Modern Engineering Thermodynamics

3.12 Thermodynamic Charts 87
vð100:°F, 95:0 psiaÞ− vð100:°F, 90 psiaÞ
v
= ð 100:°F, 100: psiaÞ− v ð 100:°F, 90:0 psia Þ
95:0 psia− 90:0 psia
100: psia − 90:0 psia
or
vð100:°F, 95:0 psiaÞ = vð100:°F, 90 psiaÞ
95:0 psia− 90:0 psia
+ ½vð100:°F, 100: psiaÞ− vð100:°F, 90:0 psiaÞŠ
100: psia − 90:0 psia
And interpolating for the specific enthalpy h gives
= 0:5751 + 0:500ð0:5086 − 0:5751Þ = 0:54185 ft 3 /lbm
hð100:°F, 95:0 psiaÞ− hð100:°F, 90:0 psiaÞ
h
= ð 100:°F, 100: psiaÞ− h ð 100:°F, 90:0 psia Þ
95:0 psia− 90:0 psia
100: psia − 90:0 psia
or
hð100°F, 95 psiaÞ = hð100:°F, 90 psiaÞ
95:0 psia− 90:0 psia
+ ½hð100:°F, 100: psiaÞ− hð100:°F, 90:0 psiaÞŠ
100: psia − 90:0 psia
= 118:39 + 0:500ð117:73 − 118:39Þ = 118:06 Btu/lbm
Exercises
15. Use Table C.1b to find the values of p sat , v f , v g , u f , u g , h f , and h g for saturated water at 100.°C. Answers: p sat = 0.1013 MPa,
v f = 0.001044 m 3 /kg, v g = 1.673 m 3 /kg, u f = 418.9 kJ/kg, u g = 2506.5 kJ/kg, h f = 419.0 kJ/kg, and h g = 2676.0 kJ/kg.
16. Use Table C.3a to find the values of v, u, and h for superheated water vapor at 2000. psia and 1000.°F. Answers:
v = 0.3945 ft 3 /lbm, u = 1328.1 Btu/lbm, and h = 1474.1 Btu/lbm.
17. Use Table C.4b to find the values of v, u, and h for compressed liquid water at 30.0 MPa and 200.°C. Answers:
v = 0.0011302 m 3 /kg, u = 831.4 kJ/kg, and h = 865.3 kJ/kg.
18. Use Table C.5a to find the values of T sat , p sat , v g , and h g for saturated ammonia when v f = 0.02446 ft 3 /lbm and
h f = 53.8 Btu/lbm. Answers: T sat = 10.0°F, p sat = 38.51 psia, v g = 7.304 ft 3 /lbm, and h g = 614.9 Btu/lbm.
19. Use Table C.8b to find the values of T and h for superheated Refrigerant-134a when the pressure is 0.500 MPa and the
specific volume is 0.06524 m 3 /kg. Answers: T = 140.°C and h = 382.42 kJ/kg.
6. Use Table C.11a to find the values of T, v, and h for saturated mercury at 1.00 psia and a quality of 50.0%. Answers:
T = T sat = 457.72°F, v = 24.211 ft 3 /lbm, and h = 77.321 Btu/lbm.
3.12 THERMODYNAMIC CHARTS
Experimental data, equations of state, and statistical thermodynamics results can be combined into very accurate
thermodynamic phase diagrams, called thermodynamic charts. These two-dimensional property diagrams can be
constructed with various useful thermodynamic properties as coordinates. For example, Figure 3.26 shows a specific
volume vs. specific internal energy chart for water. This chart also includes lines of constant pressure, temperature,
and quality. Thus, given a pair of independent properties, such as p and T (or p and x in the wet
region), the u and v values can be immediately read from the coordinate axes. Notice that, in the wet region,
where 0 < x < 1, the constant temperature and constant pressure lines lie on top of each other, since p and T are
not independent in this region.
A series of similar charts for a variety of substances can be found in the charts portion of Thermodynamics Table
to accompany Modern Engineering Thermodymanics. It must be emphasized, however, that since the physical size of
these charts is very small, the values taken from them are not as accurate as those taken from a table for the
same substance, even if interpolation must be used within the table. Therefore, small charts like these are used
only when appropriate tables are not available or a state is to be fixed without using either pressure or temperature.
For example, given values for u and v for water, it would be much easier to find the other thermodynamic
properties at that state using Figure 3.26 than to do a double interpolation within the water tables (however,
the accuracy still is not as good as using the tables).