Modern Engineering Thermodynamics

336 CHAPTER 10: Availability Analysis
EXAMPLE 10.7 (Continued )
For a steady state adiabatic, aergonic process _Q = _W ¼ 0, and the resulting energy rate balance gives (this is a type of throttling
process, see Chapter 4)
h 2 = h 1 + V2 1 − V2 2
= 1363:4 Btu
2g c
lbm + ð50:0ft/sÞ 2 − ð300: ft/sÞ 2
2 32:174 lbm
.ft
778:16
lbf .s ft .lbf
2
Btu
= 1360 Btu
lbm
The values p 2 = 14.696 psia and h 2 = 1361.7 Btu/lbm fix the exit state, and the remaining properties at the exit station can
now be determined from the steam tables (by interpolation) as T 2 = 655°F and s 2 = 1:9981 Btu/lbm.R: Then, Eq. (10.20)
gives the specific flow availability as
a f = h − h 0 − T 0 ðs − s 0 Þ + V 2
2g c
+ gZ
g c
The local environment (ground state) values are h 0 = h f (70.0°F) = 38.1 Btu/lbm, and s 0 = s f (70.0°F) = 0.0746 Btu/lbm · R,
and inserting the numerical values with Z 1 = Z 2 = 0, we get
and
a f 1 = ð1363:4 − 38:1 Btu/lbmÞ − ð70:0 + 459:67 RÞð1:3990 − 0:0746 Btu/lbm.RÞ
+
ð50:0ft/sÞ 2
2 32:174 lbm .ft
lbf .s 2
778:16 ft .lbf
Btu
+ 0 = 624 Btu
lbm
a f 2 = ð1361:7 − 38:1 Btu/lbmÞ − ð70:0 + 459:67 RÞð1:9970 − 0:0746 Btu/lbm.RÞ
+
ð300: ft/sÞ 2
2 32:174 lbm .ft
lbf .s 2
778:16 ft .lbf
Btu
+ 0 = 307 Btu
lbm
The irreversibility rate is found from Eq. (10.26), and if we divide through by _m , we have the irreversibility per unit mass of
steam flowing, or
_I / _m = I/m = 1 − T
0 _Q / _m − _W/ _m + a f 1 − a f 2
T b
and, since _Q = _W = 0 here, this equation reduces to
I
m = a f 1 − a f 2 = 624 − 307 = 317 Btu
lbm
Exercises
19. Determine the irreversibility rate in Example 10.7 if the mass flow rate out of the pipe crack is 0.100 lbm/s.
Answer: _I = 31:7 Btu/s:
20. If the crack in the pipe in Example 10.7 is horizontal at a height of 15.0 ft from the floor, determine the inlet and exit flow
availabilities of the steam relative to the floor. Answer: Only gZ/g c = 0.0193 Btu/lbm is added to a f1 and a f2 in Example 10.7.
21. If the leak in Example 10.7 is not adiabatic but has a heat transfer rate per unit flow rate of _Q / _m = 316 Btu/lbm and a
boundary temperature of 500.°F, determine the new irreversibility per unit mass of steam exiting the crack. Assume all other
variables remain unchanged (i.e., assume this small value of _Q / _m does not significantly change the values of h 2 and s 2 ).
Answer: I/m = 316 Btu/lbm.
EXAMPLE 10.8
Superheated steam at 2.80 lbm/s, 100. psia, and 500.°F enters
a horizontal, stationary, insulated nozzle with a negligible
inlet velocity and expands to 10.0 psia. The friction and
other irreversibilities within the nozzle cause the exit velocity
to be only 95.0% of that produced by an isentropic
expansion. Taking the local environment (ground state) to
be saturated liquid water at 70.0°F, determine
a. The inlet specific flow availability.
b. The exit specific flow availability.
c. The irreversibility rate inside the nozzle.
100. psia
500.°F
2.80 lbm/s
Ground state:
x 0 = 0.00
T 0 = 70.0°F
Nozzle
Insulation
10.0 psia
V = 0.95 V isentropic
Solution
First, draw a sketch of the system (Figure 10.13).
FIGURE 10.13
Example 10.8.