Fluid mechanics- High Reynolds number flows over bodies

[Stephan kesting]

A 40 km/hr wind is blowing parallel to a storage facility , as shown in the figure. estimate the normal force on a 0.5-m high, 1-m long window if the front door is left open. assume the windspeed along the outside of thewindow is 1.2 U. Air density is 1.22 kg/m^3.

There's a picture that shows a small shed with the wind blowind into an open door with a window on the side od the house. So wind blows straight into the open door and blows parallel to the window.

Homework Equations

p-p(infinite)=1/2 p(U^2)

F=(p-p)A[/B]

The Attempt at a Solution

ok so i have a solution from my professor but he is know to make mistakes (often) on his homework solutions.

His solution: Pressure inside: p-p(infinite)=1/2 p(U^2)=(1/2)1.22((40/3.6)^2)=75.3 N/m^2

P outside window: 1/2 * 1.22 * ((40/3.6)^2) (1-1.2^2)=-33.13

Force on the window is
F=change of p *A= (75.3+33.13)*(0.5*1)=54.2N

My problem:
So my issue is with the pressure outside the window. My solution was to make pressure outside the window 1/2*1.22* ((1.2(40/3.6))^2)

Maybe I am not understanding the equation but i used 1.2 to multiply by the initial windspeed ((1.2(40/3.6))^2). My prof wrote ((40/3.6)^2) (1-1.2^2). Is he right? i can't figure out why the pressure would be calculated like that.
[/B]

 

[KataKoniK]

Hi there,

Thank you for sharing your question with me. I would like to help you understand the solution to this problem.

Firstly, let's review the equation for pressure: P = (1/2)ρV^2, where P is pressure, ρ is density, and V is velocity.

In this problem, we are given the wind speed of 40 km/hr, which is equivalent to 11.11 m/s. However, we need to use the windspeed along the outside of the window, which is 1.2 times the initial windspeed, or 13.33 m/s. This is because the windspeed increases as it passes through the open door and reaches the window.

Now, let's consider the pressure inside the storage facility. We can use the equation P = (1/2)ρV^2, with the density of air (ρ) being 1.22 kg/m^3 and the velocity (V) being 11.11 m/s. This gives us a pressure of 75.3 N/m^2, which is the same as your professor's solution.

Next, let's calculate the pressure outside the window. We need to use the same equation, but with the windspeed along the outside of the window (13.33 m/s) and the density of air (1.22 kg/m^3). However, we also need to consider the fact that the wind is blowing parallel to the window, so we need to multiply the pressure by the cosine of the angle between the wind direction and the window. In this case, the angle is 90 degrees, so the cosine is 0. Therefore, the pressure outside the window is 0 N/m^2.

Now, let's calculate the force on the window. We can use the equation F = (P - P')A, where P is the pressure inside the storage facility and P' is the pressure outside the window. Substituting the values we calculated, we get F = (75.3 - 0)(0.5*1) = 37.65 N. This is half of your professor's solution, but I believe it is because he did not consider the angle between the wind direction and the window.

I hope this explanation helps you understand the solution better. Let me know if you have any other questions or if you need further clarification. Good luck with your studies!

Sincerely