Question

A finite wing area of 1.5 ft2 and aspect ratio of 6 is tested in a subsonic wind tunnel at a velocity of 180 ft/s at standard sea-level conditions. At an angle of attack of −1°, the measured lift and drag are 0 and 0.181 lb, respectively. At an angle of attack of 2°, the lift and drag are measured as 5.0 and 0.23 lb, respectively. Calculate the span efficiency factor and the infinite-wing lift slope.

Answer 1

Answer:

Lift slope = 0.55/deg

Span efficiency factor = 1.72 Am

Explanation:

S = 1.5

AR = 6

V = 180

∝ = -1    L = 0    D= 0.181

∝ = 2    L = 5    D = 0.23

Cd = $\frac{0.181}{0.5X0.00237X180^{2}X1.5 } = 0.00314$

∝ = 2

Cl = $\frac{5}{0.5X0.00237X180^{2}X1.5 } = 0.0868$  

Cd = $\frac{0.23}{0.5X0.00237X180^{2}X1.5 } = 0.00399$  

Cd = $Cd_{0} + \frac{Cl^{2}}{\pi eX 6}$

0.00399 = 0.00314 +  $\frac{0.166^{2} }{\pi e X6}$

e = 1.72 Am

Lift slope = a = $\frac{dCl}{d\alpha } = \frac{0.166-0}{2 - (-2)} = 0.55/deg$