Question

2. Industrial humidifiers has two inlet streams – 1. liquid water and 2. drier air. The entering liquid water stream flows at 5.31 mol/min. The component flow rate of air in the drier air stream is 44.3 mol/min. The exiting streams consist of a wetter air stream and a liquid water stream. The exiting liquid water stream contains 34.5% of the entering liquid water. The outlet streams are in vapor liquid equilibrium at 42.8oC and 786 mmHg.
a) Draw the process flow diagram, number the streams, and label the species in each stream.
b) Find the component mass flow rates (kg/min) of all four streams.
c) The pressure of the outlet streams decrease by 5%, will the mole fraction of water in the wetter air increase, decrease, stay the same? Justify your answer.

Answer 1

moles of water entering, Stream-1 = 5.31mol/min ,Liquid water stream contains 34.5% of entering liquid water= 0.345*5.31=1.83 moles/hr So water in wet air = 5.31-1.83=3.48 mol/min

Stream-2 contains 44.3 mol/min, total moles in humijdifed air = 44.3+3.48=47.78 mols/min

since the out let streams are in equilibrium

At 42.8 deg.c,, vapor pressure of water =64.5mm Hg

sincethe outlet streams are in vapor liquid equilibrium

moles of water vapor/ total moles= partial pressure ( = vapor pressure)/ total pressure)= 64.5/786=0.082 (1)

Moles of water vapor in humidifed air = 0.082*47.78=3.92 moles/min

moles of air in humidifed air = total mole entering-( moles of water in liquid stream+ moles of water in humidifed air )

44.3+5.31- (3.92+1.83)=43.86 mol/hr of dry air

With the moles you can fin the mass using Molecular Wight = Moles / Mass

c) When the pressure decreases by 5%, the pressure = 786*0.95=746.7 mm Hg

eq.1 now beomces, moles of water vapor/ total moles = 64.5/746.7=0.0864

moles of water vapor = 0.0864*47.78 moles/hr=4.12 moles/hr

moles of air in humidified air increases.