P7-10 (a)
E+Sk1
“ # “ E•S
E•Sk2
“ # “ E+S
E•Sk3
“ # “ P+E
P+Ek4
“ # “ E•S
r
P=k3E•S
[ ]
“P
[ ]
E
[ ]
KC
#
$
%
&
‘
(
,
KC=k3/k4
r
E•S=k1E
[ ]
S
[ ]
“k2E•S
[ ]
“k3E•S
[ ]
+k4P
[ ]
E
[ ]
E•S
[ ]
=k1E
[ ]
S
[ ]
+k4P
[ ]
E
[ ]
k2+k3
=
E
[ ]
k1S
[ ]
+k4P
[ ]
( )
k2+k3
Et
[ ]
=E
[ ]
+E•S
[ ]
=E
[ ]
1+k1S
[ ]
+k4P
[ ]
k2+k3
“
#
$
%
&
‘
E
[ ]
=Et
[ ]
k2+k3
( )
k2+k3+k1S
[ ]
+k4P
[ ]
E•S
[ ]
=
Et
[ ]
k1S
[ ]
+k4P
[ ]
( )
k2+k3+k1S
[ ]
+k4P
[ ]
r
P=
k3Et
[ ]
k1S
[ ]
+k4P
[ ]
( )
k2+k3+k1S
[ ]
+k4P
[ ]
“k4P
[ ]
Et
[ ]
k2+k3
( )
k2+k3+k1S
[ ]
+k4P
[ ]
( )
#
$
%
%
&
‘
(
(
=Et
[ ]
k1k3S
[ ]
+k3k4P
[ ]
“k2k4P
[ ]
“k3k4P
[ ]
k2+k3+k1S
[ ]
+k4P
[ ]
#
$
%
&
‘
(
#
&
7-22
P7-10 (b)
SESE k•!“!+1
SESE k+!“!•2
E•Sk3
“ # “ E•P
E•Pk4
“ # “ E•S
E•Pk5
“ # “ P+E
r
E•S=0=k3E•S
[ ]
“k4E•P
[ ]
“k5E•P
[ ]
E•P
[ ]
=k3E•S
[ ]
k4+k5
r
E•S=k1E
[ ]
S
[ ]
“k2E•S
[ ]
“k3E•S
[ ]
+k4E•P
[ ]
since E is not consumed:
ET=E+E•S+E•P
or
E=ET“E•S“E•P
E=ET“E•S“k3
k4+k5
#
$
%
&
‘
(
E•S E=ET“E•S“1+k3
k4+k5
#
$
%
&
‘
(
Insert this into the equation for rE·S and solve for the concentration of the intermediate:
E•S
[ ]
=k1S
[ ]
ET
[ ]
1+k3
k4+k5
“
#
$
%
&
‘
k1S
[ ]
+k2+k3(k4k3
k4+k5
r
P=k5E•P
[ ]
=k3k5E•S
[ ]
k4+k5
r
P=k1k3k5S
[ ]
ET
[ ]
k3+k4+k5
[ ]
k1S
[ ]
+k2k4+k3k5
P7-10 (c) No solution will be given
P7-10 (d)
1
2
1 1
k
k
E S E S
!!“
+ •
#!!
3
4
1 2 1 2
k
k
E S S E S S
!!“
• + •
#!!
5
1 2
k
E S S P E•!+
(1)
( )( ) ( ) ( )( ) ( )
11 1 2 3 2 4 1 2
0
E S
r k E S k E S k E S S k E S S= = !•!• + •
(2)
( )( ) ( ) ( )
1 2 3 2 4 1 2 5 1 2
0
E S S
r k E S S k E S S k E S S
•= = • !•!•
If we add these two rates we get:
From equation (2) we get
( ) ( )( )
3 2
1 2
4 5
k E S S
E S S
k k
•
• = +
Plug this into equation 3 and we get:
7-24
( ) ( )
( )
( )( )
( )
31
1
2
5 3
4 1 5
2
5 4 1
k E S k E S
E S k k
k P k k
k k P
! “
# $
! “
•# $
• = # $
# $
+# $
% & +
# $
+
% &
( ) ( )( )
( )
3 1
2
2 4 1 2 5 5 3
k k E S
E S
k k P k k k k
• = + +
( ) ( )( )
( )
2
1 3 5
52
2 4 1 2 5 5 3
P
k k k E S
r k E S
k k P k k k k
= • = + +
( ) ( ) ( ) ( )
1 2
T
E E E S E S= + • + •
( ) ( ) ( )
( )
( )
( )
1 3 1
5 3 2 4 1 2 5 5 3
2
5 4 1
1
T
k S k k S
E E k k k k P k k k k
k
k k P
! “
# $
# $
= + + + +
# $
+
# $
+
% &
( )( )
( )
( ) ( )
( )
( )
( )
2
1 3 5
1 3 1
2 4 1 2 5 5 3
5 3 2 4 1 2 5 5 3
2
5 4 1
1
T
P
k k k E S
r
k S k k S
k k P k k k k k k k k P k k k k
k
k k P
=! “
# $
# $
+ + + + + +
# $
+
# $
+
% &
P7-10 (f)
3
1
2
k
k
k
E S E S P
!!“
+ • “
#!!
4
5
k
k
E P E P
!!“
+ •
#!!
6
7
k
k
E S P E S P
!!“
• + • •
#!!
8
9
k
k
E P S E S P
!!“
• + • •
#!!
( )
3P
r k E S= •
(1)
( ) ( )( ) ( ) ( ) ( )( ) ( )
1 2 3 6 7
0
E S
r k E S k E S k E S k E S P k E S P
•= = !•!• + • + • •
(2)
( ) ( )( ) ( ) ( )( ) ( )
4 5 8 9
0
E P
r k E P k E P k E P S k E S P
•= = !•!• + • •
(3)
( ) ( )( ) ( ) ( )( ) ( )
6 7 8 9
0
E S P
r k E S P k E S P k E P S k E S P
• • = = • !• • + • !• •
(2) + (3):
( ) ( ) ( )( ) ( ) ( )( ) ( )
4 5 6 7
0
E P E S P
r r k E P k E P k E S P k E S P
• •
+ = = !• + • !• •
( ) ( )( ) ( )( ) ( )
1 4 5
2 3
k E S k E P k E P
E S
k k
+!•
• = +
( ) ( )( ) ( )
( )
4 9
5 8
k E P k E S P
E P
k k S
+ • •
• = +
( ) ( )( ) ( )( )
6 8
7 9
k E S P k E P S
E S P
k k
• + •
• • = +
( )
( )( ) ( )( ) ( )( )
( )
6 8
4 9
7 9
5 8
k E S P k E P S
k E P k
k k
E P
k k S
! “
• + •
+# $
+
% &
• = +
( ) ( )( ) ( )( ) ( )( ) ( )( )
( )
7 4 9 4 9 6 8
5 8
k k E P k k E P k k E S P k E P S
E P
k k S
+ + • + •
• = +
( ) ( )( ) ( )( ) ( )( )
7 4 9 4 9 6
5
k k E P k k E P k k E S P
E P
k
+ + •
• =
( )
( )( ) ( )( ) ( )( ) ( )( ) ( )( )
7 4 9 4 9 6
1 4 5
5
2 3
k k E P k k E P k k E S P
k E S k E P k
k
E S
k k
! “
+ + •
+#$ %
& ‘
• = +
( ) ( )( ) ( )( ) ( )( ) ( )( )
( )
1 4 7 4 9 4
2 3 6 9
k E S k E P k k E P k k E P
E S
k k k k P
+!+
• = +!
( ) ( ) ( ) ( ) ( ) ( )
( )
1 4 7 4 9 4
3 3
2 3 6 9
P
k S k P k k P k k P
r k E S k E
k k k k P
+!+
= • = +!
( ) ( ) ( ) ( ) ( )
T
E E E S E P E S P= + • + • + • •
( ) ( ) ( )( ) ( )( ) ( )( ) ( )( )
( )
( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )
( ) ( )
1 4 7 4 9 4
2 3 6 9
1 4 7 4 9 4
7 4 9 4 9 6
2 3 6 9
5
T
k E S k E P k k E P k k E P
E E
k k k k P
k E S k E P k k E P k k E P
k k E P k k E P k k P
k k k k P
k
+!+
= + +!
” #
+!+
+ + $ %
$ %
+!
& ‘
+
( )( ) ( )( ) ( )( ) ( )( )
( ) ( )
1 4 7 4 9 4
6
2 3 6 9
7 9
k E S k E P k k E P k k E P
k P
k k k k P
k k
! “
+#+
$ %
$ %
+#
& ‘
++
( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )
( ) ( )
( )
1 4 7 4 9 4
7 4 9 4 9 6
2 3 6 9
8
5
1 9
k E S k E P k k E P k k E P
k k E P k k E P k k P
k k k k P
k S
k
k k
! “
! “
+#+
+ +
$ %
$ %
$ %
+#
$ %
& ‘
$ %
$ %
$ %
& ‘
+
7-26
all of the terms in the numerator have (E) in it and so the (E) can be factored out and an expression for (E)
in terms of (ET), (P), and (S) can be made and plugged back into the equation for rP.
P7-10 (g)
3
1
2
k
k
k
E S E S P
!!“
+ • “
#!!
4
5
k
k
E P E P
!!“
+ •
#!!
( ) ( )( ) ( ) ( )
1 2 3
0
E S
r k E S k E S k E S
•= = !•!•
( ) ( )( )
1
2 3
k S E
E S
k k
• = +
( ) ( )( ) ( )
4 5
0
E P
r k E P k E P
•= = !•
( ) ( )( )
4
5
k E P
E P
k
• =
( ) ( )( ) ( )
3 4 5P
r k E S k E P k E P= • !+ •
( )( ) ( )( ) ( )( )
1 3
4 4
2 3
P
k k S E
r k E P k E P
k k
=!+
+
( )( )
1 3
2 3
P
k k S E
r
k k
=+
( ) ( ) ( ) ( )
T
E E E S E P= + • + •
( ) ( ) ( )( ) ( )( )
1 4
2 3 5
T
k S E k E P
E E
k k k
= + +
+
( ) ( ) ( ) ( )
1 4
1
T
k S k P
E E
k k k
! “
= + +
# $
+
( )( )
( ) ( ) ( )
1 3
1 4
2 3
2 3 5
1
T
P
k k S E
r
k S k P
k k
k k k
=! “
+ + +
# $
+
% &
( )( )
( ) ( )( )
3
4 2 3
2 3
1 1 5
T
P
k S E
r
k k k P
k k S
k k k
=! “
+
++ +
# $
% &
( )
( ) ( )
max
P
M P
V S
r
K S K P
=+ +
P7-10 (h) No solution will be given
P7-10 (i) No solution will be given
P7-10 (j) No solution will be given
7-27
P7-10 (k) No solution will be given
P7-11 (a)
The enzyme catalyzed reaction of the decomposition of hydrogen peroxide. For a batch reactor:
7-28
P7-11 (b)
P7-11 (c) Individualized solution
P7-11 (d) Individualized solution
P7-12 (a)
7-29
P7-12 (b)
7-30
7-31
P7-12 (c)
2
1 2
1
S T
S
s S
kC E
r
K C K C
!=+ +
If ET is reduced by 33%, -rS will also decrease by 33%. From the original plot, we see that if the curve –rS
X = 0.2
P7-12 (d) Individualized solution
P7-12 (e) Individualized solution
P7-13 (a)
7-32
P7-13 (b)
7-33
P7-13 (c) Individualized solution
P7-13 (d) Individualized solution
P7-14
For No Inhibition, using regression,
Equation model:
!
“
#
$
%
&
+=
‘S
aa
rs
1
10
1
a0 = 0.008 a1 = 0.0266
For Maltose,
Equation model:
!
“
#
$
%
&
+=
‘S
aa
rs
1
10
1
a0 = 0.0098 a1 = 0.33
For α-dextran,
7-34
P7-15
PEHSS rrr =!=
( )
P
r k EHS=
( ) ( )( )
M
EHS K EH S=
( )( )
P M
r kK EH S=
( ) ( )( )
2 2
EH K H EH
+ +
=
( ) ( )( )
1
EH K H E
+!
=
( ) ( )
( )
1
EH
E
K H
!
+
=
( ) ( ) ( ) ( )
2T
E E EH EH
!+
= + +
( ) ( )
( ) ( ) ( )( )
2
1
T
EH
E EH K H EH
K H
+
+
= + +
( ) ( )
( ) ( )
2
1
1
1
T
E
EH
K H
K H
+
+
=
+ +
Now plug the value of (EH) into rP
=
S
r
( )( ) ( ) ( )( )
( ) ( )
1
2
1 1 2
1
M T
P M
kK E K H S
r kK EH S
K H K K H
+
+ +
= =
+ +
At very low concentrations of H+ (high pH)
S
r
approaches 0 and at very high concentrations of H+ (low
P7-15 (a) Individualized solution
P7-15 (b) Individualized solution
7-35
P7-16 (a)
For batch reaction,
S
Sr
dt
dC =
SM
CS
SCK
CC
r
+
!=max
µ
&
SSCC rYr /
!=
( )
SSOSCCOC CCYCC !+= /
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 10 10
Cs 20 6.301E-11 20 6.301E-11
Km 0.25 0.25 0.25 0.25
Ycs 0.5 0.5 0.5 0.5
Cc 0.1 0.1 10.1 10.1
rc 0.0493827 1.273E-09 4.0390748 1.273E-09
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cs)/d(t) = rs
Explicit equations as entered by the user
P7-16 (b)
For logistic growth law:
g
Cr
dt
dC =
C
C
g
C
C
C
r!
!
#
$
$
&‘=
1
max
µ
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 7 7
ODE Report (RKF45)
Differential equations as entered by the user
Explicit equations as entered by the user
[1] umax = 1
[2] Coo = 1
P7-16 (c)
For CSTR,
SSCg rYr /
!=
CSSC
CK
CCY
+
=max/
µ
P7-16 (d)
!
!
“
#
$
$
%
&
+
‘((=
!
!
“
#
$
$
%
&
+
‘=‘
33
3
1
max/max /20/25.0
/25.0
115.01
dmgdmg
dmg
hr
CK
K
YD
SOM
M
SCprod
µ
1
max 44.0 !
=hrD prod
( )
!
!
“
#
$
$
%
&‘
+
!
!
“
#
$
$
%
&
‘
+
=D
CK
CY
DY
KCY
C
SOM
SOSC
SC
MSOSC
C
max/
max/
/
µ
µ
( ) 31
3
31
11
3
/08.944.0
/25.20
/2015.0
44.015.0
/25.0205.0 dmghr
dmg
dmghr
hrhr
dmg
CC=
!
!
“
#
$
$
%
&‘
((
!
!
“
#
$
$
%
&
‘(
+
=‘
‘
‘‘
3
max/
/83.1 dmg
DY
DK
C
SC
M
S=
!
=
µ
33
//362.1/08.915.0 dmgdmgCYC CCPP =!==
hrdmg
dmgdmg
dmgdmg
CK
CC
r
SM
CS
S./99.7
/83.1/25.0
/08.9/83.11 3
33
33
max =
+
!!
=
+
=“
µ
P7-16 (e)
If rd = kdCC
( )
VrrvCm dgOC !==
&
Divide by CCV,
( ) ( )
d
SM
SSC
C
dg k
CK
CY
C
rr
D!
+
=
!
=max/
µ
!
( )
DY
KkD
C
Md
S!
+
=
µ
Now
gCSS rYr /
=!
!
( ) ( )
dCCSSSO rDCYCCD +=!/
( )
SSOSC
CkD
CCDY
C
+
!
=/
For dilution rate at which wash out occur, CC = 0
!
CSO = CS
!
( )
DY
KkD
C
Md
SO !
+
=
µ
7-38
P7-16 (f)
Now –rm = mCC
gC rDC =
&
( ) mSSSO rrCCD !!=!
( ) VCVrvCm CgOC
µ
===
&
For dilution rate at which wash out occur, CC = 0
!
CSO = CS,
DY
DK
C
SC
M
S!
=
max/
µ
!
DY
DK
C
SC
M
SO !
=
max/
µ
!
1
33
31
max/ 494.0
/25.0/20
/2015.0 !
!
=
+
““
=
+
=hr
dmgdmg
dmghr
KC
CY
D
MSO
SOSC
MAX
µ
Now For
prod
D.max
,
( ) 0=
dD
DCd C
P7-16 (g) Individualized solution
P7-16 (h) Individualized solution
P7-17
Tessier Equation,
P7-17 (a)
For batch reaction,
S
Sr
dt
dC =
,
CCSS rYr /
=!
,
( ) C
kC
gCer S/
max 1!
!=
µ
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 7 7
Cs 20 0.0852675 20 0.0852675
Cco 0.1 0.1 0.1 0.1
Cso 20 20 20 20
Cc 0.1 0.1 10.057366 10.057366
k 8 8 8 8
RateS 0.183583 0.183583 7.7126957 0.213253
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cs)/d(t) = rs
Explicit equations as entered by the user
[1] Cco = 0.1
[2] Ycs = 0.5
[3] Cso = 20
[4] Cc = Cco+Ycs*(Cso-Cs)
[5] k = 8
[6] umax = 1
P7-17 (b) Individualized solution
P7-17 (c)
gC rDC =
( ) SSSO rCCD =!
Cg Cr
µ
=
( )
kCS
e/
max 1!
!=
µµ
7-40
Divide by CCV,
( )
kCS
eD /
max 1!
!==
µµ
!
!
#
$
$
&‘‘=
1ln
µ
D
kCS
For dilution rate at which wash out occur, CC = 0
!
CSO = CS
!
!
“
#
$
$
%
&‘‘=
max
1ln
µ
D
k
!
( ) ( ) 1/8//201
/
max 918.0111
#3
0!!!
!=!=!=hrehreD dmgdmg
kC
MAX
S
µ
P7-17 (d)
( )
SSOSCC CCDYDC !=/
!
!
“
#
$
$
%
&‘‘=
max
1ln
µ
D
kCS
!
!
#
$
$
&
!
!
#
$
$
&‘+=
/1ln
µ
D
kCDYDC SOSCC
P7-18 (a)
rg =
C
C
µ
SM
S
CK
C
+
=max
µ
µ
For CSTR,
gC rDC =
( ) SSSO rCCD !=!
gCSS rYr /
=!
( ) ( ) 33 /19.01/101 dmgdmgXCC SOS =!=!=
V
v
DO
=
P7-18 (b)
Flow of cells out = Flow of cells in
0C C
F v C=
Cell Balance:
C
C g C
dC
F r V F V dt
+!=