Blending Low-Grade Iron Ore to Meet an Export Specification
A mine near Sandur must clear a twelve-month export contract: Fe ≥ 60.0%, SiO2 + Al2O3 ≤ 8.0%, P ≤ 0.10%. Two screened stocks are available. Low-grade (L): Fe 58.5%, SiO2 6.5%, Al2O3 4.5%, P 0.12%. High-grade (H): Fe 66.0%, SiO2 1.8%, Al2O3 1.5%, P 0.05%. The mine produces 10,000 t of L per month and cannot produce more; H must be bought in from a neighbouring lease at a premium. The plant has a wet screening and scrubbing circuit never yet run on L. A lab note says most of the Al2O3 in L sits in clay coatings on lumps below 40 mm, while the SiO2 is mostly in quartz veins inside the lumps. The marketing manager proposes shipping all 10,000 t of L each month; the plant manager says the numbers will not allow it.
1. Taking x as the mass fraction of L, set up the three constraints and find the maximum x. Which constraint binds?
Read the worked answer
Let x = fraction of L, (1 − x) = fraction of H.
Iron: 58.5x + 66.0(1 − x) ≥ 60.0 66.0 − 7.5x ≥ 60.0 → x ≤ 0.800
Gangue (SiO2 + Al2O3): L has 6.5 + 4.5 = 11.0%; H has 1.8 + 1.5 = 3.3% 11.0x + 3.3(1 − x) ≤ 8.0 3.3 + 7.7x ≤ 8.0 → x ≤ 0.610
Phosphorus: 0.12x + 0.05(1 − x) ≤ 0.10 0.05 + 0.07x ≤ 0.10 → x ≤ 0.714
Answer: x_max = 0.61, i.e. 61% L : 39% H. Gangue is the binding constraint. Check the blend: Fe = 58.5(0.61) + 66.0(0.39) = 35.69 + 25.74 = 61.4% ✓ Gangue = 11.0(0.61) + 3.3(0.39) = 6.71 + 1.29 = 8.00% (exactly at the limit) P = 0.12(0.61) + 0.05(0.39) = 0.093% ✓
Note there is zero margin on gangue; routine sampling error will put cargoes over spec.
2. The marketing manager wants to ship all 10,000 t of L monthly. How much H is required, and what are the commercial implications?
Read the worked answer
At the 61:39 maximum blend ratio:
Total blend = 10,000 / 0.61 = 16,393 t/month H required = 16,393 − 10,000 = 6,393 t/month.
So each tonne of L shipped drags in about 0.64 t of purchased high-grade ore. The commercial implications are severe:
- The blend sits at exactly 8.00% gangue, so a 0.3-point analytical error or a wet cargo means rejection or a penalty at the port, with demurrage.
- H is bought at a premium, so the margin on the L tonnes may be negative once the H premium is paid.
- The plant is being run as a blending operation, not a mining operation: the value is created by the purchased ore.
A prudent target is 58–59% L, giving roughly 7–8% gangue and a working margin. The marketing manager is answering 'can we ship the tonnes?' when the real question is 'at what cost and risk?'. Also check the P limit is not breached by a switch of H supplier - 0.093% is thin cover.
3. Suppose only 4,000 t of H is available each month. Evaluate beneficiation and give the practical monthly limit without it.
Read the worked answer
Blend at 10,000 L : 4,000 H (71.4 : 28.6) Fe = (585,000 + 264,000)/14,000 = 60.6% ✓ Gangue = (110,000 + 13,200)/14,000 = 8.8% ✗ (fails by 0.8) P = (1,200 + 200)/14,000 = 0.100% - exactly on the limit, no margin ✗
Without beneficiation, solve for L with H fixed at 4,000: Gangue: (11L + 13,200)/(L + 4,000) ≤ 8 → 3L ≤ 18,800 → L ≤ 6,267 t/month (P allows L ≤ 10,000, so gangue again binds.)
Can scrubbing rescue it? To ship 10,000 t of L you must cut its gangue from 11.0% to ≤9.88%, i.e. remove 1.12 points - a quarter of the entire Al2O3 content. The lab note says Al2O3 is in clay coatings (removable by scrubbing) but SiO2 is locked in quartz veins (not removable). So removal is capped by the 4.5% Al2O3, and only the coated fraction of it. If half the alumina is intergrown, scrubbing cannot reach the target. Scrubbing also does nothing to phosphorus.
Recommendation: commission a scrubbing pilot on a 50 t sample and measure the actual alumina drop; meanwhile plan on ~6,250 t/month of L.