Refrigeration and Air-conditioning Miscellaneous


Refrigeration and Air-conditioning Miscellaneous

Refrigeration and Air-conditioning

  1. The pressure, dry bulb temperature and relative humidity of air in a room are 1 bar, 30°C and 70%, respectively. If the saturated steam pressure at 30°C is 4.25 kPa, the specific humidity of the room air in kg of water vapour/kg dry air is









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    RH = 0.7 =
    Pv
    and w =
    0.622pv
    4.25100 - pv

    Correct Option: C

    RH = 0.7 =
    Pv
    and w =
    0.622pv
    4.25100 - pv


  1. A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of air in the room are 100 kPa and 25°C respectively. Given that the saturation pressure for water at 25°C, is 3.17 kPa, the relative humidity of the air in the room is









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    ma = 35 kg, mv = 0.5 kg

    w =
    mv
    =
    0.5
    = 0.01428
    mv35

    also w = 0.622
    Pv
    Pa

    Pt = 100 kPa
    ∴ w = 0.622 ×
    Pv
    Pt - Pv

    Now, 0.01428 = 0.622 × Pv
    Pt - Pv

    Pv = 2.238 kPa
    Relative humidity
    φ =
    Pv
    =
    2.238
    × 100
    Ps3.17

    φ = 70.6% = 71%

    Correct Option: D

    ma = 35 kg, mv = 0.5 kg

    w =
    mv
    =
    0.5
    = 0.01428
    mv35

    also w = 0.622
    Pv
    Pa

    Pt = 100 kPa
    ∴ w = 0.622 ×
    Pv
    Pt - Pv

    Now, 0.01428 = 0.622 × Pv
    Pt - Pv

    Pv = 2.238 kPa
    Relative humidity
    φ =
    Pv
    =
    2.238
    × 100
    Ps3.17

    φ = 70.6% = 71%



  1. If a mass of moist air in an airtight vessel is heated to a higher temperature, then









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    R.H. decreases

    Correct Option: D

    R.H. decreases


  1. A moist air sample has dry bulb temperature of 30°C and specific humidity of 11.5 g water vapour per kg dry air. Assume molecular weight of air as 28.93. If the saturation vapour pressure of water at 30°C is 4.24 kPa and the total pressure is 90 kPa, then the relative humidity (in%) of air sample is









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    Given: DBT = 30°C, w = 11.5 g/kg
    ∴ Relative humidity

    =
    Specific humidity
    ×
    Total pressure
    0.622saturation vapour pressure

    φ =
    w
    ×
    Pa
    0.622Pvs

    φ =
    11.5 × 90
    = 0.385
    0.622 × 1000 × 4.24

    φ ≈ 38.5%

    Correct Option: B

    Given: DBT = 30°C, w = 11.5 g/kg
    ∴ Relative humidity

    =
    Specific humidity
    ×
    Total pressure
    0.622saturation vapour pressure

    φ =
    w
    ×
    Pa
    0.622Pvs

    φ =
    11.5 × 90
    = 0.385
    0.622 × 1000 × 4.24

    φ ≈ 38.5%



  1. Moist air at a pressure of 100 kPa is compressed to 500 kPa and then cooled to 35°C in an aftercooler. The air at the entry to the aftercooler is unsaturated and becomes just saturated at the exit of the aftercooler. The saturation pressure of water at 35°C is 5.628 kPa. The partial pressure of water vapour (in kPa) in the moist air entering the compressor is closest to









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    Assuming that compression is isentropic in air compressors, the process can be described on the T - s diagram. The process in the intercooler is constant pressure
    p2 = p3
    But p2 = 5p1

    p1 =
    p2
    =
    p3
    55

    p1 =
    5.628
    = 1.13kPa
    5

    Correct Option: B

    Assuming that compression is isentropic in air compressors, the process can be described on the T - s diagram. The process in the intercooler is constant pressure
    p2 = p3
    But p2 = 5p1

    p1 =
    p2
    =
    p3
    55

    p1 =
    5.628
    = 1.13kPa
    5