Sustainability and Stability - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 811

(d) How is the trajectory look like when the conditions are reversed, i.e. from the steady
state for S 0 ¼ 60 g/L back to the case where the feed concentration is S 0 ¼ 100 g/L with
all other conditions remain the same.
Solution. The flow schematic is shown in Fig. E16-2.1 .
(a) Mass balance on substrate (glucose) leads to

d ð SV Þ
d t ¼ 0

QS 0 QSþ r S V ¼ (E16-4.2)

Since V ¼ constant, Eqn (16-3.2) can be reduced to

d S
d t ¼ DðS 0 SÞþr S

(E16-4.3)

Substituting the rate

r X j total

YF X=S ¼ m G X

r S ¼ (E16-4.4)

YF X=S

We obtain

d S

d t ¼ DðS 0 SÞ m G X YF X=S ¼ MS S MC S

(E16-4.5)

At steady state, Eqn (16-4.5) is reduced to MS S ¼ MC S .
Mass balance on biomass leads to

d ð XV Þ
d t

QX 0 QXþ r X j net V ¼ (E16-4.6)

Since the feed is sterile and substitute the net growth rate in Eqn (E16-4.6) ,we
obtain

d X
d t ¼DXþðm G k d ÞX

(E16-4.7)

which can be rewritten as

m G k d X
Z

D X
Z

(E16-4.8a)

d X
d t ¼ MG X MR X

or

d X
d t ¼ Xð SMG X SMR X Þ

(E16-4.8b)