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Where does entropy generation come from?

  1. Dec 6, 2015 #1
    Let's think about two thermal reservoirs. They are internally reversible. This means they don't have entropy generation(Sgen) right? Each reservoirs' entropy change is different; one is minus(-) and the other is plus(+). But total entropy change in the isolated system has plus value(entropy generation). This means isolated system is irreversible. But reservoir's process is reversible. I think Sgen can not be in the reservoirs since these are internally reversible. And then Sgen may be in the surroundings. How can I evaluate Sgen?
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  3. Dec 6, 2015 #2


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    Depends on the interaction.
    If you have some irreversible process, you'll add more entropy to one system than you take away from the other. Numbers depend on the specific process.

    Consider exchange of heat, for example: dQ = T dS. For the same heat exchanged, a hotter system will reduce its entropy less than a colder system will increase its entropy.
  4. Dec 6, 2015 #3
    This is a really excellent question. Before addressing the issue of the reservoirs, let's take a step back and pose the following problem:
    Suppose you have two semi-infinite slabs of the same solid material, one at temperature T1 and the other at temperature T2, and, at time t = 0, you bring them together. What is the temperature at the interface between the two slabs for all times after they are brought together? Is heat conduction taking place? Is heat being transferred from the hot slab to the cold slab? Is entropy being generated? Now you separate them after a certain time, and let them self-reequilibrate. What are their final temperatures? Has a net change in entropy occurred?

  5. Dec 6, 2015 #4

    Andrew Mason

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    You are correct that there is necessarily entropy change in each reservoir. There is no entropy change in the system undergoing a cyclical process between the reservoirs since the system returns to its initial state after each cycle. So the entropy increase overall must come from the surroundings i.e. the reservoirs.

    There is zero entropy change in the surroundings in a reversible process only because the change in entropy of cold reservoir is offset by the equal and opposite change in entropy of the hot reservoir. But in an irreversible process, the gain in entropy of the cold reservoir is greater than the loss of entropy in the hot reservoir. This is because Qc/Tc > |Qh/Th|

  6. Dec 6, 2015 #5
    um... i can't understands what that means well.. Two reservoirs in internally reversible process don't match with the isolated system in irreversible process. the former doesn't have Sgen the latter has Sgen. where is Sgen created from and how ?
    Last edited: Dec 6, 2015
  7. Dec 6, 2015 #6
    For a closed system experiencing a reversible or an irreversible change (either one of your reservoirs can be regarded as a closed system), there are two contributions to the entropy change:

    1. Heat flow across the boundary of the system, occurring at the boundary temperature TB. Here, the contribution to the entropy change is Q/TB, where Q is the total heat flow during the change.

    2. Entropy generation within the system resulting from irreversible heat conduction and irreversible viscous dissipation. The rate from irreversible heat conduction is locally proportional to the square of the temperature gradient, and the rate from irreversible viscous dissipation is locally proportional to the square of the velocity gradient. To get the total rate of entropy generation in the system, you have to integrate these over the volume of the system.

    In a reversible process, item 2 is negligible, and the entropy change is totally determined by item 1. In this case also, the temperature of the system is uniform, and the boundary temperature is equal to the uniform temperature within the system.

    Now, let's turn attention to one of your thermal reservoir.

    Suppose we first consider a reversible transfer of heat to the reservoir. To accomplish this, we need to hold the boundary temperature TB only slightly higher than the reservoir temperature TR, and, under these circumstances, the rate of heat transfer is so slow that we have to wait a long time for the heat transfer to take place. Also, the capacity of the reservoir is so great that the transfer of Q to the reservoir is not sufficient to cause its temperature of to change significantly. So the change of entropy of the reservoir is Q/T = Q/TB. In addition the temperature gradients within the reservoir are not significant during the reversible change, so that mechanism #2 is negligible.

    Next let's consider an irreversible transfer of heat to the reservoir. In this case, we hold the boundary temperature TB at a value substantially higher than the bulk reservoir temperature TR. But how can we do this, if the reservoir temperature is uniform throughout at TR. The answer is that, under these circumstances, it can't be uniform throughout. The reservoir consists of a fluid, and, in a fluid, there must be a region close to the boundary where the temperature is varying rapidly with distance from the boundary (within the reservoir), from the value TB at the boundary to the value of TR in the bulk. In fluid mechanics and heat transfer, we call such a region a "boundary layer." So, when we consider mechanism #1, we find that the contribution to the change in entropy is Q/TB < Q/TR. This is less than the reversible entropy change. In the irreversible process, after Q has been transferred through the boundary, we insulate the boundary and do not allow any more heat transfer. So in both the reversible case and the irreversible case, the same amount of heat has been transferred, and the system in the end remains at TR. So the entropy change for both the reversible process and the irreversible process must be the same. But, in the irreversible case, where did the rest of the entropy change come from? Well, it came from mechanism #2. During the process, there are very high temperature gradients near the boundary, and these cause a substantial amount of entropy generation in the region near the boundary. This exactly compensates for the lower amount of entropy change from heat transfer through the boundary associated with mechanism #1.

    All that we have discussed here is related closely to the Clausius Inequality. For an irreversible process, this is given by:
    $$\Delta S > \frac{Q}{T_B}$$
    It states that the change in entropy is greater than the heat flow through the boundary divided by the boundary temperature. The difference between this being an inequality and this being an equality is just the entropy generation resulting from mechanism #2.

    Last edited: Dec 7, 2015
  8. Dec 7, 2015 #7

    Andrew Mason

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    What about the entropy change within the reservoir (due to heat flow within the reservoir caused by a temperature gradient within the reservoir)?

    Is it really necessary to consider the temperature gradient when determining entropy change? Since ##\Delta S = \frac{Q_{rev}}{T}## one does not have to take into account temperature gradients at all to determine entropy change. One simply takes the actual heat flow that occurs but done in a reversible process and divides by the reservoir temperature. In the irreversible case there is more heat flow per unit temperature into the cold reservoir than there is out of the hot reservoir. So the entropy (reversible heat flow per unit temperature) increases overall.

  9. Dec 7, 2015 #8
    That's covered by mechanism #2.

    I agree. But the OP was asking about what caused the entropy generation in the irreversible case.

    As you said, if we know the initial and final equilibrium states, the entropy change is determined, irrespective of the path.
  10. Dec 9, 2015 #9
    Hi guys,

    I have a reference that I think everyone will find interesting regarding Entropy Generation. It is in the book Transport Phenomena by Bird, R.B., Stewart, W.E., and Lightfoot, E.N., John Wiley, 2002. The analysis is in end-of-chapter problem 11D.1 "Equation of Change for Entropy." page 372. It might be fun to discuss this here, if anyone has the energy to look over the reference.

  11. Dec 9, 2015 #10

    Andrew Mason

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    "Entropy generation" can be a misleading and confusing concept, as is evident by the original question here. The phrase suggests that there is something 'physical' being produced in an ongoing process. Entropy is not exactly a physical thing. It is a mathematical construct that happens to be a useful thermodynamic potential. Thermodynamics is the study of the relationship between equilibrium states that lie at the beginning and end of thermodynamic processes. Entropy generation is really a matter of non-equilibrium thermodynamics, which is not classical thermodynamics. From my limited knowledge I am not sure it is a very clearly defined area of science.

  12. Dec 9, 2015 #11
    Please don't be judgmental before you have even had a chance to look over the analysis I suggested.

  13. Dec 9, 2015 #12

    Andrew Mason

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    Chet, I wasn't exactly being judgmental about the analysis. It is just that the analysis takes the subject to a much higher level than the OP's question. Although the title to the thread speaks about "entropy generation", the question was a simple one about thermodynamics of equilibrium states. "Entropy generation" as Bird et al. deal with it involves an analysis of non-equilibrium thermodynamics ie. dynamic processes, which is extremely complex. I am not saying it is not interesting.


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    Last edited: Dec 9, 2015
  14. Dec 10, 2015 #13
    Hi Andrew,

    If you got back and look at post #1, the focus of this post (in my judgment) was entropy generation. In my post #6, I tried to describe mechanistically how entropy generation comes into play in the irreversible transition between two thermodynamic equilibrium states (for a heat transfer situation).

    Even though in post #9, I opened with "Hi guys," I really intended to direct that Bird reference at you personally, as a respected friend who I thought might find the development in Bird very interesting. I didn't think that, at the OPs present level, he would yet have the math background to follow the analysis.

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