A frequently quoted rule of thumb in aircraft design is that wings should produc

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.) 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? (a) m/s (b) How fast (in m/s) must air move over the upper surface at a cruising speed of 240 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

Explanation / Answer

Given,

Ft = 1000 N ;

a)v = 63 m/s ; rho = 1.29 kg/m^3

from Bernoulli's theorem

P1 + 1/2 rho v1^2 = P2 + 1/2 rho v2^2

v2 = sqrt [2(P1 - P2)/rho + v1^2]

v2 = sqrt [2 x 1000/1.29 + 63^2] = 74.3 m/s

Hence, v2 = 74.3 m/s

b)v1 = 240 m/s

v2 = sqrt [2(P1 - P2)/rho + v1^2]

rho = 1.29/4= 0.323

v2 = sqrt [2 x 1000/0.323+ 250^2] = 262.1 m/s

Hence, v2 = 262.1 m/s

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