Water in a 100ft pipe

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I have a small assignment at work that I can't seem to figure out. To not bore you with the details, I have developed a simple scenario.

Say you have water in a tower that is 100ft tall that is flowing through a vertical pipe to the ground. If this pipe is instantaneously closed at the top, at what height would the water equilibrate? and what would be the pressure in inside the pipe between the top of the water and the top of the pipe?

So far, I have developed the simple, atmospheric pressure = (density of water)x(gravity)x(height of water at equilibrium) + (pressure at the top of the pipe)

Obviously, the height of the water can't be greater than 10m because then according to the previous equation, the pressure at the top of the pipe would be negative, right?

Any help would be appreciated,

Vivek
 

Answers and Replies

  • #2
boneh3ad
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If you instantaneously closed it so that no air would be in the top, you would end up with a vacuum. You would just equate the atmospheric pressure pushing on the exit plane with the weight of water and solve for the height. This, of course, ignores the inertia that the flow has, so you would probably actually lose a little bit more water than what this steady calculation predicts, but not much.
 
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boneh3ad, I agree with what you are saying, but am still confused. If you calculate the static height from the atmospheric pressure and we ignored inertia, wouldn't this give a zero pressure inside of the pipe? Isn't that impossible? and if we include inertia, the pressure inside the pipe would increase according to the equation that I posted, but that doesn't really make sense to me either.
 
  • #4
boneh3ad
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Zero pressure is possible if there are no molecules in a particular space, and in the situation you described, there would be no molecules in that space at the top.

The inclusion of inertial effects just means that the force exerted by the atmosphere wouldn't instantaneously decelerate the the column of water and some extra would leak out. Of course, this all ignores that in a vacuum, the water at the top would likely vaporize and so you would end up with pressure up there. I don't know off the top of my head how to quantify that.
 
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Okay, I neglected the fact that at lower pressures, water would vaporize and this vapor would create a positive pressure in the pipe, right? Now I just need to figure out how to actually calculate this. Thanks for your help!
 
  • #6
Andy Resnick
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I have a small assignment at work that I can't seem to figure out. To not bore you with the details, I have developed a simple scenario.

Say you have water in a tower that is 100ft tall that is flowing through a vertical pipe to the ground. If this pipe is instantaneously closed at the top, at what height would the water equilibrate? and what would be the pressure in inside the pipe between the top of the water and the top of the pipe?
If the bottom end of the pipe is at atmospheric pressure and air bubbles can't get into the pipe from the open end, at equilibrium the column of water will be about 34 feet high. The remainder space will be filled with water vapor at a pressure of about 9.4 inH2O.

If you were using mercury instead of water, the column height would be 760mm (29.9 inHg), with the 'empty' headspace filled with Hg vapor 0.28 mmHg pressure. If you like, 9.4 inH2O is 17.5 torr, and 0.28mmHg is 0.28 torr.

Torricelli got in trouble with the Vatican for doing this experiment:

http://www.eoht.info/page/Nature+abhors+a+vacuum
 
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  • #7
AlephZero
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Until the potential energy in the water column is dissipated by viscosity and turbulence, the water in the system will behave like a damped oscillator.

Consdering the pipe is 100ft high and the equlibrium position is at about 34ft, unless the pipe is very narrow it is quite likely it will completely empty. The free surface is starting at 100 - 34 = 66 feet above the equilibrium position, so with no damping it would overshoot and try to go to 66 feet below the equilibrium, or 32 feet below the bottom of the pipe.

Obviously what happens in practice will depend on the exact details of the pipe and what it is emptying into.

There is also the issue of whether the pipe will collapse by buckling under the external pressure of 1 atmosphere - again, that depends on the diameter and construction of the pipe.
 
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Andy, can you explain how you calculated the vapor pressure? I need to be able to do it for different heights. Thanks
 
  • #9
Andy Resnick
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Andy, can you explain how you calculated the vapor pressure? I need to be able to do it for different heights. Thanks
vapor pressure doesn't depend on height, it depends on temperature. There are tables with this information:

http://intro.chem.okstate.edu/1515sp01/database/vpwater.html
http://www.boulder.nist.gov/div838/SelectedPubs/NISTIR.6643.pdf [Broken]
 
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  • #10
Andy Resnick
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Obviously what happens in practice will depend on the exact details of the pipe and what it is emptying into.

There is also the issue of whether the pipe will collapse by buckling under the external pressure of 1 atmosphere - again, that depends on the diameter and construction of the pipe.
Right- for example, if the pipe were very thin like a capillary, wetting effects will contribute as well.
 
  • #11
Redbelly98
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Andy, can you explain how you calculated the vapor pressure? I need to be able to do it for different heights. Thanks
While there is a formula for calculating vapor pressure, I think most people just look it up in a table like this:

http://intro.chem.okstate.edu/1515sp01/database/vpwater.html

Andy was assuming the water was at room temperature, or 20 C (68 F), for which the vapor pressure of water is 17.5 Torr. This corresponds to a 9 inch height of water. As long as the water is at that temperature, that is the vapor pressure, and the 9 inch height will apply.
 
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Alright that definitely makes sensne.

Thanks for all the help!

Vivek
 

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