Influence of vacuum in the Conservation of energy

In summary: The external force is considered as a temporary influence that pushes the small lid down (and keeps it down) after which the energy gets transferred further within the container.
  • #1
JohnnyGui
796
51
Vacuum.png


I probably haven’t thought this through. A sideview of a closed container filled with air consisting of two vertical cylinders (with radius ##r_1## and ##r_2##) are connected by two horizontal tubes. The container is separated by a small and a large lid (red) that are circular and can move up and down freely. Furthermore, the small lid is connected to a handle that sticks out of the container to move it up and down. There is a pressure equilibrium; ##P_1 = P_2##

Pushing the handle down with a force ##F_1## makes the small lid move down by a distance of ##d_1##. The work done on the small lid would be ##E_1 = F_1 \cdot d_1##.
This would make the pressure ##P_2## increase and the large lid would undergo a force ##F_2## which makes it move upwards by a certain distance ##d_2##. According to the conservation of energy ##F_1 \cdot d_1 = F_2 \cdot d_2##.

Here’s the thing however. When pushing the small lid down, a vacuum arises in the upper container half which would counter the force ##F_1## and make the moved distance ##d_1## smaller than expected. At the same time, that same arised vacuum would support the force ##F_2## and thus make the moved distance ##d_2## of the large lid larger than expected. The work done on the smaller lid would therefore seem smaller than the work done on the large lid. Wouldn’t this vacuum thus break the conservation of energy ##F_1 \cdot d_1 = F_2 \cdot d_2##?
 
Physics news on Phys.org
  • #2
JohnnyGui said:
When pushing the small lid down, a vacuum arises in the upper container half ...
Vacuum?
JohnnyGui said:
... which would counter the force ##F_1## and make the moved distance ##d_1## smaller than expected.
Expected?
 
  • #3
A.T. said:
Vacuum?

As in ##P_1## decreases compared to ##P_2##

A.T. said:
Expected?

As in ##d_1## would be larger if ##P_1## did not decrease by the movement of the small lid.
 
  • #4
Are you considering a quasi-static situation where the pressure is uniform throughout each section or are you considering dynamic pressure waves in the container? Are the lids considered massive or massless?
 
  • #5
Dale said:
Are you considering a quasi-static situation where the pressure is uniform throughout each section or are you considering dynamic pressure waves in the container? Are the lids considered massive or massless?

Quasi-static and the lids are considered massive.
 
  • #6
First, there is an external force acting on the system. Why would energy be conserved? Second, you push the plunger and both lids start moving. Why would the net energy be the difference of the motions and not the sum? (Plus the motion of the air if you wanted to be complete) Finally, instead of one pressure and vacuum, think in terms of a pressure difference between the top half and the bottom half. Both “lids” experience the same pressure difference at all times in the quasi static model.
 
  • Like
Likes Lnewqban
  • #7
JohnnyGui said:
View attachment 257524
...
Here’s the thing however. When pushing the small lid down, a vacuum arises in the upper container half which would counter the force ##F_1## and make the moved distance ##d_1## smaller than expected.
I may be wrong, but this is the way I see it happening:
Either that, or the force ##F_1## must increase as the resultant force tends to decrease as the force ##F_{vacuum}## increases.
Think of a force being applied by the middle point of a spring, one half resists that force by increasing compression and the other half resists that force by increasing tension.
The work of that force is temporarily converted into potential energy; same happens with the pressure differential created ##(P_1-P_2)##.

JohnnyGui said:
At the same time, that same arised vacuum would support the force ##F_2## and thus make the moved distance ##d_2## of the large lid larger than expected.
Once enough pressure differential has been created (same for both pistons) to overcome inertia and friction of pistons against cylinders, both pistons will move in proportion to theirs sectional areas, displacing same volumes of air.
During this state of equilibrium, there is no reason that would make the moved distance ##d_2## of the large piston larger than it should be.

JohnnyGui said:
The work done on the smaller lid would therefore seem smaller than the work done on the large lid. Wouldn’t this vacuum thus break the conservation of energy ##F_1 \cdot d_1 = F_2 \cdot d_2##?
In any event, it would be the other way around.
A portion of the energy entering the system via work on the small piston will be used to overcome pistons-cylinders friction and to generate heat and turbulence inside the compressed and expanded volumes of air.
 
  • #8
Cutter Ketch said:
First, there is an external force acting on the system. Why would energy be conserved?

Can't the external force be considered as a temporary influence that pushes the small lid down (and keeps it down) after which the energy gets transferred further within the container?

Cutter Ketch said:
Second, you push the plunger and both lids start moving. Why would the net energy be the difference of the motions and not the sum? (Plus the motion of the air if you wanted to be complete)

I'm not sure what you mean with energy difference. What I'd expect is that the energy that is spent to push the small lid down gets transferred to pushing the large lid up. But when taking the influence of the arisen pressure difference into account, it seems that these two energies are not the same as explained in my OP.

Cutter Ketch said:
Finally, instead of one pressure and vacuum, think in terms of a pressure difference between the top half and the bottom half

Vacuum is indeed a poorly chosen word. I actually meant pressure difference.
that

Lnewqban said:
Once enough pressure differential has been created (same for both pistons) to overcome inertia and friction of pistons against cylinders, both pistons will move in proportion to theirs sectional areas, displacing same volumes of air.
During this state of equilibrium, there is no reason that would make the moved distance d2d2d_2 of the large piston larger than it should be

So what you're saying is that the pressure difference will cause both lids to rise up, and thus the ##d_2## should not be larger than expected? If that's the case, will the sum of the energies of both lids rising up be the same as the energy that pushed the small lid down in the first place? (Also, I am assuming there is no friction between the lids and the container.)
 
  • #9
JohnnyGui said:
...
So what you're saying is that the pressure difference will cause both lids to rise up, and thus the ##d_2## should not be larger than expected? If that's the case, will the sum of the energies of both lids rising up be the same as the energy that pushed the small lid down in the first place? (Also, I am assuming there is no friction between the lids and the container.)
For ideal conditions, with no energy lost, pushing the handle down distance ##d## with force ##F_1## will produce the work of moving both pistons and both masses of air inside the container.
The chain of events inside the container do not change the magnitude of input work.
##W## = ##F_1## x ##d##
 
  • Like
Likes Cutter Ketch
  • #10
If you give the plunger a quick push, in the quasi static assumption, and with no friction a pressure differential will only exist while you are pushing. When you stop pushing, both pistons will be moving as will the air. The total kinetic energy imparted by the push is the sum of all of those. The small one will be moving faster than the large one according to their diameter. They and the air will continue to move under their own inertia (the air has inertia too) until friction (ok, you said no friction) and loss in the air stops the movement or the pistons run into the corners. We could say no loss in the air, but given that the air has to squeeze down and speed up as it enters the small side, and vice versa, it is hard to imagine that the airflow won’t significantly resist the motion (in which case there will be a backwards pressure differential slowing the pistons down)
 
  • Like
Likes Lnewqban and jbriggs444
  • #11
Cutter Ketch said:
If you give the plunger a quick push, in the quasi static assumption, and with no friction a pressure differential will only exist while you are pushing. When you stop pushing, both pistons will be moving as will the air. The total kinetic energy imparted by the push is the sum of all of those. The small one will be moving faster than the large one according to their diameter. They and the air will continue to move under their own inertia (the air has inertia too) until friction (ok, you said no friction) and loss in the air stops the movement or the pistons run into the corners. We could say no loss in the air, but given that the air has to squeeze down and speed up as it enters the small side, and vice versa, it is hard to imagine that the airflow won’t significantly resist the motion (in which case there will be a backwards pressure differential slowing the pistons down)

So if I understand correctly:
$$F_1 \cdot d_1 \downarrow = (P_2 - P_1) \cdot \pi r_1^2 \cdot d_1 \uparrow + (P_2 - P_1) \cdot \pi r_2^2 \cdot d_2 \uparrow$$
Where the arrows represent the direction of the movement.
Is this equation correct?
 
  • #12
JohnnyGui said:
So if I understand correctly:
$$F_1 \cdot d_1 \downarrow = (P_2 - P_1) \cdot \pi r_1^2 \cdot d_1 \uparrow + (P_2 - P_1) \cdot \pi r_2^2 \cdot d_2 \uparrow$$
Where the arrows represent the direction of the movement.
Is this equation correct?
I see no justification for such an equation. If the applied force on piston 1 is different from the net pressure force on piston 1, the difference can only arise from the mass and acceleration of piston 1. That's Newton's second law.

The piston mass does not appear in the equation. So the equation must be wrong.

Edit: If the pistons and working fluid are massless/friction free/no viscosity then the equation holds trivially because ##P_1 = P_2## and ##F_1 = 0##.
 
Last edited:
  • Like
Likes Rolacycle
  • #13
jbriggs444 said:
I see no justification for such an equation. If the applied force on piston 1 is different from the net pressure force on piston 1, the difference can only arise from the mass and acceleration of piston 1. That's Newton's second law.

Correct me if I'm wrong but can't the difference also arise from the difference in circular area of the pistons? If the force on each piston caused by the pressure gradiënt is proportional to its circular area, and the mass of the piston is also proportional to the circular area, wouldn't the acceleration of both pistons be the same?

jbriggs444 said:
Edit: If the pistons and working fluid are massless/friction free/no viscosity then the equation holds trivially because P1=P2P1=P2P_1 = P_2 and F1=0F1=0F_1 = 0.

Sorry for the confusion, for my equation I defined ##P_2## as the pressure after the downward force on the small piston has been applied, such that ##P_2 > P_1##. In that case, without friction, would the equation then hold?
 
  • #14
JohnnyGui said:
So if I understand correctly:
$$F_1 \cdot d_1 \downarrow = (P_2 - P_1) \cdot \pi r_1^2 \cdot d_1 \uparrow + (P_2 - P_1) \cdot \pi r_2^2 \cdot d_2 \uparrow$$
Where the arrows represent the direction of the movement.
Is this equation correct?

I see where you were going with that. The work done by the force equals the work done on the two pistons is a good idea. I think significant work is also done on the air, but let’s ignore that for a second.

Let’s see: the net force on piston one is not just the pressure differential times the area. It is the applied force minus the pressure differential so you’d have to fix that. The pressure differential is the only force on the second piston, so that seems right. I’m afraid the biggest problem is that the pressure is not constant during the push. At first the pressure differential is zero, and the force accelerates piston one. This increases the pressure differential reducing the net force and acceleration of piston one while accelerating piston two. Eventually you arrive at a condition where the pressure differential equals the applied force and the pistons move at a constant speed.

If you had written your work equation with integrals over the path (given that you would still be ignoring the work done on the air and correcting the other things I) then I think you could write a work relation like that. Unfortunately it wouldn’t be terribly useful, because you don’t have any prior knowledge of the pressures as a function of time or distance moved.
 
  • #15
Cutter Ketch said:
It is the applied force minus the pressure differential so you’d have to fix that.

Thanks, but I'm not sure if I understand that part because I was considering the pressure gradient forces to occur after force ##F_1## has been applied and the first piston has finished moving the distance ##d_1 \downarrow## (holding the 2nd piston into place until it the 1st piston stops moving if you may). Not sure if this reasoning is possible.

Edit: I should have also said that the ##d_1 \downarrow## on the left hand-side of the equation is not the same as the ##d_1 \uparrow## on the right hand side of the equation since the 1st piston would not move the same distance upwards.
 
Last edited:
  • #16
JohnnyGui said:
Correct me if I'm wrong but can't the difference also arise from the difference in circular area of the pistons?
##\sum F=ma##. The equation you propose contains a term for the applied force. It contains a term for the force arising from pressure difference on the first piston. The remaining term must necessarily be equal to ##ma##. But it's not.
 
  • #17
jbriggs444 said:
##\sum F=ma##. The equation you propose contains a term for the applied force. It contains a term for the force arising from pressure difference on the first piston. The remaining term must necessarily be equal to ##ma##. But it's not.

Sorry but I still don't follow. The force ##F_1## applied to piston 1 makes ##P_2## higher than ##P_1##. Since the force of that pressure gradient can also be applied on piston 2, the work of the pressure gradiënt is spread over both pistons and thus the force on piston 1 does not have to be equal to ##F_1##.

I think it's better to describe my scenario in more detail; I am considering the pistons to move by the pressure gradient only after force ##F_`1## has been applied and piston 1 has finished moving ##d_1 \downarrow## by it. So during ##F_1## and the downward movement of piston 1, piston 2 is held in place.
 
  • #18
JohnnyGui said:
Sorry but I still don't follow. The force ##F_1## applied to piston 1 makes ##P_2## higher than ##P_1##.
No. The force applied to piston 1 makes piston 1 move. The movement of piston 1 eventually makes ##P_2## higher than ##P_1##.

If the piston is massless or if the motion is slow enough that the acceleration rate is negligible then Newton's second law has something to say:

$$F_1+(P_2-P_1)\pi r_1^2 = ma = 0$$
Since the force of that pressure gradient can also be applied on piston 2, the work of the pressure gradiënt is spread over both pistons and thus the force on piston 1 does not have to be equal to ##F_1##.
Rather than "force of that pressure gradient", I would refer to the net force on piston 1 and the net force on piston 2 arising from the pressure gradient. Yes, both forces arise from the pressure gradient.

[The term gradient actually has a slightly different meaning, but I will use it here in the same way that you have -- to mean "pressure difference"]

I do not know what "the work of the pressure gradient" is supposed to mean. One can compute a figure for work and associate it with a pressure gradient by multiplying the pressure gradient by a volume swept out. Yes, there is a work that can be computed based on the volume swept out by piston 1 and a work that can be computed based on the volume swept out by piston 2.

None of that negates Newton's second law. For negligible mass or negligible acceleration, the applied force at piston 1 is still equal to the pressure difference multiplied by the area of piston 1.
I think it's better to describe my scenario in more detail; I am considering the pistons to move by the pressure gradient only after force ##F_`1## has been applied and piston 1 has finished moving ##d_1 \downarrow## by it. So during ##F_1## and the downward movement of piston 1, piston 2 is held in place.
So you latch piston 2 in place to begin.

You force piston 1 slowly downward, applying just enough force to match the slowly increasing pressure gradient. During this process, some work is obviously done by the applied force.

Now you latch piston 1 in place. A pressure gradient now exists. Everything is at rest. Energy has been conserved. There is pressure energy in the working fluid.

You now release the latches that hold both pistons in place and let them snap back into a new equilibrium position? Or do you release them and apply a force on both pistons so that they each gently relax back into a new equilibrium position? Or do you release only piston 2 and let it snap or relax to a new equilibrium position?

Based on the original problem statement, I think that you mean to release only piston 2. You want to do it slowly, resisting the resulting motion with an external force ##F_2##. This external force will draw off the energy that had been originally injected by external force ##F_1##.

Where does a supposed discrepancy come in?
 
Last edited:
  • #19
Cutter Ketch said:
If you had written your work equation with integrals over the path (given that you would still be ignoring the work done on the air and correcting the other things I) then I think you could write a work relation like that. Unfortunately it wouldn’t be terribly useful, because you don’t have any prior knowledge of the pressures as a function of time or distance moved.

Is this basically calculating the pressure gradient forces at infinitesimally small increments of ##d_1## and substracting those pressure forces from ##F_1## until the net force is zero? And the number of times you’d have to substract from ##F_1## multiplied by the small increment of ##d_1## gives the total moved distance of the first piston?
 
  • #20
jbriggs444 said:
You now release the latches that hold both pistons in place and let them snap back into a new equilibrium position?

Yes that's exactly what I mean. I don't apply any force after releasing them so the movement of both pistons after being released is solely performed by the pressure difference.

jbriggs444 said:
What makes you think that the new equilibrium position will match the original equilibrium position?

I'm not thinking that and I'd predict that the new equilibrium position will differ because the movement of piston 2 has contributed to the new equilibrium. What I'm thinking is that the work done by the pressure difference on both pistons (after releasing them) would have to match ##F_1 \cdot d_1## that is initially applied on piston 1 while holding piston 2 in place. Please correct me if I'm wrong on this.
 
  • #21
JohnnyGui said:
Yes that's exactly what I mean. I don't apply any force after releasing them so the movement of both pistons after being released is solely performed by the pressure difference.
In the idealized case (massless pistons, massless fluid, no friction), you do not have a well defined result. Non-zero restoring force and zero mass means infinite acceleration. Two zero masses means that the mass ratio is not well defined. We cannot tell which piston will move more than the other.

If we give the pistons some mass but leave friction out of it, we will end up with simple harmonic motion. The pistons will both oscillate about some new equilibrium point. For a massless fluid, the speed of sound is infinite and the oscillations will continue forever and stay in phase with each other. The relative amplitudes of the vibrations will depend on both the mass ratio and the area ratio of the pistons. The frequency of the vibrations will depend on the masses, the areas and the volumes of the two reservoirs.

It is for this reason that I wanted to simplify the problem by slowly releasing piston 2 only against a resisting force ##F_2##. We can, if you prefer, simplify by releasing both 1 and 2 against new resisting forces ##F_{\text{1b}}## and ##F_{\text{2b}}##

In either case, we will find that the work done by original force ##F_1## is equal and opposite to the sum of the works done by the forces during relaxation.
 
Last edited:
  • #22
jbriggs444 said:
In the idealized case (massless pistons, massless fluid, no friction), you do not have a well defined result. Non-zero restoring force and zero mass means infinite acceleration. Two zero masses means that the mass ratio is not well defined. We cannot tell which piston will move more than the other.

If we give the pistons some mass but leave friction out of it, we will end up with simple harmonic motion. The pistons will both oscillate about some new equilibrium point. For a massless fluid, the speed of sound is infinite and the oscillations will continue forever and stay in phase with each other. The relative amplitudes of the vibrations will depend on both the mass ratio and the area ratio of the pistons. The frequency of the vibrations will depend on the masses, the areas and the volumes of the two reservoirs.

It is for this reason that I wanted to simplify the problem by slowly releasing piston 2 only against a resisting force ##F_2##. We can, if you prefer, simplify by releasing both 1 and 2 against new resisting forces ##F_{\text{1b}}## and ##F_{\text{2b}}##

In either case, we will find that the work done by original force ##F_1## is equal and opposite to the sum of the works done by the forces during relaxation.

How about in case of a fluid with mass that eventually fully transfers the pressure energy on both pistons. Would in this case my previous quote hold:

"What I'm thinking is that the work done by the pressure difference on both pistons (after releasing them) would have to match ##F_1 \cdot d_1## that is initially applied on piston 1 while holding piston 2 in place."

EDIT: Nvm, I see you already answered that question, thanks!
 
Last edited:
  • #23
jbriggs444 said:
It is for this reason that I wanted to simplify the problem by slowly releasing piston 2 only against a resisting force F2F2F_2. We can, if you prefer, simplify by releasing both 1 and 2 against new resisting forces F1bF1bF_{\text{1b}} and F2bF2bF_{\text{2b}}

Apologies but another question came into mind when reading this. Since the new resisting forces and the mass are both proportional to the circular area of the pistons, would this mean that both pistons would have the same acceleration by the ##F_{1b}## and ##F_{2b}##?
 
  • #24
JohnnyGui said:
Apologies but another question came into mind when reading this. Since the new resisting forces and the mass are both proportional to the circular area of the pistons, would this mean that both pistons would have the same acceleration by the ##F_{1b}## and ##F_{2b}##?
No.

We are setting ##F_\text{1b}## to be equal to the current pressure difference multiplied by the area of piston 1 and maintaining it thus. We are setting ##F_\text{2b}## to be equal to the current pressure difference multiplied by the area of piston 2 and maintaining it thus. By itself, this would mean that both pistons are at rest and remain so.

Now we apply an additional force to one or both pistons. Say we brush them with feathers so that they are gently nudged to positions where the pressure difference is reduced to zero. We measure the work absorbed by ##F_\text{1b}## and ##F_\text{2b}## during this relaxation process. The work done by the feathers is, of course, negligible. It can be made as small as we please by using fluffier feathers.

The accelerations of the pistons during this process depend on the force of the feathers. The accelerations do not depend on the pressure differences or piston areas.

That is the scenario, properly set up.
 
  • #25
jbriggs444 said:
No.

We are setting ##F_\text{1b}## to be equal to the current pressure difference multiplied by the area of piston 1 and maintaining it thus. We are setting ##F_\text{2b}## to be equal to the current pressure difference multiplied by the area of piston 2 and maintaining it thus. By itself, this would mean that both pistons are at rest and remain so.

Now we apply an additional force to one or both pistons. Say we brush them with feathers so that they are gently nudged to positions where the pressure difference is reduced to zero. We measure the work absorbed by ##F_\text{1b}## and ##F_\text{2b}## during this relaxation process. The work done by the feathers is, of course, negligible. It can be made as small as we please by using fluffier feathers.

The accelerations of the pistons during this process depend on the force of the feathers. The accelerations do not depend on the pressure differences or piston areas.

That is the scenario, properly set up.

Oh my bad, for some reason I was thinking your ##F_{1b}## and ##F_{2b}## were the pressure forces on the pistons when the pressure difference arose.

What I had in mind is when there's a pressure difference while holding onto both pistons, and then releasing them both at the same time and let them relax to the equilibrium positions without applying any resisting forces.
In that case would you still say that they'd have different initial accelerations?
 
  • #26
JohnnyGui said:
Oh my bad, for some reason I was thinking your ##F_{1b}## and ##F_{2b}## were the pressure forces on the pistons when the pressure difference arose.

What I had in mind is when there's a pressure difference while holding onto both pistons, and then releasing them both at the same time and let them relax to the equilibrium positions without applying any resisting forces.
In that case would you still say that they'd have different initial accelerations?
Yes. Certainly so. Depending on piston mass, piston area and pressure difference. In the absence of friction, one could come up with a formula for the resulting simple harmonic motion.

In the presence of [linear] friction, the differential equation is somewhat more difficult but typically resolves to the product of a decaying exponential and a sine wave. In this case kinetic/pressure energy is drained into thermal energy, of course.
 
Last edited:
  • Like
Likes Lnewqban
  • #27
jbriggs444 said:
Yes. Certainly so. Depending on piston mass, piston area and pressure difference.

Could you please tell me what I'm donig wrong here?

In the case of a pressure difference while holding the pistons and then releasing them to relax (without applying any additional forces) the pressure force on piston ##1## would be ##F_1 = (P_2 - P_1) \cdot \pi r^2_1 = m_1 \cdot a_1##. The pressure force on piston ##2## would be ##F_2 = (P_2 - P_1) \cdot \pi r^2_2 = m_2 \cdot a_2##.

I can see from this that ##F_2 = F_1 \cdot \frac{r^2_2}{r^2_1}## and that ##m_2 = m_1 \cdot \frac{r^2_2}{r^2_1}##. Substituting ##F_2## and ##m_2## with these for the force on piston ##2##:
$$F_1 \cdot \frac{r^2_2}{r^2_1} = m_1 \cdot \frac{r^2_2}{r^2_1} \cdot a_2$$
$$= F_1 = m_1 \cdot a_2$$
Therefore, I see that ##a_1 = a_2##.
 
  • #28
JohnnyGui said:
Could you please tell me what I'm donig wrong here?

In the case of a pressure difference while holding the pistons and then releasing them to relax (without applying any additional forces) the pressure force on piston ##1## would be ##F_1 = (P_2 - P_1) \cdot \pi r^2_1 = m_1 \cdot a_1##. The pressure force on piston ##2## would be ##F_2 = (P_2 - P_1) \cdot \pi r^2_2 = m_2 \cdot a_2##.

I can see from this that ##F_2 = F_1 \cdot \frac{r^2_2}{r^2_1}## and that ##m_2 = m_1 \cdot \frac{r^2_2}{r^2_1}##. Substituting ##F_2## and ##m_2## with these for the force on piston ##2##:
$$F_1 \cdot \frac{r^2_2}{r^2_1} = m_1 \cdot \frac{r^2_2}{r^2_1} \cdot a_2$$
$$= F_1 = m_1 \cdot a_2$$
Therefore, I see that ##a_1 = a_2##.
As I understand this, you are assuming that the pistons are both of the same thickness and that the material is of the same density. So the piston masses are each directly proportional to area and each with the same constant of proportionality.

Yes, under this assumption, the accelerations of the two pistons are identical to each other.

So far so good. Why do you suspect this to be wrong?
 
  • Like
Likes Lnewqban
  • #29
jbriggs444 said:
As I understand this, you are assuming that the pistons are both of the same thickness and that the material is of the same density. So the piston masses are each directly proportional to area and each with the same constant of proportionality.

Yes, under this assumption, the accelerations of the two pistons are identical to each other.

So far so good. Why do you suspect this to be wrong?

Because you said in post #26 that the initial accelerations would be different in this scenario. Perhaps I have missed your point but I'm not sure how.
 
  • #30
JohnnyGui said:
Because you said in post #26 that the initial accelerations would be different in this scenario. Perhaps I have missed your point but I'm not sure how.
That was in the scenario where we were resisting the relaxation, not the scenario where the relaxation is free.
Oops. On review, that I was not sure we were dealing with same density, same thickness at that point.
 
  • #31
jbriggs444 said:
That was in the scenario where we were resisting the relaxation, not the scenario where the relaxation is free.
Oops. On review, that I was not sure we were dealing with same density, same thickness at that point.

Ah ok. You said furthermore that it will show an oscillation when there's no friction.

In the presence of friction, would this mean that the pistons would stop at positions where the pressure difference is zero?
 
  • #32
JohnnyGui said:
Ah ok. You said furthermore that it will show an oscillation when there's no friction.

In the presence of friction, would this mean that the pistons would stop at positions where the pressure difference is zero?
Yes. At least in the case of a friction that is proportional to velocity (e.g. viscosity in the fluid).

If we are talking about a fixed retarding frictional force (e.g. piston rings against cylinder walls) then the stop positions might have a small non-zero pressure difference compensated for by the small non-zero static friction.
 
  • Like
Likes Lnewqban
  • #33
jbriggs444 said:
Yes.

Assuming both pistons would swipe the same volume to reach equlibrium, does this mean that piston 2 would move a distance that is a factor ##\frac{r^2_2}{r^2_1}## smaller than piston 1, even though they have the same initial acceleration? (again, no resisting forces added).
 
  • #34
JohnnyGui said:
Assuming both pistons would swipe the same volume to reach equlibrium, does this mean that piston 2 would move a distance that is a factor ##\frac{r^2_2}{r^2_1}## smaller than piston 1, even though they have the same initial acceleration? (again, no resisting forces added).
I thought that we had decided that pistons with equal thickness and equal density would experience equal accelerations at all times subsequent to release. Would they not then sweep out equal distances and not equal volumes?

I will assume here that any friction is from fluid viscosity and not from the pistons on walls.
 
  • Like
Likes Lnewqban
  • #35
jbriggs444 said:
I thought that we had decided that pistons with equal thickness and equal density would experience equal accelerations at all times subsequent to release. Would they not then sweep out equal distances and not equal volumes?

I will assume here that any friction is from fluid viscosity and not from the pistons on walls.

Ah, my bad. I was taking piston 2 's larger inertia to decelerate into account but I forgot that its larger area of force at pressure equilibrium would also compensate for that.

One last question; if friction from the pistons on the walls are taken into account, would the distances swept out by the pistons differ? If so, does this mean that pressure equilibrium is not reached?
 

Similar threads

Replies
2
Views
1K
Replies
30
Views
2K
Replies
3
Views
3K
Replies
5
Views
1K
Replies
4
Views
803
  • Mechanics
Replies
8
Views
22K
Replies
5
Views
962
Replies
3
Views
966
Replies
5
Views
874
Back
Top