Chapter 7

Liquid–Liquid Extraction

A solvent instead of a reboiler again — but this time the solvent dissolves part of what you were trying to keep, and the diagram gains a dimension.

Wankat, Ch. 13 4 interactive apps 3 worked examples Self-marking problem set Prerequisite: Ch. 6

By the end of this chapter you should be able to

Say when extraction beats distillation, and why.
Treat the immiscible case as Chapter 6 with new labels, and compare co-current, cross-flow and countercurrent cascades.
Read a ternary equilibrium diagram in both the equilateral and the right-triangle representation.
Use the lever rule to mix and to split streams on a ternary diagram.
Solve a single stage, finding the tie line through the mixing point with the conjugate curve.
Locate the difference point Δ, step off stages, and find the minimum solvent rate.

Separating a liquid with a liquid

Chapter 6 dissolved a solute out of a gas. This chapter does the same thing to a liquid — contact it with a second liquid that will not mix with it, and let the solute choose. The equipment is a stirred tank and a settling drum rather than a tower of trays, but the logic is the one you already have: an equilibrium relation, a mass balance, and a staircase between them.

Figure 7.1 — An extraction process. The feed carries solute B in carrier A; solvent C takes the solute away as the extract, leaving the raffinate behind. As in absorption6 §6.1, the solvent is a debt: it has to be separated from the solute again, usually by distillation, and the recovery column is where the energy goes.

Two words to fix now

The stream that leaves rich in solvent is the extract. The stream that leaves lean in solvent — mostly the original carrier — is the raffinate. Everything in this chapter is bookkeeping between those two, and half the errors made in it are the result of mixing up which is which on the diagram.

7.1 Why extraction

Distillation separates by volatility and it is very good at it. Extraction is what you reach for when volatility is the wrong handle:

Heat-sensitive material. Antibiotics, proteins, flavours and fragrances degrade at their boiling points. Extraction runs at room temperature.
A solute present in small quantity. Boiling a whole stream to recover 1 % of it is thermodynamically absurd — the same argument that put a mass separating agent1 §1.6 into a flue-gas scrubber.
An azeotrope, or components too close in volatility. A solvent can have a large selectivity where relative volatility has almost none, and an azeotrope2b §2b.6 is invisible to it.
The components are already liquid. No latent heat is paid at all.

Cooking oils, decaffeinated coffee, most antibiotic recovery, the extraction of aromatics from reformate, and the purification of rare earths are all extraction. So is leaching — the same operation with a solid feed, which is how sugar leaves beet and how metals leave ore.

7.2 The immiscible case — Chapter 6 with new labels

Take a solute B dissolved in carrier A, and contact it with solvent C.

Figure 7.2 — The solute partitions across the interface between two liquid layers, exactly as it partitioned across a gas–liquid interface in Chapter 6.

If A and C are mutually insoluble — and many useful pairs very nearly are — then the flow of A in the raffinate and the flow of C in the extract are both constant down the cascade. That is precisely the condition that made solute-free coordinates6 §6.3 work, so define

7.1

and every equation of Chapter 6 carries over unchanged. The countercurrent operating line is straight,

7.2

with A the flow of carrier and C the flow of solvent. Note the staggered subscripts, exactly as in Equation 6.6 of the absorber operating line6 §6.4: a balance over stages 1 to j has the raffinate crossing the envelope, not . and both leave stage j, so they are the equilibrium pair and sit on the equilibrium curve; and are the pair that pass each other, and they are the ones on the operating line. The extraction factor

7.3

plays the part the absorption factor6 §6.8 played there, with the same wall at and the same Kremser equation behind it.

The distribution coefficient

Extraction data are often quoted as a distribution (or partition) coefficient. Be careful which way up it is defined — both conventions are in print. The question bank uses

the weight fraction of solute in the carrier over that in the solvent. On that definition a small is a good solvent, because the solute prefers the solvent. Read the definition before you use the number; the two conventions differ by a factor of in the answer.

Three ways to cascade

Once there is more than one stage there is a choice about how to arrange them, and it matters more than students expect.

Arrangement	What happens	Ceiling
Co-current	Both phases travel together through every stage. After the first contact they are already in equilibrium, so nothing further happens.	One equilibrium stage, no matter how many vessels you buy.
Cross-flow	Fresh solvent is fed to every stage and each extract is withdrawn separately. Each stage sees a clean solvent, so each one keeps working.	Improves without limit as N grows, but the solvent is split N ways and every extract is dilute.
Countercurrent	Solvent and raffinate move in opposite directions, so the leanest raffinate meets the cleanest solvent.	Best of the three at any N, and it makes one concentrated extract instead of N dilute ones.

Think about it — why does co-current stop at one stage?

Because the two streams leaving stage 1 are already in equilibrium with each other, and stage 2 receives exactly those two streams. There is no driving force left. Countercurrent works because the stream a phase meets next is one it has not equilibrated with — it always sees something leaner, all the way to the end of the cascade. That single sentence is the reason every separation cascade in this course is countercurrent.

App 1

Three ways to cascade

The same feed, the same total solvent, the same number of stages — arranged three ways. Watch the raffinate that comes out of each, and watch what happens to the gap as you add stages or change the extraction factor. The equilibrium is straight in ratio coordinates, so this is the immiscible case of §7.2.

Equilibrium form

Feed X_F, mass B / mass A 0.0526

Equilibrium slope m 11.82

Solvent per carrier C/A 0.158

Stages 5

The construction

Countercurrent staircase against the cross-flow fan.

Raffinate against stage count

How each arrangement improves as you buy more vessels.

Countercurrent X_out

—

Cross-flow X_out

—

Co-current X_out

—

Extraction factor

—

Countercurrent recovery

—%

Best possible

—

✓—

7.3 When the solvent dissolves the carrier

Perfect immiscibility is an idealisation. In most real systems the three components are partly soluble in one another, the flows of A and C are not constant, and a two-coordinate diagram can no longer hold the state. Three components need three compositions — two of which are independent — so the equilibrium lives on a ternary diagram.

Figure 7.3 — Ternary systems come in several kinds. Type I has one partially miscible pair, and the two-phase region is a closed dome — the common case, and the only one this chapter designs with. Type II has two partially miscible pairs, so the two-phase band runs right across the diagram and there is no plait point. Type III has a larger region still. The shape of the dome is the whole design problem: it tells you which mixtures split into two layers and which do not.

Inside the dome — the binodal curve — a mixture is unstable as a single liquid and separates into two layers. The compositions of those two layers are joined by a tie line, and the tie line is the ternary equivalent of a horizontal step on an x–y diagram: it connects two phases that are in equilibrium. Tie lines are not generally parallel, and the point where they shrink to nothing is the plait point, where the two layers become identical.

7.4 The right-triangle diagram

Ternary compositions are traditionally drawn on an equilateral triangle, one pure component at each vertex. It is elegant, and it is a nuisance to plot on. Because the three fractions sum to one, any two of them fix the state, so the same information fits on a right triangle with ordinary rectangular axes.

Figure 7.4 — The same equilibrium, twice. Left: the equilateral triangle. Right: the right triangle, with the carrier on the horizontal axis and the solute on the vertical. The red lines connect matching compositions. We use the right-triangle form for every calculation in this chapter — it plots on ordinary graph paper and it reads like every other diagram in the course.

Reading the right triangle

Horizontal axis: mass fraction of the carrier (water, in our system). Vertical axis: mass fraction of the solute (acetone). The third component — the solvent — is whatever is left over, , so the hypotenuse is the solvent-free line and the origin is pure solvent. Every point in the triangle is a legitimate ternary mixture; every point outside it is not a composition at all.

Figure 7.5 — Acetone (solute) – water (carrier) – trichloroethane (solvent) at 25 °C, the system this chapter is built on. The binodal separates one liquid phase from two. Its left-hand branch carries the extract compositions, its right-hand branch the raffinate compositions, and the red tie lines say which extract belongs to which raffinate. The dashed conjugate curve on the right is a device for finding a tie line that is not one of the ones you were given — §7.6.

App 2

Reading a ternary diagram

Click anywhere in the triangle. The app reads off all three mass fractions, tells you whether that mixture is one phase or two, and — if it is two — draws the tie line through it and gives you the compositions and relative amounts of the layers that would settle out.

System

Show

Carrier fraction x 0.500

Solute fraction y 0.250

The diagram — click to move the point

Right-triangle representation; solvent is 1 − x − y.

What settles out

The two layers, drawn to scale by amount.

Carrier

—

Solute

—

Solvent

—

Extract layer

—%

Solute in extract

—

Solute in raffinate

—

✓—

7.5 Mixing and the lever rule

Two streams and are combined into a mixture . The overall balance and the two independent component balances are

7.4

Eliminating between them gives the same ratio from either component,

7.5

which says two things at once. First, lies on the straight line joining and — because the two ratios are equal, the three points are collinear. Second, it lies at the position a lever balance would put it: closer to the larger stream. This is the same lever rule you used on a T–xy diagram2a §2.9 in Chapter 2a, now working in two dimensions instead of one.

The lever arms are the wrong way round if you rush

: the amount of stream 1 is proportional to the length of the arm on the far side, next to stream 2. A mixture that is nearly all feed sits nearly on top of the feed point, so the arm to the feed is short. If your answer says the mixing point is close to the small stream, you have inverted it.

Everything else in this chapter follows from Equation 7.5 used forwards and backwards. Used forwards it mixes two streams into one. Used backwards it splits one stream into two: if a mixture inside the dome settles into an extract and a raffinate , then lies on the tie line , and the amounts follow from the arm lengths.

7.6 A single stage

Figure 7.6 — A mixer–settler. The mixer creates area for mass transfer; the settler lets gravity do the separating. Together they are one equilibrium stage — the ternary analogue of a flash drum3 §3.2.

The recipe is short:

Plot the feed and the solvent on the diagram.
Join them; find on that line by the lever rule with the known flows.
If lies outside the dome, nothing separates — you have made one homogeneous liquid, and no amount of settling will help.
If lies inside, find the tie line through . Its ends are the extract and the raffinate .
Get and from the lever rule on that tie line, or from the overall balance.

Step 4 is the only hard one, because the tie line through an arbitrary point is not one of the ones the data table gave you. That is what the conjugate curve is for.

Building and using the conjugate curve

To build it: for each tie line you were given, join the ends and . Draw a vertical through and a horizontal through ; mark where they cross, at . Do that for every tie line and join the marks — that curve is the conjugate (or auxiliary) curve, and it runs from the plait point down to the far corner.

To use it: given a point on the extract branch, run horizontally to the conjugate curve, then vertically down to the raffinate branch. You have just interpolated a tie line without interpolating a tie line.

For a point inside the dome: guess a tie line, check whether is above or below it, and correct — two or three tries is normal by hand. App 3 does the search for you and shows the trial lines.

App 3

One stage, and the conjugate curve

Set the feed and the solvent rate. The app mixes them, finds the tie line through the mixing point — showing the conjugate-curve construction that locates it — and splits the mixture into an extract and a raffinate. Push the solvent up and watch the extract get bigger and weaker; push it down until the mixing point leaves the dome and nothing separates at all.

System

Solute in the feed 0.450

Solvent / feed 0.50

Solute in the entering solvent 0.000

Show construction

The construction

Feed, solvent, mixing point, and the tie line through it.

Solute recovery against solvent rate

One stage only — note where it flattens.

Extract E/(F+S)

—

Solute in extract

—

Solute in raffinate

—

Solvent in raffinate

—

Recovery

—%

Selectivity

—

✓—

7.7 Countercurrent cascades

Figure 7.7 — The cascade and its nomenclature. Solvent enters stage 1; the extract flows towards the feed end and leaves stage N as . Feed enters stage N; the raffinate flows the other way and leaves stage 1 as . Subscripts name the stage a stream leaves, so and are the two streams in equilibrium on stage j.

The external balances are the ones you would write for any cascade:

7.6

and the same again for the second component. Read geometrically: lies on the line and on the line at the same time. Two lines, one intersection — which is how you find the extract product before you have stepped a single stage.

The external calculation, in order

Plot and ; draw .
Put on it by the lever rule with the chosen .
Plot the specified raffinate product on the raffinate branch.
Draw and extend it to the extract branch — that intersection is .
Lever rule on gives the two product flows.

7.8 The difference point

Now the internal problem. Write the overall balance as a difference rather than a sum:

7.7

and the same for each component,

7.8

The quantity is the net flow past any point in the cascade, and Equation 7.7 says something remarkable: it is the same at every point. A constant net flow with a constant composition is a single point on the diagram — and since can be rearranged to , the three points , and are collinear for every .

Δ is the operating line, collapsed to a point

In distillation each section had one operating line, and every pair of passing streams sat on it. Here each pair of passing streams sits on a line through one fixed point — so the cascade has a whole pencil of operating lines, all hinged on Δ. Finding Δ is exactly as important as finding the rectifying line was in Chapter 4a4a §4.2, and for exactly the same reason.

Figure 7.8 — Locating Δ. Because , the points , and Δ are collinear; because , so are , and Δ. Draw both lines and extend them — Δ is where they cross. It usually lands well outside the triangle, which is fine: Δ is a difference, not a mixture, so it does not have to be a real composition. Here it sits at a water fraction of 2.2.

Which side does Δ land on?

Read the sign of . If the raffinate is the net flow — more material moving towards the raffinate end than towards the extract end, so is negative — Δ lands beyond the raffinate side, at , as in Figure 7.8. Worked example 7.3 is exactly this case: and kg/h give kg/h and . Add more solvent and Δ swings out to infinity (the two lines become parallel) and then reappears on the far left, at negative , where the extract has become the net flow and is positive — as in the first long problem of §7.12, where and kg/h give kg/h and . Nothing physical happens at that transition; it is the same construction throughout, and App 4 will walk you through it.

7.9 Stepping off stages

Two relations alternate, exactly as they did on an x–y diagram:

A tie line connects the two streams that leave the same stage — the equilibrium step.
A line through Δ connects two streams that pass each other between stages — the operating step.

The construction

Start at , the specified raffinate product.
Tie line through → gives , the extract leaving stage 1.
Line from Δ through , extended to the raffinate branch → gives .
Tie line through → . And so on.
Stop when the extract reaches . The number of tie lines you drew is the number of equilibrium stages.

Note which of the two steps counts. Tie lines are stages; the lines through Δ are just bookkeeping between them. It is the same rule as McCabe–Thiele, where the horizontal steps to the equilibrium curve were the stages and the verticals to the operating line were not.

7.10 Minimum solvent

Reduce the solvent rate and the extract gets richer, which is good, and the stages multiply, which is not. Somewhere below there is a rate at which the cascade needs infinitely many stages — the minimum solvent.

The pinch happens when an operating line — a line through Δ — coincides with a tie line. Once that happens the two streams passing each other are already in equilibrium, the driving force is zero, and the staircase makes no further progress. Geometrically:

Finding (S/F)_min

Draw and extend it in both directions — Δ must lie on this line whatever the solvent rate, because is specified and is fixed.
Extend every tie line in the operating region until it meets that line. Each intersection is a candidate Δ.
If the intersections fall on the raffinate side (), the binding one is the Δ furthest from the diagram; if they fall on the solvent side (), it is the Δ closest to the diagram. Either way it is the one that demands the most solvent — every other candidate would pinch before you got there. Think of the extended line as a loop: as Δ travels outwards from the triangle to , through infinity, and back in from towards the solvent corner, the solvent rate it implies rises the whole way.
Join that to and extend to the extract branch → , the richest extract obtainable.
is where crosses ; the lever rule on then gives .

The pinch is usually not at the feed

Because the tie lines are not parallel, the tie line that binds is generally an interior one, somewhere between the feed and the raffinate product — not the tie line through the feed. Assuming it is at the feed end is the commonest error in this construction, and it always errs on the optimistic side: too little solvent, and a column that cannot make specification. It is the same trap as the interior pinch in absorption6 §6.7.

As in every other chapter, practice runs at 1.5 to 2 times the minimum. There is one extra consideration here that distillation does not have: the solvent you circulate has to be distilled back off the extract, so the cost of extra solvent shows up as reboiler duty in a different column.

The same solvent, spent two ways

App 4 will also run the cascade cross-flow: the same total solvent divided equally between the same number of vessels, each dose mixed with the raffinate from the one before and its extract drawn off separately. On the ternary the difference is easy to see. Countercurrent walks the raffinate down the binodal in long steps hinged on a single Δ; cross-flow drags the mixing point back towards the solvent corner every time, so each stage starts from a weaker driving force and each extract comes off more dilute than the last. Same money, same vessels, and — as §7.2 proved algebraically — a worse raffinate.

App 4

The countercurrent designer

The whole construction, live. Set the feed, the required raffinate and the solvent rate; the app finds , and Δ, steps the stages, and reports the count. Sweep the solvent rate down towards the minimum and watch Δ swing out to one side, off to infinity, and back in on the other; turn on the minimum-solvent search to see which tie line is doing the pinching. Switch the arrangement to cross-flow to spend the same solvent in the same number of vessels the other way, and see what it costs.

System

Solute in the feed 0.450

Required raffinate 0.100

Multiple of the minimum solvent 1.50

Arrangement

Minimum-solvent search

The construction

Δ may sit far off the triangle — the axis stretches to hold it.

Down the cascade

Solute in each phase, stage by stage.

Stages N

—

(S/F)_min

—

S/F used

—

Solute in extract

—

Recovery

—%

Δ at x =

—

Countercurrent raffinate

—

Cross-flow raffinate

—

✓—

7.11 Worked examples

Worked example 7.1One stage on a ternary diagram

1000 kg/h of a 45 wt % acetone-in-water solution is contacted once with 500 kg/h of pure 1,1,2-trichloroethane at 25 °C. Find the mixing point, the two layers that settle out, their flows, and the fraction of the acetone recovered.

Work it yourself first, then open

Plot and mix. In right-triangle coordinates (x = water, y = acetone) the feed is and the solvent is the origin, . The lever rule on with :

M = (1000 F + 500 S)/1500 = (0.3667, 0.3000)

Is it inside the dome? Yes — comfortably. So it separates.

The tie line through M. Interpolating with the conjugate curve gives

R = (0.738, 0.247) · E = (0.024, 0.349)

so the raffinate is 24.7 wt % acetone and the extract 34.9 wt %.

Flows, by the lever rule on the tie line. M sits a fraction 0.521 of the way from R to E, so

E = 0.521(1500) = 781 kg/h, R = 719 kg/h

Recovery. Acetone in the extract is 781(0.349) = 273 kg/h out of 450 fed:

recovery = 60.5 %

Read that number. One stage, with half a tonne of solvent per tonne of feed, recovers three-fifths of the acetone. Worked example 7.3 will get 87 % out of the same feed with less solvent, and leave a raffinate at 10 wt % instead of 24.7 % — because it uses the stages countercurrently instead of all at once. Arrangement beats brute force, and that is the whole content of §7.2.

App 3, acetone–water–TCE, feed 0.450, solvent/feed 0.50.

Worked example 7.2Countercurrent against cross-flow, immiscible

200 mol/h of a toluene–acid solution containing 0.05 mole fraction acid is to be extracted with water recycled from a downstream still, containing 0.002 mole fraction acid. The total water rate is 30 mol/h. Toluene and water may be taken as immiscible, and equilibrium at 1 atm and 25 °C is

(a) How many countercurrent stages take the toluene stream to 0.001 mole fraction acid? (b) If instead the water is split equally between five cross-flow stages, what leaves?

Work it yourself first, then open

Carrier flows and ratios. Acid in = 200(0.05) = 10 mol/h, toluene = 190 mol/h. Water carries 30(0.998) = 29.94 mol/h of water with 0.06 mol/h of acid.

X_F = 10/190 = 0.05263 · Y_in = 0.06/29.94 = 0.002004 · X_out = 0.001/0.999 = 0.001001

The operating line has slope A/C = 190/29.94 = 6.346 and passes through . The exit water follows from the overall balance:

Y_out = 0.002004 + 6.346(0.05263 − 0.001001) = 0.3297 → 24.8 mol % acid

(a) Stepping the staircase between that line and the equilibrium curve gives

N = 4.64 stages — five in practice

(b) Cross-flow, five stages, 6 mol/h of water each. Solve each stage in turn:

X: 0.05263 → 0.03330 → 0.02237 → 0.01550 → 0.01094 → 0.00782

so the toluene leaves at 0.0078 mole fraction — eight times the specification, using the same solvent and the same number of vessels.

And co-current? One contact of all 30 mol/h of water with the whole feed gives X = 0.0162, worse again, and adding vessels changes nothing.

Why countercurrent wins. In cross-flow every stage is fed clean solvent but only a fifth of it, so the driving force is large and the capacity small. Countercurrent gives the last stage the whole solvent flow at its cleanest, exactly where the raffinate is leanest and the driving force is hardest to find. It puts the driving force where it is scarce.

App 1 (equilibrium form y = m x), X_F = 0.0526, m = 11.82, C/A = 0.158, five stages. The app solves for the raffinate at a fixed stage count, so it reports 0.00072 at exactly five stages — consistent with 4.64 stages reaching 0.001001.

Worked example 7.3A full countercurrent design

1000 kg/h of a 45 wt % acetone-in-water solution is to be extracted countercurrently at 25 °C with pure 1,1,2-trichloroethane to give a raffinate containing 10 wt % acetone. Find the minimum solvent rate, and the number of equilibrium stages at 1.5 times that.

Work it yourself first, then open

Fixed points. , , and the raffinate product is on the raffinate branch at 10 wt % acetone: .

Minimum solvent. Draw and extend it. Extending the tie lines to meet it puts every candidate Δ on the raffinate side, beyond ; the binding one is the furthest out, at , and it belongs to an interior tie line — the one leaving a raffinate at about 39 wt % acetone, not the one through the feed. Following the construction through to and applying the lever rule on :

(S/F)_min = 0.230 → S_min = 230 kg/h

At 1.5 × minimum, S = 345 kg/h and S/F = 0.345. Then

M = (0.409, 0.335) · E_N = (0.044, 0.511) · Δ = (2.23, 0.249)

Stepping from R₁:

Stage	acetone in R	acetone in E
1	0.100	0.151
2	0.186	0.270
3	0.264	0.369
4	0.335	0.450
5	0.401	0.519

The fifth extract passes , so

N = 4.87 stages — build 5

Products and recovery. The lever rule on splits the 1345 kg/h of total flow into

E_N = 768 kg/h at 51.1 wt % acetone, R₁ = 577 kg/h at 10 wt %

acetone recovered = 768(0.511) = 392 kg/h of 450 → recovery = 87.2 %

Compare with Worked example 7.1. Same feed, less solvent (345 against 500 kg/h), and the raffinate goes from 24.7 wt % acetone to 10 wt % — recovery 87.2 % against 60.5 %. Five stages arranged countercurrently do what one stage cannot do at any solvent rate.

App 4, feed 0.450, raffinate 0.100, multiple 1.50.

7.12 Check your understanding

Five multiple-choice questions, two short problems and two long ones — 41 marks. Work them offline, then enter your numbers.

Marks earned

0 / 0

Multiple choice

Short problems

Long problems

Think about it

Why can Δ sit outside the triangle when no real composition can?

Because Δ is a difference of two streams, not a mixture of them. Mixing is a weighted average with positive weights, and averages of points inside a triangle stay inside it. A difference has one negative weight, and that throws the result outside — sometimes a long way outside, sometimes to infinity when the two streams happen to be equal. The same thing happens in distillation: the "net flow" point of a rectifying section sits at with a flow of , and if you push the reflux up far enough the operating line becomes the diagonal and the net flow vanishes. Δ is that idea in two dimensions.

Cross-flow uses the same solvent and the same vessels. Why is it ever chosen?

Two reasons, neither of them about stage efficiency. First, it needs no interstage pumping of a settled layer against the direction of the other — a batch laboratory extraction is naturally cross-flow because you simply throw away one layer and add fresh solvent to the other. Second, when you want the first extract as concentrated as possible — for analysis, or because the first fraction is the valuable one — cross-flow gives you a rich cut you can take away before the rest. For a plant that has to hit a raffinate specification at least cost, countercurrent wins every time.

What happens to this construction if the solvent and carrier become fully immiscible?

The two-phase dome stretches until it touches both ends of the base, the extract branch collapses onto the solvent axis and the raffinate branch onto the carrier axis, and the tie lines become the whole diagram. At that limit the only thing that varies is how much solute is on each axis — which is exactly two numbers, and . The ternary diagram has degenerated into the mass-ratio diagram of §7.2. Both halves of this chapter are the same construction; the immiscible one is what happens when the dome swallows the triangle.

A colleague proposes a solvent with an enormous distribution coefficient but which dissolves 20 % of the carrier. Good idea?

Usually not. Capacity is only half of the requirement; selectivity is the other half. A solvent that takes the carrier as well as the solute produces an extract you then have to separate twice — once to recover the solvent and once to remove the carrier it brought with it — and it loses carrier into the extract stream that you will never get back cheaply. On the diagram it shows up as a fat two-phase dome with tie lines that slope the wrong way, and the tell-tale is that the extract branch bends away from the solvent corner. Real solvent selection trades capacity, selectivity, density difference (for settling), interfacial tension, boiling point (for recovery), toxicity and price, and the distribution coefficient is only the first screen.

Where this chapter connects

Back to: §7.2 is Chapter 6 in solute-free coordinates6 §6.3, with the carrier liquid playing the part of the carrier gas.
The lever rule: the same tool you used on a T–xy diagram2a §2.9 in Chapter 2a, now applied in a plane.
One stage: a mixer–settler is a flash drum3 §3.2 that separates by solubility instead of by boiling.
Δ and the operating line: the pencil of lines through Δ does the job the rectifying and stripping lines4a §4.2 did in Chapter 4a.
The pinch: minimum solvent here, minimum reflux4a §4.8 there, minimum solvent6 §6.7 in absorption — one idea with three names.
Ahead: Chapter 88 §8.3 replaces the second liquid with a solid surface, equilibrium stops being a line at all, and nothing is at steady state.

Summary & key equations

Immiscible systems

Mass ratios, — carrier and solvent flows constant

Operating line — staggered: passes , while and are the equilibrium pair

Extraction factor; the wall is at , as in Chapter 6

Cascadesco-current = 1 stage, whatever you build; countercurrent beats cross-flow at equal N and equal solvent

Ternary diagrams

Right triangle = carrier, = solute, solvent

Binodalinside it two phases; the ends of a tie line are the two layers; they merge at the plait point

Conjugate curvelocus of over all tie lines — use it to interpolate a tie line

Lever rule; , and collinear

Countercurrent design

External balance lies on and on

Difference point, the same for every ; , , collinear

Locating Δintersection of with — usually outside the triangle

Steppingtie line → stage; line through Δ → between stages. Count the tie lines.

Minimum solventan operating line coincides with a tie line — search all of them, the pinch is usually interior

Practice– times the minimum

← Previous

6 · Absorption & Stripping

8 · Adsorption & Chromatography