Subtopic: Mechanics and energy requirement in size reduction of agriculture produce
Question Statement:
The work index of a material is 6.25. If 80% of the feed and 80% of the product pass through IS Sieve No. 340 (3.25 mm opening) and IS Sieve No. 40 (0.42 mm opening), respectively, the power consumed in kW to crush 5000 kg h⁻¹ of sorghum is
Options:
(A) 9.77
(B) 20.49
(C) 26.29
(D) 32.29
Official Answer Key:
A
Step-by-Step Solution & Explanation
Bond's Law: specific energy E (kWh/tonne) = 10×W_i×(1/√D_p − 1/√D_f), with particle sizes D expressed in MICRONS.
D_f = 3.25 mm = 3250 µm; D_p = 0.42 mm = 420 µm.
E = 10×6.25×(1/√420 − 1/√3250) = 62.5×(0.04879−0.01754) = 62.5×0.03125 = 1.953 kWh/tonne.
Capacity = 5000 kg/h = 5 tonne/h.
Power = E×capacity = 1.953×5 = 9.77 kW. Answer: (A).
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