μ
g
=
μ
max
S
S
D
=
μ
g
=
μ
max
μ
max
=
ln
(2)
τ
d
= 0
.
289 hr

1
However if we don’t assume this:
ln
(2)
τ
d
=
D
max
S
K
S
+
S
D
max
=
ln
(2)(
K
S
+
S
)
τ
d
S
As
S
→
0, or all the substrate is consumed to produce cell mass, indicating
the maximum growth rate,
D
max
→ ∞
.
Thus, we need some data to deter
mine when the Monod kinetics will plateau such that
μ
max
can be determined
empirically.
6.d
DX
=
DY
X/S
0

K
S
D
μ
max

D
DX
= 3
.
49 g DW/L/hr
7
Shuler 6.9
Want to maximize biomass productivity (DX) while minimizing
S
o
ut
, thus find
D such that
d
(
DX
)
dD
= 0. Since,
DX
=
DY
X/S
(
S
0

S
), maximizing
DX
also
minimizes S.
11
DX
=
DY
X/S
(
S
0

S
)
D
=
μ
g
=
μ
max
S
K
S
+
S
S
=
DK
S
μ
max

D
DX
=
DY
X/S
S
0

DK
S
μ
max

D
∴
d
(
DX
)
dD
=
Y
X/S
d
dD
DS
0

D
2
K
S
μ
max

D
Separate and apply quotient rule to get:
d
(
DX
)
dD
=
Y
X/S
S
0

K
S
D
(2
μ
max
)
(
μ
max

D
)
2
!
Setting this equal to zero to find the extrema and solving for D:
0 =
Y
X/S
S
0

K
S
D
(2
μ
max
)
(
μ
max

D
)
2
!
D
= 0
.
518 or 0
.
482
We can check whether these are maxima or minima by taking the second deriva
tive and checking the sign at each dilution rate.
d
2
(
DX
)
dD
2
=

Y
X/S
K
S
2
μ
2
max
(
μ
max

D
)
3
!
d
2
(
DX
)
dD
2
D
=0
.
518
>
0
∴
minimum
d
2
(
DX
)
dD
2
D
=0
.
482
<
0
∴
maximum
Therefore,
D
optimal
= 0
.
482 h

1
. This is the same as applying equation 6.83.
To determine whether we have sufficient
O
2
delivery, the cell concentration at
this dilution rate is needed.
12
D
=
μ
g
=
μ
max
S
K
S
+
S
S
=
K
S
D
μ
max
D
S
= 0
.
803 g/L
X
=
Y
X/S
(
S
0

S
)
X
= 10
.
6g/L
The OUR can then be compared with the OTR to determine if sufficient oxygen
can be provided.
OUR
=
q
O
2
X
=
μ
g
X
Y
X/O
2
=
D
opt
X
Y
X/O
2
Y
X/O
2
=
Y
X/S
Y
O
2
/S
= 0
.
25 g cell/g oxygen
OUR
= 20
.
4 g O
2
/L/h
As OUR is greater than the OTR of the chemostat (10 g oxygen per liter per
hour), not enough oxygen can be supplied to operate at the optimal dilution
rate.
8
Shuler 6.12
8.a
From a mass balance on cells in the CFSTR:
Accum.
= In  OUT + GEN  DESTROY
V
dX
dt
=
QX
0

QX
+
μ
g
XV

k
d
XV
dX
dt
=
DX
0

DX
+
μ
g
X

k
d
X
0 =
D
(0)

DX
+
μ
g
X

(0)
X
0 =

DX
+
μ
g
X
D
=
μ
g
13
Apply Contois equation:
μ
g
=
μ
m
S
K
SS
X
+
S
D
=
μ
m
S
K
SS
X
+
S
S
=
DK
SS
X
μ
m

D
8.b
From a mass balance on substrate in the CFSTR:
Accum.
= In  OUT + GEN  DESTROY
V
dS
dt
=
QS
0

QS

μ
g
XV
Y
X/S
dS
dt
=
D
(
S
0

S
)

μ
g
X
Y
X/S
0 =
D
(
S
0

S
)

μ
g
X
Y
X/S
D
(
S
0

S
) =