Refrigeration and Air-conditioning Miscellaneous


Refrigeration and Air-conditioning Miscellaneous

Refrigeration and Air-conditioning

  1. Refrigerant vapor enters into the compressor of a standard vapor compression cycle at –10°C (h = 402 J/kg) and leaves the compression at 50°C (h = 432 kJ/kg). It leaves the condenser at 30°C (h = 237 kJ/kg). The COP of the cycle is ________.









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    Work done = 432 – 402 = 30 kJ/kg
    Refrigerating effect = 402 – 237 = 165 kJ/kg.

    COP =
    165
    = 5.5
    30

    Correct Option: B

    Work done = 432 – 402 = 30 kJ/kg
    Refrigerating effect = 402 – 237 = 165 kJ/kg.

    COP =
    165
    = 5.5
    30


  1. The power required for compressor in kW is









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    Power required by compressor p = ṁ(h2 – h1)
    = 0.2 (276.45 – 237) = 7.9 kW

    Correct Option: C

    Power required by compressor p = ṁ(h2 – h1)
    = 0.2 (276.45 – 237) = 7.9 kW



  1. The rate at which heat is extracted, in kJ/s from the refrigerated space is









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    Rate at which heat is extracted = ṁ(h1 – h4)
    = 0.2 (237 – 95.5) = 28.3 kW

    Correct Option: A

    Rate at which heat is extracted = ṁ(h1 – h4)
    = 0.2 (237 – 95.5) = 28.3 kW


  1. In an ideal vapour compression refrigeration cycle, the specific enthalpy of refrigerant (in kJ/kg) at the following states is given as :
    Inlet of condenser : 283
    Exit of condenser : 116
    Exit of evaporator : 232
    The COP of this cycle is









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    1– 2 → work done by the compressor
    2 – 3 → condenser heat rejected at constant pressure
    3 – 4 → throttling
    4 – 1 → heat addition in evaporator
    Given: h2 = 283 kJ/ kg
    h3 = 116 kJ/kg = h4 (from ph curve)
    h1 = 232 kJ/kg
    We know

    C.O.P =
    desired effect
    =
    Cooling effect
    work inputwork done by compressor

    C.O.P =
    h1 – h4
    =
    232 - 116
    = 2.27
    h2 – h1283 - 232

    Correct Option: A


    1– 2 → work done by the compressor
    2 – 3 → condenser heat rejected at constant pressure
    3 – 4 → throttling
    4 – 1 → heat addition in evaporator
    Given: h2 = 283 kJ/ kg
    h3 = 116 kJ/kg = h4 (from ph curve)
    h1 = 232 kJ/kg
    We know

    C.O.P =
    desired effect
    =
    Cooling effect
    work inputwork done by compressor

    C.O.P =
    h1 – h4
    =
    232 - 116
    = 2.27
    h2 – h1283 - 232



  1. The vapour compression refrigeration cycle is represented as shown in the figure below, with state 1 being the exit of the evaporator. The coordinate system used in this figure is










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    T-h curve

    Correct Option: D

    T-h curve