D / F
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Chapter 5
Multicomponent Shortcut Distillation
More than two components, no diagram to draw on — Fenske, Underwood and Gilliland, and how far you can trust them.
By the end of this chapter you should be able to
- Explain why the external balances that solved a binary column cannot solve a multicomponent one.
- Choose the light and heavy keys and specify a separation by fractional recovery.
- Use the Fenske equation for and for the split of the non-key components.
- Solve the Underwood equation for , choose the correct root, and obtain .
- Read from the Gilliland correlation and estimate the feed stage.
- State what the whole method assumes, and how far to trust the answer.
When the diagram runs out
Everything since Chapter 4a has been drawn on one square of paper. That works because a binary mixture has exactly one composition variable: fix and you know both components. Add a third component and the state of a stage needs two numbers, the diagram needs a third dimension, and the staircase has nowhere to live.
There are only two honest responses. Either compute every stage numerically — which is what a simulator does, and what nobody does by hand — or accept an approximation good enough to size the column before the simulator runs. This chapter is the second: three results, from 1932, 1948 and 1940, that between them still decide the first sketch of most industrial columns.
What the shortcut is for
FUG gives you the number of stages, the reflux and the feed location in about ten minutes with a calculator. It is a first estimate — accurate to a few percent when its assumptions hold, and much worse when they do not. You use it to decide what to simulate, not what to build.
5.1 Why multicomponent is different
Compare what has to be specified. For a binary column, eight numbers; for a ternary, nine:
| Stream | Binary | Ternary |
|---|---|---|
| Feed | , , [], | , , , [], |
| Distillate | , [], quality | , quality |
| Bottoms | , [] | |
| Reflux | ||
| Feed location | ||
| Total | 8 | 9 |
Now count equations. Around the whole ternary column there are three component balances and one energy balance — four equations. The unknowns are , , the distillate compositions you did not specify, the bottoms compositions you did not specify, and — five. The external balances no longer close.
This is the whole difficulty in one sentence
In a binary you can specify and and get and from the external balances alone3 §3.7, before knowing anything about the inside of the column. In a multicomponent column the distribution of the components you did not specify is decided inside the column, by how many stages there are — so you cannot get the product compositions without already knowing the design you were trying to find.
A rigorous solution breaks that circle by iterating: guess the compositions at one end, march stage by stage down the column, check the external balances, correct the guess, repeat. That is the Lewis–Matheson and Thiele–Geddes family of methods, and it is what your simulator is doing behind its progress bar. The shortcut breaks the circle differently — by only ever asking about the two components that matter.
5.2 Keys and non-keys
A multicomponent separation is almost never asked for as "separate everything". It is asked for as a split between two adjacent components: keep this one out of the bottoms, keep that one out of the distillate. Those two are the light key (LK) and the heavy key (HK), and everything else is a non-key.
Specifications are given as fractional recoveries rather than compositions, because a recovery is something you can ask for without knowing the answer:
5.1
"Recover 98 % of the n-butane overhead and 98 % of the isopentane in the bottoms" is a complete, checkable specification. "Make the distillate 86.5 % n-butane" is a consequence, and you cannot know it in advance.
Choosing the keys
Order the components by volatility and cut between the two you care about. If the specification names a purity rather than a recovery — "no more than 2 % isopentane in the overhead" — the HK is whatever that impurity is. If two candidate cuts are possible, the cheaper one is between the pair with the largest ratio, for the same reason a large made a binary column short2a §2.7.
5.3 Fenske — the minimum stages
At total reflux the operating lines lie on the diagonal, so the vapour leaving a stage has the same composition as the liquid entering it from above. Chaining the equilibrium relation down stages for any pair of components and gives
5.2
which rearranges into the form you will actually use — written in terms of the amounts of each key in each product, and therefore in terms of the recoveries:
5.3
Two things to notice. Put the binary specifications in and equation 5.3 collapses to the Fenske equation of Chapter 4a4a §4.8 — it is the same result, written for a mixture. And appears in a logarithm on the bottom, so a separation between components of similar volatility is expensive in exactly the way a binary one was.
The Fenske equation also tells you where the non-keys go
This is the part that has no binary counterpart, and it is what makes the whole shortcut usable. Equation 5.2 holds for any pair of components, so apply it to each component against the heavy key, using the you just found:
5.4
Because is typically 6–15, is enormous for a component well above the keys and negligible for one well below — so light non-keys come out at essentially 100 % recovery overhead and heavy non-keys at essentially 0 %. Only components whose is close to a key's come out anywhere in between. Those are the ones that distribute, and they are the reason a real distillate has traces of things nobody asked about.
Keys, non-keys and Fenske
Set up the feed once — all three apps on this page share it. Pick the keys, set the two recoveries, and Fenske gives and the fate of every other component. The plot is the point: drag a non-key's towards a key and watch its recovery leave 0 or 1 and start to distribute.Where each component goes
Fractional recovery in the distillate against relative volatility. Note the log axis.
Products
What Fenske says the two products contain.
Nmin
—
αLK,HK
—
Distributing components
—
✓—
5.4 Underwood — the minimum reflux
Minimum reflux was easy to see on a binary diagram: rotate the operating line until it touches the equilibrium curve. With no diagram, Underwood's 1948 analysis does the same job algebraically. Defining
5.5
the difference between the minimum vapour rates above and below the feed can be written as a sum over the components. Underwood's contribution was to show that, under constant relative volatility and CMO, there exist common values of and — which collapses two coupled problems into one. The result is the first Underwood equation:
5.6
so, per mole of feed,
5.7
Everything in equation 5.7 is known except . Solve it, then substitute into the second Underwood equation, which uses the distillate the Fenske split just gave you:
5.8
Choosing the root
Equation 5.7 is not a polynomial you solve once. Each term has a pole at , so the function jumps from to at every relative volatility in the mixture, and there is a root in every gap between adjacent 's — roots for components.
The root you want
When the non-keys do not distribute, the physically meaningful root is the one lying between the two keys: . Pick a root in the wrong gap and every number after it is wrong, usually without looking wrong. If several components distribute, more than one root becomes active and the analysis needs all of them — a case this course notes and leaves to a simulator.
The Underwood function
Figure 5.2, made live for your feed. Drag and watch the sum blow up as you cross each . The roots are where the curve meets the horizontal line ; the one between the keys is highlighted. Change the feed quality and watch every root move at once.Σ αi zi / (αi − φ) against φ
Poles at every relative volatility; roots where the curve crosses 1 − q.
Root between the keys, φ
—
Vmin per mole feed
—
Rmin
—
All roots
—
Value at your φ
—
✓—
5.5 Gilliland — the actual stages
Fenske gives the stages at infinite reflux; Underwood gives the reflux at infinite stages. The real column is neither, and there is no exact way to interpolate between two limits that are both singular. Gilliland's answer, in 1940, was to plot a great many rigorous solutions and observe that they collapse onto a single curve when written in the right variables:
5.9
The curve is usually evaluated through Molokanov's fit, which avoids reading a log-log chart by eye:
5.10
and then follows by rearranging equation 5.9:
5.11
Where the correlation is weakest
At small — that is, at a reflux close to the minimum — the curve is steep, and a small error in moves a great deal. That is exactly the region designers like, because it is where energy is cheapest. Worked example 5.1 runs at and Gilliland overestimates the stage count by 5 %; worked example 5.2 runs at and it is right to better than 0.2 %. Know which end you are at.
The finished design
The whole chain, on the feed you set up in App 1: Fenske → Underwood → Gilliland → and the feed stage. When the feed has just two components the panel on the right also runs the Chapter 4a McCabe–Thiele stepper4a App 3 on the same numbers, so you can see for yourself how good the shortcut is.Gilliland
Your design, placed on the correlation.
How the shortcut compares
Available whenever the feed has exactly two components.
Nmin
—
Rmin
—
R used
—
Stages N
—
Feed stage
—
X
—
Y
—
N / Nmin
—
✓—
5.6 Where does the feed go?
The shortcut has no diagram, so the elegant "switch at the intersection" rule of Chapter 4a4a §4.7 is not available. Two estimates are in common use, and neither is very good.
The Fenske ratio. Apply Fenske a second time, but to the rectifying section alone — treating the feed as its "bottoms" — to get the minimum stages above the feed:
5.12
Kirkbride's correlation is the common alternative, fitted rather than derived:
5.13
Both assume the ratio of stages above and below the feed is preserved between the total-reflux column and the real one, which is not true. Expect them to agree with each other to a stage or two and with reality to rather less than that. The feed stage is the least reliable output of the whole method — and, fortunately, the cheapest to fix, since real columns are built with two or three spare feed nozzles for exactly this reason.
5.7 The recipe, and its limits
| Step | What you do | What you get |
|---|---|---|
| 1 | Order components by ; choose LK and HK; state the two recoveries | a well-posed problem |
| 2 | Fenske, equation 5.3 | |
| 3 | Fenske again, equation 5.4, for every component | the split of the non-keys, and so , and both product compositions |
| 4 | Underwood 1, equation 5.7, root between the keys | |
| 5 | Underwood 2, equation 5.8 | , hence |
| 6 | Choose –; Gilliland, equations 5.10 and 5.11 | |
| 7 | Equation 5.12 or 5.13 |
What the whole chain assumes, and what each assumption costs:
- Constant relative volatility. Both Fenske and Underwood need it. Real varies with temperature, and therefore up the column. The usual dodge is a geometric mean of the at the top, the feed and the bottom — which is a patch, not a fix. Where varies strongly, this is the dominant error.
- Constant molal overflow. Same assumption as Chapter 4a4a §4.4, with the same failure mode when the latent heats differ.
- Non-keys that do not distribute. When a component sits between the keys, one Underwood root is no longer enough.
- Gilliland is a correlation, with the scatter you can see in Figure 5.3 — several percent even where it behaves.
So what is it for?
Screening. FUG will tell you in ten minutes whether a separation needs 12 stages or 120, whether the reflux is 1.2 or 12, and therefore whether the idea is worth a simulation at all. It also gives the simulator a starting point good enough to converge from. What it will not do is size a column you intend to buy.
5.8 Worked examples
Worked example 5.1Benzene–toluene · FUG against the exact answer
A feed of 20 % benzene, 80 % toluene is separated into a distillate of 99.0 % benzene and a bottoms of 1.0 % benzene. The feed is a subcooled liquid such that one mole of vapour condenses on the feed plate for every seven moles of feed. The relative volatility is 4.13 and the column runs at . Find , , and the feed stage, then compare with a McCabe–Thiele construction.
Work it yourself first, then open
Feed quality. One mole of vapour condensed per seven of feed means — a subcooled liquid, as stated.
Recoveries. , so
FRLK,D = 0.1939(0.99)/0.20 = 0.9597 · FRHK,B = 0.8061(0.99)/0.80 = 0.9976
Fenske.
Nmin = ln[(0.9597/0.0403)(0.99758/0.00242)]/ln 4.13 = ln(9 801)/1.4183 = 6.48
Keep the full precision on FRHK,B: the two recovery ratios are just and in disguise, so their product is exactly . Rounding FRHK,B to 0.9976 first turns 411.6 into 415.7 and shifts the product by a per cent.
Underwood. With , solving 4.13(0.20)/(4.13−φ) + 1(0.80)/(1−φ) = −0.1429 for the root between 1 and 4.13 gives φ = 2.330. Then
Vmin = 4.13(0.1919)/(4.13−2.330) + 1(0.00194)/(1−2.330) = 0.4404 − 0.0015 = 0.4389
Rmin = Vmin/D − 1 = 0.4389/0.1939 − 1 = 1.264
Gilliland. , so X = (1.643−1.264)/2.643 = 0.1434. Molokanov gives Y = 0.5112 and
N = (0.5112 + 6.48)/(1 − 0.5112) = 14.30 stages
Feed stage. NF,min = ln[(0.99/0.01)(0.80/0.20)]/ln 4.13 = 4.217, so NF = 14.30(4.217/6.48) = 9.3.
Now the exact answer. A McCabe–Thiele construction on the same numbers gives Rmin = 1.2635 — identical to Underwood, as it must be for a binary at constant α — and N = 13.61 stages with the feed on stage 7.
So Gilliland overestimated the stage count by 5 % and the feed-stage estimate was out by two. That is characteristic: this design sits at X = 0.14, on the steep part of the correlation. Both errors are on the safe side, and both are small compared with the uncertainty in the VLE data.
To reproduce this in the apps: load the binary preset in App 1 — it comes loaded with the next example's system, so overwrite the table with α = 4.13 for benzene, 1.00 for toluene and z = 0.20 and 0.80. The recovery sliders move in steps of 0.0005 and q in steps of 0.01, so they cannot land on 0.9597, 0.99758 and 1.143 exactly; use the nearest reachable values, 0.9595, 0.9975 and q = 1.14. R/Rmin = 1.3 in App 3 is reachable exactly. You should then read Nmin = 6.45, Rmin = 1.2677, N = 14.25 and feed stage 9.3 — the small differences from 6.48, 1.264 and 14.30 above are the slider steps, not the method. The comparison panel does the McCabe–Thiele run for you: 13.55 stages with the feed on stage 7, the shortcut 5.1 % high.
Worked example 5.2The Chapter 4a column, done the other way
The benzene–toluene column of worked example 4a.1 — , , , saturated liquid feed, — is now run at . Design it by FUG and compare.
Work it yourself first, then open
Recoveries. , so FRLK,D = 0.9747 and FRHK,B = 0.9763.
Nmin = ln[(0.9747/0.0253)(0.9763/0.0237)]/ln 2.4 = 8.42
which is the same number Chapter 4a obtained from the Fenske equation — as it must be, since equation 5.3 is that equation.
Underwood with q = 1 gives 1 − q = 0, so 2.4(0.4402)/(2.4−φ) + 0.5598/(1−φ) = 0, whose root between 1 and 2.4 is φ = 1.4849. Then Vmin = 1.0979 per mole of feed and
Rmin = 1.0979/0.4423 − 1 = 1.4821
Chapter 4a read 1.4820 off the diagram. Underwood and the graphical pinch are the same calculation.
Gilliland. R = 3.5(1.4821) = 5.187, X = 0.5988, Y = 0.1930, so
N = (0.1930 + 8.42)/(1 − 0.1930) = 10.67 stages, NF = 10.67(4.245/8.42) = 5.4
The exact answer: McCabe–Thiele at R = 5.187 gives 10.68 stages, feed on stage 6. Gilliland is right to 0.1 %.
Why so much better than example 5.1? Because X = 0.60 here rather than 0.14 — this column runs at 3.5 times the minimum reflux, far from the steep end of the correlation. Compare the two examples in App 3 by moving the R/Rmin slider and watching the comparison bars converge and diverge.
And notice the price of that accuracy: 3.5 Rmin needs only 10.7 stages against 18.6 at 1.2 Rmin, but the vapour rate — and therefore the reboiler — is nearly three times larger. Nobody would build this column.
Worked example 5.3A five-component depropanizer
A depropanizer separates a light-hydrocarbon feed, mole fractions ethane 0.05, propane 0.35, isobutane 0.15, n-butane 0.25 and isopentane 0.20, with relative volatilities 3.50, 2.00, 1.00, 0.83 and 0.40 referenced to isobutane. Propane is the light key and isobutane the heavy key, with 98 % recovery of each in its own product. The feed is a saturated liquid and .
Work it yourself first, then open
Fenske. :
Nmin = ln[(0.98/0.02)(0.98/0.02)]/ln 2.00 = ln(2 401)/0.6931 = 11.23
The non-keys. Applying equation 5.4 with that :
| Component | α | z | FR in D | xD | Verdict |
|---|---|---|---|---|---|
| Ethane | 3.50 | 0.05 | 1.0000 | 0.126 | LNK — all overhead |
| Propane (LK) | 2.00 | 0.35 | 0.9800 | 0.865 | specified |
| Isobutane (HK) | 1.00 | 0.15 | 0.0200 | 0.008 | specified |
| n-Butane | 0.83 | 0.25 | 0.0025 | 0.002 | HNK — but not quite zero |
| Isopentane | 0.40 | 0.20 | 0.0000 | 0.000 | HNK — entirely bottoms |
giving D/F = 0.3966. Note n-butane: its α is only 17 % below the heavy key's, and it is the one component that shows any tendency to distribute. Isopentane, a factor 2.5 away, does not.
Underwood. With q = 1 the right-hand side is zero and the root between 1.00 and 2.00 is φ = 1.2884. Then
Vmin = Σ αi(Dxi)/(αi − φ) = 1.0316 per mole of feed
Rmin = 1.0316/0.3966 − 1 = 1.601
Gilliland. R = 1.2(1.601) = 1.921, X = 0.1096, Y = 0.5440, and
N = (0.5440 + 11.23)/(1 − 0.5440) = 25.8 stages, NF ≈ 12.9 by the Fenske ratio (13.5 by Kirkbride)
Reading the answer. N/Nmin = 2.3, the usual ratio. About 26 stages at 60 % tray efficiency is 40-odd real trays — a 25 m column, which is exactly what a depropanizer looks like. The design is plausible, which is all a shortcut is asked to establish. What happens next is a rigorous simulation, using these numbers as the starting guess.
This is the depropanizer preset in App 1. Try raising n-butane's α from 0.83 towards 1.0 and watch its recovery climb off the floor — that is a non-key becoming a distributor. Push it past the heavy key, to α = 1.5 say, and Apps 2 and 3 refuse to return a design at all: n-butane now sits between the keys, two roots (1.0912 and 1.6844) fall between αHK and αLK, and the one-root form of equation 5.8 no longer applies. Choosing between them is not a matter of taste — they give Rmin = 1.495 and 0.649.
5.9 Check your understanding
Five multiple-choice questions, two short problems and two long ones — 41 marks. Work them offline, then enter your numbers.
Multiple choice
Short problems
Long problems
Think about it
Think 1
Why is a multicomponent separation specified by recoveries of the keys rather than by product compositions, when the binary problem was specified by compositions?
Think first, then open
Because in a binary, specifying specifies everything — the other mole fraction is . In a multicomponent mixture the distillate composition depends on how much of every non-key came over, and that depends on , which you do not know until you have solved the problem. A recovery of a named component, by contrast, is a statement about that component alone: it can be written down before any calculation and checked after one. This is not a convention — it is forced by the degrees-of-freedom argument in §5.1.
Think 2
The Underwood function has roots. What goes wrong if you pick the one in the wrong interval?
Think first, then open
You get a number, it looks reasonable, and it is wrong. The roots in other intervals correspond to the minimum reflux for different splits — a root between two heavy components describes a column that separates those two instead of your keys. Substituted into the second Underwood equation it returns a that is usually too small, so you design a column at a reflux below the true minimum, which will never make specification no matter how many trays it has. The check costs nothing: confirm before going on.
Think 3
Both worked examples 5.1 and 5.2 are binary, and Underwood reproduced the graphical exactly, while Gilliland was out by 5 % in one case and 0.1 % in the other. What does that tell you about where the error in FUG lives?
Think first, then open
Fenske and Underwood are exact under their assumptions — constant α and CMO — so for a binary at constant α they are not approximations at all. All of the shortcut's inaccuracy is in Gilliland, which is an empirical interpolation between two exact limits. That is useful to know: if your α is genuinely constant, trust and and treat as ±10 %. If α varies strongly up the column, none of the three is safe and the geometric-mean patch is doing more work than it should.
Think 4
A sandwich component — one whose volatility lies between the two keys — breaks the single-root Underwood analysis. Why, physically?
Think first, then open
Because it genuinely leaves in both products in comparable amounts, so there is no single "pinch" that describes the column. Physically, a distributing component creates a second pinch zone: one near the feed for the key split and another where the sandwich component's profile flattens. Mathematically, more than one Underwood root becomes active and must satisfy all of them simultaneously, which turns a root-find into a small system of equations. Sandwich components are uncommon but not rare — isopentane between n-butane and n-pentane is the classic case — and they are a good reason to check a shortcut design against a simulation.
Where this chapter connects
- Back to: Fenske is the total-reflux limit of Chapter 4a4a §4.8 written for a mixture, and Underwood computes the same pinch the operating line found by touching the equilibrium curve.
- The check: for a binary feed, App 3 runs the McCabe–Thiele construction4a App 3 alongside the shortcut. The two agree on exactly.
- Same assumptions, same weaknesses: constant molal overflow4a §4.4 is required here too, and constant is required more strictly than anywhere else in the course.
- The same mixture: the depropanizer feed is the light-alkane system of the Rachford–Rice solver3 App 2 in Chapter 3, now with a column around it.
- Ahead: Chapter 6 keeps the multicomponent bookkeeping but replaces the reboiler with a solvent — absorption and stripping.
Summary & key equations
Setting up
Why the shortcut existsexternal balances do not close for — the non-key split is set inside the column
KeysLK and HK are the adjacent pair the specification cuts between
Specificationfractional recoveries and , not compositions
Fenske — total reflux
Minimum stages
Non-key split
ConsequenceLNK → all distillate, HNK → all bottoms; only components near a key distribute
Underwood — minimum reflux
First equation
Root selection — one of roots
Second equation,
Gilliland — the real column
Coordinates,
Molokanov fit
Stages
Feed stage, or Kirkbride
Accuracyworst at small ; typical –2.5
Standing assumptionsconstant · CMO · non-keys do not distribute · Gilliland is empirical