Ericsson cycle- question concerning entropy

[trelek2]

Hi!

I have trouble with the following:
1)
Consider an ericsson cycle. It is easy to calculate the change of entropy of the working substance during the isothermal parts. But assuming that all the heat transferred/removed from the working substance during the isobaric processes is done by direct thermal contact with reservoirs at constant temperatures A and B, does this imply the change of entropy of the working substances to be 0?

2)
If we don't make the above assumption, adding all the processes of the ericsson cycle together give a net change of entropy to be 0. Why does the change of entropy of a working substance taken around a closed cycle (even including irreversible processes) is always equal to 0?

3) A corrolary question: When calculating the change of entropy of the universe when the ericsson engine goes around one cycle should I make the assumption as stated in 1)? Otherwise the change of entropy of the universe would be 0, right?

 

[Chestermiller]

Hi!

I have trouble with the following:
1)
Consider an ericsson cycle. It is easy to calculate the change of entropy of the working substance during the isothermal parts. But assuming that all the heat transferred/removed from the working substance during the isobaric processes is done by direct thermal contact with reservoirs at constant temperatures A and B, does this imply the change of entropy of the working substances to be 0?

If the working fluid is in contact with constant temperature reservoirs during the isobaric segments of the process, then entropy will be generated within the working fluid. However, by the ends of these steps, the generated entropy will get transferred to the reservoirs. That is, if the end points at beginning and end of the isobaric segments are independent of whether the steps are reversible or irreversible, the entropy change for the working fluid is the same in both cases. But the entropy changes of the reservoirs will be different.

2)
If we don't make the above assumption, adding all the processes of the ericsson cycle together give a net change of entropy to be 0. Why does the change of entropy of a working substance taken around a closed cycle (even including irreversible processes) is always equal to 0?

Entropy is an inherent physical property of the working fluid, independent of any process. So if, at the end of a cycle, the working fluid is in the same thermodynamic state as at the beginning (which is guaranteed, since we are calling it a cycle), the entropy change over the cycle is zero.

3) A corrolary question: When calculating the change of entropy of the universe when the ericsson engine goes around one cycle should I make the assumption as stated in 1)? Otherwise the change of entropy of the universe would be 0, right?

If the cycle is carried out reversibly, the change in entropy of the universe will be zero. If the cycle is carried out irreversibly, the change in entropy of the universe will be positive.