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Navier Stokes EQs: Some Insight Please

  1. Feb 23, 2013 #1
    I have been doing some serious review of fluids in order to prep for some CFD. I have been re-deriving the NS Equations in all of their various forms. Something seems to have cropped up that I have worked myself in circles about. Let's take the momentum equation in Conservative Integral form:

    $$\sum \vec{F} =\left(\frac{d\vec{M}}{dt}\right)_{system} \qquad (1)$$
    $$\sum \vec{F} =
    \frac{\partial{}}{\partial{t}}\iiint_{\mathcal{V}}(\rho\vec{V})d\mathcal{V} +
    \iint_S(\rho\vec{V})\vec{V}\cdot d\vec{s} \qquad (2)$$

    The 3rd term in the equation (2) for a control volume is the net outflow of momentum from the control volume. What I am a little confused about is the sign convention here. The left-hand side of the equation takes forces acting along the positive coordinate axes as being positive. But the right-hand side takes outflow to be positive regardless of the coordinate axes.

    So it seems that an increase in force acting in the positive coordinate direction is equated to an increase of outflow of momentum from the CV regardless of what direction it flows out from.

    I hope that maybe someone can see what I am getting at. I think that my confusion might be stemming from what happens when we go from a system approach to a control volume approach (i.e. application of the Reynold's Transport Theorem (RTT)).

    Any thoughts are appreciated.
     
    Last edited: Feb 23, 2013
  2. jcsd
  3. Feb 23, 2013 #2

    AlephZero

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    Your notation of ##(\rho \vec V)\vec V.ds## is probably confusing you. The "##ds##" must be a vector quantity for ##\vec V.ds## to make any sense.

    If you write it as ##(\rho \vec V)(\vec V.\vec n)dA##, where ##\vec n## is the outwards normal of ##S## and ##dA## is the scalar area, it should be clear that the sign changes if ##\vec V## is flowing into or out of the volume.
     
  4. Feb 23, 2013 #3
    Hi AlephZero :smile: I fixed the typo, but this is not my source of confusion. I understand that inflow is negative and outflow is positive. What I don't understand is why...
     
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