Question Bank • QB_APE_008 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Size Reduction and Material Handling

Subtopic: Mechanics and energy requirement in size reduction of agriculture produce

Question Statement:

A hammer mill comminutes grain from an initial $80\%$ passing size $d_1 = 1000\text{ }\mu\text{m}$ to a product $80\%$ passing size $d_2 = 100\text{ }\mu\text{m}$. If Bond's work index of the grain is $W_i = 12.0\text{ kWh/t}$, the energy required to grind 1 tonne of grain in $\text{kWh}$ according to Bond's law ($E = 10 W_i \left[\frac{1}{\sqrt{d_2}} - \frac{1}{\sqrt{d_1}}\right]$, where $d_1, d_2$ are in $\mu\text{m}$) is ________ (round off to 2 decimal places).

Answer Key:

8.21

Step-by-Step Solution & Explanation

From Bond's third theory of comminution: $E = 10 W_i \left( \frac{1}{\sqrt{d_2}} - \frac{1}{\sqrt{d_1}} \right)$ Where: • $W_i = 12.0\text{ kWh/t}$ • $d_1 = 1000\text{ }\mu\text{m} \implies \sqrt{d_1} = 31.62277$ • $d_2 = 100\text{ }\mu\text{m} \implies \sqrt{d_2} = 10.0$ $\frac{1}{\sqrt{d_2}} = 0.10$ $\frac{1}{\sqrt{d_1}} = \frac{1}{31.62277} \approx 0.031623$ $E = 10 \times 12.0 \times (0.10 - 0.031623) = 120 \times 0.068377 = 8.205\text{ kWh/t} \approx 8.21\text{ kWh}$
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