Modern Engineering Thermodynamics

366 CHAPTER 11: More Thermodynamic Relations
11.3 GIBBS PHASE EQUILIBRIUM CONDITION
During a phase change process, the system pressure and temperature are not independent properties. This means
that, if we hold one of them constant during a phase change, the other must also remain constant. Under the
condition of constant pressure and temperature, dp = dT = 0, and Eq. (11.9) gives dg = 0. Since g = g f + xg fg ,we
can then write
dg = dg f + xdg fg + g fg dx = 0
Again, Eq. (11.8) can be used to evaluate dg f = dg fg = 0. Since x can vary during the phase equilibrium, dx
cannot be zero. Therefore, we are forced to conclude from the preceding equation that g fg = 0 at phase equilibrium,
or g f = g g . Using the definition of the Gibbs function, Eq. (11.7), we see that, at phase equilibrium,
or
g f = g g = h f − ðT sat Þs f = h g − ðT sat Þs g
or
h g − h f = h fg = ðT sat Þðs g − s f Þ = ðT sat Þs fg
s fg = h fg /T sat (11.10)
This gives us an important relation between the entropy and the enthalpy of a phase change, but we need much
more information to complete the process of determining the nonmeasurable properties u, h, ands from the
measurable properties p, v, and T. The following example illustrates the use of Eq. (11.10).
EXAMPLE 11.3
Use Eq. (11.10) to calculate the phase change entropy s fg for water at exactly 1.00 MPa and compare the result with the
value for s fg at exactly 1.00 MPa listed in Table C.2b in Thermodynamic Tables to accompany Modern Engineering
Thermodynamics.
Solution 1
The unknown is the phase change entropy of water. From (Eq. 11.10), we have
and from Table C.2b at p = 1.00 MPa, we find that
and
then, Eq. (11.10) gives
s fg = h fg
T sat
h fg = 2015:3 kJ/kg
T sat = 179:90°C
s fg = h fg 2015:3 kJ/kg
=
T sat 179:90 + 273:15 K = 4:4482 kJ/kg .K
Comparing this with the value for s fg listed in Table C.2b at p = 1.00 MPa, we find that it is exactly the same.
The following exercises illustrate some of the many uses of Eq. (11.10).
Exercises
7. Use Eq. (11.10) to compute the values of s fg for water at 0.0100 MPa and compare the result with the values listed in
Table C.2b. Find h fg and T sat at 0.01 MPa from Table C.2b. Answer: (s fg ) calc = 7.5021 kJ/kg·K.
8. Use Eq. (11.10) to compute the value of h fg for water at 100.°F and compare the result with the value listed in Table
C.1a. Find values for s fg and T sat at 100.°F from Table C.1a. Answer: (h fg ) calc = 1036.96 Btu/lbm.
9. Use the values for h fg and s fg found in Table C.2b for water at 10.0 MPa and Eq. (11.10) to calculate the value of T sat at
this state. Answer: (T sat ) calc = 584.2 K = 311.06°C.
1 To achieve the desired result, we need to carry a lot more significant figures than usual.