Modern Engineering Thermodynamics

14.9 Cascade and Multistage Vapor-Compression Systems 559
Solution
Use Figure 14.18 as the equipment schematic for this example. From Tables C.7f and C.8d for R-134a, we can find the
following data:
Loop A
Station 1A Station 2sA Station 3A Station 4hA
Compressor A inlet Compressor A outlet Condenser A outlet Expansion valve A outlet
p 1A = 500: kPa p 2sA = 1600: kPa x 3A = 0:00 h 4hA = h 3A
h 1A = 265:60 kJ/kg s 2sA = s 1A p 3A = 1600: kPa −
ðsee belowÞ = 0:9486 kJ/kg.K h 3A = 134:02 kJ/kg h 4A = 134:02 kJ/kg
s 1A = 0:9486 kJ/kg.K h 2A = 256:60 kJ/kg
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 = 500: kPa x 3B = 0:00 h 4hB = h 3B
p 1B = 100: kPa s 2sA = s 1B = 0:9395 p 3B = 500: kPa −
h 1B = 231:35 kJ/kg h 2sB = 264:25 kJ/kg h 3B = 71:33 kJ/kg h 4hB = 71:33 kJ/kg
s 1B = 0:9395 kJ/kg.K ðby interpolationÞ
The specific enthalpy of station 1A was determined from an energy rate balance on the flash chamber using Eq. (14.17) as
where h 2B was determined from Eq. (14.18) as
h 1A = x flash h g ðat p flash Þ + ð1 − x flash Þh 2B
h 2B = ðh 2sB − h 1B Þ/ðη s Þ c−B
+ h 1B
= ð264:25 − 231:35Þ/0:800 + 231:35 = 272:54 kJ/kg
The quality of the vapor exiting the flash chamber is given by Eq. (14.13) as
x flash = h f ðat the condenser pressureÞ − h f ðat the flash chamber pressureÞ
h fg ðat the flash chamber pressureÞ
= h f ðat 1600: kPaÞ − h f ðat 500: kPaÞ
h fg ðat 500: kPaÞ
=
134:02 − 71:33
184:74
= 0:339 = 33:9%
Then
h 1A = 0:339ð252:07Þ + ð1 − 0:339Þð272:54Þ = 265:60 kJ/kg
a. The mass flow rate in loop B is given by Eq. (14.14) as
_m B =
and, since for the dual-stage system _m B = _m A ð1 − x flash Þ,
b. The system COP is given by Eq. (14.16) as
COP dual
=
stage
ð _Q L Þ dual
stage ð10:0 tonsÞ½210: kJ/min/ð1 tonÞŠð1 min/60 sÞ
= = 218 kg/s
h 1B − h 4B 231:35 − 71:33 kJ/kg
_m A = _m B /ð1 − x flash Þ = ð0:218 kg/sÞ/ð1 − :0339Þ = 0:330 kg/s
=
_m ref ð1 − x flash Þðh 1B − h 4B Þ
_m ref ½ðh 2sA − h 1A Þ/ðη s Þ c−A + ð1 − x flash Þðh 2B − h 1B Þ/ðη s Þ c−B Š
ð1 − 0:339Þð231:35 − 71:33Þ
ð292:33 − 265:60Þ/0:800 + ð1 − 0:339Þð264:25 − 231:38Þ/0:800
= 1:78
(Continued )