Does pressure depend upon the observer?

[jpas]

Suppose you have a tube of moving water where bernoulli´s equation can be apllied and the water is at the same elevation all over the tube. Consider two points, 1 and 2. We have :

$$p_1 + \frac{1}{2}\rho (v_1)^2=p_2 + \frac{1}{2}\rho (v_2)^2$$

On different inertial referentials the velocities would be different, therefore, the pressures would also be different, which is absurd. For example, if on referential S, $$v_1=0$$,

$$p_1=p_2 + \frac{1}{2}\rho (v_2)^2$$

$$p_1\succ p_2$$

But on referential $$R$$ in which $$v_2=0$$

$$p_2\succ p_1$$

by the same reasoning.

Can anybody explain the paradox? Doesn´t this collide with the relativity principle which states that the laws of physics must be equal to inertial obervers?

 

[jpas]

I just realized there was another thread that dealt with the subject.
https://www.physicsforums.com/showthread.php?t=286412

However, I don´t think a convincing answer came up.

 

[russ_watters]

The velocity pressure measured by a probe is due to that probe's velocity through the fluid. There is only one relevant reference per probe...but there may be many probes moving at different velocities through one airstream.

 

[jpas]

Hi Russ,

Thanks for the reply. What you call a probe I call an observer. There is only one "relevant" reference per observer, but there may be a lot of observers. The question remains: "does pressure change with different observers?" This seems to follow from Bernoulli´s equation but it implicates that on different inertial frames the laws of physics change.

The problem remains.

 

[rcgldr]

The presssure only chages if you accelerate the air to the speed of the observer, and then only if the sensor is similar to a pitot port facing into the flow. If the port is surface mounted onto a surface pependicular to the flow, it "hides" under the boudary layer (shear layer where speed changes from surface speed to the surrounding air speed), and senses the static pressure of the surrounding air regardless of the relative speeds. If the surface included features ahead of the port that disturb the air, such as a pitot - static tube, then the static side ports need to be positioned far enough back so that they sense the static pressure and not some side effect of the disturbance of the air at the front of the pitot-static tube.

Another effect comes into play if the port is an open tube but perpendicular to the relative flow. At the leading edge of the tube, a component of air is diverted away the opening, resulting in a lower pressure vortice. This is how carburetors, perfume sprayers, and the classic "flit gun" sprayers work.

 

[blkqi]

Just think of an ideal gas for a minute. Since the ideal gas equation of state, and thus pressure, is ultimately derived from the Maxwell distribution of velocities/momenta, pressure must depend upon the observers reference frame. The Maxwell distribution holds only for observers at rest with a gas sample, no? Is this absurd?

 

[jpas]

Hi Jeff Reid,

I´m sorry but I didn´t understand what you´re saying. I´ve only begun to study fluid dynamics a few weeks ago. You talk about static pressure and I don´t know what that is nor how does it relate to my question. I´m sorry.


Hello blkqi,

I didn´t know that the gas equation of state is derived from the Maxwell distribution of velocities/momenta because I don´t even know what that is. The equation of state doesn´t bother me because I don´t see how it can be dependent of observers, since in nowhere on the equation do velocities appear.

You seem to say there´s nothing wrong with the fact that pressure may be different for different inertial frames. But notice that

$$P=\frac{dF}{dA}$$

Therefore, if pressure varies with different inertial frames, so do forces. But this contradicts the principle of relativity! Am I making sense? Oh, and thanks for the reply.

 

[russ_watters]

Thanks for the reply. What you call a probe I call an observer. There is only one "relevant" reference per observer, but there may be a lot of observers.

Ok, fine. Just so we're clear here, these observers are all in the fluid and measuring the velocity pressure of the fluid with respect to themselves, right?

The question remains: "does pressure change with different observers?"

If they are moving at different speeds through the same fluid, yes.

This seems to follow from Bernoulli´s equation but it implicates that on different inertial frames the laws of physics change.

How so? The equation doesn't change. The wording of Bernoulli's principle doesn't change. What part of the laws of physics do you see changing?

You seem to say there´s nothing wrong with the fact that pressure may be different for different inertial frames. But notice that

$$P=\frac{dF}{dA}$$

Therefore, if pressure varies with different inertial frames, so do forces. But this contradicts the principle of relativity! Am I making sense?

What part of that equation do you see being frame dependent? Note that that equation is the equation for static pressure, so there are no frame dependent components to it. In your OP, you were asking about velocity pressure.

 

[rcgldr]

Static pressure and I don´t know what that is

Wiki articles:

http://en.wikipedia.org/wiki/Static_pressure

http://en.wikipedia.org/wiki/Pressure

Pressure may be different for different inertial frames.

Static pressure is independent of inertial frame. It doesn't have a direction, and it's not related to speed. Dynamic pressure can only be "sensed" if a fluid or gas is being accelerated from one speed to another. Dynamic pressure is related to the change in speed^2 (at reasonably sub-sonic speeds), it's a bit more complex when compressability and high speeds are considered.

http://en.wikipedia.org/wiki/Dynamic_pressure

 

[jpas]

Hi,

thank you both for answering. It seems I have to study what dynamic and static pressure is. Unfortunately, it´s late here in Portugal and tomorrow I have classes. I promise to study in the afternoon and to report later.

 

[russ_watters]

You posted Bernoulli's equation in the OP. It has two components: the static pressure component p and the velocity pressure component $$\frac{1}{2}\rho (v_1)^2$$

Static pressure is pressure in a static fluid. Ie, the pressure inside a balloon. Velocity pressure is pressure due to the motion of air. It is what propels a balloon when you let it go and what you feel when you stick your hand out the window of your car, perpendicular to your direction of motion.

 

[rcgldr]

Velocity pressure is pressure due to the motion of air.

That pressure can't be sensed unless the air is accelerated though.

Ignoring how pressure differentials are created or maintained, once a pressure differential exists, then a fluid or gas will accelerate away from the higher pressure zone towards the lower pressure zone. Ignoring the work required to maintain these pressure differential zones, the internal reaction due to the pressure differentials is "work free" and the total mechanical energy remains constant. This is where Bernoulli principle applies. As the affected fluid or gas accelerates from the higher pressure zone to the lower pressure zone, it's speed increases and it's static pressure decreases, and Bernoulli equation defines the approximate relationship, ignoring factors like turbulence. If you consider the dynamic pressure of that affected fluid's or gas's speed, (1/2 density v^2), then the total pressure = dynamic pressure + static pressure is a constant, as speed increases in this case, dynamic pressure increases and static pressure decreases. At the molecular level, since total energy is constant, then the average speed of the accelerating molecules is constant, and static pressure is related to the rate of collision and the average impulse of those collisions, which is related to the average diffference in velocities of those collisions. As the flow speed increases, the average velocity of the accelerating molecules becomes more directed and less random, so the rate of collision and/or the average impulse of those collision decreases, which decreases the static pressure.

If there is vertical motion within a gravitational field, then density x g x h (gravitation potential energy per unit volume) is also included as a term, and there's may be another term to deal with compressability.

It is what propels a balloon when you let it go

What propels the balloon is an imbalance in force, the outwards force at the nozzle is less than the outwards force due to pressure at all other points within the balloon.

What you feel when you stick your hand out the window of your car, perpendicular to your direction of motion.

This is a good example of sensing dynamic pressure, since you hand accelerates the air.

 

[jpas]

Hi Russ Waters and Jeff Reid,

Thanks for the explanation. I think I got it. What I used to call "pressure" is in fact static pressure and the term $$\frac{1}{2}\rho v^2$$ is the dynamic pressure. The total pressure remains constant on an air stream when there´s no turbulence or elevation.

Still, notice that I can apply the same argument of the original post to the static pressure (which, on that post, I just called "pressure"). Static pressure does change from point 1 to point 2 (on the conditions of the OP).

Notice that, if static pressure changes, so do the forces that act on each molecule. If we change from one inertial frame to another so do the forces, which contradicts the principle of relativity.

In the OP, we just called the same thing by different names. Hope it´s clear now, since the argument is now about static pressure and not just "pressure", as I previously thought.

So, since this can´t be, where is my mistake?

 

[russ_watters]

That pressure can't be sensed unless the air is accelerated though...

Huh? I'm not following any of that. Bernoulli's equation measures the speed of the air, not the acceleration of the air. You can only get the acceleration by measuring the speed in two different places and subtracting.

Now a pitot tube measures the velocity pressure by decelerating it to stagnation in a tube - essentially converting it to static pressure. Is that what you are talking about? But you could avoid that by using a turbine flow meter and scaling the output for pressure instead of velocity, never decelerating the air. I'm not sure what the particulars of how a pressure sensor works is relevant here.

What propels the balloon is an imbalance in force, the outwards force at the nozzle is less than the outwards force due to pressure at all other points within the balloon.

That's one way to look at it. Another way is that the pressure applied to the air pushes the air out of the balloon and since forces come in pairs, the air leaving the balloon is also pushing the balloon forward. You can choose to apply bernoulli's principle to it (my way) or not (your way).

 

[russ_watters]

Thanks for the explanation. I think I got it. What I used to call "pressure" is in fact static pressure and the term $$\frac{1}{2}\rho v^2$$ is the dynamic pressure. The total pressure remains constant on an air stream when there´s no turbulence or elevation.

Yes, that is a good paraphrase of Bernoulli's principle.

Still, notice that I can apply the same argument of the original post to the static pressure (which, on that post, I just called "pressure"). Static pressure does change from point 1 to point 2 (on the conditions of the OP).

Notice that, if static pressure changes, so do the forces that act on each molecule. If we change from one inertial frame to another so do the forces, which contradicts the principle of relativity.

No, it doesn't. You're misapplying Bernoulli's equation (the form you first used). Bernoulli's principle/equation applies to steady flow along a streamline. You're applying it to two different objects moving through the pipe at different speeds at the same time. It's like saying a car driving at 20 mph and a car driving at 40 mph should feel the same apparent wind speed because the real wind speed is the same for both of them. But each car is measuring its own apparent wind speed. Each airstream also has a different energy because of the motion of the car!

What they share (in my example and yours) is that the static pressure is the same but the velocity pressure and total pressure are different.

 

[rcgldr]

That pressure can't be sensed unless the air is accelerated though.

Huh? I'm not following any of that. Bernoulli's equation measures the speed of the air, not the acceleration of the air. You can only get the acceleration by measuring the speed in two different places and subtracting.

Bernoulli's principle only applies when the acceleration is due to internal forces, pressure differentials, or gravitational potental energy per unit volume. Bernoulli relates the pressure and speeds during accelerations or changes in height. Assuming constant height, Bernoulli equation is derived based on acceleration:

wiki_derivation.html

As a common example to an exception of Bernoulli principle, the flow just aft of a propeller or fan has increased pressure and nearly the same speed, because the propeller or fan perform work on the air, changing it's total energy, violating Bernoulli principle. You need a more complicated (than Bernoulli) process to calculate the pressure jump across the virtual disk swept out by a propeller or fan. Afterwards, the flow's higher pressure causes it to accelerate towards the ambient pressure air further downstream, following Bernoulli's principle (ignoring issues like viscosity and turbulence).

balloon propulsion

Another way is that the pressure applied to the air pushes the air out of the balloon and since forces come in pairs, the air leaving the balloon is also pushing the balloon forward. You can choose to apply bernoulli's principle to it.

Bernoulli's principle would apply to the air expelled by the balloon, and Bernoulli could be used to calculate the terminal velocity of the expelled air, but you'd also need to know the mass flow rate in order to determine thrust. You could also assume that the pressure at the exit nozzle of the balloon was ambient, then use (static_pressure_in_balloon - exit_pressure_at_nozzle) x (cross_sectional_area_of_nozzle) to calculate force, but I don't know how accurate this alternate method would be.

water flow in a pipe

In the real world, because of wall friction and viscosity, in a constant diameter pipe, the flow and speed remain constant, but the pressure decreases with distance. I assume that mechanical energy is being converted into heat energy and then dissipated by the pipe, and this violates Bernoulli.


Getting back to the OP:

referential S v₁=0, referential R v₂=0

Bernoulli describes what happens when a fluid or gas accelerates from a higher pressure zone to a lower pressure zone (ignoring gravity, viscosity, turbulence, ...). Although dynamic pressure is related to velocity, it doesn't make sense to choose a reference frame that is not interacting with the flow. For example, assuming that you have a Bernoulli compliant flow in a clear pipe and observe that flow from outside that pipe while traveling at speed v₁ for case S and v₂ for case R. The point here is that the speed of the observer will not have any affect on the static pressure of the fluid in the pipe at any point in that pipe. Within the Bernoulli compliant pipe, the pressures and speeds with respect to the pipe will be compliant with Bernoulli, but Bernoulli equation will not apply when using reference frames S or R. For example, mass flow is constant within a pipe, but how can you have mass flow if the observed velocity is zero?

 

[jpas]

Hello Russ Waters and Jeff Reid,

Your conversation is interesting. However, I still think my example is valid. I´ll try to make it clearer.

Consider two points, 1 and 2, on a streamline of a fluid flowing with no turbulence (nor elevation) so that Bernoulli´s equation can be applied to the two points . Do you agree with

$$p_1 + \frac{1}{2}\rho (v_1)^2=p_2 + \frac{1}{2}\rho(v_2)^2$$

this? If you do, and I don´t see why you shouldn´t, then the rest follows.

On the referential S, $$v_1=0$$, so

$$p_1=p_2 + \frac{1}{2}\rho (v_S2)^2$$

so, $$p_1\succ p_2$$

But on referential R, in which $$v_2=0$$

$$p_2\succ p_1$$

by the same reasoning.

Any objections?

 

[jpas]

Hi Jeff Reid,

Within the Bernoulli compliant pipe

What does this mean? Bernoulli´s equation only applies on a certain frame? Or on other frames it is irrelevant?

 

[rcgldr]

Bernoulli compliant pipe flow

What does this mean? Bernoulli´s equation only applies on a certain frame?

For flow in a pipe, Bernoulli is an idealized situation. From my previous post:

In the real world, because of wall friction and viscosity, in a constant diameter pipe, the flow and speed remain constant, but the pressure decreases with distance. I assume that mechanical energy is being converted into heat energy and then dissipated by the pipe, and this violates Bernoulli.

Bernoulli is violated in a real world pipe because the interaction between the pipe and fluid reduces the total energy of the fluid or gas flowing within the pipe.

referential S v1=0, referential R v2=0

Bernoulli doesn't work if you have a zero velocity, because part of the assumption involved with Bernoulli, and also what happens in the real world once a flow stabilizes, is that mass flow is constant, which isn't possible if velocity is zero. For a flow within a pipe, the proper frame of reference is the pipe itself.

If you want to introduce new frames of references R and S, they need to interact with the fluid so that those frames of references make sense. For example if a pitot tube were moving against the flow at V, reducing the flow speed within the chamber of the pitot tube to "zero" relative to the pitot tube (note this involves some change in total mechanical energy of the affected fluid or gas), then the static pressure sensed within the chamber of the pitot tube equals the static pressure + dynamic pressure (1/2 density V^2) of the fluid that passes by unaffected by the pitot tube (plus a small amount of pressure due to the work peformed by the pitot tube).

Bernoulli considers the total pressure of the fluid to be the sum of the static and dynamic pressure, but the dynamic pressure component only makes sense when speed changes occur via interaction within the fluid itself (acceleration due to pressure gradient), or via interaction with some other object (like a pitot tube).

The reason the pitot tube can decelerate the relative flow to "zero" relative velocity and still function, is that the pitot tube doesn't significantly interfere with the surrounding flow. If you capped the end of the pipe, then the flow would stop and there would be no dynamic pressure. The pitot tube peforms work on a small portion of the fluid or gas that is accelerated to the chambers speed, but is designed so that a tapered stagnation zone builds up ahead of the opening, that distributes and minimizes the effect of the work done on the small amount of fluid or gas actually accelerated by the stagnation zone itself. Most of the fluid or gas ahead of the stagnation zone will end up being diverted and bypass the stagnation zone, with only a relatively small portion of fluid or gas being accelerated "forwards" by the stagnation zone.

So getting back to your original question, if you have two pitot tubes moving at different speeds within a fluid, then the faster moving pitot tube will sense a higher static pressure in it's inner chamber.

If you had an object that moved along with the fluid as it flowed and/or accelerated through a pipe, then the static pressure sensed by that object is the same as the static pressure sensed by the pipe in the vicinity of that object, even though the object is moving with repect to the pipe.

 

[tiny-tim]

Bernoulli's equation is only for steady flow

Hey guys!

Aren't we missing something? …

Bernoulli's equation is only for steady flow!​

Steady means always the same at the same point.

Bernoulli's equation in a frame with velocity u relative to steady flow would need an extra term ρu.v (plus a constant 1/2 ρu², which is absorbed into the equation's own constant).

Flow along a pipe that is moving is not steady (unless the pipe's velocity u is always perpendicular to the flow, u.v = 0, eg purely horizontal flow in a vertically moving pipe).

(The condition of steadiness is essential for the derivation of Bernoulli's equation, because it needs the work done on a slug of fluid to depend only on position and not on time.)

 

[jpas]

Hello Tim,

When I mentioned different frames, they don´t have to be moving pipes... just pure abstract moving frames! It can be a dog running side by side to the fluid. The static pressure it will measure will be different from the one measured on another frame.

What I got out of this discussion is that static pressure does vary with different frames but it´s irrelevant that it does so. The only relevant frame is the one of the pipe. (I´m afraid I missed that explanation about the "pitot tube" - sorry Jeff Reid).

So is this it?

 

[tiny-tim]

Hello jpas!

Hello Tim,

When I mentioned different frames, they don´t have to be moving pipes... just pure abstract moving frames! It can be a dog running side by side to the fluid.

Yes, that's what I meant …

in a frame in which the pipe is moving (for example, the frame of your running dog), the flow is not static.

 

[russ_watters]

Bernoulli's principle only applies when the acceleration is due to internal forces, pressure differentials, or gravitational potental energy per unit volume. Bernoulli relates the pressure and speeds during accelerations or changes in height.

I agree with the first sentence but not the second. In any case, I know what Bernoulli's equation/principle says, but what does that have to do with how the speed is sensed?

Assuming constant height, Bernoulli equation is derived based on acceleration:

I don't see any mention acceleration there: Bernoulli's equation is derived based on conservation of energy, comparing a starting point and an end pont, without regard to the particulars of how it got from one to the other. Since you don't know the distance between the two points, the acceleration could be literally anything.

In the real world, because of wall friction and viscosity, in a constant diameter pipe, the flow and speed remain constant, but the pressure decreases with distance. I assume that mechanical energy is being converted into heat energy and then dissipated by the pipe, and this violates Bernoulli.

 

[russ_watters]

Hello Russ Waters and Jeff Reid,

Your conversation is interesting. However, I still think my example is valid. I´ll try to make it clearer.

Consider two points, 1 and 2, on a streamline of a fluid flowing with no turbulence (nor elevation) so that Bernoulli´s equation can be applied to the two points . Do you agree with

$$p_1 + \frac{1}{2}\rho (v_1)^2=p_2 + \frac{1}{2}\rho(v_2)^2$$

this? If you do, and I don´t see why you shouldn´t, then the rest follows.

On the referential S, $$v_1=0$$, so

$$p_1=p_2 + \frac{1}{2}\rho (v_S2)^2$$

so, $$p_1\succ p_2$$

But on referential R, in which $$v_2=0$$

$$p_2\succ p_1$$

by the same reasoning.

Any objections?

Yes, I object. You need to try to relate this to reality in order to get your arms around it. Put permanently installed pito-static tubes into this device you have constructed and explain how it is possible to measure two different speeds at the same time. You can't.

I order to be in the second reference frame, you have to be in a submarine, and its engines have to be running, which imparts energy on the flow, which means you are not just traveling along a streamline.

 

[russ_watters]

Bernoulli doesn't work if you have a zero velocity, because part of the assumption involved with Bernoulli, and also what happens in the real world once a flow stabilizes, is that mass flow is constant, which isn't possible if velocity is zero. For a flow within a pipe, the proper frame of reference is the pipe itself.

If the pressure vessel generating the flow is several orders of magnitude larger than the diameter of the opening, it is close enough for a simplistic calculation. Problems from a balloon to a hydroelectric dam assumes the flow in the reservoir to be zero.

If you want to introduce new frames of references R and S, they need to interact with the fluid so that those frames of references make sense. For example if a pitot tube were moving against the flow at V, reducing the flow speed within the chamber of the pitot tube to "zero" relative to the pitot tube (note this involves some change in total mechanical energy of the affected fluid or gas)...

Reiterating for emphasis. I used a submarine to affect that change in mechanical energy in my post, but whatever you use, there has to be something.

 

[jpas]

Yes, I object. You need to try to relate this to reality in order to get your arms around it

Suppose I create two super devices that act as the two frames I imagined and can measure velocity and pressure without distrubing the fluid´s flow. In this case, could they measure different satic pressures?

 

[russ_watters]

Suppose I create two super devices that act as the two frames I imagined and can measure velocity and pressure without distrubing the fluid´s flow. In this case, could they measure different satic pressures?

No. Such a device is inherrently impossible because in order to measure velocity, you have to interact with the flow.

But that's not what I actually meant when I said you have to relate this to reality. I mean describe an actual situation where you can measure what you showed with your calculations in post #17. Let me help. You mentioned in a later post that maybe you have a dog running alongside the flow. Well I suppose the dog could have a pitostatic tube attached to it and he could be pulling it through the flow. You start with a very large tank, with water being forced out one end into a pipe and you've rigged a sealed groove that allows the dog to pull the pitostatic tube through the device.

So in the first scenario you calculated, you have the pressures measured by permanently installed pitostatic tubes. In the second, you give the dog a kick and he starts running. But wait - starts running? That takes energy! He's moving his pitostatic tube past the permanently installed one in the tank and clearly, his measurement will have more energy because they have the same static pressure, but his legs are providing velocity to give his pitostatic tube a different velocity pressure.

Your scenario here is not steady flow along a streamline, it is two completely different flow situations.

 

[jpas]

because they have the same static pressure

in your argument, I think you already assume this as a premise. After rereading a lot of times, I guess I didn´t comprehend what you said.

different flow situations

Why is the flow different? Surely the velocity of the flow is different, but that´s what´s intended.

Could you explain yourself more simply? (I´m an high school student, sorry if I´m a bit slow)

 

[rcgldr]

Assuming constant height, Bernoulli equation is derived based on acceleration:

wiki_derivation.html

I don't see any mention of acceleration there.

The first derivation in that wiki article is based on Newtons 2nd law (f = ma), it starts off with m dv/dt = F, where dv/dt represents an acceleration. A better example of a Newton based derivation can be found here, since it doesn't shortcut the integration step:

http://home.earthlink.net/~mmc1919/venturi_discuss_math.html

The static pressure it will measure will be different from the one measured on another frame.

The static pressure will measure the same regardless of the frame of reference.

This is how static ports work on civilian aircraft. These are pipes leading to a chamber inside the aircraft and to a flush mounted port (hole) on the oustide of the aircraft, perpendicular to the relative air flow, placed at a point where the flow is not being accelerated. The port hides within a boundary layer than transitions from zero speed relative to the surface to the speed of the relative flow. Regardless of the aircraft speed, from zero to perhaps mach 0.3, the static port chamber will have the same static pressure as the air outside of the aircraft, regardless of the speed of the aircraft.

The static port could be considered a frame of reference, and the sensed static pressure does not vary with the speed of the aircraft (within reason, pitot-static tubes are needed for supersonic flight, but the static port on the tubes are still just holes placed perpendicular to the relative flow).

The point here is that with pitot-static tubes, the pitot port senses the total pressure (static + dynamic), which is speed senstive, but the static port senses the static pressure, which is not speed sensitive. The pitot port senses the total pressure because it accelerates a small sample of the air to the speed of the pitot tube's chamber and the change in momentum of that air is related to dynamic pressure, but since the static port doesn't affect the air flow (since it hides inside an existing boundary layer), it can sense the static pressure regardless of speed.

Static pressure is not affected by a frame of reference. As another example, imagine two closed containers each with the same interal static pressure moving at different speeds in outer space, unaffected by any external forces. Does the difference in speed of those containers have any effect on the static pressure within those containers?