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4-21
Solving using polymath to get a table of values of X Vs T.
See Polymath program P4-9.pol.
POLYMATH Results
NLE Solution
Variable Value f(x) Ini Guess
X 0.4229453 3.638E-12 0.5
To 300
T 305.5
NLE Report (safenewt)
Nonlinear equations
[1] f(X) = (z/y)*X/((1-X)^2 - X^2/Kc) -V = 0
Explicit equations
[1] To = 300
[2] T = 305.5
0.4
0.41
0.42
0.43
X
4-22
T(in K)
X
300
0.40
301
0.4075
We get maximum X = 0.4229 at T = 305.5 K.
P4-10 (a)
For substrate:
Cell: voCC = rgV
P4-10 (b)
[ ]
SSOSCC CCYC !=/
P4-10 (c)
CC = YC/S(CSO - CS)
= 0.8(30 – 5.0)g/dm3 = 20 g/dm3 .
P4-10 (d)
4-23
P4-10 (e)
Vnew = Vo/3 = 25/3 dm3
Using equation from above, we get CS = 3.0 g/dm3 and CC = 21.6 g/dm3
P4-10 (f)
For batch reactor:
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 15 15
Cs 30 0.0382152 30 0.0382152
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cs)/d(t) = rs
Explicit equations as entered by the user
[1] Cso = 30
P4-10 (g)
P4-11
Gaseous reactant in a tubular reactor: A B
1
0.0015 min at 80 F
A A
r kC
k!
!=
=o
1000
17.21
58
B
lb
lb mol
hr
Flb hr
lb mol
= =
0
17.21
19.1
0.9
B
A
lb mol
Flb mol
hr
F
X hr
= = =
For a plug flow reactor:
At T2 = 260°F = 720°R, with k1 = 0.0015 min-1 at T1 = 80°F = 540°R,
1
2 1
1 2
1 1 25000 1 1
exp 0.0015exp 53.6 min
1.104 540 720
E
k k
R T T
!
" #
" # " #
" #
=!=!=
$ %
$ % $ %
$ %
$ % & '
& '
& '
& '
1 1
253.6 min 3219k hr
! !
= =
4-25
P4-12
A → B/2
4-26
P4-13
Given: The metal catalyzed isomerization
BA !
, liquid phase reaction
Case 1: an identical plug flow reactor connected in series with the original reactor.
Since yA = 1.0, ΘB = 0. For a liquid phase reaction
( )
A A0
C C 1 X=!
and
B A0
C C X=
4-27
( )
0 1
1
0
1 1
ln 1 1 0.853ln .355 0.883
1
1
A
A eq
eq
kC V X
F K
K
! "
# $
=% % + = %=
& '
( )
( )
& '
* +
+, -
Take advantage of the fact that two PFR’s in series is the same as one PFR with the volume of the two
combined.
VF = V1 + V2 = 2V1 and at VF X = X2
F
X
Case 2: Products from 1st reactor are separated and pure A is fed to the second reactor,
4-28
Since pure A enters both the first and second reactor CA0,2 = CA0, CB0,2 = 0, ΘB = 0
( )
C = C 1 - X
C = C X
for the second reactor.
and since V1 = V2
0 2 0 1
0 0
A A
A A
kC V kC V
F F
=
or
eq
Overall conversion for this scheme:
( ) ( )( ) ( )( )
0 0,2 2 0 0 1 2
1 2
0 0
11 1 1 1 1
A A A A
A A
F F X F F X X
X X X
F F
! ! ! ! !
= = = ! ! !
0.895X=
P4-14
Given: Ortho- to meta- and para- isomerization of xylene.
4-29
Assume that the reactions are irreversible and first order.
Then:
Check to see what type of reactor is being used.
Case 1:
Assume plug flow reactor conditions:
0M M
F dX r dV=!
or
0
0
X
M
M
dX
V F
r
=!
"
The reactor appears to be plug flow since (kV)Case 1 = (kV)Case 2
As a check, assume the reactor is a CSTR.
0 0 0M M M
F X C v X r V= = !
( )
0 0
01
M
M
C X v X
V v
r k X
= =
! !
or
0
1
v X
kV
X
=!
Again kV should be the same for both Case 1 and Case 2.
4-30
This assumes that the same hydrodynamic conditions are present in the new reactor as in the old.
P4-15
A→ B in a tubular reactor
Tube dimensions: L = 40 ft, D = 0.75 in.
nt = 50
( )
2
2
3
0.75
50
12 40 6.14
4 4
t
n D
V L ft
!
!
" #
$ %
& '
= = =
0
500
6.86
73
A
A
A
lb
mlb mol
hr
Flb
MW hr
lb mol
= = =
0
0
1
ln
1
A
A
F RT
Vky P X
! "
=# $
%
& '
or
0
0
1
ln
1
A
A
F RT
kVy P X
! "
=# $
%
& '
Assume Arrhenius equation applies to the rate constant.
E
!
1 2 1 1 2
k R T T R T T
& '
( )( )
1 2 2
1 2 1
660 760 0.740
ln ln 19,500
100 0.00152
T T k
ER
R T T k
= = =
!
o
1
1
exp E
A k
RT
! "
=# $
% &
so
1
1
1 1
exp E
k k
R T T
! "
# $
=% %
& '
( )
* +
, -
From above we have
0 1
1
A
Vy P X R T T
%
* + * +
, -
Dividing both sides by T gives:
4-32
P4-16
Reversible isomerization reaction
m-Xylene → p-Xylene
Xe is the equilibrium conversion.
Rate law:
p
m m
e
C
r k C k
! "
#=#
$ %
& '
At equilibrium,
P4-16 (a)
For batch reactor,
P4-16 (b)
For CSTR,
mo
m
F X
V
r
="
P4-16 (c)
1
1
ln
1
e
V
e e
X
X
X X
X X
!
" # " #
$% & % &
' ( % &
=% &
$
% &
' (
Following is the plot of volume efficiency as a function of the ratio (X/Xe),
See Polymath program P4-16-c.pol.
P4-16 (d)
Efficiency = VPFR / VCSTR = 1 from problem statement, which is not possible because conversion will not be
the same for the CSTR’s in series as for the PFR.
4-34
P4-17 (a)
A → ½ B
ε = -1/2, X = 0.3, W = 1 kg, yexit = 0.25
For PBR, -rA = kCA
2 and
( )
( )
X
yXC
CO
A
!
+
"
=1
1
Solving for z by trial and error in Polymath to match x and y at exit,
X = 0.3 yo = 1 and yf = 5/20 = 0.25
we get: α = 1.043 kg-1 and z = 0.7 kg-1
See Polymath program P4-17-a1.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
W 0 0 1 1
Differential equations as entered by the user
Solving we get for W = 1kg and z = 0.7 kg-1
X = 0.40
See Polymath program P4-17-a2.pol.
POLYMATH Results
NLE Solution
Variable Value f(x) Ini Guess
NLE Report (fastnewt)
P4-17 (b)
For turbulent flow:
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
ODE Report (STIFF)
Differential equations as entered by the user
Explicit equations as entered by the user
P4-17 (c) Individualized solution
P4-17 (d) Individualized solution
4-36
P4-18
P4-18 (b)
4-37
P4-18 (c)
P4-18 (d)
P4-19
Production of phosgene in a microreactor.
See Polymath program P4-19.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
W 0 0 3.5E-06 3.5E-06
X 0 0 0.7839904 0.7839904
y 1 0.3649802 1 0.3649802
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(W) = -rA/FA0
[2] d(y)/d(W) = -a*(1+e*X)/(2*y)
Explicit equations as entered by the user
[1] e = -.5 [2] FA0 = 2e-5
P4-19 (a)
P4-19 (b)
The outlet conversion of the reactor is 0.784
Therefore 10,000 kg/year / 48.95 kg/ year = 204 reactors are needed.
P4-19 (c)
Assuming laminar flow, α ~ Dp
-2, therefore
P4-19 (d)
A lower conversion is reached due to equilibrium. Also, the reverse reaction begins to overtake the
forward reaction near the exit of the reactor.
P4-19 (e) Individualized solution
P4-20 (a)
For turbulent flow:
"
=2
#
P
$
1%
&
"
o=G1#
$
( )
31.75G
( )
P4-20 (b)
See Polymath program P4-20-b.pol.
P4-20 (c)
