# the COP, The exergy destruction in each component and the... 11-34

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11-34 A vapor-compression refrigeration cycle is used to keep a space at a low temperature. The mass flow rate of R-134a,
the COP, The exergy destruction in each component and the exergy efficiency of the compressor, the second-law
efficiency, and the exergy destruction are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and
T
potential energy changes are negligible.
2
·
Analysis (a) The properties of R-134a are (Tables A-11 through A-13)
QH 2s
P2  1.2 MPa h2

T2  50C  s 2
P3  1.2 MPa  h3

x3  0
 s3
 278.27 kJ/kg
3 1.2 MPa
 0.9267 kJ/kg  K
 117.77 kJ/kg
 0.4244 kJ/kg  K
4s
4
The rate of heat transferred to the water is the energy change of the
water from inlet to exit
Q  m c (T  T )  (0.15 kg/s)(4.18 kJ/kg  C)(28  20)C  5.016 kW
H
w p
w, 2
·
Win
·
1
QL
s
w,1
The energy decrease of the refrigerant is equal to the energy increase of the water in the condenser. That is,
Q H
5.016 kW
Q H  m R (h2  h3 ) 
 m R 

 0.03125 kg/s
h2  h3 (278.27  117.77) kJ/kg
 3412 Btu/h 
Q L  Q H  W in  5.016  2.2  2.816 kW  (2.816 kW)
  9610 Btu/h
 1 kW 
The COP of the refrigerator is determined from its definition,
Q
2.816 kW
COP  L 
 1.28

2.2 kW
Win
(b) The COP of a reversible refrigerator operating between the same temperature limits is
COPCarnot 
TL
12  273

 8.156
TH  TL (20  273)  (12  273)
The minimum power input to the compressor for the same refrigeration load would be
Q L
2.816 kW

 0.3453 kW
W in,min 
COPCarnot
8.156
The second-law efficiency of the cycle is
W in,min 0.3453
 II 

 0.1569  15.7%
2.2
W in
The total exergy destruction in the cycle is the difference between the actual and the minimum power inputs:
E x
 W  W
 2.2  0.3453  1.85 kW
dest, total
in
in, min
(c) The entropy generation in the condenser is
 T w, 2
S gen,cond  m w c p ln
T
 w,1

  m R ( s 3  s 2 )


 28  273 
 (0.15 kg/s)(4.18 kJ/kg  C)ln
  (0.03125 kg/s)(0.4004  0.9267) kJ/kg  K)
 20  273 
 0.001191 kW/K
The exergy destruction in the condenser is
E x
 T S
 (293 K )(0.001191 kW/K )  0.349 kW
dest,cond
0 gen,cond
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
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