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    q3t02 Sol PDF

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    TrầnHạVy
    1) A refrigerator uses refrigerant-134a to keep a refrigerated space at -30°C. Cooling water enters the condenser at 18°C and leaves at 26°C. 2) The quality of the refrigerant at the evaporator inlet is 0.4802. The refrigeration load is 5.39 kW. 3) The COP of the refrigerator is calculated to be 2.14 based on the actual heat gain across the evaporator versus the specific work into the compressor.

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    q3t02 Sol PDF

    Uploaded by

    TrầnHạVy
    114 views2 pages
    1) A refrigerator uses refrigerant-134a to keep a refrigerated space at -30°C. Cooling water enters the condenser at 18°C and leaves at 26°C. 2) The quality of the refrigerant at the evaporator inlet is 0.4802. The refrigeration load is 5.39 kW. 3) The COP of the refrigerator is calculated to be 2.14 based on the actual heat gain across the evaporator versus the specific work into the compressor.

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    1) A refrigerator uses refrigerant-134a to keep a refrigerated space at -30°C. Cooling water enters the condenser at 18°C and leaves at 26°C. 2) The quality of the refrigerant at the evaporator inlet is 0.4802. The refrigeration load is 5.39 kW. 3) The COP of the refrigerator is calculated to be 2.14 based on the actual heat gain across the evaporator versus the specific work into the compressor.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    114 views2 pages

    q3t02 Sol PDF

    Uploaded by

    TrầnHạVy
    1) A refrigerator uses refrigerant-134a to keep a refrigerated space at -30°C. Cooling water enters the condenser at 18°C and leaves at 26°C. 2) The quality of the refrigerant at the evaporator inlet is 0.4802. The refrigeration load is 5.39 kW. 3) The COP of the refrigerator is calculated to be 2.14 based on the actual heat gain across the evaporator versus the specific work into the compressor.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 2

    ME354

    Thermodynamics 2 Name:

    Quiz #3 - T02: ID #:

    Problem: A commercial refrigerator with refrigerant- 134a as the working fluid is used to keep the
    refrigerated space at −30 ◦ C by rejecting its waste heat to cooling water that enters the condenser at
    18 ◦ C at a rate of 0.25 kg/s and leaves at 26 ◦ C. The refrigerant enters the condenser at 1.2 M P a
    and 65 ◦ C and leaves at 42 ◦ C. The inlet state of the compressor is 60 kP a and −34 ◦ C and the
    compressor is estimated to gain a net heat of 450 W from the surroundings.

    Determine:
    (a) the quality of the refrigerant at the
    evaporator inlet,
    (b) the refrigeration load, [kW ]
    (c) the COP of the refrigerator,

    Assumptions

    1. SS − SF
    2. KE = P E → 0
    3. properties are constant

    Part a)

    state T P h s comment
    (◦ C) (kP a) (kJ/kg) (kJ/kg · K)

    1 -34 60 230.05
    2 65 1200 295.13 Table A-13 1 mark
    3 42 1200 111.37
    4 60 111.37 h4 = h3

    State 1
    We first notice from Table A-12, our temperature at 1 is above the saturation temperature at 60 kP a,
    therefore state point 1 is in the superheated region. From Table A-13 at P1 = 0.06 M P a we need to
    interpolate.
    1 mark
    h1 = 0.826(227.79) + 0.174(240.76) = 230.05 kJ/kg
    State 3
    We notice from Table A-12, our temperature at 3 is below the saturation temperature 1200 kP a, there-
    fore state point 3 is in the sub cooled region. From Table A-11 we can determine

    h3 = hf (T ) + vf (T )[P − Psat (T )] = 111.26 + 0.0008786(1200 − 1072.8) = 111.37 kJ/kg


    1 mark

    State 4
    State point 4 is under the dome with pressure and entropy known, therefore from Table A-12
    111.37 − 3.841 1 mark
    x4 = = 0.4802 ⇐
    227.79 − 3.841

    Part b)
    The mass flow rate of the refrigerant can be determined by applying a 1st law energy balance across the
    condenser.
    The water properties can be obtained by using a saturated liquid at the given temperature

    hw1 = hf @18 ◦ C = 75.54 kJ/kg


    1 mark
    hw2 = hf @26 ◦ C = 109.01 kJ/kg

    ṁR (h2 − h3 ) = ṁw (hw2 − hw1 )

    ṁR (295.13 − 111.37) kJ/kg = (0.25 kg/s)(109.01 − 75.54) kJ/kg


    2 marks
    ṁR = 0.0455 kg/s

    The refrigeration load is given as


    1 mark
    Q̇L = ṁR (h1 − h4 ) = (0.0455 kg/s)(230.03 − 111.37) kJ/kg = 5.39 kW ⇐

    Note: Cengel calculates the refrigeration load as Q̇H − Ẇin. This includes the Qin of the compressor
    in addition to the enthalpy of the evaporator and provides a Q̇L = 5.85 kW . While not technically the
    cooling load, I will accept this answer.

    Part c)
    The specific work into the compressor can be determined as

    Q̇in 0.450 kJ/s


    win = (h2 − h1 ) − = (295.13 − 230.03) kJ/kg − = 55.21 kJ/kg
    ṁR 0.0455 kg/s
    1 mark
    The COP is determined as
    q̇L h1 − h4 (230.03 − 111.37) kJ/kg
    COP = = = = 2.14 ⇐ 1 mark
    ẇin win 55.21 kJ/kg
    Note: I have calculated qL as the actual heat gain across the evaporator. Cengel calculates qL =
    qH − win. Since win is biased by the heat gain at the compressor this leads to a COP = 2.33. I will
    accept this as a valid solution although, the COP should only be based on the actual energy gain which is
    h1 − h4 .

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