Question The elementary irreversible organic gas-phase reaction A + B → C is carried out adiabatically in a flow reactor. An equal molar feed in A and B enters at 27oC, and the volumetric flow rate is 2 dm3 /s. Additional information: 𝐻𝐴 𝑜 = −20 𝑘𝑐𝑎𝑙⁄𝑚𝑜𝑙 𝐻𝐵 𝑜 = −15 𝑘𝑐𝑎𝑙⁄𝑚𝑜𝑙 𝐻𝐶 𝑜 = −41 𝑘𝑐𝑎𝑙⁄𝑚𝑜𝑙 𝐶𝐴𝑜 = 0.1 𝑘𝑚𝑜𝑙 𝑚3 ⁄ 𝐶𝑝,𝐴 = 𝐶𝑝,𝐵 = 15 𝑐𝑎𝑙⁄𝑚𝑜𝑙.𝐾 , 𝐶𝑝,𝐶 = 30 𝑐𝑎𝑙⁄𝑚𝑜𝑙.𝐾 𝑘 = 0.01 𝑑𝑚3⁄𝑚𝑜𝑙. 𝑠 𝑎𝑡 300𝐾, 𝐸 = 10000 𝑐𝑎𝑙⁄𝑚𝑜𝑙 a- Calculate the PFR and CSTR volumes necessary to achieve 80% conversion. Hint: for PFR you need to set a table similar to the one in your handout and use numerical method to evaluate the integral b- What is the maximum inlet temperature one could have so that the explosion temperature (900 K) would not be exceeded even for complete conversion?

Ans (a) For adiabatically CSTRT-T_{F}=\frac{F_{m_{0}} x_{A}\left[-O H_{R}(T)\right]}{F_{p_{0}} C_{P A}+f_{B O} C_{P B}}whele f_{\text {to }}= feo _{\text {Jiven }}\begin{array}{l}F_{D_{0}}=v_{0} \times C_{A_{0}}=2 \times 0.001 \times 0.1 \times 10^{3}=0.2 \frac{\mathrm{mel}}{\mathrm{sec}} \\O H_{e}(T)=-41+35=-6 \mathrm{kcal} / \mathrm{mal} \\T-27=\frac{0.2 \times 0.8 \times 6 \times 10^{3}=160}{0.2 \times 15+0.2 \times 15} \\T=187^{\circ} \mathrm{C}\end{array}from arhenium equation:-\begin{aligned}\ln \left(\frac{K_{2}}{K_{1}}\right) & =\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) \\\ln \left(\frac{k_{2}}{0.01}\right) & =\frac{10,000 \times 4.18}{8.314}\left(\frac{1}{800}-\frac{1}{460}\right) \\k_{2} & =0.0034 \frac{m^{3}}{\text { mel.see }}\end{aligned}Fer CSTR 2^{\text {nd }} order reaction\begin{array}{l}\frac{v}{F_{n_{0}}}=\frac{x_{n}}{k C_{n_{0}^{2}}\left(-x_{n}\right)^{2}} \\\frac{v}{0.2}=\frac{0.8}{(0.2)^{2} \times 0.0034 \times 0.01 \times 10^{6}} \\V_{C S+R}=0.118 \mathrm{~m}^{3} \text { Ans. }\end{array}for adiabaticully PFRT-T_{f}=\left[\operatorname{gH}_{R}(T)\right] \text { find } \frac{d x_{n}}{Z F_{A} C_{P A}} .where \xi F_{i} C_{P_{i}}=F_{A} C_{P A}+F_{B} C_{P B}+F_{C} C_{P_{C}}\begin{array}{l}f_{A}=F_{A_{0}}\left(1-x_{n}\right) \\F_{B}=F_{\theta_{0}}\left(1-x_{n}\right) \\f_{2}=F_{\text {ono }} x_{A} \\\Sigma f_{1} C_{P_{1}}{ }^{2}=2 F_{A_{0}}\left(1-x_{A}\right) \times 15+F_{A_{0}} \times_{A} \times 30 \\=2 \times 0.2 \times 15\left(1-x_{n}\right)+30 \times 0.2 x_{A} \\6\left(1-x_{A}\right)+6 x_{A} \\=6 \\T-27=\frac{\left[-O H_{R}(T)\right] F_{A_{0}} \cdot X_{A}}{6} \\=\frac{6 \times 10^{3} \times 0.2 \times 0.8}{6}=187^{\circ} \mathrm{c} \\\end{array}80\begin{array}{l}\begin{array}{l}K_{\text {value }} \text { is }=0.0034 \frac{\mathrm{m}^{3}}{m 0} \\\frac{V}{f_{n_{0}}}=\int \frac{d x_{a}}{k c_{n_{0}}^{2}\left(1-x_{n}\right)^{2}}\end{array} \\\tau k c_{n 0}=\frac{x_{A}}{1-x_{A}} \\\frac{V K c_{s_{0}}}{v_{0}}=\frac{0.8}{0.2}=4 \\V=\frac{4 \times 2 \times 0.001}{0.0034 \times 0.1 \times 10^{3}}=0.0235 \mathrm{~m}^{3} \text { Ans } \\\end{array}\begin{array}{l}\operatorname{Ans}(b) \frac{T-T_{\max }}{}=\frac{f_{A O} \times X_{A}\left[-O X_{R}(t)\right]}{F_{A O} C_{P A}+F_{B O} C_{B}} \\ 900-T_{\max }=\frac{0.2 \times 1 \times 6 \times 10^{3}}{6} \\ T_{\text {max }}=700 \mathrm{~K} \text { Ans. } \\\end{array} ...