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Plane's wings and Bernoulli's equation 
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#1
Apr1012, 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? 


#2
Apr1012, 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. 


#3
Apr1012, 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. 


#4
Apr1012, 04:32 PM

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Plane's wings and Bernoulli's equation
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/m^{3}, 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). 


#5
Apr1012, 05:19 PM

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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. 


#6
Apr1012, 05:37 PM

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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. 


#7
Apr1012, 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 NavierStokes 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. 


#8
Apr1012, 09:19 PM

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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. 


#9
Apr1112, 12:20 AM

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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? 


#10
Apr1112, 04:13 AM

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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). 


#11
Apr1112, 06:59 AM

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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... 


#12
Apr1112, 07:08 AM

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#13
Apr1112, 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 lowspeed 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 wallnormal 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 (trailingedge 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 KuttaJoukowski Theorem. 


#14
Apr1112, 08:52 PM

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#15
Apr1112, 10:03 PM

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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 boundarylayer 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. 


#16
Apr1112, 11:48 PM

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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. 


#17
Apr1212, 10:13 AM

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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. 


#18
Apr1212, 09:05 PM

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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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