Question 2
A crusher was used to crush material of compressive strength of 22.5 MN/m². The size of feed particles was in the range 40 to 50 mm and the energy required was 13.0 kJ/kg and the
mean size of the product was 0.5mm. The same crusher was used on a different material with twice the compressive strength. Determine the power required to reduce the size of this
material, the size distribution of the feed and the product are given in Table 1, if the mass feed rate is 1kg/s; (Assume Kick law holds)
Table 1: Particle size distributions of the material before and after crushing.
| | Sieve size (mm) | 4.750 | 4.000 | 3.350 | 2.800 | 2.360 | 2.000 | 1.700 | 1.400 | 1.180 | 1.000 | 0.850 | 0.500 | 0.000 |
|-------------|-----------------|-------|-------|-------|-------|-------|-------|-------|-------|-------|-------|-------|-------|-------|
| | dp,ave(mm) | 4.750 | 4.375 | 3.675 | 3.075 | 2.580 | 2.180 | 1.850 | 1.550 | 1.290 | 1.090 | 0.925 | 0.675 | 0.250 |
| Feed | xi,feed | 0.000 | 0.100 | 0.150 | 0.250 | 0.180 | 0.090 | 0.063 | 0.040 | 0.035 | 0.032 | 0.030 | 0.020 | 0.010 |
| Product | xi,product | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.050 | 0.150 | 0.200 | 0.350 | 0.200 | 0.050 |
Addition Information
Kicks Law is given as:
$$E = K_k ln(\frac{d_{s,a}}{d_{s,b}})$$
where $K_k$ is the Kick's constant, $d_{s,a}$ is surface volume diameter of the particles before crushing and $d_{s,b}$ is the surface volume of the particles after crushing.
