How dynamic and static pressures are related

[thetexan]

I teach aerodynamics and I’m looking for a way to explain how d and s pressure are related.

I understand that as dynamic increases static decreases with the total sum remaining the same.

With dynamic being zero the total goes to static. As dynamic increases the static pressure decreases. But why? Dynamic is measured parallel to the surface and static measured normal to the surface. And dynamic is related to velocity of the air.

Is there a way to explain this relationship between velocity and dynamic pressure and between dynamic and static pressures that are relatively easy to present?

It as if when velocity across the surface (wing) increases the fluid (air) doesn’t “have time” to spend exerting static pressure perpendicular to the surface. And the slower the velocity the more normal pressure that can be exerted.

I’m looking for a good way to explain this phenomena.

Any help would be appreciated.

Thanks

Tex

 

[CWatters]

I don't teach aerodynamics so I could be wrong but I thought static pressure was independent of velocity. What I mean is for something like an aircraft or submarine its related to height/depth only not the velocity of the fluid flow.

 

[thetexan]

Well...

There is a relationship between fluid velocity and dynamic pressure...the faster the velocity the higher the dynamic pressure although I don’t completely understand how velocity and viscosity fit into the dynamic pressure formula.

And...consequently, the higher the dynamic pressure the lower the static pressure. And that is what I want to know how that works. The total pressure equals total dynamic pressure plus total stati pressure...and the total sum of dynamic and static pressure is a constant for a given situation.

I don’t understand that and how it works and it makes it hard to explain it to students.

I need help

Tex

 

[CWatters]

Ah Ok you are talking about Bernoulli's principle. It derives from conservation of energy.

A unit of water flowing into a section of pipe has some energy. It has KE and PE (which includes static pressure). If the pipe narrows, forcing the water to speed up, it gains KE. Conservation of energy means it must loose PE (which includes static pressure).

 

[bkercso]

"As dynamic increases the static pressure decreases. But why?"

When applying impact pressure to a fluid or gas, it is accelerated, after that its pressure decrases because of energy conservation (Bernoulli law). Static pressure is the pressure you could feel inside this fluid or gas. Dynamic pressure is what you feel when try to stop this flowing medium.

 

[boneh3ad]

There is a relationship between fluid velocity and dynamic pressure...the faster the velocity the higher the dynamic pressure although I don’t completely understand how velocity and viscosity fit into the dynamic pressure formula.

Viscosity has absolutely nothing to do with what you are asking here.

And...consequently, the higher the dynamic pressure the lower the static pressure. And that is what I want to know how that works. The total pressure equals total dynamic pressure plus total stati pressure...and the total sum of dynamic and static pressure is a constant for a given situation.

I don’t understand that and how it works and it makes it hard to explain it to students.

Dynamic pressure isn't really a pressure at all, but a measure of the kinetic energy of the gas due to bulk motion per unit volume. After all, look at the equation,
q = \dfrac{1}{2}\rho V^2.
It's the same as the physics 1 equation for kinetic energy, except it uses density, so it is basically a kinetic energy density. If the kinetic energy due to bulk motion changes, that had to come from somewhere. Barring a change in elevation, the only other source is the pool of kinetic energy that exists due to the random motion of air molecules, otherwise known as static pressure.

Dynamic is measured parallel to the surface and static measured normal to the surface.

What does this even mean? Both of these quantities are directionless. They are scalar.

Dynamic pressure is what you feel when try to stop this flowing medium.

Under no circumstances can you "feel" dynamic pressure. Dynamic pressure cannot be felt or directly measured. In that sense, it is a somewhat abstract quantity. If you, say, stick your hand out the window of your car and feel the pressure on it as you drive, that is total pressure or stagnation pressure, which is the sum of static and dynamic pressure. You'd have to measure total and static pressure and subtract static from total in order to calculate the dynamic pressure (and velocity). This is how a Pitot static tube works.

 

[thetexan]

Then this is what I need to understand.

On a wing the static pressure is measured as psi on a unit of area normal to the surface. That’s what we care about...the difference of the psi on top of the wing compared to the psi on the bottom of the wing.

To me that represents a force of pressure pressing against the wing. That’s easy to for me to understand. So what exactly is dynamic pressure AS A PRESSURE and how is it related to static pressure so that when it increases the pressure measured against the wing decreases. Why does increasing one reduce the other. I guess what I’m trying to understand is the mechanism of that relationship.

Tex

 

[russ_watters]

Why does increasing one reduce the other. I guess what I’m trying to understand is the mechanism of that relationship.

The mechanism is energy conversion between static and kinetic (dynamic). If you think about what happens when air is released from a balloon, the static pressure pushes the air out of the balloon, trading static pressure (potential energy) for kinetic/dynamic.

 

[thetexan]

Maybe if I ask it this way...

Let’s start with zero velocity.

There will be no dynamic pressure. So all of the pressure in the fluid is static which to me means that all of the pressure is acting in all directions. For example, an aircraft sitting on the ramp...the static pressure against every square inch of the wing is about 14.7 psi. But if we move that wing thru the air and introduce velocity to the fluid dynamic pressure begins to increase (and, by the way, I don’t quite understand how velocity is related to dynamic pressure) and the 14.7 begins to decrease. So somehow by increasing the kenetic energy of the flow parallel to the surface reduces the pressure normal to the surface.

Let me try this...

Is it this way...

14.7 at zero velocity represents the total kenetic energy available at the wing surface, which exerts pressure in all directions at that point on the surface. Then when you add velocity some of that total energy is used in the moving of the fluid which subtracts from that available to press against the wing.

Is that how it works?

Tex

 

[jbriggs444]

Maybe if I ask it this way...

Let’s start with zero velocity.

There will be no dynamic pressure. So all of the pressure in the fluid is static which to me means that all of the pressure is acting in all directions. For example, an aircraft sitting on the ramp...the static pressure against every square inch of the wing is about 14.7 psi. But if we move that wing thru the air and introduce velocity to the fluid dynamic pressure begins to increase (and, by the way, I don’t quite understand how velocity is related to dynamic pressure) and the 14.7 begins to decrease. So somehow by increasing the kenetic energy of the flow parallel to the surface reduces the pressure normal to the surface.

If you are considering Bernoulli, the relevant difference is between the flow velocity upstream and downstream from the wing compared with the flow velocity above and below the wing. The flow velocity is high above the wing and low in front and behind. This must be so in order for the flow to have accelerated in front and to have decelerated behind.

14.7 at zero velocity represents the total kenetic energy available at the wing surface, which exerts pressure in all directions at that point on the surface. Then when you add velocity some of that total energy is used in the moving of the fluid which subtracts from that available to press against the wing.

Speeding up a box containing a parcel of air does nothing to change the pressure of that parcel of air. Speeding up a parcel of air by having it rush from an area of high pressure to an area of low pressure... results in the air being at a lower pressure. (obviously!)

 

[thetexan]

I’m still in the dark.

Most pilot students including myself 40 years ago learned it this way...fast velocity over the top of the wing than below the wing results in lower pressure above the wing than below so the wing is either sucked up due to lower pressure or pushed up due to higher pressure below. We learned that it’s the difference in pressure above and below the wing that does the trick. We learned that velocity, being faster above the wing due to camber, means a difference in velocity above and below the wing, with a resulting difference in pressure above and below the wing...and all due to something called Bernullis principle. We even learned the basics of B’s principle, that is, tha velocity reduces pressure as in a Venturi tube. This is the usual extent of their knowledge.

Then...here comes an aerodynamics class where they ask “why should it be so?”

So I’m struggling in explaining why, when we increase velocity over a surface, even a flat surface, the pressure against the surface reduces because of something called dynamic pressure! Not velocity but dynamic pressure!

So, I have to be able to explain how and increase in velocity increases this thing called dynamic pressure and how increasing dynamic pressure decreases the pressure acting against the surface of the wing (static pressure) and what is this thing called dynamic pressure. Static pressure is easy to understand and visuallize, it is simply the pressure acting perpendicular against the surface of the wing. But how to visuallize dynamic pressure?

tex

 

[russ_watters]

Let’s start with zero velocity.

There will be no dynamic pressure. So all of the pressure in the fluid is static which to me means that all of the pressure is acting in all directions. For example, an aircraft sitting on the ramp...the static pressure against every square inch of the wing is about 14.7 psi.

But if we move that wing thru the air and introduce velocity to the fluid dynamic pressure begins to increase (and, by the way, I don’t quite understand how velocity is related to dynamic pressure) and the 14.7 begins to decrease. So somehow by increasing the kenetic energy of the flow parallel to the surface reduces the pressure normal to the surface...

14.7 at zero velocity represents the total kenetic energy available at the wing surface, which exerts pressure in all directions at that point on the surface. Then when you add velocity some of that total energy is used in the moving of the fluid which subtracts from that available to press against the wing.

Is that how it works?
[emphasis added]

To start with, you have to be a little careful with the airplane example because switching reference frames can be confusing. The air is stationary with respect to itself, so its dynamic pressure is zero in that frame. But by the airplane moving, using the airplane's reference frame, the air is now declared to be moving, with a non-zero dynamic pressure and increased total pressure.

...now, that highlighted bit you said isn't really accurate as stated. For the pito-static tube attached to the airplane, the static pressure of the air has not changed, but the dynamic and therefore total pressure has.

For air flowing over the wing, though, there is an additional increase in velocity (above the airspeed of the plane) in some places and the pressure does drop below ambient.

(and, by the way, I don't quite understand how velocity is related to dynamic pressure)

Can you elaborate on what is confusing you? You know the equation, right? I feel like, though, by jumping right to flow around a wing you are making the issue more complicated instead of ensuring you understand the simple versions first (a balloon, a Venturi tube, etc) and applying the solidly understood principles to the more complicated case.

 

[russ_watters]

Then...here comes an aerodynamics class where they ask “why should it be so?”

So I’m struggling in explaining why, when we increase velocity over a surface, even a flat surface, the pressure against the surface reduces because of something called dynamic pressure! Not velocity but dynamic pressure!
[emphasis added]

For emphais: the static pressure does not drop just by a flat, parallel surface gaining speed in air.

 

[thetexan]

Here's my understanding and explanation...

At zero velocity on the ramp let's say the pressure against every square inch of wing surface above and below the wing is 14.7 psi. If we are going to take advantage of Bernouli and reduce the pressure above the wing compared to below the wing we have to somehow make the velocity above the wing faster than below the wing. So we create a cambered airfoil. When we move this through the air thus increasing velocity both above and below the wing we reduce the 14.7 both above and below the wing...only we reduce it more above the wing because of the relatively increased velocity above the wing compared to below because of the camber.

So, for example, at 100 kts we may have a pressure of 13.9 above the wing and 14.5 below the wing. The pressure against both surfaces have reduced...the top surface has reduced more due to the faster velocity. There is still a greater pressure below thus giving us the upward force due to differential pressure. So far so good. The question then is why does an increase in velocity compared to some other velocity reduce the resulting static pressure to 13.9 psi against the wing? What is actually happening that causes the static pressure of 14.7 psi to be reduced to 13.9 just because the air moved at an increased velocity across the surface? The answer is dynamic pressure. Increasing velocity increased this thing called dynamic pressure which in turn reduced the static pressure from 14.7 to 13.9.

So I have struggled to understand the relationship of velocity to dynamic pressure and...how dynamic pressure is related to static pressure in that there is something about the increase in dynamic pressure the reduces the 14.7 to 13.9. The answer, I believe is in the idea of the constant total pressure...so when one is reduced the other increases. I understand that it is so, but I don't understand WHY it is so.

I know I am making a mess of the explanation and I am sorry I am not explaining myself better.

tex

 

[jbriggs444]

The question then is why does an increase in velocity compared to some other velocity reduce the resulting static pressure

Exactly what velocity do you imagine is being increased in comparison to exactly what other velocity? i.e. as @russ_watters suggested, are you frame-jumping?

 

[cjl]

Here's my understanding and explanation...

At zero velocity on the ramp let's say the pressure against every square inch of wing surface above and below the wing is 14.7 psi. If we are going to take advantage of Bernouli and reduce the pressure above the wing compared to below the wing we have to somehow make the velocity above the wing faster than below the wing. So we create a cambered airfoil. When we move this through the air thus increasing velocity both above and below the wing we reduce the 14.7 both above and below the wing...only we reduce it more above the wing because of the relatively increased velocity above the wing compared to below because of the camber.

So it's worth being careful here. When we move the airfoil through the air, it doesn't matter that above and below the wing, the velocity is higher than it was earlier when the wing was stationary. What matters is what the local velocities are relative to freestream. In the freestream, the static pressure is 14.7 psi and the velocity is some value V. Locally, around a cambered airfoil at a nonzero angle of attack, I'd tend to expect the lower surface local velocity to be slower than V, and the upper surface local velocity to be higher than V. This leads us to your next paragraph:

So, for example, at 100 kts we may have a pressure of 13.9 above the wing and 14.5 below the wing. The pressure against both surfaces have reduced...the top surface has reduced more due to the faster velocity. There is still a greater pressure below thus giving us the upward force due to differential pressure. So far so good. The question then is why does an increase in velocity compared to some other velocity reduce the resulting static pressure to 13.9 psi against the wing? What is actually happening that causes the static pressure of 14.7 psi to be reduced to 13.9 just because the air moved at an increased velocity across the surface? The answer is dynamic pressure. Increasing velocity increased this thing called dynamic pressure which in turn reduced the static pressure from 14.7 to 13.9.

As I said above, I'd actually expect a situation more along the lines of 13.9 above the wing and 15.1 below the wing, since the air on the underside of the wing will be moving slower than freestream. It is very important to remember that we're always comparing local velocity to freestream, and not velocity now to some other velocity at a different arbitrary point in time.

If you remember this, it starts to make a lot more sense why the pressure has to do what it does. Above the wing, the local airspeed is higher than freestream. As a result, the air must accelerate from far away to the point above the wing. The only way to cause the air to accelerate is for there to be a pressure gradient forcing each parcel of air to speed up. This means that where the local velocity is higher, the pressure *must* be lower than it is in the freestream, otherwise the air could not have accelerated to get to that point.

Similarly, below the wing, the airspeed is below freestream. To slow the air down, there must have been an adverse pressure gradient it was fighting against, implying the local pressure below the wing *must* be higher than out far from the wing. Trying to claim that the pressure mysteriously went to this thing called "dynamic pressure" is just increasing confusion and introducing something that isn't necessary to understand the basic principles here.

So I have struggled to understand the relationship of velocity to dynamic pressure and...how dynamic pressure is related to static pressure in that there is something about the increase in dynamic pressure the reduces the 14.7 to 13.9. The answer, I believe is in the idea of the constant total pressure...so when one is reduced the other increases. I understand that it is so, but I don't understand WHY it is so.

I know I am making a mess of the explanation and I am sorry I am not explaining myself better.

tex

Again, be very careful. Total pressure is not constant in the scenario you laid out. In the case of your static wing with no flow, total pressure is the same everywhere (and equal to the static pressure), and in the case of your moving wing, total pressure is also the same everywhere, but there's absolutely no reason why total pressure in the moving wing case needs to be equal to total pressure in the static wing case.

 

[russ_watters]

When we move this through the air thus increasing velocity both above and below the wing we reduce the 14.7 both above and below the wing...

No: the pressure on the bottom surface is generally higher than ambient, not lower.

You are indeed fame-jumping.

 

[jartsa]

It as if when velocity across the surface (wing) increases the fluid (air) doesn’t “have time” to spend exerting static pressure perpendicular to the surface. And the slower the velocity the more normal pressure that can be exerted.

A parcel of air next to the surface always pushes on the surface. Sometimes said parcel does work as it pushes, sometime it doesn't. Said work consists of the parcel pushing itself to a higher speed. (That is true in such frame where the kinetic energy of the parcel increases)

Actually parcel is not doing work, as its energy stays constant. Parcel's static pressure is doing work on parcel's dynamic pressure, maybe that's how it could be expressed.

 

[thetexan]

There is an increased flow velocity below the wing also isn't there. The flow across the lower surface of the wing is faster than before so why should it not also reduce the static pressure below...just not as much. Any fluid velocity across a surface reduces static pressure on that surface. I know enough about how wings work to know that. Every course I know of teaches that there is a reduction of pressure on all parts of the wing just more so along the top due to increased velocity due to camber. If static pressure is decreased won't it be below some starting point...namely...14.7?

How can increasing the flow along the bottom of the wing increase the pressure and not decrease the pressure?

 

[russ_watters]

There is an increased flow velocity below the wing also isn't there.

No, it's decreased.

The flow across the lower surface of the wing is faster than before...

Sure, but so what? Bernoulli's principle applies to steady flow. Is flow steady for an accelerating airplane?

Any fluid velocity across a surface reduces static pressure on that surface.

No. It. Does. Not.

 

[thetexan]

I don’t understand. You say flow along bottom of wing is decreased. Decreased from what?In the ramp the velocity across the bottom of the wing is zero...now it’s 100kts! How can you say that velocity s decreased?

 

[cjl]

Decreased from the flow velocity far away from the wing in the same scenario. You can't compare when the airplane is stationary to when it is flying because those are not the same frame of reference and are not the same flow field. You need to look at what the flow is doing in a single, steady moment in time.

It's completely valid to compare the flow next to the wing when it is on the ramp to the flow far from the wing also when it is on the ramp (though it'll be somewhat boring, since flow velocity is zero both next to the wing and far from it). It's also valid to compare the flow next to the wing and the flow far from the wing while the airplane is in flight. It's not valid, however, to compare the flow next to the wing with the wing in flight to the flow next to the wing with the plane on the ramp. There's no connection between these two flows, and no reason they would have any particular relationship to each other.

 

[thetexan]

Perhaps someone can explain how a wing reduces lift because some of these answers don’t make sense. Maybe we have different unferstandings.

Can someone Please explain how lift is generated on a wing using airstream velocity, free stream velocity, dynamic pressure and statiic pressure and Bernoulli’s principle in the definition. Somehow I think I don’t understand everything I have ever been taught.

 

[cjl]

A wing moving through air generates a flowfield in which the local velocity above the wing is higher than ambient (far from the wing), and the local velocity below the wing is lower than ambient. In steady flow, the only way for a local area to have a velocity higher than ambient is for the pressure to be lower than ambient. Similarly, the only way for a region to have a local velocity below the velocity far from the wing is for it to have a locally higher pressure. Because the pressure below the wing is higher than the pressure above the wing, upward force is generated.

Alternatively, you can say that a wing deflects airflow downwards. This creates a reaction force pushing the airfoil upwards. This is sometimes called the "Newtonian" explanation for lift, and both the above pressure-based explanation and the momentum based Newtonian explanation are correct and valid, though neither actually gets into the details of exactly why the airfoil creates the flowfield that it does with an overall downward deflection of the flow and higher velocity above it.

Note that in this explanation, I never once compare a moving airfoil to a static airfoil. The important thing to consider is the flow of air next to the airfoil vs the flow of air far from the airfoil. This is where you can apply Bernoulli.

 

[thetexan]

Well there is camber below the wing just like above just not as much. Bernouli is acting on the bottom just like the top. Just not as much since the velocity is greater on top due to greater camber above.

 

[russ_watters]

Well there is camber below the wing just like above just not as much. Bernouli is acting on the bottom just like the top. Just not as much since the velocity is greater on top due to greater camber above.

I'm not sure you realize that camber is the sum of the curvatures. There is only one camber.

And again: bernoulli deals with one flow scenario at a time. A stationary and a moving plane are different scenarios. This is your critical misunderstanding.

 

[russ_watters]

Can someone Please explain how lift is generated on a wing using airstream velocity, free stream velocity, dynamic pressure and statiic pressure and Bernoulli’s principle in the definition. Somehow I think I don’t understand everything I have ever been taught.

It's pretty simple once you fix your reference frame problem:

1. The reference velocity is freestream.
2. Velocity above the wing is higher, so pressure is lower.
3. Velocity below the wing is lower, so pressure is higher.
[in general for level flight]

That's it!

 

[rcgldr]

Well there is camber below the wing just like above just not as much. Bernouli is acting on the bottom just like the top. Just not as much since the velocity is greater on top due to greater camber above.

A flat wing can produce lift, it just needs an angle of attack. Using the wing as a frame of reference, the air is deflected off the bottom, increasing pressure, and the air above the wing is drawn into what would otherwise be a void swept out by the wing as it passes through a parcel of air, reducing pressure above the wing. The wing at an effective angle of attack causes the air to accelerate downwards, curving the relative flow, both below and above the wing. There is a pressure gradient perpendicular to the flow, higher above and lower below (the air is accelerated towards the center of radius of curvature of the flow). The higher and lower pressure zones also cause the air to accelerate in the direction of flow.

From a molecular perspective, static pressure is related to the momentum of random collisions between molecules of air. In the case of flow in a pipe, if the pipe narrows, then the velocity of the flow increases, and the flow becomes more organized, due to the change in the net velocity of the molecules, which reduces the momentum of collisions, since the amount of random movement between molecules decreases as the flow velocity increases and becomes less random. This is assuming that no external forces are involved, so the exchange of energy between pressure and kinetic doesn't involve any externally induced gains or losses in overall energy.

 

[boneh3ad]

https://www.physicsforums.com/insights/airplane-wing-work-primer-lift/

 

[thetexan]

Who said anything about an accelerating airplane? At a steady 100kt velocity there is no accelerating airplane.

I have studied many diagrams and now understand the dynamic/static relationship.

There is an acceleration then deceleration of the flow over the wings curved surface...and the same over the wings curves surface under the wing. A gentler curve but a curve none the less. If Bernouli applies above the wing it does so also under the wing and that IS a fact.

So, as I understand it the static pressure reduction is what we are interested in when considering the difference between upper and lower wing pressures that produce lift, is that right?

Tex

 

[russ_watters]

Who said anything about an accelerating airplane? At a steady 100kt velocity there is no accelerating airplane.

In order for the plane to go from 0-100 kts it has to accelerate. My point in pointing this out is that you are mixing together two different scenarios as if they were part of the same scenario. An airplane sitting stationary on a runway and an airplane flying at 100 kts are two completely different scenarios.

I have studied many diagrams and now understand the dynamic/static relationship.

There is an acceleration then deceleration of the flow over the wings curved surface...and the same over the wings curves surface under the wing. A gentler curve but a curve none the less. If Bernouli applies above the wing it does so also under the wing and that IS a fact.

Of course Bernoulli's principle applies in both surfaces. But Bernoulli's principle applies to flows that both speed up and slow down. On the top surface it speeds up and on the bottom it slows down. Here's a diagram indicating the resulting pressure profile:

As you can see, the red arrows indicate pressure above ambient on the bottom of the airfoil.

So, as I understand it the static pressure reduction is what we are interested in when considering the difference between upper and lower wing pressures that produce lift, is that right?

Static pressure change. It can be an increase (on the bottom surface) or reduction (on the top).

Here's a diagram with a velocity profile:

https://www.mh-aerotools.de/airfoils/velocitydistributions.htm

Let us follow the flow from the stagnation point, along both sides of the Eppler E 64 airfoil at 2º angle of attack, as shown above:

• Starting from v/
  
  =0 in the stagnation point, the velocity v increases rapidly to 1.38 times the velocity
  
  of the onset flow near the location x/c=0.1. Further downstream the velocity gradually decreases and reaches at the trailing edge approximately 85% of the free stream velocity
  
  .
• The velocity of the flow on the lower surface looks similar, but its level is considerably lower. In this example, it always stays below the free stream velocity.

 

[thetexan]

In classroom examples we consider a wing in flight, say at 100kts. It is not accelerating. So the free stream airmass is not accelerating. However, the local stream is accelerating as it move from the leading edge to the higher velocity point at the thickest point on the airfoil then it begins decelerating back to the free stream velocity at the trailing edge. The dynamic pressure increases and then decreases during this trip. And, accordingly, the static pressure decreases then increases.

The same is happening on the lower surface but the static pressures are greater overall on the lower surface than the upper.

All of this is going on with the aircraft in non accelerated flight.

The stagnation point in the boundary layer (one of two) on the top of the wing is where the adverse pressure gradient causes a reverse flow towards the front and reaches equilibrium with the normal favorable gradient flow toward the rear. This stagnant barrier causes the boundary layer to separate from the surface as the rearward flow goes up and over the stagnation barrier. This causes wake turbulence as the higher aft pressure rushes into fill the void. With the boundary layer separated Bernoulli and his effects are eliminated behind the separation point and that portion of lift aft of that pont is lost.

No. I don’t believe I’m changing my frame.

Remember, my original question is why dynamic pressure and static pressure work inversely to each other...how they are related and why. I believe on reply hit on it concerning the potential and kinetic energy and the conservation thereof. Please explain that more.

Tex

 

[boneh3ad]

From a molecular perspective, static pressure is related to the momentum of random collisions between molecules of air.

No it isn't. From a molecular perspective, according to kinetic theory, static pressure is related to the kinetic energy of the individual particles based on the mean of squares of the particle velocities,
p = \dfrac{1}{3}\rho\overline{v^2},
where $\overline{v^2}$ is the mean-square velocity. In other words, it is an energy density based on the Brownian motion of the particles and is not related to momentum.

Well there is camber below the wing just like above just not as much.

You misunderstand airfoil terminology. Airfoils have three generally-important properties: angle of attack, camber, and thickness. Angle of attack is the angle the chord line (straight from leading edge to trailing edge) makes with the free stream. The camber line comes from starting at the leading edge and drawing a curved line to the trailing edge in a way that exactly half the thickness is above and below the camber line at all times. Thickness is, as was just mentioned, the distance from the camber line to the airfoil surface. See, for example:

The same is happening on the lower surface but the static pressures are greater overall on the lower surface than the upper.

Not true, necessarily. As @russ_watters already pointed out with links and images, the flow under the airfoil is not necessarily accelerated at all, and is often decelerated. This is the meaning of that plot that shows the bottom side has a value of $v/v_{\infty}<1$. In other words, the local velocity, $v$, is less than the free-stream velocity, $v_{\infty}$, and therefore the pressure there is higher.

The stagnation point in the boundary layer (one of two) on the top of the wing is where the adverse pressure gradient causes a reverse flow towards the front and reaches equilibrium with the normal favorable gradient flow toward the rear. This stagnant barrier causes the boundary layer to separate from the surface as the rearward flow goes up and over the stagnation barrier.

That's not really a proper way to describe things here. In a boundary layer, the velocity at the surface is always zero relative to the surface. In effect, the entire surface is a stagnation point. The key to separation is the shape of the boundary layer. The typical criterion for defining the separation point is based on shear stress at the wall, which is proportional to the wall-normal derivative of the velocity in the boundary layer. When the shear stress is zero, the boundary layer separates and reverses downstream.

 

[thetexan]

These students don’t get that deep into this. They understand the blausian graphs of the boundary layer and the point where the adverse gradient reverses the flow.

youre right. I’ve been saying camber when I should have been using the word “curved surface”. Because of the lower surface curve Bernoulli should work just like it does on the top surface as far as dynamic/static pressures are concerned. The main difference between upper and lower surfaces is that the lower surface, because of the typical positive AOA, is exposed to direct relative wind willing strikes the lower surface directly, adding a Newtonian reactive component to the total lifting force.
There is also the downwash deflecting of the upper flow also adding a reactive component from the top side.

Tex

 

[russ_watters]

No. I don’t believe I’m changing my frame.

You were, but you are slowly appearing to correct it. Please answer this question and we can be sure:

Atmospheric pressure is 14.7 psi.

Is the total pressure of the free stream for your 100kt airplane 14.7 psi, higher than 14.7 psi or lower than 14.7 psi?

The same is happening on the lower surface but the static pressures are greater overall on the lower surface than the upper.

This feels like a way to try to avoid saying the pressure on the bottom surface of the wing is above ambient. Please say you recognize that to be true.

Remember, my original question is why dynamic pressure and static pressure work inversely to each other...how they are related and why. I believe on reply hit on it concerning the potential and kinetic energy and the conservation thereof. Please explain that more.

I gave an example of a balloon. Can you reread and respond to the description I gave?

The main difference between upper and lower surfaces is that the lower surface, because of the typical positive AOA, is exposed to direct relative wind willing strikes the lower surface directly, adding a Newtonian reactive component to the total lifting force.

Please note: that's an alternative way to describe how pressure is applied to a surface, but it does not eliminate Bernoulli's principle. Bernoulli's principle still applies fully.

 

[thetexan]

In the free stream above and outside of the boundary layer, 14.7. As well a below and outside of the lower boundary layer. In Other words, all around the wing in the free stream clear of local airfoil created flows, 14.7 ambient.

 

[russ_watters]

...I may have another way to explain the velocity change you might find simpler:

Consider a Venturi tube. The velocity increases in the constriction to maintain the constant mass flow rate. That's a "what", but not really a why/how. To get to the why/how, consider what would happen if the velocity didn't increase in the constriction. The fluid would build-up before the constriction, pressurizing itself or bursting the tube (both of these can actually happen). It's pressure - static pressure - that would build, and so it is static pressure that pushes the fluid through the constriction. By doing this, the static pressure is in a way consumed and converted to velocity pressure.

 

[russ_watters]

In the free stream above and outside of the boundary layer, 14.7. As well a below and outside of the lower boundary layer. In Other words, all around the wing in the free stream clear of local airfoil created flows, 14.7 ambient.

No. The total pressure of the free stream is above 14.7 psi/ambient. It has to be; that's how the pressure on the bottom surface of the wing can be above 14.7psi/ambient!

You haven't corrected your reference frame problem yet.

[Very late edit]
Let me complete that thought:
As you know, Bernoulli's equation is:
Static Pressure + Dynamic Pressure = Total Pressure

For stationary air, static pressure is 14.7psi, dynamic pressure is zero and total pressure is 14.7psi. This is applying the calculation to the wrong reference frame, which you keep going back to.

For an airplane moving at 100kts, static pressure is still 14.7psi, dynamic pressure has been added to the system in the amount of about 0.2psi and the total pressure is now 14.9psi.

 

[cjl]

youre right. I’ve been saying camber when I should have been using the word “curved surface”. Because of the lower surface curve Bernoulli should work just like it does on the top surface as far as dynamic/static pressures are concerned. The main difference between upper and lower surfaces is that the lower surface, because of the typical positive AOA, is exposed to direct relative wind willing strikes the lower surface directly, adding a Newtonian reactive component to the total lifting force.
There is also the downwash deflecting of the upper flow also adding a reactive component from the top side.

Tex

The way you've phrased this makes me concerned that you have some common misconceptions about the way airflow and lift work. First of all, you say a curved surface implies bernoulli. Are you under the impression that a longer path for the fluid to travel implies a greater velocity (the equal transit time fallacy)? This is not the case at all. Bernoulli applies everywhere within the flow - it is simply a relation between static pressure and local velocity, and on the lower surface, Bernoulli works just fine, but the local velocity is below ambient so Bernoulli's relation (correctly) predicts that the pressure there is higher than the ambient static pressure.

Also, the "Newtonian" component is not an independent effect. Both the upper and lower surface of the wing contribute to deflecting air downwards, and you can correctly predict the entirety of the lift based on measuring this downward deflection and the associated momentum flow. You can also correctly predict the entirety of the lift purely from knowing the velocity distribution around the airfoil, and applying the Bernoulli relation to obtain pressure. These are not independent effects - they are two different ways to approach the same fundamental physical phenomenon.

 

[rcgldr]

From a molecular perspective, static pressure is related to the momentum of random collisions between molecules of air.

No it isn't. From a molecular perspective, according to kinetic theory, static pressure is related to the kinetic energy of the individual particles based on the mean of squares of the particle velocities ...

By molecular, I meant the kinetic theory of gasses, the effect of collisions between molecules of the air and whatever the molecules of air collide with, such as the internal surface of a pipe (based on force = change in momentum per unit time, leading to force per unit area). With a fixed amount of total energy per unit volume of air, the static pressure is greatest when there is no net velocity of flow. If there is a net velocity of flow, then some of the total energy is related to the net flow velocity, with the remainder related to static pressure. With a net velocity of flow, the flow is more "organized" and less random, reducing the rate of and/or change in momentum of collisions. Link to NASA article that covers some of this:

https://www.grc.nasa.gov/www/k-12/rocket/kinth.html

 

[cjl]

In classroom examples we consider a wing in flight, say at 100kts. It is not accelerating. So the free stream airmass is not accelerating. However, the local stream is accelerating as it move from the leading edge to the higher velocity point at the thickest point on the airfoil then it begins decelerating back to the free stream velocity at the trailing edge. The dynamic pressure increases and then decreases during this trip. And, accordingly, the static pressure decreases then increases.
Thus,
The same is happening on the lower surface but the static pressures are greater overall on the lower surface than the upper.

The same doesn't happen on the lower surface though. On the lower surface, the flow decelerates, then accelerates. Thus, the static pressure increases, then decreases, and the overall average pressure on the lower side of the airfoil is actually higher than the ambient static pressure.

 

[boneh3ad]

These students don’t get that deep into this. They understand the blausian graphs of the boundary layer and the point where the adverse gradient reverses the flow.

It doesn't matter how deep they get. If you are instructing the course, it is important that you understand what is going on.

 

[thetexan]

You say that the static pressure in the 100kt freestream remains at 14.7 and since there is now 100kts of velocity there must be a higher ambient total pressure due to the ADDED dynamic pressure due to velocity. Why should the static pressure remain the same? Why not the total? And if that works in the free stream (that static remains the same) why does that not apply in the local stream near the wing?We say that there is only a fixed total amount of in the atmosphere, say 14.7 at rest. Why the. Is there, say, 14.9 at 100kts. Unless the velocity adds dynamic pressure to the rest state 14.7. If that is the case then the over-the-wing curve induced velocities and dynamic pressures add to the rest state 14.7 static also. Is that what your saying?In other words, are you saying that velocity is adding dynamic to the static, which remains the same, so in either free stream or local stream were actually starting at some number above 14.7, say 14.9 and using that number as our starting value.If that is the case, that dynamic adds to the static, then static never reduces below the starting value. Dynamic never has the effect of lowering static and then what good does the faster velocity over the wing do?Tex

 

[thetexan]

I don’t understand why you say that flow over the lower part of the wing is not accelerating/decelerating just as on the upper surface. The lower curve is much shallower that the upper but it’s there just the same.

Now the relative wind IS hitting the surface straight on due to AOA and that is different but the curve effect on velocity should act the same on top and bottom notwithstanding the difference in the way the relative wind strikes the wing.

 

[russ_watters]

Why should the static pressure remain the same? Why not the total?

Because static pressure is static pressure. The air that was stationary with respect to itself is still stationary with respect to itself. Nothing has changed for the air by - improperly - selecting another reference frame from which to view it.

There are thousands of airplanes flying around right now, all with different velocities. But static pressure is static pressure. It doesn't care.

And if that works in the free stream (that static remains the same) why does that not apply in the local stream near the wing?

Because those thousands of airplanes are all in different flow situations and not connected by streamlines. Bernoulli's principle applies in steady flow along a streamline, not across different flows on different continents.

We say that there is only a fixed total amount of in the atmosphere, say 14.7 at rest. Why the. Is there, say, 14.9 at 100kts. Unless the velocity adds dynamic pressure to the rest state 14.7. If that is the case then the over-the-wing curve induced velocities and dynamic pressures add to the rest state 14.7 static also.

It is: Bernoulli's principle says the total pressure is constant along the streamline. If it's 14.9 in free stream, its 14.9 over and under the wing even as the speeds change. That's the entire point of Bernoulli's principle.

In other words, are you saying that velocity is adding dynamic to the static, which remains the same, so in either free stream or local stream were actually starting at some number above 14.7, say 14.9 and using that number as our starting value.

If that is the case, that dynamic adds to the static, then static never reduces below the starting value. Dynamic never has the effect of lowering static and then what good does the faster velocity over the wing do?

Here you appear to be flipping back and forth between static and total pressure. 14.9psi is the total pressure, not the static pressure. The rules are:
1. The static and dynamic sum to 14.9
2. Neither can be negative.

So if dynamic goes up, static goes down, and vice versa.

 

[256bits]

Confounding problem,
The motionless wing on a tarmac with still air is easy to understand.
Total pressure = stagnation pressure = static pressure.
Dynamic pressure = 0

Adding energy to the air is also easy to understand, if one can understand the underlying principles.
And since that is the way the wing situation is explained most cases, it so is kind of ingrained in the mental process.
Such would be the case of a wing in a wind tunnel, with a fan exciting the air increasing the total pressure.
Subsequently, stagnation pressure = total pressure = dynamic pressure + static pressure.
Is static pressure the same as atmospheric pressure in the test section between the entrance plenum and exiting diffuser - probably not, but we will say that it is due to ingenious design.
But we can leave that discussion to wind tunnel designers.

For a wing moving through air, the energy is imparted to the wing to keep it in flight, and not the air.
But now we face a problem.
For one, we did say in case 1 above, still air, P_stag = P_total=P_static.
Equally problematic is that V_ambient is 0.
For two, due to drag, the air is carried along with the wing, and has a velocity that of the wing, or part thereof depending upon distance from the chord of the wing. Along the chord length, we also might have some air positive and negative velocities from ambient ( zero velocity ) as the air is subject to different pressures along the length.

we can remove ourselves from this mess, if we take an educated leap of faith and assume that a wing moving through air is the same as
air moving over a wing.
V_ambient becomes the speed of the aircraft.
P_ambient becomes the atmospheric pressure. We will assign that as P_static.
Subsequently, the analysis becomes more straight forward as to what we are accustomed to seeing for wing analysis.
Now, since we know the static pressure and the velocity of the air, we can determine the new and improved total pressure, stagnation pressure, and resume the analysis.
See post 38.

 

[cjl]

I don’t understand why you say that flow over the lower part of the wing is not accelerating/decelerating just as on the upper surface. The lower curve is much shallower that the upper but it’s there just the same.

Now the relative wind IS hitting the surface straight on due to AOA and that is different but the curve effect on velocity should act the same on top and bottom notwithstanding the difference in the way the relative wind strikes the wing.

Because the airfoil surface being curved is not, in and of itself, sufficient to accelerate the flow. There is no "curve effect on velocity" - the flow velocity will vary based on the overall geometry of the flow, and there's no simple or easy way to tell what it will be doing at any point in a given flow. In the case of almost any airfoil at a nonzero angle of attack, the lower surface will actually slow down the flow relative to ambient, despite its curvature.

EDIT: Out of curiousity, could you explain your understanding of what curvature does to velocity? I think we might be dealing with a misconception here, but I want to see where your understanding is first before just making assumptions.

 

[rcgldr]

Out of curiousity, could you explain your understanding of what curvature does to velocity? I think we might be dealing with a misconception here, but I want to see where your understanding is first before just making assumptions.

The curvature of the relative flow is coexistent with a pressure gradient perpendicular to the flow, with the pressure decreasing in the direction towards the center of curvature of the flow, and the pressure differential across the flow coexists with change in speed in the direction of flow.

A wing produces lift by diverting the relative flow downwards, and since air has momentum, it can't change direction instantly, so the relative flow is curved downwards, usually both above and below a wing. Below a wing, the pressure is greatest just outside the boundary layer and decreases with distance from the bottom surface of a wing. Above a wing, the pressure is lowest just outside the boundary layer and increases with distance from the top surface of a wing.

 

[boneh3ad]

Let's take a step back here for a moment. In aerodynamics, we generally use some terminology for various "types" of pressure: static, dynamic, and total (also called stagnation).

Static pressure is the pressure due to the random motion of molecules in the air. It is typically donated as just $p$. As I mentioned in a previous post, the static pressure can be derived entirely from molecular quantities. Specifically,
p = \dfrac{1}{3}\rho\overline{v_m^2},
where $\rho$ is the density and $\overline{v_m^2}$ is the mean of the squared velocities of the air particles (gas molecules move with random speed and direction, so taking the average before squaring would equal zero). Clearly the speed of molecules in a chunk of air, say, floating above Kansas, are not affected by a plane taking off in Chicago. The static pressure in the atmosphere is therefore constant at a given altitude (more or less, still subject to small fluctuations in the form of barometric pressure like you see on the news). It is frame-invariant.

Dynamic pressure is not really a pressure, per se. Often denoted $q$, it is a kinetic energy density due to bulk motion of the fluid, which has units of pressure. It is
q = \dfrac{1}{2}\rho v^2,
where $v$ is the velocity of the air relative to your reference frame. In other words, dynamic pressure depends on your frame or reference since it depends on flow velocity.

Finally, there is total (or stagnation) pressure. This is (for an incompressible flow) the sum of static and dynamic pressure, or
p_t = p + \dfrac{1}{2}\rho v^2.
Total pressure is therefore also frame-dependent. So, as the plane that left Chicago approaches our chunk of air from before, static pressure far from the plane is the same whether you are watching the wing from the ground or watching it from your window seat on the plane. The dynamic pressure in the air, however, changes in these two situations, and therefore so does the total pressure.

So if you look at the case of a wind tunnel where the wing is stationary and the air is moving, that air has both a static pressure, $p$, and a velocity, $v$, and its total pressure is therefore greater than the static pressure. At the leading edge stagnation point on the model, the velocity is brought down to zero, so the static pressure rises until the static pressure equals the total pressure at that point.

With a plane flying through the air, the situation is more complicated because the problem is not a steady-state one. The bottom line is that the static pressure is atmospheric pressure, and the plane passing through temporarily adds velocity and the relationship between all of the pressures is not as nice. However, but simply shifting your frame to follow along with the plane, the problem is identical to that of the wind tunnel above. Doing that frame shift, however, adds a constant free-stream velocity to the atmosphere, so the static pressure is unchanged but the total pressure now reflects that velocity.

 

[thetexan]

Let me say that I really appreciate everyone’s helping me to better understand these concepts.

I believe I now understand the relationship of dynamic and static pressure as they relate to each other.

I understand, I think, why the ambient pressure at velocity is greater than the still-air pressure due to the fact that velocity has added dynamic pressure to the unchanged static pressure. For example...if the static pressure of an air mass is 14.7 then it still is 14.7 STATIC at velocity. What has changed is the dynamic pressure added by the velocity to the 14.7 to, say, 15.0. And that 15.0 is our starting point at that velocity. For example let’s use 100kts and 15.0. That pressure may not be precise but it makes the point.

Now as the air travels across the top of the wing it accelerates as it picks up speed over the top of the airfoil. As the velocity increases the dynamic pressure increases and the static pressure decreases. But the total pressure is based on our new reference total pressure of 15.0 (14.7 static in this case). So let’s say the dynamic pressure increases .5. That means static becomes 14.5 (keeping the total at 15.0) which is .2 below 14.7 and, voila we have a net decrease in static and thus some lift.

God, I hope that’s right.

I suppose what’s left that I need to clarify is...

Why Bernoulli doesn’t apply below the wing.

At a zero lift AOA the relative wind produces as much lift above as it does below. This zero lift AOA is different for different airfoils. This AOA for a symmetrical airfoil is zero degrees where for a cambered foil it is a negative angle.

But in either case the air flows over the top and bottom of the wing and produces a positive or negative lift respectively by the same method... the acceleration of the flow as it passes across the surface. Bernoulli is working both above and below.

As the AOA is increased positively more lift is produced above than below and as the relative wind begins to strike the bottom of the wing Bernoulli tapers off to nothing and we begin to see a reactive Newtonian force adding to the total lift...induced lift on top and Newton on the bottom.

This is my current understanding of what happens. Where am I wrong in any of this please?

Tex