{"id":487,"date":"2022-12-11T23:09:08","date_gmt":"2022-12-11T23:09:08","guid":{"rendered":"https:\/\/books.gw-project.org\/introduction-to-karst-aquifers\/chapter\/exercise-12-solution\/"},"modified":"2023-01-14T20:36:19","modified_gmt":"2023-01-14T20:36:19","slug":"exercise-12-solution","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/introduction-to-karst-aquifers\/chapter\/exercise-12-solution\/","title":{"raw":"Exercise 12 Solution","rendered":"Exercise 12 Solution"},"content":{"raw":"
\r\n

Substituting 64\/R<\/em>e and the equation for the Reynolds number into Equation\u00a04<\/a> proves that this results in the Hagen-Poiseuille equation for a circular pipe.<\/p>\r\n

Reynolds Number: [latex]\\displaystyle Re=\\frac{VD\\rho }{\\mu }=\\frac{VD}{\\nu }[\/latex]<\/p>\r\n

Equation 4<\/a>: [latex]\\displaystyle \\frac{\\Delta p}{L}=f_{D}\\frac{\\rho }{2}\\frac{V^{2}}{D}[\/latex]<\/p>\r\n

Hint<\/em>: Remember that the radius of a circle is one half the diameter.<\/p>\r\n

[latex]\\displaystyle \\frac{\\Delta p}{L}=\\frac{64\\mu }{VD\\rho }\\frac{\\rho V^{2}}{2D}[\/latex]<\/p>\r\n

[latex]\\displaystyle \\frac{\\cancel{64}\\mu }{\\cancel{V}D\\rho }\\frac{\\rho V^{\\cancel{2}}}{\\cancel{2}D}=\\frac{32\\mu V}{D^{2}}[\/latex]<\/p>\r\n

[latex]\\displaystyle \\frac{32\\mu V}{D^{2}}=\\frac{32\\mu V}{2^{2}r^{2}}[\/latex]<\/p>\r\n

[latex]\\displaystyle \\frac{\\cancel{32}\\mu V}{\\cancel{2}^{\\cancel{2}}r^{2}}=\\frac{8\\mu V}{r^{2}}[\/latex]<\/p>\r\n

Return to Exercise 12<\/span><\/span><\/a><\/p>\r\n

Return to where text linked to Exercise 12<\/a><\/p>\r\n

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

\n

Substituting 64\/R<\/em>e and the equation for the Reynolds number into Equation\u00a04<\/a> proves that this results in the Hagen-Poiseuille equation for a circular pipe.<\/p>\n

Reynolds Number: \"\displaystyle Re=\frac{VD\rho }{\mu }=\frac{VD}{\nu }\"<\/p>\n

Equation 4<\/a>: \"\displaystyle \frac{\Delta p}{L}=f_{D}\frac{\rho }{2}\frac{V^{2}}{D}\"<\/p>\n

Hint<\/em>: Remember that the radius of a circle is one half the diameter.<\/p>\n

\"\displaystyle \frac{\Delta p}{L}=\frac{64\mu }{VD\rho }\frac{\rho V^{2}}{2D}\"<\/p>\n

\"\displaystyle \frac{\cancel{64}\mu }{\cancel{V}D\rho }\frac{\rho V^{\cancel{2}}}{\cancel{2}D}=\frac{32\mu V}{D^{2}}\"<\/p>\n

\"\displaystyle \frac{32\mu V}{D^{2}}=\frac{32\mu V}{2^{2}r^{2}}\"<\/p>\n

\"\displaystyle \frac{\cancel{32}\mu V}{\cancel{2}^{\cancel{2}}r^{2}}=\frac{8\mu V}{r^{2}}\"<\/p>\n

Return to Exercise 12<\/span><\/span><\/a><\/p>\n

Return to where text linked to Exercise 12<\/a><\/p>\n

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