Enzymes - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 360

Separation of variable leads to,
K m
S þ1

dS

¼r max d t (E8-2.4)

Integrating Eqn (E8-2.4) from t ¼ 0 and S ¼ S 0 , we obtain

S
S

K m ln

0 þ

S S

0 ¼r max t

(E8-2.5)

The conversion of substrate is defined as

S

0

S

f S ¼ (E8-2.6)

S

0

Substitute Eqn (E8-2.6) into Eqn (E8-2.5) , we obtain

S f S K m lnð1 f S Þ

r max

0

t ¼ (E8-2.7)

Thus, the time required for 90% conversion is

S f S K m lnð1 f S Þ

r max

¼ 0:10:90:0265 lnð10:9Þ
2:710 4

0

t ¼ s ¼ 559:3 s ¼ 9 : 322 min

Example 8-3. Consider the reaction:

P
catalyzed by enzyme E. Michaelis e Menten rate parameters are r max ¼

S /

0.15/min and

K m ¼ 0.01 mol/L when the enzyme loading 0.1 g/L. CSTR has been selected to produce

100 mol of P per hour from a stream of substrate containing 1 mol-S/L. S costs $0.1/mol,
P sells for $1/mol, and the enzyme E cost $100/g. The cost of operating the reactor is
$0.1/L/h. Assume no value or cost of disposal of unreacted S (i.e. separation or recovery
cost to S is identical to the fresh S cost).

(a) What is the relationship between rate of formation of P and enzyme loading?

(b) Perform a mole balance on the reactor to relate the concentration of exiting S with reactor
size.

(c) What is the optimum concentration of S at the outlet of the reactor? What is the optimum
enzyme loading?
(d) What is the cash flow per mole of product from the process?

Solution. A sketch of the reactor system is shown in Fig. E8-3 . Substrate and enzyme are
fed to the CSTR. The enzyme is let out unchanged, while some substrate is turned into
product P.

(a) The maximum rate is proportional to the enzyme loading, that is,

r max ¼ k E (E8-3.1)

0