Solution Exercise 2

Assume one-dimensional flow for the purpose of calculations and consider the rate of S consumption in a 1 L cube of aquifer material (10 cm x 10 cm x 10 cm).

Determine the amount of S in the 1 L aquifer cube which weighs:

\displaystyle weight\ of\ 1\ L\ cube=10cm\ 10cm\ 10cm\ 1.8\frac{g}{{cm}^3}=1800\ g

We know S is 0.02 percent of the dry material by weight, so:

\displaystyle grams\ of\ S\ in\ 1\ L\ cube=1800\ g\ \ 0.0002=\ 0.36\ g\ S\ in\ 1\ L\ cube

NO3N flux through the cube is determined as follows.

\displaystyle Darcy\ Velocity=\left(Interstital\ Groundwater\ Velocity\right)\ Porosity

\displaystyle = 2800\frac{cm}{yr}\ 0.3=840\frac{cm}{yr}

\displaystyle Volumetric\ Flow\ through\ cube=\left(Darcy\ Velocity\right)\ \left(Flow\ Area\right)

\displaystyle =\ 840\frac{cm}{yr}\ 10cm\ 10cm=84000\frac{{cm}^3}{yr}

\displaystyle NO_{3}^{-}\textup{-}N\ flux\ through\ cube=(Volumetric\ Flow) (Concentration)

\displaystyle =84000\frac{cm^{3}}{yr}\frac{50\ mg\ NO_{3}^{-}\textup{-}N}{1000\ cm^{2}}=4200\frac{mg\ NO_{3}^{-}\textup{-}N}{yr}=4.2\frac{g\ NO_{3}^{-}\textup{-}N}{yr}

\displaystyle \frac{Moles\ NO_{3}^{-}\textup{-}N}{yr}=4.2\frac{g}{yr}\ \frac{1\ M\ NO_{3}^{-}\textup{-}N}{14\ g}=0.3\frac{M\ NO_{3}^{-}\textup{-}N}{yr}

From Equation 14, 1 M of NO3N will consume 10/14 M S:

\displaystyle Moles\ S\ consumed=0.3\frac{M\ NO_{3}^{-}\textup{-}N}{yr}\frac{10\ M\ S}{14\ M\ NO_{3}^{-}\textup{-}N}=0.21\frac{M\ S}{yr}

\displaystyle \frac{g\ S\ consumed}{yr}=0.21\frac{M\ S}{yr}\ \frac{32.06g\ S}{1\ M\ S}=6.87\frac{g\ S}{yr}

Given that there is 0.36 g S in each 1 L cube of aquifer, each cube has sufficient S to support autotrophic denitrification of a plume for 0.052 years as shown below.

\displaystyle Number\ of\ years\ required\ to\ deplete\ S\ from\ cube = \displaystyle \frac{0.36\ g\ S}{6.9\frac{g\ S}{yr}} = \displaystyle 0.052\ yr

Thus, the NO3N plume will advance 10 cm every 0.052 yr, or one could say it can consume all S from 19.2 cubes each year (1/0.052=19.2), which is a rate of NO3N plume advance of 192 cm/yr. Another way to calculate the advance rate is as follows.

\displaystyle NO_{3}^{-}\textup{-}N\ plume\ advance\ rate=\frac{10\ cm}{0.052\ yr}=\frac{192\ cm}{yr}

Return to Exercise 2

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Septic System Impacts on Groundwater Quality Copyright © 2021 by William Robertson. All Rights Reserved.