Modern Engineering Thermodynamics

Summary 351
In this problem, we have p H = p C = p m = p 0 and we neglect all kinetic and potential energy terms. Then, the flow
availabilities become
a fH = a f 1 = c w ðT 1 − T 0 Þ − c w T 0 ln T
1
= 1:00 Btu
ð130: − 55:0RÞ
= 4:99 Btu/lbm
T 0
− 1:00 Btu
ð55:0 + 459:67 RÞ ln
lbm.R
lbm.R
130: + 459:67 R
55:0 + 459:67 R
a fC = a f 2 = c w ðT 2 − T 0 Þ − c w T 0 ln T
2
= 1:00 Btu
ð60:0 − 55:0RÞ
T 0 lbm.R
− 1:00 Btu
ð55:0 + 459:67 RÞ ln
60:0 + 459:67 R
lbm.R
55:0 + 459:67 R
= 0:0240 Btu/lbm
a fM = a f 3 = c w ðT 3 − T 0 Þ − c w T 0 ln T
3
= 1:00 Btu
ð88:0 − 55:0RÞ
T 0 lbm.R
− 1:00 Btu
ð55:0 + 459:67 RÞ ln
88:0 + 459:67 R
lbm.R
55:0 + 459:67 R
= 1:02 Btu/lbm
The second law availability efficiency is now given by Eq. (10.37) with y = _m H / _m m = _m 1 / _m 3 = 0:180/ð0:180+0:270Þ =
0.400. Then,
ε mixing HX
= ð1−yÞða f 3 −a f 2 Þ
yða f 1 −a f 3 Þ
ð1−0:400Þ 1:02−0:0240
Btu
=
lbm.R
0:400 4:99−1:02 Btu = 0:374 = 37:4%
lbm.R
Exercises
47. The hot water temperature in Example 10.16 is reduced from 130.°F to 120.°F. Determine the new mixture temperature
in the sink and the second law efficiency of the mixing in the sink. Assume all the other variables (except T m ) remain
unchanged. Answer: T m = 84°F, and ε HX
= 38.2%.
48. If the hot and cold mass flow rates in Example 10.16 are equalized at 0.225 lbm/s each, determine the mixture
temperature in the sink and new second law efficiency of this mixing process. Assume all the other variables (except T m )
remain unchanged. Answer: T m = 95.0°F, and ε HX
= 41.5%.
49. Explain how the second law efficiency in Example 10.16 might be increased. (Hint: Describe how the mass flow rates
and the temperatures could be altered to increase the value of ε.)
50. According to Figure 10.8, when T b > T 0 , the heat and availability transports are in the same direction, but when T b < T 0 ,
they are in opposite directions. In a heat exchanger that has T C < T H < T 0 , show that the hot flow stream gains
availability during the internal heat transfer process and the cold flow stream loses it. (Hint: Look at the difference
between the inlet and exit flow stream specific flow availabilities.)
SUMMARY
In this chapter, we study a new concept in applied thermodynamics called available energy. The importance of
this material is discussed in the Introduction, and necessary background material is presented in the sections on
scalar and vector fields, conservative fields, and conservative forces. The concept of availability is based on the
maximum reversible work possible in a system, limited by the conditions present in the local environment. Oncethe
necessary background material is presented, we are able to define availability as the maximum possible useful
reversible work that a system could supply relative to its local environment as a ground state. At this point, we
could develop a closed system availability balance and carry out the solution of several example problems. Before
we could extend this to open systems, we had to define the concept of flow availability. With this as the concept
of how available energy crosses the system boundary, we are able to develop an open system availability balance
and modify the general form of this equation for several typical conditions, such as steady state and steady
flow. The chapter concludes with a discussion of a new type of efficiency, the second law efficiency. This efficiency
tells us how efficiently the available energy within the system is used. This is a very important concept in the
design of energy conversion systems.