Two-Stage Cross-Current Liquid–Liquid Extraction: 23-Chem-A2 Unit Operations and Separation Processes, May 2018, Q6

Question B3: Two-Stage Cross-Current Liquid–Liquid Extraction (25 points)

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	Solvent (each stage)	Conjugate tie pairs — pyridine wt% (raffinate → extract)
250 kg, 50 wt% pyridine (125 kg pyridine + 125 kg water)	200 kg fresh chlorobenzene	5.02→11.05, 11.05→18.95, 18.90→24.10, 25.50→28.60, 36.10→31.55, 44.95→35.05, 53.20→40.60; plait point 49.0

The printed table mixes both phases in order of pyridine content. Rows with more than 50% chlorobenzene belong to the extract (chlorobenzene-rich) branch and the rest to the raffinate (water-rich) branch. The i-th raffinate row pairs with the i-th extract row, as in Treybal's source data for this system, and 49.0/37.80/13.20 is the plait point. Above about 27% pyridine the tie lines reverse, so the extract holds less pyridine than the raffinate it is in equilibrium with.

Find. (a) pyridine wt% in the final raffinate $R_2$; (b) overall percentage of pyridine extracted.

Figure B3 — Each stage receives fresh solvent; the raffinate is chased down a staircase of tie lines. Cross-current re-exposes the raffinate to solute-free solvent at every stage.

Approach. For each stage, locate the mixing point on the feed-to-solvent line, find the tie line through it by collinearity, and apply the lever-arm rule on the chlorobenzene axis to split into extract and raffinate; the stage-1 raffinate becomes the stage-2 feed.

Stage 1 mixing point. Mixing $F=250$ kg (125 kg pyridine, 125 kg water) with 200 kg chlorobenzene gives $M_1=450$ kg at $(P,C,W)=(27.8,\ 44.4,\ 27.8)$ wt%.
Stage 1 tie line and lever-arm split. Interpolating the tie line through $M_1$ and applying the lever-arm rule on the chlorobenzene axis: $M_1$ lies between the 25.50→28.60 and 36.10→31.55 tie lines, close to the first. $$R_1:\ 26.0\%\ \text{pyridine}\ (161.5\ \text{kg}),\qquad E_1:\ 28.8\%\ \text{pyridine}\ (288.5\ \text{kg}).$$ The pyridine balance closes as a check: $288.5(0.2875)+161.5(0.2604)=\boxed{125\ \text{kg}}.$ ✓
Stage 2 mixing point. $R_1$ (161.5 kg, 26.0% pyridine, 0.64% chlorobenzene) is the feed; mixing with a further 200 kg chlorobenzene gives $M_2\approx361$ kg at $(P,C,W)=(11.6,\ 55.6,\ 32.8)$ wt%.
(a) Stage 2 split → final raffinate. The tie line through $M_2$ and the lever-arm rule give $M_2$ lies between the 5.02→11.05 and 11.05→18.95 tie lines. $$\boxed{R_2:\ \approx7.25\ \text{wt}\%\ \text{pyridine}},\qquad R_2\approx126\ \text{kg}\ \Rightarrow\ \text{pyridine in }R_2\approx9.1\ \text{kg}.$$
(b) Percentage extracted. Of 125 kg pyridine fed, 9.1 kg remains: $$\%\,\text{extracted}=\frac{125-9.1}{125}\times100=\boxed{92.7\%}.$$ Two fresh 200-kg contacts beat a single 400-kg charge, which leaves a 13.7% raffinate and extracts only about 85%, because each stage re-exposes the raffinate to solute-free solvent.

Quantity	Value
Stage-1 raffinate $R_1$	161.5 kg, 26.0% pyridine
Stage-1 extract $E_1$	288.5 kg, 28.8% pyridine
(a) Final raffinate $R_2$	≈7.25 wt% pyridine (126 kg)
(b) Pyridine extracted	92.7%

Check: the tie-line interpolations are graphical reads on the ternary diagram (which the student plots from the tabulated binodal); a ±1–2 wt% band on the raffinate is expected. The pyridine balance closing to 125 kg at each stage is the internal consistency check.

Concept & Theory

Topic. Cross-current (fresh-solvent) liquid–liquid extraction on the ternary diagram.

Key relations. Overall mixing $M=F+S$ on the feed–solvent line; tie line through $M$ connects equilibrium $R$ and $E$; lever-arm rule $E/M=\overline{MR}/\overline{RE}$; stage-to-stage $R_n\to$ feed of stage $n{+}1$.

Why this works. Extraction exploits the two-phase region bounded by the binodal curve; tie lines pair equilibrium raffinate/extract compositions. In cross-current operation each stage is fed fresh solvent, so the solute is chased down a staircase of tie lines — mix, settle, re-extract. It uses more total solvent than counter-current for the same recovery but is simple and robust, ideal for a few batch stages.

Common pitfall. Pairing the interleaved rows on the assumption that the extract is always richer in pyridine, which forces 36.10→40.60 and misses the tie-line reversal near the plait point; or applying the lever arm on the pyridine (narrow-spread) axis instead of the chlorobenzene axis where the spread — and hence the read accuracy — is greatest.

Source. Treybal ch. 10 (liquid extraction); Geankoplis §12.5–12.6; Perry's §15.

Question 6 of 6: Two-Stage Cross-Current Liquid–Liquid Extraction

Question B3: Two-Stage Cross-Current Liquid–Liquid Extraction (25 points)