Question 1 of 6: Mine, Mill and Metallurgical-Accounting Economics

Question 1 — Mine, Mill and Metallurgical-Accounting Economics (30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) a labelled flow sheet of the circuit; (b) the copper grade of the final tailings; (c) the daily comminution power cost from Bond's equation; (d) the net operating profit in $/day; (e) the sample mass (Gy equation) needed for ±0.1% Cu accuracy at 95% confidence.

Approach. Part (b) is a two-product metallurgical-accounting balance on copper; part (c) applies Bond's third law of comminution directly to the stated F and P; part (d) is a revenue-minus-cost operating statement built from the same mass balance; part (e) applies Gy's sampling-constant equation with the crusher top size as the particle diameter and a 95%-confidence z-value to convert the stated absolute accuracy into Gy's relative-variance term s.

1. Part (a) — Flow sheet. The circuit described, stage by stage, is drawn above: primary jaw crushing → secondary/tertiary crushing → screening to a −1.5 cm (15,000 µm) product → rod mill → ball mill closed-circuit with a hydrocyclone (cyclone underflow/sands recycle to the ball mill, overflow at 210 µm to flotation) → conditioner → rougher flotation → two stages of cleaning → final concentrate to thickening/filtering/drying → rail to smelter. Rougher tails report to a scavenger stage; scavenger concentrate and cleaner-1 tails are reground in an open-circuit regrind ball mill (per the question) and returned to the conditioner, while scavenger tails leave the circuit as final tailings.
2. Part (b) — Copper content of mill tailings. Basis: 100 mass units of ore feed. Copper in feed $=100\times0.01=1.00$ unit. At 95% recovery, copper reporting to concentrate $=1.00\times0.95=0.95$ unit, so copper reporting to tailings is $1.00-0.95=0.05$ unit. Concentrate mass, from its own grade: $$C=\dfrac{0.95}{0.27}=3.5185\ \text{units (per 100 feed)}$$ Tailings mass is the remainder: $T=100-3.5185=96.4815$ units. Tailings grade: $$t=\dfrac{0.05}{96.4815}=5.183\times10^{-4}=0.0518\%\ \text{Cu}$$ \(\boxed{t \approx 0.052\%\ \text{Cu}}\) — roughly a twentieth of the head grade, consistent with a well-liberated sulphide ore recovering 95% of its copper.
3. Part (c) — Comminution power cost (Bond's equation). Take the crusher product as the grinding-circuit feed size, $F=15{,}000\ \mu\text{m}$, and the grind product as $P=210\ \mu\text{m}$ (the size stated for the rod–ball milling stage), applying Bond's equation directly over that single reduction span with the paper's own overall Work Index. $$W=10(12)\left(\dfrac{1}{\sqrt{210}}-\dfrac{1}{\sqrt{15{,}000}}\right)=120(0.06900-0.008165)=120(0.060842)$$ $$W=7.301\ \text{kWh/tonne}$$ Daily energy $=7.301\times1{,}500=10{,}951.5\ \text{kWh/day}$. At 20¢/kWh: $$\text{Power cost}=10{,}951.5\times0.20=\text{\$}2{,}190\ \text{/day}$$ \(\boxed{\approx\ \text{\$}2{,}190\text{/day}}\)
4. Part (d) — Net operating profit. From part (b)'s mass balance, concentrate produced $=1{,}500\times(3.5185/100)=52.78\ \text{t/day}$, carrying contained copper $=52.78\times0.27=14.25\ \text{t Cu/day}$ — identical to $1{,}500\times0.01\times0.95=14.25\ \text{t Cu/day}$ computed straight from head grade and recovery, which cross-checks the balance. Payable copper (90%) $=14.25\times0.90=12.825\ \text{t/day}$, so revenue $=12.825\times7{,}000=\text{\$}89{,}775\text{/day}$. Deducting the treatment charge on concentrate mass, $100\times52.78=\text{\$}5{,}278\text{/day}$, gives a net smelter return of $\text{\$}84{,}497\text{/day}$. Mining and milling costs apply to the full ore tonnage; freight applies to concentrate tonnage only: mining $=30\times1{,}500=\text{\$}45{,}000$, milling $=6\times1{,}500=\text{\$}9{,}000$, freight $=50\times52.78=\text{\$}2{,}639$/day. $$\text{Net profit}=84{,}497-45{,}000-9{,}000-2{,}639=\text{\$}27{,}858\text{/day}$$ \(\boxed{\approx\ \text{\$}27{,}858\text{/day}}\)
5. Part (e) — Gy sampling equation. The particle diameter to use is the top size entering the fine-ore bin, i.e. the crusher product, $d=1.5\ \text{cm}$. A required accuracy of ±0.1% Cu (absolute) on a 1% Cu head grade is a relative allowable error of $e=0.1/1=0.10$ (10%). At 95% confidence (two-tailed normal, $z=1.96$), Gy's fractional standard deviation is $s=e/z=0.10/1.96=0.0510$. $$M=\dfrac{Cd^3}{s^2}=\dfrac{(1.0)(1.5)^3}{(0.0510)^2}=\dfrac{3.375}{0.002604}=1{,}296\ \text{g}$$ \(\boxed{M\approx1.3\ \text{kg}}\) (using the common $z=2$ engineering approximation instead gives $s=0.05$ and $M=1{,}350\ \text{g}$ — the same order, since the exact 95% z-value only shifts the answer by about 4%).