0.4% Copper: Smelt or Leach?
A copper deposit averages 0.40% Cu, mostly chalcopyrite. Two development options are on the table. Option 1: froth flotation to a 28% Cu concentrate (88% recovery), then smelting. Smelting needs 1.4 kWh of electricity per kg Cu; the SO2 in the off-gas is 95% captured and converted to sulfuric acid, but the regional acid market is saturated at 200 t/y. Option 2: heap leaching with sulfuric acid (62% recovery) followed by solvent extraction and electrowinning, which uses 2.6 kWh/kg Cu and 5.0 kg H2SO4 per kg Cu. Grid electricity emits 0.75 kg CO2 per kWh. The company's target is 10 000 t Cu per year.
1. Calculate the ore tonnage and concentrate mass for each option at 10 000 t Cu per year, and comment on what this means for the project.
Read the worked answer
Ore per tonne of copper = 1 / (0.0040 × recovery)
- Option 1: 1 / (0.0040 × 0.88) = 284 t ore per t Cu → 2.84 Mt/y
- Option 2: 1 / (0.0040 × 0.62) = 403 t ore per t Cu → 4.03 Mt/y
Concentrate (Option 1)
10 000 t Cu at 28% Cu → 10 000 / 0.28 ≈ 35 700 t/y concentrate, leaving about 2.81 Mt/y of tailings to manage and rehabilitate.
Comment
The leach moves 42% more rock for the same copper and produces a proportionally larger spent-ore dump. At 0.4% Cu both options are bulk-materials businesses: ore handling, tailings and closure costs may well decide the project before the chemistry does. Neither option is 'cheap' merely because the process is well understood - the tonnage is the dominant number.
2. Calculate the SO2 and sulfuric acid production from smelting and compare it with the leach's acid demand. What does the comparison show about combining the two?
Read the worked answer
Per tonne of copper
- n(Cu) = 1000 / 63.55 = 15.7 kmol
- Cu2S + O2 → 2Cu + SO2, so n(SO2) = 15.7 / 2 = 7.87 kmol
- m(SO2) = 7.87 × 64.1 ≈ 505 kg; 95% captured → about 480 kg captured, 25 kg emitted
- Acid equivalent: 7.87 kmol H2SO4 × 98.1 = 772 kg, × 0.95 ≈ 730 kg H2SO4 per t Cu
Scale
At 10 000 t Cu/y: about 7 300 t/y H2SO4 produced against a market of 200 t/y. Roughly 7 100 t/y must be neutralised, stored or dumped at a cost, and the smelter may have to run below capacity.
Leach demand
5.0 kg H2SO4 per kg Cu → 50 000 t/y. The smelter's acid covers only about 15% of that.
What it shows
The neat story that 'the smelter makes the acid for the leach' does not survive the numbers at this grade and scale. A combined operation would still need to buy roughly 43 000 t/y of sulfuric acid, so the acid market is a material risk for Option 2, not a free input. The smelter's acid is better sold, or the leach is better sited near an acid supplier.
3. Compare the two options on energy, CO2 and emissions, and recommend a course of action, identifying the weakest assumption in the brief.
Read the worked answer
Annual electricity and CO2
- Option 1: 1.4 kWh/kg × 10 000 000 kg = 14 GWh → 14 000 MWh × 0.75 = 10 500 t CO2
- Option 2: 2.6 kWh/kg × 10 000 000 kg = 26 GWh → 19 500 t CO2
So the leach emits about 9 000 t/y more CO2 on this grid.
SO2
5% of 505 kg/t = 25 kg/t → 250 t/y emitted by Option 1, plus the acid surplus problem; Option 2 emits no SO2 but needs 50 000 t/y of purchased acid.
Recommendation
Option 1 looks better on energy, CO2 and acid by-product, so it is preferred if the ore behaves. But the weakest number in the brief is the 88% flotation recovery at 0.4% Cu: liberation is poor at low grade and recovery usually falls well short of the test value. At 70% recovery, Option 1 needs 357 t ore per t Cu and its ore advantage largely disappears.
Commission locked-cycle flotation testwork on a representative 0.4% composite before choosing. If recovery is confirmed, smelt and sell or neutralise the acid; if not, leach the bulk low-grade material, and consider a hybrid that smelts a higher-grade lens while the heap handles the low-grade halo.