Modern Engineering Thermodynamics

11.2 Two New Properties: Helmholtz and Gibbs Functions 365
Table 11.1 Summary of Thermodynamic Property Relations
New Property Relations
T = ∂u
∂s = ∂h
v ∂s p
v = ∂h ∂g
∂p = s ∂p T
p = − ∂u ∂f
∂v
= − s ∂v T
∂f ∂g
s = −
∂T
= − v ∂T p
Table 11.1 summarizes these results. The importance of this set of partial differential equations lies in the fact
that they relate easily measurable properties (p, v, T) to nonmeasurable properties (u, h, s, f, andg). Therefore,
accurate p, v, T data on any pure substance can be used to generate information about u, h, s, f, andg for that
substance. However, they do not provide a direct method for calculating u, h, ors from p, v, and T information.
We must look for additional information to complete this task. But, first, we take a short diversion into phase
change processes for which we can determine important results based on what we already know.
EXAMPLE 11.2
The design of a new Happy Food fast-food processing system requires the values of the specific Helmholtz and Gibbs functions
for superheated water vapor at 200. psia and 400.°F. Since most thermodynamic tables do not list these properties
directly, you are asked to calculate them for these conditions.
Solution
First, draw a sketch of the system (Figure 11.1).
Superheated water vapor
at 200. psia and 400.°F
The unknowns are the values of the specific Helmholtz and Gibbs
functions for superheated water vapor at 200. psia and 400.°F.
The specific Helmholtz and Gibbs functions are defined in the
text as
and
Specific Helmholtz function: f = u − Ts
Specific Gibbs function: g = h − Ts
New Happy Food
fast−food processing
system
FIGURE 11.1
Example 11.1.
Happy food
Therefore, we need u, h, and s information at the state defined by p = 200. psia and T = 400.°F = 860.R. From Table C.3a in
Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that, at this state,
and
Then,
and
uðp = 200: psia, T = 400:°FÞ = 1123:5 Btu/lbm
hðp = 200: psia, T = 400:°FÞ = 1210:8 Btu/lbm,
sðp = 200: psia, T = 400:°FÞ = 1:5602 Btu/lbm.R
f = 1123:5 Btu/lbm − ð400: + 459:67 RÞð1:5602 Btu/lbm.RÞ = −218 Btu/lbm
g = 1210:8 Btu/lbm − ð400: + 459:67 RÞð1:5602 Btu/lbm.RÞ = −131 Btu/lbm
The following exercises consider different pressures and temperatures and explore why the specific Helmholtz and Gibbs
functions were negative in Example 11.2.
Exercises
4. Determine the value of the specific Helmholtz function of the superheated water vapor in Example 11.2 if the pressure is
maintained at 200. psia but the temperature is increased to 1000.°F. Answer: f = −1320 Btu/lbm.
5. Determine the specific Helmholtz function of saturated liquid water at 200. psia. Answer: f = −103 Btu/lbm.
6. The Helmholtz and Gibbs functions calculated in Example 11.2 are both negative. Though this has no particular
significance at this point, use Table C.3a to determine the temperature and pressure at which the Helmholtz function is
zero for water. Answer: f = 0 for saturated liquid water at 0.0887 psia and 32.018°F (the triple point of water).