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## Q. 19.15

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

$r_{1}=k_{1} C_{A} \quad r_{2}=k_{2} C_{B}$

where the rate constants have the following temperature dependence:

\begin{aligned}&k_{1}=3.0 \times 10^{6} \exp (-5000 / T) \\&k_{2}=6.0 \times 10^{6} \exp (-5500 / T)\end{aligned}

Each rate constant has units of $h ^{-1}$, and $T$ is in $K$.

Use the MATLAB Optimization Toolbox or Excel to determine the optimum values of temperature $T( K )$ and flow rate $F_{B}( L / h )$ that maximize the steady-state production rate of component B. The allowable values are $0 \leq F_{B} \leq 200$ and $300 \leq T \leq 500$

Available Information

(i) The reactor is perfectly mixed.

(ii) The volume of liquid, $V$, is maintained constant using an overflow line (not shown in Fig. E19.14).

(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\end{aligned}

## 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{aligned}&0.3 F_{B}+3 \times 10^{6} e ^{(-5000 / T)} C_{A} V-6 \times 10^{6} e ^{(-5500 / T)} C_{B} V=\left(150+F_{B}\right) C_{B} \\&45+6 \times 10^{6} e ^{(-5500 / 7)} C_{B} V-3 \times 10^{6} e ^{(-5000 / T)} C_{A} V=\left(150+F_{B}\right) C_{A} \\&\quad F_{B} \leq 200 \\&300 \leq T \leq 500 \\&C_{A}, C_{B}, F_{B} \geq 0\end{aligned}

By using Excel- Solver, the optimum values are

\begin{aligned}&F_{B}=200 l / hr \\&C_{A}=0.104 molA / l \\&C_{B}=0.177 mol B / 1 \\&T=311.3 K\end{aligned}