Question

A four-stroke diesel engine operating at 800 rpm uses 0.10 kg of fuel in 4 minutes while developing a torque of 70 Nm. Calculate brake specific fuel consumption

(a) 0.262 kg/kwh

(b) 0.256 kg/kwh

(c) 0.242 kg/kwh

(d) 0.236 kg/kwh

This question was posed to me by my college professor while I was bunking the class.

The doubt is from Fuel Supply System in Ignition or Diesel Engine topic in chapter Ignition and Power Transmission of Farm Machinery

Answer 1

The correct answer is (b) 0.256 kg/kwh

Best explanation: Power=\(\frac{2πNT}{60}=\frac{2*π*800*70}{60}\) = 5864.31 W

Brake specific fuel consumption = \(\frac{0.10*60}{4*5.86}\) = 0.256 kg/kwh.