Question 1 of 6: Highland Valley Copper — Metallurgical Balance

Question 1: Highland Valley Copper — Metallurgical Balance (24 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.

Value mineral: bornite, $\text{Cu}_5\text{FeS}_4$. Atomic masses: Cu 63.5, Fe 55.8, S 32.0 g/mol.

Approach. All six parts follow from the classical two-product (feed = concentrate + tailings) mass and metal balance built on the three assays; the yield of concentrate to feed, $Y=C/F$, is the pivot value that (2)–(4) and (6) all build on.

1(1) — Enrichment ratio (3 marks)

1. Ratio of concentrate to feed grade. $$ER=\frac{c}{f}=\frac{41.4}{0.388}=\boxed{106.7}$$

1(2)/1(3) — Concentration ratio and weight % concentrate produced (6 marks)

1. Two-product yield. The mass fraction of feed reporting to concentrate is fixed by the three assays: $$Y=\frac{C}{F}=\frac{f-t}{c-t}=\frac{0.388-0.033}{41.4-0.033}=\frac{0.355}{41.367}=0.008582$$ so the weight % of Cu concentrate produced is $\boxed{0.858\%}$ of the feed mass.
2. Concentration ratio. $K=F/C$ is the reciprocal of the yield: $$K=\frac{1}{Y}=\frac{1}{0.008582}=\boxed{116.5}$$

1(4) — Copper recovery into the concentrate (3 marks)

1. Metal-balance recovery. $$R=\frac{Cc}{Ff}=Y\cdot\frac{c}{f}=\frac{c(f-t)}{f(c-t)}\times100=\frac{41.4\times0.355}{0.388\times41.367}\times100=\boxed{91.57\%}$$

1(5) — Bornite content of the concentrate (6 marks)

1. Molar mass and Cu fraction of bornite. $$M_{\text{Cu}_5\text{FeS}_4}=5(63.5)+55.8+4(32.0)=317.5+55.8+128.0=501.3\ \text{g/mol}$$ $$\%\text{Cu in bornite}=\frac{5\times63.5}{501.3}\times100=\frac{317.5}{501.3}\times100=63.33\%$$
2. Bornite content of the concentrate. Since Cu is assumed to be present only as bornite, the concentrate's bulk Cu assay divided by bornite's own Cu assay gives the mineral fraction: $$\%\text{bornite}=\frac{c}{63.33}\times100=\frac{41.4}{63.33}\times100=\boxed{65.4\%}$$

The remaining ≈35% of the concentrate mass is gangue reporting with the floated bornite – a normal figure for a real (imperfectly selective) copper flotation concentrate.

1(6) — Separation efficiency (6 marks)

1. Recovery of gangue mineral into the concentrate. Separation efficiency is a MINERAL balance, so "gangue" is everything that is not bornite (bornite's own Fe and S belong to the value mineral). With $m=63.34\%$ Cu in pure bornite from 1(5), the feed carries $0.388/63.34\times100=0.613\%$ bornite, i.e. $99.387\%$ gangue, and the concentrate carries $100-65.37=34.63\%$ gangue: $$R_{g}=Y\cdot\frac{100-\%\text{bornite}_c}{100-\%\text{bornite}_f}\times100=0.008582\times\frac{34.63}{99.387}\times100=0.299\%$$
2. Schuhmann separation efficiency. $SE$ is the recovery of the valuable mineral less the recovery of gangue mineral into the same product (every Cu atom is in bornite, so bornite recovery equals Cu recovery): $$SE=R_{m}-R_{g}=91.57-0.30=\boxed{91.3\%}$$ Cross-check with Wills' closed form: $SE=\dfrac{100\,m\,Y\,(c-f)}{(m-f)\,f}=\dfrac{100(63.34)(0.008582)(41.012)}{(62.95)(0.388)}=91.27\%$.