Modern Engineering Thermodynamics

326 CHAPTER 10: Availability Analysis
EXAMPLE 10.3
The instrument cooling system on an aircraft flying at 30.0 × 10 3 ft at 500. mph contains 5.00 lbm of saturated liquid
Refrigerant-22 at 50°F. As the aircraft lands, a malfunction occurs that causes the refrigerant to be heated to 400.°F and
100. psia. Determine
a. The total availability of the refrigerant before and after the aircraft lands.
b. The change in total availability during the landing.
Take the local environment (ground state) to be saturated liquid Refrigerant-22 at T 0 = 70.0°F (sothatp 0 = p sat (70.0°F) =
136.12 psia).
Solution
First, draw a sketch of the system (Figure 10.7).
R-22
Ground state:
R-22 at 70.0°F and 136.12 psia
FIGURE 10.7
Example 10.3.
The unknowns are the total availability of the refrigerant before and after the aircraft lands and the change in total availability
during the landing. The material is R-22, and the system is closed.
a. 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
Tables C.9a and C.10a give the thermodynamic properties of the refrigerant as
State 1 ðflyingÞ State 2 ðlandedÞ Ground state
x 1 = 0:00 p 2 = 100:psia x 0 = 0:00
T 1 = 50:0°F T 2 = 400:°F T 0 = 70:0°F
v 1 = 0:01281 ft 3 /lbm v 2 = 1:046 ft 3 /lbm v 0 = 0:01325 ft 3 /lbm
u 1 = 24:04 Btu/lbm u 2 = 154:77 Btu/lbm u 0 = 29:78 Btu/lbm
s 1 = 0:0519 Btu/lbm.R s 2 = 0:31464 Btu/lbm.R s 0 = 0:06296 Btu/lbm.R
V 1 = 500 mph V 2 = 0 p 0 = 136:12 psia
Z 1 = 30,000 ft Z 2 = 0
The total availability is given by Eq. (10.5) as
A = mu− u 0 + p 0 ðv − v 0 Þ − T 0 ðs − s 0 Þ + V 2
/2g c + gZ/g c