Question

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)	hf (kJ/kg)	hg(kJ/kg)
– 10	0.2191	26.85	183.1
40	0.9607	74.53	203.1

Answer 1

The cycle is shown on T-s diagram in

Given : Evaporator temperature = – 10°C 

Condenser temperature = 40°C 

Enthalpy at the end of compression, h₃ = 220 kJ/kg 

From the table given, we have 

h₂ = 183.1 kJ/kg ; h₁ = h_f4 = 26.85 kJ/kg

(i) The C.O.P. the cycle :

(ii) Refrigerating capacity : 

Refrigerating capacity = m(h₂ – h₁) 

[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(h₃ – h₂) 

= 1 × (220 – 183.1) = 36.9 kJ/min or 0.615 kJ/s 

= 0.615 kW.