Design of Fermentation Processes
This was of course just what was done in Example 6.5 when the two non-linear rates
were linearized around a given state (.
The solution of the two linear differential equations with constant coefficients is a weighted sum of
exponentials. If both exponentials have negative arguments then the perturbation will die out, and
the steady state will be exponentially approached.
J determine the exponentials of the solution.
Matrix J of dimension (N x N) has N eigenvalues. These are determined as the N zeros of the
polynomial obtained by calculating the determinant of the matrix J - AI.
For our case of N = 2 the determinant is a polynomial of degree 2 in
. ~ YxsM
stisx ~ D ~ !
is defined as j ~ j
and the subscript ss stands for the steady state that is investigated.
In the steady state
p = D
must be different from
during the transient since
With this simplification the following equation is obtained for
X(Yxsp,x + D + X) + YxsxDps =0
There is always one negative eigenvalue = -
is positive for the steady state the other
eigenvalue is also negative.
For Monod kinetics p increases monotonically with
is positive for any steady state.
For substrate inhibition kinetics, Eq (9.28), all steady states to the left of the maximum
s = (Kj Ks)'a
> 0, while
0 to the right of the maximum
The stability of the steady state is determined by the following
If all eigenvalues have a negative real part then the steady state is