Question

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

Answer 1

Answer:

$v_{1}=164.4 m/s$  

Explanation:

The Bernoulli equation is:

$P+\frac{1}{2}\rho v^{2}+\rho gh=constant$ (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:

$P_{1}+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\frac{1}{2}\rho v_{2}^{2}$ (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:

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

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

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

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

$v_{1}=\sqrt{\frac{2(11649.4)}{1.23}+90^{2}}$  

$v_{1}=164.4 m/s$  

I hope it helps you!