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28 July, 19:25

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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  1. 28 July, 19:31
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    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%
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Home » Engineering » 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.
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