Calculating Power Output of Wind Turbine

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To calculate the power output of a wind turbine with a 20m diameter and 40% efficiency at a wind speed of 10m/s, the relevant equation is P=0.5 * air density * swept area * wind velocity^3. Using an air density of 1.23 kg/m³, the initial calculation yields 12,300 Watts, leading to an output of 4,920 Watts after applying the efficiency factor. However, concerns were raised about the accuracy of the swept area calculation, as it should be based on the area of a circle (π*r²) rather than just the diameter. At sea level and 20 °C, the air density is approximately 1.2 kg/m³, which could also affect the results. The discussion emphasizes the importance of correctly calculating the swept area to ensure accurate power output estimations.
JosephF
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Homework Statement



1. A wind turbine with a swept diameter of 20m and an efficiency of extracting energy from the wind of 40% is subjected to an average wind speed of 10m/s.
(a) How much power might the turbine produce?


Please can some one point me to the right equation as everyone I can find includes cut-in speeds etc...
 
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This link might be useful:
http://www.otherpower.com/windbasics1.html
 
Last edited by a moderator:
Thanks for the link.

P=0.5 * air density * swept area * wind velocity3

Taking air density as 1.23kg per cubic meter.

P=0.5 * 1.23 * 20 * 103

P=12,300Watts

40% efficent so output power is 4,920Watts.

Is this correct as the value seems quite high? Also there was no mention of air density in the question... Thanks,
 
JosephF said:
Thanks for the link.

P=0.5 * air density * swept area * wind velocity3

Taking air density as 1.23kg per cubic meter.

P=0.5 * 1.23 * 20 * 103

P=12,300Watts

40% efficent so output power is 4,920Watts.

Is this correct as the value seems quite high? Also there was no mention of air density in the question... Thanks,

Wikipedia said:
At sea level and 20 °C, air has a density of approximately 1.2 kg/m3.

They are after all 30 foot blades. That's a big area and a hefty chunk of air at 10m/s. I haven't calculated it out, but you would expect power in the kilowatts I would think.
 
You have calculated your "swept area" incorrectly.

You have used the diameter, not the area.

Remember area of circle = pi*r^2
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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