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A diesel power station has fuel consumption of 0.215 kg per kWh, the calorific value of fuel being 10,000 kcal / kg. Determine the overall efficiency.


1. 40%
2. 30%
3. 45%
4. 55%

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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:

m= 0.215 kg / kWh = \(\frac{{0.215}}{{3600 × {{10}^3}}}\;kg/J\)

CV = 10,000 kcal / kg = 4.18 × 10J / kg

\(\% \eta = \frac{{100 × 3600 × {{10}^3}}}{{0.215 × 4.18 × {{10}^7}}} \approx 40\% \)

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