{"id":629,"date":"2020-11-10T23:50:43","date_gmt":"2020-11-10T23:50:43","guid":{"rendered":"https:\/\/books.gw-project.org\/hydrogeologic-properties-of-earth-materials-and-principles-of-groundwater-flow\/?post_type=chapter&p=629"},"modified":"2020-12-29T18:06:24","modified_gmt":"2020-12-29T18:06:24","slug":"deriving-the-tangent-law-of-refraction","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/hydrogeologic-properties-of-earth-materials-and-principles-of-groundwater-flow\/chapter\/deriving-the-tangent-law-of-refraction\/","title":{"raw":"Box 8 Deriving the Tangent Law of Refraction","rendered":"Box 8 Deriving the Tangent Law of Refraction"},"content":{"raw":"Eileen Poeter, William Woessner, and Paul Hsieh<\/strong>\r\n\r\nFigure Box 8-1 shows flow lines and equipotential lines at an interface between materials of different hydraulic conductivity. One side of the interface, Region 1 has hydraulic conductivity K<\/em>1<\/sub>; the other side of the interface, Region 2 has hydraulic conductivity K<\/em>2<\/sub>. Under steady state conditions the discharge in a flow tube formed by two parallel flowlines must be the same on both sides of an interface (Q<\/em>1<\/sub> = Q<\/em>2<\/sub>), and given that Darcy\u2019s Law must be followed, the gradient and flow area (A<\/em>) must differ on each side of the interface to accommodate the differing hydraulic conductivities. This causes the flow lines to refract at the interface as shown in Figure Box 8-1 because the gradient hdiff<\/em>1<\/sub>\/l<\/em>1<\/sub> in Region 1 and hdiff<\/em>2<\/sub>\/l<\/em>2<\/sub> in Region 2) and the flow area (indicated by the width of the flow tubes, d<\/em>1<\/sub> and d<\/em>2<\/sub>, times a unit width into the image) must adjust to carry the same volumetric flow in materials of different hydraulic conductivity.\r\n\r\n[caption id=\"attachment_632\" align=\"alignnone\" width=\"1024\"]\"Figure Figure Box 8-1 - Geometry of flow lines at an interface between materials of differing hydraulic conductivity showing the angle of refraction, the width of the flow tube d<\/em>1<\/sub> and d<\/em>2<\/sub>, and the distance between equipotential lines l<\/em>1<\/sub> and l<\/em>2<\/sub> on different sides of the interface.[\/caption]\r\n\r\nDarcy\u2019s Law can be written for either Region 1, or 2, with the subscript n<\/em> = 1 to represent region 1 or n<\/em> = 2 to represent region 2, using Equation Box 8-1.\r\n\r\n\r\n\r\n
<\/td>\r\nQ<\/em>n<\/em><\/sub> = \u2013 K<\/em>n<\/em><\/sub> i<\/em>n<\/em><\/sub> A<\/em>n<\/em><\/sub><\/td>\r\n(Box 8-1)<\/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>n<\/em><\/sub><\/td>\r\n=<\/td>\r\nvolumetric flow through a unit width into the image for region n (L3<\/sup>\/T)<\/td>\r\n<\/tr>\r\n
K<\/em>n<\/em><\/sub><\/td>\r\n=<\/td>\r\nhydraulic conductivity of region n (L\/T)<\/td>\r\n<\/tr>\r\n
i<\/em>n<\/em><\/sub><\/td>\r\n=<\/td>\r\nhydraulic gradient in region n (dimensionless) = [latex]-\\frac{head difference}{distance}=-\\frac{hdiff_{n}}{l_{n}}[\/latex]<\/td>\r\n<\/tr>\r\n
A<\/em>n<\/em><\/sub><\/td>\r\n=<\/td>\r\narea perpendicular to flow in tube of region n (L2<\/sup>), A<\/em>n<\/em><\/sub> = d<\/em>n<\/em><\/sub> w<\/em> where w<\/em> is a unit width into the image<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSetting Q<\/em>1<\/sub> = Q<\/em>2<\/sub> leads to Equation Box 8-2.\r\n\r\n\r\n\r\n
[latex]\\displaystyle -K_1\\frac{{hdiff}_1}{l_1}\\ d_1w={-K}_2\\frac{{hdiff}_2}{l_2}\\ d_2w[\/latex]<\/td>\r\n(Box 8-2)<\/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
hdiff<\/em>1<\/sub><\/td>\r\n=<\/td>\r\nhead difference between equipotential lines region 1 (L)<\/td>\r\n<\/tr>\r\n
hdiff<\/em>2<\/sub><\/td>\r\n=<\/td>\r\nhead difference between equipotential lines region 2 (L)<\/td>\r\n<\/tr>\r\n
d<\/em>1<\/sub>, d<\/em>2<\/sub>, l<\/em>1<\/sub>, l<\/em>2<\/sub><\/td>\r\n=<\/td>\r\ndistances defined in Figure Box 8-1 (L)<\/td>\r\n<\/tr>\r\n
w<\/em><\/td>\r\n=<\/td>\r\na unit width into the image (L)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nnote: in this case hdiff<\/em>1<\/sub> equals hdiff<\/em>2<\/sub>, hdiff<\/em> = hdiff<\/em>1<\/sub> = hdiff<\/em>2<\/sub> = (129m \u2013 130m = \u20131m).\r\n\r\nCanceling the equal head differences, hdiff<\/em>, and canceling the equal distances perpendicular to the flow direction, w<\/em>, Equation Box 8-2 simplifies to Equation Box 8-3.\r\n\r\n\r\n\r\n
<\/td>\r\n[latex]\\displaystyle K_1\\frac{d_1}{l_1}=K_2\\frac{d_2}{l_2}[\/latex]<\/td>\r\n(Box 8-3)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nGiven the distance b<\/em> and angles of \u03b8<\/em>\u2019s as defined in Figure Box 8-1, trigonometric relationships result in Equations Box 8-4 and Box 8-5.\r\n\r\n\r\n\r\n\r\n
<\/td>\r\nd<\/em>1<\/sub> = b<\/em> cos \u03b8<\/em>1<\/sub><\/td>\r\n(Box 8-4)<\/td>\r\n<\/tr>\r\n
<\/td>\r\nd<\/em>2<\/sub> = b<\/em> cos \u03b8<\/em>2<\/sub><\/td>\r\n(Box 8-5)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nwhere:\r\n\r\n\r\n\r\n
\u03b8<\/em>n<\/sub><\/td>\r\n=<\/td>\r\nangles shown in Figure Box 8-1 (degrees)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting Equations Box 8-4 and Box 8-5 for d<\/em>1<\/sub> and d<\/em>2<\/sub> in Equation Box 8-3 yields Equation Box 8-6.\r\n\r\n\r\n\r\n
<\/td>\r\n[latex]\\displaystyle K_1\\frac{b\\ \\cos{\\theta_1}}{l_1}=K_2\\frac{b\\ \\cos{\\theta_2}}{l_2}[\/latex]<\/td>\r\n(Box 8-6)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nUsing rules of trigonometry, the Equation Box 8-7 and Equation Box 8-8 are true.\r\n\r\n\r\n\r\n\r\n
<\/td>\r\n[latex]\\displaystyle \\frac{b\\ }{l_1}=\\frac{1}{\\sin{\\theta_1}}[\/latex]<\/td>\r\n(Box 8-7)<\/td>\r\n<\/tr>\r\n
<\/td>\r\n[latex]\\displaystyle \\frac{b\\ }{l_2}=\\frac{1}{\\sin{\\theta_2}}[\/latex]<\/td>\r\n(Box 8-8)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting Equation Box 8-7 and Equation Box 8-8 into Equation Box 8-6 and canceling b, results in Equation Box 8-9.\r\n\r\n\r\n\r\n
<\/td>\r\n[latex]\\displaystyle K_1\\frac{\\cos{\\theta_1}}{\\sin{\\theta_1}}=K_2\\frac{\\cos{\\theta_2}}{\\sin{\\theta_2}}[\/latex]<\/td>\r\n(Box 8-9)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRearranging by multiplying both sides first by sin \u03b8<\/em>1<\/sub> and then by sin \u03b8<\/em>2<\/sub>, and dividing both sides first by cos \u03b8<\/em>1<\/sub> and then by cos \u03b8<\/em>2<\/sub> results in Equation Box 8-10.\r\n\r\n\r\n\r\n
<\/td>\r\n[latex]\\displaystyle K_1\\frac{\\sin{\\theta_2}}{\\cos{\\theta_2}}=K_2\\frac{\\sin{\\theta_1}}{\\cos{\\theta_1}}[\/latex]<\/td>\r\n(Box 8-10)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nUsing the Trigonometric identity of Equation Box 8-11.\r\n\r\n\r\n\r\n
<\/td>\r\n[latex]\\displaystyle \\tan{\\theta}=\\frac{\\sin{\\theta}}{\\cos{\\theta}}[\/latex]<\/td>\r\n(Box 8-11)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nEquation Box 8-10 can be expressed as Equation Box 8-12.\r\n\r\n\r\n\r\n
<\/td>\r\n[latex]\\displaystyle K_1\\tan{\\theta_2}=K_2\\tan{\\theta_1}[\/latex]<\/td>\r\n(Box 8-12)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThen dividing both sides of Equation Box 8-12, first by K<\/em>2<\/sub> and then by tan \u03b8<\/em>2<\/sub> the tangent law of Equation Box 8-13 is obtained.\r\n\r\n\r\n\r\n
<\/td>\r\n[latex]\\displaystyle \\frac{K_1}{K_2}=\\frac{\\tan{\\theta_1}}{\\tan{\\theta_2}}[\/latex]<\/td>\r\n(Box 8-13)<\/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
K<\/em>1<\/sub><\/td>\r\n=<\/td>\r\nhydraulic conductivity of layer 1 (L\/T)<\/td>\r\n<\/tr>\r\n
K<\/em>2<\/sub><\/td>\r\n=<\/td>\r\nhydraulic conductivity of layer 2 (L\/T)<\/td>\r\n<\/tr>\r\n
\u03b8<\/em>1<\/sub><\/td>\r\n=<\/td>\r\nangle the flow line makes with a perpendicular to the boundary in stratum 1 as shown in Figure Box 8-1 (degrees)<\/td>\r\n<\/tr>\r\n
\u03b8<\/em>2<\/sub><\/td>\r\n=<\/td>\r\nangle the flow line makes with a perpendicular to the boundary in stratum 2 as shown in Figure Box 8-1 (degrees)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

Return to where text links to Box 8<\/a><\/p>","rendered":"

Eileen Poeter, William Woessner, and Paul Hsieh<\/strong><\/p>\n

Figure Box 8-1 shows flow lines and equipotential lines at an interface between materials of different hydraulic conductivity. One side of the interface, Region 1 has hydraulic conductivity K<\/em>1<\/sub>; the other side of the interface, Region 2 has hydraulic conductivity K<\/em>2<\/sub>. Under steady state conditions the discharge in a flow tube formed by two parallel flowlines must be the same on both sides of an interface (Q<\/em>1<\/sub> = Q<\/em>2<\/sub>), and given that Darcy\u2019s Law must be followed, the gradient and flow area (A<\/em>) must differ on each side of the interface to accommodate the differing hydraulic conductivities. This causes the flow lines to refract at the interface as shown in Figure Box 8-1 because the gradient hdiff<\/em>1<\/sub>\/l<\/em>1<\/sub> in Region 1 and hdiff<\/em>2<\/sub>\/l<\/em>2<\/sub> in Region 2) and the flow area (indicated by the width of the flow tubes, d<\/em>1<\/sub> and d<\/em>2<\/sub>, times a unit width into the image) must adjust to carry the same volumetric flow in materials of different hydraulic conductivity.<\/p>\n

\"Figure
Figure Box 8-1 – Geometry of flow lines at an interface between materials of differing hydraulic conductivity showing the angle of refraction, the width of the flow tube d<\/em>1<\/sub> and d<\/em>2<\/sub>, and the distance between equipotential lines l<\/em>1<\/sub> and l<\/em>2<\/sub> on different sides of the interface.<\/figcaption><\/figure>\n

Darcy\u2019s Law can be written for either Region 1, or 2, with the subscript n<\/em> = 1 to represent region 1 or n<\/em> = 2 to represent region 2, using Equation Box 8-1.<\/p>\n\n\n\n
<\/td>\nQ<\/em>n<\/em><\/sub> = \u2013 K<\/em>n<\/em><\/sub> i<\/em>n<\/em><\/sub> A<\/em>n<\/em><\/sub><\/td>\n(Box 8-1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n\n\n\n
Q<\/em>n<\/em><\/sub><\/td>\n=<\/td>\nvolumetric flow through a unit width into the image for region n (L3<\/sup>\/T)<\/td>\n<\/tr>\n
K<\/em>n<\/em><\/sub><\/td>\n=<\/td>\nhydraulic conductivity of region n (L\/T)<\/td>\n<\/tr>\n
i<\/em>n<\/em><\/sub><\/td>\n=<\/td>\nhydraulic gradient in region n (dimensionless) = \"-\frac{head difference}{distance}=-\frac{hdiff_{n}}{l_{n}}\"<\/td>\n<\/tr>\n
A<\/em>n<\/em><\/sub><\/td>\n=<\/td>\narea perpendicular to flow in tube of region n (L2<\/sup>), A<\/em>n<\/em><\/sub> = d<\/em>n<\/em><\/sub> w<\/em> where w<\/em> is a unit width into the image<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Setting Q<\/em>1<\/sub> = Q<\/em>2<\/sub> leads to Equation Box 8-2.<\/p>\n\n\n\n
\"\displaystyle -K_1\frac{{hdiff}_1}{l_1}\ d_1w={-K}_2\frac{{hdiff}_2}{l_2}\ d_2w\"<\/td>\n(Box 8-2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n\n\n\n
hdiff<\/em>1<\/sub><\/td>\n=<\/td>\nhead difference between equipotential lines region 1 (L)<\/td>\n<\/tr>\n
hdiff<\/em>2<\/sub><\/td>\n=<\/td>\nhead difference between equipotential lines region 2 (L)<\/td>\n<\/tr>\n
d<\/em>1<\/sub>, d<\/em>2<\/sub>, l<\/em>1<\/sub>, l<\/em>2<\/sub><\/td>\n=<\/td>\ndistances defined in Figure Box 8-1 (L)<\/td>\n<\/tr>\n
w<\/em><\/td>\n=<\/td>\na unit width into the image (L)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

note: in this case hdiff<\/em>1<\/sub> equals hdiff<\/em>2<\/sub>, hdiff<\/em> = hdiff<\/em>1<\/sub> = hdiff<\/em>2<\/sub> = (129m \u2013 130m = \u20131m).<\/p>\n

Canceling the equal head differences, hdiff<\/em>, and canceling the equal distances perpendicular to the flow direction, w<\/em>, Equation Box 8-2 simplifies to Equation Box 8-3.<\/p>\n\n\n\n
<\/td>\n\"\displaystyle K_1\frac{d_1}{l_1}=K_2\frac{d_2}{l_2}\"<\/td>\n(Box 8-3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Given the distance b<\/em> and angles of \u03b8<\/em>\u2019s as defined in Figure Box 8-1, trigonometric relationships result in Equations Box 8-4 and Box 8-5.<\/p>\n\n\n\n\n
<\/td>\nd<\/em>1<\/sub> = b<\/em> cos \u03b8<\/em>1<\/sub><\/td>\n(Box 8-4)<\/td>\n<\/tr>\n
<\/td>\nd<\/em>2<\/sub> = b<\/em> cos \u03b8<\/em>2<\/sub><\/td>\n(Box 8-5)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n
\u03b8<\/em>n<\/sub><\/td>\n=<\/td>\nangles shown in Figure Box 8-1 (degrees)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Substituting Equations Box 8-4 and Box 8-5 for d<\/em>1<\/sub> and d<\/em>2<\/sub> in Equation Box 8-3 yields Equation Box 8-6.<\/p>\n\n\n\n
<\/td>\n\"\displaystyle K_1\frac{b\ \cos{\theta_1}}{l_1}=K_2\frac{b\ \cos{\theta_2}}{l_2}\"<\/td>\n(Box 8-6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Using rules of trigonometry, the Equation Box 8-7 and Equation Box 8-8 are true.<\/p>\n\n\n\n\n
<\/td>\n\"\displaystyle \frac{b\ }{l_1}=\frac{1}{\sin{\theta_1}}\"<\/td>\n(Box 8-7)<\/td>\n<\/tr>\n
<\/td>\n\"\displaystyle \frac{b\ }{l_2}=\frac{1}{\sin{\theta_2}}\"<\/td>\n(Box 8-8)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Substituting Equation Box 8-7 and Equation Box 8-8 into Equation Box 8-6 and canceling b, results in Equation Box 8-9.<\/p>\n\n\n\n
<\/td>\n\"\displaystyle K_1\frac{\cos{\theta_1}}{\sin{\theta_1}}=K_2\frac{\cos{\theta_2}}{\sin{\theta_2}}\"<\/td>\n(Box 8-9)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Rearranging by multiplying both sides first by sin \u03b8<\/em>1<\/sub> and then by sin \u03b8<\/em>2<\/sub>, and dividing both sides first by cos \u03b8<\/em>1<\/sub> and then by cos \u03b8<\/em>2<\/sub> results in Equation Box 8-10.<\/p>\n\n\n\n
<\/td>\n\"\displaystyle K_1\frac{\sin{\theta_2}}{\cos{\theta_2}}=K_2\frac{\sin{\theta_1}}{\cos{\theta_1}}\"<\/td>\n(Box 8-10)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Using the Trigonometric identity of Equation Box 8-11.<\/p>\n\n\n\n
<\/td>\n\"\displaystyle \tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}\"<\/td>\n(Box 8-11)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Equation Box 8-10 can be expressed as Equation Box 8-12.<\/p>\n\n\n\n
<\/td>\n\"\displaystyle K_1\tan{\theta_2}=K_2\tan{\theta_1}\"<\/td>\n(Box 8-12)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Then dividing both sides of Equation Box 8-12, first by K<\/em>2<\/sub> and then by tan \u03b8<\/em>2<\/sub> the tangent law of Equation Box 8-13 is obtained.<\/p>\n\n\n\n
<\/td>\n\"\displaystyle \frac{K_1}{K_2}=\frac{\tan{\theta_1}}{\tan{\theta_2}}\"<\/td>\n(Box 8-13)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n\n\n\n
K<\/em>1<\/sub><\/td>\n=<\/td>\nhydraulic conductivity of layer 1 (L\/T)<\/td>\n<\/tr>\n
K<\/em>2<\/sub><\/td>\n=<\/td>\nhydraulic conductivity of layer 2 (L\/T)<\/td>\n<\/tr>\n
\u03b8<\/em>1<\/sub><\/td>\n=<\/td>\nangle the flow line makes with a perpendicular to the boundary in stratum 1 as shown in Figure Box 8-1 (degrees)<\/td>\n<\/tr>\n
\u03b8<\/em>2<\/sub><\/td>\n=<\/td>\nangle the flow line makes with a perpendicular to the boundary in stratum 2 as shown in Figure Box 8-1 (degrees)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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