Refrigerant 22 flows in a theoretical single-stage compression refrigeration cycle with a mass flow rate of 0.05 kg/s. The condensing temperature is 48 oc, and the evaporating temperature is - 16 oc. If the power input to the cycle is 2.5 kW, determine: (a) the work done by the compressor in kJ/kg, (b) the heat rejected from the condenser in kJ/kg, (c) the heat absorbed by the evaporator in kJ/kg, (d) the coefficient of performance, and (e) the refrigerating efficiency.

Question

Engineering

Wilkinson

11 January, 10:42

Refrigerant 22 flows in a theoretical single-stage compression refrigeration cycle with a mass flow rate of 0.05 kg/s. The condensing temperature is 48 oc, and the evaporating temperature is - 16 oc. If the power input to the cycle is 2.5 kW, determine: (a) the work done by the compressor in kJ/kg, (b) the heat rejected from the condenser in kJ/kg, (c) the heat absorbed by the evaporator in kJ/kg, (d) the coefficient of performance, and (e) the refrigerating efficiency.

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Answers (1)

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Ruth Mcneil · Answer 1

a. The work done by the compressor is 447.81 Kj/kg

b. The heat rejected from the condenser in kJ/kg is 187.3 kJ/kg

c. The heat absorbed by the evaporator in kJ/kg is 397.81 Kj/Kg

d. The coefficient of performance is 2.746

e. The refrigerating efficiency is 71.14%

Explanation:

According to the given data we would need first the conversion of temperaturte from C to K as follows:

Temperature at evaporator inlet = Te=-16+273=257 K

Temperatue at condenser exit=Te=48+273=321 K

Enthalpy at evaporator inlet of Te - 16=i3=397.81 Kj/Kg

Enthalpy at evaporator exit of Te 48=i1=260.51 Kj/Kg

b. To calculate the the heat rejected from the condenser in kJ/kg we would need to calculate the Enthalpy at the compressor exit by using the compressor equation as follows:

w=i4-i3

W/M=i4-i3

i4=W/M + i3

i4=2.5/0.05 + 397.81

i4=447.81 Kj/kg

a. Enthalpy at the compressor exit=447.81 Kj/kg

Therefore, the heat rejected from the condenser in kJ/kg=i4-i1

the heat rejected from the condenser in kJ/kg=447.81-260.51

the heat rejected from the condenser in kJ/kg=187.3 kJ/kg

c. Temperature at evaporator inlet = Te=-16+273=257 K

The heat absorbed by the evaporator in kJ/kg is Enthalpy at evaporator inlet of Te - 16=i3=397.81 Kj/Kg

d. To calculate the coefficient of performance we use the following formula:

coefficient of performance=Refrigerating effect/Energy input

coefficient of performance=137.3/50

coefficient of performance=2.746

the coefficient of performance is 2.746

e. The refrigerating efficiency = COP/COPc

COPc=Te / (Tc-Te)

COPc=255 / (321-255)

COPc=3.86

refrigerating efficiency=2.746/3.86

refrigerating efficiency=0.7114=71.14%