Design of Fermentation Processes
365
This can be used to calculate the optimal value of the operating variable k, a for given D and
s2f.
K2D/s2j-
= £ > § ^
1 1
-
VÏ|JC
5]
-D
(9-50)
Example 9.7 Optimal design of a single cell production.
Let S| be methane which is fed to the stirred tank reactor as natural gas (90% methane) mixed with
oxygen in volumetric ratio 1 to 1.31, and
P= \
atm total pressure.
s*
= 4.67 mg L
'1
and
Y,lX
= 0.8 g g
' 1
.
The second substrate is ammonia.
0.35 h
' 1
, K
2
= 50 mg L"1.
The biomass composition is CH|.8Oo.5No,2.
Let the dilution rate be 0.2 h"1, and
s2f
200 mg L'1.
From (9.50) one obtains
{kta)m
Q2
24.6/(0.2-17)
200
50-0.2/20Q^ _ 5 1 6 h ,1
0.8
4.67
{
0.35-0.20
J
. x
0.8-4.67-51.6
0.2
965 mg U
1
(
1
)
It would be useless to increase
kt a
above 51.6 h
' 1
since the liquid phase reaction would then become rate
limiting.
If the biomass concentration calculated in (
1
) is considered to be too low to obtain a reasonable
volumetric production rate it becomes interesting to redesign the process for e.g. x = 10 g L
' 1
From
(9.48) the required mass transfer coefficient is calculated to be
ki
a = 51.6 ' (10 / 0.965) = 535 h'1. But at
the same time the ammonia feed concentration must be increased; otherwise the higher rate of methane
consumption cannot be sustained.
Solution of (9.50) for
s2
f
when
k, a
is 535 h
'1
yields
s2
/ =
1449 mg L”1.
The conclusion is clearly that if one feed stream is “improved” then the other feed stream must also be
“improved”.
An overall maximum productivity is obtained when the volumetric rate of the bioreaction is maximized
according to Eq (9.34-9.35) and die matching
k/ a
value is calculated.
With the given data for the liquid phase reaction and with
s2
/=
1449 mg NH
3
L
"1
one obtains
Dop) =
0.2861 h'1,
x = Ys2x
( s2f- s2) = 8865 mg L"1,
qx ~ Dop,x =
2.54 g L’1
h'1
.
(2)
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