Question

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

Answer 1

Correct Answer - Option 3 : 1.6 kg/hr

Concept:

Mechanical efficiency:

\(\eta_m = \frac{Brake\ Power}{Indicated\ Power} = \frac{B.P.}{I.P.}\)

Indicated thermal efficiency:

\(\eta_{ith} = \frac{Indicated\ Power}{Heat\ added\ per\ second}\)

Heat added per second HA/s =  \(\dot m_f\ \times \) (C.V.)_f, where C.V. = Calorific value of fuel 

Calculation:

Given:

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

\(η_m = \frac{B.P}{I.P.}\)  ⇒  \(I.P = \frac{B.P}{η_m} = \frac{4.5}{0.85} \) = 5.294 kW

\(η_{ith} = \frac{I.P}{HA/s}\)

\(HA/s = \frac{I.P}{η_{ith}} = \frac{5.294}{0.3} = 17.647\ kW\) 

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

\(HA/s = \dot m_f\times {C.V}_f\) = 17.647 kJ/s

\(\dot m_f = \frac{17.647\ \times\ 3600 }{40000}\) = 1.6 kg/hr.