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Usimov [2.4K]
3 years ago
12

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
Physics
1 answer:
svetlana [45]3 years ago
7 0

Answer:

    v₂ = 63.62 m / s

Explanation:

For this exercise in fluid mechanics we will use Bernoulli's equation

         P₁ + ρ g v₁² +  ρ g y₁ = P₂ +  ρ g v₂² +  ρ g y₂

where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.

We will assume that the distance between the two parts is small, so y₁ = y₂

        P₁-P₂ =  ρ g (v₂² - v₁²)

pressure is defined by

        P = F / A

we substitute

        ΔF / A =  ρ g (v₂² - v₁²)

         v₂² = \frac{\Delta F}{A \ \rho  \ g} + v_1^2

suppose that the area of ​​the wing is A = 1 m²

we substitute

         v₂² = \frac{1000}{1 \ 1.29 \ 9.8} + 63^2

         v₂² = 79.10 + 3969

         v₂ = √4048.1

         v₂ = 63.62 m / s

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