Iron Ore Grinding Circuit: 24-MMP-A3 Mineral Processing, December 2018, Q2

Question 2: Iron Ore Grinding Circuit (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.

Grinding circuit operating data
Stage	Parameter	Value
Rod mill	Feed rate, $F$	100 t/h
Rod mill	Feed grade, $f$	35% Fe
Rod mill	Discharge size, $F_{80}$	2500 µm (80% passing)
Magnetic separator	Non-magnetic waste grade, $t$	5% Fe
Hydrocyclone	Overflow grade	45% Fe
Ball mill	Circulating load, $CLR$	250%
Ball mill	Product size, $P_{80}$	150 µm (80% passing)
Ball mill	Bond work index, $W_i$	11.5 kWh/t

Find. (1) Ball mill power rating, $P$ (kW). (2) Total ore flowrate through the ball mill, i.e. fresh feed plus circulating load (t/h).

Check – two assumptions bridge the given data: (i) the magnetic separator does not change particle size, so the ball mill circuit's feed size is the rod mill's own discharge size, $F_{80}=2500\ \mu\text{m}$; (ii) the hydrocyclone classifies by size only (not by magnetic content), so its overflow grade (45% Fe) equals the magnetic concentrate's own grade, which is the figure needed to mass-balance the magnetic separator.

Approach. Mass-balance the magnetic separator (two-product formula) to find the ball mill circuit's fresh feed rate; apply Bond's Third Theory of Comminution to that fresh feed rate for the power rating (250% is Bond's own standard test circulating load, so no correction factor is needed); then scale the fresh feed rate up by the circulating load for the total mill throughput.

Magnetic separator mass balance. Treat the separator as a two-product split of the 100 t/h, 35% Fe rod-mill product into a magnetic concentrate (grade $c$, by assumption (ii) equal to the hydrocyclone overflow grade, 45% Fe) and non-magnetic waste (5% Fe): $$C=F\cdot\frac{f-t}{c-t}=100\times\frac{35-5}{45-5}=100\times\frac{30}{40}=\boxed{75.0\ \text{t/h}}$$ This magnetic concentrate is the ball mill circuit's fresh feed rate. (Non-magnetic waste $=100-75=25.0$ t/h, for the record.)
Specific grinding energy (Bond's Third Theory). With $F_{80}=2500\ \mu\text{m}$ (assumption (i)), $P_{80}=150\ \mu\text{m}$ and $W_i=11.5\ \text{kWh/t}$: $$E=10\,W_i\left(\frac{1}{\sqrt{P_{80}}}-\frac{1}{\sqrt{F_{80}}}\right)=10(11.5)\left(\frac{1}{\sqrt{150}}-\frac{1}{\sqrt{2500}}\right)$$ $$E=115\left(0.08165-0.02000\right)=115\times0.06165=\boxed{7.090\ \text{kWh/t}}$$
Ball mill power rating. Bond's specific energy is per tonne of new (fresh) feed to the circuit – not per tonne of total mill throughput – and the given 250% circulating load is exactly Bond's own standard closed-circuit test condition, so $E$ applies directly with no correction factor: $$P=E\times C=7.090\times75.0=\boxed{531.7\ \text{kW}}$$
Total ore flowrate through the ball mill (Part 2). The mill itself processes the fresh feed plus the circulating load returned by the hydrocyclone underflow: $$Q_{\text{mill}}=C\,(1+CLR)=75.0\times(1+2.50)=75.0\times3.50=\boxed{262.5\ \text{t/h}}$$

Quantity	Value
Magnetic concentrate (ball mill fresh feed), $C$	75.0 t/h
Non-magnetic waste rejected	25.0 t/h
Specific grinding energy, $E$	7.09 kWh/t
(1) Ball mill power rating, $P$	531.7 kW
(2) Total ore flowrate through ball mill, $Q_{\text{mill}}$	262.5 t/h

Topic. Bond's Third Theory of Comminution; circulating load and mill throughput vs. net (fresh-feed) production rate.

Key relations. $E=10W_i\left(1/\sqrt{P_{80}}-1/\sqrt{F_{80}}\right)$ kWh/t (sizes in µm); $P=E\times(\text{fresh feed rate})$; total mill throughput $=$ fresh feed $\times(1+CLR)$; two-product balance $C=F(f-t)/(c-t)$.

Why this works. Bond's work index is itself calibrated from a locked-cycle laboratory test run at a 250% circulating load, so $E$ already represents the energy needed per tonne of finished (fresh) product under that standard condition – multiplying by the gross, recirculating tonnage instead would over-state the power draw by the circulating-load factor. The circuit's classification hardware (hydrocyclone) does not add net energy or net tonnage; it only determines how many extra passes a given particle makes through the mill before it reports to product.

Common pitfall. Multiplying the specific energy by the TOTAL ball mill throughput (262.5 t/h) instead of the fresh feed rate (75.0 t/h) – this over-predicts the power draw by the $(1+CLR)$ factor, here 3.5× too high. A second common slip is forgetting that particle size, not magnetic grade, sets $F_{80}$ for the ball mill circuit.

Source. Wills & Finch, Wills' Mineral Processing Technology, 8th ed., Ch. 6 (Bond's Third Theory, §6.3) and Ch. 13 (magnetic separation mass balance).

Question 2 of 4: Iron Ore Grinding Circuit

Question 2: Iron Ore Grinding Circuit (20 marks)