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Hi Steve & All, Feb 17, 2007
I guess the terse wording of e-mails, just to save keystrokes, can
be misunderstood as irritation. So I wish to assure you that we are on
the same page. I view exchanges of the elevation type as ego-free,
uncompetitive, laid-back discussions that one might engage in while
waiting for socks to dry by a campfire. Some ideas may prevail or prove
better than others, in one context or another, but in that case the
laurels go to the idea and not to the person who championed the idea.
Not all messages can appeal to all. About the only bird posts that
wake me up are those in which someone says they saw some bird on the way
to work and this leads me to pause over the delete key just long enough
to wonder whether it was carrying a lunch pail or a briefcase. So those
who have become bored by what I consider to be great fun can exercise
their delete finger.
The shape of the pressure/elevation curve in the atmosphere, the
measurement of absolute pressure using a barometer and the use of a
differential manometer to measure the difference between two pressures,
are three related but quite distinct topics. I think you may have them
mixed together in ways that don't adequately make allowance for the
different context of each.
I have not visited the Wikipedia page but don't need to because the
figure that you have calculated for 39 m above sea level, using their
equation, agrees very well with my figure of 5 cm water, i.e. using your
figure for atmospheric pressure (corrected to sea level implied) of 10.3
m, Pb (pressure at sea level) consequently = 1030 cm water and p at 39 m
is equal to 0.995*Pb. But remember, a differential manometer measures
the difference between two pressures not the absolute pressure of
either. In this case, the difference is equal to 1030- (1030*0.995) =
5.15 cm water as compared to my figure of 5 cm using water at the
temperature of maximum density and therefore minimum height at a given
pressure difference.
A differential water manometer can be easily made using routine lab
supplies (as detailed below) and it will be entirely reliable under
stable lab conditions where the two gas bodies are at the same location,
at the same temperature and where the pressure of the gas, which is to
be compared with atmospheric pressure, is maintained constant. Building
something from household scraps that can deal with field conditions,
possible temperature difference initially or over time, the lack of a
way to maintain pressure of the closed gas body constant and in fact
just the opposite, the certainty that the difference between the two
pressures with cause pressure of the closed body to change, is a very
different and much more difficult matter. So different that I didn't
initially notice my prior use of the same principles for many different
purposes under lab conditions (regulation of small positive and negative
pressures and measurement of small stable negative pressures relative to
atmospheric.
The po = h + p of a previous e-mail is essentially an expression of
forces at equilibrium and therefore self-evident. The pressure p plus
the pressure exerted by the water column h must equal po or h would
increase or decrease.
One can readily generate a pressure that is greater than atmospheric
by a known amount, say 10 cm of water, by securing a vertical glass tube
with the lower end 10 cm below water level and supplying air under
pressure to the upper end (air tap or rubber bulb with 1-way valve). As
soon as the pressure in the closed system exceeds 10 cm water, bubbles
will emerge from the lower end. If a U-shaped glass tube, partly filled
with water, is connected by a tee into the closed system then h will
oscillate between 10 cm water and slightly more due to the overpressure
required to form and release an air bubble. For more stable pressures
the lower end can be drawn out to a smaller diameter and turned up.
Minor modifications of this setup enable relatively stable negative
pressures for e.g. controlled removal of supernatant without disturbing
a pellet. And so forth.
I hope this clears up some (all ?) of the confusion.
Yt, Dave Webster
Steve Shaw wrote:
> Hi Dave and others,
> Not to flog a dead horse too long here I hope -- most NNS-ers may
> want to skip this post -- and of course I was being flippant and
> chauvinistic in the earlier reply, just to defend home turf and get
> you going a bit: apologies if you took it to heart, not intended.
> Both methods we cooked up would work in principle, the question being
> whether either would work in practice, and with what sensitivity and
> what likely error if so, always important in designing experiments.
>
> What I meant originally was that I actually hadn't tried to
> understand your method because the result seemed so obviously out of
> whack, while you don't derive your method for po = h + p (previous
> e-mail), perhaps thinking it self-evident. Looking at it, aren't you
> putting variables for water pressure and air pressure illegally in the
> same units in the same equation and that's how you come up with such
> amazing sensitivity?
>
> Despite the decimal place correction you make here, you can't say that
> you'll get anything like a 5 cm change for a 39 m rise, because (a)
> you haven't said what the absolute volume of air is that you've
> trapped in the manometer, and (b) something must be wrong with the
> derivation. On (a), logic would dictate that if you have a column of
> air of 1000 ml trapped in the fixed-diameter manometer tube, and the
> length of it increased by 5 cm when you ascended to a certain
> elevation above sea level, if you now reduced this to a volume of only
> 10 ml in the same tube (same height ascended) the change now would be
> only 1/100 of this, or 0.05 cm -- obviously, the actual change in cm
> must depend directly on the volume trapped. On (b), as I said last
> time, I simply looked this up on Wikipedia -> Atmospheric Pressure ->
> Barometric formulae, where I picked equation (2) to calculate P, the
> atmospheric pressure at different elevations (the constants are a bit
> different for the 6 altitude ranges given, and it would be appropriate
> to use the lowest one here). I've just re-checked this to see if I
> made a mistake, and so that you can check this for yourself if your
> interest is still piqued, it is
> P = Pb*exp-[g*M*(h-hb)/R*T],
> where Pb is the standard atm pressure at sea