Membranes

A wall that chooses

Every other chapter of this course creates a second phase and lets the components partition into it. A membrane does something different: it puts a barrier between two streams and lets some molecules through faster than others. There is no vapour to condense, no solvent to recover, no adsorbent to regenerate — and, usually, no phase change at all. That is the attraction, and it is also the reason membrane processes are the fastest-growing separations in industry.

9.1 The family of processes

The pressure-driven processes form a continuous family, distinguished by what size of thing they stop.

Gas separation, pervaporation and dialysis sit alongside these. Reverse osmosis alone supplies drinking water to well over a hundred million people, and gas-separation membranes now take CO₂ out of natural gas, nitrogen out of air and hydrogen out of refinery purge gas at very large scale.

9.2 Membrane structure

The point of all three is the same. Flux goes as , so the whole art is to make the selective layer as thin as it can be made without defects — and then to support it well enough that it survives 60 bar and a decade of service.

9.3 Flux, permeability and permeance

At steady state, the flux of a component through the membrane is proportional to its partial-pressure difference across it:

with and the total pressures on the retentate and permeate sides. The group — a diffusivity times a solubility — is the permeability, a property of the material alone. Divided by the thickness it becomes the permeance, a property of the finished membrane.

9.4 Selectivity and the trade-off

The ideal selectivity is the ratio of the two permeabilities,

and it is the direct analogue of relative volatility2a §2.7. Splitting it into the two factors is worth doing, because they can be engineered separately: a glassy polymer with tight, rigid chains gives a large diffusivity ratio and separates by size, while a rubbery polymer gives a large solubility ratio and can separate a larger, more condensable molecule preferentially — which is how a membrane can be made to permeate propylene ahead of hydrogen.

9.5 Designing one module

Take a module well enough mixed that the retentate leaves at the same composition as the gas everywhere inside it. Two things then fix the answer: a mass balance and the ratio of the two fluxes.

Define the stage cut — the fraction of the feed that permeates — and write the balance on the fast component:

The permeate composition is whatever ratio the two fluxes happen to be in:

Equations 9.3 and 9.4 are two equations in the two unknowns and . Equation 9.4 is a quadratic in , so the pair can be solved directly — and the two of them together are the whole design.

9.6 The pressure-ratio limit

Look again at Equation 9.1. The flux of the fast component cannot be positive unless

and nothing in that statement mentions the membrane. It is a thermodynamic ceiling set entirely by the pressures. A perfect membrane with infinite selectivity, fed a gas containing 10 % of the fast component through a pressure ratio of 5, cannot make a permeate richer than 50 %.

9.7 From a stage cut to an area

Everything in §9.5 was dimensionless. You now know the permeate composition, the retentate composition and the recovery — and not one number that a vendor could quote you a price against. Nothing so far says how big the thing is.

The step that fixes that is short. Every mole in the permeate got there by crossing the membrane, so the permeate flow and the flux must agree:

with both fluxes evaluated from Equation 9.1 at the compositions the module actually has — on the retentate side and on the permeate side. That last clause is the whole difficulty, and it is why the composition problem has to be solved first.

Getting the units right

Permeance is quoted in GPU and pressures in cmHg, so it pays to collapse the conversion once and for all. One GPU is cm³(STP)/(cm²·s·cmHg); multiply by 3600 s/h and by 10⁻² to turn cm³/cm² into m³/m², and

and with the feed rate also in m³(STP)/h, Equation 9.6 returns square metres directly. Because everything is at standard temperature and pressure, volumes here are just moles in disguise, so the balances of §9.5 carry over unchanged.

The area does not scale with the cut

Here is the base case for the rest of the chapter: 10 000 m³(STP)/h of natural gas containing 10 mol % CO₂ at 50 bar, permeate at 2 bar, on a membrane with a CO₂ permeance of 1000 GPU and a selectivity of 20 (so the methane permeance is 50 GPU). Run the ladder at a series of stage cuts:

Read the last four columns together. Going from θ = 0.10 to θ = 0.20 doubles the permeate flow but multiplies the area by 2.4, and the column headed A/θ shows why: the specific area climbs steadily because the retentate keeps getting leaner and the CO₂ driving force collapses with it — 152 cmHg down to 93. The last mole of CO₂ costs far more membrane than the first. That is the same diminishing return you met as an absorber approached its pinch and as a column approached minimum reflux; here it appears as area rather than as stages.

Notice also that barely moves — 6.11 up to 6.41 — because the methane driving force is dominated by the 3750 cmHg of feed-side partial pressure and hardly notices what the permeate is doing. Almost all of the flux variation in a gas-separation membrane is in the fast component.

Designing to a spec, not to a stage cut

No customer ever asks for a stage cut. They ask for a pipeline gas below 2 % CO₂, or a permeate above 40 % CO₂, or 80 % of the CO₂ captured. Each of those is a single equation in the single unknown θ, because every quantity in the table is a monotonic function of the cut:

Three specs, three completely different plants — and the third one is a disaster. Meeting a pipeline specification with a single well-mixed stage throws away two-fifths of the product gas, because the only way to strip the retentate that far is to permeate almost half the feed, and at α = 20 nearly four-fifths of what permeates is methane. That single line is the reason §9.9 exists.

9.8 Cross-flow: the module is not a stirred tank

The well-mixed model made a strong claim: every square metre of membrane sees retentate at , the exit composition. In the base case at θ = 0.20 that means the whole module works against 3.86 % CO₂ when the gas entering it is at 10 %. That is plainly pessimistic — the first stretch of membrane sees the feed at full strength.

A real spiral-wound or hollow-fibre module is much closer to cross-flow: the retentate travels along the module in something like plug flow, its composition falling as it goes, while the permeate is swept off the far face and never mixes back. Each element of area then works against the local composition.

The bookkeeping is a differential version of Equation 9.3. Let be the retentate flow remaining and its composition. In a slice that permeates , the material removed has the local permeate composition , so , and

where , and are all evaluated at the local using the same Equations 9.1 and 9.4 as before. Integrate from until . The permeate you actually collect is the mixed average of everything produced along the way, which the overall balance hands you for free:

Two integrations and you are done. The comparison, on the base case:

Cross-flow wins on every column, and the gap grows with the cut. At θ = 0.30 the well-mixed model predicts a residue of 2.80 % CO₂ where cross-flow gives 0.54 % — a factor of five, and comfortably the difference between meeting a pipeline spec and not meeting it.

9.9 Two stages and a recycle

Figure 9.5 and the spec table in §9.7 both end at the same wall: one membrane stage is one contact, and one contact cannot give a high purity and a high recovery. A distillation column escapes that wall by doing the separation many times over and sending product back — and a membrane plant escapes it in exactly the same way, by putting a second module in and recycling one of its streams.

Why it needs iteration

The recycle is what makes this more than two module calculations in a row. Stage 1 no longer sees the fresh feed; it sees the fresh feed plus the recycle, at a composition you cannot know until you have solved stage 2 — which needs stage 1's answer. The standard remedy is the one every recycle problem in the course uses: guess, go round, repeat.

1. Guess the recycle flow and its composition — zero and will do.
2. Mix: and .
3. Solve stage 1 at its cut, then stage 2 on whichever stream it treats.
4. Read the new recycle off stage 2 and go back to step 2. Stop when it stops moving.

On the base case this converges in about twenty passes and is entirely undramatic — the recycle is a modest fraction of the feed, so the loop is well damped. When you check your answer, do it on the overall balance, , which knows nothing about the recycle and is therefore an independent test.

(a) The permeate cascade — buying purity

Take the base case at θ₁ = 0.20 and θ₂ = 0.30. The recycle settles at 1628 m³/h of 21.1 % CO₂, which lifts the stage-1 feed from 10.0 % to 11.56 % and its permeate to 39.4 %. Stage 2 works on that 39.4 % gas and produces:

An 82 % CO₂ product at 57 % recovery, from 248 m² of membrane. The point is not that 82 % is high — it is that a single stage cannot reach it at all. Push the cut of a single stage to zero and the permeate tends to its ceiling of ; every value above that is off the map for one contact, no matter how much area you buy. The second stage does not merely improve the answer, it enlarges the set of achievable answers — which is precisely what reflux does for a column.

The bill is the compressor. Recompressing 2326 m³(STP)/h from 2 to 50 bar takes about 480 kW of ideal duty, and that single number usually decides whether a two-stage membrane plant is built or an amine unit is built instead.

(b) The retentate cascade — buying recovery

Now the other spec: the pipeline wants the gas below 2 % CO₂. §9.7 did it in one stage, at θ = 0.441 and 551 m² — and lost 39 % of the methane. Put stage 1 at θ₁ = 0.20 and let stage 2 finish the job at θ₂ = 0.130, recycling stage 2's permeate:

The cascade meets the same specification with a third less membrane (365 against 551 m²) and loses about two-fifths as much methane — 15.4 % against 39.2 %, a ratio of 0.39 — for one compressor of 238 kW. The methane saved is the difference, 3524 − 1387 = 2137 m³(STP)/h; the 1387 m³/h is what the cascade still throws away. At any plausible gas price those 2137 m³/h pay for the compressor many times over — which is why essentially every CO₂-removal membrane plant in the world has a recycle in it, and why a single-stage answer to a retentate spec should always make you suspicious.

9.10 Modules and contactors

Area is everything, so membranes are sold as modules that pack a great deal of it into a small volume. Spiral-wound modules roll flat sheets and spacers around a central permeate tube and reach roughly 1000 m² per cubic metre; hollow-fibre modules bundle tens of thousands of fibres a fraction of a millimetre across into a shell-and-tube arrangement and reach ten times that, at the price of being much harder to clean.

A membrane contactor is a different idea altogether. Here the membrane does no separating — it is a microporous wall whose only job is to hold a gas and a liquid in contact at a fixed, known interfacial area without them mixing. The separation is still absorption, exactly as in Chapter 66 §6.1; the membrane just replaces the packing and removes any possibility of flooding, entrainment or foaming. It is one of the more elegant hybrids in the subject.

9.11 Worked examples

9.12 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

Summary & key equations