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The following data is available for a steam power station:

maximum demand = 25 MW

Load factor = 0.4

Coal consumption = 0.88 kg/kWh

Boiler efficiency = 85%

Turbine efficiency = 90%

Price of coal = Rs. 55 per tones

Find the thermal efficiency of the station.


1. 65.2%
2. 76.5%
3. 99.8%
4. 62.32%

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Correct Answer - Option 2 : 76.5%

Concept:

Thermal Efficiency:

  • Thermal Efficiency (\(\eta_T\)) is the ratio of the heat utilized by a heat engine to the total heat units in the fuel consumed.
  • It is also the product of Boiler efficiency and Turbine efficiency.
  • \(\eta_T\) = Boiler efficiency × Turbine efficiency.


Load Factor:

  • It is the ratio of Average load to maximum demand.
  • Load Factor is always less than or equal to 1.


Calculation:

Given

Maximum demand = 25 MW

Load factor = 0.4

Coal consumption = 0.88 kg/kWh

Boiler efficiency = 85%

Turbine efficiency = 90%

Price of coal = Rs. 55 per tones

From concept,

\(\eta_T = 0.85 × 0.9 = 0. 765\)

= 76.5 % (In percentage)

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