Confusion on entropy change calculations for irreversible process

[Chestermiller]

Yes.. exactly...

OK. We've looked at three systems so far (each of the two slabs individually and the combination of the two slabs) and we've seen what happens when we allow an irreversible equilibration of heat between the slabs. If we want to use this as a starting point for demonstrating the closed cycle version of the Clausius inequality, we need provide an additional process step to return anyone of these three systems to its original state. Which one do you want to consider (the simplest would be the combination of the two slabs)?

Chet

 

[Signature]

OK. We've looked at three systems so far (each of the two slabs individually and the combination of the two slabs) and we've seen what happens when we allow an irreversible equilibration of heat between the slabs. If we want to use this as a starting point for demonstrating the closed cycle version of the Clausius inequality, we need provide an additional process step to return anyone of these three systems to its original state. Which one do you want to consider (the simplest would be the combination of the two slabs)?

Chet

Ok.. We will go for the simplest i.e. the combination of the two slabs.

 

[Chestermiller]

Ok.. We will go for the simplest i.e. the combination of the two slabs.

OK. For the combination of the two slabs, we've already shown that, for the irreversible leg of the cycle,

$$\int {\frac{dq}{T_I}}=0$$
and
$$ΔS=mC\ln \left(\frac{(T_H+T_C)^2}{4T_HT_C}\right)$$

Now, to complete the cycle, we are going to return each of the two individual slabs reversibly to their original equilibrium states. Do you know what the combined change in entropy for this process is, or what the change in entropy for each of the slabs individually is for this process? Do you know what the integral of dq/T_I is for this process?

Chet

 

[Signature]

OK. For the combination of the two slabs, we've already shown that, for the irreversible leg of the cycle,

$$\int {\frac{dq}{T_I}}=0$$
and
$$ΔS=mC\ln \left(\frac{(T_H+T_C)^2}{4T_HT_C}\right)$$

Now, to complete the cycle, we are going to return each of the two individual slabs reversibly to their original equilibrium states. Do you know what the combined change in entropy for this process is, or what the change in entropy for each of the slabs individually is for this process? Do you know what the integral of dq/T_I is for this process?

Chet

For the initially hot slab to this process,
Initial temperature = (TH+TC)/2
Final temperature = TH

$$ΔS_H=mC\ln \left(\frac{2T_H}{T_H+T_C}\right)$$

$$\int {\frac{dq}{T_I}}=\frac{Q}{T_I}=mC\frac{T_H-T_C}{T_H+T_C}$$

For the initially cold slab
Initial temperature = (TH+TC)/2
Final temperature = TC

$$ΔS_C=mC\ln \left(\frac{2T_C}{T_H+T_C}\right)$$

$$\int {\frac{dq}{T_I}}=\frac{-Q}{T_I}=-mC\frac{T_H-T_C}{T_H+T_C}$$


If we take hot and cold slab as a system (isolated),

the combined change in entropy $$ΔS=mC\ln \left(\frac{2T_H}{T_H+T_C}\right)+mC\ln \left(\frac{2T_C}{T_H+T_C}\right)$$

The ΔS is less than zero.

For Ex., if we take TH as 700 K and TC as 300 K it will give ΔS = 0.3365mC-0.5108mC = -0.1743 mC <0

But for isolated system ΔS should be greater than or equal to zero.

And

$$\int {\frac{dq}{T_I}}=0$$ (since the combined system is not interacting with their surroundings)Now I want to know what mistake I have done in calculating entropy of isolated system.

Please also prove the Clausius Inequality theorem i.e. closed integral of δQ / T is less than or equal to zero with this example.

 

[Chestermiller]

For the initially hot slab to this process,
Initial temperature = (TH+TC)/2
Final temperature = TH
$$ΔS_H=mC\ln \left(\frac{2T_H}{T_H+T_C}\right)$$

For the initially hot slab in the return part of the cycle, this is correct.

For the initially cold slab
Initial temperature = (TH+TC)/2
Final temperature = TC
$$ΔS_C=mC\ln \left(\frac{2T_C}{T_H+T_C}\right)$$

For the initially cold slab in the return part of the cycle, this is correct.

the combined change in entropy $$ΔS=mC\ln \left(\frac{2T_H}{T_H+T_C}\right)+mC\ln \left(\frac{2T_C}{T_H+T_C}\right)$$

The ΔS is less than zero.

Yes, for the initially hot slab and the initially cold slab in the return part of the cycle, this is correct. For the hot slab and the cold slab over the entire cycle, if we add the entropy change for the irreversible equilibration part of the cycle (positive) with the entropy change in the reversible return part of the cycle (negative), they add up to zero (as they must for a cycle).

$$\int {\frac{dq}{T_I}}=\frac{-Q}{T_I}=-mC\frac{T_H-T_C}{T_H+T_C}$$

$$\int {\frac{dq}{T_I}}=\frac{Q}{T_I}=mC\frac{T_H-T_C}{T_H+T_C}$$

These equations for the hot slab and the cold slab in the reversible part of the cycle are incorrect. They should read
$$\int {\frac{dq}{T_I}}=ΔS_H$$
and
$$\int {\frac{dq}{T_I}}=ΔS_C$$
The integral of dq/T along a reversible path determines the change in entropy.

But for isolated system ΔS should be greater than or equal to zero.

In the reversible return part of the cycle, we are free to do whatever is necessary to return the two slabs to their original state. It certainly can't happen spontaneously. So we separate them, and return them each to their original states by contacting each of them with a continuous sequence of constant temperature reservoirs at slightly different temperatures. So the system is not isolated during the return part of the cycle.

Please also prove the Clausius Inequality theorem i.e. closed integral of δQ / T is less than or equal to zero with this example.

For the irreversible part of the cycle, we already showed that the integral of dq/T_I is zero. For the reversible (return) part of the cycle, we have now shown that the integral of dq/T_I is negative. So, for the entire cycle, the entropy change of our system is zero, and the integral of dq/T_I for our system is negative.

Chet

 

[Signature]

For the initially hot slab in the return part of the cycle, this is correct.For the initially cold slab in the return part of the cycle, this is correct.Yes, for the initially hot slab and the initially cold slab in the return part of the cycle, this is correct. For the hot slab and the cold slab over the entire cycle, if we add the entropy change for the irreversible equilibration part of the cycle (positive) with the entropy change in the reversible return part of the cycle (negative), they add up to zero (as they must for a cycle).


These equations for the hot slab and the cold slab in the reversible part of the cycle are incorrect. They should read
$$\int {\frac{dq}{T_I}}=ΔS_H$$
and
$$\int {\frac{dq}{T_I}}=ΔS_C$$
The integral of dq/T along a reversible path determines the change in entropy.


In the reversible return part of the cycle, we are free to do whatever is necessary to return the two slabs to their original state. It certainly can't happen spontaneously. So we separate them, and return them each to their original states by contacting each of them with a continuous sequence of constant temperature reservoirs at slightly different temperatures. So the system is not isolated during the return part of the cycle.

For the irreversible part of the cycle, we already showed that the integral of dq/T_I is zero. For the reversible (return) part of the cycle, we have now shown that the integral of dq/T_I is negative. So, for the entire cycle, the entropy change of our system is zero, and the integral of dq/T_I for our system is negative.

Chet

Nice sir. Now I understood the concept.. It's very good in having such interactions.

Now I need to get some idea on another example.

Suppose we are having an engine whose piston and cylinder housings are made of an imaginary material whose thermal diffusivity is zero so that the heat produced during combustion of fuel will not be absorbed by the piston and cylinder housings but only some amount of heat can be taken as work from the piston.

Will this engine's efficiency can be equal to the efficiency of engine operating in carnot cycle with the same working temperatures?

 

[Chestermiller]

Nice sir. Now I understood the concept.. It's very good in having such interactions.

Great. Maybe you can now help out when others ask about these concepts.

Now I need to get some idea on another example.

Suppose we are having an engine whose piston and cylinder housings are made of an imaginary material whose thermal diffusivity is zero so that the heat produced during combustion of fuel will not be absorbed by the piston and cylinder housings but only some amount of heat can be taken as work from the piston.

Will this engine's efficiency can be equal to the efficiency of engine operating in carnot cycle with the same working temperatures?

To avoid confusion, please start a new thread with this.

Chet

 

[rajendra123]

Useful Nucleus, are you still out there? Is there anyone else out there who would like to offer answers to the questions I posed in my previous post. Is there anyone out there who would like to see me post the answers to the questions I posed my my previous post?

Chet

Hello dear Chet , I would like to join the discussion,

At equilibrium both slabs will reach a temperature (Hot slab will lose heat and cold body will gain heat)
Change in entropy of hot body = C ln(T/T_h1)
Change in entropy of hot body = C ln(T/T_c1)
Where, T_h1 and T_c1 are initial temperatures of hot and cold slab respectively.

 

[Useful nucleus]

Hi Chet, I'm sorry that I was not able to respond in due time to your detailed analysis. rajendra123 's post was a good reminder. I think my problem in the analysis you presented for the two slabs is that you insist to perform the integral of δQ/T assuming constant temperature during the transfer process. This cannot be the case. A quasi static process implies that the slab will spend sufficient time visiting the temperatures all the way from the initial temperature to the final temperature. Let me clarify this for the cold slab for example.

You assumed that T is constant and is equal to (T_H+T_C)/2 as such you obtained that ∫δQ/T =mC (T_H-T_C)/(T_H+T_C).

But in fact the temperature of the cold body is changing during the transfer process and cannot be taken out the integration sign. So if I still assume a T-independent heat capacity as you did I will get
∫δQ/T = ∫(mCdT)/T = mC ln (T_final/T_initial) =mC ln [(T_H+T_C)/2T_C]

This is in fact exactly the same expression you obtained for the change of entropy of the cold slab. Thus for the cold slab (and the same applies for the hot slab);
δQ/T=dS even though this heat transfer process is irreversible.

However, I agree with you that for the overall system made of the two slabs we have δQ=0 and dS>0.

 

[Chestermiller]

Hi Chet, I'm sorry that I was not able to respond in due time to your detailed analysis. rajendra123 's post was a good reminder.

Hi Useful Nucleus, Long time no see.

I think my problem in the analysis you presented for the two slabs is that you insist to perform the integral of δQ/T assuming constant temperature during the transfer process. This cannot be the case. A quasi static process implies that the slab will spend sufficient time visiting the temperatures all the way from the initial temperature to the final temperature. Let me clarify this for the cold slab for example.

You assumed that T is constant and is equal to (T_H+T_C)/2 as such you obtained that ∫δQ/T =mC (T_H-T_C)/(T_H+T_C).

But in fact the temperature of the cold body is changing during the transfer process and cannot be taken out the integration sign. So if I still assume a T-independent heat capacity as you did I will get
∫δQ/T = ∫(mCdT)/T = mC ln (T_final/T_initial) =mC ln [(T_H+T_C)/2T_C]

In the irreversible heating of a cold body like this, the temperature of the body is not uniform spatially within the body. So, when one writes δQ/T, there is significant ambiguity as to what value of T to use. Does one use the temperature at the portion of the system boundary where the heat transfer is occurring, the mass-average temperature of the body, or the temperature at some other location? The Clausius inequality applied to any arbitrary situation specifically calls for the use of the temperature at location on the part of the boundary of the system where the heat transfer is occurring:
$$\Delta S \geq \left(\int{\frac{δQ}{T}}\right)_b $$
Moran, Shapiro, Boettner, and Bailey, Fundamentals of Engineering Thermodynamics, use the symbol δ to represent the fact that Q is path dependent and they use the subscript b to indicate that the integral must be evaluated at the part of the boundary of the system where heat transfer is occurring. In my writings, I have been using $T_I$ to represent the temperature on the part of the boundary where the heat transfer is occurring (interface I). So, if one is going to discuss irreversible heat transfer in terms of the Clausius inequality, one must use the temperature at the boundary. This is what I did when I applied the Clausius inequality to the system we have been analyzing. In our system, a transient heat transfer analysis tells us that, immediately after the hot body is brought into contact with the cold body, the interface temperature becomes equal to the average of their initial values, and stays at this initial average value throughout the entire irreversible transfer equilibration.

Now, what you did in your analysis was use the average temperature of the cold body in your evaluation of the integral, and this give you the correct result for the entropy change. Using the average temperature to determine the entropy change is valid for cases in which no work is being done, provided one makes clear that this is not the same thing as applying the Clausius inequality. Moreover, it will not give the correct result for the entropy change for a system in which both irreversible work and irreversible heat transfer are occurring.

You may find it interesting to check out Moran et al. They give the following equation for the entropy change of a system experiencing an irreversible process:
$$\Delta S = \left(\int{\frac{δQ}{T}}\right)_b+\sigma $$
where the integral is interpreted as representing the net amount of entropy entering the system through its boundaries, and the σ is interpreted as the entropy generated within the system as a result of finite heat conduction within the body (from finite temperature gradients) and viscous dissipation of mechanical energy within the body.

This is in fact exactly the same expression you obtained for the change of entropy of the cold slab. Thus for the cold slab (and the same applies for the hot slab);
δQ/T=dS even though this heat transfer process is irreversible.

As I said above, this is only because you used the average temperature of the body, rather than the temperature at the boundary (as would be consistent with the Clausius inequality).

 

[Useful nucleus]

Thank you for elaborating on this subtle point in Clausius inequality , Chet. I have to admit that since my undergraduate thermodynamics course I never needed to deal with Clausius inequality. I will certainly have a look at the book you recommended.

 

[Chestermiller]

Thank you for elaborating on this subtle point in Clausius inequality , Chet. I have to admit that since my undergraduate thermodynamics course I never needed to deal with Clausius inequality. I will certainly have a look at the book you recommended.

It's an important point that, unfortunately, is omitted from almost all thermodynamics treatises, to the detriment and confusion of generation after generation of students. I was happy to find that Moran et al got it right. Students need this wonderful book.