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

• tracker890 Source h
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.
$$\frac{dS_{sys}}{dt}=0,\,\,\,\,dt\ne 0$$
$$\therefore dS_{sys}=0\,\,\,\,\,\,\,\,\therefore ∆Ssys=∆Sair=0$$
$$\therefore ∆\overset{\cdot}{S}sys=∆\overset{\cdot}{S}air=0$$

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.

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Chestermiller said:
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.
$$∆\overset{\cdot}{S}sys=\frac{dS_{sys}}{dt}=\left( \overset{.}{S}_{in}-\overset{.}{S}_{out} \right) +\overset{.}{S}_{gen}=\left( \sum{\frac{\overset{.}{Q}_{in\ sys}}{T_{sys}}}+\sum{\overset{.}{S}_{mass,input}-}\sum{\overset{.}{S}_{mass,output}} \right) +\overset{.}{S}_{gen}$$
$$may\ be\ \sum{\frac{\overset{.}{Q}_{in\ sys}}{T_{sys}}}\ne 0,\sum{\overset{.}{S}_{mass,input}}\ne 0,\sum{\overset{.}{S}_{mass,output}}\ne 0,\overset{.}{S}_{gen}\ne 0,$$
$$but\ ∆\overset{\cdot}{S}sys=0\ in\ steady\ flow\ system$$

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tracker890 Source h said:
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}##

Chestermiller said:
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

tracker890 Source h said:
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.

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Chestermiller said:
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?
$$\frac{dS_{sys}}{dt}=0,\,\,\,\,dt\ne 0\ \ \therefore dS_{sys}=0\,\,\,\,\,\,$$

tracker890 Source h said:
"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..."

Chestermiller said:
"the system is at steady state..."

tracker890 Source h said:
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.

Chestermiller said:
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##.

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tracker890 Source h said:
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##.

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

tracker890 Source h

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

In a steady flow open system, entropy is not necessarily zero because the system exchanges both energy and matter with its surroundings, leading to potential entropy changes due to heat transfer, work interactions, and mass flow.

## Does heat transfer affect the entropy of a steady flow open system?

Yes, heat transfer can significantly affect the entropy of a steady flow open system. When heat is added to the system, entropy increases, and when heat is removed, entropy decreases, according to the second law of thermodynamics.

## How does mass flow contribute to the entropy change in a steady flow open system?

Mass flow contributes to the entropy change because the entering and exiting mass may carry different amounts of entropy. The difference in the specific entropy of the incoming and outgoing streams will result in an overall entropy change in the system.

## Can irreversibilities within the system cause the entropy to be non-zero?

Yes, irreversibilities such as friction, turbulence, mixing, and non-ideal processes within the system can generate entropy, causing the entropy value of the system to be non-zero even in a steady state.

## Is it possible for an open system to have zero entropy change under certain conditions?

It is theoretically possible for an open system to have zero entropy change if the process is perfectly reversible and there is no net heat transfer or mass flow carrying entropy in or out. However, in practical scenarios, some degree of irreversibility and heat transfer is almost always present, leading to a non-zero entropy change.

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