# Does pressure depend upon the observer?

1. Feb 17, 2010

### 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?

2. Feb 18, 2010

### jpas

3. Feb 18, 2010

### Staff: Mentor

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.

4. Feb 18, 2010

### 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.

5. Feb 18, 2010

### 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.

6. Feb 18, 2010

### 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?

7. Feb 18, 2010

### 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.

8. Feb 18, 2010

### Staff: Mentor

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

9. Feb 18, 2010

### rcgldr

Wiki articles:

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

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

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

10. Feb 18, 2010

### 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.

11. Feb 18, 2010

### Staff: Mentor

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.

12. Feb 18, 2010

### rcgldr

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.

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.

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

Last edited: Feb 18, 2010
13. Feb 19, 2010

### 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?

14. Feb 19, 2010

### Staff: Mentor

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.
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).

15. Feb 19, 2010

### Staff: Mentor

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

16. Feb 19, 2010

### rcgldr

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).

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.

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:
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?

Last edited: Feb 19, 2010
17. Feb 19, 2010

### 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?

Last edited: Feb 19, 2010
18. Feb 19, 2010

### jpas

Hi Jeff Reid,

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

19. Feb 19, 2010

### rcgldr

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

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.

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.

Last edited: Feb 19, 2010
20. Feb 20, 2010

### 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.)