Two-column distillation-sequence material balance: 23-Chem-B6 Petroleum Refining and Petrochemicals, Undated paper, Q5

Question 5: Two-column distillation-sequence material balance (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.

Quantity	Value
Feed rate $F$	200 mol/min
Feed composition	20% A (40), 65% B (130), 15% C (30) mol/min
Volatility order	C has the highest molecular weight, then B, then A ⇒ A is the lightest, C the heaviest
Column 1 distillate $D_1$	95 mol% A, 99% recovery of A overhead
Column 2 overhead $D_2$	split fraction of B $=0.88$; purity 98 mol% B

Find. (a) the three product-stream rates; (b) the purity of C in the column-2 bottoms; (c) the schematic with the A/B/C composition of every stream.

Approach. Because A is the lightest and C the heaviest, the sequence is a direct one: column 1 takes A overhead and column 2 takes B overhead, leaving a C-rich bottoms. A recovery spec plus a purity spec fixes each overhead completely (key-component rate, then total rate); everything else follows by difference. The impurity in $D_1$ is the next-lightest component, B; the light A entering column 2 all leaves in $D_2$, so the balance of $D_2$'s non-B content is C.

Feed component rates. From $F=200$ mol/min: $$A_F=0.20(200)=40,\quad B_F=0.65(200)=130,\quad C_F=0.15(200)=30\ \text{mol/min}.$$
(a) Column 1 overhead $D_1$ (A product). 99% of the A is recovered overhead at 95 mol% purity: $$A_{D_1}=0.99(40)=39.60,\qquad D_1=\frac{A_{D_1}}{0.95}=\boxed{41.68\ \text{mol/min}}.$$ The non-A in $D_1$ is the next-lightest component, B: $B_{D_1}=41.68-39.60=2.08$ mol/min.
Column 1 bottoms $W_1$ (feed to column 2). By difference, $$W_1=F-D_1=158.32\ \text{mol/min},\quad A_{W_1}=0.40,\ B_{W_1}=127.92,\ C_{W_1}=30.00.$$
(a) Column 2 overhead $D_2$ (B product). The split fraction applies to the B entering column 2, and the overhead is 98 mol% B: $$B_{D_2}=0.88(127.92)=112.57,\qquad D_2=\frac{B_{D_2}}{0.98}=\boxed{114.86\ \text{mol/min}}.$$ All the remaining light A (0.40 mol/min) leaves overhead, so closing the 98% purity fixes the C carried over: $C_{D_2}=114.86-112.57-0.40=1.90$ mol/min.
(a) Column 2 bottoms $W_2$ (C product). By difference, $$W_2=W_1-D_2=\boxed{43.45\ \text{mol/min}},\quad A_{W_2}=0,\ B_{W_2}=15.35,\ C_{W_2}=28.10.$$
(b) Purity of C in the second column's bottoms. $$x_{C,W_2}=\frac{28.10}{43.45}=\boxed{64.7\%\ \text{C}},\qquad x_{B,W_2}=\frac{15.35}{43.45}=35.3\%\ \text{B}.$$

Each component closes on the feed: A $=39.60+0.40=40$; B $=2.08+112.57+15.35=130$; C $=1.90+28.10=30$; and $D_1+D_2+W_2=41.68+114.86+43.45=200$ mol/min. ✓

Check

The C product is only 64.7% pure, which is worth remarking on rather than glossing over: it is a direct consequence of the specifications the question gives. B is 65% of the feed, and only 88% of the B reaching column 2 is taken overhead — so 15.35 mol/min of B (12% of it) is left behind in a bottoms stream that contains just 28.1 mol/min of C. A high-purity C bottoms would need a much higher B split fraction; at the printed 0.88 the answer is genuinely an impure bottoms, not an arithmetic slip.

(c) Schematic diagram with stream compositions

Figure 5. Direct two-column sequence: column 1 removes A overhead ($D_1$, 95% A); its bottoms $W_1$ feed column 2, which takes B overhead ($D_2$, 98% B) and leaves the C-rich bottoms $W_2$. Rates are mol/min.

Stream	Rate (mol/min)	A (mol/min)	B (mol/min)	C (mol/min)	Composition
$F$ (feed)	200.00	40.00	130.00	30.00	20% A, 65% B, 15% C
$D_1$ (A product, col-1 overhead)	41.68	39.60	2.08	0	95.0% A, 5.0% B
$W_1$ (col-1 bottoms → col 2)	158.32	0.40	127.92	30.00	0.3% A, 80.8% B, 18.9% C
$D_2$ (B product, col-2 overhead)	114.86	0.40	112.57	1.90	0.3% A, 98.0% B, 1.7% C
$W_2$ (C product, col-2 bottoms)	43.45	0	15.35	28.10	35.3% B, 64.7% C

Concept & Theory

Topic: Overall material balances on a sequence of separators using recovery, split-fraction and purity specifications.

Key relations: component recovery $r_i=(\text{i in product})/(\text{i in feed})$; split fraction $\xi_i=(\text{i to overhead})/(\text{i entering that column})$; purity $x_i=(\text{i in stream})/(\text{total stream})$; overall balances $F=D_1+W_1$, $W_1=D_2+W_2$, and one balance per component.

Why this works: a recovery (or split) spec plus a purity spec on one product completely fixes that product stream — the first gives the key component's molar rate, the second divides it by a mole fraction to give the total rate — and the partner stream then follows by difference. No stage-by-stage (McCabe–Thiele) calculation is needed for an overall balance. Because A is the lightest and C the heaviest, A reports only to the overheads and C only to the bottoms, which makes each column's non-key distribution unambiguous.

Common pitfall: applying the 0.88 split fraction to the B in the original feed (130) rather than to the B actually entering column 2 (127.92) — a split fraction is always referred to that column's own feed. The second pitfall is treating the distillates as pure: the 95%/98% purities mean each overhead carries a few percent of a neighbouring component, and that carry-over must be tracked or the bottoms purity comes out wrong.

Source: Perry's Chemical Engineers' Handbook, Sec. 13 (multicomponent distillation, column sequencing); Gary, Handwerk & Kaiser, Petroleum Refining, Ch. 4 (fractionation balances).

Question 5 of 5: Two-column distillation-sequence material balance

Question 5: Two-column distillation-sequence material balance (10 marks)

(c) Schematic diagram with stream compositions