Ideal Flow Reactors - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 205

Since the conversionofAmust be between 0 (no conversion) and1 (100%conversion) inorder
for it to be physically possible, we discard the value that is greater than unit and obtain
f A ¼ 20 3

20 3

p

p

1
1199 ¼ 0:97194

(E5-4.14)

Next, we check if indeed this conversion gives the maximum gross profit. We can either
plot out the variation of GP$ vs f A to see if we indeed having a maximum GP$ at this
f A value or checking the second derivative of GP$,
d 2 GP$
d f A ¼

2 $ A F B
f A

2 $ V F B
kC A 0 1 f A 3

(E5-4.15)

Plug in the values we know up until this step, we obtain
d 2 GP$
d f A ¼

2 2ð $ = mol Þ100ð mol = h Þ

0:97194 3

2 0:03½ $ =ð lh Þ 100ð mol = h Þ

3

0:15ðmin 1 Þ60ð min = h Þ2ð mol = l Þð1 0:97194Þ

Þ< 0
A negative second-order derivative means that the value of GP$ is lower for either higher
or lower f A . Thus, we indeed have obtained a maximum value of the gross profit at the
conversion of f A ¼

¼15527ð $ = h

0.97194. This confirms that the optimum conversion is f A ¼ 0.97194.

The reactor size can be computed from Eqn (E5-4.4)

F B 100

0:15 60 2 ð1 0:97194Þ

V ¼

kC A 0 1 f A ¼

¼ 197:99 l

l

3. The cash flow based on the optimum conversion:

$ B F B
F A 0

F B
f A F A 0

F B

GP$
F A 0 ¼

kC A 0 1 f A F A 0

$ A $ V

$ V f A

¼

$ B f A $ A

kC A 0 1 f A

0:03 0:97194
0:15 60 2 ð1 0:97194Þ

¼ 5 0:97194 2 $ = mol A

¼ $ 2:802= mol A

4. 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.