If the weight of water in a pipe is 35kg's how much vacuum is required
to prevent that water from falling? What is the required volume of
pipe to create a vacuum equivalent to the first answer? To create the
vacuum I would fill the pipe with water from the bottom and let air
out the top then seal the top.

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Request for Question Clarification by livioflores-ga on 01 Nov 2006 19:55 PST

I think that we need to know the measures of the pipe, recall that
pressure of water is proportional to the heigh rather than its weight.

iang says : it is not vaccum but ..

OK. More exactly, it is the differential pressure - difference in
pressure outside of tube and inside.

The picture is here:

http://www.quicksilver-barometers.co.uk/barometer%20people.html

Nicer picture:
http://brunelleschi.imss.fi.it/museum/isim.asp?c=100144

Inside of tube  is not vaccum - pressure is vapor pressure of the
liquid, which depends on temperature and for mercury is very small.

So instruments combines the function of barometer and thermometer.

Mercury has higher density then water (which makes column shorter) but is
poisonous (which is why it is not used any more). Formula for height h is

  pressure= density * g * h

 is discussed here:


http://www.du.edu/~jcalvert/phys/mercury.htm#Baro


Note on units:
http://www.madsci.org/posts/archives/2001-03/984630300.Ch.r.html

Value of g:

http://en.wikipedia.org/wiki/Acceleration_due_to_gravity

Hedgie

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Request for Answer Clarification by hughesy-ga on 15 Dec 2006 20:35 PST

The pipe is 14.5meters long and 32mm in diameter.

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Clarification of Answer by hedgie-ga on 16 Dec 2006 22:08 PST

hughesy-ga

               The crossection A of the pipe cancels out from the equation:

pressure * A = force 
g * D * Volume = Force (weight)     where volume is V  = A *h 

so only h height matters and for water.

 Critical height for water is about 10 m   (depending ..)

http://greenbooks.theonering.net/guest/files/050504_01.html

 and for higher list one has to use another method: push rather then pull.

string to enter into the search engine is

SEARCH TERM: drilled well

http://www.wellowner.org/awaterwellbasics/typesdrilled.shtml

you can get deeper, and pump is placed IN THE HOLE.  The height of the lift
is then limited only by strenth of the pump.

Good like

Hedgie

Great answer, it has made a few things a lot clearer. I am fiddling
with windmills and home made pumps and trying to reduce head pressure
to make pumping easier. I have a few new ideas which I am keen to try.
I unfortunately have a problem though, I have run out of water in my
bore hole and the only answer to that is in the hands of nature.

It's not the vacuum in the pipe that stops the water falling, it's the
air pressure outside.  Try an experiment - fill a bottle with water,
put your thumb over the end and and hold it upside down in a basin of
water. Now take your thumb away.  Nothing happens!  You don't see
water running out until there's enough vacuum to support it - the air
pressure pushing down on the water in the basin holds it in place.  If
you could use big enough bottles (and basins!) or pipes you'd find
that nothing would change until the column of water was about 34 feet
high.  After that, no matter how long your pipe was, the water column
would always be the same height because that height of water is
pushing down with the same pressure as the atmosphere, so they're in
balance.  If you could leave your pipe in place and measure the height
of the water column you'd see it going up and down as the air pressure
changes - you've got a barometer!

Ian G.

The upward forces must balance the downward forces.

The upward force is from atmospheric pressure, multiplied by the
cross-section area of the pipe.

The downward forces are from (1) gravity, = 35kg x 9.8N/kg, and (2)
the pressure in the partial vacuum above the water, multiplied by the
cross-sectional area.

It looks like we need to know the cross-sectional area to solve the problem.

At any rate, putting it all into one equation gives:

Area x 101,325 N/m^2 = 343N + Area x Pvac

Solving for Pvac, the absolute pressure above the water, gives

Pvac = 101325 N/m^2 - (343N)/Area

Plug in the correct area, and you'll have your answer.  (You'll need
to figure out how many significant figures are appropriate, too.)

Just to follow up my earlier post:

If the area is too small, the equation gives a negative value for
Pvac, meaning that it becomes impossible to prevent the water from
falling for small cross-sections.