Why is the entropy value of this steady flow open system not equal to zero?

[tracker890 Source h]

Homework Statement

To determine system entropy in a steady flow

Relevant Equations

entropy balance in a steady flow

Q： Why the entropy value of this steady flow open system is not equal to zero?
My idea is as represented by the following equation.
$$
\frac{dS_{sys}}{dt}=0,\,\,\,\,dt\ne 0
$$
$$
\therefore dS_{sys}=0\,\,\,\,\,\,\,\,\therefore ∆Ssys=∆Sair=0
$$
$$
\therefore ∆\overset{\cdot}{S}sys=∆\overset{\cdot}{S}air=0
$$
reference. 7-25answer

 

[Chestermiller]

The system is the air. Just because the system is at steady state, that does not mean that the properties of the air exiting the compressor are the same as the properties of the air entering. Are you familiar with the open system (control volume) version of the 1st law of thermodynamics. If so, please write it down for this system operating at steady state.

 

[tracker890 Source h]

The system is the air. Just because the system is at steady state, that does not mean that the properties of the air exiting the compressor are the same as the properties of the air exiting. Are you familiar with the open system (control volume) version of the 1st law of thermodynamics. If so, please write it down for this system operating at steady state.

My thoughts are as follows, but I'm not sure if they are correct.
$$
∆\overset{\cdot}{S}sys=\frac{dS_{sys}}{dt}=\left( \overset{.}{S}_{in}-\overset{.}{S}_{out} \right) +\overset{.}{S}_{gen}=\left( \sum{\frac{\overset{.}{Q}_{in\ sys}}{T_{sys}}}+\sum{\overset{.}{S}_{mass,input}-}\sum{\overset{.}{S}_{mass,output}} \right) +\overset{.}{S}_{gen}
$$
$$
may\ be\ \sum{\frac{\overset{.}{Q}_{in\ sys}}{T_{sys}}}\ne 0,\sum{\overset{.}{S}_{mass,input}}\ne 0,\sum{\overset{.}{S}_{mass,output}}\ne 0,\overset{.}{S}_{gen}\ne 0,
$$
$$
but\ ∆\overset{\cdot}{S}sys=0\ in\ steady\ flow\ system
$$

 

[Chestermiller]

My thoughts are as follows, but I'm not sure if they are correct.
$$
∆\overset{\cdot}{S}sys=\sum{\frac{\overset{.}{Q}_{in\ sys}}{T_{sys}}}+\sum{\overset{.}{S}_{input}-}\sum{\overset{.}{S}_{output}}+\overset{.}{S}_{gen}
$$
$$
may\ be\ \sum{\frac{\overset{.}{Q}_{in\ sys}}{T_{sys}}}\ne 0,\sum{\overset{.}{S}_{input}}\ne 0,\sum{\overset{.}{S}_{output}}\ne 0,\overset{.}{S}_{gen}\ne 0,
$$
$$
but\ ∆\overset{\cdot}{S}sys=0 in steady flow system
$$

They are looking for $\dot{S}_{out}-\dot{S}_{in}$

 

[tracker890 Source h]

They are looking for $\dot{S}_{out}-\dot{S}_{in}$

Apologies, I cannot understand the content of your reply. Could you please provide more details, thank you.

And I think the key is：
$$
\frac{dS_{sys}}{dt}=0,\,\,\,\,dt\ne 0\ \ \therefore dS_{sys}=0\,\,\,\,\,\,
$$
reference

 

[Chestermiller]

Apologies, I cannot understand the content of your reply. Could you please provide more details, thank you.

And I think the key is：
$$
\frac{dS_{sys}}{dt}=0,\,\,\,\,dt\ne 0\ \ \therefore dS_{sys}=0\,\,\,\,\,\,
$$
reference

They are looking for $$\dot{S}_{out}-\dot{S}_{in}=\frac{\dot{Q}}{T_I}+\dot{\sigma}$$ where $T_I$ is the temperature at the interface between the system and surroundings. The term $\frac{\dot{Q}}{T_{system}}$ in your reference is incorrect.

 

[tracker890 Source h]

They are looking for $$\dot{S}_{out}-\dot{S}_{in}=\frac{\dot{Q}}{T_I}-T_I\dot{\sigma}$$ where $T_I$ is the temperature at the interface between the system and surroundings. The term $\frac{\dot{Q}}{T_{system}}$ in your reference is incorrect.

"What you said is correct, but how do we prove that the equation below is incorrect?
$$
\frac{dS_{sys}}{dt}=0,\,\,\,\,dt\ne 0\ \ \therefore dS_{sys}=0\,\,\,\,\,\,
$$

 

[Chestermiller]

"What you said is correct, but how do we prove that the equation below is incorrect?
$$
\frac{dS_{sys}}{dt}=0,\,\,\,\,dt\ne 0\ \ \therefore dS_{sys}=0\,\,\,\,\,\,
$$

"the system is at steady state..."

 

[tracker890 Source h]

"the system is at steady state..."

so the system is unsteady?

 

[Chestermiller]

so the system is unsteady?

No. The system is steady. At any location in the system, all thermodynamic parameters are constant in time. But the parameters vary with location through the system.

 

[tracker890 Source h]

No. The system is steady. At any location in the system, all thermodynamic parameters are constant in time. But the parameters vary with location through the system.

So, the conclusion is as follows, correct?
$$
∆\overset{\cdot}{S}sys=\frac{dS_{sys}}{dt}=\left( \overset{.}{S}_{in}-\overset{.}{S}_{out} \right) +\overset{.}{S}_{gen}=\left( \sum{\frac{\overset{.}{Q}_k}{T_k}}+\sum{\overset{.}{S}_{mass,input}-}\sum{\overset{.}{S}_{mass,output}} \right) +\overset{.}{S}_{gen}
$$
$$
may\ be\ \sum{\frac{\overset{.}{Q}_k}{T_k}}\ne or=0\ ,\sum{\overset{.}{S}_{mass,input}}\ne or=0,\sum{\overset{.}{S}_{mass,output}}\ne or=0,\overset{.}{S}_{gen}\ne or=0,$$
but in any case, when the system is a steady flow system, $∆\overset{\cdot}{S}sys=0$.

ref. Youtube teach

 

[Chestermiller]

So, the conclusion is as follows, correct?
$$
∆\overset{\cdot}{S}sys=\frac{dS_{sys}}{dt}=\left( \overset{.}{S}_{in}-\overset{.}{S}_{out} \right) +\overset{.}{S}_{gen}=\left( \sum{\frac{\overset{.}{Q}_k}{T_k}}+\sum{\overset{.}{S}_{mass,input}-}\sum{\overset{.}{S}_{mass,output}} \right) +\overset{.}{S}_{gen}
$$
$$
may\ be\ \sum{\frac{\overset{.}{Q}_k}{T_k}}\ne or=0\ ,\sum{\overset{.}{S}_{mass,input}}\ne or=0,\sum{\overset{.}{S}_{mass,output}}\ne or=0,\overset{.}{S}_{gen}\ne or=0,$$
but in any case, when the system is a steady flow system, $∆\overset{\cdot}{S}sys=0$.

ref. Youtube teach

No. The correct equation is:
$$
\frac{dS_{sys}}{dt}= \sum{\frac{\overset{.}{Q}_k}{T_k}}+\sum{\overset{.}{S}_{mass,input}-}\sum{\overset{.}{S}_{mass,output}} +\overset{.}{S}_{gen}=0
$$