PSF to MPH windspeed conversion calculation 5

[Grunchy]

I'm not a civil engineer so this is an alien topic to me.
I have a conversion calculation as follows:

MPH = sqrt( PSF / 0.00249 )

What I am doing is testing rollshutters for windload using a vacuum chamber. I am able to measure the vacuum inside the chamber in inches of water column, which converts readily to pounds per square feet (PSF). Now, somewhere or other somebody got the above conversion to determine equivalent MPH wind speed. It is clearly empirical and seems awfully simplified to me, which raises alarm flags. Even worse, nobody knows where it came from!

My questions: is this a legitimate calculation, also does anybody know its source? Are there limitations, ie the equation is accurate only for 0-10 PSF? Is there a better or more universally recognized conversion calc?

Thanks in advance,
Oyster (Mech)

 

[Ron]

I have to assume you are placing an actual vacuum AGAINST the shutter and not placing the shutter in a vacuum chamber. With that premise, the equation you are using is close to the basic equation used to convert wind speed to wind pressure as given in many texts and in ASCE 7, "Minimum Design Loads for Buildings and Structures". The basic equation is:

p=0.00256(V^2), where

p=pressure in psf
V=wind velocity in mph

the constant, 0.00256 corresponds to the mass density of air at standard atmospheric pressure and temperature, with the appropriate unit conversions applied.

Re-written in the form you are using, the equation becomes:

V= SQRT[p/0.00256]

The 0.00249 you are using is the mass density of air at something other than standard atmosphere.

Ron

 

[Ron]

One other point....

Usually there are numerous factors applied to components such as this for wind loads. These include shape factors, height factors, terrain type,structural importance and others. Consider these when you are doing your tests, as these usually make the pressures increase.

Check ASCE 7-88, 7-93, 7-95, and 7-98 for treatment of these factors. These are all evolutions of the original ANSI document (ANSI A58.1) for loads applied to buildings and structures. see

http://www.asce.org

for info on purchasing these standards.

 

[meca]

There are inexpensive spreadsheets that calculate wind loads per ASCE 7-93, 95, and 98 at

http://www.mecaconsulting.com

Chris Rosencutter
chris@mecaconsulting.com

http://www.mecaconsulting.com

 

[Grunchy]

You are correct Ron, the vacuum is against one side of the rollshutter. The vacuum figure is the pressure differential withstood by the rollshutter curtain.

I hate unit conversions. By the way, I checked my thermodynamics & fluid flow texts and didn't find this calculation. Is it fair to say, it's a basic equation for civil engineering?

Back-calculating:

V = sqrt( P / 0.00256 )

V = sqrt( (32.1719 ft/s^2) * (3600 s/hr)^2 / (5280 ft/mi)^2 x P / 0.038287 lbm/ft^3 )

Where wind speed V is in mi/hr,
vacuum P is in lbm/ft^2,
and the STP density of air would be 0.038287 lbm/ft^3.

Now if I check my air table "Properties of Air at Standard Atmospheric Pressure", the air would have to be nearly 600 degF to have density of 0.038 lbm/ft^3. ??huh??

Or is there one of those "fudge factors" so common to empirical equations. In which case the 0.00249 constant in my equation would be the constant applicable at an air temperature of -7.8 degC and standard pressure.

The units of the constant are (lbm * hr^2) / (ft^2 * mi^2), or M x T^2 / L^4 (density divided by acceleration).

 

[Ron]

Oyster,
Your equation for the constant should be....

0.00256=1/2{[0.0765lb/cu.ft.]/[32.2ft/sec^2]}X[(1mi/hr)(5280 ft/mi)}X{(1hr/3600 sec)]^2

I believe you have an extra squared term in your equation.

The value, 0.00256, is standard at 59F and 29.92 inches of mercury.

The wind speed to pressure formula is common to structural engineering.

Ron

 

[TowerEngineer]

Oyster,

The constant 0.00256 is derived in the Commentary section of ASCE 7 (formerly ANSI A58.1-1982) as noted in the equation stated by Ron in his last response. This portion of the equivalence is indisputable.

However, there is no simple conversion factor to MHP using ASCE 7 or any other code-type source because of the great quantity of parametric variables that go into determining dynamic wind force acting on an object... some of which are noted by Ron in his first response.

It is my opinion that the current codes and standards (IBC, UBC, EIA, BOCA, ASCE 7, etc.) are very conservative with respect to wind pressures acting on objects. If these codes were correct, then practically all structures built before 1982 would have experienced lateral wind forces that would have brought about their demise long before now. I have reviewed many tall, slender structures that were designed and built in the time frame from 1935 to 1982. The lateral forces derived from the historical design methodology (say 1935) for a 90 MPH wind is approximately 1/2 to 1/6th (depending on the structure height, site elevation, and topographic features surrounding the structure) as compared to the lateral forces derived from current design standards and codes.

The basic problem lies in two areas, namely the Reynold's number(viscous flow) and compressibilty of air. The design methodologies and force coefficients contained in current design standards are generally based on research performed i n the late 1950's and early 1960's. This body of research is presented in ASCE Transactions Paper No. 3269 'Wind Forces on Structures' which includes significant numbers of references to empirical drag force coefficients. The problem is that the code writing bodies have moved away from the research that was performed back then and have steadily moved toward larger and larger drag coefficients and shape factors, with none of it having been based on new experimental research.

For instance, the drag force coefficient for a smooth cylinder with height/width ratio of 25 is 0.55 in the Transaction Report, but current design standards require the use of a drag coeffiecient of 0.7 or more depending on the standard or code used as a reference. This is even worse for essentially square or flat objects. For these objects, with height/width ratio of 25, the 1961 paper indicates a drag force coefficient of 1.4, but current design standards require a drag coefficient of 1.8 or more, again depending on the standard or code used as a reference.

This alone makes it very difficult or impossible to back into a wind velocity based upon a certain negative pressure in PSF. I suggest that if you want an answer to your question, you look into design materials used by mechanical and areonautical engineers for derivation of drag coefficients. I believe they use more realistic design standards and codes with respect to drag coefficients because of the nature of their work.

Best of Luck to you in your quest. I'd be happy to give you more information on how to obtain a copy of the ASCE Transaction Paper No. 3269 if you so desire.

 

[Boothby171]

SInce we're on subject--is there any correlation between the ASCE wind pressure calculations and the C-sub-D typically used in the equation:

Force = Area * density * C-sub-D * Velocity^2

(from our old fluids/physics classes)

I'd love to just multiply the surface pressure I get through ASCE * Area * C-sub-D, to get the force on an object with a known drag coefficient, some Y feet in the air. The sense I get, though, is that ASCE incorporates (directly or indirectly) some of its own C-sub-D values by the time I've calculated a pressure, and I'd need to normalize against those.

Any ideas?

--Steve

 

[Ron]

boothby...the "Cd" term in your equation is limited to drag. In ASCE 7, the factors that would substitute for the Cd in your equation can be numerous and depend greatly on structural type, location (geographically and on the structure), shape, proximity to other structures, duration of wind, statistical factors and other considerations. Don't mix up the two.

If you are computing forces on a structure, use the entire ASCE 7 (or similar) premise as these have been shown to have specific applicability.

**Need Professional Development Hours / Continuing Education ??
**Try this course...

http://pdhonline.org/cgi-bin/quiz/courses/courselist.cgi?class_name=p112