CR4 - The Engineer's Place for News and Discussion ®
Login | Register for Engineering Community (CR4)

 Previous in Forum: What is Avionics? Next in Forum: Too Cold Downstairs
Power-User

Join Date: Jul 2008
Posts: 353

08/22/2008 10:25 AM

How do you calculate the force in PSI or KPa from the wind speed normal to a flat surface, other than look it up in tables?

__________________
The nice thing about Standards is there are so many to choose from.
Pathfinder Tags: Pressure wind speed
Interested in this topic? By joining CR4 you can "subscribe" to
Guru

Join Date: Jan 2007
Location: In the bothy, 7 chains down the line from Dodman's Lane level crossing, in the nation formerly known as Great Britain, and now disconnecting as Little England and Wales (not too sure about Wales bit, either). Kettle's on.
Posts: 24744
#1

Re: Wind Load Force

08/22/2008 10:30 AM

The pressure on the flat plate normal to the static streamline is equal to 0.5 times the density times the square of the velocity, from first principles. All one needs to do is multiply that pressure by a drag coefficient, which is related to the size and shape of the plate.

__________________
"Did you get my e-mail?" - "The biggest problem in communication is the illusion that it has taken place" - George Bernard Shaw, 1856
Guru

Join Date: Aug 2007
Posts: 15020
#2

Re: Wind Load Force

08/22/2008 12:15 PM

You could use the formula for sails:

Load in Pounds = Sail Area * (Wind Speed )2 * 0.00431

or try this link;

__________________
phoenix911
Power-User

Join Date: Sep 2006
Posts: 119
#3
In reply to #2

Re: Wind Load Force

08/22/2008 11:13 PM

Got this one from a Naval Architect:

Force on Sail

Velocity of wind squared, divided by 295 = pressure per square foot

Multiply by the number of Square Feet

Power-User

Join Date: Jul 2008
Posts: 353
#4
In reply to #3

Re: Wind Load Force

08/23/2008 12:07 AM

Thank you all for your help on this. Erich, are we looking at ft/min and lbs/ftÂ˛? I'm from Australia, but I can convert this to ISO units. Tony

__________________
The nice thing about Standards is there are so many to choose from.
Power-User

Join Date: Sep 2006
Posts: 119
#5
In reply to #4

Re: Wind Load Force

08/23/2008 12:37 AM

Just multiply by the number of square feet and it gives you the total pushing force on the sail. If you want it by the square inch; there are 144 square inches in 1 square foot.

Guru

Join Date: Aug 2007
Posts: 15020
#8
In reply to #3

Re: Wind Load Force

08/23/2008 11:21 AM

Erich

I like to point out what units of velocity, considering you got that from a naval architect, The velocity would be in knots, that would be the same from my post earlier also.

__________________
phoenix911
Guru

Join Date: Feb 2006
Location: Piney Flats, Tennessee
Posts: 1753
#6

Re: Wind Load Force

08/23/2008 3:20 AM

You will also have to consider the amoint of Humidity in the air.

__________________
If you never do anything you never have problems.
Guru

Join Date: Aug 2007
Posts: 15020
#7
In reply to #6

Re: Wind Load Force

08/23/2008 11:17 AM

You will also have to consider the amoint of Humidity in the air.

How's that?

__________________
phoenix911
Guru

Join Date: Feb 2007
Location: Cairo, Egypt
Posts: 1746
#9

Re: Wind Load Force

08/23/2008 11:21 AM

Please refer to the following thread Calculating Wind Resistance to find the wind force due to wind speed.

__________________
It is better to be defeated on principles, than to win on lies!
Guru

Join Date: Aug 2006
Posts: 4542
#10

Re: Wind Load Force

08/23/2008 4:26 PM

In general, the formula is 1/2*rho*V2*Cd*area.

In feet, etc: Rho (the mass density of air) is .0024. Cd is the coefficient of drag: 1.0 is a good figure to use for a flat plate. (Whatever is behind the flat plate, such as the rest of a building, will change that Cd)

So, in a breeze of 20 feet per second, a flat plate of 100 ft2 would experience a force of:

1/2 * .0024 * 400 * 1 * 100 = 48 lb.

If you needed to express this as a pressure rather than a force, you could say that 100 ft2 is equal to 14,400 in2, so the pressure would be 48/14,400, or .0033... psi.

Figures for sail force, incidentally, have to be taken as valid only when you read all the disclaimers, of which there should be (but rarely are) many. Sails spend only a small amount of time acting as a barn door. Most of the time (because the weather gods always dictate that the wind is coming from precisely where you want to go), sails act as airfoils, creating lift (which, because of the foil's orientation, is directed parallel to the water's surface, rather than upward as in an airplane.) Therefore, one must know the Cl (Coeff of lift) at which the sail is operating. This rarely exceeds .8, and is often more like .5, but can sometimes be 1.3 or so. Add to this the fact that the apparent wind that the sails see can be (very roughly) the boat speed plus the wind speed (upwind) or the wind speed minus the boat speed (downwind). Thus the factor that gets squared varies tremendously (and its squared value varies hugely). This variability makes rules of thumb somewhat suspect.

__________________
There is more to life than just eating mice.
Power-User

Join Date: Jul 2008
Posts: 353
#11
In reply to #10

Re: Wind Load Force

08/24/2008 3:01 AM

Thank you Ken, that was great, the units and an example, it has been copied to my useful formulas file.

__________________
The nice thing about Standards is there are so many to choose from.
Guru

Join Date: Aug 2006
Posts: 4542
#13
In reply to #11

Re: Wind Load Force

08/24/2008 3:22 PM

You're welcome -- glad to help.

__________________
There is more to life than just eating mice.
Power-User

Join Date: May 2008
Location: Aggieland, Texas
Posts: 200
#12

Re: Wind Load Force

08/24/2008 8:36 AM

Here in the US, we use the formula, .002496V^2 where V is in miles per hour. The ".002496" takes into account the density of the air and all other constants for standard conditions. For instance, a wind blowing at 100 miles per hour would exert a force on the side of a flat, vertical, rigid face of 25 pounds per square ft. If other than a flat square surface with the wind blowing normal to the surface, the formula must be modified with shape factors. For instance, a cylindical surface would be .002496V^2(.6). The result would be in pounds per square foot of projected area.

Guru

Join Date: May 2008
Location: CHENNAI, TAMIL NADU, INDIA.
Posts: 1678
#14

Re: Wind Load Force

05/09/2011 1:46 PM

Dear Mr.Tonymech,

The Formula for the pressure of Wind/Air on any surface perpendicular to the Direction of wind flow is 0.0004 V^2 PSI where V is the Velocity of Air/Wind in Ft/Sec.

For Youngsters who studied Velocity as Metres/Sec. Conversion Factor of M/Sec to Ft/Sec is

Ft/Sec = Metres/Sec.x 3.28.

DHAYANANDHAN.S

Guru

Join Date: May 2008
Location: CHENNAI, TAMIL NADU, INDIA.
Posts: 1678
#15

Re: Wind Load Force

07/27/2012 1:03 PM

Dear Mr. Tonymech,

The equation for the wind speed striking on surface PERPENDICULARLY is as follows.

P = 0.0004 V^2

where

P = Pressure in Pounds/ Sq.Inch, V = velocity of Wind in Feet/Sec.

Thanks,

DHAYANANDHAN.S

Interested in this topic? By joining CR4 you can "subscribe" to