Sm Chapter V


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Sm Chapter V

  1. 1. Chapter V Flow through Soils We know that soil consists of voids and are interconnected. As such water flows through these pores in soil from point of higher energy to point of lower energy, in a similar way it flows in pipes or open channels. Assessment of rate of flow and quantity of flow are very important in many soil engineering problems such as in filtration, flow through earthen dam, flow beneath weirs, flow into bore wells, piping failure etc. study of flow through soil constitutes understanding one of the important characteristic of soil. To get into study of flow through soils we need to recall some fundamentals of flow. As we said fluid flows from a point of higher energy to a point of lower energy. Bernoulli gives the total energy possessed by a flowing fluid, expressed as equivalent head of water as u v2 h= + +z γ w 2g Where the first term is pressure head, second term is velocity head and the third term is elevation head and also h = total head, u = pressure, v = velocity, g = acceleration due to gravity and γw = unit weight of water. In soils the velocity of flow we encounter are very small that velocity head can be neglected in comparison to other heads. As such for problems of flow through soils u h= +z γw Figure no.1 shows the relationship among pressure, elevation and total heads for the flow through soils. The pressure head at a point is the height of the vertical column of water in the peizometer installed at that point. The elevation head of a point is the vertical distance measured from any arbitrary horizontal reference datum plane to the point. Δh uA / γw uB / γw Flow hA hB zA L zB Fig. No. 1. Energy in water during flow through soil.
  2. 2. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT The loss of head between points A and B can be given by ⎛u ⎞ ⎛u ⎞ Δh = h A − h B = ⎜ A + z A ⎟ − ⎜ B + z B ⎟ ⎜γ ⎟ ⎜ ⎟ ⎝ w ⎠ ⎝γw ⎠ The head loss Δh, can be expressed in a non dimensional form as Δh i= L Where i = hydraulic gradient L= distance between points A and B, that is the length of flow over which the loss of head occurred. The variation of velocity ‘v’ with the hydraulic gradient ‘i’ is as shown in figure no.2. It consists of three zones. 1. Laminar flow zone (zone I) 2. Transition zone (zone II) 3. Turbulent zone (zone III) v Zone III Zone II Zone I i Fig.No.2. Relationship between velocity of flow and hydraulic gradient In zone I & II flow remains laminar when ‘i’ is increased, and discharge velocity bears a linear relationship with i. At a higher hydraulic gradient the flow becomes turbulent. When ‘i’ is decreased laminar flow exists only in zone I. In most soils hydraulic gradients and velocities are such that laminar conditions exist. Then vαi Darcy’s law Darcy’s law states that velocity of flow is directly proportional to hydraulic gradient. Darcy, in 1856, assuming laminar flow conditions for flow through soils, gave the relationship between discharge velocity of water through saturated soil and hydraulic gradient as v = ki _____________________________________________________________________________________________________________ 42 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  3. 3. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT Where v = discharge velocity, which is the quantity of water flowing in unit time through a unit gross cross sectional area of soil at right angles to the direction of flow. k = coefficient of permeability or hydraulic conductivity. The above equation is derived by Darcy for clean sands and its applicability to wide range of other soils is with errors. The Darcy’s equation v, is discharge velocity and is based on gross cross sectional area. However the actual velocity of water (known as seepage velocity and represented by vs) through the void spaces is higher than v. AV VV A V AS VS Fig.No.3. Relationship between discharge velocity and seepage velocity q q v= and v s = A AV vA = v s AV ⇒ vV = v sVV vV v That is v s = = VV n Coefficient of permeability The coefficient of permeability has the same unit as velocity. It can be expressed in m/sec or cm/sec. It depends on several factors viz. fluid viscosity, pore size distribution, grain size distribution, void ratio, roughness of mineral particles and degree of saturation. Coefficient of permeability varies widely for different types of soils. Some typical values of coefficients of permeability are given below in Table.No.1. Soil type k, cm/sec Clean gravel 1.0 - 100 Coarse sand 1.0 – 0.01 Fine sand 0.01 – 0.001 silts 0.001 – 0.00001 clays < 0.000001 Table.No.1. Typical values of coefficients of permeability _____________________________________________________________________________________________________________ 43 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  4. 4. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT Laboratory determination of coefficient of permeability Two methods are widely employed in laboratory to determine the coefficient of permeability. They are constant head permeability test and falling head permeability test. Constant head permeability test A typical experimental setup is shown in figure. No 4. In this type of test, water supply at the inlet is adjusted in such a way that the difference of head between the inlet and the outlet remains constant during the test. After a constant flow rate is established water is collected in a graduated flask for a known duration. Surplus inlet Overflow h L Over flow Fig.No.4. Set up for constant head permeability test Then total volume of water collected Q = Avt = Akit Where Q = Volume of water collected A = Cross sectional area of soil suspension T = time duration over which water is collected h Then i = L h ∴ Q = Ak t L QL ∴k = Aht This test is suitable for determination of coefficient of permeability of coarse grained soil. _____________________________________________________________________________________________________________ 44 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  5. 5. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT It can not be used for fine grained soils as the quantity of water collected will be very less even over a long period of time and small error in measurement of volume of water will lead to large error in the value of coefficient of permeability. Falling head permeability test A typical arrangement of equipment is shown in figure no.5. Water from a stand pipe flows through the soil sample. The initial head difference h1 at time t = t1 is recorded and water is allowed to flow through the soil specimen such that the final head difference at time t = t2 is h2. dh h2 h h1 L Over flow Fig.No.5. Set up for falling head permeability test Then the rate of flow of water through the specimen at any time‘t’ can be given by h dh q = k A = −a L dt Where q = flow rate a = cross sectional area of stand pipe A = cross sectional area of soil specimen aL ⎛ dh ⎞ ⇒ dt = ⎜− ⎟ Ak ⎝ h ⎠ aL h t= log e 1 Ak h2 This test is suitable for the determination of coefficient of permeability of fine grained soils. _____________________________________________________________________________________________________________ 45 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  6. 6. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT Effect of temperature of water on coefficient of permeability Coefficient of permeability is related to the properties of fluid flowing through it by the γw equation k = k μ Where γw = unit weight of water η = viscosity of water k = absolute coefficient of permeability (L2, cm2, m2) Therefore ⎛ η T ⎞⎛ γ w(T1 ) ⎞ k T1 = ⎜ 2 ⎟⎜ ⎟ k T2 ⎜ η T1 ⎟⎜ γ w(T2 ) ⎟ ⎝ ⎠⎝ ⎠ It is conventional to express the value of k at a temperature of 200C. We can assume γ w(T1 ) = γ w(T2 ) with in the range of temperature encountered in experimentation. Therefore ηT k 200 C = 2 k ηT T 0 C 20 Figure. No. 6. shows the relationship between viscosity and temperature for usual range of temperatures encountered in laboratory. 1.2 ηT 2 ηT20 1.0 12 20 30 T0C Fig.No.6. Relationship between viscosity and temperature _____________________________________________________________________________________________________________ 46 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  7. 7. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT Empirical relations for coefficient of permeability Many researchers have proposed empirical relationships for obtaining ‘k’ from basic properties of soil. It will be convenient to use these formulae for first approximation purposes and when the other accurate data are not available. Hazen (1930) proposed an empirical relation for the coefficient of permeability in the form k = cD10 (for loose, clean filter sand) 2 Where k = coefficient of permeability in cm/sec. c = constant that varies from 1.0 to 1.5. D10 = effective size in mm. Casagrande proposed a simple relationship for the coefficient of permeability for fine to medium clean sand in the following form k = 1.4e 2 k 0.85 Where k = Coefficient of permeability at void ratio ‘e’. K0.85 = Coefficient of permeability at void ratio e = 0.85. Equivalent permeability in stratified soils When soil is stratified the equivalent permeability will be different for flow in different directions with respect to stratification. Flow parallel to stratification In this case as shown in figure no.7, water will be flowing in all the layers under same hydraulic gradient and total rate of discharge is equal to sum of rate of discharges through all the layers. k1 H1 Flow k2 H2 H Direction k3 H3 k4 H4 Figure. No 7. Flow condition when flow is parallel to stratification Therefore q = v.1.H = v1.1.H + v2.1.H + v3.1.H + v4.1.H + ……. Where v = average discharge velocity and v1, v2, v3… are discharge velocities through layers 1, 2, 3 …. Say kH is the equivalent coefficient of permeability and k1, k2, k3 … are coefficients of permeability of layers 1, 2, 3 … . Then from Darcy’s law _____________________________________________________________________________________________________________ 47 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  8. 8. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT v = kH ieq, v1 = k1i1, v2 = k2i2, v3= k3i3 … But ieq = i1 = i2 = i3 …… Therefore kHiH = k1iH1 + k2iH2 + k3iH3 + …… → kH = (1/H)[k1H1 + k2H2 + k3H3 + ….] i=n ∑k H i i i.e. k H = i =1 i=n ∑H i =1 i Flow perpendicular to stratification In this case water will be flowing with the same velocity in all the layers and the head loss over the stratification will be equal to summation of head lost over individual layers. h4 h h3 h2 h1 H1 k1 H2 k2 H3 k3 H4 k4 Flow direction Figure. No 8. Flow condition when flow is perpendicular to stratification v = v1 = v2 = v3 = v4 …….. h = h1 + h2 + h3 + ……… Where v is discharge velocity for the stratification and v1,v2,v3 … are discharge velocities in various layers. Then by Darcy’s law v = kV (h/H), v1 = k1 (h1 /H1), v2 = k2 (h2 / H2), v3 = k3 (h3 / H3), ……. Obtaining values of h, h1, h2, h3 ….. from above equation and substituting it in equation for ‘h’ we get vH v1 H 1 v 2 H 2 v3 H 3 = + + + ..... kV k1 k2 k3 But we know from equation for ‘v’ that v = v1 = v2 = v3 = v4 …….. _____________________________________________________________________________________________________________ 48 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  9. 9. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT i=n H H1 H 2 H 3 H ∑H i Therefore = + + + ..... Which gives kV = i.e. kV = i =1 i =n kV k1 k2 k3 H1 H 2 H 3 Hi k1 + k2 + k3 ∑k i =1 i Seepage through soils The seepage case considered in the previous half of the chapter was a relatively simple one; flow considered was one dimensional and linear that is the flow velocity was presumed constant at every point in a cross section perpendicular to the flow line and the flow lines were also supposed to be parallel. However in many practical situations, such as flow around a sheet pile wall, under masonry dams and through earth dams, flow is not one dimensional but two dimensional, which means that the velocity components in the horizontal and vertical directions vary from point within the cross section of soil mass. Such problems are covered in this part. The following assumptions are made. a) The soil is fully saturated and Darcy’s law is valid. b) The soil mass is homogeneous and isotropic (except special cases). c) Both soil grains and pore fluid are incompressible. d) Flow conditions do not change with time, i.e. steady state conditions exist. Laplace equations for two dimensional flow Consider an elemental soil block of dimensions dx, dy, and dz, as shown in figure no. 9. (dy perpendicular to the plane of paper). Let vx, and vz be the components of discharge velocity in horizontal and vertical directions respectively. The rate of flow into the elemental block in the horizontal direction is equal to vxdzdy, and in the vertical direction it is vxdxdy. The rates of outflow from the block in the horizontal and vertical directions are respectively, ⎛ ∂v ⎞ dy ⎜ v z + z dz ⎟dxdy ⎝ ∂z ⎠ ⎛ ∂v ⎞ ⎜ v x + x dx ⎟dzdy ⎝ ∂x ⎠ v x dzdy dz dx v z dxdy Fig. No. 9. Two dimensional flow through an element of soil _____________________________________________________________________________________________________________ 49 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  10. 10. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT ⎛ ⎞ ⎛ ⎞ ⎜ v x + ∂v x dx ⎟dzdy and ⎜ v z + ∂v z dz ⎟dxdy ⎜ ∂v y ⎟ ⎜ ∂z ⎟ ⎝ ⎠ ⎝ ⎠ With the assumptions made earlier total rate of inflow into element should be equal to total rate of ⎛ ∂v ⎞ ⎛ ∂v ⎞ outflow. Hence ⎜ v x + x dx ⎟dzdy + ⎜ v z + z dz ⎟dxdy - [v x dzdy + v z dxdy ] = 0 ⎜ ⎟ ⎜ ∂v y ⎠ ⎟ ∂z ⎝ ⎝ ⎠ ∂v x ∂v z ⇒ + = 0 ……………………………………………..(1) ∂x ∂z From Darcy’s law, discharge velocities can be expressed as ∂h ∂h vx = k xix = k x and v z = k z i z = k z ∂x ∂z Where kx and kz are coefficients of hydraulic conductivities in horizontal and vertical directions respectively. Therefore equation (1) becomes ∂2h ∂2h k x 2 + k z 2 = 0 ………………………………………….(2) ∂x ∂z And for isotropic soils with respect to hydraulic conductivity i.e. kx=kz=k ∂ 2h ∂ 2h Equation (2) ⇒ 2 + 2 = 0 …………………….…….……………(3) ∂x ∂z Which is known as Laplace equation for flow through soil. This same equation is used by physicists to describe flow of heat in a thermal medium and flow of current in a conducting medium. In general, Laplace equation describes the loss of energy in any resistive medium. The solution of above equation gives two sets of orthogonal curves known as equipotential lines and stream lines, which describes the flow pattern and energy dissipation and constitutes flow nets. Flownets Sheet pile wall H1 H2 a' A a b B b' c flow line equipotential line d d' Impervious layer Fig.No.10. Flow around a sheet pile wall _____________________________________________________________________________________________________________ 50 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  11. 11. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT Consider the figure .No. 10 shown below. Sheet pile wall retains water to a height H1 on one side and on the other side water height is H2 ( H2 < H1). So water seeps through the soil from higher level to lower level. A water particle enters soil (pervious media) at A and flow through it along the path shown and exits it at point B. likewise all along boundary surface aa' number of water particles will enter pervious media and exits boundary surface bb', traveling along different paths. These path traced by water particles are called flow paths or flow lines. Water particles enters boundary aa' with an energy equivalent of total head of water H1 and leaves boundary bb', gradually. Along various flow lines we can imagine / identify points having equal energy for example 0.9 (H1-H2), 0.8 (H1- H2), 0.3 (H1-H2) … etc. the line joining these points of equal energy / potential are known as equipotential lines. Number of flow lines and equipotential lines can be drawn like this. A combination of families of these equipotential and flow lines is called a flow net. The solution of Laplace equation yields nothing but these families of curves. Also in an isotropic medium, the hydraulic gradient has to be the steepest possible, meaning thereby that the length of flow shall be the shortest. Thus, flow lines have to cross equipotential lines orthogonally. The space between two adjacent flow lines is called flow path and the figure formed on the flownet between any two adjacent flowlines and adjacent equipotential lines is referred to as field. Once a flow net is constructed, its graphical properties can be used in obtaining solutions for many seepage problems such as the estimation of seepage losses from reservoirs, determination of seepage pressures, uplift pressures below dams, etc. towards this end, flow nets are constructed such that a) Head lost between successive equipotentials is the same, say Δh and b) Quantity of flow in each of the flow path is the same, say Δq. Δh Flowline Flow channel Δq a b Δq Equipotential lines Field Fig.No.11. Quantity of flow from flownet _____________________________________________________________________________________________________________ 51 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  12. 12. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT Thus the total head lost H = Δh x Number of equipotential drops H = Δh(η d ) The total seepage quantity q = Δq(η f ) According to darcy’s law v = ki and q = kiA Δh H a ⎛ H⎞ Therefore Δq = kiΔA = k a.1 = k -------------- ⎜Q Δh = ⎟ b ηd b ⎜ ⎝ ηd ⎟ ⎠ Therefore total seepage = q = Δq (ηf) H a H a =k =k ηf ηd b ηd b It is convenient to construct a flow net in which all the ‘fields’ in a flow net are elementary squares, that is a = 1 . For that condition b ηf q = kH ………………………………………….(4) ηd Drawing of the flow net takes several trials. While constructing the flow net, keep the boundary conditions in mind. For the flow net shown in figure No.10 the following boundary conditions apply. a) The u/s and d/s surfaces, ‘aa'’ and ‘bb'’, of the permeable layer are equipotentials. b) Because aa' and bb' are equipotential lines, all flow lines intersect them at right angles. c) The boundary of impervious layer dd', is a flow line, and so is the surface of the sheet pile wall acb. d) The equipotential lines intersect ac and dd' at right angles. Flownets in Anisotropic soil For anisotropic soil having hydraulic conductivity kx and kz in horizontal and vertical directions Laplace equation takes the form ∂2h ∂2h kx 2 + kz 2 = 0 ∂x ∂z ∂ h 2 ∂ 2h ⇒ + kz 2 = 0 kz 2 ∂z ∂x kx kz Substituting x ' = x , we can express above equation as kx ∂ 2h ∂ 2h + 2 = 0 …………………….……….(5) ∂ ( x ' ) 2 ∂z Equation (5) is a form similar to that of equation (3) with ‘x’ replaced by ‘x'’, which is the new transformed coordinate. To construct the flow net, use the following procedure. a) Adopt a vertical scale (that is, z axis) for drawing the cross section. b) Adopt a horizontal scale (that is, x axis) such that horizontal scale = k z k x . (ver.scale) _____________________________________________________________________________________________________________ 52 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  13. 13. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT c) With scales adopted in steps 1 and 2, plot the vertical section through the permeable layer parallel to the direction of flow. d) Draw the flow net for the permeable layer on the section obtained from step 3, with flow lines intersecting equipotential lines at right angles and the elements as approximate squares. e) It can be transformed back to understand the actual pattern of flow. f) The rate of seepage per unit length can be calculated by modifying equation (4) to ηf q = kxkz H ηd Uplift pressure under hydraulic structures Flow nets can be used to determine the uplift pressure at the base of a hydraulic structure. To compute uplift pressure under a hydraulic structure, draw the required flow net. The pressure diagram at the base of the weir can be obtained from the equipotential lines from which uplift force can be calculated. Unconfined flow 0.3 BC B D C E H X(x,y) directrix Datum A F H Impervious Layer Horizontal filter d Fig.No.12. Flownet for flow through an earthen dam Consider the case of a homogeneous, isotropic earth dam resting on an impermeable foundation as shown in figure no 12. In this problem boundaries AC - an equipotential line with total head equal to H, AF – is a boundary flow line and discharge surface FH is a boundary equipotential line with potential equal to tail water head. Thus three boundaries are defined and the fourth one is not defined yet. It is however, clear that this top flow line would begin at point C in a direction at right angles to equipotential boundary AC and would end at some point, yet unknown, on the equipotential boundary FH. A. Casagrande (1937) suggested the use of Kozeny’s base parabola for different conditions in earth dams. He found that in the central portion, the top flow line coincided with the base parabola. The base parabola when continued upstream meets the water surface at the corrected entrance point D. since actual top flow line has to begin at C at right angles to AC, a short reverse curve is drawn from A to join the base parabola. Thus the following steps are used to sketch top flow line for earth dam with horizontal filter as discharge face. _____________________________________________________________________________________________________________ 53 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  14. 14. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT 1. Intersection of impervious boundary and filter is taken as focus of base parabola. 2. Point ‘D’ is marked on water surface as shown in figure. 3. Directrix is located by striking an arc with D as centre and DF as radius to cut horizontal through D at E. then draw vertical through E to meet filter at H. Then EH is the directrix. If ‘s’ is the distance from focus to the directrix and X (x,y) any point on the parabola, then x 2 + y 2 = x + s ………………………………………………(6) Point D is on this basic parabola with coordinates (d,H). ∴s = d2 + H 2 − d ∴ x 2 + y 2 = x + ( d 2 + H 2 − d ………………….…………(7) Using equation (7) points on parabola can be obtained and plotted. The short length of reverse curve can then be sketched from point C to complete the top flow line. Once the flow line is drawn, the rest of the flow net can be drawn with consideration of other boundary conditions. Seepage computations can be done using flow net as well as by the following method. Equation (6) ⇒ y = s 2 + 2 xs differentiation of which yields dy s = dx 2 xs + s 2 At x = 0, dy/dx =1 and y = s If it is assumed that the gradient of the top flow line can represent the average gradient on the vertical, dy/dx may be used in place of dy/ds. ⎛ dy ⎞ q = kiA = k ⎜ ⎟ y.1 …………………………………………..(8) ⎝ dx ⎠ At the vertical section passing through F, dy/dx =1, x = 0 and y = s Therefore q = ks…………………………………………..……………….(9) If the discharge face is not horizontal, a further correction at the exit surface may have to be made to the theoretical base parabola, depending on the conditions at the toe. Angle β is used to describe the direction of the discharge surface relative to the bottom boundary flow line. Figure No. 13 shows the top flow line details at exit. When the exit surface is horizontal i.e. β=1800, no correction is needed to the base parabola. When an angular rock toe forms the exit surface β=1800, a correction is made by relocating the exit point of top flow line by the method suggested by Casagrande. For an embankment without a drainage filter β=900, the top flow line exits tangentially along the downstream slope. The following table no. 2 can be used to locate the actual point of exit. β in degrees 30 60 90 120 150 180 Δa 0.36 0.32 0.26 0.18 0.10 0.00 a + Δa Δa Note: Intermediate values can be readout from the graph plotted between β and with the values given here. a + Δa Table.No.2. Values of Δa for different angle of inclination of downstream slope _____________________________________________________________________________________________________________ 54 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA
  15. 15. HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT Phreatic line Base parabola a Δa No correction to Toe filter Base parabola β = 1800 Horizontal filter β β F F F Δa a Fig. No. 13. Phreatic line intersection with down stream slope / filters For an embankment without filter and α<300, L = (a-Δa) can be obtained by the relation d d2 H2 − L= − cos α cos 2 α sin 2 α And the rate of seepage per unit length of dam by q = k L tanα sinα For α ≥ 300 Casagrande showed that L = d 2 + H 2 − d 2 − H 2 cos 2 α and q = kLsin2α Problems 1. Determine the equivalent coefficient of permeability of a soil stratification containing alternate ‘n’ layers of silt with k = 5x10-4cm/sec and ‘n+1’ layers of clay with k = 3x10-6 cm /sec for both flow parallel and perpendicular to the strartification. 2. A flownet for flow around a single row of sheet piles in a permeable soil layer is shown in figure. Given kx = kz = 4.2x10-3cm/sec. determine a. How high (above the ground surface) the water will rise if piezometers are placed at points a, b, c, and d. and b. The rate of seepage through flow channel II per unit length (perpendicular to the section shown). 5m 1.5m 11m 0 6 I II III C 2.3 2.3 1 A 14m 1 15m B Hor. Sand filter 1 2 3 4 5 Impervious layer 12m 3. The section of a homogeneous earth is shiown in figure. The coefficients of permeability in the x and y directions are 8x10-7 cm/sec and 3.6x10-7 respectively. Draw the flownet and estimate the seepage through the dam section. ***Hand outs are not meant to be exhaustive; students are advised to refer various books on the subject. _____________________________________________________________________________________________________________ 55 SOIL MECHANICS- I CHAPTER-V FLOW THROUGH SOILS 2008-09 MU JAGADEESHA