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More Thoughts on lift generation

  1. Aug 31, 2005 #1
    I recently read an article where the author stated that the momentum transfered to the air by a passing wing will eventually strike the earth, even from a high cruise altitude (in other words, out of ground effect)

    I have a problem with this for the following reasons, feedback would be appreciated.

    There is only two ways that a fluid can impose forces on an object immersed in it:

    1) Normal Forces (Pressure)
    2) Tangential Forces (Friction)

    From the standpoint of 2D potential flow:
    Let's say we have a fluid element floating above a point on the earth. Relative to that point on the earth, the total pressure is just the static pressure of the air, made up of the density and the translational (internal) kinetic energy of the molecules.

    However, approaching the fluid element from the right (the direction really does not matter) is an airplane with an airspeed of 100 Knots. Now, relative to that airplane the fluid element has a total pressure of static pressure + dynamic pressure. So, even before the airplane makes contact with the fluid element, it has effectively already done "work" on it, as the total pressure relative to the airplane has increased.

    When the fluid element comes in to the flow field of the wing (does not have to contact the wing, can be some distance (finite) above or below) it will go thru various velocity and "static" pressure changes. But, the mechanical energy of the flow remains constant, all accelerating and decelerating come about as a result of "internal" forces (no external forces increase or decrease the total mechanical energy- we are talking 2D potential flow).

    After the wing leaves, and the fluid element exits the flow field, it will have to readjust to the freestream static pressure and direction. The fluid element will have been displaced from its initial position, downward (down wash) but it caint keep travelling in the downward direction after it leaves the flow field of the wing, as this would represent a transfer of energy to the fluid, which would mean the energy came from the airplane, which means the airplane would expierence a "drag" which 2D potential theory (D'Alembert's Paradox) says does not exist.

    Now in 3D potential flow, energy will be imparted to the fluid (even without friction) as the wings will be a finite length, and the resluting wing tip vortices will repesent a transfer of energy to the fluid, which will be felt by the airplane as a drag force. These wingtip vortices distort the pressure distribution around the wing so that the pressure forces in the drag direction no longer cancel (D'Alembert's Paradox no longer valid).

    Now, in real life the vortices will eventually be dissipated due to viscosity. I can see sound waves, which are weak pressure waves reaching the ground, but sound waves are not a mass movement of air.

    feedback would be appreciated
    Last edited: Aug 31, 2005
  2. jcsd
  3. Aug 31, 2005 #2
    I always forget to include something

    From Newtons third law, the bending of the streamlines back up to the freestream direction and velocity, should reduce the lift slightly...and it does. This can be seen from a pressure distribution stand point also. As long as the fluid (streamlines) dont return to there original start position (there has to be a down wash) there will still be a positive lift.

    In potential flow, there will be a stagnantion point at the trailing edge of the airfoil. However, on the last 10% to 15% of the airfoil, on the top surface, there will be a positive pressure (higher than freestream). This positive pressure on an aft facing surface will decrease the drag, but will also decrease the lift slightly, so this holds from a pressure distribution standpoint and Newton's 3rd law standpoint.

    In addition, it usually takes 2 or 3 chord lengths past the trailing edge for the flow to return to the freestream static pressure and direction.

    (yes, I know in potential flow without friction, how is the stagnation point going to move to the rear, however, remember potential flow and D'Alembert's Paradox are mathematical solutions.)
    Last edited: Aug 31, 2005
  4. Sep 25, 2005 #3
    I re-read the article a few more times. I think the author is talking about a momentum flow reaching the earth..........now this makes more sense.
  5. Sep 25, 2005 #4
    Due to the inertia of air/vapor molecules and their elastic structure, I think the momentum would be dissapated and absorbed before reaching earth ffom say several miles high. Most of the momentum would be conserved as angular momentum in the constiuent air particles.

    Just my humble opinion, if you please.
  6. Sep 27, 2005 #5
    Well, you really cain't dissipate momentum (you can dissipate energy)

    However, if you are talking about the collision of two molecules with different masses, then I may see your point. However, wouldn't that just slow down the momentum flow?

    Of course, if you were to have a glancing collision I can see where some of the momentum may be conserved as angular momentum ( in diatomic molecules at least)
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