Question

An airplane has a mass of 10x10 kg, and the ar flows put the lower surface of the wings at 82 m/s - Part A the wings have a surface area of 1300 ml how fast must the air flow over the upper surface of the wing if the plane Express your answer to two significant figures and include the appropriate units. JARO? Value

Answer 1

By Bernoulli's theorem sum of pressure head and velocity head above and below the wing must be equal i.e,

Pa+3 pos = Po + 5 pula

P denotes the pressure, v denotes the velocity and $\rho$ is the density of air

The index a denotes above and b denotes below the wings.

Pa+3 pos = Po + 5 pula

or, (P - Pa) + pub= pua

0r, vk – 2[P. – P.) +

2(Po – Pa) + uz or, va = ...)

Now we consider the term Pb-Pa

This is nothing but the pressure difference between the upper and lower surface of the wings.

This pressure difference is equal to the force per unit area i.e,

Force P- PA Area

Here the force is supplied by the weight of the aircraft. So we can write

mg Po - Pa=

m is the mass of the aircraft and g is the acceleration due to gravity.

putting this value in equation (i) we have

$or,\ v_a=\sqrt{\frac{2mg}{\rho A}+v_b^2}$

Putting the values we have,

2 x 1.9 x 100 x 9.8 + 822 m/s 1.2 x 1300

= 174.92 m/s

This is the velocity of the air flow above the surface.