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

[MaxManus]

Why does the air flow faster over the wing? In my fluid mechanics course we used streamlines and a half body to show that the air flow faster over a half circle, but I do not understand why it does so.

 

[rcgldr]

An older thread asking the same question for a normal wing:

https://www.physicsforums.com/showthread.php?t=430717

In the case of a half cylinder (flat on bottom, circular on top) "airfoil", it wouldn't generate lift unless there was a large angle of attack. You'd have higher pressure, slower speed upwash at the front, then air would accelerate normal (perpendicular) to the flow if it's mostly a laminar flow and follows the upper surface near the peak to avoid creating a void (a Coanda like effect), or if's turbulent, a vortice would form in order to avoid creating a void, depending on the speed, and rate of curvature of the half cylinder (viscosity is also a factor, but I'm assuming this is air, not helium). In the laminar flow case, towards the rear of the upper surface, pressure would increase and the flow would decrease in speed.

A website with a casual explanation of how wings work:

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

The half cylinder laminar flow case would be similar to the hill case in section 4.3 of the web page linked to below, and the turbulent flow case would be similar to the roof case further down on that same web page. Note that in the case of the pipe example on that web page, even if the leading edge of the pipe creates a vortice,the pressure is still lower than ambient across the opening, and some carburetors use a tube protuding into a venturi pipe with the opening perpendicular or aft of the relative flow to reduce pressure futher still to draw more fuel into the flow, with an intended turbulent vortice to also atomize the fuel. The old Flit Gun pump sprayer used a similar method, a thin vertical pipe positioned with it's open end within the flow generated by a hand pump with a narrow outlet.

http://user.uni-frankfurt.de/~weltner/Mis6/mis6.html

 

[MaxManus]

Thanks.

 

[DrDu]

The flow at the end of the wing has to fulfill the so called Kutta Joukowski condition. I. e., the steam lines from above and below the wing have to be parallel and of equal velocity. That determines the degree of circulation around the wing and the difference of velocity below and above wing.

 

[K^2]

No, it doesn't. Kutta condition merely tells you that the streams meet, but because of the separation layer between them, it doesn't tell you anything about their relative velocities. In other words, Kutta condition doesn't tell you that the stream over the top of cambered wing has to travel faster.

On the other hand, the fact that air over the wing travels faster isn't directly related to the lift. It does tell you that you have positive circulation, which means the air is deflected downwards, which is how you really compute lift, because you can now use momentum conservation (Kutta Joukowski Theorem). But lift itself is caused by pressure at the boundary, and the air speed at the boundary is zero all around the wing. So you cannot use it as an explanation of lift.

 

[MaxManus]

On the other hand, the fact that air over the wing travels faster isn't directly related to the lift. It does tell you that you have positive circulation, which means the air is deflected downwards, which is how you really compute lift, because you can now use momentum conservation (Kutta Joukowski Theorem). But lift itself is caused by pressure at the boundary, and the air speed at the boundary is zero all around the wing. So you cannot use it as an explanation of lift.

Thanks, but now I'm not sure if I understand. It might be that I misunderstood my fluid mechanic course, but I believe we used the fact that the air flows faster over the wing and the bernoulli equation to show that there will be a lift. I have read the links and see that there are other things that create lift too, but is it wrong to say that that faster flow over the wing creates lift?

 

[DrDu]

No, it doesn't. Kutta condition merely tells you that the streams meet, but because of the separation layer between them, it doesn't tell you anything about their relative velocities. In other words, Kutta condition doesn't tell you that the stream over the top of cambered wing has to travel faster.

On the other hand, the fact that air over the wing travels faster isn't directly related to the lift. It does tell you that you have positive circulation, which means the air is deflected downwards, which is how you really compute lift, because you can now use momentum conservation (Kutta Joukowski Theorem). But lift itself is caused by pressure at the boundary, and the air speed at the boundary is zero all around the wing. So you cannot use it as an explanation of lift.

I am not a specialist in fluid mechanics.
Which separation layer? Wouldn't a difference of speed after the wing lead to turbulences which finally carry away circulation until the speeds become equal?

I cannot follow your argumentation that the lift cannot be calculated from the air speed as the air speed vanishes at the wing. The vanishing of the velocity is due to friction.
Bernoulli only applies when friction can be neglected. But this is the case already at distances small compared with the dimesnion of the wing.

 

[Andy Resnick]

No, it doesn't. Kutta condition merely tells you that the streams meet, but because of the separation layer between them, it doesn't tell you anything about their relative velocities. In other words, Kutta condition doesn't tell you that the stream over the top of cambered wing has to travel faster.

On the other hand, the fact that air over the wing travels faster isn't directly related to the lift. It does tell you that you have positive circulation, which means the air is deflected downwards, which is how you really compute lift, because you can now use momentum conservation (Kutta Joukowski Theorem). But lift itself is caused by pressure at the boundary, and the air speed at the boundary is zero all around the wing. So you cannot use it as an explanation of lift.

This is correct. The Kutta-Zhukovsky theorem states that a vortex of circulation G, moving in a uniform fluid of density p with velocity v_inf (the subscript is there for technical reasons), produces a force directed perpendicular to the direction of motion and the axis of the vortex of magnitude p*v_inf*G per unit length.

This theory neglects vortex tip shedding- the free vortices often seen behind airplanes.

Going to a finite wingspan, the problem was solved by the Prandtl-Lanchester lifting line theory, and from there, found an interesting application in Fletter's rotorship.

http://en.wikipedia.org/wiki/Lifting-line_theory

http://en.wikipedia.org/wiki/Rotor_ship

 

[DrDu]

Dear Andy,

I don't doubt the Kutta Joukowski theorem.
The presence of circulation explains different velocities above and below the wing and by Bernoulli leads also to a difference in pressures. So there is no contradiction with the statement that it is the higher velocity above the wing which is responsible for lift as stated by MaxManus.

The amount of circulation is fixed by the Kutta condition. My point is that after the wing the velocity of the fluid elements passing above and below the wing at the line (or surface) where they meet has to be equal as pressure must be a continuous function.

 

[Andy Resnick]

DrDu,

I'm having a hard time parsing your post- the Kutta condition requires the velocity to be *bounded*, not *continuous*. I'm not sure how well I can demonstrate it, but a derivation can be found in Saffman's "Vortex Dynamics". AFAIK, the relationship between a discontinuous velocity and accelerated flow past a wing pertains to shock waves and stability, but it's been a while since I went through the detailed derivation.

 

[DrDu]

DrDu,

I'm having a hard time parsing your post- the Kutta condition requires the velocity to be *bounded*, not *continuous*.

Yes, that is true.
I doubt on physical grounds that in sub-sonic stationary systems a shock wave with discontinuous velocities or pressure can pertain.
A more formal argument is the following:
For some special wing shaped bodies the velocity field around the body can be obtained from the fluid flow around a cylinder via a conformal transformation. Now the flow around a cylinder is an analytical function of r and theta for all values of these parameters and a conformal transformation cannot introduce a line of discontinuities.

For the solutions for a cylinder see 3.93 and 3.94 (alas in German only) in

The equality of the velocities behind the wing is explicitly derived in that script from the Kutta condition in node35 of that script.

 

[Cleonis]

Thanks, but now I'm not sure if I understand. It might be that I misunderstood my fluid mechanic course, but I believe we used the fact that the air flows faster over the wing and the bernoulli equation to show that there will be a lift. I have read the links and see that there are other things that create lift too, but is it wrong to say that that faster flow over the wing creates lift?

Faster flow over the wings is a necessary part of the whole, but it shouldn't be thought of as the key factor.

The most direct example of a setup producing lift is a jet engine or rocket engine that directs its exhaust downward. A man with a jetpack strapped on can hover. If the exhaust is directed in horizontal direction, to the left, then you will accelerate to the right; the action-reaction principle. If you direct the exhaust downward you are using the propulsion to hover (or climb).

(For completeness: another category of lift is buoyant lift, as in the case of a hot air balloon. In the following discusssion I refer to 'all forms of lift', meaning 'all forms of non-buoyant lift', but that's such a mouthful.)

All forms of lift have in common that mass is accelerated downward. In the case of a jetpack exhaust is accelerated downward, helicopter blades accelerate air mass downward, and in the case of an aircraft wing the air mass flowing around the wing is deflected downward. Action-reaction.

As you know, the crudest wing of all is just a flat board, at an angle of attack. That is a horribly inefficient wing design, but it will perform as a wing; it will deflect air mass downward. The reason wings are design for good flow characteristics is of course to optimize performance. As you know, with a flat board air mass will have a strong tendency to separate from the top of the surface. Performance is all about getting the air to not separate from the top surface. That is how you maximize lift given the total available wing area.Incidentially, in the case of buoyancy (balloon, airship) the lift does come fundamentally from a force difference; in a state of positive buoyancy more force is exerted on the underside of the craft than on the upper side.

Jetpacks, helicopters, and aircraft wings generate lift by accelerating air mass downwards; the lift comes from action-reaction.

 

[Andy Resnick]

The flow at the end of the wing has to fulfill the so called Kutta Joukowski condition. I. e., the steam lines from above and below the wing have to be parallel and of equal velocity. That determines the degree of circulation around the wing and the difference of velocity below and above wing.

I cannot follow your argumentation that the lift cannot be calculated from the air speed as the air speed vanishes at the wing. The vanishing of the velocity is due to friction.
Bernoulli only applies when friction can be neglected. But this is the case already at distances small compared with the dimesnion of the wing.

The amount of circulation is fixed by the Kutta condition. My point is that after the wing the velocity of the fluid elements passing above and below the wing at the line (or surface) where they meet has to be equal as pressure must be a continuous function.

Yes, that is true.
I doubt on physical grounds that in sub-sonic stationary systems a shock wave with discontinuous velocities or pressure can pertain.

DrDu,

I'm still having trouble understanding your questions/comments/points. Does this have to do with the origin of lift (generated by a wing), the origin of circulation/vortices, detailed solutions to either, general solutions to either,... ?

 

[rcgldr]

After the wing the velocity of the fluid elements passing above and below the wing at the line (or surface) where they meet has to be equal as pressure must be a continuous function.

In the real world, depending on air foil, angle of attack, wing loading, ..., the flows aren't quite equal, there's turbulence and vortices at the trailing edge, as differences in velocity and pressure exist at the trailing edge where the flows merge. Wing tip vortices also significantly interfere with the flow on a typical aircraft (high end gliders with 80+ foot wing spans would be an exception).

Getting back to the OP, even a flat airfoil with a reasonable angle of attack will end up drawing more air downwards from above than pushing air downwards from below, so the greater difference from ambient pressure above the wing results in a higher maximum speed of the air above.

On a side note, in the case of small balsa models that glide at low speeds, nearly flat air foils aren't that inefficient.

 

[DrDu]

Dear Andy,

well, for me it's also not easy to parse your posts, e.g. expressions like "vortex tip shedding".

In my first post in that thread I wanted to point out the importance of the Kutta condition in understanding the lift of a wing. I think we both can agree on that.

My further comments referred mainly to what K^2 wrote and of which you said "This is correct".
I wondered especially about the "separation layer" and how it may or may not separate sheets of air of different velocity.
You also later wrote:
"AFAIK, the relationship between a discontinuous velocity and accelerated flow past a wing pertains to shock waves and stability, but it's been a while since I went through the detailed derivation." I still cannot believe that there is a discontinuity in the velocity (below velocity of sound) and tried to find some arguments against it. I would be interested in some further explications of your point on that.

 

[Andy Resnick]

DrDu,

That's fair- ok, let's back up a bit- all the way back to the motion of a rigid object through a viscous fluid.

The motion of a sphere/ellipsoid was found a long time ago (say 1845) because the solution is well behaved: in the frame of the ellipsoid, the fluid velocity decays to a constant as 1/r^3, so the disturbance created by the ellipsoid is very localized. Not so with a cylinder- the translation of a cylinder with constant velocity (perpendicular to the cylinder axis) though an infinite fluid creates a disturbance in the fluid that grows without bound.

Oseen linearized the relevant equations and found the drag force is finite- the derivation is in Lamb's book, but we summarized the final results here:

http://www.ncbi.nlm.nih.gov/pubmed/17526573

We'll get to wings shortly- first, let's discuss 'boundary layer separation'. When fluid moves past a body, the no-slip condition ensures that the fluid in contact with the solid moves at the speed of the solid. Thus, there must be a velocity gradient to match the no-slip fluid motion with the far-field fluid motion.

Now, the no-slip condition has a very fuzzy pedigree- it's required to keep the stress finite, but it rules out observed phenomena like wetting. There's lots of material out there trying to reconcile the two, and since it's not that relevant, I'll pass over this point without comment.

Boundary layer flow, which is the region of fluid close to the object, is also a fuzzy concept. There's no clear demarcation between the boundary layer and far-field velocity. A key concept is the Reynolds number- the ratio of inertia to viscosity. Small Reynolds numbers correspond with highly viscous flow (eg Stokes flow, the result above) and high Reynolds numbers correspond to Bernoulli's equation and Euler's equation.

Prandtl was the one who thought of the boundary layer- it's a region of flow where the Reynolds number is important. For fluids with low viscosity (gases), this region of flow is found near the surface of a rigid object. Furthermore, as the Reynolds number increases, the boundary layer thickness decreases and the (local) shear rate is larger.

The usual definitions for boundary layer thickness can be found easily online, but to summarize the result 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'. Steven Vogel's "Life in Moving Fluids" covers this topic in an incredibly clear chapter.

Boundary layer separation is a feature of viscous flow past a 'bluff' body, and occurs when the velocity within the boundary layer changes sign from positive (with respect to the body) to negative. The details of where this occurs is complicated, but has to do with the stress gradient. Separation of the boundary layer leads to the phenomenon of 'wake' behind a body moving though a fluid. There is also the phenomenon of 'stall'. When the boundary layer separates, the flow pattern becomes so diffrent from laminar flow that the results from simple inviscid+boundary layer flow have to be abandoned completely.

This is all fine, but what does it have to do with an airfoil? The conformal mapping of a cylinder to an airfoil results in a solution that (excluding stall) has negligible boundary layer separation. However, the *inviscid* flow pattern- a limiting flow pattern- around a bluff body has, AFAIK, remained resistant to solution by either theory, calculation, or experiment.

So now what? Now we introduce the idea of vorticity and circulation. Consider the flow of an inviscid incompressible fluid past a body started from rest by conservative forces. The motion is not unique, unless the flow is continuous. If the flow is continuous, the Helmholtz-Kelvin theorem states that the velocity is the gradient of a single-valued potential. If discontinuous flow is allowed, by for example allowing the creation of a vortex sheet from separated flow, the unsteady flow no longer has a unique solution.

This has the best images I could find right now:

http://www.biketechreview.com/aerodynamics/misc/466-skinsuits-and-boundary-layers?start=2

Vortices have 'circulation' associated with them, and so again, by the Helmholtz-Kelvin theorem (circulation is conserved), there is circulation associated with the wing to compensate for the generation of a vortex sheet- the vortices detach from the wing, forming a "von Karmen vortex street":

http://www.simerics.com/animation/karman_vortex_street_experiment.gif

The Kutta condition comes into play simply because an airfoil cross-section is a conformal map of a circle.

There's a lot here; let me pause and ask you what makes sense/what does not make sense

 

[rcgldr]

A lot of this seems to be going beyond what the OP was asking.

For a simple answer, air is sort of like a rope, pulling it has more effect than pushing it. If an object is pushing the air (advancing), a lot of the affected air will tend to flow around the object rather than getting pushed forwards by the object. However if an object is pulling the air (retreating), the air is forced to follow the object in order to fill in what would otherwise be a void.

In the case of a wing, the upper surface is pulling air downwards, and the lower surface is pushing air downwards. The flow that is "going around" the wing occurs just ahead of the wing and around the wing tips. Because the pressure above is lower and the pressure below is higher, the air flow is diverted upwards ahead of the wing. The air "going around" the wing decreases the overall lift, but the end result is that for a typical wing, there's more pressure reduction above a wing than pressure increase below the wing, and this corresponds to a greater maximum speed of flow above the wing.

 

[DrDu]

I think I have a good understanding on how a boundary layer forms. In old times I even wrote a paper on multiple scales techniques (although in QM, not fluid mechanics).

I also saved a copy of Prandtls collected works from the library which is here on my shelf.

This is all fine, but what does it have to do with an airfoil? The conformal mapping of a cylinder to an airfoil results in a solution that (excluding stall) has negligible boundary layer separation. However, the *inviscid* flow pattern- a limiting flow pattern- around a bluff body has, AFAIK, remained resistant to solution by either theory, calculation, or experiment.

I am confused. I thought the *inviscid* flow pattern to be known, at least for a cylinder?
What is a "bluff" body? I can't find a really fitting translation into German.

I always thought that after a wing ideally there is no von Karman street, only the vortex formed when starting? But maybe that is an (over-)idealization.
In a von Karman sheet, the direction of the vortices alternates. So I would expect that at least on the mean the velocity above and below the street is equal.

Well, thank you for your pacience. I am enjoying very much this discussion.

 

[Andy Resnick]

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...

 

[DrDu]

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?

 

[Andy Resnick]

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.

 

[russ_watters]

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

 

[Andy Resnick]

in other words, it's an inside-out venturi tube.

like! :)

 

[rcgldr]

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.

 

[Andy Resnick]

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.

 

[russ_watters]

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.

 

[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

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 website 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

 

[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.

 

[Andy Resnick]

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.

 

[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.

 

[SDEric]

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.

 

[MaxManus]

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 I am 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?

 

[rcgldr]

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

 

[SDEric]

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

 

[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.

 

[SDEric]

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?

 

[rcgldr]

Thanks for the informative post. (Aside: So an airfoil would not generate lift in liquid helium?

I'm not sure if it's impossible to deflect a flow for even a near inviscid fluid if the flow speed is fast enough that momentum prevents the pressures from equalizing. If you dropped a very light (hollow) sphere in a very tall column of liquid helium would it accelerate the same as it would in a vacuum? To eliminate the issue of pressure increasing with depth, consider accelerating a charged hollow sphere with an electrical field through a long container filled with liquid helium, free of gravitational effects, and compare this to the sphere accelerating in a vacuum.

Is lift generated by an airfoil that is (1) symmetrical front to back, and (2) at a zero angle of attack?

Symmetrical means the same shape above and below. Such a shape should not generate lift at zero angle of attack (assuming that it's isn't a transitory state with angle of attack changing over time).

In a similar question, does wind blowing across a plain "lift" a small hill it encounters ... not turbulent ... is the net effect on the hill upward?

Not sure, laminar (non-turbulent) flow is rare except at low speeds. For an air foil with a flat bottom and a curved top and zero angle of attack (the flat bottom horizontal), if the peak of the curve is near the leading edge, you get positive lift, if it's near the trailing edge you get negative lift (and a lot of drag, it would be similar to an inverted wedge). At somepoint in between, there would be zero lift, but I don't know if this corresponds to having the peak exactly at the middle.

 

[RandomGuy88]

If you dropped a very light (hollow) sphere in a very tall column of liquid helium would it accelerate the same as it would in a vacuum?

No. An object that is accelerated in any fluid (even an inviscid fluid) will experience drag due to the "added mass" effect. Basically if the object is accelerating in a fluid it must also be accelerating the fluid itself. So if you apply a given force F to an object of mass m it will not accelerate at a rate of F/m because you must also consider the mass of the fluid that is accelerated.

So the sphere dropped into a column of liquid helium would a lower acceleration than a sphere dropped in a vacuum.

In addition to this there is also buoyancy to consider.

 

[daqddyo1]

MaxManus - Are you still here?
The air moves faster over the rounded top suface of the wing simply because it has to travel further than the air that travels below the wing. When the air at the front (leaing edge) of the wing separates into two parts, those parts must meet at the trailing edge of the wing at the same time. Think about what would happen if this were not the case.

I have played with a perfect airfoil in a wind tunnel. You get no lift if the bottom edge of the wing is parallel to the airflow. As soon as you present an angle of attack (tilt the wing up a bit), lift occurs. There is an optimum angle of attack for maximum lift. This depends in part to the relative windspeed. With an angle of attack and sufficient airspeed, the air passing over the top of the wing behaves as if it were trying to separate away from the upper surface causing a reduction in pressure, resulting in lift. As well, below the wing, the air srikes that surface at a slight relative angle which also contribute to the lift.

 

[RandomGuy88]

MaxManus - Are you still here?
The air moves faster over the rounded top suface of the wing simply because it has to travel further than the air that travels below the wing. When the air at the front (leaing edge) of the wing separates into two parts, those parts must meet at the trailing edge of the wing at the same time. Think about what would happen if this were not the case.

NOOOOOOOOOOOO! That is absolutely INCORRECT! There is NO reason that the two parts of air that split at the leading edge must meet again at the trailing edge.

Lift is created because the streamlines must curve due to the body. A pressure gradient is then created due to this streamline curvature. Think of a particle of air traveling in a circle. There is a centrifugal force on this particle and this centrifugal force is balanced by the pressure. So for air moving in a circle the pressure at the center of the circle is lower then the edge of the circle. So if you think air flowing over the surface of an airfoil, it curves and on the upper surface the centrifugal force is trying to push the air particle away from the surface so the pressure at the surface decreases in order to balance this centrifugal force. On the lower surface the same mechanism causes the pressure to be higher at the surface and this pressure difference causes lift.

 

[daqddyo1]

One is always in danger when referring to centrifugal force as an action. It is a reaction force and is exerted BY a particle forced to move in a curved path and NOT on that particle. The air moving in the curved path over the top of the wing has some upward momentum that it didn't have before the wing hit it. This means there is less air density close to the top of the wing.
At sufficient airspeed, this increases the effective pressure difference to provide the lift force provided by the wing.


If the air passing over the top of the wing continually takes more (or less) to time pass over the wing, then somewhere there is going to be a great buildup of of air.

P.S. I'm going to bed now.

 

[SDEric]

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.

According to one UC Davis M.S. thesis, a frisbee generates some lift at a zero angle of attack. See figure 2-3 in http://morleyfielddgc.files.wordpress.com/2009/04/hummelthesis.pdf.

I'd love to have an intuitive understanding of this lift.

According to Foilsim, there is lift on a flat-bottomed, ellipse topped shape even when the flat bottom is parallel to the airflow. It says there is positive pressure on the leading curve and on the trailing curve, and negative pressure at the top. I understand the first part; that seems obvious, but I don't get why the top would have lower pressure than ambient and I don't get why the trailing edge would have higher pressure. If anyone can help me understand this, I would be most grateful.

 

[rcgldr]

For an air foil with a flat bottom and a curved top and zero angle of attack (the flat bottom horizontal), if the peak of the curve is near the leading edge, you get positive lift, if it's near the trailing edge you get negative lift (and a lot of drag, it would be similar to an inverted wedge). At somepoint in between, there would be zero lift, but I don't know if this corresponds to having the peak exactly at the middle.

According to one UC Davis M.S. thesis, a frisbee generates some lift at a zero angle of attack. ... I'd love to have an intuitive understanding of this lift.

According to Foilsim, there is lift on a flat-bottomed, ellipse topped shape even when the flat bottom is parallel to the airflow. It says there is positive pressure on the leading curve and on the trailing curve, and negative pressure at the top.

From my quoted post, there is a flat bottom, smoothly curved top airfoil with zero lift at zero angle of attack, but the peak is probably a bit aft of the middle of the chord length. The frisbee article doesn't mention the wind tunnel air speed versus the speed of a typical throw. If the speed in the wind tunnel was signficantly slower, then tubulent effects would be reduced. Also a frisbee doesn't have a flat bottom, the bottom has a similar shape to the top, but the effects may be similar to a flat bottom, ellipse top airfoil shape.

Because of the convex curvature of relative flow, there's a low pressure area near and a bit aft of the peak. There may be places on such an airfoil where pressure above is greater than pressure below, but apparently there is sufficient area where the pressure above is less than the pressure below to result in a net upwards force (lift). Turbulent flow could reduce or eliminate this lift.

As far as an explanation for this lift, it's because there is net downwash aft of the airfoil. The airfoil's presence in a relative horizontal flow causes upwash on the leading edge, then curves this flow back downwards again. If a downwash is created, the air foil passes by the affected air before it stops such a downwash. If the total effect of the upwash, bending of flow, and then downwash results in a net downwash of air as the air foil passes through, then lift is created.

 

[boneh3ad]

The air moving in the curved path over the top of the wing has some upward momentum that it didn't have before the wing hit it. This means there is less air density close to the top of the wing.
At sufficient airspeed, this increases the effective pressure difference to provide the lift force provided by the wing.

It means no such thing. The air need not change density for lift to be generated. In fact, if it had to, the Wright brothers would never have left the ground and birds wouldn't be able to glide. In a flow below Mach 0.3, density is effectively constant. Above Mach 0.3, density plays very little role until you approach sonic conditions.

If the air passing over the top of the wing continually takes more (or less) to time pass over the wing, then somewhere there is going to be a great buildup of of air.

No there isn't. In reality, the air moving over the top of an airfoil passes over the wing much faster than the air passing over the bottom surface. This occurs because the effect of viscosity ensures that the rear stagnation point must be located at the sharp trailing edge (a concept known as the Kutta condition). The top and bottom streams of air leave the trailing edge parallel to one another but at different velocities. This tends to keep the flow turning for a little distance past the trailing edge, contributing to the downwash.

 

[SDEric]

From my quoted post, there is a flat bottom, smoothly curved top airfoil with zero lift at zero angle of attack, but the peak is probably a bit aft of the middle of the chord length. The frisbee article doesn't mention the wind tunnel air speed versus the speed of a typical throw. If the speed in the wind tunnel was signficantly slower, then tubulent effects would be reduced. Also a frisbee doesn't have a flat bottom, the bottom has a similar shape to the top, but the effects may be similar to a flat bottom, ellipse top airfoil shape.

Because of the convex curvature of relative flow, there's a low pressure area near and a bit aft of the peak. There may be places on such an airfoil where pressure above is greater than pressure below, but apparently there is sufficient area where the pressure above is less than the pressure below to result in a net upwards force (lift). Turbulent flow could reduce or eliminate this lift.

Thanks. I think the general question is, for frisbees, hills, and semi-circular cross section airfoils, how does an object, symmetrical front to rear, and oriented parallel to the air flow, generate lift? It just begs for the (incorrect) longer path explanation.

As far as an explanation for this lift, it's because there is net downwash aft of the airfoil. The airfoil's presence in a relative horizontal flow causes upwash on the leading edge, then curves this flow back downwards again. If a downwash is created, the air foil passes by the affected air before it stops such a downwash. If the total effect of the upwash, bending of flow, and then downwash results in a net downwash of air as the air foil passes through, then lift is created.

The problem I see with this explanation is that the airflow turned down at the back seems like it is exactly offset by the airflow turned up at the front.

But that view is wrong. The airflow is turned up at the front, and by the time it reaches the middle it is back to moving parallel to the direction of travel. So it has been turned up and then turned back to parallel. Then it turns down, and by the time it leaves the back, it is turned down and is left turned down. So you are exactly right. All general discussion of airfoils focuses on the more complicated situation in which the airfoil has a shape not symmetrical front to rear, and angled to the airflow. What happened to the physics principles of deconstruction? No wonder everyone is frustrated by these explanations. Even that stupid NASA website boils down to "it's too complicated to explain clearly".

I recall Feynman complaining that no one had worked out the physics of such an obvious phenomenon as wind generated waves.

 

[DrDu]

I think it should be possible to get some symmetric airfoil with the Joukowski transformation.
You can try it out here (Java does not work for me):

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

 

[olivermsun]

Thanks. I think the general question is, for frisbees, hills, and semi-circular cross section airfoils, how does an object, symmetrical front to rear, and oriented parallel to the air flow, generate lift? It just begs for the (incorrect) longer path explanation.

Hills are easy -- the flow only goes around one side of the hill.

Are you sure that frisbees and semicircular airfoils generate lift when oriented parallel to the flow?

I recall Feynman complaining that no one had worked out the physics of such an obvious phenomenon as wind generated waves.

It's a lot more complicated because the waves (which have a whole set of dynamics of their own) exert a feedback on the wind stress.

 

[SDEric]

Hills are easy -- the flow only goes around one side of the hill.

Ah! That is fallacy #1. The air flowing over the top of the wing has NOTHING to do with the air flowing under the bottom! Assuming that the two flows are somehow connected leads one to the "equal transit time" fallacy.

Are you sure that frisbees and semicircular airfoils generate lift when oriented parallel to the flow?

I am not sure. The simulations, and supposedly some wind tunnel data, say there is lift, but I have a hard time developing an intuition why it exists.

It's a lot more complicated because the waves (which have a whole set of dynamics of their own) exert a feedback on the wind stress.

Sure, but again, make it simple. A small, smooth wave (that is, not breaking), a wind speed low enough to not cause turbulence but faster than the propagation speed of the waves. For some reason the wave builds. I can understand why the wave moves (same as drag on the wing) but why does the wave build? I think it is because pressure is higher at the base of the wave and less at the top, much like how the curved top wing has some lift.

 

[olivermsun]

Ah! That is fallacy #1. The air flowing over the top of the wing has NOTHING to do with the air flowing under the bottom! Assuming that the two flows are somehow connected leads one to the "equal transit time" fallacy.

You yourself agreed in an earlier post that the net momentum change of the flow over the wing is important (the flow remains "downturned" after it passes by the wing). My point is that this condition is impossible for a symmetric hill (the flow cannot leave with net downward momentum since the ground is in the way). This places strong constraints on the net lift produced by the hill.

I am not sure. The simulations, and supposedly some wind tunnel data, say there is lift, but I have a hard time developing an intuition why it exists.

Could you link to some simulation results? The diagrams I was able to find of lift generated by a Frisbee all show an angle of attack, but my search was by no means exhaustive.

I think it is because pressure is higher at the base of the wave and less at the top, much like how the curved top wing has some lift.

I guess here it may be helpful to point out that pressure differences exist over airfoils, whether or not a net lift is generated.

 

[RandomGuy88]

For the flow over a 2D airfoil the downwash exactly equals the upwash and there is no net downward momentum added to the flow. The force on the airfoil is a result of the difference in pressure between the upper and lower surface. This pressure difference is due to streamline curvature.

Before you argue that there can be no force without a net downwash consider this example.

Imagine you have an open loop wind tunnel. So the air is pulled in from the atmosphere at one end and exhausted at the other end. Because the flow is accelerated from rest to whatever speed you want in the test section the pressure drops considerably. As a result, the pressure in the test section is less than atmospheric pressure. So if you have a door that gives you access to the test section and this door opens out it will be very difficult to open this door because of the pressure difference between the low pressure in the tunnel and the atmospheric pressure outside. The door is getting sucked into the tunnel. This is analogous to the lift of a 2D airfoil because there is a force but there is no net downwash. The force is only a result of the pressure difference.

Half a cylinder with the bottom face parallel to the flow generates lift because the pressure decreases over the rounded surface. This pressure decreases because of streamline curvature. The same is true of a hill and a frisbee. A frisbee generates lift a zero degrees angle of attack. Think about it like this, when you throw a frisbee horizontally it doesn't drop as fast as if you just dropped it straight down right? The lift is less than the weight so it still falls but just not as quickly.