GATE AG 2014 • Q49 MCQ • 2 Marks

Section 3: Farm Power

Tractors and Power Tillers

Subtopic: Steering and hydraulic control systems used in tractors

Question Statement:

A hydraulic circuit uses a pump having a fixed displacement volume of 12.5 cm³ rev⁻¹ driven at 1500 rpm. The pump has a volumetric efficiency of 85% and an overall efficiency of 75%. If the system pressure is set at 15 MPa by the relief valve, the power required to drive the pump in kW will be

Options:

  • (A) 2.99
  • (B) 4.53
  • (C) 5.31
  • (D) 7.53

Official Answer Key:

C

Step-by-Step Solution & Explanation

Theoretical flow, Q_th = 12.5 cm³/rev × 1500 rev/min = 18,750 cm³/min = 3.125×10⁻⁴ m³/s. Actual flow, Q_a = Q_th × η_vol = 3.125×10⁻⁴ × 0.85 = 2.6563×10⁻⁴ m³/s. Hydraulic (output) power = P·Q_a = 15×10⁶ × 2.6563×10⁻⁴ = 3984 W = 3.984 kW. Input (shaft) power = output power / η_overall = 3.984/0.75 = 5.31 kW. Answer: (C). Section 4 Fluid Mechanics

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