Modern Engineering Thermodynamics

572 CHAPTER 14: Vapor and Gas Refrigeration Cycles
EXAMPLE 14.11
Suppose 4.00 lbm/s of air at 530. R enters the compressor of a modern reversed Brayton ASC refrigeration unit. The
isentropic pressure ratio of the compressor is 3.00 to 1 and the inlet temperature of the expander is 600. R. Determine
a. The expander power.
b. The compressor power.
c. The coefficient of performance of the unit.
d. The refrigeration capacity of the unit in tons.
Solution
Using Figure 14.27 as the illustration for this example, station 1 is the compressor inlet, station 2 is the compressor outlet,
station 3 is the expander inlet, and station 4 is the expander outlet. For an ASC, all the processes are reversible, so the η s of
the expander and the compressor are both 1.00.
a. An energy rate balance on the expander gives
_W expander = _m ðh 3 − h 4s Þ = _mc p ðT 3 − T 4s Þ
where
Then,
T 4s = T 3 ðp 4s /p 3 Þ ðk −1Þ/k
= 600: ð1/3:00Þ 0:40/1:40 = 438 R
_W expander = ð4:00 lbm/sÞ½0:240 Btu/ ðlbm.RÞŠð600: − 438: RÞ = 155 Btu/s
b. An energy rate balance on the compressor gives
_W compressor = _m ðh 1 − h 2s Þ = _mc p ðT 1 − T 2s Þ
where
Then,
T 2s = T 1 ðp 2s /p 1 Þ ðk −1Þ/k
= 530: ð3:00Þ 0:40/1:40 = 725 R:
_W compressor = 4:00ð0:240Þð530: − 725Þ = −188 Btu/s
c. Equation (14.25) gives the COP of this unit as
−1
COP = PR ðk −1Þ/k
− 1 = 3:00 0:40/1:40 − 1 −1
= 2:71
d. Finally, the refrigeration capacity of this unit is
_Q L = refrigeration capacity = COP R/AC
× j _W in j net = 2:71jð188 − 155Þj
= ð88:0 Btu/sÞð60 s/minÞ = ð5280 Btu/minÞðl ton/200: Btu/minÞ
= 26:4 tons of refrigeration
Exercises
31. If the compressor inlet temperature of the reversed Brayton ASC refrigeration unit in Example 14.11 is reduced from
530. R to 500. R, what is the new refrigeration capacity of this unit? Answer: _Q L = 49.8 tons.
32. The pressure ratio across the compressor of the reversed Brayton ASC refrigeration unit in Example 14.11 is increased
from 3.00 to 500. Determine the new ASC coefficient of performance for the new unit. Answer: COP ASC R/AC = 1.71.
33. The reversed Brayton ASC refrigeration unit discussed in Example 14.11 is to be scaled up to handle a refrigeration capacity
of 230. tons. Determine the corresponding air mass flow rate required for the new unit. Answer: _m air = 34.9 lbm/s.
14.15 REVERSED STIRLING CYCLE REFRIGERATION
A reversed Stirling cycle refrigerator was first implemented by the Scottish engineer Alexander Carnegie Kirk
(1830–1892) in 1862. Kirk was searching for a cooling technology that was safer than the prevailing
vapor-compression machines that used explosive ether. He was aware of the engine developed by Robert Stirling
in 1816 and felt that, if he put power into the engine instead of letting it produce power, the displacer piston
would be cooled. With a compressor pressure ratio of 2.0, he reached an expander temperature of −13°C, and
when he increased the pressure ratio to 7.0, he reached −40.°C.