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The Benouilli Effect

  1. Oct 18, 2003 #1
    The Benouilli effect, used by aeroplanes, race cars and of course blowing pieces of paper.

    OK, In the case of an aeroplane wing, the air flowing over the top of the wing has a greater distance to travel than the air flowing under the wing, therefore the air travels faster over the top and has a greater Kinetic Energy.
    The gain in Kinetic energy comes from a loss of Gravitational Potential Energy, (I think it's GPE, at least that's what my teacher fobbed me off with). It is this drop in Gravitational Potential Energy that (somehow) causes lower pressure on top of the wing and therefore the aircraft is pushed up by the higher pressure from underneath the wing.

    These are the bits I need explaining:

    * Why does a drop in GPE cause lower pressure over the top of the wing.

    * If there is a drop in GPE, then surely the air will be lower than it was before, but it isn't, it's still at the same height (ie: the height of the top of the wing above the ground).

    * Why does the GPE decrease? I understand that the energy must come from somewhere, but why GPE?

    * Also, I have trouble applying this example to other situations, for example, blowing over the top of a piece of paper. Try this yourself, get a piece of A4 paper, hold it at two corners that are next to each other (it's better if the corners are close together), and blow, gently, just over the top of the paper. Unless it's very thick paper, it should rise up due to the Benoulli effect.
    Well, where is the GPE loss here? You give it the KE needed by blowing the air.

    If you can think of any other uses for the Benouilli effect, feel free to keep them to yourself. Alternatively you can write them down on a piece of paper, screw it up, then throw it away. Joke.
    But seriously, if you can answer all or part of this question, please respond.
    Last edited: Oct 18, 2003
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  3. Oct 18, 2003 #2


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    Hi lavalamp!

    It's spelled Bernoulli, FYI...

    This is actually a VERY common misconception. Yes, the air travels faster over the top of the wing. No, it is not because it has to go farther.

    Air going along the top of the wing would actually GAIN gravitational potential energy. That would lead to the conclusion that it would lose kinetic energy.

    Let me point you to one explanation on NASA's websites:


    And I'll try to explain what's really going on:

    Let me warn you ahaid of time: this explanation is not going to be the whole story. NASA's site talks about Venturi nozzles, which is essentially what I'm going to describe (I will note why it isn't entirely right). What is really going on is much more complicated, and really requires a semester or three of aero- and fluid-dynamics courses to get a good understanding. It is a decent approximation for low speed aircraft, however.

    To explain lift in low speed airflows, you need three different sets of formulas.

    The first is Bernoulli's equation, which is actually a specialized form of Newton's second law.

    p1 + 1/2*rho*V12 = p2 + 1/2*rho*V22 = constant {1}

    Where p is the pressure, rho is the density, and V is the velocity. 1 and 2 designate two points in the flow.

    What that means is that if the velocity decreases, the pressure must increase to compensate.

    The second is mass continuity. This is sometimes known as the Venturi effect.

    rho1*V1*A1 = rho2*V2*A2 = constant {2}

    What this one means is that if you decrease the area a flow travels through, the velocity and density will increase to match. You can test this by holding your finger over the tip of a garden hose (decreasing the area).

    The third formulas are the isentropic flow relations:

    The isentropic flow equations are most likely a bit beyond your mathematical capabilities, so I'll just boil it down to the result. What they give is the way that the ideal gas equation distributes: If pressure increases, how much do rho and T increase.


    So how does all this get tied together? Well, when a wing goes through the air, it disturbs the air. If you assume that all of the air is broken up into 'streamlines', then when the wing travels through them, some of the streamlines need to compress to get out of the way. As it happens, the streamlines which compress are the ones which flow over the top of the wing. If we look at equation {1} that means that either the density must increase or the velocity must increase. We don't know which at this point.

    This is where the isentropic flow relations (IFR) come in. If you take a piece of paper and run it through the air so you're pushing the air, the IFR give how much the density, temperature, and pressure change. As it turns out, for speeds less than Mach 0.3, the air acts like an incompressible flow. What that means, is that the density will not increase - the area a mass of air takes up will not compress or decrease. This incompressible flow assumption is ~95% accurate at M = 0.3, but drops off rapidly at speeds higher than that. For high speed flows, you must use different methods, since Bernoulli assumes incompressible flow (there is only one value for density listed for both positions in the flow).

    Now, we've got the result we want. We've decreased the area, and we know that the density doesn't change. Therefore, the velocity must increase.

    rho*V*A = constant

    We can now take that result for velocity, and plug it into Bernoulli's equation and obtain the pressure at the new point.

    If you or your teacher have any further questions don't hesitate to ask.

    PS. tell your teacher what you've found after class. Like I said, it's a common mistake, but no-one liked being put on the spot in front of others, especially if they're your students.
  4. Oct 18, 2003 #3
    Well I've read through what you've written and I get most of it, although a re-read sometime tomorrow when I'm awake would probably help me get the rest. I've stayed up a little past my bedtime, it's 04:44 where I am.

    I doubt that my teacher was wrong, I think it's far more likely that I remembered wrong.

    I'll have to take a look at that NASA web-site tomorrow, I'm in danger of breaking my keyboard if my head droops anymore. Thanks for responding so soon.
    The reason I ask about this, is because next year I'm going to be doing aerodynamics at University but the Bernoulli effect has been taken off the syllabus, that kind of puts me at a disadvantage straight away.
    You'll be happy to hear that I did spell Bernoulli correctly when I was searching for information on him on Yahoo and Ask Jeeves, not that I found much.

    It seems as though you have a pretty decent knowledge of this sort of stuff, who knows, maybe one day I'll get this stuff as well as you do.
    I saw something on the Venturi Effect affecting the wind velocity in mountains and canyons, but I didn't read it. Maybe that's another task for tomorrow.

    By the way, if the IFR equations are differential equations, and that's what you meant by too hard, then I have done them. If they're are something else then could you tell me what type of equation they are, so that if I come accross them I'll know.
    If they are genuinely too hard and you were right all along, then I'm sorry for bugging you about them but I do like to aim high.
  5. Oct 18, 2003 #4


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    Always good to have another aero-head on the forums . I seriously doubt that you won't see Bernoulli. It will most likely be covered either in one of your intro physics courses or in the first few weeks of an intro to aerodynamics course. It has limited application, simply because it is an approximation, but I can practically guarantee you'll see it (if you don't, you may be going to the wrong school...)

    I certainly hope so. My area of focus is in space systems, so I've had about half the aerodynamics that aeronautics focused students have had.

    I usually don't go indepth for high-school students, because their derivation does require differential equations and thermodynamics and can be confusing (and is a big pain in the butt to go through). I also didn't know the level of your physics experience, so it is better to err on the side of caution. If you have seen these already, then you are much farther along then I was when I was in high-school...

    The final results follow:

    T0/T = 1 + (gamma-1)/2*M^2

    p0/p = (1 + (gamma-1)/2*M^2)^(gamma/(gamma-1))

    rho0/rho = (1 + (gamma-1)/2*M^2)^(1/(gamma-1))

    Where gamma is the ratio of specific heats (1.4 for air), M is the mach number, T is temperature, p is pressure, rho is density, and the subscript zero denotes total (or stagnation) conditions which mean that that is the value when the flow isn't moving. The total conditions are constant for an isentropic flow - no heat added or taken away, and no energy losses (e.g. friction). If you invert the rho equation and graph it with respect to Mach number, you'll see the line cross .95 around M=0.3, and then drop off quicker from there.

    If you want to get a head start studying this, I'd recommend you check out http://search.barnesandnoble.com/textbooks/booksearch/isbnInquiry.asp?userid=2WWUAGCB2E&isbn=007109282X&TXT=Y&itm=3 [Broken] book. It is expensive, but it covers a huge amount of material, and is one of the most well-written textbooks I've ever used. It's a sophomore level book, so it will be understandable to anyone with intro level physics background.
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  6. Oct 18, 2003 #5
    When I said that the Bernouilli Effect had been taken off the syllabus, I meant the A-level course syllabus that I'm doing now, not the University course (that'd just be weird), sorry for the mis-understanding.

    I'm not sure what age a sophomore is, but I'm going to assume that it's a 17/18 year old and therefore perfect for me.

    I haven't done anything about thermodynamics, but if I start working on him on Monday, I reckon that I can break him by Friday ( ) when he'll divulge as much as he's willing to and then recommend a book in the library. So I can expect to get that covered over the half term holidays (which start next week-end :smile:).

    You were right about that book. It works out at £85.83 + shipping and handling, I hope it's a big thick book. The bits about hypersonics and space vehicles look interesting, although I'm not sure why aerodynamics would apply to things in space unless it is talking about entering and leaving the atmosphere.
  7. Oct 19, 2003 #6


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    A sophomore is a second year university student in the US. The progression is freshman, sophomore, junior, senior.

    The book covers the following:

    A history of aeronautic engineering
    A section on fundamentals: discussion of units, conversions, anatomy of aircraft & spacecraft
    The standard atmosphere
    A 100+ page chapter on basic aerodynamics
    A 100+ page chapter on wings & airfoils
    A 100+ page chapter on airplane performance
    A 50+ page chapter on stability and control
    A 50+ page chapter on space flight. Mostly orbital motion and atmospheric re-entries.
    A 50+ page chapter on propulsion, including a little bit of rocket propulsion
    A short chapter on aircraft structures
    A short chapter on hypersonic flight
    and many, many, many very useful tables and charts.

    The book also covers all the thermo you'd need to understand the isentropic flow relations.
  8. Oct 19, 2003 #7
    Hmm, a second year University student. I wouldn't say that I'm quite up to that standard yet, considering that I haven't yet applied to any universities, (the schools dead line for that is next Friday so I will have applied soon).

    I suppose I could get the book for when the time comes that I'll be able to understand it, by then newer versions of the book will be out (probably), so would it be worth it.

    You said it was aimed at sophomores, well does that mean that any previous knowledge of aerodynamics, or other kinds of physics or even maths will be required for me to understand the book fully?
    If so, do you know of any other good books that are aimed at a slightly lower level. There's no use in setting my sights on something that is out of my range.
  9. Oct 19, 2003 #8


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    If you have had some physics in high school (and understood it), as well as calculus, you should be able to follow.

    It is the very first aerodynamics book they teach from in my school's program, and they teach it during the first semester of second year.

    All the students have had at that point is integral and derivative calculus, chemistry, and Newtonian physics.

    I'd say go check it out in a library before you drop the cash.
  10. Oct 19, 2003 #9
    I doubt that they would have it in the library, but there may be a remote possibilty that I could get them to order it for me.

    There is someone else in all of my classes (much better than me at Maths and Chemistry but I think we're even on Physics), he got them to order a book about fractals some time last year.
    So far it has only been checked out twice and it was him both times. As you can imagine, they're a little bit miffed about this ( ), but like I said, there's always a chance.

    Anyway, thanks for the info. It's got me another rung up the ladder to aerodynamic heaven. :smile:
  11. Oct 19, 2003 #10
    Bernoulli's principal is utterly simple (but not simple for Bernoulli to figure out the first time) although the math can be made as complicated as you like. You can't achieve directed motion in a gas or liquid without reducing the random kinetic motion (energy) of its molecules in all other directions (predominately to the side of such directed motion). To put it in even plainer English, if the molecules are going in one direction, they can't be very busy vibrating in other directions, otherwise they couldn't be going in just one direction.

    The result of this is that if you discharge an air hose between two hanging strings, or just plain blow between them, the strings will be attracted toward the air stream. The reason for this, as can be inferred from the info above, is that the lateral kinetic energy (pressure) existing in the air jet stream is lower than the kinetic energy (pressure) in all directions in the rest of the air in the surrounding environment, therefore, this environmental, relatively higher pressure, air tends to move in on the low lateral pressure (energy) of the jet stream, and will tend to join it.

    Physics books will usually give you the formula, P + .5pV^2 = K, (where P is the lateral (static) pressure in a (horizontal) pipe or, more fuzzily, in a jet stream, p is the directed pressure, V is the relative velocity, and K stays constant), without telling you why it works, which is too bad to say the least.

    As for the air foil shape of airplane wings, the air is made to travel a little farther over the top than on the bottom in the same length of time, therefore it must travel a little faster, which reduces it effective lateral pressure, in this case up and down, compared to the static pressure of the air underneath the wing. It would appear that the air on top of the wing would be a little compressed because of the squeezing effect of the upper bulge of the wing, but at common flying speeds this is more than compensated for by the reduction in pressure achieved. At supersonic speeds this kind of air foil cannot be used.

    The importance of the "lift" airfoil design is over hyped because most airplanes can be flown upside down without losing altitude, it just takes more gas. But if you're going to build an airplane wing, you might just as well build in as much lift as you can so as to save fuel.
  12. Oct 19, 2003 #11


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    Did you read any of my posts? The air over the top does not need to travel a longer distance in the same period of time. It's a fluid. There is nothing which dictates that the two flows need to meet up at the end.

    In fact, the air passing the top surface generally takes less time to travel past the chord than the flow on the bottom.

    Ref: the link I posted above to NASA's website or the link to the book which has several pages going over that very myth.
  13. Oct 20, 2003 #12

    I got my information from a basic physics book by Isaac Asimov. I missed your link to the NASA Web site the first time I read through your reply because it looked like a title to a section of your reply. However, so far I haven't been able to see, at the NASA site, where the explanation I presented about the air foil is wrong, but the material there is rather long winded, so maybe I will eventually find it.
  14. Oct 20, 2003 #13


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  15. Oct 22, 2003 #14

    I have read you making the assertion elsewhere that the reason the air velocity increases on top of the wing has nothing to do with the fact it has farther to go. I read your above explanation, which as informed and detailed as it is, still fails to explain why the upper airstream goes faster. You merely assert that it does:"as it happens the streamlines which compress are the ones which flow over over the top of the wing". You go on from there to explain how we can rule out density and can attribute this compression to velocity which leads to lower pressure, but you still haven't explained why "it happens" that there is a compressing of streamlines on the top to begin with.

    It seems to me that the force at work creating a situation where the the top and bottom flow would seem to want to meet up is simple inertia: the air is sitting there, minding its own business, inert, and along comes an aerofoil, parts it into an upper and lower stream, then leaves. We speak of the air as flowing but as the aerofoil cuts through it, its true motion is up or down. That being the case, once the aerofoil has passed the main vector of the top stream is down and of the bottom stream, up. If you envision a purely "Bernoulli" aerofoil, free of N III angle of attack effects, this should be clearer. The camber in the upper surface represents a longer distance. By simply trying to stand still the air is pushed up this hill and falls back down the other side.
    This would certainly account for why the streamlines on top are compressed to begin with.
  16. Oct 23, 2003 #15
    Sorry, but I have to butt back in here. While your explanation does seem to make sense, I have to say that the air is not just sitting there.

    The r.m.s. (root mean square) velocity is Mach 1 (approx 750 mph). This is because while the average velocity of the air (assuming there is no wind) is 0, each atom/molecule's speed very fast indeed. There are a range of energies for the particles which "give" them Kinetic Energy.
  17. Oct 23, 2003 #16
    I don't percieve you as butting in. You started this thread.
    "...just sitting there, minding its own business..." was, of course, intended as a comically anthropomorphic characterization of its being inert, not as a scientifically exhaustive exploration of all its dynamics on all scales, macro and micro.
    This I did not know. I am not familiar with the concept of "root mean square velocity". Why root? Why square? This is a way of figuring the average velocity of molecules is a gas? Sounds interesting.
    I was familiar with this fact.
    This I didn't know. I wasn't aware there was a restricted range.

    I don't see how this interpolation of information affects anyone's arguments one way or another, and don't understand the urgency to point it out, although I do find it interesting.

    I am not out to prove Enigma wrong. For a long, long time it was taught that Bernoulli was the only thing getting planes into the air. This is what I was taught. In high school my science teacher asked everyone how they thought planes fly. A couple people (not me, thankfully) suggested it was like when you stick your hand out a car window and tilt it a little. These people were firmly castigated for approaching science with mere intuition and then he taught Bernoulli.

    Now, of course, everyone has faced the fact that Newton III is capable of getting a plane into the air all on its lonesome, if you want to design a wing that way.

    So, Enigma may be completely right about the greater distance over the top of the wing being wrong. I would rather hear him explain it, rather than assert it.
    Last edited: Oct 23, 2003
  18. Oct 23, 2003 #17
    Sound travels through air as fast as the particles in the air can "pass on the message", ie: one particles vibrates at a certain frequency which then makes another particle vibrate at a certain frequency (not neccesarily the same frequency as the second particle could be heavier or lighter than the first and not all of the energy may be passed on). This means that the average speed of all of the particles in air (at s.t.p) must be the same speed as the speed of sound, Mach 1 or approx 750 mph.

    The root mean square speed is when you get all of the velocities of the particles in the air, square them all individually (to get rid of any negatives), add them together and then take the positive square root the answer. Hence it is called the root of the mean of all the speeds squared, or r.m.s. speed for short.

    The energy that a particle has is Kinetic Energy, (assuming that all of the elctrons in their are in their lowest possible orbitals etc. blah blah all chemistry and no physics here), and the average Kinetic Energy of the particles in a gas (or liquid, solid or liquid crystal for that matter) determines the temperature of the substance. The Kinetic Energy of all the particles is obviously not just the same for each particle, there are a range of energies that form a graph, which starts at 0 KE and goes up to a peak value (not neccesarily the r.m.s. speed) and then tails off rapidly to infinity as there is no real limit to the KE a particle can have.

    I said "butt back in" because for a while I butted out, I read the posts of course, but because my question had been answered and since I couldn't answer any other points raised I didn't post anything (until now).
    Last edited: Oct 23, 2003
  19. Oct 23, 2003 #18


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    Hi zoobyshoe,

    Unfortunately, I can't really explain it in detail. It has to do with boundary layer conditions and advanced fluid dynamics topics (which I haven't studied yet). I can show you pictures of flow fields though, to back up my statement:

    http://www2.icfd.co.jp/examples/naca0012_2d/image/ma00901.jpg [Broken]

    Flowfield 2

    Well, we know that there is a compressing of streamlines. There has to be, since the airfoil is flowing through. Unfortunately, I can't explain it in any greater detail than "it is that way". Sorry.

    This depends on where you're drawing your coordinate frame. If you use the ground as a reference frame, then the wing pulls the particles with it as it flows through, and it pulls the ones on the top different from the ones on the bottom. If you're in the reference frame of the wing, and integrate the velocity as a function of distance travelled, you'll find that for the majority of cases, the air will be farther downstream if it passed over the top.

    You can't be "free" of N3 effects, just as you can't be free of Bernoulli! All of Newton's laws (Bernoulli is a specialized case of N2) always apply. But the underlying cause of a change in momentum is not described by N3. N3 doesn't say anything about how momentum is transferred, only that it has been. Every (non gravity, non electromagnetic) force is applied either by pressure or by friction. There are no "real" point forces.

    It's also damn near impossible to measure the net downdraft of air to work backwards and determine the lift.

    I hope that helps a bit. I'll try to go farther if it isn't.
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  20. Oct 24, 2003 #19


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    Actually, I think its more simple than that: its due to the incompressibility of air at low speed. If the air moving over the top surface of the wing were just pushed up and then went back down, it would compress. At low speed, the pressure is able to keep the molecules closer to the right distance apart, but to do that some of them have to move out of the way - toward the back of the airfoil.

    At higher speed (above 220mph or so) compressibility starts to play a role, but the same effect exists - its just reduced.
    You don't work with fans much, do you?
  21. Oct 24, 2003 #20
    This makes perfect sence. It cleary explains why there would be a flow of air over the top surface that was even faster than the speed of the craft cutting through inert fluid.

    It seems, though, still correct to say: It has farther to go, it has to do this without compressing, therefore its only option is to do it faster.
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