Design of Fermentation Processes
383
discussed in Section 9.3.2.
Example 9.11 Calculation of final values of
s,p
andx after a change in
Consider the situation of figure 9.3 C with
= 2 g L
'1
and
S/=
4 g L'1. Ysp=Ysll=
Vi
.
For
the parameter a = 0.2 and for
sf
the parameter is 0.1.
Likewise the parameter
sfYsp/pmBI
changes from = % to 1
'A
.
For D = 0.2 h
"1
the algebraic equation (1) is solved for Sf = 2 and 4 g L'1, respectively:
P
max
For sf = 2 g L
1
: S = 0.4 and s = 0.8 g L'1. x = 0.5 (2 - 0.8) = 0.6 g L
'1
= p.
For sf = 4 g L
'1
: S = 0.6 and s = 2.4 g L"1. x = p = 0.8 g L'1.
Due to the product inhibition the steady state value of s is much higher at
sf =
4 g L
' 1
than at
2 g L~'. The relation (9.90) is not valid anymore._____________________________________
Z> = -
s + K.
After this general discussion of the relationship between
x, s
and
p
during the transient, the time-
profile will be calculated for a few examples.
a
First consider a change in
D.
The algebraic relations between
s
and
x
and between
p
and
x
are inserted in the biomass balance (9.85 a) and this is solved by separation of variables from
x(/=
0
) -
x(i
dx
=
p{s(x)
, />(x))
x-D x
;
x{t =
0
) = x
0
at
(9.92)
0
*(*)
~ D) X
For Monod kinetics:
Mmax (Sf
f
K s + ( S f ~ Yv X)
(9.93)
(9.93) is inserted in (9.92) which is integrated to
previous page 406 Bioreaction Engineering Principles, Second Edition  read online next page 408 Bioreaction Engineering Principles, Second Edition  read online Home Toggle text on/off