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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
1
B C
C C A
C
C C
r k C
K
! "
=#
$ %
& '
A D
'
2 2D D A
r k C=
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
Fe 0 0 1.3722476 1.3722476
y 1 0.3404952 1 0.3404952
Ft 10 9.7581913 13.220737 9.7581913
Cto 0.6 0.6 0.6 0.6
Cd 0 0 0.1019635 0.0571654
Cc 0 0 0.2117037 0.1017844
k3e 5 5 5 5
r1c 1.2 0.0051635 1.2 0.0051635
r3e 0 0 0.0216828 0.0029612
rd 0.24 0.0014457 0.24 0.0014457
rb 1.2 0.0051635 1.2 0.0051635
rc 1.2 -0.0042625 1.2 0.0022023
re 0 0 0.0216828 0.0029612
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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
[6] Ca = Cto*(Fa/Ft)*y
[7] Cd = Cto*(Fd/Ft)*y
[8] Cc = Cto*(Fc/Ft)*y
[14] ra = -r1c-r2d
[15] r3e = k3e*(Cc^2)*Cd
[16] rd = r2d-(r3e/2)
[17] rb = r1c
[21] alfa = 0.008
[22] Fto = 10
P6-21 (b)
The interesting concentrations here are species C and D, both of which go through a maximum. Species C
goes through a maximum for two reasons: (1) it is an intermediate product which is formed and then
consumed, and (2) there is pressure drop along the length of the reactor and as pressure drops, so does
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
P6-22 (b) Oscillations eventually cease because the CA is decreasing and becomes the limiting factor.
P6-22 (ac) Observation 1: τ1 and τ2 decreased
Observation 2: τP1 increased
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Now,
"
1=1
#
ln
µ
0
µ
1
*
$
%
&
'
(
)
and
"
2=1
#
ln
µ
0
µ
2
*
$
%
&
'
(
)
"
P1=2
#
(
µ
1
*2 +KU)1/ 2
and
"
P2=2
#
(
µ
2
*2 +KU)1/ 2
"
=k0
k2
,
µ
1,2
*=(1"2KU)±1"8KU
2
#
$
%
&
'
(
1/ 2
From observation 2 and 3, we get
Decreasing µ1
* and increasing µ2
* => KU = (kU/k2) will increase
Also, from observation 1, ε increased => k0/k2 should be increased
Now, P A k = k0 …….(1)
A B k = kU …… (2)
P6-22 (d) Individualized solution
P6-23 Individualized solution
CDP6-24
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CDP6-25 (a)
PFR:
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CDP6-25 (b)
When θM is reduced to 1.5, it now takes a τ of 0.24 h to achieve a maximum of xylene. Increasing θM to
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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)
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d(c11)/d(V) = r1/vo # c11(0)=0.137
d(c9)/d(V) = (-r2+r3)/vo # c9(0)=0
d(c10)/d(V) = (-r1+r2)/vo # c10(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) #
S89 = c8/(c9+0.0000001) #
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
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CDP6-26 (c) Individualized solution
CDP6-27
6-70
CDP6-28
6-71
6-72
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CDP6-29 No solution will be given
CDP6-A
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b) No solution will be given.
CDP6-B
