Under pressure (doo doo doo doot doot doo doo)
December 19, 2005 7:38 AM
How much lower is the air pressure on the 23rd floor of a standard apartment building than at ground level?
For the sake of accurate calculations, I'm on the 23rd floor but it's the 22nd storey as there's no 13th floor. I get migraines, and air pressure drops seem to be one of my main triggers. It occurred to me that I may be handicapping myself (literally! ha!) by living in a high-rise. Is there a significant difference?
For the sake of accurate calculations, I'm on the 23rd floor but it's the 22nd storey as there's no 13th floor. I get migraines, and air pressure drops seem to be one of my main triggers. It occurred to me that I may be handicapping myself (literally! ha!) by living in a high-rise. Is there a significant difference?
It might be you are effected by the rapid compression/decompression, more than the absolute pressure. Possibly you could decompress for a few minutes on intermediate floors.
posted by StickyCarpet at 7:49 AM on December 19, 2005
posted by StickyCarpet at 7:49 AM on December 19, 2005
Oh yeah, keep everything in SI and realise that this is just a rough approximation based on air as an incompressible fluid.
posted by meehawl at 7:50 AM on December 19, 2005
posted by meehawl at 7:50 AM on December 19, 2005
Additional info for the math-heads: you live in Toronto, which is just a teeny bit above sea level (380 feet or so). The 22nd floor would presumably bring that to a bit over 600 feet, ish. Presuming you don't live in the Scotia Plaza or another super-tall building, which might have internal air pressure management system.
posted by RJ Reynolds at 7:56 AM on December 19, 2005
posted by RJ Reynolds at 7:56 AM on December 19, 2005
USA today has a pressure per altitude table which lists the standard pressure of sea level at 1013hPa and the standard of 1000m above sea level at 900hPa. It appears to be an exponential relationship, but taking a floor at about 4m, you would gain around 0.5hPa per floor, which would total around 11hPa difference over 22 floors. Apparently 1hPa = 100Pa which is 100 Newton per square meter, which I gather means an additional 100kg of pressure on each square meter of a body 22 floors lower.
But I have no idea what I'm talking about.
posted by scottreynen at 8:01 AM on December 19, 2005
But I have no idea what I'm talking about.
posted by scottreynen at 8:01 AM on December 19, 2005
Oh, that would be 1100kg or pressure, which seems like an awful lot. Hopefully someone will come along and correct my mistake(s).
posted by scottreynen at 8:03 AM on December 19, 2005
posted by scottreynen at 8:03 AM on December 19, 2005
Oh, that would be 1100kg or pressure, which seems like an awful lot. Hopefully someone will come along and correct my mistake(s).
No, you're actually right. There's an awful lot of atmospheric pressure out there.
If the numbers Scott mentioned above are accurate (and there's no reason to think they aren't), then there's about a 1% drop in atmospheric pressure from the ground to your apartment. This is less than the normal variations due to weather and such, so I think it's doubtful that this could be the cause. On the other hand, it could be the rate of change of pressure that is the trigger and not the change itself, in which case a way to solve the problem would be to take the stairs. (And get some exercise, too!)
posted by Johnny Assay at 8:21 AM on December 19, 2005
No, you're actually right. There's an awful lot of atmospheric pressure out there.
If the numbers Scott mentioned above are accurate (and there's no reason to think they aren't), then there's about a 1% drop in atmospheric pressure from the ground to your apartment. This is less than the normal variations due to weather and such, so I think it's doubtful that this could be the cause. On the other hand, it could be the rate of change of pressure that is the trigger and not the change itself, in which case a way to solve the problem would be to take the stairs. (And get some exercise, too!)
posted by Johnny Assay at 8:21 AM on December 19, 2005
a rough approximation based on air as an incompressible fluid.
If air is something like an "incompressible fluid," what would be a "compressible fluid?"
posted by rxrfrx at 8:38 AM on December 19, 2005
If air is something like an "incompressible fluid," what would be a "compressible fluid?"
posted by rxrfrx at 8:38 AM on December 19, 2005
If you assume the atmosphere is in hydrostatic equilibrium (good enough I think) you use
dP = - (density) g dr
You can get the density assuming air's an ideal gas (again good enough I reckon) and you'll deal with the compressibility issue. This puts density proportional to pressure, and then you do indeed have an exponential relation.
Again, solving is left as an exercise for the reader.
posted by edd at 8:49 AM on December 19, 2005
dP = - (density) g dr
You can get the density assuming air's an ideal gas (again good enough I reckon) and you'll deal with the compressibility issue. This puts density proportional to pressure, and then you do indeed have an exponential relation.
Again, solving is left as an exercise for the reader.
posted by edd at 8:49 AM on December 19, 2005
what would be a "compressible fluid?"
The stuff in shock absorbers? Perhaps he meant it as a simplifying assumption as opposed to meaning that air was really incompressible. Like the old joke about the physicist at the races: "Let's assume that all the horses are perfect spheres..."
posted by GuyZero at 8:50 AM on December 19, 2005
The stuff in shock absorbers? Perhaps he meant it as a simplifying assumption as opposed to meaning that air was really incompressible. Like the old joke about the physicist at the races: "Let's assume that all the horses are perfect spheres..."
posted by GuyZero at 8:50 AM on December 19, 2005
(oh and I assume temperature is constant, which it should again be near enough or I don't imagine people would build towers that tall)
posted by edd at 8:52 AM on December 19, 2005
posted by edd at 8:52 AM on December 19, 2005
Well, from Wikipedia: http://en.wikipedia.org/wiki/Air_pressure
Air pressure can be approximated:
log(Pressure) = 5 - h/15500
Pressure = 10^(5- h/15500)
so pressure at 0m above sea level:
P=10^5
= 100000 Pascals
= 100 kPa (which sounds right)
assuming 1 story = 4m
23stories = 4x23 = 92m
P = 10^(5-92/15500)
= 10^(4.99406...)
= 98642.601 Pascals
= 98.64 kPa
so really atmospheric pressue doesn't really change all that much (less than 2 kPa), and for most of us should be hardly noticeable.
posted by ruwan at 9:14 AM on December 19, 2005
Air pressure can be approximated:
log(Pressure) = 5 - h/15500
Pressure = 10^(5- h/15500)
so pressure at 0m above sea level:
P=10^5
= 100000 Pascals
= 100 kPa (which sounds right)
assuming 1 story = 4m
23stories = 4x23 = 92m
P = 10^(5-92/15500)
= 10^(4.99406...)
= 98642.601 Pascals
= 98.64 kPa
so really atmospheric pressue doesn't really change all that much (less than 2 kPa), and for most of us should be hardly noticeable.
posted by ruwan at 9:14 AM on December 19, 2005
Doesn't all of that stuff go out the window because of things such as ceilings and windows and ventilation? Wouldn't the only real way to tell be to take actual measurements?
posted by furtive at 9:24 AM on December 19, 2005
posted by furtive at 9:24 AM on December 19, 2005
Ceilings and windows etc won't make that much difference. The pressure will equalise with outside for the most part. There are slight issues with wind etc. There is a pressure effect that applies to buildings though, related to heating. See previously this AskMe.
posted by edd at 9:29 AM on December 19, 2005
posted by edd at 9:29 AM on December 19, 2005
GuyZero, the stuff in shock absorbers is incompressible (unless they are um, air shocks). Shocks are absorbed by forcing the incompressible fluid through apertures; that turns the energy into heat.
posted by Kirth Gerson at 9:31 AM on December 19, 2005
posted by Kirth Gerson at 9:31 AM on December 19, 2005
Atmospheric pressure at ground level changes regularly (hence "highs", "lows", and the function of a barometer.)
That said, assuming that you're using a mean atmospheric pressure at sea level of 1013.25 millibars...
At mean altitude above MSL (mean sea level), Toronto is at 858.45 mb. (Ground floor of your building.)
On the 22nd floor, the pressure is approximately 794.90 mb.
I did this by using a formula I learned in a physical oceanography class.
Po = pressure at sea level
Ph = pressure at height
M = .029 kg/mol (mass of a mol of air)
g = 9.8 m/s2 (gravity)
z = height above sea level (97 meters for Toronto, 142 for 22nd floor)
R = 8.314 J/K/mol (gas constant)
T = temperature (assumed 20 degrees, arbitrarily)
e = 2.7182
Ph = Po*e^(-Mgz/RT)
ruwan's formula is just as good (and much faster.) You can generally assume 1% decrease for every 80 meters.
posted by nekton at 9:50 AM on December 19, 2005
That said, assuming that you're using a mean atmospheric pressure at sea level of 1013.25 millibars...
At mean altitude above MSL (mean sea level), Toronto is at 858.45 mb. (Ground floor of your building.)
On the 22nd floor, the pressure is approximately 794.90 mb.
I did this by using a formula I learned in a physical oceanography class.
Po = pressure at sea level
Ph = pressure at height
M = .029 kg/mol (mass of a mol of air)
g = 9.8 m/s2 (gravity)
z = height above sea level (97 meters for Toronto, 142 for 22nd floor)
R = 8.314 J/K/mol (gas constant)
T = temperature (assumed 20 degrees, arbitrarily)
e = 2.7182
Ph = Po*e^(-Mgz/RT)
ruwan's formula is just as good (and much faster.) You can generally assume 1% decrease for every 80 meters.
posted by nekton at 9:50 AM on December 19, 2005
(incidentally, what nekton's done there is my 'exercise left for the reader' so the assumptions I used are what go in to getting that formula)
posted by edd at 10:21 AM on December 19, 2005
posted by edd at 10:21 AM on December 19, 2005
At mean altitude above MSL (mean sea level), Toronto is at 858.45 mb. (Ground floor of your building.)
On the 22nd floor, the pressure is approximately 794.90 mb.
Which is a difference of about 7.5%, right? That certainly seems detectable by the human body.
/I know nothing, I'm from Barcelona.
posted by dash_slot- at 2:56 PM on December 19, 2005
On the 22nd floor, the pressure is approximately 794.90 mb.
Which is a difference of about 7.5%, right? That certainly seems detectable by the human body.
/I know nothing, I'm from Barcelona.
posted by dash_slot- at 2:56 PM on December 19, 2005
You may be experiencing the stack effect. I used to operate 2, 35 floor high rise office towers and the airflow in the winter (cold outside, warm inside, warm air rising) was so bad elevator doors had problems closing. So when you are in the top of this building during winter you probably are under higher pressure than outside at that point.
This online calculator may help explain the effect.
My building had pressurization fans to pressurize the lobby to fix the issue of the elevator doors closing, however when it was -15F outside, freezing the sprinkler lines in the drop ceiling was a bigger issue (yes they did).
This is a great reason why when workers replace a window in a high rise you keep the office door closed or the desk/ paper gets a cleaning (i have seen this first hand) Woosh.
As an aside, lots of these buildings have a 13th floor, but it is often a mech equipment floor (IAAOE).
posted by blink_left at 3:58 PM on December 19, 2005
This online calculator may help explain the effect.
My building had pressurization fans to pressurize the lobby to fix the issue of the elevator doors closing, however when it was -15F outside, freezing the sprinkler lines in the drop ceiling was a bigger issue (yes they did).
This is a great reason why when workers replace a window in a high rise you keep the office door closed or the desk/ paper gets a cleaning (i have seen this first hand) Woosh.
As an aside, lots of these buildings have a 13th floor, but it is often a mech equipment floor (IAAOE).
posted by blink_left at 3:58 PM on December 19, 2005
Numbers are nice, but do they convey much? Maybe, maybe not.
One thing I can confirm: the pressure difference can be felt. With an ear infection, it can be extremely painful. This goes to support the notion that it isn't pressure that's the problem, but the speed of change.
posted by Goofyy at 4:16 AM on December 21, 2005
One thing I can confirm: the pressure difference can be felt. With an ear infection, it can be extremely painful. This goes to support the notion that it isn't pressure that's the problem, but the speed of change.
posted by Goofyy at 4:16 AM on December 21, 2005
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P23 + (air density)(gravitational attraction)(height) + (.5)(air density)(wind speed at 23rd floor)^2 = Pground + (air density)(gravitational attraction)(ground level) + (.5)(air density)(wind speed at ground level)^2
P23 is the atmospheric pressure that you are looking for. Pground is the atmospheric pressure 23 floors below you.
The calculation of this is left as an exercise for the reader.
posted by meehawl at 7:49 AM on December 19, 2005