Design of Fermentation Processes
discussed in Section 9.3.2.
Example 9.11 Calculation of final values of
andx after a change in
Consider the situation of figure 9.3 C with
= 2 g L
4 g L'1. Ysp=Ysll=
the parameter a = 0.2 and for
the parameter is 0.1.
Likewise the parameter
changes from = % to 1
For D = 0.2 h
the algebraic equation (1) is solved for Sf = 2 and 4 g L'1, respectively:
For sf = 2 g L
: S = 0.4 and s = 0.8 g L'1. x = 0.5 (2 - 0.8) = 0.6 g L
For sf = 4 g L
: 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
4 g L
2 g L~'. The relation (9.90) is not valid anymore._____________________________________
Z> = -
s + K.
After this general discussion of the relationship between
during the transient, the time-
profile will be calculated for a few examples.
First consider a change in
The algebraic relations between
are inserted in the biomass balance (9.85 a) and this is solved by separation of variables from
) = x
~ D) X
For Monod kinetics:
K s + ( S f ~ Yv X)
(9.93) is inserted in (9.92) which is integrated to