Design of Fermentation Frocesses
407
A corresponds to Figure 9.15 B
y
2
d0
2
dO
+
% +
(A
■ 1 ) ^ = 0
dB
(
10
)
which on integration from
0
=
0 and réintroduction of X andX yields
A
+ ln(X,) + (A - l ) l n
(X2)=K
(11)
where the value of
K
is determined from the initial condition at
t =
0". If for
t
= 0+,
Xs = X!0 + &XS
and
X? =
Xo, one obtains
(A ~ 1)ln(J^2) =
0 2xloX 2
+ X - |A
- fi2Ysxsf X l
+ ln(X,)]=
/32xl0X 2 +a
(12)
For a given value of AX/, one may calculate the value of
a-.
The straight line
fi2xl0X 2
+
a
intersects
(fi,
-l)ln(X) twice if
a
is smaller (more negative) than the value
obtained from the intersection of the tangent of (A -l)ln(X) with slope A xl0 and the ordinate axis. The
slope of (A -1 )ln(X) is A^io for
X2=
(A -1) /
(fixl0)
= Xu- At this value of X , one obtains the following
equation for
Xf.
A
~P2Ysxsf X,
+ ln (X )= A
~P2Ysxsf (XlQ
+AXl) + M X w + AXl)
(13)
One solution of Eq. (13) is obviously X =
X l0 +
AXj, The other solution is obtained by numerical solution
of
F{X,
) = ln(X,) -
ß 2Ytxsf X ]
- ln(X10
+AX]) + ß 2Ysxsf (Xl0 + AX,
) = 0
(14)
A standard Newton iteration
Y
= x
f
F
\
X UF{X
u )
u+x
u
[ d F /d X j^
U
l - ß 2Ysxsf X u
(15)
will find the other zero of Eq. (14) in a few steps. Consequently, for a given Ax, the phase-plane plot ofX
versus
X,
extends in the X direction between a value of
Xs
found by solution of Eq. (15) and
Xs -
Xo
+
AXj. The maximum extension in the X direction is found for X =
X2o
. Ax, is large and positive, the
solution of Eq. (15) is close to zero, and if A x ~ -Xo* the solution of Eq. (15) is large and positive.
The extension of the phase plane plot in the
X2
direction is found by an analogous procedure. X must be
equal to
X!0
and the two solutions of