Question

Wet air containing 60.4 mol% of pure water at 82.5˚C is fed to a dehumidifier at a rate of 185 mol/hr. Outlet of the dehumidifier consists of 2 streams – one stream of drier air and the other stream is pure liquid water. The outlet streams are in vapor-liquid equilibrium at 812 mmHg and 31.7oC.
a) Draw the process flow diagram, number the streams, and label the components in each stream.
b) Find the component mole flow rates (mol/hr) exiting the dehumidifier.
c) If the temperature of the inlet wet air increases by 10oC, will the molar flow rate of liquid water exiting the dehumidifier increase, decrease, or stay the same? Justify your answer.

Answer 1

Feed rate= 185 moles/hr   Molar flow rate of water= 185*0.604=111.74 moles/hr

Molar flow rate of dry air = 185-111.74 =73.26 moles of dry air

at 31.7 deg.c, vapor pressrue of water= 35.663 mm Hg

At equibrium

F(185)= D( dehumidifed air)+W (water) =185

D ( dehumified air)= 185-W

let moles of water vapor in dry air =x

moles of water in pure water= 111.74-x

W= moles of dry air

moles of water vapor./moles of dry air = partial pressure of water vapor/ partial pressure of dry air= 35.363/(812-35.363)=0.0455

moles of water vapor = 0.0455*73.26 moles/hr=3.33 moles/ hr

dry air contains 3.33 moles/ hr of water and rest water has to leave from pure water, pure water leaving dehumidifier= 111.74-3.33=108.41 mols/hr

so moles of Dehumidified air = 3.33+73.26= 76.59 moles/hr and water = 108.41 moles/hr

c) As long as the outelt air leaves at equilibrium at 812mm Hg and 31.7 deg.c, the saturation conditions remain the same and this will not influecne the moles leaving.