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6-61
P6-21 (a)
Isothermal gas phase reaction in a membrane reactor packed with catalyst.
A B + C
'
1 1
B C
C C A
C C
r k C
K
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=#
$ %
See Polymath program P6-21-a.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
W 0 0 100 100
Fa 10 0.349438 10 0.349438
Fb 0 0 3.2375418 0.4443151
Fc 0 0 4.9873025 4.8617029
Fd 0 0 2.7304877 2.7304877
Fe 0 0 1.3722476 1.3722476
Cd 0 0 0.1019635 0.0571654
Cc 0 0 0.2117037 0.1017844
kb 1 1 1 1
k1c 2 2 2 2
r2d 0.24 0.0029263 0.24 0.0029263
k3e 5 5 5 5
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ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(W) = ra
[2] d(Fb)/d(W) = rb-(kb*Cb)
Explicit equations as entered by the user
[1] k2d = 0.4
[2] K1c = 0.2
[3] Ft = Fa+Fb+Fc+Fd+Fe[4] Cto = 0.6
[5] Cb = Cto*(Fb/Ft)*y
[6] Ca = Cto*(Fa/Ft)*y
[7] Cd = Cto*(Fd/Ft)*y
[8] Cc = Cto*(Fc/Ft)*y
P6-21 (b)
The interesting concentrations here are species C and D, both of which go through a maximum. Species C
P6-21 (c) Individualized Solution
P6-22 (a) What factors influence the amplitude and frequency of the oscillation reaction?
Ans: k and the initial conditions
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P6-22 (d) Individualized solution
P6-23 Individualized solution
CDP6-24
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CDP6-25 (a)
PFR:
CDP6-25 (b)
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CDP6-25 (c)
To find out the reactor schemes needed, use the attainable region to get these graphs
CDP6-25 (d)
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CDP6-26 (a)
d(c11)/d(V) = r1/vo # c11(0)=0.137
d(c9)/d(V) = (-r2+r3)/vo # c9(0)=0
d(c6)/d(V) = -r5/vo # c6(0)=0
d(ch)/d(V) = (r1+r2+r3+r4+r5)/vo # ch(0)=0.389
S9o = c9/(c10+c8+c7+c6+0.0000001) #
S87 = c8/(c7+0.0000001) #
V(0)=0 # V(f)=0.8
CDP6-26 (b)
The polymath program is the same as the first, we see that the value of c11(o)=0.092 and ch(o)=0.434 and
follows that the volume would be smaller because earlier the reaction ends the less C7 is formed.
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CDP6-26 (c) Individualized solution
CDP6-27
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CDP6-28
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CDP6-29 No solution will be given
CDP6-A
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Plug those into POLYMATH:
b) No solution will be given.
CDP6-B
