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6-56 Refrigerant-134a flows through the condenser of a residential heat pump unit. For a given compressor power
consumption the COP of the heat pump and the rate of heat absorbed from the outside air are to be determined.
Assumptions 1 The heat pump operates steadily. 2 The
kinetic and potential energy changes are zero.
QH
800 kPa
x=0
Properties The enthalpies of R-134a at the condenser
inlet and exit are
800 kPa
35C
Condenser
P1  800 kPa 
h1  271.22 kJ/kg
T1  35C 
P2  800 kPa 
h2  95.47 kJ/kg
x2  0

Expansion
valve
Win
Compressor
Evaporator
Analysis (a) An energy balance on the condenser gives
the heat rejected in the condenser
QL
Q H  m (h1  h2 )  (0.018 kg/s)(271.22  95.47) kJ/kg  3.164 kW
The COP of the heat pump is
COP 
Q H 3.164 kW

 2.64
Win
1.2 kW
(b) The rate of heat absorbed from the outside air
Q L  Q H  W in  3.164  1.2  1.96 kW
6-57 A commercial refrigerator with R-134a as the working fluid is considered. The evaporator inlet and exit states are
specified. The mass flow rate of the refrigerant and the rate of heat rejected are to be determined.
Assumptions 1 The refrigerator operates steadily. 2 The kinetic and
potential energy changes are zero.
QH
Properties The properties of R-134a at the evaporator
inlet and exit states are (Tables A-11 through A-13)
Condenser
P1  100 kPa 
h1  60.71 kJ/kg
x1  0.2

P2  100 kPa 
h2  234.74 kJ/kg
T2  26C 
Expansion
valve
Compressor
Evaporator
Analysis (a) The refrigeration load is
Q L  (COP)W in  (1.2)(0.600 kW)  0.72 kW
Win
100 kPa
x=0.2
QL
100 kPa
-26C
The mass flow rate of the refrigerant is determined from
m R 
Q L
0.72 kW

 0.00414 kg/s
h2  h1 (234.74  60.71) kJ/kg
(b) The rate of heat rejected from the refrigerator is
Q H  Q L  W in  0.72  0.60  1.32 kW
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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