Pressure and Lift around a Wing

[boneh3ad]

Fog37, In reply I think you might want to look at what happens at the leading edge of the wing. I am getting close to the limits of my understanding here.
The air running into the leading edge curve is flung/forced upwards which compacts the air above the wing, at the same time making a lower pressure near the surface of the wing. The pressure layers higher up cannot exert downward force because they are being constantly compacted and separated from the wing by the up flow from the curved leading edge.
Pretty cool question you asked, as it made me reorganize what I "knew" about airfoils and think about the details more.

Air running into the leading edge is directed both over and under the wing, not simply over it. You are correct that the pressure away from the surface does not impact it directly (only the surface pressure acts on the surface), but it is related.

Consider the streamlines as they move around the airfoil over the top. Near the leading edge they are deflected upward and have a curvature whose center points away from the airfoil. Curved motion requires a centripetal force, which here must be provided by a pressure gradient. Since the pressure far from the airfoil must be atmospheric, this means that the pressure in that region must be locally higher than atmosphere. This shouldn't be surprising because this is the result of being near the stagnation point.

After that point, the curvature flips direction as the streamlines bent to follow the curve of the airfoil over the top. Now the center of curvative is in or below the wing and the pressure must therefore decrease as you approach the surface. There is therefore a low pressure near the surface at that point.

So the pressure far from the surface does not directly impact the lift force, but it definitely plays an indirect role here.

Added: Are you familiar with the triangular strips of metal that are put on the leading edge of the wings, near the fuselage to make that part of the wing stall before the outboard part where the ailerons are?
Rather interesting little detail on real-world wings.

That's not what those triangles do. Those are vortex generators and they actually delay stall. The turbulence they induce energizes the boundary layer near the wall and makes it more resistant to separation.

I think this is an excellent video. As an interested amateur, I wish that I had seen this decades ago. It really gives one a clue that any simple explanation necessarily omits a too much and is misleading when predicting lift and drag. That is why CFD programs are used in spite of the huge amount of computer power required and the limitations of the answers.

Well, CFD is quite effective for lift. It isn't terribly accurate for drag. Computers do a terrible job of predicting laminar-turbulent transition, for example, and as a result do a terrible job predicting separation.

 

[DarioC]

On the triangles, I think somehow you have misunderstood which "triangles" I meant. The ones I am talking about are very small long strips that are located on the leading edge of the inboard part of the wing. Their purpose is to disturb the airflow at high angles of attack and cause the inboard part of the wing to "stall" before the sections outboard that have the ailerons. The desired result is the you dump a lot of the lift of the wing before you loose roll control when the aileron area stalls. It allows you to get a good comfortable sink rate during the final flare. A Beachcraft Bonanza has them as I remember and some others I am sure. I suspect they are called something like leading edge spoilers.

They are completely different in appearance and function from the various vortex generator "tabs."

Of course air goes under the wing as well as over it, but I was talking about the air going over the wing specifically, I wasn't discussing the air going under the wing, that is why I didn't write anything about that part of the airflow.

Anyone who has compared the landing characteristics of a Piper cub versus a Taylorcraft will understand the significance of the shape of the wing leading edge. Interestingly enough those two planes were designed by the same person.

 

[OCR]

I suspect they are called something like leading edge spoilers.

Stall strips... eh ? .

.

 

[DarioC]

Yes that is one. Thanks for the photo.

 

[fog37]

Thank you DarioC , Boneh3ad and everyone else.

I am back to thinking about the fundamentals of pressure.

In a fluid at rest, the pressure at a hypothetical point $P$ could be measured using a Bourdon vacuum pressure gauge (the higher the spring compression the higher the pressure). The measured pressure at the infinitesimal point $P$ is the same no matter the orientation of the gauge (pressure is isotropic both in hydrostatics and fluid dynamics).

But when the fluid is moving, the pressure at a particular point along a specific streamline will be dependent on the fluid speed and on the way we measure it: if we placed Bourdon gauge facing the fluid upstream, the fluid would come to rest against the measuring surface of the gauge and the measured pressure would depend on the fluid speed (stagnation point): the higher the flow speed the higher the stagnation pressure. At that same stagnation point, with the first gauge remaining in place, the pressure measured by an extra Bourdon gauge oriented perpendicularly to the first gauge would be the same since the pressure is isotropic.

What if we don't create a stagnation point but use a Bourdon gauge oriented perpendicularly to the streamline? I guess it is called a tap if we measure that pressure at the surface of the pipe (pressure does not change across the streamlines). What pressure would the gauge measure in that case? This pressure would be smaller than the pressure at the stagnation point and the faster the fluid is moving along the streamline the smaller that pressure measured by the Bourdon gauge would be. In both case (fluid at rest or moving), the pressure $p$ in Bernoulli's theorem is the same. However, it is interesting that its value changes depending on the way we measure it (facing the fluid and creating a stagnation point or perpendicularly to a streamline).

In a pipe that has a narrowing, since the fluid must accelerate to maintain the same flow rate ($m^3 / s$), the pressure in the narrowing must be smaller to have a pressure gradient.Thanks!

 

[David Lewis]

The pressure on top of the wing is slightly lower than the free stream pressure. This pressure distribution produces a net force L_top on top of the wing directed downward.

Wouldn't the total force acting on the upper surface of the wing be pointing upward?

The pressure at the bottom of the wing is slightly larger than the free stream pressure farther away from the wing and produces a net lift force L_bottom.

The actual pressure distribution depends on the airfoil and angle of attack, but typically, while the pressure acting on the lower surface near the trailing edge may be above free stream, the rest of the bottom surface experiences pressures less than free stream. So the total force (on the lower surface) would be directed downward, but smaller than force acting on the upper surface.

(The variable L usually stands for lift, so it's better to represent total force by F, or F_upper, and F_lower).

 

[boneh3ad]

There is almost nothing about that analysis that makes sense to me. At the most fundamental level, a fluid cannot exert a pressure force on a surface in any direction other than normal to it and toward it. Any force on the upper surface points down, and any force on the lower surface points up.

 

[sophiecentaur]

Wouldn't the total force acting on the upper surface of the wing be pointing upward?

That goes against all the mechanics I ever learned. A gas can't 'suck'. The lift has to be because of the fact that there is more total (positive) pressure upwards than (positive) pressure downward.

 

[fog37]

The pressure above the surface is lower than the free stream atmospheric pressure. The pressure below the wing is higher than the free stream atmospheric pressure. A pressure gradient is directed from high pressure to low pressure which I assume that is where the net force point. would point.

We know lift is due to the pressure difference between the top and bottom surface. But, given the relative difference between free stream pressure and the pressures at the wing produce a net force point down on both the top and bottom surfaces?

 

[boneh3ad]

Again, it is literally impossible for a fluid to "pull" on a surface. The force is always directed normal to and into the surface. Always.

 

[fog37]

Ok, I finally get it. What matters is the local pressure distribution on the top and bottom surfaces of the wing to produce the net lift force.

A gradient is a vector pointing from small to large values. I guess I got confused about these negative of the pressure gradients, $- \nabla p$, one directed towards from the top surface and one directed away from the bottom surface.

 

[sophiecentaur]

a net force point down on both the top and bottom surfaces?

I suggest that you really meant to use the word towards ?

 

[David Lewis]

Diagram of lift coefficient distribution from Airplane Performance Stability and Control -- Perkins and Hage.

 

[FactChecker]

The pressure above the surface is lower than the free stream atmospheric pressure. The pressure below the wing is higher than the free stream atmospheric pressure. A pressure gradient is directed from high pressure to low pressure which I assume that is where the net force point. would point.

One needs to be careful about the difference between pressure and pressure gradient. The pressure force is always normal to the surface and directed from the gas toward the surface. The pressure gradient is derived from the difference in pressure. It is positive in the direction of increasing pressure. It is not the same as the force from the pressure and positive is in the opposite direction from the force.

We know lift is due to the pressure difference between the top and bottom surface. But, given the relative difference between free stream pressure and the pressures at the wing produce a net force point down on both the top and bottom surfaces?

The force on both surfaces is from the gas side toward and normal to the surface. If there is lift, the total force on the bottom is greater than the total force on the top.

 

[David Lewis]

The pressure gradient is derived from the difference in pressure.

The difference between which pressures?

The pressure force is always normal to the surface and directed from the gas toward the surface.

There is pressure (force/area), and then there is force. Which one are you analyzing? If it's force, then aerodynamic force has both normal and tangential components. The tangential component (skin friction) arises due to viscosity within the boundary layer.

 

[FactChecker]

The difference between which pressures?

Between the pressures at two infinitely close points. It's a derivative of the pressure.

There is pressure (force/area), and then there is force. Which one are you analyzing?

Sorry. I should have said the force due to the pressure.

If it's force, then aerodynamic force has both normal and tangential components. The tangential component (skin friction) arises due to viscosity within the boundary layer.

Good point. I was ignoring any tangential force for discussion of lift. Actually, this is not my area of expertise and I am not sure exactly how the viscosity relates to pressure in this situation.

 

[fog37]

Again, it is literally impossible for a fluid to "pull" on a surface. The force is always directed normal to and into the surface. Always.

Hi Boneh3ad,
I am still pondering about Bernoulli's equation: in Bernoulli's equation where the pressure $p$ is called static pressure and the term $\frac{\rho v^2}{2}$ is called dynamic pressure. Is the term $p$ called static because it is measured by a device that is not moving relative to the fluid? Any device moving at the same speed as the fluid would measure zero pressure. The dynamic pressure term can only be measure by bringing the fluid to rest (at a stagnation point). So it is the just the increment to $p$ when we measure $p$ exactly against the direction of fluid motion perpendicular to the streamline. We would only measure $p$ if we measured the pressure parallel to the streamline. Somehow that pressure parallel to the streamline is affected by the speed of the fluid (the faster the fluid the lower that pressure and the higher the stagnation pressure).

Does this make any sense?

 

[boneh3ad]

Is the term $p$ called static because it is measured by a device that is not moving relative to the fluid?

Static pressure is so named because it is the pressure that has to due with the fluid's state (as opposed to its motion). It is otherwise known as thermodynamic pressure or just pressure.

Any device moving at the same speed as the fluid would measure zero pressure.

This isn't true. A device moving at the same speed as a given fluid would still register a pressure due to the impact of the various fluid molecules on the sensing element.

The dynamic pressure term can only be measure by bringing the fluid to rest (at a stagnation point).

Dynamic pressure can never be measured. You can measure static pressure using a static port (i.e. a pressure port that does not disturb the flow in any way that alters its velocity, usually flush-mounted in a surface in the flow). You can measure total pressure (or stagnation pressure, if you prefer) with a Pitot probe that slows the flow down to zero at the measurement point. The different between these two allows you to calculate the dynamic pressure (in an incompressible flow).

So it is the just the increment to $p$ when we measure $p$ exactly against the direction of fluid motion perpendicular to the streamline. We would only measure $p$ if we measured the pressure parallel to the streamline. Somehow that pressure parallel to the streamline is affected by the speed of the fluid (the faster the fluid the lower that pressure and the higher the stagnation pressure).

I am having a hard time understanding what you are saying here. The first portion is very confusing, and the second portion seems to make a conceptual error in the phrase "pressure parallel to the streamline". Pressure is a scalar quantity. It has no direction, so all of this talk of perpendicular and parallel really doesn't make sense to me.

 

[russ_watters]

Any device moving at the same speed as the fluid would measure zero pressure.

Zero gauge pressure, perhaps (with respect to itself?), but it would still have absolute pressure.

Note: in practice, essentially all pressure is measured as gauge pressure. Even a supposed absolute pressure measuring device just uses a reference pressure of a vacuum. This is useful to remember for the next issue...

The dynamic pressure term can only be measure by bringing the fluid to rest (at a stagnation point).

That's total pressure. As said previously, dynamic pressure is total pressure minus static pressure. But you can measure dynamic pressure in one shot by using static pressure as the reference pressure, as in a pito-static tube.

So it is the just the increment to $p$ when we measure $p$ exactly against the direction of fluid motion perpendicular to the streamline. We would only measure $p$ if we measured the pressure parallel to the streamline.

You have some jumbled words there, but in general, its static pressure that is measured by orienting the pressure port parallel to the flow and total pressure or dynamic pressure measured by orienting it perpendicular to the flow.

Somehow that pressure parallel to the streamline is affected by the speed of the fluid (the faster the fluid the lower that pressure and the higher the stagnation pressure).

The first part isn't correct: Since static pressure is measured the same whether you are moving with the flow or not, it wouldn't be correct to say it is in general lower for higher speed fluid. For example, an airplane's static pressure port measures exactly the same static pressure regardless of its own speed. But the dynamic pressure is a function of speed.

 

[boneh3ad]

Zero gauge pressure, perhaps (with respect to itself?), but it would still have absolute pressure.

It wouldn't necessarily be zero gauge. It depends on what the gauge is measured against. For example, in wind tunnel testing, gauge pressure is usually measured against local atmosphere, which is not the same as the static pressure at the port location, so the gauge pressure would be nonzero.

Note: in practice, essentially all pressure is measured as gauge pressure. Even a supposed absolute pressure measuring device just uses a reference pressure of a vacuum. This is useful to remember for the next issue...

The definition of absolute pressure is pressure referenced to a vacuum, so in that sense, you are measuring both a gauge and absolute pressure simultaneously. They are the same in this instance.

The first part isn't correct: Since static pressure is measured the same whether you are moving with the flow or not, it wouldn't be correct to say it is in general lower for higher speed fluid. For example, an airplane's static pressure port measures exactly the same static pressure regardless of its own speed. But the dynamic pressure is a function of speed.

Be careful here. This applies to the static pressure port related to the Pitot-static system on an airplane only because those systems are designed to measure static pressure where the velocity is locally equivalent to the free stream. It would not be correct to apply this to static pressure in general. Consider a series of static pressure ports along the chord of an airfoil moving at constant velocity. They will all be registering different static pressure values owing to the varying flow speed over that portion of the 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.