Bernoulli's Principle and Static Gas Pressure


by Thomas2
Tags: bernoulli, pressure, principle, static
russ_watters
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#55
Mar24-05, 01:31 PM
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Quote Quote by Thomas2
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_st...el_final1.html
Q_Goest
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Mar24-05, 02:44 PM
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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/b...-bernoulli.htm
http://www.fluidmech.net/tutorials/b...ssumptions.htm
Q_Goest
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Mar24-05, 02:49 PM
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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
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Mar24-05, 04:14 PM
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Quote Quote by Thomas2
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
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Mar25-05, 09:21 AM
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Quote Quote by russ_watters
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
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Mar25-05, 10:19 AM
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Quote Quote by Thomas2
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 gonna 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.
Thomas2
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Mar25-05, 10:45 AM
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Quote Quote by russ_watters
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.
The question is whether the 'pressure' being measured is the pressure that the wing experiences. As I have indicated before in this thread, the 'pressure taps' in the wing will indicate a lower pressure for a tangential airflow because molecules from the air in the pressure tap are sucked into the airflow by means of viscous friction. The material surface of the airfoil will not experience this effect as its molecules can not move.
Andrew Mason
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Mar25-05, 01:51 PM
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Quote Quote by Q_Goest
Not really. The reason the gas cools upon expansion can be seen in the first law of thermodynamics.
If P decreases, the energy per unit volume of the fluid decreases, so the temperature must decrease. This was a response to Thomas2's statement:
But this would correspond to a decrease in temperature as the flow speeds up, which I don't think will be observed here
AM
Thomas2
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Mar26-05, 08:44 AM
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Quote Quote by Andrew Mason
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.
I don't actually think that a refrigerator would work with an incompressible gas (and incompressibility is the standard assumption in connection with applications of Bernoulli's principle).
Q_Goest
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Mar26-05, 09:00 AM
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I don't actually think that a refrigerator would work with an incompressible gas (and incompressibility is the standard assumption in connection with applications of Bernoulli's principle).
Correct. As stated previously, the simplified Bernoulli's equation assumes isothermal conditions.
russ_watters
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#65
Mar26-05, 10:09 AM
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Quote Quote by Thomas2
The question is whether the 'pressure' being measured is the pressure that the wing experiences. As I have indicated before in this thread, the 'pressure taps' in the wing will indicate a lower pressure for a tangential airflow because molecules from the air in the pressure tap are sucked into the airflow by means of viscous friction. The material surface of the airfoil will not experience this effect as its molecules can not move.
Yes, I know: hence your claim about most of the lift being generated on the back third of the wing (and, hence: your misunderstanding of what is going on in a venturi tube). I guess its just a coincidence to you that the results in both cases are exactly what aerodynamics predicts. But regardles, once again, wind tunnel tests prove that the pressure profile measured and predicted is real because it accurately models the lift on the wing. Once again, your model fails because it doesn't match reality.
russ_watters
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Mar26-05, 10:10 AM
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Regarding refrigerators: compressibility is largeley irrelevant as a cause of temperature fluctuations because the working fluid is undergoing a phase change. That's where most of the energy comes from.
Andrew Mason
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Mar26-05, 11:12 PM
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Quote Quote by Thomas2
I don't actually think that a refrigerator would work with an incompressible gas (and incompressibility is the standard assumption in connection with applications of Bernoulli's principle).
There is no such thing as an incompressible gas. If you are talking about a gas in a pipe, which is what I thought you were talking about, the gas compresses. Bernoulli's equation works quite well for a gas in a pipe, with a slight modification to factor in the energy that goes into the work of compressing the gas or the work done in expanding it.

If you are talking about air flow around a wing, the air is considered non-compressible because there is nothing to compress it against (like a pipe wall) so the air just moves out of the way and creates a compression wave that propagates away much faster than the wing speed.

AM
hydrologyphys
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Nov4-07, 01:03 AM
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An airfoil (wing of an airplane) is designed such that the upper surface of the wing is longer than the bottom. (a curved line is longer than a straight line of same horizontal displacement)

Now, think of a bend in a river. The water on the outer edge of the bend travels a longer distance over the same period of time as it takes the water on the innder edge to round the bend. Since velocity = distance/time, the water on the outer ede of the bend must travel at a faster velocity in order to travel a larger distance over the same period of time. Say the water travelling on the outer edge is travelling at velocity v2 and the water at the inner edge is travelling at v1. From this we have v2>v1. Similarly, v2 represents the velocity of the air flowing on top of the wing and v1 is the velocity of air flowing under the wing.

Now use Bernoulli's Equation: p1 -p2 = [1/2]rho(v2^2 - v1^2) where p is pressure, rho is the greek symbol representing density and v is velocity.
Density (rho) remains constant in a given system. (Assume the density of air does not change as it passes around the wing). THEN since v2>v1, p1>p2.

So the pressure on the bottom of the wing is greater. A resultant force is created due to the pressure different. (keep in mind pressure=force/area). So the force is acting perpendicular to the wing, thus holding the wing up.

Another way to derive this force is by considering the difference in velocities. Since
v2>v1, then V2 must be ACCELERATED from v1. There must be a net force inbetween the pressures in order to accelerate the wind.
arildno
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#69
Nov4-07, 06:52 AM
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The equal transit time principle is FALSE.
There is no natural law that says particles travelling along the outer bend must arrive at the same time as those travelling around the inner bend.

Utilization of the equal transit time principle yields totally false pressure distributions.
russ_watters
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#70
Nov4-07, 08:28 AM
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No need to open a 2.5 year old thread...


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