xN leaving
—
Chapter 6
Absorption & Stripping
Separating with a solvent instead of a reboiler — and the bookkeeping change that makes the operating line straight again.
By the end of this chapter you should be able to
- Distinguish absorption from stripping, and physical from chemical absorption.
- Explain why an absorber is almost always drawn next to a regenerator.
- Use Henry's law and say when it applies.
- Convert to solute-free (mole-ratio) coordinates and explain why the operating line is straight there and curved in mole fractions.
- Construct an absorber or a stripper stage by stage, and find or .
- Use the Kremser equation for dilute systems, and know what the absorption factor means.
Separating with a solvent
Everything from Chapter 3 to Chapter 5 created a second phase with heat. Boil part of the mixture, and the vapour is richer in whatever is more volatile. That is an energy separating agent, and it works well when the components differ enough in volatility to make it worth the reboiler.
Sometimes they do not. A flue gas is 12 % CO₂ in nitrogen at atmospheric pressure and near-ambient temperature; condensing the whole stream to distil it is absurd. What you do instead is bring the gas into contact with a liquid that dissolves the CO₂ and ignores the nitrogen — and the separation is done by a difference in solubility rather than a difference in volatility. That is a mass separating agent1 §1.6, and this chapter is how you size the equipment for it.
What is new, and what is not
The method is identical to the McCabe–Thiele construction4a §4.6: an equilibrium curve, an operating line from a mass balance, and a staircase between them. Two things change. The equilibrium relation is now Henry's law rather than relative volatility, and — because the solute physically leaves one phase and joins the other — the total flows are not constant, so the balance has to be written in different coordinates to stay straight.
6.1 Absorption and stripping
Absorption removes one or more components from a gas by contacting it with a liquid: the solute is absorbed and leaves with the liquid. Stripping is the same operation run backwards — a component is removed from a liquid by contacting it with a gas.
| Type | How the solute is held | Example |
|---|---|---|
| Physical absorption | The solute is simply more soluble in the solvent than the other components are. Nothing reacts; heating or dropping the pressure releases it again. | CO₂ into chilled methanol; H₂S into a physical solvent at high pressure |
| Chemical absorption | The solute reacts with the solvent and the product stays in the liquid. Much higher capacity at low partial pressure, and a much larger heat of absorption to pay back on regeneration. | CO₂ into aqueous amine — the reaction is why Quest can capture from a 12 % stream at all |
Both are everywhere: sweetening natural gas of H₂S and CO₂, scrubbing SO₂ and ammonia from stack gas, recovering solvent vapours, and the post-combustion capture plants of Chapter 11 §1.2. This course treats physical absorption quantitatively; chemical absorption uses the same equipment and the same diagram, with an equilibrium curve that bends sharply because of the reaction.
The regeneration loop
A mass separating agent has to be paid for twice. You buy the solvent, and then you have to get the solute back out of it — otherwise you are consuming solvent continuously and shipping the problem downstream. So an absorber is almost never alone:
Where the energy actually goes
The absorber itself uses almost no energy — it is a tower with a pump. Essentially the entire operating cost of a solvent process is in the regenerator: the reboiler that boils the solute back out, or the compression of the stripping gas. When Chapter 1 reported 972 kWh per tonne of CO₂ captured against a thermodynamic floor of 41, almost all of that gap1 App 1 was the reboiler on the right-hand column of Figure 6.1, not the absorber on the left.
6.2 Equilibrium — Henry's law
Consider a solute B distributed between a carrier gas C and a solvent A.
At low concentration the partial pressure of the solute above the liquid is proportional to its liquid mole fraction, which is Henry's law:
6.1
Counting degrees of freedom confirms what a single number can describe: , so beyond temperature and pressure, one composition fixes the state.
6.2
The size of is what decides whether a solvent is any use at all. Small means a soluble gas and an easy absorption:
Reading the ratio, not the number
What matters is not for the solute alone but the ratio of for the solute to for the carrier — the same role relative volatility2a §2.7 played in distillation. Water at 20 °C separates H₂S from air with a solubility ratio of about 140, which is why a simple water scrubber works. Separating N₂ from air by absorption, with a ratio near 1.2, would need a column nobody would build.
6.3 Why the coordinates change
Here is the one genuinely new idea in the chapter, and it is worth slowing down for.
In distillation, constant molal overflow4a §4.4 let us treat and as constant, so the balance was a straight line. In an absorber that assumption is not merely inaccurate — it is wrong by construction. The whole point of the operation is that the solute leaves the gas and joins the liquid, so the total gas flow falls from the bottom of the column to the top, and the total liquid flow grows on the way down. Neither is constant, and a balance written with them is not a straight line.
The fix is to count what does stay constant. The carrier gas is insoluble and the solvent is non-volatile, so:
6.3
are both constant through the column. Measure the solute against those constant streams instead of against the total:
6.4
X and Y can be greater than 1
They are ratios, not fractions. A liquid holding two moles of solute per mole of solvent has , which is perfectly sensible and catches everybody out once. The conversions back are and , and the equilibrium curve must be converted point by point — Henry's law is straight in and curved in .
The pay-off is that and are now molar flows of solute — moles of solute per hour, full stop — so a component balance in these variables is linear.
Mole fractions or ratios?
The same absorber, drawn both ways. On the left, in mole fractions, the balance is a curve because the total flows change down the column. On the right, in solute-free ratios, the same balance is a straight line. Drive the inlet concentration up and watch how far apart the two pictures get — and how little difference it makes when the gas is dilute.Mole fractions
The balance is not a straight line here — the total flows change.
Solute-free ratios
Same column, same data — now the balance is exactly straight.
YN+1
—
Y1
—
XN leaving
—
Gas shrinks by
—%
Max error if you use x–y
—%
✓—
6.4 The operating line
The column and the envelope look exactly like a distillation section — because they are one.
A solute balance around that envelope gives
6.5
and rearranged,
6.6
Read what this is. It relates the gas entering stage from below to the liquid leaving stage — two passing streams, not in equilibrium. It is the operating line of Chapter 4a4a §4.2, with in place of and mole ratios in place of mole fractions. There is no feed line and no second section, because there is no feed in the middle and no reboiler: an absorber is one section, one operating line.
The standing assumptions
Four, and they buy a great deal: the heat of absorption is negligible and the column is isothermal — between them these make the energy balance unnecessary; the solvent is non-volatile and the carrier gas is insoluble — between them these make and constant. Chemical absorption breaks the first two badly, which is why a real amine absorber has a temperature bulge partway up.
6.5 The construction
With both curves on axes, the staircase is the one you already know. For an absorption the operating line lies above the equilibrium curve, because the gas must always be richer than equilibrium with the liquid it is meeting for the solute to keep transferring into the liquid.
Absorber & stripper designer
One construction, two operations. In absorb mode the operating line sits above the equilibrium curve and you step up from the solvent inlet; in strip mode it sits below and you step down. The minimum solvent — or minimum gas — is marked, and the app refuses specifications the pinch makes impossible, with the reason.The construction, in mole ratios
Operating line straight; equilibrium curved by the conversion.
Down the column
Solute in each phase, stage by stage.
Stages N
—
(L/G)min
—
L/G used
—
Rich stream out
—
Recovery of solute
—%
A = L/(mG), dilute basis
—
A at the rich end, ratio basis
—
✓—
6.6 Stripping — the mirror image
Nothing in the algebra changes. Equation 6.6 is the operating line for a stripper too; what changes is which side of the equilibrium curve it falls on, and which end you know.
One rule covers both
The operating line is on the side the transfer is going. Absorption moves solute gas → liquid, so the gas must be above equilibrium: operating line above. Stripping moves solute liquid → gas, so the liquid must be above equilibrium: operating line below. Everything else — the balance, the stepping, the pinch — is the same, which is why App 2 uses one piece of code for both.
6.7 Minimum solvent, minimum gas
The limiting case is the familiar one: rotate the operating line until it touches the equilibrium curve, and the stages go to infinity.
For absorption, the slope of the operating line is , so less solvent means a shallower line pivoting about the fixed top point — it falls towards the equilibrium curve as the solvent rate drops. At some slope it touches the equilibrium curve — usually at the bottom of the column, where the gas is richest — and that is . With the pinch at the bottom,
6.7
For stripping, it is the gas that is limiting, and the same argument run the other way gives a maximum — equivalently a minimum gas rate , pinched at the top where the liquid is richest.
Practice sits at 1.2 to 1.5 times the minimum. The trade-off is the same one App 4 of Chapter 4a4a App 4 drew: more solvent means fewer stages but a bigger regenerator, and the regenerator is where the energy is. Since the reboiler duty scales roughly with solvent circulation, using more solvent than you need is paid for twice a day, every day, for thirty years.
The pinch is not always at the end
Equation 6.7 assumes the line first touches equilibrium at the bottom end. It touches partway up instead when the equilibrium curve is concave in the ratio coordinates you are actually stepping in — bending over as rises — because a line pivoting about then runs tangent to it before reaching . A convex curve cannot do this: the chord slope from the pivot only increases, so the pinch stays at the end. Note which way round that puts : in , is convex in mole fractions and pushes the pinch back to the end, while pulls the tangency inside. And the conversion to ratios is concave-making all by itself, so curvature in the equilibrium data is not even required — with a perfectly straight , App 2 pinches at the bottom for an inlet of 15 % but at an interior for an inlet of 20 %. Whenever the tangency is interior, equation 6.7 returns a minimum that is too small and a column built on it will not make specification. App 2 searches for the true tangency rather than assuming the end point, and tells you which case you are in.
6.8 Kremser — when the lines are straight
Dilute systems are the common case in absorption: a few percent of solute at most, often parts per million. Then and , Henry's law is linear, and both curves on the diagram are straight lines. When that happens the staircase between two straight lines is a geometric series, and it can be summed exactly — no drawing needed. The result is the Kremser equation:
6.8
where is the absorption factor — the ratio of the operating-line slope to the equilibrium-line slope. It is the single most useful number in the chapter:
- : the operating line is steeper than equilibrium, the lines diverge going down the column, and any recovery can be reached with enough stages.
- : the lines are parallel. Equation 6.8 is indeterminate and reduces to — the staircase makes equal steps forever.
- : the lines converge and then meet. There is a maximum possible recovery no column can beat, however tall.
For stripping the same algebra with the stripping factor gives the mirror-image result.
Why A ≈ 1.4
Below the separation is impossible; just above it the stage count is enormous; well above it you are pumping solvent you did not need and paying to regenerate it. The economic optimum for physical absorption has sat near since Colburn pointed it out in 1939, and it has not moved much. App 3 shows the curve that puts it there.
The absorption factor
Kremser in one picture. Sweep and watch the stage count fall away from the wall at — and watch the maximum achievable recovery collapse when drops below it. The right-hand panel checks Kremser against a stage-by-stage construction on the same numbers.Stages against the absorption factor
The wall at A = 1 is the whole story.
Kremser against the staircase
Same specification, stepped exactly in mole ratios.
N from Kremser
—
A
—
L/G needed
—
Outlet gas y1
—
Max removal at this A
—%
N stepped exactly
—
✓—
6.9 Worked examples
Worked example 6.1A vent scrubber · minimum water
A vent gas is 15.0 wt % Z in air, and the local pollution authority requires it to leave at no more than 4.0 wt %. You will scrub it with pure water at 30 °C, isothermally. The laboratory reports equilibrium as in weight fractions. Find , and at find the number of equilibrium stages and the outlet water concentration.
Work it yourself first, then open
Ratios first. Weight fractions work exactly like mole fractions here — the ratios are just mass ratios instead of mole ratios, and every equation is unchanged.
YN+1 = 0.15/0.85 = 0.17647 · Y1 = 0.04/0.96 = 0.041667 · X0 = 0
The pinch. With a straight equilibrium in weight fractions the pinch is at the bottom, where the gas is richest. In equilibrium with y = 0.15, x* = 0.15/0.5 = 0.30, so
X* = 0.30/0.70 = 0.42857
(L/G)min = (0.17647 − 0.041667)/(0.42857 − 0) = 0.13481/0.42857 = 0.3145
At 1.22 × minimum, L/G = 0.3837, and the outlet liquid follows from the overall balance:
XN = X0 + (YN+1 − Y1)/(L/G) = 0.13481/0.3837 = 0.3513 → xN = 0.3513/1.3513 = 0.260
so the water leaves at 26.0 wt % Z. Stepping between the converted equilibrium curve and the operating line gives N = 4.70 stages — five in practice.
Sanity checks. On the ratio basis the equilibrium slope is not m. For y = 0.5x, Y = 0.5X/(1 + 0.5X), so
dY*/dX = 0.5/(1 + 0.5X)² → 0.500 at the lean top end, 0.362 at XN = 0.3513, 0.339 at X* = 0.4286
With L/G = 0.3837 the local absorption factor therefore runs from A = 0.77 at the top to A = 1.06 at the bottom — increasing down the column, because the ratio-basis equilibrium line flattens as it climbs. The pinch is nevertheless at the bottom, and not because of A: the ratio curve is concave everywhere, and a line pivoting about (0, 0.041667) would run tangent to it at X = 0.513, which is past X* = 0.4286 and so outside the column. The steepest chord you can actually draw is the one to the bottom end, which is exactly what equation 6.7 assumes. And 4.7 stages for a 76 % removal of the solute at 1.22 × minimum is a perfectly ordinary scrubber.
Set App 2 to absorb, m = 0.5, inlet 0.15, outlet 0.04, lean 0, multiple 1.22. It reports (L/G)min = 0.3145, L/G = 0.3837, N = 4.70, an end pinch, and the two absorption factors above: 0.767 on the dilute basis (which is just L/G divided by m, the ratio-basis slope at the lean end) and 1.061 measured against the ratio-basis slope at the rich end. Then push the inlet to 0.20 and watch the pinch leave the end of the column.
Worked example 6.2Kremser · a dilute recovery
Solute Q is to be absorbed from air into a pure solvent. The entering gas is 0.0015 mole fraction Q and must leave at 0.00003; the outlet liquid is 0.00065 mole fraction Q. Equilibrium is . How many stages?
Work it yourself first, then open
The solvent rate comes from the balance, not from a specification:
L/G = (yN+1 − y1)/(xN − x0) = (0.0015 − 0.00003)/(0.00065 − 0) = 2.2615
A = L/(mG) = 2.2615/2.21 = 1.0233
Kremser. With x0 = 0 the group in the logarithm simplifies to (yN+1/y1) = 50:
N = ln[50(1 − 1/1.0233) + 1/1.0233]/ln 1.0233 = ln(1.1385 + 0.9772)/0.023032
N = 0.74933/0.023032 = 32.5 stages
Thirty-two stages for a dilute absorption? Yes — and the reason is A = 1.02. The operating line is barely steeper than the equilibrium line, so the two nearly run parallel and each stage achieves almost nothing. Raise the solvent rate by 40 % — L/G = 3.166, A = 1.4(1.0233) = 1.4326 — and the same duty needs N = ln[50(1 − 1/1.4326) + 1/1.4326]/ln 1.4326 = 7.68 stages (7.64 by an exact staircase in ratios). Thirty-two down to eight, for 40 % more solvent. That is the entire argument for the A ≈ 1.4 rule of thumb, in one comparison.
Check: stepping the staircase exactly in mole ratios gives 32.8 stages against Kremser's 32.5 — a 0.7 % difference, which is the price of the dilute approximation at these concentrations. At 15 % solute it would be much larger.
App 3, with yN+1 = 0.0015, removal 98 %, m = 2.21. Drag A and watch the wall.
Worked example 6.3Stripping with a curved equilibrium
A stripping column uses air to regenerate a liquid containing 1.0 mol % of a solute, removing 98 % of it. The air is solute-free. Equilibrium is with and in mole fractions. Find the minimum air rate, and the number of plates at 1.5 times that.
Work it yourself first, then open
Ends of the column. , , and the gas enters clean, .
X0 = 0.010101 · XN = 0.00020004
Minimum gas. For stripping the pinch is at the top, where the liquid is richest. In equilibrium with x = 0.01,
y* = 0.1(0.01) + 0.01(0.01)² = 0.001001 → Y* = 0.0010020
(L/G)max = (Y* − YN+1)/(X0 − XN) = 0.0010020/0.0099010 = 0.10120
so the minimum air rate is (G/L)min = 9.881 moles of air per mole of solvent — the stripping gas outnumbers the liquid ten to one, which is normal when the solute would rather stay in the liquid.
At 1.5 × the minimum air, L/G = 0.10120/1.5 = 0.067468 and
Y1 = 0 + 0.067468(0.0099010) = 0.000668
Stepping down from the top — horizontal to the equilibrium curve, vertical to the operating line, repeat — gives N = 7.28 stages, so eight plates.
Check: the system is dilute enough that Kremser applies. With , — 1 % below the exact staircase, the whole discrepancy being the term.
Why the curvature matters. The term is tiny at these concentrations — it changes y* by 0.1 % — so this problem behaves almost linearly and Kremser with m = 0.1 would give a close answer. Make the feed 10 % solute instead and the quadratic term contributes 10 % of y*, the equilibrium curve bends visibly, and the pinch can leave the end of the column entirely. Try it in App 2.
6.10 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 the operating line curved in mole fractions but straight in mole ratios? Both describe the same column.
Think first, then open
Because a mole fraction is measured against a denominator that changes. As solute leaves the gas, the total gas flow falls, so the same number of moles of solute is a different mole fraction depending on where you are in the column — the balance picks up a varying coefficient and stops being linear. A mole ratio is measured against the carrier, which by assumption does not change at all, so the balance keeps constant coefficients and stays straight. This is the same trick as choosing a basis in a mass balance: pick the thing that is conserved.
Think 2
An absorber is running at and failing to meet its specification. Your colleague proposes adding trays. Will it work?
Think first, then open
No, and this is the most important practical consequence of the absorption factor. With the operating line is shallower than the equilibrium line, so the two converge going down the column and meet at a pinch. Beyond that pinch no stage does anything, and the recovery is capped at a value below the specification no matter how many trays you add. The fixes are all about : raise the solvent rate, lower the temperature or raise the pressure to reduce , or change to a solvent with a smaller . App 3 shows the maximum removal collapsing as falls below 1.
Think 3
Absorbers are run cold and strippers hot, on the same solvent, in the same loop. Why does that work — and what does it cost?
Think first, then open
Because rises with temperature (equation 6.2 and Figure 6.3): the solute is more soluble cold and less soluble hot. Absorbing cold gives a small , hence a large , hence few stages; stripping hot gives a large , hence a large stripping factor, hence easy regeneration. The loop exploits the same temperature dependence twice, in opposite directions. What it costs is the heating and cooling of the entire circulating solvent between the two columns — which is why plants spend so much effort on cross-exchangers between the rich and lean streams, and why solvent circulation rate, not absorber size, dominates the operating cost.
Think 4
Chemical absorption gives far higher capacity at low partial pressure than physical absorption. Why is it not always the answer?
Think first, then open
Because the same bond that holds the solute has to be broken again. A reaction that is strongly favourable at absorber conditions is strongly unfavourable to reverse, so the regenerator must supply the heat of reaction on top of the sensible and latent heat — which is precisely why amine capture costs around 3.5 GJ per tonne of CO₂ while the thermodynamic minimum is nearer 0.15. Chemical absorption is the right answer when the solute is dilute and must be removed almost completely; physical absorption wins when the partial pressure is high enough to do the job without chemistry, because then regeneration is just a pressure let-down.
Where this chapter connects
- Back to: the construction is McCabe–Thiele4a §4.6 with different axes; the pinch is the minimum-reflux pinch4a §4.8 under another name.
- The agent: a solvent is a mass separating agent1 §1.6, and §6.1 is where the debt it creates gets paid — the regenerator.
- The number that decides everything: does for absorption what relative volatility2a §2.7 does for distillation.
- The same problem, five chapters on: the flue-gas capture of Chapter 1's minimum-work app1 App 1 is an absorber and a stripper, and its energy penalty lives almost entirely in the second column.
- Ahead: Chapter 77 §7.2 changes the second phase from a gas to another liquid — extraction. Its immiscible half is this chapter with new labels; its miscible half needs a whole extra dimension7 §7.3.
Summary & key equations
Setting up
Absorptionsolute moves gas → liquid; operating line above equilibrium
Strippingsolute moves liquid → gas; operating line below equilibrium
Assumptionsnegligible heat of absorption · isothermal · non-volatile solvent · insoluble carrier
Henry's law, valid dilute; , so rises with
Solute-free coordinates
Constant flows,
Mole ratios, — may exceed 1
Back again,
Whythe balance is linear in and non-linear in
Design
Operating line
Minimum solvent — check for an interior tangent
Practice– times the minimum
Dilute systems
Absorption factor; stripping factor
Kremser
A < 1the lines converge — there is a maximum recovery no number of stages can beat
A = 1parallel lines;
Economic optimum