A batch reactor converts component A into B, which in turn decomposes into C: k1 k2 A → B → C where k1 = k10e^−E1/RT and k2 = k20e^−E2/RT. The concentrations of A and B are denoted by x1 and x2, respectively. The reactor model is dx1/dt = −k10x1e^−E1/RT dx2/dt = k10x1e^−E1/RT − k20x2e^−E2/RT Thus,

Question

A batch reactor converts component A into B, which in turn decomposes into [latex]C[/latex] :

[latex]\stackrel{k_{1}}{ A ightarrow B } \stackrel{k_{2}}{ ightarrow C }[/latex]

where [latex]k_{1}=k_{10} e^{-E_{1} / R T}[/latex] and [latex]k_{2}=k_{20} e^{-E_{2} / R T}[/latex].

The concentrations of [latex]A[/latex] and [latex]B[/latex] are denoted by [latex]x_{1}[/latex] and [latex]x_{2}[/latex], respectively. The reactor model is

[latex]\begin{aligned}&\frac{d x_{1}}{d t}=-k_{10} x_{1} e^{-E_{1} / R T} \&\frac{d x_{2}}{d t}=k_{10} x_{1} e^{-E_{1} / R T}-k_{20} x_{2} e^{-E_{2} / R T}\end{aligned}[/latex]

Thus, the ultimate values of [latex]x_{1}[/latex] and [latex]x_{2}[/latex] depend on the reactor temperature as a function of time. For

[latex]\begin{array}{ll}k_{10}=1.335 imes 10^{10} min ^{-1}, & k_{20}=1.149 imes 10^{17} min ^{-1} \E_{1}=75,000 J / g mol , & E_{2}=125,000 J / g mol \R=8.31 J /( g mol K ) & x_{10}=0.7 mol / L , \quad x_{20}=0\end{array}[/latex]

Find the constant temperature that maximizes the amount of B, for [latex]0 \leq t \leq 6 min[/latex].

Accepted Answer

The reactor equations are:

[latex]\frac{d x_{1}}{d t}=-k_{1} x_{1} (1)[/latex]

[latex]\frac{d x_{2}}{d t}=k_{1} x_{1}-k_{2} x_{2} (2)[/latex]

Where [latex]k_{1}=1.335 * 10^{10} e^{-75000 /(8.31 * T)} ; k_{2}=1.149 * 10^{17} e^{-125000 /(8.31 * T)}[/latex]

By using MATLAB, this differential equation system can be solved using the command "ode45". Furthermore, we need to apply the command "fminsearch" ino order to optimize the temperature. In doing so, the results are:

[latex]T_{o p}=360.92 K ; x_{2, \max }=0.343[/latex]

Question 19.8: A batch reactor converts component A into B, which in turn d...

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