356
Chapter 9
It is desired to treat a feed stream v of substrate with concentration .syin a stirred tank reactor
of given volume
V
.
The kinetics follows a Monod expression with
Ks =asf
. Now the
maximum amount (in kilograms per hour) of substrate that can he converted is
-
V q „ = m * f -
*m ) -
sfDVQ.
- Sopt)
where ^ =
Sopt
is calculated from Eq. (9.34) and
D
=
from Eq. (9.35).
If a given feed stream value v is larger than
D ^ V
, the best policy is to divert some of the
feed stream— a ’’shunt” around the bioreactor—or to use cell recirculation as discussed in
section 9.1.3.
By the same reasoning, the highest cell productivity (in kilograms per hour) for the given volume
V
is calculated from Eq. (9.34). The same formula can be used to calculate
(Pp )mtx
since
Pp =Yxp'Px.
Needless to say, the term “optimal” used in Eqs. (9.34) and (9.35) would be highly misleading in
situations where the cost of substrate has to be weighted against the value of the biomass or
product. Environmental constraints might make it necessary to go to lower effluent values of the
substrate concentration than that given by Eq. (9.34), at the cost of increasing
V
,
to have enough
capacity to process the given feed stream. The above mathematical treatment is used solely to
illustrate how one of many optimization problems associated with bioreactor operation can be
solved.
Example 9.3 Optimal productivity in steady- state stirred tank bioreactors.
In figure 9.3 the problem of finding the maximum steady state productivity in the stirred tank has been
solved graphically for several typical unstructured kinetic models. Substrate inhibition, product
inhibition or maintenance added to a basic kinetic rate model always lowers the maximum productivity,
which is proportional to
qx.
It is easily seen that the area of the hatched rectangle is (
J-S) Y„ sf qx!
=
fi1
= D 1
. With the graphical representation of the problem it is also clear that a given value of
s
(or
D)
corresponds to one and only one value of
q„
while a given value of
qx
can be obtained for two different
values of
s
(or
D
) , one with low
x
and high
D
and the other with high
x
and low
D.
Hatched area : Holding time to achieve maximum productivity.
A : 3.37 h ; B : 7.17 h ; C : 4.69 h ; D : 4.29 h.
Substrate inhibition shifts
towards lower values because of the negative influence of
S
on /j at high
S
values. Product inhibition, and to a smaller extent maintenance demands shifts
Sop
,
towards higher
S
values, in the first case to avoid that too much product is formed and in the last case because the relative
amount of substrate that goes to biomass decreases for small
D
values (i.e. small
S).