11-38
11-49 A geothermal heat pump is considered. The degrees of subcooling done on the refrigerant in the condenser, the mass
flow rate of the refrigerant, the heating load, the COP of the heat pump, the minimum power input are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the refrigerant-134a tables (Tables A-11 through A-13)
T4  20C P4  572.1 kPa

x 4  0.23  h4  121.24 kJ/kg
h3  h4
Condenser
3
Expansion
valve
P1  572.1 kPa  h1  261.59 kJ/kg

x1  1 (sat. vap.) s1  0.9223 kJ/kg
P2  1400 kPa 
 h2  280.00 kJ/kg
s 2  s1

From the steam tables (Table A-4)
hw1  h f @ 50C  209.34 kJ/kg
.
QH
4
.
Win
Compressor
1
Evaporator
sat. vap.
20C
x=0.23
40C
hw 2  h f @ 40C  167.53 kJ/kg
The saturation temperature at the condenser pressure of
1400 kPa and the actual temperature at the condenser
outlet are
2
1.4 MPa
s2 = s1
Water
50C
T
·
QH
Tsat @ 1400 kPa  52.40C
2
1.4 MPa
P3  1400 kPa

 T3  48.59C (from EES)
h3  121.24 kJ/kg 
·
Win
3
Then, the degrees of subcooling is
4s
4
Tsubcool  Tsat  T3  52.40  48.59  3.81C
·
QL
1
s
(b) The rate of heat absorbed from the geothermal water in the evaporator is
Q L  m w (hw1  hw 2 )  (0.065 kg/s)(209.34  167.53)kJ/kg  2.718 kW
This heat is absorbed by the refrigerant in the evaporator
m R 
Q L
2.718 kW

 0.01936 kg/s
h1  h4 (261.59  121.24)kJ/kg
(c) The power input to the compressor, the heating load and the COP are
W in  m R (h2  h1 )  Q out  (0.01936 kg/s)(280.00  261.59)kJ/kg  0.6564 kW
Q H  m R (h2  h3 )  (0.01936 kg/s)(280.00  121.24)kJ/kg  3.074 kW
COP 
Q H
3.074 kW

 4.68

Win 0.6564 kW
(d) The reversible COP of the cycle is
COPrev 
1
1  TL / TH

1
 12.92
1  (25  273) /(50  273)
The corresponding minimum power input is
W in, min 
Q H
3.074 kW

 0.238 kW
COPrev
12.92
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