Modern Engineering Thermodynamics

10.6 Availability 325
and the total available energy is then
A = ð0:433 kgÞ½ð42:0 − 83:9 kJ/kgÞ + ð101 kN/m 2 Þð0:001000 − 0:001002 m 3 /kgÞ
− ð293 KÞð0:1510 − 0:2965 kJ/kg .KÞ + 0 + ð9:81 m/s 2 Þð0:762 mÞ/1Š = 3:56 kJ
Exercises
1. In Example 10.1, the total available energy was positive even though the water is colder than the local environment and
the contribution from the potential energy term is quite small. Determine the contribution of each term in the total
availability equation in this example and comment on its significance. Answer: m(u − u 0 ) = −18.1 kJ, mp 0 (v − v 0 ) ≈ 0 kJ,
−mT 0 (s − s 0 ) = 18.5 kJ, and mgZ/g c = 3.23 kJ. The entropy change dominates.
2. Recalculate the total availability in Example 10.1 when the water is at 20.0°C instead of 10.0°C. Answer: 3.23 kJ.
(The only nonzero term here is potential energy.)
3. If the mass of water in the glass in Example 10.1 is doubled, but all the other variables remain unchanged, determine
the new availability of the water. Answer: A = 7.12 kJ.
CRITICAL THINKING
Why is some of the energy within a system unavailable for use? Does its availability or unavailability depend on the
technology being used? Could energy that is unavailable in one system be considered as available in another? Give some
examples.
EXAMPLE 10.2
Determine the specific available energy in a stationary, rigid tank containing air (an ideal gas) at 20.0°C and 1.500 MPa.
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 10.6).
The unknown is the specific available energy in a stationary, rigid tank, The
material is air and the system is closed.
The specific availability is given by Eq. (10.4) as
p 0 = 0.101 MPa and T 0 = 20.0°C
p = 1.500 MPa
T = 20.0°C
Stationary
rigid tank
a = u − u 0 + p 0 ðv − v 0 Þ − T 0 ðs − s 0 Þ + V 2 /2g c + gZ/g c
For an ideal gas, u − u 0 = c v (T − T 0 ), v − v 0 = R(T/p − T 0 /p 0 ), and s − s 0 =
c p ln(T/T 0 ) − Rln(p/p 0 ). Then,
FIGURE 10.6
Example 10.2.
a = c v ðT − T 0 Þ + p 0 RðT/p − T 0 /p 0 Þ − T 0 c p lnðT/T 0 Þ + V 2 /2g c + gZ/g c
or
a = ð0:781 kJ/kg .KÞð20:0 − 20:0°CÞ + ð101 kN/m 2 293 K
Þð0:286 kJ/kg .KÞ
1500: kN/m 2 − 293 K
101 kN/m 2
− ð293 KÞ 1:004 kJ/kg .KÞ ln 293 K
1500: kN/m2
− ð0:286 kJ/kg .KÞ ln
293 K
101 kN/m 2 + 0 + 0 = 148 kJ/kg
Exercises
4. If the tank temperature in Example 10.2 is increased from 20.0°C to 200.°C and all the other variables remain unchanged,
determine the new specific availability of the gas in the tank. Answer: a = 140. kJ/kg.
5. If the pressure in the tank in Exercise 10.2 is reduced from 1.500 MPa to 0.500 MPa and all the other variables remain
unchanged, determine the new specific availability of the gas in the tank. Answer: a = 67.0 kJ/kg.
6. Recalculate the specific availability in Example 10.2 when the tank contains carbon dioxide at 50.0°C and 2.250 MPa.
Assume all the other variables remain unchanged. Answer: a = 115 kJ/kg.