24-MMP-B5 Mineral Processing Design and Operations · May 2016
Question 2 of 8: Crusher selection (power and capacity criteria) and preliminary cost
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 09-MMP-B5, Mill Design & Operations — May 2016, 3 hours. Candidates were instructed to answer any 6 of the 8 questions (each of equal value); all 8 are solved below as a complete study resource.
Reference texts: Wills' Mineral Processing Technology (B.A. Wills & J. Finch, 8th ed., Butterworth-Heinemann) — Ch. 4 Comminution, Ch. 8 Screening, Ch. 9 Classification, Ch. 12 Froth Flotation, Ch. 14 Solid-Liquid Separation; Mular, Halbe & Barratt (eds.), Mineral Processing Plant Design, Practice and Control (SME, 2002); Mular & Poulin, CIM Special Volume 47 (1998) preliminary capital cost estimation; SME Mining Engineering Handbook (3rd ed.).
Question 2: Crusher selection (power and capacity criteria) and preliminary cost (2/6)
Given.$W_i = 20.3\ \text{kWh/t}$; circulating load $130\%$; ore tonnage 4,500 t/day; availability 85%; reduction ratio $RR=3.5$; crusher product $P_{80}=24.4\ \text{mm}$ (crusher discharge table, 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 = 1,750.
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 daily tonnage to an hourly design rate through the crusher (new feed grossed up by the circulating load), size by capacity against the manufacturer table, cross-check against the Bond-equation power demand, then escalate the purchase cost from the base M&S index to the current one.
Design throughput through the crusher. New (fresh) feed rate, corrected for availability:
$$\dot m_{\text{new}} = \frac{4{,}500\ \text{t/day}}{24\ \text{h}\times0.85}=220.6\ \text{t/h}$$
The crusher itself handles new feed plus the 130% recirculating load:
$$\dot m_{\text{crusher}} = 220.6\times(1+1.30) = \boxed{507.4\ \text{t/h}}$$
Capacity criterion. Dividing 507.4 t/h by each candidate's rated capacity and rounding up: 120/6-8 (100 t/h) needs 6 units; 175/8-11 (150 t/h) needs 4 units; 210/9-13 (270 t/h) needs $\lceil 507.4/270\rceil = 2$ units (540 t/h ≥ 507.4 t/h).
Power criterion (Bond equation). With $P_{80}=24.4\ \text{mm} = 24{,}400\ \mu\text{m}$ and, from the stated reduction ratio, $F_{80}=3.5\times24.4=85.4\ \text{mm}=85{,}400\ \mu\text{m}$:
$$W = 10 W_i\left(\frac{1}{\sqrt{P_{80}}}-\frac{1}{\sqrt{F_{80}}}\right)=10(20.3)\left(\frac{1}{\sqrt{24{,}400}}-\frac{1}{\sqrt{85{,}400}}\right)=0.605\ \text{kWh/t}$$
Total power demand: $P_{tot}=0.605\times507.4=\boxed{306.9\ \text{kW}}$. By unit rating: 120/6-8 (110 kW) needs 3 units; 175/8-11 (150 kW) needs 3 units; 210/9-13 (220 kW) needs $\lceil306.9/220\rceil=2$ units (440 kW ≥ 307 kW).
Select the crusher. The 210/9-13 unit is the only size where the power criterion (2 units) and the capacity criterion (2 units) agree on the same fleet size, giving the smallest, most balanced installation: 2 × 210/9-13 SH cone crushers.
Preliminary cost, escalated to the current M&S index.$X = D = 210\ \text{cm} = 6.890\ \text{ft}$.
$$\text{Cost}_{1400} = aX^{1.70}=30{,}010\times(6.890)^{1.70}=\$798{,}400/\text{unit}$$
$$\text{Cost}_{1750}=\text{Cost}_{1400}\times\frac{1750}{1400}=\$798{,}400\times1.25=\$998{,}000/\text{unit}$$
$$\text{Total (2 units)} = \boxed{\$1{,}996{,}000}$$
Final Results — Question 2
Quantity
Value
Design crusher throughput
507.4 t/h
Bond specific power
0.605 kWh/t
Total power demand
306.9 kW
Selected crusher & count
2 × 210/9-13 SH cone
Preliminary cost (M&S = 1,750), total
≈ USD 2.00 million
Check: the reduction ratio (3.5) is applied to the crusher's own P80 to back-calculate its F80, per the standard definition $RR = F_{80}/P_{80}$; the crusher discharge table's 24.4 mm row (80.1% cum. passing) is read as the design P80 since it lands almost exactly on 80%.