1. Wet air containing 60.4 mol% of pure water at 82.5˚C is fed to a dehumidifier...

1. 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.

Expert Solution

Inlet to the humidifier

Flow rate of wet air = 185 mol/hr, Flow rate of water= 185*0.604=111.74 moles/hr

Moles of dry air = 185-111.74 =73.26 moles/hr

Outlet of humidifier

At 31.7 deg.c, vapor pressure of water =33.6 mm Hg, Total pressure= 812 mm Hg

At Equilibrium, partial pressure of water vapor = vapor pressure of llquid= 33.6 mmHg

Parital pressue of dry air= 812-33.6=778.4 mm Hg

Partial pressure of water vapor/ Partial pressure of dry air= 33.6/778.4= 0.0432

= moles of water vapor/moles of dry air

Mole of water vapor = 0.0432*73.26=3.165 moles/hr, moles of dry air = 73.26 moles/hr

Second stream from de humidifier

Hence molar flow rate of liquid water leaving= moles ofwater entering-( moles of water in air=)111.74- 3.165=108.575 moles/hr

C) when temperature of inlet wet air is increased and outlet conditinos remains the same at 31.7 deg.c and 812mm Hg, there will not be any change in molar flow rate of liqud water exiting the dehumidifier. Since moles of water vapor are calculated based on moles of dry air.

queen_honey_blossom answered 2 years ago

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