{"id":595,"date":"2020-11-09T15:48:57","date_gmt":"2020-11-09T15:48:57","guid":{"rendered":"https:\/\/books.gw-project.org\/hydrogeologic-properties-of-earth-materials-and-principles-of-groundwater-flow\/?post_type=chapter&p=595"},"modified":"2020-12-29T18:01:35","modified_gmt":"2020-12-29T18:01:35","slug":"equation-derivation-for-equivalent-k-and-a-4-layer-application","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/hydrogeologic-properties-of-earth-materials-and-principles-of-groundwater-flow\/chapter\/equation-derivation-for-equivalent-k-and-a-4-layer-application\/","title":{"raw":"Box 5 Equation Derivation for Equivalent K and a 4-layer Application","rendered":"Box 5 Equation Derivation for Equivalent K and a 4-layer Application"},"content":{"raw":"To calculate the equivalent hydraulic conductivity parallel to layers, K<\/em>x<\/em><\/sub>, consider flow through the system shown in Figure Box 5-1. Define \u2206h<\/em> as the hydraulic head difference over a horizontal distance \u0394L<\/em>x<\/em><\/sub> (so that \u2206h<\/em>\/\u0394L<\/em>x<\/em><\/sub> is the hydraulic gradient). The volumetric discharge Q<\/em> (L3<\/sup>\/T) through a unit width (into the image) of the system is the sum of the volumetric discharges through each layer. The specific discharge is q<\/em> = Q<\/em>\/A<\/em>. Here the area is equal to the product of the thickness d<\/em> and one unit of distance into the image, so A<\/em>=d<\/em>(1), thus q<\/em> = Q<\/em>\/d<\/em>(1), or q<\/em> = Q<\/em>\/d<\/em>. Knowing that q<\/em> = \u2212Ki<\/em>, this logic can be applied to the stack of layers as follows in Equations Box 5-1 through Equation Box 5-3. Equation Box 5-1 indicates the horizontal flow through each layer.\r\n\r\n[caption id=\"attachment_599\" align=\"alignnone\" width=\"953\"]\"Figure Figure Box 5-1 -<\/strong> Calculating the equivalent K<\/em>x<\/em><\/sub> of a layered heterogeneous system composed of isotropic and homogeneous layers. The original system is made up of four layers that have different thicknesses (d<\/em>1<\/sub>, d<\/em>2<\/sub>, d<\/em>3<\/sub>, and d<\/em>4<\/sub>) and hydraulic conductivities (K<\/em>1<\/sub>, K<\/em>2<\/sub>, K<\/em>3<\/sub>, and K<\/em>4<\/sub>). The total horizontal flux per unit thickness into the image is represented by Q<\/em>\/d<\/em>(1) where Q<\/em> is the flow through a cross sectional area that is d<\/em> long and the unit thickness is 1 unit wide. Conceptually, flow in the x<\/em> direction passes through all of the layers under the same gradient, \u2206h<\/em>\/\u0394L<\/em>x<\/em><\/sub>. The equivalent K<\/em>x<\/em><\/sub> for the entire thickness, d<\/em>, is computed using Equation Box 5-3.[\/caption]\r\n\r\n\r\n\r\n
[latex]\\displaystyle Q_{i}=-K_{i}d_{i}\\frac{\\Delta h}{\\Delta L_{x}}[\/latex]<\/td>\r\n(Box 5-1)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nwhere:\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Q<\/em>i<\/em><\/sub><\/td>\r\n=<\/td>\r\nvolumetric discharge through an individual layer of a unit width denoted by i (L3<\/sup>\/T)<\/td>\r\n<\/tr>\r\n
d<\/em>i<\/em><\/sub><\/td>\r\n=<\/td>\r\nthickness of individual layer i (L)<\/td>\r\n<\/tr>\r\n
\u0394L<\/em>x<\/em><\/sub><\/td>\r\n=<\/td>\r\nlength of the layers in the x<\/em>-direction (L)<\/td>\r\n<\/tr>\r\n
\u2206h<\/em><\/td>\r\n=<\/td>\r\nhydraulic head difference over horizontal distance \u0394L<\/em>x<\/em><\/sub> (L)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nEquation Box 5-2 provides the specific discharge for the entire section by summing the flow through every layer and dividing by the total flow area of the section, and shows that the specific discharge for the entire stack of layers is equal to the product of the equivalent K<\/em>x<\/em><\/sub> and the gradient.\r\n\r\n\r\n\r\n
[latex]\\displaystyle q=-\\sum_{i=1}^{n}\\left [ \\frac{K_{i}d_{i}}{d} \\right ]\\frac{\\Delta h}{\\Delta L_{x}}=-K_{x}\\frac{\\Delta h}{\\Delta L_{x}}[\/latex]<\/td>\r\n(Box 5-2)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nwhere:\r\n\r\n\r\n\r\n\r\n
q<\/em><\/td>\r\n=<\/td>\r\nspecific discharge through the entire section per unit width in the horizontal direction (L)<\/td>\r\n<\/tr>\r\n
K<\/em>x<\/em><\/sub><\/td>\r\n=<\/td>\r\nequivalent K<\/em>x<\/em><\/sub> of the entire section (L\/T)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThen, simplifying Equation Box 5-2 by cancelling the gradients, the equivalent K<\/em> in the x<\/em> direction is as shown in Equation Box 5-3.\r\n\r\n\r\n\r\n
[latex]\\displaystyle K_{x}=\\sum_{i=1}^{n}\\left [ \\frac{K_{i}d_{i}}{d} \\right ][\/latex]<\/td>\r\n(Box 5-3)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThis calculation of the equivalent K<\/em>x<\/em><\/sub> for the layered formation is a thickness-weighted arithmetic mean. This is the same as K<\/em>h<\/em><\/sub>, equivalent horizontal hydraulic conductivity, referred to in the main text. Some readers may find it useful to note that the equation for calculating equivalent K<\/em> parallel to layers is identical to the equation used to calculate the equivalent conductance through electrical resistors that are wired in parallel.\r\n\r\nTo calculate the equivalent the hydraulic conductivity perpendicular to the layers, K<\/em>z<\/em><\/sub>, consider vertical flow through the system shown in Figure Box 5-2. Let \u2206h<\/em> be the hydraulic head difference over a vertical distance \u0394L<\/em>z<\/em><\/sub> (so that, \u2206h<\/em>\/\u0394L<\/em>z<\/em><\/sub> is the overall hydraulic gradient). Mass must be conserved, so the volumetric inflow Q<\/em> (L3<\/sup>\/T) through a unit width (into the image) of the system at the bottom must be equal to the outflow at the top. In fact, water cannot be created nor destroyed along the flow path, so the specific discharge must be the same through each layer of the system. Given that the hydraulic conductivities differ between layers, then by Darcy\u2019s law, with a constant flow rate through each layer, the gradient will be different in each layer, as indicated by Equation Box 5-4.\r\n\r\n[caption id=\"attachment_604\" align=\"alignnone\" width=\"650\"]\"Calculating Figure Box 5-2 -<\/strong> Calculating the equivalent K<\/em>z<\/em><\/sub> of a layered heterogeneous system composed of isotropic and homogeneous layers. The vertical flux rate, q<\/em>, is constant. The original system is made up of four layers that have different thickness (d<\/em>1<\/sub>, d<\/em>2<\/sub>, d<\/em>3<\/sub>, and d<\/em>4<\/sub>) and hydraulic conductivity (K<\/em>1<\/sub>, K<\/em>2<\/sub>, K<\/em>3<\/sub>, and K<\/em>4<\/sub>). Conceptually flow in the z direction passes through all of the layers under the overall gradient, \u2206h<\/em>\/\u2206L<\/em>z<\/em><\/sub>. Because mass must be conserved the flow rate through each layer will be the same, and given their differing K<\/em> values, the gradients will differ. The equivalent K<\/em>z<\/em><\/sub> for the entire thickness, d<\/em>, is computed using Equation Box 5-6.[\/caption]\r\n\r\n\r\n\r\n
[latex]\\displaystyle q=-\\frac{K_{1}\\Delta h_{1}}{d_{1}}[\/latex] = [latex]\\displaystyle -\\frac{K_{2}\\Delta h_{2}}{d_{2}}[\/latex] = [latex]\\displaystyle -\\frac{K_{3}\\Delta h_{3}}{d_{3}}[\/latex] = [latex]\\displaystyle -\\frac{K_{4}\\Delta h_{4}}{d_{4}}[\/latex] = [latex]\\displaystyle -\\frac{K_{z}\\Delta h}{d}[\/latex]<\/td>\r\n(Box 5-4)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nwhere:\r\n\r\n\r\n\r\n\r\n\r\n
q<\/em><\/td>\r\n=<\/td>\r\nspecific discharge in the vertical direction (L\/T)<\/td>\r\n<\/tr>\r\n
\u2206h<\/em>i<\/em><\/sub><\/td>\r\n=<\/td>\r\nhydraulic head difference across each layer in the vertical direction, the values of \u2206h<\/em>i<\/em><\/sub> sum to \u2206h<\/em> (L)<\/td>\r\n<\/tr>\r\n
K<\/em>z<\/em><\/sub><\/td>\r\n=<\/td>\r\nequivalent K<\/em>z<\/em><\/sub> of the entire section (L\/T)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRearranging Equation Box 5-4 produces Equation Box 5-5, where \u2206h<\/em> is expanded into the sum of the layer \u2206h<\/em>i<\/em><\/sub> values, and, by Darcy\u2019s law, each \u2206h<\/em>i<\/em><\/sub> can be expressed as qd<\/em>i<\/em><\/sub> \/ K<\/em>i<\/em><\/sub>.\r\n\r\n\r\n\r\n
[latex]\\displaystyle K_{z}=\\frac{qd}{\\Delta h}[\/latex] = [latex]\\displaystyle \\frac{qd}{\\Delta h_{1}+\\Delta h_{2}+\\Delta h_{3}+\\Delta h_{4}}[\/latex] = [latex]\\displaystyle \\frac{qd}{\\frac{qd_{1}}{K_{1}}+\\frac{qd_{2}}{K_{2}}+\\frac{qd_{3}}{K_{3}}+\\frac{qd_{4}}{K_{4}}}[\/latex]<\/td>\r\n(Box 5-5)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nBy cancelling the q<\/em>\u2019s and summing the d<\/em>z<\/em><\/sub> \/ K<\/em>i<\/em><\/sub>, Equation Box 5-6 provides the procedure for calculating the equivalent K<\/em>z<\/em><\/sub>.\r\n\r\n\r\n\r\n
[latex]\\displaystyle K_{z}=\\frac{d}{\\sum_{i=1}^{n}\\frac{d_{i}}{K_{i}}}[\/latex]<\/td>\r\n(Box 5-6)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThis calculation of the equivalent K<\/em>z<\/em><\/sub> for the layered formation is a thickness-weighted harmonic mean. It is the same value as the K<\/em>v<\/em><\/sub>, equivalent vertical hydraulic conductivity in the main text. Some readers may find it useful to note that the equation for calculating equivalent K<\/em> perpendicular to layers is identical to the equation used to calculate conductance through electrical resistors that are wired in series.\r\n\r\nEquations Box 5-3 and Box 5-6 provide the K<\/em>x<\/em><\/sub> and K<\/em>z<\/em><\/sub> values for a homogeneous but anisotropic formation that is hydraulically equivalent to the layered heterogeneous system of homogeneous, isotropic geologic formations (Figure Box 5-3). Given that K<\/em>x<\/em><\/sub> is a thickness-weighted arithmetic mean, the high hydraulic conductivity values dominate its value, whereas the low hydraulic conductivity layers dominate the thickness-weighted harmonic mean K<\/em>z<\/em><\/sub> value. For example, suppose the materials from top to bottom are coarse sand, medium gravel, silty sand and fine gravel. The layers from top to bottom have K<\/em> values of K<\/em>1<\/sub> = 100 m\/d, K<\/em>2<\/sub> = 1000 m\/d K<\/em>3<\/sub> = 0.1 m\/d and K<\/em>4<\/sub> = 400 m\/d and corresponding thicknesses of d<\/em>1<\/sub> = 125 m, d<\/em>2<\/sub> = 58m, d<\/em>3<\/sub> = 125 m and d<\/em>4<\/sub> = 67 m. The computed horizontal and vertical hydraulic conductivities are K<\/em>x<\/em><\/sub> = 260 m\/d and K<\/em>z<\/em><\/sub> = 0.3 m\/d (using Equations Box 5-3 and Box 5-6). The anisotropy ratio<\/em>, K<\/em>x<\/em><\/sub>\/K<\/em>z<\/em><\/sub>, is on the order of 900. In the field, it is common for layered heterogeneity to lead to regional anisotropy values on the order of 10:1 and in many settings much greater values.\r\n\r\n[caption id=\"attachment_607\" align=\"alignnone\" width=\"991\"]\"Figure Figure Box 5-3 -<\/strong> Equivalent hydraulic conductivities can be calculated for layered materials using Equations Box 5-3 and Box 5-6.[\/caption]\r\n

Return to where text links to Box 5<\/a><\/p>","rendered":"

To calculate the equivalent hydraulic conductivity parallel to layers, K<\/em>x<\/em><\/sub>, consider flow through the system shown in Figure Box 5-1. Define \u2206h<\/em> as the hydraulic head difference over a horizontal distance \u0394L<\/em>x<\/em><\/sub> (so that \u2206h<\/em>\/\u0394L<\/em>x<\/em><\/sub> is the hydraulic gradient). The volumetric discharge Q<\/em> (L3<\/sup>\/T) through a unit width (into the image) of the system is the sum of the volumetric discharges through each layer. The specific discharge is q<\/em> = Q<\/em>\/A<\/em>. Here the area is equal to the product of the thickness d<\/em> and one unit of distance into the image, so A<\/em>=d<\/em>(1), thus q<\/em> = Q<\/em>\/d<\/em>(1), or q<\/em> = Q<\/em>\/d<\/em>. Knowing that q<\/em> = \u2212Ki<\/em>, this logic can be applied to the stack of layers as follows in Equations Box 5-1 through Equation Box 5-3. Equation Box 5-1 indicates the horizontal flow through each layer.<\/p>\n

\"Figure
Figure Box 5-1 –<\/strong> Calculating the equivalent K<\/em>x<\/em><\/sub> of a layered heterogeneous system composed of isotropic and homogeneous layers. The original system is made up of four layers that have different thicknesses (d<\/em>1<\/sub>, d<\/em>2<\/sub>, d<\/em>3<\/sub>, and d<\/em>4<\/sub>) and hydraulic conductivities (K<\/em>1<\/sub>, K<\/em>2<\/sub>, K<\/em>3<\/sub>, and K<\/em>4<\/sub>). The total horizontal flux per unit thickness into the image is represented by Q<\/em>\/d<\/em>(1) where Q<\/em> is the flow through a cross sectional area that is d<\/em> long and the unit thickness is 1 unit wide. Conceptually, flow in the x<\/em> direction passes through all of the layers under the same gradient, \u2206h<\/em>\/\u0394L<\/em>x<\/em><\/sub>. The equivalent K<\/em>x<\/em><\/sub> for the entire thickness, d<\/em>, is computed using Equation Box 5-3.<\/figcaption><\/figure>\n\n\n\n
\"\displaystyle Q_{i}=-K_{i}d_{i}\frac{\Delta h}{\Delta L_{x}}\"<\/td>\n(Box 5-1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n\n\n\n
Q<\/em>i<\/em><\/sub><\/td>\n=<\/td>\nvolumetric discharge through an individual layer of a unit width denoted by i (L3<\/sup>\/T)<\/td>\n<\/tr>\n
d<\/em>i<\/em><\/sub><\/td>\n=<\/td>\nthickness of individual layer i (L)<\/td>\n<\/tr>\n
\u0394L<\/em>x<\/em><\/sub><\/td>\n=<\/td>\nlength of the layers in the x<\/em>-direction (L)<\/td>\n<\/tr>\n
\u2206h<\/em><\/td>\n=<\/td>\nhydraulic head difference over horizontal distance \u0394L<\/em>x<\/em><\/sub> (L)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Equation Box 5-2 provides the specific discharge for the entire section by summing the flow through every layer and dividing by the total flow area of the section, and shows that the specific discharge for the entire stack of layers is equal to the product of the equivalent K<\/em>x<\/em><\/sub> and the gradient.<\/p>\n\n\n\n
\"\displaystyle q=-\sum_{i=1}^{n}\left [ \frac{K_{i}d_{i}}{d} \right ]\frac{\Delta h}{\Delta L_{x}}=-K_{x}\frac{\Delta h}{\Delta L_{x}}\"<\/td>\n(Box 5-2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n\n
q<\/em><\/td>\n=<\/td>\nspecific discharge through the entire section per unit width in the horizontal direction (L)<\/td>\n<\/tr>\n
K<\/em>x<\/em><\/sub><\/td>\n=<\/td>\nequivalent K<\/em>x<\/em><\/sub> of the entire section (L\/T)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Then, simplifying Equation Box 5-2 by cancelling the gradients, the equivalent K<\/em> in the x<\/em> direction is as shown in Equation Box 5-3.<\/p>\n\n\n\n
\"\displaystyle K_{x}=\sum_{i=1}^{n}\left [ \frac{K_{i}d_{i}}{d} \right ]\"<\/td>\n(Box 5-3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

This calculation of the equivalent K<\/em>x<\/em><\/sub> for the layered formation is a thickness-weighted arithmetic mean. This is the same as K<\/em>h<\/em><\/sub>, equivalent horizontal hydraulic conductivity, referred to in the main text. Some readers may find it useful to note that the equation for calculating equivalent K<\/em> parallel to layers is identical to the equation used to calculate the equivalent conductance through electrical resistors that are wired in parallel.<\/p>\n

To calculate the equivalent the hydraulic conductivity perpendicular to the layers, K<\/em>z<\/em><\/sub>, consider vertical flow through the system shown in Figure Box 5-2. Let \u2206h<\/em> be the hydraulic head difference over a vertical distance \u0394L<\/em>z<\/em><\/sub> (so that, \u2206h<\/em>\/\u0394L<\/em>z<\/em><\/sub> is the overall hydraulic gradient). Mass must be conserved, so the volumetric inflow Q<\/em> (L3<\/sup>\/T) through a unit width (into the image) of the system at the bottom must be equal to the outflow at the top. In fact, water cannot be created nor destroyed along the flow path, so the specific discharge must be the same through each layer of the system. Given that the hydraulic conductivities differ between layers, then by Darcy\u2019s law, with a constant flow rate through each layer, the gradient will be different in each layer, as indicated by Equation Box 5-4.<\/p>\n

\"Calculating
Figure Box 5-2 –<\/strong> Calculating the equivalent K<\/em>z<\/em><\/sub> of a layered heterogeneous system composed of isotropic and homogeneous layers. The vertical flux rate, q<\/em>, is constant. The original system is made up of four layers that have different thickness (d<\/em>1<\/sub>, d<\/em>2<\/sub>, d<\/em>3<\/sub>, and d<\/em>4<\/sub>) and hydraulic conductivity (K<\/em>1<\/sub>, K<\/em>2<\/sub>, K<\/em>3<\/sub>, and K<\/em>4<\/sub>). Conceptually flow in the z direction passes through all of the layers under the overall gradient, \u2206h<\/em>\/\u2206L<\/em>z<\/em><\/sub>. Because mass must be conserved the flow rate through each layer will be the same, and given their differing K<\/em> values, the gradients will differ. The equivalent K<\/em>z<\/em><\/sub> for the entire thickness, d<\/em>, is computed using Equation Box 5-6.<\/figcaption><\/figure>\n\n\n\n
\"\displaystyle q=-\frac{K_{1}\Delta h_{1}}{d_{1}}\" = \"\displaystyle -\frac{K_{2}\Delta h_{2}}{d_{2}}\" = \"\displaystyle -\frac{K_{3}\Delta h_{3}}{d_{3}}\" = \"\displaystyle -\frac{K_{4}\Delta h_{4}}{d_{4}}\" = \"\displaystyle -\frac{K_{z}\Delta h}{d}\"<\/td>\n(Box 5-4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n\n\n
q<\/em><\/td>\n=<\/td>\nspecific discharge in the vertical direction (L\/T)<\/td>\n<\/tr>\n
\u2206h<\/em>i<\/em><\/sub><\/td>\n=<\/td>\nhydraulic head difference across each layer in the vertical direction, the values of \u2206h<\/em>i<\/em><\/sub> sum to \u2206h<\/em> (L)<\/td>\n<\/tr>\n
K<\/em>z<\/em><\/sub><\/td>\n=<\/td>\nequivalent K<\/em>z<\/em><\/sub> of the entire section (L\/T)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Rearranging Equation Box 5-4 produces Equation Box 5-5, where \u2206h<\/em> is expanded into the sum of the layer \u2206h<\/em>i<\/em><\/sub> values, and, by Darcy\u2019s law, each \u2206h<\/em>i<\/em><\/sub> can be expressed as qd<\/em>i<\/em><\/sub> \/ K<\/em>i<\/em><\/sub>.<\/p>\n\n\n\n
\"\displaystyle K_{z}=\frac{qd}{\Delta h}\" = \"\displaystyle \frac{qd}{\Delta h_{1}+\Delta h_{2}+\Delta h_{3}+\Delta h_{4}}\" = \"\displaystyle \frac{qd}{\frac{qd_{1}}{K_{1}}+\frac{qd_{2}}{K_{2}}+\frac{qd_{3}}{K_{3}}+\frac{qd_{4}}{K_{4}}}\"<\/td>\n(Box 5-5)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

By cancelling the q<\/em>\u2019s and summing the d<\/em>z<\/em><\/sub> \/ K<\/em>i<\/em><\/sub>, Equation Box 5-6 provides the procedure for calculating the equivalent K<\/em>z<\/em><\/sub>.<\/p>\n\n\n\n
\"\displaystyle K_{z}=\frac{d}{\sum_{i=1}^{n}\frac{d_{i}}{K_{i}}}\"<\/td>\n(Box 5-6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

This calculation of the equivalent K<\/em>z<\/em><\/sub> for the layered formation is a thickness-weighted harmonic mean. It is the same value as the K<\/em>v<\/em><\/sub>, equivalent vertical hydraulic conductivity in the main text. Some readers may find it useful to note that the equation for calculating equivalent K<\/em> perpendicular to layers is identical to the equation used to calculate conductance through electrical resistors that are wired in series.<\/p>\n

Equations Box 5-3 and Box 5-6 provide the K<\/em>x<\/em><\/sub> and K<\/em>z<\/em><\/sub> values for a homogeneous but anisotropic formation that is hydraulically equivalent to the layered heterogeneous system of homogeneous, isotropic geologic formations (Figure Box 5-3). Given that K<\/em>x<\/em><\/sub> is a thickness-weighted arithmetic mean, the high hydraulic conductivity values dominate its value, whereas the low hydraulic conductivity layers dominate the thickness-weighted harmonic mean K<\/em>z<\/em><\/sub> value. For example, suppose the materials from top to bottom are coarse sand, medium gravel, silty sand and fine gravel. The layers from top to bottom have K<\/em> values of K<\/em>1<\/sub> = 100 m\/d, K<\/em>2<\/sub> = 1000 m\/d K<\/em>3<\/sub> = 0.1 m\/d and K<\/em>4<\/sub> = 400 m\/d and corresponding thicknesses of d<\/em>1<\/sub> = 125 m, d<\/em>2<\/sub> = 58m, d<\/em>3<\/sub> = 125 m and d<\/em>4<\/sub> = 67 m. The computed horizontal and vertical hydraulic conductivities are K<\/em>x<\/em><\/sub> = 260 m\/d and K<\/em>z<\/em><\/sub> = 0.3 m\/d (using Equations Box 5-3 and Box 5-6). The anisotropy ratio<\/em>, K<\/em>x<\/em><\/sub>\/K<\/em>z<\/em><\/sub>, is on the order of 900. In the field, it is common for layered heterogeneity to lead to regional anisotropy values on the order of 10:1 and in many settings much greater values.<\/p>\n

\"Figure
Figure Box 5-3 –<\/strong> Equivalent hydraulic conductivities can be calculated for layered materials using Equations Box 5-3 and Box 5-6.<\/figcaption><\/figure>\n

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