 Welcome to a course on advanced geotechnical engineering module 2 on permeability and see page and lecture 5. So in previous lectures we have discussed about permeability and the factors affecting the permeability and how to determine the permeability in the laboratory and also cases for construction of the flow nets for isotropic soils. But in nature we do have quite often the soils which are actually anastrophic in nature. Anastrophic in the nature means the soils are having different permeabilities in horizontal direction as well as vertical direction. This is possible for stratified soil deposits and when we construct some embankments or dams we have different materials that means that two soils or three soils having different permeabilities then what will happen how the flow lines and equipotential lines can be constructed or drawn will be seen in this lecture. So this is module 2 lecture number 5 in permeability and see page. Before discussing anastrophic and non-homogeneous soil conditions let us solve this example problem. This is an example problem where an example of a two dimensional problem where we have got a pipe which facilitates the draining of the water from the soil which is actually shown here. So this particular case there is ground surface and two dimensional problem because it is basically a plane strain structure where the length is actually perpendicular to the length of this figure it is continuous in nature. So we need to construct the flow net around the horizontal drain and at a certain point here above the center line of the pipe we need to determine what is the pore water pressure and also determine the velocity once you know the void ratio you can also determine what is the seepage velocity. So the flow net is constructed it will be like this and this being symmetry you can actually have two identical flow nets and this is also an example of radial flow nets wherein you have the eq potential lines running like this and you have flow channels running towards in the periphery in the surrounding the pipe. So the soil which is surrounding the drain is given as k is equal to pi into 10 to the power of minus 7 meter per second and this is a long horizontal drain which actually has got the capacity to drain the water from the surrounding soil. The strata which is permeable is about 6 meter from the ground surface and the datum is selected here but even the datum can be selected even also here but for convenience the datum is actually selected here. So and this is the impervious stratum so that is represented here so when we draw the flow channels so you can see this is the first flow channel second third fourth fifth and then sixth. So six flow channels which are actually there and when you look into the head loss which is actually there from the center of the pipeline is about 3 meters the drain that is from the center line to the ground water table surface it is 3 meters. So the head loss which occurs from this potential line to this potential line is about 3 meters that means that this is the potential line 9, 8, 7, 6, 5, 4 and then 3, 2, 1. So with this now we will be able to calculate discharge and we will be able to calculate what is seepage velocity and if we know the elevations we will be able to calculate what is the pore water pressure at different points. So what we are doing is that we are doing an example problem the flow net around the horizontal drain the question is that find the discharge to the drain in meter cube per day per meter length of the drain because it is a plain strain condition so per meter length and two dimensional analysis is very relevant and we consider the datum at the free surface. So this makes the total head at top of the ground is 3 meters now the head loss HL from the surface to the drain center line is about 3 meters. Now by considering the one half of the flow region the number of flow channels we counted as 6 and number of potential drops we counted as 9 by using Q is equal to k H into Nf by Nd and considering the symmetry multiplying by 2 and converting into meter cube per day we will get the discharge as 0.1426 meter cube per day per meter length. So this is the discharge which actually the pipe is capable to drain the water from the surrounding soil. Now the question is that how to determine the pore water pressure at x that is 1.5 meter into the soil directly above the drain. So we considered if you recollect the diagram which we have shown the datum is at the ground surface so the point x lies 1.5 meter below the datum so elevation head at x is minus 1.5 meters so total head at x if you wanted to write it is 5.6 by 9 why because we have said 9 8 that is 9 8 7 6 and so here we actually have we are counting between approximating between 6th equipotential line and fifth equipotential line so this is somewhat closer to the 6th equipotential line so this is actually approximated as 5.6 by 9 so this is that point which is above the drain. So with this we can actually now when we once you know the total head at x and elevation head at x then total head is equal to pressure head plus elevation head so we need to calculate what is pressure head at x so it is nothing but 1.867 meters that is nothing but 5.67 divided by 9 into 3 3 is the total head loss so with that we will get 1.867 minus of minus 1.5 you will get with that 3.67 meters now taking gamma w of the water permeating to the drain is say having 10 kilo Newton per meter cube the pore water pressure at x is given as 3.67 into 10 that is 36.7 kilo Pascal's or kilo Newton per meter square. Now next question is that estimation of the velocity of the flow at x once we know the velocity once we know the void ratio we can also determine what is the seepage velocity of that point. So we said that between each potential line the potential drop is nothing but 3 by 9 which is nothing but 0.33 meters so between each that means that from 9th equipotential line to 8th equipotential line the potential drop is 3 by 9 that is 0.333 meters and if the length of that small incremental flow line is known from the say flow net which is drawn on to the scale then we can calculate what is the hydraulic gradient that is nothing but 0.333 divided by 0.4 which gives 0.83 by using Darcy's law V is equal to Ki, K is given to us and I is actually estimated as 0.83 we will be able to calculate what is the discharge velocity or superficial velocity at that particular point. So that is actually obtained as 4.75 into 10 to the power of minus 7 meter per second and if you use Vs is equal to V by n then by using the porosity by knowing the void ratio calculating the porosity we can also calculate the seepage velocity at that point. So coming to the flow nets in anastrophic soils particularly in sedimentary soil deposits there is always a possibility that the condition of anisotropy occurs that means that the soil or permeable soil layers soil layer or soil layers are no longer isotropic in nature they have different permeabilities say for example in this case when we consider the two-dimensional cases that is x and say z direction the permeability in both x and z direction is different. So if a flow occurs parallel to the bedding planes the flow lines are also parallel to these planes and the hydraulic gradient has at every point of every bed the same value of i which is independent of the coefficient of permeability of the beds. So the k horizontal combined or equivalent we can actually get by 1 by h into k1 h1 plus k2 h2 so on to k and h1 where n number of layers can be accounted. Suppose we are having said two layers k equivalent horizontal is equal to 1 by h into k1 h1 plus k2 h2. Similarly when the flow is occurring right angles to the bedding planes every water particle passes in succession through every one of layers since the flow is continuous the discharge velocity V must be the same in every layer whereas the coefficient of permeability of the layers is different hence the hydraulic gradient must also be different. So when the coefficient of permeability is different the hydraulic gradient is different so i is equal to i1 plus i2 plus i3 by using again V is equal to k and then simplification we get k vertical equivalent of n number of layers say for example having thickness h1 h2 so on to hn we get h divided by h1 by k1 plus h2 by k2 so on to hn by kn. So let us see how we can actually construct flow nets in anastrophic material particularly having permeability k horizontal and k vertical different in both horizontal direction and vertical direction. Previously what we have assumed is that in developing the procedure for plotting flow nets we assume that the permeable layer is isotropic that is k horizontal is equal to k vertical is equal to k. Considering the case of constructing the flow nets for c-phase to soils that show anisotropicity, anisotropy with respect to permeability k horizontal is not equal to k permeability. In fact we also should know that k horizontal is more than k vertical the reasons are you know the ease with which water can flow through the horizontal direction is superior than vertical direction. The reason is that the sigma horizontal which is actually available at any given point for major types of soil deposits is less than sigma v. So this makes you know the k horizontal more than k vertical but in certain type of soils where they have been you know subjected to a very high pre consolidation pressures and in case if there is a locking of pressures takes place then in that case the k horizontal will be less than k vertical but this is not quite common. So we actually have discussed the Laplace equation when you have got kx is not equal to kz and for two dimensional problems we can write this kx dou square h by dou x square plus kz dou square h by dou z square is equal to 0 where k horizontal is represented here as kx and k vertical as kz and kx is greater than kz. Now in all the sedimentary deposits as well as in the most of the earthen dams there is pronounced tendency for the permeability to be greater in horizontal than in the vertical direction. This we have discussed and the permeability is generally of the order of 5 to 10 times that is kx is 5 to 10 times the vertical permeability. So this k is dou square h by dou x square plus kz dou square h by dou z square is equal to 0 is not a Laplace equation equation since this equation applies for a soil mass in which stratification occurs. So this results in distorted flow net this results in distorted flow net. So by rewriting that equation what we have done is that by dividing through the kz throughout we can write in the form of equation A which is kx by kz dou square h by dou x square plus dou square h by dou z square is equal to 0. A closer analysis of the above equation indicates that a simple Laplace equation may be endangered if an inadequate change of variable is used. So let us now consider a capital X is equal to c times small x and the x small x is the horizontal direction capital X is here considered here where c is a constant. So by substituting x is equal to cx that is by writing this we can write this in the plane like the equation can be x is substituted as dou square h by dou x square plus dou square h by dou z square this is nothing but a equation in xz plane that capital X and z small z plane. So by writing capital X is equal to cx we can write this dou square h by dou cx whole square plus dou square h by dou z square so by simplifying this we can write 1 by c square dou square h by dou x square plus dou square h by dou z square is equal to 0 this is termed as b. Now the constant c is defined by comparing equations a and b that is equation a is kx by kz dou square h by dou x square plus dou square h by dou z square is equal to 0 equation b finally 1 by c square dou square h by dou x square plus dou square h by dou z square is equal to 0 this gives constant c as root over kz by kx. So equation b is of the same form as dou square h by dou x square plus dou square h by dou z square is equal to 0 which governs that the flow in isotropic soils and should represent two sets of orthogonal curves in capital X and small z plane. So we can write capital X is equal to root over the ratio of the vertical and horizontal permeability is root over kz by kx into small x. Now by using this concept let us try to write the construct convert this natural anastrophic section into a transformer section by using the logic what we discussed. Now as we have said that kx is greater than kz that means that when we transform the L the L is actually multiplied with in the transformer section what we do is that we retain the vertical distance or height same as in the original plane but the horizontal distance are reduced by root over kz by kx. So the transformer flow through natural anastrophic section is given as that is direction of the flow which is shown there and which is nothing but kz and then kx and what we are actually doing is that by converting into transform section we are actually making this into the equivalent permeability in both the directions. So the anastrophic case is converted into an isotropic section by transforming from anastrophic section to a transformer section. The assumption is that anastrophe merely causes linear distortion of the flow net so that is what under that assumption and this actually has been done. Now starting with the natural section that is the flow quantity in the anastrophic section can be given as q is equal to L into 1 because L is the length and perpendicular to that plane it is 1 unit. So L into 1 is the area over which this flow is occurring into this kz, kz is the permeability in the direction into I. So while for the transformer section the same quantity of the flow is calculated as q is equal to now the L is nothing but L into root over kz by kx into 1 into kt. The kt is nothing but the equivalent permeability in transform section into I. So comparing both C and D we actually get the transform permeability is established as kt is equal to root over kx and kz. So in the transform permeability where we convert anastrophic case into the isotropic case the kt is equal to root over kx into kz that is k permeability in horizontal direction multiplied by the permeability of vertical direction and the square root of that. So how to construct flow nets for the anastrophic cases the procedure is actually given here the same expression could have been found using the flow in the horizontal direction but note that then the hydraulic gradient changes from I is equal to we have taken in the vertical direction but the same case can be applied for the horizontal direction also in that case the natural section from hydraulic gradient is H by L in the natural section and in the case of transformable section it will be H by L into root over kz by kx. So steps involved in the construction of flow net anastrophic medium is that to plot the section of the hydraulic structure adopt a vertical scale. So first you adopt a vertical scale and determine root over kz by kx which is nothing but root over k vertical by k horizontal and adopt a horizontal scale. So vertical scale is considered as it is and the horizontal scale is modified the scale horizontal is nothing but root over kz by kx into the whatever the scale which is adopted for the vertical scale. So the scale for horizontal is nothing but root over kz by kx into scale vertical. So in the fourth step when the scales adopted in step 1 and step 3 plot the cross section into the structure. So draw the flow net for the transform section plotted in step 4 in the same manner as is done for the seepage to the isotropic soils and calculate the rate of the seepage but here in case of in calculating the discharge here we need to use the kt that is the transformable section permeability which is nothing but root over kx kz into H into nm by nt. So let us try to look into this example problem wherein we actually have a particular concrete dam with butters as retaining a head of water of 5 meters and this is the downstream side and this is the upstream side and this is the vertical scale and this is the horizontal scale. The horizontal length here is 12.8 meters and at this point it is 9.6 meters and at this point is 10 meters and this is called cut off wall basically this is provided to direct the flow and then reduce the uplift pressures and all. Considering the concrete dam on a layer of a homogeneous anastrophic clay having permeability 16 into 10 to the power of minus 8 meter per second in the horizontal direction and kz is 1 into 10 to the power of 8 meter per second in the vertical direction. So the permeability for this clay is 16 times more than vertical direction so this could be a normally consolidated clay and the flow quantity seeping on the downstream side of a dam is required. So the flow takes place in this direction and this is the impervious rock stratum so this is the impervious boundary for this and now the first step is to determine the constant C which is nothing but the for the root over kz by kx so with that we will get 1 by 4 because the kz by kx is ratio is 1 by 16 root over that we get 1 by 4 and the transformed permeability kt is equal to root over kx kz so we get 4 into 10 to the power of minus 8 meter per second is the transformed permeability. Now when we do the you know without transforming and the anastrophic case when you put you have convalinear rectangles of a sides ratio of root over kx kz and this is what actually we have told because of the anastrophic the flow net in the natural section will be distorted and equipotential lines and flow lines do not cross at right angles and as the tangent to flow line does not correspond to the normal to the equipotential line as was the case for isotropic soils. So flow net is harder because in this case the flow net is ordered to sketch because the sides of every quadrangle in the must have a ratio of root over kx by kz so this type of problem where we cannot actually sketch the flow net comfortably. So for that the transformed section is one option and for that what we said is that represent the vertical scale and adopt the horizontal scale such a way that the scale is root over kz by kx times of the vertical. Now in the next slide when we see that the transformed section flow net is shown here the 12.8 meters now is shrunk by about 4 because the scale what we adopted is the 3.2 meters and the vertical heights are remain constant so with this now we constructed the flow net for the transformed section the scale for the horizontal is 1 by 4 times of the scale for the vertical so with this we constructed and by adopting the similar procedure now considering kt that is the equivalent permeability because of the horizontal and vertical permeability of the permeability layer we actually have kt into h into enough by nd and we have got 10 potential drops here the same procedure so here the this is the 10th equipotential line and this is the 0th equipotential line and what we can see is that here the head available is about 5 meters then as the water progresses it loses the head and by the time it comes here the head loss is about is completely lost so with that q can be obtained so this is the you know how the transformed sections are handled and so the important is that by adopting vertical or horizontal we actually have to construct from and convert anastrophic case into a transformed case and taking kt that is equivalent permeability that is transformed permeability better refer it as is q is equal to kt into h into nf by nd and the problem is actually handled in a similar manner as that of the isotropic soils. Now we have said that we have a case we have discussed about the anastrophic cases but now we may actually have hydraulic structures resting on non-homogeneous soils that means that most of the cases stratified soil deposits are quite common and particularly when we have soils having two different permeabilities there is a possibility that the deviations of you know flow lines flow nets occur so when a flow net is constructed across the boundary of two soils with the different permeabilities the flow net deflects at the boundary there is something like a refraction type it actually takes place and the condition where the deflection of you know the flow net happens and this is called the transfer condition. So how to handle this transfer condition particularly for flow nets for hydraulic structures on non-homogeneous soils. So in the transfer condition consider we have say on this side this is the boundary the yellow line which is actually shown is the boundary and this side is actually having a soil with permeability k1 and this side we are having a soil with a permeability k2. So here soil 1 having permeability k1 coefficient of permeability k1 and this side we are actually having soil 2 having coefficient of permeability k2 so this is the flow channel so this is the flow line what we see these are the flow lines but once it crosses the boundary it undergo deviation the deviation which is actually shown here. Now here this distance between two flow lines is represented here as b1 and distance between these are the equipotential lines which are actually shown the broken lines here and these are after deviation and here the b1 is this dimension between two flow lines and this is the l1 is the dimension between two adjacent equipotential lines on this side of the boundary. On this side of the boundary that is soil 2 side with this distance is l2 and this distance is b2 and this distance is l2 but as the discharge is same where we have got delta q enters and then delta q comes out we are using the equation of continuity. So here let alpha 1 be the angle which makes with the boundary and alpha 2 is the angle which actually makes the flow line which makes with the boundary. Similarly here this is the equipotential line so this angle is theta 2 in boundary 2 that is in the soil 2 and this angle is theta 1 in soil 1 so this angle is theta 1. Now this boundary point on the point where the deflection of flow line is occurring here is termed as a and this point is termed as b and this point is termed as c. Now by using the geometry we can say that that l1 is equal to ab sin theta 1 is equal to ab cos alpha 1. Similarly b1 is equal to that is b1 is equal to ac cos theta 1 is equal to ac sin alpha 1. Similarly we can write l2 is equal to ab sin theta 2 is equal to ab cos alpha 2 and b2 is equal to ac cos theta 2 is equal to ac sin alpha 2. So by using this we can now handle these cases now we let delta h be the loss of hydraulic head between two consecutive equipotential lines considering a unit length perpendicular section shown the rate of the seepage to the flow channel is given as delta q is equal to k1 is the permeability of the soil 1 delta h is the head loss and l1 which is actually there in the soil 1 that is the distance between any two potential lines and b1 is the distance between two flow lines into 1 is equal to k2 into delta h by l2 that is the hydraulic gradient in soil 2 into b2 into 1. This is because of the equation of continuity and by simplifying that we can write k1 by k2 is equal to b2 by l2 divided by b1 by l1. So by using this expression it is possible for us to construct the flow lines or understand about how the flow lines deviate when k1 is greater than k2 or k1 is less than k2. When k1 is equal to k2 it is which is nothing but b2 by l2 is equal to b1 by l1 is equal to 1. So what is that ratio b1 by l1 is nothing but the aspect ratio which is in for isotropic cases the mostly the aspect ratio is 1. Now the transfer condition where we have b1 by l1 is equal to by using the deliberations what we have discussed here by using this conditions and now once we substitute in the expression what we said is that k1 by k2 is equal to b2 by l2 by b1 l1 by substituting there we will actually get b1 by l1 is equal to cos theta 1 divided by sin theta 1 is equal to sin alpha 1 by cos alpha 1 and then b2 by l2 is equal to cos theta 2 by sin theta 2 is equal to sin alpha 2 by cos alpha 2. So by using k1 by k2 is equal to b2 by l2 by b1 by l1 and substituting this we will get k1 by k2 is equal to tan theta 1 by tan theta 2 and is equal to tan alpha 2 by tan alpha 1. So this will tell us how the deviations of the flow lines happen or the equipotential lines suffer when you actually have say two cases which are discussed here. One is let us assume that this is the boundary this is the boundary and this is the soil one having permeability k1 and this is the soil actually having permeability k2 that is soil 2 which is shown here. So when k1 is k2 then the b2 by l2 is actually greater than 1. So we have here you get large rectangles and not necessary that here on the soil one boundary where the square elements are ensured in one of the soils that is soil one it can be vice versa if you take soil one as soil 2 and then this thing. So here you have square elements and here you have rectangle elements when k1 is k2 you get large rectangles but similarly when you have k1 less than k2 and with the ratio as less than 1 so b2 by l2 is less than 1. So you get here the deviation will become narrow here and where you actually get small rectangles but on the soil one side again it maintains the square elements which is equal to b1 is equal to l1 the aspect ratio is maintained as 1. In fact we will see this through a numerical simulation also. So what the case we discussed is that the flow channel at the boundary between two soils with different coefficients of permeability we can write k1 by k2 is equal to b2 by l2 by b1 by l1, b2 by l2 is in soil 2 region where b2 is the distance between two flow lines after deflection in soil 2, l2 is the distance between two equipotential lines and similarly b1 by l1 is nothing but the aspect ratio in soil 1. If k1 is k2 that is if you are having two soils where the permeability of the soil 1 is greater than soil 2 we may plot square flow elements in layer 1 this means that l1 is equal to b1 and so k1 by k2 is equal to b2 by l2. So thus the flow elements in layer 2 will be rectangles and their width to length ratios will be equal to k1 by k2 that is what it says because in this when we put b1 by l1 is equal to 1 the k1 by k2 is equal to b2 by l2. So whatever the ratio is there by having two different permeabilities we can actually plot this as two ratio which is actually b2 by l2 is equal to in the ratio of k1 by k2. If say k1 is less than k2 we may plot the square flow elements in the layer 1 that is similar and this means that again the aspect ratio is 1 in soil 1 and so k1 by k2 is equal to b2 by l2 so thus the flow elements in layer 2 will be rectangles again and the width to length ratios will be equal to k1 by k2. So now having discussed and so we actually have two types of conditions we discussed anastrophic condition and other one is non-homogeneous soils and whenever we actually construct embankment dams and this condition is actually referred as unconfined seepage. Suppose if you are having a hydraulic structure in the form of a sheet pile wall or a coffered dam the flow surrounding that is actually referred as a confined seepage. So example of an unconfined seepage is nothing but one boundary of the flow region being priatic surface on which the pressure is atmospheric. So the cross sections of the top flow line are at atmospheric so because of this condition is referred as unconfined seepage condition. So the seepage through embankment dams the example is an example for unconfined seepage condition. The priatic surface constitutes the top flow line and its position must be estimated before drawing the flow net. Suppose if you are having an embankment dams for example in practical cases these structures are like levees or dams and nowadays the modern structures like ash pond dams or you have some tailing dams these are the examples of these embankment dams which are actually constructed for different purposes. So in which case when you have got water retained on the upstream size, upstream surface and downstream side if it is actually having say tail water level and if it is actually constructed with earth then it is called as earth and that is the major construction material and so the priatic surface which constitutes the top flow line and it must be estimated before the drawing the flow net. So important requirement is that how to estimate the priatic line when you are actually have dam which is retaining a certain head of water. So for that in order to calculate the idea is that once you construct and then to estimate the seepage through an earthen dam resting on impervious base. Suppose impervious base because otherwise the flow can actually lead it to the subsoil. So several solutions have been proposed for determining the quantity of seepage through a homogeneous earthen dam. To name a few we actually have we discuss here Dupont solution and the second one being Schaeffernach solution and third one is Kasekrande solution. All actually have given methods which are actually for constructing the uppermost flow line that is the priatic surface which actually can be formed and this can be also estimated from the numerical simulation which are actually inbuilt and also all the physical simulation. So in this particular slide we are actually discussing going to discuss about the Dupont solution for flow through an earthen dam, Dupont solution for flow through an earthen dam. So here we actually have an embankment which is constructed on an impervious base and the value of the X axis is shown here and H1 is the head of water upstream water level and this is the head of water that is tile stream water level. So AB this part this is an equipotential line and again this is an equipotential line with a head that is H2. So the portion AB is the uppermost flow line or these called as priatic surface AB. So this is the point where is a conflict point where you actually have a flow net portion a flow line portion and then equipotential line portion which actually commences at this level. So this ordinate being Z here the axis is shown like this vertical axis Z and X and this is a two dimensional case. So the slope which is actually indicated as dz by dx that is nothing but a I hydraulic gradient. So assumption is that hydraulic gradient I is equal to the slope of the free surface and is constant with the depth. Now when we consider the perimeter section if you are actually having height z then the permeability this the seepage can be written as Q is nothing but k into dz by dx into z into 1. Now by integrating this X that is from X is equal to 0 here and to the point D that is the point where the upstream water level meets the or where the water of the priatic surface meets this point and this distance is D by integrating up to that and the z is actually integrated with H1 and H2 because this is H1 minus H2 is nothing but the differential water level. So 0 to dx is equal to H2 to H1 kz dz by simplifying that we get Q is equal to k by 2D into H1 square minus H2 square. So if you look into this, this is the equation of the parabolic surface which is actually estimated by Dupit in the solution with an assumption that the hydraulic gradient is nothing but the slope of the free surface and is constant with the depth. So that is what actually assumption which has been made in the Dupit's solution for estimating the flow through the net than that but one thing we must observe here is that no attention for the entrance and exit conditions for the priatic surface were considered and if H2 is equal to 0 let us say that this tail water level is at this point that is if the tail water level as is at this point the priatic line will actually intersect the impervious base, the priatic line is actually intersecting the impervious base. In the coming to the next solution that is the Schaffernag's solution for flow through an attendant and where the priatic surface is assumed like line AB that is actually which is shown here line AB and will intersect at a distance L along the point which is from O to B that is the L here which is actually given from the impervious base and by considering the triangle OBD the slope which is actually given an I is equal to dz by dx and this angle which is actually subtended by this OB with OD is beta so we can write dz by dx is equal to tan beta. So the hydraulic gradient is considered as dz by dx is equal to tan beta. So we can write q is equal to kz into dz by dx and now that we can write now by substituting for dz by dx is tan beta and z we can write in terms of L sin beta by taking the triangle OBD that is here OBD and with that by integrating from the distance L sin beta to L sin beta is this vertical distance that is this distance to this that is H and then this one is that L cos beta L cos beta to D that is the horizontal distance from this point to this point where it meets the upstream water surface. So by simplifying that we will actually get L that is the L is the length along the downstream slope surface which is nothing but L is equal to D by cos beta minus root over D square by cos beta minus H square by sin square beta. Once we knew L by using the expression now here we can calculate the flow which is nothing but q is equal to k L sin beta tan beta. So this in the Schaeffenach solution pre-erotic surface was assumed to be like an line AB and will intersect at a distance L from the impervious base. So to some extent what Jupit could not consider has been improvised in this particular solution. The graphical construction for determining L is also given previously we have seen the analytical solution and the Schaeffenach actually has given the graphical solution also. So here in this the graphical solution goes like this the extend BC downwards that means that the along the slope line this line is to be extended and draw a vertical line AE and they will intersect the projection line BC at F so this is that point F with C as the diameter draw a semicircle FHC and the draw a horizontal line AG that is the point with C as the center CG as the diameter sorry CG as the radius draw an arc GH so it cuts at H. So with F as an center F as center FH as radius draw an arc HB so B is the point by measuring the distance graphically we can determine BC is equal to L. So how to look determine L once we get the L then there is a possibility that we will be able to estimate again the flow through an attendant by using Schaeffenach solution. Now third solution what we discussed and put forwarded is the Casagrande solution in Casagrande in 1937 is shown experimentally that the parabola AB actually starts from point A dash which is that is AA dash is equal to 0.3 delta. So the parabola was assumed in the Casagrande solution as that the parabola starts instead of at this point it actually so the entrance correction which is actually given here with 0.3 delta and the slope of this line again d dz by dx and the ds which is shown and this length is L which is considered and this angle is beta and this point is the this is the vertical axis and this is the horizontal axis and this delta is nothing but from the upstream point where it meets this distance to and this distance this is delta. So it is one third of this distance was considered by Casagrande and with the modification with this modification value of d is equal to A dash C that is the value of d will be equal to A dash C that is this particular distance. So considering now the triangle BCD we can write I is equal to dz by ds is equal to sin beta considering BCD instead of dz by dx. So here he has considered dz by ds is equal to sin beta and using q is equal to ka where flow is actually happening perpendicular to the plane and where q is equal to k sin beta into that is sin beta is nothing but dz by ds and then A that is area of the cross section which is nothing but the vertical distance is nothing but L sin beta. So with that we can write q is equal to k L square sin beta by solving for L we get L is equal to s minus root over s square minus h square by sin square beta where s is equal to root over d square plus h square where the s is nothing but the length of the curve A dash BC length of the curve is obtained nothing but d square plus root over d square plus h square. Once L is known the rate of seepage q can be calculated once the L is estimated the rate of seepage can be estimated. Because Kasekrande solution further the plotting of the periodic line for the seepage to earthen for construction of the flow nets for seepage to earthen dams the periodic line needs to be established first. So the curve A dash E B dash C is a parabola with its focus at C that is the focus at C so when the parabola runs like this this is the way it actually erupts and the periodic line coincides with the parabola with some deviation upstream and downstream faces and this is the deratrix so all point at point A at point A the periodic line starts at an angle 90 degrees. So with this Kasekrande solution the procedure for constructing parabola is given as follows. So parabola here is nothing but A dash E F B dash C so here C C dash is equal to P that is C C dash is equal to P the from the distance which is equal distance from the this side is again P so AC which is nothing but root over x square plus z square A is a point with coordinates x and z AD is nothing but AD is nothing but 2P plus x so AD distance horizontal distance is nothing but 2P that is from the deratrix it is 2P plus x so AC is equal to AD based on the properties of the parabola so we can write root over x square plus root over x square plus z square is equal to 2P plus x by using the properties of parabola we can write root over x square plus z square is equal to 2P plus x so at x x is equal to D that is x is equal to D and z is equal to h we can write P is equal to 1 by 2 root over this square plus h square minus D by knowing D and h P can be calculated. So when P with P value the value of the x various values of z can be calculated with by knowing the P value of the P and with various values of z can be calculated so for different values of x various values of z can be calculated when beta less than 30 L can be calculated by using the expression which is L is equal to D by cos beta minus root over D square by cos square beta minus h square by sin square beta, otherwise the table which is actually given also allows to estimate the length delta L and for beta this is for a beta less than 30 degrees in case when the beta is greater than 30 degrees Kasekara and I propose this table when you have got beta 30 degrees this is actually 0.36 but when you have 60 degrees 90 degrees delta L by 1 plus delta L plus delta L so after locating b dash b is equal to delta L bc is equal to L the curve of b can be approximated. So this is the condition what will happen is that how the estimate the periodic surface which is given by from the Kasekara and I solutions. So when we construct flow net construction through an earthen dam basically what we need to do is that we need to construct the periodic surface so for that three methods have been discussed one is Jupitt solution other one is Shaffernack but Kasekara and I solution is widely use and note that Ag is an equipotential line and Gc is an flow line and pressure head at any point on the periodic line is 0 and the difference of the total head at any two equipotential line should be equal to the difference in the elevation because the total head between any two should be equal to is nothing but pressure at any point on the because it is on the atmosphere the pressure head is 0 so just what we need to do is that by knowing the elevations we will be able to calculate what is the total head between any two equipotential lines. So delta H is nothing but you know in the similar way we once we construct the upper more identify the upper most periodic surface or the flow line and then construct these flow channels so here channel one channel two channel three and this is the equipotential line say one, two, three, four, five, six so these are the potential drops and we can actually draw the head lines which are nothing but delta H is nothing but H by Nd so by knowing the number of drops we can actually calculate what is delta H in this case being very easy and we can construct the flow net and then determine Q is equal to which can be C page or it can be leakage and suppose leakage can be in the form of reservoir where K into H is equal to Nf by D, Nf by D. So in this particular lecture we actually have discussed case for anisotropic case and when you have actually have got then we said that it has to be changed to transformed condition and when you have got say non homogenous soils and how the flow nets deviate and all we have discussed and then when we have say embankment dams or unconfined C page conditions we have discussed about how a flow net can be constructed. So in the next lecture we will try to look into some numerical and physical simulations of these examples and then we will try to see how the use of filters or the chimney drains helps in changing the course of these priatic surfaces which are priatic surface which will be there within the at hand dam or when you actually have a concrete dam with or without cut-offs how there is a diversion of how the flow nets will change and then we will also try to see the examples of how to estimate an upstream uplift pressures and then calculate the factor of safety against uplift pressure. In case if there is a case of uplift then how this can be solved by using some case studies we will discuss in the next lecture of this particular permeability and C page module.