Question 2 of 7: Rod Mill – Ball Mill Work-Index Efficiency and Reduction Ratio

Question 2: Rod Mill – Ball Mill Work-Index Efficiency and Reduction Ratio (20%)

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.

a) Work-index efficiency and rod-mill reduction ratio

Given. Blend above; circuit rate 250 t/h; RM feed F₈₀=1.2 cm (12,000 μm); Wio_RM=12.0 kWh/t at 500 kW; Wio_BM=9.61 kWh/t at 1250 kW.

Find. Rowland work-index efficiency of the RM and the BM, and the rod mill's reduction ratio (F₈₀/P₈₀).

Approach. Weight-average each mill's laboratory Wi over the feed blend to get the design (target) work index, compare it with the given operating Wio via Rowland's efficiency ratio; separately back-calculate the rod mill's product P₈₀ from its own specific energy through the Bond equation, then take F₈₀/P₈₀.

1. Blended (design) work indices. $$Wi_{RM}=0.25(16.0)+0.35(14.0)+0.40(11.0)=\boxed{13.30\ \text{kWh/t}}$$ $$Wi_{BM}=0.25(15.0)+0.35(13.5)+0.40(10.0)=\boxed{12.48\ \text{kWh/t}}$$
2. Rowland work-index efficiency. $$\eta_{RM}=\frac{Wi_{RM}\times100}{Wio_{RM}}=\frac{13.30\times100}{12.0}=\boxed{110.8\%}$$ $$\eta_{BM}=\frac{Wi_{BM}\times100}{Wio_{BM}}=\frac{12.48\times100}{9.61}=\boxed{129.8\%}$$ Both mills are running more efficiently than the ore's own laboratory Wi would predict (η>100%), the ball mill markedly so.
3. Rod mill specific energy and P₈₀ from Bond's equation. $$w=\frac{P}{T}=\frac{500\ \text{kW}}{250\ \text{t/h}}=2.00\ \text{kWh/t}$$ $$w=Wio_{RM}\left(\frac{10}{\sqrt{P_{80}}}-\frac{10}{\sqrt{F_{80}}}\right) \ \Rightarrow\ \frac{10}{\sqrt{P_{80}}}=\frac{w}{Wio_{RM}}+\frac{10}{\sqrt{F_{80}}} =\frac{2.00}{12.0}+\frac{10}{\sqrt{12{,}000}}=0.2580$$ $$P_{80}=\left(\frac{10}{0.2580}\right)^2=\boxed{1{,}503\ \mu\text{m}}$$
4. Reduction ratio. $$RR=\frac{F_{80}}{P_{80}}=\frac{12{,}000}{1{,}503}=\boxed{7.98\approx8.0:1}$$

b) New North-zone-only rod mill tonnage

Given. North zone alone: Wi_RM=11.0 kWh/t; same 110.8% work-index efficiency as part (a); same 500 kW power draw and same P₈₀=1,503 μm as part (a); new F₈₀=1.0 cm (10,000 μm).

Find. The new rod-mill throughput (t/h).

Approach. "Same operating conditions of work index efficiency" fixes the ratio Wi/Wio at the 110.8% found in (a), so the new operating work index follows directly from the North zone's own laboratory Wi; with Wio_new, the unchanged P₈₀ and the new F₈₀, Bond's equation gives the new specific energy, and tonnage follows from the fixed 500 kW power draw.

1. New operating work index. $$Wio_{new}=\frac{Wi_{North,RM}\times100}{\eta_{RM}}=\frac{11.0\times100}{110.8}=\boxed{9.92\ \text{kWh/t}}$$
2. New specific energy (Bond, same P₈₀, new F₈₀). $$w_{new}=Wio_{new}\left(\frac{10}{\sqrt{P_{80}}}-\frac{10}{\sqrt{F_{80,new}}}\right) =9.92\left(\frac{10}{\sqrt{1{,}503}}-\frac{10}{\sqrt{10{,}000}}\right)$$ $$=9.92(0.2580-0.1000)=\boxed{1.568\ \text{kWh/t}}$$
3. New tonnage at the fixed 500 kW power draw. $$T_{new}=\frac{P}{w_{new}}=\frac{500\ \text{kW}}{1.568\ \text{kWh/t}}=\boxed{319\ \text{t/h}}$$

The finer North-zone-only F₈₀ (1.0 vs. 1.2 cm) combined with its lower laboratory Wi both push toward a lower specific energy demand at the same P₈₀, so – holding power constant – the mill can process appreciably more tonnage (319 vs. 250 t/h) than the blended-feed case.