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

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