Modern Engineering Thermodynamics

14.16 Miscellaneous Refrigeration Technologies 575
35. The microcooler described in Example 14.12 can also produce 0.250 J/s of cooling at 120. K with a power input of only
1.70 W when the ambient temperature is 22.0°C. Determine the ASC and actual coefficients of performance under these
conditions. Answer: COP reversed Stirling ASC R/AC = 0.685 and COP actual Stirling R/AC = 0.147.
36. Note that the Stirling ASC coefficient of performance of a refrigerator vanishes as T L → 0. What happens to the power
required to drive these systems as you approach lower and lower temperatures? If a Stirling ASC has the same efficiency as
a Carnot ASC and a Carnot ASC is the most efficient cycle possible, then what is the likelihood that we will ever be able to
reach absolute zero temperature in the laboratory? Answer: AsT L → 0, _W comp → ∞, consequently we will never be able to
reach absolute zero temperature in the laboratory (although temperatures in the range of 0.001 K have been reached).
14.16 MISCELLANEOUS REFRIGERATION TECHNOLOGIES
14.16.1 Joule-Thomson Expansion Cooling
The cooling that results from the expansion or throttling of a gas from a high to a low pressure is called
Joule-Thomson cooling. A refrigeration ASC analysis predicts that no cooling occurs in this type of throttling,
because an ideal gas throttling process is both isenthalpic and isothermal. However, real gases and vapors do
undergo a temperature change during isenthalpic processes, as dictated by their Joule-Thomson coefficient. This
coefficient is discussed in Chapter 6 and defined in Eq. (6.25) as
so that, approximately,
μ J =
∂T
∂p
h
(6.25)
ðΔTÞ h
= μ J ðΔpÞ h
(14.28)
The value of μ J can be either positive or negative, and it is usually larger at lower temperatures. 6 For example, the
Joule-Thomson coefficient for air at 20°C and several atmospheres is only about 0.3°C/atm (see Figure 6.6).
Therefore, throttling air from 100 psig (~6 atm) down to atmospheric pressure produces a temperature drop in
the air of only about 2°C. Such a system is shown in Figure 14.32.
The refrigeration or air conditioning COP of a Joule-Thomson expansion throttling device is given by
COP J
T
R/AC
and if T 1 = T 3 , p 2 = p 3 , and p 4 = p 1, then this reduces to
Q
= _ L T
= 1 − T 4h
j _Wj c
ðjT 1 − T 2s jÞ/ðη s Þ c
COP J
T
R/AC
=
μ J ðp 2 − p 1 Þ
h
i
(14.29)
T 1 ðp 2 /p 1 Þ ðk−1Þ/k
− 1 / ðη s Þ e
The vortex tube discussed in Chapter 9 is a variation on this technique. It does not appear to depend on the
Joule-Thomson effect, and it can produce cold temperatures in only part of the outlet flow. The remainder of
the outlet flow is quite warm (see Figure 9.15).
Q H
3
4
Warm air
Expansion
(throttling)
valve
Cool air
Q L
2
1
W comp
T
3
s = c
p = c
p = c
4h
s
2s
s = c
1
FIGURE 14.32
A Joule-Thomson refrigeration system.
6 If μ J is negative, then you have Joule-Thomson heating of the gas on throttling.