Formulae for Atm. Pressure w.r.to height

[quark]

The ambient pressure can be calculated, when we know the height above mean sea level, by the simplified formula,

P = 29.92-0.001*H where, P is ambient pressure in inches Hg, H is height above mean sea level in ft.

I recently came across with another formula which can be valid upto 11km height, with +/-3% climatic variation.

Pamb = Pmsl[(288-6.5H)/288]^5.255 where, Pamb is the ambient pressure at height in bar, Pmsl is the atmospheric pressure at mean sea level(1.0132bar) and H is height in km.

The difference in both the formulae varies from +0.46% to -1.55% upto 2.5km and is significant beyond that(3% at 3km, 5% at 3.5km and 8% at 4km etc.)

Anybody got any other formula?

Regards,

 

[Tunalover]

Quark-
My formula looks like yours a little:
Palt=Psl*[(288-.00198z)/288]**4.256
where Psl=14.69psia and z=feet above sea level.



Tunalover

 

[vzeos]

Try:
P = 10 ^ (1.1682 - (H / 63800))
where P = ambient pressure in psia and H=feet above sea level.

 

[25362]

The first formula shown by quark, is apparently a ROT. Namely, for every 1000 ft ascent there is a drop in barometric pressure of ~1 in Hg.

Another formula relating altitude to barometric pressure attributed to the CRC handbook:

Z = 44,331.5 - 4,946.624 * P^0.190263​

Z: altitude above sea level, m
P: pressure, Pa (1 Pa = 0.01 millibar)

It appears to be correct when compared with the U.S. Standard Atmosphere (1976) published values up to 10 km altitude.

In general, the published relations between altitude and pressure appear to be logarithmic, giving a straight line in a semi-log graph paper up to, at least, the tropopause.

 

[quark]

Thanks all for the responses. I fitted all these formulae in excel table and got the following results.
[tt]km ft ROT quark tuna rgas 25362[/tt]
[tt]0 0 1.0132 1.0132 1.0128 1.0155 1.0132[/tt]
[tt]1 3280 0.9021 0.8987 0.9191 0.9022 0.8987[/tt]
[tt]2 6560 0.7910 0.7948 0.8322 0.8014 0.7949[/tt]
[tt]3 9840 0.6799 0.7009 0.7517 0.7120 0.7010[/tt]
[tt]4 13120 0.5689 0.6162 0.6773 0.6325 0.6164[/tt]
[tt]5 16400 0.4578 0.5400 0.6087 0.5619 0.5402[/tt]
[tt]6 19680 0.3467 0.4716 0.5455 0.4991 0.4718[/tt]
[tt]7 22960 0.2356 0.4104 0.4875 0.4434 0.4106[/tt]
[tt]8 26240 0.1246 0.3558 0.4343 0.3939 0.3560[/tt]
[tt]9 29520 0.0135 0.3072 0.3857 0.3499 0.3074[/tt]
[tt]10 32800 -0.0975 0.2642 0.3413 0.3108 0.2643[/tt]

The values are chopped off after the 4th digit. At lower altitudes, all formulae seem to be good. Can you guys give me the range in which they work? In future(about 5 years), I am planning to come up with engineering tips book(particularly for pharma mechanical engineers). The main intent is to provide the data, formulae and procedures which are not generally available. I do acknowledge you guys help

Tunalover,

Can you once again check the exponential coefficient? 6.5H in km should equal 0.00198H in feet.

Regards,

 

[vzeos]

Quark

I guess the range in which the equation works depends on the accuracy you’re willing to accept. My equation is based on the equation below which was cited in the U. S. Bureau of Mines Monograph No 6, (Flow of Natural Gas through High Pressure Transmission Lines by Johnson and Berwald, 1935.)

h=62,900Log(Po/Ph)
h=altitude
Ph=pressure at altitude
Po=pressure at sea level

Johnson and Berwald attributed this equation to Lichty, Lester Clyde, Measurement, Compression, and Transmission of Natural Gas, 1924.

I just made some minor adjustment to fit the data I had available to me at the time. I guess it could still be modified to better fit the U.S. Standard Atmosphere (1976) published values that 25362 pointed out.