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Chapter 19

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}

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{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}