Modern Engineering Thermodynamics

3.8 Quality 75
Although Eq. (3.26) was developed using specific volume, an identical argument can be used to expand it to all
other intensive properties (except pressure and temperature), resulting in equations of the form
x = v − v f
v fg
= u − u f
u fg
= h − h f
h fg
(3.27)
In addition, the term m f /m =1− x represents the relative amount of liquid present in the mixture, called the
moisture of the mixture.
EXAMPLE 3.4
Saturated water at 14.696 psia and 212°F has the following
properties:
v f = 0.01672 ft 3 /lbm v g = 26.80 ft 3 /lbm
u f = 180.1 Btu/lbm u g = 1077.6 Btu/lbm
h f = 180.1 Btu/lbm h g = 1150.5 Btu/lbm
If 0.200 lbm of saturated water at 14.696 psia is put into a
sealed rigid container whose total volume is 3.00 ft 3 (Figure
3.18), determine the following properties of the system:
a. The specific volume v.
FIGURE 3.18
b. The quality, x, and moisture, 1 − x.
Example 3.4.
c. The specific internal energy u.
d. The specific enthalpy h.
e. The mass of water in the liquid and vapor phases, m f and m g .
Solution
The system is a closed rigid container.
∀ = 3.00 ft 3
m = 0.200 lbm
p = 14.696 psia
System
boundary
a. The specific volume can be calculated directly from its definition, Eq. (3.4). as
v = V/m = 3:00/0:200 = 15:0ft 3 /lbm
b. The quality can be calculated from Eq. (3.26) or (3.27) and Eq. (3.25) as
x = v − v f 15:0 − 0:01672
=
v fg 26:80 − 0:01672 = 0:559
or x = 55.9% vapor. Therefore, the amount of moisture present is 1 − x = 0.441, or the mixture consists of 44.1% moisture.
c. The specific internal energy can be obtained by combining Eq. (3.27) with the definition u fg = u g − u f to give u = u f + xu fg =
u f + x(u g − u f ), or
u = 180:1 + ð0:559Þð1077:6 − 180:1Þ = 682 Btu/lbm
d. The specific enthalpy can be obtained by combining Eq. (3.27) with the definition h fg = h g − h f to give h = h f + xh fg =
h f + x(h g − h f ), or
h = 180:1 + ð0:559Þð1150:5 − 180:1Þ = 722 Btu/lbm
e. To obtain the mass of water in the liquid and vapor phases, we can use the original definition of quality given in Eq.
(3.22) to get m g = xm = 0.559(0.2) = 0.112 lbm of saturated water vapor and then m f =(1− x)m = m − m g = 0.088 lbm
of saturated liquid water.
EXAMPLE 3.5
What total mass of saturated water (liquid plus vapor) should be put into a 0.500 ft 3 sealed, rigid container at 14.696 psia so that
the water passes exactly through the critical state when the container is heated? Also, determine the initial quality in the vessel.
Solution
Processes carried out in sealed rigid containers are constant volume (or isochoric) processes. Therefore, the process path on a
p-v diagram is a vertical straight line, as shown in the p-v diagram of Figure 3.19. In this problem, we are given the final
(Continued )