{"id":106,"date":"2022-12-30T18:15:40","date_gmt":"2022-12-30T18:15:40","guid":{"rendered":"https:\/\/books.gw-project.org\/variable-density-groundwater-flow\/chapter\/solution-exercise-5\/"},"modified":"2023-01-08T04:12:59","modified_gmt":"2023-01-08T04:12:59","slug":"solution-exercise-5","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/variable-density-groundwater-flow\/chapter\/solution-exercise-5\/","title":{"raw":"Solution Exercise 5","rendered":"Solution Exercise 5"},"content":{"raw":"
\r\n

It can be seen that the density does not vary between the screens, so this means the point water hydraulic heads can be used directly in the calculation of q<\/em>z <\/em><\/sub>with Darcy\u2019s law. Given the negative numbers, the smallest absolute value is the highest hydraulic head so the hydraulic head increases with depth, thus flow is upward from screen\u00a03 to screen\u00a02. The factor of 1000 in the equation\u00a0converts meters to millimeters.<\/p>\r\n

[latex]\\displaystyle q_{z}=-20\\frac{(-3.08--3.04)}{(-96.44--118.93)}\\times 1000=36\\;\\textrm{mm}\\;\\textrm{d}^{-1}[\/latex]<\/p>\r\n

While from screen 2 to 1 it is<\/p>\r\n

[latex]\\displaystyle q_{z}=-20\\frac{(-3.13--3.08)}{(-76.41--96.44)}\\times 1000=50\\;\\textrm{mm}\\;\\textrm{d}^{-1}[\/latex]<\/p>\r\n

For the more general case where the density varies, one would first convert the hydraulic heads to freshwater heads using [latex]h_{f}=z_{i}+(h_{i}-z_{i})\\frac{\\rho _{i}}{\\rho _{f}}[\/latex], which yields (three decimals are provided to prevent large rounding errors in the calculation of q<\/em>z<\/em><\/sub>):<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Screen<\/strong><\/td>\r\nh<\/em><\/strong>f<\/em><\/strong><\/sub><\/td>\r\n<\/tr>\r\n
1<\/td>\r\n\u22122.815<\/td>\r\n<\/tr>\r\n
2<\/td>\r\n\u22122.679<\/td>\r\n<\/tr>\r\n
3<\/td>\r\n\u22122.542<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

Using [latex]q_{z}=-K_{f}\\left ( \\frac{\\Delta h_{f}}{\\Delta z}+\\frac{\\rho _{mean}-\\rho _{f}}{\\rho _{{f}}} \\right )[\/latex] one finds for the flow between piezometer\u00a03 and 2:<\/p>\r\n

[latex]\\displaystyle q_{z}=-20\\left ( \\frac{(-2.679--2.542)}{(-96.44--118.93)}+\\frac{1004.3-1000}{1000} \\right )\\times 1000=36\\;\\textrm{mm}\\;\\textrm{d}^{-1}[\/latex]<\/p>\r\n

And for the flow between piezometer\u00a02 and 1<\/p>\r\n

[latex]\\displaystyle q_{z}=-20\\left ( \\frac{(-2.815--2.679)}{(-76.41--96.44)}+\\frac{1004.3-1000}{1000} \\right )\\times 1000=5\\;\\textrm{mm}\\;\\textrm{d}^{-1}[\/latex]<\/p>\r\n

Which is the same answer as before (note that the unrounded result has a 0.43 percent error because K<\/em>f<\/em><\/sub> was equated to the hydraulic conductivity).<\/p>\r\n

Return to <\/span>Exercise <\/span>5<\/span><\/a><\/p>\r\n

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

\n

It can be seen that the density does not vary between the screens, so this means the point water hydraulic heads can be used directly in the calculation of q<\/em>z <\/em><\/sub>with Darcy\u2019s law. Given the negative numbers, the smallest absolute value is the highest hydraulic head so the hydraulic head increases with depth, thus flow is upward from screen\u00a03 to screen\u00a02. The factor of 1000 in the equation\u00a0converts meters to millimeters.<\/p>\n

\"\displaystyle q_{z}=-20\frac{(-3.08--3.04)}{(-96.44--118.93)}\times 1000=36\;\textrm{mm}\;\textrm{d}^{-1}\"<\/p>\n

While from screen 2 to 1 it is<\/p>\n

\"\displaystyle q_{z}=-20\frac{(-3.13--3.08)}{(-76.41--96.44)}\times 1000=50\;\textrm{mm}\;\textrm{d}^{-1}\"<\/p>\n

For the more general case where the density varies, one would first convert the hydraulic heads to freshwater heads using \"h_{f}=z_{i}+(h_{i}-z_{i})\frac{\rho _{i}}{\rho _{f}}\", which yields (three decimals are provided to prevent large rounding errors in the calculation of q<\/em>z<\/em><\/sub>):<\/p>\n\n\n\n\n\n\n
Screen<\/strong><\/td>\nh<\/em><\/strong>f<\/em><\/strong><\/sub><\/td>\n<\/tr>\n
1<\/td>\n\u22122.815<\/td>\n<\/tr>\n
2<\/td>\n\u22122.679<\/td>\n<\/tr>\n
3<\/td>\n\u22122.542<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Using \"q_{z}=-K_{f}\left ( \frac{\Delta h_{f}}{\Delta z}+\frac{\rho _{mean}-\rho _{f}}{\rho _{{f}}} \right )\" one finds for the flow between piezometer\u00a03 and 2:<\/p>\n

\"\displaystyle q_{z}=-20\left ( \frac{(-2.679--2.542)}{(-96.44--118.93)}+\frac{1004.3-1000}{1000} \right )\times 1000=36\;\textrm{mm}\;\textrm{d}^{-1}\"<\/p>\n

And for the flow between piezometer\u00a02 and 1<\/p>\n

\"\displaystyle q_{z}=-20\left ( \frac{(-2.815--2.679)}{(-76.41--96.44)}+\frac{1004.3-1000}{1000} \right )\times 1000=5\;\textrm{mm}\;\textrm{d}^{-1}\"<\/p>\n

Which is the same answer as before (note that the unrounded result has a 0.43 percent error because K<\/em>f<\/em><\/sub> was equated to the hydraulic conductivity).<\/p>\n

Return to <\/span>Exercise <\/span>5<\/span><\/a><\/p>\n

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