Closed-Circuit Crushing and the Screen Partition Curve: 21-Mat-B1 Hydrometallurgy and Electrometallurgy, Undated paper, Q2

Question 2 — Closed-Circuit Crushing and the Screen Partition Curve (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. Fresh feed $F=600\ \text{t/h}$; circulating load ratio $=150\%$; oversize/undersize size-retained tables above (each column sums to 100%).

Find. (1) Every stream tonnage on the crushing-circuit flowsheet. (2) The partition (Tromp) curve and its 50% cut size $d_{50}$.

Fig. Q2(1) — closed-circuit crushing flowsheet with tonnage rates.

Approach. Part (1) uses the circulating-load-ratio definition to size the recycle and product streams; part (2) converts each size class's wt % retained into an actual tonnage on the oversize and undersize streams, computes each size class's partition number (% of that size reporting to oversize), plots the resulting curve on semi-log axes, and reads $d_{50}$ where the curve crosses 50%.

Part (1) — Circuit tonnages. Circulating load ratio $C=L/F=1.50$, so the recycled screen oversize is $$L = 1.50\times600 = 900\ \text{t/h}$$ Only the screen undersize leaves the circuit as final product, so at steady state it equals the fresh feed: $$\text{Undersize} = F = 600\ \text{t/h}$$ The crusher receives fresh feed plus recycle, and (crushing changes size, not mass) discharges the same tonnage to the screen: $$\text{Screen feed} = \text{crusher product} = F+L = 600+900 = \boxed{1500\ \text{t/h}}$$
Part (2), step A — Convert wt % retained to tonnage. Multiply each size class's retained fraction by its stream's total tonnage (oversize 900 t/h, undersize 600 t/h): $$q_{O,i} = \left(\dfrac{\text{wt\%}_{O,i}}{100}\right)\times900,\qquad q_{U,i} = \left(\dfrac{\text{wt\%}_{U,i}}{100}\right)\times600$$ giving oversize tonnages $412.2,\ 427.5,\ 47.7,\ 5.4,\ 2.7,\ 4.5$ t/h (sum 900.0) and undersize tonnages $4.2,\ 42.0,\ 289.2,\ 153.6,\ 64.2,\ 46.8$ t/h (sum 600.0) for size classes $+32,+16,+8,+4,+2,-2$ mm respectively.
Part (2), step B — Partition numbers. Each size class's partition number is the fraction of that size (from both streams combined) that reports to the oversize: $$P_i = \dfrac{q_{O,i}}{q_{O,i}+q_{U,i}}\times100$$ Evaluating: +32 mm → 99.0%, +16 mm → 91.1%, +8 mm → 14.2%, +4 mm → 3.4%, +2 mm → 4.0%, −2 mm (pan) → 8.8%. The steep drop between +16 mm and +8 mm is where the screen does almost all of its separating work; the small upturn at the finest (−2 mm) class is a real "fish-hook" effect — a small fraction of very fine particles is carried mechanically with the oversize (near-size misplacement/entrainment) rather than passing cleanly, which is common on an inclined vibrating screen and does not indicate a data error.
Part (2), step C — Plot and read $d_{50}$. Plotting $P_i$ (linear, 0–100%) against particle size (log scale) gives the curve in the figure below. The curve crosses 50% between the $+8$ mm point (14.2%) and the $+16$ mm point (91.1%); interpolating linearly in $\log_{10}(\text{size})$ between those two bracketing points, $$\log_{10}d_{50} = \log_{10}(8)+\dfrac{50-14.2}{91.1-14.2}\big[\log_{10}(16)-\log_{10}(8)\big] = 1.0434$$ $$\boxed{d_{50} \approx 11.1\ \text{mm}}$$

Fig. Q2(2) — screen partition (Tromp) curve; d₅₀ read at the 50% crossing. Pan point (−2 mm, open-ended class) plotted at a nominal 1 mm for reference only.

Item	Result
Recycled oversize (150% circulating load)	900 t/h
Screen feed = crusher product	1500 t/h
Final undersize product	600 t/h
Cut size d₅₀	≈ 11.1 mm

Check: d₅₀ is read by linear interpolation of the partition number against log(size) between the two data points bracketing 50% — the standard hand-plotting convention on semi-log graph paper (which is exactly the blank sheet the exam supplies for this question). A smooth-curve fit through all six points would shift the reading by well under 1 mm here, since the two bracketing points already dominate the interpolation.

Topic: Closed-circuit comminution (circulating load) and the screen partition (Tromp) curve.

Key relations: Circulating load ratio $C=L/F$; screen feed $=F(1+C)$; partition number $P_i=\dfrac{q_{O,i}}{q_{O,i}+q_{U,i}}\times100$; cut size $d_{50}$ read where $P=50\%$.

Why this works: In a closed circuit at steady state, whatever mass leaves via the undersize must equal what enters as fresh feed (nothing else exits), which fixes both the undersize tonnage and, via the stated circulating load ratio, the recycle tonnage directly from the feed rate — no iteration is needed. The partition curve simply asks, size class by size class, "of all the material of this size in the system, what fraction ends up in the oversize?" — a perfect screen would give a step function (100% below $d_{50}$'s aperture, near it 0% above, ideal 0/100 split), so how close the actual data comes to a step function, i.e. how narrow the S-shaped transition is, is itself the measure of screening efficiency (see the probable-error definition in Question 4(9)).

Common pitfall: Confusing $d_{50}$ (read from the PARTITION curve, a measure of screen performance) with $P_{80}$ or $F_{80}$ (read from a cumulative SIZE-DISTRIBUTION curve of a single stream, a measure of product fineness) — the two curve types share axes but answer different questions. Also, forgetting that circulating load is defined relative to NEW feed, not to the screen's own total throughput, is a common source of a 2.5× error in the recycle tonnage.

Source: Wills & Finch, Mineral Processing Technology, 8th ed., §6.6 (closed-circuit grinding, circulating load) and §8.3 (screen partition curves).

Question 2 of 4: Closed-Circuit Crushing and the Screen Partition Curve

Question 2 — Closed-Circuit Crushing and the Screen Partition Curve (20 marks)