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In many books of Engineering thermodynamics, the entropy balance equation is wrritten as

$\frac{dS}{dt}=\Pi_S+I_S$

$I_S=\sum_{j=1}\frac{\dot{Q}_{j}}{T_j}+\sum_{i=1}\dot{m}_is_i-\sum_{e=1}\dot{m}_es_e$

$\frac{dS}{dt}=\Pi_S+\sum_{j=1}\frac{\dot{Q}_{j}}{T_j}+\sum_{i=1}\dot{m}_is_i-\sum_{e=1}\dot{m}_es_e$

Where the term "$\Pi_S$" represents the entropy rate production, which is always postive or equal to zero, $\Pi_S\ge0$ , and the term $I_S$ means the entropy exchange rate. $\sum_{i=1}\dot{m}_is_i$, $\sum_{e=1}\dot{m}_es_e$ are the incoming and outgoing entropy respectively due to mass flow. Where $s_i$, $s_e$ are the intensive entropy per unit mass.

Where does the $\sum_{i=1}\dot{m}_is_i-\sum_{e=1}\dot{m}_es_e$ come from? How to derived it in a formal way? Other authors like Ansermet, can derived that in fact $I_S=\frac{\dot{Q}}{T}$ but no the entropy exchanged by mass flows

In the book "Principles of thermodynamics" by Jean Phillipe Ansermet, says that the entropy evolution in a macroscopic scale, can be described as follows:

$\frac{dS}{dt}=\Pi_S+I_S$

A few pages back, the author says that the internal energy evolution for a system can be described as

$\dot{U}=T\dot{S}-P\dot{V}+\sum_{i=1}^C\mu_i\dot{N_i}=P_Q+P_W+P_C$

Where $P_Q , P_W, P_C$ are the heat power, mechanical power and chemical power respectively. The heat power $P_Q$ is the same variable that $\dot{Q}_j$, so $P_Q=\dot{Q}_j$. We can re write the internal energy evolution equation in terms of $\dot{S}$.

$\dot{S}=\frac{1}{T}(P_Q+P_W+P_C+P\dot{V}-\sum_{i=1}^C\mu_i\dot{N_i})$

Entropy exchanges happens when exist heat transfer. Due to the above, we can rewrite the equation as follows:

$\dot{S}=\frac{1}{T}(P_W+P_C+P\dot{V}-\sum_{i=1}^C\mu_i\dot{N_i})+\frac{P_Q}{T}$

Therefore, when the process is reversible there is no entropy production $\Pi_S=0$ and there is no entropy exchage $I_S=0$. Then, we can deduce the reversible heat, mechanical work and chemical work

$\frac{1}{T}(P_W+P_C+P\dot{V}-\sum_{i=1}^C\mu_i\dot{N_i})=\Pi_S=0$ y $\frac{P_Q}{T}=I_S=0$

$P_W=-P\dot{V}\Rightarrow\delta W=P_Wdt=-PdV\Rightarrow W_{if}=\int_i^f\delta W=-\int_{V_i}^{V_f}PdV$

$P_C=\sum_{i=1}^C\mu_i\dot{N_i}\Rightarrow\delta C=P_Cdt=\sum_{i=1}^C\mu_idN_i\Rightarrow C_{if}=\int_i^f\delta C=\int_{N_{i_i}}^{N_{i_f}}\sum_{i=1}^C\mu_idN_i$

$P_Q=TI_S$

In reversible process, we can rewrite the internal energy equation in order to obtain another expression for $P_Q$

$\dot{U}=T\dot{S}-P\dot{V}+\sum_{i=1}^C\mu_i\dot{N_i}=P_Q+P_W+P_C\Rightarrow T\dot{S}-P\dot{V}+\sum_{i=1}^C\mu_i\dot{N_i}=P_Q-P\dot{V}+\sum_{i=1}^C\mu_i\dot{N_i} \Rightarrow P_Q=T\dot{S}$

$P_Q=T\dot{S}\Rightarrow \delta Q=P_Qdt=TdS\Rightarrow Q_{if}=\int_i^f\delta Q=\int_{S_i}^{S_f}TdS$

But, Ansermet cannot show where does the $\sum_{i=1}\dot{m}_is_i-\sum_{e=1}\dot{m}_es_e$ come from.

Also, how can we fit the term $\sum_{i=1}\dot{m}_is_i-\sum_{e=1}\dot{m}_es_e$ when we take the differential form of entropy $dS$?

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  • $\begingroup$ Are you familiar with the open system (control volume) version of the 1st law of thermodynamics? If not, please look that over and get back to us. If so, do you have any issues with the convective terms in that equation? $\endgroup$ Jul 2 '21 at 11:50
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I'm not familiar with this equation exactly, but I think it makes sense if I understand your definitions correctly.

As a general rule (see, for example, Wikipedia's "continuity equation") the rate at which the amount of stuff in a region is changing per unit time is the rate per unit time that it's being produced within the region, plus the net rate at which it's flowing into the region.

Taking that latter term into account, it would be the entropy per unit time flowing in, minus the entropy per unit time flowing out. If we chose to split these up and index multiple inflows and outflows, this would be

$$\sum_i \dot{S}_i - \sum_e \dot{S}_e.$$

But by definition, the relationship between an entropy, a mass and an entropy per unit mass is $S_i = m_i s_i$. Making this substitution yields the terms about which I believe you're inquiring.

There's one last issue. In the case of entropy, we know that flows of heat themselves carry entropy. So in addition to the internal entropy production rate, we must include heat flows in addition to mass flows, yielding the $\dot{Q}_j/T_j$ terms. As to why these correspond to entropies well: the way I teach the subject, at least, this serves as the definition for entropy and the consistency of this definition follows from the Second Law of Thermodynamics.

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