Batch Reactor - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 158

Since the flow rate of product B is fixed, not a function of the conversion f A , we have

ln 1 f A
k

$ A F B

V F B

F B
C A0 f A

dGP$
d

$

kC A0 f A 1 f A þ

0 ¼ ¼ 0 þ

f A

$ t L (E4-5.10)

V

f A

which can be reduced to,

$ A kC A0
$

f A
1 f A þ kt L lnð1 f A Þ

0 ¼ (E4-5.11)

V

Since

$ A kC A0
$

2ð $ = mol Þ0:15=min 2ð mol = l Þ

þ kt L ¼ þ 0:15=min 10 h

0:03 $ =ð L $ h Þ

V

¼ 1290

Equation (E-5.11) is reduced to

f A
1 f A þ lnð1 f A Þ

1290 ¼ (E4-5.12)

This nonlinear equation can be solved to give

f A ¼ 0:9992297
Thus, the optimum conversion is f A ¼ 0.9992297. The reactor size can be computed from

Eqn (E4-5.6)

ln 1 f A
k

F B lnð1 0:9992297Þ

0:15 60

100
2 0:9992297

V ¼ t L

C A0 f A ¼

10

L

¼ 540:24 L

c. The cash flow based on the optimum conversion:

ln 1 f A
k

GP$
F A0 ¼

$ B F B
F A0

F B
F A0 f A

F B
F A0 C A0 f A

$ A $ t L

V

ln 1 f A
k

$

V
C A0

¼ $ B f A $ A t L

lnð1 0:99922297Þ
0:15 60

0:03
2

¼ 5 0:9992297 2 10 $ = mol A

¼ $ 2:8342 = mol A

d. At breaking even point, the net profit (or in this case the gross profit) is zero. Therefore, at
the break-even point, operating cost and the value of product are equal.
The value of product is,

$ B F B ¼ 5 100 $ = h ¼ 500 $ = h

which is also the operating cost at break even.