Design of Fermentation Processes
391
Figure 9.9 gives the time profiles of
S = s /sf
after perturbation of an unstable steady state with
substrate inhibition kinetics. 5 tends to either 5 = 1 (for a positive perturbation of 5) or (for a
negative perturbation of
S)
to the asymptotically stable steady state which accompanies the
D
value below
pex)r
of Eq (9.29).
Those two stable steady states are the
attractors
for the solution of the dynamic mass balances.
In Figure 9.11 a much more general representation of the dynamics is given in terms of a phase
diagram for the same example as discussed in Figure 9.9. Here the track of
(SyX)
is followed
from any starting point ( 5 ^ ;) , and the diagram clearly shows which initial conditions will lead
to a transient that ends up in each o f the two attractors. The diagram also shows that a large
positive 5-perturbation of the asymptotically stable steady state will bring the transient “over the
hill” and force it towards the stable (but trivial) steady state at 5 = 1.
Maintenance substrate demands coupled to more complex kinetics is easily included in the
analysis if numerical solution of the dynamic mass balances is acceptable. Only in a few
exceptional cases will a more complicated rate expression
p(x,p)
We shall briefly discuss the solution of the mass balances when
qs =
-(Fa1™
pis) + ms) x
as in Eqs
(9.12)-(9.13). In this case, and in complete analogy with the derivation of (9.104) one obtains:
2A = -
+ D)± ^ { Y ^ x + D)2
-4p,xm,
(9.105)
The solution of Eq. (9.105) is
H
f
-D-Ö
[~YX
sPsX
+
^
for
Ps >
0
at
(*o>so)
(9.106a)
x=<
\
-D +
Ô
[-YxsMsX-ô
for
M
t <
o at
(*0>50)
(9.106b)
where the value of the positive constant
8
can be calculated for a given maintenance constant
ms,
and it is assumed that
(Y^PsX-D)2 >4psxms
when
ps >0
(9.107)
With reference to the previous example of substrate-inhibition kinetics it appears that maintenance
destabilizes the nontrivial steady states when ^ >
0
, while an unstable steady state would be
somewhat stabilized. We shall give a little more detail for the very simplest case where
p
is given
by the Monod expression. For a given
S-slsf ,D= p mmSI
(5 +
a),
and
X - xl
(T
„ 5
f ) given by
Eq. (1) of Example 9.3:
P* =
1
dp
s j dS
D
sf
5(5
+ a)
(9.108)