VAWT Swept Area Calculation

[Yeddid]

I Build My Blade Like A Picture Below



How I can Calculate the swept area.. BTW Some one has done to explain but, iam not sure that explanation can be applied in my turbine or not

Thank You.. This Picture below is his explanation by "IntegEner"



THANK YOU

[electrondady1]

height x width is the traditional way of calculating swept area
i have never seen the method used in your attachment.

[XXLRay]

I understand the shown calculation but it is dfinitely not the way it is usually done. To keep values comparable to other turbines you should just multiply height by width.

[RP]

I don't know.  That 4/3 radius instead of 2X radius just looks like a way to make a VAWT seem 50% more efficient to me.

[Bruce S]

I don't know.  That 4/3 radius instead of 2X radius just looks like a way to make a VAWT seem 50% more efficient to me.

IT would be better to read the entire post found here :http://fieldlines.com/board/index.php?topic=138783.0
It explains the calculations showing the VAWTS have less or worse numbers not more efficient.

Its the last line on the calculations page that summaries it.

Hope this helps clear that up.
Bruce S
 

[fabricator]

I don't know.  That 4/3 radius instead of 2X radius just looks like a way to make a VAWT seem 50% more efficient to me.

They have to do something to make vawts seem worth building.

[Bruce S]

I don't know.  That 4/3 radius instead of 2X radius just looks like a way to make a VAWT seem 50% more efficient to me.

They have to do something to make vawts seem worth building.

I disagree.
VAWTs are very much worth building. Windstuff Ed has proven that time and time again.
Go tell him they're not worth building

When building VAWTs they like HAWTs need to be built with correct expectations, NOT wild ideas of running your whole house off one for much less and have enough to "sell" back to the grid to retire off of.
That nonsense should be saved for shysters and used car salesmen
 
As I said in my previous post... that is taken out of context the corrected part is the very last line that says only 2/3rds.

TRY READING THE ENTIRE SECTION.

Incomplete information is just bad. and serves no good.


 

[REdiculous]

If your vawt has 3 blades but only 1 is "in the wind" at any given moment then it seems to me that the effective swept area is equal to the width*height of one blade, not all 3.

[XXLRay]

... it seems to me that the effective swept area is equal to the width*height of one blade, not all 3.

No. Important is the amount of wind that hits the turbine. Otherwise you would calculate the the swept area by the blade area for HAWTs too. That is nonsense. It is of interest how much energy can be fixed by a "tunnel" of wind. This one is determined by the complete area covered by the turbine blades at any time facing the wind. For HAWTs this is a circle and for VAWTs this is a rectangle (or oval). It does not matter that a blade does not cover a part of the area at all times. When it passes once it counts.

[Kevortex]

"It does not matter that a blade does not cover a part of the area at all times. When it passes once it counts. ~ XXLRay"

I believe XXLRay to be correct. This is the way I have calulated my own "Kevonious" wind turbines and it works quite well.

Do keep in mind that the distance from the axis of rotation to the VAWT blade must be taken into consideration.

Greater distance (swept diameter) = more torque. Now add in your actual surface area of the blade and you'll have a more accurate
calculation.

On a HAWT this would be the equivelant of the difference with using long skinny blades or shorter wider blades... or more blades.

I hope this makes sence ! LOL