Question 1 of 6: The Bell Concentrator — Flowsheet, Metallurgical Balance and Operating Profit

Question 1: The Bell Concentrator — Flowsheet, Metallurgical Balance and Operating Profit (34 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.

(a) Identifying the fourteen unit operations

Given. The printed flowsheet carries fourteen unlabelled blocks (A to H, then J to N and P; the letters I and O are skipped, as is normal on engineering drawings) and four blocks that are already named: primary crusher, primary screen, regrind cyclone and third-stage cleaner flotation cells. The candidate list contains exactly fourteen unit operations, so the answer is a one-to-one assignment.

Find. The unit operation belonging at each lettered position, justified by the stream connections drawn on the flowsheet rather than by guesswork.

Approach. Read the diagram in three blocks — the closed crushing circuit, the closed grinding circuit and the flotation and dewatering train — and in each block use the recycle loops to fix the identity of the equipment, because a closed loop only makes sense for one pairing of machines.

Working from the primary crusher down, the primary screen sends its oversize sideways to A and its undersize downwards. A machine that takes screen oversize and discharges to another screen, C, is a secondary crusher, and the screen it feeds is the secondary screen. That screen's oversize goes right to B and B returns its product to the same screen: a closed-circuit third stage of crushing, so B is the tertiary crusher. The secondary screen's undersize joins the primary screen undersize and passes to the grinding circuit — the standard three-stage crushing arrangement of a 1970s porphyry copper mill.

In the grinding block, D takes the crushing product directly and discharges to E; E splits, sending one stream to F and one stream on to flotation, and F returns to E. Only a classifier can split a stream two ways, so E is the bank of hydrocyclones, F is the ball mill grinding the cyclone underflow, and D, which is fed dry crushed ore and needs no classifier of its own, is the rod mill. The same logic identifies the regrind loop lower down: the regrind cyclone underflow goes to J and J returns to the cyclone feed, so J is the regrind ball mill.

The flotation block is read from the tailings side. G receives the cyclone overflow, which is fresh flotation feed, so it is the rougher flotation cells; its tailing goes to H, whose concentrate returns to the rougher feed and whose tailing leaves as final mill tailings, which is exactly the duty of the scavenger flotation cells. The rougher concentrate is reground and then upgraded in a countercurrent cleaner train, so K and L are the first-stage and second-stage cleaner flotation cells feeding the named third-stage cells. Finally the concentrate is dewatered in three steps of decreasing water content: M is the thickener, N the filter and P the dryer.

(b) Tonnes per day of copper concentrate

Given. The mill treats a fixed tonnage of ore and produces one concentrate and one tailing, with all three copper assays measured.

Find. The tonnes per day of copper concentrate produced, and the copper recovery that goes with it.

Approach. Write a solids balance and a copper balance around the whole plant and eliminate the tailing tonnage; this is the two-product formula.

1. Write the two balances around the plant. Solids in equals solids out, and copper in equals copper out: $$\begin{aligned} F &= C + T \\ Ff &= Cc + Tt \end{aligned}$$ where $F$, $C$ and $T$ are the tonnages of feed, concentrate and tailing.
2. Eliminate the tailing tonnage. Substituting $T = F - C$ into the copper balance and collecting the terms in $C$: $$Ff = Cc + (F-C)t \;\Longrightarrow\; C(c-t) = F(f-t)$$ which rearranges to the two-product formula $$\frac{C}{F} = \frac{f-t}{c-t}$$
3. Substitute the assays. The numerator is the upgrade the plant has to achieve and the denominator is the upgrade one tonne of concentrate represents: $$\frac{C}{F} = \frac{0.50 - 0.04}{28.0 - 0.04} = \frac{0.46}{27.96} = 0.016452$$ so one tonne of concentrate is made from just over sixty tonnes of ore. The ratio of concentration is $K = 1/0.016452 = 60.78$.
4. Scale to the daily tonnage. Multiplying by the ore rate: $$C = 12\,000 \times 0.016452 = 197.4\ \text{t/d}$$ $$\boxed{C = 197.4\ \text{tonnes per day of copper concentrate}}$$
5. Check by recovery. Copper recovery to the concentrate follows from the same three assays, and it is the number a metallurgist would quote to check the answer: $$R = \frac{c(f-t)}{f(c-t)} \times 100 = \frac{28.0 \times 0.46}{0.50 \times 27.96} \times 100 = 92.13\ \%$$ Recomputing it from the tonnages instead, $197.4 \times 28.0 / (12\,000 \times 0.50) = 92.13\ \%$, gives the same figure, which confirms the arithmetic.

A recovery of just over 92 per cent at a concentrate grade of 28 per cent copper is entirely typical of a British Columbia porphyry copper operation of that era, and the tailing assay of 0.04 per cent copper is the metallurgical evidence for it: only about eight per cent of the contained copper is being lost.

(c) Comminution power cost by Bond's equation

Given. A Bond work index of 16 and electricity at 6 cents per kilowatt hour, applied to the same 12 000 t/d of ore. The feed and product sizes are not stated, so the question asks explicitly for the assumptions to be listed.

Find. The daily cost of the electricity used to crush and grind the ore.

Approach. Bond's third theory gives the specific energy for the whole size reduction from run-of-mine to final grind; multiply by the tonnage and by the tariff.

1. State Bond's third theory of comminution. The energy needed is proportional to the difference of the reciprocal square roots of the product and feed sizes: $$W = 10\,W_i\left(\frac{1}{\sqrt{P_{80}}} - \frac{1}{\sqrt{F_{80}}}\right)$$ with $W$ in kWh per tonne, $W_i$ the work index and both sizes in micrometres.
2. Evaluate the two size terms separately. Keeping them apart makes the structure of the answer obvious — almost all of the energy is spent making fines: $$\begin{aligned} \frac{10 W_i}{\sqrt{P_{80}}} &= \frac{160}{\sqrt{150}} = \frac{160}{12.247} = 13.064\ \text{kWh/t} \\ \frac{10 W_i}{\sqrt{F_{80}}} &= \frac{160}{\sqrt{150\,000}} = \frac{160}{387.30} = 0.4131\ \text{kWh/t} \end{aligned}$$ The feed term is only about 3 per cent of the product term, which is why crushing is cheap and grinding is not.
3. Take the difference to get the specific energy. Subtracting: $$W = 13.064 - 0.4131 = 12.65\ \text{kWh per tonne of ore}$$ $$\boxed{W = 12.65\ \text{kWh/t}}$$
4. Scale to the plant and to money. Multiplying by the daily tonnage gives the energy, and by the tariff the cost: $$\begin{aligned} E &= 12.65 \times 12\,000 = 151\,810\ \text{kWh/d} \\ \text{cost} &= 0.06 \times 151\,810 = 9\,109\ \text{per day} \end{aligned}$$ $$\boxed{\text{power cost} \approx 9\,100\ \text{dollars per day, or } 0.76\ \text{dollars per tonne of ore}}$$ Spread over 24 hours this is an installed comminution draw of about 6 325 kW, which is a sensible figure for a 12 000 t/d rod-mill and ball-mill plant and is a useful sanity check on the whole calculation.

If instead the work index is read in its original units of kilowatt hours per short ton, the 12 000 tonnes become 13 228 short tons, the daily energy becomes 167 338 kWh and the cost rises to about 10 040 dollars per day. Either answer is defensible provided the assumption is stated; what would not be defensible is quoting a number without saying which tonne is meant.

(d) Specific gravity of the copper concentrate

Given. The concentrate assays 28.0 per cent copper. The only two mineral species present are chalcopyrite, which carries 34.6 per cent copper and has a specific gravity of 4.2, and siliceous gangue of specific gravity 2.7.

Find. The specific gravity of the concentrate.

Approach. Convert the copper assay into a chalcopyrite content by stoichiometry, then combine the two minerals on a volume basis, because it is volumes and not densities that add.

1. Convert the copper assay to a mineral assay. All the copper reports to chalcopyrite, so the mass fraction of chalcopyrite in the concentrate is the copper assay divided by the copper content of pure chalcopyrite: $$x_{cp} = \frac{28.0}{34.6} = 0.8092$$ That is 80.92 per cent chalcopyrite. The balance, $x_g = 0.1908$ or 19.08 per cent, is siliceous gangue. This is worth a moment: a 28 per cent copper concentrate is a 81 per cent chalcopyrite concentrate, which is a much better result than the copper number alone suggests.
2. Combine the minerals on a volume basis. One kilogram of concentrate occupies the sum of the volumes of its two components, and dividing through by that kilogram gives the reciprocal-density rule: $$\frac{1}{\rho_c} = \frac{x_{cp}}{\rho_{cp}} + \frac{x_g}{\rho_g}$$
3. Substitute and evaluate. Working the two contributions separately: $$\begin{aligned} \frac{x_{cp}}{\rho_{cp}} &= \frac{0.8092}{4.2} = 0.19268 \\ \frac{x_g}{\rho_g} &= \frac{0.1908}{2.7} = 0.07065 \\ \frac{1}{\rho_c} &= 0.19268 + 0.07065 = 0.26333 \end{aligned}$$ Inverting, $$\rho_c = \frac{1}{0.26333} = 3.797$$ $$\boxed{\text{SG of the copper concentrate} = 3.80}$$

The result sits much closer to the chalcopyrite value than to the gangue value, as it must, because the heavier mineral is also the more abundant one. Note the common trap: a mass-weighted average of the two specific gravities would give $0.8092 \times 4.2 + 0.1908 \times 2.7 = 3.91$, which is nearly 3 per cent high. Densities never average by mass.

(e) Net operating profit

Given. The 197.4 t/d of concentrate calculated in part (b), assaying 28.0 per cent copper, sold on the stated smelter schedule against the stated operating costs.

Find. Income minus operating costs, in dollars per day.

Approach. Build the income from the payable copper only, then subtract the two ore-based costs and the two concentrate-based costs. Keeping those two families apart is what prevents the commonest error in this question.

1. Find the copper actually in the concentrate. At 28.0 per cent copper: $$m_{Cu} = 197.4 \times 0.280 = 55.28\ \text{t/d of contained copper}$$
2. Apply the payable term. The smelter pays for only 90 per cent of the contained copper; the remaining 10 per cent is the smelter's deduction and is simply lost to the mine: $$m_{pay} = 0.90 \times 55.28 = 49.75\ \text{t/d of payable copper}$$
3. Value the payable copper at the LME price. Multiplying by 2 500 dollars per tonne of copper: $$\text{income} = 49.75 \times 2\,500 = 124\,378\ \text{dollars per day}$$
4. Build the cost side. Mining and milling are charged on the ore fed to the plant, while freight and treatment are charged on the concentrate shipped: $$\begin{aligned} \text{mining} &= 4 \times 12\,000 = 48\,000 \\ \text{milling} &= 3 \times 12\,000 = 36\,000 \\ \text{freight} &= 30 \times 197.4 = 5\,923 \\ \text{treatment} &= 100 \times 197.4 = 19\,742 \end{aligned}$$ Adding the four gives a total operating cost of $48\,000 + 36\,000 + 5\,923 + 19\,742 = 109\,665$ dollars per day.
5. Take the difference. Income less operating cost: $$\text{profit} = 124\,378 - 109\,665 = 14\,712\ \text{dollars per day}$$ $$\boxed{\text{net operating profit} \approx 14\,700\ \text{dollars per day}}$$

Two features of the answer are worth reading. First, the margin is thin: 14 712 dollars on 124 378 dollars of revenue is a margin of about 12 per cent, or 1.23 dollars per tonne of ore against an operating cost of 9.14 dollars per tonne. Second, mining and milling together are 77 per cent of the total cost, so the plant's profitability is far more sensitive to ore grade and to the mining cost than it is to the smelter terms. A drop of 0.05 percentage points in head grade — from 0.50 to 0.45 per cent copper — would remove roughly 12 400 dollars per day of income and take the operation to break-even. This is exactly why the Bell mine, like most low-grade porphyry operations, lived and died by the copper price.