Answer:
Explanation:
a.) To find Secrional liftSectional lift Coefficient [tex](c_{I})=\frac{W}{(0.5*Rho*V^{2})}[/tex]
where Rho = 1.225Kg/m³ (density of air at sea level)
[tex](c_{I})=\frac{1000}{(0.5*1.225*55^{2})}=0.5397[/tex]
b.) To obtain the sectional induced coefficient:
[tex]C_{D} = \frac{(C_{L})^2}{π(AR)}[/tex]
[tex]C_{D} = \frac{(0.5397)^2}{π(5)}=0.0185[/tex]
C.) For the effective angle of attack ∝0
∝0 = [tex]\frac{C_{l}}{m_{0}}\\=\frac{0.5397}{5.7}\\=0.0947rad[/tex]
For induced angle of attack ∝i:
∝i= [tex]\frac{-C_{D}}{C_{L}}\\=\frac{-0.0185}{0.5397}\\=-0.0343rad[/tex]
For absolute angle of attack ∝a:
∝a = ∝0 - ∝i = 0.0947 - (-0.0343) = 0.1290 rad
For geometric angle of attack ∝g:
∝g = ∝a + ∝L given ∝L = -2°≅-0.0349 rad
= 0.1290-0.0349 = 0.0941rad
