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I'm having a hard time trying to understand a few concepts of the second law of thermodynamics.

Let's take the example bellow:

A heat source at 800K loses 2000kJ of heat to a sink at 500K. Determine the total entropy generation.

SOLUTION

The case involve heat transfer through a finite temperature difference, and therefore it is irreversible. The magnitude of the entropy generation (irreversibility) associated with each process can be determined by calculating the total entropy change for each case. The total entropy change for a heat transfer process involving two reservoirs (a source and a sink) is the sum of the entropy changes of each reservoir since the two reservoirs form an adiabatic system.

Or do they?

The problem statement gives the impression that the two reservoirs are in direct contact during the heat transfer process. But this cannot be the case since the temperature at a point can have only one value, and thus it cannot be 800 K on one side of the point of contact and 500 K on the other side. In other words, the temperature function cannot have a jump discontinuity.

Therefore, it is reasonable to assume that the two reservoirs are separated by a partition through which the temperature drops from 800 K on one side to 500 K on the other. In that case, the entropy change of the partition should also be considered when evaluating the total entropy change for this process. However, considering that entropy is a property and the values of properties depend on the state of a system, we can argue that the entropy change of the partition is zero since the partition appears to have undergone a steady process and thus experienced no change in its properties at any point. We base this argument on the fact that the temperature on both sides of the partition and thus throughout remains constant during this process. Therefore, we are justified to assume that the $\Delta S_{partition} = 0$ since the entropy (as well as the energy) content of the partition remains constant during this process. The entropy change for each reservoir can be determined since each reservoir undergoes an internally reversible, isothermal process.

$$\Delta S_{source} = \frac{Q_{source}}{T_{source}}=\frac{-2000kJ}{800K}=-2.5kJ/K$$

$$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}=\frac{2000kJ}{500K}=4.0kJ/K$$

$$ S_{gen} = \Delta S_{total} = \Delta S_{source} + \Delta S_{sink} + {\Delta S_{partition}} = 1.5$$

My question is: If both reservoirs are internally reversible, since their state don't change and the "partition" is also internally reversible for the same reason. What is causing the entropy generation (irreversibilities)?

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The entropy of each of the reservoirs changes, because one loses heat and the other gains heat. So their states change, but, since they are internally reversible, there is no entropy generation within the reservoirs. So, where does the entropy generation occur? It occurs within the partition, where finite heat conduction is occurring via a temperature gradient. But, if the partition returns to its original temperature in the end, its entropy doesn't change. The entropy generated within the partition is transferred to the combination of the two reservoirs. So, even though the reservoirs are internal reversible, they receive the generated entropy from the partition.

Incidentally, this was a great question.

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  • $\begingroup$ Thank you for the great answer, Chester Miller. I have another question that you might be able to help me with. Why are thermal reservoirs considered to be internally reversibles? $\endgroup$ Sep 19 '17 at 20:14
  • $\begingroup$ They are taken to be ideal, and thus, internally reversible. In thermodynamics, they are employed to represent ideal heat transfer media. $\endgroup$ Sep 19 '17 at 22:06
  • $\begingroup$ Hi Chet. Came across this while researching irreversible isothermal processes. When you say "The entropy generated within the partition is transferred to the combination of the two reservoirs" is there a way to determine how much of the entropy generated transfers to the two reservoirs? Im thinking how this may apply to an irreversible isothermal expansion of an ideal gas that is irreversible due to heat transfer across a finite temperature gradient and how to determine the entropy generated in the gas. Couldn't find this example in Grandpa Chet's entropy recipe. $\endgroup$
    – Bob D
    Feb 13 '20 at 13:04
  • $\begingroup$ @BobD It seems to me that, in the case of the two reservoirs, all the generated entropy within the partition is flowing into the cold reservoir along with the entropy flowing from the hot reservoir. In the case of the irreversible isothermal expansion, the reservoir is assumed to be ideal, so all the generated entropy remains in the gas. If you can determine the final state of the system from knowing the external work, then the generated entropy within the gas during the process is equal to the total entropy change minus the entropy flow from the reservoir $Q_{irrev}/T_{res}$ $\endgroup$ Feb 13 '20 at 15:29
  • $\begingroup$ @BobD. For more details on the problem involving entropy generation with two reservoirs and a conductive partition, see post #27 of the following thread: physicsforums.com/threads/… $\endgroup$ Feb 13 '20 at 16:18

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