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Forum Index : Windmills : swept area of savonius rotor

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ewells13
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Joined: 07/05/2008
Location: Australia
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Posted: 09:29am 07 May 2008

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Chasing some info on how to calculate the "swept area" of a simple two-blade savonius rotor. Anything would be much appreciated, thanks

Elliott

Gizmo

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Joined: 05/06/2004
Location: Australia
Posts: 5116

Posted: 11:22am 07 May 2008

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Do you want swept area or wind frontage?
On a HAWT windmill, the swept area is also the area facing the wind. This is the area of wind the turbine can extract power from. Area would be Pi*R*R ( Pi R squared ).
The swept area of a VAWT ( eg Savonius ) isn't really relevant. The wind frontage, that is the area facing the wind, would be calculated with diameter by height. So a 2 meter diameter Savonius that was 3 meters high would have a area facing the wind of 6 square meters. But a VAWT could be making power from the back as well as the front. A high speed Darius windmill makes power for over 300degress of its rotation, even when the blades are heading upwind they are making power.
Glenn

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Wells

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Joined: 05/05/2008
Location: Australia
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Posted: 03:44am 08 May 2008

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Thanks Glenn,

I thought I was chasing swept area. The reason I want it is I have a BASIC equation for the calculation of power for a rotor (P=0.5*rho*A*V^3), where A is swept area; but I wasnt sure if you use the frontal area for a Savonius rotor due to the fact that at any one time only one rotor (half the diameter) is catching the wind... Reading your reply I am guessing this assumption is incorrect?
Elliott

EWells

GWatPE

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Joined: 01/09/2006
Location: Australia
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Posted: 06:59am 08 May 2008

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Hi Elliott,

The equation you have presented is the energy of the moving air related to the velocity and the density and the cross section that is moving. The Rotor power will have additional vane/blade efficiency factors and then there may be gearing losses and alternator/generator efficiency factors.

For a fair comparison with other turbine types, the total area that is in the wind should be included. The efficiency would be the power extracted from the total cross section in the wind, divided by the calculated power in the wind at that moment for that same cross section... .. Gordon.

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Wells

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Joined: 05/05/2008
Location: Australia
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Posted: 07:08am 08 May 2008

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Thanks very much Gordon, well explained

Elliott

EWells

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