# 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 =

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The following data is available for a steam power station:

maximum demand = 25 MW

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.

• 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

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)