Posted by *Curbie* on October 1, 2009, 9:09 pm

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

Posted by *Robert Copcutt* on October 2, 2009, 3:54 pm

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???*

You need to check consistency of units. In the SI system force is

measured in Newtons. Dividing .0748 by the gravitational constant (about

32 ft/s^2) gets you almost there.

When the turbine (assuming HAWT) is facing the wind and spinning at

optimum speed then (Area) will be the swept area of the blades and the

result will be the maximum possible force on the tower. This force gets

huge in strong winds so the turbine needs to do something to protect

itself and make the effect area facing the wind smaller.

*> *

*> 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*

*> *

Cd is the drag coefficient and you need to be careful that you are using

a coefficient that applies to your situation.

Have a look at http://www.scoraigwind.com/wpNotes/index.htm and the

bladedesign.pdf.

Posted by *daestrom* on October 2, 2009, 9:44 pm

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

Posted by *Curbie* on October 3, 2009, 12:05 am

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

> 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???