Another contradiction in thermodynamics?

[Amin2014]

Consider an ideal gas operating in a quasi-static (very slow) cycle that is identical to the heat engine version of the carnot cycle in every aspect, except that friction is present. So even though the cycle is quasi-static, it is irreversible due to friction.

Now the question is: How does the efficiency of this quasi-static cycle compare to ideal carnot efficiency? Let η_rev be the efficiency of the carnot cycle and η_irrev be the efficiency of the quasi-static cycle:
Is η_irrev=η_rev?

 

[mfb]

Its efficiency is worse. You convert mechanical work to heat with a 1:1 ratio instead of the thermodynamic optimum.

 

[Amin2014]

You convert mechanical work to heat with a 1:1 ratio.

I'm guessing you're talking about work lost due to friction, correct?
But how does this affect our equations? Take the adiabatic step for instance. Since the process is quasi-static, we could still use dW=-PdV with p referring to the system's pressure. dq=0 right? It's an ideal gas, so dU=C_vdT. I'm taking the gas confined in the cylinder as my system. Where does friction between piston and cylinder come into the picture? If C_vdT = -PdV, then we're going to get the same PV^gamma= Const equation.

 

[Chestermiller]

I can help you work this out. There are two ways of doing this, but the easiest way is to treat the piston and cylinder as a combined system. Assume that all the "frictional heat" goes into the gas. To get things started, let F be the frictional force acting on the piston by the cylinder wall. Do a free body diagram on the piston, and then write a force balance on the piston. Don't forget the supplemental force you need to apply to the piston to keep the conditions quasistatic. What do you obtain?

Chet

 

[Amin2014]

I can help you work this out. There are two ways of doing this, but the easiest way is to treat the piston and cylinder as a combined system. Assume that all the "frictional heat" goes into the gas. To get things started, let F be the frictional force acting on the piston by the cylinder wall. Do a free body diagram on the piston, and then write a force balance on the piston. Don't forget the supplemental force you need to apply to the piston to keep the conditions quasistatic. What do you obtain?

Chet

Thnx Chet. I attached a diagram with three forces/pressures acting on the piston: internal pressure of the gas P, external pressure P_ext, and f_k due to kinetic friction. I've assumed the gas is undergoing expansion. I'm not sure about the supplemental force that you've mentioned though. I'm also confused about your choice of system.

treat the piston and cylinder as a combined system

Is it gas + piston, with the cylinder as surroundings? or is it piston+ cylinder?

Assume that all the "frictional heat" goes into the gas

Why all of the heat?

And what ratio are we trying to calculate here?

 

[Chestermiller]

Thnx Chet. I attached a diagram with three forces/pressures acting on the piston: internal pressure of the gas P, external pressure P_ext, and f_k due to kinetic friction. I've assumed the gas is undergoing expansion. I'm not sure about the supplemental force that you've mentioned though. I'm also confused about your choice of system.

Yes. The diagram is correct. I see that you had the cylinder sitting in a bath. I was assuming you were going to be taking the system to be adiabatic when you wrote PV^γ=const.

The "supplemental" force I was referring to is what you called P_ext. So the force balance on the piston is, as you indicated,

$PA - F = P_{ext}A$

Is it gas + piston, with the cylinder as surroundings? or is it piston+ cylinder?

Sorry. I meant to say, treat the gas and piston as a combined system, and adiabatic cylinder as part of the surroundings.

Why all of the heat?
And what ratio are we trying to calculate here?

To do this problem, you have to decide where the frictional heat is going. If it all somehow goes into the surroundings, then the gas experiences exactly the same treatment as in an adiabatic reversible expansion, and your PV^γ is correct. On the other hand, if part or all of the heat goes into the gas, it experiences a different treatment, equivalent to a non-adiabatic reversible expansion. I was assuming that you wanted to consider the case in which all the frictional heat goes into the gas.

Chet

 

[Chestermiller]

Continuing: If we multiply your force balance by dx (the displacement of the piston) we obtain:
$$PdV-\frac{F}{A}dV=P_{ext}dV=dW$$
where dV = Adx and dW is the work done by the system on the surroundings. If you combine this equation with the differential version of the first law and assume that the cylinder is adiabatic, what do you get?

Chet

 

[Amin2014]

Yes. The diagram is correct. I see that you had the cylinder sitting in a bath. I was assuming you were going to be taking the system to be adiabatic when you wrote PV^γ=const.

The "supplemental" force I was referring to is what you called P_ext. So the force balance on the piston is, as you indicated,

$PA - F = P_{ext}A$Sorry. I meant to say, treat the gas and piston as a combined system, and adiabatic cylinder as part of the surroundings.To do this problem, you have to decide where the frictional heat is going. If it all somehow goes into the surroundings, then the gas experiences exactly the same treatment as in an adiabatic reversible expansion, and your PV^γ is correct. On the other hand, if part or all of the heat goes into the gas, it experiences a different treatment, equivalent to a non-adiabatic reversible expansion. I was assuming that you wanted to consider the case in which all the frictional heat goes into the gas.

Chet

So what ratio do you have in mind as efficiency? Are you considering the Work done by P_ext as the numerator? But what if we choose the gas as our system, then it should be work done by the gas (PdV), no?

 

[Amin2014]

Continuing: If we multiply your force balance by dx (the displacement of the piston) we obtain:
$$PdV-\frac{F}{A}dV=P_{ext}dV=dW$$
where dV = Adx and dW is the work done by the system on the surroundings. If you combine this equation with the differential version of the first law and assume that the cylinder is adiabatic, what do you get?

Chet

I understand you're writing the wok-energy theorem for the massless piston.
PdV-P_extdV-f_kdx=dk_pist=0

dU=dW+dq+dW_*= -P_extdV-Fdx

 

[Amin2014]

where F= (P-P_ext)A
Substituting for f yields:
dU= -P_extdV+ (P_extdV-PdV)= -PdV

 

[Chestermiller]

So what ratio do you have in mind as efficiency? Are you considering the Work done by P_ext as the numerator? But what if we choose the gas as our system, then it should be work done by the gas (PdV), no?

Let's not worry about the efficiency yet. If we chose the gas as our system, then, yes, this would be the work. We'll do the problem that way after we do it this way, OK?

 

[Chestermiller]

I understand you're writing the wok-energy theorem for the massless piston.
PdV-P_extdV-f_kdx=dk_pist=0

dU=dW+dq+dW_*= -P_extdV-Fdx

I don't know exactly what you did here, but the result should be with a plus sign in front of the F and it should be P, not P_ext:
$$dU=\frac{F}{A}dV-PdV$$

Now, the next step is to substitute the ideal gas relationships into this.

Chet

 

[Amin2014]

I don't know exactly what you did here, but the result should be with a plus sign in front of the F and it should be P, not P_ext:
$$dU=\frac{F}{A}dV-PdV$$

Now, the next step is to substitute the ideal gas relationships into this.

Chet

my equation is dU= -P_extdV - Fdx. The first term on the right side of my equation equals the entire right side of your equation, i.e. -P_extdV= F/AdV - PdV.
The difference between my equation and yours is that I have added the term -Fdx, which is the work done by the cylinder on the piston. Considering that the cylinder is part of the surroundings and the piston is part of the system, therefore the force of friction, F, is external to our system, and since its point of application is moving, we must include a term for its work.

I added -Fdx because I believe it's part of the boundary work for our choice of system.

 

[Amin2014]

dW is the work done by the system on the surroundings

Remember that our system is working against TWO forces, not one. dW = P_extdV is only PART of the the work done by the system on its surroundings. The other half is done against friction.

 

[Chestermiller]

my equation is dU= -P_extdV - Fdx. The first term on the right side of my equation equals the entire right side of your equation, i.e. -P_extdV= F/AdV - PdV.
The difference between my equation and yours is that I have added the term -Fdx, which is the work done by the cylinder on the piston. Considering that the cylinder is part of the surroundings and the piston is part of the system, therefore the force of friction, F, is external to our system, and since its point of application is moving, we must include a term for its work.

I added -Fdx because I believe it's part of the boundary work for our choice of system.

How many times do you want to include it? We already included it once.

The cylinder is stationary, so our system does no work on the cylinder. The cylinder is considered part of the surroundings, so our system does no work on that part of the surroundings.

Chet

 

[Chestermiller]

Remember that our system is working against TWO forces, not one. dW = P_extdV is only PART of the the work done by the system on its surroundings. The other half is done against friction.

No, P_extdV is all the work done by our system on its surroundings.

 

[Amin2014]

No, P_extdV is all the work done by our system on its surroundings.

You are right, the piston does no work on the cylinder as the cylinder isn't moving. So P_extdV is the only work done BY the system ON the surroundings. So I made a mistake here. But I don't think you should use the engineering convention of the first law of thermodynamics here. Start with dU=dW +dq, where dW is the work done by the surroundings on the system. This I believe is a generalization of conservation of mechanical energy, and is consistent with the conventions used there.

 

[Amin2014]

In other words, in this particular problem, the work done on the surroundings by the system is not equal to the negative of the work done on the system by surroundings, and thus you cannot use dU= dq-dW where dW is the work done by the system as a valid form of the first law.

 

[Chestermiller]

You are right, the piston does no work on the cylinder as the cylinder isn't moving. So P_extdV is the only work done BY the system ON the surroundings. So I made a mistake here. But I don't think you should use the engineering convention of the first law of thermodynamics here. Start with dU=dW +dq, where dW is the work done by the surroundings on the system.

Actually, the version of the first law that I am accustomed to is dU = dq - dW, where dW is the work done by the system on the surroundings. So, from that perspective, what I said was OK.

So, it's a bit paradoxical, huh? But I am confident that I can resolve the paradox for you, so that we are both in agreement. But, for sure, handling the friction effect at the interface between the piston and the cylinder wall is a bit tricky.

I am going to analyze the friction effect fusing a different approach. Are you familiar with viscous fluids and Newtonian viscosity?

Please stay tuned. But right now, my wife an I are going out to dinner. So, see you later.

Chet

 

[Amin2014]

Actually, the version of the first law that I am accustomed to is dU = dq - dW, where dW is the work done by the system on the surroundings. So, from that perspective, what I said was OK.

So, it's a bit paradoxical, huh? But I am confident that I can resolve the paradox for you, so that we are both in agreement. But, for sure, handling the friction effect at the interface between the piston and the cylinder wall is a bit tricky.

I am going to analyze the friction effect fusing a different approach. Are you familiar with viscous fluids and Newtonian viscosity?

Please stay tuned. But right now, my wife an I are going out to dinner. So, see you later.

Chet

Ok take your time thank you and I hope you have a great time.

 

[Amin2014]

Are you familiar with viscous fluids and Newtonian viscosity?

Two solid objects, grinding against each other, no lubrication. What does this have to do with fluid mechanics? There was a time when I appreciated this branch of physics but I definitely need to refresh my memory.

 

[Amin2014]

I think we need to clarify something here. The cylinder wall is adiabatic right? That means no heat can cross the wall. But the rubbing is occurring between the piston and INNER surface of the wall, which means any heat produced will be confined within the cylinder. So even though the cylinder wall is adiabatic, depending on our choice of system, we may or may not have heat flow to the system.

If we choose our system to be just the gas and piston, that means the heat produced from friction will cross the boundary of our system. All of it, since it can't escape the adiabatic cylinder. So we have to account for dq=Fdx as well : dU= dW + dq= -P_extdV +Fdx -Fdx= -P_extdV, where dW= =P_extdV +Fdx.

On the other hand, if we had taken the boundary of our system to include the frictional force, then dq would be zero, and the frictional work would become internal. So in this case our equation would be dU=dW + dq= -P_extdV. Am I correct?

 

[Chestermiller]

I think we need to clarify something here. The cylinder wall is adiabatic right? That means no heat can cross the wall. But the rubbing is occurring between the piston and INNER surface of the wall, which means any heat produced will be confined within the cylinder. So even though the cylinder wall is adiabatic, depending on our choice of system, we may or may not have heat flow to the system.

Yes.

if we had taken the boundary of our system to include the frictional force, then dq would be zero, and the frictional work would become internal. So in this case our equation would be dU=dW + dq= -P_extdV. Am I correct?

Yes. This is the case I was trying to get us to consider. The equations for this were presented in posts #9 and #12:
$$dU=-PdV+\frac{F}{A}dV$$

If we choose our system to be just the gas and piston, that means the heat produced from friction will cross the boundary of our system. All of it, since it can't escape the adiabatic cylinder. So we have to account for dq=Fdx as well : dU= dW + dq= -P_extdV +Fdx -Fdx= -P_extdV, where dW= =P_extdV +Fdx.

No. If were going to do this type of thing then we would choose as the system the gas. The heat generated at the piston interface with the cylinder would flow into the piston, and then into the gas. So the heat flow into the gas would be dq = Fdx. The surroundings in this case (as reckoned by the gas) would be the piston, and the work done by the gas on the piston (its surroundings) would be dW=PdV, where P is the gas pressure (i.e., the pressure at the interface between our system, the gas, and its surroundings). So, in this case, the first law would give us:
$$dU=dq - dW=Fdx-PdV=\frac{F}{A}dV-PdV$$
Note that both methods give us exactly the same result.

Chet

 

[Amin2014]

Thanks Chet, I understand what you are saying, your analyeses and everything. But they aren't quite addressing my own personal source of confusion. For one, the frictional force DOES do work on the moving piston after all. So there has to be some choice of system where this becomes external work. I was thinking we could choose the system as either gas alone (which you analyzed above), or gas + piston, or somehow, gas + piston + friction. That's how I tried to view the problem. As you see, there is still some source of confusion for me; I can't quite "visualize" everything here.

Why can't we just consider the microscopic details of friction as another "unit" and decide to include or exclude it in our system?

 

[Andrew Mason]

Just a comment here: if the friction results in heat flow into the system, the cycle will not be a Carnot cycle. The Carnot cycle operates between two reservoirs each at different temperatures, Th and Tc. The cycle requires all heat flow to occur where the surroundings and the system are infinitesimally close to thermal equilibrium. Heat flow from the hot reservoir at Th occurs with the system at temperature Th. In order to have heat flow from the system to the cold reservoir the system temperature has to change from being infinitesimally close to Th to infinitesimally close to Tc. This change in temperature has to occur, therefore, without any heat flow occurring (adiabatic). If friction causes heat flow into the system during the adiabatic expansion, it ceases to be adiabatic. So it is not a Carnot cycle.

AM

 

[Chestermiller]

Thanks Chet, I understand what you are saying, your analyeses and everything. But they aren't quite addressing my own personal source of confusion. For one, the frictional force DOES do work on the moving piston after all. So there has to be some choice of system where this becomes external work. I was thinking we could choose the system as either gas alone (which you analyzed above), or gas + piston, or somehow, gas + piston + friction. That's how I tried to view the problem. As you see, there is still some source of confusion for me; I can't quite "visualize" everything here.

Why can't we just consider the microscopic details of friction as another "unit" and decide to include or exclude it in our system?

OK. Let's consider the microscopic gap between the piston and the cylinder as an additional "unit" in the picture. The outer surface of this unit is stationary, since it is the inner surface of the cylinder. The inner surface of this unit is moving with the velocity of the piston. The unit has negligible mass, so there are no changes in its internal energy. From the frame of reference of the laboratory, since the outer surface is not moving, no work is done by the "unit" on the cylinder at this surface; but the inner surface is moving, and the work done by the unit on the piston at this surface is -Fdx. If we apply the first law to this "unit," we obtain:

$-dq=-Fdx$

where dq is the heat flow into the piston from the unit at the piston interface with the "unit" (there is no heat flow at the cylinder surface of the unit).

Now, we can either include the "unit" as part of our gas/piston system as a third element, or we can omit it from our gas/piston system. I think that the case you are interested in is if we don't include it as a third element. If we do this, then the work done by the gas/piston system on its surroundings is:

$P_{ext}dV+Fdx$

where Fdx is the work done on the "unit" (which is now outside our system). There is also heat transfer into the gas/piston system from the unit. So, with this choice of system,

$dW=P_{ext}dV+Fdx=PdV-Fdx+Fdx=PdV$

$dq = Fdx$

And so, from the first law, $dU=Fdx-PdV$

If we were to include the "unit" as part of our system, then no work is done at the cylinder surface by this system, since the cylinder is stationary. And no heat enters this system from the surroundings, because the cylinder is insulated. So, for this choice of the system, we have

$dW = P_{ext}dV$

$dq = 0$

And so, from the first law, $dU=-P_{ext}dV = - PdV + Fdx$

So we get the same answer either way.Chet

 

[Amin2014]

if the friction results in heat flow into the system, the cycle will not be a Carnot cycle.

Yes thankyou. I never implied the cycle is carnot. If friction exists, the cycle is not carnot, period. Friction causes irreversibility, while the carnot is defined to be a reversible cycle between two (constant temperature) reservoirs.

In other words, reversibility is embedded in the definition of carnot cycle; there's' no such thing as an irreverible carnot.

 

[Amin2014]

OK. Let's consider the microscopic gap between the piston and the cylinder as an additional "unit" in the picture. The outer surface of this unit is stationary, since it is the inner surface of the cylinder. The inner surface of this unit is moving with the velocity of the piston. The unit has negligible mass, so there are no changes in its internal energy. From the frame of reference of the laboratory, since the outer surface is not moving, no work is done by the "unit" on the cylinder at this surface; but the inner surface is moving, and the work done by the unit on the piston at this surface is -Fdx. If we apply the first law to this "unit," we obtain:

$-dq=-Fdx$

where dq is the heat flow into the piston from the unit at the piston interface with the "unit" (there is no heat flow at the cylinder surface of the unit).

Now, we can either include the "unit" as part of our gas/piston system as a third element, or we can omit it from our gas/piston system. I think that the case you are interested in is if we don't include it as a third element. If we do this, then the work done by the gas/piston system on its surroundings is:

$P_{ext}dV+Fdx$

where Fdx is the work done on the "unit" (which is now outside our system). There is also heat transfer into the gas/piston system from the unit. So, with this choice of system,

$dW=P_{ext}dV+Fdx=PdV-Fdx+Fdx=PdV$

$dq = Fdx$

And so, from the first law, $dU=Fdx-PdV$

If we were to include the "unit" as part of our system, then no work is done at the cylinder surface by this system, since the cylinder is stationary. And no heat enters this system from the surroundings, because the cylinder is insulated. So, for this choice of the system, we have

$dW = P_{ext}dV$

$dq = 0$

And so, from the first law, $dU=-P_{ext}dV = - PdV + Fdx$

So we get the same answer either way.Chet

This is an awesome analysis Chet! And I'm glad that we're both saying the same thing (I had come to dU= -P_extdV in each case as well). A few things come to mind though: First of all, If we take the "unit" as our system, I believe the fist law leads to dq= -Fdx, which means that the heat is flowing out of this unit (into the gas))
Secondly, in order to clarify the problem further more, we should state that the piston itself is also adiabatic, so no heat can flow into the piston, which means all the heat created will flow directly into the gas. In this case, dq for gas is the negative of dq= -Fdx calculated above.
Again thanks for your patience and depth of analysis.

 

[Amin2014]

If we apply the first law to this "unit," we obtain:

$-dq=-Fdx$

where dq is the heat flow into the piston from the unit at the piston interface with the "unit" (there is no heat flow at the cylinder surface of the unit).
Chet

This is also correct since you are taking dq as heat flowing OUT of the "unit".

 

[Andrew Mason]

Yes thankyou. I never implied the cycle is carnot. If friction exists, the cycle is not carnot, period. Friction causes irreversibility, while the carnot is defined to be a reversible cycle between two (constant temperature) reservoirs.

In other words, reversibility is embedded in the definition of carnot cycle; there's' no such thing as an irreverible carnot.

You can have a Carnot cycle with friction. It is just that you can't have the friction causing heat flow into the system.

The reversibility of the Carnot cycle relates to the thermodynamic cycle. It does not depend on how the work output of the engine is used.

AM

 

[Amin2014]

OK. Let's consider the microscopic gap between the piston and the cylinder as an additional "unit" in the picture. The outer surface of this unit is stationary, since it is the inner surface of the cylinder. The inner surface of this unit is moving with the velocity of the piston.

So Chet does this mean that we have a velocity profile between the cylinder surface and piston, where the velocity at the cylinder surface is zero while velocity at piston is maximum, so the work done by system on surroundings would always be negative to the work done by surroundings on system when analyzing contact forces, am I correct?

I am asking this because I'm used to the scientific convention of the first law dU= dq + dW

 

[Chestermiller]

So Chet does this mean that we have a velocity profile between the cylinder surface and piston, where the velocity at the cylinder surface is zero while velocity at piston is maximum, so the work done by system on surroundings would always be negative to the work done by surroundings on system when analyzing contact forces, am I correct?

We are not really concerning ourselves with the details of the geometry and contents of the frictional interface. But, because of Newton's third law, this relationship between the work that the system does on the surroundings and the work that the surroundings do on the system must always be true. It's just more difficult to visualize in the frictional case we've been looking at. I like to picture the surfaces at the microscale as having surface asperities that rub against the opposite surface. However, the asperities associated with the piston move with the velocity of the piston, and the asperities associated with the cylinder move with zero velocity. There also must be some substance in the interstices between the asperities, like air, gas, or whatever. So the substance within the interstices must have a velocity profile, but not the asperities. Hope that addresses your question.

I am asking this because I'm used to the scientific convention of the first law dU= dq + dW

We should be able to come up with exactly the same final equations using your sign convention on the work.

Chet

 

[Amin2014]

We are not really concerning ourselves with the details of the geometry and contents of the frictional interface. But, because of Newton's third law, this relationship between the work that the system does on the surroundings and the work that the surroundings do on the system must always be true. It's just more difficult to visualize in the frictional case we've been looking at. I like to picture the surfaces at the microscale as having surface asperities that rub against the opposite surface. However, the asperities associated with the piston move with the velocity of the piston, and the asperities associated with the cylinder move with zero velocity. There also must be some substance in the interstices between the asperities, like air, gas, or whatever. So the substance within the interstices must have a velocity profile, but not the asperities. Hope that addresses your question.We should be able to come up with exactly the same final equations using your sign convention on the work.

Chet

Just to further clarify what you are saying in case anybody else may be reading this and also to reinforce my own understanding, I believe you are saying that if we have CONTACT forces (as opposed to long distance forces) acting between two POINTS on two different bodies, then their works would always be equal in magnitude and negative in sign, because for the two points two remain in contact, they'd have to move together (and of course as soon as they depart there would be no work, since we are assuming contact forces).

This friction work is kinda funny as you mentioned; In mechanics we take the kinetic friction between a (fixed) incline and the sliding block to be action-reaction pairs, and yet the work on the incline is zero while the work on the block is not.

 

[Amin2014]

But if we take the friction forces to be (long distance) electromagnetic adhesions between the molecules of the cylinder and the molecules of the piston, here the action-reaction forces aren't necessarily doing equal and opposite work, since they aren't moving together. But this brings up another question: aren't the surface molecules of the cylinder re-arranged during contact? Oh my this whole friction work has caused so much confusion for me even though I can calculate it now.

 

[Amin2014]

I guess we'll have to assume the cylinder has a perfectly rigid surface. I'm still finding it confusing though.

 

[Chestermiller]

Hi Amin2014.

The reason that kinetic friction work is so tricky to quantify is because of the idealized macroscopic model we are using to describe it. The macroscopic model features a velocity discontinuity at the boundary, but the rate of doing work is determined by the shear force times the velocity. But, with a velocity discontinuity, which velocity do you use? There's to rub.

You yourself envisioned a way to work around this by imagining an intermediate material layer (on the microscale) between the two surfaces in which the velocity varies continuously (velocity profile). Unfortunately we do not know all the microscopic details of dry kinetic friction, but, what we can do is conceive of a microscopic model that, on the macroscopic scale behaves exactly the same as the idealized kinetic friction model, but without the velocity discontinuity. This is what I was talking about also when I proposed introducing a microscopic fluid layer between the surfaces to mimic the kinetic friction. Such a fluid would have a viscosity that varies inversely with the velocity gradient (non-Newtonian fluid) so that the shear force is always constant irrespective of the magnitude of the velocity difference between the surfaces. So, on the macroscopic scale, it would behave exactly the same as our macroscopic model of dry kinetic friction, except that it would eliminate the velocity discontinuity by providing a linear velocity profile across the intermediate microscopic layer. Thinking of it in this way would totally eliminate the confusion that we have been experiencing, and would be consistent with what we were discussing in several of our last few posts. And, most importantly, it would enable us to correctly determine the work done by whom on what, and by what on whom.

Thoughts?

Chet

 

[Amin2014]

Hi Amin2014.

The reason that kinetic friction work is so tricky to quantify is because of the idealized macroscopic model we are using to describe it. The macroscopic model features a velocity discontinuity at the boundary, but the rate of doing work is determined by the shear force times the velocity. But, with a velocity discontinuity, which velocity do you use? There's to rub.

You yourself envisioned a way to work around this by imagining an intermediate material layer (on the microscale) between the two surfaces in which the velocity varies continuously (velocity profile). Unfortunately we do not know all the microscopic details of dry kinetic friction, but, what we can do is conceive of a microscopic model that, on the macroscopic scale behaves exactly the same as the idealized kinetic friction model, but without the velocity discontinuity. This is what I was talking about also when I proposed introducing a microscopic fluid layer between the surfaces to mimic the kinetic friction. Such a fluid would have a viscosity that varies inversely with the velocity gradient (non-Newtonian fluid) so that the shear force is always constant irrespective of the magnitude of the velocity difference between the surfaces. So, on the macroscopic scale, it would behave exactly the same as our macroscopic model of dry kinetic friction, except that it would eliminate the velocity discontinuity by providing a linear velocity profile across the intermediate microscopic layer. Thinking of it in this way would totally eliminate the confusion that we have been experiencing, and would be consistent with what we were discussing in several of our last few posts. And, most importantly, it would enable us to correctly determine the work done by whom on what, and by what on whom.

Thoughts?

Chet

Hi Chet, last night (after I had posted here), something exciting happened: I was suddenly able to view the problem from what I think is your point of view. I think I can now explain your view, and compare it to mine, and hopefully reach a cool conclusion, so here goes:

I wish to discuss two things here, one is comparing dU= dq - dW with dU = dq + dW, so I'll start with this first. The reason that you are CONSISTENTLY getting correct results using the engineering convention is because you are calculating dW based on the displacement of the boundary of the system. In other words, dx for you is always the displacement of the boundary of system. This way you are guaranteeing that this "work" that you are calculating is always negative the work done by SURROUNDINGS on system, and since you have changed the plus sign in front of it to a minus sign, you are going to always get results similar to the scientific convention.

What you have to realize is this quantity that you are calculating and calling work is not always going to be the work done by SYSTEM on surroundings. In the case of contact forces, it makes no difference and you may call dW in the engineering convention "work done by system on surroundings". However, if you are going to calculate work done by long distance forces, this "virtual" work that you are calculating will in general differ from the work done by system on surroundings, do you see why?There is no law stating that the WORKS done by action-reaction forces are equal and opposite. This in general is not true. Maybe
someday in physics all forces in nature will be convincingly reduced to contact forces, but with current models, taking forces such as gravity to be long
distance, then we can't use that convention. In other words, for the engineering convention to be consistently correct, dW has to be the negative
of work done on the system, which may or may not be equal to the work done BY the system.

You could still use the engineering convention the way you do, but you should stop calling dW "work done by system on surroundings"
when you are dealing with work done by long distance forces.

In the case of long distance forces, using displacement of SYSTEM BOUNDARY to calculate dx does NOT yield work done on surroundings, as the displacement of surroundings may be different from displacement of system boundary.

I still have more to say on this if you are interested.

 

[Amin2014]

Hi Amin2014.

The reason that kinetic friction work is so tricky to quantify is because of the idealized macroscopic model we are using to describe it. The macroscopic model features a velocity discontinuity at the boundary, but the rate of doing work is determined by the shear force times the velocity. But, with a velocity discontinuity, which velocity do you use? There's to rub.

You yourself envisioned a way to work around this by imagining an intermediate material layer (on the microscale) between the two surfaces in which the velocity varies continuously (velocity profile). Unfortunately we do not know all the microscopic details of dry kinetic friction, but, what we can do is conceive of a microscopic model that, on the macroscopic scale behaves exactly the same as the idealized kinetic friction model, but without the velocity discontinuity. This is what I was talking about also when I proposed introducing a microscopic fluid layer between the surfaces to mimic the kinetic friction. Such a fluid would have a viscosity that varies inversely with the velocity gradient (non-Newtonian fluid) so that the shear force is always constant irrespective of the magnitude of the velocity difference between the surfaces. So, on the macroscopic scale, it would behave exactly the same as our macroscopic model of dry kinetic friction, except that it would eliminate the velocity discontinuity by providing a linear velocity profile across the intermediate microscopic layer. Thinking of it in this way would totally eliminate the confusion that we have been experiencing, and would be consistent with what we were discussing in several of our last few posts. And, most importantly, it would enable us to correctly determine the work done by whom on what, and by what on whom.

Thoughts?

Chet

Let's think of kinetic friction as gum sticking to your shoe. So when you're walking on the ground, there's a strip of gum attaching your shoe to the ground which makes it harder to move forwards (analogous to kinetic friction). As the shoe moves forwards, negative
work is being done on the shoe due to the CONTACT force at the shoe, and equal and opposite this work is being done on the end
of the gum attached to the shoe. At the same time, the other end of the gum is applying force to the ground, but it can't move
the ground, so there's no work being done there. The ground doesn't do any work on the gum either. As you can see, by considering
a massive "medium" between the surface and the shoe, we have been able to reduce all forces to contact forces, and thus the work
done by any force is opposite the work done by its reaction pair. As you've mentioned, we've eliminated velocity discontinuity.

Now, If we introduce a further assumption that the mass*acceleration of the gum is negligible compared to the magnitudes of these forces, then the forces on either end of the gum (and their corresponding reactions), will be equal in magnitude, and we will have cleared up all confusions.

I came up with this model last night, and I find it very similar to your non-Newtonian fluid description. Very nice ;)

 

[Chestermiller]

Hi Chet, last night (after I had posted here), something exciting happened: I was suddenly able to view the problem from what I think is your point of view. I think I can now explain your view, and compare it to mine, and hopefully reach a cool conclusion, so here goes:

I wish to discuss two things here, one is comparing dU= dq - dW with dU = dq + dW, so I'll start with this first. The reason that you are CONSISTENTLY getting correct results using the engineering convention is because you are calculating dW based on the displacement of the boundary of the system. In other words, dx for you is always the displacement of the boundary of system. This way you are guaranteeing that this "work" that you are calculating is always negative the work done by SURROUNDINGS on system, and since you have changed the plus sign in front of it to a minus sign, you are going to always get results similar to the scientific convention.

What you have to realize is this quantity that you are calculating and calling work is not always going to be the work done by SYSTEM on surroundings. In the case of contact forces, it makes no difference and you may call dW in the engineering convention "work done by system on surroundings". However, if you are going to calculate work done by long distance forces, this "virtual" work that you are calculating will in general differ from the work done by system on surroundings, do you see why?There is no law stating that the WORKS done by action-reaction forces are equal and opposite. This in general is not true. Maybe
someday in physics all forces in nature will be convincingly reduced to contact forces, but with current models, taking forces such as gravity to be long
distance, then we can't use that convention. In other words, for the engineering convention to be consistently correct, dW has to be the negative
of work done on the system, which may or may not be equal to the work done BY the system.

You could still use the engineering convention the way you do, but you should stop calling dW "work done by system on surroundings"
when you are dealing with work done by long distance forces.

In the case of long distance forces, using displacement of SYSTEM BOUNDARY to calculate dx does NOT yield work done on surroundings, as the displacement of surroundings may be different from displacement of system boundary.

I still have more to say on this if you are interested.

I think we are on the same wavelength. In thermo, when we calculate work on the surroundings or done by the surroundings, I always envisioned that it was implicit that we are dealing exclusively with contact forces.

Chet

 

[Chestermiller]

Let's think of kinetic friction as gum sticking to your shoe. So when you're walking on the ground, there's a strip of gum attaching your shoe to the ground which makes it harder to move forwards (analogous to kinetic friction). As the shoe moves forwards, negative
work is being done on the shoe due to the CONTACT force at the shoe, and equal and opposite this work is being done on the end
of the gum attached to the shoe. At the same time, the other end of the gum is applying force to the ground, but it can't move
the ground, so there's no work being done there. The ground doesn't do any work on the gum either. As you can see, by considering
a massive "medium" between the surface and the shoe, we have been able to reduce all forces to contact forces, and thus the work
done by any force is opposite the work done by its reaction pair. As you've mentioned, we've eliminated velocity discontinuity.

Now, If we introduce a further assumption that the mass*acceleration of the gum is negligible compared to the magnitudes of these forces, then the forces on either end of the gum (and their corresponding reactions), will be equal in magnitude, and we will have cleared up all confusions.

I came up with this model last night, and I find it very similar to your non-Newtonian fluid description. Very nice ;)

Excellent. I like your gum model. If I had thought you would relate to gum better than fluid, I would have introduced it instead. But I think that it's better that you thought of all this on your own. What you said here is definitely what I was trying to convey.

You have the soul of a modeler. Not everyone can dope things out like this. Welcome to the club.

Do you have the energy to continue with the solution to the differential equation and get (a) the final temperature, (b) the total work done, (c) the work done by the gas and (d) the efficiency, or would you just like to skip the rest?

Chet

 

[Amin2014]

I think we are on the same wavelength. In thermo, when we calculate work on the surroundings or done by the surroundings, I always envisioned that it was implicit that we are dealing exclusively with contact forces.

Chet

But in the case of kinetic friction between the piston and cylinder for instance, we aren't dealing with contact forces. For two reasons: both the realistic model, and because as you see, the work done on the cylinder surface is zero, but the work done on the piston surface is not. This was really confusing me, and I think it could be entirely avoided with the scientific convention. Anyway

 

[Amin2014]

Excellent. I like your gum model. If I had thought you would relate to gum better than fluid, I would have introduced it instead. But I think that it's better that you thought of all this on your own. What you said here is definitely what I was trying to convey.

You have the soul of a modeler. Not everyone can dope things out like this. Welcome to the club.

Do you have the energy to continue with the solution to the differential equation and get (a) the final temperature, (b) the total work done, (c) the work done by the gas and (d) the efficiency, or would you just like to skip the rest?

Chet

I didn't suggest that my gum model was better than the fluid model, and I don't think it is. It's just that you didn't explain your fluid model until later, and thus I was confused at first. Also, and more importantly, according to my own understanding of physics, you kept misusing the term work on surroundings by system, which got me all confused. So it took me some time to figure out what you were doing there, and to do so I had to step out of what I thought and still think to be the correct terminology in physics.

Your model makes perfect sense; The only advantage my model might have is that people can relate to it even if they don't know about fluids, especially since gum and friction sort of go hand in hand; you can get "Stuck" with both of them, lol.

With regards to solving the differential equation, I don't think that should be much of a problem; One way to do it is to write T in terms of P and V ( T=PV/nR), differentiate to find dT= (PdV +VdP)/nR, and substitute for dT in dU= nC_vdT and then separate the variables in the differential equation we had obtained: dU= -PdV +FdV/A. All the terms are written in terms of P and V of gas, and the rest is a bunch of constants.

 

[Chestermiller]

With regards to solving the differential equation, I don't think that should be much of a problem; One way to do it is to write T in terms of P and V ( T=PV/nR), differentiate to find dT= (PdV +VdP)/nR, and substitute for dT in nC_vdt and then separate the variables in the differential equation we had obtained: dU= -PdV +FdV/A. All the terms are written in terms of P and V of gas, and the rest is a bunch of constants.

For what it's worth, I would do it a little differently. I would substitute P = nRT/V, and solve for T rather than P.

Chet

 

[Amin2014]

With my way, the separation of variables is straightforward. With your way, I get nC_vdT= -nRT dV/V +(F/A)dV.
While I recognize this as a familiar differential equation that should be solvable with Laplace transformation, I can't separate the variables, and don't recall the alternative solution. What's the simple solution (without Laplace)?

 

[Chestermiller]

With my way, the separation of variables is straightforward. With your way, I get nC_vdT= -nRT dV/V +(F/A)dV.
While I recognize this as a familiar differential equation that should be solvable with Laplace transformation, I can't separate the variables, and don't recall the alternative solution. What's the simple solution (without Laplace)?

This is a first order linear ordinary differential equation. One method of solving it is by using an integrating factor.

Chet

 

[Amin2014]

This is a first order linear ordinary differential equation. One method of solving it is by using an integrating factor.

Chet

Yeah I think we have to multiply both sides by some factor to make it a complete differential. Also you could check my solution. Based on the form of the equation, I find it easier to use T=PV/nR than the conventional P=nRT/V

 

[Chestermiller]

Yeah I think we have to multiply both sides by some factor to make it a complete differential. Also you could check my solution. Based on the form of the equation, I find it easier to use T=PV/nR than the conventional P=nRT/V

Are you saying that you would like me to provide my solution for the temperature?

Chet

 

[Amin2014]

Are you saying that you would like me to provide my solution for the temperature?

Chet

No it's ok I can do it. I just said that if you write T in terms of P and V in this particular equation, then you can separate all the variables in one step and integrate ;)

 

[Chestermiller]

No it's ok I can do it. I just said that if you write T in terms of P and V in this particular equation, then you can separate all the variables in one step and integrate ;)

Yes. I tried it your way, and I liked a lot better. Much simpler.

Chet

 

[Amin2014]

Do you have the energy to continue with the solution to the differential equation and get (a) the final temperature, (b) the total work done, (c) the work done by the gas and (d) the efficiency, or would you just like to skip the rest?

The work done by the gas is ∫PdV. We have (P- P_c)V ^gamma = Const, so we substitute P in terms of V to evaluate the integral. By "total work". I assume you mean ∫P_extdV, which can readily be found from -PdV + (F/A)dV = -P_extdV. For part d, I think we can define two different efficiencies, one with the work done by the gas as numerator and one with the total work as numerator.