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Potential flow, inviscid flow, incompressible flow

  1. Nov 24, 2015 #1
    I have a couple of questions regarding several key areas of fluid mechanics:
    1. My first question deals with the Navier-Stokes equations. Does incompressible&irrotational flow imply inviscid flow? My answer is yes and here is my thought process.
    In the incompressible form of the Navier-stokes equation, the viscous term is (assuming constant transport properties):
    ##\nu \nabla^2\vec{V}##

    Using some vector calculus identities, this becomes:
    ##\nu [\nabla(\nabla\cdot\vec{V}) - \nabla \times \vec{\omega}]##
    The first time of this equation (divergence of velocity) is zero because we have assumed incompressible flow. The second term is zero because we have assumed irrotational flow. Thus, the viscous term is equal to zero, and therefore irrotational&incompressible flow implies inviscid flow.


    2) When comparing potential and viscous flow for, say, across a sphere or cross flow over a cylinder:
    In the case of potential flow, the streamline upstream of the frontal stagnation point will actually reach the stagnation point. In the case of viscous flow, the streamline upstream of the frontal stagnation point will only reach some finite distance away from this stagnation point and then will be deflected close to the body where viscous effects are present. Am I understanding this correctly?
     
  2. jcsd
  3. Nov 25, 2015 #2
    This is an incorrect mathematical assessment. If the viscosity is zero, it doesn't matter what either of those terms is equal to.

    No. In viscous flow, the streamline goes right to the surface also.

    Chet
     
  4. Nov 25, 2015 #3

    boneh3ad

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    I guess I will take the side of @pyroknife here. The vector Laplacian of the velocity is defined as ##\nabla^2\vec{v} = \nabla(\nabla \cdot \vec{v}) - \nabla \times (\nabla \times \vec{v})##, so the viscous stress term in the Navier-Stokes equations is, as he said,
    [tex]\mu\left[ \nabla(\nabla \cdot \vec{v}) - \nabla \times (\nabla \times \vec{v}) \right].[/tex]
    If the flow is incompressible, then ##\nabla \cdot \vec{v} \equiv 0##. If the flow is known to be irrotational, then ##\nabla \times \vec{v} \equiv 0##. That would mean that the entire viscous stress term is zero and the flow can be treated as being inviscid in those regions. I am not 100% sure where you are seeing the error here, @Chestermiller. Did I miss something?
     
  5. Nov 25, 2015 #4
    The words "inviscid flow" means that the viscosity is zero. Now an incompressible irrotational flow my behave like an inviscid flow, but the viscosity of the fluid is not necessarily zero. I guess I'm just quibbling with terminology.

    Chet
     
  6. Nov 25, 2015 #5

    boneh3ad

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    But it doesn't mean that. It means that the effects of viscosity on the flow field are zero (or negligible). It sort of like how airfoils can be analyzed using the tenets of inviscid flow even though the air certainly still has a finite viscosity. The effects of viscosity are confined only to the boundary layer. As long as that small region is treated as viscous or the shape of the airfoil is altered to simulate that effect (based on the displacement thickness), the entire rest of the flow field can be calculated with inviscid theory.

    I suppose I view it a lot like I view the term "incompressible flow". If you have a flow with zero divergence, then you call the flow incompressible, but that doesn't mean that the compressibility of the fluid is zero. You can have incompressible flow of air and yet the air is still highly compressible.
     
  7. Nov 25, 2015 #6
    Sure, of course I know all that. Maybe I'm just not up on the terminology.

    Speaking of airfoils (now that you brought it up), .........., what's the latest? Is it a "go," or is the mission scrubbed?

    Chet
     
  8. Nov 25, 2015 #7

    boneh3ad

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    It's "in progress." Just got back from an APS meeting and have a few days off for the holiday so hopefully that should be more than enough free time.
     
  9. Nov 25, 2015 #8
    In either cases, I thought theoretically the streamline can never actually reach the stagnation point?
    Using potential flow theory, I did a calculation where I followed a fluid element along its trajectory and found that the time it would take to reach the frontal stagnation point is infinity, implying that it can never reach the stagnation point?

    In viscous flow, such as Stoke's creeping flow, I thought all the streamlines are deflected and thus, no streamlines can reach the frontal stagnation?
    http://ocw.mit.edu/courses/earth-at...-structures-fall-2006/course-textbook/ch3.pdf
    The way the streamlines in Figure 3-1 is drawn gives me this impression.
     
  10. Nov 25, 2015 #9

    boneh3ad

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    The streamlines in that figure certainly aren't all of the streamlines. There are infinitely many and exactly one of them will impact the stagnation point. Now, in any flow, you are correct, a point moving on the stagnation streamline will take infinite time to reach the stagnation point. Of course a point is infinitely small and therefore does not adhere to the continuum approximation and streamlines don't make sense at that level anyway. In essence, a streamline is a theoretical construct. A single molecule moving along the stagnation streamline, for example, will not actually perfectly follow that and would be knocked off of the stagnation path by its Brownian motion before it came to rest anyway.
     
  11. Nov 25, 2015 #10
    Oh I see. I typically only see the stagnation streamline (not sure if this is the correct terminology, but you know what I mean) drawn for potential flow, and not for others. Do you know why this is the case?
     
  12. Nov 25, 2015 #11
    At the continuum level, a streamline is a trajectory over which the stream function is constant. If you look up the solution for the Stokes flow stream function (e.g. in Bird, Stewart, and Lightfoot, Transport Phenomena), you will find that the stream function has the same value over the the entire trajectory leading up to the forward stagnation point, over the entire sphere, and over the continuation of the trajectory departing from the rear stagnation point. And there are streamlines infinitely close to this streamline. If you check the stream function for flow over a cylinder (either potential flow or slow viscous flow), you will find the same thing. And if you check the stream function for inviscid flow past a sphere, you will find the same thing.

    Chet
     
  13. Nov 25, 2015 #12
    Usually, more than just the stagnation streamline are drawn.

    Chet
     
  14. Nov 25, 2015 #13
    Yes, what I mean is, when I see streamlines drawn for potential flow, I always see the stagnation streamline drawn, whereas for viscous flow, I never see the stagnation streamline drawn. I was wondering if there was any significance in this.

    I mean in potential inviscid flow, I always see the stagnation streamline included in the drawing, whereas for viscous flow, I have not yet seen this stagnation streamline drawn. I wasn't sure if there was any significance to why it is drawn in the inviscid case, but not in the viscous.
    I actually don't know very much about continuum flow. Most of what I do is rarefied. I'm a UM student btw.
     
  15. Nov 25, 2015 #14
    http://farside.ph.utexas.edu/teaching/336L/Fluidhtml/node116.html
    GO BLUE. Beat Duh Ohio State University.
     
  16. Nov 25, 2015 #15

    boneh3ad

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    Rarefied? More like BORE-ified, amirite!?!?
     
  17. Nov 25, 2015 #16
    I do a lot of high temperature reentry simulations. I actually enjoy it, though the rarefied part doesn't seem to have much applicability outside of academia/NASA.

    Do you use the Crocco-busemann compressible boundary layer in any of your work?
     
  18. Nov 25, 2015 #17
  19. Nov 26, 2015 #18

    boneh3ad

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    Nope. I never had to deal with it as an experimentalist. I measured the wall temperature at any given time anyway.
     
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