# Liquid helium and curved piston

1. Dec 12, 2013

### Gh778

It's a theoretical study. Like the image showing, the curved piston move in container and need energy for that. In the same time, another piston move up and recover energy. The pressure inside container is always the same (need to control speed of pistons, but there is a pressure). No gravity here. Like one piston is curved it seems pressure on curved piston is reduced and atoms of helium has a up force in the same time. So, the image is false ? where ?

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2. Dec 12, 2013

### Staff: Mentor

The picture seems fine to me, but I am not sure what you are asking about. The only thing that seems odd is your comment about "reduces pressure". Since you have already specified that the pressure inside the piston is constant then the pressure clearly cannot be reduced.

3. Dec 12, 2013

### wreckemtech

The force from the curved object being moved into will be distributed evenly over all surfaces inside the piston as pressure. Also, the curved object itself is likely to develop an area of higher pressure in front of it as it is moved into the cylinder. That's what happens in real pistons anyway, and just one of the many things that cause irreversibilities. The drawing works on a quasi-static assumption that doesn't hold for real pistons.

4. Dec 13, 2013

### Gh778

Sure, the pressure must be same everywhere, but I "see" a reduced pressure on curved piston maybe this force is canceled by something I gorgot ?

5. Dec 13, 2013

### Staff: Mentor

How can you "see" a reduced pressure if the pressure is not reduced? That doesn't make sense. I don't get what you are asking.

6. Dec 13, 2013

### Gh778

On my picture, forces are lower on curved piston than it must be. I think the pressure must be the same, but I don't understand where is my error in the picture. Like I drawn, the curved piston receive less force due to the position of atoms (black forces).

7. Dec 13, 2013

### Staff: Mentor

I'm sorry, but you keep on asking about an error in the picture, but I still don't understand what you think that the picture means and why.

What do the black forces have to do with the force on the piston? The black forces aren't even acting on the piston. You haven't identified any forces on the curved piston at all, so I have a hard time guessing what there is in the picture that you think suggests a reduction in the force on the piston.

By the way, the way that you have drawn it is a fairly poor way to think about pressure mechanically. You are much better off thinking about it in terms of collisions with moving particles rather than static forces transmitted through small rigid bodies. Perhaps that it the primary source of your confusion?

8. Dec 14, 2013

### Staff: Mentor

I have been thinking about what you could possibly mean by the drawing and your statements. You seem to believe that a curved section of fluid will have different forces acting on it than a straight section. Specifically, you seem to believe that a concave fluid section will have less force.

9. Dec 15, 2013

### Gh778

Yes, it's that. If I study last layer, the pressure is different than the straight surface. Why ? because I think there is not the same number of contact on a straight line compared to a curved line. Take a 1/4 circle of radius 1, with atoms with diameter 0.5. in the straight line the force is 2 units, always. With a curved line it's 3.92 (imagine it's 4, an integer). But the pressure come from upper layer with more contact it's not an integer. And according the position of that extra contact the angle don't work with the same force. So, in this case the force from curved shape is greater than straight line. For me, it a problem of quantification of the number of contacts.

Edit: and with black forces all these forces must done the same pressure. The different position of contact give an angle different so I can't imagine the same pressure. Maybe the mean about a long distance ?

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10. Dec 15, 2013

### Staff: Mentor

It isn't just the number of contacts, it is also the direction of the contacts. The force on a small surface element is given by $d\vec{f}=P\;dA\;\hat{f}$ where $\hat{f}$ is a unit vector normal to the surface element.

Now, for a concrete example, let's take the drawing you posted in post 9 above with 1/4 of a circle arc and calculate the total force from that surface which we can then compare with the force from the flat face.

$\vec{f}=\int d\vec{f}= \int P\;\hat{f}\;dA = \int P\;(\cos(\theta),\sin(\theta)) \; r \; d\theta=(P\;r,P\;r)$

It is pretty simple to see that the force from the flat face is $(-P\;r,0)$, so the horizontal component of the force from the curved wall balances exactly with the force from the flat face. So although there are more contacts, they don't all add together, and if you sum the contributions from all contacts (while accounting for the direction) then you get the forces properly balancing.

Last edited: Dec 15, 2013
11. Dec 15, 2013

### Gh778

You're right the direction must take in account. You consider the size of atom like 0 in your formula ?

12. Dec 15, 2013

### Staff: Mentor

If you wish to consider a finite size atom then you simply replace the integral by a sum. This is a standard numerical approximation technique. You may introduce some numerical loss of precision in that manner, but the limit will be as shown.

13. Dec 15, 2013

### Gh778

Maybe it's for that I don't find the solution. For you, with a finite size atom, the pressure is increased by upper layer or not ?

14. Dec 15, 2013

### Staff: Mentor

The pressure is not increased by the upper layer. Let's go ahead and work it out.

Let's consider a pressure vessel with a corner, but instead of a 90º bend it has this 1/4 circular arc. The pressure at the wall is P, and the pressure from the upper layer of fluid is P'. The finite size atoms form a layer of thickness $\delta r$. Therefore, they are bounded by a vertical segment, the arc, and a horizontal segment of vessel wall (P) as well as by an arc of the fluid (P').

The force on the fluid due to the vertical segment of vessel wall is $(0,\delta r \; P)$.

The force on the fluid due to the horizontal segment of vessel wall is $(\delta r \; P,0)$.

The force on the fluid due to the arc of vessel wall is $\int P\;(\cos(\theta),\sin(\theta)) \; r \; d\theta=(P\;r,P\;r)$

The force on the fluid due to the upper layer is $\int P'\;(-\cos(\theta),-\sin(\theta)) \; (r+\delta r) \; d\theta=-(P'\;(r+\delta r),P'\;(r+\delta r))$

Now, because the fluid is static we know that $(0,\delta r \; P)+(\delta r \; P,0)+(P\;r,P\;r)-(P'\;(r+\delta r),P'\;(r+\delta r))=0$ which is true for P=P'. Therefore the pressure is not increased (nor decreased) by the upper layer even for a layer of finite sized atoms.

Last edited: Dec 15, 2013
15. Dec 15, 2013

### Gh778

Formulas must works in every cases I think. Why I can't draw this image ? Where my forces are wrong ?

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16. Dec 15, 2013

### Staff: Mentor

What makes you think they are wrong?

I still don't understand what you think the drawings show that is problematic. And I have already shown how the math works. What is the relationship between the drawing and the math?

Do you understand and agree with the math?

17. Dec 15, 2013

### Gh778

Yes, I understand math. But I thought these formulas can be used in every case and this one for example. Like I drawn forces, I find a force greater than with a straight line. Even I take in account angles. The force must be the same, no ?

18. Dec 15, 2013

### Staff: Mentor

How do you find that? This is what I don't see in your drawings.

It is clear from the math that the forces are the same. It is not clear from your drawing that the forces are not the same.

19. Dec 15, 2013

### sophiecentaur

Is your problem simply due to the difference between treating the problem as a continuum and treating it as quantised in some way? Concepts like pressure are classical and are based on a continuous medium. If you want to change the simple model into one involving infinitely slippery snooker balls then I would not be surprised if you were to get a few discrepancies due to quantisation. But, if the billiard balls are of atomic size, there would be other issues which would introduce other errors into any real or imagined measurement.
You may well be 'right' but is it relevant?

20. Dec 15, 2013

### Staff: Mentor

Gh778, the other mentors have let me know that this is a re-opening of a closed thread, so I am closing this one down also. I hope that the math is helpful. It shows conclusively that the pressure is the same on both sides of a curved surface.

If you wish to consider the finite size of an atom in a tangential direction then you can simply change the integrals to discrete sums. You may get some numerical imprecision in that process, but the discrepancy should always be less than the force on one atom, a truly negligible loss of precision, and nothing more than an artifact of the calculation method anyway.

Furthermore, your drawings do not contradict the math in any way. In order to actually use the drawings as you seem to intend to use them you would need to write an expression for the force from each atom as a function of the positions of the neighboring atoms. Atoms are not small rigid balls, and the distances between atoms will vary as will the forces. You would then need to solve the resulting system of equations in order to determine the forces involved. Simply drawing a picture can, at best, tell you the directions of the many forces. It is not sufficient to claim anything about the magnitude of the net force, as you are attempting to do.

Given that the math is so clear and simple and given the immense number of degrees of freedom available in your drawing, I have a hard time understanding why you would even think that there is a possibility of a significant discrepancy. But in any case, having presented the physics there is nothing more that can be accomplished here.

Last edited: Dec 15, 2013