# Compressing an incompressible fluid.

1. Oct 17, 2009

### dE_logics

I was reading the proof of how the work done on an ideal fluid which has been pressurized to a pressure 'P' is P.

There I read that the work done will be the product of the area of cross section of the piston, the pressure applied on the piston and the distance moved by the piston.

If the piston got moved, the fluid should have been compressed

2. Oct 17, 2009

### Staff: Mentor

Work done by a piston is indeed P A dx. A is finite, but for an incompressible fluid the pressure required to compress it is infinite and dx is 0, so the result of this formula is undefined. You will have to use some other equation.

3. Oct 17, 2009

### dE_logics

Yes, I agree, that will mean although pressure can be made in an incompressible fluid, the work done on it will be 0...in an ideal fluid, there can't be any energy in the fluid by virtue of pressure.

4. Oct 17, 2009

### dE_logics

So the P in Bernoulli's equation is wrong.

5. Oct 17, 2009

### Staff: Mentor

No, it is not that there can't be any energy in the fluid by virtue of pressure, just that you have to use a different calculation to find it.

You are looking at two very specific equations, the piston equation and Bernoulli's equation. These two equations are not general laws but apply for specific situations. Bernoulli's principle applies along a streamline in an inviscid flow, and the piston equation is valid for a piston. There are not any inviscid flow streamlines in a piston and there are not any pistons in an inviscid flow streamline, the two equations describe completely different physical situations.

6. Oct 17, 2009

### dE_logics

So we can safely eliminate the P in Bernoulli's equation in an ideal fluid.

7. Oct 17, 2009

### rcgldr

The inviscid fluid also causes another problem with the classic example of Bernoulli flow in a pipe of varying diameters. Since there is zero viscosity, a high speed stream exiting from a narrow section of the pipe would does not interact with the surrounding fluid, and would continue moving at that same high speed, without affecting the surrounding fluid.

With an incompressable fluid, you also get things like instant changes in speed (infinite accelerations), and the speed of sound is infinite.

8. Oct 18, 2009

### ideasrule

9. Oct 18, 2009

### Staff: Mentor

Only if you don't believe in the conservation of energy and think that the kinetic energy of a fluid can spontaneously and magically increase without and decrease in any other form of energy.

10. Oct 18, 2009

### amannikus26

haha i love physics <3

11. Oct 18, 2009

### Staff: Mentor

I suspect there is more to that problem than you are telling us. For example, is the fluid being moved by a pump, like a hydraulic cylinder? With a hydraulic cylinder, there is a pressure, a cross sectional area and a distance moved, from which you can calculate work. But there is more to it than that - the ideal version of Bernoulli's equation doesn't deal with internal work from compression, but that's not a flaw, it just isn't what it is for.

Ie, consider a mercury barometer. The column of mercury has a height, weight and cross sectional area. Is work (energy) being done? No. Stored in the compression of the mercury. Maybe, but not enough to bother worrying about. So what does Bernoulli have to say about this then? Bernoulli's equation in this case helps you explore is the potential energy - potential work - of the column of mercury.
Bernoulli's equation is all about "P". That's its entire purpose/point. So obviously, you can't throw out the "P". You just haven't understood Bernoulli's principle/equation yet.

Last edited: Oct 18, 2009
12. Oct 19, 2009

### dE_logics

Yes, I know, there's hardly any use in an ideal fluid.

But we discussed that the energy possessed by virtue of pressure is negligible...ok, I get it, if the fluid is maintained at constant pressure, then it will result in work being done continuously on that fluid...but then again a negligible part of the energy will be by virtue of pressure (cause the fluid does not get compressed).

That means there can be considerable energy by virtue of pressure, but we do not agree with that do we?

Another thing that can happen is that, since this is a flow, continuous energy can be delivered by virtue of pressure, and cause of this continuous work done, the K.E of the individual particles will increase in the narrow cross section cause of this continuous energy given.

Yes, that's exactly what I'm talking about above, but in the book the situation is not like this....it's a closed cylinder...so there's a direct relation between the work done on the fluid and pressure.

Ok then, how do we prove in an ideal fluid that the potential energy in it = P?

13. Oct 19, 2009

### Staff: Mentor

No.
No.
That isn't what is being described by the equation.
Correct: there can be considerable energy there.
Correct.
It doesn't have to be work done it may just be potential work.
Punch a hole in the bottom of the cylinder and attach a turbine! http://en.wikipedia.org/wiki/Hoover_Dam

14. Oct 19, 2009

### Staff: Mentor

NO! That is not at all what I said. I don't know if you are deliberatly misunderstanding or if there is a language barrier but look at what I said.
There IS energy due to pressure in an incompressible fluid, but you cannot use the piston equation to find it. The piston equation is completely irrelevant if you don't have a piston. When is the last time you heard of a piston using water or mercury or some other incompressible fluid?

There is pressure energy in an incompressible fluid.

The piston equation doesn't apply.

15. Oct 19, 2009

### rcgldr

This assumes the fluid isn't being allowed to move. Assume fluid is contained between two pistons, perhaps the first piston is in a horizontal section of pipe, and the second pistion is in a vertical section of pipe. Assuming no losses, the work done by the first piston would end up as worked peformed onto the second piston and whatever that second piston supported.

Hydraulics, commonly used in some types of elevators, tractor shovels, ...

16. Oct 19, 2009

### Staff: Mentor

Just to close this gap, there are (at least) three separate concepts being discussed here and the OP's problem is in distinguishing between them.

Example 1 is the idea of energy stored in a fluid due to compression. The typical example is an inverted piston-cylinder arrangement with air in the cylinder. A weight is placed on top of the piston, compressing the air and storing some potential energy (and releasing some entropy...). This example has nothing to do with Bernoulli.

But then the OP made blanket statements about the uslessnes of "P" in Bernoulli's equation. Why Bernoulli's equation is even beind discussed here is beyond me, since from the first 3 posts it doesn't appear to apply, but in any case, obviously, the ideal (incompressible) form of Bernoulli's equation doesn't help us examine the above situation.

Example 2 is a hydraulic cylinder with an (assumed) incompressible fluid inside. This example covers a gap in the OP as the situation in the OP is described vaguely enough that this example applies. Ie, the power of the hydraulic pump (or piston) is the flow times the total pressure. However, this is not, strictly speaking, a Bernoulli's situation either (though it can be modified a bit to be). So...

Example 3, a hydroelectric dam, is related to example two in that you have a column of (assumed) incompressible fluid under some pressure. It is different from Example 2, in that its own height provides the pressure instead of a large weight put on top of it (like in the piston). This is a Bernoulli's example straight out of the first week of most intro fluid dynamics textbooks/courses.

17. Oct 19, 2009

### Staff: Mentor

Hmm, you are right. I have always used the term "hydraulic cylinder" but it is in fact a piston, isn't it.

In any case, in a hydraulic cylinder piston the dx is not equal to 0 so hopefully dE_logics will see how the piston equation could work even with an incompressible fluid in such cases.

18. Oct 20, 2009

### dE_logics

@russ_watters

But how?...ok in a piston cylinder arrangement, if work done on an ideal fluid then it's negligible...that means that the fluid absolutely does not store any energy by virtue of the force the molecules apply on each.

So what happens when the molecules are in motion and at a certain pressure?...for pressure to do work, the fluid should expand, an ideal fluid does not.

Ok, so this is actually happening...so pressure is working as a ram.

Potential work?...never heard of it.

I meant theoretical proof.

Calculating the work done that way is not an easy task!!...you need to take in the efficiency of the turbine, then by someway measure the power delivery by it.

Not heard, I did with an injection.........a few days ago.

And that is the work done by pressure!

Pressure's working like an infinitely small RAM.

The energy is delivered through the pressure.

I guess this clears the doubt for the mean time...but I think it's not that much clear.

And yes, the doubt appeared while I was revising the post.

In Bernoulli's equation, we directly make use of this P...which, from what what I've concluded above, should be related from the energy transfered through pressure.

So, suppose in a flowing fluid work is being done (that's why it's flowing), if the flowing fluid is made to climb a height, then this pressure energy will turn to potential continuously.

If the column suddenly narrows, then the pressure will be converted to kinetic...or actually the work being done on the flowing fluid will be continuously used to increase the K.E of the fluid instead of imparting pressure to it (that's according to the law...I do not know why does this happen).

Ok, the problem appears to be solved then...hopefully.

I was just giving an opinion on what was giving in the book.

From my point of view I was right.

Although the dx will remain dx, the F can increase...so F dx might just increase...but it will eventually result in dw.

19. Oct 20, 2009

### rcgldr

The only means to for that pressure to be utilized is via internal acceleration of the fluid, for example, when the fluid flows through a narrower section of pipe, the speed of the fluid increases and the pressure decreases. What happens with the pipe diameter increases is complicated. There's no viscosity, so streamlines don't interact via friction at the edges. The stream line would have to be decelerated at the front of the stream as it impacts with the higher pressure, slower moving, surrounding fluid. Without visocity or compressability, I'm not sure what the interaction would be between a high speed low pressure stream within a surrounding, lower pressure, initially non moving stream. The leading edge of the high speed stream ends up colliding with the slower moving fluid just ahead of the stream, and I'm not sure how to calculate the reaction.

Last edited: Oct 21, 2009
20. Oct 21, 2009

### dE_logics

Ok then...thanks everyone!