Mineral Processing and Non-Ferrous Extraction: Comminution, Beneficiation, Mass Balances, and Aluminium, Copper, Titanium and Zinc
2. Size classification and the concentration methods
Size classification separates particles by the rate at which they settle rather than by aperture. For a small sphere in the laminar (Stokes) regime the terminal velocity is v = g d²(ρ_s − ρ_f)/(18μ), valid while the particle Reynolds number is below about 1; larger particles settle in the Newton regime, where v ∝ √d. Screens separate coarse sizes by aperture; mechanical classifiers and hydrocyclones separate fine sizes in water, the cyclone’s centrifugal field multiplying g many times. A classifier’s performance is its partition (Tromp) curve and its cut size d₅₀, the size with equal probability of reporting to overflow and underflow. In closed-circuit grinding the classifier underflow returns to the mill, and the circulating load is the ratio of that returned stream to the fresh feed.
Froth flotation separates on surface chemistry. Air bubbles attach to particles whose surfaces are hydrophobic (a finite contact angle) and carry them into a froth, while hydrophilic particles stay in the pulp. The reagents: collectors adsorb on the chosen mineral and make it hydrophobic (xanthates for sulphides, fatty acids for oxides and salt-type minerals, amines for silicates); frothers stabilise the froth (pine oil, methyl isobutyl carbinol); activators restore floatability (copper sulphate activates sphalerite, which xanthate does not collect on its own); depressants prevent it (cyanide depresses sphalerite and pyrite, lime depresses pyrite, sodium silicate depresses silicate gangue); and pH regulators set the conditions (lime, soda ash). Differential flotation of a lead-zinc ore floats galena first with sphalerite depressed, then activates and floats the sphalerite.
| Method | Property exploited | Typical use |
|---|---|---|
| Gravity: jigs, shaking tables, spirals, dense-medium separation | Density difference; easy when (ρ_heavy − ρ_f)/(ρ_light − ρ_f) > 2.5 | Coal washing, cassiterite, gold, iron ore, beach sands |
| Low-intensity magnetic (wet or dry drum) | Ferromagnetism | Magnetite ores; recovering magnetite from dense-medium circuits |
| High-intensity / high-gradient magnetic | Paramagnetism | Hematite, ilmenite, removing iron-bearing impurities from kaolin |
| Electrostatic (high-tension roll) | Electrical conductivity | Separating conducting rutile and ilmenite from non-conducting zircon and monazite in beach sands |
| Froth flotation | Surface hydrophobicity | Sulphides of Cu, Pb, Zn, Mo; fine particles below about 0.2 mm |
3. Mass balance: grade, recovery and the two-product formula
A separation splits a feed F of grade f (fraction of the valuable metal or mineral) into a concentrate C of grade c and a tailing T of grade t. Two balances hold: total mass, F = C + T, and valuable component, Ff = Cc + Tt. Eliminating T gives the two-product formula C/F = (f − t)/(c − t), the mass yield. The recovery is the fraction of the valuable component that reports to the concentrate, R = Cc/(Ff) = c(f − t)/(f(c − t)). Other figures of merit follow: the ratio of concentration F/C (tonnes of feed per tonne of concentrate), the enrichment ratio c/f, and separation efficiency, which subtracts the gangue that was recovered along with the value. Grade and recovery trade against each other along a grade-recovery curve: cleaning a concentrate raises its grade and loses recovery.
Worked: a copper ore of 2.0% Cu gives a concentrate of 25% Cu and a tailing of 0.2% Cu. C/F = (2.0 − 0.2)/(25 − 0.2) = 1.8/24.8 = 0.0726, so 72.6 kg of concentrate come from every tonne of feed and the ratio of concentration is 13.8. Recovery = 0.0726 × 25/2.0 = 0.907, i.e. 90.7%. As a check on the tailing side, 927.4 kg of tailing at 0.2% carry 1.85 kg of copper, which is the 9.3% of the 20 kg in the feed that was lost.
4. Principles of pyro-, hydro- and electro-metallurgy
Pyrometallurgy uses high temperature. Roasting treats a solid with gas below its melting point: an oxidising (dead) roast converts sulphide to oxide, a sulphating roast to a water-soluble sulphate, a chloridising roast to a chloride — the choice read from the predominance diagram. Calcination drives off CO₂ or water. Smelting melts the charge and separates a metal or matte from a slag; matte smelting of copper produces a molten sulphide mixture (Cu₂S-FeS) because copper has a lower affinity for oxygen and a higher affinity for sulphur than iron, so iron is preferentially oxidised into the slag. Converting blows air through the molten matte to oxidise first the remaining iron and then the sulphur. Refining removes the last impurities, by fire, by distillation, or by electrolysis.
Hydrometallurgy works in aqueous solution at modest temperature: leaching (dissolving the metal with acid, alkali, cyanide or ammonia, in heaps, vats, agitated tanks or autoclaves), solution purification and concentration (solvent extraction with an organic extractant, ion exchange, precipitation of impurities), and metal recovery (cementation by a more electropositive metal, precipitation, hydrogen reduction, or electrowinning). It suits low-grade and complex ores and avoids SO₂ emissions. Electrometallurgy uses electrical energy: electrowinning deposits metal from a solution with an inert anode, at which oxygen is evolved, and so needs the whole decomposition voltage; electrorefining dissolves an impure anode and deposits pure metal at the cathode, needing only a fraction of a volt to overcome resistance and polarization, while impurities nobler than the metal fall as anode slime and less noble ones stay in solution. Fused-salt electrolysis is used for metals too reactive to be deposited from water (Al, Mg, Na). Faraday’s law gives the mass: m = ItM/(nF) × current efficiency.
5. Aluminium, copper, titanium and zinc: the unit processes and their balances
Aluminium. Bauxite is refined to alumina by the Bayer process: digestion in hot caustic soda dissolves the aluminium hydroxide minerals as sodium aluminate, the insoluble iron oxide, silica and titania residue is separated as red mud, the cooled and seeded liquor precipitates Al(OH)₃, and calcination gives Al₂O₃. Alumina is then reduced by the Hall-Héroult process: it is dissolved in molten cryolite (Na₃AlF₆) at about 950-980 °C, which lowers the working temperature far below alumina’s melting point of about 2050 °C, and electrolysed between a carbon cathode lining and consumable carbon anodes. Aluminium collects at the cathode; the anodes burn to CO₂ by the overall reaction 2Al₂O₃ + 3C = 4Al + 3CO₂, so the theoretical carbon consumption is 3 × 12.01/(4 × 26.98) = 0.334 kg per kg of Al (practical consumption is higher). The metal carries three electrons per atom, so it needs 3F per mole — 2980 A h per kg — which is why smelters are sited next to cheap power.
Copper. Sulphide ores (chalcopyrite, CuFeS₂) are concentrated by flotation to about 25-30% Cu, then smelted — today mostly in flash smelters, where dried concentrate burns in oxygen-enriched air, using the heat of sulphide oxidation — to a matte of Cu₂S and FeS and an iron silicate slag. The matte is converted in two stages: the slag-forming blow, 2FeS + 3O₂ + SiO₂ → 2FeO·SiO₂ + 2SO₂, then the copper-making blow, Cu₂S + O₂ → 2Cu + SO₂, which gives blister copper of about 98.5-99.5% Cu. Fire refining removes sulphur and oxygen, the metal is cast into anodes, and electrorefining in acidified copper sulphate gives cathode copper of 99.99%, with gold, silver and selenium recovered from the anode slime. Oxide and low-grade ores follow the hydrometallurgical route instead: heap leaching with sulphuric acid, solvent extraction, and electrowinning (SX-EW).
Titanium. TiO₂ cannot be reduced by carbon — titanium’s oxide line lies low on the Ellingham diagram and titanium forms a stable carbide — and titanium dissolves oxygen and nitrogen, which embrittle it. So the Kroll process goes through the chloride. Rutile or upgraded ilmenite slag is carbochlorinated, TiO₂ + 2Cl₂ + 2C → TiCl₄ + 2CO (or with CO₂); the TiCl₄, a volatile liquid, is purified by distillation; and it is reduced by molten magnesium in a sealed steel retort under argon, TiCl₄ + 2Mg → Ti + 2MgCl₂. The product is titanium sponge, from which MgCl₂ and excess Mg are removed by vacuum distillation or leaching; the MgCl₂ is electrolysed to recover both Mg and Cl₂ for reuse. The sponge is consolidated by vacuum arc remelting. The process is batch, which is the root of titanium’s cost.
Zinc. Sphalerite (ZnS) concentrate is first roasted to ZnO (ZnS + 1.5O₂ → ZnO + SO₂, the SO₂ going to sulphuric acid). About 80-90% of zinc is then made by the roast-leach-electrowin route: the calcine is leached in spent electrolyte (sulphuric acid); iron, which forms zinc ferrite in the roast, is removed by precipitation as jarosite, goethite or hematite; copper, cadmium, cobalt and nickel are cemented out with zinc dust, because even traces of them ruin electrowinning; and zinc is electrowon onto aluminium cathodes with lead-silver anodes. Zinc can be deposited from acid solution only because hydrogen evolution has a very large overpotential on zinc — the Tafel effect of the previous chapter doing industrial work. The pyrometallurgical alternative, the Imperial Smelting Process, reduces a sintered zinc-lead charge with coke in a blast furnace; zinc leaves as vapour (it boils at 907 °C) and is shock-condensed in a splash of molten lead so that it does not reoxidise, while lead bullion is tapped from the hearth.
Key takeaways
- Rittinger E ∝ (1/x₂ − 1/x₁) for fine grinding, Kick E ∝ ln(x₁/x₂) for coarse crushing, Bond W = 10W_i(1/√P₈₀ − 1/√F₈₀) with sizes in µm; critical mill speed 42.3/√(D − d) rpm.
- Flotation: collectors make the mineral hydrophobic, frothers hold the froth, CuSO₄ activates and cyanide depresses sphalerite. Magnetic separation needs low intensity for magnetite and high intensity for hematite; electrostatic separation splits conductors from non-conductors.
- Two-product formula C/F = (f − t)/(c − t); recovery R = c(f − t)/(f(c − t)); ratio of concentration F/C.
- Electrowinning needs the full decomposition voltage and electrorefining only a fraction of a volt; Faraday: m = ItM/(nF) × current efficiency.
- Al: Bayer then Hall-Héroult in cryolite (0.334 kg C per kg Al in theory). Cu: flotation, matte smelting, converting, fire refining, electrorefining. Ti: Kroll, TiCl₄ reduced by Mg. Zn: roast, leach, purify with zinc dust, electrowin — or the Imperial Smelting Process.
Practice questions (13)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
An ore with a Bond work index of 12 kWh/t is ground from a feed of F₈₀ = 10 000 µm to a product of P₈₀ = 100 µm. What specific energy does Bond’s law predict, in kWh/t (to one decimal place)?
Numerical answer — type the value.
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Answer: 10.8
W = 10 W_i (1/√P₈₀ − 1/√F₈₀) = 10 × 12 × (1/10 − 1/100) = 120 × 0.09 = 10.8 kWh/t. Sanity check against the definition: grinding from infinite size to 100 µm would cost exactly W_i = 12 kWh/t, and starting at 10 mm saves the 1/√10000 term, 10% of it. Entering sizes in mm gives 120 × (1/0.316 − 1/3.16) = 341.5, a thousandfold error in the root.Grinding a material from 10 mm to 1 mm requires 10 kWh/t. If Rittinger’s law holds, how much energy, in kWh/t, is required to grind the same material from 1 mm to 0.1 mm?
Numerical answer — type the value.
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Answer: 100
Rittinger: E = K_R(1/x₂ − 1/x₁). First stage: 10 = K_R(1/1 − 1/10) = 0.9K_R, so K_R = 11.11 kWh mm/t. Second: E = 11.11 × (1/0.1 − 1/1) = 11.11 × 9 = 100 kWh/t. By ratio: (10 − 1)/(1 − 0.1) = 9/0.9 = 10 times the first stage. Kick’s law would give the same 10 kWh/t for the same reduction ratio, and Bond’s law √10 × 10 = 31.6 kWh/t.What is the critical speed, in rpm, of a ball mill 2.0 m in internal diameter charged with 100 mm balls (to one decimal place)? Take g = 9.81 m/s².
Numerical answer — type the value.
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Answer: 30.7
At critical speed a ball at the top has ω²(D − d)/2 = g, so ω = √(2 × 9.81/1.9) = 3.213 rad/s and N_c = 60ω/(2π) = 30.7 rpm. The shortcut 42.3/√(D − d) = 42.3/√1.9 = 42.3/1.378 = 30.7 rpm agrees. Ignoring the ball diameter gives 42.3/√2 = 29.9 rpm.A copper ore assaying 2.0% Cu is floated to give a concentrate of 25% Cu and a tailing of 0.2% Cu. What is the copper recovery in the concentrate, in percent (to one decimal place)?
Numerical answer — type the value.
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Answer: 90.7
Two-product formula: C/F = (f − t)/(c − t) = (2.0 − 0.2)/(25 − 0.2) = 0.07258. Recovery R = (C/F) × c/f = 0.07258 × 25/2.0 = 0.9073 = 90.7%. Directly: R = c(f − t)/(f(c − t)) = 25 × 1.8/(2 × 24.8) = 45/49.6 = 90.7%. Stopping at C/F gives 7.3%, the mass yield, not the recovery.For the same separation (feed 2.0% Cu, concentrate 25% Cu, tailing 0.2% Cu), what is the ratio of concentration, i.e. tonnes of feed per tonne of concentrate (to one decimal place)?
Numerical answer — type the value.
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Answer: 13.8
F/C = (c − t)/(f − t) = 24.8/1.8 = 13.78. Check with a copper balance on 13.78 t of feed: 0.2756 t of Cu, of which 0.25 t is in the 1 t of concentrate and 12.78 × 0.002 = 0.0256 t in the tailing — the balance closes. The enrichment ratio c/f = 12.5 is a different figure.A quartz particle 50 µm in diameter (density 2650 kg/m³) settles in water (density 1000 kg/m³, viscosity 0.001 Pa·s). What is its Stokes terminal velocity, in mm/s (to two decimal places)? Take g = 9.81 m/s².
Numerical answer — type the value.
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Answer: 2.25
v = g d²(ρ_s − ρ_f)/(18μ) = 9.81 × (50 × 10⁻⁶)² × 1650/(18 × 0.001) = 9.81 × 2.5 × 10⁻⁹ × 1650/0.018 = 2.248 × 10⁻³ m/s = 2.25 mm/s. Validity check: Re = ρvd/μ = 1000 × 2.248 × 10⁻³ × 50 × 10⁻⁶/0.001 = 0.11 < 1, so Stokes applies. Using the particle density instead of the density difference gives 3.61 mm/s.Which of the following reagent roles in sulphide froth flotation are correctly stated?
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Answer: A — Potassium amyl xanthate acts as a collector; B — Pine oil acts as a frother; D — Sodium cyanide depresses sphalerite and pyrite
Xanthates are the standard sulphide collectors (A), pine oil is a frother (B), and cyanide depresses sphalerite and pyrite, which is how galena is floated first in lead-zinc differential flotation (D). C is backwards: copper sulphate ACTIVATES sphalerite, depositing copper on its surface so that xanthate can adsorb — the step that lets the zinc be floated after the lead.Which of the following pairings of a mineral separation with a suitable method are correct?
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Answer: A — Magnetite from silica: low-intensity magnetic separation; B — Rutile from zircon: electrostatic (high-tension) separation; D — Coal from shale: dense-medium (gravity) separation
Magnetite is ferromagnetic and is taken by a low-intensity drum (A). Rutile conducts and zircon does not, so a high-tension roll separates them (B). Coal is much lighter than shale, the textbook case for dense-medium separation (D). Hematite is only paramagnetic, so it needs HIGH-intensity or high-gradient magnetic separation — a low-intensity drum barely holds it (C is wrong).A copper electrowinning tankhouse passes 20 000 A for 24 h at 90% current efficiency. How much copper, in kg, is deposited (to the nearest whole number)? Take M(Cu) = 63.55 g/mol and F = 96 485 C/mol.
Numerical answer — type the value.
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Answer: 512
Charge Q = 20 000 × 86 400 = 1.728 × 10⁹ C. Moles of Cu = ηQ/(nF) = 0.9 × 1.728 × 10⁹/(2 × 96 485) = 8059 mol, and m = 8059 × 63.55 g = 512.2 kg. Check through the electrochemical equivalent: 63.55/(2 × 96 485) = 0.3293 mg/C, times 1.555 × 10⁹ effective coulombs = 512 kg. Using n = 1 doubles it to 1024 kg.In the Hall-Héroult cell, taking the overall reaction as 2Al₂O₃ + 3C = 4Al + 3CO₂, what is the theoretical carbon anode consumption per kg of aluminium, in kg (to three decimal places)? Take M(C) = 12.01 and M(Al) = 26.98 g/mol.
Numerical answer — type the value.
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Answer: 0.334
3 mol C per 4 mol Al: 3 × 12.01/(4 × 26.98) = 36.03/107.92 = 0.334 kg C per kg Al. Electron check: 4 Al need 12 electrons, and each C going to CO₂ gives 4, so 3 C — the same stoichiometry. If the anode gas were all CO the carbon per Al would double to 0.668 kg; real cells produce some CO and burn some anode in air, so practical consumption is higher than 0.334.In the Hall-Héroult process the electrolyte in which alumina is dissolved is essentially molten:
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Answer: A — Cryolite, Na₃AlF₆
Alumina melts at about 2050 °C; dissolved in molten cryolite it can be electrolysed at about 950-980 °C, which is the whole point of the process. Sodium aluminate solution is the Bayer-process liquor that precedes it, bauxite is the ore, and aluminium chloride electrolysis is a different route that is not the Hall-Héroult cell.In the Kroll process for titanium, TiCl₄ is reduced by:
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Answer: A — Molten magnesium
TiCl₄ + 2Mg → Ti + 2MgCl₂ in a sealed retort under argon gives titanium sponge (sodium does the same job in the older Hunter process). Carbon is ruled out because titanium forms a carbide and its oxide lies too low on the Ellingham diagram; titanium is far too reactive to be deposited from water; and hydrogen is not a strong enough reductant for TiCl₄.Which of the following statements about zinc extraction are correct?
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Answer: A — Zinc dust is added to the leach liquor to cement out copper and cadmium before electrowinning; B — Zinc can be electrowon from acid sulphate solution because hydrogen evolution has a high overpotential on zinc; D — Sphalerite concentrate is roasted to zinc oxide before leaching
Zinc dust, more electropositive than Cu and Cd, cements them out (A); a high hydrogen overpotential on zinc is what lets the cathode deposit zinc rather than evolve hydrogen from an acid bath (B); roasting ZnS to ZnO precedes leaching (D). C is wrong: in the ISP zinc leaves as VAPOUR with the furnace gas and is condensed in a lead splash condenser; it is lead bullion that is tapped from the hearth.