Subtopic: Mechanics and energy requirement in size reduction of agriculture produce
Question Statement:
Feed grain of average particle size $D_1 = 8.0\text{ mm}$ is comminuted to a product size of $D_2 = 2.0\text{ mm}$ consuming $15.0\text{ kWh/tonne}$ of specific energy. Assuming Kick's law of comminution applies ($E = K_K \ln(D_1/D_2)$), the specific energy required in $\text{kWh/tonne}$ to grind the same feed grain from $D_2 = 2.0\text{ mm}$ to a final product size of $D_3 = 0.5\text{ mm}$ is ________ (answer in integer).
Answer Key:
15
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Direct Constant Evaluation**
1. By Kick's law for Stage 1 ($D_1 = 8.0\text{ mm} \to D_2 = 2.0\text{ mm}$):
$E_1 = K_K \ln\left(\frac{D_1}{D_2}\right) = K_K \ln\left(\frac{8.0}{2.0}\right) = K_K \ln(4) \approx 1.38629 K_K$
$K_K = \frac{15.0}{1.38629} = 10.8202\text{ kWh/tonne}$
2. For Stage 2 ($D_2 = 2.0\text{ mm} \to D_3 = 0.5\text{ mm}$):
$E_2 = K_K \ln\left(\frac{D_2}{D_3}\right) = 10.8202 \times \ln\left(\frac{2.0}{0.5}\right) = 10.8202 \times \ln(4) = 15.0\text{ kWh/tonne}$
**Method 2: Reduction Ratio Invariance**
Kick's law asserts that the energy required is directly proportional to the logarithm of the size reduction ratio:
$E \propto \ln(R), \quad \text{where } R = \frac{D_{\text{feed}}}{D_{\text{product}}}$
- Stage 1 ratio: $R_1 = \frac{8.0}{2.0} = 4$
- Stage 2 ratio: $R_2 = \frac{2.0}{0.5} = 4$
Since $R_1 = R_2$, the energy expended in both stages is identical: $E_2 = E_1 = 15\text{ kWh/tonne}$.