Question

An airplane has a mass of 4 space x space 10 to the power of 6kg and the air flows past the lower surface of the wings at 400 m/s. If the wings have a surface area of 1200 m squared, how fast must the air flow over the upper surface of the wings if the plane is to stay in the air? Consider only the Bernoulli’s effect

Answer 1

Answer:

Ok I will answer on start

Answer 2

Given:

Aeroplane has mass of 4 × 10^6 kg.

Air travels along lower surface of wing at 400 m/s. Surface area of wing is 1200 m².

To find:

Velocity of air along upper surface of wing.

Concept:

For the aircraft to remain steady in air , its weight must be balanced by the up-thrust provided by Bernoulli's Effect.

Velocity of air above the wing surface is always greater than that of below the wing surface.

Calculation:

Let the required velocity be v ;

Since the aeroplane is in translational equilibrium , the forces will be equal and opposite :

$\therefore \: weight = F_{b}$

$= > \: mg = F_{b}$

$= > \: mg = (area) \times \Delta P$

$= > (4 \times {10}^{6}) \times 10 = 1200 \times \bigg \{ \dfrac{1}{2} \times \rho \times ( {v}^{2} - {400}^{2}) \bigg \}$

$= > 4 \times {10}^{7} = 600 \times \bigg \{ 1 \times ( {v}^{2} - {400}^{2}) \bigg \}$

$= > \dfrac{4}{6} \times {10}^{5} = ( {v}^{2} - {400}^{2})$

$= > 6.67 \times {10}^{4} = {v}^{2} -( 16 \times {10}^{4} )$

$= > {v}^{2} = 22.67 \times {10}^{4}$

$= > v = 4.76 \times {10}^{2}$

$= > v = 476 \: m {s}^{ - 1}$

So , final answer is :

Air travels along upper surface at a speed of 476 m/s.