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An engine working on Otto cycle consumes 8 litres of fuel per hour and develops 25 kW. The specific gravity of fuel is 0.75 and its calorific value is 44000 kJ/kg. The indicated thermal efficiency of the engine is:

1. 54.5%
2. 24.2%
3. 34.1%
4. 28.4%

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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:

f = 8 lit/hr = \(\frac{8\;×\;10^{-3}}{3600}\)m3/sec, CV = 44000 kJ/kg, SG = 0.75, IP = 25 kW

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

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

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

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