ChE 335 · Lecture 4 · Mass Transfer and Equipment Design
Binary Distillation
One flash drum gives you a small separation. Stack the flashes inside one column with internal recycle and you can split almost anything. This guide teaches the McCabe-Thiele method by drawing it, one line and one dot at a time, with the reason for every move.
Part 1 · The idea
Part 2 · Build the diagram
Part 3 · Work it out
Part 4 · Real columns
Part 5 · Use it
The idea
Why we build a column
You already know the flash drum. Boil a mixture once and the vapour is richer in the light component. The problem is that once is not enough. This page shows what happens when you try to fix that with more drums, and why the fix is one tall column instead.
GoalBe able to say, in your own words, what a distillation column actually is: a stack of flash stages with the intermediate streams recycled inside the shell instead of piped around outside it.
Watch the problem, then watch the fix
vapour, rich in benzeneliquid, poor in benzeneequilibrium curve45° line
The sentence to rememberA distillation column is a cascade of flash stages in one shell. The liquid falling down is the recycle that a cascade of separate drums would have to pipe around, and the vapour rising up is the heat that a cascade would have to supply drum by drum. One reboiler at the bottom and one condenser at the top serve every stage.
What you gain, in three numbers
One flash of the feed
0.61
Benzene mole fraction in the vapour, starting from a feed at 0.50. That is the best a single drum can do.
Three flashes in series
0.80
Better, but you now have three heaters, three coolers, three drums and two side streams you did not want.
One column, 9 stages
0.95
One reboiler, one condenser, one shell, and the intermediate streams never leave the equipment.
What a cascade costs you
Each extra drum needs its own heat input and its own cooling. Worse, each drum throws off a liquid side stream that is neither product nor feed. By the third drum most of your original material is sitting in those side streams and only a trickle reaches the top.
The counting is brutal. To reach a high purity you might need ten stages. Ten drums, ten heaters, ten coolers, nine side streams, and a very small amount of product.
What the column does instead
Send every liquid side stream down to the stage below and every vapour up to the stage above. Nothing is thrown away. The side streams become the internal traffic of the column.
Now you only need heat at one place, the reboiler, and cooling at one place, the condenser. The vapour carries the heat up through every stage on the way, and the liquid carries the recycle down through every stage on the way back.
The hardware
Inside the column
Before any graph, you should be able to point at a column drawing and name every stream and every number on it. This page builds the drawing piece by piece and puts the Example 4.1 numbers on each arrow.
vapourliquidproductfeed
The two ratios that control everything
Reflux ratio, set at the top
R = LD
How much condensed liquid you send back down for every mole you take off as product. Then L = R·D and V = (R + 1)D.
Boilup ratio, set at the bottom
VB = V̄B
How much liquid you boil back up for every mole you draw off as bottoms. Then V̄ = VB·B and L̄ = (1 + VB)B.
Bar notationA bar over a symbol always means below the feed. So L and V are the liquid and vapour in the rectifying section, and L̄ and V̄ are the liquid and vapour in the stripping section. The feed is what makes them different.
Constant molar overflow, and why it saves you
Look at any one tray. Vapour arrives from below and some of it condenses. Liquid arrives from above and some of it evaporates. Condensing releases heat, evaporating absorbs it.
If the two components have the same molar enthalpy of vaporization, then every mole that condenses supplies exactly the heat needed to evaporate one mole. The two swaps cancel, so the same number of moles leaves the tray as vapour as arrived, and the same number leaves as liquid.
Constant molar overflow
Vn = Vn+1 = VLn = Ln−1 = L
Molar flows do not change from tray to tray within a section. That is the whole reason the operating lines are straight.
This is why we work in mole fractions and molar flows all through this lecture, never mass. In mass units the cancellation does not happen.
The four McCabe-Thiele assumptions
1. Equal molar heats of vaporization. The two components take about the same energy per mole to boil. Benzene and toluene do, which is why they are the classic teaching pair.
2. Sensible heat changes are negligible. Compared with the latent heat of vaporization, the heat needed to warm or cool a stream by a few degrees is small.
3. No heat loss through the wall. The column is insulated, so no energy leaks out along the way.
4. Uniform pressure. The pressure drop from tray to tray is small enough to ignore, so one equilibrium curve serves the whole column.
What these buy youWith these four assumptions you never write an energy balance inside the column. Material balances alone are enough, and material balances are straight lines. That is what turns a hard simultaneous problem into a drawing you can do with a ruler.
The core
The two rules that make the staircase
If you understand only one page of this lecture, make it this one. Every McCabe-Thiele diagram ever drawn is these two rules used alternately.
Rule 1 · leaving streamsThe vapour and the liquid that leave the same stage are in equilibrium with each other. On the chart, that pair of compositions sits on the equilibrium curve.
Rule 2 · passing streamsThe vapour going up and the liquid coming down between two stages are tied by a mass balance. On the chart, that pair sits on the operating line.
You are holding a vapour composition yn. The liquid leaving the same stage must be in equilibrium with it, so slide horizontally from the operating line across to the equilibrium curve. Where you land is xn.
The horizontal move is the stage doing its work. One horizontal move equals one theoretical stage.
The vertical move
Now you are holding a liquid composition xn flowing down out of stage n. The vapour rising past it is coming up from stage n+1. Those two are passing streams, so drop vertically from the curve back to the operating line. Where you land is yn+1.
The vertical move is just bookkeeping. It carries you to the next stage but it separates nothing.
Common mistakeStudents often count the vertical moves as well, and end up with twice the number of stages. Count only the horizontal moves, or equivalently count the corners that touch the equilibrium curve. One corner on the curve equals one theoretical stage.
Build the diagram
Where the equilibrium curve comes from
The curve answers one question: if the liquid on a tray has composition x, what is the vapour above it? Everything about how hard your separation will be is hidden in the shape of that curve.
From the boiling diagram to the y-x plot
bubble point line, liquiddew point line, vapourequilibrium curve, y against x45° line
Read the 45° line as the do-nothing lineOn the diagonal, y = x, so the vapour has the same composition as the liquid and the stage separates nothing. The vertical gap between the curve and the diagonal is your separation per stage. A fat curve means few stages. A curve that hugs the diagonal means many stages, or no separation at all where it touches.
Three ways to get the curve, and when to use each
Constant relative volatility. If α barely changes with composition, one number gives you the whole curve. This is what Example 4.1 does with α = 2.47.
y = αx1 + x(α − 1)
Tabulated data. For non-ideal pairs such as methanol and water, α changes a lot across the range. You plot measured points and draw a smooth curve through them.
Methanol and water: α falls from about 7.5 in dilute solution to about 2.5 near pure methanol.
Vapour pressures. For similar molecules that follow Raoult law, get the bubble point at each x from Antoine equations, then compute y from the partial pressures.
yi = xi PisatP
Compare the three systems in this guide
System
How much does α matter
Drag through the values and watch the curve move away from the diagonal. The stage count is for a separation from xB = 0.05 to xD = 0.95, at total reflux, which is the easiest the separation can ever be.
2.47
The design lessonBelow about α = 1.2 the stage count explodes and distillation stops being sensible. That is the region where you look at extraction, adsorption, membranes, or adding a third component to move the curve.
Build the diagram
The rectifying operating line
This line is one mass balance around the top of the column, nothing more. Once you see where it comes from, you never have to memorise it.
Slope L/V, and it passes through the point (xD, xD) on the 45° line.
In terms of reflux ratio, the useful form
yn+1 = RR + 1 xn + xDR + 1
Substituting L = RD and V = (R + 1)D and dividing through by D.
Why it must pass through (xD, xD)At the very top of the column the vapour going into the condenser has composition y1. A total condenser condenses all of it, so the reflux and the distillate both come out at that same composition: x0 = xD = y1. A point whose x and y are equal lies on the 45° line. Put xn = xD into the equation and you get y = xD as well, which is the same statement in algebra.
Move the reflux ratio and watch the slope
3.35
Two ends of the sliderPush R up and the slope R/(R+1) approaches 1, so the line lies on the 45° line and the gap to the curve is as wide as it can be. Pull R down and the slope falls, the line swings up toward the curve, and the gap you have to step through gets thin.
Build the diagram
The stripping operating line
Same idea, other end of the column. Draw the envelope around the bottom instead of the top and you get the second straight line.
Because L̄ = V̄ + B and B is positive. So the stripping line is always steeper than the 45° line, while the rectifying line is always flatter than it.
A useful check on your drawingRectifying line: below the diagonal, slope less than 1, anchored at xD. Stripping line: above the diagonal, slope greater than 1, anchored at xB. If your two lines do not look like that, something is wrong before you step off a single stage.
The bar flows, tied back to the feed
You almost never get told L̄ and V̄ directly. You work them out from what the feed does when it enters.
The feed splits itself between the two sections
L̄ = L + qFandV̄ = V − (1 − q)F
The liquid part of the feed, qF, joins the liquid running down. The vapour part, (1 − q)F, joins the vapour running up, so it never reaches the stripping section below.
Quantity
Formula
Example 4.1 value (mol/h)
Where it comes from
D
F(zF − xB) / (xD − xB)
50.0
Overall balance on the light component
B
F − D
50.0
Overall balance on total moles
L
R·D
167.5
Definition of reflux ratio
V
(R + 1)D
217.5
Balance around the condenser
L̄
L + qF
217.5
Liquid part of the feed joins the downflow
V̄
V − (1 − q)F
167.5
Vapour part of the feed joins the upflow
VB
V̄ / B
3.35
Definition of boilup ratio
Check it closes: L̄ − V̄ = 217.5 − 167.5 = 50.0 = B. Whatever goes down and does not come back up has to leave as bottoms.
Build the diagram
The q-line, or what state the feed is in
Cold liquid, boiling liquid, half vapour, saturated vapour, superheated vapour. One number q covers all five, and that one number sets where the two operating lines meet.
What q means
q = heat to turn 1 mol of feed into saturated vapourmolar enthalpy of vaporization
Equivalently q = (hV − hF) / Δhvap. For a feed that is already a boiling liquid you need the full latent heat, so q = 1. For a feed that is already saturated vapour you need nothing, so q = 0.
The line it draws
y = qq − 1 x − zFq − 1
Slope q/(q − 1), always through the point (zF, zF) on the 45° line. Your lecture writes the same line as y = zF/(1 − q) − qx/(1 − q).
The five feed states, drawn
equilibrium curveq-line45° line
Feed state
q
Slope q/(q−1)
Line looks like
What it does to the column
Cold liquid, below its bubble point
q > 1
positive, greater than 1
leans forward, steep
Extra liquid load below the feed. The reboiler has to work harder.
Saturated liquid, at its bubble point
q = 1
infinite
vertical
The standard textbook case. All of the feed joins the liquid going down.
Partly vaporized
0 < q < 1
negative
leans backward
The feed splits. Fraction q goes down as liquid, fraction 1 − q goes up as vapour.
Saturated vapour, at its dew point
q = 0
0
horizontal
All of the feed joins the vapour going up. Nothing added to the liquid below.
Superheated vapour
q < 0
between 0 and 1
leans forward, shallow
The hot feed evaporates some of the liquid coming down. The condenser has to work harder.
Why the two operating lines have to meet on itSubtract the stripping section balance from the rectifying section balance. Everything cancels except the feed terms, and what survives is exactly the q-line equation. So the point where the rectifying line and the stripping line cross is always sitting on the q-line. That is why you draw the q-line before the stripping line: the q-line tells you where the stripping line has to aim.
Preheating the feed is not freeMaking q smaller, by vaporizing some of the feed before it enters, moves work from the reboiler to the feed preheater. The total energy does not disappear. It usually only pays when you have a cheap hot stream available to do the preheating.
Work it out
Example 4.1, drawn step by step
Benzene and toluene at 101.3 kPa. Every line and every dot appears one at a time, with the reason it goes where it goes. Press play, or walk through with Next at your own pace.
ProblemFeed at zF = 0.50 benzene with feed quality q = 0.50. Take xD = 0.95 overhead and allow only xB = 0.05 in the bottoms. Relative volatility is constant at α = 2.47 and the column runs at R = 3.35. Find the number of theoretical stages.
Take the basis as F = 100 mol/h. Two balances, two unknowns.
F = D + BF zF = D xD + B xB
Solve
D = F zF − xBxD − xB = 100 0.50 − 0.050.95 − 0.05 = 50.0
and B = 100 − 50 = 50.0 mol/h. A symmetric split, because the feed sits exactly halfway between the two product compositions.
Step B · the four internal flows
Reflux ratio 3.35 fixes the top. Feed quality 0.50 fixes how the feed splits.
L = RD = 3.35(50) = 167.5
V = (R+1)D = 4.35(50) = 217.5
L̄ = L + qF = 167.5 + 50 = 217.5
V̄ = V − (1−q)F = 217.5 − 50 = 167.5
All in mol/h. Half the feed enters as liquid and half as vapour, so exactly 50 mol/h moves from the vapour traffic into the liquid traffic as it crosses the feed tray.
Step C · the three lines, as numbers
Line
Slope
Anchor point
Equation
q-line
q/(q−1) = −1.000
(0.500, 0.500)
y = 1.000 − x
Rectifying
R/(R+1) = 0.7701
(0.950, 0.950)
y = 0.7701x + 0.2184
Stripping
L̄/V̄ = 1.2985
(0.050, 0.050)
y = 1.2985x − 0.0149
The two operating lines cross at (0.4416, 0.5584), which sits on the q-line as it must.
Step D · read the answer off the staircase
Graphical total
9.02
Corners on the curve, counting the partial reboiler as one of them.
Stages in the column
8.02
Take the reboiler out of the count. Round up to 9 trays of theoretical duty.
Feed stage
5
Counting down from the top. Switch lines here.
R / Rmin
2.20
Rmin is 1.52 for this separation, so 3.35 is a generous reflux.
The last step lands just past xB, which is where the 0.02 fraction comes from. A fraction of a stage is not something you can buy, so in practice you round up.
If you are comparing with the lecture slideThe chart printed on slide 37 of the lecture is not drawn for the Example 4.1 numbers. Working backwards from the plotted stages, that chart uses zF = 0.45 with a saturated liquid feed (q = 1) and R = 3.0, which gives 9.4 total. Example 4.1 as written on slide 33 uses zF = 0.50, q = 0.50 and R = 3.35, which gives 9.02. Both are correct for their own numbers. The next page reproduces the slide 37 chart exactly so you can see them side by side.
Work it out
Counting the stages correctly
This is where marks get lost. Is the reboiler a stage? Is the condenser? What does 8.4 stages mean when you can only buy whole trays? This page is drawn on the same numbers as the chart in your lecture slides.
Drawn for zF = 0.45, saturated liquid feed, xD = 0.95, xB = 0.05, R = 3.00. These are the numbers behind the chart on slide 37, and the construction reproduces every plotted stage to within 0.002 in x.
The three questions, answered once
Is the partial reboiler a stage?
Yes. Liquid enters it, it boils, and the vapour it sends back up is in equilibrium with the bottoms liquid leaving it. That is exactly the definition of an equilibrium stage, so it appears as one corner on the curve.
It counts as a stage, but it is not a tray you install in the shell.
Is the total condenser a stage?
No. A total condenser condenses everything it receives, so it changes phase but not composition: xD = y1. No separation happens, so no corner on the curve.
A partial condenser is different. It only condenses part of the vapour, so it does separate, and then it does count as one stage.
What is 0.4 of a stage?
The staircase almost never lands exactly on xB. The last step overshoots, and the fraction is how far along that last step you actually needed to go.
You cannot install 0.4 of a tray. Report the fraction to show your work, then round up when you specify the equipment.
Interpolating the last stage
N + 1 = 9 + x9 − xBx9 − x10
x9 is the composition on the last stage that had not yet reached the target and x10 is where the next full step would land. The fraction is the part of that final step you actually used.
How your lecture states the answerSlide 38 reads it as 8.4 theoretical stages in the column plus the partial reboiler, and it numbers the partial reboiler as stage 1 counting from the bottom. So the graphical count of 9.4 is N + 1, and N = 8.4 is what goes inside the shell. Class Exercise 2 uses the same convention, with its answer footnoted as stages in the column.
Choosing the feed stageSwitch from the rectifying line to the stripping line at the first stage whose step crosses the intersection point. Switching there gives the smallest total number of stages. Switch too early or too late and the staircase has to squeeze through a narrower gap, so you pay for it with extra stages. On a real column you install feed nozzles on two or three trays around the calculated one, because the feed composition drifts over the life of the plant.
Numbering from the top
Stage 1 is the top tray, the one whose vapour goes to the condenser. Numbers increase downward, and the partial reboiler is the last number. This is the convention used when you step off the diagram starting at (xD, xD), which is what this guide does.
Numbering from the bottom
Stage 1 is the partial reboiler and numbers increase upward. This is the convention on slide 38 of the lecture. Either is fine as long as you say which one you are using, and as long as the reboiler is included exactly once.
Work it out
The two limits of any column
Every real design sits between two impossible ones. At one end you get the fewest stages but no product. At the other you get product but need an infinite column. Understanding both tells you what your R is buying.
Limit 1 · total reflux, R → ∞Send everything back. D = 0, so both operating lines collapse onto the 45° line. The gap between line and curve is as wide as it can ever be, so you need the fewest possible stages, Nmin. You also make no product at all, so nobody runs a column this way except during startup and test runs.
Limit 2 · minimum reflux, R = RminDrop R until the operating lines just touch the equilibrium curve. At that touching point, called the pinch, the driving force is zero and the staircase takes infinitely many tiny steps to get past it. You need an infinite number of stages. Below Rmin the separation is simply impossible.
Move R and watch both limits appear
equilibrium curverectifying linestripping linepinch point
3.35
Jump to
Working reflux
Finding Rmin without guessing
Start at the point (xD, xD) and swing the rectifying line downward until it first touches the equilibrium curve. For a well-behaved curve the touch happens right where the q-line crosses the curve. Read the intersection point (x', y') and the slope follows.
Minimum reflux from the pinch point
RminRmin + 1 = xD − y′xD − x′
For Example 4.1 the q-line meets the curve at (0.3889, 0.6111), giving a slope of 0.6039 and Rmin = 1.524.
Watch out for the bulgeIf the equilibrium curve has an S shape, as methanol and water does, the rectifying line can touch the curve somewhere else before it reaches the q-line. Then the tangent point is the real pinch, and Rmin is larger than the q-line construction would suggest. Always check by eye that your line does not cut through the curve anywhere.
Finding Nmin without drawing
At total reflux both operating lines are the 45° line, so the staircase is between the curve and the diagonal only. For constant α that staircase can be summed exactly, and the result is the Fenske equation.
For Example 4.1: ln(19 × 19) / ln 2.47 = 5.889 / 0.9042 = 6.51, against 6.61 read off the drawing. The small gap is drawing error, not a mistake in either method.
Nmin, graphical
6.61
Corners counted at total reflux, including the reboiler.
Nmin, Fenske
6.51
Same answer from the closed formula.
Rmin
1.52
Below this, no column of any height can do the job.
At R = 3.35
9.02
Roughly 40 percent more stages than the absolute floor.
A note on slide 45The minimum reflux chart in the lecture is drawn for xF = 0.45 with a slanted q-line, while the numbering chart on slide 37 uses a vertical q-line at the same feed composition. The construction on this page uses the Example 4.1 numbers throughout, so the pinch sits at 0.389 rather than where the slide shows it. The method is identical either way.
Real columns
Choosing the reflux ratio
R is the one number the designer really chooses. Everything else follows from the specification. Push it up and you buy a shorter column with a bigger steam bill. Push it down and you buy a taller column that is cheap to run.
capital cost, more traysoperating cost, more steamtotal annual costcheapest point
3.35
The shape is the pointCapital cost falls steeply as R rises from Rmin, because the stage count is dropping fast. Operating cost rises in a straight line, because every extra mole of reflux is another mole to boil and another mole to condense. The sum has a flat bottom, and that flat bottom is good news: a design that sits anywhere near the optimum costs almost the same, so you can move R around to suit other constraints without paying much.
What you actually pay for
Reboiler duty, the steam bill
QR = [ D(1 + R) − (1 − q)F ] Δhvap
Every mole of vapour you raise in the reboiler needs its latent heat. More reflux means more vapour, so a straight line in R.
Condenser duty, the cooling water bill
| QC | = (R + 1) D Δhvap
Everything that comes overhead has to be condensed, whether it leaves as product or returns as reflux.
Example 4.1 duties, scaled to a plant
Take Δhvap as roughly 32 kJ/mol for benzene and toluene near their boiling points. On the 100 mol/h basis of the example the duties are small, so scale the feed up to 100 kmol/h, which is a modest but realistic column.
Reflux ratio
Stages needed
V̄ (kmol/h)
Reboiler duty (MW)
Condenser duty (MW)
Going from R = 2.0 to R = 8.0 saves about three stages and roughly triples the energy bill, every hour, for the next twenty years. That is the trade in one sentence.
What designers actually pick
Rule of thumb. Most columns are designed at R = 1.1 to 1.3 times Rmin. The cost curve is flat there and the extra reflux gives you room to handle upsets.
When energy is expensive, or when the column is huge, designers push closer to 1.05 Rmin and accept the taller column. Propylene splitters live here.
When trays are expensive, or headroom is limited, or the mixture fouls, designers go higher, 1.5 Rmin or more, and accept the energy bill.
Column height, a quick estimateOnce you know the actual number of trays, a common first estimate of the shell height is 4 ft of clearance at the top, 2 ft of tray spacing for each tray gap, and 10 ft of liquid sump at the bottom. A 30-tray column then stands roughly 72 ft, about 22 m, before you add the skirt.
Real columns
From theoretical stages to real trays
A theoretical stage assumes the vapour and the liquid leave in perfect equilibrium. Real trays never quite manage it. The gap between the two is efficiency, and it is what decides how many trays you actually buy.
The number you order from the fabricator
actual trays = NEO
N is the theoretical stages inside the column, with the reboiler already taken out. EO is the overall column efficiency, typically 0.4 to 0.8 for ordinary hydrocarbon and solvent systems.
In mPa·s, taken at the average column temperature and averaged over the feed composition.
O'Connell correlation
EO = 0.503 [ α ⟨μ⟩ ]−0.226
μ in mPa·s. Both a high α and a thick liquid pull efficiency down. Fitted to a large set of commercial columns.
Drickamer and Bradford correlation
EO = 13.3 − 66.8 log10 μ
Gives EO as a percentage, with μ in cP. Older and simpler. It ignores α entirely, which is why O'Connell replaced it.
Why efficiency dropsThe vapour is only in contact with the liquid for a second or two as it bubbles through. A thick liquid slows diffusion, so less of the possible transfer happens in that second. A large α means the target composition is further away, so the same contact time gets you a smaller fraction of the way there. Both effects push EO down.
Murphree efficiency, tray by tray
Overall efficiency lumps the whole column into one number. Murphree efficiency looks at a single tray and asks how far it got, compared with how far it could have gone.
Murphree vapour efficiency
EMV = yn − yn+1y*n − yn+1
The vapour arrives at yn+1 and leaves at yn. If the tray were perfect it would leave at y*n, in equilibrium with the liquid on the tray. Actual rise divided by possible rise.
On the diagram this shifts the whole equilibrium curve downward, toward the operating line. You then step off stages on that lower effective curve, and because the gaps are smaller, you need more of them.
true equilibrium curveeffective curve at EMVoperating lines
0.60
Do it in the right orderSolve the McCabe-Thiele diagram for theoretical stages first. Take the reboiler out of the count, because the reboiler is a real equilibrium stage and does not need an efficiency correction. Then divide the remaining stages by EO and round up. Never apply efficiency to the reboiler or to a total condenser.
Real columns
When the curve touches the diagonal
Everything on the last few pages assumed the equilibrium curve stays above the 45° line all the way across. Some mixtures break that assumption. Where the curve crosses the diagonal, distillation stops working, and no amount of reflux or trays will get you past it.
What an azeotrope is
At one particular composition the vapour leaving a stage has exactly the same composition as the liquid it left. So y = x, the point sits on the 45° line, and boiling the mixture changes nothing. The mixture behaves like a pure substance: it boils at one fixed temperature and the distillate comes over unchanged.
The composition where this happens is xaz. It is fixed by the chemistry of the pair and by the pressure, not by anything you do to the column.
Why the staircase stops
A step of the staircase is only possible when there is a vertical gap between the operating line and the equilibrium curve. At the azeotrope the curve meets the diagonal, and any operating line drawn from (xD, xD) also starts on the diagonal.
Push xD toward xaz and that gap shrinks to nothing. The steps get smaller and smaller and never arrive. xD can never pass xaz.
The system on slide 36: isopropyl ether and 2-propanol
This pair boils at 68.3 °C and 82.3 °C, so it looks like an easy separation. It is not. The two form a minimum-boiling azeotrope at x = 0.780 that boils at 66.2 °C, which is lower than either pure component. That low-boiling point is the first thing to leave the top of any column, so it is the ceiling on your distillate purity.
dew point, vapourbubble point, liquidazeotrope, 66.2 °C
Read the boiling diagram first
Both curves dip down and meet at x = 0.780. To the left of that point isopropyl ether is the lighter component and rises. To the right the roles swap and 2-propanol becomes the lighter one. The lowest boiling point on the whole diagram is not a pure component, it is the mixture.
The give-awayAny time the bubble point curve has a minimum or a maximum inside the diagram rather than at one of the two ends, you have an azeotrope. Check this before you draw a single operating line.
Move xD and watch the column become impossible
equilibrium curveazeotrope wall at 0.780rectifying linestripping line
0.750
Jump to
Feasible
What the slider is showingThe feed is 40 mol% ether, saturated liquid, bottoms 5 mol%, and the reflux ratio is held at R = 0.96. Drag xD upward and the number of stages climbs slowly, then very fast, then runs away completely. There is no cliff edge you can see coming from the specification sheet. It only shows up on the diagram.
One honest note on the numbers
Slide 36 reports N + 1 = 10.2 stages for this separation. The curve on this page is built from published van Laar constants fitted so the azeotrope lands exactly at x = 0.780 and 66.2 °C, and stepping it off gives 10.8. The difference is which set of equilibrium data was used, not a mistake in either construction. Different sources for the same pair routinely disagree by half a stage or more.
This is worth knowing in itself: a McCabe-Thiele answer is only as good as the equilibrium curve underneath it. When your answer and someone else's differ by a fraction of a stage, check the VLE data before you check the drawing.
So what do you actually do about it
Change the pressure
The azeotrope composition moves with pressure. Run two columns at different pressures and pass the material between them, and each column pushes past the other one's ceiling. This is pressure-swing distillation, and it works when the azeotrope shifts enough, roughly 5 mol% or more, over a practical pressure range.
Add a third component
A heavy solvent that clings to one component changes the relative volatility and can remove the azeotrope entirely. That is extractive distillation, used industrially to make dry ethanol with ethylene glycol. A light entrainer that forms a new low-boiling azeotrope taken overhead is azeotropic distillation.
Stop using distillation
Get as close to the azeotrope as economics allow, then switch method. Ethanol plants take the overhead at about 95% and finish with molecular sieves or a pervaporation membrane. Neither of those cares about relative volatility, so the azeotrope is simply not their problem.
Reading the diagram before you trust itNotice that on the ether-rich side of 0.780 the curve sits below the 45° line. Over there the vapour is poorer in ether than the liquid, so the staircase runs the other way and the roles of light and heavy component are reversed. If your feed starts on that side, the component you can take overhead is 2-propanol, not the ether.
Real columns
The same diagram, in real plants
Everything here is from operating industrial columns and published plant data, not from textbook exercises. The point is that the numbers you have been drawing all afternoon are the same numbers that decide how tall a real tower is and how much steam it burns.
Share of world energy
10 to 15%
Thermal separations, distillation dominant. Sholl and Lively, Nature 2016.
US industrial energy
~ half
Separations account for roughly half of all industrial energy use in the United States.
Refinery throughput
90 Mbbl/d
Crude processed worldwide, drawing on the order of 230 GW continuously.
Second law efficiency
< 10%
Typical for distillation, which is why the choice of R matters so much.
A benzene and toluene column you could walk up to
This is the plant in Figure 7.1 of Seader, the same pair used in Example 4.1, drawn with the flows a real design gives. Compare its reflux ratio with the one you calculated.
Item
Value
Feed
620 lbmol/h, 46% benzene
Feed condition
bubble point liquid, 55 psia, 294 °F
Distillate
281 lbmol/h, 99 mol% benzene
Bottoms
339 lbmol/h, 98 mol% toluene
Reflux
623 lbmol/h, so R = 2.215
Boilup
708 lbmol/h
Rmin
1.708, giving R / Rmin = 1.297
Theoretical stages
21, against Nmin = 10.7
Actual trays
25 sieve trays, feed at tray 13
Overall efficiency
20 / 25 = 80%
Column
5 ft diameter, 24 in tray spacing
Reboiler duty
10,030,000 Btu/h
Condenser duty
11,820,000 Btu/h
Three numbers worth memorisingThe ratio of actual stages to minimum stages is 21 / 10.7 = 1.96, close to 2, which is typical of operating columns. The ratio of actual to minimum reflux is 1.30, right in the middle of the usual design window. And the overall tray efficiency is 80%, close to the average for trayed distillation. If your own design lands far from these, go back and check it.
Twenty-three real binary columns
These are commercial distillations in decreasing order of difficulty. Notice how sharply the tray count climbs as α falls toward 1, and how tightly every plant sits in the R / Rmin band of 1.06 to 1.71.
hover or tap a dot to name the pair
What the second chart saysNot one of these plants runs anywhere near minimum reflux, and not one runs anywhere near total reflux. Every single one sits in a narrow band just above Rmin. That band is the flat bottom of the cost curve, found on real plants rather than on paper.
Binary mixture
αaverage
Trays
Pressurepsia
R / Rmin
Read the first row againButadiene and vinylacetylene have α = 1.16 and need 130 trays. Water and ethylene glycol have α = 81.2 and need 16. Same method, same diagram, same equations. The only thing that changed is how far the equilibrium curve sits from the 45° line.
The propylene splitter, the hardest column in the plant
Why it is so tall
Propylene and propane differ by one double bond. Relative volatility is about 1.40 at 280 psia, and polymer-grade propylene must be 99.5 mol% or better. Table 7.1 gives 138 trays at R / Rmin = 1.06, which is almost at the minimum-reflux limit because the steam bill on a column that size is enormous.
Many plants split the duty into two towers in series simply because a single tower would be too tall to build or to support against wind.
A real one, shipped in one piece
The propane dehydrogenation plant at Kallo in Belgium, run by Borealis, uses a propylene splitter about 105 m long and 10 m wide, weighing roughly 1,600 tonnes, serving a unit rated at 750,000 tonnes of propylene a year.
A column that size is one of the largest single pieces of equipment on any chemical site, and its whole height exists because α is 1.40 instead of 2.47.
When a real splitter stops workingA published troubleshooting case on a C3 splitter, 28 ft internal diameter at 105 psig with four-pass valve trays, found overhead propane at 3.4 mol% against a 0.5% specification. The equilibrium data had not changed and the reflux ratio was on target. The overall tray efficiency had fallen from a design 80 to 90% down to 40 to 50% because of tray damage and maldistribution across the four passes. The McCabe-Thiele diagram was right the whole time. The hardware was not.
Efficiency measured on a working column
Eastman Kodak in Rochester, New York ran a methylene chloride and ethylene chloride column at total reflux specifically to measure how good its trays really were. This is the data set behind the efficiency page.
Column detail
Value
Diameter
5.5 ft, 65.5 in I.D.
Trays
60, spaced 18 in
Tray type
115 bubble caps per tray, 3 in diameter
Bubbling area
20 ft²
Liquid rate
24.5 gal/min per ft of weir
Percent of flooding
85%
Top tray pressure
33.8 psia
Bottom tray pressure
42.0 psia
Sample point
mol% MC in liquid
Tray 33
89.8
Tray 32
72.6
Tray 29
4.64
How this becomes an efficiencyAt total reflux the operating line is the 45° line, so the whole diagram is just the curve and the diagonal. Step off the ideal stages needed to move from 89.8% to 4.64%, compare with the 6 real trays actually used from 35 to 29, and the ratio is the overall efficiency. The pressure drop of 8.2 psi over 60 trays is why α changes down the column.
Two more you will meet
Ethanol from fermentation
A distillery train takes about 30,000 kg/h of fermented beer at roughly 10% ethanol by volume and produces 95% v/v rectified spirit, using an analyzer column followed by a rectifier. Steam consumption runs near 2.80 kg of steam per kg of ethanol.
The 95% ceiling is not a design choice. It is the ethanol and water azeotrope, and everything above it is done by molecular sieve.
Why refineries look the way they do
A crude tower is not binary, but every side stream drawn off it is designed with light key and heavy key thinking that starts here. Table 7.1 has ethylbenzene and styrene at 1 psia, run under vacuum only because styrene polymerises in a hot reboiler.
Ethylene and ethane at 230 psia go the other way and need refrigerant overhead, because cooling water cannot condense ethylene at all.
The thread through all of itPressure is chosen so the overhead can be condensed with the cheapest available coolant and the bottoms will not decompose. That choice sets the temperatures, the temperatures set α, α sets the equilibrium curve, and the curve sets the number of stages. Every plant number on this page comes out of that chain, and the McCabe-Thiele diagram is where you see it happen.
Use it
Design solver
Every specification in one panel. Change any of them and the operating lines, the q-line and the whole staircase redraw immediately. Use it to build the feel for what each number actually controls, which is something no formula will give you.
Watch the stage count creep up, then jump. The last stretch before Rmin costs you far more stages than the first. This is the shape of the whole economics argument.
Move q from 1 to 0
The q-line swings from vertical to horizontal. Notice that the stage count changes, but the boilup and the reboiler duty change more. Vaporising the feed moves work from the reboiler to the preheater.
Switch to methanol and water
The curve bulges. The same specifications now need a different number of stages, and near the top the curve hugs the diagonal, so the last bit of purity is expensive.
Drive xD to 0.995
Stages pile up at the top corner of the diagram. Purity is not linear in cost. Going from 95 to 99 percent costs more stages than going from 50 to 95.
One rule the solver obeysThe solver refuses to draw when the specification is impossible, that is when R is at or below Rmin for the numbers you set, or when xB is above zF, or xD is below it. The status strip tells you which. A design that will not draw is not a bug in the tool, it is a mass balance telling you something.
Use it
Class exercises, step by step
Both exercises from the lecture, solved the same way every time: balance first, flows second, lines third, staircase last. Play each build and read what every line means as it appears.
Exercise 1A methanol and water mixture, 100 mol/h at 40 mol% methanol, enters as a saturated liquid. The distillate must be 95 mol% methanol and the bottoms 5 mol%. The reflux ratio is R = 2.5 with a total condenser and a partial reboiler. Find the number of stages and the feed stage.
Graphical total
5.75
Corners counted from xD down to xB, reboiler included.
Stages in the column
4.75
Take the reboiler out, so five real trays are needed.
Feed stage
4
The first corner that crosses the operating line intersection.
R / Rmin
3.71
Rmin = 0.674, so this specification is generous with reflux.
The arithmetic, in order
Quantity
Working
Value
D
100 (0.40 − 0.05) / (0.95 − 0.05)
38.89
B
100 − 38.89
61.11
L
R D = 2.5 × 38.89
97.22
V
(1 + R) D
136.11
L̄
L + q F = 97.22 + 100
197.22
V̄
V − (1 − q) F
136.11
VB
V̄ / B
2.227
Check
L̄ − V̄ − B = 197.22 − 136.11 − 61.11
0
The three lines
Rectifying
y = 0.7143 x + 0.2714
Slope L / V = 97.22 / 136.11, through (0.95, 0.95).
q-line
x = 0.40, vertical
q = 1 for a saturated liquid feed, so the slope q / (q − 1) is infinite.
Stripping
y = 1.4490 x − 0.0224
Slope L̄ / V̄, through (0.05, 0.05) and through the intersection at (0.400, 0.557).
Exercise 2A hexane and octane mixture, 100 kmol/h at 21 mol% hexane, enters half vaporised so q = 0.5. The distillate must be 98 mol% hexane and the bottoms 2 mol%. The reflux ratio is R = 4.0. Find the stages in the column and the feed stage.
Graphical total
7.13
Everything the drawing gives you, reboiler included.
Stages in the column
6.13
This is the answer the exercise asks for, so seven trays.
Feed stage
5
Switch operating lines here, not before.
R / Rmin
1.44
Rmin = 2.785. This one is a realistic industrial choice.
Why this one feels harderThe feed is dilute, only 21% hexane, but the distillate must be 98%. That is a long way to travel on the diagram, and the feed enters low down where the curve is steep. Notice the stripping line is very steep, slope 2.64, because L̄ is large and V̄ is small. The boilup ratio is only 0.61, which is unusual, and it comes straight from the half-vaporised feed carrying its own vapour into the column.
Check yourself
One page to take into the exam
Overall balances
F = D + B
F zF = D xD + B xB
Internal flows, top
L = R D
V = (1 + R) D
Internal flows, bottom
L̄ = L + q F
V̄ = V − (1 − q) F
V̄ = VB B
Rectifying line
y = RR + 1 x + xDR + 1
Through (xD, xD).
Stripping line
y = L̄V̄ x − B xBV̄
Through (xB, xB), slope = 1 + 1 / VB.
q-line
y = qq − 1 x − zFq − 1
Through (zF, zF).
Equilibrium, constant α
y = α x1 + (α − 1) x
Fenske and pinch
Nmin = ln [ (xD/(1−xD))((1−xB)/xB) ]ln α
RminRmin+1 = xD − y′xD − x′
Duties and trays
QR = [D(1+R) − (1−q)F] Δhvap
trays = N / EO
The six moves, every timeDraw the equilibrium curve and the 45° line. Mark xD, zF and xB on the diagonal. Draw the q-line from (zF, zF). Draw the rectifying line from (xD, xD) with slope R/(R+1) until it hits the q-line. Join that intersection to (xB, xB) for the stripping line. Step off from (xD, xD), horizontal to the curve, vertical to whichever line applies, until you pass xB.
