{"id":923,"date":"2020-12-09T05:09:26","date_gmt":"2020-12-09T05:09:26","guid":{"rendered":"https:\/\/books.gw-project.org\/hydrogeologic-properties-of-earth-materials-and-principles-of-groundwater-flow\/?post_type=chapter&p=923"},"modified":"2020-12-10T21:24:29","modified_gmt":"2020-12-10T21:24:29","slug":"solution-to-exercise-15","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/hydrogeologic-properties-of-earth-materials-and-principles-of-groundwater-flow\/chapter\/solution-to-exercise-15\/","title":{"raw":"Solution to Exercise 15","rendered":"Solution to Exercise 15"},"content":{"raw":"15) A map view of a heterogeneous and isotropic field site containing a complex confined aquifer system is presented in the accompanying figure. If a contaminant is released at A and each zone of hydraulic conductivity is isotropic and homogeneous:\r\n\r\n15a) Map the flowline from A to the row of houses. Would wells at any of the houses be impacted?<\/em>\r\n\r\nA copy of the base map is made and drawn on below. To determine if any of the houses (1-5) would be impacted by the flow from A, a flow line crossing equipotential lines at right angles (isotropic and homogeneous) is constructed from A to the north boarder of the zone with the lower K<\/em>. At the point of intersection, a black line bisector is constructed at right angles to the boundary and the angle the flow line makes with the boundary is measured (with a protractor). As the flow will be refracted into the lower K<\/em> zone the tangent law needs to be applied, Equation 91 of this book. Figure 71<\/a> illustrates the application of this principle.\r\n\r\n\r\n\r\n
[latex]\\displaystyle \\frac{K_1}{K_2}=\\frac{{\\tan{\\ \\theta}}_1}{{\\tan{\\ \\theta}}_2}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFor this problem, the flowline from A makes an angle of 40\u00b0 with the boundary. With the given hydraulic conductivity values the refraction angle of the flow line is computed.\r\ntan 40\u00b0 \/ tan \u03b8<\/em>2<\/sub> = 10 m\/d \/ 1 m\/d\r\nRe-arranging tan \u03b8<\/em>2<\/sub> = tan 40\u00b0\/(10 m\/d \/ 1 m\/d) = 0.084\r\narctan 0.084 = 4.8\u00b0 or about 5\u00b0\r\nA 5\u00b0 angle is constructed that represents the flow path though the low K<\/em> zone.\r\nWhere the flow line intersects the southern boundary, a perpendicular bisector is constructed and since the K<\/em> in the third zone is the same as the first zone, 10 m\/d, the flow refracts at an angle of 40\u00b0.\r\nResults show that the well at house 2 would be impacted by the contamination that entered the aquifer at A.\r\n\r\n\"\"\r\n\r\n15b) Now assume that only the northern portion of the aquifer is anisotropic Kx<\/sub> = 10 m\/d and Ky<\/sub> = 1 m\/d. Starting at A construct the anisotropic flow path and show which, if any, of the houses (blue rectangles) will be affected by this new flow path. Present your construction and show how you determined the flow path in the northern portion of the aquifer.<\/em>\r\n\r\nThis part of the problem asks that if A is located in an anisotropic groundwater system, which house might be impacted by the flow from A. The anisotropy of the area where A is found is K<\/em>x<\/em><\/sub> =10 m\/d and K<\/em>y<\/em><\/sub> = 1 m\/d. The flow line from A can be constructed using the inverse hydraulic conductivity ellipse method illustrated in Figure 69<\/a> of this book.\r\n\r\nAn inverse hydraulic conductivity ellipse is created by using a piece graph paper and plotting an ellipse with an x axis of 1\/K<\/em>x<\/em><\/sub>0.5<\/sup> and y axis equal to 1\/K<\/em>y<\/em><\/sub>0.5<\/sup>. The ellipse size can be any size; however, it is kept at a manageable size for the base map presented with this problem. An example is presented on the solution below. Following the instructions in Figure 69<\/a> and keeping the axis parallel to x and y of the base map, the ellipse is placed on location A and a flow line for the anisotropic conditions is plotted (red arrow).\r\n\r\nThe anisotropy causes the flow to be deflected in the K<\/em>x<\/em><\/sub> direction. In this problem the flow from A does not reach the houses south of the low K<\/em> zone.\r\n\r\n\"\"\r\n

Return to Exercise 15<\/a><\/p>","rendered":"

15) A map view of a heterogeneous and isotropic field site containing a complex confined aquifer system is presented in the accompanying figure. If a contaminant is released at A and each zone of hydraulic conductivity is isotropic and homogeneous:<\/p>\n

15a) Map the flowline from A to the row of houses. Would wells at any of the houses be impacted?<\/em><\/p>\n

A copy of the base map is made and drawn on below. To determine if any of the houses (1-5) would be impacted by the flow from A, a flow line crossing equipotential lines at right angles (isotropic and homogeneous) is constructed from A to the north boarder of the zone with the lower K<\/em>. At the point of intersection, a black line bisector is constructed at right angles to the boundary and the angle the flow line makes with the boundary is measured (with a protractor). As the flow will be refracted into the lower K<\/em> zone the tangent law needs to be applied, Equation 91 of this book. Figure 71<\/a> illustrates the application of this principle.<\/p>\n\n\n\n
\"\displaystyle \frac{K_1}{K_2}=\frac{{\tan{\ \theta}}_1}{{\tan{\ \theta}}_2}\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

For this problem, the flowline from A makes an angle of 40\u00b0 with the boundary. With the given hydraulic conductivity values the refraction angle of the flow line is computed.
\ntan 40\u00b0 \/ tan \u03b8<\/em>2<\/sub> = 10 m\/d \/ 1 m\/d
\nRe-arranging tan \u03b8<\/em>2<\/sub> = tan 40\u00b0\/(10 m\/d \/ 1 m\/d) = 0.084
\narctan 0.084 = 4.8\u00b0 or about 5\u00b0
\nA 5\u00b0 angle is constructed that represents the flow path though the low K<\/em> zone.
\nWhere the flow line intersects the southern boundary, a perpendicular bisector is constructed and since the K<\/em> in the third zone is the same as the first zone, 10 m\/d, the flow refracts at an angle of 40\u00b0.
\nResults show that the well at house 2 would be impacted by the contamination that entered the aquifer at A.<\/p>\n

\"\"<\/p>\n

15b) Now assume that only the northern portion of the aquifer is anisotropic Kx<\/sub> = 10 m\/d and Ky<\/sub> = 1 m\/d. Starting at A construct the anisotropic flow path and show which, if any, of the houses (blue rectangles) will be affected by this new flow path. Present your construction and show how you determined the flow path in the northern portion of the aquifer.<\/em><\/p>\n

This part of the problem asks that if A is located in an anisotropic groundwater system, which house might be impacted by the flow from A. The anisotropy of the area where A is found is K<\/em>x<\/em><\/sub> =10 m\/d and K<\/em>y<\/em><\/sub> = 1 m\/d. The flow line from A can be constructed using the inverse hydraulic conductivity ellipse method illustrated in Figure 69<\/a> of this book.<\/p>\n

An inverse hydraulic conductivity ellipse is created by using a piece graph paper and plotting an ellipse with an x axis of 1\/K<\/em>x<\/em><\/sub>0.5<\/sup> and y axis equal to 1\/K<\/em>y<\/em><\/sub>0.5<\/sup>. The ellipse size can be any size; however, it is kept at a manageable size for the base map presented with this problem. An example is presented on the solution below. Following the instructions in Figure 69<\/a> and keeping the axis parallel to x and y of the base map, the ellipse is placed on location A and a flow line for the anisotropic conditions is plotted (red arrow).<\/p>\n

The anisotropy causes the flow to be deflected in the K<\/em>x<\/em><\/sub> direction. In this problem the flow from A does not reach the houses south of the low K<\/em> zone.<\/p>\n

\"\"<\/p>\n

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