Correct Answer - Option 3 : 39.2%
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
Calculation:
mf = 0.2 kg / kWh = \(\frac{{0.2}}{{3600 × {{10}^3}}}\;kg/J\)
CV = 11, 000 k cal / kg = 4.18 × 11 × 106 J/kg = 45.98 × 106 J/kg {1 Cal = 4.18 J)
\(\% \eta = \frac{{100 × 3600 × {{10}^3}}}{{0.2 × 45.98 × {{10}^6}}} = 39.14\%\simeq39.2\% \)