Entropy balance of surroundings in open systems

Question

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?

Answer 1

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.

Answer 2

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 $$\frac{dS_{surr}}{dt}=-\frac{Q}{T_I}$$

Answer 3

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.

Answer 4

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.