{"id":1425,"date":"2026-05-12T21:47:49","date_gmt":"2026-05-12T21:47:49","guid":{"rendered":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/chapter\/solution-exercise-24\/"},"modified":"2026-05-13T03:39:59","modified_gmt":"2026-05-13T03:39:59","slug":"solution-exercise-24","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/chapter\/solution-exercise-24\/","title":{"raw":"Solution Exercise 24","rendered":"Solution Exercise 24"},"content":{"raw":"
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Calculate the concentration for part (a) using Equation (98)<\/span><\/a> and for part (b) using Equation (99)<\/span><\/a>.<\/blockquote>\r\n<\/div>\r\n
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Average linear velocity<\/strong> is given by Equation (93)<\/span><\/a>:<\/blockquote>\r\n<\/div>\r\n
\r\n\r\n[latex]\\overline{v_{x}} = \\frac{K\\frac{\\partial h}{\\partial x}}{\\varphi_{eff}} = \\frac{0.1\\ \\frac{cm}{s}\\frac{10\\ cm}{100\\ cm}}{0.2} = \\mathbf{0.05\\ }\\frac{\\mathbf{cm}}{\\mathbf{s}}[\/latex]\r\n\r\n<\/div>\r\n
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Dispersion coefficient<\/strong> is given by Equation (96)<\/span><\/a>:<\/blockquote>\r\n<\/div>\r\n
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[latex]D_{x} = \\alpha_{x}\\overline{v_{x}} + D_{mol} = 5\\ cm\\ 0.05\\ \\frac{cm}{s} + {1x10}^{- 9}\\ \\frac{m^{2}}{s}\\frac{{10,000\\ cm}^{2}}{m^{2}} = \\mathbf{0.25}\\frac{\\mathbf{cm}^{\\mathbf{2}}}{\\mathbf{s}}[\/latex]<\/blockquote>\r\n<\/div>\r\n
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Concentration is given by Equation (98)<\/span><\/a>:<\/blockquote>\r\n<\/div>\r\n
\r\n\r\n[latex]C = \\frac{C_{o}}{2}\\left\\lbrack erfc\\left( \\frac{x - \\overline{v_{x}}t}{2\\sqrt{D_{x}t}} \\right) + \\exp\\left( \\frac{x\\overline{v_{x}}}{D_{x}} \\right)erfc\\left( \\frac{x + \\overline{v_{x}}t}{2\\sqrt{D_{x}t}} \\right) \\right\\rbrack[\/latex]\r\n\r\n<\/div>\r\n
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a) For 25 cm the concentration is<\/u><\/strong> calculated as follows:<\/blockquote>\r\n<\/div>\r\n
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[latex]C = \\frac{1000\\frac{mg}{L}}{2}\\left\\lbrack erfc\\left( \\frac{25\\ cm - 0.05\\ \\frac{cm}{s}1000\\ s}{2\\sqrt{0.25\\frac{{cm}^{2}}{s}1000\\ s}} \\right) + exp\\left( \\frac{25\\ cm\\ 0.05\\ \\frac{cm}{s}}{0.25\\frac{{cm}^{2}}{s}} \\right)erfc\\left( \\frac{25\\ cm + 0.05\\ \\frac{cm}{s}\\ 1000\\ s}{2\\sqrt{0.25\\frac{{cm}^{2}}{s}1000\\ s}} \\right) \\right\\rbrack[\/latex]<\/blockquote>\r\n<\/div>\r\n
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[latex]C = 500\\frac{mg}{L}\\left\\lbrack erfc( - 0.79057) + \\exp(5)erfc(2.3717) \\right\\rbrack[\/latex]<\/blockquote>\r\n<\/div>\r\n
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[latex]erfc( - 0.79057) = 1 - \\left( - {erf}(0.79057) \\right) = 1 + 0.7364[\/latex] = 1.7364<\/blockquote>\r\n<\/div>\r\n
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[latex]\\exp(5) = 148.4132[\/latex]<\/blockquote>\r\n<\/div>\r\n
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[latex]erfc(2.3717) = 1 - \\left( {erf}(2.3717) \\right) = 1 - 0.999204[\/latex] = 0.000796<\/blockquote>\r\n<\/div>\r\n
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[latex]C = 500\\ \\frac{mg}{L}\\left\\lbrack (1.7364) + 148.4132(0.000796) \\right\\rbrack[\/latex]<\/blockquote>\r\n<\/div>\r\n
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[latex]C = 500\\ \\frac{mg}{L}\\lbrack 1.854537\\rbrack = \\mathbf{927\\ }\\frac{\\mathbf{mg}}{\\mathbf{L}}[\/latex]<\/blockquote>\r\n<\/div>\r\n
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Following similar steps:<\/blockquote>\r\n<\/div>\r\n
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b) at 50 cm<\/u>, C=500 mg\/L<\/strong><\/blockquote>\r\n<\/div>\r\n
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c) at 75 cm,<\/u> C=132 mg\/L<\/strong><\/blockquote>\r\n<\/div>\r\n
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  1. These values make sense.<\/u><\/strong> We could have deduced the concentration at 50 cm because it takes 1000 seconds to reach 50 cm so the concentration would be half of the source concentration. The concentration at 25 cm should be higher than half the source concentration because it is closer than the location the contaminant would travel in 1000 seconds, while the concentration at 75 cm should be less because it is further than the average travel distance at 10000 seconds.<\/li>\r\n \t
  2. Field systems will deviate from the conditions described by analytical solutions. Among other reasons, deviation from values predicted by analytical solutions will occur because:<\/u><\/strong><\/li>\r\n<\/ol>\r\n