Answer:
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
