1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lift more complex than Bernoulli?

  1. Sep 21, 2012 #1
    Hello

    Over the past year I've taken up the hobby of RC model airplane flying. During that time I've had many discussions with fellow modelers about lift. I've additionally read a bit about it as well. Most explanations I see like to explain it on Bernoulli's principle. That the wing construction being such that the flat bottom provides a shorter distance for the air to travel. that the air passing over the rounded top must go faster to get to the trailing edge at the same time. this more rapid flow results in lower pressure on top and there in comes the lift.

    Some things that occur to me is that not all wings are flat bottomed. In modeling many, especially the higher performance models, have symetrical wings which would result in zero lift at an attack angle of zero. I can see that increasing the angle of attack would cause the flow on top to again have to take a longer path and again bournoulli's principle would again be an explanation.

    One thing about bernoulis principle that I have trouble with is "why must the air traveling over the top exit the trailing edge at the same time as the air traveling over the bottom of the wing?"

    I keep having this image of the wing advancing on the air and since air has a density it will offer resistance, that the wing will compress it and climb up on top of it. I am told by some that it does not work this way and by some that it does but the effect is minimal.

    I also consider that just about any angle of attack above the horizontal will result in some of the air being forced downard which in my mind would push the wing in the opposite direction that is to say up and a bit forward.

    Can someone discuss lift here with an eye towards helping the nerdy layman grasping the principles? I did search the forum but couldn't find a thread on point.

    Regards,
    Mike
     
  2. jcsd
  3. Sep 21, 2012 #2

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

  4. Sep 22, 2012 #3
    Alphazero

    thanks very much for the links. I found the allstar link especially thought provoking. I knew I was missing something.

    Mike
     
  5. Sep 22, 2012 #4

    rcgldr

    User Avatar
    Homework Helper

    Another link about wings and lift with animated diagram of how the air is affected.

    http://www.avweb.com/news/airman/183261-1.html

    In some cases, the "longer path" is on the bottom, but lift is still generated, as shown in this pre shuttle lifting body prototype (m2-f2 glider next to F104 jet):

    m2f2.jpg
     
  6. Sep 23, 2012 #5

    cjl

    User Avatar

    One thing to keep in mind - bernoulli's principle is correct. The air flowing over the top of the wing is indeed moving faster than the air moving under the bottom, and this speed difference is correlated with a pressure difference which can be calculated using the bernoulli relation. The part of the popular explanation which is wrong (as you suspected) is the equal transit time assumption - the cause of the speed difference is much more complicated than that.

    Basically, when you work out the math, the view that the wing gets its lift from a pressure differential between the top and bottom surface, and the view that the wing is pumping air downwards are equivalent - either way, the lift is the same.
     
  7. Sep 23, 2012 #6

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    And, of course, all this only really applies to subsonic flows. Supersonic flows follow several fundamentally different rules.
     
  8. Sep 23, 2012 #7

    rcgldr

    User Avatar
    Homework Helper

    If the concept of faster moving air focuses on the mostly horizontal flow relative to the wing in the immediate vicinity of the wing, than note that relative to the air (the unaffected air), that horizontal component of flow above a wing is slower than the flow below. The actual flow relative to the air, is mostly downwards (lift) and somewhat forwards (drag). The accelerations are independent of frame of reference (air or wing), and are mostly downwards (lift) and somewhat forwards (drag).

    Bernoulli is violated somewhat in the process because the air affected by a wing ends up with a downwards and somewhat forwards velocity at the moment it's pressure returns to ambient, meaning that wing has performed work on the air, increasing it's overall mechanical energy (using the unaffected air as a frame of reference). In the case of propellers or rotors, this velocity is called "exit velocity", while for wings, a more generic term, "downwash" is used.

    From what I've read, in a typical situation and using the air as a frame of reference, the fastest moving air occurs just aft and below a wing (the moment where the affected air's higher pressure returns to ambient, it's "exit velocity").
     
    Last edited: Sep 24, 2012
  9. Sep 24, 2012 #8

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    For starters, you ought to really use a more descriptive phrase than "using the air as a reference frame." The air is moving at any number of different velocities in this flow field. I believe what you are referring to is the Eulerian specification of the flowfield, which is where you essentially are a fixed observer watching the foil move through the initially quiescient air. Is this what you are referring to?

    At any rate, the concept of Bernoulli's principle still holds in this situation. After all, static pressure does not depend on the frame of reference. It is a thermodynamic state variable. Bernoulli's principle is essentially an energy balance cast in terms of pressure, stating that all applicable variables are constant. By changing your reference frame, you are essentially just changing your constant, which is just the stagnation pressure when dealing when neglecting gravity.
     
  10. Sep 24, 2012 #9

    rcgldr

    User Avatar
    Homework Helper

    I forgot to include "unaffected air" in my last post. I updated my last post to include that. The observer could be on the ground in a no wind situation, or in a balloon moving at the same speed as the air unaffected by the aircraft passing through the air.

    Yes, after the foil passes through the air, the work performed on the air shows up as a downwash. For propellers or rotors, a term called "exit velocity" is used to describe the velocity of the affected air at the moment the affected air's pressure returns to ambient. Link to NASA article:

    NASA_propeller_analysis.htm

    Bernoulli doesn't account for the work performed by the wing that results in downwash. This is mentioned in that NASA article about propellers:

    We can apply Bernoulli's equation to the air in front of the propeller and to the air behind the propeller. But we cannot apply Bernoulli's equation across the propeller disk because the work performed by the engine (propeller) violates an assumption used to derive the equation.

    The amount of pressure jump and work performed by a wing is much less than a propeller (ignoring cases like an F-16 fighter in a 400 knot 9 g turn), but there is still significant work being peformed in order to create the downwash.
     
    Last edited: Sep 24, 2012
  11. Sep 25, 2012 #10

    K^2

    User Avatar
    Science Advisor

    Bernoulli Effect explanation is also technically wrong. It simply happens to be a good approximation in many cases.

    Bernoulli Effect only applies within the bulk of the fluid. At the boundary, the speed of the fluid is always zero. So Bernoulli Effect at the surface of the wing, where the force is actually applied between medium and the aircraft, is precisely zero.

    You have to make an assumption that the fluid has low viscosity and almost constant density in order to make use of Bernoulli Equation. Former tells you that you can get very close to the surface of the wing without running into boundary layer effects, and later tells you that the pressure gradient is zero in direction normal to the surface. So the pressure at the surface of the wing is almost identical to pressure at a contour following the wing where you can use Bernoulli Equation.

    These assumptions do hold fairly well for an aircraft at air speeds significantly lower than speed of sound. The closer you get to speed of sound, the worse that approximation gets, which is one of the reasons why in practice you'll rarely see people use Bernoulli Equation to compute lift. Circulation is easier to compute numerically and gives correct values for higher air speeds. Circulation can be related to lift via Kutta-Joukowski Theorem.


    By the way, Kutta-Joukowski Theorem also tells you that if equal transit hypothesis was actually valid, lift of the wing would be zero. Equal transit would result in zero circulation around the wing, meaning there is no lift either. For a wing to generate lift, the transit time for streams above and bellow the wing actually have to be different.

    In most supersonic aircraft, the wing operates within the subsonic stream behind the shockwave generated by the nose of the aircraft. If you look at geometry of supersonic fighters, for example, you can tell the design cruise speed by the angle of the cone within which the airplane fits.
     
  12. Sep 26, 2012 #11

    cjl

    User Avatar

    K^2: You're neglecting one of the assumptions stated when using the bernoulli equation: namely, that the flow is inviscid. In most aircraft-related flows, this is true (to a pretty good approximation) everywhere except in the boundary layer, and the boundary layer is thin enough that it can be fairly safely neglected for the purposes of lift and pressure calculations (it is important for drag though). Another of those assumptions (as you mentioned) is that the flow is incompressible -this is the one which is violated as the flow approaches the speed of sound, which is why the bernoulli equation doesn't apply to many modern aircraft. It's a perfectly valid relationship within the realm of flows for which it was designed though.

    As for supersonic wings? They most definitely are operating within supersonic air (which is why they tend to be extremely thin with sharp leading and trailing edges). It's true that they are within the mach cone generated by the nose of the aircraft, but the nose of the aircraft is generating an oblique shock, not a normal shock, and as a result, the air after the shock is still traveling supersonically relative to the aircraft. You could probably design an aircraft which would push a large normal shock in front of it, and have the aircraft operating within the subsonic bubble generated by the normal shock, but this would cause a phenomenal amount of drag.
     
  13. Sep 26, 2012 #12

    cjl

    User Avatar

    Very true. This is part of why it is often much more convenient, both when visualizing the flow and when calculating the details of the flow, to imagine the wing as stationary with the flow moving over it. The two situations are of course physically equivalent, but the situation with the wing stationary has the wing performing no work, and has a time-independent solution (ignoring turbulence and such things). In my opinion, looking at it from the frame of the air is a much more difficult way to envision the problem, and potentially more misleading, though it is certainly mathematically a possible way to do it.
     
  14. Sep 26, 2012 #13

    cjl

    User Avatar

    This depends entirely on the wing. Some wings, especially those with a very high Cl (high camber and angle of attack) will have a much higher flow velocity over the top of the wing, even from the reference frame of the unaffected air (although I do still think it's more useful to think of it from the wing frame of reference). It's also true that the accelerations are independent of the frame of reference, but in the wing frame of reference, the air's energy is conserved (and it is just redirected as it flows over the wing), allowing conservation of energy to contribute to the problem's solution (which is really all bernoulli is - a statement of conservation of energy in an inviscid, incompressible flow). In the frame of reference of the stationary air, the wing is imparting energy to the air, which complicates the solution, since an energy balance is substantially more difficult. A momentum balance is relatively straightforwards in either frame of course, but it alone does not provide sufficient information to really determine the flow characteristics.

    As I said, Bernoulli is basically a statement of conservation of energy, and it is assuming that nothing is adding or subtracting energy to the flow. This is easy enough to do in an ordinary flow around an aircraft - by using the frame of reference in which the aircraft is stationary, it is unable to perform any work (since it is not moving), and thus the work done on the flow by the aircraft is zero. If viscosity is also neglected, the flow must always contain the same energy as it entered the control volume with, since there is no means of transferring the energy either to an object or to any other parcel of fluid, and this simple principle leads directly to the bernoulli equation. In the case of propellers or rotors, there is no single reference frame which can be chosen which makes all portions of the craft stationary, so there is no frame in which the bernoulli relation can be used without accounting for the energy added. However, for just a wing, the easiest way to treat the problem is simply to have the wing stationary and thus enabling the statement of conservation of energy to apply to the entirety of the flow.

    Again, this will depend on the geometry of the wing. A highly cambered wing at a high angle of attack will probably have the largest absolute flow velocity (relative to the stationary air) somewhere around the quarter chord point (or possibly in front of it), on the top surface of the wing. A more ordinary wing at a low angle of attack will be different - it could be just behind the wing, it could be on top, or it might even be below the wing. A supercritical airfoil will be different still. This all depends on the specific shape.
     
  15. Sep 26, 2012 #14

    K^2

    User Avatar
    Science Advisor

    Note that fluid that is perfectly inviscid cannot generate lift. Kutta condition relies on existence of boundary layer which requires some viscosity. But yes, it's low, meaning the boundary layer is thin.

    I'm not sure what exactly you are trying to correct in this entire paragraph, because I've stated these exact things in my post.

    The only time you get supersonic stream over a wing is in transonic region, and that results in separation, turbulence, wave drag, loss of control, and many other unpleasant things. During supersonic flight, the flow over the wing is subsonic. That's what allows supercruise in the first place.

    The difference between oblique and normal shock waves is that a normal shock wave guarantees subsonic flow behind it at any air speed. The oblique shock wave slows the flow down as well, but whether the flow behind it is subsonic will depend on the speed. Simple check. Take an object at Mach 1+. It generates a 45° oblique shock. Can the flow behind that be supersonic? No. Because if it's just supersonic before the shock and just supersonic behind, there cannot be a shock in between. You need a sharp change in velocity which you can't get. So for low Mach numbers, somewhere within Mach 1-2 region, you do get a good range of air speeds where the flow is entirely subsonic within the cone.

    So what happens if you go even faster? Naturally, at high enough speed the flow within the oblique shock becomes supersonic. But the flow over the wing is still going to be subsonic. Why? Because the wing is going to generate its own bow shock with a subsonic flow behind it.

    This might break down for hypersonic flights. I know nothing about hypersonic aerodynamics. But a typical modern supersonic jet does not deal with supersonic flows over the surfaces except for transition regions.

    Oh, and the wings are made thin primarily to reduce drag in transonic regions, providing for better control and making it easier to punch through to supersonic.
     
  16. Sep 26, 2012 #15

    rcgldr

    User Avatar
    Homework Helper

    Absent gravity, the frame of an aircraft would be accelerating and non-inertial, which creates issues. An external force, normally gravity, is required to prevent the aircraft from accelerating when diverting the air flow. The source of that force needs to be included to end up with a closed system, one where momentum and energy are conserved (in order to use the aircraft as an inertial frame of reference).

    From what I recall, in addition to friction drag, induced drag also contributes to slowing down the air's speed when the flow is diverted (using the aircraft as a frame of reference). Friction drag converts the energy into heat, but it's not clear to me where the induced drag energy loss goes. My guess is that the energy loss related to induced drag corresponds to an increase in energy of the source of the force that keeps the aircraft from accelerating. In the normal case where the earth is the source of the external force via gravity, the energy could be going into the earth, since the earth would accelerate towards the aircraft, ignoring interactions between air and earth.

    As an alternative example, imagine an ice boat on a huge flat block of ice, that rests on a frictionless surface. Say the ice boat is moving perpendicular to the wind, accelerating the huge block of ice in the direction of the wind, in order to keep the ice boat moving at constant velocity with respect to some inertial frame (the "angle of attack" of the skates on the ice boat would be increasing over time as the huge block of ice accelerates). From the ice boats frame of reference, the energy of the block of ice is increasing, and the source of that energy is the energy being extracted from the air.
     
    Last edited: Sep 26, 2012
  17. Sep 26, 2012 #16
    Really?

    I thought it was commonplace for hydraulic engineers to apply Bernoulli to the free surface of a liquid, particularly in reservoir engineering.
     
  18. Sep 26, 2012 #17

    K^2

    User Avatar
    Science Advisor

    Gravity also does zero work, so including an external force does not change the argument. There is no work being done by aircraft on the medium in frame of reference attached to the aircraft. Total energy in the medium is definitely conserved. The only source of error is heat generated due to viscosity, and it's a minor correction at low speeds.

    As discussed above, it's an approximation. If fluid is well approximated as inviscid and incompressible in given problem, it's a good approximation. But the actual, physical Bernoulli Effect precisely at the surface is always zero.
     
  19. Sep 26, 2012 #18
    I've no idea what that means, bu I do know that I can assign precise and exact values to all the variables at the free surface in Bernoulli's Theorem. There are no approximations there.
     
  20. Sep 26, 2012 #19

    rcgldr

    User Avatar
    Homework Helper

    My point was that gravity would be doing work on the earth (not the aircraft) in the situation where the aircraft was not accelerating.

    What about the ice boat analogy (using the acceleration of a second medium to offset the force related to the first medium (as opposed to using gravity))? In the frame of reference of the ice boat, the huge block of ice is gaining energy; what is the source of that energy if not the diverted air flow?
     
  21. Sep 26, 2012 #20

    cjl

    User Avatar

    Sorry, but this is incorrect. The flow over a wing in supersonic flight is supersonic, aside from the fluid inside the (thin) boundary layer.

    No, it will depend on the flow turning angle. In almost all cases, an oblique shock has two possible solutions, a strong oblique shock and a weak oblique shock. In the vast majority of cases, the weak oblique shock solution is the one which will be physically realized within a flow, and the weak oblique shock (by definition) has M2 > 1. This is a good thing too, as this is the weakest possible shock wave for the flow turning angle and incoming mach number, which means the smallest losses within the flow. Smaller flow losses mean less drag.

    At M1 = 1, you don't actually get a shock, you get a mach wave, and it is at a 90° angle to the flow (not 45). To generate a 45 degree oblique shock, you must have the proper combination of incoming mach number and flow deflection angle, which could range from the weakest possible shock (a mach wave) with M1 = sqrt(2), or you could have an incoming flow at M1 = 3, with a 25° flow deflection, just to name a couple of examples. You can see more possibilities on the chart http://upload.wikimedia.org/wikipedia/en/c/c7/ObliqueShockAngleRelation.png [Broken]. Interestingly enough, a 45 degree shock will always be a weak shock, and thus in every case, a 45 degree shock will have M2 > 1 (the flow out the back of the shock will still be supersonic). Strong oblique shocks have shock angles greater than 60°, even for very high M1 and small turning angles.


    Only if it is an extremely poorly designed wing. This would cause an immense amount of drag for no real benefit. The only subsonic flow over a modern supersonic wing should be in the boundary layer.

    Note also that there is such a thing as a supersonic leading edge vs a so-called subsonic leading edge on a supersonic aircraft. This does not mean that the flow is subsonic on the wing, even for a subsonic leading edge. A subsonic leading edge means that the leading edge of the wing is swept back sufficiently far that the normal component of the flow velocity is subsonic (or, in other words, the leading edge is swept back more steeply than the mach angle). This allows spanwise disturbances from the root of the wing to propagate in front of the wing itself, which means the flow ahead of the wing is affected by the wing. On a supersonic leading edge, the flow is unaffected until it hits the wing itself. In either case, the flow over the wing is supersonic though - this terminology is referring only to the component of the flow velocity normal to the leading edge of the wing.
    The wings are thin primarily to reduce drag in supersonic regions - airfoils made to reduce transonic drag look quite different. This is why, for example, the wings on a modern commercial airliner (made to minimize drag in the transonic region of around M = 0.75 to M = 0.95) look very different from the wings on an F-104 starfighter (made for M = 2 or so). Interestingly, if you look at a modern fighter, the wings are somewhat thicker than older fighters. This is largely for transonic and low supersonic flight - aircraft optimized for purely supersonic operation (SR-71, D-21, XB-70, TU-144, concorde, etc) have extremely thin wings which give substantially poorer performance at low speeds, but for a fully supersonic aircraft, the very thin wing gives much better drag characteristics.
     
    Last edited by a moderator: May 6, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Lift more complex than Bernoulli?
Loading...