Answer:
a) at 2° => 0.22
6° => 0.66
10° => 1.1
b) at 2° => 0.363
6°=>0.803
10° => 1.243
c) at 2° => 0.0275
6° => 0.4675
10° => 0.9075
Explanation:
We are given:
Lift curve slope = 0.11 per degree
[tex] a_1=2degrees; a_2=6degrees; a_3=10degrees[/tex]
a) to find symmetric airfoil, we use the formula;
[tex]C_L = C_ha * a[/tex]
[tex] at a_1 = 2 => C_L = 0.11*5[/tex]
= 0.22
[tex] at a_2 = 6=> C_L = 0.11*6[/tex]
=0.66
[tex] at a_3 =10=> C_L = 0.11*10[/tex]
= 1.1
b) to find cambered airfoil, with a zero lift angle of attack at -1.3°, we use:
[tex] \frac{C_L-0}{a+1.3} = 0.11[/tex]
Making C_L subject of the formula, we have:
[tex] C_L = 0.11(a +1.3)[/tex]
At a = 2 => 0.11(2+1.3) = 0.363
At a = 6 => 0.11(6+1.3) = 0.803
At a = 10 => 0.11(10+1.3)= 1.243
c) to calculate for a eflexed airfoil with a zero lift angle of attack of 1.75 degrees, we use:
[tex] \frac{C_L-0}{a-1.75}[/tex]
[tex] C_L = 0.11(a-1.75)[/tex]
At a= 2 => 0.11(2-1.75) = 0.0275
At a = 6 => 0.11(6-1.75) = 0.4675
At a = 10 => 0.11(10-1.75) = 0.9075
