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In my text book, "Chemical, Biochemical, and Engineering Thermodynamics" by Sandler. The entropy balance equation for an open system is defined as, $$\frac{dS}{dt}_{sys} = \sum_k M_kS_k + \frac{Q}{T} + S_{gen},$$ which can help us to calculate the entropy change of in the system and predict the reversibility of the process. However, if we need to calculate the total entropy change of a process, that is ,we need to calculate the entropy change of the system and surroundings, what is the entropy balance of surroundings in an open system?

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  • $\begingroup$ please, use mathjax $\endgroup$
    – hyportnex
    Commented Feb 2, 2023 at 19:09
  • $\begingroup$ The website automatically change the equation to a more readable one. If it's not clear enough, I could try to fix it. $\endgroup$
    – Richard Wu
    Commented Feb 2, 2023 at 19:13
  • $\begingroup$ The website does not change anything automatically, I converted your text into mathjax. Anyhow, although I do not quite understand your question neither am I familiar with the book this may help you authors.library.caltech.edu/4392/1/TOLrmp48.pdf $\endgroup$
    – hyportnex
    Commented Feb 2, 2023 at 19:15
  • $\begingroup$ This is my first time to use this website, so very grateful to your help! As for my question, the equation (3.1) in the paper you offered is the entropy balance equation for the system (which is what I type for my question statement). But my question is that if there is a corresponding equation for the surroundings? $\endgroup$
    – Richard Wu
    Commented Feb 2, 2023 at 19:21
  • $\begingroup$ Because there are some textbook using the total entropy change, which is the entropy change of system plus that of surroundings, but the conditions are usually in a closed system of mass. But if the mass will enter and exit a system, it becomes an open system of mass, and the equation for the system becomes eq(3.1). However I don't see there is some related information about that one of the surroundings. $\endgroup$
    – Richard Wu
    Commented Feb 2, 2023 at 19:29

4 Answers 4

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The underlying assumption in Tolman-Fine is that anything that happens in the "environment" happens in a reversible manner. That assumption is valid as long as all exchanges of entropy, energy, mass, volume, etc., are infinitesimally small relative to the respective amount in the environment. As a consequence, all changes in the environment are assumed to be reversible. That is the change of entropy due to heat exchange is exactly $-\frac{Q}{T}$, the change of entropy due to mass transfer is excatly $-\sum_kM_kS_k$, etc.

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  • $\begingroup$ Thank you for your answer, will the generation or irreversibility term appear in the equation for the surroundings? $\endgroup$
    – Richard Wu
    Commented Feb 2, 2023 at 19:32
  • $\begingroup$ Oh I see, because all changes in the surroundings is reversible, the irreversibility term will vanish? $\endgroup$
    – Richard Wu
    Commented Feb 2, 2023 at 19:34
  • $\begingroup$ As long as the system is coupled to the environment and is not isolated from it then whatever entropy is generated inside the system by irreversibilities it will be exchanged as any entropy is exchanged with the environment. The environment cannot tell the difference from what source the entropy comes from, it is just entropy. $\endgroup$
    – hyportnex
    Commented Feb 2, 2023 at 19:36
  • $\begingroup$ What do you mean by "vanish"? It is added to the internal entropy, and it is just entropy for any future interaction. For a engine performing an irreversible cycle it will also be dumped into the environment but from the environment's point of view in a reversible manner. $\endgroup$
    – hyportnex
    Commented Feb 2, 2023 at 19:38
  • $\begingroup$ Thank you for your answer, so even if the mass transfer of the system is in steady state, the accumulation term can be zero, but the accumulation term in the surrounding one may not? $\endgroup$
    – Richard Wu
    Commented Feb 2, 2023 at 19:40
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The equation you wrote inherently assumes that the rate of heat transfer from the surroundings to the system Q takes place at the interface temperature between the system and surroundings $T=T_I$. So T is not the average temperature of the system, but, rather, the interface temperature. If the surroundings is modeled as an ideal reservoir (having negligible entropy generations), then $$\frac{dS_{surr}}{dt}=-\frac{Q}{T_I}$$

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  • $\begingroup$ Doesn't the entropy change from the mass transfer term should be included in this equation for surroundings? Because the mass would enter and exist from the point of view of surroundings. $\endgroup$
    – Richard Wu
    Commented Feb 2, 2023 at 19:51
  • $\begingroup$ Well, yes, I can see it being interpreted from that perspective. Then you're back to having a closed combined system. But, when looking at the open system behavior, the focus is usually exclusively on the system. $\endgroup$ Commented Feb 2, 2023 at 21:32
  • $\begingroup$ Thanks for your comment. Some other textbooks use the "the total change of entropy" term which consists of change of entropy both from system and surrounding to predict the spontaneity (i.e. plust the two together to check if it's > 0, if so, it's spontaneous). But most cases are in closed system. Then I'm confused about how this concept works in an open system. $\endgroup$
    – Richard Wu
    Commented Feb 2, 2023 at 23:36
  • $\begingroup$ In an open system, the condition is that, for an open system operating at steady state in contact with an ideal constant temperature reservoir at the same temperature T as the entering streams (and exit streams), the maximum shaft work is equal to $-\Delta G$ of the exiting streams relative to the entering streams. $\endgroup$ Commented Feb 3, 2023 at 0:44
  • $\begingroup$ Yes, I know that. But here is a example of my problem. For an open steady system, the term of entropy change in the CV per time is zero. And if the process is adiabatic, then the overall result will be "the generation term = mass transfer term". But when I consider that equation of surrounding, if the accumulation term still be zero because the process is in steady state? Or we should not make such assumption because the irreversibility of the system will contribute on the surrounding. $\endgroup$
    – Richard Wu
    Commented Feb 3, 2023 at 1:05
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I can see what your issue is now. It is in establishing the relationship between the closed system version of the equations and the open system version.

To do this, we first have to precisely identify the specific closed system used in the derivation of the open system equations. This closed system consists of the mass contained within the control volume at time t plus the mass entering the control volume between times t and t + $dt$. It is also equal to the mass contained within the control volume at time t + $dt$ plus the mass which exited the control volume between times t and t + dt. So our mass balance on this closed system reads: $$m_{closed}(t+dt)-m_{closed}(t)=m_{CV}(t+dt)+\dot{m}_{out}dt-m_{CV}(t)-\dot{m}_{in}dt$$or, dividing by dt, $$\frac{dm_{closed}}{dt}=\frac{dm_{CV}}{dt}+\dot{m}_{out}-\dot{m}_{in}$$But, since the mass of a closed system is constant, we obtain:$$\frac{dm_{CV}}{dt}=\dot{m}_{in}-\dot{m}_{out}$$

Using this same closed system for the entropy balance, we have $$S_{closed}(t+dt)-S_{closed}(t)=S_{CV}(t+dt)+\dot{m}_{out}s_{out}dt-S_{CV}(t)-\dot{m}_{in}s_{in}dt=\frac{Q}{T_I}dt+\sigma dt$$sor, dividing by dt, we have $$\frac{dS_{closed}}{dt}=\frac{dS_{CV}}{dt}-\dot{m}_{in}s_{in}+\dot{m}_{out}s_{out}=\frac{Q}{T_I}+\sigma $$So the interpretation is that the mass flow terms combine with the rate of change within the control volume to give the rate of change within the closed system. And finally, $$\frac{(dS_{closed}+dS_{surroundings})}{dt}=\frac{dS_{CV}}{dt}-\dot{m}_{in}s_{in}+\dot{m}_{out}s_{out}-\frac{Q}{T_I}=\sigma$$Hope this helps.

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  • $\begingroup$ Thank you for your explanation! I am still figuring it out but I think it's sufficient for me to understand. $\endgroup$
    – Richard Wu
    Commented Feb 3, 2023 at 14:31
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I will elaborate on the answer by hyportnex. The balance for the system and the surroundings (the universe) is $$ \frac{dS_\text{sys}}{dt} + \frac{dS_\text{surr}}{dt} = \dot S_\text{gen} \geq 0 $$ which simply states the second law: the entropy in an isolated system can only increase. Since you have an equation for the entropy change of the system, back-calculate the entropy of the surroundings. The result is $$ \frac{dS_\text{surr}}{dt} % = \dot S_\text{gen} - \frac{dS_\text{sys}}{dt} = -\frac{Q}{T}-\sum_k M_k S_k $$ which is the answer by hyportnex.

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    $\begingroup$ Thanks. This helps a lot $\endgroup$
    – Richard Wu
    Commented Feb 3, 2023 at 14:32

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