Question 4 of 7: Ethylbenzene Reactor with a Di-Ethylbenzene Side Reaction

Question 4: Ethylbenzene Reactor with a Di-Ethylbenzene Side Reaction (Part B — 25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Reactor feed (basis) 100 mol/s: 80 mol/s benzene + 20 mol/s ethylene, delivered at 400 °C, 5 bar. Ethylene is limiting; its conversion is 90%; $\xi_1/\xi_2=5$. Heat-capacity data (kJ/mol·K, $T$ in °C — Felder Table B.2 form): $C_{p,\text{Bz,liq}}=126.5\times10^{-3}+23.4\times10^{-5}T$; $C_{p,\text{Bz,vap}}=74.06\times10^{-3}+32.95\times10^{-5}T-25.20\times10^{-8}T^2+77.57\times10^{-12}T^3$; $C_{p,\text{C}_2\text{H}_4,\text{vap}}=40.75\times10^{-3}+11.47\times10^{-5}T-6.891\times10^{-8}T^2+17.66\times10^{-12}T^3$.

Find. (a) reactor-effluent flows of benzene, ethylene, ethylbenzene and di-ethylbenzene; (b) the heater/mixer duty.

Approach. Part (a) is stoichiometry with two extents: the 90% ethylene conversion and the 5:1 extent ratio give two equations for $\xi_1,\xi_2$. Part (b) is an energy balance on the heater/mixer, taking liquid benzene through vaporisation (benzene boils at ~143 °C at 5 bar) and superheating both feeds to 400 °C.

1. Extents from ethylene consumption. Ethylene reacted $=0.90(20)=18$ mol/s $=\xi_1+2\xi_2$; with $\xi_1=5\xi_2$, $$7\xi_2=18\;\Rightarrow\;\xi_2=2.571,\quad \xi_1=12.857\ \text{mol/s}.$$
2. Reactor-effluent component flows. Benzene consumed $=\xi_1+\xi_2$; ethylene left $=20-18$: $$\dot n_{\text{Bz}}=80-15.43=\boxed{64.57},\quad \dot n_{\text{C}_2\text{H}_4}=2.00,\quad \dot n_{\text{EB}}=\xi_1=12.86,\quad \dot n_{\text{DEB}}=\xi_2=2.571\ \text{mol/s}.$$ (Total 82.0 mol/s — the two moles lost are the net ethylene+benzene combined into product rings.) That is part (a).
3. Benzene boiling point and latent heat at 5 bar. Antoine (NIST) gives $T_b(5\ \text{bar})=415.9$ K (142.8 °C); Watson-scaling the normal $\Delta H_{vap}=30.72$ kJ/mol ($T_c=562$ K) to 415.9 K: $$\Delta H_{vap}(5\ \text{bar})=30.72\left(\tfrac{562-415.9}{562-353.2}\right)^{0.38}=26.82\ \text{kJ/mol}.$$
4. Heater/mixer duty. Benzene: liquid 25 °C → 142.8 °C, vaporise, superheat → 400 °C; ethylene: gas 25 °C → 400 °C. All three inlet streams (fresh benzene, fresh ethylene and the recycle) enter at 25 °C, 5 bar, so the duty is the same as heating 80 mol/s benzene and 20 mol/s ethylene from 25 °C. Integrating the $C_p$ polynomials with $T$ in °C: $$\int_{25}^{142.8}\!C_{p,\text{Bz,liq}}\,dT=17.21,\quad \int_{142.8}^{400}\!C_{p,\text{Bz,vap}}\,dT=37.41,\quad \int_{25}^{400}\!C_{p,\text{C}_2\text{H}_4}\,dT=23.06\ \text{kJ/mol}$$ $$Q=80\,(17.21+26.82+37.41)+20\,(23.06)=6515+461=\boxed{6980\ \text{kW}}.$$ That is part (b).
   Check

   The exam prints the units as kJ/mol·K, but these coefficient sets are the Felder Table B.2 entries, which take $T$ in °C. The check is physical: in °C the liquid-benzene expression gives $C_p(25^\circ\text{C})=0.132$ kJ/mol·K and ethylene 0.0436, both close to measured values (0.136 and 0.043). Substituting $T$ in kelvin gives 0.196 and 0.069, which is 45–60 % too high, and would inflate the duty to about 8660 kW.