Power in the Wind

[richhagen]

This is just some basic information that may be of use to someone just starting out with wind power.  It goes over how much power is available in the wind.  Feel free to copy, post or forward it if you find it useful. I uploaded the file as HTML to the board here.  

http://www.otherpower.com/images/scimages/742/02_power_in_the_wind.htm

If you copy it you may need the below files from my uploads as well. Just copy the pictures and place them in the same folder as the html document.  Just maybe this will be of use or interest to someone.  Rich

[scottsAI]

I think the right side is labeled upside down. 2 feet at bottom and 24 at top.

[richhagen]

You have to match the color of the curve with the line in the legend since the legend is listed in numerical order (red line in legend matches to red curve on chart).  I didn't think about it at the time, but I guess I could shift it so that the vertical order of the legend matches the order of the curves by power if that makes it simpler to read.  Rich Hagen

[FrankG]

Certainly handy to have such a reference on hand... But one obvious caution is that I assume that the curves apply to ideal blade designs...

As a relative newcomer to "Home-brew" Wind I'm wary of such power curves as the numerous variables in blade design, generator efficency/inefficency etc... make such values hard to attain...

Personally I view such data as Theoretical "Maximums" to aim toward, but not as the expected outcome... Don't get me wrong, I'm grateful for the posting, but have woken-up on the "Conservative Side of the Bed" (It's Election Day in Canada...) Get it??? Conservative Side of the Bed!!!, Election day...

... never mind...

[wind4Reg]

Hey FrankG, I get it, course I'm Canadian too.

[ghurd]

"Theoretical Maximums". The second one is 35%, quite achievable.

G-

[DanB]

Nice charts there!

Here is another good page along the same lines with some explanation.

http://www.ndsu.nodak.edu/ndsu/klemen/Perfect_Turbine.htm

[erichtopp]

Hummmmm, interesting, not until you reach 15 mph wind speed does the blade diameter make much difference at all in the output of power. Rgds, Erich

[richhagen]

Actually it does still make a huge difference, it's just that the scale of the chart makes it difficult to see, because since the power is going up with the cube of the windspeed, the difference, although the exact same percentage wise between any two curves, is difficult to see because the absolute values are so much smaller.  Since the power available is determined by the mass of air moving through the swept area, the power is proportionate to the swept area.  The swept area increases proportionate to the square of the blade diameter.  For example, at any given wind speed, the power available to a 4 foot diameter rotor is 4 times greater than the power available to a 2 foot diameter rotor.  This is because the swept area is 4 times greater.  Swept area is pie multiplied by the square of the radius, and the radius is 1/2 of the rotor diameter. The ratio of the 4 foot diameter rotor swept area to the 2 foot diameter rotor swept area would be:

   (pi*(2 feet)^2)/(pi*(1 foot)^2) = 4

Regardless of the wind speed, the available power will always be 4 times greater for the 4 foot diameter rotor than it is for the 2 foot diameter rotor.

Rich Hagen

[richhagen]

The top chart shows the theoretical value.  The bottom chart shows 35% of this value, which I believe should be obtainable within a narrow range of wind speeds around which the turbine is designed for. (not the whole curve) I base this on the performance data of machines posted here, also, I think I have heard from Dan or someone that this is a reasonable number, and additionally it is about the same as the similar web page that DanB links to in his response to this post.  I hadn't seen that page before but I am happy to see a similar estimate, it seems to me that we should all come to about the same results anyway.  

Unfortunately, in practice, my devices do not always achieve what they possibly could.  Some do pretty good, and some fail miserably.  

Frank, I like to look at the theory as a guess, or model, of what I expect should happen, and then if it doesn't, try to figure out why not.  Rich

[Texas Al]

Do the lines in your graph represent diameter (2x blade length), radius (blade length), or swept area? I really don't expect it to be swept area, because the largest one, 24 feet (square feet presumably), would come out to 2.76 foot blades.

[richhagen]

The curves are labeled for blade diameter, two times the length of an individual blade.  I should re-caption the chart to make that more clear.  Rich