LEM
304S12
Fogler
CHAPTER 8
HOMEWORK SOLUTION
#8.6
P8-6
ABC
Since the feed is equimolar, CA0 = CB0 = .1 mols/dm3
CA = CA0(1-X)
CB = CB0(1-X)
1
2
3
4
5
6
7
8
9
P8-6 (a)
0
0
PFR A
A
A
CSTR
A
dX
VF
r
FX
V
r





For the PFR, FA0 = CA0v0 = (.1)(2) = .2 mols/dm3
See Polymath program P8-6-a.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 X 0 0 0.85 0.85
2 V 0 0 308.2917 308.2917
3 Ca0 0.1 0.1 0.1 0.1
4 Fa0 0.2 0.2 0.2 0.2
5 T 300. 300. 470. 470.
6 k 0.01 0.01 4.150375 4.150375
7 ra -0.0001 -0.0018941 -0.0001 -0.0009338
Differential equations
1 d(V)/d(X) = -Fa0 / ra
Explicit equations
1 Ca0 = .1
2 Fa0 = .2
3 T = 300 + 200 * X
4 k = .01*exp((10000 / 2) * (1 / 300 - 1 / T))
5 ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2)
V = 308.2917dm3
8-24
For the CSTR,
X = .85, T = 300+(200)(85) = 470 K.
k = 4.31 (Using T = 470K in the formula).
-rA
= .000971 mol/dm3/s
03
-4
.1 2 .85
175 dm
9.71 10
A
A
10
FX
V
r



The reason for this difference is that the temperature and hence the rate of reaction
remains
constant throughout the entire CSTR (equal to the outlet conditions), while for a PFR,
the rate
increases gradually with temperature from the inlet to the outlet, so the rate of increases
with
length.
P8-6 (b)
0
[]
i
R
iP
XH
TT
C



For boiling temp of 550 k,
550 = T0 + 200
T0 = 350K
P8-6 (c)
P8-6 (d)
0
0
()
A
CSTR
A
CSTR
A
A
FX
V
r
V
Xr
F



For V = 500 dm3, FA0=.2
222
0
(1 ) .01 (1 ) A A
11
r k C X k X
T 300 200 X
Now use Polymath to solve the non-linear equations.
See Polymath program P8-6-d-1.pol.
8-25
Calculated values of NLE variables
Variable Value f(x) Initial Guess
1 T 484.4136 0 480.
2 X 0.9220681 -2.041E-09 0.9
Variable Value
1 k 6.072856
2 ra 0.0003688
Nonlinear equations
1 f(T) = 300 + 200 * X - T = 0
2 f(X) = 500 - .2 * X / ra = 0
Explicit equations
1 k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T))
2 ra = 0.01 * k * (1 - X) ^ 2
Hence, X = .922 and T = 484.41 K
For the conversion in two CSTR’s of 250 dm3 each,
For the first CSTR, using the earlier program and V = 250 dm3,
Calculated values of NLE variables
Variable Value f(x) Initial Guess
1 T 476.482 1.137E-13 480.
2 X 0.88241 -5.803E-09 0.9
Variable Value
1 k 5.105278
2 ra 0.0007059
Nonlinear equations
1 f(T) = 300 + 200 * X - T = 0
2 f(X) = 250 - .2 * X / ra = 0
Explicit equations
1 k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T))
2 ra = 0.01 * k * (1 - X) ^ 2
T = 476.48 ad X = .8824
Hence, in the second reactor,
8-26
01
1
0
()
()
A
CSTR
A
CSTR
A
A
FXX
V
r
V
XrX
F



12

See Polymath program P8-6-d-2.pol.
Calculated values of NLE variables
Variable Value f(x) Initial Guess
1 T 493.8738 0 480.
2 X 0.9693688 -1.359E-09 0.8824
Variable Value
1 k 7.415252
2 ra 6.958E-05
3 X1 0.8824
Nonlinear equations
1 f(T) = 476.48 + 200 * (X - X1) - T = 0
2 f(X) = 250 - .2 * (X - X1) / ra = 0
Explicit equations
1 k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T))
2 ra = 0.01 * k * (1 - X) ^ 2
3 X1 = .8824
Hence, final X = .9694
P8-6 (e) Individualized solution
P8-6 (f) Individualized solution
P8-7 (a)
For reversible reaction, the rate law becomes
C
AAB
C
C
rkCC
K



, 1 1 200( ) out CSTR
T T X X
8-27
300 200
11
(300) exp
300
11
(450) exp
450
Rxn
CC
TX
E
kk
RT
H
13
KK
RT







Stoichiometry:
0
0
0
(1 )
(1 )
CA
AA
BA
CCX
CCX
CCX



See Polymath program P8-7-a.pol.
POLYMATH Results
No Title 03-21-2006, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 10 10
X 0 0 0.0051176 0.0051176
T 300 300 301.02352 301.02352
k 0.01 0.01 0.010587 0.010587
Fa0 0.2 0.2 0.2 0.2
Ca0 0.1 0.1 0.1 0.1
Kc 286.49665 276.85758 286.49665 276.85758
ra -1.0E-04 -1.048E-04 -1.0E-04 -1.048E-04
Xe 0.8298116 0.827152 0.8298116 0.827152
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
Explicit equations as entered by the user
[1] T = 300+200*X
[2] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] Kc = 10 * exp(-6000 /1.987 * (1 / 450 - 1 / T))
[6] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)
[7] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
P8-8 (a)
ABC
Species Balance:
0
3
0
14
0
20 /
10
A
A
dX r
dW F
v dm s
P atm




Stoichiometry:
0
0
0
0
1
,1
1
1
1
AA
AA
XT
C C where
XT
XT
CC
XT








Rate Law is:
11
, 0.133exp
450
31400
20,000 /
AA
Rxn
E
r kC with k
RT
15
E
H J mol





Energy Balance:
0
0
[ ( )]
15 24 40 0
20,000
450 450 500
40
i
R
iPP
P
XHT
TT
CXC
C
X
TX






% %
See Polymath program P8-8-a.pol.
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 X 0 0 0.8 0.8
2 W 0 0 43.13711 43.13711
3 T 450. 450. 850. 850.
4 v0 20. 20. 20. 20.
5 T0 450. 450. 450. 450.
6 k 0.133 0.133 6.904332 6.904332
Differential equations
1 d(W)/d(X) = v0 * (1 + X) * T / k / (1 - X) / T0
8-36
Explicit equations
1 T = 450 + 500 * X
2 v0 = 20
3 T0 = 450
4 k = .133 * exp(31400 / 8.314 * (1 / T0 - 1 / T))
P8-8 (b)
Species Balance for CSTR:
0
16
450 500 450 500(.8) 850
31400 1 1
.133exp 6.9
8.314 450 850
39.42
A
CSTR
A
CSTR
FX
W
r
TXK
k
W kg







P8-8 (c) Individualized solution
P8-8 (d)
For pressure drop, an extra equation is added

0
00
0
00
1
2(/)
1
1
AA
dP T P
X
dW T P P
XTP
CC
XTP








17
See Polymath program P8-8-d.pol.
Using POLYMATH program CRE_8_8d.pol
For .019
8-37
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 W 0 0 0.8 0.8
2 X 0 0 0.0544753 0.0544753
3 P 1.013E+06 1.002E+06 1.013E+06 1.002E+06
4 T 450. 450. 850. 850.
5 v0 20. 20. 20. 20.
6 T0 450. 450. 450. 450.
7 k 0.133 0.133 6.904332 6.904332
8 P0 1.013E+06 1.013E+06 1.013E+06 1.013E+06
9 alpha 0.019 0.019 0.019 0.019
Differential equations
1 d(X)/d(W) = k / v0 * (1 - X) / (1 + X) * T0 / T * P / P0
2 d(P)/d(W) = -alpha / 2 * (T / T0) * P0 ^ 2 / P * (1 + X)
Explicit equations
1 T = 450 + 500 * W
2 v0 = 20
3 T0 = 450
4 k = .133 * exp(31400 / 8.314 * (1 / T0 - 1 / T))
5 P0 = 1013250
6 alpha = .019
18