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Velocity of fluid at rear of airfoil

  1. Nov 29, 2004 #1
    Hello, we've just watched a few movies about drag, reynold's number, etc., as part of our fluid dynamics course. In one of the movies, they claimed that given a laminar flow around an arifoil, the velocity of the fluid at both ends of the airfoil is zero. Now, I understand why the velocity is zero at the front of the foil (at the stagnation point), but why is it also zero at the rear end?
     
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  3. Nov 30, 2004 #2

    Clausius2

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    I'm not sure of what I'm saying. In laminar flow (I mean potential flow) there is a paradox called D'Alembert Paradox. That states all symmetric aerodynamic body has no drag inmersed in a fluid stream. Why? The flow is deccelerated until the body nose, where v=0 and the static pressure P=Pmax. As the flow goes over the body shoulders it losses static pressure at the same time gaining velocity. When the flow reaches the rear part, the velocity becomes zero, and the pressure is P=Pmax because in potential flow there are not any losses of pressure (isentropic flow). So that, the integral of the pressure over the body (drag coefficient) is zero.

    Why is v=0 at the rear?. Think of it. There are two stream lines (one travelling at the higher part and another travelling at the lower part of the body). When both streamlines crash into each other at the rear part, the velocity must be zero at that point. That's because of the definition of streamline. A streamline is a line tangential to the velocity vector. If two streamlines are defined just at one space point, the velocity has to be zero at that point.

    Of course, it only happens at potential flow. Turbulent regimes are quite different. Surely it is formed a vortex at the rear due to the Kelvin's Theorem.

    What do you think of this?
     
  4. Nov 30, 2004 #3
    Yes, this is what was shown in that movie (among other things). Although they didn't use any integrals, and didn't mention the term "potential flow".
    Ok, that makes sense, I guess. But that's a purely mathematical explanation. What is the physical reason for the velocity to drop back to zero? I mean you've got these fluid particles coming in from the top and from the bottom, but they're coming in at an angle. I can see why their vertical velocities may cancel out, but why do the horizontal also zero out? Or, looking from a slightly different angle, why does the pressure rise back to Pmax? For that matter, why does it rise at all?? (Don't tell me it rises because the volcity drops - it's the other way around if I'm not mistaken).
    Yes, turbulent flow is quite a different story. There you get backflow, boundary layer separation, and all sorts of other mess...
     
  5. Nov 30, 2004 #4

    russ_watters

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    Think of it this way - unless, the trailing edge of an airfoil is perfectly sharp, there will be room behind it for air molecules to sit undistrubed by the straight flow of air above and below.
     
  6. Nov 30, 2004 #5
    Are you implying that if the rear end was sharp to within atomical precision, the velocity at the rear end would not be zero? How does that sit with the mathematical definition of a streamline (see Clausius2's post)?
     
  7. Dec 1, 2004 #6

    Clausius2

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    It is consistent with russ_watter said. If the airfoil is perfectly sharp, both ends of the rear are infinitesimally tangential, so that when both streamlines come one into other they do not cross, but they collide tangentially.

    In my opinion, mathematics and reality are the same thing at Fluid Mechanics. I have given to you a mathematical explanation, well, if you take it carefully you will see it has a physical explanation. The velocity drops to zero because fluid particles are braked because of the positive pressure gradient. If the gradient is not enough to reach the leading stagnation pressure (there are pressure drops due to irreversibilities) then you won't recover the rear stagnation point, and v is not 0 at the rear.
     
  8. Dec 1, 2004 #7

    russ_watters

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    Actually, I left out that second part: infinitessimally tangential. You got what I meant though. In practice, actually getting the streamlines parallel at the trailing edge would require an infinitely long (chord) wing, with a symmetric/hyperbolic cross-section toward the trailing edge.
     
    Last edited: Dec 1, 2004
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