Modern Engineering Thermodynamics

330 CHAPTER 10: Availability Analysis
EXAMPLE 10.4 (Continued )
Solution
First, draw a sketch of the system (Figure 10.9).
The unknown is the irreversibility of this process. The material is water,
and the system is closed.
The irreversibility for this system can be computed from Eq. (10.14) using
an entropy balance to calculate the entropy production or directly from
Eq. (10.18). We use Eq. (10.18) for this example. Solving this equation
for the irreversibility gives
1I 2 = 1 − T 0
T bi
ð 1
Q 2 Þ − 1 W 2 + p 0 mðv 2 − v 1 Þ − ½mða 2 − a 1 ÞŠ system
and, since the entire process takes place at constant pressure, we can write
Z
1W 2 = m pdv = mpðv 2 − v 1 Þ
Applying the closed system energy balance equation to this system gives
x 1 = 0.00
T 1 = 120.°C
State 1 State 2
FIGURE 10.9
Example 10.4.
Ground state:
T 0 = 20.0°C
p 0 = 0.101 MPa
T s = 130.°C
x 2 = 0.500
1Q 2 = mðu 2 − u 1 Þ + 1 W 2
The state properties are
State 1 State 2 Ground State
x 1 = 0 x 2 = 0:500 T 0 = 20:0°C = 293 K
T 1 = 120:°C p 2 = p 1 = p sat ð120:°CÞ = 198:5 kN/m 2 p 0 = 0:101 MPa
v 1 = v f ð120:°CÞ
v 2 = v f ð120:°CÞ + x 2 v fg ð120:°CÞ
= 0:001060 m 3 /kg = 0:44648 m 3 /kg
u 1 = u f ð120:°CÞ = 503:5 kJ/kg u 2 = u f + x 2 u fg = 1516:4 kJ/kg
s 1 = s f ð120:°CÞ = 1:5280 kJ/kg .K s 2 = s f + x 2 s fg = 4:3292 kJ/kg .K
The change in specific availability of the system is calculated from Eq. (10.4) with V 1 = V 2 = 0 and Z 1 = Z 2 as
a 2 − a 1 = u 2 − u 1 + p 0 ðv 2 − v 1 Þ − T 0 ðs 2 − s 1 Þ
= ð1516:4 − 503:5 kJ/kgÞ+ ð101 N/m 2 Þð0:44648 − 0:001060 m 3 /kgÞ
− ð293 KÞð4:3292 − 1:5280 kJ/kg .KÞ = 237 kJ/kg
Then, we can compute the work using p 2 = p 1 = p as
and the heat transfer becomes
1W 2 = mpðv 2 − v 1 Þ
= ð1:00 kgÞð198:5 kN/m 2 Þð0:44648 − 0:001060 m 3 /kgÞ = 88:4kJ
1Q 2 = mðu 2 − u 1 Þ + 1 W 2
= ð1:00 kgÞð1516:4 − 503:5 kJ/kgÞ+ 88:4kJ= 1100 kJ
Then, Eq. (10.15) gives the irreversibility of this process as
1I 2 = 1 − 293 K
ð110 kJÞ − 88:4kJ+ ð101 kN/m 2 Þð1:00 kgÞð0:44648 − 0:001060 m 3 /kgÞ
403 K
− ð1:00 kgÞð236:9 kJ/kgÞ = 20:3kJ
Exercises
10. Suppose the initial state in Example 10.4 is a saturated vapor at 120.°C instead of a saturated liquid at 120.°C and the
system boundary temperature is lowered from 130.°C to 100.°C. Determine the irreversiblilty in the process under these
conditions, assuming all the other variables remain unchanged. Answer: 1 I 2 = 44.0 kJ.
11. Determine the irreversibility of the process in Example 10.4 if the system boundary temperature is (a) 150.°C, and
(b) 110.°C instead of 130.°C. Assume all the other variables remain unchanged. Answer: (a) 58.3 kJ, (b) −21.4 kJ. (The fact
that 1 I 2 < 0 in part b means that it would be impossible to have a surface temperature of only 110.°C for this process.)
12. What is the minimum possible system boundary temperature for the process described in Example 10.4? (Hint: Find T b
that produces 1 I 2 = 0.) Answer: T b = 120.°C.