Question 13 of 13: Net Smelter Return for Four Metals

Question 7: Net Smelter Return for Four Metals (15 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.

Stope sample assay: 0.2% Cu, 0.8% Zn, 0.30 g/t Ag, 0.02 g/t Au. 1 troy ounce = 31.1035 g.

Find. An NSR formula ($ per unit grade per tonne of ore) for each metal, then the individual-metal and total NSR of the stope sample.

Approach. For the two base metals (Cu, Zn), each tonne of ore's payable value is netted against the smelting/transport charge on the concentrate tonnage that one tonne of ore actually produces; for the precious metals (Ag, Au), which carry no separate smelting charge, NSR is simply the payable metal value. Deriving each result as a $/unit-grade factor lets it be applied directly to any assay, including the stope sample.

1. Copper NSR factor (per 1% Cu grade). Payable Cu value per tonne of ore at 1% grade: 0.01×0.60×$\$7,500$ = $\$45.00$/t ore. Concentrate produced per tonne of ore at 1% grade: (0.01×0.60)/0.20 = 0.03000 t conc./t ore. Smelting/transport charge: 0.03000×$\$185$ = $\$5.55$/t ore. $$\text{NSR}_{Cu}=\$45.00-\$5.55=\boxed{\$39.45\ \text{per \%Cu, per t ore}}$$
2. Zinc NSR factor (per 1% Zn grade). Payable Zn value: 0.01×0.80×$\$1,875$ = $\$15.00$/t ore. Concentrate produced: (0.01×0.80)/0.52 = 0.015385 t conc./t ore. Charge: 0.015385×$\$350$ = $\$5.3846$/t ore. $$\text{NSR}_{Zn}=\$15.00-\$5.3846=\boxed{\$9.6154\ \text{per \%Zn, per t ore}}$$
3. Silver NSR factor (per 1 g/t Ag grade). Payable Ag at 1 g/t and 14% recovery: 1×0.14 = 0.14 g/t ore = 0.14/31.1035 = 0.0045022 troz/t ore. No smelting charge. $$\text{NSR}_{Ag}=0.0045022\times\$28=\boxed{\$0.12603\ \text{per g/t Ag, per t ore}}$$
4. Gold NSR factor (per 1 g/t Au grade). Payable Au at 1 g/t and 55% recovery: 1×0.55 = 0.55 g/t ore = 0.55/31.1035 = 0.017682 troz/t ore. No smelting charge. $$\text{NSR}_{Au}=0.017682\times\$1{,}600=\boxed{\$28.293\ \text{per g/t Au, per t ore}}$$
5. Applying the factors to the stope sample. Cu: 39.45×0.2=$\$7.890$/t ore. Zn: 9.6154×0.8=$\$7.692$/t ore. Ag: 0.12603×0.30=$\$0.0378$/t ore. Au: 28.293×0.02=$\$0.5659$/t ore.
6. Total NSR of the sample. $$\text{NSR}_{\text{total}}=7.890+7.692+0.038+0.566=\boxed{\$16.186\ \text{per t ore}}$$