Design of Fermentation Processes
371
one obtains the following:
v(f) = 0.25'0.02114 ‘ 1 '
V0exp(p.25 t)
(2)
V(t)= V0
( 1 -
b
+
b exp{
0.25 0)
(3)
jc(0 =
exp(0.2St) V0 /V
for
x0= \g
L
' 1
(4)
At the time
tfmai
when
V = V
fi„a
, = 4 V(!
one obtains from (3) that (1 -
b
+
b exp{
0.25 f)) = 4 and
tfmai
is
calculated to 19.84 h.
From (4) the corresponding x-value is determined to 35.7 g L"1.
(5)
Assume that the largest attainable value of
kt
a is 650 h
'1
and that the oxygen tension in the medium
needs to be
10
%
of the saturation value.
( q ') ^
= 650( 1.16 10'3‘ 0.2096) ■
0.9 = 0.1422 mol 0
2
h
' 1
L
’ 1
.
This oxygen uptake can support a volumetric biomass growth rate
(qx)max ~
(0.6836)'1
24.6
{q!
= 5.116 g L
1
h
"1
corresponding to x =
x* ~
20.46 g L 1,
Solving (4) for x = 20.46 g L
"1
yields /• = 14.26 h, and from (3)
V - V*
= 1.73
Va.
From (9.65) *= 5.116/ (0.4730 ■
100) = 0.1082 h 1.
Thus from
t = t*
(
0
= 0) to
tfinal
the reactor volume increases as
V=
V*
exp
(0.1082
t),
and for
V - 4V„
(i.e.
V! V
* ~ 2.31) one obtains
0
= 7.73 h and
tfinai
= 22.0 h.
xfmai
is calculated from (9.64) to 35.62 g L'1, and apart from the permissible approximation in (9.62) this
is the same as the concentration reached at the end of a constant
—p
fed batch fermentation. The increase
in production time from 19.8 to 22 h is not large.
The optimal design of a fed batch fermentation that gives the maximum productivity and yet satisfies the
constraint imposed by a limited oxygen transfer should follow the lines illustrated in this example. One
should, however, not be misled to believe that this is the over all best production policy. A continuous
steady state fermentation has a far greater productivity.
Let the reactor volume be 4
V0
since this volume must be available at the end of the fed batch process. If
a continuous production of biomass with
x =
35.62 g L
"1
is to be maintained in the reactor then sf should
be 35.62/
Ysx
= 61.75 g L
' 1
when the miniscule effluent glucose concentration is neglected. If D = 0.25 h"\
the highest D value for which no ethanol is produced, then
qx
= 0.25 ' 35.62 = 8.905 g L
’ 1
h
'1
which
cannot be supported by the available mass transfer coefficient. To obtain
qx
=
qxmwi =
5.116 g L
/ 1
h
"1
the
dilution rate must be lower, namely
D
= 0.1436 h‘\ But still, a much higher volume of glucose can be
processed to give
x =
35.62 g L
' 1
: v = 4 V
0
‘ 0.1436 L h
’1
or in 22 hours -
tfiaai
a total volume of 12.64
V0
compared to only 4
VD
by the optimal fed batch process._________________________________ ______
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