Sustainability and Stability - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 797

which can be reduced to

DðS 0 SÞ¼r S (E16-2.3)

The rate of consumption of substrate is given by

r X j total

YF X=S ¼ m G X

r S ¼ (E16-2.4)

YF X=S

Thus, mass balance on substrate leads to

DðS 0 SÞ¼ m G X
YF X=S

(E16-2.5a)

or

X
YF X=S

m max S
K S þSþS 2

DðS 0 SÞ¼ (E16-2.5b)

= K I

which is equivalent to MS S ¼ MC S . However, the right-hand side still contains an
unknown quantity X that needs to be determined. Therefore, we cannot solve this equa-
tion alone.
Mass balance on biomass leads to

d ð XV Þ
d t ¼ 0

QX 0 QXþ r X j net V ¼ (E16-2.6)

Since the feed is sterile and substitute the net growth rate in Eqn (E16-2.6) , we obtain

QXþðm G k d ÞXV ¼ 0 (E16-2.7)

which can be reduced to

D ¼ m G k d (E16-2.8)

0. One should note that X ¼ 0 is also a solution for Eqn (E16-2.7) , which corresponds
to the wash out conditions.
Eqn (16-2.8) can also rearranged to give

If X s

Dþ k d ¼ m G (E16-2.9)

which is in direct analogy to what we have learned so far:

SMR X ¼ SMG X (E16-2.10)
i.e. specific mass removal rate of biomass (by reactor effluent and cell death) ¼ specific mass
generation rate of biomass (by growth). This equation contains only one unknown and thus
can be solved directly.
Substituting Eqn (E16-2.1) into Eqn (E16-2.9) , we obtain

m max S
K S þSþS 2

Dþ k d ¼ (E16-2.11)

= K I