GATE AG 2007 • Q25 MCQ • 1 Mark

Section 3: Farm Power

Farm Power

Subtopic: Power efficiencies and measurement, engine performance curves

Question Statement:

The mechanical efficiency of a power tiller engine developing 7.5 kW is 80%. The calorific value of diesel is 45 MJ kg⁻¹. If the indicated thermal efficiency is 35%, the brake specific fuel consumption of the engine is

Options:

  • (A) 0.135 kg kW⁻¹h⁻¹
  • (B) 0.228 kg kW⁻¹h⁻¹
  • (C) 0.245 kg kW⁻¹h⁻¹
  • (D) 0.286 kg kW⁻¹h⁻¹

Official Answer Key:

D

Step-by-Step Solution & Explanation

Indicated power, IP = BP/η_mech = 7.5/0.80 = 9.375 kW. Indicated thermal efficiency = IP/(fuel energy rate) ⟹ fuel rate = IP/(η_ith × CV) = 9.375/(0.35×45000 kJ/kg) = 5.952×10⁻⁴ kg/s = 2.143 kg/h. BSFC = fuel rate / BP = 2.143/7.5 = 0.286 kg kW⁻¹h⁻¹.

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