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6 December, 18:18

A room contains 50 kg of dry air and 06 kg of water vapor at 25°C and 95 kPa total pressure. The saturation p ressure of water at 25 "C is 3.1698 kPa. The molar masses for air and water are 29 kg/kmol and 18 kg/kmol, respectively Determine (a) The specific humidity of air in the room; (b) The relative humidity of air in the room.

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  1. 6 December, 18:30
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    a) w = 0.12 kg / kg of dry air

    b) d = 484.69 %

    Explanation:

    Given:

    Mass of dry air, m = 50 kg

    mass of water vapor = 6 kg

    saturation pressure of water = 3.1698 kPa

    Molar mass of the air = 29 kg/mol

    Molar mass of the water = 18 kg/mol

    a) The specific humidity (w) is calculated as:

    w = (mass of water vapor) / (mass of air)

    on substituting the values, we get

    w = 6 kg / 50 kg

    or

    w = 0.12 kg / kg of dry air

    b) Now,

    w = (0.622Pv) / (Pt - Pv)

    where,

    Pv is the actual vapor pressure

    Pt is the vapor pressure at room temperature

    on substituting the values, we have

    0.12 = (0.622 * Pv) / (95 - Pv)

    or

    11.4 - 0.12Pv = 0.622Pv

    or

    Pv = 15.36 kPa

    also,

    Relative humidity is given as:

    d = (actual vapor pressure) / (saturated vapor pressure)

    saturated vapor pressure for water = 3.1698 kPa

    on substituting the values, we have

    d = (15.36 / 3.1698) * 100

    or

    d = 4.846 * 100

    or

    d = 484.69 %
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