Batch and differential distillation, Rayleigh equation

Why it matters

Pharmaceutical, fine-chemical, flavour and solvent-recovery plants often process small or variable batches, where a continuous column would be uneconomic. Batch and differential (simple) distillation are used there, and the Rayleigh equation – a differential material balance – tells you how much residue is left and what composition the collected distillate has. It is also a favourite GATE numerical.

Key ideas

Differential (simple, Rayleigh) distillation. A charge of liquid is boiled in a still; the vapour is removed as soon as it forms and condensed, with no reflux. At every instant the vapour leaving is in equilibrium with the liquid then in the still. Because the vapour is richer in the light component, the still liquid becomes steadily leaner, the boiling temperature rises, and the instantaneous vapour composition falls. The collected distillate is a mixture of all vapour produced, so its average composition lies between the first and last vapour.

Assumptions behind the Rayleigh equation.

• The liquid in the still is perfectly mixed.
• The vapour leaving is in equilibrium with the still liquid at every instant (one equilibrium stage).
• The vapour is removed immediately – no condensation back into the still, no entrainment, no reflux.
• Vapour holdup is negligible. (No assumption of ideality or constant α is needed for the general integral form; those enter only in the closed-form solutions.)

Rayleigh equation. A solute balance over a small amount dW vaporised gives d(W·x) = y·dW, which rearranges and integrates to ln(F/W) = ∫ dx/(y − x) between x_W and x_F. With tabulated VLE, evaluate the integral graphically or by Simpson's rule; with constant α or a linear equilibrium it integrates in closed form.

Comparison with flash. For the same fraction vaporised, differential distillation gives a richer distillate and a leaner residue than a flash, because early vapour is produced from richer liquid. It is still only roughly one stage.

Batch rectification. Adding a column and reflux to the still gives a much sharper separation. Two operating policies:

• Constant reflux ratio: distillate composition falls with time; analysed by Rayleigh with x_D in place of y, using McCabe–Thiele at each instant.
• Constant distillate composition: reflux ratio is increased progressively as the still empties of the light component. Batch columns handle variable feeds, several products from one column (successive cuts), and small throughputs, at the cost of higher energy use per kg and operator attention.

Formulas

Rayleigh equation (general): ln(F/W) = ∫ from x_W to x_F of dx/(y − x)

• F – initial charge, kmol; W – residue remaining, kmol; x_F, x_W – initial and final liquid mole fractions of the light component; y – vapour in equilibrium with x.

Constant relative volatility: ln(F/W) = [1/(α − 1)]·[ ln(x_F/x_W) + α·ln((1 − x_W)/(1 − x_F)) ] Equivalent form: ln(F·x_F/(W·x_W)) = α·ln(F(1 − x_F)/(W(1 − x_W)))

Linear equilibrium y = K·x (dilute): ln(F/W) = [1/(K − 1)]·ln(x_F/x_W)

Overall balances for the distillate collected: D = F − W, x_D,avg = (F·x_F − W·x_W)/D

Worked examples

Example 1 – residue and distillate from a given final composition (standard). Given: F = 100 kmol, x_F = 0.5, constant α = 2.5. The still is run until x_W = 0.3. Find W, D and x_D,avg.

1. ln(F/W) = [1/(α − 1)][ln(x_F/x_W) + α·ln((1 − x_W)/(1 − x_F))].
2. ln(0.5/0.3) = 0.5108; ln(0.7/0.5) = 0.3365; α × 0.3365 = 0.8412.
3. ln(F/W) = (0.5108 + 0.8412)/1.5 = 0.9013 → F/W = 2.463.
4. W = 100/2.463 = 40.6 kmol; D = 59.4 kmol.
5. x_D,avg = (100 × 0.5 − 40.6 × 0.3)/59.4 = (50 − 12.18)/59.4 = 0.637.

W = 40.6 kmol, D = 59.4 kmol, x_D,avg = 0.637.

Example 2 – final composition when a fixed fraction is distilled (GATE level). Given: F = 100 kmol, x_F = 0.4, α = 2.5; 40 % of the charge is vaporised (W = 60 kmol). Find x_W and x_D,avg.

1. Target: ln(F/W) = ln(100/60) = 0.5108; x_W must be found by trial.
2. Try x_W = 0.30: [ln(0.4/0.3) + 2.5 ln(0.7/0.6)]/1.5 = (0.2877 + 0.3854)/1.5 = 0.4487 → W = 63.8 kmol (too much residue, so x_W must be lower).
3. Try x_W = 0.28: [ln(0.4/0.28) + 2.5 ln(0.72/0.6)]/1.5 = (0.3567 + 0.4558)/1.5 = 0.5417 → W = 58.2 kmol (too little).
4. Interpolating and refining gives x_W = 0.2866, for which W = 60.0 kmol. ✓
5. x_D,avg = (100 × 0.4 − 60 × 0.2866)/40 = (40 − 17.20)/40 = 0.570.

x_W ≈ 0.287, x_D,avg ≈ 0.570.

Common mistakes

• Writing the Rayleigh equation as ln(F/W) = (x_F − x_W)/(y − x) – the integral cannot be replaced by a single difference unless y − x is constant.
• Using the average distillate composition as the equilibrium y in the integral.
• Inverting the log ratio (ln(W/F)) and getting a negative result.
• Forgetting to check that F·x_F ≥ D·x_D,avg (light component cannot be created).
• Mixing mass and mole bases – Rayleigh is written in moles and mole fractions.
• Assuming simple distillation gives the same result as flash distillation.

For GATE CH

Expect direct Rayleigh calculations with constant α or linear equilibrium: find W given x_W, or x_D,avg from balances, sometimes x_W by trial when the amount distilled is given. Conceptual questions test the assumptions and the comparison with flash distillation. Practise the logarithm arithmetic – answers are sensitive to rounding.

Quick check

1. Write the general Rayleigh equation.
2. During differential distillation, does the still temperature rise or fall?
3. For y = 3x, F = 100 kmol, x_F = 0.1, x_W = 0.05, what is W?
4. Which gives a richer distillate for the same fraction vaporised – flash or differential distillation?

Answers: 1. ln(F/W) = ∫ dx/(y − x) from x_W to x_F; 2. it rises, as the liquid becomes richer in the heavy component; 3. ln(F/W) = ln 2/2, so W = 70.7 kmol; 4. differential distillation.