408
Chapter 9
< A
-\)\n{X2) - p 2x[0X 1
+ ln
X '°
} + 0 2Y„sf àXx
+P2xi0X 20
- (A - 1) ln(X20) = 0
(16)
+AXy J
span the range of the phase plane plot in the
direction.
In summary: For AA)
-
0.4250
[X:{t
= 0^) = 0.55]—the situation depicted in Fig. 9.15B—the phase plane
curve is delimited by the values listed in the following table:
X,
0.55
0.00715
0.125
0.125
X
0.25
0.25
1.68642
0.001999
For any
Xt
value between 0.55 and 0.00715, the corresponding two values of
X2
can be found by iterative
solution of Eq. (11) at the given
&Xj
= 0.4250. The closed curve A in Fig. 9.16 is the result of these
calculations. For large AA), the point
(Xh X2)
spends most of the very large cycle time
T
in the close vicinity
of (0, 0), while
1. An obvious question is why the transient is not caught by the steady-state solution
X}-
X2
= 0. The answer is quite simply that this steady state is not stable when A > 1. as is easily seen by
insertion in Eq. (5). The eigenvalues are zeros of
U + 1)2(A -1 -> 1 ) = 0
(17)
and one eigenvalue is real and positive for
1.
The third possible steady state with
Xs
> 0 and
X2
= 0 is not stable either, since one eigenvalue is positive as
long as A > 1. As discussed further in Problem 9.9, a steady state with 0 <
X}<
1 and
X2 ~
0 can also be
obtained for A > 1
if the growth kinetics for
Xs
is changed to include substrate limitation
(3).
Now the steady state is determined by
A /(A > ) = 1 ; -^2o= ®>
* 10= 1 -A >
(IB)
for which the eigenvalues are solutions of
{(32YsxSfX w - \- X ) \X 2
+
A
- 1
x in+i
a
+ A
*)
àS)Sn
(19)
i.e.,
-1
|
-*io
So
(20)
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