{"id":65,"date":"2021-06-21T21:09:53","date_gmt":"2021-06-21T21:09:53","guid":{"rendered":"https:\/\/books.gw-project.org\/septic-system-impacts-on-groundwater-quality\/chapter\/solution-exercise-2\/"},"modified":"2021-06-23T02:05:16","modified_gmt":"2021-06-23T02:05:16","slug":"solution-exercise-2","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/septic-system-impacts-on-groundwater-quality\/chapter\/solution-exercise-2\/","title":{"raw":"Solution Exercise 2","rendered":"Solution Exercise 2"},"content":{"raw":"
\r\n

Assume one-dimensional flow for the purpose of calculations and consider the rate of S<\/em> consumption in a 1 L cube of aquifer material (10 cm x 10 cm x 10 cm).<\/p>\r\n

Determine the amount of S<\/em> in the 1 L aquifer cube which weighs:<\/p>\r\n

[latex]\\displaystyle weight\\ of\\ 1\\ L\\ cube=10cm\\ 10cm\\ 10cm\\ 1.8\\frac{g}{{cm}^3}=1800\\ g[\/latex]<\/p>\r\n

We know S<\/em> is 0.02 percent of the dry material by weight, so:<\/p>\r\n

[latex]\\displaystyle grams\\ of\\ S\\ in\\ 1\\ L\\ cube=1800\\ g\\ \\ 0.0002=\\ 0.36\\ g\\ S\\ in\\ 1\\ L\\ cube [\/latex]<\/p>\r\n

NO<\/em>3<\/em><\/sub>-<\/em><\/sup>-<\/em>N<\/em> flux through the cube is determined as follows.<\/p>\r\n

[latex]\\displaystyle Darcy\\ Velocity=\\left(Interstital\\ Groundwater\\ Velocity\\right)\\ Porosity[\/latex]<\/p>\r\n

[latex]\\displaystyle = 2800\\frac{cm}{yr}\\ 0.3=840\\frac{cm}{yr}[\/latex]<\/p>\r\n

[latex]\\displaystyle Volumetric\\ Flow\\ through\\ cube=\\left(Darcy\\ Velocity\\right)\\ \\left(Flow\\ Area\\right)[\/latex]<\/p>\r\n

[latex]\\displaystyle =\\ 840\\frac{cm}{yr}\\ 10cm\\ 10cm=84000\\frac{{cm}^3}{yr}[\/latex]<\/p>\r\n

[latex]\\displaystyle NO_{3}^{-}\\textup{-}N\\ flux\\ through\\ cube=(Volumetric\\ Flow) (Concentration) [\/latex]<\/p>\r\n

[latex]\\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}[\/latex]<\/p>\r\n

[latex]\\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}[\/latex]<\/p>\r\n

From Equation 14<\/a>, 1 M of NO<\/em>3<\/em><\/sub>-<\/em><\/sup>-<\/em>N<\/em> will consume 10\/14 M S<\/em>:<\/p>\r\n

[latex]\\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}[\/latex]<\/p>\r\n

[latex]\\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}[\/latex]<\/p>\r\n

Given that there is 0.36 g S<\/em> in each 1 L cube of aquifer, each cube has sufficient S<\/em> to support autotrophic denitrification of a plume for 0.052 years as shown below.<\/p>\r\n

[latex]\\displaystyle Number\\ of\\ years\\ required\\ to\\ deplete\\ S\\ from\\ cube[\/latex] = [latex]\\displaystyle \\frac{0.36\\ g\\ S}{6.9\\frac{g\\ S}{yr}}[\/latex] = [latex]\\displaystyle 0.052\\ yr[\/latex]<\/p>\r\n

Thus, the NO<\/em>3<\/em><\/sub>-<\/em><\/sup>-<\/em>N<\/em> plume will advance 10 cm every 0.052 yr, or one could say it can consume all S<\/em> from 19.2 cubes each year (1\/0.052=19.2), which is a rate of NO<\/em>3<\/em><\/sub>-<\/em><\/sup>-<\/em>N<\/em> plume advance of 192 cm\/yr. Another way to calculate the advance rate is as follows.<\/p>\r\n

[latex]\\displaystyle NO_{3}^{-}\\textup{-}N\\ plume\\ advance\\ rate=\\frac{10\\ cm}{0.052\\ yr}=\\frac{192\\ cm}{yr}[\/latex]<\/p>\r\n

Return to Exercise 2<\/span><\/a><\/p>\r\n

<\/p>\r\n\r\n<\/div>","rendered":"

\n

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

Determine the amount of S<\/em> in the 1 L aquifer cube which weighs:<\/p>\n

\"\displaystyle weight\ of\ 1\ L\ cube=10cm\ 10cm\ 10cm\ 1.8\frac{g}{{cm}^3}=1800\ g\"<\/p>\n

We know S<\/em> is 0.02 percent of the dry material by weight, so:<\/p>\n

\"\displaystyle grams\ of\ S\ in\ 1\ L\ cube=1800\ g\ \ 0.0002=\ 0.36\ g\ S\ in\ 1\ L\ cube \"<\/p>\n

NO<\/em>3<\/em><\/sub>–<\/em><\/sup>–<\/em>N<\/em> flux through the cube is determined as follows.<\/p>\n

\"\displaystyle Darcy\ Velocity=\left(Interstital\ Groundwater\ Velocity\right)\ Porosity\"<\/p>\n

\"\displaystyle = 2800\frac{cm}{yr}\ 0.3=840\frac{cm}{yr}\"<\/p>\n

\"\displaystyle Volumetric\ Flow\ through\ cube=\left(Darcy\ Velocity\right)\ \left(Flow\ Area\right)\"<\/p>\n

\"\displaystyle =\ 840\frac{cm}{yr}\ 10cm\ 10cm=84000\frac{{cm}^3}{yr}\"<\/p>\n

\"\displaystyle NO_{3}^{-}\textup{-}N\ flux\ through\ cube=(Volumetric\ Flow) (Concentration) \"<\/p>\n

\"\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}\"<\/p>\n

\"\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}\"<\/p>\n

From Equation 14<\/a>, 1 M of NO<\/em>3<\/em><\/sub>–<\/em><\/sup>–<\/em>N<\/em> will consume 10\/14 M S<\/em>:<\/p>\n

\"\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}\"<\/p>\n

\"\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}\"<\/p>\n

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

\"\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\"<\/p>\n

Thus, the NO<\/em>3<\/em><\/sub>–<\/em><\/sup>–<\/em>N<\/em> plume will advance 10 cm every 0.052 yr, or one could say it can consume all S<\/em> from 19.2 cubes each year (1\/0.052=19.2), which is a rate of NO<\/em>3<\/em><\/sub>–<\/em><\/sup>–<\/em>N<\/em> plume advance of 192 cm\/yr. Another way to calculate the advance rate is as follows.<\/p>\n

\"\displaystyle NO_{3}^{-}\textup{-}N\ plume\ advance\ rate=\frac{10\ cm}{0.052\ yr}=\frac{192\ cm}{yr}\"<\/p>\n

Return to Exercise 2<\/span><\/a><\/p>\n

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