Modern Engineering Thermodynamics

556 CHAPTER 14: Vapor and Gas Refrigeration Cycles
EXAMPLE 14.6 (Continued )
For this design, determine
a. The mass flow rate of refrigerant in loops A and B.
b. The system’s coefficient of performance.
c. The pressure ratios across each of the compressors.
Solution
Use Figure 14.17 as the equipment schematic for this example. From Tables C.9b and C.10b for R-22, we can find the
following property values:
Loop A
Station 1A Station 2sA Station 3A Station 4hA
Compressor A in Compressor A out Condenser A out Expansion valve A out
x 1A = 1:00 p 2sA = 1500: kPa x 3A = 0:00 h 4hA = h 3A
T 1A = − 20:0°C s 2sA = s 1A T 3A = 25:0°C
h 1A = 242:05 kJ/kg h 2sA = 289:08 kJ/kg h 3A = 74:91 kJ/kg h 4hA = h 3A = 74:91 kJ/kg
s 1A = 0:95927 kJ/kg.K ðby interpolationÞ
p 1A = 244:8 kPa T 2sA = 71:07°C
ðby interpolationÞ
Loop B
Station 1B Station 2sB Station 3B Station 4hB
Compressor B inlet Compressor B outlet Condenser B outlet Expansion valve B outlet
x 1B = 1:00 p 2sB = 300: kPa x 3B = 0:00 h 4hB = h 3B
T 1B = −50:0°C s 2sB = s 1B T 3B = −20:0°C
h 1B = 228:51 kJ/kg h 2sB = 264:05 kJ/kg h 3B = 21:73 kJ/kg h 4hB = h 3B = 21:73 kJ/kg
s 1B = 1:02512 kJ/kg.K ðby interpolationÞ
p 1B = 63:139 kPa T 2sB = 15:0°C
a. The mass flow rate in loop B can be found from an energy rate balance on the evaporator as
_m B =
_Q L ð40:0 tonsÞ½210: kJ/min/ð1 tonÞŠð1 min/60 sÞ
= = 0:677 kg/s
h 1B − h 4hB 228:51 − 21:73 kJ/kg
Equation (14.10) can now be used to find the actual compressor outlet state in loop B as
h 2B = ðh 2sB − h 1B Þ/ðη s Þ c − B
+ h 1B = ð264:05 − 228:51Þ/0:80 + 228:51 = 272:9 kJ/kg
Then Eq. (14.9) can be used to find the mass flow rate in loop A as
h
_m A = _m 2B − h 3B
272:94 − 21:73 kJ/kg
B
= ð0:677 kg/sÞ = 1:02 kg/s
h 1A − h 4hA
242:05 − 74:91 kJ/kg
b. Equation (14.12) provides the system COP as
_m B ðh 1B − h 4hB Þ
COP dual
=
_m A ðh 2sA − h 1A Þ/ðη
cascade
s Þ c−A
+ _m B ðh 2sB − h 1B Þ/ðη s Þ c−B
0:677ð228:51 − 21:73Þ
=
1:02ð289:08 − 242:05Þ/0:80 + 0:677ð264:05 − 228:51Þ/0:80
= 1:55
c. The compressor pressure ratios are obtained form these data as
PR compressor A = p 2SA /p 1A = 1500:/244:8 = 6:13
PR compressor B = p 2SB /p 1B = 300:/63:139 = 4:75