Leaching and solid-liquid extraction

Why it matters

Leaching (solid–liquid extraction) recovers sugar from beet and cane, edible oil from oilseeds, caffeine from coffee, active ingredients from plants, and copper, gold and uranium from ores. Washing a precipitate free of mother liquor is the same operation. Design questions are always the same: how much solvent, how many stages, and how much solute is lost with the solids that leave.

Key ideas

What happens in leaching. The solvent penetrates the solid, dissolves the soluble constituent (solute), and the solute diffuses out through the pores to the bulk liquid. The insoluble residue is the inert solid. Rates are usually limited by diffusion inside the particle, so:

• Particle size: smaller particles shorten diffusion paths and speed leaching, but very fine particles drain and filter badly. Oilseeds are flaked; sugar beet is sliced into cossettes.
• Solvent: selective, low viscosity, easily recovered (hexane for oil, water for sugar, dilute acid or cyanide for metals).
• Temperature: higher solubility and diffusivity, lower viscosity, but watch for dissolving unwanted material or degrading the product.
• Agitation: thins the external film and keeps fine solids suspended; it does not speed internal diffusion.

Ideal (equilibrium) stage in leaching. After mixing and settling, the solution retained by the solids (underflow) has the same composition as the clear solution leaving (overflow) – all solute dissolved and concentrations equalised. Leaching "equilibrium" is not a partition coefficient; the real non-ideality is how much solution the solids carry with them.

Underflow and overflow.

• Underflow = inert solids + the solution they retain (the solids cannot be drained dry).
• Overflow = clear solution, normally free of solids.
• Retention is expressed as kg solution per kg inert (or its inverse, N = kg inert/kg solution, on Ponchon–Savarit-type diagrams). It may be constant (constant underflow) or vary with concentration (then use the right-triangle or N-versus-y diagram with tie lines that are vertical for ideal stages).

Contacting schemes.

• Single-stage batch: simple, poor recovery.
• Cross-current: fresh solvent to every stage; good recovery but dilute, expensive-to-recover extract.
• Counter-current (continuous counter-current decantation, CCD): solids move one way, solution the other; best recovery with the most concentrated extract. With constant underflow, the stages after the first behave like a linear "washing" cascade.

Equipment. Percolation tanks for coarse solids; Bollman (bucket elevator), Rotocel and Kennedy extractors for oilseeds; screw/diffuser extractors for sugar beet; agitated tanks (Pachuca, Dorr) for fine ores; thickeners in series for CCD washing; heap and in-situ leaching for low-grade ores.

Stage efficiency. Real stages fall short of equilibrium if contact time is too short for internal diffusion; an overall stage efficiency or longer residence time is then used.

Formulas

Ideal-stage condition: concentration of solute in underflow solution = concentration in overflow, y_n = x_n (mass fraction of solute in solution).

Solute balance on stage n (constant underflow U kg solution, overflow V between stages): U·x_n−1 + V·x_n+1 = U·x_n + V·x_n

Single batch stage, dry feed solids (solute S_A, inerts B, solvent W, retention r kg solution/kg inert): x = S_A/(S_A + W); solute lost with underflow = r·B·x; recovery = 1 − r·B·x/S_A

Counter-current washing with constant underflow and pure solvent at the last stage: unrecovered fraction falls roughly by a factor R = V/U per additional washing stage (exact values from stage-by-stage balances or the Kremser form with factor V/U).

• U, V – kg/h solution; x, y – mass fraction of solute in solution; B – kg inert; r – kg solution retained per kg inert.

Worked examples

Example 1 – single batch stage (standard). Given: 100 kg inert solids containing 20 kg soluble solute are mixed with 200 kg water. After settling, the underflow retains 0.5 kg solution per kg inert. Ideal stage.

1. Total solution = 20 + 200 = 220 kg; concentration x = 20/220 = 0.0909.
2. Underflow solution = 0.5 × 100 = 50 kg, carrying 50 × 0.0909 = 4.55 kg solute.
3. Overflow = 220 − 50 = 170 kg, carrying 15.45 kg solute.
4. Recovery = 15.45/20 = 0.773.

Overflow 170 kg at 9.09 wt % solute; 77.3 % of the solute recovered.

Example 2 – counter-current leaching with constant underflow (GATE level). Given: 100 kg/h inerts with 25 kg/h solute (dry feed) enter stage 1; 150 kg/h fresh water enters the last stage N. Every underflow carries 50 kg/h solution. Ideal stages. Find the recovery with N = 2, and the number of stages for at least 98 % recovery.

1. Solution balance: underflow solution U = 50 kg/h everywhere; overflow between stages V = 150 kg/h; overflow product from stage 1 = 150 + 25 − 50 = 125 kg/h.
2. N = 2, stage 2 (solute): U·x₁ + 0 = (U + V)·x₂ → 50x₁ = 200x₂ → x₂ = x₁/4.
3. Stage 1: 25 + V·x₂ = (125 + 50)x₁ → 25 + 37.5x₁ = 175x₁ → x₁ = 0.1818, x₂ = 0.0455.
4. Solute lost = U·x₂ = 50 × 0.0455 = 2.27 kg/h → recovery = 1 − 2.27/25 = 0.909.
5. Repeating the balances for more stages: N = 3 → 97.0 %; N = 4 → 99.0 %.

N = 2 gives 90.9 % recovery; 4 ideal stages are needed for ≥ 98 % (extract 125 kg/h at 18.2 wt % with N = 2, rising towards 20 wt %).

Common mistakes

• Assuming the underflow leaves dry – retained solution carries solute out with the solids.
• Using a distribution coefficient as in liquid extraction; for an ideal leaching stage, underflow and overflow solutions have the same composition.
• Forgetting that the first stage's overflow differs from the inter-stage overflow when the feed brings in solute.
• Mixing kg of solution with kg of solvent in retention data.
• Grinding as finely as possible without considering drainage and filtration.
• Thinking agitation speeds internal pore diffusion.

For GATE CH

Expect single-stage and multistage balance problems with constant underflow, recovery fractions, overflow concentrations, and the number of counter-current stages for a target recovery. Conceptual questions cover ideal-stage definition, underflow and overflow, effects of particle size, temperature and agitation, and equipment such as Bollman and Rotocel extractors and thickeners.

Quick check

1. What defines an ideal stage in leaching?
2. What is the underflow?
3. Why is counter-current leaching preferred to cross-current?
4. In Example 1, how much solute leaves with the solids?

Answers: 1. the solution retained by the solids has the same composition as the overflow; 2. inert solids plus the solution they retain; 3. higher recovery with less solvent and a more concentrated extract; 4. 4.55 kg.