Subtopic: Mechanics and energy requirement in size reduction of agriculture produce
Question Statement:
In grinding of black pepper, 80% of the feed passes through a 6 mm screen, and 80% of the product passes through a 0.5 mm screen. The gross energy required is 10 kW·h per ton of material. Assuming Bond’s law holds good, and for the same feed conditions, if 80% of the product passes through a 0.2 mm screen, the energy required, in kW·h per ton of material, is nearest to
Options:
(A) 18.17
(B) 9.09
(C) 5.50
(D) 11.00
Official Answer Key:
A
Step-by-Step Solution & Explanation
Bond’s law: E = K(1/√D_p − 1/√D_f), where D_f and D_p are the 80%-passing feed and product sizes (in mm) and K is the Bond constant for the material.
Step 1 — find K from the given case (D_f = 6 mm, D_p = 0.5 mm, E = 10):
1/√0.5 = 1.4142; 1/√6 = 0.4082
10 = K(1.4142 − 0.4082) = K(1.0060)
K = 9.940
Step 2 — apply to the finer grind (D_f = 6 mm unchanged, D_p = 0.2 mm):
1/√0.2 = 2.2361; 1/√6 = 0.4082
E = 9.940 × (2.2361 − 0.4082) = 9.940 × 1.8279
E = 18.17 kW·h per ton
Physical sense: finer grinding requires more energy, so the answer must exceed 10 — eliminating (B) and (C) immediately.
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