Question 2 of 8: Crusher selection (power and capacity criteria) and preliminary cost

Question 2: Crusher selection (power and capacity criteria) and preliminary cost (2/6)

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. $W_i=16.3$ kWh/t; circulating load 130%; ore tonnage 9,000 t/day; availability 82%; reduction ratio $RR=3.5$; crusher product $P_{80}=24.4$ mm (from the crusher-discharge table, Question 1, cum. % passing 24.4 mm = 80.1% ≈ 80%); three candidate crusher sizes with rated kW/t-h. Cost law $\text{Cost}=aX^{1.70}$, $a=30{,}010$ at M&S = 1,400; current M&S = 1,800.

Find. Crusher size and number of units by (i) power and (ii) capacity criteria, and the preliminary current cost of the selected fleet.

Approach. Convert the daily tonnage to an hourly design rate through the crusher (fresh feed grossed up by the circulating load, since the crusher physically re-processes the recirculating oversize as well as new ore), cross-check the total throughput against the manufacturer table's capacity rating and against the Bond-equation power demand, then escalate the purchase cost from the base M&S index to the current one.

1. Feed size from the stated reduction ratio. With $P_{80}=24.4$ mm (Question 1's crusher-discharge table): $$F_{80}=RR\times P_{80}=3.5\times24.4=\boxed{85.3\ \text{mm}}$$
2. Design throughput through the crusher. Fresh feed rate, corrected for availability: $$\dot m_{\text{new}}=\frac{9{,}000\ \text{t/day}}{24\ \text{h}\times0.82}=457.3\ \text{t/h}$$ The crusher itself handles new feed plus the 130% recirculating load through the bin (Fig. 1): $$\dot m_{\text{crusher}}=457.3\times(1+1.30)=\boxed{1{,}051.8\ \text{t/h}}$$
3. Power criterion (Bond equation). With $P_{80}=24{,}400\ \mu\text{m}$, $F_{80}=85{,}300\ \mu\text{m}$: $$W=10W_i\left(\frac{1}{\sqrt{P_{80}}}-\frac{1}{\sqrt{F_{80}}}\right)=10(16.3)\left(\frac{1}{\sqrt{24{,}400}}-\frac{1}{\sqrt{85{,}300}}\right)=0.486\ \text{kWh/t}$$ Total power demand (on the crusher's own throughput, since the machine expends energy on every tonne that passes through it, recirculating or not): $$P_{tot}=0.486\times1{,}051.8=\boxed{511.3\ \text{kW}}$$ By unit rating: 120/6-8 (110 kW) needs $\lceil511.3/110\rceil=5$ units; 175/8-11 (150 kW) needs 4 units; 210/9-13 (220 kW) needs $\lceil511.3/220\rceil=3$ units.
4. Capacity criterion. Dividing 1,051.8 t/h by each candidate's rated capacity and rounding up: 120/6-8 (100 t/h) needs 11 units; 175/8-11 (150 t/h) needs 8 units; 210/9-13 (270 t/h) needs $\lceil1{,}051.8/270\rceil=\boxed{4}$ units (1,080 t/h ≥ 1,051.8 t/h).
5. Select the crusher. The 210/9-13 unit needs the fewest units under EITHER criterion (3 by power, 4 by capacity); the capacity criterion governs, so 4 × 210/9-13 SH cone crushers are selected (the 4-unit fleet also comfortably clears the 511.3 kW power demand at 880 kW installed).
6. Preliminary cost, escalated to the current M&S index. $X=D=210\ \text{cm}=6.90$ ft: $$\text{Cost}_{1400}=aX^{1.70}=30{,}010\times(6.90)^{1.70}=\$798{,}400/\text{unit}$$ $$\text{Cost}_{1800}=\text{Cost}_{1400}\times\frac{1800}{1400}=\$798{,}400\times1.286=\$1{,}026{,}500/\text{unit}$$ $$\text{Total (4 units)}=\boxed{\$4{,}106{,}000}$$