Bernoulli's Principle and Static Gas Pressure

[Thomas2]

According to the popular interpretation of 'Bernoulli's Principle', moving air should always be associated with a lower static gas pressure than resting air, but clearly this can not be correct:

If one considers a pipe with air resting in it, then the static pressure on the inside of the wall (due to the random motion of molecules) is given through the ideal gas law as p=n*k*T (where n is the volume number density of air, k the Boltzmann constant and T the temperature).
Now consider the air in the pipe not resting but moving uniformly through it, i.e. assume that the amount of air leaving the pipe at one end is exactly replaced by the amount of air entering the pipe at the other end. Because the total amount of air is the same as for the stationary case, the density n is the same as well and hence also to static gas pressure p=n*k*T (assuming that the gas temperature is unchanged).
The walls of the pipe experience therefore the same static pressure whether or not the air is moving in it (in contradiction to the popular view based on Bernoulli's Principle).

 

[dextercioby]

I don't know where and how you came up with this "popular" obviously incorrect interpretation of Bernoulli's principle...

Daniel.

 

[Thomas2]

I don't know where and how you came up with this "popular" obviously incorrect interpretation of Bernoulli's principle...

Well, just do a corresponding Google search and you will find a lot of sites claiming this (e.g. http://nasaui.ited.uidaho.edu/curriculum/velocity.htm ).

 

[FredGarvin]

Look up the definition of total pressure. Static pressure is only one component of total pressure. For anything to flow in an enclosed vessel, i.e. a pipe, there must be a pressure differential.

 

[Andrew Mason]

According to the popular interpretation of 'Bernoulli's Principle', moving air should always be associated with a lower static gas pressure than resting air, but clearly this can not be correct:

You are quite right. The popular interpretation is wrong or, at best incomplete, as are most explanations of Bernouilli's principle in explaining airplane wing lift.

Bernouilli's principle is simply a statement about conservation of energy in a dynamic fluid. Pressure is a measure of energy/unit volume and represents potential or stored energy. If the pressure changes, then the energy density changes. If the quantify of fluid is constant, the energy has to go somewhere. It does. It results in change in kinetic energy of the fluid.

Furthermore, motion is relative. One cannot say that a volume of gas is 'moving' and another 'at rest' without defining a reference frame. So it cannot at all be correct to say that moving air has a lower static gas pressure than resting air, because we can't say, in absolute terms, which one is moving.

One can only speak about changes in pressure and velocity. Essentially Bernouilli's law says that for a given quantity of fluid in thermal isolation and not subject to external forces (ie. external pressures), if that fluid experiences an increase in kinetic energy (ie. speed), the pressure of that fluid must decrease so that the increase in kinetic energy of fluid is equal to the drop in pressure x volume.

AM

 

[arildno]

Andrew Mason:
Add to that the all-important condition that this relation between pressure&velocity holds between two points on the SAME streamline.

 

[Thomas2]

Look up the definition of total pressure. Static pressure is only one component of total pressure. For anything to flow in an enclosed vessel, i.e. a pipe, there must be a pressure differential.

If the flow is uniform as assumed (i.e. it has the same speed at both ends of the pipe) there can not be a pressure differential as the latter would correspond to a force and hence accelerate the flow (assuming the air to be non-viscous). As an example, imagine for instance a pipe (with a sufficiently large diameter to prevent friction) moving longitudinally through the stationary air. Clearly no pressure differential is necessary here for the air to flow through the pipe, and obviously it would be wrong to assume that in this case the static pressure on the walls of the pipe would be reduced (as Bernoulli's Principle would suggest).

 

[FredGarvin]

Good post AM.

Thomas2,

I agree that the examples you cited on that website do over simplify the explanation by saying flow equals lower static pressure than non flow conditions. All of those examples do require viscous flow to work, which is a limitation of Bernoulli. If that is what you are saying, then I misread your original post.

 

[Thomas2]

I agree that the examples you cited on that website do over simplify the explanation by saying flow equals lower static pressure than non flow conditions. All of those examples do require viscous flow to work, which is a limitation of Bernoulli.

It is true that not all sources explicitly state the connection between pressure and flow speed in this blunt way, but even serious scientific references dealing with aerodynamics imply in fact the same thing through the simplifying assumptions that are being made for the corresponding physical models:
all corresponding derivations actually assume not only zero viscosity but also incompressibility. See for instance the NASA webpage www.grc.nasa.gov/WWW/K-12/airplane/bern.html and note the statement (halfway down the page) 'Assuming that the flow is incompressible, the density is a constant' (which obviously must be the case). Now have a look at the page dealing with the concept of stream lines around an airfoil http://www.grc.nasa.gov/WWW/K-12/airplane/stream.html . I quote from this page: 'Since the streamline is traced out by a moving particle, at every point along the path the velocity is tangent to the path. Since there is no normal component of the velocity along the path, mass cannot cross a streamline. The mass contained between any two streamlines remains the same throughout the flowfield. We can use Bernoulli's equation to relate the pressure and velocity along the streamline. Since no mass passes through the surface of the airfoil (or cylinder), the surface of the object is a streamline'.
Now taking these statements together, this obviously results in the logical conclusion that the density n along the whole of the airfoil is constant i.e. equal to original density of the undisturbed air. Since one certainly can assume the temperature T to be constant as well, this means the static pressure p=n*k*T is constant along the airfoil. According to these assumptions, aerodynamic lift should therefore be impossible altogether (obviously, the dynamic pressure could not contribute either with this 'model' as the velocity is always tangential to the airfoil).

 

[Clausius2]

Now taking these statements together, this obviously results in the logical conclusion that the density n along the whole of the airfoil is constant i.e. equal to original density of the undisturbed air. Since one certainly can assume the temperature T to be constant as well, this means the static pressure p=n*k*T is constant along the airfoil. According to these assumptions, aerodynamic lift should therefore be impossible altogether (obviously, the dynamic pressure could not contribute either with this 'model' as the velocity is always tangential to the airfoil).

I think density remains constant along the airfoil. Lifting is not an exclusive propierty of compressible flows. In incompressible flow there is a decoupling of energy transport and momentum transport. In fact one can solve

$$\nabla \cdot \overline{v}=0$$

$$\overline{v}\nabla\overline{v}=-\frac{1}{\rho}\nabla P+\nu\nabla^2\overline{v}$$

$$\overline{v}\cdot\nabla T=\alpha\nabla^2 T$$

and obtain a lift component with constant density. The temperature will reshape itself according to the momentum transport. At low Mach numbers making use of the equation of state has little sense.

 

[Thomas2]

I think density remains constant along the airfoil.
...
The temperature will reshape itself according to the momentum transport.

I can't really see how (in the absence of any heat sources or sinks) the temperature of the air could change along a streamline given the assumption of incompressibility (which is usually made for small velocities). Since therefore neither the density n nor the temperature T changes along a streamline, the pressure p=n*k*T should consequently be constant as well.

 

[Clausius2]

I can't really see how (in the absence of any heat sources or sinks) the temperature of the air could change along a streamline given the assumption of incompressibility (which is usually made for small velocities). Since therefore neither the density n nor the temperature T changes along a streamline, the pressure p=n*k*T should consequently be constant as well.

I can't really see how the temperature of the air is NOT going to change. As I have said before, the expression P=nkT has NO sense for incompressible aerodynamics. Your substance is incompressible, and this formula is given for compressible substances. In incompressible flow, the pressure reshapes itself into the fluid field in order to yield the continuity of mass, and it is decoupled of temperature variations, which are a CONSEQUENCE of the transport of Internal Energy due to convective and diffusive phenomena (See the last equation I wrote: it is the Navier-Stokes Internal Energy Equation).

Various effects enhance a change of temperature (spatial variations):

i) Although initial temperatures of surrounding air and airfoil were the same, heat transfer by convective effects will change the temperature of the airfoil. If you're so lucky to have a thermal resistance which makes possible to control the airfoil temperature to be the same of the surrounding air (namely $$T_{\infty}$$) then the energy equation would yield a constant temperature along the flow field. Let's see. Try to solve

$$\overline{v}\cdot\nabla T=\alpha\nabla^2 T$$

with $$T=T_{\infty}$$ at $$x\rightarrow\infty$$ and $$x=x_{airfoil}$$

The solution of this problem is $$T=T_{\infty}$$ everywhere.

ii) Of course this last result is far away from reality, because I have neglected viscous dissipation, which surely has a strong effect inside the boundary layer and will change the temperature field.

Pay attention: No matter the temperature will be constant in our "imaginary" experiment I have modeled theoretically above, the pressure is NOT the same along the flow field. The proper deflection of the flow over the airfoil is going to reorder pressure gradients to yield the conservation of mass.

As a conclusion, the equation of state in an incompressible fluid is NOT P=nkT, but $$\rho=constant$$, which is very different. The density is NOT a function of the thermodynamic state. Moreover, fluid-mechanical variables (velocity & pressure) are DECOUPLED of fluid-thermal variables (Temperature).

 

[Q_Goest]

Hey Thomas, maybe this will help a bit. The assumptions for Bernoulli's are:

- points 1 and 2 lie on a streamline,
- the fluid has constant density,
- the flow is steady, and
- there is no friction

Ref: http://www.princeton.edu/~asmits/Bicycle_web/Bernoulli.html

Also:

Incompressible Flow:
The question of whether or not the density of the air flowing around an airplane will change is crucial in Aerodynamics. Generally the electro-chemical forces between the air molecules attempt to keep the molecules a certain distance apart (that distance depends on the temperature and pressure.) Thus, the molecules will move as required to maintain a constant Air Density, if possible. This will only be possible if the flow is at a speed well below the speed of sound.
Thus, if the velocities involved are well below the speed of sound the flow can be thought of as incompressible. That means that when we apply Bernoulli's equation to a given flow we will assume that the Air Density does not change. As a result when Velocity increases, Static Pressure decreases.

Ref: http://142.26.194.131/aerodynamics1/Basics/Page3.html

So the incompressible (constant density) assumption is only valid for velocity below the speed of sound. (Still, that gives us plenty of velocity to play with!)

From Cambridge University Press:

Physics
With respect to the Bernoulli equation, the main difference between a compressible and incompressible flow is that the variations in the pressure in a compressible flow will result in compressions and expansions of the fluid blob as it moves along its own path. If the blob were stationary, the work done by the compressions and expansions would be converted to and stored as internal energy, at least if there is no energy loss during the compressions and expansions. In general, the internal energy is comprised of both thermal energy and energy associated with the intermolecular forces, although it is simply proportional to the temperature in the perfect gas model. Thus, in a compressible flow, energy must be exchanged not only among the kinetic energy and the potential energies due to gravity and pressure, but also with the internal energy. Because the energy principle used to derive the incompressible form of the Bernoulli equation only accounts for the mechanical energy, the (independent) law of conservation of energy must be employed to completely describe the energy exchange in a compressible flow.

Ref: http://www.fluidmech.net/tutorials/bernoulli/compressible-bernoulli.htm

What all this says is that for the incompressible case, internal energy and density remain constant, so you are safe to say that temperature does not change for the incompressible flow case. You are correct in saying the temperature does not change. This is true for a gas, and it's true for a liquid of course as well. For the compressible flow case, temperature does change.

But that isn't your question. The real question is, "Why does the flow across a wing result in a lower pressure?" (hope I got that right! lol)

Imagine a given mass of gas moving through a converging/diverging nozzle. Imagine just one gram of air and the volume it would take up, and you are traveling along with this one gram of air. When static, you might imagine this volume as being a cube shape.

As this gram of air accelerates past the smallest point in the nozzle, the volume (in order to maintain constant density) stretches in the direction of motion. In other words, the flow accelerates and the cube shape this volume took before must take on a rectangular shape as it accelerates. So:
i) the length of the cube in the direction of motion increases, and
ii) the length of the cube in the direction perpendicular to the motion decreases.

The total pressure of this deformed cube stays constant, as does the density. That's what Bernoulli's tells us. But the total pressure is made up of 3 different forms, the static, dynamic, and head pressure. For a streamline going past a wing, we can ignore changes in head pressure since we will assume horizontal flight.

The static pressure then decreases as dynamic pressure increases as the small cube of air accelerates. But this is with respect to a stationary observer. If one were to stand in the center of the converging/diverging nozzle, you might sense the velocity increase because the dynamic pressure on you could be felt just as you feel the push of the wind. Similarly, that stationary observer would sense a lower static pressure (ie: perpendicular to the direction of motion).

A physical explanation for the lower static pressure for the stationary observer might best be explained by examining the cause of the pressure. Pressure is caused by the change in momentum of the molecules. As the molecules bounce off the surface of a wing or internal surface of the nozzle, they impart a small force, and it's the sum of these forces which results in pressure. In the case for the moving fluid, the impacts in the direction of motion perpendicular to the direction of flow are spread out. I don't know how to explain that any better, I guess it's a visualization you have to think about for a while. But anyway, the static pressure decreases, and static pressure is the pressure in the direction perpendicular to the direction of motion.

______________________________________________________

"Beware of those who stand on soap boxes. They are not there to help you but to impress themselves."
~ Anonymous.

 

[russ_watters]

Look up the definition of total pressure.

This is the answer to the OP. Static pressure isn't the only kind of pressure present in the pipe when the air is moving, Thomas2. When the air is stationary, it is. So the total pressure when the air is moving is equal to the static pressure when the air is stationary (which is kinda the whole point of Bernoulli's equation).

Thomas2, you are falling into a common trap where you assume that if an equation has all the variables you are looking for (the ideal gas law in this case), you can apply it wherever you want. This is not the case. In this case, the ideal gas law may provide all the variables you want - but it doesn't have all the variables you need.

Andrew_Mason, for a question about Bernouli's principle in a pipe, lift on a wing is not relevant (though I do know that that's where Thomas2 is going). Let's try to keep lift out of it for now.

edit: Its also interesting that you're making separate, simultaneous claims, Thomas2: that Bernoulli's equation is wrong and also just that its interpreted (or applied) wrong.

 

[Clausius2]

To say the truth I am a bit lost in this thread. I haven't got the point about what is being discussed here. So don't take my previous comments seriously. I think I have misunderstood what Thomas meant. Anyway, all what I posted I think is true (although maybe not related with the subject of this thread by the way).

 

[Thomas2]

...

Of course this last result is far away from reality, because I have neglected viscous dissipation, which surely has a strong effect inside the boundary layer and will change the temperature field.
...
As a conclusion, the equation of state in an incompressible fluid is NOT P=nkT, but n=constant which is very different. The density is NOT a function of the thermodynamic state. Moreover, fluid-mechanical variables (velocity & pressure) are DECOUPLED of fluid-thermal variables (Temperature).

First of all, the 'incompressibility' of a gas is clearly a very relative property. All gases are in principle compressible as there is a lot of empty space between the molecules. It is just that for certain densities and temperatures one needs large forces to compress it, i.e. the compression will in many cases only be small and consequently one may treat it then as 'incompressible'. Obviously, the density can in this case be taken as constant, but the fact that the ideal gas law p=nkT is restricted by the condition n=const. does not mean that it it is not valid anymore (see for instance http://www.grc.nasa.gov/WWW/K-12/airplane/eqstat.html where the ideal gas law is in fact used in connection with aerodynamics).
You mentioned that viscous dissipation should lead to a variation of the temperature along the streamline of an airfoil, but a) the air is usually assumed to be non-viscous for these basic considereations of aerodynamic lift and b) dissipation of energy would anyway lead to an increase of the temperature and hence the static gas pressure, whereas what you want to produce lift is a pressure decrease.

 

[Clausius2]

First of all, the 'incompressibility' of a gas is clearly a very relative property. All gases are in principle compressible as there is a lot of empty space between the molecules. It is just that for certain densities and temperatures one needs large forces to compress it, i.e. the compression will in many cases only be small and consequently one may treat it then as 'incompressible'. Obviously, the density can in this case be taken as constant, but the fact that the ideal gas law p=nkT is restricted by the condition n=const. does not mean that it it is not valid anymore (see for instance http://www.grc.nasa.gov/WWW/K-12/airplane/eqstat.html where the ideal gas law is in fact used in connection with aerodynamics).
You mentioned that viscous dissipation should lead to a variation of the temperature along the streamline of an airfoil, but a) the air is usually assumed to be non-viscous for these basic considereations of aerodynamic lift and b) dissipation of energy would anyway lead to an increase of the temperature and hence the static gas pressure, whereas what you want to produce lift is a pressure decrease.

I repeat again I think I am out of the aim of this thread.

If you are treating with Incompressible substances, the equation of state is not valid and not needed to solve the flow field. If you are treating with Gas dynamic at low Mach Numbers but allowing a short compressibility, the equation of state is another constitutive equation in order to solve the Navier Stokes equations.

I don't know how basic is the lifting force, but it is CLOSELY related to the viscous behavior of the fluid.

 

[arildno]

Clausius2:
You might consider viewing the following thread prior to engaging in a discussion with Thomas2:
https://www.physicsforums.com/showthread.php?t=64384

 

[Clausius2]

Clausius2:
You might consider viewing the following thread prior to engaging in a discussion with Thomas2:
https://www.physicsforums.com/showthread.php?t=64384

Ok. Message received. Thanks for forewarning me.

 

[Thomas2]

In the meanwhile I have looked into some of the issues that have come up in the course of this discussion in some more detail. Despite looking through several aerodynamics theory books , I could actually not find any indication that the pressure profile along an airfoil should be temperature related (as Clausius suggested above). As mentioned by me already, this raises the question how the static pressure can change if the density stays the same (which it should under the assumption of an incompressible flow).

I also found some apparent confusion in the literature regarding the definition of 'viscosity' and 'friction' in this context. In aerodynamics theory, an inviscous fluid is claimed to result in a flow without drag on objects (see http://www.centennialofflight.gov/essay/Theories_of_Flight/Ideal_Fluid_Flow/TH7.htm , http://astron.berkeley.edu/~jrg/ay202/node95.html ). This assumption actually equates friction with viscosity, which is clearly incorrect. Viscosity is generally defined as the presence of 'shear stress' i.e. a friction between parallel streamlines. However, even if the viscosity is zero in this sense (i.e. if the molecules of a gas are assumed not to be interacting with each other), there will be a friction and a resulting momentum loss by molecules hitting the object and being elastically reflected (this is easy to show from the energy and momentum conservation laws for an elastic collision of two objects). Even in an ideal non-viscous gas, an object will therefore lose momentum due to friction and hence experience drag (a plane surface whose normal has an angle alpha to the gas flow experiences a normal force proportional to cos(alpha) and thus the drag is proportional to cos^2(alpha) whereas the lift is ~sin(alpha)*cos(alpha) ).

 

[Nenad]

I don't like how people say that the pressure decreases as kinetic energy increases. Total Pressure stays the same, where as with greater speed, dynamic pressure increases while static pressure decreases.

Regards,

Nenad

 

[Antiphon]

According to the popular interpretation of 'Bernoulli's Principle', moving air should always be associated with a lower static gas pressure than resting air, but clearly this can not be correct:

[example of pipe moving through air]

The pipe does not contradict B's law because both inside and outside of
the pipe measure an equal (and oppositely directed) pressure drop.

At the risk of being presumptuous, let me try to reivigorate the original
spirit of the question:

Does anyone have a good "kinetic gas" explanation for how Bernoulli's
law actuall works at a fluid/material interface? The energy conservation
arguments are sound and elegant but don't leave the amateur fluid
dynamicists like myself with an obvious mental picture of the actual
mechanism of pressure drop...

 

[Andrew Mason]

Does anyone have a good "kinetic gas" explanation for how Bernoulli's law actuall works at a fluid/material interface? The energy conservation arguments are sound and elegant but don't leave the amateur fluid
dynamicists like myself with an obvious mental picture of the actual
mechanism of pressure drop...

The pressure drop is the result of potential energy (pressure x volume = energy) stored in the fluid being converted into kinetic energy of the fluid. Pressure is really a measure of the potential energy density of the fluid. If some of that is converted into kinetic energy, the potential energy density of the moving fluid has to decrease. The mechanism is 'work'. Pressure decreases because it does work in accelerating the fluid.

AM

 

[Antiphon]

The pressure drop is the result of potential energy (pressure x volume = energy) stored in the fluid being converted into kinetic energy of the fluid. Pressure is really a measure of the potential energy density of the fluid. If some of that is converted into kinetic energy, the potential energy density of the moving fluid has to decrease. The mechanism is 'work'. Pressure decreases because it does work in accelerating the fluid.

AM

Ahh thanks. I think I can translate your accurate description into the
following qualitative picture now. Please correct it wherever it's erroneous.

The fluid is caused to speed up (by shaping an airfoil or pipe constriction
etc) but no little or no work is done on the (compressible or not) fluid in
the act of constriction. This means the gas must speed up as it is
contricted, and this in turn means the average momentum of the fluid
molecules is "vectored" more parallel to the surface. But because we
are not adding additional energy, this reoriented momentum comes with
a decrease in the fluid momentum normal to the surface, which
translates directly into a decrease in pressure.

 

[russ_watters]

You pretty much have it, Antiphon. The consevation of energy also explains why in the ideal case (a Venturi tube with no friction), the velocities are equal ahead of and behind the "pinch" in the tube: potential (pressure) energy is converted to kinetic and then back to potential while mass flow rate/momentum are equal at all points in the tube.

 

[Thomas2]

This means the gas must speed up as it is
contricted, and this in turn means the average momentum of the fluid
molecules is "vectored" more parallel to the surface. But because we
are not adding additional energy, this reoriented momentum comes with
a decrease in the fluid momentum normal to the surface, which
translates directly into a decrease in pressure.

The average momentum of the molecules is parallel to the walls of the pipe before and after the constriction, so there is no 'vectoring' of the flow. As I mentioned above already, the molecular pressure on the walls is given by the density and the temperature of the gas or fluid and if these are the same (which they should be for an incompressible medium) the pressure on the walls should also be the same. In this sense,it is clear that the observed apparent decrease of pressure for faster moving fluids is only due to its viscosity: a moving medium will pull molecules of the neigbouring medium into the flow due to viscosity and thus reducing the number of molecules (and hence the pressure) in the neigbouring medium. This is why corresponding devices measure a lower pressure adjacent to faster moving flows. It does not mean that the pressure in the flow itself is also reduced.

 

[Andrew Mason]

The average momentum of the molecules is parallel to the walls of the pipe before and after the constriction, so there is no 'vectoring' of the flow. As I mentioned above already, the molecular pressure on the walls is given by the density and the temperature of the gas or fluid and if these are the same (which they should be for an incompressible medium) the pressure on the walls should also be the same. In this sense,it is clear that the observed apparent decrease of pressure for faster moving fluids is only due to its viscosity: a moving medium will pull molecules of the neigbouring medium into the flow due to viscosity and thus reducing the number of molecules (and hence the pressure) in the neigbouring medium. This is why corresponding devices measure a lower pressure adjacent to faster moving flows. It does not mean that the pressure in the flow itself is also reduced.

The pressure change is unrelated to viscosity. A non-viscous fluid will have the same change in pressure. The decrease in pressure in the flow is real, not merely apparent.

AM

 

[russ_watters]

...it is clear that the observed apparent decrease of pressure for faster moving fluids is only due to its viscosity... [emphasis added]

Ehh, no - the Bernouli effect actually depends on inviscid flow. But you already know that since we've discussed it before...

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

 

[arildno]

Besides, I would like to add, what is pertinent in a flight discussion is the pressure distribution NORMAL to the wing, i.e, the typical vertical pressure distribution.

Since Bernoulli's equation relates quantities along a streamline, rather than across them, I do not find Bernoulli's equation as the most natural starting point for the discussion of the flight phenomenon.

 

[russ_watters]

Since Bernoulli's equation relates quantities along a streamline, rather than across them, I do not find Bernoulli's equation as the most natural starting point for the discussion of the flight phenomenon.

Well, that's true, but its clear that Thomas2 just plain doesn't understand how air works - and Bernoulli's is about as simple and basic of a concept as you can get to start off with in fluids. Its where any intro to aero or fluids course starts.

 

[arildno]

This wasn't meant to be taken as a criticism of either your or AM's good efforts in this thread.

It was more like a sigh from me over the unjustified importance Bernoulli's equation has gained in the common/popular explanations of flight.

 

[Andrew Mason]

Since Bernoulli's equation relates quantities along a streamline, rather than across them, I do not find Bernoulli's equation as the most natural starting point for the discussion of the flight phenomenon.

I have never really understood how Bernouilli's law explains wing lift. If the air is stationary and the wing just passes through it, the airflow (relative to the earth) is the same above and below the wing. So the pressure can't be different.

It seems to me that wing lift is a combination of a few things, but mainly the upward deflection of air by the leading edge to create an upward airflow over the wing, thereby creating a partial vacuum over the wing.

AM

 

[russ_watters]

I have never really understood how Bernouilli's law explains wing lift. If the air is stationary and the wing just passes through it, the airflow (relative to the earth) is the same above and below the wing. So the pressure can't be different.

Motion is motion: whether its the air moving toward the wing or the wing moving toward the air, the effect is exactly the same.

 

[PBRMEASAP]

This wasn't meant to be taken as a criticism of either your or AM's good efforts in this thread.

It was more like a sigh from me over the unjustified importance Bernoulli's equation has gained in the common/popular explanations of flight.

Then what makes an airplane fly?

Also, doesn't Bernoulli's law hold between any two points (not just along streamlines) if you have irrotational flow? That is a reasonable approximation (i think) outside the boundary layer, so wouldn't that make Bernoulli's law applicable?

 

[Andrew Mason]

Motion is motion: whether its the air moving toward the wing or the wing moving toward the air, the effect is exactly the same.

I agree. The equivalence of the Earth and wing frames makes the point.

I was thinking of the explanation one often sees of the air speed relative to the wing surface being greater on top than the bottom of the wing because the path over the wing is longer. Since the air covers that longer distance in the same time, it is said to be moving faster (in the wing frame) so its pressure is less. But in the rest frame of the earth, it is not moving horizontally at all so its pressure cannot possibly be less due to horizontal flow. That is why I don't think Bernouilli's law applies to wing lift. It has a superficial appeal but upon closer examination it does not make sense.

AM