An industrial wastewater flowing at a rate of 37,000 m3 /day is contaminated by
ID: 942474 • Letter: A
Question
An industrial wastewater flowing at a rate of 37,000 m3 /day is contaminated by 187 mg/L of propanol – CH3CH2CH2OH. As an environmental consultant, you propose to treat the water in an activated sludge process in which propanol is oxidized by oxygen into carbon dioxide – CO2. The reaction is highly favorable and assumed to be complete.
a) Equilibrate the half-reactions and overall reaction that are expected to occur during the treatment process.
b) Determine the theoretical quantity of oxygen (in mol/day) required for the complete oxidation of propanol
c)Determine the volume of pressurized air (2 atm, 15°C) to be provided (in m3 /day) for the complete oxidation of propanol. The partial pressure of O2 in air is 0.2095 atm and the efficiency of oxygen transfer is estimated to be 15%.
Explanation / Answer
The reaction is CH3CH2CH2OH+ 4.5 O2---->3CO2+ 4H2O
Flow of waste water= 37000 m3/day =37000*1000l/day
concentration of propanol =37000*1000 l/day*187 mg/L= 6.92*109 mg/day=6.92*106 g/day
moles/ day of propanol= mass/Molecular weight= 6.92*106/60=115333.33 moles
moles of oxygen required= 4.5 time moles of oxygen = 4.5*115333.33=519000/day
c) eficiency of oxygen tranfer= 15%, moles of oxygen required= 519000/0.15=3460000 mol/day
Mass of air required =3460000/0.2095=16515513 moles/day
Volume of air / day,V = nRT/P= 16515513*0.08206*(15+273.15)/2=16022996 L/day= 16022.996 m3/day