Mass Balance (20 marks): 21-Mat-A3 Structure and Characterization of Materials, Dec-10-Met-A3 2018, Q2

Question 2 — Mass 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. A closed-circuit ball mill/cyclone grinding loop: fresh ore enters the mill at 5 % moisture, dilution water is added at the mill discharge, the cyclone feed is 25 % solids, the cyclone underflow (recycled to the mill) is 80 % solids, and the cyclone overflow (10 t/h dry, product to flotation) is 12.5 % solids.

Given data
Quantity	Symbol	Value
Cyclone overflow, dry solids (= flotation feed)	$O$	10 t/h
Fresh ore moisture (ore bin feed)	—	5 %
Cyclone feed, % solids	$s_{CF}$	25 %
Cyclone underflow, % solids	$s_U$	80 %
Cyclone overflow, % solids	$s_O$	12.5 %

Find. (a) The circulating load (underflow, dry t/h) and (b) the water that must be added to the ball mill discharge stream ahead of the cyclone.

Figure 2.1 — The closed-circuit grinding loop. Underflow recycles to the mill as the circulating load; overflow, fixed by the steady-state solids balance at 10 t/h dry, is the only stream that leaves the loop.

Approach. Convert each stream's percent-solids into a water:solids dilution ratio, then close a water balance first around the cyclone (to get the circulating load $U$) and then around the dilution point ahead of it (to get the water to add). No reagent or solids are lost anywhere in the loop, so both balances are exact.

Set the dilution ratios. For a stream at percent solids $s$, the water-to-solids mass ratio is $\mathrm{dil}(s)=(1-s)/s$: $$A=\mathrm{dil}(0.25)=\frac{0.75}{0.25}=3.000,\quad B=\mathrm{dil}(0.125)=\frac{0.875}{0.125}=7.000,\quad C=\mathrm{dil}(0.80)=\frac{0.20}{0.80}=0.250$$ ($A$ = cyclone feed, $B$ = overflow, $C$ = underflow.) The overflow's dry solids are fixed by the steady-state solids balance around the whole circuit: with no other solids exit, $O=10$ t/h, the fresh feed rate.
Part (a) — water balance around the cyclone to find the circulating load. All solids entering the cyclone (the mill-discharge/cyclone-feed stream, dry mass $CF=O+U$) split into overflow $O$ and underflow $U$; the cyclone itself adds or removes no water, so the water entering equals the water leaving: $$(O+U)A=OB+UC$$ Solving for $U$, $$U=\frac{O(B-A)}{A-C}=\frac{10\times(7.000-3.000)}{3.000-0.250}=\frac{40.00}{2.750}$$ $$\boxed{U\approx14.5\ \text{t/h dry (circulating load)}}$$ As a fraction of fresh feed, the circulating-load ratio is $U/O\times100=145.5\ \%$ — the mill sees roughly 2.45 times the fresh feed tonnage once recycle is included, which is a realistic figure for a cyclone closed circuit.
Part (b) — water balance at the dilution point to find the water to add. The cyclone feed stream must carry $CF\!\cdot\!A=(O+U)\times3.000=24.545\times3.000=73.64$ t/h of water. Before any dilution water is added, the ball-mill discharge carries only the water that entered the mill: the fresh ore's own moisture, plus the water already riding on the recycled underflow. $$\text{fresh ore, wet}=\frac{O}{1-0.05}=\frac{10}{0.95}=10.53\ \text{t/h}\ \Rightarrow\ \text{water}_{fresh}=10.53-10=0.53\ \text{t/h}$$ $$\text{water}_U=U\times C=14.545\times0.250=3.64\ \text{t/h}$$ $$\text{water present before dilution}=0.53+3.64=4.16\ \text{t/h}$$ The dilution water added at the discharge makes up the shortfall to the 73.64 t/h the cyclone feed requires: $$\text{water to add}=73.64-4.16$$ $$\boxed{\text{water to add}\approx69.5\ \text{t/h}}$$ This is by far the larger of the two water additions in the loop — the closed circuit recycles most of its own water via the underflow, but still needs a substantial fresh water make-up to hit a dilute 25 % cyclone feed.

Final results — Question 2
Quantity	Value
(a) Circulating load, dry t/h	14.5 t/h ($\approx$145.5 % of fresh feed)
(b) Dilution water to add	69.5 t/h

Topic. Closed-circuit grinding: solids and water balances around a classifying cyclone.

Key relations. Dilution ratio $\mathrm{dil}(s)=(1-s)/s$; cyclone solids balance $CF=O+U$; cyclone water balance $CF\!\cdot\!A=OB+UC$.

Why this works. A classifying cyclone splits one dry-solids stream into two products of different particle size (here proxied by their percent-solids), but every kilogram of water and solid entering must leave — there is no accumulation at steady state. Writing the water balance in terms of dilution ratios rather than raw tonnages turns a two-unknown problem ($U$ and its water content) into one linear equation, because water content is tied to dry solids through a single known ratio at each point.

Common pitfall. The most common error is balancing solids alone and forgetting that percent-solids specifications are only meaningful through a water balance — the circulating load cannot be found from the solids balance ($CF=O+U$) by itself, since $U$ is unknown on both sides. A second common slip is applying the fresh-ore moisture correction to the wrong stream, or forgetting it entirely when computing the water already present at the dilution point.

Source. Wills' Mineral Processing Technology, 8th ed., Ch. 9 (classification, circulating load); Habashi, Textbook of Pyrometallurgy, Ch. 3 (mass balance conventions).

Question 2 of 7: Mass Balance (20 marks)

Question 2 — Mass Balance (20 marks)