Question 1 of 3: Copper Flotation Circuit – Flowsheet, Metallurgical Balance and Economics
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Exam — Mining and Mineral Processing Engineering, 09-MMP-A3 Mineral Processing (National Exams May 2019). 3 hours duration, closed book; only an approved Casio or Sharp calculator permitted. Three questions constitute a complete exam paper (100 marks total).
Reference texts: Wills & Finch, Wills' Mineral Processing Technology, 8th ed. (comminution and Bond's Third Theory – Ch. 6; classification and hydrocyclones – Ch. 9; froth flotation, cells and flotation columns – Ch. 12; metallurgical balances, recovery/economic efficiency – Ch. 1 & 12; particle characterization, sedimentation and gravity concentration – Ch. 1 & 10); Taggart, Handbook of Mineral Dressing (classical definition of metallurgical/economic efficiency); SME Mining Engineering Handbook, 3rd ed. (porphyry copper mill flowsheets); BC Health, Safety and Reclamation Code for Mines for the Canadian regulatory context.
Approach. Trace the description arrow by arrow: three parallel rougher banks combine on concentrate, feed a closed-circuit regrind mill/cyclone, gravity-feed a column surge tank, then two columns in series with a scavenger stage recycling its own concentrate back to the surge tank ahead of the columns.
Fig. 1 – Copper flotation circuit: three rougher banks combine on concentrate into a closed-circuit regrind mill/cyclone, gravity to a column surge tank, two columns in series to the concentrate thickener, and a scavenger stage recycling its concentrate to the surge tank while its tails report to final tailings.
Check – the source text does not state where rougher-bank TAILINGS report; only rougher CONCENTRATE routing is described. The sketch above follows the only tailings path the text actually specifies (the column-scavenger loop); rougher tailings are the other rejection point in a real circuit of this type and would typically join the same final tailings line.
(2)–(6) Metallurgical balance and economics
Given. Feed grade $f=0.31\%\text{ Cu}$; recovery $R=80\%$; concentrate grade $c=28\%\text{ Cu}$ (stated in the circuit description); mill throughput $30{,}000$ t/d of ore; chalcopyrite CuFeS₂ with atomic masses Cu 63.5, Fe 55.8, S 32; mining $\$3.00$/t ore, milling $\$4.00$/t ore, concentrate freight $\$150$/t, smelting $\$250$/t, 350 operating days/yr, copper payable at $\$7$/kg contained in the concentrate.
Given data
Quantity
Symbol
Value
Feed grade
$f$
0.31% Cu
Recovery
$R$
80%
Concentrate grade
$c$
28% Cu
Mill throughput
$F$
30,000 t/d
Atomic masses
Cu, Fe, S
63.5, 55.8, 32
Mining / milling cost
\$3.00 / \$4.00 per t ore
Freight / smelting
\$150 / \$250 per t conc.
Operating days/yr
350
Copper payment
\$7/kg contained Cu
Find. (2) tailings grade $t$; (3) concentrate tonnes/day; (4) %Cu in pure chalcopyrite; (5) % chalcopyrite in the concentrate; (6)(i) Economic Efficiency (%); (6)(ii) operating profit ($M/yr).
Approach. Apply the two-product (recovery) formula to get the mass yield and tailings grade, scale by the daily throughput for the concentrate tonnage, use chalcopyrite stoichiometry for the mineral content, then build a full revenue/cost statement and compare it (per Taggart's classical definition) against a theoretically perfect concentration to obtain the Economic Efficiency.
Mass yield and tailings grade. Recovery $R=\dfrac{c(f-t)}{f(c-t)}$ rearranges, using the mass-yield form $Y=\dfrac{Rf}{c}$:
$$Y=\frac{Rf}{c}=\frac{0.80\times0.31}{28}=0.008857\ (0.886\%\text{ of feed mass})$$
$$t=\frac{f(1-R)}{1-Y}=\frac{0.31(1-0.80)}{1-0.008857}=\boxed{0.0626\%\ \text{Cu}}$$
Cross-check with the standard two-product form $Y=(f-t)/(c-t)=(0.31-0.0626)/(28-0.0626)=0.00886$ – matches.
Concentrate produced per day.
$$C=F\cdot Y=30{,}000\times0.008857=\boxed{265.7\ \text{t/d}}$$
%Cu in pure chalcopyrite (CuFeS₂). Molar mass $M=63.5+55.8+2(32)=183.3$ g/mol:
$$\%\text{Cu}=\frac{63.5}{183.3}\times100=\boxed{34.64\%}$$
% chalcopyrite in the concentrate. The concentrate assays 28% Cu; if chalcopyrite is the only copper mineral, the concentrate's copper is diluted pure-chalcopyrite:
$$\%\text{CuFeS}_2=\frac{c}{\%\text{Cu}_{\text{ccp}}}\times100=\frac{28}{34.64}\times100=\boxed{80.8\%}$$
(6)(ii) Operating profit, million \$/yr.
$$\$204{,}514/\text{d}\times350\ \text{d/yr}=\$71.58\times10^6/\text{yr}\approx\boxed{\$71.6\ \text{million/yr}}$$
(6)(i) Economic Efficiency (Taggart). E.E. compares the value actually realized per tonne of ore (net of freight and smelting, but not mining/milling, per Taggart's classical definition) against the value obtainable from a theoretically perfect concentration (100% recovery into pure, gangue-free chalcopyrite, same smelter terms). Actual case, per tonne of ore:
$$V_{\text{actual}}=\frac{\text{Revenue}-\text{Freight}-\text{Smelting}}{F}$$
$$=\frac{520{,}800-39{,}857-66{,}429}{30{,}000}=\boxed{\$13.82/\text{t ore}}$$
Perfect-concentration case: all 0.31% Cu reports to a pure-chalcopyrite concentrate ($Y_{\text{perfect}}=f/\%\text{Cu}_{\text{ccp}}=0.31/34.64=0.008948$, $C_{\text{perfect}}=30{,}000\times0.008948=268.5$ t/d, Cu recovered $=93.0$ t/d $=93{,}000$ kg/d):
$$V_{\text{perfect}}=\frac{93{,}000\times7-150\times268.5-250\times268.5}{30{,}000}=\boxed{\$18.12/\text{t ore}}$$
$$\text{E.E.}=\frac{V_{\text{actual}}}{V_{\text{perfect}}}\times100=\frac{13.82}{18.12}\times100=\boxed{76.3\%}$$
