calculating lift/Bernoulli forces on hydrofoil for windsurfing board

Discussion in 'Hydrodynamics and Aerodynamics' started by Ittiandro, Mar 20, 2018.

  1. Ittiandro
    Joined: Sep 2016
    Posts: 6
    Likes: 0, Points: 1, Legacy Rep: 10
    Location: montreal

    Ittiandro Junior Member

    I know little about physics and I am planning to make a prototype of a hydrofoil wing for my windsurfing board,

    Knowing the weight ( board, rig, and myself, 110 kg), the velocity of the board and the wing surface is it possible to calculate the lift force on the wing, to see if it can offset the 110 Kg downward weight?

    Using the lift force formula L= 1/2ρ ( density) x V2 x A x CL, I have come up with
    L=4,650,125 N=474 kg, which is almost 5 times greater than 110 Kg., hence sufficient to lift the board off the water.

    Here are the parameters I used:

    ρ ( rho, water density) = 1000 kg/m3 ( water) V=10 kmh, A ( area) =1200 cm2 and Cl= 1.005215,( Lift coefficient)

    Note, from what I read : Cl=L/(q x A) where q= 1/2ρ x V2

    I read that the angle of attack affects the Cl, but I can’t see how to fit it in the calculation.

    The best angle of attack of the wing is given 10°-15 ° for the smallest drag.

    I don’t know if the hydrostatic pressure from below further increases the lift and how to calculate it, but if 474 kg of Bernoulli force alone is correct, this should ensure the lift., regardless of the hydrostatic pressure.

    I may be wrong, because I don’t much I physics.Could you comment on this?


    Thanks


    Ittiandro
     
    Last edited: Mar 20, 2018
  2. Erwan
    Joined: Oct 2005
    Posts: 460
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    Location: France

    Erwan Senior Member

    Interesting "homework" you have to do.

    At first glance hydrofoils work better within a Cl (Lift coef) range below 0.6 / 0.5, so for AoA below 4°

    The drag you have is probably the 2D (section) drag as a function of Cl.

    I think you should consider the induced drag too, using the following formula which is probably only a proxy, but accurate enought.

    Cx(induced)= Cl^2/(Pi*AR) while you effective lift coef (Cl3d)is also dependent of the AR =(Aspect Ratio)

    With Cl(3d)= Cl2d/(1+AR/2)

    So with a little hand made itération you should make it.

    As section drag and other friction drags are function of (V^2)
    and
    Induced drag coef is function of (1/V^2)
    So the minimum of their sum is when V is such that Sum of Induced drags = Sum of friction drags

    It works for an airplane, for a boat with aero drag in addition to hydro drag it is a bit more complicated, But I feel confident you will find on this forum a more academic answer likely to complement my intuitive one.

    Cheers

    Erwan
     
  3. Doug Lord
    Joined: May 2009
    Posts: 16,679
    Likes: 349, Points: 93, Legacy Rep: 1362
    Location: Cocoa, Florida

    Doug Lord Flight Ready

    ==========================
    The book : "Hydrofoils Design Build Fly" by Ray Vellinga is an excellent reference for foiler design(and includes some Moth data). It would help you immensely and be of value for a long time to come. https://www.amazon.com/Hydrofoils-Design-Build-Ray-Vellinga/dp/0982236115
    Good Luck!
    PS-if you don't have it already buy the "Theory of Wing Sections Including A Summary of Airfoil Data" because used with Vellinga's book you can determine angle of attack(incidence) ,Coefficient of Lift(and drag) and all the pertinent info you need to come very close to the foil you need. The two together are not very expensive and will serve you well:
    https://www.amazon.com/Theory-Wing-Sections-Aeronautical-Engineering/dp/0486605868
     
    Last edited: Mar 28, 2018
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