{"id":274,"date":"2020-11-19T23:44:29","date_gmt":"2020-11-19T23:44:29","guid":{"rendered":"https:\/\/books.gw-project.org\/introduction-to-isotopes-and-environmental-tracers-as-indicators-of-groundwater-flow\/?post_type=part&p=274"},"modified":"2022-09-20T16:09:42","modified_gmt":"2022-09-20T16:09:42","slug":"exercise-solutions","status":"publish","type":"part","link":"https:\/\/books.gw-project.org\/introduction-to-isotopes-and-environmental-tracers-as-indicators-of-groundwater-flow\/part\/exercise-solutions\/","title":{"raw":"8 Exercise Solutions","rendered":"8 Exercise Solutions"},"content":{"raw":"
<\/a>Exercise 1 Solution<\/span><\/h5>\r\nAnswer:<\/strong> The proportion of river water in piezometer P2 is calculated to be 25 percent using 2<\/sup>H values, and 20 percent using chloride concentrations. The corresponding proportions of regional groundwater in piezometer P2 are then 75 percent and 80 percent using 2<\/sup>H and chloride, respectively.\r\n\r\nDetails:<\/strong> Equation 7<\/a> can be re-arranged to [latex]\\displaystyle f=\\frac{c_m-c_2}{c_1-c_2}[\/latex]\r\n\r\nwhere f<\/em> is the proportion of end-member 1, c<\/em>1<\/em><\/sub> and c<\/em>2<\/em><\/sub> are the respectively end-member concentrations and c<\/em>m<\/em><\/sub> is the concentration of the mixed sample. Using c<\/em>m<\/em><\/sub> = 380 mg\/L, c<\/em>1<\/em><\/sub> = 25 mg\/L and c<\/em>2<\/em><\/sub> = 500 mg\/L gives f<\/em> = (-120)\/(-475) = 0.25. Using c<\/em>m<\/em><\/sub> = -26 \u2030, c<\/em>1<\/em><\/sub> = -10 \u2030 and c<\/em>2 <\/em><\/sub>= -30 \u2030 gives f<\/em> = 4\/20 = 0.2.\r\n

Return to exercise 1<\/a><\/p>\r\n\r\n

<\/a>Exercise 2 Solution<\/span><\/h5>\r\nAnswer:<\/strong> The velocity is 4.6 meters per year and the hydraulic conductivity is 16.3 meters per day.\r\n\r\nDetails<\/strong>:<\/strong> First use Equation 1<\/a> to calculate the groundwater age at each well. Using c<\/em>0<\/em><\/sub> = 90 pmC and \u03bb<\/em> = 1.21 \u00d7 10-<\/sup>4<\/sup>, gives t<\/em> = 1077 years for the upstream well and 3921 years for the downstream well. The age gradient is then (3921 \u2013 1077) \/ 13 000 years per meter = 0.219 y m-<\/sup>1<\/sup>. From Equation 9, the horizontal velocity is the inverse of the age gradient, which equals 4.57 m y-<\/sup>1<\/sup>.\r\n\r\nDarcy\u2019s Law gives the relationship between the seepage velocity and the hydraulic conductivity as:\r\n\r\nv<\/em> = Ki<\/em>\/n<\/em>e<\/em><\/sub>\r\n\r\nso\r\n\r\nK<\/em> = vn<\/em>e<\/em><\/sub>\/<\/em>i<\/em>\r\n\r\ni<\/em> = (h<\/em>1<\/sub>-h<\/em>2<\/sub>)\/(distance between h<\/em>1<\/sub> and h<\/em>2<\/sub>) = 2 m \/ 13000 m = 0.000154\r\n\r\nK<\/em> = 4.57 m\/y \u00d7 0.2 \/ 0.000154 = 5943 m\/y = 16.3 m\/day\r\n

Return to exercise 2<\/a><\/p>\r\n\r\n

<\/a>Exercise 3 Solution<\/span><\/h5>\r\nAnswer:<\/strong> The recharge rate is 60 millimeters per year.\r\n\r\nDetails:<\/strong> Equation 9<\/a> can be re-arranged to give [latex]R=\\frac{H\\theta}{t}\\ln \\left(\\frac{H}{H-z}\\right)[\/latex]\r\n\r\nUsing H<\/em> = 50 m, z<\/em> = 10 m, \u03b8<\/em> = 0.2 and t<\/em> = 39 years, gives R<\/em> = 0.057 m\/y or 57 mm\/y or rounding 60 mm\/yr.\r\n

Return to exercise 3<\/a><\/p>\r\n\r\n

<\/a>Exercise 4 Solution<\/span><\/h5>\r\nAnswer:<\/strong> The groundwater age is 7.3 years.\r\n\r\nWorking<\/strong>: In Equation 3<\/a>, c<\/em>p<\/em><\/sub> = 10 TU, c<\/em>d<\/em><\/sub> = 5 TU and \u03bb= 0.0558 y-<\/sup>1<\/sup>. The groundwater age is therefore calculated to be 7.3 years.\r\n

Return to exercise 4<\/a><\/p>","rendered":"

<\/a>Exercise 1 Solution<\/span><\/h5>\n

Answer:<\/strong> The proportion of river water in piezometer P2 is calculated to be 25 percent using 2<\/sup>H values, and 20 percent using chloride concentrations. The corresponding proportions of regional groundwater in piezometer P2 are then 75 percent and 80 percent using 2<\/sup>H and chloride, respectively.<\/p>\n

Details:<\/strong> Equation 7<\/a> can be re-arranged to \"\displaystyle f=\frac{c_m-c_2}{c_1-c_2}\"<\/p>\n

where f<\/em> is the proportion of end-member 1, c<\/em>1<\/em><\/sub> and c<\/em>2<\/em><\/sub> are the respectively end-member concentrations and c<\/em>m<\/em><\/sub> is the concentration of the mixed sample. Using c<\/em>m<\/em><\/sub> = 380 mg\/L, c<\/em>1<\/em><\/sub> = 25 mg\/L and c<\/em>2<\/em><\/sub> = 500 mg\/L gives f<\/em> = (-120)\/(-475) = 0.25. Using c<\/em>m<\/em><\/sub> = -26 \u2030, c<\/em>1<\/em><\/sub> = -10 \u2030 and c<\/em>2 <\/em><\/sub>= -30 \u2030 gives f<\/em> = 4\/20 = 0.2.<\/p>\n

Return to exercise 1<\/a><\/p>\n

<\/a>Exercise 2 Solution<\/span><\/h5>\n

Answer:<\/strong> The velocity is 4.6 meters per year and the hydraulic conductivity is 16.3 meters per day.<\/p>\n

Details<\/strong>:<\/strong> First use Equation 1<\/a> to calculate the groundwater age at each well. Using c<\/em>0<\/em><\/sub> = 90 pmC and \u03bb<\/em> = 1.21 \u00d7 10–<\/sup>4<\/sup>, gives t<\/em> = 1077 years for the upstream well and 3921 years for the downstream well. The age gradient is then (3921 \u2013 1077) \/ 13 000 years per meter = 0.219 y m–<\/sup>1<\/sup>. From Equation 9, the horizontal velocity is the inverse of the age gradient, which equals 4.57 m y–<\/sup>1<\/sup>.<\/p>\n

Darcy\u2019s Law gives the relationship between the seepage velocity and the hydraulic conductivity as:<\/p>\n

v<\/em> = Ki<\/em>\/n<\/em>e<\/em><\/sub><\/p>\n

so<\/p>\n

K<\/em> = vn<\/em>e<\/em><\/sub>\/<\/em>i<\/em><\/p>\n

i<\/em> = (h<\/em>1<\/sub>–h<\/em>2<\/sub>)\/(distance between h<\/em>1<\/sub> and h<\/em>2<\/sub>) = 2 m \/ 13000 m = 0.000154<\/p>\n

K<\/em> = 4.57 m\/y \u00d7 0.2 \/ 0.000154 = 5943 m\/y = 16.3 m\/day<\/p>\n

Return to exercise 2<\/a><\/p>\n

<\/a>Exercise 3 Solution<\/span><\/h5>\n

Answer:<\/strong> The recharge rate is 60 millimeters per year.<\/p>\n

Details:<\/strong> Equation 9<\/a> can be re-arranged to give \"R=\frac{H\theta}{t}\ln \left(\frac{H}{H-z}\right)\"<\/p>\n

Using H<\/em> = 50 m, z<\/em> = 10 m, \u03b8<\/em> = 0.2 and t<\/em> = 39 years, gives R<\/em> = 0.057 m\/y or 57 mm\/y or rounding 60 mm\/yr.<\/p>\n

Return to exercise 3<\/a><\/p>\n

<\/a>Exercise 4 Solution<\/span><\/h5>\n

Answer:<\/strong> The groundwater age is 7.3 years.<\/p>\n

Working<\/strong>: In Equation 3<\/a>, c<\/em>p<\/em><\/sub> = 10 TU, c<\/em>d<\/em><\/sub> = 5 TU and \u03bb= 0.0558 y–<\/sup>1<\/sup>. The groundwater age is therefore calculated to be 7.3 years.<\/p>\n

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