A textile finishing process involves drying a fabric that has been treated with

Question

A textile finishing process involves drying a fabric that has been treated with a volatile solvent. The drying process involves evaporation of solvent and removal of solvent vapor by air. The wet fabric entering the dryer contains 50% solvent. Solvent-free air enters the dryer at a rate of 8 kilograms per kilogram of dry (solvent-free) fabric. The drying process is 96% efficient (i.e., 96% of the entering solvent is removed from fabric).

(a) Calculate the percentage of solvent in the dried fabric by weight.

(b) What is the concentration of solvent in the dryer exhaust (in g/m3) if the exhaust is at 80 degrees C and 1 atm? The solvent has a molecular weight of 86.

Explanation / Answer

degrees C and 1 atm? The solvent has a molecular weight of 86.

Basis : 100 kg of wet fabric,

Solvent in the fabric =50 kgs

Amount of dry fabric = 100-50= 50 kgs

The drying process is 96% efficient, i.e 96% of solvent is removed

Solvent present in the fabric = 50*0.04= 2 kgs

Weight of dried fabric =50 ( dry fabic)+ 2 kg (solvent)= 52 kgs

a)Percentage of solvent in the dried fabric = 100*2/52=3.84%

b)

b) air is 8 kg/kg of solvent free fabric

1 kg of solvent free fabric requires 8 kg of air

50 kg of solvent free fabric requires =200 kg of air

So air leaving the dryer contains =200 kg of air and 48 kgs of solvent

So total mass leaving the dryer= 200+48=248 kgs

Moles of air and solvent leaving the dryer= 200/29+48/86 ( 29 and 86 are molar masses of air and solvent)= 6.9+0.56= 7.46 kgmoles=7460 gmoles

From PV= nRT where P =1 atm V = volume in L to be calculated , n = number of moles and R=0.08206 L.atom/mole.K T= 80+273.15= 353.15 K

V= 7460*0.08206*353.15/1 =216187 L = 216187*0.001=216.187 m3

Concentration of solvent = 48/ 216.187=0.22203 kg/m3 =222.03 g/m3

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