Subtopic: Measurement of force, stress, torque, speed, displacement and acceleration on machine elements – shafts, couplings, keys, bearings and knuckle joints
Question Statement:
A vertical hydraulic cylinder having an inside diameter of 100 mm is used in a hoist for lifting load. A constant flow rate pump capable of delivering 12 litre per minute at a maximum pressure of 18 MPa is used for supplying the fluid. Volumetric and mechanical efficiencies of the pump are 91% and 85% respectively. The rate of lifting of load in m s⁻¹ and maximum amount of load that can be lifted in kN are
Options:
(A) 0.020, 180
(B) 0.025, 141
(C) 0.064, 45
(D) 0.080, 565
Official Answer Key:
B
Step-by-Step Solution & Explanation
Cylinder area: A = π(0.100)²/4 = 7.854×10⁻³ m². The pump's rated delivery of 12 L/min = 2×10⁻⁴ m³/s is already the actual (delivered) flow to the cylinder, so rate of lifting v = Q/A = 2×10⁻⁴/7.854×10⁻³ = 0.0255 ≈ 0.025 m/s. Maximum load is set by the cylinder force at maximum pressure, F = P×A = 18×10⁶×7.854×10⁻³ = 141,372 N ≈ 141 kN (mechanical efficiency governs the pump's driving power, not the hydraulic force delivered by pressure×area). Both figures match option (B). A common error is to multiply F by the mechanical efficiency (giving ≈120 kN, not an option) or to apply volumetric efficiency again to the already-actual flow rate.
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