An Overview of Chemical Reaction Analysis - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 131

Unfortunately, A also decomposes into Q at the condition where R is formed:
A

(E3-8.2)
A costs $0.05/lb, R sells for $0.10/lb, and Q costs $0.03/lb to dispose. Assume that cost to
the reactor and operation is negligible.

/ Q ; r

2 ¼

k C A

2

(a) What relation between the k's will give a positive cash flow?
(b) What relation between the k's will give a 30% profit on sales?

Solution:
This is a system of first-order parallel reactions. The instantaneous selectivity of R is given
by, Equation (3.103) ,

r A j forming R

d ð #moles of A converted to R Þ r k

1 1

s R = A ¼ ¼ ¼

2 ¼

(E3-8.3)

d ð #moles of A reacted Þ r A r

1 þ

r k

1 þ

k

2

which is constant. Therefore, one can conclude that the conversion of A into R is always a frac-
tion of the total A reacted. This can simplify our computations (before we even know how to
perform reactor analysis). The fraction of A converted to R is s R/A and the fraction of A con-
verted to Q is 1

s R/A . Therefore, for every pound of A reacted,

k k

1 1

Profit per lb of A reacted ¼ $ 0:10 $ 0:03 1 $ 0:05 (E3-8.4)

k

1 þ

k k

1 þ

k

2 2

(a) For the process to give a positive cash flow, it is equivalent to require:

k k

1 1

0 < Profit per lb of A reacted ¼ $ 0:10 $ 0:03 1 $ 0:05 (E3-8.5)

k

1 þ

k k

1 þ

k

2 2

That is to say

10 k k

1 1

3 1 > 5 (E3-8.6)

k

1 þ

k k

1 þ

k

2 2

or

13 k

1

> 8 (E3-8.7)

k

1 þ

k

2

which can be reduced to

k

< 5

2

8 ¼ 0:625

(E3-8.8)

k

1

(b) To make 30% profit on sales is equivalent to say that the profit is 30% of the value of R
produced. Thus,

k k k

1 1 1

$ 0:10 $ 0:03 1 $ 0:05 ¼ 30 % $ 0:10 (E3-8.5)

k

1 þ

k k

1 þ

k k

1 þ

k

2 2 2