Two-Stage Grinding Circuit Mass Balance and Bond Power: 21-Mat-B1 Hydrometallurgy and Electrometallurgy, Undated paper, Q3

Question 3 — Two-Stage Grinding Circuit Mass Balance and Bond Power (25 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. Circuit feed $=50\ \text{t/h}$ dry solids; ore $SG=3.0$; $W_i=13.5\ \text{kWh/t}$; GGS distribution modulus $m=0.6$; the stream table above.

Find. (1)(i) dilution water added to the classifier; (1)(ii) solids tonnage ground in the ball mill; (2) rod mill power rating.

Fig. Q3 — grinding circuit with computed steady-state tonnages annotated on the exam's own flowsheet.

Approach. Since the rod mill is open circuit, its feed and discharge solids tonnage both equal the circuit feed; only the classifier overflow leaves the whole circuit, so it too equals the circuit feed. A −100 µm (fines) balance around the classifier alone then solves for the unknown sands/ball-mill tonnage, after which a water balance gives the dilution addition. Part (2) fits a Gates-Gaudin-Schuhmann line through each stream's single known point to recover $F_{80}$/$P_{80}$ for Bond's equation.

Part (1)(ii) — Solids tonnage ground in the ball mill. The rod mill is open circuit, so its discharge $R$ equals the circuit feed: $R=50\ \text{t/h}$. Only the overflow $O$ leaves the whole circuit (sands recycle through the ball mill indefinitely), so $O=50\ \text{t/h}$ too. In the ball mill's own closed loop, feed (sands, $S$) equals discharge ($X$) at steady state, so $X=S$. A −100 µm mass balance around the classifier (in: rod discharge + ball discharge; out: overflow + sands) gives $$R(0.20)+X(0.475)=O(0.75)+S(0.20)$$ Substituting $R=O=50$ and $X=S$: $$50(0.20)+S(0.475)=50(0.75)+S(0.20)\ \Rightarrow\ 0.275S=27.5\ \Rightarrow\ \boxed{S=X=100\ \text{t/h}}$$ i.e. the ball mill grinds 100 t/h of solids — twice the fresh feed rate, a circulating load of 200% on the ball mill.
Part (1)(i) — Dilution water to the classifier. Convert each stream's solids tonnage and %solids to a water tonnage via $\text{water}=\text{solids}\times\dfrac{1-\%\text{solids}}{\%\text{solids}}$: rod discharge $12.5\ \text{t/h}$, ball discharge $33.33\ \text{t/h}$, sands $33.33\ \text{t/h}$, overflow $=50\times\dfrac{1-0.333}{0.333}=100.15\ \text{t/h}$. A water balance around the classifier (in: rod discharge water + ball discharge water + dilution water $W_d$; out: overflow water + sands water) gives $$12.5+33.33+W_d=100.15+33.33 \ \Rightarrow\ \boxed{W_d\approx 87.65\ \text{t/h}}$$ (equivalently, since the ball mill and rod mill add no water themselves, this equals the overflow's own water minus the circuit feed's own water, $100.15-12.5=87.65$ — the same answer from a whole-circuit check.)
Part (2), step A — Recover $F_{80}$ and $P_{80}$ from GGS. The Gates-Gaudin-Schuhmann distribution is $Y=(x/k)^m$; with $m=0.6$ fixed and one known point per stream ($x=100\ \mu\text{m}$), solve for that stream's own size modulus $k=x/Y^{1/m}$, then evaluate at $Y=0.80$ for the 80%-passing size. For the circuit feed ($Y=0.05$ at $100\ \mu\text{m}$, i.e. the rod mill's OWN feed): $$F_{80}=100\times\left(\dfrac{0.80}{0.05}\right)^{1/0.6}\approx 10{,}159\ \mu\text{m}$$ For the rod mill discharge ($Y=0.20$ at $100\ \mu\text{m}$): $$P_{80}=100\times\left(\dfrac{0.80}{0.20}\right)^{1/0.6}\approx 1{,}008\ \mu\text{m}$$
Part (2), step B — Bond's equation. Applying the given form directly with $W_i=13.5\ \text{kWh/t}$: $$W=10(13.5)\left(\dfrac{1}{\sqrt{1008}}-\dfrac{1}{\sqrt{10{,}159}}\right)=135(0.03150-0.00991)=135(0.02160)$$ $$W\approx 2.91\ \text{kWh/t}$$ The rod mill's own feed rate is the circuit feed, $50\ \text{t/h}$, so $$\boxed{\text{Power} = 2.91\times50 \approx 146\ \text{kW}}$$

Item	Result
(1)(i) Dilution water to classifier	≈ 87.65 t/h
(1)(ii) Solids ground in ball mill	100 t/h (200% circulating load)
(2) Rod mill power rating	≈ 146 kW

Check: the GGS fit is extrapolated far beyond the single calibration point (100 µm) used to define each stream's own size modulus — a real distribution can deviate from a pure power law over 1–2 orders of magnitude, so $F_{80}/P_{80}$ found this way are approximate. Other assumptions requested by the question: Bond's equation is applied here across the rod mill's own feed-to-discharge span in isolation (not the whole circuit); the standard Bond efficiency correction factors (mill diameter, wet open-circuit rod milling, feed size beyond the standard test range) are not applied, since the question supplies a single overall $W_i$ and asks only for an "approximate" rating; and no rod mill mechanical/no-load power allowance is added.

Topic: Multi-stream grinding-circuit mass balance and Bond's third law of comminution.

Key relations: Open-circuit mill: feed = discharge. Closed-circuit mill at steady state: feed = discharge (whatever enters must leave, since nothing accumulates). Component (fines) balance: $\sum(\text{stream}\times\text{fraction}_\text{in})=\sum(\text{stream}\times\text{fraction}_\text{out})$. GGS distribution $Y=(x/k)^m$. Bond's equation $W=10W_i\!\left(\tfrac{1}{\sqrt{P_{80}}}-\tfrac{1}{\sqrt{F_{80}}}\right)$.

Why this works: Every stream in a mineral-processing circuit obeys its own steady-state mass balance for total solids AND for any conserved sub-quantity (here, the mass fraction finer than 100 µm); with five streams and two of them determined immediately by the open/closed-circuit topology, only one true unknown ($S$) remains, recoverable from a single extra balance equation. Bond's power law is empirically calibrated so that $\sqrt{1/P_{80}}$ terms capture the (much larger) energy cost of creating new surface area as particles get finer, which is why the equation is written in terms of 80%-passing sizes rather than a mean size.

Common pitfall: Assuming the ball mill's feed rate equals the fresh circuit feed (50 t/h) rather than solving for the true circulating tonnage (100 t/h) — the ball mill's throughput, and therefore its own power draw, is set by the CIRCULATING load, not the new feed rate. Also, using the ROD MILL's own %-100µm values (20% at its discharge) as if they were $P_{80}$ directly, rather than first converting them to an 80%-passing size via the stated size-distribution law.

Source: Wills & Finch, Mineral Processing Technology, 8th ed., §4.2 (particle-size distributions, Gates-Gaudin-Schuhmann) and §6.3 (Bond's law).

Question 3 of 4: Two-Stage Grinding Circuit Mass Balance and Bond Power

Question 3 — Two-Stage Grinding Circuit Mass Balance and Bond Power (25 marks)