Ideal Flow Reactors - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 203

change rate of A” vs “concentration of A” (or “fractional conversion of A”), the steady-state
solution is the intercept between the “molar supply rate” and the “molar consumption rate”.
That is,

MC A ¼ MS A (5.57)

Example 5-4. Consider the reaction:

A / B ; k ¼ 0:15=min

in a CSTR. A costs $2/mol, and B sells for $5/mol. The cost of operating the reactor is $0.03/
L/h. We need to produce 100 mol of B/h using C A0 ¼

2 mol/L. Assume no value or cost of
disposal of unreacted A (i.e. separation or recovery cost to A is identical to the fresh A cost).

1. Perform a mole balance on the reactor to relate the conversion with reactor size.
2. What is the optimum conversion and reactor size?
3. What is the cash flow (or profit per mole of A feed to the reactor) from the process?
4. At what operating cost do we break-even?

Solution. A sketch of the reactor system is shown in Fig. E5-4 .

1. It is understood that the CSTR is operating under constant temperature (isothermal) and
steady-state conditions, i.e. no accumulation of any sort of materials inside the reactor.
Mole balance around the reactor for B yields

0 F B þ r B V ¼ 0 (E5-4.1)

The rate law gives

r B ¼ n B r ¼ kC A (E5-4.2)

It is understood that the reaction mixture density is not changing, i.e. the volumetric flow
rates at the reactor inlet and the outlet are nearly unchanged. We have

F A
Q ¼

F A 0 ð 1 f A Þ
Q

C A ¼ ¼ C A 0 ð1 f A Þ (E5-4.3)

F A0 , C A0 = 2 M

FIGURE E5-4 Well-mixed reaction vessel.