Another contradiction in thermodynamics?

[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.

 

[Chestermiller]

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 in the numerator and one with the total work as numerator.

I have another method for you to consider (not involving further integration).

We have nC_vdT=-P_extdV, so the total work done on the surroundings is -nC_vΔT.
The integral of (F/A)dV is the work to overcome friction, (F/A)ΔV. So the work done by the gas is -nC_vΔT+(F/A)ΔV
For the efficiency, I get the total work done on the surroundings divided by the work done by the gas.

Chet