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 [latex]A[/latex] into [latex]B[/latex], which in turn decomposes into [latex]C[/latex] :

[latex]\begin{gathered}k_{1} \quad k_{2} \A ightarrow B ightarrow C\end{gathered}[/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{aligned}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{aligned}[/latex]

Find the constant temperature in [latex]K[/latex] that maximizes the amount of B, for [latex]0 \leq t \leq 8 min[/latex]. Next allow the temperature to change as a cubic function of time

[latex]T(t)=a_{0}+a_{1} t+a_{2} t^{2}+a_{3} t^{3}[/latex]

Find the values of [latex]a_{0}, a_{1}, a_{2}, a_{3}[/latex] that maximize [latex]x_{2}[/latex] by integrating the model and using a suitable optimization method.

Accepted Answer

The reactor equations are:

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

where [latex]k_{1}=1.335 imes 10^{10} e ^{-75,000 /(8.31 imes T)}[/latex] and [latex]k_{2}=1.149 imes 10^{17} e ^{-125,000 /(8.31 imes T)}[/latex]

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

a)

Isothermal operation to maximize conversion [latex]\left(x_{2}(8) ight)[/latex] :

[latex]T_{o p}=357.8 K \quad[/latex] and [latex]\quad x_{2 \max }=0.3627[/latex]

b) Cubic temperature profile: the values of the parameters in [latex]T=a_{0}+[/latex] [latex]a_{1} t+a_{2} t^{2}+a_{3} t^{3}[/latex] that maximize [latex]x_{2}(8)[/latex] are:

[latex]\left\{\begin{array}{l}a_{0}=372.78 \a_{1}=-10.44 \a_{2}=2.0217 \a_{3}=-0.1316\end{array} \quad ext { and } \quad x_{2 \max }=0.3699 ight.[/latex]

The optimum temperature profile and the optimum isothermal operation are shown in Fig. S22.10.

Question 22.10: A batch reactor converts component A into B, which in turn d...

Related Answered Questions