CHAPTER TWO
2.1 (a)
(b)
(c)
2.2 (a)
(b)
(c)
3 wk
7d
24 h 3600 s 1000 ms
1 wk 1 d
1 h 1 s
38.1 ft / s 0.0006214 mi 3600 s
3.2808
554 m 4
1d
ft
h
1
1000 g
m
1 h
0.0006214 mi 3600 s
1
kg
35.3145 ft 3
5.37 × 10 3 kJ 1 min 1000 J
min
60 s
1
m
4
= 3.85 × 10 4 cm 4 / min⋅ g
= 340 m / s
1 m3
921 kg 2.20462 lb m
m3
= 25.98 mi / h ⇒ 26.0 mi / h
1 kg 10 8 cm 4
1h
d ⋅ kg 24 h 60 min
760 mi
1 h
= 1.8144 × 10 9 ms
= 57.5 lb m / ft 3
1.34 × 10 -3 hp
1 kJ
1
J/s
= 119.93 hp ⇒ 120 hp
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 =
= 5.18 × 10 6 ≈ 5 million balls
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 1 d
3600 s 1.86 × 10 5 mi
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
1 report
0.0006214 mi
0.001 m
= 4 × 10 11 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
$1.25
1 gal
gal 44.7 mi
x (mi)
Equate the two costs ⇒ x = 4.3 × 10 5 miles
2-1
= 14,500 + 0.04464 x
x (mi)
= 21,700 + 0.02796x
2.7
106 cm 3
220.83 imp. gal
5320 imp. gal 14 h 365 d
plane ⋅ h 1 d 1 yr
0.965 g
1 cm 3
1 kg 1 tonne
1000 g 1000 kg
tonne kerosene
plane ⋅ yr
9
4.02 ×10 tonne crude oil 1 tonne kerosene
plane ⋅ yr
5
yr
7 tonne crude oil 1.188 ×10 tonne kerosene
= 1.188 ×105
= 4834 planes ⇒ 5000 planes
2.8 (a)
(b)
(c)
2.9
2.10
2.11
32.1714 ft / s2
25.0 lb m
1
lb f
32.1714 lb m ⋅ ft / s 2
25 N
1 kg ⋅ m / s2
1
9.8066 m / s 2
10 ton
1N
1 lb m
5 × 10
50 × 15 × 2 m 3
-4
980.66 cm / s2
1 g ⋅ cm / s2
35.3145 ft 3
85.3 lb m
3
3
1
1 dyne
2.20462 lb m
1
500 lbm
= 2.55 kg ⇒ 2.6 kg
1000 g
ton
= 25.0 lb f
kg
2.20462 lb m
m
1 ft
1 m3
11.5 kg
32.174 ft
1 s
2
1 lb f
32.174 lb m / ft ⋅ s
F 1I F 1 I
JG J
H 2 KH10 K
≈ 5 × 10 2 G
= 9 × 10 9 dynes
2
= 4.5 × 10 6 lb f
≈ 25 m 3
(a)
mdisplaced fluid = mcylinder ⇒ ρ f V f = ρ cVc ⇒ ρ f hπr 2 = ρ c Hπr 2
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 =
= 1.71 g / cm 3
h
(30 cm - 20.7 cm)
2-2
ρc
H
ρf
h
2.12
Vs =
πR 2 H
πR 2 H πr 2 h
R r
R
; Vf =
−
;
= ⇒r= h
3
3
3
H h
H
h
2
πR 2 H πh F Rh I
πR 2 F
h3 I
⇒ Vf =
−
H
−
G
J =
G
J
3
3 HHK
3 H
H2 K
ρ f V f = ρ s Vs ⇒ ρ f
H
⇒ ρ f = ρs
H−
2.13
h3
r
H
πR 2 F
h3 I
πR 2 H
H
−
=
ρ
s
J
3 G
3
H
H2 K
= ρs
H3
R
1
= ρs
H 3 − h3
ρs
hI
J
HH K
3
F
1− G
H2
Say h( m ) = depth of liquid
y
y= 1
dA
y=y=1––1+h
h
x
⇒
A(m 2 )
h
1m
x = 1– y 2
y= –1
dA
1− y 2
dA = dy ⋅
2
z dx = 2 1 − y dy
− 1− y
2
−1+ h
⇒ Adm
2
i
=2
2
z 1 − y dy
−1
E Table of integrals or trigonometric substitution
Adm 2 i = y 1 − y 2 + sin−1 y
W bN g =
h −1
−1
= bh − 1g 1 − bh − 1g + sin −1 bh − 1g +
2
4 m × A( m 2 ) 0.879 g 10 6 cm 2
cm
3
1m
3
1 kg
9.81 N
3
kg
{
10 g
= 3.45 × 10 4 A
g g0
E
Substitute for A
L
W bN g = 3.45 × 10 4 Mbh − 1g 1 − bh − 1g + sin −1 bh − 1g +
2
N
πO
2 PQ
2.14 1 lb f = 1 slug ⋅ ft / s2 = 32 .174 lb m ⋅ ft / s2 ⇒ 1 slug = 32.174 lb m
1
1 poundal = 1 lb m ⋅ ft / s2 =
lb f
32.174
2-3
π
2
ρf
2.14(cont’d)
(a) (i) On the earth:
175 lb m
M=
W=
175 lb m
(ii) On the moon
175 lb m
M=
W=
175 lb m
1 slug
= 5.44 slugs
32.174 lbm
32.174 ft
1 poundal
3
2
2 = 5.63 × 10 poundals
s 1 lb m ⋅ ft / s
1 slug
= 5.44 slugs
32.174 lb m
32.174 ft
1 poundal
= 938 poundals
2
6
s 1 lb m ⋅ ft / s2
(b ) F = ma ⇒ a = F / m =
1 lb m ⋅ ft / s 2
1 poundal
355 poundals
25.0 slugs
1 slug
32.174 lb m
1m
3.2808 ft
= 0.135 m / s2
F 1I
2.15 (a) F = ma ⇒ 1 fern = (1 bung)(32.174 ft / s 2 )G J = 5.3623 bung ⋅ ft / s2
H6 K
1 fern
⇒
5.3623 bung ⋅ ft / s2
3 bung 32.174 ft
1 fern
= 3 fern
2
6 s 5.3623 bung ⋅ ft / s2
On the earth: W = (3)(32.174 ) / 5.3623 = 18 fern
(b) On the moon: W =
2.16 (a) ≈ (3)(9 ) = 27
≈
(d)
≈ 50 × 10 3 − 1 × 10 3 ≈ 49 × 10 3 ≈ 5 × 10 4
( 2.7)(8.632 ) = 23
(c) ≈ 2 + 125 = 127
2.365 + 125.2 = 127 .5
2.17 R ≈
4.0 ×10−4
≈ 1× 10−5
40
(3.600 ×10−4 ) / 4 5 = 8.0 × 10−6
(b)
4.753 × 10 4 − 9 × 10 2 = 5 × 10 4
(7 × 10−1 )(3 × 105 )(6)(5 ×104 )
≈ 42 ×10 2 ≈ 4 ×103 (Any digit in range 2-6 is acceptable)
6
(3)(5 × 10 )
Rexact = 3812.5 ⇒ 3810 ⇒ 3.81 ×103
2-4
2.18 (a)
A: R = 731
. − 72.4 = 0.7 o C
72.4 + 731
. + 72.6 + 72.8 + 73.0
= 72.8 o C
5
X=
( 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
o
= 0.3 C
s=
B: R = 1031
. − 97.3 = 58
. oC
X=
97.3 + 1014
. + 98.7 + 1031
. + 100.4
= 100.2 o C
5
s=
(97.3 − 1002
. ) 2 + (1014
. − 1002
. ) 2 + (98.7 − 1002
. ) 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.
2.19 (a)
12
X =
∑
C min=
12
Xi
i =1
= 735
.
s=
12
= X − 2 s = 73.5 − 2 (1.2) = 711
.
∑ ( X − 73.5)
i =1
12 − 1
2
= 12
.
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
2-5
2.20 (a),(b)
(a) Run
1
2
X
134 131
Mean(X) 131.9
Stdev(X) 2.2
Min
127.5
Max
136.4
(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
4
5 6
7 8
9 10 11 12 13 14 15
133 135 131 134 130 131 136 129 130 133 130 133
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
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
h
2.22 N Pr =
N Pr ≈
Cp µ
k
1 h
2
3600 s
kg
m
−4
(2 × 10 )(2 )(9 )
≈ 12 × 10 ( −4− 3) ≈ 12
. × 10 −6 lb ⋅ ft 2 / s
3 × 10 3
= 1.48 × 10 −6 lb ⋅ ft 2 / s = 0.00000148 lb ⋅ ft 2 / s
(b) Q' approximate ≈
Q' exact
2.10462 lb 3.2808 2 ft 2
=
0.583 J / g ⋅ o C 1936 lb m
1 h 3.2808 ft
1000 g
o
0.286 W / m ⋅ C
ft ⋅ h 3600 s
m 2.20462 lbm
(6 × 10 −1 )(2 × 10 3 )(3 × 10 3 ) 3 × 10 3
≈
≈ 15
. × 10 3 . The calculator solution is 1.63 × 10 3
(3 × 10 −1 )(4 × 10 3 )(2 )
2
2.23
Duρ 0.48 ft
1 m
2.067 in
1 m
=
−3
µ
s 3.2808 ft 0.43 × 10 kg / m ⋅ s 39.37 in
Re ≈
(5 × 10 −1 )(2 )(8 × 10 − 1 )(10 6 ) 5 × 101−( −3)
≈
≈ 2 × 10 4 ⇒ the flow is turbulent
(3)(4 × 10)(10 3 )(4 × 10 −4 )
3
2-6
0.805 g
cm 3
1 kg 10 6 cm 3
1000 g 1 m3
Re =
2.24
(a)
kgdp y
D
1/3
µ I
J
H ρD K
F
= 2.00 + 0.600G
= 2.00 +
L
0.600M
.
N(100
= 44.426 ⇒
1/2
F d p uρ I
G
J
H µ K
1/ 3
O
1.00 × 10 −5 N ⋅ s / m2
P
3
−5
2
kg / m )(100
. × 10 m / s) Q
k g (0.00500 m)(0.100)
1.00 × 10 −5 m 2 / s
L (0.00500 m)(10.0 m / s)(100
. kg
M
−5
(100
. × 10 N ⋅ s / m 2 )
N
1/2
/ m3 ) O
P
Q
= 44.426 ⇒ k g = 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 D (m2/s) µ (N-s/m 2) ρ (kg/m 3) u (m/s) kg
0.005 0.1 1.00E-05 1.00E-05
1
10
0.889
0.010 0.1 1.00E-05 1.00E-05
1
10
0.620
0.005 0.1 2.00E-05 1.00E-05
1
10
1.427
0.005 0.1 1.00E-05 2.00E-05
1
10
0.796
0.005 0.1 1.00E-05 1.00E-05
1
20
1.240
2.25 (a) 200 crystals / min ⋅ mm; 10 crystals / min ⋅ mm 2
(b) r = 200 crystals
0.050 in 25.4 mm
min ⋅ mm
in
= 238 crysta ls / min ⇒
(c) Dbmmg =
10 crystals
−
(25.4) 2 mm 2
min ⋅ mm2
238 crystals 1 min
min
0.050 2 in 2
in 2
= 4.0 crystals / s
60 s
crystals 60 s
D ′ bing 25.4 mm
F crystals I
= 60r ′
= 25.4 D ′ ; r G
J = r′
H
K
min
s
1 min
1 in
2
⇒ 60r ′ = 200b25.4 D ′g − 10b25.4 D ′g ⇒ r ′ = 84 .7 D ′ − 108bD ′g
2
2.26 (a) 70.5 lb m / ft 3 ; 8.27 × 10 -7 in 2 / lb f
−7
2
9 × 10 6 N
14.696 lb f /in 2 
(b) ρ = (70.5 lb / f t 3) exp 8.27 × 10 in


m
lbf
m 2 1.01325 × 10 5 N/m 2 

70.57 lb m 35.3145 ft 3 1 m 3
1000 g
=
= 1.13 g /cm 3
3
3
6
3
ft
m 10 cm 2.20462 lb m
g
1 lb m
28,317 cm 3
cm 3
453.593 g
1 ft 3
N
F lb I
P G f2 J = P ' 2
Hin K
m
0.2248 lb f
12
F lb m I
J =
H ft 3 K
(c) ρ G
ρ′
⇒ 62.43 ρ ′ = 70.5 exp d8.27 × 10
m2
2
1N
39.37 in
−7
= 62.43ρ ′
2
. × 10
i d145
−4
= 1.45 × 10 −4 P'
P 'i ⇒ ρ ′ = 113
. expd1.20 × 10 −10 P'i
P' = 9.00 × 10 6 N / m 2 ⇒ ρ ' = 113
. exp[(1.20 × 10 −10 )(9.00 × 10 6 )] = 113
. g / cm 3
2-7
2.27 (a) V dcm 3 i =
V ' din 3 i
28,317 cm 3
1728 in 3
= 16.39V ' ; t bsg = 3600t ′bhr g
⇒ 16.39V ' = expb3600t ′g ⇒ V ' = 0.06102 expb3600t ′ g
(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:
Cint =
( 0.6 − 0) + 3.00 = 1.4 mol / L
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
(0.10 − 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)
p* =
2.29 (a)
(b)
c
1
902
*
903
2
60 − 20
(185 − 166.2) + 20 = 42 mm Hg
1998
. − 1662
.
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)
CONTINUE
WRITE (5, 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
FORMAT (10X, F5.1, 10X, F5.1)
CONTINUE
END
2-8
2.29 (cont’d)
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
DATA
OUTPUT
98.5
1.0
TEMPERATURE
VAPOR PRESSURE
131.8
5.0
(C)
(MM HG)
100.0
1.2
M
M
215.5
100.0
105.0
1.8
M
M
215.0
98.7
2.30 (b) ln y = ln a + bx ⇒ y = ae bx
b = (ln y 2 − ln y 1 ) / ( x 2 − x 1 ) = (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 y 1 ) / (ln x 2 − ln x 1 ) = (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 = ae by / x ⇒ y = ( a / x )e by / 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 − 3 ln(2.0) ⇒ a = 2 ⇒ xy = 2e 3 y / x ⇒ y = (2 / x )e 3y / x
(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 ]
= (ln 807.0 − ln 40.2 ) / (ln 2.0 − ln 1.0) = 4.33
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)
(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
2-9
x [rect. axes], slope =
a
1
, intercept =
b
b
2.31 (cont’d)
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)
1
b
x
x
= ax + ⇒ = ax 2 + b ⇒ Plot
vs. x 2 [rect.], slope = a , intercept = b
y
x
y
y
OR
1
b
1
b
1
1
= ax + ⇒
= a + 2 ⇒ Plot
vs. 2 [rect.] , slope = b, intercept = a
y
x
xy
x
xy
x
2.32 (a) A plot of y vs. R is a line through ( R = 5 , y = 0.011 ) and ( R = 80 , y = 0.169 ).
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
0.169 − 0.011
U
= 2.11 × 10 −3
|
−3
−4
80 − 5
V ⇒ y = 2 .11 × 10 R + 4.50 × 10
−3
−4
b = 0.011 − d2.11 × 10 i b5g = 4.50 × 10 |W
a=
(b) R = 43 ⇒ y = d2.11 × 10 −3 i b43g + 4.50 × 10 −4 = 0.092 kg H 2O kg
b1200
kg h gb0.092 kg H 2 O kgg = 110 kg H 2 O h
2-10
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) T = 9677.6φ −1.19 ⇒ φ = b9677.6 / T g
0.8403
T = 85 o C ⇒ φ = b9677.6 / 85g
0.8403
= 53.5 L / s
0.8403
T = 175 o C ⇒ φ = b9677.6 / 175g
0.8403
T = 290 o C ⇒ φ = b9677.6 / 290g
= 29.1 L / s
= 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.
2-11
ln ((C A-CAe )/(CA0-C Ae))
2.34 (a) Yes, because when ln[(CA − C Ae ) / (C A0 − CAe )] 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 (min)
Slope = -0.0093 ⇒ k = 9.3 × 10 -3 min −1
(b) ln[(CA − C Ae ) / (C A0 − C Ae )] = −kt ⇒ C A = (C A 0 − CAe )e − kt + CAe
−3
CA = (0.1823 − 0.0495)e− (9.3×1 0
)(120)
+ 0.0495 =9.300 ×10-2 g/L
9.300 ×10-2 g 30.5 gal 28.317 L
C =m/ V ⇒ m=CV =
= 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(3.53 × 10 −2 ) ; or
V(logarithmic axis) vs. t2 in semilog coordinates, slope=2, intercept= 3.53 × 10−2
(c) V ( m 3 ) = 100
. × 10 −3 exp(15
. × 10− 7 t 2 )
2.36 PV k = C ⇒ P = C / V k ⇒ ln P = ln C − k ln V
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 = e 12.736 = 3.40 × 10 5 mm Hg ⋅ cm 4.719
G − GL
1
G −G
G −G
=
⇒ 0
= K L Cm ⇒ ln 0
= ln K L + mln C
m
G0 − G K L C
G − GL
G − GL
ln(G 0 -G)/(G-G L )= 2.4835lnC - 10.045
3
ln(G 0-G)/(G-G L)
2.37 (a)
2
1
0
-1
3.5
4
4.5
5
lnC
2- 12
5.5
2.37 (cont’d)
m = slope = 2.483 (dimensionless)
Intercept = ln KL = −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 ln V& + c ln p
ln(3.5) = ln a + b ln(1.02 ) + c ln(9.1)
ln(2.58) = ln a + b ln(102
. ) + c ln(11.2 )
ln(3.72 ) = ln a + b ln(175
. ) + c ln(11.2 )
b = 0.68
⇒ c = −1.46
a = 86.7 volts ⋅ kPa 1.46 / (L / s) 0.678
& . 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
0.5
1
1.5
lnV
lnZ = 0.5199lnV + 1.0035
b = slope = 0.52
Intercept = lna + c ln P = 1.0035
&
When V is constant (runs 5 to 7), plot lnZ vs. lnP. Slope=c, Intercept= ln a + c ln V&
2
lnZ
1.5
1
0.5
0
1.5
1.7
1.9
2.1
2.3
lnP
lnZ = -0.9972lnP + 3.4551
c = slope = −0.997 ⇒ 1.0
Intercept = lna + b ln V& = 3.4551
Z
Plot Z vs V& b P c . Slope=a (no intercept)
7
6
5
4
3
2
1
0.05
0.1
0.15
b
b c
Z = 31.096V P
VP
0.2
c
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- 13
2.39 (a)
sxy =
sxx =
n
1
n
∑x y
1
n
∑x
i
i
= [( 0.4 )(0.3) + (2.1)(1.9) + (3.1)(3.2 )] / 3 = 4.677
i =1
n
2
i
= (0.32 + 1.9 2 + 3.2 2 ) / 3 = 4 .647
i =1
1 n
1
xi = (0.3 + 1.9 + 3.2 ) / 3 = 1.8; s y =
n i =1
n
sxy − sx sy 4.677 − (1.8)(1.867)
a=
=
= 0.936
2
4.647 − (18
. )2
sxx − bsx g
∑
sx =
b=
sxx s y − sxy sx
2
sxx − bsx g
n
∑y
i
= (0.4 + 2.1 + 3.1) / 3 = 1.867
i =1
(4.647 )(1867
. ) − (4.677 )(18
.)
= 0182
.
2
4.647 − (18
.)
=
y = 0.936x + 0.182
(b) a =
sxy
sxx
=
4.677
= 10065
.
⇒ y = 1.0065 x
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
t
1
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- 14
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 = 0168
.
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
Lab 3
k = 0.0063 s-1
t
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- 15
2.43 y i = ax i ⇒ φ (a ) =
n
∑
d i2 =
i =1
n
⇒a =
∑
i =1
b yi
2
− axi g ⇒
dφ
= 0=
da
n
∑
i =1
2b yi − axi gxi ⇒
n
∑
i =1
yi xi − a
n
∑x
2
i
i =1
n
∑y x /∑x
2
i
i i
i =1
2.44
n
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- 16
=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 cm 2 / 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
ln D
-10.0
-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- 17
(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
16 × 6 × 2 m 3 1000 kg
(a) m =
m3
(b) m& =
≈ b2 × 10gb5gb2 gd103 i ≈ 2 × 105 kg
8 oz 1 qt
10 6 cm3
1g
4 × 106
≈
≈ 1 × 10 2 g / s
3
3
2 s 32 oz 1056.68 qt cm b3 × 10gd10 i
(c) Weight of a boxer ≈ 220 lb m
12 × 220 lb m 1 stone
Wmax ≥
≈ 220 stones
14 lb m
dictionary
(d)
V=
≈
πD L 314
.
4.5 ft
=
4
4
2
2
2
800 miles 5880 ft 7.4805 gal 1 barrel
1 mile
1 ft 3
42 gal
3 × 4 × 5 × d8 × 10 2 i × d5 × 10 3 i × 7
4 × 4 × 10
(e) (i) V ≈
6 ft × 1 ft × 0.5 ft 28,317 cm 3
(ii) V ≈
1 ft 3
≈ 1 × 10 7 barrels
≈ 3 × 3 × 10 4 ≈ 1 × 105 cm 3
1 ft 3
28,317 cm 3
62.4 lb m
1 ft 3
150 lb m
150 × 3 × 104
≈ 1 × 105 cm 3
60
≈
(f) SG ≈ 105
.
3.2
995 kg
(a) (i)
m
(ii)
3
0.028317 m 3
1 lb m
0.45359 kg
1 ft
995 kg / m 3 62.43 lb m / ft 3
1000 kg / m 3
3
= 62.12 lb m / ft 3
= 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 × 10 3 kg
1 m3
m3 10 3 L
m 3 1000 L 1 min
1150 kg
min
10 gal
= 35 kg
0.7 × 1000 kg
1 ft 3
2 min 7.481 gal
1 m3
0.70 × 62.43 lb m
1 ft 3
60 s
= 27 L s
≅ 29 lb m / min
3- 1
3.3 (cont’d)
(d) Assuming that 1 cm 3 kerosene was mixed with Vg (cm 3 ) gasoline
Vg dcm 3gasoline i ⇒ 0.70Vg dg gasoline i
1dcm3 kerosene i ⇒ 0.82 dg kerosene i
SG =
d0.70Vg
+ 0.82 i dg blendi
Vg + 1dcm blend i
3
Volumetric ratio =
3.4
In France:
In U.S.:
3.5
50.0 kg
= 0.78 ⇒ V g =
0.82 − 0.78
3
= 0.5 0 cm
0.78 − 0.70
Vgasoline 0.50 cm 3
=
= 0.50 cm 3 gasoline / cm3 kerosene
Vkerosene
1 cm3
L
5 Fr
$1
= $68.42
1L 5.22 Fr
1 gal
$1.20
= $22.64
0.70 × 1.0 kg 3.7854 L 1 gal
0.7 × 1.0 kg
50.0 kg
L
V&B ( ft 3 / h ), m& B ( lb m / h )
V& ( ft 3 / h ), SG = 0.850
V& H ( ft 3 / h ), m& H ( lb m / h )
700 lb m / h
700 lb m
ft 3
(a) V& =
= 13.19 ft 3 / h
h
0.850 × 62.43 lb m
3
&
V dft i 0.879 × 62.43 lb m
& B bkg / hg
m& B = B
= 54.88 V
3
ft
bh g
m& H = dV&H hb0.659 × 62.43g = 4114
. V&H bkg / h g
V&B + V&H = 1319
. ft 3 / h
m& B + m& H = 54.88V&B + 4114
. V&H = 700 lb m
⇒ V&B = 114
. ft 3 / h ⇒ m& B = 628 lb m / h benzene
& H = 71.6 lb m / h hexane
V&H = 1.74 ft 3 / h ⇒ m
(b) – No buildup of mass in unit.
– ρ B and ρ H at inlet stream condit ions 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 1.2563 × 1.000 kg
= 445 L
(b)
L
18255
.
× 1.00 kg
195.5 kg H 2SO 4
0.65 kg H 2O
+
0.35 kg H 2SO 4
470 − 445
% error =
× 100% = 5.6%
445
V ideal =
3.7
195.5 kg H 2SO 4
L
= 470 L
1.000 kg
Buoyant force bup g= Weight of block bdowng
E
Mass of oil displaced + Mass of water displaced = Mass of block
ρ oil b0.542 gV + ρ H
2O
b1 − 0.542 gV
= ρc V
From Table B.1: ρ c = 2.26 g / cm3 , ρ w = 100
. g / cm 3 ⇒ ρ oil = 3.325 g / cm3
moil = ρ oil × V = 3.325 g / cm3 × 35.3 cm 3 = 117.4 g
moil + flask = 117.4 g + 124.8 g = 242 g
3.8
Buoyant force bup g = Weight of block bdowng
⇒ Wdisplaced liquid = Wblock ⇒ ( ρVg ) disp. Liq = ( ρVg ) block
Expt. 1: ρ w b15
. Agg = ρ B b2 Agg ⇒ ρ B = ρ w ×
ρ w =1.00 g/cm 3
15
.
2
ρ B = 0.75 g / cm 3 ⇒ bSG gB = 0.75
Expt. 2: ρ soln b Agg = ρ B b2 Agg ⇒ ρ soln = 2 ρ B = 15
. g / cm3 ⇒ bSG gsoln = 15
.
3.9
Let ρ w = density of water. Note: ρ A > ρ w (object sinks)
Volume displaced: Vd 1 = Abhsi = Ab dh p1 − hb1 i
hs 1
WA + WB
Archimedes ⇒
h ρ1
hb 1
h p1 − hb1 =
ρ wVd 1g
= WA + WB
123
weight of displaced water
Subst. (1) for Vd 1 , solve for dh p1 − hb1 i
Before object is jettisoned
WA + WB
p w gAb
(2)
bi g
Volume of pond water: Vw = Ap h p1 − Vd1 ⇒ Vw = Ap h p1 − Ab dh p1 − hb1 i
subst. ( 2 )
for b p1 − hb 1
W A + WB
V
W + WB
⇒ hp1 = w + A
ρw g
Ap
ρ w gAp
(3)
Vw (W A + WB )  1
1
+
− 

Ap
ρ wg
 Ap Ab 
(4)
Vw = Ap h p1 −
subst. ( 3for
) hp1 in
(2, )solve for
hb1
hb1 =
(1)
3- 3
3.9 (cont’d)
hs 2
WB
WA
WA
ρ Ag
(5)
Volume displaced by boat: Vd 2 = Ab dh p2 − hb 2 i
(6)
Let V A = volume of jettisoned object =
hρ 2
h b2
After object is jettisoned
Archimedes ⇒ ρ WVd 2 g = WB
E
Subst. for Vd 2 , solve for dh p2 − hb 2 i
h p 2 − hb2 =
WB
ρ w gAb
(7)
Volume of pond water: Vw = Ap h p2 − Vd 2 − VA
solve for
⇒ h p2 =
h p 21
subst. ( 8 )
⇒
for h p 2 in (7, )solve for hb 2
(5), (6 ) & ( 7 )
Vw = Ap h p 2 −
Vw
WB
WA
+
+
Ap ρw gAp ρ A gAp
hb2 =
WB
W
− A
ρwg ρ A g
(8)
Vw
WB
WA
WB
+
+
−
Ap ρ w gAp ρ A gAp ρ w gAb
(9)
(a) Change in pond level
( 8)− (3)
h p2 − hp1 =
WA
Ap g
 1
1  W A ( ρw − ρ A )
 ρ − ρ  = ρ ρ gA < 0 (since ρw < ρ A )
W 
A w
p
 A
⇒ the pond level falls
(b)
Change in boat level
>0
>0
}
 644744
8
5) 

( 9)− (4 ) W  1
(



A
 
1
1
V
ρ  p
A
h p 2 − hp1 =
−
+
− 1   > 0

 =  A  1 +  A 
Ap g  ρ A Ap ρW A p ρW A b   Ap    ρW  Ab
 




⇒ the boat rises
3.10 (a) ρ bulk =
2.93 kg CaCO 3 0.70 L CaCO3
L CaCO3
(b) Wbag = ρ bulkVg =
L total
= 2.05 kg / L
2.05 kg 50 L 9.807 m / s2
1N
= 1.00 × 10 3 N
L
1 kg ⋅ m / s2
Neglected the weight of the bag itself and of the air in the filled bag.
(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
Wb − WI
(1202 N - 44.0 N)
1 kg ⋅ m / s2
=
= 119 L
ρwg
0.996 kg / L × 9.807 m / s2
1N
m
122 .5 kg
ρb = b =
= 1.03 kg / L
Vb
119 L
Vb =
m f + mnf = mb
(b)
xf =
mf
mb
(1)
⇒ m f = mb x f
(1),( 2) ⇒ mnf = mb d1 − x f
V f + Vnf = Vb ⇒
b2 g,b3g
F
⇒ mb G
xf
Hρ f
(c) x f =
+
1− xf
I
ρ nf
J
K
1 / ρ b − 1 / ρ nf
=
1 / ρ f − 1 / ρ nf
mf
ρf
=
+
(2)
(3)
i
mnf
ρ nf
=
mb
ρb
F 1
1 / ρ b − 1 / ρ nf
mb
1 I
1
1
⇒ xf G
−
−
⇒ xf =
J =
ρb
ρ nf K ρ b ρ nf
1 / ρ f − 1 / ρ nf
Hρ f
1 / 1.03 − 1 / 1.1
= 0.31
1 / 0.9 − 1 / 1.1
(d) V f + Vnf + Vlungs + Vother = Vb
mf
ρf
+
mnf
ρ nf
+ Vlungs + Vother =
m =m x
f
b f
mnf = mb (1 − x f )
F
1
1
−
ρ nf
Hρ f
⇒ xf G
mb
ρb
xf 1− x f
−
ρ nf
Hρ f
F
mb G
I
J =
K
I
J + (Vlungs
K
1
1 I FVlungs + Vother I
−
−G
J
G
ρ nf JK H
mb
K
Hρ b
F
1
1
−
ρ nf
Hρb
I
J
K
1
1 Vlungs + Vother
−
−
ρb ρ nf
mb
F
⇒ xf =
F
+ Vother ) = mb G
1
1
−
G
ρ nf
Hρ f
I
=
F 1
G
H1.03
J
K
3- 5
1 I F 12
. + 0.1I
J −G
J
K
H
11
.
122.5 K
= 0.25
1I
F 1
− J
G
H 0.9 11
.K
−
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)
From the plot above, r = 5455
. ρ − 539.03
(b) For ρ = 0.9940 g / cm 3 ,
m& Ile =
150 L
0.994 g
h
3
cm
r = 3.197 g Ile / 100g H 2O
1000 cm 3
3.197 g Ile
L
1 kg
103.197 g sol 1000 g
= 4.6 kg Ile / h
(c) The density of H2 O increases as T decreases, therefore the density was higher than it
should have been to use the calibration formula. The valve of r and hence the Ile mass
flow rate calculated in part (b) would be too high.
3.13 (a)
Mass Flow Rate (kg/min)
1.20
1.00
y = 0.0743x + 0.1523
R2 = 0.9989
0.80
0.60
0.40
0.20
0.00
0.0
2.0
4.0
6.0
8.0
10.0
12.0
Rotameter Reading
& = 0.0743 b5.3g + 0.1523 = 0.55 kg / min
From the plot, R = 5.3 ⇒ m
3- 6
3.13 (cont’d)
(b)
Rotameter Collection Collected
Reading
Time
Volume
(min)
(cm3)
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
Difference
Duplicate
(Di)
Mean Di
0.004
0.006
0.004
0.0104
0.012
0.026
1
b0.004 + 0.006 + 0.004 + 0.012 + 0.026g = 0.0104 kg / min
5
95% confidence limits: ( 0.610 ± 174
. Di ) kg / min = 0.610 ± 0.018 kg / min
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 C6 H 6
15.0 kmol C6 H 6
15,000 mol C 6H 6
78.114 kg C 6H 6
= 117
. × 10 3 kg C6 H 6
kmol C6 H 6
1000 mol
= 15
. × 10 4 mol C6 H 6
kmol
lb - mole
453.6 mol
15,000 mol C6 H 6
= 33.07 lb - mole C 6H 6
6 mol C
1 mol C 6H 6
15,000 mol C 6H 6
6 mol H
1 mol C6 H6
90,000 mol C 12.011 g C
mol C
90,000 mol H 1.008 g H
mol H
15,000 mol C 6H 6
= 90,000 mol C
= 90,000 mol H
= 1.08 × 10 6 g C
= 9.07 × 10 4 g H
6.022 × 10 23
mol
= 9.03 × 1027 molecules of C6 H 6
3- 7
3.15 (a) m& =
(b) n& =
175 m3
1000 L
0.866 kg
3
h
m
2526 kg
min
L
1000 mol 1 min
92.13 kg
60 s
1h
60 min
= 2526 kg / min
= 457 mol / s
(c) Assumed density (SG) at T, P of stream is the same as the density at 20o C and 1 atm
3.16 (a)
200.0 kg mix 0150
.
kg CH 3OH
kg mix
(b) m& mix =
3.17
M=
100.0 lb - mole MA
h
0.25 mol N 2
m& N 2 =
3000 kg
h
kmol CH3OH 1000 mol
32.04 kg CH3OH
74.08 lb m MA
1 kmol
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
kmol 0.25 kmol N 2
8.52 kg
kmol feed
= 936 mol CH3OH
= 8715 lb m / h
2.02 g H2
= 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 CaCO3 = 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 CaCO3 / 500 g suspension = 0.300 g CaCO 3 g suspension
3.19
Assume 100 mol mix.
mC 2 H5 OH =
10.0 mol C2 H5OH
46.07 g C 2 H5OH
= 461 g C2 H5OH
mol C 2H 5OH
75.0 mol C4 H8 O2 88.1 g C 4 H8O2
mC 4 H8 O2 =
= 6608 g C 4 H8O2
mol C 4 H8O2
15.0 mol CH 3COOH 60.05 g CH 3COOH
mCH 3COOH =
= 901 g CH 3COOH
mol CH3COOH
461 g
x C 2 H5OH =
= 0.0578 g C 2 H 5OH / g mix
461 g + 6608 g + 901 g
6608 g
x C4 H8 O2 =
= 0.8291 g C4 H 8O 2 / g mix
461 g + 6608 g + 901 g
901 g
x CH 3COOH =
= 0.113 g CH3COOH / g mix
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
(b) m gypsum = 1 L slurry
0.35 kg CaSO 4 ⋅ 2 H 2 O
= 0. 35 kg CaSO 4 ⋅ 2 H 2 O
L slurry
L CaSO4 ⋅ 2H2 O
= 0151
. L CaSO 4 ⋅2 H2O
2.32 kg CaSO4 ⋅ 2 H2O
0.35 kg gypsum 136.15 kg CaSO 4
CaSO 4 in gypsum: m =
= 0.277 kg CaSO 4
172.18 kg gypsum
. g L sol 1.05 kg 0.209 kg CaSO 4
b1 − 0151
CaSO 4 in soln.: m =
= 0.00186 kg CaSO4
L 100.209 kg sol
Vgypsum =
(c) m =
0.35 kg CaSO4 ⋅ 2H2 O
0.35 kg gypsum
0.05 kg sol
0.209 g CaSO 4
= 384
. × 10 -5 kg CaSO4
0.95 kg gypsum 100.209 g sol
0.277 g + 3.84 × 10 -5 g
% recovery =
× 100% = 99.3%
0.277 g + 0.00186 g
3.21
CSA:
FB:
45.8 L
0.90 kg
min
L
55.2 L
0.75 kg
min
L
kmol
kmol U
|
0.5496
mol CSA
75 kg
min |
⇒
= 12
.
kmol
kmol V|
0.4600
mol FB
= 0.4600
90 kg
min |W
= 0.5496
She was wrong.
The mixer would come to a grinding halt and the motor would overheat.
3.22 (a)
150 mol EtOH
46.07 g EtOH
= 6910 g EtOH
mol EtOH
6910 g EtO H
0.600 g H 2 O
= 10365 g H 2 O
0.400 g Et OH
6910 g EtOH
L
10365 g H 2 O
L
V=
+
= 19.123 L ⇒ 19.1 L
789 g EtOH
1000 g H 2 O
(6910 +10365) g
L
SG =
= 0903
.
191
. L
1000 g
(b) V ′ =
(6910 + 10365) g mix
% error =
L
= 18.472 L ⇒ 18.5 L
935.18 g
(19.123 − 18.472 ) L
× 100% = 3.5%
18.472 L
3-9
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
0.95 kmol air
h
0.05 kmol CH 4
= 43.01 kmol air h
Dilution air required: b43.01 - 22.89 g kmol air 1000 mol = 20200 mol air h
h
1 kmol
Product gas: 700 kg + 20.20 kmol Air 29 kg Air = 1286 kg h
h
h
kmol Air
43.01 kmol Air 0.21 kmol O2 32.00 kg O2
h
kg O2
= 0.225
h
1.00 kmol Air 1 kmol O2 1286 kg total
kg
3.24
xi =
mi
m
M
, ρi = i , ρ =
M
Vi
V
m m
1
A: ∑ x i ρ i = ∑ i i =
M Vi
M
B:
∑
Not helpful.
mi Vi
1
V
1
=
Vi =
=
Correct.
∑
mi
M
M
ρ
i
xi
0.60
0.25
0.15
=
+
+
= 1.091 ⇒ ρ = 0.917 g / cm 3
ρ i 0.791 1.049 1.595
xi
∑ρ
1
=
ρ
mi2
∑ Vi ≠ ρ
=
∑M
3.25 (a) Basis: 100 mol N 2 ⇒ 20 mol CH 4 ⇒
N total = 100 + 20 + 64 + 32 = 216 mol
R
20 ×
|
S
| 20 ×
T
80
= 64 mol CO 2
25
40
= 32 mol CO
25
32
64
= 0.15 mol CO / mol , x CO 2 =
= 0.30 mol CO 2 / mol
216
216
20
100
=
= 0 .09 mol CH 4 / mol , x N 2 =
= 0.46 mol N 2 / mol
216
216
xC O =
xC H 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
1
CH4
C2H6
C3H8
C4H10
16.04
30.07
44.09
58.12
0.150
0.287
0.467
0.583
Peak
Area
3.6
2.8
2.4
1.7
Mole
Fraction
0.156
0.233
0.324
0.287
Mass
Fraction
0.062
0.173
0.353
0.412
moles
mass
0.540
0.804
1.121
0.991
8.662
24.164
49.416
57.603
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, /,
3-11
3.26 (cont’d)
∗ ' SPECIES MOLE FR. MASS FR.', /,
∗ 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
1
0.156
0.062
2
0.233
0.173
3
0. 324
0. 353
0. 287
0.412
4
SAMPLE: 2
(ETC.)
3.27 (a)
MASS FR
(8.7 × 10 6 × 0.40) kg C
44 kg CO 2
= 1.28 × 10 7 kg CO 2 ⇒ 2 .9 × 10 5 kmol CO 2
12 kg C
(11
. × 10 6 × 0.26) kg C 28 kg CO
= 6.67 × 10 5 kg CO ⇒ 2.38 × 10 4 kmol CO
12 kg C
( 3.8 × 10 5 × 0.10) kg C
16 kg CH 4
12 kg C
= 5.07 × 10 4 kg CH 4 ⇒ 3.17 × 10 3 kmol CH 4
(1.28 × 10 + 6 .67 × 10 + 5.07 × 10 4 ) kg 1 metric ton
metric tons
m=
= 13,500
yr
1000 kg
7
M =
∑y M
i
i
5
= 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
gal
(c) u = V =
87 L
10 3 mL 1.03 g
1L
mL
m3
A
= 144 s
0.0500 g H 2 SO 4
g
1 min
min 1000 L
60 s
L
45 m
t= =
= 88 s
u 0.513 m / s
(π × 0.06 2 / 4) m 2
1 lbm
= 23.6 lb m H 2 SO 4
453.59 g
= 0.513 m / s
3.29 (a)
n&1 (mol/min)
n& 2 (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)
n& 3 =
150
. L 0.659 kg 1000 mol
min
L
86.17 kg
= 1147
. mol / min
Hexane balance: 0.180n&1 = 0050
. n&2 + 1147
. (mol C6 H14 / min)U solve R
n& = 838
. mol / min
| 1
V ⇒S
&
Nitrogen balance: 0.820n&1 = 0950
. n&2 (mol N2 / min)
| n2 = 72.3 mol / min
W
T
(b) Hexane recovery =
30 mL
3.30
n&3
1147
.
× 100% =
× 100% = 76%
n&1
0180
. b838
.g
1 L 0.030 mol 172 g
= 0155
. g Nauseum
103 mL
lL
1 mol
3-13
ln(CA)
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
1
0
-1
-2
-3
-4
-5
y = -0.4137x + 0.2512
R2 = 0.9996
0.0
5.0
t (min)
10.0
k = 0.414 min −1
ln CAO = 02512
.
⇒ CAO = 1286
.
lb - moles ft 3
mol 28.317 liter 2.26462 lb- moles
F 1b - moles I
(c) C A G
= 0.06243C′A
J = C A′
3
H
K
liter
1 ft3
1000 mol
ft
t ′bsg 1 min
t bming =
= t ′ 60
60 s
C = C exp( − kt )
A
A0
0.06243C ′A = 1334
.
exp b−0.419 t ′ 60g
drop primes
⇒
C A bmol / Lg = 214
. expb−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 2O 101.325 kPa
= 822.0 kPa
33.9 ft H 2 O
3.00 atm 101325
.
× 105 N m 2
12 m2
(c)
= 30.4 N cm 2
2
2
1 atm
100 cm
(d)
280 cm Hg 10 mm 101325
.
× 10 6 dynes cm 2 1002 cm 2
(e) 1 atm −
1 cm
760 mm Hg
2
1 m
20 cm Hg 10 mm
1 atm
= 0.737 atm
1 cm 760 mm Hg
3-14
2
= 3.733 × 1010
dynes
m2
3.32 (cont’d)
(f)
(g)
25.0 psig 760 mm Hg bgaugeg
= 1293 mm Hg bgauge g
14.696 psig
b25.0 + 14.696gpsi
760 mm Hg
= 2053 mm Hg babsg
14.696 psi
(h) 325 mm Hg − 760 mm Hg = −435 mm Hg bgaugeg
(i)
Eq. (3.4-2) ⇒ h =
=
P
ρg
35.0 lbf
in
2
144 in 2
1 ft
ft3
1.595 × 62.43 lbm
2
s2 32.174 lbm ⋅ ft
32.174 ft
lbf ⋅ s
2
100 cm
3.2808 ft
= 1540 cm CCl4
0.92 × 1000 kg 9.81 m / s2
m3
⇒ h (m) = 0.111Pg (kPa)
3.33 (a) Pg = ρgh =
h (m)
1N
1 kg ⋅ m / s2
1 kPa
10 N / m 2
3
h
Pg
Pg = 68 kPa ⇒ h = 0.111 × 68 = 7.55 m
F
moil = ρV = G0.92 × 1000
H
kg I F
16 2 3 I
×
7
.
55
×
π
×
m J = 14
. × 106 kg
J
G
m3 K H
4
K
(b) Pg + Patm = Ptop + ρgh
68 + 101 = 115 + b0.92 × 1000g × b9.81g / 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 ) Aρb g = h1 Aρ1 g + h2 Aρ 2 g ⇒ ρb = 1 1
h1 + h2
3-15
3.34 (cont’d)
(b)
Ptop = Patm + ρ1gh0 , Pbottom = Patm + ρ1g(h0 + h1) + ρ2gh2 , Wb = ρb (h1 + h2 ) A
⇒ Fdown = ( Patm + ρ1gh0 ) A + ρb (h1 + h2 ) A , Fup = [Patm + ρ1g(h0 + h1 ) + ρ2gh2 ]A
Fdown = Fup ⇒ ρb (h1 + h2 ) A = ρ1gh1 A + ρ2 gh2 A ⇒ Wblock =Wliquid displaced
3.35
∆ P = b Patm + ρghg − Pinside
= 1 atm − 1 atm +
F=
3.36
154 N 65 cm2
2
cm
. g1000
b105
m3
kg 9.8066 m 150 m 12 m 2
1N
2
2
2
s
100 cm 1 kg ⋅ m / s2
F 0.22481 lb f I
= 100
. × 104 N × G
J = 2250 lb f
H
1N K
1.4 × 62.43 lb m
1 ft 3
2.3 × 106 gal
= 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 14
= 14.7 f2 +
in
ft 3
s2
32.174 lb m ⋅ ft / s2 12 2 in2
= 32.9 psi
m = ρV =
— Structural flaw in the tank.
— Tank strength inadequate for that much force.
— Molasses corroded tank wall
π × 24 2 × 3 in 3
1 ft 3
8.0 × 62.43 lb m
= 392 lb m
3
3
4
12 in
ft 3
392 lb m
32.174 ft / s 2
1 lb f
W = mhead g =
= 392 lb f
32.174 lb m ⋅ ft / s2
3.37 (a) mhead =
( 30 + 14.7 )  lb f π × 202 in 2
Fnet = Fgas − Fatm − W =
2
in
4
−
14.7 lbf
in
2
π × 24 2 in 2
4
− 392 lb f = 7.00 ×10 3 lb f
The head would blow off.
3-16
3.37 (cont’d)
7.000 × 10 lbf
Fnet
=
392 lb m
mhead
3
Initial acceleration: a =
32.174 lbm ⋅ ft/s 2
1 lbf
= 576 ft/s
2
(b) Vent the reactor through a valve to the outside or a hood before removing the head.
3.38 (a)
a
2m
1m
b
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) 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
&
Vtub = ≈
= 2.5 ft 3 / min ⇒ V& = 5 × 2.5 = 125
. ft 3 / min
t
10 min
(i)
For a full room, h = 10 m
1000 kg 9.81 m
1N
10 m 2 m 2
⇒ F > 2.0 × 105 N
3
2
2
m
s 1 kg ⋅ m / s
The door will break before the room fills
⇒F>
(ii)
If the door holds, it will take
3
35.3145 ft 3
1h
V
b5 × 15 × 10g m
t fill = room
=
= 31 h
V&
12.5 ft 3 / min
1 m3
60 min
He will not have enough time.
3.39 (a) dPg i tap =
d Pg i
25 m H2 O
junction
101.3 kPa
= 245 kPa
10.33 m H 2O
101.3 kPa
b25 + 5g m H 2O
=
= 294 kPa
10.33 m H 2O
(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.
3-17
3.40
Pabs = 800 mm Hg
Pgauge = 25 mm Hg
Patm = 800 − 25 = 775 mm Hg
3.41 (a) P1 + ρ A g bh1 + h2 g = P2 + ρ B gh1 + ρ C gh2
⇒ P1 − P2 = bρ B − ρ A ggh1 + bρ C − ρ A ggh2
Lb1.0 −
(b) P1 = 121 kPa + M
N
×
3.42
0.792 g g 981 cm 30.0 cm
+
cm 3
s2
F 1 dyne
IF
101.325 kPa
G
2 JG
6
H1 g ⋅ cm / s KH1.01325 × 10 dynes /
I
cm
2 J
.
b137
− 0.792 g g 981 cm 24.0 cmO
P
cm 3
s2
Q
= 123.0 kPa
K
(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
(i) Hg: ρ t = 0.866, ρ m = 13.6, h = 150 cm ⇒ R = 238
. cm
(ii) H 2 O: ρ t = 0.866, ρ m = 1.00, h = 150 cm ⇒ R = 2260 cm
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.
P
3.43
Patm = ρ f gb7.23 m g ⇒ ρ f = atm
7.23 g
F P
I
Pa − Pb = dρ f − ρ w i gb26 cmg = G atm − ρ w gJ b26 cm g
H 7.23 m
K
F756mmHg
=G
H 7.23 m
1 m 1000 kg 9.81 m/s2
1N
760mmHg
1m I
−
Jb26 cmg
100 cm
m3
1 kg⋅ m/s2 1.01325×105 N m2 100 cmK
⇒ Pa − Pb = 81
. mm Hg
3-18
3.44 (a) ∆h = 900 − hl =
75
. psi 760 mm Hg
= 388 mm Hg ⇒ hl =900 − 388 =512 mm
14.696 psi
(b) ∆h = 388 − 25 × 2 = 338 mm ⇒ Pg =
338 mm Hg
14.696 psi
= 6.54 psig
760 mm Hg
3.45 (a) h = L sin θ
(b) h = b8.7 cmg sinb15°g = 2.3 cm H 2 O = 23 mm H 2 O
3.46 (a) P = Patm − Poil − PHg
= 765 − 365 −
9.81 m / s2 0.10 m
920 kg
m3
1N
760 mm Hg
2
1 kg ⋅ m/ s 1.01325 × 105 N / m2
= 393 mm Hg
(b) — Nonreactive with the vapor in the apparatus.
— Lighter than and immiscible with mercury.
— Low rate of evaporation (low volatility).
3.47 (a) Let ρ f = manometer fluid density c110
. g cm 3 h , ρ ac = acetone density c0.791 g cm 3 h
Differential manometer formula: ∆P = dρ f − ρ ac i gh
∆P ( mm Hg ) =
(1.10 − 0.791) g
981 cm h (mm)
3
cm
s
1 cm
1 dyne
10 mm 1 g ⋅ cm/s
2
= 0.02274 h ( mm )
V& ( mL s )
62
87
107
123
138
151
5
10
15
20
25
30
h ( mm )
∆P ( mm Hg )
0.114 0.227 0.341 0.455 0.568 0.682
(b) lnV& = n lnb∆P g + ln K
6
ln(V)
5.5
y = 0.4979x + 5.2068
5
4.5
4
-2.5
-2
-1.5
-1
-0.5
0
ln( P)
3-19
760 mm Hg
2
1.01325×106 dyne/cm2
3.47 (cont’d)
.
lnb∆Pg + 52068
.
From the plot above, ln V& = 04979
⇒n = 04979
.
≈ 05
. , ln K = 5.2068 ⇒ K = 183
ml s
0 .5
bmm Hgg
0. 5
(c) h = 23 ⇒ ∆P = b0.02274gb23g = 0.523 mm Hg ⇒ V& = 183b0.523g = 132 mL s
132 mL 0.791 g
104 g 1 mol
= 104 g s
= 180
. mol s
s
mL
s 58.08 g
. = 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) ∆T =
(d)
85° C 10
. °K
85° C 18
. °F
85° C 1.8° R
= 85° K;
= 153° F;
= 153° R
10
. °C
1° C
1.0° C
150° R 1° F
1° R
= 150° F;
150° R 1.0o K
1.8° R
= 83.3° K;
150° R 1.0° C
1.8° R
= 83.3° C
3.49 (a) T = 0.0940 × 1000o FB + 4.00 = 98.0o C ⇒ T = 98.0 × 1.8 + 32 = 208o F
(b) ∆T (o C) = 0.0940 ∆T (o FB) = 0.94o C ⇒ ∆T (K) = 0.94 K
0.94o C 1.8o F
o
∆T ( F) =
= 1.69o F ⇒ ∆T (o R) = 1.69o R
o
1.0 C
o
o
(c) T1 = 15o C ⇒ 100o L ; T2 = 43 C⇒1000 L
T ( o C) = aT (o L) + b
o
F oCI
C
a=
= 0.0311G J ;
o
o
1000 - 100 L
H LK
b43 − 15g
b
b = 15 − 0.0311 × 100 = 119
. oC
g
⇒ T ( o C) = 0.0311T ( o L) + 11.9 and T ( o L) = 32.15T ( o C) − 382.6
(d) Tbp = −88.6o C ⇒ 184.6 K ⇒ 332.3o R ⇒ -127.4o F ⇒ −9851
. o FB ⇒ −3232o L
(e) ∆T = 50.0o L ⇒ 1.56o C ⇒ 16.6o FB ⇒ 156
. K ⇒ 2.8o F ⇒ 2.8o R
3-20
3.50
bTb gH
2O
= 100° C
bTm gAgCl
= 455° C
(a) V bmV g = aT b° Cg + b
5.27 = 100a + b
a = 0.05524 mV ° C
⇒
24.88 = 455a + b
b = −0.2539 mV
V bmVg = 0.05524T b° Cg − 0.2539
⇓
T b° Cg = 1810
. V bmVg + 4.596
. mV→136
. mV ⇒1856
. ° C→2508
. °C ⇒
(b) 100
3.51 (a) ln T = ln K + n ln R
n=
. − 1856
. g°C
dT b2508
=
= 326
. °C / s
dt
20 s
T = KR n
lnb250.0 110.0g
= 1184
.
lnb40.0 20.0g
.
ln K = ln 1100
. −1184
. (ln 200
. ) = 1154
. ⇒ K = 3169
. ⇒T = 3169
. R1184
(b) R =
F 320 I
G
J
H 3169
K
.
1/ 1.184
= 49.3
(c) Extrapolation error, thermocouple reading wrong.
3.52 (a) PV = 0.08206nT
Pbatmg=
P′bpsigg + 14696
.
14696
.
nbmolg = n′blb - molesg×
⇒
b P′ + 14.696g
14.696
28317
.
ft 3
, VbLg = V ′dft i ×
L
3
453.59 mol
T ′( o F) − 32
, T(o K) =
+ 27315
.
lb − moles
1.8
× V ′ × 28.317 = 008206
.
× n′ ×
⇒ bP ′ + 14.696g × V ′ =
453.59 L(T ′ − 32)
O
×M
+ 27315
. P
1
N 1.8
Q
0.08206 × 14.696 × 453.59
× n′ × bT ′ + 459.7 g
28.317 × 18
.
⇒bP′ + 14.696gV ′ = 1073
. n′bT ′ + 459.7g
3-21
3.52 (cont’d)
b500 + 14.696g × 3.5
(b) ntot
′ =
10.73 × b85 + 459.7g
mCO =
(c) T ′ =
= 0.308 lb - mole
0308
. lb- mole 0.30 lb- mole CO 28 lbm CO
= 2.6 lb m CO
lb- mole
lb - mole CO
.
b3000 + 14.696g × 35
10.73 × 0.308
− 459.7 = 2733o F
3.53 (a) T b° Cg = a × r bohmsg + b
0 = 23624
. a +b U
a = 10634
.
⇒ Tb°Cg = 10634
. rbohmsg− 25122
.
V ⇒
100 = 33028
. a + bW
b = −25122
.
n& ′ (kmol) 1 min n& ′
=
min
60 s 60
P ′bmm Hgg
1 atm
P′
Pbatmg =
=
760 mm Hg 760
F kmol I
J
H s K
(b) n& G
3
Fm
V&G
H s
=
3
I
& m
J =V ′
K
min
,
TbKg = T ′b°Cg + 27316
.
1 min V& ′
=
60 s 60
0.016034P′bmm HggV& ′dm3 mini
V& ′
n&′ 12.186 P ′
=
⇒ n& ′ =
60
T ′b° Cg + 27316
.
760 T ′ + 27316
.
60
(c) T = 10.634 r − 251.22
r1 = 26.159 ⇒ T1 = 26.95° C
⇒ r2 = 26157
.
⇒ T2 = 26.93° C
r3 = 44.789 ⇒ T3 = 2251
. °C
F 760 mm Hg I
P (mm Hg) = h + Patm = h + ( 29.76 in Hg) G
J = 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)
(d) n&1 =
b0.016034 gb987.9gb947
60g
= 0.8331 kmol CH 4 min
26.95 + 27316
.
. gb195g
b0.016034gb9119
n& 2 =
= 9.501 kmol air min
26.93 + 27316
.
n& 3 = n&1 + n& 2 = 10.33 kmol min
(e) V3 =
(f)
n&3 bT2 + 27316
. g
0.016034 P3
=
. + 27316
. g
b10.33gb2251
b0.016034gb829.9g
= 387 m3 min
0.8331 kmol CH 4 16.04 kg CH 4
kg CH 4
= 13.36
min
kmol
min
0.21× 9.501 kmol O2 32.0 kg O2 0.79 × 9.501 kmol N2
+
min
kmol O2
min
xC H4 =
13.36 kg CH 4 min
= 0.0465 kg CH4 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 b5, ∗g MW, NT
DO 10 IT=1, NT
READ b5, ∗g TC, N
TK(IT) = TC + 273.15
READ b5, ∗g (TIME (J), CA (J), J = 1 , N)
DO 1 J=1, N
CAbJ g = CAbJ g / MW
3.54
XbJ g = TIMEbJ g
YbJg = 1./CAbJg
1
CONTINUE
CALL LS (X, Y, N, SLOPE, INTCPT)
KbITg = 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
XbJ g= 1./TKbJg
YbJg = LOGcKbJgh
3-23
28.0 kg N 2
kg air
= 274
kmol N2
min
3.54 (cont’d)
4
2
3
5
10
CONTINUE
CALL LS (X, Y, NT, SLOPE, INTCPT)
KO = EXPbINTCPTg
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
4
6
8.1
4.3
[OUTPUT]
TEMPERATURE (K): 367.15
TIME CA
(MIN) (MOLS/L)
10.00 0.1246
3-24
3.54 (cont’d)
30.0
40.0
50.0
60.0
3.0
2.2
1.8
1.5
20.00
30.00
40.00
50.00
60.00
0.0662
0.0462
0.0338
0.0277
0.0231
KbL / MOL ⋅ MINg: 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
bat 94° Cg
TEMPERATURE (K): 383.15
M
KbL / MOL ⋅ MINg: 1.758
M
M
K0bL / MOL − MINg: 0.2329E + 10
M ETC
E bJ / MOLg: 0.6690E + 05
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=
kg
kg dn
dn
kg
− 3.00 =
⇒
= 3.00
s
s
dt
dt
s
1.00 m3 1000 kg 1s
= 333 s
1 m3 3.00 kg
a.
Continuous, Steady State
b.
k = 0 ⇒ CA = CA0
c.
Input – Output – Consumption = 0
Steady state ⇒ Accumulation = 0
A is a reactant ⇒ Generation = 0
k = ∞ ⇒ CA = 0
3
3
Fm I
Fm I
C A0
F mol I
F mol I
F mol I
&
&
V G J CA 0 G 3 J = V G J C A G 3 J + kVC A G
J ⇒ CA =
kV
Hm K
Hm K
H s K
H s K
H s K
1+
V&
4.3
a.
m& v bkg / h g
100 kg / h
0.850kg B / kg
0.550kg B / kg
0.150kg T / kg
0.450kg T / kg
m& l bkg / h g
Input – Output = 0
Steady state ⇒ Accumulation = 0
No reaction ⇒ Generation = 0, Consumption = 0
0.106kg B / kg
0.894kg T / kg
(1) Total Mass Balance: 100.0 kg / h = m& v + m& l
& v + 0106
&l
(2) Benzene Balance: 0.550 × 100.0 kg B / h = 0.850 m
. 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.
c.
n(mol)
0 .500 mol N 2 mol
0 .500 mol CH 4 mol
100.0 g / s
x E bg C 2 H6 gg
0.500n bmol N 2 g 28 g N 2 1 kg
= 0.014 n bkg N 2 g
mol N 2 1000 g
n& E =
100x E bg C2H 6 g 1 lb m lb - mole C2H 6 3600 s
s
453.593 g 30 lb m C2H 6
h
= 26.45 x E blb - mole C2H 6 / h g
x P bg C3H8 g g
x B bg C4H10 g g
d.
n& 1 blb - mole H2 O sg
Rn& 2 blb - mole DA sg
|
. lb - moleO2
S 021
| 079
. lb- moleN 2
T
U
|
lb - mole DA V
lb - mole DA|W
n& O2 = 0.21n& 2 blb - mole O2 / sg
x H2 O =
x O2 =
e.
n ( mol)
n&1 F lb - mole H2O I
G
J
n&1 + n& 2 H lb - mole K
0.21n&2 F lb - mole O 2 I
G
J
n&1 + n&2 H lb - mole K
nN 2O4 = n  0.600 − y NO2  ( mol N 2O 4 )
0.400mol NO mol
yNO 2 ( mol NO 2 mol)
0.600 − yNO 2 ( mol N 2 O4 mol )
4- 2
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 C 3H 8 / h
n& 6 blb m / h g
002
. lb m C3H 8 / lb m
n& 7 blb m / hg
098
. lb m C3 H 6 / lbm
097
. lb m C3H 8 / lb m
Still
0.03 lb m C3 H 6 / lb m
Compressor
Reactor
n&1 blb m C3 H8 / h g
n& 1 blb m C3H 8 / h g
n& 2 blb m C3 H 6 / h g
n& 2 blb m C3H 6 / hg
n& 3 blb m CH4 / h g
n& 3 blb m CH 4 / h g
n& 4 blb m H 2 / hg
n& 4 blb m H 2 / h g
Note: the compressor and the off gas from
the absorber are not mentioned explicitly
in the process description, but their presence
should be inferred.
n& 5 blbm / hg
Stripper
Absorber
n&1 blb m C3 H8 / h g
n& 2 blb m C3 H 6 / h g
n& 5 blbm oil / h g
4.6
b.
Overall objective : To produce C3 H6 from C3 H8 .
Preheater function: Raise temperature of the reactants to raise the reaction rate.
Reactor function: Convert C3 H8 to C3H6 .
Absorption tower function: Separate the C3H8 and C3 H6 in the reactor effluent from the other
components.
Stripping tower function: Recover the C3H8 and C3H6 from the solvent.
Distillation column function: Separate the C3 H5 from the C3 H8.
a.
3 independent balances (one for each species)
b.
7 unknowns ( m& 1 , m
& 3 , m& 5 , x2 , y2 , y 4 , z4 )
– 3 balances
– 2 mole fraction summations
2 unknowns must be specified
c.
y2 = 1 − x2
F kg A I
J =
H h K
A Balance: 5300 x2 G
Overall Balance: m& 1 + 5300
B Balance: 0.03m
& 1 + 5300 x2
F kg A I
J
H h K
& 3 + b1200gb0.70g G
m
F kg I
G
J =
Hh K
m& 3 + 1200 + m& 5
F kg B I
G
J =
H h K
z4 = 1 − 0.70 − y 4
4- 3
F kg I
G
J
H h K
1200 y4 + 0.60m& 5
F kg B I
G
J
H h K
4.7
a.
3 independent balances (one for each species)
b.
Water Balance:
400 g 0.885 g H2O m& R bg g 0.995 g H 2O
& R = 356 g min
=
⇒m
min
g
g
bming
F g CH 3OOH I
J
H
K
min
Acetic Acid Balance: b400gb0115
. gG
& R + 0.096m
&E
= 0.005m
F g CH 3 OOH I
G
J
H
K
min
& E = 461g min
⇒m
g I
& R + m& E
J = m
H min K
F
Overall Balance: m& C + 400 G
c.
F g I
. gb400g − b0.005gb356g G
b0115
J
H min K
F g I
G
J ⇒
H min K
& C = 417 g min
m
F g I
= b0.096gb461g G
J ⇒ 44 g min = 44 g min
H min K
d.
CH3 COOH
H 2O
someCH 3COOH
CH3COOH
H 2O
Extractor
C4 H 9 OH
C 4H 9OH
CH 3COOH
Distillation
Column
C 4 H9 OH
4.8
a.
120 eggs/min
0.30 broken egg/egg
0.70 unbroken egg/egg
b.
Large: n1 broken eggs/min
n2 unbroken eggs/min
120 = 25 + 35 + n1 + n2 beggs ming ⇒ n1 + n2 = 50U
b0.30gb120g =
c.
X-large: 25 broken eggs/min
35 unbroken eggs/min
|
V⇒
|W
25 + n1
n1 = 11
n2 = 39
n1 + n2 = 50 large eggs min
n1 large eggs broken/50 large eggs = b11 50g = 0.22
d.
22% of the large eggs (right hand) and b25 70g ⇒ 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.
4- 4
4.9
a.
m3 blb m W evaporated g
m1 blb m strawberries g
015
. lb m S / lb m
0.85 lb m W / lb m
m2 c lb m S sugar
1.00 lb m jam
0.667 lb m S / lb m
h
0.333 lb m W / lb m
b.
3 unknowns ( m1 , m2 , m3 )
– 2 balances
– 1 feed ratio
0 DF
c.
Feed ratio: m1 / m2 = 45/55
(1)
S balance: 0.15m1 + m2 = 0.667 (2)
Solve simultaneously ⇒ m1 = 0.49 lb m strawberries, m 2 = 0.59 lb m sugar
4.10
a.
300 gal
m1 blb m g
0.750 lb m C 2 H 5OH / lb m
0.250 lb m H 2O / lb m
m3 blb m g
0.600 lb m C 2 H 5OH / lb m
0.400 lb m H 2O / lb m
4 unknowns ( m1, m2 ,V40 , m3 )
– 2 balances
– 2 specific gravities
0 DF
V40 bgal g
m2 blb m g
0.400 lb m C 2 H 5 OH / lb m
0.600 lb m H 2O / lb m
b.
m1 =
300gal
1 ft3
0.877 × 62.4 lb m
= 2195 lb m
7.4805 gal
ft 3
Overall balance: m1 + m2 = m3
C2 H5OH balance: 0.750m1 + 0.400m2 = 0.600m3
Solve (1) & (2) simultaneously ⇒ m2 = 1646 lb m, , m3 = 3841lb m
V40 =
1646 lb m
ft 3 7.4805gal
= 207 gal
0.952 × 62.4lb m
1ft 3
4- 5
(1)
(2)
4.11
3 unknowns ( n&1 , n&2 , n&3 )
– 2 balances
1 DF
a.
n&1 bmol / sg
0.0403 mol C3H 8 / mol
0.9597 mol air / mol
n&3 bmol / s g
n&2 bmol air / sg
0.0205 mol C 3H 8 / mol
0.9795 mol air / mol
0.21 mol O2 / mol
0.79 mol N 2 / mol
b.
Propane feed rate: 0.0403n&1 = 150 ⇒ n&1 = 3722 bmol / sg
Propane balance: 0.0403n&1 = 0.0205n&3 ⇒ n&3 = 7317 bmol / sg
Overall balance: 3722 + n&2 = 7317 ⇒ n&2 = 3600 bmol / sg
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.
4.12
a.
m
& bkg / h g
1000 kg / h
0.500 kg CH3OH / kg
0.960 kg CH3OH / kg
0.040 kg H2O / kg
0.500 kg H 2O / kg
2 unknowns ( m
& ,x )
– 2 balances
0 DF
673 kg / h
x bkg CH3OH / kg g
1 − x bkg H 2O / kg g
b.
& + 673 ⇒ m
& = 327 kg / h
Overall balance: 1000 = m
Methanol balance: 0.500b1000g = 0.960b327 g + x b673g ⇒ x = 0.276 kg CH3OH / kg
Molar flow rates of methanol and water:
673 kg 0.276 kg CH3OH 1000 g molCH3OH
= 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 × 10 4 mol H 2O / h
h
kg
kg 18 g H 2O
Mole fraction of Methanol:
5.80 × 10 3
= 0176
.
mol CH3OH / mol
5.80 × 103 + 2.71 × 104
c.
Analyzer is wrong, flow rates are wrong, impurities in the feed, a reaction is taking place, the
system is not at steady state.
4- 6
4.13
a.
Product
Feed
Reactor effluent
Reactor
2 2 5 3 kg
Purifier
2253k g
R = 388
1239 k g
R = 583
Waste
mw bk g g
R = 140
Analyzer Calibration Data
1
xp
x p = 0.000145R
1.364546
0.1
0.01
100
b.
1.3645
Effluent: x p = 0.000145b388g
1.3645
Product: x p = 0.000145b583g
1.3645
Waste: x p = 0.000145b140g
Efficiency =
c.
1000
R
= 0.494 kg P / kg
= 0.861 kg P / kg
= 0123
. kg P / kg
0.861b1239g
× 100% = 95.8%
0.494b2253g
Mass balance on purifier: 2253 = 1239 + mw ⇒ mw = 1014 kg
P balance on purifier:
Input: b0.494 kg P / kggb2253 kg g = 1113 kg P
Output: b0.861 kg P / kg gb1239 kg g + b0123
.
kg P / kg gb1014 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- 7
4.14
a.
n1 blb - mole/ hg
&
00100
.
lb -mole H2O/ lb -mole
09900
.
lb- mole DA / lb -mole
n3 blb- mole/ hg
&
0100
. lb -mole H2O/ lb- mole
0900
.
lb- mole DA/ lb - mole
n2 blb- mole HO/
hg
2
&
v2 dft 3 / hi
&
4 unknowns ( n&1 , n&2 , n& 3 , v& ) – 2 balances – 1 density – 1 meter reading = 0 DF
Assume linear relationship: v& = aR + b
v& − v&
96.9 − 40.0
Slope : a = 2 1 =
= 1626
.
R 2 − R1
50 − 15
Intercept: b = v&a − aR1 = 40.0 − 1.626b15g = 15.61
v&2 = 1.626b95g + 15.61 = 170 cft 3 / hh
n& 2 =
170 ft 3 62 .4 lb m lb - mol
= 589 blb - moles H 2 O / h g
h
ft 3
18.0 lb m
DA balance: 0.9900n&1 = 0.900n& 3
(1)
Overall balance: n&1 + n&2 = n& 3
(2)
&
&
Solve (1) & (2) simultaneously ⇒ n1 = 5890 lb - moles / h , n 3 = 6480 lb - moles / h
b.
4.15
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.
a.
m& bkg / s g
100 kg / s
0.900 kg E / kg
0100
.
kg H 2 O / kg
0.600 kg E / kg
0.050 kg S / kg
0.350 kg H 2 O / kg
& bkg / s g
m
x E bkg E / kg g
x S bkgS / kg g
1 − x E − x S bkg H 2 O / kgg
b.
Overall balance: 100 = 2m
& ⇒ m& = 50.0 bkg / sg
S balance: 0.050b100g = xS b50g ⇒ xS = 0100
. bkg S / kg g
E balance: 0.600b100g = 0.900b50g + x E b50g ⇒ x E = 0.300 kg E / kg
0.300b50 g
kg Ein bottom stream
kg Ein bottom stream
=
= 0.25
kg E in feed
0.600b100g
kg E in feed
4- 8
3 unknowns ( m
& , xE , xS )
– 3 balances
0 DF
4.15 (cont’d)
c. x = aR b ⇒ lnbxg = lnbag + b lnbR g
b=
lnbx2 / x1 g lnb0.400 / 0100
. g
=
= 1491
.
lnbR2 / R1 g
lnb38 / 15g
lnbag = lnbx1 g − b ln bR1 g = lnb0100
. g − 1491
. lnb15g = −6340
.
⇒ a = 1764
.
× 10−3
x = 1764
.
× 10−3 R1.491
1
R
d.
F xI b
=G J
Ha K
1
0900
.
F
I 1.491
=G
−3 J
H1764
.
× 10 K
= 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- 9
4.16
a.
b.
4.00 mol H 2SO 4 0.098 kg H2SO 4 L of solution
= 0.323 bkg H2SO 4 / kg solution g
L of solution
molH2SO 4 1.213kg solution
5 unknowns ( v1 , v2 , v3 , m2 , m3 )
– 2 balances
– 3 specific gravities
0 DF
v1 bL g
100 kg
v3 bL g
0.200 kg H 2 SO 4 / kg
m3 bkg g
0.800 kg H 2 O / kg
0.323 kg H 2SO 4 / kg
0.677 kg H 2 O / kg
SG = 1213
.
SG = 1139
.
v 2 bL g
m 2 bkg g
0.600 kg H 2SO 4 / kg
0.400 kg H 2 O / k g
SG = 1.498
Overall mass balance: 100 + m2 = m3
Water balance: 0.800b100 g + 0.400 m2 =
v1 =
100 kg
v2 =
44.4 kg
U
V⇒
0.677 m3 W
m2 = 44.4 kg
m3 = 144 kg
L
= 87.80 L20% solution
1139
. kg
L
= 29.64 L 60%solution
1498
.
kg
v1 87.80
L 20% solution
=
= 2.96
v 2 29.64
L 60% solution
c.
4.17
1250 kg P 44.4 kg 60% solution
L
= 257 L / h
h
144 kg P
1.498 kg solution
m1 bkgg@$18 / kg
0.25 kg P / kg
0.75 kg H2O / kg
100
. kg
017
. kg P/ kg
0.83 kg H2O / kg
m2 bkg g@$10 / kg
012
. kg P / kg
0.88 kg H2O / kg
Overall balance: m1 + m2 = 100
.
(1)
Pigment balance: 0.25m1 + 0.12m2 = 0.17b1.00g
(2)
Solve (1) and (2) simultaneously ⇒ m1 = 0.385 kg 25% paint , m2 = 0.615kg12% 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
4- 10
4.18
a.
m1 bkg H 2 O gb85%of enteringwaterg
100 kg
0.800 kgS / kg
0.200 kg H 2O / kg
m2 bkgSg
m3 bkg H 2 Og
85% drying: m1 = 0.850b0.200 gb100 g = 17.0 kg H2O
Sugar balance: m2 = 0.800b100g = 80.0 kg S
Overall balance: 100 = 17 + 80 + m3 ⇒ m3 = 3 kg H 2O
xw =
3 kg H2O
= 0.0361 kg H 2O / kg
b3 + 80 gkg
m1
17 kg H2O
=
= 0.205 kg H2O / kg wet sugar
m2 + m3 b80 + 3g kg
b.
1000 tonswet sugar
3 tonsH 2 O
= 30 tons H 2 O / day
day
100 tonswet sugar
1000 tons WS 0.800 tons DS 2000 lb m $0.15 365days
= $8.8 × 107 per year
day
ton WS
ton
lb m
year
c.
1
bx w1 + x w 2 +...+ x w10 g = 0.0504 kg H 2 O / kg
10
1
2
2
SD =
bx w1 − x w g +...+ bx w10 − x w g = 0.00181 kg H 2 O / kg
9
Endpoints = 0.0504 ± 3b0.00181g
xw =
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 cm 3 h
m1 bkg H 2 O g
SG = 1.00
v 3 dm 3 i
m 3 b kg suspension g
v 2 dm 3 i
SG = 1.48
5 unknowns ( v1 , v2 , v3 , m1 , m3 )
– 1 mass balance
– 1 volume balance
– 3 specific gravities
0 DF
400 kg galena
S G = 7 .44
Total mass balance: m1 + 400 = m3
(1)
4- 11
4.19 (cont’d)
Assume volume additivity:
m1 bkg g
m3
400 kg m 3
m bkg g m 3
+
= 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.
n&1 bmol / h g
n&2 bmol/ h g
0.040 mol H2O / mol
0.960 mol DA / mol
x bmol H 2O / molg
1 − x bmol DA / molg
n&3 bmol H 2O adsorbed / h g
97% of H 2O in feed
Adsorption rate: n& 3 =
b3.54 − 3.40gkg
5h
molH2O
= 1556
.
molH2O / h
0.0180 kg H2O
97% adsorbed: 156
. = 0.97b0.04n&1g ⇒ n&1 = 401
. mol/ h
Total mole balance: n&1 = n& 2 + n& 3 ⇒ n&2 = 401
. − 1556
.
= 38.54 mol / h
Water balance: 0.040 ( 40.1) = 1.556 + x ( 38.54 ) ⇒ x = 1.2 × 10−3 ( molH 2O/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.
300lb m / h
0.55 lb m H 2SO 4 / lb m
0.45 lb m H 2O / lb m
& C blb m / h g
m
& B blb m / h g
m
0.75 lb m H2SO 4 / lb m
0.25 lb m H2O / lb m
0.90 lb m H2SO 4 / lb m
0.10 lb m H2O / lb m
Overall balance: 300 + m& B = m& C
(1)
H2 SO4 balance: 0.55b300g+ 0.90m
& B = 0.75m& C
& B = 400lb m / h , m
& C = 700 lb m / h
Solve (1) and (2) simultaneously ⇒ m
(2)
4- 12
4.21 (cont’d)
b.
500 − 150
& A = 7.78 R A − 44.4
b RA − 25g ⇒ m
70 − 25
800 − 200
& B − 200 =
& B = 15.0 RB − 100
m
bRB − 20g ⇒ m
60 − 20
ln 100 − ln 20
0.2682 Rx
ln x − ln 20 =
bRx − 4g ⇒ ln x = 0.2682 Rx + 1.923 ⇒ x = 6.841e
10 − 4
300 + 44.4
400 + 100
mA = 300 ⇒ RA =
= 44.3, m B = 400 ⇒ RB =
= 33.3,
7.78
15.0
1
F 55 I
x = 55% ⇒ Rx =
ln G
J = 7.78
0.268 H 6.841K
c.
Overall balance: m& A + m& B = m& C
& A − 150 =
m
H2 SO4 balance: 0.01xm
& A + 0.90m& B = 0.75m& C = 0.75bm
& A + m& B g ⇒ m& B =
⇒ 15.0 RB − 100 =
0.75 − 0.01d6.841e
⇒ RB = d2.59 − 0.236 e
0.2682 Rx
0.2682 Rx
i b7.78 RA
&A
b0.75 − 0.01x gm
0.15
− 44.4 g
015
.
. e 0.2682 Rx − 813
.
i RA + 135
Check: RA = 44.3, Rx = 7.78 ⇒ RB = e2.59 − 0.236e 0.2682b7.78g j 44.3 + 135
. e 0.2682b7.78 g − 813
. = 33.3
4.22
a.
n& A bkmol / h g
0.10 kmolH2 / kmol
0.90 kmolN 2 / kmol
100 kg / h
n& P bkmol / h g
0.20 kmolH 2 / kmol
n& B bkmol / h g
0.80 kmol N 2 / kmol
0.50 kmolH2 / kmol
0.50 kmolN 2 / kmol
MW = 0.20b2.016g + 0.80b28.012g = 22.813 kg / kmol
100 kg kmol
= 4.38 kmol / h
h 22.813 kg
Overall balance: n& A + n& B = 4.38
H2 balance: 0.10 n& A + 0.50n& B = 0.20b4.38g
⇒ n& P =
Solve (1) and (2) simultaneously ⇒ n& A = 3.29 kmol / h , n& B = 110
. kmol / h
4- 13
(1)
(2)
4.22 (cont’d)
b.
n& P =
m& P
22.813
m
&P
22.813
x m
&
H2 balance: x An& A + xB n& B = P P
22.813
m& P b xB − x P g
⇒ n& A =
22.813 b xB − x A g
Overall balance: n& A + n& B =
c.
Trial
1
2
3
4
5
6
7
8
9
10
11
12
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
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
n& B =
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
m& P
bx P
− xAg
22.813 bx B − x A g
mP
100
100
100
100
100
100
250
250
250
250
250
250
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.
d.
4.23
Results are the same as in part c.
Arterialblood
200.0 ml / min
190
. mg urea / ml
Venous blood
1950
. ml / min
1.75 mg urea / ml
Dialysate
Dialyzing fluid
1500 ml / min
a.
v bml / ming
&
c bmg urea / mlg
Water removal rate: 200.0 − 195.0 = 5.0 ml / min
Urea removal rate: 1.90b200.0g − 1.75b1950
. g = 38.8 mg urea / min
b.
v& = 1500 + 5.0 = 1505 ml / min
c=
38.8mgurea/min
= 0.0258mgurea/ml
1505ml/min
4- 14
4.23 (cont’d)
c. b2.7 − 11
. g mg removed
1 min
10 3 ml 5.0 L
= 206 min (3.4 h)
ml
38.8 mg removed
1L
4.24
a.
n&1 bkmol / min g
20.0 kg CO2 / min
n&3 bkmol / min g
0.023 kmol CO2 / kmol
n& 2 bkmol / min g
0.015kmol CO2 / kmol
20.0 kg CO2
kmol
= 0.455 kmolCO2 / min
min
44 .0 kg CO2
Overall balance: 0.455 + n&2 = n& 3
CO2 balance: 0.455 + 0.015 n& 2 = 0.023n&3
Solve (1) and (2) simultaneously ⇒ n&2 = 55.6 kmol / min , n&3 = 561
. kmol / min
n&1 =
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 0.123 kmol 60 s 8.33 m
Spectrophotometer calibration: C = kA ====> C bµg / Lg = 3.333 A
4.25
A = 0.9
C =3
Dye concentration: A = 018
. ⇒ C = b3333
. gb018
. g = 0.600 µg / L
Dye injected =
0.60 cm 3
1L
10 3 cm 3
5.0 mg 10 3 µ g
1L
1 mg
= 3.0 µ g
⇒ b3.0 µ g g V bLg = 0.600 µ g / L ⇒ V = 5.0 L
4.26
a.
1000 LB / min
n&3 bkmol / min g
& 2 bkg B / min g
m
y 3 bkmol SO 2 / kmol g
V&1 dm 3 / min i
1 − y3 bkmol A / kmol g
n&1 bkmol / min g
& 4 bkg / min g
m
y1 bkmol SO 2 / kmolg
x4 bkg SO 2 / kg g
1 − y1 bkmol A / kmolg
1 − x 4 bkg B / kg g
4- 15
(1)
(2)
4.26 (cont’d)
–
–
–
–
–
b.
8 unknowns ( n&1 , n&3 , v&1, m& 2 , m
& 4 , x4 , y1 , y3 )
3 material balances
2 analyzer readings
1 meter reading
1 gas density formula
1 specific gravity
0 DF
Orifice meter calibration:
A log plot of V& vs. h is a line through the points dh1 = 100, V&1 = 142 i and dh2 = 400, V&2 = 290i .
ln V& = b ln h + ln a ⇒ V& = ah b
b=
ln dV&2 V&1 h
lnbh2 h1 g
=
lnb290 142g
lnb400 100g
= 0.515
ln a = ln V&1 − b ln h1 = lnb142g − 0.515ln 100 = 2.58 ⇒ a = e 2.58 = 13.2 ⇒ V& = 13.2 h 0.515
Analyzer calibration:
ln y = bR + ln a ⇒ y = ae bR
b=
ln by 2 y 1 g
R2 − R1
=
lnb0.1107 0.00166g
90 − 20
= 0.0600
ln a = ln y 1 − bR1 = lnb0.00166g − 0.0600b20 g =
E
a = 5.00 × 10 −4
c.
U
|
|
|
−7.60V ⇒
|
|
|
W
y = 5.00 × 10 −4 e 0.0600R
0.515
h1 = 210 mm ⇒ V&1 = 13.2b210g
= 207.3 m 3 min
ρ feed gas =
b12.2g b150 + 14.7 g
b75 + 460g
14.7 batm g
18
. bK g
= 0.460 mol / L = 0.460 kmol / m 3
E
n&1 =
207.3 m 3 0.460 kmol
= 95.34 kmol min
min
m3
R1 = 82.4 ⇒ y1 = 5.00 ×10−4 expb0.0600 × 82.4g = 0.0702 kmol SO 2 kmol
R3 = 116
. ⇒ y3 = 500
. × 10−4 exp b00600
.
×116
. g = 0.00100 kmol SO2 kmol
&2 =
m
1000 L B 130
. kg
= 1300 kg / min
min
LB
4- 16
4.26 (cont’d)
A balance: b1 − 0.0702gb95.34g = b1 − 0.00100gn3 ⇒ n3 = 88.7 kmol min
& 4 x4
SO2 balance: b0.0702gb9534
. g(64.0 kg / kmol) = b0.00100 gb88.7g(64 ) + m
(1)
& 4 (1 − x4 )
B balance: 1300 = m
(2)
& 4 = 1723 kg / min, x4 = 0.245 kg SO2 absorbed / kg
Solve (1) and (2) si multaneously ⇒ 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 SO 2 from the gas to the liquid phase.
a.
V&2 dm 3 / min i
n& 3 bkmol / min g
& 2 bkg B / min g
m
y 3 bkmolSO 2 / kmolg
1 − y 3 bkmol A / kmolg
V&1 dm 3 / min i
R3
n&1 bkmol / min g
& 4 bkg / min g
m
y1 bkmolSO 2 / kmolg
x 4 bkgSO 2 kg g
1 − y1 bkmol A / kmolg
1 − x 4 bkg B / kg g
P1 , T1 , R1 , h1
b.
& 2, n&3, y3, R3 , m
& 4 , x4 )
14 unknowns ( n&1,V&1, y1, P1, T1, R1, h1,V&2, m
– 3 material balances
– 3 analyzer and orifice meter readings
– 1 gas density formula (relates n&1 andV&1 )
& and V& )
– 1 specific gravity (relates m
2
2
6 DF
A balance: b1 − y1 gn&1 = b1 − y3 gn&3
&4
x4 m
SO2 balance: y1n&1 = y3n& 3 +
64 kgSO 2 / kmol
& 2 = b1 − x 4 gm
&4
B balance: m
Calibration formulas:
(1)
(2)
(3)
−4 0.060 R1
y1 = 5.00 × 10 e
(4)
y 3 = 5.00 × 10 −4 e 0.060 R3
V& = 13.2 h 0.515
(5)
1
Gas density formula : n&1 =
1
12.2 bP1 + 14.7 g / 14.7 &
V1
.
bT1 + 460g / 18
& bkg g m 3
m
Liquid specific gravity: SG = 130
. ⇒ V&2 = 2
h 1300 kg
4- 17
(6)
(7)
(8)
4.27 (cont’d)
c.
T1
75 °F
y1
P1
150 psig
V1
207 m3/h
h1
210 torr
n1
95.26 kmol/h
R1
0.07 kmol SO2 /kmol
82.4
x4 (kg SO 2/kg) y3 (kmol SO 2/kmol) V2 (m3/h) n3 (kmol/h)
0.10
0.050
0.89
93.25
0.10
0.025
1.95
90.86
0.10
0.010
2.56
89.48
0.10
0.005
2.76
89.03
0.10
0.001
2.92
88.68
0.20
0.050
0.39
93.25
0.20
0.025
0.87
90.86
0.20
0.010
1.14
89.48
0.20
0.005
1.23
89.03
0.20
0.001
1.30
88.68
Trial
1
2
3
4
5
6
7
8
9
10
m4 (kg/h)
m2 (kg/h)
1283.45 1155.11
2813.72 2532.35
3694.78 3325.31
3982.57 3584.31
4210.72 3789.65
641.73
513.38
1406.86 1125.49
1847.39 1477.91
1991.28 1593.03
2105.36 1684.29
3
V 2 (m /h)
V2 vs. y 3
3.50
3.00
2.50
2.00
1.50
1.00
0.50
0.00
0.000
0.020
0.040
0.060
y 3 (kmol SO 2 /kmol)
x4 = 0.10
x4 = 0.20
For a given SO 2 feed rate removing more SO 2 (lower y3 ) requires a higher solvent feed
rate (V&2 ).
For a given SO 2 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
Overall mass balance ⇒ m& 3
Mass balance - Unit 1 ⇒ m& 1
A balance - Unit 1 ⇒ x1
Mass balance - mixing point ⇒ m& 2
A balance - mixing point ⇒ x 2
C balance - mixing point ⇒ y2
4- 18
4.29
a.
100 mol / h
n&2 bmol / hg
n&4 bmol / h g
0.300 mol B / mol
0.250 mol T / mol
x B 2 bmol B / mol g
0.940 mol B / mol
xT 2 bmol T / molg
Column 2 0.060 molT / mol
Column 1
1 − x B 2 − xT 2 bmol X / molg
0.450 mol X / mol
n&3 bmol / hg
n&5 bmol / h g
0.020 molT / mol
0980
.
molX / mol
xB 5 bmolB / molg
xT 5 bmol T / molg
1 − xB 5 − xT 5 bmol X / molg
Column 1
4 unknowns ( n& 2 , n&3, x B 2, x T 2 )
–3 balances
– 1 recovery of X in bot. (96%)
0 DF
Column 2:
4 unknowns ( n& 3 , n&4 , n&5 , y x )
– 3 balances
– 1 recovery of B in top (97%)
0 DF
Column 1
96% X recovery: 0.96b0.450gb100g = 0.98 n&3
Total mole balance: 100 = n& 2 + n&3
(2)
B balance: 0.300b100g = x B 2n&2
(3)
T balance: 0.250b100g = x T 2 n&2 + 0.020 n& 3
(4)
Column 2
97% B recovery: 0.97 xB 2n&2 = 0.940 n&4
b.
(1)
(5)
Total mole balance: n& 2 = n&4 + n&5
(6)
B balance: xB 2n&2 = 0.940 n&4 + xB5n&5
(7)
T balance: xT 2 n&2 = 0.060n& 4 + xT 5n&5
(8)
(1) ⇒ n& 3 = 44 .1 mol / h
(2) ⇒ n& 2 = 55.9 mol / h
( 3) ⇒ x B 2 = 0.536 molB / mol
( 4) ⇒ x T 2 = 0.431 molT / mol
(5) ⇒ n& 4 = 30.95 mol / h
( 6) ⇒ n& 5 = 24.96 mol / h
( 7) ⇒ x B5 = 0.036 mol B / mol
(8) ⇒ x T 5 = 0.892 mol T / mol
0.940b30.95g
Overall benzene recovery:
× 100% = 97%
0.300b100 g
Overall toluene recovery:
0.892b24.96g
× 100 = 89%
0.250b100g
4- 19
4.30
a.
100 kg / h
0.035 kg S / kg
0.965 kg H 2 O / kg
& 3 bkg / h g
m
1
1 − x3 bkg H2 O / kg g
0100
. m& w b kg H 2O / h g
m
& w bkg H 2O / h g
b.
m& 4 bkg / h g
x3 bkg S / kgg
4
x4 bkg S / kgg
1 − x4 bkg H2 O / kg g
& w bkg H 2 O / h g
0100
. m
Overall process
m& 10 (kg / h)
0.050 kg S/kg
0.950 kg H2 O/kg
100 kg/h
0.035 kg S/kg
0.965 kg H2 O/kg
m& w ( kg H 2 O / h )
Salt balance: 0.035b100g = 0.050m& 10
& w + m& 10
Overall balance: 100 = m
H2 O yield: Yw =
& w bkg H2O recovered g
m
96.5 bkg H2Oin freshfeed g
First 4 evaporators
m4 b kg/ hg
&
x 4 bkg S/ kgg
1 − x4 bkg H2 O / kgg
100 kg/ h
0.035 kg S/ kg
0965
.
kg H2 O / kg
4 × 0100
. mw bkg H2 O / hg
&
&4
Overall balance: 100 = 4b0100
. gm& w + m
&4
Salt balance: 0.035b100g = x4 m
c.
Yw = 0 .31
x4 = 0.0398
4- 20
& 10 b kg / h g
m
10
0.050 kg S / kg
0.950 kg H2 O / kg
& w bkg H 2 O / h g
0.100 m
4.31
a.
2 n&1 bmolg
Condenser
0.97 mol B / mol
0.03 mol T / mol
100 mol
0.50 mol B / mol
0.50 mol T / mol
n&1 bmol g
n&1 bmol g( 89.2% of Bin feed )
0.97 mol B / mol
0.03 mol T / mol
0.97 mol B / mol
0.03 mol T / mol
Still
n&4 bmol gb45% of feed to reboiler g
y B bmol B / mol g
1 − y B bmol T / mol g
n&3 bmolg
n& 2 bmolg
Reboiler
z B bmolB / molg
1 − zB bmolT / molg
Overall process:
Condenser:
x B bmol B / mol g
1 − x B bmol T / molg
3 unknowns ( n&1 , n&3 , xB )
Still: 5 unknowns ( n&1 , n&2 , n&4 , y B , zB )
– 2 balances
– 2 balances
– 1 relationship (89.2% recovery)
3 DF
0 DF
1 unknown ( n&1 )
– 0 balances
1 DF
Reboiler:
6 unknowns ( n& 2 , n& 3 , n& 4 , 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: 0.892b0.50gb100g = 0.97 n&1
Overall balance: 100 = n&1 + n& 3
B balance: 0.50b100g = 0.97n&1 + x B n& 3
Reboiler
y B / e1 − y B j
Composition relationship:
x B / b1 − x B g
= 2.25
Percent vaporized: n& 4 = 0.45 n& 2
(1)
Mole balance: n& 2 = n&3 + n& 4
(2)
(Solve (1) and (2) simultaneously.)
B balance: z Bn&2 = x Bn&3 + yB n&4
4- 21
4.31 (cont’d)
c. B fraction in bottoms : xB = 0100
.
mol B / mol
Moles of overhead: n&1 = 46.0 mol
Recovery of toluene:
4.32
b1 − x B gn&3
0.50b100g
Moles of bottoms : n& 3 = 54.0 mol
× 100% =
b1 − 0.10gb54.02g
0.50b100 g
× 100% = 97%
a.
m3 bkg H2Og
Bypass
Basis: 100 kg
100 kg
0.12 kg S / kg
0.88 kg H2O / kg
Mixing point
Evaporator
m1 bkg g
0.12 kg S / kg
0.88 kg H2 O / kg
m4 bkgg
m5 bkgg
0.58 kg S / kg
0.42 kg H2O / kg
0.42 kg S / kg
0.58 kg H2O / kg
m2 bkgg
0.12 kg S / kg
0.88 kg H2 O / 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
Overall S balance: 0.12b100g = 0.42m5
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 = 71.4 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- 22
4.33
a.
m& 4 bkg Cr / h g
m
& 1 bkg / hg
m& 2 bkg / h g
0.0515 kgCr / kg
09485
.
kg W / kg
0.0515 kgCr / kg
0.9485 kg W / kg
Treatment
Unit
& 5 bkg / h g
m
m& 6 bkg / h g
x 5 bkg Cr / kg g
x6 bkg Cr / kg g
1 − x 5 bkg W / kgg
1 − x 6 bkg W / kgg
m& 3 bkg / hg
0.0515 kgCr / kg
0.9485 kg W / kg
b.
&1 = 6000 kg / h ⇒ m
& 2 = 4500 kg / h bmaximum allowed value g
m
Bypass point mass balance: m& 3 = 6000 − 4500 = 1500 kg / h
& 4 = 0.95b0.0515gb4500g = 2202
95% Cr removal: m
. kg Cr / h
Mass balance on treatment unit : m& 5 = 4500 − 220.2 = 4279.8 kg / h
0.0515b4500 g − 220.2
= 0.002707 kg Cr / kg
47798
.
& 6 = 1500 + 42798
Mixing point mass balance: m
. = 5779.8 kg / h
Cr balance on treatment unit : x 5 =
Mixing point Cr balance: x 6 =
c.
0.0515b1500 g + 0.0002707b4279.8g
= 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- 23
x5
m 6 (kg/h)
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 productio n in coming years, capacity of waste lagoon,
regulatory limits on Cr emissions.
a.
175 kg H 2 O / s b45% of water fed to evaporator g
& 1 bkg / s g
m
m
& 4 bkg K2SO 4 / s g
0196
.
kg K 2SO4 / kg
0.804 kg H 2 O / kg
& 5 bkg H2O / s g
m
m
& 6 bkg K2SO 4 / s g
Evaporator
Crystallizer
Filter
& 7 bkg H2 O / s g
m
Filter cake
& 2 bkg K2SO4 / sg
10m
m
& 2 bkg soln / sg
R0.400
S
T 0.600
Filtrate
& 3 bkg / s g
m
kg K2SO4 / kgU
kg H2O / kg
0.400 kg K2SO 4 / kg
0.600 kg H2 O / kg
Let K = K2 SO4 , W = H2 Basis: 175 kg W evaporated/s
Overall process: 2 unknowns ( m& 1, m& 2 )
- 2 balances
0 DF
Evaporator:
4 unknowns ( m& 4 , m& 5 , m& 6 , m& 7 )
– 2 balances
– 1 percent evaporation
1 DF
Mixing point:
Crystallizer:
4 unknowns ( m&1 , m& 3, m& 4 , m& 5 )
- 2 balances
2 DF
4 unknowns ( m& 2 , m& 3 , m& 6 , m& 7 )
– 2 balances
2 DF
&1, m
&2
Strategy: Overall balances ⇒ m
U verify that each
|
% evaporation ⇒ m
&5
| chosen subsystem involves
V
Balances around mixing point ⇒ m
& 3, m
& 4 | no more than two
& 7 |Wunknown variables
Balances around evaporator ⇒ m& 6 , m
4- 24
V
W
4.34 (cont’d)
& 1 = 175 + 10m
&2 + m
&2
U
Overall mass balance: m
|
V
Overall K balance:
0196
. m& 1 = 10m& 2 + 0.400m
&2 |
W
&2
Production rate of crystals = 10 m
45% evaporation: 175 kg evaporated min = 0.450m& 5
& 1 + 0.600m
& 3 = m& 5
W balance around mixing point: 0.804 m
& 3 = m& 4 + m& 5
Mass balance around mixing point: m& 1 + m
&4
K balance around evaporator: m& 6 = m
W balance around evaporator: m& 5 = 175 + m& 7
Mole fraction of K in stream entering evaporator =
b.
m& 4
m
&4 + m
&5
&1 = 221 kg / s
Fresh feed rate: m
& 2 = 416
Production rate of crystals = 10 m
. kg K bsg s
Recycle ratio:
c.
& 3bkg recycle sg
m
352.3
kg recycle
=
= 160
.
&
m1bkg fresh feed sg 220.8
kg fresh feed
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- 25
4.35
a.
Overall objective : Separate components of a CH 4 -CO2 mixture, recover CH4 , and discharge
CO2 to the atmosphere.
Absorber function: Separates CO 2 from CH4.
Stripper function: Removes dissolved CO2 from CH3 OH 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.
n& 1 bmol / hg
n& 5 bmol N 2 / hg
0.010 molCO 2 / mol
0.990 mol CH 4 / mol
100 mol / h
0.300 molCO 2 / mol
0.700 molCH 4 / mol
n& 6 bmolCO 2 / h g
n& 2 bmol / h g
Absorber
Stripper
0005
.
molCO 2 / mol
0995
.
molCH3 OH / mol
n& 5 bmol N 2 / hg
n& 3 bmol CO 2 / h g
n& 4 bmol CH3 OH / h g
Overall:
Stripper:
3 unknowns ( n&1, n&5 , n& 6 )
– 2 balances
1 DF
Absorber:
4 unknowns ( n&1, n& 2 , n&3 , n&4 )
– 3 balances
1 DF
4 unknowns ( n& 2 , n& 3 , n&4 , n& 5 )
– 2 balances
– 1 percent removal (90%)
1 DF
Overall CH4 balance: b0.700gb100g bmol CH4 / h g = 0.990 n&1
Overall mole balance: 100bmol / h g = n&1 + n&6
Percent CO2 stripped: 0.90 n&3 = n& 6
Stripper CO2 balance: n& 3 = n& 6 + 0.005 n& 2
Stripper CH3 OH balance: n& 4 = 0.995n&2
d.
n&1 = 70.71 mol / h , n&2 = 651.0 mol / h , n&3 = 32.55 mol CO2 / h, n&4 = 6477
. mol CH 3OH / h ,
n& 6 = 29.29 mol CO 2 / h
30.0 − 0.010n&1
Fractional CO2 absorption: f CO2 =
= 0.976 molCO 2 absorbed / mol fed
30.0
4- 26
4.35 (cont’d)
Total molar flow rate of liquid feed to stripper and mole fraction of CO2 :
n&3
n& 3 + n&4 = 680 mol / h , x3 =
= 0.0478 molCO 2 / mol
n&3 + n&4
e.
Scale up to 1000 kg/h (=106 g/h) of product gas:
MW1 = 0.01b44 g CO2 / molg + 0.99b16 g CH 4 / molg = 16.28 g / mol
bn&1 gnew
= d1.0 × 10 6 g / hi b16.28 g / molg = 6.142 × 10 4 mol / h
bn&feed gnew
4.36
= b100 mol / hg (6142
.
× 10 4 mol / h) / (70.71 mol / h) = 8.69 × 10 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
Condenser
m ekg C H j
5
6 14
m ekg C H j
1
6 14
300 kg C6H14
Ex
m2 bkgg
F
x2 bkg S / kgg
m4 bkgg
y 4 bkg oil / kgg
Ev
m6 bkg oilg
1 − y4 bkg C6 H14 / kgg
y 2 bkg oil / kgg
1 − x2 − y 2 bkg C 6H14 / kgg
13.0 kg oil
87.0 kg S
m3 bkgg
0.75 kg S / kg
y3 bkg oil / kgg
0.25 − y3 bkg C6 H14 / kgg
Overall:
4 unknowns ( m1 , m3 , m6 , y3 )
– 3 balances
1 DF
Extractor:
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- 27
4.36
(cont’d)
Extractor S balance: 87.0 kg S = x2m2
Extractor oil balance: 13.0 kg oil = y2 m2
Filter S balance: 87.0 kg S = 0.75m3
Filter mass balance: m2 bkg g = 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
Evaporator hexane balance: b1 − y 4 gm4 = m5
Mixing pt. Hexane balance: m1 + m5 = 300 kg C 6H14
Evaporator oil balance: y4m4 = m6
b.
Yield =
c.
Lower heating cost for the evaporator and lower cooling cost for the condenser.
m6
11.8 kg oil
=
= 0118
. bkg oil / kg beans fed g
100 100 kg beans fed
m
28 kg C6H 14
Fresh hexanefeed = 1 =
= 0.28bkg C 6 H14 / kg beans fed g
100 100 kg beans fed
m
272 kg C 6H14 recycled
Recycle ratio = 5 =
= 9.71bkg C6H14 recycled / kg C6H14 fed g
m1
28 kg C6H14 fed
4.37
m blb m dirt g
1
98 lb m dry shirts
3 lb m Whizzo
100 lbm
2 lbm dirt
98 lb m dry shirts
Filter
Tub
m blb m Whizzo g
2
m blb m g
3
0.03 lb m dirt / lb m
m blb m g
4
0 .13 lb m dirt / lb m
0.97 lb m Whizzo / lb m
0 .87 lb m Whizzo / lb m
m blb m g
5
0 .92 lb m dirt / lb m
0 .08 lb m Whizzo / lb m
m6 blb m g
1− x blbm dirt / lb m g
x blb m Whizzo/ lbm g
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- 28
4.37
(cont’d)
Balances around the mixing point involve 3 unknowns bm3 , m6 , x g , as do balances
around the filter bm4 , m6 , x g , but the tub only involves 2 bm3, m4 g and 2 balances are
allowed for each subsystem. Balances around tub ⇒ m3, m4
Balances around mixing point ⇒ m6 , x (solves Part (b))
a.
95% dirt removal: m1 = b0.05gb2.0g = 010
. lb m dirt
Overall dirt balance: 2.0 = 010
. + b0.92gm5 ⇒ m5 = 2.065 lb m dirt
Overall Whizzo balance: m2 = 3 + b0.08 gb2.065g blb m Whizzog = 317
. lb m Whizzo
b.
Tub dirt balance: 2 + 0.03m3 = 010
. + 013
. m4
Tub Whizzo balance: 0.97 m3 = 3 + 0.87 m4
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:
(1)
(2)
3.17 + x (17.3) = ( 0.97 )( 20.4) ⇒ x = 0.961 lbm 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
Filter 3
C 2L kg L
C 2S kg S
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:
0.01b620g = F1L ⇒
F1L = 6.2 kg L
balance:
620 = 6.2 + C1L ⇒ C1L = 6138
. kg L
U
0
.
01
6138
.
+
F
=
F
mixer filter 2 :
b
F2 L = 6.2 kg L
3L g
2L
|
balance: 6138
. + F3L = F2 L + C3L V ⇒ C2 L = 613.7 kg L
|
mixer filter 3: 0.01C2 L = F3L
F3L = 61
. kg L
W
balance:
613.7 = 6.1+ C3L ⇒ C3L = 6076
. kg L
4- 29
4.38 (cont’d)
Solvent
m f 1:
balance:
m f 2:
balance:
m f 3:
balance:
015
. b3300g = C1S ⇒
3300 = 495 + F1S ⇒
015
. b495 + F3S g = C2 S U
C1S = 495 kg S
F1S = 2805 kg S
C2S = 482.6 kg S
|
495 + F3S = C2S + F2S |
F2 S = 2734.6 kg S
V⇒
C3S = 480.4 kg S
015
. b2720 + C2 S g = C3S |
|
F3S = 2722.2 kg S
2720 + C2S = F3S + C3S W
Holding Tank Contents
6.2 + 6.2 = 12.4 kg leaf
2805 + 27346
. = 5540 kg solvent
b.
5540 kg S
0165
. kg E / kg
0.835 kg W / kg
Q R bkgg
. kg E / kg
Extraction 013
0.15 kg F / kg
Unit
QD bkg D g
0.855 kg W / kg
QF bkg Fg
QE bkg Eg
QD bkg D g
QF bkg Fg
Q0 bkgg
Steam
Stripper
0.200 kg E / kg
0.026 kg F / kg
0.774 kg W / kg
QB bkg g
0.013 kg E / kg
0.987 kg W / kg
Q3 bkg steam g
1 kg D
620 kg leaf
= 0.62 kg D = QD
1000 kg leaf
Water balance around extraction unit: 0.835b5540g = 0.855QR ⇒ QR = 5410 kg
Ethanol balance around extraction unit:
0.165b5540g = 013
. b5410 g + QE ⇒ QE = 211 kg bethanol in extract g
Mass of D in Product:
c.
F balance around stripper
0.015b5410g = 0.026 Q0 ⇒ Q0 = 3121 kg bmass of stripper overhead product g
E balance around stripper
0.13b5410g = 0.200b3121g + 0.013QB ⇒ QB = 6085 kg bmass of stripper bottom productg
W balance around stripper
0.855b5410g + QS = 0.774b3121g + 0.987b6085g ⇒ QS = 3796 kg steam fed to stripper
4.39
a.
C 2 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- 30
4.39 (cont’d)
b.
nH 2
nC 2 H2
= 1.5 < 2.0 ⇒ H2 is limiting reactant
15
. molH2 fed ⇒ 1.0 mol C2 H2 fed ⇒ 0.75 molC 2H2 required (theoretical)
1.0 mol fed − 0.75 mol required
% excess C2H 2 =
× 100% = 333%
.
0.75 mol required
c.
4 × 10 6 tonnes C2 H 6 1 yr 1 day 1 h 1000 kg 1 kmol C2 H 6 2 kmolH 2 2.00 kg H 2
yr
300 days 24 h 3600 s tonne 30.0 kg C2 H 6 1 kmolC 2H 6 1 kmol H 2
= 20.6 kg H2 / s
4.40
d.
The extra cost will be involved in separating the product from the excess reactant.
a.
4 NH3 + 5 O 2 → 4 NO + 6 H2O
5 lb - mole O2 react
= 125
. lb - mole O 2 react / lb - mole NO fo rmed
4 lb - mole NO formed
b.
dn O2 i
theoretical
=
100 kmol NH3 5 kmol O2
= 125 kmol O2
h
4 kmol NH3
40% excess O 2 ⇒ dn O2 i
c.
b100.0 kg O 2 gb1 kmol O 2
nO 2
I
G
J
H n NH3 Kfed
=
= 1.40b125 kmol O2 g = 175 kmol O2
kmol NH3 / 17 kg NH3 g = 2.94 kmol NH3
b50.0 kg NH 3 gb1
F
fed
/ 32 kg O2 g = 3125
.
kmol O2
F nO I
3125
.
= 1.06 < G 2 J
2.94
H nNH3 K
=
stoich
5
= 1.25
4
⇒ O2 is the limiting reactant
Required NH3 :
3125
. kmol O2 4 kmol NH3
= 2.50 kmolNH3
5 kmol O2
2.94 − 2.50
× 100% = 17.6% excess NH3
2.50
Extent of reaction: nO2 = dnO2 i − vO2 ξ ⇒ 0 = 3125
.
− b−5gξ ⇒ ξ = 0.625kmol = 625 mol
% excess NH3 =
0
Mass of NO:
4.41
a.
3125
.
kmol O2 4 kmol NO 30.0 kg NO
= 75.0 kg NO
5 kmol O2 1 kmol NO
By adding the feeds in stoichometric proportion, all of the H2 S and SO 2 would be consumed.
Automation provides for faster and more accurate response to fluctuations in the feed stream,
reducing the risk of release of H2 S and SO 2 . It also may reduce labor costs.
4- 31
4.41 (cont’d)
b.
n& c =
3.00 × 10 2 kmol 0.85 kmol H 2S 1 kmol SO 2
= 1275
. kmol SO 2 / h
h
kmol
2 kmol H 2S
c.
Calibration Curve
1.20
X (mol H 2S/mol)
1.00
0.80
0.60
0.40
0.20
0.00
0.0
20.0
40.0
60.0
80.0
100.0
Ra (mV)
X = 0.0199Ra − 0.0605
d.
n& c bkmol SO2 / h g
n& f bkmol / h g
Blender
x bkmol H 2S / kmolg
Flowmeter calibration:
n& f = aR f
U
20
n& =
R
&n f = 100 kmol / h , R f = 15 mV VW f
3 f
Control valve calibration:
n&c = 25.0 kmol / h, R c = 10.0 mV U
7
5
Vn& = R +
n& c = 60.0 kmol / h , Rc = 25.0 mV W c 3 c 3
1
7
5 1 F 20 I
n& f x ⇒ Rc + = G R f J b0.0119 Ra − 0.0605g
K
2
3
3 2H 3
10
5
⇒ Rc =
R f b0.0119 Ra − 0.0605 g −
7
7
Stoichiometric feed: n& c =
n& f = 3.00 × 10 2 kmol / h ⇒ R f =
3
n& f = 45 mV
20
4- 32
4.41 (cont’d)
10
5
= 53.9 mV
b45g b0.0119 gb76.5g − 0.0605 −
7
7
7
5
⇒ n&c = b53.9 g + = 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 =
e.
4.42
165 mol / s
n& bmol / sg
x bmol C2H 4 / molg
0.310 mol C 2 H 4 / mol
1 − x bmol HBr / molg
0173
.
mol HBr / mol
0.517 mol C 2H 5Br / mol
C 2 H 4 + HBr → C 2H 5Br
C balance:
165 mol x bmol C 2H4 g 2 mol C
= n&b0.310 gb2 g + n&b0.517gb2g
s
mol
mol C2 H4
Br balance: 165 (1 − x )( 1) = n& ( 0.173 )( 1) + n& ( 0.517 )(1)
(1)
(2)
Solve (1) and (2) simultaneously ⇒ n& = 108.77 mol / s, x = 0.545mol C2H 4 / mol
⇒ b1 − x g = 0.455mol HBr / mol
Since the C2 H4 /HBr feed ratio (0.545/0.455) is greater than the stoichiometric ration (=1),
HBr is the limiting reactant .
bn& HBr gfed
= b165mol / sgb0.455mol HBr / molg = 75.08 mol HBr
Fractional conversion of HBr =
dn& C2 H4 i
stoich
dn& C2 H4 i
fed
75.08 − b0173
. gb108.8 g
× 100% = 0.749 mol HBr react / molfed
75.08
= 75.08 molC2 H4
= b165 mol / sgb0.545mol C2H 4 / molg = 89.93 mol C2 H4
89.93 − 75.08
= 19.8%
75.08
Extentof reaction: n&C 2H5 Br = dn& C2 H5 Br i + v C2 H5 Brξ ⇒ b1088
. gb0.517g = 0 + b1gξ ⇒ ξ = 56.2 mol / s
% excess of C2H 4 =
0
4- 33
4.43
a.
2HCl +
1
O 2 → Cl 2 + H 2 O
2
Basis: 100 mol HCl fed to reactor
100 mol HCl
n2 bmol HClg
n3 bmolO 2 g
n1 bmol air g
n4 bmol N 2 g
0.21 mol O 2 / mol
n5 bmol Cl 2 g
0.79 mol N2 / mol
n6 bmol H2Og
35% excess
stoic =
bO 2 g
100 mol HCl 0.5 mol O 2
2 mol HCl
= 25 mol O 2
35% excess air: 0.21n1bmol O 2 fedg = 1.35 × 25 ⇒ n1 = 160.7 mol air fed
85% conversion ⇒ 85 mol HCl react ⇒ n2 = 15 mol HCl
n5 =
85 mol HCl react
1 mol Cl 2
2 mol HCl
= 42.5 mol Cl 2
n6 = b85gb1 2g = 42.5 mol H2O
N 2 balance:
b160.7gb0.79g =
n 4 ⇒ n4 = 127 mol N2
O balance:
b160.7 gb0.21g mol
O2
2 mol O
1 mol O 2
= 2 n3 +
42.5 mol H 2 O
1 mol O
1 mol H2O
⇒ n3 = 12.5 mol O 2
Total moles:
5
∑ n j = 239.5 mol ⇒
j =2
15 mol HCl
mol HCl
molO 2
mol N 2
= 0.063
, 0.052
, 0.530
,
239.5 mol
mol
mol
mol
0177
.
b.
molCl 2
mol H 2 O
, 0177
.
mol
mol
As before, n1 = 160.7 mol air fed, n2 = 15 mol HCl
1
2HCl + O2 → Cl2 + H2O
2
n i = bn i g0 + v i ξ
E
HCl: 15 = 100 − 2ξ ⇒ ξ = 42.5 mol
4- 34
4.43 (cont’d)
1
ξ = 12.5 mol O 2
2
N 2 : n4 = 0.79b160.7g = 127 mol N 2
O 2 : n 3 = 0.21b160.7 g −
Cl 2 : n 5 = 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.
FeTiO3 + 2H 2SO 4 → bTiOgSO 4 + FeSO 4 + 2H 2O
Fe 2O3 + 3H 2SO 4 → Fe 2 bSO 4 g3 + 3H2O
bTiOgSO 4
+ 2H 2O → H2TiO3bsg + H 2SO 4
H2TiO3bsg → TiO 2bsg + H2O
Basis: 1000 kg TiO 2 produced
1000 kg TiO 2
kmol TiO 2
1 kmol FeTiO 3
79.90 kg TiO2
1 kmol TiO 2
12.52 kmol FeTiO3 dec.
1 kmol FeTiO3 feed
0.89 kmol FeTiO3 dec.
14.06 kmol FeTiO3
= 12.52 kmol FeTiO 3 decomposes
= 14.06 kmol FeTiO3 fed
1 kmol Ti
47.90 kg Ti
1 kmol FeTiO3
kmol Ti
= 6735
. kg Ti fed
673.5 kg Ti / M bkg oreg = 0.243 ⇒ M = 2772 kg ore fed
Ore is made up entirely of 14.06 kmol FeTiO 3 + nbkmol Fe 2O3 g (Assumption!)
n = 2772 kg ore −
638.1 kg Fe2 O3
14.06 kmol FeTiO3 151.74 kg FeTiO3
kmol FeTiO3
kmol Fe 2O3
159.69 kg Fe 2O3
= 6381
. kg Fe 2O3
= 4.00 kmol Fe2 O3
14.06 kmol FeTiO3 2 kmol H2SO4 4.00 kmol FeTiO3 3 kmol H2SO4
+
= 4012
. kmol H2SO4
1 kmol FeTiO3
1 kmol Fe2O3
50% excess: 15
. b4012
. kmol H2SO 4 g = 6018
. kmol H2SO 4 fed
Mass of 80% solution:
60.18 kmol H 2SO 4 98.08 kg H2SO 4
1 kmol H2SO 4
= 59024
. kg H2SO 4
5902.4 kg H 2SO 4 / M a bkg soln g = 0.80 ⇒ M a = 7380 kg 80% H 2SO 4 feed
4- 35
4.45
a.
Plot C (log scale) vs. R (linear scale) on semilog paper, get straight line through
d R1
= 10, C1 = 0.30 g m 3 i and
FR
H 2
= 48, C2 = 2.67 g m3 I
K
ln C = bR + ln a ⇔ C = ae br
b=
ln b2.67 0.30g
= 0.0575 , ln a = ln b2.67 g − 0.0575b48g = −178
. ⇒ a = e −1.78 = 0.169
48 − 10
⇒ C = 0169
. e 0.0575 R
Cdg m 3 i =
C ′( lb m ) 4536
. g 35.31 ft 3
ft 3
1 lb m
1 m3
= 16,020C ′
E
16 ,020C' = 0.169e 0.0575R ⇒ C ′dlb m SO 2 ft 3 i = 1.055 × 10 −5 e 0.0575 R
b.
d2867
ft 3 si b60 s min g
1250 lb m min
= 138 ft 3 lb m coal
R = 37 ⇒ C ′dlb m SO 2 ft 3 i = 1055
.
× 10 −5 e b0.0575 gb37g = 8.86 × 10
8.86 × 10 −5 lb m SO 2
ft
c.
3
138 ft 3
1 lb m coal
= 0.012 < 0.018
−5
lb m SO 2 ft 3
lb m SO 2
compliance achieved
lb m coal
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
= 1249
. lb m SO 2 generated min
2867 ft3 60 s 8.86 ×10−5 lb m SO 2
= 15.2 lb m SO 2 min in scrubbed gas
s
1 min
ft3
air
1250 lbm coal/min
62.5 lb m S/min
% removal =
d.
scrubbing fluid
furnace stack gas
124.9 lbm SO2 /min
ash
scrubber scrubbed gas
15.2 lb m SO2 /min
liquid effluent
(124.9 – 15.2)lbm SO2 (absorbed)/min
b124.9 − 15.2 g lb m
SO 2 scrubbed min
× 100% = 88%
1249
. lb m SO 2 fed to scrubber min
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 SO 2 emitted per mass of coal
burned is independent of the flow rate of air in the stack.
4- 36
4.46
a.
A + B ===== C + D
nA = nA − ξ
0
n B = nB − ξ
y A = en A − ξ j n T
nC = nC + ξ
y B = en B − ξj nT
n D = nD + ξ
y C = en C
0
0
0
0
0
nI = nI
Total nT = ∑ ni
At equilibrium:
0
+
ξ j nT
y D = en D + ξj nT
0
0
yC y D bnC0 + ξ c gbnD0 + ξ c g
=
= 487
. (nT ’s cancel)
y A y B bn A0 − ξ c gbn B0 − ξ c g
387
. ξ 2c − cnC0 + nD0 + 487
. bnA0 + nB0 ghξ c − bnC0 nD0 − 487
. nA0 nB0 g = 0
[aξ 2c + bξ c + c = 0]
a = 387
.
1
2
∴ξc = e−b ± b − 4ac j where b = − nC0 + nD 0 + 4.87bnA0 + nB0 g
2a
c = − nC0nD0 − 4.87nA0nB0
b.
Basis: 1 mol A feed
nA0 = 1 nB0 = 1 nC0 = nD0 = nI0 = 0
Constants: a = 3.87 b = −9.74
ξe =
(
1
9.74 ±
2 ( 3.87 )
c = 4.87
( 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
nA 0
nA 0
nA0 = 80, nC0 = nD 0 = nJ 0 = 0
nC 0 = 0
nC = 70 = nC 0 + ξ c =======> ξ c = 70 mol
n A = nA 0 − ξ c = n A0 − 70 mol
nB = n B0 − ξ c = 80 − 70 = 10 mol
nC = nC 0 + ξ c = 70 mol
nD = nD 0 + ξ c = 70 mol
4.87 =
b70gb70g
yC y D nC nD
=
⇒
= 4.87 ⇒ n A0 = 170.6 mol methanol fed
y A y B n An B
bn A0 − 70 gb10g
4- 37
4.46 (cont’d)
Product gas n A = 170.6 − 70 = 100.6 mol U y A = 0.401 mol CH3OH mol
|
y B = 0.040 mol CH3COOH mol
n B = 10 mol
|
V⇒
y C = 0.279 mol CH3COOCH3 mol
nC = 70 mol
|
|
y D = 0.279 mol H 2O mol
n D = 70 mol
W
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 + H2O ←
→ CO 2 + H2
(A)
(B)
(C)
(D)
1.00 mol
n A bmol COg
0.20 mol CO / mol
n B bmol H 2O g
010
. mol CO 2 / mol
0.40 mol H 2O / mol
nC bmol CO 2 g
n D bmol H 2 g
0.30 mol I / mol
n I bmol Ig
Degree of freedom analysis :
6 unknowns ( n A , nB , nC , n D , n I ,ξ )
– 4 expressions for ni bξ g
– 1 balance on I
– 1 equilibrium relationship
0 DF
b.
Since two moles are prodcued for every two moles that react,
bn total gout = bntotal gin = 1.00 bmol g
n A = 0.20 − ξ
nB = 0.40 − ξ
nC = 010
. +ξ
nD = ξ
(1)
(2)
(3)
(4)
n I = 0.30
(5)
ntot = 1.00 mol
At equilibrium:
. + ξ gbξ g
yC yD nC n D
b010
F 4020 I
=
=
= 0.0247exp G
.
mol
J ⇒ ξ = 0110
H1123K
y A yB n A nB b0.20 − ξ gb0.40 − ξ g
y D = n D = ξ = 0110
. bmol H 2 / molg
c.
The reaction has not reached equilibrium yet.
4- 38
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)
0
0
0
0
0
0
0
0
0
0
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
0.1
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
ln Ke = ln A0 + E T bK g
E=
ln bKe1 / K e2 g
1 T1 − 1 T2
=
ln d10.5 / 2.316 × 10 −4 i
1 373 − 1 573
= 11458
ln A0 = ln K e1 − 11458 T1 = ln 10.5 − 11458 373 = −28.37 ⇒ A0 = 4.79 × 10 −13
K e = 4.79 × 10 −13 expc11458 T bKgh atm −2 ⇒ K e (450 K ) = 00548
.
atm −1
b.
n A = n A0 − ξ
nB =
nC =
nT =
y A = bn A0 − ξ g bnT 0 − 2ξ g
U
n B0 − 2ξ ||
V
nC 0 + ξ |
nT 0 − 2ξ |W
⇒
y B = bn B0 − 2ξ g bnT 0 − 2ξ g
yC = bnC 0 + ξ g bnT 0 − 2ξ g
bnT 0 = n A0 + n B0 + nC 0 g
At equilibrium,
2
yC 1 bnC 0 + ξ e gbnT 0 − 2ξ e g 1
=
= Ke bT g (substitute for K bT g from Part a.)
e
y A yB2 P 2 bn A0 − ξ e gbnB 0 − 2ξ e g2 P 2
c.
Basis: 1 mol A (CO)
n A0 = 1 n B0 = 1 nC 0 = 0 ⇒ nT 0 = 2 , P = 2 atm , T = 423K
2
ξ e b2 − 2ξ e g
2
b1 − ξ e gb1 − 2ξ e g
1
-2
2
.
=0
2 = Ke b423g = 0.278 atm ⇒ ξ e − ξ e + 01317
4 atm
4- 39
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 n B .
y A = b1 − 0156
. g c2 − 2b0156
. gh ⇒ y A = 0.500
y B = c1 − 2b0156
. gh c2 − 2b0156
. gh ⇒ y B = 0.408
y C = b0 + 0.156 g c2 − 2b0156
. gh ⇒ y C = 0.092
Fractional Conversion of CO b Ag =
d.
n A0 − n A
ξ
=
= 0.156 mol A reacted / mol A feed
n A0
n A0
Use the equations from part b.
i)
ii)
iii)
iv)
Fractional conversion decreases with increasing fraction of CO.
Fractional conversion decreases with increasing fraction of CH3 OH.
Fractional conversion decreases with increasing temperature.
Fractional conversion increases with increasing pressure.
*
1
2
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 ξ e = 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
GO TO 1
NT = 1. – 2. * EKPI
YA = (YA0 – EKPI)/NT
YB = (YB0 – 2. + EKPI)/NT
YC = (YC0 + EKPI)/NT
4- 40
4.48 (cont’d)
CON = EKPI/YA0
WRITE (6, 3) YA, YB, YC, CON
STOP
4
WRITE (6, 5) INMAX, EKPI
3
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
Note: This will only find one root — there are two others that can only be found by
choosing different initial values of ξ a
4.49
a.
CH4 + O 2 
→ HCHO + H 2O
(1)
CH4 + 2O 2 → CO2 + 2H 2O
(2)
100 mol / s
0.50 mol CH4 / mol
n&1 bmol CH4 / sg
0.50 mol O 2 / mol
n&2 bmol O2 / sg
n&3 bmol HCHO / sg
n&4 bmol H 2O / sg
n&n&5 bmol
(molCO
CO2 g2 /s)
5
7 unknowns ( n&1, n& 2 , n&3, n&4 , n&5 , ξ& 1,ξ& 2 )
– 5 equations for n&i eξ& 1 , ξ& 2 j
2 DF
b.
n&1 = 50 − ξ& 1 − ξ& 2
n& = 50 − ξ& − 2ξ&
2
1
n& 3 = ξ& 1
n& = ξ& + 2ξ&
4
1
(1)
(2)
2
(3)
(4)
2
n& 5 = ξ& 2
c.
(5)
Fractional conversion:
Fractional yield :
( 50 − n&1 ) = 0.900 ⇒ n&
1
50
= 5.00 mol CH /s
4
n&3
= 0.855 ⇒ n&3 = 42.75 mol HCHO/s
50
4- 41
4.49 (cont’d)
y CH = 0.0500 mol CH 4 /mol
4
Equation 3 ⇒ ξ1 = 42.75 
y O = 0.0275 mol O 2 /mol
Equation 1 ⇒ ξ2 = 2.25 
2
 y
Equation 2 ⇒ n&2 = 2.75  ⇒ HCHO = 0.4275 mol HCHO/mol
Equation 4 ⇒ n&4 = 47.25 y H O = 0.4725 mol H 2O/mol

2
Equation 5 ⇒ n&5 = 2.25 
y CO = 0.0225 mol CO 2 /mol
2
Selectivity: [(42.75mol HCHO/s)/(2.25molCO 2 /s) = 19.0 mol HCHO/mol CO2
4- 42
4.50
a.
Design for low conversion and feed ethane in excess. Low conversion and excess ethane
make the second reaction unlikely.
b.
C2 H6 + Cl2 → C2 H5 Cl + HCl, C2 H5 Cl + Cl2 → C2H4 Cl2 + HCl
Basis: 100 mol C2 H5 Cl produced
c.
n 1 (mol C2 H6 )
100 mol C2 H5 Cl
n 2 (mol Cl2 )
n 3 (mol C2 H6 )
n 4 (mol HCl)
n 5 (mol C2 H5 Cl2 )
5 unknowns
–3 atomic balances
2 D.F.
Selectivity: 100 mol C 2 H 5Cl = 14 n5 (mol C 2 H 4 Cl 2 ) ⇒ n5 = 7.143 mol C 2 H 4 Cl 2
15% conversion: b1 − 0.15gn1 = n 3
2 n1 = 2b100g + 2 n 3 +
C balance:
U
n1
|
V⇒
2b7.143g|W n3
= 714.3 mol C 2 H 6 in
= 114.3 mol C 2 H 6 out
H balance:
6b714.3g = 5b100g + 6b114.3g + n4 + 4b7 .143g ⇒ n4 = 607 .1 mol HCl
Cl balance:
2 n2 = 100 + 607.1 + 2b7 .143g ⇒ n2 = 114 .3 mol Cl 2
. mol Cl 2 / mol C 2 H 6
Feed Ratio : 114.3 mol Cl 2 / 714.3 mol C 2 H 6 = 016
Maximum possible amount of C2 H5 Cl:
114.3 mol Cl 2 1 mol C 2 H 5Cl
n max =
= 114.3 mol C 2 H 5Cl
1 mol Cl 2
Fractional yield of C2 H5 Cl:
4.51
nC2 H5 Cl
n max
=
100 mol
= 0.875
114.3 mol
d.
Some of the C2 H4 Cl2 is further chlorinated in an undesired side reaction:
C2 H5 Cl2 + Cl2 → C2H4 Cl3 + HCl
a.
C2 H4 + H2O → C2 H5OH, 2 C2H5 OH → (C2 H5 )2 O + H2O
Basis: 100 mol effluent gas
100 mol
n1 (mol C 2 H 4 )
n [mol H O (v)]
2
2
n 3 (mol I)
0.433 mol C 2 H 4 / mol
0.025 mol C H OH / mol
2 5
0.0014 mol (C H ) O / mol
2 5 2
0.093 mol I / mol
3 unknowns
-2 independent atomic balances
-1 I balance
0 D. F.
0.4476 mol H O (v) / mol
2
(1) C balance: 2 n1 = 100b2∗0.433 + 2∗0.025 + 4∗0.0014 g
(2) H balance: 4 n1 + 2n 2 = 100b4∗0.433 + 6∗0.025 + 10∗0.0014 + 2∗0.4476 g
(3) O balance: n 2 = 100b0.025 + 0.0014 + 0.4476g
Note; Eq. (1)∗2 + Eq. (3)∗2 = Eq. (2 ) ⇒2 independent atomic balances
(4) I balance: n3 = 9.3
4-43
4.51 (cont'd)
b.
(1) ⇒ n1 = 46.08 mol C 2 H 6 U
|
(3) ⇒ n 2 = 47.4 mol H 2 O V ⇒ Reactor feed contains 44.8% C 2 H 6 , 46.1% H 2 O, 9.1% I
|
(4) ⇒ n 3 = 9.3 mol I
W
46.08 − 43.3
× 100% = 6.0%
46.08
If all C2 H4 were converted and the second reaction did not occur, dn C2 H5OH i
% conversion of C2 H4 :
⇒ Fractional Yield of C2 H5 OH: n C2 H5OH / dnC2 H5 OH i
max
max
= 46.08 mol
= b2.5 / 46.08g = 0.054
Selectivity of C2 H5 OH to (C2 H5)2 O:
2.5 mol C 2 H 5OH
= 17.9 mol C 2 H 5OH / mol (C2 H 5 ) 2 O
0.14 mol (C 2 H 5 ) 2 O
c.
4.52
Keep conversion low to prevent C2 H5OH from being in reactor long enough to form
significant amounts of (C2 H5)2 O. Separate and recycle unreacted C2H4 .
CaF2 bsg + H 2SO 4 bl g → CaSO 4 bsg + 2HFbg g
1 metric ton acid
1000 kg acid
0.60 kg HF
1 metric ton acid
1 kg acid
= 600 kg HF
Basis: 100 kg Ore dissolved (not fed)
100 kg Ore dissolved
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
n1
n2
n3
n4
(kg CaSO4)
(kg HF)
(kg H 4SiF6 )
(kg H 2 SO 4)
n5 (kg H2 O)
Atomic balance - Si:
0.04b100g kg SiO2
28.1 kg Si
60.1 kg SiO 2
=
n 3 (kg H 4 SiF6 )
28.1 kg Si
146.1 kg H 4 SiF6
Atomic balance - F:
n (kg HF) 19.0 kg F
38.0 kg F
= 2
20.0 kg HF
78.1 kg CaF2
9.72 kg H 4SiF6
114.0 kg F
+
⇒ n 2 = 41.2 kg HF
146.1 kg H 4 SiF6
0.96b100g kg CaF2
600 kg HF 100 kg ore diss.
41.2 kg HF
1 kg ore feed
0.95 kg ore diss.
4-44
= 1533 kg ore
⇒ n 3 = 9.72 kg H 4 SiF6
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 3Cl 3 + HCl
Convert output wt% to mol%: Basis 100 g output
species
C6H 6
C 6 H 5Cl
C 6 H 4 Cl 2
C 6 H 3Cl 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
total 1.136
mol %
73.2
25.0
1.5
0.3
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 mol C6 H6
73.2 mol C6 H6
32.0 mol C 6 H 5 Cl
25.0 mol C6 H5 Cl
2.5
Cl 2
1.5mol
molCC66H
H44 Cl
2
0.5
mol
C
H
Cl 3
0.3 mol C6 H 3Cl
6
b.
3
4
-3
-1
0
unknowns
atomic balances
wt% Cl 2 in feed
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: 2 n2 = 28.9 + 25.0 + 2 (1.5 ) + 3 ( 0.3) ⇒ n2 = 28.9 mol Cl 2
Theoretical C 6 H 6 = 28.9 mol Cl 2 (1 mol C6H 6 1 mol Cl 2 ) = 28.9 mol C6 H6
(100 − 28.9 ) 28.9 × 100% = 246% excess C6H6
Fractional Conversion: (100 − 73.2) 100 = 0.268 mol C6 H6 react/mol fed
Excess C 6 H 6 :
Yield: (25.0 mol C 6 H 5Cl) (28.9 mol C 6 H 5Cl maximum)=0.865


g gas

 ⇒ 0.268
g liquid
 78.11 g C6 H6 

Liquid feed: (100 mol C6 H6 ) 
 = 7811 g liquid 
 mol C6H6 

Gas feed:
28.9 mol Cl2 70.91 g Cl2 1 g gas
= 2091 g gas
mole Cl 2 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.
e.
Dissolve in water to produce hydrochloric acid.
Reagent grade costs much more. Use only if impurities in technical grade mixture affect the
reaction rate or desired product yield.
4-45
4.54
a.
2CO 2 ⇔ 2CO + O 2
2A ⇔ 2B + C
O 2 + N 2 ⇔ 2NO
C + D ⇔ 2E
n A = n A 0 − 2ξ e1
y A = bn A0 − 2ξ e1 g bn T 0 + ξ e 1 g
n B = n B0 + 2ξ e2
y B = bn B 0 + 2ξ e1 g bnT 0 + ξ e1 g
n C = n C 0 + ξ e1 − ξ e2 ⇒ y C = bn C 0 + ξ e1 − ξ e 2 g bn T 0 + ξ e1 g
n D = n D 0 − ξ e2
y D = bn D0 − 1ξ e 2 g bn T 0 + ξ e1 g
n E = n E 0 + 2ξ e2
y E = bn E 0 + 2ξ e 2 g bnT 0 + ξ e1 g
ntotal = n T 0 + ξ e1
bn T 0
= n A0 + n B0 + n C 0 + n D0 + n E 0 g
Equilibrium at 3000K and 1 atm
y 2B y C
y 2A
=
bn B 0
2
+ 2ξ e 1 g bn C 0 + ξ e1 − ξ e2 g
bn A 0
2
− 2ξ e1 g bn T 0 + ξ e 1 g
= 0.1071
yE2
( nE 0 + 2ξe 2 )
=
= 0.01493
yC y D ( nA0 + ξe 1 − ξ e 2)( nD0 − ξe 2 )
2
E
Defines functions
f 1 = 0.1071bn A0 − 2ξ e1 g bn T 0 + ξ e 1 g − bn B 0 + 2ξ e1 g bn C 0 + ξ e1 − ξ e 2 g = 0U
|
V f 1bξ 1 , ξ 2 g and
2
f 2 = 0.01493bnC 0 + ξ e1 − ξ e 2 gbn D 0 − ξ e2 g − bn E 0 + 2ξ e2 g = 0
|
f 2 bξ 1 , ξ 2 g
W
2
2
b.
Given all n io ’s, solve above equations for ξ e1 and ξ e2 ⇒ n A , n B, n C, n D , n E ⇒ yA , yB, yC, yD , yE
c.
n A0 = n C0 = nD0 = 0.333, n B0 = n E0 = 0 ⇒ ξ e1 =0.0593, ξ e2 = 0.0208
⇒ yA = 0.2027, yB = 0.1120, yC = 0.3510, yD = 0.2950, yE = 0.0393
d.
a 11d 1 + a 12 d 2 = − f 1
a f − a 22 f 1
d 1 = 12 2
a 11a 22 − a 12 a 21
a 21d 1 + a 22 d 2 = − f 2
a f − a 11 f 2
d 2 = 21 1
a11a 22 − a12 a 21
bξ e1 gnew
bξ e 2 gnew
= ξ e1 + d 1
= 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
4-46
4.54 (cont’d)
2
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
3
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
100
CONTINUE
WRITE (6, 4) NMAX
4
FORMAT('0', 10X, 'PROGRAM DID NOT CONVERGE IN', I2, 'ITERATIONS'/)
STOP
120
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
4-47
0.33
X1C,
X1C,
X1C,
0.00
X2C = 0.06418
X2C = 0.05969
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
X1C, X2C = 0.05931
X1C, X2C = 0.05930
X1C, X2C = 0.05930
YA, YB, YC, YD, YE =
2.0270E − 01
1.1197 E − 01
2 .9501E − 01
3.9319 E − 02
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
0.20
0.20
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.
m
& B 0 (kg A/h)
1 kg B/kg A fed to reactor
( P)
( A)
& A0 (kg A/h)
m
m& B 0 (kg A/h)
xRA (kg R/kg A)
x RA (kg R/kg A)
R → S
& 3 (kg A/h)
m
m
& P (kg P/h)
xR 3 (kg R/kg)
0.0075 kg R/kg P
99% conv.
f m& A0 (kg A/h)
xRA (kg R/kg A)
Splitting point: 1 allowed material balance
Reactor: 1 mass balance + 99% conversion of R (=> 2 equations)
Mixing point: 2 allowed material balances (1 mass, 1 on R)
& A0 , f , x RA ,m& B0 , m
& 3 , x R 3 , m& P ) − 5 equations = 2 degrees of freedom
⇒ 7 unknowns ( m
b.
Mass balance on splitting point: mA0 = mB0 + f mA0
(1)
Mass balance on reactor: 2 mB0 = m3
99% conversion of R: xR3 m3 = 0.01 xRA mB0
Mass balance on mixing point: m3 + f mA0 = mP
R balance on mixing point: xR3 m3 + xRA f mA0 = 0.0075 mP
Given xRA and mP , solve simultaneously for mA0 , mB0 , f, m3 , xR3
(2)
(3)
(4)
(5)
4-48
4.55(cont’d)
c. mA0 = 2778 kg A/h
mB0 = 2072 kg B/h
fA = 0.255 kg bypass/kg fresh feed
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 vs. x RA
f (kg bypass/kg fresh feed)
d.
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 (kg R/kg A)
4-49
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 CH3 OH/h (fresh feed)
Single pass conversion:
c.
4.57
a.
30.0
= 0.60 ⇒ n 3 = 20.0 kmol CH 3 OH / h
n1 + n3
n 1 + n 3 = 50.0 kmol CH3 OH fed to reactor/h
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%.
Convert effluent composition to molar basis. Basis: 100 g effluent:
10.6 g H2
1 mol H2
2.01 g H 2
64.0 g CO
= 5.25 mol H 2
1 mol CO
28.01 g CO
25.4 g CH3 OH
= 2.28 mol CO
⇒
H : 0.631 mol H / mol
2
2
CO: 0.274 mol CO / mol
CH 3 OH: 0.0953 mol CH 3 OH / mol
1 mol CH 3OH
32.04 g CH3 OH
= 0.793 mol CH3O H
4-50
4.57 (cont’d)
n4 (mol / min)
0.004 mol CH 3OH(v)/ mol
x (mol CO /mol)
(0.896 - x) (mol H 2 / mol)
Cond.
Reactor
350 mol/ min
n1 (mol CO/ min)
n 2 (mol H 2 / min)
0.631 mol CH 3OH(v)/ mol
n 3 (mol CH 3OH(l) / min)
0.274 mol CO/ mol
CO+ H 2 → CH 3OH 0.0953 mol H / mol
2
Condenser
3 unknowns (n3 , n4 , x)
-3 balances
0 degrees of freedom
Overall process
2 unknowns (n1 , n2 )
-2 independent atomic balances
0 degrees of freedom
Balances around condenser
n = 32.1 mol CH 3OH(l) / min
U
CO: 350∗0.274 = n ∗ x
3
4
|
H : 350∗0.631 = n ∗ (0.996 − x ) V ⇒ n = 318.7 mol recyc le / min
2
4
4
CH OH: 350∗0.0953 = n + 0.004∗ n |
x =.301 molCO / mol
3
3
4W
Overall balances
n = 32.08 mol / min CO in feed
C: n = n U
1
3 | ⇒ 1
H: 2n = 4n V|
n = 64.16 mol / min H 2 in feed
2
2W
2
Single pass conversion of CO:
(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.)
Overall conversion of CO:
b.
–
–
–
4-51
4.58
a.
Basis: 100 kmol reactor feed/hr
n3 (kmol CH4 /h)
100 kmol /h
Reactor
n1 (kmol CH4 /h) 80 kmol CH4 /h
n2 (kmol Cl2 /h) 20 kmol Cl
2 /h
n3 (kmol CH4 /h)
n4 (kmol HCl /h)
5n5 (kmol CH3 Cl /h)
n5 (kmol CH2 Cl 2 /h)
Cond.
Solvent
Absorb
n3 (kmol CH4 /h)
n4 (kmol HCl/h)
n4 (kmol HCl/h)
5n5 (kmol CH3Cl /h)
Still
5n5 (kmol CH3Cl /h)
n5 (kmol CH2Cl 2 /h)
n5 (kmol CH2Cl 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 )
Cl2 balance: n2 = 20
b.
c.
Solve for n1
For a basis of 100 kmol/h into reactor
n1 = 17.1 kmol CH4 /h
n2 = 20.0 kmol Cl2 /h
n3 = 62.9 kmol CH4 /h
n4 = 20.0 kmol HCl/h
5n5 = 14.5 kmol CH3 Cl/h
(1000 kg CH3 Cl/h)(1 kmol/50.49 kg) = 19.81 kmol CH3 Cl/h
Scale factor =
19 .81 kmol CH 3 Cl/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
4-52
4.59
a.
Basis: 100 mol fed to reactor/h ⇒ 25 mol O2 /h, 75 mol C2 H4 /h
n1 (mol C2H 4 //h)
n2 (mol O2 /h)
Seperator
reactor
nC2H4 ( mol C2 H 4 /h)
nO2 (mol O2 /h)
75 mol C2H 4 //h
25 mol O2 /h
n1 (mol C 2H 4 //h)
n2 (mol O2 /h)
n3 (mol C2H 4O /h)
n4 (mol CO2 /h)
n5 (mol H2O /h)
n3 (mol C2 H 4O /h)
n4 (mol CO2 /h)
n5 (mol H2O /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 (productio n 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) C2 H4 balance (mix pt): nC2H4 = 75 – n1
Overall conversion of C2 H4 : 100%
b.
n1 = 60.0 mol C2 H4 /h
n2 = 13.75 mol O2 /h
n3 = 13.5 mol C2 H4 O/h
n4 = 3.00 mol CO2 /h
c.
Scale factor =
n5 = 3.00 mol H2 O/h
nO2 = 11.25 mol O2 /h
nC2H4 = 15.0 mol C2 H4 /h
100% conversion of C2 H4
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 C2 H4 /h
nO2 = (3.363)(11.25) = 37.8 lb-mol O2 /h
4-53
4.60
a.
Basis: 100 mol feed/h
100 mol/h
&11 (mol /h)
nn
32 mol CO/h
64 mol H 2 / h
4 mol N 2 / h
.13 mol N 2 /mol
reactor
n&n23 (mol CH 3 OH / h)
cond.
500 mol / h
x1 (mol N 2 /mol)
x2 (mol CO / mol)
1-x1-x 2 (mol H 2 / h)
n3 (mol / h)
(mol/h)
x1n&(mol
N 2 /mol)
3
x2x(mol
CO
mol)
(mol
N2/ /mol)
1
1-x1-x 2 (mol H 2 / h)
Purge
x 2 (mol CO/mol)
1 − x1 − x2 (mol H2 /mol)
Mixing point balances:
total: (100) + 500 = n&1 ⇒ n&1 = 600 mol/h
N2 : 4 + x1 * 500 = .13 * 600 ⇒ x1 = 0.148 mol N2 /mol
Overall system balances:
N2 : 4 = n&3 (0.148)
⇒ n&3 = 27 mol/h
Atomic C:32(1) = n& 2 (1) + 27 x 2 (1)
Atomic H:64(2) = n& 2 (4) + 27(1 − 0.148 − x2 )(2)
=>
n& 2 = 24.3 mol CH3 OH/h
x 2 = 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.
3H2NH
-> NH 3
2
N2 + 2N
3H22 +→
3
(1-yp) (1-fsp) n1 (mol N 2)
(1-yp) (1-fsp) 3n1 (mol H 2)
(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 N 2)
3n1 (mol H2)
n2 (mol I)
Reactor
4-54
(1-fsp) n1 (mol N 2)
(1-fsp) 3n1 (mol H 2)
n2 (mol I)
nr (mol)
(1-fsp) n1 (mol N2)
(1-fsp) 3n1 (mol H 2)
n2 (mol I)
2 fsp n1 (mol NH 3)
yp (1-fsp) n1 (mol N 2)
yp (1-fsp) 3n1 (mol H 2)
yp n2 (mol I)
Condenser
np (mol)
2 fsp n1 (mol NH 3)
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 = 2fsp n1
Overall N2 conversion:
b.
(1 − X I 0 ) / 4 − y p (1 − f sp ) n 1
(1 − X I 0 ) / 4
XI0 = 0.01 fsp = 0.20 yp = 0.10
n1 = 0.884 mol N2
n2 = 0.1 mol I
× 100 %
nr = 3.636 mol fed
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
4-55
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 4H10)
n 3 (i-C 4H 10)
n 1 (C 8H 18)
m4 (91% H 2SO 4)
E
Units of n: kmol
Units of m: kg
reactor
C
B
P
n 1 (C 8H 18)
n 2 (n-C4 H10)
F
decanter
n 1 (C 8H 18)
n 2 (n-C 4H 10)
n 3 (i-C 4H 10)
still
n 5 (n-C 4H 10)
n 6 (i-C 4H 10)
n 7 (C 8H 18)
m8 (91% H 2SO 4)
m 4 (kg 91% H 2SO 4)
40000 kg
A
n 0 kmol
0.25 i-C4 H 10
0.50 n-C4 H10
0.25 C4 H 8
n 3 (i-C 4H 10)
Calculate moles of feed
M = 0.25 M L− C 4 H10 + 0.50 M n− C4 H10 + 0.25 M C 4 H8 = b0.75gb58.12 g + b0.25gb56.10g
= 57 .6 kg kmol
n 0 = b40000 kg gb1 kmol 57.6 kgg = 694 kmol
Overall n - C4 H10 balance: n 2 = b0.50gb694 g = 347 kmol n - C4 H 10 in product
C 8 H 18 balance:
n1 =
b0.25gb694 g kmol
C4 H 8 react 1 mol C8 H 18
1 mol C 4 H 8
= 1735
. kmol C 8 H 8 in product
At (A), 5 mol i - C 4 H 10 1 mole C 4 H 8 ⇒ n bmol i - C 4 H 10 gA = b5gb0.25gb694g = 867.5 kmol
14
4244
3
i -C 4 H10 at
moles C 4 H8 at
A =173.5
Note: nbmol C 4 H 8 g = 173.5 at (A), (B) and (C) and in feed
i - C 4 H 10 balance around first mixing point ⇒ b0.25gb694g + n3 = 867 .5
⇒ n3 = 694 kmol i - C 4 H 10 recycled from still
At C, 200 mol i - C4 H 10 mol C4 H 8
⇒ nbmol i - C4 H 10 gC = b200gb1735
. g = 34,700 kmol i - C4 H10
4-56
bA g
and bB g
4.62 (cont’d)
i - C 4 H 10 balance around second mixing point ⇒ 8675
. + 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,
n 5 bmoles n - C 4 H 10 gE
n2 bmoles n - C 4 H 10 gD
n7 bmoles C8 H18 gE
n1 bmoles C8 H 18 gD
=
=
n 6 bmoles i - C 4 H10 gE
n3 bmoles i - C 4 H 10 gD
⇒ n5 = 16,900 kmol n - C 4 H10
n6
⇒ n 7 = 8460 kmol C 4 H 18
n3
Hydrocarbons entering reactor:
kg I
J
kmol K
kg I
kg I
F
F
+ b867 .5 + 33800gbkmol i - C 4 H 10 g G58.12
J + 173.5 kmol C 4 H 8 G56.10
J
H
K
H
kmol
kmolK
kg I
F
6
+ 8460 kmol C8 H 18 G114.22
J = 4.00 × 10 kg .
H
kmol K
b347
F
+ 16900gbkmol n - C 4 H 10 g G58.12
H
H 2SO 4 solution entering reactor
band
leaving reactor g
=
4.00 × 10 6 kg HC 2 kg H 2SO 4 baq g
1 kg HC
= 8.00 × 10 6 kg H 2 SO 4 baq g
m8 bH 2SO 4 in recycle g
8.00 × 10 bH 2 SO 4 leaving reactor g
6
=
n5 bn - C 4 H 10 in recycle g
n2 + n5 bn - C 4 H 10 leaving reactorg
⇒ m8 = 7 .84 × 10 6 kg H 2 SO 4 baq g in recycle E
m4 = H 2 SO 4 entering reactor − H 2 SO 4 in E
= 1.6 × 10 5 kg H 2 SO 4 baq g recycled from decanter
⇒ d1.6 × 10 5 i b0.91gkg H 2SO 4 b1 kmol 98.08 kg g = 1480 kmol H 2SO 4 in recycle
d1.6 × 10 i b0.09 gkg
5
H 2 O b1 kmol 18.02 kg g = 799 kmol H 2 O from decanter
Summary: (Change amounts to flow rates)
Product: 173.5 kmol C8 H 18 h , 347 kmol n - C 4 H10 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 - C4 H10 h , 8460 kmol C 8 H18 h,
7.84 × 10 6 kg h 91% H 2 SO 4 ⇒ 72,740 kmol H 2SO 4 h , 39,150 kmol H 2 O h
4-57
4.63
a.
A balance on ith tank (input = output + consumption)
v&bL min gC A , i−1 bmol Lg = vC
& Ai + kC Ai C Bi bmol liter ⋅ min gV bLg
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 = CBi + k τ C Ai CBi
Subtract equations ⇒ CBi − C Ai = CB, i−1 − C A, i−1
=
CB , i −2 − C A, i − 2 =K = CB 0 − C A0
A
from balances on
st
bi −1g tank
b.
C Bi − C Ai = CB 0 − C A0 ⇒ C Bi = C Ai + C B0 − C A 0 . Substitute in A balance from part (a).
C A, i−1 = C Ai + kτ C Ai C Ai + bCB0 − C A0 g . Collect terms in C 2Ai , C 1Ai , C 0Ai .
2
C Ai
k τ + C AL 1 + kτ bC B0 − C A0 g − C A, i−1 = 0
⇒ α C 2AL + β C AL + γ = 0 where α = kτ , β = 1 + k τ bCB 0 − C A0 g, 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.)
Solution: C Ai =
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), (x min = 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)
4-58
4.64
a.
Basis: 1000 g gas
Species
m (g)
MW
n (mol)
mole % (wet)
mole % (dry)
C3 H8
800
44.09
18.145
77.2%
87.5%
C4 H10
150
58.12
2.581
11.0%
12.5%
H2 O
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.
C3 H8 + 5 O2 → 3 CO2 + 4 H2 O, C4 H10 + 13/2 O2 → 4 CO2 + 5 H2O
Theoretical O2 :
C3H8:
C 4 H 10 :
100 kg gas 80 kg C 3 H 8 1 kmol C 3 H 8
5 kmol O 2
= 9.07 kmol O 2 / h
h
100 kg gas 44.09 kg C3 H 8 1 kmol C 3 H8
100 kg gas 15 kg C 4 H10 1 kmol C 4 H 10
6.5 kmol O 2
= 1.68 kmol O 2 / h
h
100 kg gas 58.12 kg C4 H 10 1 kmol C4 H10
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 H14
= 38.3 mol C6 H14
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 C7 H16
= 27.36 mol C 7 H 16
L C 7 H 16
100 kg C 7 H16
C6 H14 +19/2 O2 → 6 CO2 + 7 H2 O
C6 H14 +13/2 O2 → 6 CO + 7 H2 O
C7 H16 + 11 O2 → 7 CO2 + 8 H2 O
C7 H16 + 15/2 O2 → 7 CO + 8 H 2 O
Theoretical oxygen:
38.3 mol C6 H14
9.5 mol O 2
27.36 mol C7 H 16
+
mol C 6 H14
11 mol O 2
= 665 mol O2 required
mol C 7 H16
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
4-59
4.66
CO +
1
O 2 → CO 2
2
H2 +
1
O2 → H2O
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 C O . We include its calculation for illustrative
purposes.
A plot of x vs. R on log paper is a straight line through the points bR1 = 10.0 , x1 = 0.05g and
bR2
= 99.7 , x 2 = 1.0g .
b = lnb1.0 0.05g ln b99.7 10.0g = 1.303
ln x = b ln R + ln a
ln a = ln b1.0g − 1.303lnb99.7 g = −6.00 ⇒ x = 2.49 × 10 −3 R1.303
a = exp −6.00 = 2.49 × 10 −3
@
x = a Rb
b
R = 38.3 ⇒ x = 0.288
g
moles CO
mol
Theoretical O 2 : 175 kmol 0.288 kmol CO 0.5 kmol O2
h
kmol
kmol CO
+
Air fed:
4.67 a.
175 kmol
h
0.212 kmol H 2
kmol
kmol O2
0.5 kmol O2
= 43.75
kmol H 2
h
43.75 kmol O 2 required
1 kmol air
1.2 kmol air fed
h
0.21 kmol O 2
1 kmol air required
CH 4 + 2O 2 → CO 2 + 2H 2 O
7
O → 2CO 2 + 3H 2 O
2 2
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
13
O → 4CO 2 + 5H 2 O
2 2
Theoretical O2 :
0.944b100gkmol CH 4
h
kmol air
h
100 kmol/h
0.944 CH4
0.0340 C2H6
0.0060 C3H8
0.0050 C4H10
C2H6 +
C 4 H 10 +
= 250
17% excess air
n a (kmol air/h)
0.21 O2
0.79 N2
2 kmol O 2 0.0340b100gkmol C 2 H 6 3.5 kmol O 2
+
1 kmol CH 4
h
1 kmol C 2 H 6
0.0060b100gkmol C 3 H 8
5 kmol O 2
0.0050b100gkmol C 4 H 10 6.5 kmol O 2
+
h
1 kmol C 3 H 3
h
1 kmol C 4 H 10
= 207.0 kmol O 2 h
+
4- 60
4.67 (cont’d)
Air feed rate: n f =
207.0 kmol O 2
1 kmol air
1.17 kmol air fed
h
0.21 kmol O2
b.
n a = n f b2 x1 + 3.5 x2 + 5 x3 + 6.5 x4 gb1 + Pxs 100gb1 0.21g
c.
n& f = aR f , (n& f = 75.0 kmol / h, R f = 60) ⇒ n& f = 1.25 R f
kmol air req.
= 1153 kmol air h
n& a = bRa , ( n&a = 550 kmol / h, Ra = 25) ⇒ n&a = 22 .0 Ra
x i = kAi ⇒
∑x
i
=k
i
∑A
i
=1 ⇒ k =
i
1
∑A
i
i
⇒ xi =
Ai
, i = CH 4 , C 2 H 4 , C 3 H 8 , C 4 H 10
Ai
∑
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
n CO2 (mol CO2)
n H2O (mol H2O)
n C4H10 (mol C4H10)
n O2 (mol O2)
n N2 (mol N2)
Pxs (% excess air)
n air (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- 61
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 , n O2 = 0
U 73.1% N 2
n N2 = b0.79 gb3095 mol g = 2445 mol
|
n CO2 = b100 mol C 4 H 10 reactgb4 mol CO 2 mol C 4 H 10 g = 400 mol CO 2 V12.0% CO2
n H2O = b100 mol C 4 H10 reactgb5 mol H 2O mol C 4 H 10 g = 500 mol H 2 O |W14.9% H 2 O
ii) 100% conversion ⇒ n C4H10 = 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 ⇒ n C4H10 = 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:
10 mol C4H10
0.3% 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
n 1 (mol C3H8)
n 2 (mol H2)
n 3 (mol CO2)
n 4 (mol CO)
n 5 (mol H2O)
n 6 (mol O2)
n 7 (mol N2)
n 0 (mol air)
0.21 mol O2/mol
0.79 mol N2/mol
Theoretical oxygen:
75 mol C 3H 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- 62
4.69 (cont’d)
Air feed rate: n 0 =
387.5 mol O 2 1 kmol air 1.25 kmol air fed
= 2306.5 mol air
h
0.21 kmol O2 1 kmol air req'd.
90% propane conversion ⇒ n1 = 0.100(75 mol C3 H 8 ) = 7.5 mol C 3 H 8
(67.5 mol C 3 H 8 reacts)
85% hydrogen conversion ⇒ n 2 = 0150
. (25 mol C 3 H 8 ) = 3.75 mol H 2
95% CO 2 selectivity ⇒ n 3 =
0.95(67.5 mol C 3 H 8 react) 3 mol CO 2 generated
mol C3 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
= 10.1 mol CO
mol C 3 H 8 react
F
H balance: (75 mol C 3H 8 )G8
H
mol H I
J + ( 25 mol H 2 )(2)
mol C 3 H 8 K
= (7.5 mol C 3 H 8 )(8 ) + ( 3.75 mol H 2 )(2 ) + n 5 ( mol H 2 O)(2) ⇒ n5 = 291.2 mol H 2 O
mol O
) = (192.4 mol CO2 )( 2)
mol O 2
+ (10.1 mol CO)(1) + (2912
. mol H 2 O)(1) + 2 n6 ( mol O 2 ) ⇒ n6 = 1413
. mol O2
O balance: (0.21 × 2306.5 mol O 2 )(2
N 2 balance: n 7 = 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.
10.1 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- 63
4.70
a.
C5H12 + 8 O2 → 5 CO2 + 6 H2O
Basis: 100 moles dry product gas
n 1 (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
n 3 (mol H2O)
Excess air
n 2 (mol O2)
3.76n 2 (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.
n 1 (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
n 3 (mol H2O)
Excess air
n 2 (mol O2)
3.76n 2 (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 √
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
Fractional conversion of C5H12:
% excess air:
22.23 mol O 2 fed - 18.75 mol O 2 required
× 100 % = 18 .6% excess air
18.75 mol O2 required
4- 64
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
n&1 (mol O 2 / h)
3.76n&1 (mol N 2 / h)
4 unknowns ( n&1 , 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 + 2n 1 = 2627[0.0045 + 2(0.0903) + 0.0181 + 2(0.8871 - x)] + 569.6
N2 balance: 3.76 n& 1 = x ( 2627)
Solving simultaneously ⇒ n&1 = 574.3 mol O 2 / h, x = 0.822 mol N 2 / mol DG
Fractional conversion:
296 .6 − 2627 (0.0045 )
= 0.960 mol CH3 OH react/mol fed
296 .6
574 .3 − 444 .9
× 100 % = 29.1%
444 .9
569 .6 mol H 2 O
Mole fraction of water:
= 0.178 mol H 2 O/mol
(2627 + 569 .6 ) mol
% excess air:
4.72
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 xCH4 and xC 2 H 6 are
the mole fractions of methane and ethane in the fuel, then
ns bmol gx C 2 H6 bmol C 2 H 2 mol gb2 mol C 1 mol C2 H 6 g
ns bmol gx CH 4 bmol CH4 mol gb1 mol C 1 mol CH 4 g
=
20
85
E
x C 2 H6 bmol C 2 H 6 mol fuel g
xC H4 bmol CH 4 mol fuelg
= 0.1176 mole C 2H 6 mole CH 4 in fuel gas
4- 65
4.72 (cont’d)
g H 2Ogb1 mol 18.02 gg
mole H 2O
= 0.126
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).
Condensation measurement:
b1.134
CH 4 + 2O 2 → CO2 + 2H 2 O
7
C 2 H6 + O2 → 2CO2 + 3H 2O
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
C balance ⇒ n 2 ;
N 2 balance U
O balance
V⇒
W
n3 , x
H balance: 4 n1 + b6gb0.1176n1 g = b100gb0.126 gb2 g ⇒ n1 = 5.356 mol CH 4 in fuel
⇒ 0.1176(5.356) = 0.630 mol C2H6 in fuel
C balance: 5.356 + b2 gb0.630g + n2 = b100gb0.874gb0.119 g ⇒ n2 = 3.784 mol CO2 in fuel
Composition of fuel: 5.356 mol CH 4 , 0.630 mol C 2 H 6 , 3.784 mols CO2
⇒ 0.548 CH 4 , 0.064 C 2 H 6 , 0.388 CO 2
N 2 balance: 3.76n3 = b100gb0.874gx
O balance: b2 gb3.784 g + 2 n3 = b100gb0126
. g + b100gb0.874gb2 g 0119
.
+ b0.881 − xg
Solve simultaneously: n3 = 18.86 mols O2 fed , x = 0.813
5.356 mol CH 4 2 mol O2 0.630 mol C2 H 6 3.5 mol O2
Theoretical O2 :
+
1 mol CH 4
1 mol CH 4
= 12 .92 mol O2 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- 66
4.73
a.
C3H8 +5 O2 → 3 CO2 + 4 H2O, C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
Basis 100: mol product gas
n 1 (mol C3H8)
n 2 (mol C4H10)
100 mol
0.474 mol H2O/mol
x (mol CO2/mol)
(0.526–x) (mol O2/mol)
n 3 (mol O2)
Dry product gas contains 69.4% CO2 ⇒
x
69 .4
=
⇒ x = 0.365 mol CO2 /mol
0.526 − x 30 .6
3 unknowns (n 1, n 2, n 3) – 3 balances (C, H, O) = 0 D.F.
O balance: 2 n 3 = 152.6 ⇒ n 3 = 76.3 mol O2
C balance : 3 n 1 + 4 n 2 = 36.5 
n = 7.1 mol C 3 H8
⇒ 1
⇒ 65 .1 % C 3 H8 , 34.9% C 4 H10

H balance : 8 n1 + 10 n 2 = 94.8 n 2 = 3.8 mol C 4 H10
b.
n c=100 mol (0.365 mol CO2/mol)(1mol C/mol CO2) = 365 mol C
n h = 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 C3 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- 67
× 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
n 1 (kmol O2)
3.76 n 1 (kmol N2)
n 2 (kmol N2)
n 3 (kmol O2)
n 4 (kmol CO2)
(8/92) n 4 (kmol CO)
n 5 (kmol SO2)
n 6 (kmol H2O)
C + O2 → CO2
C + 1/2 O2 → CO
2H + 1/2 O2 → H2O
S + O2 → SO2
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 C
2 kmol H
1 kmol S
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 SO 2
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 + O 2 --> 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 CO 2 / h)
0.1 n3 (kmol CO / h)
n4 (kmol SO2 / h)
n5 (kmol H2O / h)
mo kg slag / h
Theoretical O 2:
C:
0.75b5000g kg C
1 kmol C
1 kmol O2
h
12.01 kg C
1 kmol C
4- 68
= 312.2 kmol O 2 h
4.75 (cont’d)
H:
S:
0.17b5000g kg H 1 kmol H 1 kmol H 2O
h
1.01 kg H
2 kmol H
0.02b5000g kg S
1 kmol S
1 kmol O2
h
32.06 kg S
1 kmol S
1 kmol O 2
2 kmol H2 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
O 2 fed = 0.21b3000g = 630 kmol O2 h
Excess air:
630 − 525.7
× 100% = 19.8% excess air
525.7
b. Balances:
1 kmol C
b0.94 gb0.75gb5000g kg C react
C:
= n& 3 + 0.1n&3
h
12 .01 kg C
⇒ n&3 = 266.8 kmol CO 2 h , 0.1n& 3 = 26.7 kmol CO h
H:
b0.17 gb5000g kg
H 1 kmol H 1 kmol H 2 O
h
1.01 kg H
2 kmol H
= n5 ⇒ n5 = 420.8 kmol H 2O h
3.1 kmol O 2 bfor SO2 g 1 kmol SO2
S:
(from part a)
N2 :
b0.79 gb3000g kmol
O:
b0.21g(3000)b2 g = 2 n&1
h
1 kmol O 2
. kmol SO 2 h
= n& 4 ⇒ n&4 = 31
N 2 h = n& 2 ⇒ n&2 = 2370 kmol N 2 h
+ 2b2668
. g + 1b26.68g + 2 b31
. g + b1gb420.8g
⇒ n&1 = 136.4 kmol O 2 / h
Stack gas total = 3223 kmol h
Mole fractions:
xC O = 26.7 3224 = 8.3 × 10 − 3 mol CO mol
xSO 2 = 31
. 3224 = 9.6 × 10 −4 mol SO 2 mol
c.
1
SO 2 + O2 → SO 3
2
SO 3 + H 2O → H2SO 4
3.1 kmol SO 2 1 kmol SO 3 1 kmol H2SO 4
h
1 kmol SO 2
1 kmol SO 3
4- 69
98.08 kg H 2SO 4
kmol H2SO 4
= 304 kg H2SO 4 h
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
b1.234
= 95.03 g air - dried coal; 4.97 g H 2 O lost by air drying
− 1.204 g g H 2 O
1.234 g a.d. c.
= 2.31 g H 2 O lost in second drying step
Total H 2 O = 4.97 g + 2 .31 g = 7.28 g moisture
95.03 g a. d. c
b1.347 −
0.811g g bv. m.+ H 2 O g
1.347 g a. d.c.
95.03 g a. d.c
0.111 g ash
1.175 g a. d.c.
− 2 .31 g H 2 O = 35.50 g volatile matter
= 8.98 g ash
Fixed carbon = b100 − 7.28 − 3550
. − 8.98gg = 48.24 g fixed carbon
7.28 g moisture
48.24 g fi xed carbon
35.50 g volatile matter ⇒
8.98 g ash
100 g coal as received
7.3% moisture
48.2% fixed carbon
35.5% volatile matter
9.0% ash
b. Assume volatile matter is all carbon and hydrogen.
C + CO 2 → CO 2 :
2H +
1 mol O 2
1 mol C
10 3 g
1 mol air
1 mol C
12.01 g C
1 kg
0.21 mol O 2
= 396.5 mol air kg C
1
0.5 mol O 2 1 mol H 10 3 g
1 mol air
O2 → H 2O :
= 1179 mol air kg H
2
2 mol H
1.01 g H 1 kg 0.21 mol O 2
Air required:
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
kg
7 kg v. m.
4- 70
kg H
= 3.72 × 105 mol air
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 + O 2 --> SO2
0.720 mol CO2 / mol
0.0257 mol CO / mol
0.000592 mol SO2 / mol
0.254 mol O2 / mol
n4 (mol O2)
(20% excess)
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
⇒ n 2 = 183.6 mol H, n 4 = 144.6 mol O2, n 5 = 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, N2, O 2, CO ,2H O2
(612.5 ppm SO )2
x n3 mol SO 2
(N2 , O 2, CO 2, H O)
2
100 g oil
0.87 g C/g
0.10 g H/g
0.03 g S/g
n1 mol O2
3.76 n1 mol N2
(25% excess)
0.10 (1 – x) n 5 mol SO 2
(N2 , O 2, CO 2, H O)
2
furnace
Alkaline solution
(1 – x) n5 mol SO 2
(N2 , O 2, CO 2, H O)
2
n2
n3
n4
n5
n6
CO 2:
H2 O:
0.87b100gg C
mol N 2
mol O 2
mol CO 2
mol SO 2
mol H 2O
1 mol C
12.01 g C
0.90 (1 – x) n5 mol SO 2
F 7.244 mol O 2 I
⇒ n 4 = 7.244 mol CO 2 G
J
1 mol C
Hconsumed
K
1 mol CO 2
F 2 .475 mol O 2 I
⇒ n6 = 4.95 mol H 2 OG
J
2 mol H
Hconsumed
K
0.10b100gg H 1 mol H 1 mol H 2 O
1.01 g H
scrubber
4- 71
4.78 (cont’d)
SO 2 :
0.03b100gg S
1 mol S
32.06 g S
F 0.0956 mol O 2 I
⇒ n5 = 0.0936 mol SO2 G
J
1 mol S
Hconsumed
K
1 mol SO 2
25% excess O 2 : n1 = 1.25b7.244 + 2.475 + 0.0936g ⇒ 12.27 mol O 2
O 2 balance: n 3 = 12.27 mol O 2 fed − b7.244 + 2.475 + 0.0936g mol O 2 consumed
= 2.46 mol O2
N 2 balance: n 2 = 3.76b12.27 molg = 46.14 mol N 2
SO 2 in stack bSO 2 balance around mixing point g:
x F0.0936I + 010
. b1 − x gb0.0936g = 0.00936 + 0.0842x bmol SO 2 g
H
n5
K
Total dry gas in stack (Assume no CO2 , O 2 , or N 2 is absorbed in the scrubber)
7.244 + 2.46 + 46.14 + b0.00936 + 0.0842 x g = 55.85 + 0.0842 x bmol dry gasg
bC O2 g
bO 2 g
bN 2 g
bSO2 g
612.5 ppm SO 2 bdry basisg in stack gas
0.00936 + 0.0842 x
612 .5
=
⇒ x = 0.295 ⇒ 30% bypassed
5585
. + 0.0842x
1.0 × 10 6
Basis: 100 mol stack gas
4.79
n 1 (mol C)
n 2 (mol H)
n 3 (mol S)
n 4 (mol O 2)
3.76 n 4 (mol O 2)
a.
C + O 2 → CO 2
1
2H + O 2 → H 2O
2
S + O 2 → SO 2
100 mol
0.7566 N 2
0.1024 CO 2
0.0827 H O
2
0.0575 O 2
0.000825 SO
2
C balance: n1 = b100gb0.1024 g = 10.24 mol C
10.24 mol C
mol C
⇒
= 0.62
H balance: n 2 = b100gb0.0827gb2g = 16.54 mol H
16.54 mol H
mol H
The C/H mole ratio of CH 4 is 0.25, and that of C2H 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.
S balance: n 3 = b100gb0.000825g = 0.0825 mol S
122.88
b10.24 mol Cgb12.0 g 1 mol g = 122.88 g CU
= 7.35 g C g H
|
16.71
⇒ No. 4 fuel oil
b16.54 mol H gb1.01 g 1 mol g = 16.71 g HV ⇒
2.65
|
×
100%
=
1
.9%
S
b0.0825 mol Sgb32.07 g 1 mol g = 2.65 g S W
142.24
4- 72
4.80
a.
Basis: 1 mol CpHqOr
1 mol CpHqO r
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 CO 2)
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 - Xs ) (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 ⇒ n 1 = (1 + P/100) (p +1/4 q – ½ r) mol O2 fed
(2)
C balance: n 2 = p
(3)
H balance: n 5 = q/2
(4)
S balance: n 3 = n 0
(5)
O balance: r + 2n 1 = 2n 2 + 2n 3 + 2n 4 + n 5 ⇒ n 4 = ½ (r+2n 1-2n 2-2n 3-n 5)
(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 SO 2
(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 SO 2
4- 73
CHAPTER FIVE
5.1
Assume volume additivity
Av. density (Eq. 5.1-1):
m
a.
A
& + m0 ⇒ m
& =
= mt
A
mass of tank
at time t
mass of
empty tank
& / min) =
⇒ V(L
b.
5.2
1
0.400
0.600
=
+
⇒ ρ = 0.719 kg L
ρ 0.703 kg L 0.730 kg L
A
A
ρO
ρD
b250 − 150gkg
b10 − 3g min
= 14.28 kg min
& =
bm
mass flow rate of liquid g
14.28 kg
1L
&
m(kg
/ min)
& =
⇒ V
= 19.9 L min
min
0.719 kg
ρ (kg / L)
& = 150 − 14.28b3g = 107 kg
m0 = m(t) - mt
void volume of bed: 100 cm3 − b2335
. − 184 gcm 3 = 50.5 cm 3
porosity: 50.5 cm3 void 184 cm3 total = 0.274 cm 3 void cm3 total
bulk density: 600 g 184 cm 3 = 3.26 g cm 3
absolute density: 600 g b184 − 50.5gcm 3 = 4 .49 g cm 3
5.3
C 6 H 6 ( l)
m
& B (kg / min)
& = 20.0 L / min
V
B
C 7 H 8 (l )
& (kg / min)
m
& (L / min)
V
m
& T (kg / min)
& (L / min)
V
T
2
2
& = ∆ V = πD ∆h = π (5.5 m) 0.15 m = 0.0594 m3 / min
V
∆t
4 ∆t
4
60 min
Assume additive volumes
&T = V
& -V
& B = b59.4 − 20.0g L / min = 39.4 L / min
V
& B + ρT ⋅ V
& T = 0.879 kg 20.0 L + 0.866 kg 39.4 L = 51.7 kg / min
& = ρB ⋅V
m
L
min
L
min
xB =
m
& B ( 0.879 kg / L)(20.0 L / min)
=
= 0.34 kg B / kg
m
&
(51.7 kg / min)
5-1
a.
1
x
b1 − x c g
= c +
⇒ check units!
ρ sl ρ 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
∆P
2775
c. i.) ρ sl =
=
= 1415 kg / m 3
gh b9.8066gb0.200g
b.
ii.)
F 1
1
x
b1 - x c g
1I F 1
1I
= c +
⇒ xc G − J = G − J
ρ sl ρ c
ρl
ρ l K H ρ sl ρ l K
Hρc
F
xc =
G
G1415
H
I
1
1
−
J
kg / m3 12
. d1000 kg / m 3 i JK
F
I
1
1
−
G
J
G2.3 1000 kg / m 3
1.2d1000 kg / m3 i JK
i
H d
iii.) Vsl =
= 0.316 kg crystals / kg slurry
m sl
175 kg
1000 L
=
= 123.8 L
ρ sl 1415 kg / m 3 m 3
iv.) mc = x c m sl = b0.316 kg crystals / kg slurry gb175 kg slurry g = 55.3 kg crystals
v.) mCuSO4 =
55.3 kg CuSO 4 ⋅ 5H 2O 1 kmol
1 kmol CuSO4
159 .6 kg
= 35.4 kg CuSO 4
249 kg 1 kmol CuSO4 ⋅ 5 H 2 O 1 kmol
vi.) ml = b1 − x c gm sl = b0.684 kg liquid / kg slurrygb175 kg slurry g = 120 kg liquid solution
vii.)
Vl =
h(m)
ρl(kg/m^3)
ρc(kg/m^3)
∆P(Pa)
xc
ρsl(kg/m^3)
0.2
1200
2300
2353.58
0
1200.00
ml
120 kg
1000 L
=
= 100 L
ρ l b1.2gd1000 kg / m3 i m 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
P1 = P0 + ρ slgh 1 U
F 1 N I F 1 Pa I
|
P2 = P0 + ρ slgh 2 V ⇒ ∆ P = P1 − P2 = ρ sl e mkg3 j gesm2 j hbm gG kg⋅m J G N J = ρ slgh
H1 s2 KH 1 m2 K
|
h = h1 − h 2
W
0.5
0.4
0.3
∆P = 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.
Basis: 1 kg slurry ⇒ x c bkg crystalsg, Vc dm 3 crystalsi =
b1 - x c gbkg liquid g,
ρ sl =
5.5
x c bkg crystalsg
Vl dm 3 liquid i =
ρ c dkg / m3 i
b1- x c gbkg
liquid g
ρ l dkg / m 3 i
1 kg
1
=
3
x c b1 − x c g
bVc + Vl gdm i
+
ρc
ρl
Assume Patm = 1 atm
3
$ = RT ⇒ V
$ = 0.08206 m ⋅ atm 313.2 K 1 kmol = 0.0064 m 3 mol
PV
kmol ⋅ K 4.0 atm 103 mol
ρ=
5.6
a.
1 mol
0.0064 m air
V=
1 kg
mol 10 3 g
= 4.5 kg m3
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
b3.06L - 2.8Lg
2.8L
× 100% = 9.3%
Assume Patm = 1.013 bar
a.
PV = nRT ⇒ n =
b.
kmol ⋅ K 28.02 kg N2
= 249 kg N 2
3
kmol
b25 + 273.2 gK 0.08314 m ⋅ bar
20.0 m 3
PV
nRT
T P n
=
⇒ n = V⋅ s ⋅ ⋅ s
Ps Vs n s RTs
T Ps Vs
n=
5.8
b10 + 1.013gbar
20.0 m 3
273K
298.2K
b10 + 1.013gbar
1.013 bar
1 kmol
28.02 kg N 2
= 249 kg N 2
kmol
22.415 m bSTP g
3
a.
R=
Ps Vs
1 atm 22.415 m 3
atm ⋅ m 3
=
= 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 o R
lb - mole ⋅o R
5-3
5.9
P = 1 atm +
10 cm H 2O
1m
1 atm
= 1.01 atm
10 cm 10.333 m H 2O
2
3
& = 2.0 m = 0.40 m 3 min = 400 L min
T = 25o C = 298.2 K , V
5 min
& = n& bmol / ming ⋅ MWbg / mol g
m
a.
& =
m
b.
m
& =
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 LbSTP g
F mI
J
HsK
5.10 Assume ideal gas behavior: uG
=
T P D 2 60.0 m 333.2K
u 2 = u1 2 1 12 =
T1 P2D 2
sec
300.2K
5.11 Assume ideal gas behavior: n =
& dm 3 si
V
Adm 2 i
g
mol
=
g
mol
= 458 g min
= 458 g min
&
&
nRT
P
u 2 nR
T2 P1 D12
⇒
=
⋅
⋅ ⋅
u1 nR
&
T1 P2 D22
π D2 4
(1.80 + 1.013 ) bar ( 7.50 cm )
(1.53 + 1.013) bar ( 5.00 cm )2
2
= 165 m sec
PV b1.00 + 1.00 g atm 5 L
=
= 0.406 mol
L⋅ atm
0.08206 mol
300 K
RT
⋅K
MW = 13.0 g 0.406 mol = 32.0 g mol ⇒ Oxygen
5.12 Assume ideal gas behavior: Say m t = mass of tank, n g = mol of gas in tank
N2 :
37.289 g = m t + n g b28.02 g molgU
n g = 0.009391 mol
|
V⇒
CO2 : 37.440 g = m t + n g b44.1 g mol g |W m t = 37.0256 g
unknown: MW =
5.13 a.
b.
b37.062
− 37.0256gg
0.009391 mol
= 3.9 g mol ⇒ Helium
3
3
∆V
&Vstd cm3 bSTPg min = ∆ V liters 273K 763 mm Hg 10 cm = 9253
.
∆t
∆ t min 296.2K 760 mm Hg
1L
φ
& cm3 bSTPg min
V
std
5.0
9 .0
12.0
139
268
370
U
|
|
V straight line plot
E
|
& std + 0.93
φ
=
0
.
031
V
|
W
3
3
& std = 0.010 mol N 2 22.4 litersbSTP g 10 cm = 224 cm3 / min
V
min
1 mole
1L
φ = 0.031d224 cm 3 / mini + 0.93 = 7 .9
5-4
nbkmolgM(kg / kmol)
5.14 Assume ideal gas behavior ρbkg Lg =
V bLg
n P
=
V RT
PM
====> RT
12
V2 dcm si = V1dcm
3
3
Fρ I
si ⋅ G 1 J
Hρ 2 K
= V1 P1M 1T2 P2 M 2 T1
12
12
cm3 L 758 mm Hg 28.02 g mol 323.2K O
3
= 350
M
P = 881 cm s
s N1800 mm Hg 2.02 g mol 298.2K Q
a.
VH 2
b.
M = 0.25MCH4 + 0.75MC3 H8 = b0.25gb16.05g + b0.75gb44.11g = 37 .10 g mol
12
cm 3 L b758 gb28.02 gb323.2 g O
Vg = 350
M
P
s Nb1800gb37.10 gb2982
. gQ
= 205 cm3 s
5.15 a.
Reactor
∆h
soap
b.
n& CO2
2
&
PV
πR 2∆ h π d0.012 m i
&
=
⇒ V=
=
RT
∆t
4
n& C O2 =
2
1.2 m 60 s
= 11
. × 10− 3 m 3 / min
7.4 s min
755 mm Hg
1 atm
1.1 ×10-3 m 3 /min 1000 mol
= 0.044 mol/min
3
⋅atm
0.08206 mkmolK
760 mm Hg
300 K
1 kmol
⋅
5.16
& air = 10.0 kg / h
m
n& air (kmol / h)
n& (kmol / h)
yCO 2 (kmol CO 2 / kmol)
3
&
V
CO2 = 20.0 m / h
n& CO (kmol / h)
o
150 C, 1.5 bar
2
Assume ideal gas behavior
10.0 kg 1 kmol
n& air =
= 0.345 kmol air / h
h
29.0 kg air
n& CO2 =
&
PV
1.5 bar
100 kPa 20.0 m 3 / h
=
= 0.853 kmol CO 2 / h
3
RT 8.314 mkmol⋅kPa
1
bar
423.2
K
⋅K
y CO2 × 100% =
0.853 kmol CO 2 / h
× 100% = 71.2%
b0.853 kmol CO 2 / h + 0.345 kmol air / hg
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
o
167 C, - 40 cm H 2O gauge
m
& 4 (kg S / min)
a.
n& 3 =
1 atm
311 m3
356.2K
min
kmol ⋅ K
0.08206 m 3 ⋅ atm
H 2O balance: 0.70 m1 =
= 10.64 kmol min
10.64 kmol 0.12 kmol H 2O 18.02 kg
kmol
kmol
min
& 1 = 32.9 kg min milk
⇒m
&4⇒m
& 4 = 9.6 kg S min
Sbolids g balance: 0.30b32.2 kg min g = 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 m 3 ⋅ atm
kmol ⋅ K
min
440K
(1033 − 40 ) cm H2O
1033 cm H 2O
1 atm
= 352 m air min
3
u air (m/min)=
& air (m3 /s) 352 m3 1 min
V
=
A (m2 )
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 CO2 =
5.19 a.
ρ CO2
ρ air
=
x CO2 = 0.75
PM CO 2
RT
PM air
RT
=
M CO2
M air
=
44 kg / kmol
= 152
.
29 kg / kmol
x air = 1 − 0.75 = 0.25
Since air is 21% O 2 , x O2 = (0.25)(0.21) = 0.0525 = 5.25 mole% O2
b.
mC O2 = n ⋅ x CO2 ⋅ M CO2 =
3
1 atm
b2 × 1.5 × 3gm 0.75 kmol CO2 44.01 kg CO 2
=12 kg
m3 ⋅atm
298.2 K
kmol
kmol CO 2
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 CO2 =
15.7 kg
1 kmol
= 0.357 kmol CO2
44.01 kg
Assume ideal gas behavior, negligible temperature change bT = 19° C = 292.2 Kg
a.
P1V
n 1RT
n1
P
102kPa
=
⇒
= 1 =
P2 V bn 1 + 0.357gRT
n 1 + 0.357 P2 3.27 × 103 kPa
⇒ n1 = 0.0115 kmol air in tank
b.
Vtank =
n1RT 0.0115 kmol 292.2 K 8.314 m3 ⋅ kPa 10 3L
=
= 274 L
P1
102kPa
kmol ⋅ K
m3
15700 g CO2 + 11.5 mol air ⋅ (29.0 g air / mol)
= 58.5 g / L
274 L
c. 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.
ρf =
5.21 At point of entry, P1 = b10 ft H 2 Ogb29.9 in. Hg 33.9 ft H 2Og + 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 (1.2 )(1.00 g / cm ) (100
. g / cm 3 )
= 0.967
cm 3
1.03 g 2.20 lb 5 × 10 −4 ton 106 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
−3
o
hr
4.3 × 10 ton 1000 gal
492 R 37 .1 in Hg
b.
4
π D2
P2 V2 nRT
V2 P1
3 e 2 j
37.1
=
⇒
=
⇒
=
⇒ D32 = 1.31D13
3
D1
P1V1 nRT
V1 P2
28
.
3
4
π
3 e2 j
3
% change =
b2.2 - 2.0g mm
2.0 mm
× 100 = 10%
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 = 306 K
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 = 306 K
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
P3V3 = n 3RT3 (3 ≡ final condition in lab): P3 = 400 kPa, T3 = 296K
n3 =
400 kPa 2 L
mol ⋅ K
= 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
× 10 6 =
× 106 = 0.950 × 10 6 ⋅ m BGC
n1
0.078
9 am: y B = (0.950 × 10 6 )(0.656 × 10− 6 ) = 0.623 ppm U
1 pm:
5 pm:
|
|
y B = ( 0.950 × 10 )(0.788 × 10 ) = 0.749 ppm VThe
|
y B = ( 0.950 × 10 6 )(0.910 × 10 −6 ) = 0.864 ppm|
W
6
−6
avg. is below the PEL
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
4
3
5.23 Volume of balloon = π b10 m g = 4189 m 3
3
Moles of gas in balloon
nbkmolg =
a.
4189 m 3 492° R 3 atm
1 kmol
535° R 1 atm 22.4 m 3 bSTP g
= 5159
. kmol
He in balloon:
m = b5159
. kmolg⋅ b4.003 kg kmolg = 2065 kg He
mg =
b.
2065 kg 9.807 m
1N
= 20,250 N
2
s 1 kg ⋅ m / s2
dPgas in balloon i V
dPair displaced i V
= n gasRT
= n air RT
Fbuoyant
⇒ n 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 1 N
= 48,920 N
1 kmol
s2 1 kg2⋅m
s
Since balloon is stationary,
Wtotal
Fcable
∑F = 0
1
Fcable = Fbuoyant − Wtotal = 48920 N −
b2065 + 150 gkg
9.807 m 1 N
= 27,200
s2 1 kg⋅2m
s
c. When cable is released, Fnet dA i = 27200 N = M tota
⇒a=
27200 N
1 kg ⋅ m / s2
b2065 + 150gkg
N
= 12 .3 m s2
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
3
PN VN 5.7 atm 400 m / h
3
a. PN VN = Pc Vc ⇒ Vc =
=
= 240 m h
9.5 atm
Pc
b.
Mass flow rate before diversion:
400 m 3 273 K 5.7 atm
1 kmol
44.09 kg
kg C 3 H 6
= 4043
3
h
303 K 1 atm 22.4 m ( STP )
kmol
h
5- 9
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 m 3
hr
273 K
303 K
2.8 atm
1 atm
1 kmol
22.4 m
3
44.09 kg
kmol
= 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
F 4.004 gI
0.35 mol He G
J = 1.40 g He
H mol K
0.20 mol
0.45 mol
U
|
|
|
F 16.05 gI
CH 4 G
J = 3.21 g CH 4 V17.22
H mol K
|
|
F 28.02 gI
N2 G
J = 12 .61 g N 2 |
H mol K
W
g ⇒ mass fraction CH 4 =
3.21 g
= 0.186
17.22 g
g of gas
= 17.2 g / mol
mol
m ndMWi P dMWi b2.00 atm gb17.2 kg / kmolg
= =
=
=
= 115
. kg / m 3
3
m ⋅atm
V
V
RT
0.08206
b
363
.
2
K
g
e
kmol ⋅K j
c.
MW =
d.
ρ gas
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
C3 H8 mole fraction can drop below the UFL, thereby producing a fire hazard.
b.
fuel-air mixture
n& 1 ( mol / s)
y C3 H8 = 0.0403 mol C3 H8 / mol
n& C3 H8 = 150 mol C3H 8 / s
n& 3 (mol / s)
0.0205 mol C3 H 8 / mol
diluting air
n& 2 ( mol / s)
n& 1 =
150 mol C 3H8
mol
= 3722 mol / s
s
0.0403 mol C 3H8
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.
n& 2 = 1.3bn& 2 gmin = 4674 mol / s
3
U
& = 4674 mol / s 8.314 m ⋅ Pa 398.2 K = 118 m 3 / s|
V
2
&
mol ⋅ K 131,000 Pa
m 3 diluting air
| V
2
=
141
.
V &
3
V
m 3 fuel gas
&V = 3722 mol 8.314 m ⋅ Pa 298.2 K = 83.9 m 3 / s | 1
1
|
s
mol ⋅ K 110000 Pa
W
y2 =
150 mol / s
150 mol / s
=
× 100% = 1.8%
n& 1 + &n 2
3722
mol
/ s + 4674 mol / sg
b
d. The incoming propane mixture could be higher than 4.03%.
If n& 2 = bn& 2 gmin , fluctuations in the air flow rate would lead to temporary explosive
conditions.
5.27
Basis: (12 breaths min )( 500 mL air inhaled breath ) = 6000 mL inhaled min
o
24 C, 1 atm
6000 mL / min
lungs
n& in (mol / min)
0.206 O 2
0.774 N 2
0.020 H 2O
a.
n& in =
blood
37o C, 1 atm
n& out (mol / min)
0.151 O 2
0.037 CO2
0.750 N 2
0.062 H 2 O
6000 mL
1L
273K
1 mol
= 0.246 mol min
3
min
10 mL 297K 22.4 L bSTPg
N 2 balance: b0.774gb0.246g = 0.750 &n out ⇒ n& out = 0.254 mol exhaled min
O 2 transferred to blood:
b0.246gb0.206g − b0.254 gb0.151g bmol
O2 min g 32.0 g mol
= 0.394 g O2 min
CO2 transferred from blood:
b0.254 gb0.037g bmol CO2
min g 44.01 g mol
= 0.414 g CO2 min
H2 O transferred from blood:
b0.254 gb0.062g − b0.246gb0.020g bmol
= 0.195 g H 2 O min
5- 11
H 2O min g 18.02 g mol
5.27 (cont’d)
PVin
n RT
= in in
PVout n out RTout
⇒
b.
Vout F n out I F Tout I F 0.254 mol min I F 310KI
=
J = 1.078 mL exhaled ml inhaled
JG
J =G
JG
Vin G
H n in KH Tin K H0.246 mol min KH 297KK
b0.414
g CO2 lost min g + b0.195 g H 2O lost ming − b0.394 g O 2 gained min g = 0.215 g min
STACK
5.28
Ts (K)
M s (g/mol)
Ps (Pa)
LL(m)
( M)
PM
RT
Ideal gas: ρ =
a.
Ta (K)
Ma (g/mol)
Pc (Pa)
D = bρgLgcombust. − bρgLgstack =
Pa M a
PM
P gL L M
M O
gL − a s gL = a M a − a P
RTa
RTs
R N Ta
Ts Q
b. M s = b0.18gb44.1g + b0.02 gb32 .0g + b0.80 gb28.0g = 31.0 g mol , Ts = 655K ,
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
kmol - K
2
760 mm Hg
s
0.08206 m3 − atm
I
. kg kmol O F
1N
323 N 1033 cm H2 O
L 29.0 kg kmol 310
×M
−
×G
=
2J
P
294K
655K
m 2 1.013 × 105 N m 2
N
Q H1 kg ⋅ m / s K
= 3.3 cm H 2 O
5.29 a.
MWCCl 2 O = 98.91 g /mol
ρ CCl 2 O 98.91
PbMWg
=
= 3.41
RT
ρ air
29.0
Phosgene, which is 3.41 times more dense than air, will displace air near the ground.
=======>
ρ=
2
b.
Vtube =
π bDin g L π
2
= 0.635 cm - 2b0.0559 cmg b15.0 cmg = 3.22 cm 3
4
4
m CCl2 O = Vtube ⋅ ρ CCl2 O =
c.
n CCl 2 O(l) =
3.22 cm 3
1L
1 atm
98.91 g / mol
= 0.0131 g
3
3
L⋅atm
10 cm 0.08206 mol⋅K
296.2 K
3.22 cm 3 1.37 × 1.000 g mol
= 0.0446 mol CCl 2 O
cm 3 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 CCl2 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 H6 + O2 → 2 CO 2 + 3H 2O
2
C 3H 8 + 5O 2 → 3CO 2 + 4 H 2 O
3
o
1450 m / h @ 15 C, 150 kPa
n& 1 (kmol / h)
0.86 CH 4 , 0.08 C2 H 6 , 0.06 C 3 H 8
n& 2 (kmol air / h)
8% excess, 0.21 O 2 , 0.79 N2
n& 1 =
1450 m 3 273.2K
h
288.2K
b101.3 + 150gkPa
1 kmol
101.3 kPa
22.4 m 3 bSTPg
= 152 kmol h
Theoretical O 2:
F 2 kmol O 2 I
F 3.5 kmol O 2 I
F 5 kmol O2 I O
152 kmol L
M0.86G
J + 0.08G
J + 0.06G
J P = 349.6 kmol h O2
h
H kmol CH 4 K
H kmol C 2 H 6 K
H kmol C3 H 8 K P
M
N
Q
& air = 1.08b349.6g kmol O2
Air flow: V
h
1 kmol Air
22.4 m 3bSTP g
0.21 kmol O 2
kmol
5- 13
= 4.0 × 104 m 3 bSTPg h
5.31 Calibration formulas
bT =
25.0; RT = 14 g , bT = 35.0, R T = 27 g ⇒ Tb° Cg = 0.77R T + 14.2
dPg
= 0; R p = 0i , dPg = 20.0, R r = 6 i ⇒ Pgauge bkPag = 3.33R p
&F
dV
& F = 2.0 × 103 , R F = 10i ⇒ V
& F dm3 h i = 200R F
= 0; R p = 0i , dV
&A
dV
& A = 1.0 × 10 5 , RA = 25i ⇒ V
& A dm 3 h i = 4000R A
= 0; R A = 0h , dV
& (m 3 / h), T, P
V
F
g
CH 4 + 2O 2 → CO2 + 2 H 2 O
7
C 2 H 6 + O 2 → 2CO2 + 3H 2 O
2
C 3 H 8 + 5O 2 → 3CO2 + 4H 2 O
13
C 4 H 10 + O 2 → 4CO 2 + 5H 2 O
2
n& F (kmol / h)
xA (mol CH 4 / mol)
xB (mol C 2 H 6 / mol)
xC (mol C 3 H8 / mol)
xD (mol n - C 4 H 10 / mol)
xE (mol i - C 4 H10 / mol)
& ( m3 / h) (STP)
V
A
& A (kmol / h)
n
0.21 mol O2 / mol
0.79 mol N2 / mol
n& F =
& F dm 3 h i
V
=
273.2K
bT + 273.2 gK
dPg
+ 101.3i kPa
101.3 kPa
1 kmol
22.4 m 3 bSTP g
& F dPg + 101.3i kmol
0.12031V
F
I
G
J
H
h K
bT + 273g
Theoretical O 2:
& o2 i
dn
Th
= n& F c2x A + 3.5x B + 5x C + 6.5bx D + x E ghkmol O 2 req. h
n&
kmol O 2 req.
1 kmol air
Air feed: n& A = d o2 i Th
h
0.21 kmol O 2
F
= 4.762G1 +
H
b1 +
Px 100gkmol feed
1 kmol req.
Px I
J dn o i
100 K 2 Th
& A = n& a bkmol air hg d22.4 m 3 bSTPg kmol i = 22.4n& A m3 bSTPg h
V
RT T(C) Rp Pg(kPa) Rf xa xb xc xd xe PX(%) nF nO2, th nA
Vf(m3/h) Va (m3/h) Ra
23.1 32.0 7.5
25.0 7.25 0.81 0.08 0.05 0.04 0.02
15 72.2 183.47 1004.74
1450 22506.2 5.63
7.5 20.0 19.3
64.3 5.8 0.58 0.31 0.06 0.05 0.00
23 78.9 226.4 1325.8
1160 29697.8 7.42
46.5 50.0 15.8
52.6 2.45 0.00 0.00 0.65 0.25 0.10
33 28.1 155.2 983.1
490 22022.3 5.51
21 30.4
3
10.0
6 0.02 0.4 0.35 0.1 0.13
15 53.0 248.1 1358.9
1200 30439.2 7.6
23 31.9
4
13.3
7 0.45 0.12 0.23 0.16 0.04
15 63.3 238.7 1307.3
1400 29283.4 7.3
25 33.5
5
16.7
9 0.5 0.3 0.1 0.04 0.06
15 83.4 266.7 1460.8
1800 32721.2 8.2
27 35.0
6
20.0 10 0.5 0.3 0.1 0.04 0.06
15 94.8 303.2 1660.6
2000 37196.7 9.3
5- 14
5.32 NO + 12 O2 ⇔ NO2
1 mol
0.20 mol NO / mol
0.80 mol air / mol U
R
|
|
0.21 O 2
S
V
|
|
0.79 N 2
T
W
P0 = 380 kPa
a.
n1 (mol NO)
n2 (mol O 2 )
n3 (mol N2 )
n4 (mol NO2 )
Pf (kPa)
Basis: 1.0 mol feed
90% NO conversion: n1 = 0.10 (0.20) = 0.020 mol NO ⇒ NO reacted = 0.18 mol
0.18 mol NO 0.5 mol O 2
= 0.0780 mol O2
mol NO
N2 balance: n 3 = 0.80(0.79 ) = 0.632 mol N 2
O 2 balance: n 2 = 0.80(0.21) −
n4 =
y NO
0.18 mol NO 1 mol NO2
= 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 O2 = 0.086
mol O2
mol N 2
mol NO2
y N2 = 0.695
y NO2 = 0.198
mol
mol
mol
Pf V n f RT
n
F 0.91 mol I
=
⇒ Pf = P0 f = 380 kPa G
J = 346 kPa
H 1 mol K
P0 V n 0RT
n0
b.
Pf
360 kPa
= (1 mol)
= 0.95 mol
P0
380 kPa
n i = n i0 + υ iξ
nf = n0
E
n1 (mol NO) = 0.20 − ξ
n 2 ( mol O2 ) = ( 0.21)(0.80 ) − 0.5ξ
n 3 (mol N 2 ) = (0.79 )(0.80)
n 4 ( mol NO2 ) = ξ
n f = 1 − 0.5ξ = 0.95 ⇒ ξ = 010
.
⇒ n 1 = 0.10 mol NO , n 2 = 0.118 mol O 2 , n 3 = 0.632 mol N 2 ,
n 4 = 0.10 mol NO 2 ⇒ y NO = 0.105, y O2 = 0.124, y N 2 = 0.665, y NO2 = 0.105
NO conversion =
P (atm) =
Kp =
b0.20- n1 g
0.20
× 100% = 50%
360 kPa
= 3.55 atm
101.3 kPa
atm
(y NO 2 P )
( y NO P)( y O2 P )
0.5
=
(y NO2 )
0.5
( y NO )( y O2 ) P
0.5
5- 15
=
0.105
0.5
(0.105)b0.124 g
1
0.5
b3.55g
= 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 C6 H 6 fed
3
o
V1 (m ) @ 40 C, 120 kPa
n1 (kmol)
0.920 HCl
0.080 Cl2
1 kmol C6 H 6 ( 78.12 kg)
n 0 (kmol Cl2 )
n 2 (kmol)
0.298 C6 H 6
0.481 C 6 H5Cl
0.221 C 6 H4 Cl 2
C 6H 6 + Cl2 → C 6H 5Cl + HCl
C balance:
1 kmol C6 H 6
C 6H 5Cl + Cl 2 → C6 H 5Cl 2 + HCl
6 kmol C
= n 2 0.298 × 6 + 0.481 × 6 + 0.221 × 6
1 kmol C6 H 6
⇒ n 2 = 100
. kmol
H balance:
1 kmol C6 H6
6 kmol H
= n 1b0.920 g(1)
1 kmol C6 H6
+ n 2 0.298 × 6 + 0.481 × 5 + 0.221 × 4 ⇒ n1 = 1.00 kmol
V1 =
n1RT 1.00 kmol 101.3 kPa 0.08206 m3 ⋅ atm 313.2 K
=
= 21.7 m3
P
120 kPa
1 atm
kmol ⋅ K
⇒
b.
V1
217
. m3
=
= 0.278 m 3 / kg B
m B 78.12 kg B
2
&
& ( m3 / s) = u(m / s) ⋅ A(m 2 ) = u ⋅ πd ⇒ d 2 = 4 ⋅ Vgas
V
gas
4
π ⋅u
d2 =
& B0 (kg B) 0.278 m3
4m
s 1 min 10 4 cm 2
= 5.90 m
& B0 (cm2 )
min
kg B π (10) m 60 s
m2
1
& B0 g 2
⇒ d(cm) = 2.43⋅ bm
c. Decreased use of chlorinated products, especially solvents.
5- 16
5.34
Vb ( m 3 / min) @900o C, 604 mtorr
60% DCS conversion
n& 1 (mol DCS / min) U
n& 2 (mol N 2 O / min) |
& b (mol / min)
Vn
n& 3 (mol N 2 / min)
|
n& 4 (mol HCl(g) / min) W
3.74 SCMM
& a ( mol / min)
n
0.220 DCS
0.780 N 2 O
SiH 2Cl 2(g) + 2 N 2O(g) → SiO2(s) + 2 N 2(g) + 2 HCl (g)
a.
n& a =
3.74 m 3 (STP)
10 3 mol
=167 mol / min
min
22.4 m3 (STP)
F 0.220 mol DCS I
60% conversion: n& 1 = b1 - 0.60gG
J b167 mol / ming = 14.7 mol DCS / min
H
K
mol
mol DCS
DCS reacted: b0.60gb0.220gb167 g
= 22.04 mol DCS reacted / min
min
mol N 2O
min
22.04 mol DCS 2 mol N 2O
−
= 86.18 mol N 2O / min
min
mol DCS
N2 O balance: &n 2 = 0.780b167g
N2 balance: n& 3 =
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
min
mol DCS
n& B = n& 1 + n& 2 + n& 3 + n& 4 = 189 mol / min
& =
⇒V
B
n& B RT 189 mol 62.36 L ⋅ torr 0.001 m3 1173 K
=
= 2.29 × 104 m 3 / min
P
min
mol ⋅ K
L
0.604 torr
n&
14.7 mol DCS/min
⋅ P= 1 P=
⋅ 604 mtorr=47.0 mtorr
DCS
DCS
n&
189 mol/min
B
n&
86.2 mol N 2O/min
p N2 O = x N 2O ⋅ P= 2 P=
⋅ 604 mtorr=275.5 mtorr
&n B
189 mol/min
b. p
=x
r = 3.16 × 10−8 p DCS ⋅ p N2 O0.65 = 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 SiO2 60 s 120 min 60.09 g/mol 1010 A
&
c. h(A)=r
⋅t ⋅
=
ρSiO2
min
m2 ⋅ s
2.25 × 10 6 g/m 3 1 m
(Table B.1)
&
=1.1 ×105 A
The films will be thicker closer to the entrance where the lower conversion yields higher
pDCS and p N 2O 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 )
& (m3 )
V
2
n2 (kmol air)
0.21 O 2
0.79 N 2
R100 kmol
| 0.105 CO 2
S0.053 O
| 0.842 N2
T
2
o
n 3 (kmol H2 O)
30 C, 98 kPa
a.
dry gasU
|
V
|
W
N2 balance: 0.79n 2 = 0.842 (100) ⇒ n 2 = 106.6 kmol air
O balance: 2b0.21n 2 g = 100 2b0.105g + 2b0.053g + n 3 ⇒ n 3 = 1317
. kmol H 2O
C balance:
n 1 dkmol C x H y i
x bkmol Cg
dkmol C x H y i
= 100b0105
. g ⇒ n 1x = 10.5
n 3 =13.17
H balance: n 1y = 2n 3 ====> n1y = 26.34
b1g
b2g
y 26.34
=
= 2 .51 mol H / mol C
x 10.5
O 2 fed: 0.21b106.6 kmol air g = 22.4 kmol
Divide b2g by b1g ⇒
O 2 in excess = 5.3 kmol ⇒ Theoretical O2 = b22.4 - 5.3g kmol = 17.1 kmol
% excess =
b.
V2 =
m1 =
5.3 kmol O2
× 100% = 31% excess air
17.1 kmol O 2
106.6 kmol N 2 22.4 m 3 (STP) 101.3 kPa 303 K
= 2740 m 3
kmol
98 kPa 273 K
n1x bkmol Cg 12.0 kg
kmol
+
n1y bkmol H g 101
. kg
kmol
n 1x=10.5
=====> m 1 = 152.6 kg
n 1y =26.34
V2
2740 m3 air
m3 air
=
= 18.0
m1 152 .6 kg fuel
kg fuel
5.36 3N 2 H 4 → 6 xH 2 + (1 + 2 x)N 2 + (4 − 4 x)NH3
a. 0 ≤ x ≤ 1
b.
n N 2 H4 =
50 L 0.82 kg 1 kmol
= 1.28 kmol
L
32.06 kg
L 6 x kmol H 2
b1 + 2 xg kmol N 2
b4 − 4 xg kmol NH3 O
n product = 1.28 kmol N2 H 4 M
+
+
P
3 kmol N 2 H4
3 kmol N2 H 4 Q
N3 kmol N 2H 4
=
1.28
b6x + 1 + 2x + 4 − 4x g = 1.707 x + 2.13 kmol
3
5- 18
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
3
& cm3 / hh
V
air
C A (g A / m )
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.
F kg A I
J
H h K in
F kg A I
J
H h Kout
=m
& AG
b.
m
& AG
c.
CA e m3 j ⋅ V
mol A
yA =
=
gA
mol air MA e mol
j ⋅ nair
d.
yA = 50 × 10−6
3
& F m I C FGkg A IJ
=V
air G
J A
H m3 K
H h K
gA
===================>
m
&
PV
CA = A ; nair =
k⋅Vair
RT
yA =
& A RT
m
k ⋅ Vair M A P
m
& A = 90
. g/h
3
&
dVair h
min
&
8.314 mmol⋅Pa
m
RT
9.0 g / h
293 K
⋅K
= A
=
= 83 m3 / h
3
−6
kyA MA P 0.5d50 × 10 i 101.3 × 10 Pa 104.14 g / mol
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
C 3H 6 + H 2 ⇔ C3H 8
5.38 Basis: 2 mol feed gas
U
np (mol C3 H8 )
|
(1- np )(mol C 3 H 6 ) V n 2 = n p + 2(1 − n p ) = 2 − n p
(1- np )(mol H 2 ) |W
1 mol C3 H 6
1 mol H 2
25o C, 32 atm
235o C, P2
a. At completion, n p = 1 mol , n 2 = 2 − 1 = 1 mol
1 mol 508K
P2 V n 2 RT2
n T
=
⇒ P2 = 2 2 P1 =
2 mol 298K
P1V n 1RT1
n 1 T1
b.
32.0 atm
= 27.3 atm
P2 = 35.1 atm
n2 =
35.1 atm 298K
P2 T1
n1 =
32.0 atm 508K
P1 T2
2 mol
= 1.29 mol
1.29 = 2 − n p ⇒ n p = 0.71 mol C3H 8 produced
⇒ b1 - 0.71g = 0.29 mol C 3H 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
60
%conv.
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 ⇒
95 g CH4 b1 mol 16.04 gg = 5.92 mol CH 4 U 97.2 mol % CH 4
V⇒
5 g C 2 H6 b1 mol 30.07 gg = 017
. mol C2 H 6 W 2.8 mol % C 2 H6
500 m3 / h
n& 2 (kmol CO2 / h)
n& 3 (kmol H2 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
40 o C, 1.1 bar
& (SCMH)
V
air
25% excess air
n& 1 =
&
P1V
11
. bar 500 m3
1
=
RT1 313K
h
kmol ⋅ K
0.08314 m3 ⋅ bar
7
C2 H 6 + O 2 → 2 CO2 + 3H 2O
2
CH 4 + 2O 2 → CO 2 + 2 H2 O
Theoretical O 2 =
21.1 kmol L0.972 kmol CH 4
M
kmol
h
N
+
Air Feed:
= 211
. kmol h
0.028 kmol C 2 H 6
kmol
2 kmol O 2
1 kmol CH 4
3.5 kmol O 2
1 kmol
O
C 2 H 6 PQ
= 43.1
125
. b431
. kmol O2 g
1 kmol Air
22.4 m3 bSTPg
h
0.21 kmol O 2
1 kmol
kmol O 2
h
= 5700 SCMH
5.40 Basis: 1 m3 gas fed @ 205° C, 1.1 bars Ac = acetone
1 m 3 @205 o C, 1.1 bar
n1 (kmol)
y1 (kmol Ac / kmol)
(1 - y1 )(kmol air / kmol)
pAC = 0100
.
bar
n3 (kmol), 10o C, 40 bar
condenser
y 3 (kmol Ac / kmol)
(1- y 3 )(kmol air / kmol)
p AC = 0.379 bar
n 2 (kmol Ac(l))
a.
n1 =
y1 =
100
. m 3 273K
110
. bars
1 kmol
478K 10132
.
bars 22.4 m3bSTP g
= 0.0277 kmol
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
Air balance: b0.0277 gb0.910 g = (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
1 kmol Ac
5- 21
= 0.133 kg acetone condensed
5.40 (cont’d)
Product gas volume =
b.
20.0 m 3 effluent
0.0254 kmol 22.4 m 3 bSTPg 283K 1.0132 bars
273K
0.0277 kmol feed
0.0909 kmol Ac 58.08 kg Ac
3
h
0.0149 m effluent
5.41 Basis: 1.00 × 10 6 gal. wastewater day.
kmol feed
kmol Ac
= 196 kg Ac h
Neglect evaporation of water.
1.00 × 106 gal / day
Effluent gas: 68o F, 21.3 psia(assume)
n1 (lb - moles H2 O / day)
0.03n1 (lb - moles NH3 / day)
a.
40.0 bars
= 0.0149 m 3
n2 (lb - moles air / day)
n3 (lb - moles NH3 / day)
300 × 106 ft 3 air / day
Effluent liquid
68o F, 21.3 psia
n 2 (lb-moles air/day)
n1 (lb - moles H2 O / day)
n4 (lb - moles NH 3 / day)
Density of wastewater: Assume ρ = 62.4 lb m ft 3
Ln1
M
N
lb - moles H 2O 18.02 lbm 0.03 n1 lb m NH3 17.03 lbm O
1 ft 3
+
P×
day
1 lb - mole
day
1 lb - mole Q 62.4 lb m
= 1.00 × 10 6
7.4805 gal
1 ft 3
gal
day
⇒ n 1 = 450000 lb - moles H 2 O fed day , 0.03n 1 = 13500 lb - moles NH 3 fed day
n2 =
300 × 106 ft 3
day
492 o R
21.3 psi
o
527.7 R 14.7 psi
1 lb - mole
359 ft 3bSTP g
= 113
. × 106 lb - moles air day
93% stripping: n 3 = 0.93 × 13500 lb - moles NH 3 fed day = 12555 lb - moles NH 3 day
Volumetric flow rate of effluent gas
PVout n out RT
n
300 × 10 6 ft 3
=
⇒ Vout = Vin out =
PVin
n in RT
n in
day
d1.13 × 10
6
+ 12555i lb - moles day
1.13 × 106 lb - moles day
= 303 × 10 6 ft 3 day
Partial pressure of NH 3 = y NH 3 P =
12555 lb - moles NH 3 day
d1.129
× 10 6 + 12555i lb - moles day
= 0.234 psi
5- 22
× 21.3 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 @ 23o C, 510 mm H 2O
n& 1 (mol / min)
y (mol Cl2 / mol)
(1 - y)(mol air / mol)
n 2 (mol air / mol)
0.013 mol Cl 2 / min
Assume Patm = 10.33 m H 2O ⇒ bPabs gin = b10.33 + 0.510g m H 2 O = 10.84 m H2 O
n& 1 =
2L
min
273K 10.84 m H 2O
1 mol
= 0.0864 mol min
296K 10.33 m H 2O 22.4 LbSTP g
Cl balance: 0.0864y = 0.013 ⇒ y = 0150
.
5.43
mol Cl 2
,∴ specification is wrong
mol
& (L / min) @ 65o C, 1 atm
V
3
125 L / min @ 25o C, 105 kPa
n& C 2 H6 (mol C 2 H 6 / min)
n& C 2 H4 (mol C2 H 4 / min)
n& air (mol air / min)
n& H 2 O (mol H 2 O / min)
n& 1 ( mol / min)
y1 (mol H2 O / mol)
Rb1- y 1 g( mol dry gas / mol)
U
|
|
0.235 mol C2 H 6 / mol DGV
S
0.765 mol C2 H 4 / mol DG|W
|
T
355 L / min air @ 75o C, 115 kPa
n& 2 (mol / min)
y2 ( mol H2 O / mol)
(1- y2 )( mol dry air / mol)
a.
Hygrometer Calibration ln y = bR + ln a
b=
ln by 1 y 2 g
R2 − R1
=
lnd0.2 10 −4 i
90 − 5
dy =
ae bR i
= 0.08942
ln a = ln y1 − bR 1 = ln 10 −4 − 0.08942b5g ⇒ a = 6.395 × 10 −5 ⇒ y = 6.395 × 10 −5 e 0.08942R
b.
n& 1 =
125 L
min
273K 105 kPa
298K 101 kPa
1 mol
= 5.315 mol min wet gas
22.4 L bSTPg
n& 2 =
355 L 273K 115 kPa
1 mol
= 14.156 mol min wet air
min 348K 101 kPa 22.4 L bSTPg
R1 = 86.0 → y 1 = 0.140 , R 2 = 12.8 → y 2 = 2.00 × 10 −4 mol H 2O mol
5- 23
5.43 (cont’d)
F
C 2 H6 balance: n& C 2 H6 = b5.315 mol min gGb1 − 0.140g
H
mol DG I F
mol C 2 H6 I
J G0.235
J
mol KH
mol DG K
= 1.07 mol C2 H 6 min
C 2 H 4 balance: n& C 2 H 4 = b5.315gb0.860gb0.765g = 3.50 mol C2 H 4 min
Dry air balance: n& air = b14.156gd1 − 2.00 × 10 −4 i = 14.15 mol DA min
Water balance: n& H 2 O = b5.315gb0.140g + b14.156gd1.00 × 10 −4 i = 0.746 mol H 2O min
n& dry product gas = b1.07 + 3.50 + 14.15g mol min = 18.72 mol min ,
n& total = b18.72 + 0.746g = 19.47 mol min
& 3 = 19.47 mol min 22.4 L bSTPg 338K = 540 liters min
V
mol
273K
Dry basis composition:
c.
p H 2O = y H 2 Ol ⋅ P =
F 1.07 I
G
J × 100%
H18.72K
= 5.7% C 2 H6 , 18.7% C2 H 4 , 75% dry air
0.746 mol H 2O
× 1 atm = 0.03832 atm
19.47 mol
y H 2O = 0.03832 ⇒ R =
1
F 0.03832 I
ln G
J = 71.5
0.08942 H6.395 × 10 −5 K
5.44 CaCO 3 → CaO + CO 2
n& CO2 =
1350 m3
273K
1 kmol
1273K 22.4 m 3 bSTPg
h
= 12.92 kmol CO2 h
12.92 kmol CO2 1 kmol CaCO3 100.09 kg CaCO3
h
1 kmol CO2
1 kmol CaCO 3
1362 kg limestone
0.17 kg clay
h
0.83 kg limestone
= 279 kg clay h
Weight % F e 2O3
kg Fe2 O 3 kg clay
279
b0.07 g
1362 + 279 − 12.92 b44.1g
kg clay 14243
× 100% = 18%
.
Fe 2 O 3
kg limestone
kg CO2 evolved
5- 24
1 kg limestone
0.95 kg CaCO 3
= 1362 kg limestone h
5.45
R864.7 g C b1 mol 12.01 g g = 72 .0 mol C
|
. mol H
| 116.5 g H b1 mol 1.01 gg = 1153
S
| 13.5 g S b1 mol 32.06 gg = 0.4211 mol S
| 5.3 g I
T
Basis: 1 kg Oil ⇒
53
. gI
n1 (mol CO 2 )
n 2 (mol CO)
n 3 (mol H 2 O)
n 4 (mol SO 2 )
n5 (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 → CO2
1
C + O 2 → CO
2
S + O 2 → SO 2
1
2H + O 2 → H 2O
2
na (mol), 0.21 O 2 , 0.79 N 2
15% excess air
o
175 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:
)
1 mol Air
0.21 mol O
554 mol Air
1 kg oil
2
22.4 liter ( STP )
mol
1 m3
3
10 liter
= 554 mol Air = n
2
448K
760 mm Hg
273K
940 mm Hg
a
= 16.5 m air kg oil
3
b. S balance: n 4 = 0.4211 mol SO 2
H balance: 115.3 = 2n 3 ⇒ n 3 = 57.6 mol H 2O
C balance: 0.95b72.0 g = n 1 ⇒ n1 = 68.4 mol CO 2 ⇒ 0.05(72.0) = n 2 = 3.6mol CO
N2 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 ) wet: 585 mols dry: 527 mols
dry basis:
wet basis:
3.6 mol CO
527 mol
3.6 mol CO
585 mol
= 6.8 ×10 −3
mol CO
mol
,
× 10 6 = 6150 ppm CO ,
5- 25
0.4211 mol SO2
527 mol
= 7.2 × 10 −4
0.4211 mol SO 2
585 mol
mol SO 2
mol
× 10 6 = 720 ppm SO 2
5.46 Basis: 50.4 liters C 5H 12 bl g min
50.4 L C5 H12 (l) / min
n& 1 (kmol C5 H12 /min)
heater
n& 1, n& 2
Combustion
chamber
15% excess air, V& air (L / min)
&n 2 kmol air
0.21 O 2
0.79 N 2
336 K, 208.6 kPa (gauge)
& 3 (kmol C5 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)
n
Condenser
C5 H 12 + 80 2 → 5CO 2 + 6 H 2O
a.
n& 1 =
n& 3 =
50.4 L
0.630 kg
1 kmol
min
L
72.15 kg
&n 4 (kmol O2 / min)
& 5 (kmol N 2 / min)
n
&n 6 (kmol CO 2 / min)
& (L/min)
V
liq
&
m=3.175
kg C5 O12 /min
n& 3 (kmol C5O12 /min)
n& 7 (kmol H 2O(l) / m i n )
= 0.440 kmol min
3175
.
kg 1 kmol
= 0.044 kmol / min
min 72.15 kg
frac. convert =
n& 2 =
V& gas (L/min), 275 K, 1 atm
0.440 - 0.044 kmol
× 100 = 90% C5 H12 converted
0.440
0.440 kmol C5 H12 1.15 (8 kmol O2 )
1 mol air
= 19.28 kmol air min
min
kmol C5 H12
0.21 mol O2
& air = 19.28 kmol 22.4 LbSTP g 336K 101 kPa 1000 mol = 173000 L min
V
min
mol
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
19.28 kmol air 0.79 kmol N2
= 15.23 kmol N2 /min
min
kmol air
0.90(0.440 kmol C5H12 ) 5 kmol CO2
n& 6 =
= 1.98 kmol CO2 / min
min
kmol C 5 H12
n& 5 =
& = 0.882+15.23+1.98 kmol 22.4 L(STP) 275 K 1000 mol = 4.08 × 10 5 L/min
V
gas
min
mol
273 K kmol
5- 26
5.46 (cont’d)
n& 7 =
0.9( 0.440 kmol C5H 12 ) 6 kmol H 2 O
= 2.38 kmol H2 O( l ) / min
min
kmol C5H 12
Condensate:
&C H =
V
5 12
&H O =
V
2
0.044 kmol 72.15 kg
min
kmol
2.38 kmol 18.02 kg
min
L
kmol
0.630 kg
L
1 kg
= 5.04 L min
= 42 .89 L min
Assume volume additivity (liquids are immiscible)
& liq = 5.04 + 42.89 = 47.9 L min
V
b.
C5 H12 ( l)
C5 H12 blg
H2 O blg
H2 O blg
5.47
o
n& air (kmol / min), 25 C, 1 atm
0.21 O 2
0.79 N 2
n& 0 (kmol / min)
n& 1 (kmol H 2 S / min) Furnace
H 2 S + 32 O 2 → SO 2 + H 2 O
0.20 kmol H 2S / mol
0.80 kmol CO 2 / mol
Reactor
&n 2 (kmol H 2S / min)
2H 2 S +SO 2 → 3S(g) + 2 H 2 O
10.0 m 3 / min @ 380o 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 m 3 / min
=
= 0.377 kmol / min
m3 ⋅ kPa
RT 8.314 kmol
653 K
⋅K
n& 1 = b0.20gn& 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 2S 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
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
Overall S balance: n& 6 =
= 0.200 n& 0 (kmol S / min)
(min)
1 kmol H 2S
Overall N2 balance: n& 3 =
Overall CO 2 balance: n& 5 = 0.800n& 0 (kmol CO2 / 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 2S
min
1 kmol H 2 O
⇒ n& 4 = 0.200n& 0 (kmol H 2 O / min)
n& exit = n& 0 b0.376 + 0.200 + 0.200 + 0.800g = 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 FeS 2 (s), 18.0 kg I.
n FeS 2 fed = b82.0 kg FeS2 gb1 kmol / 120.0 kgg = 0.6833 kmol FeS 2
100 kg ore
0.6833 kmol FeS2
18 kg I
Vout m3 (STP)
40% excess air
n 1 (kmol)
0.21 O 2
0.79 N 2
V1 m 3 (STP)
n2
n3
n4
n5
(kmol SO2 )
(kmol SO 3 )
(kmol O 2 )
(kmol N2 )
m6 (kg FeS2 )
m7 (kg Fe 2O 3 )
18 kg I
2 FeS2(s) + 11
O2(g) → Fe 2 O3(s) + 4SO 2(g)
2
2 FeS2(s) + 152 O2(g) → Fe 2O3(s) + 4SO 3(g)
a.
n1 =
0.6833 kmol FeS2 7 .5 kmol O2 1 kmol air req' d 1.40 kmol air fed
= 17.08 kmol air
2 kmol FeS 2 0.21 kmol O2
kmol air req'd
V1 = b17.08 kmolgb22.4 SCM / kmolg = 382 SCM / 100 kg ore
n2 =
( 0.85)(0.40)0.6833 kmol FeS 2 4 kmol SO 2
= 0.4646 kmol SO2
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 FeS 2
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
n5 = b0.79 × 17.08g kmol N2 = 13.49 kmol N 2
Vout = (0.4646+0.6970+1.641+13.49 ) kmol [ 22.4 SCM (STP)/kmol]
= 365 SCM/100 kg ore fed
ySO 2 =
0.4646 kmol SO 2
× 100% = 2.9%; y SO 3 = 4.3%; y O2 = 10.1%; y N 2 = 82.8%
16.285 kmol
b.
Product gas, Te o Cj
Converter
0.4646 kmol SO2
0.697 kmol SO 3
1633
.
kmol O 2
1349
. kmol N 2
nSO2 (kmol)
nSO3 (kmol)
nO2 (kmol)
nN 2 (kmol)
Let ξ (kmol) = extent of reaction
n SO 2 = 0.4646 − ξ
n SO3 = 0.697 + ξ
n O2 = 1.641 − 12 ξ
n N 2 = 13.49
n=16.29- 12 ξ
K p (T)=

0.4646 − ξ
0.697 + ξ
 ySO =
, y SO3 =
2
1
16.29- 2 ξ
16.29- 12 ξ

⇒
1.641 − 12 ξ
13.49

yO 2 =
, y N2 =
1

16.29- 2 ξ
16.29- 12 ξ

(0.697 + ξ ) (16.29 − 12 ξ ) 2
1
P ⋅ y SO3
1
2
P ⋅ y SO2 (P ⋅ yO2 )
⇒
(0.4646 − ξ ) (1.641 − 12 ξ )
1
2
-1
⋅ P 2 = K p (T)
P=1 atm, T=600 o C, K p = 9.53 atm - 2 ⇒ ξ = 0.1707 kmol
1
⇒ n SO2 = 0.2939 kmol ⇒ fSO2 =
( 0.4646 − 0.2939 ) kmol SO2
0.4646 kmol SO 2 fed
reacted
= 0.367
P=1 atm, T=400 o C, Kp = 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 SO 2 .
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 2SO 4 98 kg H 2 SO 4
= 1133
. kg H 2SO 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 FeS 2 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 H2SO 4 98 kg
= 1339
. kg H2SO 4
kmol FeS 2
1 kmol S
kmol
133.9 kg H 2SO 4
kg H2SO 4
= 3.06
43.8 kg S
kg S
The sulfur is not completely converted to H2 SO4 because of (i) incomplete oxidation of
FeS2 in the roasting furnace, (ii) incomplete conversion of SO 2 to SO 3 in the converter.
5.49 N 2 O 4 ⇔ 2 NO 2
dPgauge
+ 1.00i V
atmgb2.00 L g
= 0.103 mol NO2
b473K gb0.08206 L ⋅ atm mol - Kg
n0 =
b.
n1 = mol NO 2 , n 2 = mol N 2 O 4
RT0
=
b2.00
a.
F n2 I
n1 I
n12
P
,
p
=
P
⇒
K
=
P
J
N 2O 4
G
J
p
n 2 bn1 + n 2 g
H n1 + n2 K
H n1 + n2 K
F
p NO2 = y NO 2 P = G
Ideal gas equation of state ⇒ PV = bn 1 + n 2 gRT ⇒ n 1 + n 2 = PV / RT
b1g
Stoichiometric equation ⇒ each mole of N 2O 4 present at equilibrium represents a loss
of two moles of NO 2 from that initially present ⇒ n1 + 2n 2 = 0.103
Solve (1) and (2) ⇒ n1 = 2(PV / RT) − 0.103
b3g ,
b2 g
n 2 = 0.103 − (PV / RT)
b4g
Substitute (3) and (4) in the expression for Kp , and replace P with Pgauge + 1
2
24.37 dPg + 1i
dPgauge + 1i V
− 0.103g
Kp =
⇒ nt =
dPgauge + 1i where n t =
V =2 L
n t b0.103 − n t g
RT
T
b2n t
Pgauge(atm)
0.272
0.111
-0.097
-0.224
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
Variation of Kp with Temperature
ln Kp
T(K)
350
335
315
300
2
1
y = -7367x + 22.747
R2 = 1
0
-1
-2
0.0028
0.003
0.0032
1/T
5- 30
0.0034
5.49 (cont’d)
c.
A semilog plot of Kp vs.
1
is a straight line. Fitting the line to the exponential law
T
yields
ln K p = −
7367
a = 7.567 × 109 atm
F −7367 I
+ 22.747 ⇒ K p = 7.567 × 10 9 expG
J ⇒
H T K
T
b = 7367K
10.00 atm
5.50
& 1 (kmol A / h)
n
&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)
V&rcy (SCMH)
A + H2
S
5.00 kmol S 1 kmol A react
= 5.00 kmol A / h
h
1 kmol S form
5.00 kmol S 1 kmol H 2 react
Overall H 2 balance: n 2 =
= 5.00 kmol H 2 / h
h
1 kmol S form
Overall A balance: n1 =
Extent of reaction equations: &n i = n& i0 + ν iξ&
A + H2 ↔ S
n& 4 = 3 n& 3 − ξ&
H 2 : n& 5 = n& 3 − ξ&
S: 5.00 = ξ& =====> n& 4 = 3n& 3 − 5.00 U
|
n& 5 = n& 3 − 5.00 |
n& 4
3n& - 5.00
P= 3
10.0
V⇒ pA = yA P =
&n S = 5.00
n& tot
4 n& 3 − 5.00
|
A:
n& tot = 4 n& 3 − 5.00 |W
n& 5
n& - 5.00
P= 3
10.0
&n tot
4 n& 3 − 5.00
5.00
p S = yS P =
10.0
4 n& 3 − 5.00
p H 2 = yH 2 P =
Kp =
5.00b4n& 3 − 5.00g
pS
=
= 0.100 ⇒ n& 3 = 1194
. kmol H 2 / h
p A p H 2 10.0b3n& 3 − 5.00 gbn& 3 − 5.00 g
n& 4 = 3(11.94 ) - 5.00 = 30.82 kmol A / h
n& 5 = 11.94 − 5.00 = 6.94 kmol H 2 / h
& = b30.82 + 6.94 g kmol / h d22.4 m 3 (STP ) / kmoli = 846 SCMH
V
rcy
5- 31
5.51
n& 4 (kmol CO / h)
n& 5 (kmol H 2 / h)
Reactor
100 kmol CO / h
& 1 (kmol CO / h)
n
&n 2 (kmol H 2 / h)
n& 4 (kmol CO/h)
n& 5 (kmol H 2 /h)
n& 6 (kmol CH 3OH/h)
T, P
&n 3 (kmol H2 / h)
T (K), P (kPa)
H xs (% H 2 excess)
a.
Separator
n& 6 (kmol CH 3 OH/h)
%XS H 2 , 2 atomic balances, Eq. relation ⇒ four equations in n& 3 , n& 4 , n& 5 , and n& 6
5% excess H2 in reactor feed:
n&3 =
100 mol CO 2 mol H2 req'd 1.05 mol H2 fed
mol H 2
= 210
h
mol CO
1 mol H2 req'd
h
C balance: 100(1) = n&4 (1) + n&6 (1) ⇒ n&4 = 1 − n&6
H balance: 210(2) = n&5 (2) + n&6 (4) ⇒ n&5 = 210 − 2n&6
(1)
(2)
n& T = n& 4 + &n5 + &n6 = (100 − n& 6 ) + ( 210 − 2n& 6 ) + n& 6 = 310 − 2n& 6
9143.6
I
21.225 +
− 7.492lnb500K g
G
J
500 K
exp
G
J
2
-3
-8
H+4.076 × 10 b500K g - 1.161 × 10 b500 K g K
F
K p bT = 500K g = 1390
.
× 10
−4
= 9.11 × 10 −7 kPa-2
Kp =
yM P
(
y CO P y H2 P
K p P = 9.11 × 10
2
)
−7
2
(1) − (3)
yM
⇒ Kp P 2 =
( )
y CO y H2
kPa
-2
( 5000 kPa )
2
2
====>
= 22.775 =
n& 6
( 310 − 2n& 6 )
(100 − n& 6 ) ( 210 − 2n& 6 )
( 310 − 2n& 6 ) ( 310 − 2n& 6 )2
2
n& 6 (310 − 2n& 6 )
2
(100 − n& 6 )( 210 − 2n& 6 )
2
(3)
Solving for n& 6 ⇒ n& 6 = 75.7 kmol M/h
⇒ n& 4 = 100 − n& 6 = 24.3 kmol CO/h , n& 5 = 210 − 2n& 6 = 58.6 kmol H2 / h
1 kmol CO
2 kmol H 2
n& 6 = 75.7 kmol CO/h , n& 2 =
n& 6 = 151 kmol H2 /h
1 kmol M
1 kmol M
3
& = ( n& + n& ) 22.4 m (STP) = 1860 SCMH
V
rec
4
5
kmol
n& 1 =
5-32
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
KpcP^2
CO/h)
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)
CO/h)
H2/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
Vrec
H2/h) (SCMH)
51.10
5227
18.00
6339
173.44
1116
197.85
296
151.36
1858
29.16
5964
146.70
1791
151.36
1858
155.55
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 + 12 O 2
1.0 mol CO 2
1.0 mol O 2
1.0 mol N 2
T = 3000 K, P = 5.0 atm
1
A ⇔ B+ C
2
1
1
C+ D = E
2
2
K1 =
dp CO p O2
1/ 2
i
= 0.3272 atm1/ 2
p CO 2
1
O2 + 12 N 2 ⇔ NO
2
p NO
K2 =
= 01222
.
1/ 2
dp N 2 p O2 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
U
|
nB = ξ1
| yA
1
1
|
nC = 1 + ξ1 − ξ2
yB
||
2
2
1
V yC
nD = 1 − ξ 2
| y
2
D
|
nE = ξ 2
| yE
1
6 + ξ1 |
n tot = 3 + ξ 1 =
2
2 |W
K1 =
p CO p1O22
p CO2
= 2 ξ 1 b6 + ξ 1 g
= b2 + ξ 1 − ξ 2 g b 6 + ξ 1 g
= b2 − ξ 2 g b 6 + ξ 1 g
= 2ξ 2
b6
p i = y iP
+ ξ1g
12
=
y B y 1C 2 b1+ 12 −1g 2ξ 1b2 + ξ 1 − ξ 2 g
12
p
=
5 = 0.3272
12 b g
yA
2b1 − ξ 1gb6 + ξ 1 g
12
⇒ 0.3272b1 − ξ 1 gb6 + ξ 1g
K2 =
= n A n tot = 2 b1 − ξ 1 g b6 + ξ 1 g
p NO
dp O2 p N 2 i
12
=
yE
12 12
yC yD
⇒ 0.1222b2 + ξ 1 − ξ 2 g
12
12
= 2.236ξ 1b2 + ξ 1 − ξ 2 g
p 1−1 2 −1 2 =
b2 − ξ 2 g
12
(1)
2ξ 2
b2 + ξ 1
12
− ξ2 g
b2
12
− ξ 2g
= 01222
.
= 2ξ 2
(2)
Solve (1) and (2) simultaneously with E-Z Solve ⇒ ξ 1 = 0.20167 , ξ 2 = 012081
.
,
y A = 2 b1 − ξ 1 g b6 + ξ 1 g = 0.2574 mol CO2 mol y D = 0.3030 mol N 2 mol
y B = 0.0650 mol CO mol
y E = 0.0390 mol NO mol
y C = 0.3355 mol O2 mol
5.53 a.
n& 4 (kmol / h)
0.04 O2
0.96 N2
PX=C8 H10 , TPA=C6H 10O 4 , S=Solvent
& (m 3 / h) @105o C, 5.5 atm
V
3
n& 3O (kmol O2 / h)
n& 3N (kmol N 2 / h)
n& 3W (kmol H 2O(v) / h)
& (m 3 / 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 H2 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
m
& s (kg S / h)
separator
100 mol TPA / s
5.53 (cont’d)
b. Overall C balance:
F kmol PXI 8 kmol C
J
H
K kmol PX
h
n& 1 G
c.
=
100 kmol TPA 8 kmol C
⇒ n& 1 = 100 kmol PX / h
h
kmol TPA
100 kmol TPA 3.0 kmol O 2
= 300 kmol O2 /h
h
1 kmol TPA
kmol O 2

Overall O2 balance: 0.21n& 2 = 300
+0.04n& 4 
n& 2 = 1694 kmol air/h
h
 ⇒
n& 4 = 1394 kmol/h

Overall N2 balance: 0.79n& 2 = 0.96n& 4

100 kmol TPA 2 kmol H2 O
Overall H2 O balance: n& 3W =
= 200 kmol H 2O / h
h
1 kmol TPA
O2 consumed =
3
& = n& 2RT = 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
&V3W = 200 kmol H 2O (l) 18.0 kg 1 m = 3.60 m 3 H 2O(l) / h leave condenser
h
kmol 1000 kg
n& 1 =100
d. 90% single pass conversion ⇒ n& 3p = 0.10dn& 1 + n& 3p i ====> n& 3p = 111
. kmol PX / h
m& recycle = m& S + m& 3 P
=
(100+11.1) kmol PX
h
3 kg S 11.1 kmol PX
+
h
1 kmol PX kg PX
106 kg PX
106 kg
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
0.90n& 2
2 kmol N2 / h
n& 1 , n& 2 , n& 3
2 kmol N2 / 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 CO2 / 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)
&n5 (kmol H 2 O / h)
5.54 (cont’d)
CO + 2H 2 ⇔ CH3OH(M)
CO2 + 3H2 ⇔ CH 3OH + H 2O
a.
Let ξ 1 ( kmol / h) = extent of rxn 1, ξ 2 ( kmol / h) = extent of rxn 2
CO: n& 1 = 30 - ξ 1
H2 : n& 2 = 63 - 2ξ 1 − 3ξ 2
CO2 : n& 3 = 5 - ξ 2
U
|
n& 4 = ξ 1 + ξ 2
|
|
n& 5 = ξ 2
V ⇒ dK p i
1
|
n& N 2 = 2
|
&n tot = 100 - 2ξ 1 − 2ξ 2 |W
M:
H2 O:
N2 :
2
dK p i ⋅ P =
1
dK p i
⋅P =
2
2
n& 4
&n tot
n& 1 F n& 2 I
n& tot G
& tot JK
Hn
2
=
bξ 1
P⋅ yM
P ⋅ y CO dP ⋅ y H 2 i
3
2
, dK p i =
2
bP ⋅ y M gdP ⋅ y H2 O i
dP ⋅ y CO2 i dP ⋅ y H 2 i
2
+ ξ 2 gb100 − 2ξ 1 − 2 ξ 2 g
b30 − ξ 1 gb63 − 2ξ 1
F n
& 4 I F n& 5 I
G
& tot JKG
& tot JK
Hn
Hn
F n
& 3 I F n& 2 I
G
JG
n
&
& tot JK
H tot KH n
=
2
− 3ξ 2 g
= 84 .65
(1)
2
=
ξ 2 bξ 1 + ξ 2 gb100 − 2ξ 1 − 2ξ 2 g
b5 − ξ 2 gb63 − 2ξ 1
2
− 3ξ 2 g
= 1.259
(2)
Solve (1) and (2) fo r 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 CO2 / 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
⇒
53.4% M
0.03% H 2 O
n total = 49 .4 kmol / h
n& 6 = 25.4 kmol CO / h
C balance: n& 4 = 25.3 kmol / h
U
⇒
V
O balance: n& 6 + 2 n& 8 = n& 4 + n& 5 = 25.44 mol / sW n& 8 = 0.02 kmol CO 2 / h
H balance: 2n& 7 = 2 (0.9 n& 2 ) + 4 n& 4 + 2 n& 5 = 1237
. ⇒ 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
3
5.54 (cont’d)
3
F 237 kmol / h I F 22.4 m (STP)I
Process feed: b25.4 + 618
. + 0.02 + 2.0gG
JG
J = 18, 700 SCMH
H25.3 kmol / h KH
kmol
K
Reactor effluent flow rate:
b
F 237 kmol / h I
49.4 kmol / h gG
J = 444 kmol / h
H25.3 kmol / sK
kmol I F 22.4 m 3 (STP) I
F
⇒ V&std G444
J
J = 9946 SCMH
H
h KG
kmol
H
K
c.
9950 m 3 (STP )
⇒ V&actual =
h
3
&
$ = V = 354 m / h 1000 L
V
n& 444 kmol / h m 3
4732
. K 1013
. kPa
= 354 m 3 / h
273.2 K 4925 kPa
1 kmol
= 0.8 L / mol
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.
$
PV
B
RT
= 1 + $ ⇒ B = c bBo + ωB1 g
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
Bo = 0.083 − 1.6 = 0.083 −
= −0.333
1.6
Tr
3082
.
K
e
305.4 Kj
0.172
0.172
B1 = 0.139 − 4.2 = 0.139 −
= −0.0270
4 .2
Tr
308
.
2
K
e
305.4K j
RTc
0.08206 L ⋅ atm 305.4 K
B(T) =
−0.333 − b0.098g0.0270
bBo + ωB1 g =
Pc
mol ⋅ K 48.2 atm
= −01745
.
L / mol
$2
PV
mol ⋅ K I $ 2 $
$ - B = FG10.0 atm
−V
J V − V + 0.1745 = 0
RT
H 308.2K 0.08206 L ⋅ atm K
$ =
⇒V
1 ± 1 - 4b0.395 mol / Lgb0.1745 L / molg
2b0.395 mol / Lg
= 2.343 L / mol, 0.188 L / mol
V$ideal = RT / P = 0.08206 × 3082
. / 10.0 = 2.53, so the second solution is
b.
c.
likely to be a mathematical artifact.
$
PV
10.0 atm 2.343 L / mol
z=
=
= 0.926
L ⋅atm
RT 0.08206 mol
308.2K
⋅K
& =
m
&
V
1000 L
mol 30.0 g 1 kg
MW =
= 12.8 kg / h
$
h 2.343 L
mol 1000 g
V
5-37
5.56
$
PV
B
RT
= 1 + $ ⇒ B = c bBo + ωB1 g
RT
Pc
V
From Table B.1 Tc bCH3OHg = 513.2 K, Pc = 78.50 atm
Tc bC3H 8 g = 369.9 K, Pc = 42 .0 atm
From Table 5.3-1 ω bCH3OHg = 0.559, ω bC 3H8 g = 0.152
0.422
Bo (CH 3OH) = 0.083 −
Bo (C 3H 8 ) = 0.083 −
0.422
Tr 1.6
B1 (CH 3OH) = 0.139 −
B1 (C3 H8 ) = 0.139 −
B(CH 3OH) =
Tr
1.6
= 0.083 −
= 0.083 −
0.422
1.6
373.2K
e
513.2Kj
0.422
373.2K
e
1.6
0.172
Tr 4.2
= 0.139 −
4 .2
513.2K j
0.172
373.2K
e
= −0.333
369.9K j
0172
.
0.172
= 0.139 −
Tr 4.2
373.2K
e
4 .2
= −0.619
= −0.516
= −0.0270
369.9Kj
RTc
bBo + ω B1 g
Pc
0.08206 L ⋅ atm 513.2K
c−0.619 − b0.559g0.516h = −0.4868
mol ⋅ K 78.5 atm
RT
B(C3 H8 ) = c bBo + ωB1 g
Pc
=
=
Bmix =
0.08206 L ⋅ atm 369.9 K
c −0.333 − b0.152 g0.0270h = −0.2436
mol ⋅ K 42.0 atm
∑∑ y y B
i
i
j
ij
L
mol
L
mol
⇒B ij = 0.5dBii + B jj i
j
Bij = 0.5b−0.4868 − 0.2436gL / mol = -0.3652 L / mol
Bmix = b0.30gb0.30gb−0.4868g + 2b0.30gb0.70gb−0.3652 g + b0.70gb0.70gb−0.2436 g
= −0.3166 L / mol
$2
PV
mol ⋅ K I $ 2 $
$ - B = FG10.0 atm
−V
J V − V + 0.3166 = 0
mix
RT
H 373.2K 0.08206 L ⋅ atm K
$ $ =
Solve for V:V
1 ± 1 - 4b0.326 mol / Lgb0.3166 L / molg
2b0.326 mol / Lg
= 2.70 L / mol, 0.359 L / mol
RT 0.08206 L ⋅ atm 3732
. 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 m 3
=135 m3 / h
0.30 kmol CH3OH / kmol 1 kmol 1000 L
5-38
5.57 a.
van der Waals equation: P =
RT
a2
− 2
$
$
V
dV - b i
$ 2 dV
$ - bi ⇒ PV
$ 3 − PV
$ 2 b = RTV
$ 2 − aV
$ + ab
Multiply both sides by V
$ 3 + b-Pb - RTgV
$ 2 + aV
$ - ab = 0
PV
c 3 = P = 50.0 atm
c 2 = b-Pb - RTg = b−50.0 atmgb0.0366 L / molg − c0.08206
L⋅ atm
223 Kg =
mol⋅ K hb
−20.1 L ⋅ atm / mol
c 1 = − a = 1.33 atm ⋅ L2 / mol 2
c 0 = −ab = -d1.33 atm ⋅ L2 / mol2 i b0.0366 L / molg
atm ⋅ L3
mol 3
= −0.0487
b.
RT 0.08206 L ⋅ atm 223 K
$
V
=
= 0.366 L / mol
ideal =
P
mol ⋅ K 50.0 atm
c.
T(K)
P(atm)
223
223
223
223
223
1.0
10.0
50.0
100.0
200.0
c3
c2
1.0
10.0
50.0
100.0
200.0
-18.336
-18.6654
-20.1294
-21.9594
-25.6194
c1
c0
1.33
1.33
1.33
1.33
1.33
-0.0487
-0.0487
-0.0487
-0.0487
-0.0487
V(ideal)
V
f(V)
% error
(L/mol) (L/mol)
18.2994 18.2633 0.0000
0.2
1.8299
1.7939 0.0000
2.0
0.3660
0.3313 0.0008
10.5
0.1830
0.1532 -0.0007
19.4
0.0915
0.0835 0.0002
9.6
d. 1 eq. in 1 unknown - use Newton-Raphson.
$i
b1g ⇒ gdV
$ 3 + b-20.1294 gV
$ 2 + b1.33gV-.0487
$
= 50.0V
=0
∂g
$ 2 − 40.259 V
$ +1.33
= 150 V
$
∂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
b
5.58 C 3H 8 : TC = 369 .9 K
Specific Volume
PC = 42.0 atmd4.26 × 10 6 Pa i
5.0 m3
44.09 kg
1 kmol
75 kg
1 kmol
10 3 mol
ω = 0.152
= 2.93 × 10 −3 m 3 mol
Calculate constants
a=
0.42747
b=
0.08664
d8.314
m 3 ⋅ Pa mol ⋅ Ki
2
b369.9
4.26 × 10 Pa
6
d8.314
m 3 ⋅ Pa mol ⋅ Ki
b369.9
4.26 × 10 Pa
6
2
Kg
Kg
= 0.949 m6 ⋅ Pa mol 2
= 6.25 × 10 −5 m 3 mol
m = 0.48508 + 1.55171b0.152 g − 0.15613b0.152 g = 0.717
2
α = 1 + 0.717 e1 − 298.2 369.9 j
2
= 115
.
SRK Equation:
P=
d8.314
m 3 ⋅ Pa mol ⋅ Ki b298.2 Kg
d2.93 × 10
−3
− 6.25 × 10 −5 i m 3 mol
−
115
. d0.949 m 6 ⋅ Pa mol 2 i
2.93 × 10 −3 m3 mold2.93 × 10 −3 + 6.25 × 10 −5 i m 3 mol
⇒ P = 7.40 × 10 6 Pa ⇒ 7.30 atm
P=
Ideal:
3
. Kg
RT d8.314 m ⋅ Pa mol ⋅ Ki b2982
=
= 8.46 × 10 6 Pa ⇒ 8.35 atm
−
3
3
$V
2.93 × 10 m mol
(8.35 − 7.30) atm
× 100% = 14.4%
7.30 atm
Percent Error:
TC = 304.2 K
5.59 CO2 :
PC = 72.9 atm ω = 0.225
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 )
CO 2 : a = 3.65
Ar:
a = 1.37
f bTg =
L2 ⋅ atm
mol2
L ⋅ atm
2
mol2
, m = 0.826 , b = 0.0297
L
, α = 1 + 0.826e1 − T 304.2 j
mol
2
, m = 0.479 , b = 0.0224
L
, α = 1 + 0.479e1 − T 1512
. j
mol
2
RT
a
−
1 + me1 − T TC j
$
$ dV
$ + bi
V−b V
2
−P =0
Use E-Z Solve. Initial value (ideal gas):
L I F
L ⋅ atm I
F
Tideal = b51.0 atmgG0.70
. K
J G0.08206
J = 4350
H
mol K H
mol ⋅ K K
E - Z Solve ⇒ bTmax gCO = 455.4 K ,
2
bTmax gAr
= 431.2 K
5-40
5.60 O2 : TC = 154.4 K ; PC = 49.7 atm ; ω = 0.021 ; T = 208.2 Kb65° Cg ; P = 8.3 atm ;
m
& = 250 kg h ; R = 0.08206 L ⋅ atm mol ⋅ K
SRK constants: a = 1.38 L2 ⋅ atm mol2 ; b = 0.0221 L mol ; m = 0.517 ; α = 0.840
$i =
f dV
SRK equation:
E-Z Solve
RT
aα
$ = 2.01 L / mol
−
− P = 0 =====> V
$
$
$
V
−
b
V
V
+
b
d
i
d
i
& =
⇒V
5.61
W
∑F
y
250 kg
kmol
h
32.00 kg
= PCO 2 ⋅ A - W = 0
103 mol 2.01 L
1 kmol
mol
= 15,700 L h
where W = mg = 5500 kge9.81
m
s2 j
= 53900 N
PCO2 ⋅ A
a.
PCO2 =
W
A piston
=
53900 N
π
0.15
4b
2
mg
b. SRK equation of state: P =
1 atm
= 30.1 atm
1.013 × 10 5 N / m2
RT
αa
−
$
$ dV
$ + bi
V
dV - b i
For CO 2: Tc = 304 .2, Pc = 72.9 atm , ω = 0.225
a = 3.654 m 6 ⋅ atm / kmol2 , b = 0.02967 m 3 / kmol, m = 0.8263, α (25o C) = 1.016
m 3 ⋅atm
301
. atm =
e0.08206 kmol⋅ K j b298.2
$ - 0.02967 m3 j
kmol
eV
Kg
−
b1.016 ge3.654
m 6 ⋅atm
kmol2 j
$ dV
$ + 0.02967 i
V
m6
2
kmol
E-Z Solve
$ = 0.675 m3 / kmol
=====> V
Vbbefore expansiong = 0.030 m 3
2
Vbafter expansiong = 0.030 m 3 + π4 b0.15 mg b15
. mg = 0.0565 m 3
mC O2 =
V
0.0565 m3
44.01 kg
MW =
= 3.68 kg
3
$V
0.675 m / kmol kmol
mC O2 ( initially) =
PV
1 atm
0.030 m 3 44.01 kg
MW =
= 0.0540 kg
3
RT
298.2 K kmol
0.08206 mkmol⋅atm
⋅K
mCO2 (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 CO2
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
Initial: n o =
Vo
RT
Final: V = Vo +
Vo +
bPo
= 1g
πd 2 h
3.63 (kg)
V
, n = no +
= o + 0.0825
4
44 (kg / kmol) RT
πd 2 h
4
$ =V=
V
n Vo
+ 0.0825
RT
W
RT
αa
53,900
RT
αa
P=
=
−
⇒ 2
=
−
$
$
$ dV
$ + bi
$ dV
$ + bi
A piston V - b V
πd / 4 V - b V
b1g
$ in b1g ⇒ one equation in one unknown. Solve for d .
Substitute expression for V
5.62 a. Using ideal gas assumption:
Pg =
nRT
35.3 lb m O 2 1 lb - mole 10.73 ft 3 ⋅ psia 509.7 o R
− 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 =
RT
αa
−
$
$ dV
$ + bi
V
dV - b i
3
ft 3
$ = 2.5 ft 32.0 lb m / lb - mole = 2.27
V
35.3 lb m
lb - mole
For O2: Tc = 277.9 o R, Pc = 7304
. psi, ω = 0.021
a = 5203.8
ft 6 ⋅ psi
ft 3
,
b
=
0
.
3537
, m = 0.518, α d50o Fi = 0.667
lb - mole 2
lb - mole
3
ft ⋅ psi
b2400 + 14.7 g psi =
e10.73 lb-mole⋅o R j d509.7
$ - 0.3537 i
dV
o
Ri
ft 3
lb -mole
−
ft ⋅ psi
lb-mole2 j
$ dV
$ + 0.3537 i
V
ft 6
2
lb-mole
$ = 2.139 ft 3 / lb - mole
E - Z Solve ⇒ V
V
2.5 ft 3
32.0 lb m
mO 2 = MW =
= 37.4 lb m
3
$
2.139 ft / lb - mole lb - mole
V
5-42
6
.
b0.667 ge52038
5.62 (cont’d)
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 =
RT
αa
−
$
$
$ + bi
VdV
dV - b i
$ dV
$ - b i dV
$ + bi :
⇒ multiply both sides of the equation by V
$ i = PV
$ dV
$ - bi dV
$ + b i − RTV
$ dV
$ + b i + α adV
$ - bi = 0
f dV
$ i = PV
$ 3 − RTV
$ 2 + dαa - b 2 P - bRTi V
$ - α ab = 0
f dV
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
READ (5, *) T, P
IF (T.LT.Q.) STOP
5-43
5.63 (cont’d)
R = 0.42747 *R*R/PC*TC*TC
B = 0.08664 *R*TC/PC
M = 0.48508 + W = b155171
.
− W∗015613
.
g
ALP = d1.+ M∗ c1 − bT / TCg∗∗0.5hi ∗∗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
Tr = b40 + 273.2 g 126.2 = 2.48
U
N2 : TC = 126.2 K
| Fig. 5.4-4
⇒
.
40 MPa
10 atm
V ⇒ z = 12
PC = 33.5 atm Pr =
= 11.78|
33.5 atm 1.013 MPa
W
He: TC = 5.26 K
PC = 2.26 atm
⇒
Fig. 5.4-4
Tr = b−200 + 2732
. g b5.26 + 8g = 552
. U
|
V ⇒ z = 1.6
Pr = 350 b2.26 + 8 g = 34.11
|
W
↑
Newton’s correction
5-44
5.65 a. ρ dkg / m3 i =
=
b.
m (kg)
3
V (m )
=
(MW)P
RT
30 kg kmol
9.0 MPa
10 atm
= 69 .8 kg m3
3
m
⋅
atm
465 K
0.08206 kmol ⋅ K 1.013 MPa
Tr = 465 310 = 1.5U Fig. 5.4-3
V ⇒ z = 0.84
Pr = 9.0 4.5 = 2.0 W
ρ=
(MW)P 69.8 kg m3
=
= 83.1 kg m 3
zRT
0.84
100 lb m CO2
1 lb - mole CO 2
= 2.27 lb - moles
44.01 lb m CO2
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 
5.66 Moles of CO 2:
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 = 1.507 , Vr = 0.80 ⇒ z = 0.85
PV 1614.7 psi
10.0 ft 3
lb - mole⋅° R
1 atm
T=
=
= 779° R = 320 ° F
3
znR
0.85
2.27 lb - moles 0.7302 ft ⋅ atm 14.7 psi
5.67 O : T = 154.4 K
2
C
PC = 49.7 atm
Tr1 = 298 154.4 = 1.93|U
Pr1 = 1 49.7 = 0.02
V z1
|W
Tr2 = 358 154.4 = 2.23
Pr2 = 1000
V2 = V1
V2 =
= 1.00 (Fig. 5.4 - 2)
U
|
Vz
49.7 = 20.12|W 2
= 1.61 bFig. 5.4 - 4g
z 2 T2 P1
z 1 T1 P2
127 m3 1.61 358 K
1 atm
= 0.246 m3 h
h
1.00 298 K 1000 atm
5.68 O 2: TC = 154.4 K
PC = 49.7 atm
Tr = b27 + 273.2g 154.4 = 1.94
Pr1 = 175 49.7 = 3.52 ⇒ z1 = 0.95
(Fig. 5.3-2)
Pr2 = 1.1 49.7 = 0.02 ⇒ z 2 = 1.00
n1 − n 2 =
10.0 L
V F P1 P2 I
− J=
G
RT H z1 z2 K 300.2 K
mol ⋅ K
. atm I
F175 atm 11
−
G
J = 74.3 mol O2
0.08206 L ⋅ atmH 0.95
1.00 K
5- 45
5.69 a.
b.
$ = V = 50.0 mL 44.01 g = 440.1 mL / mol
V
n
5.00 g
mol
RT 82.06 mL ⋅ atm
1000 K
P= $ =
= 186 atm
mol ⋅ K 440.1 mL / mol
V
For CO 2: Tc = 304.2 K, Pc = 72.9 atm
T 1000 K
=
= 3.2873
Tc 304.2 K
$
VP
4401
. mL 72.9 atm
mol ⋅ K
c
Vr ideal =
=
= 1.28
RTc
mol 304.2 K 82.06 mL ⋅ atm
Tr =
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 / mol2 , b = 29 .67 mL / mol, m = 0.8263, α (1000 K) = 0.1077
mL ⋅atm
P=
c82.06 mol ⋅ K hb1000
Kg
mL
b440.1 - 29.67g mol
−
.
b01077
ge3.654 × 10
6 mL2 ⋅atm
mol 2 j
2
mL
440.1b440.1+ 29.67 g mol
2
= 198 atm
5.70 a. The tank is being purged in case it is later filled with a gas that could ignite in the
presence of O 2 .
b. Enough N2 needs to be added to make x O2 = 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 = 33.5 atm, Tr = 2.36
n initial = n 1 =
PV
1 atm
5000 L
=
= 204.3 mol
L⋅ atm
RT 0.08206 mol⋅ K 298.2 K
F 0.21
n O2 = 204.3 mol airG
H
n O2
n2
mol O2 I
J = 42.9 mol O2
mol air K
= 10 × 10 −6 ⇒ n 2 = 4.29 × 10 −6 mol
5000 L
= 116
. × 10 -3 L / mol
4.29 × 10 6 mol
$
. × 10 −3 L
mol ⋅ K 33.5 atm
$ ideal = VPc = 116
V
= 3.8 × 10 −3
r
RTc
mol 0.08206 L ⋅ atm 126.2 K
$ =
V
⇒ not found on compressibility charts
Ideal gas: P =
RT 0.08206 L ⋅ atm
2982
. K
=
= 2.1 × 10 4 atm
−3
$
mol ⋅ K 116
V
. × 10 L / mol
The pressure required will be higher than 2.1 × 10 4 atm if z ≥ 1, which fro m
Fig. 5.3- 3 is very likely.
n added = 4.29 × 106 − 204.3 ≅ d4.29 × 106 mol N 2 i b0.028 kg N 2 / mol g = 1.20 × 105 kg N2
5- 46
5.70 (cont’d)
c.
143
. kmol N 2
143
. kmol N 2
n initial = 0.204 kmol
y O2 = 0.21 kmol O 2 / kmol
1.43 kmol N 2
y1
1.43 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
=
= 1.633 kmol
L ⋅atm
zRT
0.99 0.08206 mol
298.2 K
⋅K
y1 =
y init n init b0.21g0.204
=
= 0.026
1.634
1634
.
y2 =
y 1n init
F n
I
= y init G init J = 0.0033
H1.634 K
1.634
2
yn I
J
H y init K
F
n
F n
I
y n = y init G init J ⇒ n =
H1.634 K
ln G
F n
I
ln G init J
H1.634K
= 4.8 ⇒ Need at least 5 stages
Total N 2 = 5b143
. kmol N 2 gb28.0 kg / kmolg = 200 kg N 2
d. Multiple cycles use less N2 and require lower operating pressures. The disadvantage is
that it takes longer.
5.71
& = MW
a. m
&
& F 44.09 lb m / lb - mol I SPV
&
&
PV
SPV
SPV
& = MW
⇒ Cost ($ / h) = mS
=G
=
60
.
4
3
J
RT
RT H 0.7302 ft ⋅ atmo
T
K T
lb-mol ⋅ R
b.
Tc = 369.9 K = 665.8 R ⇒ Tr = 0.85U Fig. 5.4-2
V ⇒ z = 0.91
Pc = 42.0 atm
⇒ Pr = 0.16 W
o
& = 60.4
m
& 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 N2 : Tc = 12620
. K = 227.16 o R, Pc = 335
. atm
U
609.7 o R
= 2.68
|
o
22716
. R
|
V ⇒ z = 1.02
600 psia 1 atm
Pr =
= 1.2 |
|
33.5 atm 14.7 psia
W
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
o
P
min
600
psia
lb - mole ⋅ R
b.
tank =
0.418 lb - mole 28 lb m / lb - mole 60 min 24 h 7 days 2 weeks
min b0.81g62.4 lb m / ft 3
h
day week
= 4668 ft 3 = 34,900 gal
5.73 a.
For CO: Tc = 133.0 K, Pc = 34 .5 atm
300 K
U
= 2.26
| Fig. 5.4-3
133.0 K
|
V ⇒ z = 1.02
2514.7 psia 1 atm
Pr1 =
= 5.0|
34.5 atm 14.7 psia
|W
Initially: Tr1 =
n1 =
2514.7 psia 150 L 1 atm
mol ⋅ K
= 1022 mol
1.02
300 K 14.7 psia 0.08206 L ⋅ atm
300 K
U
= 2 .26
| Fig. 5.4-3
133.0 K
|
V ⇒ z = 1.02
2258.7 psia 1 atm
Pr1 =
= 4 .5|
34.5 atm 14.7 psia
|W
After 60h: Tr1 =
n2 =
2259.7 psia 150 L 1 atm
mol ⋅ K
= 918 mol
1.02
300 K 14.7 psia 0.08206 L ⋅ atm
n& leak =
b.
n1 − n2
= 1.73 mol / h
60 h
n 2 = y 2 n air = y 2
t min =
PV 200 × 10 −6 mol CO
1 atm
30.7 m 3 1000 L
=
= 0.25 mol
L ⋅atm
RT
mol air
0.08206 mol
300 K
m3
⋅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 = 45.8 atm
C 2 H6 : Tc = 305.4 K , Pc = 48.2 atm
C 2H 4 : Tc = 2831
. K , Pc = 50.5 atm
Pseudocritical temperature: Tc′ = b0.20gb1907
. g + b0.30gb305.4g + b0.50gb2831
. g = 2713
. K
Pseudocritical pressure: Pc′ = b0.20gb45.8g + b0.30gb48.2g + b0.50gb50.5g = 48.9 atm
Reduced temperature:
Tr =
b90 + 273.2 gK
= 1.34
2713
. K
200 bars
1 atm
Reduced pressure:
Pr =
=
48.9 atm 1.01325 bars
Mean molecular weight of mixture:
U
| Figure 5.4 -3
|
⇒ z
V
|
4.04
|
W
= 0.71
M = b0.20gM CH4 + b0.30gM C2 H 6 + b0.50gM C 2 H4
= b0.20gb16.04g + b0.30gb30.07 g + b0.50gb28.05g
= 26.25 kg kmol
V=
znRT 0.71 10 kg 1 kmol
=
P
26.25 kg
0.08314 m3 ⋅ bar
b90 + 273g K
kmol ⋅ K
200 bars
= 0.041 m 3 (41 L)
5.75 N 2 : Tc = 126.2 K, PC = 33.5 atm U T ′ = 0.10b309.5g + 0.90b126.2g = 144.5 K
c
N 2 O: Tc = 309.5 K, PC = 71.7 atm VWPc′ = 0.10b71.7g + 0.90b33.5g = 37.3 atm
M = 0.10b44.02 g + 0.90b28.02g = 29.62
n = 5.0 kgb1 kmol 29.62 kg g = 0.169 kmol = 169 mol
a.
Tr = b24 + 273.2g 144.5 = 2.06
37.3 atm
mol ⋅ K
$ = 30 L
V
=
r
169 mol 144.5 K 0.08206 L ⋅ atm
P=
b.
U
|
V⇒
0.56|
W
0.97 169 mol 297.2 K 0.08206 L ⋅ atm
30 L
mol ⋅ K
z = 0.97bFig. 5.4 - 3g
= 133 atm ⇒ 132 atm gauge
Pr = 273 37.3 = 7.32 |U
$ = 0.56 bfrom a.g V| ⇒ z = 1.14 bFig. 5.4 - 3g
V
r
W
T=
273 atm 30 L
mol ⋅ K
= 518 K ⇒ 245° C
1.14 169 mol 0.08206 L ⋅ atm
5- 49
5.76 CO: Tc = 133.0 K, Pc = 34.5 atmU Tc′ = 0.60b133.0g + 0.40b33 + 8g = 96.2 K
V
H 2 : Tc = 33 K, Pc = 12 .8 atm
WPc′ = 0.60b34.5g + 0.40b12.8 + 8 g = 29.0 atm
Tr = b150 + 273.2g 96.2 = 4.4
Turbine inlet:
Pr =
2000 psi 1 atm
29.0 atm 14.7 psi
Turbine exit: Tr = 373.2 96.2 = 3.88
U
|
Fig. 5.4-1
→ z
V 
= 4.69 |
W
≈ 1.01
⇒ 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
&
Pout V
z out n RTout
Pin z out Tout
min 2000 psia 1.00 373.2
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
5.77 CO: Tc = 133.0 K, Pc = 34.5 atm UTc′ = 0.97b133.0g + 0.03b304.2g = 138.1 K
V
CO2 : Tc = 304.2 K, Pc = 72.9 atm WPc′ = 0.97b34.5g + 0.03b72.9g = 35.7 atm = 524.8 psi
Initial: Tr = 303.2 138.1 = 2.2
Pr = 2014.7
U
V
524.8 = 3.8 W
5.4-3
Fig.

→ z 1 = 0.97
Final: Pr = 1889.7 524.8 = 3.6 ⇒ z 1 = 0.97
Total moles leaked:
F P1
n1 − n 2 = G
H z1
−
P2 I V b2000 − 1875gpsi 30.0 L 1 atm
=
J
z 2 K RT
0.97
303 K 14.7 psi
mol ⋅ K
0.08206 L ⋅ atm
= 10.6 mol leaked
Moles CO leaked: 0.97b10.6g = 10.3 mol CO
Total moles in room:
Mole% CO in room =
24.2 m 3 10 3 L 273 K
1m
3
1 mol
303 K 22.4 LbSTP g
10.3 mol CO
× 100% = 1.0% CO
9734
. mol
5- 50
= 9734
. mol
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
545
. kmol CH 3OH( l) / h
a.
n& 1 =
54.5 kmol CH 3OH 1 kmol CO react
h
1 kmol CO fed
1 kmol CH 3OH
0.25 kmol CO react
= 218 kmol h CO
2n& 1 = 2 b218g = 436 kmol H 2 h ⇒ b218 + 436g = 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
Tc′ =
1
2
. g + b33 + 8g = 71.7 K
b1330
3
3
Pc′ =
1
2
b34.5g + b12.8 + 8g = 25.4 atm
3
3
Tr = 644 71.7 = 8.98
U
|
Fig. 5.4 -4
34.5 MPa
10 atm
→ z 1 = 1.18
V  
Pr =
= 13.45|
24.5 atm 1.013 MPa
W
644 K
& feed = 1.18 654 kmol
V
h
34.5 MPa
Vcat =
120 m 3 h
0.08206 m 3 ⋅ atm 1.013 MPa
kmol ⋅ K
1 m3 cat
25,000 m3 / h
10 atm
= 120 m 3 h
= 0.0048 m 3 catalyst (4.8 L)
b.
CO, H 2
n& 4 kmol CO / h
2n& 4 kmol H 2 / h
54.5 kmol CH3OH (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
644 K
& feed = 1.18 163.5 kmol
V
h
34.5 MPa
0.08206 m 3 ⋅ atm 1.013 MPa
kmol ⋅ K
5- 51
10 atm
= 29.9 m 3 h
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(41.3 K) + 0.85(419.6 K) = 362.8 K
Pc ' = 0.15(20.8 atm) + 0.85(39.7 atm) = 36.9 atm
Tr ' = 0.89 U Fig. 5.4-2
V ⇒ z = 0.86
Pr ' = 0.27 W
&
0.86 35 kmol 0.08206 m3 ⋅ atm 323 K 1 h
& = znRT
V
=
= 133
. m3 / min
P
h
kmol ⋅ K 10 atm 60 min
3
2
&
& F m I = u FG m IJ Adm 2 i = u × πd ⇒ d = 4V =
V
G
J
H min K
4
πu
H min K
5.80
CH4 :
4d1.33 m 3 / min i F100 cm I
G
J = 10.6 cm
π b150 m / ming H m K
Tc = 190.7 K Pc = 45.8 atm
C2 H4 : Tc = 283.1 K Pc = 50.5 atm
C2 H6 : Tc = 305.4 K Pc = 48.2 atm
T=90 o C
Tc ' = 0.15(1907
. K) + 0.60(283.1 K) + 0.25(305.4 K) = 274.8 K ====>
U
Tr ' = 1.32 |
V
P=175 bar
Pc ' = 0.15(45.8 atm) + 0.60(50.5 atm) + 0.25(48.2 atm) = 49.2 atm =====> Pr ' = 35
. |W
5.4-3
Fig.

→ z = 0.67
& Fm
V
G
H s
n& =
5.81
3
I
F mI
2
J = uG J A dm i
H
K
s
K
F
= G10
H
mI F
s I π
m3
2
b0.02 mg = 0.188
J G60
J
s KH
min K 4
min
&
PV
175 bar 1 atm
kmol ⋅ K 0.188 m 3 / min
=
= 1.63 kmol / min
zRT
0.67 1.013 bar .08206 m3 ⋅ atm
363 K
N2 :
Tc = 126.2 K = 227.16o R Pc = 33.5 atm
acetonitrile: Tc = 548 K = 986.4o R
Pc = 47.7 atm
Fig. 5.4-3
Tank 1 (acetonitrile): T1 = 550o F, P1 = 4500 psia ⇒ Tr1 = 1.02 Pr1 = 6.4 ⇒ z 1 = 0.80
⇒ n& 1 =
P1V1
306 atm 0.200 ft 3
=
z 1RT1
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)
F 0.104 I
F 0.027 I
o
o
o
Final: Tc ' = G
J 986.4 R + G
J 227.16 R = 830 R
H 0.131 K
H 0.131 K
.
F 0104
I
J 47.7
H 0.131 K
Pc ' = G
$
dV r i
P=
ideal
=
o
T=550

F→
Tr ' = 1.22
F 0.027 I
J 33.5 atm = 44.8 atm
H 0.131 K
atm + G
$ '
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 0.131 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 Ca H bO c , 26.8o C, 499.9 kPa
n c (mol C), n H (mol H), n O (mol O)
1 L @483.4o C, 1950 kPa
n p (mol)
0.387 mol CO2 / mol
0.258 mol O 2 / mol
0.355 mol H 2O / mol
n O2 (mol O 2 )
26.8o C, 499.9 kPa
a.
Volume of sample: 3.42 gd1 cm3 159
. g i = 2.15 cm 3
O 2 in Charge:
n O2 =
1.000 L − 2.15 cm 3 d10 −3 L km 3 i
0.08206
L ⋅ atm
mol ⋅ K
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
0.08206
756.6 K 101.3 kPa
mol ⋅ K
Balances:
np =
O: 2b0.200g + n O = 0.310 2b0.387 g + 2b0.258g + 0.355 ⇒ n O = 0.110 mol O in sample
C: n C = 0.387b0.310g = 0.120 mol C in sample
H: n H = 2 b0.355gb0.310g = 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.
MW = 12.01a + 1.01b + 16.0c = 12.01b1.1c g + 1.01b2c g + 16.0c = 31.23c
300 < MW < 350 ⇒ c = 10 ⇒ C11H 20 O10
5- 53
C5 H10 +
5.83 Basis: 10 mL C5H10 bl g charged to reactor
15
O 2 → 5CO2 + 5H 2O
2
10 mL C 5H10 blg
n1 (mol C5H10 )
n 2 (mol air)
0.21 O 2
0.79 N 2
27 o C, 11.2 L, Po (bar)
a.
n 3 (mol CO 2 )
n 4 bmol H 2 O(v)g
n 5 (mol N 2 )
75.3 bar (gauge), Tad d o Ci
10.0 mL C5H 10 blg 0.745 g
n1 =
mL
Stoichiometric air: n 2 =
Po =
1 mol
70.13 g
= 0.1062 mol C5 H 10
0.1062 mol C5H10
7.5 mol O2
1 mol air
1 mol C 2 H10 0.21 mol O 2
= 3.79 mol air
nRT 3.79 mol 0.08314 L ⋅ bar 300K
=
= 8.44 bars
V
11.2 L
mol ⋅ K
(We neglect the C5H 10 that may be present in the gas phase due to evaporation)
Initial gauge pressure = 8.44 bar − 1 bar = 7.44 bar
b.
5 mol CO2
U
= 0.531 mol CO 2 |
1 mol C 5 H10
|
0.531 mol CO 2 1 mol H 2 O
|
n4 =
= 0.531 mol H 2 O
V ⇒ 4.052 mol product gas
1 mol CO 2
|
n 5 = 0.79 b3.79 g = 2.99 mol N 2
|
n3 =
0.1062 mol C5 H 10
|
W
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
Tc ' = 0.131(304.2 K) + 0.131(647.4 K) + 0.738(126.2 K) = 217.8 K
Pc ' = 0.131(72.9 atm) + 0.131(218.3 atm) + 0.738(33.5 atm) = 62.9 atm ⇒ Pr ' = 1.21
$
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=
PV b75.3 + 1gbars 11.2 L
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 increas es, V increas es .
Point B: Neglect the variation of the density of liquid water with temperature, so ? = 1.00 g/mL
and VB = 10 mL
Point C: H2 O (v, 100°C)
b.
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,
(3.000 - 0.010) L
mol ⋅ K 243 mm Hg
1 atm
119.39 g
m(vapor) =
= 4.59 g
(30 + 273.2) K 0.08206 L ⋅ atm
760 mm Hg
mol
m(liquid) =
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
log10 p ∗ = 7.09808 −
1238.71
= 2.370 ⇒ p * = 10 2.370 = 2345
. mm Hg
45 + 217
*
*
∆H$ v 1
∆H$ v lnd p2 / p1 i
b. ln p = −
+ B⇒−
= 1 1 =
R T
R
T2 − T1
∗
B = ln( p1* ) +
lnb760 / 118.3g
1
b77 .0 + 273.2 gK
1
− b29.5+ 273.2
gK
∆H$ v / R
4151 K
= ln b118.3g +
= 18.49
T1
b29.5 + 273.2 gK
6-1
= −4151K
6.3 (cont’d)
ln p ∗ (45 o C) = −
4151
+ 18.49 ⇒ p ∗ = 231.0 mm Hg
b45 + 273.2 g
231.0 − 234.5
× 100% = −1.5% error
2345
.
c.
. − 760 I
F 1183
p∗ = G
J b45 − 29.5g + 118.3 = 327.7 mm Hg
H 29.5 − 77 K
327.7 − 234.5
× 100% = 39.7% error
2345
.
1
(rect. scale) on semilog paper
T + 273.2
⇒ straight line: slope = −7076 K , intercept = 21.67
Plot p ∗ blog scaleg vs
ln p∗ bmm Hgg =
L
O
−7076
−7076
+ 2167
. ⇒ p∗ bmm Hgg = exp M o
+ 2167
. P
T ( C) + 2732
.
.
NT ( C) + 2732
Q
o
7076 K
∆ Hv
= 7076K ⇒ ∆ H$ v =
R
1 kJ
mol ⋅ K 10 3 J
= 58.8 kJ mol
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
1/T
6-2
0.003
0.0026
0.0024
0.0022
0
y = -7075.9x + 21.666
0.002
T( oC)
79.7
105.8
120.0
141.8
178.5
197.3
ln(p*)
6.5
8.314 J
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.
T(°C)
1/T(K)
42.7
58.9
68.3
77.9
88.6
98.3
105.8
3.17×10 -3
3.01×10 -3
2.93×10 -3
2.85×10 -3
2.76×10 -3
2.69×10 -3
2.64×10 -3
p * (mm Hg)
=758.9 + hright -hleft
34.9
78.9
122.9
184.9
282.9
404.9
524.9
b. Plot is linear, ln p∗ = −
∆H$ v
−51438
. K
+ B ⇒ ln p∗ =
+ 19 .855
RT
T
At the normal boiling point, p∗ = 760 mmHg ⇒ Tb = 116° C
8.314 J 5143.8 K 1 kJ
∆ 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.
ln p∗ = a bT + 273.2g + b ⇒ y = ax + b
y = ln p∗ ; x = 1 bT + 273.2g
Perry' s Handbook, Table 3 - 8:
T1 = 39.5° C , p1 ∗ = 400 mm Hg ⇒ x 1 = 31980
.
× 10 − 3 , y1 = 5.99146
T2 = 56.5° C , p 2 ∗ = 760 mm Hg ⇒ x 2 = 3.0331 × 10 −3 , y 2 = 6.63332
T = 50° C ⇒ x = 3.0941 × 10 −3
F x − x1 I
6.39588
y = y1 + G
= 599 mm Hg
J by 2 − y1 g = 6.39588 ⇒ p∗ b50° Cg = e
H x 2 − x1 K
b. 50° C = 122° F
Cox Chart ⇒ p∗ =
c.
6.8
log p∗ = 7.02447 −
12 psi 760 mm Hg
14.6 psi
1161.0
= 2.7872 ⇒ p∗ = 10 2.7872 = 613 mm Hg
50 + 224
Estimate p ∗ b35° Cg: Assume ln p∗ =
a=
ln bp 2 ∗ p1 ∗g
1
T2
−
1
T1
b = ln p1 ∗−
=
= 625 mm Hg
lnb200 50g
1
45+ 273.2
−
1
25 + 273.2
a
+ b , interpolate given data.
T bK g
U
= − 65771
.|
a
6577.1
= lnb50g +
= 25.97
T1
25 + 273.2
|
V⇒
|
|W
6-3
ln p∗ b35° Cg = −
6577.1
+ 25.97 = 4.630
35 + 273.2
p∗ b35° Cg = e 4.630 = 102.5 mm Hg
6.8 (cont’d)
Moles in gas phase: n =
150 mL
273 K
102.5 mm Hg
1L
1 mol
3
760 mm Hg 10 mL 22.4 LbSTP g
b35 + 273.2g K
= 8.0 × 10 −4 mol
6.9
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
a.
⇒ 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. VH2 O =
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.
T1 = 58.3° C , p1 ∗ = 755 mm Hg − b747 − 52gmm Hg = 60 mm Hg
T2 = 110°C , p 2 ∗ = 755 mm Hg − b577 − 222 gmm Hg = 400 mm Hg
ln p∗ =
a=
a
+b
T bK g
ln bp 2 ∗ p1 ∗g
1
T2
−
b = ln p1 ∗−
1
T1
=
ln b400 60g
1
110+ 273.2
−
1
58.3 + 273 .2
= −46614
.
a
4661.4
= lnb60g +
= 18.156
T1
58.3 + 273.2
T=130o C=403.2 K
−46614
.
+ 18.156
T
ln p∗ b130° Cg = 6.595 ⇒ p∗ b130° Cg = e 6.595 = 7314
. mm Hg
ln p∗ =
b.
Basis: 100 mol feed gas CB denotes chlorobenzene.
n 1 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)
n 2 mol CB (l)
Saturation condition at inlet: y o P = p CB ∗ b130° Cg ⇒ y o =
731 mm Hg
= 0.962 mol CB mol
760 mm Hg
Saturation condition at outlet: y 1 P = p CB ∗ b58.3° Cg ⇒ y 1 =
60 mm Hg
= 0.0789 mol CB mol
760 mm Hg
Air balance: 100b1 − y o g = n1 b1 − y1 g ⇒ n1 = b100gb1 − 0.962g b1 − 0.0789g = 4.126 mol
Total mole balance: 100 = n1 + n 2 ⇒ n 2 = 100 − 4.126 = 95.87 mol CBbl g
% condensation:
95.87 mol CB condensed
× 100% = 99 .7%
b0.962 × 100g mol CB feed
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%
y H 2O P = 0.87 p∗ b25.56° Cg
Table B.3
yH 2O =
0.87 ( 24.559 mm Hg)
( )
759.5 mm Hg
Dew Point: p ∗ Tdp = yH2 O P = 0.0281( 759.5 ) = 21.34 mm Hg
6-5
= 0.0281 mol H 2 O mol air
Table B.3
Tdp = 23.2 °C
6.13 (cont’d)
0.0281
= 0.0289 mol H2 O mol dry air
1 − 0.0281
0.0289 mol H2O 18.02 g H2O mol dry air
ha =
= 0.0180 g H2 O g dry air
mol dry air
mol H2O
29.0 g dry air
hm =
hp =
hm
p ∗ ( 25.56 °C ) P − p ∗ ( 25.56 °C )
×100% =
0.0289
× 100 = 86.5%
24.559 [759.5 − 24.559 ]
6.14 Basis I : 1 mol humid air @ 70° F (21.1° C), 1 atm, h r = 50%
h r = 50% ⇒ y H 2 O P = 0.50 p H 2 O ∗ b21.1° Cg
Table B.3
Mass of air:
y H 2O =
0.50 × 18.765 mm Hg
mol H 2 O
= 0.012
760.0 mm Hg
mol
0.012 mol H 2 O 18.02 g 0.988 mol dry air 29.0 g
+
= 28.87 g
1 mol
1 mol
Volume of air:
1 mol
22.4 L bSTPg
b273.2
1 mol
+ 21.1gK
273.2K
= 24.13 L
28.87 g
= 1196
.
g L
24.13 L
Basis II : 1 mol humid air @ 70° F (21.1° C), 1 atm, hr = 80%
Density of air =
h r = 80% ⇒ y H 2 O P = 0.80 p H 2 O ∗ b21.1° Cg
Table B.3
Mass of air:
y H 2O =
0.80 × 18.765 mm Hg
mol H 2 O
= 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
22.4 L bSTPg
+
0.980 mol dry air 29.0 g
1 mol
b273.2
1 mol
+ 21.1gK
273.2K
= 24.13 L
28.78 g
= 1193
.
g L
24.13 L
Basis III: 1 mol humid air @ 90° F (32.2° C), 1 atm, h r = 80%
h r = 80% ⇒ y H2 O P = 0.80 p H2 O ∗ b32.2° Cg
Table B.3
y H 2O =
0.80 × 36.068 mm Hg
mol H 2 O
= 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
Volume of air:
Density =
1 mol
1 mol
+
22.4 L bSTP g
0.962 mol dry air
29.0 g
1 mol
b273.2
1 mol
+ 32.2gK
273.2K
= 28.58 g
= 25.04 L
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 i n 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.
6.15 a. h r = 50% ⇒ y H2 O P = 0.50 p H2 O ∗ b90° Cg
Table B.3
y H 2O =
0.50 × 52576
. mm Hg
= 0.346 mol H 2 O / mol
760.0 mm Hg
Dew Point: y H2 O p = p∗ dTdp i = 0.346b760 g = 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 m3 feed gas 10 3 L 273K
m
3
363K
mol
22.4 L bSTPg
= 33.6 mol
n 1 mol @ 25°C, 1atm
y1 (mol H2 O (v)/mol) (sat’d)
(1-y1 ) (mol air/mol)
33.6 mol @ 90°C, 1atm
0.346 H2 O mol /mol
0.654 mol air/mol
n 2 mol H2 O (l)
p *H 2O b25° Cg
23.756
= 0.0313 mol H 2 O mol
P
760
Dry air balance: 0.654 (33.6 ) = n1 (1 − 0.0313 ) ⇒ n1 = 22.7 mol
Saturation Condition: y 1 =
=
Total mol balance: 33.6=22.7+n2 ⇒ n2 = 10.9 mol H 2 O condense/m 3
c. y H 2O P = p∗ b90° Cg ⇒ P =
p * ( 90° C) 525.76 mmHg
=
= 1520 mm Hg = 2.00 atm
yH 2 O
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
n&condensed =
1 ft 3
10 gal H2O
min
62.43 lb m
7.4805 gal
ft
1 lb-mol
3
18.02 lb m
= 4.631 lb-mole/min
n& 2 (lb - moles / min)
V&1 (ft 3 / min)
n&1 (lb - moles / min)
y2 (lb-mol H2 O (v)/lb-mol) (sat’d)
(1-y2 ) (lb-mol DA/lb-mol)
40o F (4.4o C), 754 mm Hg
y1 (lb-mol H2 O (v)/lb-mol)
(1-y1 ) (lb-mol DA/lb-mol)
h r =95%, 90o F (32.2o C),
29.7 in Hg (754 mm Hg)
4.631 lb-moles H2 O (l)/min
95% hr at inlet: y H2 O P = 0.95 p ∗ b32.2° Cg
y H 2O =
Table B.3
0.95b36.068 mm Hgg
= 0.045 lb - mol H 2O lb - mol
754.4 mm Hg
Raoult's law: y2 P = p* ( 4.4° C)
Table B.3
y2 =
6.274
= 0.00817 lb-mol H 2O lb-mol
754.4
Mole balance: n&1 = n&2 + 4.631
 n&1 = 124.7 lb-moles/min
⇒
Water balance: 0.045 n&1 = 0.00817 n&2 + 4.631 n& 2 = 120.1 lb-moles/min
Volume in: V& =
124.7 lb-moles 359 ft 3 (STP) (460+90) oR
min
lb-moles
o
492 R
760 mm Hg
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 H2 O ) final = 0.20 × 462.7 = 92.6 mm Hg
p * (15° C) = 12.79 mm Hg < p H2 O . Impossible ⇒ condensation occurs.
Tfinal
288 K
= ( 0.80 × 760) mm Hg ×
= 370.2 mm Hg
Tinitial
473 K
= 370.2 + 12.79 = 383 mm Hg
( p air ) final = ( p air ) initial
P = p H 2 O + p air
b. Basis:
1 L 273 K
473 K
mol
22.4 L (STP)
= 0.0258 mol
6-8
6.17 (cont’d)
n 1 mol @ 15°C,
383.1 mm Hg
y1 (mol H2 O (v)/mol) (sat’d)
(1-y1 ) (mol dry air/mol)
0.0258 mols @ 200°C,
760 mm Hg
0.20 H2 O mol /mol
0.80 mol air/mol
n 2 mol H2 O (l)
Saturation Condition: y1 =
c.
p *H2 O b15° Cg
P
=
12.79 mm Hg
= 0.03339 mol H 2 O mol
383.1 mm Hg
Dry air balance: 0.800b0.0258g = n1 b1 − 0.03339 g ⇒ n1 = 0.02135 mol
Total mole balance: 0.0258 = 0.02135 + n 2 ⇒ n 2 = 0.00445 mol
Mass of water condensed =
0.00445 mol
18.02 g
mol
= 0.0802 g
6.18 Basis: 1 mol feed
3
n2 (mol), 15.6°C, 3 atm
y 2 (mol H 2O (v)/mol)(sat'd)
(1 – y 2) (mol DA/mol)
V1(m )
1 mol, 90°C, 1 atm
0.10 mol H O
2 (v)/mol
0.90 mol dry air/mol
heat
100°C, 3 atm
n2 (mol)
3
V2(m )
n3 (mol) H 2O( l), 15.6°C, 3 atm
Saturation: y2 =
p *H2 O b15.6° Cg
P
Table B.3
y2 =
13.29 mm Hg
atm
3 atm
760 mm Hg
= 0.00583
Dry air balance: 0.90b1g = n 2 b1 − 0.00583g ⇒ n 2 = 0.9053 mol
H 2 O mol balance: 0.10b1g = 0.00583b0.9053g + n 3 ⇒ n3 = 0.0947 mol
Fraction H 2 O condensed:
hr =
0.0947 mol condensed
= 0.947 mol condense mol fed
0100
.
mol fed
y2 P × 100% 0.00583b3 atm g
=
× 100% = 1.75%
p∗ b100° Cg
1 atm
6-9
6.18 (cont’d)
V2 =
0.9053 mol 22.4 L bSTPg 373K 1 atm
V1 =
1 mol 22.4 L bSTPg 363K
= 9.24 × 10 −3 m 3 outlet air @ 100° C
3
mol
273K 3 atm 10 L
1 m3
= 2 .98 × 10 −2 m 3 feed air @ 90° C
273K 10 3 L
mol
1 m3
V2 9.24 × 10 −3 m 3 outlet air
=
= 0.310 m 3 outlet air m 3 feed air
−2
3
V1
2.98 × 10 m feed air
25 L 1.00 kg
6.19 Liquid H 2 O initially present:
Saturation at outlet: y H2 O =
⇒
1 kmol
L
p *H2 O b25° Cg
P
=
= 1.387 kmol H 2 O bl g
18.02 kg
23.76 mm Hg
= 0.0208 mol H 2 O mol air
1.5 × 760 mm Hg
0.0208
= 0.0212 mol H 2 O mol dry air
1 − 0.0208
Flow rate of dry air:
Evaporation Rate:
15 L bSTPg
1 mol
= 0.670 mol dry air min
min
22.4 LbSTP g
0.670 mol dry air
0.0212 mol H 2 O
min
mol dry air
Complete Evaporation:
1.387 kmol 10 3 mol
min
kmol
6.20 a. Daily rate of octane use =
= 0.0142 mol H 2 O min
1h
0.0142 mol 60 min
= 1628 h
b67.8
daysg
7.069 × 103 ft 3 7.481 gal
π
⋅ 302 ⋅ (18 − 8) =
= 5.288 × 104 gal / day
4
day
ft 3
5.288 × 10 4 gal
1 ft 3
0.703 × 62.43 lb m
day
7.481 gal
ft 3
( SG ) C8 H18 = 0.703 ⇒
= 3.10 × 10 5 lb m C 8 H 18 / day
b.
∆p =
0.703 × 62 .43 lb m
32.174 ft
ft 3
c. Table B.4: p *C8 H18 (90 o F) =
s2
1 lb f
lb ⋅ ft
32.174 m2
s
20.74 mm Hg
(18 - 8) ft
14.696 psi
29.921 in Hg
14.696 lb f / in 2
= 6.21 in Hg
= 0.40 lb f / in 2 = p octane = y octane P
760 mm Hg
Octane lost to environment = octane vapor contained in the vapor space displaced by liquid
during refilling.
Volume:
5.288 × 10 4 gal
1 ft 3
7.481 gal
= 7069 ft 3
6-10
6.20 (cont’d)
pV
(16.0 + 14.7) psi
7069 ft 3
=
= 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 H18 : y = 8 18 =
= 0.0130 lb - mole C 8 H18 / 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.
*
*
6.21 a. Antoine equation ⇒ p tol
(85o F) = p tol
(29.44 o C) = 35.63 mmHg = p tol
Mole fraction of toluene in gas: y =
p tol 35.63 mmHg
=
= 0.0469 lb - mole toluene / lb - mole
P
760 mmHg
yPV
RT
0.0469 lb - mole tol
Toluene displaced = yntotal =
=
1 atm
ft ⋅ atm
3
lb - mole
0.7302
lb - mole ⋅ o R
1 ft 3
900 gal
(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 C7 H8 (v)/mol
0.9531 mol G/mol
n V (mol)
y (mol C7 H8 (v)/mol)
(1-y) (mol G/mol)
T(o F), 5 atm
Assume G is
noncondensable
n L [mol C7 H8 (l)]
90% of C7 H8 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:
*
p tol = yP = (0.004907 )(5 × 760) = 18.65 mmHg = p tol
(T )
Antoine equation:
T=
B − C ( A − log 10 p* ) 1346.773 − 219.693(6.95805 − log10 18.65)
=
= 17.11o C=62.8 o F
A − log10 p *
6.95805 − log10 18.65
6-11
6.22 a. Molar flow rate: n& =
&
VP
100 m 3
kmol ⋅ K
2 atm
=
= 6.53 kmol / h
-3
3
RT
h
82.06 × 10 m ⋅ atm (100 + 273) K
b. Antoine Equation:
1175.817
= 3.26601
100+224.867
⇒ p * = 1845 mm Hg
log 10 p *Hex (100°C)=6.88555-
p Hex = y Hex ⋅ P =
0.150(2.00) atm 760 mm Hg
atm
p *Hex (T ) = 228 mm Hg ⇒ log10 228=6.88555-
c.
6.53 kmol/h
0.15 C6 H14 (v)
0.85 N2
= 228 mm Hg < p *Hex ⇒ not saturated
1175.817
= 2.35793 ⇒ T = 34.8°C
T+224.867
n V (kmol/h)
y (kmol C6 H14 (v)/kmol), sat’d
(1-y) (kmol N2 /kmol)
T (o C), 2 atm
n L (kmol C6 H14 (l)/h)
80% of C6 H14 in feed
80% condensation:
n L = 0.80(0.15)(6.53 kmol / h) = 0.7836 kmol C 6 H 14 ( l) / h
Mole balance: 6.53 = nV + 0.7836 ⇒ nV = 5.746 kmol / h
Hexane balance:
0.15(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 )
1171530
.
Antoine equation: log10 5182
. = 6.87776 −
⇒ T = 2.52 o C
T + 224.366
Raoult’s law:
6.23 Let H=n-hexane
a.
n& 0 ( kmol / min)
y 0 (kmol H(v)/kmol
(1-y 0 ) (kmol N2 /kmol)
80o C, 1 atm, 50% rel. sat’n
Condenser
n&1 ( kmol / min)
0.05 kmol H(v)/kmol, sat’d
0.95 kmol N2 /kmol
T (o C), 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
6-12
6.23 (cont’d)
Antoine equation: log 10 38 = 6.88555 −
Mole balance: n&0 = n&1 + 150
.
N 2 balance: (1 − 0.703)n&0 =
1175.817
⇒ T1 = −3.26o C
T1 + 224.867
U
Rn&0
V⇒S
0.95n&1 W Tn&1
= 2.18 kmol / min
= 0.682 kmol / min
(0.95)0.682 kmol 22.4 m 3 (STP )
N2 volume: V&N 2 =
= 14.5 SCMM
min
kmol
b.
Assume no condensation occurs during the compression
2.18 kmol/min
0.703 H(v)
0.297 N2
80o C, 1 atm
Compressor
V&0 ( m 3 / min )
2.18 kmol/min
0.703 H(v)
0.297 N2
T0 (o C), 10 atm, 50% R.S.
V&1 (m3 / min)
0.682 kmol/min
0.05 H(v), sat’d
0.95 N2
T1 (o C), 10 atm
Condenser
1.5 kmol H(l)/min
50% relative saturation at condenser inlet:
0.500 p *H (T0 ) = 0.703(7600 mmHg) ⇒ p *H (T0 ) = 1.068 × 10 4 mmHg
Saturation at outlet: 0.050(7600 mmHg) = 380 mmHg = p *H (T1 )
Volume ratio :
Antoine
Antoine
T0 = 187 o C
T1 = 48.2° C
V&1 n1R T1 / P n1 (T1 + 273.2) 0.682 kmol/min 321 K
m 3 out
=
=
=
×
= 0.22 3
V&0 n0 RT0 / P n0 (T0 + 273.2) 2.18 kmol/min 460 K
m in
c. The cost of cooling to −3.24 o 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
kmol
L C 9 H 20
128.25 kg
Assume T = 25o C, P = 1 atm
n gas =
2 × 10 4 L 273 K
1
kmol
298 K 22.4 × 10 L(STP)
3
6-13
= 0.818 kmol
= 0.084 kmol C 9 H 20
6.24 (cont’d)
y max =
n max 0.084 kmol C 9 H 20
=
= 0.10 kmol C 9 H 20 / kmol (10 mole%)
n gas
0.818 kmol
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, p 2* = 40.0 mmHg
ln( 40.0 / 5.00)
5269
5269
⇒ A = 5269, B = ln(5.00) +
= 19.23 ⇒ p * = exp(19.23 −
)
1
1
299
T ( K)
−
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 H 2O)
23°C, 1 atm
n1 (mol) @ hr = 10%
0.79 mol N 2/mol
y1 (mol H 2O/mol)
+ O2 , CO2
Lungs
O2
Air inhaled: n 1 =
37°C, 1 atm
n2 (mol), saturated
0.75 mol N /mol
2
y 2 (mol H 2O/mol)
+ O2 , CO 2
CO2
12 breaths 500 ml 1 liter
min
breath 10 3 ml
273K
1 mol
60 min 24 hr
1 hr
1 day
b23 + 273gK 22.4 literbSTPg
= 356 mol inhaled day
Inhaled air - -10% r. h.: y1 =
Inhaled air - -50% r. h.: y1 =
0.10 p∗ H2 O b23° Cg
P
=
0.10b2107
. mm Hgg
760 mm Hg
= 2.77 × 10 −3
mol H 2 O
mol
0.50 p∗ H 2O b23° Cg 0.50b2107
. mm Hg g
mol H 2 O
=
= 1.39 × 10 − 2
P
760 mm Hg
mol
6-14
6.25 (cont’d)
H 2 O balance: n 0 = n 2 y 2 − n1 y 1 ⇒ ( n0 ) 10%
F
= G356
H
rh
− ( n0 ) 50%
rh
= ( n1 y 1 ) 50% − (n1 y 1 ) 10%
mol H 2 O OF 18.0 g I
molI L
G
J = 71 g / day
J M(0.0139 − 0.00277)
day K N
mol PQH1 mol K
Although the problem does not call for it, we could also calculate that n 2 = 375 mol exhaled/day,
y2 = 0.0619, and the rate of weight loss by breathing at 23o C and 50% relative humidity is
n 0 (18) = (n 2y2 - n 1 y1 )18 = 329 g/day.
6.26 a. To increase profits and reduce pollution.
b. Assume condensation occurs. A=acetone
n 1 mol @ To C, 1 atm
1 mol @ 90o C, 1 atm
y1 mol A(v)/mol (sat’d)
(1-y1 ) mol N2 /mol
0.20 mol A(v)/mol
0.80 mol N2 /mol
n 2 mol A(l)
For cooling water at 20o C
(
)
log 10 p *A 20 o C = 7.11714 −
(
)
1210.595
= 2.26824 ⇒ p*A 20 o C = 184.6 mmHg
20.0 + 229.664
Saturation: y1 ⋅ P = p *A d20 o Ci ⇒ y1 =
184.6
= 0.243 > 0.2 , so no saturation occurs.
760
For refrigerant at –35o C
(
)
log 10 p *A −35 o C = 7.11714 −
(
)
1210.595
= 0.89824 ⇒ p A* −35 o C = 7.61 mmHg
−35.0 + 229.644
Note: –35o C is outside the range of Antoine equation coefficients in Table B.4. If the
correct vapor pressure of acetone at that temperature is looked up (e.g., in Perry’s
Handbook) and used, the final result is almost identical.
Saturation: y1 ⋅ P = p *A d−35 o Ci ⇒ y1 =
7.61
= 0.0100
760
N2 mole balance: 1b0.8g = n1 b1 − 0.01g ⇒ n1 = 0.808 mol
Total mole balance: 1 = 0.808 + n2 ⇒ n 2 = 0192
.
mol
0.192
×100% = 96%
2
c. Costs of acetone, nitrogen, cooling tower, cooling water and refrigerant
d. 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.
Percentage acetone recovery:
6-15
6.27
Basis:
12500 L
1 mol
273 K 103000 Pa
= 5285
. mol / h
h
22.4 L(STP) 293 K 101325 Pa
n o (mol/h) @ 35o C, 103 KPa
y0 [mol H2 O(v)/mol]
yo1–mol
H2 O(v)/mol
y0 (mol
DA/mol)
(1-y
o ) mol DA/mol
h r=90%
h r =90%
528.5 (mol/h) @ 20o C, 103 KPa
y0 [mol H2 O(v)/mol]
y11–mol
H2 O(v)/mol
(sat’d)
y0 (mol
DA/mol)
(1-y1 ) mol DA/mol
H2 O(l)/h]
2 [mol
n 2nmol
H2O(l)/h
Inlet: y o =
hr ⋅ p H* 2 O d35 o Ci
P
Outlet: y1 =
p *H 2O d20 o Ci
P
=
=
0.90 × 42.175 mmHg 101325 Pa
= 0.4913 mol H 2 O / mol
103000 Pa
760 mmHg
17 .535 mmHg 101325 Pa
= 0.02270 mol H 2 O / mol
103000 Pa 760 mmHg
Dry air balance: b1 − 0.04913gn o = b1 − 0.02270 gb528.5g ⇒ n o = 5432
. mol / h
5432
. mol 22.4 L(STP) 308 K 101325 Pa
= 13500 L / h
h
mol
273 K 103000 Pa
Total balance: 543.2 = 528.5 + n 2 ⇒ n 2 = 14.7 mol / h
Inlet air :
Condensatio n rate:
6.28
Basis:
14.7 mol 18.02 g H 2 O 1 kg
= 0.265 kg / h
h
1 mol H 2 O 1000 g
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
n 1 lb -mole/min
40o F, 29.8 in.Hg
y1 [lb-mole H2 O(v)/lb-mole]
1- y1 (lb-mole DA/mol)
24.82 lb -mole/min
90o F, 29.8 in.Hg
y0 [lb-mole H2 O(v)/mol
1- y0 (lb-mole DA/mol)
h r = 88%
n 1 lb-mole/min
65o F, 29.8 in.Hg
y1 [lb-mole H2 O(v)/lb-mole]
1- y1 (lb-mole DA/lb-mole)
n 2 [lb-mole H2 O(l)/min]
Inlet: y o =
hr ⋅ p *H2 O d90 o Fi
Outlet: y1 =
P
p H* 2 O d40 o Fi
P
=
=
0.88b36.07 mmHgg 1 in Hg
= 0.0419 lb - mol H 2O / lb - mol
29 .8 in Hg
25.4 mmHg
6.274 mmHg 1 in Hg
= 0.00829 lb - mol H 2 O / lb - mol
29 .8 in Hg 25.4 mmHg
Dry air balance: 24.82b1 − 0.0419 g = n1 b1 − 0.00829 g ⇒ n1 = 23.98 lb - mol / min
Total balance: 24.82 = 23.98 + n 2 ⇒ n 2 = 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 @ 65o F:
23.98 lb - mol 359 ft 3 (STP) 525 o R 29.92 in Hg
= 9223 ft 3 / min
min
1 lb − mol 492 o 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))
nn2 2lb-mol
O(l)/min
Outlet: y1 =
h r ⋅ p H* 2 O d25o Ci
0.55b23.756 g
= 0.0172 mol H 2 O / mol
P
760
Saturation at T1 : 0.0172 b760g = 13.07 = p *H2 O bT1 g ⇒ T1 = 15.3 o C
Inlet: yo =
hr ⋅ p *H2 O d32 o Ci
P
=
=
0.70b35.663g
760
= 0.0328 mol H 2 O / mol
Dry air balance: n o b1 − 0.0328 g = 100b1 − 0.0172 g ⇒ no = 1016
. mol
Total balance: 101.6 + n 2 = 100.0 ⇒ n 2 = −1.6 mol (i.e. removed)
kg H 2 O removed :
kg dry air :
Ratio:
16
. mol 18.02 g 1 kg
= 0.0288 kg H 2 O
1 mol 1000 g
100b1 − 0.0172g mol 29.0 g 1 kg
= 2.85 kg dry air
1 mol 1000 g
0.0288
= 0.0101 kg H 2O 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∗ H 2 O b22° Cg ⇒ y 1 =
b0.40g19.827
mm Hg
= 0.01044 mol H 2 O mol
760 mm Hg
Second sample − T = 50° C , P = 839 mm Hg , saturated:
y2 P = p ∗H 2 O (50 °C ) ⇒ y2 =
92.51 mm Hg
= 0.1103 mol H2 O mol
839 mm Hg
ln y = bH + ln a ⇔ y = ae bH , y 1 = 0.01044 , H1 = 5 , y 2 = 01103
.
, H 2 = 48
b=
ln by 2 y 1 g
H2 − H1
=
lnb0.1103 0.01044 g
48 − 5
= 0.054827
ln a = ln y1 − bH 1 = lnb0.01044g − b0.054827gb5g = −4.8362 ⇒ a = expb−4.8362g = 7.937 × 10 −3
⇒ y = 7.937 × 10 −3 expb0.054827 H g
b.
Basis:
1 m3 delivered air
273K
b22
1 k mol
+ 273gK 22.4m 3 bSTP g
10 3 mol
1 kmol
o
41.31
mol,
1 atm
41.4
mol,
22o22
C,1C,atm
41.31 mol, T, 1 atm
n o mol, 35o C, 1 atm
0.0104 mol H22 O(v)/mol, (sat’d)
sat’d mol DA/mol
0.09896
0.9896 mol DA/mol
yo mol H2 O(v)/mol
(1-yo ) mol DA/mol
H=30
= 41.31 mol air delivered
0.0104
0.0104mol
molH2HO(v)/mol
2 O(v)/mol
0.09896
0.9896 mol
mol DA/mol
DA/mol
n 1 mol H2 O(l)
Saturation condition prior to reheat stage:
yH 2 O P = p*H2 O (T ) ⇒ ( 0.01044 )( 760 mm Hg ) = 7.93 mm Hg
⇒ T = 7.8°C (from Table B.3)
Part ba g
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 H2 O condensed
Mass of condensed water =
1.32 mol H 2 O 18.02 g H 2 O 1 kg
1 mol H 2 O 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
n 1 (mol) @ T f (°C), P4 (mm Hg)
y 1 [mol S(v)/mol] (sat’d)
(1–y 1 ) (mol G/mol)
n 0 (mol) @ T0 (°C), P0 (mm Hg)
y 0 (mol S/mol)
(1-y 0 ) (mol G/mol)
Td0 (°C) (dew point)
Inlet dew point = T0 ⇒
n 2 (mol S (l))
y o Po = p ∗ bTdo g ⇒ y o =
p ∗ bTdo g
Saturation condition at outlet: y 1 Pf = p∗ dT f i ⇒ y1 =
p∗ dT f i
(2)
Pf
(1)

→ n 2 = n0 fp∗ bT0 g P0
Fractional condensation of S = f ⇒ n 2 = n0 y 0 f
Eq. b3 g for n1
Total mole balance: n& 0 = n1 + n 2 ⇒ n1 = n&0 − n2
(1)
Po
⇒
n1 = n& 0 −
n&0 fp∗ bTdo g
Po
S balance: bn0 gby 0 g = n1 y 1 + n2
(1) - (4)
n& 0 p∗ bTdo g L
n& 0 fp∗ bTdo gOF p ∗ dT f
= Mn& 0 −
PG
G
Po
Po
M
P
N
QH Pf
⇒
b.
b1 −
f g p∗ bTdo g
Po
L
= M1 −
fp∗ bTdo gO p∗ dT f
M
N
P
P
Q
Po
iI
J
J
K
i
Pf
+
n&0 fp∗ bTdo g
po
⇒ Pf =
L
p∗ dT f i M1 −
M
M
N
b1 −
fp∗ eT j O
do P
P
P
o
P
Q
p∗ eT
fg
do j
Po
Condensation of ethylbenzene from nitrogen
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 : 120m 3 min feed @ 1000 o C(1273K), 35 atm . Use Kay’s rule.
Cmpd. Tc ( K )
Pc ( atm )
(T c )corr . ( Pc )corr ( Apply Newton's corrections for H2 )
H2
33.2
12.8
41.3
20.8
CO
133.0
34.5
−
−
CO 2
304.2
72.9
−
−
CH 4
190.7
45.8
−
−
∑y T
P′ = ∑ y P
Tc′ =
c
i ci
i
ci
= 0.40b41.3g + 0.35b133.0g + 0.20b304.2 g + 0.05b1907
. g = 133.4 K
= 0.40b20.8g + 0.35b34.5g + 0.20b72.9 g + 0.05b458
. g = 37.3 atm
Feed gas to heater
Tr = 1273 K 133.4 K = 9.54 
 Fig. 5.3-2 ⇒ z = 1.02
Pr = 35.0 atm 37.3 atm = 0.94
Vˆ =
1.02
8.314 N ⋅ m
1273 K
1 atm
mol ⋅ K 35 atm 101325 N m
⇒ n&gas feed =
120 m 3
min
−3
3.04 ×10
mol
1 kmol
3
103 mol
m
2
= 3.04 × 10 −3 m 3 mol
= 39.5 kmol min
Feed gas to absorber
Fig. 5.4-1
Tr = 283K/133.4K = 2.12, Pr = 35.0 atm/37.3 atm = 0.94 
→ z = 0.98
znRT 0.98 39.5 kmol
V& =
=
P
min
8.314 N ⋅ m
283 K
1 atm
mol ⋅ K 35.0 atm 101325 N/m 2
103 mol
1 kmol
n 1 (kmol/min), 261 K, 35atm
1.2(39.5) kmol/min MeOH(l)
y NaOH sat’d
y H2
y CH4 (2% of feed)
y CO
39.5 kmol/min, 283K, 35 atm
0.40 mol H2 /mol
0.35 mol CO/mol
0.20 mol CO2 /mol
0.05 mol CH4 /mol
n 2 (kmol/min), liquid
y MeOH
y CO2
y CH4 (98% of feed)
6-20
= 25.7
m3
min
6.32 (cont’d)
Saturation at Outlet: y McOH
7 .87863 −1473 .11 b−12 + 2300g
mm Hg
p∗ MeOH b261K g 10
=
=
P
35 atmb760 mm Hg atmg
= 4.97 × 10 −4 mol MeOH mol
y McOH =
nMeOH
n MeOH
n
+ H 2 + n CH4
A
= input
b.
+ n CO
=
nMeOH
A
A
=0.02 of input
=input
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.
6.33
n& 0 (kmol/min wet air) @ 28°C, 760 mmHg
y 1 (mol H2 O/mol)
(1-y 1 ) (mol dry air/mol)
50% rel. sat.
y 2 (mol H2 O/mol)
(1-y 2 ) (mol dry air/mol)
Tdew point = 40.0o C
1500 kg/min wet pulp
m& 1 (kg/min wet pulp)
0.75 /(1 + 0.75) kg H2 O/kg
1/1.75 kg dry pulp/kg
Dry pulp balance: 1500 ×
n& 1 (kmol/min wet air) @ 80°C, 770
0.0015 kg H2 O/kg
0.9985 kg dry pulp/kg
1
=m
& 1 (1 − 0.0015) ⇒ m
& 1 = 858 kg / min
1 + 0.75
50% rel. sat’n at inlet: y1 P = 0.50 p*H2 O (28 o 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 *H2 O ( 40 o C)
⇒ y 2 = (55.324 mm Hg) / (770 mm Hg)
= 0.0718 mol H 2O / mol
Mass balance on dry air:
n& 0 (1 − 0.0187 ) = n&1 (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, n&1 = 658.4 kmol / min
Mass of water removed from pulp : [1500(0.75/1.75)–858(.0015)]kg H2 O = 642 kg / min
622.8 kmol 22.4 m3 (STP) (273 + 28) K
Air feed rate : V&0 =
= 1.538 × 10 4 m 3 / min
min
kmol
273 K
6-21
6.34
Basis: 500 lb m hr dried leather (L)
n&1 (lb - moles / h)@130o F, 1 atm
n& 0 ( lb - moles dry air / h)@140o F, 1 atm
y1 (lb - moles H2 O / lb - mole)
(1 - y1 )(lb - moles dry air / lb - mole)
m& 0 (lb m / h)
500 lb m / h
0.61 lb m H 2O(l) / lb m
0.39 lb m L / lb m
0.06 lb m H 2 O(l) / lb m
0.94 lb m L/ lb
Dry leather balance: 0.39 m0 = b0.94 gb500g ⇒ m0 = 1205 lb m wet leather hr
Humidity of outlet air: y 1 P = 0.50 p∗ H 2O b130° Fg ⇒ y 1 =
0.50 (115 mm Hg)
mol H 2 O
= 0.0756
760 mmHg
mol
H 2 O balance: b0.61gb1205 lb m hrg = ( 0.06) a500 lb m hr f +
b0.0756n1 glb -
moles H 2 O
hr
18.02 lb m
1 lb - mole
E
n1 = 517.5 lb - moles hr
Dry air balance: n 0 = b1 − 0.0756g(5175
. ) lb - moles hr = 478.4 lb - moles hr
Vinlet =
6.35 a.
478.4 lb - moles 359 ft 3 bSTPg
hr
b140 +
460 g° R
492 ° R
1 lb - mole
= 2.09 × 105 ft 3 hr
Basis: 1 kg dry solids
n1 (kmol)N 2, 85°C
n 2 (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
n 3 (kmol) 28°C, 5.0 atm
y 3 (mol Hex/mol) sat'd
(1 – y)3 (mols N 2/mol)
n 4 (kmol) Hex(l)
0.05 kg Hex
1.00 kg solids
Mol Hex in gas at 80° C
b0.78
− 0.05gkg
kmol
86.17 kg
= 8.47 × 10 −3 kmol Hex
Antoine eq.
↓
70% rel. sat.: y 2 =
0.70 p ∗hex ( 80°C )
P
6.885551175.817
−
(80+ 224.867)
0.70 )10
(
=
760
6-22
= 0.984 mol Hex mol
6.35 (cont’d)
n2 =
8.47 × 10 −3 kmol Hex
1 kmol
0.984 kmol Hex
= 0.0086 kmol
N 2 balance on dryer: n1 = b1 − 0.984g0.0086 = 1376
.
× 10 −4 kmol
Antoine Eq.
↓
−
( 28+224.867)
p ∗hex ( 28 °C ) 106.885551175.817
Saturation at outlet: y3 =
=
= 0.0452 mol Hex mol
P
5 ( 760)
Overall N 2 balance: 1.376 × 10 -4 = n 3 b1 − 0.0452 g ⇒ n 3 = 1.44 × 10 −4 kmol
Mole balance on condenser: 0.0086 = 1.44 × 10 −4 + n 4 ⇒ n 4 = 0.0085 kmol
Fractional hexane recovery:
b.
0.0085 kmol cond. 86.17 kg
0.78 kg feed
kmol
= 0.939 kg cond. kg feed
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 – y3) (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 + 0.9n 3 (0.0452) = n 2 (0.984)
(1)
N2 balance on dryer: n1 + 0.9n 3 (1 − 0.0452 ) = n 2 (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.
6-23
6.36 b.
m& 1 (lb m/h)
300 lb m/h wet product
0.2
= 0167
.
lb m T(l) / lb m
1 + 0.2
0.833 lb m D / lb m
0.02 / (1.02 ) = 0.0196 lb m T(l) /
0.9804 lb m D / lb m
Dryer
@ 200O F,
y3 (lb-mole T/lb-mole)
(1-y3 )( lb-mole N2 /lb-mole)
n&3 ( lb-mole/h)
n& 1 (lb-mole/h)
y1 (lb-mole T(v)/lb-mole)
(1–y1 ) (lb-mole N2 /lbmole)
T=toluene
70% r.s.,150o F, 1.2 atm
D=dry solids
Heater
n&3 ( lb-mole/h)
y3 (lb-mole T(v)/lb-mole)
(1-y3 ) (lb-mole N2 /lb-mole)
Condenser
Eq.@ 90O F,
1atm
n& 2 ( lb-mole T(l)/h
)T(l)
Strategy: Overall balance⇒ m& 1 & n& 2 ;
Relative saturation⇒y1; , Gas and liquid equilibrium⇒y3
Balance over the condenser⇒ n&1 & n& 3
Toluene Balance: 300 × 0.167 = m& 1 × 0.0196 + n&2 × 92.13U Rm
& 1 = 255 lb m / h
V⇒ S
& 1 × 0.9804
Dry Solids Balance: 300 × 0.833 = m
&
W T n2 = 0.488 lb - mole / h
70% relative saturation of dryer outlet gas:
pC*7 H8 (150 O F=65.56 O C)=10
(6.95805 −
y1 P = 0.70 pC* 7 H8 (150 O F) ⇒ y 1 =
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
1.2 × 760
Saturation at condenser outlet:
1346.773
(6.95805)
*
o
o
32.22 + 219.693
pC7 H8 (90 F=32.22 C)=10
y3 =
p*C7 H8
P
=
= 40.90 mmHg
40.90
= 0.0538 mol T(v)/mol
760
Condenser Toluene Balance: n&1 × 0.1324 = 0.488 + n& 3 × 0.0538 U
= 5.875 lb - mole / h
Condenser N 2 Balance: n&1 × (1 − 0.1324) = n& 3 × (1 − 0.0538)
= 5.387 lb - mole / h
6-24
Rn&1
V⇒S
T n&3
W
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
= 0.182 lb m / h
Vinlet =
6.37
5.387 lb - moles 359 ft 3 bSTPg (200 + 460)° R
hr
492° R
1 lb - mole
C 6 H 14 +
Basis: 100 mol C 6 H 14
100 mol C 6H
19
O2 → 6CO 2 + 7H 2 O
2
n1 (mol) dry gas, 1 atm
0.821 mol N /mol
D.G.
2
0.069 mol CO
2 /mol D.G.
0.069 mol CO /mol
D.G.
2
0.021 mol CO/mol D.G.
0.021 mol CO/mol D.G.
0.00265 mol C6 H14 /mol
0.086
molOO/mol)
/mol
D.G.
2
x (mol
2
0.00265
mol
C
H
/mol
D.G.
6
14/mol)
(0.907–x) (mol N
2
n2n (mol
H
O)
2 O)
(mol H
14
n 0 (mol) air
0.21 mol O 2/mol
0.79 mol N 2/mol
2
C balance:
= 2590 ft 3 h
2
L
O
6b100g = n1 M0.069 + 0.021+ 6b0.00265gP ⇒ n1
bC O2 g
bCOg
bC6 H 14 g
M
P
N
Q
= 5666 mol dry gas
100 − 0.00265b5666g mol reacted
× 100% = 85.0%
100 mol fed
H balance: 14b100g = 2 n 2 + 5666b14 gb0.00265g ⇒ n 2 = 595 mol H 2 O
Conversion:
Dew point: y H 2 O =
p∗ dTdp i
Table B.3
595
=
⇒ p∗ dTdp i = 72.2 mm Hg ⇒ Tdp = 451
. °C
595 + 5666 760 mm Hg
N2 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 n 0 = 5888 mol air, x = 0.086 mol O2 /mol
Theoretical air:
Excess air:
100 mol C 2 H14
19 mol O 2
1 mol air
2 mol C 2 H 14
0.21 mol O 2
5888 − 4524
× 100% = 30.2% excess air
4524
6-25
= 4524 mol air
6.38 Basis: 1 mol outlet gas/min
n& 0 ( mol / min)
y0 ( mol CH 4 / mol)
(1 − y0 ( mol C 2 H 6 / mol)
1 mol / min @ 573K, 105 kPa
y1 (mol CO2 / mol)
n&1 (mol O2 / min)
y 2 (mol H 2 O / mol)
(1 − y1 − y2 ) mol N 2 / mol
3.76n& 1 (mol N 2 / min)
CH 4 + 2O 2 → CO 2 + 2H 2 O
p CO2 = 80 mmHg ⇒ y1 =
C2 H6 +
7
O 2 → 2CO 2 + 3H 2 O
2
80 mmHg 101325 Pa
= 0.1016 mol CO2 / mol
105000 Pa 760 mmHg
7
2
100% O2 conversion : 2no y o + no b1 − y o g = n1
(1)
.
C balance: no yo + 2no b1 − yo g = 01016
(2)
N2 balance: 376
. n1 = 1 − y1 − y2
(3)
H balance: 4 no y o + 6no b1 − y o g = 2 y2
(4)
Solve equations 1 to 4 ⇒
Rn o = 0.0770 mol
|
| y o = 0.6924 mol CH 4
S
| n1 = 0.1912 mol O2
| y = 01793
.
mol H 2 O
T 2
/ mol
/ mol
Dew point:
p *H2 O dTdp i =
01793
.
b105000g Pa 760 mmHg
= 141.2 mmHg ⇒ Tdp = 58.8 o C bTable B.3g
101325 Pa
6.39 Basis: 100 mol dry stack gas
n P (mol C 3H 8)
n B (mol C 4 H10 )
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 C3 H 8/mol
0.000527 mol C 4 H 10/mol
0.0148 mol CO/mol
0.0712 mol CO 2/mol
+ O2, N 2
nw (mol H2 O)
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
C 4 H10 +
6-26
13
O 2 → 4CO 2 + 5H 2 O
2
6.39 (cont’d)
Dew point = 46.5° C ⇒ y w P = p ∗ w b46.5° Cg ⇒ y w =
But yw =
77.6 mm Hg
mol H 2 O
= 0.0995
780 mm Hg
mol
nw
= 0.0995 ⇒ nw = 11.05 mol H2 O (Rounding off strongly affects the result)
100 + n w
C balance: 3n p + 4 n B = b100g b0.000527 gb3g + b0.000527 gb4g + 0.0148 + 0.0712
⇒
3n p + 4 n B = 8.969
b1g
H balance: 8 n p + 10 nB = (100 ) ( 0.000527 )(8 ) + ( 0.000527 )(10 )  + (11.05 )( 2 )
( 2)
⇒ 8n p + 10 nB = 23.047
 49% C3 H8
 n p = 1.25 mol C3H8 

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.
L&1 (lb - mole C10 H 22 / h)
L& 2 (lb - mole / h)
x 2 (lb - mole C 3 H 8 / lb - mole)
1 − x 2 (lb - mole C10 H 22 / lb - mole)
G& 1 (lb - mole / h)
G& 2 = 1 lb - mole / h
y1 (lb - mole C 3 H8 / lb - mole)
0.07 (lb - mole C 3 H 8 / lb - mole)
1 − y 1 (lb - mole N 2 / lb - mole)
0.93 (lb - mole N 2 / lb - mole)
Basis: G& 2 = 1 lb - mole h feed gas
N 2 balance: b1gb0.93g = G& 1 b1 − y1 g ⇒ G& 1 b1 − y 1 g = 0.93
b1g
98.5% propane absorption ⇒ G& 1 y1 = b1 − 0.985gb1gb0.07 g ⇒ G& 1 y1 = 105
. × 10
& = 0.93105 lb - mol h , y = 1128
.
× 10 −3 mol C H mol
b1g & b2 g ⇒ G
1
1
3
−3
b2 g
8
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
Raoult's law: x 2 p∗ C3 H8 b80° Fg = 0.07 p ⇒ x 2 =
b0.07 gb1.0
atmg
10.89 atm
Propane balance: b0.07 gb1g = G& 1 y1 + L&2 x 2 ⇒ L& 2 =
= 0.006428
mol H 2 O
mol
0.07 − b0.93105gd1128
.
× 10 −3 i
0.006428
= 10.726 lb - mole h
Decane balance: L&1 = b1 − x 2 gd L& 2 h = b1 − 0.006428gb10.726g = 10.66 lb - mole h
⇒
& / G& h
2
dL1
min
= 10.7 mol liquid feed / mol gas feed
6-27
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& C 3H 8 =
10.726 lb - mole 0.006428 lb - mole C 3H 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 =
0.06895 lb - mole C 3 H 8 h
= 0.00536 lb - mole C 3 H 8 / lb - mole
b0.06895 + 12.8g lb - moles h
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
100 mol/s @ 30o C, 1 atm
n& 4 (mol/s) @ 30°C, 1 atm
y 4 (mol B/mol)
(1-y 4 ) (mol N2 /mol)
xB =12.5 mol B/s
87.5 mol other hydrocarbon/s
n& 3 (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) 0.125 × 2120
=
= 0.3487
P
760
95% n-butane stripped: n& 4 ⋅ b0.3487g = b12.5gb0.95g ⇒ n& 4 = 34.06 mol / s
Raoult's law: y 4 P = x B p B* (30 o C) ⇒ y 4 =
Total mole balance: 100 + n&3 = 34.06 + 88.125 ⇒ n&3 = 22.18 mol/s
⇒ 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),
95% n-butane is stripped: n& 4 ⋅ b0.2790g = b12.5gb0.95g ⇒ n&4 = 42 .56 mol / s
Total mole balance: 100 + n&3 = 42.56 + 88.125 ⇒ 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.
6-28
6.42 Basis: 100 mol NH 3
Preheated
air
100 mol NH 3
780
kPasat’d
sat'd
820
kPa,
820
kPa, sat’d
N2
O2
converter
n3
n4
n5
n6
n 1 (mol) O2
3.76 n 1 (mol) N2
n 2 (mol) H2 O
1 atm, 30°C
h r= 0.5
a. i)
hydrator
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)
NH 3 feed: P = P∗ bTsat g = 820 kPa = 6150 mm Hg = 8.09 atm
Antoine :
log10 b6150g = 7 .55466 − 1002.711 bTsat + 247885
. g ⇒ Tsat = 18.4° C = 291.6 K
Table B.1 ⇒
V NH 3 =
Pc = 1113
. atm ⇒ Pr = 8.09 / 1113
. = 0.073U
V⇒z
W
Tc = 405.5 K ⇒ Tr = 291.6 / 405.5 = 0.72
0.92b100 molg 8.314 Pa
291.6 K
820 × 10 3 Pa
mol - K
= 0.92
(Fig. 5.3-1)
= 0.272 m 3 NH 3
Air feed: NH 3 + 2 O 2 → HNO 3 + H 2 O
n1 =
100 mol NH 3
2 mol O 2
mol NH 3
= 200 mol O 2
hr ⋅ p * b30° Cg 0.500 × 31824
.
Water in Air: y H2 O =
=
= 0.02094
p
760
n2
⇒ 0.02094 =
⇒ n 2 = 20.36 mol H 2O
n 2 + 4.76( 200)
A
( 4 .76 mol air mol O2 )
Vair =
4.76b200 g + 20.36 mol 22.4 L bSTPg 303K
1 mol
1 m3
3
273K 10 L
= 24.2 m 3 air
ii) Reactions: 4 NH 3 + 5O 2 → 4 NO + 6 H 2 O , 4 NH 3 + 3O 2 → 2 N 2 + 6 H 2 O
Balances on converter
NO: n 3 =
97 mol NH 3
4 mol NO
4 mol NH 3
= 97 mol NO
6- 29
6.42 (cont’d)
N 2 : n 4 = 3.76b2.00g mol +
O 2 : n 5 = 200 mol −
3 mol NH 3
2 mol N 2
4 mol NH 3
97 mol NH 3
= 7535
. mol N 2
5 mol O 2
4 mol NH 3
−
3 mol NH 3
3 mol O 2
4 mol NH 3
H 2 O: n 6 = 20.36 mol +
100 mol NH 3
= 76.5 mol O2
6 mol H 2 O
4 mol NH 3
= 1704
. 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 absorb er: n8 =
H 2 O in product: n9 =
97 mol NO react 4 mol HNO 3
4 mol NO
97 mol HNO3
63.02 g HNO 3
mol
= 97 mol HNO 3
45 g H 2 O
1 mol H 2 O
55 g HNO 3 18.02 g H 2 O
= 27756
. mol H 2 O
H balance on absorber: b1704
. gb2g + 2n 7 = 97 + b277 .6gb2 gbmol H g
⇒ n7 = 1557
. mol H 2 O added
VH 2 O =
b.
155.7 mol H 2 O 18.02 g H 2 O 1 cm 3
1 mol
1g
1 m3
10 6 cm 3
= 2.81 × 10 −3 m 3 H 2 Obl g
63.02 g HNO3 277.6 mol H 2 O 18.02 g HNO3
+
mol
mol
= 11115 g = 11115
.
kg
M acid in old basis =
Scale factor =
97 mol HNO 3
b1000
metric tons gb1000 kg metric ton g
11.115 kg
= 8.997 × 10 4
VNH3 = d8.997 × 10 4 i d0.272 m 3 NH 3 i = 2.45 × 10 4 m 3 NH 3
Vair = d8.997 × 10 4 i d24.2 m 3 air i = 2 .18 × 10 6 m 3 air
VH 2 O = d8.997 × 10 4 i d2.81 × 10 −3 m 3 H 2Oi = 253 m 3 H 2 Obl g
6- 30
6.43 a. Basis: 100 mol feed gas
100 mol
0.10 mol NH3 /mol
0.90 mol G/mol
G = NH 3 -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 = 10o C (50o F), ρ = 0.9534 g/mL ⇒ 0.120 g NH3 /g solution
⇒
12.0 g NH 3
88.0 g H 2 O
= 0.706 mol NH 3 ,
= 4.89 mol NH 3
(17.0 g / 1 mol)
(18.0 g / 1 mol)
⇒ 12.6 mole% NH 3 (aq), 87.4 mole% H 2 O(l)
Composition of gas effluent
T = 50 F, x A = 0.126 →
o
Perry
p NH 3 = 121
. psia bTable 2 - 23g
p H 2 O = 0.155 psia bTable
p total = 14 .7 psia
U
|
2 - 21gV
|
W
y A = 1.21 / 14.7 = 0.0823 mol NH 3 mol
⇒ y W = 0.155 / 14.7 = 0.0105 mol H 2 O mol
y G = 1 − y A − yW = 0.907 mol G mol
G balance: b100gb0.90g = n 2 y G ⇒ n 2 = b100gb0.90g b0.907 g = 99.2 mol
NH 3 absorbed = b100gb0.10 gin − b99.2gb0.0823gout = 184
. mol NH 3
% absorption =
1.84 mol absorbed
× 100% = 18.4%
b100 gb0.10gmol fed
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 2SO 4
1 kmol H 2SO 4
1 kmol SO 3
80.07 kg SO 3
98.08 kg H 2SO 4
1 kmol H 2 SO 4
1 kmol SO 3
⇒ 84.4% SO 3
6- 31
= 84 .4 kg
6.44 (cont’d)
b.
Basis 1 kg liquid feed
n o (mol), 40oC, 1.2 atm
n1 (mol), 40oC, 1.2 atm
0.90 mol SO3 /mol
0.10 mol G/mol
y1 mol SO3/mol
(1-y 1) mol G/mol
Equilibrium @ 40o C
1 kg 98% H2SO4
m1 (kg) 15% oleum
0.98 kg SO3
0.02 kg H2O
0.15 kg SO3 /kg
0.85 kg H2SO4 /kg
p SO3 b40° C, 84.4% g
115
.
= 151
. × 10 −3 mol SO 3 mol
P
760
ii)
0.98 kg H 2SO 4
2.02 kg H
0.02 kg H 2 O
2.02 kg H
H balance:
+
98.08 kg H 2SO 4
18.02 kg H 2 O
0.85 m 1 H 2 SO 4
2.02 kg H
=
⇒ m1 = 1.28 kg
98.08 kg H 2SO 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 = b1.28 − 1.0g kg =
= 3.50 mol
kg 80.07 g
⇒ 3.5 mol = n 0 − n1
G balance: 0.10n 0 = d1 − 151
. × 10 −3 i n1
1444444444444442444444444444443
i)
y1 =
=
E
n 0 = 3.89 mol
n1 = 0.39 mol
V=
3.89 mol
1 kg liquid feed
22.4 L bSTPg 313K
1 atm
1 m3
273K 1.2 atm 10 3 L
mol
= 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 p ∗B (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: y B P = x B p ∗B ⇒
yB =
yT =
0.40 (1350.1)
10 (760 )
0.60 ( 556.3)
10 ( 760)
= 0.0711 mol Benzene mol
= 0.0439 mol Toluene mol
y N2 = 1 − 0.0711 − 0.0439 = 0.885 mol N 2 mol
6- 32
6.47 N 2 - Henry' s law: Perry' s Chemical Engineers' Handbook, Page. 2 - 127, Table 2 -138
⇒ H N 2 b80° Cg = 12 .6 × 10 4 atm mole fraction
⇒ p N 2 = x N 2 H N2 = b0.003gd12 .6 × 10 4 i = 378 atm
H 2 O - Raoult's law: p H∗ 2 O b80° Cg =
3551
. mm Hg
1 atm
760 mm Hg
= 0.467 atm
⇒ p H 2O = dx H 2 O i d p ∗H2 O i = b0.997gb0.467 g = 0.466 atm
Total pressure: P = p N 2 + p H2 O = 378 + 0.466 = 378.5 atm
Mole fractions:
y H 2O = p H2 O P = 0 / 466 / 3785
. = 1.23 × 10 −3 mol H 2 O mol gas
y N 2 = 1 − y H2 O = 0.999 mol N 2 mol gas
6.48 H 2 O - Raoult's law: p ∗H2 O b70° Cg =
2337
. mm Hg
1 atm
760 mm Hg
= 0.3075 atm
⇒ p H 2O = xH 2 O p H∗ 2 O = b1 − xm gb0.3075g
Methane − Henry' s law: p m = x m ⋅ Hm
Total pressure: P = p m + p H2 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 2 O =
1000 cm 3
1g
cm
3
mol
18.02 g
= 55.49 mol
Moles of nitrogen:
n N2 =
(1 - 0.334) × 14.1 cm 3 (STP )
1 mol
1L
= 4.192 × 10 −4 mol
3
22.4 L (STP) 1000 cm
Moles of oxygen:
n O2 =
(0.334) ⋅ 14.1 cm3 (STP)
mol
L
= 2.102 × 10 −4 mol
22.4 L (STP) 1000 cm3
Mole fractions of dissolved gases:
n N2
4 .192 × 10 −4
x N2 =
=
n H 2O + n N 2 + n O2 55.49 + 4.192 × 10 −4 + 2.102 × 10 −4
x O2
= 7.554 × 10 −6 mol N 2 / mol
nO2
2.102 × 10 −4
=
=
n H2 O + n N2 + nO2 55.49 + 4.192 × 10 −4 + 2 .102 × 10 −4
= 3.788 × 10 −6 mol O 2 / mol
6- 33
6.49 (cont’d)
Henry' s law
Nitrogen: HN 2 =
x N2
=
0.79 ⋅ 1
= 1046
.
× 10 5 atm / mole fraction
7.554 × 10 −6
0.21 ⋅ 1
= 5544
.
× 10 4 atm / mole fraction
x O2 3.788 × 10 −6
Mass of oxygen dissolved in 1 liter of blood:
Oxygen: HO2 =
b.
p N2
p O2
=
2.102 × 10 -4 mol 32.0 g
= 6.726 × 10 −3 g
mol
0.4 g O2
1 L blood
Mass flow rate of blood: m& blood =
= 59 L blood / min
min
6.72 × 10 -3 g O 2
c. 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.
m O2 =
6.50 a.
Basis: 1 cm 3 H 2 Obl g
H2O =1.0
(SG)

→
= 0.0901
1 g H 2 O 1 mol
= 0.0555 mol H 2 O
18.0 g
CO2
 
→
( SC)
0.0901 cm3 bSTPgCO 2
p CO2 = 1 atm ⇒ x CO 2 =
d4.022
× 10 −6 i mol CO 2
d0.0555 + 4.022
p CO2 = x CO2 HCO2 ⇒ H CO2 b20° Cg =
b.
1 mol
= 4.022 × 10 − 6 mol CO 2
3
22,400 cm bSTP g
× 10
−6
i
mol
= 7.246 × 10 −5 mol CO 2 mol
1 atm
= 13800 atm mole fraction
7.246 × 10 −5
For simplicity, assume n total ≈ n H 2 O bmolg
xCO2 = p CO2 H = b3.5 atm g b13800 atm mole fractiong = 2.536 × 10 −4 mol CO 2 mol
1L
10 3 g H 2 O 1 mol H 2 O 2.536 × 10 − 4 mol CO 2
33.8 oz
1L
18.0 g H 2 O
1 mol H 2 O
= 0.220 g CO 2
n CO2 =
c. V =
12 oz
0.220 g CO 2
1 mol CO2
22.4 L bSTP g
b273 + 37 gK
44.0 g CO 2
1 mol
273K
6- 34
44.0 g CO 2
1 mol CO 2
= 0127
.
L = 127 cm 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. Calculate xbmol SO 2 molg in terms of rbg SO 2 100 g H 2 Og
Basis: 100 g H 2Ob1 mol 18.02 gg = 5.55 mol H2 O
r (g SO2 )b1 mol 64.07 gg = 0.01561r (mol SO 2 )
0.01561r
F mol SO 2 I
G
J
5.55 + 0.01561r H mol K
From this relation and the given data,
pS O2 (torr)
0
42
⇒ x SO2 =
xSO2 (mol SO2 /mol)
85
–3
0
1.4x10
129
–3
176
–3
2.8x10
5.6x10–3
4.2x10
A plot of pSO 2 vs. xS O2 is a straight line. Fitting the line using the method of least squares
(Appendix A.1) yields
mm Hg
, H SO 2 = 3.136 × 10 4
d pSO2 = H SO2 xS O2 i
mole fraction
c. 100 ppm SO 2 ⇒ ySO = 100 mol SO 2 = 1.00 × 10 −4 mol SO 2 /mol gas
2
106 mols gas
(
)
⇒ pSO2 = ySO2 P = 1.0 ×10−4 ( 760 mm Hg ) = 0.0760 mm Hg
Henry' s law ⇒ xSO 2 =
Since xS O2
H SO2
=
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
n final ≈ ninitial =
⇒ nS O2 =
pSO 2
140 L 103 g H 2 Obl g 1 mol
1L
18 g
= 7.78 × 10 3 moles
7.78 × 10 3 mol solution 2.40 × 10 −6 mol SO 2
1 mol solution
= 0.0187 mol SO 2 dissolved
0.0187 mol SO2 dissolved
= 1.34 × 10 −4 mol SO 2 L
140 L
Gas-phase composition
ySO2 = 1.0 × 10
−4
mol SO 2
mol
yH 2 O =
xH2 O p*H2O (30o C)
P
=
(1)(31.824 torr)
760 torr
= 4.19 × 10 −2
mol H 2 O(v)
mol
yair = 1 − ySO2 − yH2O = 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 SO 2.
6- 35
6.52 Raoult’s law + Antoine equation (S = styrene, T = toluene):
−
( T +214.985)
y S P = xS p∗S ⇒ x S = 0.650(150 mm Hg)/10 7.066231507.434
yT P = xT pT∗ ⇒ xT = 0.350(150 mm Hg)/106.95805−1346.773 (T + 219.693 )
0.65(150)
1 = x S + xT =
+
7.066231507.434
−
(T + 214.985)
0.35(150)
6.958051346.773
−
( T +219.693)
10
10
⇒ T = 86.0°C (Determine using E-Z Solve or a spreadsheeet)
xS =
0.65(150)
10
7.066231507.434
−
( 86.0 +214.985)
= 0.853 mol styrene/mol ⇒ xT = 0.147 mol toluene/mol
6.892721203.531
−
( 85+ 219.888)
6.53 PB∗ ( 85°C ) = 10
= 881.6 mm Hg
6.95805−1346.773 (85.0+ 219.693 )
PT∗ ( 85°C ) = 10
= 345.1 mm Hg
yB = 0.35 (881.6 ) /[10(760)] = 0.0406 mol Benzene mol
Raoult's Law: y B P = x B PB∗ ⇒
yT = 0.65 (345.1) /[10(760)] = 0.0295 mol Toluene mol
y N2 = 1 − 0.0406 − 0.0295 = 0.930 mol N2 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 ⋅pnB
+x iB ⋅ piB
= 0.50(140) + 0.30(3 5)+ 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 , p *nB = 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 140o F unless a fire breaks out, so the container
is adequate.
6.844711060.793
−
(120 +231.541)
6.55 a. Antoine: Pnp∗ (120 °C ) = 10
Pip∗ (120°C) = 10
6.73457 −992.019 (120 + 231.541)
= 6717 mm Hg
= 7883 mm Hg
Note: We are using the Antoine equation at a temperature well above the validity ranges in
Table B.4, so that all calculated values must be considered rough estimates.
When the first bubble of vapor forms,
x np = 0.500 mol n - C5H12 (l) / mol
xip = 0.500 mol i -C5H 12 (l)/mol
Total pressure: P=xnp ⋅ p *np +x ip ⋅ pip* = 0.50(6717) + 0.50(7883) = 7300 mm Hg
6- 36
6.55 (cont’d)
ynp =
xnp ⋅ p*np
P
=
0.500(6725)
= 0.46 mol n-C5 H12 (v)/mol
7342
yip = 1 − ynp = 1 − 0.46 = 0.546 mol i-C5H 12 (v)/mol
When the last drop of liquid evaporates,
ynp = 0.500 mol n - C5H 12 (v) / mol
x np + xip =
xnp =
y np P
*
pnp (120 o C)
+
yip P
*
pip (120o C)
yip = 0.500 mol i-C5H12 (v)/mol
=
0.500 P 0.500P
+
= 1 ⇒ P = 7250 mm Hg
6717
7883
0.500*7250 mm Hg
= 0.54 mol n-C5H12 (l)/mol
6717 mm Hg
xip = 1 − x np = 1 − 0.54 = 0.46 mol i -C5H12 (l)/mol
b. When the first drop of liquid forms,
ynp = 0.500 mol n - C5H 12 (v) / mol
yip = 0.500 mol i - C12 H12 (v) / mol
P = (1200 + 760) = 1960 mm Hg
x np + xip =
0.500 P 0.500 P
980
980
+ *
= 6.844711060.793/(
+ 6.73457− 992.019/(T + 231.541) = 1
*
−
Tdp + 231.541)
dp
pnp (Tdp ) pip (Tdp ) 10
10
⇒ Tdp = 63.1o C
p ∗np = 10
pip∗
6.844711060.793
−
( 63.1+ 231.541 )
6.73457 −992.019 ( 63.1+ 231.541 )
= 10
x np =
= 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,
x np = 0.500 mol n - C5H12 (l) / mol
x ip = 0.500 mol i - C5H 12 (l) / mol
Total pressure: P= xnp ⋅ p *np +x ip ⋅ p*ip
−
( T + 231.541)
⇒ 1960 = (0.5)106.844711060.793
+ (0.5)10
6.73457 − 992.019/(Tb p + 231.541)
⇒ T = 62.6°C
(Obtained with E-Z Solve)
xnp ⋅ p*np (62.6o C)
0.5(1731)
= 0.44 mol n -C5H12 (v)/mol
P
1960
yip = 1 − y np = 1 − 0.44 = 0.56 mol i -C5 H12 (v)/mol
ynp =
=
6- 37
6.56 B = benzene, T = toluene
n&v (mol / min) at 80o C, 3 atm
n& N2 =
10 L(STP)/min
xB [mol B(l)/mol]
yN2 (mol N2 /mol)
yB [mol B(v)/mol]
n& N2 ( mol / min)
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
6.958051346.773
−
(80+ 219.693 )
= 291.2 mm Hg
Antoine: p ∗B (80 °C ) = 10
pT∗ ( 80°C ) = 10
a. Initially, xB = 0.500, xT = 0.500.
Raoult's law:
yB =
x B p∗B (80o C) 0.500 ( 757.6 )
=
= 0.166 mol B(v) mol
P
3(760)
yT =
xT p∗T (80o C) 0.500 ( 291.2 )
=
= 0.0639 mol T(v) mol
P
3(760)
N 2 balance: 0.4464 mol N 2 / min = n&v (1 − 0.166 − 0.0639) ⇒ n&v = 0.5797 mol / min
mol I F
mol BI
mol B(v)
.
J G0166
J = 0.0962
min K H
mol K
min
F
⇒ n&B 0 = G0.5797
H
F
n&T0 = G0.5797
H
mol I F
mol BI
mol T(v)
J G0.0639
J = 0.0370
min KH
mol K
min
b. Since benzene is evaporating more rapidly than toluene, xB decreases with time and xT (=
1–xB) increases.
c. Since xB decreases, yB (= xBp B*/P) also decreases. Since xT increases, yT (= xTp T*/P) also
increases.
6.57 a. P =
∗
xhex phex
dTbp i
+
x hep p∗hep dTbp i
, yi =
x i pi∗ dTbp i
P
760 mm Hg = 0.500 10

6.885551175.817/(
−
Tbp + 224.867)
, Antoine equation for p ∗i
 + 0.500 10 6.90253−1267.828 /(Tbp + 216.823) 



E-Z Solve or Goal Seek ⇒ Tbp = 80.5° C ⇒ yhex = 0.713, y hep = 0.287
b.
xi =
yi P
∗
pi dTdp i
⇒
∑x
i i
∑
=P
yi
∗
i
pi dTdp i
=1
6- 38
6.57 (cont’d)

0.30
0.30

760 mmHg  6.88555−1175.817/(T + 224.867) + 6.902531267.828
 =1
−
/(
T
+
216.823)
dp
dp
10
10

E-Z Solve or Goal Seek ⇒ Tdp = 711
. ° C ⇒ x hex = 0.279 , x hep = 0.721
6.58
a.
f (T ) = P −
N
∑
xi pi* (T )
= 0 ⇒ T , where
pi* (T )
F
G Ai
H
= 10
−
Bi I
J
T + Ci K
i =1
yi (i = 1,2,L , N ) =
x i pi* (T )
P
b.
Calculation of Bubble Points
A
B
Benzene
6.89272 1213.531
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
xT
0.331
0.331
0.548
T bp (o C)
108.09
96.47
104.48
When x B = 1bpure benzene g, Tbp = dTbp i
C6 H 6
= 80.1o C
When x EB = 1bpure ethylbenzeneg, Tbp = dTbp i
When x T = 1bpure tolueneg, Tbp = dTbp i
C 7 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
C 8H 10
= 136.2 o C ⇒ Tbp , EB > Tbp ,T > Tbp, B
= 1106
. oC
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- 39
a. Basis: 150.0 L/s vapor mixture
6.59
n& 1 (mol/s) @ T(o C), 1100 mm Hg
0.600 mol B(v)/mol
0.400 mol H(v)/mol
&n2 (mol/s)
x2 [mol B(l)/mol]
(1- x2 ) [mol H(l)/mol]
n& 0 (mol/s)@120°C, 1 atm
1atm
0.500 mol B(v)/mol
0.500 mol H(v)/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 ⋅10 6.82485 −943.453/(T +239.711)
(1)
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 )
(2)
(3)
(4)
6.88555− 1175.817/(T + 224.867)
c. From (1) and (2), 1=
⇒ T = 57.0 ° C
x2 =
1100(0.6)
1100(0.4)
+
943.453
1175.817
10**(6.82485 −
) 10**(6.88555 −
)
T + 239.711
T + 224.867
1100(0.6)
6.82485943.453/(57.0
−
+ 239.711)
10
= 0.149 mol butane /mol
Solving (3) and (4) simultaneously ⇒ n&1 = 3.62 mol C4 H10 /s; n&2 = 1.03 mol C6 H14 /s
d.
Assumptions: (1) Antoine equation is accurate for the calculation of vapor pressure;
(2) Raoult’s law is accurate;
(3) Ideal gas law is valid.
6.60 P = n-pentane, H = n-hexane
170.0 kmol/h, T1a (o C), 1 atm
85.0 kmol/h, T1b (o C), 1
0.98 mol P(l)/mol
0.02 mol H(l)/mol
&n0 (kmol/h)
0.45 kmol P(l)/kmol
0.55 kmol H(l)/kmol
&n2 (kmol/h) (l),
o
x2 (kmol P(l)/kmol)
(1- x2 ) (kmol H(l)/kmol)
6- 40
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)
10
6.844711060.793/(
−
T1a + 231.541)
+
0.02(760)
10
6.88555−1175.817 /( T1a + 224.687)
= 1 ⇒ T1a = 37.3o C
Flow rate of column overhead vapor effluent. Assuming ideal gas behavior,
170 kmol 0.08206 m3 ⋅ atm (273.2 + 37.3) K
&
Vvapor =
= 4330 m 3 / h
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
V&distillate =
h
kmol P 0.621 kg P
0.02 (85) kmol H 86.17 kg H
L
= 9 .9 × 10 3 L / h
h
kmol H 0.659 kg H
c. Reboiler temperature.
+
6.84471−1060.793/(T2 + 231.541)
0.04 ⋅10
+ 0.96 ⋅106.88555−1175.817/(T2 + 224.867) = 760 ⇒ T2 =66.6°C
Boilup composition.
y2 =
−
+ 231.541)
x2 pP* (66.6 o C) 0.04 ⋅ 106.844711060.793/(66.6
=
= 0.102 mol P(v)/mol
P
760
⇒ (1 - y 2 ) = 0.898 mol H(v) / mol
d. Minimum pipe diameter
Fm
V& G
H s
3
I
J
K
2
πD min
FmI
= u max G J ×
( m2 )
Hs K
4
⇒ D min =
4V&vapor
π ⋅ u max
=
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- 41
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 5H 12 vapor mixture at
P = Pmin
Total condenser: 40° C is the bubble point of a 96% C 4 H 10 - 4% C5 H 12 liquid mixture at
P = Pmin
Dew Point: 1 =
(Raoult's Law)
∑x = ∑ p
i
yi P
∗
i b40° Cg
Antoine Eq. for
pi∗
⇒ Pmin =
i
pi∗ b40° Cg

943.453 
 6.82485 −

40 +239.711 

( C4 H10 ) = 10
Antoine Eq. for pi∗ ( C5 H12 ) = 10
⇒ Pmin =
∑y
1

1060.793 
 6.84471−

40 + 231.541 

= 2830.7 mm Hg
= 867.2 mm Hg
1
= 2596 mm Hg ( partial condenser )
0.96 2830.7 + 0.04867.2
Bubble Point: P =
∑ y P = ∑ x p b40° Cg
i
i
∗
i
P = 0.96 ( 2830.7 ) + 0.04 (867.2 ) = 2752 mm Hg ( total condenser )
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 kmolbutane kmol
Total balance: F& = 75 + 112.5 = 187.5 kmol / h
c.
Total balance as in b.
R& = 1125
. kmol / h
F& = 187.5 kmol / h
Equilibrium: 0.96 P = x1 b2830.70g
U P = 2596 mm Hg
V
bRaoult' s lawg 0.04 P = b1 − x1 gb867.22gWx 1 = 0.8803 mol butane mol
Butane balance: 187 .5 x 0 = 112.5b0.8803g + 0.96b75g ⇒ x 0 = 0.9122 mol butane mol reflux
6.62 a.
y i p ∗i
y x
p ∗ P p ∗A
=
⇒ α AB = A A = ∗A
=
= α AB
xi
P
y B xB
p B P p ∗B
Raoult's law:
b.
(
)

1507.434 
 7.06623 −

85 + 214.985 

pS 85 C = 10
*
o
1423.543 

 6.95650 − 85 + 213.091 


pEB (85 C) = 10
*
= 109.95 mm Hg
o

1213.531 
 6.89272−

85 + 219.888 

p*B (85o C) = 10
= 151.69 mm Hg
= 881.59 mm Hg
6- 42
6.62 (cont’d)
α S,EB
p*S 109.95
p * 881.59
= * =
= 0.725 , α B,EB = *B =
= 5.812
pEB 151.69
pEB 151.69
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 = 5.810 ⇒ y B =
xB
yB
P
6.63 a.
=
α ij x i
yi xi
⇒ yi =
(1 − y i ) b1 − xi g
1 + dα ij − 1i x i
x Bα B, EB
1 + (α B , EB − 1) x B
=
5.81 x B
*
, P = x B p B* + (1 − x B ) p EB
1 + 4.81x B
0.0
0.2
0.4
0.6
0.8
1.0 mol B blg mol
0.0 0.592 0.795 0.897 0.959 1.0 mol B bv g 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 ; x 1 = 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
−
T + 214.985)
P1 = (0.400 × 760) mmHg = ( 0.55 )10 6.89272−1203.531/(T + 219.888) + ( 0.45 )107.066231507.434/(
E-Z Solve

→ T1 = 67.6 o C
⇒ y1 =
x1 p ∗B bT g
=
0.55b508g
= 0.920 mol B mol ⇒ 0.080 mol S mol
P
0.400 × 760
B balance: y 0 n& v + x 2 n& l = y 1n& v + x 1n&l ⇒ x 2 = 0.910 mol B mol ⇒ 0.090 mol S mol
E-Z Solve
Stage 2: (0.400 × 760) mmHg = 0.910 p *B (T2 ) + 0.090 p *S ( T 2) 
→T2 = 55.3o C
y2 =
0.910b3310
. g
= 0.991 mol B mol ⇒ 0.009 mol S mol
760 × 0.400
B balance: y 1n& v + x 3n& l = y 2n&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- 43
6.64 Basis: 100 mol/s gas feed. H=hexane.
200 mol oil/s
n F (mol/s)
y F (mol H/mol)
1 – y F (mol N 2/mol)
100 mol/s
0.05 mol H/mol
0.95 mol N /mol
2
a.
n 2 (mol/s)
x 2 (mol H/mol)
1 – x 2 (mol Oil/mol)
n l (mol/s)
x +i 1 (mol H/mol)
n v (mol/s)
y i (mol H/mol)
Stage i
n l (mol/s)
x i (mol H/mol)
n v (mol/s)
y i– 1 (mol H/mol)
99.5% of H in feed.
n F = 95.025 mol s
N 2 balance: 0.95b100g = b1 − y F gn F
U
|
V⇒
99.5% absorption: 0.05b100 gb0.005g = y F n F |W y F = 2 .63 × 10 −4 mol H(v) mol
Mole Balance: 100 + 200 = 95.025 + n 2 ⇒ n2 = 205 mol s
Hexane Balance: 0.05b100g = 2.63 × 10 −4 b95.025g + x 1 b204.99g ⇒ x 1 = 0.0243 mol H(l) mol
n& L =
1
1
b200 + 205g ⇒ n& L = 202.48 mol s , n&G = b100 + 95.025g ⇒ n&G = 97 .52 mol s
2
2
Antoine
B
b.
y1 = x1 p ∗H b50° Cg / P = 0.0243b403.73g / 760 = 0.0129 mol H(v) mol
H balance on 1 st Stage: y0 n&v + x2 n& l = y1 n&v + x1n& l ⇒ x2 = 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=
nGf=
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
ye= 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- 44
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)
2.43E-02
1.24E-02
6.43E-03
3.44E-03
1.94E-03
...
4.38E-04
y(i)
5.00E-02
2.53E-02
1.29E-02
6.69E-03
3.58E-03
2.02E-03
...
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 70o C, 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
70o C, 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 ⇒ x B = 0.24 mol B mol bliquid g, y B = 0.46 mol B mol bliquid g
Basis: 1 mol
0.30 mol B(v)/mol
n V (mol vapor)
0.46 mol B(v)/mol
n L (mol liquid)
0.24 mol B(l)/mol
Balances
Total moles: 1 = nV + n L
n L = 0.727 mol nV
U
mol vapor
⇒
= 0.375
V⇒
B:
0.30 = 0.46 nV + 0.24 n L W nV = 0.273 mol n L
mol liquid
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, y B = 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
nV = 3.16 mol
U
n
mol vapor
⇒ v = 0.46
V⇒
B bal.:
3 = 0.43nV + 0.24 n L W n L = 6.84 mol n l
mol liquid
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- 45
mol T
10 −3 L
0.866 g T
= 1.0 L
6.66 (cont’d)
V=
3.16 mol vapor 0.08206 L ⋅ atm
373 K
760 mm Hg
mol ⋅ K 750 mm Hg
1 atm
− 0.6 L = 97.4 L
(Liquid volume is about 0.6 L)
iii) 600 mm Hg
v=
10 mol vapor
0.08206 L ⋅ atm
373K
760 mm Hg
mol ⋅ K 600 mm Hg
1 atm
= 388 L
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
U
nV
x −x
= F
V ⇒ x F nV + x F n L = ynV + xn L ⇒ f =
MeOH balance: x F n f = ynV + xn L W
nL
y−x
x F = 0.4 , x = 0.23, y = 0.62 ⇒ 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
o
T( C)
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- 46
0.8
1
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 + 0.34 nL 
⇒ 76.2 mole% vapor
(iii) ρ A (l ) = 0.791 g/cm 3 , ρ E(l) = 0.789 g/cm 3 ⇒ ρ l ≈ 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
⇒ Ml = ( 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
(1 mol)(50.15 g / mol)
Liquid volume: Vl =
= 63.48 cm 3
3
(0.790 g / cm )
Vapor volume:
3.2 mol 22400 cm 3 (STP) (65 + 273)K
= 88 ,747 cm 3
mol
273K
88,747
Volume percent of vapor =
× 100% = 99.9 volume % vapor
88747 + 63.48
Vv =
d. For a basis of 1 mol fed, guess T, calculate n V as above; if n V ≠ 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 = xi p i* ⇒ 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 = 66.16 o C
xp*A 0.5 ×107.11714 −1210.595/(66.25+ 229.664)
=
= 0.696 mol acetone/mol
P
760
∆Tbp
66.25 − 61.8
The actual Tbp = 61.8o C ⇒
=
×100% = 7.20% error in Tbp
Tbp (real)
61.8
y=
y A = 0.674 ⇒
∆y A
0.696 − 0.674
=
×100% = 3.26% 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- 47
6.69 a. B = benzene, C = chloroform. At 1 atm, (Tbp )B = 80.1o C, (Tbp)C = 61.0o C
The Txy diagram should look like Fig. 6.4-1, with the curves converging at 80.1o C when
xC = 0 and at 61.0o C when xC = 1. (See solution to part c.)
b.
Txy Diagram for an Ideal Binary Solution
A
B
C
Chloroform
6.90328
1163.03
227.4
Benzene
6.89272 1203.531
219.888
P(mmHg)=
760
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
Txy diagram
(P=1 atm)
85
75
Vapor
o
T( C)
80
70
Liquid
65
60
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Mole fraction of chloroform
6- 48
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
0.2 0.3 0.4
0.5
65
60
0
0.1
0.6
0.7
0.8 0.9
1
Mole fraction of choloroform
∆T
71 − 75.3
=
× 100% = −5.7% error in Tbp
Tactual
75.3
Raoult’s law: Tbp = 71o C, y = 0.58 ⇒
∆y
0.58 − 0.60
=
× 100% = −3.33% error in y
y actual
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.
6.70 P ≈ 1 atm = 760 mm Hg = x m p m* dTbp i + b1 − x m gp *P dTbp i
7.878631473.11/(
−
Tbp + 230)
760 = 0.40 ×10
7.74416 −1437.686/(Tbp + 198.463)
+ 0.60 × 10
E-Z Solve

→T = 79.9 o 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- 49
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.
•
Calculate molar flow rate of product dn& p i 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 U
•
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 U
•
•
•
•
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.
V ⇒ n& c , n&r
W
A balance about still
V ⇒ n& 0 , n&3
Overall H balanceW
Molar flow rate of product
M = 0.97 M A + 0.03 M E = b0.97 gb44.05g + b0.03gb46.05g = 44.11 g mol
n& p =
1000 kg 1 kmol
= 22.67 kmol h
h
44.11 kg
Table B.4 (Antoine) ⇒ p *A ( −40 °C ) = 44.8 mm Hg
p *E ( −40 °C ) = 0.360 mm Hg
Note: We are using the Antoine equation at a temperature below the ranges of validity in Table
B.4, so that all calculated values must be considered rough estimates.
Raoult’s law ⇒ yA3 =
0.550 p*A ( −40 °C )
P
=
0.550(44.8)
= 0.03242 kmol A/kmol
760
6- 50
6.71 (cont’d)
yE3 =
0.450 p*E ( −40 °C)
yH3
P
= 1 − y A3 − y E3
0.450(0.360)
= 2.13 × 10 −4 kmol E kmol
760
= 0.9674 kmol H 2 kmol
=
Mole balance about still: n& c = n& p + n& r ⇒ n&c = 22.67 + n&r U
n& r = 29.5 kmol / h recycle
A balance about still: 0.550n&c = 0.97 (22 .67) + 0.05n& r
n& c = 52.1 kmol / h
V⇒
W
A balance about scrubber: n& A4 = n& 3 y A3 = 0.02815n& 3
(1)
E balance about scrubber: n& E4 = n&3 y E3 = 2.03 × 10 −4 n& 3
(2)
H 2 balance about scrubber: n& H4 = n&3 y H3 = 0.9716n&3
(3)
Overall C balance:
n& 0 (mol E) 2 mol C
h
1 mol E
= bn& A4 gb2 g + bn& E4 gb2 g + d0.97n& p i b2 g + d0.03n& p i b2 g
⇒ n& 0 = n& A 4 + n& E 4 + 22.67
(4)
Overall H balance:
6n& 0 = 2 n& H4 + 4 n& A4 + 6n& E4 + n& p b0.97gb4g + b0.03gb6g
(5)
Solve (1)–(5) simultaneously (E-Z Solve):
n&0 = 23.4 kmol E/h (fresh feed), n&H 4 = 22.8 kmol H2 /h (in off-gas)
n&3 = 23.5 kmol/h, n&A 4 = 0.76 kmol A/h, n&E 4 = 0.0050 kmol E/h
A balance about feed mixing point: n& A1 = 0.05n& r = 1.47 kmol A h
E balance about feed mixing point: n& E1 = n&0 + 0.95 n&r = 51.5 kmol E h
E balance about condenser: n& E2 = n&3 y E3 + 0.450 n&c = 23.5 kmol E h
Ideal gas equation of state:
Vreactor feed =
(1.47 + 51.5 ) kmol
b. Overall conversion =
h
n&0 − 0.03 n& p
22.4 m 3 (STP ) ( 273+280 ) K
= 2.40 × 10 3 m 3 h
1 kmol
273K
23.4 − ( 0.03 )( 22.67 )
× 100% = 97%
n&0
23.4
n& − n&
51.5 − 23.5
Single-pass conversion = E1 E2 × 100% =
× 100% = 54%
n&E1
51.5
Feed rate of A to scrubber: n&A4 =0.76 kmol A/h
Feed rate of E to scrubber:
× 100% =
n& E4 = 0.0050 kmolE h
6- 51
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 /106 SCF gas
90 o F, 500 psia
Absorber
n& 7 ( lb - mole W / d)
4.0 × 106 SCF / d
4 × 80 = 320 lb m W / d
n& 1 ( lb - mole G / d)
n& 2 [lb - mole W(v) / d]
lb - mole TEG I
J
K
d
F lb - mole W I
n& 6 G
J
H
K
d
F
n& 5 G
Distillation
Column
H
F lb - mole TEG I
n&5 G
J
H
K
d
F lb- mole W I
n&8 G
J
H
K
d
Overall system D.F. analysis:
Water feed rate : n& 2 =
5 unknowns (n&1 , 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
d
18.0 lb m
= 17 .78 lb - moles W / d
Dry gas feed rate:
4.0 × 106 SCF 1 lb- mole
lb - moles W
&n1 =
− 17.78
= 1112
. × 104 lb - moles G/ d
d
359 SCF
d
Overall G balance: n&1 = n& 3 ⇒ n&3 = 1112
.
× 10 4 lb - moles G / d
Flow rate of water in dried gas:
n& 4 =
(n& 3 + n&4 ) lb - moles 359 SCF gas 10 lb m W 1 lb - mole W
d
lb - mole 10 6 SCF
18.0 lb m
n& =1.112 ×104
3
 
→ n& 4 = 2.218 lb - mole W(l) / d
Overall W balance:
n& 7 =
(17.78 − 2.218) lb - moles W
18.0 lb m
d
1 lb - mole
6-52
= 280
lb m W
×
d
F 1 ft 3 I
G
J
H62.4 lb m K
= 4.5
ft 3 W
d
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
n&3 + n& 4 (2.218 + 1.112 × 10 ) lb - moles / d
lb - mole
Henry’s law: yw P = Hwxw ⇒
(1.99 × 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
( x w ) max =
37 lb m TEG 1 lb - mole TEG 18.0 lbm W
n&5
lb - mole TEG
=
= 4.434
lb m W 150.2 lb m TEG 1 lb m W
n&2 − n&4
lb - mole W absorbed
⇒ n&5 = 4.434(17.78 − 2.22) = 69.0 lb - moles TEG / d
xw = 0.80(0.0170) = 0.0136
lb - mole W
n&6
&5 = 69 .0
=
n
→ n&6 = 0951
. lb - mole W/ d
lb - mole
n&5 + n&6
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
n&8 = (17.78 + 095
. − 2.22) lb -moles W/ d = 16.51 lb - moles W/ d
16.51 lb - moles W / d
⇒ xw =
= 019
. lb - mole W / lb - mole
(16.51 + 69.9) lb - moles / d
c. The distillation column recovers the solvent for subsequent re-use in the absorber.
6.73 Basis: Given feed rates
G1
G2
G3
100 mol/h
200 mol air/h
n1 (mol/h)
0.96 H2
0.999 H 2
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 2S/mol)
(1 – x 3) (mol solvent/mol)
0.998 solvent
0°C
heater
6-53
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)
Equilibrium condition: At G1, p H 2S = b0.04gb18
. atmg = 0.072 atm
⇒ x3 =
p H 2S
H H2 S
=
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 ⇒ n&1 , 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
Overall H 2 balance: b100gb0.96g = 0.999 n1 ⇒ n&1 = 96.1 mol h
Overall H 2S balance:
b100gb0.04 g = 0.001n&1
n&1 = 96.1
+ n&2 ⇒ n&2 = 3.90 mol H 2S h
Volumetric flow rate at stripper outlet
b200 + 3.90gmol 22.4 litersbSTP g
V&G4 =
h
1 mol
b273 +
40gK
273 K
= 5240 L hr
H 2 S and solvent balances around absorber:
b100gb0.04 g + 0.002n& 4
= 0.001n&1 + n& 3 x 3 ⇒ n& 4 = 1.335n& 3 − 1952U
|
V ⇒ n&3 ≈ n& 4 = 5830 mol h
0.998n&4 = n& 3 d1 − 2.67 × 10 −3 i
|
W
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))
NaHCO 3 balance ⇒ 16.4 = 111
. + ms ⇒ m s = 5.3 g NaHCO 3 bsg
% crystallization =
5.3 g cryst allized
× 100% = 32.3%
16.4 g fed
6.75 Basis: 875 kg/h feed solution
m1 (kg H2 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
6-54
6.75 (cont’d)
Analysis of feed: 2KOH + H 2 SO 4 → K 2 SO 4 + 2H 2 O
22.4 mL H 2 SO 4 bl g
1L
0.85 mol H 2 SO 4 2 mol KOH
3
5 g feed soln
10 mL
L
1 mol H 2 SO 4
= 0.427 g KOH g feed
x0 =
56.11 g KOH
1 mol KOH
60% recovery: 875 (0.427 )( 0.60 ) = 224.2 kg KOH h
m3 =
224.2 kg KOH 92.15 kg KOH ⋅ 2H2 O
= 368.2 kg KOH ⋅ 2H2 O h (143.8 kg H 2 O h )
h
56.11 kg KOH
KOH balance: 0.427 (875 ) = 224.2 + 1.03m2 ⇒ m2 = 145.1 kg h
Total mass balance: 875 = 368.2 + 2.03 (145.1) + m1 ⇒ m1 = 212kg H 2 O h evaporated
6.76 a.
R
0 30
45
g A dissolved
mL solution
CA 0 0.200 0.300
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 = 41.0 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 =
= 0106
.
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
= 0.118 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 bsg
100 g S
b. Mass of solution:
c.
A remaining in sol n
6 4 g44
47444 4
8
g A initia l
6 44
7448 0.5 × 390 g S 11.8 g A
= 23.0 g A bsg
b160 − 114 gg A −
100 g S
6.77 a. Table 6.5-1 shows that at 50o F (10.0o F), the salt that crystallizes is MgSO 4 ⋅ 7 H 2 O , which
contains 48.8 wt% MgSO 4.
b. Basis: 1000 kg crystals/h.
m& 0 (g/h) sat’d solution @ 130o F
m& 1 (g/h) sat’d solution @ 50o F
0.35 g MgSO4 /g
0.65 g H2 O/g
0.23 g MgSO4 /g
0.77 g H2 O/g
6-55
1000 kg MgSO4 ·7H2 O(s)/h
6.77 (cont’d)
& 0 = 2150 kg feed / h
m
Mass balance: m& 0 = m& 1 + 1000 kg / h
MgSO 4 balance: 0.35m 0 = 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
kg anhydrous MgSO 4
h
6.78 Basis: 1 lb m feed solution.
Figure 6.5-1 ⇒ a saturated KNO3 solution at 25o C contains 40 g KNO3 /100 g H2 O
⇒ x KNO3 =
40 g KNO3
= 0.286 g KNO3 / g = 0.286 lb m KNO3 / lb m x
(40 + 100) g solution
1 lb m solution @ 80o C
0.50 lb m KNO3 /lb m
0.50 lb m H2 O/lb m
m1 (lb m) sat’d solution @ 25o C
0.286 lb m KNO3 /lb m soln
0.714 lb m H2 O/lb m soln
m2 [lb m KNO3 (s)]
m1 = 0.700 lb m solution / lb m feed
Mass balance: 1 lb m = m1 + m2
⇒
m2 = 0.300 lb m crystals/ lb m feed
KNO3 balance: 0.50 lb m KNO 3 = 0286
. m1 + m2
Solid / liquid mass ratio =
0.300 lb m crystals/ lb m feed
= 0.429 lb m crystals/ lb m solution
0.700 lb m solution / lb m feed
6.79 a. Basis: 1000 kg NaCl(s)/h.
Figure 6.5-1 ⇒ a saturated NaCl solution at 80o C contains 39 g NaCl/100 g H2 O
⇒ 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& 1 (kg/h) sat’d solution @ 80o C
m& 0 (kg/h) solution
0.100 kg NaCl/kg
0.900 kg H2 O/kg
0.281 kg NaCl/kg soln
0.719 kg H2 O/kg soln
1000 kg NaCl(s)/h
& 0 = m&1 + m
&2
Mass balance: m
NaCl balance: 0.100 kg NaCl = 0281
. m& 1 + m
&2
Solid / liquid mass ratio =
⇒
& 1 =0.700 lbm solution / lb m feed
m
m& 2 = 0.300 lbm crystals/ lb m feed
0.300 lbm crystals/ lb m feed
=0.429 lbm crystals / lb m solution
0.700 lbm 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
6-56
6.79 (cont’d)
0100
. (m& 0 ) min = 1000 kg NaCl / h ⇒ ( m
& 0 ) min = 10,000 kg / min
& 2 = 9000 kg H 2O / h
Evaporation rate: m
&1 = 0
Exit solution flow rate: m
b.
m& 2 [kg H 2 O(v) / h]
m& 1 (kg/h) sat’d solution @ 80o C
0.281 kg NaCl/kg soln
0.719 kg H2 O/kg soln
1000 kg NaCl(s)/h
m& 0 (kg/h) solution
0.100 kg NaCl/kg
0.900 kg H2 O/kg
40% solids content in slurry ⇒ 1000
kg NaCl
kg
& 1 ) max ⇒ (m& 1 ) max = 2500
= 0.400(m
h
h
NaCl balance: 0.100m
& 0 = 0.281(2500) ⇒ m
& 0 = 7025 kg / h
Mass balance: m
& 0 = 2500 + m
&2 ⇒ m
& 2 = 4525 kg H2O evaporate / h
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 b1 + 0.20g kg soln
n& 2 (mol/ h)
y2 (mol W(v) / mol)
m& e [kg W(v) / h)
(1 − y2 )(mol A / mol)
90 C, 1 atm, Tdp = 392
. C
o
& f (kg/ h)
m
m& f + m& r (kg / h)
CRYSTALLIZERCENTRIFUGE
0.210 kg K / kg
0.790 kg W(l)/ kg
m& 1 ( kg / h)
0.90 kg K(s) / kg
DRYER
o
1000 kg K(s) / h
0.10 kg soln / kg
0.1667 kg K/ kg
0.8333 kg W/ kg
n& a (mol A / h)
& r (kg recycle / h)
m
0.1667 kg K / kg
0.8333 kg W / kg
Dryer outlet gas: y2 P = p *W b39.2° Cg ⇒ y2 =
53.01 mm Hg
= 0.0698 mol W mol
760 mm Hg
6-57
& f = 1000 kg K h ⇒ m
& f = 4760 kg h feed solution
Overall K balance: 0.210m
6.80 (cont’d)
.
. m
& 1 g = 1000 kg h ⇒ m& 1 = 1090 kg h
K balance on dryer: 0.90m& 1 + b01667
gb010
Mass balance around crystallizer-centrifuge
m& f + m& r = m
& e + m& 1 + m
& r ⇒ me = 4760 − 1090 = 3670 kg h water evaporated
b0.10 × 1090g kg h not recycled
95% solution recycled ⇒ m& r =
95 kg recycled
5 kg not recycled
= 2070 kg h recycled
Water balance on dryer
. gb1090g
b0.8333gb010
1801
. × 10
−3
kg W h
kg mol
= 0.0698n&2 ⇒ n&2 = 7.225 × 104 mol h
Dry air balance on dryer
1 − 0.0698g7.225 × 10 4 mol 22.4 L bSTPg
na = b
= 151
. × 10 6 LbSTPg h
h
1 mol
6-58
6.81. Basis : 100 kg liquid feed. Assume P atm=1 atm
100 kg Feed
0.07 kg Na 2CO 3 / kg
0.93 kg H 2 O / kg
n2w (kmol H2 O )(sat' d)
Reactor
Reactor
e
n1(kmol)
0.70 kmol CO 2 / kmol
0.30 kmol Air / kmol
n2c (kmol CO2 )
n2a (kmol Air)
70o C, 3 atm(absolute)
m3 ( kg NaHCO 3( s))
Rm 4 (kg solution)
|
S0.024 kg NaHCO3
|
T0.976 kg H 2 O / kg
/
U
|
kgV
|
W
Filtrate
m5 (kg)
Filter
0.024 kg NaHCO 3 / kg
0.976 kg H 2 O / kg
Filter cake
m6 (kg)
0.86 kg NaHCO 3 (s) / kg
R0.14 kg solution
|
S0.024 kg NaHCO3
|
T0.976 kg H 2 O / kg
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
46 kg Na
(m + 0.024m4 ) kg NaHCO3
23 kg Na
= 3
kg
106 kg Na2 CO3
84 kg NaHCO3
⇒ 3.038 = 0.2738(m3 + 0.024m4 )
(1)
Air balance: 0.300 n1 = n2a
(2 )
C balance on reactor :
n1 (kmol) 0.700 kmol CO2
12 kg C
100 kg 0.07 kg Na 2CO3
12 kg C
+
kmol
1 kmol CO2
kg
106 kg Na 2CO3
12
= (n2c )(12) + (m3 + 0.024m2 )( ) ⇒ 8.40n1 + 0.7924 = 12n2c + 01429
.
(m3 + 0.024m4 )
(3)
84
H balance :
2
1
2
) = ( n2 w )(2 ) + (m3 + 0.024 m4 )( ) + 0.976m4 ( )
18
84
18
⇒ 10.33 = 2 n2 w + 0.01190( m3 + 0.024 m4 ) + 0.1084 m4
(4 )
(100)(0.93)(
6-59
U
|
/ kgV
|
W
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
= (n 2w )(16) + n2c (32) + (m3 + 0.024 m4 )( ) + 0.976m4 ( )
84
18
⇒ 22.4 n1 + 85.84 = 16n 2w + 32n 2c + 0.5714(m 3 + 0.024 m4 ) + 0.8676 m4
(5)
Raoult's Law :
y w P = p *w (70 o C) ⇒
⇒ n2 w = 0.1025( n2 w
n2 w
233.7 mm Hg
=
n 2w + n 2c + n 2a (3 * 760) mm Hg
+ n 2c + n 2a )
(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,
n 2w = 0.0848 kmol H 2 O(v),
n2c = 0.500 kmol CO 2 ,
m3 = 8.874 kg NaHCO3 (s),
m4 = 92.50 kg solution
NaHCO3 balance on filter:
m3 + 0.024 m4 = 0.024 m5 + m6 [0.86 + (0.14 )( 0.024 )]
m3 =8.874
11.09 = 0.024m 5 + 0.8634m6
(7 )
m4 =92 .50
Mass Balance on filter: 8.874 + 92.50 = 1014
. = m5 + m6
Solve (7) & (8) ⇒
Scale factor =
m5 = 91.09 kg filtrate
m6 = 10.31 kg filter cake
(8)
⇒ (0.86)(10.31) = 8.867 kg NaHCO3 (s)
500 kg / h
= 56.39 h −1
8.867 kg
(a) Gas stream leaving reactor
R46.7kmol / h
n& 2w = (0.0848)(56.39) = 4.78 kmol H 2 O(v) / h U |
|
| 0.102 kmol H 2 O(v) / kmol
n& 2c = (0.500)(56.39) = 28.2 kmol O 2 / h V ⇒ S
0.604 kmol CO 2 / kmol
n& 2a = (0.2426)(56.39) = 13.7 kmol air / h |W |
| 0.293 kmol Air / kmol
T
n& RT
V&2 = 2
=
P
(46.7 kmol / h)(0.08206
3 atm
m 3atm
)(343 K)
kmol ⋅ K
= 438 m 3 / h
56.39 × 0.8086 kmol 22.4 m 3 (STP)
1h
(b) Gas feed rate : V&1 =
= 17.0 SCMM
h
kmol
60 min
6-60
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 Na 2 CO3 solution.
(d) If T dropped in the filter, more solid NaHCO 3 would be recovered and the residual
solution would contain less than 2.4% NaHCO 3.
(e)
Benefit: Higher pressure ⇒ greater pCO2
Henry's law
higher concentration of CO2 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
m
& 1 ( lb m H 2O / h)
m
& 6 (lb m / h)
0.23 lb m MgSO 4 / lb m
0.77 lbm H 2O / lb m
Filter II
Filter I
& 2 (lbm soln / h) U
Rm
|
|
0.32
kg MgSO 4 / kgV
S
| 0.68 kg H O / kg
|
T
2
W
6000 lb m I / h
110o F
& 4 ( lb m MgSO 4 ⋅ 7 H 2 O / h)
m
6000 lbm I / h
R300 lbm soln /
|
S0.32 MgSO4
|
T0.68 H 2 O
m
& 3 ( lb m so ln/ h)
0.32 MgSO4
0.68 H 2 O
Crystallizer
& 5 (lb m soln )
Rm
U
|
|
| 0.23 lb m MgSO 4 / lb m |
S
V
| 0.77 lb m H 2 O / lb m
|
|
|W
T
& 4 ( lb m MgSO 4 ⋅ 7H 2 O)
m
& 4 ( lb m soln )
R0.05 m
U
|
|
S0.23 lb m MgSO 4 / lbm V
|
|
T0.77 lbm H 2 O / lb m
W
a. Heating the solution dissolves all MgSO 4 ; filtering removes I, and cooling recrystallizes
MgSO 4 enabling subsequent recovery.
(b) Strategy: Do D.F analysis.
6-61
hU
|
V
|
W
6.82(cont'd)
Overall mass balance
Overall
Diss. tank overall mass balanceU
U
& 1 , m& 4
V⇒m
MgSO 4 balanceW
V⇒
W
Diss. tank MgSO 4 balance
&2,m
&6
m
( MW) MgSO4 = (2431
. + 32.06 + 6400
. ) = 12037
. , ( MW) MgSO4 ⋅7H2 O = (12037
. + 7 *1801
. ) = 24644
.
Overall MgSO 4 balance:
60,000 lb m
0.90lb m MgSO4 ⋅ 7H 2 O
h
lb m
= (300 lb m / h)(0.32 lb m
120.37 lb m MgSO 4
246.44 lb m MgSO 4 ⋅ 7H 2 O
MgSO 4 / lb m ) + m& 4 (12037
. / 246.44 ) + 0.05m& 4 (0.23)
& 4 = 5.257 x10 4 lb m crystals / h
⇒ m
& 4 = 5.257 x10 4 lb m / h
m
&4
Overall mass balance: 60,000 + m& 1 = 6300 + 1.05m
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
U
54 ,000(120.37 / 24644
. ) + 0.23m
& 6 = 0.32 m& 2 VW
m
& 2 = 1512
.
x10 5 lb m / h
& 6 = 9.575x10 4 lb m / h recycle
m
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 2SO 4 / h (10 wt%)
1000 kg HNO 3 / h
m
& w (kg H 2 O / h)
m
& 2 (kg CaSO4 / h)
m
& 3 (kg Ca(NO3) 2 / h)
m
& 4 (kg H2O / h)
Filter cake
& 5 (kg / h)
m
0.96 kg CaSO 4 (s) / kg
0.04 kg soln / kg
& 0 (kg CaCO3 / h)
m
& 0 (kg solution / h)
2m
m
& 0 (kg CaCO 3 / h)
2m
& 0 (kg solution / h)
Solution composition:
6-62
m
& 8 (kg soln / h)
X a (kg CaSO 4 / kg)
R
|
500 X a (kg H 2 O / kg)
S
|
T (1 − 501X a )(kg Ca(NO 3 ) 2
/
U
|
V
kg)|W
6.83 (cont’d)
b. Acid is corrosive to pipes and other equipment in waste water treatment plant.
c. Acid feed:
1000 kg H 2SO 4 / h
= 0.10 ⇒ m
& w = 8000 kg H 2 O / h
(2000 + m& w ) kg / h
Overall S balance:
1000 kg H 2SO 4
32 kg S
h
98 kg H 2SO 4
+
=
& 5 (kg / h) (0.96 + 0.04 X a ) (kg CaSO 4 )
m
kg
32 kg S
136 kg CaSO 4
& 8 (kg / h) X a (kg CaSO 4 )
m
32 kg S
kg
136 kg CaSO 4
⇒ 3265
. = 0.2353m
& 5 (0.96 + 0.04 X a ) + 0.2353m& 8 X a
(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 ) + 0.171m& 8 (1 − 501X a )
(2)
Overall Ca balance:
& 5 (kg / h) (0.96 + 0.04X a ) (kg CaSO 4 )
40 kg Ca
m
40 kg Ca
=
100 kg CaCO 3
kg
136 kg CaSO 4
& 5 (kg / h)
(1 − 501X a ) (kg Ca(NO 3 ) 2 ) 0.04m
40 kg Ca
+
kg
164 kg Ca(NO 3 ) 2
& 8 (kg / h) X a (kg CaSO 4 )
m
40 kg Ca
+
kg
136 kg CaSO 4
& 8 (kg / h) (1 − 501X a ) (kg Ca(NO 3 ) 2 )
m
40 kg Ca
+
kg
164 kg Ca(NO 3 ) 2
& 0 (kg / h)
m
& 0 = 0.294m& 5 (0.96 + 0.04 X a ) + 0.00976m
& 5 (1 − 501 X a )
⇒ 0.40m
& 8 X a + 0.244m& 8 (1 − 501X a )
+ 0.294 m
( 3)
Overall C balance :
m
& 0 (kg / h)
12 kg C
100 kg CaCO 3
⇒ 0.01m
& 0 = n&1
=
n& 1 (kmol CO 2 / h)
(4 )
6-63
1 kmol C
12 kg C
1 kmol CO 2
1 kmol C
6.83 (cont’d)
Overall H balance :
1000 (kg H 2SO4 )
2 kg H
+
1000 kg HNO3
h
98 kg H 2SO 4
⇒ 92517
. = 2.22m
& 5 X a + 5556
. m
&8 Xa
(5)
1 kg H
h
+
m
& w (kg / h)
2 kg H
63 kg HNO 3
18 kg H 2O
& 5 (kg / h) 500 X a (kg H 2 O)
& (kg / h) 500 X a (kg H 2 O) 2 kg H
0.04m
2 kg H
m
=
+ 8
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,
& 8 = 9584.9 kg soln / h,
m
n&1 = 18.1 kmol CO 2 / h(v),
X a = 0.00173 kg CaSO 4 / kg
Recycle stream = 2 * m& 0 = 3625 kg soln / h
0.00173(kg CaSO 4 / kg)
R
|
500 * 0.00173(kg H 2 O / kg)
S
| (1 − 501 * 0.00173)(kg Ca(NO )
T
3 2
d.
/
U
|
V⇒
kg) |W
R0173%
.
CaSO 4 U
|
|
|
|
S86.5% H 2 O
V
|
|
| 13.3% Ca(NO 3 ) 2 |
T
W
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):
z = 0.86 ⇒ V$ =
zRT 0.86 0.08206 L ⋅ atm 313.2 K
L
=
= 0.737
mol
⋅
K
30
atm
P
mol CO2
Volumetric flow rate of CO2 :
18.1kmol CO 2
V& = n&1 * V$ =
h
e.
0.737 L
1000 mol
mol CO 2
1 kmol
= 1.33x104 L / h
Solution saturated with Ca(NO3 )2 :
⇒
1 − 501X a (kg Ca(NO 3 ) 2 / kg)
= 1.526 ⇒ X a = 0.00079 kg CaSO4 / kg
500Xa (kg H 2 O / kg)
Let m& 1 (kg HNO3 /h) = feed rate of nitric acid corresponding to saturation without
crystallization.
6-64
6.83 (cont’d)
Overall S balance:
1000kg H 2SO 4
32 kg S
h
98 kg H 2SO 4
+
=
m
& 5 (kg / h) (0.96 + (0.04)(0.00079)) (kg CaSO 4 )
32 kg S
kg
136 kg CaSO 4
& 8 (kg / h) 0.00079(kg CaSO4 )
m
kg
32 kg S
136kg CaSO 4
& 5 + 0.000186m
&8
⇒ 326.5 = 0.226m
(1' )
Overall N balance:
m
&1 (kg HNO3 )
14 kg N
h
63 kg HNO3
=
0.04m
& 5 (kg / h) (1 − (501)(0.00079)) (kg Ca(NO 3) 2 )
kg
+
28 kg N
164kg Ca(NO 3) 2
& 8 (kg / h) (1 − (501)(0.00079)) (kg Ca(NO3 )2 )
m
kg
28 kg N
164 kg Ca(NO3 )2
&1 = 000413
& 5 + 0103
&8
⇒ 0222
. m
.
m
. m
(2')
Overall H balance:
1000 (kg H 2SO 4 )
2 kg H
h
98 kg H 2SO 4
+
8000 (kg / h)
2 kg H
18 kg H2 O
=
+
+
m& 1 kg HNO3
1 kg H
h
63 kg HNO3
& 5 (kg / h) 500(0.00079) (kg H 2 O)
0.04m
kg
18 kg H2 O
m& 8 (kg / h) 500(0.00079) (kg H 2O)
kg
2 kg H
2 kg H
18 kg H2 O
& 1 = 0.00175m& 5 + 0.0439m& 8
⇒ 909.30 + 0.0159m
(3')
Solve eqns (1')-(3') simultaneously using E-Z solve:
m& 1 = 1155
. x10 4 kg / h;
m
& 5 = 1.424 x10 3 kg / h;
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
6-65
m
& 8 = 2.484x10 4 kg / h
6.84
Moles of diphenyl (DP):
Moles of benzene (B):
⇒ x DP =
56.0 g
U
= 0.363 mol
|
154.2 g mol
|
V
550.0 ml 0.879 g 1 mol
= 6.19 mol|
|
ml
78.11 g
W
0.363
= 0.0544 mol DP mol
6.19 + 0.363
p *B bT g = (1 − x DP ) p *B bT g = 0.945b120.67 mm Hgg = 114.0 mm Hg
2
8.314b2732
. + 5.5g
RTm0
∆Tm =
x DP =
$
9837
∆H
2
b0.0554 g = 3.6
K = 3.6o C ⇒ Tm = 5.5 − 3.6 = 1.9 ° C
m
2
∆Tbp =
8.314b273.2 + 801
.g
RTb02
x DP =
. K = 1.85o C
b0.0554 g = 185
$
30
,
765
∆H
v
⇒ Tb = 80.1 + 185
. = 82.0 ° C
6.85
Tm0 = 0.0 o C, ∆Tm = 4.6 o C=4.6K
∆Tm∆Hˆ m
(4.6K)(600.95 J/mol)
Eq. 6.5-5

→
x
=
=
= 0.0445 mol urea/mol
u
Table B.1
2
R(Tm 0 )
(8.314 J/mol ⋅ K)(273.2K)2
Eq. (6.5-4) ⇒ ∆Tb =
RTb 0 2
(8.314)(373.2) 2
xu =
0.0445 = 1.3K = 1.3o C
∆Hˆ v
40,656
1000 grams of this solution contains mu (g urea) and (1000 – mu ) (g water)
nu 1 (mol urea) =
mu1 (g)
60.06 g/mol
nw 1 (mol water) =
(1000 − mu1 )(g)
18.02 g/mol
mu1
(mol urea)
60.06
xu 1 = 0.0445 =
⇒ mu 1 = 134 g urea, mw1 = 866 g water
 mu1 + (1000 − mu1 )  (mol solution)
 60.06

18.02
∆Tb = 3.0 o C = 3.0K ⇒ xu 2 =
∆Tb ∆Hˆ 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
6-66
6.86 x aI =
.
b0.5150 gg b1101
.
b0.5150 g g b1101
∆Tm =
⇒
b1 −
g mol g
g molg+ b100.0 gg b94.10 g mol g
= 0.00438 mol solute mol
RTm20
∆T I
xI
∆T II
0.49 ° C
mol solute
x s ⇒ mII = IIs ⇒ xsII = x Is mI = 0.00438
= 0.00523
$
0.41° C
mol solution
∆Tm
xs
∆Tm
∆Hm
0.00523g mol solvent 94.10 g solvent
0.00523 mol solute
1 mol solvent
8.314b2732
. − 5.00g
RTm20
xs =
∆Tm
0.49
0.4460 g solute
95.60 g solvent
= 8350
. g solute mol
2
∆H$ m =
6.87 a.
ln ps* bTb0 g = −
b0.00523g = 6380
J mol = 6.38 kJ / mol
∆H vI
∆H vII
+ B , ln ps* bTbs g = −
+B
RTb0
RTbs
Assume ∆HvI ≅ ∆HvII ; T0 Ts ≅ T02
⇒ ln Ps* bTb 0 g − ln P0* bTbs g = −
b. Raoult’s Law:
p *s bTb0 g = b1 −
∆Hv
R
F 1
G
H Tb0
x g p *0 bTbs g
−
1 I ∆Hv Tbs − Tb 0
≅
Tbs JK
R
Tb20
∆Hv ∆Tb
RTb20
⇒ ln b1 − x g ≈ − x = −
⇒ ∆Tb =
x
∆Hv
RTb20
6.88
m1 (g styrene)
90 g ethylbenezene
100 g EG
90 g ethylbenzene
30 g styrene
m2 (g styrene)
100 g EG
Styrene balance: m1 + m2 = 30 g styrene
Equilibrium relation:
F m1 I
m2
= 0.19G
J
100 + m2
H90 + m1 K
solve simultaneously
m1 = 25.6 g styrene in ethylbenzene phase
m 2 = 4.4 g styrene in ethylene glycol phase
6-67
6.89 Basis: 100 kg/h.
A=oleic acid; C=condensed oil; P=propane
100 kg / h
95.0 kg C / h
m
& 2 kg A / h
0.05 kg A / kg
0.95 kg C / kg
& 3 kg A / h
m
& 1 kg P / h
m
m
& 1 kg P / h
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:
0.15 =
0.5 / ( n&1 + 0.5)
⇒ n&1 = 73.2 kg P / h
4.5 / (4 .5 + 95)
b. Operating pressure must be above the vapor pressure of propane at T=85o C=185o F
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=H 2 O, H=Hexane, B=Benzene
100 (kg)
0.30 kg A / kg
0.70 kg W / kg
m1 (kg)
0.10 kg A / kg
0.90 kg W / kg
m2 (kg A)
m H (kg H) or m B (kg B)
m H (kg H)
or mB (kg B)
Balance on W:
100 * 0.70 = m1 * 0.90 ⇒ m1 = 77.8 kg
Balance on A:
100 * 0.30 = m2 + 77.8 * 0.10 ⇒ m 2 = 22.2 kg
Equilibrium for H:
KH =
m2 / ( m2 + mH ) 22.2 / (22 .2 + mH )
=
= 0.017 ⇒ mH = 1.30 x10 4 kg H
xA
010
.
Equilibrium for B:
KB =
m2 / (m2 + mB ) 22.2 / (22.2 + mB )
=
= 0.098 ⇒ mB = 2.20 x10 3 kg B
xA
0.10
(b) Other factors in picking solvent include cost, solvent volatility, and health, safety, and
environmental considerations.
6-68
6.91
a. Basis: 100 g feed ⇒ 40 g acetone, 60 g H 2 O. A = acetone, H = n - C6 H 14 , W = water
40 g A
60 g W
e1 (g A)
60 g W
25°C
100 g H
100 g H
r 1 (g A)
25°C
75 g H
e 2 (g A)
60 g W
75 g H
r 2 (g A)
xA in H pha se / xA in W phase = 0.343 bx = mass fraction g
Balance on A − stage 1:
Equilibrium condition − stage 1:
Balance on A − stage 2:
Equilibrium condition − stage 2:
% acetone not extracted =
40 = e1 + r1
r1 b100 + r1 g
e1 b60 +
U
e1
|
V⇒
= 0.343|
r1
e1 g
W
27.8 = e2 + r2
r2 b75 + r2 g
e 2 b60 +
U
r2
|
V⇒
= 0.343|
e2
e2 g
W
= 27.8 g acetone
= 12 .2 g acetone
= 7 .2 g acetone
= 20.6 g acetone
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.
40.0 = e1 + r1
r1 b175 + r1 g
e 1 b60 + e 1 g
= 17.8 g acetone
= 22.2g acetone
22.2 g A remaining
× 100% = 55.5%
40 g A fed
40 g A
60 g W
19.4 g A
60 g W
20.6 g A
m (g H)
m (g H)
Equilibrium condition:
=
U
r1
|
V⇒
0.343|
e1
W
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.
6-69
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 ⇒ m 3P = 0.90(15
. ) = 135
. kg P
P balance:
1.5 = m2 P + 1.35 ⇒ m 2 P = 0.15 kg P
1.35 / (1.35 + m1 )
pH=2.1 ⇒ K = 25.0 =
⇒ m1 = 34.16 kg BA
015
. / ( 015
. + 98.5)
In Unit II, 90% transfer: m5 P = 0.90(m3 P ) = 1.215 kg P
P balance:
m 3P = 1.215 + m6 P ⇒ m 6 P = 0135
.
kg P
m / ( m6 P + 34.16)
pH=5.8 ⇒ K = 0.10 = 6 P
⇒ m4 = 29.65 kg Alk
1.215 / (1215
.
+ m4 )
m1 34.16 kg BA
=
= 0.3416 kg butyl acetate / kg acidified broth
100 100 kg broth
m 4 29.65 kg Alk
=
= 0.2965kg alkaline solution / kg acidified broth
100 100 kg broth
Mass fraction of P in the product solution:
m5P
1.215 P
xP =
=
= 0.394 kg P / kg
m4 + m5P (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.
6-70
6.93
W = water, A = acetone, M = methyl isobutyl ketone
x W = 0.20 U Figure 6.6-1
Phase 1: x W = 0.07 , x A = 0.35, x M = 0.58
|
x A = 0.33 V
Phase 2: x W = 0.71, x A = 0.25, x M = 0.04
x M = 0.47 |W
⇒
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:
m = 0.95 kg in MIBK - rich phase
R
| 1
⇒S
1.2 * 0.33 = 0.35m1 + 0.25m 2
| m 2 = 0.24 kg in water - rich phase
1.2 * 0.20 = 0.07m1 + 0.71m2
T
6.94 Basis: Given feeds: A = acetone, W = H2 O, M=MIBK
Overall system composition:
5000 g b30 wt% A, 70 wt% W g ⇒ 1500 g A, 3500 g WU
|
3500 g b20 wt% A, 80 wt% Mg ⇒ 700 g A, 2800 g M
2200 g A U
|
⇒ 3500 g WV ⇒ 25.9% A, 41.2% W, 32.9% M
2800 g M |W
V
|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:
m = 4200 g in MIBK - rich phase
R
| 1
⇒S
2200 = 0.31m1 + 0.21m 2
| m2 = 4270 g in water - rich phase
3500 = 0.06 m1 + 0.76m 2
T
6.95 A=acetone, W = H 2 O, M=MIBK
41.0 lb m / h
32 lb m / h
x AF (lb m A / lb m )
x WF (lb m W / lb m )
xA,1 , x W,1, 0.70
& 2 lbm / h
m
xA ,2 , xW ,2 , x M ,2
m
& 1 (lb m M / h)
Figure 6.6-1⇒ Phase 1: x M = 0.700 ⇒ xw ,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:
32.0 lb m / h + m& 1 = 41.0 lb m h + m& 2 U m& 1 = 28.1 lb m MIBK / h
V⇒
& 1 = 41.0 * 0.7 + m& 2 * 0.03
m
m& 2 = 19.1lb m h
W
6-71
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
maq,1 = 417
. kg
Mass balance:
maq ,1 + morg,1 = 100
U
⇒
Acetone balance: maq ,1 * 0.275 + morg ,1 * 0.375 = 33.33VW 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
= 0.925 mol W
m aq,2 = 22 .2 kg
Mass balance:
maq ,2 + morg, 2 = 100
U
V⇒
Acetone balance: maq ,2 * 0.055 + morg , 2 * 0.100 = 9W morg,2 = 77.8 kg
b. K a,1 =
xa ,org,1
xa ,aq,1
=
0.375
= 136
. ;
0.275
K a, 2 =
xa, org ,2
x a, aq , 2
=
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
=
xa, org / x w ,org
=
xa ,aq / x w ,aq
0.375 / 0.075
0.100 / 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 aw → ∞
d.
Organic phase= extract phase; aqueous phase= raffinate phase
β a, w =
( xa / x w ) org
( xa / x w ) aq
=
( x a ) org / ( xa ) aq
( xw ) 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& 2 (kg / h)
x2A (kg A / kg)
x2W (kg W / kg)
x2M (kg M / kg)
e& 1 (kg / h)
x1A (kg A / kg)
x1W (kg W / kg)
x1M (kg M / kg)
200 kg / h
0.30 kg A / kg
0.70 kg M / kg
Stage I
&r1 (kg / h)
y1A (kg A / kg)
y1W (kg W / kg)
y1M (kg M / kg)
300 kg W / h
6-72
Stage II
Stage
IIStage
300 kg W / h
&r2 (kg / h)
y2A (kg A / kg)
y2W (kg W / kg)
y2M (kg M / kg)
6.97(cont'd)
Overall composition of feed to Stage 1:
b200gb0.30 g = 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 = 0.15, y1W = 0.035, y1M = 0.815
e& = 273 kg / h
R
| 1
⇒S&
60 = 0.095e&1 + 015
. r&1 | r1 = 227 kg / h
500 = e&1 + r&1
Mass balance
Acetone balance:
T
Overall composition of feed to Stage 2:
b227gb0.15g = 34
kg A h
U
527 kg h
|
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: x2 A = 0.04, x 2W = 0.94, x2 M = 0.02
Raffinate: y2 A = 0.085, y2 W = 0.025, y2 M = 0.89
Mass balance:
Acetone balance:
e& = 240 kg / h
R
| 2
⇒ S&
34 = 0.04 e2 + 0.085r2
| r2 = 287 kg / h
527 = e&2 + r&2
T
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 = e&1 + e&2 = 273 + 240 = 513 kg / h
Acetone:
( x1 A e&1 + x 2 A e&2 ) kg A
=
0.095 * 273 + 0.04 * 240
= 0.069 kg A / kg
513
Water :
( x1w e&1 + x 2w e&2 ) kg W 0.88 * 273 + 0.94 * 240
=
= 0.908 kg W / kg
e&1 + e&2
513
MIBK:
( x1 M e&1 + x 2 M e&2 ) kg M 0.025 * 273 + 0.02 * 240
=
= 0.023 kg M / kg
(e&1 + e& 2 )kg
513
6-73
6.98. a.
1.50 L / min
25o C, 1atm, rh = 25%
n&0 (mol / min)
y 0 (mol H2O / mol)
(1- y 0 ) (mol dry air / mol)
n& 0 =
M (g gel)
M a (g H2 O)
(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)
= 025
.
Silica gel saturation condition: X * = 12 .5
Water feed rate :
⇒ m& H 2O =
y0 =
0.25 p *H2 O ( 25 o C)
p
p H2 O
p *H2 O
=
= 12 .5 * 0.25 = 3125
.
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 18.01g H 2 O
min
mol
g H 2 O ads
100 g silica gel
mol H 2 O
= 0.00863 g H 2 O / min
Adsorption in 2 hours = (0.00863 g H 2 O / min)(120min) = 1.035 g H 2 O
Saturation condition:
1.035 g H 2O
3.125 g H 2 O
=
⇒ M = 33.1g 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
dehumidific ation 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
*
p c (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
P0V 0
=
= 1.985 mol
RT0 0.08206 L ⋅ atm / mol ⋅ K 307 K
Initial moles of CCl4 in tank:
n c 0 = yc 0 n0 =
p c0
50.7 mm Hg
n0 =
× 1.985 mol = 0.1324 mol CCl 4
P0
760 mm Hg
6-74
6.99 (cont’d)
50% CCl4 adsorbed ⇒ n c = 0.500n c0 = 0662 mol CCl 4 (= n ads)
Total moles in tank: n tot = n 0 − n ads = (1.985 − 0.0662) mol = 1.919 mol
Pressure in tank. Assume T = T0 and V = V0 .
P=
n tot RT0 F (1919
.
)( 0.08206)(307)
I F 760 mm Hg I
=G
atmJ G
J = 735 mm Hg
H
K H
K
V0
50.0
atm
yC =
nc
0.0662 mol CCl 4
mol CCl 4
=
= 0.0345
n tot
1.919 mol
mol
⇒ p c = 0.0345(760 mm Hg) = 26.2 mm Hg
c. Moles of air in tank: n a = n0 − nc 0 = (1.985 − 01324
.
) mol air = 1.853 mol air
yc =
nc
mol CCl 4
= 0.001
⇒ nc = 1.854 × 10 −3 mol CCl 4
n c + 1853
.
mol
⇒ n tot = nc + n air = 1.854 mol
L n RT O 1854
.
× 10 −3 mol 0.08206 L ⋅ atm 307 K
p c = y c P = 0.001M tot 0 P =
50.0 L
mol ⋅ K
N V0
Q
760 mm
1 atm
= 0.710 mm Hg
g CCl 4 I
0.0762 p c
0.0762( 0.710)
g CCl 4 adsorbed
⇒ X* =
= 0.0506
J =
1 + 0.096 (0.710)
g carbon
H g carbon K 1 + 0.096 p c
F
X *G
Mass of CCl4 adsorbed
m ads = (n c0 − nc )( MW ) c =
(0.1324 − 0.001854 ) mol CCl 4
153.85 g
1 mol CCl 4
= 20.3 mol CCl 4 adsorbed
20.3g CCl 4 ads
Mass of carbon required: m c =
= 400 g carbon
g CCl 4 ads
0.0506
g carbon
β
X * = K F p βNO2 ⇒ ln X * = ln K F + β ln p NO
2
ln(PNO2)
6.100 a.
y = 1.406x - 1.965
2
1.5
1
0.5
0
-0.5
-1
-1.5
0
1
2
ln(X*)
6-75
3
6.100 (cont’d)
.406
.406
ln X * = 1406
.
ln p NO2 − 1.965 ⇒ X * = e −1.965 p 1NO
= 0.140p 1NO
2
2
K F = 0.140 (kg NO 2 / 100 kg gel)(mm Hg)−1.406 ; β = 1.406
b. Mass of silica gel : m g =
π * (0.05m) 2 (1 m) 10 3 L 0.75kg gel
1m 3
L
= 5.89 kg gel
Maximum NO2 adsorbed :
p NO2 in feed = 0.010(760 mm Hg) = 7.60 mm Hg
m ads =
0.140(7.60) 1.406 kg NO 2
5.89 kg gel
100 kg gel
= 0.143 kg NO 2
Average molecular weight of feed :
MW = 0.01( MW ) N O2 + 0.99 ( MW ) air = ( 0.01)(46.01) + ( 0.99)(29.0) = 29.17
kg
kmol
Mass feed rate of NO 2 :
m& =
8.00 kg
1 kmol
0.01 kmolNO 2
46.01 kg NO 2
h
29.17 kg
kmol
kmol NO2
Breakthrough time :
tb =
= 0.126
kg 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
6-76
3.39
4.89
5.65
6.78
Let S=sucrose, I=trace impurities, A=activated carbon
Add m A (kg A)
m S (kg S)
mS (kg S)
mI (kg I)
R (color units / kg S)
V (L)
m I0 (kg I)
R0 (color units / kg S)
Come to equilibrium
V (L)
mA (kg A)
mI A (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
•
k
⇒ ∆R = k ( Ci 0 − Ci ) = ( mI 0 − mI )
V
mIA = mI 0 −mI
∆R =
kmIA
V
kmIA / V
m
∆R
x100% =
x100 = 100 IA
R0
kmI 0 / V
m I0
m
Equilibrium adsorption ratio : X i* = I A
mA
Normalized percentage color removal:
% removal of color =
υ=
m m
% removal ( 3) 100 m IA / mI 0
=
= 100 IA S
m A / mS
mA / mS
mA mI 0
m
mI 0
⇒ υ = 100X *i S ⇒ X i* =
υ
mI 0
100 mS
( 1),(5)
Freundlich isotherm X i* = K F Ci β
⇒ υ=
100 mS 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
6-77
intercept = lnK'F
6.101 (cont’d)
ln υ = 0.4504 ln p NO2 + 8.0718 ⇒ υ = e8.0718 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
% color reduction
97.5
⇒ 2344 =
=
⇒ m A = 20.0 kg carbon
m A / mS
m A / 480
6-78
CHAPTER SEVEN
7.1
0.80 L 3.5 × 10 4 kJ 0.30 kJ work
h
L
1 kJ heat
7.2
1 kW
3600 s 1 k J s
2.33 kW 10 3 W 1.341 × 10 −3 hp
1 kW
1h
1W
= 2.33 kW ⇒ 2.3 kW
= 312
. hp ⇒ 3.1 hp
All kinetic energy dissipated by friction
mu 2
2
5500 lbm 552 miles2
=
2
h2
= 715 Btu
(a) E k =
52802 ft 2
1 2 mile 2
12 h 2
36002 s2
1 lbf
9 .486 × 10 − 4 Btu
32.174 lbm ⋅ ft / s2 0.7376 ft ⋅ lb f
(b)
3 ×108 brakings 715 Btu 1 day
day
braking
24 h
1h
1W
−4
3600 s 9.486 × 10
1 MW
Btu/s 1 06 W
= 2617 MW
⇒ 3000 MW
7.3
(a) Emissions:
1000 sacks
Paper ⇒
Plastic ⇒
2000 sacks
1000 sacks
(0.0045 + 0.0146) oz
(724 + 905) Btu
sack
Plastic ⇒
2000 sacks
1 lbm
sack 16 oz
sack
Energy:
Paper ⇒
(0.0510 + 0.0516) oz
= 6.41 lb m
1 lbm
16 oz
= 2 .39 lb m
= 1.63 × 10 6 Btu
(185 + 464 ) Btu
sack
= 1.30 × 10 6 Btu
(b) For paper (double for plastic)
Materials
for 400 sacks
Raw Materials
Acquisition and
Production
Sack
Production and
Use
7- 1
1000 sacks
Disposal
400 sacks
7.3 (cont’d)
Emissions:
Paper ⇒
400 sacks
Plastic ⇒
0.0510 oz
1 lb m 1000 sacks
+
sack 16 oz
800 sacks
0.0045 oz
sack
0.0516 oz
1 lb m
= 4.5 lb m
sack 16 oz
⇒ 30% reduction
1 lb m 2000 sacks
+
16 oz
0.0146 oz
1 lb m
= 2.05 lb m
sack 16 oz
⇒ 14% reduction
Energy:
Paper ⇒
400 sacks
Plastic ⇒
(c) .
724 Btu
sack
800 sacks
3 × 10 8 persons
+
185 Btu
sack
1000 sacks
+
905 Btu
= 119
. × 10 6 Btu; 27% reduction
sack
2000 sacks
464 Btu
= 1.08 × 10 6 Btu; 17% reduction
sack
1 sack
1 day
1h
person - day 24 h 3600 s
649 Btu
1J
1 MW
-4
1 sack 9.486 × 10 Btu 10 6 J / s
= 2 ,375 MW
Savings for recycling: 0.17 (2 ,375 MW) = 404 MW
(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
(0.792)(62.43) lb m
1 ft
3
2
2
mu 2 0.330 lb m
=
2
s
. g
b1225
60 s
1
7.4805 gal Π b0.5g in
2
ft2 1
2
s
1 min
1
1 ft
1 min
12 in
60 s
lb f
2 32.174 lb m ⋅ ft / s2
= 0.330 lb m s
= 1.225 ft s
= 7.70 × 10 −3
F 1341
. × 10−3 hp I
= d7.70 × 10 −3 ft ⋅ lb f / si G
. × 10−5 hp
J = 140
H0.7376 ft ⋅ lb f / s K
(b) Heat losses in electrical circuits, friction in pump bearings.
7- 2
ft ⋅ lb f
s
7.5
(a) Mass flow rate:
m& =
42.0 m π ( 0.07 m )
s
2
103 L 673 K
1 m3
4
1 mol
29 g
273 K 101.3 kPa 22.4 L (STP )
42.0 2 m 2
1N
1J
2
s
1 kg ⋅ m / s2 N ⋅ m
mu
& 2 127.9 g 1 kg
E& k =
=
2
2
s 1000 g
(b)
130 kPa
mol
= 127.9 g s
= 113 J s
127.9 g 1 mol 22.4 L (STP ) 673 K 101.3 kPa 1 m3
4
= 49.32 m s
s 29 g
1 mol
273 K 130 kPa 10 3 L π (0.07)2 m2
& 2 127.9 g 1 kg
mu
E& k =
=
2
2
s 1000 g
49.32 2 m 2
s
2
1N
1J
= 1558
. J/s
1 kg ⋅ m / s2 N ⋅ m
∆E& k = E& k (400 o C) - E& k (300 o 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)
∆E p = mg∆z =
1 gal
1 ft 3
62.43 lbm 32.174 ft −10 ft
1 lbf
= −83.4 ft ⋅ lb f
3
2
7.4805 gal
1 ft
s
32.174 lbm ⋅ ft / s2
12
mu 2
12
L F
ft I
O
(b) E k = − ∆E p ⇒
= mg b− ∆zg ⇒ u = 2 g b− ∆zg = M2G32.174 2 J b10 ft gP
H
K
2
s
N
Q
= 25.4
ft
s
(c) False
7.7 (a)
∆E& k ⇒ positive When the pressure decreases, the volumetric flow rate increases, and
hence the velocity increases.
∆E& ⇒ negative The gas exits at a level below the entrance level.
p
5 m π b1.5g cm 2
s
2
(b)
&=
m
1 m3
10 4 cm 2
273 K
10 bars
1 kmol
16.0 kg CH 4
3
303 K 1.01325 bars 22.4 m bSTP g
1 kmol
= 0.0225 kg s
2
&
PoutV&out nRT
V&
P
u (m/s) ⋅ A(m)
P
=
⇒ out = in ⇒ out
= in
&
&
2
&
PinVin
nRT
Vin Pout
uin (m/s) ⋅ A ( m ) Pout
P
10 bar
⇒ u out = uin in = 5 ( m s )
= 5.555 m s
Pout
9 bar
1
2
∆E&k = m& (uout
− u in2 ) =
0.5(0.0225) kg (5.5552 − 5.000 2 )m 2
s
2
s
1N
1W
1 kg ⋅ m/s
2
2
1 N ⋅ m/s
= 0.0659 W
& ( zout − zin ) =
∆E& p = mg
0.0225 kg 9.8066 m -200 m
s
s
= − 44.1 W
7- 3
1N
kg ⋅ m/s
1W
2
1 N ⋅ m/s
7.8
∆E& p = mg
& ∆z =
105 m3 103 L 1 kg H2O 981
. m −75 m
1 m3
h
1J
2.778 × 10−7 kW ⋅ h
1 kg ⋅ m / s2 1 N⋅ m
s2
1L
1N
1J
= −204
. × 10 kW ⋅ h h
4
The maximum energy to be gained equals the potential energy lost by the water, or
2.04 × 10 4 kW ⋅ h
h
7.9
24 h
7 days
1 day 1 week
= 3.43 × 10 6 kW ⋅ h week (more than sufficient)
(b) Q − W = ∆U + ∆E k + ∆E p
∆E k = 0 bsystem is stationary g
∆E p = 0 bno height change g
Q − W = ∆U , Q < 0,W > 0
(c) Q − W = ∆U + ∆E k + ∆E p
Q = 0 badiabaticg, W = 0bno moving parts or generated currentsg
∆E k = 0 bsystem is stationary g
∆E p = 0 bno height changeg
∆U = 0
(d). Q − W = ∆U + ∆E k + ∆E p
W = 0 bno moving parts or generated currentsg
∆E k = 0 bsystem is stationary g
∆E p = 0 bno height change g
Q = ∆U , Q < 0 Even though the system is isothermal, the occurrence of a chemical
reaction assures that ∆U ≠ 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:
∆U + ∆E k + ∆E p = Q − W
R ∆E k
S
|T ∆E p
= 0 binitial / final states stationaryg
= 0 bby assumption g
∆U = Q − W
(b)
Constant T ⇒ ∆U = 0 ⇒ Q = W =
−7.65 L ⋅ bar
8.314 J
0.08314 L ⋅ bar
= −765 J
(c) Adiabatic ⇒ Q = 0 ⇒ ∆U = −W = 7.65 L ⋅ bar > 0, Tfinal > 30° C
7- 4
transferred from
gas to
surroundings
2
π b3g cm 2
1 m2
= 2.83 × 10 −3 m 2
10 4 cm 2
(a) Downward force on piston:
7.11 A =
Fd = Patm A + mpiston+weight g
=
1 atm 1.01325 × 10 5 N / m2 2.83 × 10 −3 m2
atm
+
24.50 kg 9.81 m
s
2
1N
1 kg ⋅ m / s2
= 527 N
Upward force on piston: Fu = APgas = d2.83 × 10 −3 m 2 i Pg dN m 2 i
Equilibrium condition:
Fu = Fd ⇒ 2.83 × 10 −3 m2 ⋅ P0 = 527 ⇒ P0 = 1.86 × 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 N2
=
= 0.677 L
P0
28.02 g 1.86 × 105 Pa
1 atm
mol ⋅ K
(b) For any step, ∆U + ∆E k + ∆E p = Q − W ⇒ ∆U = Q − W
∆Ek = 0
∆E p = 0
Step 1: Q ≈ 0 ⇒ ∆U = −W
Step 2: ∆U = 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 ⇒ ∆U = 0 ⇒ Q − W = 0
In step 1, the gas expands ⇒ W > 0 ⇒ ∆U < 0 ⇒ T decreases
(c) Downward force Fd = b100
. gd101325
.
× 10 5 i d2.83 × 10 −3 i + b4 .50 gb9.81gb1g = 331 N (units
as in Part (a))
F
331 N
=
= 116
. × 10 5 N m 2
A 2.83 × 10 − 3 m 2
P
1.86 × 10 5 Pa
Since T0 = T f = 30° C , Pf V f = P0V0 ⇒ V f = V0 0 = b0.677 Lg
= 108
. L
Pf
116
. × 105 Pa
Final gas pressure Pf =
∆V b1.08 − 0.677g L
Distance traversed by piston =
=
A
1 m3
103 L
2.83 × 10 −3 m2
= 0142
.
m
⇒ W = Fd = b331 Ngb0.142 mg = 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
Ref state dU$ = 0i ⇒ liquid Bromine @ 300 K, 0.310 bar
(a)
(b) ∆U$ = U$ final − U$ initial = 0.000 − 28.24 = −28.24 kJ mol
∆ H$ = ∆U$ + ∆ dPV$ i = ∆U$ + P∆V$ (Pressure Constant)
∆ Hˆ = −28.24 kJ mol +
0.310 bar
(0.0516 − 79.94) L
8.314 J
1 kJ
mol 0.08314 L ⋅ bar 103 J
= −30.7 kJ mol
∆ H = n∆ H$ = b5.00 molgb−30.7 kJ / molg = −15358
. kJ ⇒ − 154 kJ
(c) U$ independent of P ⇒ U$ b300 K, 0.205 bar g = U$ b300 K , 0.310 bar g = 28.24 kJ mol
U$ d340 K, P i = U$ b340 K, 1.33 bar g = 29.62 kJ mol
f
∆U$ = U$ final − U$ initial
E
∆U$ = 29.62 − 28.24 = 1.380 kJ mol
$ = 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
∆U = n∆U$ = b0.0414 mol gb1.38 kJ / molg = 0.0571 kJ
∆U + ∆E k + ∆E 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
U$ bT , Pg = U$ bT g ⇒ H$ bT , P g = U$ bT g + RT = H$ bT g independent of P
.
cal 50 K
cal 1987
(b) ∆ H$ = ∆U$ + R∆T = 3500
+
= 3599 cal mol
mol
mol ⋅ K
∆ H = n∆ H$ = b2.5 molgb3599 cal / molg = 8998 cal ⇒ 9.0 × 10 3 cal
7.15 ∆U + ∆E k + ∆E p = Q − Ws
∆ E k = 0 bno change in m and ug
∆ E p = 0 bno elevation change g
Ws = P∆V bsince energy is transferred from the system to the surroundingsg
∆U = Q − W ⇒ ∆U = Q − P∆V ⇒ Q = ∆U + P∆V = ∆(U + PV ) = ∆H
7- 6
7.16. (a) ∆ E k = 0 bu1 = u 2 = 0g
∆ E p = 0 bno elevation changeg
.
∆P = 0 (the pressure is constant since restraining force is constant, and area is constrant)
Ws = P∆V bthe only work done is expansion work g
H$ = 34980 + 35.5T (J / mol), V1 = 785 cm3, T1 = 400 K, P = 125 kPa, Q = 83.8 J
125 × 103 Pa
785 cm3
1 m3
PV
=
= 0.0295 mol
RT 8.314 m3 ⋅ Pa / mol ⋅ K 400 K 10 6 cm3
$ -H
$ ) = 0.0295 mol 34980 + 35.5T - 34980 - 35.5(400K) (J / mol)
Q = ∆H = 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 m 3 ⋅ Pa 10 6 cm3 480 K
=
= 941 cm 3
125 × 105 Pa
mol ⋅ K
1 m3
P
125 × 105 N (941- 785)cm3 1 m 3
ii ) W = P∆V =
= 19.5 J
m2
10 6 cm 3
iii ) Q = ∆U + P∆V ⇒ ∆U = Q − ∆PV = 83.8 J − 19.5 J = 64.3 J
i) V =
(b) ∆Ep = 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) ∆U + ∆ E p + ∆ E R = Q& ∆t − W& ∆t
∆ E p = 0, ∆ 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 ) = 1.26 t
Moles in tank: n = PV RT =
1 atm
2.10 L
b25 + 273gK
1
mol ⋅ K
= 0.0859 mol
0.08206 L ⋅ atm
U
1.26 t (J)
U$ = =
= 14.67 t
n 0.0859 mol
Thermocouple calibration: T = aE + b
T = 0 , E =−0.249
T =100 , E =5 .27
T b° Cg = 181
. E bmVg + 4.51
U$ = 14.67 t
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 bT g only.
Ideality could be tested by repeating experiment at several initial pressures ⇒ same
results.
7- 7
7.18 (b) ∆H& + ∆E& k + ∆E& p = Q& − W&s (The system is the liquid stream.)
∆ E& k =0 cno change in m and uh
∆ E& p =0 c no elevation change h
W& s = 0 c no moving parts or generated currentsh
∆ H& = Q& , Q& > 0
(c) ∆H& + ∆E& k + ∆E& p = Q& − W&s (The system is the water)
∆ H& =0 c T a nd P ~ constant h
∆ E& k =0 c no change in m and uh
Q& =0 c no ∆ T between system and surroundings h
∆E& p = −W& s , W& s > 0 bfor water system g
(d) ∆H& + ∆E& k + ∆E& p = Q& − W&s (The system is the oil)
∆ E& k =0 c no velocity change h
∆H& + ∆ E& p = Q& − W& s Q& < 0 (friction loss); W& s < 0 (pump work).
(e) ∆H& + ∆E& k + ∆E& p = Q& − W&s (The system is the reaction mixture)
∆ E& k = ∆ E& p =0 c given h
∆W& s = 0 c no moving parts or generated current h
∆ H& = Q& , Q& pos. or neg. depends on reaction
7.19 (a) molar flow:
1.25 m3 273 K 122 kPa
min
423 K 101.3 kPa
1 mol
103 L
= 43.4 mol min
22.4 L bSTPg 1 m3
∆ H& + ∆ E& k + ∆ E& p = Q& − W& s
∆ E& k = ∆ E& p =0 cgiven h
W& s = 0 c no moving partsh
43.37 mol 1 min
Q& = ∆ H& = n&∆ H$ =
min 60s
3640 J
kW
mol 10 3 J / s
= 2.63 kW
(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
7.20 (a) H$ = 1.04 T b° Cg − 25 H$ in kJ kg
H$ out = 1.04 34.0 − 25 = 9.36 kJ kg
H$ = 104
. 30.0 − 25 = 5.20 kJ kg
n&
(m
ol/
s)
N2
30
o
C
in
∆ H$ = 9.36 − 5.20 = 4.16 kJ kg
∆ H& + ∆ E& k + ∆ E& p = Q& − W& s
∆ E& k = ∆ E& p =0 c assumed h
W& s = 0 c no moving partsh
Q& = ∆ H& = n& ∆ H$
⇒ n& =
P=
11
0
kP
a
34
o
C
Q&
=1
.25
k
W
Q&
1.25 kW
kg
1 kJ / s 10 3 g 1 mol
=
= 10.7 mol s
∆ H&
4.16 kJ
kW
1 kg 28.02 g
10.7 mol 22.4 L bSTPg 303 K 101.3 kPa
⇒ V& =
= 245.5 L / s ⇒ 246 L s
s
mol
273 K 110 kPa
(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
7.21 (a) H$ = aT + b a = H 2 − H 1 = 129.8 − 25.8 = 5.2
|
$
.
T2 − T1
50 − 30
V ⇒ H bkJ kgg = 5.2 T b° Cg − 1302
|
$
b = H 1 − aT1 = 258
. − b5.2gb30g = −1302
. W
1302
.
H$ = 0 ⇒ Tref =
= 25° C
5.2
1 m3
Table B.1 ⇒ bS . G.gC H blg = 0.659 ⇒ V$ =
= 1.52 × 10 −3 m 3 kg
6 14
659 kg
U$ bkJ kg g = H$ − PV$ = b5.2 T − 1302
. gkJ / kg
−
1 atm 1.0132 × 105 N / m2 1.52 × 10 −3 m3
1J
1 kJ
1
atm
1
kg 1 N ⋅ m 103 J
⇒ U$ bkJ kgg = 5.2 T − 1304
.
(b) Energy balance: Q = ∆U =
∆Ek , ∆E p , W =0
20 kg [(5.2 × 20 - 130.4) - (5.2 × 80 -130.4)] kJ
= −6240 kJ
1
kg
Average rate of heat removal =
6240 kJ 1 min
5 min
60 s
7- 9
= 20.8 kW
7.22
m& (kg/s)
260°C, 7 bars
H$ = 2974 kJ/kg
u0 = 0
& (kg/s)
m
200°C, 4 bars
H$ = 2860 kJ/kg
u (m/s)
∆H& + ∆E& k + ∆E& p = Q& − W&s
∆ E& p = Q& = W&s = 0
& 2
mu
∆E& k =− ∆H& ⇒
=− m
& dH$ out − H$ in i
2
(2)b2974 − 2860gkJ 10 3 N ⋅ m 1 kg ⋅ m / s2
m2
u 2 = 2dH$ in − H$ out i =
= 2.28 × 10 5 2 ⇒ u = 477 m / s
kg
1 kJ
1N
s
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
&
& ∆ dPV$ i + ∆du 2 i + mg
& ∆z = Q& − W& s
m& ∆U$ + m
2
∆U$ = 0 (given )
& $ aPout − Pin f = V&∆P
m& ∆PV$ = mV
∆ u 2 = 0 (assume for incompressible fluid )
∆z = 0
W& s = 0 (all energy other than flow work included in heat terms )
Q& = Q& in − Q& out
V&∆P = Q& in − Q& out
5 L b100 − 0gmm Hg
1 atm
8.314 J
(b) Flow work: V&∆P =
= 66.7 J min
min
760 mm Hg 0.08206 liter ⋅ atm
5 ml O2 20.2 J
Heat input: Q& in =
= 101 J min
min
1 ml O 2
Efficiency:
V& ∆ P 66.7 J min
=
×100% = 66%
Q& in
101 J min
7.24 (a) ∆H& + ∆E& k + ∆E& p = Q& − W&s ; ∆E& k , ∆E& p , W& s = 0 ⇒ ∆H& = Q&
H$ b400° C, 1 atm g = 3278 kJ kg (Table B.7)
H$ b100° C, sat' d ⇒ 1 atm g = 2676 kJ kg (Table B.5)
7- 10
7.24 (cont’d)
100 kg H 2 O(v) / s
100 kg H 2 O(v) / s
o
o
100 C, saturated
400 C, 1 atm
Q& (kW)
100 kg
Q& =
s
b3278 − 2676.0gkJ
kg
10 3 J
1 kJ
= 6.02 × 10 7 J s
(b) ∆U + ∆E k + ∆E p = Q − W ; ∆E k , ∆E p , W = 0 ⇒ ∆U = 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 V̂ =1.673 at 400o C, and then interpolate again to
find Û at 400o C and that pressure:
 3.11 − 1.673 
3
o
ˆ
Vˆ = 1.673 m /g ⇒ Pfinal = 1.0 + 4.0 
 = 3.3 bar , U (400 C, 3.3 bar) = 2966 kJ/kg
 3.11 − 0.617 
⇒ Q = ∆U = m ∆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 insignif icant.)
7.25 H$ cH 2 Obl g, 20° Ch = 83.9 kJ kg (Table B.5)
H$ bsteam, 20 bars, sat'd g = 2797.2 kJ kg (Table B.6)
m& [kg H 2 O(l) / h]
m& [kg H 2 O(v) / h]
o
20 C
20 bar (sat'd)
Q& = 0.65(813 kW) = 528 kW
(a) ∆H& + ∆E& k + ∆E& p = Q& − W&s ; ∆E& k , ∆E& p , W& s = 0 ⇒ ∆H& = Q&
∆H& = m& ∆H$
m& =
528 kW
Q&
=
∆ H$
kg
1 kJ / s 3600 s
= 701 kg h
1 kW
1 h
b2797.2 − 83.9gkJ
(b) V& = b701 kg h gd0.0995 m 3 kgi = 69.7 m 3 h sat'd steam @ 20 bar
A
Table B.6
701 kg / h
10 3 g / kg 485.4 K 0.08314 L ⋅ bar 1 m3
&
nRT
(c) V& =
=
= 78.5 m 3 / h
P
18.02 g / mol
20 bar
mol ⋅ K
103 L
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
7.26 H$ cH 2 Obl g, 24° C, 10 bar h = 100.6 kJ kg (Table B.5 for saturated liquid at 24o C; assume H$
independent of P).
H$ b10 bar, sat'd steamg = 27762
. kJ kg (Table B.6) ⇒ ∆ H$ = 2776.2 − 100.6 = 2675.6 kJ kg
& [kg H2O(l)/h]
m
& [kg HO(v)/h]
m
2
3
o
24 C, 10 bar
15,000 m /h @10 bar (sat'd)
&(kW)
Q
15000 m 3
m& =
kg
4
0.1943 m 3 = 7.72 × 10 kg h
h
A
b Table
8.6g
Energy balance d∆E& p , W& s = 0i : ∆H& + ∆E& k = Q&
∆E& k = E&k final − E& k initial
∆E& k =
mu
& f
2
E& kinitial ≈0
7.72 × 10 4 kg
2
=
∆ E& k = E& kfinal
d15,000
m 3 hi
2
2
0.15 2 π 4 m 2
h
1
1
h3
1J
1 kg ⋅ m 2 / s2
2 3600 3 s3
A
A=π D2 4
= 5.96 × 10 5 J / s
7.72 × 10 4 kg 2675.6 kJ
1h
5.96 × 10 5 J 1 kJ
& ∆H$ + ∆E& k =
Q& = m
+
h
kg 3600 s
s 10 3 J
= 57973 kJ s = 5.80 × 10 4 kW
228 g/min
25o C
7.27 (a)
228 g/min
T(o C)
Q& ( kW)
=0
Energy balance: Q& = ∆H& ⇒ Q& bWg =
∆ E& x , ∆ E& p , W&s =0
228 g 1 min ( H$ out − H$ in) J
min
60 s
g
⇒ H$ out bJ g g = 0.263Q& bW g
T b° Cg
H$ bJ g g = 0.263Q& bW g
(b) H$ = bbT − 25g
25 26.4 27.8 29.0 32.4
0
4.47 9.28 13.4 24 .8
Fit to data by least squares (App. A.1)
b=
⇒ H$ bJ gg = 3.34 T b° Cg − 25
7.27 (cont’d)
7- 12
∑ H$
i
i bTi
− 25g
∑ bT
i
i
2
− 25g = 3.34
350 kg 10 3 g 1 min 3.34 b40 − 20gJ
(c) Q& = ∆ 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.
7.28
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]
o
16 C, 2.5 bar
m& e [ kg C 2 H 6 / min]
o
93 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
= 2.487 × 103 kg min
1 K - mol 30.01 g
1 kg
0.08314 L - bar
mol 1000 g
H$ ei = 941 kJ kg , H$ ef = 1073 kJ kg
Energy Balance on C 2 H 6 : ∆E& p , W& s = 0, ∆E& k ≅ 0 ⇒ Q& = ∆H&
kJ O 2.487 × 103 kJ 1 min
kg L
Q& = 2.487 × 103 min Mb1073 − 941g P =
= 5.47 × 103 kW
min 60 s
kg Q
N
(b) H$ s1 b3.00 bar, sat' d vapor g = 2724.7 kJ kg (Table B.6)
H$ bliquid, 27° Cg = 1131
. kJ kg (Table B.5)
s2
Assume that heat losses to the surroundings are negligible, so that the heat given up by the
condensing steam equals the heat transferred to the ethane d5.47 × 10 3 kW i
Energy balance on H 2O: Q& = ∆H& = m
& dH$ s2 − H$ s1 i
&=
⇒m
−5.47 × 10 3 kJ
Q&
=
s
H$ s2 − H$ s1
kg
= 2.09 kg s steam
. − 2724.7 gkJ
b1131
⇒ V&s = b2.09 kg / sgd0.606 m 3 kg i = 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)
Turbine
250 kg/min
5 bar, T 2 (°C), H$ 2(kJ/kg)
Heat
exchanger
250 kg/min
5 bar, 500°C
H$3 (kJ/kg)
Q(kW)
W s =1500 kW
H 2 Obv , 40 bar, 500° Cg: H$ 1 = 3445 kJ kg (Table B.7)
H Obv , 5 bar, 500° Cg: H$ = 3484 kJ kg (Table B.7)
2
3
(a) Energy balance on turbine: ∆E& p = 0, Q& = 0, ∆E& k ≅ 0
&
∆H& = −W&s ⇒ m& dH$ 2 − H$ 1 i = −W& s ⇒ H$ 2 = H$ 1 − W&s m
=
3445 kJ 1500 kJ
−
kg
s
min
60 s
250 kg
1 min
= 3085 kJ kg
H$ = 3085 kJ kg and P = 5 bars ⇒ T = 310° C (Table B.7)
(b) Energy balance on heat exchanger: ∆E& p = 0, W&s = 0, ∆E& k ≅ 0
250 kg
Q& = ∆H& = m& dH$ 3 − H$ 2 i =
min
b3484
− 3085gkJ 1 min 1 kW
= 1663 kW
kg 60 s 1 kJ / s
(c) Overall energy balance: ∆E& p = 0 , ∆E& k ≅ 0
∆H& = Q& − W& s ⇒ m& s dH$ 3 − H$ 1 i = Q& − W&s
250 kg
Q& = ∆H& + ∆W& s =
min
b3484 −
3445gkJ 1 min 1 kW 1500 kJ 1 kW
+
kg
60 s 1 kJ / s
s
1 kJ / s
= 1663 kW √
(d) H 2 Obv , 40 bar, 500° Cg: V$1 = 0.0864 m 3 kg (Table B.7)
H 2 Obv , 5 bar, 310° Cg: V$2 = 0.5318 m 3 kg (Table B.7)
u1 =
u2 =
250 kg 1 min 0.0864 m3
min
250 kg
min
60 s
kg
1
0.5 π 4 m 2
2
min 0.5318 m3
60 s
kg
1
0.5 π 4 m 2
2
m& 2
250 kg 1 1 min
∆E& k =
u 2 − u12 =
2
min
2 60 s
= 0.26 kW << 1500 kW
= 183
. ms
= 11.3 m s
2
b11.3g
7- 14
2
− b183
. g m2
s
2
1 kW ⋅ s
1N
1 kg ⋅ m / s
2
10 3 N ⋅ m
kJ
7.30 (a) ∆E& p , ∆E& k , W& s = 0 ⇒ Q& = ∆H& ⇒ −hAbTs − To g = −300 kJ h ⇒ 18
. h bTs − To g = 300
h
(b) Clothed:
h = 8 ⇒ To = 13.4° C
Ts =34 .2
Nude, immersed: h = 64 ⇒ To = 31.6° C (Assuming Ts remains 34.2°C)
Ts = 34.2
(c) The wind raises the effective heat transfer coefficient. (Stagnant air acts as a thermal
insulator —i.e., in the absence of wind, h is low.) For a given To , the skin temperature
must drop to satisfy the energy balance equation: when Ts drops, you feel cold.
7.31 Basis: 1 kg of 30°C stream
1 kg H2 O(l)@30o C
3 kg H2 O(l)@Tf(o C)
2 kg H2 O(l)@90o C
1
2
o
o
o
b30 Cg + b90 Cg = 70 C
3
3
(b) Internal Energy of feeds: U$ b30° C, liq.g = 1257
. kJ kg U
|
V
U$ b90° C, liq.g = 376.9 kJ kg |W
(Table B.5 - neglecting effect of P on H$ )
(a) T f =
Energy Balance: Q - W = ∆U + ∆E p + ∆E k
Q= W= ∆E = ∆E =0
p
k
∆U = 0
⇒ 3U$ f − (1 kg)b125.7 kJ / kgg − (2 kg)b376.9 kJ / kg g = 0
⇒ U$ = 293.2 kJ kg ⇒ T = 70.05° C (Table B.5)
f
Diff.=
7.32
f
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 = 27475
. kJ kg
52.5 m3
$ bar, sat'd) = 0.375 m 3 / kg ⇒ m
V(5
& =
h
1 kg
= 140 kg h
0.375 m 3
(b) H2 O evaporated = b015
. gb140 kg hg = 21 kg h (Q = energy needed to vaporize this much water)
Energy balance:Q& = ∆H&
=
21 kg
h
( 2747.5 − 640.1) ] kJ
1h
1 kW
= 12 kW
kg 3600 s 1 kJ s
7- 15
7.33 (a) P = 5 bar
Table B.6
Tsaturation = 1518
. o C . At 75°C the discharge is all liquid
(b) Inlet: T=350°C, P=40 bar
Outlet: T=75°C, P=5 bar
u in =
u out =
H$ in = 3095 kJ / kg , V$in = 0.0665 m3 / kg
Table B.7
Table B.7
H$ out = 314.3 kJ / kg , V$out = 1.03 × 10 -3 m 3 / kg
3
V&in 200 kg 1 min 0.0665 m / kg
=
= 50.18 m / s
min 60 s
π (0.075) 2 / 4 m 2
Ain
V&out 200 kg 1 min
=
min 60 s
Aout
0.00103 m 3 / kg
= 1.75 m / s
π (0.05)2 / 4 m 2
m&
Energy balance: Q& − W&s ≈ ∆H& + ∆E& k = m
& ( H$ 2 − H$ 1 ) + (u 22 − u12 )
2
200 kg 1 min (314-3095) kJ
200 kg 1 min (1.752 -50.18 2 ) m 2
Q& − W&s =
+
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 H 2O(v)/s)
25 bars, sat'd
100 kg oil/min
185°C
m& (kg H2O(l)/s)
25 bars, sat'd
& dH$ out − H$ in i
Energy balance on H 2O: Q& = ∆H& = m
∆E& p , ∆ E& 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:
b962.0
kg
= 0.091 kg s
− 2800.9gkJ
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
trap ⋅ s
kg stream
= $7.4 × 10 5 / year
7-16
2.6 × 10 −3$ 3600 s 24 h 360 day
kg lost
h
day
year
7.35 Basis: Given feed rate
200 kg H2 O(v)/h
10 bar, sat’d, H$ = 2776.2 kJ / kg
n& 3 [kg H 2 O(v) / h]
10 bar, 250o C, 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)
Energy balance: Q& = ∆H& = n&3 b2943g− 200b2776.2g − n&2 b3052 g, Q& in kJ h
(2)
& =0
∆ E& K , ∆ E& p , W
(a) n& 3 = 300 kg h
(b) Q& = 0
(1),(2 )
(1)
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 o 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, 250o C) = 1085.8 kJ / kg
Mass balance: mv + ml = 100 kg
Energy balance: ∆H& = 0
(1)
& , ∆ E& , W& = 0
∆ E& K , Q
p
⇒ mv H$ v + ml H$ l − m1 H$ 1 = mv H$ v + m l 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, ∆E&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)
7-17
7.36 (cont’d)
(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.
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 H2 O (v)
10.0 m3
m in (kg)
275 oC, 1.5
, 15bar
bar
mv [kg H2 O (v)]
ml [kg H 2O (l)]
Q
(a) P=1.2 bar, saturated,
1.2 bar, saturated
Table B.6
(b) Total mass of water: min =
10 m3
T2 = 104.8 o C
1 kg
= 55 kg
0.1818 m 3
Mass Balance: m v + m l = 55.0
Volume additivity:
Vv + Vl = 10.0 m 3 = m v (1.428 m 3 / kg) + m l (0.001048 m 3 / kg)
⇒ mv = 7.0 kg, ml = 48.0 kg condensed
(c) Table B.7 ⇒ U$ in = 2739.2 kJ / kg; V$in = 0.1818 m3 / kg
R$
3
$
|
Table B.6 ⇒ SU l = 439.2 kJ / kg; Vl = 0.001048 m / kg
|T U$ v = 2512.1 kJ / kg;
V$v = 1.428 m3 / kg
Energy balance: Q = ∆U = mvU$ v + ml U$ l − minU$ in
∆E p ,∆Ek , 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 + 1 5 0 o C
m& l [ kg H 2 O (l ) / s]
m& v [ kg H 2 O( v ) / s]
1.0 bar, saturated
7-18
7.38 (cont’d)
(b) Table B.6 ⇒ Tsat'n (15 bar) = 198.3o C ⇒ Tin = 3483
. oC
Table B.7 ⇒ H$ in = H$ ( 348.3o 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
Energy balance: ∆H& = 0 ⇒ m
& H$ + m& H$ − m& H$ = 0
&s =0
∆E& p ,∆E& k ,Q& , W
⇒m
& in H$ in = m
& l H$ l + m
& v H$ v
l
l
v
m& v + m& l
in
v
in
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.
& in = m& out = 1
m
(c) Energy balance ⇒ m
& out H$ out = m& in H$ in
3149 kJ / kg = H$ (1 bar, Tout )
Table B.7
Tout ≈ 337o C
(This answer is only approximate, since ∆E& k is not zero in this process).
7.39 Basis: 40 lb m min circulation
(a) Expansion valve
R = Refrigerant 12
40 lb m R(l)/min
93.3 psig, 86°F
H$ = 27.8 Btu/lb m
40 lb m / min
x v lb m R (v ) / lb m
(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: ∆E& p , W& s , Q& = 0 , neglect ∆E& k ⇒ ∆H& =
∑ n& H$ − ∑ n& H$
i
i
out
40 X v lb m Rbv g 77.8 Btu 40b1 − X v glb m Rbl g
+
min
lb m
min
9.6 Btu 40 lb m
−
min
lb m
i
i
=0
in
27.8 Btu
=0
lb m
E
X v = 0.267 b26.7% evaporates g
(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: ∆E& p , W&s = 0, neglect ∆E& k ⇒ Q& = ∆H&
40 lb m
Q& =
min
77.8 Btu b40gb0.267 glb m Rbv g
−
lb m
min
= 2000 Btu min
7-19
77.8 Btu b40 gb0.733glb m R bl g 9.6 Btu
−
lb m
min
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
closed (no mass flow in or out). Consider a time interval ∆tbming . Since the system is at
steady state, the changes ∆U , ∆E k and ∆E p over this time interval all equal zero. The
total heat input is Q& ∆t , the work input is W& ∆t , and (Eq. 8.3-4) yields
c
−500 Btu 1 min
1.341 × 10 −3 hp
Q& ∆t − W&c ∆t = 0 ⇒ W&c = Q& =
= 11.8 hp
min
60 s 9.486 × 10 −4 Btu s
7.40 Basis: Given feed rates
n&1 (mol / h)
n&C 3 H8 (mol C 3 H 8 / h)
n&C 4 H10 (mol C 4 H 10 / h)
227o C
0.2 C3 H 8
0.8 C 4 H 10
0 o C, 1.1 atm
n& 2 (mol / h)
0.40 C3 H 8
0.60 C 4 H 10
25 o C, 1.1 atm
Q& (kJ / h)
Molar flow rates of feed streams:
n&1 =
300 L 1.1 atm
1 mol
= 14.7 mol h
hr 1 atm 22.4 LbSTP g
n& 2 =
200 L 273 K 1.1 atm
1 mol
hr 298 K 1 atm 22.4 LbSTP g = 9.00 mol h
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 C3 H 8 h
Total mole balance: n&C 4 H10 = (14.7 + 9 .00 − 6.54 ) mol C 4 H 20 h = 17.16 mol C 4 H 20 h
Propane balance ⇒ n&C 3 H8 =
Energy balance: ∆E& p , W&s = 0, neglect ∆E& k ⇒ Q& = ∆H&
Q& = ∆H& =
∑ N& H$ − ∑ N& H$
i
out
−
b0.40 ×
i
i
in
9.00g mol C3 H 8
h
i
=
6.54 mol C 3H 8
h
20.685 kJ 17.16 mol C4 H 10
+
mol
h
1.772 kJ b0.60 × 9.00g mol C 4 H10
−
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 m3 273 K 10 3 L
1 mol
1 kmol
= 21.4 kmol min
min 291 K m 3 22.4 LbSTP g 10 3 mol
7.41 Basis:
(a)
Q& (kJ/min)
n. 0 ( kmol/min)
38°C, h r = 97%
y 0 ( mol H 2 O(v)/mol )
(1 – x 0) ( mol dry air/ mol)
21.4 kmol/min
18°C, sat'd
y 1 ( mol H 2 O(v)/ mol
)
(1 – y 1) (mol dry air)
.
n 2 ( kmol H 2O(l )/ mol)
18°C
Inlet condition: yo =
hr PH∗2O ( 38°C )
P
PH∗2 O
(18 °C )
=
0.97 ( 49.692 mm Hg )
= 0.0634 mol H2 O mol
760 mm Hg
15.477 mm Hg
= 0.0204 mol H2 O mol
P
760 mm Hg
Dry air balance: b1 − 0.0634 gn& o = b1 − 0.0204g214
. ⇒ n& o = 22.4 kmol min
Outlet condition: y1 =
=
Water balance: b0.0634 g22.4 = n& 2 + b0.0204 g21.4 ⇒ n&2 = 0.98 kmol min
0.98 kmol 18.02 kg
= 18 kg / min H 2 O condenses
min
kmol
(b). Enthaphies: H$ air b38° Cg = 0.0291b38 − 25g = 0.3783 kJ mol
H$ b18° Cg = 0.0291b18 − 25g = −0.204 kJ mol
air
2570.8 kJ 1 kg 18.02 g
U
H$ H2 O bv , 38° Cg =
= 46.33 kJ mol |
kg 10 3 g
mol
|
25345
. kJ 1 kg 18.02 g
|
H$ H2 O bv , 18° Cg =
=
45
.
67
kJ
mol
VTable B.5
kg 10 3 g
mol
|
75.5 kJ
1 kg 18.02 g
$
|
H H2 O bl , 18° Cg =
= 136
. kJ mol
kg 10 3 g
mol
|W
Energy balance:
∆ E& , W& = 0, ∆E& ≅0
p
s
k
Q& = ∆H& =
∑ n& H$ − ∑ n& H$
i
out
i
i
i
⇒ Q& = b1 − 0.0204 gd21.4 × 10 3 i b−0.204 g
in
+b0.0204gd21.4 × 10 3 i b45.67g + d0.98 × 10 3 i b136
. g − b1 − 0.0634gd22.4 × 10 3 i b0.3783g
−b0.0634gd22.4 × 10 3 i b46.33g = −5.67 × 10 4 kJ min
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
0.5n 2 (mol)
0.98 A(l )
0.02 B(l )
100 mol, 67.5°C
0.65 A(l )
0.35 B(l )
Qc (cal)
56.8°C
n 5 (mol), 98.7°C
0.544 A(v )
0.456 B(v )
0.5n 2 (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 = 0.5n 2 + n 5
U n 2 = 120 mol
A: 0.65b100g = 0.98b0.5n 2 g + 0.155n 5 VWn 5 = 40 mol
Product flow rates: Overhead
0.5b120g0.98 = 58.8 mol A
0.5b120g0.02 = 1.2 mol B
Bottoms 0.155b40g = 6.2 mol A
0.845b40g = 33.8 mol B
Overall energy balance: Q = ∆H =
n H$ −
n H$
∑
∆ E , W =0 , ∆E ≅ 0
p
2
i
out
x
i
∑
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) Flow through condenser: 2b58.8g = 117.6 mols A
2b12
. g = 2 .4 mols B
Energy balance on condenser: Q c = ∆H
∆E , W = 0 , ∆E ≅ 0
p
3
k
Qc = 117 .6b0 − 7322 g + 2.4b0 − 6807 g = −8.77 × 10 5 cal heat removed from condenser
Assume negligible heat transfer between system & surroundings other than Q c & Q r
(
)
Qr = Q −Qc = 1.82× 10 4 − −8.77 ×10 5 = 8.95 × 105 cal heat added to reboiler
7.43
1.96 kg, P1= 10.0 bar, T1
2.96 kg, P 3= 7.0 bar, T3=250o C
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$ (H O( v), P = 7.0 bar, sat'd) = 2760 kJ kg ( Table B.6)
2
2
Energy balance
∆E , Q, W , ∆E ≅0
p
s
k
∆H = 0 = 2.96 H$ 3 − 196
. H$ 1 − 1.0 H$ 2 ⇒ 1.96 H$ 1 = 2 .96 kg(2954 kJ / kg) -1.0 kg(2760 kJ / kg)
⇒ H$ (10.0 bar, T ) = 3053 kJ / kg ⇒ T ≅ 300 o 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. ) = 133.5o C
(a)
Vapor
V$l ( P = 3.0 bar, sat' d. ) = 0.001074 m3 / kg
V$ ( P = 3.0 bar, sat'd.) = 0.606 m 3 / kg
P=3 bar
Liquid
v
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
1 m3
1 kg
= 0.0376 kg
1000 L 0.606 m 3
(b) P = Pmax = 20.0 bar;
m total = 165.0 + 0.0376 = 165.04 kg
T1 = T ( P = 20.0 bar, sat'd.) = 212.4o C
V$l ( P = 20.0 bar, sat' d.) = 0.001177 m3 / kg; V$v ( P = 20.0 bar, sat'd.) = 0.0995 m3 / kg
V
= m V$ + m V$ ⇒ m V$ + (m
− m )V$
total
l
l
v v
l
l
total
l
v
1 m 3 = 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
⇒ 200.0 L
Vl =
0.001177 m 3
kg
m evaporated =
1000 L 164.98 kg
= 194.2 L;
m3
V space = 200.0 L - 194.2 L = 5.8 L
( 0.06 - 0.04) kg 1000 g
= 20 g
kg
(c) Energy balance Q = ∆U = U ( P = 20.0 bar, sat'd) − U ( P = 3.0 bar, sat'd)
∆E , W , ∆E ≅ 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
1 kg H 2 O,(v) 1 atm
0.97 kg H2 O(v)
0.03 kg H2 O(l)
H$ 1 (kJ/kg)
1 kg H2 O
Tamb , 1 atm
H$ 2 (kJ/kg)
Q=0
Q
Enthalpies: H$ bv , 20 bars, sat'd g = 2797 .2 kJ kgU
|
VbTable B.7g
$
H bl , 20 bars, sat' d g = 908.6 kJ kg |
W
Energy balance on condenser: ∆H = 0 ⇒ H$ 2 = H$ 1 = 0.97b2797.2 g + 0.03b9086
. g
∆E , ∆E , Q, W =0
p
K
3
Table B.7
⇒ H$ 2 = 2740 kJ / kg
T ≈ 132o 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.
7.46 Basis:
8 oz H 2O bl g 1 quart
1 m3
1000 kg
32 oz 1057 quarts
m3
(For simplicity, we assume the beverage is water)
0.2365 kg H 2O (l)
18°C
m (kg H 2O (s))
32°F (0°C)
= 0.2365 kg H 2O bl g
(m + 0.2365) (kg H 2O (l))
4°C
Assume P = 1 atm
Internal energies (from Table B.5):
U$ bH 2 O(l ), 18° Cg = 755
. kJ / kg; U$ bH 2 O(l ), 4° Cg = 16.8 kJ / kg; U$ bH 2 O(s), 0° Cg = -348 kJ / kg
Energy balance bclosed system g: ⇒ ∆U =
∆E p , ∆E k , Q , W = 0
∑ n U$ − ∑ n U$
i
out
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: ∆U = 0
∆E , ∆E , Q, W = 0
k
p
25 g Fe, 175°C
25 g Fe
1000 g H 2O
T f (°C)
1000 g H 2O(l)
20°C
U Fe dT f i + U H2 O dT f i − U Fe b175° Cg − U H 2 O b20° C, 1 atmg = 0 or ∆U Fe + ∆U H2 O = 0
∆U Fe =
25.0 g 4.13dT f − 175i cal 4.184 J
= 432 Tf − 175 J
g
cal
1.0 L 10 3 g eU$ H2 O dT f i − 83.9 j J
= 1000eU$ H 2 O dT f i − 83.9 j J
1 L
g
5
⇒ 432T f + 1000U$ H 2 O dT f i − 1.60 × 10 = f dT f i = 0
Table B.5 ⇒ ∆U H 2O =
⇒
Tf ° C
30
f dT f
−2.1 × 10
i
40
4
+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
H 2 O( l), 100 °C
⇒ 1.08 bar sat'd
⇒ Tf = 101.8°C (Table 8.5)
H 2 O( l), Tf
T0
Tf
Energy balance - closed system:
∆E p , ∆ E K , W , Q = 0
∆U = 0 =
mvIIU$ vII
+
mlIIU$ lII
+
mbII U$ bII
−
mvI U$ vI
−
mlIU$ lI
−
mbI U$ bI
v-vapor
l -liquid
b-block
V$l bL kg g
V$v bL kg g
U$ l bL kg g
U$ v bL kg g
I b1.01 bar, 100° Cg II b1.08 bar, 101.8° Cg
1.044
1046
.
1673
419.0
2506.5
1576
426.6
2508.6
Initial vapor volume: VvI = 20.0 L − 5.0 L −
50 kg
1L
8.92 kg
= 14.4 L H 2 Obv g
Initial vapor mass: mvI = 14.4 L b1673 L kgg = 8.61 × 10 −3 kg H 2 Obvg
Initial liquid mass: mlI = 5.0 L b1.044 L kg g = 4.79 kg H 2 Obl g
Final energy of bar: U$ bII = 0.36b101.8g = 36.6 kJ kg
Assume negligible change in volume & liquid ⇒ VvII = 14.4 L
Final vapor mass: mvII = 14.4 L b1576 L kg g = 9.14 × 10 − 3 kg H 2 Obvg
Initial energy of the bar:
1
−3
. b36.6g − 8.61 × 10 −3b25065
. g − 4.79b419.0gi
d9.14 × 10 b2508.6g + 4.79b426.6g + 50
5.0 kg
= 44.1 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 = 45.8 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
+ atmospheri c pressure
area of piston
9.807 N
400.0 cm 2
2
b100 cmg
kg
2
12 bmg
10
. bar
10 5 N m 2
+
1 atm 1.013 bar
atm
= 108
. bar
⇒ Tsat = 1018
. °C
Heat required to bring the water and block to the boiling point
Q = ∆U = mw dU$ wl b1.08 bar, sat'd g − U$ wl bl , 20° Cgi + mAl dU$ Al bTsat g − U$ Al b20° Cgi
. − 20) ]kJ
− 83.9 gkJ 3.0 kg [0.94 (1018
+
= 2630 kJ
kg
kg
2630 kJ < 3310 kJ ⇒ Sufficient heat for vaporization
=
7.0 kg
b426.6
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 2O(l )
T ≡ 101.8°C
P ≡ 1.08 bars
H$ = 426.6 kJ / kg
V
$ = 1.046 L / kg
mv (kg H2 O( v ))
1576 L/kg, 2508.6 kJ/kg
1.046 L/kg, 426.6 kJ/kg
m l (kg H 2 O( l))
Q (kJ)
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 = m l + mv (1)
Work done by the piston: W = F ∆ z = w piston + Patm A ∆ z
=
Lw
MA
N
O
+ Patm Pb A ∆ z g = P ∆ V ⇒ W = b108
. bar g 1576mv + 1046
.
m l − b1.046gb7.0g L
Q
×
8.314 J / mol ⋅ K
1 kJ
0.08314 liter - bar / mol ⋅ K 10 3 J
= b170.2m v + 0.113ml − 0.7908gkJ
Energy balance: ∆U = Q − W
Q
∆U444448
W
6444447
6447
44
8 6444447
444448
⇒ 25086
. mv + 426.6mL − 426.6b7 g = ( 3310 − 2630 ) − (1702
. mv + 0113
. mL − 0.7908)
⇒ 2679mv + 4267
. mL − 3667 = 0 (2)
Solving (1) and (2) simultaneously yields m v = 0.302 kg , m l = 6.698 kg
Liquid volume = b6.698 kggb1.046 L kgg = 7.01 L liquid
Vapor volume = b0.302 kg gb1576 L kgg = 476 L vapor
. g L 10 3 cm 3
1
∆V 7.01 + 476 − b7.0gb1046
Piston displacement: ∆z =
=
= 1190 cm
A
1 L 400 cm 2
(c) Tupper ⇒ All 3310 kJ go into the block before a measurable amount is transferred to the
water. Then ∆U AL = Q ⇒ b3.0 kg g 0.94bTu − 20g kJ kg = 3310 ⇒ Tu = 1194° C if melting is
neglected. In fact, the bar would melt at 660o C.
7-27
7.50 1.00 L H 2 O( v ), 25 o C
m v1 (kg)
m v2 [kg H2 O( v)]
= m v1 + me
4.00 L H 2 O(l ), 25 o C
m L1 (kg)
m L2 [kg H 2O( l) ]
= m L1 + m e
Assume not all the liquid
vaporized. Eq. at
T f , Pf . me = kg H 2 O vaporized.
U
V is
W
Q=2915 kJ
Initial conditions: Table B.5 ⇒ U$ L1 = 104.8 kJ kg , V$L1 = 1.003 L kg P = 0.0317 bar
T = 25° C, sat' d ⇒ U$ = 2409.9 kJ kg , V$ = 43,400 L kg
v1
−5
m v1 = b1.00 l g b43400 l kg g = 2.304 × 10
L1
kg , m LI = b4.00 l g b1.003 l kgg = 3.988 kg
Energy balance:
∆U = Q ⇒ d2.304 × 10 −5 + me i U$ v dT f i + b3.988 − me gU$ L dT f i − d2.304 × 10 −5 i b2409.9 g
−b3.988g(1048
. ) = 2915 kJ
⇒ d2.304 × 10 −5 + me i U$ v dT f i + b3.988 − me gU$ v dT f i = 3333
E
−5
3333 − d2.304 × 10 i U$ v − 3.988U$ L
⇒ me =
U$ − U$
v
F
(1)
L
I
V L + Vv = Vtan k ⇒ G2.304 × 10 −5 + m e J V$L dT f i + b3.988 − me gV$L dT f i = 5.00 L
G
H
A
kg
J A
K liters kg
5.00 − d2.304 × 10 − 5 i V$v − 3.988V$L
⇒ me =
V$ − V$
v
b1g − b2 g ⇒
f dT f i =
b2g
L
3333 − d2.304 × 10 −5 i U$ v dT f i − 3.988U$ L dT f
i
U$ v − U$ L
5.00 − d2.304 × 10 −5 i V$v − 3.988V$L
−
=0
V$ − V$
v
L
Table 8.5
Procedure: Assume T f
Tf
U$ v
U$ L
201.4 25938
. 856.7
198.3 2592.4 842.9
195.0 2590.8 828.5
196.4 25915
. 834 .6
Eqb1 g
⇒ U$ v , U$ L , V$v , V$L ⇒ f dT f i Find T f such that f dT f i = 0
V$v
123.7
1317
.
140.7
136.9
V$L
f
1159
.
− 512
. × 10 −2
1154
.
−193
. × 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 Eqb2 g
7-28
Basis: 1 mol feed
7.51.
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)
(a)
in equilibrium
at T(°C), P(mm Hg)
nL (mol liquid)
x B(mol B(l)/mol)
(1 – x B ) (mol T(l)/mol)
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 − n 2
Benzene balance: z B = nV y B + n L x B
(1)
(2)
C 6 H 6 bl g: dT = 0, H$ = 0i , dT = 80, H$ = 10.85i ⇒ H$ BL = 0.1356 T
C 6 H 6 bv g:
dT
= 80, H$ = 41.61i , dT = 120, H$ = 45.79 i ⇒ H$ BV = 01045
.
T + 33.25
(3)
(4)
C 7 H 8 bl g: dT = 0, H$ = 0i , dT = 111, H$ = 18.58i ⇒ H$ TL = 0.1674 T
(5)
C 7 H 8 bv g: dT = 89 , H$ = 49.18 i , dT = 111, H$ = 52 .05i ⇒ H$ TV = 0.1304T + 37 .57
(6)
Energy balance: ∆E p , Ws = 0, neglect ∆E k
Q = ∆H = nV y B H$ BV + nV b1 − y B gH$ TV + n L x B H$ BL + n L b1 − x B gH$ TL − b1gz B H$ BL bTF g
− b1gb1 − zB gH$ TL bTF g
Raoult's Law:
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:
p *B (90o C) = 10[6.89272−1203.531/(90+ 219.888)] = 1021 mmHg
pT* (90 o C) = 10[6.95805−1346.773/(90+219.693)] = 406.7 mmHg
Adding equations (8) and (9) ⇒
P = x B p *B + (1 − x B ) pT* ⇒ x B =
yB =
P − p*T
p *B
−
x B p *B
P
(7)
pT*
=
=
P − p*T
p *B
−
pT*
=
652 − 4067
.
= 0.399 mol B(l) / mol
1021 - 406.7
0.399(1021 mmHg)
= 0.625 mol B(v ) / mol
652 mmHg
z B − xB
0.5 − 0.399
=
= 0.446 mol vapor
y B − x B 0.625 − 0.399
n L = 1 − nV = 1 − 0.446 = 0.554 mol liquid
Solving (1) and (2) ⇒ nV =
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)[0.1304( 90) + 37.57 ]
+ 0.554 (0.399 )[0.1356(90)] + 0.554 (1 − 0.399 )[0.1674( 90)] − 0.5[0.1356(130)]
− 0.5[0.1674 (130)] ⇒ Q = 8.14 kJ / mol
(c). If P<P min, all the output is vapor. If P>P max, 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
7.52 (a). Bernoulli equation:
∆P ∆u 2
+
+ g∆z = 0
ρ
2
∆P d0.977 × 10 −5 − 1.5 × 10 5 i Pa 1 N / m 2
=
ρ
Pa
m3
1.12 × 10 3 kg
= −46.7
m2
s2
g∆z = (9.8066 m / s2 )b6gm = 58.8 m 2 / s 2
Bernoulli ⇒
∆u 2
= b46.7 − 58.8 g m 2 / s2 ⇒ u 22 = u12 + 2d−12.1 m 2 / s2 i
2
2
=b5.00g m 2 / s2 − (2)(12.1) m 2 / s2 = 0.800 m 2 / s 2 ⇒ u 2 = 0.894 m / s
(b). Since the fluid is incompressible, V& dm 3 si = π d 12 u1 4 = π d 22 u 2 4
u2
0.894 m s
= b6 cmg
= 2 .54 cm
u1
5.00 m s
⇒ d1 = d 2
A
7.53 (a). V& dm 3 si = A1 dm 2 i u 1bm sg = A2 dm 2 i u 2 bm sg ⇒ u 2 = u1 1
A2
(b). Bernoulli equation ( ∆z = 0)
A1 =4 A2
u2 = 4u 1
ρdu 22 − u 12 i
∆ P ∆u 2
+
= 0 ⇒ ∆ P = P2 − P1 = −
ρ
2
2
Multiply both sides by − 1
Substitute u 2 = 16u1
2
2
2
Multiply top and bottom of right - hand side by A1
note V& 2 = A12 u12
P1 − P2 =
(c) P1 − P2 = dρ Hg − ρ H2 O i gh =
15ρ H 2 OV& 2
2 A12
⇒ V& 2 =
I
2 A12 gh F ρ Hg
− 1J
G
15 H ρ H 2O
K
2
2 π b7.5g cm4 1
m4
V& 2 =
15
108 cm 4
⇒ V& = 0.044 m 3 s = 44 L s
2
15ρV& 2
2 A12
9.8066 m 38 cm
1m
s2
102 cm
7-31
b13.6 − 1g
= 1955
.
× 10−3
m6
s2
7.54 (a). Point 1- surface of fluid . P1 = 31
. bar , z1 = +7 m , u1 = 0bm sg
Point 2 - discharge pipe outlet . P2 = 1 atm , z2 = 0bmg , u 2 = ?
b=1.013
∆ρ b1.013 − 3.1gbar
=
ρ
g∆z =
9.8066 m
s2
10 N
1
m3
= −263.5 m 2 s 2
m 2 ⋅ bar 0.792 × 10 3 kg
−7 m
Bernoulli equation ⇒
bar g
5
= −68.6 m 2 s2
∆u 2
∆P
=−
− g ∆z = b263.5 + 68.6g m 2 s2 = 332.1 m 2 s 2
2
ρ
∆u 2 = u 22 − 0 2
u 22 = 2 (332.1 m 2 s2 ) = 664.2 m 2 s 2 ⇒ u2 = 25.8 m / s
π (1.00 2 ) cm 2
V& =
4
2580 cm 1 L
60 s
= 122 L / min
3
3
1
s 10 cm 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
Point 2 - pipe outlet . P2 = 1 atm , z2 = zbft g
u2 =
V& 95 gal
=
A
min
1 ft 3
1
144 in 2 1 min
= 35.3 ft s
2
7.4805 gal π b0.5 × 1.049g in 2
1 ft 2
60 s
Pressure drop: ∆ P ρ = 0
b P1
= P2 g
Friction loss: F = 0.041b2 zg ft ⋅ lb f lb m = 0.0822 z (ft ⋅ lb f lb m )
Z
F
I
= 2zJ
GL =
H
K
sin 30 °
Shaft work:
W& s -8 hp 0.7376 ft ⋅ lb f / s
=
m&
1.341 × 10 − 3 hp
1 min 7.4805 gal
1 ft 3
95 gal
1 ft 3
60 s
62.4 lb m
1 min
= −333 ft ⋅ lb f lb m
2
Kinetic energy: ∆ u 2 =
2
Potential energy: g ∆z =
Eq. b7.7 - 2 g ⇒
b35.3g
− 0 2 ft 2
2
2
s
32.174 ft
s2
zbft g
1
lb f
32.174 lb m ⋅ ft / s2
= 19.4 ft ⋅ lb f lb m
1 lb f
= zbft ⋅ lb f lb m g
32.174 lb m ⋅ ft / s 2
∆P ∆u 2
−W& s
+
+ g∆z + F =
⇒ 19.4 + z + 0.082 z = 333 ⇒ z = 290 ft
&
ρ
2
m
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), u 2 = ? , z2 = 0
∆P ρ = 0
2
&2 6 2
&
∆u 2 u22 dV Ah V (m / s )
=
=
=
2
2
2
(2)
1
2 2
π b35g
10 8 cm 4
1 N
1 m4
1 kg ⋅ m / s2
cm 4
= 3.376V& 2 bN ⋅ m kg g
g∆z =
9.8066 m −65 m
1N
= −637 N ⋅ m kg
2
s
1 kg ⋅ m / s2
s
1 m3
W&s 0.80 × 10 6 W 1 N ⋅ m / s
=
= 800 V& bN ⋅ m kgg
W
V& dm 3 i 1000 kg
m
&
Mechanical energy balance: neglect F bEq. 7.7 - 2g
∆P ∆u 2
−W&s
800 T + E
127
. m 3 60 s
+
+ g ∆z =
⇒ 3.376V& 2 − 637 = − & ⇒ V& =
= 76.2 m 3 min
&
ρ
2
m
V
s
1 min
Include friction (add F > 0 to left side of equation) ⇒ V& increases.
7.57 (a). Point 1: Surface at fluid in storage tank, P1 = 1 atm , u1 = 0 , z1 = Hbmg
Point 2 (just within pipe): Entrance to washing machine. P2 = 1 atm , z2 = 0
u2 =
600 L
10 3 cm 3 1 min
1m
= 7.96 m s
2
min π b4.0 cmg 4 1 L
60 s 100 cm
∆u 2 u22 b7.96 m sg2
∆P
=0;
=
=
2
2
ρ
2
g∆z =
9.807 m
s
c0 −
2
Bernoulli Equation:
H bmgh
1 J
= 31.7 J / kg
1 kg ⋅ m 2 / s2
1
J
1 kg ⋅ m / s2
2
= −9 .807H (J / kg)
∆P ∆u 2
+
+ g ∆z = 0 ⇒ H = 3.23 m
ρ
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 , u 2 = 7.96 m s , z2 = 3.23 m
∆u 2
J
J
J
∆P
=0;
= 31.7
; g∆z = 9.807b3.23 − 0g = 31.7
; F = 72
2
kg
kg
kg
ρ
L ∆P
O
∆u 2
Mechanical energy balance: W& s = −m& M +
+ g∆z + F P
2
Nρ
Q
600 L 0.96 kg 1 min
⇒ W&s = −
min
L
60 s
b31.7 + 31.7 +
Rated Power = 1.30 kW 0.75 = 1.7 kW
7-33
72 g J
1 kW
= −1.30 kW
kg 10 3 J s
(work applied to the system)
7.58 Basis: 1000 liters of 95% solution . Assume volume additivity.
x i 0.95 0.05
1
l
Density of 95% solution:
=
=
+
= 0.804
⇒ ρ = 1.24 kg liter
ρ
ρ i 1.26 1.00
kg
bEq. 6.1-1g
∑
1 0.35 0.65
l
=
+
= 0.9278
⇒ ρ = 1.08 kg liter
ρ 126
.
100
.
kg
1000 liters 1.24 kg
Mass of 95% solution:
= 1240 kg
liter
Density of 35% solution:
G = glycerol
W = water
1240 kg (1000 L)
0.95 G
0.05 W
m 2 (kg)
0.60 G
0.40 W
23 m
m1 (kg)
0.35 G
0.65 W
5 cm I.D.
Mass balance: 1240 + m1 = m2
U
m1 = 1740 kg 35% solution
V⇒
Glycerol balance: b0.95gb1240g + b0.35gbm1 g = b0.60gbm2 gW m2 = 2980 kg 60% solution
Volume of 35% solution added =
1740 kg
1L
= 1610 L
1.08 kg
⇒ Final solution volume = b1000 + 1610g 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 , z2 = 23 m
u2 =
1610 L
13 min
1 m3 1 min
1
2
3
10 L
60 s π b2.5g cm 2
10 4 cm 2
= 1.051 m s
1 m2
∆u 2 ∆u 22 b1.051g2 m 2 / s 2
1 N
∆P ρ =0,
=
=
= 0.552 N ⋅ m kg
2
2
(2)
1 kg ⋅ m / s 2
g∆z =
9.8066 m
s2
Mass flow rate: m& =
23 m
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
Mechanical energy balance bEq. 7.7 - 2 g
2 .23 kg
L ∆P
O
∆u 2
W& s = −m& M +
+ g∆z + F P = −
2
s
Nρ
Q
b0.552 +
= −0.62 kW ⇒ 0.62 kW delivered to fluid by pump.
7-34
225.6 + 50gN ⋅ m
kg
1J
1 kW
1 N ⋅ m 10 3 J s
CHAPTER EIGHT
8.1
a.
U$ ( T ) = 25.96T + 0.02134 T 2 J / mol
U$ ( 0o C) = 0 J / mol U$ (100 o C) = 2809 J / mol
$ o C) = 0)
Tref = 0o 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 = ∆U + ∆E k + ∆E p
∆E k = 0, ∆E p = 0, W = 0
Q = ∆ U = ( 30
. mol )[( 2809 − 0) J / mol] = 8428 J ⇒ 8400 J
d.
F ∂U$ I
Cv = G
J
H ∂T K $
V
=
dU$
= [ 25.96 + 004268
.
T ] J / (mol⋅ o C)
dT
T2
∆U$ =
100
F
z Cv (T ) dT =
. + 0.04268T ) dT = G2596
. T + 004268
.
z (2596
T1
0
G
H
T2
2
100 I
O
P
P
Q0
J
J
K
J / mol
∆U = ( 30
. mol) ⋅ ∆U$ ( J / mol)
= ( 30
. mol) ⋅ [2596
. (100 − 0) + 0.02134(1002 − 0)] (J / mol) = 8428 J ⇒ 8400 J
8.2
a.
Cv = C p − R ⇒ Cv = b353
. + 0.0291T g[ J / (mol⋅° C)] − b8.314 [J / (mol ⋅ K)]gb1 K 1° Cg
⇒ C v = 27 .0 + 0.0291T [ J / (mol⋅° C)]
100
b.
∆Hˆ =
∫
C p dT = 35.3T ]25 + 0.0291
100
25
100
c.
d.
8.3
a.
∆U$ =
100
100
T2 
 = 2784 J mol
2 25
100
$
z Cv dT =
z C p dT −
z RdT = ∆H − R∆T = 2784 − b8.314 gb100 − 25g = 2160
25
25
25
H$ is a state property
C v [ kJ / (mol ⋅ o C) ] = 0.0252 + 1547
.
× 10 −5 T − 3012
.
× 10 −9 T 2
PV
( 2.00 atm)( 3.00 L )
n=
=
= 0245
.
mol
RT ( 0.08206[ atm⋅ L / (mol ⋅ K) ]( 298 K)
1000
Q1 = n∆U$ 1 = ( 0.245 mol) ⋅
.
dT ( kJ / mol) = 6.02 kJ
z 00252
25
1000
Q2 = n∆U$ 2 = ( 0.245) ⋅
.
× 10
z[ 0.0252 + 1547
−5
T ] dT = 7.91 kJ
−5
T − 3012
.
× 10 −9 T 2 ] dT = 7.67 kJ
25
1000
Q3 = n∆U$ 3 = ( 0245
. )⋅
.
× 10
z[ 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
J mol
8.3 (cont’d)
b.
C p = Cv + R
C p [ kJ / (mol⋅ o C)] = ( 0.0252 + 1547
.
× 10 −5 T − 3012
.
× 10 −9 T 2 ) + 0.008314
= 0.0335 + 1547
.
× 10 −5 T − 3.012 × 10 −9 T 2
T2
Q = ∆ H = n z C P dT
T1
1000
= ( 0.245 mol) ⋅
.
× 10
z[ 0.0335 + 1547
−5
T − 3012
.
× 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.
dC p i
b.
dC p i
C 6 H 6 bl g
C 6 H 6 bvg
b313 K g =
0.06255 + 23.4 × 10 −5 b313g = 01360
.
[kJ / (mol ⋅ K)]
b40° Cg = 0.07406 + 32.95 × 10
−5
b40g − 25.20 × 10
−8
2
b40 g
+ 77.57 × 10− 12 b40g
3
= 0.08684 [kJ / (mol⋅ o C)]
c.
dC p i
C bs g
b313
−2
Kg = 001118
.
+ 1095
.
× 10−5 b313g − 4.891 × 10 2 b313 g
= 0.009615 [ kJ / (mol ⋅ K)]
300
d.
∆H$ C H
6
6 bvg
= 0.07406T +
32.95 × 10 −5 2 25.20 × 10−8 3 77.57 × 10−12 4 O
T −
T +
T P
2
3
4
P
Q
= 31.71 kJ mol
40
573
e.
∆H$ C bsg = 0.01118 T +
O
1.095 × 10− 5 2
T + 4.891 × 10 2 T −1 P
2
P
Q
= 3.459 kJ / mol
313
8.5
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
350
b.
H$ =
.
+ 06886
.
× 10
z 003346
−5
T + 0.7604 × 10− 8 T 2 − 3593
.
× 10 −12 T 3 dT
100
= 8845
.
kJ mol ⇒ 491.4 kJ kg
Difference results from assumption in (b) that H$ is independent of P. The numerical difference
is ∆H$ for H 2 Ob v, 350 ° C, 1 atmg → H 2 Obv, 350° C, 100 bar g
80
8.6
b.
dC p i
= 02163
.
kJ / (mol⋅ C) ⇒ ∆H$ = z[ 0.2163 ] dT = 11.90 kJ / mol
o
n− C6 H14 (l)
25
The specific enthalpy of liquid n-hexane at 80o C relative to liquid n-hexane at 25o C is 11.90 kJ/mol
c.
dC p i
−5
o
n− C6 H14 (v)
−8
2
[ kJ / (mol⋅ C)] = 013744
.
+ 40.85 × 10 T − 2392
. × 10 T + 57.66 × 10
−12
T
3
0
∆H$ =
z[ 0.13744 + 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 500o C relative to hexane vapor at 0o C is 110.7 kJ/mol. The
specific enthalpy of hexane vapor at 0o C relative to hexane vapor at 500o C is –110.7 kJ/mol.
8- 2
8.7
1
T ′b° Fg − 32 = 05556
.
T ′ b° Fg − 17.78
18
.
C p bcal mol⋅° Cg = 6.890 + 0.001436 0.5556 T ′ b° Fg − 17 .78 = 6.864 + 0.0007978 T ′b° Fg
T b° Cg =
C ′p bBtu lb - mole⋅° F g
=
Cp
E
drop primes
cal
453.6 mol 1 Btu
1° C
= b100
. gC p
mol ⋅° C 1 lb - mole 252 cal 1.8° F
C p bBtu lb - mole⋅° Fg = 6.864 + 0.0007978T b° Fg
8.8
dC p i
C H3 CH2 OH(l)
bT g =
01031
.
+
.
− 01031
.
b01588
g
100
T = 01031
.
+ 0000557
.
T [kJ / (mol⋅ o C)]
78 .5
Q = ∆H =
550
. L 789 g 1 mol F
0.000557 2 O
.
T+
T P
G01031
s
1 L 46.07 g H
2
Q20
144444244444
3
kJ mol
= 941.9 × 7.636 kJ / s = 7193 kW
8.9
a.
k J mol
6444444444444474444444444444
8
200
Q& = ∆ H& = b5,000 mol sg⋅
.
× 10
z 0.03360 + 1367
−5
T − 1607
.
× 10−8 T 2 + 6.473 × 10−12 T 3 dT
100
= 17,650 kW
b.
Q = ∆ U = ∆H − ∆ PV = ∆H − nR∆T = 17,650 kJ − b5.0 kmolg⋅ b8.314 [kJ / (kmol ⋅ K)]g ⋅ b100 K g
= 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
m∆T
Q
(16.73- 6.14) kJ 1 L 86.17 g 10 3 J
Cp =
=
= 0.223 kJ / (mol ⋅ K)
m∆T (2.00 L)(3.10 K) 659 g 1 mol
1 kJ
Q = mC p ∆ T ⇒ C p =
Table B.2 ⇒ C p = 0.216 kJ / (mol⋅ o C) = 0.216 kJ / (mol ⋅ K)
8.11
a∂ ∂T f P
PV$ = RT
F ∂H$ I
F ∂U$ I
F ∂U$ I
H$ = U$ + PV$ =====> H$ = U$ + RT =====> G J = G
J + R ⇒ Cp = G
J +R
H ∂T K
H ∂T K
H ∂T K
p
p
p
But since U$ depends only on T,
F ∂ U$ I
G
J
H ∂T K
=
p
dU$ F ∂ U$ I
=
≡ Cv ⇒ C p = Cv + R
J
dT G
H ∂T K $
V
8- 3
8.12 a.
o
dC p i
n=
H2 O(l)
= 754
. kJ / (kmol ⋅ C) =75.4 kJ/(kmol.o C) V = 1230 L ,
Vρ 1230 L 1 kg 1 kmol
=
= 68.3 kmol
M
1 L 18 kg
T2
n ⋅ zdC p i
Q
Q& = =
t
b.
H2 O(l)
dT
T`
=
t
683
. 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
40
Q& to water
n ⋅ z C P( H2 O) dT
Q
= to water =
t
=
29
t
68.3 kmol 754
. kJ / (kmol⋅ o C) 11 o C
= 5245
.
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 = $4.72
d.
If the lid is removed, more heat will be transferred into the surroundings and lost, resulting in higher
cost.
8.13 a.
∆H$ N
o
o
2 (25 C) → N 2 (700 C)
b.
∆H$ H
2 (800
c.
∆H$ CO
d.
∆H$ O
8.14 a.
o
F)→ H 2 (77 o F)
= H$ N
= H$ H
o
o
2 (300 C)→ CO2 (1250 C)
2 (970
o
F) →O 2 (0o F)
= H$ O
& = 300 kg / min n& =
m
2 (700
2 (77
o
o
F)
= H$ CO
2 (0
o
F)
C)
− H$ N
− H$ H
2 (1250
− H$ O
o
2 (25
2 (800
C)
o
o
= b2059
. − 0g = 2059
. kJ mol
C)
F)
= b0 − 5021g = − 5021 Btu / lb - mol
− H$ C O
o
2 (970 F)
o
2 (300 C)
= b63.06 − 1158
. g = 5148
. kJ mol
= b−539 − 6774 g = −7313 Btu / lb- mol
300 kg 1 min 1000 g 1 mol
= 1785
. mol / s
min 60 s 1 kg 28.01 g
T
2
Q& = n& ⋅ ∆H$ = n& ⋅ z C p dT
T1
50
= (178.5 mol / s) ⋅
z[0.02895 + 0.411 × 10
−5
T + 0.3548 × 10 −8 T 2 − 2.22 × 10− 12 T 3 ] dT [kJ / mol]
450
= (178.5 mol / s)b− 12.076 [kJ / mol]g = −2,156 kW
b.
8.15 a.
Q& = n& ⋅ ∆H$ = 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& = n& ∆H$ =
= −1.278 kW
h
1 kg
1000 g 3600 s 1 mol
i)
T
2
Q& = n& ∆H$ = n& ⋅ z Cp dT
T1
ii)
=
250 mol 1 h 100
[0.03346 + 0.6880 × 10−5T + 0.7604 × 10−8 T 2 − 3.593× 10−12 T 3 ] = −1.274 kW
h
3600 s z600
8- 4
8.15 (cont’d)
250 mol
Q& =
⋅ b2.54 − 20.91g [kJ / mol] = − 1276
.
kW
3600 s
Method (i) is most accurate since it is not based on ideal gas assumption.
The work done by the water vapor.
iii)
b.
c.
8.16 Assume ideal gas behavior, so that pressure changes do not affect ∆H$ .
200 ft 3 492 o R 12
. atm 1 lb - mol
= 0.6125 lb- mole / h
o
h
537 R 1 atm 359 ft 3 (STP)
lb - mole
Q& = n& ∆H$ = ( 0.6125
) ⋅ b( 2993 − 0) [Btu / lb - mole]g = 1833 Btu / h
h
n& =
8.17 a.
50 kg 1.14 kJ
kg⋅° C
b50 − 10 g° C
= 2280 kJ
b.
(C )
p Na C O
2
3
≈ 2 (C p ) Na + ( C p)C + 3 ( C p )O = 2 ( 0.026 ) + 0.0075 + 3 ( 0.017 ) = 0.1105 kJ mol ⋅°C
50,000 g 0.1105 kJ 1 mol
b50 − 10g° C
= 2085 kJ
mol ⋅° C 105.99 g
2085 − 2280
% error =
× 100% = −8.6% error
2280
8.18
dC p i
dC p i
o
C 6 H14 O(l)
= 6b0.012g + 14b0.018g + 1b0025
. g = 0.349 kJ / (mol⋅ C) (Kopp’s Rule)
−5
C H3 COCH3 (l)
= 01230
.
+ 18.6 × 10 T kJ (mol⋅° C)
Assume ∆Hmix ≅ 0
↓ CH 3 COCH 3
C pm =
↓ C 6 H 14 O
0.30 ( 0.1230+18.6 × 10 −5T ) kJ 1 mol 0.70 ( 0.349 ) kJ 1 mol
+
mol ⋅°C
58.08 g
mol ⋅°C
102.17 g
= [0.003026 + 9.607 ×10−7 T] kJ (g ⋅ °C)
20
∆Hˆ = ∫ [0.003026 + 9.607 ×10−7 T] dT = −0.07643 kJ g
45
8.19 Assume ideal gas behavior, ∆Hmix ≅ 0
Mw =
1
2
g
b16.04 g + b32 .00 g = 26.68
3
3
mol
350
∆H$ O2 = z
25
dC p i
O2
dT = 10.08 kJ / mol, ∆H$ CH4 =
350
z25
dC p i
C H4
dT = 14.49 kJ / mol
2
L1
OF 1000 g I F 1 mol I
H$ = M b14.49 kJ / mol g + b10.08 kJ / molgPG
JG
J = 433 kJ kg
3
N3
QH 1 kg KH 26.68 g K
8- 5
8.20
1000 m 3 1 min 273 K
1 kmol
= 0.6704 kmol s = 670.4 mol / s
min
60 s 303 K 22.4 m3 bSTP g
Energy balance on air:
Table B.8 for ∆ H$
670.4 mol 0.73 kJ 1 kW
Q = ∆ H = n∆H
Q=
= 489.4 kW
s
mol
1 kJ s
n=
Solar energy required =
Area required =
8.21
489.4 kW heating 1 kW solar energy
= 1631 kW
0.3 kW heating
1627 kW 1000 W 1 m2
= 1813 m 2
1 kW 900 W
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
n& fuel =
n& air =
1.35 × 105 SCFH 1 lb - mol
lb - mol
= 376
h
h
359 ft 3
376 lb − mol 5 lb - mol O 2
1 lb - mol air 115
.
lb − mol
= 103
. × 104
h
1b - mol C3 H 8 0.211b - mol O2
h
T2
Q = ∆H = n& ⋅ zC p dT
T1
F
= G1.03 × 104
H
lb − mol I
J⋅
K
h
302
.
+ 0.4147 × 10
z[ 002894
−5
T + 0.3191 × 10 −8 T 2 − 1965
.
× 10 −12 T 3 ] dT
0
103
. × 104 lb - mol 8.954 kJ 453.593 mol 9.486 × 10 -1 Btu
=
= 3.97 × 10 7 Btu / h
h
mol
lb - mol
kJ
8.22 a.
Basis : 100 mol feed (95 mol CH4 and 5 mol C2 H6 )
7
CH 4 + 2O 2 → CO 2 + 2H 2 O
C 2 H 6 + O 2 → 2CO 2 + 3H 2 O
2
L 95 mol CH
5 mol C2 H 6 35
. mol O 2 O
4 2 mol O2
nO 2 = 125
. ⋅M
+
P = 259.4 mol O 2
1 mol CH 4
1 mol C 2 H 6 PQ
M
N
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)
$
$
∆H
=H
− H$
= 18.845 − 42.94 = −24.09 kJ / mol
o
o
CO2
(CO2 , 450 C)
$
$
∆H
H 2 O = H (H
2 O,
o
450 C)
(CO2 , 900 C)
− H$ (H
$
$
∆H
O2 = H (O
2,
450 o C)
− H$ (O
$
$
∆H
N 2 = H (N
2,
450o C)
$
−H
(N
2,
2 O,
o
900 C)
= 1512
. − 3332
. = −18.20 kJ / mol
900 o C)
= 13375
.
− 2889
. = −15.51 kJ / mol
900 o C)
= 12.695 − 2719
. = −14.49 kJ / mol
2,
Q = ∆H = 105(-24.09) + 205(-18.20) + 51.9(-15.51) + 975(-14.49)
Q = 21,200 kJ / 100 mol feed
8- 6
8.22 (cont’d)
$ (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 ⋅ ∆H
c.
From part (b), 8.07 kg steam is produced per 100 mol feed
n& feed =
d.
1250 kg steam 01
. kmol feed 1 h
= 4.30 × 10−3 kmol / s
h
8.07 kg steam 3600 s
4.30 mol feed 1336.9 mol product gas 8.314 Pa ⋅ m 3
723 K
V&product gas =
= 341
. m3 / s
5
s
100 mol feed
mol ⋅ K
1.01325 × 10 Pa
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 ∆Hmix ≅ 0 ⇒ ∆H = ∆H C10 H12 O2 + ∆H C6 H6
8.23
Kopp’s rule: dC p i
∆H C10 H12 O2 =
∆H C6 H6 =
o
C10 H12 O2
o
= 10(12) + 12(18) + 2( 25) = 386 J emol ⋅ Cj = 2.35 J eg⋅ Cj
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 348
O
⋅ z [ 0.06255 + 23.4 × 10 −5 T] dTP = 1166 kJ
L 78.11 g M
N 298
Q
∆H = 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 )
b.
Vw1 (m3 )
References: H2 O (l, 0.01 o C), C3 H8 (gas, 40 o C)
240
C3 H8 : H$ in = 0 kJ / mol; H$ out = zC pC H dT = 19.36 kJ mol (Cp from Tabl e B.2)
3 8
40
H 2 O : Hˆ in = 3065 kJ/kg (Table B.7); Hˆ out = 640.1 kJ/kg (Table B.6)
c.
∆Hˆ C3H 8 = 19.36 kJ/mol, ∆Hˆ w = (640.1 − 3065) kJ/kg = −2425 kJ/kg
Q = ∆ H = 100 ∆H$ C 3 H8 + m w ∆H$ w = 0 ⇒ mw = 0.798 kg
From Table B.7: V$steam b50
. bar, 300 ° Cg = 0522
.
m 3 kg
0.008314 m 3 ⋅ kPa (mol ⋅ K) 313 K
V$C3 H8 b40° C, 250 kPa g =
= 0.0104 m 3 mol C 3 H 8
250 kPa
0.798 kg steam 0.522 m3 steam
100 mol C3 H 8
1 kg steam
1 mol C3 H8
0.0104 m 3 C 3 H 8
= 0.400 m 3 steam m 3 C 3H 8
d.
Q = mw ∆Hˆ w = 0.798 kg × (-2425 kJ/kg)=-1935 kJ
e.
A lower outlet temperature for propane and a higher outlet temperature for steam.
8- 7
8.25
a.
5500 L(STP)/min CH 3 OH (v) 65o C
n 2 mol/min CH 3 OH (v) 260 o C
n 2 (mol/min)
mw kg /min H 2 O(l, sat‘d) @ 90o C
mw kg /min H 2 O(v, sat‘d) @ 300o C
Vw 2 ( m3 /min )
n2 =
Vw 1 ( m3 /min )
5500 L(STP)
1 mol
= 245.5 mol CHOH(v)/min
3
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 H2 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)/mol
0.100 mol CO/mol
0.800 mol CO2 /mol
n 2 mol/s (30o C)
0.020 mol H2 O(v)/mol
y 2 mol
(molCO/s
CO/mol)
(molCO
CO
(0.980-y 2 ) mol
2/mol)
2 /s
m3 kg humid air/s (50o C)
m4 kg humid air/s (30
(48o oC)
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: n 2 , m3 , m4 , y2 , y4
– 4 independent material balances, H2 O(v), CO, CO2 , dry air
– 1 energy balance equation
0 degrees of freedom (all unknowns may be determined)
b.
(1) CO balance:
(100)(0.100) = n& 2 y 2
(2) CO 2 balance: (100)(0.800) = n2 (1 −
U
|
⇒ n& 2
y2 ) V|
W
= 9184
. mol / s, x 2 = 0.1089 mol CO / mol
1.000
= m 4 (1 − y 4 )
1002
.
(100)(0.100)(18)
0.002
( 0.020)(18)
(4) H 2 O balance:
+ m& 3
= 91.84
+ m& 4 y 4
1000
1.002
1000
References: CO, CO2 , H2 O(v), air at 25o C ( H$ values from Table B.8 )
substance
n& in ( mol / s)
n& out ( mol / s)
H$ (kJ / mol)
(3) Dry air balance: m 3
0.169
0.146
91.84(0.020)
10
0.193
80
H$ out (kJ / mol)
0.169
0.146
0.193
0.847
0.727
m4 y 4 (1000/18 )
m4 (1-y 4 ) (1000 /29 )
0.779
0.672
in
H2 O(v)
CO
10
10
CO2
H2 O(v)
dry air
80
0.002
1000
m3 (
/1.002 )( /18 )
m3 (1.000 /1.002 ) (1000/29 )
8- 8
8.26 (cont’d)
(5) Energy balance:
F 0.002 I F 1000I
F 1.000 I F1000 I
JG
J ( 0.847 ) + m 3 G
JG
J (0.727 )
H 1.002 K H 18 K
H1.002 K H 29 K
10(0.169) + m 3 G
F 1000I
F 1000I
= 91.84 ( 0.020 )( 0.169) + m4 y4 ( 0.779 ) G
J + m 4 (1 − y 4 )( 0.672 ) G
J
H 18 K
H 29 K
Solve Eqs. (3)–(5) simultaneously ⇒ m3 = 2.55 kg/s, m4 = 2.70 kg/s, y4 = 0.0564 kg H2 O/kg
2.55 kg humid air / s
kg humid air
= 0.0255
100 mol gas / s
mol gas
Mole fraction of water :
00564
.
kg H2 O
c.
8.27 a.
=.0963
(1-.0564) kg dry air kmol DA 18 kg H 2 O
⇒
0.0963 kmol H2 O
(1 + 00963
.
) kmol humid air
pH 2 O
Relative humidity:
29 kg DA 1 kmol H2 O
p*H2 O e48 o Cj
=
= 0.0878
kmol H2 O
kmol DA
kmol H2 O
kmol humid air
( 0.0878)( 760 mm Hg )
× 100% = 79.7%
83.71 mm Hg
The membrane must be permeable to water, impermeable to CO, CO2 , O2 , and N2 , and both durable
and leakproof at temperatures up to 50o C.
p* b57° Cg
12982
. mm Hg
= 0171
.
mol H 2 O mol
P
760 mm Hg
↓
3
28.5 m bSTP g
1 mol
= 1270 mol h ⇒ 217.2 mol H 2 O h
h
0.0224 m 3 bSTPg
y H2 O =
=
1270 − 217 .2 = 1053
mol dry
h
R 89.5
|
gas
| 110.5
=======> S
percentages
5.3
|
| 847.6
T
given
.
b391
kg H 2 O hg
mol CO h
mol CO 2 h
mol O2 h
mol N 2 h
1270 mol/h, 620°C
425°C
m (kg H2 O( l )/h), 20°C
References for enthalpy calculations:
CO, CO 2 , O 2 , N 2 at 25°C (Table B.8); H 2 Oel, 0.01 o Cj (steam tables)
substance
CO
CO 2
O2
N2
H 2 Obv g
H 2 Oblg
nin
89.5
110.6
5.3
847.6
3.91
m
H$ in
18.22
27.60
19.10
18.03
3749
83.9
nout
89.5
110.6
5.3
847.6
H$ out
12.03
17.60
12.54
11.92
3.91 + m 3330
---
8- 9
U
| n in mol h
V $
H in kJ mol
|
W
U n in kg
V $
W H in kJ
h
kg
8.27 (cont’d)
∆H =
∑ n H$ − ∑ n H$
i
i
i
out
b.
i
= 0 ⇒ − 8504 + 3246m = 0 ⇒ m = 2.62 kg h
in
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.
8.28 2°C, 15% rel. humidity ⇒ pH 2 O = b015
. gb5294
.
mm Hg g = 0.7941 mm Hg
d y H2 O i
inhaled
n& inhaled =
b0.7941g b760 g = 1.045 × 10
−3
mol H 2 O mol inhaled air
5500 ml 273 K 1 liter
1 mol
= 0.2438 mol air inhaled min
3
min
275 K 10 ml 22.4 litersbSTPg
Saturation at 37 °C ⇒ y H2 O =
p * b37° Cg
760 mm Hg
=
47.067
= 0.0619 mol H 2 O mol exhaled dry gas
760
0.2438 mol/min 2o C
n2 kmol/min 37oC
1.045 x 10-3 H2O
0.999 dry gas
0.0619 H2 O
0.9381 dry gas
n 1 mol H2O(l)/min 22o C
Mass of dry gas inhaled (and exhaled) =
b0.2438 gb0.999 gmol dry gas
29.0 g
min
mol
= 7.063 g min
Dry gas balance: b0.999 gb0.2438g = 0.9381 n&2 ⇒ n&2 = 0.2596 mols exhaled min
H 2 O balance:
.
.
× 10
b02438
ge1045
−3
j
+ n& 1 = b0.2596 gb0.0619 g ⇒ n& 1 = 0.0158 mol H 2O min
References for enthalpy calculations: H 2 Oblg at triple point, dry gas at 2 °C
substance
m& in
Dry gas
H 2 Obv g
7.063
0.00459
0.285
H 2 Oblg
Q = ∆H =
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 = 18.02n&H 2O
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
= 1.39 × 10 6 J day
min
1 hr
1 day
8- 10
8.29 a.
75 liters C 2 H 5OH blg 789 g
liter
1 mol
46.07 g
= 1284 mol C 2 H 3OH bl g
( C p ) C H3 OH = 01031
.
+ 0557
.
× 10 −3 T ekJ / (mol⋅ o C)j (fitting the two values in Table B.2)
55 L H 2 Oblg 1000 g
liter
1 mol
18.01 g
= 3054 mol H 2 O blg ( C p ) H2 O = 0.0754 bkJ mol⋅° Cg
1284 mol C2 H5 OH(l) (70.0oC)
1284 mol C2H5 OH (l) (To C)
3054 mol H2 O(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
70
Q = ∆ U ≅ ∆ H ( liquids )
 ⇒ ⇓ Integrate, solve quadratic equation
Q = 0 (adiabatic )

T=44.3 o C
b.
1.
2.
3.
4.
5.
6.
7.
25
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
Mistakes in measured volumes & initial temperatures of feed liquids
Thermometer is wrong
8.30 a.
1515 L/s air
500oC, 835 tor,
Td p=30 o C
1515 L/s air , 1 atm
110 g/s H2O(v)
110 g/s H2O, T=25oC
Let n& 1 (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 H2 O in the air stream.
1515 L 835 mm Hg
mol ⋅ K
n& 1 + n& 2 =
= 262
. mol / s
s
773 K
62.36 L ⋅ mm Hg
n& 2
p * (30 o C) 31824
.
mmHg
= y=
=
= 00381
.
mol H 2 O / mol air
&n1 + n& 2
Ptotal
835 mmHg
⇒ n& 1 = 252
. mol dry air / s; n& 2 = 10
. mol H 2 O / s
8- 11
8.30 (cont’d)
References: H2 O (l, 25o C), Air (v, 25o C)
substances
n& in (mol / s)
H$ in (kJ / mol)
dry air
25.2
H2 O(v)
1.0
14.37
100
dC p i
z25
H2O (l )
25.2
T
z25 dC p i air dT
100
7.1
dT + H$ vap
z25
500
z100
H2 O(l)
H$ out (kJ / mol)
n& out (mol / s)
dC p i
H2O (l )
dT + H$ vap
T
dC p i
H2 O( v )
6.1
dT
z100 dC p i H O( v ) dT
2
0
--
--
∆H = 0 = n& out ⋅ H$ out − n& in ⋅ H$ in
F T
b25.2 gG
dC p i
air
H 25
z
I
F 100
K
H 25
dT J + b71
. gGz
F 100
−b25.2gb14.37g − b100
. gGz
H 25
T
dC p i
H2 O( l )
dT + H$ vap + z
100
dC p i
500
dC p i
H2 O( l )
dT + H$ vap + z
100
dC p i
I
H2 O( v)
dT J
K
I
H2 O( v )
dT J = 0
K
Integrate, solve : T = 139 C
o
b.
139
139
Q& = − ( 25.2 ) ∫ ( C p )air dT − (1.00 ) ∫ ( C p )H O( v) dT = −290 kW
500
500
2
This heat goes to vaporize the entering liquid water and bring it to the final temperature of 139o C.
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
Basis:
520 kg NH 3 103 g 1 mol
1h
= 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
Q = –7 kW
n2 (mol/s)
0.100 NH 3
0.900 air
600°C
NH 3 balance: 848
. = 0100
. n2 ⇒ n2 = 848
. mol s
Air balance: n1 = b0.900gb84.8g = 76.3 mol air s
References for enthalphy calculations: NH 3bgg , air at 25°C
NH 3
H$ in = 0.0
H$ out =
600
25
Cp from
dC p i
NH3
dT
⇒
Table B.2
H$ out = 2562
. kJ mol
Air: C p bJ mol ⋅° Cg = 28.94 + 0.4147 × 10 −2 T b° Cg
H$ in =
T
25
L
FT 2
M
N
H2
C p dT = M28.94bT − 25g + 0.004147 G
−
25 2 I O J
1 kJ
×
P
2 JK PQmol 103 J
= e2.0735 × 10 −6 T 2 + 0.02894T − 0.7248j bkJ molg
H$ out =
100
25
Cp dT = 17.39 kJ mol
Energy balance: Q = ∆ H = ∑ n i H$ i − ∑ ni H$ i
E
out
in
−7 kJ s = b8.48 mols NH 3 sgb25.62 kJ mol g + b763
. mols air sgb17.39 kJ molg
− b8.48gb00
. g − b76.3ge2.0735 × 10−6 T 2 + 0.02894 T − 07248
.
j
E
1582
.
× 10−4 T 2 + 2.208 T − 1606 = 0 ⇒ T = 693° C (–14,650°C)
8.32 a.
Basis: 100 mol/s of natural gas. Let M represent methane, and E for ethane
100 mol/s
0.95 mol M/mol
0.05 mol E/mol
Furnace
Stack gas (900oC)
Stack gas (ToC)
n3
n4
n5
n6
n3
n4
n5
n6
mol CO2/s
mol H2O/s
mol O2/s
mol N2/s
Heat
Exchanger
mol CO2/s
mol H2O/s
mol O2/s
mol 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 → CO2 + 2H 2O
C 2 H 6 + b7 / 2gO 2 → 2CO2 + 3H 2 O
8- 13
8.32 (cont’d)
 95 mol M 2 mol O2 4.76 mol air 5 mol E 3.5 mol O2 4.76 mol air 
n&air = 1.2 
+

s
1 mol M
mol O 2
s
1 mol E
mol O2 

n&air = 1185 mol air/s
n&1 = 0.21× 1185 = 249 mol O 2 /s, n&2 = 0.79 ×1185 = 936 mol N 2 /s
n&3 =
95 mol M 1 mol CO 2 5 mol E 2 mol CO 2
+
= 105 mol CO2 /s
s
1 mol M
s
1 mol E
n&4 =
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 O2 5 mol E 3.5 mol O2
+
= 41.5 mol O 2 /s
s
1 mol M
s
1 mol E
n&6 = n&2 = 936 mol N 2 /s
n&5 = 249 −
Energy balance on air:
Q& = n&air
∫
o
245 C
20 o C
(C )
p
air
mol air 
kJ 
kJ

dT = 1185
6.649
= 7879
⇒ 7879 kW


s 
mol 
s

Energy balance on stack gas:
6
∑  n& ∫ ( C ) dT 
Q& = − ∆H = −
T
i
i= 3
−7879 = n&3
900
∫ (C )
T
p CO
2
900
p i
dT + n&4
∫ (C )
T
900
p H O (v )
2
dT + n&5
∫ (C )
T
900
p O
2
dT + n&6
∫ (C )
T
900
p N
2
dT
Substitute for the heat capacities (Table B.2), integrate, solve for T using E-Z Solve⇒ T = 732 C
o
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& ′ = 00434
.
b7851g = 341 kW
Scale factor =
8.33 a.
b.
100
335
. + 4b351
. + 38.4 + 42.0g + 2b367
. + 40.2g439
. = 23100 J mol
3
150 mol 23100 J 1 kW
Q = ∆H = n∆H$ =
= 3465 kW
s
mol 1000 J / s
∆H$ =
600
0
C p dT =
The method of least squares (Equations A1-4 and A1-5) yields (for X = T , y = C p )
Cp = 0.0334 + 1732
.
× 10 −5 Tb° Cg 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.
ln C p = bT 1 2 + ln a ⇒ C p = a expebT 1 2 j ,
b=
ln C p2 C p1
= 00473
.
T2 − T1
ln a = ln C p1 − b T1 = −14475
.
⇒a
T1 = 71
. , C p1 = 0.329 ,
U
|
|
V⇒ Cp
−1.4475
=e
= 0.235 |
|W
8- 14
T2 = 17.3 , C p2 = 0533
.
= 0235
.
expe0.0473T 1 2 j
8.34 (cont’d)
b.
150
0235
. expe0.0473T
12
1800
1
20
30
40
2
200
j dT
=
b0.235gb2g R
L
. T 1 2 j MT 1 2
S exp e0473
00473
.
N
T
150
1 OU
−
= −1730 cal g
V
.0473PQW1800
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 INTEGRATION bbbDELTA(H)b= ', E11.4,'bCAL/G')
CONTINUE
STOP
END
Solution: N = 11 ⇒ ∆H$ = −1731 cal g
N = 101 ⇒ ∆H$ = −1731 cal g
Simpson's rule with N = 11 thus provides an excellent approximation
8.35 a.
U
m& = 175 kg / min
|
175 kg 1000 g 1 mol 56.9 kJ 1 min
M . W. = 62.07 g / molV ⇒ Q& = ∆H =
= 2670 kW
min
kg 62.07 g mol 60 s
|
$
∆H = 56.9 kJ / mol
W
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 fro m state A to state B as shown
in the following phase diagram.
P
B
A
T
8- 15
8.36 a.
Table B.1 ⇒ Tb = 68.74o C, ∆H$ v (Tb ) = 28.85 kJ / mol
Assume: n - hexane vapor is an ideal gas, i.e. ∆H$ is not a function of pressure
$
∆H
Total
 
→
bC6 H 14 g
l, 20 o C
bC 6 H 14 g
v, 200o C
$
∆H
1
B
A
$ bT
∆H
bC6 H 14 g
l, 68.74o C
∆H$ 1 =
∆H$ 2 =
68 .74
20
200
v
b
 

→
g
∆H$ 2
bC6 H 14 g
v, 68.74 o C
0.2163 dT = 10.54 kJ / mol
68.74
013744
.
+ 4085
. × 10 −5 T − 2392
. × 10 −8 T 2 + 57.66 × 10−9 T 3 dT
∆H$ 2 = 24.66 kJ / mol
∆H$ Total = ∆H$ 1 + ∆ H$ 2 + ∆ H$ v bTb g = 1054
. + 24.66 + 28.85 = 64.05 kJ / mol
b.
∆H$ = −64.05 kJ / mol
c.
U$ e200o C, 2 atmj = H$ − PV$
Assume ideal gas behavior ⇒ PV$ = RT = 393
. kJ / mol
U$ = 64.05 − 3.93 = 6012
. kJ / mol
8.37
Tb = 100.00° C
∆H$ v btb g = 40.656 kJ mol
H2 O el, 50o Cj
 
→
B
$ 50 o C
∆H
ve
j
∆H$ 1
A
100
C pH
2 O bl g
∆H$ 2
$ 100o C
∆H
ve
j
 → H2 O ev, 100 o Cj
H2 O el, 100 o Cj
∆H$ 1 =
H 2 O ev, 50o Cj
dT = 377
. kJ mol
25
∆H$ 2 =
25
C pH
2 O bv g
dT = − 169
. kJ mol
100
Table B.1
B
∆H$ v b50° Cg = 377
. + 40.656 − 169
. = 42.7 kJ mol
Steam table:
( 2547.3 − 104.8 ) kJ
kg
18.01 g 1 kg
= 44.0 kJ mol
1 mol 1000 g
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 50o C in Table B.5 is for a pressure of 0.1234 bar (0.12 atm). The difference is ? H for
liquid water going from 50o C and 0.1234 bar to 50o C and 1 atm plus ? H for water vapor going from
50o C and 1 atm to 50o C and 0.1234 bar.
8.38
n& =
1.75 m3
879 kg
2.0 min
3
m
1 kmol
78.11 kg
103 mol 1 min
1 kmol
Tb = 801
. ° C , ∆H$ v bTb g = 30765
.
kJ mol
8- 16
60 s
= 164.1
mol
s
8.38 (cont’d)

→
C 6 H 6 ev, 580 o Cj
B
$
∆H
1
80 .1
C pC
6 H 6 bvg
A
−∆ µ
Hv

→
C 6 H 6 ev, 80.1o Cj
∆H$ 1 =
C6 H 6 el, 25 o Cj
∆H$ 2
C6 H 6 el, 80.1o Cj
dT = −77.23 kJ mol
580
∆H$ 2 =
298
Cp C
dT = −7.699 kJ mol
6 H 6 bl g
3531
.
∆H$ = ∆H$ 1 − ∆H$ v d80.1o Ci + ∆H$ 2 = −1157
. kJ / mol
Q = ∆H = n ∆H$ = b164 .1 mol / sgb−115.7 kJ / mol g = −1.90 x 10 −4 kW
Antoine
B
8.39
35° C
15% relative
( ∆ H$ v ) CCl 4
U
V⇒
saturation W
Table B.1
=
300
.
y CCl 4 = 0.15
PV∗ b25° Cg
1 atm
= 0.15
176.0 mm Hg
= 0.0347 mol CCl4 mol
760 mm Hg
10 mol 0.0347 mol CCl 4
kJ
⇒ Q = ∆H =
mol
min
mol
30.0 kJ
= 104
. kJ min
mol CCl 4
Time to Saturation
6 kg carbon 0.40 g CCl 4
g carbon
8.40 a.
1 mol CCl 4
153.84 g CCl 4
1 mol gas
0.0347 mol CCl 4
CO2 bg, 20° Cg → CO2 bs, − 78.4 ° Cg: ∆H$ =
−78.4
20
dC p i
1 min
= 45.0 min
10 mol gas
CO2 bg g
dT − ∆H$ sub b− 78.4° Cg
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
F kJ I
∆H$ ≈ z
.03611 + 4.233 × 10−5 T − 2.887 × 10 −8 T 2 + 7.464 × 10 −12 T 3 dT G
J
H mol K
20
− 6030
cal 4184
.
× 10 −3 kJ
= − 28.66 kJ mol
mol
1 cal
300 kg CO 2
Q = ∆ H = n∆H$ =
h
10 3 g 1 mol 28.66 kJ removed
= 195
. × 105 k J 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, T fusion =-56o C
Q& = ∆ H = n& ∆H$
where, ∆H$ =
Q& = n& L
−56
M
N 20
−56
20
dCp i
dC p i
CO2 (v)
CO2 (v)
dT +∆H$ v e− 56o Cj +
dT +∆ H$ v e− 56o Cj +
−78 .4
−56
8- 17
−78. 4
−56
dCp i
dC p i
CO2 (l)
CO2 (l)
dT O
P
Q
dT
C p = a + bT
8.41 a.
b=
5394
. − 50.41
= 0.01765
500 − 300
a = 5394
. − b0.01765gb500g =
U
|
|
V ⇒ C p bJ
|
4512
. |W
mol ⋅ Kg = 4512
. + 001765
.
T bKg
NaCl bs, 300 Kg → NaClbs, 1073 Kg → NaClbl , 1073 Kg
1073
1073
O J
b4512
. + 0.01765T gdT P
N 300
Q mol
L
∆H$ = z C ps dT + ∆H$ m b1073 K g = Mz
300
+
30.21 kJ 103 J
mol
1 kJ
= 7.44 × 104 J mol
1073
Q = ∆U = nz C v dT + ∆U$ m b1073 K g
b.
300
Cv ≈ Cp
∆ U m ≅ ∆H m
Q ≈ ∆H = n∆H$ =
t=
c.
8.42
200 kg 10 3 g
1 kg
2.55 × 10 8 J
s
1 mol
74450 J
58.44 g
mol
1 kJ
0.85 × 3000 kJ 103 J
= 2.55 × 108 J
= 100 s
∆H$ v = 35.98 kJ mol , Tb = 136.2° C = 409.4 K , Pc = 37.0 atm , Tc = 619.7 K (from Table B.1)
Trouton's rule: ∆H$ v ≈ 0.088Tb = b0.088 gb409.4 Kg = 360
. kJ molb01%
.
errorg
Chen's rule:
L
F Tb I
J − 0.0327 + 0.0297 log 10
H Tc K
Tb M0.0331G
∆H$ v ≈
M
N
FT I
107
. −G bJ
H Tc K
O
Pc P
P
Q
= 35.7 kJ mol (–0.7% error)
F 619.7 − 373.2 I
J
H 619 .7 − 409.4 K
Watson’s correlation : ∆H$ v b100° Cg ≈ 3598
. G
8.43
0.38
= 38.2 kJ mol
C 7 H 2 N : Kopp's Rule ⇒ C p ≈ 7b0.012g + 12b0.018 g + 0.033 = 0.333 k J (mol ⋅° C)
Trouton's Rule ⇒ ∆H$ v b200 ° Cg = 0.088 b200+ 273.2 g = 41.6 kJ mol
C 7 H12 Nbl, 25° Cg → C 7 H12 Nbl, 200° Cg → C 7 H 12 Nbv , 200° Cg
200
∆Hˆ =
kJ
∫ C dT + ∆Hˆ ( 200 °C ) ≈ 0.333(200 − 25) mol
p
v
25
8- 18
+ 41.6
kJ
= 100 kJ mol
mol
8.44 a.
Antoine equation: Tb b° Cg =
1211.033
− 220.790 = 261
. °C
6.90565 − logb100g
F 562 .6 − 299 .3I
Watson Correction: ∆H$ v b261
. ° Cg = 30.765G
J
H 562.6 − 353.1 K
b.
= 336
. kJ mol
Antoine equation: Tb b50 mm Hgg = 118
. ° C ; Tb b150 mm Hgg = 352
. °C
$
lnb p2 p1 g
∆H v
Clausius-Clapeyron: ln p = −
+ C ⇒ ∆H$ v = − R
RT
1 T2 − 1 T1
∆H$ v = −0.008314
c.
0. 38
lnb150 50g
U
kJ R
|
|
S
V = 34.3 kJ mol
mol ⋅ K T| 1 3084
. K − 1 285.0 K W
|
C6 H 6 ( l , 26.1°C)
C6 H 6 (v, 26.1°C)
∆ H$1
∆H$ v (80.1°C)
C6 H 6 ( l , 80.1°C)
∆H$ 1 =
80 .1
dC p i
26 .1
∆H$ 2 =
l
dC p i
C6 H 6 (v, 80.1°C)
dT = 7.50 kJ mol
26 .1
80.1
∆ H$2
v
dT = −4.90 kJ mol
∆H$ v b261
. ° Cg = 7.50 + 30.765 − 4.90 = 334
. kJ mol
8.45 a. Tout = 49.3o C. 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 49.3o C 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 = 0550
. ( 44.66 mol / s) = 24.56 mol C 5 H10 ( l) / s
Cyclopentane balance ⇒ n& v = ( 44.66 − 24.56) mol C5H 10 / s = 20.10 mol C 5H 10 (v) / s
Reference: C5 H10 (l) at 49.3o C
n& in
(mol/s)
H$ in
(kJ/mol)
n& out
(mol/s)
H$ out
(kJ/mol)
C5 H10 (l)
—
—
24.56
0
C5 H10 (v)
44.66
H$ f
20.10
H$ v
Substance
T
i
Hi = ∆H$ v + z
o
49. 3 C
8- 19
C p dT
8.45 (cont’d)
Substituting for ∆H$ v from Table B.1 and for C p from Table B.2
⇒ H$ f = 38.36 kJ / mol, H$ v = 27.30 kJ / mol
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 dry air/mol)
100 mol
y 1 (mol H2 O/mol)
1-y 1 (mol dry air/mol)
50o C, 1 atm, 2o superheat
n 3 (mol H2 O(l))
There are five unknowns (n 2 , n 3 , y 1 , y 2 , Q) and five equations (two independent material
balances, 2o C superheat, saturation at outlet, energy balance). The problem can be solved.
b.
2° C superheat ⇒ y1 =
p∗b48° Cg
p
saturation at outlet ⇒ y2 =
dry air balance:
p∗b20° Cg
p
b100gb1 − y1g = n2 b1 − y2 g
H2 O balance: b100gby1 g = bn2 gby 2 g + n3
c.
References: Air b25° Cg , H 2Obl, 20° Cg
H$ 1 =
H$ 2 =
=
100 ⋅ b1 − y1 g
H$ in
H$
H 2 Obv g
100 ⋅ y1
H 2 Obl g
−
Substance
nin
Air
n2 ⋅ b1 − y2 g
H$ out
H$
n in mol
H$ 2
n2 ⋅ y2
H$ 4
H$ in kJ mol
−
n3
0
nout
1
3
50
50
dT =
0.02894 + 0.4147 × 10−5 T + 0.3191× 10−8 T 2
dC p i
25
air
25
100
50
dT + ∆ H$ v e100o Cj +
dT
dC p i
dC p i
20
100
H 2O(l)
H 2 O(v)
100
20
50
100
0.0754 dT + 40.656 +
0.03346 + 0.688 × 10−5 T + 0.7604 × 10− 8 T 2 − 3593
.
× 10 −12 T 3 dT
20
H$ 3 =
25
H$ 2 =
20
dCp i
100
− 1965
.
× 10 −12 T 3 dT
air
dC p i
dT
H 2 O(l)
dT + ∆H$ v e100 o Cj +
20
100
dC p i
8- 20
H 2O(v)
dT
8.46 (cont’d)
c.
Q = ∆ H = ∑ ni H$ i − ∑ ni H$ i
out
Vair =
in
100 mol 8314
.
Pa ⋅ m 3
323 K
5
mol ⋅ K
101325
.
× 10 Pa
∑ n H$ − ∑ n H$
i
⇒
d.
i
i
i
Q
out
in
=
V air 100 mol 8.314 Pa ⋅ m 3
323 K
mol ⋅ K
101325
.
× 105 Pa
2° C superheat ⇒ y1 =
p∗b48° Cg 8371
. mm Hg
=
= 0110
.
mol H 2 O mol
p
760 mm Hg
p∗b20° Cg 17.535 mm Hg
=
= 0023
.
mol H 2O mol
p
760 mm Hg
saturation at outlet ⇒ y2 =
. g=
b100 gb1 − 0110
dry air balance:
n2 b1 − 0.023g ⇒ n2 = 9110
. mol
H2 O balance: b100 gb0110
. g = b9110
. gb0.023 g + n3 ⇒ n3 =
890
. mol H 2O 0.018 kg
1 mol
= 0160
.
kg H2 O condensed
Q = ∆ H = ∑ n i H$ i − ∑ ni H$ i = − 4805
. kJ
out
Vair =
100 mol 8.314 Pa ⋅ m 3
323 K
= 2.65 m 3
mol ⋅ K
1.01325 × 105 Pa
⇒
⇒
e.
f.
in
0.160 kg H 2 O condensed
2.65 m 3 air fed
− 4805
. kJ
3
2.65 m air fed
= 0.0604 kg H 2 O condensed / m 3 air fed
= −181 kJ / m 3 air fed
Solve equations with Maple.
Q=
8.47 Basis:
−181 kJ 250 m 3 air fed 1 h 1 kW
= −12.6 kW
h
3600 s 1 kJ / s
m 3 air fed
226 m3
273 K
10 3 mol
min
309 K
22.415 m 3 bSTPg
= 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)
n& 1 ( 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
8- 21
n& 2 [ mol H 2 O(l) / min], 10 o C
8.47 (cont’d)
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.
Table B.3
B
098
. p *w b36° Cg ⇒
b. Inlet air: y0 P =
y0 =
0.98( 44.563 mm Hg)
= 0.0575 mol H 2 O(v) mol
760 mm Hg
Outlet air: y1 = p ∗ (10 o C) / P = b9.209 mm Hg g b760 mm Hgg = 0.0121 mol H 2 O(v) mol
b1 − 0.0575g(8908
Air balance:
F
H 2 O balance: 0.0575G8908
H
mol / min) = b1 − 00121
.
gn& 1 ⇒ n& 1 = 8499 mol / min
mol I
mol
) + n& 2 ⇒ n& 2 = 409 mol H 2 O(l) min
J = 0.0121(8499
min K
min
References: H 2 Obl, triple point g, air b77° Fg
n& in
Substance
Air
H$ in
n& out
8396 0.3198 8396
H 2 Obv g
512
462
103
H 2 Obl g
−
−
409
H$ out
−0.4352 n& in mol min
453
H$ in kJ / mol
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 = ∆H =
∑
ni H$ i −
out
∑
ni H$ i =
−196
. × 105 kJ
60 min 9486
. × 10−4 Btu
min
in
1h
0.001 kJ
1 ton
−12000 Btu h
= 930 tons
8.48
Basis:
746.7 m3 outlet gas / h 3 atm
1 kmol
1 atm 22.4 m3 bSTP g
= 100.0 kmol / h
100 kmol/h at 0°C, 3 atm
yout (kmol C 6 H 14( v)/kmol), saturated
(1 – yout) (kmol N 2/kmol)
n 2 kmols/h nC 6 H 14( v), 0°C
nn&11 (kmol/h) at 75°C, 3 atm
yin (kmol C 6 H 14( v)/kmol), 90% sat'd
(1 – yin) (kmol N 2/kmol)
o
n&2 [kmol n-CH
( v)/h],0 C
6 14
Antoine:
log p∗v = 6.88555 −
1175.817
224.867 + T
p∗v (0 °C ) = 45.24 mm Hg, pv∗ (75°C) = 920.44 mm Hg
8- 22
yout =
yin =
p∗v b0° Cg
P
=
45.24
= 0.0198 kmol C 6 H 14 kmol ,
3b760 g
0.90 p∗v b75° Cg
=
P
b0.90gb920.44 g
3b760g
= 0.363
kmol C 6 H 14
kmol
8.48 (cont’d)
N 2 balance: n& 1 b1 − 0.363g = 100b1 − 00198
.
g ⇒ n& 1 = 153.9 kmol h
C 6 H 14 balance: b1539
. gb0363
. g = b100gb0.0198g + n& 2 ⇒ n& 2 = 5389
. kmol C 6 H 14 bl g h
Percent Condensation: b5389
. kmol h condenseg b0363
.
× 1539
. gbkmol h in feedg × 100% = 96.5%
References: N2 (25o C), n-C6 H14 (l, 0o C)
Substance
n
H$
n
in
N2
98000
in
146
.
out
98000 −0.726
n - C 6 H 14 br g 55800 44.75
−
n - C 6 H 14 blg
−
H$ out
2000
33.33
53800
0.0
n& in mol h
H$ in kJ mol
68 .7
N 2 : H$ = C p bT − 25g, n − C 6 H 14 (v): H$ =
T
$
z C pl dT + ∆ Hv b68.7g +
0
z C pv dT
68.7
Energy balance: Q = ∆ H = ( −2.64 × 10 6 kJ h)(1 h / 3600 s) ⇒ −733 kW
∑ n H$ − ∑ n H$
i
i
out
i
i
in
8.49 Let A denote acetone.
Q& ( kW)
W& s = − 25.2 kW
n& 1 (mol / s) @ − 18o C, 5 atm
y1[mol A(v) / mol], sat'd
(1 − y 1 )( 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
a.
Degree of freedom analysis :
6 unknowns ( n& 0 , n& 1 , n& 2 , y0 , 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
8- 23
b.
Ideal gas equation of state
Raoult’s law
P0V&0
RT0
(1) n& 0 =
(2) y1 =
p*A ( −18o C)
5 atm
(Antoine equation for p *A )
Feed stream analysis
(3)
F mol A I
J
H mol K
y0 G
=
[(4.973 − 4.017) g A][1 mol A /58.05 g]
[( 300
. L) P0 / RT0 ] mol feed gas
8.49 (cont’d)
Air balance
(4) n& 1 =
Acetone balance
n& 0 (1 − y0 )
(1− y1 )
(5) n& 2 = n& 0 y 0 − n&1 y1
o
o
Reference states : A(l, –18 C), air(25 C)
Hˆ in
n&in
(mol/s)
Substance
A(l)
−
A(v)
air
(kJ/mol)
(kJ/mol)
n&0 y0
−
Hˆ A0
n&1 y1
0
Hˆ A1
n&0 (1 − y 0 )
Hˆ a 0
n&1 (1 − y1 )
Hˆ a1
56 o C
(6) H$ A(v) ( T ) = z
−18 C
o
n&2
( C p ) A(l) dT + ( ∆H$ v ) A +
Table B.2
(7)
Hˆ out
n&out
(mol/s)
T
z56
Tab le B.1
o
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&
c.
(1) ⇒ n&0 = 5.32 mol feed gas/s
$
in H in
(W& s = −252
. kJ / s)
(2) ⇒ y1 = 6.58 × 10 −3 mol A(v)/mol outlet gas
(3) ⇒ y 0 = 0.147 mol A(v)/mol feed gas
(4) ⇒ n&1 = 4.57 mol outlet gas/s
(5) ⇒ n&2 = 0.75 mol A(l)/s
(6) ⇒ Hˆ A 0 = 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- 24
8.50 a.
2
3 m π × b35g cm 2
s
1 m2
273 K
104 cm 2
b273 + 40gK
850 mmHg
1 kg ⋅ mol
10 3 mol
760 mmHg 22.4 m 3 bSTP g 1 kg ⋅ mol
= 50.3 mol s
H = n-hexane
50.3 mol/s, 850 mmHg
assume P=850 mmHg
n2 mol H(v)/mol, sat’d @ To C
n3 mol air/mol
n 1 (mols H( l )/s)
(90%
feed)
(60% of
of H in feed)
x0 mol H/mol
(1-x0) mol air/mol
40o C, Tdp =20o C
8.50 (cont’d)
Degree-of-freedom analysis
5 unknowns (n 1 , n 2 , n 3 , x0 and T)
– 2 independent material balances
– 1 saturation condition
– 1 60% recovery equation
– 1 energy balance
0 degrees of freedom
All unknowns can be calculated.
b.
Antoine equation, Table B.4
dTdp i
feed
= 25 ° C ⇒ x 0 =
60% recovery
p*H b25° Cg
⇒ n1 =
P
0.600
151 mm Hg
= 0178
.
mol H mol
850 mm Hg
b50.3gb0.178g mols H feed
= 5.37 mols H blg s
s
=
n2 = b0.400gb50.3gb0178
. g = 358
. mols Hbv g s
Air balance: n3 = b50.3gb1 − 0178
. g = 413
. mols air s
Mole fraction of hexane in outlet gas:
p*H bT g
n2
3.58
=
=
⇒ p*H bT g = 67.8 mm Hg
n2 + n3 b3.58 + 41.3g 850 mm Hg
Antoine equation: p *H = 67 .8 mm Hg ⇒ T = 7 .8 ° C
Reference states: C 6 H 14 bl, 7.8° Cg , 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
n& out
8- 25
n& in mol/s
$
H in kJ/mol
68 . 74
T
C p from Table B.2
C6 H14 ( v ) : H$ = z C pl dT + ∆ H$ v b6 8 .74 ° C g + z C pv d T ,
∆ H$ v from Table B.1
7 .8
68 .7 4
Air: H$ from Table B.8
Energy balance: Q = ∆ H =
∑ n& H$ − ∑ n& H$
i
out
c.
u ⋅ A = u'⋅ A '; A =
i
i
i
=
−257 kJ s 1 kW cooling
in
π ⋅ D2
1 U
|
; D' = DV ⇒ u' = 4 ⋅ u = 12.0 m / s
4
2 |W
8- 26
−1 kJ s
= 257 kW
8.51
n& v ( mol / min) @ 65o C, P0 ( atm)
y[ mol P(v) / mol], sat'd
(1- y) (mol H(v) / mol)
100 mol / s @80o 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 ( 65o C) = 1851 mm Hg, p *H ( 65o C) = 675 mm Hg
Raoult' s law for pentane and hexane
0.410 p *P (65o C) = yP0
y = 0.656 mol P(v) / mol
⇒
0.590 p *H ( 65o 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 =
Fractional vaporization: f =
nv RT 36.6 mol
=
P0
s
⇒
n& v = 36.6 mol vapor / s
n& l = 634
. mol liquid / s
0.08206 L ⋅ atm
mol ⋅ K
36.6 mol vapor / s
mol vaporized
= 0.366
100 mol / s
mol fed
References: P(l), H(l) at 65 o C
H$ in
Substance n& in
P(v)
−
P(l)
50
H(v)
−
H(l)
50
H$ out
n& out
−
24.0 24.33
2.806 26.0
−
0
12.6 29.05
3.245 37.4
Vapor: H$ ( T ) =
Liquid: H$ ( T) =
n& in mol s
H$ in kJ / mol
0
Tb
$
T
z65 C C pl dT + ∆Hv (Tb ) + zT
o
C pv dT
b
T
z65
o
C
C pl dT
Tb and ∆H$ v from Table B.1, C p from Table B.2
Energy balance:
Q& =
∑ n&
$
out H out
−
∑ n&
$
in H in
8- 27
= 1040 kW
(65 + 273)K
152
. atm
= 667 L / s
8.52 a.
B=benzene; T=toluene
n 2 mol/s 95o C
1320 mol/s 25o C
0.735 mol B/mol
0.265 mol T/mol
0.500 mol B/mol
0.500 mol T/mol
n 3 mol/s 95o C
0.425 mol B/mol
0.575 mol T/mol
Q
Total mole balance: 1320 = n2 + n3
U
Rn 2 = 319 mol / s
V⇒ S
Benzene balance: 1320(0.500) = n2 ( 0.735) + n3 ( 0.425) W Tn 3 = 1001 mol / s
References: B(l, 25o C), T(l, 25o C)
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)
Q=
660
-660
--
∑ n Hˆ − ∑ n Hˆ
i
i
out
b.
i
0
-0
-i
425
234
576
85
9.838
39.91
11.78
46.06
= 2.42 × 10 4 kW
in
Antoine equation (Table B.4) ⇒ p *B e95o Cj = 1176 torr , pT* e95o Cj = 476.9 torr
Raoult's law
Benzene: b0.425gb1176g = b0.735gP ⇒ P = 680 torr
Toluene: b0.575gb476.9g = b0.265gP ' ⇒ P ' = 1035
|U
V⇒
torr|W
P ≠ P'
⇒ 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).
8.53 Kopp’s rule (Table B.10):
C 5 H12 Obs g — C p = b5gb7.5g + b12gb9.6g + 17 = 170 J mol
C 5 H12 Oblg — C p = b5gb12g + b12gb18g + 25 = 301 J mol
Trouton’s rule — Eq. (8.4-3): ∆Hv = b0109
. gb113 + 273g = 421
. kJ mol
Eq. (8.4-5) ⇒ ∆H$ m = b0.050 gb52 + 273g = 16.25 k J mol
Basis :
235 m3
h
273 K
1 kmol
3
389 K 22.4 m bSTP g
103 mol
1h
1 kmol
3600 s
= 2.05 mol s
Neglect enthalpy change for the vapor transition from 116°C to 113°C.
C 5 H 12 Obv , 113° C g → C 5H 12 Obl, 113° Cg → C 5 H 12 Obv, 52° Cg
→ C 5 H 12 Obs, 52° Cg → C 5 H12 Obs, 25° Cg
8- 28
8.53 (cont’d)
∆H$ = − ∆H$ v + C pl b52 − 113g − ∆H$ m + C ps b25 − 52g
kJ
kJ
J
1 kJ
= −421
.
− 16.2
− b301gb61g + b170gb27g
× 3 = −813
. kJ mol
mol
mol
mol 10 J
Required heat transfer: Q = ∆H = n∆H$ =
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
90% A evaporation
5 kg acetone
95 kg DF
5 kg C3 H 6O( l)
Tf 1 = 35°C
n 1 mol air
Ta1 , 1.01 atm
0.5 kg acetone remain in film
4.5 kg acetone exit in gas phase
95 kg DF
0.5 kg C 3 H 6O( l)
Tf 2
n 1 mol air
4.5 kg C3 H 6O( v) (40% sat'd)
Ta2 = 49°C, 1.0 atm
Antoine equation (Table B.4) ⇒ p *C3 H 6O = 59118
. mm Hg
1 kmol
10 3 mol
58.08 kg
kmol
4.5 kg C3 H 6 O
⇒y=
= 77.5 mol C 3 H 6Obv g in exit gas
171.6 mol 22.4 LbSTPg
040
. b59118
. mm Hgg
LbSTPg
775
.
=
⇒ n1 =
= 405
.
775
. + n1
760 mm Hg
kg DF
mol
95 kg DF
References: Airb25° Cg, C 3 H 6Obl , 35° Cg, DFb35° Cg
b.
Substance
nin
H$ in
nout
H$ out
DF
95
0
95
1.33 dT f 2 − 35i
C 6 H 14 Oblg
86.1
0
8.6 0.129 dT f 2 − 35i
C 6 H 14 Obvg
—
—
77.5
32.3
dT 171.6
0.70
171.6
Air
86
H$ A(v) =
Ta1
z25
dC p i
air
n in kg
H$ in kJ/kg
n in mol
H$ in kJ/mol
49
$
$
zdC p i dT + ∆H v + zdC p i dT , HDF = C p bT − 35g
l
v
35
86
Energy balance
∆H =
∑ n H$ − ∑ n H$
i
i
i
out
⇒
c.
= 126.4dT f 2 − 35i + 111
. (T f 2 − 35) + 26234
. − 171.6
in
Ta1
z25
dC p i
air
Ta1
Ta1 = 120° C ⇒ z
25
dT =
dC p i
Ta 1
z25
i
1275
. dT f 2 − 35i + 2623.4
air
1716
.
dT = 2.78 kJ mol ⇒ dT f 2 − 35i ° C = −168
. °C
8- 29
dC p i
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.
Tset b p = 200 psiag ≈ 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 C3 H8 (v)/h
200 psia, 100o F
8.357 lb-mole C3 H8 (l)/h
200 psia, 100o F
Q&
m(lb
& - mole H 2O(l) / h
70o F
m(lb
& - mole H 2O(l) / h
85o F
The outlet water temperature is 85o F. 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
& = − n& ∆H$ = 8.357 lb − moles −18.77 kJ 0.9486 Btu 453.593 mol = −6.75 × 104 Btu
Q& = ∆ H
v
h
h
mol
kJ
1 lb ⋅ mole
Energy balance on cooling water: Assume no heat loss to surroundings.
& p ∆T ⇒ m& =
Q& = ∆H& = mC
6.75 ×104 Btu
lb m ⋅°F
1.0 Btu 15 °F
h
= 4500
lbm cooling water
h
8.56
m& 2 [kg H2 O(v)/h]@100o C, 1 atm
o
1000 kg/h, 30 C
0.200 kg solids/kg
0.800 kg H2 O(l)/kg
m& 3 (kg/h) @ 100 o C
0.350 kg solids/kg
0.650 kg H2 O(l)/kg
m& 1 [ kg H 2 O(v) / h], 1.6 bar, sat'd
a.
m& 1 [ kg H 2 O(l) / h], 1.6 bar, sat'd
Solids balance: 200 = 0.35m3
⇒ m3 = 571.4 kg h slurry
H 2 O balance: 800 = m2 + 0.65b5714
. g
⇒ m2 = 428.6 kg h H 2 Obvg
8- 30
8.56 (cont’d)
References: Solids (0.01°C), H 2 O (l, 0.01o C)
200
800
—
H$ in
62.85
125.7
—
200
571.4
428.6
H$ out
209.6
419.1
2676
m
&1
2696.2
m
&1
475.4
Substance
m
& in
Solids
H 2 Oblg
H 2 Obv g
H 2 O , 1.6 bar
E.B. Q = ∆H =
m
& out
∑ m& Hˆ − ∑ m& Hˆ
i
i
i
out
i
& ( kg h ) H$ H 2O from steam tables
m
H$ bkJ kgg
& 1 = 592 kg steam h
= 0 ⇒ 1.315 × 106 − 2221m& 1 = 0 ⇒ m
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)
n (kg A(l )/h)
condenser 1
329 K
303 K
15000 kg/h
0.46 A
0.27 B
0.27 C
348 K, 1 atm
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.
n& 2 = b001
. gb0.46gb15,000 kg hg = 69 kg A h
Acetic acid balance: n& 3 = b0.27gb15,000 g = 4050 kg B h
Acetic anhydride balance: n& 4 = b027
. gb15, 000g = 4050 kg h
Acetone balance:
. gb15,000g = n1
b046
+ 69 ⇒ n1 = 6831 kg h `
⇓
Distillate product: 6831 kg acetone h
8169 kg h
0.8% acetone
Bottoms product: b69 + 4050 + 4050g kg h =
49.6% acetic acid
49.6% acetic anhydride
b.
Energy balance on condenser
8- 31
8.57 (cont’d)
C 3 H 6 O bv , 329 K g → C 3 H 6 Obl, 329 K g → C 3 H 6 Obl, 303 K g
303
∆H$ = − ∆H$ v b329 K g +
z C pl dT = −520.6 + b2.3gb−26g = −580.4
kJ kg
329
b2 × 6831gkg − 580.4 kJ
Q& c = ∆H& = n& ∆H$ =
= −7.93 × 10 6 kJ h
h
kg
c.
Overall process energy balance
Reference states : A(l), B(l), C(l) at 348 K (All H$ m = 0 )
Substance
n& in
H$ in
A bl, 303 Kg
—
—
—
0
0
0
6831 –103.5
69
115.0
4050 109.0
—
0
4050
A bl, 398 Kg
B bl, 398 Kg
C bl, 398 Kg
Acetic anhydride (l):
n& out
C p ≈ b4 × 12g + b6 × 18g + b3 × 25g
H$ out
n& in kg/h
H$ in kJ/kg
113
J
1 mol 103 g 1 kJ
mol⋅° C 102.1 g 1 kg 103 J
= 2.3 kJ kg⋅° C
$
H bT g = C p bT − 348 g (all substances)
Q& = ∆H& ⇒ Q& c + Q& r =
∑ n& H$ − ∑ n& H$
i
out
i
i
i
⇒ Q& r = −Q& c +
in
A
∑ n& H$
i
i
= e7.93 × 106 + 2.00 × 105 j kJ h
out
=0
= 813
. × 106 kJ h
(We have neglected heat losses from the still.)
d.
H 2 O (saturated at ≈ 11 bars): ∆H$ v = 1999 kJ kg (Table 8.6)
813
. × 106 kJ h
Q& r = n& H2 O ∆H$ v ⇒ n& H 2O =
= 4070 kg steam h
1999 kJ kg
8.58
Basis : 5000 kg seawater/h
a.
S = Salt
n 3 (kg H2 O(l )/h @ 4 bars)
2738 kJ/kg
5000 kg/h @ 300 K
0.035 S
0.965 H 2O( l)
113.1 kJ/kg
n 5 (kg H2 O(l )/h @ 4 bars)
605 kJ/kg
b.
S balance on 1st effect:
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 2O( l)
360 kJ/kg
. gb5000g =
b0035
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 2O(l )/h @ 0.6 bars)
360 kJ/kg
0055
. n&1 ⇒ n& 1 = 3182 kg h
Mass balance on 1st effect: 5000 = 3182 + n& 2 ⇒ n& 2 = 1818 kg h
8- 32
8.58 (cont’d)
Energy balance on 1st effect:
∆H& = 0 ⇒ bn& 2 gb2654g + bn& 1 gb360 g + bn& 5 gb605 − 2738g − b5000 gb1131
. g=0
n& 5 = 2534 kg H 2 Obv g h
n&1 = 3182
n&2 =1818
c.
Mass balance on 2nd effect: 3182 = n& 3 + n& 4 (1)
Energy balance on 2nd effect:
b∆H
= 0g
bn 4 gb2610g + bn 3 gb252 g + bn 2 gb360 − 2654g − bn 1 gb360g =
E
0
n1 = 3182, n2 = 1818
5.316 × 10 = 252 n3 + 2610n4
6
(2)
Solve (1) and (2) simultaneously:
n& 3 = 1267 kg h brine solution
n& 4 = 1915 kg h H 2 Obv g
Production rate of fresh water = n& 2 + n& 4 = b1818 + 1915g = 3733 kg h fresh water
Overall S balance: b0035
. gb5000g = 1267 x ⇒ x = 0138
.
kg salt kg
d.
e.
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.
n 5 (kg H 2 O(v )/h)
2738 kJ/kg
3733 kg/h H 2 O(v ) @ 0.2 bar
2610 kJ/kg
5000 kg/h
0.035 S
0.965 H 2O( l)
113.1 kJ/kg
n 1 (kg brine/h @ 0.2 bar
252 kJ/kg
Q3
n 5 (kg H 2 O(l )/h)
605 kJ/kg
Mass balance: 5000 = 3733 + n& 1 ⇒ n&1 = 1267 kg h
Energy balance:
& = 0i
d∆H
b3733gb2610 g + b1267gb252 g + n& 5 b605 − 2738g − b5000gb113.1g =
0
⇒ n& 5 = 4452 kg H 2 O bv g h
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.
Salt balance: x L7 n& L 7 = x L1n& L1 ⇒ n& L1 =
= 583 kg h
030
.
= 5000 − 583 = 4417 kg fresh water h
Fresh water produced: n L 7 − n L1
b.
Final result given in Part (d).
c.
Salt balance on i th effect:
b0.035gb5000g
n& Li x Li = bn& L gi + 1b x L gi +1 ⇒ x Li =
&L g
bn
i +1
bx L g
n& θ Li
i +1
(1)
Energy balance on i th effect:
∆H& = 0 ⇒ n& vi H$ vi + bn& v g L −1 e H$ v j
⇒
bn& v g
L −1
=
L −1
+ n& Li H$ Li − bn& L g L +1 e H$ L j
n& vi H$ vi + n& Li H$ Li − bn& L gi +1 eH$ L j
Mass balance on bi − 1g
$
e Hv j
th
i −1
− e H$ L j
L +1
− bn& v g L −1 e H$ v j
L −1
=0
(2)
i +1
L −1
effect:
n& Li = bn& v gi −1 + bn& L gi −1 ⇒ bn& L gi −1 = n& Li − bn& v g i −1
(3)
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 p i
v
= dC p i = 20 cal (mol⋅° C) ; bCv gv ≈ dC p i − R ≈ b10 − 2g
v
l
cal
= 8 cal (mol ⋅° C)
mol⋅° C
b.
n 0 (mol N2 )
n 0 (mol N2 )
o
3.00 L@ 93 C, 1 atm
n 2 [mol A(v)]
85o C, P(atm)
n 1 (mol A(l)
n 3 [mol A(l)]
o
85o C, P(atm)
0.70 mL, 93 C
n0 =
3.00 L
n1 =
70.0 mL
273 K
1 mol
= 0100
.
mol N 2
b273 + 93gK 22.4 LbSTP g
0.90 g 1 mol
= 15
. mol Ablg
mL 42 g
Energy balance ⇒ ∆U = 0 ⇒
∑ n U$ − ∑ n U$
i
i
i
out
c.
i
=0
in
References: N 2 bgg, Ablgb85° C, 1 atm g
nin U$ in nout U$ out
010
.
398
. 010
.
0
n in mol
15
.
160 n3
0
U$ in cal mol
−
−
n2 20050
Substance
N2
Abl g
Abv g
Abl, 93° Cg and N 2 b g, 93° Cg: U$ = Cv b93 − 85g
Abv, 85° Cg: U$ A( v ) = 20b90 − 85g + 20,000 + 10b85 − 90g = 20050 cal mol
∆U = 0 ⇒ nv1 b20050 g − b010
. gb398
. g − b15
. gb160 g = 0 ⇒ nv1 = 0.012 mol A evaporate
0.012 mol A 42 g A
⇒
d.
mol A
= 051
. g evaporate
Ideal gas equation of state
P=
bn 0
+ n2 gRT
V
=
0.112 mol
3.00 liters
b273 + 85gK
0.08206 L ⋅ atm
= 1.097 atm
mol ⋅ K
Raoult’s law
p ∗A b85° Cg = y A P =
n2
0.012 mol 1097
.
atm
P=
= 0.117 atm
0.112 mol
n0 + n2
8- 35
b= 89.3 mmHgg
8.61 (a) i)
b4.4553 − 3.2551gkg
kg
F mI
Expt 1 ⇒ G J
=
= 0.600 ⇒ bSG gliquid = 0.600
HV K liquid
2.000 L
L
ii) Expt 2 ⇒ Mass of gas = b3.2571 − 3.2551gkg = 0.0020 kg = 2.0 g
Moles of gas =
b763 − 500gmm
2.000 L 273 K
363 K
Hg
1 mol
= 0.0232 mol
22.4 litersbSTPg
760 mm Hg
Molecular weight = b2.0 g g b0.0232 mol g = 86 g mol
iii) Expt. 1 ⇒ n =
bliquidg
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 = ∆U = nC v ∆T
⇒ bCv gliquid =
Q
800 J
=
= 24 J mol ⋅ K@ 284.2 K
n∆T b14 molsgb2.4 Kg
=
800 J
= 24 J mol ⋅ K@331.2 K
b14 molsgb2.4 Kg
⇒ bCv gliquid ≡ 24 J mol ⋅ K
Expt. 2 ⇒ n = 0.0232 mol from biig
bvapor g
T
L
1
N
Cv = a + bT ⇒ Q = 0.0232 T 2 (a + bT ) dT = 0.0232Ma ( T2 − T1) +
b 2
O
(T2 − T12 )P
2
Q
b
OU
(3669
. 2 − 363.0 2 ) P|
a = −4.069
2
N
Q|
V⇒
b
b = 0.05052
L
O
1.30 J = 0.0232 Ma(492.7 - 490.0) + (492.7 2 − 490.02 )P|
2
N
Q|W
L
1.30 J = 0.0232 Ma(366.9 - 363.0) +
⇒ bCv gvapor (J / mol ⋅ K) = −4.069 + 0.05052 T bKg
iv) Liquid: C p ≈ Cv ≡ 24 J mol ⋅ K
Vapor: Assuming ideal gas behavior, C p = Cv + R = Cv + 8.314 J mol ⋅ K
⇒ C p bJ mol ⋅ Kg = 4.245 + 0.05052 T bKg
v)
Expt. 3 ⇒ T
T
T
T
= 315K , p ∗ = b763 − 564gmm Hg = 199 mm Hg
= 334K , p ∗ = 401 mm Hg
= 354K , p ∗ = 761 mm Hg
= 379K , p ∗ = 1521 mm Hg
8-36
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
.
× 10 7 expb− 3770 T g
T bK g
1 17 .28 − lnb760g
=
= 2 .824 × 10 − 3 K −1 ⇒ Tb = 354 K
A
T
3770
Part v b
vi) p ∗ = 760 mm Hg ⇒
vii)
∆H$ v
= 3770bKg ⇒ ∆H$ v = b3770 K gb8.314 J mol ⋅ Kg ⇒ ∆H$ v = 31,300 J mol
R PartA v
3.5 L feed 273 K
1 mol
= 0.0836 mol s feed gas
s
510 K 22.4 lbSTP g
Let A denote the drug
(b) Basis:
.
0.0836 mol/s @ 510 K
0.20 A
0.80 N 2
Q(kW)
n 1 [mol A(v)/s]
n. 2 [mol N 2/s]
T(K), saturated with A
.
n 3 (mols A(l )/s), 90% of A in feed
T(K)
N 2 balance: n&2 = b0.800gb0.0836 mol sg = 0.0669 mol N 2 s
90% condensation: n&3 = b0.900gb0.200 × 0.0836 g = 0.01505 mol Abl g s
n&1 = b0.100gb0.200 × 0.0836g = 1.67 × 10− 3 mol Abv g s
Partial pressure of A in outlet gas:
pA =
n&1
1.67 × 10 −3 mol
P=
( 760 mm Hg) = 18.5 mm Hg = p∗A bT g
n
&
+
n
&
0.0686
mol
b 1
2g
E
Part (a) - (v)
1 17.28 − lnb18.5g
=
= 3.81 × 10 −3 K −1
T
3770
⇓
T = 262 K
(c) Reference states: N2 , Abl g at 262 K
substance
n&in
H$ in
n& out
H$ out
N2
0.0669 7286
0.0669
0
n& in mol s
−3
Abv g
0.0167 37575 1.67 × 10
31686 H$ in J mol
Abl g
−
−
0.01505
0
8-37
8.61 (cont’d)
N 2 b510 Kg: H$ N2 (510 K) - H$ N 2 ( 262 K) = H$ N 2 (237o C) - H$ N 2 (−11o C)
Table B.8
B
= [6.24 - (-1.05)] kJ / mol = 7.286 kJ / mol = 7286 J / mol
A(v, 262K): H$ = C pl bTb − 262g + ∆H$ v b359K g +
262
Tb
Cpv dT
Part (a) results for Tb , C pl , C pv , ∆H$ v
H$
2
262
L
T O
= 24b354 − 262g + 31300 + M4.245 + 0.05052 P
2 Q354
N
A(v, 510K): H$ = C pl bTb − 262g + ∆H$ v b354K g +
510
Tb
= 31686 J mol
C pv dT = 37575 J mol
−1060 J s 1 kW cooling
Energy balance: Q& = ∆H& = ∑ ni H$ i − ∑ ni H$ i =
= 1.06 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 = 0.96b0.72 × 50 kg min g = 34 .56 kg H 2 O bv g min
m& 3 = 0.04b0.72 × 50 kg ming = 1.44 kg H 2 O bl g min
Dry meat balance:
m& 2 = b0.28gb50g = 14.0 kg D.M. min
Reference states: Dry meat at −26° C , H 2 Obl, 0° Cg
substance
m& in
H$ in
m
& out
H$ out
dry meat
14.0
0
14.0
105
−
−
209
H 2 O bs, − 26° Cg 36.0 −390
H 2 O bl , 50° Cg
−
−
1.44
H 2 Obv , 60° Cg
−
−
34.56 2599
1.38 kJ
Dry meat: H$ b50° Cg = C p 50 − b− 26g =
kg ⋅ C°
m
& in kg min
H$ in kJ kg
76° C
= 105 kJ kg
H 2 Obs, − 26° Cg: H 2 Obl , 0° Cg → H 2 Obs, 0° Cg → H 2 Obs, − 26° Cg
8-38
8.62 (cont’d)
−26
∆H$ = − ∆H$ m b0° Cg + z C p dT =
−6.01 kJ
1 mol 10 3 g 2.17 kJ −26° C
= −390 kJ kg
18.02 g 1 kg + kg⋅° C
mol
0
A
Table B.1
H 2 Obl, 50° Cg: H 2 Obl , 0° Cg → H 2 Obl , 50° Cg
0.0754 kJ
50
$
∆H = zC pdT = mol ° C
b50 −
0g° C
1 mol
18.02 g
1000 g
1 kg = 209 kJ kg
A
0
Table B.2
H 2 Obv , 60° Cg: H 2 Obl , 0° Cg → H 2 Obl , 100° Cg → H 2 Obv , 100° Cg → H 2Obv, 60° Cg
0.0754 kJ
∆H$ = mol⋅° C
b100
− 0g° C
+ 40.656
A
Table B.2
=
60
kJ
+ z dC p i H O(v) dT
2
mol 100
A
A
Table B.1 d ∆H$ v
46.830 kJ
1 mol
1000 g
mol
18.02 g
1 kg
Table B.2
i
= 2599 kJ kg
Energy balance:
Q = ∆H =
∑ m H$ − ∑ m H$
i
out
i
i
i
=
1.06 × 105 kJ 1 min
min
in
60 s
1 kW
1 kJ s
= 1760 kW
8.63 Basis: 20,000 kg/h ic e crystallized. S = solids in juice. W = water
.
Qf
preconcentrate
.
.
m 1 (kg/h) juice
m 2 (kg/h)
freezer
0.12 solids(S)
x 2 (kg S/kg)
0.88 H 2O( l )(W) (1 – x 2) (kg W/kg)
20°C
.
Slurry(10% ice), –7°C
m 5 (kg/h) product
filter
20,000 kg W(s )/h
0.45 kg S/kg
.
m4 kg residue/h
0.55 kg W/kg
0.45 kg S/kg
20,000 kg W(s )/h
.
0.55 kg W( l )/kg
m 4 (kg/h), 0.45 S, 0.55 W
.
m 3 (kg/h), 0°C
separator
0.45 kg S/kg
20,000 kg W( s )/h
0.45 kg W( l )/kg
20000 10
=
⇒m
& 4 = 180000 kg h concentrate leaving freezer
&4
m
90
& 1 = 27273 kg h feed
m
Overall S balance: 012
. m& 1 = 0.45m
&5
U
⇒
V
& 5 = 7273 kg h concentrate product
m
&1 = m
& 5 + 20000
Overall mass balance: m
(a) 10% ice in slurry ⇒
W
Mass balance on filter: 20000 + m& 4 + m& 5 + 20000 + m& 6
⇒
& 4 =180000
m
m
& 5 = 7273
m& 6 = 172730 kg h recycle
Mass balance on mixing point:
& 2 = 2.000 × 105 kg h preconcentrate
27273 + 172730 = m& 2 ⇒ m
8-39
8.63 (Cont’d)
S balance on mixing point:
5
b0.12 gb27273g + b0.45gb172730g = 2.000 × 10 X 2 ⇒ X 2 ⋅ 100% = 40.5% S
(b) Draw system boundary for every balance to enclose freezer and mixing point (Inputs:
fresh feed and recycle streams; output; slurry leaving freezer)
Refs: S, H2 Obl g at −7° C
& out
substance
m& in
H$ in
m
H$ out
& bkg h g
12% soln 27273 108
−
−
m
$
45% soln 172730 28 180000
0
HbkJ kg g
H 2Obsg
−
−
20000 −337
Solutions : H$ bT g = 4.00 T − b−7g kJ kg
Ice: H$ = − ∆H$ b−T ° Cg ≈ − ∆H$ b0° Cg
m
m
= −6.0095 kJ mol ⇒ −337 kJ kg
D
Table B.1
−1.452 × 10 7 kJ
1h
1 kW
E.B. Q& c = ∆H& = ∑ m
& i H$ i − ∑ m
& i H$ i =
= −4030 kW
h
3600 s 1 kJ s
out
in
8.64 a. B=n-butane, I=iso-butane, hf=heating fluid. ( C ) = 2.62 kJ / kg⋅o C
d
i
p hf
24.5 kmol/h @ 10o C, P (bar)
0.35 kmol B(l)/h
24.5 kmol/h @ 180o C
0.35 kmol B(l)/h
Q& ( kW)
o
m& (kg HF / h), T( C)
m& (kg HF / h), 215 o C
From the Cox chart (Figure 6.1-4)
p *B d10 o Ci = 22 psi, pI* d10 o Ci = 32 psi
F 1.01325 bar I
p min = p B + p I = x B p B* + x I p I* = 28.5 psiG
. bar
J = 196
H 14.696 psi K
b.
$
$
Hv
H1
B dl, 10 o Ci ∆

→ B dv, 10 o Ci ∆

→ B dv, 180 o Ci
$
$
Hv
H2
I dl, 10 o Ci ∆

→ I dv, 10 o Ci ∆

→ I dv, 180o Ci
Assume temperature remains constant during vaporization.
Assume mixture vaporizes at 10o C i.e. won’t vaporize at respective boiling points
as a pure component.
8-40
8.64 (cont’d)
References: B(l, 10o C), I(l, 10o C)
substance
n& in bmol / h g H$ in bkJ / molg
B (l)
8575
0
B (v)
--I (l)
15925
0
I (v)
--$
dH out i
B
$
dH out i
∆H =
= d∆H$ v i
I
180
+z
B
10
180
= d∆H$ v i + z
I
10
dC p i
∑ n H$ − ∑ n H$
i
i
out
i
dC p i
i
B
I
n& out bmol / hg
-8575
-15925
H$ out bkJ / mol g
-42.21
-41.01
= 42.21 kJ / mol
= 41.01 kJ / mol
= 8575b42.21g − 15825b41.01g
in
∆H& = 1.015 × 10 6 kJ / h
c.
& hf 2.62 kJ / dkg⋅o Ci
Q = 1015
.
× 10 6 kJ / h = m
b215 −
45go C
m& hf = 2280 kg / h
d.
b2540
kg / hg 2.62 kJ / dkg⋅o Ci
b215 −
45go C = 1131
.
× 10 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 215o C.
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: SG Toluene=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 cm 3
3
3
0.879 g / cm
0.866 g / cm
n total =
Vtotal
dx f i
C6 H 6
=
0.640 mol C 6H 6
= 0.541 mol C6 H6 mol
1.183 mol
Actual feed:
32.5 m 3 106 cm 3
h
1 m
3
1183
.
mol mixture
3
1h
114.6 cm mixture 3600 s
= 93.19 mol / s
T = 90° C ⇒ p ∗C6 H 6 = 1021 mm Hg , pC∗ 7 H8 = 407 mm Hg (from Table 6.1-1)
Raoult's law: p tot = x C 6 H6 pC∗ 6 H6 + x C7 H8 p C∗ 7 H8 = b0.541gb1021g + b0.459 gb407 g
=
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 H6 = 648 mm Hg , pC∗ 7 H8 = 244 mm Hg (from Table 6.1-1)
Raoult's law ⇒ p tank = xC 6 H6 pC∗ 6 H6 + xC7 H 8 p C∗ 7 H 8 = b0.439 gb648g + b0.561gb244g
= b284 + 137 gmm Hg = 421 mmHg ⇒ Ptank = 0.554 atm
yC 6 H6 =
284 mm Hg
= 0.675 mol C6 H 6 bv g 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
nv = 40.27 mol vapor s
Mole balance: 93.19 = nv + n L
U
⇒
C 6H 6 balance: b0.541gb93.19 g = 0.675nv + 0.439 n L VW n L = 52.92 mol liquid s
(c) Reference states: C 6H 6 bl g, C6 H6 bl g at 75° C
Substance
n&in
H$ in
n& out
H$ out
C 6 H 6 bv g
−
−
27.18
31.0
23.23
0
13.09
35.3
C 6 H 6 bl g
C 7 H 8 bv g
C 7 H 8 bl g
50.41 2.16
−
−
42.78 2.64 29.69
n& in mol s
H$ in kJ mol
0
C 6H 6 bl , 90° Cg: H$ = b0144
. gb90 − 75g = 2.16 kJ mol
C 7H 8 bl , 90° Cg: H$ = b0176
. gb90 − 75g = 2.64 kJ mol
C 6H 6 bv , 75° Cg: H$ = b0.144gb801
. − 75g + 30.77 +
A
∆H$ v b 80 .1 °C g
75
80.1
0.074 + 0.330 × 10− 3 T dT
= 31.0 kJ mol
C 7 H 8 bv , 75° Cg: H$ = b0.176gb110.6 − 75g + 33.47 +
75
110.6
0.0942 + 0.380 × 10 −3 T dT
= 35.3 kJ mol
Energy balance: Q& = ∆H& =
∑ 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 equilibr ium.
(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
nV mol vapor/s @ T, P
1 mol/s @ TFo C
y mol A/mol
(1-y) mol B/mol
xF mol A/mol
(1-xF) mol B/mol
nL mol vapor/s @ T, P
vapor and liquid streams
in equilibrium
x mol A/mol
(1-x) mol B/mol
Raoult's law ⇒ x ⋅ p ∗A bT g + b1 − xg ⋅ p B∗ bT g = P ⇒ x =
pA = y ⋅ P = x ⋅
p ∗A bT g ⇒
y=
P − p ∗B bT g
(1)
p ∗A bT g − p B∗ bT g
x ⋅ p ∗A bT g
(2)
P
Mole balance: 1 = n& L + n&V ⇒ n&V = 1 − n& L
(4)
for n v from (4)
A balance: bx F gb1g = y ⋅ n&V + x ⋅ n& L Substitute

→ n& L =
Energy balance: ∆H& =
∑ n& H$ − ∑ n& H$
i
i
i
out
in
A
6.84471
6.88555
B
1060.79
1175.82
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
-0.01
60.0
1609
573
0.412
0.663
0.649
0.351
6.8
7.6
32.5
43.7
-0.01
62.3
1714
617
0.349
0.598
0.394
0.606
7.3
8.0
32.8
44.1
-0.01
i
y − xF
y− x
(3)
=0
(5)
b.
Tref(deg.C) = 25
Compound
n-pentane
n-hexane
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
ABRITARY 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'EVAPORATOR bPRESSUREb=', 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 (1H b, 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 (1H b, 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
$DATA (Fields of 10 Columns)
6.84471 1060.793 231.541 6.88555 1175.817 224.867
25.0
25.0
0.195
36.07
25.77
0.115 0.000341
0.216
68.74
28.85
0.137 0.000409
0.500
110.0
760.0
Solution:
Tevaportor = 52.2° C
n L = 0.552 mol, dxC 5H 12 i
liquid
= 0.383 mol C5 H 12 mol liquid
n v = 0.448 mol, dx C5 H12 i
vapor
= 0.644 mol C5 H 12 mol liquid
8-45
8.67 Basis:
2500 kmol product 1 kmol condensate
= 10,000 kmol h fed to condenser
h
.25 kmol product
.
m
m& 11(kg/h)
, (kg/h)atatT1 T1
1090 kmol/h C3 H8 (v )
7520 kmol/h i -C 4H10 ( v)
1390 kmol/h n -C 4H10 (v )
saturated vapor at T f, P
1090 kmol/h C 3 H 8 ( l )
7520 kmol/h i -C H ( l )
P (mm Hg) 1390 kmol/h -C 4H10 ( )
n 4 10 l
T out
.
m1 (kg/h) at
2 T2
(a) Refrigerant: Tout = 0 o C , T1 = T2 = −6o C .
Antoine constants
C 3H 8
i − C 4H 10
n − C 4 H 10
A
7.58163
6.78866
6.82485
B
1133.65
899.617
943.453
C
283.26
241.942
239.711
Calculate P for Tout = Tbubble pt.
P=
∑ x p ( 0°C ) = 0.109( 3797 mm Hg ) + 0.752(1176 mm Hg ) + 0.139( 775 mm Hg)
i
*
i
i
⇒ P = 1406 mm Hg
Dew pt. T f = Tdp ⇒ f (T f ) = 1 − P
∑ p (T ) = 0
yi
i
*
i
trial & error to find T f
f
Substitute Antoine expressions, use E-Z Solve ⇒ T f = 5.00°C
Refs: C 3H8 bl g , C 4H10 bl g at 0 °C, Refrigerant @ –6°C
substance
C 3H 8
i − C 4H 10
n − C 4 H 10
Refrigerant
n&in
H$ in
n&out
H$ out
Assume: ∆H$ v bTb g , Table B.1
↓
1090 19110 1090
7520 21740 7520
1390 22760 1390
0
0
0
n& (kmol/h)
H$ (kJ/kmol)
0
151
m& (kg/h)
H$ (kJ/kmol)
m& 1
m& 1
$ bvapor g = ∆H$ b0° Cg +
UH
2
v
|
4 .95
V
C p dT bTable B.2g
|
W
0
z
U $
VH
W
= ∆H$ v
E.B.:
∆H = ∑ n&i H$ i − ∑ n&i H$ i = 0 ⇒ 151m& 1 − 2.16 × 106 = 0 ⇒ m& 1 = 1.43 × 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=
∑ x p ( 40°C) = 0.109 (11877 ) + 0.752( 3961) + 0.139( 2831) = 4667 mm Hg
i
*
i
i
( )
f Tf = 1 − P
∑ ( )
i
E-Z Solve
yi
=
0
⇒ T f = 45.7° C
p*i T f
Refs: C 3H8 bl g , C 4H10 bl g @ 40°C, H2 Obl g @ 25°C.
& 1 = 5.74 × 10 6 kg H 2 O / h
∆H& = 0 ⇒ 37 .7m& 1 − 2.17 × 108 = 0 ⇒ m
(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
n3 (mol O 2 )
3.76n 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
n4 (mol H 2 O(v ))
0.0834 mol CH 3OH/mol
0.303 mol N 2/mol
n 1 (mol CH 3 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 2 O( v)/mol 3.1 bars, sat'd
30°C
Q(kJ)
CH 3 OH(l ), 1 atm, sat'd
2.5n8 (mol CH3 OH)
n 6a (mol HCHO)
distillation
absorption
(l )
n 6b (mol CH3 OH(l ))
sat'd, 1 atm
n 6c (mol H2 O(l ))
Product solution
88°C, 1 atm
n 7 (mol)
0.37 g HCHO/g ( x 1 mol/min)
Absorber off-gas
m w3 (kg H 2 O( l ))
0.01 g CH 3OH/g ( x 2mol/min)
n5 a (mol N 2 )
30°C
20oC
0.82 g H 3 O/g (x 3 mol/min)
n5 b (mol O 2 )
n5 c (mol H 2 )
n5 d (mol H 2 O(v )), sat'd
n5 e (mol HCHO(v )), 200 ppm
27°C, 1 atm
a. Strategy
C balance on conversion reactor ⇒ n 2 , N 2 balance on conversion reactor ⇒ n 3
H balance on conversion reactor ⇒ n 4 , (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
H 2 O saturation of absorber off - gas U
V ⇒ n5 d , n 5b
200 ppm HCHO in absorber off - gas W
8-47
8.68 (cont’d)
HCHO balance on absorber ⇒ n6a , CH 3OH balance on absorber ⇒ n6b
Wt. fractions of product solution ⇒ x 1 , x 2 , x 3
HCHO balance on distillation column ⇒ n 7
CH 3OH balance on distillation column ⇒ n8
CH 3OH balance on recycle mixing point ⇒ n1
Energy balance on waste heat boiler ⇒ m w1 , E.B. on cooler ⇒ mw2
Energy balance on reboiler ⇒ Q
C balance on conversion reactor:
n 2 = 19.9 mol HCHO + 8.34 mol CH 3 OH = 28.24 mol CH 3OH
N 2 balance on conversion reactor:
3.76n 3 = 30.3 ⇒ n 3 = 8.06 mol O 2 , 3.76 × 8.06 = 30.3 mol N 2 feed
H balance on conversion reactor:
n 4 b2g + 28.24b4 g − 19.9b2 g + 8.34b4 g + 5b2 g + 35.6b2 g ⇒ n 4 = 20.7 mol H 2O fed
O balance: 65.1 mol O in, 65.5 mol O out. Accept (precision error)
N 2 balance on absorber: 30.3 = n 5a ⇒ 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:
p w* b27° Cg L 26.739 mm Hg
n5d
O
yw =
=M
=
P
30.3 + 0.83 + 5.00 + n5d + n 5e PQ
N 760 mm Hg
⇒ n5d = 0.03518b3613
. + n5d + n5e g 1 U
200 ppm HCHO in off gas:
⇒
n5e
200
= 6 2
36.13 + n5d + n5e 10
|
| solve
|
V ⇒
|
|
|W
n5d = 1.318 mol H 2 O
n5e = 7.49 × 10
−3
mol HCHO
Moles of absorber off-gas = n5a + n5b + n 5c + n 5e = 37.46 mol off - gas
HCHO balance on absorber: 19.9 = n 6a + 7 .49 × 10 −3 ⇒ n 6a − 19.89 mol HCHO
CH 3 OH balance on absorber: 8.34 = n 6b ⇒ n 6b = 8.34 mol CH 3OH
Product solution
%MW
U
Basis-100 g ⇒ 37.0 g HCHO ⇒ 1.232 mol HCHO|
x 1 = 0.262 mol HCHO mol
|
1.0 g CH 3 OH ⇒ 0.031 mol CH 3OH V ⇒ x 2 = 0.006 mol CH 3OH mol
|
62.0 g H 2 O ⇒ 3.441 mol H 2 O
x 3 = 0.732 mol H 2 O mol
|W
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 = 0.262 n 7 ⇒ n7 = 75.9 mol product
CH 3 OH balance on distillation column:
8.34 = 0.006b75.9g + n8 ⇒ n 8 = 7.88 mol CH 3OH
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:
n1 = 20.4 mol CH 3 OHbl g fresh feed
n 2 = 75.9 mol p roduct solution
n 3 = 7.88 mol CH 3 OHbl g recycle
n 4 = 37.5 mol a bsorber off - gas
Waste heat boiler:
Refs: HCHObv, 145° Cg , CH 3 OHbv ,145° Cg ; N 2 , O 2 , H 2 , H 2 Obv g at 25°C for product
gas, H 2 Obl, triple point g for boiler water
substance
n in
H$ in
n out
H$ out
HCHO
CH 3OH
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
n (mol)
3.51
3.60 H$ (kJ/mol)
3.47
4.09
m w1
566.2
m w1 2726.32 m (kg)
H$ (kJ/kg)
N2
O2
H2
H2O
H2O
(boiler)
E.B. ∆H =
∑ n H$ − ∑ n H$
i
out
i
i
i
T
U $
VH
W
=
U
|
| $
VH
|
|
W
= C p bT g T − 25
U $
VH
W
zC p dT
145
from steam tables
= 0 ⇒ −1814 + 2160mw1 = 0 ⇒ mw1 = 0.84 kg 3.1 bar steam
in
8-49
8.68 (cont’d)
Gas cooler: Same refs. as above for product gas, H 2 Obl, 30° Cg for cooling water
substance
n in
H$ in
n out
H$ out
HCHO
CH 3OH
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
m w2
0
m w2
62.76
N2
O2
H2
H2O
H2O
(coolant)
E.B. ∆H =
∑ n H$ − ∑ n H$
i
out
i
i
i
n (mol)
H$ (kJ/mol)
m (kg)
H$ (kJ/kg)
H$ = 4.184
kJ
bT − 30g° C
kg⋅° C
= 0 ⇒ −1581
. + 62 .6mw 2 = 0 ⇒ mw 2 = 2.52 kg cooling water
in
Condenser: CH 3OH condensed = n8 + 2.5n8 = b35
. gb7.88g = 27 .58 mol CH 3 OH condensed
E.B.:
b.
Q = −n ∆H$ v b1 atm g = − b27.58 molgb35.27 kJ molg
= −973 kJ (transferred from condenser)
3.6 × 10 4 tonne / y
10 6 g
1 yr
1d
1 metric ton 350 d 24 h
b0.37 gd4.286
= 4.286 × 10 6 g h product soln
× 10 6 i = 1586
.
× 10 6 g HCHO h ⇒ 5.281 × 10 4 mol HCHO h U
⇒ b0.01gd4.286 × 10 6 i = 4.286 × 10 6 g CH 3 OH h ⇒ 1338 mol CH 3OH h
b0.62 gd4.286
× 10 6 i = 2.657 × 10 6 g H 2 O h ⇒ 1.475 × 10 5 mol H 2 O h
⇒ 2.016 × 10 5 mol h ⇒ Scale factor =
||
V
|
|W
2.016 × 10 5 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 ≈ 13 o 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
= 10.1 lb m DA
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
10.1 lb m DA
From the psychrometric chart, Tdb = 90 o F, ha = 0.0903
h r = 67%
Twb = 80.5 o F
Tdew point = 77 .3o F
8.71
Tdb = 35° C
Tab = 27 ° C
8.72 a.
$ = 44.0 − 0.11 ≅ 43.9 Btu / lb DA
H
m
⇒ h r = 55% He wins
Tdb = 40° C, Tdew point = 20° C
b. Mass of dry air: m da =
Mass of water:
c.
$ = 14.3 ft 3 / lb DA
V
m
hr = 33%, h a = 0.0148 kg H 2 O kg dry air
Fig. 8.4-1
2.00 L
⇒
Twb = 25.5° C
1 m3
1 kg dry air
= 2.2 × 10 −3 kg dry air
0.92 m3
↑ from Fig. 8.4-1
3
10 L
2.2 × 10 −3 kg dry air 0.0148 kg H 2O 10 3 g
1 kg dry air
1 kg
= 0.033 g H 2 O
H$ b40° C, 33% relative humidity g ≈ b78.0 − 0.65g kJ kg dry air = 77.4 kJ kg dry air
H$ b20° C, saturated g ≈ 57.5 kJ kg dry air (both values from Fig. 8.4-1)
∆H 40→ 20 =
2.2 × 10 −3 kg dry air
b57 .5 − 77.4g kJ
10 3 J
kg dry air
1 kJ
= −44 J
d. Energy balance: closed system
n=
2.2 × 10 −3 kg dry air 10 3 g 1 mol
1 kg
29 g
+
0.033 g H 2 O 1 mol
18 g
= 0.078 mol
Q = ∆U = n∆U$ = nd∆H$ − R∆T i = ∆H − nR∆T
= −44 J −
0.078 mol 8.314 J
b20
− 40 g° C 1 K
mol ⋅ K
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 2O min
(b) ha =
= 0.025 kg H2 O kg dry air , Tdb = 50° C
400 kg dry air min
8.73 (a)
min
Fig. 8.4-1
H$ = b116 − 11
. g = 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 2O kg dry air , H$ = 29.5 kJ kg dry air
(d)
b0.0250 − 0.0077g kg
400 kg dry air
min
H 2O
kg dry air
= 6.92 kg H2 O min condense
References: Dry air at 0° C, H 2Obl g at 0° C
substance
m& in
m& out
H$ in
H$ out
Air
400
115
400
29.5
H2 Obl g
—
—
6.92
42
m& air in kg dry air/min, m& H 2 O in kg/min
H$ air in kJ/kg dry air, H$ H 2 O in kJ/kg
H2 Obl, 0° Cg → H 2Obl , 20° Cg :
H$ =
75.4
Q = ∆H =
J
1 mol
mol⋅° C
18 g
∑ m& Hˆ − ∑
i
i
out
(e)
in
b10 − 0g° C
1 kJ
103 g
= 42 kJ kg
10 3 J 1 kg
−34027.8 kJ 1 min 1 kW
m& i Hˆ i =
= −565 kW
min 60 s 1 kJ/s
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 D.A.
m
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 = 21.4 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.
min
13.07 ft 2
= 76.5 lb m D.A. min
H 2 O condensed:
76.5 lb m D.A.
min
b0.0226
− 0.0075g lb m H 2 O
lb m D.A.
8-52
= 1.2 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
References: Same as Fig. 8.4-2 [including H 2 Obl, 32° Fg ]
substance
m& in H$ in m& out H$ out
Air
76.5 45.5 76.5 21.4 m& air in lb m D.A./min
H 2 Obl, 49° Fg
—
—
1.2 17.0 H$ in Btu/ lb D.A.
air
m
m&
in lb /min, H$
H2 O
m
H2 O
in Btu/ lb m
− 455
. + 1.2(17.0) (Btu) 60 min 1 ton cooling
min
1h
−12 ,000 Btu h
= 9.1 tons cooling
Q = ∆H =
.
b76.5g 214
b.
6
7
(76.5 lb m DA/min)
hr = 40%, ha = 0.0075 lb m H 2O/lbm DA
o
75F,
26.2 Btu/lb m DA
1
7
(76.5 lb m DA/min)
hr = 80%, ha = 0.0226 lb m H 2O/lb m DA
Coolerreheater
o
87F,
45.5 Btu/lb m DA
76.5 lb m DA/min
ha = 0.0075 lbm H 2O/lb m DA
Lab
o
55F,
21.4 Btu/lbm DA
Q& lab
m& H2 O (kg H2 O(l)/min)
Q& (tons)
Water balance on cooler-reheater (system shown as dashed box in flow chart)
1  76.5
7

lb m H 2O  6
lb m DA 
 0.0226
 + ( 76.5

min 
lbm DA  7
)( 0.0075 ) = (76.5)(0.0075) + m& H O
⇒ m& H2 O = 0.165 kg H2 O condensed/min
Energy balance on cooler-reheater
References: Same as Fig. 8.4-2 [including H2 O(l, 32o F)]
8-53
2
Substance
m& in
Hˆ in
10.93 45.5
65.57 26.2
Fresh air feed
Recirculated air feed
Delivered air
o
Condensed water (49 F)
& i Hˆ i − ∑ m& i Hˆ i =
Q& = ∆H& = ∑ m
out
in
m& out
—
—
Hˆ out
—
—
21.4
—
—
76.5
—
—
0.165 17.0
m& DA in lb m dry air/min
Hˆ air in Btu/lb m dry air
m& H2O ( l ) in lb m /min
Hˆ HO(l)
in Btu/lb m
2
−575.3 Btu 60 min 1 ton cooling
= 2.9 tons
min
1h
-12,000 Btu/h
Percent saved by recirculating =
(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 =38o C, Twb =29o C
m2a (kg DA)
m2w [kg H2 O(v)]
Inlet air: 11.6 m3 (STP), Tdb =100o C
m1a (kg DA)
1 kg wet chips, 19o C
0.40 kg H2 O(l)/kg
0.60 kg DC/kg
m3c (kg dry chips)
m3w [kg H2 O(l)]
T (o C)
(a) Dry air: m1a =
11.6 m3 bSTPg DA
1 kmol
3
29.0 kg
22.4 m bSTP g 1 kmol
= 15.02 kg DA = m 2a
Outlet air:
Fig. 8.4-1
→ Hˆ 2 = (95.3 − 0.48) = 94.8 kJ kg D.A.
(Tdb = 38°C, Twb = 29°C) 
ha2 = 0.0223 kg H 2O kg D.A.
Water in outlet air: m2 w = ha2 m2a = 0.0223b15.02 g = 0.335 kg H 2 O
(b) H2 O balance: 0.400 kg = 0.335 kg + m3w ⇒ m3w = 0.065 kg H 2O
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
8-54
8.75 (cont’d)
(c) References: Dry air, H 2 Obl g , dry chips @ 0°C.
H$ in
substance
min
Air
H2 Obl g
dry chips
15.02
0.400
0.600
H$ out
mout
100.2 15.02 94.8 mair in kg DA, H$ air in kJ/kg DA
79.5 0.065 4.184T m in kg DC, H$ in in kJ/kg DC
39.9
0.6 2.10T
Energy Balance:
∆H =
Tdb = 45° C
hr = 10%
b. Twb = 21.0° C
hr = 60%
∑m
ˆ
out Hout
8.76 a.
H 2 O added:
8.77
in
= 0 ⇒ −136.8 + 1.532T = 0 ⇒ T = 89.3°C
ha = 0.0059 kg H 2 O kg DA
Tdb = 26.8° C h a = 0.0142 kg H 2 O kg DA
Fig. 8.4-1
15 kg air
1 kg D. A.
min
1.0059 kg air
0.92 m 3
Fig. 8.4-1
− 0.0059g kg H 2O
1 kg D.A.
= 0.12 kg H 2 O min
V$ = 0.92 m 3 kg D.A. , Twb = 22 ° C
= 12.3 kg D.A. min
saturated
12.3 kg D.A.
Tdb = 45° C
U
Tdew point = 4° CVW
b0.0142
h a = 0.0050 kg H 2 O kg D.A.
Outlet air: Twb = Tas = 22° C
8.78 a.
in
Tas = Twb = 210
. °C
11.3 m3 1 kg D.A.
Evaporation:
∑ m Hˆ
Fig. 8.4 -1
Inlet air: Tdb = 50° C
Tdew pt. = 4° C
min
−
⇒ T = 22° C ha = 0.0165 kg H 2O kg D. A.
b0.0165 − 0.0050g kg
min
kg D.A.
bha gin
Fig. 8.4-1
H 2O
= 0.14 kg H2 O min
= 0.0050 kg H 2 O kg D. A.
Twb = 20.4 ° C, V$ = 0.908 m 3 kg D.A.
Twb = Tas = 20.4° C, saturated ⇒ bha gout = 0.0151 kg H 2 O kg D.A.
8-55
8.78 (cont’d)
b. Basis: 1 kg entering sugar (S) solution
m1 (kg D.A.)
0.0050 kg H2 O/kg DA
m1 (kg D.A.)
0.0151 kg H2 O(v)/kg
1 kg
0.05 kg S/kg
0.95 kg H2 O/kg
m2 (kg)
0.20 kg S/kg
0.80 kg H2 O/kg
Sugar balance: b0.05gb1g = b0.20gm2 ⇒ m2 = 0.25 kg
Water balance: bm1 gb0.0050 g + b1gb0.95g = bm1 gb0.0151g + b0.25gb0.80g
m1 = 74 kg dry air
⇒
8.79
A
1 lb mD.A.
h a1 (lb mH 2O)
T d = 20°F
h r = 70%
Inlet air (A):
Coil
bank
Outlet air (D):
1 kg D. A.
B
C
1 lb mD.A.
Spray 1 lb mD.A.
h a2 (lb mH 2O) chamber h a3 (lb mH 2O)
Td = 75°F
H2 O
Tdb = 20° FU
hr = 70%
V=
0.908 m3
74 kg dry air
Fig. 8.4-2
V
W
Tdb = 70° FU
hr = 35%
Fig. 8.4-2
V
W
a. Inlet of spray chamber (B):
= 67 m 3
D
1 lb mD.A.
h a3 (lb m H 2O)
T d = 70°F
h r = 35%
Coil
bank
ha1 ≈ 0.0017 lb m H 2 O lb m D.A.
V$ ≈ 12.2 ft 3 lb m D.A.
h a3 = 0.0054 lb m H 2 O lb m D.A.
ha = 0.0017 lb m H 2 O lb m D.A.U
Tdb = 75° F
V ⇒ Twb
W
= 49.5° 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,
h = 0.0054 lb m H 2 O lb m D.A.U
Outlet of spray chamber (C): a
V ⇒ hr = 52%
Twb = 49.5° F
W
At point C, Tdb = 58.5° F
b.
bha 3
− ha1 g lb m H 2O evaporate
lb m DA
lb m DA
=
V$A dft 3 inlet airi
8-56
b0.0054 − 0.0017 g
12.2
= 3.0 × 10 −4
lb m H 2 O
ft 3 air
8.79 (cont’d)
(20 - 6.4) Btu / lb m dry air
c. QBA = ∆H = H$ B − H$ A ≅
= 1.1 Btu / ft 3
3
12.2 ft / lb m dry air
(23 - 20) Btu / lb m dry air
QDC = ∆H = H$ D − H$ C ≅
= 0.25 Btu / ft 3
12.2 ft 3 / lb m dry air
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.
h a1 (kg H2 O/kg D.A.)
Tdb= 40°C, Tab = 18°C
1 kg D.A.
ha2 (kg H2 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
Outlet air:
⇒ h a2 = 0.0122 kg H 2O kg D.A.
Twb = 18° C badiabatic humidification g
Inlet air:
Overall H 2 O balance: mw + b1gbha1 g = b1gbh a2 g ⇒ m n = b0.0122 − 0.0039gkg H 2 O kg D. A.
= 0.0083 kg H 2 O kg D. A.
b.
1250 kg/h
T=37 o C, h r=50%
ma (lb m H2 O/h)
T=15 o C, sat’d
mc (lb m H2O/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 =
Fig. 8.4-1
⇒
= 0.0198 kg H 2 O kg DA
= b88.5- 0.5g kJ kg DA = 88.0 kJ kg DA
Rha1
S $
T H1
1250 kg
1 kg DA
h
1.0198 kg
Rha = 0.0106 kg H 2 O
S
$
T H 2 = 42.1 kJ kg DA
Fig. 8.4-1
Outlet air: Tdb = 15° C, sat'd
⇒
Overall water balance ⇒ m
&c =
= 1226 kg DA h
1226 kg DA
b0.0198 −
h
kg DA
0.0106 g kg H 2 O
kg DA
= 11.3 kg H 2 O h withdrawn
Reference states for enthalpy calculations: H 2 Obl g , dry air at 0o C.
kJ
⇒ H2O ( l , 12°C ): Hˆ =
kg ⋅o C
Overall system energy balance:
(C p )H O2 ( l ) = 4.184
Q& c = ∆H& =
∑ m& H$ − ∑ m& H$
i
i
out
i
∫
12
0
C p dT = 50.3 kJ/kg
i
in
L 11.3 kg H 2 O 50.3 kJ
1226 kg DA b42.1 − 88g kJ OF 1 h I F 1 kW I
=M
+
JG
J
PG
h
kg H 2 O
h
kg DA QH3600 s KH1 kJ / sK
N
= −155
. kW
∆H =
8.81
8.82 a.
b.
400 mol NH 3
− 78.2 kJ
mol NH3
= −31,280 kJ
HClbg , 25° Cg, H 2 Obl , 25° Cg → HClb25° C, r = 5g .
Table B.11
∆H$ = ∆H$ s b25° C, r = 5g  → ∆H$ = −64.05 kJ mol HCl
HClbaq, r = ∞ g → HClbr = 5g, H 2 Obl g
∆H$ = ∆H$ s b25° C, n = 5g − ∆H$ s b25° C, n = ∞ g
= b−64.05 + 75.14g kJ mol HCl = 11.09 kJ / mol HCl
8.83 Basis: 100 mol solution ⇒ 20 mol NaOH, 80 mol H2 O
⇒r=
80 mol H2O
mol H2O
= 4.00
20 mol NaOH
mol NaOH
8-58
8.83 (cont’d)
Refs: NaOH(s), H2 Obl g@25° C
H$ in
substance
nin
NaOH bsg
20.0 0.0
H 2 Obl g
80.0 0.0
NaOH br = 4.00g −
−
n out
H$ out
−
−
n in mol
−
−
H$ in kJ mol
20.0 − 34.43 ← n in mol NaOH
H$ bNaOH, r = 4.00g = −34.43 kJ mol NaOH (Table B.11)
−688.6 kJ 9.486 × 10 −4 Btu
∆H = ∑ ni H$ i − ∑ ni H$ i = (20)(−34.43) =
= −6532
. Btu
10 −3 kJ
out
in
−6532
. Btu
103 g
Q=
= −132.3 Btu lb m product solution
20.0b40.00g + 80.0b18.01g g 2.20462 lb m
8.84 Basis: 1 liter solution
n H 2SO 4 =
mtotal =
nH 2O =
1 L 8 g - eq
L
1 mol
2 g - eq
1 L 1.230 kg
L
H 2 O 1000 mol H 2 O
= 46.5 mol H 2 O
18.02 kg H 2 O
n H2 O
n H 2SO4
=
0.392 kg H 2 SO 4
= 1230
.
kg solution
b1.230 − 0.392gkg
⇒r=
F 0.09808 kg I
J =
H 1 mol
K
= 4 mol H 2 SO 4 × G
46.49 mol H 2 O
mol H 2 O
= 11.6
4 mol H 2 SO 4
mol H 2 SO 4
H 2SO 4 daq, r = ∞ ,25 o Ci → H 2 SO 4 daq, r = 11.6, 25 o Ci + H 2 Odl , 25 o Ci
∆H$ 1 = ∆H$ s (r = 11.6) − ∆H$ s (r = ∞ )
H$ bH 2 SO 4 , r = 11.6, 60° Cg =
Table B.11
=
L
n H S O ∆H 1
M
N 2 4
( −67.6 + 96.19 ) = 28.6
kJ
mol H 2 SO 4
60
+ mz C pdT O
kJ
P
25
Q
n H2 SO ( mol H 2SO 4 )
4
=
R4 mol H 2 SO 4
1
S
4 mol H 2 SO 4 T
28.6 kJ
mol H 2 SO 4
= 60.9 kJ mol H 2 SO 4
8-59
+
1.230 kg
3.00 kJ
kg⋅° C
b60 − 25g° CU
V
W
8.85 2 mol H2SO 4 = 0.30d2.00 + nH 2 O i ⇒ n H2 O = 4.67 mol H2 O ⇒ r =
a.
For this closed constant pressure system,
2 mol H 2SO 4
Q = ∆H = n H2 SO4 ∆H$ s b25° C, r = 2.33g =
b. msolution =
2 mol H 2SO 4
98.08 g H 2SO 4
mol
+
4.67
mol H 2O
= 2.33
2
mol H 2SO 4
−44.28 kJ
mol H 2SO 4
= −88.6 kJ
4.67 mol H2 O 18.0 g H 2 O
mol
= 2802
. g
T
∆H = 0 ⇒ n H2 SO4 ∆H$ s b25° C, r = 2.33g + m 25 C pdT = 0
−88.6 kJ +
8.86 a.
Basis:
b280.6 + 150gg
3.3 J
bT
− 25g° C
g⋅° C
1 kJ
1000 J
= 0 ⇒ T = 87° C
1 L product solution 1.12 e103 gj
= 1120 g solution
L
1 L 8 mol HCl 36.47 g HCl
= 292 g HCl
L
mol HCl
46.0 mol H2O(l, 25°C)
8.0 mol 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 2O
18.0 g
46.0 mol H 2O
= 575
. mol H 2O mol HCl
8.0 mol HCl
Assume all HCl is absorbed
Volume of gas:
8 mol 293 K
273 K
760 mm Hg 22.4 L bSTPg
790 mm Hg
mol
= 185 liter bSTP g gas feed L HCl solution
b. Ref: 25°C
substance
nin
H$ in nout
H 2Obl g
46.0 00
.
−
HCl b gg
8.0 −015
.
−
HClbn = 5.75g −
−
8.0
H$ out
−
−
−59.07
n in mol
$
H in kJ mol
8-60
8.86 (cont’d)
1
H$ b HCl, n = 5.75g = ∆H$ s b25° C, n = 575
. g+
nHCl
= −64.87 kJ mol +
H$ e HCl, 20o Cj =
20
25
40
mC pdT
25
1120 g 0.66 cal
8 mols
g⋅° C
b40 − 25g° C
4184
.
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 = ∆ H = − 471 kJ L product
c.
Q = 0 = ∆H = 8e H$ j − 8b−015
. g
o
1120 g 0.66 cal bT − 25g C 4.184 J 1 kJ
−015
. = H$ = − 64.87 +
8 mol g⋅ o C
cal 1000 J
T = 192 o C
8.87 Basis: Given solution feed rate
.
n. a (mol air/min)
.
n1 (mol H 2O( v)/min)
saturated @ 50°C, 1 atm
na (mol air/min)
200°C, 1.1 bars
150 mol/min solution
0.001 NaOH
0.999 H 2O
25°C
.
n2 (mol/min) @ 50°C
0.05 NaOH
0.95 H 2O
NaOH balance: b0.001gb150g = 0.05n&2 ⇒ n&2 = 3.0 mol min
H 2 O balance: b0.999gb150g = n&1 + 0.95b3.0g ⇒ n&1 = 147 mol H2 O min
Raoult’s law: y H 2O P =
Table B.4
n&1
mol air
P = pH∗ 2 O b50° Cg = 92.51 mm Hg ⇒ n& a = 1061
&
n1 =147
n&1 + n& a
min
P = 760
1061 mol 22.4 LbSTP g 473 K 1.013 bars
V&inlet air =
= 37 ,900 L min
min
1 mol
273 K 1.1 bars
References for enthalpy calculations: H 2Obl g, NaOHbsg, air @ 25° C
999 mol H2 O Table B.11 $
⇒ ∆Hs b25° Cg = −42.47 kJ mol NaOH
1 mol NaOH
95 mol H 2O 19 mol H 2O
kJ
5% solution @ 50°C: r =
=
⇒ ∆H$ s b25° Cg = −42 .81
5 mol NaOH
mol NaOH
mol NaOH
0.1% solution @ 25°C: r =
Solution mass: m=
1 mol NaOH 40.0 g 19 mol H 2 O 18.0 g
g solution
+
= 382
1 mol
1 mol
mol NaOH
50
H$ b50° Cg = ∆H$ s b25° Cg + m 25 C pdT
= −42.81
382 g
4.184 J
kJ
+
mol NaOH mol NaOH 1 g⋅° C
8-61
b50 − 25g° C
1 kJ
= −2.85 kJ
10 3 J
8.87 (cont’d)
Air @ 200°C: Table B.8 ⇒ H$ = 5.15 kJ mol
Air (dry) @ 50°C: Table B.8 ⇒ H$ = 0.73 kJ mol
1 kg 18.0 g
b2592 − 104.8g kJ
H2 Obv , 50° Cg: Table B.5 ⇒ H$ =
= 44.81 kJ mol
kg
10 3 g 1 mol
substance
n&in
H$ in
n&out
H$ out
NaOHbaq g 015
.
−42.47 0.15 −2.85 n& in mol min
H2 Obv g
−
−
147 44 .81 H$ in kJ mol
Dry air
1061 5.15 1061 0.73
Energy balance: Q& = ∆H& = ∑ ni H$ i − ∑ ni H$ i = 1900 kJ min transferred to unit
bneglect
∆E n g
out
in
8.88 a. Basis: 1 L 4.00 molar H2 SO4 solution (S.G. = 1.231)
4.00 mol H 2 SO 4
1231 − 392.3 = 8387
. g H2 O
1 L 1231 g
= 1231 g ⇒
⇒
= 392.3 g H 2SO 4
L
= 46.57 mol H 2 O
B.11
⇒ r = 11.64 mol H 2 O / mol H 2SO 4 Table

→ ∆H$ s = −67.6 kJ / mol H 2 SO 4
Ref: H 2 Obl , 25° Cg , H 2SO 4 b25° Cg
substance
n in
H$ in
n out H$ out
H 2 Obl g
46.57 0.0754bT − 25g −
−
H 2 SO 4 bl g
4.00
0
−
−
H 2 SO 4 b25° C, n = 11.64 g
−
−
4.00 −67.6
n in mol
$
H in kJ mol
Q = ∆H = 0 = 4.00b−67.6g − 46.57 b0.0754 gbT − 25g ⇒ T = −52° C
(The water would not be liquid at this temperature ⇒ impossible alternative!)
b. Ref: H 2 Obl , 25° Cg , H 2SO 4 b25° Cg
substance
nin
H$ in
n out
H$ out
H 2 O bl g
nl
0.0754b0 − 25g
−
−
H 2O bsg
ns
−6.01 + 0.0754b0 − 25g
−
−
H 2SO 4 ( l )
4.00
0
−
−
H 2 SO 4 b25° C, n = 11.64 g
n in mols
$
H in kJ mol
4.00 −67.61
∆H$ m bH 2 O, 0° Cg = 6.01 kJ mol
A
Table B.1
n l + n s = 46.57
∆H = 0 = 4 .00b−67.61g − nl b−1885
. g − b46.57
⇒ 291.4 g H 2 Obl g + 547 .3 g H 2 Obsg@0° C
8-62
U
nl
V⇒
− n l gb−7.895gW n s
= 1618
. mol liquid H 2O
= 30.39 mol ice
8.89 P2O 5 + 3H 2O → 2H 3PO 4
mol H 3 PO 4
a.
wt% P2 O 5 =
nb14196
. g
mt
× 100% , wt% H 3 PO 4 =
B
}
2n
g H3 PO 4 mol
B
b98.00g
mc
× 100%
A
g total
where n = mol P2 O 5 and mt = total mass .
wt% H 3 PO 4 =
2b98.00g
wt% P 2 O 5 = 1381
.
wt% P2O 5
14196
.
b. Basis: 1 lb m feed solution 28 wt% P2O 5 ⇒ 38.67 wt% H 3PO 4
m1 (lb m 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
H3 PO 4 balance: 0.3867 = 0.5800m2 ⇒ m2 0.667 lb m solution
Total balance: 1 = m1 + m2 ⇒ m1 = 0.3333 lb m H 2Obr g
Evaporation ratio: 0.3333 lb m H 2 O bvg lb m feed solution
c. Condensate:
P = 3.7 psia b0.255 bar g
Table B.6
⇒ Tsat = 654
. o C= 149 o F, Vliq =
m& =
100 tons feed
day
46.3 lb m
V& =
min
000102
.
m3 353145
.
ft 3 / m 3
ft 3
= 0.0163
kg 2.205 lb m / kg
lb m H 2O(l)
2000 lb m 1 lb m H 2O
1 ton
1 day
(24 × 60) min
3 lb m
0.0163 ft 3
7.4805 gal
lb m
ft 3
= 46.3 lb m / min
= 5.65 gal condensate / min
Heat of condensation process:
46.3lbm H2O(v)/min
46.3 lbm H2O(l)/min
(149+37)°F, 3.7 psia
149°F, 3.7psia
.
Q (Btu/min)
8-63
8.89 (cont’d)
Table B.6 ⇒
R
| $
H H2 O (v ) (186o F = 85.6 o C) =
|
|
S
|
o
o
$
| H
H2 O (l ) (149 F = 65.4 C) =
|
T
Btu
F
(2652 kJ /
kg)G0.4303
G
H
kJ
lb m
kg
I
J
J
K
= 1141 Btu / lb m
(274 kJ / kg)b0.4303 g = 118 Btu / lb m
lb L
Btu
Q& = m
& ∆H$ = ( 46.3 m ) M(118 − 1141)
min N
lb m
O
P = −47, 360
Q
Btu / min
⇒ 4.74 × 10 4 Btu min available at 149 o F
d. Refs: H3 PO 4 bl g, H 2Oblg@77° F
substance
min
H$ in
mout
H$ out
H 3PO 4 b28% g 1.00 13.95
−
−
m in lb m
$
H 3PO 4 b42% g −
−
0.667 34.13 H in Btu lb m
H 2Obv g
−
−
0.3333 1099
H$ bH 3 PO 4 , 28% g =
+
0.705 Btu
lb m ⋅° F
b125 − 77g° F
H$ bH 3 PO 4 , 42% g =
+
0.705 Btu
lb m ⋅° F
−5040 Btu
lb - mole H 3PO 4
0.3867 lb m H 3 PO 4
1.00 lb m soln
= 13.95 Btu lb m soln
−5040 Btu
lb - mole H 3 PO 4
b186.7 −
1 lb - mole H 3 PO 3
98.00 lb m H 3 PO 4
77 g° F
1 lb - mole H 3PO 4
98.00 lb m H 3 PO 4
0.5800 lb m H 3 PO 4
1.00 lb m sol.
= 34.13 Btu lb m soln
H$ bH 2O g = H$ b3.7 psia, 186° Fg − H$ bl , 77° Fg = b2652 − 104.7 g kJ kg ⇒ 1096 Btu lb m
At 27.6 psia (=1.90 bar), Table B.6 ⇒ ∆H$ v = 2206 kJ / kg = 949 Btu / lb m
∆H = ∑ ni H$ i − ∑ ni H$ i = 375 Btu = m steam ∆H$ v ⇒ m steam =
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 3PO 4
24 h
day
3292 lb m steam
(46.3 × 60) lb m H 2 O evaporated / h
= 118
.
8-64
= 3292 lb m steam / h
lb m steam
lb m H2 O evaporated
8.90 Basis: 200 kg/h feed solution. A = NaC2 H 3O 2
.
n1 (kmol H 2O( v)/h)
50°C, 16.9% of H O
2
in feed
200 kg/h @ 60°C
.
n0 (kmol/h)
0.20 A
0.80 H O
2
Product slurry @ 50°C
.
n2 (kmol A-3H 2O( v)/h)
.
n3 (kmol solution/h)
0.154 A
0.896 H 2O
Q (kJ/hr)
a. Average molecular weight of feed solution: M = 0.200 M A + 0.800 M H 2 O
= b0.200gb82.0g + b0.800gb18.0g = 30.8 kg k
Molar flow rate of feed: n 0 =
200 kg
1 kmol
h
30.8 kg
= 6.49 kmol h
b. 16.9% evaporation ⇒ n1 = b0.169gb0.80 gb6.49 kmol h g = 0.877 kmol H 2 O bv g h
A balance: b0.20gb6.49 kmol hg =
n2 bkmol A ⋅ 3 H 2 O g
1 mole A
h
1 mole A ⋅ 3 H 2 O
E
+ 0154
. n3
⇒ n2 + 0.154 n3 = 1.30
H 2 O balance: b0.80gb6.49 kmol h g = 0.877 +
(1)
n 2 bkmol A ⋅ 3 H 2 O g
3 moles H 2 O
h
1 mole A ⋅ 3 H 2 O
+ 0.846n 3 ⇒ 3n 2 + 0.846n 3 = 4.315
b2g
Solve b1g and b2g simultaneously ⇒ n 2 = 113
. kmol A ⋅ 3H 2 O bsg h
n 3 = 1.095 kmol solution h
Mass flow rate of crystals
1.13 kmol A ⋅ 3H 2 O 136 kg A ⋅ 3H 2 O 154 kg NaC2 H 3O 2 ⋅ 3H 2 Obsg
=
h
1 kmol
h
Mass flow rate of product solution
200 kg feed 154 kg crystals b0.877 gb18.0 gkg H 2 O bv g
−
−
= 30 kg solution h
h
h
h
c.
References for enthalpy calculations: NaC2 H 3O 2 bsg, H 2 Obl g@25° C
60
Feed solution: nH$ = n A ∆H$ s b25° Cg + mz C p dT (form solution at 25° C , heat to 60° C )
25
4
b0.20g6.49 kmol A −1.71 × 10 kJ 200 kg 3.5 kJ
nH$ =
+
h
kmol A
hr
kg⋅° C
8-65
b60 − 25g° C
= 2300 kJ h
8.90 (cont’d)
50
Product solution: nH$ = n A ∆H$ s b25° Cg + mz C p dT
25
kmol A −1.71 × 10 4 kJ 30 kg 3.5 kJ
+
h
kg⋅° C
h
kmol A
= −259 kJ h
=
b0.154 g1.095
b50
− 25g° C
50
Crystals: nH$ = n A ∆H$ hydration + mz C p dT (hydrate at 25° C , heat to 50° C )
25
1.13 kmol A ⋅ 3H 2 Obsg −3.66 × 10 4 kJ 154 kg 1.2 kJ
+
h
kg⋅° C
h
kmol
= −36700 kJ h
=
b50 −
25g° C
50
H 2 Obv , 50° Cg: n∆H = n LM∆H$ v + z C p dT O
(vaporize at 25° C , heat to 50° C )
P
25
N
=
Q
4.39 × 10 4 + b32.4 gb50 − 25g kJ
0.877 kmol H 2 O
h
Energy balance: Q = ∆H =
bneglect ∆ E R g
∑ n H$ − ∑ n H$
i
i
out
i
i
= 39200 kJ h
= b−259 − 36700 + 39200g − b2300g kJ h
in
= −60 kJ h (Transfer heat from unit)
8.91 50 mL H2SO4
1834
.
g
mL
84.2 mL H2 Oblg
U
= 917
. g H2SO4 ⇒ 0.935 mol H2SO4 |
|
V ⇒ r = 5.00
100
. g
= 84.2 g H2 Oblg ⇒ 4.678 mol H2Oblg|
|
mL
W
mol H 2O mol H 2SO 4
Ref: H 2 O , H 2SO 4 @ 25 °C
H$ ( H 2 O(l ), 15 o C) = [0.0754 kJ / (mol ⋅o C)](15 − 25) o C = − 0.754 kJ / mol
(91.7 + 84.2) g
2.43 J
kJ
H$ bH 2 SO 4 , r = 5.00 g = −58.03
+
mol 0.935 mol H 2 SO 4 g⋅° C
= ( −69.46 + 0.457 T )( kJ / mol H 2SO 4 )
substance
nin
H$ in
bT
− 25g° C
1 kJ
10 3 J
H$ out
nout
4.678 –0.754 —
—
0.935
0.0
—
—
—
0.935 b−69.46 + 0.457 T g
H2SO 4 br = 4 .00g —
H2 Obl g
H 2SO 4
n in mol
H$ in kJ/mol
n bmol H 3SO 4 g
Energy Balance: ∆H = 0 = 0.935b−69.46 + 0.457T g − 4.678b−0.754 g ⇒ T = 144 ° C
Conditions: Adiabatic, negligible heat absorbed by the solution container.
8-66
8.92 a.
mA (g A) @ TA0 (o C)
n A (mol A)
n S (mol solution) @ Tmax (o C)
mB (g B) @ TB0 (o C)
n B (mol B)
Refs: A(l), B(l) @ 25 °C
substance n in H$ in n out
H$ out
—
—
n in mol
B
n A H$ A
n B H$
B
—
—
H$ in J / mol
S
— —
nA
H$ S (J mol A )
A
Moles of feed materials: n A (mol A) =
mA (g A)
m
, nB = B
M A (g A / mol A)
MB
Enthalpies of feeds and product
H$ A = m A C pA ( TA 0 − 25o C), H$ B = m B C pB ( TB 0 − 25o C)
r (mol B mol A) = n B n A =
H$ S
J
F
G
H mol
A
I
J
K
=
1
n A (mol A)
mB / M B
mA / M A
L
J I
F
$
J
Mn A ( mol A) × ∆ H m ( r ) G
H mol A K
M
M
F
J
M+ ( m A + m B )( g soln) × C ps G
o
H g soln ⋅
M
N
I
J
CK
× (Tmax −
O
P
P
P
25)( o C) P
P
Q
1
⇒ H$ S =
n A ∆ H$ m (r ) + ( m A + m B )C ps (Tmax − 25)
nA
Energy balance
∆H = n A H$ S − n A H$ A − n B H$ B = 0
mA $
∆Hm brg + (mA + mB )C ps (Tmax − 25) − mA CpA bTA0 − 25g − mB CpB bTB0 − 25g = 0
MA
m
m AC pA bTA0 − 25g + mB C pB bTB0 − 25g − A ∆H$ m br g
MA
⇒ Tmax = 25 +
(mA + mB )Cps
⇒
Conditions for validity: Adiabatic mixing; negligible heat absorbed by the solution container,
negligible dependence of heat capacities on temperature between 25o C and TA0 for A, 25o C
and TB0 for B, and 25o C and Tmax for the solution.
b.
m A = 100.0 g
M A = 40.00 TA0 = 25° C
m B = 2250
. g
M B = 18.01 TB 0 = 40° C C pB = 4.18 J
C ps = 3.35 J (g⋅° C)
C pA = ?birrelevantg U
V⇒r
(g⋅° C) |W
= 5.00
∆H$ m bn = 5.00g = −37,740 J mol A ⇒ Tmax = 125° C
8-67
mol H 2 O
mol NaOH
8.93 Refs: Sulfuric acid and water @ 25 °C
b.
substance
nin
H$ in
nout
H2 SO4
H2 O
H 2SO 4 baq g
1
r
—
M A CpA bT0 − 25g
M w C pw bT0 − 25g
—
—
—
1
H$ out
—
n in mol
—
H$ in J/mol
∆H$ m brg + b M A + rM w gC ps bTs − 25g
(J/mol H2 SO4 )
∆H = 0 = ∆H$ m br g + b M A + rM w gC ps bTs − 25g − M A C pa bT0 − 25g − rM w C pw bT0 − 25g
= ∆H$ m br g + b98 + 18rgC ps bTs − 25g − (98C pa + 18 rC pw )bT0 − 25g
⇒ Ts = 25 +
1
(98C pa + 18rC pw )bT0 − 25g − ∆H$ m br g
(98 + 18r) C ps
c.
H2 O(l)
H2 SO4
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
∆H$ 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 A0 = P0V g / RT0
Total moles of B: n B0 (mol B) =
(1)
Vl ( L) × bSG B × 1 kg / Lgd10 3 g / kgi
(2)
M B (g / mol B)
Total moles of A: n Ao = n Av + n Al
(3)
n Al
n RT
A(l)I
= bc 0 + c1T g Av
J = ks pA ⇒
mol B K
nB 0
Vg
F mol
Henry’s Law: r G
H
(4)
Solve (3) and (4) for nAl and n Av.
n Al =
n Av =
n B0 RT
bc0 + c1 T g
Vg
L
M1 +
M
N
(5)
O
n B0 RT
bc0 + c1T gP
Vg
P
Q
n Ao
L
M1 +
M
N
(6)
O
n B0 RT
bc0 + c1T gP
Vg
P
Q
Ideal gas equation of state
P=
n Av RT ( 6)
n A 0 RT
=
Vg
Vg + nB 0 RT bc0 + c1T g
(7)
Refs: Abg g, Bbl g @ 298 K
U$ in
U$ eq
substance
n in
Abg g
n Ao
M A CvA bT0 − 298g n Av
Bbl g
n B0
M B Cv B bT0 − 298g
—
—
Solution
—
—
n Al
U$ 1 (kJ/mol A)
n eq
M A CvA bT − 298g
n in mol
$
U in kJ/mol
1
U$ 1 = ∆U$ s +
bn Al M A + n B0 M B gCvs bT − 298g
n Al
E.B.: ∆U = 0 =
∑ n U$ − ∑ n U$
i
out
i
i
i
in
0 = cn Av CvA + bn Al M A + n B M B gCvs hbT − 298g + n Al ∆U$ s − bn AoCvA + nB CvB gbT0 − 298g
⇒ T = 298 +
n Al d− ∆U$ s i + bn Ao CvA + n B CvB gbT0 − 298g
n Av CvA + bn Al M A + n B M B gCvs
8-69
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
350 mL 85% H2 SO 4
ma(g), 60 oF, ρ=1.78
Q=0
30% H2 SO 4
ms(g), T( oF)
H2O, Vw(mL),
mw(g), 60 oF
a.
Vw =
350 mL feed
178
.
g
0.85(70 / 30) − 0.15 g H 2O added 1 mL water
1 mL feed
g feed
1 g water
= 1140 mL H 2O
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:
m Hˆ + mw Hˆ w
∆H = 0 = mp 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
c. 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 2SO 4
$ = −10 Btu / lb
@ 77 o F ⇒ H
1
m2 (lb m ) 80.0 wt% H 2SO 4
$ = −120 Btu
@ 60 o F ⇒ H
2
Total mass balance:
U
|
m
|
|
adiabatic mixing
→
V  
|
|
/ lb m |W
2.30 + m2 = m3
H 2SO 4 mass balance: 2.30b0.150g + m2 b0.800g =
$
m3 ( lb m ) 60.0 wt% H 2SO 4 @ T o F, H
3
U
|
⇒
m3 (0.600 )VW
|
m
R
| 2
Sm
|T 3
= 5.17 lb m (80%)
= 7.47 lb m (60%)
b. Adiabatic mixing ⇒ Q = ∆H = 0
$
b7.47 gH
3
− b2.30 gb−10g − b517
. gb−120g = 0 ⇒ H$ 3 = −861
. Btu / lb m
E
Figure 8.5 - 1
T = 140o F
c.
H$ d60 wt%, 77 o Fi = −130 Btu / lb m
Q = m3 H$ d60 wt%, 77 o Fi − H$ 3 = b7 .475gb−130 + 86.1g = −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
xNH 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 H2 O lb m
0.90 lb m liquid
−25 Btu
0.10 lb m vapor
670 Btu
Enthalpy: H$ =
+
= 44 Btu lb m
1 lb m
1 lb m liquid
1 lb m
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), m L (g liquid)
.14
.60
.80 xNH3
Mass Balance: mv + m L = 250
NH3 Balance: 0.80m g + 014
. mL = (0.60)( 250) ⇒ mv = 175 g, mL = 75 g
Vapor: m NH3 = b0.80 gb175 gg = 140 g NH 3 , 35 g H 2 O
Liquid: mNH 3 = b0.14 gb75 g g = 10.5 g NH 3 , 64.5 g H 2O Liquid
8.99 Basis: 200 lb m feed h
m& v (lb m h)
xv(lb m NH3 (g)/lb m)
H$ v ( Btu lb m )
200 lb m/h
0.70 lb m NH3 (aq)/lb m
0.30 lb m H2 O(l)/lb m
m& l (lb m h)
H$ f = −50 Btu lb m
xl [lb m NH3 (aq)/lb m]
in equilibrium
at 80o F
H$ l ( Btu l b 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 NH3 in liquid: x l = 0.30 lb m NH 3 lb m
Specific enthalpies: H$ v = 650 Btu lb m , H$ l = −30 Btu lb m
& v = 120 lb m h vapor
m
&l
Mass balance:
200 = m& v + m
U
⇒
& l = 80 lb m h liquid
& v + 0.30m& l VW m
Ammonia balance: b0.70gb200g = 0.96 m
Energy balance: Neglect ∆E& k .
Q& = ∆H& =
∑ m& H$
i
out
i
120 lb m
& f H$ f =
−m
h
= 86,000
650 Btu 80 lb m
+
lb m
h
Btu
h
8-73
−30 Btu 200 lb m
−
lb m
h
−50 Btu
lb m
CHAPTER NINE
4 NH 3 ( g) + 5O2 ( g) → 4NO(g) + 6H 2O(g)
∆H$ 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
2 NH 3 ( g) + O 2 ( g) → 2NO(g) + 3H 2O(g)
2
Reducing the stoichiometric coefficients of a reaction by half reduces the heat of reaction by half.
904.7
∆H$ ro = −
= −452.4 kJ / mol
2
3
5
NO(g) + H 2O(g) → NH 3 ( g) + O 2 ( g)
2
4
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)
∆H$ ro = −
= +2262
. kJ / mol
4
d.
e.
& NH3 = 340 g / s
m
340 g 1 mol
= 20.0 mol / s
s
17.03 g
$o
n&
∆H
r
20.0 mol NH 3 −904.7 kJ
& = ∆H
& = NH3
Q
=
s
4 mol NH 3
ν NH
n& NH 3 =
= −4.52 × 10 4 kJ / s
3
The reactor pressure is low enough to have a negligible effect on enthalpy.
f.
9.2
Yes. Pure water can only exist as vapor at 1 atm above 100°C, but in a mixture of gases, it can
exist as vapor at lower temperatures.
C 9 H20 ( l) + 14O2 ( g) → 9CO 2 (g) + 10H2 O(l)
∆H$ o = −6124 kJ / mol
r
a.
b.
c.
When 1 g-mole of C9 H20 (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.
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.
& =
Q& = ∆ H
$0
n& C9 H 20 ∆ H
r
ν C 9 H 20
=
25.0 mol C9 H 20
s
−6124 kJ
1 mol C 9 H20
9- 1
1 kW
= −153
. × 105 kW
1 kJ / s
9.2 (cont'd)
Heat Output = 1.53×105 kW.
The reactor pressure is low enough to have a negligible effect on enthalpy.
d.
C 9 H 20 ( g) + 14O2 ( g) → 9CO 2 (g) +10H 2O(l)
∆H$ o = −6171 kJ / mol
(1)
C 9 H 20 ( l) + 14O2 (g) → 9CO 2 (g) +10H 2O(l)
∆H$ o = −6124 kJ / mol
(2)
r
r
(2) − (1) ⇒ C9 H 20 ( l) → C 9 H 20 (g)
∆H$ o (C H , 25o 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.
19
O 2 bgg → 6CO2 bgg + 7 H2 Obgg b1g ∆H$ ro = ?
2
19
C 6H 14 blg + O 2 bgg → 6CO2 bgg + 7 H 2Oblg b2g ∆H$ 2 = ∆H$ or = −1791
.
× 106 Btu lb- mole
2
C 6H 14 bgg → C6 H14 blg b3g ∆H$ 3 = − e∆H$ v j
= −13,550 Btu lb - mole
C 6H 14 bgg +
C 2 H14
H2 Oblg → H 2Obgg
b4g
b1g = b2g + b3g + 7 × b4g
c.
m& = 120 lb m / s
∆H$ 4 = e∆H$ v j
Hess's law
⇒
MO2 =32.0
⇒
H2 O
= 18,934 Btu lb - mole
∆H$ 1 = ∆ H$ 2 + ∆ H$ 3 + 7 ∆H$ 4 = −1672
.
× 106 Btu lb - mole
n& = 375
. lb - mole / s.
n&O ∆Hˆ ro 3.75 lb-mole/s −1.672 ×106 Btu
Q& = ∆H& = 2
=
= −6.60 × 105 Btu/s (from reactor )
9.5
1 lb-mole O 2
v O2
CaC 2 bsg + 5H 2Oblg → CaObsg + 2CO 2 bgg + 5H 2 bgg , ∆H$ 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.
∆U$ ro
=
∆H$ ro
L
M
− RT M
νi
Mgaseous
M
Nproducts
∑
O
P
−
ν i P = 69.36
gaseous P
reactants P
Q
∑
= 52.0 kJ mol
9- 2
kJ 8.314 J 1 kJ 298 K
−
mol mol ⋅ K 103 J
b7 − 0 g
9.4 (cont’d)
∆U$ ro is the change in internal energy when 1 g- mole of CaC2 (s) and 5 g - moles of H 2O(l) at 25o C and
1 atm react to form 1 g - mole of CaO(s), 2 g- moles of CO 2 (g) and 5 g - moles of H 2 (g) at 25o C
and 1 atm.
c.
Q = ∆U =
nCaC 2 ∆ U$ ro
vCaC 2
=
150 g CaC2
1 mol
52.0 kJ
= 121.7 kJ
64.10 g 1 mol CaC2
Heat must be transferred to the reactor.
9.5
a.
Given reaction = (1) – (2)
Hess's law
⇒
∆ H$ ro = ∆H$ ro1 − ∆H$ ro2 = b1226 − 18, 935g Btu lb- mole
= −17 ,709 Btu lb - mole
b.
Given reaction = (1) – (2)
Hess's law
⇒
∆H$ ro = ∆H$ ro1 − ∆ H$ ro2 = b−121740
,
+ 104,040g Btu lb - mole
= −17, 700 Btu lb - mole
Hess's law
9.6
9.7
a.


Reaction (3) = 0.5× (1) − ( 2 ) ⇒ ∆Hˆ r = 0.5  −326.2
o
kJ  
kJ 
kJ
−  −285.8
= 122.7


mol  
mol 
mol
b.
Reactions (1) and (2) are easy to carry out experimentally, but it would be very hard to decompose
methanol with only reaction (3) occurring.
a.
N2 bgg + O2 bgg → 2NO bgg , ∆H$ ro = 2e∆H$ of j
b.
n − C5 H12 bgg +
∆H$ ro = 5e∆H$ fo j
Table B.1
F
NO(g)
B
I
= 2G90.37 kJ mol J = 180.74 kJ mol
G
H
J
K
11
O 2 bgg → 5CObgg + 6H 2Oblg
2
+ 6 ∆ H$ o
− ∆H$ o
CO(g)
e
f j
H 2 Obl g
e
f j
n − C5 H12 bg g
= b5gb−11052
. g + b6gb−285.84g − b−1464
. g kJ mol = −21212
. kJ mol
c.
C 6H 14 blg +
(
19
O 2 bgg → 6CO2 bgg + 7H 2Obgg
2
∆Hˆ ro = 6 ∆Hˆ fo
)
CO 2
(
+ 7 ∆ Hˆ fo
)
H2 O( g)
(
− ∆ Hˆ fo
)
C6 H 14 (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)
∆H$ o = ∆H$ o
+ 4 ∆H$ o
− ∆H$ o
r
e
f j
Na 2S( l )
e
f j
CO2 (g )
e
f j
Na 2S O4 (l )
− 4e∆H$ fo j
CO(g)
= ( − 3732
. + 6.7) + b4gb−393.5g − b− 13845
. + 24.3g − 4( − 11052
.
kJ mol = − 138.2 kJ mol
9- 3
9.8
∆H$ ro1 = e∆H$ fo j
a.
C 2 H 2 Cl 4 (l )
∆H$ ro2 = e∆H$ of j
C 2 HCl3 bl g
− e∆H$ fo j
+ e∆H$ fo j
C 2 H4 ( g)
HCl bg g
⇒ e∆H$ fo j
− e∆H$ fo j
C2 H 2 Cl 4 (l )
C2 H 2 Cl 4 (l )
= −385.76 + 52.28 = −333.48 kJ mol
= − 276.2 − 92.31 + 33348
. = −3503
. kJ mol
Given reaction = b1g + b2g ⇒ −385.76 − 35.03 = −420.79 kJ mol
b.
c.
9.9
a.
b.
300 mol C2 HCl 3 − 420.79 kJ
Q& = ∆ H& =
= −126
. × 105 kJ h b= − 35 kWg
h
mol
Heat is evolved.
5
C 2 H2 ( g) + O 2 ( g) → 2CO 2 ( g) + H 2O(l)
∆H$ co = −1299.6 kJ mol
2
The enthalpy change when 1 g-mole of C2 H2 (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 H2 O(l) at 25°C and 1 atm is -1299.6 kJ.
∆H$ co = 2e∆H$ of j
Table B.1
B
=
c.
CO2 ( g )
H 2 O bl g
− e∆H$ fo j
2b− 3935
. g + b−28584
. g − b226.75g
( i) ∆H$ ro = d∆H$ fo i
Table B.1
B
=
C2 H6 ( g )
Table B.1
B
=
− d∆H$ fo i
b−84.67g − b226.75g
( ii) ∆H$ ro = d∆H$ co i
d.
+ e∆H$ fo j
C 2 H2 ( g )
C 2 H 2 bg g
kJ
kJ
= −1299.6
mol
mol
C2 H2 bg g
kJ
kJ
= −3114
.
mol
mol
+ 2 d∆H$ co i
H2 (g )
− d∆H$ co i
C 2 H 6 bg g
b−1299.6g + 2 ( −285.84 ) − b−1559.9 g
5
C 2 H2 ( g) + O 2 ( g) → 2CO2 ( g) + H 2 O( l)
2
1
H2 ( g) + O 2 ( g) → H 2 O( l)
2
7
C 2 H6 (g) + O 2 ( g) → 2CO 2 ( g) + 3H 2 O( l)
2
kJ
kJ
= −3114
.
mol
mol
(1) ∆H$ co1 = − 1299.6 kJ mol
(2) ∆H$ c 2 = − 28584
. kJ mol
o
(3) ∆H$ co3 = − 1559.9 kJ mol
The acetylene dehydrogenation reaction is (1) + 2 × (2) − (3)
Hess's law
⇒
∆ H$ ro = ∆H$ co1 + 2 × ∆H$ co2 − ∆H$ co3
= b−1299.6 + 2( − 28584
. ) − ( − 1559.9)g kJ mol = −3114
. kJ / mol
9- 4
9.10
25
O2 (g) → 8CO 2 bgg + 9H 2 Obgg
2
∆H$ ro = −4850 kJ / mol
a.
C 8H18 blg +
b.
When 1 g-mole of C8 H18 (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 H2 O(g), the change in enthalpy equals -4850 kJ.
Energy balance on reaction system (not including heated water):
∆E k , ∆E p , W = 0 ⇒ Q = ∆U = nbmol C8 H 18 consumed g∆U$ co bkJ molg
( Cp ) H 2 O(l) from Table B.2 = 75.4 × 10 −3 kJ / mol.o C
−Q = mH2 O ( Cp ) H 2 O(l) ∆T =
Q = ∆ U ⇒ − 89.4 kJ =
1.00 kg
1 mol
75.4 × 10−3 kJ 21.34° C
= 89.4 kJ
18.0 × 10−3 kg
mol.o C
2.01 g C 8H18 consumed 1 mol C8H 18
114.2 g
∆U$ co (kJ)
1 mol C8 H18
⇒ ∆U$ co = −5079 kJ mol
L
M
O
P
νi −
νiP
Mgaseous
gaseous P
M
reactants P
Nproducts
Q
∆H$ co = ∆U$ co + RT M
∑
=− 5079 kJ mol +
∑
8.314 J
1 kJ
298 K
b8 + 9 − 12.5g
mol ⋅ K 10 J
3
⇒ ∆Hˆ co = −5068 kJ mol
c.
% difference =
∆H$ co = 8e∆H$ of j
(
⇒ ∆Hˆ
(−5068) −( −4850)
×100 = − 4.3 %
−5068
CO2 bg g
o
f
)
+ 9e∆H$ off j
C 8H18 ( l )
H 2 Obg g
− e∆H$ fo j
C8 H18 bl g
= 8 ( −393.5) + 9 ( −241.83 ) + 5068 kJ/mol = −256.5 kJ/mol
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.
n − C 4H10 bgg → i − C 4H10 bgg
Basis : 1 mol feed gas
0.930 mol n -C4 H10
(n n- C4H10 )out
0.050 mol i-C4 H10
( n i-C4H10)out
0.020 mol HCl
0.020 mol HCl
149°C
Q(kJ/mol)
149°C
(n n-CH4 H10 ) out = 0.930(1 − 0.400) = 0.560 mol
(n i-C H4 H10 ) out = 0.050 + 0. 930 × 0.400 = 0.420 mol
ξ =
(n n-C4 H10 ) out − ( n n-C4 H10 ) in
ν n-C
=
0.560 − 0.930
= 0.370 mol
1
4 H 10
⇒
∆H$ ro = e∆H$ fo j
c.
References: n − C 4H10 bgg, i − C 4 H10 bgg at 25° C
i − C 4 H10
substance
− e∆H$ fo j
Table B.1
b.
n − C4 H10
H$ in
n in
(mol)
H$ out
n out
n − C 4 H 10
1
mol g (mol)
H$ 1
0.600
i − C 4 H 10
−
−
L 149
H$ 1 = M
z
M 25
N
∆ H$ ro = − 134.5 − b− 124.7g kJ mol = −9.8 kJ mol
bkJ
Table B.2
B
O
kJ
C p dT P
P mol
Q
bkJ
0.400
molg
H$ 1
H$ 2
H$ 2 =
= 14.29 kJ mol
L 149
M
M 25
N
z
Table B.2
B
O
C p dT P
kJ
P mol
Q
= 1414
. kJ mol
ˆ ro + ∑ ni H
ˆ i − ∑ ni Hˆ i ] = 0.370 −9.8 + (1)(14.142 ) − (1)(14.287 ) kJ
Q = ∆H = ξ [ ∆ H


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
∆Hˆ 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
n 0 (mol)
0.111 mol SiH4 /mol
0.8889 mol O2 /mol
n 1 (mol O2 )
n 2 (mo l SiO2 )
n 3 (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
298 K
760 torr
1 mol
22.4 × 10-3 m 3
= 0.1614 mol
ni = nio + ν iξ
SiH4 : 0=0.1111(0.1614 mol) − ξ ⇒ ξ = 0.0179 mol
O2 : n1 =0.8889(0.1614 mol) − ξ = 0.1256 mol O 2
SiO2 : n2 = ξ = 0. 0179 mol SiO2
H 2 : n3 =2ξ =0.0358 mol H 2
b.
∆H$ ro = ( ∆H$ fo ) SiO 2 ( s) − ( ∆H$ fo ) SiH 4 (g)
= [− 851 − ( −619
. )] kJ mol = −789.1 kJ / mol
References : SiH 4 ( g),O2 (g), SiO2 (g), H 2 ( g) at 298 K
Substance
SiH 4
O2
nin
nout
Hˆ in
Hˆ out
(mol h) (kJ mol) (mol h) (kJ mol)
0.0179
0
−
−
0.1435
0
0.1256
Hˆ 1
SiO2
−
−
0.0179
H2
−
−
0.0358
Hˆ 2
Hˆ
3
Table B.8
B
O 2 (g,1375K): H$ 1 = H$ O2 (1102 o C) = 3614
. kJ / mol
1375
SiO 2 (s,1375K): H$ 2 =
z( C p ) SiO ( s) dT = 79.18 kJ / mol
2
298
Table B.8
B
H 2 (g,1375K): H$ 3 = H$ H 2 (1102o C) = 32.35 kJ / mol
c.
Q = ∆H = ξ ∆Hˆ ro + ∑ ni Hˆ i − ∑ ni Hˆ i = −7.01 kJ/m 3 feed
out
−7.01 kJ 27.5 m 3
&
Q=
m3
h
in
1h
1 kW
= −0.0536 kW (transferred from reactor)
3600 s 1 kJ/s
9- 7
9.13
a.
Fe 2O3 bsg + 3Cbsg → 2 Febsg + 3CObgg , ∆H$ r ( 77 o F) = 2111
. × 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 Fe2 O3 fed
53.72 lb - moles C fed
17.9 lb -moles Fe2 O3 (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.
References: Fe 2O3 bsg, Cbsg, Febsg, CObgg at 77° F
Substance
Fe2 O3 bs,77° F g
Cbs,77° Fg
Febl,2800° Fg
CObg,570° Fg
nin
nout
H$ in
H$ out
(lb - moles) (Btu lb - mole) (lb - moles) (Btu lb - mole)
17.91
0
−
−
5372
.
0
−
−
−
−
3581
.
H$ 1
−
−
5372
.
H$ 2
Fe(l,2800o F): H$ 1 =
2794
z77
dC p i
dT + ∆H$ m b2794° Fg +
Febs g
CO(g,570 o F): H$ 2 = H$ C O (570o F)
=
2800
z2794 dC p i Fe l dT = 28400
Btu lb - mole
b g
3486 Btu lb - mole
A
F interpolating I
H from Table B.9K
c.
Q = ∆H =
nFe ∆H$ ro
+
ν Fe
∑ n H$ − ∑ n H$
i
out
i
i
i
in
. ge2111
. × 10 j
b3581
5
=
d.
+ b3581
. gb28400g + b53.72gb3486g − 0 = 4.98 × 106 Btu / ton Fe produced
2
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.
C 7 H16bgg → C 6H 5CH3 ( g) + 4 H 2 ( g)
Basis : 1 mol C7 H16
1 mol C7 H16
1 mol C6 H5 CH3
400°C
4 mol H2
400° C
Q (kJ/mol)
References: Cbsg, H 2 bgg at 25° C
b.
substance nin
bmolg
H$ in
nout
H$ out
bkJ
bmolg
bkJ
C7 H 16
1
mol g
$
H1
C7 H 8
−
−
1
H2
−
−
4
−
mol g
−
H$ 2
H$
3
 400 ↓

C7 H16 ( g,400°C ): Hˆ 1 = (∆Hˆ fo ) C7 H16 (g) + 
C p dT 
 25



= ( − 187.8 +91.0) kJ/mol= −96.8 kJ/mol
0.2427
∫
L 400
C 6 H 5CH 3 bg,400° Cg: H$ 2 = ( ∆H$ fo ) C6 H 5CH 3 ( g) + Mz
M 25
N
Table B.2
B
O
C p dT P
P
Q
= (+50 + 60.2) kJ / mol = 110.2 kJ / mol
Table B.8
B
H2 bg,400 ° Cg: H$ 3 = H$ H 2 ( 400 o C) = 1089
. kJ mol
c.
Q = ∆H =
∑ n Hˆ − ∑ n Hˆ
i
out
i
i
i
in
= [(1)(110.2) + (4)(10.89) - (1)(-96.8)] kJ = 251 kJ (transferred to reactor)
d.
∆Hˆ r (400o C)=
251 kJ
= 251 kJ/mol
1 mol C7 H16 react
9- 9
9.15
a.
bCH 3 g2 Obg g → CH 4 bgg + H 2 bg g + CObg g
Moles charged: (Assume ideal gas)
2.00 liters 273 K 350 mm Hg
1 mol
873 K 760 mm Hg 22.4 litersbSTPg
= 0.01286 mol bCH 3 g2 O
Let x = fraction bCH 3 g2 O decomposed (Clearly x<1 since Pf < 3P0 )
0.01286 mol
(CH 3)2 O
600°C, 350 mm Hg
0.01286(1 – x ) mol (CH3 )2O
0.01286 x mol CH4
0.01286 x mol H2
0.01286 x mol CO
600°C
875 mm Hg
Total moles in tank at t = 2 h = 0.01286 b1 − x g + 3 x = 0.01286b1 + 2 x gmol
Pf V
P0V
b.
=
n f RT
n0 RT
⇒
nf
n0
=
Pf
P0
⇒
0.01286b1 + 2 xg
0.01286
=
875
⇒ x = 0.75 ⇒ 75% decomposed
350
References: C ( s ), H 2 ( g ) , O 2 ( g ) at 25 C
o
substance
bCH 3 g
2
Obgg
CH 4 bgg
H 2 bgg
CObgg
nin
nout
H$ in
H$ out
(mol) ( kJ / mol)
( mol)
(kJ / mol)
$
0.01286
H1
0.25 × 0.01286
H$ 1
−
−
0.75 × 0.01286
H$ 2
−
−
0.75 × 0.01286
H$ 3
−
−
0.75 × 0.01286
H$
4
o
$
$o
bCH 3 g O(g,600 C): H1 = ( ∆H f ) bCH
2
L
3 g2 O
+M
given
873 B
z
M 298
N
O
C p dT P
J
P mol
Q
×
1 kJ
= ( − 18016
. + 62.40 ) kJ / mol
103 J
= − 118 kJ mol
 600 Table↓ B.2 
o
ˆ
ˆ
CH4 (g,600C):H 2 = ( ∆H f )CH 4 +  ∫ C p dT  = −74.85 + 29.46 = −45.39 kJ mol
 25

o
Table B.8
B
H 2 ( g,600 C): H$ 3 = H$ H2 ( 600o C) = 16.81 kJ mol
o
Table B.1
Table B.8
Table B.8
B
B
CO(g,600o C): H$ 4 = ( ∆H$ fo ) C O + H$ CO (600o C) = − 110.52 + 17.57 kJ mol = −92.95 kJ mol
c.
For the reaction of parts (a) and (b), the enthalpy change and extent of reaction are :
∆H = ∑ nout Hˆ out − ∑ nin Hˆ in = [ −1.5515 −( −1.5175)] kJ = −0.0340 kJ
ξ=
(n CH 4 )out − (n CH 4 )in
ν CH 4
=
0.75 × 0.01286
mol = 0.009645 mol
1
∆H = ξ ∆Hˆ r ( 600°C) ⇒ ∆Hˆ r ( 600°C ) =
−0.0340 kJ
= −3.53 kJ/mol
0.009645
9- 10
9.15 (cont’d)
∑ν
∆U$ r b600° Cg = ∆H$ r b600° Cg − RT [
i
= −3.53 kJ mol −
9.16
∑ν
−
gaseous
products
i
]
gaseous
reactants
(1 + 1 + 1 − 1)
8.314 J 1 kJ 873 K
mol ⋅ K 10 3 J
= −18.0 kJ mol
d.
Q = ξ ∆Uˆ r ( 600 °C ) = (0.009645 mol)( − 18.0 kJ/mol) = −0.174 kJ (transferred from reactor)
a.
1
SO 2 ( g) + O 2 ( g) → SO3 (g)
2
Basis :
10 3 mol SO3
80.07 kg SO3
l00 kg SO 3
min
= 1249 mol SO 3 min
n&0 (mol SO2 /min)
1249 mol SO3 /min
n& 2 (mol SO2 /min)
n&3 (mol O2 /min)
3.76 n&1 (mol N 2 /min)
o
450C
100% excess air
n&1 (mol O 2 / min)
3.76n&1 (mol N 2 /min)
5 5 0o C
o
450C
m& w (kg H 2 O(l) /min)
m& w (kg H 2 O(l) /min)
o
40 C
o
25C
Assume low enough pressure for H$ to be independent of P.
SO 3 balance :
bGeneration =outputg
n& 0 (mol SO 2 fed) 0.65 mol SO 2 react 1 mol SO3 produced
mol SO3
= 1249
min
1 mol SO2 fed
1 mol SO 2 react
min
⇒ n&0 = 1922 mol SO 2 / min fed
100% excess air: n&1 =
1922 mol SO 2
0.5 mol O2 reqd
min
1 mol SO 2
b1 + 1g mol
O 2 fed
1 mol O 2 reqd
= 1922 mol O 2 min fed
N2 balance : 376
. b1922g = 7227 mol / min bin & outg
65% conversion : n&2 = 1922b1 − 0.65g mol s = 673 mol SO2 min out
O balance:
b2gb1922 g + b2gb1922g = b3gb1249 g + b2gb673g + 2n& 3
.
b.
.
Extent of reaction : ξ =
⇒ n&3 = 1298 mol / min out
.
(n SO2 ) out − ( nS O2 ) in
ν SO2
Table B.1
B
=
673 − 1922
= 1249 mol / min
1
∆H$ ro = ( ∆H$ fo )S O3 ( g) − ( ∆ H$ fo ) SO2 ( g) = − 395.18 − ( −296.9) = −99.28 kJ / mol
9- 11
9.16 (cont’d)
References : SO 2 bgg, O 2 bg g, N 2 bg g, SO 3 bgg at 25 o C
Substance
SO 2
O2
N2
SO 3
n& in
n& out
H$ in
H$ out
( mol / min) ( kJ / mol) ( mol / min) ( kJ / mol)
1922
H$ 1
673
H$ 4
1922
H$ 2
1298
H$ 5
7227
H$ 3
7227
H$ 6
−
−
1249
H$ 7
450
SO 2 (g,450 C) : H$ 1 = z
o
25
Table B.2
B
C p dT = 19.62 kJ / mol
Table B.8
B
O 2 ( g,450 C) = H$ 2 = H$ O2 (450 C) = 1336
. kJ / mol
o
o
Table B.8
B
N 2 ( g,450 o C) = H$ 3 = H$ N2 ( 450o C) = 12.69 kJ / mol
Out :
Table B.2
∫
SO2 (g,550 C) : Hˆ 4 =
o
↓
550
C p dT = 24.79 kJ/mol
25
Table B.8
B
O 2 ( g,550 C) = H$ 5 = H$ O2 (550 C) = 1671
. kJ / mol
o
o
Table B.8
B
N2 ( g,550 C) = H$ 6 = H$ N2 ( 550 o C) = 1581
. kJ / mol
o
550
SO 3 (g,550 C) : H$ 7 = z
o
25
Q& = ∆ H& = ξ&∆Hˆ ro +
Table B.2
B
C p dT = 3534
. kJ / mol
∑ n& Hˆ − ∑ n& Hˆ
i
out
i
i
i
in
= (1249 )( −98.28 ) + ( 673)( 24.796 ) + (179.8)(16.711) + ( 7227 )(15.808 ) + (1249 )(35.336 )
− (1922 )(19.623 ) − 1922 (13.362 ) − ( 7227 )(12.691)
−8.111 × 10 4 kJ 1 min 1 kW
=
= −1350 kW
min 60 s 1 kJ/s
c.
&
Assume system is adiabatic, so that Q& lost from reactor = Q
gained by cooling water
L
O
M
P
Q& = ∆ H& = m& w M H$ w el, 40o Cj − H$ w el, 25o Cj P
A
MTable B.5
N
⇒ 8111
. × 104
d.
A
Table B.5
P
Q
kJ
kJ
F kg I
= m& w G
⇒ m& w = 1290 kg min cooling water
J 167.5 − 104 .8
H min K
min
kg
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 ∆H& .
9- 12
CO(g) + H 2Obvg → H2 ( g) + CO 2 (g) ,
9.17
Table B.1
∆H$ ro = e∆H$ fo j
a.
Basis :
− e∆ H$ fo j
CO2 ( g)
CO(g)
− e∆H$ fo j
B
H 2 O bv g
= − 4115
.
kJ
mol
2.5 m 3 bSTP g product gas h 1000 mol 22.4 m 3 bSTP g = 111.6 mol h
nn&33 (mol CO2 /h)
nn&44 (mol H 2/h)
111.6 mol/h
nn& (mol H 2O(v)/h), sat'd
condenser 55
0.40 mol H 2/mol
15°C, 1 atm
0.40 mol CO 2/mol
nn&66 (mol H 2O( l )/h)
0.20 mol H 2O( v)/hmol
15°C, 1 atm
500°C
n&10 (mol CO/h)
25°C
nn&22 (mol H 2 O(v)/h)
150°C
reactor
Q& rr (kW)
Q
Q&cc (kW)
C balance on reactor : n&1 = b0.40gb111.6 mol hg = 44.64 mol CO h
H balance on reactor : 2 n& 2 = 111.6 b2 gb0.40 g + b2gb0.20g mol h ⇒ n& 2 = 66.96 mol H 2 O bv g h
Steam theoretically required =
% excess steam
=
44.64 mol CO 1 mol H2 O
= 44.64 mol H 2O
h
1 mol CO
b66.96 − 44 .64 g
mol h
44.64 mol h
× 100% = 50% excess steam
CO2 balance on condenser : n& 3 = b0.40gb111.6 mol hg = 44.64 mol CO2 h
H 2 balance on condenser: n&4 = b0.40gb111.6 mol hg = 44.64 mol H 2 h
Saturation of condenser outlet gas:
yH 2O =
p∗w b15° Cg
p
⇒
n&5 bmol H 2 O hg
b44.64 + 44.64 + n&5 gbmol h g
=
12.788 mm Hg
⇒ n& 5 = 153
. mol H 2O bvg h
760 mm Hg
H 2 O balance on condenser: b111.6 gb0.20gmol H 2 O h = 1.53 + n& 6
⇒ n& 6 = 20.8 mol H 2O h condensed = 0.374 kg / h
b.
Energy balance on condenser
o
References : H 2 ( g) , CO 2 (g) at 25 C, H 2 O at reference point of steam tables
Substance
CO 2 ( g)
H 2 (g )
H 2 Obv g
H 2 Oblg
n&in
n& out
H$ in
H$ out
mol / h kJ / mol mol / h kJ / mol
44.64
H$ 1
44.64
H$ 4
44.64
H$ 2
44.64
H$ 5
22.32
H$ 3
1.53
H$ 6
−
−
20.80
H$ 7
Enthalpies for CO2 and H2 from Table B.8
CO 2 (g,500o C) : H$ 1 = H$ CO2 (500o C) = 2134
. kJ / mol
9- 13
9.17 (cont’d)
H 2 (g,500o C) : H$ 2 = H$ H2 (500o C) = 1383
. kJ / mol
o
H 2O(v,500C)
: Hˆ 3 = 3488
kJ  18 kg 
×
= 62.86 kJ mol
kg  10 3 mol 
o
CO 2 (g,15C)
: Hˆ 4 = Hˆ CO2 (15 o C) = −0.552 kJ/mol
H 2 (g,15 o C) : H$ 5 = H$ H 2 (15 o C) = −0.432 kJ / mol
kJ F 18.0 kg I
H 2 O(v,15o C) : H$ 6 = 2529
×G
J = 45.52 kJ mol
kg H10 3 molK
kJ F 18.0 kg I
H 2 O(l,15 o C) : H$ 7 = 62.9 × G 3
. kJ mol
J = 113
kg H10 mol K
Q& = ∆H& =
∑ n& H$ − ∑ n& H$
i
i
i
out
=
− 2971.8g kJ
b49.22
1 kW
3600 s 1 kJ s
= −0.812 kW
from condenser g
Energy balance on reactor :
References : H2 ( g) , C(s), O2 (g) at 25° C
Substance
CO( g)
H2 O( v)
H2 bgg
CO2 bgg
n&in
n&out
H$ in
H$ out
( mol / h) ( kJ / mol) ( mol / h) ( kJ / mol)
44.64
H$ 1
−
−
$
6696
.
H2
22.32
H$ 3
−
−
44.64
H$ 4
−
−
44.64
H$ 5
CO(g,25 o C) : H$ 1 = (∆H$ fo ) CO
Table B.1
=
−110.52 kJ / mol
H 2 O(v,150 o C) : H$ 2 = ( ∆H$ fo ) H2 O(v) + H$ H 2 O (150o C)
H 2 O(v,500 o C) : H$ 3 = ( ∆H$ fo ) H 2O(v) + H$ H2 O (500o C)
H 2 (g,500 o C) : H$ 4 = H$ H 2 (500 o C)
Q = ∆H =
∑ n H$ − ∑ n H$
i
out
bheat
i
i
i
=
in
Tables B.1, B.8
=
Tables B.1, B.8
=
−237.56 kJ mol
−224.82 kJ mol
Table B.8
=
13.83 kJ / mol
CO 2 (g,500o C) : H$ 5 = ( ∆H$ of ) CO2 + H$ CO2 (500 o C)
d.
1h
h
in
bheat transferred
c.
i
Tables B.1, B.8
=
− 21013.83 − ( −20839.96) kJ
h
−372.16 kJ / mol
1h
1 kW
3600 s 1 kJ s
= −0.0483 kW
transferred from reactorg
Benefits
Preheating CO ⇒ more heat transferred from reactor (possibly generate additional steam for plant)
Cooling CO ⇒ lower cooling cost in condenser.
9- 14
9.1 8
b.
References : FeO(s), CO(g), Fe(s), CO2 ( g) at 25 o C
CO
Fe
−
CO2
∑n
$
out H out
H$ out
nout
( mol) ( kJ / mol) ( mol) ( kJ / mol)
1.00
0
n1
H$ 1
n0
H$ 0
n2
H$ 2
−
−
n
H$
FeO
Q = ξ ∆H$ ro +
H$ in
nin
Substance
−
−
∑n
3
3
n4
H$ 4
$
in H in
⇒ Q = ξ ∆H$ ro + n1 H$ 1 + n 2 H$ 2 + n 3 H$ 3 + n 4 H$ 4 − n0 H$ 0
(1.00 − n1 )
Fractional Conversion : X =
⇒ n1 = 1 − X
1.00
CO consumed :
1 mol CO
(1 − n1 ) mol FeO consumed
1 mol FeO consumed
= (1− n1 ) mol CO
⇒ n2 = n0 − (1 − n1 ) = n0 − X
Fe produced : n3 =
(1 − n1) mol FeO consumed
1 mol Fe
1 mol FeO consumed
CO 2 produced : n4 =
= (1 − n1 ) mol Fe = X
1 mol CO 2
(1 − n1 ) mol FeO consumed
= (1 − n1 ) mol CO 2 = X
1 mol FeO consumed
Extent of reaction : ξ =
( nCO ) out − ( nC O ) in
ν CO
=
n2 − n0
1
=X
T
H$ i = z Cpi dT
for i = 0,1,2,3,4
25
H$ 0 = 0.02761 (T0 − 298) + 2.51 × 10 −6 (T0 2 − 2982 )
⇒ H$ 0 = ( −8.451 + 0.02761 T0 + 2.51 × 10 −6 T02 ) kJ / mol
H$ 1 = 0.0528 (T − 298) + 31215
.
× 10 −6 (T 2 − 2982 ) + 3188
.
× 10 2 (1 / T − 1 / 298 )
⇒ H$ = ( −17.0814 + 0.0528 T + 3.1215 × 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) + 1335
.
× 10 −5 (T 2 − 2982 )
⇒ H$ 3 = ( −6.335 + 0.01728 T + 1335
.
× 10 −5 T 2 ) kJ / mol
H$ 4 = 0.04326(T − 298) + 0.573 × 10 −5 ( T 2 − 2982 ) + 8.18 × 10 2 (1 / T − 1 / 298)
⇒ H$ = ( −16.145 + 0.04326 T + 0.573 × 10 −5 T 2 + 8.18 × 10 2 / T ) kJ / mol
4
9-15
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
∆Hˆ 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
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
T
500
500
500
500
500
500
500
500
500
500
500
Xi
no To
0.5 400
0.6 400
0.8 400
1.0 400
1.2 400
X
0.5
0.5
0.5
0.5
0.5
T
400
400
400
400
400
1
1
1
1
1
1
1
1
1
1
1
n1
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
H4
0
0
0
0
0
0
0
0
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
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
0
0
0
0
0
0
0
0
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
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
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
Xi
0.5
0.5
0.5
0.5
0.5
n1
0.5
0.5
0.5
0.5
0.5
n2 n3 n4
0 0.5 0.5
0.1 0.5 0.5
0.3 0.5 0.5
0.5 0.5 0.5
0.7 0.5 0.5
H0
2.995
2.995
2.995
2.995
2.995
H1
5.335
5.335
5.335
5.335
5.335
H2
2.995
2.995
2.995
2.995
2.995
H3
2.713
2.713
2.713
2.713
2.713
H4
4.121
4.121
4.121
4.121
4.121
Q
-3.653
-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
9-16
9.18 (cont'd)
1.4
1.6
1.8
2
9.19
a.
400
400
400
400
0.5
0.5
0.5
0.5
400
400
400
400
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.9
1.1
1.3
1.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
2.995
2.995
2.995
2.995
5.335
5.335
5.335
5.335
2.995
2.995
2.995
2.995
2.713
2.713
2.713
2.713
4.121
4.121
4.121
4.121
Fermentor capacity : 550,000 gal
Solution volume : (0.9 × 550,000) = 495,000 gal
R0.071 lb m
|
C 2 H 5OH / lb m solution
Final reaction mixture : S 0.069 lb m (yeast, other species) / lb m solution
|
T 0.86
Mass of tank contents :
lb H 2O / lb m solution
1 ft 3
65.52 lb m
= 4335593 lb m
7.4805 gal
1 ft 3
495,000 gal
Mass of ethanol produced :
4.336 × 10 6 lb m solution
0.071 lb m C 2 H 5 OH
lb m solution
⇒
⇒
b.
3.078 × 105 lb m C 2 H 5OH
= 3.078 ×105 lb m C2 H5 OH
1 lb-mole C 2H 5OH
46.1 lb m C 2 H 5 OH
= 6677 lb-mole C2 H 5OH
1 ft 3 C2 H5OH
7.4805 gal
49.67 lb m C 2 H 5 OH
1 ft3
307827 lb m C2 H5 OH
= 46,360 gal C2 H 5OH
C12 H 22 O11 (s) + 12O2 (g) → 12CO2 (g ) + 11H2 O(l)
∆H$ co = −56491
. kJ / mol
o
o
o
o
∆H$ c = 12 ∆H$ f (CO2 ) + 11∆H$ f (H 2O) − ∆H$ f (C12 H 22 O11 )
⇒ ∆H$ fo (C12 H 22 O11 ) = −2217.14 kJ / mol
C12 H 22 O11 (s) + H 2 O(l) → 4C2 H 5OH(l ) + 4CO2 (g)
c.
∆H$ ro = 4 ∆H$ fo (C2 H5OH) + 4 ∆H$ fo (CO2 ) − ∆H$ fo (C12 H 22 O11 ) − ∆H$ fo (H 2 O) = − 184.5kJ / mol
−1815
. kJ 453.6 mol 0.9486 Btu
⇒ ∆H$ ro =
= −7.811 × 104 Btu / lb - mole
1 mol
1 lb - mole
1 kJ
Moles of maltose :
4.336 × 10 6 lb m solution 0.071 lb C2 H5 OH 1 lb - mole C 2 H 5OH 1 lb - mole C12 H22O11
1 lb m solution
46.1 lb C2 H 5OH
4 lb - mole C 2 H 5OH
= 1669 lb - moles C 12 H 22 O11 ⇒ ξ = nC10 H22 O11 = 1669 lb - moles
Q = ξ∆H$ r + mC p (95o F - 85o F)
Btu
Btu
) + (4.336 × 106 lb m )(0.95 o )(10 o F)
lb - mole
lb - F
7
= −8.9 × 10 Btu ( heat transferred from reactor)
= (1669 lb - moles)( −7.811 × 10 4
d.
Brazil has a shortage of natural reserves of petroleum, unlike Venezuela.
9-17
-3.653
-3.653
-3.653
-3.653
9.20
a.
4NH 3 + 5O2 → 4NO + 6H2 O,
3
2NH3 + O2 → N 2 + 3H2 O
2
References: N2 bgg, H2 bgg, O2 (g), at 25° C
n& in
n& out
H$ in
H$ out
(mol min ) (kJ mol) (mol min) (kJ mol)
100
H$ 1
−
−
$
900
H2
−
−
−
−
90
H$ 3
−
−
150
H$ 4
−
−
716
H$ 5
−
−
69
H$ 6
Substance
NH 3
Air
NO
H 2O
N2
O2
T
H$ i = ∆ H$ foi + z Cpi dT
25
Table B.1
NH 3 bg, 25° Cg: H$ 1 = ( ∆H$ fo ) NH 3
B
=
− 46.19 kJ mol
Table B.8
B
Airbg, 150° Cg: H$ 2
=
3.67 kJ mol
Table B.1,Table B.2
700
B
NObg, 700° Cg: H$ 3 = 90.37 + z Cp dT
=
25
111.97 kJ mol
Table B.1,
Table B.8
B
H 2O bg, 700° Cg: H$ 4
=
− 216.91 kJ mol
Table B.8
N 2 bg, 700 ° Cg: H$ 5
B
=
20.59 kJ mol
Table B.8
O 2 bg, 700° Cg: H$ 6
b.
Q& = ∆ H& =
B
=
21.86 kJ mol
∑ n& H$ − ∑ n& H$
i
out
i
i
i
= −4890 kJ min × (1 min / 60s) = − 815
. kW
in
(heat transferred from the reactor)
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 ∆H& . The value of Q&
would remain unchanged.
9-18
9.21
a.
Basis : 1 mol feed
I=Inert
1 mol at 310°C
0.537 C2 H4 (v)
0.367 H2 0 (v)
0.096 N2 (g)
Products at 310°C
n 1 (mol C2 H4 (v))
n 2 (mol H2 O(v))
0.096 mol N2 (g)
n 3 (mol C2 H5 OH (v))
n 4 (mol (C2 H5 )2 O) (v))
C 2 H4 ( v) + H 2O(v) ⇔ C2 H 5OH(v)
2C2 H 5OH(v) ⇔ bC 2 H5 g2 O(v) + H 2 O(v)
5% ethylene conversion:
. gb0.05g =
b0537
002685
.
mol C2 H 4 consumed
⇒ n1 = b0.95gb0537
. g = 0510
.
mol C2 H 4
90% ethanol yield:
0.02685 mol C 2 H 4 consumed 0.9 mol C2 H 5OH
n3 =
= 0.02417 mol C 2 H5OH
1 mol C 2 H4
C balance :
⇒ n 4 = 1.415 × 10 −3 mol bC 2 H 5 g2 O
b2 gb0.537g = b2 gb0.510g + b2 gb0.02417g + 4 n4
O balance : 0.367 = n2 + 0.02417 + 1415
.
× 10−3 ⇒ n2 = 0.3414 mol H 2O
References: Casf , H 2 bg g, O 2 bg g at 25o C, I bg g at 310o C
substance
C 2 H4
H2 O
H$ in
nin
(mol) (kJ / mol)
0.537
H$ 1
0.367
H$
nout
H$ out
(mol)
(kJ / mol)
H$ 1
H$
0.510
0.3414
2
2
I
0. 096
0
0.096
C 2 H5OH
−
−
0. 02417
bC 2 H 5 g
2
−
O
−
C2 H 4 ( g, 310° C ) : Hˆ 1 = (∆Hˆ fo )C2 H4 +
1.415 × 10
∫
310
C p dT
25
bC2 H 5 g
2
Obg, 310 ° Cg: H$ 4 = e∆H$ fo j
H$ 4
⇒
⇒
( 52.28 + 16.41) = 68.69
b−241.83 + 9.93g =
Table B.1
Table B.8
C 2 H5OHbg, 310 ° Cg : H$ 3 = ( ∆H$ fo ) C2 H5OH(g) + z C pdT ⇒
25
3
−3
Table B.1 for ∆Hˆ fo
Table B.2 for Cp
H2 Obg, 310 ° Cg : H$ 2 = ( ∆H$ fo ) H 2 O(v) + H$ H 2 O(v) ( 310 o C)
310
0
$
H
−23190
. kJ mol
. + 24.16g =
b− 23531
Table B.1
Table B.2
kJ mol
− 21115
. kJ mol
310
(C 2 H 5 )O(l)
+ ∆H$ v b25° Cg + z Cp dT = b−272 .8 + 26.05 + 42.52g
25
= −204.2 kJ mol
Energy balance: Q = ∆H =
∑ n H$ − ∑ n H$
i
out
b.
i
i
i
=− 1.3 kJ ⇒ 1.3 kJ transferred from reactor mol feed
in
To suppress the undesired side reaction. Separation of unconsumed reactants from products and
recycle of ethylene.
9-19
9.22
C 6H 5CH3 + O 2 → C 6 H5CHO + H 2O
C 6H 5CH3 + 9O 2 → 7CO2 + 4H 2 O
Basis : 100 lb-mole of C 6H 5CH3 fed to reactor.
100 lb-moles C 6 H 5CH 3
n0 (lb-moles O 2 )
3.76n 0 (lb-moles N2 )
350°F, 1 atm
V0 (ft3 )
Vp (ft3 ) at 379°F, 1 atm
n1 (lb-moles C6 H 5 CH3 )
n2 (lb-moles O 2)
3.76n0 (lb-moles N 2 )
n3 (lb-moles C 6H 5 CHO)
n4 (lb-moles CO 2 )
n5 (lb-moles H2 O)
reactor
Q(Btu)
jacket
mw(lb m H 2 O( l )), 80°F
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 6H 5CH3 . The calculated quantities will then be scaled to the known flow rate of water in
the product gas b29.3 lb m 4 hg .
Plan of attack: % excess air ⇒ n0
13% C 6H5CHO formation ⇒ n3
0.5% CO2 formation ⇒ n4
C balance ⇒ n1
Ideal gas equation of state ⇒ V0
Ideal gas equation of state ⇒ Vp
E.B. on reactor ⇒ Q
E.B. on jacket ⇒ mw
Scale V0 , Vp , Q, mw by bn&5 gactual / bn5 gbasis
H balance ⇒ n5
O balance ⇒ n2
100% excess air:
n0 =
100 lb-moles C6 H 5CH3
1 mole O2 reqd
1 mole C 6H 5CH3
(1 + 1) mole O2
fed
= 200 lb-moles O2
1 mole O2 reqd
N 2 feed b& output g = 3.76b200glb - moles N 2 = 752 lb - moles N 2
13% → C6H5CHO ⇒ n3 =
100 lb-moles C6H5CH3
0.13 mole C6H5 CH3 react 1 mole CHCHO
form
6 5
1 mole C6 H5 CH3 fed
1 mole C6H5 CH3 react
=13 lb-moles C6H5CHO
0.5% → CO2 ⇒ n4 =
b100gb0.005glb - moles C 6 H5CH3
react
7 moles CO2
1 mole C6 H5CH3
= 35
. lb - moles CO2
mol C mole C 7H8
B
C balance:
a100f a7f
lb - moles C = 7n1 + a13f a7f + a3.5f a1f ⇒ n1 = 86.5 lb - moles C 6 H5CH 3
H balance:
b100 gb8glb - moles H
O balance:
b200gb2glb - moles
= b865
. gb8g + b13gb6g + 2n5 ⇒ n5 = 150
. lb - moles H 2Obvg
O = 2n2 + b13gb1g + b35
. gb2g + b15gb1g ⇒ n2 = 182 .5 lb- moles O 2
9-20
9.22 (cont’d)
Ideal gas law − inlet:
V0 =
b100 +
200 + 752glb - moles 359 ft 3 bSTPg
1 lb - moles
o
b350 +
460g R
= 6.218 × 105 ft 3
o
492 R
Ideal gas law – outlet:
O2
C 7 H 8 O CO2 H 2 O
N2 I
F C 7 H8
G86.5 + 182.5+ 13 + 3.5 + 15 + 752 J lb - moles
Vp = H
K
359 ft 3
o
b379 + 460g
1 lb - mole
R
= 6.443 × 105 ft 3
492 o R
Energy balance on reactor (excluding cooling jacket)
References : Cbsg, H 2 bgg, O 2 bgg, N 2 bgg at 25o Ce77o F j
H$ in
nin
substance
blb - molesg
C 6H 5CH3
100
O2
200
N2
bBtu
H$ out
nout
lb- moleg
H$
blb- molesg
86.5
1
bBtu
lb - mole g
H$
4
182.5
752
H$ 2
H$
C6 H 5CHO
−
−
13
CO 2
−
−
3.5
H$ 7
H$
H2O
−
−
15
H$ 9
752
3
H$ 5
H$
6
8
Enthalpies:
Table B.1
L
O
M
N
P
Q
B
430.28 Btu lb - mole
Btu
o P
C 6H 5CH3 ( g,T): H$ bT g = M∆ H$ fo bkJ molg ×
+ 31
bT − 77 g F
M
P
1 kJ mol
1b - mole⋅° F
C 6H 5CH3 ( g,350 F): H$ 1 = 2.998 × 104 Btu lb - mole
C 6H 5CH3 ( g,379 o F): H$ 4 = 3.088 × 104 Btu lb - mole
o
o
C6 H 5CHO(g, T): H$ bT g = −17200 + 31bT − 77g F Btu lb - mole
⇒ H$ 7 = −7.83 × 10 3 Btu lb - mole
Table B.9
B
O 2 eg,350 F j : H$ 2 = H$ O2 ( 350o F ) = 1972
.
× 103 Btu / lb − mole
o
Table B.9
B
N2 eg,350 Fj : H$ 3 = H$ N2 ( 350 o F) = 1911
.
× 103 Btu / lb − mole
o
Table B.9
B
O 2 eg,379 Fj : H$ 5 = H$ O2 ( 379 F) = 2186
.
× 103 Btu / lb − mole
o
o
Table B.9
B
N2 eg,379 Fj : H$ 6 = H$ N 2 ( 379 o F ) = 2.116 × 103 Btu / lb − mole
o
CO 2 eg,379 Fj : H$ 8 = ( ∆H$ of ) CO2 ( g) + H$ CO2 ( 379 o F )
o
Table B.1 and B.9
B
H2 Obg,379 ° Fg: H$ 9 = ( ∆H$ of ) H 2 O( g) + H$ H 2 O ( 379 o F)
9-21
=
− 1664
.
× 105 Btu / lb − mole
Table B.1 and B.9
B
=
− 1016
.
× 105 Btu / lb − mole
9.22 (cont’d)
Energy Balance :
Q = ∆H =
∑ n H$ − ∑ n H$
i
i
i
out
i
= −2.376 × 106 Btu
in
Energy balance on cooling jacket:
105
Q = ∆H = m w z
80
dC p i H O l dT
2 b g
Q = +2 .376 × 10 4 Btu , C p = 1.0 Btu (lb m ⋅ o F)
2.376 × 10 6 Btu = mw blb m g × 1.0
Scale factor:
bn&5 g
actual
bn5 g
basis
=
Btu
o
× b105 − 80g F ⇒ m w = 9.504 × 10 4 lb m H 2 Oblg
lb m ⋅° F
29.3 lb m H 2O 1b- mole H 2 O
1
= 0.02711 h −1
4h
18.016 lb m H2 O 15.0 lb - moles H2 O
V0 = d6.218 × 10 5 ft 3 i d0.02711 h −1 i = 1.69 × 10 4 ft 3 h bfeed g
V p = d6.443 × 10 5 ft 3 i d0.02711 h −1 i = 175
. × 10 4 ft 3 h bproduct g
Q = d−2 .376 × 10 6 Btui d0.02711 h − 1 i = −6 .44 × 10 4 Btu / h
& w = d9.504 × 104 Btui d0.02711 h −1 i =
m
2577 lb m
1 ft 3
7.4805 gal 1 h
= 515
. gal H 2O min
h
62.4 lb m
1 ft 3 60 min
9-22
9.23
a.
CaCO3 ( 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 CaCO3 =
= 10.0 kmol CaCO3 ⇒ 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
Substance (mol)
(kJ/mol) (mol) (kJ/mol)
Hˆ1
−
−
−
10 4
Hˆ
4
CaCO3
10
CaO
−
CO2
−
Hˆ out
nout
2
−
10
Hˆ 3
4
Table B.1
B
CaCO3 ( s, 25 C) : H$ 1 = ( ∆H$ fo ) CaCO 3 ( s)
=
o
− 1206.9 kJ / mol
Table B.1,
Table B.2
1173
CaO( s, 900 o C) : H$ 2 = ( ∆H$ fo ) CaO(s) +
B
=
C p dT
z
( −6356
. + 4854
. ) kJ / mol = −587 .06 kJ / mol
298
Table B.1,
Table B.8
CO2 ( g, 900o C) : H$ 3 = ( ∆H$ fo ) C O2 (g) + H$ CO2 ( 900 o C)
∑ n H$ − ∑ n H$ JK =
H
F
I
Energy balance: Q = ∆H = G
i
out
b.
i
i
i
B
=
( −3935
. + 42.94) kJ / mol = −350.56 kJ / mol
2.7 × 106 kJ
in
Basis : 1000 kg CaCO3 fed ⇒ 10.0 kmol CaCO3
CaCO3 ( s) → CaO(s) + CO2 ( g)
2CO + O2 → 2CO2
10 kmol CaCO3
25 o C
Product gas at 900o C
200 kmol at 900o C
0.75 N2
0.020 O2
0.090 CO
0.14 CO2
n 1 [kmol CaO(s)]
10 kmol CaCO 3 react ⇒ n1 = 100
. kmol CaO
9-23
n 2 (kmol CO2 )
n 3 (kmol N2 )
n 4 (kmol CO)
9.23 (cont'd)
n2 = ( 014
. )( 200 ) +
10.0 kmol CaCO3 react 1 kmol CO 2 4 kmol O2 react
+
1 kmol O 2
2 kmol CO2
= 46 kmol CO2
1 kmol O 2
n3 = ( 075
. )(200) = 150 kmol N2
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), O2 (g), N 2 ( g) at 25o C
nin
nout
Hˆ in
Hˆ out
Substance (mol) (kJ/mol) (mol) (kJ/mol)
CaCO3
10.0
Hˆ 1
−
−
CaO
−
−
10
− 587.06
CO2
28
− 350.56
46
− 350.56
ˆ
CO
18
H
10
Hˆ
2
O2
4.0
N2
150
CaCO3 (s, 25 C) : Hˆ 1 =
o
Hˆ 3
Hˆ
4
(∆Hˆ fo )CaCO(s)
3
2
−
150
= − 1206.9 kJ/mol
Table B.8
↓
O2 (g, 900 C) : Hˆ 3 = Hˆ O2 (900 C)
=
o
N2 (g, 900 C) : Hˆ 4 = Hˆ N2 (900 C)
o
o
∑ n H$ − ∑ n H$ JK =
H
F
I
Q = ∆H = G
i
out
i
i
i
Table B.1,
Table B.8
↓
=
(−110.52 + 27.49) kJ/mol = −83.03 kJ/mol
28.89 kJ/mol
Table B.8
↓
=
27.19 kJ/mol
0.44 × 106 kJ
in
% reduction in heat requirement =
c.
4
Table B.1
↓
CO(g, 900 oC) : Hˆ 2 = (∆Hˆ fo ) CO(g) + Hˆ CO (900 o C)
o
−
Hˆ
2.7 × 106 − 0.44 × 106
2.7 × 106
× 100 = 83.8%
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.
9-24
9.24
a.
A+B→ C
2C → D + B
(1)
(2)
Basis : 1 mol of feed gas
1.0 mol
x AO (mol A/mol)
x BO (mol B/mol)
x IO (mol I/mol)
Fractional conversion:
C generated: n0 =
fA =
n A (mol A)
nB (mol B)
nC (mol C)
n D (mol D)
n I (mol I)
T (o C)
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 AYC
D generated: nD = 0.5 × mol C consumed = (1 2) × (mol A consumed − mol C out)
⇒ n D = (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.
Species Formula
DHf
a
A
C2H4(v)
52.28 0.04075
B
H2O(v)
-241.83 0.03346
C
C2H5OH(v) -235.31 0.06134
D
(C4H10)O(v) -246.75 0.08945
I
N2(g)
0 0.02900
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
xA0
0.537
H(in)
(kJ/mol)
68.7
-231.9
-211.2
-204.2
9.4
xB0
0.367
n(out)
(mol)
0.510
0.341
0.024
0.001
0.096
o
b
1.15E-04
6.88E-06
1.57E-04
4.03E-04
2.20E-05
xI0
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
c. For Tf = 125 C, Q = 7.90 kJ . Raising Tp , lowering fA , and raising YC all increase Q.
9-25
9.25
a.
CH 4 ( g) + O2 ( g) → HCHO(g) + H 2O(g)
10 L, 200 kPa
n 0 (mol feed gas) at 25°C
n 3 (mol HCHO)
n 4 (mol H2 O)
0.851 mol CH4 /mol
0.15 mol O2 /mol
n 5 (mol CH4 )
T (°C), P(kPa), 10L
Q (kJ)
Basis : n0 =
200 kPa 1000 Pa 10 L 10 −3 m 3
1 mol K
1 kPa
1L
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
1 mol O 2 fed
= 0.1211 mol CH 4
⇒ n 5 = (0.6861 − 01211
.
) mol CH 4 = 0.5650 mol CH 4
HCHO produced : n3 =
H2 O produced : n4 =
1 mol HCHO
01211
.
mol CH 4 consumed
= 01211
.
mol HCHO
1 mol CH 4 consumed
1 mol H 2O
01211
.
mol CH 4 consumed
= 01211
.
mol H 2O
1 mol CH 4 consumed
Extent of reaction :ξ =
( nO 2 ) out − ( nO 2 ) in
ν O2
=
0 − 01211
.
1
= 01211
.
mol
References : CH4 ( g), O 2 ( g), HCHO(g), H 2O(g), at 25 o C
Substance
CH4
O2
HCHO
H 2O
U$ i =
nin
nout
U$ in
U$ out
mol kJ mol mol
kJ mol
0.6861
0
05650
.
U$ 1
01211
.
0
−
−
−
−
01211
.
U$ 2
−
−
01211
.
U$ 3
T
T
z(Cv ) i dT =
z( C p − R ) i dT
25
25
i = 1,2,3
Using ( Cp ) i from Table B.2 and R = 8.314 × 10−3 kJ / mol ⋅ K:
U$ 1 = ( 0.02599 T + 2.7345 × 10−5 T 2 + 01220
.
× 10− 8 T 3 − 2.75 × 10−12 T 4 − 0.6670) kJ / mol
U$ 2 = (0.02597 T + 21340
.
× 10−5 T 2 − 21735
.
× 10− 12 T 4 − 0.6623) kJ / mol
U$ 3 = ( 0.02515 T + 03440
.
× 10−5 T 2 + 0.2535 × 10−8 T 3 − 08983
.
× 10−12 T 4 − 0.6309) kJ / mol
9-26
9.25 (cont'd)
Q=
100 J 85 s
s
1 kJ
1000 J
= 8.5 kJ
Table B.1
B
∆H$ ro = ( ∆H$ fo ) HCHO + (∆H$ fo ) H2 O − ( ∆H$ fo ) CH4
=
. ) + (−24183
. ) − (−74.85)g kJ
b(−11590
/ mol
= −282.88 kJ / mol
∆U$ ro = ∆H$ ro − RT (
∑
∑ν )
νi −
gaseous
products
i
gaseous
reactants
= −282.88 kJ / mol −
8.314 J
mol K
(1 + 1 − 1 − 1)
298 K
1 kJ
= −282.88 kJ / mol
10 3 J
Energy Balance :
Q = ξ∆U$ 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 − 4329
. kJ / mol
Solve for T using E - Z Solve ⇒ T = 1091o C = 1364 K
⇒ P = nRT / V =
0.8072 mol 8.314 m3 ⋅ Pa 1364 K
mol ⋅ K
10 L
1L
10 −3 m3
= 915 × 10 3 Pa = 915 kPa
Add heat to raise the reactants to a temperature at which the reaction rate is significant.
b.
c.
9.26
Side reaction : CH 4 + 2O2 → CO2 + 2H 2O. 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.
1
O 2 ( g) → C 2 H 4 Obgg
2
C 2 H 4 + 3O 2 → 2CO 2 + 2H 2 O
a.
C 2 H 4 bgg +
Basis: 2 mol C 2 H 4 fed to reactor
Qr (kJ)
heat
n 1 (mol C 2H 4)
n 2 (mol O 2 )
25°C
2 mol C2 H4
1 mol O2
450°C
reactor
n 3 (mol C 2H 4)
n 4 (mol O 2 )
n 6 (mol CO2 )
n 7 (mol H 2O( l ))
25°C
separation
n3 ( mol C 2H 4)
process
n4 ( mol O 2 )
n5 ( mol C 2 H 4O)
n6 ( mol CO2 )
n7 ( mol H2 O)
450°C
25% conversion ⇒ 0.500 mol C 2 H 4 consumed ⇒ n 3 = 1.50 mol C 2 H 4
9-27
n5 (mol C 2H 4O(g ))
25°C
9.26 (cont'd)
70% yield ⇒ n5 =
0.500 mol C 2 H 4 consumed 0.700 mol C 2 H 4 O
= 0.350 mol C 2 H 4 O
1 mol C 2 H 4
C balance on reactor:
. g + b2gb0350
. g + n6
b2gb2g = b2gb150
⇒ n 6 = 0.300 mol CO 2
1 mol H 2 O
= 0.300 mol H 2 O
1 mol CO 2
O balance on reactor: b2gb1g = 2n4 + 0.350 + b2gb0.300g + 0300
.
⇒ n4 = 0.375 mol O 2
Water formed: n7 =
0.300 mol CO 2
Overall C balance: 2n1 = n6 + 2n5 = 0300
.
+ b2gb0.350 g ⇒ n1 = 0500
.
mol C2 H 4
Overall O balance: 2n2 = 2n6 + n7 + n5 = b2gb0.300 g + b0300
. g + b0.350g ⇒ n2 = 0.625 mol O 2
Feed stream: 44.4% C2 H 4 , 556%
.
O2
Reactor inlet: 66.7% C 2 H 4 , 333%
.
O2
Recycle stream: 80.0% C 2 H 4 , 20.0% O2
Reactor outlet: 53.1% C 2 H 4 , 13.3% O2 , 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 4O
44.05 g 1 kg
= 00154
.
kg
1 mol 10 3 g
References for enthalpy calculations :Cbsg, H 2 bgg, O 2 bgg at 25° C
T
H$ i bT g = ∆H$ ofi + z C p dT for C 2 H 4
25
T +273
= ∆H$ 0f + z C p dT for C 2 H 4 O
298
= ∆H$ ofi + H$ i ( table B.8)
= ∆H$ of for H 2 Oblg
for O 2 , CO 2 , H 2 Obgg
Overall Process
nin
H$ in
(mol) ( kJ / mol)
C2 H4 0.500
5228
.
O2 0625
.
0
C2 H 4 O
−
−
CO2
−
−
H2Oblg
−
−
Substance
Reactor
nout
H$ out
( mol) ( kJ / mol)
−
−
−
−
0350
.
−5100
.
0300
.
−3935
.
0300
.
−28584
.
Energy balance on process: Q = ∆H =
nin
nout
H$ in
H$ out
( mol) ( kJ / mol) ( mol) ( kJ / mol)
C2 H 4
2
7926
.
150
.
79.26
O2
1
1337
.
0375
.
1337
.
C2 H 4 O
−
− 0350
.
−1999
.
CO2
−
− 0300
.
−374.66
H2Obgg
−
− 0300
.
−22672
.
substance
∑ n H$ − ∑ n H$
i
i
i
out
Energy balance on reactor: Q = ∆H =
∑ n H$ − ∑ n H$
i
out
c.
i
= − 248 kJ
in
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
103 mol
1500 kg day
= 9.73 × 104 day −1
0.01542 kg
9-28
) = 0.01542 kg C2 H 4 O
9.26 (cont'd)
In initial basis, fresh feed containsU
|
M = b0.500gb28.05 g C 2 H 4 molg + b0.625 gb32.0 g O 2 molg
0.500 mol C 2 H 4
V
|
0.625 mol O2
W
= 34.025 × 10−3 kg
Fresh feed rate = e34.025× 10 −3 kg j e9 .73 × 10 4 day −1 j = 3310 kg day (44.4% C 2 H 4 , 55.6% O 2 )
Qprocess =
Qreactor =
b− 248
kJ ge9.73 × 104 day −1 j
1 day 1 hr
1 kW
= − 279 kW
24 hr 3600 s 1 kJ s
b− 236
kJ ge9.73 × 10 4 day −1 j
1 day 1 hr
1 kW
= −265 kW
24 hr 3600 s 1 kJ s
9-29
9.27
a.
1200 lb m C 9 H12
h
Overall process :
Basis :
1 lb - mole
= 10.0 lb - moles cumene produced h
120 lb m
n 1 (lb-moles/h)
0.75 C 3H 6
0.25 C 4H 10
n 3 (lb-moles C 3 H6 /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
∆H$ r b77° Fg = −39520 Btu lb- mole
C 3H 6 blg + C6 H 6 blg → C9 H12 blg,
Benzene balance: n& 2 =
binput =consumption g
=
10.0 lb - moles C 9H12 produced 1 mole C 6 H6 consumed
h
1 mole C 9H12 produced
10.0 lb - moles C 6H 6
h
0.75n& 1 = n& 3 +
Propylene balance:
binput =output+ consumption g
781
. lb m C6 H 6
= 781 lb m C 6H 6 h
1 lb - mole
10.0 lb - moles C 9H12 1 mole C 3H 6
h
1 mole C 9H12
⇒ 0.75n&1 = n& 3 + 10
n&1 = 16.67 lb - moles h
n& 3 = 2.50 lb - moles C 3H 6 h
U
V⇒
20% C 3H 6 unreacted⇒ n& 3 = 0.20b0.75n&1 gW
Mass flow rate of C3 H6 / C 4 H10 feed =
+
b0.75gb16.67 glb - moles
b0.25gb16.67 glb - moles
h
h
C3 H 6 42.08 lb m C 3H 6
1 lb - mole
C4 H10 58.12 lb m C 4 H10
= 768 lb m h
1 lb - mole
Reactor :
Benzene feed rate =
10.0 lb - moles fresh feed
h
10.0 lb-moles C9 H12 /h
2.50 lb-moles C3 H6 /h
4.17 lb-moles C4 H10 /h
30.0 lb-moles C6 H6 /h
400o F
40.0 lb-moles C6 H6 /h
b.
fed to reactor
1 mole fresh feed
16.67 lb-moles/h @ 77o F
0.75 C3 H6
0.25 C4 H10
Overhead from T1 ⇒
a3 + 1f moles
2.50 lb - moles C 3H 6 h U
6.67 lb - moles h
V ⇒ 37.5% C 3 H 6
4.17 lb- moles C4 H 10 hW
62.5% C 4H 10
Heat exchanger :
Reactor effluent at 400°F
10.0 lb-moles C9H12 /h
2.50 lb-moles C3H6 /h
4.17 lb-moles C4H10 /h
30.0 lb-moles C6H6 /h
200°F
40.0 lb-moles C6H6 /h
77°F
T (°F)
9- 30
= 40 lb - moles C 6 H 6 h
46.7 lb-moles/h
21.4% C9 H12
5.4% C3 H6
8.9% C4 H10
64.3% C6 H6
9.27 (cont'd)
Energy balance: ∆H = 0 ⇒
∑ n eH$
i
i , out
− H$ i , in j =
∑n C
i
pi bTout
− Tin gi = 0
(Assume adiabatic)
C 3H 6
L10 lb - moles C 9 H 12
M
N
120 lb m
0.40 BtuO
1 lb - mole 1bm ⋅
o
h
Pe200
FQ
o
B
F − 400o F j + b2.50gb42.08gb0.57ge200 o F − 400
C4 H10
B
+ b4.17gb58.12gb0.55ge200 o F − 400o F j + b30.0gb78.11gb0.45ge200o F − 400o Fj
A
C6 H6 in
effluent
+ b40.0g b78.11gb0.45geT − 77o F j = 0 ⇒ T = 323° F
A
C6 H6 fed
to reactor
(Refer to flow chart of Part b: T = 323° F )
References : C 3H 6 blg, C 4 H10blg, C6 H 6 blg, C 9H 12 blg at 77° F
H$ i bBtu lb - moleg = C pi bBtu lb m ⋅° Fg M i blb m lb - molegbT − 77 gb° F g
Substance
C3 H 6
H$ in
n& in
n& out
H$ out
(lb - mole / h) (Btu / lb - mole) (lb - mole / h) (Btu / lb - mole)
12.0
0
2.50
7750
C 4 H 10
4.17
0
4.17
10330
C6 H 6
40.0
8650
30. 0
11350
C9 H 12
−
−
10.0
15530
Energy balance on reactor :
n& C9 H12 ∆H$ ro
Q = ∆H =
+
n& i H$ i −
vC 9 H1 2
out
∑
=
b10.0gb−39520g
b1g
∑ n& H$
i
i
in
+ b2.50gb7750g + b417
. gb10330 g + b30.0gb11350g + b10.0gb15530g
−b40.0gb8650g = −183000 Btu h bheat removalg
9.28
Basis :
a.
100 kg C 8H 8 103 g 1 mol
= 960 mol h styrene produced
h
1 kg 104.15 g
C 8H10 ( g) → C 8H 8 (g) + H 2 ( g)
Overall system
n 2 (mol H2 /h)
Fresh feed
n 1 (mol C8H 10/h)
960 mol C8H 8 /h
960 mol C 8H8 1 mol C8 H10
= 960 mol C8 H10 h fresh feed
h
1 mol C 8H8
bC 8 H10 balanceg
960 mol C8 H10
1 mol H 2
H 2 balance : n& 2 =
= 960 mol H 2 h
h
1 mol C8 H10
Fresh feed rate: n&1 =
9- 31
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)
0.35n3 bmol C8 H 10 reactg 1 mol C 8H 8
= 960 mol C8 H 8 h
h
1 mol C8 H 10
⇒ n& 3 = 2740 mol C8 H10 h fed to reactor
35% 1-pass conversion ⇒
⇒ Recycle rate = a2740 − 960 f = 1780 mol C 8H10 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
Energy balance: ∆H = 2740 ∆H$ C H + n& 4∆ H$ H O = 0bkJ h g
8 10
2
bNeglect
Q , ∆E k g
L
O
J
P
b118 + 0.30T gdT
M 500 14
4244
3 P mol ⋅o
M
Cp
P
N
Q
600
∆H$ C8 H10 = Mz
∆H$ H 2O
a2740 f a28.3f
b.
Table B.8
⇒
P =1 bar
C
×
1 kJ
= 28.3 kJ mol
103 J
= − 39
. kJ mol
+ n& 4 a−3.9 f = 0 ⇒ n& 4 = 1.99 × 10 4 mol H2 O / h
Ethylbenzene preheater bA g :
960 mol fresh feed 1780 mol recycled 2740 mol EBblg
+
=
at 25° C
h
h
h
2740 mol EBbvg
⇒
at 500° C
h
136
500
∆H$ = z C pi dT + ∆H$ v a136° Cf + z C pv dT = a20.2 + 36.0 + 77.7f kJ mol = 133.9 kJ mol
25
2740 mol C8 H10
Q& A = ∆H& =
h
Steam generator bFg :
136
133.9 kJ
= 367
. × 10 5 kJ h bpreheaterg
mol C8 H10
19400 mol h H 2Obl, 25° Cg → 19400 mol h H 2Obv, 700 ° C, 1 atmg
Table B.5 ⇒ H$ bl, 25° Cg = 104.8 kJ kg ;
Table B.7 ⇒ H$ bv, 700° C, 1 atm ≈ 1 bar g = 3928 kJ kg
9- 32
9.28 (cont'd)
19400 mol H2 O 18.0 g 1 kg
Q& F = ∆ H& =
h
1 mol 10 3 g
a3928 − 104.8 f kJ
kg
= 134
. × 10 kJ h bsteam generator g
6
Reactor bCg :
References: C 8H 8 bvg, C8 H10 bvg, H 2 bgg, H2 Obvg at 600° C
560
H$ i e560o Cj = z
600
dC pv i i dT
for C 8H10 , C8 H8
≈ H$ (T) for H 2 , H 2 O (interpolating from Table B.8)
Substance
n& in
n& out
H$
H$
in
C8 H10
out
(mol h) (kJ mol) (mol h) (kJ mol)
2740
0
1780
−11.68
H2 O
19900
0
19900
−1.56
C8 H 8
−
−
960
−10.86
H2
−
−
960
− 119
.
Energy balance :
960 mol C 8H 8 produced
124.5 kJ
Q& c = ∆H& =
+
h
1 mol C8 H 8
∑ n& H$ − ∑ n& H$
i
out
i
i
i
in
= 5.61 × 10 4 kJ h areactor f
c.
This is a poorly designed process as shown. The reactor effluents are cooled to 25o 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
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 O 2/mol air
0.79 mol N 2/mol air
n s mol H2 O(v )/h
saturated at 145°C
reactor
reactor
product gas, 600°C
n 1 (mol CH 3 OH/h)
n 2 (mol O 2 /h)
waste
n 3 (mol N 2 /h)
heat
n 4 (mol HCHO/h)
boiler
0.37 kg HCHO/h
n 5 (mol H 2 /h)
0.63 kg H 2O/h
n 6 (mol H 2 O/h)
mb(kg
(kgHHO(l)/h)
mb (kg H2 O(v )/h)
2 O(v )/h)
m
b
2
30°C
sat'd at 3.1 bars
o
30 C
b.
In the absence of data to the contrary, we assume that the separation of methanol from
formaldehyde is complete.
Methanol vaporizer:
The product stream, which contains 42 mole % CH 3OHbvg , is saturated at Tm e o Cj and 1 atm.
9- 33
9.29 (cont'd)
ym P = pm∗ bTm g ⇒ b0.42gb760 mmHgg = 319 .2 mmHg = pm∗ bTm g
 → p∗m = 319.2 mmHg ⇒ Tm = 44.1 o C
Antoine equation
c.
Moles HCHO formed :
=
36 × 106 kg solution 0.37 kg HCHO
1 kmol
1 day
kmol HCHO
= 52.80
h
350 days
1 kg solution 30.03 kg HCHO 24 h
but if all the HCHO is recovered, then this equals n& 4 , or n& 4 = 5280
. kmol HCHO h
70% conversion :
52.80 kmol HCHO 1 kmol CH 3OH react
1 kmol CH 3OH fed
1 kmol feed gas
= n& f
h
1 kmol HCHO formed 0.70 kmol CH3OH react 0.42 kmol CH 3OH
⇒ n& f = 179.59 kmol h
Methanol unreacted:
n&1 =
b0.42 gb179.59 gkmol
CH 3OH fed
h
b1 − 0 .70g kmol
CH 3OH fed
1 kmol CH 3OH fed
= 22.63
kmol CH 3OH
h
N 2 balance: n&3 = b179.6 kmol hgb058
. gb079
. g = 82.29 kmol N 2 h
Four reactor stream variables remain unknown — n& s , n&2 , n&5 , and n& 6 — and four relations are
available — H and O balances, the given H2 content of the product gas (5%), and the energy
balance. The solution is tedious but straightforward.
H balance:
b179 .6 gb0.42 gb4 g + 2n s
= b22 .63gb4 g + b52.8 gb2 g + 2n& 5 + 2 n& 6
⇒ n& s = n& 5 + n& 6 − 52.80
O balance:
(1)
. gb1g + b179.6g(0.58)b0.21gb2g + n& s
b179 .6gb042
= ( 22.63)(1) + 2n& 2 + (52.80)(1) + n& 6
⇒ n& s = 2n& 2 + n& 6 − 4375
.
H 2 content:
(2)
n& 5
= 0.05 ⇒ 19n&5 − n& 2 − n& 6 = 157 .72
22.63 + n& 2 + 82.29 + 5289
. + n& 5 + n& 6
References : Cbsg, H 2 bgg, O 2 bgg, N 2 bgg at 25° C
Table B.2
H=
∆H$ fo
T
B
+ z C pdT
25
or Table B.8 for O 2 , N 2 and H 2
9- 34
(3)
9.29 (cont'd)
substance
n& in
kmol / h
n& out
H$ in
kmol
/h
kJ / kmol
H$ out
kJ / kmol
CH 3 OH
O2
N2
H2O
75.43
21.88
82.29
ns
−195220
3620
3510
−237740
22 .63
n2
82.29
n6
−163200
18410
17390
−220920
HCHO
H2
−
−
−
−
52.80
n5
−88800
16810
Energy Balance :
∆H = ni H$ i − ni H$ i = 0 ⇒ 18410n2 + 16810n5 − 220920n6 + 237704n s = −7.406 × 10 6
∑
∑
out
(4)
in
We now have four equations in four unknowns. Solve using E-Z Solve.
n& s =
58.8 kmol H 2 Obvg 18.02 kg
h
1 kmol
= 1060 kg steam fed h
n& 2 = 2.26 kmol O2 h , n& 5 = 1358
. kmol H 2 h , n& 6 = 9800
. kmol H2 O h
Summarizing, the product gas component flow rates are 22.63 kmol CH3 OH/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 H2 O/h
⇒
d.
272 kmol h product gas
8% CH 3OH, 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
Reference States: Cbsg, H 2 bgg, O 2 bgg, N 2 bgg at 25° C for reactor gas
H2 Oblg at triple point for boiler water
Substance
n&i n
Hˆ in
n& out
Hˆ out
kmol/h
mol
CH 3O H
22.63
kJ/kmol
−163200
22.63
kJ/mol
− 195220
O2
N2
2.26
82.29
18410
17390
2.26
82.29
3620
3510
H2O
98.02
−220920
98.02
− 237730
HCHO
H2
52.80
13.58
m& b
− 88800
16810
52.80
13.58
m& b
− 111350
3550
H2O
(kg/h)
125.7
(kJ/kg)
Energy Balance :
∆H =
∑ n H$ − ∑ n H$
i
out
i
i
i
=0
in
⇒ mb b27261
. − 1257
. g − 4.92 × 106 = 0
⇒ mb = 1892 kg steam h
9- 35
(kg/h)
2726.1
(kJ/kg)
9.30
a.
C 2 H 4 + HCl → C 2 H5Cl
Basis:
1600 kg C 2 H5Clblg 103 g 1 mol
= 24800 mol h C 2 H5Cl
h
1 kg 64.52 g
n 3 (mol HCl( g)/h)
n 4 (mol C 2H 4( g)/h)
n 5 (mol C 2H 6( g)/h)
n 6 (mol C 2H 5Cl( g)/h)
50°C
A
n 1 (mol HCl( g)/h)
0°C
B
C
condenser
n 3 (mol HCl( g)/h)
n 4 (mol C 2H 4( g)/h)
n 5 (mol C 2H 6( g)/h)
0°C
[mol
Cl(l)/h]
n 6 n(mol
C C2H2 H5Cl(
g)/h)
6
5
reactor
n 2 (mol/h) at 0°C
0.93 C 2H 4
0.07 C 2H 6
( n 6 – 24,800) (mol C 2H 5Cl( l)/h)
0°C
D
24,800 mol C2 H 5Cl( l )/h
Product composition data:
(1)
( 2)
( 3)
n&3 = 0.015n&1
n&4 = 0.015 ( 0.93n&2 ) = 0.01395 n&2
n&5 = 0.07 n&2
Overall Cl balance :
n&1 ( mol HCl h )
1 mol Cl
= ( n&3 ) (1) + ( 24800 )(1)
1 mol HCl
( 4)
Solve (4) simultaneously with (1) ⇒ n&1 = 25180 mol h = 25.18 kmol HCl fed/h
n&3 = 378 mol HCl ( g ) h
Overall C balance :
n& 2 ( 0.93 )( 2 ) + n&2 ( 0.07 )( 2 ) = 2n&4 + 2 n&5 + ( 2 )( 24800)
From Eqs. (2) and (3) ⇒ 2n&2 0.93 + 0.07 − 0.0139 − 0.07  = ( 2 )( 24800 )


n& 2 = 27070 mol fed h = 27.07 kmol h of Feed B
b.
c.

n&3 = 378 mol HCl h
 2.65 kmol / h of Product C
n& 4 = 0.01395 ( 27070 ) =378 mol C 2H4 h 
14.3% HCl, 14.3% C2 H 4 , 71.4% C 2 H 6
n&5 = 0.07 ( 27070 ) =1895 mol C2 H6 h 
References : C 2 H 4 bg g, C 2 H 6 bg g, C 2 H 5 Clbg g, HClbg g at 0 o C
50
C 2 H4 eg, 50o Cj : H$ = zC p dT
0
Table B.2
⇒
2181
.
kJ mol
9- 36
9-30 (cont’d)
50
o
C2 H6 ( g, 50C
) : Hˆ = ∫ C p dT ⇒ 2.512 kJ mol
Table B.2
0
HCleg, 50o Cj : H$ =
Table B.2
50
zC pdT
0
⇒ 1.456 kJ mol
C 2 H5Clel, 0o Cj : H$ = − ∆ H$ v e0o Cj = −24.7 kJ mol
50
C 2 H5Cleg, 50o Cj : H$ = zC pv dT = 2.709 kJ mol
0
n& out
Hˆ i n
mol/h
kJ/mol mol/h
HCl
25180
0
378
C2 H 4
25175
0
378
C 2H 6
1895
0
1895
C 2 H 5Cl n&6 − 24800 −24.7
n&6
Energy balance:
substance
∆H = 0 ⇒
⇒
n&in
( )+
n&A ∆Hˆ r 0o C
νA
Hˆ out
kJ/mol
1.456
2.181
2.512
2.709
∑ n& Hˆ − ∑ n& Hˆ
i
i
i
out
( 25180 − 378) mol HCl react
h
i
=0
in
−64.5 kJ
+ (378 )(1.456 ) + ( 378)( 2.181) + (1895)( 2.512 )
1 mol HCl
+ 2.627n&6 − ( n&6 − 24800 )( −24.7 ) = 0 ⇒ n&6 = 80732 mol C2H5Cl h in reactor effluent
80732 mol condensed 24800 mol product
mol
−
= 55932
h
h
h
kmol recycled
= 55.9
h
C2 H 5Cl recycled =
d.
C p is a linear function of temperature.
∆H$ 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 C2 H4 or C2 H6 is absorbed in the ethyl chloride product.
9.31
a.
4NH3 (g) + 5O2 (g) à 4NO(g) +6H2 O(g)
∆H$ ro = −904 .7 kJ / mol
Basis : 10 mol/s Feed gas
4 mol / s NH3
6 mol / s O 2
n& 3 (mol O2 )
n& 4 (mol NO)
Tin = 200 o C
n&5 (mol H2 O)
Tout
9- 37
9.31 (cont'd)
5 mol O 2 4 mol NH 3 fed
= 5 mol / s ⇒ n& 3 = ( 6 − 1) mol O 2 / s = 1 mol O 2 /
4 mol NH 3
s
4 mol NO produced 4 mol NH 3 fed
NO produced : n& 4 =
= 4 mol NO / s
4 mol NH 3
s
O 2 consumed :
H 2 O produced : n& 5 =
6 mol H 2O produced 4 mol NH 3 fed
4 mol NH 3
Extent of reaction : ξ& =
b.
s
( n& NH 3 ) out − (n& NH3 ) in
ν NH 3
=
0 −4
4
= 6 mol H 2 O / s
= 1 mol / s
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.
References : NH3 ( g), O 2 ( g), NO(g), H2 O(g) at 25o C, 1atm
c.
Substance
NH 3 ( g)
O 2 ( g)
NO(g)
H$ out
n& out
( mol / s) ( kJ / mol) ( mol / s) ( kJ / mol)
4.00
H$ 1
−
−
$
6.00
H2
1.00
H$ 3
−
−
4.00
H$
4
−
H 2O(g)
−
H$ 5
6.00
Table B.2
200
H$ 1 =
H$ in
n& in
Table B.8
B
z (C p ) NH
3
dT
=
6.74 kJ / mol,
H$ 2 = H$ O2 ( 200 o 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: ∆H& = 0
∆H& = ξ& ∆H$ ro +
5
∑
( ni ) out ( H$ i ) out −
i =3
2
∑ (n
$
i ) in ( Hi ) in
i =1
⇒ ∆H& = ξ&∆Hˆ 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& , ∆Hˆ r , and Hˆ1 through Hˆ 6
∆H& = (0.3479 Tout + 4.28 × 10 −5 Tout 2 + 0.9285 × 10 −8 T out 3 − 4.697 × 10 −12 Tout 4 )
− 972.24 kJ/mol = 0
E-Z Solve ⇒ Tout = 2223 o C
9- 38
9-31 (cont’d)
T
If only the first term from Table B.2 is used, H$ i =
d.
z (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 = 00295
.
( Tout − 25) , H$ 5 = 0.03346( Tout − 25)
ˆ2 =0
E.B. ∆H& = ξ&∆Hˆ ro + (1.00) Hˆ 3 + (4.00) Hˆ 4 + (6.00)Hˆ 5 − (4.00)Hˆ 1 − (6.00)H
⇓
o
Substitute for ξ& (=1 mol/s), ∆Hˆ r ( = − 904.7 kJ/mo l) and Hˆ1 through Hˆ 6
0=0.3479 Tout − 969.86 ⇒ Tout = 2788 o C ⇒ % error=
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).
e.
9.32
Basis : 100 lb m coke fed
⇒ 84 lb m C ⇒ 7.00 lb - moles C fed ⇒ 7.00 lb - moles CO2 fed
7.00 lb-moles CO 2
400°F
n1 (lb-moles CO)
n2 (lb-moles CO 2)
1830°F
n3 lb-moles C( s)/hr
16 lb mash/hr
1830°F
7.00 lb-moles(84 lb )C/hr
m
16 lb mash/hr
77°F
585,900 Btu
a.
Cbsg + CO 2 bgg → 2CObgg ,
∆H$ ro e77o Fj = e∆ H$ co j
= 25° C
=
CO 2 bg g
− 2e∆H$ co j
CObg g
−393.50 − b2gb−282 .99g kJ 0.9486 Btu 453.6 mols
mol
1 kJ
1 lb - mole
= 74,210 Btu lb - mole
Let x = fractional conversion of C and CO2 :
E
n1 =
7.00xblb - moles C reactedg 2 lb - moles CO formed
= 14.0 x lb- moles CO
1 lb - mole C reacted
n2 = 7.00b1 − x g lb- moles CO 2
n3 = 7.00b1 − x g lb- moles Cbsg
o
References for enthalpy calculations: Cbsg, CO2 bgg, CObgg, ash at 77 F
o
CO2 bg ,400° Fg: H$ = H$ CO2 ( 400 F)
Table B.9
⇒ 3130 Btu lb - mole
o
CO 2 bg,1830° Fg: H$ = H$ CO2 (1830 F)
CObg,1830° Fg: H$ = H$ C O (1830o F)
Table B.9
⇒
Table B.9
20,880 Btu lb - mole
⇒ 13,280 Btu lb - mole
9- 39
9.32 (cont'd)
b1830 − 77 g° F
0.24 Btu
Solid b1830° F g: H$ =
lb m ⋅ o F
= 420 Btu lb m
Mass of solids (emerging)
=
7.00b1 − xg lb - moles C
12.0 lb m
1 lb - mole
+ 16 lb m = b100 − 84 xg lb m
nin
nout
H$ in
H$ out
(lb − moles) ( Btu lb - mole) (lb − moles) ( Btu lb- mole)
7.00
3130
7.00b1 − xg
20,890
−
−
14.0x
13,280
(lbm )
(Btu lbm )
(lb m )
(Btu lbm )
100
0
100 − 84x
420
substance
CO2
CO
solid
Extent of reaction: n CO = ( n CO ) o + ν CO ξ ⇒ 14.0 x = 2ξ ⇒ ξ ( lb - moles) = 7.0 x
Energy balance:
Q = ∆H = ξ ∆H$ ro +
∑ n H$ − ∑ n H$
i
i
i
out
585,900 Btu =
i
in
7.0 x (lb - moles)
74,210 Btu
+ 7.00a1 − x f a20,880 f
lb - mole
+ a14. 0x f a13,280f + a100 − 84 x f a420f − a7.00f a3130 f
E
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.
9.33
17.1 m3 103 L 273 K 5.00 atm
1 mol
= 3497 mol h feed
3
h
1m
298 K 1.00 atm 22.4 LaSTP f
CObgg + 2H 2 bgg → CH 3OHbgg ,
Basis :
∆H$ ro = e∆H$ fo j
CH 3OHbg g
− e∆H$ fo j
CO(g)
3497 mol/h
0.333 mol CO/mol
0.667 mol H /mol
2
25°C, 5 atm
= −9068
. kJ mol
n1 (mol CH 3OH /h)
n2 (mol CO/h)
n3 (mol H 2/h)
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).
9- 40
9.33 (cont’d)
CO reacted : =
( 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
CH 3OH produced : n&1 =
= 1166 f mol CH 3OH h
1 mol CO
CO remaining : n& 2 = 1166 a1 − f f mol CO h
H2 remaining : n& 3 = b3497gb0.667 g mol H 2 fed −
1166 f mol CO react 2 mol H 2 react
1 mol CO react
= 2332 b1 − f g mol H 2 h
Reference states : CO(g), H 2 bgg , CH 3OHbgg at 25°C
Substance
bmol
CO
H2
CH3OH
H$ in
n& in
hg
bkJ
1166
2332
−
H$ out
n& out
bmol h g
mol g
0
1166a1 − f f
0
2332a1 − f f
−
1166 f
bkJ
mol g
H$ 1
H$ 2
H$ 3
Table B.8
B
COeg,127 Cj : H$ 1 = H$ CO (127o C) = 2.99 kJ mol
o
Table B.8
B
H2 eg,127 Cj : H$ 2 = H$ H2 (127o C) = 2.943 kJ mol
o
122
CH 3OH(g,127 C): H$ 3 =
o
Table B.2
B
z C pdT
25
& = ξ& ∆H$ o +
Energy balance : Q& = ∆ H
r
= 5.009 kJ / mol
∑ n& H$ − ∑ n& H$
i
out
⇒
i
i
i
in
−17.05 kJ 3600 s
kJ
= (1166 f )( −90.68)
+ 1166b1 − f g b2.99g
s
1h
h
+ 2332b1 − f
g b2 .993g +
1166 f b5009
. g bkJ hg
⇒ 1102
.
× 10 f = 7173
.
× 10 ⇒ f = 0.651 mol CObor H 2 g converted mol fed
5
4
n&1 = 1166b0.651g = 759 .1 mol h
n& 2 = 1166b1− 0.651g = 406.9 mol h
n& 3 = 2332b1 − 0.651g = 8139
. mol h
E
n& tot = 1980
9.34
a.
mol
1980 mol
⇒ Vout =
h
h
22.4 LbSTP g 400 K 1.00 atm 1 m 3
= 13.0 m 3 h
1 mol
273 K 5.00 atm 103 L
CH 4 bgg + 4Sbgg → CS 2 bgg + 2 H2 Sbgg , ∆H$ r ( 700 ° C ) = −274 kJ mol
Basis : 1 mol of feed
9- 41
9.34 (cont'd)
1 mol at 700°C
0.20 mol CH /mol
Reactor
4
0.80 mol S/mol
Q = –41 kJ
Product gas at 800°C
n 1 (mol CS2)
n 2 )(mol H 2S)
n 3 (mol CH 4 )
n 4 (mol S (v))
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
n3 = 0.20b1 − f g mol CH 4
n4 = 0.80 mol S fed −
0.20 f bmol CH 4 react g
4 mol S react
= 080
. b1 − f g mol S
1 mol CH 4 react
1 mol CS 2
= 020
. f mol CS2
1 mol CH 4
0.20 f mol CH 4 react 2 mol H 2 S
n2 =
= 0.40 f mol H 2S
1 mol CH 4
n1 =
0.20 f mol CH 4 react
References: CH 4 (g), Sbgg, CS 2 (g), H2S(g) at 700°C (temperature at which ∆H$ r is known)
H$ in
substance nin
bmolg
H$ out
nout
CH4
0.20
b molg
molg
0
0.20b1 − f g
S
0.80
0
0.80b1 − f g
CS2
−
−
0.20 f
H2S
−
−
0.40 f
bkJ
H$ out = Cpi ( 800 − 700) ⇒
b kJ
molg
$
H1
H$
2
H$ 3
H$
4
CH 4 bg, 800 ° Cg: H$ 1 = 7.14 kJ / mol
Sbg, 800° Cg: H$ 2 = 3.64 kJ / mol
CS bg, 800° Cg: H$ = 3.18 kJ / mol
2
3
H 2Sbg, 800° Cg: H$ 4 = 4.48 kJ / mol
Energy balance on reactor:
Q& = ∆ H& = ξ& ∆H$ r +
∑ n& H$ − ∑ n& H$
i
out
b.
=
. f gb− 274.0g
b020
b1g
i
i
i
= 41
in
kJ
s
+ 0.20b1− f gb7140
. g + 080
. b1 − f gb3640
. g + 0.20 f b3180
. g + 0.40 f b4.480g
⇒ f = 0.800
9- 42
9.34 (cont'd)
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.
References : CH 4 (g), Sblg, CS 2 (g), H 2S(g) at 200°C
H$ in
nin
Substance
bmolg
nout
bmolg
bCH 4 g
0.20
bCH 4 g
0.04
mol g
$
H1
H$
2
0.04
0
080
.
016
.
016
.
0.32
H$ 3
H$ 4
H$ 5
H$
016
.
0.80
016
.
0.32
0
$
H8
0
0
150°, 700 °
800°,200 °
Sblg
Sbgg
CS 2
H 2S
bkJ
H$ out
6
0.20
bkJ
molg
$
H7
H$ i = C pi aT − 200 f for all substances but S
= dC p i
= dC p i
S al f
aT − 200 f
for Salf
F
I
∆ H$ v bTb g + dCp i
aT
G444.6 − 200J +
Sbg g
Tb
K = 83. 7 kJ mol
S al f H
− 444.6f for Sbgg
c.
CH 4 bg, 150° Cg: H$ 1 = − 3.57 kJ / mol
CH 4 bg, 800 ° Cg: H$ 2 = 42.84 kJ / mol
Sbl, 150° Cg: H$ = − 1.47 kJ / mol
CS2 bg, 800 ° Cg: H$ 5 = 19.08 kJ / mol
H 2Sbg, 800° Cg: H$ 6 = 26.88 kJ / mol
CH bg, 700 ° Cg: H$ = 35.7 kJ / mol
3
4
Sbg, 800° Cg: H$ 4 = 103.83 kJ / mol
Energy balance: Q bkJ g =
∑ n H$ − ∑ n H$
i
out
7
Sbg, 700° Cg: H$ 8 = 100.19 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- 43
9.35
Basis : 1 mol C 2 H 6 fed to reactor
1 mol C H
2
6
1273 K, P atm
a.
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)
C2 H6 ⇔ C2 H4 + H 2 , K p =
x C2 H 4 x H 2
P = 7.28 × 106 exp[−17,000 / T ( K)]
xC 2 H 6
(1)
Fractional conversion = f bmols C 2 H 6 react mol fed g
ξ(mol) = f
U
|
nC 2 H 6 = b1 − f gbmol C 2 H 6 g|
|
nC 2 H 4 = f bmol C 2 H 4 g
V⇒
|
n H2 = f b mol H 2 g
|
n = 1 + f b mols g
|
W
f
Kp =
e1 −
b.
f
x C2 H 4 x H 2
x C 2 H4
2
j Kp
F
= f P⇒ f =G
2
2
b1 + f
P⇒ Kp =
1 − f mol C2 H 6
1+ f
mol
f mol C2 H 4
x C2 H 4 =
1+ f
mol
f mol H 2
xH2 =
1 + f mol
x C2 H6 =
2
g
b1−
fg
b1+
fg
Kp
P
=
f 2P
f2
=
P
2
b1 − f gb1 + f g 1 − f
12
I
b2 g
J
HP + K p K
References : C 2 H 6 bgg, C 2 H 4 bgg, H 2 bgg at 1273 K
Energy balance:
∆H = 0 ⇒ ξ ∆ H$ r b1273 K g +
ni H$ i −
ni H$ i
∑
∑
out
$
eH i j
in
= 0 binlet temperature = reference temperatureg
T
$
eH i j
in
=z
out
1273
C pi dT
⇓ energy balance
f ∆H$ r b1273 K gkJ + b1 − f
T
z1273
g
dC p i
dT + f
C2H 6
T
z1273
T
dC p i
C2 H4
dT + f z
1273
rearrange, reverse limits and change signs of integrals
1273
1− f
=
f
∆H$ r b1273K g − z
T
1273
dC p i
C2 H 4
dT − z
T
dC p i
1273
H2
dT
Cp i
dT
zT d4
C2 H 6
1444444444
2444444444
4
3
φ bT g
1− f
1
= φ bT g ⇒ 1 − f = fφ bT g ⇒ f =
f
1 + φbT g
9- 44
b4 g
b3 g
dC p i
H2
dT = 0
9.35 (cont'd)
φbT g =
F
Kp
G
H1 +
1273
T
T
e26.90 + 4.167 × 10
−3
T j dT
1273
zT b11.35 + 0.1392T gdT
⇒ φbT g =
c.
1273
145600 − z b9.419 + 01147
.
T gdT − z
12
I
J
Kp K
3052 + 36.2T + 0.05943T 2
127240 − 113
. T − 0.0696T 2
F Kp I
1
=
⇒G
J
1 + φ bT g H1 + K p K
12
−
1
= ψ bT g = 0
1 + φ bT g
φbT g given by expression of Part b. K p bT g given by Eq. (1)
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- 45
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 = 120000
.
PSI = 08226
.
E + 00
T = 115000
.
T = 110000
.
T = 105000
.
T = 100000
.
T = 950.00
T = 959.80
T = 960.25
T = 960.27
Solution: T = 960.3 K ,
PSI = 0.7048E + 00
PSI = 05551
.
E + 00
PSI = 0.3696E + 00
PSI = 01619
.
E + 00
PSI = − 03950
.
E − 01
PSI = − 01824
.
E − 02
PSI = − 07671
.
E − 04
PSI = − 03278
.
E − 05
f = 0.360 mol C 2 H 6 reacted mol fed
9- 46
2CH 4 → C 2 H 2 + 3H 2
C 2 H 2 → 2C(s) + H 2
9.36
a.
?
?
?
?
?
?
?
?
?
?
9- 47
9.46
a.
H2 SO4 baqg + 2NaOH baqg → Na 2SO4 baq g + 2H2 Oblg
H 2SO 4 solution::
4 mol H 2SO 4
1L
75 mL
= 0.30 mol H 2SO 4
3
1 L acid soln 10 mL
b75 mL gb1.23 g mLg = 92.25 g, (0.3 mol H 2 SO 4 ) b98.08 g mol g = 29 .42 g H 2 SO 4
75 ml of 4M H 2SO 4 solution ⇒
⇒ b92.25 − 29.42g g H 2 O ⇒ b62.83 g H 2 O gb1 mol 18.02 gg = 3.49 mol H 2 O
⇒ r = 3.49 mol H 2 O 0.30 mol H 2 SO 4 = 11.63 mol H 2 O / mol H 2SO 4
Table B.1,
Table B.11
$o
e∆H f j
soln
= e∆H$ fo j
+ e∆H$ of j
H2 SO4 bl g
B
=
H 2SO 4 baq., r =11. 63 g
b−811.32 − 67 .42 g
kJ
mol
= −87874
. kJ mol H 2SO 4
NaOH solution required:
0.30 mol H 2S O 4
b50.00
2 mol NaOH
1 mol H 2 SO 4
10 3 m L
= 50.00 mL NaOH b aq g
1L
1 L NaOH(aq )
12 mol NaOH
mLgb137
. g mLg = 685
. g
40 g/ mol NaOH
12 mol NaOH
1L
50 mL
=
0
.
60
mol
NaOH
24.00 g NaOH
1 L NaOH(aq) 103 mL
⇒
⇒ b68.5 − 24.00g g H2 O ⇒ b44.5 g H 2Ogb1 mol 18.02 gg = 2.47 mol H 2 O
⇒ r = 2.47 mol H 2O 0.6 mol NaOH =
$o
e∆H f j
soln
= e∆H$ fo j
NaOH bs g
+ e∆H$ so j
4.12 mol H 2 O
mol NaOH
NaOHbsgbaq., r =4.12 g
= b−426.6 − 3510
. g
kJ
mol
= −46170
. kJ mol NaOH
Na 2SO 4 baqg:
$o
kJ
= −13857
. kJ mol Na 2SO 4
soln
Na 2SO4 bsg
Na 2SO 4 baqg
mol
= total mass of reactants or products = (92.25g H 2SO 4 soln + 68.5g NaOH) = 160.75g = 0.161 kg
e∆H f j
mtotal
= e∆H$ fo j
+ e∆H$ fo j
= b−1384.5 − 117
. g
Extent of reaction: (nH 2S O ) final = ( nH2 SO4 ) fed + ν H2 SO4 ξ ⇒ 0 = 0.30 mol − (1)ξ ⇒ ξ = 0.30 mol
4
Standard heat of reaction
∆H$ ro = e∆H$ of j
+ 2e∆H$ fo j
Na 2SO 4 baqg
H 2 O bl g
− e∆H$ of j
H 2SO 4 baqg
− 2e∆H$ fo j
NaOH baq g
Energy Balance : Q = ∆H = ξ∆H$ ro + mtotal C p ( T − 25) o C
F
kJ I
H
kg o C K
= ( 030
. mol) (1552
. kJ / mol) + (0161
. kg)G4.184
b.
Volumes are additive.
Heat transferred to and through the container wall is negligible.
9- 55
J (T − 25)
o
C = 0 ⇒ T = 94o C
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
90% SO 2 removal: n& 4 = 0100
. a0.00300 f b50, 000 mol hg = 15.0 mol SO2 h
n&5 = a0.997f b50, 000 mol hg = 49,850 mol N2 h
N 2 balance:
. gbn&1 g =
b2gb0100
NH +4 balance:
n&2 + b15
. gb2gn&2 ⇒ n&1 = 20n&2
U
n&1
|
V⇒
n&2
0100
. n&1 + b000300
.
. + n& 2 + 15
. n&2 W
gb50,000 g = 150
|
S balance:
H2 O balance: n& 3 = b0900
. gb5400g−
= 5400 mol h
= 270 mol NH 4 HSO 3 h
270 mol NH 4 HSO3 produced
1 mol H 2O consumed
h
2 mol NH4 HSO3 produced
= 4725 mol H 2 Oblg h
Heat of reaction:
∆H$ ro = 2e∆H$ fo j
NH4 HSO4 baqg
− e∆H$ fo j
bNH 4 g
SO3 baq g
2
− e∆H$ fo j
SO2 ( g )
− e∆ H$ fo j
H 2 O(l)
= 2b−760g − b−890g − b−29690
. g − b− 28584
. g kJ mol = − 47.3 kJ mol
References : N 2 bgg, SO 2 bgg, bNH4 g2 SO 3baqg, NH4 HSO 3baqg, H 2Oblg at 25°C
50
o
SO 2 eg, 50 Cj : H$ = z dC p i
dT = 101
. kJ mol ( C p from Table B.2)
SO 2
25
35
o
SO 2 eg, 35 Cj : H$ = z dC p i
dT = 0.40 kJ mol
SO2
25
N 2 eg, 50 Cj : H$ = 073
. kJ mol (Table B.8)
o
o
N 2 eg, 35 Cj : H$ = 0292
.
kJ mol
Entering solution: H$ = 0
Effluent solution at 35°C
m& bg h g =
+
270 mol NH4 HSO 3
h
99 g
mol
1.5 × 270 mol bNH 4 g2 SO 3 116 g 4725 mol H 2O 18 g
g
+
= 159 ,000
h
1
mol
h
h
mol
159,000 g 4 J
nH
& $ = mC p∆ T =
h
g⋅° C
a35 − 25f ° C
1 kJ
103 J
= 6360 kJ / h
Extent of reaction:
( n& NH 4 HSO3 ) out = (n& NH4 HSO3 ) in + ν NH 4 HSO3 ξ& ⇒ 270 mol / h = 0 + 2ξ& ⇒ ξ& = 135 mol / h
9- 56
9.47 (cont'd)
Energy balance: Q& = ∆H& = ξ& ∆H$ ro +
Q=
+
9.48
a.
135 mol
h
−47.3
mol
effluent solution
6360
∑n& H$ − ∑ n& H$
i
i
i i
in
SO2 out
N 2 out
kJ
+ a15f a0.40f + a49,850 f a0.292f
out
− a50,000f a0.003 f a1.01f − a49,850f a0.73f =
−22, 000 kJ
1h
1 kW
= − 611
. kW
h
3600 s 1 kJ s
CH 4 bgg + 2O2 bgg → CO2 ( g) + 2H 2Obvg
Table B.1 HHV
− ∆H$ co
HHV = 890.36 kJ / mol, LHV =
at 25 °C
B
B
− 2 e∆H$ v j
H2 O
= 890.36 − 2b44.01g kJ mol
= 802 .34 kJ mol CH 4
7
C 2 H4 bgg + O2 ( g) → 2CO 2 ( g) + 3H 2O(v)
2
HHV = 1559.9 kJ / mol, LHV = 1559.9 − 3a44.01f kJ mol = 142787
. kJ mol C 2 H6
C 3H 8 bgg + 5O2 (g) → 3CO2 ( g) + 4H 2O(v)
HHV = 2220 .0 kJ / mol, LHV = 2220.0 − 4a44. 01f kJ mol = 2043.96 kJ mol C 3H 8
bHHV gnatural gas
= b0.875gb890.36 kJ molg + b0.070 gb1559.9 kJ mol g + b0.020gb2200.00 kJ mol g
= 933 kJ mol
b LHV gnatural gas
= b0.875gb802.34 kJ molg + b0.070 gb1427.87 kJ molg + b0020
. gb204396
. kJ molg
= 843 kJ mol
b.
F
1 mol natural gas ⇒ [b0.875 mol CH 4 gG16.04
H
F
+ b0.020 mol C3 H8 gG44.09
H
⇒
c.
843 kJ
mol
g I
g I
F
.
J + b0 .070 mol C2 H 6 gG3007
J
H
mol K
mol K
g I
g I
1 kg
F
.
mol N 2 gG28.02
= 0.01800 kg
J + b0035
J ]×
H
mol K
mol K 103 g
1 mol
= 46800 kJ kg
0.01800 kg
The enthalpy change when 1 kg of the natural gas at 25o C is burned completely with oxygen at
25o C and the products CO2 (g) and H2 O(v) are brought back to 25o C.
9.49
Table B.1
B
Cbsg + O 2 (g) → CO2 ( g),
∆ H$ co
= e∆H$ fo j
CO2 ( g)
=
−393.5 kJ
1 mol
103 g
mol
12.01 g
1 kg
Table B.1
B
Sbsg + O 2 (g) → SO 2 ( g), ∆H$ co = e∆H$ fo j
SO2
= −296.90 kJ mol
MSO = 32 . 064
2
B
⇒
− 9261 kJ / kg S
Table B.1
B
1
H 2 bgg + O 2 ( g) → H 2 Oblg, ∆H$ co = e∆H$ fo j
= −285.84 kJ mol H 2
H 2 O bl g
2
9- 57
= − 32,764 kJ kg C
M H =1. 008
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
A
in water
F
Eq. (9.6-3) ⇒ HHV = 32,764 C + 141, 790GH −
H
OI
J + 9261S
8K
This formula does not take into account the heats of formation of the chemical constituents of
coal.
b.
C = 0. 758 , H = 0. 051 , O = 0. 082 , S = 0. 016 ⇒ b HHV gDulong = 31, 646 kJ kg coal
1 kg coal ⇒
φ=
c.
0.016 kg S
64 .07 kg SO 2 formed
32. 06 kg S burned
= 0. 0320 kg SO 2 kg coal
0.0320 kg SO 2 kg coal
= 101
. × 10−6 kg SO 2 kJ
31,646 kJ kg coal
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.
CH 4 + 2O2 → CO2 + 2H 2Oblg , HHV = − ∆H$ co = 89036
. kJ mol bTable B.1g
9.50
7
C 2 H 6 + O2 → 2CO2 + 3H 2Oblg , HHV = 1559 . 9 kJ mol
2
1
CO + O2 → CO 2 , HHV = 282 . 99 kJ mol
2
2.000 L
273.2K
2323 mm Hg
1 mol
Initial moles charged:
= 0.25 mol
a
25
+
273.2
f
K
760
mm
Hg
22
.4
LaSTP f
(Assume ideal gas)
Average mol. wt.: ( 4.929 g) (0.25 mol) = 19.72 g / mol
Let x1 = mol CH 4 mol gas , x2 = mol C 2 H 6 mol gas c⇒ b1 − x1 − x2 gmol CObmol gasgh
MW = 19.72 ⇒ x1 b16.04 g mol CH 4 g + x2 b3007
. g + b1− x1 − x2 gb28.01g = 19.72
HHV = 9637
. kJ mol ⇒ x1 b890.36g + x 2 b1559.9g + b1 − x1 − x 2 gb282.99g = 963.7
b1g
b2 g
Solving (1) & (2) simultaneously yields
x1 = 0.725 mol CH 4 mol, x 2 = 0.188 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) + 2O2 (g) → CO 2 (g) + 2H2 O(v), ∆H$ co = −890.36 kJ / mol
7
C 2 H 6 (g) + O2 (g) → 2CO2 (g) + 3H2 O(v), ∆ H$ co = −1559.9 kJ / mol
2
Excess O2 , 25°C
n& 2 , mol CO2
n& 3 , mol H 2O
n& 4 , mol O2
25°C
1 mol/s fuel gas, 25°C
85% CH4
15% C2 H6
9- 58
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 C2 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
C balance : n& 2 = b0.85gb1g + b015
. gb2g ⇒ n& 2 = 115
. mol CO 2 / s
H balance : 2n& 3 = b0.85gb4g + b015
. gb6g ⇒ n& 3 = 215
. mol H 2 O / s
10% excess O 2 ⇒ n& 4 = b01
. gb2.225g mol O2 / s = 0.223 mol O 2 / s
Extents of reaction: ξ& 1 = n& CH 4 = 0.85 mol / s, ξ& 2 = n& C2 H 6 = 015
. mol / s
Reference states: CH4 bgg, C2 H 6 bgg, N 2 bgg, O2 bgg, H 2 Oblg, CO 2 (g) at 25o C
(We will use the values of ∆H$ co given in Table B.1, which are based on H2 Oblg as a
combustion product, and so must choose the liquid as a reference state for water)
Substance
CH4
C2 H 6
O2
CO2
H 2Obvg
n& in
n&out
H$ in
H$ out
mol kJ mol mol kJ mol
085
.
0
−
−
015
.
0
−
−
2.225
0
0.223
0
−
−
115
.
0
−
−
215
.
H1
H$ 1 = ∆H$ v e25o Cj = 44.01 kJ / mol
Energy Balance :
Q& = n&
∆H$ o
CH4 e
c j
C H4
+ n& C 2 H6 e∆H$ co j
C2 H 6
+
∑ n& H$ − ∑ n& H$
i
out
i
i
i
in
= b0.85 mol / s CH 4 gb−890.36 kJ mol g + b0.15 mol / s C 2 H 6 gb− 1559.9 kJ molg
+ b2.15 mol / s H 2 O gb44.01 kJ / molg = − 896 kW
⇒ − 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 H6
Theoretical oxygen = 2.225 mol O 2
Assume 10% excess O 2 ⇒ O 2 fed = 1.1× 2.225 = 2.448 mol O2
C balance : n2 = b0.85gb1g + b015
. gb2g ⇒ n2 = 115
. mol CO2
H balance : 2n3 = b085
. gb4g + b015
. gb6g ⇒ n3 = 215
. mol H2 O
10% excess O2 ⇒ n4 = b01
. gb2.225g mol O 2 = 0.223 mol O 2
9- 59
9.51 (cont' d)
Reference states: CH4 bgg, C2 H 6 bgg, N 2 bgg, O2 bgg, H 2 Oblg, 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
CH4
085
.
0
−
−
C2 H 6
O2
CO2
H 2Obvg
015
.
2.225
−
−
−
0.223
115
.
215
.
0
0
−
−
−
0
0
$
U1
8.314 J
U$ 1 = ∆U$ v e25o Cj = ∆H$ v e25o Cj − RT = 44.01 kJ / mol −
Eq. (9.1-5) ⇒
∆U$ co
=
∆H$ co
− RT (
∑ν
i
gaseous
products
⇒ e∆U$ co j
CH 4
= b− 890.36 kJ molg −
$o
e∆U c j
C2 H6
∑ν
−
i
1 kJ
298 K
mol K 1000 J
= 4153
.
kJ
mol
)
gaseous
reactants
298 K (1 + 2 − 1 − 2)
8.314 J
1 kJ
3
= − 890.36
kJ
mol
mol K
10 J
8.314 J 298 K (3 + 2 − 35
. − 1) 1 kJ
kJ
= b− 1559.9 kJ mol g −
= −156114
.
3
mol
mol K
10 J
Energy balance:
Q = ∆ U = nCH 4 e∆U$ co j
CH 4
+ nC2 H6 e∆U$ co j
C2H 6
+
∑ n U$ − ∑ n& U$
i
out
i
i
i
in
= b0.85 mol / s CH 4 gb−890.36 kJ molg + b015
. mol / s C 2 H 6 gb−156114
. kJ molg
+ b2.15 mol / s H 2 Ogb4153
. kJ / mol g = −902 kJ
⇒ −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 ∆H& and ∆ U .
a.
n& fuel ( − ∆H$ co ) = W& s − Q& l (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 103 g
L
1 J/s
1.341 × 10− 3 hp
49 kJ
1 kg
1 kJ
103 J
g
3600 s
h
+
15 × 106 kJ
⇒ V& = 25
. 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- 60
9.53
a.
Energy balance: ∆U = 0 ⇒
nblb m fuel burnedg ∆U$ co (Btu)
lb m
+ mCv bTout − 77° F g = 0
⇒ a0.00215f ∆U$ co +b4.62 lbm gb0.900 Btu lb m ⋅° F ga87.06° F − 77.00° F f = 0
⇒ ∆U$ co = −19500 Btu lb m
b.
The reaction for which we determined ∆ U$ co is
1 lb m oil + aO2 ( g) → bCO2 (g) + cH2 O(v)
(1)
The higher heating value is ∆H$ r for the reaction
1 lb m oil + aO 2 ( g) → bCO2 ( g) + cH2 O(l)
(2)
o
Eq. (9.1-5) on p. 441 ⇒ ∆H$ co1 = ∆U$ c1
+ RT ( b + c − a)
$ o + c∆H$ ( H O, 77o F)
Eq. (9.6-1) on p. 462 ⇒ − ∆H$ co2 = − ∆H
c1
v
2
( 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 CO2 formed when 1 lb m fuel oil is burned
c = lb - moles of H 2O formed when 1 lbm fuel oil is burned
9.54
a.
3
CH 3 OHbvg + O 2 ( g) → CO 2 ( g) + 2H 2 Oblg
2
∆H$ ro = e∆H$ co j
CH3 OHbv g
= −764 .0
kJ
mol
Basis : 1 mol CH 3OH fed and burned
1 mol CH 3OH( l)
25°C, 1.1 atm
vaporizer
Q2 (kJ)
1 mol CH 3OH( v)
100°C
1 atm
Q 1 (kJ)
reactor
n 0 (mol O 2)
3.76 n 0 (mol N 2)
100°C
Overall C balance:
1 mol CH 3OH
Effluent at 300°C, 1 atm
n p (mol dry gas)
0.048 mol CO /mol
D.G.
2
0.143 mol O /mol
D.G.
2
0.809 mol N /mol
D.G.
2
n w (mol H 2O)
1 mol C
= n p b0048
. gb1g ⇒ n p = 20.83 mol dry gas
1 mol CH3OH
N2 balance: 3.76n0 = b2083
. gb0.809g ⇒ n0 = 4.482 mol O 2
Theoretical O 2:
% excess air =
H balance:
b1
b1 mol
CH 3OH gb1.5 mol O 2 mol CH 3OH g = 15
. mol O 2
(4.482 − 15
. ) mol O 2
× 100% = 200% excess air
15
. mol O2
mol CH 3OHgb4 mol H 1 mol CH 3OH g = nw b2g ⇒ nw = 2 mol H 2O
(An atomic O balance ⇒ 9.96 mol O = 9.96 mol O , so that the results are consistent.)
p∗w =
Table B.3
nw
2 mol H 2 O
×P=
× 760 mm Hg = 66.58 mm Hg = pw∗ dTdp i ⇒ Tdp = 44.1° C
nw + n p
a2 + 20.83f mol
9- 61
9.54 (cont'd)
b.
Energy balance on vaporizer:
L
64.7
M
Q1 = ∆H = n∆H$ = 1 mol Mz
M
N
25
C pl
A
O
kJ
Cpv dT P
P
64.7
mol
A
P
Table B.2
Q
100
dT + ∆ H$ v + z
A
Table B.2
Table B.1
= 40.33 kJ
References : CH 3 OH av f , N 2 (g), O 2 (g), CO 2 (g), H 2 O alf at 25° C
n in
n out
H$ in
H$ out
Substance
(mol)
(mol)
(kJ / mol)
(kJ / mol)
CH 3OH
1.00
3.603
−
−
N2
16.85
2.187
16.85
8.118
4.482
2.235
2.98
8.470
CO 2
−
−
1.00
11.578
H 2O
−
−
2.00
53.58
O2
H$ aT f = H$ i for N 2 , O 2 , CO 2 (Table B.8)
= ∆H$ v d25o Ci + H$ i for H2 Oav f (Eq. 9.6 - 2a on p. 462, Table B.8)
T
= z C p dT for CH3OHavf (Table B.2)
25
(Note: H2 Oblg was chosen as the reference state since the given value of ∆H$ co presumes liquid
water as the product.)
Extent of reaction: ( nCH 3OH ) out = ( nCH3 OH ) in + ν CH 3OHξ ⇒ 0 = 1 mol − ξ ⇒ ξ = 1 mol
Energy balance on reactor: Q2 = ξ∆H$ co +
∑ n H$ − ∑ n H$
i
i
out
i
i
in
= b1gb− 764.0g+ b1685
. gb8118
. g+K− b4.482gb2.235g kJ
bTable
B.1 g
= −534 kJ ⇒ 534 kJ transferred from reactor
9.55
a.
CH 4 + 2O 2 → CO 2 + 2H 2 O
3
CH 4 + O2 → 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 ⇒ n&1 = 0.10 (1000 ) = 100 mol CH 4 s
Theoretical O2 required = 2000 mol/s
10% excess O2 ⇒ O 2 fed=1.1(2000 mol/s)=2200 mol/s
9- 62
9.55 (cont’d)
C balance:
(1000 mol CH4 s )(1 mol C mol CH 4 ) = (100 )(1) + n&3 (1) + 10 n&3 (1) ⇒ n&3 = 81.8 mol CO
s
⇒ 10 n&3 = 818 mol CO2 s
H balance:
(1000 )( 4 ) = (100 )( 4 ) + 2 n&4 ⇒ n&4 = 1800 mol H 2O
O balance:
( 2200 )( 2 ) = 2n&2 + (81.8)(1) + (818 )( 2 ) + (1800 )(1) ⇒ n&2 = 441 mol O2
s
s
References :Casf , H 2 bg g, O 2 bg g, N 2 bg g at 25o C
Hˆ in
n&in
Substance
( mol s ) ( kJ
Table B.1
B
T
n&out
mol )
Hˆ out
( mol s ) ( kJ
mol)
CH 4
1000
−74.85
100
−57.62
O2
2200
2.24
441
11.72
N2
8272
2.19
8272
11.15
CO
−
−
81.8
−99.27
CO2
−
−
818
−377.2
H2O
−
−
1800
−228.63
Table B.2
B
H$ = ∆H$ fo + z Cp dT for CH 4
25
=
∆H$ fo
Table B.8
B
+ H$ i (T) for others
& = ∆H& =
Energy balance: Q
∑ n& Hˆ − ∑ n& Hˆ
i
i
out
b.
9.56
i
i
= −5.85 ×105 kJ s ⇒ −5.85 × 105 kW
in
(increases) ⇒ − Q& A
(i)
Tair
(ii)
(iii)
%XS A ⇒ −Q& B (more energy required to heat additional O2 and N2 to 400o C, therefore
less energy transferred.)
S CO2 CO A ⇒ − Q& A (reaction to form CO2 has a greater heat of combustion and so releases
(iv)
more thermal energy)
& B (more energy required to heat combustion products)
Tstack A ⇒ − Q
A
CH 4 + 2O 2 → CO 2 + 2H2 O, C 2 H 6 +
7
O2 → 2CO 2 + 3H2 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
9- 63
9.56 (cont’d)
a.
N2 balance: 3.76n3 = b100gb0.7352gmol N2 ⇒ n3 = 19.55 mol O2 fed
C balance: n1b1g + n2 b2g = b100gb0.0532gb1g + b100 gb0.0160 gb1g|U
H balance: n1 b4g + n 2 b6g = b100gb01224
.
gb2 g
Vf =
. + 160
. gmol
b372
Theoretical O 2 =
= 160
. mol C 2 H 6
fuel gas 22.4 LbSTPg 298.2 K 1 m 3
= 0130
.
m3
1 mol
273.2 K 103 L
3.72 mol CH4
2 mol O 2 1.60 mol C 2 H 6 3.5 mol O 2
+
= 1304
. mol O 2
1 mol CH 4
1 mol C 2 H 6
3.72 mol CH 4 U 69.9 mole% CH 4
V⇒
. mole% C2 H 6
1.60 mol C 2 H 6 W 301
Fuel composition:
b19.55 − 13.04 gmol
% Excess air:
n1 = 372
. mol CH 4
V⇒
n2
|W
O2 in excess
13.04 mol O 2 required
× 100% = 50% excess air
References : Cbsg, H 2 bgg, O 2 bgg, N 2 bgg at 25° C
b.
CH 4
nin
mol
3.72
H$ in
kJ / mol
− 74.85
nout
mol
−
H$ out
kJ / mol
−
C2H6
160
.
−84.67
−
−
O2
19.55
5.31
7.32
2535
.
N2
7352
.
513
.
7352
.
2386
.
CO
−
−
160
.
− 86.39
CO 2
−
−
532
.
− 3561
.
H 2O
−
−
12.24 −212.78
Substance
Table B.1
B
T
H$ = ∆H$ of + z
Table B.2, for
CH 4 , C2 H6
B
C p dT
25
Table B.8
B
= ∆H$ fo + H$ i (T) for O 2 , N 2 , CO, CO2 , H 2Obvg
Energy balance:
Q = ∆H =
∑ n H$ − ∑ n H$
i
out
i
i
in
i
=
− 2764 kJ
0.130 m 3 fuel
9- 64
= −2.13 × 104 kJ m 3 fuel
9.57
Basis : 50000 lb m coal fed h ⇒
1b - mole C
h
12.01 lbm
= 3039 1b - mole C h
. gb50000g
b0047
101
. = 2327 lb - moles H h (does not include H in water)
. gb50000g
b0037
32.07 = 57.7 lb - moles S h
. gb50000g
b0068
18.02 = 189 lb - moles H2 O h
. gb50000 g = 5900
b0118
a.
b0.730gb50000 glb mC
lb m ash h
50,000 lb mcoal/h
3039 lb-moles C/h
2327 lb-moles H/h
57.7 lb-moles S/h
189 lb-moles H O/h
2
5900 lb mash/h
77°F, 1 atm (assume)
Stack gas at 600°F, 1 atm (assume)
n 2 (lb-moles CO 2/h)
n 3 (lb-moles H 2O/h)
n 4 (lb-moles SO 2/h)
n 5 (lb-moles O 2/h)
n 6 (lb-moles N 2/h)
m 7(lb mfly ash/h)
n1 (lb-moles air/h)
0.210 O 2
0.790 N 2
77°F, 1 atm (assume)
m 8(lb mslag/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 :
O2 required to oxidize carbon bC + O2 → CO 2 g =
3039 lb - moles C 1 lb - mole O2
h
1 lb - mole C
= 3039 lb - moles O 2 h
Air fed: n& 1 =
1.5 × 3039 lb- moles O 2 fed
1 mole air
h
0.210 mole O 2
= 21710 lb - moles air h
& 8 = 2540 lb m slag / h
30% ash in coal emerges in slag ⇒ 0.697m& 8 = 0.30b5900 lb m hg ⇒ m
& 7 = 0.700b5900 g = 4130 lb m fly ash h
⇒ m
C balance: 3039 blb - moles C h g = n& 2 + b0.287 gb2540g 12.01
M CO 2 = 44 .01
⇒ n& 2 = 2978 lb - moles CO 2 h
131
. × 105 lb m CO 2 h
H balance: 2327 blb - moles H hg + b189 gb2g = 2n& 3
⇒ n& 3 = 1352.5 lb - moles H 2 O h
M H2 O =18.02
2.44 × 10 4 lb m H 2 O h
N2 balance: n& 6 = b0.790g21710 lb - moles h = 17150 lb - moles N 2 h
MN = 28 .02
2
481
. × 105 lbm N 2 h
S balance: 57.7blb- moles S hg = b1gn& 4 + 0016
.
b2540g 32.06
⇒ n& 4 = 564
. lb - moles SO 2 h
O balance:
MSO2 = 64 .2
3620 lb m SO 2 h
b189 gb1g + b0.21gb21710gb2g = b2978 gb2g + b1352.5gb1g + b56.4gb2g + 2n& 5
bcoal g
bair g
bC O2 g
⇒ n& 5 = 943 lb - moles O 2 h ⇒ 30200 lb m O 2 h
9- 65
eH 2 O j
bS O2 g
bO 2 g
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 CO2 h
1352.5 lb - moles H 2 O h ⇒ 24400 lb m H 2O h
56.4 lb - moles SO 2 h ⇒
3620 lb m SO 2 h
943 lb - moles O2 h ⇒
30200 lb m O 2 h
17150 lb - moles N 2 h ⇒
48100 lb m N 2 h
674,350 lb m stack gas/h
4130 lb m fly ash h
Check: 50000 + b21710gb29g
⇒
b679600gin
⇔ 674350 + 2540
in
out
⇔ b676900 gout (0.4% roundoff error)
Total molar flow rate = 22480 lb - moles h at 600° F , 1 atm (excluding fly ash)
⇒V=
b.
22480 lb- moles 359 ft 3 aSTPf 1060° R
= 1. 74 × 107 ft 3 h
h
1 lb- mole
492 ° R
References: Coal components, air at 77°F ⇒ ∑ ni H$ i = 0
in
Stack gas: nH$ =
674350 lbm
7.063 Btu
lb - mole ⋅° F 28.02 lbm
h
2540 lb m
Slag: nH$ =
h
b600 − 77 g° F
1 lb - mole
b600 − 77 g° F
0.22 Btu
lb m ⋅° F
= 2.92 × 105 Btu h
Energy balance: Q = ∆H = n coal burned ∆H$ co b77 ° Fg +
∑ n H$ − ∑ n H$
i
i
i
out
=
= 8.90 × 107 Btu h
5 × 104 lbm
−18
. × 104 Btu
h
lb m
i
in
+ e8.90 × 107 + 2.92 × 105 j Btu h
= −8.11 × 10 Btu h
8
Power generated =
c.
b035
. ge811
. × 10 j Btu
Q$ = e−811
. × 108 Btu hj
⇒
8
h
b5000
1 hr
1W
3600 s 9.486 × 10
−4
1 MW
Btu s 106 W
= 831
. MW
lb m coal hg = − 162
. × 104 Btu lbm coal
− Q$
1. 62 × 104 Btu lb m
=
= 0. 901
HHV 1. 80 × 10 4 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- 66
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& O (mol O 2 / s)
Ta ( o C)
2
3.76n& 0 (mol N 2 / s)
n& CO (mol CO / s)
rn& CO (mol CO 2 / s)
1 mol / s @ 25 o C
xm (mol CH 4 / mol)
n& H2 O (mol H 2 O / s)
n& Ar (mol Ar / s)
xa (mol Ar / mol)
(1 − xm − xa ) (mol C 2 H 6 / mol)
CH 4 + 2O 2 → CO 2 + 2H 2 O
C2 H6 +
7
2
O 2 → 2CO2 + 3H 2 O
Pxs
) 2 xm + 35
. (1 − x m − xa )
100
x + 2(1 − x m − xa )
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& C O + 2 r n& CO + n& H2 O ⇒ n& O2 = 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
C 2 H6
A
xm
(1 − xm − x A )
xA
O2
no
N2
CO
CO 2
H2 O
376
. no
−
−
−
−
−
xA
0
0
0
H$ 1
H$
nO2
3.76 no
nCO
r nCO
nH 2 O
2
−
−
−
H$ out
−
−
$
H3
H$ 4
H$
5
H$ 6
H$ 7
H$
8
Ta or Ts Table B.2
c.
H$ i = ( ∆H$ f ) i +
B
z
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 = 700o C
⇒ no = 2153
. , nCO = 0.0955, n H2 O = 2.00, nO2 = 01500
.
H$ 1 (kJ / mol) = 8.091, H$ 2 = 29.588, H$ 3 = 0.702, H$ 4 = 3.279,
H$ 5 = 166.72, H$ 6 = − 8567
. , H$ 7 = − 345.35, H$ 8 = − 433.82
Energy balance: Q& =
∑ n&
out
H$ out −
∑ n&
$
in H in
9- 67
= − 655 kW
9.58 (cont'd)
Xa
Pxs
r
Ta
Ts
Q
1087
996
-905
-813
-722
631
-905
893
-869
-799
682
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
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
-905
-893
-882
-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- 68
-200 0
0.2
0.4
0.6
0.8
1
1.2
80
100
120
0.4
0.5
-400
Q
-600
-800
-1000
-1200
Xm
0
0
200
Q 400
600
-800
1000
20
40
60
Pxs
-850
-860 0
0.1
0.2
0.3
0.6
-870
Q -880
-890
-900
-910
x
0
0
20
40
60
80
100
120
200
250
300
800
1000
1200
-200
-400
Q
10
10
10
10
10
10
-600
-800
-1000
r
-800
-850
Q
0.0
5
1
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
d.
-1000
-1500
Ts
9.59
a.
Basis:
207.4 liters 273.2 K 1.1 atm
s
1 mol
278.2 K 1.0 atm 22.4 litersbSTP g
= 10.0 mols s fuel gas to furnace
H = C6 H14 ; M = CH 4
mw (kg H 2O( l)/s)
25°C
Qc (kW)
n0 (mol/s)
y0 (mol C 6H 14/mol)
(1 – y0) (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
nb (mol C 6H 14( l)/s)
mw (kg H 2 O( v)/s)
10 bars, sat'd
reactor
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)
n a (mol air/s) @ 200°C
0.21 mol O /mol
2
0.79 mol N /mol
2
100% excess
Antoine Eq.
Tdp = 55° C ⇒ y 0 P =
⇒ y0 =
p αH b55° Cg
B
=
483.3 mm Hg
4833
. 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 b5° Cg
5889
. mm Hg
y2 =
=
= 0.070 mol C 6 H 14 mol = 0.93% mol CH 4 mol
P
1.1 × 760 mm Hg
y0 =0 .530
Methane balance on condenser: n& 0 b1 − y0 g = 10.0b1 − y2 g ⇒
n& 0 = 19.78 mol s
y 2 =0 .070
Hexane balance on condenser: n& 0 y 0 = n& b + 10.0y 2
⇒
n&0 =19 .78
y 0 =0 .530
y 2 =0 .070
n& b = 9.78 mol C6 H14 s condensed
9.78 mol C6 H14 blg 86.17 g
cm 3
1L
3600 s
3
3
Volume of condensate =
s
0.659
g
10
cm
1h
mol
A
Table B.1
A
Table B.1
= 4600 L C 6 H 14 ( l ) h
b.
References : CH 4 eg, 5o Cj , C 6H14 el, 5o Cj
Substance
CH4
C6 H14 bvg
C 6H 14 blg
n& in
n& out
H$ in
H$ out
(mol / s) (kJ / mol) (mol / s) (kJ / mol)
9.30
1985
.
9.30
0
10.48
41212
.
0.70
32.940
−
−
9.78
0
T
Table B.2
B
Tb
CH 4 bgg: H$ = z C p dT
Table B.1
B
B
T
C 6H 14 bvg: H$ = z C pR dT + ∆H$ v + z Cpv dT
5
Condenser energy balance: Q& c = ∆H& =
Table B.1
5
∑
n& i H$ i −
out
9- 69
∑
in
n& i H$ i = −427 kW
Tb
9.59 (cont'd)
CH 4 + 2O2 → CO2 + 2H 2 O , C 6 H 14 +
9.30 mol CH 4
s
Theoretical O 2:
19
2
O 2 → 6CO 2 + 7H 2 O
2 mol O 2 0.70 mol C 6H14 9.5 mol O 2
+
= 25.3 mol O2 s
1 mol CH 4
s
1 mol C 6H 14
100% excess ⇒ bO2 gfed = 2 × bO2 gtheor. ⇒ 0.21n& a = 2 × 253
. ⇒ n& a = 240.95 mol air s
N 2 balance: 0.79b240.95g = n& 4 ⇒ n& 4 = 190.35 mol N 2 s
C balance:
9.30 mol CH 4
1 mol C
s
1 mol CH4
+
0.70 mol C6 H 14
6 mol C
1 mol C6 H 14
=
n5bmol CO 2 g
1 mol C
1 mol CO2
⇒ n&5 = 13.5 mol CO2 s
H balance:
b9.30
mol CH4 sgb4 mol H mol CH 4 g + b0.70gb14g = n& 6 b2g ⇒ n& 6 = 235
. mol H 2O
1
bO2 g
⇒ n& 3 = 253
. mol O 2 s
fed
2
References : Cbsg, H 2 bgg, O 2 bgg, N 2 bgg at 25° C for reactor side, H2 Oblg at triple point for
steam side (reference state for steam tables)
Since combustion is complete, bO2 gremaining = bO2 gexcess =
CH 4
n& in
mol / s
9.30
n& out
H$ in
kJ / mol mol / s
−75553
.
−
H$ out
kJ / mol
−
C 6 H 14 bvg
0.70
−170.07 −
−
O2
506
.
531
.
253
.
1172
.
N2
190.35
513
.
190.35
1115
.
CO 2
−
−
135
.
−37715
.
H 2 Obvg
−
−
235
.
−228.60
Substance
H 2 Obboiler waterg m& w (kg / s) 104.8
Table B.1 and B.2
B
H$ bT g
=
m& w ( kg / s) 2776.2
T
∆H$ fo + zC p dT
for CH4 , C 6H 14
25
Table B.1 and B.8
B
=
∆H$ fo + H$ i bT g for O2 , N 2 , CO2 , H 2 Obvg
Energy balance on reactor (assume adiabatic):
∆H& =
∑ n& H$ − ∑ n& H$
i
out
i
i
i
= 0 ⇒ − 8468 + m& w b2776.2 − 104 .8g = 0 ⇒ m& w = 3.2 kg steam s
in
9- 70
9.60
a.
CH 4 + 2O 2 → CO 2 + 2 H 2 O
Basis: 450 kmol CH 4 fed h
n a ( kmol air / h)@25 o C
Stack gas@300 o C
n& 1 (kmol CO2 / h)
n& 2 (kmol H 2 O / h)
n& 3 (kmol O 2 / h)
n& 4 (kmol N 2 / h)
0.21 kmol O 2 / kmol
0.79 kmol N 2 / kmol
450 kmol CH 4 / h @ 25 o C
Q& ( kJ / h)
m& w [kg H 2 O(l) / h]
m& w [kg H 2 O(v) / h]
o
17 bar, 250 o C
25 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 ⇒ n&1 = 450 kmol CO 2 h , n& 2 = 900 kmol H 2 O h
N 2 balance: n& 4 = b0.79ge5143
.
× 10 6 mol hj = 4060 kmol N 2 h
Molecular O2 balance:
n& 3 = b0.21gb5143g
mol O2 fed
450 kmol CH 4 react 2 mol O2
−
= 180 kmol O 2 h
h
1 mol CH 4
h
450 kmol CO 2 h U
900 kmol
4060 kmol
180 kmol
|
H 2 O h|
|
V⇒
N2 h |
|
O2 h |
W
y CO2 = 00805
.
y H2 O = 0161
.
y N2 = 0.726
y O2 = 0.0322
5590 kmol / h
Mean heat capacity of stack gas
Cp =
∑y C
i
pi
= b0.0805gb0.0423g + b0.161gb0.0343g + b0.726gb0.0297g + b0.0322gb0.0312g
= 0.0315 kJ mol ⋅ o C
Energy balance on furnace (combustion side only)
References: CH 4 bgg , CO 2 bgg, O 2 bgg, N 2 bgg, H 2 Oblg at 25o 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& CH4 = 450 kmol / h
9- 71
9.60 (cont’d)
H& p = n& 2 ( ∆H$ v ) H
o
2 O(25 C)
+ n& stack gas ( C p ) stack gas (Tstack gas − 25o C)
3
5590 kmol 103 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& = ∆ H& = ξ& ( ∆H$ co ) CH 4 +
∑ n H$ − ∑ n H$
i
i
out
=
i
0.0315 kJ (300- 25)o C
mol ⋅ o C
i
in
kmol I F
mol I F
kJ I
kJ
kJ
.
. × 10 7
= −3.44 × 108
J G1000
J G−89036
J + 563
h KH
kmol KH
mol K
h
h
F
G450
H
Energy balance on steam boiler
Table B.7 Table B.6
kJ L F kg I OL
kJ O
Q& = m& w ∆H$ w ⇒ + 3.44 × 108
= Mm& w G J P Mb2914 − 105g P
h N H h K QN
kg Q
5
⇒ m& w = 123
. × 10 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 2O/h)
n 2 (mol H 2O/h)
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 150o C instead of 300o C.
References: CH 4 bgg , CO 2 bgg, O 2 bgg, N 2 bgg, H 2 Oblg at 25 o C bfurnace sideg
H 2 O alf at triple point (steam table reference) (steam tube side)
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 ( ∆H$ 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 103 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- 72
0.0315 kJ (150- 25)o C
mol ⋅ o C
9.60 (cont’d)
∆H& = ξ& ( ∆ H$ co ) CH 4 +
∑ n H$ − ∑ n H$
i
i
out
⇒
F
G450
H
i
i
=0
in
kmol I F
mol I F
kJ I
7 kJ
J G1000
J G−890.36
J + 2 .99 × 10
K
H
K
H
K
h
kmol
mol
h
L
kJ O
F kg I OL
+ Mm& w G J PMb2914 − 105g P = 0 ⇒ mw = 1.32 × 105 kg steam / h
H h KQN
kg Q
N
Energy balance on preheater: ∆H& = d∆H& i
b∆H g
stack gas
= nC p ∆T =
b− ∆H gstack gas
+ d∆H& i
air
=0
5590 kmol 10 3 mol 0.0315 kJ
h
1 kmol mol ⋅ o C
o
b150 − 300g
C
= − 2.64 × 10 7
kJ
h
7
kJ
= b∆ H g air = na H$ air ( Ta ) ⇒ H$ air ( Ta ) = 2.64 × 10 kJ / h 1 3kmol = 5.133
5143 kmol / h 10 mol
mol
H$ air = 5133
.
kJ / mol
c.
stack gas
Table B.8
Ta = 199 o C
∑n
The energy balance on the furnace includes the term −
$
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
076
. × 40000gkg C 103 g 1 mol C
Basis: 40000 kg coal h ⇒ b
= 2.531 × 106 mol C h
h
1
kg
12.01
g
Assume coal enters at 25°C
. × 4000gkg
b005
H h e10 3 101
. j = 198
. × 10 6 mol H h
. × 4000 gkg
b008
O h e10 3 16.0j = 2.00 × 105 mol O h
. × 40000 g =
b011
4400 kg ash 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
furnace
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)
Preheated air at Ta (°C)
a.
air at 30°C, 1 atm, hr = 30%
n1 (mol O2 /h)
3.76n1 (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 × 106 = 0.078n 3 + 0.012 n3 ⇒ n3 = 2.812 × 107 mol h dry flue gas
N 2 balance: 376
. n1 = b0.796 ge2.812 × 10 7 j ⇒ n1 = 5.95 × 106 mol O 2 h b376
. ge595
. × 10 6 j
= 224 × 107 mol N 2 h
9- 73
9.61 (cont'd)
30% relative humidity (inlet air):
y H2 O P =
∗
030
. pH
b30° Cg ⇒
2O
Table B.3
B
n& 2
595
. × 106 + 2.24 × 10 7 + n2
b760 mm
Hgg = 0.300 b31824
.
mm Hgg
⇒ n& 2 = 3.61× 105 mol H 2 O h
Volumetric flow rate of inlet air:
6
7
. × 105 j mol
e5.95 × 10 + 224 × 10 + 361
V& =
h
Air/fuel ratio:
22.4 litersbSTP g
1 m3
1 mol
10 3 liters
= 6.43 × 105 SCMH
643
. × 105 m 3 air h
= 161
. SCM air kg coal
40000 kg coal h
H balance: 1.98 × 10 6 mol H h + 2e3.61 × 10 5 j mol H h = 2n& 4 ⇒ n& 4 = 1.351 × 10 6 mol H 2 O h
144424443 1444
424444
3
H in coal
H 2 O content of stack gas =
b.
H in water vapor
1.357 × 106 mol H 2 O h
e1.357 × 10
+ 2.812 × 107 j mol h
6
× 100% = 4.6% H 2 O
Energy balance on stack gas in preheater
References : CO 2 , CO, O 2 , N 2 , H 2 Obvg at 25o C
Substance
CO 2
CO
O2
N2
H 2O
nin
mol h
2.193 × 10 6
0.337 × 106
3.706 × 10 6
22.38 × 106
1.357 × 10 6
H$ in
kJ mol
4.942
3669
3758
3655
4266
H$ out
kJ mol
9.738
6.961
7193
.
6918
.
8135
nout
mol h
2.193 × 106
0.337 × 10 6
3.206 × 106
72.38 × 10 6
1.351× 106
Table B.2
H$ i (T ) from Table B.8 for inlet
Q=
∑ n H$ − ∑ n H$
i
i
out
i
i
B
H$ i (T ) = z C p
dT for outlet
= − 1.01 × 108 kJ h bHeat transferred from stack gasg
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
Ta (°C)
(We assume preheater is adiabatic, so that Qstack gas = −Qair )
Energy balance on air:
Q = ∆ H ⇒ 101
. × 108 kJ hr =
∑
Ta
Ta
z ni (C p ) i dT =
z ndry air (C p ) dry air dT + zn H2 O (C p ) H 2 O dT
30
30
9- 74
Table B.2
Ta
B
Table B.2
B
30
9.61 (cont'd)
⇒ 101
. × 10 8 = 8.31 × 105 ( Ta − 30) + 59.92( Ta2 − 302 ) + 0.031( Ta3 − 30 3 ) − 142
. × 10 −5 ( Ta4 − 304 )
o
⇒ 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 mol CO/h
3.206 × 106 mol O2 /h
22.38 × 106 mol N2 /h
1.351 × 106 mol H2 O( v)/h
5.95 ×10 6 mol O2 /h
2.24 ×10 7 mol N2 /h
3.61 ×10 5 mol H2 O( v)/h
150°C(= 149.8°C)
m (kg H2 O(l )/h)
50°C
m (kg H 2 O(v )/h)
30 bars, sat'd
References for energy balance on furnace: CO 2 , CO, O 2 , N 2 , H 2 Oblg, coal at 25° C
(Must choose H 2 Oblg since we are given the higher heating value of the coal.)
substance
nin
H$ in
nout
H$ out
Coal
40000
0
−
−
nbkg hg
Ash
−
−
4400
412.25 H$ bkJ kgg
O2
595
. × 10 6 3.758 3.206 × 10 6 7193
.
N2
2.24 × 10 6 3.655 2.24 × 10 7
6.918
nbmol hg
6
$
CO 2
−
−
2193
. × 10
9.738 HbkJ molg
6
CO
−
−
0.337 × 10
6.961
H2O
3.61 × 105 4828
.
1351
.
× 10 6 5214
.
(Furnace only — exclude boiler water)
Heat transferred from furnace
Q = ncoal ∆H$ io +
∑ n H$ − ∑ n H$
i
out
i
i
i
in
F
F
= G4 × 10 4
H
I
kg I F
kJ
4 kJ I
3
G
J G−2 .5 × 10
J + 2.74 × 10 − 122
. × 10 8 J
h KH
kg K G
kg
J
A
= − 876
. × 10
H
8
H$ of preheated air
K
kJ h
8
8
Heat transferred to boiler water: 0.60(8.76x10 kJ/h) = 5.25x10 kJ/h
F kg I
& G J H$ cH 2 Oblg, 30b, sat'dh − H$ eH 2 Oblg, 50 o Cj
Energy balance on boiler: Q& bkJ hg = m
H h K
L
O
⇒ 525
. × 108 kJ h = m& M2802 .3 − 209.3P
M A
N Table B.6
9- 75
A
P
Table B.5 Q
kJ
⇒ m& = 2.02 × 105 kg steam h
kg
9.62
1
CO + O 2 → CO 2 , ∆H$ co = −282.99 kJ mol
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
1 mol CO react 0.5 mol O2
= n0 − 0.5
1 mol CO
References: CO, CO2 , O2 , N2 at 25o C
n in
nout
H$ in
H$ out
Substance bmol g bkJ molg bmol g bkJ molg
Oxygen in product gas: n1 = n0 bmol O2 fedg −
CO
O2
N2
CO 2
1
n0
3.76n0
−
0
0
0
−
−
n0 − 0.5
376
. n0
1
−
H$ 1
H$ 2
H$ 3
Table B.8
B
O 2 bg,1400° Cg: H$ 1 = H$ O2 (1400 C) = 47.07 kJ mol
o
Table B.8
B
N 2 bg,1400 ° Cg: H$ 2 = H$ N 2 (1400o C) = 44.51 kJ mol
Table B.8
B
CO 2 bg,1400 ° Cg: H$ 3 = H$ CO 2 (1400o C) = 7189
. kJ mol
E.B.:
∆H = nCO ∆H$ co +
∑n H$ − ∑ n H$
i
out
i
⇒ n0 = 1094
.
mol O2
i
i
= −282.99 + 47.07bn0 − 0.5g + 44.51b3.76n0 g + 7189
. =0
in
Theoretical O 2 = b1 mol COgb0.5 mol O2 mol COg = 0500
.
mol O 2
1094
.
mol fed − 0.500 mol reqd.
× 100% = 119% excess oxygen
0.500 mol
Increase %XS air ⇒ Ta d 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 C2 H 6
CH 4 (g) + 2O2 (g) → CO 2 (g) + 2H2 O(v), ∆H$ co = −890.36 kJ / mol
7
C 2 H 6 (g) + O2 (g) → 2CO2 (g) + 3H2 O(v), ∆ H$ co = −1559.9 kJ / mol
2
82 mol CH4
18 mol C2 H6
298 K
Stack gas at T(°C)
n 2 (mol CO2 )
n 3 (mol H2 O (v))
n 4 (mol O2 )
n 5 (mol N2 )
n 0 (mol air) at 423 K
0.21 O2 (20% XS)
0.79 N2
9- 76
9.63 (cont’d)
Theoretical oxygen =
Air fed : n1 =
2 mol O 2
1 mol CH 4
1.2 × 227 mol O 2
82 mol CH4
+
3.5 mol O 2
1 mol C 2 H6
18 mol C2 H 6
= 227 mol O 2
1 mol air
= 129714
. mol air
0.21 mol O 2
C balance : n2 = b82.00gb1g + b18.00gb2g ⇒ n2 = 118.00 mol CO 2
H balance : 2n3 = b82.00gb4g + b1800
. gb6g ⇒ n3 = 21800
. mol H 2 O
20% excess air, complete combustion ⇒ n4 = b0.2gb227g mol O2 = 4540
. mol O 2
N 2 balance : n5 = b079
. gb129714
. g = 1024 .63 mol N 2
Extents of reaction: ξ1 = nCH4 = 82 mol, ξ1 = nC 2 H 6 = 18 mol
Reference states: CH 4 bgg, C 2H 6 bgg, N 2 bgg, O2 bgg, H 2 Oblg at 298 K
(We will use the values of ∆H$ co given in Table B.1, which are based on H2 Oblg as a combustion
product, and so must choose the liquid as a reference state for water.)
H$ i bT g = C pi bT − 298 K g for all species but water
= ∆H$ v,H2 O b298 K g + C p, H2 O bT − 298 K g for water
Substance
nin
mol
H$ in
kJ mol
CH 4
C2H6
O2
N2
CO 2
H 2 Obvg
82.00
18.00
272.40
1024.63
−
−
0
414
.
3.91
−
−
H$ out
kJ mol
n out
mol
−
−
−
−
4540
.
0.0331bT − 298 g
1024.63
00313
.
bT − 298g
11800
.
0.0500bT − 298g
218.00 44.013 + 0.0385bT − 298 g
Energy balance : ∆H = 0
ξ 1 e∆H$ co j
CH 4
+ ξ 2 e∆H$ co j
C2H 6
+
∑ n H$ − ∑ n H$
i
i
out
i
i
=0
in
⇒ b82.00 mol CH 4 gb−890.36 kJ mol g + b1800
. mol C 2 H 6 gb−1559.90 kJ mol g
+ b45.40gb00331
.
g + b1024.63gb0.0313g + b118 .00gb0.0500g + b218.00gb0.0385g bT − 298 g
+ b218.00gb44.01g − ( 272 .40)( 414
. ) − (1024 .63)(3.91) = 0
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- 77
C 3 H 6 Obg g + 4O 2 bg g → 3CO2 bg g + 3H 2 O alf, ∆H$ io = −1821.4 kJ mol
9.64
Basis :
1410 m 3 bSTPg feed gas
103 mol
1 min
= 1049 mol s feed gas
3
min
22.4 m bSTP g 60 s
Stochiometric proportion:
1 mol C3H 6O ⇒ 4 mol O 2 ⇒ 4 × 3.76 = 15.04 mol N2 ⇒ b1+ 4 + 1504
. g = 2004
. mol
yC 3 H 6 O =
1 mol C 3 H 6 O
mol C 3H 6O
4
= 0. 0499
, yO 2 =
= 0.1996 mol O 2 mol
20.04 mol
mol
20. 04
Preheat
Feed gas
1049 mol/s
0.0499 C 3H 6O
0.1996 O
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 CO2 /s)
n 2 (mol H 2 O/s)
n 3 (mol N 2 /s)
T a (°C)
Q1 (kW)
Relative saturation = 12.2% ⇒ yC 3 H6 O P = 0122
.
p∗C3 H 6O dT f
⇒ p∗ =
b.
Reaction
.
b00499
gb1500
mm Hg g
= 61352
. mm Hg
0.122
Product gas
350°C
Q2 (kW)
i
Table B.4
T f = 50.0o C
Feed contains b1049 mol sgb0.0499 C3 H 6O molg = 52.34 mol C 3H 6O s
b1049gb0.1996g = 209.4
b1049gb0.7505g =
mol O 2 s
787.3 mol N 2 s
n1 = b52.34gb3g = 157.0 mol CO2 s U 14.25 mole% CO 2
|
⇒ Product contains n2 = b52.34gb3g = 157.0 mol H 2 O sV ⇒ 14.25% H 2 O
|
n3 = 787.3 mol N 2 s
71.5% N
W
2
o
References : C3 H6 Obgg, O 2 , N 2 , H 2Oblg, CO2 at 25 C
Hˆ in
n&in
Substance
(mols) (kJ/mol)
n&out
Hˆ out
(mols)
(kJ/mol)
o
Ta
(562 C)
52.34
67.66
−
−
O2
N2
209.4
787.3
17.72
17.18
−
787.3
−
0.032 ( Ta − 25)
CO 2
−
−
157.0
H2 O
−
−
157.0
C3 H 6O
0.052 ( Ta − 25)
44.013 + 0.040( Ta − 25 )
H$ H2 O = ∆H$ v d25 o Ci + H$ H2 O ( T)
Energy balance on reactor:
∆H = nC 3H 6O ∆H$ co +
∑ n H$ − ∑ n H$
i
out


⇒ (5234 mol s )  − 1821.1
i
i
i
= 0 bkJ sg
in
kJ 
4
 + 39.638 (Ta − 25 ) + 157.0 ( 44.013 ) − 2.078 × 10 = 0 ⇒ Ta = 2780 °C
mol 
9- 78
9.64 (cont'd)
c.
Preheating step: References: C 3H 6 bgg, O 2 , N 2 at 25° C
n& in
H$ in
n& out
H$ out
( mol / s) (kJ / mol) (mol / s) ( kJ / mol)
(50 o C)
(562 o C)
52.34
315
.
52.34
67.66
209.4
0826
.
209.4
17.72
787 .3
0.775
787 .3
16.65
Substance
C 3H 6O
O2
N2
E.B. ⇒ Q& 1 =
∑ n& H$ − ∑ n& H$
i
i
i
out
= 194
. × 104 kW
i
in
Cooling step. References: CO2 (g), H2 Oavf , N2 (g) at 25 o C
nin
H$ in
nout
H$ out
Substance
( mol) (kJ / mol) (mol) (kJ / mol)
o
e2871
CO 2
H 2O
N2
E.B. ⇒ Q 2 =
157.0
157.0
787 .3
142 .3
10815
.
88.23
∑ n& H$ − ∑ n& H$
i
i
out
Cj
i
i
e350
157.0
157.0
787.3
o
Cj
16.25
12.35
1008
.
= −9.64 × 104 kW
in
Exchange heat between the reactor feed and product gases.
9.65
a.
Basis : 1 mol C5 H12 (l)
C 5H12 (l) + 8O 2 (g) → 5CO2 (g) + 6H 2 O(v),
1 mol C5 H12 (l)
n 2 (mol CO2 )
n 3 (mol H2 O (v))
n 4 (mol O2 )
Tad(o C)
n 0 (mol O2 ) , 75°C
30% excess
Theoretical oxygen =
∆H$ co = −3509.5 kJ / mol
1 mol C5 H12
8 mol O 2
= 8 mol O 2
1 mol C 5H12
30% excess ⇒ n0 = 13
. × 8 = 10.4 mol O 2
C balance: n2 = b1gb5g ⇒ n2 = 5 mol CO2
H balance: 2n3 = b1gb12g ⇒ n3 = 6 mol H 2O
30% excess O2 , complete combustion ⇒ n4 = b0.3gb8g mol O2 = 2.4 mol O 2
Reference states: C5H 12 blg, O2 bgg, H 2 Oblg, CO 2 (g) at 25o C
(We will use the values of ∆H$ c0 given in Table B.1, which are based on H2 Oblg as a
combustion product, and so must choose the liquid as a reference state for water)
9- 79
9.65 (cont'd)
substance
C5H 12
O2
CO2
H 2O
nin
mol
100
.
10.40
−
−
H$ in
kJ mol
0
H$ 1
−
−
nout
H$ out
mol kJ mol
−
−
2.40
H$ 2
500
.
H$ 3
6.00
H$ 4
T
H$ i = z ( C p ) i dT
i = 2,3
25
T
= ∆H$ v e25o Cj + z( C p ) H2 O(v) dT for H 2 O(v)
25
Table B.8
H$ 1 = H$ O 2 ( 75o C)
B
=
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
H$ 3 = ( 0.03611 Tad + 2.1165 × 10 −5 Tad 2 − 0.9623 × 10 −8 Tad 3 + 1.866 × 10 −12 Tad 4 − 0.9158 )
mol
kJ
2
3
4
H$ 4 = 4401
. + ( 003346
.
Tad + 03440
.
× 10 −5 Tad + 02535
.
× 10 −8 Tad − 08983
.
× 10−12 Tad − 0838
. )
mol
kJ
2
3
4
⇒ H$ 4 = 4317
. + (0.03346 Tad + 03440
.
× 10 −5 Tad + 0.2535 × 10 −8 Tad − 08983
.
× 10 −12 Tad )
mol
Energy balance :∆H = 0
nC5 H12 e∆H$ co j
C5 H 12 ( l)
+
∑ n H$ − ∑ n H$
i
out
i
i
i
=0
in
(1 mol C5H12 )( − 35095
. kJ / mol) + ( 2.40) H$ 2 + (5.00) H$ 3 + ( 6.00) H$ 4 − (10.40)( H$ 1) = 0
$ through H
$
Substitute for H
1
4
∆H& = (04512
.
Tad + 14036
. × 10−5 Tad 2 − 3777
.
× 10−8 Tad 3 + 4727
.
× 10 −12 Tad 4 ) − 3272.20 kJ / mol = 0
⇒ f (Tad ) = −3272 .20 + 04512
.
Tad + 14.036 × 10−5 Tad 2 − 3777
.
× 10−8 Tad 3 + 4727
.
× 10 −12 Tad 4 = 0
Check :
−3272.20
4727
.
× 10 −12
= −6.922 × 1014
Solving for Tad using E - Z Solve ⇒ Tad = 4414o C
b.
Terms
1
2
3
Tad
% Error
7252
64.3%
3481 –21.1%
3938 –10.8%
9- 80
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)
n&3 (mol O2 /s)
3.76 n&2 (mol N2 /s)
n&4 (mol CO2 /s)
25% excess air
n&2 (mol O2 /s)
Adiabatic
Reactor
3.76 n&2 (mol N2 /s) n& 4 (mol
CO2 /s)
n&5 (mol H2 O/s)
150°C, 1.1 atm
T(°C), 1.05 atm
2CH 4 + 2O 2 → CO2 + 2H 2O
Fuel feed rate : =
550
. L 273 K 1.1 atm
mol
= 0.247mol CH 4 / s
s
298 K 1.0 atm 22.4 L(STP)
Theoretical O 2 = 2 × 0.247 = 0.494 mol O 2 / s
25% excess air ⇒ n& 2 = 1.25( 0.494) = 0.6175 mol O 2 / s ,
⇒ 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 O2 fed / s − 0.494 mol consumed / s
= 0.124 mol O 2 / s
References: CH 4, O ,2N , 2C O , 2H O2( l ) a t 2 5o C
Substance
n&in
n&out
Hˆ in
Hˆ out
(mol/s) (kJ/mol) (mol/s) (kJ/mol)
CH 4
0.247
O2
0.6175
−
−
Hˆ
N2
2.32
0
ˆ
H1
Hˆ
CO 2
−
−
0.247
Hˆ 4
Hˆ
H 2O
−
−
0.497
Hˆ 6
H$ 1 = H$ (O 2, 150o C)
Table B.8
H$ 2 = H$ ( N 2, 150 o C)
Table B.8
2
3.78 kJ / mol
366
. kJ / mol
( ∆ H$ co ) CH 4 = −890 .36 kJ / mol
T
H$ i = z C pi dT , i = 3 − 5
25
9- 81
0.124
2.32
3
5
9.66 (cont'd)
T
H$ b = ( ∆ H$ v ) H
2 O(25
o
C)
+ z ( C p ) H 2 O(v) dT
25
a.
Energy Balance
∆H& = ξ& ( ∆H$ o )
c
CH 4
+
∑ n&
$
out H out
−
∑ n&
$
in H in
=0
$ )
Table B.2 for C p i ( T ), ( ∆H
v H O = 44 .01 kJ / mol
2
0.247 ( − 890.36) + 0494
. (44.01) + 0.0963( T − 25) + 174
. × 10 −5 (T 2 − 252 ) + 0305
.
× 10 −8 ( T 3 − 253 )
−161
. × 10 −12 ( T 4 − 25 4 ) − 0.6175( 378
. ) − 2.32( 366
. )=0
⇒ −211.4 + 0.0963Tad + 1.74 × 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 = 1832 o C, P = 1.05× 760 = 798 mmHg
0.494 mol/s
y H2 O =
= 0.155 mol H 2O/mol
(0.124 + 2.32 + 0.247 + 0.494) mol/s
Raoult's law: y H2O P = pH* 2O (Tdp ) ⇒ pH* 2O = (0.155)(798) = 124 mmHg
Table B.3
⇒T
dp
= 56 o C
Degrees of superheat = 1832o C − 56o C = 1776o C superheat
9.67
a.
CH 4 ( l) + 2O 2 ( g) → CO 2 (g) + 2H 2O(v)
Basis : 1 mol CH 4
2 mol O 2
= 2.00 mol O2
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 CH4
1 mol CH4
2.60 mol O2
9.78 mol N2
25° C, 4.00 atm
n 2 (mol CO2 )
n 3 (mol H2 O)
n 4 (mol O2 )
Complete combustion ⇒ n2 = 1.00 mol CO2 , n3 = 2.00 mol H 2 O
2.00 mol O 2 consumed ⇒ n4 = ( 2.60 − 2.00) mol O 2 = 0.60 mol O 2
F
G
Internal energy of reaction: Eq. (9.1-5) ⇒ ∆U$ co = ∆H$ co − RT G
I
∑ν
Ggaseous
Hproducts
⇒ e∆U$ co j
F
CH 4
T
U$ =
z( Cv )dT
25
= G− 890.36
H
Ideal Gas
⇒
i
−
∑ ν JJ
i
gaseous J
reactants K
kJ I 8.314 J 298 K (1 + 2 − 1 − 2) 1 kJ
kJ
= −890.36
J −
3
K
mol
K
10
J
mol
mol
T
z( C p − Rg ) dT
25
If C p is independent of T ⇒ U$ = ( C p − Rg )( T − 25o C)
9- 82
9.67 (cont’d)
b.
Reference states: CH 4 bgg, N 2 bgg, O 2 bgg, H 2Oblg, CO2 (g) at 25 o C
(We will use the values of ∆H$ c0 given in Table B.1, which are based on H2 Oblg as a
combustion product, and so must choose the liquid as a reference state for water.)
Substance
CH4
O2
N2
CO2
H 2Obvg
nin
U$ in
mol kJ mol
100
.
0
2.60
0
9.78
0
−
−
−
−
nout
U$ out
mol kJ mol
−
−
0.60
U$ 1
9.78
U$ 2
100
.
U$ 3
2.00
U$ 4
Part a
B
U$ i = ( C p − Rg )( T − 25) for all species except H 2 O(v)
= ∆U$ v e25 o Cj + ( C p − Rg )(T − 25) = ∆H$ v e25o Cj − Rg Tref + (C p − Rg )( T − 25) for H 2 O(v)
Substituting given values of ( C p ) i and R g = 8.314 × 10 −3 kJ / mol yields
U$ 1 = (0.033 − 8.314 × 10− 3 )( T − 25) kJ / mol = ( 002469
.
T − 0.6172) kJ / mol
U$ 2 = ( 0.032 − 8.314 × 10 −3 )(T − 25) kJ / mol = ( 0.02369 T − 0.5922) kJ / mol
U$ 3 = ( 0.052 − 8.314 × 10−3 )(T − 25) kJ / mol = ( 004369
.
T − 10922
.
) kJ / mol
L
kJ F
kJ I
O
kJ
−3
U$ 4 = M44.01
− G8.314 × 10 −3
J (298 K)P + ( 0.040 − 8.314 × 10 )( T − 25)
H
K
mol
mol ⋅ K
mol
N
Q
kJ
kJ
kJ
⇒ U$ 4 = 4153
.
+ ( 0.052 − 8314
.
× 10 −3 )( T − 25)
= ( 0.03167 T − 40.74)
mol
mol
mol
Energy Balance
Q = nCH 4 e∆U$ co j
CH 4
+
∑ n U$ − ∑ n U$
i
i
out
i
i
=0
in
⇒ Q = (100
. ) b−890.36 kJ / molg + ( 0.60)U$ 1 + (9.87)U$ 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.
Pf Tf
F ( 2295 + 273) K I
=
⇒ Pf = G
J × 4.00 atm = 34 .5 atm
Pi
Ti
H (25 + 273) K K
– Heat loss to and through reactor wall
– Tank would expand at high temperatures and pressures
9- 83
9.68
b.
1 mol natural gas
yCH4 (mol CH 4 / mol)
nCO2 (mol CO 2 )
yC2 H6 (mol C 2 H 6 / mol)
nH2 O (mol H 2O)
yC3 H8 (mol C3 H 8 / mol)
nN 2 (mol N 2 )
nO2 mol O 2 )
Humid air
n a (mol air)
y wo (mol H2 0(v)/mol)
(1-y wo ) (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) + 2O2 (g) → CO 2 (g) + H 2 O(v)
7
C 2 H6 (g) + O2 (g) → 2CO2 (g) + 3H 2O(v)
2
C 3H 8 (g) + 5O 2 (g) → 3CO2 (g) + 4H 2O(v)
Theoretical oxygen :
2 mol O2
1 mol CH 4
y CH4 (mol CH4 )
+
3.5 mol O2
1 mol C2 H6
yC2 H6 (mol C2 H6 )
+
5 mol O2
yC 3H8 (mol C3H8 )
1 mol C3H 8
= ( 2 yCH4 + 35
. yC2 H 6 + 5 yC3 H8 )
F
Excess oxygen: 0.21na (1 − y wo ) = G1 +
H
F
⇒ na = G1 +
H
Pxs I
J ( 2 y CH4 + 3.5 y C 2 H 6 + 5 y C 3H 8 ) mol O 2
100K
Pxs I
1
. y C 2 H6 + 5 y C 3 H8 )
mol air
J ( 2 y CH4 + 35
100K
0.21(1− y w 0 )
Feed components
( nO 2 ) in = 021
. na (1 − y wo ), ( n N2 ) in = 0.79na (1 − y wo ), (n H2 O ) in = na y wo
N 2 in product gas: nN2 = ( nN 2 ) in mol N 2
CO2 in product gas :
nCO2 =
1 mol CO2 nCH4 (mol CH4 ) 2 mol CO2 nC2H6 (mol C2 H6 ) 3 mol CO2 nC3 H8 (mol C3 H8 )
+
+
1 mol CH4
1 mol C2H 6
1 mol C3H8
= (nCH4 + 2nC2H6 + 3nC3H8 ) mol CO2
H 2 O in product gas :
nH2 O =
1 mol H2O nCH4 (mol CH4 ) 3 mol H 2O nC2 H6 (mol C2 H 6 ) 4 mol O2 nC3 H8 (mol C3H 8 )
+
+
1 mol CH 4
1 mol C2 H 6
1 mol C3H 8
= [2nCH4 + 3nC2H6 + 4nC3H8 + na (1- ywo )] mol H 2O
O 2 in product gas : nO2 =
Pxs
( 2 nCH 4 + 35
. n C2 H 6 + 5 nC 3H 8 ) mol O 2
100
9- 84
9.68 (cont’d)
c.
References : C(s), H 2 ( g) at 25o C
T
$
H
( T) = ( ∆ H of ) CH 4 + ( C p ) CH4 dT
CH
z
4
25
Using ( ∆ Hfo )CH4 from Table B.1 and ( C p )CH4 from Table B.2
FT
I
H$ CH (T) = −7485
. kJ / mol+ G (0.03431+5.469 × 10−5 T + 03661
.
× 10−8T 2 − 1100
. × 10−12 T 3 ) dT J kJ / mol
z
G
H25
4
J
K
⇒ H$ CH (T ) = [ −7572
. + 3.431 × 10 −2 T + 2.734 × 10 −5 T 2 + 0122
. × 10 −8 T 3 − 2.75× 10 −12 T 4 ] kJ / mol
4
nin
mol
n1
n2
n3
n4
n5
n6
−
Substance
CH 4
C2H6
C3H8
O2
N2
CO 2
H 2O
7
∆H =
nout H$ out
H$ in
kJ / mol mol kJ / mol
H$ 1
−
−
H$ 2
−
−
$
H3
−
−
H$ 4
n7
H$ 7
H$ 5
n8
H$ 8
−
n9
H$ 9
−
n10
H$ 10
6
∑(n )
i out ( H i ) out
−
i= 4
∑ (n )
i in ( H i ) in
i =1
H$ i = ai + bi T + ci T 2 + di T 3 + ei T 4
6
3
∑ (n )
=
i in ( Hi ) in
i =1
∑ (n )
6
$
i in Hi (Tf
)+
i =1
∑ (n )
$
i in Hi (Ta )
i =4
7
⇒ ∆H =
3
∑
( ni ) out (ai + bi T + ci T 2 + di T 3 + ei T 4 ) out −
i =4
7
⇒ ∆H =
7
∑(n )
i out a i
+
i =1
i out bi T +
i =4
3
−
∑ (n )
∑ (n )
i out ci T
$
2
i =1
6
) −
i in H i ( Tf
i =1
7
7
∑(n )
i =1
∑ (n )
∑
( ni ) in H$ i ( Tf ) −
+
∑ (n )
i out d i T
i =1
$
i in H i (Ta )
i =4
= α 0 + α 1T + α 2T 2 + α 3T 3 + α 4 T 4
7
where α 0 =
∑
3
(ni ) out ai −
i =1
∑
( ni ) in H$ i ( Tf ) −
i =1
7
α1 =
∑(n )
∑
α2 =
∑ (n )
i out ci
i =1
7
7
∑ (n )
i out d i
i =1
α4 =
∑ (n )
i out ei
i =1
.
9- 85
$
i in Hi ( Ta )
i =4
7
( ni ) out bi
i =1
α3=
6
6
∑(n )
$
i in Hi ( Ta )
i =4
7
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- 86
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
n14 (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
T ad (°C)
Feed gas, 650°C
n 14 (mol CH 4 /h)
0.96n 15 (mol O2 /h)
0.04n 15 (mol N2 /h)
Converter
product
38°C
0.917 n1 (mol DMF/h)
Lean solvent
filter
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 CO 2 /h)
n 11 (mol H 2 O/h)
n 12 (mol N 2/h)
n 13 (mol C(s )/h)
n15 (mol/h)
0.96 mol O2 /mol
0.04 mol N2 /mol
25°C
Basis:
5000 kg/h Product gas
n1 (mol/h)
0.991 mol C2 H2 ( g)/mol
0.00059 mol H2 O/mol
0.00841 mol CO2 /mol
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)
stripper
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
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)
Average M.W. of product gas:
M = 0.991b2604
. g + 0.00059b18016
. g + 000841
.
. g mol
b44.01g = 2619
Molar flow rate of p roduct gas: n0 =
5000 kg 103 g 1 mol 1 day
= 7955 mol h
day
1 kg 26.19 g 24 h
Material balances -- plan of attack (refer to flow chart):
Stripper balances: C 2 H 2 ⇒ n1 , CO ⇒ n2 , CH 4 ⇒ n3 , H 2O ⇒ n4 , CO2 ⇒ n5
Absorber balances: CH 4 ⇒ n6 , C 2 H2 ⇒ n7 , CO ⇒ n9 , CO2 ⇒ n10 , H2 O ⇒ n11
R5.67% soot formationU
S
V ⇒ n13 , n14 ,
Tconverter C balance W
converter H balance ⇒ n8
Converter O balance ⇒ n15 , converter N 2 balance ⇒ n12
Stripper balances:
C 2 H 2: 0.0155n1 = 0991
. b7955 mol hg ⇒ n1 = 5086
.
× 105 mol h
CO: b0.00055ge5.086 × 105 j = n2 ⇒ n2 = 79.7 mol CO h
CH 4 : b0.00055ge5086
.
× 105 j = n3 ⇒ n3 = 79.7 mol CH4 h
H2 O: b0.0596 ge5086
.
× 105 j = n4 + b0.00059gb7955g ⇒ n4 = 30308 mol H2 O h
CO2 : b0.0068ge5086
.
× 105 j = n5 + b0.00841gb7955g ⇒ n5 = 3392 mol CO 2 h
Absorber balances
CH 4 : n6 = 0.950n6 + b0.00055ge5086
.
× 105 j = n6 ⇒ 5595 mol CH 4 h
9- 87
9.69 (cont'd)
C 2 H2 : n7 = b0.0155ge5.086 × 105 j + 0.006 n7 ⇒ n7 = 7931 mol C2 H 2 h
CO: n9 = 0.988 n9 + b0.00055ge5.086 × 105 j ⇒ n9 = 23311 mol CO h
CO2 : n10 = b0.0068ge5086
.
× 105 j = 3458 mol CO2 h
H2 O: n11 = b0.0596ge5.086 × 105 j = 30313 mol H 2 O h
Soot formation:
n13 = b0.0567gn14 (mol CH 4 )
1 mol C
h
1 mol CH 4
⇒ n13 = 0.0567n14
b1g
Converter C balance:
n14 = b5595 mol CH 4 hgb1 mol C mol CH 4 g + b7931gb2g + b23311gb1g + b3458gb1g + n13
⇒ n14 = n13 + 48226
b2 g
Solve (1) & (2) simultaneously ⇒ n13 = 2899 mol Cbsg h , n14 = 51120 mol CH 4 h
Converter H balance:
51120 mol CH 4
4 mol H
h
1 mol Ch4
CH 4
C2H 2
H2 O
H2
= b5595gb4g + b7931gb2g + 2n8 + b30313gb2g
⇒ n8 = 52816 mol H2 h
b0.96 n15 gb2 g =
Converter O balance:
23311 mol CO
1 mol O
h
1 mol CO
C O2
H 2O
+ b3458 gb2g + b30313gb1g
⇒ n15 = 31531 mol h
Converter N 2 balance: b0.04gb31531gn12 ⇒ n12 = 1261 mol N 2 h
a.
Feed stream flow rates
VC H4 =
VO 2 =
b.
51120 mol CH 4
h
0.0244 m 3 bSTP g
= 1145 SCMH CH 4
1 mol
31531 mol bO 2 + N 2 g 0.0244 m 3 bSTPg
= 706 SCMH O 2 b+ N 2 g
h
1 mol
Gas feed to absorber
5595 mol
7931 mol
23311 mol
3458 mol
30313 mol
52816 mol
1261 mol
CH 4 h U
C 2 H 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 V ⇒ 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
|
W
Absorber off-gas
52816 mol H 2 h U
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 V ⇒ 82.5 kmol h, 6.4% CH , 0.06% C H
4
2 2
41.6 mol C 2 H 2 h|
|
8.2471× 104 mol h|W
9- 88
9.69 (cont'd)
Stripper off-gas
279.7 mol CO h U
279.7 mol CH 4 h |
30308 mol H 2 O h |V ⇒ 34.3 kmol h, 0.82% CO, 0.82% CH 4 , 88.5% H 2 O, 9.9% CO2
3392 mol CO h |
3.4259 × 10 4 mol h |W
c.
F
DMF recirculation rate = 0.917 G5.086 × 105
H
d.
Overall product yield =
b0.991gb7955g
mol I F 1 kmol I
JG
J = 466 kmol DMF h
h K H10 3 mol K
mol C2 H 2 in product gas
51120 mol CH 4 in feed h
= 0154
.
mol C2 H 2
mol CH 4
The theoretical maximum yield would be obtained if only the reaction 2CH 4 → C 2 H2 + 3H2
occurred, the reaction went to completion, and all the C 2 H2 formed were recovered in the
product gas. This yield is (1 mol C2 H2 /2 mol CH4 ) = 0.500 mol C2 H2 /2 mol CH4 .
The ratio of the actual yield to the theoretical yield is 0.154/0.500 = 0.308.
e.
Methane preheater
Table B.2
B
650
Q& CH 4 = ∆H& = n& 14 z
25
dC p i
dT =
CH4
51120 mol 32824 J
1h
1 kJ
= 466 kW
h
mol
3601 s 103 J
Oxygen preheater
Table B.8
Table B.8
B
B
Q& O2 = ∆H& = 0.96n&15 H$ ( O 2 ,650o C) + 0.04n& 15 H$ ( N 2 ,650 o C)
F
= G31531
H
f.
mol I L
kJ OF 1 h I
. + 0.04 × 18.99g
J Mb0.96 × 20135
J = 176 kW
PG
h KN
mol ⋅ o C QH 3600 s K
References : Cbsg, H 2 bgg, O 2 bgg, N 2 bgg at 25° C
Substance
n& in
H$ in b650° Cg
n& out
CH 4
51120
−42.026
5595
O2
30270
20.125
−
H$ out bTout g
− 74.85 +
Ta
z25 C p dT
−
Ta
N2
1261
18.988
1261
z35 C p dT
C2 H2
−
−
7931
+226.75z C p dT
H2
−
−
52816
zC p dT
CO
−
−
23311 −110.52 + z C p dT
CO 2
−
−
3458
H 2O
−
−
30313
Cbsg
−
−
2899
9- 89
Ta
25
−393.5 + C p dT
z
−24183
. + z C p dT
zC p dT
nbmol hg
H$ bkJ molg
9.69 (cont’d)
T
H$ i = ∆H$ i0 + zC pidT
kJ mol
∑ n& H$
i
i
25
kJ mol ⋅° C
= − 1575
.
× 10 6 kJ h
in
∑ n& H$
i
Tout
i
out
= −9.888 × 106 kJ h + z
L
5595dC p i
M
CH4
N
25
+52816dC p i
Tad + 273
+z
298
H2
dC p i
+ 23311dC p i
×
C bs g
1 kJ
CO
+ 1261dC p i
+3458dC p i
CO2
N2
+ 7931dC p i
+ 3013dC p i
O
C 3 H2
1 kJ
3
H 2 O bvg P
Q 10
J
dT
dT
103 J
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
Tad
i
out
= −9.888 × 106 kJ h + z
25
.
− 5.9885 × 10
e3902 + 12185
Tad +273 F
+z
298
∑ n& H$
i
i
G32 .411 + 0.031744 T −
−4
T 2 − 10162
.
× 10 −7 T 3 j dT
14179
.
× 106 I
T2
H
JdT
K
= −1.000 × 107 + 3943 Ta + 0.6251Ta2 − 1.996 × 10 −4 Ta3 − 2.5405 × 10−8 Ta4 +
out
Energy balance: ∆H& =
∑ n& H$ − ∑ n& H$
i
out
i
i
i
=0
in
⇒ f bTc g = −8.485 × 10 6 + 3943Tc + 0.6251Tc2 − 1996
.
× 10 −4 Tc3 − 2.5405 × 10 −8 Tc4 +
E-Z Solve
1.418 × 106
Ta + 273
Tc = 2032 o C.
9- 90
1.418 × 10 6
=0
Tc + 273
9.70 a.
m& 1[kg W(v)/d]
W = H2 O
1 0 0o C
F
o
24 ,000 kg sludge / d, 22 C
0.35 solids, 0.65 W(l)
DRYER
Q&2
& 2 (kg conc. sludge/d), 100o C
m
INCINERATOR
Waste gas
0.75 solids, 0.25 W(l)
& 3 [kg W(v)/d]
m
4B, sat'd
C
& 3 [kg W(l)/d]
m
BOILER
m& 3 [kg W(l)/d]
Q&3 (kJ / d)
4B, sat'd
o
20C
D
110oC
0.90 kmol CH4 /kmol
0.10 kmol C2 H6 / kmol
Q&4 (kJ / d)
Q&1
m& 4 (kg oil/d)
0.87 C
0.10 H
0.0084 S
0.0216 ash
& 6 (kg gas/d)
m
Stack gas
CO 2 , H2 O(v)
o
125 C
SO2
O 2, N 2
ash
m
& 7 (kg air/d)
25 oC
Q& 0 (kJ/d)
E m& 5 (kg air/d)
2 5o C
9-90
9.70 (cont'd)
Solids balance on dryer:
&2 ⇒ m
& 2 = 11200 kg/d ⇒ F :11.2 tonnes concentrated sludge/d
0.35 × 24,000 kg/d = 0.75 m
Mass balance on dryer: 24,000 = m& 1 + 11200 ⇒ m& 1 = 12,800 kg/d
Energy balance on sludge side of dryer:
References : H 2O(l,22o C), Solids(22o C)
m& in
Substance
Hˆ in
(kg d) (kJ kg)
m& out
Hˆ out
(kg d) (kJ kg)
8400
Hˆ 1
Solids
8400
0
HO(l)
2
15600
0
2800
H 2 O(v)
−
−
12800
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
from Table B.5)
water
Q& 2 =
∑ m& H$ − ∑ m& H$ ⇒ Q&
i
i
out
Q& steam
i
2
i
= 356
. × 107 kJ day
in
356
. × 107
=
= 647
. × 107 kJ / d ⇒ Q& 3 = 2.91 × 107 kJ / d
055
.
Energy balance on steam side of dryer:
∆ Hˆ v for
HO(sat'd,
)
2
6.47 × 107
↓
 kJ   1 tonne 
kJ
kg
& 3   × 2133    3
=m
 ⇒ m& 3 = 30.3 tonnes boiler feedwater/d
d
 d 
 kg  10 kg 
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 =
&4 =
⇒m
8.04 × 107
= 13
. × 108 kJ / d
0.62
1.30 × 10 8 kJ/d
= 3458 kg/d ⇒ D :3.5 tonnes fuel oil/day
3.75 × 10 4 kJ/kg
Air feed to boiler furnace: C + O 2 → CO 2 , 4H + O2 → 2H 2 O,
( nO2 )theo = 3458
S + O2 → SO 2
kg 
kgC 1 kmol C 1 kmol O 2
1 1
1 1
)(
)(
)+(0.10)( )( ) + (0.0084)( )( ) 
(0.87
d 
kg
12 kg
1 kmol C
1 4
32 1 
= 338 kmol O2 /d
9- 91
9.70 (cont’d)
Air fed (25% excess)=1.25(4.76
⇒
2011 kmol 29 kg 1 tonne
kmol 103 kg
d
kmol air
kmol O2
kmol air
)(338
) = 2011
kmol O2
d
d
⇒ E :58.3 tonnes air to boiler/d
Energy balance on boiler air preheater:
2011 kmol 103 mol
kJ
Table B.8 ⇒ Hˆ a i r , 1 2o5 C = 2.93
⇒ Q& 0 =
d
1 kmol
mol
2.93 kJ
kJ
= 5.89 × 10 6
mol
d
Supplementary fuel for incinerator:
11.2 tonne sludge 195 SCM
1 kmol
n&6 =
= 97.5kmol d
d
tonne
22.4 SCM
MWgas = 0.90 MWCH 4 + 010
. MWC 2 H6 = ( 0.90)(16) + ( 010
. )(30) = 17.4 kg kmol
m& gas = (97.5
kmol
kg
)(17.4
) ⇒ G :1.7 tonne natural gas/d
d
kmol
Air feed to incinerator:
CH 4 + 2O 2 → CO2 + 2H2 O, C2 H 6 +
7
O2 → 2CO2 + 3H 2O
2
11200 kg conc. sludge
(air)th for sludge:
(air)t h for gas: 97.5
d
kmol 
d
 0.90

0.75 kg solids
1 kg conc. sludge
kmol CH 4
kmol
×
2 kmol O2
kmol CH 4
19000 kJ
2.5 SCM air
1 kmol
1 kg solids
4
22.4 SCM
10 kJ
= 1781
kmol air
d
 4.76 kmol air 
kmol air
 = 997.8
1
kmol
O
d


2
+ (0.10)(3.5)  
kmol air
kmol air
= 5558
d
d
5558 kmol air 29 kg air 1 tonne
tonnes incinerator air
⇒
⇒ H :161
d
1 kmol 103 kg
d
100% XS: n&7 = 2(1781 + 997.8)
Energy balance on incinerator air preheater
5558 kmol 103 mol 2.486 kJ
kJ
kJ
Table B.8 ⇒ Hˆ a i r , 1 1o0 C = 2.486
⇒ Q& 4 =
= 1.38 ×107
d
1 kmol
mol
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- 92
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
bn1 , n2 , n3 , x 1 , x 2 ,
y 1 , y 2 , y 3 , T1 , T2 , Qg
−5 relations
b4 material balances and 1 energy balance g
6 degrees of freedom
A feasible set of design variables: l n1 , n 2 , x1 , x2 , T1 , T2 q
Calculate n 3 from total mole balance, y1 , y 2 , and y 3 from component balances,
Q from energy balance.
An infeasible set: l n1 , n 2 , n 3 , x1 , x 2 , T1q
Specifying n1 and n 2 determines n 3 (from a total mole balance)
n 2 (mol gas), T 2 , P
y 2 (mol C 6H 14/mol)
1 – y 2 (mol N 2/mol)
c.
n 1 (mol gas), T 1 , P
y 1 (mol C 6H 14/mol)
1 – y 1 (mol N 2/mol)
n 3 (mol C 6H 14( )/mol),
l
T2, P
Q (kJ)
9 variables
bn1 , n 2 , n 3 , y1 , y 2 , T1 , T2 , Q, P g
*
−4 relations
d2 material, 1 energy, and 1 equilibrium: y 2 P = PC 6 H14 bT2 gi
5 degrees of freedom
A feasible set: l n , y1 , T1 , P, n3 q
Calculate n 2 from total balance, y 2 from C 6 H 14 balance, T2 from Raoult’s law:
[ y2 P = PC∗6H 4 bT2 g ], Q from energy balance
An infeasible set: l n 2 , y 2 , n 3 , P, T2 q
Once y 2 and P are specified, T2 is determined from Raoult’s law
10- 1
10.2 10 variables bn 1 , n 2 , n 3 , n4 , x1 , x 2 , x 3 , x 4 , T , Pg
−2 material balances
−2 equilibrium relations: [ x 3 P = x 4 PB* bT g, b1 − x 3 gP = b1 − x 4 gPC* bT g]
6 degrees of freedom
a. A straightforward set: l n1 , n 3 , n4 , x1 , x 4 , T q
Calculate n 2 from total material balance, P from sum of Raoult's laws:
P = x 4 p ∗B bT g + b1 − x 4 gPc∗ bT g
x 3 from Raoult's law, x 2 from B balance
b. An iterative set: l n1 , n 2 , n 3 , x1 , x 2 , x3 q
Calculate n 4 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: l n1 , n 2 , n 3 , n4 , T , Pq
Once n1 , n 2 , and n 3 are specified, a total mole balance determines n 4 .
10.3 2BaSO 4 bsg + 4Cbsg → 2BaSbsg + 4CO2 bg g
a.
100 kg ore, T 0 (K)
xb (kg BaSO 4/kg)
n 1 (kg C)
n 2 (kg BaS)
n 3 (kg CO 2)
n 4 (kg other solids)
T f (K)
n 0 (kg coal), T 0 (K)
xc (kg C/kg)
P ex (% excess coal)
Q (kJ)
11 variables dn0 , n1 , n 2 , n 3 , n4 , x b , x c , T0 , Tf , Q, Pex i
−5 material balancesbC, BaS, CO 2 , BaSO 4 , other solidsg
−1 energy balance
+1 reaction
−1 relation defining Pex in terms of n0 , x b , and x c
5 degrees of freedom
b. Design set: n x b , x c , T0 , T f , Pe x s
Calculate n 0 from x b , x c , and Pex ; n1 through n 4 from material balances,
Q from energy balance
10- 2
10.3 (cont’d)
c. Design set: l x B , xc , T0 , n 2 , Qq
Specifying x B determines n 2 ⇒ impossible design set.
d. Design set: l x B , xc , T0 , Pex , Qq
Calculate n 2 from x B , n 3 from x B
n 0 from x B , x c and Pex
n1 from C material balance, n 4 from total material balance
T f from energy balance (trial-and-error probably required)
10.4 2C 2 H 5OH + O 2 → 2CH 3CHO + 2H 2 O
2CH 3COH + O 2 → 2CH 3CHOOH
n f (mol solution), T 0
x ef (mol EtOH/mol)
1 – x ef (mol H 2O/mol)
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.
n e (mol EtOH), T
n ah (mol CH 3CHO)
n ea (mol CH 3COOH)
n w (mol H 2O)
n ax (mol O 2)
n n (mol N 2)
Q (kJ)
13 variables dn f , naw , ne , n eh , n ea , n w , nex , n 0 , x ef , T0 , T , Q, Pxs i
−6 material balances
−1 energy balance
−1 relation between Pxs , n f , x ef , and nair
+2 reactions
7 degrees of freedom
b. Design set: nn f , x ef , Pxs , n e , n ah , T0 , Ts
Calculate n air from n f , x ef and Pxs ; n n from N 2 balance;
n aa and n w from n f , x ef , ne , n ah and material balances;
n ex from O atomic balance; Q from energy balance
c. Design set: nn f , x ef , T0 , n air , Q, ne , n w s
Calculate Pxs from n f , x ef and n air ; n’s from material balances; T from energy
balance (generally nonlinear in T)
d. Design set: l n air , n n , Kq . Once n air is specified, an N 2 balance fixes n n
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
n 10 (kg catalyst)
n 11(mol C3 H6 )
n 16(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:
10 variables bn1 − n 16 g
−6 material balances
+2 reactions
6 degrees of freedom
Flash Tank:
12 variables bn 4 − n15 g
−6 material balances
6 degrees of freedom
Separation:
10 variables bn11 − n 20 g
−5 material balances
5 degrees of freedom
Hydrogenator:
5 variables bn19 − n 23 g
−3 material balances
+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 n 9 = 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
(n 9 = n 10). It may therefore not be determined unless either n 9 or n 10 is specified.
10- 4
10.6 n − C 4 H10 → i − C 4 H 10 bn − B = i − Bg
n 1 (mol n-B)
mixer
n 2 (mol n-B)
n 3 (mol i-B)
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)
a. Mixer:
5 variables bn1 , n2 , n3 , n r , x r g
−2 material balances
3 degrees of freedom
Reactor:
4 variables bn 2 , n 3 , n 3 , n5 g
−2 material balances
+1 reaction
3 degrees of freedom
Still:
6 variables bn 4 , n5 , n 6 , x 6 , n r , 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 − C4 H 10 , x 6 = 0.115 mol n − C 4 H 10 mol , x r = 0.85 mol n − C 4 H 10 mol
Overall C balance: b100gb4g = n6 b0.115gb4g + b0.885gb4g mol C ⇒ n 6 = 100 mol overhead
Overall conversion =
100 mol n - B fed − b100gb0.115gmol n - B unreacted
100 mol n - B fed
Mixer n-B balance: 100 + 0.85nT = n 2
× 100% = 88.5%
b1g
b1g
35% S.P. conversion: n 4 = 0.65n 2 ⇒ n 4 = 65 + 0.5525n r
b2g
Still n – B balance:
b2 g
n 4 = n6 x6 + n r xr ⇒ 65 + 0.5525n r = b0.115gb100 g + 0.85n r ⇒ n r = 17983
. mol
Recycle ratio = b179 .83 mol recycleg b100 mol fresh feed g = 179
.
10- 5
mol recycle
mol fresh feed
10.6 (cont’d)
c.
nr
n 2 = 100 + 0.85n r
n 3 = n r b1 − 0.85g
n 4 = 0.65n 2
n 5 = n2 + n 3 − n4
n6 =
n 4 = 0115
. n6 +
nr =
Error:
d.
k =2
k=3
100.0
185.0
132.3 151.5
212.5 228.8
15.0 19.85 22.73
120.25 1381
. 148.7
79.75 94.21 102 .8
n 4 + n5 = n 6 + n r
U
V⇒
0.85n r W
k =1
67.69
132.3
80.76 88.55
1515
. 163.0
179.83 − 163.0
× 100 = 9.3% error
17983
.
w=
1515
. − 132.3
= 0.595
132.3 − 100.0
q=
0.595
= −1470
.
0.595 − 1
n rb g = −1.470b132.3g + c1 − b−1.470ghb1515
. g = 179.8
3
Error:
179.8 − 1798
.
× 100 = < 0.1% error
179.8
e. Successive substitution, Iteration 32: n r = 179.8319 à n r = 179.8319
Wegstein, Iteration 3: n r = 179.8319 à n r = 179.8319
S1
10.7
SF
Split
S2
a.
1
2
3
4
5
6
7
8
A
X1 =
B
C
0.6
Molar flow rates (mol/h)
SF
S1
nA
85.5
51.3
nB
52.5
31.5
nC
12.0
7.2
nD
0.0
0.0
T(deg.C)
315
315
Formula in C4: = $B$1*B4
Formula in D4: = B4-C4
10- 6
D
S2
34.2
21.0
4.8
0.0
315
10.7 (cont’d)
b.C
**CHAPTER 10 -- PROBLEM 7
DIMENSION SF(8), S1(8), S2(8)
FLOW = 150.
N=3
SF(1) = 0.35*FLOW
SF(2) = 0.57*FLOW
SF(3) = 0.08*FLOW
SF(8) = 315.
X1 = 0.60
CALL SPLIT (SF, S1, S2, X1, N)
WRITE (6, 900)' STREAM 1', S1(1), S1(2), S1(3), S1(B)
WRITE (6, 900)' STREAM 2', S2(1), S2(2), S2(3), S2(B)
900
FORMAT (A10, F8.2,' mols/h n-octane', /,
*10X, F8.2,' mols/h iso-octane', /,
*
*10X, F8.2,' mols/h inerts', /,
10X, F8.2,' K')
END
C
C
C
100
SUBROUTINE SPLIT
SUBROUTINE SPLIT (SF, S1, S2, X1, N)
DIMENSION SF(8), S1(8), S2(8)
D0 100 J = 1, N
S1(J) = X1*SF(J)
S2(J) = SF(J) – S1(J)
S1(8) = SF (8)
S2(8) = SF (8)
RETURN
END
Program Output: Stream 1 3150
. mols h n-octane
51.30 mols h iso-octane
7.20 mols h inerts
315.00 K
Stream 2 21.00 mols h n-octane
34.20 mols h iso-octane
4.80 mols h inerts
315.00 K
10- 7
10.8
a. Let Bz = benzene, Tl = toluene
Antoine equations: p*Bz = 106.90565−1211.033/( T +220.790) (=1350.491)
pTl* = 106.95334−1343 .943/( T +219.377 ) (= 556.3212)
Raoult's law: x Bz = ( P − p*Tl ) / ( p*Bz - pTl* ) (= 0.307) , yBz = x Bz p *Bz / P ( = 0518
. )
Total mole balance: 100 = nv + nl
U
V
Benzene balance:
40 = yBz nv + x Bz nl W
⇒ nv =
40 − 100x Bz
(= 44.13) , nl = 100 − nv (= 55.87)
y Bz − xBz
Fractional benzene vaporization: f B = nv y Bz / 40 (= 0.571)
Fractional toluene vaporization: f T = nv (1 − y Bz ) / 60 (= 0.354)
The specific enthalpies are calculated by integrating heat capacities and (for vapors)
adding the heat of vaporization.
Q = ∑ n out H$ out − ∑ n in 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 n v=0) and the dew-point
temperature (find the temperature for which n l =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
10.9 a. Mass Balance: NF = NL + NV
b1g
XF bI g∗ NF = XLbI g∗ NL + XV bI g∗ NV
N
I = 1,2Kn − 1
Energy Balance: Q = bT − TF g∗ ∑ CPbI g∗ c XLbI g∗ NL + XV bI g∗ NV h
I=1
N
+ NV ∗ ∑ HV bI g∗ XV b1g
b3g
I =1
where: XLb N g = 1 −
N−1
∑ XLbI g
I =1
XV b N g = 1 −
N
Raoult’s law: P = ∑ XL bI g∗ PV bI g
b4g
I =1
10- 9
N −1
∑ XV bI g
I =1
b2 g
10.9 (cont’d)
XV bI g∗ P = XLbI g∗ PV bI g I = 1,2 ,K N − 1
where: PV bI g = 10∗∗ d AbI g − BbI g cCbI g + T hi
b5g
I = 1,2,K N − 1
3 + 3bN − 1g + N + 4 variables b NF , NL , NV , XF ( I ), XL( I ), XV ( I ), PV ( I ), TF , T , P , Qg
− N mass balance
−1 energy balances
− N equilibrium relations
− N Antoine equations
N + 3 degrees of freedom
Design Set mTF , T , P, NF , XF bI gr
Eliminate NL form (2) using (1)
Eliminate XV(I) form (2) using (5)
Solve (2) for XL(I)
XLbI g = XF bI g∗ NF dNF + NV ∗ c PV bI g P − 1hi
b6g
Sum (6) over I to Eliminate XL(I)
N
f bNV g = −1 + NF ∗ ∑ XF bI g dNF + NV ∗ c PV bI g P − 1hi = 0
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
b7 g
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)
n&v (mol / s)
x v ( mol A(v) / mol)
1 − x v ( mol B(g) / mol)
T (K), P (mm Hg)
n& F (mol / s)
x F (mol A(v) / mol)
1 − x F (mol B(g) / mol)
TF (K), P(mm Hg)
n& l (mol A(l) / s)
10- 12
10.10 (cont’d)
10 variables ( n& F , x F , TF , P , n& v , x v , T , n&l , p *,A , Q& )
–2 material balances
–1 Antoine equation
–1 Raoult’s law
–1 energy balance
5degrees of freedom
b.
References: A(l), B(g) at 25oC
Substance
n& in
H$ in
n& out
A(l)
—
—
n& l
A(v)
n& F x F
n& v x v
B(g)
n& F (1 − x F )
H$ 1
H$
2
n& v (1 − x v )
H$ out
H$
3
H$ 4
H$
5
Given n& F and x F (or n& AF and n& BF ), TF , P, y c (fractional condensation),
Fractional condensation ⇒ n& l = y c n& F x F
Mole balance ⇒ n& v = n& F − n&l
A balance ⇒ x v = (n& F x F − n& l ) / n& v
Raoult's law ⇒ p *A = x v P
B
−C
A − log 10 p *A
Enthalpies: H$ 1 = ∆H$ v + C pv ( TF − 25), H$ 2 = C pg (TF − 25), H$ 3 = C pl (T − 25),
Antoine' s equation ⇒ T =
H$ 4 = ∆H$ 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
η1 A1 + η 2 A2 + η 3 A3 +K η m Am = 0
a. Extent of reaction equations:
ξ = −[ SF bIX g∗ X ] NU bIX g
SPbI g = SF bI g + NU bI g∗ ξ
I = 1,2,K N
Energy Balance: Reference states are molecular species at 298K.
TF = SF b N + 1g TP = SPbN + 1g
N
∆H$ r = ∑ HF bI g∗ NU bI g
I =1
N
N
I =1
I =1
Q = ξ ∗ ∆H$ r + bTP − 298g∗ ∑ SPbI g∗ CPbI g − (TF − 298)∗ ∑ SF bI g∗ CPb I g
b. C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
Subscripts: 1 = C3H8, 2 = O2, 3 = N2, 4 = CO2, 5 = H2O
270 m 3
mol ⋅ K
1 atm
273K 0.08206 liter ⋅ atm
h
3.348 mol C3 H 8 1.2b5 mol O 2 g
sec
mol C3H 8
1000 liter
m
3
h
3600 s
= 3.348 mol C 3 H 8 s [=SF(1)]
= 20.09 mol O 2 s [= SF(2)] ⇒ 7554
. mol N 2 s [= SF(3)]
X C3 H8 = 0.90 ⇒ n& C3 H8 = 010
. (3.348) = 0.3348 mol C 3 H 8 / s in product gas [= SP(1)]
ξ = −[ SF bIX g∗ X ] NU bIX g = –(3.348 mol/s)(0.90)/(–1) = 3.013 mol/s
Nu
nin (SF)
X
Xi
nout (SP)
Cp
Tin
Hin
Tout
Hout
HF
DHr
Q
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
10- 16
10.11 (cont’d)
300
10.12
HP = HP + SP(I)*CP(I)
HP = HP + (SP(N + 1) – SF (N + 1))
Q = EXT * HR + HP
RETURN
END
a.
Extent of reaction equations:
ξ = − SF bIX g∗ X NU bIX g
SPbI g = SF bI g + NU bI g∗ ξ
I = 1, N
Energy Balance: Reference states are molecular species at feed stream temperature.
N
N
T
i =1
I =1
Tfeed
Q = ∆H = ξ∆H$ r + ∑ n out H$ out = 0 ⇒ 0 = ξ ∑ NU bI gHF bI g + ∑ SPbI g CPbI gdT
2
CP(I) = ACP(I) + BCP(I)*T + CCP(I)*T + DCP(I)*T
N
f bT g = ξ ∗ ∑ NU bI g * HF bI g + AP∗ (T − Tfeed ) +
I =1
+
3
BP
2
∗ (T 2 − Tfeed
)
2
CP
DP
3
4
∗ ( T 3 − Tfeed
)+
∗ (T 4 − Tfeed
) =0
3
4
N
where: AP = ∑ SPbI g∗ ACPbI g , 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
2 mol CO 1.25b1 mol O2 g
2 mol CO
= 1.25 mol O 2 ⇒ 4.70 mol N 2
⇒ Total moles fed = (2.00 + 1.25 + 4.70) mol = 7.95 mol
Scale to given basis:
(23.0
kmol
1 h 10 3 mol
SF (1) = 1.607 mol CO fed s
)(
)(
)
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.
21
10.14C
C
C
100
110
200
C
300
400
500
900
**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
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
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)
XCM(I) = XC(I)
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
CONTINUE
Convergence
RETURN
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)
XA(I) = Q = XAM(I) + (1. – Q)*XCM(I)
GOTO 110
WRITE (6, 900)
FORMAT (' CONVG did not converge')
STOP
END
10- 22
Elementary Principles of Chemical Processes
23
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
1
0.150000E + 01 0.115578E + 01
2
2
0.275000E + 01 0.282353E + 01
3
1
0.395135E + 00 0.482384E + 00
3
2
0.283152E + 01 0.245041E + 01
8
1
0.113575E + 01 0.113289E + 01
8
2
0.269023E + 01 0.269315E + 01
4
1
0.113199E + 01 0.113180E + 01
9
2
0.269186E + 01 0.269241E + 01
M
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 m& 1 M p / M
b.
Balance on mass: Accumulation = input – output
E
dM
& 0 − m& 1
=m
dt
t = 0, M = M 0 (mass in tank when leakage begins)
Balance on H 2O 2: Accumulation = input – output – consumption
E
dM p
dt
F
MpI
H
M
= m& 0 x p 0 − m& 1G
J − kM p
K
t = 0, M p = M p0
11.2
a.
Balance on H3 PO4 : Accumulation = input
Density of H3 PO4 : ρ = 1834
.
g / ml .
Molecular weight of H3 PO4 : M = 9800
. g / mol .
Accumulation =
Input =
dn p
(kmol / min)
dt
20.0 L 1000 ml 1.834 g
min
L
ml
mol
1 kmol
98.00 g 1000 mol
= 0.3743 kmol / min
E
dn p
dt
= 0.3743
t = 0, n p0 = 150 × 0.05 = 7.5 kmol
np
b.
t
dn p = 03743
.
dt ⇒ n p = 75
. + 0.3743t (kmol H3PO 4 in tank )
7 .5
xp =
c.
0
np
015
. =
n
=
np
n0 + n p − n p0
=
7.5 + 0.3743t
150 + 0.3743t
kmol H 3PO 4
kmol
7.5 + 03743
.
t
⇒ t = 471
. min
150 + 03743
.
t
11-1
11.3
a.
& w = 750g & bt = 5, m
& w = 1000g ⇒ m
& w bkg h g = 750 + 50 tb h g
m& w = a + bt bt = 0, m
Balance on methanol: Accumulation = Input – Output
M = kg CH3OH in tank
dM
= m& f − m& w = 1200 kg h − b750 + 50tg kg h
dt
E
dM
= 450 − 50tbkg hg
dt
t = 0, M = 750 kg
M
b.
dM =
750
t
b450 − 50t gdt
0
E
M − 750 = 450t − 25t 2
E
M = 750 + 450 t − 25t 2
Check the solution in two ways:
(1) t = 0, M = 750 kg ⇒ satisfies the initial condition;
(2)
c.
dM
= 450 − 50t
dt
⇒ reproduces the mass balance.
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 ±
2
b450 g
+ 4b25gb750 g
2b−25g
3.40 m 3 103 liter 0.792 kg
1 m3
1 liter
⇒ t = –1.54 h, 19.54 h
= 2693 kg (capacity of tank)
M = 2693 = 750 + 450t − 25t 2
t=
−450 ±
2
b450g
+ 4b25gb750 − 2693g
2b−25g
⇒ t = 719
. h ,1081
. h
Expressions for M(t) are:
M(t) =
R750 + 450t - 25t 2 b0 ≤ t ≤ 719
. and 1081
. ≤t
|
( 719
. ≤ t ≤ 10.81)
S2693
|
(19.54 ≤ t ≤ 2054
. )
|T0
≤ 1954
. g (tank is filling or draining)
(tank is overflowing)
(tank is empty, draining
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
= 0.0258 lb - mole
532° R 359 ft 3 bSTP g
Air in tank after 15 s:
Pf V
P0V
=
N f RT
N0 RT
⇒ N f = N0
Rate of addition: n& =
b.
Pf
P0
=
0.0258 lb - mole 114.7 psia
14.7 psia
.
− 0.0258g lb - mole
b02013
15 s
air
= 0.2013 lb - mole
= 0.0117 lb - mole air s
Balance on air in tank: Accumulation = input
dN
= 0.0117 blb - moles sg ; t = 0 , N = 0.0258 lb - mole
dt
N
c.
Integrate balance:
t
dN = n& dt ⇒ N = 0.0258 + 0.0117t blb - mole airg
0 .0258
0
Check the solution in two ways:
(1) t = 0, N = 0.0258 lb - mole ⇒ satisfies the initial condition
dN
( 2)
= 0.0117 lb - moleair / s ⇒ reproduces the mass balance
dt
d.
t = 120 s ⇒ N = 0.0258 + b0.0117gb120 g = 143
. lb - moles air
O 2 in tank = 0.21b143
. g = 030
. lb - mole O 2
11-3
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 ⇒
dV 540 m
=
dt
h
3
(
1h
− ν& m 3 min
60 min w
)
t = 0, V = 3.00 × 10 3 m 3 ( t = 0 corresponds to 8:00 AM )
V
t
∫
( )
∫
3.00×10 3
b.
t
dV = (9.00 −ν&w ) dt ⇒ V m 3 = 3.00 × 10 3 + 9.00t − ν&w dt t in minutes
∫
0
0
Let ν& w i = tabulated value of ν& w at t = 10bi − 1g
240
ν& w dt ≅
0
i = 1, 2, K , 25
24
24
O 10
10 L
& w1 + ν& w 25 + 4 ∑ ν& w i + 2 ∑ ν& w i P =
114
. + 98
. + 4b1246
. g + 2b113.4g
Mν
3M
3
i =2 , 4 , K
i = 3, 5, K P
N
Q
= 2488 m 3
V = 300
. × 103 + 9.00b240g − 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
M
Results:
TIME (MIN)
0.00
10.00
20.00
M
VOLUME (CUBIC M ETERS)
3000.
2974.
2944.
M
230.00
240.00
2683.
2674.
Vtrapezoid = 2674 m 3 ; VSimpson = 2672 m 3 ;
2674 − 2672
× 100% = 0.07%
2672
Simpson’s rule is more accurate.
11-4
a.
ν& out bL ming = kV bLg ⇒ ν& out = 0.200V ν& out = 20.0 L min ⇒ Vs = 100 L
b.
Balance on water: Accumulation = input – output (L/min).
(Balance volume directly since density is constant)
V = 300
ν& out = 60
dV
= 20.0 − 0200
. V
dt
t = 0, V = 300
c.
dV
= 0 = 200 − 0200
. Vs ⇒ Vs = 100 L
dt
The plot of V vs. t begins at (t=0, V=300). When t=0, the slope (dV/dt) is
200
. − 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.
V
11.6
t
V
d.
t
dV
= dt
20.0 − 0.200V 0
300
⇒−
1
F 20.0 − 0.200V I
lnG
J =t
K
0.200 H
− 40.0
⇒ −0.5 + 0.005V = expb− 0.200t g ⇒ V = 100.0 + 200.0 expb−0.200tg
V = 101
. b100g = 101 L b1% from steady stateg ⇒
101 = 100 + 200 expb− 0.200t g ⇒ t =
lnb1 200 g
−0.200
= 26.5 min
11-5
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=
ln D2 D1 lnb755 2385g
=
= − 0230
.
t 2 − t1
6−1
ln a = ln D1 − bt1 = lnb2385g + b0.230 gb1g = 8.007 ⇒ a = e 8. 007 = 3000
E
D = 3000e −0.230 t
b.
Inventory balance: Accumulation = –output
dI
= −3000 e −0.230t b kg week g
dt
t = 0, I = 18,000 kg
I
18, 000
11.8
t
dI = −3000e −0.230 t dt ⇒ I − 18,000 =
0
c.
t = ∞ ⇒ I = 4957 kg
a.
Total moles in room: N =
1100 m3
Molar throughput rate: n& =
3000 −0.230t
e
0.230
273 K
t
0
⇒ I = 4957 + 13,043e −0.230 t
103 mol
295 K 22.4 m 3bSTP g
700 m 3
min
273 K
103 mol
295 K 22.4 m3bSTPg
= 45, 440 mol
= 28,920 mol min
SO 2 balance ( t = 0 is the instant after the SO 2 is released into the room):
N bmolgx bmol SO 2 molg = mol SO2 in room
Accumulation = –output.
d
dx
&
⇒
= −0.6364x
bNxg = − nx
N =45, 440 dt
dt
n& = 28 ,920
t = 0, x =
b.
15
. mol SO 2
= 330
. × 10−5 mol SO2 mol
45, 440 mol
The plot of x vs. t begins at (t=0, x=3.30×10-5 ). When t=0, the slope (dx/dt) is
− 06364
.
× 330
. × 10 −5 = − 210
. × 10 −5. As t increases, x decreases.⇒
dx dt = −0.6364x 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:
x
t
3.30 ×10 −5
dx
x
= − 06364
.
dt ⇒ ln
= − 06364
.
t ⇒ x = 330
. × 10−5 e −0. 6364t
−5
x 0
330
. × 10
Check the solution in two ways:
(1) t = 0, x = 3.30 × 10-5 mol SO2 / mol ⇒ satisfies the initial condition;
dx
(2)
= −0.6364 × 330
. × 10−5 e− 0.6364 t = −0.6364x ⇒ reproduces the mass balance.
dt
d.
e.
CS O2 =
45,440 moles
1100 m 3
i)
t = 2 min ⇒ CSO2
ii)
x = 10 −6 ⇒ t =
x mol SO2 1 m 3
= 4131
. × 10−2 x = 13632
.
× 10−6 e −0.6364t mol SO2 / L
mol
103 L
mol SO2
= 382
. × 10−7
liter
lne10 −6 3.30 × 10−5 j
−0.6364
= 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
Molar flow rate of entering and leaving gas: n& (
)=
h
RT
&
P
ν
kmol F kmol CO I
p
Rate at which CO leaves: n& (
) xG
x
J =
H kmol K
h
RT
CO balance: Accumulation = -output
Pν& p
d( Nx )
dx
F P I
& px
=−
x⇒
= −G
Jν
H NRT K
dt
RT
dt
PV = NRT
ν& p
dx
=−
x
dt
V
kmol CO
t = 0, x = 0.01
kmol
E
ν& p r
dx
V
=−
dt ⇒ tr = −
lnb100 xg
x
V 0
ν& p
0 .01
x
b.
t
c.
V = 350 m 3
350
tr = −
lne100 × 35 × 10 −6 j = 283
. hrs
700
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
dM
& bkg ming = 0 ⇒∴ M is a constant = 200 kg
= m& − m
dt
b.
Sodium nitrate balance: Accumulation = - output
x = mass fraction of NaNO3
d bxM g
dt
= − xm& bkg ming
E
&
dx
m
m&
=−
x=−
x
dt
M
200
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
& increases ⇒
m
becomes more negative ⇒ x decreases faster.
dt
x
d.
t
dx
m&
& I
x
m&
F mt
=−
dt ⇒ ln
=−
t ⇒ x = 0.45 expG−
J
H 200 K
0
.
45
200
x
M
0 .45
0
Check the solution:
(1) t = 0, x = 0.45 ⇒ satisfies the initial condition;
&
dx
m&
mt
m&
(2)
= −0.45 ×
exp(−
)= −
x ⇒ satisfies the mass balance.
dt
200
200
200
0.45
0.4
& = 50 kg / min
m
0.35
m& = 100 kg / min
0.3
& = 200 kg / min
m
x
0.25
0.2
0.15
0.1
0.05
0
0
5
10
15
20
t(min)
e.
m& = 100 kg min ⇒ t = − 2 lndx f 0.45i
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.
Mass of tracer in tank: V em 3 j Cekg m 3 j
Tracer balance: Accumulation = –output. If perfectly mixed, C out = C tank = C
d bVCg
dt
C
b.
zm
0
c.
V
= −ν& C bkg min g
dC
ν&
=− C
dt
V
m
t = 0, C = 0
V
V is constant
t ν
&
&t I
F C I
m
dC
ν& t
F ν
=−
dt ⇒ lnG
⇒ C = 0 expG− J
J =−
z
0V
H VK
C
V
V
H m0 V K
Plot C (log scale) vs t (rect. scale) on semilog paper: Data lie on straight line (verifying assumption
of perfect mixing) through et = 1, C = 0223
.
× 10 −3 j & et = 2, C = 0.050 × 10 −3 j .
−
ν& lnb0.050 0.223g
=
= −1.495 min −1
V
2 −1
E
V = e30 m 3 minj
11.12 a.
e1.495
min − 1 j = 201
. m3
In tent at any time, P=14.7 psia, V=40.0 ft 3 , T=68°F=528°R
14.7 psia
PV
⇒N=
= m(liquid) =
ft 3 ⋅ psia
RT
10.73
lb - mole ⋅ o R
b.
40.0 ft 3
528 o R
= 01038
.
lb - mole
Molar throughout rate:
n& in = n& out = n& =
60 ft 3
min
492° R 16.0 psia
528° R 14.7 psia
1 lb - mole
= 01695
.
lb - mole min
359 ft 3 bSTP g
F lb - mole O2
Moles of O2 in tank= N(lb - mole) × G
H lb - mole
Balance on O2 : Accumulation = input – output
db Nx g
dt
= 0.35n& − xn& ⇒ 0.1038
I
J
K
dx
= 1.63b0.35 − xg
dx
= 0.1695b0.35 − xg ⇒ dt
dt
t = 0, x = 0.21
t
b0.35 − x g
dx
= 163
= 163
. t
z0. 21 0.35 − x z0 . dt ⇒ − ln b0.35 − 021
. g
x
c.
035
. −x
= e −1.63t ⇒ x = 0.35 − 014
. e −1.63t
014
.
1 L
F 0.35 − 0.27 I O
x = 0.27 ⇒ t =
− lnG
J P = 0.343 min ( or 20.6 s)
H 0.35 − 0.21 KQ
1.63 M
N
⇒
11-10
11.13 a.
Mass of is otope at any time = V blitersgCbmg isotope liter g
Balance on isotope: Accumulation = –consumption
dC
= − kC
dt
t = 0, C = C0
Cancel V
d
F mg I
bVCg = − kC G
JV bL g
HL⋅ sK
dt
Separate variables and integrate
C
zC
0
t
− lnbC C0 g
F C I
dC
= − kdt ⇒ lnG J = − kt ⇒ t =
C z0
k
H C0 K
C = 0.5C0 ⇒ t1 2 =
b.
t 1 2 = 2.6 hr ⇒ k =
C = 0.01C0
− lnb0.5g
k
⇒ t1 2 =
ln 2
k
ln 2
= 0267
.
hr − 1
2.6 hr
t=
t=-ln(C/C0)/k
− lnb0.01g
0.267
= 17.2 hr
11.14 A → products
a.
Mole balance on A: Accumulation = –consumption
dbC AV g
dt
= − kC AV
bV
constant; cancelsg
t = 0, C A = C A0
CA
t
FC I
dC A
= z − kdt ⇒ lnG A J = − kt ⇒ C A = C A0 expb− kt g
CA0 C A
0
H C A0 K
⇒z
b.
Plot C A (log scale) vs. t (rect. scale) on semilog paper. The data fall on a straight line (verifies
assumption of first-order) through bt = 213
. , C A = 00262
.
g & bt = 120.0, C A = 0.0185g .
ln C A = − kt + ln C A0
−k =
lnb0.0185 00262
.
g
120.0 − 213
.
= −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
dbC AV g
dt
= − kC 2AV
bV
constant; cancelsg
t = 0, C A = C A0
⇒
CA
zC
A0
dC A
C 2A
=
t
z0
− kdt ⇒ −
−1
L 1
O
1
1
+
= − kt ⇒ C A = M
+ kt P
C A C A0
NC A 0
Q
11-11
11.15 (cont’d)
b.
C A = 0.5C A0 ⇒ −
1
1
1
n
P
RT
+
= − kt1 2 ⇒ t1 2 =
; but C A0 = A0 = 0 ⇒ t1 2 =
0.5C A0 C A0
kC A0
V
RT
kP0
n A = 0.5n A0
n B = b0.5n A0 mol A react.gb2 mol B 2 mol A react.g = 05
. n A0
nC = b0.5n A0 mol A react.gb1 mol C 2 mol A react.g = 0.25n A0
total moles = 125
. n A0 ⇒ P1 2 = 125
.
c.
n A0 RT
= 125
. P0
V
Plot t1 2 vs. 1 P0 on rectangular paper. Data fall on straight line (verifying 2nd order
decomposition) through dt1 2 = 1060, 1 P0 = 1 0135
. i & dt1 2 = 209 , 1 P0 = 1 0.683 i
Slope:
RT
1060 − 209
=
= 1432
. s ⋅ atm
k
1 0135
. − 1 0683
.
⇒k =
d.
t1 2 =
b1015
Kgb008206
.
L ⋅ atm mol ⋅ Kg
143.2 s ⋅ atm
= 0582
.
L mol ⋅ s
F t1 2 P0 I
RT
1 E 1
F E I
expG J ⇒ lnG
+
J = ln
H RT K
k 0 P0
RT
k
R T
H
K
0
Plot t1 2 P0 RT (log scale) vs. 1 T (rect. scale) on semilog paper.
t1 2 bsg, P0 = 1 atm, R = 0.08206 L ⋅ atm / (mol ⋅ K) , TbKg
Data fall on straight line through dt1 2 P0 RT = 74.0, 1 T = 1 900i &
dt1 2 P0
RT = 0.6383, 1 T = 1 1050i
R=8.314 J/ (mol ·K)
E lnb0.6383 74.0g
=
= 29,940 K
R 1 1050 − 1 900
ln
e.
E = 2.49 × 105 J mol
1
29,940
12
= lnb06383
.
g−
= − 28.96 ⇒ k 0 = 3.79 × 10 L (mol ⋅ s)
k0
1050
F
T = 980 K ⇒ k = k 0 expG−
H
C A0 =
E I
J = 0.204 L (mol ⋅ s)
RT K
070
. b120
. atm g
b0.08206
L ⋅ atm mol ⋅ Kgb980 Kg
= 1045
.
× 10 −2 mol L
90% conversion
1L 1
1 O
1 L
1
1
−
−
M
P=
M
−
3
k NC A C A0 Q 0.204 N1.045 × 10
1.045 × 10 −2
= 4222 s = 70.4 min
C A = 0.10C A0 ⇒ t =
11-12
O
P
Q
11.16 A → B
a.
Mole balance on A: Accumulation = –consumption(V constant)
dC A
kC
=− 1 A
dt
1+ k2C A
t = 0, C A = C A0
CA
zC
b.
1 + k 2C A
A0
k1 C A
dC A =
t
CA
1
z0− dt ⇒ k 1 ln C A0
+
k2
k1
bC A
− C A0 g = −t ⇒ t =
k2
k1
bC A 0
− C A g−
C
1
ln A
k1 C A 0
Plot t bC A − C A0 g vs. lnbC A / C A0 g bC A 0 − C A g on rectangular paper:
x 44
y
64
47
8
647
48
ln
C
C
b
t
1
k
A
A0 g
=−
+ 2
k
C
−
C
k
C
−
C
b A0
Ag
1
A0
A
;
11
slope
intercept
F
y1
x1
I
F
K
H
y2
x2
I
Data fall on straight line through G116.28, − 02111
.
J & G130.01, −0.2496J
H
−
K
1
13001
. − 116.28
=
= − 356.62 ⇒ k1 = 2.80 × 10 −3 L (mol ⋅ s)
k 1 −0.2496 − b−0.2111g
k2
= 130.01 + 356.62b− 02496
.
. ⇒ k 2 = 0115
.
L mol
g = 4100
k1
11.17 CO + Cl 2 ⇒ COCl 2
a.
3.00 L
273 K
1 mol
303.8 K 22.4 LbSTP g
bCCO g
i
dCCl 2 i
i
= 0.12035 mol gas
= 0.60b012035
.
molg 3.00 L = 0.02407 mol L CO
= 0.40b012035
.
molg 3.00 L = 0.01605 mol L
U
|
Vinitial
Cl 2 |
W
concentrations
CC O bt g = 0.02407 − C p bt g U
|
V Since
CCl 2 bt g = 0.01605 − C p bt g |
1 mol COCl 2 formed requires 1 mol of each reactant
W
b.
Mole balance on Phosgene: Accumulation = generation
ddVC p i
dt
c.
=
. CCl 2
d1 + 586
+ 34.3C p i
dC p
V=3.00 L
875
. CCO C Cl 2
dt
2
=
2.92d0.02407 − C p i d0.01605 − C p i
.
− 24.3C p i
d1941
t = 0, C p = 0
Cl 2 limiting; 75% conversion ⇒ C p = 075
. b0.01605g = 0.01204 mol L
1
t=
2.92
0 .01204
z0
.
− 24.3C p i
d1941
2
d0 .02407 − C p i d0.01605 − C p i
dC p
11-13
2
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 = 904
. minutes
Moles of CO2 in liquid phase at any time = V ecm 3 j C A emols cm 3 j
Balance on CO 2 in liquid phase: Accumulation = input
dC A kS *
=
d
F mols I
*
eC A − C A j
⇒
bVC A g = kS eC A − C A j G
J
dt
V
H s K ÷V
dt
t = 0, C A = 0
Separate variables and integrate. Since p A = y A P is constant, C *A = p A H is also a constant.
dC A
CA
z0
C *A
⇒ ln
−CA
t
=z
C *A − C A
C *A
1− C A C A*
0
kS
dt ⇒ − lneC *A − C A j
V
=−
CA
C A =0
=
kS
t
V
kS expb g C A
t ⇒ 1 − * = e − kSt V ⇒ C A = C *A e1 − e − kSt V
V
CA
11-14
j
11.18 (cont’d)
b.
V L CA
ln M1 − *
kS M
CA
N
t=−
O
P
P
Q
V = 5 L = 5000 cm 3 , k = 0.020 cm s , S = 785
. cm 2 , C A = 0.62 × 10 − 3 mol / cm 3
C *A = y A P H = a0.30f a20 atm f d9230 atm⋅ cm 3 mol i = 0.65 × 10 − 3 mol cm 3
e5000
t=−
.
b002
cm 3 j
cm sge785
. cm 2 j
F
0.62 × 10 −3 I
H
0.65 × 10−3 K
lnG1 −
J
= 9800 s ⇒ 2.7 hr
(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 A0v& − ( kC A )V C =N /V
A
A
dt
dN A
= C Ao v& − kN A
dt
t = 0, N A = 0
dN A
C v&
= 0 ⇒ N A = A0
dt
k
b.
Steady State:
c.
& ⇒ V = vt
&
z0dV = z0vdt
V
NA
z0
⇒−
t
t
dN A
= z dt
C A0 v& − kN A
0
C A0 v& − kN A
1 F C A0 v& − kN A I
lnG
= e − kt
J =t⇒
k H C A0 v&
C A0 v&
K
⇒ NA =
CA =
C A0v&
1 − exp b− kt g
k
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.
lim C A = lim
t →∞
t →∞
C A0 [1 − exp(− kt )]
C
= lim A0 = 0
t →∞ kt
kt
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) = 300
. 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
100o C
b.
c.
11.21 a.
z18dTC
o
240 s
=z
0
0.0797 Q& 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 ⋅ o C)(1 mol / 0.0180 kg) = 4.184 kJ / (kg ⋅o C)
Q& = a0.97f (2.50) = 2.425 kJ s
W& = 0
dT
= 0.0290 b° C sg , t = 0 , T = 25° C
dt
The other 3% of the energy is used to heat the vessel or is lost to the surroundings.
T
b.
z
o
t
dT =
25 C
c.
.
dt ⇒ T = 25° C + 0.0290t bsg
z 00290
0
T = 100° C ⇒ t = b100 − 25g 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
dTb & &
= Q − W = −UAbTb − Tw g
dt
Table B.1
B
3
3
M = e60 cm j e7. 7 g cm j = 462 g
Cv = 0.46 kJ (kg ⋅° C), Tw = 25° C
2
U = 0.050 J (min ⋅ cm ⋅° C)
A = 2 a2 fa3f + a2 fa10f + a3fa10 f cm2 = 112 cm 2
dTb
= −0.02635bTb − 25g b° C ming
dt
t = 0, Tb = 95° C
dTb
= 0 = −0.02635dTbf − 25i ⇒ Tbf = 25° C
dt
b.
95
85
75
Tb(o C)
65
55
45
35
25
15
5
0
t
Tb
c.
t
dTb
= −0.02635dt
T − 25 0
95 b
F T − 25 I
⇒ lnG b
.
t
J = − 002635
H95 − 25 K
⇒ Tb btg = 25 + 70 exp b−0.02635t g
Check the solution in three ways:
(1) t = 0, Tb = 25 + 70 = 95o C ⇒ satisfies the initial condition;
dTb
(2)
= −70 × 0.02635e −0. 02635 t = − 002635
.
( Tb − 25) ⇒ reproduces the mass balance;
dt
(3) t → ∞ , Tb = 25o C ⇒ confirms the steady state condition.
Tb = 30° C ⇒ t = 100 min
11-17
11.23
12.0 kg/min
25o C
12.0 kg/min
T (o C)
Q (kJ/min) = UA (Tsteam-T)
a.
Energy Balance: MCv
dT
& p b25 − T g + UAbTsteam − T g
= mC
dt
M = 760 kg
m& = 12 .0 kg min
dT / dt = 150
. − 0.0224T ( o C min), t = 0, T = 25o C
Cv ≈ Cp = 2.30 kJ (min ⋅° C)
UA = 11.5 kJ (min ⋅° C)
Tsteam asat'd; 7.5barsf = 167.8° C
b.
Steady State:
dT
= 0 = 150
. − 0.0224 Ts ⇒ Ts = 67° C
dt
T(o C)
67
25
0
t
Tf
c.
dT
1
. − 0.0224T I
150
. − 094
. exp( −0.0224t )
F150
= dt ⇒ t = −
ln G
J ⇒T =
H
K
150
. − 00224
.
T 0
00224
.
0.94
0.0224
25
t
t = 40 min. ⇒ T = 498
. °C
d.
U changed. Let x = (UA) new . The differential equation becomes:
dT
= 0.3947 + 0.096x − ( 0.01579 + 5.721x ) T
dt
55
z
25
dT
=
0.3947 + 0.096x − ( 0.01579 + 5721
.
× 10−4 x )T
⇒
−
L 0.3947
1
M
ln
0.01579 + 5.721 × 10−4 x M0.3947
M
N
40
zdt
0
+ 0.096 x − e0.01579 + 5721
.
× 10−4 x j × 55 O
P
+ 0.096x − e0.01579 + 5721
. × 10−4 x j × 25 PP
⇒ x = 14.27 kJ / (min⋅o C)
∆U
∆(UA)
1427
. − 115
.
=
=
× 100% = 241%
.
U initial (UA) initial
115
.
11-18
Q
= 40
11.24 a.
Energy balance: MCv
dT & &
= Q −W
dt
W& = 0, Cv = 1.77 J g ⋅° C
M = 350 g, Q& = 40.2W = 40.2 J s
dT
U
T = 20 + 0.0649tbs g
= 0.0649 b° C sg|
V⇒
dt
. min
t = 0, T = 20° C |W T = 40° C ⇒ t = 308 s ⇒ 51
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 bneglect evaporation g: t =
= 926 s
0.0649
Time remaining: 40 minutes b60 s ming − 926 s = 1474 s
Evaporation: ∆H$ v = b30.765 kJ mol gb1 mol 78.11 g gb1000 J kJ g = 393 J g
Evaporation rate = b40.2 J sg / b393 J gg = 0102
.
g s
Benzene remaining = 350 g − b0102
.
g sgb1474 sg = 200 g
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.
Moles of air in room: n =
60 m3
273 K
1 kg - mole
= 2.58 kg - moles
283 K 22.4 m 3 bSTP g
dT & &
Energy balance on room air: nCv
= Q −W
dt
Q& = m& ∆H$ bH O, 3bars, sat'dg − 30.0bT − T
s
v
2
0g
W& = 0
nCv
dT
= m& s∆ H$ v − 30.0bT − T0 g
dt
N = 2.58 kg - moles
Cv = 20.8 kJ (kg- mole⋅° C)
∆H$ = 2163 kJ kg bfrom Table B.6 g
v
T0 = 0° C
dT
= 40.3m& s − 0.559T b° C hr g
dt
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)
b.
& s − 0.559T = 0 ⇒ m& s =
At steady-state, dT dt = 0 ⇒ 40.3m
0559
. T
40.3
T = 24° C ⇒ m& s = 0.333 kg hr
c.
Separate variables and integrate the balance equation:
dT
=
40.3m& s − 0.559T
Tf
10
t
0
dT
23
dt
10 13.4 − 0.559T
m& s = 0.333
T f = 23 ° C
E
t =−
11.26 a.
Integral energy balance
Q = ∆U = MCv ∆T =
bt
=t
L13.4 − 0559
. b23g O
1
ln M
P = 4.8 hr
0559
.
. − 0.559b10g PQ
M
N134
= 0 to t = 20 ming
250 kg
4.00 kJ
b60 − 20g° C
kg⋅° C
= 400
. × 104 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
M = 250 kg
Cv = 4 .00 kJ kg ⋅° C
T
Integrate:
t
t
0
0
dT
= 0.001Q& bt g
dt
t = 0, T = 20° C
&
dT = 0001
.
Q& dT ⇒ T = 20o C + Qdt
20 o C
Evaluate the integral by Simpson's Rule (Appendix A.3)
600 s
0
& = 30 33 + 4b33 + 35 + 39 + 44 + 50 + 58 + 66 + 75 + 85 + 95g
Qdt
3
+ 2b34 + 37 + 41 + 47 + 54 + 62 + 70 + 80 + 90g + 100 = 34830 kJ
⇒ T b600 sg= 20o C + e0.001 oC / kJj b34830 kJg = 54.8° C
c.
10 kW
Past 600 s, Q& = 100 +
bt − 600 sg = t 6
60 s
t
&
T = 20 + 0.001 Qdt
z
0
⇒ T = 548
. +
L
O
M600
t
P
t P
M &
= 20 + 0001
. M Qdt +
dt
6 P
M0
P
600
123
M
P
N 34830
Q
z
z
F t2
0001
.
6002 I
−
⇒ t bsg = 12000 bT − 248
. g
G
6 H6
2 JK
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
ρ=constant
dM tot
d( ρV)
&o ⇒
= m& i − m
= 800
. ρ − 4.00ρ
dt
dt
dV
= 4.00 L / s
dt
t = 0, V0 = 400 L
KCl Balance:
dM KCl
d( CV )
& o, KCl ⇒
= m& i , KCl − m
= 100
. × 8.00 − 400
. C
dt
dt
dC 8 − 8C
dV dt = 4
=
dC
dV
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
V
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 + 05
.t
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
When the tank overflows, V = 400 + 4t = 2000 ⇒ t = 400 s
1
C = 1= 0.96 g / L
2
b1+ 0.01 × 400g
11.28 a. Salt Balance on the 1st tank:
Accumulation=-Output
E
d(C S1V1 )
dC
v&
= − CS1v& ⇒ S1 = − CS1 = −0.08CS1
dt
dt
V1
CS1( 0) = 1500 500 = 3 g / L
Salt Balance on the 2nd tank:
Accumulation=Input-Output
E
d(C S2V2 )
dCS2
v&
= CS1v& − CS 2v& ⇒
= ( CS1 − CS 2 ) = 0.08( CS1 − CS 2 )
dt
dt
V2
CS 2 ( 0 ) = 0 g / L
Salt Balance on the 3rd tank:
Accumulation=Input-Output
E
d(C S3V3 )
dC
v&
= CS 2v& − CS 3v& ⇒ S3 = ( CS 2 − CS 3 ) = 0.04( CS 2 − CS 3 )
dt
dt
V3
CS 3 ( 0 ) = 0 g / L
b.
C S1, C S2, C S3
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 )⇒ d CS2 /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 )⇒ d CS3 /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 = 2 r1 = 0.2 CA
(ii) Rate of consumption of B in the 2nd reaction: −rB2 = r2 = 0.2CB2
b. Mole Balance on A:
Accumulation=-Consumption
E
d( C AV )
dC
= −01
. C AV ⇒ A = −01
. CA
dt
dt
t = 0, C A0 = 100
. mol / L
Mole Balance on B:
Accumulation= Generation-Consumption
E
d ( CB V )
dC
= 0.2CAV − 0.2CB2V ⇒ B = 0.2CA − 0.2CB2
dt
dt
t = 0, CB0 = 0 mol / L
11-23
11.29 (cont’d)
c.
2
C A, 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 2B < CA )⇒ d CB/dt =0.2(CA - C 2B ) becomes less positive
until dCB/dt changes to negative ( C 2B > 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 2B becomes positive also increases with increasing t
⇒ CC increases faster until CB decreases with increasing t ⇒ dCc/dt =0.2 C 2B 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=
ax x =1, y=1
a
⇒ 1=
⇒ a =1+ b
x +b
1+ b
pC5 H12 = yP = xp *C5 H12 ( 46o C) ⇒ y =
b. Raoult’s Law:
p *C5 H12 ( 46 o C ) = 10
Antoine Equation:
⇒ y=
xp *C5 H12 ( 46o C )
P
=
(6 .85221−
1064 .63
)
46 + 232. 00
xp *C5 H12 ( 46o C )
P
= 1053 mm Hg
0.7 × 1053
= 0.970
760
ax x=0.70, y=0.970
0.70a
R
.
0970
.
=
LL (1) |Ra = 1078
| y =
⇒S
x+b
0.70 + b
S
|T b = 0.078
| From part (a), a = 1+ bLLLLLLLLLLL ( 2)
T
c. Mole Balance on Residual Liquid:
Accumulation=-Output
E
dN L
= − n&V
dt
t = 0, N L = 100 mol
Balance on Pentane:
Accumu lation=-Output
E
d( N L x )
dN L
dx
ax
= − n&V y ⇒ x
+ NL
= − n&V
dt
dt
dt
x+b
E dN L / dt = − n&V
dx
n& F ax
I
=− V G
− xJ
dt
NL Hx + b K
t = 0, x = 0.70
d. Energy Balance: Consumption=Input
E
)
n&V ∆H vap = Q&
From part (c),
)
∆H vap =27.0 kJ/mol
t = 0, N L
dN L
= − n& V
dt
n&V
Q& 27.0
=
&
NL
Qt
100 27.0
= 100 mol
Q&
. kJ / molg
b270
&
Qt
N L = 100 − n&V t = 100 −
27.0
n&V =
Substitute this expression into the equation for dx/dt from part (c):
11-25
11.30 (cont’d)
& 270
dx
n& F ax
Q
. F ax
I
I
=− V G
− xJ = −
− xJ
& G
Hx + b
K
dt
N L Hx + b K
Qt
100 27.0
x(0) = 0.70
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
CHAPTER THIRTEEN
Problem
13.1
Methanol
Production
Rate
430,000 metric tons/year
51,114 kg/h
1,597 kmol/h
Process
stoichiometry:
CH4 + H20 ---> CH30H + H2
So that the required feed rates (with given assumptions) are
CH4 Feed Rate =
Steam Feed Rate =
Problem
1,597 kmol/h
1,597 kmol/h
cubic
meters/min
13.2
1
P =
r.h. = (pH20/p*H20
p*H20 =
0.0424
pH20 =
0.02968
yH20
0.029292
Basis:
1 mole of
component moles
N2
0.79
0.21
02
0.030176
Water
Total
1.03
component
N2
02
Total
Basis:
atm
@ 30C) x 100% =
bar
bar
70%
dry air (79 mole% N2, 21 mole% 02)
Mw
mass
mole frac
28
22.12
32
6.72
0.0293
18
0.54
29.38
1.0000
moles
Mw
0.79
0.21
1.00
mass
22.12
6.72 .
28.84 Difference in avg molecular weight
is due to presence of water; the
difference is slight.
1 km01 of CH4 burned
flow rate of air/km01
Problem
596 standard
28,75lkg/h
28
32
nat gas burned
13.3
Composition of effluent gas from burners
mole frac
mass frac
Component
km01
kg
3.2
0.0103
02
0.100
0.0088
N2
7.900
0.6990
221.2
0.7139
44.0
0.1420
co2
1.000
0.0885
0.1337
H20
2.302
0.2037
41.4
1.0000
309.8
1.0000
total
11.302
p*H20 @ 150C =
4.74 bar
< pH20
Therefore, there is no condensation in cooling the exhaust gases
to 15OC, which means the effluent gas and stack gas have the same
composition.
Volumetric
flow rate
effluent gas
stack gas
density of air =
density of stack gas =
specific gravity =
1,143 m3/kmol CH4 burned
392 m3/kmol CH4 burned
1.1471 kg/m3
0.7899 kg/m3
1 0.6886 relative to ambient air
13-1
I
Problem
13.4
Reformer
Temperature
Reformer
Temperature
Reformer
Pressure
Equilibrium
Temperature
855
1128
15.8
1128
C
K
atm
K
1.6 Mpa
CH4 + Hz0 ---> CO + 3Hz
Production
rates
CH4
H20
CO
CO2
H2
feed rate CH4 x (1 - fractional conversion)
feed rate H20 - fractional conversion x feed rate CH4
fractional conversion x feed rate CH4
feed rate CO2
feed rate + 3 x feed rate CH4 x fractional conversion
(a) methane:steam of 3:l
Stoichiometric
Feed
I
(kmol/h)(
1600
4800
0
0
0
6400
Table:
CH4
H20
co
co2
H2
Total
(kmol/h)
170
3,370
1,430
0
4,291
9,261
574.3668
KP~
Ratio1 574.366846
fractional conversion of CH4* =
=
Product
MolFrac
(kg/h)
0.0183
2,716
0.3639
60,655
0.1544
40,047
0.0000
0
0.4633
8,582
1.0000
112,000
MassFrac
0.0242
0.5416
0.3576
0.0000
0.0766
1.0000
lO"(-(11,769/T(K))+l3.1927)
( (Yco x YH2A3) / (YCH4 x YH20)
6.8939
-3.57E-07 Converge* = ((Kpl/Ratiol)-1)lOO
* the Goal Seek tool is used to adjust the fractional
conversion until Converge is close to zero
Methane Conversion =
) p2
1
Product Flow Rate =-I
(b)
methane:steam of 1:l
Stoichiometric
Table:
CH4
H20
co
co2
H2
Total
fractional
Feed
I
(kmol/h)l
1600
1600
0
0
0
3200
574.3668
Kpl
Ratio1 1653.61852
conversion of CH4 =
(kmol/h)
471
471
1,129
0
3,387
5,458
=
lO"(-(11,769/T(K))+l3.1927)
=
( kc0 x YH2A3)
/ (YCH4
x YH20)
Jp
2
0.7055
-65.266061Converge = ((Kpl/Ratiol)-1)lOO
Methane Conversion --I-[
Product Flow Rate ='
Product
MolFrac
(kg/h)
MassFrac
0.0863
7,538
0.1386
0.0863
8,480
0.1559
0.5810
0.2068
31,608
0.0000
0
0.0000
0.6205
6,773
0.1245
54.400
1.0000
1.0000
5,458 kmol/h
54,400 kg/h
13-2
Problem 13.4 (cont'd)
methane:steam
of 2:l
Stoichiometric
Table:
CH4
H20
co
co2
H2
Total
Feed
I
(kmol/h)l,
1600
3200
0
0
0
4800
Product
(kmol/h) MolFrac
MassFrac
(kg/h)
0.0476
248
0.0330
3,960
1,848
0.2462
33,256
0.3997
0.1802
37,869
0.4552
1,352
0
0.0000
0
.o.oooo
4,057
0.5406
8,115
0.0975
7,505
1.0000
83,200
1.0000
574.3668
=
lO"(-(11,769/T(K))+l3.1927)
Kpl
Ratio1
874.51201
=
( (ycO x YH2^3) / (YCH4 x YH20) )p2
fractional conversion of CH4 = 0.84529344
-34.321446 Converge = ((Kpl/Ratiol)-I)100
Methane
Conversion
=1
Product Flow Rate =
methane:steam
of
Stoichiometric
7,505 kmol/h
83,200 kg/h
4:l
Table:
CH4
H20
co
co2
H2
Total
Feed
I
(kmol/h)l
1600
6400
0
0
0
8000
Product
MassFrac
(kmol/h) MolFrac
(kg/h)
0.0147
130
0.0118
2,074
4,930
0.4506
88,733
0.6302
1,470
0.1344
41,171
0.2924
0.0000
0
0.0000
0
0.4032
8,822
0.0627
4,411
10,941
1.0000
140,800
1.0000
=
574.3668
lO"(-(11,769/T(K))+l3.1927)
Kpl
=
Ratio1 411.494126
( (Yco x YH2^3) / (YCH4 x YH~o) jp2
fractional conversion of CH4 = 0.91899524
39.5808125 Converge = ((Kpl/Ratiol)-I)100
Methane
Conversion = 1
Product Flow Rate =I[
Summary
Moles
Steam:Mole
CH4
1
2
3
4
Methane
H2
Conversion Produced
6,773
70.6%
84.5%
8,115
89.4%
8,582
91.9%
8,822
13-3
H2:CO
3.0000
3.0000
3.0000
3.0000
Problem
13.5
Reformer Temperature
Reformer Temperature
Reformer Pressure
Equilibrium
Temperature
Production
1
855
1128
15.79
1128
C
K
atm
K
1.6 Mpa
CH, + Hz0 ---> CO + 3H,
CO f Hz0 ---> COz f H,
rates
CH4 feed rate CH4 x (1 - fractional conversion)
H20 feed rate H20 - fractional conversion x feed rate CH4
- production rate of CO2
co fractional conversion x feed rate CH4
- production rate of CO2
co2 feed rate CO2 + production rate of CO2
H2 feed rate + 3 x feed rate CH4 x fractional conversion
+ production rate of CO2
methane:steam of 3:l
Stoichiometric
Feed
I
(kmol/h)[
CH4
1600
H20
4800
co
0
co2
0
H2
0
Total
6400
(kmol/h)
176
3,156
1,205
219
4,492
9,249
Product
MolFrac
(kg/h)
0.0190
2,811
0.3413
56,814
0.1303
33,740
0.0237
9,650
0.4857
8,985
1.0000
112,000
MassFrac
0.0251
0.5073
0.3013
0.0862
0.0802
1.0000
574.4
=
lO"(-(11,769/T(K))+l3.1927)
Kpl
Ratio1
574.4
=
( (Yco x YHZh3) / (YCHQ x YH20) )P2
0.2590
=
10"(1,197.8/T(K)-1.6485)
KP~
0.2590
=
Ratio2
(Y co2 x zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQP
Y t I 2 ) / (Yco x YH20)
fractional conversion of CH4*
0.89021022
Converge 1" -4.3538E-05
= ((Kpl/Ratiol)-1)*100
moles of CO2 formed**
219.322711
= ((Kp2/Ratio2)-l)*lOO
Converge 2"" -0.00028801
* Use Goal Seek tool to adjust CH4 conversion so that Converge 1 goes to zero.
** Use Goal Seek tool to adjust CO2 formation so that Converge 2 goes to zero.
CH4 Conv =
CH4 to CO =
-1
Product Flow Rate =mI
CO with water-gas shift reaction:CO
without water-gas shift reaction =
184.381
13-4
Problem 13.5 (cont'd)
methane:steam of 1:l
Stoichiometric
Feed
I
(Iunol/h)l
CH4
1600
H20
1600
co
0
co2
0
H2
0
Total
3200
*(kmol/h)
482
445
1,082
37
3,392
5,437
Product
MolFrac
(kg/h)
0.0886
7,706
0.0818
8,007
0.1990
30,286
0.0068
1,617
0.6239
6,784
1.0000
54,400
MassFrac
0.1416
0.1472
0.5567
0.0297
0.1247
1.0000
=
KP~
574.4
lO"(-(11,769/T(K))+l3.1927)
=
Ratio1
1662.1
(YCH4 x YH20) ) p2
( (Yco x Y,2^3 11 zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQ
=
KP~
0.2590
10A(1,197.8/T(K)-1.6485)
=
Ratio2
0.2590
(Y co2 x Y,,) / (Yco x YIi20)
fractional conversion of CH4*
0.69900166
Converge 1" -65.4436238
= ((Kpl/Ratiol)-1)*100
moles of CO2 formed**
36.7477842
Converge 2** -5.79043-05
= ((Kp2/Ratio2)-I)*100
CH4 Conv =
CH4 to CO =
Product Flow Rate =I1
methane:steam of 2:l
Stoichiometric
Feed
I
(kmol/h)l
CH4
1600
3200
H20
co
0
0
co2
H2
0
Total
4800
(kmol/h)
257
1,726
1,213
130
4,160
7,487
=
574.4
Kpl
=
876.6
Ratio1
=
0.2590
KP~
=
Ratio2
0.2590
fractional conversion of CH4*
Converge l* -34.4757385
moles of CO2 formed**
Converge 2** -1.52263-05
5:;4: g;Fh
Product
MolFrac
(kg/h)
0.0343
4,108
0.2306
31,074
0.1620
33,961
0.0174
5,737
0.5557
8,320
1.0000
83,200
lO^(-(11,769/T(K))+13.1927)
( (yco x ~~2~3) 1 (Y c H 4 x YHZO) ) P2
10"(1,197.8/T(K)-1.6485)
(Y c o2 x YH2) / (Yco x YH20)
0.83954105
= ((Kpl/Ratiol)-I)*100
130.381595
= ((Kp2/Ratio2)-l)*lOO
CH4 Conv =
CH4 to CO =
Product Flow Rate -Im[
13-5
MassFrac
0.0494
0.3735
0.4082
0.0690
0.1000
1.0000
Problem 13.5 (cont'd)
methane:steam of 4:l
Stoichiometric
Feed
I
(kmol/h)
CH4
H20
co
co2
H2
Total
1600
6400
0
0
0
8000
1
,(kmol/h)
133
4,634
1,169
299
4,700
10,934
Product zyxwvutsrqponmlkjihgfedcbaZYXWVUTSR
MolFrac
MassFrac
(kg/h)
0.0122
0.0151
2,126
0.4238
83,418
0.5925
0.1069
32,722
0.2324
0.0273
13,134
0.0933
0.4298
0.0668
9,400
140,800
1.0000
1.0000
574.4
=
lO"(-(11,769/T(K))+l3.1927)
Kpl
Ratio1
410.9
=
~~2 ~x 1 ! (YCH4 x YH20) ) p2
( (Yco x zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQP
0.2590
=
10"(1,197.8/T(K)-1.6485)
Kp2
Ratio2
0.2590
=
(Y co2 x YH2) / (Yco x YH20)
fractional conversion of CH4*
0.91695885
Converge 1* 39.77214561
= ((Kpl/Ratiol)-l)*lOO
moles of CO2 formed**
298.507525
Converge 2**
-l.O4E-05
= ((Kp2/Ratio2)-l)*lOO
CH4 Conv =
CH4 to CO =
Product Flow Rate =mI
Moles
Methane
H2
H2:CO
Steam:Mole CH4 Conversion Production Production
CO:H2
1
69.9%
3,392
3.1359 0.3188883
84.0%
4,160
3.4300 0.2915462
2
91.7%
4,492
3.7280 0.2682379
3
4,700
4.0217 0.2486487
4
89.0%
13-6
Problem
13.6
Reformer
Reformer
Temperature
Temperature
Pressure
855 C
1128 K
,15.79 atm
1.6 MPa
Production
rates
CH4 feed rate CH4 x (1 - fractional conversion)
H20 feed rate H20 - fractional conversion x feed rate CH4
- production rate of CO2
CO fractional conversion x feed rate CH4
- production rate of CO2
CO2 feed rate CO2 + production rate of CO2
H2 feed rate + 3 x feed rate CH4 x fractional conversion
+ production rate of CO2
Stoichiometric
Table:
COMPONENT
FEED
PRODUCT
MolFrac
CH4
1600
176
0.0190
4800
H20
3,156
0.3413
co
0
1,205
0.1303
co2
0
219
0.0237
H2
0
4,492
0.4857
Total
6400
9,249
1.0000
574.4
=
Kpl
574.4
=
Ratio1
0.2590
=
Kp2
Ratio2
0.2590
=
fractional conversion of CH4*
Converge 1* -4.35353-05
moles of CO2 formed**
Converge 2** -0.00028798
Reformer
Reformer
Pressure
Stoichiometric
CH4
H20
co
co2
H2
Total
T
T
lO"(-(11,769/T(K))+l3.1927)
( (Yco x Y,,^3) / (YCHI x YHZO) )P2
10"(1,197.8/T(K)-1.6485)
(Y CO2 x YH2) / (YCO x YH20)
0.8902102
((Kpl/Ratiol)-l)*lOO
219.=32271
=
((Kp2/Ratio2)-l)*lOO
750 c
1023 K
15.79 atm
1.6 MPa
Table:
FEED
1600
4800
0
0
0
6400
PRODUCT
576
3,510
759
266
3,338
8,448
MOLFRAC
0.0681
0.4155
0.0898
0.0314
0.3951
Kpl
48.79
Ratio1
48.79
Kp2
0.3329
Ratio2
0.3329
fractional conversion of CH4*
0.64015
Converge 1" - 6 . 4 5 1 9 3 - 0 5 =
moles of CO2 formed**
265.59039
Converge 2** 0.000608333
=
13-7
nCO/nCH4
nH2/nCH4
nCO/nC02
0.474156
2.086444
2.856465
((Kpl/Ratiol)-l)*lOO
((Kp2/Ratio2)-1)*100
Problem 13.6 (cont'd)
Reformer
Reformer
Stoichiometric
Temperature
Temperature
Pressure
800 C
1073 K
.15.79 atm
1.6 MPa
Table:
FEED
1600
4800
0
0
0
6400
CH4
H20
co
co2
H2
Total
PRODUCT
357
3,312
999
244
3,975
8,887
MOLFRAC
0.0401
0.3727
0.1124
0.0275
0.4472
Kpl
167.6
Ratio1
167.6
Kp2
0.2936
Ratio2
0.2936
fractional conversion of CH4*
0.7771058
Converge l* 0.000238464
((Kpl/Ratiol)-l)*lOO
moles of CO2 formed**
244: 4398
Converge 2** -2.48073-05
=
((Kp2/Ratio2)-l)*lOO
Reformer
Reformer
Stoichiometric
CH4
H20
co
co2
H2
Total
Temperature
Temperature
Pressure
900 c
1173 K
15.79 atm
1.6 MPa
Table:
FEED
1600
4800
0
0
0
6400
PRODUCT
88
3,086
1,311
201
4,738
9,424
MOLFRAC
0.0093
0.3275
0.1391
0.0214
0.5027
1.0000
Kpl
1443.6
Ratio1
1443.6
KP~
0.2359
Ratio2
0.2359
fractional conversion of CH4*
0.9451022
Converge l* -4.1853-05
=
moles of CO2 formed**
201.39252
Converge 2"" -0.00020159
=
13-8
((Kpl/Ratiol)-I)*100
((Kp2/Ratio2)-l)*lOO
Problem 13.6 (cont'd)
Reformer
Reformer
Stoichiometric
Temperature
Temperature
Pressure
950 c
' 1223 K
15.79 atm
1.6 MPa
Table:
FEED
1600
4800
0
0
0
6400
CH4
H20
co
co2
H2
Total
PRODUCT
39
3,054
1,377
185
4,869
9,523
MOLFRAC
0.0040
0.3207
0.1445
0.0194
0.5113
1.0000
Kpl
3712.3
Ratio1
3712.3
KP~
0.2142
Ratio2
0.2142
fractional conversion of CH4*
0.9759079
Converge l* 0.000662212
=
moles of CO2 formed**
184.93694
Converge 2** -2.0226E-05
=
1.60 MPa
T (Cl
nCO/nCH4
nH2/nCH4
nCO/nCO2
750
0.474
2.086
2.856
800
0.624
2.484
4.087
13-9
855
0.753
2.808
5.494
((Kpl/Ratiol)-l)*lOO
((Kp2/Ratio2)-l)*lOO
900
0.819
2.961
6.509
950
0.860
3.043
7.443
Problem 13.6 (cont'd)
Reformer
Reformer
Temperature
Temperature
Pressure
855 C
1128 K
*11.84 atm
1.2 MPa
Production
rates
CH4 feed rate CH4 x (1 - fractional conversion)
H20 feed rate H20 - fractional conversion x feed rate CH4
- production rate of Co2
CO fractional conversion x feed rate CH4
- production rate of CO2
CO2 feed rate CO2 + production rate of CO2
H2 feed rate + 3 x feed rate CH4 x fractional conversion
+ production rate of CO2
Stoichiometric
Table:
COMPONENT
FEED
PRODUCT
MolFrac
CH4
1600
116
0.0124
H20
4800
3,098
0.3307
co
0
1,266
0.1352
co2
0
218
0.0232
H2
0
4,670
0.4985
6400
9,368
Total
1.0000
574.4
=
Kpl
Ratio1
574.4
=
0.2590
=
Kp2
Ratio2
0.2590
=
fractional conversion of CH4*
Converge 1* 5.08455E-05
moles of CO2 formed**
Converge 2** -4.7046E-05
Reformer T
Pressure
Stoichiometric
CH4
H20
co
co2
H2
Total
lO"(-(11,769/T(K))+13.1927)
zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPON
x YH20) )P2
( (Yco x YH2&3) / (YCH4
10"(1,197.8/T(K)-1.6485)
(Y CO2 x yi-12) / (YCO x YH20)
0.9275941
((Kpl/Ratiol)-1)*100
217.=65358
=
((Kp2/Ratio2)-1)*100
1023 K
11.84 atm
1.2 MPa
Table:
FEED
1600
4800
0
0
0
6400
PRODUCT
472
3,405
861
267
3,651
8,656
MOLFRAC
0.0545
0.3934
0.0994
0.0309
0.4218
KP~
48.79
Ratio1
48.79
0.3329
KP~
0.3329
Ratio2
fractional conversion of CH4*
0.7049568
Converge 1* -0.00022618
moles of CO2 formed**
267 .=23789
Converge 2** 0.000779089
=
13-10
nCO/nCH4 0.537933
nH2/nCH4 2.281894
3.2207
nCO/nCO2
((Kpl/Ratiol)-l)*lOO
((Kp2/Ratio2)-l)*lOO
Problem 13.6 (cont'd)
Reformer Temperature
Reformer Temperature
Pressure
Stoichiometric
800 c
1073 K
*11.84 atm
Table:
FEED
1600
4800
0
0
0
6400
CH4
H20
co
co2
H2
Total
PRODUCT
265
3,221
1,092
243
4,249
9,071
MOLFRW
0.0292
0.3551
0.1204
0.0268
0.4685
KP~
167.6
Ratio1
167.6
KP~
0.2936
Ratio2
0.2936
fractional conversion of CH4*
0.8346471
Converge 1* -0.00043805
moles of CO2 formed**
243.14362
Converge 2** -2.3364E-05
=
Reformer Temperature
Reformer Temperature
Pressure
Stoichiometric
CH4
H20
co
co2
H2
Total
1.2 MPa
900 c
1173 K
11.84 atm
((Kpl/Ratiol)-l)*lOO
((Kp2/Ratio2)-l)*lOO
1.2 MPa
Table:
FEED
1600
4800
0
0
0
6400
PRODUCT
54
3,054
1,346
200
4,839
9,492
MOLFRAC
0.0057
0.3217
0.1418
0.0211
0.5098
1.0000
KP~
1443.6
Ratio1
1443.6
Kp2
0.2359
Ratio2
0.2359
fractional conversion of CH4*
0.966348
Converge l* -0.00083839
=
moles of CO2 formed**
200.31077
Converge 2** -2.46013-05
=
13-11
((Kpl/Ratiol)-l)*lOO
((Kp2/Ratio2)-l)*lOO