The outside air temperature is 30°C, the pressure is 103 kPa, and the relative h

Question

The outside air temperature is 30°C, the pressure is 103 kPa, and the relative humidity is 85%. An air conditioner draws in the outside air, cools it to 20°C,and some water condenses. The gas stream leaves the condenser at a volumetric flow rate of 11.8 SCMH. This stream is then mixed with 250 mol/h of dry air. The temperature of the stream leaving the mixer is also 20°C. You may assume a constant pressure of 103 kPa throughout the system. 250 mol/h dry air V1 (m3/h) Air Conditioner 11.8 SCMH 30°C, 103 kPa (Condenser) 20°C 20°C Mixer 85% relative humidity VH20 (L/h) a) Calculate the volumetric flow rate of the water that is condensed by the air conditioner (Vi 2o in Uh). [7 mar b) Calculate the volumetric flow rate of the air drawn in by the air conditioner (Vi in m3/h). [1 mark] c) Calculate the relative humidity of the gas stream leaving the mixer (h). 12 marksl

Explanation / Answer

Vapor pressure of water at 30 deg.c = 31.8 mm Hg

Relative humidity = 100% partial pressure of water vapour/vapour pressure of liquid

=0.85= partial pressure of vapour/31.8

Partial pressure of vapour= 31.8*0.85= 27 mm Hg                             

103 Kpa =(103/101.3)*760 =773 mmHg

        Partial pressure of water vapor/partial pressure of dry air = 27/(773-27) =0.036        

At the exit of air conditioner, where the air is assumed to be saturated, partial pressure of water vapour= vapour pressure of liquid = 17.5 mm Hg

Volumetric flow rate of gas at the exit = 11.8 m3/hr at STP= 11.8*1000 L/hr

1 mole of any gas at STP occupies 22.4 L

11.8*1000 L/hr occupies 11.8*1000/22.4=527 moles/hr

At the exit , partial pressure of water vapour/ total pressure = 17.5/773 =moles of water vapour/ total moles

Moles of water vapour at exit = 527*17.5/773=12 moles/hr

Moles of dry gas remains same throughout the process , hence moles of dry air = 527-12= 515 moles/hr

Moles of water vapour at the inlet = 0.036*515 =19 moles/hr

Moles of water vapour condensed = 19-12= 7 moles/hr

Volumetric flow rate of air condensed at STP = 7*22.4 L/hr=156.8 L/hr         =0.1568 m3/hr

At the given conditions of 20 deg.c (20+273= 293K ) and 760 mm Hg ( 1 atm pressure)

=nRT/P= 7*0.0821*293/1 =168 L/hr=0.168 m3/hr

Volumetric flow rate of air at the inlet to air conditioner = 515 ( dry air )+19 ( water vapour)= 534 moles/hr

Volumetric flow rate of wet air = 534*0.0821*(30+273)/1 = 13283 L/hr= 13.3 m3/hr

After mixer

Moles of dry air = 250+515= 765 moles/hr

Moles of water vapour = 12 moles/hr

Partial pressure of water vapour = mole fraction* total pressure = (12/765)*773 = 12 mm Hg

Relative humidity =100* partial pressure of water vapour/ vapor pressure of liquid =

100* 12/17.5                       =68.57%

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