Chapitre 10. Flow Nets 1

Soil Mechanics Flo Nets page 1 Contents of this chapter : CHAPITRE 10. FLOW NETS...1 10.1 INTRODUCTION...1 10.2 REPRESENTATION OF SOLUTION...2 10.3 SOME GEOMETRIC PROPERTIES OF FLOW NETS...3 10.4 COMMON BOUNDARY CONDITIONS...4 10.4.1 SUBMERGED SOIL BOUNDARY = EQUIPOTENTIAL...4 10.4.2 IMPERMEABLE BOUNDARY = FLOW LINE...5 10.4.3 LINE OF CONSTANT PORE WATER PRESSURE...6 10.5 PROCEDURE FOR DRAWING FLOW NETS...7 10.6 CALCULATION OF QUANTITIES OF INTEREST FROM FLOW NETS...7 10.6.1 CALCULATION OF FLOW...7 10.6.2 CALCULATION OF PORE WATER PRESSURE...8 10.6.3 EXERCISE...8 10.6.4 EXAMPLE STRANDED VESSEL RESCUE...13 10.6.5 EXERCISE...14 10.7 FLOW NETS FOR ANISOTROPIC MATERIALS...15 10.7.1 INTRODUCTION...15 10.7.2 FLOW NETS FOR SOIL WITH ANISOTROPIC PERMEABILITY...15 10.8 EFFECT OF WATER FLOW IN A SOIL MASS...17 10.8.2 SEEPAGE FORCE : GENERAL FORMULATION...19 10.9 HYDRAULIC HEAVE OR PIPING...20 10.9.1 EXERCISE...21 Chapitre 10. Flo Nets 1 10.1 Introduction Let us consider a state of plane seepage as for example in the earth dam shon in Figure 1. 1 Réseaux d'écoulement

Soil Mechanics Flo Nets page 2 Phreatic line Unsaturated Soil Drainage blanket z Flo of ater x Fig. 1 Flo through an earth dam For an isotropic material, the head h of the ater floing into the soil satisfies Laplace's equations, thus analysis involves the solution of: 2 2 h h + 0 2 z 2 = x subject to certain boundary conditions. 10.2 Representation of Solution At every, point (x,z) here there is flo there ill be a value of head h(x,z). In order to represent these values e dra contours of equal head (red lines) as shon on Figure 2. Flo line (FL) Equipotential (EP) Fig.2 Flo lines and equipotentials These lines are called equipotentials 2. On an equipotential (EP). by definition: h ( x, z) = constant 2 équipotentielles

Soil Mechanics Flo Nets page 3 It is also useful in visualising the flo in a soil to plot the flo lines 3 (blue lines noted FL), these are lines that are tangential to the flo at a given point and are illustrated in Figure 2. It can be seen from Fig. (2) that the flo lines and equipotentials are orthogonal. 10.3 Some Geometric Properties of Flo Nets Let us consider another example of seepage under a sheet pile all. The flo net is represented on figure 3. Observation pipes Figure 3 On Fig. 3, each interval beteen to equipotentials corresponds to a head loss h equal to 1/N d of the total head loss h through the soil : h = h / N d Where N d = Total number of equipotentials. Consider a pair of flo lines, clearly the flo through this flo tube must be constant and so as the tube narros the velocity must increase. Let us consider the flo through abcd delimited by to flo lines and to equipotentials. The hydraulic gradient is : i = h/l 1 = h / (N d. l 1 ) Where l 1 is the distance beteen the to equipotentials. 3 Lignes de courant

Soil Mechanics Flo Nets page 4 Let l 2 be the distance beteen the to flo lines. Applying Darcy's la, the velocity in the tube abcd is: v = k. i = k. h / (N d. l 1 ) The flo passing into abcd, per m idth of soil, is : q abcd = Area x velocity = l 2.1. k.h / (N d. l 1 ) If e dra the flo net taking l 2 = l 1 (a "squared" mesh net is more convenient to dra : it is possible to dra an inscribed circle, see Fig.4), q abcd = k.h / N d (the flo through any quadrilateral of the flo net is thus the same as the one through abcd ). The total flo ill then be equal to : Figure 4 Inscribing Circles in a Flo Net Q = N f. q = k.h. N f / N d Where N f = Number of flo tubes. From that equation, one can see that the flo is function of the ratio N f / N d and thus if the flo net is refined by dividing each cell in four smaller cells, the ratio ill remain unchanged. That means that Q is independent of the refinement of the flo net! It is thus easy to determine quickly an estimate of the flo of ater passing under a dam or all. To calculate quantities of interest, that is the flo and pore ater pressures, a flo net must be dran. The flo net must consist of to families of orthogonal lines that ideally define a square mesh, and that also satisfy the boundary conditions. The three most common boundary conditions are discussed belo. 10.4 Common boundary conditions 10.4.1 Submerged Soil Boundary = Equipotential Consider the submerged soil boundary shon in Figure 5

Soil Mechanics Flo Nets page 5 H Water Datum z H-z Soil Figure 5 Equipotential boundary The head at the indicated position is calculated as follos: u h = + z γ no so u h = = γ ( H z) ( H z) γ γ + z = H That is, the head is constant for any value of z, hich is by definition an equipotential. Alternatively, this could have been determined by considering imaginary observation pipes placed at the soil boundary, as for every point the ater level in the standpipe ould be the same as the ater level. A consequence of this is that all the flo lines arrive perpendicularly to a submerged soil boundary. The upstream face of the dam shon in Figures 1 and 2 is an example of this situation. 10.4.2 Impermeable Boundary = Flo Line At a boundary beteen permeable and impermeable material the velocity normal to the boundary must be zero since otherise there ould be ater floing into or out of the impermeable material, this is illustrated in Figure 6.

Soil Mechanics Flo Nets page 6 Perm eable Soil Flo Line v n =0 v t Im perm eable M aterial Figure 6 Flo line boundary A consequence of this is that all the equipotentials arrive perpendicularly to an impermeable soil boundary. The phreatic surface shon in Figures 1 and 2 is also a flo line marking the boundary of the flo net. A phreatic surface is also a line of constant (zero) pore ater pressure as discussed belo. 10.4.3 Line of Constant Pore Water Pressure Sometimes a portion of saturated soil is in contact ith air and so the pore ater pressure of the ater just beneath that surface is atmospheric. The phreatic surface shon in Figure 7 belo is an example of such a condition. We can sho from the expression for head in terms of pore ater pressure that equipotentials intersecting a line of constant pore ater pressure do so at equal vertical intervals as follos: u h = + z γ thus u h = + z γ no u = 0 and so h = z

Soil Mechanics Flo Nets page 7 Figure 7 Constant pore ater pressure boundary 10.5 Procedure for Draing Flo Nets 1. Mark all boundary conditions. Determine the head at the inlet and outlet of the flo net. 2. Dra a coarse net hich is consistent ith the boundary conditions and hich has orthogonal equipotential and flo lines. (It is usually easier to start by draing the flo lines). 3. Modify the net so that it meets the conditions outlined above and so that the mesh located beteen adjacent flo lines and equipotentials are square (you could dra an inscribed circle). Refine the flo net by repeating step 3. 10.6 Calculation of Quantities of Interest from Flo Nets 10.6.1 Calculation of flo We have seen at 10.3 that, hen the flo net has been dran so that the elemental rectangles are approximately square, the total flo is equal to : Q = N f. q = k.h. N f / N d It should be noted in the development of this formula it as assumed that each flo tube as of unit idth and so the flo is given per unit idth (into the page). Let us come back to the earth dam example.

Soil Mechanics Flo Nets page 8 15 m 5m h = 15m P h = 0 h = 12m h = 9m h = 6m h = 3m Figure 8 Value of Head on Equipotentials In this example, e have N f = the number of flo tubes = 5, and N d = the number of equipotential drops = 5. Suppose that the permeability of the underlying soil is k=10-5 m/sec (typical of a fine sand or silt) then the flo per unit idth of dam is: 5 3 Q = 15 10 m /sec (per m idth) and if the dam is 25m ide the total flo under the dam: 5 3 Q = 25 15 10 m /sec 10.6.2 Calculation of Pore Water Pressure The pore ater pressure at any point can be found using the expression u h = + z γ No referring to Fig. 8 suppose that e ish to calculate the pore ater pressure at the point P. Taking the datum to be at the base of the dam it can be seen that z = - 5m and so: u = [ 12 ( 5)]γ = 17γ 10.6.3 Exercise 1. A sheet pile all is driven to a depth of 6m into a permeable sand layer (k = 6.10-3 mm/s) of 13.5 m thickness lying on an impermeable layer. The ater on one side of the all is at a height of 4.5m, hile on the other side, pumps maintain the ater level at ground level. To design the pumping system, dra, by hand, the flo net and estimate the flo under the all in m³/day.

Soil Mechanics Flo Nets page 9 4.5 m 6 m 13.5m (sand layer) Impermeable soil Figure 9 To check your flo net dran by hand, you can use FDSOLVER.XLA, a excel macro using the Finite Difference Method. That macro is available on the course Moodle ebsite, together ith an install note 4 and an help file containing examples. That macro can be also be useful in other domains like heat diffusion, electric potential, fluid flo, vibrations, The steps to follo are explained in the help file. You should "dra" the model like this : 4 An update to the install note has been added by a 2011-1012 student, François Diffels, for Excel 2007 and 2010. Thanks François!

Soil Mechanics Flo Nets page 10 The light green cells (left and bottom) are the y and x scales (0.5m step) (not mandatory, but it helps to dra the model to scale. The yello cells are either head specified boundaries (the top lines at either side of the all), or an impermeable boundary (the all itself). To solve the system and get the head values at any point (cell), just select the area ith the hite and yello cells, then select Tools, FDSOLVER, 2D-LAPLACE, then specify the scale if necessary, then press "run". The numerical results appear after a hile :

Soil Mechanics Flo Nets page 11 No, select the result area and choose the good graph options (Excel 2003) : Then by clicking on the left vertical axis :

Soil Mechanics Flo Nets page 12 And by clicking on the right vertical axis (otherise the graph appears upside don!): Finally, one obtains the head graph : S1 S3 S5 S7 4.00-4.50 3.50-4.00 S9 S11 S13 S15 3.00-3.50 2.50-3.00 S17 S19 S21 S23 2.00-2.50 1.50-2.00 S25 S27 1.00-1.50 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 0.50-1.00 Of course that macro only plots the head lines, as it is the head that follos the Laplace's equation. The flo lines must be dran by hand, trying to dra "square" meshes.

Soil Mechanics Flo Nets page 13 S1 S3 S5 S7 4.00-4.50 3.50-4.00 S9 S11 S13 S15 3.00-3.50 2.50-3.00 S17 S19 S21 S23 2.00-2.50 1.50-2.00 S25 S27 1.00-1.50 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 0.50-1.00 Q = k.h. N f / N d = 6. 10-3 x10-3 x4,5x 4/9 x 60x60x24= 1.04 m³/day / m of all 10.6.4 Example Stranded 5 Vessel Rescue The figure 10 shos a long vessel, 20 metres ide, stranded on a sand bank. It is proposed to inject ater into a ell point, 10 metres don, under the centre of the vessel to assist in toing the vessel off. The ater depth is 1 metre. 20m (not to scale) 20m 10m Figure 10 Stranded Vessel Example The sand has a permeability of 3 10-4 m/sec. Assuming that a pressure head of 50 m can be applied at the ell point : 1. dra the flo net by hand (and check eventually ith the Excel macro) : 5 échoué

Soil Mechanics Flo Nets page 14 2. calculate: The pore ater pressure distribution across the base of the vessel The total upthrust due to this increase in pore ater pressure The rate at hich ater must be pumped into the ell point. 10.6.5 Exercise 2. The figure 11 shos a concrete dam lying on a permeable soil layer (k = 12.5 10-3 mm/s). a) dra the flo net b) calculate the total flo, per m idth of dam, passing under the dam c) calculate the upthrust due to the ater flo under the dam. Figure 11

Soil Mechanics Flo Nets page 15 10.7 Flo Nets For Anisotropic Materials 10.7.1 Introduction Many soils are formed in horizontal layers as a result of sedimentation through ater. Because of seasonal variations such deposits tend to be horizontally layered and this results in different permeabilities in the horizontal and vertical directions. 10.7.2 Flo nets for soil ith anisotropic permeability It can be demonstrated that in the case of plane flo in an anisotropic material having a horizontal permeability k H and a vertical permeability kv, the solution can be reduced to that of flo in an isotropic material by doing a variable change ith and x z = = x α z α = k kh V So the flo in anisotropic soil can be analysed using the same methods (including sketching flo nets) that are used for analysing isotropic soils. Example - Seepage in an anisotropic soil Suppose e ish to calculate the flo under the dam shon in Figure 12; H Im perm eable 1 dam H 2 z x L Soil layer Z Im perm eable bedrock Fig. 12 Dam on a permeable soil layer over impermeable rock (natural scale)

Soil Mechanics Flo Nets page 16 For the soil shon in Fig. 12 it is found that k H = 4 k and therefore V α 4 kv = = 2 k V so x x = 2 z = z In terms of transformed co-ordinates this becomes as shon in Figure 13 z x H 1 H 2 L/2 Soil layer Z Impermeable bedrock Fig. 13 Dam on a permeable layer over impermeable rock (transformed scale) The flo net can no be dran in the transformed co-ordinates and this is shon in Fig.14 5m Impermeable bedrock Fig. 14 Flo net for the transformed geometry

Soil Mechanics Flo Nets page 17 It is possible to use the flo net in the transformed space to calculate the flo underneath the dam by introducing an equivalent permeability k = k k eq H V A rigorous proof of this result ill not be given here Example Suppose that in Figure 12 H1 = 13m and H2 = 2.5m, and that kv = 10-6 m/sec and k H =4 10-6 m/sec The equivalent permeability is: k = eq ( 4 10 6 ) ( 10 6 ) = 2 10 6 m / sec The total head drop is 10.5 m. There are 14 head drops and thus Nd=14. There are 6 flo tubes and thus Nf=6 The flo underneath the dam is, Q = k eq h N f /N d = (2 10-6 ) 10.5x6/14 = 9.0 10-6 m 3 /sec(/m idth of dam) For a dam ith a idth of 50 m, Q = 450 10-6 m 3 /sec = 41.47 m 3 /day 10.8 Effect of ater flo in a soil mass Introduction Fig. 15 illustrates equilibrium conditions in a column of soil. The left-hand tank contains ater and is connected to the right-hand tank containing soil and ater. When the ater level is the same in both tanks. there ill be no flo of ater through the soil. Figure 15 At level xx : σ = H. γ + Z. γ sat (total stress) u = (H+Z). γ (pore ater pressure) σ = Z. (γ sat - γ ) (contrainte effective = poids de la colonne de sol déjaugé)

Soil Mechanics Flo Nets page 18 Any change from equilibrium conditions ill cause ater to flo through the soil and this ill alter the effective stress and pore ater pressure. As it flos, the ater exerts a frictional drag on the soil particles and the effect of this force is knon as the seepage pressure. (a) Flo donards through the soil. If the left-band tank is loered, and the level of ater in the right-hand tank is maintained, ater ill flo donards through the soil (Fig. 16). At level x-x σ = H. γ + Z. γ sat u = (H+Z-h). γ σ = Z. (γ sat - γ ) + h. γ Figure 16 Thus the effective pressure is increased by h.γ. This quantity h.γ is the seepage pressure exerted by the floing ater. (b) Flo upards through the soil. If the left-hand tank is raised, ater ill flo upards through the soil (Fig. 17). At level x-x σ = H. γ + Z. γ sat u = (H+Z+h). γ σ = Z. (γ sat - γ ) - h. γ Figure 17

Soil Mechanics Flo Nets page 19 Thus the effective pressure is decreased by h.γ, the amount of seepage pressure. If h is increased, it may happen that the effective stress σ vanishes : it is the heave due to seepage of ater in the ground also called piping 6. This kind of phenomenon may be catastrophic and ill be developed in the next paragraph. Figure 18 Piping 10.8.2 Seepage Force : general formulation It can be demonstrated that the seepage force acting on a soil volume dv, is a force acting in the direction of the flo given by the formula: r J = γ Where : r. i. dv i r is the hydraulic gradient in the direction of the flo. dv is a soil volume subjected to three forces : Figure 19 Seepage Force - its on eight W = γ sat. dv - the buoyancy (Archimede's thrust) A = γ.dv ρ ρ - the seepage force J j dv. ρ =. i dv =γ. 6 «renard»

Soil Mechanics Flo Nets page 20 10.9 Hydraulic Heave or Piping Many dams on soil foundations have failed because of the sudden formation of a piped shaped discharge channel. As the store ater rushes out, the channel idens and catastrophic failure results. This results from erosion of fine particles due to ater flo. Another situation here flo can cause failure is in producing quicksand conditions. This is also often referred to as piping failure (Fig. 20). Figure 20 Piping Failure : 1) initiation and first deterioration, 2) regressive erosion, 3) formation of flo channel, 4) liquefaction and collapse We ill no sho ho to check the danger of piping in an example.

Soil Mechanics Flo Nets page 21 10.9.1 Exercise Verify the safety against hydraulic heave in the folloing case. GWL 7.0m H = 2.5 m Water 1.0m Sand γ k = 20kN/m 3 3.0m The flo net, obtained ith the Visual AEM freeare, gives the folloing equipotentials around the base of the all. (The datum is at the ground level at the right of the all.) h z m m A 1.6-3 B 1.5-2.81 C 1.4-2.46 D 1.3-1.99 E 1.2-1.36 F 1.1-0.69 1 0

Soil Mechanics Flo Nets page 22 As already mentioned in the chapter 8 about Eurocode 7, hen considering a limit state of failure due to heave by seepage of ater in the ground (HYD), it shall be verified, for every relevant soil column, that : 1. the design value of the destabilising total pore ater pressure (u dst;d ) at the bottom of the column is less than or equal to the stabilising total vertical stress (σ stb;d ) at the bottom of the column : u dst;d σ stb;d, or 2. the design value of the seepage force (S dst;d ) in the column is less than or equal to the submerged eight 7 (G dst;d ) of the same column S dst;d G stb;d The partial factor set for this case are: Parameter Symbol HYD - Partial factor set Permanent action (G) Unfavourable γ G, dst 1.35 Favourable γ G, stb 0.9 Variable action (Q) Unfavourable γ Q, dst 1.5 Favourable - 0.0 z u/γ u=(h-z).γ.1.35 σtot=γsat*z*0.9+γ.1.35 u<σtot? i=dh/ds Seepage Force per unit m m kn/m² kn/m² kn /m² -3 4.6 62.1 67.5 YES 0.526315789-1.35-2.81 4.31 58.185 64.08 YES 0.285714286-1.35-2.46 3.86 52.11 57.78 YES 0.212765957-1.35-1.99 3.29 44.415 49.32 YES 0.158730159-1.35-1.36 2.56 34.56 37.98 YES 0.149253731-1.35-0.69 1.79 24.165 25.92 YES 0.144927536-1.35 0 1 13.5 13.5 YES S dst;d Total Seepage Force (kn/m²) -8.1 G stb;d Total eight (kn/m²) -13.5 The all is thus safe against piping, folloing Eurocode 7. 7 Poids déjaugé