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Author: Subject: does anyone know how much water 1 atm can support in a column
g3nius427
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[*] posted on 17-11-2004 at 14:14
does anyone know how much water 1 atm can support in a column


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Saerynide
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[*] posted on 17-11-2004 at 14:19


Maybe you should stop asking to be spoonfed your homework.



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HNO3
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[*] posted on 17-11-2004 at 14:54


Search GOOGLE, HMMMMMM?
"water column atmospheric pressure"
fourth hit
then read the page

[Edited on 11-17-04 by HNO3]




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g3nius427
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[*] posted on 17-11-2004 at 15:26


rofl ok no more spoonfeeding sorry
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thalium
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[*] posted on 18-11-2004 at 11:44


Saerynide is right...I'm new here too and I don't need the people here to do my homework...pV=(number of moles)RT
and R=0,082 and the temperature has to be in K (C plus 273)




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Dodoman
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[*] posted on 18-11-2004 at 17:01


Quote:
Originally posted by thalium
pV=(number of moles)RT
and R=0,082 and the temperature has to be in K (C plus 273)


I don't get how are these info useful in solving the problem.
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Hermes_Trismegistus
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[*] posted on 18-11-2004 at 20:12


Take the density of water at your given T and P.. 1 g per ml?

Density of mercury something like 13 g per ml?

get ratio....1:13 (or something like it.)

take column of mercury absolute vacuum (~29.92 inches)

convert.

Works for all substances.

Mmmm......Dimensional Analysis.....(drooling)
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Dodoman
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[*] posted on 19-11-2004 at 05:27


The way i see it is as follows. The pressure at a point in a liquid= dgh (d is the density , g is the free fall acceleration and h is the hight of the liquid column).

Now 1 atm = 76 cmHg. dHg=13579.04 kg/m 3 (13600 Kg/m3). dH2O=1000 kg/m 3.

d(Hg) g h(Hg) = d(H2O) g h(H2O)

d(Hg) h(Hg) = d(H2O) h(H2O)

h(H2O)=13600x76/1000 cm

Now i don't mean to spoon feed anyone. I'm just tring to prove a point. This approach is so much easier don't you think.
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Hermes_Trismegistus
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[*] posted on 20-11-2004 at 00:01
no. NO! that's not easier at all...


In fact it makes not one whit of sense! It must be a deliberate joke.....

It's really easy.

How much water can an evacuated column (perfect vacuum) support....

Ok....29.92 inches mercury.....mercury is 13.546 times more dense than water.

Multiply your figure (29.92) for mercury by 13.546....and you have the column of water that could be supported by the "vacuum"

405.30 inches....thats 33.77 feet....good luck with that.

of course, this doesn't work in reality. The water vapourises in a vacuum until a certain pressure is reached.

You may have heard of it......Vapour Pressure

And I believe that's how the figures for vapour pressure were originally obtained.

By floating some of the liquid into the 30 plus inch column of mercury and seeing how much the murcury level dropped.


[Edited on 20-11-2004 by Hermes_Trismegistus]




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Dodoman
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[*] posted on 20-11-2004 at 07:53


:o You must be the one joking Hemes.

Just how exactly did you get the specific gravity of mercury ?

Is it not from 13546/1000 And what did you did then didn't you multiply it by the hight of the mercury column ?

Doesn't your equation look like this:

h(H2O)=d(Hg)/d(H2O) x h(Hg)

........?

where did that equation come from ? Hint Hint ;)

Both methods come from the same equation but the easy thing about using the general formula not the specific gravity method is that the problem may have other liquids than water.
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Hermes_Trismegistus
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[*] posted on 20-11-2004 at 12:04
It's like you have a GRANT for this or something the way you're complicating the hell out of it.


Specific gravity???

Free Fall accelleration???

What the hell.

It's like you have a GRANT for this or something the way you're complicating the hell out of it.

We don't need calculus for this, it is a simple conversion.

The question, a common first year chemistry problem basically boils down to this.....

Take torricelli's barometer, fill it with water instead of mercury....how high is the column....


BLAH!!!!

EASY AS PIE!!!................PIE!!!

BLAH!




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Magpie
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[*] posted on 20-11-2004 at 20:29


I will make my stab at simplicity and grant money:

Assume atmospheric pressure = 14.7 lbs/in2. A cubic inch of water weighs 0.0361 lbs. Therefore a column of water able to counterbalance this pressure would be = 14.7/0.0361 = 407.1 inches high.

or, H = 33.9 ft = 1034.0 cm = 10.34 m.




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JohnWW
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[*] posted on 21-11-2004 at 00:00


The practical consequences of this:
(a)water cannot be siphoned over a height of more than 33.9 ft or 10.34 m (at sea level) and down the other side to below the water level on the original side (but the limit, at least at temperatures much above freezing, would be less than this due to vapor formation with cavitation), and less at above sea level;
and (b) that a Torricelli-type barometer employing water as a liquid instead of Hg would have to be at least this height.
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Oxydro
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[*] posted on 21-11-2004 at 20:21


You neglected, JohnWW, to mention that this also means that if you want to pump water more than that height you need to put the pump at the bottom :D. Actually, this is pretty similar to syphoning, so I guess you had it pretty well covered ;).

Oh yeah, that reminds me, let's not forget that for every 33.9 ft you go under water, the pressure increases by 1 atm. Useful stuff, huh? :) Especially if you're building a submarine :D!

Fresh water of course, that is. :cool:

Edit: Well, regarding the pump, I should say, it has to be within that distance of the surface of the water, assuming atmospheric pressure on the water's surface. Got to specify, you know:P!

[Edited on 22-11-2004 by Oxydro]
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neutrino
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[*] posted on 22-11-2004 at 03:28


How does atmospheric pressure have any effect on this? At 40 ft, wouldn't the water at the top of the column boil at room temp due to the vacuum exerted by the column(s) of water?
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Oxydro
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[*] posted on 22-11-2004 at 06:23


Neutrino, I assume you're referring to the last section of my post, saying that it's assumed the pressure outside the tube is atmospheric. It does have an effect, because it is atmospheric pressure which forces the water up the column. Remember, this is how a manometer works. If the pressure outside drops, the column will fall, while if it rises, the column will also rise.

Anything that uses suction to pump a liquid, etc., is actually just generating a difference between internal and atmospheric pressure.

I'm not sure that I interpreted you correctly, but I can't think of any other meaning. Sorry if that isn't what you mean, I don't mean to offend you by explaining something you probably know.
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neutrino
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[*] posted on 22-11-2004 at 14:03


Sorry about that, my brain was obviously not thinking about the force of the atmosphere (I always leave some force out, which is why I’m failing physics…) The height is calculated by subtracting the vapor pressure of the liquid from the atmospheric pressure.
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[*] posted on 26-11-2004 at 11:21


Offtopic:
A little question of mine: How do you survive using feet, inches, lbls/inch^2 etc.?
Let's face it: the metric system is superior? Or is it not? Feel free to criticize my point of view.

[Edited on 26-11-2004 by TheBear]
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[*] posted on 26-11-2004 at 21:28


Mr Bear:

It's not easy. I hate the English system but in the country I happen to live in that is the predominant system. So in my studies and in my work I would be continually converting between English and metric. The force/mass units are especially tedious in English. But when the equipment you have been supplied is in English units you must take your readings in English units.

The weather forecast each day is in degrees F. I think in degrees F. If I get a Canadian weather report in deg C I have to do a conversion to get an accurate feel for the predicted temperature.




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[*] posted on 27-11-2004 at 04:27


Quote:
Originally posted by TheBear
the metric system is superior? Or is it not? Feel free to criticize my point of view.


Neither is superior, it only matters what easier for the person using the units.

Personally I like the metric system the best because it makes more sense to me (everything is in multiples of ten, exponential notation is easier, etc.) and most the formulas i use are based on metric or metric derived units.
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