TITLE.PM5

752 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th14-3.pm5

(i)The C.O.P. of the cycle.
(ii)The refrigerating capacity and the compressor power assuming a refrigerant flow rate
of 1 kg/min. You may use the extract of Freon-12 property table given below :

t(°C) p(MPa) hf(kJ/kg) hg(kJ/kg)

• 10 0.2191 26.85 183.1
  40 0.9607 74.53 203.1
  (GATE 1997)
  Solution. The cycle is shown on T-s diagram in Fig. 14.29.
  Given : Evaporator temperature = – 10°C
  Condenser temperature = 40°C
  Enthalpy at the end of compression, h 3 = 220 kJ/kg
  From the table given, we have
  h 2 = 183.1 kJ/kg ; h 1 = hf 4 = 26.85 kJ/kg

T

s

3

3 ′

40°C

4

(^1) – 10°C 2
Evaporation
Compression
Condensation
Throttling
Fig. 14.29
(i)The C.O.P. the cycle :
C.O.P. =
R
W
hh
hh
n= −
−
21
32
= 183.1 .53
220 183.1
−
−
(^74) = 2.94. (Ans.)
(ii)Refrigerating capacity :
Refrigerating capacity = m(h 2 – h 1 )
[where m = mass flow rate of refrigerant = 1 kg/min ...(Given)]
= 1 × (183.1 – 74.53) = 108.57 kJ/min. (Ans.)
Compressor power :
Compressor power = m(h 3 – h 2 )
= 1 × (220 – 183.1) = 36.9 kJ/min or 0.615 kJ/s
= 0.615 kW. (Ans.)