382
Chapter 9
Xq
-
Ypxp0
Xg - YpX
YSp
( Sf Sq
)
Xg - YX
I (sj-
Sg)
0
B.
b
Feed substrate concentration changed from s® to sf ; D is unchanged:
x0+Y„s0=A +
s°f + Ysx {Sf
- j J ) or
A
= -
Ya (sf - s°).
X0
-
YptPo = Xg-YSI
( i /
-S0) = 0 - B .
c
A pulse of substrate added at
t
= 0+. D and
sf
are unchanged.
x0 + Ysx s0 + A s = A + YSI sf ,
or
A = As.
x0
-
Yptp 0 = Xg-Ysx(s/ -sg) = 0 - B.
In all three situations
B
is zero, and during the whole transient
p
= Yxp x.
The substrate- and
biomass balances are also uncoupled. For a step change in
D
the relation between s and x is an
algebraic equation
s
=
sf -Y„
x. When
sf
is changed (case b) or when a pulse of substrate is
added (case c) then
x + Y sxs= Y„ Sf - (sf - s°f)Ysz exp(-D t
)
x +
Y^s
=
As exp(-D t
)
(9.88 a, b)
For
Sf
>
the sum of x and
Ysx s
is always below its final value
sf
In case c the sum is
always positive.
Since D is unchanged during the transient then at the end of the transient when s and p have
reached their final values
sm
and p^
D = p
(
s0
,po, s
f
) =
fi(svp„sj)
(9.89)
If
p
does not depend on
p
then
s
must return to its initial value, and
,= s„.
Then from (9.88 a):
=
Ysxsf - Yn sx = Ysx( S f-sg) - xg
+
Ysx
(
Sf-s/)
(9.90)
When
p
depends on
p
then
sx.
is not equal to
sg
since
sf
will appear in
p
when
p
is eliminated
using a mass balance relation between
s
and
p
.The final value of x is obtained from
x r =Ypxp
x , where
p, = Ysp (sf - s j
and
s*,
is determined by solution of (9.89).
(9.90)
In the pulse experiment both x and s return to their initial values following (9.88 b
)provided that
the initial steady state is stable,
and that the added pulse is not too large. This aspect will be
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