McCabe-Thiele method for continuous distillation

Graphical design of binary distillation columns: constant molal overflow, rectifying and stripping operating lines, the q-line, stepping off stages, optimum feed stage, total and minimum reflux.

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Why it matters

The McCabe–Thiele construction is the standard way to size a binary distillation column: it gives the number of ideal stages, the best feed location and the minimum reflux from nothing more than an x–y diagram and three straight lines. Even when simulators do the real design, engineers use it to understand pinches, feed conditions and reflux trade-offs, and GATE tests it every year in one form or another.

Key ideas

Column layout. Feed enters part-way up. Above it is the rectifying (enriching) section; below it the stripping section. A condenser at the top returns reflux L = R·D (R = reflux ratio, D = distillate); a reboiler at the bottom produces boil-up vapour. A partial reboiler (and a partial condenser) acts as an equilibrium stage; a total condenser does not.

Constant molal overflow (CMO). If the molar latent heats of the two components are nearly equal, heat losses and heats of mixing are negligible, then each mole of vapour condensed on a tray vaporises one mole of liquid. Liquid and vapour molar flows are then constant within each section (L, V above the feed; L′, V′ below). This turns both operating lines into straight lines. When CMO fails badly, use the Ponchon–Savarit enthalpy method.

Operating lines come from component balances around the top or bottom of the column and relate the vapour rising from a stage to the liquid falling into it:

  • Rectifying line: slope L/V = R/(R + 1), passes through (x_D, x_D), intercept x_D/(R + 1).
  • Stripping line: slope L′/V′ (> 1), passes through (x_W, x_W).

Feed condition, q. q is the fraction of the feed that adds to the liquid flow below the feed: q = (heat to convert 1 mol feed to saturated vapour)/(molar latent heat).

  • Subcooled liquid q > 1; saturated liquid q = 1 (vertical q-line); two-phase 0 < q < 1; saturated vapour q = 0 (horizontal q-line); superheated vapour q < 0.
  • The q-line passes through (z_F, z_F) with slope q/(q − 1); the two operating lines and the q-line meet at one point.

Stepping off stages. Start at (x_D, x_D), draw horizontally to the equilibrium curve (a stage: vapour y_n in equilibrium with liquid x_n), then vertically to the operating line (find y_n+1 from x_n), and repeat. Switch from the rectifying to the stripping line at the step that crosses the intersection point – this is the optimum feed stage, giving the fewest stages. Stop when x ≤ x_W. Count includes the reboiler if partial.

Limiting conditions.

  • Total reflux (R → ∞, D = 0): operating lines coincide with the 45° line; the number of stages is the minimum, N_min (Fenske equation for constant α).
  • Minimum reflux R_min: the operating lines touch the equilibrium curve (a pinch), usually where the q-line meets the equilibrium curve; for curved non-ideal curves a tangent pinch may occur above it. Infinite stages are needed.
  • Economic optimum is typically R = 1.2–1.5 R_min. Gilliland's correlation links N, N_min, R and R_min for quick estimates.

Real trays. Divide ideal stages (excluding the reboiler) by an overall efficiency, or step off with a pseudo-equilibrium curve drawn using the Murphree efficiency.

Formulas

Overall balances: F = D + W, F·z_F = D·x_D + W·x_W Rectifying line: y_n+1 = (R/(R + 1))·x_n + x_D/(R + 1), with L = R·D, V = (R + 1)·D Stripping line: y_m+1 = (L′/V′)·x_m − W·x_W/V′ Flows below the feed: L′ = L + q·F, V′ = V − (1 − q)·F q-line: y = (q/(q − 1))·x − z_F/(q − 1) q from enthalpies: q = (H_V − H_F)/(H_V − H_L); subcooled liquid q = 1 + c_p,L (T_bp − T_F)/λ Minimum reflux from pinch point (x′, y′): R_min/(R_min + 1) = (x_D − y′)/(x_D − x′), i.e. R_min = (x_D − y′)/(y′ − x′) Fenske (total reflux, constant α): N_min = ln[(x_D/(1 − x_D))·((1 − x_W)/x_W)] / ln α (includes the reboiler)

  • F, D, W, L, V – molar flows, kmol/h; x, y, z – mole fractions of the light component; R – reflux ratio L/D; q – dimensionless; H – molar enthalpies, kJ/kmol; λ – molar latent heat, kJ/kmol.

Worked examples

Example 1 – saturated-liquid feed (standard). Given: F = 100 kmol/h, z_F = 0.5, saturated liquid (q = 1); x_D = 0.95, x_W = 0.05; α = 2.5; total condenser, partial reboiler; R = 1.5 R_min.

  1. D = F(z_F − x_W)/(x_D − x_W) = 100 × 0.45/0.90 = 50 kmol/h; W = 50 kmol/h.
  2. Pinch on the vertical q-line at x′ = 0.5: y′ = 2.5 × 0.5/(1 + 1.5 × 0.5) = 0.7143.
  3. R_min = (0.95 − 0.7143)/(0.7143 − 0.5) = 1.10; R = 1.65.
  4. Rectifying line: y = (1.65/2.65)x + 0.95/2.65 = 0.6226x + 0.3585.
  5. L = 82.5, V = 132.5; L′ = 82.5 + 100 = 182.5, V′ = 132.5 kmol/h. Stripping line: y = 1.3774x − 0.01887.
  6. Stepping from the top (x_n = y_n/(α − (α − 1)y_n)): x₁ = 0.884, x₂ = 0.799, x₃ = 0.704, x₄ = 0.611, x₅ = 0.531, x₆ = 0.470 (crosses x = 0.5, so stage 6 is the feed stage), then on the stripping line x₇ = 0.404, x₈ = 0.317, x₉ = 0.223, x₁₀ = 0.139, x₁₁ = 0.077, x₁₂ = 0.037 < 0.05.
  7. Fenske check: N_min = ln[(0.95/0.05)(0.95/0.05)]/ln 2.5 = 5.889/0.916 = 6.43.

R_min = 1.10, R = 1.65; 12 ideal stages including the reboiler (11 ideal trays), feed on stage 6 from the top; N_min = 6.4.

Example 2 – two-phase feed (GATE level). Given: same column but the feed is 50 % vaporised (q = 0.5); R = 2.25.

  1. q-line: slope q/(q − 1) = −1, intercept −z_F/(q − 1) = 1 → y = 1 − x.
  2. Pinch: 2.5x/(1 + 1.5x) = 1 − x → 1.5x² + 2x − 1 = 0 → x′ = (−2 + √10)/3 = 0.3874, y′ = 0.6126.
  3. R_min = (0.95 − 0.6126)/(0.6126 − 0.3874) = 0.3374/0.2252 = 1.50 (larger than for a liquid feed).
  4. With R = 2.25: L = 112.5, V = 162.5 kmol/h; L′ = 112.5 + 0.5 × 100 = 162.5; V′ = 162.5 − 0.5 × 100 = 112.5 kmol/h.
  5. Stripping line: y = (162.5/112.5)x − 50 × 0.05/112.5 = 1.444x − 0.0222.

q-line y = 1 − x; R_min = 1.50; L′ = 162.5 kmol/h, V′ = 112.5 kmol/h; stripping line y = 1.444x − 0.0222.

Common mistakes

  • Writing the rectifying slope as R instead of R/(R + 1).
  • Getting the q-line sign wrong: saturated liquid is vertical, saturated vapour horizontal.
  • Using V′ = V for a vapour or two-phase feed (V′ = V − (1 − q)F).
  • Counting the total condenser as a stage, or forgetting the partial reboiler is one.
  • Taking R_min from the q-line intersection when the equilibrium curve has an inflection (tangent pinch).
  • Dividing the reboiler by tray efficiency.

For GATE CH

Expect: D and W from balances; operating-line equations and intercepts; q from feed enthalpy or state; q-line equation; R_min from the pinch with constant α; N_min from Fenske; flows L′ and V′ in the stripping section; and occasionally counting stages for a short column. Practise reading an x–y diagram quickly and recognising which line is which.

Quick check

  1. What is the slope of the rectifying line for R = 3?
  2. What is the q value and q-line slope for a saturated-vapour feed?
  3. How many stages are needed at minimum reflux?
  4. For α = 2, x_D = 0.9, x_W = 0.1, what is N_min?

Answers: 1. 3/4 = 0.75; 2. q = 0, slope 0 (horizontal); 3. infinite; 4. ln 81/ln 2 = 6.34.

Interview questions

All Mass Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What are the key assumptions of the McCabe–Thiele method?Concept

    It treats a binary mixture, assumes each stage is an ideal equilibrium stage and, most importantly, assumes constant molal overflow: the molar liquid and vapour flows are constant within each section. CMO holds when the components have similar molar latent heats and heat losses, heats of mixing and sensible-heat effects are small. Under these assumptions both operating lines are straight, so the column can be designed graphically on an x–y diagram. When the latent heats differ strongly, the enthalpy-based Ponchon–Savarit method is used instead.

  2. 2.What is the q-line and how does feed condition affect it?Concept

    q is the fraction of the feed that joins the liquid flowing down below the feed point, q = (H_V − H_F)/(H_V − H_L). The q-line, y = (q/(q − 1))x − z_F/(q − 1), passes through (z_F, z_F) and is the locus where the rectifying and stripping lines intersect. A saturated-liquid feed gives a vertical line, a saturated vapour a horizontal one, a two-phase feed a negative slope, a subcooled liquid a steep positive slope (q > 1) and a superheated vapour a shallow positive slope (q < 0).

  3. 3.Explain minimum reflux and total reflux.Concept

    At total reflux no product is withdrawn, the operating lines coincide with the 45° line and the number of ideal stages is the minimum, given for constant α by the Fenske equation. At minimum reflux the operating lines touch the equilibrium curve, usually at the q-line intersection, creating a pinch where infinite stages would be needed. Real columns operate between the two, typically at 1.2–1.5 times R_min, which balances the capital cost of stages against the energy cost of reboiler and condenser duty.

  4. 4.How do you locate the optimum feed stage on a McCabe–Thiele diagram?Concept

    Step off stages from the top on the rectifying line, and switch to the stripping line at the first step that crosses the intersection of the two operating lines (on the q-line). That stage is the optimum feed stage and gives the fewest total stages. Introducing the feed earlier or later forces some steps to be taken on the line further from the equilibrium curve, so more stages are needed; on a real column several feed nozzles are often provided to allow for this.

  5. 5.For a column with R = 2, x_D = 0.96 and a total condenser, write the rectifying operating line.Concept

    The rectifying line is y = (R/(R + 1))x + x_D/(R + 1). With R = 2 the slope is 2/3 = 0.667 and the intercept is 0.96/3 = 0.32, so y = 0.667x + 0.32. It passes through (0.96, 0.96) on the 45° line, which is a quick check, and the intercept on the y-axis is often used to draw it.

  6. 6.Why does a two-phase or vapour feed change the flows in the stripping section?Concept

    Only the liquid part of the feed, q·F, joins the downflowing liquid, so L′ = L + qF. The vapour part, (1 − q)F, joins the rising vapour above the feed, so the vapour below the feed is V′ = V − (1 − q)F. A vapour feed therefore reduces the boil-up the reboiler must provide for the same reflux, but it raises the minimum reflux because the pinch moves to a leaner, less favourable point on the equilibrium curve.

  7. 7.Is the reboiler counted as an ideal stage in McCabe–Thiele, and what about the condenser?Concept

    A partial reboiler produces vapour in equilibrium with the bottoms liquid, so it counts as one ideal stage; the number of trays is the stage count minus one. A total condenser only condenses the vapour without changing its composition, so it is not a stage. A partial condenser, which returns liquid reflux in equilibrium with a vapour product, does count as a stage. Tray efficiency is applied to the trays only, not to the reboiler.

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