Zet Posted February 9, 2015 Author Posted February 9, 2015 ... but to say the energy comes from the fluid …seems misleading to me. Are you comfortable with “The increase in vertical kinetic energy and in gravitational potential energy is offset by an equal decrease in the kinetic energy of the fluid” (if that, in fact, is how it is offset)? --- This has become complicated ... If there is more force on an object in one direction (more force from the pressure pushing up on the bottom of the solid) and if there is less force on the same object in the other direction (the force from the pressure pushing down on the top of the object plus its weight) then it will move (upwards) if it is free to do so (and it will not move upwards, even though there is an overall upward force on it, if it is fixed in place). When it moves upwards there is an increase in vertical kinetic energy (in the form of the moving solid upwards and displaced fluid downwards) and there is an increase in gravitational potential energy (and if it is prevented from moving then there will not be these same energy increases). And so … for energy to be conserved … there must be a decrease (an equal decrease) in another form (or forms) of energy (that does not also occur when the airfoil is prevented from rising). When MigL suggested adding an angle of attack (in his thought experiment of a tilted flat hand outside the window of a moving car) that could be addressed in and of itself (the bottom of the airfoil will collide with the fluid and the airfoil will be redirected upwards and the air will be redirected downward due to this. There is an increase in vertical kinetic energy and there will be an equal decrease in horizontal kinetic energy. Energy is conserved.) There was no need to address and include every other aspect of this system to understand this part of it. The same thing here. If there is more upward force than downward force then the solid will rise if it is free to do so and it will not if is prevented from doing so. And if it rises then there is an increase in vertical kinetic energy and in gravitational energy. And so there must be a decrease in another form of energy that does not occur when it does not. What is it? --- I agree that seems to be the correct answer to your question. J.C.MacSwell’s logic works. There is an increase in vertical kinetic energy and in gravitational potential energy in the one case that does not occur in the other. And so, if there is an equal decrease in the kinetic energy of the fluid (overall in all of its various motions) in the one case that does not occur in the other then (… ta-da! …) energy is conserved. The logic works, but we still need a link to a reputable web site showing that this in fact physically the case. (Just because this logically works out, does not mean that this necessarily is what is happening in the real physical world.) And (assuming for the moment that this is right, that there is a greater decrease in the overall kinetic energy of the moving fluid in the rising case than in the non-rising case), there is still a problem then with the next step in this conservation of energy analysis. If (in the thought experiment in post #83) there are differently shaped neutrally buoyant airfoils all of equal masses and of equal volumes in the horizontal part of the L shaped container, then when the fluid drains out of the L shaped container there will a greater decrease in pressure pushing down of the top of each one and a lesser decrease in pressure pushing up on the bottom each one. If they are free to rise then they will do so. In the end (since each differently shaped airfoil is of the same mass and volume) there will be the same increase in gravitational potential energy. (There is not only an increase in gravitational potential energy when the airfoils rise but also an increase in vertical kinetic energy. And so, the odds are that when each differently shaped airfoil rises they will do so at differently velocities (thus different vertical kinetic energies). If what brings the rising airfoils to a stop (when the reach the top of the frictionless track that allows them to move vertically but not horizontally) is a perfectly elastic collision (such as compressing a hypothetically perfect spring) then all of the vertical kinetic energy of the rising solid will become an equal amount of elastic potential energy. Energy is conserved. This will balance out in and of itself. And the vertically moving fluid displaced downwards becomes part of the rest of the overall movement of the fluid.) If (… if …) energy is conserved when the airfoil rises because there is a greater decrease in the overall kinetic energy of the fluid than when it is prevented from rising, then this additional decrease in the kinetic energy of the fluid must be the same in every rising case and regardless of the shape of the airfoil. It could be. The logic works. But I find it doubtful. When each differently shaped airfoil is held in place and the fluid passes over it (the odds are) there will different amounts of drag and turbulence in each case. And when each differently shaped airfoil rises, it stands to reason, that if there is an lesser increase in the turbulence that it would be different lesser increases in each case due to them having different shapes (and not regardless of their shapes). The logic works. But I find it physically doubtful. ? --- ... I haven't read it closely ... (Modesty? Your post was very detailed.) I understand that the rise of an airfoil can be explained in terms of Newton’s Third Law of Motion (or also in terms of the Law of Conservation of Momentum). But this still doesn’t get around the fact that when a fluid and an airfoil are in relative motion then there is an overall upward force on an otherwise neutrally buoyant body and it will rise if free to do so. And so, regardless of whatever else is causing this rising airfoil to rise (and regardless of the conservation of energy analyses involved in those other aspects), there must be a decrease in another form of energy to offset the increase in gravitational potential energy from this contributing factor. (And I’m guessing you and I are in agreement about this.) However, my question (and I suspect Zet's ) is what happens when there's 0 angle of attack and the only lift is provided by the difference in the static presssure between the upper and lower surfaces of the airfoil. Yep. (I keep using the terms “dynamic pressure” and “dynamic buoyant force” but I just made those up because they seem descriptively accurate but I don’t know if they are, in fact, the correct terms I should be using. You said “static pressure.” Is that right? Typo?) What about the difference in pressure between the upper surface of the air foil and the lower surface of the airfoil? Yep. All of your questions are also my questions. Thanks for the links! --- ? --- (And while J.C.MacSwell is the only one who has come up with a workable (logically workable) answer in this thread (... as far as I can tell ...), I’m still hesitant about it, and not so sure this is, in fact, how the real physical world works (based on the next step in the analysis where then differently shaped rising airfoils of equal mass and volume would all have to have the same lesser increase in turbulence … for energy to be conserved). We need some links!) --- ?
studiot Posted February 9, 2015 Posted February 9, 2015 Zet but I just made those up because they seem descriptively accurate but I don’t know if they are, in fact, the correct terms I should be using. You said “static pressure.” Is that right? Typo?) How much else have you made up? Zet You said “static pressure.” Is that right? Typo?) Do you mean to say that you have invoked Bernoulli's theorem, but you don't know what static pressure is? Zet If there is more force on an object in one direction (more force from the pressure pushing up on the bottom of the solid) and if there is less force on the same object in the other direction (the force from the pressure pushing down on the top of the object plus its weight) then it will move (upwards) if it is free to do so (and it will not move upwards, even though there is an overall upward force on it, if it is fixed in place). Codswallop. Weightless objects only exist in fantasy.
Essay Posted February 10, 2015 Posted February 10, 2015 (edited) Are you comfortable with “The increase in vertical kinetic energy and in gravitational potential energy is offset by an equal decrease in the kinetic energy of the fluid” (if that, in fact, is how it is offset)? =-= J.C.MacSwell’s logic works. There is an increase in vertical kinetic energy and in gravitational potential energy in the one case that does not occur in the other. And so, if there is an equal decrease in the kinetic energy of the fluid (overall in all of its various motions) in the one case that does not occur in the other then (… ta-da! …) energy is conserved. The logic works, but we still need a link to a reputable web site showing that this in fact physically the case. (Just because this logically works out, does not mean that this necessarily is what is happening in the real physical world.) =-=> [edited ] When each differently shaped airfoil is held in place and the fluid passes over it (the odds are) there will different amounts of drag and turbulence in each case. And when each differently shaped airfoil rises, it stands to reason, that if there is a lesser increase in the turbulence that it would be a different "lesser increase" in each case due to them having different shapes (and not regardless of their shapes). The logic works. But I find it physically doubtful. =-=> [edit] "But I find it physically doubtful." Why? Not me; it seems very straightforward, about the extra turbulence, that each would be different, though very difficult to measure accurately enough to see the relatively tiny and chaotic differences. This seems valid, so I'm comfortable with it: "The increase in vertical kinetic energy and in gravitational potential energy is offset by an equal decrease in the kinetic energy of the fluid." I'm comfortable expecting that the forces offset and balance each other, and I think you have the correct point made for airfoils of different shapes also. With a poor design, much more ineffective turbulence will be generated, so if any lift could be achieved, I'd expect it would require much much more KE overall. As for finding a citable site "showing that this in fact physically the case," that may be problematic, since we aren't using the standard metrics and concepts that science uses for measuring and describing the physics. What we've described as vorticity or swirl or turbulence or "ineffective turbulence" are not easily measured or quantified into a useful metric for calculating the relevant forces. All we've done here is to gain a deeper intuitive understanding about how the forces are distributed. "Understanding about how those forces are distributed" (in reality) is a nice thing; but if you want to model the forces (in theory) so you can make predictions, then you'll need measurable effects and observations. Istm, Bernoulli uses quantities that are more easily measurable, in metrics relevant to the problem at hand, which work well enough to make predictions with. As with Boltzmann, who allows us to measure overall temperature without examining each particle's unique temperature, Bernoulli allows us to look at overall flow without examining every unique change in flow. ....imho. But I'm no expert in physics, so I may be 'mixing metaphors' and assuming way too much; and though comfortable with this, still looking forward to further considerations, clarifications, or corrections. ~ Edited February 10, 2015 by Essay
studiot Posted February 10, 2015 Posted February 10, 2015 Looking more closely at Zet's words I find he has included the weight of the airfoil in the pale typeface so apologies for missing this. However these words are at variance with his diagrams since Zet does not include the weight or any fixing force in his diagrams so they are all flawed force balances.
Essay Posted February 10, 2015 Posted February 10, 2015 (edited) Looking more closely at Zet's words I find he has included the weight of the airfoil in the pale typeface so apologies for missing this. However these words are at variance with his diagrams since Zet does not include the weight or any fixing force in his diagrams so they are all flawed force balances. Right, when I was talking about balancing the forces, I meant in general rather than from the specified diagram. As I pointed out post #85, "Regardless of the numbers in the examples that you posted, I think you’d find small differences in the fluid exiting the tube," I wasn't expecting to find the balance in the listed forces. Mainly I've been trying to help Zet understand what you said in post #18: "Some of this work goes into the kinetic energy of vertical motion. (did you remember that?) Some goes into the increased gravitational energy." ...and especially, "Similarly if the body is in level flight the drag will be doing work on the air, leaking kinetic energy away to the air. This KE is maintained by the power of the engine supplying the thrust, which does work on the body." In trying to see this on a very non-mathematical level, more than ever before thanks to Zet's questions, the terminology got loose. In the end though, I think we're all saying the same thing, except mostly with different languages. N'est-ce pas? ~ Edited February 10, 2015 by Essay
studiot Posted February 10, 2015 Posted February 10, 2015 (edited) Hello Essay, The problem is that I have never been comfortable with the following statements I'm not sure who made this one Are you comfortable with “The increase in vertical kinetic energy and in gravitational potential energy is offset by an equal decrease in the kinetic energy of the fluid” (if that, in fact, is how it is offset)? Zet There is an increase in vertical kinetic energy and in gravitational potential energy in the one case that does not occur in the other. And so, if there is an equal decrease in the kinetic energy of the fluid (overall in all of its various motions) in the one case that does not occur in the other then (… ta-da! …) energy is conserved. A simple treatment of energy regards objects as "point particles", and it is the view taken by Zet. This view has a special definition in Physics such that it regards all the mass of a body as being concentrated at a single point. Whilst this view can suffice in the case of the mechanics of simple solid objects is is entirely inadequate fot fluid bodies as a whole. In particular the mass of a fluid body cannot be considered as a point particle, but is distributed throught the entire body. Solid objects can possess mechanical energy in four forms Kinetic Energy, Rotational Energy, Field Energy (eg gravitational), Internal Energy The children with their hoops are demonstrating rotational energy. When they are bowling these along that brings in kinetic energy. If they are bowling them uphill they are adding gravitational field energy Here is the repository for the fourth type internal energy. The springs can twist and flap, extend and contract. All of which changes the internal energy whilst the object can be regarded as the same entity. Internal energy is so much more important for fluids. So when considering energy balances all four forms need to be considered. Often we need only consider two for solid objects, but we always need to consider at least three for fluids. I said before that Bernoulli represents an energy accounting for a fluid and as such you have four terms in the full expression. If we limit discussion to an irrotational fluid for the moment the four terms have particular names in physics The kinetic energy is called the kinetic head or kinetic pressure The internal energy is called the static head or static pressure The field energy is called the gravitational head and is the gravitational potential energy per unit volume. Now I explained, back in posts# 93, 98 & 100, that Zet has not provided enough information to distribute energy added to or taken from the fluid between these three, using Bernoulli. A possible key missing variable is the volumetric flowrate. I have tried to keep this discussion nonmathematical as requested. A final point the static pressure in a fluid should not be confused with the stagnation pressure. The stagnation pressure is the pressure of the fluid if it is brough to rest so all the kinetic energy is turned into internal energy, or if you like the static pressure when the velocity is zero. Edited February 10, 2015 by studiot
cxxLjevans Posted February 11, 2015 Posted February 11, 2015 (edited) [snip] Now I explained, back in posts# 93, 98 & 100, that Zet has not provided enough information to distribute energy added to or taken from the fluid between these three, using Bernoulli. A possible key missing variable is the volumetric flowrate. [snip] Couldn't the missing variables simply be given a name, such as VolFlowRate, and the a solution found using in terms of that name? If not, then pick some value for the missing varable and solve the problem. Maybe you mean the volumetric flow rate at the start of the channel before encountering the airfoil. I guess you could just set that to 1 in some aribitrary units and set the density to 1 (of both airfoil and fluid), and viscosity to 0. IOW, for every term in the Bernoulli equatoin, pick a value and solve the problem. Why wouldn't that work? OOPS. Bernoulli s for steady state. Hmm. So maybe that wouldn't work. Maybe just using some CFD code, such as: http://su2.stanford.edu/ could answer the problem; however I suspect it would require a lot of time to learn how to do that ; however, since this thread seems to be going on for so long, maybe resorting to using a CFD to answer the question would be desirable. -regards, Larry Edited February 11, 2015 by cxxLjevans
Zet Posted February 11, 2015 Author Posted February 11, 2015 (edited) Essay: ""But I find it physically doubtful." Why? Not me; it seems very straightforward, about the extra turbulence, that each would be different, though very difficult to measure accurately enough to see the relatively tiny and chaotic differences." This can be argued a different way … at this point … and in the inverse. In this forum we must assume all of the Laws of Physics as axioms. One of those is energy can change forms but the total amount of energy always remains. (The Law of Conservation of Energy). And so, your intuition must be right and mine must be wrong. It seems strange to me that each one of the differently shaped airfoils all of equal mass and volume all produce the same lesser increase in the overall amount of kinetic energy in the decrease in the swirl in the rising case than in the non-rising case in the fluid in the end, but to you it does not. However, if energy is conserved (… and it must …) and if the only energy change option available is the same decrease in the overall amounts of kinetic energy in the fluid in the end when differently shaped airfoils of equal mass and volume all rise the same distance then they must all have the same lesser increase in the swirl in the rising case than in the non-rising case. Accepting the premises of Physics, one of which is the total amount of energy must always remain the same, then the only way for this to happen is your intuition must be right and my intuition must be wrong. And we can know this to a logical certainty. I’m sure I’ll never be fully comfortable with this answer ( … just to honest … ) but I accept the logic. Thank you all for helping me out with this! Cheers! --- A possible test. Sheldon on tv told a story about Archimedes measuring the amount of gold in the king’s crown by submersing it in a fluid. In a real friction filled world , such as ours , the motion of the fluid in both cases will come to a stop. There will be a decrease in kinetic energy and an equal increase in thermal energy. So, we could measure the temperatures of the two cases in the end (after all of the motion of the fluid has come to a stop) and the one should be higher than the other. Or … mass energy equivalence … after the end the of thought experiment the one airfoil is lowered half the distance and the other airfoil is raised half the distance between them , and so different actual motions but similar that any energy change difference between the two will be negligible if not zero , and then weight the two cases and the one should weight more than the other. So, two possible tests. ? Edited February 11, 2015 by Zet
cxxLjevans Posted February 11, 2015 Posted February 11, 2015 Why does it not occur anyway? (it occurs more so...but baby steps) The airfoil is slowing the wind. Drag is being created. What makes you think it slows the wind more by allowing it to rise? If I do a brake stand in my car, revving the engine at full power, and finally take my foot off the brake, are you going to ask "I wonder where the energy came from to accelerate the car?" Would there be drag if there was no viscosity and no turbulence and no vortices created? I think in this thought experiment, those were the assumptions.
studiot Posted February 11, 2015 Posted February 11, 2015 Blind misapplication of Bernoulli can lead you wildly astray. Here are some more conundrums to ponder. With reference to the diagram. An airflow is proceeding from left to right with steady velocity V and pressure P until it encounters an obstacle. Consider two parcels of air at A and B with such velocity V and pressure P. The 'Bernoulli Explanation' is supposed to run as follows: Parcel A passes over the top of the obstruction, gaining velocity over the stream velocity, Which I have shown as V+ at C, D, and E then finally returning to match the stream at F at velocity V. As a result it is claimed that Bernoulli's theorem says the static pressure must fall below stream pressure at C, D, and E (shown as P-) returning to p at F. In mirror fashion parcel B passes under the obstruction and looses velocity, denoted V- but gains pressure at G, H and I (denoted P+) , returning to V and P at J. So the pressure difference exerts a net upward force on the obstruction, causing it to rise and extract energy from the stream. Question 1) If parcels A and B between them loose some energy to the obstruction, where does the energy come from to rejoin the stream at stream pressure and velocity? Question 2) When parcel A has velocity V+ and pressure P-, then its pressure is less than the pressure of the air immediately above it in the undisturbed stream. So why does it not rise. Question3) Similarly when parcel B has velocity V- and pressure P+ its pressure is greater than the air immediately below it so why does it not fall?
cxxLjevans Posted February 11, 2015 Posted February 11, 2015 Blind misapplication of Bernoulli can lead you wildly astray. Here are some more conundrums to ponder. With reference to the diagram. An airflow is proceeding from left to right with steady velocity V and pressure P until it encounters an obstacle. Consider two parcels of air at A and B with such velocity V and pressure P. The 'Bernoulli Explanation' is supposed to run as follows: Parcel A passes over the top of the obstruction, gaining velocity over the stream velocity, Which I have shown as V+ at C, D, and E then finally returning to match the stream at F at velocity V. As a result it is claimed that Bernoulli's theorem says the static pressure must fall below stream pressure at C, D, and E (shown as P-) returning to p at F. In mirror fashion parcel B passes under the obstruction and looses velocity, denoted V- but gains pressure at G, H and I (denoted P+) , returning to V and P at J. So the pressure difference exerts a net upward force on the obstruction, causing it to rise and extract energy from the stream. Question 1) If parcels A and B between them loose some energy to the obstruction, where does the energy come from to rejoin the stream at stream pressure and velocity? Question 2) When parcel A has velocity V+ and pressure P-, then its pressure is less than the pressure of the air immediately above it in the undisturbed stream. So why does it not rise. Question3) Similarly when parcel B has velocity V- and pressure P+ its pressure is greater than the air immediately below it so why does it not fall? bern3.jpg If the angle of attach is positive, I can see why the velocity below would decrease and the pressure would increase. However, if the angle of attach was 0, and the airfoil was symmetrical, I would think the pressure above would exactly mirror the pressure below. At: http://www.grc.nasa.gov/WWW/k-12/airplane/foil3.html when a symmetric airfoll (an ellipse) is specified with 0 angle of attack, the pressures above an below look the same. The velocities both look same too. Could you explain more the reasoning behind your pressures? Is your airfoil not symmetric or is the angle of attack positive? TIA. -Larry
studiot Posted February 11, 2015 Posted February 11, 2015 Larry, this is not an angle of attack question. You can have any angle you like and any shape you like. This is to try to get folks to think about the fluid as well as the obstruction. Note, I have not said there are no pressure changes. We can measure pressure changes so the logic presented must be flawed.
J.C.MacSwell Posted February 12, 2015 Posted February 12, 2015 Would there be drag if there was no viscosity and no turbulence and no vortices created? I think in this thought experiment, those were the assumptions. There would be no drag, and no lift, if that was the case.
Zet Posted February 12, 2015 Author Posted February 12, 2015 (edited) Larry, The link to SU2 was fascinating (post #108). I was a c (… not a c++ …) programmer back in the mid-nineties. And … I’m guessing some things have probably changed by now. (If you ever need a mean DOS batch file programmer, I’m your guy.) I couldn’t get the NASA link to work (post #112). I reloaded Java and downloaded the Java Control Panel and followed their instructions under “security settings” for Windows 8 and greater but I couldn’t get the Java Control Panel to let me type or paste in the web site URL in the excepted list box. I guess I’m getting old. It looks cool, and perhaps very promising for perhaps resolving this thread. I’ll keep trying to get it to work on my (new but very low end) laptop. Could this piece of software (even if only in beta and even if only approximately) answer/address whether or not there would be the same decrease in the swirl for differently shaped airfoils of equal mass and volume all rising the same distance as opposed to when they don’t rise? That would be exciting. --- And, whether or not it is okay to stipulate a perfectly incompressible fluid in a thought experiment, I don’t know. I trust you that it is. But, if it’s not, this doesn’t change the logic of the issue. In post #83 I wrote: “The fluid is incompressible and has zero viscosity.” But … perhaps … I should have written: “The fluid is as theoretically close to incompressible as possible which is something greater than zero and the viscosity of the fluid is as low as theoretically possible which is greater than zero.” (Again, assuming it’s not simply okay to just stipulate zero compressibility and zero viscosity.) And then I could have gone on (in a thus modified post #83) to say, “And with the compressibility of the fluid so miniscule and so therefore with any internal pressure energy changes so minor, any pressure changes in the fluid itself in this thought experiment can reasonably (or “justifiably”) be ignored for the ease of analysis … and ditto for the impacts of a something greater than zero amount of viscosity.” No? ? --- J.C.MacSwell, Are you saying whether not there is an upward “dynamic buoyant” force on an otherwise neutrally buoyant body that this is not a factor in whether or not it rises? (Or, are you saying it is not part of “lift” … that word? If so, and if the presence of a “dynamic upward buoyant” force is a factor in whether or not the body rises (along with the rest of the factors falling under “lift”), then the body rises, in part, due to the “non-lift” factor (or whatever you want to call it) of an increased upward “dynamic buoyant” force. No?) ? ? Edited February 12, 2015 by Zet
cxxLjevans Posted February 12, 2015 Posted February 12, 2015 (edited) There would be no drag, and no lift, if that was the case. The Euler equation: http://www.grc.nasa.gov/WWW/k-12/airplane/eulereqs.html assume viscosity can be neglected: The Euler equations neglect the effects of the viscosity of the fluid So are you saying the Euler equatons would not predict lift for an airfoil. But then why does a further quote from same page: For some problems, like the lift of a thin airfoil at low angle of attack, a solution of the Euler equations provides a good model of reality. indicates that it would? Hmm, maybe your saying that with Euler turbulence and vortices would be present and they would produce lift? Sorry, I don't understand. Could you explain more? -regards, Larry Larry, The link to SU2 was fascinating (post #108). I was a c (… not a c++ …) programmer back in the mid-nineties. And … I’m guessing some things have probably changed by now. (If you ever need a mean DOS batch file programmer, I’m your guy.) I couldn’t get the NASA link to work (post #112). I reloaded Java and downloaded the Java Control Panel and followed their instructions under “security settings” for Windows 8 and greater but I couldn’t get the Java Control Panel to let me type or paste in the web site URL in the excepted list box. I guess I’m getting old. It looks cool, and perhaps very promising for perhaps resolving this thread. I’ll keep trying to get it to work on my (new but very low end) laptop. Could this piece of software (even if only in beta and even if only approximately) answer/address whether or not there would be the same decrease in the swirl for differently shaped airfoils of equal mass and volume all rising the same distance as opposed to when they don’t rise? That would be exciting. [snip] Hi Zet, I also, initially, could not get the java applet to work. I had to make some adjustments to my FileFox browser prefserences (or something having to do with enableing java applets, but I forget what exaclty) but eventually I got it to work. I've looked at the code (which you can if you download) briefly, and I found code that mentioned viscosity, but viscosity of different forms. I've yet to figure out how to allow the airfoil to rise; hence, I don't know whether it can answer the question. I tried emailing the author listed on the web page (I think it was Thomas Benson) ;however, my mailer said there was no longer such an email address I've also looked at the SU2 docs, but, agian, it willl talk some time to figure out how to use it and graph the results to make the more understandable. I did ask for help here: http://www.cfd-online.com/Forums/cfd-freelancers/148415-effect-rising-hydrofoil-fluid.html and got a "private" reply where some guy offered help for a "reasonable price". I'm not sure I want to pay, but paying might be less work than going on forever here I probably try my hand using su2 (since I've been a c++ programmer for some time) and if it gets too hard, I might pay the money. -regards, Larry Edited February 12, 2015 by cxxLjevans
J.C.MacSwell Posted February 12, 2015 Posted February 12, 2015 The Euler equation: http://www.grc.nasa.gov/WWW/k-12/airplane/eulereqs.html assume viscosity can be neglected: The Euler equations neglect the effects of the viscosity of the fluid So are you saying the Euler equatons would not predict lift for an airfoil. But then why does a further quote from same page: For some problems, like the lift of a thin airfoil at low angle of attack, a solution of the Euler equations provides a good model of reality. indicates that it would? Hmm, maybe your saying that with Euler turbulence and vortices would be present and they would produce lift? Sorry, I don't understand. Could you explain more? -regards, Larry The way I understand it: I think that if the Euler Equations for inviscid flows are set (certain assumptions are made with regard to the wake, or division of flow at the trailing edge) up to allow lift, they invariably include a net drag force on the airfoil. You have circulation about the airfoil, with faster moving flow from above the wing converging with slower moving air from below. In all the set ups that have zero drag (D'alemberts paradox), there is no circulation and the flows converge at zero speed wrt the airfoil...they meet at a second stagnation point and there is no net lift force
studiot Posted February 12, 2015 Posted February 12, 2015 In all the set ups that have zero drag (D'alemberts paradox), there is no circulation and the flows converge at zero speed wrt the airfoil...they meet at a second stagnation point and there is no net lift force The first part is correct, there is no lift without circulation, but I'm not convinced about the second part. There are a lot of possible combinations of compressible/incompressible and viscid / inviscid flows, as well as laminar / turbulent. I'm not sure what you mean by 'they' meet at a stagnation point, streamlines cannot intersect and end at a surface.
J.C.MacSwell Posted February 12, 2015 Posted February 12, 2015 (edited) The first part is correct, there is no lift without circulation, but I'm not convinced about the second part. There are a lot of possible combinations of compressible/incompressible and viscid / inviscid flows, as well as laminar / turbulent. I'm not sure what you mean by 'they' meet at a stagnation point, streamlines cannot intersect and end at a surface. It would look the same as when the streams divide at the stagnation point at the front. There is a point (same point in each stream) that comes to rest (wrt the airfoil) before reaccelerating on it's path/s. There is no boundary layer in this (inviscid) case, so "they" (the streams) meet at a point after going above or below the airfoil. Edited February 12, 2015 by J.C.MacSwell
studiot Posted February 12, 2015 Posted February 12, 2015 (edited) Sorry, your terminology is imprecise. Are you referring to streamlines, pathlines or what? Can you post a link or better a picture.? Incidentally, Bernoulli is blown wide open of you allow stream lines to divide, or join, since one way of stating the theorem is The total energy is constant along a streamline. So if two join the total energy becomes indeterminate. and then it becomes impossible to solve for dynamic and static pressures, velocities etc. If you like I will show how Bernoulli and streamlines should be applied to the example in post 83 Edited February 12, 2015 by studiot
J.C.MacSwell Posted February 13, 2015 Posted February 13, 2015 (edited) Sorry, your terminology is imprecise. Are you referring to streamlines, pathlines or what? Can you post a link or better a picture.? Incidentally, Bernoulli is blown wide open of you allow stream lines to divide, or join, since one way of stating the theorem is The total energy is constant along a streamline. So if two join the total energy becomes indeterminate. and then it becomes impossible to solve for dynamic and static pressures, velocities etc. If you like I will show how Bernoulli and streamlines should be applied to the example in post 83 By stream I mean the flow bounded by two streamlines, or the airfoil surface and a streamline when no boundary later is present (as in the hypothetical of inviscid flow). Edit: trying to post an image but keep getting You are not allowed to use that image extension on this community. Edited February 13, 2015 by J.C.MacSwell
studiot Posted February 13, 2015 Posted February 13, 2015 OK so you mean pathlines, which are a function of time. Streamlines are at one particular instant in time.
J.C.MacSwell Posted February 13, 2015 Posted February 13, 2015 OK so you mean pathlines, which are a function of time. Streamlines are at one particular instant in time. In this case it is steady flow, so they are the same thing.
Essay Posted February 13, 2015 Posted February 13, 2015 (edited) Essay: ""But I find it physically doubtful." Why? Not me; it seems very straightforward, about the extra turbulence, that each would be different, though very difficult to measure accurately enough to see the relatively tiny and chaotic differences." It seems strange to me that each one of the differently shaped airfoils all of equal mass and volume all produce the same lesser increase in the overall amount of kinetic energy in the decrease in the swirl in the rising case than in the non-rising case in the fluid in the end, but to you it does not. However, if energy is conserved (… and it must …) and if the only energy change option available is the same decrease in the overall amounts of kinetic energy in the fluid in the end when differently shaped airfoils of equal mass and volume all rise the same distance then they must all have the same lesser increase in the swirl in the rising case than in the non-rising case. .... It would be strange. I don't think different airfoils produce the "same" effect (lesser increase in ...the decrease in the swirl), even if no lifting occurs. I think you should be able to see the differences "in the exit fluid" for each different airfoil. I'd expect some combination of extra turbulence and decreased velocity would be found in the cases with poor airfoil design ...which could be converted (theoretically) into lifting force in the case of a well-designed airfoil. I wouldn't expect that you could get various poorly-designed airfoils to rise, by using the same amount of 'input' energy, so your second point quoted above should make sense after accounting for the change in input energy needed to get different airfoils to rise. ~[edit] p.s. I think of these statements, “The increase in vertical kinetic energy and in gravitational potential energy is offset by an equal decrease in the kinetic energy of the fluid” (if that, in fact, is how it is offset) & There is an increase in vertical kinetic energy and in gravitational potential energy in the one case that does not occur in the other. And so, if there is an equal decrease in the kinetic energy of the fluid (overall in all of its various motions) in the one case that does not occur in the other then (… ta-da! …) energy is conserved. ...that while these statements are valid, they may raise objections since the statements can't be used to define workable parameters for describing lift, which Bernoulli does do. But for a general understanding how the forces (or the energy) will balance, they seem fairly good. Edited February 14, 2015 by Essay
studiot Posted February 14, 2015 Posted February 14, 2015 In this case it is steady flow, so they are the same thing. With respect I see nothing steady about the flow in the post83 example, which I have offered to discuss.
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