화학공정계산-4판-솔루션-elementary-principles-of-chemical-processes-4th-solution
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화학공정계산 4판 솔루션 elementary principles of chemical
processes, 4th solution
화학공학양론 (Kangwon National University)
Studocu is not sponsored or endorsed by any college or university
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CHAPTER TWO
2.1
(a)
(b)
(c)
2.2
(a)
(b)
(c)
2.3
(a)
Assume that a golf ball occupies the space equivalent to a 2 in x 2 in cube.
For a classroom with dimensions 40 ft x 40 ft x 15 ft:
The estimate could vary by an order of magnitude ore more, depending on
the assumptions made.
2.4
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2.5
2.6
2.7
2.8
2.9
Say h
depth of liquid
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2.10
(a)
(i) On the earth:
(ii) On the moon:
(b)
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2.11
(a)
(b)
2.12
(a)
(b)
(c)
2.13
(a)
(b)
2.14
-1
The t in the exponent has a coefficient of s .
(a)
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(b)
(a)
Yes, because when ln [( C A C A e ) / ( C A 0 C A e )] is plotted vs. t in rectangular
coordinates, the plot is a straight line.
0
ln ((CA-CAe)/(CA0-CAe))
2.15
50
100
150
200
0
-0.2
-0.4
-0.6
-0.8
-1
-1.2
-1.4
-1.6
t (min)
S lo p e = -0 .0 0 9 3 k = 9 .3 1 0
-3
m in
1
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(b)
ln [( C A C A e ) /( C A 0 C A e )] k t C A ( C A 0 C A e ) e
C A ( 0 .1 8 2 3 0 .0 4 9 5 ) e
C = m /V m = C V
3
9 .3 0 0 1 0
)(1 2 0 )
-2
0 .0 4 9 5 = 9 .3 0 0 1 0
g
3 0 .5 g a l
L
(a)
1/C vs. t. Slope= b, intercept=a
(b)
b s lo p e = 0 .4 7 7 L / g h ;
2
C
1.5
1
0.5
0
1
2
1/C = 0.4771t + 0.0823
3
2 8 .3 1 7 L
g /L
1 0 .7 g
4
5
6
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
1
2
t
3
4
5
t
C
(c)
-2
a In te rc e p t = 0 .0 8 2 L / g
3
0
C Ae
7 .4 8 0 5 g a l
2.5
1/C
2.16
( 9 .3 1 0
kt
C-fitted
C 1 / ( a b t ) 1 / [ 0 .0 8 2 0 .4 7 7 ( 0 )] 1 2 .2 g / L
t ( 1 / C a ) / b ( 1 / 0 .0 1 0 .0 8 2 ) / 0 .4 7 7 2 0 9 .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.
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2.17
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(a)
(b)
(c)
E(cal/mol), D0 (cm2/s)
ln D vs. 1/T, Slope=-E/R, intercept=ln D0.
In te rc e p t = ln D 0 = -3 .0 1 5 1 D 0 = 0 .0 5 c m
2
/s.
3.0E-03
2.9E-03
2.8E-03
2.7E-03
2.6E-03
2.5E-03
-10.0
-10.5
-11.0
-11.5
-12.0
-12.5
-13.0
-13.5
-14.0
ln D = -3666(1/T) - 3.0151
(d)
2.4E-03
2.3E-03
2.2E-03
2.1E-03
2.0E-03
S lo p e = E / R = -3 6 6 6 K E = (3 6 6 6 K )(1 .9 8 7 cal / m o l K ) = 7 2 8 4 cal / m o l
ln D
2.18
1/T
Spreadsheet
T
D
1/T
lnD (1/T)*(lnD)
347 1.34E-06 2.88E-03 -13.5
-0.03897
374.2 2.50E-06 2.67E-03 -12.9
-0.03447
396.2 4.55E-06 2.52E-03 -12.3
-0.03105
420.7 8.52E-06 2.38E-03 -11.7
-0.02775
447.7 1.41E-05 2.23E-03 -11.2
-0.02495
471.2 2.00E-05 2.12E-03 -10.8
-0.02296
Sx
Sy
Syx
Sxx
-E/R
ln D0
2.47E-03
-12.1
-3.00E-02
6.16E-06
-3666
-3.0151
D0
7284
E
0.05
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(1/T)**2
8.31E-06
7.14E-06
6.37E-06
5.65E-06
4.99E-06
4.50E-06
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CHAPTER THREE
3.1
(a)
(b)
(c)
(d)
(e)
(f)
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3.2
3.3
(a)
(b)
– No buildup of mass in unit.
– B and H at inlet stream conditions are equal to their tabulated values (which are
o
strictly valid at 20 C and 1 atm.)
– Volumes of benzene and hexane are additive.
– Densitometer gives correct reading.
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3.4
(a)
(b)
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3.5
3.6
(a)
(b)
(c)
(d)
(e)
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(f)
(g)
(h)
3.7
(a)
(b)
3.8
M s u s p e n s io n 5 6 5 g 6 5 g 5 0 0 g
,
M C aC O
3
215 g 65 g 150 g
(a)
V 4 5 5 m L m in , m 5 0 0 g m in
(b)
m / V 5 0 0 g / 4 5 5 m L 1.1 0 g m L
(c)
1 5 0 g C a C O 3 / 5 0 0 g s u s p e n s io n 0 .3 0 0 g C a C O 3 g s u s p e n s io n
3.9
Assume 100 mol mix.
mC H OH
2
1 0 .0 m o l C 2 H 5 O H
4 6 .0 7 g C 2 H 5 O H
5
mC H O
4
8
m o l C 2 H 5O H
7 5 .0 m o l C 4 H 8 O 2
mCH COOH
3
8 8 .1 g C 4 H 8 O 2
2
m ol C 4 H 8O 2
1 5 .0 m o l C H 3 C O O H
4 6 1 g C 2 H 5O H
6608 g C 4 H 8O 2
6 0 .0 5 g C H 3 C O O H
m o l C H 3C O O H
9 0 1 g C H 3C O O H
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xC H OH
2
3.10
5
461 g
461 g + 6608 g + 901 g
0 .0 5 7 8 g C 2 H 5 O H / g m ix
(a)
Unit
Crystallizer
Filter
Dryer
Function
Form solid gypsum particles from a solution
Separate particles from solution
Remove water from filter cake
(b)
(c)
% reco very =
0 .2 7 7 g + 3 .8 4 1 0
-5
g
1 0 0 % 9 9 .3 %
0 .2 7 7 g + 0 .0 0 1 8 6 g
3.11
(a)
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V
6 9 1 0 g E tO H
L
10365 g H 2O
7 8 9 g E tO H
SG
(6 9 1 0 + 1 0 3 6 5 ) g
1 9 .1 L
L
1 9 .1 2 3 L 1 9 .1 L
1000 g H 2O
L
0 .9 0 3
1000 g
(b)
V
( 6 9 1 0 1 0 3 6 5 ) g m ix
L
1 8 .4 7 2 L 1 8 .5 L
9 3 5 .1 8 g
% e rro r
( 1 9 .1 2 3 1 8 .4 7 2 ) L
1 0 0 % 3 .5 %
1 8 .4 7 2 L
3.12
3.13
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3.14
(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
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(b)
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3.15
(a)
6
(8.7 1 0 0 .4 0 ) k g C
44 kg C O 2
1. 2 8 1 0
7
k g C O 2 2 .9 1 0 k m o l C O 2
6 .6 7 1 0
5
k g C O 2 .3 8 1 0
4
5 .0 7 1 0
4
k g C H 4 3 .1 7 1 0
3
12 kg C
6
( 1.1 1 0 0 .2 6 ) k g C
28 kg CO
5
kmol CO
12 kg C
5
( 3 .8 1 0 0 .1 0 ) k g C
16 kg CH 4
12 kg C
7
m
5
4
( 1.2 8 1 0 6 .6 7 1 0 5 .0 7 1 0 ) k g
1 m e tric t o n
1 3 ,5 0 0
kmol CH 4
m e tric to n s
yr
1000 kg
M y i M i 0 .9 1 5 4 4 0 .0 7 5 2 8 0 .0 1 1 6 4 2 .5 g / m o l
3.16
(a)
Basis: 1 liter of solution
1000 m L
1 .0 3 g
5 g H 2SO 4
mL
m ol H 2SO 4
100 g
0 .5 2 5 m o l / L 0 .5 2 5 m o la r s o lu tio n
9 8 .0 8 g H 2 S O 4
(b)
t
V
V
55 gal
5 5 gal
3 .7 8 5 4 L
m in
60 s
gal
87 L
m in
mL
1 .0 3 g
0 .0 5 0 0 g H 2 S O 4
1 lb m
g
4 5 3 .5 9 g
3 .7 8 5 4 L
10
gal
3
1 L
144 s
mL
2 3 .6 lb m H 2 S O 4
(c)
u
V
A
t
L
u
3
87 L
m
m in
1000 L
45 m
1 m in
60 s
( 0 .0 6
2
/ 4) m
2
0 .5 1 3 m / s
88 s
0 .5 1 3 m / s
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3.17
(a)
n 3
1.5 0 L
0 .6 5 9 k g
1000 m ol
m in
L
8 6 .1 7 k g
1 1.4 7 m o l / m in
(b)
3.18
3.19
(a)
(b)
A se m ilo g p lo t o f C A v s. t is a stra ig h t lin e ln C A ln C A O kt
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k 0 .4 1 4 m in
1
ln C A O 0 .2 5 1 2 C A O 1.2 8 6 lb - m o le s ft
3
(c)
t 2 0 0 s C A 5 .3 0 m o l / L
3.20
(a)
(b)
(c)
(d)
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(e)
(f)
(g)
(h)
(i)
3.21
3.22
(a)
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(b)
(i)
Assume an average bathtub 5 ft long, 2.5 ft wide, and 2 ft high takes about 10 min to
fill.
For a full room, h 1 0 m
F
1000 kg
9 .8 1 m
3
2
m
s
1N
1 kg m / s
10 m
2
2m
2
The door will break before the room fills
(ii)
If the door holds, it will take
He will not have enough time.
3.23
(a)
h = L sin
(b)
3.24
5
F 2 .0 1 0 N
(a)
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(b)
3.25
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CHAPTER FOUR
4.1
(a)
Because the mass flow rate out is less than the mass flow rate in, there is a net
accumulation in the tank. Thus, transient.
(b)
Because the flow rate in and the flow rate out are consistent for the duration of the
process, the process is continuous.
(c)
4.2
(a)
Because the flow rates in and out are equal and unchanged throughout the process, the
process is steady-state.
(b)
Because the flow rates are consistent for the duration of the process, the process is
continuous.
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(4.2 cont’d)
(c)
4.3
(a)
Input – Output = 0
Steady state Accumulation = 0
No reaction Generation = 0,
Consumption = 0
(1) Total Mass Balance: 1 0 0 .0 k g / h m v m l
(2) Benzene Balance: 0 .5 5 0 1 0 0 .0 k g B / h 0 .8 5 0 m v 0 .1 0 6 m l
Solve (1) & (2) simultaneously m v 5 9 .7 k g h , m l 4 0 .3 k g h
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4.4
(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.
(a)
X-large:
25 broken eggs/min
35 unbroken eggs/min
120 eggs/min
0.30 broken egg/egg
Large:
0.70 unbroken egg/egg
n 1 broken eggs/min
n 2 unbroken eggs/min
(b)
(c)
n 1 n 2 5 0 la rg e e g g s m in
(d)
4.5
(a)
(b)
3 unknowns ( m 1 , m 2 , m 3 )
– 2 balances
– 1 feed ratio
0 DF
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(c)
F e e d r a tio : m 1 / m 2 4 5 / 5 5
(1 )
S b a la n c e : 0 .1 5 m 1 m 2 0 .6 6 7 ( 2 )
S o lv e s im u lta n e o u s ly m 1 0 .4 9 lb m s tr a w b e r r i e s , m 2 0 .5 9 lb m s u g a r
4.6
(a)
4 unknowns ( m 1 , m 2 , V 4 0 , m 3 )
– 2 balances
– 2 specific gravities
0 DF
(b)
m1
1 ft
3
0 .8 7 7 6 2 .4 lb m
7 .4 8 0 5 g a l
ft
3
2 1 9 5 lb m
Overall balance: m 1 m 2 m 3
(1)
C2H5OH balance: 0 .7 5 0 m 1 0 .4 0 0 m 2 0 .6 0 0 m 3
Solve (1) & (2) simultaneously m 2 1 6 4 6 lb m , , m 3 3 8 4 1 lb m
(2)
V 40
4.7
300 gal
1 6 4 6 lb m
ft
3
0 .9 5 2 6 2 .4 lb m
7 .4 8 0 5 g a l
1 ft
3
207 gal
(a)
3 unknowns ( n 1 , n 2 , n 3 )
– 2 balances
1 DF
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(b)
(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.
4.8
(a)
4 unknowns ( n 1 , n 2 , n 3 , v ) – 2 balances – 1 density – 1meter reading = 0 DF
Assume linear relationship: v a R b
Slope: a
v 2 v 1
R 2 R1
9 6 .9 4 0 .0
50 15
1. 6 2 6
DA balance: 0 .9 9 0 0 n 1 0 .9 0 0 n 3
(1)
Overall balance: n 1 n 2 n 3
(2)
Solve (1) & (2) simultaneously n 1 5 8 9 0 lb - m o le s / h , n 3 6 4 8 0 lb - m o le s / h
(b)
Bad calibration data, not at steady state, leaks, 7% value is wrong, v R relationship is
not linear, extrapolation of analyzer correlation leads to error.
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4.9
4.10
(a)
(b)
1 0 0 0 to n s w e t s u g a r
3 to n s H 2 O
day
1 0 0 to n s w e t s u g a r
3 0 to n s H 2 O / d a y
1 0 0 0 to n s W S 0 .8 0 0 to n s D S 2 0 0 0 lb m $ 0.1 5 3 6 5 d a y s
day
to n W S
to n
lb m
7
$ 8.8 1 0 p e r y e a r
year
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(c)
(d)
4.11
The evaporator is probably not working according to design specifications since
x w 0 .0 3 6 1 0 .0 4 5 0 .
(a)
Total mole balance: n 1 n 2 n 3 n 2 4 0 .1 1.5 5 6 3 8 .5 4 m o l / h
Water balance: 0 .0 4 0 4 0 .1 1 .5 5 6 x 3 8 .5 4 x 1 .2 1 0 3 m o l H 2 O / m o l
(b)
The calcium chloride pellets have reached their saturation limit. Eventually the mole
fraction will reach that of the inlet stream, i.e. 4%.
4.12
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(a)
Water removal rate: 2 0 0 .0 1 9 5 .0 5 .0 m l / m in
(b)
v 1 5 0 0 5 .0 1 5 0 5 m l / m in
c
3 8 .8 m g u re a /m in
0 .0 2 5 8 m g u re a /m l
1 5 0 5 m l/m in
(c)
4.13
(a)
(b)
Overall balance: 1 0 0 m w m 1 0
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First 4 evaporators
(c)
Y w 0 .3 1
x 4 0 .0 3 9 8
4.14
(a)
Overall process: 2 unknowns ( m 3 , m 5 )
Bypass:
– 2 balances
0 DF
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2 unknowns ( m 1 , m 2 )
– 1 independent balance
1 DF
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Evaporator: 3 unknowns ( m 1 , m 3 , m 4 )
Mixing point:
– 2 balances
1 DF
3 unknowns ( m 2 , m 4 , m 5 )
– 2 balances
1 DF
Overall mass balance: 1 0 0 m 3 m 5
Mixing point mass balance: m 4 m 2 m 5
(1)
Mixing point S balance: 0 .5 8 m 4 0 .1 2 m 2 0 .4 2 m 5
(2)
Solve (1) and (2) simultaneously
Bypass mass balance: 1 0 0 m 1 m 2
(b)
m 1 9 0 .0 5 k g , m 2 9 .9 5 k g , m 3 7 1.4 k g , m 4 1 8 .6 5 k g , m 5 2 8 .6 k g p ro d u c t
Bypass fraction:
(c)
4.15
m2
0 .0 9 5
100
Over-evaporating could degrade the juice; additional evaporation could be uneconomical;
a stream consisting of 90% solids could be hard to transport.
(a)
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(b)
Bypass point mass balance: m 3 6 0 0 0 4 5 0 0 1 5 0 0 k g / h
Mass balance on treatment unit: m 5 4 5 0 0 2 2 0 .2 4 2 7 9 .8 k g / h
(c)
m 1 ( k g /h )
m 2 ( k g /h )
m 3 ( k g /h )
m 4 ( k g /h )
m 5 ( k g /h )
1000
1000
0
4 8 .9
951
0 .0 0 2 7 1
951
0 .0 0 2 7 1
2000
2000
0
9 7 .9
1902
0 .0 0 2 7 1
1902
0 .0 0 2 7 1
3000
3000
0
147
2853
0 .0 0 2 7 1
2853
0 .0 0 2 7 1
4000
4000
0
196
3804
0 .0 0 2 7 1
3804
0 .0 0 2 7 1
5000
4500
500
220
4280
0 .0 0 2 7 1
4780
0 .0 0 7 8 1
6000
4500
1500
220
4280
0 .0 0 2 7 1
5780
0 .0 1 5 4
7000
4500
2500
220
4280
0 .0 0 2 7 1
6780
0 .0 2 0 7
8000
4500
3500
220
4280
0 .0 0 2 7 1
7780
0 .0 2 4 7
9000
4500
4500
220
4280
0 .0 0 2 7 1
8780
0 .0 2 7 7
10000
4500
5500
220
4280
0 .0 0 2 7 1
9780
0 .0 3 0 1
x5
m 1 vs. x6
0 .0 3 5 0 0
x 6 (k g C r / k g )
0 .0 3 0 0 0
0 .0 2 5 0 0
0 .0 2 0 0 0
0 .0 1 5 0 0
0 .0 1 0 0 0
0 .0 0 5 0 0
0 .0 0 0 0 0
0
2000
4000
6000
m
1
8000
10000
(k g /h )
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12000
m 6 ( k g /h )
x6
lOMoARcPSD|32610952
(d)
4.16
Cost of additional capacity – installation and maintenance, revenue from additional
recovered Cr, anticipated wastewater production in coming years, capacity of waste
lagoon, regulatory limits on Cr emissions.
(a)
(b)
Basis: 1 mol A feed
n A0 1
Constants: a 3 .8 7
b 9 .7 4
e
nB0 1
nC 0 n D 0 n I 0 0
c 4 .8 7
9 .7 4 9 .7 4 4 3 .8 7 4 .8 7 0 .6 8 8
2 3 .8 7
1
2
e1
e 2 1 .8 3 is a ls o a s o lu tio n b u t le a d s to a n e g a tiv e c o n v e r s io n
Fractional conversion: X A X B
nA0 nA
n A0
e1
0 .6 8 8
n A0
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4.17
(c)
n A 0 80, nC 0 n D 0 n J 0 0
(d)
Cost of reactants, selling price for product, market for product, rate of reaction, need for
heating or cooling, and many other items.
(a)
A 2B C
ln A 0 ln K e 1 1 1 4 5 8 T1 ln 1 0 .5 1 1 4 5 8 3 7 3 2 8 .3 7 A 0 4 .7 9 1 0
(b)
At equilibrium,
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13
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(c)
Basis: 1 mol A (CO)
n A0 1
nB0 1
n C 0 0 n T 0 2 , P 2 atm , T 4 2 3K
(For this particular set of initial conditions, we get a quadratic equation. In general, the
equation will be cubic.)
e 0 .1 5 6 , 0.844
(d)
Reject the second solution, since it leads to a negative n B .
Use the equations from part b.
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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 a
4.18
(a)
C H 4 O 2 H C H O H 2 O
(1)
C H 4 2 O 2 C O 2 2 H 2 O
(2)
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(b)
n 1 5 0 1 2
(1)
n 2 5 0 1 2 2
(2)
n 3 1
(3)
n 4 1 2 2
(4)
n 5 2
(c)
(5)
Fractional conversion:
5 0 n1
50
Fractional yield:
n3
50
0 .9 0 0 n1 5 .0 0 m o l C H
4
/s
0 .8 5 5 n 3 4 2 .7 5 m o l H C H O /s
y C H 0 .0 5 0 0 m o l C H 4 /m o l
4
E q u a tio n 3 1 4 2 .7 5
y O 0 .0 2 7 5 m o l O 2 /m o l
2
E q u a tio n 1 2 2 .2 5
E q u a tio n 2 n 2 2 .7 5 y H C H O 0 .4 2 7 5 m o l H C H O /m o l
E q u a tio n 4 n 4 4 7 .2 5
y H O 0 .4 7 2 5 m o l H 2 O /m o l
2
E q u a tio n 5 n 5 2 .2 5
y C O 0 .0 2 2 5 m o l C O 2 /m o l
2
Selectivity: [( 4 2 .7 5 m o l H C H O /s )/( 2 .2 5 m o l C O /s ) 1 9 .0 m o l H C H O /m o l C O
2
4.19
(a)
2C O 2 2C O O 2
2A 2B C
O 2 N 2 2N O
C D 2E
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2
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(b)
(c)
nA0 = nC0 = nD0 = 0.333, nB0 = nE0 = 0 e1 =0.0593, e2 = 0.0208
yA = 0.2027, yB = 0.1120, yC = 0.3510, yD = 0.2950, yE = 0.0393
(d)
(Solution given following program listing.)
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4.20
(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
7 u n k n o w n s ( m A 0 , f , x R A , m B 0 , m 3 , x R 3 , m P ) 5 e q u a tio n s 2 d e g re e s o f fre e d o m
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(b)
Mass balance on splitting point: mA0 = mB0 + f mA0
(1)
Mass balance on reactor: 2 mB0 = m3
(2)
99% conversion of R: xR3 m3 = 0.01 xRA mB0
(3)
Mass balance on mixing point: m3 + f mA0 = mP
(4)
R balance on mixing point: xR3 m3 + xRA f mA0 = 0.0075 mP
(5)
Given xRA and mP, solve simultaneously for mA0, mB0, f, m3, xR3
(c)
mA0 = 2778 kg A/h
mB0 = 2072 kg B/h
fA = 0.255 kg bypass/kg fresh feed
(d)
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f (kg bypass/kg fresh feed)
f vs. xRA
4.21
(a)
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0.00
0.02
0.04
0.06
0.08
0.10
0.12
xRA (kg R/kg A)
Basis: 100 mol feed/h
1 0 0 m o l/h
3 2 m o l C O /h
n 1 ( m o l /h )
r e a cto r
co n d .
n3 (m ol C H 3 O H / h)
.1 3 m o l N 2 /m o l
64 m ol H 2 / h
4 mol N 2 / h
5 00 m ol / h
x 1 ( m o l N 2 /m o l)
n3 (m ol / h)
x 2 ( m o l C O / m o l)
x 1 ( m o l N 2 /m o l)
1 - x1 - x2 ( m o l H 2 / h )
x 2 ( m o l C O / m o l)
1 - x1 - x2 ( m o l H 2 / h )
P urg e
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 .1 4 8 )
n 3 = 27 mol/h
A to m ic C : 3 2 (1) n 2 (1) 2 7 x 2 (1)
A to m ic H : 6 4 ( 2 ) n 2 ( 4 ) 2 7 (1 0 .1 4 8 x 2 )( 2 )
n 2 2 4 .3 m o l C H 3 O H / h
=>
x 2 0 .2 8 4 m o l C O / m o l
Overall CO conversion: 100*[32-0.284(27)]/32 = 76%
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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.
4.22
(a)
Basis: 1000 g gas
Species
m (g)
MW
n (mol)
mole % (wet)
mole % (dry)
C3H8
800
44.09
18.145
77.2%
87.5%
C4H10
150
58.12
2.581
11.0%
12.5%
H2O
50
18.02
2.775
11.8%
Total
1000
23.501
100%
100%
Total moles = 23.50 mol, Total moles (dry) = 20.74 mol
Ratio: 2.775 / 20.726 = 0 .1 3 4 m o l H 2 O / m o l d ry g a s
(b)
C3H8 + 5 O2 3 CO2 + 4 H2O, C4H10 + 13/2 O2 4 CO2 + 5 H2O
Theoretical O2:
g C not
H change
1 k m o lfor
C incomplete
H
5 k m combustion
ol O
1 0 0 The
k g ganswer
a s 8 0 kdoes
C 3H 8:
C 4 H 10 :
3
h
8
100 kg gas
1 0 0 k g g as 1 5 k g C 4 H 10
h
100 kg gas
3
8
2
4 4 .0 9 k g C 3 H 8 1 k m o l C 3 H 8
1 k m o l C 4 H 10
9 .0 7 k m o l O 2 / h
6 .5 k m o l O 2
5 8 .1 2 k g C 4 H 1 0 1 k m o l C 4 H 1 0
1 .6 8 k m o l O 2 / h
Total: (9.07 + 1.68) kmol O2/h = 10.75 kmol O2/h
Air feed rate:
1 0 .7 5 k m o l O 2
h
1 k m o l A ir
1 .3 k m o l a ir fe d
.2 1 k m o l O 2 1 k m o l a ir re q u ire d
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6 6 .5 k m o l a ir / h
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4.23
(a)
C4H10 + 13/2 O2 4 CO2 + 5 H2O
(b)
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4.24
(a)
C3H8 + 5 O2 3 CO2 + 4 H2O
H2 +1/2 O2 H2O
C3H8 + 7/2 O2 3 CO + 4 H2O
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(b)
4.25
(a)
C5H12 + 8 O2 5 CO2 + 6 H2O
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(b)
(c)
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4.26
(a)
Basis: 5000 kg coal/h; 50 kmol a ir m in 3 0 0 0 k m o l a ir h
5 0 0 0 kg c o a l / h
0 .7 5 kg C / k g
n 1 ( km o l O 2 / h )
0 .1 7 k g H / k g
n 2 ( km o l N 2 / h )
0 .0 2 k g S / k g
C + 0 2 - -> C O 2
0 .0 6 k g a s h / k g
2 H + 1 /2 O 2 - - > H 2 O
3 0 0 0 km o l a ir / h
n 3 ( km o l C O 2 / h )
0 .1 n 3 ( km o l C O / h )
S + O 2 - -> S O 2
n 4 ( km o l S O 2 / h )
C + 1 /2 O 2 - - > C O
n 5 ( km o l H 2 O / h )
0 .2 1 k m o l O 2 / km o l
0 .7 9 k m o l N 2 / km o l
m o k g s la g / h
T h e o re tic a l O 2 :
Total = (312.2+210.4 + 3.1) kmol O2/h = 5 2 5 .7 k m o l O 2 h
Excess air:
(b)
6 3 0 5 2 5 .7
1 0 0 % 1 9 .8 % e x c e s s a ir
5 2 5 .7
Balances:
n 3 2 6 6 .8 k m o l C O 2 h , 0 .1 n 3 2 6 .7 k m o l C O h
n 4 3 .1 k m o l S O 2 h
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n 1 1 3 6 .4 k m o l O 2 / h
Stack gas total 3 2 2 3 k m o l h
Mole fractions:
x C O 2 6 .7 3 2 2 4 8 .3 1 0
x SO
2
3 .1 3 2 2 4 9 .6 1 0
3
4
m ol C O m ol
m ol SO 2 m ol
(c)
SO 2
1
2
O 2 SO 3
S O 3 H 2 O H 2S O 4
4.27
3 .1 k m o l S O 2
1 km ol SO 3
1 km ol H 2S O 4
9 8 .0 8 k g H 2 S O 4
h
1 km ol SO 2
1 km ol SO 3
km ol H 2S O 4
304 kg H 2S O 4 h
Basis 100 mol dry fuel gas. Assume no solid or liquid products!
n1 (m ol C )
1 00 m ol d ry g as
n2 (m ol H )
n3 (m ol S)
C + 0 2 - -> C O 2
C + 1 /2 O 2 - - > C O
2 H + 1 /2 O 2 - - > H 2 O
S + O 2 - -> S O 2
n 4 ( m o l O 2)
: n2 2 n5
O balance
: 2 n 4 100 [ 2(0.720)
20 % excess
O 2 : (1.20) (74.57
0 .0 0 0 5 9 2 m o l S O 2 / m o l
0 .2 5 4 m o l O 2 / m o l
n 5 ( m o l H 2O ( v) )
( 2 0% excess)
H balance
0 .7 2 0 m o l C O 2 / m o l
0 .0 2 5 7 m o l C O / m o l
0.0257
0.0592
2 (0.000592)
2 (0.254)]
0.25 n 2 ] n 4
n5
n2 = 183.6 mol H, n4 = 144.6 mol O2, n5 = 91.8 mol H2O
Total moles in feed: 258.4 mol (C+H+S) 28.9% C, 71.1% H, 0.023% S
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CHAPTER FIVE
5.1
5.2
(a)
(b)
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5.3
(a)
(b)
n CO
2
5.4
755 m m H g
0 .0 8 2 0 6
3
m a tm
k m o lK
1 a tm
760 m m H g
1 .1 1 0
-3
m
3
/ m in 1 0 0 0 m o l
300 K
0 .0 4 4 m o l/m in
1 km ol
Basis: Given flow rates of outlet gas. Assume ideal gas behavior
3
o
3 1 1 m / m in , 8 3 C , 1 a tm
n 3 ( k m o l / m in )
1 ( k g / m in )
m
0 .7 0 k g H 2 O / k g
0 .1 2 k m o l H 2 O / k m o l
0 .3 0 k g S / k g
0 .8 8 k m o l d r y a ir / k m o l
n 2 ( k m o l a i r / m i n )
3
V ( m / m in )
4 ( k g S / m in)
m
2
o
1 6 7 C , - 4 0 cm H 2 O gauge
(a)
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(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.
5.5
(a)
(b)
5.6
(c)
CO2 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.
(a)
It is safer to release a mixture that is too lean to ignite.
If a mixture that is rich is released in the atmosphere, it can diffuse in the air and the
C3H8 mole fraction can drop below the UFL, thereby producing a fire hazard.
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(b)
fuel-air mixture
n 1 ( m o l / s )
y C H 0 .0 4 0 3 m o l C 3 H 8 / m o l
3
n C H
3
8
8
n 3 ( m o l / s )
0 .0 2 0 5 m o l C 3 H 8 / m o l
150 m ol C 3H 8 / s
diluting air
n 2 ( m o l / s )
(c)
(d)
5.7
o
2 4 C , 1 a tm
6 0 0 0 m L / m in
n in (m o l / m in )
0 .2 0 6 O 2
0 .7 7 4 N 2
0 .0 2 0 H 2 O
o
lungs
3 7 C , 1 a tm
n o u t ( m o l / m in )
0 .1 5 1 O 2
blood 0 .0 3 7 C O 2
0 .7 5 0 N 2
0 .0 6 2 H 2 O
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(a)
(b)
5.8
(a)
B a s is : 1 .0 m o l fe e d
9 0 % N O c o n v e rs io n : n 1 0 .1 0 ( 0 .2 0 ) 0 .0 2 0 m o l N O N O re a c te d = 0 .1 8 m o l
O 2 b a la n c e: n 2 0 .8 0 ( 0 .2 1 )
0 .1 8 m o l N O 0 .5 m o l O 2
m ol N O
0 .0 7 8 0 m o l O 2
N 2 b a la n c e : n 3 0 .8 0 ( 0 .7 9 ) 0 .6 3 2 m o l N 2
n4
0 .1 8 m o l N O 1 m o l N O 2
1 m ol N O
0 .1 8 m o l N O 2 n f n 1 n 2 n 3 n 4 0 .9 1 m o l
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y NO
yO
2
0 .0 2 0 m o l N O
0 .0 2 2
m ol N O
0 .9 1 m o l
0 .0 8 6
m ol O 2
m ol
m ol
yN
2
0 .6 9 5
m ol N 2
m ol
y NO
2
0 .1 9 8
(b)
nf = n0
Pf
( 1 m o l)
P0
P (a tm ) =
0 .9 5 m o l
380 kPa
360 kPa
1 0 1 .3
5.9
360 kPa
kPa
a tm
3 .5 5 a tm
B a s is : 1 0 0 k m o l d ry p ro d u c t g a s
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m ol N O 2
m ol
lOMoARcPSD|32610952
(a)
N 2 b a la n c e : 0 .7 9 n 2 0 .8 4 2 ( 1 0 0 ) n 2 = 1 0 6 .6 k m o l a ir
(b)
5.10
B a s is: 1 .0 0 1 0
1. 0 0 1 0
6
6
g a l. w a s te w a te r d a y .
gal / day
E fflu e n t g a s : 6 8 F , 2 1 .3 p s ia (a s s u m e )
n 1 (lb -m o le s H 2 O /d a y)
0 .0 3 n 1 (lb -m o le s N H 3 /d a y)
300 10
6
ft
3
N e g le c t e v a p o ra tio n o f w a te r .
a ir / d a y
n 2 (lb -m o le s a ir/d a y)
n 3 (lb -m o le s N H 3 /d a y)
300 10
6
ft
3
a ir / d a y
6 8 F , 2 1 .3 p s ia
n 1 (lb -m o le s H 2 O /d a y)
n 2 (lb -m o le s a ir/d a y)
n 4 (lb -m o le s N H 3 /d a y)
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(a)
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5.11
n e x it
PV
RT
205 kPa
1 0 .0 m
3
8 .3 1 4
m k P a
k m o lK
3
/ m in
0 .3 7 7 k m o l / m in
653 K
A ir fe e d to fu rn a c e : n a ir
0 .0 6 6 7 n 0 (k m o l H 2 S fe d ) 1.5 k m o l O 2
1 k m o l a ir
1 k m o l H 2 S 0 .2 1 k m o l O 2
(m in )
0 .4 7 6 4 n 0 k m o l a ir / m in
O v e ra ll N 2 b a la n c e : n 3
O v e ra ll S b a la n c e : n 6
0 .4 7 6 4 n 0 (k m o l a ir) 0 .7 9 k m o l N 2
(m in )
m in
0 .2 0 0 n 0 (k m o l H 2 S )
(m in )
1 km ol S
1 km ol H 2S
0 .3 7 6 4 n 0 ( k m o l N 2 / m in )
0 .2 0 0 n 0 (k m o l S / m in )
O v e ra ll C O 2 b a la n c e : n 5 0 .8 0 0 n 0 (k m o l C O 2 / m in )
O v e ra ll H b a la n c e :
0 .2 0 0 n 0 (k m o l H 2 S )
2 km ol H
(m in )
1 km ol H 2S
n 4 = 0 .2 0 0 n 0 (k m o l H 2 O / m in )
n a ir = 0 .4 7 6 4 (0 .2 4 k m o l a ir / m in ) 0 .1 1 4 k m o l a ir / m in
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n 4 k m o l H 2 O
2 km ol H
m in
1 km ol H 2 O
lOMoARcPSD|32610952
5.12
B a s is : 1 0 0 k g o re fe d 8 2 .0 k g F e S 2 ( s ), 1 8 .0 k g I.
V out m
1 0 0 kg ore
3
(S T P )
n 2 (k m o l S O 2 )
n 3 (k m o l S O 3 )
0 .6 8 3 3 k m o l F e S 2
18 kg I
n 4 (k m o l O 2 )
n 5 (k m o l N 2 )
4 0 % ex cess air
m 6 (k g F e S 2 )
m 7 (k g F e 2 O 3 )
18 kg I
n 1 ( k m o l)
0 .2 1 O 2
0 .7 9 N 2
V1
m
3
(S T P )
2 F e S 2 (s ) 121 O 2 (g ) F e 2 O 3 (s ) 4 S O 2 (g )
2 F e S 2 (s ) 125 O 2 (g ) F e 2 O 3 (s ) 4 S O 3 (g )
(a)
n1
n2
n3
n
4
0 .6 8 3 3 k m o l F e S 2 7 .5 k m o l O 2 1 k m o l a ir re q ' d 1.4 0 k m o l a ir fe d
2 km ol FeS 2
0 .2 1 k m o l O 2
( 0 .8 5 )( 0 .4 0 ) 0 .6 8 3 3 k m o l F e S 2 4 k m o l S O 2
2 km ol FeS 2
( 0 .8 5 )( 0 .6 0 ) 0 .6 8 3 3 k m o l F e S 2 4 k m o l S O 2
2 km ol FeS 2
0 .2 1 1 7 .0 8 k m o l O
.6 9 7 k m o l S O
.4 6 4 6 k m o l S O
2
3
fe d
0 .4 6 4 6 k m o l S O 2
0 .6 9 7 0 k m o l S O 3
2
5 .5 k m o l O
4 km ol SO
7 .5 k m o l O
4 km ol SO
2
1 7 .0 8 k m o l a ir
k m o l a ir r e q ' d
1 .6 4 1 k m o l O
2
2
2
3
V o u t = 0 .4 6 4 6 + 0 .6 9 7 0 + 1 .6 4 1 + 1 3 .4 9 k m o l 2 2 .4 S C M (S T P )/k m o l
3 6 5 S C M /1 0 0 k g o re fe d
y SO
2
0 .4 6 4 6 k m o l S O 2
1 6 .2 8 5 k m o l
1 0 0 % 2 .9 % ; y S O 4 .3 % ; y O 1 0 .1 % ; y N 8 2 .8 %
3
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2
2
lOMoARcPSD|32610952
(b)
converter
n SO
0 .4 6 4 6 k m o l S O 2
0 .6 9 7 k m o l S O 3
1. 6 3 3 k m o l O 2
1 3 .4 9 k m o l N 2
n SO
nO
2
nN
2
2
3
( k m o l)
( k m o l)
( k m o l)
( k m o l)
L e t (k m o l) = e x te n t o f re a c tio n
0 .4 6 4 6
0 .6 9 7
y SO
, y SO
2
3
1
1 6 .2 9 - 2
1 6 .2 9 - 12
1 .6 4 1 12
1 3 .4 9
yO
, yN
2
2
1
1 6 .2 9 - 2
1 6 .2 9 - 12
n S O 0 .4 6 4 6
2
n S O 0 .6 9 7
3
n O 1 .6 4 1 12
2
nN
1 3 .4 9
2
n = 1 6 .2 9 - 12
K p (T )=
P y SO
3
P y SO ( P y O )
2
1
2
2
( 0 .6 9 7 ) 1 6 .2 9 12
1
2
( 0 .4 6 4 6 ) 1 .6 4 1 12
-1
1
2
P 2 K p (T )
-1
P = 1 a tm , T = 6 0 0 C , K p 9 .5 3 a tm 2 0 .1 7 0 7 k m o l
o
n S O 0 .2 9 3 9 k m o l f S O
2
0 .4 6 4 6 0 .2 9 3 9 k m o l S O 2 r e a c te d
2
0 .3 6 7
0 .4 6 4 6 k m o l S O 2 f e d
-1
P = 1 a tm , T = 4 0 0 C , K p 3 9 7 a tm 2 0 .4 5 4 8 k m o l
o
n S O 0 .0 0 9 8 k m o l f S O 0 .9 7 9
2
2
The gases are initially heated in order to get the reaction going at a reasonable rate. Once
the reaction approaches equilibrium the gases are cooled to produce a higher equilibrium
conversion of SO2.
(c)
S O 3 le a v in g c o n v e rte r: (0 .6 9 7 0 + 0 .4 6 8 7 ) k m o l = 1 .1 5 6 k m o l
1 .1 5 6 k m o l S O 3 1 k m o l H 2 S O 4 9 8 k g H 2 S O 4
m in
S u lfu r in o re:
1 km ol SO 3
0 .6 8 3 k m o l F e S 2
1 1 3 .3 k g H 2 S O 4
2 .5 9
2 km ol S
3 2 .1 k g S
km ol FeS 2
km ol
0 .6 8 3 k m o l F e S 2
1 3 3 .9 k g H 2 S O 4
4 3 .8 k g S
4 3 .8 k g S
kg S
2 km ol S
km ol FeS 2
1 1 3 .3 k g H 2 S O 4
kg H 2S O 4
4 3 .8 k g S
100% conv.of S :
km ol
3 .0 6
1 km ol H 2S O 4 98 kg
1 km ol S
kg H 2S O 4
kg S
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km ol
1 3 3 .9 k g H 2 S O 4
lOMoARcPSD|32610952
The sulfur is not completely converted to H2SO4 because of (i) incomplete oxidation of
FeS2 in the roasting furnace, (ii) incomplete conversion of SO2 to SO3 in the converter.
5.13
n 4 (k m o l C O / h )
n 5 (k m o l H 2 / h )
Reactor
n 1 ( k m o l C O / h )
n 2 ( k m o l H 2 / h )
separator
100 km ol C O / h
n 4 ( k m o l C O /h )
n 3 (k m o l H 2 / h )
T (K ), P (k P a )
n 6 (k m o l C H 3 O H /h )
n 5 ( k m o l H 2 /h )
n 6 ( k m o l C H 3 O H /h )
H x s (% H 2 e x c e s s )
T, P
(a)
% X S H 2 , 2 a to m ic b a la n c e s , E q . re la tio n f o u r e q u a tio n s in n 3 , n 4 , n 5 , a n d n 6
5 % e x c e s s H 2 in re a c to r fe e d :
n3
100 m ol C O
2 m o l H 2 re q 'd
1 .0 5 m o l H 2 fe d
h
m ol C O
1 m o l H 2 re q 'd
210
m ol H 2
h
C b a la n c e: 1 0 0 (1) n 4 (1) n 6 (1) n 4 1 n 6
(1 )
H b a la n c e: 2 1 0 ( 2 ) n 5 ( 2 ) n 6 ( 4 ) n 5 2 1 0 2 n 6
(2 )
n T n 4 n 5 n 6 1 0 0 n 6 2 1 0 2 n 6 n 6 3 1 0 2 n 6
S o lv in g fo r n 6 n 6 7 5 .7 k m o l M /h
n 4 1 0 0 n 6 2 4 .3 k m o l C O /h , n 5 2 1 0 2 n 6 5 8 .6 k m o l H 2 / h
n1
1 km ol C O
1 km ol M
n 6 7 5 .7 k m o l C O /h , n 2
3
V re c n 4 n 5
2 2 .4 m (S T P )
1860 SC M H
km ol
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2 km ol H 2
1 km ol M
n 6 1 5 1 k m o l H 2 /h
lOMoARcPSD|32610952
(b)
(c)
(d)
(e)
n 3 (k m o l
n 4 (k m o l
P (k P a )
T (K )
H x s (% )
K p (T)E 8
K p P ^2
H 2/h)
C O /h)
n 5 (k m o l
H 2/h)
1000
500
5
9.1E + 01
0.91
210
74.45
158.90
5000
500
5
9.1E + 01
22.78
210
91.00
192.00
10000
500
5
9.1E + 01
91.11
210
13.28
36.56
5000
400
5
3.1E + 04
7849.77
210
1.07
12.15
5000
500
5
9.1E+ 01
22.78
210
24.32
58.64
5000
600
5
1.6E + 00
0.41
210
85.42
180.84
5000
500
0
9.1E + 01
22.78
200
26.65
53.30
5000
500
5
9.1E + 01
22.78
210
24.32
58.64
5000
500
10
9.1E + 01
22.78
220
22.23
64.45
n 6 (k m o l
ntot
K p P ^2 -
n 1 (k m o l
n 2 (k m o l
V re c
M /h)
(k m o l/ h )
K pc E 8
K p c P ^2
C O /h)
H 2/h)
(S C M H )
25.55
258.90
9 . 1 E -0 1
1 . 3 E -0 5
25.55
51.10
5227
9.00
292.00
2 . 3 E -0 1
2.3E + 01
9.00
18.00
6339
86.72
136.56
9.1E + 01
4 . 9 E -0 3
86.72
173.44
1116
98.93
112.15
7.8E + 03
3 . 2 E -0 8
98.93
197.85
296
75.68
158.64
2.3E+ 01
3 . 4 E -0 3
75.68
151.36
1858
14.58
280.84
4 . 1 E -0 1
-2 . 9 E -0 4
14.58
29.16
5964
73.35
153.30
2.3E + 01
9 . 8 E -0 3
73.35
146.70
1791
75.68
158.64
2.3E + 01
3 . 4 E -0 3
75.68
151.36
1858
77.77
164.45
2.3E + 01
-3 . 1 E -0 3
77.77
155.55
1942
Increase yield by raising pressure, lowering temperature, increasing Hxs. Increasing the
pressure raises costs because more compression is needed.
If the temperature is too low, a low reaction rate may keep the reaction from reaching
equilibrium in a reasonable time period.
Assumed that reaction reached equilibrium, ideal gas behavior, complete condensation of
methanol, not steady-state measurement errors.
5.14 (a)
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lOMoARcPSD|32610952
(b)
RT
0 .0 8 2 0 6 L a tm 2 2 3 K
0 .3 6 6 L / m o l
V id e a l
P
m o l K 5 0 .0 a tm
(c)
T(K )
(d)
P (a t m )
c3
c2
c1
V (id e a l)
V
(L / m o l)
(L / m o l)
f(V )
% e rro r
223
1.0
1.0
-1 8 . 3 3 6
1.33
-0 . 0 4 8 7
18.2994
18.2633
0.0000
223
10.0
10.0
-1 8 . 6 6 5 4
1.33
-0 . 0 4 8 7
1.8299
1.7939
0.0000
2.0
223
50.0
50.0
-2 0 . 1 2 9 4
1.33
-0 . 0 4 8 7
0.3660
0.3313
0.0008
10.5
223
100.0
100.0
-2 1 . 9 5 9 4
1.33
-0 . 0 4 8 7
0.1830
0.1532
-0 . 0 0 0 7
19.4
223
200.0
200.0
-2 5 . 6 1 9 4
1.33
-0 . 0 4 8 7
0.0915
0.0835
0.0002
9.6
1 eq. in 1 unknown - use Newton-Raphson.
Eq. (A.2-13) a
g
V
2
1 5 0 V 4 0 .2 5 9 V + 1 .3 3
s o lv e
Eq. (A.2-14) a d g d
Then V (k + 1 ) V (k ) d
g
a
Guess V (1 ) V id e a l 0 .3 6 6 0 L / m o l .
(k )
V
1
2
3
4
5.15
c0
0.3660
0.33714
0.33137
0.33114
(k + 1 )
V
0.33714
0.33137
0.33114
0.33114 converged
(a)
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0.2
lOMoARcPSD|32610952
(b)
P ro b le m 5 . 6 3 -S R K E q u a t io n S p re a d s h e e t
S p e c ie s
CO2
Tc (K )
304.2
P c (a t m )
72.9
R = 0 . 0 8 2 0 6 m ^3 a t m / k m o l K
0.225
a
3 . 6 5 3 9 2 4 m ^6 a t m / k m o l^2
b
0 . 0 2 9 6 6 8 m ^3 / k m o l
m
0.826312
f(V )= B 1 4 * E 1 4 ^3 -0 . 0 8 2 0 6 * A 1 4 * E 1 4 ^2 + ($ B $ 7 * C 1 4 -$ B $ 8 ^2 * B 1 4 -$ B $ 8 * 0 . 0 8 2 0 6 * A 1 4 )* E 1 4 -C 1 4 * $ B $ 7 * $ B $ 8
(c)
T(K )
P (a t m )
a lp h a
V (id e a l)
V (S R K )
f(V )
200
6.8
1.3370
2.4135
2.1125
0.0003
250
12.3
1.1604
1.6679
1.4727
0.0001
300
6.8
1.0115
3.6203
3.4972
0.0001
300
21.5
1.0115
1.1450
1.0149
0.0000
300
50.0
1.0115
0.4924
0.3392
0.0001
E-Z Solve solves the equation f(V)=0 in one step. Answers identical to VSRK values in
part b.
(d)
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lOMoARcPSD|32610952
5.16 (a)
(b)
5.17
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lOMoARcPSD|32610952
5.18
(a)
5 0 .0 m L 4 4 .0 1 g
V
V =
4 4 0 .1 m L / m o l
5 .0 0 g
n
m ol
RT
P =
(b)
V
For C O 2 :
Tr
Vr
T
Tc
id e a l
8 2 .0 6 m L a tm
m o l K 4 4 0 .1 m L / m o l
1000 K
(c)
5.19
(a)
3 .2 8 7 3
3 0 4 .2 K
V P c
zR T
m o l 3 0 4 .2 K 8 2 .0 6 m L a tm
id e a l
m ol K
6
1. 2 8
1 .2 8 a n d T r 3 .2 9 z = 1 .0 2
1 .0 2 8 2 .0 6 m L a tm
a = 3 .6 5 4 1 0
m ol K
4 4 0 .1 m L 7 2 .9 a tm
R Tc
Vˆ
1 8 6 a tm
T c 3 0 4 .2 K , P c 7 2 .9 a tm
F ig u re 5 .4 -3 : V r
P=
1000 K
2
m ol 1000 K
1 9 0 a tm
4 4 0 .1 m L
2
m L a tm /m o l , b = 2 9 .6 7 m L /m o l, m 0 .8 2 6 3,
(1 0 0 0 K ) 0 .1 0 7 7
F o r C O : T c 1 3 3 .0 K , P c 3 4 .5 a tm
n1
n2
2 5 1 4 .7 p s ia 1 5 0 L
1 .0 2
2 2 5 9 .7 p s ia 1 5 0 L
n le a k =
1 .0 2
n1 n 2
1 a tm
m ol K
3 0 0 K 1 4 .7 p s ia 0 .0 8 2 0 6 L a tm
1 a tm
m ol K
3 0 0 K 1 4 .7 p s ia 0 .0 8 2 0 6 L a tm
1022 m ol
918 m ol
1.7 3 m o l / h
60 h
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(b)
n 2 y 2 n a ir y 2
t m in
n2
n le a k
PV
200 10
RT
6
m o l a ir
0 .2 5 m o l
1 a tm
3 0 .7 m
0 .0 8 2 0 6 mL oaltm
K
300 K
m ol C O
3
1000 L
m
3
0 .2 5 m o l
0 .1 4 h
1 .7 3 m o l / h
t m in w o u ld b e g re a te r b e c a u s e th e ro o m is n o t p e rfe c tly s e a le d
(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.
5.20
P in V in
P o u t V o u t
z in n R T in
z out n R Tout
V in V o u t
P o u t z in
P in
T in
1 5 ,0 0 0
z out Tout
ft
3
1 4 .7 p s ia 1 .0 1 4 2 3 .2 K
m in 2 0 0 0 p s ia 1 .0 0
1 2 6 ft
3
/ m in
If the ideal gas equation of state were used, the factor 1.01 would instead be 1.00
1 % e rro r
5.21
Final:
P r 1 8 8 9 .7 5 2 4 .8 3 .6 z 1 0 .9 7
Total moles leaked:
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3 7 3 .2
lOMoARcPSD|32610952
Mole% CO in room =
1 0 .3 m o l C O
1 0 0 % 1. 0 % C O
9 7 3 .4 m o l
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lOMoARcPSD|32610952
CHAPTER SIX
6.1
(a)
P fin a l 2 4 3 m m H g . Since liquid is still present, the pressure and temperature must lie
(b)
on the vapor-liquid equilibrium curve, where by definition the pressure is the vapor
pressure of the species at the system temperature.
Assuming ideal gas behavior for the vapor,
m (v a p o r)
(3 .0 0 0 - 0 .0 1 0 ) L
m ol K
(3 0 + 2 7 3 .2 ) K
0 .0 8 2 0 6 L a tm
m (liq u id )
10 m L
1 .4 8 9 g
243 m m H g
1 a tm
1 1 9 .3 9 g
760 m m H g
m ol
1 4 .8 9 g
mL
m to ta l m (v a p o r) + m (liq u id ) = 1 9 .5 g
x vapor =
6.2
4 .5 9
0 .2 3 5 g v a p o r / g to ta l
1 9 .4 8
(a) lo g 1 0 p 7 .0 9 8 0 8
1 2 3 8 .7 1
45 217
2 .3 7 0 p
*
10
2 .3 7 0
2 3 4 .5 m m H g
(b)
(c)
6.3
Hv
R
7 0 7 6 K H v
7076 K
8 .3 1 4 J
m ol K
1 kJ
10
3
5 8 .8 k J m o l
J
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4 .5 9 g
lOMoARcPSD|32610952
6.4
6.5
T1 3 9 .5 C , p 1 4 0 0 m m H g x 1 3 .1 9 8 0 1 0
3
T 2 5 6 .5 C , p 2 7 6 0 m m H g x 2 3 .0 3 3 1 1 0
T 5 0 C x 3 .0 9 4 1 1 0
, y 1 5 .9 9 1 4 6
3
, y 2 6 .6 3 3 3 2
3
6.6
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lOMoARcPSD|32610952
(6.6 cont’d)
6.7
(a)
At the dew point,
p ( H 2 O ) = p ( H 2 O ) = 5 0 0 0 .1 = 5 0 m m H g T = 3 8 .1 C fro m T a b le B .3 .
(b)
VH O
3 0 .0 L
2
(c)
6.8
273 K
500 m m H g
1 m ol
0 .1 0 0 m o l H 2 O
1 8 .0 2 g 1 c m
(5 0 + 2 7 3 ) K
760 m m H g
2 2 .4 L (S T P )
m ol
m ol
1. 3 4 c m
g
( iv ) ( th e g a u g e p r e s s u r e )
T 7 8 F = 2 5 .5 6 C , Pb a r 2 9 .9 in H g = 7 5 9 .5 m m H g , h r 8 7 %
D ew P o in t: p T d p y H O P 0 .0 2 8 1 7 5 9 .5 2 1 .3 4 m m H g
hm
ha
hp
6.9
3
0 .0 2 8 1
1 0 .0 2 8 1
T d p 2 3 .2 C
2
0 .0 2 8 9 m o l H 2 O m o l d ry a ir
0 .0 2 8 9 m o l H 2 O
1 8 .0 2 g H 2 O
m o l d ry a ir
m o l d ry a ir
m ol H 2O
2 9 .0 g d ry a ir
hm
p 2 5 .5 6 C P p 2 5 .5 6 C
100%
0 .0 1 8 0 g H 2 O g d ry a ir
0 .0 2 8 9
2 4 .5 5 9 7 5 9 .5 2 4 .5 5 9
1 0 0 8 6 .5 %
B a s is I : 1 m o l h u m id a ir @ 7 0 F (2 1 .1 C ), 1 a tm , h r 5 0 %
yH O
0 .5 0 1 8 .7 6 5 m m H g
2
M a s s o f a ir:
0 .0 1 2
7 6 0 .0 m m H g
0 .0 1 2 m o l H 2 O
1 8 .0 2 g
0 .9 8 8 m o l d ry a ir
1 m ol
D e n s ity o f a ir
2 8 .8 7 g
m ol H 2 O
m ol
2 9 .0 g
1 m ol
1.1 9 6 g L
2 4 .1 3 L
B a s is II : 1 m o l h u m id a ir @ 7 0 F (2 1 .1 C ), 1 a tm , h r 8 0 %
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2 8 .8 7 g
3
lOMoARcPSD|32610952
yH O
0 .8 0 1 8 .7 6 5 m m H g
2
0 .0 2 0
m ol H 2 O
7 6 0 .0 m m H g
m ol
(6.9 cont’d)
M a s s o f a ir:
0 .0 2 0 m o l H 2 O
1 8 .0 2 g
0 .9 8 0 m o l d ry a ir
1 m ol
D e n s ity o f a ir
2 8 .7 8 g
2 9 .0 g
2 8 .7 8 g
1 m ol
1.1 9 3 g L
2 4 .1 3 L
B a s is III: 1 m o l h u m id a ir @ 9 0 F (3 2 .2 C ), 1 a tm , h r 8 0 %
yH O
0 .8 0 3 6 .0 6 8 m m H g
2
M a s s o f a ir:
0 .0 3 8
m ol H 2 O
7 6 0 .0 m m H g
0 .0 3 8 m o l H 2 O
1 8 .0 2 g
m ol
0 .9 6 2 m o l d ry a ir
1 m ol
D e n s ity
2 8 .5 8 g
2 9 .0 g
2 8 .5 8 g
1 m ol
1.1 4 1 g L
2 5 .0 4 L
In c r e a s e in T in c r e a s e in V d e c r e a s e in d e n s ity
In c r e a s e in h r m o r e w a te r ( M W = 1 8 ) , le s s d r y a ir ( M W = 2 9 )
d e c r e a s e in m d e c r e a s e in d e n s ity
S in c e th e d e n s ity in h o t, h u m id a ir is lo w e r th a n in c o o le r , d r y e r a ir , th e b u o y a n c y f o r c e
o n th e b a ll m u s t a ls o b e lo w e r . T h e r e f o r e , th e s ta te m e n t is w r o n g .
6.10
T 9 0 F = 3 2 .2 C , p 2 9 .7 in H g = 7 5 4 .4 m m H g , h r 9 5 %
Basis: 10 gal water condensed/min
n condensed
3
10 gal H 2O
1 ft
m in
7 .4 8 0 5 g a l
3
V1 ( f t / m in )
6 2 .4 3 lb m
ft
3
1 lb -m o l
4 .6 3 1 lb -m o le /m in
1 8 .0 2 lb m
n 2 (lb - m o le s / m in )
n 1 ( lb - m o le s / m in )
y2 (lb-mol H2O (v)/lb-mol) (sat’d)
y1 (lb-mol H2O (v)/lb-mol)
(1-y1) (lb-mol DA/lb-mol)
hr=95%, 90oF (32.2oC),
29.7 in Hg (754 mm Hg)
(1-y2) (lb-mol DA/lb-mol)
40oF (4.4oC), 754 mm Hg
4.631 lb-moles H2O (l)/min
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(6.10 cont’d)
R a o u lt's la w : y 2 P p
*
4 .4 C
y2
6 .2 7 4
7 5 4 .4
0 .0 0 8 1 7 lb -m o l H 2 O lb -m o l
M o le b a la n c e : n 1 n 2 4 .6 3 1
n 1 1 2 4 .7 lb -m o le s /m in
W a te r b a la n c e : 0 .0 4 5 n 1 0 .0 0 8 1 7 n 2 4 .6 3 1
n 2 1 2 0 .1 lb -m o le s /m in
1 2 4 .7 lb -m o le s
3 5 9 ft
3
o
(S T P )
(4 6 0 + 9 0 ) R
V o lu m e in : V =
m in
5 .0 4 1 0
lb -m o le s
4
760 m m H g
o
492 R
754 m m H g
3
ft / m in
6.11
0 .0 2 0 8
1 0 .0 2 0 8
0 .0 2 1 2 m o l H 2 O m o l d ry a ir
0 .6 7 0 m o l d ry a ir
0 .0 2 1 2 m o l H 2 O
m in
m o l d ry a ir
E v a p o ra tio n R a te :
6.12
(a)
D a ily ra te o f o c ta n e u s e =
30
2
(1 8 8 )
7 .0 6 9 1 0
4
8
18
0 .7 0 3
ft
3
7 .4 8 1 g a l
day
5 .2 8 8 1 0
(SG ) C H
3
0 .0 1 4 2 m o l H 2 O m in
day
4
gal
1 ft
3
ft
5
5 .2 8 8 1 0
0 .7 0 3 6 2 .4 3 lb m
7 .4 8 1 g a l
3 .1 0 1 0
3
ft
lb m C 8 H 1 8 / d a y
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3
4
gal / day
lOMoARcPSD|32610952
(6.12 cont’d)
0 .7 0 3 6 2 .4 3 lb m
(b)
p
ft
3 2 .1 7 4 ft
3
s
1 lb f
2
3 2 .1 7 4
(1 8 - 8 ) ft
lb m ft
s
(c)
*
o
Table B.4: p C H ( 9 0 F )
8
2 0 .7 4 m m H g
18
2 9 .9 2 1 in H g
1 4 .6 9 6 lb f / in
2
1 4 .6 9 6 p s i
760 m m H g
0 .4 0 lb f / in
2
2
6 .2 1 in H g
p o c ta n e y o c ta n e P
Octane lost to environment = octane vapor contained in the vapor space displaced by
liquid during refilling.
V o lu m e:
5 .2 8 8 1 0
4
gal
1 ft
3
7 0 6 9 ft
3
7 .4 8 1 g a l
pV
T o ta l m o le s: n
RT
(1 6 .0 + 1 4 .7 ) p s i
1 0 .7 3 ft
M o le fra c tio n o f C 8 H 1 8 : y =
3
pC H
8
p s i / (lb - m o le
18
o
R)
0 .4 0 p s i
P
7 0 6 9 ft
(1 6 .0 + 1 4 .7 ) p s i
(9 0 + 4 6 0 )
3
o
3 6 .7 7 lb - m o le s
R
0 .0 1 3 0 lb - m o le C 8 H 1 8 / lb - m o le
O c ta n e lo s t 0 .0 1 3 0 ( 3 6 .7 7 ) lb - m o le 0 .4 7 9 lb - m o le ( 5 5 lb m 2 5 k g )
(d)
6.13
A mixture of octane and air could ignite.
Let H=n-hexane
(a)
1.50 kmol H(l)/min
50% relative saturation at inlet: y o P 0 .5 0 0 p H* ( 8 0 o C )
yo
( 0 .5 0 0 )( 1 0 6 8 m m H g )
= 0 .7 0 3 k m o lH / k m o l
760 m m H g
Saturation at outlet: 0 .0 5 P p H* ( T1 ) p H* ( T1 ) 0 .0 5 ( 7 6 0 m m H g ) = 3 8 m m H g
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Antoine equation: lo g 1 0 3 8 6 .8 8 5 5 5
1 1 7 5 .8 1 7
T1 2 2 4 .8 6 7
o
T1 3 .2 6 C
(6.13 a cont’d)
N2 volume: V N
(b)
2
3
2 2 .4 m ( S T P )
(0 .9 5 )0 .6 8 2 k m o l
m in
1 4 .5 S C M M
km ol
Assume no condensation occurs during the compression
50% relative saturation at condenser inlet:
Volume ratio:
V1
V0
6.14
n 1 R T1 / P
n 0 R T0 / P
n 1 ( T1 2 7 3 .2 )
n 0 ( T 0 2 7 3 .2 )
0 .6 8 2 k m o l/m in
2 .1 8 k m o l/m in
321 K
0 .2 2
m
460 K
(c)
The cost of cooling to 3.2 4 o C (installed cost of condenser + utilities and other
operating costs) vs. the cost of compressing to 10 atm and cooling at 10 atm.
(a)
Maximum mole fraction of nonane achieved if all the liquid evaporates and none
escapes.
Assume T 2 5 o C , P = 1 a tm
n gas
2 10
4
L
273 K
298 K
1
2 2 .4 1 0
km ol
3
0 .8 1 8 k m o l
L (S T P )
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3
m
out
3
in
lOMoARcPSD|32610952
y m ax
n m ax
0 .0 8 4 k m o l C 9 H 2 0
0 .8 1 8 k m o l
n gas
0 .1 0 k m o l C 9 H 2 0 / k m o l (1 0 m o le % )
(6.14 a cont’d)
As the nonane evaporates, the mole fraction will pass through the explosive range (0.8%
to 2.9%). The answer is therefore y e s .
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
o
*
o
*
T1 2 5 .8 C = 2 9 9 K , p 1 5 .0 0 m m H g
T
T 2 6 6 .0 C = 3 3 9 K , p 2 4 0 .0 m m H g
A
ln ( 4 0 .0 / 5 .0 0 )
1
339
(c)
6.15
1
A 5 2 6 9 , B = ln ( 5 .0 0 ) +
5269
1 9 .2 3 p
299
*
e x p ( 1 9 .2 3
5269
)
T (K )
299
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.
Basis: 24 hours of breathing
n 0 (mol H O)
2
23°C, 1 atm
n 1 (mol) @
37°C, 1 atm
h r = 10%
0.79 mol N /mol
2
Lungs
n 2 (mol), sat urat ed
0.75 mol N /mol
2
y 1 (mol H O/mol)
2
y 2 (mol H O/mol)
2
+
+
O 2 , CO 2
O2
O 2 , CO 2
CO 2
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lOMoARcPSD|32610952
(6.15 cont’d)
Although the problem does not call for it, we could also calculate that n2 = 375 mol
exhaled/day, y2 = 0.0619, and the rate of weight loss by breathing at 23oC and 50%
relative humidity is n0 (18) = (n2y2 - n1y1)18 = 329 g/day.
6.16
(a)
B a s is: n 0 m o l fe e d g a s .
S so lv en t ,
G s o lv e n t - fre e g a s
n1 (mol) @ Tf (C), P4 (mm Hg)
y1 [mol S(v)/mol] (sat’d)
n0 (mol) @ T0 (C), P0 (mm Hg)
(1–y1) (mol G/mol)
y0 (mol S/mol)
(1-y0) (mol G/mol)
n2 (mol S (l))
Td0 (C) (dew point)
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(6.16 a cont’d)
C o n d e n s a tio n o f e th y lb e n z e n e f r o m n itr o g e n
A n to in e c o n s ta n ts f o r e th y lb e n z e n e
A=
6 .9 5 6 5
B=
1 4 2 3 .5
C=
2 1 3 .0 9
R un
T0
P0
Td0
f
Tf
p * (T d 0 )
p *(T f)
Pf
C re fr
C com p
107027 109702
1
50
765
40
0 .9 5
45
2 1 .4 7 2
2 7 .6 0
19139
2675
2
50
765
40
0 .9 5
40
2 1 .4 7 2
2 1 .4 7
14892
4700
83329
88029
3
50
765
40
0 .9 5
35
2 1 .4 7 2
1 6 .5 4
11471
8075
64239
72314
4
50
765
40
0 .9 5
20
2 1 .4 7 2
7 .0 7
4902
26300
27582
53882
(b)
6.17
C to t
(c)
When Tf decreases, Pf decreases. Decreasing temperature and increasing pressure both
to increase the fractional condensation. When you decrease Tf, less compression is
required to achieve a specified fractional condensation.
(d)
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.
(a)
B a s is : 1 2 0 m
3
o
m in fe e d @ 1 0 0 0 C (1 2 7 3 K ), 3 5 a tm .
T c c o r r . Pc c o r r
C m pd.
Tc K
Pc a tm
H2
3 3 .2
1 2 .8
4 1 .3
2 0 .8
CO
1 3 3 .0
3 4 .5
CO 2
3 0 4 .2
7 2 .9
CH 4
1 9 0 .7
4 5 .8
Use Kay’s rule.
A p p ly N e w to n 's c o rre c tio n s fo r H 2
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(6.17 a cont’d)
Feed gas to heater
T r 1 2 7 3 K 1 3 3 .4 K 9 .5 4
F ig . 5 .3 -2 z 1 .0 2
Pr 3 5 .0 a tm 3 7 .3 a tm 0 .9 4
Vˆ
1 .0 2
n g a s fe e d
8 .3 1 4 N m
1273 K
1 a tm
m ol K
3 5 a tm
101325 N m
120 m
3
m ol
3 .0 4 1 0
m in
3
m
3
1 km ol
10
3
2
3 .0 4 1 0
3
m
3
m ol
3 9 .5 k m o l m in
m ol
Feed gas to absorber
F ig . 5 .4 -1
T r 2 8 3 K /1 3 3 .4 K = 2 .1 2 , Pr 3 5 .0 a tm /3 7 .3 a tm = 0 .9 4
z 0 .9 8
V
znR T
0 .9 8
3 9 .5 k m o l
m in
P
1.2(39.5) kmol/min MeOH(l)
8 .3 1 4 N m
m ol K
283 K
1 a tm
3 5 .0 a tm
1 0 1 3 2 5 N /m
10
2
3
m ol
1 km ol
n1 (kmol/min), 261 K, 35atm
yNaOH sat’d
39.5 kmol/min, 283K, 35
yH2
0.40 mol H2/mol
yCH4 (2% of feed)
0.35 mol CO/mol
0.20 mol CO2/mol
0.05 mol CH4/mol
yCO
n2
(kmol/min), liquid
yMeOH
yCO2
yCH4 (98% of feed
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2 5 .7
m
3
m in
lOMoARcPSD|32610952
y M eO H
n M eO H
n M eO H + n H 2
in p u t
nCH
nCO
4
in p u t
0 .0 2 o f in p u t
n M eO H
n M e O H 3 9 .5 0 .4 0 0 .0 2 0 .0 5 0 .3 5
n M e O H 0 .0 1 4 8 k m o l m in M e O H in g a s
(6.17 cont’d)
(b)
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.
6.18
Dry pulp balance: 1 5 0 0
1
1 0 .7 5
m 1 ( 1 0 .0 0 1 5 ) m 1 8 5 8 k g / m in
50% rel. sat’n at inlet: y 1 P 0 .5 0 p H* O ( 2 8 o C ) y 1 0 .5 0 ( 2 8 .3 4 9 m m H g )/( 7 6 0 m m H g )
2
= 0 .0 1 8 7 m o l H 2 O /m o l
40oC dew point at outlet: y 2 P p H* O ( 4 0 o C ) y 2 ( 5 5 .3 2 4 m m H g ) / ( 7 7 0 m m H g )
2
= 0 .0 7 1 8 m o l H 2 O / m o l
Mass balance on dry air: n 0 (1 0 .0187 ) n 1 (1 0 .0718 )
( 1)
Mass balance on water:
n 0 ( 0 .0 1 8 7 )( 1 8 .0 k g / k m o l ) 1 5 0 0 ( 0 .7 5 / 1.7 5 ) n 1 ( 0 .0 7 1 8 )( 1 8 ) 8 5 8 ( 0 .0 0 1 5 ) ( 2 )
Solve (1) and (2) n 0 6 2 2 .8 k m o l / m in , n 1 6 5 8 .4 k m o l / m in
Mass of water removed from pulp: 1500(0.75/1.75)–858(.0015)]kg H2O = 6 4 2 k g / m in
Air feed rate: V0
6.19
6 2 2 .8 k m o l
3
2 2 .4 m ( S T P )
(2 7 3 + 2 8 ) K
km ol
273 K
m in
B a s is: 5 0 0 lb m h r d rie d le a th e r (L)
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1 .5 3 8 1 0
4
m
3
/ m in
lOMoARcPSD|32610952
o
n 1 ( lb - m o le s / h )@ 1 3 0 F , 1 a tm
o
n 0 ( lb - m o le s d ry a ir / h )@ 1 4 0 F , 1 a tm
y 1 ( lb - m o le s H 2 O / lb - m o le )
(1 - y 1 )( lb - m o le s d ry a ir / lb - m o le )
m 0 ( lb m / h )
5 0 0 lb m / h
0 .6 1 lb m H 2 O (l) / lb m
0 .0 6 lb m H 2 O (l) / lb m
0 .3 9 lb m L / lb m
0 .9 4 lb m L / lb
(6.19 cont’d)
= 478.4 lb – moles/hr
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lOMoARcPSD|32610952
6.20
(b)
300 lbm/h wet product
0 .2
1 0 .2
m 1 (lbm/h)
0 .0 2 / ( 1.0 2 ) 0 .0 1 9 6 lb m T (l) / lb m
0 .1 6 7 lb m T (l) / lb m
0 .9 8 0 4 lb m D / lb m
Dryer
0 .8 3 3 lb m D / lb m
(lb-mole/h) @ 200OF,
n 1 (lb-mole/h)
n 3
y1 (lb-mole T(v)/lb-mole)
y3 (lb-mole T/lb-mole)
(1–y1) (lb-mole N2/lb-mole)
(1-y3)( lb-mole N2/lb-mole)
70% r.s.,150oF, 1.2 atm
T=toluene
D=dry solids
heater
(lb-mole/h)
3n
y3 (lb-mole T(v)/lb-mole)
Eq.@ 90OF,
(1-y3) (lb-mole N2/lb-mole) 1atm
n 2 ( lb-mole T(l)/h )T(l)
condenser
Strategy: Overall balance m 1 & n 2 ;
Relative saturationy1, Gas and liquid equilibriumy3
Balance over the condenser n 1 & n 3
70% relative saturation of dryer outlet gas:
( 6 .9 5 8 0 5
*
O
O
p C H (1 5 0 F = 6 5 .5 6 C ) = 1 0
7
8
*
*
y 1 P 0 .7 0 p C H ( 1 5 0
7
8
O
F) y1
0 .7 0 p C H
7
P
8
1 3 4 6 .7 7 3
6 5 .5 6 2 1 9 .6 9 3
( 0 .7 0 )( 1 7 2 .4 7 )
1. 2 7 6 0
)
1 7 2 .4 7 m m H g
0 .1 3 2 4 lb - m o le T (v ) / lb - m o le
Saturation at condenser outlet:
*
pC H
7
8
y3
(9 0
o
*
pC H
7
8
P
F = 3 2 .2 2
o
4 0 .9 0
(6 .9 5 8 0 5
C )= 1 0
1 3 4 6 .7 7 3
3 2 .2 2 2 1 9 .6 9 3
0 .0 5 3 8 m o l T (v )/m o l
760
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)
4 0 .9 0 m m H g
lOMoARcPSD|32610952
C ir c u la tio n r a te o f d r y n itr o g e n = 5 .8 7 5 ( 1 - 0 .1 3 2 4 ) =
5 .0 9 7 lb - m o le
lb - m o le
h
2 8 .0 2 lb m
0 .1 8 2 lb m / h
6.21
Basis: 1 mol outlet gas/min
n 0 ( m o l / m in )
y 0 ( m o l C H 4 / m o l)
(1 y 0 ( m o l C 2 H 6 / m o l)
1 m o l / m in @ 5 7 3 K , 1 0 5 k P a
y 1 (m o l C O 2 / m o l)
n 1 (m o l O 2 / m in )
y 2 (m o l H 2 O / m o l)
(1 y 1 y 2 ) m o l N 2 / m o l
3 .7 6 n 1 (m o l N 2 / m in )
C H 4 2O 2 C O 2 2H 2 O
p CO
2
80 m m H g y1
80 m m H g
C2H 6
7
2
O 2 2C O 2 3H 2 O
101325 Pa
105000 Pa 760 m m H g
0 .1 0 1 6 m o l C O 2 / m o l
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6.22
B a s is: 1 0 0 m o l N H 3
P reh eat ed
a ir
N2
O2
100 m ol N H 3
7 8 0 k P a s a t 'd
co n v ert er
n 1 ( m o l) O 2
3 .7 6 n 1 ( m o l) N 2
n 2 ( m o l) H 2 O
ab s o rb er
n 3 (m o l N O )
n 8 (m o l H N O 3 )
n 5 (m o l O 2 )
n 9 ( m o l H 2O )
n 6 ( m o l H 2O )
n 7 (m o l H 2O )
1 at m , 3 0 ° C
h r = 0 .5
(a)
A ir fe e d : N H 3 2 O 2 H N O 3 H 2 O
n1
100 m ol N H 3
5 5 w t % H N O 3 (a q )
n 4 (m o l N 2)
2 m ol O 2
m ol N H 3
200 m ol O 2
Vair =
Balances on converter
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N O: n3
H N O 3 b a l. in a b s o rb e r: n 8
97 m ol N H 3
4 m ol N O
97 m ol N O
4 m ol N H 3
9 7 m o l N O re a c t
4 m ol H N O 3
4 m ol N O
H 2 O in p ro d u c t: n 9
97 m ol H N O 3
97 m ol H N O 3
6 3 .0 2 g H N O 3
45 g H 2 O
1 m ol H 2 O
m ol
55 g H N O 3
1 8 .0 2 g H 2 O
2 7 7 .5 6 m o l H 2 O
(b)
M a c id in o ld b a s is
97 m ol H N O 3
6 3 .0 2 g H N O 3
2 7 7 .6 m o l H 2 O
m ol
1 1 1 1 5 g 1 1.1 1 5 k g
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1 8 .0 2 g H N O 3
m ol
lOMoARcPSD|32610952
6.23
6.24
(a)
(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)
(torr)
0
42
85
129
176
(mol SO2/mol)
0
1.4x10–3
2.8x10–3
4.2x10–3
5.6x10–3
pSO
xS O
2
2
From this relation and the given data,
A plot of p S O vs. x S O is a straight line. Fitting the line using the method of least
2
2
squares (Appendix A.1) yields
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(c)
1 0 0 p p m S O 2 ySO
pSO
ySO P
2
2
H e n ry ' s la w x S O
2
2
100 m ol SO 2
10
1 .0 1 0
pSO
2
H SO
2
6
1 .0 0 1 0
4
m o ls g a s
4
m o l S O 2 /m o l g a s
7 6 0 m m H g 0 .0 7 6 0 m m H g
0 .0 7 6 0 m m H g
4
3 .1 3 6 1 0
2 .4 0 1 0
6
m m H g m o le fra c t io n
m ol SO 2 m ol
Since x S O is so low, we may assume for simplicity that V fin a l V in itia l 1 4 0 L
2
nSO
2
7 .7 8 1 0
3
m o l s o lu tio n
2 .4 0 1 0
6
m ol SO 2
1 m o l s o lu tio n
0 .0 1 8 7 m o l S O 2 d is s o lv e d
1.3 4 1 0
4
0 .0 1 8 7 m o l S O 2 d is s o lv e d
m ol SO 2 L
140 L
Gas-phase composition
*
ySO
1 .0 1 0
2
y a ir 1 y S O
6.25
4
m ol SO 2
yH O
2
2
2
m ol
2
o
xH O p H O (3 0 C )
(1) ( 3 1 .8 2 4 to r r )
4 .1 9 1 0
2
m o l H 2 O (v )
7 6 0 to r r
P
m ol
y H O 0 .9 5 8 m o l d r y a ir /m o l
2
(d)
Agitate/recirculate the scrubbing solution, change it more frequently. Add a base to the
solution to react with the absorbed SO2.
(a)
F ro m th e C o x c h a rt, a t 7 7 F , p P 1 4 0 p s ig , p n B 3 5 p s ig , p iB 5 1 p s ig
*
*
*
*
*
*
T o ta l p r e s s u r e P = x p p p + x n B p n B + x iB p iB
0 .5 0 (1 4 0 ) 0 .3 0 ( 3 5 ) 0 .2 0 ( 5 1) 9 1 p s ia 7 6 p s ig
P 2 0 0 p s ig , s o th e c o n ta in e r is te c h n ic a lly s a fe .
(b)
*
*
*
F ro m th e C o x c h a rt, a t 1 4 0 F , p P 3 0 0 p s ig , p n B 9 0 p s ig , p iB 1 2 0 p s ig
T o ta l p re s s u re P = 0 .5 0 ( 3 0 0 ) 0 .3 0 ( 9 0 ) 0 .2 0 ( 1 2 0 ) 2 0 0 p s ig
The temperature in a room will never reach 140oF unless a fire breaks out, so the
container is adequate.
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6.26
(a)
*
y i ( i 1,2 , , N )
(b)
xi pi (T )
P
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
Tbp(oC)
108.09
96.47
104.48
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
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
6.27
Basis: 100 mol/s gas feed. H=hexane.
n l ( m o l/s )
2 0 0 m o l o il/s
n F ( m o l/s )
x +i 1 ( m o l H /m o l)
y F ( m o l H /m o l)
1 –
n l ( m o l/s )
n 2 ( m o l/s )
0 .0 5 m o l H /m o l
x 2 ( m o l H /m o l)
0 .9 5 m o l N /m
ol
2
1 –
x 2 ( m o l O il/m o l)
y i ( m o l H /m o l)
S t a ge
y F ( m o l N 2/m o l)
1 0 0 m o l/s
n v ( m o l/s )
x i ( m o l H /m o l)
9 9 .5 % o f H in f e e d .
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i
n v ( m o l/s )
y i– 1 ( m o l H /m o l)
lOMoARcPSD|32610952
(a)
(b)
H b a la n c e o n 1
(c)
st
S ta g e : y 0 n v x 2 n l y 1 n v x 1 n l x 2 0 .0 0 6 4 3 m o l H (l) m o l
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
y(i)
5.00E-02
1.30E-02
3.45E-03
1.00E-03
3.74E-04
2.13E-04
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202.48
T
70
p*(T)
790.5546
i
0
1
2
3
4
5
...
21
x(i)
y(i)
5.00E-02
2.43E-02 2.53E-02
1.24E-02 1.29E-02
6.43E-03 6.69E-03
3.44E-03 3.58E-03
1.94E-03 2.02E-03
...
...
4.38E-04 4.56E-04
lOMoARcPSD|32610952
6.28
(e)
If the column is long enough, the liquid flowing down eventually approaches equilibrium
with the entering gas. At 70oC, the mole fraction of hexane in the exiting liquid in
equilibrium with the mole fraction in the entering gas is 4.56x10–4 mol H/mol, which is
insufficient to bring the total hexane absorption to the desired level. To reach that level
at 70oC, either the liquid feed rate must be increased or the pressure must be raised to a
value for which the final mole fraction of hexane in the vapor is 2.63x10–4 or less. The
solution is Pm in 1 0 3 7 m m H g .
(a)
M = methanol
n V ( m o l)
y ( m o l M ( v ) m o l)
n f ( m o l)
x F ( m o l M ( l) /m o l)
n L ( m o l)
x ( m o l M ( l) /m o l)
x F 0 .4 , x 0 .2 3 , y 0 . 6 2 f
(b)
0 .6 2 0 .2 3
0 .4 3 6
o
T m in 7 5 C , f 0 , T m a x 8 7 C , f 1
(a)
Txy diagram
(P=1 atm)
80
75
Vapor
70
T(oC)
6.29
o
0 .4 0 .2 3
65
liquid
60
55
50
0
0.2
0.4
0.6
Mole fraction of Acetone
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0.8
1
lOMoARcPSD|32610952
(b)
x A 0 .4 7 ; y A 0 .6 6
(c)
(i) x A 0 .3 4 ; y A 0 .5 5
( ii) M o le b a l.:
A b a l.:
1 nV n L
n V 0 .7 6 2 m o l v a p o r , n L 0 .2 3 8 m o l liq u id
0 .5 0 0 .5 5 n V 0 .3 4 n L
7 6 .2 m o le % v a p o r
(iii) A ( l ) 0 .7 9 1 g /c m 3 , E (l) 0 .7 8 9 g /c m 3 l 0 .7 9 0 g /c m 3
(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 5 8 .0 8 g /m o l, M E 4 6 .0 7 g /m o l
M l 0 .3 4 5 8 .0 8 1 0 .3 4 4 6 .0 7 5 0 .1 5 g /m o l
B a s is: 1 m o l liq u id (0 .7 6 2 m o l v a p o r/0 .2 3 8 m o l liq u id ) = 3 .2 m o l v a p o r
L iq u id v o lu m e :
Vl
(1 m o l)( 5 0 .1 5 g /m o l)
3 .2 m o l
22400 cm
V a p o r v o lu m e : V v =
(e)
3
(S T P )
m ol
V o lu m e p e rc e n t o f v a p o r
(d) For a
T, calculate nV as above;
6 3 .4 8 c m
3
3
( 0 .7 9 0 g /c m )
T
65 C
64.5 C
8 8 7 4 7 6 3 .4 8
*
8 8, 7 4 7 cm
3
273K
8 8, 7 4 7
xA
0.34
0.36
(6 5 + 2 7 3 )K
1 0 0 % 9 9 .9 v o lu m e % v a p o r
yA
0.55
0.56
*
fV
0.333
0.200
basis of 1 mol fed, guess
if nV 0.20, pick new T.
*
R a o u lt' s la w : y i P = x i p i P x A p A x E p E
7 6 0 0 .5 1 0
*
y
xp A
P
7 .1 1 7 1 4 1 2 1 0 .5 9 5 /( T b p 2 2 9 .6 6 4 )
0 .5 1 0
0 .5 1 0
8 .1 1 2 2 0 1 5 9 2 .8 6 4 /( T b p 2 2 6 .1 8 4 )
7 .1 1 7 1 4 1 2 1 0 .5 9 5 /( 6 6 .2 5 2 2 9 .6 6 4 )
0 .6 9 6 m o l a c e to n e /m o l
760
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o
T b p 6 6 .1 6 C
lOMoARcPSD|32610952
o
T h e a c tu a l T b p 6 1 .8 C
y A 0 .6 7 4
Tb p
T b p (re a l)
yA
y A (re a l)
6 6 .2 5 6 1 .8
6 1 .8
0 .6 9 6 0 .6 7 4
0 .6 7 4
1 0 0 % 7 .2 0 % e rro r in T b p
1 0 0 % 3 .2 6 % e rro r in y A
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.
6.30
(a)
B = benzene, C = chloroform. At 1 atm, (Tbp)B = 80.1oC, (Tbp)C = 61.0oC
The Txy diagram should look like Fig. 6.4-1, with the curves converging at 80.1oC when
xC = 0 and at 61.0oC when xC = 1. (See solution to part c.)
(b)
Txy Diagram for an Ideal Binary Solution
A
B
C
6.90328 1163.03
227.4
Chloroform
6.89272 1203.531
219.888
Benzene
760
P(mmHg)=
x
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
T
80.10
78.92
77.77
76.66
75.58
74.53
73.51
72.52
71.56
70.62
69.71
68.82
67.95
67.11
66.28
65.48
64.69
63.93
63.18
62.45
61.73
y
0
0.084
0.163
0.236
0.305
0.370
0.431
0.488
0.542
0.593
0.641
0.686
0.729
0.770
0.808
0.844
0.879
0.911
0.942
0.972
1
p1
0
63.90
123.65
179.63
232.10
281.34
327.61
371.15
412.18
450.78
487.27
521.68
554.15
585.00
614.02
641.70
667.76
692.72
716.27
738.72
760
p2
760
696.13
636.28
580.34
527.86
478.59
432.30
388.79
347.85
309.20
272.79
238.38
205.83
175.10
145.94
118.36
92.17
67.35
43.75
21.33
0
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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
lOMoARcPSD|32610952
Txy diagram
(P=1 atm)
85
T(oC)
80
Vapor
75
70
Liquid
65
60
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.9
1
Mole fraction of chloroform
(d)
Txy diagram
(P=1 atm)
85
T(oC)
80
yc
xc
75
70
65
x
y
60
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Mole fraction of choloroform
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lOMoARcPSD|32610952
(6.30 d cont’d)
T
o
Raoult’s law: T b p = 7 1 C , y = 0 .5 8
7 1 7 5 .3
7 5 .3
T a c tu a l
y
1 0 0 % 5 .7 % e rro r in T b p
0 .5 8 0 .6 0
1 0 0 % 3 .3 3 % e rro r in y
0 .6 0
y a c tu a l
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.
6.31
7 6 0 0 .4 0 1 0
7 .8 7 8 6 3 1 4 7 3 .1 1 /( T b p 2 3 0 )
0 .6 0 1 0
7 .7 4 4 1 6 1 4 3 7 .6 8 6 /( T b p 1 9 8 .4 6 3 )
E -Z S o lv e
o
T 7 9 .9 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.
6.32
Basis: 1000 kg/h product
nH4
E = C2 H 5 O H ( M =
4 6 .0 5 )
A = CH 3CH O ( M =
4 4 .0 5 )
P = 760 m m H g
( m o l H 2 /h )
H y d ro carb o n s
s cru b b er
n3
( m o l/h )
yA3
( m o l A /m o l) , s a t 'd
y E3
( m o l E /m o l) , s a t 'd
F res h feed
yH3
( m o l H 2 /h )
n 0 ( m o l E /h )
v ap o r, – 4 0 ° C
nA1
( m o l A /h )
nE1
( m o l E /h )
280°C
react o r
Scru b b ed
co n d en s er
nA2
( m o l A /h )
nE2
( m o l E /h )
0 .5 5 0 A
nH2
( m o l H 2/h )
0 .4 5 0 E
nC
( m o l/h )
liq u id , – 4 0 ° C
n r ( m o l/h )
0 .0 5 A
0 .9 5 E
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nA4
( m o l A /h )
nE4
( m o l E /h )
P ro d u ct
1 0 0 0 k g/h
s t ill
n p ( m o l/h )
0 .9 7 A
0 .0 3 E
lOMoARcPSD|32610952
(6.32 cont’d)
(a)
Molar flow rate of product
n p
1000 kg
1 km ol
h
4 4 .1 1 k g
2 2 .6 7 k m o l h
Table B.4 (Antoine)
p A 4 0 C 4 4 .8 m m H g
*
p E 4 0 C 0 .3 6 0 m m H g
*
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.
0 .5 5 0 p A 4 0 C
*
Raoult’s law y A 3
0 .4 5 0 p E 4 0 C
P
0 .4 5 0 ( 0 .3 6 0 )
0 .5 5 0 ( 4 4 .8 )
0 .0 3 2 4 2 k m o l A /k m o l
760
P
*
y E3
2 .1 3 1 0
4
km ol E km ol
760
y H 3 1 y A 3 y E 3 0 .9 6 7 4 k m o l H 2 k m o l
A balance about scrubber: n A 4 n 3 y A 3 0 .0 2 8 1 5 n 3
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lOMoARcPSD|32610952
(6.32 a cont’d)
E balance about scrubber: n E 4 n 3 y E 3 2 .0 3 1 0 4 n 3
H 2 b a la n c e a b o u t s c ru b b e r:
n H 4 n 3 y H 3 0 .9 7 1 6 n 3
Overall C balance:
Solve (1)–(5) simultaneously (E-Z Solve):
n 0 2 3 .4 k m o l E /h (fre s h fe e d ), n H 4 2 2 .8 k m o l H 2 / h (in o ff-g a s )
n 3 = 2 3 .5 k m o l/h , n A 4 = 0 .7 6 k m o l A /h , n E 4 = 0 .0 0 5 0 k m o l E /h
A balance about feed mixing point: n A 1 0 .0 5 n r 1.4 7 k m o l A h
E balance about feed mixing point: n E 1 n 0 0 .9 5 n r 5 1 .5 k m o l E h
E balance about condenser: n E 2 n 3 y E 3 0 .4 5 0 n c 2 3 .5 k m o l E h
Id e a l g a s e q u a tio n o f s ta te :
1 .4 7 5 1 .5 k m o l 2 2 .4 m S T P 2 7 3 + 2 8 0 K
3
V re a c to r fe e d
(b)
h
Overall conversion
1 km ol
n 0 0 .0 3 n p
100%
2 3 .4 0 .0 3 2 2 .6 7
nE1 nE2
nE1
Feed rate of A to scrubber:
100%
3
100% 97%
2 3 .4
n0
Single-pass conversion
2 .4 0 1 0
273K
5 1 .5 2 3 .5
5 1 .5
n A 4 = 0 .7 6 k m o l A /h
Feed rate of E to scrubber: n E 4 0 .0 0 5 0 k m o l E h
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100% 54%
m
3
h
lOMoARcPSD|32610952
6.33
Basis: Given feed rates
G2
G1
G3
G4
2 0 0 m o l a ir /h
2 0 0 m o l a ir /h
1 0 0 m o l/h
n 1 ( m o l/h )
0 .9 6 H 2
0 .9 9 9 H 2
n 2 m o l H 2S /m o l
0 .0 4 H 2 S , s a t 'd
0 .0 0 1 H 2 S
0 .4 0 ° C , 1 a t m
1 .8 a t m
ab s o rb er
L2
s t r ip p e r
0°C
L1
40°C
n 3 ( m o l/h )
n 4 ( m o l/h )
x 3 ( m o l H 2 S /m o l)
0 .0 0 2 H 2 S
n 3 ( m o l/h )
x 3 ( m o l H 2S /m o l)
( 1 – x 3) ( m o l s o lv e n t /m o l)
0 .9 9 8 s o lv e n t
( 1 – x 3 ) ( m o l s o lv e n t /m o l)
0°C
40°C
h eat er
Equilibrium condition: At G1,
x3
pH S
2
HH S
2
0 .0 7 2 a tm
2 .6 7 1 0
3
2 7 a tm m o l fra c tio n
m o le H 2 S m o le
Strategy: Overall H 2 and H 2 S balances n 1 , n 2
n 2 air flow rate volumetric flow rate at G4
and solvent balances around absorber n 3 , n 4
0 .9 9 8 n 4 solvent flow rate
H 2S
Volumetric flow rate at stripper outlet
H 2 S and solvent balances around absorber:
Solvent flow rate 0 .9 9 8 n 4 5 8 2 0 m o l s o lv e n t h
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lOMoARcPSD|32610952
6.34
Basis: 875 kg/h feed solution
m 1 ( k g H 2 O (v ) /h )
8 7 5 k g/h
S a t 'd s o lu t io n 1 0 ° C
x 0 ( k g K O H /k g)
m 2 ( k g H 2 O ( 1 ) /h )
( 1 – x0) ( k g H 2O /k g)
1 .0 3 m 2 ( k g K O H /h )
m 3 ( k g K O H - 2 H 2O ( s ) /h )
6 0 % o f K O H in f e e d
Analysis of feed: 2 K O H H 2 S O 4 K 2 S O 4 2 H 2 O
60% recovery: 8 7 5 0 .4 2 7 0 .6 0 2 2 4 .2 k g K O H h
m3
2 2 4 .2 k g K O H
9 2 .1 5 k g K O H 2 H 2 O
h
5 6 .1 1 k g K O H
3 6 8 .2 k g K O H 2 H 2 O h
1 4 3 .8 k g H 2 O h
KOH balance: 0 .4 2 7 8 7 5 2 2 4 .2 1 .0 3 m 2 m 2 1 4 5 .1 k g h
Total mass balance: 8 7 5 3 6 8 .2 2 .0 3 1 4 5 .1 m 1 m 1 2 1 2 k g H 2 O h e v a p o ra te d
6.35 (a) Table 6.5-1 shows that at 50oF (10.0oF), the salt that crystallizes is M g S O 4 7 H 2 O , which
contains 48.8 wt% MgSO4
(b) Basis: 1000 kg crystals/h.
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lOMoARcPSD|32610952
6.36
(a)
Heating the solution dissolves all MgSO4; filtering removes I, and cooling
recrystallizes MgSO4 enabling subsequent recovery.
(b)
Strategy:
( M W ) M gS O ( 2 4 .3 1 3 2 .0 6 6 4 .0 0 ) 1 2 0 .3 7 , ( M W ) M gS O 7 H O ( 1 2 0 .3 7 7 * 1 8 .0 1 ) 2 4 6 .4 4
4
4
2
Overall MgSO4 balance:
6 0 ,0 0 0 lb m
0 .9 0 lb m M g S O 4 7 H 2 O
1 2 0 .3 7 lb m M g S O 4
h
lb m
2 4 6 .4 4 lb m M g S O 4 7 H 2 O
(3 0 0 lb m / h )(0 .3 2 lb m M g S O 4 / lb m ) m 4 ( 1 2 0 .3 7 / 2 4 6 .4 4 ) 0 .0 5 m 4 ( 0 .2 3 )
m 4 5 .2 5 7 x 1 0
4
lb m c ry s ta ls / h
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lOMoARcPSD|32610952
(6.36 b cont’d)
Overall mass balance:
4 5 .2 5 7 x 1 0
m
4
lb m / h
1 6 3 0 0 1. 0 5 m
4
6 0 ,0 0 0 m
1 1 4 9 4 lb m H 2 O / h
m
(c)
Recycle/fresh feed ratio
9 .5 7 5 x 1 0
4
lb m / h
64
1 4 9 4 lb m / h
6.37
m 1 (g styrene)
90 g ethylbenezene
100 g EG
90 g ethylbenzene
m 2 (g styrene)
30 g st yrene
100 g EG
Styrene balance: m 1 m 2 3 0 g s tyren e
m 1 2 5 .6 g s ty re n e in e th y lb e n z e n e p h a s e
m 2 4 .4 g s ty re n e in e th y le n e g ly c o l p h a s e
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lb m re c y c le / lb m fre s h fe e d
lOMoARcPSD|32610952
6.38
(a)
P--penicillin; Ac--acid solution; BA--butyl acetate; Alk--alkaline solution
(b)
In Unit I, 90% transfer m 3 P 0 .9 0 ( 1.5 ) 1.3 5 k g P
P balance:
pH=2.1 K 2 5 .0
1.5 m 2 P 1.3 5 m 2 P 0 .1 5 k g P
1.3 5 / ( 1.3 5 m 1 )
0 .1 5 / ( 0 .1 5 9 8 .5 )
m 1 3 4 .1 6 k g B A
In Unit II, 90% transfer: m 5 P 0 .9 0 ( m 3 P ) 1.2 1 5 k g P
P balance:
pH=5.8 K 0 .1 0
m1
100
m4
m 3 P 1.2 1 5 m 6 P m 6 P 0 .1 3 5 k g P
m 6 P / ( m 6 P 3 4 .1 6 )
1. 2 1 5 / ( 1. 2 1 5 m 4 )
3 4 .1 6 k g B A
m 4 2 9 .6 5 k g A lk
0 .3 4 1 6 k g b u ty l a c e ta te / k g a c id ifie d b ro th
1 0 0 k g b ro th
100
2 9 .6 5 k g A lk
0 .2 9 6 5 k g a lk a lin e s o lu tio n / k g a c id ifie d b ro th
1 0 0 k g b ro th
Mass fraction of P in the product solution:
xP
m5P
m4 m5P
1. 2 1 5 P
0 .3 9 4 k g P / k g
(2 9 .6 5 + 1 .2 1 5 ) k g
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lOMoARcPSD|32610952
(6.38 cont’d)
(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.
6.39
(a)
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lOMoARcPSD|32610952
(6.39 a cont’d)
(b)
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lOMoARcPSD|32610952
(6.39 b cont’d)
6.40
(a)
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lOMoARcPSD|32610952
(6.40 a cont’d)
(b)
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lOMoARcPSD|32610952
(6.40 b cont’d)
6.40
(c)
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lOMoARcPSD|32610952
CHAPTER SEVEN
7.1
4
0 .8 0 L
3 .5 1 0 k J
0 .3 0 k J w o rk
1 h
1 kW
h
L
1 kJ heat
3600 s
1 kJ s
2 .3 3 k W
7.2
3
W
1 .3 4 1 1 0
1 kW
1 W
10
3
hp
2 .3 3 k W 2 .3 k W
3 .1 2 h p 3 .1 h p
All kinetic energy dissipated by friction
(a)
(b)
3 10
8
b ra k in g s
day
7 1 5 B tu
b ra k in g
1 day
24 h
1 h
1 W
3600 s
9 .4 8 6 1 0
1 MW
4
B tu /s
3000 M W
7.3
(a)
Emissions:
P aper
P la s tic
1000 sacks
2000 sacks
( 0 .0 5 1 0 0 .0 5 1 6 ) o z
1 lb m
sack
16 oz
( 0 .0 0 4 5 0 .0 1 4 6 ) o z
1 lb m
sack
16 oz
6 .4 1 lb m
2 .3 9 lb m
Energy:
P aper
1000 sacks
( 7 2 4 9 0 5 ) B tu
1. 6 3 1 0
6
B tu
sack
P la s tic
2000 sacks
( 1 8 5 4 6 4 ) B tu
1. 3 0 1 0
6
B tu
sack
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6
10 W
2617 M W
lOMoARcPSD|32610952
(7.3 cont’d)
(b)
For paper (double for plastic)
R a w M a te ria ls
Sack
A c q u is itio n a n d
P ro d u c tio n a n d
P ro d u c tio n
U se
M a te ria ls
400 sacks
1000 sacks
D is p o s a l
fo r 4 0 0 s a c k s
Emissions:
P aper
400 sacks
0 .0 5 1 0 o z
1 lb m
sack
16 oz
1000 sacks
0 .0 5 1 6 o z
1 lb m
sack
16 oz
4 .5 lb m
3 0 % re d u c tio n
P la s tic
800 sacks
0 .0 0 4 5 o z
1 lb m
sack
16 oz
2000 sacks
0 .0 1 4 6 o z
1 lb m
sack
16 oz
2 .0 5 lb m
1 4 % re d u c tio n
Energy:
P aper
400 sacks
7 2 4 B tu
1000 sacks
9 0 5 B tu
sack
P la s tic
800 sacks
1 8 5 B tu
1.1 9 1 0
6
B tu ; 2 7 % re d u c tio n
sack
2000 sacks
4 6 4 B tu
sack
1. 0 8 1 0
6
B tu ; 1 7 % re d u c tio n
sack
(c)
3 10
8
p e rs o n s
1 sack
p e rs o n - d a y
1 day
1 h
24 h
6 4 9 B tu
3600 s
1 sack
1 J
9 .4 8 6 1 0
1 MW
-4
B tu
10
6
J / s
2 ,3 7 5 M W
Savings for recycling: 0 .1 7 ( 2 , 3 7 5 M W ) = 4 0 4 M W
7.4
(d)
Cost, toxicity, biodegradability, depletion of nonrenewable resources.
(a)
M a s s flo w ra te : m
3 .0 0 g a l
m in
1 ft
3
7 .4 8 0 5 g a l
(0 .7 9 2 )(6 2 .4 3 ) lb m
1 ft
3
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1 m in
60 s
0 .3 3 0 lb m s
lOMoARcPSD|32610952
(7.4 a cont’d)
7.5
(b)
Heat losses in electrical circuits, friction in pump bearings.
(a)
E k p o s itiv e
When the pressure decreases, the volumetric flow rate increases, and
hence the velocity increases.
E p n e g a tiv e
The gas exits at a level below the entrance level.
Po u t V o u t
V out
Pin V in
nRT
nRT
V in
u o u t u in
Ek
1
2
2
Po u t
m ( u o u t u in )
u o u t (m /s ) A (m 2 )
u in (m /s ) A (m )
2
Po u t
Pin
2
Pin
5m s
10 bar
Pin
Po u t
5 .5 5 5 m s
9 bar
0 .5 ( 0 .0 2 2 5 ) k g
2
2
( 5 .5 5 5 5 .0 0 0 ) m
s
s
2
2
1 N
1 W
1 k g m /s
2
1 N m /s
0 .0 6 5 9 W
E p m g ( z o u t z in )
0 .0 2 2 5 k g
9 .8 0 6 6 m
-2 0 0 m
k g m /s
s
s
1 N
1 W
2
4 4 .1 W
4.00 L, 30 °C, 5.00 bar V (L), T (°C), 8.00 bar
7.6
(a)
(b)
Constant T U 0 Q W
7 .6 5 L b a r
8 .3 1 4 J
0 .0 8 3 1 4 L b a r
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765 J
1 N m /s
lOMoARcPSD|32610952
(7.6 cont’d)
(c)
Adiabatic Q 0 U W 7 .6 5 L b a r > 0 , T fin a l 3 0 C
(a)
Downward force on piston:
7.7
F d P a tm A m p is to n + w e ig h t g
1 a tm
1 .0 1 3 2 5 1 0
5
N / m2
2 .8 3 1 0
3
m
2
2 4 .5 0 k g
9 .8 1 m
1 N
2
1 kg m / s2
s
a tm
527 N
Equilibrium condition:
F u F d 2 .8 3 1 0
V0
(b)
nRT
1 .4 0 g N 2
P0
3
m 2 P 0 5 2 7 P 0 1.8 6 1 0
1 m ol N 2
2 8 .0 2 g
For any step, U E k E p Q W
N m
1 .0 1 3 2 5 1 0
303 K
1 .8 6 1 0
5
5
Pa
5
2
1.8 6 1 0
Pa
0 .0 8 2 0 6 L a tm
Pa
m ol K
1 a tm
5
0 .6 7 7 L
U Q W
E k 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 n R T P g must vary. This implies that the piston moves, so that W is not
zero.
Overall: T in itia l T fin a l U 0 Q W 0
In step 1, the gas expands W 0 U 0 T d e c re a s e s
(c)
Final gas pressure P f
F
A
331 N
2 .8 3 1 0
3
m
2
1.1 6 1 0
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5
N m
2
lOMoARcPSD|32610952
(7.7 c cont’d)
Since work is done by the gas on its surroundings, W 4 7 J
Q 47 J
Q W 0
(heat transferred to gas).
V
7.8
3 2 .0 0 g
4 .6 8 4 c m 3
m ol
g
103 L
106 cm 3
H U P V 1 7 0 6 J m o l
0 .1 4 9 9 L m o l
4 1 .6 4 a tm
0 .1 4 9 9 L
m ol
7.9
8 .3 1 4
0 .0 8 2 0 6 L a tm / (m o l K )
(a)
Ref state
(b)
U U fin a l U in itia l 0 .0 0 0 2 8 .2 4 2 8 .2 4 k J m o l
Hˆ 2 8 .2 4 k J m o l
0 .3 1 0 b a r
0 .0 5 1 6 7 9 .9 4 L
m ol
(c)
J / (m o l K )
8 .3 1 4 J
1 kJ
0 .0 8 3 1 4 L b a r
10
3
2338 J m ol
3 0 .7 k J m o l
J
U independent of
V c h a n g e s w ith p re s s u re . A t c o n s ta n t te m p e ra tu re P V = P ' V ' V ' = P V / P '
(0 .3 1 0 b a r )(7 9 .9 4 L / m o l)
1 2 0 .8 8 L / m o l
V ' (T = 3 0 0 K , P = 0 .2 0 5 b a r) =
0 .2 0 5 b a r
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lOMoARcPSD|32610952
(7.9 c cont’d)
n
5 .0 0 L
1 m ol
0 .0 4 1 4 m o l
1 2 0 .8 8 L
(d)
7.10
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.
(a)
H E k E p Q W s
Q H n H
(b)
7.11
4 3 .3 7 m o l
1 m in
3640 J
m in
60s
m ol
kW
10
3
2 .6 3 k W
J / s
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.
(a)
1 3 0 .2
H 0 T re f
2 5 C
5 .2
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lOMoARcPSD|32610952
(7.11 a cont’d)
⇒
(b)
E n e rg y b a l a n c e : Q U
Ek , E p, W 0
1
A v e ra g e ra te o f h e a t re m o v a l
7.12
(a)
[ (5 .2 2 0 - 1 3 0 .4 ) - (5 .2 8 0 - 1 3 0 .4 )] k J
20 kg
6240 kJ
1 m in
5 m in
60 s
6240 kJ
kg
2 0 .8 k W
H E k E p Q W s ; E k , E p , W s 0 H Q
kg (Table B.7)
kg (Table B.5)
1 0 0 k g H 2 O (v ) / s
1 0 0 k g H 2 O (v ) / s
o
o
1 0 0 C , s a tu ra te d
4 0 0 C , 1 a tm
Q (k W )
(b)
U E k E p Q W ; E k , E p , W 0 U Q
T a b le B .5 Uˆ 1 0 0 C , 1 a tm 2 5 0 7
kJ
, Vˆ 1 0 0 C , 1 a tm 1 .6 7 3
m
kg
3
kg
Vˆ 4 0 0 C , Pfin a l
Interpolate in Table B.7 to find P at which Vˆ =1.673 at 400oC, and then interpolate again to
find Uˆ at 400oC and that pressure:
3
Vˆ 1 .6 7 3 m /g Pfin a l 1 .0 4 .0
3 .1 1 1 .6 7 3
ˆ
o
3 .3 b a r , U ( 4 0 0 C , 3 .3 b a r ) = 2 9 6 6 k J /k g
3 .1 1 0 .6 1 7
3
Q U m Uˆ 1 0 0 k g 2 9 6 6 2 5 0 7 k J k g 1 0 J k J
4 .5 9 1 0 J
7
The difference is the net energy needed to move the fluid through the system (flow work).
(The energy change associated with the pressure change in Part (b) is insignificant.)
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lOMoARcPSD|32610952
7.13
m [ k g H 2 O (l) / h ]
m [ k g H 2 O (v ) / h ]
o
20 C
2 0 b a r (s a t' d )
Q = 0 . 6 5 ( 8 1 3 k W ) 5 2 8 k W
(a)
(b)
(c)
V
n R T
701 kg / h
103 g / kg
1 8 .0 2 g / m o l
20 bar
P
4 8 5 .4 K
0 .0 8 3 1 4 L b a r
1 m3
m ol K
103 L
7 8 .5 m 3 / h
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.
7.14
m [k g H 2 O (l)/h ]
m [k g H 2 O (v )/h ]
3
o
24 C , 10 bar
1 5 ,0 0 0 m /h @ 1 0 b a r (sa t'd )
Q (k W )
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lOMoARcPSD|32610952
(7.14 cont’d)
E k E k fin a l E k in itia l
7.15
(a)
E k in itia l 0
E k E k fin a l
228 g/min
228 g/min
25oC
T(oC)
Q ( k W )
=0
(b)
(c)
(d)
Heat is absorbed by the pipe, lost through the insulation, lost in the electrical leads.
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lOMoARcPSD|32610952
7.16
(a)
(b)
C lo th e d :
h 8 T o 1 3 .4 C
T s = 3 4 .2
N u d e , im m e rse d :
(c)
h 6 4 T o 3 1.6 C
T s = 3 4 .2
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 T o , the skin temperature
must drop to satisfy the energy balance equation: when T s drops, you feel cold.
7.17
.
3
m ( k g /h )
5 2 .5 m H 2 O ( v ) /h
.
m ( k g /h )
0 ..8 5 k g H 2 O ( v) /k g
o
5 b a r, T ( C )
0 .1 5 k g H 2 O ( l ) /k g
.
o
5 b a r , s a tu r a te d , T ( C )
P 5 b a rs
(a)
T a b le B .6
Q (k W )
T 1 5 1.8 C , H L 6 4 0 .1 k J k g , H V 2 7 4 7 .5 k J k g
3
V ( 5 b a r , s a t' d ) = 0 .3 7 5 m / k g m
5 2 .5 m
h
3
1
kg
0 .3 7 5 m
3
140 kg h
(b)
E n e rg y b a la n c e : Q H
7.18
T a b le B .6
21 kg
2 7 4 7 .5 6 4 0 .1 ] k J
h
kg
(a)
P 5 bar
(b)
Inlet: T=350°C, P=40 bar
1 h
1 kW
3600 s
1 kJ s
12 kW
o
T s a tu ra tio n 1 5 1.8 C . At 75°C the discharge is all liquid.
Outlet: T=75°C, P=5 bar
T a b le B .7
T a b le B .7
3
H in = 3 0 9 5 k J / k g , Vin = 0 .0 6 6 5 m / k g
-3
3
H o u t = 3 1 4 .3 k J / k g , Vo u t = 1 .0 3 1 0 m / k g
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lOMoARcPSD|32610952
(7.18 b cont’d)
u in
u out
Vin
200 kg
m in
A in
Vo u t
200 kg
1 m in
3
/ kg
2
60 s
( 0 .0 7 5 )
1 m in
0 .0 0 1 0 3 m
m in
A out
0 .0 6 6 5 m
60 s
( 0 .0 5 )
2
2
5 0 .1 8 m / s
2
1.7 5 m / s
/ 4 m
3
/ kg
/ 4 m
m
2
2
E n e rg y b a la n c e : Q W s H E k m ( H 2 H 1 )
(u 2 u1 )
2
Q Ws
200 kg
1 m in
(3 1 4 -3 0 9 5 ) k J
m in
60 s
kg
2
2
200 kg
1 m in
(1 .7 5 -5 0 .1 8 ) m
2 m in
60 s
s
1 3 , 4 6 0 k W ( 1 3 ,4 6 0 k W tra n s f e rre d fro m th e tu rb i n e )
7.19
(a)
Assume all heat from steam transferred to oil
1 .0 0 1 0
Q
4
kJ
m in
1 m in
167 kJ s
60 s
1 0 0 k g o il/m in
1 0 0 k g o il/m in
135°C
m ( k g H O (v )/s )
185°C
m ( k g H 2 O (l)/s )
2 5 b a r s , s a t'd
2 5 b a r s , s a t'd
2
T im e b e tw e e n d is c h a rg e s :
1200 g
1 s
1 kg
d is c h a rg e
0 .0 9 1 k g
10
3
g
1 3 s d is c h a rg e b
(b)
Y e a rly c o s t:
1 0 0 0 tra p s
3
0 .0 9 1 k g s tre a m
0 .1 0 k g la s t
2 .6 1 0
tra p s
k g s tre a m
k g lo s t
$ 7 .4 1 0
5
$
3600 s
24 h
360 day
h
day
year
/ year
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2
2
lOMoARcPSD|32610952
10 m3, n moles of steam(v), 275°C, 15 bar 10 m3, n moles of water (v+l), 1.2 bar
7.20
1 0 .0 m 3 H 2 O ( v )
1 0 .0 m 3
m in ( k g )
m v [k g H 2O (v )]
2 7 5 o C , 1 .5 b a r
m l [ k g H 2 O ( l) ]
Q
1 .2 b a r , s a tu r a te d
T a b le B .6
(a)
P=1.2 bar, saturated,
(b)
Total mass of water: m in =
10 m
T 2 1 0 4 .8 C
3
1 kg
0 .1 8 1 8 m
3
55 kg
M a s s B a la n c e : m v m l 5 5 .0
V v V l 1 0 .0 m
V o lu m e a d d itiv ity :
3
m v ( 1. 4 2 8 m
3
/ k g ) m l ( 0 .0 0 1 0 4 8 m
3
/ kg)
m v 7 .0 k g , m l 4 8 .0 k g c o n d e n s e d
(c)
E n e rg y b a la n c e : Q = U = m v U v m l U l m in U in
E p , E k , W 0
[ ( 7 .0 ) (2 5 1 2 .1 k J / k g ) + ( 4 8 .0 )( 4 3 9 .2 ) - 5 5 k g (2 7 3 9 .2 )] k J
= 1 .1 2 1 0
7.21
(a)
5
kJ
Assume both liquid and vapor are present in the valve effluent.
1 kg H 2 O (v ) / s
1 5 b a r, Tsat 1 5 0
o
m l [ k g H 2 O ( l ) / s ]
C
m v [ k g H 2 O ( v ) / s ]
1 .0 b a r , s a t u r a t e d
(b)
o
o
T a b le B .6 T s a t' n ( 1 5 b a r) = 1 9 8 .3 C T in 3 4 8 .3 C
T a b le B .7 H H ( 3 4 8 .3 C , 1 5 b a r) 3 1 4 9 k J / k g
in
T a b le B .6 H l ( 1 .0 b a r, s a t' d ) = 4 1 7 .5 k J / k g ;
H v ( 1 .0 b a r, s a t' d ) = 2 6 7 5 .4 k J / k g
E n e rg y b a l a n c e : H 0 m l H l m v H v m in H in 0
E p , E k , Q , W s 0
m in H in m l H l m v H v
m v m l
3 1 4 9 k J / k g m l ( 4 1 7 .5 ) ( 1 m l )( 2 6 7 5 .4 )
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lOMoARcPSD|32610952
(7.21 b cont’d)
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 twophase assumption is therefore incorrect; the effluent must be pure vapor.
(c)
Energy balance m o u t H o u t m in H in
m in m o u t 1
T a b le B .7
7.22
3 1 4 9 k J / k g = H ( 1 b a r, T o u t )
Tout 3 3 7 C
B a s is : 4 0 lb m m in c irc u la tio n
(a)
Expansion valve
R = Refrigerant 12
4 0 lb m R ( l) /m in
9 3 .3 p s ig , 8 6 ° F
H = 2 7 .8 B tu /lb
4 0 lb m / m in
x v lb m R ( v ) / lb m
( 1 x v ) lb m R ( l ) / lb m
m
H v 7 7 .8 B tu / lb m ,
E n e rg y b a la n c e : E p , W s , Q 0 , n e g le c t E k H
H l 9 .6 B tu / lb m
n H n H 0
i
i
out
(b)
i
i
in
Evaporator coil
4 0 lbm /m in
4 0 lb m R ( v ) /m in
0 .2 6 7
R (v )
1 1 .8 p s ig , 5 ° F
0 .7 3 3
R(l )
H = 7 7 .8 B tu /lb m
1 1 .8 p s ig , 5 ° F
H v = 7 7 .8 B tu /lb m ,
H l = 9 .6 B tu /lb m
E n e rg y b a l a n c e : E p , W s 0 , n e g le c t E k Q H
(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
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lOMoARcPSD|32610952
(7.22 c cont’d)
Q Q e v a p o ra to r Q c o n d e n s e r 2 0 0 0 2 5 0 0 5 0 0 B tu m in
5 0 0 B tu
Q t W c t 0 W c Q
m in
7.23
1. 3 4 1 1 0
1 m in
60 s
9 .4 8 6 1 0
4
3
hp
1 1.8 h p
B tu s
B a s is : G iv e n fe e d ra te s
n 1 ( m o l / h )
n C H ( m o l C 3 H 8 / h )
3
8
n C H ( m o l C 4 H 1 0 / h )
0 .2 C 3 H 8
0 .8 C 4 H 1 0
4
10
o
o
227 C
0 C , 1 .1 a tm
n 2 ( m o l / h )
0 .4 0 C 3 H 8
0 .6 0 C 4 H 1 0
Q ( k J / h )
o
2 5 C , 1 .1 a tm
P ro p a n e b a la n c e
n C H
3
8
1 4 .7 m o l
0 .2 0 m o l C 3 H 8
h
m ol
9 .0 0 m o l
h
0 .4 0 m o l C 3 H 8
m ol
6 .5 4 m o l C 3 H 8 h
T o ta l m o le b a la n c e : n C H
4
10
( 1 4 .7 9 .0 0 6 .5 4 ) m o l C 4 H 2 0 h 1 7 .1 6 m o l C 4 H 2 0 h
E n e rg y b a l a n c e : E p , W s 0 , n e g le c t E k Q H
( H i 0 for components of 1st feed stream)
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lOMoARcPSD|32610952
7.24
(a)
Q (k J /m in )
.
n 0 ( k m o l / m in )
2 1 . 4 k m o l / m in
38°C , h r = 97%
1 8 ° C , s a t'd
y 0 ( m o l H 2 O ( v ) /m o l )
y 1 ( m o l H 2 O ( v ) / m o l)
( 1 – y 1 ) ( m o l d r y a ir /m o l )
( 1 – y 0) ( m o l d r y a ir / m o l )
.
n 2 ( k m o l H 2 O ( l ) / m in )
18°C
h r PH O 3 8 C
In le t c o n d itio n : y o
2
760 m m H g
P
PH O 1 8 C
O u tle t c o n d itio n : y 1
2
P
0 .9 7 4 9 .6 9 2 m m H g
1 5 .4 7 7 m m H g
760 m m H g
0 .0 6 3 4 m o l H 2 O m o l
0 .0 2 0 4 m o l H 2 O m o l
(b)
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lOMoARcPSD|32610952
(7.24 b cont’d)
5 .6 7 1 0
4
kJ
6 0 m in
0 .9 4 8 6 B tu
m in
h
kJ
1 to n c o o lin g
1 2 0 0 0 B tu
2 7 0 to n s o f c o o lin g
Basis: 100 mol feed
7.25
n 2 ( m o l ), 6 3 . 0 ° C
A - A ceto n e
0 . 9 8 A (v )
B - A c e t ic A c id
0 . 0 2 B (v )
Qc (c a l)
0. 5 n 2 ( mo l)
1 0 0 m o l, 6 7 . 5 ° C
0 . 9 8 A (l )
0 . 6 5 A (l )
0. 5 n 2 ( mo l)
5 6 .8 ° C
0 . 9 8 A (l )
0 . 0 2 B (l )
0 . 3 5 B (l )
0 . 0 2 B (l )
n 5 ( m o l ), 9 8 . 7 ° C
0 . 5 4 4 A (v )
0 . 4 5 6 B (v )
n 5 ( m o l ), 9 8 . 7 ° C
0 . 1 5 5 A (l )
0 . 8 4 5 B (l )
Qr (c a l)
(a)
O v e ra ll b a la n c e s :
O v erall en erg y b alan c e : Q H
E , W 0, E 0
p
2
x
n H n H
i
out
i
i
i
in
in te rp o la te in ta b le
in te rp o la te in ta b le
Q 5 8 .8 0 1 .2 0 6 .2 1 3 8 5 3 3 .8 1 3 1 2 6 5 3 5 4 3 5 3 3 5 1 .8 2 1 0 4 c a l
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(7.25 cont’d)
(b)
E n e rg y b a la n c e o n c o n d e n s e r: Q c H
E , W 0, E 0
3
p
k
Assume negligible heat transfer between system & surroundings other than Q c & Q r
Q r Q Q c 1 .8 2 1 0
4
8 .7 7 1 0
5
8 .9 5 1 0 c a l heat added to reboiler
5
7.26
1 .9 6 k g , P 1 = 1 0 .0 b a r , T 1
o
2 .9 6 k g , P 3 = 7 .0 b a r , T 3 = 2 5 0 C
1 .0 0 k g , P2 = 7 .0 b a r , T 2
Q= 0
(a)
T 2 T ( P 7 .0 b a r, s a t' d s te a m ) = 1 6 5 .0
o
C
o
H 3 ( H 2 O ( v ), P = 7 .0 b a r, T = 2 5 0 C ) 2 9 5 4 k J k g (T a b le B .7 )
H ( H O ( v ), P = 7 .0 b a r, s a t' d ) 2 7 6 0 k J k g
( T a b le B .6 )
2
2
E n e rg y b a l a n c e
E
p
,
Q , W , E 0
s
k
H 0 2 .9 6 H 3 1.9 6 H 1 1.0 H 2 1.9 6 H 1 2 .9 6 k g (2 9 5 4 k J / k g ) - 1 .0 k g (2 7 6 0 k J / k g )
H 1 ( 1 0 .0 b a r, T 1 ) 3 0 5 3 k J / k g T 1 3 0 0 C
(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.
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lOMoARcPSD|32610952
7.27
(a)
(b)
P Pm a x 2 0 .0 b a r;
m to ta l 1 6 5.0 0 .0 3 7 6 1 6 5.0 4 k g
T 1 T ( P 2 0 .0 b a r, s a t' d . ) = 2 1 2 .4 C
3
3
Vl ( P 2 0 .0 b a r, s a t' d . ) = 0 .0 0 1 1 7 7 m / k g ; Vv ( P 2 0 .0 b a r, s a t' d . ) = 0 .0 9 9 5 m / k g
V to ta l m l Vl m v Vv m l Vl ( m to ta l m l )Vv
3
2 0 0 .0 L
1 m
3
3
m l k g ( 0 .0 0 1 1 7 7 m / k g ) + (1 6 5 .0 4 - m l ) k g ( 0 .0 9 9 5 m / k g )
1000 L
m l 1 6 4 .9 8 k g ; m v 0 .0 6 k g
Vl
0 .0 0 1 1 7 7 m
kg
1000 L
m
1 6 4 .9 8 k g
3
( 0 .0 6 - 0 .0 4 ) k g
m e v a p o r a te d
(c)
3
1000 g
kg
E n e rg y b a la n c e
1 9 4 .2 L ;
V s p a c e 2 0 0 .0 L - 1 9 4 .2 L = 5 .8 L
20 g
Q = U U ( P 2 0 .0 b a r, s a t' d ) U ( P 3 .0 b a r, s a t' d )
E , W , E 0
p
s
k
U l ( P 2 0 .0 b a r, s a t' d . ) = 9 0 6 .2 k J / k g ; U v ( P 2 0 .0 b a r, s a t' d . ) = 2 5 9 8 .2 k J / k g
U l ( P 3 .0 b a r, s a t' d . ) = 5 6 1 .1 k J / k g ; U v ( P 3 .0 b a r, s a t' d . ) = 2 5 4 3 k J / k g
Q 0 .0 6 k g (2 5 9 8 .2 k J / k g ) + 1 6 4 .9 8 k g ( 9 0 6 .2 k J / k g ) - 0 .0 4 k g (2 5 4 3 k J / k g )
1 6 5 .0 k g ( 5 6 1 .1 k J / k g ) = 5 .7 0 1 0
4
kJ
Heat lost to the surroundings, energy needed to heat the walls of the tank
(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.
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(7.27 cont’d)
(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.
7.28
(a)
When T 0 C , H 0 , T re f 0 C
(b)
Energy Balance-Closed System: U 0
o
E
,
k
E
o
,
p
Q,W 0
25 g F e, 175°C
25 g Fe
1 0 0 0 g H 2 O ( l)
1 0 0 0 g H 2O
20°C
T f (°C )
7.29
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lOMoARcPSD|32610952
(7.29 cont’d)
In itia l c o n d itio n s : T a b le B .5 U L 1 1 0 4 .8 k J k g , V L 1 1.0 0 3 L k g
P 0 .0 3 1 7 b ar T 2 5 C , sat' d U v 1 2 4 0 9 .9 k J k g , V L 1 4 3 , 4 0 0 L k g
Tf
U v
U L
Vv
V L
f
2 0 1.4
2 5 9 3 .8
8 5 6 .7
1 2 3 .7
1.1 5 9
5 .1 2 1 0
2
1 9 8 .3
2 5 9 2 .4
8 4 2 .9
1 3 1.7
1.1 5 4
1.9 3 1 0
2
2
1 9 5 .0
2 5 9 0 .8
8 2 8 .5
1 4 0 .7
1.1 4 9
1.3 4 1 0
1 9 6 .4
2 5 9 1.5
8 3 4 .6
1 3 6 .9
1.1 5 1
4 .0 3 1 0
4
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T f 1 9 6 .4 C , P f 1 4 .4 b a rs
lOMoARcPSD|32610952
7.30
(a)
P o in t 1 - s u rfa c e o f flu id . P1 3 .1 b a r , z 1 7 m ,
P o in t 2 - d is c h a rg e p ip e o u tle t .
gz
(b)
7.31
7m
9 .8 0 6 6 m
s
2
6 8 .6 m
,
2
s
m/s
m, u 2 ?
2
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.
P o in t 1 - s u rfa c e o f la k e . P1 1 a tm , z 1 0 , u 1 0
P o in t 2 - p ip e o u tle t . P2 1 a tm ,
S h a ft w o rk :
W s
m
-8 h p
0 .7 3 7 6 ft lb f / s
1 .3 4 1 1 0
3
hp
1 m in
7 .4 8 0 5 g a l
95 gal
1 ft
3 3 3 ft lb f lb m
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3
1
ft
3
6 2 .4 lb m
60 s
1 m in
lOMoARcPSD|32610952
(7.31 cont’d)
7.32
P o in t 1 - s u rfa c e o f re s e rv o ir . P1 1 a tm (assume), u 1 0 , z 1 6 0 m
P o in t 2 - d is c h a rg e p ip e o u tle t . P2 1 a tm (assume), u 2 ? , z 2 0
P 0
gz
P
9 .8 0 6 6 m
s
u
2
65 m
2
2
gz
1 N
1 kg m / s
W s
m
2
637 N m kg
2
3 .3 7 6V 6 3 7
800 T E
V
V
1. 2 7 m
s
3
60 s
7 6 .2 m
1 m in
Include friction (add F 0 to left side of equation) V increases.
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3
m in
lOMoARcPSD|32610952
CHAPTER EIGHT
8.1
(a)
2
U ( T ) 2 5.9 6 T 0 .0 2 1 3 4 T J / m o l
o
U ( 0 C ) 0 J / m o l
(b)
o
U ( 1 0 0 C ) 2 8 0 9 J / m o l
T r e f 0 C ( s in c e U ( 0 C ) = 0 )
o
We can never know the true internal energy. U ( 1 0 0 o C ) is just the change from U ( 0 o C ) to
o
U ( 1 0 0 C ) .
(c)
Q W U E k E p
E k 0, E p 0, W 0
Q U ( 3 .0 m o l )[( 2 8 0 9 0 ) J / m o l] 8 4 2 8 J 8 4 0 0 J
(d)
8.2
(a)
C v 2 7 .0 0 .0 2 9 1 T [ J / (m o l C )]
100
(b)
Hˆ
C d T 3 5 .3 T
p
25
100
25
100
0 .0 2 9 1
2784 J m ol
2 25
T
2
(c)
(d)
o
H is a s ta t e p r o p e r ty
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8.3
(a)
(b)
(c)
(d)
(e)
8.4
o
o
H 2 O (v , 1 0 0 C , 1 a tm ) H 2 O (v , 3 5 0 C , 1 0 0 b a r)
(a)
H 2 9 2 6 k J k g 2 6 7 6 k J k g 2 5 0 k J k g
(b)
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8.5
8.6
(a)
(b)
The difference is the flow work done on the gas in the continuous system.
(c)
Q a d d itio n a l heat needed to raise temperature of vessel wall + heat that escapes from
wall to surroundings.
8.7
(a)
(b)
(c)
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lOMoARcPSD|32610952
(d)
8.8
(a)
(b)
Q n H n H
8.9
o
(50 C )
H
o
(450 C)
( 1 7 8 .5 m o l / s )(0 .7 3 - 1 2 .8 1 5 [ k J / m o l] ) = 2 ,1 5 7 k W
Assume ideal gas behavior, so that pressure changes do not affect H .
n
2 0 0 ft
h
3
o
4 9 2 R 1.2 a tm
o
1 a tm
537 R
1 lb - m o l
3
0 .6 1 2 5 lb - m o le / h
3 5 9 ft (S T P )
8.10
Assume H m ix 0
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8.11
8.12
(a)
o
5 5 0 0 L (S T P )/ m in C H 3 O H (v ) 6 5 C
o
n 2 m o l/ m in C H 3 O H (v ) 2 6 0 C
n 2 (m o l/ m in )
o
m w k g / m in H 2 O ( l, s a t ‘d ) @ 9 0 C
o
m w k g / m in H 2 O (v , s a t ‘d ) @ 3 0 0 C
3
3
V w 2 ( m / m in )
n2
5 5 0 0 L (S T P )
1 m ol
m in
2 2 .4 L (S T P )
V w 1 ( m / m in )
2 4 5 .5 m o l C H 3 O H (v )/m in
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 .1 3 k g H 2 O /m in .
(b)
kg
k J 1 m in 1 k W
Q 1 .1 3
2 3 7 3 .9
4 4 .7 k W
m in
k g 6 0 s e c 1 k J /s
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8.13
(a)
1270 mol/h, 620°C
425°C
m (kg H O(
)/h),
l 20°C
2
(b)
When cold water contacts hot gas, heat is transferred from the hot gas to the cold water
lowering the temperature of the gas (the object of the process) and raising the
temperature of the water.
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8.14
8.15
(a)
1 5 1 5 L / s a ir
o
5 0 0 C , 8 3 5 t o r,
o
Tdp= 3 0 C
1 5 1 5 L / s a ir , 1 a t m
1 1 0 g /s H 2O ( v)
o
1 1 0 g /s H 2O , T= 2 5 C
Let n 1 (m o l / s ) be the molar flow rate of dry air in the air stream, and n 2 (m o l / s) be
the molar flow rate of H2O in the air stream.
n 1 + n 2
n 2
n 1 + n 2
1515 L 835 m m H g
s
773 K
o
= y =
p * (3 0 C )
P to ta l
m ol K
6 2 .3 6 L m m H g
3 1.8 2 4 m m H g
835 m m H g
2 6 .2 m o l / s
0 .0 3 8 1 m o l H 2 O / m o l a ir
n 1 2 5 .2 m o l d r y a ir / s ; n 2 1.0 m o l H 2 O / s
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(b)
Q 2 5 .2
(c)
139
500
C
p
a ir
d T 1 .0 0
139
500
C
p
dT 290 kW
H 2O (v )
This heat goes to vaporize the entering liquid water and bring it to the final temperature
of 139oC.
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 air.
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8.16
8.17
(a)
(b)
H 6 4 .0 5 k J / m o l
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(c)
8.18
n
1 .7 5 m
3
2 .0 m in
879 kg
m
3
3
1 km ol
10
m ol
1 m in
7 8 .1 1 k g
1 km ol
60 s
1 6 4 .1
m ol
s
8.19
Time to Saturation
6 k g c a rb o n
0 .4 0 g C C l 4
1 m ol C C l 4
1 m ol gas
1 m in
g c a rb o n
1 5 3 .8 4 g C C l 4
0 .0 3 4 7 m o l C C l 4
10 m ol gas
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4 5 .0 m in
lOMoARcPSD|32610952
H v 3 5 .9 8 k J m o l , Tb 1 3 6 .2 C 4 0 9 .4 K , Pc 3 7 .0 a tm , Tc 6 1 9 .7 K
8.20
8.21
(a)
Antoine equation:
Watson Correction:
(b)
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(from Table B.1)
lOMoARcPSD|32610952
(c)
8.22
(a)
Tout = 49.3oC. The only temperature at which a pure species can exist as both vapor and
liquid at 1 atm is the normal boiling point, which from Table B.1 is 49.3oC for
cyclopentane.
(b)
Ideal gas equation of state
n f
1550 L
273 K
1 m ol
s
423 K
2 2 .4 L (S T P )
4 4 .6 6 m o l C 5 H 1 0 (v ) / s
55% condensation: n l 0 .5 5 0 ( 4 4 .6 6 m o l / s) = 2 4 .5 6 m o l C 5 H 1 0 ( l ) / s
Cyclopentane balance n v ( 4 4 .6 6 2 4 .5 6 ) m o l C 5 H 1 0 / s = 2 0 .1 0 m o l C 5 H 1 0 (v ) / s
Reference: C5H10(l) at 49.3oC
n in
H in
n o u t
Substance
H o u t
(mol/s)
(kJ/mol)
(mol/s)
(kJ/mol)
C5H10 (l)
—
—
24.56
0
C5H10 (v)
44.66
H f
20.10
H v
S u b s titu ti n g f o r H v f r o m T a b le B .1 a n d f o r C p f r o m T a b le B .2
H f 3 8 .3 6 k J / m o l, H v 2 7 .3 0 k J / m o l
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Energy balance: Q n o u t H o u t n in H in 1.1 6 1 0 3 k J / s = 1.1 6 1 0 3 k W
8.23
100 kmol/h at 0°C, 3 atm
y out (kmol C H6 (14)/kmol),
v
saturated
n 1 (kmol/h) at 75°C, 3 atm
(1 – y out
) (kmol N /kmol)
2
90% sat'd
v
y in (kmol C H6 (14)/kmol),
(1 – y in)
n 2 kmols/h
(kmol N /kmol)
2
n C 6H 14( ),v 0°C
Antoine:
lo g p v 6 .8 8 5 5 5
1 1 7 5 .8 1 7
2 2 4 .8 6 7 T
p v 0 C 4 5 .2 4 m m H g , p v 7 5 C 9 2 0 .4 4 m m H g
Percent Condensation:
References: N2(25oC), n-C6H14(l, 0oC)
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8.24
(a)
H = n -h e xa n e
5 0 . 3 m o l/ s , 8 5 0 m m H g
a ss u me P = 8 50 m mH g
x0 m o l H / m o l
n 2 m o l H (v )/ m o l , s a t ’d @ T C
(1 - x0 ) m o l a ir/ m o l
o
o
40 C , T d p = 20 C
o
n 3 m o l a ir/ m o l
n 1 (m o ls H ( l )/ s )
(9 0 % o f H in fe e d )
(b)
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Antoine equation:
*
p H 6 7 .8 m m H g T 7 .8 C
Air: H fro m T a b le B .8
Energy balance: Q H n i H i n i H i
out
257 kJ s
in
1 k W c o o lin g
1 kJ s
257 kW
(c)
8.25
o
n v ( m o l / m in ) @ 6 5 C , P 0 ( a tm )
y [ m o l P (v ) / m o l] , s a t' d
(1 - y ) (m o l H (v ) / m o l)
o
1 0 0 m o l / s @ 8 0 C , 5 .0 a tm
0 .5 0 0 m o l P (l) / m o l
Q ( k J / s )
0 .5 0 0 m o l H (l) / m o l
o
n l ( m o l / m in ) @ 6 5 C , P 0 ( a tm )
0 .4 1 m o l P (l) / m o l
0 .5 9 m o l H (l) / m o l
(a)
Degree of freedom analysis
5 unknowns
– 2 material balances
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– 2 equilibrium relations (Raoult’s law) at outlet
– 1 energy balance
0 degrees of freedom
Antoine equation (Table B.4) p *P ( 6 5 o C ) = 1 8 5 1 m m H g , p *H ( 6 5 o C ) = 6 7 5 m m H g
T o ta l m o le b a la n c e : 1 0 0 m o l = n v n l
n v 3 6 .6 m o l v a p o r / s
n l 6 3 .4 m o l liq u id / s
P e n ta n e b a la n c e : 5 0 m o le P = 0 .6 5 6 n v + 0 .4 1 0 n l
Id e a l g a s e q u a tio n o f s ta te : V v
nv RT
P0
Fractional vaporization:
0 .0 8 2 0 6 L a tm
3 6 .6 m o l
m ol K
s
f
(6 5 + 2 7 3 )K
3 6 .6 m o l v a p o r / s
0 .3 6 6
100 m ol / s
1.5 2 a tm
m o l v a p o riz e d
m o l fe d
References: P (l), H (l) a t 6 5 o C
S u b s ta n c e
n in
H in
n o u t
H o u t
P (v )
2 4 .0
2 4 .3 3
n in m o l s
P (l)
50
2 .8 0 6
2 6 .0
0
H in k J / m o l
H (v )
1 2 .6
2 9 .0 5
H (l)
50
3 .2 4 5
3 7 .4
0
T b a n d H v fro m T a b le B .1 , C p fro m T a b le B .2
Energy balance:
Q
n
out H out
n H
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in
in
1040 kW
667 L / s
lOMoARcPSD|32610952
8.26
Neglect enthalpy change for the vapor transition from 116C to 113C.
8.27
(a)
Solids balance: 2 0 0 0 .3 5 m 3
m 3 5 7 1.4 k g h slurry
H 2 O b a la n c e :
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(b)
5 9 2 .0 4 2 8 .6 1 6 3 k g h a d d itio n a l s te a m
(c)
The cost of compressing and reheating the steam vs. the cost of obtaining it externally.
8.28
Basis: 5000 kg seawater/h
(a)
S = Salt
n 3 (kg H O(
)/h
l @ 4 bars)
2
2738 kJ/kg
n 4 kg H O(
)/h
v @ 0.2 bars
2
n 2 (kg H O(
)/h
v @ 0.6 bars)
2
2610 kJ/kg
2654 kJ/kg
(b)
5000 kg/h @ 300 K
n 1 (kg/h @ 0.6 bars)
0.035 S
0.055 S
x (kg S/kg)
0.965 H O(
2 ) l
0.945 H O(
)l
2
(1 –
113.1 kJ/kg
360 kJ/kg
252 kJ/kg
n 3 (kg/h @ 0.2 bars)
n 5 (kg H O(
)/h
l @ 4 bars)
2
n 2 (kg H O(
)/h
l @ 0.6 bars)
2
605 kJ/kg
360 kJ/kg
S balance on 1st effect:
Mass balance on 1st effect: 5 0 0 0 3 1 8 2 n 2 n 2 1 8 1 8 k g h
(c)
)x (kg H 2O( )/hr)
l
Mass balance on 2nd effect: 3 1 8 2 n 3 n 4
(1)
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Energy balance on 2nd effect:
Solve (1) and (2) simultaneously:
n 3 1 2 6 7 k g h
n 4 1 9 1 5 k g h
(d)
brine solution
H2O (v)
The entering steam must be at a higher temperature (and hence a higher saturation
pressure) than that of the liquid to be vaporized for the required heat transfer to take
place.
(e)
n 5 (kg H O(
)/h)
v
2
2738 kJ/kg
3733 kg/h H O(2 ) @
v 0.2 bar
2610 kJ/kg
5000 kg/h
n 1 (kg brine/h @ 0.2 bar
0.035 S
252 kJ/kg
0.965 H O(
)l
2
113.1 kJ/kg
Q3
l
n 5 (kg H O(
)/h)
2
605 kJ/kg
Mass balance: 5 0 0 0 3 7 3 3 n 1 n 1 1 2 6 7 k 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?
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8.29
(a)
(b)
Final result given in Part (d).
(c)
S a lt b a la n c e o n i
th
e ffe c t:
(d)
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8.30
(a)
Basis: 50 kg wet steaks/min
D.M. = dry meat
m 1 (kg H O(
)/min)
v
2
(96% of H 2 O in feed)
60°C
50 kg/min @ –26°C
0.72 H O(
2 ) s
m 2 (kg D.M./min)
0.28 D.M.
m 3 (kg H O(
)/min)
l
2
Q (kW)
50°C
Dry meat:
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Energy balance:
Q H
m i H i
out
8.31
m i H i
1 .0 6 1 0
5
m in
in
kJ
1 m in
1 kW
60 s
1 kJ s
1760 kW
Basis: 20,000 kg/h ice crystallized. S = solids in juice. W = water.
.
.
Q f
preconcent rate
.
m 1 (kg/h) juice
m 2 (kg/h)
0.12 solids(S)
x 2 (kg S/kg)
0.88 H O(
l
2 )(W)
(1 – x )2 (kg W/kg)
.
Slurry(10% ice), –7°C
freezer
20,000 kg W ( )/hs
.
0.55 kg W /kg
0.45 kg S/kg
20,000 kg W ( )/hs
0.55 kg W ( )/kg
l
m4
m 3 (kg/h), 0°C
0.45 kg S/kg
.
separat or
0.45 kg W ( )/kg
l
10% ice in slurry
20000
m 4
10
90
m 4 1 8 0 0 0 0 k g h
concentrate leaving freezer
Mass balance on filter:
2 0 0 0 0 m 4 m 5 2 0 0 0 0 m 6
0.45 kg S/kg
.
m 4 kg residue/h
20°C
(a)
m 5 (kg/h) product
filter
m 4 1 8 0 0 0 0
m 5 7 2 7 3
m 6 1 7 2 7 3 0 k g h re c y c le
Mass balance on mixing point:
2 7 2 7 3 1 7 2 7 3 0 m 2 m 2 2 .0 0 0 1 0
5
k g h p r e c o n c e n tr a te
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(kg/h), 0.45 S, 0.55 W
20,000 kg W ( )/hs
lOMoARcPSD|32610952
(b) Draw system boundary for every balance to enclose freezer and mixing point (Inputs:
fresh feed and recycle streams; output; slurry leaving freezer)
8.32
(a)
Basis: 1 mol feed/s
n V m o l v a p o r/ s @ T , P
y mo l A /mo l
o
1 m o l/ s @ T F C
(1 -y ) m o l B / m o l
xF m o l A / m o l
(1 - xF ) m o l B / m o l
n L m o l v a p o r/ s @ T , P
x mo l A /mo l
(1 - x ) m o l B / m o l
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v a p o r a n d liq u id s t re a m s
in e q u il ib r iu m
lOMoARcPSD|32610952
(b)
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lOMoARcPSD|32610952
(c)
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lOMoARcPSD|32610952
8.33
(a)
For 24°C and 50% relative humidity, from Figure 8.4-1,
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(b)
(c)
(d)
8.34
3
V ro o m 1 4 1 f t . D A = d r y a ir .
m DA =
ha
1 4 0 ft
3
o
lb - m o l R 2 9 lb m D A
0 .7 3 0 2
0 .2 0 5 lb m H 2 O
1 0 .1 lb m D A
ft
3
a tm
1 a tm
lb - m o l 5 5 0
o
R
1 0 .1 lb m D A
0 .0 2 0 3 lb m H 2 O / lb m D A
8.35
Td b 3 5 C
Tab 2 7 C
8.36
h r 5 5 % H e w in s
(a)
F ig . 8 .4 -1
T d b 4 0 C , T d e w p o in t 2 0 C
h r 3 3 % , h a 0 .0 1 4 8 k g H 2 O k g d ry a ir
T w b 2 5 .5 C
(b)
Mass of water:
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2 .2 1 0
3
k g d ry a ir
0 .0 1 4 8 k g H 2 O
1 k g d ry a ir
10
3
g
1 kg
0 .0 3 3 g H 2 O
(c)
(d)
8.37
(a)
400 kg
2 .4 4 k g w a te r
m in
9 7 .5 6 k g a ir
1 0 .0 k g w a te r e v a p o ra te s / m in
(b)
(c)
Td b 1 0 C , saturated
h a 0 .0 0 7 7 k g H 2 O k g d r y a ir , H 2 9 .5 k J k g d r y a ir
(d)
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substance
m in
H in
m o u t
H o u t
Air
400
115
400
29.5
m a ir in kg dry air/min, m H O
2
—
H2O (l)
—
6.92
42
in kg/min
H a ir in kJ/kg dry air, H H O in kJ/kg
2
H a ir in kJ/kg dry air, H H O in kJ/kg
2
H2O (l, 00C)
Q H
H2O (l, 200C):
m Hˆ m Hˆ
i
i
out
(e)
8.38
i
3 4 0 2 7 .8 k J
1 m in
1 kW
m in
60 s
1 k J /s
i
in
565 kW
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.
Basis: 1 kg wet chips. DA = dry air, DC = dry chips
(a)
Outlet air: Tdb=38oC, Twb=29oC
Inlet air: 11.6 m3(STP), Tdb=100oC
m2a (kg DA)
m2w [kg H2O(v)]
m1a (kg DA)
1 kg wet chips, 19oC
0.40 kg H2O(l)/kg
0.60 kg DC/kg
m3c (kg dry chips)
m3w [kg H2O(l)]
T (oC)
Dry air: m1a =
Outlet air:
T d b 3 8 C , T w b 2 9 C Hˆ 2 (9 5 .3 0 .4 8 ) 9 4 .8 k J k g D .A .
F ig . 8 .4 -1
h a 0 .0 2 2 3 k g H 2 O k g D .A .
2
Water in outlet air:
(b)
H 2 O b a la n c e :
0 .4 0 0 k g = 0 .3 3 5 k g + m 3 w m 3 w 0 .0 6 5 k g H 2 O
Moisture content of exiting chips:
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0 .0 6 5 k g w a te r
1 0 0 % 9 .8 % 1 5 % m e e ts d e s ig n s p e c ific a tio n
0 .6 0 0 k g d ry c h ip s + 0 .0 6 5 k g w a te r
(c)
References: Dry air, H2O (l), dry chips @ 0°C.
H in
substance
m in
Air
H2O (l)
dry chips
15.02 100.2
0.400 79.5
0.600 39.9
m out
H o u t
15.02 94.8 m a ir in kg DA, H a ir in kJ/kg DA
0.065 4.184T m in kg DC, H in in kJ/kg DC
0.6 2.10T
Energy Balance: H m o u t Hˆ o u t m in Hˆ in 0 1 3 6 .8 1 .5 3 2 T 0 T 8 9 .3 C
8.39
(a)
Td b 4 5 C
T a s T w b 2 1. 0 C
hr 10%
h a 0 .0 0 5 9 k g H 2 O k g D A
F ig . 8 .4 -1
(b)
8.40
In le t a ir : T d b 5 0 C
F ig . 8 .4 -1
h a 0 .0 0 5 0 k g H 2 O k g D . A .
T d e w p t. 4 C
1 1 .3 m
m in
3
1 kg D .A .
0 .9 2 m
3
O u tle t a ir : T w b T a s 2 2 C
s a tu ra te d
3
V 0 .9 2 m k g D . A . , T w b 2 2 C
1 2 .3 k g D . A . m in
T 22 C
h a 0 .0 1 6 5 k g H 2 O k g D . A .
Evaporation:
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8.41
(a)
(b)
Basis: 1 kg entering sugar (S) solution
m1 (kg D.A.)
0.0050 kg H2O/kg DA
1 kg
0.05 kg S/kg
0.95 kg H2O/kg
m1 (kg D.A.)
0.0151 kg H2O(v)/kg
m2 (kg)
0.20 kg S/kg
0.80 kg H2O/kg
Sugar balance:
Water balance:
8.42
Basis: 1 kg D.A.
(a)
Inlet air:
Tdb 4 0 C
Twb 1 8 C
h a 1 0 .0 0 3 9 k g H 2 O k g D . A .
Outlet air:
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(b)
12 5 0 kg /h
m a (lb m H 2 O / h )
o
o
T = 1 5 C , s a t ’d
T = 3 7 C , h r= 5 0 %
m c (lb m H 2 O / h )
liq u id , 1 2 ° C
Q c ( Btu /h )
Inlet air:
Moles dry air:
m a
1250 kg
1 kg D A
h
1 .0 1 9 8 k g
1226 kg D A h
Outlet air:
1 1.3 k g H 2 O h w ith d ra w n
Reference states for enthalpy calculations: H2O (l), dry air at 0o C.
( C p ) H O (l) = 4.184
2
kJ
o
kg C
H 2 O l , 1 2 C : Hˆ
12
C d T 5 0 .3 k J/k g
0
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p
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Overall system energy balance:
Q c H
m H m H
i
i
out
i
i
in
8.43
H
400 m ol N H 3
7 8 .2 k J
3 1,2 8 0 k J
m ol N H 3
8.44
Basis: 100 mol solution 20 mol NaOH, 80 mol H2O
r
80 m ol H 2O
4 .0 0
20 m ol N aO H
m ol H 2O
m ol N aO H
Refs: NaOH(s), H2O (l) @ 25oC
NaOH (Table B.11)
H n i H i n i H i ( 2 0 ) ( 3 4 .4 3 )
out
8.45
6 8 8 .6 k J
9 .4 8 6 1 0
in
Basis:
1 L
8 m ol H Cl
3 6 .4 7 g H C l
L
m ol H Cl
292 g H Cl
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10
3
4
kJ
B tu
6 5 3 .2 B tu
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4 6 . 0 m o l H 2 O (l , 2 5 ° C )
8 . 0 m o l H C l (g , 2 0 ° C , 7 9 0 m m H g )
1 L H C l (a q )
1120 g 292 g 828 g H 2 O ]
828 g H 2O
m ol
1 8 .0 g
n
4 6 .0 m o l H 2 O
8 .0 m o l H C l
4 6 .0 m o l H 2 O
5 .7 5 m o l H 2 O m o l H C l
Assume all HCl is absorbed
Volume of gas:
(b)
Ref: 25 C
Q H 4 7 1 k J L p ro d u c t
(c)
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8.46
Basis: Given solution feed rate
.
.
n a ( m o l a ir / m i n )
n a ( m o l a ir / m i n )
.
n 1 ( m o l H 2 O ( v) /m in )
2 0 0 ° C , 1 .1 b a r s
s a tu r a te d @ 5 0 ° C , 1 a t m
1 5 0 m o l/ m in s o lu tio n
.
0 .0 0 1 N a O H
n 2 ( m o l/ m in ) @ 5 0 ° C
0 .9 9 9 H 2O
0 .0 5 N a O H
2 5 °C
0 .9 5 H 2 O
NaOH balance:
min
S o lu tio n m a s s : m
1 m ol N aO H
4 0 .0 g
19 m ol H 2 O
1 m ol
Air @ 200C: Table B.8 H 5 .1 5 k J m o l
Air (dry) @ 50C: Table B.8 H 0 .7 3 k J m o l
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1 8 .0 g
1 m ol
382
g s o lu tio n
m ol N aO H
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8.47
(a)
Average molecular weight of feed solution:
M 0 .2 0 0 M A 0 .8 0 0 M H O
2
Molar flow rate of feed:
n0
(b)
200 kg
1 km ol
h
3 0 .8 k g
6 .4 9 k m o l h
16.9% evaporation:
A balance:
(1)
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Mass flow rate of crystals:
Mass flow rate of product solution:
(c)
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8.48
Ref: H 2 O , H 2 S O 4 @ 25 C
o
o
o
H ( H 2 O ( l ), 1 5 C ) [ 0 .0 7 5 4 k J / (m o l C )](1 5 2 5 ) C =
0 .7 5 4 k J / m o l
Energy Balance:
Conditions: Adiabatic, negligible heat absorbed by the solution container.
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8.49
(a)
Enthalpies of feeds and product:
Energy balance:
Conditions for validity: Adiabatic mixing; negligible heat absorbed by the solution
container, negligible dependence of heat capacities on temperature between 25oC and
TA0 for A, 25oC and TB0 for B, and 25oC and Tmax for the solution.
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(b)
C p s 3 .3 5 J (g C )
8.50
Ideal gas equation of state: n A 0 P0 V g / R T 0
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T m ax 1 2 5 C
(1)
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E.B.: U 0 n i U i n i U i
out
in
(b)
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(c)
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8.51
(a)
(b)
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(c)
(d)
A d d th e c o n c e n tra te d s o lu tio n to th e d ilu te s o lu tio n . 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.
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CHAPTER NINE
C 9 H 2 0 ( l ) 1 4 O 2 ( g ) 9 C O 2 ( g ) + 1 0 H 2 O ( l)
9.1
H r 6 1 2 4 k J / m o l
o
(a)
When 1 g-mole of C9H20(l) and 14 g-moles of O2(g) at 25C and 1 atm react to form 9 g-moles of
CO2(g) and 10 g-moles of water vapor at 25C and 1 atm, the change in enthalpy is -6124 kJ.
(b)
Exothermic at 25C. 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)
Q H =
n C H
9
H r
0
20
C H
9
2 5 .0 m o l C 9 H 2 0
6124 kJ
1 kW
s
1 m o l C 9 H 20
1 kJ / s
20
1.5 3 1 0
5
kW
5
Heat Output = 1.5310 kW.
The reactor pressure is low enough to have a negligible effect on enthalpy.
(d)
C 9 H 2 0 ( g ) 1 4 O 2 ( g ) 9 C O 2 ( g ) + 1 0 H 2 O ( l)
(1 )
H r 6 1 7 1 k J / m o l
o
C 9 H 2 0 ( l ) 1 4 O 2 ( g ) 9 C O 2 ( g ) + 1 0 H 2 O ( l)
(2 )
H r 6 1 2 4 k J / m o l
o
( 2 ) (1) C 9 H 2 0 ( l ) C 9 H 2 0 ( g )
H v ( C 9 H 2 0 , 2 5 C )
o
9.2
6 1 2 4 k J / m o l ( 6 1 7 1 k J / m o l) = 4 7 k J / m o l
(e)
Yes. Pure n-nonane can only exist as vapor at 1 atm above 150.6C, but in a mixture of gases, it
can exist as a vapor at lower temperatures.
(a)
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)
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(c)
M O = 3 2 .0
2
m 1 2 0 lb m / s
n O Hˆ r
o
Q H
2
vO
n 3 .7 5 lb - m o le / s .
1 .6 7 2 1 0
3 .7 5 lb -m o le /s
9 .5
6
B tu
1 lb -m o le O 2
2
5
6 .6 0 1 0 B tu /s
fro m re a c to r
CaC2 (s) + 5H2O (l) CaO (s) + 2CO2 (g) + 5H2 (g), H ro 6 9 .3 6 k J k m o l
9.3
(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 r is th e c h a n g e in in te rn a l e n e rg y w h e n 1 g - m o le o f C a C 2 (s ) a n d 5 g - m o le s o f H 2 O (l) a t 2 5 C a n d
o
1 a tm re a c t to fo rm 1 g - m o le o f C a O (s ), 2 g - m o le s o f C O 2 (g ) a n d 5 g - m o le s o f H 2 (g ) a t 2 5 C
a n d 1 a tm .
(c)
Q U
o
n C a C U r
2
v C aC
150 g C aC 2
1 m ol
5 2 .0 k J
6 4 .1 0 g
1 m ol C aC 2
2
Heat must be transferred to the reactor.
9.4
(a)
N2 (g) + O2 (g) 2NO (g),
(b)
(c)
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1 2 1 .7 k J
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Hˆ r 6 Hˆ f
o
o
CO2
7 Hˆ f
o
H 2O g
Hˆ f
o
C 6 H 14 l
6 3 9 3 .5 7 2 4 1 .8 3 1 9 8 .8 k J m o l 3 8 5 5 k J m o l
(d)
9.5
N a 2S O 4 (l) 4 C O (g ) N a 2S (l) 4 C O 2 (g )
(a)
Given reaction = (1) + (2) ⇒ -385.76 – 35.03 = -420.79 kJ/mol
C 2 H 2 (g)
9.6
(a)
5
2
O 2 (g ) 2 C O 2 (g ) + H 2 O (l)
H c 1 2 9 9 .6 k J m o l
o
The enthalpy change when 1 g-mole of C2H2(g) and 2.5 g-moles of O2(g) at 25C and 1 atm react
to form 2 g-moles of CO2(g) and 1 g-mole of H2O(l) at 25C and 1 atm is -1299.6 kJ.
(b)
(c)
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(d)
C 2 H 2 (g)
H 2 (g)
1
2
C 2 H 6 (g)
9.7
(a)
5
2
O 2 (g ) 2 C O 2 (g ) + H 2 O (l)
O 2 (g ) H 2 O (l)
7
2
O 2 (g ) 2 C O 2 (g ) + 3 H 2 O (l)
(1 )
H c 1 1 2 9 9 .6 k J m o l
(2 )
H c 2 2 8 5 .8 4 k J m o l
(3 )
H c 3 1 5 5 9 .9 k J m o l
o
o
o
C7H16 (g) C6H5CH3 (g) + 4H2 (g)
Basis: 1 mol C7H16
1 mol C7H16
1 mol C6H5CH3
400C
4 mol H 2
400 C
Q (kJ/mol)
(b)
o
References: C (s), (2 (g) @ 25 C
C 7 H 1 6 g , 4 0 0 C : Hˆ 1 ( Hˆ f ) C H ( g )
7
16
C p dT
0 .2 4 2 7
400
25
( 1 8 7 .8 + 9 1 .0 ) k J /m o l= 9 6 .8 k J /m o l
(c)
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Q H
n Hˆ n Hˆ
i
i
out
i
i
in
= [ ( 1 ) ( 1 1 0 .2 ) + ( 4 ) ( 1 0 .8 9 ) - ( 1 ) ( -9 6 .8 ) ] k J = 2 5 1 k J ( tr a n s f e r r e d to r e a c to r )
(d)
9.8
(a)
Hˆ r ( 4 0 0 C )=
251 kJ
2 5 1 k J /m o l
1 m o l C 7 H 1 6 re a c t
(CH3)2 O (g) CH4 (g) + H2 (g) + CO (g)
Moles charged: (Assume ideal gas)’
Let x + fraction (CH3)2O decomposed (Clearly x<1 since P f 3 P0 )
0 .0 1 2 8 6 m o l
0 .0 1 2 8 6 ( 1 – x ) m o l ( C H 3 ) 2 O
( C H 3 )2 O
0 .0 1 2 8 6 x
m ol C H 4
6 0 0 °C
6 0 0 °C , 35 0 m m H g
0 .0 1 2 8 6 x
m ol H 2
875 m m H g
0 .0 1 2 8 6 x
m ol C O
Total moles in tank at t = 2h = 0.01286[(1-x) + 3x] = 0.01286 (1 + 2x) mol
(b)
References: C s , H 2 g , O 2 g a t 2 5 C
600
C p d T 7 4 .8 5 2 9 .4 6 4 5 .3 9 k J m o l
C H 4 ( g ,6 0 0 C ) : Hˆ 2 ( Hˆ f ) C H
4
25
T a b le B .2
o
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(c)
For the reaction of parts (a) and (b), the enthalpy change and extent of reaction are:
H n o u t Hˆ o u t n in Hˆ in 1 .5 5 1 5 ( 1 .5 1 7 5 ) k J 0 .0 3 4 0 k J
( n C H ) o u t ( n C H ) in
4
4
CH
0 .7 5 0 .0 1 2 8 6
m o l 0 .0 0 9 6 4 5 m o l
1
4
0 .0 3 4 0 k J
H Hˆ r 6 0 0 C Hˆ r 6 0 0 C
3 .5 3 k J /m o l
0 .0 0 9 6 4 5
3 .5 3 k J m o l
(d)
8 .3 1 4 J
1 kJ
m ol K
3
10
873 K
1 1 1 1
J
18 .0 k J m o l
Q Uˆ r 6 0 0 C ( 0 .0 0 9 6 4 5 m o l)( 1 8 .0 k J /m o l) 0 .1 7 4 k J (tra n s fe rre d fro m re a c to r)
CO (g) + H2O (v) H2 (g) + CO2 (g),
9.9
(a)
3
3
Basis: [2.5 m (STP) product gas/h][1000 mol/22.4 m (STP)] = 111.6 mol/h
C balance on reactor: ṅ1 = (0.40)(111.6 mol/h) = 44.64 mol CO/h
H balance on reactor:
Steam theoretically required:
4 4 .6 4 m o l C O
1 m ol H 2O
h
1 m ol CO
4 4 .6 4 m o l H 2 O
% excess steam:
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Saturation of condenser outlet gas:
(b)
Energy balance on condenser
References: H 2 ( g ) , C O 2 ( g ) a t 2 5 C , H 2 O at reference point of steam tables
Enthalpies for CO2 and H2 from Table B.8
C O 2 ( g ,5 0 0 C ) : H 1 H C O ( 5 0 0 C ) 2 1.3 4 k J / m o l
2
H 2 ( g ,5 0 0 C ) : H 2 H H ( 5 0 0 C ) 1 3 .8 3 k J / m o l
2
kJ 18 kg
H 2 O (v ,5 0 0 C ) : Hˆ 3 3 4 8 8
6 2 .8 6 k J m o l
3
kg 10 m ol
C O 2 ( g ,1 5 C ) : Hˆ 4 Hˆ C O (1 5 C ) 0 .5 5 2 k J /m o l
2
H 2 ( g ,1 5 C ) : H 5 H H ( 1 5 C ) 0 .4 3 2 k J / m o l
2
Qc H
n Hˆ n Hˆ
i
out
i
i
4 9 .2 2 2 9 7 1 .8 k J
1 h
1 kW
h
3600 s
1 kJ s
i
in
h e a t tra n s fe rre d fro m c o n d e n s e r
(c)
Energy balance on reactor:
References: H 2 ( g ) , C (s), O 2 (g ) a t 2 5 C
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0 .8 1 2 k W
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C O ( g ,2 5 C ) : H 1 ( H f ) C O
T a b le B .1
1 1 0 .5 2 k J / m o l
H 2 O (v ,1 5 0 C ) : H 2 = ( H f ) H O (v ) H H O ( 1 5 0 C )
2
H 2 O (v ,5 0 0 C ) : H 3 = ( H f ) H O (v ) H H O ( 5 0 0 C )
2
Qr H
n Hˆ n Hˆ
i
i
i
out
T a b le s B .1 , B .8
2 2 4 .8 2 k J m o l
1 3 .8 3 k J / m o l
H C O ( 5 0 0 C )
2
2 3 7 .5 6 k J m o l
T a b le B .8
2
2
H 2 ( g ,5 0 0 C ) : H 4 H H ( 5 0 0 C )
C O 2 ( g ,5 0 0 C ) : H 5 ( H f ) C O
T a b le s B .1 , B .8
2
2
T a b le s B .1 , B .8
3 7 2 .1 6 k J / m o l
2 1 0 1 3 .8 3 ( 2 0 8 3 9 .9 6 ) k J
1 h
1 kW
h
3600 s
1 kJ s
i
in
0 .0 4 8 3 k W
h e a t tra n s fe rre d fro m re a c to r
(d)
Benefits
Preheating CO more heat transferred from reactor (possibly generate additional steam for
plant)
Cooling CO lower cooling cost in condenser.
9.10
C 6 H 5C H 3 O 2 C 6 H 5C H O H 2 O
C 6 H 5C H 3 9O 2 7C O 2 4H 2 O
Basis: 100 lb-mole of C 6 H 5 C H 3 fed to reactor.
3
r e a c to r
1 0 0 lb - m o le s C 6 H 5 C H 3
n 0 ( lb - m o le s O 2 )
3 .7 6 n 0 ( lb - m o le s N 2 )
3 5 0 ° F , 1 a tm
Q ( B tu )
ja c k e t
3
V 0 ( ft )
m w ( lb m H 2 O ( l ) ), 8 0 ° F
V p ( ft ) a t 3 7 9 ° F , 1 a tm
n 1 ( lb - m o le s C 6 H 5 C H 3 )
n 2 ( lb - m o le s O 2 )
3 .7 6n 0 ( lb - m o le s N 2 )
n 3 ( lb - m o le s C 6 H 5 C H O )
n 4 ( lb - m o le s C O 2 )
n 5 ( lb - m o le s H 2 O )
m w ( lb m H 2 O ( l ) ), 1 0 5 ° F
Strategy:
All material and energy balances will be performed for the assumed basis of 100 lb-mole
C 6 H 5 C H 3 . The calculated quantities will then be scaled to the known flow rate of water in the
product gas (23.3 lbm/4 h).
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100% excess air:
n0
1 0 0 lb -m o le s C 6 H 5 C H 3
1 m o le O 2 re q d
1 1 m o le O 2 f e d
1 m o le C 6 H 5 C H 3
1 m o le O 2 re q d
2 0 0 lb -m o le s O 2
N2 feed (& output) = 3.76(200)lb-moles N2 = 752 lb-moles N2
1 3 % C 6 H 5C H O n3
1 0 0 lb -m o le s C 6 H 5 C H 3
0 .1 3 m o le C 6 H 5 C H 3 re a c t
1 m o le C 6 H 5 C H O fo rm
1 m o le C 6 H 5 C H 3 fe d
1 m o le C 6 H 5 C H 3 re a c t
=1 3 lb -m o le s C 6 H 5 C H O
C balance:
H balance: (100)(8)lb-moles H = (86.5)(8) + (13)(6) + 2n5 ⇒ n5 = 15.0 lb-moles H2O (v)
O balance: (200)(2)lb-moles O = 2n2 + (13)(1) + (3.5)(2) + (15)(1) ⇒ n2 = 182.5 lb-moles O2
Ideal gas law inlet:
Ideal gas law – outlet:
Energy balance on reactor (excluding cooling jacket)
2
o
o
References: C (s), H2 (g), O (g), N2 (g) at 25 C (77 F)
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lOMoARcPSD|32610952
Enthalpies:
o
C6H5CHO (g, T): Ĥ(T) = [-17200 = 31(T – 77) F] Btu/lb-mole
3
⇒ Ĥ7 = -7.83x10 Btu/lb-mole
Energy Balance:
Q H
n H n H 2 .3 7 6 1 0 B tu
6
i
i
out
i
i
in
Energy balance on cooling jacket:
Q 2 .3 7 6 1 0
4
B tu , C p 1.0 B tu ( lb m
F)
Scale factor:
5
3
-1
4
3
5
3
-1
4
3
V0 = (6.218x10 ft )(0.02711 h ) = 1.69x10 ft /h (feed)
VP = (6.443x10 ft )(0.02177 h ) = 1.75x 0 ft /h (product)
4
-1
4
Q = (9.504x10 Btu)(0.02177 h ) = -6.44x10 Btu/h
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lOMoARcPSD|32610952
9.11
(a)
C a C O 3 ( s) C a O (s) +C O 2 ( g )
CaO(s)
900C
CaCO3(s)
25C
CO2(g)
900C
Q (kJ)
1 0 .0 k m o l C a O ( s ) p r o d u c e d
B a s is : 1 0 0 0 k g C a C O 3
1000 kg
1 m ol
0 .1 0 0 k g
1 0 .0 k m o l C a C O 3 1 0 .0 k m o l C O 2 ( g ) p r o d u c e d
1 0 .0 k m o l C a C O 3 ( s ) f e d
References: Ca(s), C(s), O2(g) at 25C
n in
Hˆ in
nout
Hˆ o u t
S u b s ta n c e
(m o l)
(k J /m o l)
(m o l)
(k J /m o l)
C aC O 3
10
Hˆ 1
4
C aO
10
CO 2
10
4
Hˆ 2
4
Hˆ 3
Energy balance:
(b)
Basis : 1000 kg CaCO3 fed 10.0 kmol CaCO3
C a C O 3 ( s ) C a O (s ) + C O 2 ( g )
2CO + O 2 2CO 2
10 kmol CaCO3
25 oC
Product gas at 900 oC
200 kmol at 900oC
n2 (kmol CO2 )
0.75 N2
0.020 O2
0.090 CO
0.14 CO2
n3 (kmol N2 )
n4 (kmol CO)
n1 [kmol CaO(s)]
1 0 k m o l C a C O 3 re a c t n 1 1 0 .0 k m o l C a O
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lOMoARcPSD|32610952
n 2 ( 0 .1 4 )( 2 0 0 )
1 0 .0 k m o l C a C O 3 re a c t
1 km ol C O 2
4 k m o l O 2 re a c t
1 km ol O 2
2 km ol C O 2
1 km ol O 2
46 km ol C O 2
n 3 ( 0 .7 5 )( 2 0 0 ) 1 5 0 k m o l N 2
C b a la n c e : (1 0 .0 )(1 ) + (2 0 0 )(0 .0 9 )(1 ) + (2 0 0 )(0 .1 4 )(1 ) = 4 6 (1 ) + n 4 ( 1) n 4 1 0 .0 k m o l C O
References: C a (s), C (s), O 2 ( g ), N 2 ( g ) a t 2 5 C
n in
Hˆ in
nout
Hˆ o u t
S u b s ta n c e
(m o l)
(k J /m o l)
(m o l)
(k J /m o l)
C aC O 3
1 0 .0
Hˆ 1
C aO
10
5 8 7 .0 6
CO 2
28
3 5 0 .5 6
46
3 5 0 .5 6
CO
18
Hˆ 2
10
Hˆ 2
O2
4 .0
Hˆ 3
150
Hˆ 4
150
Hˆ 4
N2
o
C a C O 3 (s , 2 5 C ) : Hˆ 1 ( Hˆ f ) C a C O ( s )
T a b le B .1
o
1 2 0 6 .9 k J /m o l
3
o
o
o
C O (g , 9 0 0 C ) : Hˆ 2 ( Hˆ f ) C O ( g ) Hˆ C O (9 0 0 C )
o
o
O 2 (g , 9 0 0 C ) : Hˆ 3 Hˆ O (9 0 0 C )
T a b le B .1 ,
T a b le B .8
( 1 1 0 .5 2 2 7 .4 9 ) k J /m o l 8 3 .0 3 k J /m o l
T a b le B .8
2 8 .8 9 k J /m o l
2
N 2 (g , 9 0 0 C ) : Hˆ 4 Hˆ N (9 0 0 C )
o
o
T a b le B .8
2 7 .1 9 k J /m o l
2
% r e d u c tio n in h e a t r e q u ir e m e n t
(c)
2 .7 1 0
6
0 .4 4 1 0
2 .7 1 0
6
6
1 0 0 8 3 .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.
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lOMoARcPSD|32610952
9.12
(a)
Fractional conversion: f A
C g e n e ra te d : n 0
m ol A consum ed
m o l A fe e d
x A 0 (m o l A fe d )
x AO n A
x AO
n A x A O (1 f A )
f A (m o l A c o n s u m e d )
Y C (m o l C g e n e ra te d )
m o l A fe d
m ol A consum ed
n C x AO f A YC
D g e n e ra te d : n D = 0 .5 m o l C c o n s u m e d = (1 2 ) (m o l A c o n s u m e d m o l C o u t)
n D ( 1 2 )( x A O f A n C )
B a la n c e o n B: m o l B o u t = m o l B in m o l B c o n s u m e d in (1 ) + m o l B g e n e ra te d in ( 2 )
= m o l B in m o l A c o n s u m e d in (1 ) + m o l D g e n e ra te d in (2 )
n B x BO x AO f A n D
B a la n c e o n I: m o l I o u t = m o l I in n I x IO
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lOMoARcPSD|32610952
(b)
9.13
o
(c)
For T f
(a)
C H 4 ( g ) O 2 ( g ) H C H O (g ) + H 2 O (g )
= 1 2 5 C , Q = 7 .9 0 k J . Raising Tp, lowering fA, and raising YC all increase Q.
10 L, 200 kPa
n3 (mol HCHO)
n0 (mol feed gas) at 25C
n4 (mol H2O)
0.851 mol CH4/mol
n5 (mol CH4)
0.15 mol O2 /mol
T (C), P(kPa), 10L
Q (kJ)
B a s is : n 0
200 kPa
1000 Pa
10 L
10
1 kPa
3
m
1 L
3
1 m ol K
8 .3 1 4 m
3
Pa
298 K
0 .8 0 7 2 m o l f e e d g a s m ix tu r e
0 .8 0 7 2 m o l fe e d g a s m ix tu re
(0 .8 5 )(0 .8 0 7 2 ) = 0 .6 8 6 1 m o l C H 4 ,
(0 .1 5 )(0 .8 0 7 2 ) = 0 .1 2 1 1 m o l O 2
C H 4 consum ed :
1 m ol C H 4
0 .1 2 1 1 m o l O 2 fe d
1 m o l O 2 fe d
0 .1 2 1 1 m o l C H 4
n 5 ( 0 .6 8 6 1 0 .1 2 1 1 ) m o l C H 4 0 .5 6 5 0 m o l C H 4
H C H O p ro d u c e d : n 3
1 m ol H C H O
0 .1 2 1 1 m o l C H 4 c o n su m e d
1 m o l C H 4 c o n su m e d
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0 .1 2 1 1 m o l H C H O
lOMoARcPSD|32610952
1 m ol H 2O
H 2 O p ro d u c e d : n 4
0 .1 2 1 1 m o l C H 4 c o n s u m e d
0 .1 2 1 1 m o l H 2 O
1 m o l C H 4 co n su m ed
E x te n t o f r e a c tio n :
( n O ) o u t ( n O ) in
2
2
O
0 0 .1 2 1 1
0 .1 2 1 1 m o l
1
2
o
R e fe re n c e s : C H 4 ( g ) , O 2 ( g ), H C H O (g ), H 2 O (g ), a t 2 5 C
n in
U in
n out
U o u t
m ol
kJ m ol
m ol
CH 4
0 .6 8 6 1
0
0 .5 6 5 0
kJ m ol
U
S u b s ta n c e
1
O2
0 .1 2 1 1
0
HCHO
0 .1 2 1 1
U 2
H 2O
0 .1 2 1 1
U 3
Q
100 J
85 s
1 kJ
s
8 .5 k J
1000 J
o
o
U r H r R T (
i
gaseous
p r o d u c ts
)
i
gaseous
r e a c ta n ts
2 8 2 .8 8 k J / m o l
8 .3 1 4 J
( 1 + 1 1 1)
298 K
1 kJ
m ol K
10
3
2 8 2 .8 8 k J / m o l
J
E n e rg y B a la n c e :
o
Q U r
(n
i ) o u t (U i ) o u t
(n
i ) in ( U i ) in
(0 .1 2 1 1 ) ( 2 8 2 .8 8 k J / m o l) + 0 .5 6 5 0 U 1 0 .1 2 1 1 U 2 0 .1 2 1 1 U 3
S u b s titu te fo r U 1 th ro u g h U 3 a n d Q
0 0 .0 2 0 8 8 T 1.8 4 5 1 0
5
T
2
0 .0 9 9 6 3 1 0
8
T
3
1.9 2 6 1 0
12
T
4
4 3 .2 9 k J / m o l
o
S o lv e fo r T u s in g E - Z S o lv e T 1 0 9 1 C 1 3 6 4 K
P nRT / V
(b)
0 .8 0 7 2 m o l
8 .3 1 4 m
3
Pa
m ol K
1364 K
10 L
1 L
10
3
m
3
915 10
3
Pa 915 kPa
Add heat to raise the reactants to a temperature at which the reaction rate is significant.
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(c)
S id e re a c tio n : C H 4 2 O 2 C O 2 2 H 2 O . T w o u ld h a v e b e e n h ig h e r (m o re n e g a tiv e h e a t o f
re a c tio n fo r c o m b u s tio n o f m e th a n e ), v o lu m e a n d to ta l m o le s w o u ld b e th e s a m e , th e re fo re
P n R T / V w o u ld b e g re a te r .
9.14
(a)
B a s is : 2 m o l C 2 H 4 fe d to re a c to r
n 6 (m ol C O 2 )
Qr ( k J )
n 7 (m ol H 2 O (l ))
2 5 °C
h e at
re a ctor
se p aration
n 1 (m ol C 2 H 4)
2 m ol C 2 H4
n 3 (m ol C 2 H 4)
n 2 (m ol O 2 )
1 m ol O2
n 4 (m ol O 2 )
p roce ss
2 5 °C
4 5 0 °C
n 5 (m ol C 2 H 4O )
n 5 ( m o l C 2 H 4 O (g ) )
2 5 °C
n 6 (m ol C O 2 )
n 7 (m ol H 2 O )
n 3 (m ol C 2 H 4)
4 5 0 °C
n 4 (m ol O 2 )
25% conversion 0 .5 0 0 m o l C 2 H 4 c o n s u m e d n 3 1.5 0 m o l C 2 H 4
70% yield n 5
0 .5 0 0 m o l C 2 H 4 c o n s u m e d
0 .7 0 0 m o l C 2 H 4 O
1 m ol C 2 H 4
0 .3 5 0 m o l C 2 H 4 O
C balance on reactor: (2)(2) = (2)(1.50) + (2)(0.350) = n6 ⇒ n6 = 0.300 mol CO2
Water formed: n 7
0 .3 0 0 m o l C O 2
1 m ol H 2 O
1 m ol C O 2
0 .3 0 0 m o l H 2 O
O balance on reactor: (2)(1) = 2n4 + 0.350 + (2)(0.300) + 0.300 ⇒ n4 = 0.375 mol O2
Overall C balance: 2n1 = n6 + 2n5 = 0.300 + (2)(0.350) ⇒ n1 = 0.500 mol C2H4
Overall O balance: 2n2 = 2n6 + n7 + n5 = (2)(0.300) + (0.300) + (0.350) ⇒ n2 = 0.625 mol O2
F e e d s tre a m : 4 4 .4 % C 2 H 4 , 5 5 .6 % O 2
R e a c to r in le t: 6 6 .7 % C 2 H 4 , 3 3 .3 % O 2
R e c y c le s tre a m : 8 0 .0 % C 2 H 4 , 2 0 .0 % O 2
R e a c to r o u tle t: 5 3 .1 % C 2 H 4 , 1 3 .3 % O 2 , 1 2 .4 % C 2 H 4 O , 1 0 .6 % C O 2 , 1 0 .6 % H 2 O
Mass of ethylene oxide:
0 .3 5 0 m o l C 2 H 4 O
4 4 .0 5 g
1 kg
1 m ol
10
3
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g
0 .0 1 5 4 k g
lOMoARcPSD|32610952
(b)
o
References for enthalpy calculations: C (s), H2 (g), O2 (g) at 25 C
Energy balance on process: Q H n i H i n i H i 2 4 8 k J
out
in
(c)
S c a le to 1 5 0 0 k g C 2 H 4 O d a y :
C 2 H 4 O p r o d u c ti o n f o r in itia l b a s i s ( 0 .3 5 0 m o l) (
S c a le f a c t o r
1500 kg day
9 .7 3 1 0
4
4 4 .0 5 k g
10
3
day
m ol
) 0 .0 1 5 4 2 k g C 2 H 4 O
1
0 .0 1 5 4 2 k g
-3
4
-1
Fresh feed rate = (34.025x10 kg)(9.73x10 day ) = 3310 kg/day (44.4% C2H4, 55.6% O2
9.15
(a)
Basis:
1 2 0 0 lb m C 9 H 1 2
1 lb - m o le
h
1 2 0 lb m
1 0 .0 lb - m o le s c u m e n e p ro d u c e d h
Overall process:
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n 1 ( lb - m o le s /h )
n 3 ( lb - m o le s C 3 H 6 /h )
0 .7 5 C 3H 6
n 4 ( lb - m o le s C 4 H1 0 /h )
0 .2 5 C 4H 1 0
n 2 ( lb - m o le s C 6 H 6 /h )
1 0 .0 lb -m o le s C9 H1 2 /h
Reactor:
16.67 lb-moles/h @ 77oF
0.75 C3H6
0.25 C4H10
40.0 lb-moles C6H6/h
10.0 lb-moles C9H12/h
2.50 lb-moles C3H6/h
4.17 lb-moles C4H10/h
30.0 lb-moles C6H6/h
400oF
Overhead from T1
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46.7 lb-moles/h
21.4% C9H12
5.4% C3H6
8.9% C4H10
64.3% C6H6
lOMoARcPSD|32610952
(b)
Heat exchanger:
R e a c t o r e ffl u e n t a t 4 0 0 °F
2 0 0 °F
1 0 .0 l b -m o l e s C 9 H 1 2 / h
2 .5 0 l b -m o l e s C 3 H 6 / h
4 .1 7 l b -m o l e s C 4 H 1 0 / h
3 0 .0 l b -m o l e s C 6 H 6 / h
4 0 .0 l b -m o l e s C 6H 6 / h
T (°F )
7 7 °F
Energy balance:
(Refer to flow chart of Part b: T 323 F )
References: C3H6(l), C4H10 (l), C6H6 (l), C9H12 (l) at 770C
H in
n in
S u b s ta n c e
( lb - m o le / h )
n o u t
H o u t
1 2 .0
( B tu / lb - m o le )
0
( lb - m o le / h )
C 3H 6
2 .5 0
( B tu / lb - m o le )
7750
C 4 H 10
4 .1 7
0
4 .1 7
10330
C 6H 6
4 0 .0
8650
3 0 .0
11350
C 9 H 12
1 0 .0
15530
Energy balance on reactor:
Q H
n C H
9
H r
o
12
vC H
9
12
n H n H
i
out
i
i
i
in
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lOMoARcPSD|32610952
9.16
(a)
C H 3O H H C H O H 2 ,
H2
1
2
O 2 H 2O
CH OH
3
O 2, N
reactor
n f (mol/h) at 145°C, 1 atm
product gas, 600°C
0.42 mol CH OH/mol
3
n 1 (mol CH OH/h)
3
0.58 mol air/mol
n 2 (mol O /h)
2
waste
n 3 (mol N /h)
2
heat
n 4 (mol HCHO/h)
boiler
0.21 mol O /mol
air
2
reactor
0.79 mol N /mol
air
2
product gas
145°C
separation
units
0.37 kg HCHO/h
n 5 (mol H /h)
2
n s mol H O(
)/h
v
2
(b)
0.63 kg H O/h
2
n 6 (mol H O/h)
2
saturated at 145°C
2
H2
m b (kg H O(
)/h)
v
2
m b (kg H O(
)/h)
v
2
30°C
sat'd at 3.1 bars
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 % CH3OH (v), is saturated at Tm (°C) and 1 atm.
A n to in e e q u a tio n
p m 3 1 9 .2 m m H g T m 4 4 .1 C
(c)
Moles HCHO formed:
36 10
6
k g s o lu tio n
350 days
0 .3 7 k g H C H O
1 km ol
1 day
1 k g s o lu tio n
3 0 .0 3 k g H C H O
24 h
5 2 .8 0
km ol H C H O
h
but if all the HCHO is recovered, then this equals n 4 , or n 4 5 2 .8 0 k m o l H C H O h
70% conversion:
5 2 .8 0 k m o l H C H O
1 k m o l C H 3 O H re a c t
1 k m o l C H 3 O H fe d
1 k m o l fe e d g a s
h
1 k m o l H C H O fo rm e d
0 .7 0 k m o l C H 3 O H re a c t
0 .4 2 k m o l C H 3 O H
n f 1 7 9 .5 9 k m o l h
Methanol unreacted:
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n f
lOMoARcPSD|32610952
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 H 2 content of the product gas (5%), and the energy
balance. The solution is tedious but straightforward.
H balance:
n s n 5 n 6 5 2 .8 0
(1)
n s 2 n 2 n 6 4 3 .7 5
(2)
O balance:
n 5
H 2 c o n te n t:
2 2 .6 3 n 2 8 2 .2 9 5 2 .8 9 n 5 n 6
0 .0 5 1 9 n 5 n 2 n 6 1 5 7 .7 2
(3)
References: C (s), H2 (g), O2 (g), N2 (g) at 25oC
or Table B.8 for O 2 , N 2 a n d H 2
H in
n in
s u b s ta n c e
CH 3OH
H o u t
n o u t
km ol / h
kJ / km ol
km ol / h
kJ / km ol
7 5 .4 3
195220
2 2 .6 3
163200
O2
2 1.8 8
3620
n2
18410
N 2
8 2 .2 9
3510
8 2 .2 9
17390
H 2O
ns
237740
n6
220920
HCHO
5 2 .8 0
88800
H 2
n5
16810
Energy Balance
H
n H n H 0 1 8 4 1 0 n 1 6 8 1 0 n 2 2 0 9 2 0 n 2 3 7 7 0 4 n 7 .4 0 6 1 0
i
out
i
i
i
2
5
6
s
6
(4)
in
We now have four equations in four unknowns. Solve using E-Z Solve.
n 2 2 .2 6 k m o l O 2 h , n 5 1 3 .5 8 k m o l H 2 h , n 6 9 8 .0 0 k m o l H 2 O h
Summarizing, the product gas component flow rates are 22.63 kmol CH 3OH/h, 2.26 kmol O2/h, 82.29 kmol
N2/h, 52.80 kmol HCHO/h, 13.58 kmol H2/h, and 98.02 kmol H2O/h
2 7 2 k m o l h p ro d u c t g a s
8 % C H 3 O H , 0 .8 % O 2 , 3 0 % N 2 , 1 9 % H C H O , 5 % H 2 , 3 7 % H 2 O
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lOMoARcPSD|32610952
(d)
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 600C), and for all but two of them at the
boiler outlet (at 145C), we will use the same reference states for the boiler calculation.
Reference States: C (s), H2 (g), O2 (g), N2(g) at 25oC for reactor gas
H2O (l) at triple point for boiler water
n in
Hˆ i n
nout
Hˆ o u t
k m o l/ h
k J /k m o l
m ol
k J /m o l
C H 3O H
2 2 .6 3
163200
2 2 .6 3
195220
O2
2 .2 6
18410
2 .2 6
3620
N2
8 2 .2 9
17390
8 2 .2 9
3510
S u b s ta n c e
9.17
H 2O
9 8 .0 2
220920
9 8 .0 2
237730
HCHO
5 2 .8 0
88800
5 2 .8 0
111350
H2
1 3 .5 8
16810
1 3 .5 8
3550
H 2O
mb
1 2 5 .7
mb
2 7 2 6 .1
( k g /h )
(k J /k g )
(k g /h )
( k J /k g )
Basis:
3497 mol/h
n 1 (mol CH OH
/h)
3
0.333 mol CO/mol
n 2 (mol CO/h)
0.667 mol H /mol
2
n 3 (mol H /h)
2
25°C, 5 at m
127°C, 5 at m
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).
CO reacted:
3497 0 . 333 mol CO feed
f mol react
1166 f mol CO react
mol feed
C H 3 O H p ro d u c e d : n 1
1 1 6 6 f m o l C O re a c t
1 m ol C H 3O H
1 m ol C O
CO remaining:
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1166 f m ol C H 3O H h
lOMoARcPSD|32610952
Reference states: CO(g), H2 (g), CH3OH (g) at 25C
Energy balance: Q H H ro n i H i n i H i
out
9.18
(a)
in
CH4 (g), 4S (g), → CS2 (g) + 2H2S (g), ∆ Ĥr (700°C) = -274 kJ/mol
1 m o l at 7 00 °C
0 .2 0 m o l C H /m o l
R e a c to r
4
0 .8 0 m o l S /m o l
P ro d u c t g a s a t 8 0 0 ° C
n 1 (m o l C S 2 )
) ol H S)
n (m
2
Q = –41 kJ
2
n 3 (m o l C H 4 )
n 4 (m o l S (v ))
Let f fractional conversion of CH 4 (which also equals fractional conversion of S, since the
species are fed in stoichiometric proportion).
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lOMoARcPSD|32610952
References: CH4 (g), S (g), CS2 (g), H2S (g) at 700C (temperature at which H r is known)
H out C pi 800 700
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lOMoARcPSD|32610952
(b)
0 .0 4 m o l C H 4
0 .1 6 m o l S ( l )
0 .1 6 m o l C S 2
0 .3 2 m o l H 2 S
Q (k J)
2 0 0 C
p r e h e a te r
0 .2 0 m o l C H 4
0 .2 0 m o l C H 4
0 .8 0 m o l S ( l )
0 .8 0 m o l S ( g )
0 .2 0 m o l C H 4
0 .8 0 m o l S ( l )
1 5 0 °C
T (°C )
7 0 0 °C
0 .0 4 m o l C H 4
0 .1 6 m o l S ( g )
0 .1 6 m o l C S 2
0 .3 2 m o l H 2 S
8 0 0 °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.
(c)
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lOMoARcPSD|32610952
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.
9.19
B a s is : 1 m o l C 2 H 6 fe d to re a c to r
1 m ol C H
2
6
1 2 7 3 K , P at m
n ( m o ls ) @
n C 2H 6
T ( K ) , P at m
( m o l C 2H 6)
n C 2H 4 ( m o l C 2H 4)
n H 2 ( m o l H 2)
(a)
C2H 6 C2H 4 H 2,K p
xC H xH
2
4
xC H
2
2
P 7 .2 8 1 0
6
e x p [ 1 7 , 0 0 0 / T ( K )]
6
(b)
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(1)
lOMoARcPSD|32610952
(c)
(d)
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lOMoARcPSD|32610952
(e)
C **PROGRAM FOR PROBLEM 9-35
WRITE (5, 1)
FORMAT ('1', 20X, 'SOLUTION TO PROBLEM 9-35'//)
1
T 1200.0
TLAST 0.0
PSIL 0.0
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.
CONTINUE
10
IF (T.GE.0.0) GO TO 45
40
WRITE (3, 2)
FORMAT (1X, 'T LESS THAN ZERO -- ERROR')
STOP
C **APPLY REGULA-FALSI
2
45
50
3
93
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
CONTINUE
IF (I.EQ.20) WRITE (3, 3)
FORMAT ('0', 'REGULA-FALSI DID NOT CONVERGE IN 20 ITERATIONS')
STOP
END
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lOMoARcPSD|32610952
1
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
FORMAT (6X, 'T ', F6.2, 4X, 'PSI ', E11,4)
RETURN
END
OUTPUT: SOLUTION TO PROBLEM 9-35
T 1 2 0 0 .0 0
P S I 0 .8 2 2 6 E 0 0
T 1 1 5 0 .0 0
P S I 0 .7 0 4 8 E 0 0
T 1 1 0 0 .0 0
P S I 0 .5 5 5 1 E 0 0
T 1 0 5 0 .0 0
P S I 0 .3 6 9 6 E 0 0
T 1 0 0 0 .0 0
P S I 0 .1 6 1 9 E 0 0
T 9 5 0 .0 0
P S I 0 .3 9 5 0 E 0 1
T 9 5 9 .8 0
P S I 0 .1 8 2 4 E 0 2
T 9 6 0 .2 5
P S I 0 .7 6 7 1 E 0 4
T 9 6 0 .2 7
P S I 0 .3 2 7 8 E 0 5
Solution: T 9 6 0 .3 K , f 0 .3 6 0 m o l C 2 H 6 re a c te d m o l fe d
9.20
2 C H 4 C 2 H 2 3H 2
C 2 H 2 2 C (s ) + H 2
Basis: 10 mol CH4 (g) fed/s
1 0 .0 m o l C H 4 ( g ) /s
1500
o
n 1 (m o l C H 4 / s )
C
n 2 (m o l C 2 H 2 / s )
n 3 (m o l H 2 / s )
n 4 (m o l C (s )/s )
o
1500 C
975 kW
(a)
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lOMoARcPSD|32610952
Yield of acetylene
2 .5 0 m o l C 2 H 2 s
6 .0 0 m o l C H 4 c o n s u m e d s
0 .4 1 7 m o l C 2 H 2 m o l C H 4 c o n s u m e d
(b)
If no side reaction,
n 1 1 0 .0 ( 1 0 .6 0 0 ) 4 .0 0 m o l C H 4 / s
n 3 0 n 2 3 .0 0 m o l C 2 H 2 / s , n 4 9 .0 0 m o l H 2 / s
Yield of acetylene
3 .0 0 m o l C 2 H 2 s
0 .5 0 0 m o l C 2 H 2 m o l C H 4 c o n s u m e d
6 .0 0 m o l C H 4 c o n s u m e d s
Reactor Efficiency
0 .4 1 7
0 .8 3 4
0 .5 0 0
9.21
B a sis : 1 m o l C 3 H 8 fe d
H e a t in g ga s
4 .9 4 m 3 a t 1 4 0 0 ° C , 1 a t m
n g ( m o l) , 9 0 0 ° C
n g ( m o l)
1 m o l C H (g )
3
8
a
P r o d u c t ga s , 8 0 0 ° C
n 1 (m o l C H ) = 0
3
6 m ol H O ( g )
n 2 (m o l H O )
125°C
n 3 (m o l C O )
2
8
2
n 4 (m o l C O )
2
n 5 (m o l H )
2
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lOMoARcPSD|32610952
4 .9 4 m
ng
3
10
3
1 m
L
3
273 K
1 m ol
1673 K
2 2 .4 L
3 5 .9 9 m o l h e a tin g g a s
Let 1 and 2 be the extents of the two reactions.
n1 0
n1 1 1 1 1 m ol
n4 2
1 1
1 1
n 2 6 3 1 2 n 2 3 2
n5 7 1 2 n5 7 2
1 1
n 3 3 1 2 n 3 3 2
o
References: C (s), H2 (g), O2 (g) at 25 C
S u b s ta n c e
n in
H in
n out
H o u t
m ol
kJ / m ol
m ol
kJ / m ol
C 3H 8
1
9 5 .3 9
0
H 2O
6
2 3 8 .4 3
32
2 1 2 .7 8
CO
32
8 6 .3 9
CO 2
2
3 5 6 .1 5
H 2
7 2
2 2 .8 5
h e a tin g g a s
3 5 .9 9
2 0 0 .0 0
3 5 .9 9
0
Energy Balance:
n H n H 0
i
out
i
i
i
2
2 .0 0 m o l n 2 1 m o l H 2 O , n 3 1 m o l C O ,
in
n 4 1 m o l C O 2 , n 5 9 m o l H 2 7 .7 m o l % H 2 O , 7 .7 % C O , 1 5 .4 % C O 2 , 6 9 .2 % H 2
9.22
M o le s o f H 2 O ( 3.1 9 2 1 0
6
5
2 .9 4 1 1 0 ) g H 2 O (
1 m ol
1 8 .0 2 g
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) 1.6 1 1 0
5
m ol H 2 O
lOMoARcPSD|32610952
H ( 3 0 0 0 m o l H 2 S O 4 )(-8 8 4 .7 k J / m o l H 2 S O 4 ) = -2 .6 5 1 0
9.23
8 5 0 0 k to n n e C l 2
10
yr
10
3
J
1 kJ
3
to n n e
10
1 k to n n e
2 .7 7 8 1 0
7
1 J
kW h
3
kg
1 to n n e
1 MW h
10
3
kW h
10
3
g
1 m ol C l 2
2 2 2 .4 4 k J
1 kg
7 0 .9 1 g C l 2
0 .5 m o l C l 2
1. 4 8 1 0
7
M W h / yr
(a)
(b)
9.25
kJ
(a)
(b)
9.24
6
(a)
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lOMoARcPSD|32610952
(b)
(c)
o
The enthalpy change when 1 kg of the natural gas at 25 C is burned completely with oxygen at
o
o
25 C and the products CO2(g) and H2O(v) are brought back to 25 C.
9.26
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lOMoARcPSD|32610952
9.27
(a)
(b)
The reaction for which we determined U co is
1 lb m o il + a O 2 ( g ) b C O 2 ( g ) + c H 2 O (v )
(1 )
The higher heating value is H r for the reaction
1 lb m oil + a O 2 ( g) b C O 2 ( g) + c H 2 O (l)
(2)
Eq. (9.1-5) on p. 441 H co1 U co1 R T ( b c a )
Eq. (9.6-1) on p. 462 H co2 H co1 c H v ( H 2 O , 7 7 F )
(HHV )
( LH V )
To calculate the higher heating value, we therefore need
a lb - m o le s o f O 2 th a t r e a c t w ith 1 l b m f u e l o il
b lb - m o le s o f C O 2 f o r m e d w h e n 1 lb m f u e l o il is b u r n e d
c lb - m o le s o f H 2 O f o r m e d w h e n 1 lb m f u e l o il is b u r n e d
9.28
(a)
% excess air
( 4 .4 8 2 1.5 ) m o l O 2
1 0 0 % 2 0 0 % e x c e s s a ir
1.5 m o l O 2
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lOMoARcPSD|32610952
(An atomic O balance 9 .9 6 m o l O 9 .9 6 m o l O , so that the results are consistent.)
(b)
Energy balance on vaporizer:
Extent of reaction: ( n C H 3 O H ) o u t ( n C H 3 O H ) in C H 3 O H 0 1 m o l 1 m o l
Energy balance on reactor: Q 2 H co n i H i n i H i
out
9.29
CH 4 2 O 2 CO 2 2 H 2 O , C 2 H 6
7
2
in
O 2 2 CO 2 3 H 2 O
Basis: 100 mol stack gas. Assume ideal gas behavior.
n 1 ( m o l C H4 )
n2 ( m o l C 2 H6 )
3
V f ( m a t 2 5 ° C , 1 a tm )
1 0 0 m o l a t 8 0 0 ° C , 1 a tm
0 .0 5 3 2 m o l C O 2 / m o l
n3 ( m o l O2 )
0 .0 1 6 0 m o l C O / m o l
3 .7 6 n 3 ( m o l N 2 )
0 .0 7 3 2 m o l O 2 / m o l
2 0 0 ° C , 1 a tm
0 .1 2 2 4 m o l H 2 O /m o l
0 .7 3 5 2 m o l N 2 / m o l
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lOMoARcPSD|32610952
(a)
T h e o re tic a l O 2
3 .7 2 m o l C H 4
2 m ol O 2
1 .6 0 m o l C 2 H 6
1 m ol C H 4
3 .5 m o l O 2
1 m ol C 2 H 6
(b)
Energy balance:
Q H
2764 kJ
n H n H 0 .1 3 0 m fu e l 2 .1 3 1 0
i
out
i
i
i
4
3
in
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kJ m
3
fu e l
1 3 .0 4 m o l O 2
lOMoARcPSD|32610952
9.30
50,000 lb coal/h
m
St ack gas at 600°F, 1 atm
3039 lb-moles C/h
n 2 (lb-moles CO /h)2
2327 lb-moles H/h
n 3 (lb-moles H O/h)
2
57.7 lb-moles S/h
n 4 (lb-moles SO /h)2
189 lb-moles H O/h2
n 5 (lb-moles O /h)2
5900 lb ash/h
m
n 6 (lb-moles N /h)2
77°F, 1 at m
m 7 (lb mfly ash/h)
n 1 (lb-moles air/h)
m 8 (lb mslag/h) at 600°F
0.210 O
2
0.287 lb C/lb
m
m
0.790 N
2
0.016 lb S/lb
m
m
77°F, 1 at m
(a)
(assume)
(assume)
(assume)
0.697 lb ash/lb
m
m
Feed rate of air:
Air fed: n 1
1 .5 3 0 3 9 lb - m o le s O 2 fe d
1 m o le a ir
h
0 .2 1 0 m o le O 2
M CO
n 2 2 9 7 8 lb - m o le s C O 2
2 1 7 1 0 lb - m o le s a ir h
M H
n 3 1 3 5 2 .5 lb - m o le s H 2 O h
M SO
n 4 5 6 .4 lb - m o le s S O 2
4 4 .0 1
2
1.3 1 1 0
h
h
2O
2
5
lb m C O 2
h
1 8 .0 2
2 .4 4 1 0
4
lb m H 2 O h
6 4 .2
3 6 2 0 lb m S O 2
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h
lOMoARcPSD|32610952
Summary of component mass flow rates
(b)
References: Coal components, air at 77F n i H i 0
in
(c)
Q
HHV
1 . 62 10
4
1 . 80 10
4
Btu lb m
0 . 901
Btu lb m
Some of the heat of combustion goes to vaporize water and heat the stack gas.
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lOMoARcPSD|32610952
9.31
(a)
B a sis: 4 5 0 k m o l C H 4 fe d h
A ir f e d : n a
CH 4 2 O 2 CO 2 2 H 2 O
450 km ol C H 4
2 k m o l O 2 re q ' d
1 .2 k m o l O 2 f e d
1 k m o l a ir
h
1 km ol C H 4
1 k m o l O 2 re q ' d
0 .2 1 k m o l O 2
5 1 4 3 k m o l a ir h
4 5 0 k m o l / h C H 4 re a c t n 1 4 5 0 k m o l C O 2 h , n 2 9 0 0 k m o l H 2 O h
Energy balance on furnace (combustion side only)
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lOMoARcPSD|32610952
Extent of reaction: n C H 4 4 5 0 k m o l / h
H p n 2 ( H v )
=
o
o
H 2 O (2 5 C )
180 km ol H 2 O
h
= 5 .6 3 1 0
7
n s ta c k g a s ( C p ) s ta c k g a s ( T s ta c k g a s 2 5 C )
3
m ol
4 4 .0 1 k J
1 km ol
m ol
10
5590 km ol
h
10
3
m ol
0 .0 3 1 5 k J
1 km ol
o
(3 0 0 - 2 5 ) C
o
m ol C
kJ / h
(b)
n a ( m o l a ir /h ) a t
T a (° C )
S t a c k ga s
n 1 ( m o l C O 2/h )
45 km ol C H
4
/h
fu rn ace
25°C
a ir
n 2 ( m ol H O
/h )
2
n 3 ( m o l O 2/h )
p reh eat er
n 4 ( m o l N 2/h )
300°C
m w ( k g H 2O /h )
n 1 ( m o l C O 2/h )
n 2 ( m ol H O
/h )
2
n 3 ( m o l O 2/h )
n 4 ( m o l N 2/h )
150°C
m w ( k g H 2O /h )
L iq u id , 2 5 ° C
v ap o r, 1 7 b ars
n a ( m o l a ir /h ) a t 2 5 ° C
250°C
0 .2 1 O
0 .7 9 N
2
2
E.B. on overall process: The material balances and the energy balance are identical to those of part (a),
except that the stack gas exits at 150oC instead of 300oC.
S u b s ta n c e
H in
n in
n o u t H o u t
(k m o l / h )
( k J / k m o l)
(k J / h )
CH 4
450
0
A ir
S ta c k g a s
5143
0
H p
H 2O
m w ( k g / h )
105 kJ / kg
m w ( k g / h )
2914 kJ / kg
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lOMoARcPSD|32610952
H p n 2 ( H v )
=
o
180 km ol H 2 O
h
= 2 .9 9 1 0
(c)
o
H 2 O (2 5 C )
n s ta c k g a s ( C p ) s ta c k g a s ( T s ta c k g a s 2 5 C )
10
3
m ol
4 4 .0 1 k J
1 km ol
7
m ol
5590 km ol
h
10
3
m ol
1 km ol
0 .0 3 1 5 k J
o
(1 5 0 - 2 5 ) C
o
m ol C
kJ / h
The energy balance on the furnace includes the term n in H in . If the air is preheated and
the stack gas temperature remains the same, this term and hence Q become more negative,
meaning that more heat is transferred to the boiler water and more steam is produced. The
stack gas is a logical heating medium since it is available at a high temperature and costs nothing.
9.32
CO
Basis: 1 mol CO burned.
1
2
o
O 2 C O 2 , H c 2 8 2 .9 9 k J m o l
1 m ol C O
1 m ol C O2
n 0 m o l O2
( n 0 – 0 .5 ) m o l O 2
3 .7 6 n 0 m o l N 2
3 .7 6 n 0 m o l N 2
25°C
1400°C
(a)
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lOMoARcPSD|32610952
Excess oxygen:
1.0 9 4 m o l fe d 0 .5 0 0 m o l re q d .
1 0 0 % 1 1 9 % e x c e ss o x y g e n
0 .5 0 0 m o l
9.33
(b)
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 ).
(a)
Basis : 1 mol C5H12 (l)
C 5 H 1 2 (l) 8 O 2 (g ) 5 C O 2 (g ) 6 H 2 O (v ),
1 mol C5H12 (l)
H c 3 5 0 9 .5 k J / m o l
o
n2(mol CO2)
n3 (mol H2O (v))
n4 (mol O2)
Tad(oC)
n0 (mol O2) , 75C
30% excess
T h e o re tic a l o x y g e n
1 m o l C 5 H 12
8 m ol O 2
1 m o l C 5 H 12
8 m ol O 2
3 0 % e x c e ss n 0 1.3 8 1 0 .4 m o l O 2
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s u b s ta n c e
n in
H in
n out
H o u t
kJ m ol
m ol
kJ m ol
m ol
C 5 H 12
1.0 0
0
O2
1 0 .4 0
H 1
2 .4 0
H 2
CO 2
5 .0 0
H 3
H 2O
6 .0 0
H 4
S u b s titu tin g ( C p ) i fro m T a b le B .2 :
kJ
2
3
4
5
8
12
H 2 ( 0 .0 2 9 1 T a d 0 .5 7 9 1 0
T a d 0 .2 0 2 5 1 0
T a d 0 .3 2 7 8 1 0
T a d 0 .7 3 1 1 )
m ol
kJ
2
3
4
5
8
12
H 3 ( 0 .0 3 6 1 1 T a d 2 .1 1 6 5 1 0
T a d 0 .9 6 2 3 1 0
T a d 1.8 6 6 1 0
T a d 0 .9 1 5 8 )
m ol
2
3
4
5
8
12
H 4 4 4 .0 1 ( 0 .0 3 3 4 6 T a d 0 .3 4 4 0 1 0
T a d 0 .2 5 3 5 1 0
T a d 0 .8 9 8 3 1 0
T a d 0 .8 3 8 )
2
3
4
5
8
12
H 4 4 3.1 7 ( 0 .0 3 3 4 6 T a d 0 .3 4 4 0 1 0
T a d 0 .2 5 3 5 1 0
T a d 0 .8 9 8 3 1 0
Tad )
kJ
m ol
kJ
m ol
( 1 m o l C 5 H 1 2 )( 3 5 0 9 .5 k J / m o l ) ( 2 .4 0 ) H 2 ( 5 .0 0 ) H 3 ( 6 .0 0 ) H 4 ( 1 0 .4 0 )( H 1 ) 0
S u b s titu te fo r H 1 th ro u g h H 4
H ( 0 .4 5 1 2 T a d 1 4 .0 3 6 1 0
5
Tad
2
3 .7 7 7 1 0
f ( T a d ) 3 2 7 2 .2 0 0 .4 5 1 2 T a d 1 4 .0 3 6 1 0
C heck :
3 2 7 2 .2 0
4 .7 2 7 1 0
12
6 .9 2 2 1 0
5
8
Tad
Tad
2
3
4 .7 2 7 1 0
3 .7 7 7 1 0
14
S o lv in g f o r T a d u s in g E - Z S o lv e
o
Tad 4 4 1 4 C
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8
12
Tad
4
T a d ) 3 2 7 2 .2 0 k J / m o l = 0
3
4 .7 2 7 1 0
12
Tad
4
0
lOMoARcPSD|32610952
(b)
Terms
1
2
3
Tad
7252
3481
3938
% Error
64.3%
–21.1%
–10.8%
(c)
T
(d)
f(T )
f '( T )
Tnew
7252
6 .0 5 E + 0 3
3 .7 4
5634
5634
1 .7 3 E + 0 3
1 .8 2
4680
4680
3 .1 0 E + 0 2
1 .2 2
4426
4426
1 .4 1 E + 0 1
1 .1 1
4414
4414
3 .1 1 E -0 2
1 .1 1
4414
The polynomial formulas are only applicable for T 1500C
9.34
F u e l fe e d ra te :
5 .5 0 L
273 K
1 .1 a tm
m ol
s
298 K
1 .0 a tm
2 2 .4 L (S T P )
0 .2 4 7 m o l C H 4 / s
T h e o r e tic a l O 2 2 0 .2 4 7 0 .4 9 4 m o l O 2 / s
2 5 % e x c e s s a ir n 2 1.2 5 ( 0 .4 9 4 ) 0 .6 1 7 5 m o l O 2 / s
,
3 .7 6 0 .6 1 7 5 2 .3 2 m o l N 2 / s
C o m p le te c o m b u s tio n n 1 = 0 .2 4 7 m o l / s , n 4 0 .2 4 7 m o l C O 2 / s , n 5 0 .4 9 4 m o l H 2 O / s
n 3 0 .6 1 7 5 m o l O 2 fe d / s 0 .4 9 4 m o l c o n s u m e d / s
0 .1 2 4 m o l O 2 / s
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lOMoARcPSD|32610952
o
R e f e r e n c e s: C H 4 , O 2 , N 2 , C O 2 , H 2 O ( l) a t 2 5 C
n in
Hˆ in
nout
Hˆ o u t
( m o l/s )
( k J /m o l)
( m o l/s )
( k J /m o l)
CH 4
0 .2 4 7
0
O2
0 .6 1 7 5
Hˆ 1
0 .1 2 4
Hˆ 3
N2
2 .3 2
Hˆ 2
2 .3 2
Hˆ 4
CO2
0 .2 4 7
Hˆ 5
H 2O
0 .4 9 7
Hˆ 6
S u b s ta n c e
(a)
E n e rg y B a la n c e
o
H ( H c ) C H
4
n
out H out
T a b le B .2 fo r C p i ( T ), ( H v ) H
2
O
n H
in
in
0
4 4 .0 1 k J / m o l
0 .2 4 7 ( 8 9 0 . 3 6 ) 0 .4 9 4 ( 4 4 .0 1 ) 0 .0 9 6 3 ( T 2 5 ) 1.7 4 1 0
1.6 1 1 0
12
(T
4
5
(T
2
2
2 5 ) 0 .3 0 5 1 0
8
(T
3
3
25 )
4
2 5 ) 0 .6 1 7 5 ( 3 .7 8 ) 2 .3 2 ( 3 .6 6 ) 0
2 1 1 .4 0 .0 9 6 3 T a d 1 .7 4 1 0
5
Tad
2
0 .3 0 5 1 0
8
Tad
3
1 .6 1 1 0
12
Tad
4
0
o
T 1832 C
(b)
In p ro d u c t g a s ,
o
T 1 8 3 2 C , P 1 .0 5 7 6 0 7 9 8 m m H g
yH O
2
0 .4 9 4 m o l/s
( 0 .1 2 4 2 .3 2 0 .2 4 7 0 .4 9 4 ) m o l/s
0 .1 5 5 m o l H 2 O /m o l
T a b le B .3
*
*
R a o u lt's la w : y H O P p H O ( T d p ) p H O ( 0 .1 5 5 )( 7 9 8 ) 1 2 4 m m H g
2
2
2
o
o
o
D e g re e s o f s u p e rh e a t = 1 8 3 2 C 5 6 C = 1 7 7 6 C s u p e rh e a t
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T
dp
56 C
lOMoARcPSD|32610952
9.35
(b)
1 mol natural gas
yCH
yC H
2
yC H
3
( m o l C H 4 / m o l)
nCO
6
( m o l C 2 H 6 / m o l)
n H O (m o l H 2 O )
8
( m o l C 3 H 8 / m o l)
nN
2
(m o l N 2 )
nO
2
m ol O 2 )
4
2
(m o l C O 2 )
2
Humid air
na (mol air)
ywo (mol H20(v)/mol)
(1-ywo) (mol dry air/mol)
0.21 mol O2/mol DA
0.79 mol N2/mol DA
Basis: 1 g-mole natural gas
C H 4 (g ) 2 O 2 (g ) C O 2 (g ) H 2 O (v )
C 2 H 6 (g )
7
2
O 2 (g ) 2 C O 2 (g ) 3 H 2 O (v )
C 3 H 8 (g ) 5O 2 (g ) 3C O 2 (g ) 4 H 2 O (v )
T h e o re tic a l o x y g e n :
2 m ol O 2
y CH
(m o l C H 4 )
4
1 m ol C H 4
( 2 yCH
4
3 .5 y C H
3 .5 m o l O 2
2
6
+ 5 yC H )
n CH
4
(m o l C H 4 )
3
yC H
2
6
(m o l C 2 H 6 )
1 m ol C 2 H 6
5 m ol O 2
yC H
3
(m o l C 3 H 8 )
8
1 m ol C 3 H 8
8
C O 2 in p ro d u c t g a s :
n CO
2
1 m ol C O 2
1 m ol C H 4
(n CH
4
2 m ol C O 2
nC H
2
6
(m o l C 2 H 6 )
1 m ol C 2 H 6
2 n C H 3n C H ) m o l C O 2
2
6
3
8
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3 m ol C O 2
1 m ol C 3 H 8
nC H
3
8
(m o l C 3 H 8 )
lOMoARcPSD|32610952
H 2 O in p ro d u c t g a s :
nH O
2
1 m ol H 2 O
n CH
(m o l C H 4 )
4
1 m ol C H 4
[2 n C H
4
3 m ol H 2 O
nC H
2
6
(m o l C 2 H 6 )
1 m ol C 2 H 6
4 m ol O 2
nC H
3
8
(m o l C 3 H 8 )
1 m ol C 3 H 8
3 n C H 4 n C H + n a (1 - y w o )] m o l H 2 O
2
6
3
O 2 in p ro d u c t g a s : n O
2
8
P xs
100
( 2n CH
4
3.5 n C H
2
6
+ 5 n C H ) m ol O 2
3
8
(c)
o
U s in g ( H f )C H
4
fro m T a b le B .1 a n d ( C p )C H
4
fro m T a b le B .2
2
5
8
12
2
3
4
H C H ( T ) [ 7 5 .7 2 + 3 .4 3 1 1 0 T + 2 .7 3 4 1 0 T 0 .1 2 2 1 0 T 2 .7 5 1 0
T ] kJ / m ol
4
S u b s ta n c e
n in
H in
n out
H o u t
m ol
kJ / m ol
H
m ol
kJ / m ol
H 2
H
H 7
H
CH 4
n1
C2H 6
n2
1
C 3H 8
n3
O2
n4
N2
n5
H 4
H
CO 2
n6
n9
H 2O
n 10
3
5
n7
n8
8
H 9
H
10
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lOMoARcPSD|32610952
7
H
6
(n )
i
out ( H i ) out
(n ) ( H )
i4
i
in
i
in
i1
2
3
4
H i a i b i T c i T d i T e i T
6
3
( n i ) in ( H i ) in
i1
6
( n i ) in H i ( T f )
i1
( n ) H ( T )
in
i
a
ei T
4
) out
i
i4
7
H
2
diT
3
i4
7
(n i ) out a i
i1
7
(n i ) out bi T
i1
i
in
f
i
a
7
(n i ) out d i T
3
i1
(n )
i
out e i T
4
i1
2
3T
3
4T
4
6
( n i ) in H i ( T f )
( n ) H ( T )
i
in
i
a
i4
i1
7
7
(n i ) out bi
2
i1
(n )
i
out ci
i1
7
a
i
3
(n i ) out a i
i1
in
i
i4
7
3
i
( n ) H ( T ) ( n ) H ( T )
0 1T 2 T
1
2
in
i
6
i1
0
(n i ) out c i T
( n ) H ( T )
i4
7
i4
3
w h e re
( n i ) in H i ( T f )
i1
7
H
6
3
(n i ) out (a i bi T c i T
7
(n i ) out d i
i1
4
(n )
i
out ei
i1
(d)
yCH4
yC2H6
yC3H8
Tf
Ta
Pxs
ywo
nO2i
nN2
nH2Oi
HCH4
HC2H6
HC3H8
HO2i
HN2i
HH2Oi
nCO2
nH2O
nO2
nN2
Tad
alph0
alph1
alph2
alph3
alph4
Delta H
Run 1
Run 2
Run 3
Run 4
Run 5
Run 6
0.75
0.21
0.04
40
150
25
0.0306
3.04
11.44
0.46
-74.3
-83.9
-102.7
3.6
3.8
-237.6
1.29
2.75
0.61
11.44
1743.1
-1052
0.4892
0.0001
-3.00E-08
3.00E-12
3.00E-07
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
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
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
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
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
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lOMoARcPSD|32610952
Species
a
CH4
C2H6
C3H8
O2
N2
H20
CO2
-75.72
-85.95
-105.6
-0.731
-0.728
-242.7
-394.4
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
9.36
A b s o r b e r o f f- g a s
n 1 4 ( m o l C H 4 /h )
n 1 3 ( m o l C (s ) /h )
25°C
B a s is :
n 8 ( m o l H 2 /h )
5 0 0 0 k g /h P r o d u c t g a s
n 1 2 ( m o l N 2 /h )
n 1 ( m o l/h )
0 .9 8 8 n 9 ( m o l C O /h )
0 .9 9 1 m o l C 2 H 2 ( g) /m o l
0 .9 5 0 n 6 ( m o l C H 4 /h )
0 .0 0 0 5 9 m o l H 2 O /m o l
0 .0 0 6 n 7 ( m o l C 2 H 2 /h )
0 .0 0 8 4 1 m o l C O2 / m o l
0 .9 1 7 n 1 ( m o l D M F /h )
P r eh e a ter s
C o n v e r te r
c o n v e r te r
F eed g a s, 65 0 °C
p rod u ct
C o n v e r te r
q u e n ch
T ad (°C )
p rod u ct
L e a n s o lv e n t
filte r
38°C
a bsorb er
n 1 4 ( m o l C H 4 /h )
0 .9 6 n 1 5 ( m o l O 2 /h )
n 6 ( m o l C H 4 /h )
n 6 ( m o l C H 4 /h )
0 .0 4 n 1 5 ( m o l N 2 /h )
n 7 ( m o l C 2 H 2 /h )
n 7 ( m o l C 2 H 2 /h )
n 8 ( m o l H 2 /h )
n 8 ( m o l H 2 /h )
n 9 ( m o l C O /h )
n 9 ( m o l C O /h )
n 1 0 ( m o l C O 2 /h )
n 1 0 ( m o l C O 2 /h )
n 1 1 ( m o l H 2 O /h )
n 1 1 ( m o l H 2 O /h )
n 1 2 ( m o l N 2 /h )
n 1 2 ( m o l N 2 /h )
n 1 5 ( m o l/h )
0 .9 6 m o l O2 /m o l
0 .0 4 m o l N2 /m o l
25°C
s tr ip p er
R ic h s o lv e n t
S tr ip p er o ff- g a s
n 2 ( m o l C O /h )
n 1 3 ( m o l C (s ) /h )
n 3 ( m o l C H 4 /h )
n 1 ( m o l/h )
0 .0 1 5 5 m o l C 2 H 2 / m o l
0 .0 0 6 3 m o l C O 2 / m o l
n 4 ( m o l H 2 O (v )/h )
n 5 ( m o l C O 2 /h )
0 .0 0 0 5 5 m o l C O / m o l
0 .0 0 0 5 5 m o l C H 4 / m o l
0 .0 5 9 6 m o l H 2 O /m o l
0 .9 1 7 m o l D M F / m o l
Average M.W. of product gas:
Molar flow rate of product gas: n 0
5000 kg
day
10
3
g
1 m ol
1 day
1 kg
2 6 .1 9 g
24 h
7955 m ol h
Material balances -- plan of attack (refer to flow chart):
Stripper balances: C 2 H 2 n 1 , C O n 2 , C H 4 n 3 , H 2 O n 4 , C O 2 n 5
Absorber balances: C H 4 n 6 , C 2 H 2 n 7 , C O n 9 , C O 2 n 1 0 , H 2 O n 1 1
C o n v e rte r O b a la n c e n 1 5 , c o n v e rte r N 2 b a la n c e n 1 2
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Absorber balances
Converter C balance:
(a)
Feed stream flow rates
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lOMoARcPSD|32610952
(b)
Gas feed to absorber
Absorber off-gas
Stripper off-gas
(c)
(d)
The theoretical maximum yield would be obtained if only the reaction
2 C H 4 C 2 H 2 3 H 2 occurred, the reaction went to completion, and all the C 2 H 2
formed were recovered in the product gas. This yield is (1 mol C2H2/2 mol CH4) = 0.500
mol C2H2/2 mol CH4.
The ratio of the actual yield to the theoretical yield is 0.154/0.500 = 0.308.
(e)
Methane preheater
Oxygen preheater
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(f)
n H 1.5 7 5 1 0
i
i
6
kJ h
in
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
out
1. 4 1 8 1 0
7
2
4
3
8
4
n i H i 1.0 0 0 1 0 3 9 4 3 T a 0 .6 2 5 1 T a 1.9 9 6 1 0 T a 2 .5 4 0 5 1 0 T a
Ta 2 7 3
Energy balance:
H
n H n H 0
i
out
i
i
i
in
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6
lOMoARcPSD|32610952
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lOMoARcPSD|32610952
CHAPTER TEN
10.1
(a)
Balance on H3PO4: Accumulation = input
Density of H3PO4: 1.8 3 4 g / m l .
Molecular weight of H3PO4: M 9 8 .0 0 g / m o l .
(b)
(c)
10.2
(a)
Air initially in tank:
Air in tank after 15 s:
Pf V
P0 V
N f RT
N 0 RT
N f N0
Pf
P0
0 .0 2 5 8 lb - m o le
1 1 4 .7 p s ia
1 4 .7 p s ia
Rate of addition:
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0 .2 0 1 3 lb - m o le
lOMoARcPSD|32610952
(b)
Balance on air in tank: Accumulation = input
; t = 0, N = 0.0258 lb-mole
(c)
Integrate balance:
C h e c k th e s o lu tio n i n tw o w a y s:
( 1 ) t = 0 , N = 0 .0 2 5 8 lb - m o le s a tis f ie s th e in itia l c o n d itio n
(2)
dN
0 .0 1 1 7 lb - m o le a ir / s r e p r o d u c e s th e m a s s b a la n c e
dt
(d)
O2 in tank = 0.30 lb-mole O2
10.3
(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).
A c c u m u la tio n = in p u t o u tp u t
dV
dt
t 0 , V 3 .0 0 1 0
V
(b)
m
3
540 m
3
1 h
h
6 0 m in
dV
3
w m
3
m in
t 0 c o rre s p o n d s to 8 :0 0 A M
t
3 .0 0 1 0
3
t
9 .0 0 d t V m 3 .0 0 1 0 9 .0 0 t d t t in m in u te s
3
3
w
w
0
0
Let w i tabulated value of w at t = 10 (i -1)
i = 1, 2, …, 25
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lOMoARcPSD|32610952
(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)
10
1
2
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
CONTINUE
FORMAT ('TIME (MIN) VOLUME (CUBIC METERS)')
FORMAT (F8.2, 7X, F6.0)
END
$DATA
11.4
11.9
Results:
TIME (MIN)
0.00
10.00
20.00
230.00 2683.
240.00 2674.
12.1
11.8
11.5
11.3
VOLUME (CUBIC METERS)
3000.
2974.
2944.
V tra p e z o id 2 6 7 4 m
3
; V S im p s o n 2 6 7 2 m 3 ;
2674 2672
1 0 0 % 0 .0 7 %
2672
Simpson’s rule is more accurate.
10.4
(a)
(b)
Balance on water: Accumulation = input – output (L/min).
(Balance volume directly since density is constant)
dV
2 0 .0 0 .2 0 0 V
dt
t 0, V 300
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lOMoARcPSD|32610952
(c)
dV
dt
0 2 0 0 0 .2 0 0 V s V s 1 0 0 L
The plot of V vs. t begins at (t=0, V=300). When t=0, the slope (dV/dt) is
20 .0 0 .200 ( 300 ) 40 .0 . As t increases, V decreases. dV / dt 20 .0 0 .200V
becomes less negative, approaches zero as t . The curve is therefore concave up.
(d)
10.5
(a)
A plot of D (log scale) vs. t (rectangular scale) yields a straight line through the points
( t 1 week, D 2 3 8 5 k g w e e k ) and ( t 6 weeks, D 7 5 5 k g w e e k ).
(b)
Inventory balance: Accumulation = –output
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lOMoARcPSD|32610952
10.6
(c)
t I 4957 kg
(a)
Balance on CO: Accumulation=-output
(b)
3
(c)
V 350 m
(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:
P u rg e th e la b o ra to ry lo n g e r th a n th e c a lc u la te d p u rg e tim e . U s e a C O d e te c to r
to m e a s u re th e re a l c o n c e n tra tio n o f C O in th e la b o ra to ry a n d m a k e s u r e it is
lo w e r th a n th e s a fe le v e l e v e ry w h e re in th e la b o r a to ry .
10.7
(a)
Total mass balance: Accumulation = input – output
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(b)
Sodium nitrate balance: Accumulation = - output
x = mass fraction of N a N O 3
(c)
x
0.45
0
t(min)
dx
x 0 , x d e c re a s e s w h e n t in c re a s e s
200
dt
dx
m
b e c o m e s le s s n e g a tiv e u n til x re a c h e s 0 ;
dt
E a c h c u rv e is c o n c a v e u p a n d a p p ro a c h e s x = 0 a s t ;
m in c re a s e s
dx
b e c o m e s m o re n e g a tiv e x d e c re a s e s fa s te r.
dt
(d)
Check the solution:
( 1 ) t = 0 , x = 0 .4 5
(2 )
dx
dt
0 .4 5
m
200
s a tis f ie s th e in iti a l c o n d iti o n ;
exp(
m t
200
)
m
x s a tis f ie s th e m a s s b a la n c e .
200
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x
lOMoARcPSD|32610952
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
5
10 t(min) 15
20
25
(e)
90% x f 0 .0 4 5 t 4 .6 m in
99% x f 0 .0 0 4 5 t 9 .2 m in
99.9% x f 0 .0 0 0 4 5 t 1 3.8 m in
10.8
(a)
Mass of tracer in tank:
Tracer balance: Accumulation = –output. If perfectly mixed, C o u t C ta n k C
dC
C
V is constant
dt
V
t 0, C
m0
V
(b)
(c)
Plot C (log scale) vs t (rect. scale) on semilog paper: Data lie on straight line (verifying
assumption of perfect mixing) through
-3
10 )
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10.9
(a)
In tent at any time, P=14.7 psia, V=40.0 ft3, T=68F=528R
N
PV
1 4 .7 p s ia
m (liq u id )
1 0 .7 3
RT
(b)
ft
3
4 0 .0 ft
p s ia
lb - m o le
o
528
o
3
0 .1 0 3 8 lb - m o le
R
R
Molar throughout rate:
Moles of O2 in tank =
Balance on O2: Accumulation = input – output
(c)
0 .3 5 x
e
1. 6 3 t
x 0 .3 5 0 .1 4 e
1. 6 3 t
0 .1 4
10.10
A B
(a)
Mole balance on A: Accumulation = –consumption (V constant)
(b)
Plot
on rectangular paper:
Slope
intercept
Data fall on straight line through
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10.11
CO Cl2 COCl2
(a)
(b)
Mole balance on Phosgene: Accumulation = generation
(c)
Cl2
limiting; 75% conversion
(d)
20
30
40
10
1
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)
CONTINUE
SUM1 = 0.
DO 30 J = 2, NM1, 2
SUM = SUM1 + F(S)
CONTINUE
SUM2 = 0.
DO 40 J = 3, NM2, 2
SUM2 = SUM2 + F(J)
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
CONTINUE
FORMAT (I4, 'POINTS —', 2X, F7.1, 'MINUTES')
END
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RESULTS
5 POINTS —
21 POINTS —
51 POINTS —
91.0 MINUTES
90.4 MINUTES
90.4 MINUTES
t 9 0 .4 m in u te s
10.12
(a)
Accumulation = input
Separate variables and integrate. Since p A y A P is constant, C *A p A H is also a
constant.
(b)
3
2
V 5 L 5 0 0 0 c m , k 0 .0 2 0 c m s , S 7 8 .5 c m , C A 0 .6 2 1 0
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3
m ol / cm
3
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10.13
(a)
MCv
dT
Q W
dt
M ( 3 . 0 0 L ) ( 1. 0 0 k g / L ) = 3 . 0 0 k g
C v C p ( 0 .0 7 5 4 k J / m o l
o
C )( 1 m o l / 0 .0 1 8 k g ) = 4 .1 8 4 k J / k g
o
C
W 0
(b)
(c)
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).
10.14
(a)
Energy balance:
MCv
dT
Q W
dt
The other 3% of the energy is used to heat the vessel or is lost to the surroundings.
(b)
(c)
No, since the vessel is closed, the pressure will be greater than 1 atm (the pressure at
the normal boiling point).
10.15
(a)
Moles of air in room:
n=
= 258 kg-moles
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Energy balance on room air: n C v
dT
Q W
dt
t 0, T 1 0 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.)
(b)
At steady-state, d T d t 0 4 0 .3 m s 0 .5 5 9 T 0 m s
0 .5 5 9 T
4 0 .3
T 2 4 C m s 0 .3 3 3 k g h r
(c)
Separate variables and integrate the balance equation:
23
𝑑𝑇
∫10 13.4−0.559𝑇 = t
𝑡= −
10.16 (a)
13.4−0.559(23)
1
ln [
] = 4.8 hr
13.4−0.559(10
0.559
Total Mass Balance:
Accumulation=Input– Output
KCl Balance:
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Accumulation=Input-Output
V
dM KCl
dt
dC
dt
m i , K C l m o , K C l
C
dV
8 4C
d(C V )
dV dt 4
dt
1.0 0 8 .0 0 4 .0 0 C
dt
dV
4 .0 0 L / s
dt
t 0, V0 400 L
(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
1
C
(b)
0
t
therefore concave down.
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dC
dt
8 8C
V
V 400 4 t
dC
dt
1 C
5 0 0 .5 t
When the tank overflows, V 4 0 0 4 t 2 0 0 0 t 4 0 0 s
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