388 | Thermodynamics
Substituting, the rate of entropy generation is determined to be
Discussion Note that entropy is generated during this process at a rate of
8.65 Btu/min · R. This entropy generation is caused by the mixing of two
fluid streams (an irreversible process) and the heat transfer between the
mixing chamber and the surroundings through a finite temperature difference
(another irreversible process).
8.65 Btu/min#R
¬
180 Btu>min
530 R
1 322.7 0.18174 300 0.0360922.7 1.7406 2 Btu>min#R
S
#
genm
#
3 s 3 m
#
1 s 1 m
#
2 s 2
Q
#
out
Tb
EXAMPLE 7–21 Entropy Generation Associated with Heat
Transfer
A frictionless piston–cylinder device contains a saturated liquid–vapor
mixture of water at 100C. During a constant-pressure process, 600 kJ of
heat is transferred to the surrounding air at 25C. As a result, part of the
water vapor contained in the cylinder condenses. Determine (a) the entropy
change of the water and (b) the total entropy generation during this heat
transfer process.
Solution Saturated liquid–vapor mixture of water loses heat to its surround-
ings, and some of the vapor condenses. The entropy change of the water and
the total entropy generation are to be determined.
Assumptions 1 There are no irreversibilities involved within the system
boundaries, and thus the process is internally reversible. 2 The water
temperature remains constant at 100C everywhere, including the
boundaries.
Analysis We first take the water in the cylinderas the system (Fig. 7–69).
This is a closed systemsince no mass crosses the system boundary during
the process. We note that the pressure and thus the temperature of water in
the cylinder remain constant during this process. Also, the entropy of the
system decreases during the process because of heat loss.
(a) Noting that water undergoes an internally reversible isothermal process,
its entropy change can be determined from
(b) To determine the total entropy generation during this process, we
consider the extended system, which includes the water, the piston–cylinder
device, and the region immediately outside the system that experiences a
¢Ssystem
Q
Tsystem
600 kJ
1100 273 K 2
1.61 kJ/K
T = 100°C
H 2 O
600 kJ
Tsurr = 25°C
FIGURE 7–69
Schematic for Example 7–21.