Engineering & Technology · Reaction Engineering, Separations & Transport

Distillation, absorption, extraction, membranes and adsorption

A separator can produce a concentrated product while losing much of the desired component, or recover most of that component into an impure stream. The useful process must meet the needed combination of composition, amount, energy and material destinations.

Use Process Balances & Design for amount, flow, energy and boundary accounts. Review material structure and properties. Chemistry’s reaction rates and equilibrium explain the molecular and thermodynamic ideas further. Exponentials are evaluated explicitly; each model states its scope before calculation.

Back to the chapter

After reading, you can

Distinguish the mechanisms of common separations and track purity, recovery and the destination of rejected material.

Choose a mechanism that matches the mixture

Distillation uses different volatility and repeated contact between vapour and liquid, with heat addition and removal. Absorption transfers selected gas components into a liquid; stripping transfers components out of liquid into a gas. Liquid–liquid extraction distributes a solute between suitable liquid phases. The added solvent must later be recovered or otherwise accounted for.

Adsorption accumulates species on an available solid surface, including surfaces inside pores. It differs from absorption into the body of a liquid. Finite capacity and transport can make outlet concentration rise during use, called breakthrough. Regeneration or replacement changes the whole material and energy account; disappearance from the feed is not destruction.

A membrane produces at least two accounts

A membrane separates through differences in passage under an applicable pressure, concentration, electrical or other driving force. Passage can depend on pores, partitioning and diffusion rather than a simple size sieve. The permeate passes through and the retentate remains on the feed side. Rejection of one component does not mean it cannot pass at all under every condition.

Consider an original steady mass balance with feed 100 kg/h:10 kg/h A and 90 kg/h B. Suppose supplied permeate data are 20 kg/h containing 8 kg/h A and 12 kg/h B. With no reaction, accumulation or other outlet, retentate is 80 kg/h containing 2 kg/h A and 78 kg/h B. These are imaginary measured stream specifications for a balance exercise, not membrane selectivity data.

Feed 100 kg/h contains A 10 and B 90. Permeate 20 contains A 8 and B 12; retentate 80 contains A 2 and B 78, in kg/h throughout. All bar components use 4 drawing units per kg/h. Permeate purity is 40 mass% and A recovery into permeate is 80%. In every bar the left segment is A (orange) and the right segment is B (blue).

Purity and recovery ask different questions

Permeate A purity is the A mass flow divided by total permeate mass flow:8/20=40%. A recovery into permeate is its A flow divided by feed A flow:8/10=80%. Retentate A fraction is 2/80=2.5%. The plotted component widths preserve all feed material instead of showing only the attractive product stream.

Actual process selection also needs flux or stage capacity, operating conditions, energy, solvent or adsorbent use, fouling, product sensitivity and final handling of each stream. A high theoretical separation factor can coexist with a very small transfer rate. A high initial flux can decline as deposits develop. Compare service over time and relevant configurations rather than ranking techniques by one favourable number.

Use a mechanism to predict a bounded outcome

Compare two original ways to remove at least 1.5 mol of a fictitious nonreactive solute A from 1.00 m³ of liquid initially containing 2.00 mol. This is a separate service from the mass-flow example above. Treat liquid volumes as unchanged and the two extraction liquids as immiscible, with no reaction or other loss. The target is a supplied amount-account objective, not a water-quality or product criterion.

Option E makes one equilibrium contact with 0.50 m³ of fresh receiving liquid. Supply its distribution relation C_R = 4 C_F, where C_R and C_F are final molar concentrations in receiving and original feed liquids. The solute balance is 2 = 1 C_F + 0.5 C_R = 3 C_F. Hence C_F = 0.6667 and C_R = 2.6667 mol/m³. The receiving liquid captures 0.5×2.6667 = 1.3333 mol, or two thirds of the input. Even this assumed equilibrium result misses the 1.5 mol target; better mixing alone cannot exceed this one-contact equilibrium amount under these assumptions.

Option S contacts the same feed with 2.00 kg of a solid. Supply the toy equilibrium loading q = min(κ C_F, qmax), where q is adsorbed amount per solid mass, κ = 4 m³/kg and qmax = 0.80 mol/kg. The rule is initially linear in the remaining liquid concentration but cannot exceed the stipulated loading cap. Assume sufficient contact to attain it. If the cap is active, captured amount is 2×0.80 = 1.60 mol, leaving C_F = 0.40 mol/m³. Check that branch: κ C_F = 1.60 mol/kg exceeds qmax, so the capped value is consistent. It meets the target under these assumptions. Merely multiplying solid mass by capacity would not be enough without checking the remaining equilibrium state; finite contact could also prevent attaining it.

The limits differ. E depends on a distribution relation and the amount of receiving phase; S depends on a concentration-dependent loading and its cap. If the solid is not saturated, the balance is 2 mol = (1.00 m³)C_F + Mκ C_F, so the remaining concentration still controls how much can be captured. Neither method destroys A. E leaves a loaded liquid; S leaves a loaded solid to regenerate or manage. Compare useful output, residual handling and relevant energy/material service rather than declaring a universal winner. The earlier volatility and diffusion examples explain why equilibrium composition and rate capability remain separate questions.

Original comparison under supplied equilibrium relations. Concentrations, distribution and capped-linear loading are fictional model inputs, not material or equipment data.
Option	Transferred A	A left in feed liquid	Meets 1.5 mol capture?
E: receiving volume 0.50 m³	1.3333 mol	0.6667 mol	No, even at supplied equilibrium
S: solid mass 2.00 kg	1.6000 mol	0.4000 mol	Yes, if the stipulated equilibrium state is attained

All quantities invented. Liquid capture is (2 mol) × 4V/(1 m³ + 4V), with receiving volume V, fixed 1 m³ feed and ratio 4. At V = 0.5 m³ capture 1.333 mol fails the 1.5 mol target; at V = 1 m³ capture 1.6 mol passes. Solid loading q = min(κ C_F, qmax), with κ = 4 m³/kg and qmax = 0.8 or 0.6 mol/kg. Combining this rule with the 2 mol solute balance gives capture as the smaller of (2 mol)κ M/(1 m³ + κ M) and qmax M. At M = 2 kg capture is 1.6 or 1.2 mol. The capped-linear rule is a constructed equilibrium approximation, not real sorption data or a rate guarantee. In the solid panel the orange dashed curve uses qmax = 0.8 mol/kg and the violet dotted curve uses qmax = 0.6 mol/kg.

Expose a recovery–purity trade-off

For a separate operating case on the same 100 kg/h feed, supply permeate 30 kg/h containing 9 kg/h A and 21 kg/h B. Its A recovery becomes 90%, while purity falls to 30%. Retentate contains 1 kg/h A and 69 kg/h B, total 70 kg/h. Both components and the overall mass balance still close.

The two cases demonstrate that recovering more A can require accepting more B in the chosen stream. No physical claim is made that changing a particular valve produces those values. A device model and relevant evidence would be needed to connect controllable inputs to the supplied outcomes. Subsequent purification could add energy, losses and another residual stream.

For a variable fresh receiving volume V_R, keep units in the solute balance: 2 mol = (1.00 m³)C_F + V_R(4C_F). Thus C_F = (2 mol)/(1.00 m³+4V_R). To capture 1.5 mol, only 0.5 mol may remain in the 1.00 m³ feed phase, so C_F = 0.5 mol/m³ and C_R = 2 mol/m³. The required receiving volume is then (1.5 mol)/(2 mol/m³) = 0.75 m³. This is the amount threshold in the supplied model, not a contact-time prediction or solvent-selection instruction.

Try the reasoning

Is adsorption the same as gas absorption into a liquid?
Show the answer and reasoning
No. Adsorption concerns accumulation at surfaces; absorption involves uptake into a bulk phase.
Does 80% A recovery mean that the permeate is 80% A?
Show the answer and reasoning
No. Here recovery is 8/10, while purity is 8/20=40%.
Where must the A that does not reach permeate appear in this model?
Show the answer and reasoning
In retentate, because the stated steady boundary has no reaction, accumulation or other loss route.
The receiving volume is increased to 1.00 m³, while S’s loading cap falls to 0.60 mol/kg at the same 2.00 kg mass and κ. Keep all other equilibrium assumptions. Which option now meets the capture target, and what consequence remains?
Show the answer and reasoning
E gives C_F = (2 mol)/(1 m³+4×1 m³) = 0.40 mol/m³ and C_R = 1.60 mol/m³, so the receiving phase captures 1.60 mol and meets the target. For S, a capped loading gives 2×0.60 = 1.20 mol captured and C_F = 0.80 mol/m³ remaining. Check κ C_F = 3.20 mol/kg, above the 0.60 cap, so that branch is consistent and S fails. The conditional outcome reverses. E uses twice the original receiving-liquid volume and leaves a loaded liquid for recovery or management. Equilibrium attainability, rate, regeneration, energy and costs still need evidence; this amount comparison is not a universal ranking.

What to carry forward

Compare the separation mechanism and complete duty, and state the numerator and denominator of every purity or recovery claim.

References

Teaching resources and research supporting this lesson’s explanations and examples.