SSE JE Electrical Engineering 10 Dec 2020 Papers

Overall efficiency can be calculated as $\%\mathrm{\eta }=\frac{100}{{\mathrm{m}}_{\mathrm{f}}\times \mathrm{CV}}$
Where,
${\mathrm{m}}_{\mathrm{f}}=$ mass flow rate of fuel or fuel consumption in kg / Joule
${\mathrm{C}}_{\mathrm{V}}=$ calorific value of fuel in Joule / kg
1 cal = 4.18 Joule
${\mathrm{m}}_{\mathrm{f}}=0.215\mathrm{kg}∕\mathrm{kWh}=\frac{0.215}{3600\times {10}^3}\mathrm{kg}∕\mathrm{J}$
$\mathrm{CV}=10,000\mathrm{kcal}∕\mathrm{kg}=4.18\times {10}^7\mathrm{J}∕\mathrm{kg}$
$\%\mathrm{\eta }=\frac{100\times 3600\times {10}^3}{0.215\times 4.18\times {10}^7}\approx 40\%$