Question 10.4: Determination of the Volume Requirements for Adiabatic Opera...

Determination of the Volume Requirements for Adiabatic Operation of a Tubular Reactor with Exothermic Reaction

Butadiene will react with ethylene in the gas phase at temperatures above 500°C. This reaction is a simple example of a Diels–Alder reaction.

Determination of the Volume Requirements for Adiabatic Operation of a Tubular Reactor with Exothermic Reaction Butadiene will react with ethylene in the gas phase at temperatures above 500°C. This reaction is a simple example of a Diels–Alder reaction. If an equimolar mixture of butadiene and

If an equimolar mixture of butadiene and ethylene at 450°C and 1 atm is fed to a tubular reactor, determine the space times required to convert 10% of the butadiene to cyclohexene for isothermal and for adiabatic modes of operation.

Wasserman (6) reported the following data for this reaction:

k=10^{7.5} e^{-27,500 / R T}  L /( mol \cdot s )

\Delta H_R=-30,000  cal / g - mol

The reverse reaction may be neglected. The following values of gas-phase heat capacities may be assumed to be constant over the temperature range of interest.

C_{p, C_4 H _6}=36.8  cal /( g - mol \cdot K )

C_{p, C_2 H _4}=20.2  cal /( g - mol \cdot K )

C_{p, C_6 H _{10}}=59.5  cal /( g - mol \cdot K )

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From the units on the reaction rate constant, the reaction is second-order. There is a volume change on reaction and δ_{a} = (−1/2). Thermal expansion will also occur, so equations (3.1.44) and (3.1.47) must be combined to obtain the reactant concentrations. Because equimolar concentrations of reactants are used, the design equation becomes

V=V_0\left(1+\delta_{ A } f_{ A }\right) \frac{T}{T_0} \frac{P_0}{P}          (3.1.44)

C_{ A }=\frac{n_{ A }}{V}=\frac{n_{ A 0}\left(1-f_{ A }\right)}{V_0\left(1+\delta_{ A } f_{ A }\right)}=C_{ A 0}\left(\frac{1-f_{ A }}{1+\delta_{ A } f_{A }}\right)                    (3.1.47)

\begin{aligned}\tau & =C_{ A 0} \int_0^{0.1} \frac{d f_{ A }}{k C_{ A 0}^2\left\{\left(1-f_{ A }\right)^2 /\left[1-\left(f_{ A } / 2\right)\right]^2\left(T_0 / T\right)^2\right\}} \\& =\int_0^{0.1}\left(\frac{T}{T_0}\right)^2 \frac{\left[1-\left(f_{ A } / 2\right)\right]^2 d f_{ A }}{k C_{ A 0}\left(1-f_{ A}\right)^2}\end{aligned}                     (A)

The initial reactant concentration can be determined from the ideal gas law:

C_{ A 0}=\frac{Y_{ A } P_{\text {tot }}}{R T_0}=\frac{0.5(1)}{0.08206(723)}=8.43 \times 10^{-3}  g – mol / L

For isothermal operation at 450°C the rate constant is equal to 0.156 liters/mol-sec and

k C_{A 0} \tau=\int_0^{0.1} \frac{\left[1-\left(\frac{f_A}{2}\right)\right]^2 df_A}{\left(1-f_A\right)^2}

With Ψ = 1 − f_{A} as a dummy variable, integration of the above equation gives

k C_{A 0} \tau=\left.\left\{\frac{1}{\Psi}-\left[\frac{1}{\Psi}+\ln (\Psi)\right]-\frac{1}{4}\left[\Psi-2 \ln (\Psi)-\frac{1}{\Psi}\right]\right\}\right|_{\Psi=1} ^{\Psi=0.9 .}

or

\tau=\left\{\left.\frac{-\frac{1}{2} \ln (\Psi)+\left[\frac{(1-\Psi)(1+\Psi)}{4 \Psi}\right]}{0.156\left(8.43 \times 10^{-3}\right)}\right|_{\Psi=1} ^{\Psi=0.9}\right\}=80.19  sec

For adiabatic operation the temperature must be related to the fraction conversion so that equation (A) can be integrated.

Equation (10.4.7) is appropriate for use if we set the heat transfer term equal to zero.

\begin{aligned}& \int_{f_{ A \text { in }}}^{f_{ A \text { out }}} U\left(T_m-T\right) \frac{4}{D} F_{ A 0} \frac{d f_{ A }}{\left(-r_{ A }\right)} \\& \quad=\sum\left(F_i \int_{T_0}^T \bar{C}_{p i} d T\right)-\frac{F_{ A 0} \Delta H_{R \text { at } T_0}}{\nu_{ A }}\left(f_{ A \text { out }}-f_{ A \text { in }}\right) \end{aligned}                 (10.4.7)

\sum\left(F_i \int_{T_0}^T \bar{C}_{p i} d T\right)=\frac{F_{ A 0} \Delta H_{R \text { at } T_0}\left(f_{ A \text { out }}-f_{ A \text { in }}\right)}{\nu_{ A }}                          (B)

From the reaction stoichiometry and the tabulated values of the heat capacities the contributions to the summation may be written as:

Butadiene:

F_i C_{p i}\left(T-T_0\right)=F_{ A 0}\left(1-f_{ A }\right) 36.8\left(T-T_0\right)

Ethylene:

F_i C_{p i}\left(T-T_0\right)=F_{ A 0}\left(1-f_{ A }\right) 20.2\left(T-T_0\right)

Cyclohexene:

F_i C_{p i}\left(T-T_0\right)=F_{ A 0} f_{ A }(59.5)\left(T-T_0\right)

Hence,

\sum\left(F_i \int_0^T C_{p i} d T\right)=\left(57.0 F_{ A 0}+2.5 f_{ A } F_{ A 0}\right)\left(T-T_0\right)

Substitution of numerical values into equation (B) and rearrangement then gives

T=723+\frac{30,000 f_{ A }}{57.0+2.5 f_{ A }}              (C)

where T is expressed in K. At a fraction conversion of 0.1, the gas temperature will have risen to 775 K. The design equation then becomes

\tau=\int_0^{0.1}\left(\frac{T}{723}\right)^2 \frac{\left[1-\left(f_{ A } / 2\right)\right]^2 d f_{ A }}{10^{7.5} e^{-27,500 / 1.987 T}\left(8.43 \times 10^{-3}\right)\left(1-f_{ A }\right)^2}

where T is given by equation (C).

Numerical evaluation of this integral gives T = 47.11 s. This space time is about 59% of that required for isothermal operation. This number may be somewhat low because the reaction is exothermic and the rate of the reverse reaction may be appreciable at the highest temperatures involved in our calculation.

We now wish to examine the case where we allow for heat exchange with a substantially constant temperature heat sink (e.g., an evaporating or condensing fluid or a material flowing at a velocity such that its temperature change over the reactor length is quite small compared to the driving force for heat transfer).

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