Wind speed (100mph) to Pressure (psi)

[Ultrasonic2]

ok guys I am 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 I am to thick to work it out even with an equation.
https://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 I am at sea level

Please help.

Thanks for your help in advance Michael D

 

[tiny-tim]

Welcome to PF!

Hi Michael! Welcome to PF!

The pressure difference (on either side of the window) is 1/2 ρv², where the density ρ of dry air at sea level is about 1.25 kg / m³, and v is the speed of the wind.

 

[Ultrasonic2]

Hi Michael! Welcome to PF!

The pressure difference (on either side of the window) is 1/2 ρv², where the density ρ of dry air at sea level is about 1.25 kg / m³, and v is the speed of the wind.

Thanks so that means 100mph to PSI is 0.17 PSI ? cos that doesn't sound like much

That would mean a 1000mph would only result in 17.4 PSI

Maybe I am 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

 

[tiny-tim]

Hi Michael!

(just got up :zzz: …)

Maybe I am 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

ah, that thread was about the wind blowing past the box …

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)

 

[Mensanator]

Thanks so that means 100mph to PSI is 0.17 PSI ? cos that doesn't sound like much

That would mean a 1000mph would only result in 17.4 PSI

Maybe I am 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

Only? According to "The Effects of Nuclear Weapons" a mere 5 PSI will level a brick house.

 

[russ_watters]

Hi Michael!

(just got up :zzz: …)


ah, that thread was about the wind blowing past the box …

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)

No. The assumption for a problem like this is that you achieve stagnation and for a big, unaerodynamic object, the assumption is pretty reasonable. So it is a straight-up Bernoulli's equation problem and your first post included the proper approach.

 

[russ_watters]

Thanks so that means 100mph to PSI is 0.17 PSI ? cos that doesn't sound like much

That would mean a 1000mph would only result in 17.4 PSI

Nope, you got it! Surprising, isn't it?

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.

Maybe I am 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

Nope - you asked the question right.

 

[tiny-tim]

No. The assumption for a problem like this is that you achieve stagnation and for a big, unaerodynamic object, the assumption is pretty reasonable.

Hi russ!

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 wind-tunnels)

 

[mgb_phys]

Thanks so that means 100mph to PSI is 0.17 PSI ? cos that doesn't sound like much

The famous Tay bridge disaster was partly caused by someone working this out and then deciding that they should use a value half this (10lbs/sqft = 0.07 psi) - the bridge fell down at a wind speed of 105mph

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

 

[varr]

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

 

[Alan Larsen]

Thanks so that means 100mph to PSI is 0.17 PSI ? cos that doesn't sound like much

That would mean a 1000mph would only result in 17.4 PSI

Maybe I am 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

Hi, this got me thinking, what would be the pressure at the end of a funnel, say built inside an old trailer 30 feet long pointed into 12-50mph prevailing winds, with fences to funnel air into the trailer, and the air at the end, some or all of it, blown into a tube say half an inch in diameter? Then it would go into a vortex tube to make free a.c. and heat. You can get vortex tubes for less than $200, they're used for spot cooling in industry, just blow in high pressure air and get instant cold air out one side and hot out the other. The ones they have use at least 20psig, 8cfm, for only like 100btu of cooling, the more air the better. They don't use them for heat, simple electric heaters work, but you could use the heat when the winter wind whips. It's very inefficient, I read you multiply the hp of a compresser by 5 to get the cfm at 100psig, so a 7hp compressor would make 35 cfm and make I think it was 380 btu. So, 7 hp is about 5.2kw, that's like two big home wind turbines, a lot of wind. But if you could take say 100 square feet of concentrated wind down to a half inch diameter to get the psi, maybe it could work? And yes, you do need a.c. when the wind blows, its a hot wind. Thanks, love the site!

 

[AlephZero]

Thanks so that means 100mph to PSI is 0.17 PSI ? cos that doesn't sound like much

It's not the pressure that does the damage, it's the total force when the pressure acts over a large area.

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.

 

[Tolma]

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...

 

[gary007]

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

 

[gary007]

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

 

[rcgldr]

Using wiki's numbers and english units:

ρ = 0.074887 lbm / ft³ = 0.00232756 slug / ft³

v = 100 mph = 146.667 ft / s

1/2 ρ v² = 1/2 (0.00232756 slug / ft³) (21511.1 ft² / s²)

1/2 ρ v² = 25.0342 slug / (ft s²) = 25.0342 (slug ft / s²) (1 / ft² ) = 25.0342 lb / ft²

1/2 ρ v² = (25.0342 lb / ft²) (ft² / 144 in²) = 0.17385 lb / in²

 

[jeffrey c mc.]

ok guys I am 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 I am to thick to work it out even with an equation.
https://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 I am at sea level

Please help.

Thanks for your help in advance Michael D

Well not sure if the OP is still interested in his quarries, yet I will assume he is and speak directly to his musings.

So, while his question is an important aspect of wind shear, and loading as applied to standing structures--ie. Buildings, Windmills, ect sic--I 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 materials--ie, 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

 

[rcgldr]

(wind ... ) 100mph, it strikes, say a 150 foot square billboard ... it will shear 90 degrees, which will then radiate equally in all direction on the surface of the standing structure.

You'll end up with a somewhat cone shaped stagnation zone in front of the billboard with the air flowing around the stagnation zone and edges of the billboard. I'm not sure how much of the billboard will experience an increase of about .17 psi in pressure. The billboard slows the air affected by the billboard, extracting energy from the wind (converting the energy into heat) so Bernoulli is violated. What happens at the edges of the billboard is complicated because of turbulence like vortices. The downwind side of the billboard also has a stagnation zone with lower than ambient pressure, but I don't know how to calculate this.

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

 

[jeffrey c mc.]

[?You'll end up with a somewhat cone shaped stagnation zone in front of the billboard with the air flowing around the stagnation zone and edges of the billboard.?] I'm not sure how much of the billboard will experience an increase of about .17 psi in pressure. The billboard slows the air affected by the billboard, extracting energy from the wind (converting the energy into heat) so Bernoulli is violated. What happens at the edges of the billboard is complicated because of turbulence like vortices. The downwind side of the billboard also has a stagnation zone with lower than ambient pressure, but I don't know how to calculate this.

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

I would agree that a stagnation zone would be present. Presumably at the geographic center of the rectangular object in question.

I would propose that it would be insignificant, in the wind shear being generated by the continued application of the oncoming wind.

The object in question, in the case, a billboard is not absorbing the energy, so much as it is changing the force direction, the force vector; wind shear; which would flow outward from the central stagnation point; radially, in all directions.

I'm unclear as to why you say Bernoulli is violated; is this to mean, it can not be applied in the specific instance?

 

[rcgldr]

I'm unclear as to why you say Bernoulli is violated; is this to mean, it can not be applied in the specific instance?

The billboard is going to slow down the wind, extracting energy from the air. I was thinking that all of the energy was being converted into heat, but much of the energy is probably going into accelerating the Earth by a very tiny amount in the direction of the wind. Bernoulli is violated when energy is extracted or added to a stream line. Another way to look at this is to note that one set of stream lines end on the upwind side of the billboard and another set (with reduced energy) starts on the downwind side of the billboard.

 

[Dale]

The billboard is going to slow down the wind, extracting energy from the air. I was thinking that all of the energy was being converted into heat, but much of the energy is probably going into accelerating the Earth by a very tiny amount in the direction of the wind.

I am not convinced that significant amounts of energy are extracted. Simply slowing the wind doesn't extract energy if the pressure goes up correspondingly. The turbulence and drag would cause frictional heating, which I would guess would be much greater than losses to turning the earth, but that is merely a guess.

 

[rcgldr]

I am not convinced that significant amounts of energy are extracted. Simply slowing the wind doesn't extract energy if the pressure goes up correspondingly.

The issue is that pressure is below ambient on the downwind side of the billboard and initially at zero speed in the stagnation zone. There's a large amount of drag on the billboard.

If you switch the frame of reference to the air (wind), then the drag force times distance (wrt to the air) equals the work done by the billboard. In this frame of reference, the energy of the air is increased.

As an analogy, imagine a bus (or if it was possible, a vehicle similar in shape to a flat plate with extended wheel base to keep it stable), moving along a highway in a no wind condition. To anyone that has experienced what it is like when a bus passes by, or witnessed the reaction of leaves on the road, after the bus passes by, there's a significant wind created aft of the bus. Similar to the billboard in a wind situation, the bus experiences a net drag force, most of which is due to reduced pressure at the aft end of the bus, since the air fore of the bus will mostly tend to separate and flow around the bus. I don't know what the relative higher and lower pressures are for a flat plate. The trailing edges of the bus are too sharp for the air to follow what what otherwise be a void if the air wasn't accelerated to follow the bus, so a reduced pressure zone is created due to the combination of acceleration and momentum of the air. This is why high speed bonneville type vehicles mostly rely on a long tapered tail to gradually introduce the trailing "void" to reduce drag.

 

[stoneman5000]

Hi russ!

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 wind-tunnels)

Looking for answer to see if wind speed gives enough hyperbaric like pressure to give health benefits.

 

[stoneman5000]

Looking for answer to see if wind speed gives enough hyperbaric like pressure to give health benefits.

Ex motorcyle riding.

 

[ian richards]

For an 85 yo amateur could you young wizards tell me approx what would be the pressure in kilopascals on a flat plate 1400MM x 700MM in a wind of 42 M/S. You would have my undying gratitude.
Tks Ian Richards.

 

[Fearnoth]

I also have a question that goes along somewhat with this conversation. If i took the example of a hélicopter, what would be the formula to calculate the possible lift if has. Let's say it has a 10 feet by 5 inch surface (lets say 5000 square inch) on a 12 degree angle (no idea if that is a good angle or not, just putting a numer out there), rotating at 500 rpm (so let's say 12500 feet per min, or 63m/sec since I've seen you guys are metric). No idea if my numbers are correct just putting them to be understood. (My goal is to figure out how many fans i would need, if possible, to lift myself of the ground). Thank in advance

 

[Jeff Boor]

Does anyone have this in an excel spreadsheet they could share? I need this for work to convert wind speed in MPH to PSI for structures.

 

[DentJ]

As a car travels through the air, the air pressure on the front of the car increase. As the MPH increase so does the PSI on the front of the car. Yet Bernoulli says pressure decreases as speed increases. How can both things be correct?

 

[jbriggs444]

As a car travels through the air, the air pressure on the front of the car increase. As the MPH increase so does the PSI on the front of the car. Yet Bernoulli says pressure decreases as speed increases. How can both things be correct?

The reduced pressure (if we model a car with Bernoulli in a wind tunnel with no viscosity) is on the top, bottom and sides of the car, not on its front.

 

[Dan Thomas]

My name is Dan Thomas and I have a question about the solution of 'Wind speed (100mph) to Pressure (psi)
discussion.
I wanted to know if the calculation was based on 1 atm?
If it does, then will the .17 psi double as each additional 1atm is used for the math?

1 atm = .17 psi
2 atm = .34 psi
3 atm = .51 psi
etc

I appreciate this discussion.
I hope someone can/will answer my question.

Dan Thomas. 9/14/21

 

[jack action]

I wanted to know if the calculation was based on 1 atm?
If it does, then will the .17 psi double as each additional 1atm is used for the math?

The Bernoulli equation (adapted for this problem) states the following:
$$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$$
This means that along a streamline:

{The static pressure + the dynamic pressure} at state 1 equals {The static pressure + the dynamic pressure} at state 2.

The static pressure is the atmospheric pressure when there is no wind.

For example, in a wind tunnel when the air doesn't move the static pressure $p_1$ is, say, 15 psi. Obviously, the dynamic pressure is 0 since the wind speed is zero. This is state 1. Total pressure is thus 15 + 0 = 15 psi.

Then somebody starts the fan and the wind speed increases to 100 mph. This is state 2. The dynamic pressure is now at 0.17 psi, as calculated in previous posts. This means that the new static pressure is now 15 psi - 0.17 psi = 14.83 psi, such that:
$$\{15\ psi + 0\ psi\}_1 = \{14.83\ psi + 0.17\ psi\}_2$$
That is in a closed wind tunnel. In the atmosphere that is so vast, a gust of wind will not be in a solid tunnel but surrounded by stagnant air. This would be like if our wind tunnel walls were made of stretchy rubber, like a ballon. This means that the walls would be subjected to the outside pressure and thus, the static pressure will always remain at 15 psi because the wall will move and compress the air in the tunnel until the inside pressure equals the outside pressure. In such a case, with a wind speed of 100 mph, the total pressure will be 15 psi + 0.17 psi = 15.17 psi.

If the atmospheric pressure would be 30 psi, it wouldn't change anything to the dynamic pressure, which would remain at 0.17 psi. Only the total pressure would change.