Connect with us

Wind-Turbine Tower Forces

Discussion in 'Home Power and Microgeneration' started by Curbie, Oct 1, 2009.

Scroll to continue with content
  1. Curbie

    Curbie Guest

    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
     
  2. daestrom

    daestrom Guest


    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...

    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
     
  3. Curbie

    Curbie Guest

    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
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-