Modern Engineering Thermodynamics

Summary 273
Note that this example does not include the entropy production due to the heat transfer required to cool the mixture down
to a drinkable temperature.
Exercises
29. Recalculate the entropy production in Example 8.15 when the average specific heat of the mixture of coffee and cream of
3856 J/(kg·K) is used instead of the value for pure water. Keep the values of all the other variables the same as they are
in Example 8.15. Answer: 1 (S P ) 2 = 0.259 J/K.
30. Determine the entropy production in the mixing process described in Example 8.15 when the temperature of the cream is
20.0°C and the temperature of the coffee is 90.0°C. Keep the vales of all the other variables the same as they are in
Example 8.15. Answer: 1 (S P ) 2 = 0.265 J/K.
31. (a) Determine the entropy produced in the process described in Example 8.15 when the mass of the cream added is
increased from 3.00 g to 10.0 g. Keep the values of all the other variables the same as they are in Example 8.15. (b) Plot
1(S P ) 2 /(m a + m b ) vs. the mass fraction y over the range 0 ≤ y ≤ 1 for the values of c, T a , and T b used in Example 8.15.
(c) What value of y maximizes 1 (S P ) 2 in part b? Answers: (a) 1 (S P ) 2 = 0.911 J/K, (b) See Figure 8.20, (c) [ 1 (S P ) 2 ] max
occurs at y = 0.52.
Coffee entropy production per unit mass
of mixture, J/kg.K
30
24
18
12
6
0
0.0
0.2 0.4 0.6 0.8 1.0
y = m a /(m a + m b )
FIGURE 8.20
Example 8.15, Exercise 31, part b.
Equations (8.12) and (8.13) show that the amount of entropy production in this type of mixing process
depends on the amounts mixed and their initial states. The larger the property differences between the initial
states, the larger is entropy production on mixing. That is, the farther the initial states are from the final equilibrium
state, the larger the associated entropy production is. This is a general characteristic of the second law.
SUMMARY
In this chapter, the energy and entropy balance and rate balance equations are used to investigate various
closed system applications. The indirect method of calculating values for S P and _S P occurs when they are calculated
from the entropy balance or the entropy rate balance equations. But, when these values are calculated
from the applicable entropy production rate density equations (σ W ) defined in Chapter 7, we called this the
direct method.
Reversible processes are defined as processes that occur with no internal irreversibilities, such as friction, viscosity,
heat transfer, or diffusion. Then, S P and _S P = 0, and the closed system entropy balance and entropy rate
balance equations for a reversible process reduce to
Entropy balance ðSBÞ: S 2 − S 1 = ms ð 2 − s 1 Þ = 1 Q 2
(8.7)
T b
and
Q
Entropy rate balance ðSRBÞ: _S = m_s =
_
T b
rev
rev
(8.8)