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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
Differential equations
1
d(X)/d(W) = -rA / v0 / CA0
Explicit equations
1
T0 = 450
2
v0 = 20
P8-9 (e) Individualized solution
8-42
P8-10 (a)
A
A T
A B C
F
C C
F
"+
=
P8-10 (b)
Mole balance:
0
A
A
r
dX
dV F
!
=
Rate law:
A A
r kC!=
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
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
P8-10 (c)
There is a maximum at θ = 8. This is because when θ is small, adding inerts keeps the
P8-10 (d)
The only change to the Polymath code from part (b) is that the heat of reaction changes sign. The
See Polymath program P8-10-d.pol.
The maximum conversion occurs at low values of theta (θ < 8) because the reaction is now
inerts as there was in the endothermic case.
P8-10 (e)
We need to alter the equations from part (c) such that
2
A A
r kC!=
and CA0 = 1
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
r k C
! "
#=#
$ %
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 =! ! " ! "
&&
( ) 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
and note that FB0 = FC0 = 0
0 0 0 0 0 0 0
A A A A B A C A A A
H F H F X H F X H F X H F!+ + !=
and combining and substituting terms
gives:
( )
0 0 0 0
A A A A RX
F H H F X H!+"=
( )
0 0 0 0
A PA A RX
F C T T F X H!+"=
Combine that with the mole balance and rate law:
C
A B
A A A C C
A A
dF
dF dF
r r r k C
dV dV dV
r kC
= = !=! !
!=
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
8-48
11
Ca
2.710027
2.610831
2.710027
2.610831
12
Kc
0.0006905
0.0002635
0.0006905
0.0002635
Differential equations
1
d(Fc)/d(V) = -ra - kc * Cc
Explicit equations
1
Ft = Fa + Fb + Fc
2
dHrx = -20000
8-49
P8-11 (b)
Now, the hear balance equation needs to be modified.
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
!
" #
$ $ % +% $ =$
& ' (
8-50
Now go back to the isothermal case:
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! ! ! " = + !
This equation also brings in another unknown: k. We know that the specific reaction rate is
dependant on temperature and if we have the activation energy, we can make an implicit equation
for k as a function of T. To calculate the activation energy we will use the isothermal and
Solving for k at 300 and 350 K gives:
k(300 K) = 0.00015625
k(350 K) = 0.0005555
8-51
POLYMATH Results
NLES Solution
Variable Value f(x) Ini Guess
T 327.68712 -2.274E-13 340
X2 0.4214731 -6.666E-12 0.4
UA 4
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
P8-12(c)
Now we need the differential form of the energy balance
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
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] Cao = 1000
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
! "
#=#
$ %
& '
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
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
P8-12(e) Individualized solution
P8-12(f)
For the gas phase the only the stoichiometry changes.
( )
00.5 1 1 1 0.5
A
From Polymath we see the exiting conversion is X = 0.365
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
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
P8-13
( )
2
2
1
C D e
C
A B e
C C X
K
C C X
= =
!
8-55
See Polymath program P8-13.pol.
Calculated values of NLE variables
Variable
Value
f(x)
Initial Guess
Nonlinear equations
1
f(Xe) = Xe - (1 - Xe) * Kc ^ 0.5 = 0
Explicit equations
1
T = 300
T
X
300
1
320
0.999
340
0.995
Xe
0.7
0.8
0.9
1
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
( ) ( )
( )( )
01
2
0
02 0 02
0
.2 .12
1.7
0
i
A
B
A
A P i A R
pA B pB
R
F
F
F C T T F X H
C C T T
XH
!
!
!
= =
" # " +"$ =
+"
="$
Slope is now negative
3rd reactor:
X = 0.3
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
R
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
mass balance equations are the same as in part (b) and so the plot of equilibrium conversion will
decrease from reactor 1 to reactor 2, and, likewise, from reactor 2 to reactor 3.
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.
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
=
+
Canceling and combining gives:
Now solve this expression for X:
0
2
0 0 0
1( )
S S
S S S
F K
X
T VC F C
µ
=!!
( ) ( )
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
[5] mu = mumax*(.0038*T*exp(21.6-
6700/T))/(1+exp(153-48000/T))
[6] Ks = 5
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
the slope of the R(T). What we are looking to do is get R(T) to intersect with G(T) at 310 K.
Since we now have a limited coolant flow rate we will use a different value for Q.
And now plug in the known values. Assume the maximum coolant flow rate and that will give the
minimum heat exchange area.
hr
g
g
J
10015600
=
"
#
$
%
&
'
P8-15 (c)
There are two steady states for this reaction. There is an unstable steady state at about 294.5 K
and a stable steady state at 316 K.
P8-15 (d)
