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7-21
P7-10 (a)
E+Sk1
" # " E•S
E•Sk2
" # " E+S
E•Sk3
" # " P+E
P+Ek4
" # " E•S
E
#
&
[ ]
[ ]
[ ]
[ ]
k2+k3
#
$
&
'
E
[ ]
=Et
[ ]
k2+k3
( )
k2+k3+k1S
[ ]
+k4P
[ ]
E•S
[ ]
=
Et
[ ]
k1S
[ ]
+k4P
[ ]
( )
k2+k3+k1S
[ ]
+k4P
[ ]
r
P=
Vmax S
[ ]
"P
[ ]
K
#
$
%
&
'
(
Km+S
[ ]
+KPP
[ ]
Vmax =k3Et
[ ]
,K=k1k3/k2k4,Km=k2+k3
( )
/k1,KP=k4/k1
7-22
P7-10 (b)
SESE k•!"!+1
SESE k+!"!•2
E•Sk3
" # " E•P
E•Pk4
" # " E•S
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
P7-10 (c) No solution will be given
P7-10 (d)
1
2
1 1
k
k
E S E S
!!"
+ •
#!!
7-23
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:
( ) ( ) ( )
( )
( )( )
( )
1 1 1 3 1 2
3 2 2 4 5 2 3 2
2
4 5
1
T
k S k k S S
E E
k S k k k k k S
k
k k
! "
# $
# $
= + +
# $
+ +
+
# $
+
% &
P7-10 (e)
( )
1
21
k
k
E S E S
!!"
+ •
#!!
( ) ( )
3
4
1
1 2
k
k
E S E S P
!!"
• • +
#!!
( )
5 4 1
k k P
7-24
( ) ( ) ( )
( )
( )
( )
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
! "
# $
# $
= + + + +
# $
+
# $
+
% &
( )( )
1 3 5
T
k k k E S
r
=! "
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
k
k
E P S E S P
!!"
• + • •
#!!
7-25
( ) ( )( ) ( )( )
6 8
7 9
k E S P k E P S
E S P
k k
• + •
• • = +
( )( ) ( )( ) ( )( )
6 8
4 9
7 9
k E S P k E P S
k E P k
k k
! "
• + •
+# $
+
% &
5
( )
( )( ) ( )( ) ( )( ) ( )( ) ( )( )
7 4 9 4 9 6
1 4 5
5
k k E P k k E P k k E S P
k E S k E P k
k
E S
! "
+ + •
+#$ %
& '
• = +
( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )
( ) ( )
1 4 7 4 9 4
7 4 9 4 9 6
2 3 6 9
5
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
! "
+#+
$ %
$ %
+#
& '
++
P7-10 (g)
3
1
2
k
k
k
E S E S P
!!"
+ • "
#!!
( ) ( )( ) ( )
4 5
0
E P
r k E P k E P
•= = !•
( ) ( )( )
4
k E P
E P
• =
( )( )
1 3
2 3
P
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
2 3 5
1
T
k S k P
E E
k k k
! "
= + +
# $
+
% &
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:
P7-11 (b)
P7-11 (c) Individualized solution
P7-11 (d) Individualized solution
P7-12 (a)
P7-12 (b)
7-30
P7-12 (c)
P7-12 (d) Individualized solution
P7-13 (a)
P7-13 (b)
7-33
P7-13 (c) Individualized solution
P7-14
For No Inhibition, using regression,
Equation model:
!
"
#
$
%
&
+=
'S
aa
rs
1
10
1
a0 = 0.008 a1 = 0.0266
P7-15
PEHSS rrr =!=
( )
P
r k EHS=
( ) ( )( )
M
EHS K EH S=
( ) ( ) ( ) ( )
2T
E E EH EH
!+
= + +
Now plug the value of (EH) into rP
At very low concentrations of H+ (high pH)
S
r
approaches 0 and at very high concentrations of H+ (low
pH)
S
r
also approaches 0. Only at moderate concentrations of H+ (and therefore pH) is the rate much
greater than zero. This explains the shape of the figure.
P7-15 (b) Individualized solution
P7-16 (a)
For batch reaction,
S
Sr
dt
dC =
CS
SCK
CC
r
+
!=max
µ
&
SSCC rYr /
!=
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
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cs)/d(t) = rs
Explicit equations as entered by the user
[1] Cso = 20
P7-16 (b)
For logistic growth law:
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
[1] d(Cc)/d(t) = rg
Explicit equations as entered by the user
P7-16 (c)
For CSTR,
SSCg rYr /
!=
SM
CSSC
CK
CCY
+
=max/
µ
7-37
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
=hrD prod
P7-16 (e)
If rd = kdCC
( )
VrrvCm dgOC !==
&
Divide by CCV,
( ) ( )
d
SM
SSC
C
dg k
CK
CY
C
rr
D!
+
=
!
=max/
µ
1
33
3131
max/ 493.0
/25.0/20
/25.002.0/2015.0 !
!!
=
+
"!""
=
+
!
=hr
dmgdmg
dmghrdmghr
KC
KkCY
D
MSO
MdSOSC
MAX
µ
There is not much change in Dilution rate while considering cell death to one where cell death is neglected.
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
P7-16 (g) Individualized solution
P7-16 (h) Individualized solution
P7-17
P7-17 (a)
For batch reaction,
7-39
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
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
P7-17 (b) Individualized solution
P7-17 (c)
7-40
Divide by CCV,
( )
kCS
eD /
!==
µµ
P7-17 (d)
( )
SSOSCC CCDYDC !=/
!
!
#
$
$
&''=
1ln
µ
D
kCS
P7-18 (a)
rg =
C
C
µ
SM
S
CK
C
+
=max
µ
µ
P7-18 (b)
Flow of cells out = Flow of cells in
