Question

An airplane has a mass of 1.9E6 kg and the air flow past the

lowersurface of the wings at 90 m/s. If the wings have surfacearea
of 1.6E3 m^3, how fast must the air over the upper surface ofthe
wing if the plane is to stay in the air. Consider onlythe Bernoulli
effect.

Answer 1

`v_(1)=164.4 m/s`

Step-by-step explanation:

The Bernoulli equation is:

`P+(1)/(2)\rho v^(2)+\rho gh=constant` (1)

• P is pressure related to the fluid
• ρ is the density of the fluid (ρ(air)=1.23 kg/m³)
• v is the speed of the fluid
• h is the displacement from one position to the other

Now let's apply the equation (1), for our case:

`P_(1)+(1)/(2)\rho v_(1)^(2)=P_(2)+(1)/(2)\rho v_(2)^(2)` (2)

here we assume that h is the same in both cases so it canceled out.

• subscript 1 is related to the upper surface of the wings
• subscript 2 is related to the low surface of the wings

Solving the equation for v₁ we have:

`v_(1)=\sqrt{(2(P_(2)-P_(1)))/(\rho)+v_(2)^(2)}` (3)

Now, we know that pressure P=F/A (force over area)

`\Dela P=(W)/(A)=(mg)/(A)=(1.9\cdot 10^(6)\cdot 9.81)/(1.6\cdot 10^(3)) =11649.4 N/m^(2)` (4)

Combining (3) and (4), we can find v1.

`v_(1)=\sqrt{(2(11649.4))/(1.23)+90^(2)}`

`v_(1)=164.4 m/s`

I hope it helps you!