Question Bank • QB_APE_SIZE_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:

Corn is crushed in a hammer mill from an initial size where $80\%$ passes $D_{pa} = 3000\,\mu\text{m}$ to a product size where $80\%$ passes $D_{pb} = 300\,\mu\text{m}$. The Bond work index of the corn is $W_i = 12.5\text{ kWh/tonne}$. According to Bond's comminution law, $W = 10 W_i \left( \frac{1}{\sqrt{D_{pb}}} - \frac{1}{\sqrt{D_{pa}}} \right)$, the power required to grind corn at a throughput of $5.0\text{ tonnes/h}$ is ________ $\text{kW}$ (round off to two decimal places).

Answer Key:

24.67

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Bond's Energy Equation** 1. Bond's work formula: $W = 10 W_i \left( \frac{1}{\sqrt{D_{pb}}} - \frac{1}{\sqrt{D_{pa}}} \right)$ Where: - $W_i = 12.5\text{ kWh/tonne}$ - $D_{pb} = 300\,\mu\text{m} \implies \sqrt{300} \approx 17.3205\,\mu\text{m}^{1/2}$ - $D_{pa} = 3000\,\mu\text{m} \implies \sqrt{3000} \approx 54.7723\,\mu\text{m}^{1/2}$ 2. Specific energy consumed: $\Delta = \frac{1}{17.3205} - \frac{1}{54.7723} = 0.057735 - 0.018257 = 0.039478\,\mu\text{m}^{-1/2}$ $W = 10 \times 12.5 \times 0.039478 = 125 \times 0.039478 = 4.93475\text{ kWh/tonne}$ 3. Total mill power for throughput $\dot{m} = 5.0\text{ tonnes/h}$: $P = \dot{m} \times W = 5.0\text{ tonnes/h} \times 4.93475\text{ kWh/tonne} = 24.6738\text{ kW} \approx 24.67\text{ kW}$ **Method 2: Generalized Comminution Law** Walker's general differential equation: $\frac{dE}{dL} = -C L^{-n}$. For Bond's law ($n = 1.5$): $E = 2 C \left( \frac{1}{\sqrt{L_2}} - \frac{1}{\sqrt{L_1}} \right) = 10 W_i \left( L_2^{-0.5} - L_1^{-0.5} \right)$ Multiplying mass flow rate gives $P = 24.67\text{ kW}$.
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