Chapter 4b

Advanced Binary Distillation

Several feeds, side draws, a partial condenser and plates that are not equilibrium stages — the same method, with the bookkeeping done honestly.

Wankat, Ch. 4–6 2 interactive apps 3 worked examples Self-marking problem set Prerequisite: Ch. 4a

By the end of this chapter you should be able to

Divide a column into sections and write the operating line for any of them from a single balance.
Construct a McCabe–Thiele diagram for a column with two feeds or with a side draw.
Explain what changes when the condenser is partial, and count stages correctly.
Distinguish overall from Murphree efficiency, and use each in the right place.
Draw and use a pseudo-equilibrium curve.
Name the three routes past an azeotrope and say when each is used.

Real columns are not two-section columns

Chapter 4a4a §4.6 built the whole method on a column with one feed, one distillate and one bottoms — two sections, two operating lines. Real columns rarely look like that. A refinery has several streams of similar composition and would rather feed them to one column than build three. A plant that needs an intermediate-purity product takes it off the side rather than building a second column to make it. And no plate is an equilibrium stage.

None of this changes the method. It changes the number of operating lines, and the curve you step against. This chapter is the general case, and it is shorter than 4a because there is only one new idea in it.

The one new idea

A section is the part of a column between two places where material enters or leaves. Within a section the flows are constant, so the balance is one straight line. Number of operating lines = number of sections. Everything else you already know.

4b.1 Sections and the net flow Δ

Take any envelope that cuts the column horizontally somewhere in section , and includes everything above the cut. Two streams cross the cut — the vapour rising and the liquid falling — plus whatever leaves the top of the envelope. Mass and component balances give

4b.1

where is the net flow leaving the envelope — the sum of everything drawn off above the cut minus everything fed in above it — and is the net flow of the light component. Rearranged, the operating line for any section is

4b.2

That is the only equation in this chapter. The rectifying line of Chapter 4a is the case , . The stripping line is the case where the net flow points downward, so and the intercept goes negative. Everything else is bookkeeping.

A rule that will save you every time

Work from the top down, and keep a running total. Start with . Every time you pass a draw-off, add its ; every time you pass a feed, subtract its . When you reach the bottom the running total must equal . If it does not, you have made an arithmetic error — and you have found it before drawing anything.

4b.2 Multiple feeds

Two feeds divide the column into three sections. Numbering from the top, with the richer feed entering higher up:

Figure 4b.1 — A two-feed column. Section 1 lies above the upper feed, section 2 between the feeds, section 3 below the lower feed. Each section carries its own pair of flows and therefore its own operating line.

The flows accumulate as you pass each feed. For saturated-liquid feeds the liquid grows and the vapour is unchanged; in general the feed quality splits it:

4b.3

and the running total of light component gives the three intercepts:

4b.4

Drawing it

The construction is a chain, and each link is fixed by the previous one:

Mark on the diagonal and draw the section 1 line with slope .
Draw the feed line for through . Mark where it crosses the section 1 line.
The section 2 line must pass through that crossing. Draw it — its slope is .
Draw the feed line for and mark where it crosses the section 2 line.
Join that crossing to on the diagonal. That is the section 3 line.

Figure 4b.2 — The three operating lines, each starting where the previous one meets a feed line. Step as usual, switching lines at each intersection; the switch stages are the optimum feed locations for both feeds at once.

Feed the right stream to the right place

Two feeds of different composition should enter at different heights — the richer one higher up, where the liquid on the trays is already about as rich as it is. Feeding both at the same point throws away the separation you were given for nothing, and shows up on the diagram as a section 2 that has collapsed to zero length. Worked example 4b.1 puts a number on the cost.

4b.3 Side streams

A draw-off does the same thing in reverse: it removes material, so the flow below it is smaller and the section changes.

Figure 4b.3 — A column with a liquid side draw of composition , taken above the feed. Again three sections, again three operating lines.

For a saturated-liquid draw the vapour is untouched and the liquid falls by :

4b.5

and below the feed, as before, , , with intercept .

The construction is the same chain, except that the draw-off gets a pseudo-feed line — vertical at for a liquid draw, horizontal at for a vapour draw — because a draw-off is a feed of negative flow with its own thermal condition.

Figure 4b.4 — The side-draw construction. Section 1 runs from to the pseudo-feed line at ; section 2 from there to the feed line; section 3 from there to .

Why a side draw costs stages

Taking liquid off the side lowers below the draw, so falls and the section 2 line lies closer to the equilibrium curve than section 1 did. Less driving force per stage means more stages for the same separation. A side draw is never free — you are stealing reflux from the section below it, and the column tells you so.

App 1

Section builder

Configure the column, and each section's operating line is generated from its own balance — with the running Δ total shown so you can see the bookkeeping close at . Move the second feed's composition through the first and watch the middle section vanish; take a side draw and watch section 2 flatten towards the equilibrium curve.

Column

Relative volatility α 2.40

x_D 0.950

x_B 0.050

Reflux ratio R 2.00

Lower feed F₁ 100

z_F1 0.300

q₁ 1.00

Upper feed F₂ 60

z_F2 0.600

q₂ 1.00

The column

One operating line per section, coloured to match the diagram.

The construction

Each line starts where the previous one meets a feed or pseudo-feed line.

Stages N

—

Distillate D

—

Bottoms B

—

Switch stages

—

Closing Δx_Δ

—

Should equal −Bx_B

—

✓—

4b.4 The partial condenser

A total condenser condenses everything the column sends up, so the reflux and the distillate have the same composition and the construction starts at on the diagonal. A partial condenser condenses only enough to provide reflux and takes the distillate off as a vapour. Vapour and liquid leave it in contact, so it is an equilibrium stage in its own right:

4b.6

where is the reflux returned to the top tray. Three things follow, and only the first is obvious:

The construction starts at on the equilibrium curve, not on the diagonal.
The condenser counts as stage 1 — you get one stage of separation free from a piece of equipment you were going to install anyway.
The operating line is unchanged. The balance around the top of the column does not care how the reflux was produced, only how much of it there is. This is worth pausing on: the equipment changed, the mass balance did not.

Partial condensers are used when the distillate is wanted as a vapour, or when the overhead is so volatile that condensing it entirely would need refrigeration. Toggle the condenser in the Chapter 4a app — App 3 there4a App 3 — and watch the first step move onto the equilibrium curve.

4b.5 Stage efficiency

Every construction so far assumed the streams leaving a plate are in equilibrium. They are not. Vapour bubbles through a shallow pool of liquid in a second or two; equilibrium takes longer than that. How much longer depends on diffusivity, on the vapour and liquid loads, on the tray design, on how well the liquid is mixed across the tray, and on how much vapour bypasses the liquid altogether.

Real tray efficiencies run from about 30 % to 90 %, and the design cannot ignore the difference: at 60 % efficiency you buy two-thirds more trays than the equilibrium calculation says.

Overall efficiency

4b.7

One number for the whole column. It is the easiest thing to use — do the McCabe–Thiele construction, divide by , order that many trays — and the hardest to predict, because it lumps together every effect in the column into a single ratio. In practice it comes from data on a similar column, or from a correlation such as O'Connell's, which relates to the product of relative volatility and liquid viscosity.

The reboiler is usually not included

A partial reboiler is a large, well-mixed, long-residence-time vessel and is normally taken as a genuine equilibrium stage. So when you convert, convert the trays: . Forgetting this is a small error on a tall column and an embarrassing one on a short column.

Murphree efficiency

The overall efficiency is a whole-column average; the Murphree vapour efficiency is defined stage by stage, as the fraction of the possible composition change that a plate actually achieves:

4b.8

where is the vapour entering the plate from below, the vapour actually leaving it, and the vapour that would leave if it reached equilibrium with the liquid . A liquid-phase version, , is defined the same way on the other phase.

Graphically this is the useful one, because it turns into a curve you can step against. At each composition, instead of stepping all the way to the equilibrium curve, step a fraction of the vertical distance from the operating line towards it. Joining those points gives the pseudo-equilibrium curve, and stepping against it counts real plates directly.

Figure 4b.5 — The pseudo-equilibrium curve lies between the operating lines and the true equilibrium curve, a fraction of the way up. Stepping against it gives real stages (black); the equilibrium stages (red) are what the ideal construction would have given. The gap between the two staircases is the efficiency.

App 2

Efficiency explorer

Left: what means on one plate — the vapour enters at , could reach , actually reaches . Right: the pseudo-equilibrium curve that follows, and the two staircases side by side. The readouts show that the overall efficiency you would quote is not equal to the Murphree efficiency you put in.

Murphree efficiency E_MV 60 %

Relative volatility α 2.40

Feed z_F 0.440

x_D 0.950

x_B 0.050

R / R_min 1.30

One plate, close up

The vertical bar is the composition change the plate could make; the filled part is what it does make.

Both staircases

Equilibrium stages against the true curve; real plates against the pseudo-curve.

Equilibrium stages

—

Real plates needed

—

Overall efficiency E_O

—%

E_MV you set

—%

Extra plates

—

Column height at 0.6 m spacing

—m

✓—

4b.6 When distillation alone will not do

Everything in Chapters 4a and 4b assumes the equilibrium curve stays clear of the diagonal. When it does not — at an azeotrope2b §2b.6 — the staircase runs out of room and no reflux, no stages and no efficiency will get you past. There are three standard escapes, and all of them work by changing the property difference rather than by trying harder with the same one.

Route	How it works	Classic example
Extractive distillation	Add a high-boiling solvent that is not itself very volatile. It alters the activity coefficients of the two components unequally, pulling away from 1 and destroying the azeotrope. The solvent is fed near the top and leaves in the bottoms, then goes to its own recovery column.	Ethanol–water with ethylene glycol; separating close-boiling aromatics with sulfolane.
Heterogeneous azeotropic distillation	Add an entrainer that forms a heterogeneous azeotrope — one whose condensate splits into two liquid phases. The decanter does the separation that the column cannot, and each phase is returned where it does most good.	Ethanol–water with cyclohexane or benzene; drying solvents with toluene.
Pressure-swing distillation	Azeotropic composition usually moves with pressure. Run two columns at different pressures and route each one's distillate to the other; between them they walk the composition past the azeotrope with no added chemical at all.	Tetrahydrofuran–water; ethanol–water (only marginally, since its azeotrope moves little).

There is a fourth answer, and increasingly it is the one industry picks: stop distilling. Ethanol is taken to about 0.89 mole fraction by distillation and then dried on a molecular sieve, because a solid separating agent1 §1.6 exploits molecular size instead of volatility, and size is a property difference the azeotrope leaves completely untouched. That is Chapter 8.

Self-study — this section is not examinable in detail

Wankat, Separation Process Engineering, covers all three routes with worked designs; the extractive and heterogeneous cases are treated with residue-curve maps, which are the ternary version of the diagrams in this chapter. If you want one idea to take away, take this: every route above works by adding a third component or a second pressure, because with two components at one pressure the azeotrope is a genuine thermodynamic wall.

4b.7 Worked examples

Worked example 4b.1Two feeds · where the middle section earns its keep

A column with a total condenser and partial reboiler is fed two saturated-liquid streams: mol/h at entering lower down, and mol/h at entering higher up. The products are and , the reflux ratio is , and with CMO. Find the product flows, the three operating lines, and the number of stages.

Work it yourself first, then open

External balances. mol/h and mol/h of light component, so

D = (66.0 − 160 × 0.05)/(0.95 − 0.05) = 58.0/0.90 = 64.44 mol/h, B = 95.56 mol/h

Flows by section. , . Both feeds are saturated liquids, so the vapour never changes and the liquid grows:

Section	L	V	L/V	Δx_Δ	Operating line
1 — above F₂	128.9	193.3	0.667	61.22	y = 0.6667x + 0.3167
2 — between	188.9	193.3	0.977	25.22	y = 0.9770x + 0.1305
3 — below F₁	288.9	193.3	1.494	−4.78	y = 1.4943x − 0.0247

The closing check: ✓ — the running total lands exactly where it must.

Construction. Both feeds are saturated liquids so both feed lines are vertical, at 0.60 and 0.30, and the operating lines cross there. Stepping from gives N = 11.73 stages including the reboiler, with the upper feed on stage 5 and the lower feed on stage 8.

What the middle section is worth. Mix the two feeds instead and put the combined stream (160 mol/h at z = 0.4125) into a single-feed column at the same reflux, and the same construction needs 13.49 stages. Splitting the feeds saves 1.8 stages — worth having, and worth more the further apart the two compositions are. Push to 0.85 in App 1 and the saving grows sharply: the further apart the feeds, the more you lose by mixing them.

Set App 1 to these numbers and read the table off the screen.

Worked example 4b.2A side draw · and what it costs

A column separates 100 mol/h of a saturated-liquid feed at into and , with and . A saturated-liquid side stream of 20 mol/h is drawn off at . Find the products, the section 2 operating line, and the stage count.

Work it yourself first, then open

Balances. Now three products: and .

D = [(100 × 0.50 − 20 × 0.70) − (100 − 20)(0.05)]/(0.95 − 0.05) = (36.0 − 4.0)/0.90 = 35.56 mol/h

B = 100 − 20 − 35.56 = 44.44 mol/h

Sections. , . The draw is a saturated liquid, so and :

section 1: y = 0.7143x + 0.2714 · section 2: y = 0.5536x + 0.3839 · section 3: y = 1.3571x − 0.0179

with the running total going 33.78 → 47.78 → −2.22, and ✓.

Construction. Vertical pseudo-feed line at , vertical feed line at 0.50. Stepping gives N = 10.41 stages, with the draw on stage 3 and the feed on stage 5.

Notice the slopes. Section 2 is flatter than section 1 — 0.554 against 0.714 — because taking liquid off reduced the reflux flowing below the draw. A flatter line sits closer to the equilibrium curve, so those stages work harder for less. Without the draw the same separation needs 9.38 stages; the side product costs 1.03 stages, and that is the price of not building a second column to make it.

Switch App 1 to "Feed + side draw", enter these numbers, and drag S from 0 upward to watch section 2 rotate towards the equilibrium curve.

Worked example 4b.3Efficiency · trays you actually order

A McCabe–Thiele construction gives 18.6 equilibrium stages including the partial reboiler. Plant data on a similar service suggests . How many real trays should be ordered, and how tall is the column at 0.6 m tray spacing?

Work it yourself first, then open

The reboiler is an equilibrium stage in its own right and is not a tray, so take it out before dividing:

N_{trays, equilibrium} = 18.6 − 1 = 17.6

N_{real trays} = 17.6/0.55 = 32.0 → 32 trays

At 0.6 m spacing the trayed section is 32 × 0.6 = 19.2 m, plus disengagement space top and bottom — call it 22 m of column, before the skirt.

The error worth avoiding: dividing all 18.6 stages by 0.55 gives 33.8 → 34 trays, two more than needed. On a small column that is a real cost; on a 100-stage column the same slip is invisible. Know which case you are in.

And a caution: is not . Run App 2 at and the overall efficiency it reports is not 55 % — the two are equal only when the operating line and the equilibrium curve are parallel, which never quite happens. Use for stage counts and for the construction, and do not swap them.

4b.8 Check your understanding

Five multiple-choice questions, two short problems and two long ones — 41 marks. Work them offline, then enter your numbers; everything is marked in your browser.

Marks earned

0 / 0

Multiple choice

Short problems

Long problems

Think about it

Think 1

A colleague proposes feeding both streams of Worked example 4b.1 at the same tray to save the cost of one nozzle. What do you say?

Think first, then open

That the nozzle is the cheapest thing in the conversation. Mixing two streams of different composition destroys separation work that has already been done — it is the reverse of everything the column is for, and it happens before the column can object. On the diagram, section 2 vanishes and the construction reverts to two lines through the mixed composition. In this example the cost is about 1.8 stages — 13.49 instead of 11.73; with feeds at 0.30 and 0.85 it becomes several more. Rule of thumb: if two streams differ enough that you would not deliberately mix them, feed them separately.

Think 2

Where along the column should a side draw be taken to get the purest possible side product?

Think first, then open

You cannot choose the purity independently of the location — the tray composition is whatever the construction gives at that height, so picking the stage picks the purity. What you can do is put the draw where the composition profile is changing slowly, so that small upsets do not move the product specification much. Those regions are exactly the pinched parts of the diagram, near the feed or near a pinch point, where successive stages differ little. That is also why side products are rarely high purity: to make one, you would have to draw near the top, where the profile changes fastest and the draw steals the most reflux.

Think 3

Why is the overall efficiency not simply equal to the Murphree efficiency?

Think first, then open

Because is a fraction of a local distance, and that distance varies down the column. Where the operating line runs close to the equilibrium curve — near a pinch — a plate at 60 % efficiency loses very little in absolute terms, because there was little to gain. Where the two curves are far apart, the same 60 % throws away a lot. The overall efficiency is a weighted average of those local effects, weighted by how the construction actually walks the diagram. The two coincide only if the operating line and equilibrium curve are parallel everywhere, which cannot happen for a real curve. App 2 shows the two numbers side by side for exactly this reason.

Think 4

Extractive distillation adds a solvent that leaves in the bottoms. Where does the solvent enter, and why not with the feed?

Think first, then open

Near the top, a few trays below the condenser, and above the feed. The solvent only helps where it is present, so to change the relative volatility on the trays that do the difficult separation it must be flowing down through them — which means entering above them. It cannot go in at the very top or it would leave with the distillate, so the few trays above the solvent entry act as a rectifying section for the solvent itself. Feeding it with the main feed would leave the entire section above the feed unhelped, which is the part of the column where the azeotrope actually bites.

Where this chapter connects

Back to: every operating line here is the same balance as Chapter 4a4a §4.2 with a different net flow; the construction rules are unchanged.
Feed lines: a side draw gets a pseudo-feed line4a §4.5 of its own, because a draw is a feed with a negative flow.
The limits still apply: total and minimum reflux4a §4.8 bracket a multi-section column exactly as they do a two-section one — each section simply has its own pinch to avoid.
The wall: §4b.6 exists because of azeotropes2b §2b.6, and the cleanest way past one is to stop using volatility — the argument made in Chapter 11 §1.6.
Ahead: Chapter 55 §5.7 drops the binary assumption entirely, and with it the diagram — Fenske, Underwood and Gilliland take over.

Summary & key equations

Sections

Definitiona section is the part of the column between two points where material enters or leaves

General operating line

Δ bookkeepingstart at ; add at a draw, subtract at a feed; must close at

Flows across a feed,

Flows across a liquid draw,

Constructioneach line starts where the previous one meets that stream's feed or pseudo-feed line

Condensers

Totalreflux and distillate share ; start at on the diagonal; not a stage

Partial in equilibrium with ; start on the equilibrium curve; counts as stage 1

Either waythe operating line is unchanged

Efficiency

Overall, typically 0.3–0.9

Trays to order — the reboiler is a stage, not a tray

Murphree, vapour

Graphicallystep against a pseudo-equilibrium curve drawn a fraction from the operating line towards equilibrium

Warning — they coincide only if operating and equilibrium lines are parallel

Past an azeotrope

Extractivehigh-boiling solvent alters unequally; leaves in the bottoms

Heterogeneous azeotropicentrainer forms a two-liquid-phase azeotrope; a decanter does the separation

Pressure swingtwo columns at two pressures; the azeotrope moves, no chemical added

Orchange the property difference entirely — molecular sieve, membrane

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5 · Multicomponent Shortcut