Atmospheric pressure at high altitude

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I was a bit surprised to find that the average recorded atmospheric pressures at such locations as Mexico City (2.2 km above sea level) are roughly equivalent to those *at* sea level, i.e. in the 99-102 kPa range. I find this a bit odd, since the air density at this location is just under 80% that of sea level. So, one would expect the barometric pressure to be correspondingly lower as well.

Are there any atmospheric physicists who can help shed light on this conundrum?
 

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  • #2
mathman
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Your information sounds wrong. Pressure does decrease with altitude. That is why airplane cabins have to be pressurized nowadays.
 
  • #3
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mathman said:
Your information sounds wrong. Pressure does decrease with altitude. That is why airplane cabins have to be pressurized nowadays.

No, I'm aware of the "correct" interpretation. If you look at weather forecasts, for example, the pressure is *not* as low as you would expect. For example, currently in Mexico City it is 100.76 KPa at 17 C (77% rel. humidity). If you calculate the density altitude which corresponds to those figures, you find a value about 10x smaller than you would expect.

I'm curious if there's a meteorological aspect I'm missing.
 
  • #4
HallsofIvy
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Mexico City is not nearly as high as an airplane would fly! All of the cities on earth are at altitudes of less than 5 miles. The atmosphere extends roughly 100 miles. Changes in altitude between cities are relatively small.
 
  • #5
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HallsofIvy said:
Mexico City is not nearly as high as an airplane would fly! All of the cities on earth are at altitudes of less than 5 miles. The atmosphere extends roughly 100 miles. Changes in altitude between cities are relatively small.

Actually, most cities are well below about 4 km, let alone 5 miles. The point remains: Mexico City's air density is about 80% that of sea level (an elevation of 2.2 km). A barometer should read a lower pressure. So why does a weather forecast show a pressure of about 100 kPa? This doesn't make sense.
 
  • #6
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Meteorologocal pressures are corrected to sea level. That is, the surface measurement at Mexico City is adjusted to represent the pressure that would be actually measured at the bottom of a one mile deep shaft, or sea level.
 
  • #7
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Bystander said:
Meteorologocal pressures are corrected to sea level. That is, the surface measurement at Mexico City is adjusted to represent the pressure that would be actually measured at the bottom of a one mile deep shaft, or sea level.

So are they adjusted according to the Standard Atmosphere? That would make sense (sort of). For example, the conditions I cited yesterday in Mexico City would correspond to a density altitude of about 200m (not 2000 m, as it should be).

Actually, do you have an official reference for this? It would be helpful, as I'm currently writing a paper in which I will be discussing it.
 
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  • #9
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Not "standard atmosphere," the "adjustment" is to the value that would exist at sea level. Actual pressure readings are useless for meteorological purposes for just the reason you asked the question --- the altitudes at which barometric readings are taken vary. Charting isobars on a weather map is done to give an idea of high and low atmospheric potentials for air movement, not the altitudes of reporting stations. The barometric equation is applied to whatever reading is available at whatever altitude, and a sea level value is calculated for such purposes. Reference? That's a little like asking for a reference for the alphabet --- you might find references to original conventions on standard methods of reporting in the CRC Handbook of Chemistry and Physics --- should be in a library in Mexico City. Try looking in the older editions first --- check the index for "barometric corrections, meteorological tables, sea level barometer," and permutations.

edit: Try googling "reduction of barometric pressure."
 
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  • #10
Clausius2
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I feel that 2 Km of altitude is not worth decreasing pressure:

Pa=1bar (h=0)

P/Pa=exp(-2000m/288*298K)=0,976 (Boltzmann)

(P-Pa)/Pa=-2,4% of variation

Surely a change of local temperature would cause your doubt, because pressure decreases with h inside isothermal atmosphere. In a low ranges of h, pressures of higher sites might be larger than in lower ones.

Saludos, colegas.
 
  • #11
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Clausius2,

Where does this come from?

P/Pa=exp(-2000m/288*298K)=0,976 (Boltzmann)

I thought an approximate rule of thumb is that pressure drops about 1 inch of Hg for each 1000 ft of elevation (at least for the first few miles above sea level). That would be more like a drop by 3% for 300m. I don't know if that's right, but your equation seems to give a drop of only 20% at 20,000m (about 12 miles). That can't be right, can it?
 
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  • #12
russ_watters
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Bystander said:
Meteorologocal pressures are corrected to sea level. That is, the surface measurement at Mexico City is adjusted to represent the pressure that would be actually measured at the bottom of a one mile deep shaft, or sea level.
I saw this thread before and didn't respond because I'd forgotten this. But how could I - I was in navigation in the Navy and one of my duties was weather forcasting. We even corrected our baometric pressure readings to account for the height of the ship - which wasn't even that high. D'oh.
 
  • #13
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Bystander and russ_watters,

Why is this correction made? If I want to know what the weather is like in Mexico City, don't I want to know the actual barometric pressure at that altitude? Or does the weather depend only on changes in pressure and not on absolute pressure?
 
  • #14
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Clausius2 said:
I feel that 2 Km of altitude is not worth decreasing pressure:

Pa=1bar (h=0)

P/Pa=exp(-2000m/288*298K)=0,976 (Boltzmann)

You're missing a factor of g in your formula. It should read:

P(z) = Pa exp(-zg/RT)

So, in this case you would find P/Pa = 0.796, or about 80%. If you want to introduce temperature variations, you would include the temperature lapse function, T(z) = T0 - Lz, in the original differential form (with L the lapse coefficient, and T0 the sea level temperature), and then re-integrate to obtain P(z). However, temperature effects would not explain the discrepancy.
 
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  • #15
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jdavel said:
Bystander and russ_watters,

Why is this correction made? If I want to know what the weather is like in Mexico City, don't I want to know the actual barometric pressure at that altitude? Or does the weather depend only on changes in pressure and not on absolute pressure?

This was my basic question. It's easy enough to correct the reported pressure to the absolute pressure, but why do I care if the corrected pressure is 101 kPa if I'm experiencing 88 kPa of pressure?
 
  • #16
Clausius2
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GRQC said:
You're missing a factor of g in your formula. It should read:

P(z) = Pa exp(-zg/RT)

So, in this case you would find P/Pa = 0.796, or about 80%. If you want to introduce temperature variations, you would include the temperature lapse function, T(z) = T0 - Lz, in the original differential form (with L the lapse coefficient, and T0 the sea level temperature), and then re-integrate to obtain P(z). However, temperature effects would not explain the discrepancy.

:surprise: Thanks, I've no memory at all!
 
  • #17
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"Why reduce actual station values for atmospheric pressure to sea level?"

Think hydrostatics. Air is a fluid, and seeks its own level. If you're interested in predicting weather, it's useful to know how far from hydrostatic equilibrium an air mass, or the air in the vicinity of a weather station is. We gotta throw in temperature, relative humidity, and who knows what all else, to predict weather, but the barometric reduction is an important bit of information.

Verstehen?
 
  • #18
Nereid
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Besides, weather maps - the ones with isobars that is - would look really, really ugly if they showed the actual pressure. I mean, how could any Californian make sense of such a map??? :rolleyes: :tongue2: :smile:
 
  • #19
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Moreover, the aviation standards require sea-level-corrected air pressure to calibrate the good old barometric altimeters. Although it seems nice to see zero altitude when departing from Mexico city but that would lead to big surprises when landing on LA or Colorado Springs
 
  • #20
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What would be the atmospheric pressure in Kg on 1cm square?
 
  • #21
russ_watters
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tumor said:
What would be the atmospheric pressure in Kg on 1cm square?
That's a strange choice of units, but about 1.0.
 
  • #22
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"kg/cm2 strange?" Check gauges on whatever regulators are handy --- output side usually includes this rather peculiar "metric engineering unit."
 
  • #23
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russ_watters said:
That's a strange choice of units, but about 1.0.

Actually, measuring density in mass/area isn't that uncommon, especially in radiation and particle physics. Usually this is referred to as "mass thickness", and e.g. with radiation shielding is a method of determining the amount of matter needed to attenuate gamma rays (independent of the density of the specific material, be it lead, iron, wood, etc...).

It frequently comes up in particle physics in any atmospheric experiments, e.g. cosmic ray showers and/or atmospheric neutrino expts.
 
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  • #24
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I'd ask the same question from the tech point: at what temperature water boils in Mexico city?
 
  • #25
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Here is code I use. It gives pressure = 78,500 Pascals at 2000 meters.
REM ===============================
Rzero=6378100 ! Earth radius in meters
R=Rzero
g=9.81
Pzero=100760 ! pressure in Pascals at sea level
P=Pzero
Dzero=1.28 !Density of air at 0 deg C, kilograms per cubic meter
D=Dzero
dR=+1 ! going up 1 meter
DO
D=Dzero*P/Pzero ! ideal gas, const temperature
R=R+dR
g=9.81*(Rzero/R)^2
P=P-dR*D*g
LOOP while R<Rzero+2000
PRINT R,P,D
REM ====================================
 
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