Question Bank • QB_APE_ADV_002 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 grinding corn from a feed diameter $d_1 = 6\text{ mm}$ to a product diameter $d_2 = 1.5\text{ mm}$ requires $12\text{ kW}$ of power for a throughput of $2\text{ tonnes/hour}$. Assuming Rittinger's law ($P = K_R \cdot \dot{m} \left( \frac{1}{d_2} - \frac{1}{d_1} \right)$) holds, the power required to grind the same corn feed to a finer product of diameter $0.75\text{ mm}$ at the same throughput is ________ $\text{kW}$ (round off to two decimal places).

Answer Key:

28.00

Step-by-Step Solution & Explanation

According to Rittinger's law: $P \propto \left( \frac{1}{d_2} - \frac{1}{d_1} \right)$ For Case 1 ($d_1 = 6\text{ mm}, d_2 = 1.5\text{ mm}, P_1 = 12\text{ kW}$): $\Delta_1 = \frac{1}{1.5} - \frac{1}{6} = 0.6667 - 0.1667 = 0.500\text{ mm}^{-1}$ For Case 2 ($d_1 = 6\text{ mm}, d_2' = 0.75\text{ mm}$): $\Delta_2 = \frac{1}{0.75} - \frac{1}{6} = 1.3333 - 0.1667 = 1.1667\text{ mm}^{-1}$ Computing $P_2$: $P_2 = P_1 \times \frac{\Delta_2}{\Delta_1} = 12\text{ kW} \times \frac{1.1667}{0.500} = 12 \times 2.3333 = 28.00\text{ kW}$
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