Modern Engineering Thermodynamics

570 CHAPTER 14: Vapor and Gas Refrigeration Cycles
WHO INVENTED MECHANICAL REFRIGERATION FIRST, BRAYTON
OR GORRIE?
If Gorrie developed his refrigeration cycle in 1844 and Brayton developed his power cycle in 1873, why do we call Gorrie’s
cycle a reversed Brayton cycle? The reversed name order is just tradition. It probably occurred because power cycles have
always been more important to the development of societal goals than refrigeration cycles.
The original reversed Brayton cycle using a reciprocating piston compressor and expander (sometimes called gas expansion
with external work), developed by Dr. Gorrie in 1844, was not commercially successful. He received a patent in 1851 but
could not raise capital to produce his refrigerator, and he died a disappointed man.
The reversed Brayton cycle was finally established as a viable refrigeration technology by Sir William Siemens (1823–1883)
in 1857. Carl von Linde (1842–1934) unsuccessfully tried to liquefy air using this cycle in 1894. Then, he switched to a
Joule-Thompson expansion technique that allowed him to liquefy air at −190°C in 1895. In 1902, the French engineer
Georges Claude (1870–1960) finally succeeded in liquefying air using a reversed Brayton cycle. The reciprocating piston
technology was finally replaced by turbine technology around 1935 in Germany.
and
T 3 = T 4s ðp 3 /p 4s
Þ ðk−1
Þ/k = T 4s PR ðk−1
where PR is the isentropic pressure ratio. Hence, T 2s /T 3 = T 1 /T 4s , and it can be shown that, for an ASC
(i.e., (η s ) c =(η s ) e = 1.0), Eqs. (14.22) and (14.23) reduce to
and
COP reversed
=
Brayton ASC
HP
COP reversed
=
Brayton ASC
R=AC
T 3
T 3 − T 4s
=
T 4s
T 3 − T 4s
=
Þ/k
1 − PR (14.24)
−1
−1
PR ðk−1Þ/k
− 1
(14.25)
It is easy to show that these equations can also be obtained directly from Eqs. (13.23), (14.4), and (14.5). These
results are illustrated in the following example.
EXAMPLE 14.10
Determine the COP and cycle minimum cooling temperature of Gorrie’s 1844 reversed Brayton cycle refrigerator
if it has a pressure ratio of 2.00 to 1, a compressor inlet temperature of 70.0°F, and an expander inlet temperature
of 80.0°F, using
a. An ASC analysis.
b. An ideal gas analysis that includes typical mid 19th century compressor and expander isentropic efficiencies of
65.0% each.
Solution
Use Figure 14.27 as the system illustration for this example.
a. From the problem statement, we have PR = 2.00, T 1 = 70.0°F = 530. R, and T 3 = 80.0°F = 540. R. Then, Eq. (14.25)
gives the COP for a reversed Brayton R/AC ASC as
COP reversed Brayton
ASC R=AC
= ð2:00 0:40/1:40 − 1Þ −1 = 4:57
and the minimum temperature within the cycle is the cooling temperature T 4s , which is given by
T 4s = T 3 /PR ðk
−1Þ/k
= ð540: RÞ/2:00 0:40/1:40 = 443 R = −17:0°F