Refrigeration and Air-conditioning Miscellaneous


Refrigeration and Air-conditioning Miscellaneous

Refrigeration and Air-conditioning

  1. Moist air is treated as an ideal gas mixture of water vapor and dry air (molecular weight of air = 28.84 and molecular weight of water = 18). At a location, the total pressure is 100 kPa, the temperature is 30°C and the relative humidity is 55%. Given that the saturation pressure of water at 30°C is 4246 Pa, the mass of water vapor per kg of dry air is ______ grams.









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    w = 0.622
    Pv
    P - Pv

    φ =
    Pv
    Ps

    ⇒ 0.55 =
    Pv
    Ps

    ⇒ Pv = 2235.3
    w = 0.622 ×
    2235.3
    - 2235.3
    (100 × 103)

    w = 0.0142 kg/kg of dry air
    w = 14.2 gm/kg of dry air

    Correct Option: A

    w = 0.622
    Pv
    P - Pv

    φ =
    Pv
    Ps

    ⇒ 0.55 =
    Pv
    Ps

    ⇒ Pv = 2235.3
    w = 0.622 ×
    2235.3
    - 2235.3
    (100 × 103)

    w = 0.0142 kg/kg of dry air
    w = 14.2 gm/kg of dry air


  1. A stream of moist air (mass flow rate = 10.1 kg/s) with humidity ratio of 0.01 kg/kg dry air mixes with a second stream of superheated water vapour flowing at 0.1 kg/s. Assuming proper and uniform mixing with no condensation, the humidity ratio of the final stream in (kg/kg dry air) is ______.









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    ωnew =
    m1ω1 + m2ω2
    m1 + m2

    0.1 × 10.1 × + .1 × 1
    = 0.02kg/kg dry air
    10.1 + .1

    Correct Option: A

    ωnew =
    m1ω1 + m2ω2
    m1 + m2

    0.1 × 10.1 × + .1 × 1
    = 0.02kg/kg dry air
    10.1 + .1



  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