In a feed milling plant, it has been observed that 80 % of the feed passes through IS sieve 340 (3.25 mm opening) and 80 % of the ground feed passes through IS sieve 40 (0.42 mm opening). The power requirement to crush the material with a feed rate of 3 ton h⁻¹ is 6 kW. Power requirement to crush 2 ton h⁻¹ of the same feed using the above system so that 80 % of the ground feed pass through IS sieve 15 (opening 0.157 mm) is _________ kW.
Official Answer Key:
7.90 to 8.10
Step-by-Step Solution & Explanation
Bond's law: P/ṁ = K (1/√L₂ − 1/√L₁), with L in mm.
Case 1 (calibration): 6/3 = 2 = K (1/√0.42 − 1/√3.25) = K (1.5430 − 0.5547) = 0.9883 K
⇒ K = 2.0236 kW h ton⁻¹ mm^{1/2}.
Case 2: 1/√0.157 = 2.5237; (2.5237 − 0.5547) = 1.9690.
P/ṁ = 2.0236 × 1.9690 = 3.984 kW h ton⁻¹.
P = 2 × 3.984 = 7.97 kW — within the official range 7.90–8.10.
Note the throughput drops from 3 to 2 ton h⁻¹ yet the power rises — the finer grind more than compensates.
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