2
E.B. ⇒ Q 1 =
∑ n H − ∑ n H
i
i
i
out
i
= 194
. × 10 4 kW
in
af
Cooling step. References: CO 2 (g), H 2 O v , N 2 (g) at 25D C
n
H in
n out
H out
Substance in
( mol) (kJ / mol) (mol) (kJ / mol)
e2871 Cj
e350 Cj
D
CO 2
H 2O
N2
E.B. ⇒ Q2 =
157.0
157.0
787.3
142.3
10815
.
88.23
∑ n H − ∑ n H
i
i
out
i
i
D
157.0
157.0
787.3
16.25
12.35
10.08
= −9.64 × 10 4 kW
in
Exchange heat between the reactor feed and product gases.
9.65
a.
Basis : 1 mol C5H12 (l)
C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(v),
1 mol C5H12 (l)
n2(mol CO2)
n3 (mol H2O (v))
n4 (mol O2)
Tad(oC)
n0 (mol O2) , 75°C
30% excess
Theoretical oxygen =
ΔH co = −3509.5 kJ / mol
1 mol C 5 H 12
8 mol O 2
= 8 mol O 2
1 mol C5 H 12
30% excess ⇒ n0 = 13
. × 8 = 10.4 mol O 2
b gb g
= b1gb12g ⇒ n
C balance: n2 = 1 5 ⇒ n2 = 5 mol CO 2
H balance: 2n3
3
= 6 mol H 2 O
b gb g
30% excess O2, complete combustion ⇒ n4 = 0.3 8 mol O 2 = 2.4 mol O 2
bg b g
bg
Reference states: C5 H 12 l , O 2 g , H 2 O l , CO 2 (g) at 25o C
(We will use the values of ΔH c0 given in Table B.1, which are based on H 2 O l as a
bg
combustion product, and so must choose the liquid as a reference state for water)
9-82
9.65 (cont'd)
substance
C5 H 12
O2
CO 2
H2O
nin
H in
mol kJ mol
100
0
.
10.40
H 1
−
−
−
−
z
nout
H out
mol kJ mol
−
−
2.40
H 2
5.00
H 3
6.00
H 4
T
H i =
i = 2,3
(C p ) i dT
25
e
j
z
T
= ΔH v 25 o C + (C p ) H 2 O(v) dT for H 2 O(v)
25
Table B.8
B
H 1 = H O 2 (75 o C)
=
148
. kJ / mol
Substituting (C p ) i from Table B.2 :
kJ
H 2 = (0.0291 Tad + 0.579 × 10 −5 Tad 2 − 0.2025 × 10 −8 Tad 3 + 0.3278 × 10 −12 Tad 4 − 0.7311)
mol
kJ
2
3
4
−5
−8
−12
H 3 = (0.03611 Tad + 2.1165 × 10 Tad − 0.9623 × 10 Tad + 1866
Tad − 0.9158)
.
× 10
mol
kJ
H 4 = 44.01 + (0.03346 Tad + 0.3440 × 10 −5 Tad 2 + 0.2535 × 10 −8 Tad 3 − 08983
.
. )
× 10 −12 Tad 4 − 0838
mol
kJ
. + (0.03346 Tad + 0.3440 × 10 −5 Tad 2 + 0.2535 × 10 −8 Tad 3 − 08983
.
⇒ H 4 = 4317
× 10 −12 Tad 4 )
mol
Energy balance : ΔH = 0
e j
nC5 H12 ΔH co
C5 H 12 ( l)
+
∑ n H − ∑ n H
i
out
i
i
i
=0
in
(1 mol C5 H 12 )( −3509.5 kJ / mol) + (2.40) H 2 + (5.00) H 3 + (6.00) H 4 − (10.40)( H 1 ) = 0
Substitute for H 1 through H 4
ΔH = (0.4512 Tad + 14.036 × 10 −5 Tad 2 − 3777
× 10 −8 Tad 3 + 4.727 × 10 −12 Tad 4 ) − 3272.20 kJ / mol = 0
.
⇒ f (Tad ) = −3272.20 + 0.4512 Tad + 14.036 × 10 −5 Tad 2 − 3777
× 10 −8 Tad 3 + 4.727 × 10 −12 Tad 4 = 0
.
Check :
−3272.20
4.727 × 10 −12
= −6.922 × 1014
Solving for Tad using E - Z Solve ⇒ Tad = 4414 o C
b.
Terms
1
2
3
Tad
% Error
7252
64.3%
3481 –21.1%
3938 –10.8%
9-83
9.65 (cont’d)
c.
d.
T
7252
5634
4680
4426
4414
f(T)
6.05E+03
1.73E+03
3.10E+02
1.41E+01
3.11E-02
f'(T) Tnew
3.74 5634
1.82 4680
1.22 4426
1.11 4414
1.11 4414
The polynomial formulas are only applicable for T ≤ 1500°C
9.66
5.5 L/s at 25°C, 1.1 atm
n 1(mol CH4/s)
n4 (mol CO 2 /s)
Adiabatic
Reactor
25% excess air
n 2 (mol O2/s)
3.76 n 2 (mol N2/s) n 4 (mol CO2/s)
150°C, 1.1 atm
n 3 (mol O2/s)
3.76 n 2 (mol N2/s)
n 5 (mol H2O/s)
T(°C), 1.05 atm
2CH 4 + 2O 2 → CO 2 + 2 H 2 O
Fuel feed rate : =
550
. L 273 K 1.1 atm
s
mol
298 K 1.0 atm 22.4 L(STP)
= 0.247 mol CH 4 / s
Theoretical O 2 = 2 × 0.247 = 0.494 mol O 2 / s
. (0.494) = 0.6175 mol O 2 / s ,
25% excess air ⇒ n 2 = 125
⇒ 3.76 × 0.6175 = 2.32 mol N 2 / s
Complete combustion ⇒ ξ = n1 = 0.247 mol / s, n 4 = 0.247 mol CO 2 / s, n 5 = 0.494 mol H 2 O / s
n 3 = 0.6175 mol O 2 fed / s − 0.494 mol consumed / s
.
mol O 2 / s
= 0124
Re ferences: CH 4 , O 2 , N 2 , CO 2 , H 2 O at 25 o C
n in
n out
H in
H out
Substance
( mol / s) ( kJ / mol) ( mol / s) ( kJ / mol)
CH 4
O2
N2
CO 2
H2O
Hˆ 1 = Hˆ (O 2,150o C)
Hˆ 2 = Hˆ (N 2,150o C)
0.247
0.6175
2.32
−
−
0
H 1
H 2
−
−
Table B.8
3.78 kJ/mol
Table B.8
3.66 kJ/mol
(ΔHˆ co )CH4 = −890.36 kJ/mol
T
Hˆ i = ∫ C pi dT , i = 3 − 5
25
9-84
−
0124
.
2.32
0.247
0.497
−
H 3
H 4
H 5
H
6
9.66 (cont'd)
H b = ( ΔH v ) H
z
T
D
2 O(25 C)
+ (C p ) H 2 O(v) dT
25
a.
Energy Balance
ΔH = ξ(ΔHˆ co )CH4 + ∑ nout Hˆ out −∑ nin Hˆ in = 0
Table B.2 for C pi (T ), ( ΔHˆ v ) H O = 44.01 kJ/mol
2
0.247(−890.36) + 0.494(44.01) + 0.0963(T − 25) + 1.02 ×10−5 (T 2 − 252 ) + 0.305 ×10−8 (T 3 − 253 )
−1.61×10−12 (T 4 − 254 ) − 0.6175(3.78) − 2.32(3.66) = 0
⇒ −211.4 + 0.0963Tad + 1.02 × 10−5 Tad 2 + 0.305 × 10−8 Tad 3 − 1.61× 10−12 Tad 4 = 0 ⇒ T = 1832o C
b.
In product gas,
T = 1832o C, P = 1.05 × 760 = 798 mmHg
y H2O =
0.494 mol/s
= 0.155 mol H 2 O/mol
(0.124 + 2.32 + 0.247 + 0.494) mol/s
Raoult's law : y H2O P = pH* 2O (Tdp ) ⇒ pH* 2 O = (0.155)(798) = 124 mmHg
Table B.3
⇒T
dp
= 56D C
Degr. superheat = 1832D C − 56D C = 1776D C
9.67
a.
CH 4 (l) + 2O 2 (g) → CO 2 (g) + 2H 2 O(v)
Basis : 1 mol CH 4
2 mol O 2
= 2.00 mol O 2
1 mol CH 4
30 % excess air ⇒ 130
. (2.00) = 2.60 mol O 2 , ⇒ 3.76 × 2.60 = 9.78 mol N 2
Theoretical oxygen =
1 mol CH 4
1 mol CH4
2.60 mol O2
9.78 mol N2
25° C, 4.00 atm
n2 (mol CO2)
n3 (mol H2O)
n4 (mol O2 )
Complete combustion ⇒ n 2 = 100
. mol CO 2 , n 3 = 2.00 mol H 2 O
2.00 mol O 2 consumed ⇒ n 4 = (2.60 − 2.00) mol O 2 = 0.60 mol O 2
Internal energy of reaction: Eq. (9.1-5) ⇒
ΔU co
=
ΔH co
F
I
G
J
− RT G ∑ ν − ∑ ν J
GH
JK
i
gaseous
products
e
⇒ ΔU co
U =
z
j
CH 4
T
25
FG
H
= −890.36
Ideal Gas
(Cv )dT
⇒
z
IJ
K
i
gaseous
reactants
8.314 J 298 K (1 + 2 − 1 − 2) 1 kJ
kJ
kJ
−
= −890.36
mol K
10 3 J
mol
mol
T
(C p − Rg )dT
25
If C p is independent of T ⇒ U = (C p − Rg )(T − 25o C)
9-85
9.67 (cont’d)
b.
bg bg bg
bg
Reference states: CH 4 g , N 2 g , O 2 g , H 2 O l , CO 2 (g) at 25o C
(We will use the values of ΔH c0 given in Table B.1, which are based on H 2 O l as a
bg
combustion product, and so must choose the liquid as a reference state for water.)
nin
U in
nout
CH 4
mol
100
.
kJ mol
0
mol
−
O2
2.60
0
N2
9.78
0
kJ mol
−
0.60
U 1
9.78
U
CO 2
−
−
100
.
H 2O v
−
−
2.00
Substance
bg
U out
2
U 3
U
4
Part a
B
U i = (C p − Rg )(T − 25) for all species except H 2 O(v)
= ΔU v 25 o C + (C p − Rg )(T − 25) = ΔH v 25 o C − Rg Tref + (C p − Rg )(T − 25) for H 2 O(v)
e
j
e
j
Substituting given values of (C p ) i and Rg = 8.314 × 10 −3 kJ / mol yields
U 1 = (0.033 − 8.314 × 10 −3 )(T − 25) kJ / mol = (0.02469T − 0.6172) kJ / mol
U 2 = (0.032 − 8.314 × 10 −3 )(T − 25) kJ / mol = (0.02369T − 0.5922) kJ / mol
U 3 = (0.052 − 8.314 × 10 −3 )(T − 25) kJ / mol = (0.04369T − 10922
.
) kJ / mol
LM
N
FG
H
OP
Q
IJ
K
kJ
kJ
kJ
U 4 = 44.01
− 8.314 × 10 −3
(298 K) + (0.040 − 8.314 × 10 −3 )(T − 25)
mol
mol ⋅ K
mol
kJ
kJ
kJ
⇒ U 4 = 4153
+ (0.052 − 8.314 × 10 −3 )(T − 25)
= (0.03167T − 40.74)
.
mol
mol
mol
Energy Balance
e
j + ∑ n U − ∑ n U = 0
⇒ Q = (100
. )b −890.36 kJ / molg + (0.60)U + (9.87)U
Q = n CH 4 ΔU co
CH 4
i
i
out
i
i
in
1
2
+ (100
. )U 3 + (2.00)U 4 = 0
Substituting U 1 through U 4
0.3557 T − 81619
. = 0 ⇒ T = 2295 o C
Ideal Gas Equation of State ⇒
c.
FG
H
IJ
K
Pf Tf
(2295 + 273) K
× 4.00 atm = 34.5 atm
=
⇒ Pf =
Pi
Ti
(25 + 273) K
– Heat loss to and through reactor wall
– Tank would expand at high temperatures and pressures
9-86
9.68
b.
1 mol natural gas
yCH 4 (mol CH 4 / mol)
nCO 2 (mol CO 2 )
yC 2 H 6 (mol C 2 H 6 / mol)
nH 2 O (mol H 2 O)
yC 3H 8 (mol C 3 H 8 / mol)
nN 2 (mol N 2 )
nO 2 mol O 2 )
Humid air
na (mol air)
ywo (mol H20(v)/mol)
(1-ywo) (mol dry air/mol)
0.21 mol O2/mol DA
0.79 mol N2/mol DA
Basis : 1 g-mole natural gas
CH 4 (g) + 2O 2 (g) → CO 2 (g) + H 2 O(v)
7
C 2 H 6 (g) + O 2 (g) → 2CO 2 (g) + 3H 2 O(v)
2
C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4 H 2 O(v)
Theoretical oxygen :
2 mol O2
1 mol CH 4
yCH4 (mol CH 4 )
+
3.5 mol O2
1 mol C2 H 6
= ( 2 yCH 4 + 35
. y C 2 H 6 + 5 yC 3 H 8 )
FG
H
Excess oxygen: 0.21n a (1 − y wo ) = 1 +
FG
H
⇒ na = 1 +
IJ
K
yC2H6 (mol C2 H 6 )
+
5 mol O2
1 mol C3 H 8
yC3H8 (mol C3 H8 )
IJ
K
Pxs
( 2 y CH 4 + 35
. y C 2 H 6 + 5 y C3 H 8 ) mol O 2
100
Pxs
1
( 2 y CH 4 + 35
. y C2 H 6 + 5 y C3H 8 )
mol air
100
0.21(1 − y w 0 )
Feed components
(n O 2 ) in = 0.21n a (1 − y wo ), (n N 2 ) in = 0.79n a (1 − y wo ), (n H 2 O ) in = n a y wo
N 2 in product gas: n N 2 = (n N 2 ) in mol N 2
CO2 in product gas :
nCO2 =
1 mol CO2 nCH4 (mol CH4 ) 2 mol CO2 nC2H6 (mol C2 H 6 ) 3 mol CO2 nC3H8 (mol C3 H8 )
+
+
1 mol CH4
1 mol C2 H 6
1 mol C3 H8
= (nCH4 + 2nC2H6 + 3nC3H8 ) mol CO2
H 2 O in product gas :
nH2O =
1 mol H 2 O nCH4 (mol CH 4 ) 3 mol H2 O nC2H6 (mol C2 H6 ) 4 mol O2 nC3H8 (mol C3 H8 )
+
+
1 mol CH4
1 mol C2 H 6
1 mol C3 H8
= [2nCH4 + 3nC2H6 + 4nC3H8 + na (1- ywo )] mol H2 O
O 2 in product gas : n O 2 =
Pxs
( 2n CH 4 + 35
. n C2 H 6 + 5 n C 3H 8 ) mol O 2
100
9-87
9.68 (cont’d)
c.
References : C(s), H 2 ( g) at 25o C
z
T
o
H
CH 4 ( T) = ( Δ H f ) CH 4 + ( C p ) CH 4 dT
25
Using ( Δ H fo )CH 4 from Table B.1 and (C p )CH 4 from Table B.2
H CH (T) = −7485
. kJ / mol +
4
F
GG
H
z
I
JJ
K
T
(0.03431+5.469 × 10−5 T + 03661
.
× 10−8 T 2 − 1100
. × 10−12 T 3 ) dT kJ / mol
25
. × 10 −8 T 3 − 2.75 × 10 −12 T 4 ] kJ / mol
⇒ H CH (T ) = [ −75.72 + 3.431 × 10 −2 T + 2.734 × 10 −5 T 2 + 0122
4
7
∑
H out
n out
mol kJ / mol mol kJ / mol
−
−
n1
H 1
n2
H 2
−
−
n3
H3
−
−
n4
H 4
n7
H 7
n5
H 5
n8
H 8
n6
n9
H 9
−
n10
H 10
−
−
CH 4
C2 H 6
C3H8
O2
N2
CO 2
H2O
ΔH =
H in
n in
Substance
6
(ni ) out ( Hi ) out −
i =4
∑ (n )
i in ( Hi ) in
i =1
3
H i = ai + bi T + ci T 2 + di T + ei T 4
6
∑
3
(ni ) in ( Hi ) in =
i =1
∑
(ni ) in H i (Tf ) +
i =1
6
∑ (n )
i in Hi (Ta )
i =4
7
⇒ ΔH =
∑
3
(ni ) out (ai + bi T + ci T 2 + d i T 3 + ei T 4 ) out −
i =4
7
⇒ ΔH =
7
∑ (n )
i out ai
+
i =1
i out bi T +
i =4
3
−
∑ (n )
∑ (n )
i out ci T
i =1
6
i in Hi (Tf
2
)−
i =1
∑ (n )
+
∑ (n )
i out d i T
i =1
i in Hi (Ta )
i =4
= α 0 + α 1T + α 2 T + α 3 T + α 4 T 4
2
where α 0 =
α1 =
α3 =
7
∑
3
3
(ni ) out ai −
i =1
7
∑
(ni ) in H i (Tf ) −
i =1
∑(n )
α2 =
∑(n )
α4 =
i out bi
i =1
7
i out d i
i =1
6
∑ (n )
i =4
7
∑ (n )
i out ci
i =1
7
∑ (n )
i out ei
i =1
.
9-88
i in Hi (Ta )
6
∑(n )
i in Hi (Ta )
i =4
7
i =1
7
7
∑ (n )
∑
(ni ) in H i (Tf ) −
3
+
∑ (n )
i out ei T
i =1
4
9.68 (cont’d)
d.
Run 1
yCH4
0.75
yC2H6
0.21
yC3H8
0.04
Tf
40
Ta
150
Pxs
25
ywo 0.0306
nO2i
3.04
nN2
11.44
nH2Oi
0.46
HCH4
-74.3
HC2H6
-83.9
HC3H8
-102.7
HO2i
3.6
HN2i
3.8
HH2Oi
-237.6
nCO2
1.29
nH2O
2.75
nO2
0.61
nN2
11.44
Tad 1743.1
alph0
-1052
alph1 0.4892
alph2 0.0001
alph3 -3.00E-08
alph4 3.00E-12
Delta H 3.00E-07
Species
CH4
C2H6
C3H8
O2
N2
H20
CO2
a
-75.72
-85.95
-105.6
-0.731
-0.728
-242.7
-394.4
Run 2
0.86
0.1
0.04
40
150
25
0.0306
2.84
10.67
0.43
-74.3
-83.9
-102.7
3.6
3.8
-237.6
1.18
2.61
0.57
10.67
1737.7
-978.9
0.4567
1.00E-04
-3.00E-08
3.00E-12
9.00E-06
Run 3
0.75
0.21
0.04
150
150
25
0.0306
3.04
11.44
0.46
-70
-77
-93
3.6
3.8
-237.6
1.29
2.75
0.61
11.44
1750.7
-1057
0.4892
0.0001
-3.00E-08
3.00E-12
-4.00E-07
Run 4
0.75
0.21
0.04
40
250
25
0.0306
3.04
11.44
0.46
-74.3
-83.9
-102.7
6.6
6.9
-234.1
1.29
2.75
0.61
11.44
1812.1
-1099
0.4892
0.0001
-3.00E-08
3.00E-12
-1.00E-04
Run 5
0.75
0.21
0.04
40
150
100
0.0306
4.87
18.31
0.73
-74.3
-83.9
-102.7
3.6
3.8
-237.6
1.29
3.02
2.44
18.31
1237.5
-1093
0.7512
0.0001
-4.00E-08
4.00E-12
-1.00E-05
b
x 10^2
3.431
4.937
6.803
2.9
2.91
3.346
3.611
c
x 10^5
2.734
6.96
11.3
0.11
0.579
0.344
2.117
d
x 10^8
0.122
-1.939
-4.37
0.191
-0.203
0.254
-0.962
e
x 10^12
-2.75
1.82
7.928
-0.718
0.328
-0.898
1.866
9-89
Run 6
0.75
0.21
0.04
40
150
25
0.1
3.04
11.44
1.61
-74.3
-83.9
-102.7
3.6
3.8
-237.6
1.29
3.9
0.61
11.44
1633.6
-1058
0.5278
0.0001
-2.00E-08
2.00E-12
6.00E-04
9.69
n 14 (mol CH4 /h)
25°C
Preheaters
Absorber off-gas
n 8 (mol H 2/h)
n 12 (mol N 2/h)
0.988 n 9 (mol CO/h)
0.950 n 6 (mol CH4 /h)
0.006 n 7 (mol C 2H 2/h)
n 13 (mol C(s )/h)
Converter
converter product quench
Tad (°C)
Feed gas, 650°C
n 14 (mol CH4 /h)
0.96n 15 (mol O2 /h)
0.04n 15 (mol N2 /h)
Converter
product
38°C
0.917 n 1 (mol DMF/h)
Lean solvent
filter
absorber
n 6 (mol CH4 /h)
n 7 (mol C 2H 2/h)
n 8 (mol H 2/h)
n 9 (mol CO/h)
n 10 (mol CO2 /h)
n 11 (mol H 2O/h)
n 12 (mol N 2/h)
n 13 (mol C(s )/h)
n 15 (mol/h)
0.96 mol O2 /mol
0.04 mol N2 /mol
25°C
Basis:
5000 kg/h Product gas
n 1 (mol/h)
0.991 mol C2 H2 ( g)/mol
0.00059 mol H2 O/mol
0.00841 mol CO2 /mol
n 6 (mol CH4 /h)
n 7 (mol C 2H 2/h)
n 8 (mol H 2/h)
n 9 (mol CO/h)
n 10 (mol CO2 /h)
n 11 (mol H 2O/h)
n 12 (mol N 2/h)
Rich solvent
n 1 (mol/h)
0.0155 mol C2 H2 /mol
0.0063 mol CO2 /mol
0.00055 mol CO/mol
0.00055 mol CH4 /mol
0.0596 mol H2 O/mol
0.917 mol DMF/mol
Average M.W. of product gas:
b
g
b
g
b
stripper
Stripper off-gas
n 2 (mol CO/h)
n 3 (mol CH4 /h)
n 4 (mol H 2O(v )/h)
n 5 (mol CO2 /h)
g
M = 0.991 26.04 + 0.00059 18.016 + 0.00841 44.01 = 26.19 g mol
Molar flow rate of product gas: n0 =
5000 kg 103 g
day
1 kg
1 mol
1 day
26.19 g
24 h
= 7955 mol h
Material balances -- plan of attack (refer to flow chart):
Stripper balances: C 2 H 2 ⇒ n1 , CO ⇒ n2 , CH 4 ⇒ n3 , H 2 O ⇒ n4 , CO 2 ⇒ n5
Absorber balances: CH 4 ⇒ n6 , C 2 H 2 ⇒ n7 , CO ⇒ n9 , CO 2 ⇒ n10 , H 2 O ⇒ n11
RS5.67% soot formationUV ⇒ n
Tconverter C balance W
13 , n14 ,
converter H balance ⇒ n8
Converter O balance ⇒ n15 , converter N 2 balance ⇒ n12
Stripper balances:
b
g
C 2 H 2 : 0.0155n1 = 0.991 7955 mol h ⇒ n1 = 5.086 × 105 mol h
b
ge
j
CH : b0.00055ge5.086 × 10 j = n ⇒ n = 79.7 mol CH h
H O: b0.0596ge5.086 × 10 j = n + b0.00059gb7955g ⇒ n = 30308 mol H O h
CO : b0.0068ge5.086 × 10 j = n + b0.00841gb7955g ⇒ n = 3392 mol CO h
CO: 0.00055 5.086 × 105 = n2 ⇒ n2 = 79.7 mol CO h
5
3
4
3
4
5
4
2
4
2
5
5
2
Absorber balances
b
5
ge
j
CH 4 : n6 = 0.950n6 + 0.00055 5.086 × 105 = n6 ⇒ 5595 mol CH 4 h
9-90
2
9.69 (cont'd)
b
ge
j
CO: n = 0.988n + b0.00055ge5.086 × 10 j ⇒ n = 23311 mol CO h
CO : n = b0.0068ge5.086 × 10 j = 3458 mol CO h
H O: n = b0.0596ge5.086 × 10 j = 30313 mol H O h
n = b0.0567gn (mol CH )
1 mol C
Soot formation:
⇒ n = 0.0567n
b1g
h
1 mol CH
C 2 H 2 : n7 = 0.0155 5.086 × 105 + 0.006n7 ⇒ n7 = 7931 mol C 2 H 2 h
5
9
9
9
5
2
10
2
5
2
11
2
13
14
4
13
14
4
Converter C balance:
b
gb
g b gb g b
gb g b gb g
n14 = 5595 mol CH 4 h 1 mol C mol CH 4 + 7931 2 + 23311 1 + 3458 1 + n13
⇒ n14 = n13 + 48226
b2 g
bg
= b5595gb4g + b7931gb2g + 2n + b30313gb2g
Solve (1) & (2) simultaneously ⇒ n13 = 2899 mol C s h , n14 = 51120 mol CH 4 h
Converter H balance:
51120 mol CH 4 4 mol H
h
1 mol Ch 4
CH 4
C2 H 2
H 2O
H2
8
⇒ n8 = 52816 mol H 2 h
b
gb g
Converter O balance: 0.96n15 2 =
23311 mol CO
1 mol O
h
1 mol CO
b gb g b
CO 2
H 2O
gb g
+ 3458 2 + 30313 1
⇒ n15 = 31531 mol h
b gb
g
Converter N 2 balance: 0.04 31531 n12 ⇒ n12 = 1261 mol N 2 h
a.
Feed stream flow rates
VCH 4 =
VO 2 =
b.
51120 mol CH 4
h
b
31531 mol O 2 + N 2
h
Gas feed to absorber
5595 mol
7931 mol
23311 mol
3458 mol
30313 mol
52816 mol
1261 mol
b g
0.0244 m 3 STP
= 1145 SCMH CH 4
1 mol
g
b g
b
0.0244 m 3 STP
= 706 SCMH O 2 + N 2
1 mol
g
U|
||
|V
||
||
W
CH 4 h
C2H 2 h
CO h
CO 2 h
4.5 mole% CH 4 , 6.4 % C 2 H 2 , 18.7% CO ,
H 2 O h ⇒ 125 kmol h , 2.8% CO 2 , 24.3% H 2 O , 42.4% H 2 , 1.0% N 2
H2 h
N2 h
1.2469 × 10 5 mol h
Absorber off-gas
52816 mol H 2 h
1261 mol N 2 h
23031 mol CO h
64.1 mole% H 2 , 1.5% N 2 , 27.9% CO,
5315 mol CH 4 h ⇒ 82.5 kmol h, 6.4% CH , 0.06% C H
4
2 2
41.6 mol C 2 H 2 h
8.2471 × 10 4
U|
||
V|
|
mol h |W
9-91
9.69 (cont'd)
Stripper off-gas
279.7 mol CO h
279.7 mol CH 4 h
30308 mol H 2 O h ⇒ 34.3 kmol h, 0.82% CO, 0.82% CH 4 , 88.5% H 2 O, 9.9% CO 2
3392 mol CO h
3.4259 × 10 4
c.
d.
U|
|V
||
mol h W
mol I F 1 kmol I
FG
JG
J = 466 kmol DMF h
H
h K H 10 mol K
b0.991gb7955g mol C H in product gas = 0154
mol C H
Overall product yield =
.
DMF recirculation rate = 0.917 5.086 × 10 5
3
2
2
2
51120 mol CH 4 in feed h
2
mol CH 4
The theoretical maximum yield would be obtained if only the reaction 2CH 4 → C 2 H 2 + 3H 2
occurred, the reaction went to completion, and all the C 2 H 2 formed were recovered in the
product gas. This yield is (1 mol C2H2/2 mol CH4) = 0.500 mol C2H2/2 mol CH4.
The ratio of the actual yield to the theoretical yield is 0.154/0.500 = 0.308.
e.
Methane preheater
Q CH 4 = ΔH = n14
z
Table B.2
650
B
d i
Cp
25
dT =
CH 4
51120 mol 32824 J
1h
1 kJ
= 466 kW
h
mol
3601 s 10 3 J
Oxygen preheater
Table B.8
Table B.8
B
B
Q O 2 = ΔH = 0.96n15 H (O 2 ,650 D C) + 0.04n15 H ( N 2 ,650 D C)
FG
H
kJ OF 1 h I
IJ LMb0.96 × 20135
+ 0.04 × 18.99g
.
PG J = 176 kW
KN
mol ⋅ C QH 3600 s K
Cbsg, H bgg, O bgg, N bgg at 25° C
n
H b650° Cg n
H bT g
= 31531
f.
References :
Substance
mol
h
D
2
in
CH 4
51120
O2
30270
2
in
2
out
−42.026
.
20125
18.988
5595
−
−74.85 +
z
−
out
Ta
25
z
z
z
z
z
z
z
Ta
C p dT
C p dT
N2
1261
C2 H 2
−
H2
−
−
52816
CO
−
−
23311 −110.52 + C p dT
CO 2
−
−
3458
H2O
−
−
30313 −24183
. + C p dT
Cs
−
−
2899
bg
1261
out
35
−
7931
+226.75
Ta
25
9-92
C p dT
C p dT
−3935
. + C p dT
C p dT
b g
H b kJ molg
n mol h
9.69 (cont’d)
z
T
H i = ΔH i0 + C pi dT
kJ mol
∑ n H
i
i
= −1575
.
× 10 6 kJ h
i
= −9.888 × 10 6 kJ h +
in
∑ n H
i
25
kJ mol⋅°C
out
z
LM5595dC i + 1261dC i + 7931dC i
N
OP 1 kJ dT
+52816dC i + 23311dC i +3458dC i
+ 3013dC i
b g Q 10 J
1 kJ
+
dC i b g × 10 J dT
Tout
p CH
4
25
z
p H
2
p CO
p N
2
p CO
2
p C H
3 2
p H O v
2
3
Tad + 273
298
p C s
3
We will apply the heat capacity formulas of Table B.2, recognizing that we will probably
push at least some of them above their upper temperature limits
∑ n H
i
i
out
+
z
out
Tad
−4
Tad + 273
j
T 2 − 10162
.
× 10 −7 T 3 dT
25
298
∑
z
.
− 5.9885 × 10
e3902 + 12185
F 32.411 + 0.031744T − 14179
.
× 10 I
GH
JKdT
T
= −9.888 × 10 6 kJ h +
6
2
1418
.
× 10
n i H i = −1000
.
× 10 7 + 3943Ta + 0.6251Ta2 − 1996
.
× 10 −4 Ta3 − 2.5405 × 10 −8 Ta4 +
Ta + 273
6
Energy balance: ΔH =
∑ n H − ∑ n H
i
out
i
i
i
=0
in
b g
⇒ f Tc = −8.485 × 106 + 3943Tc + 0.6251Tc2 − 1996
. × 10 −4 Tc3 − 2.5405 × 10 −8 Tc4 +
E-Z Solve
Tc = 2032 o C.
9-93
1418
. × 106
=0
Tc + 273
9.70 a.
m 1 [ k g W (v)/d ]
W = H2O
100oC
o
24,000 kg sludge / d, 22 C
0.35 solids, 0.65 W(l)
DRYER
F
Q 2
m 2 (kg conc. sludge/d), 100 o C
INCINERATOR
0.75 solids, 0.25 W (l)
Waste gas
m 3 [kg W(v)/d]
4B, sat'd
C
m 3 [kg W(l)/d]
Q3 (kJ / d)
m 3 [kg W(l)/d] BOILER
4B, sat'd
20 o C
0.90
km ol C H 4
110oC
kmol
km ol C 2 H 6
0.10
kmol
Q4 (kJ / d)
Q1
D
m 6 (kg gas/d)
m 4 (kg oil/d)
Stack gas
0.87 C
CO 2 , H 2 O(v)
o
0.10 H
125 C
SO 2
0.0084 S
O2 , N2
0.0216 ash
ash
m 7 (kg air/d)
25 o C
Q 0 (kJ/d)
E
m 5 (kg air/d)
25 o C
9-94
9.70 (cont'd)
Solids balance on dryer:
0.35 × 24,000 kg / d = 0.75n 2 ⇒ n2 = 11200 kg / d ⇒
F
11.2 tonnes / d (conc. sludge)
Mass Balance on dryer: 24,000 = n1 + 11200 ⇒ n1 = 12,800 kg / d
Energy balance on sludge side of dryer:
References : H 2 O(l,22 D C), Solids(22 D C)
nin
Substance
(kg d)
Solids
nout
Hˆ in
(kJ kg) (kg d)
8400
0
8400
H 2 O(l)
15600
0
2800
H 2 O(v)
−
−
12800
Hˆ out
(kJ kg)
Hˆ
1
Hˆ 2
Hˆ
3
Hˆ 1 = 2.5(100 − 22) = 195.0 kJ/kg
Hˆ = (419.1 − 92.2) = 326.9 kJ/kg
2
Hˆ 3 = (2676 − 92.2) = 2584 kJ/kg
( Hˆ water from Table B.5)
Q 2 =
∑ m H − ∑ m H ⇒ Q
i
i
out
i
i
2
= 356
. × 107 kJ day
in
7
. × 10
356
Q steam =
= 6.47 × 107 kJ / d ⇒ Q 3 = 2.91 × 107 kJ / d
0.55
Energy balance on steam side of dryer:
6.47 × 107
FG IJ
H K
ΔH v for
H 2 O(sat'd, )
B
kJ
kg
= n3
× 2133
d
d
FG kJ IJ F 1 tonne I ⇒ n
H kg K GH 10 kg JK
3
3
= 30.3 tonnes / d (boiler feedwater)
Energy balance on steam side of boiler:
Q1 = (30300
kg
kJ
)(2737.6 − 83.9)
= 8.04 × 107 kJ / d
d
kg
62% efficiency ⇒ Fuel heating value needed =
⇒ n4 =
130
. × 108 kJ / d
3.75 × 104 kJ / kg
8.04 × 107
= 13
. × 108 kJ / d
0.62
= 3458 kg / d ⇒ D = 3.5 tonnes / day (fuel oil)
Air feed to boiler furnace: C + O 2 → CO 2 , 4H + O 2 → 2H 2 O,
(nO2 ) theo = 3458
S + O 2 → SO 2
kg ⎡
kgC 1 kmol C 1 kmol O 2
1 1
1 1⎤
(0.87
)(
)(
)+(0.10)( )( ) + (0.0084)( )( ) ⎥
⎢
d ⎣
kg
12 kg
1 kmol C
1 4
32 1 ⎦
= 338 kmol O 2 /d
9-95
9.70 (cont’d)
Air fed (25% excess) = 1.25(4.76
⇒
kmol air
kmol O2
kmol air
)(338
) = 2011
kmol O2
d
d
2011 kmol 29 kg 1 tonne
⇒ E = 58.3 tonnes/ d (air to boiler)
d
kmol 103 kg
Energy balance on boiler air preheater:
2011 kmol 103 mol 2.93 kJ
kJ
⇒ Q 0 =
= 5.89 × 106 kJ/d
Table B.8 ⇒ Hˆ air (125o C) = 2.93
d
1 kmol
mol
mol
Supplementary fuel for incinerator:
n6 =
11.2 tonne sludge 195 SCM
d
tonne
1 kmol
22.4 SCM
= 97.5 kmol d
MWgas = 0.90 MWCH 4 + 010
. MWC 2 H 6 = (0.90)(16) + (010
. )(30) = 17.4 kg kmol
= 1.7 tonne / d (natural gas)
M gas = (97.5)(17.4) ⇒ G
CH 4 + 2O 2 → CO 2 + 2H 2 O, C 2 H 6 +
7
O 2 → 2CO 2 + 3H 2 O
2
Air feed to incinerator:
(air)th, sludge :
11200 kg sludge 0.75 kg sol 19000 kJ 2.5 m3 (STP) air
(air)th , gas : 97.5
d
kg sludge
1 kg sol
4
10 kJ
1 kmol
22.4 m3 (STP)
= 1781
kmol air
d
⎤ ⎛ 4.76 kmol air ⎞
kmol CH 4 2 kmol O2
kmol ⎡
kmol air
×
+ (0.10)(3.5) ⎥ ⎜
⎟ = 998
⎢0.90
d ⎣
kmol
kmol CH 4
d
⎦ ⎝ 1 kmol O2 ⎠
kmol air
= 5558 kmol air/d
d
5558 kmol air 29.0 kg air 1 tonne
⇒
= 161 tonne air/d (incinerator air)
d
1 kmol air 103 kg
100% excess air: n7 = 2(1781 + 998)
Energy balance on air preheater :
Table B.8 ⇒ Hˆ air (110o C) = 2.486
5558 kmol 103 mol
kJ
⇒ Q 4 =
d
1 kmol
mol
2.486 kJ
kJ
= 1.38 × 107
mol
d
b.
Cost of fuel oil, natural gas, fuel oil and air preheating, pumping and compression, piping, utilities,
operating personnel, instrumentation and control, environmental monitoring. Lowering environmental
hazard might justify lack of profit.
c.
Put hot product gases from boiler and/or incinerator through heat exchangers to preheat both air streams.
Make use of steam from dryer.
d.
Sulfur dioxide, possibly NO2, fly ash in boiler stack gas, volatile toxic and odorous compounds in gas
effluents from dryer and incinerator.
9-96
CHAPTER TEN
10.1 b. Assume no combustion
n 1 (mol gas),T1 (°C)
x 1 (mol CH4 /mol)
x 2 (mol C2 H6 /mol)
1 – x 1 – x 2 (mol C3 H8 /mol)
n 3 (mol), 200°C
y 1 (mol CH4 /mol)
y 2 (mol C2 H6 /mol)
y 3 (mol C3 H8 /mol)
1 – y 1 – y 2 – y 3 (mol air/mol)
n 2 (mol air), T2 (°C)
Q (kJ)
11 variables
bn , n , n , x , x , y , y , y , T , T , Qg
b4 material balances and 1 energy balanceg
1
−5 relations
6 degrees of freedom
2
3
1
2
1
2
3
1
2
ln , n , x , x , T , T q
A feasible set of design variables:
1
2
1
2
1
2
Calculate n3 from total mole balance, y1 , y 2 , and y 3 from component balances,
Q from energy balance.
An infeasible set:
ln , n , n , x , x , T q
1
2
3
1
2
1
Specifying n1 and n2 determines n3 (from a total mole balance)
c.
n 2 (mol gas), T 2 , P
y 2 (mol C 6 H 14/mol)
1 – y 2 (mol N 2 /mol)
n 1 (mol gas), T 1 , P
y 1 (mol C 6 H 14/mol)
1 – y 1 (mol N 2 /mol)
n 3 (mol C 6 H 14( l )/mol), T 2 , P
Q (kJ)
b
d
g
n1 , n2 , n3 , y1 , y 2 , T1 , T2 , Q, P
9 variables
2 material, 1 energy, and 1 equilibrium: y 2 P = PC*6 H14 T2
−4 relations
5 degrees of freedom
A feasible set:
b gi
ln, y , T , P, n q
1
1
3
Calculate n2 from total balance, y 2 from C 6 H 14 balance, T2 from Raoult’s law:
b g
[ y 2 P = PC∗6 H 4 T2 ], Q from energy balance
An infeasible set:
ln , y , n , P, T q
2
2
3
2
Once y 2 and P are specified, T2 is determined from Raoult’s law
10- 1
b
10.2 10 variables n1 , n2 , n3 , n4 , x1 , x 2 , x 3 , x 4 , T , P
−2 material balances
g
bgb
g b
g bg
−2 equilibrium relations: [ x 3 P = x 4 PB* T , 1 − x 3 P = 1 − x 4 PC* T ]
6 degrees of freedom
ln , n , n , x , x , Tq
a. A straightforward set:
1
3
4
1
4
Calculate n2 from total material balance, P from sum of Raoult's laws:
bg b
g bg
P = x 4 p B∗ T + 1 − x 4 Pc∗ T
x 3 from Raoult's law, x 2 from B balance
b. An iterative set:
ln , n , n , x , x , x q
1
2
3
1
2
3
Calculate n4 from total mole balance, x 4 from B balance.
Guess P, calculate T from Raoult's law for B, P from Raoult’s law for C, iterate until
pressure checks.
c. An impossible set:
ln , n , n , n , T , Pq
1
2
3
4
Once n1 , n2 , and n3 are specified, a total mole balance determines n4 .
bg
bg
bg
bg
10.3 2BaSO 4 s + 4C s → 2BaS s + 4CO 2 g
a.
100 kg ore, T 0 (K)
xb (kg BaSO4 /kg)
n 1 (kg C)
n 2 (kg BaS)
n 3 (kg CO2 )
n 4 (kg other solids)
T f (K)
n 0 (kg coal), T 0 (K)
xc (kg C/kg)
Pex (% excess coal)
Q (kJ)
d
i
11 variables n0 , n1 , n2 , n3 , n4 , x b , x c , T0 , T f , Q, Pex
−5 material balances C, BaS, CO 2 , BaSO 4 , other solids
−1 energy balance
b
g
+1 reaction
−1 relation defining Pex in terms of n0 , x b , and x c
5 degrees of freedom
n
b. Design set: x b , x c , T0 , T f , Pex
s
Calculate n0 from x b , x c , and Pex ; n1 through n4 from material balances,
Q from energy balance
10- 2
10.3 (cont’d)
l
q
c. Design set: x B , x c , T0 , n2 , Q
Specifying x B determines n2 ⇒ impossible design set.
l
q
d. Design set: x B , x c , T0 , Pex , Q
Calculate n2 from x B , n3 from x B
n0 from x B , x c and Pex
n1 from C material balance, n4 from total material balance
T f from energy balance (trial-and-error probably required)
10.4 2C 2 H 5 OH + O 2 → 2CH 3 CHO + 2H 2 O
2CH 3 COH + O 2 → 2CH 3 CHOOH
n f (mol solution), T 0
x ef (mol EtOH/mol)
1 – x ef (mol H 2 O/mol)
n e (mol EtOH), T
n ah (mol CH 3 CHO)
n ea (mol CH 3 COOH)
n w (mol H 2O)
n ax (mol O 2)
n n (mol N 2)
n w (mol air), Pxs , T 0
0.79 n air (mol N 2 )
0.21 n air (mol O 2 )
(Pxs = % excess air)
a.
Q (kJ)
d
13 variables n f , naw , ne , neh , nea , n w , nex , n0 , x ef , T0 , T , Q, Pxs
−6 material balances
−1 energy balance
−1 relation between Pxs , n f , x ef , and nair
+2 reactions
i
7 degrees of freedom
n
b. Design set: n f , x ef , Pxs , ne , nah , T0 , T
s
Calculate nair from n f , x ef and Pxs ; nn from N 2 balance;
naa and nw from n f , x ef , ne , nah and material balances;
nex from O atomic balance; Q from energy balance
n
c. Design set: n f , x ef , T0 , nair , Q, ne , n w
s
Calculate Pxs from n f , x ef and nair ; n’s from material balances; T from energy
balance (generally nonlinear in T)
l
q
d. Design set: nair , nn , … . Once nair is specified, an N 2 balance fixes nn
10- 3
10.5
a. n 1 (mol CO)
n 2 (mol H2 )
reactor
n 3 (mol C3 H6 )
n 4 (mol C3 H6 )
n 5 (mol CO)
n 6 (mol H2 )
n 7 (mol C7 H8 O)
n 8 (mol C4 H7 OH)
n 9 (kg catalyst)
Flash
tank
n10 (kg catalyst)
n 16(mol C3 H6 )
n 11(mol C3 H6 )
n 12(mol CO)
n 17(mol CO)
n 13(mol H2 )
Separation n 18(mol H2 )
n 14(mol C7 H8 O)
n 15(mol C4 H7 OH)
n 19(mol C7 H8 O)
n 20(mol C4 H7 OH)
n 21(mol H2 )
Hydrogenator
n 22(mol H2 )
n 20(mol C4 H7 OH)
Reactor:
b
10 variables n1 − n16
−6 material balances
g
+2 reactions
6 degrees of freedom
Flash Tank:
b
12 variables n4 − n15
g
−6 material balances
6 degrees of freedom
Separation:
b
10 variables n11 − n20
g
−5 material balances
5 degrees of freedom
Hydrogenator:
b
5 variables n19 − n23
−3 material balances
g
+1 reaction
3 degrees of freedom
Process:
20 Local degrees of freedom
−14 ties
6 overall degrees of freedom
The last answer is what one gets by observing that 14 variables were counted two times
each in summing the local degrees of freedom. However, one relation also was counted
twice: the catalyst material balances on the reactor and flash tank are each n9 = n10 . We
must therefore add one degree of freedom to compensate for having subtracted the same
relation twice, to finally obtain 7 overall degrees of freedom (A student who gets this one
has done very well indeed!)
b. The catalyst circulation rate is not included in any equations other than the catalyst balance
(n9 = n10). It may therefore not be determined unless either n9 or n10 is specified.
10- 4
b
10.6 n − C 4 H 10 → i − C 4 H 10 n − B = i − B
n 1 (mol n-B)
mixer
n 2 (mol n-B)
n 3 (mol i-B)
g
reactor
n 4 (mol n-B)
n 5 (mol i-B)
still
n 6 (mol)
x 6 (mol n-B/mol)
(1 – x 6)(mol i-B/mol)
n r (mol)
x r (mol n-B/mol)
(1 – x r)(mol i-B/mol)
b
a. Mixer:
5 variables n1 , n2 , n3 , nr , x r
g
−2 material balances
3 degrees of freedom
b
Reactor:
4 variables n2 , n3 , n3 , n5
g
−2 material balances
+1 reaction
3 degrees of freedom
b
Still:
6 variables n4 , n5 , n6 , x 6 , nr , x r
g
−2 material balances
4 degrees of freedom
Process:
10 Local degrees of freedom
− 6 ties
4 overall degrees of freedom
b. n1 = 100 mol n − C 4 H 10 , x 6 = 0.115 mol n − C 4 H 10 mol , x r = 0.85 mol n − C 4 H 10 mol
b gb g
b
gb g b gb g
100 mol n - B fed − b100gb0115
. gmol n - B unreacted
× 100% = 88.5%
Overall conversion =
Overall C balance: 100 4 = n6 0.115 4 + 0.885 4 mol C ⇒ n6 = 100 mol overhead
100 mol n - B fed
Mixer n-B balance: 100 + 0.85nT = n2
b1g
b1g
35% S.P. conversion: n4 = 0.65n2 ⇒ n4 = 65 + 0.5525nr
Still n – B balance:
b2 g
b
b2g
gb g
n4 = n6 x 6 + nr x r ⇒ 65 + 0.5525nr = 0115
.
100 + 0.85nr ⇒ nr = 179.83 mol
b
Recycle ratio = 179.83 mol recycle
mol recycle
.
g b100 mol fresh feedg = 179
mol fresh feed
10- 5
10.6 (cont’d)
c.
k =1 k = 2 k = 3
100.0 132.3 1515
.
185.0 212.5 228.8
nr
n2 = 100 + 0.85nr
n3 = nr 1 − 0.85
b
g
n4 = 0.65n2
n5 = n 2 + n 3 − n 4
n 4 + n5 = n 6 + n r
n4 = 0.115n6 + 0.85n r
Error:
d. w =
q=
UV
W
15.0
120.25
79.75
n6 = 67.69
⇒
n r = 132.3
19.85
138.1
94.21
80.76
1515
.
22.73
148.7
102.8
88.55
163.0
179.83 − 163.0
× 100 = 9.3% error
179.83
1515
. − 132.3
= 0.595
132.3 − 100.0
0.595
= −1470
.
0.595 − 1
b
g c b
ghb g
nrb 3g = −1470
.
132.3 + 1 − −1470
.
1515
. = 179.8
Error:
179.8 − 179.8
× 100 = < 01%
.
error
179.8
e. Successive substitution, Iteration 32: nr = 179.8319 Æ nr = 179.8319
Wegstein, Iteration 3: nr = 179.8319 Æ nr = 179.8319
S1
10.7
SF
Split
S2
a.
1
2
3
4
5
6
7
8
A
X1 =
nA
nB
nC
nD
T(deg.C)
B
C
0.6
Molar flow rates (mol/h)
SF
S1
85.5
51.3
52.5
31.5
12.0
7.2
0.0
0.0
315
315
Formula in C4: = $B$1*B4
Formula in D4: = B4-C4
10- 6
D
S2
34.2
21.0
4.8
0.0
315
10.7 (cont’d)
b. C
**CHAPTER 10 -- PROBLEM 7
DIMENSION SF(8), S1(8), S2(8)
FLOW = 150.
N=3
SF(1) = 0.35*FLOW
SF(2) = 0.57*FLOW
SF(3) = 0.08*FLOW
SF(8) = 315.
X1 = 0.60
CALL SPLIT (SF, S1, S2, X1, N)
WRITE (6, 900)' STREAM 1', S1(1), S1(2), S1(3), S1(B)
WRITE (6, 900)' STREAM 2', S2(1), S2(2), S2(3), S2(B)
900
FORMAT (A10, F8.2,' mols/h n-octane', /,
*10X, F8.2,' mols/h iso-octane', /,
*10X, F8.2,' mols/h inerts', /,
*
10X, F8.2,' K')
END
C
C
C
100
SUBROUTINE SPLIT
SUBROUTINE SPLIT (SF, S1, S2, X1, N)
DIMENSION SF(8), S1(8), S2(8)
D0 100 J = 1, N
S1(J) = X1*SF(J)
S2(J) = SF(J) – S1(J)
S1(8) = SF (8)
S2(8) = SF (8)
RETURN
END
. mols h n-octane
Program Output: Stream 1 3150
51.30 mols h iso-octane
7.20 mols h inerts
315.00 K
Stream 2 21.00 mols h n-octane
34.20 mols h iso-octane
4.80 mols h inerts
315.00 K
10- 7
10.8
a. Let Bz = benzene, Tl = toluene
*
Antoine equations: pBz
= 106.89272−1211.033/(T + 220.790) (=1350.491)
pTl* = 106.95805−1346.773/(T + 219.693) (=556.3212)
*
*
/ P ( = 0.518)
Raoult's law: xBz = (P − pTl* )/(pBz
-pTl* ) (=0.307) , yBz = xBz pBz
Total mole balance: 100 = nv + nl
Benzene balance:
⎫
⎬
40 = yBz nv + xBz nl ⎭
40 − 100 xBz
⇒ nv =
(=44.13), nl = 100 − nv (=55.87)
yBz − xBz
Fractional benzene vaporization : f B = nv yBz / 40 (=0.571)
Fractional toluene vaporization : fT = nv (1 − yBz ) / 60 (=0.354)
The specific enthalpies are calculated by integrating heat capacities and (for vapors)
adding the heat of vaporization.
Q = ∑ nout H out − ∑ nin H in (= 1097.9)
b. Once the spreadsheet has been prepared, the goalseek tool can be used to determine the
bubble-point temperature (find the temperature for which nv=0) and the dew-point
temperature (find the temperature for which nl =0). The solutions are
Tbp = 96.9 o C, Tdp = 103.2 o C
c.
C **CHAPTER 10 PROBLEM B
DIMENSION SF(3), SL(3), SV(3)
DATA A1, B1, C1/6.90565, 1211.033, 220.790/
DATA A2, B2, C2/6.95334, 1343.943, 219.377/
DATA CP1, CP2, HV1, HV2/ 0.160, 0.190, 30.765, 33.47/
COMMON A1, B1, C1, A2, B2, C2, CP1, CP2, NV1, NV2
FLOW = 1.0
SF(1) = 0.30*FLOW
SF(2) = 0.70*FLOW
T = 363.0
P = 512.0
CALL FLASH2 (SF, SL, SV, T, P, Q)
WRITE (6, 900) 'Liquid Stream', SL(1), SL(2), SL(3)
WRITE (6, 900) 'Vapor Stream', SV(1), SV(2), SV(3)
900
FORMAT (A15, F7.4,' mol/s Benzene',/,
* 15X, F7.4, mol/s Toluene',/,
* 15X, F7.2, 'K')
WRITE (6, 901) Q
10- 8
10.8 (cont’d)
901
FORMAT ('Heat Required', F7.2,' kW')
END
C
C
C
C
C
C
SUBROUTINE FLASN2 (SF, SL, SV, T, P, Q)
REAL NF, NL, NV
DIMESION SF(3), SL(3), SV(3)
COMMON A1, B1, C1, C2, CP1, CP2, NV1, NV2
Vapor Pressure
PV1 = 10.**(A1 – B1/(T – 273.15 + C1))
PV2 = 10.**(A2 – B2/(T – 273.15 + C2))
Product fractions
XL1 = (P – PV2)/(PV1 – PVS)
XV1 = XL1*PM/P
Feed Variables
NF = SF(1) + SF(2)
XF1 = SF(1)/NF
Product flows
NL = NF*(XF1 – XV1)/(XL1 – XV1)
NV = NF – NL
SL(1) = XL1*NL
SL(2) = NL – SL(1)
SY(1) = XY1*NY
SY(2) = NV – SY(1)
SL(3) = T
SV(3) = T
Energy Balance
Q = CP1*SF(1)*SF(1) + CP2*SF(2)
Q = Q*(T – SF(3)) + (NV1*XV1 + HV2*(1 – XV1))*NV
RETURN
END
b1g
XF b I g∗ NF = XLb I g∗ NL + XV b I g∗ NV I = 1,2… n − 1 b2g
Energy Balance: Q = bT − TF g∗ ∑ CPb I g∗ c XLb I g∗ NL + XV b I g∗ NV h
10.9 a. Mass Balance: NF = NL + NV
N
I =1
bg
N
b g b3g
+ NV ∗ ∑ HV I ∗ XV 1
I =1
b g
where: XL N = 1 −
∑ XLb I g
N −1
b g
XV N = 1 −
I =1
b g b4 g
XV b I g∗ P = XLb I g∗ PV b I g
N
∑ XV b I g
N −1
I =1
bg
Raoult’s law: P = ∑ XL I ∗ PV I
I =1
10- 9
I = 1,2, … N − 1
b5g
10.9 (cont’d)
bg
d b g b g cCb I g + T hi I = 1,2,… N − 1
3 + 3b N − 1g + N + 4 variables b NF , NL, NV , XF ( I ), XL( I ), XV ( I ), PV ( I ), TF , T , P , Qg
where: PV I = 10∗∗ A I − B I
− N mass balance
−1 energy balances
− N equilibrium relations
− N Antoine equations
N + 3 degrees of freedom
b gr
m
Design Set TF , T , P , NF , XF I
Eliminate NL form (2) using (1)
Eliminate XV(I) form (2) using (5)
Solve (2) for XL(I)
XL I = XF I ∗ NF NF + NV ∗ PV I P − 1
bg
bg
c bg
d
hi b6g
Sum (6) ove I to Eliminate XL(I)
b g
N
b g d NF + NV ∗ c PV b I g P − 1hi = 0 b7g
f NV = −1 + NF ∗ ∑ XF I
I =1
Use Newton's Method to solve (7) for NV
Calulate NL from (1)
XL(I) from (2)
XV(I) from (5)
Q from (3)
b.
C **CHAPTER 10 - - PROBLEM 9
DIMENSION SF(8), SL(8), SV(8)
DIMENSION A(7), B(7), C(7), CP(7), HV(7)
COMMON A, B, C, CP, NV
DATA A/6.85221, 6.87776, 6.402040, 0., 0., 0., 0./
DATA B/1064.63, 1171.530, 1268.115, 0., 0., 0., 0./
DATA C/232.00, 224.366, 216.900, 0., 0., 0., 0./
DATA CP/0.188, 0.216, 0.213, 0., 0., 0., 0./
DATA NV/25.77, 28.85, 31.69, 0., 0., 0., 0./
FLOW = 1.0
N*3
SF(1) = 0.348*FLOW
SF(2) = 0.300*FLOW
SF(3) = 0.352*FLOW
SF(4) = 363
SL(4) = 338
SV(4) = 338
P*611
CALL FLASHN (SF, SL, SV, N, P, Q)
WRITE (6, 900)' Liquid Stream', (SL(I), I = 1, N + 1)
WRITE (6, 900)' Vapor Stream', (SV(I), I = 1, N + 1)
10- 10
10.9 (cont’d)
900
901
C
C
100
200
C
300
C
500
400
900
C
FORMAT (A15, F7.4,' mols/s n-pentane', /,
*15X, F7.4,' mols/s n-hexane', /,
*15X, F7.4,' mols/s n-hephane', /,
*
15X, F7.2,' K')
WRITE (6, 901) Q
FORMAT ('Heat Required', F7.2, 'kW')
END
SUBROUTINE FLASHIN (SF, SL, SV, N, P, Q)
REAL NF, NL, NV, NVP
DIMENSION SF(8), SL(8), SV(8)
DIMENSION XF(7), XL(7), XV(7), PV(7)
DIMENSION A(7), B(7), C(7), CP(7), HV(7)
COMMON A, B, C, CP, HV
TOL = 1,5 – 6
Feed Variables
NF = 0.
DO 100 I = 1, N
NF = NF + SF(I)
DO 200 I = 1, N
XF(I) = SF(I)/NF
TF = SF (N + 1)
T = SL (N + 1)
TC = T – 273.15
Vapor Pressures
DO 300 I = 1, N
PV(I) = 10.**(A(I) – B(I)/(TC + C(I)))
Find NV -- Initial Guess = NF/2
NVP = NF/2
DO 400 ITER = 1, 10
NV = NVP
F = –1.
FP = 0.
DO 500 I = 1, N
PPM1 = PV(I)/P – 1.
F = F + NF*XF(I)/(NF + NV*PPM1)
FP = FP – PPM1*XF(I)/(NF + NV*PPM1)**2.
NVP = NV – F/FP
IF (ABS((NVP – NV)/NVP).LT.TOL) GOTO 600
CONTINUE
WRITE (6, 900)
FORMAT ('FLASHN did not converge on NV')
STOP
Other Variables
10- 11
10.9 (cont’d)
600
700
800
NL = NF – NVP
DO 700 I = 1, N
XL(I) = XF(I)*NF/(NF + NV**(PV(I)/P – 1))
SL(I) = XL(I)*NL
XV(I) = XL(I)*PV(I)/P
SV(I) = SF(I) – SL(I)
Q1 = 0.
Q2 = 0.
DO 800 I = 1, N
Q1 = Q1 + CP(I)*SF(I)
Q2 = Q2 + HV(I)*XV(I)
Q = Q1*(T – TF) + Q2*NVP
RETURN
END
Program Output: Liquid Stream 0.0563 mols
0.1000 mols
0.2011 mols
338.00 K
Vapor Stream 0.2944 mols
0.2000 mols
0.1509 mols
338.00 K
Heat Required 13.01 kW
s n-pentane
s n-hexane
s n-heptane
s n-pentane
s n-hexane
s n-heptane
10.10
a.
Q(kW)
nv (mol / s)
x v ( mol A(v) / mol)
1 − x v ( mol B(g) / mol)
T (K), P(mm Hg)
n F (mol / s)
xF
(mol A(v) / mol)
1 − x F (mol B(g) / mol)
TF (K), P(mm Hg)
nl (mol A(l) / s)
10- 12
10.10 (cont’d)
10
–2
–1
–1
–1
5
variables (n F , x F , TF , P , nv , x v , T , nl , p *,A , Q)
material balances
Antoine equation
Raoult’s law
energy balance
degrees of freedom
b.
References: A(l), B(g) at 25oC
Substance
nin
H in
nout
H out
A(l)
—
—
nl
H3
A(v)
nF x F
H1
nv x v
H4
B(g)
n F (1 − x F )
H2
nv (1 − x v )
H5
Given n F and x F (or n AF and n BF ), TF , P , y c (fractional condensation),
Fractional condensation ⇒ nl = y c n F x F
Mole balance ⇒ nv = n F − nl
A balance ⇒ x v = (n F x F − nl ) / nv
Raoult' s law ⇒ p *A = x v P
Antoine' s equation ⇒ T =
B
−C
A − log 10 p *A
Enthalpies: H1 = ΔH v + C pv (TF − 25), H 2 = C pg (TF − 25), H 3 = C pl (T − 25),
H 4 = ΔH v + C pv (T − 25), H 5 = C pg (T − 25)
Energy balance: Q = ∑ n out H out − ∑ n in H in
c.
nAF
0.704
nV
0.3664
Cpv
0.050
nBF
0.296
xV
0.1921
Cpg
0.030
nF
1.00
A
7.87863
H1
37.02
xF
0.704
B
1473.11
H2
1.05
TF
333
C
230
H3
0.2183
P
760
pA*
146.0
H4
35.41
yc
0.90
T
300.8
H5
0.0839
nL
0.6336
Cpl
0.078
Q
–23.7
Greater fractional methanol condensation (yc) ⇒ lower temperature (T). (yc = 0.10 ⇒
T = 328oC.)
10- 13
10.10 (cont’d)
e.
C **CHAPTER 10 -- PROBLEM 10
DIMENSION SF(3), SV(3), SL(2)
COMMON A, B, C, CPL, HV, CPV, CPG
DATA A, B, C / 7.87863, 1473.11, 230.0/
DATA CPL, HV, CPV, CPG,/ 0.078, 35.27, 0.050, 0.029/
FLOW = 1.0
SF(1) = 0.704*FLOW
SF(2) = FLOW – SF(1)
YC = 0.90
P = 1.
SF(3) = 333.
CALL CNDNS (SF, SV, SL, P, YC, Q)
WRITE (6, 900) SV(3)
WRITE (6, 401) 'Vapor Stream', SV(1), SV(2)
WRITE (6, 401) 'Liquid Stream', SL(1)
WRITE (6, 902)Q
900
FORMAT ('Condenser Temperature', F7.2,' K')
901
FORMAT (A15, F7.3,' 'mols/s Methyl Alcohol', /,
*15X, F7.3, 'mols/s air')
902
FORMAT ('Heat Removal Rate', F7.2,' kW')
END
C
SUBROUTINE CNDNS (SF, SV, SL, P, YC, Q)
REAL NF, NL, NV
DIMENSION SF(3), SV(3), SL(2)
COMMON A, B, C, CPL, HV, CPV, CPG
C
Inlet Stream Variables
NF = SF(1) + SF(2)
TF = SF(3)
XF = SF(1)/NF
C
Solve Equations
NL = YC * XF * NF
NV = NF - NL
XV = (XF*NF - NL)/NV
PV = P * XV * 760.
T = B/(A - LOG(N)/LOG (10.)) - C
T = T + 273.15
Q = ((CPV * XV + CPG * (1 - XY)) * NV + CPL * NL) * (T - TF) - NL * HV
C
Output Variables
SL(1) = NL
S2(2) = T
SV(1) = XV*NV
SV(2) = NV - SV(1)
SV(3) = T
RETURN
END
10- 14
10.11
η 1 A1 + η 2 A2 + η 3 A3 +…η m Am = 0
a. Extent of reaction equations:
b g
b g
SPb I g = SF b I g + NU b I g∗ ξ I = 1,2, … N
ξ = −[ SF IX ∗ X ] NU IX
Energy Balance: Reference states are molecular species at 298K.
b
g TP = SPb N + 1g
= ∑ HF b I g∗ NU b I g
TF = SF N + 1
ΔH r
N
I =1
b
g
N
bg bg
N
bg bg
Q = ξ∗ ΔH r + TP − 298 ∗ ∑ SP I ∗ CP I − (TF − 298)∗ ∑ SF I ∗ CP I
I =1
I =1
b. C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
Subscripts: 1 = C3H8, 2 = O2, 3 = N2, 4 = CO2, 5 = H2O
mol ⋅ K
270 m 3 1 atm
h
273K 0.08206 liter ⋅ atm
b
1000 liter
m
3
h
3600 s
= 3.348 mol C 3 H 8 s [=SF(1)]
g
3.348 mol C3 H 8 1.2 5 mol O2
= 20.09 mol O2 s [= SF(2)] ⇒ 7554
. mol N 2 s [= SF(3)]
sec
mol C3 H8
X C3 H8 = 0.90 ⇒ nC3 H8 = 010
. (3.348) = 0.3348 mol C 3 H 8 / s in product gas [= SP(1)]
b g
b g
ξ = −[ SF IX ∗ X ] NU IX
Nu
nin (SF)
X
Xi
nout (SP)
Cp
Tin
Hin
Tout
Hout
HF
DHr
Q
= –(3.348 mol/s)(0.90)/(–1) = 3.013 mol/s
1-C3H8
-1
3.348
2-O2
-5
20.09
3-N2
0
75.54
4-CO2 5-H2O(v)
3
4
0.3348
0.1431
5.024
0.033
75.54
0.0308
9.0396
0.0495
12.0528
0.0375
17.9
4.1
3.9
6.2
4.7
107.6
-103.8
24.8
0
23.2
0
37.2
-393.5
28.2
-241.83
0.90
3.01
423
1050
-2044
-4006
For the given conditions, Q = −4006 kJ / s . As Tstack increases, more heat goes into the
stack gas so less is transferred out of the reactor: that is, Q becomes less negative.
10- 15
10.11 (cont’d)
C **CHAPTER 10 PROBLEM 11
DIMENSION SF(8), SP(8), CP(7), HF(7)
REAL NU(7)
DATA NU/–1., –5, 0., 3., 4., 0., 0./
DATA CP/0.1431, 0.0330, 0.0308, 0.0495, 0.0375, 0., 0./
DATA HF/–103.8, 0., 0., –393.5, –241.83, 0., 0./
COMMON CP, HF
SF(1) = 3.348
SF(2) = 20.09
SF(3) = 75.54
SF(4) = 0.
SF(5) = 0.
SF(6) = 423.
SP(6) = 1050.
IX = 1
X = 0.90
N=5
CALL REACTS (SF, SP, NU, N, X, IX, Q)
WRITE (6, 900) (SP(I), I = 1, N + 1), Q
900
FORMAT ('Product Stream', F7.3, ' mols/s propane', /,
*15X, F7.3,' mols/s oxygen', /,
*15X, F7.3,' mols/s nitrogen', /,
*
*15X, F7.3,' mols/s carbon dioxide', /,
*15X, F7.3,' mols/s water', /,
*15X, F7.2,'K', /,
Heat required', F8.2, 'kW')
END
C
SUBROUTINE REACTS (SF, SP, NU, N, X, IX, Q)
DIMENSION SF(8), SP(8), CP(7), HF(7)
REAL NU(7)
COMMON CP, HF
C
Extent of Reaction
EXT = –SF(IX)*X/NU(IX)
C
Solve Material Balances
DO 100 I = 1, N
100
SP(I) = SF(I) + EXT = NU(I)
C
Heat of Reaction
HR = 0
DO 200 I = 1, N
200
HR = HR + NF(I)*NU(I)
C
Product Enthalpy (ref * inlet)
HP = 0.
DO 300 I = 1, N
300
HP = HP + SP(I)*CP(I)
HP = HP + (SP(N + 1) – SF (N + 1))
Q = EXT * HR + HP
RETURN
END
10- 16
10.12 a. Extent of reaction equations:
ξ = − SF IX ∗ X NU IX
b g
b g
SPb I g = SF b I g + NU b I g∗ ξ
I = 1, N
Energy Balance: Reference states are molecular species at feed stream temperature.
N
bg bg
N
b g z CPb I gdT
Q = ΔH = ξΔH r + ∑ nout H out = 0 ⇒ 0 = ξ ∑ NU I HF I + ∑ SP I
i =1
I =1
T
Tfeed
CP(I) = ACP(I) + BCP(I)*T + CCP(I)*T2 + DCP(I)*T3
bg
bg
N
bg
f T = ξ ∗ ∑ NU I * HF I + AP∗ (T − Tfeed ) +
I =1
+
N
BP
2
∗ (T 2 − Tfeed
)
2
CP
DP
3
4
∗ (T 3 − Tfeed
)+
∗ ( T 4 − Tfeed
)=0
3
4
bg
bg
where: AP = ∑ SP I ∗ ACP I , and similarly for BP, CP, & DP
I =1
Use goalseek to solve f (T ) = 0 for T [= SP(N+1)]
b. 2CO + O 2 → 2CO 2
Temporary basis: 2 mol CO fed
b
g
2 mol CO 1.25 1 mol O2
= 125
. mol O2 ⇒ 4.70 mol N 2
2 mol CO
⇒ Total moles fed = (2.00 + 1.25 + 4.70) mol = 7.95 mol
Scale to given basis:
(23.0
kmol
1h
10 3 mol
mol CO fed s
.
SF (1) = 1607
)(
)
)(
h
3600 s 1 kmol = 0.8036 ⇒ SF (2) = 1.004 mol O fed s
2
7.95 mol
SF (3) = 3.777 mol N 2 fed s
10- 17
10.12 (cont’d)
Solution to Problem 10.12
Nu
nin (SF)
X
Xi
nout (SP)
ACP
BCP
CCP
DCP
AP
BP
CP
DP
Tfeed
DHF
DHr
T
f(T)
1-CO
-2
1.607
2-O2
-1
1.004
3-N2
0
3.777
4-CO2
2
0
0.45
0.36
0.88385
0.02895
4.11E-06
3.55E-09
-2.22E-12
0.642425
3.777 0.72315
0.0291
0.029 0.03611
1.16E-05 2.20E-06 4.23E-05
-6.08E-09 5.72E-09 -2.89E-08
1.31E-12 -2.87E-12 7.46E-12
0.1799
5.00E-05
-2.90E-11
-6.57E-12
650
-110.52
0
0
-393.5
-566
1560
-4.7E-08
The adiabatic reaction temperature is 1560 o C .
As X increases, T increases. (The reaction is exothermic, so more reaction means
more heat released.)
d.
C **CHAPTER 10 -- PROBLEM 12
DIMENSION SF(8), SP(B), NU(7), ACP(7), BCP(7), CCP(7), DCP(7), HF(7)
COMMON ACP, BCP, CCP, DCP, NF
DATA NU / –2., –1., 0., 2., 0., 0., 0./
DATA ACP/ 28.95E-3, 29.10E-3, 29.00E-3, 36.11E-3, 0., 0., 0./
DATA BCP/ 0.4110E-5, 1.158E-5, 0.2199E-5, 4.233E-6, 0., 0., 0./
DATA CCP/ 0.3548E-B, –0.6076E-8, 0.5723E-8, –2.887E-8, 0., 0., 0./
DATA DCP/ –2.220 E-12, 1.311E-12, –2.871E-12, 7.464E-12, 0., 0., 0./
DATA HF / –110.52, 0., 0., –393.5, 0., 0., 0./
SF(1) = 1.607
SF(2) = 1.004
SF(3) = 3.777
SF(4) = 0.
SF(5) = 650.
IX = 1
X = 0.45
N=4
CALL REACTAD (SF, SP, NU, N, X, IX)
WRITE (6, 900) (SP(I), I = 1, N + 1)
10- 18
10.12 (cont’d)
900
C
C
C
100
C
200
C
300
C
400
900
FORMAT ('Product Stream', F7.3, ' mols/s carbon monoxide', /,
*15X, F7.3, 'mols/s oxygen', /.
*15X, F7.3, 'mols/s nitrogen', /.
*
*15X, F7.3, 'mols/s carbon dioxide', /,
15X, F7.2, 'C')
END
SUBROUTINE REACTAD (SF, SP, NU, N, X, IX)
DIMENSION SF(8), SP(8), NU(7), ACP(7), BCP(7), CCP(7), DCP(7), HF(7)
COMMON ACP, BCP, CCP, DCP, NF
TOL = 1.E-6
Extent of Reaction
EXT = –SF(IX)*X/NU(IX)
Solve Material Balances
DO 100 I = 1, N
SP(I) = SF(I) + EXT*NU(I)
Heat of Reaction
HR = 0
DO 200 I = 1, N
HR = HR + HF(I) * NU(I)
HR = HR * EXT
Product Heat Capacity
AP = 0.
BP = 0.
CP = 0.
DP = 0.
DO 300 I = 1, N
AP = AP + SP(I)*ACP(I)
BP = BP + BP(I)*BCP(I)
CP = CP + SP(I)*CCP(I)
DP = DP + SP(I)*DCP(I)
Find T
TIN = SF (N + 1)
TP = TIN
D0 400 ITER = 1, 10
T = TP
F = HR
FP = 0.
F = F +T*(AP + T*(BP/2. + T*(CP/3. + T*DP/4.)))
*–TIN*(AP + TIN*(BP/2. + TIN*(CP/3. + TIN*DP/4.)))
FP = FP + AP + T *(BP + T*(CP + T*DP))
TP = T – F/FP
IF(ABS((TP – T)/T).LT.TOL) GOTO 500
CONTINUE
WRITE (6, 900)
FORMAT ('REACTED did not converge')
STOP
10- 19
10.12 (cont’d)
500
SP(N + 1) = T
RETURN
END
Program Output:
0.884 mol/s carbon monoxide
0.642 mol/s oxygen
3.777 mol/s nitrogen
0.723 mol/s carbon dioxide
T = 1560.43 C
10- 20
10.13
37.5 mol C2H4O
a.
Separator
50 mol C2H4
50 mol O2
208.3333 mol C2H4
50 mol O2
Reactor
166.6667
18.75
37.5
8.333333
8.333333
mol C2H4
mol O2
mol C2H4O
mol CO2
mol H2O
8.333333
18.75
8.333333
8.333333
mol C2H4
mol O2
mol CO2
mol H2O
Xsp = 0.2
Ysp = 0.9
158.3333 mol C2H4 (Ra)
158.3333 mol C2H4 (Rc)
Rc-Ra =
0
Procedure: Assume Ra, perform balances on mixing point, then reactor, then separator. Rc is recalculated recycle rate.
Use goalseek to find the value of Ra that drives (Rc-Ra) to zero.
b.
Xsp
0.2
0.2
0.3
0.3
Ysp
0.72
1
0.75333
1
Yo
0.6
0.833
0.674
0.896
no
158.33
158.33
99.25
99.25
The second reaction consumes six times more oxygen per mole of ethylene consumed. The lower the single pass ethylene oxide
yield, the more oxygen is consumed in the second reaction. At a certain yield for a specified ethylene conversion, all the oxygen in
the feed is consumed. A yield lower than this value would be physically impossible.
10-21
21
10.14 C **CHAPTER 10 -- PROBLEM 14
DIMENSION XA(3), XC(3)
N=2
EPS = 0.001
KMAX = 20
IPR = 1
XA(1) = 2.0
XA(2) = 2.0
CALL CONVG (XA, XC, N, KMAX, EPS, IPR)
END
C
SUBROUTINE FUNCGEN(N, XA, XC)
DIMENSION XA(3), XC(3)
XC(1) = 0.5*(3. – XA(2) + (XA(1) + XA(2))**0.5
XC(2) = 4. – 5./(XA(1) + XA(2))
RETURN
END
C
SUBROUTINE CONVG (XA, XC, N, KMAX, EPS, IPR)
DIMENSION XA(3), XC(3), XAH(3), XCM(3)
K=1
CALL FUNCGEN (N, XA, XC)
IF (IPR.EQ.1) CALL IPRNT (K, XA, XC, N)
DO 100 I = 1, N
XAM(I) = XA(I)
XA(I) = XC(I)
100
XCM(I) = XC(I)
110
K=K+1
CALL FUNCGEN (N, XA, XC)
IF (IPR.EQ.1) CALL IPRNT (K, XA, XC, N)
D0 200 I = 1, N
IF (ABS ((XA(I) - XC(I))/XC(I)).GE.EPS) GOTO 300
200
CONTINUE
C
Convergence
RETURN
300
IF(K.EQ.KMAX) GOTO 500
DO 400 I = 1, N
W = (XC(I) – XCM(I))/(XA(I) – XAM(I))
Q = W/(W – 1.)
IF (Q.GT.0.5) Q = 0.5
IF (Q.LT.–5) Q = –5.
XCM(I) = XC(I)
XAM(I) = XA(I)
400
XA(I) = Q = XAM(I) + (1. – Q)*XCM(I)
GOTO 110
500
WRITE (6, 900)
900
FORMAT (' CONVG did not converge')
STOP
END
10- 22
10.14 (cont’d)
C
SUBROUTINE IPRNT (K, XA, XC, N)
DIMENSION XA(3), XC(3)
IF (K.EQ.1) WRITE (6, 400)
IF (K.NE.1) WRITE (6, *)
DO 100 I = 1, N
100
WRITE (6, 901) K, I, XA(I), XC(I)
RETURN
900
FORMAT (' K Var Assumed Calculated')
901
FORMAT (I4, I4, 2E15.6)
END
Program Output: K Var
Assumed
Calculated
1
1 0.200000E + 01 0.150000E + 01
1
2
0.200000E + 01 0.275000E + 01
2
2
1
2
0.150000E + 01 0.115578E + 01
0.275000E + 01 0.282353E + 01
3
3
1
2
0.395135E + 00 0.482384E + 00
0.283152E + 01 0.245041E + 01
8
1
0.113575E + 01 0.113289E + 01
8
2
0.269023E + 01 0.269315E + 01
4
9
1
2
0.113199E + 01 0.113180E + 01
0.269186E + 01 0.269241E + 01
10- 23
CHAPTER ELEVEN
11.1 a.
The peroxide mass fraction in the effluent liquid equals that in the tank contents, which is:
xp =
Mp
M
Therefore, the leakage rate of hydrogen peroxide is m1 M p / M
b. Balance on mass: Accumulation = input – output
E
dM
= m0 − m1
dt
t = 0, M = M 0 (mass in tank when leakage begins)
Balance on H 2 O 2 : Accumulation = input – output – consumption
E
dM p
dt
= m0 x p 0 − m1
FG M IJ − kM
HMK
p
p
t = 0, M p = M p 0
11.2 a.
Balance on H3PO4: Accumulation = input
.
g / ml .
Density of H3PO4: ρ = 1834
Molecular weight of H3PO4: M = 98.00 g / mol .
dn p
Accumulation =
(kmol / min)
dt
20.0 L 1000 ml 1.834 g
mol
1 kmol
= 0.3743 kmol / min
Input =
min
L
ml
98.00 g 1000 mol
E
dn p
dt
= 0.3743
t = 0, n p0 = 150 × 0.05 = 7.5 kmol
z
z
np
b.
t
dn p = 0.3743 dt ⇒ n p = 7.5 + 0.3743t (kmol H 3PO 4 in tank )
7.5
xp =
c.
0
np
015
. =
n
=
np
n0 + n p − n p 0
=
7.5 + 0.3743t
150 + 0.3743t
kmol H 3PO 4
kmol
7.5 + 0.3743t
⇒ t = 471
. min
150 + 0.3743t
11-1
11.3 a.
g b
b
g
b
g
bg
mw = a + bt t = 0, mw = 750 & t = 5, mw = 1000 ⇒ mw kg h = 750 + 50t h
Balance on methanol: Accumulation = Input – Output
M = kg CH 3OH in tank
dM
= m f − mw = 1200 kg h − 750 + 50t kg h
dt
b
g
E
b
dM
= 450 − 50t kg h
dt
t = 0, M = 750 kg
z zb
M
b.
dM =
750
E
t
g
g
450 − 50t dt
0
M − 750 = 450t − 25t 2
E
M = 750 + 450t − 25t 2
Check the solution in two ways:
(1) t = 0, M = 750 kg ⇒ satisfies the initial condition;
dM
(2)
= 450 − 50t ⇒ reproduces the mass balance.
dt
c.
dM
= 0 ⇒ t = 450 50 = 9 h ⇒ M = 750 + 450(9) − 25(9)2 = 2775 kg (maximum)
dt
M = 0 = 750 + 450t − 25t 2
t=
d.
−450 ±
b450g + 4b25gb750g ⇒ t = –1.54 h, 19.54 h
2b −25g
2
3.40 m 3 103 liter 0.792 kg
= 2693 kg (capacity of tank)
1 m3
1 liter
M = 2693 = 750 + 450t − 25t 2
t=
−450 ±
b450g + 4b25gb750 − 2693g ⇒ t = 719
. h,10.81 h
2b −25g
2
Expressions for M(t) are:
R|750 + 450t - 25t b0 ≤ t ≤ 719
. and 10.81 ≤ t ≤ 19.54g (tank is filling or draining)
( 719
. ≤ t ≤ 10.81)
M(t) = S2693
(tank is overflowing)
||T0
(19.54 ≤ t ≤ 20.54)
(tank is empty, draining
2
as fast as methanol is fed to it)
11-2
11.3 (cont’d)
3000
2500
M(kg)
2000
1500
1000
500
0
0
5
10
15
20
t(h)
11.4 a.
Air initially in tank: N 0 =
10.0 ft 3
492° R
1 lb - mole
b g = 0.0258 lb - mole
532° R 359 ft 3 STP
Air in tank after 15 s:
Pf V
P0V
=
N f RT
N 0 RT
⇒ N f = N0
Rate of addition: n =
Pf
P0
=
0.0258 lb - mole 114.7 psia
= 0.2013 lb - mole
14.7 psia
b0.2013 − 0.0258g lb - mole air = 0.0117 lb - mole air s
15 s
b. Balance on air in tank: Accumulation = input
b
g
dN
= 0.0117 lb - moles s ; t = 0, N = 0.0258 lb - mole
dt
z z
N
c.
Integrate balance:
t
b
dN = n dt ⇒ N = 0.0258 + 0.0117t lb - mole air
0.0258
0
Check the solution in two ways:
(1) t = 0, N = 0.0258 lb - mole ⇒ satisfies the initial condition
( 2)
d.
dN
= 0.0117 lb - mole air / s ⇒ reproduces the mass balance
dt
b
gb g
t = 120 s ⇒ N = 0.0258 + 0.0117 120 = 143
. lb - moles air
b g
O 2 in tank = 0.21 143
. = 0.30 lb - mole O 2
11-3
g
11.5 a.
Since the temperature and pressure of the gas are constant, a volume balance on the gas
is equivalent to a mole balance (conversion factors cancel).
Accumulation = Input – Output
1h
dV 540 m 3
=
− ν w m 3 min
dt
h
60 min
e
j
b
t = 0, V = 3.00 × 103 m 3 t = 0 corresponds to 8:00 AM
z zb
V
dV =
3.00×103
b.
t
g
0
0
b g
z
ν w dt ≅
0
z
t
e j
9.00 − ν w dt ⇒ V m 3 = 3.00 × 103 + 9.00t − ν w dt t in minutes
Let ν w i = tabulated value of ν w at t = 10 i − 1
240
g
LM
MN
i = 1, 2, … , 25
OP
PQ
b
g b
24
24
10
10
. + 9.8 + 4 124.6 + 2 113.4
ν w1 + ν w 25 + 4 ∑ ν w i + 2 ∑ ν w i =
114
3
3
i = 3, 5, …
i = 2, 4, …
= 2488 m 3
b g
V = 3.00 × 103 + 9.00 240 − 2488 = 2672 m 3
c.
Measure the height of the float roof (proportional to volume).
The feed rate decreased, or the withdrawal rate increased between data points,
or the storage tank has a leak, or Simpson’s rule introduced an error.
d.
REAL VW(25), T, V, V0, H
INTEGER I
DATA V0, H/3.0E3, 10./
READ (5, *) (VW(I), I = 1, 25)
V= V0
T=0.
WRITE (6, 1)
WRITE (6, 2) T, V
DO 10 I = 2, 25
T = H * (I – 1)
V = V + 9.00 * H – 0.5 * H * (VW(I – 1) + VW(I))
WRITE (6, 2) T, V
10 CONTINUE
1 FORMAT ('TIME (MIN) VOLUME (CUBIC METERS)')
2 FORMAT (F8.2, 7X, F6.0)
END
$DATA
11.4 11.9
12.1
11.8
11.5
11.3
Results:
TIME (MIN)
0.00
10.00
20.00
VOLUME (CUBIC METERS)
3000.
2974.
2944.
230.00
240.00
2683.
2674.
Vtrapezoid = 2674 m3 ; VSimpson = 2672 m 3 ;
2674 − 2672
× 100% = 0.07%
2672
Simpson’s rule is more accurate.
11-4
g
11.6 a.
b
g
bg
ν out L min = kV L
⇒ ν out = 0.200V ν out = 20.0 L min ⇒ Vs = 100 L
V = 300
ν out = 60
b. Balance on water: Accumulation = input – output (L/min).
(Balance volume directly since density is constant)
dV
= 20.0 − 0.200V
dt
t = 0, V = 300
c.
dV
= 0 = 200 − 0.200Vs ⇒ Vs = 100 L
dt
V
The plot of V vs. t begins at (t=0, V=300). When t=0, the slope (dV/dt) is
20.0 − 0.200(300) = −40.0. As t increases, V decreases. ⇒ dV / dt = 20.0 − 0.200V
becomes less negative, approaches zero as t → ∞ . The curve is therefore concave up.
t
z
z
V
d.
t
dV
= dt
20.0 − 0.200V 0
300
⇒−
FG
H
IJ
K
1
20.0 − 0.200V
ln
=t
−40.0
0.200
b
g
b
⇒ −0.5 + 0.005V = exp −0.200t ⇒ V = 100.0 + 200.0 exp −0.200t
b g
b
g
. 100 = 101 L 1% from steady state ⇒
V = 101
b
g
101 = 100 + 200 exp −0.200t ⇒ t =
b
ln 1 200
−0.200
g = 26.5 min
11-5
g
11.7 a.
A plot of D (log scale) vs. t (rectangular scale) yields a straight line through the points ( t = 1 week,
D = 2385 kg week ) and ( t = 6 weeks, D = 755 kg week ).
ln D = bt + ln a ⇔ D = ae bt
b=
b
g
ln D2 D1 ln 755 2385
=
= −0.230
6−1
t 2 − t1
b g b
gb g
ln a = ln D1 − bt1 = ln 2385 + 0.230 1 = 8.007 ⇒ a = e 8.007 = 3000
E
D = 3000e −0.230t
b. Inventory balance: Accumulation = –output
b
dI
= −3000e −0.230t kg week
dt
t = 0, I = 18,000 kg
g
z z
I
18,000
c.
11.8 a.
t
dI = −3000e −0.230t dt ⇒ I − 18,000 =
0
3000 −0.230t
e
0.230
t
0
⇒ I = 4957 + 13,043e −0.230t
t = ∞ ⇒ I = 4957 kg
Total moles in room: N =
1100 m 3
Molar throughput rate: n =
273 K
103 mol
b g = 45,440 mol
295 K 22.4 m 3 STP
700 m 3
min
273 K
103 mol
b g = 28,920 mol min
295 K 22.4 m 3 STP
SO 2 balance ( t = 0 is the instant after the SO 2 is released into the room):
b gb
g
N mol x mol SO 2 mol = mol SO 2 in room
Accumulation = –output.
b g
d
dx
= −0.6364 x
Nx = − nx ⇒
N = 45, 440 dt
dt
n = 28,920
t = 0, x =
15
. mol SO 2
= 330
. × 10 −5 mol SO 2 mol
45,440 mol
b. The plot of x vs. t begins at (t=0, x=3.30×10-5). When t=0, the slope (dx/dt) is
−0.6364 × 330
. × 10 −5 = −210
. × 10 −5 . As t increases, x decreases. ⇒
dx dt = −0.6364 x becomes less negative, approaches zero as t → ∞ . The curve
is therefore concave up.
11-6
x
11.8 (cont’d)
0
t
c.
Separate variables and integrate the balance equation:
z
x
z
t
3.30×10 −5
dx
x
= −0.6364dt ⇒ ln
= −0.6364t ⇒ x = 330
. × 10 −5 e −0.6364t
x 0
330
. × 10 −5
Check the solution in two ways:
(1) t = 0, x = 3.30 × 10-5 mol SO 2 / mol ⇒ satisfies the initial condition;
dx
(2)
= −0.6364 × 330
. × 10 −5 e −0.6364t = −0.6364 x ⇒ reproduces the mass balance.
dt
d.
e.
CSO 2 =
45,440 moles x mol SO 2
1100 m 3
mol
i)
t = 2 min ⇒ CSO 2 = 382
. × 10 −7
ii)
x = 10 −6 ⇒ t =
e
1 m3
103 L
mol SO 2
liter
ln 10 −6 3.30 × 10 −5
−0.6364
. × 10 −2 x = 13632
.
= 4131
× 10 −6 e −0.6364t mol SO 2 / L
j = 55. min
The room air composition may not be uniform, so the actual concentration of the SO2
in parts of the room may still be higher than the safe level. Also, “safe” is on the average;
someone would be particularly sensitive to SO2.
11-7
11.9 a.
Balance on CO: Accumulation=-output
N ( mol ) x ( mol CO / mol) = total moles of CO in the laboratory
Pν p
kmol
)=
h
RT
Pν p
kmol CO
kmol
)x
Rate at which CO leaves: n (
=
x
h
kmol
RT
CO balance: Accumulation = -output
Molar flow rate of entering and leaving gas: n (
FG
H
FG
H
IJ
K
IJ
K
Pν p
d ( Nx )
dx
P
x⇒
=−
=−
ν px
dt
RT
dt
NRT
E PV = NRT
νp
dx
=−
x
dt
V
kmol CO
kmol
t = 0, x = 0.01
z
z
νp r
dx
V
=−
ln 100 x
dt ⇒ tr = −
νp
x
V 0
0.01
t
x
b.
c.
b g
V = 350 m 3
350
tr = −
ln 100 × 35 × 10 −6 = 2.83 hrs
700
e
j
d. The room air composition may not be uniform, so the actual concentration of CO
in parts of the room may still be higher than the safe level. Also, “safe” is on the
average; someone could be particularly sensitive to CO.
Precautionary steps:
Purge the laboratory longer than the calculated purge time. Use a CO detector
to measure the real concentration of CO in the laboratory and make sure it is
lower than the safe level everywhere in the laboratory.
11.10 a.
Total mass balance: Accumulation = input – output
b
g
dM
= m − m kg min = 0 ⇒∴ M is a constant = 200 kg
dt
b. Sodium nitrate balance: Accumulation = - output
x = mass fraction of NaNO 3
b g
E
b
d xM
= − xm kg min
dt
g
dx
m
m
=−
x=−
x
200
dt
M
t = 0, x = 90 200 = 0.45
11-8
11.10 (cont’d)
c.
0.45
m = 50 kg / min
m = 100 kg / min
x
m = 200 kg / min
0
t(min)
dx
m
=−
x < 0 , x decreases when t increases
dt
200
dx
becomes less negative until x reaches 0;
dt
Each curve is concave up and approaches x = 0 as t → ∞;
dx
becomes more negative ⇒ x decreases faster.
dt
m increases ⇒
z
z
x
d.
FG
H
t
dx
m
x
m
mt
dt ⇒ ln
=−
=−
t ⇒ x = 0.45 exp −
0.45
200
200
x
M
0.45
0
IJ
K
Check the solution:
(1) t = 0, x = 0.45 ⇒ satisfies the initial condition;
(2)
dx
m
mt
m
= −0.45 ×
exp( −
)=−
x ⇒ satisfies the mass balance.
dt
200
200
200
0.45
0.4
m = 50 kg / m in
0.35
m = 100 kg / m in
0.3
m = 200 kg / m in
x
0.25
0.2
0.15
0.1
0.05
0
0
5
10
15
20
t(m in)
e.
d
i
m = 100 kg min ⇒ t = −2 ln x f 0.45
90% ⇒ x f = 0.045 ⇒ t = 4.6 min
99% ⇒ x f = 0.0045 ⇒ t = 9.2 min
99.9% ⇒ x f = 0.00045 ⇒ t = 138
. min
11-9
25
11.11 a.
e je
Mass of tracer in tank: V m 3 C kg m 3
j
Tracer balance: Accumulation = –output. If perfectly mixed, Cout = C tank = C
b g = −ν C bkg ming
d VC
dC
ν
=− C
dt
V
m
t = 0, C = 0
V
V is constant
dt
b.
z
C
m0 V
c.
dC
=−
C
z
t
ν
0
V
dt ⇒ ln
FG C IJ = − νt ⇒ C = m
V
Hm VK V
0
FG νt IJ
H VK
exp −
0
Plot C (log scale) vs t (rect. scale) on semilog paper: Data lie on straight line (verifying assumption
e
j e
j
of perfect mixing) through t = 1, C = 0.223 × 10 −3 & t = 2, C = 0.050 × 10 −3 .
−
ν
V
=
b
2 −1
E
V = e30 m
11.12 a.
g = −1495
min
.
ln 0.050 0.223
3
min
−1
min j = 201
.
. m
j e1495
−1
3
In tent at any time, P=14.7 psia, V=40.0 ft3, T=68°F=528°R
14.7 psia
40.0 ft 3
PV
3
⇒N=
= m(liquid) =
= 01038
.
lb - mole
ft ⋅ psia
RT
528 o R
10.73
o
lb - mole ⋅ R
b. Molar throughout rate:
60 ft 3 492° R 16.0 psia
1 lb - mole
lb - mole min
= 01695
n in = n out = n =
.
min 528° R 14.7 psia 359 ft 3 STP
Moles of O2 in tank= N (lb - mole) ×
FG lb - mole O IJ
H lb - mole K
b g
2
Balance on O2: Accumulation = input – output
dx
b g = 0.35n − xn ⇒ 01038
. b0.35 − x g
dx
.
= 01695
.
b0.35 − xg ⇒ dt = 163
d Nx
dt
c.
z
dt
z
t = 0, x = 0.21
b
g
t
0.35 − x
dx
= 163
. t
= 163
. dt ⇒ − ln
0.21 0.35 − x
0
0.35 − 0.21
x
b
g
0.35 − x
. e −1.63t
⇒
= e −1.63t ⇒ x = 0.35 − 014
014
.
1
0.35 − 0.27
x = 0.27 ⇒ t =
− ln
= 0.343 min (or 20.6 s)
163
.
0.35 − 0.21
LM FG
N H
IJ OP
KQ
11-10
11.13 a.
b gb
Mass of isotope at any time = V liters C mg isotope liter
g
Balance on isotope: Accumulation = –consumption
FG IJ b g
H K
b g
dC
= − kC
dt
t = 0, C = C0
Cancel V
mg
d
VC = − kC
V L
L⋅s
dt
Separate variables and integrate
z z
C
C0
dC
=
C
FG C IJ = − kt ⇒ t = − lnbC C g
k
HC K
− lnb0.5g
ln 2
=
⇒t =
t
0
0
C = 0.5C0 ⇒ t 1 2
b.
0
− kdt ⇒ ln
t 1 2 = 2.6 hr ⇒ k =
C = 0.01C0
12
k
k
ln 2
= 0.267 hr −1
2.6 hr
t=-ln(C/C0)/k
t=
b g = 17.2 hr
− ln 0.01
0.267
11.14 A → products
a.
Mole balance on A: Accumulation = –consumption
b g = − kC V
d C AV
A
dt
bV constant; cancelsg
t = 0, C A = C A0
⇒
z
CA
CA0
dC A
=
CA
z
t
− kdt ⇒ ln
0
FG C IJ = − kt ⇒ C
HC K
A
A
b g
= C A0 exp − kt
A0
b. Plot C A (log scale) vs. t (rect. scale) on semilog paper. The data fall on a straight line (verifies
b
g b
g
assumption of first-order) through t = 213
. , C A = 0.0262 & t = 120.0, C A = 0.0185 .
ln C A = − kt + ln C A0
−k =
b
ln 0.0185 0.0262
120.0 − 213
.
g = −353
. × 10
−3
min −1 ⇒ k = 35
. × 10 −3 min −1
11.15 2 A → 2 B + C
a.
Mole balance on A: Accumulation = –consumption
b g = − kC V
d C AV
2
A
dt
bV constant; cancelsg
t = 0, C A = C A0
⇒
z
CA
dC A
CA0
C A2
z
LM
N
t
1
1
1
= − kdt ⇒ −
+
= − kt ⇒ C A =
+ kt
0
C A C A0
C A0
11-11
OP
Q
−1
11.15 (cont’d)
b.
C A = 0.5C A0 ⇒ −
n A = 0.5n A0
b
= b0.5n
n
P
RT
1
1
1
+
= − kt 1 2 ⇒ t 1 2 =
; but C A0 = A0 = 0 ⇒ t 1 2 =
0.5C A0 C A0
kC A0
kP0
V
RT
gb
g
mol A react.gb1 mol C 2 mol A react.g = 0.25n
n B = 0.5n A0 mol A react. 2 mol B 2 mol A react. = 0.5n A0
nC
A0
total moles = 125
. n A0 ⇒ P1 2 = 125
.
c.
A0
n A0 RT
= 125
. P0
V
Plot t 1 2 vs. 1 P0 on rectangular paper. Data fall on straight line (verifying 2nd order
d
i d
i
.
& t 1 2 = 209, 1 P0 = 1 0.683
decomposition) through t 1 2 = 1060, 1 P0 = 1 0135
RT
1060 − 209
=
= 143.2 s ⋅ atm
k
1 0135
.
− 1 0.683
Slope:
⇒k =
d.
t1 2 =
b1015 Kgb0.08206 L ⋅ atm mol ⋅ Kg = 0.582 L mol ⋅ s
143.2 s ⋅ atm
I
JK
F
GH
FG IJ
H K
t 1 2 P0
RT
E
1 E 1
exp
⇒ ln
= ln
+
k 0 P0
RT
RT
k0 R T
Plot t 1 2 P0 RT (log scale) vs. 1 T (rect. scale) on semilog paper.
bg
bg
t 1 2 s , P0 = 1 atm, R = 0.08206 L ⋅ atm / (mol ⋅ K), T K
d
i
Data fall on straight line through t 1 2 P0 RT = 74.0, 1 T = 1 900 &
dt
i
RT = 0.6383, 1 T = 1 1050
1 2 P0
b
g
E ln 0.6383 74.0
=
= 29,940 K
R 1 1050 − 1 900
ln
e.
b
R=8.314 J/ (mol ·K)
E = 2.49 × 10 5 J mol
g
1
29,940
= ln 0.6383 −
= −28.96 ⇒ k 0 = 3.79 × 1012 L (mol ⋅ s)
k0
1050
FG
H
T = 980 K ⇒ k = k 0 exp −
C A0 =
b
IJ
K
E
= 0.204 L (mol ⋅ s)
RT
g
0.70 120
. atm
b0.08206 L ⋅ atm mol ⋅ Kgb980 Kg
= 1045
.
× 10 −2 mol L
90% conversion
LM
N
OP
Q
LM
N
1 1
1
1
1
1
−
=
−
−
3
k C A C A0
0.204 1045
.
× 10
.
× 10 −2
1045
= 4222 s = 70.4 min
C A = 010
. C A0 ⇒ t =
11-12
OP
Q
11.16 A → B
a.
Mole balance on A: Accumulation = –consumption(V constant)
dC A
k C
=− 1 A
dt
1+ k2CA
t = 0, C A = C A0
z
1+ k2CA
dC A =
C A0
k 1C A
CA
b
z
t
− dt ⇒
0
g
b
b
b. Plot t C A − C A0 vs. ln C A / C A0
b
g
b
g
k
C
C
k
1
1
ln A + 2 C A − C A0 = − t ⇒ t = 2 C A0 − C A −
ln A
k1
k 1 C A0
k 1 C A0 k 1
g bC
A0
g
− C A on rectangular paper:
x
y
k
t
1 ln C A C A0
=−
+ 2
k 1 C A0 − C A
k1
C A0 − C A
g ;
b
g
slope
1
FG
H
intercept
y1
IJ FG
K H
x1
y2
x2
IJ
K
Data fall on straight line through 116.28, −0.2111 & 130.01, −0.2496
−
1
130.01 − 116.28
=
= −356.62 ⇒ k 1 = 2.80 × 10 −3 L (mol ⋅ s)
k 1 −0.2496 − −0.2111
b
g
b
g
k2
= 130.01 + 356.62 −0.2496 = 4100
. ⇒ k 2 = 0115
.
L mol
k1
11.17 CO + Cl 2 ⇒ COCl 2
a.
3.00 L
273 K
1 mol
.
mol gas
b g = 012035
.
molg 3.00 L = 0.02407 mol L CO U
bC g = 0.60b012035
|Vinitial concentrations
.
molg 3.00 L = 0.01605 mol L Cl |
dC i = 0.40b012035
W
C bt g = 0.02407 − C bt g U
| Since 1 mol COCl formed requires 1 mol of each reactant
C bt g = 0.01605 − C bt g V|
W
303.8 K 22.4 L STP
CO i
Cl 2
2
i
CO
p
Cl 2
p
2
b. Mole balance on Phosgene: Accumulation = generation
d i=
d VC p
dt
c.
d1 + 58.6C
Cl 2
dC p
V=3.00 L
8.75CCO CCl 2
+ 34.3C p
i
dt
2
=
d
id
2.92 0.02407 − C p 0.01605 − C p
− 24.3C i
.
d1941
p
t = 0, C p = 0
b
g
Cl 2 limiting; 75% conversion ⇒ C p = 0.75 0.01605 = 0.01204 mol L
1
t=
2.92
z
.
− 24.3C i
d1941
dC
d0.02407 − C id0.01605 − C i
2
0.01204
0
p
p
p
11-13
p
2
i
11.17 (cont’d)
d.
11.18 a.
REAL F(51), SUM1, SUM2, SIMP
INTEGER I, J, NPD(3), N, NM1, NM2
DATA NPD/5, 21, 51/
FN(C) = (1.441 – 24.3 * C) ** 2/(0.02407 – C)/(0.01605 – C)
DO 10 I = 1, 3
N = NPD(I)
NM1 = N – 1
NM2 = N – 2
DO 20 J = 1, N
C = 0.01204 * FLOAT(J – 1)/FLOAT(NM1)
F(J) = FN(C)
20
CONTINUE
SUM1 = 0.
DO 30 J = 2, NM1, 2
SUM = SUM1 + F(S)
30
CONTINUE
SUM2 = 0.
DO 40 J = 3, NM2, 2
SUM2 = SUM2 + F(J)
40
CONTINUE
SIMP = 0.01204/FLOAT(NM1)/3.0 * (F(1) + F(N) + 4.0 * SUM1 + 2.0 * SUM2)
T = SIMP/2.92
WRITE (6, 1) N, T
10 CONTINUE
1 FORMAT (I4, 'POINTS —', 2X, F7.1, 'MINUTES')
END
RESULTS
5 POINTS — 91.0 MINUTES
21 POINTS — 90.4 MINUTES
51 POINTS — 90.4 MINUTES
t = 90.4 minutes
e j e
Moles of CO 2 in liquid phase at any time = V cm 3 C A mols cm 3
j
Balance on CO 2 in liquid phase: Accumulation = input
b g
e
d
VC A = kS C *A − C A
dt
FjG molsIJ ⇒ dCdt = kSV eC
H s K t = 0, C = 0
A
*
A
− CA
j
÷V
A
Separate variables and integrate. Since p A = y A P is constant, C *A = p A H is also a constant.
z
dC A
CA
0
C *A
⇒ ln
− CA
=
C *A − C A
C *A
z
t
0
e
kS
dt ⇒ − ln C *A − C A
V
=−
j
CA
CA =0
=
kS
t
V
C
kS expb g
t ⇒ 1 − *A = e − kSt V ⇒ C A = C *A 1 − e − kSt V
V
CA
e
1− C A C *A
11-14
j
11.18 (cont’d)
b.
LM
MN
C
V
ln 1 − *A
kS
CA
t=−
OP
PQ
V = 5 L = 5000 cm , k = 0.020 cm s , S = 78.5 cm , C A = 0.62 × 10
3
2
−3
mol / cm
a fa f d9230 atm ⋅ cm moli = 0.65 × 10 mol cm
e5000 cm j lnF1 − 0.62 × 10 I = 9800 s ⇒ 2.7 hr
t=−
b0.02 cm sge78.5 cm j GH 0.65 × 10 JK
C *A = y A P H = 0.30 20 atm
−3
3
3
3
3
−3
−3
2
(We assume, in the absence of more information, that the gas-liquid interfacial surface area equals
the cross sectional area of the tank. If the liquid is well agitated, S may in fact be much greater than
this value, leading to a significantly lower t than that to be calculated)
11.19 A → B
a.
Total Mass Balance: Accumulation = input
dM d ( ρV )
=
= ρv
dt
dt
E
dV
= v
dt
t = 0, V = 0
A Balance: Accumulation = input – consumption
dN A
= C A0 v − ( kC A )V C =N /V
A
A
dt
dN A
= C Ao v − kN A
dt
t = 0, N A = 0
b. Steady State:
c.
z z
z
V
t
0
0
dV =
NA
0
⇒−
dN A
C v
= 0 ⇒ N A = A0
dt
k
⇒ V = vt
vdt
dN A
=
C A0 v − kN A
z
t
dt
0
FG
H
IJ
K
C v
=
1 − expb− kt g
k
C v − kN A
C v − kN A
1
ln A0
= t ⇒ A0
= e − kt
k
C A0 v
C A0 v
⇒ NA
CA =
A0
t →∞⇒ NA =
N A C A0 [1 − exp( − kt )]
=
V
kt
11-15
C A0 v
k
11.19 (cont’d)
When the feed rate of A equals the rate at which A reacts, NA reaches a steady value.
NA would never reach the steady value in a real reactor. The reasons are:
⇒ t → ∞, V → ∞.
(1) In our calculation, V = vt
But in a real reactor, the volume is limited by the reactor volume;
(2) The steady value can only be reached at t → ∞. In a real reactor, the reaction time is finite.
d.
C
C A0 [1 − exp( − kt )]
= lim A0 = 0
t →∞
t →∞ kt
kt
lim C A = lim
t →∞
From part c, t → ∞, N A → a finite number, V → ∞ ⇒ C A =
11.20 a.
MCv
NA
→0
V
dT
= Q − W
dt
M = (3.00 L)(100
. kg / L) = 3.00 kg
Cv = C p = (0.0754 kJ / mol ⋅ o C)(1 mol / 0.018 kg) = 4.184 kJ / kg ⋅ o C
W = 0
dT
= 0.0797Q (kJ / s)
dt
t = 0, T = 18 o C
b.
c.
11.21 a.
z z
100o C
240 s
18 C
0
dT
=
o
0.0797Q dt ⇒ Q =
100 − 18
kJ
= 4.287
= 4.29 kW
240 × 0.0797
s
Stove output is much greater.
Only a small fraction of energy goes to heat the water.
Some energy heats the kettle.
Some energy is lost to the surroundings (air).
Energy balance: MCv
dT
= Q − W
dt
M = 20.0 kg
C v ≈ C p = ( 0.0754 kJ / mol ⋅ C)(1 mol / 0.0180 kg) = 4.184 kJ / (kg ⋅ C)
Q = 0.97 ( 2.50) = 2.425 kJ s
o
o
a f
W = 0
b g
dT
= 0.0290 ° C s , t = 0, T = 25° C
dt
The other 3% of the energy is used to heat the vessel or is lost to the surroundings.
z z
T
b.
t
dT =
o
25 C
c.
bg
0.0290dt ⇒ T = 25° C + 0.0290t s
0
b
g
T = 100° C ⇒ t = 100 − 25 0.0290 = 2585 s ⇒ 43.1 min
No, since the vessel is closed, the pressure will be greater than 1 atm (the pressure at the normal
boiling point).
11-16
11.22 a. Energy balance on the bar
MCv
b
dTb = Q − W = −UA Tb − Tw
dt
g
B
Table B.1
e
M = 60 cm
3
je7.7 g cm j = 462 g
3
Cv = 0.46 kJ (kg ⋅° C), Tw = 25° C
2
U = 0.050 J (min ⋅ cm ⋅° C)
a fa f a fa f a fa f
2
A = 2 2 3 + 2 10 + 3 10 cm = 112 cm
b
gb
dTb
= −0.02635 Tb − 25 ° C min
dt
2
g
t = 0, Tb = 95° C
d
i
dTb
= 0 = −0.02635 Tbf − 25 ⇒ Tbf = 25° C
dt
b.
95
85
75
Tb( oC)
65
55
45
35
25
15
5
0
t
z
z
Tb
c.
t
dTb
= −0.02635dt
T − 25 0
95 b
FG T − 25 IJ = −0.02635t
H 95 − 25K
⇒ T bt g = 25 + 70 expb−0.02635t g
⇒ ln
b
b
Check the solution in three ways:
(1) t = 0, Tb = 25 + 70 = 95o C ⇒ satisfies the initial condition;
dTb
= −70 × 0.02635e −0.02635t = −0.02635(Tb − 25) ⇒ reproduces the mass balance;
dt
(3) t → ∞, Tb = 25o C ⇒ confirms the steady state condition.
(2)
Tb = 30° C ⇒ t = 100 min
11-17
11.23
12.0 kg/min
25oC
12.0 kg/min
T (oC)
Q (kJ/min) = UA (Tsteam-T)
a.
b
g
b
dT
p 25 − T + UA Tsteam − T
= mC
dt
Energy Balance: MCv
g
M = 760 kg
m = 12.0 kg min
dT / dt = 150
. − 0.0224T ( o C min), t = 0, T = 25o C
Cv ≈ C p = 2.30 kJ (min ⋅° C)
UA = 115
. kJ (min ⋅° C)
a
f
Tsteam sat' d; 7.5bars = 167.8° C
b. Steady State:
dT
= 0 = 150
. − 0.0224Ts ⇒ Ts = 67° C
dt
T(oC)
67
25
0
t
z
c.
IJ
K
FG
H
z
Tf
dT
1
150
. − 0.0224T
150
. − 0.94 exp( −0.0224t )
⇒T =
= dt ⇒ t = −
ln
150
. − 0.0224T 0
0.0224
0.94
0.0224
25
t
t = 40 min. ⇒ T = 49.8° C
d. U changed. Let x = (UA) new . The differential equation becomes:
dT
= 0.3947 + 0.096 x − ( 0.01579 + 5.721x )T
dt
z
55
25
0.3947 + 0.096 x − (0.01579 + 5.721 × 10−4 x )T
⇒
−
1
0.01579 + 5.721 × 10−4
z
40
dT
=
dt
0
L 0.3947 + 0.096x − e0.01579 + 5.721 × 10 xj × 55 OP
= 40
ln M
x M 0.3947 + 0.096 x − e0.01579 + 5.721 × 10 x j × 25 P
PQ
NM
−4
−4
⇒ x = 14.27 kJ / (min⋅o C)
ΔU
Δ(UA)
14.27 − 115
.
=
=
× 100% = 241%
.
U initial (UA)initial
115
.
11-18
11.24 a.
Energy balance: MCv
dT
= Q − W
dt
. J g ⋅° C
W = 0, Cv = 177
M = 350 g, Q = 40.2W = 40.2 J s
b gU|V ⇒ T = 20 + 0.0649tbsg
|W T = 40° C ⇒ t = 308 s ⇒ 51. min
dT
= 0.0649 ° C s
dt
t = 0, T = 20° C
b. The benzene temperature will continue to rise until it reaches Tb = 801
. ° C ; thereafter the heat
input will serve to vaporize benzene isothermally.
801
. − 20
Time to reach Tb neglect evaporation : t =
= 926 s
0.0649
Time remaining: 40 minutes 60 s min − 926 s = 1474 s
Evaporation: ΔH = 30.765 kJ mol 1 mol 78.11 g 1000 J kJ = 393 J g
b
v
g
b
b
b
gb
g
gb
g
gb
g
Benzene remaining = 350 g − b0102
.
g sgb1474 sg = 200 g
Evaporation rate = 40.2 J s / 393 J g = 0102
.
g s
c.
11.25 a.
1. Used a dirty flask. Chemicals remaining in the flask could react with benzene. Use a clean flask.
2. Put an open flask on the burner. Benzene vaporizes⇒ toxicity, fire hazard.
Use a covered container or work under a hood.
3. Left the burner unattended.
4. Looked down into the flask with the boiling chemicals. Damage eyes. Wear goggles.
5. Rubbed his eyes with his hand. Wash with water.
6. Picked up flask with bare hands. Use lab gloves.
7. Put hot flask on partner’s homework. Fire hazard.
60 m 3
273 K 1 kg - mole
= 2.58 kg - moles
283 K 22.4 m 3 STP
dT
Energy balance on room air: nCv
= Q − W
dt
Q = m ΔH H O, 3bars, sat' d − 30.0 T − T
Moles of air in room: n =
b g
W = 0
nCv
s
v
b
g
2
b
dT
s ΔH v − 30.0 T − T0
=m
dt
N = 2.58 kg - moles
b
0
g
g
Cv = 20.8 kJ (kg - mole⋅° C)
ΔH v = 2163 kJ kg from Table B.6
T0 = 0° C
b
b
dT
= 40.3m s − 0.559T ° C hr
dt
g
g
t = 0, T = 10° C
(Note: a real process of this type would involve air escaping from the room and a constant pressure
being maintained. We simplify the analysis by assuming n is constant.)
11-19
11.25 (cont’d)
0.559T
40.3
b. At steady-state, dT dt = 0 ⇒ 40.3m s − 0.559T = 0 ⇒ m s =
T = 24° C ⇒ m s = 0.333 kg hr
c.
Separate variables and integrate the balance equation:
z
Tf
10
dT
=
40.3m s − 0.559T
z
t
0
dt
z
m s = 0.333
E
T f = 23°C
t=−
11.26 a.
b
Integral energy balance t = 0 to t = 20 min
Q = ΔU = MCv ΔT =
dT
23
10 13.4 − 0.559T
=t
LM
MN
b g OP = 4.8 hr
b g PQ
13.4 − 0.559 23
1
ln
0.559 13.4 − 0.559 10
g
b60 − 20g° C = 4.00 × 10
250 kg 4.00 kJ
kg⋅° C
4
kJ
4.00 × 104 kJ 1 min 1 kW
Required power input: Q =
= 333
. kW
20 min
60 s 1 kJ s
b. Differential energy balance: MCv
dT
= Q
dt
bg
dT
= 0.001Q t
dt
M = 250 kg
Cv = 4.00 kJ kg⋅°C
z z
T
Integrate:
t = 0, T = 20° C
z
t
t
dT = 0.001 Q dT ⇒ T = 20 o C + Qdt
20o C
0
0
Evaluate the integral by Simpson's Rule (Appendix A.3)
600 s
= 30 33 + 4 33 + 35 + 39 + 44 + 50 + 58 + 66 + 75 + 85 + 95
Qdt
3
0
z
b
g
b
g
+2 34 + 37 + 41 + 47 + 54 + 62 + 70 + 80 + 90 + 100 = 34830 kJ
b g
jb
e
g
⇒ T 600 s = 20o C + 0.001 oC / kJ 34830 kJ = 54.8° C
c.
b
g
10 kW
Past 600 s, Q = 100 +
t − 600 s = t 6
60 s
LM
= 20 + 0.001M Qdt
T = 20 + 0.001 Qdt
MM +
MN
0.001 F t
600 I
⇒ T = 54.8 +
−
⇒ t bsg =
G
6 H6
2 JK
z
z z
600
t
0
0
34830
2
2
t
OP
t P
dt
6 P
PP
Q
600
b
12000 T − 24.8
g
T = 85° C ⇒ t = 850 s = 14 min, 10 s ⇒ explosion at 10:14 + 10 s
11-20
11.27 a. Total Mass Balance:
Accumulation=Input– Output
E
dM tot
d( ρV)
i − m
o ⇒
=m
= 8.00ρ − 4.00ρ
dt
dt
ρ=constant
dV
= 4.00 L / s
dt
t = 0, V0 = 400 L
KCl Balance:
dM KCl
d( CV )
i, KCl − m
o, KCl ⇒
. × 8.00 − 4.00C
=m
= 100
dt
dt
dC 8 − 8C
=
dV dt = 4
dV
dC
dt
V
⇒V
+C
= 8 − 4C
dt
dt
t = 0, C 0 = 0 g / L
Accumulation=Input-Output ⇒
b. (i)The plot of V vs. t begins at (t=0, V=400). The slope (=dV/dt) is 4 (a positive constant).
V increases linearly with increasing t until V reaches 2000. Then the tank begins to overflow
and V stays constant at 2000.
V
2000
400
0
t
(ii) The plot of C vs. t begins at (t=0, C=0). When t=0, the slope (=dC/dt) is (8-0)/400=0.02.
As t increases, C increases and V increases (or stays constant)⇒ dC/dt=(8-8C)/V becomes
less positive, approaches zero as t→ ∞. The curve is therefore concave down.
C
1
0
t
c.
dV
=4⇒
dt
z
V
z
t
dV = 4 dt ⇒ V = 400 + 4t
400
0
11-21
11.27 (cont’d)
dC 8 − 8C
dC
1− C
=
=
V = 400 + 4 t
dt
V
dt 50 + 0.5t
C dC
t
dt
C
t
=
⇒ − ln(1 − C ) 0 = 2 ln(50 + 0.5t ) 0
0 1− C
0 50 + 0.5t
50 + 0.5t
⇒ ln(1- C)-1 = 2 ln
= ln(1 + 0.01t )2
50
1
1
⇒
= (1 + 0.01t )2 ⇒ C = 1 −
1- C
(1 + 0.01t )2
z
z
When the tank overflows, V = 400 + 4t = 2000 ⇒ t = 400 s
1
= 0.96 g / L
C = 12
1+ 0.01 × 400
b
g
11.28 a. Salt Balance on the 1st tank:
Accumulation=-Output
E
v
d(CS1V1 )
dC
= − CS1v ⇒ S1 = − CS1 = −0.08CS1
V1
dt
dt
CS1 ( 0) = 1500 500 = 3 g / L
Salt Balance on the 2nd tank:
Accumulation=Input-Output
E
v
d(CS2V2 )
dC
= CS1v − CS 2 v ⇒ S2 = ( CS1 − CS 2 ) = 0.08( CS1 − CS 2 )
dt
V2
dt
CS 2 ( 0) = 0 g / L
Salt Balance on the 3rd tank:
Accumulation=Input-Output
E
v
d(CS3V3 )
dC
= CS 2 v − CS 3v ⇒ S3 = ( CS 2 − CS 3 ) = 0.04( CS 2 − CS 3 )
dt
V3
dt
CS 3 ( 0) = 0 g / L
b.
CS1, CS2, CS3
3
CS1
CS2
CS3
0
t
11-22
11.28 (cont’d)
The plot of CS1 vs. t begins at (t=0, CS1=3). When t=0, the slope (=dCS1/dt) is −0.08 × 3 = −0.24 .
As t increases, CS1 decreases ⇒ dCS1/dt=-0.08CS1 becomes less negative, approaches zero as
t→ ∞. The curve is therefore concave up.
The plot of CS2 vs. t begins at (t=0, CS2=0). When t=0, the slope (=dCS2/dt) is 0.08(3 − 0) = 0.24 .
As t increases, CS2 increases, CS1 decreases (CS2 < CS1)⇒ dCS2/dt =0.08(CS1-CS2) becomes less
positive until dCS2/dt changes to negative (CS2 > 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