Wind speed and pressure
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I saw a post online asking about the pressure at a wind speed of 15 m/s. Here’s how they calculate it: Suppose we take an area S that is perpendicular to the wind. Since the wind speed is V, the volume of air that hits this area S in a time interval t is V = Svt. Using the concept of density, we have m = ρSvt. When the wind encounters an obstacle, its speed drops to 0, and the resulting pressure is F. According to the law of momentum, Ft = (ρSvt)v ； After sorting, the resulting average force F = ρSv^2 ； The pressure formula: P = F/S = ρv^2 ； The density of air is approximately 1.3 kilograms per cubic meter; using this value in the calculations, the resulting pressure is about 300 pascals. There’s one thing I don’t understand: in the unit conversion, was the 9.8 coefficient omitted somewhere? I am a beginner. I hope the experts here can help me figure this out. Thank you!
This post was last edited by Lantian on 2017-4-21 at 16:44. This is a conversion of the relationship between wind pressure and wind speed; it’s easy to think, upon hearing about it, that atmospheric pressure has changed.
It has already been multiplied by g; g was simply canceled out, as it was originally in the denominator
I understand now, thank you! But another problem arises. Later, I saw elsewhere that the formula was written as: p=0.5·r·v^2/g – it just lacks a 1/2 in it
Look at Bernoulli’s equation, and you’ll naturally understand everything
The original poster is really great; thanks for sharing this. I don’t quite understand this Dongben either, and I’m not good at doing the calculations