{"id":109,"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-8\/"},"modified":"2023-01-08T04:37:45","modified_gmt":"2023-01-08T04:37:45","slug":"solution-exercise-8","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/variable-density-groundwater-flow\/chapter\/solution-exercise-8\/","title":{"raw":"Solution Exercise 8","rendered":"Solution Exercise 8"},"content":{"raw":"
\r\n

From the photograph it can be inferred that H<\/em>\u00a0=\u00a01\u00a0m, so:<\/p>\r\n

[latex]\\displaystyle Ra=\\frac{\\Delta \\rho gkH}{\\mu D_{C}}=\\frac{235\\times 9.81\\times 1.25\\times 10^{-10}\\times 1}{1\\times 10^{-3}\\times 5.0\\times 10^{-10}}=6\\times 10^{5}[\/latex]<\/p>\r\n

Return to <\/span>Exercise <\/span>8<\/span><\/a><\/p>\r\n

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

\n

From the photograph it can be inferred that H<\/em>\u00a0=\u00a01\u00a0m, so:<\/p>\n

\"\displaystyle Ra=\frac{\Delta \rho gkH}{\mu D_{C}}=\frac{235\times 9.81\times 1.25\times 10^{-10}\times 1}{1\times 10^{-3}\times 5.0\times 10^{-10}}=6\times 10^{5}\"<\/p>\n

Return to <\/span>Exercise <\/span>8<\/span><\/a><\/p>\n

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