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Plane's wings and Bernoulli's equation

by mathsciguy
Tags: bernoulli, equation, plane, wings
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mathsciguy
#1
Apr10-12, 06:28 AM
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It's just a simple question about the pressure under and over the plane's wing problem that I'm trying to answer. Well, actually I've already answered it, but one just keeps bugging me. Why don't we consider the difference in altitude of the lower and upper points of the fluid (air) when we use Bernoulli's equation on the fluid on the plane's wings?

A possible answer I've thought of is that the fluid is in horizontal motion so there is no work done by the gravity, but I'm not sure if that's a sufficient explanation, it might not even be correct.

Maybe a more general question is, given a fluid traversing some object horizontally, and we are considering the upper and lower portion of the fluid in contact with the object, why do we omit the terms (rho)*g*y in Bernoulli's equation?
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AlephZero
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Apr10-12, 08:21 AM
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The ##\rho g y## term represents the upwards buoyancy force on the wing from the air.

This is negligible compared with the weight of the plane. It is the difference between the weight of the plane in air and its weight in a vacuum.
K^2
#3
Apr10-12, 12:44 PM
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You really shouldn't be using Bernoulli to compute lift. You cannot use the Bernoulli principle to explain pressure differential at the wing surface, which is where you would need it to compute the force on the wing. The pressure differential in areas where you can use Bernoulli can actually be quite a bit different.

What you should be looking at is momentum transfer to the air flow, which you evaluate via circulation. It's a simpler thing to look at analytically, and more precise if you perform finite element analysis numerically.

cjl
#4
Apr10-12, 04:32 PM
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Plane's wings and Bernoulli's equation

Quote Quote by K^2 View Post
You really shouldn't be using Bernoulli to compute lift. You cannot use the Bernoulli principle to explain pressure differential at the wing surface, which is where you would need it to compute the force on the wing. The pressure differential in areas where you can use Bernoulli can actually be quite a bit different.

What you should be looking at is momentum transfer to the air flow, which you evaluate via circulation. It's a simpler thing to look at analytically, and more precise if you perform finite element analysis numerically.
Bernoulli works just fine as a method of determining the pressure around an airfoil if you already know the velocity profile, and the flow is incompressible (Mach <0.3 or so). You can't use the bernoulli principle to determine the velocity profile - it must be determined by some other method, but once the velocity profile is known, bernoulli is completely valid.


As for why the gravitational term is omitted? Simple - it will give a differential pressure that is equal to ρgΔy, where Δy is the thickness of the wing. Since ρ is of the order 1 kg/m3, the pressure differential will be about 10 pascals per meter of wing thickness. This means that for pretty much any airplane in existence, the pressure difference will be on the order of a few pascals. This is tiny compared to the pressure differential due to the different velocity of air above and below the wing, which will be several thousand or even tens of thousands of pascals.

Basically, it's negligible (unless you're in a blimp).
K^2
#5
Apr10-12, 05:19 PM
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Quote Quote by cjl View Post
Bernoulli works just fine as a method of determining the pressure around an airfoil if you already know the velocity profile, and the flow is incompressible (Mach <0.3 or so). You can't use the bernoulli principle to determine the velocity profile - it must be determined by some other method, but once the velocity profile is known, bernoulli is completely valid.
Of course you know velocity profile at the wing boundary. Velocity of the flow at the boundary is zero. That's one of your boundary condition for solving flow equation in the first place. So Bernoulli effect at the wing surface is precisely zero.

Bernoulli gives you pressure in the flow. Pressure within the flow itself doesn't affect the wing. It only affects the flow. To find lift, you need pressure at boundary.
cjl
#6
Apr10-12, 05:37 PM
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Quote Quote by K^2 View Post
Of course you know velocity profile at the wing boundary. Velocity of the flow at the boundary is zero. That's one of your boundary condition for solving flow equation in the first place. So Bernoulli effect at the wing surface is precisely zero.

Bernoulli gives you pressure in the flow. Pressure within the flow itself doesn't affect the wing. It only affects the flow. To find lift, you need pressure at boundary.
I can't tell if you're deliberately misunderstanding me, or if you simply haven't studied very much aerodynamics.

Yes, the velocity profile at the surface of the wing is zero. However, the lift and the induced drag acting on a wing can be modeled quite accurately with the assumption that the flow is inviscid (and thus the velocity at the wing surface is nonzero). Using potential flow, the velocity profile can be obtained around the wing, and this inherently assumes the validity of bernoulli for incompressible, inviscid flows. This works very well for flows up to a mach number of approximately 0.3.

The reason that this works well even when viscous effects are added back in is because, to a decent approximation at least, the pressure gradient across a boundary layer is zero. As a result, the flow just outside the boundary layer has the same pressure as the point at the surface just below the boundary layer. This allows you to use the bernoulli principle with the velocity of air just outside the boundary layer to obtain the pressure on the surface of the wing.

Oh, and since the pressure gradient across the boundary layer is zero, the pressure within the flow itself does indeed affect the wing.
K^2
#7
Apr10-12, 06:23 PM
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So what's your boundary condition for tangential component, then? Because even with inviscid and incompressible assumption, I should be able to get a family of solutions to Navier-Stokes that differ by circulation around the boundary. And naturally, all of these will produce different lift.

I do know that pressure gradient near boundary is low, which is why you can get a descent estimate using Bernoulli equation. And yes, that estimate improves as you drop the air speed, but you aren't going to get around the fact that pressure at the boundary can't be explained with Bernoulli principle. It's a completely wrong approach both from perspective of underlying physics and because even as an estimate, it provides no advantage. If you know the flow, circulation gives you an exact solution for lift even at speeds where compressibility becomes a factor.

People are wrongly taught that Bernoulli effect is actually responsible for lift. It is not. It is partly responsible for shaping the flow around the wing, which does result in lift, but it's not the direct source of lift.



By the way, the actual answer to the boundary condition question is that in true inviscid flow, Kutta condition breaks down, and you get zero circulation. So viscosity is actually a necessary condition to generate lift.
AlephZero
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Apr10-12, 09:19 PM
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Quote Quote by K^2 View Post
So what's your boundary condition for tangential component, then? Because even with inviscid and incompressible assumption, I should be able to get a family of solutions to Navier-Stokes that differ by circulation around the boundary. And naturally, all of these will produce different lift.
That is irrelevant to the OP's question. If you choose ANY of those flow conditions, Bernouilli's equation will correctly calculate the lift that corresponds to it.

You are right that Bernouilli's principle can not tell you the "true" velocity distribution around the wing, and that the physical cause of lift is the circulation caused by viscosity, but don't throw the baby out with the bathwater.
K^2
#9
Apr11-12, 12:20 AM
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Quote Quote by AlephZero View Post
That is irrelevant to the OP's question. If you choose ANY of those flow conditions, Bernouilli's equation will correctly calculate the lift that corresponds to it.
Is it a physics problem or an exercise in mathematics? If it's the later, I would suggest that it's not the place for it. Using a meaningful formula with meaningless setup does not lead to a meaningful result. The problem is undefined if you don't fix circulation at boundary, it is not physical if that circulation is fixed at anything but zero, and it produces zero lift, which does not represent physical reality, when the circulation is fixed at zero.

When OP is asking about specifics of computing lift using Bernoulli equation, something, somewhere went horribly wrong, and that needs to be made clear. Is that not so?
rcgldr
#10
Apr11-12, 04:13 AM
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Quote Quote by cjl View Post
Bernoulli works just fine as a method of determining the pressure around an airfoil if you already know the velocity profile.
How large an area is included in this velocity profile?

Also isn't part of the pressure differential related to work done by a wing onto the air? The work done results in a downwards (lift) and somewhat forwards (drag) flow of air after a wing passes through a volume of air (wrt air) (the exit velocity of the affected air when it's pressure returns to ambient). That work done on the air involves a mechanical interaction that violates Bernoulli, but is responsible for part of the lift (but I don't know by how much).
cjl
#11
Apr11-12, 06:59 AM
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Quote Quote by rcgldr View Post
How large an area is included in this velocity profile?
I'm not really sure what you're asking here...


Quote Quote by rcgldr View Post
Also isn't part of the pressure differential related to work done by a wing onto the air? The work done results in a downwards (lift) and somewhat forwards (drag) flow of air after a wing passes through a volume of air (wrt air) (the exit velocity of the affected air when it's pressure returns to ambient). That work done on the air involves a mechanical interaction that violates Bernoulli, but is responsible for part of the lift (but I don't know by how much).
You're trying to split up a single problem here, and the truth is that it's really just multiple ways of looking at the same answer. If you know the velocity distribution around a low speed airfoil (< mach 0.3), bernoulli can give you a very accurate lift and induced drag calculation, and it does indeed account for the work done on the air. That isn't a separate term, it's merely a different way of looking at the same problem (and it certainly doesn't "violate" bernoulli). You can get exactly the same answer for the lift by either taking a control volume around the airfoil and looking at the momentum flux into and out of the control volume (effectively a newtonian analysis), and by looking at the pressure distribution on the airfoil surface itself. In oversimplified explanations, these two approaches are often stated as conflicting, or different, but in reality, they are both valid.

The real flaw in popular belief about how airfoils work isn't in the fact that high velocity air has a low pressure - this is both true, and a valid way to analyze the problem. The problem with popular belief is the explanation for why the air is traveling faster over the top of the airfoil. This is commonly explained with the equal transit time assumption, which is fatally flawed in many ways. The true reason has to do with the generation of circulation around the airfoil due to the need to have the flow attached at the trailing edge (known as the kutta condition). However, when this circulation is calculated and the velocity profile around the airfoil is obtained, the bernoulli relation can absolutely be used to transform that velocity profile into a pressure distribution, and the force on the airfoil is (as would be expected) simply the pressure integrated around the outside surface of the airfoil.

If you plot the pressure distribution around the airfoil, you will also notice a large low pressure bubble above the top surface of the airfoil, and a less substantial high pressure bubble below the airfoil. Both of these will tend to turn the freestream flow farther from the airfoil downwards, so the downwash generated by the airfoil is not an independent effect from the pressure distribution - rather the two effects are interlinked and inseparable from each other.

Hopefully this helps clear things up a bit...
cjl
#12
Apr11-12, 07:08 AM
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Quote Quote by K^2 View Post
So what's your boundary condition for tangential component, then? Because even with inviscid and incompressible assumption, I should be able to get a family of solutions to Navier-Stokes that differ by circulation around the boundary. And naturally, all of these will produce different lift.
Yes, they will. However, only one of them will also satisfy the condition of having the flow attached at the trailing edge, and that is the only one that makes physical sense (I'm assuming an unstalled airfoil here - flow separation is a different, and quite challenging topic).

Quote Quote by K^2 View Post
I do know that pressure gradient near boundary is low, which is why you can get a descent estimate using Bernoulli equation. And yes, that estimate improves as you drop the air speed, but you aren't going to get around the fact that pressure at the boundary can't be explained with Bernoulli principle. It's a completely wrong approach both from perspective of underlying physics and because even as an estimate, it provides no advantage. If you know the flow, circulation gives you an exact solution for lift even at speeds where compressibility becomes a factor.



People are wrongly taught that Bernoulli effect is actually responsible for lift. It is not. It is partly responsible for shaping the flow around the wing, which does result in lift, but it's not the direct source of lift.
You seem to be under the impression that the bernoulli relation says something about the way the flow will be shaped. This might be some of the reason for the confusion. Bernoulli actually says nothing at all about the streamlines or the flow velocity. It simply relates the local flow velocity along a streamline in an inviscid, incompressible flow to the pressure along that same streamline. Other assumptions or methods are required to obtain the shape of the streamlines and the flow velocity at each point. Bernoulli just lets you take the flow velocity (once it is known) and then obtain the pressure as well. As such, you can't really say anything about how bernoulli "shapes" the flow, or how it causes incorrect solutions. As long as the flow is effectively incompressible and no significant viscous dissipation is present, bernoulli is very accurate in relating pressure and velocity. If the velocity profile is wrong, then the pressure profile will be wrong as well, but that's not a problem with the bernoulli relation - that's a problem with the method used to obtain the velocity profile.

Quote Quote by K^2 View Post
By the way, the actual answer to the boundary condition question is that in true inviscid flow, Kutta condition breaks down, and you get zero circulation. So viscosity is actually a necessary condition to generate lift.
Viscosity is necessary, yes, but simply through the application of the kutta condition without any other viscous effects (no boundary layers, skin drag, or viscous loss of any kind), a surprisingly accurate solution for lift and induced drag can be found. Obviously, the overall drag will be wrong (as viscous drag is not negligible), but the effects of viscosity don't really affect lift to a significant effect (so long as the kutta condition is applied to the inviscid case).
boneh3ad
#13
Apr11-12, 07:30 AM
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I'll add half a bit here. To expand on what cjl is talking about, the OP asked about calculating lift, not the nature of lift. For any low-speed airfoil (M<0.3) the inviscid velocity at any point on the surface almost exactly approximates the velocity at the boundary layer edge in the viscous case.

Because of the negligible wall-normal pressure gradient in an incompressible boundary layer, this means that any pressure calculated at the boundary layer edge from the edge velocity and Bernoulli's equation accurately reflects the pressure at the surface.

As long as one knows the velocity profile over a slow, inviscid wing (and the OP implies that he does) then one can validly use Bernoulli to get a very accurate value for lift and induced drag. After all, the assumptions made for Bernoulli to be valid are not violated anywhere except inside the boundary layer, and you are effectively ignoring the negligible effects of the boundary layer here. Of course if you want viscous drag that's a whole different story.

To answer the question about boundary conditions in an inviscid case, all one needs are the no penetration condition ([itex]\vec{V}\cdot\hat{n}=0[/itex]) and the Kutta condition (trailing-edge velocity is zero. That will be a unique solution that can utilize Bernoulli for lift and induced drag, though by its nature you would solve lift anyway by solving for the circulation and using the Kutta-Joukowski Theorem.
K^2
#14
Apr11-12, 08:52 PM
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Quote Quote by cjl View Post
Yes, they will. However, only one of them will also satisfy the condition of having the flow attached at the trailing edge, and that is the only one that makes physical sense (I'm assuming an unstalled airfoil here - flow separation is a different, and quite challenging topic).

Viscosity is necessary, yes, but simply through the application of the kutta condition without any other viscous effects (no boundary layers, skin drag, or viscous loss of any kind), a surprisingly accurate solution for lift and induced drag can be found. Obviously, the overall drag will be wrong (as viscous drag is not negligible), but the effects of viscosity don't really affect lift to a significant effect (so long as the kutta condition is applied to the inviscid case).
So lets see. In order to make an estimate using Bernoulli principle for lift, you are suggesting to treat gas as incompressible, and furthermore, treat it as viscous and inviscid fluid at the same time. You are getting further and further from the physics of lift to try and make an equation work. That's bad. That's kind of like saying Bohr's model is a good description of atom because for hydrogen atom you can get a good estimate fro the radius and ground state energy if you take a whole bunch of assumptions. None of it changes the fact that the very approach is wrong. It simply happens to work out at low air speeds for a high quality wing.


Quote Quote by cjl
You seem to be under the impression that the bernoulli relation says something about the way the flow will be shaped. This might be some of the reason for the confusion. Bernoulli actually says nothing at all about the streamlines or the flow velocity.
You do realize that for inviscid flow in steady state Navier-Stokes is just differential form of Bernoulli, right?
boneh3ad
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Apr11-12, 10:03 PM
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Quote Quote by K^2 View Post
So lets see. In order to make an estimate using Bernoulli principle for lift, you are suggesting to treat gas as incompressible, and furthermore, treat it as viscous and inviscid fluid at the same time. You are getting further and further from the physics of lift to try and make an equation work. That's bad. That's kind of like saying Bohr's model is a good description of atom because for hydrogen atom you can get a good estimate fro the radius and ground state energy if you take a whole bunch of assumptions. None of it changes the fact that the very approach is wrong. It simply happens to work out at low air speeds for a high quality wing.
At no point did he suggest that Bernoulli's equation explained the physics behind lift. Other than perhaps the OP, I think everyone here is pretty clear on that. However, Bernoulli's equation is absolutely valid at the edge of the boundary layer, where the pressure is effectively the same as at the surface, meaning that for a low-speed airfoil, you can, without violating any physical principles, get very nearly the exact lift of an airfoil from Bernoulli's equation provided you already know the velocity field around the airfoil from other means.

As far as the comment about treating the fluid as viscous and inviscid at the same time goes, you are missing the point. You can solve the boundary-layer edge velocity about an airfoil using potential flow if you simply take into account the viscous effects using the Kutta condition. This has been done for years successfully. Boeing, Airbus, Lockheed, Dassault and every other airframer you can name does a fair bit of airfoil design using this exact concept when initially designing their planes' wings. Of course, it then would require more robust methods such as the incorporation of a boundary layer solver and eventually a full, 3D CFD package, but the lift and induced drag on an airfoil can be calculated using potential flow using the Kutta condition for viscous correction with a very great degree of accuracy.
K^2
#16
Apr11-12, 11:48 PM
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Quote Quote by boneh3ad View Post
At no point did he suggest that Bernoulli's equation explained the physics behind lift. Other than perhaps the OP, I think everyone here is pretty clear on that.
That's kind of my whole point. OP's post leads me to conclude that he's probably confused. Therefore, talking about approximations we can make to force Bernoulli principle to be workable for computation of lift isn't helping.
You can solve the boundary-layer edge velocity about an airfoil using potential flow if you simply take into account the viscous effects using the Kutta condition.
I know. And if you are doing it to simplify your computations, it's entirely acceptable. But when you use zero circulation at boundary to get Kutta condition so that you can later say that you'll assume flow at boundary to use Bernoulli... Something went wrong in that logic. Incompressibility argument is much better, because that's the actual reason you can use Bernoulli as an approximation, even if the flow is viscous. However, even for a true incompressible fluid, it's an approximation.

But again, this is not where you start the conversation with the person who tries to use Bernoulli principle to compute lift. You start it by telling them to compute lift by other means.
mathsciguy
#17
Apr12-12, 10:13 AM
P: 132
Wow, my question seems to have spawned a lot of answers and an interesting discussion (though, most of those are all over my head for now).

All I want to know is that why do we omit [tex] \rho g \Delta y [/tex] when we calculate the pressure on the wings.

Yes, the question plane question I've been working on seems simple that the speed of the fluid over and under the wings are given.

Thanks for your help everyone.
cjl
#18
Apr12-12, 09:05 PM
P: 1,018
I thought I answered that in my first post. That term will only cause a difference of a few pascals between the top and bottom surface, which is completely negligible compared to the overall pressure difference (which will likely be thousands or tens of thousands of pascals).


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