Thermodynamics: Gas Expansion with Piston Friction

[Chestermiller]

A friend of mine and I have been discussing how to apply the first law of thermodynamics to analyze the quasi static expansion of an ideal gas in a cylinder featuring a piston having both mass and friction (with the cylinder). We have identified two different systems that can be used in the analysis:

1. gas alone as system, with piston as part of the surroundings
2. gas and piston (and cylinder) as system

We wanted to see how the analysis using the first law of thermodynamics would play out in each of these approaches, knowing that, ultimately, the answer would have to come out the same in each case. So here is my specification of the problem:

There is a vertical insulated cylinder cross sectional area A containing an ideal gas. There is a piston of mass m situated above the gas. The piston is insulated at the top and has negligible heat capacity so that any frictional heat generated by piston contact with the cylinder is ultimately transferred to the gas (rather than the surroundings). The frictional force F is always constant and present both initially and finally. This can be achieved if we say that the coefficient of static friction is equal to the coefficient of kinetic friction, and that, in the initial and final states, the piston is at the verge of slipping (This is an idealization which does not detract from what we are trying to achieve in our analysis). We have a force f applied downward on the piston which decreases gradually (quasi statically) from $f_i$ in the initial state of the system to $f_f$ in the final state of the system; this is how we bring about the desired volume increase. There are n moles of ideal gas in the cylinder, initially at temperature $T_i$, and initially at mechanical equilibrium with the mass of the piston, the frictional force F, and the outside downward force $f_i$.

I am now going to stop and allow my friend to make comments and suggestions about the problem description. Is this basically what we had in mind? (Others are invited to participate).

Chet

 

[Delta2]

You are saying that in the initial and final state the piston is at the verge of slipping. Doesn't that imply that the net force (and hence pressure) on the piston is the same in the initial and final states?

EDIT:NVM I now noticed about the external force F acting on the piston. Is the external force present in the initial and final states or is it present only during the movement of the piston?

 

[Chestermiller]

You are saying that in the initial and final state the piston is at the verge of slipping. Doesn't that imply that the net force (and hence pressure) on the piston is the same in the initial and final states?

EDIT:NVM I now noticed about the external force F acting on the piston. Is the external force present in the initial and final states or is it present only during the movement of the piston?

I think I made that clear in the problem statement.

This problem is not about how the friction is handled. The focus of the problem is on the thermodynamics of the system. The friction force is as I say it is: F in the initial state, F during the gas expansion, and F in the final state.

 

[CWatters]

Perhaps I misunderstand the question but consider..

A Champaign cork is held in by a combination of things, tension in the wire cage, friction between cork and bottle and the weight of the cork.

If you shake it up, increasing the pressure inside so the cork starts moving slightly the gas is allowed to expand. As it expands it does work loosing energy to all of the things that are opposing the motion of the cork... eg The energy stored in stretching the wire, friction, gravity and inertia. Conservation of energy basically says this lot must add up to the energy lost by the gas.

The slight complication is you say..

any frictional heat generated by piston contact with the cylinder is ultimately transferred to the gas (rather than the surroundings).

That only slightly changes the sum you do when applying conservation of energy. I believe that ultimately it means you can ignore energy lost to friction because it appears on both sides of the equation.

 

[Chestermiller]

Perhaps I misunderstand the question but consider..

A Champaign cork is held in by a combination of things, tension in the wire cage, friction between cork and bottle and the weight of the cork.

If you shake it up, increasing the pressure inside so the cork starts moving slightly the gas is allowed to expand. As it expands it does work loosing energy to all of the things that are opposing the motion of the cork... eg The energy stored in stretching the wire, friction, gravity and inertia. Conservation of energy basically says this lot must add up to the energy lost by the gas.

The slight complication is you say..
That only slightly changes the sum you do when applying conservation of energy. I believe that ultimately it means you can ignore energy lost to friction because it appears on both sides of the equation.

This is incorrect. The energy lost to friction cannot be ignored in the problem as I specified it. I refuse to change the problem description. It is exactly the way I want. The objective is not to decide what the exact model for frictional force should be. The friction model is what I said it is. The objective is to evaluate, using the first law of thermodynamics, the effect of specifying two different "systems" in the problem, and seeing if the different analyses lead to the same end result (as they must). So please, everyone, take your focus off the friction force model.

 

[Robert Davidson]

Thank you inviting me to participate in the forum. Sorry for not jumping in sooner.

The subject for discussion is, as you said, to apply the first law in the analysis of a the quasi-static adiabatic expansion of an ideal gas in a cylinder having both mass and friction. I agree with your specification of the problem, although my original question during our prior discussion on the Physics Stack Exchange had to do with a non quasi-static, constant external pressure adiabatic expansion. I might still like to follow up on that after this analysis, if I still have issues.

With regard to the present problem description, I do have some questions on how to handle the piston mass when it is included as part of the system, as discussed later below.

As to the application of the first law, I think the system definition would only influence the manner in which energy transfers are accounted for, that is, how heat and/or work is determined, as defined by the first law equation. Of course the choice of system also determines where entropy is generated (internal or external to the system). In any case, I would like to use this a starting point for my comments on the initial analysis below.

Please refer to the diagrams abov, which are intended to help (at least me) visualize the two system definition approaches. The approach is similar to the one you introduced, regarding identification of system work based on system definition, in our previous discussion at the Physics Stack Exchange. (Note, however, I am using f for the imposed force on the outside face of the piston and F for the friction force in accordance with the current problem description, whereas they were reversed in our prior discussion).

PISTON MASS:

First you will note that I have left out the mass of the piston in this initial analysis, for the following reasons:

In the case of System 1, since the piston is part of the surroundings, one could simply add the force, mg, to the variable external force, f. This would simply modify the initial and final external force by a constant factor.

It is the case of System 2, where the piston is part of the system, where my questions arise. Here the gas is doing work in raising the piston (increasing its potential energy), but since the piston is part of the system the work does not cross the boundary, i.e. the work is not done on the surroundings. So technically this would not be W in the first law equation. On the other hand, there is no change in internal energy of the system since the decrease in the internal energy of the gas portion of the system equals the increase in potential energy of the mass portion of the system.

QUESTION: Given that the increase in the potential energy portion of the system equals the decrease kinetic energy portion (gas), can we simply ignore the piston and apply the first law equation as if it were massless? Your thoughts?

The following pertains to the case where we ignore the mass of the piston in the two systems.

SYSTEM 1:

Friction, and therefore entropy, is generated outside the system. Since the cylinder walls and piston are part of the surroundings, the friction work occurs in the surroundings. Since the cylinder walls are part of the surroundings, and we know that friction work will increase the temperature of the cylinder walls, I am showing heat transfer (quasi-static) from the walls to the gas. But I am also showing the friction force F as part of the surroundings.

QUESTION: Is this depiction correct? Should both the friction work and heat transfer be shown? I think so because in doing so the final conclusion about the change in internal energy will be consistent with that of System 2 which I think should be. Also, is it correct to interpret equation 7 to mean the work done by the system is equal to the reversible work, pdV, minus the irreversible (friction) work, dW F?

SYSTEM 2:

For system 2, there is no heat transfer to the surroundings, so dQ=0. Friction work is internal to the system. So the only system work is dW f and we have the same as system 1. The change in internal energy is the same as System 1 and the system work is, once again, the reversible work minus the friction work.

That is a far as I can take it now, without resolving the questions asked. Any help or suggestions will be appreciated.

 

[Robert Davidson]

After my post and upon further thinking I believe I can answer some of my questions. This post is therefore a revision to my original post. (Not sure how to revise a post being new at the Forum). See revised diagrams. Regarding the following questions concerning System 1:

Is this depiction correct? Should both the friction work and heat transfer be shown? I think so because in doing so the final conclusion about the change in internal energy will be consistent with that of System 2 which I think should be. Also, is it correct to interpret equation 7 to mean the work done by the system is equal to the reversible work, pdV, minus the irreversible (friction) work, dW F?

I believe the depiction is incorrect. I can show the effect of friction as either negative work transfer (equivalent of gas compression) or as equivalent heat transfer but not both. Since the cylinder walls (and friction work) are outside the system (gas only) I am able to show it as heat transfer. See the revised diagram of System 1.

Regarding the work done I think the interpretation is correct for System2. In System 1 you can view the process as reversible since the friction and thus entropy is generated is outside the system. But in System 2, the friction occurs within the system. So for System2 only the work is the reversible work minus the irreversible (friction) work, as shown in the updated depiction of System 2.

Your thoughts?

 

[Chestermiller]

After my post and upon further thinking I believe I can answer some of my questions. This post is therefore a revision to my original post. (Not sure how to revise a post being new at the Forum). See revised diagrams. Regarding the following questions concerning System 1:

Is this depiction correct? Should both the friction work and heat transfer be shown? I think so because in doing so the final conclusion about the change in internal energy will be consistent with that of System 2 which I think should be. Also, is it correct to interpret equation 7 to mean the work done by the system is equal to the reversible work, pdV, minus the irreversible (friction) work, dW F?

I believe the depiction is incorrect. I can show the effect of friction as either negative work transfer (equivalent of gas compression) or as equivalent heat transfer but not both. Since the cylinder walls (and friction work) are outside the system (gas only) I am able to show it as heat transfer. See the revised diagram of System 1.

Regarding the work done I think the interpretation is correct for System2. In System 1 you can view the process as reversible since the friction and thus entropy is generated is outside the system. But in System 2, the friction occurs within the system. So for System2 only the work is the reversible work minus the irreversible (friction) work, as shown in the updated depiction of System 2.

Your thoughts?

View attachment 236873

Hi Bob.

First of all, WELCOME TO PHYSICS FORUMS!

Secondly, if you want to edit a previous post, go to the bottom of the page and hit the edit button. This will reopen the editing window. Afterwards, you can hit the SAVE EDITS button. If you want to reply to a post in sections, you hit the reply button at the bottom of the post.

I had some trouble following the details of your analysis, so I am going to present my own short development. I think this will answer most of your questions.

As in your analysis, everything starts with the FORCE BALANCE ON THE PISTON:
$$PA=mg+F+f$$or equivalently
$$P=\frac{mg}{A}+\frac{F}{A}+\frac{f}{A}$$
And, if there is a differential change in volume dV, we have
$$PdV=\frac{mg}{A}dV+\frac{F}{A}dV+\frac{f}{A}dV\tag{1}$$

FIRST LAW OF THERMODYNAMICS APPLIED TO SYSTEM 1, THE GAS:
$$dU=dQ-PdV\tag{2}$$
where dQ is the differential heat transferred from the piston to the gas.

FIRST LAW OF THERMODYNAMICS APPLIED TO SYSTEM 2, THE GAS AND PISTON:
In this case, we need to use the more general form of the first law which includes the change in potential energy of whatever is present with the boundaries of the system (in this case, the piston).
$$dE=dU+d(PE)=dU+\frac{mg}{A}dV=-\frac{f}{A}dV\tag{3}$$where $\frac{F}{A}dV$ is the work done by system 2 on its surroundings. Note that, in this case, there is no heat transfer because the combined system is adiabatic. If we substitute Eqns. 1 into this relationship, we obtain:
$$dU=\frac{F}{A}dV-PdV\tag{4}$$From Eqns. 2 and 4, it follows that the heat transferred from the piston to the gas dQ is given by:
$$dQ=\frac{F}{A}dV\tag{5}$$

FIRST LAW OF THERMODYNAMICS APPLIED TO THE PISTON ALONE:

We can obtain the equation for the first law applied to the piston alone by subtracting Eqn. 2 from Eqn. 3 to yield:
$$\frac{mg}{A}dV=-dQ-\left(\frac{f}{A}-P\right)dV\tag{6}$$
The left hand side is the change in potential energy of the piston. Again, dQ is the heat transferred from the piston to the gas. The work done by the piston on its surroundings is $\left(\frac{f}{A}-P\right)dV$. Note that the piston does not do any work on the cylinder, since the displacement of the cylinder is zero.

FURTHER DISCUSSION
For an ideal gas, Eqn. 4 becomes:
$$nC_vdT=\frac{F}{A}dV-\frac{nRT}{V}dV\tag{7}$$This equation does not have a simple analytic solution that I am aware of, and it would probably have to be integrated numerically. In any event, from this equation, it follows from this equation that the change in entropy of the gas is obtained from:
$$dS=\frac{F}{A}\frac{dV}{T}$$

 

[Robert Davidson]

Thanks Chester for your analysis and yes, it did help to answer some of my questions. I do have a question regarding your equation 1:

$$PdV=\frac{mg}{A}dV+\frac{F}{A}dV+\frac{f}{A}dV$$

Shouldn’t the friction work be negative? In which case, shouldn’t

$$PdV = \frac{mg}{A}dV-\frac{F}{A}dV+\frac{f}{A}dV$$

Then

$$PdV_{without friction}>PdV_{with friction}$$

Which would give us

$$PdV_{reversible}>PdV_{irreversible}$$

 

[Chestermiller]

Thanks Chester for your analysis and yes, it did help to answer some of my questions. I do have a question regarding your equation 1:

$$PdV=\frac{mg}{A}dV+\frac{F}{A}dV+\frac{f}{A}dV$$

Shouldn’t the friction work be negative? In which case, shouldn’t

$$PdV = \frac{mg}{A}dV-\frac{F}{A}dV+\frac{f}{A}dV$$

The force balance from the free body diagram on the piston makes pretty clear that the sign on the friction term is correct. And, if you would rather I had written Eqn. 4 with a negative sign on the friction term, that would be consistent with moving the three terms on the right hand side of Eqn. 4 to the left-hand side, to yield:
$$PdV-\frac{mg}{A}dV-\frac{F}{A}dV-\frac{f}{A}dV=0$$For the pressure term, the displacement is in the same direction as the force, and, for the other three terms, the displacement is in the opposite direction of the force (as you pointed out for the friction term).

There is more that I can discuss on this problem with regard to the friction work and friction heat, with specific focus on the interface region between the cylinder and the piston. This is very tricky because velocity is discontinuous at the interface and the rate of doing work involves velocity. If you want to hear more about this, I would be glad to pursue it.

 

[Robert Davidson]

I agree that the force balance would seem to indicate the friction work should be positive. I’m sure you’re right about the sign of the friction work, but consider the following.

We know it is the nature of friction that work against friction is negative. And we already acknowledged that for system 1 friction work is equivalent to heat transfer to the system. Given friction work increases internal energy and that

$$dU=-dW$$

Then shouldn’t the friction work component of dW be negative?

Perhaps we need to revisit the force balance equation on the piston, keeping in mind that the externally applied forces always exist but the friction force only exists when opposing motion.

For system 1, during the expansion process, the externally applied force (mg + f) is gradually reduced so that there is always a small net force on the piston in the same direction as the displacement. The system work against mg + f positive and the contribution to the change in internal energy is negative. But during this time the sliding friction force is constant in the opposite direction of the displacement. The friction work is negative and the contribution to the internal energy is positive.

Where am I going wrong?

 

[Chestermiller]

I agree that the force balance would seem to indicate the friction work should be positive. I’m sure you’re right about the sign of the friction work, but consider the following.

We know it is the nature of friction that work against friction is negative. And we already acknowledged that for system 1 friction work is equivalent to heat transfer to the system. Given friction work increases internal energy and that

$$dU=-dW$$

Then shouldn’t the friction work component of dW be negative?

Perhaps we need to revisit the force balance equation on the piston, keeping in mind that the externally applied forces always exist but the friction force only exists when opposing motion.

1. If you feel that the force balance obtained from the free body diagram is an issue, please show your free body diagram for comparison. You can UPLOAD it using the UPLOAD button. Also, regarding friction, for simplicity, in my problem statement, I was very careful to specify "The frictional force F is always constant and present both initially and finally. This can be achieved if we say that the coefficient of static friction is equal to the coefficient of kinetic friction, and that, in the initial and final states, the piston is at the verge of slipping (This is an idealization which does not detract from what we are trying to achieve in our analysis)."

2. When talking about "work," it is very important to specify what is doing work on what. Please specify this in your wording.

 

[Robert Davidson]

Thanks Chester. I will deliberate on this further.

There must be some way to demonstrate that, for system 1 and/or system 2, the work done without friction (reversible adiabatic process) will be greater than the work done with friction (irreversible adiabatic process).

This goes to the heart of what I would like to see demonstrated by our example.

Be back with you later.

 

[Chestermiller]

Thanks Chester. I will deliberate on this further.

There must be some way to demonstrate that, for system 1 and/or system 2, the work done without friction (reversible adiabatic process) will be greater than the work done with friction (irreversible adiabatic process).

Before we start dealing with that, let's first make sure that we reach consensus on the actual analysis. However, you are aware that, if you have an adiabatic irreversible process on a closed system, it is impossible to define an adiabatic reversible process through the same two end states, correct?

I'm going to go back and add to my analysis, providing much more detail, to help us reach consensus. This will include an idealized model of the gap/interface between the cylinder and piston to help us better quantify what is happening there with regard to the friction.

Chet

 

[Chestermiller]

IDEALIZED MODEL OF FRICTIONAL INTERFACE/"GAP" BETWEEN PISTON AND CYLINDER

I'm introducing this idealized model of the interface/"gap" between the piston and cylinder so that I can more easily describe the application of the first law of thermodynamics to the piston, including its interaction with the cylinder (via the interface/"gap") and its mechanical and thermal interaction with the gas.

In the figure, the cylinder is exerting a downward force F at one side of the interface/gap, while the piston is exerting an upward force F at the other side of the interface/gap. As a result of the work being done at the interface/gap, heat Q is being transferred from the gap to the piston. On the cylinder side of the interface/gap, no heat is being transferred because the cylinder is assumed to be non-conductive.

Let's take a look at the work. Since the cylinder is stationary, the work done by the cylinder on the interface and by the interface on the cylinder are zero.

At the piston side of the interface/gap, the piston is moving upward with a displacement dV/A, where dV is the change in volume of the gas and A is the area of the piston. So, the work done by the piston on the interface/gap is $+\frac{F}{A}dV$, and the work done by the interface/gap on the piston is $-\frac{F}{A}dV$; the latter is the work done by the interface/gap on its surroundings. If we apply the first law of thermodynamics to the interface/gap, we have:
$$dU=0=-dQ+\frac{F}{A}dV$$or equivalently $$dQ=\frac{F}{A}dV\tag{1}$$. The change in internal energy of the interface/gap is zero because, in the limit, the interface/gap has no mass. According to this equation, the heat generated due to the friction at the interface is transferred to the piston.

In summary, all this idealized model of the interface really shows us is what we already knew:

1. If the gas is expanding so that the piston moves upward, the interface exerts a downward friction (reaction) force on the piston
2. the heat generated by friction at the interface is transferred to the piston.

Let's next turn attention to how this analysis of the frictional interface/gap plays out in terms of applying the first law of thermodynamics specifically to the piston.

APPLICATION OF FIRST LAW OF THERMODYNAMICS TO PISTON
The figure below contains the information necessary to apply the first law of thermodynamics to the piston.

First let's consider the work.

For the "external" force f, dV is in the opposite direction of f. So, work done by the surroundings on the piston = $-\frac{f}{A}dV$ and work done by the piston on the surroundings = $+\frac{f}{A}dV$.

For the gas force PA, dV is in the same direction of PA. So, work done by surroundings (the gas) on piston = $PdV$ and work done by the piston on the surroundings (the gas) = $-PdV$.

For the frictional force F from the interface/gap, dV is in the opposite direction of F. So, frictional work done by the surroundings on the piston = $-\frac{F}{A}dV$ and frictional work done by the piston on the surroundings = $+\frac{F}{A}dV$

In terms of heat transfer, the piston receives dQ from the frictional interface/gap, and delivers dQ to the gas below.

Based on all of this, the application of the first law of thermodynamics to the piston yields the following:
$$dE=dU+\frac{mg}{A}dV=\frac{mg}{A}dV=dQ-dQ-\frac{f}{A}dV+PdV-\frac{F}{A}dV$$or, equivalently:
$$\frac{mg}{A}dV=-\frac{f}{A}dV+PdV-\frac{F}{A}dV$$where the internal energy change of the piston is assumed to be negligible. This is just the same as Eqn. 1 of post #8. If we combine this with Eqn. 1 of the present post, we obtain:
$$\frac{mg}{A}dV=-dQ+PdV-\frac{f}{A}dV$$This is the same as Eqn. 6 of post #8.

 

[Chestermiller]

So, based on your analysis,

$$\frac{mg}{A}dV+\frac{f}{A}dV=PdV-\frac{F}{A}dV$$

Then is it not correct to say:

$$dW_{surroundings}=\frac{mg}{A}dV+\frac{f}{A}dV$$

$$dW_{gas}=PdV$$

$$dW_{friction}=-\frac{F}{A}dV$$

$$dW_{surroundings}=dW_{gas}-dW_{friction}$$

From which we can conclude the work done on the surroundings is greater without friction than with friction?

What are you specifically referring to when you write these equations for work, (a) the combination of piston, cylinder, and gas or (b) the combination of piston and gas, (c) the piston alone, or (d) the gas alone?

The work done by the combination of piston, gas, and cylinder on its surroundings is just
$$dW=\frac{f}{A}dV$$
This does not include the change in potential energy of the piston which is included in $dE_{piston}=\frac{mg}{A}dV$

The work done by the gas on its surroundings (the piston) is $$dW=PdV$$

The work done by the piston on its surroundings is $$dW=\frac{f}{A}dV+\frac{F}{A}dV-PdV$$Here again, this does not include the change in potential energy of the piston which is included in $dE_{piston}=\frac{mg}{A}dV$

 

[Robert Davidson]

When trying to make a sign correction in my post, I inadvertently deleted it. Sorry about that.

In any event, to answer your question, the work I am interested is work done by the combination of the piston, cylinder, and gas (as the system), since I want friction to be included within the system. Under these conditions (and I will ignore the change in potential energy in all of the following as you did) the work done on the surroundings is, as you indicated

$$dW_{surroundings}=\frac{f}{A}dV$$

Which I believe should equal the sum of work done by the gas and the friction work, or

$$\frac{f}{A}dV=PdV + \frac{F}{A}dV$$Since the friction force is in the opposite direction of dV, F in this equation will be negative. Which means the friction work will be negative.

Based on the above, the work done by the system on the surroundings will be a maximum when the friction work is or approaches zero (the process is reversible). This would then be consistent with the principle

$$W_{reversible}>W_{irreversible}$$

Incidentally, I came across an analysis titled “Irreversible Work and Internal Friction” on the website chem.libretexts.org. The system they analyzed is shown in the diagram below. The friction is between the piston rod R and the bushing B and is included as part of the system along with the gas. They concluded

$$w=-\int_{V1}^{V2}pdV-\int_{V1}^{V2}\frac{F}{A}dV$$

Where w on the left is the work done on the surroundings, the first term on the right is the work of expanding or compressing the gas and the second term is the frictional work where the friction force will be negative and thus the friction work will be positive. (Note that the negative signs on the work terms is due to their use the chemists version of the first law, dU = dQ + dW.).

They then state:

“As a result, the irreversible work of expansion is less negative than the reversible work for the same volume increase, and the irreversible work of compression is more positive than the reversible work for the same volume decrease. These effects of piston velocity on the work are consistent with the minimal work principle.”

The different signs not withstanding, I believe this is consistent with my above conclusion.

They also make the following very interesting statement:

Consider the situation when the piston moves very slowly in one direction or the other. In the limit of infinite slowness the friction force and friction work vanish, and the process is reversible with expansion work given by

$$-\int pdV$$

It seems to me this is tantamount to saying a quasi-static process with friction is reversible, or, alternatively, there is no friction force or friction work if the process is quasi-static. This is contrary to the premise that a quasi-static process with friction is irreversible.

 

[Robert Davidson]

Forgot to add the diagram below

 

[Chestermiller]

When trying to make a sign correction in my post, I inadvertently deleted it. Sorry about that.

In any event, to answer your question, the work I am interested is work done by the combination of the piston, cylinder, and gas (as the system), since I want friction to be included within the system. Under these conditions (and I will ignore the change in potential energy in all of the following as you did) the work done on the surroundings is, as you indicated

$$dW_{surroundings}=\frac{f}{A}dV$$

Which I believe should equal the sum of work done by the gas and the friction work, or

$$\frac{f}{A}dV=PdV + \frac{F}{A}dV$$Since the friction force is in the opposite direction of dV, F in this equation will be negative. Which means the friction work will be negative.

It is very unconventional to treat the frictional force in this way. Since frictional force is a vector quantity, it is typically treated as a positive magnitude times an appropriately oriented unit vector (in this case, a unit vector in the negative z direction). It really goes against my comfort to say that the friction force is negative. The net result of all this is that your parameter F is the negative of my parameter F. Any chance you might consider going back and editing your posts so that our F's are the same?

Based on the above, the work done by the system on the surroundings will be a maximum when the friction work is or approaches zero (the process is reversible). This would then be consistent with the principle

$$W_{reversible}>W_{irreversible}$$

I really don't like this interpretation. Even for the same volume change, the temperature and pressure histories are going to be different, so the reversible work the gas does with friction is going to be different from the reversible work it does without friction. It doesn't make sense to me to compare reversible and irreversible work in this way. Usually, one would compare them for processes between the same to end points for the same identical system. I' m thinking of going back and integrating the following equation numerically (after reducing it to dimensionless form) to enable us to make more rational comparisons between cases with and without friction:
$$nC_vdT=\frac{F}{A}dV-\frac{nRT}{V}dV$$
Meanwhile, based on the equations we are discussing here , the only interpretation I'm willing to make is that the work that gas has to do to overcome the frictional force and the external force is greater than the work the combined system has to do to overcome the external force alone.

Incidentally, I came across an analysis titled “Irreversible Work and Internal Friction” on the website chem.libretexts.org. The system they analyzed is shown in the diagram below. The friction is between the piston rod R and the bushing B and is included as part of the system along with the gas. They concluded

$$w=-\int_{V1}^{V2}pdV-\int_{V1}^{V2}\frac{F}{A}dV$$

Where w on the left is the work done on the surroundings, the first term on the right is the work of expanding or compressing the gas and the second term is the frictional work where the friction force will be negative and thus the friction work will be positive. (Note that the negative signs on the work terms is due to their use the chemists version of the first law, dU = dQ + dW.).

They then state:

“As a result, the irreversible work of expansion is less negative than the reversible work for the same volume increase, and the irreversible work of compression is more positive than the reversible work for the same volume decrease. These effects of piston velocity on the work are consistent with the minimal work principle.”

The different signs not withstanding, I believe this is consistent with my above conclusion.

They also make the following very interesting statement:

Consider the situation when the piston moves very slowly in one direction or the other. In the limit of infinite slowness the friction force and friction work vanish, and the process is reversible with expansion work given by

$$-\int pdV$$

It seems to me this is tantamount to saying a quasi-static process with friction is reversible, or, alternatively, there is no friction force or friction work if the process is quasi-static. This is contrary to the premise that a quasi-static process with friction is irreversible.

Bob, I looked over what you have written here, but was unable to follow, and didn't understand the comment that, when the process is quasi-static, the process is reversible. Maybe you can provide the exact statement of the problem.

 

[Chestermiller]

I've started to do calculations with the model I discussed in my previous post. The dimensionless parameters are as follows:
$$\bar{V}=\frac{V}{V_0}$$
$$\bar{T}=\frac{T}{T_0}$$
$$\bar{F}=\frac{F}{AP_0}$$
$$\bar{W_f}=\frac{\int_{V_0}^V{\frac{f}{A}dV}}{nC_vT_0}$$
$$\bar{W_P}=\frac{\int_{V_0}^V{PdV}}{nC_vT_0}$$where the subscript 0 refers to the initial state of the gas. The calculations are being performed for the case of $\bar{F}=0.5$, meaning that the frictional force and the external force initially contribute equally to supporting the initial pressure $P_0$, or equivalently, the frictional force is always equal to half the initial pressure force. The calculations are also performed for $\gamma=\frac{C_p}{C_v}=1.4$. So far, I've calculated the dimensionless temperature as a function of the dimensionless volume. This is shown in the figure below:

Note the significant difference between the cases with and without friction. Note also that, with friction, the temperature first decreases with increasing volume, but then passes through a minimum and then starts increasing. The minimum corresponds to the point where the external force f has dropped to zero, and friction is supporting the entire pressure of the gas. Further increases in volume correspond to actually applying tension (rather than compression) to the upper surface of the piston (i.e., f < 0). In this situation, the gas pressure has dropped to the point where it is insufficient alone to overcome the friction.

The figure below shows the calculated work for (a) the gas acting on the piston (green line), (b) the external work done by the combined system (blue line), and (c) the gas and the external work for the case of no friction (where both curves coincide).

The difference between the green curve and the blue curve is the work done to overcome friction. Note that the system does much less work on the surroundings than the gas does on the piston. Note also that the work done by the combined system increases for small volume increases, passes through a maximum, and then begins to decrease with further volume increases. The maximum work corresponds to the point where one needs to begin pulling on the piston from the outside to overcome friction; the gas no longer has enough pressure to do this on its own. The reversible case with no friction is only provided for comparison, and is not really directly related to the frictional case.

Chet

 

[Chestermiller]

Incidentally, I came across an analysis titled “Irreversible Work and Internal Friction” on the website chem.libretexts.org. The system they analyzed is shown in the diagram below. The friction is between the piston rod R and the bushing B and is included as part of the system along with the gas. They concluded

$$w=-\int_{V1}^{V2}pdV-\int_{V1}^{V2}\frac{F}{A}dV$$

Where w on the left is the work done on the surroundings, the first term on the right is the work of expanding or compressing the gas and the second term is the frictional work where the friction force will be negative and thus the friction work will be positive. (Note that the negative signs on the work terms is due to their use the chemists version of the first law, dU = dQ + dW.).

They then state:

“As a result, the irreversible work of expansion is less negative than the reversible work for the same volume increase, and the irreversible work of compression is more positive than the reversible work for the same volume decrease. These effects of piston velocity on the work are consistent with the minimal work principle.”

The different signs not withstanding, I believe this is consistent with my above conclusion.

They also make the following very interesting statement:

Consider the situation when the piston moves very slowly in one direction or the other. In the limit of infinite slowness the friction force and friction work vanish, and the process is reversible with expansion work given by

$$-\int pdV$$

It seems to me this is tantamount to saying a quasi-static process with friction is reversible, or, alternatively, there is no friction force or friction work if the process is quasi-static. This is contrary to the premise that a quasi-static process with friction is irreversible.

I'm in disagreement with almost all aspects of this analysis. Let's take a look at what the force balance on the piston looks like when the gas is expanding and when it is being compressed. If $\mathbf{i_x}$ represents a unit vector in the positive x direction, then for expansion, the friction force on the piston shaft is $F(-\mathbf{i_x})$, and, if the gas is being compressed, the friction force on the piston shaft $F(+\mathbf{i_x})$, where, in each case, F is the (positive) magnitude of the friction force. So, for expansion, the vector force balance on the piston is $$P_{ext}A(-\mathbf{i_x})+F(-\mathbf{i_x})+pA(+\mathbf{i_x})=\mathbf{0}$$
On the other hand, for compression, the vector force balance on the piston is $$P_{ext}A(-\mathbf{i_x})+F(+\mathbf{i_x})+pA(+\mathbf{i_x})=\mathbf{0}$$Note that the force balance equation, and thus the equations for the work are different, depending on whether there is expansion or compression. The component of each of these equations in the x-direction is $$P_{ext}A=pA-F\tag{expansion}$$$$P_{ext}A=pA+F\tag{compression}$$In both these equations, the magnitude F of the friction force is positive. If follows from these equations that the work done by the system on the surroundings is $$W=\int_{V_1}^{V_2}{P_{ext}dV}=+\int_{V_1}^{V_2}{\left(p-\frac{F}{A}\right)dV}\tag{expansion}$$and $$W=\int_{V_1}^{V_2}{P_{ext}dV}=+\int_{V_1}^{V_2}{\left(p+\frac{F}{A}\right)dV}\tag{compression}$$Based on these equations, in each case, $$\Delta U=-W\tag{expansion and compression}$$

I assert that all these equations that I have written have the correct signs.

They then state:

“As a result, the irreversible work of expansion is less negative than the reversible work for the same volume increase, and the irreversible work of compression is more positive than the reversible work for the same volume decrease. These effects of piston velocity on the work are consistent with the minimal work principle.”

The different signs not withstanding, I believe this is consistent with my above conclusion.

They also make the following very interesting statement:

Consider the situation when the piston moves very slowly in one direction or the other. In the limit of infinite slowness the friction force and friction work vanish, and the process is reversible with expansion work given by

$$-\int pdV$$

It seems to me this is tantamount to saying a quasi-static process with friction is reversible, or, alternatively, there is no friction force or friction work if the process is quasi-static. This is contrary to the premise that a quasi-static process with friction is irreversible.

None of this makes any sense to me. It seems to me that, in the first part, they are comparing apples and oranges. In the second part, clearly, no matter how slowly the piston is being moved, the friction term is present and the process is irreversible. Maybe, in the problem they are solving, they are dealing with viscous fluid between the bearing and shaft, so that the frictional force varies with the velocity of shaft advance.

I might also point out that there is another fundamental difference between our problem and this one. In our problem, the friction force and gas pressure force are in series, while in this problem, they are in parallel.

 

[Chestermiller]

Second Thoughts. Based on the results I presented in post #20, I think I can provide and apples-apples comparison/interpretation that will satisfy the intuition of my friend @Robert Davidson.

For the same change in volume, the work done by the gas in the overall adiabatic irreversible expansion involving piston friction will be greater than the work done by the gas in the adiabatic reversible expansion not involving piston friction; this is because of the higher temperatures of the gas in the frictional expansion, and the correspondingly higher pressures. However, the work done by the combined system on the external surroundings will be less with piston friction than without piston friction.

 

[Robert Davidson]

Chet,

Didn't forget about this. Just a lot to digest. Be back soon.

Bob

 

[Robert Davidson]

Hi Chet,

Thanks for proving by mathematical analysis what I was only able to express qualitatively. After first pass, I must admit I didn’t follow the post 20 analysis completely (I will study further). Clearly, you are at a higher level than I.

Regarding the following conclusion of your last post:

For the same change in volume…the work done by the combined system on the external surroundings will be less with piston friction than without piston friction.

Am I correct that this conclusion is based on the comparison of the red and blue curves of dimensionless work from post 20? I only ask because the red curve says “gas & combined system” whereas the word “gas” is missing in the caption for the blue curve. I am assuming, however, that both curves are for the combined system (our system 2) which includes the gas.

Regarding the following conclusion:

For the same change in volume, the work done by the gas in the overall adiabatic irreversible expansion involving piston friction will be greater than the work done by the gas in the adiabatic reversible expansion not involving piston friction; this is because of the higher temperatures of the gas in the frictional expansion, and the correspondingly higher pressures.

Can I interpret this as well to mean that only part of the work done by the gas is, for want of a better term, “productive”, whereas the other part (against friction) is returned to the gas, the end result being a smaller decrease in internal energy (higher final temperature and pressure)? Or, to put it another way, only part of the work done by the gas is done on the surroundings of the combination system.

In post 14 you asked me the following question:

However, you are aware that, if you have an adiabatic irreversible process on a closed system, it is impossible to define an adiabatic reversible process through the same two end states, correct?

The answer is yes, I was (and am) aware of that. In fact, that was the reason for my original example (that led us to the forum) to compare two constant pressure adiabatic expansions, one with and one without piston/cylinder friction, in which the question is which will result in a greater increase in height of the weighted piston (increase in volume of gas). The answer to that should clearly demonstrate whether more work is done on the weighted piston with or without friction, without the need to have the same initial and final end points (even taking into account that the piston with friction will get "stuck" when the net force on the piston equals the maximum static friction). Do you think it would be work while to pose this question in a new post? Or have we, as they say, beat this to death.

Again, thanks for your help. I feel I learned a lot.

Bob

 

[Chestermiller]

Hi Chet,

Thanks for proving by mathematical analysis what I was only able to express qualitatively. After first pass, I must admit I didn’t follow the post 20 analysis completely (I will study further). Clearly, you are at a higher level than I.

I didn't include all the analysis that led up to the results in post # 20. If you like, I can provide that. The graphs include only the final results of the analysis.

Regarding the following conclusion of your last post:

For the same change in volume…the work done by the combined system on the external surroundings will be less with piston friction than without piston friction.

Am I correct that this conclusion is based on the comparison of the red and blue curves of dimensionless work from post 20? I only ask because the red curve says “gas & combined system” whereas the word “gas” is missing in the caption for the blue curve. I am assuming, however, that both curves are for the combined system (our system 2) which includes the gas.

Correct.

Regarding the following conclusion:

For the same change in volume, the work done by the gas in the overall adiabatic irreversible expansion involving piston friction will be greater than the work done by the gas in the adiabatic reversible expansion not involving piston friction; this is because of the higher temperatures of the gas in the frictional expansion, and the correspondingly higher pressures.

Can I interpret this as well to mean that only part of the work done by the gas is, for want of a better term, “productive”, whereas the other part (against friction) is returned to the gas, the end result being a smaller decrease in internal energy (higher final temperature and pressure)? Or, to put it another way, only part of the work done by the gas is done on the surroundings of the combination system.

Yes. Exactly.

In post 14 you asked me the following question:

However, you are aware that, if you have an adiabatic irreversible process on a closed system, it is impossible to define an adiabatic reversible process through the same two end states, correct?

The answer is yes, I was (and am) aware of that. In fact, that was the reason for my original example (that led us to the forum) to compare two constant pressure adiabatic expansions, one with and one without piston/cylinder friction, in which the question is which will result in a greater increase in height of the weighted piston (increase in volume of gas). The answer to that should clearly demonstrate whether more work is done on the weighted piston with or without friction, without the need to have the same initial and final end points (even taking into account that the piston with friction will get "stuck" when the net force on the piston equals the maximum static friction). Do you think it would be work while to pose this question in a new post? Or have we, as they say, beat this to death.

Again, thanks for your help. I feel I learned a lot.

Bob

For the process to be quasi-static, we cannot let the piston rise on its own. So it would have to involve gradual removal of weights from the top of the piston. This would entail problems with specifying the initial state, because, in one case, friction would be helping to balance the gas pressure initially, while, in the other case, it would not be helping. So, in the non-frictional case, we would have to start with a heavier piston (or more weights). If you want to take a shot at specifying this problem, we can proceed in the present thread. Here, instead of matching the final volumes, we would be doing something like matching the final amounts of weight removed (I guess).

 

[Chestermiller]

I'm thinking of the following comparison:

System 1: massless piston with friction force F, and a pile of small masses totaling M sitting on top of the piston initially

System 2: frictionless piston having mass m, such that mg for this system is equal to F for system 1, and the same pile of small masses totaling M sitting on top of the piston initially

The initial states of these two systems will be the same, but System 2 will experience an adiabatic reversible expansion and System 1 will experience an adiabatic irreversible expansion when the same pile of small masses is removed quasi-statically from the top of each piston.

 

[Robert Davidson]

I didn't include all the analysis that led up to the results in post # 20. If you like, I can provide that. The graphs include only the final results of the analysis.

Yes, I would very much appreciate it if you could provide all the analysis.

On your last post:

For the process to be quasi-static, we cannot let the piston rise on its own…..

I was not thinking of a quasi-static process. What I had in mind was to compare the final volume for two irreversible constant pressure adiabatic expansions, one with sliding friction and one without.

 

[Chestermiller]

I didn't include all the analysis that led up to the results in post # 20. If you like, I can provide that. The graphs include only the final results of the analysis.

Yes, I would very much appreciate it if you could provide all the analysis.

I'll post this in a little while.

On your last post:

For the process to be quasi-static, we cannot let the piston rise on its own…..

I was not thinking of a quasi-static process. What I had in mind was to compare the final volume for two irreversible constant pressure adiabatic expansions, one with sliding friction and one without.

So you would consider the same comparison as in my previous post, except with sudden removal of mass M at time zero?

 

[Chestermiller]

The starting equations for the frictional analysis I alluded to in post #20 is:
$$dU=nC_vdT=\frac{F}{A}dV-\frac{nRT}{V}dV\tag{1}$$and $$dU=nC_vdT=-\frac{f}{A}dV\tag{2}$$Dividing Eqn. 1 by $nC_vT$ yields:
$$\frac{dT}{T}=\frac{FdV}{AnC_vT}-\frac{R}{C_v}\frac{dV}{V}=\left[\frac{FV}{AnC_vT}-\frac{R}{C_v}\right]\frac{dV}{V}\tag{3}$$
In Eqn. 3, $$\frac{FV}{AnC_vT}=\frac{FV_0}{nC_vAT_0}\frac{(V/V_0)}{(T/T_0)}=\frac{FR}{AC_v}\frac{V_0}{nRT_0}\frac{(V/V_0)}{(T/T_0)}$$
But, from the ideal gas law, $$\frac{nRT_0}{V_0}=P_0$$
Therefore, $$\frac{FV}{AnC_vT}=\frac{F}{P_0A}\frac{C_v}{R}\frac{(V/V_0)}{(T/T_0)}\tag{4}$$
Substituting Eqn. 4 into Eqn. 3 then yields:
$$\frac{dT}{T}=\frac{C_v}{R}\left[\frac{F}{P_0A}\frac{(V/V_0)}{(T/T_0)}-1\right]\frac{dV}{V}=(\gamma -1)\left[\frac{F}{P_0A}\frac{(V/V_0)}{(T/T_0)}-1\right]\frac{dV}{V}\tag{5}$$
We now make the following dimensionless substitutions:
$$\bar{T}=T/T_0$$
$$\bar{V}=V/V_0$$
This then yields:
$$\frac{d\ln{\bar{T}}}{d\ln{\bar{V}}}=-(\gamma -1)\left[1-\frac{F}{P_0A}e^{(ln{\bar{V}}-\ln{\bar{T}})}\right]\tag{6}$$
This is the differential equation I integrated to get the results in post #20.
To get the work terms, I did the following: From Eqn. 2, if I divide by $nC_vT_0$ and integrate, I obtain:
$$\frac{\left[\int_{V_0}^V{\frac{f}{A}dV}\right]}{nC_vT_0}=1-\bar{T}$$
Similarly, for the integral of the gas pressure, I get:
$$\frac{\left[\int_{V_0}^V{PdV}\right]}{nC_vT_0}=(\gamma-1)\frac{F}{P_0A}(\bar{V}-1)+(1-\bar{T})\tag{7}$$

 

[Robert Davidson]

Chet,

Thanks for the analysis.

Regarding the new example, I had in mind the following:

System 1: massless piston with friction between the piston and cylinder walls, and a pile of small masses sitting on top of the piston, such that the total downward force due to the masses plus the downward force due to atmospheric pressure equals the upward force due to the gas pressure. Or, the pressure upward exerted by the gas equals the pressure downward exerted by the weights plus atmospheric pressure.

System 2: Same as system 1 except there is no friction.

An equal amount of weight is abruptly removed from each system so that the external downward pressure exerted by the weights plus the atmosphere is ½ the original external downward pressure.

To solve: What will be the final volume of the two systems?

If we can agree on the problem statement, I will give it a try on my own, with the benefit of the analysis you did on our previous example, and then post it.

Bob

 

[Chestermiller]

Chet,

Thanks for the analysis.

Regarding the new example, I had in mind the following:

System 1: massless piston with friction between the piston and cylinder walls, and a pile of small masses sitting on top of the piston, such that the total downward force due to the masses plus the downward force due to atmospheric pressure equals the upward force due to the gas pressure. Or, the pressure upward exerted by the gas equals the pressure downward exerted by the weights.

System 2: Same as system 1 except there is no friction.

An equal amount of weight is abruptly removed from each system so that the external downward pressure exerted by the weights plus the atmosphere is ½ the original external downward pressure.

To solve: What will be the final volume of the two systems?

If we can agree on the problem statement, I will give it a try on my own, with the benefit of the analysis you did on our previous example, and then post it.

Bob

Bob,

Although I don't think that the two cases are really going to be comparable, I think each of them is interesting on its own. So please proceed and let's see what you come up with. (Please note that, in both these cases, the gas experiences a non-quasistatic deformation, and entropy is generated within the gas in each case).

Chet

 

[Chestermiller]

I've been able to analytically solve the differential equation (Eqn. 6) of post #29 for the dimensionless temperature of the gas as a function of the dimensionless volume ratio. The result, which agrees with the numerical solution I presented in post #20 is:
$$\bar{T}=\frac{(1-\xi)}{\bar{V}^{(\gamma-1)}}+\xi \bar{V}$$where $$\xi=\frac{(\gamma-1)}{\gamma}\frac{F}{P_0A}$$

 

[Robert Davidson]

Hi Chet,

Haven’t forgot this. So far only tackled the easy case (system 2). Here’s what I have so far.

Going to assume dealing with ideal gas (air) with the following initial conditions and properties:

$P_{i}=10atm$, $T_{i}=300K$, $n=1$, $C_{v}=0.2053\frac{l.atm}{mol.K}$, $R_{u}=0.08205\frac{l.atm}{mol.K}$, $k=1.4$

Applying the ideal gas equation to get the initial volume:

$$V_{i}=\frac{nRT_{i}}{P_{i}}=\frac{(1)(0.08205)(300)}{10}=2.46 l$$

Weight is abruptly removed to reduce the external pressure by one half, or to 5 atm, and the gas allowed to expand non-quasi-statically at constant pressure until equilibrium is reached.

Since the final pressure is known, we can (1) determine the relationship between the final temperature and volume by applying the general gas equation to the initial and final states and (2) substitute the relationship into the first law equation for a constant pressure adiabatic process.

Applying the general gas equation:

$$\frac{P_{i}V_{i}}{T_{i}}= \frac{P_{f}V_{f}}{T_{f}}$$

$$T_{f}=\frac{P_{f}V_{f}T_{i}}{P_{i}V_{i}}$$

$$T_{f}=\frac{(5)(V_{f})(300)}{(10)(2.46)}=61V_{f}$$

Applying the first law for an adiabatic process:

$$\Delta U=-W$$

$$nC_{v}(T_{f}-T_{i})=-P_{f}(V_{f}-V_{i})$$

Substituting $T_{f}=61V_{f}$ and the other known values

$$(1)(0.2053)[61V_{f}-300]=-5(V_{f}-2.46)$$

Which gives us $V_{f}=4.22l$ and $T_{f}=257K$, and the work done is

$$W=5(4.22-2.46)l.atm=892j$$

Comparing to the final volume and work done for a reversible adiabatic expansion with the same initial conditions and same final pressures:

$$P_{i}V_{i}^{k}= P_{f}V_{f}^{k}$$

$$V_{f}=6l$$

$$W_{rev}=\frac{(P_{f}V_{f}-P_{i}V_{i})}{1-k}$$

$$W_{rev}= 13.5 l.atm=1368j$$

Which, as expected, is greater than the irreversible work.

The next step is to tackle system 1, which adds piston friction. I would like to bring system 2 to the same final pressure and then compare the work on the surroundings to system 2. I recall that in your analysis of the quasi-static expansion with friction it was necessary to apply an external tensile force to overcome static friction in order to reach the same volume as the non-friction system. It appears that will again be the case here, except it would be in order to reach the same final pressure.

I was thinking about a possible alternative approach, such as assuming the surface of the cylinder at some point changes to a smooth service prior to reaching the maximum static friction force so that the expansion could be completed, albeit without friction work done at the end. But that would probably create more complications than it's worth.

 

[Chestermiller]

I approached this problem differently but, of course, arrived at the same results. My focus was on the final temperature, rather than the final volume.
$$nC_v(T_f-T_i)=-P_f(V_f-V_i)=P_f\left(\frac{nRT_f}{P_f}-\frac{nRT_i}{P_i}\right)$$
Solving for $T_f$ then yields:
$$T_f=\left[\frac{1}{\gamma}+\left(1-\frac{1}{\gamma}\right)\frac{P_f}{P_i}\right]T_i$$For $\gamma=1.4$ and $P_f/P_i=0.5$, this yields $T_f=0.8571T_i$. And, for $T_i=300\ K$, $T_f=257.1K$.

The work is given by: $$W=-\Delta U=-nC_v(T_f-T_i)=-(1)(2.5)(8.314)(257.1-300)=892\ J$$
It'll be interesting to compare the approach and results you get for system 2 with those that I obtained.

 

[Chestermiller]

Here's an interesting paradox for you. As $P_f/P_i\rightarrow 0$, $$\frac{T_f}{T_i}\rightarrow\frac{1}{\gamma}$$and then $$W\rightarrow nC_vT_i\left(1-\frac{1}{\gamma}\right)>0$$But how can this be if $P_f/P_i\rightarrow 0$ implies free expansion.