Batch Reactor - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 159

The operating conditions for the break-even point can be obtained by setting the gross
profit to zero.

ln 1 f A
k

$ A F B

F B
C A0 f A

0 ¼ GP$ ¼ $ B F B

f A

$ V t L (E4-5.13)

which can be rearranged to

ln 1 f A
k

$ $ B $ 1

C A0 ¼ 0

$ B

f A t L (E4-5.14)

or

0:001
3

f A 0:43 þ lnð1 f A Þ¼0 (E4-5.15)

Solving Eqn (E4-5.15) , we obtain

f A ¼ 0:430187

4.3. NON-ISOTHERMAL BATCH REACTORS

The solution of a batch reactor problem involves the solution of sets of ordinary differen-
tial equations from the species mole balance in the reactor as we have noted above. Mole
balance leads to,

d n j

d

r j V ¼ (4.2)

t

or

d n j

d

t ¼ r j V

(4.23)

with initial conditions set in the reactor as t C j 0 . If the reactor is not oper-

ated isothermally, an energy balance equation is needed as well, Eqn (3.134), or

¼ 0, n j ¼ n j 0, or C j ¼

N S N R

d ðpVÞ

d

d d

¼ Q W s V

C Pj n j

t

r i DH Ri (4.24)

t

j¼1 i¼1

T 0 . For liquids, d

ð Þ

d t z

pV

with t 0 , which further simplifies Eqn (4.24) .

The reaction rates r j 's given as functions of the concentrations (and temperature):

¼ 0, T ¼

N R

d n j

d t ¼ r j V ¼ V

n ij r i (4.25)

i¼1

r i ¼ f ðC k ; TÞ i ¼ 1; 2; 3; . :; N R (4.26)