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7-1
Solutions for Chapter 7 – Reaction Mechanisms,
Pathways, Bioreactions and Bioreactors
P7-1 (a) Example 7-1
The graph of Io/I will remain same if CS2 concentration changes. If concentration of M increases the slope
of line will decrease.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 12 12
C1 0.1 2.109E-04 0.1 2.109E-04
C2 0 0 1.311E-09 1.311E-09
C6 0 0 3.602E-09 3.602E-09
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(C1)/d(t) = -k1*C1-k2*C1*C2-k4*C1*C6
[4] d(C4)/d(t) = k2*C1*C2-k3*C4+k4*C6*C1-
k5*C4^2
[5] d(C7)/d(t) = k4*C1*C6
7-2
P7-1 (c) Example 7-3
The inhibitor shows competitive inhibition.
See Polymath program P7-1-c.pol.
P7-1 (d) Example 7-4
1)Now Curea = 0.001mol/dm3 and t = 10 min = 600 sec.
Solving, we get X = 0.9974.
2) For CSTR,
sec7.461== t
!
urea
urea
r
XC
!
=
"
ureaM
ureaMAX
urea CK
CV
r
+
=!
( )
sec7.461
1/1.0/0266.0
Xdmmoldmmol
!+
Solving, we get X = 0.675
See Polymath program P7-1-c.pol.
POLYMATH Results
NLE Solution
7-3
P7-1 (e) Example 7-5
( ) gg
C
C
Y
P
S
PS /18.2
14.203.5
2457.238
/=
!
!!
=
"
"!
=
P7-1 (f) Example 7-6
1) if we go for 24 hrs, fermentation will stop at 13.2 hrs as CP = CP
* .
See Polymath program P7-1-f-1.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 13.2 13.2
Cc 1 1 16.558613 16.550651
ODE Report (RKF45)
Differential equations as entered by the user
Explicit equations as entered by the user
[1] rd = Cc*0.01
2)Semi-Batch reactor:
See Polymath program P7-1-f-2.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 24 24
Cc 1.0E-04 1.0E-04 0.0474697 0.0474697
Cs 1.0E-04 1.0E-04 12.206266 12.206266
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg-rd
Explicit equations as entered by the user
[1] rd = Cc*0.01
[2] Ysc = 1/0.08
7-5
3) Changes from part(2)
I
S
SS
SC
P
P
g
K
C
CK
CC
C
C
r2
52.0
*
max 1
++
!
!
"
#
$
$
%
&'=
µ
See Polymath program P7-1-f-3.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 24 24
Cc 1.0E-04 1.0E-04 1.514E-04 1.514E-04
Cs 1.0E-04 1.0E-04 12.823709 12.823709
Cp 0 0 4.669E-04 4.669E-04
7-6
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg-rd
[2] d(Cs)/d(t) = vo*Cso/V + Ysc*(-rg)-rsm
[3] d(Cp)/d(t) = rg*Ypc
Explicit equations as entered by the user
4) After 9.67 hrs, CS = 0.
P7-1 (g) Individualized solution
P7-1 (h) Individualized solution
P7-2 Solution is in the decoding algorithm given with the modules.
P7-3
Burning:
7-8
7-9
P7-3 (d) Individualized solution
P7-3 (e) Individualized solution
P7-4
P7-5 (a)
7-12
P7-5 (b)
Cl2
k1
" # "
k2
$ " " 2Cl
Cl +CO
k3
" # "
k4
$ " " COCl
( )
4 5 2
( ) ( ) ( )( ) ( ) ( )( )
2
1 2 2 3 4 5 2
0
Cl
r k Cl k Cl k Cl CO k COCl k COCl Cl= = ! ! + +
add rCOCl to rCl
( ) ( )2
1 2 2
0 0
Cl COCl
r r k Cl k Cl+ = + = !
( )
( )
4 5 2 4 5 2
k k Cl k k Cl
+ +
P7-5 (c) Individualized solution
P7-5 (d) Individualized solution
P7-6 (a)
NO2 + hv
1
K
!
NO + O
P7-6 (b)
NOOMOO
OCCkCCCk
dt
dC
32
3
32 !=
NOONO
NO CCkCk
dt
dC
32
2
31 +!=
P7-6 (c)
P7-6 (d) Individualized solution
P7-6 (e) Individualized solution
P7-7(a)
P7-7(b)
Low temperatures with anti-oxidant
7-16
P7-7(c)
If the radicals are formed at a constant rate, then the differential equation for the concentration of the
and
[ ] [ ]
0
i
k
I
k RH
=
The substitution in the differential equation for R· also changes. Now the equation is:
1 2
P
Now we are ready to look at the equation for the motor oil.
[ ] [ ][ ] [ ][ ]
2 2i P
d RH
k I RH k RO RH
dt =! !
and making the necessary substitutions, the rate law for the degradation of the motor oil is:
P7-7(d)
Without antioxidants With antioxidants
7-18
P7-7 (e) Individualized solution
P7-8 (a)
P7-8 (b)
P7-8 (c)
7-19
P7-8 (d)
See Polymath program P7-8-d.pol.
P7-8 (e) Individualized solution
P7-8 (f) Individualized solution
P7-9 (a)
Starting with the design equation for a batch reactor
7-20
Using the data at 40oF and 45oF the following graphs are made.
P7-9 (b)
P7-9 (c)
P7-9 (d)
The data appears that it may fit the Monod equation for substrate consumption at the stationary phase.
