Chapter 4a

Binary Distillation

How many stages, where does the feed go, and how much reflux — the McCabe–Thiele method, built from mass balances rather than asserted.

Wankat, Ch. 4 4 interactive apps 3 worked examples Self-marking problem set Prerequisite: Ch. 3 Flash

By the end of this chapter you should be able to

Derive the rectifying and stripping operating lines from a balance envelope of your own choosing — not recall them.
State what constant molal overflow assumes, and recognise a system where it fails.
Compute the feed quality , draw the feed line, and predict how it rotates with feed condition.
Carry out a complete McCabe–Thiele construction: stages, feed stage, and the reading of the diagram.
Explain why the optimum feed stage is the one that straddles the operating-line intersection.
Find at total reflux and at the pinch, and explain why a real column sits between them.

From one stage to a column

Chapter 3 ended with a cascade3 §3.5: a stack of flash drums, each fed by the products of the one before, with the liquid returning down and the vapour rising. That picture was qualitatively right and quantitatively useless — every drum needed its own energy balance, and the bookkeeping grew faster than the separation.

This chapter replaces the bookkeeping with a diagram. The trick, due to McCabe and Thiele in 1925, is to notice that the streams passing each other between two stages are related by a mass balance, and that under one assumption that balance is a straight line on the same axes the equilibrium data already lives on. Once both curves are on one plot, stepping between them counts the stages, and a hundred coupled equations become a staircase you can draw by hand.

The two relations, and only two

Every McCabe–Thiele construction alternates between exactly two statements. Equilibrium — the streams leaving a stage are in equilibrium, so lies on the equilibrium curve. Mass balance — the streams passing between two stages satisfy the operating line, so lies on it. Horizontal moves are equilibrium; vertical moves are mass balance. That is the entire method.

4.1 The design problem

A column is fed a binary mixture and must deliver a distillate rich in the light component and a bottoms lean in it. Three things are wanted:

the number of equilibrium stages needed,
where along the column the feed should enter,
the condenser and reboiler duties.

The third we can already do — the external balances of Chapter 33 §3.7 give , , and from the specifications alone, without knowing anything about the inside of the column. The first two are what this chapter adds.

Figure 4.1 — The column marked up for design. Above the feed is the rectifying (or enriching) section; below it the stripping section. The circled quantities are what a design problem normally specifies: feed rate, composition and thermal condition; the reflux ratio and its temperature; the pressure. The blue and red envelopes are the two balances this chapter is built on.

The standing assumptions, which we relax later:

Two components only.
Every plate is an equilibrium stage. Real plates are not, and Chapter 4b4b §4b.5 deals with the correction.
VLE data for the system is available.
Negligible pressure drop up the column, so one pressure describes the whole thing.
The column is well insulated — adiabatic.
A total condenser and a partial reboiler, unless said otherwise.

Conventions — worth ten minutes now

Half the errors in this chapter are notation errors. We track only the light component, so and always mean its mole fraction in liquid and vapour. Stages are numbered from the top down: stage 1 is the top tray, stage the bottom. Rectifying-section flows are and ; stripping-section flows carry a bar, and . A subscript names the stage a stream leaves, so is the vapour rising off stage 2 and the liquid falling off stage 1. Liquid enthalpy is , vapour enthalpy . A total condenser makes the reflux and the distillate the same composition, ; a partial reboiler is itself an equilibrium stage and counts as one.

4.2 The rectifying operating line

Draw a boundary around the condenser and the top stage. Four streams cross it: the vapour rising into stage 1 from below, the liquid falling out of stage 1, the distillate , and the condenser duty .

Figure 4.2 — The smallest useful envelope: the condenser plus stage 1. Only , , and cross it.

Writing the balances for that envelope:

4.1

Three balances, plus equilibrium on stage 1 and two enthalpy relations, is six equations in the six unknowns . The system closes — but only for that one stage, and doing this times is exactly the bookkeeping we are trying to escape.

So enlarge the envelope. Take it around the condenser and the top stages. Nothing changes structurally: the same four kinds of stream cross the boundary, only with different subscripts.

Figure 4.3 — The general rectifying envelope, around the top stages. The vapour enters, the liquid leaves, and and leave as before.

4.2

Eliminating between the two gives the rectifying operating line in its general form:

4.3

Read what this says. It relates the vapour coming up to stage to the liquid going down from stage — two streams that pass each other in the gap between the stages, and which are emphatically not in equilibrium. Equation 4.3 is a mass balance and nothing else.

It is not yet a straight line, because may change from stage to stage. Fix that in §4.4 and it becomes one.

App 1

Draw the envelope

Choose where to cut the column. The balances for that envelope write themselves, reduce, and — where they can — become a line on the diagram. The point of this app is that the operating line is not a formula to memorise: it is whatever falls out of the boundary you drew.

Balance envelope

Reflux ratio R = L/D 2.00

x_D 0.95

x_B 0.05

z_F 0.50

Constant molal overflow

Where you cut

Streams crossing the boundary are the only ones that appear in the balance.

What the balance gives you

Written out, then reduced.

Slope

—

Intercept

—

L/V

—

Passes through

—

Unknowns

—

Equations

—

✓—

4.3 The stripping operating line

Nothing new happens at the bottom; only the labels change. Cut around stage down through stage and the partial reboiler. The liquid enters from above, the vapour leaves upward, and the bottoms leaves at the foot with the reboiler duty entering.

Figure 4.4 — The stripping envelope. Note the direction of the net flow: in the rectifying section material moves up and out as ; here it moves down and out as , which is why the sign of the intercept flips.

4.4

and therefore

4.5

The stripping line has a slope greater than one and a negative intercept; the rectifying line has a slope less than one and a positive intercept. Both facts follow from the direction of the net flow, and both are worth being able to reconstruct rather than remember.

4.4 Constant molal overflow

Equations 4.3 and 4.5 are exact, and useless as drawn, because and still carry stage subscripts. The constant molal overflow assumption removes them:

4.6

and the operating lines become genuinely straight:

4.7

In terms of the reflux ratio and the boil-up ratio , which is how specifications usually arrive:

4.8

Two consequences drop straight out and are worth noting now, because the whole graphical method leans on them. Setting in the rectifying line gives : the rectifying line always meets the diagonal at . The same operation on the stripping line gives . Each line is therefore fixed by one point on the diagonal plus a slope — exactly like the flash operating line3 §3.2, which pivots about .

What CMO actually assumes

CMO is not a law. It holds when, per mole of vapour condensed, exactly one mole of liquid vaporises — so the vapour flow neither grows nor shrinks as it climbs. That needs all of:

Equal molar latent heats. The dominant requirement. If , condensing a mole of one component releases just enough to vaporise a mole of the other.
Negligible sensible-heat changes compared with latent heat — true when the column's temperature span is modest.
Negligible heat of mixing — safe for near-ideal systems, less safe once activity coefficients2b §2b.2 depart badly from one.
An adiabatic column at essentially constant pressure.

Benzene–toluene satisfies all four to a fraction of a percent. Methanol–water does not: water's molar latent heat is about 40.7 kJ/mol against methanol's 35.2, a 16 % difference, so the vapour flow really does change up the column. The app below lets you see how much that costs.

App 2

Does CMO hold?

Constant molal overflow is usually asserted and never tested. Here it is tested. The model keeps the heat flow up the column constant rather than the molar flow — so — and lets the molar flows do whatever that implies. Watch the operating line stop being a line.

System

λ_heavy / λ_light 1.08

Relative volatility α 2.40

R / R_min 1.30

x_D 0.95

The operating locus

Straight if CMO holds; curved if it does not. Same specification either way.

Vapour flow up the column

Normalised to the flow leaving the top stage.

Stages, CMO assumed

—

Stages, heat balance

—

Error in N

—

V at the bottom / V at the top

—

λ spread

—%

✓—

4.5 The feed stage and the quality q

The two operating lines describe two different sections. Something has to join them, and that something is the feed stage. Cut an envelope around it alone: the feed enters, the liquid from above and the vapour from below enter, and the vapour to the section above and the liquid to the section below leave.

Figure 4.5 — The feed-stage envelope. The feed's thermal condition decides how it splits between the liquid running down and the vapour running up.

4.9

Under CMO the subscripts drop, and combining the two gives , which defines the feed quality:

4.10

Two readings of the same number, and both are useful. On the left, is the fraction of the feed that joins the liquid running down the column. On the right, it is the heat needed to bring the feed to a saturated vapour, divided by the latent heat — which is how you actually compute it from enthalpies.

Figure 4.6 — What the feed does, for each of the five thermal conditions. A subcooled liquid () condenses extra vapour on the feed tray, so the liquid below exceeds the feed plus the liquid above. A superheated vapour () does the reverse, vaporising liquid on its way in.

Feed condition	Feed-line slope q/(q−1)	On the diagram
Subcooled liquid	positive, > 1	leans back over the liquid region
Saturated (bubble-point) liquid	infinite	vertical
Partially vaporised	negative	into the two-phase wedge
Saturated (dew-point) vapour	0	horizontal
Superheated vapour	positive, < 1	leans forward under the diagonal

The feed line

Where do the two operating lines cross? Subtract equation 4.7's two forms at the crossing point and substitute the feed-stage balance, and the locus of intersections is itself a straight line:

4.11

Setting gives . So the feed line pivots about on the diagonal and rotates as the feed's thermal condition changes — vertical for a bubble-point liquid, horizontal for a dew-point vapour, and through the two-phase wedge in between.

Figure 4.7 — The feed line for all five conditions, all pivoting about . The two operating lines and the feed line meet at a single point; fix any two and the third follows.

Three lines, one point — and why that saves you work

The rectifying line, the stripping line and the feed line are concurrent. In practice you draw the rectifying line from with slope , draw the feed line from , mark where they cross, and then simply join that crossing to — that is the stripping line. You never need to compute to draw it.

4.6 The McCabe–Thiele construction

Everything is now on one diagram. Starting at the top with , alternate the two relations:

Move horizontally to the equilibrium curve — this applies equilibrium on a stage, taking to the that leaves with it.
Move vertically to the operating line — this applies the mass balance between stages, taking to the passing it.

Each corner touching the equilibrium curve is one equilibrium stage. Continue until you step past ; a partial fraction of the last step is a fractional stage, which you round up when you order trays.

Figure 4.8 — A complete construction. The rectifying line has slope , the stripping line slope , and the staircase between the equilibrium curve and the operating lines counts the stages from down to .

App 3

McCabe–Thiele designer

The full construction, live. Drag any specification and the staircase redraws; use step to build it one stage at a time and watch each horizontal move apply equilibrium and each vertical move apply the balance. The feed-stage control lets you override the optimum and pay the price. Equilibrium is at constant relative volatility.

This app is drawn by the browser — if you are reading this, JavaScript is switched off.

4.7 The optimum feed location

The construction switches from the rectifying line to the stripping line at the feed stage. When you switch is a design decision, and only one choice is best.

Switch too late — keep stepping on the rectifying line past the intersection — and the steps crowd into the narrowing gap between that line and the equilibrium curve, each one achieving less. Switch too early and the same thing happens on the stripping side. Switch at the stage that straddles the intersection of the two operating lines, and each step is taken against whichever line is furthest from the equilibrium curve at that composition, which is the most any step can achieve.

(A) Feed below the optimum — 12 stages.

(B) Feed above the optimum — 11 stages.

(C) Feed at the optimum — fewest stages.

The rule, and the reason

Switch operating lines at the first stage whose liquid composition passes the intersection point. The reason is not a convention: the vertical distance between the operating line and the equilibrium curve is what a stage converts into separation, and since both operating lines run below the equilibrium curve, that distance is greatest for whichever line is lower at the composition in question. Switching at the intersection is precisely what keeps you on the lower envelope of the two lines all the way down. Anything else wastes stages you have paid for.

Set the feed stage manually in App 3 and step through it — the crowding in figures (A) and (B) appears on screen exactly as drawn here.

4.8 Total reflux and minimum reflux

Two limits bracket every real column, and both are read off the same diagram.

Total reflux — the fewest stages

Take : everything condensed is returned, everything boiled is returned. Then and both operating lines collapse onto the diagonal. The staircase between the equilibrium curve and the line is as wide as a step can possibly be, so this gives the minimum number of stages .

Figure 4.9 — Total reflux. Both operating lines lie on the diagonal and the construction gives — here 7 stages for a separation that needed 12 at a working reflux.

For constant relative volatility the construction can be done analytically, giving the Fenske equation — which Chapter 5 derives properly5 §5.3, and which you can already use as a check:

4.12

Total reflux is not a fantasy: columns are started up on total reflux to build a composition profile before feed is introduced, and are put back on it when downstream units go down.

Minimum reflux — the most stages

Now rotate the rectifying line the other way. As falls, the line's slope falls, and the line swings down towards the equilibrium curve. At some reflux it touches. The point of contact is the pinch point: there, operating line and equilibrium curve coincide, successive stages produce no change in composition, and an infinite number of them would be needed.

Figure 4.10 — Minimum reflux. The operating line has been rotated down until it meets the equilibrium curve; the contact is the pinch. For a well-behaved (concave) equilibrium curve the pinch sits at the feed line, and follows from the geometry.

For an ordinary concave equilibrium curve — and with has everywhere, so it always is — the pinch occurs where the feed line meets the equilibrium curve, at , and the minimum reflux follows from the slope of the line joining that point to :

4.13

When the pinch is not at the feed line

If the equilibrium curve has an inflection — common in strongly non-ideal systems2b §2b.1 — the operating line can touch it before reaching the feed line. Then the tangent point, not the feed-line intersection, sets , and equation 4.13 gives an answer that is too small. Always look at the diagram. And if the curve touches the diagonal at all, you have an azeotrope2b §2b.6, and no reflux whatsoever will carry you past it.

And in between

Real columns run at somewhere between the two limits, typically to times . Below that the column grows without bound; above it, the extra vapour costs reboiler duty, condenser duty and column diameter for very little saving in trays. App 4 draws the trade-off explicitly.

App 4

The reflux trade-off

The two limits are the two ends of a single curve. Sweep the reflux and watch fall towards at one end and run away to infinity at the pinch at the other. The cost panel puts a price on both axes, and the last readout compares the reboiler duty you are spending with the thermodynamic minimum work1 §1.4 for this same separation.

Relative volatility α 2.40

Feed z_F 0.44

x_D 0.97

x_B 0.02

Feed quality q 1.00

R / R_min 1.20

Stages against reflux

Two asymptotes: as , and infinity at the pinch.

What it costs

Trays are capital, vapour is energy. The sum has a shallow minimum.

R_min

—

R used

—

Stages N

—

N_min (Fenske)

—

Feed stage

—

Vapour per mole feed

—

Reboiler duty

—kJ/mol feed

Second-law efficiency

—%

✓—

4.9 Worked examples

Worked example 4a.1Benzene–toluene · the full design

A column at 101 kPa separates 30 kg/h of a benzene–toluene mixture containing 40 wt % benzene into a distillate of 0.97 and a bottoms of 0.02 mole fraction benzene. The feed is a saturated liquid, the reflux is returned saturated, and the column has a total condenser and a partial reboiler. Take and assume CMO. Find and , , and the number of stages and feed stage at .

Work it yourself first, then open

Feed composition in mole fractions. On a 100 g basis, benzene 40/78.11 = 0.5121 mol and toluene 60/92.14 = 0.6512 mol, so

z_F = 0.5121/(0.5121 + 0.6512) = 0.440

and the molar feed rate is 30 kg/h × 11.633 mol per 1000 g = 349.0 mol/h.

External balances — the Chapter 3 result, unchanged by anything inside the column:

D/F = (z_F − x_B)/(x_D − x_B) = (0.440 − 0.02)/(0.97 − 0.02) = 0.4423

so D = 154.4 mol/h and B = 194.6 mol/h.

Minimum reflux. The feed is a saturated liquid, so the feed line is vertical at x = z_F and the pinch sits at x_p = 0.440, where

y_p = 2.4(0.440)/[1 + 1.4(0.440)] = 1.0565/1.6163 = 0.6537

R_min = (0.97 − 0.6537)/(0.6537 − 0.440) = 0.3164/0.2135 = 1.482

At R = 1.2 R_min = 1.778, the rectifying line is y = 0.6401x + 0.3491 and the operating lines cross at (0.440, 0.6309). Stepping from x_D = 0.97 gives

N = 18.6 stages including the partial reboiler → 19 in practice, with the feed on stage 9.

Sanity check. Fenske gives N_min = ln[(0.97/0.03)(0.98/0.02)]/ln 2.4 = ln(1584)/0.8755 = 8.42. A working column needing about 2.2 × N_min at 1.2 R_min is entirely normal — and if your stepping had given, say, 10 or 40, the Fenske number would have caught it.

Set App 3 to α = 2.4, x_D = 0.97, z_F = 0.44, x_B = 0.02, q = 1, R/R_min = 1.2 and confirm all of it.

Worked example 4a.2The same column, different feed

The same separation is offered a choice of feed condition: saturated liquid (), 50 % vaporised (), or saturated vapour (). Everything else is unchanged and the reflux is still in each case. Which feed do you want, and why is the answer not obvious?

Work it yourself first, then open

Rotating the feed line changes where it meets the equilibrium curve, and therefore :

Feed	q	R_min	R = 1.2R_min	Stages	Feed stage
Saturated liquid	1.0	1.482	1.778	18.6	9
Half vaporised	0.5	1.997	2.396	17.6	9
Saturated vapour	0.0	2.739	3.287	16.1	8

The vapour feed needs the fewest stages — and the most reflux. Nothing was gained for free: pre-vaporising the feed does work that the reboiler would otherwise have done, so the heat has simply moved from inside the column to upstream of it. The right choice depends on where that heat is cheapest and on whether the feed arrives hot already.

Note also that the stage count changes by only 14 % across the whole range while changes by 85 %. Stage count is remarkably insensitive; vapour rate is not. That asymmetry is why energy, not trays, dominates the operating cost of a distillation column — and it is the quantitative version of the claim in Chapter 11 §1.2 that separations are an energy problem.

Drive the q slider in App 4 and watch move while barely does.

Worked example 4a.3Reading a specification that will not work

A colleague specifies and for a system with , feed as a saturated liquid, at . Before doing any stepping, estimate whether this is a sensible column.

Work it yourself first, then open

Fenske first, because it costs one line:

N_min = ln[(0.995/0.005)(0.995/0.005)]/ln 1.15 = ln(39 601)/0.1398 = 10.59/0.1398 = 75.7 stages

That is the number at infinite reflux. A real column at 1.1 R_min typically needs somewhere around 2–2.5 times N_min, so expect 150 to 190 stages. At 0.6 m tray spacing that is a column roughly 100 m tall — taller than anything ever built for this duty, so it would have to be split into two or three columns in series, with the associated pumps and reboilers.

The lesson is not that the specification is impossible; it is that close to 1 is expensive in a way that no amount of reflux fixes. Doubling the reflux buys a handful of stages here; changing the property difference you exploit — a different pressure, an entrainer, a membrane — is the real design move. This is the p-xylene/m-xylene case from Chapter 2a2a §2.7, met again with a column attached.

Set App 4 to α = 1.15, x_D = 0.995, x_B = 0.005 and look at the vertical scale on the N-versus-R plot.

4.10 Check your understanding

Five multiple-choice questions, two short problems and two long ones — 41 marks in total. Work them offline with a calculator and the diagram, then enter your numbers. Everything is marked in your browser and nothing is sent anywhere; your answers are remembered on this device until you reset them.

Marks earned

0 / 0

Multiple choice

Short problems

Long problems

Think about it

Think 1

Everything above assumed a total condenser. What changes if the condenser is partial — condensing only enough vapour to provide reflux, and taking the distillate off as a vapour?

Think first, then open

A partial condenser is an equilibrium stage: the vapour leaving it, , is in equilibrium with the liquid it returns as reflux. So the construction starts at on the equilibrium curve rather than at on the diagonal, and the condenser counts as stage 1 — you get one stage free. The operating line is unchanged, because the mass balance around the top does not care how the reflux was made. Chapter 4b4b §4b.4 works this through, and App 3 has a condenser toggle you can experiment with now.

Think 2

What actually sets the optimum reflux ratio for a given separation — why 1.1 or 1.2 times and not 1.01 or 3?

Think first, then open

Two costs pulling opposite ways. As the stage count blows up, so capital cost — trays, shell, height — goes to infinity. As grows, the vapour rate grows linearly, so reboiler duty, condenser duty and column diameter all grow with it. The total has a minimum, and because both curves are shallow near it the minimum is broad: anywhere in 1.05–1.3 is usually within a few percent of optimal. Energy prices push the optimum down towards ; expensive steel pushes it up. App 4 plots exactly this.

Think 3

If the plates are not equilibrium stages — the streams leaving are not quite in equilibrium — would you need more plates or fewer?

Think first, then open

More. Every real plate achieves less than a full equilibrium step, so it takes more of them to cover the same ground: , with overall efficiencies typically 0.4–0.8. On the diagram the effect is a pseudo-equilibrium curve drawn a fraction of the way from the operating line up to the true equilibrium curve, and the staircase stepped against that instead. Chapter 4b develops both measures; App 3 already has the Murphree efficiency control if you want to see the curve move.

Think 4

What happens to the McCabe–Thiele diagram if there are two feeds, or a side stream drawn off partway up?

Think first, then open

Another section, and therefore another operating line. Each point where material enters or leaves the column changes the net flow, so the balance — and hence the line — changes there. A column with two feeds has three sections and three operating lines; a side draw adds a fourth. The method does not change at all: you still step between the equilibrium curve and whichever operating line governs the section you are currently in, switching at each draw or feed point. That is Chapter 4b4b §4b.2.

Where this chapter connects

Back to: the staircase is the flash cascade3 App 3 done properly — the same repeated equilibrium contact, with the bookkeeping replaced by a diagram.
The same two relations: every step alternates a mass balance2a §2.9 with equilibrium, exactly as in a single flash. Only the geometry is new.
The property difference: from Chapter 2a2a §2.7 sets how far the equilibrium curve bows from the diagonal, and therefore how much a single stage can achieve. closes the gap and the column grows without bound.
The wall you cannot climb: at an azeotrope2b §2b.6 the equilibrium curve meets the diagonal, the staircase has nowhere to step, and neither reflux nor stages will help.
Ahead: Chapter 4b adds multiple feeds, side draws, partial condensers and plate efficiency; Chapter 5 replaces the diagram with Fenske, Underwood and Gilliland when there are more than two components.

Summary & key equations

Operating lines — from balances

Rectifying, general

Rectifying, CMO, through

Stripping, CMO, through

Reflux and boil-up ratios,

CMO requiresequal molar latent heats · negligible sensible heat · no heat of mixing · adiabatic

Feed

Feed quality

Feed line, through

Conditions subcooled · sat. liquid (vertical) · two-phase · sat. vapour (horizontal) · superheated

Concurrencyrectifying line, stripping line and feed line meet at one point

Construction

Horizontal moveequilibrium on a stage: on the equilibrium curve

Vertical movemass balance between stages: on the operating line

Optimum feed stagethe first stage whose liquid passes the operating-line intersection

Partial reboilercounts as one equilibrium stage

Limits

Total reflux; both lines on the diagonal; gives

Fenske

Minimum reflux at the pinch

Practice;

Vapour rate — the quantity that sets duty and diameter

← Previous

3 · Flash Distillation

4b · Advanced Binary Distillation