Question 3 of 6: Material Balance on a Two-Stage Water-Only Cyclone Coal Circuit

Question 3: Material Balance on a Two-Stage Water-Only Cyclone Coal Circuit (17 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. The circuit of Figure 1: fresh feed and the secondary cyclone overflow join in the first sump and are pumped to the primary cyclone; the primary overflow is clean coal; the primary underflow falls into a second sump where dilution water is added and is pumped to the secondary cyclone, whose underflow is the rejects and whose overflow recirculates to the first sump. Every stream has been sampled for both per cent solids and per cent ash.

Find. (a) the clean coal tonnage, (b) the dilution water added to the second sump, and (c) the solids in the primary cyclone underflow.

Approach. Ash behaves exactly like an assay, so a two-product ash balance around the whole circuit gives the clean coal without ever touching the recycle. The internal streams then follow from a second ash balance around the primary cyclone alone, and the water is obtained last, from the per cent solids figures, because water can only be balanced once every solids tonnage is known.

(a) Clean coal produced by the circuit

1. Draw the boundary round the whole circuit. Across that boundary there is one feed and two products — clean coal and rejects — and the recycle stream is entirely internal, so it cannot appear: $$\begin{aligned} F &= C + R \\ F a_F &= C a_C + R a_R \end{aligned}$$ where $a$ is the ash content of each stream.
2. Apply the two-product formula to ash. Eliminating $R$ exactly as in Problem 1: $$\frac{C}{F} = \frac{a_F - a_R}{a_C - a_R} = \frac{25 - 70}{10 - 70} = \frac{-45}{-60} = 0.750$$
3. Scale to the circuit feed. With 100 t/h of solids entering: $$\begin{aligned} C &= 0.750 \times 100 = 75\ \text{t/h} \\ R &= 100 - 75 = 25\ \text{t/h} \end{aligned}$$ $$\boxed{\text{clean coal} = 75\ \text{t/h at }10\ \%\ \text{ash}}$$
4. Check the ash balance closes. Ash out is $0.10 \times 75 + 0.70 \times 25 = 7.5 + 17.5 = 25.0$ t/h, exactly matching the $0.25 \times 100 = 25.0$ t/h of ash in. The yield is 75 per cent, and on a combustible (ash-free) basis the recovery is $75 \times 0.90 / (100 \times 0.75) = 90.0$ per cent, which is the number a coal preparation engineer would quote.

(c) Solids in the primary cyclone underflow

Part (c) is taken next, because part (b) needs its answer. The circulating load cannot be read from the circuit boundary; it needs a boundary drawn around the primary cyclone alone.

1. Draw the boundary round the primary cyclone. Its feed splits into the clean coal overflow, already known, and the underflow: $$\begin{aligned} P &= C + U \\ P a_P &= C a_C + U a_U \end{aligned}$$ with $a_P = 25$, $a_C = 10$ and $a_U = 47.5$ per cent ash.
2. Substitute the known clean coal tonnage and solve for the underflow. Putting $P = 75 + U$ into the ash balance: $$25(75 + U) = 10(75) + 47.5\,U$$ $$1\,875 + 25U = 750 + 47.5U \;\Longrightarrow\; 1\,125 = 22.5\,U$$ $$\boxed{U = 50\ \text{t/h of solids in the primary cyclone underflow}}$$
3. Complete the internal streams. The primary cyclone feed is therefore $P = 75 + 50 = 125$ t/h, and the secondary cyclone, fed by those 50 t/h, must send $50 - 25 = 25$ t/h to its overflow, since the rejects are 25 t/h from part (a). The circulating load is 50 t/h, or 50 per cent of new feed.
4. Cross-check against the printed ash assays. Around the secondary cyclone, $0.25 \times 25 + 0.70 \times 25 = 6.25 + 17.5 = 23.75$ t/h of ash out against $0.475 \times 50 = 23.75$ t/h in. And because the recycle carries 25 per cent ash, the same ash as the fresh feed, the combined primary cyclone feed also assays 25 per cent ash, exactly as the survey reports. The whole balance is self-consistent.

(b) Dilution water added to the sump

1. Convert every per cent solids figure into a water tonnage. If a stream carries $S$ tonnes per hour of solids at $p$ per cent solids, the water it carries is $$W = S\,\frac{100 - p}{p}$$
2. Apply it to the four streams that bracket the second sump. The sump receives the primary cyclone underflow and the dilution water, and discharges the secondary cyclone feed, which in turn leaves as the secondary overflow and the rejects: $$\begin{aligned} \text{primary underflow} &: 50 \times \tfrac{75}{25} = 150\ \text{t/h of water} \\ \text{secondary overflow} &: 25 \times \tfrac{83.33}{16.67} = 125\ \text{t/h of water} \\ \text{rejects} &: 25 \times \tfrac{66.67}{33.33} = 50\ \text{t/h of water} \end{aligned}$$
3. Balance the water across the sump and the secondary cyclone together. Water is neither made nor destroyed in a cyclone, so the water leaving in the two secondary cyclone products equals the water entering the sump: $$W_{dil} = (125 + 50) - 150 = 25\ \text{t/h}$$ $$\boxed{\text{dilution water added to the sump} = 25\ \text{t/h}}$$
4. Check against the rest of the circuit. The circuit feed carries $100 \times 87.5/12.5 = 700$ t/h of water and the recycle adds 125 t/h, so the primary cyclone feed carries 825 t/h of water with 125 t/h of solids — that is $125/950 = 13.16$ per cent solids, precisely the surveyed value, which independently confirms both the circulating load and the water figures. The primary cyclone's own water balance also closes, $675 + 150 = 825$ t/h. Finally, on the whole circuit, $700 + 25 = 725$ t/h of water enters and $675 + 50 = 725$ t/h leaves.

The secondary cyclone feed therefore runs at 50 t/h of solids in 175 t/h of water, or 22.2 per cent solids. The purpose of the dilution is exactly this: a water-only cyclone separates on density difference in a shear field, and it will only make a sharp cut if the feed pulp is thin enough for particles to move relative to the fluid. Feeding the secondary cyclone at the 25 per cent solids of the primary underflow would crowd the apex and push clean coal into the rejects.