Two-column separation of a light hydrocarbon mixture: 23-Chem-B6 Petroleum Refining and Petrochemicals, May 2018, Q4

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 $F=100\ \text{mol/s}$, composition:

Component	Feed (mol/s)
Propane (C₃)	20
n-Butane (nC₄)	40
n-Pentane (nC₅)	20
n-Hexane (nC₆)	20

P1 is 95 mol% propane with $n\text{C}_4:n\text{C}_5=50:1$ (no hexane). Per the exam diagram, the C1 bottoms is stream P2 and is fed in full to C2; the n-pentane split ratio in C2 is 0.65, i.e. 65% of the n-pentane entering C2 leaves overhead in P3.

Find. (a) the molar flow rate and composition of P1–P4; (b) the recovery of n-butane in its product stream, relative to the feed.

[Figure not reproduced: Two-column train as printed in the exam: C1 takes propane overhead (P1); its bottoms P2 is routed in full to C2, which sends all n-butane and 65% of the n-pentane overhead (P3) and the rest of the n-pentane with all n-hexane to the bottoms (P4). See the official exam paper.]

Check — stated assumptions (well-posed reading). The printed diagram shows the C1 bottoms leaving as stream P2 and being routed, without any branch, into column C2; so P2 is the whole C1 bottoms and the C2 feed. The exam leaves some splits unstated, so per exam note 1 the following are adopted: (i) all feed propane is recovered overhead in P1, and P1 contains no n-hexane; (ii) "the split ratio of n-pentane in C2 = 0.65" is the fraction of the n-pentane entering C2 that reports to the overhead P3 (the remaining 0.35 goes to the bottoms P4); (iii) n-pentane is the only distributed component in C2, so all n-butane (light) goes overhead to P3 and all n-hexane (heavy) goes to the bottoms P4.

Approach. Fix P1 from its purity and impurity ratio with a total propane recovery; take P2 (the C1 bottoms) by component difference; apply the 0.65 n-pentane split and the sharp n-butane/n-hexane splits in C2; then tabulate.

Stream P1 from the 95% purity and 50:1 impurity ratio. All 20 mol/s propane report to P1, which is 95% propane: $$P1=\frac{n_{\text{C}_3}}{0.95}=\frac{20}{0.95}=\boxed{21.05\ \text{mol/s}}.$$ The $21.05-20=1.053\ \text{mol/s}$ of impurity splits as nC4:nC5 = 50:1, giving $$n\text{C}_4^{P1}=1.053\tfrac{50}{51}=1.032,\qquad n\text{C}_5^{P1}=1.053\tfrac{1}{51}=0.021\ \text{mol/s}.$$
Stream P2 (C1 bottoms) by component difference. Propane is exhausted; the rest descends: $$n\text{C}_4=40-1.032=38.97,\quad n\text{C}_5=20-0.021=19.98,\quad n\text{C}_6=20,$$ so $P2=\boxed{78.95\ \text{mol/s}}$, all of which is fed to C2.
Column C2 split. With 65% of the n-pentane overhead, all n-butane overhead and all n-hexane to the bottoms: $$P3=38.97+0.65(19.98)=38.97+12.99=\boxed{51.95\ \text{mol/s}},$$ $$P4=0.35(19.98)+20=6.99+20.00=\boxed{26.99\ \text{mol/s}}.$$ Check: $P3+P4=78.95=P2$, and the external balance $F=P1+P3+P4=21.05+51.95+26.99=100.0\ \text{mol/s}$ closes (P2 is an intermediate stream, not a final product).
(b) Recovery of n-butane. n-Butane's product is the C2 overhead P3; relative to the 40 mol/s in the feed, $$R_{n\text{C}_4}=\frac{38.97}{40}=\boxed{97.4\%}.$$ The only loss is the 1.03 mol/s of n-butane that slips overhead with the propane in P1.

Component (mol/s)	Feed	P1	P2	P3	P4
Propane	20	20.00	—	—	—
n-Butane	40	1.03	38.97	38.97	—
n-Pentane	20	0.02	19.98	12.99	6.99
n-Hexane	20	—	20.00	—	20.00
Total	100	21.05	78.95	51.95	26.99

n-Butane recovery in P3 (based on feed): 97.4%.

Concept & Theory

Topic. Multi-column separation sequencing; distributed-component split ratios; component recovery from overall balances.

Key relations. Overall balance $F=P1+P3+P4$; component balances $z_iF=\sum x_{i,j}P_j$; split ratio of component $i$ in a column $=$ (moles of $i$ overhead)/(moles of $i$ fed); recovery $R_i=(\text{component in product})/(\text{component in feed})$.

Why this works. With sharp key splits, each column's separation is fixed by the light/heavy key assignment, so the whole train can be solved by sequential component balances without tray-by-tray calculation. Components lighter than the light key go overhead, heavier than the heavy key go to the bottoms, and only a key or intermediate component (here n-pentane in C2) needs a stated split ratio.

Common pitfall. Reading a column split ratio as a stream-splitter fraction (it applies to one component inside a column, not to the whole stream); and double-counting the propane impurity — the small nC4/nC5 in P1 must be subtracted from the feed before the bottoms balance.

Source. Felder & Rousseau, Elementary Principles ch. 4 (splitters, recovery); Wankat, Separation Process Engineering (column sequencing).

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

Question 4 — Two-column separation of a light hydrocarbon mixture (10 marks)