Pressure and Lift around a Wing

[fog37]

Hello and thanks for the replies.

The red surface is a rigid surface that measures the pressure.

Case 1: The fluid is inviscid and traveling at speed $V_1$. The pressure $P_A$ measured by the red surface is less than the pressure $P_B$ measured at the stagnation point.

Case 2: The fluid is inviscid and traveling at speed $V_2 = 2 V_1$. The pressure $P_C$ measured by the red surface is less than the pressure $P_D$ measured at the stagnation point. But pressure $P_C < P_A$. Why? if the fluid was not moving at all, i.e. $V=0$, would the pressure measured by the horizontal red surface be larger than $P_A$?

 

[FactChecker]

The comparison is not true unless you have related the two diagrams so that their total energy is the same. The typical way to do that would be to make the second tube follow the first with a restrictive venturi between and the area of the second tube is 1/2 that of the first. That would make $V_2 = 2 V_1$. One way to intuit the reduced pressure $P_C < P_A$ is to realize that the venturi would have resisted the pressure from the first tube so that a higher static pressure, $P_A$, was required to "squirt" the fluid through at twice the velocity into the second tube.

 

[256bits]

The red surface is a rigid surface that measures the pressure.

How did you solve for the problem of water coming out of a hole in a tank?
One has to take into account the pressure differential between upstream and downstream, and in your two cases the second one has a higher ΔP, hence the greater velocity. If the exit pressures are the same in both cases, what does that say about P_A and P_C, and the exit pressure? What is the pressure in the line?
Maybe you are thinking about a venturi where the static pressure within the venture will drop for same entrance and exit pressures.

 

[russ_watters]

Case 2: The fluid is inviscid and traveling at speed $V_2 = 2 V_1$. The pressure $P_C$ measured by the red surface is less than the pressure $P_D$ measured at the stagnation point. But pressure $P_C < P_A$.

That isn't necessarily true (it would only be true if something in the way you created the airflow made it true). Please note my airplane example where the static pressure measured by the plane (atmospheric pressure) is independent of the speed of the plane.

 

[fog37]

Thanks.

well, I am assuming two different cylindrical pipes in which the fluid flows at two different speeds (due to different conditions upstream and downstream that we don't worry about) But It could well be a Venturi tube in which case the pipes diameters are different.

I think it is fair to say that the stagnation pressure $P_B > P_D$ since the fluid in the 2nd figure will impact the red surface at a higher speed.

And I still think that along the streamline, the pressure $P_C$ would be less than the pressure $P_A$ but I am not sure, conceptually, why the liquid molecules would exert a lesser pressure if the pass by the horizontal surface faster. Do they have less time to impact? But if the go faster it also means more liquid molecules would end up impacting with the horizontal red surface and that could make up for the fact that the impact less in that direction.

Then I wonder if the liquid thermodynamic pressure at that same point would be higher, if we stopped the liquid and the liquid parcels were not flowing to the right of the red surface...

 

[jbriggs444]

I am not sure, conceptually, why the liquid molecules would exert a lesser pressure if the pass by the horizontal surface faster. Do they have less time to impact?

Trying to think in terms of molecules and impacts usually distracts from the mathematical facts of the situation. A high speed liquid exerts less pressure on the passing surface because the liquid has less pressure. It has less pressure because there must have been a pressure gradient in order to accelerate it to the higher speed.

If you must think in terms of molecules, think in terms of almost rigid balls that are not packed so tightly together in the area where their speed is higher.

 

[FactChecker]

well, I am assuming two different cylindrical pipes in which the fluid flows at two different speeds (due to different conditions upstream and downstream that we don't worry about)

Sorry that I didn't make my point clearly. Do the two have the same total energy or not? The only reason that I mentioned the setup with the venturi is because that guarantees equal total energy. Without saying something about total energy, no conclusions can be reached.

 

[russ_watters]

well, I am assuming two different cylindrical pipes in which the fluid flows at two different speeds (due to different conditions upstream and downstream that we don't worry about)

I'm sorry, but that just isn't tightly enough constrained. With what you're specifying there, the static pressures in the two tubes could be almost anything and there is no way to know without specifying something, which one will be higher. E.G.:

But It could well be a Venturi tube in which case the pipes diameters are different.

That helps. That satisfies @FactChecker's constraint that the total pressures be equal. In that case, Pc<Pa.

I think it is fair to say that the stagnation pressure $P_B > P_D$ since the fluid in the 2nd figure will impact the red surface at a higher speed.

It's not. You're mixing-up stagnation (total) pressure and dynamic pressure. The dynamic pressure is higher and per your constraint (a Venturi tube), the total/stagnation pressures are equal.

And I still think that along the streamline, the pressure $P_C$ would be less than the pressure $P_A$ but I am not sure, conceptually, why the liquid molecules would exert a lesser pressure if the pass by the horizontal surface faster. Do they have less time to impact?

No. Again, static pressure in and of itself is not a function of speed. In a Venturi you make the static pressure drop by using it to accelerate the fluid.

But if the go faster it also means more liquid molecules would end up impacting with the horizontal red surface.

This isn't true. Think about running or driving in the rain. A horizontal surface can't gather more rain because the total amount of rain reaching the ground is fixed. But a vertical surface "sweeps up" rain as it travels. That's what happens here with the two different pressure port orientations.

Then I wonder if the liquid thermodynamic pressure at that same point would be higher, if we stopped the liquid and the liquid parcels were not flowing to the right of the red surface...

I don't know what "thermodynamic pressure" is, but we're using a low-speed; incompressible, constant density, constant temperature flow assumption here.

 

[FactChecker]

Intuitively:
Assume that the total energy is identical in the two figures. Look at a molecule in each and their velocity vectors and assume that they have the same energy (same vector length). If the vector in the second figure points more in the direction of flow, then it must point less toward the side. In other words, the faster that molecule flows downstream, the less it impacts the side. The relationship is a simple Pythagorean theorem.

 

[fog37]

Thank again. The more correct situation would be this:

Pressure $P_B >P_C$ and $P_D>P_B$. My dilemma was about $P_A$ and $P_C$ and why would one be smaller from the other from the conceptual point of view. These pressures are called "static" but was is static is the red measuring surface, not the fluid. If the fluid was stopped, the pressures at these two locations would be different than when the fluid is in motion and be about equal (neglecting the difference in depth).

 

[FactChecker]

It's almost trivial. No energy is added as the fluid flows through the system. So the average magnitude of the velocity vectors of the molecules remain constant. At position C, the velocity vectors point more in the direction of flow than they do at A. Because their average length has not increased, there are smaller velocity components in the direction pressing on the red horizontal lines at $P_C$ than at $P_A$. That means static pressure $P_C < P_A$.

 

[jbriggs444]

My dilemma was about $P_A$ and $P_C$

In addition to the reasoning used by @FactChecker...

In order for the flow to be as you assert with V₂ > V₁, it must be the case that there is a pressure gradient causing the flow to accelerate rightward (or decellerate leftward in the case of a flow moving right to left). Either way, $P_a > P_c$

 

[fog37]

Ok, thanks. So you are saying that the velocity vector magnitude at B is the same as the magnitude at point A and point C

"so the average velocity vectors of the molecules remain constant magnitude."

From the figure below, the pressure $P_B < P_A$ but the velocity vector at B is angle downward with a nonzero component towards the horizontal red surface A increasing the pressure while at point A the fluid particles don't have a bulk fluid velocity component perpendicular to the red surface. But that is not what happens since $P_B < P_A$ . I would argue that may the fact the fluid is incompressible and has to pass through a smaller cross-section it has to speed up hence the magnitudes of the velocity vectors at points A,B,C are different, with the fluid being the fastest at point C.

what about the pressure at point C where the tube is not sloped anymore? The pressure $P_C <P_B$.

 

[FactChecker]

I should have been more careful in my wording. To have constant energy, the magnitude of the velocity vectors, averaged over all molecules at a point, should remain constant. They will all point in random directions if there is no flow. If there is flow, they will tend to point toward the direction of flow. In any case, the amount of flow does not change the average magnitude -- only the direction they tend to point.

You have drawn the red line at position B horizontal, so it is not a static pressure. For it to be a static pressure, it should be tilted along a flow line. Otherwise, it will include some dynamic pressure. Because $P_B$ includes some dynamic pressure, you can not be sure that $P_B < P_A$

 

[boneh3ad]

Thank again. The more correct situation would be this:

View attachment 226925

Pressure $P_B >P_C$ and $P_D>P_B$. My dilemma was about $P_A$ and $P_C$ and why would one be smaller from the other from the conceptual point of view. These pressures are called "static" but was is static is the red measuring surface, not the fluid. If the fluid was stopped, the pressures at these two locations would be different than when the fluid is in motion and be about equal (neglecting the difference in depth).

"Static" pressure is so named because it is the pressure associated with the state of the fluid rather than the motion of it. In other words, it is the thermodynamic pressure associated with the fluid's equation of state. It has nothing to do with what is or isn't moving, and it is not frame-dependent.

 

[fog37]

thanks boneh3ad.

So, "static" pressure is essentially the thermodynamic pressure which does not depend on the fluid state of motion but only on the fluid temperature, density and fluid composition.

That said, how do we call that gauge pressure that we measure with vertical piezometric tubes attached to the walls of a pipe having a fluid flowing inside?

I get what the stagnation pressure is: it is juts the pressure that we measure when a flowing fluid is brought to rest by a surface oriented perpendicular to the fluid direction. The stagnation pressure is alway larger than the pressure measured by a surface oriented in any other direction.

When the surface is horizontal, like at point A and point C in my sketch in #103, the pressure should be the same (but it is not) if does not depend on the state of motion of the fluid. The fluid at C is faster hence the pressure is lower...

 

[russ_watters]

That said, how do we call that gauge pressure that we measure with vertical piezometric tubes attached to the walls of a pipe having a fluid flowing inside?

Gauge pressure is just any pressure measured with a gauge, having a readng with respect to a reference. It doesn't denote a specific type of pressure (static, dynamic, steam, water, whatever). The alternative is absolute pressure, which is measured against or converted to a zero reference pressure.

You can measure/express static or total pressure in gauge or absolute form. Velocity pressure is by nautre a gauge pressure (achieved by subtraction).

 

[fog37]

Those vertical piezometers measure gauge pressure relative to the atmosphere: the height reached by the fluid in the vertical tube linearly correlates with the gauge pressure: zero height means the pressure inside the fluid has the same value as atmospheric pressure. I think I am correct on this.

The height in those piezometric tubes varies depending on the pressure and in regions where the pressure is lower the height is lower. The pressure is not the stagnation pressure but the pressure measured when the red surface is oriented parallel to the flow like $P_A$ in thread#103.

 

[boneh3ad]

So, "static" pressure is essentially the thermodynamic pressure which does not depend on the fluid state of motion but only on the fluid temperature, density and fluid composition.

It certainly does depend on the fluid's state of motion. If a fluid accelerates through a nozzle, the static pressure decreases. This is what Bernoulli's equation is essentially saying. It can be derived from a momentum balance or an energy balance, but either way, it shows an inverse relationship between pressure and velocity. It is still a state variable, and as it changes, so to do the fluid density and temperature according to an equation of state (e.g. the ideal gas law).

When the surface is horizontal, like at point A and point C in my sketch in #103, the pressure should be the same (but it is not) if does not depend on the state of motion of the fluid. The fluid at C is faster hence the pressure is lower...

No they should not be the same. At those two points, the fluid velocity is different, so the static pressures will be different. What I meant by the comment about static pressure being a state variable and not an artifact of the fluid's velocity relative to some frame of reference. It fundamentally arises from the random motion of molecules, not the bulk motion of the fluid. A fish tank sitting on a truck driving by you will have the same pressure inside as a fish tank sitting still that you drive by in your car. The dynamic pressure, such as the concept even makes sense in this context, would be different.

However, static pressure must still obey the laws of thermodynamics, which means if a fluid accelerates without changing its total energy, that increase in kinetic energy had to come from somewhere. Static pressure represents one of the possible pools of energy from which kinetic energy (dynamic pressure) can be drawn.

 

[FactChecker]

So, "static" pressure is essentially the thermodynamic pressure which does not depend on the fluid state of motion but only on the fluid temperature, density and fluid composition.

I think "does not depend" is not exactly correct since there is often a trade-off between static and dynamic pressure if no energy is added. It's better to say that static pressure represents the remaining energy after the dynamic energy of fluid velocity has been accounted for. When the trade-off between static and dynamic pressure changes, such as a gas going through a venturi, the temperature instantly changes accordingly. The temperature in the throat of a venturi is lower.

One way of looking at this is the following: Suppose there is a fluid whose molecules have a given set of velocity vectors. You can determine the average motion and velocity, $V_{avg}$, of the fluid by averaging all the velocity vectors of the molecules. That tells you how much energy can be represented by the dynamic motion of the fluid. Now suppose that we subtract $V_{avg}$ from every molecular velocity vector. The result is a set of vectors with no direction. That represents the static pressure. The temperature is directly associated with the static pressure. The more fluid flow, the greater the magnitude of $V_{avg}$, and (given constant total energy) the lower the remaining static pressure is.

There are other ways to look at the same thing and they all tell the same story -- Bernouli is correct. You can look at it as the pressure differential causing the increased velocity in a venturi. You can also look at it as conservation of total energy. When something is true, there are often many ways to convince yourself of the truth.

PS. I doubt that my explanation is very formally correct. I like the intuative, geometric aspect of it. But for the real formal explanation, I would recommend the posts of @boneh3ad ,@jbriggs444 , and @russ_watters

 

[fog37]

Thanks FactChecker.

So, for the same overall kinetic energy, when the fluid is moving faster, the fluid trades some of the chaotic kinetic energy that was first used by the colliding molecules to produce static pressure for kinetic energy that is now used by the molecules to move in a more organized fashion in a specific direction. For a compressible fluid like a gas, when the static pressure decreases, also the fluid temperature decreases (this does not happen in incompressible fluids).

In the extreme case when the fluid is not moving, i.e. when $V_{avg} = 0$, all the pressure is static (hydrostatic).

 

[FactChecker]

Thanks FactChecker.

So, for the same overall kinetic energy, when the fluid is moving faster, the fluid trades some of the chaotic kinetic energy that was first used by the colliding molecules to produce static pressure for kinetic energy that is now used by the molecules to move in a more organized fashion in a specific direction. For a compressible fluid like a gas, when the static pressure decreases, also the fluid temperature decreases (this does not happen in incompressible fluids).

In the extreme case when the fluid is not moving, i.e. when $V_{avg} = 0$, all the pressure is static (hydrostatic).

That's how I like to look at it. I doubt that my explanation is very formally correct. I like the intuative, geometric aspect of it. But for the real formal explanation, I would recommend the posts of @boneh3ad ,@jbriggs444 , and @russ_watters

 

[fog37]

I like the intuitive perspective.

I think the splitting of pressure, defined as the normal compressive force per unit area on an infinitesimal surface, into static and dynamic terms is confusing and confused me. Pressure is one and always isotropic and can be affected by the speed of the flow and by how we measure it. For example the dynamic pressure term becomes a contribution to pressure when the flow is brought to rest at a stagnation point at which the isotropic pressure, being the in that case the sum of the static and dynamic pressures, is increased. But even the fluid is moving and not stopped, its pressure is affected by the fluid speed via the term $0.5 \rho v^2$.

 

[FactChecker]

For example the dynamic pressure term becomes a contribution to pressure when the flow is brought to rest at a stagnation point at which the isotropic pressure, being the in that case the sum of the static and dynamic pressures, is increased.

That seems wrong. I think the total sum of the static and dynamic pressures is a constant unless energy is added.

 

[fog37]

I agree that the sum of the two pressures is constant. But there is really only one pressure, which is $p$. The dynamic pressure is the extra pressure that would build up in addition to the static pressure if the fluid was brought to a stop. The dynamic pressure at some point in a flow is a hypothetical pressure, i.e. the pressure if the fluid was brought to rest. It is the pressure increase obtained from integrating the deceleration along a streamline so that the pressure stops the fluid from moving.

 

[arydberg]

Not sure i should even get involved here but here goes. I am a x glider pilot. One of the maneuvers we did when being towed up to altitude was a called "boxing the wake" Starting from 200 feet directly behind the tow plane we would descend until we encountered the turbulence of the wake from the tow plane. We would keep going through the turbulent area until we were below it. We would then move to the left of the wake rise up then move to the right of the wake and descend to the bottom of the wake. Then rise up in the center of the wake.

Having been 200 feet behind a tow plane and getting bounced around by the extreme turbulence my take is that it is the air coming off the bottom of the wing and being forced down that develops a upward force on the plane.

You might look for a gliding club and ask about it or even take a rede.

 

[arydberg]

The shape of the wing forces air to go 'down', so therefore the wing obtains a force going 'up' (lift).

It is the angle of attack of the wing that forces the air to go down.

 

[FactChecker]

Not sure i should even get involved here but here goes. I am a x glider pilot. One of the maneuvers we did when being towed up to altitude was a called "boxing the wake" Starting from 200 feet directly behind the tow plane we would descend until we encountered the turbulence of the wake from the tow plane. We would keep going through the turbulent area until we were below it. We would then move to the left of the wake rise up then move to the right of the wake and descend to the bottom of the wake. Then rise up in the center of the wake.

Having been 200 feet behind a tow plane and getting bounced around by the extreme turbulence my take is that it is the air coming off the bottom of the wing and being forced down that develops a upward force on the plane.

You might look for a gliding club and ask about it or even take a rede.

Very interesting comment from pilot experience. But I think it would be practically impossible for a pilot to distinguish between air pushed down by impacting the bottom of the wing versus air directed down after going above the wing. In fact, calculations of force from air hitting the bottom of the wing usually come up way short of the actual lift force. So the complete story is more complicated than that.

 

[arydberg]

Interesting. Where can i find some of those calculations.

 

[FactChecker]

Interesting. Where can i find some of those calculations.

Do you mean the "correct" calculations or just the calculations of the impact on the bottom?
I have never seen any calculations only of the impact on the bottom. Just imagine ignoring all the air flow over the top that gets redirected in this photo. Obviously that would ignore a lot. I don't think there is a reason to do a serious calculation like that.

The reason for all that downward redirection of the flow over the wing is complicated. Any simple calculation is seriously flawed. The most "correct" calculations are done using Computational Fluid Dynamics (CFD). (see https://en.wikipedia.org/wiki/Computational_fluid_dynamics) Those calculations are massively complicated because everything effects everything else. Even the CFD results are very limited. They must be adjusted to agree with wind tunnel results and then adjusted again as each flight test is done.