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31 May, 22:52

Consider an NACA 2412 airfoil with a 2-m chord in an airstream with a velocity of 50 m/s at standard sea level conditions. If the lift per unit span is 1353 N, what is the angle of attack

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  1. 31 May, 23:10
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    Explanation

    Given that,

    c=2m at sea level

    V=50m/s at infinity

    Lift per unit span is

    L=1353N

    The dynamic pressure is given as

    Density of water at infinity p∞=1.23kg/m³

    q=½p∞V∞²

    q=½*1.23*50²

    q=1537.5 N/m²

    Now,

    Lift coefficient is given as

    Cl = L/qc

    Cl = 1353 / (1537.5 * 2)

    Cl = 0.44

    Then, coefficient is given as

    Cl=2πα

    Where α is the required angle

    α = Cl/2π

    α = 0.44/2π

    α=0.007rad

    Now, 1 rad = 57.296°

    Then, α=0.007rad = 0.007*57.296°

    α=4.01°

    The angle attack is 4.01°

    Or Cl can also be

    Cl=2πSinα

    Sinα=Cl/2π

    Sinα=0.44/2π

    Sinα=0.007

    Then, α=arcsin (0.007)

    α=4.02°

    Both are correct
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