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Wind speed (100mph) to Pressure (psi) 
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#1
Dec2109, 02:43 AM

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ok guys im an old guy that is useless at maths so i'd like your help.
I have searched the net and this forum and i have found some formula but im to thick to work it out even with an equation. http://www.physicsforums.com/showthread.php?t=117105 This isn't home work as i haven't been to school for a long time now .. probably if i was at school i could have understood peoples replies Basically what i want to know is for a 100mph wind what is that in psi ? i obviously want to be able to change the mph and see the result in psi is but i need REAL basic maths. From what i've read so far it's important to know im at sea level Please help. Thanks for your help in advance Michael D 


#2
Dec2109, 05:24 PM

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Hi Michael! Welcome to PF!
The pressure difference (on either side of the window) is 1/2 ρv^{2}, where the density ρ of dry air at sea level is about 1.25 kg / m^{3}, and v is the speed of the wind. 


#3
Dec2109, 10:13 PM

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That would mean a 1000mph would only result in 17.4 PSI Maybe im not asking the right question .. what i want to know is if i blew 100mph wind into a sealed box what would be PSI inside the box Thanks for your help 


#4
Dec2209, 02:18 AM

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Wind speed (100mph) to Pressure (psi)
Hi Michael!
(just got up …) blowing against the box (how can it blow into a sealed box? ), you need pressure = force/area, and (good ol' Newton's second law …) force = rate of loss of momentum. (actually, it's quite complicated, because of course the wind doesn't stop dead, and everything depends on exactly how it it flows round the box) 


#5
Dec2209, 02:02 PM

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#6
Dec2209, 05:56 PM

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#7
Dec2209, 06:02 PM

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Note, though, that above a couple hundred mph the air compresses enough for that equation to beome inaccurate. And above the speed of sound, the rules change again. So you can't use it at 1000 mph. 


#8
Dec2309, 04:47 AM

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I thought that stagnation points only occurred at limited places on an object, such as the centre line of a building? (and is a building unaerodynamic? … ok, I know they don't fly, but you do see them a lot in windtunnels) 


#9
Dec2309, 10:13 AM

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For the fourth bridge, which is still there, they used 56lbs/sqft ( = 0.39psi) 


#10
Feb2010, 09:24 PM

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pressure for us workers is uselly in cubic feet so just under 25 pounds psi if they are correct



#11
Sep3011, 10:56 PM

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#12
Oct111, 08:41 AM

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Suppose the window was 5ft x 4ft = 20 sq ft = 2880 sq in. The total force from the 0.17 psi pressure is 0.17 x 2880 = about 500 lb. If this is still surprising, think about the fact that the entire weight of a car is supported by the air pressure inside the tires (at about 25 or 30 psi), with only a few square inches of tire in contact with the ground. 


#13
Jun1112, 05:18 AM

P: 1

ok a different approach, I am no math whiz so a simple basic number will do.
Ignoring sea level, wind funneling, massice blades, or such, what is the basic Pounds per square inch if I were to hand crank this thing would it take to drive a wind turbine desinged to optimize at a 10mph or 20mph wind? A simple number please I have not been to high school math class in over 30 years. So a bunch of equations will simply cause my eyes to cross and start to water. LOl.. No Seriously I mean it... 


#14
May2613, 08:37 PM

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I believe the end calculation result is incorrect above of .17psi
First it is important to covert the wind speed into meters /sec. If my calculations are correct the answer should be 1.698psi My best 


#15
May2713, 10:28 AM

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If the above initial calculation was correct then one can assume a 1000mph wind would only produce 1.7psi which I'm sure that one could see this would be a lot less than actual. I have included what I hope is all of my calculation to derive at my revised PSI figure. Please advise if I am inaccurate in my calculation.
100mph = .447 (100) =44.7m/s 44.7m/s2 = 1998 .0017 (P) * 1998 (v2) = 3.396 3.396/2 = 1.698 PSI 


#16
May2713, 02:04 PM

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Using wiki's numbers and english units:
ρ = 0.074887 lbm / ft^{3} = 0.00232756 slug / ft^{3} v = 100 mph = 146.667 ft / s 1/2 ρ v^{2} = 1/2 (0.00232756 slug / ft^{3}) (21511.1 ft^{2} / s^{2}) 1/2 ρ v^{2} = 25.0342 slug / (ft s^{2}) = 25.0342 (slug ft / s^{2}) (1 / ft^{2} ) = 25.0342 lb / ft^{2} 1/2 ρ v^{2} = (25.0342 lb / ft^{2}) (ft^{2} / 144 in^{2}) = 0.17385 lb / in^{2} 


#17
May2713, 02:09 PM

P: 45

So, while his question is an important aspect of wind shear, and loading as applied to standing structuresie. Buildings, Windmills, ect sicI believe, and so assert and retort; that a direct and constant that will convert wind velocity to applied force, can not be made without considering the 'area', in square proportions, of the structure under analysis. With that said, mind you; the general principles that must come under study; are those set forward by Sir Newton, Messrs Pascal, Bernoulli in their discourses of physical minutia. So Sir Newton would of course consider the kinetic forces of the moving wind, and of course would need to know the general atmospheric pressure, at the point(s) in question, ie, sea level, versus, at some specified elevation, he would must also be of interest in the moisture content of the wind mass moving at the stated rate; along with any other dejecta materialsie, sand, dirt, objects de art, Dorthy and Toto, to name a few. So after Sir Newton felt he had stated his principles sufficiently he would sit down in triumph on his laurels, and allow Mr Pascal to step up the the podium. Cough several times and retort; 'A change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid' and then, after glaring at all the bright minds before him, gesture to Mr Bernoulli, and return to his sedan. Now of course Mr Bernoulli having had the problem so well quoted would glance up and to the left, cock his left eyebrow and utter; "that for an inviscid flow, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy. " So now with the issues so dramatically made know, an analysis of the question at hand can be sought First; we have the mass of the atmospheric fluid; yes a fluid, if a gas isn't fluid then I don't wish to consider living any longer; so, the mass of the gaseous medium, considered at what elevation it is at, and the moisture content, along with any other stray objects picked up in the course of it's travels; this force is Kinetic; yet described in terms of itself only, is potential. If this mass is moving at the rate of 100mph, for example we will say 44.704 m/s AND if, it strikes, say a 150 foot square billboard, say 1,614.60 m sq As this force (sic) strikes square direct, on the structure, it will shear 90 degrees, which will then radiate equally in all direction on the surface of the standing structure. This force acting on this surface are which creates the pressure differential, between the wind face; and the area behind this face. The pressure behind the object does not change; the change in vector forces, which create a 'LOWERED' pressure on the wind side area of the structure that will at first be pulled in the inverse direction of the wind (the billboard). The window is not being 'blown' in, it's being 'sucked out' The higher force, 'static air pressure; as compared to the lower pressure on the wind loading surface, caused by the vector change in conjunction with the change in velocity in the new direction, cause a lowered state of pressure, as noted by Pascal, and Bournelli. Thus, the wind doth truly suck. Respectfully submitted for discussion jeffrey 


#18
May2713, 02:28 PM

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The overall force on the billboard could be approximated using the coefficient of drag for a rectangular flat pate. 


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