Continuous Cultivation - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 629

effluent volumetric flow rates: Q m Q m 1 is the volumetric fresh feed rate to stage m . The

specific growth rates:

m G m ¼ m max S m

K S þ S m ¼ m net m þ k d

(12.52)

The substrate and biomass concentrations can be solved one stage after another.

s

b 2 m 4D m ðm max D m k d ÞðD m þ k d Þ

Q m1
Q m

1 Q m1
Q m

b m D m S m1 þ D m S 0m

S m ¼

2D m ðm max D m k d Þ

(12.53)

where

b m ¼ðD m þ k d ÞD m K S þ m max Q m1

D m S m1
YF X = S þðm max D m k d Þ

Q m

(12.54)

1 Q m1
Q m

D m S 0m þ Q m1
Q m

D m S m1

Substituting Eqn (12.52) into Eqn (12.50) , one obtain

Q m1
Q m

D m X m1

YF X = S þ Q m1

1 Q m1
Q m

D m S m1 þ D m S 0m D m S m

Q m

X m ¼ YF X = S (12.55)

D m þ k d

Example 12-3. Data for the production of a secondary metabolite from a small-scale batch
reactor are shown in Fig. E12-3.1 . There is a stream of substrate available at 120 L/h of the
same medium as that generating the batch fermentation data. Assume that two reactors,
each with 800-L working volume, are available. You will use exactly the same culture condi-
tions (medium, pH, temperature, and so on) as in the batch reactor.

(a) Determine the outlet concentration of the product.
(b) Compare that obtained in part (a) to the value predicted if a single 1600-L reactor was
used.

Solution. This is a chemostat problem. Mass balance for biomass over the i th chemostat at
steady state yields:

QX i1 QXiþ r X i V i ¼ 0 (E12-3.1)

Dividing through by V i , we obtain

r Xi ¼ D i ðX i X i1 Þ (E12-3.2)
where r X i is the rate of biomass production inside the i th chemostat and X i is the biomass
concentration in the effluent stream of the i th chemostat, which is the same as that inside
the i th chemostat. Therefore, to solve the chemostat problem, we will need the rate of biomass
production.