Question

In a typical grinding operation, 80% of the feed material passes through a sieve opening of 4.75 mm; whereas, 80% of the ground product passes through 0.5 mm opening. If the power required to grind 2 tonnes/h of the feed material is 3.8 kW, the work index of the material is ........ (rounded off to 2 decimal places)

Hint:

The work index is a measure of the energy required to grind a material. It can be calculated using the power consumption in the grinding process and the material properties such as feed and product sizes.

Answer 1

The work index (Wi) is related to the power required for grinding by the following formula: \[ P = \frac{W_i}{\sqrt{P_1} - \sqrt{P_2}} \times \frac{F}{t} \] Where:
- \(P\) = Power required (kW)
- \(W_i\) = Work index (kWh/ton)
- \(P_1\) = Sieve opening size for feed material (in microns)
- \(P_2\) = Sieve opening size for ground product (in microns)
- \(F\) = Feed rate (tonnes/h)
- \(t\) = time (h)
Given:
- \(P = 3.8\) kW
- \(F = 2\) tonnes/h
- \(P_1 = 4.75 \, {mm} = 4750 \, \mu m\)
- \(P_2 = 0.5 \, {mm} = 500 \, \mu m\)
Rearranging the formula to solve for the work index \(W_i\): \[ W_i = \frac{P \times \left( \sqrt{P_1} - \sqrt{P_2} \right) \times t}{F} \] Substituting the known values: \[ W_i = \frac{3.8 \times \left( \sqrt{4750} - \sqrt{500} \right) \times 1}{2} \] Calculating the square roots: \[ W_i = \frac{3.8 \times \left( 68.95 - 22.36 \right) \times 1}{2} \] \[ W_i = \frac{3.8 \times 46.59}{2} = \frac{176.10}{2} = 6.29 \] Thus, the work index of the material is approximately 6.29 kWh/ton.