Modern Engineering Thermodynamics

8.2 Systems Undergoing Reversible Processes 255
For an incompressible material, u 2 – u 1 = c(T 2 – T 1 ) and s 2 – s 1 = c ln(T 2 /T 1 ). Then,
This is the required final formula.
ðT b Þ min = T 2 − T 1
ln ðT 2 /T 1 Þ
If you look closely at the last equation in Example 8.3, you see that (T b ) min always falls between T 1 and T 2 . This
means that, to change the temperature of an incompressible substance by a heat transfer process from an external
source without producing entropy (i.e., reversibly), the system must be constructed with a well-defined
boundary that is somehow maintained isothermal at the temperature given by this equation. This causes all the
entropy generated by the heat transfer process to be produced outside the system.
EXAMPLE 8.4
Show that a reversible and adiabatic process carried out in a closed system results in the system having a constant entropy;
that is, that a closed system reversible adiabatic process is also isentropic.
Solution
First, draw a sketch of the system (Figure 8.4).
Here, we are to show that a reversible and adiabatic process carried out in a closed
system results in the system having a constant entropy. We begin with just the
entropy balance equation and note that
Z Z
ms ð 2 − s 1 Þ =
τ Σ
_q
dA dt
T b
0 ðadiabaticÞ
+ 1 ðS P Þ 2
= 0
0 ðreversibileÞ
and therefore s 2 = s 1 and we have a constant entropy or isentropic process. Similarly, the entropy rate balance equation
gives
Z
_q
_S = dA + _S P = 0
and therefore S (and, of course, s) is again constant.
Σ
T b
0 ðadiabaticÞ
0 ðreversibileÞ
Exercises
4. If the size of the solar collectors in Example 8.2 is doubled and the collectors capture 200. × 10 3 Btu/h of solar energy,
determine the new maximum steady state electrical power output of the power plant. Answer: _W electrical
rev = 14:2kW:
5. Using the result of Example 8.3, determine the minimum isothermal boundary temperature as liquid water is heated
from 50.0°F to 70.0°F at atmospheric pressure. Answer: (T b ) min = 520°F.
6. Example 8.4 shows that a reversible and adiabatic process carried out in a closed system is also an isentropic process.
Use a similar argument to show that an isentropic Z process in a closed system is not necessarily reversible and adiabatic.
Answer: s 2 – s 1 = 0 simply implies that _S P = − ð _q /T b ÞdA, not that _S P = 0 and _q = 0:
Σ
S 1
1 Q 2 = 0
S 2 = ?
1 (S P ) 2 = 0
State 1
Reversible and
adiabatic process
State 2
FIGURE 8.4
Example 8.4.
Isentropic processes are an important new category that we add to our list of constant property processes
(isothermal, isobaric, isochoric, and isenthalpic).
ARE ALL ISENTROPIC PROCESSES REVERSIBLE AND ADIABATIC?
Example 8.4 shows that all reversible and adiabatic processes are isentropic but not all isentropic processes are necessarily
reversible and adiabatic. It does not automatically go both ways. For example, heat loss from a system results in an entropy
loss for that system; and if this entropy loss exactly balances the entropy production for the process the system is
undergoing, then the process also is isentropic without being either reversible or adiabatic.