Question

A R-12 refrigerator works between the temperature limits of – 10°C and + 30°C. The compressor employed is of 20 cm × 15 cm, twin cylinder, single-acting compressor having a volumetric efficiency of 85%. The compressor runs at 500 r.p.m. The refrigerant is sub-cooled and it enters at 22°C in the expansion valve. The vapour is superheated and enters the compressor at – 2°C. Work out the following : 

(i) Show the process on T-s and p-h diagrams ; 

(ii) The amount of refrigerant circulated per minute ; 

(iii) The tonnes of refrigeration ; 

(iv) The C.O.P. of the system.

Answer 1

(i) Process on T-s and p-h diagrams : 

The processes on T-s and p-h diagrams are shown in

(ii) Mass of refrigerant circulated per minute : 

The value of enthalpies and specific volume read from p-h diagram are as under : 

h₂ = 352 kJ/kg ; h₃ = 374 kJ/kg 

h₄ = h₁ = 221 kJ/kg ; v₂ = 0.08 m3/kg 

Refrigerants effect per kg = h₂ – h₁ = 352 – 221 = 131 kJ/kg 

Volume of refrigerant admitted per min.

= \(\cfrac{\pi}4\) D²L × r.p.m. × 2 × η_vol, for twin cylinder, single acting

= \(\cfrac{\pi}4\) (0.2)² × 0.15 × 500 × 2 × 0.85 = 4 m³/min

Mass of refrigerant per min = \(\cfrac{4}{0.08}\) = 50 kg/min.

(iii) Cooling capacity in tonnes of refrigeration : 

Cooling capacity = 50(h₂ – h₁) = 50 × 131 

= 6550 kJ/min or 393000 kJ/h

= \(\cfrac{393000}{14000}\) = 28.07 TR.

\(\because\) 1 tonne of refrigeration TR = 14000 kJ/h)

(iv) Work per kg = (h₂ – h₁) = 374 – 352 = 22 kJ/kg

C.O.P. = \(\cfrac{131}{22}\) = 5.95