Ram Pressure of Standard Sea Level Air

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larry cottrill
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Ram Pressure of Standard Sea Level Air

Post by larry cottrill » Fri Aug 12, 2005 9:31 pm

EDIT: All constants now corrected (see explanation in following post)!

Sometimes I get to wondering just how to calculate air speed vs ram pressure. I think I've found the method.

First, we sort of need to define the air we're working with. Standard sea level air is defined as follows, from my old Introduction to Gas-Turbine and Jet-Propulsion Design textbook (Harper & Brothers, 1948, but I don't think air has basically changed much except the degree of pollution):
Specific Volume = 13.33 cu.ft/lb
(Density then would be 0.075 lb/cu.ft, but Specific Volume is handier for the gas flow equations)
Temperature = 520 degF, absolute
Pressure = 14.7 PSI, absolute

According to the University of Denver site's page on fans and windmills ( http://www.du.edu/~jcalvert/tech/fluids/fan.htm ), the relation between airspeed and pressure can be expressed in English units as:
V = 66.8 x sqrt( pdiff )
where pdiff is the difference in measured impact pressure (pressure due to airspeed) and static pressure. The difference is handy to use because that's what you get from a standard Pitot tube - the static port is balanced against the impact port, so the measured pressure relates only to the airspeed. The units are ft/sec for V and the highly convenient inches H2O for pdiff.

Nicer units:

OK - I did this quickly, and need for you to check me out on this:

The site says that an inch of water head = .036 PSI, 1.489 mb or 1489 dynes/sq.cm - take your pick. Since the pressure difference is under the radical sign, we need the square root of each of these factors so we can move it out to the front and combine it with the factor already shown. These square roots are.1897, 1.220 and 38.59, respectively. I have also included a value for pressure in lb/sq.ft, since this is a commonly used pressure unit in the flow equations. So, for airspeed still in ft/sec, I get the following set of equivalents:
For V in ft/sec (edited to correct):
V = 66.8 x sqrt( pdiff ) for pdiff in inches of water
V = 352.1 x sqrt( pdiff ) for pdiff in PSI
V = 29.34 x sqrt( pdiff ) for pdiff in PSF
V = 54.74 x sqrt( pdiff ) for pdiff in millibars
V = 1.731 x sqrt( pdiff ) for pdiff in dynes/sq.cm

Carrying metric equivalency to its logical conclusion, I determined that one foot = 0.3048 metres. So, if we want our airspeed in metres/sec, we would simply multiply the factors above by .3048:
For V in m/sec (edited to correct):
V = 20.36 x sqrt( pdiff ) for pdiff in inches of water
V = 107.3 x sqrt( pdiff ) for pdiff in PSI
V = 8.943 x sqrt( pdiff ) for pdiff in PSF
V = 16.68 x sqrt( pdiff ) for pdiff in millibars
V = .5276 x sqrt( pdiff ) for pdiff in dynes/sq.cm

EDIT -----
Note that to express airspeed in 'lay persons' units, we can apply the following factors:
To get MPH (miles per hour), multiply ft/sec by 3600 (yielding ft/hr) and divide by 5280, i.e. simply multiply by the factor .6818
To obtain Km/hr, multiply metres/sec by 3600 (yielding m/hr) and divide by 1000, i.e. simply multiply by 3.6

Now, we can turn the equations around, so we can derive pressure from the relative airspeed. (Sort of what I said in the first place ;-) Transposing, we initially get sqrt( pdiff ) = V / f ; we can square both sides, obtaining pdiff = V^2 / f^2 with f of course being the coefficient we found earlier for each set of units. It is interesting to note that ram pressure, like aerodynamic drag, is directly related the the square of the relative airspeed. So, we now have:

For V in ft/sec (edited to correct):
pdiff inches of water = V^2 / 4462
pdiff PSI = V^2 / 123951
pdiff PSF = V^2 / 860.8
pdiff millibars = V^2 / 2996.8
pdiff dynes/sq.cm = V^2 / 2.997

For V in m/sec (edited to correct):
pdiff inches of water = V^2 / 414.5
pdiff PSI = V^2 / 11515
pdiff PSF = V^2 / 79.96
pdiff millibars = V^2 / 278.39
pdiff dynes/sq.cm = V^2 / .2784

Again, we are often confronted with speeds in common units. For MPH airspeeds, we must multiply the V^2 x .6818^2, that is, multiply by .46485; However, we would rather not introduce a second constant into the equation, so we take 1/.46485 = 2.151, and multiply the denominator of each equation for ft/sec by that value. So, we obtain:
For V in MPH (edited to correct):
pdiff inches of water = V^2 / 2074
pdiff PSI = V^2 / 57608
pdiff PSF = V^2 / 400.05
pdiff millibars = V^2 / 1392.8
pdiff dynes/sq.cm = V^2 / 1.3928

Similarly, for V expressed in Km/hr, we square the factor 3.6, obtaining 12.96, and multiply denominators by the reciprocal, .07716:
For V in Km/hr(edited to correct):
pdiff inches of water = V^2 / 5372.3
pdiff PSI = V^2 / 149231
pdiff PSF = V^2 / 1036.3
pdiff millibars = V^2 / 3608
pdiff dynes/sq.cm = V^2 / 3.608
----- End of EDIT

Please let me know if you find any error in my procedure; I sometimes make very silly mistakes! But, if my error is pointed out, I will edit my post to correct, so it can be of use for calculation, for working with UFLOW1D, or whatever.

L Cottrill
Last edited by larry cottrill on Sat Aug 13, 2005 9:00 pm, edited 2 times in total.

larry cottrill
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CORRECTION of Silly Mistake

Post by larry cottrill » Sat Aug 13, 2005 8:49 pm

I awoke this morning certain that I had made a basic error. Sure enough, I had (I think) calculated the square root factors correctly, but I applied them by multiplying rather than dividing. Here are the corrections, which I will apply to my original post as an edit.

What I showed as:
For V in ft/sec:
V = 66.8 x sqrt( pdiff ) for pdiff in inches of water
V = 12.67 x sqrt( pdiff ) for pdiff in PSI
V = 1.056 x sqrt( pdiff ) for pdiff in PSF
V = 81.51 x sqrt( pdiff ) for pdiff in millibars
V = 2578 x sqrt( pdiff ) for pdiff in dynes/sq.cm
should be:
For V in ft/sec:
V = 66.8 x sqrt( pdiff ) for pdiff in inches of water
V = 352.1 x sqrt( pdiff ) for pdiff in PSI
V = 29.34 x sqrt( pdiff ) for pdiff in PSF
V = 54.74 x sqrt( pdiff ) for pdiff in millibars
V = 1.731 x sqrt( pdiff ) for pdiff in dynes/sq.cm

What I showed as:
For V in m/sec:
V = 20.36 x sqrt( pdiff ) for pdiff in inches of water
V = 3.862 x sqrt( pdiff ) for pdiff in PSI
V = .3218 x sqrt( pdiff ) for pdiff in PSF
V = 24.84 x sqrt( pdiff ) for pdiff in millibars
V = 785.5 x sqrt( pdiff ) for pdiff in dynes/sq.cm
should be:
For V in m/sec:
V = 20.36 x sqrt( pdiff ) for pdiff in inches of water
V = 107.3 x sqrt( pdiff ) for pdiff in PSI
V = 8.943 x sqrt( pdiff ) for pdiff in PSF
V = 16.68 x sqrt( pdiff ) for pdiff in millibars
V = .5276 x sqrt( pdiff ) for pdiff in dynes/sq.cm

What was given as:
For V in ft/sec:
pdiff inches of water = V^2 / 4462
pdiff PSI = V^2 / 160.5
pdiff PSF = V^2 / 1.115
pdiff millibars = V^2 / 6645
pdiff dynes/sq.cm = V^2 / 6645000

For V in m/sec:
pdiff inches of water = V^2 / 414.5
pdiff PSI = V^2 / 14.92
pdiff PSF = V^2 / .1036
pdiff millibars = V^2 / 617.2
pdiff dynes/sq.cm = V^2 / 617200
should be:
For V in ft/sec:
pdiff inches of water = V^2 / 4462
pdiff PSI = V^2 / 123951
pdiff PSF = V^2 / 860.8
pdiff millibars = V^2 / 2996.8
pdiff dynes/sq.cm = V^2 / 2.997

For V in m/sec:
pdiff inches of water = V^2 / 414.5
pdiff PSI = V^2 / 11515
pdiff PSF = V^2 / 79.96
pdiff millibars = V^2 / 278.39
pdiff dynes/sq.cm = V^2 / .2784

And what I showed as:
For V in MPH:
pdiff inches of water = V^2 / 9599
pdiff PSI = V^2 / 345.3
pdiff PSF = V^2 / 2.399
pdiff millibars = V^2 / 14295
pdiff dynes/sq.cm = V^2 / 14295000

For V in Km/hr:
pdiff inches of water = V^2 / 31.98
pdiff PSI = V^2 / 1.15
pdiff PSF = V^2 / .007994
pdiff millibars = V^2 / 47.62
pdiff dynes/sq.cm = V^2 / 47620
should really be:
For V in MPH:
pdiff inches of water = V^2 / 2074
pdiff PSI = V^2 / 57608
pdiff PSF = V^2 / 400.05
pdiff millibars = V^2 / 1392.8
pdiff dynes/sq.cm = V^2 / 1.3928

For V in Km/hr:
pdiff inches of water = V^2 / 5372.3
pdiff PSI = V^2 / 149231
pdiff PSF = V^2 / 1036.3
pdiff millibars = V^2 / 3608
pdiff dynes/sq.cm = V^2 / 3.608


My apologies for the confusion my mental lapse may have caused. I hope the revised numbers are provably correct; I have tried at least one example with each revised factor in place. My original post will be corrected with the new numbers as soon as I can accomplish it.

L Cottrill

Mike Everman
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re: Ram Pressure of Standard Sea Level Air

Post by Mike Everman » Thu Aug 25, 2005 7:05 am

Oooh, thanks again. I just got around to reading this. I like psi per mph, but that's just me... If my car had kilometers per second on the speedometer, everything might be different for me. ;-)
Mike
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