Question

A hammer mill is used to grind blackgram. The size distribution is such that 80% passes through a 6-mesh (3.36 mm) screen. The power required to produce a powder where 80% passes through a 45-mesh (0.354 mm) screen is 4.5 kW. Find the power in kW required to produce a finer powder, 80% of which passes through a 60-mesh (0.25 mm) screen. Use Bond’s law of size reduction.

Hint:

Bond’s law is most accurate for intermediate particle sizes. Use \( P \propto \left( \tfrac{1}{\sqrt{D_p}} - \tfrac{1}{\sqrt{D_f}} \right) \) for quick calculations.

Answer 1

Correct Answer: 5.2

Step 1: Recall Bond’s law.
\[ E = k \left( \frac{1}{\sqrt{D_p}} - \frac{1}{\sqrt{D_f}} \right) \] where, \( D_f = \) feed size (mm), \( D_p = \) product size (mm), \( E = \) energy (or power, proportional).
Step 2: Given data.
- Feed size \( D_f = 3.36 \, \text{mm} \)
- Case 1 product size \( D_{p1} = 0.354 \, \text{mm} \), Power = 4.5 kW
- Case 2 product size \( D_{p2} = 0.25 \, \text{mm} \), Power = ?
Step 3: Compute energy terms.
For Case 1: \[ E_1 \propto \left( \frac{1}{\sqrt{0.354}} - \frac{1}{\sqrt{3.36}} \right) \] \[ \frac{1}{\sqrt{0.354}} = 1.681, \frac{1}{\sqrt{3.36}} = 0.545 \] \[ E_1 = 1.681 - 0.545 = 1.136 \] For Case 2: \[ E_2 \propto \left( \frac{1}{\sqrt{0.25}} - \frac{1}{\sqrt{3.36}} \right) \] \[ \frac{1}{\sqrt{0.25}} = 2.000, \frac{1}{\sqrt{3.36}} = 0.545 \] \[ E_2 = 2.000 - 0.545 = 1.455 \]
Step 4: Ratio of powers.
\[ \frac{P_2}{P_1} = \frac{E_2}{E_1} = \frac{1.455}{1.136} = 1.281 \]

Step 5: Compute final power.
\[ P_2 = 4.5 \times 1.281 = 5.77 \, \text{kW} \] Final Answer: \[ \boxed{5.77 \, \text{kW}} \]