Question 1 of 6: Propane-Pyrolysis Plant — Flowsheet and Material Balance

Question 1: Propane-Pyrolysis Plant — Flowsheet and Material Balance (20 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. Feed gas of fixed composition; absorber recovers 75 % of the feed propane and the total absorbed stream is 50 mol/hr; single reactor reaction C₃H₈→C₂H₄+CH₄, per-pass conversion 60 %, k = 0.28 s⁻¹; first fractionator recovers 95 % of the unconverted propane (3 % ethane impurity) for recycle; reactor feed diluted with an equal volume (hence equal moles) of steam.

Find. A labelled process flow diagram, and the calculable stream quantities: feed gas rate and composition, propane lost in the lean gas, propane fed to the reactor, ethylene product, recycle stream, steam and total reactor-inlet flow, plus the reactor space-time implied by k and X.

Approach

Take a 1-hour basis. The absorber and its 75 % recovery fix the whole feed; a steady-state propane balance around the reactor–fractionator recycle loop fixes the reactor throughput; reaction stoichiometry then gives the ethylene product, and the equal-volume steam dilution and the first-order rate law close the reactor inlet and space-time.

1. Fix the feed from the absorber recovery. The 50 mol/hr absorbed is the recovered propane (the heaviest, most-absorbable component), i.e. 75 % of the propane in the feed: $$n_{C_3,\text{feed}} = \frac{50}{0.75} = 66.67 \text{ mol/hr}$$ Because propane is 10 mol % of the feed, the total feed gas is $$F = \frac{n_{C_3,\text{feed}}}{y_{C_3}} = \frac{66.67}{0.10} = 666.7 \text{ mol/hr}$$ $\boxed{\text{Feed gas } F = 666.7\ \text{mol/hr}}$
2. Feed component flows and propane leaving in the lean gas. Multiplying $F$ by each mole fraction gives N₂ 33.33, CH₄ 433.33, C₂H₆ 133.33, C₃H₈ 66.67 mol/hr. The 25 % of propane not recovered leaves overhead in the lean gas: $$n_{C_3,\text{lost}} = 66.67 - 50 = 16.67 \text{ mol/hr (with essentially all the N}_2\text{, CH}_4\text{, C}_2\text{H}_6\text{)}$$
3. Propane balance around the reactor–fractionator loop. All 50 mol/hr of stripped propane enters the loop as fresh feed $F_p$. Each pass converts 60 %; of the unconverted 40 %, the fractionator recovers 95 % for recycle. At steady state the propane fed to the reactor $P_r$ satisfies $$P_r = F_p + 0.95\,(1-X)\,P_r = 50 + 0.95(0.40)P_r$$ Solving, $P_r\,[1 - 0.38] = 50$, so $$P_r = \frac{50}{1-0.38} = 80.65 \text{ mol/hr}$$ $\boxed{\text{Propane to reactor } P_r = 80.65\ \text{mol/hr}}$
4. Per-pass conversion products. Converted propane $= X P_r = 0.60(80.65) = 48.39$ mol/hr; unconverted $= 32.26$ mol/hr, of which recycled $= 0.95(32.26) = 30.65$ and purge-lost $= 1.61$ mol/hr. The stoichiometry C₃H₈→C₂H₄+CH₄ makes one ethylene and one methane per converted mole: $$n_{C_2H_4} = n_{CH_4,\text{made}} = 48.39 \text{ mol/hr}$$ $\boxed{\text{Ethylene product} = 48.39\ \text{mol/hr}}$ A whole-loop check closes: fresh in (50) = converted (48.39) + purge loss (1.61). ✓
5. Recycle stream (with 3 % ethane). The recycled propane is 97 % of the bottoms, so $$R = \frac{30.65}{0.97} = 31.59 \text{ mol/hr} \;\;(30.65\text{ C}_3\text{H}_8 + 0.95\text{ C}_2\text{H}_6)$$ $\boxed{\text{Recycle} = 31.59\ \text{mol/hr}}$ The combined hydrocarbon feed to the reactor is $F_p + R = 50 + 31.59 = 81.59$ mol/hr, whose propane content $50 + 30.65 = 80.65 = P_r$ — consistent with Step 3.
6. Steam dilution and reactor inlet. "Equal volume of steam" means equal moles for an ideal gas, so steam $= 81.59$ mol/hr and $$n_{\text{reactor in}} = 81.59 + 81.59 = 163.2 \text{ mol/hr}$$
7. Reactor space-time. For a first-order reaction in an (assumed) plug-flow reactor, the space-time for 60 % conversion is $$\tau = \frac{-\ln(1-X)}{k} = \frac{-\ln(0.40)}{0.28} = 3.27 \text{ s}$$ $\boxed{\tau \approx 3.3\ \text{s}}$ The steam both supplies pyrolysis endotherm and lowers hydrocarbon partial pressure (favouring the cracking).