Modern Engineering Thermodynamics

7.10 Differential Entropy Balance 227
7.9 ENTROPY TRANSPORT MECHANISMS
Several conceptual problems occur when Eq. (7.22) is used to define the entropy of a system. First of all, this
equation is limited to closed systems, and at this point, we do not know how it must be altered to accommodate
open systems. Second, it does not indicate how system entropy may be influenced by the work transport of
energy; third, it deals only with hypothetical “reversible” processes. The third point is particularly bothersome,
since all of our auxiliary formulae for heat transfer have been developed from an empirical basis and therefore
always give the actual rather than the reversible heat transport of energy (see Chapter 4).
From the work of Carnot, Thomson, and Clausius discussed earlier in this chapter, it seems clear that energy and
entropy are related in some way. In this chapter, we investigate the possibility that the energy transport mechanisms
of heat transfer and work modes are also mechanisms for entropy transport. In Chapter 9, we expand this
investigation to include mass flow transport of entropy.
First, we investigate heat and work transports of entropy by again restricting our analysis to closed systems. In
Chapter 4, we note a modern definition of heat transfer: It is an energy transport mechanism that is neither a
work mechanism nor a mass flow mechanism. It is often conveniently viewed as a nonwork, nonmass flow
mechanism for transporting internal energy.
7.10 DIFFERENTIAL ENTROPY BALANCE
In a reversible process, the production of entropy is always zero, by definition. Therefore, if Eq. (7.22) is viewed
as a differential entropy balance for a closed system undergoing a reversible heat transport of energy, then from
a differential form of Eq. (7.2), it is clear that the heat transport of entropy is given by
ðdS G
Þ Q
= ðdS
Þ Q
= ðdS T Þ rev
+ ðdS P
Q
0
Þ rev
Q
= dQ T
rev
So, the differential entropy transport due to a hypothetical reversible heat transfer is simply
ðdS T Þ rev = dQ
Q T
rev
(7.42)
Unfortunately, the two sides of Eq. (7.42) do not have the same differential form. The left side is the total differential
of S T , whereas the right side is a differential divided by an absolute temperature. To integrate this equation
over the surface area of the system, we need to know the exact mathematical relationship between Q and the
temperature of the boundary T b at the point where this heat transfer occurs. For a reversible Carnot cycle, this
relation is very simple, since a reversible heat transfer occurs only during an isothermal process; so T b must be a
constant. In this case, Eq. (7.42) can be integrated to give
ð Þ 2 rev Q and
1 S T
isothermal
= 1 Q 2
T b
rev Q and
isothermal
(7.43)
However, Eq. (7.42) cannot be as easily integrated for common nonisothermal heat transfer processes. This problem
can be solved by changing the form of Eq. (7.42) by introducing the following mathematical identity
dQ
T
= d Q
+ Q dT (7.44)
T T2 Then, Eq. (7.42) becomes
ðdS T Þ rev = d Q
+ Q
Q T
rev
T dT 2 rev
Recall that heat transfer irreversibility is simply due to the heat transport of energy through a finite temperature
difference, so that for all reversible heat transfers, we must have dT = 0, or
Q
T 2 dT = 0
rev