son hong chang said:
um... i can't understands what that means well.. Two reservoirs in internally reversible process don't match with the isolated system in irreversible process. the former doesn't have Sgen the latter has Sgen. where is Sgen created from and how ?
For a closed system experiencing a reversible or an irreversible change (either one of your reservoirs can be regarded as a closed system), there are two contributions to the entropy change:
1. Heat flow across the boundary of the system, occurring at the boundary temperature T
B. Here, the contribution to the entropy change is Q/T
B, where Q is the total heat flow during the change.
2. Entropy generation within the system resulting from irreversible heat conduction and irreversible viscous dissipation. The rate from irreversible heat conduction is locally proportional to the square of the temperature gradient, and the rate from irreversible viscous dissipation is locally proportional to the square of the velocity gradient. To get the total rate of entropy generation in the system, you have to integrate these over the volume of the system.
In a reversible process, item 2 is negligible, and the entropy change is totally determined by item 1. In this case also, the temperature of the system is uniform, and the boundary temperature is equal to the uniform temperature within the system.
Now, let's turn attention to one of your thermal reservoir.
Suppose we first consider a reversible transfer of heat to the reservoir. To accomplish this, we need to hold the boundary temperature T
B only slightly higher than the reservoir temperature T
R, and, under these circumstances, the rate of heat transfer is so slow that we have to wait a long time for the heat transfer to take place. Also, the capacity of the reservoir is so great that the transfer of Q to the reservoir is not sufficient to cause its temperature of to change significantly. So the change of entropy of the reservoir is Q/T = Q/T
B. In addition the temperature gradients within the reservoir are not significant during the reversible change, so that mechanism #2 is negligible.
Next let's consider an irreversible transfer of heat to the reservoir. In this case, we hold the boundary temperature T
B at a value substantially higher than the bulk reservoir temperature T
R. But how can we do this, if the reservoir temperature is uniform throughout at T
R. The answer is that, under these circumstances, it can't be uniform throughout. The reservoir consists of a fluid, and, in a fluid, there must be a region close to the boundary where the temperature is varying rapidly with distance from the boundary (within the reservoir), from the value T
B at the boundary to the value of T
R in the bulk. In fluid mechanics and heat transfer, we call such a region a "boundary layer." So, when we consider mechanism #1, we find that the contribution to the
change in entropy is Q/T
B < Q/T
R. This is less than the reversible entropy change. In the irreversible process, after Q has been transferred through the boundary, we insulate the boundary and do not allow any more heat transfer. So in both the reversible case and the irreversible case, the same amount of heat has been transferred, and the system in the end remains at T
R. So the entropy change for both the reversible process and the irreversible process must be the same. But, in the irreversible case, where did the rest of the entropy change come from? Well, it came from mechanism #2. During the process, there are very high temperature gradients near the boundary, and these cause a substantial amount of entropy generation in the region near the boundary. This exactly compensates for the lower amount of entropy change from heat transfer through the boundary associated with mechanism #1.
All that we have discussed here is related closely to the Clausius Inequality. For an irreversible process, this is given by:
$$\Delta S > \frac{Q}{T_B}$$
It states that the
change in entropy is greater than the heat flow through the boundary divided by the boundary temperature. The difference between this being an inequality and this being an equality is just the entropy generation resulting from mechanism #2.
Chet
