# Fluid mechanics- High Reynolds number flows over bodies

1. Nov 28, 2014

### 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.

2. Relevant equations p-p(infinite)=1/2 p(U^2)

F=(p-p)A

3. 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 im 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.

2. Dec 4, 2014

### Staff: Admin

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

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