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Hydrostatic pressure in a submerged pipe

  1. Nov 25, 2009 #1
    What would be the hydrostatic pressure near the top of a vertical pipe closed at the top and open on the bottom? Would you calculate the pressure density*gravity*depth measuring depth from the top of the pipe or from its opening at the bottom or the difference between the point of measurement and the top of the pipe?

    say the pipe was 50 feet tall, and submerged underwater 10 feet, what would the calculation for the pressure 1 foot from the top of the pipe be?

    d*g*11 or d*g*60? or d*g*1

    this isnt homework question. i never did my homework in school anyway. (and it probably shows). im trying to build this. help please?
    Last edited: Nov 25, 2009
  2. jcsd
  3. Nov 28, 2009 #2
    If I understand your configuration correctly, it should be d*g*11. As long as the water inside the pipe is not isolated from the outside, the pressure at any point inside the pipe is exactly the same as water pressure outside at the same level. I assume there is no air column inside the pipe. If there is, the above won't be the case.

    Wai Wong
  4. Nov 28, 2009 #3
    Ah wonderful, thank you. I assumed that would be the case, but then I thought about it too long and got myself confused again.

    How does the pressure measurement change with a column of air inside the pipe? Would you need to use a ratio of air/water density in the pipe? Or does it have more to do with the compressibility of air?

    What would be the method of calculation?

    Thanks again!
  5. Nov 29, 2009 #4
    The pressure at the top of an air column is practically the same as the pressure at the bottom of the column. When compared with a water column where the pressure change is l*d*g, an air column introduces a difference of that magnitude. So, to work out the pressure at any level, you first assume there are no air columns, and then add l*d*g for each air column *below* the desired level. Or you can add the l's and then multiply by d*g. For example if the pipe contains all air, then at the bottom end the pressure is 60*d*g, and so the top should also be 60*d*g. My suggested method gives the following result:

    P inside pipe at 10ft = P outside pipe at 10 ft + air column difference
    = 10*d*g + 50*d*g = 60*d*g (the same)

    Wai Wong
  6. Nov 29, 2009 #5
    Ok, thanks!

    ...only I still don't quite understand why that is exactly. Why doesn't the standard l*g*density apply?
    My guess is it's because the water above the column is denser than the air column, correct? And so the water pressure dictates the air pressure, since the weight of the air is more or less negligible compared to the water surrounding it?
    An so basically, the pressure at any height within a column of air in the pipe is about the same as the pressure of the water level at the bottom of that column of air, and the pressure of the water is equal to its absolute depth*density*gravity. Am I following you correctly?
  7. Nov 29, 2009 #6
    The formula l*d*g applies to the water pressure in an open container because the water pressure at the surface is zero (or should I say atmospheric pressure). That is not necessarily the case in a closed container.

    In the absence of air columns, the formula still applies because the pressure is the same at a certain level (the bottom end) and the pressure gradient is the same inside and outside the pipe all the way to the top end.

    From my understanding of your argument, it sounds correct.
  8. Dec 1, 2009 #7


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    Just treat it like a manometer...start at one point (the surface of the liquid exposed to atmosphere) and work your way down the outside of the tube and then back up the inside summing the pressure changes as you go.

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