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Navier stokes question

  1. Feb 3, 2014 #1

    joshmccraney

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    hey pf!

    so when deriving navier stokes we have, from newton's second law, [itex]\sum \vec{F} = m\frac{d \vec{V}}{dt}[/itex] when deriving the full navier stokes (constant density) the acceleration term can be thought of as two pieces: a body change of velocity within the control volume and a mass flow leaving the control volume through the a small surface element [itex]d\vec{S}[/itex]. my question is, when momentum leaves we have: [tex]\iint \rho \vec{V}\vec{V}\cdot d\vec{S}[/tex] and through the divergence theorem we have, as an velocity [itex]\vec{V}_x[/itex] component, [tex]\nabla \cdot
    (\rho V_x \vec{V})[/tex]

    can someone explain how? also, why is there no negative sign (momentum leaving, not entering)?

    thanks!
     
  2. jcsd
  3. Feb 3, 2014 #2
    If you are asking how the area integral translates into the divergence over the volume, you need to study the divergence theorem. If you are asking how the component in the x direction is obtained from the divergence of the momentum flux, just write it out in component form (using cartesian coordinates) and you will see.

    Regarding the integral of the momentum flux over the surface area, you are dotting the momentum flux tensor with the outwardly directed normal. If the velocity vector is directed inward through the surface, then the dot product of the velocity vector with the outwardly directed normal is negative. This accounts for your missing negative sign. The integral over the area includes both momentum coming in as well as momentum going out.
     
  4. Feb 3, 2014 #3

    joshmccraney

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    i understand the divergence theorem, but i don't understand it with a dyad (intuitively or computationally); can you explain it?

    also, shouldn't there be a "negative sign" in front of the integral? like you pointed out, momentum entering should be positive, and by the orientation of the normal surface, incoming momentum yields a negative.

    please help and thanks
     
  5. Feb 4, 2014 #4
    I'm not quite sure how to answer this question. My inclination would be to suggest trying to apply it in component form to a simple shape like a cube. Something like this will show how the math works out. There is no reason mathematically why the divergence theorem should apply just for vectors and not second order tensors.
    This is easier to answer. The rate of accumulation of momentum within the control volume is equal to the rate of momentum flow in minus the rate of momentum flow out. The latter is the same as the minus the net rate of momentum flow out. And this is the same as minus your integral, since your integral is plus the net rate of momentum flow out. So, minus your integral should go on the other side of the equation. When you bring it over to the same side of the equation as the rate of accumulation of momentum, the sign in front of the integral changes to positive.

    Chet
     
  6. Feb 9, 2014 #5

    joshmccraney

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    what if i was adding the net rate of momentum flow in? would anything change (other than the sign of this integral)? i mean, it seems something is wrong here (with my intuition). if we are talking about the sum of forces and rate of change of momentum, shouldn't defining change of momentum in or out change everything? in other words, how would the entire navier stokes equations change if we consider (as i believe we should) rate of change of momentum flow in?

    sorry to keep this post going, but obviously im still confused

    thanks chestermiller for your time!
     
  7. Feb 10, 2014 #6
    You are already automatically including the momentum flow in when you do your integral over the entire surface. The differential area vector dS has attached to it an outwardly directed normal from your control volume. So, if the normal component of the velocity vector at the surface is also outwardly directed, you have flow out, and the contribution to the integral is positive. But, if the normal component of the velocity vector at the surface is directed into the control volume, you have flow in, and the contribution to the integral is negative. If the entire integral turns out to be negative, then you have net flow in.
     
  8. Feb 10, 2014 #7

    joshmccraney

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    But isn't this counter intuitive? Shouldn't momentum in be considered positive?
     
  9. Feb 10, 2014 #8

    olivermsun

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    Momentum (mv) is a signed quantity. Think of your classic billiard-balls problem. Balls can end up with forward or backward momentum after the collision.
     
  10. Feb 10, 2014 #9

    BruceW

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    well, the equation is like this:
    [tex]\frac{\partial \rho \vec{v}}{\partial t} + \nabla \cdot ( \rho \vec{v} \otimes \vec{v} ) = \nabla \cdot \sigma + \vec{f} [/tex]
    (where ##\otimes## is the dyadic product, ##\sigma## is the Cauchy stress tensor and ##\vec{f}## is the body force). So, the change in momentum inside a volume due to flow of momentum will be
    [tex]- \iint \rho \vec{v}\vec{v}\cdot d\vec{S}[/tex]
    I hope this helps.

    edit: p.s. this is of course, when the surface of that volume is moving with the same velocity as the fluid. (otherwise, you get a dependence on the velocity of the surface also).

    edit again: no... wait, this is when the velocity of the surface is zero. I think. Anyway, the differential equation is always true, since it doesn't depend on any kind of volume of integration.
     
  11. Feb 10, 2014 #10

    BruceW

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    [tex]\iiint \frac{\partial \rho \vec{v}}{\partial t} dV + \iint \rho \vec{v}\vec{v}\cdot d\vec{S} = \iint \sigma \cdot d\vec{S} + \iiint \vec{f} dV [/tex]
    And due to Reynold's transport theorem, we have:
    [tex] \iiint \frac{\partial \rho \vec{v}}{\partial t} dV = \frac{d}{dt} ( \iiint \rho \vec{v} dV ) - \iint \rho \vec{v} \vec{v}_s \cdot d\vec{S} [/tex]
    (where ##\vec{v}_s## is the velocity of the surface). And therefore:
    [tex]\frac{d}{dt} ( \iiint \rho \vec{v} dV ) = \iint \rho \vec{v} (\vec{v}_s-\vec{v}) \cdot d\vec{S} + \iint \sigma \cdot d\vec{S} + \iiint \vec{f} dV [/tex]
    (where the left-hand side is simply the total change of momentum inside the volume).
     
  12. Feb 10, 2014 #11
    You could set it up the other way by using an inward directed normal to the control volume surface, but then you would need to change the sign in the divergence theorem. By convention, the normal is outwardly directed. Sorry if that doesn't satisfy your intuition, but that's the way that it's been done for a long time. Have you learned about the Cauchy stress relationship yet? That also uses an outwardly directed normal.
     
  13. Feb 10, 2014 #12

    joshmccraney

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    i have learned about the cauchy stress tensor (if that's what you mean) and i am comfortable with normal directed outward. what is confusing me is, if we have [itex]\sum \vec{F}=m \vec{a}[/itex] shouldn't momentum change be on the R.H.S and shouldn't momentum added to the control volume be positive? thanks a ton for you patience!
     
  14. Feb 10, 2014 #13

    joshmccraney

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    im considering the momentum flux, and if [itex]\vec{V} \cdot d\vec{S}[/itex] is positive, this means momentum is leaving (as you have already stated. but wouldn't we put a "negative" sign up, since momentum leaving is negative?
     
  15. Feb 10, 2014 #14

    BruceW

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    it depends what side of the equation you want to put it on. For a volume with static boundary, and no body forces or stresses, the equation is:
    [tex]\frac{d\vec{P}}{dt} + \iint \rho \vec{v}\vec{v}\cdot d\vec{S} = 0[/tex]
    Or, taking it over to the other side:
    [tex]\frac{d\vec{P}}{dt} = - \iint \rho \vec{v}\vec{v}\cdot d\vec{S}[/tex]
     
  16. Feb 10, 2014 #15
    Yes. If net momentum is leaving, then, as you said, there should be a negative sign in front of it on the right hand side of the momentum balance equation (when the rate of change of momentum within the control volume is on the left hand side of the equation). But, when you bring the momentum flux integral over to the left hand side of the equation, it changes sign to positive.
     
  17. Feb 11, 2014 #16

    joshmccraney

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    doesn't this seem weird? i mean, if we set it up as you have said, we have [itex]m \frac{d \vec{V}}{dt}=\sum \vec{F}[/itex] then shouldn't the [itex]-\iint \rho\vec{V}\vec{V}\cdot d\vec{S}[/itex] term be derived on the [itex]m \frac{d \vec{V}}{dt}[/itex] side, thus giving us [tex]\frac{\partial}{\partial t}\iiint \rho \vec{V} dv -\iint \rho\vec{V}\vec{V}\cdot d\vec{S}=\sum \vec{F}[/tex] with the negative, showing that this is change of momentum, not a force.
     
  18. Feb 11, 2014 #17

    joshmccraney

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    i know what i wrote is incorrect, for the record. (not trying to cause waves here)
     
  19. Feb 11, 2014 #18

    BruceW

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    something Roger Penrose once said: "The best thing about physical intuition, is that it can be changed to fit the facts." (or roughly something like that). In this case, the real equation is:
    [tex]\frac{\partial}{\partial t} \iiint \rho \vec{v} dV + \iint \rho \vec{v} \vec{v} \cdot d\vec{S} = \sum \vec{F} [/tex]
    so try to fit your intuition around this equation.

    edit: also, the Forces are inside the volume, so maybe it is best to think of them as sources of momentum. So, for example if the first term on the left-hand-side is zero (i.e. the momentum inside the volume doesn't change), and if the forces inside the volume are radially outward, then this means there is a source of outwardly radial momentum. And therefore, on the surface of that volume, there must be fluid flowing outwards, since this radial momentum must go somewhere, and if the momentum inside the volume isn't changing, this would mean the radial momentum must be leaving the volume. In other words, momentum must go somewhere.
     
    Last edited: Feb 11, 2014
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