You are to design a reactor for removal of reduced iron (Fe2+) from water. The i
ID: 811156 • Letter: Y
Question
You are to design a reactor for removal of reduced iron (Fe2+) from water. The influent water has an iron concentration of 10mg/L, and this must be reduced to 0.1mg/L. The water has a pH of 6.5 and the plan is to oxidize the iron to Fe3+ using pure oxygen gas, then removing the resulting particulate matter in a sedimentation basin. It has been found that the reduction of in Fe2+ concentration over time equals Kapparent x [Fe2+], where Kapparent equals 8x10^13x[partial pressure of O2]xKw^2/[H+]^2. The units of Kapparent determined from this experiment are min^-1 and the partial pressure of oxygen is 0.21atm and the dissociation constant for water, Kw, equals 10^-14. Determine the volume (m^3) of a plug flow reactor to treat 1 MGD of water.
Explanation / Answer
The rate law is d[Fe2+]/dt = Kapparent*[Fe2+]
Therefore, ln[Fe2+] = ln[Fe2+]o -Ka*t
From pH value [H+] is 3.162x10-7 and Kw2/[H+]2 = 1x10-15
Ka = 8x1013*0.21*1x10-15 = 0.0168
Therefore, using the rate law, time taken to decrease [Fe2+] from 10 mg/L to 0.1 mg/L is,
t = 1/Ka * ln(10/0.1) = 0.0774 min
That is it takes 0.0774 min to treat 1L water to bring the [Fe2+] to 0.1 mg/L or 12.92 L/min
1 MGD corresponds to 3785400L/1440min or 2628.75 L/min
Therefore, the volume of reactor should be 2628.75/12.92 = 203.47 L or 53.75 gallons.
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