I Entropy generation during irreversible isothermal expansion

1. Nov 7, 2018

dRic2

Sometimes I go back and think about this stuff, and I always find something I don't understand very well.

Consider an irreversible isothermal expansion of an ideal gas from state $A$ to state $B$ and suppose I know the amount of heat given to the system to perform the expansion - I'll simply call it $Q$.

First of all I need to evaluate the change in entropy between the two states. In order do this I forget about the irreversibility and I evaluate the change in entropy like this:

$$T dS = de + pdV$$

I suppose the process is a reversible thermal expansion so, at constant temperature the change in internal energy $de$ is zero:

$$dS = \frac p T dV = \frac {nR} V dV$$
$$\Delta S_{A→B} = nR \ln \frac {V_{B}} {V_{A}}$$

This result should be correct because the entropy is a function of state so it does not care about the path taken (reversible or irreversible) to go from $A$ to $B$ as long as the initial and final states are the same.

I also know that, for a closed system, the following is true:

$$dS = dS_e + dS_i$$

where $dS_e$ is the entropy received by the system from the surroundings and it is equal to $Q/T$ where $T$, in our case, is a constant. $dS_i$ is the entropy generated inside the system due to irreversibility.

Integrating and re-arranging the above equation:

$$\Delta S_i = \Delta S_{A→B} - \Delta S_e = \frac {Q_{rev}} T - \frac Q T = nR \ln \frac {V_{B}} {V_{A}} - \frac Q T$$

Since the generation of entropy should be greater than zero then $Q_{rev} - Q > 0$ or $Q_{rev} > Q$ which is very strange to me.

Did I miss something ? Is It wrong ?

2. Nov 7, 2018

Andrew Mason

Qrev > Q is not a problem. It will always be the case in an isothermal expansion unless heatflow occurs to the surroundings..

Consider an isothermal free expansion (no heat flow, no work done). The reversible process between those two states would be a quasi-static adiabatic expansion (doing work on surroundings) with heat flow into the system. So the entropy increase of the system would be Qrev/T > 0. Qrev > Qactual = 0

In your case where the irreversible isothermal expansion does work and, therefore, requires heat flow, the work will be less than a quasi-static work so the heatflow will be less. Either that or heatflow must occur to the surroundings which would result in an entropy increase to the surroundings that you have not accounted for.

AM

Last edited: Nov 7, 2018
3. Nov 7, 2018

Lord Jestocost

There is a useful link "A detailed explanation of the concept of "reversibility" as it affects changes in entropy" on http://pollux.chem.umn.edu/4501/ which addresses some questions.

4. Nov 7, 2018

dRic2

I can't Read the article right now because I'm using my smartphone but I think I figured it out. I have Just one more question:
Is it possibile to have $Q_{rev} < Q$?

To test my intuition, I invented a stupid exercise with numebrs and I found this other result that I can not interpret very well.

5. Nov 7, 2018

Andrew Mason

Certainly. That is the difference between a Carnot engine and a real one.

One can see this in the adiabatic part of the cycle. Consider a reversible (quasi-static) adiabatic expansion of an ideal gas between state A(P0, V0, T0) and state B (P1, V1, T1). Qrev = 0. An irreversible path between those two states could be a path in which that same amount of work is done but where the expansion is done more quickly by adding heat flow Q to the gas and then letting the gas cool with heat flow to the surroundings to get to T1.

AM

6. Nov 8, 2018

dRic2

If $Q_{rev} < Q$ does it means that entropy is generated only in the surroundings while the system undergoes a reversible transformation?
Because I really can't get how $S_i$ (the entropy generated inside the system) could be negative.

PS: I read the articles, very nice! :)

7. Nov 8, 2018

Andrew Mason

Please clarify. Change in entropy is $\int Q_{rev}/T$. Change in entropy of a system can easily be negative.

AM

8. Nov 8, 2018

dRic2

Sorry again for the late reply.

If $Q > Q_{rev}$ then

$$dS_i = dS - dS_e = \frac {dQ_{rev}} T - \frac {dQ} T < 0$$

So the entropy generated in the system $S_i$ is negative (??) What does it means physically ?

I suppose that, if such a situation occurs (when the surroundings exchange more heat than the heat required by the reversible process), the system is forced to undergo a transformation without generation of "new entropy" (aka no irreversibility), while the generation of entropy takes place entirely in the surroundings...

9. Nov 8, 2018

Andrew Mason

First of all, you seem to be assuming T is the same for all parts of the reversible and irreversible processes. In the example I gave you of a rapid expansion due to large quantity of heat flow into the system followed by rapid cooling, there are a large differences in temperatures between the system and surroundings in both parts. Since heat flow can only occur from hotter to cooler temperatures, total entropy change of the system and surroundings at every place where heatflow occurs must be positive. At each location where heatflow occurs, Qin is the same magnitude as Qout = |Q| but differs in sign so ΔS = -|Q|/Th + |Q|/Tc > 0. As Th → Tc the ΔS can get arbitrarily close to 0 but never reach 0 (which is the reversible case).

A loss of entropy to the system will occur for example if the system is a bottle of hot water placed on a bed of ice. Since the heatflow occurs at more than an infinitesimal temperature difference ΔS > 0 : dS = dShw + dSice = -dQ/Th + dQ/Tc > 0 because Th > Tc.

To do this reversibly, one could operate a Carnot engine between the hot water bottle and the ice. This would cause the same amount of heat to flow out of the hot water bottle but much less heat to flow into the ice (the difference being the work done by the Carnot engine). So the total heat flow into the ice in the irreversible example, Q, is greater than the heat flow into the ice in the reversible case: Q > Qrev.

AM

10. Nov 9, 2018

Staff: Mentor

Getting back to your original example, if there is just one single ideal reservoir at temperature T used for the reversible path, then the reversible heat added is always greater than the irreversible heat added. Here is a breakdown of how it plays out:

For the irreversible path, if the system were adiabatic, the gas temperature would cool during the expansion. But, in the actual irreversible process, since the boundary of the system is held at the temperature T, the interior temperatures of the gas throughout the irreversible expansion will be cooler than T. This means that, during the irreversible process, there will be significant temperature gradients within the gas, which is one of the mechanisms causing entropy generation.

In addition, during the irreversible process, the gas is deforming very rapidly, and thus is experiencing large velocity gradients. This leads to viscous dissipation of mechanical energy (to internal energy), which is a second mechanism for entropy generation.

So, in the irreversible process, there is significant entropy generation within the system resulting from large temperature- and velocity gradients. Since the entropy change for both the reversible path and the irreversible path is the same, less entropy needs to be transferred from the surroundings to the system during the irreversible process. This means that the net amount of heat absorbed from the surroundings during the irreversible path must be less than during the reversible path (for a reversible path featuring a single ideal reservoir at T).

One of the questions is whether the heat transferred from the surroundings must always be greater for a reversible path. This depends on whether only one ideal reservoir can be used. If more than one ideal reservoir can be used (say, at a different temperature than T), then, at the end of the reversible path, we can always append an ideal Carnot cycle which either does positive or negative work (and heat transfer) using the original reservoir and a second reservoir at a different temperature. If the Carnot cycle does negative work, the total net amount of heat received by the gas during the new overall reversible path can be less than that during the original irreversible path. However, the entropy change of the gas will be the same.

In reality, there are an infinite number of reversible paths between the initial and final states of the gas, and they can differ in the amount of heat transferred to the system. However, the entropy change for each and every one of these reversible paths will be the same, and equal to that of the irreversible path.

One final point. In studying thermodynamics, when we focus on a system experiencing an irreversible change, we always model the surroundings as containing one or more ideal reservoirs at constant temperatures. Because of this, none of the irreversibility (i.e. entropy generation) takes place in the surroundings, and all the entropy generation is implicitly assumed to be confined to the system.

Last edited: Nov 10, 2018
11. Nov 10, 2018

dRic2

Hi thanks a lot for the replies, I'll be without access to my computer for the next 2 days and I also need some time to think about it. I'll be back as soon as possibile. Thank you again :)

12. Nov 12, 2018

dRic2

Hi @Chestermiller your explanation was very clear but I have 3 questions left:

1) For this irreversible isothermal expansion if I write an entropy balance on the whole universe I should get $$\Delta S_{universe} = \Delta S_{system}^{A→B} + \Delta S_{surroundings} = \Delta S_{system}^{A→B} - \frac Q {T_{ambient}}$$ On the other hand if I write the entropy balance for the system alone I get $$\Delta S_{system}^{A→B} = \frac Q {T_{system}} + S_{generated}$$ So $$S_{generated} = \Delta S_{system}^{A→B} - \frac Q {T_{system}} ≠ \Delta S_{universe}$$... Where is the flaw ?

2) Can you recommend a book or some chapters of book to study the mechanism of entropy generation.

3)
Here $Q > Q_{rev}$ because I have to take into the account the work I'm doing on the system, right?

13. Nov 12, 2018

Staff: Mentor

This equation is where the flaw occurs. It should read:
$$\Delta S_{system}^{A→B} = \frac Q {T_{ambient}} + S_{generated}$$Whenever entropy is transferred, the transfer occurs at the boundary between the system and its surroundings, at the boundary temperature. In this case, the boundary temperature is $T_{ambient}$. For more on how to determine the entropy change for a system that has experienced an irreversible process, see my Physics Forums Insights article: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Bird, Stewart, and Lightfoot, Transport Phenomena, Chapter 11, Problem 11D.1, Equation for change for entropy.
I'm not sure I fully understand this question. In a reversible Carnot cycle, the working fluid either receives a net amount of heat from the surroundings and does a net amount of work, or it discharges a net amount of heat to the surroundings and receives a net amount of work. In either case, the entropy change of the working fluid over the cycle is zero.

14. Nov 12, 2018

dRic2

I always have trouble to determine the boundary Temperature at which heat is transfered. I Read again the definition of entropy and it seems to me that T is always the temperature of the reservoir - so it doesn't really matter if the system undergoes an isothermal process or not because even if the temperature of the systen changes, as long as the reservoir remains at constant T, the entropy associated with the heat transfer is always $\frac {dQ} {T_{surroundings}}$.
Am I correct?

Here I assumed that $T_{boundary} = T_{surroundings}$

I apologize for errors because I'm writing with my phone.

15. Nov 12, 2018

Staff: Mentor

In most cases, you are correct, especially for problems in thermodynamics books where an ideal reservoir is involved. But sometimes the surroundings is not a reservoir, for example, if two fluid streams are exchanging heat in a heat exchanger. In such a case, there would be entropy generation not only within the system (say the main process stream) but also In the surroundings (the cooling or heating stream). Here, you would have to determine the temperature at the boundary, taking into account the heat transfer coefficients on both sides of the boundary.

16. Nov 13, 2018

dRic2

"Example 1" of your recipe is an other of this cases, right? And to take into account the change in temperature of the surroundings (the other object) you assumed the heat was transferred to an infinite number of reservoir at gradually higher (or lower, for the other solid) temperature.

I think I'm almost there but I have a huge problem now... Consider my first post:

There is a very important thing going on here. The temperature figuring in the denominator of the fraction $\frac p T$ is not (by definition) the temperature of the system, but it is the temperature at the boundary so it is the temperature of the reservoir! So the equality $\frac p T = \frac {nR} v$ should not be true because $T$ is not the temperature of the gas!!

17. Nov 13, 2018

dRic2

The only thing that comes to my mind is that, for a reversible process, there should not be temperature gradients inside the system so the temperature at the boundary should be equal at the temperature of the whole system. But this means that a reversible thermal expansion can occur only at the temperature of the reservoir (assuming $T_{boundary} = T_{reservoir}$)

18. Nov 13, 2018

Staff: Mentor

Nice.
For the reversible path, the temperature of the gas is virtually equal to the temperature of the reservoir.

19. Nov 13, 2018

dRic2

So is my assunption correct? (I wrote an other post just after the first one that you have quoted here, you might have not seen it)

Anyway thank you very much, you really helped me a lot!

Btw Just for the sake of curiosity: is there a particular need to evalute The entropy chnage in a heat Exchanger or is it Just a textbook problem?

20. Nov 13, 2018

Staff: Mentor

Yes, this is a correct assessment.