Question 12.18: In a standard vapour compression refrigeration cycle, operat...
In a standard vapour compression refrigeration cycle, operating between an evaporator temperature of – 10°C and a condenser temperature of 40°C, the enthalpy of the refrigerant, Freon-12, at the end of compression is 220 kJ/kg. Show the cycle diagram on T-s plane. Calculate :
(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) | h_{f} (kJ/kg) | h_{g} (kJ/kg) |
| – 10 40 |
0.2191 0.9607 |
26.85 74.53 |
183.1 203.1 |
(GATE)
The cycle is shown on T-s diagram in Fig. 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} = h_{f_4} = 26.85 kJ/kg
(i) The C.O.P. the cycle :
C.O.P. = \frac{R_{n}}{W} = \frac{ h_{2} – h_{1}}{ h_{3} – h_{2}}
= \frac{183.1 – 74.53}{220 – 183.1} = 2.94 .
(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.
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.
