11-47
11-61 A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The
cooling rate of the high-temperature evaporator, the power required by the compressor, and the COP of the system are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Condenser
2
T
3
·
QH
Compressor
1
Expansion
valve
Expansion
valve
3 800 kPa
4
·
Win
5
-26.4C
Expansion
valve
7
0C
4
Evaporator 1
5
2
6
Evaporator 2
· 7 1
QL
s
6
Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),
P3  800 kPa 
 h3  h f
sat. liquid

@ 800 kPa
 95.47 kJ/kg
h4  h6  h3  95.47 kJ/kg ( throttling)
T5  0C 
 h  h g @ 0C  250.45 kJ/kg
sat. vapor  5
T7  26.4C 
 h7  h g @  26.4C  234.44 kJ/kg
sat. vapor

The mass flow rate through the low-temperature evaporator is found by
Q L
8 kJ/s
Q L  m 2 (h7  h6 ) 
 m 2 

 0.05757 kg/s
h7  h6 (234.44  95.47) kJ/kg
The mass flow rate through the warmer evaporator is then
m 1  m  m 2  0.1  0.05757  0.04243 kg/s
Applying an energy balance to the point in the system where the two evaporator streams are recombined gives
m h  m 2 h7 (0.04243)(250.45)  (0.05757)(234.44)
m 1 h5  m 2 h7  m h1 
 h1  1 5

 241.23 kJ/kg
m
0.1
Then,
P1  Psat @  26.4C  100 kPa 
 s1  0.9789 kJ/kg  K
h1  241.23 kJ/kg

P2  800 kPa 
 h2  286.26 kJ/kg
s 2  s1

The cooling rate of the high-temperature evaporator is
Q  m (h  h )  (0.04243 kg/s)(250.45  95.47) kJ/kg  6.58 kW
L
1
5
4
The power input to the compressor is
W  m (h  h )  (0.1 kg/s)(286.26  241.23) kJ/kg  4.50 kW
in
2
1
The COP of this refrigeration system is determined from its definition,
Q
(8  6.58) kW
COPR  L 
 3.24

4.50 kW
Win
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