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Navier-stokes control volume

  1. Jan 26, 2007 #1
    navier-stokes smoothness problem almost solved

    Penny Smith has made progress with showing that smooth conditions exist for all time in a domain for the Navier Stokes equations

    However a flaw was found in the mathematics - hopefully it can be sorted out soon.There's still the problem though of finding actual solutions to the equations!

    Thanks for the info on control volume.I was just wondering if in reality there
    is a real,if very small size,to the differential element.
    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. Jan 26, 2007 #2


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    It's not a control volume but a differential element (infinitely small). Imagine a cube with sides that measure, in cartesian coordinates [tex]\delta x[/tex], [tex]\delta y[/tex] and [tex]\delta z[/tex].
  4. Jan 26, 2007 #3


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    Actually Fred, the NS equations have 4 forms:

    Lagrangian (moving frame of reference)
    Eulerian (stationary)
    ..and then...

    So for example, looking simply at continuity (sorry again for my lack of latex):
    Differential Lagrangian:
    Dp/Dt + rho*del•V =0
    Where D/Dt is the substantial derivative with respect to time

    Differential Eulerian:
    dp/dt + del•(rho*V) = 0
    (note this form is now strongly conservative as all variables are inside of a derivative)

    Integral Lagrangian:
    D/Dt [Volume Integral] rho dV = 0

    Integral Eulerian:
    d/dt [Volume Integral] rho dV + [surface integral] rho*V dS = 0

    The entire equations can be derived any of the four ways. It's easiest (at least for me) to remember one form, and then how to go from one form to another.

    To answer the question, the control volume for an integral approach is simply V, there is no need to know anything else besides that.
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