Re: O2


"Doug Semler" <dougsemler@xxxxxxxxx> wrote in message
news:v4edna2Uy6-5pCTbnZ2dnUVZ_qmlnZ2d@xxxxxxxxxxxxxxxxxxx
"Jim Logajan" <JamesL@xxxxxxxxx> wrote in message
news:Xns9985BCD277AF0JamesLLugojcom@xxxxxxxxxxxxxxx
Doug <dougsemler@xxxxxxxxx> wrote:
[ Very thoughtful, but long, article elided for brevity. ]

An impressive analysis Doug! But I believe you made a few errors (which
I would have made too - but somehow reading another person's explanation
with a critical eye makes them more evident; odd psychology, I know):

It was contrived, as I admitted. That's part of the problem. (I kinda
wish
the original had been cross-posted to r.a.p, because I responded in that
thread as well.) After thinking about it, I realized there is one more
issue with the problem as stated, and that is that it assumes a constant
density and temperature vertically along the measured column of air.
Which,
of course, is absolute baloney in the real world.


1) You can't use the ideal gas law for a liquid. It produces very wrong
numbers (whereas the ideal gas law produces "good enough" numbers for
real gases near STP). There are other equations of state that do better;
see for example the Peng-Robinson equation, among others:
http://en.wikipedia.org/wiki/Equation_of_state

I don't remember using the ideal gas law for liquid, I used hydrostatic
pressure calculations, and the combined gas law. While for the purposes
of
the example, the laws are sufficient, they aren't in the real world (yet
another hand wave here). Just like the Jeppesen book, these are
simplifications aimed at non-physicists. I totally handwaved the phase
transitions, heat transfer caclulations, energy transfers, energy
expendetures at the molecular level, etc. I admit, using "water" was
probably bad in the sense that you can't really apply the gas laws to it,
however I was only attempting to use it as a visualization technique.
(Using it as a visualization allows me to also demonstrate that the
atmosphere is subject to fluid dynamics). I thought I had said that you
need to "picture" the atmosphere as water in a pool. <shrug> Maybe I
should
have been more clear...

The people in this group are more familiar with density =
pressure/temperature (which is really just a simplified derivation of the
ideal gas law). Or do you REALLY want me to get into nth order
differential
equations, Euler integrations, Legendre transforms, Maxwell relations and
stuff like that? If so, we should be in sci.physics <g>


2) When you set P1*V1/T1 = P2*V2/T2 you overlooked the fact that P is
dependent on height h above the ground. That is P is a function P(h). So
for example at 0.5 m above ground the pressure P1(0.5) is not equal to
P2(0.5). Only P1(0) = P2(0).

mmmm, it's a function of the height of the column above it, not the height
above ground. Anyway, the equation is only relevant if the gas has the
same mass over the same surface area. In other words, it is exerting
exactly the same pressure. However, because P1(0) = P2(0), we can
calculate
the density change, which allows us to calculate the volume in height of
the
entire system, which allows us to calculate the pressure differential when
"moving" the surface up half a meter. (note, again, this is simplified)

I don't know that it's particularly relevant, however. The "issue" as I
see
it is that there seems to be an intermixture of the pressure measured at
the
surface (what we know as "atmospheric" pressure) with that of gauge
pressure, which is the pressure measured at an altitude. What you also
have
to remember, is that at "height", you have "moved" your gauge, and have
therefore reduced the mass of the atmosphere that is above it and also the
volume. Remember, pressure is a measurement of force applied to an area.
For example, if you have an instantaneous change in altitude, you have
reduced the amount of air above it (in the atmosphere, you reduce the
force
by about half give or take). Note, I am ignoring the change in
acceleration
due to gravity from 9.81 m/s^2 at a height of 18,000 ft, because it is
less
significant than the change in mass in the air. I am also ignoring
temprature differences at the different altitudes. This is pretty much
how pitot-static altimiters work. you set the altimiter to the measured
pressure at mean sea level, and by measuring the pressure at height h, you
can get the "height" above the ground.


3) An ideal gas is considered compressible. An ideal liquid is not. So
for a gas P(h) will decrease exponentially while for a liquid P(h) will
decrease linearly. The volume of the ideal gas would extend to infinity
whereas the volume of the ideal liquid would be finite. This presents
obvious problems with regard to V1 and V2 (both being equal to infinity
for a gas!) :-) (And no, I don't know, off hand, how to address that
problem!)

Gravity constrains the volume of the atmosphere. Think of it as the very
top of the sides of the pool. Once the height of the water reaches that
point, it spills over and is no longer in the system. <g>


4) When you wrote "I am also not addressing the localized differential
of adding heat," I think you sidestepped the problem of most interest to
real-world piloting.

The reason I handwaved it was because the temperature of an air column of
atmosphere is not constant throughout the column. As I said in r.a.p.,
the
key thing that a pilot is really interested is how pressure, temperature,
and density relate to each other. 2000 ft of density altitude is 2000 ft
density altitude, whether the conditions are at 200 ft actual elevation or
at 2000 ft actual. And the key way for a pilot to know how to get there
"close enough" is to understand the general relationship between
temperature, density, and pressure. There are certain aspects of piloting
that can be considered "close enough." You'll hear them often as "rules
of
thumb," and there are several of them in piloting. Knowing whether the
density altittude is exactly 2039 feet or "about" 2000 feet is really
irrelevant unless you are really anal, because what you are really trying
to
do is make informed decisions based on the information that you are
provided. What you know is that your performance is going to be affected
as if your plane was about 2000 ft higher than it actually is, and that
adjustments should be made accordingly. Your decision to fly may not be
affected if you are in San Diego; however, you may decide to wait another
day if you are in Santa Fe, NM. There are similar ones for calculating
distance to start descent, true airspeed calculation, wind correction,
weight and balance calculations, and so forth. The key is learning which
ones can be considered "close enough," which ones should be calculated
precisely, and when a value that you got that's "close enough" tells you
to
recalculate it precisely.

--
Doug Semler
a.a. #705, BAAWA. EAC Guardian of the Horn of the IPU (pbuhh).
The answer is 42; DNRC o-
Gur Hfrarg unf orpbzr fb shyy bs penc gurfr qnlf, abbar rira
erpbtavmrf fvzcyr guvatf yvxr ebg13 nalzber. Fnq, vfa'g vg?


Since the thread seems to have changed to how pressure differentials can
exist in the atmosphere, here is some additional food for thought--with the
disclaimer that I am neither a physicist not a meteorologist, nor do I play
either of them on TV...

Local heating near the surface might cause expansion which would result in a
local area of decreased weight and therefore locally decreased pressure--and
that much has been alluded to above.

In addition, I have read that most of the weather which we observe is
closely related to the presence of water vapor in the atmosphere. The
average molecular weight of dry air appears to be approximately 29; since O2
is approximately 32, N2 is approximately 28, and CO2 is approximately 44.
Water, which is a major variable in the warmer atmosphere nearer the ground,
has a molecular weight of approximately 18. Therefore, wherever there is
strong local heating concurrent with moisture to be absorbed into the
atmosphere, the resulting local pressure changes and vertical motion can be
greatly amplified.

I regret that I am unable to recall and provide any citations at this time.

Peter


.

Relevant Pages

  • Re: O2
    ... I don't remember using the ideal gas law for liquid, I used hydrostatic pressure calculations, and the combined gas law. ... I thought I had said that you need to "picture" the atmosphere as water in a pool. ... pressure is a measurement of force applied to an area. ...
    (rec.aviation.student)
  • Re: The weather and the many different factors that drive it...
    ... > winds travel upwards and suck air away from the surface of the Earth ... > like a giant vacuum cleaner, decreasing the air pressure above the ... > the levels in the atmosphere involved. ... > Troposphere where in reality the Stratosphere has a part to play. ...
    (sci.geo.meteorology)
  • Re: world with less air pressure - survivable?
    ... provides a great pressure garment - you don't "explode in a vacuum", ... At 37 deg-C, the vapor pressure of water is 0.07 Atm, ... Why do you think a lower-density or lower-pressure atmosphere would ... but also the atmospheric temperature gradient. ...
    (rec.arts.sf.science)
  • Re: Is it this easy to live on Earth?
    ... and its atmospheric pressure is 101.3 kpa. ... and Mars' atmosphere obtains a lesser figure - the sum of all figures ... also means that Venus already has nitrogen and mineral salts to work ... For Venus you need 278 tonnes for every 10 tonnes useful payload ...
    (sci.space.policy)
  • Re: Reply to Eric Rowley -- Ozone
    ... pushing all the heat in a large area into a smaller area. ... temperature is higher, but there is no actual new heat energy in the ... If this is correct, then pressure is not a heat source, as no ... We can move it around the atmosphere. ...
    (talk.origins)