In a simple vapour compression cycle, following are the properties of the refrigerant R-12 at various points : Compressor inlet : [latex] h_{2} = 183.2 kJ/kg v_{2} = 0.0767 m³/kg[/latex] Compressor discharge : [latex] h_{3} = 222.6 kJ/kg v_{3} = 0.0164 m³/kg[/latex] Compressor exit : [latex] h_{4} = 84.9 kJ/kg v_{4} = 0.00083 m³/kg[/latex] The piston displacement volume for compressor is 1.5 litres per stroke and its volumetric efficiency is 80%. The speed of the compressor is 1600 r.p.m. Find : (i) Power rating of the compressor (kW) ; (ii) Refrigerating effect (kW). (GATE)

Piston displacement volume = [latex] \frac{π}{4} d² × l[/latex] = 1.5 litres = 1.5 × 1000 × [latex] 10^{–6}[/latex] m³/stroke = 0.0015 m³/revolution. (i) Power rating of the compressor (kW) : Compressor discharge = 0.0015 × 1600 × 0.8 ([latex]η_{vol.} [/latex] ) = 1.92 m³/min. Mass flow rate of compressor, m = [latex] \frac{Compressor discharge }{v_{2}} [/latex] = [latex] \frac{1.92}{0.0767} [/latex] = 25.03 kg/min. Power rating of the compressor = [latex] \dot{m} (h_{3} – h_{2})[/latex] = [latex] \frac{25.03}{60} [/latex] (222.6 – 183.2) = 16.44 kW. (ii) Refrigerating effect (kW) : Refrigerating effect = [latex] \dot{m} (h_{2} – h_{1}) = \dot{m} (h_{2} – h_{4}) (∵ h_{1} = h_{4})[/latex] = [latex] \frac{25.03}{60} [/latex] (183.2 – 84.9) = 41 kW.

Question 12.22: In a simple vapour compression cycle, following are the prop...

Applied Thermodynamics

In a simple vapour compression cycle, following are the properties of the refrigerant R-12 at various points :
Compressor inlet : $h_{2} = 183.2 kJ/kg v_{2} = 0.0767 m³/kg$
Compressor discharge : $h_{3} = 222.6 kJ/kg v_{3} = 0.0164 m³/kg$
Compressor exit : $h_{4} = 84.9 kJ/kg v_{4} = 0.00083 m³/kg$
The piston displacement volume for compressor is 1.5 litres per stroke and its volumetric efficiency is 80%. The speed of the compressor is 1600 r.p.m.
Find : (i) Power rating of the compressor (kW) ;
(ii) Refrigerating effect (kW). (GATE)

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