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Homework Help: Theoretical power from wind turbine

  1. Sep 24, 2008 #1
    We honestly haven't gone over this in class and I have no idea why my prof wants us to do this question... so I'm given it a go and come up empty. NEED HELP!

    1. The problem statement, all variables and given/known data
    Calculate the maximum theoretical power from a wind turbine if the radius of the vane area is 18.1 m and the wind speed is 8.3 m/s. Express the answer in kilowatts.

    2. Relevant equations
    I'm thinking, from what I've found on the internet, that the correct equation would be either
    P = 0.5 x rho x A x V3
    P = 0.5 x rho x A x Cp x V3 x Ng x Nb
    Assuming: Rho=1.225 kg/m3, Cp=.59, Ng=1, and Nb=1

    and of course A=pi*r2

    3. The attempt at a solution
    I've tried each method and I've gotten the wrong answer each time.
    For the first I get 360.451 kW
    for the second I get 212.66 kW

    Anyone want to give me some help
    It'd be much appreciated!
  2. jcsd
  3. Sep 25, 2008 #2
    Anyone? I'm really struggling here and my Prof just said to go find help and wouldn't help me at all... which leaves me here.
  4. Sep 25, 2008 #3


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    I'm guessing the maximum possible power is to capture all the kinetic energy of the wind hitting the area of the blades. So E = 1/2mv2

    To get the mass/s you need the density of air and th evolume per second.
    The volume is just the area of the circle swept by the blades and the lenght of the
    column of air that passes through each second.
  5. Sep 25, 2008 #4
    Your approach is correct. What are you using for the air density? I found a solution of 330kw assuming total energy conversion. Using this equation;

    P = 0.5*p*A*V^3

    In the real world power of a wind turbine is limited by the Betz limit, which in that one equation is represented by Cp. So the actual maximum efficiency of a wind turbine is;

    Pmax = 0.5*p*A*V^3*0.59

    This equation is derived from newtons second law. In a nutshell you are just finding the total kinetic energy that the wind poses.
  6. Sep 25, 2008 #5
    I think this is all wrong... just disregard this...
    ok, so in that case...

    -Circ=pi*18.12=1029.217 m2
    -Vol=1029.217 m2*8.3 m/sec=8542.503 m3/sec
    -Density=1.225 kg/m3 or perhaps 1.29 kg/m3 or maybe 1.275 kg/m3 I can't find one exact number online... Anyone know for sure? We're talking STP I believe.
    -Mass=Vol/Density=6973.47 kg (1.225) 6700.002 kg (1.275) or 6622.095 kg (1.29)
    -And, P=.5*m*8.32

    1.225=240201.24 w=240.20 kw
    1.275=230781.58 w=230.78 kw
    1.29=228098.07 w=228.10 kw

    Each answer comes back wrong! I'm going out of my mind here! What am I doing wrong?
    Last edited: Sep 25, 2008
  7. Sep 25, 2008 #6
    So your rho is approximately 1.1215 kg/m3?

    I'm guessing this is not counting Betz limit as this is a Phy101 class. What frustrates the hell out of me is we haven't discussed this at all and it's not even in the readings. We're doing the buoyancy right now, which IS fluids I suppose, but it's not anywhere near this problem...

    BTW, it said that was wrong too... So far I've used 14 of my 25 guesses at this one. Needless to say I'm getting frustrated....
  8. Jan 11, 2011 #7
    Excuse me for being PLANE; but, the wind mill does not produce any electricity (perhaps static)--that is the job of the attached device--generator/ alternator. These devices are dependent on some sort of transmission to deal with the variables encountered with the wind. Sun up and down's transition in wind and if there is any at all. Be it prevailing at 5 M.P.H. or 35--with 55 M.P.H. gusts. A very heavy generator will not turn in a light wind whereas a small on may burn up in a gale. Gary.Once again excuse me for avoiding theory.
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