Question 4 of 5: Two-column separation of a light hydrocarbon mixture

Question 4 — Two-column separation of a light hydrocarbon mixture (10 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 $F=500\ \text{mol/s}$; product purities as tabulated.

Find. (a) flows P1–P4; (b) composition of P2; (c) key components of each column; (d) split fractions/ratios of the light keys; (e) recovery of n-butane based on the feed.

Approach. Fix P1 from the methane recovery and its 96% purity; obtain P2 (=C1 bottoms) by component difference; solve the two coupled C2 balances for P3 and P4; then read off keys, split ratios and the n-butane recovery.

1. (a) Stream P1 from complete methane recovery at 96% purity. All 100 mol/s of methane report to P1, which is 96% methane: $$P1=\frac{n_{\text{CH}_4}}{0.96}=\frac{100}{0.96}=\boxed{104.17\ \text{mol/s}}.$$ Its n-butane is 1%, $n\text{C}_4^{P1}=0.01(104.17)=1.04\ \text{mol/s}$, and the balance is propane, $n\text{C}_3^{P1}=104.17-100-1.04=3.13\ \text{mol/s}$ (3.0%).
2. (b) Stream P2 (C1 bottoms) by component difference. Methane is exhausted; the rest descends: $$n\text{C}_3=200-3.13=196.88,\quad n\text{C}_4=200-1.04=198.96,\quad n\text{CH}_4=0,$$ giving $P2=\boxed{395.83\ \text{mol/s}}$ with composition $$x_{\text{C}_3}=\frac{196.88}{395.83}=49.7\%,\qquad x_{n\text{C}_4}=\frac{198.96}{395.83}=50.3\%.$$
3. (a, cont.) Column C2 — solve for P3 and P4. C2 is fed by P2. With P3 = 99% C₃ (1% nC₄) and P4 = 92% nC₄ (8% C₃), write the propane balance and the total balance ($P3+P4=395.83$): $$0.99\,P3+0.08\,P4=196.88.$$ Substituting $P4=395.83-P3$ gives $0.91\,P3=165.21$, so $$P3=\boxed{181.55\ \text{mol/s}},\qquad P4=395.83-181.55=\boxed{214.29\ \text{mol/s}}.$$ Check on n-butane: $0.01(181.55)+0.92(214.29)=1.82+197.14=198.96$ — matches P2, and $P1+P3+P4=104.17+181.55+214.29=500.0\ \text{mol/s}$, so the overall balance closes.
4. (c) Key components. The keys are the adjacent pair straddling each column's cut: Column C1 (de-methaniser) separates methane from propane, so the light key is methane (CH₄) and the heavy key is propane (C₃) (n-butane is a heavy non-key). Column C2 separates propane from n-butane, so the light key is propane (C₃) and the heavy key is n-butane (nC₄).
5. (d) Split fractions and split ratios of the light keys. Using $\text{SF}=\dfrac{\text{LK in distillate}}{\text{LK in that column's feed}}$ and $\text{SR}=\dfrac{\text{LK in distillate}}{\text{LK in bottoms}}$: $$\text{C1, LK}=\text{CH}_4:\quad \text{SF}=\frac{100}{100}=\boxed{1.00}\ (100\%),\quad \text{SR}\to\infty,$$ a complete split (no methane in the bottoms). For C2 the light key propane (feed to C2 = 196.88 mol/s, of which $0.99(181.55)=179.73$ goes overhead, $196.88-179.73=17.14$ to bottoms): $$\text{C2, LK}=\text{C}_3:\quad \text{SF}=\frac{179.73}{196.88}=\boxed{0.913}\ (91.3\%),\quad \text{SR}=\frac{179.73}{17.14}=\boxed{10.5}.$$
6. (e) Recovery of n-butane (based on the process feed). The n-butane product is the C2 bottoms P4 ($0.92\times214.29=197.14\ \text{mol/s}$); relative to the 200 mol/s in the feed, $$R_{n\text{C}_4}=\frac{197.14}{200}=\boxed{98.6\%}.$$ The 2.86 mol/s shortfall is the n-butane lost overhead — 1.04 in P1 and 1.82 in P3.

P2 composition: 49.7% propane, 50.3% n-butane. Keys — C1: CH₄/C₃; C2: C₃/nC₄. Light-key splits — C1 methane SF = 1.00; C2 propane SF = 0.913, SR = 10.5. n-Butane recovery = 98.6%.