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3 years ago

14

When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing

is 273 mph relative to the airplane. (a) What suction pressure is developed on the wing at that point? (b) What is the pressure at the leading edge (a stagnation point) of the wing?

1 answer:

Sedaia [141]3 years ago

5 0

Answer:

P1 = 0 gage

P2 = 87.9 lb/ft³

Explanation:

Given data

Airplane flying = 200 mph = 293.33 ft/s

altitude height = 5000-ft

air velocity relative to the airplane = 273 mph = 400.4 ft/s

Solution

we know density at height 5000-ft is 2.04 × $10^{-3}$ slug/ft³

so here P1 + $\frac{\rho v1^2}{2}$ = P2 + $\frac{\rho v2^2}{2}$

and here

P1 = 0 gage

because P1 = atmospheric pressure

and so here put here value and we get

P1 + $\frac{\rho v1^2}{2}$ = P2 + $\frac{\rho v2^2}{2}$

0 + $\frac{2.048 \times 10^{-3} \times 293.33^2}{2}$ $= P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}$

solve it we get

P2 = 87.9 lb/ft³

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$\boxed{q_{B}=+2q}$

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$\boxed{q_{C}=+1.5q}$

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First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

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