Question 1 of 5: Gibraltar Mine Copper Flotation Circuit

Question 1: Gibraltar Mine Copper Flotation Circuit (32 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.

1(i) — Flow sheet of the flotation circuit (8 marks)

Approach. Trace the material path exactly as narrated: three parallel grinding circuits each feed their own rougher bank; the three rougher concentrates join into one regrind (ball mill + cyclone) circuit; the regrind cyclone overflow reports to a column surge tank feeding two column cells in series; the two columns' concentrates combine to the thickener (final Cu concentrate); the second column's tail is scavenged by a Denver bank whose concentrate recycles to the surge tank and whose tail is the final plant tailings.

[Figure not reproduced: Figure 1.1 — Gibraltar Mine copper flotation flow sheet, redrawn from the text description (3 grinding circuits → rougher banks → regrind ball mill/cyclone → column flotation in series → column scavenger, with the scavenger concentrate recycled to the column feed surge tank. See the official exam paper.]

Rougher flotation runs at 42–44% solids, pH 9.6–10.2 (lime added ahead of the rod mill for pH control); sodium isopropyl xanthate (collector) and Orform F-2 (frother) are dosed to the regrind cyclone overflow, i.e. into the column feed, not the roughers. The scavenger loop (column tail → 16×300H Denver → concentrate recycled to the surge tank) is the circuit's cleaner-scavenger stage: it recovers copper that reported to the column tailings before that material is finally rejected.

1(ii) — Tailings grade (5 marks)

Given. Feed grade $f = 0.31\%\ \text{Cu}$; copper recovery $R = 80\%$; concentrate grade $c = 28\%\ \text{Cu}$ (stated in the circuit description).

Find. Tailings grade $t$ (% Cu).

Approach. Combine the recovery definition with the two-product (feed = concentrate + tailings) mass and metal balance to solve for $t$.

1. Yield to concentrate from the recovery definition. Recovery is the fraction of feed metal reporting to concentrate, $R = Cc/(Ff)$, so the concentrate yield (mass fraction of feed) is $$Y=\frac{C}{F}=\frac{Rf}{c}=\frac{0.80\times0.31}{28}=0.008857\ \ (0.8857\%\ \text{of feed mass})$$
2. Metal balance on the tailings stream. Per unit feed mass, metal in $=f$, metal to concentrate $=Yc=Rf$, so metal to tailings is $f(1-R)$ over a tailings mass fraction $(1-Y)$: $$t=\frac{f(1-R)}{1-Y}=\frac{0.31\times(1-0.80)}{1-0.008857}=\frac{0.0620}{0.991143}=\boxed{0.0626\%\ \text{Cu}}$$

Substituting back through the standard two-product formula $Y=(f-t)/(c-t)$ as a check gives the same $Y = 0.008857$, confirming the tailings grade.

1(iii) — Concentrate produced per day (4 marks)

Given. Ore throughput $F = 30{,}000\ \text{t/d}$; concentrate yield $Y = 0.008857$ (from 1(ii)).

Find. Concentrate produced, tonnes/day.

1. Apply the yield to the daily ore feed. $$C = FY = 30{,}000 \times 0.008857 = \boxed{265.7\ \text{t/d concentrate}}$$

1(iv) — % chalcopyrite in the copper concentrate (2 marks)

Given. Concentrate grade $c = 28\%\ \text{Cu}$; chalcopyrite (CuFeS₂) is the only copper mineral present and is stated to itself assay $34.6\%\ \text{Cu}$.

Find. Mass % chalcopyrite in the concentrate.

1. Divide the bulk assay by the mineral assay. All the Cu in the concentrate is locked in chalcopyrite, so the chalcopyrite mass fraction is the ratio of the two grades: $$\%\text{CuFeS}_2 = \frac{28}{34.6}\times100=\boxed{80.9\%}$$

The remaining ≈19% of the concentrate mass is gangue (silicate/pyrite) that reports with the floated chalcopyrite – consistent with a real, imperfectly selective flotation concentrate rather than a pure mineral product. The stated 34.6% Cu assay is itself consistent with the CuFeS₂ formula unit (atomic weights Cu 63.5, Fe 55.8, S 32.1: $63.5/(63.5+55.8+2\times32.1)\times100=34.60\%$), confirming the given figure.

1(v) — Economic efficiency and net operating profit (9 marks)

Given.

Find. (a) Economic efficiency, %; (b) net operating profit, USD million/year.

Approach. (b) is a direct revenue-minus-costs calculation on the actual operation. (a) uses Taggart's classical economic (recovery) efficiency: the value of the products actually realized per tonne of crude ore, against the value obtainable per tonne of ore by a theoretically perfect concentration (100% recovery, zero gangue – i.e. a pure-chalcopyrite concentrate), both valued net of the concentrate-handling charges (freight, smelting) that scale with concentrate tonnage.

1. (b) Actual daily revenue. Copper paid on is the copper actually recovered to concentrate, $RFf$: $$\dot m_{Cu}=0.80\times30{,}000\ \text{t/d}\times0.0031 = 74.4\ \text{t/d} = 74{,}400\ \text{kg/d}$$ $$\text{Revenue}=74{,}400\times\$7=\$520{,}800/\text{d}$$
2. (b) Actual daily costs. Mining and milling scale with ore; freight and smelting scale with the 265.7 t/d concentrate from 1(iii): $$\$90{,}000\ (\text{mining}) + \$120{,}000\ (\text{milling}) + \$39{,}857\ (\text{freight}) + \$66{,}429\ (\text{smelting}) = \$316{,}286/\text{d}$$
3. (b) Net operating profit. $$\text{Profit/d}=520{,}800-316{,}286=\$204{,}514/\text{d}\ \Rightarrow\ \boxed{\$71.6\ \text{million/year}}\ (\times 350\ \text{d/yr})$$
4. (a) Value of products actually realized, per tonne of ore. Net of freight and smelting only (mining/milling are incurred regardless of metallurgical performance and are common to both scenarios compared below): $$\frac{520{,}800-39{,}857-66{,}429}{30{,}000}=\$13.82/\text{t ore}$$
5. (a) Value by perfect concentration, per tonne of ore. A theoretically perfect separation recovers all the feed copper ($R=100\%$) into a gangue-free, pure-chalcopyrite concentrate ($c=34.6\%\ \text{Cu}$, from 1(iv)); the same USD 7/kg payment and freight/smelting tariffs apply: $$Y_{\text{perfect}}=\frac{f}{34.6}=0.008958,\quad C_{\text{perfect}}=30{,}000\times0.008958=268.7\ \text{t/d}$$ $$\text{Cu paid}=1.0\times30{,}000\times0.0031\times1000=93{,}000\ \text{kg/d}\ \Rightarrow\ \text{Revenue}=\$651{,}000/\text{d}$$ $$\text{Value}=\frac{651{,}000-150(268.7)-250(268.7)}{30{,}000}=\$18.12/\text{t ore}$$
6. (a) Economic efficiency. $$E=\frac{13.82}{18.12}\times100=\boxed{76.3\%}$$

1(vi) — Sample weight by Gy's equation (4 marks)

Given. Ore ground to 95% passing $d = 200\ \mu\text{m} = 0.02\ \text{cm}$ (taken as the top particle size in Gy's formula); sampling constant $C = 60\ \text{g/cm}^3$; allowable sampling error $\pm0.01\%\ \text{Cu}$ at 95% confidence; feed grade $f=0.31\%\ \text{Cu}$ (1(ii)). Gy's equation: $M = C\,d^3/s^2$, $M$ in grams, $d$ in cm, $s$ the fractional (relative) standard deviation of the sampling error.

Find. Minimum sample mass $M$.

1. Convert the confidence interval to a fractional standard deviation. "95% of the time" is taken as the usual $\pm2\sigma$ engineering approximation, so the 1-sigma absolute error is $0.01\%/2=0.005\%\ \text{Cu}$; expressed as a fraction of the feed grade, $$s=\frac{0.005}{0.31}=0.01613$$
2. Apply Gy's equation. $$M=\frac{C\,d^3}{s^2}=\frac{60\times(0.02)^3}{(0.01613)^2}=\frac{4.80\times10^{-4}}{2.60\times10^{-4}}=\boxed{1.85\ \text{g}}$$

A sample of under 2 g is required at this fine top size (200 μm) – consistent with Gy sampling theory, where the required mass falls with the cube of particle size, so a fully ground assay pulp needs only a small sub-sample while a run-of-mine sample at cm-scale top size would need kilograms to tonnes for the same relative precision.