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Entropy Balance

  • Thread starter danago
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  • #1
danago
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Im working on some problem for a chemical thermodynamics course, and ive solved a question (maybe incorrectly) which is making me question my understanding of entropy.

Basically, the system in question is a turbine, with one stream in and one stream out. The turbine is insulated, hence no heat transfer occurs. The system is also in a steady state.

Because the system is in steady state, the rate of change of entropy within the system should be zero -- Am i right in saying this? Also, since there is no heat transferred to the surroundings, the entropy change here should also be zero? This would imply that the total entropy generation in the universe due to this process is zero?

Now that is a nice simple answer, however in the way i understand the second law of thermodynamics, entropy change will only be zero if a process is carried out reversibly. Does the result above imply that the process was carried out reversibly? Or have i missed something in calculating the entropy generation of the universe?

Thanks in advance,
Dan.
 

Answers and Replies

  • #2
Andrew Mason
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Im working on some problem for a chemical thermodynamics course, and ive solved a question (maybe incorrectly) which is making me question my understanding of entropy.

Basically, the system in question is a turbine, with one stream in and one stream out. The turbine is insulated, hence no heat transfer occurs. The system is also in a steady state.
The turbine may be insulated but that does not mean that no heat transfer occurs with the surroundings. If no heat was released to the surroundings the temperature of the system would keep increasing. Heat is transferred to the surroundings from the exit stream.

Because the system is in steady state, the rate of change of entropy within the system should be zero -- Am i right in saying this? Also, since there is no heat transferred to the surroundings, the entropy change here should also be zero? This would imply that the total entropy generation in the universe due to this process is zero?
The system's change in entropy is 0 since its state does not change. But the state of the surroundings DOES change because heat IS transferred to the surroundings. One cannot generate work from heat without transferring some of that heat to the surroundings.

Now that is a nice simple answer, however in the way i understand the second law of thermodynamics, entropy change will only be zero if a process is carried out reversibly. Does the result above imply that the process was carried out reversibly? Or have i missed something in calculating the entropy generation of the universe?
In order to determine whether the process is reversible you have to calculate the change in entropy. You have to determine:

[tex]\Delta S = \Delta S_{sys} + \Delta S_{surr} = 0 + \int dQ_h/T_h + \int dQ_c/T_c[/tex]

In the real world, it will not be 0.

AM
 
  • #3
danago
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Thanks very much for the reply Andrew!

I have revised my thinking, and come up with something like this:

[tex]
\Delta S = \Delta S_{surr} = \dot{n}(s_{out} - s_{in})
[/tex]

Where [tex]\dot{n}[/tex] is the molar flow rate through the turbine. The question states that the gas behaves as an ideal gas, and i am also given information about the state of the gas at the inlet and outlet. Based on this, i could construct a hypothetical reversible thermodynamic path between the two state points and calculate the change in entropy for this path. Doing so yields:

[tex]
s_{out} - s_{in} = \int^{T_2}_{T_1} \frac{c_p}{T} \:dT + Rln (\frac{P_1}{P_2})
[/tex]

Which, when i substitute the relevent numbers, should give me the required change in entropy of the universe per mole of gas flowing through?

Thanks again for your help.
 
  • #4
Andrew Mason
Science Advisor
Homework Helper
7,563
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Thanks very much for the reply Andrew!

I have revised my thinking, and come up with something like this:

[tex]
\Delta S = \Delta S_{surr} = \dot{n}(s_{out} - s_{in})
[/tex]

Where [tex]\dot{n}[/tex] is the molar flow rate through the turbine. The question states that the gas behaves as an ideal gas, and i am also given information about the state of the gas at the inlet and outlet. Based on this, i could construct a hypothetical reversible thermodynamic path between the two state points and calculate the change in entropy for this path. Doing so yields:

[tex]
s_{out} - s_{in} = \int^{T_2}_{T_1} \frac{c_p}{T} \:dT + Rln (\frac{P_1}{P_2})
[/tex]

Which, when i substitute the relevent numbers, should give me the required change in entropy of the universe per mole of gas flowing through?

Thanks again for your help.
You will be able to construct a reversible path between the two states but it will not be one that fits the parameters of the steam turbine. It will be a Carnot cycle in which the expansion and compression of the gas is adiabatic and the heat flow into and out of the gas occurs at constant temperature.

AM
 

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