Question 3 of 7: SABC Circuit – Cone Crusher Selection and Costing

Question 3: SABC Circuit – Cone Crusher Selection and Costing (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) Cone crusher size and number

Given. Table above, plus the crusher capacity table at R=10 mm (4 ft: 75 t/h/120 kW; 6 ft: 130 t/h/160 kW; 7 ft: 230 t/h/280 kW) and the 100(x/R) product-size table (80% passing → 100(x/R)=162).

Find. A cone crusher size/count meeting both the required instantaneous throughput and power draw.

Approach. Convert the average daily tonnage into the instantaneous rate the crusher must actually clear: correct the nameplate rate for plant availability to get the true operating-hour feed rate, take 30% of that as the circulating pebbles load, then divide by the 60% duty-cycle fraction (the crusher must clear a whole shift's circulating load in only 60% of the time it is available) to get the required crusher capacity. Cross- check the resulting duty against the crusher's installed power via Bond's equation using the CSS-derived P₈₀.

1. Required instantaneous crusher throughput. $$\dot{m}_{nameplate}=\frac{25{,}000\ \text{t/d}}{24\ \text{h}}=1{,}041.7\ \text{t/h}$$ $$\dot{m}_{operating}=\frac{1{,}041.7}{0.93}=1{,}120.1\ \text{t/h}\ \text{(true rate during running hours)}$$ $$\dot{m}_{CL}=0.30\times1{,}120.1=336.0\ \text{t/h (circulating pebbles load)}$$ $$\dot{m}_{crusher}=\frac{336.0}{0.60}=\boxed{560\ \text{t/h}}\ \text{(instantaneous duty, 60\% frequency of use)}$$
2. Select crusher(s) by capacity at CSS=10 mm. A single crusher size cannot clear 560 t/h (largest unit, 7 ft, gives only 230 t/h); with identical units (standard practice for spares/interchangeability), $$N=\left\lceil\frac{560}{230}\right\rceil=3\ \text{crushers, capacity}=3(230)=\boxed{690\ \text{t/h}\ge560\ \text{t/h}}$$
3. Check power via Bond's equation at CSS=10 mm. From the 100(x/R) table at 80% passing, 100(x/R)=162, so $$P_{80}=1.62\times R=1.62(10\ \text{mm})=16.2\ \text{mm}=16{,}200\ \mu\text{m}$$ $$F_{80}=6.5\ \text{cm}=65{,}000\ \mu\text{m}$$ $$w=W_{iC}\left(\frac{10}{\sqrt{P_{80}}}-\frac{10}{\sqrt{F_{80}}}\right) =17\left(\frac{10}{\sqrt{16{,}200}}-\frac{10}{\sqrt{65{,}000}}\right)=\boxed{0.669\ \text{kWh/t}}$$ $$P_{required}=w\times\dot{m}_{crusher}=0.669(560)=\boxed{375\ \text{kW}}$$ Three 7 ft crushers install 3(280)=840 kW – comfortably above the 375 kW actually drawn, confirming capacity (not power) is the binding constraint.

b) Current cost via Mular & Poulin

Given. cost=aX^b, a=30,010, b=1.7 (at M&S index 1400); current M&S index=1675; X=mantle diameter=7 ft (the selected crusher size).

Find. Current (index-escalated) cost of one 7 ft cone crusher, and of the full 3-unit installation.

Approach. Evaluate the base-year cost-capacity equation at X=7 ft, then escalate by the ratio of the current to base Marshall & Swift index.

1. Base-year (M&S=1400) cost. $$\text{cost}_{1400}=aX^{b}=30{,}010(7)^{1.7}=30{,}010(27.33)=\boxed{\$820{,}200}$$
2. Escalate to current M&S=1675. $$\text{cost}_{1675}=\text{cost}_{1400}\times\frac{1675}{1400}=820{,}200(1.1964)=\boxed{\$981{,}300\ \text{per crusher}}$$
3. Total for the 3-crusher installation. $$\text{cost}_{total}=3\times981{,}300=\boxed{\$2{,}944{,}000}$$