364
Chapter 9
and 13.5 m
3
h’1, respectively.
A conservatively designed ultrafilter supports a flux
F ~
10 L m
'2
h'!, and consequently the required filter
area is 1350 m2.
From an ultrafilter brochure one obtains the information that the necessary volumetric flow to the filter
in order to sustain a flux ir = 10 L m
'2
h
'1
through the filter is 0.74 m
3
(m
2
filter)'1. Hence a recirculation
flow of 0.74'1350 = 1000 m
3
h
1
is needed
Assume that a pressure drop of 4 bar across the filter is needed to support the flux, and that the pump has
a mechanical efficiency of 65 %. Then the energy input is 4 10
5
(0.65 ‘ 3600)'1
= 171 kW.
The reactor capacity has been increased by a factor 10 by cell recirculation, but at the cost of a
considerable energy input. A filter must be installed anyhow in order to separate the product stream from
the biomass and remaining proteins, but a conventional drum filter working on only 15 m
3
h
'1
is bound to
be cheaper than an ultrafilter treating 1350 m
3
h'‘.
A 190 000 t lactic acid per year plant (Dow -Cargill) has recently been commissioned in Omaha,
Nebraska. Economic considerations of the kind illustrated in this example are typical for the design of
large, integrated production plants.
9.1.4 The Stirred Tank with a Substrate Extracted From a Gas Phase
When one of the substrates is admitted with a gas phase that is sparged to the reactor there are, as
already indicated in Chapter 3 two possible candidates for the rate limiting process, the gas-to-
liquid transport and the bioreaction in the liquid phase.
If the transport process is rate limiting then:
kla(s'-st) - YUy
fi(sj x = D(s,-slf)
(9.47)
where
s,
is the liquid phase concentration of the reactant from the gas phase and
pi(s,)
is the
specific growth rate, expressed as a function of the limiting substrate concentration
s2.
As usual
for steady state continuous reactors
D
=
p(s,).
Both
Si
and
slf
can safely be set equal to zero compared with
s',
and (9.47) degenerates to
qx = Dx
= Yv s* kj a
(9.48)
If on the other hand a substrate
S2
in the liquid phase is rate limiting then (for a Monod rate
expression):
x = YS2X(s2f- s2) and qx = D x
(9.49)
The obvious choice is to work at conditions where
qx
calculated from either (9.48) or (9.49) is the
same.
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