C
g
dC r
dt =
Substrate Balance:
0 0 0
S
S S S g
C
dC v C v C Y r
dt =! !
7-42
[8] V = 5000
[9] vo = 1000
[10] D = vo/V
CC1 = 4.33 g/dm3 X = 0.867
CS1 = 1.33 g/dm3 CP1 = YP/CCC1 =0.866 g/dm3
For 2nd CSTR,
( ) gCC rCCD =!12
( ) SSS rCCD !=!
1
See Polymath program P7-18-c-2cstr.pol.
NLES Solution
Variable Value f(x) Ini Guess
Cc 4.9334151 3.004E-10 4
Cs 0.1261699 6.008E-10 5
Km 4
Ysc 2
rg 0.120683
V 5000
X 0.987383
NLES Report (safenewt)
Nonlinear equations
[1] f(Cc) = D*(Cc-Cc1)-rg = 0
[2] f(Cs) = D*(Cs1-Cs)+rs = 0
Explicit equations
[1] umax = 0.8
[2] Km = 4
[3] Csoo = 10
[4] Cs1 = 1.333
[9] vo = 1000
CC2 = 4.933 g/dm3 X = 0.987
CS2 = 1.26 g/dm3 CP1 = YP/CCC1 =0.9866 g/dm3
P7-18 (d)
For washout dilution rate, CC = 0
7-43
!
1
33
31
max
max 57.0
/108.0 !
!
=
“
=
=hr
dmghr
C
D
SO
µ
1
33
3
1
max. 37.0
/10/4
/4
18.01 !! =
“
“
#
$
%
%
&
‘
+
!=
“
“
#
$
%
%
&
‘
+
!=hr
dmgdmg
dmg
hr
CK
K
D
SOM
M
PRODMAX
µ
Production rate = CCvO(24hr) = 4.85
3
/dmg
x1000dm3/hrx24hr = 116472.56g/day
P7-18 (e)
For batch reactor,
V = 500dm3,
g
Cr
dt
dC =
S
Sr
dt
dC =
CCO = 0.5 g/dm3 CSO = 10g/dm3
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 6 6
Ysc 2 2 2 2
umax 0.8 0.8 0.8 0.8
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg
[2] d(Cs)/d(t) = rs
Explicit equations as entered by the user
[1] Km = 4
[2] Ysc = 2
[3] umax = 0.8
For t = 6hrs, CC = 5.43g/dm3.
So we will have 3 cycle of (6+2)hrs each in 2 batch reactors of V = 500dm3.
P7-18 (g) Individualized solution
P7-18 (f) Individualized solution
7-44
P7-19
Given constants:
Balance on cells
0=dCC
dt =in “out +generation
0=“D(0.1)CC+
µ
CC
D=10
µ
(1)
Balance on Substrate
0
/ /
1 1
0 ( ) ( )
s
C C
X S P S
dC D S S C C
dt Y Y
µ!µ“
= = # # # +
(2)
Balance on product ( lactic acid)
/
1
0 ( )
P
P C
P S
dC r C
dt Y
!µ“
= = #+ +
(3)
Rate of production = rP
/
1( )
P C
P s
r C
Y
!µ“
= +
(4)
From (2)
0
/ /
1 1
( ) ( ) C
X S P S
D S S C
Y Y
µ!µ“
# $
%= + +
& ‘
( )
;
µ
=fD
0
/ / /
1
( ) ( ) C
X S P S P S
D S S fD C
Y Y Y
! “
# $
%= + +
& ‘
( )
From (1)
Recycled cells
fD =
µ
=
µ
max S
KS+S
S=fDKS
µ
max “fD
0
max
/ / /
1 1
S
C
X S P S P S
fDK
D S fD
C
fD Y Y Y
µ
!
” #
$
% &
$
‘ (
=” #
+ +
% &
‘ (
from (4)
[ ] 0
/ max
/ / /
1
1 1
s
P S
P
X S P S P S
fDK
fD D S
Y fD
r
fD Y Y Y
! “ µ
“
# $
+%
& ‘
%
( )
=# $
+ +
& ‘
( )
Plot rP versus D or
( ) 0
P
d r
dD =
to obtain the optimum D for maximizing (rP) the rate of lactic acid
production by utilizing the constants mentioned above.
Recycle CSTR
0
2
4
6
8
10
12
14
012345
Dilution rate, hr–1
Production Rate
DP
Cell conc
The maximum cell concentration occurs at a dilution rate of D = 0.195/hr
The maximum production rate occurs at a dilution rate of D = 2.5/hr
P7-19 (a,b) Individualized solutions.
7-46
P7-20 (a)
X1 + S → More X1 + P1
( ) 11
/gXXSSSO
SrYCCD
dt
dC !!=
( ) 2
1
2111
1
/gXXXgXX
XrYrCD
dt
dC !+!=
( ) 22
2
gXX
XrCD
dt
dC +!=
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 1 1
Cs 10 1.2366496 10 1.2366496
Cx1 25 25 25.791753 25.456756
Km2 10 10 10 10
u1max 0.5 0.5 0.5 0.5
u2max 0.11 0.11 0.11 0.11
rgx1 6.25 1.4008218 6.25 1.4008218
Yx1s 0.14 0.14 0.14 0.14
Yx2x1 0.5 0.5 0.5 0.5
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cs)/d(t) = D*(Cso-Cs)-Ysx1*rgx1
Explicit equations as entered by the user
[1] Km1 = 10
[2] Km2 = 10
[3] u1max = 0.5
[4] u2max = 0.11
[10] Yx1x2 = 1/Yx2×1
7-47
P7-20 (b) When we increase D, CS increases, CX1 decreases, and CX2 has very little decrease.
P7-20 (c) When CSO decreased, CS and CX1 both decreases, CX2 has no noticeable change.
When CSi increased, CX1 increases, CX2 has no noticeable change for large t
P7-20 (d) Individualized solution
P7-20 (e) Individualized solution
P7-21 (a) and (b)
7-48
P7-21 (c)
P7-21 (d) Production starts at the end of the exponential (for both runs)
P7-21 (e)
P7-21 (f) Individualized solution
P7-21 (g) Individualized solution
7-49
P7-22 (a)
P7-22 (c)
7-50
For D = 0.27hr-1 , CC = 0 if CCO = 0.5 g/dm3.
For D = 0.314hr-1 , CC = 0 if CCO = 0 g/dm3.
And for maximum production rate, D = 0.876hr-1.
P7-22 (d)
For batch reactor,
g
Cr
dt
dC =
gCS
SrY
dt
dC
/
!=
( )
CS
gKCCK
CC
r++
=1
max
µ
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 2 2
Cc 0.5 0.5 2.9 2.9
Cs 30 2.01E-07 30 2.01E-07
Ycs 0.08 0.08 0.08 0.08
rateS 5.7397959 1.093E-05 36.27396 1.093E-05
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg
Explicit equations as entered by the user
[1] Ki = 50
[2] umax = 1.5
[3] Ks = 1
[7] rateS = Ysc*rg
7-51
P7-22 (e)
For semi-batch reactor,
V
Cv
r
dt
dC CO
g
C!=
( )
V
CCv
rY
dt
dC SSinO
gCS
S!
+!=/
( )
ISSS
CS
gKCCK
CC
r++
=1
max
µ
tvVV OO +=
See Polymath program P7-22-e.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 4.5 4.5
Cc 0.5 0.2329971 2.2593341 2.2593341
Cs 2 0.8327919 24.016878 0.8327919
Csin 30 30 30 30
umax 1.5 1.5 1.5 1.5
Ks 1 1 1 1
rg 0.487013 0.2329022 2.2464622 1.5283423
Ysc 12.5 12.5 12.5 12.5
VCc 5 5 530.94351 530.94351
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg-vo*Cc/V
Explicit equations as entered by the user
[1] Ki = 50
[2] vo = 50
[7] Ks = 1
[8] rg = umax*Cs*Cc/(Ks+Cs*(1+Cs/Ki))
7-52
P7-22 (f) Individualized solution
P7-22 (g) Individualized solution
P7-23 (a)
P7-23 (b)
Answer correct, typos in Equation to get answer.
Rearranging Equation (7-89)
Y
C S =CC
CSO “DKs
µmax “D
#
$
%
&
‘
(
Inserting values from Row 4
Y
C S =4
50 “1.8
0.97
1.98 “1.8
#
$
%
&
‘
(
=4
50 “9.7
7-53
Y
C S =1
Y
C S
=10.075
Answer correct solution manual
P7-23 (c)
Plot CS, CC, rg and –rS as a function of time for a batch reactor when CC0 = 10–4 g/dm3 and CS0 = 50
P7-23 (d)
Vary CC0 and CS0 in Part (c) and describe what you find.
7-54
7-55
P7-24 No solution will be given.
P7-25 No solution will be given
P7-26 See Professional Reference Shelf 7.5 on the website for a sample solution.
CDP7-A No solution will be given.
CDP7-B
7-56
CDP7-C
Assumptions
7-57
CDP7-C (b)
To increase the growth rate, you could:
CDP7-C (c)
RC=yO2RO2
From equation (4), have:
7-58
CDP7-C (d)
Assumptions: