Biomedical Engineering Reference
In-Depth Information
F
P
S
0
F
S
0
¼
(E8-3.7a)
S
0
S
F
P
E
0
S
F
E
0
¼
(E8-3.7b)
0
S
Substituting Eqns
(E8-3.7)
and
(E8-3.4)
into Eqn
(E8-3.5)
, we obtain
F
P
S
$
E
F
P
E
0
S
$
V
F
P
ðK
m
þ
S
Þ
0
GP$
¼
$
P
F
P
$
S
S
S
(E8-3.8)
S
0
0
k
E
S
0
To find the optimum reactor conditions, we maximize the gross profit by varying the
effluent substrate concentration and enzyme loading. This task can easily be achieved by
an optimizer like the Excel Solver. Substituting the known parameters into Eqn
(E8-3.8)
,
we obtain
0:1
1001
1
S
100
100
E
S
0:1
100ð0:01þ
S
Þ
0
GP$
¼
1
100
$
=
h
1
1:560
E
S
0
which is reduced to
10
1
S
10000 E
0
S
0:01þ
S
GP$
¼
100
$
=
h
(E8-3.9)
1
9E
0
S
Maximizing the gross profit GP$, we obtain: S
¼
0.075969 mol/L, E
0
¼
0.0024102 g/L, and
GP$
$10.9256/h.
Therefore, we have found the optimum enzyme loading and the effluent substrate
concentration.
¼
(d) The cash flow based on Eqn
(E8-3.8)
,
GP$
F
P
¼
S
E
$
V
ð
K
m
þ
S
Þ
0
0
$
P
$
S
S
$
E
S
(E8-3.14)
S
0
S
0
k
E
S
0
Or since we have GP$ value already,
GP$
F
P
¼
10:9256
$
=
mol-P
¼
$0
:
109256
=
mol-P
100
This concludes the example. One must be careful of the units whenever numbers are
inserted into the equation.
Example 8-3 shows that enzyme cost can be significant. Therefore, separate and reuse of
enzyme is very important for bioprocess systems.
Example 8-4. It has been observed that substrate inhibition occurs in the following enzymatic
reaction:
E
þ
S
P
þ
E
/
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