9-41
To 530 530 530 530
E 3.0E+04 3.0E+04 3.0E+04 3.0E+04
k 0.3478362 0.3478362 164.49672 164.49672
Ca 0.479966 0.0241085 0.479966 0.0241085
Gt 2.404E+05 2.404E+05 5.711E+06 5.711E+06
Rt -5.25E+05 -5.25E+05 6.825E+06 6.825E+06
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(t) = 1
[4] vo = 400
[5] UA = 250*150
[6] Ta = 530
[7] rho = 50
[8] To = 530
[9] E = 30000
[10] k = 2*7.08e11*exp(-E/1.987/T)
[14] Gt = -dHr*V*k*Ca
[15] Rt = UA*(T-Ta)+rho*vo*Cp*(T-To)
P9-16 (b)
From part (a), we find the concentration and temperatures at the points where G(T) and R(T)
intersect.
The extinction temperature is T0 = 506°R. At this point R(T) is tangent to G(T) at the upper
steady state.
P9-16 (c)
9-42
See Polymath program P9-16-c.pol
POLYMATH Report
Calculated values of DEQ variables
Variabl
e
Initial
value
Minimal
value
Maximal
value
1
t
0
0
6.
2
Ca
0.0681
0.04847
0.4290427
3
T
628.
542.6046
638.1791
4
Ca0
0.5
0.5
0.5
5
Fa0
200.
200.
200.
8
R
1.987
1.987
1.987
9
U
150.
150.
150.
1
A
250.
250.
250.
1
1
v0
400.
400.
400.
1
2
Ta
530.
530.
530.
1
3
V
48.
48.
48.
1
E
3.0E+04
3.0E+04
3.0E+04
1
5
k
51.27669
1.165967
75.24179
1
6
tau
0.12
0.12
0.12
1
7
rA
-3.491943
-3.971954
-0.4433316
1
delH
-3.0E+04
-3.0E+04
-3.0E+04
1
9
Na0
24.
24.
24.
2
Cpa
75.
75.
75.
9-43
0
Differential equations
1
d(Ca)/d(t) = rA + Ca0 / tau – Ca / tau
2
d(T)/d(t) = (U * A * (Ta – T) – Fa0 * Cpa * (T – T0) + (rA * V) * delH) /
1
Ca0 = 0.5
2
Fa0 = 200
3
T0 = 530
4
Ar = 1.416 * 10 ^ 12
7
A = 250
8
v0 = 400
9
Ta = 530
1
0
V = 48
1
E = 30000
1
5
delH = -30000
1
Na0 = Ca0 * V
9-44
When the upper steady state is used as the initial conditions, the unsteady-state mole balance
P9-16 (d)
When T0 or Ta is increases slightly, the upper steady state becomes stable. At these elevated inlet
P9-16 (e)
Starting at the lower steady state, if T0 is increased to 550 °R, the lower steady state is no longer
stable, but the upper steady state is
If we plot the concentration-temperature phase-plane trajectory, we see that increasing Ta will
shift the trajectory to the left. However, the final steady state is shifted to the right. This means
that from the initial conditions, at any temperature CA is lower for the larger Ta until the minimum
9-45
P9-16 (f) Individualized solution
P9-16 (g)
The following Polymath program gives the linear analysis of the problem
A = 1.175, B = 6.03, J = 1600, τ = .12
See Polymath program P9-16-g.pol
POLYMATH Report
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
t
0
0
6.
6.
2
y
2.
0.0176788
4.159719
0.0176788
3
x
0.2
-0.0225558
0.2
-0.0001258
4
Ca
0.0681
0.04847
0.4290427
0.4253139
5
T
628.
542.6046
638.1791
547.0711
6
Cpa
75.
75.
75.
75.
7
Ca0
0.5
0.5
0.5
0.5
8
Fa0
200.
200.
200.
200.
9
T0
530.
530.
530.
530.
10
Ar
1.416E+12
1.416E+12
1.416E+12
1.416E+12
11
R
1.987
1.987
1.987
1.987
12
U
150.
150.
150.
150.
13
Area
250.
250.
250.
250.
15
Ta
530.
530.
530.
530.
16
V
48.
48.
48.
48.
17
E
3.0E+04
3.0E+04
3.0E+04
3.0E+04
18
k
51.27669
1.165967
75.24179
1.463353
19
tau
0.12
0.12
0.12
0.12
9-46
21
delH
-3.0E+04
-3.0E+04
-3.0E+04
-3.0E+04
22
Na0
24.
24.
24.
24.
23
J
800.
800.
800.
800.
24
C
6.
6.
6.
6.
25
B
6.03
6.03
6.03
6.03
26
A
1.175
1.175
1.175
1.175
Differential equations
1
d(y)/d(t) = (-J * (1 – A) * x + (B – C) * y)
2
d(x)/d(t) = (-A * x – B * y / J) / tau
3
d(Ca)/d(t) = rA + Ca0 / tau – Ca / tau
Explicit equations
1
Cpa = 37.5 * 2
2
Ca0 = 0.5
3
Fa0 = 200
4
T0 = 530
5
Ar = 1.416 * 10 ^ 12
9
v0 = 400
10
Ta = 530
11
V = 48
12
E = 30000
13
k = Ar * exp(-E / R / T)
14
tau = V / v0
15
rA = -k * Ca
16
delH = -30000
9-47
P9-16 (f)
Only the lower steady state plot of x1 and y1 will be shown.
P9-17
9-48
P9-17 (a)
See Polymath program P9-17-a.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 0.2 0.2
Ca 0.3 1.34E-65 0.3 1.34E-65
Cc 0 0 0.3 0.3
Cb 0 -2.02E-44 0.2895784 -1.864E-65
T 283 283 915.5 915.5
UA 0 0 0 0
V 10 10 10 10
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra1
[2] d(Cc)/d(t) = –rb2
[3] d(Cb)/d(t) = rb1+rb2
Explicit equations as entered by the user
[1] UA = 0
[2] V = 10
[3] k1 = 3.03*exp((9900/1.987)*(1/300-1/T))
[4] k2 = 4.58*exp((27000/1.987)*(1/500-1/T))
9-49
P9-17 (b)
9-50
P9-17 (c)
P9-18
9-51
9-52
P9-18 (a)
The selectivity is proportional to CA/CB and therefore a small concentration of B in the reactor is
P9-18 (b)
Now it would be best to slowly add A to B in a semibatch reactor.
Now it would be best to slowly add A to B in a semibatch reactor to maintain high concentrations
of B and low concentrations of A.
P9-18 (c)
P9-19
Plotting the data of t vs. T gives the initial and final temperature of the reaction, T0 = 52.5 ˚C and
Tf =166.8 ˚C.
Recalling equation (8-29) with ΔCp = 0, X = 1 and T = Tf gives:
See Polymath program P9-19.pol
POLYMATH Report
Model: lnTdot = a0 + a1*T_inverse_kelvin
Variable
Value
95% confidence
a0
18.88162
2.425175
a1
-6224.846
880.596
R^2
0.9755663
Rmsd
0.0597988
Variance
0.0413783
( )
!“#$=%” 0
TTCpH fiiRx
( ) mol
cal
K
Kmol
cal
HRx 4.20575.528.16618 !=!
“
!=#
9-54
.6224
ln 18.88
s
T
T
!
= +
6224 E
R
=
6224 1.987 12367
cal cal
E K
mol K mol
=!=“
!
P9-20 No solution will be given
CDP9-A
CDP9-B
CDP9-C
9-55
9-56
9-57
CDP9-D
9-58
9-59
9-60