Observations: Calculation: Carnot COP, Cycle COP and Actual COP of the refrigeration system (i) Carnot CoP: CoP = Tevp / (Tcond Tevp) – = (273+6.223)/ (53.03-6.223) = 5.966 Here, Tevp and Tcond are the saturation temperatures corresponding to evaporator pressure and condenser pressure. (ii) Cycle CoP: CoP = (h1 h3) / (h2– h1) – Where h1 = enthalpy of refrigerant at evaporator outlet = kJ/kg 196.1 h2 = enthalpy of refrigerant vapor at compressor outlet = kJ/kg 220.5 h3 = enthalpy of refrigerant liquid at condenser outlet = kJ/kg 83.96 CoP = 4.802 (iii) Actual CoP: CoP = actual refrigerant effect / actual energy input to compress = energy meter reading of heater/ energy meter reading of compressor 1.05/0.45 = = 2.33 Overall heat transfer coefficients of evaporator and condenser (Uevp, Ucond) : For the evaporator:- Uevp = Where, Acoil is the area of the evaporator coil (to be obtained from the evaporator tube diameter, coil diameter and number of immersed in water) Twater & Te are the water and refrigerant temperatures in the evaporator. Te = (8+15)/2 = 11.5 oC Twater = 19.5 oC Acoil = (pi*de )*(pi*De*Nte) = 0.384 m2 Actual refrigeration effect = 1050 KJ
Observations: Calculation: Carnot COP, Cycle COP and Actual COP of the refrigeration system (i) Carnot CoP: CoP = Tevp / (Tcond – Tevp) = (273+6.223)/ (53.03-6.223) = 5.966 Here, Tevp and Tcond are the saturation temperatures corresponding to evaporator pressure and condenser pressure. (ii) Cycle CoP: CoP = (h1 – h3) / (h2– h1) Where h1 = enthalpy of refrigerant at evaporator outlet = 196.1 kJ/kg h2 = enthalpy of refrigerant vapor at compressor outlet = 220.5 kJ/kg h3 = enthalpy of refrigerant liquid at condenser outlet = 83.96 kJ/kg CoP = 4.802 (iii) Actual CoP: CoP = actual refrigerant effect / actual energy input to compress = energy meter reading of heater/ energy meter reading of compressor = 1.05/0.45 = 2.33 Overall heat transfer coefficients of evaporator and condenser (Uevp, Ucond) : For the evaporator:- Uevp = Where, Acoil is the area of the evaporator coil (to be obtained from the evaporator tube diameter, coil diameter and number of immersed in water) Twater & Te are the water and refrigerant temperatures in the evaporator. Te = (8+15)/2 = 11.5 oC Twater = 19.5 oC Acoil = (pi*de )*(pi*De*Nte) = 0.384 m2 Actual refrigeration effect = 1050 KJ
tfl_rac_sam3.docx