Modern Engineering Thermodynamics

10.7 Closed System Availability Balance 327
so
A 1 = ð5:00 lbmÞ 24:04 − 29:78 Btu
+ 136:12 lbf
144 in 2 /ft 2
bm
in 2
0:01218 − 0:01325
778:16 ft .lbf/Btu
ft3
lbm
− ð5:00 lbmÞð530 RÞ 0:0519 − 0:06296
Btu
lbm.R
½ð500: mphÞð5280 ft/miÞð1 h/3600 sÞŠ 2
+ ð5:00 lbmÞ
2ð32:174 lbm.ft/lbf .s 2 Þð778:16 ft .lbf/BtuÞ + ð32:174 ft/s 2 Þð30:0 × 10 3
ftÞ
ð32:174 lbm.ft/lbf .s 2 Þð778:16 ft .lbf/BtuÞ
= 247 Btu
and
"
A 2 = ð5:00 lbmÞ 154:77 − 29:78 Btu
+ 136:12 lbf
144 in 2 /ft 2
bm
in 2
1:046 − 0:01325
778:16 ft .lbf/Btu
ft3
lbm
#
− ð530 RÞ 0:31464 − 0:06296
Btu
+ 0 + 0 = 88:1 Btu
lbm.R
b. Now, from part a, we have A 2 − A 1 = 88.1 − 247 = −159 Btu. Note that the availability is higher in the first state.
Exercises
7. Determine the change in total availability in Example 10.3 if the initial (flying) state of the R-22 is a saturated vapor
rather than a saturated liquid. Assume all the other variables remain unchanged. Answer: A 2 − A 1 = −161 Btu.
8. If the state of the R-22 in the final (landed) state of Example 10.3 is a saturated liquid at 100.°F rather than a superheated
vapor, recompute the change in total availability for this system. Answer: A 2 − A 1 = −246 Btu.
9. To illustrate the impact of choosing the local environment (ground state), compute the total availabilities and the
change during landing in Example 10.3 if the ground state is changed from a saturated liquid at 70.0°F, where
p 0 = 136.122 psia, to a saturated liquid at −40.0°F, where p 0 = p sat (−40.0°F) = 15.222 psia, which is close to atmospheric
pressure. Answer: A 1 = 258 Btu, A 2 = 129 Btu, and A 2 − A 1 = −129 Btu.
10.7 CLOSED SYSTEM AVAILABILITY BALANCE
Since availability is a function of the system’s thermodynamic properties, it is therefore a thermodynamic property
itself. Perhaps, we can gain more insight into its engineering use if we carry out an availability balance for a
closed system. Using the general balance equation, Eq. (2.11), we can write
A transport + A production = A gain (10.7)
If availability were conserved like energy, we would be able to set A production = 0, but since availability is defined
for only reversible processes and most systems undergo irreversible processes, we can expect that A production ≠ 0
and availability is not conserved.
Since it is difficult to decide heuristically how total availability is transported across a system boundary, it is
easier to develop Eq. (10.7) from the definition of the gain in total availability given by Eq. (10.5) as
A gain = A 2 − A 1 = E 2 − E 1 + p 0 ðV 2 − V 1Þ − T 0 ðS 2 − S 1 Þ
From an energy balance on a system undergoing an actual irreversible process from state 1 to state 2, we have
(recall that we can choose to follow either an actual irreversible path or a hypothetical reversible path to evaluate
the change in the total system energy E, because energy is a point function whose integral is independent of
the integration path)
E 2 − E 1 = ð 1
Q 2 Þ act
− ð 1
W 2 Þ act
= 1 Q 2 − 1 W 2
where we drop the subscripts on heat and work transfer and allow them to represent the actual (irreversible)
process values from here on. The entropy balance for this situation is
Z 2
dQ
S 2 − S 1 = + 1 ðS P Þ
1 T 2 b