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

[boneh3ad]

You won't get lift on any hill, even this "ideal" type you concocted. You will get a heck of a lot of drag and therefore eventually some erosion, but certainly no lift.

Just as an aside, the only way you wouldn't get turbulence is if the wind was pretty much not blowing. Between Görtler vortices on the concave portion, the separation on the convex portion or just plain old' Tollmien-Schlichting waves over the whole thing, it will be nearly impossible to have laminar flow over such a shape at any reasonable velocity.

 

[RandomGuy88]

Boneh3ad is right, there won't be any lift on a hill because there isn't really another side for there to be a pressure difference. For an airfoil there are two sides, the upper surface has lower pressure then the bottom surface and this pressure difference creates lift. On a hill there really is no lower surface is there?

The reason for having a coordinate system that follows the streamlines is so that you can easily determine the forces that are tangential and perpendicular to the the streamlines. This is how you derive the expression relating the streamline curvature and the pressure gradient.

If you have curved streamlines (which are curved because of the solid body) the pressure decreases towards the center of the curvature. If the streamlines were circular the pressure in the center is lowest.

For a 2D hill you will have two areas where the curvature causes the pressure to decreases as you go away from the surface (curvature is concave up). This is at the bottom of the hill on both sides. (I am ignoring the viscosity, so I am not considering the separated wake region). So at the bottom of the hill the pressure increases relative to the freestream static pressure. As the air goes up the hill there is an inflection point and then the curvature is in the other direction (concave down). So now the lowest pressure is at the surface and the pressure increases as you go up from the surface.

So from the flat ground to the top of the hill the pressure first increases then decreases and is lowest at the peak.

 

[boneh3ad]

Don't forget in this "ideal" hill there is a major curvature discontinuity so there would be major pressure spikes at those locations.

 

[SDEric]

Boneh3ad is right, there won't be any lift on a hill because there isn't really another side for there to be a pressure difference. For an airfoil there are two sides, the upper surface has lower pressure then the bottom surface and this pressure difference creates lift. On a hill there really is no lower surface is there?

The reason for having a coordinate system that follows the streamlines is so that you can easily determine the forces that are tangential and perpendicular to the the streamlines. This is how you derive the expression relating the streamline curvature and the pressure gradient.

If you have curved streamlines (which are curved because of the solid body) the pressure decreases towards the center of the curvature. If the streamlines were circular the pressure in the center is lowest.

For a 2D hill you will have two areas where the curvature causes the pressure to decreases as you go away from the surface (curvature is concave up). This is at the bottom of the hill on both sides. (I am ignoring the viscosity, so I am not considering the separated wake region). So at the bottom of the hill the pressure increases relative to the freestream static pressure. As the air goes up the hill there is an inflection point and then the curvature is in the other direction (concave down). So now the lowest pressure is at the surface and the pressure increases as you go up from the surface.

So from the flat ground to the top of the hill the pressure first increases then decreases and is lowest at the peak.

Understand and appreciate the discussion.

The thing is, you don't need the bottom to have higher pressure for the top to have lower pressure. You can demonstrate this with various flat-bottomed airfoils; with the right (mildly convex) curvature of the bottom, you can have net lift of the bottom being zero. The top still has net negative pressure, or net lift. The bottom of the airfoil isn't necessary to generate lift.

I can really understand why the hill shouldn't have lift; I can also understand why this curvature of the airflow is a minor component to overall airfoil lift, but just like wondering whether a flat-bottomed and front-to-back symmetrical airfoil has lift, I wonder whether a hill has lift.

It's a completely symmetrical situation, but here's an aspect that is does not net to zero: the airflow being turned by the concave curvature at the front of the hill and at the back of the hill is slower than the airflow being turned at the top of the hill. Higher speed, greater acceleration, greater force. Thus the sum of the forces is net upward (and leeward, counting drag).

Are the walls of a venturi sucked toward each other?

 

[boneh3ad]

Why SHOULD a hill have lift? Even if the pressure on the hill is lower than atmospheric, it isn't like it is going to suck the hill upward. That isn't how pressure works. There will still be zero pressure on any imaginary surface of the hill not exposed to the air, and the air will exert a force everywhere it touches normal to the surface. That means a hill, in fact, sees a net DOWNWARD aerodynamic force (though zero net force since it doesn't move).

Pressure does not suck. It only pushes. An object (be it a hill or airfoil or anything else) can never experience a be aerodynamic force towards a fluid without a higher pressure fluid on the other side. It is Newton's law.

 

[olivermsun]

Boneh3ad is right, there won't be any lift on a hill because there isn't really another side for there to be a pressure difference.

If there is a pressure difference relative to "ambient" (usu. something like hydrostatic) then will there will be any confusion in calling the net force due to the pressure anomaly "lift"?

Even if the pressure on the hill is lower than atmospheric, it isn't like it is going to suck the hill upward. That isn't how pressure works. There will still be zero pressure on any imaginary surface of the hill not exposed to the air, and the air will exert a force everywhere it touches normal to the surface.

If it is a physical argument about the phenomenon that will be observed under these conditions, then imagine this:

The hill was in equilibrium with the Earth (or whatever was below it, be it trapped gases or underground water if you insist it must be fluid). If there is a flow speedup above the hill, reducing the downward pressure on the hill, and the Earth is still pushing upward, then the hill will probably relax upward a bit. Do you call this lift? Isostatic adjustment?

 

[boneh3ad]

With no wind, the hill exerted a force of X on the ground below it. This force was a combination of the weight of the hill, W, and the force of the atmosphere pushing on the hill, P1. If the wind were to blow, there will be a pressure drop over the hill compared to the pressure in slower moving wind, so the force exerted by the hill will now be Y, the same weight of the hill, W, plus the force exerted on the now very slightly lower pressure surrounding air, P2. The net force is still downward, still dominated by the weight of the hill, and still not lift. Lift is an upward force, and it is impossible to have lift if the top side of an object is surrounded by air and the bottom side is resting against a solid, such as the Earth under the hill.

 

[olivermsun]

The net force is still downward, still dominated by the weight of the hill, and still not lift. Lift is an upward force, and it is impossible to have lift if the top side of an object is surrounded by air and the bottom side is resting against a solid, such as the Earth under the hill.

I'm not sure this answers my question. For the hill surface to "rebound" when the downward pressure of its own weight is decreased by the pressure drop, there has to be some finite period of time when there was a net upward force on the hill surface, causing (allowing?) the hill to rise.

Suppose we modify the scenario and say that the dynamic pressure drop caused by wind blowing over some bump in the ocean surface causes the bump to rise slightly. Was this caused by lift?

Alternatively, suppose we have an airfoil (which we agree has lift) in level flight. Is the weight of the airfoil (and aircraft) canceling the entire upward force due to the flow, in which case there is no net force at all? Is it still lift?

 

[boneh3ad]

No, it would be caused by just a shrinking of the downward force. There is no situation ever where pressure can suck something towards it. If you really look at a situation where you think that is happening, you will find that somewhere is is actually something pushing the object.

Think of it this way: you grab a bowling ball and then stand on a bathroom scale. You then set the bowling ball down and watch your weight (i.e. the force you exert on the scale) lessen. Did you experience lift? No, you just shed a bi of weight and therefore decreased your downward force.

The same is true of a hill. Wind blows over it and whatever minute pressure drop results just ends up being a drop in the effective weight of the hill (and a very, very, very small drop at that), not any sort of upward force. The effective weight of the hill falls and therefore the ground the hill sits on exerts less upward force on the hill since the sum of forces is zero.

There can't be lift because there is nothing under the hill to push on it that is affected in any way by airspeed. The ground merely reacts to the weight of the hill. The same affect could be achieved by you standing on the hill and then leaving it so it weighs less. Certainly the hill didn't experience lift there, no?

 

[olivermsun]

In the case of an airfoil, which we agree can experience lift, there is a pressure force "pushing" from the bottom and creating the net lift. Clearly, pushing vs. pulling is not the source of disagreement here.

Your point is that in the hill example, upward forces in the (solid) medium are unaffected by the external flow, and hence the net force cannot be called "lift."

How about the "intermediate" example of flow over a bump on the ocean surface? Couldn't it be argued that any upward force due to the shear flow (even if the lower layer is prescribed to have zero flow) is a special case of "lift"?

Just wondering where you draw the line, that's all.

 

[boneh3ad]

The ocean example would not be called lift by anyone I know in the fluids community. Instead that is a wave dynamics problem.

The definition of lift is an upward aerodynamic force on a body as a result of the movement of the body through that fluid. In other words, it has to do with the flow of that fluid around a body. In your wave example, it is an entirely different fluid below the "bump" and any upward force is as a result of the dynamics of the Kelvin-Helmholtz instability rather than lift.

It may be easier to think about in terms of other descriptions of lift. Lift requires a net circulation about the body moving through the fluid. This can't happen unless the body has the same fluid on all sides.

 

[SDEric]

It seems to me that requiring the hill to have the working fluid underneath it to generate lift is unnecessary. I understand that there is no such thing as "negative pressure", only less pressure than somewhere else. We can work around that requirement, however. The question is not whether the net forces on the hill are in any particular direction (I know the hill is not able to move), but whether the net pressure on the hill is less than ambient.

Suppose the hill were like an iceberg, floating in water. If there were lift on the hill, it is free to move up out of the water. With no wind, the air pressure on the hill is balanced by the air pressure on the water (and, of course, the mass of the hill displacing mass of water, etc.). There is enough of the hill below the water that the curved section begins exactly at the water surface.

Then, when the wind is blowing, is the net force on the hill up? Down? Parallel to the wind and surface?

The air encounters the hill and is turned up to perpendicular to its original direction. Then it turns and follows the same radius of curvature back down until it is again parallel, then again perpendicular, and then turns back the other direction until it is again parallel to the original flow.

I think the air, generally speaking, slows and turns, then speeds up and turns in the other direction, then slows and turns again in the first direction.

I can vaguely see that there might be net force upward because the speed (and hence acceleration) is lower at the concave curvature sections than at the convex curvature section at the top. Thus the integral of the pressure over the hill is less than ambient.

(Nothing about that construction explains drag, however.)

 

[boneh3ad]

There would be a net upward force in that case... as a result of buoyancy. Of course then the berg would rise again, the forces would all be equal and there would be no net force.

Despite how you seem to be interpreting things, lift is a force on a body arising due to its relative motion through a fluid. Forit to exist, it has to be immersed in that fluid, not partially immersed. That is just the definition of lift.

 

[SDEric]

There would be a net upward force in that case... as a result of buoyancy. Of course then the berg would rise again, the forces would all be equal and there would be no net force.

Despite how you seem to be interpreting things, lift is a force on a body arising due to its relative motion through a fluid. Forit to exist, it has to be immersed in that fluid, not partially immersed. That is just the definition of lift.

Understand. However, you can measure lift on an airplane wing in a wind tunnel, supported by thin rods or wires, and measure the force on those rods or wires. The wing does not deflect in the wind, yet you can measure the force imposed by the lift and drag. So it is definitely not necessary that the body be freely moving through the air. It can be rigidly supported.

So what is the integral of the pressure over the hill ("integral P dHill")? Is it less when the air is moving? Specifically, at each point on the curve, the pressure is orthogonal to the curve. At each point that orthogonal pressure has an x (horizontal) component and a y (vertical) component. Is the y component of the pressure less when the air is moving than when it is still?


A related thought experiment: If the hill is simply mated to its mirror image, the net force would obviously be zero (or, more to the point, straight back due to drag). If you attached the two halves by a force meter, would the meter measure tension when the wind blew? Would the two halves pull apart?


Thanks for lending your expertise. I am amazed that after more than a hundred years, explanations of lift do not simply take apart the various aspects of lift and discuss them independently.

 

[boneh3ad]

I never said it couldn't be rigidly mounted, only that it had to be immersed in such a way that a net circulation in the flus can form around the body.

The y component of the integral of the pressure should be less when the wind is moving.

If the hill was mated to a mirror image by a force sensor, it would measure zero force assuming a perfect mate with no air seeping into it.

 

[olivermsun]

The ocean example would not be called lift by anyone I know in the fluids community. Instead that is a wave dynamics problem.

While the development of a pressure gradient over a disturbance on a surface is a subject of interest for waves people, the dynamics are very similar (at least at the outset) to flow around an airfoil (and not surprisingly so).

The definition of lift is an upward aerodynamic force on a body as a result of the movement of the body through that fluid. In other words, it has to do with the flow of that fluid around a body.

To take the Kutta-Joukowsky theorem literally, it seems there would be a "lift" associated with circulation at an interface. However, if your feeling is that the proper usage is only associated with circulation around a solid object (an airfoil) then I can certainly accept that convention.

In your wave example, it is an entirely different fluid below the "bump" and any upward force is as a result of the dynamics of the Kelvin-Helmholtz instability rather than lift.

Look closer at the dynamics of the K-H instability -- similarities between that and flow around a thin wing are quite interesting. In the first case, you have vortices bound to a wing. In the second, the vortices are bound to the interface, with the net pressure distribution contributing to growth of the instability (raising the bumps and pushing down the troughs).

It may be easier to think about in terms of other descriptions of lift. Lift requires a net circulation about the body moving through the fluid. This can't happen unless the body has the same fluid on all sides.

Are you sure that the same fluid on all sides is a requirement? What about lift in a parallel stratified flow?

 

[boneh3ad]

The Kelvin-Helmholtz has little similarity to the flow around a wing. It is the result of the interaction between two immiscible fluids flowing parallel to one another with different densities. It is not a vorticity-based phenomenon or circulation-based phenomenon.

Kutta-Joukowski was developed for solid objects in a single fluid. While there may be analogous phenomenon in other situations, there are no corollaries to Kutta-Joukowski that apply to those situation. The concept of lift applies to a solid object immersed in a fluid.

Stratified flows are still all the same gas.

 

[olivermsun]

The Kelvin-Helmholtz has little similarity to the flow around a wing. It is the result of the interaction between two immiscible fluids flowing parallel to one another with different densities.

I find there to be some resemblance between the mathematics of the growth term in K-H instability and the lift theorem. There is also similar phenomenology between the flow around a wing and the flow around the K-H disturbances (which are often seen to develop into a roll of vortices). Again, I just find it to be an interesting parallel (no pun intended). If you maintain that there is no similarity whatsoever, well, you are certainly entitled to your view and I accept that.

As an aside: we observe K-H instabilities between miscible fluids all the time. In fact, K-H instabilities are believed to be quite important for mixing in a variety of environmental flows.

It is not a vorticity-based phenomenon or circulation-based phenomenon.

There is a velocity discontinuity (in the idealized limit) but there is no vorticity at the surface?

Stratified flows are still all the same gas.

Perhaps in the stratified flows you have looked at.

 

[boneh3ad]

I find there to be some resemblance between the mathematics of the growth term in K-H instability and the lift theorem. There is also similar phenomenology between the flow around a wing and the flow around the K-H disturbances (which are often seen to develop into a roll of vortices). Again, I just find it to be an interesting parallel (no pun intended). If you maintain that there is no similarity whatsoever, well, you are certainly entitled to your view and I accept that.

As an aside: we observe K-H instabilities between miscible fluids all the time. In fact, K-H instabilities are believed to be quite important for mixing in a variety of environmental flows.

There is a velocity discontinuity (in the idealized limit) but there is no vorticity at the surface?

Sure there is vorticity, but that isn't the nature of the growth of the instability. If you had a situation where the Kelvin-Helmholtz instability would ordinarily occur that was in equilibrium, you could see a completely lack of any waves regardless of the vorticity inherent in the shear layer. You would still need some kind of initial perturbation to cause the instability to manifest.

The Kelvin-Helmholtz instability early in its life can be fully described without any viscosity terms and need not roll over to still be an example of this instability. It is an inherently irrotational problem. Treating the governing equations linearly (by dropping viscous terms that are the only source of vorticity) still results in observation of the instability. Therefore it is not a vortical phenomenon. It starts out as linear waves that grow according to the dynamics of the linearized Euler equations. Upon reaching a certain amplitude, only then to they start exhibiting any sort of nonlinear (and vortical) component. At that point they create the familiar rolls that are highly nonlinear and fractal in nature.

Lift on an airfoil, however, is an inherently viscous phenomenon and depends on the effect of viscosity in generating circulation about a solid body. I suppose I fail to see the parallel here because lift on an airfoil is intimately related to viscosity and vorticity and cannot exist without it. The Kelvin-Helmholtz instability is still present in the inviscid limit.

You can certainly find Kelvin-Helmholtz phenomena occurring in miscible fluids, so I overstepped with that description. In that you are correct. All you really need is a velocity discontinuity.

Perhaps in the stratified flows you have looked at.

I read it (and clearly typed it) wrong. In my mind, I was thinking rarefied. You are, of course, correct. My mistake.

 

[RandomGuy88]

The Kelvin-Helmholtz has little similarity to the flow around a wing. It is the result of the interaction between two immiscible fluids flowing parallel to one another with different densities. It is not a vorticity-based phenomenon or circulation-based phenomenon.

There are examples of KH instabilities without two immiscible fluids. Jets issuing into a motionless fluid are an example. It can occur in shear layers with an inflection point. There were a few other requirements but I don't remember them.

 

[boneh3ad]

There are examples of KH instabilities without two immiscible fluids. Jets issuing into a motionless fluid are an example. It can occur in shear layers with an inflection point. There were a few other requirements but I don't remember them.

Already covered in my previous post. I erred when stating the immiscible part. All that is necessary is a discontinuity in velocity in a flow. Most examples that are visible in nature involve two different fluids (or the same fluid of different properties such that the two parts stand out from one another). It is the velocity discontinuity that is important. If the velocities were the same, it would either be stable or unstable under the Rayleigh-Taylor instability, not Kelvin-Helmholtz.

The inflection point is an entirely different phenomenon. An inflection point in the shear layer profile is the criterion for inviscid instability, and I am not 100% sure that it is valid for shear layers in general, but definitely for boundary layers. It is known as the Rayleigh criterion.

 

[Stan Butchart]

Just found this thread and thought I would throw out a bananna peel just for fun.

The flow (ideal) next to a cylinder is accelerated to the velocity of the cylinder at the fwd stag point. It remains at constant 1V speed until the aft stag point.

The flow (ideal) next to a common wing is accelerated to the velocity of the wing at the fwd stag point. It SLOWS down over the top before speeding up again tothe TE.

 

[boneh3ad]

Just found this thread and thought I would throw out a bananna peel just for fun.

The flow (ideal) next to a cylinder is accelerated to the velocity of the cylinder at the fwd stag point. It remains at constant 1V speed until the aft stag point.

The flow (ideal) next to a common wing is accelerated to the velocity of the wing at the fwd stag point. It SLOWS down over the top before speeding up again tothe TE.

That isn't true at all. Maybe you have your reference frames mixed up or something. If you use the same frame of reference for both objects, say, stationary object with moving fluid for simplicity, you see qualitatively the same behavior. For example, if you place a cylinder in an inviscid flow with free stream velocity U_{\infty} moving left to right in x, the maximum velocity will be \vec{V}=2U_{\infty}\hat{\imath} at the top and bottom of the cylinders.

If you similarly place a NACA 0012 airfoil (for simplicity) into the same flow, the maximum flow speed is roughly \vec{V}=1.2U_{\infty}\hat{\imath}.

Both flows are accelerated over the wing.

If you then want to look at it in the other frame of reference with the object moving with velocity U_{\infty} through a stagnant medium, you need only add \vec{V} = -U_{\infty}\hat{\imath} to all quantities. In other words, in both cases, the air is still moving the same direction and is still accelerated over the object, accelerating at it reaches a max (whose location varies depending on the object) before decelerating back to the stagnation point. I think you must have looked at two different reference frames. The cylinder and the airfoil are qualitatively similar. The cylinder just accelerates the air to a greater degree. This is, as previously mentioned, all inviscid.

 

[Stan Butchart]

T the maximum velocity will be \vec{V}=2U_{\infty}\hat{\imath} at the top and bottom of the cylinders.QUOTE]

I'm sorry but I must stick to my guns.
For a moving cylinder, there is a movingsurface and a moveing fluid.
IF i wish to identify the inertial velocity of the fluid following the surface, as if I were to calculate Bernoulli, I need to relate path and relative to its original position (or better the remote still field)
A particle located on the fwd stagnation is accelerated to1V right to left by a cylinder moveing 1V right to left.
Using the the accepted v=2sinA*V, a little high school calculating will allow graphing of its path and velocity.

At the top and bottom the particle velocity is 1V left to right. The cylinder surface velocity is 1V righ to left.

by the same token , over the0012,the particle velocity is .2V left to right while while the wing surface is 1V right to left.

A particle is used since a following particle is on a different inertial flowpath.

Maybe we are just talking the importance of reference.

 

[boneh3ad]

Either way, you said that the particle slows down in the case of an airfoil. That isn't true. What you just said in this post is exactly what my previous post said.

 

[Stan Butchart]

Either way, you said that the particle slows down in the case of an airfoil. That isn't true. What you just said in this post is exactly what my previous post said.

Yah, I realized tha halfway through my post but it was too late to stop!

But tell me, it is moving at 1V at the stag point and .2V at the top ( all relative to the remote air.) How is that not slowing down?

 

[boneh3ad]

Well, -0.2V. It is slowing down in that frame and the speeding back up until it reaches that -0.2V, but in your earlier post you said it sped up over te cylinder, which does not happen in that frame.

 

[Stan Butchart]

in your earlier post you said it sped up over te cylinder, which does not happen in that frame.

I certainly hope that it did not come across that way. If it did I apologise.

The cylinder is constant speed, stag point to stag point.

 

[boneh3ad]

No it isn't. It is qualitatively the same as the airfoil. Both stagnation points are obviously moving at the same speed as the body and at the top/bottom the flow is moving at twice that speed. It is not constant.

 

[Stan Butchart]

No it isn't. It is qualitatively the same as the airfoil. Both stagnation points are obviously moving at the same speed as the body and at the top/bottom the flow is moving at twice that speed. It is not constant.

Oohh yes it is! This is all in friendship but communication is difficult.

I am looking at the inertial path relative to the remote still air only. The flowpath is as a lower case script "e".
Look up the source/sink pattern for the cylinder. While it is an instantaneous picture, all of the surface intercepts are at 1V. The "e" is an integration of those intercepts.