Consider two separate columns of pure water at stp. One column is a
pipe with x-section of .049 inches (.250 diameter). Second column is a
pipe with x-section of 3.1416 (2.000 diameter). Top of pipes closed
off (vacuum). Are water columns equal height when water stream pulls
apart? Would like formulas to figure
mathematically(height,x-section,lbs. vacuum). I think this relates to
how high you can siphon water. Thanks! 3rrotec

---

Request for Question Clarification by jeremymiles-ga on 28 Apr 2003 08:30 PDT

The answer is "Yes, they are the same".  Do you want the proof of that?

jeremymiles-ga

---

Request for Question Clarification by dogbite-ga on 28 Apr 2003 11:41 PDT

Hello 3rrotec-ga,

  What do you mean by "When water stream
  pulls apart?"  And, you want a forumula
  to compute the height given the x-section
  and lbs of the vacuum:

       height = f(x-section, vacuum)

  Is that right?

           dogbite-ga

---

Clarification of Question by 3rrotec-ga on 28 Apr 2003 12:58 PDT

To Jeremymiles-ga, Yes I would like mathematical proof. You may make
assumptions on barometer, specific gravity of water, and temp.
Interested in mathematical calculations of lbs. absolute in vacuum
chamber when column separates. First of many questions on this
subject. Thanks! Robert Williams 3rrotec

---

Clarification of Question by 3rrotec-ga on 30 Apr 2003 11:00 PDT

Please clarify one thing and answer will be complete. Use answer mode
so I can proceed with more questions. What will be the negative
pressure in the top void chamber if column is say 40" tall? Does the
water vaporize because of the pressure drop thus filling top chamber
with water vapor thus letting the column lower or fall?

Hello Robert,

          Since none of the previous interlocutors came back
          after your last clarification, I suppose it is OK for
          me to answer - and provide any follow-up clarifications.

 I am imagining your  two tubes to resemble this classical experiment:
  http://galileo.imss.firenze.it/multi/torricel/etorat34.html
    http://www.royal-met-soc.org.uk/weatherclub/secondary/barometer.html

 Commenter    leoj-ga   is  correct pointing out that the
cross-section
 of the tube cancels out  form the formula (below) so that it is
 the density of liquid and atmospheric pressure, which will determine
 the height of the column.

  The instrument  (inverted closed tube) is really first  barometer
 and as you correctly note, it has to do with maximal height to
 which one can 'draw'  a liquid:
  http://www.newadvent.org/cathen/14784a.htm

 Roughly: 760mm of mercury to about 10m  of water should be
            inverse of density of Hg to density of H2O (about 13)

To this day people measure pressure in mm of Hg (instead  of using
 SI units and say: Atmospheric pressure is about 100 kPa )
 http://www.taftan.com/thermodynamics/PRESSURE.HTM

You are having good physical intuition, noting that some liquid will
evaporate (as nature does abhors vacuum, after all) and for
this reason, early instruments measured combination of both:
pressure and temperature. They nevertheless were useful for weather
prediction.
  http://www.zedds.net/WeatherStations/Barometer2.htm

The mathematics:
  I was surprised when I did not find a page with calculation ready
made.
  Closest was a page with some numbers, like density of Hg and nice
picture:
  http://www.du.edu/~jcalvert/phys/mercury.htm#Baro

 So, here is the 'original' derivation :

   Force= Weight of column = g * Volume * Density = g * Area * height
* Density

   Pressure = 100 kPa = Force/Area = g * Density * height

 That is the formula for height, D=Density is  1.0  for H2O, 13.5 for
Hg
 in g/cc which we need to convert to kg/ m-cubed, as  we do use SI
units.

 Oh! What about the vapor pressure?
 Depending on the temperature and volatility of the liquid
 (Hg is good here) it will counterbalance the atmospheric 
pressure.

  So, in the formula above, you will use as Pressure
   Atmospheric Pressure - Vapor Pressure

   The tables of the later are on the Internet, 
(see search termss below), for example, google
search:

://www.google.com/search?hl=en&ie=ISO-8859-1&q=vapor+pressure%2C+water%2C+mercury

 will reveal:
     The vapor pressure of water is 20 millimeters of mercury
  (room temperature) , so, effect does exist, as a small correction.

 But, at the 100 C, it will reach the atmospheric pressure of 760 mm,
(which is the definition of the boiling point)

          http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/vappre.html

Search Terms
     Torricelli Experiment
     History of Barometer
      atmospheric pressure, Pa
      vapor pressure, water, mercury

  This looks like a complete answer to me,
 but feel free to ask for clarifications
 and please do rate the answer after it is 
all  clear.

hedgie

Thanks Hedgie!  Now go to next one.

The easiest way to think of this is to ask why the water is held in
the tube in the first place.  The answer is that air pressure from the
outside is pushing against the water.  The water will be held in place
until the weight of the water exceeds the force created by the air
pressure.  Simply put, there is about 14.7 lbs/sq. in. (psi) pushing
the water up, so you need a height of water that pushes back with that
much for the water to fall.

Since the weight of the water goes with the cross sectional area, and
so does the force created by the air pressure, the cross sectional
area does not enter into the problem (except for very small diameters
when capillary force matter).

Further, it is interesting to note that the air pressure is caused by
the height of the column of air that extends up quite a few miles in
height - the density of the fluid matters.  For the same reason, we
don't use water in barometers too much, we use a more dense liquied: 
Mercury.

Cheers