How dynamic and static pressures are related

[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

 

[cjl]

As I said earlier, "Bernoulli" vs "Newtonian" aren't two different things. They're two different explanations for the same phenomenon, and both can explain it in full. You can explain the entirety of the lift in any configuration through a Newtonian momentum transfer and associated downwash, and you can also explain the entirety of the lift by looking at local velocity distributions and using the bernoulli relation. Bernoulli never "tapers off to nothing", and the reactive Newtonian component is there from the instant the airfoil starts making lift.

 

[russ_watters]

God, I hope that’s right.

Good so far!

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

Why Bernoulli doesn’t apply below the wing.

Bernoulli's principle most certainly does apply below the wing.

At a zero lift AOA the relative wind produces as much lift above as it does below.

This is poorly worded. Lift is an upward force. It is improper to say that lift from the lower surface pulls the airfoil down. I know what you meant, but that's not the right way to say it.

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.

Yes. Acceleration or deceleration.

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.

No. As the AOA rises, the airflow under the wing decelerates more and more, causing the lift produced to increase more and more, as Bernoulli tells us.

 

[cjl]

No. As the AOA rises, the airflow under the wing decelerates more and more, causing the lift produced to increase more and more, as Bernoulli tells us.

It's worth noting that, while this is correct, most of the additional lift as you increase AoA actually comes from the suction peak on the upper surface increasing in magnitude, and not from the increasing pressure on the underside.

 

[rcgldr]

It's worth noting that, while this is correct, most of the additional lift as you increase AoA actually comes from the suction peak on the upper surface increasing in magnitude, and not from the increasing pressure on the underside.

True for a conventional airfoil, but in the case of the M2-F2 and M2-F3 prototype reentry vehicles, which would have to operate at reentry speeds, it appears that most of the lift is due to the bottom surface. The shape is similar to a cone split in half with a tapered tail. The trailing edge is blunt since that where the rocket engines went on the M2-F3.

Getting back to the main topic, think of wings as being designed to curve (divert) the relative airflow downwards, and the coexistent differences in speeds and pressures above and below a wing being consequences of that design (due to pressure gradients perpendicular to flow being coexistent with curvature of flow).

 

[cave man]

Seems to me that this is similar to potential and kinetic energy. Simply-a stretched rubber band contains potential energy. When you release it it becomes kinetic. Could you explain dynamic verses static in this way?

 

[Lee Duke]

The lift on an aircraft (wings, body, tail) is a force that is normal to the aircraft velocity vector. No velocity no lift.

The key to measuring the dynamic pressure on a vehicle is a Pitot tube which returns STATIC and TOTAL presure. The Pitot tube should be reasonable distant from the wings. On research aircraft, you often see a nose boom. Besides weather vanes for measuring angle of attack and angle of sideslip (yes, I know the measure isn't of sideslip exactly, but I wanted to give all the trivia nerds a place to show their "amazing brilliance.") Anyway, a Pitot tube is generally located at the tip of that nose boom to get it as far away from the aircraft disturbances in the air. It's called the freestream (or more correctly, tstagnation) pressure.

I think boneh3ad tried to explain all this and, to mind, did a credible job.

The pressure profile at some location on the wing is measured using either piezoelectric sensors or holes on the wing surface containing the equivalent of Pitot tubes. If an aircraft is going to fly their needs to be lower pressure beneath the "lifting portion" of that vehicle be it wings, the body, or thousands of unseen angels holding the aircraft aloft.