T-S Diagram for an Irreversible Process

[Robert Davidson]

I recently came across this concerning a T-S diagram: https://learnthermo.com/T1-tutorial/ch07/lesson-C/pg04.php

The author states that the area under the curve represents the heat transferred reversibly. The author didn't state what actual process the curve represents, but I couldn't think of a reversible process where there is an increase in entropy associated with a decrease in temperature. Is there any such process?

In the slide that follows the author uses the same diagram to discuss the case of an irreversible process and concludes that, for the irreversible process, the area under the curve does not equal the heat transfer, but the sum of the heat transfer and entropy generated. My question is this; since the process is irreversible it seems to me the temperature of the system, as a function of entropy would not be defined. Is it legitimate to use the same curve to represent the reversible and irreversible process as the author did? It seems to me all we know is the entropy at state 2 will be the same for the reversible and irreversible process, but the temperature at state 2 need not be.

Is it even legitimate to represent the irreversible process as a continuous (differentiable and integrable) function since the system is not in equilibrium going from state 1 and 2? There seems to be some disagreement among people as to whether or not you can even draw a T-S diagram for an irreversible process.

 

[Chestermiller]

I recently came across this concerning a T-S diagram: https://learnthermo.com/T1-tutorial/ch07/lesson-C/pg04.php

The author states that the area under the curve represents the heat transferred reversibly. The author didn't state what actual process the curve represents, but I couldn't think of a reversible process where there is an increase in entropy associated with a decrease in temperature. Is there any such process?

Sure. If a gas is expanded isothermally and reversibly, we need to add heat to the process tp keep its temperature constant, and its entropy increases. All we need to do is add a little less heat to the process, and its temperature will decrease while its entropy increases. Another example is an isothermal expansion followed by an adiabatic expansion.

In the slide that follows the author uses the same diagram to discuss the case of an irreversible process and concludes that, for the irreversible process, the area under the curve does not equal the heat transfer, but the sum of the heat transfer and entropy generated.

I'm not able to see that next slide at the link. How do I get it?

 

[Robert Davidson]

In the upper right area, click on "next page"

 

[Chestermiller]

I recently came across this concerning a T-S diagram: https://learnthermo.com/T1-tutorial/ch07/lesson-C/pg04.php

In the slide that follows the author uses the same diagram to discuss the case of an irreversible process and concludes that, for the irreversible process, the area under the curve does not equal the heat transfer, but the sum of the heat transfer and entropy generated. My question is this; since the process is irreversible it seems to me the temperature of the system, as a function of entropy would not be defined. Is it legitimate to use the same curve to represent the reversible and irreversible process as the author did? It seems to me all we know is the entropy at state 2 will be the same for the reversible and irreversible process, but the temperature at state 2 need not be.

Is it even legitimate to represent the irreversible process as a continuous (differentiable and integrable) function since the system is not in equilibrium going from state 1 and 2? There seems to be some disagreement among people as to whether or not you can even draw a T-S diagram for an irreversible process.

None of the next slide is valid. First of all, the author has mis-stated and mis-applied the Clausius relationship. For an irreversible process, it should read $$\Delta S=\int{\frac{dq}{T_B}}+\sigma$$where dq is the differential heat transferred to the system at its boundary with the surroundings and $T_B$ is the boundary temperature (not the system temperature), which is determinate. The system temperature is often not even uniform for an irreversible process. So what temperature could the guy mean?

In my judgment, it is typically not valid to represent an irreversible process using differentials. But it might be valid if you are assuming that local equilibrium can exist spatially in a system undergoing an irreversible process. You could then integrate over the volume of the system to get the total entropy at a given state. But, in the Clausius relationship, you would still need to use the temperature at the boundary.

To get a really good understanding of all this,, including explicit expressions for local entropy generation rate in terms of temperature gradients, velocity gradients, concentration gradients, and transport properties (thermal conductivity, viscosity, and diffusion coefficient), in a substance experiencing an irreversible process, see the following (WONDERFUL) reference:

Problem 11.D.1, Equation of change of entropy, Chapter 11, Transport Phenomena by Bird, Stewart, and Lightfoot.

This reference is extremely enlightening. The continuum mechanics guys have advanced this approach even further, and include cases where the heat flux and boundary temperature can vary with time and location on the boundary.

 

[Robert Davidson]

Sure. If a gas is expanded isothermally and reversibly, we need to add heat to the process tp keep its temperature constant, and its entropy increases. All we need to do is add a little less heat to the process, and its temperature will decrease while its entropy increases. Another example is an isothermal expansion followed by an adiabatic expansion.

I'm not able to see that next slide at the link. How do I get it?

Thanks for your reply.

I guess what I had in mind is a single reversible process where the function T(S) involves a temperature decrease and entropy increase as shown in the link. Such would not be the case for reversible isothermal, isobaric, isochoric, or adiabatic process.

For your first example, the reversible isothermal process is altered by adding less heat. But then isn't it no longer a reversible isothermal process? Also, during the period where less heat is added wouldn't the rate of entropy increase decrease?

The second example involves a combination of two processes, each of which by itself does not involve a decrease in temperature accompanied by an increase in entropy.

I am still thinking about the rest of your answer.

 

[Chestermiller]

Thanks for your reply.

I guess what I had in mind is a single reversible process where the function T(S) involves a temperature decrease and entropy increase as shown in the link. Such would not be the case for reversible isothermal, isobaric, isochoric, or adiabatic process.

For your first example, the reversible isothermal process is altered by adding less heat. But then isn't it no longer a reversible isothermal process? Also, during the period where less heat is added wouldn't the rate of entropy increase decrease?

I was thinking of a non-isothermal process, using a continuous sequence of reservoirs at decreasing temperatures. This would allow it to be reversible. As long as the temperature decrease did not offset the entropy increase, it would still involve an increase in entropy and a decrease in temperature.

The second example involves a combination of two processes, each of which by itself does not involve a decrease in temperature accompanied by an increase in entropy.

So what?

 

[Robert Davidson]

Could you please explain what you mean by "as long as the temperature decrease did not offset the entropy increase". In terms of our other conversation, if we used a continuous sequence of reservoirs at decreasing temperature to bring the temperature of our body H from TH to (TH+TC)/2, wouldn't the change in entropy of body H be negative (ΔSH =C ln (TH+TC)/2TH)?

 

[Chestermiller]

Could you please explain what you mean by "as long as the temperature decrease did not offset the entropy increase". In terms of our other conversation, if we used a continuous sequence of reservoirs at decreasing temperature to bring the temperature of our body H from TH to (TH+TC)/2, wouldn't the change in entropy of body H be negative (ΔSH =C ln (TH+TC)/2TH)?

I'm not talking about a Carnot cycle type process. I'm just talking about starting with a plain old vanilla isothermal reversible expansion. For such an expansion, the entropy increases while the temperature remains constant. So if you do the same expansion, but, this time remove just a small amount of heat (using a sequence of reservoirs to keep the process reversible), you will still get an increase in entropy, but this time with a small decrease in temperature.

 

[Robert Davidson]

I'm not talking about a Carnot cycle type process. I'm just talking about starting with a plain old vanilla isothermal reversible expansion. For such an expansion, the entropy increases while the temperature remains constant. So if you do the same expansion, but, this time remove just a small amount of heat (using a sequence of reservoirs to keep the process reversible), you will still get an increase in entropy, but this time with a small decrease in temperature.

OK. So you still get an increase in entropy with the removal of a small amount of heat and slight lowering of temperature because of the cumulative increase in entropy prior to the reduction in heat, as long as the reduction in heat is not too great to turn the change in entropy negative. Kind of like "entropy inertia".

 

[Chestermiller]

OK. So you still get an increase in entropy with the removal of a small amount of heat and slight lowering of temperature because of the cumulative increase in entropy prior to the reduction in heat, as long as the reduction in heat is not too great to turn the change in entropy negative. Kind of like "entropy inertia".

Yes. You can take the heat added all the way down to the isentropic limit before the change in entropy becomes zero.

 

[Robert Davidson]

Got it. Thanks.

 

[Robert Davidson]

None of the next slide is valid. First of all, the author has mis-stated and mis-applied the Clausius relationship. For an irreversible process, it should read $$\Delta S=\int{\frac{dq}{T_B}}+\sigma$$where dq is the differential heat transferred to the system at its boundary with the surroundings and $T_B$ is the boundary temperature (not the system temperature), which is determinate. The system temperature is often not even uniform for an irreversible process. So what temperature could the guy mean?

In my judgment, it is typically not valid to represent an irreversible process using differentials. But it might be valid if you are assuming that local equilibrium can exist spatially in a system undergoing an irreversible process. You could then integrate over the volume of the system to get the total entropy at a given state. But, in the Clausius relationship, you would still need to use the temperature at the boundary.

To get a really good understanding of all this,, including explicit expressions for local entropy generation rate in terms of temperature gradients, velocity gradients, concentration gradients, and transport properties (thermal conductivity, viscosity, and diffusion coefficient), in a substance experiencing an irreversible process, see the following (WONDERFUL) reference:

Problem 11.D.1, Equation of change of entropy, Chapter 11, Transport Phenomena by Bird, Stewart, and Lightfoot.

This reference is extremely enlightening. The continuum mechanics guys have advanced this approach even further, and include cases where the heat flux and boundary temperature can vary with time and location on the boundary.

I agree with you when you say "So what temperature could the guy mean?" I don't see how he can come up with the function T(S) for an irreversible process.

Regarding your reference, I don't have (free) access to it, but statistical mechanics (statistical thermodynamics) is above my head. Never took a course in it. Only macrothermodynamics.

 

[Chestermiller]

I agree with you when you say "So what temperature could the guy mean?" I don't see how he can come up with the function T(S) for an irreversible process.

Regarding your reference, I don't have (free) access to it, but statistical mechanics (statistical thermodynamics) is above my head. Never took a course in it. Only macrothermodynamics.

It’s not statistical mechanics. Strictly classical thermo, fluid dynamics, and heat transfer.

Later, I'll come back and write down an equation summarizing the outcome of the analysis for a closed system.

 

[Chestermiller]

For a closed system, $$\frac{dS}{dt}=-\int_A{\frac{\mathbf{q}\centerdot \mathbf{n}}{T}dA}+\int_V{\dot{\sigma} dV}$$ where $\mathbf{q}$ is the heat flux vector, $\mathbf{n}$ is a unit outwardly directed normal to the (moving and deforming) interface A between the system and surroundings, and $\dot{\sigma}$ is the rate of generation of entropy per unit volume within the system volume V at time t. BSL give an explicit expression for $\dot{\sigma}$ in terms of the local temperature gradients, velocity gradients, thermal conductivity, and viscosity within the system.

 

[Robert Davidson]

Yes. You can take the heat added all the way down to the isentropic limit before the change in entropy becomes zero.

Just so I understand correctly. See the figures below.

Fig 1 shows what I originally thought you meant when you said "Another example is an isothermal expansion followed by an adiabatic expansion". You subsequently correctly surmised I was thinking you were talking about the expansion portion of a Carnot cycle. T(S) here is not continuously differentiable, which is why I balked.

Correct me if I'm wrong, but Fig 2 shows what I think you mean by "You can take the heat added all the way down to the isentropic limit before the change in entropy becomes zero." My original concern was for the counter intuitive idea that a decrease in temperature could produce an increase in entropy. Your example shows that, for the system, it can (though entropy is decelerating in the transition between the isothermal and isentropic limits). However, would I be correct in saying that the overall change in entropy of the system plus the change in entropy of the "continuous sequence of reservoirs" which are removing the heat from the system, is zero for the reversible process?

Fig 3 is essentially what was given in the link we are discussing. Is there a physical interpretation we can give for this curve?

 

[Chestermiller]

I was thinking more like this: For an isothermal reversible expansion, $PV^n=C$ with n=1, heat is added, entropy increases, and temperature is constant. For an adiabatic reversible expansion, $PV^n=C$ with $n = \kappa$ (say, 1.4), no heat is added, entropy is constant, and temperature decreases. Now, for, say, a polytropic expansion with $PV^n=C$ with $1<n<\kappa$, heat is added (but not as much as in the isothermal case), entropy increases (but not as much as in the isothermal case), and temperature decreases (but less than in the isentropic case). So the isothermal case and the isentropic case are the limiting cases I would consider for polytropic processes featuring an increase in entropy and a decrease in temperature. A polytropic process of this type would be an example of one with a graph like in the link. Of course, it all depends on how one controls Q as a function of V, and there are many other examples that are possible.

 

[Robert Davidson]

Yes. You can take the heat added all the way down to the isentropic limit before the change in entropy becomes zero.

Just so I understand correctly. See the figures below.

Fig 1 shows what I originally thought you meant when you said "Another example is an isothermal expansion followed by an adiabatic expansion". You subsequently correctly surmised I was thinking you were talking about the expansion portion of a Carnot cycle. T(S) here is not continuously differentiable, which is why I balked.

Correct me if I'm wrong, but Fig 2 shows what I think you mean by "You can take the heat added all the way down to the isentropic limit before the change in entropy becomes zero." My original concern was for the counter intuitive idea that a decrease in temperature could produce an increase in entropy. Your example shows that, for the system, it can (though entropy is decelerating in the transition between the isothermal and isentropic limits). However, would I be correct in saying that the overall change in entropy of the system plus the change in entropy of the "continuous sequence of reservoirs" which are removing the heat from the system, is positive?

Fig 3 is essentially what was given in the link we are discussing. Can we interpret this as the gradual removal of heat starting with an isentropic process and ending with an isothermal process?

I was thinking more like this: For an isothermal reversible expansion, $PV^n=C$ with n=1, heat is added, entropy increases, and temperature is constant. For an adiabatic reversible expansion, $PV^n=C$ with $n = \kappa$ (say, 1.4), no heat is added, entropy is constant, and temperature decreases. Now, for, say, a polytropic expansion with $PV^n=C$ with $1<n<\kappa$, heat is added (but not as much as in the isothermal case), entropy increases (but not as much as in the isothermal case), and temperature decreases (but less than in the isentropic case). So the isothermal case and the isentropic case are the limiting cases I would consider for polytropic processes featuring an increase in entropy and a decrease in temperature. A polytropic process of this type would be an example of one with a graph like in the link. Of course, it all depends on how one controls Q as a function of V, and there are many other examples that are possible.

Excellent! So we are talking about a polytropic process with n being a variable (i.e. constantly changing) in the range between 1 (isothermal limit) and k (isentropic limit, e.g., 1.4 for ideal gas)? But doesn't that apply to Fig 2 since Fig 3 (the link) starts at the isentropic limit?

 

[Chestermiller]

Just so I understand correctly. See the figures below.

Fig 1 shows what I originally thought you meant when you said "Another example is an isothermal expansion followed by an adiabatic expansion". You subsequently correctly surmised I was thinking you were talking about the expansion portion of a Carnot cycle. T(S) here is not continuously differentiable, which is why I balked.

Correct me if I'm wrong, but Fig 2 shows what I think you mean by "You can take the heat added all the way down to the isentropic limit before the change in entropy becomes zero." My original concern was for the counter intuitive idea that a decrease in temperature could produce an increase in entropy. Your example shows that, for the system, it can (though entropy is decelerating in the transition between the isothermal and isentropic limits). However, would I be correct in saying that the overall change in entropy of the system plus the change in entropy of the "continuous sequence of reservoirs" which are removing the heat from the system, is positive?

Fig 3 is essentially what was given in the link we are discussing. Can we interpret this as the gradual removal of heat starting with an isentropic process and ending with an isothermal process?

Excellent! So we are talking about a polytropic process with n being a variable (i.e. constantly changing) in the range between 1 (isothermal limit) and k (isentropic limit, e.g., 1.4 for ideal gas)? But doesn't that apply to Fig 2 since Fig 3 (the link) starts at the isentropic limit?

No, n is not changing. It is constant for the entire expansion. It is just a fixed parameter for each expansion.

 

[Chestermiller]

I have a set of calculations for you to try. Derive the equations for an arbitrary polytropic expansion, deriving expressions for P, T, Q, and S as functions of V. Carry out calculations for a diatomic ideal gas with a specified starting state of P, V, and T, and for values of n = 1, 1.2, and 1.4, in each case with values of V running from the initial volume to twice the initial volume. Then, make plots of Q(V) vs V and T(V) vs S(V) showing the results for the 3 cases n = 1, n=1.2, and 1.4. I think that you will find these two plots to be very enlightening.

 

[Robert Davidson]

No, n is not changing. It is constant for the entire expansion. It is just a fixed parameter for each expansion.

OK. but which figure (1 or 2) applies? Your example started with an i

I have a set of calculations for you to try. Derive the equations for an arbitrary polytropic expansion, deriving expressions for P, T, Q, and S as functions of V. Carry out calculations for a diatomic ideal gas with a specified starting state of P, V, and T, and for values of n = 1, 1.2, and 1.4, in each case with values of V running from the initial volume to twice the initial volume. Then, make plots of Q(V) vs V and T(V) vs S(V) showing the results for the 3 cases n = 1, n=1.2, and 1.4. I think that you will find these two plots to be very enlightening.

Before I even try to follow this, I need to go back to your previous answer, that n is a constant for the entire process. I'm afraid I'm losing you (please be patient) and starting to feel stupid. Fig 4 below shows, as I understand it, TS diagrams of different polytropic expansion processes with different, but constant values of n. None of them associates a decrease in temperature with an increase in entropy.

 

[Chestermiller]

OK. but which figure (1 or 2) applies? Your example started with an i

Before I even try to follow this, I need to go back to your previous answer, that n is a constant for the entire process. I'm afraid I'm losing you (please be patient) and starting to feel stupid. Fig 4 below shows, as I understand it, TS diagrams of different polytropic expansion processes with different, but constant values of n. None of them associates a decrease in temperature with an increase in entropy.

View attachment 267594

The graph doesn't show any cases between n = 1 and n = gamma.

 

[Robert Davidson]

Oops! Sorry. Didn't look at it closely enough. Probably something like added red curve below.

 

[Chestermiller]

We have $$\ln{P}+n\ln{V}=\ln{C}$$
and, from the ideal gas law, $$\ln{P}+\ln{V}=\ln{(mR)}+\ln{T}$$where m is the number of moles of ideal gas. So,$$d\ln{P}+nd\ln{V}=0\tag{1}$$and$$d\ln{P}+d\ln{V}=d\ln{T}\tag{2}$$Subtracting Eqn.1 from Eqn. 2 then yields $$d\ln{T}=-(n-1)d\ln{V}\tag{3}$$Also, for an ideal gas, we have for the entropy variation:
$$dS=m(C_vd\ln{T}+Rd\ln{V})$$Substituting Eqn. 3 into this relationship then yields: $$d\left(\frac{S}{mR}\right)=\left[\frac{1}{(\gamma-1)}-\frac{1}{(n-1)}\right]d\ln{T}$$For $\gamma=1.4$ and $n=1.2$, this gives: $$\frac{d\ln{T}}{d\left(\frac{S}{mR}\right)}=-0.4$$

 

[Chestermiller]

Oops! Sorry. Didn't look at it closely enough. Probably something like added red curve below. View attachment 267624

Nice. The polytropic equation is just one example of an expansion path where dT/dS is negative.

 

[Robert Davidson]

We have $$\ln{P}+n\ln{V}=\ln{C}$$
and, from the ideal gas law, $$\ln{P}+\ln{V}=\ln{(mR)}+\ln{T}$$where m is the number of moles of ideal gas. So,$$d\ln{P}+nd\ln{V}=0\tag{1}$$and$$d\ln{P}+d\ln{V}=d\ln{T}\tag{2}$$Subtracting Eqn.1 from Eqn. 2 then yields $$d\ln{T}=-(n-1)d\ln{V}\tag{3}$$Also, for an ideal gas, we have for the entropy variation:
$$dS=m(C_vd\ln{T}+Rd\ln{V})$$Substituting Eqn. 3 into this relationship then yields: $$d\left(\frac{S}{mR}\right)=\left[\frac{1}{(\gamma-1)}-\frac{1}{(n-1)}\right]d\ln{T}$$For $\gamma=1.4$ and $n=1.2$, this gives: $$\frac{d\ln{T}}{d\left(\frac{S}{mR}\right)}=-0.4$$

Wow. It took me a while, but I was able to follow your derivation. What I don't understand is how you determined all the mathematical manipulations that you needed (the "road map", so to speak) to demonstrate that dT/dS is negative. I couldn't have done it.

 

[Chestermiller]

Wow. It took me a while, but I was able to follow your derivation. What I don't understand is how you determined all the mathematical manipulations that you needed (the "road map", so to speak) to demonstrate that dT/dS is negative. I couldn't have done it.

I've been playing around with the first few equations in terms of natural logs for a while now as an easy way that I dreamt up for getting the exponents in the adiabatic relationships between T and V, and T and P. And, of course, I knew the equation for $\Delta S$ as a function of ln(T/To) and ln(V/Vo). So the rest was easy.

 

[vanhees71]

Yes, and all this has to be taken with a big grain of salt. Whenever you have dimensionful arguments of a log you should think deeper!

 

[Chestermiller]

Yes, and all this has to be taken with a big grain of salt. Whenever you have dimensionful arguments of a log you should think deeper!

I was intimately familiar with all three of the equations that I used. Is there something wrong with my mathematical manipulations? I get the idea that you just don't like analyses in which the log of a dimensional quantity is used. Did you think I didn't know how to handle this?

 

[vanhees71]

I'm sure that you know how to handle this, and one should always do so!

 

[Robert Davidson]

I've been playing around with the first few equations in terms of natural logs for a while now as an easy way that I dreamt up for getting the exponents in the adiabatic relationships between T and V, and T and P. And, of course, I knew the equation for $\Delta S$ as a function of ln(T/To) and ln(V/Vo). So the rest was easy.

In short: Experience.