Question 7 of 8: Gold flotation-roast-cyanidation flowsheet sizing

Question 7: Gold flotation-roast-cyanidation flowsheet sizing (bonus)

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. Plant feed 40,000 t/day, ore SG 3.1, calcine SG 2.93. Rougher feed = plant feed × (1 + 25% scavenger circulating load); rougher slurry 40% solids by wt.; rougher retention 12 min; cell size 100 m³, 20% dead volume. Calcine (cyanidation feed) = 22.5% of plant feed by wt. = 9,000 t/day (consistent with 30% flotation recovery × 75% roasting mass retained); cyanidation slurry 50% solids by wt.; retention 25 h; cylindrical tank 14 m diameter × 12 m height, 20% dead volume, operated at 85% of full volume.

Find. (a) Number of rougher flotation cells. (b) Number of cyanidation tanks.

Approach. For each vessel, compute the slurry volumetric flow rate from the solids tonnage, SG and % solids, multiply by the stated retention time for the required active (working) volume, then divide by each vessel's own usable (post-dead-space) volume and round up.

1. Rougher feed and slurry volume. Rougher feed tonnage includes the scavenger circulating load: $$\dot m_{\text{rougher}}=40{,}000\times(1+0.25)=\boxed{50{,}000\ \text{t/day (solids)}}$$ At 40% solids by weight (SG$_\text{ore}$ = 3.1, water SG = 1.0), solids and water volumes: $$V_{\text{solids}}=\frac{50{,}000}{3.1}=16{,}129\ \text{m}^3/\text{day}\qquad V_{\text{water}}=50{,}000\left(\frac{1}{0.40}-1\right)=75{,}000\ \text{m}^3/\text{day}$$ $$V_{\text{slurry}}=16{,}129+75{,}000=91{,}129\ \text{m}^3/\text{day}=63.28\ \text{m}^3/\text{min}$$
2. Required active rougher volume and cell count. $$V_{\text{active}}=63.28\ \text{m}^3/\text{min}\times12\ \text{min}=759.4\ \text{m}^3$$ Usable volume per 100 m³ cell (20% occupied by agitator/aeration): $100\times0.80=80\ \text{m}^3$. $$n_{\text{cells}}=\left\lceil\frac{759.4}{80}\right\rceil=\left\lceil9.49\right\rceil=\boxed{10\ \text{cells}}$$
3. Cyanidation feed and slurry volume. Calcine feed $=40{,}000\times0.225=9{,}000\ \text{t/day}$ (consistent with 30% flotation recovery less 25% roasting mass loss: $40{,}000\times0.30\times(1-0.25)=9{,}000\ \text{t/day}$). At 50% solids by weight (SG$_\text{calcine}$ = 2.93): $$V_{\text{solids}}=\frac{9{,}000}{2.93}=3{,}071.7\ \text{m}^3/\text{day}\qquad V_{\text{water}}=9{,}000\left(\frac{1}{0.50}-1\right)=9{,}000\ \text{m}^3/\text{day}$$ $$V_{\text{slurry}}=3{,}071.7+9{,}000=12{,}071.7\ \text{m}^3/\text{day}=503.0\ \text{m}^3/\text{h}$$
4. Required active cyanidation volume. $$V_{\text{active}}=503.0\ \text{m}^3/\text{h}\times25\ \text{h}=\boxed{12{,}574.7\ \text{m}^3}$$
5. Tank usable volume and count. Full cylindrical tank volume, then remove BOTH the 20% agitator/aeration dead space and operate at only 85% of full volume: $$V_{\text{tank}}=\pi\left(\frac{14}{2}\right)^2(12)=1{,}847.3\ \text{m}^3$$ $$V_{\text{usable}}=1{,}847.3\times(1-0.20)\times0.85=\boxed{1{,}256.1\ \text{m}^3}$$ $$n_{\text{tanks}}=\left\lceil\frac{12{,}574.7}{1{,}256.1}\right\rceil=\left\lceil10.01\right\rceil=\boxed{11\ \text{tanks}}$$ (The ratio lands at 10.01 — just barely over 10 — so a whole 11th tank is genuinely required; ten tanks alone (12,561 m³) fall 14 m³ short of the 12,575 m³ needed.)