Tranforming a PSI test result to MPH wind force 1

[britbrat]

I have a bomb mitigation test result in PSI that I would like to equate to an equivelent MPH of wind - any suggestions??

I have the charge of the bomb blast and the distance from the test piece and the pressure at the test piece in PSI, can I use these to equate a mph wind load??

 

[CoryPad]

You can use the pressure (psi or Pascals) and multiply it by the area (square inches or square meters) it acts against and get a force (pounds or newtons). Then, you can compare that with the force generated by wind at certain speed, temp, relative humidity, etc. I can't help with that part, but someone with fluid dynamics/aerodynamics background can.

 

[poetix99]

Not so fast; the overpressure of a shock wave (whether from a bomb blast or something else) is not related via Bernoulli's equation to a "wind speed". The bomb blast is compressible fluid mechanics.

Review "The Dynamics and Thermodynamics of Compressible Fluid Flow" - Ascher Shapiro. It is a classic textbook in the field of compressible flow phenomena.

You may in fact express any differential pressure as equivalent to a "wind speed", but that will not realistically describe the phenomena of the blast, nor will it be correct for pressure differentials much greater than 4 or 5 psi, because compressibility effects are no longer negligible.

Having said all of that, if you still wish to make that sort of simple "equivalence", consult an undergraduate level fluid mechanics text, or Crane Technical Paper #410, as example references. The basic equation, Bernoulli's equation, is an expression of conservation of energy. For incompressible flows, with little change in elevations between initial and final states:

Pressure + 1/2(density)*(velocity)^2 = constant

Ensure that the units are dimensionally consistent (or just work in SI - it's much easier).

 

[arto]

Maybe look @ the ANSI/ASCE building design Standard & fiddle with their equation??
The old one gave an equation for velocity pressure of:

qz = *Kz*[IV]^2
qz = velocity pressure @ height z
V = Wind speed, mph
Kz = Velocity Pressure Exposure coef. [based on height above ground & exposure (terrain)]
I = Importance Factor [ ~risk]

 

[arto]

Oh yeah - qz is in pounds/sq foot

 

[arto]

It's early
CORRECT EQUATION:

qz = 0.00256*Kz*[IV]^2