Bernoulli's Principle and Static Gas Pressure

[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

 

[FredGarvin]

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'll take that as a nod to me and my fellow rotary wing bretheren Russ.

 

[Q_Goest]

I see the point you're getting at Thomas2, applying Bernoulli's like you would inside a converging/diverging nozzle seems to leave us with a paradox when applied to an aircraft wing.

I did find some interesting things on the web though, here at NASA:

Arguments arise because people mis-apply Bernoulli and Newton's equations and because they over-simplify the description of the problem of aerodynamic lift. The most popular incorrect theory of lift arises from a mis-application of Bernoulli's equation. The theory is known as the "equal transit time" or "longer path" theory which states that wings are designed with the upper surface longer than the lower surface, to generate higher velocities on the upper surface because the molecules of gas on the upper surface have to reach the trailing edge at the same time as the molecules on the lower surface. The theory then invokes Bernoulli's equation to explain lower pressure on the upper surface and higher pressure on the lower surface resulting in a lift force. ...

The real details of how an object generates lift are very complex and do not lend themselves to simplification. For a gas, we have to simultaneously conserve the mass, momentum, and energy in the flow.

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

And at this at Aeronautic Learning Lab for Science, Technology And Research (ALLSTAR)

The lift of a wing is equal to the change in momentum of the air it is diverting down. Momentum is the product of mass and velocity. The lift of a wing is proportional to the amount of air diverted down times the downward velocity of that air. Its that simple. (Here we have used an alternate form of Newton’s second law that relates the acceleration of an object to its mass and to the force on it; F=ma) For more lift the wing can either divert more air (mass) or increase its downward velocity. This downward velocity behind the wing is called "downwash".

Ref: http://www.allstar.fiu.edu/aerojava/airflylvl3.htm

 

[russ_watters]

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.

It may not be moving horizontally relative to the ground, but it is moving vertically relative to the ground. So there's still an increase in total speed and a corresponding decrease in static pressure. Trouble is, there is more to it than that (as you say and as Q_Goest discusses)...

The simplest explanation of lift may be downwash (if the plane is pushed up, some air must be pushed down), but that doesn't explain how the air gets pushed down, which is the main question here.

 

[Thomas2]

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

I know that the application of Bernoulli's principle assumes an inviscid and usually also an incompressible fluid. But as mentioned above, exactly these conditions would not result in a pressure drop for a moving fluid if you consider the problem on a microscopic level. I think this apparent inconsistency is simply due to a insufficient definitions of the physical terms. Consider for instance the notion 'static pressure': for a gas this is given by the random thermal motion of the molecules, but at the bottom of a water tank the pressure has nothing to do with the molecular motion but with the total weight of the water column above the point considered ( the pressure is here transmitted statically through intermolecular forces). The air pressure does also correspond to the weight of the air column, but here the density of the air decreases towards the bottom, so in this sense, one can't really say that air is incompressible and neither is the pressure really static but a gas dynamic effect. The density of water however is surely the same at everywhere in the water tank, and thus we have true incompressibility but on the other hand it can not be considered as inviscous anymore because it is just the static molecular interacation that transmits the pressure from the top to the bottom and throughout the tank. I think that these differences need to be kept in mind when applying Bernoulli's principle.

 

[russ_watters]

I know that the application of Bernoulli's principle assumes an inviscid and usually also an incompressible fluid. But as mentioned above, exactly these conditions would not result in a pressure drop for a moving fluid if you consider the problem on a microscopic level.

Well, since Bernoulli's equation accurately predicts what is seen in experiments and your idea does not, clearly your idea conflicts with reality.

 

[Q_Goest]

I know that the application of Bernoulli's principle assumes an inviscid and usually also an incompressible fluid. But as mentioned above, exactly these conditions would not result in a pressure drop for a moving fluid if you consider the problem on a microscopic level.

Thomas2, please correct me if I'm wrong, but I think what you're (still) trying to suggest is that if a fluid has a given density, the pressure should be the same in all directions regardless of how that fluid's velocity changes in relationship to a surface along one of the fluid flow's streamlines. If that's not what you're suggesting, my appologies. But you also stated that earlier:

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.

Antiphon had it right (as do a few others here.):

... 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 (normal to the flow velocity).

(emphasis mine)

In other words, the average velocity of all the molecules does not change. The bell curve shape of the velocity distribution of the molecules would not change between the inlet of a venturi and the throat for a fluid flowing at subsonic speed. But that requires the direction of all the molecules to go from 'random direction' (ie: the sum of all the vectors representing the velocity of all the molecules is zero) to some preferred direction which is a function of the velocity of the fluid as a whole. The 'average' velocity of all the molecules goes from zero to the velocity in the direction the molecules had to accelerate. Therefore, the velocity perpendicular to this direction must necessarily be reduced. That's why the pressure in this direction is also reduced.

 

[Thomas2]

Well, since Bernoulli's equation accurately predicts what is seen in experiments and your idea does not, clearly your idea conflicts with reality.

But the question (which has not been answered so far) is then why does Bernoulli's equation apparently reproduce experiments although on a microscopic level the conditions of an incompressible and inviscous flow lead to the logical conclusion that the pressure should not change if the flow is parallel to a surface.
I am not trying to promote any idea, but merely to understand the consequences of Bernoulli's equation on a microscopic level. It may be sufficient for an engineer to accept Bernoulli's principle as a macroscopic law, but for me as a physicist this is not enough.

 

[Thomas2]

But that requires the direction of all the molecules to go from 'random direction' (ie: the sum of all the vectors representing the velocity of all the molecules is zero) to some preferred direction which is a function of the velocity of the fluid as a whole. The 'average' velocity of all the molecules goes from zero to the velocity in the direction the molecules had to accelerate. Therefore, the velocity perpendicular to this direction must necessarily be reduced. That's why the pressure in this direction is also reduced.

I am generally not very fond of Thermodynamics arguments, but this would effectively mean you could create order from chaos. I just can't see it happening.

 

[FredGarvin]

I am generally not very fond of Thermodynamics arguments, but this would effectively mean you could create order from chaos. I just can't see it happening.

Please explain how you arrive at that conclusion.

 

[russ_watters]

But the question (which has not been answered so far) is then why does Bernoulli's equation apparently reproduce experiments although on a microscopic level the conditions of an incompressible and inviscous flow lead to the logical conclusion that the pressure should not change if the flow is parallel to a surface.
I am not trying to promote any idea, but merely to understand the consequences of Bernoulli's equation on a microscopic level. It may be sufficient for an engineer to accept Bernoulli's principle as a macroscopic law, but for me as a physicist this is not enough.

Simple: Bernoulli's equation does not apply on a molecular level. If you are only looking at one molecule, there is no such thing as "pressure". Thus, it is possible to ignore pressure and just consider the individual particles and the way they bounce off the wing, arriving at the erroneous conclusion you promote.

Please explain how you arrive at that conclusion.

Yes, please do - from that quote it would appear that we can add thermodynamics to the list of scientific fields you do not understand...

Q_Goest's objection is actually very similar to mine (excellent presentation, btw) - by ignoring the random motion and concentrating on linear motion and momentum, you're ignoring the entire concept of "pressure".

 

[Andrew Mason]

I am generally not very fond of Thermodynamics arguments, but this would effectively mean you could create order from chaos. I just can't see it happening.

I don't see why not. It happens all the time with a refrigerator. There is no principle that energy has to become more dispersed. It is just that it can't become less dispersed. (Entropy change > or = to 0).

Absent friction, the flow of a contained fluid is conservative of energy so it is analagous to a reversible thermodynamic process. There is no thermodynamic principle that prevents a fluid under pressure from doing work and then converting all that work back into potential energy. Think of a Carnot engine in which the flow of heat from the hot to cold reservoir is used to perform work that lifts a weight. The potential energy of the weight is then converted to back to work to reverse the heat flow from the cold to the hot reservoir. The result: a completely reversible process that loses no energy. Yet during part of that cycle, heat flows from cold to hot.

AM

 

[Thomas2]

Bernoulli's equation does not apply on a molecular level. If you are only looking at one molecule, there is no such thing as "pressure". Thus, it is possible to ignore pressure and just consider the individual particles and the way they bounce off the wing, arriving at the erroneous conclusion you promote

The only way air can exert a pressure on the wing is obviously by molecules bouncing off it. Of course it does not make much sense to use the pressure concept if you have only one molecule, but still each molecule contributes to the pressure. In this sense, it should be allowed to ask how the normal momentum transfer to a surface (i.e. the pressure) can possibly change if you impart to all molecules an additional velocity which is merely tangential to the surface.

 

[Thomas2]

Well, since Bernoulli's equation accurately predicts what is seen in experiments and your idea does not, clearly your idea conflicts with reality.

Apart from the theoretical arguments given in this thread already, I would actually question your assertion that Bernoulli's equation accurately predicts what is seen in experiments: if you have a horizontally symmetric wing profile as schematically shown in Fig.1 of my webpage http://www.physicsmyths.org.uk/bernoulli.htm , then Bernoulli's principle would predict a lift even for a zero angle of attack. However, I would not expect a lift here. I have not been able to do a corresponding experiment myself, but if somebody has the opportunity, I would challenge him to do it. Putting a correspondingly shaped object on a sensitive scales and blowing with a hairdryer over it might be sufficient to show it. One would have to be careful however that the airflow is homogeneous over the whole width of the object, because otherwise viscosity effects will arise as the adjacent resting air will be forced to move with the airflow (the latter aspect explains by the way most of the home experiments that are usually used to demonstrate Bernoulli's equation; I tried this myself for the case of two sheets of paper being attracted by blowing between them; if you cut down the paper sheets to two narrow strips about 1 cm in width (so that the whole width is covered by the flow), the effect vanishes in fact and the strips stay parallel to each other).

 

[Thomas2]

In other words, the average velocity of all the molecules does not change. The bell curve shape of the velocity distribution of the molecules would not change between the inlet of a venturi and the throat for a fluid flowing at subsonic speed. But that requires the direction of all the molecules to go from 'random direction' (ie: the sum of all the vectors representing the velocity of all the molecules is zero) to some preferred direction which is a function of the velocity of the fluid as a whole. The 'average' velocity of all the molecules goes from zero to the velocity in the direction the molecules had to accelerate. Therefore, the velocity perpendicular to this direction must necessarily be reduced. That's why the pressure in this direction is also reduced.

If you extract energy of random motion from the molecules and turn this into a systematic energy of flow, then in a reference frame moving with the flow, the shape of the velocity distribution function must obviously change because of energy conservation. But this would correspond to a decrease in temperature as the flow speeds up, which I don't think will be observed here (see also my reply to Andrew Mason below).

 

[Thomas2]

Absent friction, the flow of a contained fluid is conservative of energy so it is analagous to a reversible thermodynamic process. There is no thermodynamic principle that prevents a fluid under pressure from doing work and then converting all that work back into potential energy. Think of a Carnot engine in which the flow of heat from the hot to cold reservoir is used to perform work that lifts a weight. The potential energy of the weight is then converted to back to work to reverse the heat flow from the cold to the hot reservoir. The result: a completely reversible process that loses no energy. Yet during part of that cycle, heat flows from cold to hot.

I don't think this comparison can actually be made, because I am not aware that the pressure differences observed in a Venturi tube for example are associated with any heat exchange. Heat is anyway exchanged over much longer time periods than pressure differences as they rely on the relaxation of the velocity distribution function whereas pressure differences are exchanged with the speed of sound. Pressure equalization happens regardless of the temperature distribution and will have established itself way before any temperature changes occur, so these two can hardly be linked to each other.
Of course, if one assumes an incompressible medium i.e. a constant density (as is usually done in this context) the only way that the pressure can change should be the temperature according to the ideal gas law, but again, I am not aware that there is any temperature change associated with the speeding up of the gas flow in connection with Bernoulli's principle.

 

[Q_Goest]

If you extract energy of random motion from the molecules and turn this into a systematic energy of flow, then in a reference frame moving with the flow, the shape of the velocity distribution function must obviously change because of energy conservation.

Sorry, but I disagree with the assumption we've extracted any type of energy from the flow. First, conservation of energy still applies regardless of how that energy manifests itself. Second, we're not talking about a real flow, just a close approximation. Bernoulli's assumes only a conservation of mechanical energy, and disregards thermal. That's a close approximation so long as velocity is not too high and the overall length of the streamline is short.

In this case, the kinetic energy of any given molecule remains the same. It is simply being directed or steered in a given direction. A car traveling on a frictionless set of railroad tracks that makes a turn does not have any change in it's kinetic energy. Similarly, there is no change in the kinetic energy of the individual molecules as they are steered in a given direction. The overall kinetic energy of the flow is changed, but only at the expense of loosing it in another direction. Note that Bernoulli's is "frictionless" such that no mechanical energy is lost, and this is not a real case. In reality, some energy is converted.

But this would correspond to a decrease in temperature as the flow speeds up, which I don't think will be observed here …

You're correct in saying the temperature will NOT decrease, but that's because the kinetic energy of the molecules is assumed to stay constant.

Imagine gas in a box with molecules in random motion. The pressure is the same in all directions. But given an infinite amount of time, there can be a state where all the molecules just happen to be moving in the same direction at the same time. Needless to say, that's highly unlikely, but it's not statistically impossible. When that happens, we find the dynamic pressure in the direction of motion is higher while the pressure perpendicular to the motion decreases. A venturi simply creates that affect for molecules flowing through it.

Also, what Andrew said:

Absent friction, the flow of a contained fluid is conservative of energy so it is analagous to a reversible thermodynamic process.

… is correct, except it's not analagous to a reversible process. It IS a reversible process. There is no dQ/dT so entropy change = 0 The entire process is isothermal as you've mentioned before and I agree. But that's only because of the assumptions made, not because it really is purely isothermal. Real fluid flows experience real pressure drops, but given the assumptions made for Bernoulli's the result is a constant entropy process. The entropy throughout the flow is isentropic. If you Google 'Bernoulli's isentropic' you'll find the isentropic assumption is valid.
http://www.google.com/search?hl=en&q=bernoulli+isentropic

I just wanted to also mention that you raise a lot of very incitefull questions, and I think we all learn more from that type of questioning than we can learn without it. So I sincerely would like to thank you for flushing out all these considerations, I certainly feel I've had to learn things better in order to provide a valid argument.

 

[Andrew Mason]

The only way air can exert a pressure on the wing is obviously by molecules bouncing off it. Of course it does not make much sense to use the pressure concept if you have only one molecule, but still each molecule contributes to the pressure. In this sense, it should be allowed to ask how the normal momentum transfer to a surface (i.e. the pressure) can possibly change if you impart to all molecules an additional velocity which is merely tangential to the surface.

The point is that there is no source of energy for the additional velocity in the direction of the flow. The energy has to come from a reduction in the velocity perpendicular to the direction of the flow (ie. the pressure - or potential energy / volume of the fluid)

AM

 

[Andrew Mason]

If you extract energy of random motion from the molecules and turn this into a systematic energy of flow, then in a reference frame moving with the flow, the shape of the velocity distribution function must obviously change because of energy conservation. But this would correspond to a decrease in temperature as the flow speeds up, which I don't think will be observed here (see also my reply to Andrew Mason below).

That is exactly why a refrigerator works: By allowing a gas at high pressure to pass through a constricted space and accelerate! The accelerated gas has lower temperature and draws heat from the surroundings.

AM

 

[Andrew Mason]

Also, what Andrew said:

… is correct, except it's not analagous to a reversible process. It IS a reversible process. There is no dQ/dT so entropy change = 0 The entire process is isothermal as you've mentioned before and I agree.

I really mean a "contained and thermally isolated system", so energy does not change. If it is isothermal, the internal energy will change with changes in pressure and then B's law will not strictly apply.

Real fluid flows experience real pressure drops, but given the assumptions made for Bernoulli's the result is a constant entropy process. The entropy throughout the flow is isentropic. If you Google 'Bernoulli's isentropic' you'll find the isentropic assumption is valid.
http://www.google.com/search?hl=en&q=bernoulli+isentropic

Isentropic means that there is no overall change in entropy. There can be + and - changes in entropy in parts of the cycle. If the system is isothermal, there will be a net increase in entropy. But if it is adiabatic, there can be 0 change in entropy (with ideal conditions). So I think Isentropic and Isothermal are incompatible.

AM

 

[russ_watters]

The only way air can exert a pressure on the wing is obviously by molecules bouncing off it. Of course it does not make much sense to use the pressure concept if you have only one molecule, but still each molecule contributes to the pressure. In this sense, it should be allowed to ask how the normal momentum transfer to a surface (i.e. the pressure) can possibly change if you impart to all molecules an additional velocity which is merely tangential to the surface. [emphasis added]

Yes, indeed - that is the nature of your misunderstanding. Your premise (in bold) is wrong. You are ignoring how that tangential component affects the pressure.

Apart from the theoretical arguments given in this thread already, I would actually question your assertion that Bernoulli's equation accurately predicts what is seen in experiments: if you have a horizontally symmetric wing profile as schematically shown in Fig.1 of my webpage http://www.physicsmyths.org.uk/bernoulli.htm , then Bernoulli's principle would predict a lift even for a zero angle of attack. However, I would not expect a lift here. I have not been able to do a corresponding experiment myself, but if somebody has the opportunity, I would challenge him to do it.

Here is something you need to get onboard with: all cambered airfoils produce lift at 0 AoA. On short notice, THIS is the best I could do, but if you scroll halfway down the page, you'll see a flat-bottom airfoil that produces lift down to about -3* aoa. Please note, since there is a slight curve in the leading edge, the chord line is not parallel to the airfoil bottom: so -3* aoa is actually about -5* for the bottom surface of the wing.

Every 1st year aerospace engineering student has measured the lift on a flat-bottomed wing in a wind tunnel. That's why I'm so incredulous about this whole thing: in the hundred years since the Wright brothers did it, literally millions of people have done wind tunnel tests. I even own an RC plane with a flat-bottom wing: the construction manual is very clear in saying that you need to set the nose gear height for a negative angle of attack, otherwise the plane will take off on its own. How can you be so arrogant as to think that every one of those millions of people missed something so basic when you haven't even looked yourself? You yanked something out of the air and you guessed wrong. Deal with it! (learn from it)

Here is something you can do yourself: http://hsc.csu.edu.au/engineering_studies/aero_eng/aeronautical/Wind_Tunnel_final1.html

 

[Q_Goest]

If it is isothermal, the internal energy will change with changes in pressure and then B's law will not strictly apply.

From Cambridge University Press:

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/...e-bernoulli.htm

In other words, when applying the Bernoulli equation with the incompressible assumption, only kinetic and potential energies are exchanged. The incompressible form of Bernoulli's equation only accounts for this mechanical energy. The internal energy does NOT change along any streamline.

You could also derive this knowing the density is constant and the enthalpy is constant.
Since H = U + PdV,
where H = Enthalpy
U = Internal Energy
PdV = Density change
and since H and PdV is constant, then U is also constant.

If the system is isothermal, there will be a net increase in entropy.

I disagree. If the system is incompressible (ie: constant density), and there is no change in internal energy as described above, the following fall out:
1) The system is isothermal
2) The system is isentropic
3) The system is isenthalpic
4) The system is adiabatic

Note that this version of Bernoulli's equation is very simple in terms of the assumptions it makes. Perhaps that's why it seems so counterintuitive, because it is not very realistic in its assumptions.

Check the following references, they all refer to this type of flow as isentropic:
http://www.grc.nasa.gov/WWW/K-12/airplane/isentrop.html
http://astron.berkeley.edu/~jrg/ay202/node87.html
http://astron.berkeley.edu/~jrg/ay202/node91.html
http://www.fluidmech.net/tutorials/bernoulli/compressible-bernoulli.htm
http://www.fluidmech.net/tutorials/bernoulli/assumptions.htm

 

[Q_Goest]

Andrew:

That is exactly why a refrigerator works: By allowing a gas at high pressure to pass through a constricted space and accelerate! The accelerated gas has lower temperature and draws heat from the surroundings.

Not really. The reason the gas cools upon expansion can be seen in the first law of thermodynamics.

The flow is isenthalpic.
The flow undergoes irreversible expansion given by PdV. Note this is for a real fluid, and is not predicted by the Bernoulli equation we're talking about here.
If H = U + PdV
and if H stays constant
and if PdV changes,
The internal energy of the fluid must change which results in a cooling (or heating) of the fluid.

 

[FredGarvin]

The only way air can exert a pressure on the wing is obviously by molecules bouncing off it. Of course it does not make much sense to use the pressure concept if you have only one molecule, but still each molecule contributes to the pressure. In this sense, it should be allowed to ask how the normal momentum transfer to a surface (i.e. the pressure) can possibly change if you impart to all molecules an additional velocity which is merely tangential to the surface.

You are not taking into account turbulent boundary layer theory. Look at theory for a flat plate. In particular, look at momentum transfer across the boundary layer. It is solely due to the mixing involved ("random transport of finite-sized fluid particles associated with turbulent eddies.). It stands to reason that an increase in tangential velocity increases boundary layer turbulence and thus increases the net transfer of momentum across the boundary layer.

 

[Thomas2]

Yes, indeed - that is the nature of your misunderstanding. Your premise (in bold) is wrong. You are ignoring how that tangential component affects the pressure. Here is something you need to get onboard with: all cambered airfoils produce lift at 0 AoA. On short notice, THIS is the best I could do, but if you scroll halfway down the page, you'll see a flat-bottom airfoil that produces lift down to about -3* aoa. Please note, since there is a slight curve in the leading edge, the chord line is not parallel to the airfoil bottom: so -3* aoa is actually about -5* for the bottom surface of the wing.

It is not about the flat bottom but about the horizontal (front/back) symmetry of the airfoil profile. This airfoil in your reference is still asymmetric having a larger surface of wing downstream (i.e. in the shadow of th.e airflow) than upstream (exposed to the airflow). Show me one that is symmetric (i.e. has the highest point of the camber in the middle) or maybe even one with usual profile reversed and still produces a lift.

 

[russ_watters]

It is not about the flat bottom but about the horizontal (front/back) symmetry of the airfoil profile. This airfoil in your reference is still asymmetric having a larger surface of wing downstream (i.e. in the shadow of th.e airflow) than upstream (exposed to the airflow). Show me one that is symmetric (i.e. has the highest point of the camber in the middle) or maybe even one with usual profile reversed and still produces a lift.

Sorry, you're going to halfta make that one yourself. It is your claim, after all. I will reiterate though: all cambered airfoils produce lift. And remember your previous claim: that the front part is pushed down while the back part is pulled up? That one was refuted by the force/pressure diagrams. The lift is not distributed the way you think it is, for the same reason you think a horizontally symmetrical and flat wing will not produce a net lift.