Subtopic: Design and selection of machine elements – gears, pulleys, chains and sprockets and belts
Question Statement:
The size of motor required for operating the pump in kW is
Options:
(A) 3.60
(B) 3.96
(C) 4.24
(D) 4.65
Official Answer Key:
D
Step-by-Step Solution & Explanation
Hydraulic power actually delivered to the load = P×Q(actual) = 18×10⁶ Pa × 2×10⁻⁴ m³/s = 3600 W = 3.6 kW. The motor must supply more than this to cover both the volumetric loss (leakage) and mechanical loss (friction) inside the pump, i.e. the overall pump efficiency is η_v×η_m = 0.91×0.85 = 0.7735. Motor power = 3.6/0.7735 ≈ 4.65 kW, matching option (D).
Section 3 — Farm Power
7 question(s) in this section: Q.5, Q.14, Q.15, Q.17, Q.30, Q.52, Q.53
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.