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Chapter 19

Q. 19.14

A reversible chemical reaction, A \rightleftarrows B, occurs in the isothermal continuous stirred-tank reactor shown in Fig. E19.14. The rate expressions for the forward and reverse reactions are

\begin{aligned}&r_{1}=k_{1} C_{A} \\&r_{2}=k_{2} C_{B}\end{aligned}

For the information given below, use a numerical search procedure to determine the value of F_{B}( L / h ) that maximizes the production rate of C_{B} (i.e., the amount of C_{B} that leaves the reactor, mol  B / h ). The allowable values of F_{B} are 0 \leq F_{B} \leq 200  L / h

Available Information

(i) The reactor is perfectly mixed.

(ii) The volume of liquid, V, is maintained constant using an overflow line (not shown in the diagram).

(iii) The following parameters are kept constant at the indicated numerical values:

\begin{aligned}V &=200 L & F_{A} &=150  L / h \\C_{A F} &=0.3  mol A / L & C_{B F} &=0.3  mol B / L \\k_{1} &=2  h ^{-1} & k_{2} &=1.5  h ^{-1}\end{aligned}

Step-by-Step

Verified Solution

Material balance:

Overall : \quad F_{A}+F_{B}=F

Component B: \quad F_{B} C_{B F}+V K_{1} C_{A}-V K_{2} C_{B}=F C_{B}

Component A: \quad F_{A} C_{A F}+V K_{2} C_{B}-V K_{1} C_{A}=F C_{A}

Thus the optimization problem is:

\max \left(150+F_{B}\right) C_{B}

Subject to:

\begin{gathered}0.3 F_{B}+400 C_{A}-300 C_{B}=\left(150+F_{B}\right) C_{B} \\45+300 C_{B}-400 C_{A}=\left(150+F_{B}\right) C_{A} \\F_{B} \leq 200 \\C_{A}, C_{B}, F_{B} \geq 0\end{gathered}

 By using Excel- Solver, the optimum values are

\begin{aligned}&F_{B}=200  l / hr \\&C_{A}=0.129  mol A / l \\&C_{B}=0.171  mol B / 1\end{aligned}