Question
Download Solution PDFThe 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.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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 tone
From concept,
\(\eta_T = 0.85 × 0.9 = 0. 765\)
= 76.5 % (In percentage)
Last updated on Jun 16, 2025
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