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I Pascal's law and kinetic energy in pipes

  1. Mar 18, 2017 #1
    If you have a U-shaped pipe like the attached image, with more fluid in the left column than the right, then the fluid will level out basically because of Pascal's Law. Force on the fluid is exerted on every part of it in equal direction etc..

    So when the fluid levels itself it flows across to the right hand side to fill the empty area. The fluid inside the bottom part of the U-shape flows to the right with a certain velocity or kinetic energy. The kinetic energy of that flow comes from the potential energy of the fluid in the tall left column.

    My question is what happens when you make the bottom of the U-shape very long? Now you are moving a lot more fluid with that little bit of potential energy. How is that energy being 'created from nothing'?

    I understand flow is not that simple and there are eddies etc.. But it must be true that fluid is basically flowing across the whole pipe, otherwise you couldn't fill that empty space in the right column.

    Attached Files:

  2. jcsd
  3. Mar 18, 2017 #2
    Interesting question. Assuming inviscid flow, one can show that, if the total length of liquid contained within the U-shaped pipe is ##L## and that the system is subject to a uniform gravitational field ##g##, then the differential equation that governs the oscillations is given by
    $$\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + \frac{2g}{L}x = 0.$$
    This means that the period of oscillation is
    $$T = 2 \pi \sqrt{\frac{L}{2g}},$$
    which means that the longer the fluid is, that is, the longer the bottom part of the tube is, the period of oscillation increases proportionally to ##\sqrt{L}##. You can see then that for longer tubes, the motion becomes slower and slower, and for very long tubes, the motion of the fluid will be really slow. In the limit of an infinitely long tube, the period also goes to infinity, so there's no energy conservation issues.

    If you wish to see the derivation of the above differential equation of motion, have a look at this page.

    It's in Spanish but I think the main ideas can be understood.

    Thinking out loud now, this derivation is fine and dandy for an ideal inviscid fluid... however, at very slow speeds, I'm not sure that kind of approximation would be valid, as it would be creeping flow and the viscous terms are no longer negligible in the Navier-Stokes equation. But this first argument should make you see why energy conservation doesn't get violated.
    Last edited: Mar 18, 2017
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