Question 18 of 29: Porphyry Copper NSR Chain Calculation

Question 4.5: Porphyry Copper NSR Chain Calculation (6 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. 4.5.1–4.5.6 as listed above (1 mark each).

Approach. Deduct the 1-unit treatment loss from the assayed concentrate grade to get the PAYABLE grade, price the payable copper content to get gross value, subtract refining/treatment/freight charges (converting the treatment charge from a per-tonne to a per-short-ton basis) to get NSR per short ton of concentrate, then convert to an ore basis via the concentration factor derived from the recovered ore grade.

1. 4.5.1 Recovered ore grade. Head grade × mill recovery: $$\%Cu_{recovered}=0.70\%\times0.90=\boxed{0.63\%\,Cu}$$ (the copper mass that actually reports to concentrate, per tonne of ORE milled).
2. 4.5.2 Gross value of concentrate. Deduct the 1-unit treatment loss from the 25% assay to get the PAYABLE grade: $25\%-1\%=24\%$. Payable copper per short ton (2000 lb) of concentrate: $0.24\times2000=480\ \text{lb Cu}$. Gross value at the USD 2.75/lb metal price: $$480\ \text{lb}\times2.75\ \text{USD/lb}=\boxed{1320\ \text{USD/short ton concentrate}}$$
3. 4.5.3 Refining charge (paid metal content). Refining is charged only on the 480 lb of PAID (payable) copper per short ton: $$480\ \text{lb}\times0.24\ \text{USD/lb}=\boxed{115.20\ \text{USD/short ton concentrate}}$$
4. 4.5.4 Net Smelter Return of the mine (per short ton concentrate). The treatment charge is quoted per TONNE (metric); converting to a short-ton basis (1 short ton = 0.90718 tonne): $255\ \text{USD/t}\times0.90718\ \text{t/ST}=231.33\ \text{USD/ST}$. NSR = gross value − treatment charge − refining charge − freight: $$\text{NSR}=1320-231.33-115.20-60=\boxed{913.47\ \text{USD/short ton concentrate}}$$
5. 4.5.5 Concentration factor. Copper mass balance: 0.63% recovered grade × 2000 lb/ST ore = 12.6 lb Cu per short ton of ORE; concentrate carries 25%×2000 = 500 lb Cu per short ton of CONCENTRATE (full assay basis, mass-balance not payable-deduction basis). Tons of concentrate produced per ton of ore: $12.6/500=0.0252$; inverting gives the concentration factor: $$\frac{1}{0.0252}=\boxed{39.68\ \text{short tons ore per short ton concentrate}}$$
6. 4.5.6 Net Smelter Return of the ore. Divide the per-concentrate-tonne NSR by the concentration factor: $$\text{NSR}_{ore}=\frac{913.47}{39.68}=\boxed{23.02\ \text{USD/short ton ore}}$$