Modern Engineering Thermodynamics

346 CHAPTER 10: Availability Analysis
EXAMPLE 10.13 (Continued )
Heat pump
30.0°F
70.0°F
1.50 hp
electric motor
30.0 ×10 3
Btu/h
FIGURE 10.18
Example 10.13.
b. The second law availability efficiency is given by Eq. (10.32) with T 0 = T L = 30.0°F as
ε HP
= 1 − T
0
T H
COP actual
heat pump
!
= 1 −
30:0 + 459:67 R
70:0 + 459:67 R
ð7:86Þ = 0:594 = 59:4%
Note that, since T 0 = T L here, Eq. (10.33) could have been used with
to find
T
COP Carnot = H
=
70:0 + 459:67
heat pump T H − T L 70:0 − 30:0 = 13:241
ε HP
=
COP actual
heat pump
COP Carnot
heat pump
= 7:86 = 0:594 = 59:4%
13:24
Exercises
38. If the house temperature in Example 10.13 increases from 70.0°F to 85.0°F and all the other variables remain
unchanged, determine the new second law efficiency of the heat pump. Answer: ε HP
= 79.4%.
39. When the outside temperature in Example 10.13 increases from 30.0°F to 40.0°F, determine the new second law thermal
efficiency of the heat pump. Assume all the other variables remain unchanged. Answer: ε HP
= 44.5%.
40. The heat provided to the house in Example 10.13 is suddenly reduced from 30.0 × 10 3 Btu/h to 25.0 × 10 3 Btu/h.
What is the new second law thermal efficiency of the heat pump? Assume all the other variables remain unchanged.
Answer: ε HP
= 49.5%.
In the case of a refrigeration or air conditioning system, the net available energy input rate is also j _W in j, but
now the available energy rate of the “desired result” (cooling) is jð1 − T 0 /T L Þ _Q L j, where the absolute value has
been used to keep the efficiency a positive number. Then, the second law availability efficiency of a refrigeration
or air conditioning system is
ε R/AC
=
1 − T
0 _Q
T L
L
j _W in j
= 1 − T
0
T L
_Q L
j _W in j
!
= 1 − T
0
COP actual
T L ref or air cond
(10.34)