Question

Brake thermal efficiency of a 15 kW IC engine is 30%. If the fuel used has CV 40 MJ/kg, then the fuel consumption rate in kg/h is:

1. 4.5
2. 1
3. 3
4. 2.5

Answer 1

Correct Answer - Option 1 : 4.5

Concept:

Brake thermal efficiency\(\left( {{\eta _{bth}}} \right) = \frac{{BP}}{{{m_f}\; × \;CV}}\)

where, BP = Brake power of IC engine, m_f = fuel consumption, CV = Calorific value of the fuel

Calculation:

Given:

ηbth = 0.3, CV = 40 × 10³ kJ/kg, BP = 15 kJ/s

Now,  

∴ \(0.3 = \frac{{15}}{{40 ~\times~ 10^3\; × \;{m_f}}}\)

\({m_f} = \frac{{15}}{{0.3}} × \frac{1}{{40000}} × 3600\frac{{kg}}{{hr}}\)

∴ mf = 4.5 kg/hr