Modern Engineering Thermodynamics

236 CHAPTER 7: Second Law of Thermodynamics and Entropy Transport and Production Mechanisms
EXAMPLE 7.10 (Continued )
where
Then,
p 1 = 14:7 psia = ð14:7 lbf/in 2 Þ 144 in 2 /ft 2 = 2117 lbf/ft 2
V 1 = mRT 1
= ð 1:00 lbm Þ ½ 53:34 ft .lbf/ ðlbm.RÞŠð70 + 459:67 RÞ
p 1 2117 lbf/ft 2 = 13:35 ft 3
Now, for an isothermal process, Eq. (7.68a) gives
Equation (7.59) gives us W irr as
Therefore,
W rev = 2117 lbf/ft 2 13:35 ft 3 ln 14:7
50:0 = −34,600 ft .lbf
ðS P Þ W
=
Z 2
1
dW irr
T
= 1 T
Z 2
1
W
dW irr = irr
1 T 2
W irr = W rev − W act = −34,600 − ð−42,000Þ = 7,400 ft.lbf
ðS P Þ W
= 7400 ft .lbf
70 + 459:67 R = 13:97 ft .lbf/R
= ð13:97 ft.lbf/R
Þ½1 Btu/ ð778:16 ft.lbfÞŠ = 0:0179 Btu/R
This example illustrates that you must know the relation between W irr and T before Eq. (7.68) can be integrated. The
simplest possible case occurs if T is a constant throughout the system volume, as in this example.
Exercises
16. Determine the work mode production of entropy in Example 7.10 if carbon dioxide gas (an ideal gas) is used instead of
air. Answer: (S P ) W = 122.3 ft·lbf/R.
17. If the losses in the electric motor in Example 7.7, Exercise 15, were interpreted as a work mode irreversibility with a
work transport energy efficiency of η W = 95.0%, then Eq. (7.46) gives W irr = (1 − η W )W rev = 0.050(800. W) = 40.0 W.
Determine the work mode production rate of entropy for this system if it has a uniform internal temperature of 30.0°C.
Answer: (S P ) W = 0.132 W/K.
18. The actual work to compress a spring was measured and found to be 3.00 J. The reversible work required to compress
the same spring was calculated to be 2.90 J. Determine the work mode production of entropy in the spring if the
internal temperature of the spring is uniform at 20.0°C. Answer: ðS P Þ W = 3:41 × 10 −4 J=K.
Note that you must also know the exact relation between Q and T or T b before Eqs. (7.60), (7.61), (7.65), and
(7.66) can be integrated. These relations are often empirically derived auxiliary formulae known as constitutive
equations. Fourier’s law,Newton’s law of cooling, and Planck’s radiation law are three such constitutive
equations, briefly discussed in Chapter 4, that relate heat transfer rate and temperature.
An alternative approach to evaluating the entropy production due to work mode irreversibilities is to attempt to
identify the sources of the irreversibilities and to mathematically model them with appropriate equations. This
is normally done by deriving relations for the work mode entropy production per unit time per unit volume σ W .
Since σ W has a value at every point within the system, the total entropy production is determined by integrating
σ W over time and the system volume as
Work mode entropy production:
Z Z
ðS P Þ w
= σ w dt dV (7.69)
V τ
and the entropy production rate is determined by integrating σ W over the system volume as
Work mode entropy production rate:
Z
ð_S P Þ w
= σ w dV (7.70)
V
These results are available for various work mode dissipation mechanisms, such as viscous dissipation within a
Newtonian fluid, electrical energy resistive dissipation, diffusion of dissimilar chemicals, and so on. However,
these mathematical models are normally quite complex, and therefore only the viscous dissipation and the electrical
resistance mechanism models are introduced at this point. Dissipation resulting from diffusion is discussed
in Chapter 9 in the section on mass flow production of entropy.