Two-Stage Grinding Circuit – Circulating Load, Water Balance, Bond Power: 24-MMP-A3 Mineral Processing, Undated paper, Q2

Question 2: Two-Stage Grinding Circuit – Circulating Load, Water Balance, Bond Power (26 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 feed $F=100$ t/h solids enters the rod mill (open circuit); its discharge is the new feed to the closed ball-mill/hydrocyclone loop. Sump dilution water contains no solids, so it changes each stream's %-solids but not its dry size-distribution weight fraction. Circuit data as tabulated above.

Fig. 2 – Grinding circuit: rod mill discharge (fresh feed) and cyclone underflow (recycle) combine ahead of the ball mill; ball mill discharge reports to the sump with dilution water, then to the cyclone via the pump.

Find. (1) Circulating load, $CL$ (%); (2) dilution water added at the sump (t/h); (3) ball mill power (kW); (4) SG of the cyclone underflow slurry.

Approach. Dilution water carries no solids, so the ball mill discharge's own wt%-passing-74µm equals the cyclone FEED's size split; apply the two-product formula around the cyclone for the circulating load, then convert every stream's solids tonnage to slurry/water tonnage via its own %-solids for the sump water balance. For the Bond power, use the circuit's overall fresh feed size ($F_{80}=$ rod mill discharge) and overall product size ($P_{80}=$ cyclone overflow) – Bond's specific energy applies to the fresh feed rate, not the gross recirculating tonnage.

(1) Circulating load. Two-product formula around the hydrocyclone, using %-74µm as the size tracer (feed $=$ ball mill discharge $=40\%$, overflow $=75\%$, underflow $=25\%$): $$CL=\frac{o-f}{f-u}\times100=\frac{75-40}{40-25}\times100=\boxed{233.3\%}$$ So underflow $U=2.333\times100=233.3$ t/h solids (overflow $O=100$ t/h $=$ fresh feed, confirming steady state).
(2) Dilution water. Ball mill discharge (= cyclone feed) solids $=F+U=100+233.3=333.3$ t/h; convert each stream to slurry/water using its sampled %-solids: $$\text{Ball mill discharge: }\frac{333.3}{0.77}=432.9\ \text{t/h slurry}\ \Rightarrow\ 99.6\ \text{t/h water}$$ $$\text{Overflow: }\frac{100}{0.35}=285.7\ \text{t/h slurry}\ \Rightarrow\ 185.7\ \text{t/h water}$$ $$\text{Underflow: }\frac{233.3}{0.75}=311.1\ \text{t/h slurry}\ \Rightarrow\ 77.8\ \text{t/h water}$$ The sump combines ball mill discharge with dilution water to form the cyclone feed (overflow $+$ underflow); water balance around the sump/cyclone loop gives: $$\text{Dilution water}=(185.7+77.8)-99.6=\boxed{163.9\ \text{t/h}}$$
(3) Ball mill power (Bond's Third Theory). The circuit's overall fresh feed size is the rod mill discharge ($F_{80}=900\ \mu\text{m}$) and its overall product size is the cyclone overflow ($P_{80}=81\ \mu\text{m}$); Bond's specific energy is per tonne of FRESH feed, applied with no circulating-load correction: $$W=10W_i\left(\frac{1}{\sqrt{P_{80}}}-\frac{1}{\sqrt{F_{80}}}\right)=10(14)\left(\frac{1}{\sqrt{81}}-\frac{1}{\sqrt{900}}\right)$$ $$=140(0.1111-0.03333)=\boxed{10.89\ \text{kWh/t}}$$ $$P=W\times F=10.89\times100=\boxed{1089\ \text{kW}}$$
(4) SG of the cyclone underflow slurry. At 75% solids by weight (ore SG 3.0, water SG 1.0), the slurry SG follows from the reciprocal (volume-additive) mixing rule: $$\frac{1}{SG_{\text{slurry}}}=\frac{0.75}{3.0}+\frac{0.25}{1.0}=0.25+0.25=0.50\ \Rightarrow\ SG_{\text{slurry}}=\boxed{2.00}$$

Quantity	Value
(1) Circulating load	233.3%
(2) Dilution water added	163.9 t/h
(3) Ball mill power	1089 kW
(4) SG of cyclone underflow slurry	2.00

Topic. Circulating load, sump/cyclone water balance, Bond's Third Theory of Comminution, slurry specific gravity.

Key relations. $CL=100(o-f)/(f-u)$ (two-product, size tracer); slurry $=$ solids$/$(wt% solids); $W=10W_i(1/\sqrt{P_{80}}-1/\sqrt{F_{80}})$; $1/SG_{\text{slurry}}=x_s/SG_s+(1-x_s)/SG_w$.

Why this works. Water is inert to the size-classification tracer, so any stream's dry %-passing value survives dilution unchanged – this is what lets the ball mill discharge's assay stand in for the cyclone feed assay even though the sump adds water in between. Bond's energy law is calibrated to fresh (net) production, not gross mill throughput, because the extra tonnage recirculating through the mill has already "paid" its grinding energy on a previous pass.

Common pitfall. Multiplying $W$ by the gross ball-mill throughput (333.3 t/h) instead of the fresh feed rate (100 t/h) – over-predicts power by the $(1+CL)$ factor; or using the ROD MILL FEED size (4000 µm) instead of its DISCHARGE size (900 µm) as $F_{80}$ for the ball-mill/cyclone loop, which is the loop's own fresh feed, not the whole-circuit raw feed.

Source. Wills & Finch, Wills' Mineral Processing Technology, 8th ed., Ch. 6 (Bond's Third Theory) and Ch. 9 (classification, circulating load).

Question 2 of 3: Two-Stage Grinding Circuit – Circulating Load, Water Balance, Bond Power

Question 2: Two-Stage Grinding Circuit – Circulating Load, Water Balance, Bond Power (26 marks)