Question 1 of 7: Highland Valley milling circuit: flow sheet, copper recovery and grinding power

Problem 1 — Highland Valley milling circuit: flow sheet, copper recovery and grinding power (25 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 flow sheet of the described circuit; (b) the copper concentrate production rate in tpd; (c) the net grinding power in kW, stating the assumptions used to apply Bond's equation.

Approach. (a) is answered directly from the plant description as a block flow sheet; (b) uses the two-product (feed/concentrate/tailings) metal balance; (c) applies Bond's third theory of comminution between the crusher product (taken as the mill circuit feed size $F$) and the final cyclone-overflow grind size $P$, converted to a net power draw using the plant's daily throughput.

1. (a) Flow sheet. Ore is trucked from the pit to two gyratory crushers (-7 in product) and conveyed to a coarse-ore stockpile. Five parallel grinding lines each reclaim ore through a SAG mill in circuit with a 0.5 in screen (oversize recycles to the SAG mill) and two ball mills operating in closed circuit with 30 in cyclones (underflow recycles to the ball mills; overflow, 80 % passing 200 microns, is the final grind). Combined cyclone overflow feeds bulk copper-molybdenum rougher/scavenger flotation (scavenger concentrate recycled to the circuit head, tailings to the impoundment). Bulk concentrate is thickened (125 ft thickener), conditioned with NaHS to depress copper minerals, and floated in a Cu/Mo separation circuit: the copper stream reports to rougher/scavenger cells and on to the final copper concentrate (thickened to 65 % solids, filtered, dried to 7 % moisture, stored at 41.4 % Cu); the molybdenum rougher concentrate is reground, classified by 6 in cyclones, upgraded in a cleaner column (tails recirculated to the 125 ft thickener), then leached with ferric chloride to strip residual copper before final filtering and drying. See Fig. 1.
2. (b) Copper concentrate tonnage — two-product formula. With fresh feed $F$, concentrate $C$ and tailings $T=F-C$, a metal balance on copper gives $Ff=Cc+(F-C)t$, i.e. $$C=F\cdot\dfrac{f-t}{c-t}.$$ Substituting the given grades: $$C=130{,}000\times\dfrac{0.388-0.033}{41.4-0.033}=130{,}000\times\dfrac{0.355}{41.367}=\boxed{1{,}116\ \text{tpd copper concentrate}}\ (\approx1115.6\ \text{tpd}).$$
3. (c) Grinding power — Bond's equation. Bond's law needs $P$ and $F$ as 80 %-passing sizes in microns. The exam gives the crusher's top size (-7 in) rather than its own 80 % passing size, so a reasonable engineering assumption is required: treat the crusher product top size as the mill-circuit feed size $F_{80}\approx7\ \text{in}=177{,}800\ \text{microns}$ (a conservative choice — the true $F_{80}$ of the crusher product is somewhat finer than its top size, which would make the true power draw slightly higher than computed here). The final grind size is the cyclone overflow, $P_{80}=200$ microns. Substituting into Bond's equation with $W_i=15$ kWh/mt: $$W=\dfrac{10(15)}{\sqrt{200}}-\dfrac{10(15)}{\sqrt{177{,}800}}=10.607-0.356=10.25\ \text{kWh/mt}.$$ Converting the daily throughput to an hourly rate, $\dot{m}=130{,}000/24=5{,}416.7$ mt/hr, so the net grinding power is $$P_{net}=W\times\dot{m}=10.25\times5{,}416.7=\boxed{55{,}500\ \text{kW}\ (\approx55.5\ \text{MW})}.$$ (Using $F_{80}=0.8\times7\ \text{in}$ instead changes the answer by under half a percent, confirming the result is not sensitive to the exact $F_{80}$ assumption within a reasonable range.)