Allied & ICAR Exam Bank • ALLIED_0629 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Size Reduction and Milling

Subtopic: Size Reduction and Milling

Question Statement:

A hammer mill crushes grain from $L_1 = 8.0\text{ mm}$ to $L_2 = 2.0\text{ mm}$ (reduction ratio $L_1 / L_2 = 4.0$, with $\ln 4.0 \approx 1.3863$). If Kick's constant is $K_K = 10.0\text{ kJ/kg}$, the energy consumed $E = K_K \ln(L_1/L_2)$ in kJ/kg is (rounded off to one decimal place):

Answer Key:

13.9

Step-by-Step Solution & Explanation

$E = 10.0 \times 1.38629 = 13.86\text{ kJ/kg} \approx 13.9\text{ kJ/kg}$.
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