{"id":1392,"date":"2026-05-12T21:47:44","date_gmt":"2026-05-12T21:47:44","guid":{"rendered":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/chapter\/solution-exercise-4\/"},"modified":"2026-05-13T16:57:42","modified_gmt":"2026-05-13T16:57:42","slug":"solution-exercise-4","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/chapter\/solution-exercise-4\/","title":{"raw":"Solution Exercise 4","rendered":"Solution Exercise 4"},"content":{"raw":"
\r\n
\r\n\r\nPart 1:\r\n\r\n<\/div>\r\n
\r\n
[latex]pBulk\\ = \\ \\left( p\\min\\ \\ \\lbrack 1 - n\\rbrack \\right) + \\left( pH_{2}O\\ \\ n \\right); [\/latex]<\/blockquote>\r\n<\/div>\r\n
\r\n
[latex]p\\min\\ = \\ pBulk - \\frac{pH_{2}O\\ n}{1 - n} [\/latex]<\/blockquote>\r\n<\/div>\r\n
\r\n
[latex]\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\left( 2.61\\frac{g}{{cm}^{3}} \\right) - \\ \\frac{1\\ (0.12)}{0.88}\\ = \\ 2.83\\ \\frac{g}{cm^{3}} [\/latex]<\/blockquote>\r\n<\/div>\r\n
\r\n\r\nPart 2:\r\n\r\n<\/div>\r\n
\r\n\r\na)\r\n
[latex]pBulk\\ = \\ \\left( p\\min\\ \\ \\lbrack 1 - n\\rbrack \\right)[\/latex]<\/blockquote>\r\n<\/div>\r\n
\r\n
[latex]n\\ = \\ 1 - \\left( \\frac{pBulk}{p\\min} \\right)[\/latex]<\/blockquote>\r\n<\/div>\r\n
\r\n
[latex]1 - \\frac{2,320\\frac{kg}{m^{3}}}{2,650\\frac{kg}{m^{3}}} = \\ 0.125[\/latex]<\/blockquote>\r\nb)\u00a025\u00a0percent:<\/span>\r\n
[latex]\\ 0.125\\ \\left( 1,000\\ \\frac{kg}{m^{3}}1\\ m^{3}\\ (0.25) \\right)\\ = \\ 31.1\\ kg\\ + \\ 2,320\\ kg\\ = \\ 2,351\\ kg[\/latex]<\/span><\/blockquote>\r\n<\/div>\r\n
\r\n
75\u00a0percent: [latex]0.125\\ \\left( 1,000\\ \\frac{kg}{m^{3}}1\\ m^{3}\\ (0.75) \\right)\\ = \\ 93.4\\ kg\\ + \\ 2,320\\ kg\\ = \\ 2,413\\ kg[\/latex]<\/blockquote>\r\n<\/div>\r\n
\r\n
100\u00a0percent: [latex]0.125\\ \\left( 1000\\ \\frac{kg}{m^{3}}1\\ m^{3}\\ (1.00) \\right) = \\ 124.5\\ kg\\ + \\ 2,320\\ kg\\ = \\ 2,445\\ kg[\/latex]<\/blockquote>\r\n<\/div>\r\n
\r\n\r\nPart 3:\r\n\r\n<\/div>\r\n
\r\n\r\na) [latex]pBulk\\ = \\ \\frac{mass}{volume} = \\ \\frac{753.7\\ kg}{0.36\\ m^{3}}\\ = \\ 2093.5\\ \\frac{kg}{m^{3}}[\/latex]\r\n\r\n<\/div>\r\n
\r\n
[latex]\\ \\ pBulk\\ = \\ p\\min\\ \\lbrack 1 - n\\rbrack[\/latex]<\/blockquote>\r\n<\/div>\r\n
\r\n
[latex]\\ \\ n\\ = \\ 1 - \\frac{pBulk}{p\\min}[\/latex]<\/blockquote>\r\n<\/div>\r\n
\r\n
[latex]1 - \\frac{2093.5\\frac{kg}{m^{3}}}{2650\\frac{kg}{m^{3}}}\\ = \\ 0.21[\/latex]<\/blockquote>\r\n<\/div>\r\n
\r\n\r\nb) [latex]\\ \\left( (0.21)\\ 1000\\ \\frac{kg}{m^{3}}\\ 0.36\\ m^{3}\\ (0.62) \\right) + 753.7\\ kg = \\ 46.87\\ kg\\ 753.7\\ kg\\ = \\ 800.6\\ kg[\/latex]\r\n\r\n<\/div>\r\n
\r\n\r\nReturn to Exercise\u00a04<\/strong><\/span><\/a>\u2ba5\r\n\r\n<\/div>\r\n
\r\n\r\nReturn to where text linked to Exercise\u00a04<\/strong><\/span><\/a>\u2ba5\r\n\r\n<\/div>\r\n<\/div>","rendered":"
\n
\n

Part 1:<\/p>\n<\/div>\n

\n

[latex]pBulk\\ = \\ \\left( p\\min\\ \\ \\lbrack 1 - n\\rbrack \\right) + \\left( pH_{2}O\\ \\ n \\right);[\/latex]<\/p><\/blockquote>\n<\/div>\n

\n

[latex]p\\min\\ = \\ pBulk - \\frac{pH_{2}O\\ n}{1 - n}[\/latex]<\/p><\/blockquote>\n<\/div>\n

\n

[latex]\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\left( 2.61\\frac{g}{{cm}^{3}} \\right) - \\ \\frac{1\\ (0.12)}{0.88}\\ = \\ 2.83\\ \\frac{g}{cm^{3}}[\/latex]<\/p><\/blockquote>\n<\/div>\n

\n

Part 2:<\/p>\n<\/div>\n

\n

a)<\/p>\n

[latex]pBulk\\ = \\ \\left( p\\min\\ \\ \\lbrack 1 - n\\rbrack \\right)[\/latex]<\/p><\/blockquote>\n<\/div>\n

\n

[latex]n\\ = \\ 1 - \\left( \\frac{pBulk}{p\\min} \\right)[\/latex]<\/p><\/blockquote>\n<\/div>\n

\n

[latex]1 - \\frac{2,320\\frac{kg}{m^{3}}}{2,650\\frac{kg}{m^{3}}} = \\ 0.125[\/latex]<\/p><\/blockquote>\n

b)\u00a025\u00a0percent:<\/span><\/p>\n

[latex]\\ 0.125\\ \\left( 1,000\\ \\frac{kg}{m^{3}}1\\ m^{3}\\ (0.25) \\right)\\ = \\ 31.1\\ kg\\ + \\ 2,320\\ kg\\ = \\ 2,351\\ kg[\/latex]<\/span><\/p><\/blockquote>\n<\/div>\n

\n

75\u00a0percent: [latex]0.125\\ \\left( 1,000\\ \\frac{kg}{m^{3}}1\\ m^{3}\\ (0.75) \\right)\\ = \\ 93.4\\ kg\\ + \\ 2,320\\ kg\\ = \\ 2,413\\ kg[\/latex]<\/p><\/blockquote>\n<\/div>\n

\n

100\u00a0percent: [latex]0.125\\ \\left( 1000\\ \\frac{kg}{m^{3}}1\\ m^{3}\\ (1.00) \\right) = \\ 124.5\\ kg\\ + \\ 2,320\\ kg\\ = \\ 2,445\\ kg[\/latex]<\/p><\/blockquote>\n<\/div>\n

\n

Part 3:<\/p>\n<\/div>\n

\n

a) [latex]pBulk\\ = \\ \\frac{mass}{volume} = \\ \\frac{753.7\\ kg}{0.36\\ m^{3}}\\ = \\ 2093.5\\ \\frac{kg}{m^{3}}[\/latex]<\/p>\n<\/div>\n

\n

[latex]\\ \\ pBulk\\ = \\ p\\min\\ \\lbrack 1 - n\\rbrack[\/latex]<\/p><\/blockquote>\n<\/div>\n

\n

[latex]\\ \\ n\\ = \\ 1 - \\frac{pBulk}{p\\min}[\/latex]<\/p><\/blockquote>\n<\/div>\n

\n

[latex]1 - \\frac{2093.5\\frac{kg}{m^{3}}}{2650\\frac{kg}{m^{3}}}\\ = \\ 0.21[\/latex]<\/p><\/blockquote>\n<\/div>\n

\n

b) [latex]\\ \\left( (0.21)\\ 1000\\ \\frac{kg}{m^{3}}\\ 0.36\\ m^{3}\\ (0.62) \\right) + 753.7\\ kg = \\ 46.87\\ kg\\ 753.7\\ kg\\ = \\ 800.6\\ kg[\/latex]<\/p>\n<\/div>\n

\n

Return to Exercise\u00a04<\/strong><\/span><\/a>\u2ba5<\/p>\n<\/div>\n

\n

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