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A diesel engine develops a Brake power of 4.5 kW. Its indicated thermal efficiency is 30% and the mechanical efficiency is 85%. Considering calorific value of the fuel as 40000 kJ/kg, the fuel consumption will be

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  1. 2.1 kg/hr
  2. 3.9 kg/hr
  3. 1.6 kg/hr
  4. 4.6 kg/hr

Answer (Detailed Solution Below)

Option 3 : 1.6 kg/hr
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Detailed Solution

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Concept:

Mechanical efficiency:

Indicated thermal efficiency:

Heat added per second HA/s =   (C.V.)f, where C.V. = Calorific value of fuel 

Calculation:

Given:

Brake power, B.P = 4.5 kW, Indicated thermal efficiency ηith = 30% = 0.3, Mechanical efficiency ηm = 85%, Calorific value (C.V)f = 40000 kJ/kg

  ⇒   = 5.294 kW

 

where I.P = indicated power, HA/s = Heat added per second

 = 17.647 kJ/s

 = 1.6 kg/hr.

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