Show that entropy is a state function

In summary, I think it would be helpful to distinguish between the entropy of the system before the process and the entropy of the system after the process.
  • #1
Philip Koeck
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In a (reversible) Carnot cycle the entropy increase of the system during isothermal expansion at temperature TH is the same as its decrease during isothermal compression at TC. We can conclude that the entropy change of the system is zero after a complete Carnot cycle.
The mentioned textbook now states that any reversible cyclic process can be constructed from Carnot cycles (towards the end of chapter 20 in the 14th global edition of Young and Freedman.)

The conclusion is that any reversible cyclic process has zero entropy change for the system.

As I see it, this does not show that entropy is a state function.
One would also have to show that the entropy change of the system is zero in any irreversible cyclic process.

Is it okay to simply generalise the argument in the textbook and say that any reversible or irreversible cyclic process can be approximated by (reversible) Carnot processes and therefore the entropy of the system is always unchanged after one complete cycle, no matter whether the process is reversible or irreversible?
 
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  • #2
If the process is not reversible, you look at an open system, but entropy (and the other thermodynamical variables) are only state functions for closed systems!
 
  • #3
Philip Koeck said:
Summary:: I'm trying to show that entropy is a state function based on an analysis of the Carnot cycle and without using advanced mathematics. I'm not satisfied with the presentation in the textbook "University Physics" by Young and Freedman and would like some feedback.

Maybe, chapter "B. The Entroy" in Richard Becker's book "The Theory of Heat" might be of help:
https://books.google.de/books?id=wS...de&source=gbs_toc_r&cad=3#v=onepage&q&f=false
 
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  • #4
Lord Jestocost said:
Maybe, chapter "B. The Entroy" in Richard Becker's book "The Theory of Heat" might be of help:
https://books.google.de/books?id=wS...de&source=gbs_toc_r&cad=3#v=onepage&q&f=false
Thanks for the reference.
I'm trying to work only with Young and Freedman at the moment since that is what we use in our courses.
Unfortunately I'm missing some step that would make the presentation logically consistent, at least for me.
I'll discuss it more in my reply to vanHees.
 
  • #5
vanhees71 said:
If the process is not reversible, you look at an open system, but entropy (and the other thermodynamical variables) are only state functions for closed systems!
In that case I have a serious, didactic problem later on in the same chapter.
Young and Freedman proceed to calculating entropy changes for irreversible processes such as free expansion and irreversible heat transfer.
The argument they offer is that for such calculations, any irreversible process can be replaced by a reversible process between the same initial and final state, because entropy is a state function.
Now, if entropy is not a state function for irreversible processes, how do I know that I will get the correct value for its change after making the replacement.
Do you see my problem?
 
  • #6
If an irreversible process is cyclic, this means that it starts and ends in the same state. It is possible to devise an infinite number of alternate reversible processes that also start and end in this same state. The change in entropy for all these reversible processes is zero. Therefore the change in entropy for the irreversible process is zero.
 
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  • #7
Philip Koeck said:
Summary:: I'm trying to show that entropy is a state function based on an analysis of the Carnot cycle and without using advanced mathematics. I'm not satisfied with the presentation in the textbook "University Physics" by Young and Freedman and would like some feedback.

Thinking about this question sent me down a rabbit hole... not sure if I have a good response, but here goes:

First, it's important to note the difference between the entropy of a system S and *changes* to the system entropy ΔS that occurs during a process.

Regarding 'S', since S is a property of a system, the entropy value associated with a particular *equilibrium* state can be expressed in terms of other state variables such as P, V, and T. Then, the entropy can be simply considered as another variable specifying the state and the state can be specified in terms of, for example, S and V or T and S, etc. Thus, S is a state variable. [There is an unstated assumption that S is uniquely determined in any particular equilibrium state, not sure if we need to 'prove' that.]

I got hung up (and honestly, and still uncertain) on both the difference between S and ΔS and the requirement for the state to be in equilibrium. For example:

Consider the copy of Young and Freedman on your desk- put the book into a box. Your system is the book. What is S? I honestly don't know how to calculate it and I believe it's actually an open question:

Xiong W, Faes L, Ivanov PC. Entropy measures, entropy estimators, and their performance in quantifying complex dynamics: Effects of artifacts, nonstationarity, and long-range correlations. Phys Rev E. 2017;95(6-1):062114. doi:10.1103/PhysRevE.95.062114 , available at https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6117159/

So, in frustration, I open the container and rip up the book. Or, maybe I open the box and burn the book. Maybe I fill the container with sulfuric acid and destroy the book. The thing is, I can 'easily' calculate ΔS for each of those three irreversible processes!

For an irreversible process, as long as the start and end states are equilibrium states ('thermostatics'), I can calculate ΔS in terms of any contrived reversible process that connects the two states. It's also possible to compute ΔS if the two states are *stationary* rather than equilibrium ('thermokinetics'), but a full-on nonequilibrium thermodynamic calculation is (AFAIK) not possible at this time.

Not sure if this helps... clarification is welcome!
 
  • #8
Chestermiller said:
If an irreversible process is cyclic, this means that it starts and ends in the same state. It is possible to devise an infinite number of alternate reversible processes that also start and end in this same state. The change in entropy for all these reversible processes is zero. Therefore the change in entropy for the irreversible process is zero.

Yes, that is very helpful.
So essentially I have to correct the textbook a little.
Any cyclic process, both reversible and irreversible, can be approximated by a series of Carnot processes.

An interesting example is a process that's almost identical to the Carnot-cycle, but the reversible, isothermal expansion is replaced by a free expansion.
It can be replaced by a single Carnot-cycle for calculating the entropy change of the system.
 
  • #9
Chestermiller said:
If an irreversible process is cyclic, this means that it starts and ends in the same state. It is possible to devise an infinite number of alternate reversible processes that also start and end in this same state. The change in entropy for all these reversible processes is zero. Therefore the change in entropy for the irreversible process is zero.
Have you now proven that there are no irreversible processes?

Of course not. The reason is that in general when you consider time-dependent situations of many-body systems they are usually not described by local thermal equilibrium at any instance of time. That's only the case if the changes on the macroscopic relevant observables are slow compared to the thermalization time of the system. Only then the entropy stays constant over all time. In other words for general non-equilibrium processes the entropy is not constant but increasing.
 
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  • #10
Philip Koeck said:
Yes, that is very helpful.
So essentially I have to correct the textbook a little.
Any cyclic process, both reversible and irreversible, can be approximated by a series of Carnot processes.
Absolutely not. To determine the entropy change for an irreversible process, we need to devise an alternative reversible process between the same two thermodynamic end states, and calculate the integral of dq/T for that reversible process. The alternative reversible process does not need to bear any resemblance whatsoever to the actual irreversible process, as long as it passes through the same two end states. And there are an infinite number of reversible processes that will do this.
 
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  • #11
Chestermiller said:
Absolutely not. To determine the entropy change for an irreversible process, we need to devise an alternative reversible process between the same two thermodynamic end states, and calculate the integral of dq/T for that reversible process. The alternative reversible process does not need to bear any resemblance whatsoever to the actual irreversible process, as long as it passes through the same two end states. And there are an infinite number of reversible processes that will do this.
Then I wonder whether I have gained anything. I want to show that S is a state function.
In your recipe you use the fact that S is a state function.
The problem I have with our textbook is that it shows that S is a state function only for reversible processes, but then it uses this finding even for irreversible processes.
I'm missing one step in the argument, so to say.
 
  • #12
I really don't know how to answer this question. I guess we just accepted it when we learned ti. My main strength lies more in applying the fundaments to solve specific problems.
 
  • #13
A general change of state is described by off-equilibrium many-body theory, which can be addressed on various levels: At the most fundamental level you can use, e.g., non-equilibrium quantum-field theory methods (relativistic or non-relativistic), e.g., the Schwinger-Keldysh real-time-contour formalism, which leads to the Kadanoff-Baym equations in the 2PI (kown under the various names of Luttinger-Ward, Kadanoff-Baym, or Jackiw-Tomboulis) formalism. This is a very sophisticated approach and the solution of the equations is usually pretty difficult. It has been done mostly for toy models like the ##\phi^4##, linear $\sigma$ model, often in lower space-time dimensions.

The next level of description are Boltzmann-type (quantum) transport equations, which is found formally by using a gradient expansion or equivalently a formal expansion in powers of ##\hbar## leading to a Markovian description, usually taking into account only ##2 \rightarrow 2## scattering processes but also higher-order inelastic processes like ##2 \leftrightarrow 3## scattering etc.

Through the assumption of "molecular chaos" to cut the typical "BBGKY hierarchy" one introduces a "thermodynamical arrow of time" which is by construction identical with the "causal arrow of time" underlying all physics. The principle of detailed balance, guaranteed by the unitarity of the S-matrix, ensures that the (coarse-grained) entropy does not decrease ("H-theorem"), and that in the long-time limit thermal equilibrium is reached.

This final equilibrium state is uniquely determined by the temperature, chemical potential(s) of conserved quantitities, and external paramaters (like volume, presence of external fields, etc.), and for this equilibrium state the entropy is maximal and a "state function", which means that the equilibrium state, reached under the applied constraints is the state of "minimal information", and you cannot say from the knowledge of this state, how the system came into this state, i.e., you have no information about the "history" of the system before it reached this equilibrium state.
 
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  • #14
Philip Koeck said:
The problem I have with our textbook is that it shows that S is a state function only for reversible processes,

When you once have proven that the entropy of a system is a state function which depends only on the values of some state variables of the system, it doesn’t matter how you bring the system into the state where it has these values of these state variables.
 
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  • #15
Yes, you only have to wait until everything is in thermal equilibrium. On its way to thermal equilibrium entropy is never decreasing (H-theorem).
 
  • #16
I think I'm getting there.
What the book really shows is that every reversible cyclic process has entropy change zero.
That obviously means that the entropy change between two states can be calculated using any reversible path between these states and the result will be the same. So far everything is clear.

The step I need is that the only explanation for the above is that the system's entropy is a state function.
So it doesn't really matter that only reversible paths are considered.
Does that sound right?

With this step I can conclude that an irreversible process between the same two states will also have the same entropy change.
 
  • #17
Philip Koeck said:
I think I'm getting there.
What the book really shows is that every reversible cyclic process has entropy change zero.
That obviously means that the entropy change between two states can be calculated using any reversible path between these states and the result will be the same. So far everything is clear.

The step I need is that the only explanation for the above is that the system's entropy is a state function.
So it doesn't really matter that only reversible paths are considered.
Does that sound right?

With this step I can conclude that an irreversible process between the same two states will also have the same entropy change.
How are you defining the entropy change of a process?
 
  • #18
Chestermiller said:
How are you defining the entropy change of a process?
The whole discussion in the book starts with the Carnot cycle, so the entropy change during a reversible isotherm is Q/T, whereas for the adiabatic processes it's zero.
 
  • #19
Philip Koeck said:
The whole discussion in the book starts with the Carnot cycle, so the entropy change during a reversible isotherm is Q/T, whereas for the adiabatic processes it's zero.
How do you define it for an arbitrary process?
 
  • #20
Philip Koeck said:
With this step I can conclude that an irreversible process between the same two states will also have the same entropy change.

A technicality about "irreversible process":
Is a "process" defined as physical phenomena that can be represented as a path on a P-V diagram? If so, then is "free expansion" a process? If a gas is confined to one side of a cylinder by a partition and the partition is removed then as the gas expands without being in equilibrium, does the gas have a defined volume and pressure?
 
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  • #21
A quotation from "Extended Irreversible Thermodynamics" by Jou, Casas-Vazquez and Lebon

Using "CIT" to mean "classical irreversible thermodynamics":

The fundamental hypothesis underlying CIT is that of local equilibrium. It postulates that the local and instantaneous relations between the thermal and mechanical properties of a physical system are the same as for a uniform system at equilibrium. It assumes that the system under study can be mentally split into a series of cells sufficiently large to allow them to be treated as macroscopic thermodynamic systems, but sufficiently small that equilibrium is very close to being realized in each cell.
 
  • #22
Chestermiller said:
How do you define it for an arbitrary process?
As I see it, dS is always dQ/T for a reversible process. For an irreversible process it's not defined or at least can't be calculated directly.
 
  • #23
Stephen Tashi said:
A technicality about "irreversible process":
Is a "process" defined as physical phenomena that can be represented as a path on a P-V diagram? If so, then is "free expansion" a process? If a gas is confined to one side of a cylinder by a partition and the partition is removed then as the gas expands without being in equilibrium, does the gas have a defined volume and pressure?
Yes, I see that problem with processes where p and V are not defined, because they happen too quickly for example. They have two end-points on the pV-diagram but nothing in between so to say. That should mean they can't be approximated by reversible processes.
 
  • #24
Philip Koeck said:
As I see it, dS is always dQ/T for a reversible process. For an irreversible process it's not defined or at least can't be calculated directly.
It can be calculated directly for an irreversible process, but it isn't as simple a calculation as one might think. It involves solving the complicated partial differential equations in space and time within the system that include the transport processes of viscous fluid dynamics and heat conduction. This enables one to calculate the local rate of entropy generation within the system and to integrate that over the volume of the system. For the basics of how this is done, see Chapter 11, Transport Phenomena, Bird, Stewart, and Lightfoot, Problem 11D.1.

For pure irreversible heat conduction problems, say involving steady state conduction in a rod or even transient heat conduction in a rod, the calculation is much more straightforward to apply in practice, and I can show how it is done, if anyone is interested.
 
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  • #25
Every time you predict the direction of a constant-pressure chemical reaction or other process with Gibbs free energy arguments, you assume that entropy is a state function. Otherwise the system could have more than one value of S and G in the same apparent macroscopic state.
 
  • #26
Chestermiller said:
For pure irreversible heat conduction problems, say involving steady state conduction in a rod or even transient heat conduction in a rod, the calculation is much more straightforward to apply in practice, and I can show how it is done, if anyone is interested.
Yes, that would be interesting.
In basic textbooks irreversible heat exchange is often treated as if it was reversible without further explanation.
 
  • #27
Consider the irreversible process of steady state heat conduction occurring along a rod in contact with a reservoir of temperature ##T_H## at x = 0 and a second reservoir of temperature ##T_C## at x = L. The temperature profile along the rod is linear, and given by $$T=T_H-(T_H-T_C)\frac{x}{L}$$and the heat flux q along the rod is constant, and given by: $$q=-k\frac{dT}{dx}=k\frac{(T_H-T_C)}{L}$$ where k is the thermal conductivity of the rod metal.

The temperature profile has been determined by satisfying the differential heat balance equation, given by:
$$k\frac{d^2T}{dx^2}=0$$
The differential entropy balance equation on the rod can be obtained by dividing the differential heat balance equation by the absolute temperature, to yield:
$$\frac{k}{T}\frac{d^2T}{dx^2}=\frac{d}{dx}\left(\frac{k}{T}\frac{dT}{dx}\right)+\frac{k}{T^2}\left(\frac{dT}{dx}\right)^2=0$$or, equivalently $$-\frac{d}{dx}\left(\frac{q}{T}\right)+\frac{q^2}{kT^2}=0$$The second term in this equation represents physically the local rate of entropy generation per unit volume within the rod. If we integrate this equation between x = 0 and x = L, we obtain: $$\frac{q_H}{T_H}-\frac{q_C}{T_C}+\frac{q^2}{kT_HT_C}L=0\tag{1}$$where ##q_H=q_C=q##. The first term on the LHS represents the rate of entropy entering the rod per unit area at x = 0, and the second term represents the rate of entropy exiting the rod per unit area at x = L. The coefficient of L in the third term is positive definite, and represents the average rate of entropy generation per unit volume within the rod. Since the rod is operating at steady state, we know that the entropy of the rod is not changing with time. So ##L\frac{dS}{dt}=0##, where dS/dt is the average rate of change of entropy per unit volume of the rod. So we can write:
$$0=L\frac{dS}{dt}\gt \frac{q_H}{T_H}-\frac{q_C}{T_C}=-\frac{q^2}{kT_HT_C}L\tag{2}$$Eqn. 2 is equivalent to the Clausius inequality applied to the irreversible heat conduction process in the rod.

Note that, in this very simple case, we have been able to precisely determine the rate of entropy generation within the system. Note also that, Eqn. 1 tells us that the rate of entropy exiting the rod at x = L is just equal to the rate of entropy entering the rod at x = 0, plus the total rate of entropy generation within the rod.
 
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  • #28
Chestermiller said:
Consider the irreversible process of steady state heat conduction occurring along a rod in contact with a reservoir of temperature ##T_H## at x = 0 and a second reservoir of temperature ##T_C## at x = L. The temperature profile along the rod is linear, and given by $$T=T_H-(T_H-T_C)\frac{x}{L}$$and the heat flux q along the rod is constant, and given by: $$q=-k\frac{dT}{dx}=k\frac{(T_H-T_C)}{L}$$ where k is the thermal conductivity of the rod metal.

The temperature profile has been determined by satisfying the differential heat balance equation, given by:
$$k\frac{d^2T}{dx^2}=0$$
The differential entropy balance equation on the rod can be obtained by dividing the differential heat balance equation by the absolute temperature, to yield:
$$\frac{k}{T}\frac{d^2T}{dx^2}=\frac{d}{dx}\left(\frac{k}{T}\frac{dT}{dx}\right)+\frac{k}{T^2}\left(\frac{dT}{dx}\right)^2=0$$or, equivalently $$-\frac{d}{dx}\left(\frac{q}{T}\right)+\frac{q^2}{kT^2}=0$$The second term in this equation represents physically the local rate of entropy generation per unit volume within the rod. If we integrate this equation between x = 0 and x = L, we obtain: $$\frac{q_H}{T_H}-\frac{q_C}{T_C}+\frac{q^2}{kT_HT_C}L=0\tag{1}$$where ##q_H=q_C=q##. The first term on the LHS represents the rate of entropy entering the rod per unit area at x = 0, and the second term represents the rate of entropy exiting the rod per unit area at x = L. The coefficient of L in the third term is positive definite, and represents the average rate of entropy generation per unit volume within the rod.
Just some comments and questions to start with.

Your equation (1) can be rewritten as $$T_C = T_H - \frac{q L}{k}$$
This also follows directly by combining the first two equations in your text and then letting x = L.
Isn't equation (1) just another way of writing the temperature distribution then?
But maybe that's as it should be.

When you interpret equation (1) you say that the first term is the entropy entering the rod at x=0, for example.
Q is not transported into the rod in a reversible process in this case, so I don't understand how you can make this interpretation.
 
  • #29
Philip Koeck said:
Just some comments and questions to start with.

Your equation (1) can be rewritten as $$T_C = T_H - \frac{q L}{k}$$
This also follows directly by combining the first two equations in your text and then letting x = L.
Isn't equation (1) just another way of writing the temperature distribution then?
But maybe that's as it should be.
Yes, this equation is correct. So?

When you interpret equation (1) you say that the first term is the entropy entering the rod at x=0, for example.
Q is not transported into the rod in a reversible process in this case, so I don't understand how you can make this interpretation.
The difference between reversible and irreversible is not how entropy is transported into- and out of the system at its boundary with its surroundings. That transport is always described by the same equation (q/T). The difference between reversible and irreversible is that, in an irreversible process, entropy is also generated within the system and in a reversible process, it is not.
 
  • #30
Chestermiller said:
The difference between reversible and irreversible is that, in an irreversible process, entropy is also generated within the system and in a reversible process, it is not.
I'll try to apply this idea to the free expansion of a gas, versus the reversible isothermal expansion.

In a free expansion the entropy change of the system is exactly the same as for a reversible isothermal expansion at the same temperature. The difference between the two processes seems to lie in what happens in the surroundings. This is what you see in summary, so to say.

In fact entropy is generated inside the system during the free expansion, whereas entropy flows from the surroundings to the system in a reversible isothermal expansion.
So, although the entropy change in the system has the same value for both processes they are fundamentally different.

Does that sound about right?
 
  • #31
Philip Koeck said:
I'll try to apply this idea to the free expansion of a gas, versus the reversible isothermal expansion.

In a free expansion the entropy change of the system is exactly the same as for a reversible isothermal expansion at the same temperature. The difference between the two processes seems to lie in what happens in the surroundings. This is what you see in summary, so to say.

In fact entropy is generated inside the system during the free expansion, whereas entropy flows from the surroundings to the system in a reversible isothermal expansion.
So, although the entropy change in the system has the same value for both processes they are fundamentally different.

Does that sound about right?
It has nothing to do with the surroundings and everything to do with the system. In the alternative reversible process, you need to add heat reversibly and isothermally to the system, so $$Q_{rev}=nRT\ln{(V_2/V_1)}$$and$$\Delta S=\frac{Q_{rev}}{T}=nR\ln{(V_2/V_1)}$$
In the irreversible free expansion process between the same two states, $$Q=0$$and $$\Delta S=nR\ln{(V_2/V_1)}=\sigma+\frac{Q}{T}=\sigma$$where ##\sigma## is the entropy generated within the system during the irreversible free expansion as a result of viscous dissipation.

To give you a feel for how viscous dissipation comes into play in generating entropy, consider the following homework problem I've dreamed up for you: You have an incompressible viscous Newtonian fluid situated between two infinite insulated parallel plates in which the bottom plate is stationary and the upper plate is moving tangentially with constant velocity V. The velocity profile within the fluid between the plates is given by $$v=V\frac{y}{h}$$where h is the distance between the plates. The transient thermal energy balance on the fluid is given by$$\frac{dU}{dt}=\dot{W}$$ (where ##\dot{W}## is the rate at which the surroundings do work on the system), or, per unit volume, $$\rho C \frac{dT}{dt}=\mu\left(\frac{V}{h}\right)^2$$where ##\rho## is the fluid density, C is its heat capacity, and ##\mu## is its viscosity. If we assume that the fluid density, heat capacity, and viscosity are independent of temperature, and that the initial temperature (at time t = 0) is ##T_0##, what is the temperature at time T?

If we divide this heat balance equation by the temperature T, we obtain the entropy balance equation on the fluid. What is that entropy balance equation (in terms of the entropy per unit volume), and, by integrating that equation, what is the change in entropy per unit volume of the fluid between time t = 0 and time t?

I'll continue after you complete this.

Notice that we are doing all this without even looking at an alternative reversible path yet.
 
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  • #32
Chestermiller said:
To give you a feel for how viscous dissipation comes into play in generating entropy, consider the following homework problem I've dreamed up for you: You have an incompressible viscous Newtonian fluid situated between two infinite insulated parallel plates in which the bottom plate is stationary and the upper plate is moving tangentially with constant velocity V. The velocity profile within the fluid between the plates is given by $$v=V\frac{y}{h}$$where h is the distance between the plates. The transient thermal energy balance on the fluid is given by$$\frac{dU}{dt}=\dot{W}$$ (where ##\dot{W}## is the rate at which the surroundings do work on the system), or, per unit volume, $$\rho C \frac{dT}{dt}=\mu\left(\frac{V}{h}\right)^2$$where ##\rho## is the fluid density, C is its heat capacity, and ##\mu## is its viscosity. If we assume that the fluid density, heat capacity, and viscosity are independent of temperature, and that the initial temperature (at time t = 0) is ##T_0##, what is the temperature at time T?

If we divide this heat balance equation by the temperature T, we obtain the entropy balance equation on the fluid. What is that entropy balance equation (in terms of the entropy per unit volume), and, by integrating that equation, what is the change in entropy per unit volume of the fluid between time t = 0 and time t?

For the temperature I just get a linear increase: $$T(t)=T_0+\frac{μV^2t}{ρCh^2}$$

For the entropy increase per unit volume I would integrate dU/T from T0 to T(t), which gives me:
$$ΔS=ρC⋅ln\frac{T(t)}{T_0}$$ with the above expression for T(t).
Does that look right?
 
  • #33
Philip Koeck said:
For the temperature I just get a linear increase: $$T(t)=T_0+\frac{μV^2t}{ρCh^2}$$

For the entropy increase per unit volume I would integrate dU/T from T0 to T(t), which gives me:
$$ΔS=ρC⋅ln\frac{T(t)}{T_0}$$ with the above expression for T(t).
Does that look right?
These results are both correct. I was also looking for the equation for the instantaneous rate of change of entropy per unit volume which is $$\frac{dS}{dt}=\rho C \frac{d\ln{T}}{dt}=\frac{\mu}{T}\left(\frac{V}{h}\right)^2$$This equation tells us that the instantaneous rate of change of entropy per unit volume during this process is just equal to the rate of entropy generation (since there is no entropy transfer across the boundaries of the system), which, in turn, is equal to the rate of viscous dissipation divided by the absolute temperature.

If we eliminate the product of density and heat capacity (##\rho C##) from your two equations, we obtain the average rate of entropy change (and average rate of entropy generation) between time zero and time t:: $$\frac{\Delta S}{t}=\frac{\mu}{T_{lm}}\left(\frac{V}{h}\right)^2$$where ##T_{lm}## is called the logarithmic mean temperature of. the system over the time interval: $$T_{lm}=\frac{(T-T_0)}{\ln{(T/T_0)}}$$It turns out mathematically that the log-mean temperature is very close to the arithmetic mean of the temperatures for all but very large differences in the temperatures. We know this from engineering experience with temperature differences in the design of heat exchangers.

These results illustrate the intimate relationship between the rate of viscous dissipation and the rate of entropy generation within a system experiencing an irreversible process. This same concept applies directly to irreversible process of free expansion, although, in free expansion, the gas velocities and viscous dissipation rates are much more complicated to calculate (and typically require the use of computational fluid dynamics). Fortunately, in the case of free expansion, we can easily establish the final state from macroscopic thermodynamics considerations, which allow us to circumvent the detailed fluid mechanics calculations. This, of course, is, in most cases, not possible.

Comments?

I have another example to offer after this one, which involves entropy generation from both viscous dissipation and finite heat conduction simultaneously.
 
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  • #34
Chestermiller said:
I have another example to offer after this one, which involves entropy generation from both viscous dissipation and finite heat conduction simultaneously.
Yes, that would be interesting.
 
  • #35
Chestermiller said:
These results are both correct. I was also looking for the equation for the instantaneous rate of change of entropy per unit volume which is $$\frac{dS}{dt}=\rho C \frac{d\ln{T}}{dt}=\frac{\mu}{T}\left(\frac{V}{h}\right)^2$$This equation tells us that the instantaneous rate of change of entropy per unit volume during this process is just equal to the rate of entropy generation (since there is no entropy transfer across the boundaries of the system), which, in turn, is equal to the rate of viscous dissipation divided by the absolute temperature.

If we eliminate the product of density and heat capacity (##\rho C##) from your two equations, we obtain the average rate of entropy change (and average rate of entropy generation) between time zero and time t:: $$\frac{\Delta S}{t}=\frac{\mu}{T_{lm}}\left(\frac{V}{h}\right)^2$$where ##T_{lm}## is called the logarithmic mean temperature of. the system over the time interval: $$T_{lm}=\frac{(T-T_0)}{\ln{(T/T_0)}}$$It turns out mathematically that the log-mean temperature is very close to the arithmetic mean of the temperatures for all but very large differences in the temperatures. We know this from engineering experience with temperature differences in the design of heat exchangers.

These results illustrate the intimate relationship between the rate of viscous dissipation and the rate of entropy generation within a system experiencing an irreversible process. This same concept applies directly to irreversible process of free expansion, although, in free expansion, the gas velocities and viscous dissipation rates are much more complicated to calculate (and typically require the use of computational fluid dynamics). Fortunately, in the case of free expansion, we can easily establish the final state from macroscopic thermodynamics considerations, which allow us to circumvent the detailed fluid mechanics calculations. This, of course, is, in most cases, not possible.

Comments?
I've started thinking about the mechanism of entropy generation in your example and also for free expansion.
In your example the upper plate is moved against the friction due to the viscosity of the liquid. The work that's done is converted to inner energy and increases the temperature of the liquid and generates entropy.

In free expansion there's no work being done from the outside, no heat is transferred.
The average kinetic energy of these molecules is unchanged since the inner energy is constant (at least for an ideal gas).
When the gas expands it does no work on the surroundings.
The only thing that really happens is that molecules fly around chaotically and move further from each other.
For a real gas there's also viscous friction between the molecules, is that right?

For a real gas: The expansion should increase the potential energy of the molecules since there are attractive forces between them. I'm wondering about the energy balance in that case. Which energy decreases?
That should mean that the temperature has to decrease a bit during the free expansion of a real gas.

For an ideal gas: There is no potential energy and also no friction between the molecules. The only reason I can see for the entropy increase is the expansion.
 

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