Irrigation Water Conveyance and Application Methods
Subtopic: Irrigation efficiencies
Question Statement:
The pump is driven by an electric motor. If the pump, drive and motor efficiencies are 80%, 82% and 90% respectively, the required size of electric motor in horse-power will be
Options:
(A) 20.0
(B) 24.5
(C) 30.0
(D) 33.5
Official Answer Key:
D
Step-by-Step Solution & Explanation
Hydraulic (water) power delivered by the pump = ρgQh = 1000×9.81×(270/3600)×20 = 14,715 W. Overall efficiency (pump×drive×motor) = 0.80×0.82×0.90 = 0.5904. Required electrical input power = 14,715/0.5904 ≈ 24,924 W = 24.92 kW. Converting to horsepower (1 hp = 745.7 W): 24,924/745.7 ≈ 33.4 ≈ 33.5 hp, matching option (D).
Section 6 — Agricultural Process Engineering
7 question(s) in this section: Q.23, Q.42, Q.44, Q.46, Q.47, Q.54, Q.55
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