Question

1000 m3 of moist air at 101 kPa and 25 °C with a dew point of 11 °C enters a process. The air leaves the process at 98 kPa with a dew point of 58 °C. Assume ideal gases.

(a) Draw and label a flowsheet for the process.

b) What is the mole fraction of water in the air entering the process? Leaving the process?

(c) How many moles of water are added to or removed from the 1000 m3 of moist air?

Answer 1

Dew point is the temperature at which the partial pressure of vapour =vapour pressure of liquid at the given temperature

Vapor pressure of water

11 deg.c =9.8 mm Hg

55 deg.c =118 mmHg

101 Kpa= 0.9967 atm

98 Kpa =0.9672 atm

Partial pressure of water =9.8 mm Hg

Mole fraction of water vapour= Moles of water vapor/ total moles

Partial pressure of water vapour= Mole fraction * Total pressure of water vapor

At inlet

Partial pressure of water vapour= 9.8/760 mm Hg

Mole fraction of water vapour =(9.8/760)/ 0.9967 =0.013

At out let.

Mole fraction of water vapour= (118/760)/0.9672 =0.1605

Since there is an increase in mole fraction of water vapor at the outlet, Moisture is added to air.

Total moles of wet air entering per 1000m3 of wet air can be calculated from

PV= nRT, R=0.08206 L.atm/mole.K

n= PV/RT= 1000*1000L*0.9967 atm/(0.08206*(25+273.15)=40737 moles =40.737 Kg moles

Moles of water/ Moles of wet air =Mole fraction

At inlet

Moles of water = Mole fraction* Moles of wet air= 0.013* 40.737 kg moles=0.53 kg moles

Moles of dry air =Moles of wet air-moles of water vapour= 40.737-0.530= 40.207 Kg moles

These moles remain the same at the outlet as well

Mole fraction of dry air at the out let= 1-Mole fraction of water vapour

= 1-0.1605 =0.8395 moles

Mole fraction= Moles of dry air/ Total moles

Moles of dry air =40.207

Total moles of wet air at the outlet= 40.207/0.8395=48.53Kg moles

Moles of water vapour added = Moles of air at outlet- Moles of air at inlet= 48.53-40.737 =7.793Kg moles