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Chapter 18: Groundwater and Seepage - Free

Chapter 18: Groundwater and Seepage - Free

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<strong>18</strong>-<strong>18</strong> The Civil Engineering H<strong>and</strong>book, Second Edition8 ftA<strong>Free</strong> surface(a)10 ftB1 2DrainC25 ftQk8(Q/k) 1(b)642(Q/k) 2B =14 ft0 0.5 1.0 1.5BTFIGURE <strong>18</strong>.<strong>18</strong> Example <strong>18</strong>.5.10 ftyb =20 fta 1a 23020a 2 = <strong>18</strong>.3 ftEq. (17.36)h 1 = 70 ft31ah1 2(a)L13b3a 2100x 50h = 52.7 ftEq. (17.38)55 60h(b)FIGURE <strong>18</strong>.19 Example <strong>18</strong>.6.With tailwater absent, h 2 = 0, the flow in region 3, a type IX fragment, with cot b = 3 producesQ---k=a---- 23(<strong>18</strong>.33)Finally, from the geometry of the section, we haveL = 20 + cot b [ h d – a 2 ] = 20 + 3 [ 80 – a 2 ](<strong>18</strong>.34)The four independent equations contain only the four unknowns, h, a 2 , Q/k, <strong>and</strong> L, <strong>and</strong> hence providea complete, if not explicit, solution.© 2003 by CRC Press LLC

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