Question

Air is dried from a partial pressure of 50 mm Hg of water vapor to a partial pressure of 10 mm Hg. The temperature of the entering air is 524°F and the pressure is constant at 760 mm Hg. How many gallons of liquid water are removed per 1000 ft3 of entering air?

Answer 1

mole fraction = volume fraction for air = partial pressure/total pressure

volume of water in entering air/volume of enetring air = partial pressure of water in entering air/total pressure

Vol. of water in entering air/1000 = 50/760

or vol of water in entering air = 65.789 ft3

volume of dry air in entering air = 1000-65.789 = 934.21 ft3

volume of water in outgoing air/volume of outgoing total air = partial pressure of water in entering air/total pressure

lets assume volume of water in outgoing air = V

V/(V+934.21) = 10/760 =

or V = 12.456 ft3

so volume of water condensed = inlet water volume - outlet water volume = 65.789-12.456 = 53.333 ft3

1 ft3 = 7.48052 so 53.333 ft3 = 398.96 gallons

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