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Answer :
The given conditions for the room are: Dimensions : length = 4.7 m, width = 3.6 m, and height = 2.9 m,Pressure = 98.3 kPa,Temperature (dry bulb) = 40°CTemperature (wet bulb) = 22°CMass of moist air in the room:We will use the formula for specific volume of air to calculate the mass of moist air in the room.
specific volume = m/V
where,m = mass of moist air in the room,V = volume of the room.
Since the air is moist, we need to consider the vapor pressure of water in air (which is given by the wet bulb temperature) to find the specific volume of moist air in the room.
specific volume of moist air = specific volume of dry air + specific volume of water vapor in air
From the psychrometric chart, at 98.3 kPa pressure,40°C dry bulb and 22°C wet bulb,
specific volume of dry air = 0.854 [tex]m^{3}[/tex]/kg and specific volume of water vapor = 0.0328 [tex]m^{3}[/tex]/kg
Therefore,specific volume of moist air = 0.854 + 0.0328
= 0.8868 [tex]m^{3}[/tex]/kg
Now,Volume of the room = (4.7 m) (3.6 m) (2.9 m)
= 47.7648 [tex]m^{3}[/tex]
Therefore,mass of moist air in the room,m = specific volume × Volume of the room
= 0.8868 × 47.7648
= 42.409 kg.
Therefore, the mass of moist air in the room is 42.409 kg/s.
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