Modern Engineering Thermodynamics

86 CHAPTER 3: Thermodynamic Properties
3.11 HOW DO YOU DETERMINE THE “THERMODYNAMIC STATE”?
First, remember that you need the values of only two independent intensive thermodynamic properties to fix the
state of a homogeneous material, and the problem statement always provides these values. Usually, you are
given the values of pressure, temperature, or specific volume in the problem statement. Sometimes specific internal
energy or specific enthalpy also is one of the values given.
Second, choose one of the two given values and look up the corresponding saturation values (f and g) ofthe
other given property from the saturation tables in Thermodynamic Tables to accompany Modern Engineering Thermodynamics,
Tables C.1 and C.2, for your material and compare them with the given value. You can then determine
the state of your material by following the rules in Table 3.9.
For example, if you are given water at a pressure of 1.0 MPa and a temperature of 200.°C, then you could
choose p given = 1.0 MPa and look up T sat at that value of p given . From Table C.1 at p given = 1.0 MPa, you find that
T sat =179.9°C. Since your value of T given =200.°C > T sat =179.9°C, the water must be in a superheated vapor
state. Similarly, if you choose T given = 200.°C instead of p given , then you would look up p sat at that value of T given .
From Table C.1 at T given = 200.°C, you find that p sat = 1.554 MPa. Since your value of p given =1.0 MPa < p sat =
1.554 MPa, you again conclude that the water must be in a superheated vapor state.
While it is true that any pair of independent properties fix the state of a simple substance subjected to only one
work mode, you must be able to deduce the system’s thermodynamic state (compressed liquid, saturated liquid
or vapor, liquid-vapor mixture, or superheated vapor) from the data given in a problem statement to know in
which table to find the other properties required in the analysis. It is important to remember that thermodynamic
states are unique and a given pair of independent properties fix the state at only one point in the tables. It is therefore
essential to understand how to determine which table to use in the solution of a thermodynamics problem.
In Example 3.9, we introduce notation of the form v(X°F, Y psia), which represents the value of the specific
volume evaluated at X°F and Y psia. For example, v(100.°F, 50. psia) means the value of the specific volume at
100.°F and 50. psia. This is a convenient way of recording the pair of independent intensive properties used to
determine the value of v. The same notation is used with the intensive properties u and h.
Table 3.9 How to Find the Thermodynamic “State”
Then You Have a
Properties Given in the
Problem Statement
Choose
Look Up in
Appropriate Table
Compressed
Liquid If
Mixture of Liquid
and Vapor If
Superheated
Vapor If
p given , T given p given T sat at p given T given < T sat T given = T sat T given > T sat
p given , T given T given p sat at T given p given > p sat p given = p sat p given < p sat
p given , υ given
p given
υ f and υ g
υ given < υ f
υ f < υ given < υ g
υ given > υ g
p given , u given
p given
u f and u g
u given < u f
u f < u given < u g
u given > u g
p given , h given
p given
h f and h g
h given < h f
h f < h given < h g
h given > h g
T given , υ given
T given
υ f and υ g
υ given < υ f
υ f < υ given < υ g
υ given > υ g
T given , u given
T given
u f and u g
u given < u f
u f < u given < u g
u given > u g
T given , h given
T given
h f and h g
h given < h f
h f < h given < h g
h given > h g
EXAMPLE 3.9
Find the specific volume and specific enthalpy of Refrigerant-134a at 100.°F
and 95.0 psia (Figure 3.25).
Solution
AcheckofTableC.7aofThermodynamic Tables to accompany Modern Engineering
Thermodynamics reveals that the saturation pressure of Refrigerant-134a at
100.°F is 138.83 psia. Since our actual pressure is less than the saturation pressure,
we must have superheated vapor. A check of Table C.8a reveals that
100.°F and 95.0 psia is indeed in the superheated region. However, 95.0 psia
is not a direct entry into this table, so we must use linear interpolation to find
the needed values. This is how a linear interpolation for v is carried out:
FIGURE 3.25
Example 3.9.
R-134a
T = 100.°F
p = 95.0 psia
v = ?
h = ?