# Fluid mechanics, why does the air flow faster over the wing?

by MaxManus
Tags: faster, flow, fluid, mechanics, wing
P: 5,513
 Quote by DrDu Well, thank you for your pacience. I am enjoying very much this discussion.
Me, too!

I only have a moment to respond, I can do more later:

1) 'Bluff' bodies are the opposite of slender bodies, 'blunt' may be a more descriptive term:

http://www.princeton.edu/~asmits/Bicycle_web/blunt.html

2) I am unaware of a complete and general solution for Re -> 00 flow past a solid body. I haven't pored over the literature lately, tho.

3) We haven't considered airfoils of finite length yet- it turns out the dominant mechanism of vortex shedding from a finite airfoil is from the ends, not along the length.

http://pilotsweb.com/principle/art/v_sheet.jpg

More later...
P: 3,593
 Quote by Andy Resnick 1) 'Bluff' bodies are the opposite of slender bodies, 'blunt' may be a more descriptive term: http://www.princeton.edu/~asmits/Bicycle_web/blunt.html 2) I am unaware of a complete and general solution for Re -> 00 flow past a solid body. I haven't pored over the literature lately, tho. 3) We haven't considered airfoils of finite length yet- it turns out the dominant mechanism of vortex shedding from a finite airfoil is from the ends, not along the length. http://pilotsweb.com/principle/art/v_sheet.jpg More later...
ad 1) ok, so the contrary of streamlined
ad 2) naively, I thought that inviscible is the same as an ideal fluid. However, the limit Re-> infinity (or nu->0) is a singular limit and so it will not coincide with the nu=0 case.
However, I have no good idea how this will lead to turbulence.
ad 3) that I know. In 3d, the vortex lines are closed.

But you still avoided my primary question. Is the velocity behind a streamlined object continuous or not?
P: 5,513
 Quote by DrDu But you still avoided my primary question. Is the velocity behind a streamlined object continuous or not?
I don't know about 'avoided', but in general, no. The velocity is not continuous in the general case.

Edit: Schlichting's book "Boundary-layer theory" has some detailed information. To summarize, for flow past a circular cylinder, boundary layer separation occurs at 4 < Re < 40, and the separated flow remains laminar until around Re > 10^5, at which point it becomes turbulent separated flow.

After laminar boundary layer flow has separated from the body, the flow evolves into free shear layers downstream (the wake). In the limit Re -> 00 , the free shear layers resolve to line and point discontinuities.
Mentor
P: 22,280
My own understanding proved somewhat misguided in a recent thread and through that, I figured a simpler way to explain it:

The airfoil provides an obstruction that the air has to travel around, which squeezes it - but due to conservation of mass and the fact that it doesn't compress much, it has to travel a lot faster to get past the bigger obstruction (the top surface of the airfoil). In other words, it's an inside-out Venturi tube.
 According to the laws governing fluid dynamics, a fluid's velocity must increase as it passes through a constriction to satisfy the conservation of mass, while its pressure must decrease to satisfy the conservation of energy. Thus any gain in kinetic energy a fluid may accrue due to its increased velocity through a constriction is negated by a drop in pressure. An equation for the drop in pressure due to the Venturi effect may be derived from a combination of Bernoulli's principle and the continuity equation.
http://en.wikipedia.org/wiki/Venturi_effect
P: 5,513
 Quote by russ_watters in other words, it's an inside-out venturi tube.
like! :)
HW Helper
P: 7,107
 Quote by russ_watters airfoil ... it's an inside-out Venturi tube.
Part of an ongoing debate between web sites. Nasa link explaining that an airfoil does not behave like a Venturi, but doesn't offer an alternate explanation.

http://www.grc.nasa.gov/WWW/K-12/airplane/wrong3.html

In a Venturi, other than transitions through constriction points, the flow is uni-directional. In the case of an airfoil the direction and speed of flow varies with distance from the airfoil, following the boundary layer in the immediate vicnity of the airfoil, and as vertical distance from the airfoil increases, the flow is more downwards and perpendicular to the surface of the airfoil.

In a Venturi, the pipe restricts the flow of air. On a cambered airfoil (or the cambered stagnation zone on a flat airfoil), there's a Coanda like effect where the air tends to follow the cambered surface to fill in what would otherwise be a void.
P: 5,513
 Quote by rcgldr http://www.grc.nasa.gov/WWW/K-12/airplane/wrong3.html
I think this model is too simple, and gives an incomplete answer.

First, they only consider half of the flow: the accelerating part, as the flow is constricted. They neglect the second half, when the flow expands. Second, this is inviscid flow.

Adding viscosity back makes the second half of the flow region more important, because the second half of the flow region takes fluid from a low pressure region back to a high pressure region- the air is moving *against* the pressure gradient.

For inviscid flow, this is no problem- no energy is lost during the acceleration phase, and so all the energy is recovered as the flow slows down. Viscosity results in the loss of some energy. Viscous effects are important primarily in the boundary layer.

Thus, any fluid that has lost too much energy by viscous dissipation is then unable to move back to the high pressure region behind the airfoil, resulting in the boundary layer separating from the airfoil. And this is what is observed- boundary layer separation at some angle of attack.
Mentor
P: 22,280
[QUOTE=rcgldr;2943946]Part of an ongoing debate between web sites. Nasa link explaining that an airfoil does not behave like a Venturi, but doesn't offer an alternate explanation.

http://www.grc.nasa.gov/WWW/K-12/airplane/wrong3.html
 In a Venturi, other than transitions through constriction points, the flow is uni-directional. In the case of an airfoil the direction and speed of flow varies with distance from the airfoil, following the boundary layer in the immediate vicnity of the airfoil, and as vertical distance from the airfoil increases, the flow is more downwards and perpendicular to the surface of the airfoil.
I've never heard that before. Do you have any links to diagrams that show flow that isn't nearly parallel to the nearest airfoil surface?
 In a Venturi, the pipe restricts the flow of air. On a cambered airfoil (or the cambered stagnation zone on a flat airfoil), there's a Coanda like effect where the air tends to follow the cambered surface to fill in what would otherwise be a void.
You can make a venturi tube have flow separation if the curve isn't right and the coanda effect would play a role in that as well. Either way, though, the basic Venturi effect isn't the only thing affecting the lift on a wing, but it does explain a lot of it and the question in the OP wasn't about lift, it was just about the speeding up of the air. Anyway, from the link you posted:
 The theory is based on an analysis of a Venturi nozzle. But an airfoil is not a Venturi nozzle. There is no phantom surface to produce the other half of the nozzle. In our experiments we've noted that the velocity gradually decreases as you move away from the airfoil eventually approaching the free stream velocity. This is not the velocity found along the centerline of a nozzle which is typically higher than the velocity along the wall. [emphasis added]
The fact that it is inside-out is obviously a difference and since it is just an analogy, it can't be expected to perform perfectly, but I'm wondering if the size of the Venturi has any impact on the velocity profile.....I think it must. Basically, it seems to me that the further away the two halves of a two-dimensional Venturi, the more like an airfoil it would behave. "Typically" doesn't imply "always" to me.
 The Venturi analysis cannot predict the lift generated by a flat plate. The leading edge of a flat plate presents no constriction to the flow so there is really no "nozzle" formed.
Agreed. The Venturi analogy would not seem to work for a flat plate. The flat plate minimizes/eliminates that component and I would think makes downward deflection and coanda effect the only active parts.
 This theory deals with only the pressure and velocity along the upper surface of the airfoil. It neglects the shape of the lower surface.
Agreed - but the OP only asked about an example (a half circle) where the lower surface is essentially a zero angle of attack flat plate.

In short, there are few if any scientific theories that can be explained in 5 words. They are always more complicated than that and in the case of lift, there are several different principles at work. I don't think it is fair to say that the idea is wrong, just incomplete for fully explaining lift. More importantly, few people have a problem with the idea that a flat plate with a positive angle of attack can deflect air downards and create lift regardless of what is going on above it. It seems to me that understanding the velocity increase is the biggest problem and this explanation isolates and deals with only that piece.
HW Helper
P: 7,107
 Quote by rcgldr In a Venturi, other than transitions through constriction points, the flow is uni-directional. In the case of an airfoil the direction and speed of flow varies with distance from the airfoil
 Quote by russ_watters Do you have any links to diagrams that show flow that isn't nearly parallel to the nearest airfoil surface?
If you look at any diagram of streamlines over an airfoil, from the point of upwash to near the peak of the cambered surface, they constrict as velocity increases and pressure decreases (mass flow within streamline is constant). The inner streamline follows the airfoil surface (actually the boundary layer), but as you get further away from the airfoil surface, the outer streamlines have less of a upwash component and more of a component perpendicular to the leading portion of the airfoil surface where the streamlines contrict, otherwise a void between streamlines would be created. Once past the peak, velocity decreases, pressure increases, the streamlines expand, and as you get further away from the airfoil, the less the streamlines follow the surface. The net result is that the higher above the airfoil, the less the flow follows the surface of the airfoil, eventually reaching a point above the airfoil where the air is virtually unaffected.

This web site includes a photo as well as an animation of how a parcel of air is affected by a wing passing through it from an air based frame of reference.

http://www.avweb.com/news/airman/183261-1.html
 Sci Advisor P: 3,593 Andi, just a question. When you say inviscid, do you mean lim Re -> infinity or nu=0? I think the case nu=0 gives Euler equations which are rather trivial to solve in comparison with the Navier Stokes Equations for low Re. I see no reason why the limit should converge to the latter. However, sometimes inviscid and ideal are used interchangeably.
P: 5,513
 Quote by DrDu Andi, just a question. When you say inviscid, do you mean lim Re -> infinity or nu=0? I think the case nu=0 gives Euler equations which are rather trivial to solve in comparison with the Navier Stokes Equations for low Re. I see no reason why the limit should converge to the latter. However, sometimes inviscid and ideal are used interchangeably.
That's a very insightful question.

Hopefully you can see that setting $\mu = 0$ is very different than lim Re -> $\infty$. I am unaware of what the 'official convention' is, and in fact one way I distinguish good texts and bad is based on how clearly they make this distinction.

AFAIK, "Inviscid" simply means that viscosity effects are unimportant. While this may be true for the far-field velocity distribution of a fluid, it neglects the boundary layer. Setting the viscosity = 0 negates the boundary layer, while lim Re -> $\infty$ maintains it.
P: 5,513
 Quote by Andy Resnick for a flat surface, the thickness 'd' of the boundary layer a distance 'x' from the leading edge is given by d/x =5* Re^0.5, where Re is the local Reynolds number at the distance 'x'.
I erred- the thickness goes as Re^-0.5, not Re^0.5.
 P: 15 Could one of you physics gurus answer this one. There was an answer to this question posted early in this thread, but I'd like to make sure it reflects the consensus opinion. If you had an airfoil that was half a cylinder, with the flat bottom at a zero angle of attack (that is, parallel to the direction of flow), would lift be generated? It wouldn't even have to be an entire half a cylinder; any section of a cylinder would work for the thought experiment. The point being, if you had an airfoil that was symmetrical front to back, and at a zero angle of attack, would lift be generated? The NASA program "foilsim" says lift would be generated. If this is true, I would love an explanation. I understand the mechanics of air being deflected down by both the top and the bottom of a wing, and the whole idea of "turning" air flow. I am curious whether the Bernoulli effect alone creates lift.
 P: 297 My post is a year old, and I don't remember the equations but yes the there will be generated lift with zero angle of attach. But Im not sure of your last question. If you agree on the bernoulli effect, then why not agree in an ideal example where you haven't taken the mass of the into the problem?
 HW Helper P: 7,107 One of the links from one of my earlier posts is broken now. There's an archived version of it, but most of the images are gone: archive_/~weltner/Mis6/mis6.html As far as a true half cylinder goes, you might be able to generate the 2d coordiantes of a half cylinder based air foil and use XFOIL to calculate the polars (lift and drag versus AOA, Re, ...). http://web.mit.edu/drela/Public/web/xfoil A true half cylinder would generate a lot of drag. If the air foil was made of a thin slice of the upper part of a cylinder, a flat bottom and a circular arc of a few degrees on the top, it might work in theory, but actual examples appear to need a non-zero angle of attack to produce lift. A frisbee is similar to this, except the bottom surface is hollow so is similar in shape to the upper surface, and from what I recall, a frisbee needs an angle of attack to generate lift. An aerobie is a flying ring that combines a special triangular outer spoiler rim with an airfoil section similar to a thin slice of a cylinder for the inner shape, but it still needs some non-zero angle of attack to produce lift. The spoiler rim is designed to keep the center of lift near the center of the ring over a reasonable range of angle of attacks to eliminate pitch torque which would result in a roll due to gyroscopic precession. http://aerobie.com/about/ringscientificpaper.htm
 P: 15 Thanks! I can easily see how a slice of a cylinder would create lift if it were at a non zero angle of attack. The angled bottom would deflect air down, and the curved top would turn the upper air stream down. The net effect would absolutely be to push air down. Probably the optimum angle of attack would be with the initial portion of the upper surface tangential to the airflow, but that's just a guess. I just cannot see how lift would be generated at a zero angle of attack. Sure, the air following the rear portion of the curve would be turned down and would lift the back portion of the cylinder section. But the exact opposite effect would take place at the front of the cylinder section. It seems to me that the net effect would be to only apply torque to the cylinder section. NASA's Foilsim program seems to say that a cylinder section would have lift at a zero angle of attack. But that is a simulator, and may or may not be completely correct. Here's the "Misinterpretations of Bernoulli's Law" paper you were attempting to link to above: http://plato.if.usp.br/2-2007/fep0111d/Bernoulli.pdf
 Engineering Sci Advisor HW Helper Thanks P: 7,101 Bernouilli's equation is just Newton's laws of motion applied to a fluid. It tells you how the changes of fluid velocity around the body are related to the changes in pressure over the surface of the body, but it doesn't tell you anything about why the velociity and pressure distributions are the way they are. The thing that "causes" the lift is the fluid viscosity and the effect it has on the boundary layer of the flow. Bernouilli's equation doesn't included any viscosity terms, so it can't possibly tell you about that. In fact it there was no viscosity, there would be no lift and drag forces on a body of any shape, at any angle of attack (and there would also be no boundary layer, and no turbulence in the flow). I had a quick look at the NASA foilsim web pages. There was some theory there but I couldn't find a complete explanation of what the program does. But from what it did say, you are quite right to question whether the results would be correct for something that doesn't "look like a normal aerofoil". Many real-world computer methods in fluid dynamics only work well in particular situations. Competely "general purpose" computational fluid dynamics software may take too long to run. It is true the half-cylinder is symmetrical front-to-back, but the airflow pattern around it is not symmetrical, because of the air viscosity. The boundary layer becomes thicker as the air flows over the body, and at some point it will probably separate from the surface. The idea of "angle of attack" is not obvious for something like a half cylinder. If the flat surface was parallel to the far-field airflow, the stagnation point that defines the "leading edge position" would not be at the corner between the flat and curved surfaces, it would be some point along the curved top surface. In that sense, the airflow has a non-zero angle of attack. Alternatively, if you bisected the 90-degree angle between the curved and flat surfaces, you could argue that the angle of attack was actually -45 degrees.
P: 15
 Quote by AlephZero Bernouilli's equation is just Newton's laws of motion applied to a fluid. It tells you how the changes of fluid velocity around the body are related to the changes in pressure over the surface of the body, but it doesn't tell you anything about why the velociity and pressure distributions are the way they are. The thing that "causes" the lift is the fluid viscosity and the effect it has on the boundary layer of the flow. Bernouilli's equation doesn't included any viscosity terms, so it can't possibly tell you about that. In fact it there was no viscosity, there would be no lift and drag forces on a body of any shape, at any angle of attack (and there would also be no boundary layer, and no turbulence in the flow). I had a quick look at the NASA foilsim web pages. There was some theory there but I couldn't find a complete explanation of what the program does. But from what it did say, you are quite right to question whether the results would be correct for something that doesn't "look like a normal aerofoil". Many real-world computer methods in fluid dynamics only work well in particular situations. Competely "general purpose" computational fluid dynamics software may take too long to run. It is true the half-cylinder is symmetrical front-to-back, but the airflow pattern around it is not symmetrical, because of the air viscosity. The boundary layer becomes thicker as the air flows over the body, and at some point it will probably separate from the surface. The idea of "angle of attack" is not obvious for something like a half cylinder. If the flat surface was parallel to the far-field airflow, the stagnation point that defines the "leading edge position" would not be at the corner between the flat and curved surfaces, it would be some point along the curved top surface. In that sense, the airflow has a non-zero angle of attack. Alternatively, if you bisected the 90-degree angle between the curved and flat surfaces, you could argue that the angle of attack was actually -45 degrees.
Thanks for the informative post. (Aside: So an airfoil would not generate lift in liquid helium?!?)

I chose a half cylinder for the thought experiment, but really it is a general question: Is lift generated by an airfoil that is (1) symmetrical front to back, and (2) at a zero angle of attack?

In a similar question, does wind blowing across a plain "lift" a small hill it encounters, assuming that the wind speed and hill size and shape are such as to create no turbulence? I have seen discussion that claims that there is higher pressure on the windward and leeward sides of the hill, and lower pressure on the peak. If true, this would help explain why sand dunes exist, or at least why waves are formed when air blows across water. But is the net effect on the hill upward?

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