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Confirming fact about potential energy of each particle in a fluid

  1. Nov 24, 2009 #1
    I just want to confirm this fact -

    Referring to a point in a cylinder on earth, filled with a fluid; the potential energy of each particle in that cylinder is identical.

    Reason being the weight experienced by each particle increases as we goto more depths.
    Last edited: Nov 24, 2009
  2. jcsd
  3. Nov 24, 2009 #2
    The gravitational potential increases with height. Am I missing something?
  4. Nov 24, 2009 #3


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    What do you mean by the "weight experienced by each particle"? The weight of the particles above it? That has nothing to do with potential energy.
  5. Nov 24, 2009 #4
    So I think I'm wrong about this.

    There's lots of pressure on the lower part of the cylinder so each particle should contain lots of potential energy due to the pressure.

    So potential energy of the particle decreases as we move towards the bottom part of the cylinder?

    If I'm wrong in the main question I think this will be it, but this will again generate a series of doubts...
  6. Nov 24, 2009 #5
    There is no potential energy due to pressure. Pressure is a thermodynamic variable, not a force, and you can only define potentials in terms of a conservative force field (for example, gravity gives you the gravitational potential).

    The potential energy does decrease, but only because it is getting closer to the Earth.
  7. Nov 25, 2009 #6
    If the potential energy does decrease, then the velocity of the particle which's x cm (perpendicular) from the orifice should be less than the particle at a height x + a (where a is a reasonable positive integer) since only the potential energy of the particle gets converted to kinetic (velocity).

    But this does not happen...in ideal conditions the velocity of all particles are the same.
  8. Nov 25, 2009 #7
    The relation between potential energy and velocity is applicable for a single particle moving under only the interaction of the field responsible for that potential energy, so PE + KE = constant in this case.

    In the case of many particles in a fluid the kinetic energy of a particle (therefore velocity) will be changed many times a second by collisions, essentially interactions with a different field (electric), to which the gravitational potential is independent, so you can no longer say KE + PE = constant because you have a second potential field due to electrostatic repulsion.

    The gravitational potential is still purely a function of distance from the centre of mass of the Earth, and does not couple to the electric potential which dominates particle velocity distributions in a fluid, especially in hydrostatic equilibrium where there is no net movement up or down of the particles due to gravity.

    This is what I think, anyway, there may well be a better explanation out there.
  9. Nov 25, 2009 #8


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    Pressure energy = pressure x volume. Gravitational potential energy = gravitational potential x mass. If the volume and mass for each particle of fluid is known, you can multiply Bernoulli equation by mass where the total energy per particle in an ideal fluid (incompressable) is constant regardless of height. If the fluid is compressable, then the relationship between volume and mass changes and Bernoulli equation has to be modified to use an integral form for the pressure term.

    Bernoulli equation for ideal fluid: pressure/density + g h + 1/2 v2 = constant

    (pressure x volume) + m g h + 1/2 m v2 = total mechanical energy = constant

    Assuming v = 0 you get:

    (pressure x volume) + m g h = constant

    Last edited: Nov 25, 2009
  10. Nov 27, 2009 #9
    Consider only a situation where the fluid is under influence of a field which is not a function of distance (can be considered sorts of equal to the situation on earth where the height of the apparatus is not reaching the sky...).

    If orifice is opened, the particles at the lower part of the cylinder will initially accelerate more in comparative to the particles above them (cause the weight of the parciles above will act on the particles below), then the acceleration should reduce to a constant since G will pull each particle in an identical way...this is what I think will happen.

    All this should result in all particles gaining a constant velocity regardless of their initial position in the cylinder...and we practically see this.

    Am I right about this in the assumed field?

    Yes, exactly what I was saying.
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