Modern Engineering Thermodynamics

380 CHAPTER 11: More Thermodynamic Relations
11.7.2 Superheated vapor tables
Superheated vapor tables are somewhat easier to construct.
1. Begin by choosing a pair of pressure-temperature (p, T) values and calculate the corresponding specific
volume v from Eq. (11.31b).
2. Values are then calculated for u, h, and s from Eqs. (11.32), (3.17), and (11.33) utilizing (∂p/∂T) v
determined from Eq. (11.31b).
The compilation of v, u, h, ands for each set of p and T values chosen forms a superheated vapor table like
Table C.3a or C.3b.
Many tables do not list both u and h values, since these properties are simply related to each other through
h = u + pv. Therefore, if only one is listed in a table (it is usually h), the other can be easily calculated.
11.8 THERMODYNAMIC CHARTS
When accurate values for p, T, v, u, h, ands have been determined for the construction of saturated and
superheated property tables, it is a relatively simple task to plot these values to form thermodynamic charts.
Two-dimensional plots allow only two independent properties to be plotted, and the remaining properties have
to be added to the plot as parametric families of lines representing constant property values (isotherms, isobars,
etc.). For example, the Mollier diagram (see Figure 7.15a) has h and s as coordinate axes. This means that all the
remaining property information must be displayed as families of lines of constant p, constant T, constant v, and
so forth.
Because of the large number of variables to choose from and the lack of any standard thermodynamic chart
format, the charts found in Thermodynamic Tables to accompany Modern Engineering Thermodynamics have many
coordinate axes (h-s, T-s, p-h, p-v, v-u, etc.).
Although all thermodynamic tables and charts up to about 1950 were generated from manual calculations, the
use of a modern digital computer can substantially reduce the amount of human labor involved. Most of the
required software programming is straightforward, simply by following the steps outlined previously. However,
oneaspectofthisprocessthatisnotsoobvious involves solving Eq. (11.31b) for v when p and T are known.
These equations are often so algebraically complex that v cannot be determined explicitly in terms of p and T.
EXAMPLE 11.11
A new substance has the following equations of state corresponding to Eqs. (11.31a) through (11.31d). Here, we just letter
the equations (a) through (d) to avoid any confusion. That is, Eq. (11.31a) is just called (a) here.
p sat = exp A 1 − A
2
T sat
p = RT
v − T h
v 2 exp B 1 − B i
2
T
v f =
1
C 1 + C 2 T sat
(a)
(b)
(c)
c 0 v = D 1 = constant
(d)
where A 1 , A 2 , B 1 , B 2 , C 1 , C 2 ,andD 1 are all empirical constants. Determine the equations for u g , u f , h g , h f , s g ,ands f for this
material in the saturated region.
Solution
For the saturation tables, let A = A 1 – A 2 /T sat and B = B 1 – B 2 /T, then p sat = exp[A] and
p = RT
v − T
v 2 exp½BŠ
then,
dp
= A
2
dT
sat
Tsat
2 exp½AŠ