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A diesel power station has fuel consumption of 0.2 kg per kWh. If the calorific value of diesel is 11000 kcal per kg, overall efficiency of power station is
1. 43.3%
2. 65.5%
3. 39.2%
4. 25.6%

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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:

m= 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 × 10J/kg   {1 Cal = 4.18 J)

\(\% \eta = \frac{{100 × 3600 × {{10}^3}}}{{0.2 × 45.98 × {{10}^6}}} = 39.14\%\simeq39.2\% \)

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