Correct Answer - Option 1 : 40%
Concept:
Overall efficiency can be calculated as
\(\% \eta = \frac{{100}}{{{m_f} × CV}}\)
Where,
mf = mass flow rate of fuel or fuel consumption in kg / Joule
CV = calorific value of fuel in Joule / kg
1 cal = 4.18 Joule
Calculation:
mf = 0.215 kg / kWh = \(\frac{{0.215}}{{3600 × {{10}^3}}}\;kg/J\)
CV = 10,000 kcal / kg = 4.18 × 107 J / kg
\(\% \eta = \frac{{100 × 3600 × {{10}^3}}}{{0.215 × 4.18 × {{10}^7}}} \approx 40\% \)