{"id":96,"date":"2022-07-15T15:06:13","date_gmt":"2022-07-15T15:06:13","guid":{"rendered":"https:\/\/books.gw-project.org\/domestic-wells-introduction-and-overview\/chapter\/solution-exercise-1\/"},"modified":"2022-07-15T21:19:48","modified_gmt":"2022-07-15T21:19:48","slug":"solution-exercise-1","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/domestic-wells-introduction-and-overview\/chapter\/solution-exercise-1\/","title":{"raw":"Solution Exercise 1","rendered":"Solution Exercise 1"},"content":{"raw":"
\r\n

Determine if the well can supply the daily target water supply volume for the household by following the example calculations in Box 1<\/a>. The target water supply volume is 860\u00a0L\/day, based on four people using 215\u00a0L\/day per person.<\/p>\r\n

[latex]\\displaystyle target\\ water\\ volume = 4\\ persons * \\frac{215\\frac{L}{day}}{person}=860\\ L\/day[\/latex]<\/p>\r\n

Therefore, the well must be able to provide 860\u00a0L\/day and must be able to provide this amount on both an ongoing daily basis and within a two-hour period to meet peak water demands. This can be achieved with a combination of well yield and water storage. The volume of water that can be supplied from the two-hour well yield and daily well yield is 240\u00a0L and 2,880\u00a0L, respectively.<\/p>\r\n

[latex]\\displaystyle 2\\ hour\\ well\\ yield = 2\\ hr * 2\\frac{L}{min} * 60\\ min\/hr=240\\ L[\/latex]<\/p>\r\n

[latex]\\displaystyle 24\\ hour\\ well\\ yield = 24\\ hr * 2\\frac{L}{min} * 60\\ min\/hr[\/latex] =<\/strong> 2,880<\/strong> L<\/em><\/strong><\/p>\r\n

The amount of available water stored in the well is calculated from the well depth (50\u00a0m), well diameter (152\u00a0mm), static water level (5\u00a0m), and pump setting allowance (5\u00a0m), and is calculated to be 726\u00a0L.<\/p>\r\n

well storage\u00a0volume<\/em><\/strong> =<\/strong> (well\u00a0depth<\/em><\/strong> \u2212\u00a0static\u00a0level<\/em><\/strong> \u2212 pump setting<\/em><\/strong>) *<\/strong>\u00a0\u03c0r<\/em>2<\/sup><\/strong> = [latex]\\left (50\\ m-5\\ m-5\\ m \\right ) * \\pi \\left ( \\frac{0.152\\ m}{2} \\right )^{2}[\/latex] *<\/strong> 1,000\u00a0L<\/em>\/m<\/em>3<\/sup><\/strong> =<\/strong> 726<\/strong>\u00a0L<\/strong><\/em><\/p>\r\n

The total amount of water that this well can provide within a two-hour period is 966\u00a0L (i.e., by adding the two-hour well yield (240\u00a0L) and the well storage (726\u00a0L) together).<\/p>\r\n

total\u00a0water\u00a0available\u00a0in\u00a02\u00a0hour\u00a0period<\/em><\/strong> = 2\u00a0hour\u00a0well\u00a0yield<\/em><\/strong> + well\u00a0storage<\/em><\/strong> = 240\u00a0L<\/em><\/strong> + 726\u00a0L<\/em><\/strong> = 996\u00a0L<\/em><\/strong><\/p>\r\n

It is concluded that the well can meet this household\u2019s needs because the target water supply of 860\u00a0L\/day is exceeded by both the daily well yield (2,880\u00a0L\/day) and the two-hour yield combined with well storage (966\u00a0L).<\/p>\r\n

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

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

\n

Determine if the well can supply the daily target water supply volume for the household by following the example calculations in Box 1<\/a>. The target water supply volume is 860\u00a0L\/day, based on four people using 215\u00a0L\/day per person.<\/p>\n

[latex]\\displaystyle target\\ water\\ volume = 4\\ persons * \\frac{215\\frac{L}{day}}{person}=860\\ L\/day[\/latex]<\/p>\n

Therefore, the well must be able to provide 860\u00a0L\/day and must be able to provide this amount on both an ongoing daily basis and within a two-hour period to meet peak water demands. This can be achieved with a combination of well yield and water storage. The volume of water that can be supplied from the two-hour well yield and daily well yield is 240\u00a0L and 2,880\u00a0L, respectively.<\/p>\n

[latex]\\displaystyle 2\\ hour\\ well\\ yield = 2\\ hr * 2\\frac{L}{min} * 60\\ min\/hr=240\\ L[\/latex]<\/p>\n

[latex]\\displaystyle 24\\ hour\\ well\\ yield = 24\\ hr * 2\\frac{L}{min} * 60\\ min\/hr[\/latex] =<\/strong> 2,880<\/strong> L<\/em><\/strong><\/p>\n

The amount of available water stored in the well is calculated from the well depth (50\u00a0m), well diameter (152\u00a0mm), static water level (5\u00a0m), and pump setting allowance (5\u00a0m), and is calculated to be 726\u00a0L.<\/p>\n

well storage\u00a0volume<\/em><\/strong> =<\/strong> (well\u00a0depth<\/em><\/strong> \u2212\u00a0static\u00a0level<\/em><\/strong> \u2212 pump setting<\/em><\/strong>) *<\/strong>\u00a0\u03c0r<\/em>2<\/sup><\/strong> = [latex]\\left (50\\ m-5\\ m-5\\ m \\right ) * \\pi \\left ( \\frac{0.152\\ m}{2} \\right )^{2}[\/latex] *<\/strong> 1,000\u00a0L<\/em>\/m<\/em>3<\/sup><\/strong> =<\/strong> 726<\/strong>\u00a0L<\/strong><\/em><\/p>\n

The total amount of water that this well can provide within a two-hour period is 966\u00a0L (i.e., by adding the two-hour well yield (240\u00a0L) and the well storage (726\u00a0L) together).<\/p>\n

total\u00a0water\u00a0available\u00a0in\u00a02\u00a0hour\u00a0period<\/em><\/strong> = 2\u00a0hour\u00a0well\u00a0yield<\/em><\/strong> + well\u00a0storage<\/em><\/strong> = 240\u00a0L<\/em><\/strong> + 726\u00a0L<\/em><\/strong> = 996\u00a0L<\/em><\/strong><\/p>\n

It is concluded that the well can meet this household\u2019s needs because the target water supply of 860\u00a0L\/day is exceeded by both the daily well yield (2,880\u00a0L\/day) and the two-hour yield combined with well storage (966\u00a0L).<\/p>\n

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

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