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Why is flow unsymmetric over a delta wing?

  1. Jan 13, 2016 #1
    I understand that there are vortices formed on the delta wing for moderate angles and low speeds. I observed recently 2 asymmetric vortices forming on a small delta wing model for a moderate angle and a low speed. But, why would there be asymmetriic flow over a delta wing? If the flow is approaching the wing symmetrically and the wing itself is symmetric about the same plane, the flow should be symmetric, right?

    And ofcourse, i'm talking about the plane at right angles to the wing and containing the root chord in it.
     
  2. jcsd
  3. Jan 13, 2016 #2
    But it isn't due to the angle of attack.
    But it isn't - even if the camber is symmetric (which it isn't usually) the control surfaces create asymmetry.
     
  4. Jan 13, 2016 #3
    The case i'm talking about, there are no control surfaces, just a triangular slender sharp edged delta wing. The model which i tested in is also very small and simple, and its basically a triangular prism which tapers at the leading edge.

    WHen it comes to angle of attack, i'm talking about symmetry ACROSS the plane PERPENDICULAR to the wing and CROSSING the root chord.
    You are talking about symmetry across the plane in which the delta wing itself lies, and i'm not talking about that plane.
     
  5. Jan 14, 2016 #4
    Which?

    If you are saying that the vortices off each tip of the wing pair are not symmetrical then this may be down to swirl in the wind tunnel flow: how sophisticated is the wind tunnel?
     
  6. Jan 14, 2016 #5

    boneh3ad

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    Yeah my first impression here is that this is experimental error either due to the measurement technique, the mode, or the tunnel. For a delta wing that is, in fact, symmetric at zero yaw in a uniform free stream, the two sides should be symmetric.
     
  7. Jan 14, 2016 #6
    Alright, thank you for your clarifications. So i guess either the wind tunnel had some asymettricities in it or our delta wing might've been placed at some yaw. My doubts have been cleared.
     
  8. Jan 14, 2016 #7

    boneh3ad

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    What sort of wind tunnel are you using? Is it homemade? Is it one used for research at a university? What is its configuration?
     
  9. Jan 14, 2016 #8

    David Lewis

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    If the model is light enough, and the CG is in the right place, you can test glide it to discover flaws in the model and trim out asymmetries. At very low Reynolds number, the delta wing's boundary layer is chaotic, unpredictable and sensitive to minor deviations in surface contour or texture.
     
  10. Jan 16, 2016 #9
    Its one of the smallest wind tunnels in the institute, open suction and max 15 m/s. it's got a small test section cross section, like 1 ft squared or something, and its got a 9 contraction ratio.

    So, my professor said that its because the navier sokes equation isn't symmetric for high Re flows, now im confused again :(
     
  11. Jan 19, 2016 #10

    boneh3ad

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    Well for one, given the tunnel parameters, this is nowhere near a situation that could be considered "high Re." At room temperature the unit are is only in the ballpark of ##10^6\;\mathrm{m}^{-1}##. Even a model that's a long as your test section is wide would only have ##Re_c \approx 3\times 10^5##, which is quite small. It almost certainly wouldn't even be turbulent.

    Second, there's nothing about the Navier-Stokes equations that says they would become asymmetric in a symmetric domain. It honestly just sounds like your professor doesn't know the answer and is making up reasons.

    What are the dimensions of your model? Is it mounted in the center of your tunnel or off-center?
     
    Last edited: Jan 19, 2016
  12. Jan 19, 2016 #11

    cjl

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    I agree with boneh3ad here - there is nothing about the Navier-Stokes equations that would indicate a symmetric model at zero yaw or roll would have any asymmetries. It is likely a problem with the experimental setup - what's your model size, and how is it mounted in the tunnel?
     
  13. Jan 19, 2016 #12

    FactChecker

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    It's very possible that the right and left flows can not coexist symmetrically and one will dominate or be different from the other. They may alternate in their behavior. The situation of symmetry would require that the flow is statically stable. It is similar to the situation when a flag is being blown on. It doesn't go straight back. It flaps.
     
  14. Jan 19, 2016 #13

    boneh3ad

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    The flag example isn't really relevant to your point. In fact, the flapping of a flag (or any similar phenomenon such as von Kármán vortex shedding) may not appear symmetric when time-resolved, but they actually are statistically symmetric. The flow structures on each side will develop identically and at the same rate; they will just alternate back and forth. The net result is statistical symmetry.

    In other words, even if there was some shedding/alternating type phenomenon occurring, it would almost certainly still be statistically symmetric. Of course, it's highly unlikely that is happening anyway. There is nothing on the OPs geometry that should give rise to that.

    The more likely culprits are things like non-uniformity in the free stream flow, a blockage effect due to the model being too large, upstream effects of the mounting system, or other such systemic errors.
     
  15. Jan 20, 2016 #14

    FactChecker

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    Are you saying that the asymmetric behavior must be cyclic and "statistically symmetric"? I don't see why.
     
  16. Jan 20, 2016 #15

    boneh3ad

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    Because the domain is symmetric, and therefore so are the forces on the flow. If it overshoots one way it will tend to move its way back to symmetry again. Just look at the vortex shedding behind a cylinder. This is all assuming that the domain and incoming flow are symmetric. If not then there's no reason to expect the flow to be symmetric.
     
  17. Jan 20, 2016 #16
    The test section dimensions:
    16x12x39 inches, 39 is the length of section.
    The dimensions of delta wing:
    Root Chord: 220 mm, Span 160 mm..
    At a velocity of 4 m/s and AOA of 27 degrees, we observed one vortex bursting a bit upstream of mid chord and the other vortex bursting at a point just before TE.
     
  18. Jan 20, 2016 #17

    boneh3ad

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    Ah, so the issue is the vortices bursting at different points then. Phenomena such as this, especially in low-Re (and likely laminar) situations, are highly-dependent on initial conditions. For example, minute differences in surface roughness or blemishes can play a major role in the development of the vortices. It seems rather likely then that the character of the surface on either side of your model is probably a bit different, so while the macroscopic shape of your model is symmetric, the surface characteristics are not, and that is causing your vortices to start with slightly different conditions and burst at different locations.
     
  19. Jan 20, 2016 #18

    FactChecker

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    That is a statement about static stability that I question. There are certainly planes that do not have lateral stability if there is no tail, but that is much more complicated. I don't know about stable airflow in the wind tunnel. I am not a stability and controls expert, so I will withdraw.
     
  20. Jan 20, 2016 #19

    boneh3ad

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    Well my point was that even if it was dynamically unstable, since the domain is symmetric, the restorative force would tend to bring the flow back toward the symmetric state. It may very well overshoot and repeat the process over and over again, but it is still going to oscillate about some basic, average state. That mean state would be symmetric. This is precisely the case in the example of the vortex shedding from a cylinder (or the flag). You have a perfectly symmetric mean flow plus some unsteady component on top of that. That unsteady component itself will have a symmetry to it, though, in that each shed vortex is the exact inverse of the one that passed before it.

    In essence, if you have any kind of phenomenon like the flag you cited, the culprit is vortex shedding. The process is unsteady but still inherently symmetric so long as the domain is symmetric (the body shape and the incoming flow). If you isolated the domain, the mean flow would look identical on both sides (mirrored, of course) and if you looked at the unsteady flow, it would also look identical but one side would be out of phase with the other by π radians. That's still symmetric.

    On the other hand, the situation that @Algren describes is one where one side of the flow field looks fundamentally different than the other. Specifically, the vortex on one side bursts well upstream of the vortex on the other side. This is not an unsteady phenomenon, and is wholly unrelated to von Kármán vortex shedding. It is instead more of a stability problem (hydrodynamic stability, not aircraft stability) where the initial conditions on one side are certainly at least slightly different than the other, and in this case, the difference was large enough to lead to noticeable asymmetry. It is a very similar problem to that of boundary-layer transition on one wing of a plane compared to the other if you glue a sheet of sandpaper to the leading edge of one and polish the other. The plane's shape as a whole is symmetric, but the transition location is not because of the minute (relative to the scale of the plane) differences introduced by the sandpaper.
     
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