8-41
9
P0
1.013E+06
1.013E+06
1.013E+06
1.013E+06
10
CA0
270.8283
270.8283
270.8283
270.8283
11
CA
270.8283
258.1071
270.8283
258.1071
12
kr
0.2
0.076962
0.2
0.076962
13
CC
0
0
15.77362
15.77362
14
CB
0
0
15.77362
15.77362
15
rA
-36.02017
-36.02017
-0.0062421
-0.0062421
Differential equations
1
d(X)/d(W) = –rA / v0 / CA0
2
d(T)/d(W) = (Uarho * (Ta – T) + rA * 20000) / v0 / CA0 / 40
3
d(Ta)/d(W) = Uarho * (T – Ta) / .2 / 5000
Explicit equations
1
T0 = 450
2
v0 = 20
3
k = 0.133 * exp(31400 / 8.314 * (1 / T0 – 1 / T))
5
P0 = 1013250
6
CA0 = P0 / 8.314 / T0
7
CA = CA0 * (1 – X) / (1 + X) * T0 / T
9
CC = CA0 * X / (1 + X) * T0 / T
10
CB = CA0 * X / (1 + X) * T0 / T
11
rA = -(k * CA – kr * CB * CC)
P8-9 (e) Individualized solution
P8-10 (a)
A
A T
T
I
I
A
A B C
F
C C
F
F
F
!
“+
=
=
T A I
T A I
C C C
F F F
= +
= +
( ) 0
01 0 0
0 0
0 0
01 1
A
A A I
A I
A I
A
I
F
C C C
F F
C C
C
!
= + +
+
=+
P8-10 (b)
Mole balance:
0
A
A
r
dX
dV F
!
=
Rate law:
A A
r kC!=
Stoichiometry:
0
01
1
1
A A
T
X
C C
X T
!
“
=+
0A
y
! “
=
1 1 1 1
!
= + “=
0 0
0
0 0 0
1
1
A A
A
T A i i
F F
yF F F
!
= = =
+ +
1
1i
!“
=+
( ) 0RX PA i Pi
PA i Pi
X H C C T
T
C C
!
!
” # + +
=+
Enter these equations into Polymath
See Polymath program P8-10-b.pol.
8-43
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 500 500
X 0 0 0.417064 0.417064
Cao 0.0221729 0.0221729 0.0221729 0.0221729
Cio 0.0221729 0.0221729 0.0221729 0.0221729
theta 100 100 100 100
To 1100 1100 1100 1100
dHrx 8.0E+04 8.0E+04 8.0E+04 8.0E+04
ra -0.0110894 -0.0110894 -0.0061524 -0.0061524
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra/Fao
Explicit equations as entered by the user
[1] Cao = 2/.082/1100
[2] Cio = Cao
[5] Cao1 = (Cao+Cio)/(theta+1)
[6] e = 1/(1+theta)
[7] To = 1100
[8] dHrx = 80000
[9] Cpa = 170
[10] Cpi = 200
[11] T = (X*(-dHrx)+(Cpa+theta*Cpi)*To)/(Cpa+theta*Cpi)
8-44
P8-10 (c)
There is a maximum at θ = 8. This is because when θ is small, adding inerts keeps the
temperature low to favor the endothermic reaction. As θ increases beyond 8, there is so much
more inert than reactants that the rate law becomes the limiting factor.
P8-10 (d)
The only change to the Polymath code from part (b) is that the heat of reaction changes sign. The
new code is not shown, but the plots are below.
See Polymath program P8-10-d.pol.
8-45
The maximum conversion occurs at low values of theta (θ < 8) because the reaction is now
P8-10 (e)
We need to alter the equations from part (c) such that
2
A A
r kC!=
and CA0 = 1
A plot of conversion versus theta shows a maximum at about θ = 5.
See Polymath program P8-10-e.pol.
P8-10 (f)
We need to alter the equations from part (c) such that
B C
A A
C
C C
r k C
K
! “
#=#
$ %
& ‘
We already know that
0
0
1
1
A A
T
X
C C
X T
!
“
=+
. Now weed need expressions for CB and CC. From
stoichiometry we can see that CB = CC. In terms of CA0 we find that:
0
01
B C A
T
X
C C C
X T
!
= = +
We also need an equation for KC:
1
1
1 1 80000 1 1
exp 2exp 8.314 1100
RX
C C
H
K K
R T T T
! “
# $ ! “
%# $
=&=&
‘ (
) * ) *
‘ (
+ ,
– .
+ ,
– .
When we enter these into Polymath we find that the maximum conversion is achieved at
approximately θ = 8.
See Polymath program P8-10-f.pol.
8-46
P8-10 (g)
See Polymath program P8-10-g.pol.
P8-11 (a)
Start with the complete energy balance:
ˆ
S i i in i i out
dE Q W E F E F
dt =! ! “ ! “
&&
The following simplifications can be made:
It is steady state.
In part a, there is no heat taken away or added
There is no shaft work
Simplifying,
0 0 0 0 0 0
( ) ( ) ( ) 0
A A A B B B C C C A A B B C C
H F R V H F R V H F R V H F H F H F+ + + + + ! ! ! =
If only C is diffusing out of the reactor we get:
8-47
( ) 0 0 0 0 0 0 0
A A B B C B B A A B B C C
H F H F H F R V H F H F H F+ + + ! ! ! =
Now we evaluate Fi
0 0A A A
F F F X=!
0 0B B A
F F F X= +
0 0C C A C
F F F X R V= + !
Inserting these into our equation gives:
0 0 0 0 0 0 0 0 0 0 0 0 0
A A A A B B B A C C C A A A B B C C
H F H F X H F H F X H F H F X H F H F H F!+ + + + ! ! ! =
( )
0 0 0 0
A PA A RX
F C T T F X H!+“=
Differentiating with respect to V with ΔCP = 0
( )
( )
0 0 0
A p A Rx
dT dX
F C F H T
dV dV
+!=
( )
( )
A Rx
i p
r H T
dT
dV C
!
“
# $
% &
=‘
Combine that with the mole balance and rate law:
0 0 0
0 0 0
C
A B
A A A C C
A A
C
A B
A T B T C T
T T T
dF
dF dF
r r r k C
dV dV dV
r kC
T T F T
F F
C C C C C C
F T F T F T
= = !=! !
!=
= = =
For kc = 10
See Polymath program P8-11-a.pol.
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
V
0
0
50.
50.
2
Fc
0
0
0.0012968
0.0005261
3
Fb
0
0
0.1978837
0.1978837
4
Fa
5.42
5.222116
5.42
5.222116
5
T
450.
450.
548.9418
548.9418
6
Ft
5.42
5.42
5.421297
5.420526
7
dHrx
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
8
k
0.133
0.133
0.6036997
0.6036997
9
kc
10.
10.
10.
10.
10
Cto
2.710027
2.710027
2.710027
2.710027
8-48
11
Ca
2.710027
2.610831
2.710027
2.610831
12
Kc
0.0006905
0.0002635
0.0006905
0.0002635
13
Cc
0
0
0.0006482
0.000263
14
Cb
2.710027
2.610831
2.710027
2.610831
16
Cpa
40.
40.
40.
40.
Differential equations
1
d(Fc)/d(V) = -ra – kc * Cc
2
d(Fb)/d(V) = –ra
3
d(Fa)/d(V) = ra
Explicit equations
1
Ft = Fa + Fb + Fc
2
dHrx = –20000
3
k = .133 * exp((31400 / 8.314) * (1 / 450 – 1 / T))
4
kc = 10
5
Cto = 100 / .082 / 450
6
Ca = Cto * Fa / Ft
7
Kc = .01 * exp((dHrx / 8.314) * (1 / 300 – 1 / T))
10
ra = -k * (Ca – Cb * Cc / Kc)
11
Cpa = 40
vary kc to see how the concentration profiles change.
8-49
P8-11 (b)
Now, the hear balance equation needs to be modified.
( )
0
0
( ) ( )
a A A Rx
A i p
Ua T T F r H T
dT
dV F C
!
“+#
$ %
& ‘
=(
See Polymath program P8-11-b.pol.
P8-12(a)
To find the necessary heat removal, we start with the isothermal case of the energy balance
( ) ( )
0
0 0
S
RX P R i Pi
A A
W
QX H C T T C T T
F F
!
” #
$ $ % +% $ =$
& ‘ (
( )( )
25 35 350 300
7500
0.4
RX
kJ
H
mol
+!
“= = !
!
Now go back to the isothermal case:
( )
3
03
1000 .5 0.2 7500
min
A RX
mol m kJ
Q C vX H m mol
! “
! “ ! “
=#=$
% &
% & % &
‘ ( ‘ (
‘ (
750000
min
kJ
Q=!
P8-12(b)
We start with the energy balance for the second CSTR (already simplified):
( ) ( ) ( )( )
2 1 0
0
a RX PA PB
A
UA T T X X H C C T T
F! ! ! “ = + !
8-51
POLYMATH Results
NLES Solution
Variable Value f(x) Ini Guess
T 327.68712 -2.274E-13 340
UA 4
X1 0.2
dHrx -7500
Cao 1000
NLES Report (safenewt)
Nonlinear equations
[1] f(T) = (UA)*(Ta-T)/Fao-(X2-X1)*dHrx-60*(T-To) = 0
[2] f(X2) = V-Fao*(X2-X1)/(-ra) = 0
Explicit equations
[1] UA = 4
[2] Ta = 350
[3] vo = 0.5
[4] X1 = 0.2
[5] dHrx = -7500
[6] To = 300
[7] V = 1
P8-12(c)
Now we need the differential form of the energy balance
( ) ( )
( )
0 0
a A RX a A RX
A i Pi A PA PB
Ua T T r H Ua T T r H
dT
dV F C F C C
!
” “ # ” “ #
= = +
$
we also need the mole balance for a PFR. For this case it simplifies to:
A B
A
dC dC r
dt dt
= = !
with
A A B
r kC C!=
and we can use the same equation for k as in part (b).
When we put these equations into Polymath we get an outlet conversion of X = 0.33
See Polymath program P8-12-c.pol.
POLYMATH Results
Calculated values of the DEQ variables
8-52
Variable initial value minimal value maximal value final value
V 0 0 1 1
T 300 283.98681 300 283.98681
X 0.2 0.2 0.3281298 0.3281298
Cao 1000 1000 1000 1000
Ua 10 10 10 10
Fao 500 500 500 500
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(V) = (Ua*(Ta-T)-ra*dHrx)/(Fao*(Cpa+Cpb))
[2] d(X)/d(V) = -ra/Fao
[1] Cao = 1000
[2] Ua = 10
[3] Ta = 300
[4] dHrx = -7500
[5] Cb = Cao*(1-X)
[6] v = 0.5
[7] Fao = Cao*v
P8-12(d)
In this case we need to replace the rate law we used in part (c)
C
A A B
C
C
r k C C
K
! “
#=#
$ %
& ‘
We also need an equation to calculate Kc at different temperatures.
1
1
1 1
exp RX
C C
H
K K
R T T
! “
# $
%
=&
‘ (
) *
+ ,
– .
be careful of the units when entering Kc1 into Polymath. Also note that the initial temperature is
different than in part (c)
8-53
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 1 1
T 350 314.93211 350 314.93211
X 0.2 0.2 0.4804694 0.4804694
R 0.0083144 0.0083144 0.0083144 0.0083144
Ua 10 10 10 10
Ta 300 300 300 300
Fao 500 500 500 500
Ca 800 519.53058 800 519.53058
Cb 800 519.53058 800 519.53058
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(V) = (Ua*(Ta-T)-ra*dHrx)/(Fao*(Cpa+Cpb))
[2] d(X)/d(V) = -ra/Fao
Explicit equations as entered by the user
[1] R = 8.3144/1000
[2] Ua = 10
[5] Cao = 1000
[6] v = 0.5
[7] Fao = Cao*v
[8] Cpa = 25
[9] Cpb = 35
[10] k = 0.00015625*exp(2664*(1/300-1/T))
P8-12(e) Individualized solution
P8-12(f)
For the gas phase the only the stoichiometry changes.
0
0
1
1
A A
T
X
C C
X T
!
“# $
# $
=% &% &
+
‘ (‘ (
and
( )
00.5 1 1 1 0.5
A
y
! “
= = # # =#
8-54
See Polymath program P8-12-f.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 1 1
T 300 279.3717 300 279.3717
X 0.2 0.2 0.3650575 0.3650575
To 300 300 300 300
Ua 10 10 10 10
Fao 500 500 500 500
Ca 888.88889 834.06666 888.88889 834.06666
ra -123.45679 -123.45679 -56.423752 -56.423752
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(V) = (Ua*(Ta-T)-ra*dHrx)/(Fao*(Cpa+Cpb))
[2] d(X)/d(V) = -ra/Fao
Explicit equations as entered by the user
[1] To = 300
[2] Ua = 10
[3] Ta = 300
[6] vo = 0.5
[7] Fao = Cao*vo
[8] Cpa = 25
[9] Cpb = 35
[10] k = 0.00015625*exp(2664*(1/300-1/T))
P8-13
( )
2
2
1
1
C D e
C
A B e
C
e
C
C C X
K
C C X
K
X
K
= =
!
“=+
( )
( )
0
30000
300 300 600
25 25
A B
R
P P
H X
T T X X
C C
!
“
=!=!= +
+ +
8-55
See Polymath program P8-13.pol.
Calculated values of NLE variables
Variable
Value
f(x)
Initial Guess
1
Xe
0.9997644
3.518E-11
0.5 ( 0 < Xe < 1. )
Variable
Value
1
T
300.
2
Kc
1.8E+07
Nonlinear equations
1
f(Xe) = Xe – (1 – Xe) * Kc ^ 0.5 = 0
Explicit equations
1
T = 300
2
Kc = 500000 * exp(-30000 / 1.987 * (1 / 323 – 1 / T))
T
X
300
1
320
0.999
340
0.995
380
0.935
420
0.76
460
0.4
500
0.1529
540
0.035
Xe
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
300 350 400 450 500 550
Temp (K)
Xe (Eqm Conversion)
Xe
P8-14
For first reactor,
1
1
1
or
1 1
e C
C e
e C
X K
K X
X K
= =
!+
For second reactor
For 3rd reactor
3 3 3
3
3
1 1
B e C B
C e
e C
X K
K orX
X K
! !
+“
= =
“+
1st reactor: in first reaction Xe = 0.3
So, FB = FA01(.3)
( ) ( )
( )( )
01
2
0
02 0 02
0
1.7
0
i
A
B
A
A P i A R
pA B pB
R
F
F C T T F X H
C C T T
XH
!
!
!
= =
“ # ” +“$ =
+“
=“$
Slope is now negative
3rd reactor:
Say
e2
B 01 02 01
03 01 02 2 01 01 2 01
03 01
X = 0.3
(.2 ) .3 (.2 .3 1.8)
(1 ) 1.8 (1 ) 1 1.8(1 .3)
2.26
A A A
A A A e A A e A
A A
F F F
F F F X F F X F
F F
!
= + = + “
= + #= + #= + #
=
Feed to the reactor 3:
8-57
P8-14 (b)
The same setup and equations can be used as in part (a). The entering temperature for reactor 1 is
now 450 K and the outlet is 520 K. When the two streams are joined prior to entering reactor 2
the temperature is (520+450)/2 = 485 K
Say that the outlet temperature for reactor 2 is 510 K. Then the entering temperature for reactor 3
For any reactor j,
( ) ( )
( )( )
0 0 0
0
0
i
A j P i A j R
pA B pB
R
F C T T F X H
C C T T
XH
!
!
“ # ” +“$ =
+“
=“$
and θB for reactor 1 = 0. For reactor 2, θB > 0. This means that the slope of the conversion line
from the energy balance is larger for reactor 2 than reactor 1. And similarly θB for reactor 3 > θB
for reactor 2. So the line for conversion in reactor 3 will be steeper than that of reactor 2. The
8-58
P8-15 (a)
Substrate More cells + Product
S C + P
G(T) = X*–∆HRX
To solve for G(T) we need X as a function of temperature, which we get by solving the mass
balance equation.
/ 0c s S
g
Y F X
Vr
=
g C
r C
µ
=
and
( ) S
S S
C
T
K C
µ µ
=+
where
1max
6700
0.0038* exp 21.6
( ) 48000
1 exp 153
T
T
T
T
µ µ
! “
# $
%
& ‘
( )
* +
( )
=# $
+%
& ‘
( )
* +
, –
if we combine these equations we get:
/ 0
( )
c s S
C S
S S
Y F X
VT C C
K C
µ
=
+
( )
/ 0
( )
c s S S S
C S
Y F X K C
V
T C C
µ
+
=
Canceling and combining gives:
Now solve this expression for X:
8-59
0
2
0 0 0
1( )
S S
S S S
F K
X
T VC F C
µ
=!!
Now that we have X as a function of T, we can plot G(T).
To get R(T) we must calculate the heat removed which is the sum of the heat absorbed by
reactants to get to the reaction temperature and the heat removed from any heat exchangers.
( ) ( )
0
0
( ) PS a
S
UA
R T C T T T T
F
=!+!
Now enter the equations into polymath and specify all other constants. The adiabatic case is
shown below. The non-adiabatic case would be with explicit equation [12] as A = 1.1.
See Polymath program P8-15-a.pol.
Differential equations as entered by the user
[1] d(T)/d(t) = 1 Adiabatic Case
Explicit equations as entered by the user
[1] mumax = .5
[2] Ycs = .8
[3] vo = 1
[4] Ta = 290
[10] Cps = 74
[11] dH = -20000
[12] UA = 0*300
[13] kappa = UA/(Cps*Fso)
[14] To = 280 Non-Adiabatic Case
8-60
P8-15 (b)
To maximize the exiting cell concentration, we want to maximize the conversion of substrate. If
we look at G(T) from part A, we see that it is at a maximum at about 310 K. This corresponds to
the highest conversion that can be achieved. By changing the values of UA and mc we can change
Since we now have a limited coolant flow rate we will use a different value for Q.
( ) !
!
“
#
$
$
%
&
!
!
“
#
$
$
%
&‘
‘‘=
PCC
aPCC Cm
UA
TTCmQ exp1
and so,
( ) ( )
0
exp1)( TTC
Cm
UA
TTCmTR PS
PCC
aPCC !+
“
“
#
$
%
%
&
‘
“
“
#
$
%
%
&
‘!
!!=
Now we set R(T) equal to the maximum value of G(T) which is 15600 J/h
And now plug in the known values. Assume the maximum coolant flow rate and that will give the
minimum heat exchange area.
( ) ( )
KK
Kg
J
h
g
Kg
J
hr
g
UA
KK
Kg
J
hr
g
hr
g
g
J
28831074100
7460000
exp12903107460000
10015600
!
“
“
#
$
%
%
&
‘
“
#
$
%
&
‘
+
“
“
“
“
“
#
$
%
%
%
%
%
&
‘
“
“
“
“
“
#
$
%
%
%
%
%
&
‘
“
“
#
$
%
%
&
‘
“
#
$
%
&
‘
!
!!
“
“
#
$
%
%
&
‘
“
#
$
%
&
‘
=
“
#
$
%
&
‘
!
“
#
$
%
&
+
!
!
!
!
!
“
#
$
$
$
$
$
%
&
!
!
!
!
!
“
#
$
$
$
$
$
%
&
!
!
“
#
$
$
%
&
‘
‘
!
“
#
$
%
&
=hr
J
Khr
J
UA
hr
J
hr
J162800
4440000
exp1888000001560000
Khr
J
UA 70415=
A=1.4 m3
P8-15 (c)
There are two steady states for this reaction. There is an unstable steady state at about 294.5 K
P8-15 (d)
Increasing T0 enough will eliminate the lower temperature steady state point. It will also lower