Question

A site has groundwater contaminated with 1 mg/L hexavalent chromium, Cr(VI). A permeable reactive barrier (PRB) measuring 8 m × 8 m is placed across the path of the groundwater flow and loaded with 30 kg of iron metal Fe0 , which can reduce Cr(VI) to the much less toxic trivalent chromium, Cr(III), by this reaction: Cr(VI) + Fe0 → Cr(III) + Fe(III) In the direction of flow, the hydraulic head declines by 8 m per 1000 m. The soil has a hydraulic conductivity of 1.5 × 10-4 m/s, a porosity of 0.35, and a solids density of 1.8 kg/L. How long will the PRB last before the Fe0 needs to be replaced?

Answer 1

the first part will be absorbed by the soil and after this alcanse the absorption maximum. It s when enter/out are the same it reaches out to say stationary conditions
on the other hand the reduction reaction is
3Cr(+4 )+ Fe = 3Cr(+3) + Fe(+3) the atomic weight of Fe is 55.8g / mol Cr and 51.99g / mol
30kg of Fe must calculating how many moles of iron are present in that amount of mass have to 30kg * 1000g / kg * 55.8g / mol = 537.63mol of Fe that have the ability to absorb 1612.69mol Cr(+4
flow THROUGH soil it is equal to the hydraulic conductivity by the area 1.5 × 10-4 m / s * 8m * 1000m = 1.2m3 / s
have the concentration of Cr(+4 ) in water is 1 mg / l which equals
1mg / l * 1g / 1000mg * 1000l / m3 = 1 g / m3 of Cr(+4 )
as we know that the amount of Cr(+4 ) which can be reduced is 1612.69mol Cr(+4 ) we take the grams
1612.69mol * 51.99g / mol Cr(+4 ) = Cr(+4 ) 83843.75g
as we know that the flow is 1.2m3 / s equivalent to 1.2g ofCr(+4 ) / s m / t = flow 83843.75g Cr(+4 ) / 1.2g / s = 69869.79s =19:41h

this could take a few minutes more if we think in the porosity and the amount absorbed by the soil but for this we need the soil absorption coefficient besides this the minimum time needed to change the iron