Modern Engineering Thermodynamics

576 CHAPTER 14: Vapor and Gas Refrigeration Cycles
EXAMPLE 14.13
Determine the outlet temperature and COP of a Joule-Thomson expansion throttling device using air when the inlet
temperature and pressure are 70.0°F and 300. psia and the outlet pressure is 14.7 psia. Assume the Joule-Thomson coefficient
for air in this range is 0.0300°F/psi, and that the isentropic efficiency of the air compressor is 90.0%.
Solution
Using Figure 14.32 as the illustration for this example, from Eq. (14.28) we have
T 2 − T 1 = μ J ðp 2 − p 1 Þ = 0:0300ð14:7 − 300: Þ = −8:56°F
Then, T 2 = 70.0 − 8.56 = 61.4°F, and Eq. (14.29) gives the COP as
COP J
T
R/AC
=
0:0300ð14:7 − 300: Þ
14:7 0:40/1:40 . = 0:0252
ð70:0 + 459:67Þ
− 1 0:90
300:
Therefore, the thermal efficiency of this type of Joule-Thomson expansion throttling refrigeration or air conditioner is only
2.52%. 7
Exercises
37. The air compressor in Example 14.13 is replaced by one with an isentropic efficiency of 95%. Determine the new
Joule-Thomson coefficient of performance of the device. Assume all the other variables remain unchanged. Answer:
COP J−T = 0.027.
38. If the inlet air temperature in Example 14.13 is reduced from 70.0°F to 50.0°F, determine the outlet air temperature and
the Joule-Thomson coefficient of performance. Assume all the other variables remain unchanged. Answer: T 2 = 41.4°F
and COP J−T = 0.0260.
39. If the initial pressure in Example 14.13 is increased from 300. psia to 3000. psia, determine the new outlet temperature
and the Joule-Thomson coefficient of performance of the device. Assume all the other variables remain unchanged.
Answer: T 2 = −19.6°F and COP J−T = 0.195.
40. The Joule-Thomson coefficient for CO 2 at 2.00 MPa is 0.0150°C/kPa. Carbon dioxide initially at 20.0°C is throttled
from 2.00 MPa to atmospheric pressure. Determine the outlet temperature and the Joule-Thomson coefficient of
performance. Answer: T 2 = −8.50°C and COP J−T = 0.179.
Though the Joule-Thomson refrigerator is not very effective by itself, it is often used in conjunction with other refrigeration
systems. Figure 14.33 shows it being used with a reversed Brayton cycle to liquefy the working fluid. This was the basic
technique used by Karl von Linde (1842–1934) to produce liquid air on a large commercial scale in 1895.
Q H
3
4
Exp.
Hot gas
2
Comp.
1
(W in ) net
T
5
p = c
6h
p = c
3
4s
p = c
2s
1
Regenerator
Q regen
Makeup gas
Throttling valve
5
6
Liquid
Q L
Drain
s
FIGURE 14.33
The basic Linde process for liquefying air.
7 Note that the thermal efficiency of a heat pump is always greater than 100% (i.e., its COP > 1.0), but the thermal efficiency of a refrigerator or air
conditioner can be less than 100%.