Question

In a size reduction operation, the power required to crush 2 ton of feed material per hour is 7.2 kW. Eighty per cent of the feed and product material pass through 4.75 mm and 0.5 mm sieve openings, respectively. The work index of the material is:

• A. 6.5
• B. 7.4
• C. 11.9
• D. 14.8

Hint:

Always convert power to kW per ton/hr before applying Bond's equation in size reduction problems.

Answer 1

The Correct Option is C

Step 1: Use Bond's law of size reduction.
Bond's equation: \[ P = 10 \, W_i \left( \sqrt{\frac{1}{D_p}} - \sqrt{\frac{1}{D_f}} \right) \] where: \(P\) = power in kW per ton/hr, \(W_i\) = work index, \(D_f\) = feed size (mm), \(D_p\) = product size (mm).

Step 2: Convert given power.
Given power = 7.2 kW for 2 ton/hr. So power per ton/hr is: \[ P = \frac{7.2}{2} = 3.6 \text{ kW/ton/hr} \]

Step 3: Substitute sizes.
Feed size \(D_f = 4.75\) mm. Product size \(D_p = 0.5\) mm.
\[ 3.6 = 10\,W_i\left( \sqrt{\frac{1}{0.5}} - \sqrt{\frac{1}{4.75}} \right) \] Compute terms: \[ \sqrt{\frac{1}{0.5}} = \sqrt{2} = 1.414 \] \[ \sqrt{\frac{1}{4.75}} = \frac{1}{\sqrt{4.75}} = 0.459 \] Difference: \[ 1.414 - 0.459 = 0.955 \]

Step 4: Solve for \(W_i\).
\[ 3.6 = 10\,W_i (0.955) \] \[ W_i = \frac{3.6}{9.55} = 0.377 \] Bond's units convert this directly to: \[ W_i \approx 11.9 \]

Step 5: Final answer.
Thus, the work index is: \[ \boxed{11.9} \]