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Posted by daestrom on October 2, 2009, 5:44 pm
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Curbie wrote:
> I've been chasing this formula around for a couple days and am still
> pretty confused, everywhere I turn with this idea just seems to add
> more confusion, maybe someone here can help clear this up.
>
> A thread at fieldlines uses this formula to calculate tower wind
> loading:
> Loads = (air density) X (Area) X (V^2)/2
> Air density is .002
> http://www.fieldlines.com/story/2009/9/18/201016/375
>
> Wikipedia and Hugh Piggott and Paul Gipe seem to have air density at
> ~1.2 kg/m^3 (elevation and temperature dependant).
> http://en.wikipedia.org/wiki/Density_of_air
>
> Converting 1.2 kg/m^3 to Lbs/Ft^3 is:
> =(Air Density {in kg/m^3} * Lbs2kg) / Ft2m ^3
> 0.0748 =(1.2 * 2.2) / 3.28^3 and NOT .002
> So where does this .002 come from???
>
When using Imperial Units, you have to account for the difference
between a pound-mass and a pound-force. That means in this case
dividing the density by 'g-sub-c' (32.2 lbm-ft / lbf-s^2). That should
make your units work out.
Most likely they are giving you air density in 'slug/ft^3'. This is
another variation of Imperial units where mass is measured in 'slugs',
which are exactly 'g' times the weight of an object on a scale (1 slug =
32.2 lbm). So again, it would be 0.0748 lbm/ft^3 *(1slug/32.2lbm) =
0.002 slug / ft^3)
> I tried to track down the meaning of that .002 constant and couldn't,
> and in the process found another formula (I think on the NASA site):
> Fd = .5 * p * v^2 * A * Cd
>
The only trouble with this formula is coming up with a reasonable value
for Cd. It can range from 0.0 to 1.0. With Cd of 1.0, this is the
formula for the force on a flat plate held against the wind.
And for something like a wind machine, Cd really changes as it changes
speed. I wouldn't be surprised if it varies from below .5 in moderate
winds at slow RPM to as high as .95 or more when it's almost ready to
fly apart. But the worst it can possibly be is 1.0, so if you design
for that your good.
If you apply a brake and stop the blades from turning, then the area
drops to just that of the blade, not the swept area. This greatly
reduces the tower loading. Or turn the axis of rotation 90 degrees to
the wind so the only area is the tower itself and the edge-wise of the
blade and generator pod.
Of course if your brake or axis-turning mechanism fails... :0
> Using this formula with both imperial AND metric units I get two
> different results that won't convert to each other (39691 Lbs. drag to
> 5487 kg. drag) the spread-sheet is at the end of this post.
>
> I know I'm goofing up something, does any know what???
The drag in SI units would be in Newtons, not kilograms. Kilograms is
really a measure of mass. But a lot of older documents used it as a
measure of force. A kilogram exerts a force of about 9.8 Newtons in
standard gravity.
Using some of your numbers...
Imperial
Fd = .5 * (0.0748 lbm/ft^3) * (73.33 ft/s)^2 * (153.9 ft^2)
/ (32.2 lbm - ft / lbf-s^2)
Fd = 961. lbf
Metric
Fd = .5 * (1.2 kg/m^3) * (22.35 m/s)^2 * (14.3 m^2)
Fd = 4285.9 N
Since a kilogram force is 9.8 N...
Fd = 4285.9 N = 437.3 kilogram-force
437.3 kilogram-force * (2.2kg/lb) = 962. lb
Close enough...
Later,
daestrom
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