386 | Thermodynamics
(c) The entropy generated during this process can be determined by applying
an entropy balance on an extended systemthat includes the iron block and
its immediate surroundings so that the boundary temperature of the extended
system is at 285 K at all times:
or
Discussion The entropy generated can also be determined by taking the iron
block and the entire lake as the system, which is an isolated system, and
applying an entropy balance. An isolated system involves no heat or entropy
transfer, and thus the entropy generation in this case becomes equal to the
total entropy change,
which is the same result obtained above.
Sgen¢Stotal¢Ssystem¢Slake12.6516.974.32 kJ>K
Sgen
Qout
Tb
¢Ssystem
4838 kJ
285 K
1 12.65 kJ>K 2 4.32 kJ/K
Qout
Tb
Sgen¢Ssystem
SinSout¬ ¬ Sgen¬¬¢Ssystem
EXAMPLE 7–20 Entropy Generation in a Mixing Chamber
Water at 20 psia and 50F enters a mixing chamber at a rate of 300 lbm/min
where it is mixed steadily with steam entering at 20 psia and 240F. The
mixture leaves the chamber at 20 psia and 130F, and heat is lost to the
surrounding air at 70F at a rate of 180 Btu/min. Neglecting the changes in
kinetic and potential energies, determine the rate of entropy generation
during this process.
Solution Water and steam are mixed in a chamber that is losing heat at a
specified rate. The rate of entropy generation during this process is to be
determined.
Assumptions 1 This is a steady-flow process since there is no change with
time at any point and thus mCV0, ECV0, and SCV0. 2 There are
no work interactions involved. 3 The kinetic and potential energies are
negligible, ke pe 0.
Analysis We take the mixing chamberas the system (Fig. 7–68). This is a
control volumesince mass crosses the system boundary during the process.
We note that there are two inlets and one exit.
T 3 = 130°F
T 2 = 240°F
T 1 = 50°F
300 lbm/min Mixing
chamber
P = 20 psia
180 Btu/min
FIGURE 7–68
Schematic for Example 7–20.
123
Net entropy transfer
by heat and mass
123
Entropy
generation
123
Change
in entropy