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Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.

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- 1. 8.4 Seepage Calculation from a Flow Net 205 deﬁnition of ﬂow and equipotential lines for ﬂow in the permeable soil layer around the row of sheet piles shown in Figure 8.1 (for kx ϭ kz ϭ k). A combination of a number of ﬂow lines and equipotential lines is called a ﬂow net. As mentioned in the introduction, ﬂow nets are constructed for the calculation of ground- water ﬂow and the evaluation of heads in the media. To complete the graphic construc- tion of a ﬂow net, one must draw the ﬂow and equipotential lines in such a way that 1. The equipotential lines intersect the ﬂow lines at right angles. 2. The ﬂow elements formed are approximate squares. Figure 8.3b shows an example of a completed ﬂow net. One more example of ﬂow net in isotropic permeable layer are given in Figure 8.4. In these ﬁgures, Nf is the number of ﬂow channels in the ﬂow net, and Nd is the number of potential drops (deﬁned later in this chapter). Drawing a ﬂow net takes several trials. While constructing the ﬂow net, keep the boundary conditions in mind. For the ﬂow net shown in Figure 8.3b, the following four boundary conditions apply: Condition 1: The upstream and downstream surfaces of the permeable layer (lines ab and de) are equipotential lines. Condition 2: Because ab and de are equipotential lines, all the ﬂow lines intersect them at right angles. Condition 3: The boundary of the impervious layer—that is, line fg—is a ﬂow line, and so is the surface of the impervious sheet pile, line acd. Condition 4: The equipotential lines intersect acd and fg at right angles. Toe ﬁlter kx ϭ kz ϭ k Nf ϭ 5 Nd ϭ 9 H1 H2 H Figure 8.4 Flow net under a dam with toe ﬁlter 8.4 Seepage Calculation from a Flow Net In any ﬂow net, the strip between any two adjacent ﬂow lines is called a ﬂow channel. Figure 8.5 shows a ﬂow channel with the equipotential lines forming square elements. Let h1, h2, h3, h4, . . ., hn be the piezometric levels corresponding to the equipotential lines. The rate of seepage through the ﬂow channel per unit length (perpendicular to the vertical sec- tion through the permeable layer) can be calculated as follows. Because there is no ﬂow across the ﬂow lines, (8.17)¢q1 ϭ ¢q2 ϭ ¢q3 ϭ p ϭ ¢q
- 2. 206 Chapter 8: Seepage From Darcy’s law, the ﬂow rate is equal to kiA. Thus, Eq. (8.17) can be written as (8.18) Eq. (8.18) shows that if the ﬂow elements are drawn as approximate squares, the drop in the piezometric level between any two adjacent equipotential lines is the same. This is called the potential drop. Thus, (8.19) and (8.20) In Figure 8.3b, for any ﬂow channel, H ϭ H1 Ϫ H2 and Nd ϭ 6. If the number of ﬂow channels in a ﬂow net is equal to Nf , the total rate of ﬂow through all the channels per unit length can be given by (8.21) Although drawing square elements for a ﬂow net is convenient, it is not always nec- essary. Alternatively, one can draw a rectangular mesh for a ﬂow channel, as shown in Figure 8.6, provided that the width-to-length ratios for all the rectangular elements in the ﬂow net are the same. In this case, Eq. (8.18) for rate of ﬂow through the channel can be modiﬁed to (8.22) If b1/l1 ϭ b2/l2 ϭ b3/l3 n (i.e., the elements are not square), Eqs. (8.20) and (8.21) can be modiﬁed to (8.23)¢q ϭ kHa n Nd b ϭ p ϭ ¢q ϭ ka h1 Ϫ h2 l1 bb1 ϭ ka h2 Ϫ h3 l2 bb2 ϭ ka h3 Ϫ h4 l3 bb3 ϭ p q ϭ k HNf Nd Nd ϭ number of potential drops where H ϭ head difference between the upstream and downstream sides ¢q ϭ k H Nd h1 Ϫ h2 ϭ h2 Ϫ h3 ϭ h3 Ϫ h4 ϭ p ϭ H Nd ¢q ϭ ka h1 Ϫ h2 l1 bl1 ϭ ka h2 Ϫ h3 l2 bl2 ϭ ka h3 Ϫ h4 l3 bl3 ϭ p h1 h2 h3 h4 ⌬q l3 l2 l1 ⌬q ⌬q2 ⌬q3 ⌬q1 l3 l2 l1 Figure 8.5 Seepage through a ﬂow channel with square elements
- 3. 8.4 Seepage Calculation from a Flow Net 207 h1 h2 h3 h4 ⌬q l3 l2 l1 ⌬q ⌬q2 ⌬q3 ⌬q1 b3 b2 b1 Figure 8.6 Seepage through a ﬂow channel with rectangular elements and (8.24) Figure 8.7 shows a ﬂow net for seepage around a single row of sheet piles. Note that ﬂow channels 1 and 2 have square elements. Hence, the rate of ﬂow through these two channels can be obtained from Eq. (8.20): However, ﬂow channel 3 has rectangular elements. These elements have a width-to-length ratio of about 0.38; hence, from Eq. (8.23) ¢q3 ϭ k Nd H10.382 ¢q1 ϩ ¢q2 ϭ k Nd H ϩ k Nd H ϭ 2kH Nd q ϭ kHa Nf Nd bn Impervious layer Water level Water table 5 m Flow channel 1 ϭ 1 l b Flow channel 2 ϭ 1 l b Ground surface Scale Flow channel 3 l b 1 0.38 Ϸ 5.6 m 2.2 m a 4.1 m d c H e b Figure 8.7 Flow net for seepage around a single row of sheet piles
- 4. So, the total rate of seepage can be given as (8.25)q ϭ ¢q1 ϩ ¢q2 ϩ ¢q3 ϭ 2.38 kH Nd Example 8.2 A ﬂow net for ﬂow around a single row of sheet piles in a permeable soil layer is shown in Figure 8.7. Given that kx ϭ kz ϭ k ϭ 5 ϫ 10Ϫ3 cm/sec, determine a. How high (above the ground surface) the water will rise if piezometers are placed at points a and b. b. The total rate of seepage through the permeable layer per unit length c. The approximate average hydraulic gradient at c. Solution Part a From Figure 8.7, we have Nd ϭ 6, H1 ϭ 5.6 m, and H2 ϭ 2.2 m. So the head loss of each potential drop is At point a, we have gone through one potential drop. So the water in the piezome- ter will rise to an elevation of (5.6 Ϫ 0.567) ϭ 5.033 m above the ground surface At point b, we have five potential drops. So the water in the piezometer will rise to an elevation of [5.6 Ϫ (5)(0.567)] ϭ 2.765 m above the ground surface Part b From Eq. (8.25), Part c The average hydraulic gradient at c can be given as (Note: The average length of flow has been scaled.) ■ i ϭ head loss average length of flow between d and e ϭ ¢H ¢L ϭ 0.567m 4.1m ϭ 0.138 ϭ 6.74 ϫ 10Ϫ5 m3 /sec/m q ϭ 2.38 k1H1 Ϫ H2 2 Nd ϭ 12.38215 ϫ 10Ϫ5 m/sec215.6 Ϫ 2.22 6 ¢H ϭ H1 Ϫ H2 Nd ϭ 5.6 Ϫ 2.2 6 ϭ 0.567m 208 Chapter 8: Seepage

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