Sustainability and Stability - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 840

Note that YF B =P A is the biomass yield factor that requiring P A as one of its substrates, and
YF P A is the amount of P A made per unit mass of A. Similar definitions apply to YF A/P B and
YF P B If we consider the pure mutualistic state, then we ignore Eqn (16.71) . For a coexistent
state to exist, D ¼m A k dA ¼m B k dB . It is also clear that the rate of production of P A and
P B must exceed their consumption (by the other species). Thus,

YF P A m A X A > m B X B

r PA > 0 0 (16.76)

YF B=P A

YF P B m B X B > m A X A
YF A=P B

r PB > 0 0 (16.77)

Since all the quantities in the equalities (16.76) and (16.77) are greater than zero, left-hand side
multiply by left-hand side and right-hand side multiply by right-hand side leads to
YF P A m A X A YF P B m B X B > m A X A

m B X B
YF B=P A

(16.78)

YF A=P B

Eliminating the identical terms, we obtain

1
YF A=P B YF B=P A

YF P A Y P B > (16.79)

It is also clear that the specific growth rates are less than their maximum values, that is

D < min ðm A max k dA; m B max k dB Þ (16.80)

Eqns (16.79) and (16.80) determine whether Eqns (16.69) to (16.71) allow the potential exis-
tence of a purely mutualistic steady state. The stability of such a coexistent state has been
examined, where m B were represented by various growth functions. Using a linear

stability analysis, it can be shown that this pure mutualistic state results in a saddle point
( Fig. 16.11 c), and the system is unstable for all physically accessible values of D. If, however,
the growth-rate-limiting substrate for either A or B is S, then a stable coexistent state can be
found.

m A and

16.6.4. Predator and Prey Interactions
The growth of a protozoa (predator) on bacteria (prey) in a chemostat is a classic stability
problem. In a chemostat culture, the following balances can be written for substrate (S), prey
(b), and predator (p).

d ðX b VÞ
d t

Qð0X b Þþr b V ¼ (16.81)

d ð X p V Þ
d t

Qð0X p Þþr p V ¼ (16.82)

d ðSVÞ
d t

QðS 0 SÞþr S V ¼ (16.83)