Question 2 of 7: Two-stage grinding circuit material balance

Problem 2 — Two-stage grinding circuit material 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.

Fresh ore feed 100 t/h (SG 3.0); rod mill in open circuit (no return stream to the rod mill), so RMD = fresh feed = 100 t/h solids; ball mill fed by RMD + recycled cyclone underflow $U$; ball mill discharge (BMD) is the cyclone feed.

Find. (a)(i)–(ii) solids flow in the cyclone overflow and underflow; (b) dilution water added at the sump; (c) SG of the cyclone underflow slurry; (d) two operator actions that would sharpen the product.

Approach. Because the rod mill is open-circuit, its discharge equals the fresh feed rate, and because the cyclone underflow never leaves the ball-mill/cyclone loop, the cyclone overflow must equal the fresh feed at steady state. The circulating load (cyclone underflow) is then found from a two-product balance on the "% -200 mesh" assay across the cyclone. Water flows follow from each stream's own percent-solids figure.

1. (a)(i)-(ii) Solids balance and circulating load. Steady-state solids balance on the whole circuit: fresh feed = final product, so the cyclone overflow $O=100$ t/h. The cyclone feed is the ball mill discharge (BMD); using the "% -200 mesh" assay as the balance component across the cyclone split, with feed assay 20 %, overflow assay 30 %, underflow assay 15 %: $$(O+U)(20)=O(30)+U(15)\ \Rightarrow\ U(20-15)=O(30-20)\ \Rightarrow\ U=2O.$$ With $O=100$ t/h, $$U=\boxed{200\ \text{t/h (cyclone underflow)}},\qquad O=\boxed{100\ \text{t/h (cyclone overflow)}}.$$ Check: BMD solids $=O+U=300$ t/h must equal RMD $+U=100+200=300$ t/h — consistent, confirming the balance.
2. (b) Dilution water at the sump. Water entering the sump with the ball mill discharge: BMD solids 300 t/h at 75 % solids $\Rightarrow$ total slurry $=300/0.75=400$ t/h, so water in BMD $=400-300=100$ t/h. Water leaving the sump (i.e. entering the cyclone) is found from the two product streams' own percent solids: overflow slurry $=100/0.40=250$ t/h (water $=150$ t/h); underflow slurry $=200/0.75=266.7$ t/h (water $=66.7$ t/h). Total water after the split $=150+66.7=216.7$ t/h. A water balance around the sump gives the dilution addition: $$W_{dilution}=216.7-100=\boxed{116.7\ \text{t/h}}.$$
3. (c) SG of the cyclone underflow slurry. With solids SG $=3.0$, water SG $=1.0$, and the underflow's own 75 % solids / 25 % water by weight: $$\dfrac{1}{SG_{slurry}}=\dfrac{0.75}{3.0}+\dfrac{0.25}{1.0}=0.25+0.25=0.50\ \Rightarrow\ SG_{slurry}=\boxed{2.0}.$$
4. (d) Producing a finer product. Two independent operator actions that shift the cyclone's cut point finer (and hence sharpen/finer the overflow product): reduce the feed (throughput) rate to the ball mill, increasing residence time and grinding each particle longer before it reports to the cyclone; and reduce the cyclone feed density (add more dilution water) or feed pressure/vortex-finder diameter, which lowers the cyclone's $d_{50}$ cut size and sends more fines to the overflow at the expense of a coarser, larger-volume underflow recycle.