Develop an unsteady-state model for a stirred batch reactor, using the nonlinear continuous reactor model presented in Example 4.8 as a starting point. For the parameter values given below, compare the dynamics of the linearized models of the batch reactor and the continuous reactor, specifically

Question

Develop an unsteady-state model for a stirred batch reactor, using the nonlinear continuous reactor model presented in Example 4.8 as a starting point. For the parameter values given below, compare the dynamics of the linearized models of the batch reactor and the continuous reactor, specifically the time constants of the open-loop transfer function between [latex]c_{A}^{\prime}[/latex] and [latex]T_{c}^{\prime}[/latex], the concentration of A, and the jacket temperature, respectively. Assume constant physical properties and the following data:

Initial steady-state conditions and parameter values for the continuous case are

[latex]\begin{aligned}&\bar{T}=150^{\circ} F , \quad c_{A i}=0.8 mol / ft ^{3}, \quad \bar{q}=26 ft ^{3} / min , \&U A=142.03 \frac{ kJ }{\min ^{\circ} F }, \quad V=1336 ft ^{3}, \quad T_{c}=77^{\circ} F\end{aligned}[/latex]

The physical property data are

[latex]C=0.843 Btu / lb { }^{\circ} F , ho=52 lb / ft ^{3},-\Delta H_{R}=500 kJ / mol .[/latex]

The reaction rate is first order with a rate constant (in [latex]\min ^{-1}[/latex] )

[latex]k=2.4 imes 10^{15} e^{-20,000 / T}\left(T ext { in }{ }^{\circ} R ight) ext {. }[/latex]

For the batch case, linearize the model around [latex]T=\overline{T} [/latex].

Accepted Answer

1.- Because there is no steady state for a batch reactor, a new linearization point is selected at [latex]t=0[/latex]. Then,

Linearization point for batch reactor: [latex]t=0 \equiv t^{*}[/latex]

2.- Available information:

[latex]k=2.4 \times 10^{15} e^{-20000 / T}\left(\min ^{-1}\right)[/latex] where [latex]T[/latex] is in [latex]{ }^{\circ} R[/latex]

[latex]\begin{array}{ll}C=0.843 \frac{ BTU }{ lb ^{\circ} F } & V=1336 ft ^{3} \\rho=52 \frac{ lb }{ ft ^{3}} & q=26 \frac{ ft ^{3}}{\min } \(-\Delta H)=500 \frac{ kJ }{ mol } & C_{A i}=0.8 \frac{ mol }{ ft ^{3}} \T_{i}=150^{\circ} F & T_{s}=25^{\circ} C \U A=142.03 \frac{ kJ }{\min ^{\circ} F } &\end{array}[/latex]

For continuous reactor, [latex]\bar{T}=150^{\circ} F[/latex]

Physical properties are assumed constant.

Problem solution:

A stirred batch reactor has the following material and energy balance equations:

[latex]-k C_{A}=\frac{d C_{A}}{d t}[/latex] (1)

[latex](-\Delta H) k V C_{A}+U A\left(T_{s}-T\right)=V \rho C \frac{d T}{d t}[/latex] (2)

where [latex]k=k_{0} e^{-E / R T}[/latex]

From Eqs. 1 and 2, linearization gives:

[latex]\begin{aligned}&-\left[k^{*} C_{A}^{*}+k^{*} C_{A}^{\prime}+C_{A}^{*} k_{0} e^{-E / R T^{*}} \frac{E}{R T^{* 2}} T^{\prime}\right]=\frac{d C_{A}^{\prime}}{d t} (3) \&(-\Delta H) V\left[k^{*} C_{A}^{*}+k^{*} C_{A}^{\prime}+C_{A}^{*} k_{0} e^{-E / R T^{*}} \frac{E}{R T^{* 2}} T^{\prime}\right] \&+U A\left(T_{s}^{\prime}-T^{\prime}\right)=V \rho C \frac{d T^{\prime}}{d t} (4)\end{aligned}[/latex]

Rearranging, the following equations are obtained:

[latex]\begin{aligned}&b_{11} C_{A}^{\prime}+b_{12} T^{\prime}=\frac{d C_{A}^{\prime}}{d t} (5)\&b_{21} C_{A}+b_{22} T^{\prime}+b_{23} T_{s}^{\prime}=\frac{d T^{\prime}}{d t} (6)\end{aligned}[/latex]

where

[latex]\begin{aligned}&b_{11}=-k_{0} e^{-E / R T^{*}}=-13.615 \&b_{12}=-k_{0} e^{-E / R T^{*}} C_{A}^{*}\left(\frac{E}{R T^{* 2}}\right)=-0.586 \&b_{21}=\frac{(-\Delta H) k_{0} e^{-E / R T^{*}}}{\rho C}=155.30 \&b_{22}=\frac{1}{\rho C}(-\Delta H) k_{0} e^{-E / R T^{*}} C_{A}^{*}\left(\frac{E}{R T^{* 2}}\right)-\frac{U A}{\rho V C}=6.66 \&b_{23}=\frac{U A}{\rho V C}=2.43 \times 10^{-3}\end{aligned}[/latex]

From Example 4.8, substituting values for continuous reactor

[latex]\begin{aligned}&a_{11}=-13.636 \&a_{12}=-8.35 \times 10^{-4} \&a_{21}=155.27 \&a_{22}=-0.0159 \&b_{2}=2.43 \times 10^{-3}\end{aligned}[/latex]

(Note that, from material balance, [latex]\bar{C}_{A}=0.00114[/latex] )

Hence the transfer functions relating the steam jacket temperature [latex]T_{s}^{\prime}(s)[/latex] and the tank outlet concentration [latex]C_{A}^{\prime}(s)[/latex] are:

Continuous reactor:

[latex]\frac{C_{A}^{\prime}(s)}{T_{s}^{\prime}(s)}=\frac{-2.03 \times 10^{-6}}{s^{2}+13.651 s+0.3464}=\frac{-5.86 \times 10^{-6}}{2.887 s^{2}+39.4 s+1}[/latex]

then [latex]\tau_{\text {dom }} \approx 35 min[/latex]

Batch reactor:

[latex]\frac{C_{A}^{\prime}(s)}{T_{s}^{\prime}(s)}=\frac{-1.424 \times 10^{-3}}{s^{2}+6.931 s+0.26}=\frac{-5.47 \times 10^{-3}}{3.84 s^{2}+26.65 s+1}[/latex]

then [latex]\tau_{\text {dom }} \approx 25 min[/latex]

As noted in transfer functions above, the time constant for the batch is smaller than the time constant for the continuous reactor, but the gain is much larger.

Question 22.9: Develop an unsteady-state model for a stirred batch reactor,...

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