Modern Engineering Thermodynamics

398 CHAPTER 11: More Thermodynamic Relations
EXAMPLE 11.17
Suppose 1.00 lbm of superheated steam at 1.00 psia and 200.°F is mixed adiabatically and aergonically with 5.00 lbm of
superheated steam at 5.00 psia and 400.°F in a closed, rigid system. Determine the final temperature and pressure.
Solution
The unknowns here are T 2 and p 2 , and since this is a closed system, the energy balance (neglecting any changes in system
kinetic and potential energy) is
1Q 2
.
0
ðadiabaticÞ
− 1 W 2 = mðu 2 − u 1 Þ = U 2 − U 1
↘ 0
ðaergonicÞ
or U 2 = U 1 ,oru 2 = u 1 = U 1 /m 1 = (m A u A + m B u B )/(m A + m B ). Also, for a closed, rigid system, the total volume and mass are constant, so
v 2 = v 1 = ðm A v A + m B v B
Þ/ ðm A + m B Þ
This problem is very difficult to solve using the steam tables or the Mollier diagram. It requires the construction of lines of constant
u and constant v for various combinations of pressure and temperature, then finding the intersection of the u = u 1 = u 2 and
v = v l = v 2 lines. However, the steam states given in the problem statement fall within the ideal gas region of Figure 11.12, so we
can solve this problem with reasonable accuracy using the ideal gas equations of state. Since the reference state values ultimately
cancel out here, we can simplify the algebra by taking T o = 0 R and u o = 0 Btu/lbm, then we can write u A = c v T A , u B = c v T B , and
u 2 = c v T 2 . Also, from the Boyle-Charles ideal gas equation, we have that v A = RT A /p A , v B = RT B /p B ,andv 2 = RT 2 /p 2 .Then,
u 2 = c v T 2 = u 1 = c v ðm A T A + m B T B
Þ/ ðm A + m B Þ
or
and
or
T 2 = ðm A T A + m B T B
v 2 = RT 2
p 2
Þ/ ðm A + m B Þ
= Rm ð AT A /p A + m B T B /p B Þ
m A + m B
p 2 =
For the values given in the problem statement, we get
and
ðm A + m B ÞT 2
m A T A /p A + m B T B /p B
1:00 lbm
T 2 =
ð Þð659:67
R Þ+ ð 5:00 lbm Þð859:67
RÞ
= 827 R = 367°F
6:00 lbm
ð6:00 lbmÞð827 RÞ
p 2 =
= 3:26 psia
ð1:00 lbmÞð659:67 RÞ/1:00 ð psiaÞ+ ð5:00 lbmÞð859:67 RÞ/5:00 ð psiaÞ Note that, since the specific heat and gas constant cancel out in the equations for the final temperature and pressure in
Example 11.17, they are independent of the material being mixed. The following exercises illustrate the use of this material.
Exercises
48. Determine the final temperature and pressure in Example 11.17 if the 1.00 lbm of superheated steam is at 5.00 psia and
200.°F instead of 1.00 psia and 200.°F. Answer: T 2 = 367°F and p 2 = 5.00 psia.
49. Suppose we increase the mass and temperature of one of the components in Example 11.17 so that we are mixing
5.00 lbm of superheated steam at 1.00 psia and 800.°F with 5.00 lbm of superheated steam at 5.00 psia and 400.°F.
Determine the final temperature and pressure of this mixture. Answer: T 2 = 600.°F and p 2 = 1.48 psia.
50. Suppose the superheated steam components in Example 11.17 are replaced with superheated R-22 at exactly the same
temperatures and pressures. Determine the final temperature and pressure of this mixture. Answer: T 2 = 367°Fandp 2 = 3.26 psia.
SUMMARY
In this chapter, we discuss a series of generalized thermodynamic property relations. We also introduce two
thermodynamic properties, the Helmholtz and Gibbs functions, and develop a series of differential property
relations, known as the Maxwell thermodynamic property equations. The Clapeyron equation, Gibbs phase equilibrium