Correct Answer - Option 3 : 34.1%

**Concept**:

Indicated thermal efficiency:

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

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

__Calculation:__

__Given:__

V̇_{f} = 8 lit/hr = \(\frac{8\;×\;10^{-3}}{3600}\)m^{3}/sec, CV = 44000 kJ/kg, SG = 0.75, IP = 25 kW

ṁf = ρV̇f = 0.75 × 1000 × \(\frac{8\;×\;10^{-3}}{3600}\) kg/sec.

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

\(\eta_{ith}= \frac{{25 × 60 × 60}}{{\left( {8 × {{10}^{ - 3}} × 0.75 × 1000} \right) × 44000}} × 100 = 34.1\%\)