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?

Community Answer

Honor CodeSolved 1 Answer

See More Answers for FREE

Enhance your learning with StudyX

Receive support from our dedicated community users and experts

See up to 20 answers per week for free

Experience reliable customer service

Get Started

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} ...