Question

A refrigerator operating on standard vapour compression cycle has a coefficiency performance of 6.5 and is driven by a 50 kW compressor. The enthalpies of saturated liquid and saturated vapour refrigerant at the operating condensing temperature of 35°C are 62.55 kJ/kg and 201.45 kJ/kg respectively. The saturated refrigerant vapour leaving evaporator has an enthalpy of 187.53 kJ/kg. Find the refrigerant temperature at compressor discharge. The c_p of refrigerant vapour may be taken to be 0.6155 kJ/kg°C.

Answer 1

Given : C.O.P. = 6.5 ; W = 50 kW, 

h₃′ = 201.45 kJ/kg, 

h_f4 = h₁ = 69.55 kJ/kg ; 

h₂ = 187.53 kJ/kg 

c_p = 0.6155 kJ/kg K

Temperature, t₃ : 

Refrigerating capacity = 50 × C.O.P. 

= 50 × 6.5 = 325 kW

Heat extracted per kg of refrigerant 

= 187.53 – 69.55 = 117.98 kJ/kg

Refrigerant flow rate = \(\cfrac{325}{117.98}\) = 2.755 kg/s

Compressor power = 50 kW 

∴ Heat input per kg = \(\cfrac{50}{2.755}\) = 18.15 kJ/kg

Enthalpy of vapour after compression 

= h₂ + 18.15 = 187.53 + 18.15 

= 205.68 kJ/kg 

Superheat = 205.68 – h₃′ = 205.68 – 201.45 

= 4.23 kJ/kg 

But 4.23 = 1 × c_p (t₃ – t₃′) 

= 1 × 0.6155 × (t₃ – 35)

∴ t₃ = \(\cfrac{4.23}{0.6155}\) + 35 = 41.87°C.