 Welcome to lecture number 15 under permeability and seepage 4 and this is under module 2. So in this lecture we are going to discuss about different types of fluid flow in soils and seepage phenomenon Laplace equation of continuity and solution of those Laplace equations in one dimensional and two dimensional and three dimensional flow conditions. And then we will try to solve some typical problems. So what are the different types of fluid flows which are possible in soils? It can be one dimensional or it can be two dimensional and it can be three dimensional. A one dimensional flow condition is the one where the velocity vectors are all parallel and are equal in magnitude. In the conventional laboratory if you are doing a constant head test where the flow which occurs is one dimensional nature. Because the flow occurs in a rigid container having a particular boundaries the flow happens in the vertical direction. A one dimensional flow condition is the one where the velocity vectors are all parallel and are equal in magnitude. In other words water always moves parallel to some axis and through a constant cross section area. So a flow through a confined aquifer, a aquifer where two clay layers are there suppose if there is a sand layer in between and if there is a head which is actually driving the flow from left to right the flow can actually happen either from left to right or right to left. So steady downward flow occurs when the water is pumped from an underground aquifer or steady upward flow occurs as a result of artesian pressure when a less permeable layer is underlined by a permeable layer which is connected through the ground to a water source providing pressures higher than the local hydrostatic pressures. Suppose if you are having a less permeable layer and below that there is a permeable layer and if the permeable layer is actually receiving a water because of some artesian conditions and the water can flow in vertical direction or can have higher pressures compared to the hydrostatic pressure. So in this in such situations upward flow is possible that is a steady upward flow occurs as a result of artesian pressure when a less permeable layer is underlined by a permeable layer which is connected through the ground to a water source providing the pressures higher than the local hydrostatic pressures. What is two dimensional flow mostly flow through attend amps or flow through a sheet pile wall if you are hydraulic structures or a particular canal amendments where when you are retaining the water on both sides either in cutting or in filling the flow which actually happens is two dimensional nature. It occurs when all the velocity vectors are confined to a single plane but vary in the direction and magnitude within that plane. So if you are actually assuming x and z are the x is the horizontal play horizontal direction and z is the vertical direction if y is along the length of the canal or an attend amp then we can say that the flow which actually happens in x and z direction that means that the flow happens in x z plane. So in that situation in the slide which is actually shown here a sheet pile wall can actually have a two dimensional flow water flows downward here through this porous medium and then water actually moves upward in this region and by the time the water reaches here the head which is actually driving the flow dissipates. So this is the upstream side and this is the downstream side. In case of long excavation or a attend amp way there is a possibility that because of the upstream water level water flows in this fashion. So at different planes if you take these are the you know this it resembles completely but the flow which actually happens in x direction and z direction and this is the plane x z plane. So these are the two examples for the two dimensional flow conditions then what is three dimensional flow? In what situations the three dimensional flow can occur? Three dimensional flow the example is that most general flow condition is that flow towards a water well. Suppose if you are having a water well there is a depletion curve water table here and the flow happens in three dimensionally. So this is an example for three dimensional flow is flow towards a water well. So if the water flows in all the three direction the directions that is in x direction, z direction and y direction and that is very similar in case of flow towards a water well. In case of some ground improvement projects when we are actually using free fabricated vertical drains or sand drains the flow actually happens also in three dimensional that is because of the presence of the drains in the impermeable soil the flow happens in x and z directions as well as in y direction that is also another example for the three dimensional flow condition. So the multi dimensional flow in soils as of now we have considered one dimensional flow in soils where all fluid is actually flowing in the same directions. In most cases however the fluid is actually different fluid in different regions will be flowing in different directions. So the multi dimensional flows are actually possible. So it is required to learn how to solve the multi dimensional problems. So for this in order to develop these capabilities we use equations of continuity, equation of continuity multi dimensional form of Darcy's law application. So if you are having a cofferdam within the middle of the river the situation is that the there is a situation where you have got the vertical sheet pile walls all around the periphery of the water. So here water actually because of this the water tends to flow in this direction. This is the direction of the flow this is downward and this upward and with the symmetry also maintains in the same direction. So it is required to ensure and to know and assess the stability of this type of structures when constructed along with the water. So the seepage and we have discussed about seepage is a phenomenon which occurs because of the prevalence of a head which is actually driving from higher head to the higher potential to the downward potential. The equation of continuity and the Laplace equation is the what we are actually going to discuss. In many practical cases the nature of the flow through the soil is such that the velocity and the gradient vary throughout the medium. So for these problems the calculation of flow is generally made by the use of the graphs represented as or referred to as flow nets. So the flow nets are nothing but the graphical representation of the system along the direction of the flow and in the direction perpendicular to the flow which are represented by the set of lines or a nest of lines which are actually in the direction of the flow or in the perpendicular to the direction of the flow. From these problems the calculation of flow is generally made by the use of graphs referred to as flow nets. So the concept of the flow net is based on the Laplace equation of continuity which describes the study flow condition for a given point in the soil mass and this is actually has got applications in heat flow and electrical the current flow and as in other allied applications particularly in heat transfer or the current electricity transfer from because of the high potential to the low potential. So the concept of flow net is based on the Laplace equation of continuity. So in this particular figure a particular hydraulic structure which is actually shown here and the derivation or the use of the Laplace equation of the continuity is the one which we are going to discuss. So we are having a structure a concrete hydraulic structure which is retaining the water, H1 is the upstream water level and H2 is the downstream water level and so this is called the tail water level and this is the upstream water level. And at the point A let us assume that a small element which is actually having dimensions dx along the x direction and dz along the y direction and dy along the y direction. So the unit volume or the volume of the element is dx dy dz and that particular element is actually experiencing a flow in and flow out that means that qx can be the flow in and qx out is the flow out of that element. So the main assumption is that when the process of the flow when the water is actually flowing through this particular element or any portion of the soil it is assumed that there is no volumetric change or no change in the effective stress. So in this particular situation the enlarged view of details of the small element at point A is shown here the head at point A let us say is small h and h1 – h2 is the delta h which is the potential drop which is actually happening from upstream level to downstream level and this top surface of this downstream level is assumed to be as the datum and this is the previous soil and here is the impervious bottom where the flow can actually happen along this direction only flow cannot actually perpendicular to penetrate through this layer and there is a certain thickness for this and the flow is actually happening from upstream level to downstream level. So the purpose of this structure is to retain the water and maintain this particular condition. So here in this particular detail of the element which is actually shown here qx is the flow entering the dz and dy area and qx plus dqs is the flow coming out. So the net flow is dqx which is actually entering the plane dz dy area and coming out of the flow which is actually coming out of that particular plane is qx plus dqx. Similarly in vertical direction qz is the flow which is occurring along area dx dz dy and the qz plus dqz is the flow which is coming out of that particular area. Similarly qz in the bi direction. So we have got in the three dimensional condition condition we have considered where in x direction and y direction and z direction. So flows entering the soil prism in x and y z directions can be given by from the Darcy's law as follows. We knew that q is equal to kia and ax is nothing but the cross sectional area through which the flow is occurring. So kx is the permeability in the x direction, ky is the permeability in the z direction and kz is the permeability in the z direction, ix, iy, iz are the hydraulic gradients along x and y and z directions. So we can write qx is equal to kx, ix, aax which can be written as kx into dow h by dow x into dy by dz and h is nothing but the hydraulic headed point a which was shown in the previous slide. Similarly qy is equal to ky, iy, ay. So ay is nothing but the cross sectional area along y direction and which is nothing but dx dz. So ky into dow h by dow y into dx dz. Similarly qz is equal to kz, iz, az where kz into dow h by dow z into dx dy. So where qx, qy, qz is equal to flow entering in the directions x, y, z respectively and where kx, ky, kz are the flow entering the directions x, y, z respectively. Now when the water enters this element, let us assume that qx which is nothing but qx in entering this particular area cross sectional area that is dy by dz and qx plus dqs that is nothing but the qx out which is the one which is actually leaving the element. So x1 is at this point and x2 is at this point, x2 minus x1 is nothing but the dx. So we can write for flow in x direction, Ax is equal to dy dz and h is nothing but the total head at point A where the element is best considered. So we can write qx is equal to qx in as kx dow h by dow xx1 that is at this point into dy dz. Similarly qx out we can write it as qx plus dqx is equal to kx dow h by dow xx2 dy dz. Now the difference of these two is nothing but the net flow which actually has taken place that is nothing but dqx is equal to kx into dow h by dow xx2 minus dow h by dow xx1 into dy dz. So the dow h by dow xx2 minus dow h by dow xx1 this particular term represents the change in gradient over a distance dx. So this can be written as dqx is equal to kx into dow square h by dow x square into dx into dy by dz. So dx dy by dz is nothing but a element volume which was considered in this particular derivation. So using that we can actually now reduce the respective flows leaving the soil prism in x and y z directions and which can be given again by applying the Darcy's law as follows. qx plus dqx is equal to qx into ix plus dix that is into ax which is nothing but kx into dow h by dow x plus dow square h by dow x square dx into dy dz. Similarly in the y direction and z direction for the respective flows leaving the soil prism can be computed. Now using the principle of conservation of the fluid mass for study flow through an incompressible medium the flow entering in the soil element or flow entering in the element is equal to the flow leaving the element. So by equating the inflow is equal to outflow we can get like this which is nothing but qx plus qy plus qz is equal to qx plus duqx which is the outflow flow leaving the along the x direction and qa plus duqy in the y direction and qz plus duqz is in the z direction. By simplifying and we get and substituting the previous terms which we have discussed we get that the net flow into or out of the element per unit time. So nothing but kx into dow square h by dow x square plus ky into dow square h by dow y square plus kz into dow square h by dow z square is equal to 0. Here the 0 represents that the volume change per unit volume which is nothing but 1 by 1 plus e into dow e by dow t is equal to 0 that means that is dow e by dow t is equal to 0 so because of that this particular term volume change per unit volume is 0. But however in case of consolidation phenomenon where there is a volume change happens the term will remain. So in the steady state seepage conditions when no volume change takes place no volumetric changes takes place this particular term it becomes 0. So this particular term is the equation of continuity when kx ky kz when they are not equal to 0 then kx is equal to kx then and then they are actually a permeability in x y z direction. So for three dimensional flow condition the generalized Laplace equation of continuity is kx into dow square h by dow x square plus ky into dow square h by dow y square plus kz into dow square h by dow z square is equal to 0. And for two dimensional flow in x z plane that means that for flow through Shearpell war or flow through an earthen dam or through a long excavation or a canal embankment problem kx dow square h by dow x square and plus kz dow square h by dow z square is equal to 0. So if the soil is isotropic with respect to permeability then kx is equal to ky is equal to kz the permeability is equal that is kx is equal to ky is equal to kz is equal to kk and the continuity equation can be simplified to dow square h by dow x square plus dow square h by dow y square plus dow square h by dow z square is equal to 0 because kx is equal to ky is equal to kz is equal to k which is not equal to 0. So because of that dow square h by dow x square plus dow square h by dow y square plus dow square h by dow z square is equal to 0. So this is the simplified Laplace equation of continuity for a soil having identical permeabilities in all the directions that means that in kx is equal to ky is equal to kz is equal to k in xy and z direction when the permeability is identical that means that isotropic with respect to the permeability. In case of a two dimensional flow for the xz plane when kx not equal to kz we write it as kx into dow square h by dow z square by dow x square plus kz into dow square h by dow z square is equal to 0. If the permeability is identical in x and z direction then we can write the simplified Laplace equation for equation of continuity for two dimensional case as dow square h by dow x square plus dow square h by dow z square is equal to 0. So flow through a constant head perimeter in the laboratory as we discussed that this represents the one dimensional flow. So here the water flow happens from vertically in the direction. So this is the direction of the flow and these are the points where the head dissipation is actually taking place. This is the available head, this is the point where the higher head is there. Suppose if the head which is driving the flow is say h and by the time it reaches here the head is dissipated. So the hydraulic gradient is nothing but h by L which is nothing but the hydraulic gradient. So for one dimensional flow in the z direction the simplified equation of Laplace equation of continuity is dow square h by dow z square is equal to 0 where k is the permeability in the vertical direction. Change in the hydraulic gradient per unit distance in the z direction. So dow square h by dow z square represents change in the hydraulic gradient per unit distance in the z direction. Similarly if you are having in x direction it represents the change in hydraulic gradient per unit distance in the x direction. If you are considering for the one dimensional flow condition vx is equal to v, vy is equal to 0, velocity in x and y directions are 0 and ix and iz is equal to 0 and iz is equal to constant. So because of that for the one dimensional flow condition so these are the boundaries of the container and the flow which actually happens, confined flow happens vertically in the downward direction. So possible so having discussed about the Laplace equation of continuity what are the possible methods which are available for solving the Laplace equation, analytical closed form or a series of solutions of the partial differential equations quite mathematical and not very general. Solution methods are available, typically the finite element method or finite difference method very powerful and easy to apply and can deal with heterogeneity, nsotropy and two dimensional and three dimensional conditions. So nowadays the software's finite element method based on softwares are available which actually enable to deal with heterogeneities nsotropies and two dimensional and three dimensional cases. And graphical techniques which are actually popular and which are known as the flow net methods commonly used in engineering practice to solve two dimensional problems. So at present in this particular lecture we are going to discuss about the flow net method in the forthcoming lectures we will be discussing about some numerical solution method particularly for two dimensional cases with and without isotropic conditions. The flow net method is a straight forward graphical method to solve two seepage problems. The solutions of Laplace equation consists of two families of orthogonal curves in x and z plane, x is a particular plane over which the flow is actually happening and that is the plane x and z are the plane over which the flow is happening. So the family of these two sets of curves in x and z direction is known as the flow net. So if you look into the figure which is actually shown here there is a system of the red lines which are actually shown here and there is a set of yellow lines which are actually shown here. And of this dimension which is actually shown as EA and this dimension is B. So this can be also indicated as B and length L. This is called the ratio A by B is called aspect ratio or ratio A by B or B by L is called as the aspect ratio and these are actually called as flow lines and the space between these two is called as flow channel. So in this case in the partial flow rate which is shown here they are the two flow channels are there and the lines which are drawn within a yellow color they are called equi-potential line which is the line joining the equal heads. So if I take a head along this the total head will be identical. Similarly here the total head will be identical but this is the direction of the flow but from this point to this point if H1 is the head here H2 is the head here the H1 is actually greater than H2 so H1 minus H2 which is nothing but the delta H that is the potential drop takes place between any two equi-potential lines. The flow line is basically defined as the path which a particle of water flows in its course of seepage through a saturated soil mass this is called as the flow line. So we have said that the flow net is nothing but the nest of system of flow lines and equi-potential lines. So in this particular slide schematically a partial flow net is shown here where this is the direction of the flow and H1, H2, H3 are the heads at this particular equi-potential line 1, 2 and 3 that means that along this line the total head is H3, along this line total head is H2, along this line total head is H1 that is why it is called as equi-potential lines. And L is this length and B the B by L or previous slide we have discussed about A by B is called aspect ratio. This ratio of B by L should be equal to approximately constant and should be equal to 1 for squares for convenience we select B by L is equal to you know as 1 a curvilinear squares and H1 greater than H2 greater than H3 indicates that as the flow actually happens the dissipation of the energy takes place. So the flow net solution is a graphical method for solving the two dimensional Laplace equation and for two dimensional condition del square H is equal to 0 that is nothing but dou square H by dou x square plus dou square H by dou z square is equal to 0. For one dimensional condition if the flow is actually happening in along the x direction it is called as dou square H by dou x square is equal to 0 if the flow is actually happening only in the vertical direction that is say z direction then dou square H by dou z square is equal to 0. So it describes that the energy loss associated with the flow through the medium and is used to solve many kinds of the flow problems including those involving the heat, electricity and seepage. So this particular Laplace equation of continuity what is used in geotechnical engineering is also used in other allied areas like heat flow or electricity flow and of course what we are using for the seepage. So now having actually said now we actually have seen that the flow lines and the equipotential lines what is the condition which actually it maintains the orthogonality. So in order to prove that let us actually try to look into this particular solution. So flow nets are based on the two mathematical functions one is potential function and that is nothing but the phi and the flow of the stream function that is psi. So now let us select say a potential function phi x z that is for the two dimensional condition x and z dou pi by dou x is equal to that is nothing but the velocity in the x direction is equal to minus k dou H by dou x minus is given because the head decreases in the direction of the flow dou pi by dou z is equal to V z is equal to minus k into dou H by dou z. So let us put them as equation A and equation B dou pi by dou x is equal to V x is equal to minus k dou H by dou x dou pi by dou z is equal to V z is equal to minus k dou H by dou z. So by differentiating and substituting the Laplace equation of continuity when we substitute the previous the equations A and B and we get when we substitute in the Laplace equation of continuity for two dimensional flow condition we get dou square pi by dou x square plus dou square pi by dou z square is equal to 0. So this indicates that the potential function pi x z in the two dimensional case satisfies also the Laplace equation. So dou square pi by dou x square plus dou square pi by dou z square is equal to 0. Similarly now when we take when we take this integrating A and B that is the velocity functions we have written in the x direction and z direction that is the previously we named it as A and B which is dou pi by dou x is equal to V x is equal to minus k dou H by dou x dou pi by dou z is equal to V z is equal to minus k dou H by dou z. By integrating this particular functions we get pi x z is equal to minus k into h of x z plus function of z that is from A that is this and from B we get pi x z is equal to minus k H x z plus g x. Since x and z can be varied independently function of z is equal to g x is equal to constant. So with that we can write that is a C so we can write as H x z is equal to minus 1 by k into C minus pi x z. So if H x z represents a constant H then the equation C represents a curve in the x z plane. So if H x z represents a constant H equation C represents a curve in x z plane. So equipotential line the curve in the cross section such that pi is constant along the curve that is why we call it as equipotential line. So the curve in the cross section we basically it is a curve in the cross section such that pi is constant along the curve. So the total head is constant along the equipotential thus the family of the curve family of curves is similar to the contour lines are a topographical map. So except drawn in vertical section and lines representing equal heads not equal elevations. So what we are actually talking is that not equal elevations the elevations are different but total head is equal. So the total head is constant along the equipotential line and equipotential line is a curve in the cross section such that pi is constant along the curve. Now along such contours of say equal total head say d pi is equal to 0 from the definition of the partial differentiation and combining this equation we can write now d pi by dx dou phi by dou x is equal to vx and dou phi by dou z is equal to pz. So vx is the direction of the this is the x direction this is the z direction and this is the element dx and this is the dz in vertical direction dx in the y direction and this is the flow line. So d pi is equal to dou phi by dou x into dx plus dou phi by dou z into dz so which can be written as d pi is equal to dou phi by dou x is nothing but vx dx plus vz dz vz is nothing but dou phi by dou z. Along an equipotential line phi is constant so when phi is constant the differential of that is constant so d phi is equal to 0 by putting this 0 what we get is that the slope of this line as nothing but this is an equipotential line dz by dx phi is equal to minus vx by vz. So the slope of this equipotential line dz by dx phi is equal to minus vx by vz. So what we have deduced is that from the definition of from the along a particular contour when you take a particular equipotential line having the slope of the equipotential line we would deduced as minus vx by vz. Similarly considering the flow or the stream function psi consider the in a function of psi xz such that dou psi by dou z is equal to so psi xz is nothing but the flow function or the stream function in the x and z direction and vx dou psi by dou z is equal to vx is equal to minus k dou h by dou x minus dou psi by dou x is equal to vz is equal to minus k dou h by dou z so this is inverse of the potential function. So combining the equations and substituting in the Laplace equation of continuity we again will get that dou square psi by dou x square plus dou square high by dou z square is equal to 0 when we substitute this particular deliberations A say D and E in Laplace equation of continuity we get dou square psi by dou x square plus dou square psi by dou z square is equal to 0. So this represents that psi xz also satisfies the Laplace equation. Now this indicates that both phi xz and psi xz the satisfies the Laplace equation of continuity the system of these lines containing the flow function and potential function they form a flow net solution which actually satisfies the Laplace of the equation of continuity. So again now to deduce the slope of this stream function or flow line from the definition of the partial differentiation and combining equation we can actually get minus dou psi by dou x is equal to vz and so this is the flow line and dou psi by dou z is equal to vx. So total differentiation differential of psi xz is given by d psi is equal to dou psi by dou x into dx plus dou psi by dou z into dz. So d psi is equal to minus vz into dx plus vx dz this we have substituted here for a given flow line if psi is constant. So along a given flow line if the psi is constant the differential of that constant is 0 so d psi is equal to 0 when we equate that we actually get you know dz by dx psi is equal to vz by vx. So the slopes are identical the product of that is equal to minus 1 that means that both this flow lines and equipotential lines are orthogonal in nature that is the reason why while drawing the flow nets it has to be remembered that always the equipotential line intersects the flow line orthogonally. So it is apparent that the flow lines and equipotential lines intersect each other at right angles are orthogonal to each other that is from the deliberations what we discussed the product of this slopes say m1 m2 equal to minus 1 that represents the orthogonality condition. So here a particular system which is actually shown so this particular angle which is actually you know have to be 90 degrees so this is the flow line and this is the flow channel this is the equipotential line 1 and equipotential line 2. So delta H is actually is the pressure drop which is actually occurring from this equipotential line 2 this equipotential line delta Q is the flow which is actually happening through this particular channel and delta L is the length over which this flow is actually happening delta B is the so the area if you are actually taking delta Q with the flow is actually happening delta B into 1 the 1 is actually is the in the other direction a unit area is a unit width is actually considered. So equation of use of continuity equation for solution of simple flow problems let us say in this particular one dimensional flow through two layers of soils or you know if we can actually reduce for two layers and then we can actually use for say in principle if H1 is the head here and by the time it reaches here the head has to be 0 that means that we have discussed that if H is the head which is actually available and over a length L then the hydraulic gradient here the head available is H and here this point is 0 that means that 50% of length L by 2 it has to be H by 2 but when you are actually having two soils then the boundary conditions actually the different. So in this particular slide what we have considered is that we have got two soils soil A and soil B having two different permeabilities but the flow occurs in along the x direction and z is the direction which is actually shown here and LA is the length of the soil A and LB is the length of the soil A and LA plus LB put together is the total length and at LA the head which is actually available is H2 and at point 2 that is the end of soil A the head is actually say H2 and H1 is actually greater than H2 at this point it is equal to 0. So this if it is assumed as datum so H1 H2 and then head available is here is 0. So this is the along the x direction and the point 1 is at beginning of soil A point 2 as the interface of at the interface of soil A and soil B and point 3 is at the end of that is soil B. So in the case one that one dimensional flow through two layers of soil the flow is in the one direction only that is the x axis the length of the two soil areas which are actually described as LA and LB and the coefficient of permeabilities are given as KA and KB in the direction of x axis. The total head at sections 1 and 3 are known required is that total head at any other section for length between 0 and LA plus LB. So integration of Laplace equation of for the one dimensional flow which is nothing but along the x direction which is nothing but dou square H by dou x square is equal to 0 which actually gives the H is equal to C2 x plus C1 where C1 C2 are the constants for the flow through the soil A the boundary conditions what we can actually use is that at x is equal to 0 head is H1 because at that is at the point 1 and at x is equal to LA that is at the end of soil A head is equal to H2. When you substitute this in this particular H is equal to C2 x plus C1 C1 C2 can be obtained and then we get actually H is equal to minus of H1 minus H2 by LA into x plus H1 for 0 this is valid between 0 and LA. Now for the flow through the soil B the boundary condition can be given as at x is equal to LA H is H2 that is beginning of soil B and by again by using H is equal to C2 x plus C1 but using the new boundary conditions and at x is equal to LA plus LB that is the end of soil B H is equal to 0 that is head available is 0. So when we substitute and determine C1 and C2 and simplify we will get H is equal to minus of H2 by LB into x plus H2 into 1 plus LA by LB this is actually valid between from LA to LA plus LA B that is at the end of soil A and to the end of soil B the total length. So Q is nothing but the rate of flow through the soil A and rate of flow through the soil B by the but flows are equal because the same head which is actually driving so because of that Q is equal to K into H1 minus H2 into LA H1 minus H2 by LA into A, A is the cross section area of soil A and cross section area of soil B is also same. So the rate of flow through the soil A can be written as KA into H1 minus H2 by LA LA is nothing but the length of soil A into area A which is the cross section area of the perpendicular to the direction of the flow and KB is equal to KB into H2 by LB H2 is the head available at 0.2 where the beginning of soil B into A. So by equating and then simplifying we will be able to get H2 in terms of KA H1 divided by LA into KA by LA plus KB by LB. So the H2 if you look into this when you substitute say at LA is equal to LB and KA is equal to KB H2 is obtained as H1 by 2 that is correct because at the half length of the soil if thus both soils are actually having same permeability the head dissipated is will be 50% of the head over which the flow is actually happening. So further when you come use this one we actually have got two things for X is equal to 0 to LA we have got H is equal to H1 into 1 minus KB X KA LB plus KB LA and for X is equal to LA to LA plus LB then H is obtained as H1 plus H1 into KA by KA LB plus KB LA into LA plus LB minus X. Say here when we substitute LA is equal to LB is equal to L, KA is equal to KB is equal to KA H is nothing but H1 into 1 minus X by L for say at X is equal to 0 H is equal to H1 when X is equal to L H is equal to 0 that means that the head available at the end of the soil is 0 soil sample is 0 at mid length of the sample that is X is equal to L by 2 H is equal to H1 by 2 but this gives actually for the length up to X is equal to 0 to LA for X is equal to LA to LA plus B the head at any point along the direction of the flow can be determined this is by using the equation of continuity. The pore pressure in the steady state CPS conditions so here in this particular slide where you are actually having this is the equipotential line and this is the direction of the flow that is the flow line what we call so between two equipotential lines let us say that delta H is the drop and the elevation which is nothing but this HQ and this is the mid atom so H I is nothing but the head drop per unit length for steady state CPS I is constant so change in pore water between points P and Q is given by per unit width which is given by delta U is equal to I gamma W into delta S delta U is equal to I gamma W into delta S so the flow net solutions particularly when water flows through a porous medium such as soil head is lost to the friction similar head losses occur when water seeps through an earthen dam that is unconfined flow an example for an unconfined flow is earthen dam or under a sheet pile wall coffer dam or a concrete masonry dam so that is actually given as a confined flow. As the particle of water proceeds from A to B it exerts a pressure drag on the soil particles that is the drag in turn produces a CPS pressure in the soil structure so as the flow happens from higher potential to the lower potential the it exerts a pressure drag on the soil particles in turn it actually produces the CPS pressures so the hydraulic structures are required to be checked against the CPS pressures if the CPS pressures are high then there is a susceptibility of the different types of failures. So in this particular slide an example problem of flow net construction is shown this is an example of a sheet pile wall which is retaining a water of 3 meter in the upstream side and the downstream side this is called as a tail water level which is 0.5 meter is the you know the tail water level which is maintained so the drop or potential drop total potential head the drop which is actually given as 3 minus 0.5 which is nothing but the 2.5 meters is the total drop. Now here first of all the boundary conditions need to be identified before you know commencing the writing of the or drawing these flow nets or constructing the flow nets impervious layers and previous layers let us assume in the given problem this is the permeable layer and this is the depth of penetration of sheet pile wall and here this is the impermeable layer and it can be a clay or it can be a rock stratum so where the flow actually happens along this plane only so in identifying the boundary conditions AB which actually represents an equipotential line again EF which represents the equipotential line and similarly here when it comes to this particular line this actually represents the flow line because the flow actually happens along this boundary along this impervious layer. So the first flow line is actually said as also line is the creeping line is called as BCDE is also called as a creeping line which actually creeps along the penetrated sheet pile wall. So you can see here when it meets this equipotential line the orthogonality is actually maintained so in drawing the flow nets it has to be ensured that the orthogonality is actually maintained properly here. So this is the first flow line subsequently depending upon the convenience like it can be divided into the configuration or geometry it can be divided into 5 or 6 flow lines so here we have got flow line 1 and flow line 2, flow line 3, flow line 4 and then the flow line 5. So the space between that is actually called as a flow channel so this is one flow channel 1, flow channel 2, flow channel 3 and 1, 2, 3, 4 this can be approximated as 4.5 or so. So the number of flow channels here are 4.5 and now this is an equipotential line and this is an equipotential line and these are flow lines. Now here there is a flow line so the line which is actually if it is dropped here this being a flow line orthogonality here so this is an equipotential line this is a flow line so orthogonality by maintaining the orthogonality we can draw another line. So similarly here so these are drawn such a way or selected such a way that the orthogonality is actually maintained between a flow line and equipotential line. So in this case if you look into this there are the numbering is done such a way for convenience it is numbered as 0, 1, 2, 3, 4, 5 and 6. So this represents there are 6 potential drops so 6 potential drops in the sense at 6 potential that is here the head available is say 2.5 meters at this point the head available is 0 nothing but 2.5 into 0 by 6 is equal to 0. Similarly here 6 potential line means that is the 6 by 6 into 2.5 that means that the full head is available at this point. So the equipotential line AB is with a potential head of 2.5 meters equipotential line EF is having total head as 0 that is the potential available head available is 0 here. So here second potential drop potential line this is 1, 2, 3, 4, 5 and 6. So by knowing this number of flow channels and number of potential drops we can calculate the flow through a sheet pile wall and if it is an earthen dam we can also get what is the seepage actually happening through an earthen dam and are leakage through a reservoir. So this is another problem where the flow net construction for a wear and wear there is upstream water level which is actually retained here the tull storm water level is at the base which is actually located here. So this is the impermeable layer and this is the permeable layer the water flows in this direction so these are the direction of the flows. So if you look into this 1, 2, 3, 4 and 5 so you can say that the 5 flow channels are there and this is the last flow line so this is the equipotential line and this is the equipotential line and this is the first flow line where the creeping water flow actually happens along this and then perpendicular by maintaining the orthogonality we write the number of potential lines. So here 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 so that means that there is a 12 potential drops are there. So by knowing say 9 by 12 that is the 9 by 12 into 10 meters is the head available at this point. So by knowing this total head and by knowing the elevation from the if this is the datum if this point is datum let us say that this is located at 2 meters that is by knowing this elevation we can actually calculate the pressure available at this particular point. So in the process of seepage there will be some dissipation by the time it comes here the head which is actually takes place 1, 2, 3 drops it undergoes so 9 by 12 into 10 and then if you take the elevation head with respect to the datum which is selected here we will be able to calculate the pressures at each point. So the calculation of this total heads and this pressure is important for assessing the uplift pressure distribution or the pore water pressure distribution along these things and the in the process in turn which can be used for designing these hydraulic structures. So the seepage particularly the flow net solution the computation of discharge which what we discussed is that the aspect ratio which is nothing but a by b which is constant basically they are curvilinear squares. So hydraulic gradient i is equal to delta h by b so this delta h is given by for one potential drop which is hl by nd by b so we can get this as a delta h by b. So equipotential drops between two flow lines is delta h by hl by a delta h is equal to hl by nd so from Darcy's flow flow in each channel is given by delta q is equal to q into hl by nd by b into a. If you wanted to purchase the total discharge per unit width then it has to be multiplied with the we have calculated this for per flow channel now let us say that there are n number of flow channels then it is multiplied by n or here the total number of flow channels are indicated by n suffix f so we can write for a case where the flow is actually occurring through isotropic medium then q is equal to k hl the hl is nothing but the potential head into nf by nd into a by b if the ratio a by b is equal to 1 then the a by b will not be represented. So consider this particular example problem where the flow net method which is required to be adopted so based on this consideration the delta h is equal to 2 meters and this is the point where the datum has been selected so here based on the discussions here we can actually say that this is an equipotential line and this is an equipotential line and this is a flow line and flow channel 1, flow channel 2, flow channel 3 and flow channel 4 so we have taken nf is equal to 4, nd is equal to 12 because there is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 so if you wanted to measure say pressure at point b we can actually calculate total head at b is equal to pressure at b plus elevation at b suppose here it is given that 1.5 meters above this particular statum so that means that it is 3.5 meters below this level that is 6.5 meters below this level if this is selected as datum then this will be minus 6.5 meters so the pressure available here the total head is nothing but it is nothing but 0, 1 and 2 and the 2 meters is the delta h that is the head loss. So 2 into that is 2 by 12 is equal to pb into this particular minus 6.5 meters so with that we will be able to get pressure at b is equal to 6.33 meters and if the unit weight of water is taken as 10 kilo newton per meter cube we can calculate the pressure at b as 68.33 kilo Pascal's. So what we have done is that by with the total head which is actually available at this particular point and by knowing the elevation head we calculated the pressure at b like that each and every point it can be estimated and this particular data is useful for calculating and assessing failure against the heave and piping failures for these hydraulic structures which we will be discussing in the future course of deliberations.