4-21
Now using:
( )
( )
( )
2 2
2 2
0
1 1
C C
Z X Z X
V f X V
z z
X X
X X
K K
! “ ! “
=#= = $
% & % &
‘ ( ‘ (
) * ) *
$ $ $ $
+ , + ,
– . – .
where
1 1
exp
k E
z
k R T T
! “
# $
= = %
& ‘
( )
& ‘
and
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
V 3785.4
Kco 3
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
[3] z = 2902.2
[9] Hrx = -25000
0.39
0.4
0.42
0.43
Temperature
X
4-22
P4-10 (a)
For substrate:
Cell: voCC = rgV
P4-10 (b)
[ ]
SSOSCC CCYC !=/
( ) /0
MAX S
SO S O S C C
C
C C v VY C
K C
µ
! “
# # =
$ %
+
!
( ) ( )
/0
MAX S
SO S O C S SO S
M S
C
C C v VY C C
K C
µ
! “
# # # =
$ %
+
& ‘
( ) ( )
0.5
30 5 25 0.8 30 0
5
S
S S
C
C C
C
! “
#
$ $ # # $ =
% &
+
P4-10 (c)
CC = YC/S(CSO – CS)
P4-10 (d)
vnew = vo/2 = 2.5 dm3/h
4-23
P4-10 (e)
Vnew = Vo/3 = 25/3 dm3
P4-10 (f)
For batch reactor:
CSO = 30 g/dm3 CCO = 0.1 g/dm3
CC = CC0 + YC/S(CSO – CS)
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
Umax 0.5 0.5 0.5 0.5
Cc 0.1 0.1 24.069428 24.069428
rs -0.0535714 -6.8055436 –0.0535714 -0.1141052
ODE Report (RKF45)
Differential equations as entered by the user
Explicit equations as entered by the user
[1] Cso = 30
[2] Ycs = 0.8
[3] Km = 5
[4] Umax = 0.5
4-24
P4-10 (g)
Graphs should look the same as part (f) since reactor volume is not in the design equations for a constant
volume batch reactor.
P4-11
Gaseous reactant in a tubular reactor: A B
1
0.0015 min at 80 F
A A
r kC
k!
!=
=o
25, 000 cal
Eg mol
=
0.90X=
1000
B
lb
M
hr
=
58
A B
lb
MW MW
lb mol
= =
1 inch (I.D.)
t
D=
10L ft=
132 146.7P psig psia= =
260 720T F R= =
o o
number of tubes
t
n=
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:
0.9
2
0
0
4
t t
A
A
n D L dX
V F
r
!
= = “
#
1 1 0
!
=“=
1.0
A
y=
0
A
y
! “
= =
( )
01
A A
r kC X!=!
0
A
A
PP
C
RT RT
= =
( )
0.9 0.9
0 0
0 0
0 0
0 0
1
ln ln10
1 1 0.9
A A
A A
A A A
F F RT
dX dX
V F F
r kC X kC kP
! “
= = = =
# $
% % %
& ‘
( (
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
! !
= =
( )
( )( )
3
0
1
19.1 10.73 720
ln10 ln10
3219 146.7
A
lb mol ft psia R
hr lb mol R
F RT
VkP hr psia
!
” #
” #$ %
$ %
& ‘& ‘
= =
o
o
3
0.72V ft=
2
4
t t
n D L
V
!
=
4-25
( )
( )
3
2
2
4 0.72
413.2
110
12
t
t
ft
V
nD L ft ft
!!
= =
” #
$ %
& ‘
Therefore 14 pipes are necessary.
P4-12
A → B/2
AO AO TO
C y C=
2 2 2
2(1 1/ 4) 0.8
2( 1/ 4)(1 1/ 4) ln(1 0.8) ( 1/ 4) 0.8 2.9
1 0.8
PFR AO TO
AO
V ky C
F
!
=! ! ! +!+ =
!
……….(6)
4-26
2 2
( )
8 8 2.9 2.58
9 9
PFR AO TO
AO
V ky C
F= = =
P4-13
Given: The metal catalyzed isomerization
BA !
, liquid phase reaction
1
B
A A
eq
C
r k C K
! “
#=#
$ %
$ %
& ‘
with Keq = 5.8
For a plug flow reactor with yA = 1.0, X1 = 0.55
Case 1: an identical plug flow reactor connected in series with the original reactor.
( )
01
A A
eq
X
r kC X K
! “
#=# #
$ %
$ %
& ‘
For the first reactor,
( )
1 1
1 0 0
0 0
01
X X
A A
A
A
eq
dX dX
V F F
rX
kC X K
= =
!” #
! !
$ %
$ %
& ‘
( (
or
1
1
0 1
00
0
1 1
ln 1 1
1
11
1 1
X
X
A
A eq
eq
eq
kC V dX X
F K
XK
K
! “
# $
= = % % +
& ‘
( )
( )
# $ & ‘
* +
+, –
%+
( )
( )
* +
.
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.
0
00
0
1 1
ln 1 1
1
11
1 1
F
X
A F
A eq
eq
eq
kC V dX X
F K
XK
K
! “
# $
= = % % +
& ‘
( )
( )
# $ & ‘
* +
+, –
%+
( )
( )
* +
.
0 0 1
2
0 0
1 1
2 ln 1 1
1
1
A F A
A A eq
eq
kC V kC V X
F F K
K
! “
# $
= = % % +
& ‘
( )
( )
& ‘
* +
+, –
( )
0 1
0
2 2 0.883 1.766
A
A
kC V
F= =
2
1 1
1.766 ln 1 1
15.8
1
5.8
X
! “
# $
=% % +
& ‘
( )
* +
, –
+
X2 = 0.74
Case 2: Products from 1st reactor are separated and pure A is fed to the second reactor,
The analysis for the first reactor is the same as for case 1.
0 1
1
0
1 1
ln 1 1
1
1
A
A eq
eq
kC V X
F K
K
! “
# $
=% % +
& ‘
( )
( )
& ‘
* +
+, –
By performing a material balance on the separator, FA0,2 = FA0(1-X1)
4-28
Since pure A enters both the first and second reactor CA0,2 = CA0, CB0,2 = 0, ΘB = 0
( )
A A0
C = C 1 – X
B A0
C = C X
for the second reactor.
( )
( )
2 2
0
2 0,2
0
0 0
1
1
X X
A
A
A A
eq
F X
dX dX
V F X
r kC XK
!
= =
!! !
” “
or
1
1 2
1
1 1 1
ln 1 1 ln 1 1
1 1
1 1
eq eq
eq eq
X
X X
K K
K K
! ” ! “
# $ # $
%
% % + = % % +
& ‘ & ‘
( ) ( )
( ) ( )
& ‘ & ‘
* + * +
+, – +, –
1
1
1
2 1
1 1
1 1 1 1
X
eq eq
X X
K K
!
” #
$ % $ %
!+ = !+
& ‘
( ) ( )
( ) ( )
& ‘
* + * +
, –
( )
1
1
1
1
1
0.45
2
1
1 1 1
1 0.356
0.766
11.174
1
X
eq
eq
X
K
X
K
!
” #
$ %
! ! +
& ‘
( )
( ) !
& ‘
* +
, –
= = =
+
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.
Mk1
“ # “ P
Mk2
“ # “ O
O“ # “ P (neglect)
Pressure = 300 psig
T = 750°F
V = 1000 ft3 cat.
4-29
Assume that the reactions are irreversible and first order.
Then:
1 2M M M M
r k C k C kC!= + =
1 2
0
k k k
!
= +
=
Check to see what type of reactor is being used.
Case 1:
02500 0.37
gal
v X
hr
= =
Case 2:
01667 0.50
gal
v X
hr
= =
Assume plug flow reactor conditions:
0M M
F dX r dV=!
or
0
0
X
M
M
dX
V F
r
=!
“
( ) ( )
0 0 0
0
0 0
ln 1
1
X X
M
M
C v dX v
dX
V v X
r k X k
= = = !
! !
” “
CM0, k, and V should be the same for Case 1 and Case 2.
Therefore,
( ) ( ) ( ) [ ]
0 1
1 1 ln 1 2500 ln 1 0.37 1155
Case
Case Case
gal gal
kV v X hr hr
=! ! =! ! =
( ) ( ) ( ) [ ]
0 2
2 1 ln 1 1667 ln 1 0.50 1155
Case
Case Case
gal gal
kV v X hr hr
=! ! =! ! =
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.
( ) ( ) ( )
0 1
1
1
1
2500 0.37
1468
1 1 0.37
Case
Case
Case
Case
gal
v X gal
hr
kV X hr
= = =
! !
( ) ( ) ( )
0 2
2
2
2
1667 0.50
1667
1 1 0.50
Case
Case
Case
Case
gal
v X gal
hr
kV X hr
= = =
! !
kV is not the same for Case 1 and Case 2 using the CSTR assumption, therefore the reactor must be
modeled as a plug flow reactor.
4-30
1155 gal
kV hr
=
3 3
1155
1.55
1000 .
gal
gal
hr
kft cat hr ft cat
= =
For the new plant, with v0 = 5500 gal / hr, XF = 0.46, the required catalyst volume is:
3
1.155
hr ft cat
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
X
A
A
dX
V F
r
=!
“
( ) ( )
0
0
1
1
1
A
A A
kC X
r kC X
X
!
“
“= = “
+
( )
0
0 0
0 0
0 0
1
ln
1 1
X X
A
A A
A A A
F
dX dX
V F F
r kC X kC X
! “
= = = # $
% % %
& ‘
( (
with
0
0
A
A
y P
P
CRT RT
= =
4-31
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.
At T1 = 600°R, k1 = 0.00152
1
E
RT
Ae
!
=
At T2 = 760°R, k2 = 0.0740
2
E
RT
Ae
!
=
2
1 2 1
1 1
exp
kE
k R T T
! “
# $
%
=%
& ‘
( )
* +
, –
2 2 1
1 2 1 1 2
1 1
ln k T T
E E
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
0
1
ln
1
A
A
F RT
kVy P X
! “
=# $
%
& ‘
so
0
1
1 1 1
ln exp
1
A
F RT E
k
Vy P X R T T
! “
# $
# $ =% %
& ‘
( )
( )
%
Dividing both sides by T gives:
1
1
0
0
1 1
exp
1
ln 1
A
A
E
kR T T
F R
Vy P X T
! “
# $
% %
& ‘
( )
# $ * +
, –
=
( )
%
* +
( )( )
.00152 3600 6.14 114.7
sec
ft psia
hr
& ‘& ‘
* +* +
Evaluating and simplifying gives:
1
1 1
exp 19500 660
0.0308 T R
R
T
!
” #
$ %
! !
& ‘
( )
* +
, –
=
o
o
Solving for T gives:
T = 738°R = 278°F
P4-16
Reversible isomerization reaction
m-Xylene → p-Xylene
Xe is the equilibrium conversion.
Rate law:
p
m m
C
r k C k
! “
#=#
$ %
At equilibrium,
-rm = 0 =>
p
m
e
C
Ck
=
( )
1mo e
mo e
e
C X
C X
K
!=
1
e
e
e
X
K
X
=!
1 1
1 1
1 1 e e e
e e e e
X X X
K X X X
! “
#+#
+ = + = =
$ %
& ‘
01
m A
e
X
r kC
X
! “
#=#
$ %
& ‘
P4-16 (a)
For batch reactor,
Mole balance:
0
0
1
m A
mO A e
r V kC
dX X
dt N C X
! “
#
= = #
$ %
& ‘
ln
e e
X X
k X X
!
” #
=$ %
&
For PFR,
1
1 1
1
1
1 1
ln
AO
m
o
e
PFR
e
e e
PFR
e
dX
V F
r
vdX
kX
K
dX
kX
K
X X
k X X
!
!
=“
=# $
“+
% &
‘ (
=# $
“+
% &
‘ (
=“
)
)
)
P4-16 (b)
For CSTR,
1
1 1
mo
m
CSTR
e
F X
V
r
X
k X
K
!
=“
=# $
% &
“+
‘ (
) *
+ ,
– .
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
( )
2
2
1
yXkC
F
r
dW
dX
O
A
“
=“=
let
Ao
v
kC
z=
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
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
W 0 0 1 1
x 0 0 0.302004 0.302004
alfa 1.043 1.043 1.043 1.043
Differential equations as entered by the user
[1] d(x)/d(W) = Z*((1-x)*y/(1+esp*x))^2
[2] d(y)/d(W) = -alfa*(1+esp*x)/(2*y)
Explicit equations as entered by the user
[1] esp = -0.5
[2] alfa = 1.043
Now for CSTR:
( )
2
1
XX
XF
W
o
+
=
=
“
POLYMATH Results
NLE Solution
Variable Value f(x) Ini Guess
x 0.396566 –1.142E-13 0.5
4-35
NLE Report (fastnewt)
Nonlinear equations
[1] W = 1
[2] esp = -0.5
P4-17 (b)
For turbulent flow:
P
D
G2
)(constant =
!
!
“
#
$
%
&
!
“
#
$
%
&
=
!
!
“
#
$
$
%
&
!
!
“
#
$
$
%
&
=‘2
1
4
12
2
1
2
1
2
1
2
P
P
D
D
G
G
(
(
0326.0
32
1
2==
!
!
and
4
2=
z
z
=>
8.27.04
2=!=“z
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
w 0 0 1 1
x 0 0 0.8619056 0.8619056
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(x)/d(w) = Z*((1-x)*y/(1+esp*x))^2
Explicit equations as entered by the user
[1] esp = -0.5
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)
For turbulent flow
( ) 1
3
2
2
1
3
2
1
2
1
12 00316.0
5.1
1
1
2
018.0 !! =
“
“
#
$
%
%
&
‘
“
#
$
%
&
‘
=
“
“
#
$
%
%
&
‘
“
“
#
$
%
%
&
‘
=kgkg
A
A
D
D
c
c
P
P
((
81.0
2
=
X
P4-19
Production of phosgene in a microreactor.
CO + Cl2
!
COCl2 (Gas phase reaction)
A + B
!
C
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
e -0.5 -0.5 –0.5 -0.5
FB0 2.0E-05 2.0E-05 2.0E-05 2.0E-05
Fb 2.0E-05 4.32E-06 2.0E-05 4.32E-06
v 2.83E-07 2.444E-07 4.714E-07 4.714E-07
Cb 70.671378 9.1638708 70.671378 9.1638708
rA -19.977775 -19.977775 –0.3359061 -0.3359061
Cc 0 0 53.532416 33.259571
ODE Report (RKF45)
Differential equations as entered by the user
Explicit equations as entered by the user
[1] e = -.5 [2] FA0 = 2e-5
[3] FB0 = FA0 [4] Fa = FA0*(1-X)
P4-19 (a)
P4-19 (b)
The outlet conversion of the reactor is 0.784
The yield is then MW*FA*X = 99 g/mol * 2 e-5 mol/s * 0.784 = .00155 g/s = 48.95 g/ year.
P4-19 (c)
Assuming laminar flow, α ~ Dp
-2, therefore
( )
2
5 1 5 1
1
2 1 2
3.55 10 4 14.2 10
P
Dkg kg
! !
” “
= = #=#
4-39
P4-19 (e) Individualized solution
P4-19 (f) Individualized solution
P4-19 (g) Individualized solution
P4-20 (a)
For turbulent flow:
“
=2
#
P
oAo
$
1%
&
( )
“
o=G1#
$
( )
g
%
oDo
$
31.75G
( )
“
#
=constant
DP
“
=
“
oDPo
DP1
4-40
( ) ( ) ( )
0
0
2
0
3
2 0.001
2
10.82126 1 0.35 2.35 20
C C
atm
dm
kg
A P dm atm
dm
!
“# $
% &
‘ (
) *
= =
+% &
+‘ (
) *
5 1
08.0 *10 kg
!
” “
=
P4-20 (b)
See Polymath program P4-20-b.pol.
P4-20 (c)
See Polymath program P4-20-c.pol.