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

Lift slope = 0.55/deg

Span efficiency factor = 1.72 Am

Step-by-step explanation:

S = 1.5

AR = 6

V = 180

∝ = -1 L = 0 D= 0.181

∝ = 2 L = 5 D = 0.23

Cd =
`(0.181)/(0.5X0.00237X180^(2)X1.5 ) = 0.00314`

∝ = 2

Cl =
`(5)/(0.5X0.00237X180^(2)X1.5 ) = 0.0868`

Cd =
`(0.23)/(0.5X0.00237X180^(2)X1.5 ) = 0.00399`

Cd =
`Cd_(0) + (Cl^(2))/(\pi eX 6)`

0.00399 = 0.00314 +
`(0.166^(2) )/(\pi e X6)`

e = 1.72 Am

Lift slope = a =
`(dCl)/(d\alpha ) = (0.166-0)/(2 - (-2)) = 0.55/deg`