Two-Stage Grinding Circuit Material Balance: 24-MMP-A3 Mineral Processing, December 2017, Q2

Question 2: Two-Stage Grinding Circuit Material Balance (18 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.

Figure 2.1 — Circuit layout for Question 2 (Figure 1 of the source paper): rod mill (open circuit) feeds the ball mill together with the recycled cyclone underflow; ball mill discharge is diluted in the sump, pumped, and classified – overflow is the finished product, underflow recycles.

2(i) — Solids flowrate in the cyclone underflow (7 marks)

Given. %−200 mesh: rod mill discharge 5%, ball mill discharge 20%, cyclone overflow 30%, cyclone underflow 15%; new (fresh) feed $F_{new}=100\ \text{t/h}$.

Find. Solids flowrate in the cyclone underflow, t/h.

Approach. Grinding does not change solids mass, so the cyclone's own feed carries the same %−200 mesh as the ball mill discharge (dilution water added afterwards, in the sump, changes % solids but not the size distribution of the solids). Apply the two-product formula around the cyclone using %−200 mesh as the tracer, then close the overall solids balance around the whole circuit (new feed in = product out, in steady state) to get the underflow tonnage.

Two-product balance around the cyclone. With cyclone feed $f=20\%$ (= ball mill discharge), overflow $o=30\%$, underflow $u=15\%$, and circulating load ratio $U/F_{new}$: $$\frac{U}{F_{new}}=\frac{o-f}{f-u}=\frac{30-20}{20-15}=\frac{10}{5}=2.00\ \ (200\%\ \text{circulating load})$$
Solids flowrate in the underflow. The overall circuit is at steady state, so the cyclone overflow (final product) equals the new feed, $O = F_{new} = 100\ \text{t/h}$, and $$\boxed{U = 2.00\times100 = 200\ \text{t/h solids}}$$ The cyclone feed (= ball mill discharge, solids basis) is then $F_{new}+U = 300\ \text{t/h}$.

2(ii) — Dilution water added at the sump (5 marks)

Given. Ball mill discharge (= cyclone feed solids) $=300\ \text{t/h}$ at 75% solids; cyclone overflow $=100\ \text{t/h}$ solids at 40% solids; cyclone underflow $=200\ \text{t/h}$ solids at 75% solids (2(i)).

Find. Water added to the sump, t/h.

Approach. Convert each stream's solids tonnage to slurry tonnage via its %solids, get the water content of each, then close a water balance around the sump (water in from the mill + dilution water = water leaving in the cyclone feed = water in overflow + water in underflow).

Water leaving the ball mill (entering the sump). $$\text{Slurry}=\frac{300}{0.75}=400\ \text{t/h},\qquad \text{Water}=400-300=100\ \text{t/h}$$
Water in the two cyclone products. $$\text{O/F: slurry}=\frac{100}{0.40}=250\ \text{t/h},\ \text{water}=150\ \text{t/h}$$ $$\text{U/F: slurry}=\frac{200}{0.75}=266.7\ \text{t/h},\ \text{water}=66.7\ \text{t/h}$$ Total water leaving via the cyclone (= water entering it) $=150+66.7=216.7\ \text{t/h}$.
Water balance on the sump. Water in (mill discharge + dilution) = water out (to the pump/cyclone): $$W_{dil}=216.7-100=\boxed{116.7\ \text{t/h}}$$

2(iii) — Ball mill net power by Bond's equation (4 marks)

Given. Work index $W_i=14$; new feed to the closed circuit (rod mill discharge) $F_{80}=3600\ \mu\text{m}$; final circuit product (cyclone overflow) $P_{80}=289\ \mu\text{m}$; new feed rate $100\ \text{t/h}$.

Find. Net power drawn by the ball mill, kW.

Approach. For a ball mill closed with a classifier, Bond's law is applied across the whole closed-circuit duty: new feed entering the circuit down to the classifier's overflow (the actual finished product), not the mill's own single-pass discharge, since the recirculating load already accounts for repeated regrinding of oversize.

Specific grinding energy, Bond's equation. $$W=10W_i\left(\frac{1}{\sqrt{P_{80}}}-\frac{1}{\sqrt{F_{80}}}\right)=10(14)\left(\frac{1}{\sqrt{289}}-\frac{1}{\sqrt{3600}}\right)$$ $$=140\left(\frac{1}{17}-\frac{1}{60}\right)=140(0.05882-0.01667)=\boxed{5.90\ \text{kWh/st}}$$
Net power at the given throughput. Applied to the 100 t/h (short-ton basis, matching the Bond work-index convention) of new feed: $$P=5.90\times100=\boxed{590\ \text{kW}}\ (\approx 791\ \text{hp})$$

Check. $F_{80}$ is taken as the rod mill discharge (material entering the ball-mill/cyclone closed circuit) and $P_{80}$ as the cyclone overflow (the circuit's actual finished product) – the standard Bond convention for a closed-circuit mill, and the only pairing of the four surveyed streams that represents the full size-reduction duty the ball mill/classifier combination performs. Using the ball mill's own single-pass feed/discharge instead would understate the work actually done, since it would ignore that most of the ball mill's feed is coarser recycled underflow, not fresh rod-mill product.

2(iv) — Specific gravity of the cyclone underflow slurry (2 marks)

Given. Underflow 75% solids by weight; ore SG 3.0; water SG 1.0.

Find. Slurry SG.

Slurry SG from the weight-fraction mixing rule. $$\frac{1}{SG_{slurry}}=\frac{0.75}{3.0}+\frac{0.25}{1.0}=0.25+0.25=0.50\ \Rightarrow\ SG_{slurry}=\boxed{2.00}$$

Question 2 — summary
Part	Result
(i) Cyclone underflow solids (circulating load)	200 t/h (200%)
(ii) Dilution water	116.7 t/h
(iii) Ball mill net power	590 kW
(iv) Cyclone underflow SG	2.00

Topic. Closed-circuit grinding material balance; solids circulating load; Bond's third theory of comminution; slurry density.

Key relations. Two-product circulating load $U/F=(o-f)/(f-u)$; Bond's law $W=10W_i(1/\sqrt{P_{80}}-1/\sqrt{F_{80}})$; slurry SG $1/SG_{slurry}=x_s/SG_s+x_w/SG_w$.

Why this works. A tracer that is conserved through the mill (here %−200 mesh, a size-distribution property unaffected by simple water dilution) lets a two-product split be solved with survey data alone, with no need to measure tonnages directly. Bond's law is empirical but scale-consistent: applying it feed-to-final-product across a closed circuit captures the extra grinding work performed on recirculated coarse material without having to track the internal recirculating tonnage explicitly.

Common pitfall. Using the ball mill's own discharge (rather than the cyclone overflow) as $P_{80}$ double-counts the classifier's contribution and understates the true energy-to-final-product; using %solids-diluted streams (rather than %−200 mesh, which dilution does not alter) as the circulating-load tracer silently mixes a water effect into a solids balance.

Source. Wills & Finch, Wills' Mineral Processing Technology, 8th ed., Ch. 3 (comminution, Bond's law) & Ch. 9 (classification, circulating load).

Question 2 of 5: Two-Stage Grinding Circuit Material Balance

Question 2: Two-Stage Grinding Circuit Material Balance (18 marks)

2(i) — Solids flowrate in the cyclone underflow (7 marks)

2(ii) — Dilution water added at the sump (5 marks)

2(iii) — Ball mill net power by Bond's equation (4 marks)

2(iv) — Specific gravity of the cyclone underflow slurry (2 marks)