Batch Reactor - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 157

Solution. Let the reaction time be t and the total preparation time (reactor loading, prepara-
tion, unloading, and cleaning) be t L . The total time required for each batch is then t B ¼

t þ t L .

a. Mole balance of A in the reactor leads to

d n A

d

0 0 þ r A V ¼ (E4-5.1)

t

The rate law gives

r A ¼ v A r ¼k C A (E4-5.2)

It is understood that the reaction mixture density is not changing. We have

V ¼ n A0 1 f A

C A ¼ n A ¼ C A0 ð1 f A Þ (E4-5.3)

V

Substituting Eqns (E4-5.2) and (E4-5.3) into Eqn (E4-5.1) and integrating with the initial
point of f A ¼ 0at t ¼ 0, we obtain

ln 1 f A
k

t ¼ (E4-5.4)

We know the production rate of B, F B0 ¼ 100 mol/h. From stoichiometry,

F A0 ¼ F B
f A

(E4-5.5)

Therefore, the reactor volume can be obtained from Eqn (4.22) as

ln 1 f A
k

V ¼ t B Q ¼ðt L þ tÞ F B

F B
C A0 f A C A0 f A ¼

t L (E4-5.6)

b. Based on the economic information at hand, we can estimate the gross profit for the
reaction process as

¼

$ B F B $ A F A 0 $ V V

(E4-5.7)

GP$

where $ B ,$ A , and $ V are the molar value of product B, molar cost of reactant A, and the unit
operating cost of reactor based on the reactor volume and time, respectively.
Substituting Eqns (E4-5.5) and (E4-5.6) into Eqn (E5-4.7) , we obtain

ln 1 f A
k

$ A F B

F B
C A0 f A

GP$ ¼ $ B F B

f A

$ t L (E4-5.8)

V

To find the optimum conversion, we maximize the gross profit. That is starting by
setting

dGP$
d

¼ 0 (E4-5.9)

f A