8-21
P8-5
2A B C+!
A
B
C
Fio
lb mole
hr
!
” #
$ %
& ‘
10
10
0.0
Tio(F)
80
80
–
~
Pio
Btu
C
lb mole F
! “
# $
°
% &
51
44
47.5
,lb
MW
lb mol
! “
# $
% &
128
94
222
,3
i
lb
!
” #
$ %
63
67.2
65
20, 000
R
Btu
H
lb mol A
!=
,
Energy balance with work term included is:
[ ]
0
0
0
10
1, 1, 1
10
( )
S
A R i Pi o
A
B
A B AF
A
Q W X H C T T
F
FX
F
Q UA Ts T
!
! !
“” # =$ “
= = = = =
=“
&&%
&
Substituting into energy balance,
[ ]
{ }
[ ]
0 0 0
0 0 0
0
0
0
( )
( )
( )
63525
199
S S A R AF A pA pB
S S A R A pA pB
S S A R
A pA pB
s
UA T T W F H X F C C T T
UA T T W F H F C C UA T T
UA T T W F H
T T F C C UA
Btu
Whr
T F
! “
# # # $ = + #
% &
! “
‘ # # # $ = + + #
% &
# # # $
= + ! “
+ +
% &
#=
(= °
P8-6
A B C+!
Since the feed is equimolar, CA0 = CB0 = .1 mols/dm3
Adiabatic:
0
0
[ ( )]
30 15 15 0
i
R
i P P
P pC pB pA
X H T
T T C X C
C C C C
!
“#
= + +#
#=” “ =” “ =
$% %
( )
R C B A
H T H H H!=” “
= -41000-(-15000-(-20000) = -6000 cal/mol A
15 15 30
i i pA B pB
cal
C C C mol K
! !
= + = + =
“%
6000
300 300 200
30
X
T X= + = +
2 2 2
0(1 ) .01 (1 )
A A
r k C X k X!=!=!
P8-6 (a)
0
0
PFR A
A
A
CSTR
A
dX
V F
r
F X
V
r
=!
=!
“
For the PFR, FA0 = CA0v0 = (.1)(2) = .2 mols/dm3
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
X
0
0
0.85
0.85
2
V
0
0
308.2917
308.2917
3
Ca0
0.1
0.1
0.1
0.1
4
Fa0
0.2
0.2
0.2
0.2
1
d(V)/d(X) = -Fa0 / ra
Explicit equations
1
Ca0 = .1
2
Fa0 = .2
3
T = 300 + 200 * X
V = 308.2917dm3
8-24
For the CSTR,
k = 4.31 (Using T = 470K in the formula).
-rA
= .000971 mol/dm3/s
03
-4
.1 2 .85 175 dm
A
F X
V
! !
= = =
The reason for this difference is that the temperature and hence the rate of reaction remains
constant throughout the entire CSTR (equal to the outlet conditions), while for a PFR, the rate
P8-6 (b)
0
[ ]
i
R
i P
X H
T T
C
!
“#
= + $
For boiling temp of 550 k,
550 = T0 + 200
T0 = 350K
P8-6 (c)
P8-6 (d)
0
0
( )
A
CSTR
A
CSTR
A
A
F X
V
r
V
X r
F
=!
“=!
For V = 500 dm3, FA0=.2
Now use Polymath to solve the non-linear equations.
See Polymath program P8-6-d-1.pol.
8-25
Calculated values of NLE variables
Variable
Value
f(x)
Initial Guess
1
T
484.4136
0
480.
2
X
0.9220681
-2.041E-09
0.9
Variable
Value
1
k
6.072856
2
ra
0.0003688
Nonlinear equations
1
f(T) = 300 + 200 * X – T = 0
2
f(X) = 500 – .2 * X / ra = 0
Explicit equations
1
k = .01 * exp(10000 / 1.98 * (1 / 300 – 1 / T))
2
ra = 0.01 * k * (1 – X) ^ 2
Hence, X = .922 and T = 484.41 K
For the conversion in two CSTR’s of 250 dm3 each,
For the first CSTR, using the earlier program and V = 250 dm3,
Calculated values of NLE variables
Variable
Value
f(x)
Initial Guess
1
T
476.482
1.137E-13
480.
2
X
0.88241
-5.803E-09
0.9
Variable
Value
1
k
5.105278
2
ra
0.0007059
Nonlinear equations
1
f(T) = 300 + 200 * X – T = 0
2
f(X) = 250 – .2 * X / ra = 0
Explicit equations
1
k = .01 * exp(10000 / 1.98 * (1 / 300 – 1 / T))
2
ra = 0.01 * k * (1 – X) ^ 2
8-26
0 1
1
0
( )
( )
A
CSTR
A
CSTR
A
A
F X X
V
r
V
X r X
F
!
=!
“=!+
See Polymath program P8-6-d-2.pol.
Calculated values of NLE variables
Variable
Value
f(x)
Initial Guess
1
T
493.8738
0
480.
2
X
0.9693688
-1.359E-09
0.8824
Variable
Value
1
k
7.415252
2
ra
6.958E-05
3
X1
0.8824
Nonlinear equations
1
f(T) = 476.48 + 200 * (X – X1) – T = 0
Explicit equations
1
k = .01 * exp(10000 / 1.98 * (1 / 300 – 1 / T))
2
ra = 0.01 * k * (1 – X) ^ 2
P8-6 (e) Individualized solution
P8-6 (f) Individualized solution
P8-7 (a)
For reversible reaction, the rate law becomes
C
A A B
C
C
r k C C
K
! “
#=#
$ %
& ‘
, 1 1
200( )
out CSTR
T T X X= + !
8-27
300 200
1 1
(300) exp 300
1 1
(450) exp 450
Rxn
C C
T X
E
k k
R T
H
K K
R T
= +
! “
! “
=#
$ %
$ %
& ‘
& ‘
(
) *
! “
=#
$ %
+ ,
& ‘
– .
Stoichiometry:
0
0
0
(1 )
(1 )
C A
A A
B A
C C X
C C X
C C X
=
=!
=!
See Polymath program P8-7-a.pol.
POLYMATH Results
No Title 03-21-2006, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 10 10
X 0 0 0.0051176 0.0051176
T 300 300 301.02352 301.02352
k 0.01 0.01 0.010587 0.010587
Ca0 0.1 0.1 0.1 0.1
ra -1.0E-04 -1.048E-04 -1.0E-04 -1.048E-04
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
Explicit equations as entered by the user
[1] T = 300+200*X
[2] k = .01 * exp((10000 / 1.987) * (1 / 300 – 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] Kc = 10 * exp(-6000 /1.987 * (1 / 450 – 1 / T))
8-28
299.8
300
300.2
300.4
300.6
300.8
301
301.2
0 1 2 3 4 5 6 7 8 9 10
V [m3]
T [K]
0
0.001
0.002
0.003
0.004
0.005
0.006
0 1 2 3 4 5 6 7 8 9 10
V [m3]
X
0.827
0.8275
0.828
0.8285
0.829
0.8295
0.83
012345678910
V [m3]
Xe
P8-7 (b)
When heat exchanger is added, the energy balance can be written as
[ ]
0
( ) ( ) ( )
ˆ
( )
a A Rxn
A i pi P
Ua T T r H T
dT
dV F C C
!
“+” “#
=+#
$
So with
ˆ
P
C!
= 0,
30
i pi
C
!
=
“
,
Rxn
H!
= -6000 cal/mol
[ ]
0
( ) ( ) 6000
(30)
a A
A
Ua T T r
dT
dV F
!+!
=
Where Ua = 20 cal/m3/s/K, Ta = 450 K
See Polymath program P8-7-b.pol.
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 10 10
X 0 0 0.3634806 0.3634806
T 300 300 455.47973 450.35437
k 0.01 0.01 3.068312 2.7061663
Fa0 0.2 0.2 0.2 0.2
ra -1.0E-04 -0.0221893 -1.0E-04 -0.0010758
DH -6000 -6000 -6000 -6000
Ua 20 20 20 20
8-29
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 1.987) * (1 / 300 – 1 / T))
[2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 – 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 – X) ^ 2 – X /Ca0/ Kc)
[6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
[7] DH = -6000
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 10 10
X 0 0 0.3611538 0.3611538
T 300 300 442.15965 442.15965
Ta 450 434.90618 450 441.60853
k 0.01 0.01 2.1999223 2.1999223
Kc 286.49665 11.263546 286.49665 11.263546
Ca0 0.1 0.1 0.1 0.1
Xe 0.8298116 0.4023362 0.8298116 0.4023362
Ua 20 20 20 20
Fao 0.2 0.2 0.2 0.2
sumcp 30 30 30 30
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 1.987) * (1 / 300 – 1 / T))
[2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 – 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 – X) ^ 2 – X /Ca0/ Kc)
[6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
[7] DH = -6000
250
300
350
400
450
500
0 1 2 3 4 5 6 7 8 9 10
V [m3]
Ta
T
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0 1 2 3 4 5 6 7 8 9 10
V [m3]
Xe
X
8-31
P8-7 (d)
For counter-current flow,
See Polymath program P8-7-d.pol.
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 10 10
X 0 0 0.3647241 0.3647241
T 300 300 463.44558 450.37724
Ta 440.71 440.71 457.98124 450.00189
Fa0 0.2 0.2 0.2 0.2
ra -1.0E-04 -0.0256436 -1.0E-04 -9.963E-04
Xe 0.8298116 0.3488462 0.8298116 0.381006
DH -6000 -6000 -6000 -6000
sumcp 30 30 30 30
mc 50 50 50 50
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 1.987) * (1 / 300 – 1 / T))
[2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 – 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 – X) ^ 2 – X /Ca0/ Kc)
[11] mc = 50
[12] Cpc = 1
250
300
350
400
450
500
0 1 2 3 4 5 6 7 8 9 10
V [m3]
Ta
T
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0 1 2 3 4 5 6 7 8 9 10
V [m3]
Xe
X
8-32
P8-7 (e)
We see that it is better to use a counter-current coolant flow as in this case we achieve the
P8-7 (f)
If the reaction is irreversible but endothermic, we have
2 2 2
0(1 ) .01 (1 )
A A
r k C X k X!=!=!
as obtained in the earlier problem.
6000 /
Rxn
H cal mol!=
See Polymath program P8-7-f-co.pol.
we use 8- 7f cocurrent.pol
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 10 10
X 0 0 0.4016888 0.4016888
T 300 300 428.84625 424.16715
Ta 450 425.45941 450 425.45941
k 0.01 0.01 1.4951869 1.314808
Fa0 0.2 0.2 0.2 0.2
DH 6000 6000 6000 6000
sumcp 30 30 30 30
mc 50 50 50 50
Cpc 1 1 1 1
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 2) * (1 / 300 – 1 / T))
[2] Ca0 = 0.1
[3] Fa0 = 0.2
[10] Cpc = 1
8-33
For counter-current flow,
See Polymath program P8-7-f-counter.pol.
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 10 10
X 0 0 0.3458817 0.3458817
T 300 300 449.27319 449.27319
Ta 423.8 423.8 450.01394 450.01394
Fa0 0.2 0.2 0.2 0.2
ra -1.0E-04 -0.0141209 -1.0E-04 -0.0019877
DH 6000 6000 6000 6000
sumcp 30 30 30 30
mc 50 50 50 50
Cpc 1 1 1 1
8-34
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 2) * (1 / 300 – 1 / T))
[2] Kc = 10 * exp(6000 / 2 * (1 / 450 – 1 / T))
[3] Fa0 = 0.2
P8-7 (g)
For a runaway reaction, the following must be true:
2
r
r C
RT
T T
E
!>
and
0300 3* 450 412.5
1 1 3
a
C
T T
T
!
!
++
= = =
+ +
So if we plug this value into the original equation we get:
2
1.987 450 0
10000 r r
T T!+ >
Tr = 499 K
P8-8 (a)
A B C+!
Species Balance:
0
3
0
0
20 /
10
A
A
r
dX
dW F
v dm s
P atm
!
=“
=
=
Stoichiometry:
0
0
0
0
1, 1
1
1
1
A A
A A
X T
C C where
X T
X T
C C
X T
!
!
“
# $
= =
% &
+
‘ (
“
# $
)=% &
+
‘ (
Rate Law is:
Explicit equations
1
T = 450 + 500 * X
P8-8 (b)
Species Balance for CSTR:
0
450 500 450 500(.8) 850
31400 1 1
.133exp 6.9
8.314 450 850
39.42
A
CSTR
A
CSTR
F X
Wr
T X K
k
W kg
=!
“
= + = + =
# $
% &
=“=
‘ (
) *
+ ,
– .
=
P8-8 (c) Individualized solution
P8-8 (d)
For pressure drop, an extra equation is added
( )
0
0 0
0
0 0
1
2 ( / )
1
1
A A
P
dP T X
dW T P P
X T P
C C
X T P
!“
# $
=%+
& ‘
( )
%
# $
=& ‘
+
( )
See Polymath program P8-8-d.pol.
Using POLYMATH program CRE_8_8d.pol
For
.019
!
=
8-37
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
W
0
0
0.8
0.8
2
X
0
0
0.0544753
0.0544753
3
P
1.013E+06
1.002E+06
1.013E+06
1.002E+06
4
T
450.
450.
850.
850.
5
v0
20.
20.
20.
20.
7
k
0.133
0.133
6.904332
6.904332
8
P0
1.013E+06
1.013E+06
1.013E+06
1.013E+06
9
alpha
0.019
0.019
0.019
0.019
Differential equations
1
d(X)/d(W) = k / v0 * (1 – X) / (1 + X) * T0 / T * P / P0
2
d(P)/d(W) = –alpha / 2 * (T / T0) * P0 ^ 2 / P * (1 + X)
Explicit equations
1
T = 450 + 500 * W
2
v0 = 20
3
T0 = 450
5
P0 = 1013250
6
alpha = .019
P8-9 (a)
We use the same equations as problem P8-8, except that the energy balance changes as:
( )
0
( )( )
a A Rxn
A pA
Ua T T r H
dT
dW F C
!
“
#+# #$
=
Where
Rxn
H!“
= 20,000 J/mol, Ta=323 K, CpA=40 J/mol/K
8-38
See Polymath program P8-9-a.pol.
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
W
0
0
50.
50.
2
X
0
0
0.1376181
0.1376181
3
T
450.
381.1888
450.
381.1888
4
T0
450.
450.
450.
450.
6
k
0.133
0.0292331
0.133
0.0292331
7
Uarho
0.08
0.08
0.08
0.08
8
Ta
293.
293.
293.
293.
9
P0
1.013E+06
1.013E+06
1.013E+06
1.013E+06
11
CA
270.8283
242.3648
270.8283
242.3648
12
rA
-36.02017
-36.02017
-7.085084
-7.085084
Differential equations
1
d(X)/d(W) = k * (1 – X) / (1 + X) * T0 / T / v0
2
d(T)/d(W) = (Uarho * (Ta – T) + rA * 20000) / v0 / CA0 / 40
Explicit equations
1
T0 = 450
2
v0 = 20
3
k = 0.133 * exp(31400 / 8.314 * (1 / T0 – 1 / T))
6
P0 = 1013250
7
CA0 = P0 / 8.314 / T0
8
CA = CA0 * (1 – X) / (1 + X) * T0 / T
8-39
If
b
UA
!
was increased by a factor of 3000, we use the same program with the new value. The
profiles are in the graphs below.
P8-9 (b)
For non-constant jacket temperature, the equation for incorporating the flow needs to be
introduced.
co-current: Ta0 = 50 °C
( )
a
a
C pC
Ua T T
dT
dW m C
!
“
=&
countercurrent:
( )
a
a
C pC
Ua T T
dT
dW m C
!
“
=“&
Taf = 50 °C
P8-9 (c)
For a fluidized CSTR with W = 80 kg, UA = 500 J/s/K,
Species balance:
XMB =
“
k
1+
“
k,
“
=
W
#
b
CA0
FA0