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

i would say the answer is food.

hope this !

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