Question 5 of 6: Copper Flotation Circuit — Rougher/Cleaner Mass Balance

Question 5: Copper Flotation Circuit — Rougher/Cleaner Mass Balance (20 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. Circuit feed $F=100\ \text{t/h}$ dry solids at $0.5\%\ \text{Cu}$; assays and % solids for all five sampled streams as tabulated above.

Find. Circuit and Cleaner Cu recovery; Cleaner Tailings (recycle) solids tonnage; Rougher Cu recovery; dilution water tonnage.

Approach. Work outward from the whole-circuit two-product balance (Feed = Final Concentrate + Final Tailings) to fix the two unknown product tonnages, then close a second two-product balance around the Cleaner node alone (Rougher Concentrate = Final Concentrate + Cleaner Tailings) to fix the internal recycle stream; the Rougher recovery and dilution-water rate then follow from mass and Cu balances written explicitly around each node. Every intermediate result is cross-checked by closing the Rougher node's own mass and Cu balance.

[Figure not reproduced: Figure 5.1 — Rougher/Cleaner flotation circuit with Cleaner Tailings recycled to the Rougher feed, redrawn from the described layout. See the official exam paper.]

5(1) — Copper recovery in the circuit (2 marks)

1. Whole-circuit two-product balance. Mass: $F=FC+FT$; copper: $F\,c_F=FC\,c_{FC}+FT\,c_{FT}$, where $FC,FT$ are the Final Concentrate/Tailings dry-solids tonnages. Substituting $FT=F-FC$ and solving: $$FC=\frac{F(c_F-c_{FT})}{c_{FC}-c_{FT}}=\frac{100(0.5-0.1)}{25.0-0.1}=\frac{40.0}{24.9}=1.606\ \text{t/h}$$ $$FT=F-FC=100-1.606=98.39\ \text{t/h}$$
2. Circuit recovery. $$R_{\text{circuit}}=\frac{FC\cdot c_{FC}}{F\cdot c_F}\times100=\frac{1.606\times25.0}{100\times0.5}\times100=\boxed{80.3\%}$$

5(2) — Copper recovery in the Cleaner (2 marks)

1. Cleaner-node two-product balance. Mass: $RC=FC+CT$; copper: $RC\,c_{RC}=FC\,c_{FC}+CT\,c_{CT}$, where $RC$ = Rougher Concentrate and $CT$ = Cleaner Tailings tonnage. Substituting $RC=FC+CT$: $$CT=\frac{FC(c_{FC}-c_{RC})}{c_{RC}-c_{CT}}=\frac{1.606(25.0-13.9)}{13.9-5.0}=\frac{17.83}{8.9}=2.004\ \text{t/h}$$ $$RC=FC+CT=1.606+2.004=3.610\ \text{t/h}$$
2. Cleaner recovery. $$R_{\text{Cleaner}}=\frac{FC\cdot c_{FC}}{RC\cdot c_{RC}}\times100=\frac{1.606\times25.0}{3.610\times13.9}\times100=\boxed{80.0\%}$$

5(3) — Solids circulated back to the Rougher, i.e. the Cleaner Tailings (8 marks)

1. Read directly from the Cleaner-node balance in 5(2). $CT$ was solved above from the mass and Cu balance around the Cleaner alone: $$\boxed{CT=2.00\ \text{t/h dry solids}}$$
2. Consistency check at the Rougher node. Mass balance: Rougher feed $=F+CT=100+2.00=102.00\ \text{t/h}$, which should equal Rougher outputs $RC+FT=3.610+98.39=102.00\ \text{t/h}$ – balances. Cu balance: Rougher feed Cu $=F\,c_F+CT\,c_{CT}=100(0.005)+2.00(0.05)=0.600\ \text{t/h Cu}$, versus Rougher output Cu $=RC\,c_{RC}+FT\,c_{FT}=3.610(0.139)+98.39(0.001)=0.600\ \text{t/h Cu}$ – also balances, confirming the two two-product solutions (5(1) and 5(2)) are mutually consistent.

5(4) — Copper recovery in the Rougher (3 marks)

1. Rougher recovery from the node balance in 5(3). Recovery is Rougher-concentrate Cu over total Rougher-feed Cu (Circuit Feed + recycled Cleaner Tailings): $$R_{\text{Rougher}}=\frac{RC\cdot c_{RC}}{F\,c_F+CT\,c_{CT}}\times100=\frac{3.610\times13.9}{0.600}\times100=\boxed{83.6\%}$$

Note that $R_{\text{Rougher}}\times R_{\text{Cleaner}}/100=83.6\times80.0/100=66.9\%\ne R_{\text{circuit}}=80.3\%$ – stage recoveries do not multiply directly in a circuit with an internal recycle, because the Cleaner Tailings gets a second pass through the Rougher rather than being permanently lost; only the closed-loop node balances above give the correct circuit-level figure.

5(5) — Dilution water added to the Cleaner (5 marks)

1. Convert each Cleaner-node stream's dry solids to accompanying water using $W=\text{solids}\times(100-\%\text{solids})/\%\text{solids}$: $$W_{RC}=3.610\times\frac{100-50.0}{50.0}=3.610\ \text{t/h}\qquad W_{FC}=1.606\times\frac{100-40.0}{40.0}=2.410\ \text{t/h}$$ $$W_{CT}=2.004\times\frac{100-8.6}{8.6}=21.30\ \text{t/h}$$
2. Water balance around the Cleaner. Water in (Rougher Conc. + Dilution) = water out (Final Conc. + Cleaner Tailings): $$W_{\text{dilution}}=W_{FC}+W_{CT}-W_{RC}=2.410+21.30-3.610=\boxed{20.1\ \text{t/h}}$$

Cross-check on the whole circuit: Circuit Feed carries $W_F=100\times(100-33.3)/33.3=200.3\ \text{t/h}$ water, so total water in $=W_F+W_{\text{dilution}}=200.3+20.1=220.4\ \text{t/h}$; total water out $=W_{FC}+W_{FT}=2.41+98.39\times(100-31.1)/31.1=2.41+217.98=220.4\ \text{t/h}$ – the two match, confirming the dilution-water figure.