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Divergence of velocity of incompressible fluid in uniform gravity

  1. Feb 25, 2013 #1

    The velocity field as a function of poisition of an incompressible fluid in a uniform acceleration field, such as a waterfall accelerated by gravity can be found as follows:
    The position is [itex]\vec{x}[/itex].
    The velocity field is [itex] \vec{v} = \frac{d\vec{x}}{dt}[/itex].
    The constant acceleration field is [itex]\vec{a}=\frac{d\vec{v}}{dt} =\frac{d\vec{v}}{dt}\frac{d\vec{x}}{d\vec{x}}= \frac{d\vec{v}}{d\vec{x}}\frac{d\vec{x}}{dt} = \frac{d\vec{v}}{d\vec{x}}\vec{v}[/itex].
    Now we can find the velocity as a function of position by rearranging the above and integrating:
    [itex]\int\vec{a}\cdot d\vec{x} = \int\vec{v}\cdot d\vec{v}[/itex]

    [itex]\vec{a}\cdot \vec{x} = \frac{\vec{v}^2}{2}[/itex]

    [itex]\vec{v} = (2\vec{a}\vec{x})^{\frac{1}{2}}[/itex]

    The divergence of the velocity field is then

    [itex]\vec\nabla\cdot\vec{v} = \vec\nabla\cdot(2\vec{a}\vec{x})^{\frac{1}{2}} = (2\vec{a})^{\frac{1}{2}}\vec\nabla\cdot \frac{\vec{x}^{-\frac{1}{2}}}{2} =\sqrt{\frac{\vec{a}}{2\vec{x}}}[/itex]

    But shouldn't [itex]\vec\nabla\cdot\vec{v} = 0[/itex] in an incompressible fluid? The stramlines are all parallel to one another, as they follow the gravitational field, so they shouldn't diverge.

    Where has my thinking gone wrong?
  2. jcsd
  3. Feb 25, 2013 #2


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    You can't take the assumptions behind the ideal flow too literally.

    You're assuming that that the field has continuity (it's "full of water") and also that it's non divergent. You're also assuming the acceleration due to, e.g., gravity is the only acceleration.

    Eventually you will have to pick. Either the water has to move together horizontally to maintain continuity by some unspecified mechanism (such as self-attraction, which would cause an acceleration), so that the stream gets narrower but faster as it falls, or else you have to give up continuity and allow the stream to break up into pieces or droplets with spaces in between.
  4. Feb 25, 2013 #3
    olivermsun alluded to the answer to your question. As the water falls and the x-velocity increases with x, the streamlines get closer together, and the sheet of water gets thinner. There will be a component of velocity in the thickness direction. Check out the "curtain coating" literature.

    In the process of casting polymer sheets and in man-made fiber spinning operations, we encounter this type of deformation routinely. As the parcels of polymer stretch axially, they get thinner in the direction normal to the axis. For example, in fiber spinning, you also have an inward radial component of velocity as the polymer stretches axially, in order to satisfy continuity.
  5. Feb 26, 2013 #4
    Thank you for your answers, everybody!

    olivermsun, Chestermiller, would the thinning of the sheet also occur for an inviscid fluid, such as superfluid liquid helium?
  6. Feb 26, 2013 #5
    Even superfluid liquid helium has viscosity, so yes it would happen. But you would also have to do a hydrodynamic stability analysis on the curtain of fluid (including surface tension effects, of course) to determine under what conditions the sheet breaks up. In curtain coating operations, surfactants are added to the fluid to inhibit breakup.

  7. Feb 26, 2013 #6


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    For an incompressible flow, the continuity equation is just a statement of mass conservation. It doesn't really depend on there being viscosity (or not).

    There are lots of other examples of "inviscid" flows that depend on deforming and stretching. For example, you can think of surface water waves as being a result of the local thinning/thickening of a sheet of water as it flows horizontally away/toward any given point.

    As Chestermiller points out, surface tension will become important once you care about the details of whether the stream stays together as it thins or whether it breaks up into smaller droplets (thereby breaking continuity).
    Last edited: Feb 26, 2013
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