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Wind-Turbine Tower Forces

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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
Loads = (air density) X (Area) X (V^2)/2
Air density is .002

Wikipedia and Hugh Piggott and Paul Gipe seem to have air density at
~1.2 kg/m^3 (elevation and temperature dependant).

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???



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() *
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 *
Drag in Kilograms (Fkg)    5487.50735    kg    =0.5 * Pm * Vm^2 * Am *
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:

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.

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

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

Curbie wrote:

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)

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

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...

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

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...



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.


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