Wind-Turbine Tower Forces

[Curbie]

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???

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

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???

Thanks.

Curbie

Calculating Drag Fd = .5 * p * v^2 * A * Cd
Speed of Wind in MPH (Sm) 50 MPH <<< Input
Speed of Wind in kPH (Sk) 80.46744269 kPH =Sm * (Ft2Mi /
Ft2m) / m2km
Velocity of Wind in MPH (Vf) 73.33333333 Ft/s =Sm * (Ft2Mi /
min2h / s2min)
Velocity of Wind in kPH (Vm) 22.35206741 m/s =Vf / Ft2m
Diameter of Rotor in feet (Diaf) 14 Feet <<< Input
Diameter of Rotor in meters (Diam) 4.26721287 Meters =Diaf *
(1 / Ft2m)
Radius of Rotor in feet (Rf) 7 Feet =Diaf / 2
Radius of Rotor in meters (Rm) 2.133606435 Meters =Diam / 2
Area of Rotor in feet (Af) 153.93804 Feet^2 =PI() * Rf^2
Area of Rotor in meters (Am) 14.30139816 Meters^2 =PI() *
Rm^2
Density of Air imperial (Pi) 0.074914141 Lbs/Ft^3 =(Pm *
Lb2kg) / Ft2m^3 Altitude and temperature dependant
Density of Air metric (Pm) 1.2 kg/m^3
Altitude and temperature dependant
Drag Coefficient (Cd) 1.28 Coefficient
Drag in Pounds (Flbs) 39691.04976 Lbs =0.5 * Pi * Vf^2 * Af *
Cd
Drag in Kilograms (Fkg) 5487.50735 kg =0.5 * Pm * Vm^2 * Am *
Cd
Constants
Feet to a Mile (Ft2Mi) 5280 Feet
Meters to Kilometer (m2km) 1000 Meters
Feet to a Meter (Ft2m) 3.28083 Feet
Meters to Foot (m2Ft) 0.3048 Meters
Pound to Kilogram (Lb2kg) 2.20462 Pounds
Seconds to a Minute (s2min) 60 Seconds
Minutes to Hour (min2h) 60 Minutes

 

[daestrom]

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

 

[Curbie]

Thanks both daestrom and Robert for your time and help!
I'll try to straighten out my spread-sheet tonight.
I greatly appreciate it.

Curbie