Modern Engineering Thermodynamics

324 CHAPTER 10: Availability Analysis
WHERE DID THE NAME “AVAILABILITY” COME FROM?
The potential we call availability today has been known by various names over the years. In 1873, Gibbs called it the available
energy of a body and a medium; in 1889, Gouy called it the utilizable energy; and in 1953, Rant named it exergy. Joseph
Keenan introduced the term availability in 1941, and it is still used.
where the kinetic and potential energy terms have been moved to the end of the last expression for convenience.
The specific availability, a, is now defined as
Specific availability, a
a = A/m = u − u 0 + p 0 ðv − v 0 Þ − T 0 ðs − s 0 Þ + V 2 /2g c + gZ/g c
(10.4)
Most processes do not take the system to its ground state. In that case, as a closed system undergoes a process
that carries it from state 1 to state 2, the change in total and specific availabilities are
A 2 − A 1 = E 2 − E 1 + p 0 ðV 2 − V 1 Þ − T 0 ðS 2 − S 1 Þ
= mu 2 − u 1 + p 0 ðv 2 − v 1 Þ − T 0 ðs 2 − s 1 Þ + ðV 2 2 − V 2 (10.5)
1 Þ/2g c + gðZ 2 − Z 1 Þ/g c
a 2 − a 1 = u 2 − u 1 + p 0 ðv 2 − v 1 Þ − T 0 ðs 2 − s 1 Þ + ðV 2 2 − V 1 2 Þ/2g c + gðZ 2 − Z 1 Þ/g c (10.6)
The following examples illustrate the calculation of these availability functions for various closed systems.
EXAMPLE 10.1
A cylindrical drinking glass 0.0700 m in diameter and 0.150 m high is three-fourths full of cold liquid water at 10.0°C and
is placed at 0.762 m above the floor on a table in a room. Determine the total availability of the water in the glass relative
to the floor. Take the local environment (ground state) to be at p 0 = 0.101 MPa and T 0 = 20.0°C = 293 K.
Solution
First, draw a sketch of the system (Figure 10.5).
3/4 full
7.00 × 10 −2 m
0.762 m
0.150 m
p 0 = 0.101 MPa
T 0 = 20.0°C
FIGURE 10.5
Example 10.1.
The unknown is the total availability of the water in the glass relative to the floor. The material is water, and the system is
closed.
The total availability of a system is given by Eq. (10.3) as
A = m u− u 0 + p 0 ðv − v 0 Þ − T 0 ðs − s 0 Þ + V2 + gZ
2g c g c
Water at 10.0°C and atmospheric pressure is a slightly compressed liquid. However, the amount of compression is very
small, and we can use the u, v, s, andu 0 , v 0 , s 0 values of saturated liquid water at 10.0°C and20.0°C, respectively from
Table C.1b of Thermodynamic Tables to accompany Modern Engineering Thermodynamics in the calculation. The glass is not
moving, so V = 0, but it does have a potential energy defined by its height at Z = 0.762 m. The mass of water in the glass is
m = πR 2 Lρ = πð0:0350 mÞ 2 ð3/4 × 0:150 mÞð1000: kg=m 3 Þ = 0:433 kg