Modern Engineering Thermodynamics

3.10 Thermodynamic Tables 85
Critical point
Saturated vapor (x = 1)
Pressure, p
Compressed
liquid
region
Saturated liquid (x = 0)
0 < x < 1
All properties
in this region are
to be calculated
using the quality, x
Superheated
vapor
region
Specific volume, v
FIGURE 3.24
Regions of application of thermodynamic tables.
When thermodynamic data are given in problem statements, you normally are not told whether the state of
the system is compressed, saturated, or superheated. To decide which table to use, you must be able to deduce
the state of the system from the information given. This can be done by comparing the given properties with
the saturation properties at the same temperature or pressure. For example, suppose you are given water at
500°F and 1000. psia. How can you tell if it is a compressed liquid, saturated liquid, a mixture of liquid plus
vapor (i.e., wet), a saturated vapor, or a superheated vapor? The answer is obtained from the saturation data in
Table C.1a or C.2a of Thermodynamic Tables to accompany Modern Engineering Thermodynamics. Thesetables
tell you that, at 500°F, the saturation pressure is 680.8 psia, and at 1000. psia, the saturation temperature is
544.61°F. First of all, we could use the saturation pressure of 680.8 psia as a guide and note that the actual state
(500°F, 1000. psia) is at a pressure greater than that required to produce a saturated liquid at 500°F, consequently
the water must be in a compressed liquid state. Alternatively, we could use the saturation temperature
of 544.61°F as a guide and note that the actual state has a temperature (500.°F) that is less than that required
for a saturated liquid at 1000. psia (544.61°F), so again the water must be in a compressed (or subcooled)
liquid state. Consequently, we obtain all other desired property information from Table C.4a, the compressed
water table.
Similarly, in metric units, if you have water at 1.00 MPa and 200°C, a check of the saturation data in Table C.1b
reveals that, at 200.°C, the saturation pressure is 1.554 MPa, which is greater than the actual pressure of
1.00 MPa. Therefore, the actual state of the water must be in the superheated vapor region. A check of
Table C.3b reveals that this state can be easily found in the table.
How do you decide which table to use when you are given properties other than pressure and temperature? You use
the same basic technique. For example, suppose you are given 3.00 lbm of water in a 15.0 ft 3 closed, rigid container
at 14.696 psia. The specific volume of the system, then, is v = 15.0/3.00 = 5.00 ft 3 /lbm. A check of Table C.2a reveals
that, at 14.696 psia, v f = 0.01672 ft 3 /lbm, and v g =26.80ft 3 /lbm. Since the actual specific volume (5.00 ft 3 /lbm)
falls between these two values (v f < v < v g ), the state of the water must be in the liquid plus vapor (wet) region, and
it therefore has a quality of x = (5.00 − 0.01672)/(26.8 − 0.01672) = 0.186, or 18.6%. To get more familiar with
these tables, it is recommended that you verify the states given in Table 3.8 for water.
Table 3.8 The States of Water Fixed by Various Combinations of Property Pairs
Pair of Independent Properties
T = 500.°F, p = 1000. psia
p = 1.00 MPa, T = 200.°C
T = 170.°F, x = 1.0
p = 14.696 psia, v = 5.00 ft 3 /lbm
u = 500. Btu/lbm, p = 100. psia
h = 1192.6 Btu/lbm, T = 300.°F
p = 0.100 MPa, h = 200. kJ/kg
T = 100.°C, v = 8.585 m 3 /kg
ν = 0.10 m 3 /kg, x = 1.0
h = 3157.7 kJ/kg, u = 2875.2 kJ/kg
State (Correct Table to Use)
Compressed or subcooled liquid (C.4a)
Superheated vapor (C.3b)
Saturated vapor (C.1a)
Liquid-vapor mixture (C.2a)
Liquid-vapor mixture (C.2a)
Superheated vapor (C.3a)
Compressed or subcooled liquid (C.4b)
Superheated vapor (C.3b)
Saturated vapor (C.1b or C.2b)
Superheated vapor (C.3b)