Crystallisation and membrane separation basics

Solubility, supersaturation, metastable zone, nucleation and growth, crystalliser types and yield balances with hydrates; membrane processes from MF to RO, osmotic pressure, flux, rejection, polarisation, fouling and modules.

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Why it matters

Crystallisation produces most solid chemicals in pure, free-flowing form – sugar, salt, fertilisers, pharmaceuticals, ammonium sulphate. Membranes desalinate seawater, purify water, concentrate whey, separate hydrogen and nitrogen from gas mixtures and dehydrate solvents. Both appear in GATE as yield and flux calculations and as concept questions, and both are growing industrial alternatives to energy-intensive distillation and evaporation.

Key ideas

Crystallisation – equilibrium and driving force.

  • Solubility (usually kg solute/100 kg water, from tables or curves) is the equilibrium; most salts become more soluble with temperature, some (Na₂SO₄ above ~32 °C, Ca(OH)₂) less.
  • Supersaturation – concentration above solubility – is the driving force for both nucleation and growth. Expressed as ΔC = C − C*, ratio S = C/C*, or relative supersaturation (C − C*)/C*.
  • Metastable zone (Miers): between the solubility curve and the supersolubility curve, existing crystals grow but new nuclei hardly form. Good practice keeps the crystalliser in this zone, often with seeding, to get large, uniform crystals.
  • Nucleation: primary (homogeneous or on foreign particles, needs high supersaturation) and secondary (from existing crystals by contact with impellers and walls – dominant in industry).
  • Growth: solute diffuses to the crystal face and is integrated into the lattice. McCabe's ΔL law: geometrically similar crystals of the same material in the same solution grow by the same linear amount regardless of size.

How supersaturation is created. Cooling (solubility strongly temperature-dependent: KNO₃), evaporation of solvent (solubility nearly flat: NaCl), adiabatic vacuum cooling (flash evaporation does both), adding an anti-solvent, or reaction (precipitation).

Crystallisers. Tank and scraped-surface coolers; forced-circulation evaporative crystallisers; draft-tube baffle (DTB) units that remove fines; Oslo (Krystal) crystallisers that grow crystals in a fluidised suspension; vacuum crystallisers.

Yield. Found by material balances on solute and water, using the solubility at the final temperature and, if crystals are hydrates, the water of crystallisation they remove. Mother liquor leaving is saturated (ideal) at the final temperature.

Membrane separations – classification by driving force and size.

  • Pressure-driven, porous: microfiltration (0.1–10 µm, bacteria, suspended solids, ~0.1–2 bar), ultrafiltration (macromolecules, proteins, ~1–10 bar).
  • Pressure-driven, dense or nearly dense: nanofiltration (divalent ions, sugars) and reverse osmosis (all ions, 10–80 bar), working by solution–diffusion – size alone does not explain them.
  • Concentration-driven: dialysis (haemodialysis); electrical: electrodialysis with ion-exchange membranes.
  • Partial-pressure driven: gas permeation (H₂ recovery, N₂ from air, CO₂ removal) and pervaporation (dehydrating alcohols, breaking azeotropes).

Key concepts.

  • Osmotic pressure opposes RO: water flows only when the applied ΔP exceeds Δπ.
  • Rejection (retention) measures selectivity for solutes; gas membranes use the selectivity ratio of permeabilities.
  • Concentration polarisation: rejected solute accumulates at the membrane surface, raising local osmotic pressure and lowering flux; cross-flow velocity reduces it.
  • Fouling (deposits, biofilm, scaling) reduces flux over time; controlled by pretreatment, cross-flow and cleaning.
  • Modules: spiral-wound (RO, NF), hollow-fibre (gas, RO, dialysis), tubular and plate-and-frame (dirty feeds, food).

Formulas

Crystallisation balances (anhydrous crystals, no evaporation): Crystals = solute in feed − (solubility at final T × water remaining/100) With hydrate crystals of mass C (solute mass fraction f = M_anhydrous/M_hydrate): Solute: S_F = f·C + S_ML, Water: W_F − E = (1 − f)·C + W_ML, S_ML/W_ML = solubility (kg/kg water)

  • S – solute mass, W – water mass, E – water evaporated, ML – mother liquor; all kg.

Growth rate (diffusion-limited): G = dL/dt, ΔL law: ΔL same for all sizes.

Membranes: Water flux (RO/NF, solution–diffusion): J_w = A·(ΔP − Δπ) Van't Hoff osmotic pressure (dilute): π = i·C·R·T Solute flux: J_s = B·(C_f − C_p) Observed rejection: R = 1 − C_p/C_f Porous membrane (MF/UF) flux, resistance form: J = ΔP / (μ·(R_m + R_f)) Gas permeation: J_i = (P_i/l)·(p_i,feed − p_i,perm); ideal selectivity α_AB = P_A/P_B

  • J_w – m³/(m²·s); A – water permeability, m/(s·Pa); ΔP – applied pressure difference, Pa; π – osmotic pressure, Pa; i – van't Hoff factor (2 for NaCl); C – mol/m³; R = 8.314 J/(mol·K); μ – Pa·s; R_m, R_f – membrane and fouling resistances, m⁻¹; P_i/l – permeance.

Worked examples

Example 1 – crystal yield with a hydrate (standard). Given: 1000 kg of a 30 wt % Na₂CO₃ solution is cooled to 20 °C, where solubility is 21.5 kg Na₂CO₃/100 kg water. Crystals are Na₂CO₃·10H₂O (M = 286; Na₂CO₃ = 106). No evaporation.

  1. Feed: 300 kg Na₂CO₃, 700 kg water. f = 106/286 = 0.3706.
  2. Mother liquor: Na₂CO₃ = 300 − 0.3706C; water = 700 − 0.6294C.
  3. Saturation: (300 − 0.3706C)/(700 − 0.6294C) = 0.215.
  4. 300 − 0.3706C = 150.5 − 0.1353C → 149.5 = 0.2353C → C = 635 kg.
  5. Check: ML = 64.5 kg Na₂CO₃ in 300.1 kg water → 0.215 ✓.

About 635 kg of decahydrate crystals; 78.5 % of the Na₂CO₃ is recovered (300 − 64.5 = 235.5 kg in crystals). Hydrate formation removes water and boosts yield.

Example 2 – reverse-osmosis flux and area (GATE level). Given: seawater with 35 g/L NaCl (M = 58.44) at 298 K; applied pressure difference 6.0 MPa; water permeability A = 3.0 × 10⁻¹² m/(s·Pa); permeate salt negligible; ignore concentration polarisation. Find flux and membrane area for 100 m³/h of permeate.

  1. C = 35/58.44 = 0.599 mol/L = 599 mol/m³.
  2. π = i·C·R·T = 2 × 599 × 8.314 × 298 = 2.97 × 10⁶ Pa (29.7 bar).
  3. J_w = A(ΔP − Δπ) = 3.0 × 10⁻¹² × (6.0 − 2.97) × 10⁶ = 9.10 × 10⁻⁶ m/s = 32.7 L/(m²·h).
  4. Area = (100/3600)/9.10 × 10⁻⁶ = 3050 m².

Flux ≈ 32.7 L/(m²·h); about 3050 m² of membrane. In practice the feed concentrates along the module, so Δπ rises and recovery is limited (around 40–50 % for seawater).

Common mistakes

  • Forgetting water of crystallisation when crystals are hydrates.
  • Using solubility per 100 kg solution instead of per 100 kg water.
  • Expecting a cooling crystalliser to work for a solute whose solubility hardly changes with temperature (use evaporation).
  • Writing RO flux as A·ΔP without subtracting osmotic pressure.
  • Forgetting i = 2 for NaCl in van't Hoff's equation.
  • Saying all membranes separate by size – RO, gas permeation and pervaporation work by solution–diffusion.

For GATE CH

Typical questions: crystal yield from solubility data (anhydrous or hydrate, with or without evaporation), supersaturation concepts, ΔL law, metastable zone; osmotic pressure, RO flux and area, rejection, MF/UF/NF/RO classification and gas-permeation selectivity. Practise setting up the two-equation hydrate balance quickly.

Quick check

  1. What drives both nucleation and crystal growth?
  2. State McCabe's ΔL law.
  3. A feed of 35 g/L gives permeate of 0.35 g/L. What is the rejection?
  4. Which membrane process uses an electrical potential as driving force?

Answers: 1. supersaturation; 2. geometrically similar crystals in the same solution grow by the same length increment regardless of size; 3. 1 − 0.35/35 = 0.99 (99 %); 4. electrodialysis.

Interview questions

All Mass Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is crystallization in the context of chemical engineering?Concept

    Crystallization is a separation and purification process where solid crystals are formed from a homogeneous solution. It is used to obtain pure substances from a solution by controlling the conditions such as temperature and concentration, allowing the solute to form solid crystals while leaving impurities in the solution.

  2. 2.Explain the basic principle of membrane separation.Concept

    Membrane separation is a process that uses a semi-permeable membrane to separate components in a mixture based on their size, charge, or chemical affinity. The membrane allows certain molecules or ions to pass through while retaining others, enabling the separation of substances such as gases, liquids, or dissolved solids.

  3. 3.Why is temperature control important in the crystallization process?Application

    Temperature control is crucial in crystallization because it affects the solubility of the solute in the solvent. By carefully controlling the temperature, the solubility can be manipulated to promote the formation of crystals. Rapid cooling can lead to small, impure crystals, while slow cooling allows for the growth of larger, purer crystals.

  4. 4.What are the advantages of using membrane separation over traditional separation methods?Application

    Membrane separation offers several advantages, including lower energy consumption, the ability to operate at ambient temperatures, and the potential for continuous operation. It is also highly selective, allowing for the separation of specific components without the need for additional chemicals, making it environmentally friendly.

  5. 5.What happens if the membrane in a separation process becomes fouled?Application

    If the membrane becomes fouled, its permeability decreases, leading to reduced efficiency and flow rates. Fouling can cause increased pressure drop across the membrane, higher energy consumption, and may require frequent cleaning or replacement of the membrane, increasing operational costs.

  6. 6.Explain how supersaturation is related to crystallization.Concept

    Supersaturation is a state where a solution contains more dissolved solute than it would under normal equilibrium conditions. It is a driving force for crystallization, as the excess solute tends to precipitate out of the solution to form crystals. Controlling supersaturation levels is key to achieving desired crystal size and purity.

  7. 7.Why is membrane selectivity important in separation processes?Application

    Membrane selectivity is important because it determines the membrane's ability to differentiate between different molecules or ions. High selectivity ensures that the desired component is efficiently separated from the mixture, improving the purity and yield of the product while minimizing losses and contamination.

  8. 8.The solubility of a solute is 50 g/L at 25 °C. What does it mean for 2 L of solution at 25 °C to be supersaturated, and how is such a state produced?Numerical

    At equilibrium 2 L can hold 50 × 2 = 100 g of solute; a supersaturated solution holds more than 100 g dissolved at 25 °C. It cannot usually be made by simply stirring in extra solid; it is produced by cooling a solution saturated at a higher temperature, evaporating solvent, adding an anti-solvent or by reaction. The excess over 100 g is the driving force for nucleation and growth, and within the metastable zone it can persist until seeds or disturbance trigger crystallisation.

  9. 9.A membrane has a permeability of 0.5 m³/m²·day·atm. Calculate the flow rate through a membrane area of 10 m² under a pressure difference of 2 atm.Numerical

    Flow rate (Q) can be calculated using the formula Q = permeability × membrane area × pressure difference. Here, Q = 0.5 m³/m²·day·atm × 10 m² × 2 atm = 10 m³/day.

  10. 10.What factors can influence the rate of crystallization in a solution?Application

    The rate of crystallization can be influenced by factors such as temperature, concentration of the solute, degree of supersaturation, presence of impurities, and agitation or mixing. Each of these factors can affect the nucleation and growth rates of crystals, impacting the overall crystallization process.

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