Question

A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing. (The fact that a wing has a top and bottom surface does not double its area.)

a)At takeoff the aircraft travels at 63.0 m/s, so that the air speed relative to the bottom of the wing is 63.0 m/s. Given the sea level density of air to be 1.29 kg/m³, how fast (in m/s) must it move over the upper surface to create the ideal lift?

m/s?

b)How fast (in m/s) must air move over the upper surface at a cruising speed of 250 m/s and at an altitude where air density is one-fourth that at sea level? (Note that this is not all of the aircraft's lift--some comes from the body of the plane, some from engine thrust, and so on. Furthermore, Bernoulli's principle gives an approximate answer because flow over the wing creates turbulence.)

m/s?

Answer 1

Given that

Sea level density of air to ρ = 1.29 kg/m^3

takeoff speed od aircraft v= 63 m/s

Lift required for wing P = 1000 N/m^2

Part(a)

Bernoulli's equation for pressure and velocity of fluid

P + 1/2ρv^2 + ρgh = k (constant)

for two different points let's consider v1 and v2 are speed of air at bottom and upper surface of wing respectively

P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2

In our situation air is flowing to the upper and bottom side of wing (h1= h2)

P1 + 1/2ρv1^2 = P2 + 1/2ρv2^2

1/2 ρv2^2  = (P1-P2)+1/2ρv1^2

v2^2 = (2(P1-P2)+1/2ρv1^2 ) / ρ

v2 = Sqrt (2(P1-P2)+1/2ρv1^2 ) / ρ

v2 = Sqrt (2(ΔP)+1/2ρv1^2 ) / ρ

v2 = Sqrt [ 2(ΔP)/ρ + v1^2]

v2 = Sqrt [ 2(1000) /1.29 + (63)^2]

v2 = 74.29 m/s

Part(b)

in this case we have

ρ = 1.29/4 =0.3225 kg/m^3

v1 = 250m/s

v2 = Sqrt [ 2(1000) /0.3225 + (250)^2]

v2 = 262.10 m/s

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