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The thermal and electrical efficiencies of a 100 MW steam station are respectively 30% and 92%. The coal used has a calorific value of 6400 kcal / kg. For the supply of full-load rated capacity the coal consumption in kg / hour would be approximately
1. 24340
2. 32450
3. 48690
4. 64910

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Correct Answer - Option 3 : 48690

Concept:

Consumption in a Thermal Plant:

  • Overall efficiency = product of all given efficiencies.
  • The heat produced / hour, H = (Units Generated per hour / overall efficiency)
  • Coal Consumption / hour = (H / Calorific value)

 

\({\eta _{overall}} = {\eta _{thermal\;}} \times {\eta _{electrical}}\)

Calculation:

Given that, thermal efficiency = 30%

Electrical efficiency = 92%

\({\eta _{overall}} = 0.30 \times 0.92 = 0.276\)

Units Generated / Hour = 100 × 1000 kWh

Therefore, \(H = \frac{{100 \times 1000}}{{0.276}}\)

Using the conversion 1 kwh = 860 kcal

H = 311.6 × 106 kcal

Calorific Value = 6400 kcal/kg

Coal consumption / Hour = (H / Calorific Value)

Coal consumption / Hour = \(\frac{{311.6 \times {{10}^6}}}{{6400}}\)

= 48690 kg

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