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.**