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Wing Lift Energy

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  1. Sep 15, 2015 #1
    I would much appreciate any critique to this line of thinking.
    Draw a circle around the airplane to see it as a free body diagram. The airplane engine provides horizontal force to move the plane. Force (F=ma) is a vector quantity with a horizontal direction. The wing lift is a force in the vertical direction. These two forces are 90 degrees apart. From an energy point of view, what is the energy source for the vertical lift force? Is this energy being extracted from the surrounding compressible gas? Theoretically from a thermodynamics view, is the atmosphere gas behind the airplane being cooled slightly? Possibly the example of a glider plane instead of a powered plane would be better.
     
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  3. Sep 15, 2015 #2
    This is not a homework question. I am familiar with the descriptions of what creates lift. However, this idea of the source of the energy for the lift force is what I am troubled by, that is the question. Thanks for any help or direction.
     
  4. Sep 16, 2015 #3

    rcgldr

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    The issue is that the drag component related to generating lift is being ignored. If drag is not ignored, then the direction of the net force on the air is not perpendicular to the force from the engine. Consider the case of an ideal wing: from the aircrafts frame of reference, the relative flow is diverted without any change in speed, only a change in direction. If the diverted flow is separated into downwards and backwards components, the speed of the backwards component is decreased (air is accelerated forwards), and the speed of the downwards component is increased from zero to the downwash speed (air is accelerated downwards). The induced drag is related to the forwards acceleration of air, and even absent all other forms of drag, the engine has to at least provide the force that opposes induced drag.

    From the air's frame of reference, as a glider descends at a steady rate, the air is accelerated from zero velocity to some non-zero velocity, mostly downwards and a bit forwards. There's a pressure jump from below ambient to above ambient in the air as it flows downwards through the plane swept out by the wings forward and slightly downwards movment. This generates an impulse that eventually reaches the ground, and the net downforce over time on the ground equals the total weight of the air and the weight of the glider. The temperature is increased due to the net increase in pressure during the pressure jump, friction between glider and air, and viscosity within the air.
     
    Last edited: Sep 16, 2015
  5. Sep 16, 2015 #4
    "The energy source for a force" is not a meaningful concept. You can have forces acting between object without any change in energy.
    You only "need" energy if work is done. And then what you call the "source" may depend on your point of view. Of course, for the plane the energy come from the fuel. Which actually comes from the sun, if you go a little back on the chain. If you want to know how the energy get transferred from the chemical energy of fuel into kinetic energy of the plane (and of the surrounding air). heat and whatever, you can go to various levels of detail.
     
  6. Sep 16, 2015 #5

    A.T.

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    In the rest frame of the airmass that force does no work, so it transfers no energy.
     
  7. Sep 16, 2015 #6

    rcgldr

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    From the frame of reference of the air, as a wing passes through a volume of air, the affected air which was initially at rest, is accelerated downwards and a bit forwards, resulting in a non-zero velocity when it's pressure returns to ambient (called exit velocity in the case of propellers). The air starts off with zero KE and ends up with non-zero KE (at the moment it's pressure returns to ambient), so work is done on the air.
     
  8. Sep 16, 2015 #7
    Yes, but the air is not accelerated by the force acting vertically on the wing.
    Of course you need some energy to maintain the plane in the air. As you need for a hovering helicopter.
    But this does not mean that spending some energy is a necessary condition for the existence of a force.
    This was discussed on the forum several times, I believe.
     
  9. Sep 16, 2015 #8

    rcgldr

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    There is a Newton third law pair of forces, the upwards force that the air exerts onto the wing, and the downwards force that the wing exerts onto theair. As for the downwards force that wing exerts onto the the air, then Newton's second law somewhat applies, force = mass x acceleration. The air's pressure is increased, and it accelerates downwards (some losses due to heat). This assumes that downwash isn't prevented by something like flying in ground effect mode.
     
  10. Sep 16, 2015 #9

    FactChecker

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    Consider the case of level, unaccelerated flight. The zero-work path of an airplane is it's glide path, which slopes downward. The engine supplies enough energy to counteract that downward slope and maintain level flight. So there is work being done in level flight. The shape and orientation of the wing causes the lift force required for the glide path and also for the level flight. There are several ways to calculate the lift force, but none are very accurate. If you could sum up all the acceleration of the air forced downward both directly and indirectly, that would give the correct answer. Bernoulli's formula can provide an estimate of the lift. It gets complicated very quickly. There are computer computational fluid dynamics (CFD) programs that track millions of small parts of air motion around an airplane. Even those are far from perfect.
     
  11. Sep 16, 2015 #10

    rcgldr

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    That's not a zero work path, the work done onto the air by an aircraft following a steady (non-accelerating) glide path corresponds to the change in gravitational potential energy. The decrease in gravitational potential energy ends up equal to the increased mechanical and thermal energy of the affected air.
     
  12. Sep 16, 2015 #11

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    CORRECTION: I stand corrected. The OP was specifically asking about gliders and the energy creating lift without an engine. The following comment did not address that question.
    Trade-offs between kinetic and potential energy is not work done by the engine. The glide path is the zero work path from the point of view of the engine and airplane components. The work requirements for the engine to maintain flight are usually what people want to know.
     
    Last edited: Sep 16, 2015
  13. Sep 16, 2015 #12

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    I stand corrected. The OP was not asking about the work done by the engine.
     
  14. Sep 16, 2015 #13
    Thanks so much for the concise response, especially rcgldr, the responses are very much appreciated, Thank You. I see that the engine is supplying the additional energy for the lift force. Makes perfect sense since a heavily loaded plane would require more fuel to fly the same distance. I suppose the wing attack angle would also need to change for the heavier payload.

    What sent me down this road thinking about the energy source, was a paper about thrust augmenting ejectors from Paul Bevilaqua, 1978. He makes the analogy of the thrust increase via the ejector shroud, versus wing lift and drag. He states that the ejector can be seen as a wing shape that has been wrapped around into a pipe wall shape.

    It seems to me this analogy leans heavily on Bernoulli, which is not exactly the whole story with regards to wing lift. It is entirely possible that I am missing something here, because Bevilaqua is certainly an accomplished jet designer.

    If the wing lift energy source is due to the engine, then does it follow that a thrust augmenting ejector will require additional energy to be added to the inlet main primary flow, if the secondary, entrained flow is increased? I mean to say that if you were to stop the entrained flow, there should still be a reduced pressure area within the ejector. But then if the entrained flow were allowed to again flow, then would the inlet primary jet flow now require higher energy to maintain an equal inlet flow compared to the stopped entrainment situation? This would appear to be similar to the wing lift requiring higher energy from the airplane's engine.

    The pdf file size is 1.2M and do not see how to post here. I think you will find it an interesting read.
    Google search will bring it up:
    LIFTING SURFACE THEORY FOR THRUST AUGMENTING EJECTORS, Dr. P. M. Bevilaqua
     
  15. Sep 17, 2015 #14

    A.T.

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    The question was about the lift component, which per definition acts perpendicularly to the wings velocity, so it does no work on the wing.
     
    Last edited: Sep 17, 2015
  16. Sep 17, 2015 #15

    boneh3ad

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    But you can't separate lift from drag, and the drag does work on the plane. It's fun to treat them separately, but any lift-generating shape will generate lift-induced drag as well. The resultant force vector does work, ehich must be overcome by the engines. After all, a wing is simply a device that converts horizontal motion into an upward force.
     
  17. Sep 17, 2015 #16

    russ_watters

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    Why can't you separate them? Seems like a good idea to me too, in the context of the OP's question..
     
  18. Sep 17, 2015 #17

    boneh3ad

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    You can separate them mathematically, but physically you can't have one without the other.
     
  19. Sep 17, 2015 #18

    rcgldr

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    My previous post was about work performed (by the wing) on the air, not work performed (by the air) on the wing. The energy for the work performed on the air has to come from somewhere.

    In that case, you have lift and zero drag, but even in this case, from the air's frame of reference, work is performed on the air.
     
  20. Sep 17, 2015 #19

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    If you could add up the work done on every air molecule after the wing passes, it would give you the total work done. The energy must come from a combination of reduced airplane velocity, reduced airplane altitude, and engine. There are many scenarios where different amounts of energy would come those three sources. The nearest thing we have today to adding up the molecule-by-molecule work is computational fluid dynamics (CFD), which tracks millions of packages of air. The results of CFD calculations can be far from perfect, so wind tunnel experiments are usually done to measure forces. Those are also far from perfect, so the first test flights are very cautious.
     
    Last edited: Sep 17, 2015
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