Question Bank • QB_APE_EXP_026 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:

Using Rittinger's law $E = K_R \left(\frac{1}{d_2} - \frac{1}{d_1}\right)$, the energy required to crush a material from an initial feed diameter $d_1 = 20.0\text{ mm}$ to product diameter $d_2 = 5.0\text{ mm}$ is $12.0\text{ kWh/tonne}$. If the material is crushed from $d_1 = 20.0\text{ mm}$ to a finer product $d_3 = 2.0\text{ mm}$, the energy required in $\text{kWh/tonne}$ is ________ (answer in integer).

Answer Key:

36

Step-by-Step Solution & Explanation

1. For first reduction: $\Delta \left(\frac{1}{d}\right)_1 = \frac{1}{5.0} - \frac{1}{20.0} = 0.20 - 0.05 = 0.15\text{ mm}^{-1}$ $E_1 = K_R (0.15) = 12.0 \implies K_R = \frac{12.0}{0.15} = 80.0\text{ kWh}\cdot\text{mm/tonne}$ 2. For second reduction: $\Delta \left(\frac{1}{d}\right)_2 = \frac{1}{2.0} - \frac{1}{20.0} = 0.50 - 0.05 = 0.45\text{ mm}^{-1}$ $E_2 = K_R (0.45) = 80.0 \times 0.45 = 36.0\text{ kWh/tonne}$
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