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)?
Best Answer
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.