Modern Engineering Thermodynamics

7.8 Numerical Values for Entropy 225
The tables and charts in Thermodynamic Tables to accompany Modern Engineering Thermodynamics list specific
entropy along with the specific properties v, u, and h. Specific entropy values are obtained from these sources in
the same way that any of the other specific properties are obtained. In particular, the quality x of a liquid-vapor
mixture is computed using the same type of lever rule relation as was used with v, u, and h; that is,
x = v − v f
v fg
= u − u f
u fg
= h − h f
h fg
= s − s f
s fg
(7.41)
EXAMPLE 7.6
Determine the change in total entropy of 3.00 lbm of steam at 100.°F and 80.0% quality when it is heated in an unknown
process to 200. psia and 800.°F.
Solution
First, draw a sketch of the system (Figure 7.14).
Steam
m = 3.00 lbm
T 1 = 100.°F
x 1 = 80.0%
State 1
Unknown process
s 2 − s 1 = ?
State 2
m = 3.00 lbm
p 2 = 200. psia
T 2 = 800.°F
FIGURE 7.14
Example 7.6.
The unknown is the change in total entropy of the steam. Since we are given two independent properties in each state in
this problem, we do not need to know how the heating process (i.e., the path) took place. We have a closed system consisting
of 3.0 lbm of water, for which
State 1
T 1 = 100:°F
x 1 = 0:800
s 1 = s f ð100:°FÞ + x 1 s fg ð100:°FÞ
= 0:1296 + 0:800ð1:8528Þ
= 1:6118 Btu/lbm.R
Unknown process path
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State 2
p 2 = 200: psia
T 2 = 800:°F
s 2 = 1:7662 Btu/lbm.R
where the specific entropy values have been found in Tables C.1a and C.3a. Then,
S 2 − S 1 = mðs 2 − s 1 Þ
= ð3:00 lbmÞ½ð1:7662 − 1:6118Þ Btu/ ðlbm.RÞŠ
= 0:463 Btu/R
Exercises
10. Determine the change in specific entropy of steam as it is cools from a saturated vapor (x = 100.%) at 100.°F toa
saturated liquid (x = 0%) at 100.°F. Answer: s 2 − s 1 = s f (100.°F) − s g (100.°F) = −s fg (100.°F) = −1.8528 Btu/lbm· R.
11. Saturated liquid ammonia at 0.00°C is heated in a constant pressure process until it has a quality of 50.0%. Determine
the change in specific entropy of the ammonia. Answer: s 2 − s 1 = 2.311 kJ/kg·K.
12. Determine the magnitude of the change in specific entropy of a water liquid-vapor mixture at 100.°C as its quality
decreases from 1.000 by (a) 1.00%, (b) 10.0%, (c) 50.0%. Answers: (a) |s 2 − s 1 | = 0.0605 kJ/kg·K, (b) |s 2 − s 1 | =
0.605 kJ/kg·K, and (c) |s 2 − s 1 | = 3.024 kJ/kg·K.
Figure 7.15 is an h-s plot for water. It is called a Mollier diagram after the German engineer Richard Mollier
(1863–1935), who developed it in 1904. States 1 and 2 of Example 7.6 are shown on this chart to illustrate its
use. Small charts like this are usually inaccurate for engineering problem solving. Professional engineers used
much larger charts like this before thermodynamic property software became available.
At this point, we must expand the classical concepts presented thus far so that they fit into the more general
balance equations introduced at the beginning of this chapter. We must now look for a set of general entropy
transport mechanisms, valid for both open and closed systems, that are consistent with Eq. (7.28) when applied
to closed systems.