GATE AG 2014 • Q41 MCQ • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Power efficiencies and measurement, engine performance curves

Question Statement:

A four-stroke, four-cylinder diesel engine running at 2000 rpm develops brake power of 60 kW and the fuel consumption is 0.30 kg kW⁻¹ h⁻¹. The engine has a bore of 120 mm and stroke of 100 mm. If air-fuel ratio is 15:1 and air density is 1.15 kg m⁻³, the volumetric efficiency of the engine in percent is

Options:

  • (A) 43.25
  • (B) 66.32
  • (C) 75.22
  • (D) 86.50

Official Answer Key:

D

Step-by-Step Solution & Explanation

Fuel consumption rate = sfc × BP = 0.30 × 60 = 18 kg/h. Air consumption = 18 × 15 (A/F ratio) = 270 kg/h. Actual air volume flow = 270 / 1.15 = 234.78 m³/h = 0.065217 m³/s. Swept volume/cylinder = (π/4)D²L = (π/4)(0.12)²(0.10) = 1.1310×10⁻³ m³; total for 4 cylinders = 4.5239×10⁻³ m³. 4-stroke engine: 1 intake stroke per cylinder every 2 revolutions ⇒ intake events/s (all cylinders) = (N/2)×4/60 = (2000/2)×4/60 = 66.667 /s. Theoretical (swept) volumetric flow rate = 1.1310×10⁻³ × 66.667 = 0.075398 m³/s. η_vol = actual/theoretical = 0.065217/0.075398 = 86.50%. Answer: (D). Tractors and Power tillers

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