Question 2 of 5: Two-Stage Grinding Circuit Material Balance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Exam — Mining and Mineral Processing Engineering, 09-MMP-A3 Mineral Processing, 2013-May. 3 hours duration, closed book; only an approved Casio or Sharp calculator permitted. Question 3 allows a choice of any six (6) of nine sub-terms. Question 5 and its bonus (page 5) were to be handed in with the exam booklet; Question 4's log-log plot (page 6) is reproduced here as a computed inline figure.
Reference texts: Wills & Finch, Wills' Mineral Processing Technology, 8th ed. (flotation circuit design and metallurgical balances – Ch. 12; comminution, Bond's law and circulating load – Ch. 3 & 6; particle size analysis – Ch. 4; sampling theory, Gy's equation – Ch. 3; gravity concentration, dense medium separation, magnetic/electrostatic separation – Ch. 10, 11 & 13); Taggart, Handbook of Mineral Dressing (classical economic-recovery/economic-efficiency formula used in Question 1(vi)).
Question 2: Two-Stage Grinding Circuit Material Balance (18 marks)
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.
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.
Two-product balance around the cyclone. With cyclone feed $f=20\%$ (= ball mill discharge), overflow $o=30\%$, underflow $u=15\%$, and circulating load ratio $U/F_{new}$:
$$\frac{U}{F_{new}}=\frac{o-f}{f-u}=\frac{30-20}{20-15}=\frac{10}{5}=2.00$$
Circulating load.
$$\text{C.L.}=200\%\ \Rightarrow\ \boxed{U = 2.00\times100 = 200\ \text{t/h}}$$
The overflow (final product) equals the new feed in steady state, $O = F_{new} = 100\ \text{t/h}$, and the cyclone feed (= ball mill discharge, solids basis) is $F_{new}+U = 300\ \text{t/h}$.
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.
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.
$$\frac{1}{SG_{slurry}}=\frac{0.75}{3.0}+\frac{0.25}{1.0}=0.25+0.25=0.50\ \Rightarrow\ SG_{slurry}=\boxed{2.00}$$
