 Welcome to lecture number 17 of Geotechnical Engineering 1. Having done the seepage analysis, now this is the right time to introduce you to the concept of seepage theory which is a quite a vast topic to discuss. So, until now we have talked about simple problems where the seepage is occurring through the porous media and it so happened that this porous media was you know of defined geometry. Now, most of the time when we deal with the porous media which is a well defined geometry we assume the variation of N, V and I you know easy to determine. But in most of the situations which we come across in the real life the geometry of the porous media is not so well defined and hence it is becoming difficult or it becomes difficult to find out the variation of V and I in the real life situation. Suppose if I say that there is a dam body, now this dam could be made up of RCC or this could be earthen dam alright. Now, this system is sitting on the ground in such a manner, so this happens to be the ground surface and we have to show that there are two sides which are defined as upstream and downstream. So, suppose this is the downstream water and this happens to be the upstream water and if I demonstrate the hard strata by drawing this line, so this is the impervious layer. This is how we define the impervious layer that means there is no flow which is going to take place across this line. If I define this as H1 and if I say that the height of the downstream water column is H2, now I hope you realize that it would be very difficult for me to find out the value of V and I and for that matter even the H value where H happens to be the head at different points. So, this could be point number 1, point number 2, point number 3, point number 4, 5 and so on. Now, this is a typical cross section of the earthen dam or a RCC dam, now my interest is to find out how much seepage is taking place through the foundations of this system. So, this part is defined as the foundation soil or the foundation of the earthen dam, this is the body of the dam. So, very soon we will realize that the difference of head between H1 and H2, now this is responsible for causing the flow through the foundations. So, what I have to find out is I have to find out the values of three parameters V, I and H. Now, as I said when you have complicated systems like this, it becomes very difficult to obtain the parameters as compared to the situations which we have already dealt with. So, what we are supposed to do, we take help of the seepage theory or the flow net. So, if I consider a control volume, so this is the control volume and if I define y, x and z, these three directions, what I am interested in is finding out how much flow is entering into all the three directions. So, if I say that the flow is Q which is entering the control volume. So in the x direction, the Qx enters and what comes out is Qx plus delta Qx. I am just using delta Qx to demonstrate that there is some change in the discharge. In the z direction, this is Qz and what comes out of the system is Qz plus delta Qz. Similarly, in the y direction, I can show that Qy or for the sake of signs, you can say this is the Qy which enters the control volume and what comes out is Qy plus delta Qy. Now, I know that Q is equal to k into i into a, where a is the area of cross section across which the flow is taking place. So, for me it is easy to write, if I say Qx equal to and if I consider the soil mass to be anisotropic, so let us generalize the situation, anisotropic porous media, so the Qx will be equal to kx into ix into area through which the x is flowing. So, x is flowing through, let us say if I consider the dimensions as y and z, so this is the area of cross section through which the Qx is taking place. Similarly, I can write Qy equal to ky iy into zx and Qz equal to kz into iz xy, alright. So, I have the basic equations of the discharge or the flow which is entering into the control volume. So, this is the control volume and what I am assuming is that the dimension of these control volumes are x, y and z. So, I can find out the incremental form also, the deltas of Qx will be equal to, if I assume kx to be constant, you know what will be the function for this. I will skip these steps and you can save some time, we can write this as or what we can do is, we can write as Q plus delta x as equal to kx into Ax ix plus dix, which will be equal to kx into Ax, what is ix minus delh upon delx, where delh is considered to be equal to h. So, if I consider this as h, the hydraulic gradient is delh upon delx minus delh square h upon delx square into dx. Similarly, I can write for the other terms also that is Qy plus delta Qy and Qz plus delta Qz. Now, if I use the continuity equation, I can say that the total discharge which is passing through the system is flow ingress is equal to the flow x-race, alright. So, that means all these functions can be written as your Q terms are nothing but this where this is Ax, this is Ay, this is Az. So, I can skip some steps to save time and what I will be getting is, I will be getting as kx into delh square h upon delx square plus ky delh square h upon dely square plus kz delh square h upon delz square equal to 0. Now, this is what is known as the Laplace equation for anisotropic system, anisotropic porous media. If I am talking about the two dimensional plane, y can be eliminated and this becomes, we can eliminate this or we can assume this to be 0 and hence we have kx delh square h upon delx square plus kz delh square h upon delz square equal to 0. To make this system isotropic, isotropic porous media, I can assume that kx is equal to kz equal to k and hence I will be getting this equation k delh square h upon delx square plus delh square h upon delz square equal to 0. What this indicates is delh square h equal to 0 and this is again the Laplace equation. Now, if I play with this equation, there are two ways of interpreting this. Now, h happens to be the total head which is causing flow to occur in this case. So, suppose if I write it as if I consider one dimensional case, this will be equal to delh upon delx equal to c1 or if I further integrate it, I can say h equal to c1 x plus c2 alright. So, using the Laplace equation, I can also solve these problems. Let us take a simple case. Suppose there is a system like this. There is a composite soil system, soil A and soil B. This is the length of the sample La, LB. I am considering kA and kB as their hydraulic conductivities. Now, if I connect this soil mass with let us say a water tub or a water bath alright. So, these are the piezometric cubes. Somewhere here I can take the datum and with respect to this datum, if I say that this is h2, this is h1 alright. Now, try to find out the discharge and the hydraulic gradients which are existing in the soil mass A and soil mass B of different lengths. So, let me introduce the concept of equipotential lines here and flow lines. I can assume 1 1 as the equipotential line because at each and every point on this line 1 1, the potential is constant and this is equal to h1. At 3 and 3, the total potential is 0. Now, somewhere in between that is at 2 and 2, we are assuming this to be h2. Now, suppose if I ask you to solve this problem, establish the discharge seepage velocity through this system by using the Laplace equation is fairly simple. So, let me consider first a distance x starting from 0.1 and x is such that that this is less than La and greater than 0. If I consider x to be this should be my reference point. So, what is going to happen? Now, at this point at x equal to 0, h is equal to h1 alright. So, h1 equal to c2. Is this correct? What will be the discharge? So, discharge can be obtained. The moment h1 is gone, what I should be doing is this h is corresponding to somewhere in between alright. So, all these lines are going to be parallel to each other that means all the equipotential lines where the h is acting is going to be parallel to 1 1 2 2 and 3 3. The lines which are going to cut them perpendicular to each other would be the flow lines. So, the blue ones are the flow lines and the white ones which I have drawn are the equipotential lines. Now, suppose if I say that h if this function is done I can substitute the value of h1 c2. So, h is equal to c1 x plus h1. So, basically c1 comes out to be h minus h1 over x, x equal to La, this will be equal to h is h2 minus h1 upon La. So, I have the composite function now. I can substitute the value of this over here and I can solve this expression. Now, similarly what you should be doing is try to find out the values of h in the second domain that is when the value of x is less than equal to Lb and La. Suppose this is the point which I have considered here let us say this is the x value all right. Now, I will give you the final expressions and please try it yourself to save some time. You can show that in this case the h will be equal to minus h2 upon Lb x plus h2 upon Lb La plus Lb. So, c1 I can compute as minus h2 upon Lb and c2 can be computed as h2 over Lb La plus Lb. So, this is the head distribution which we are trying to find out. Now, another thing which I would like to do is I will like to find out the discharge which is taking place through the system. So, the q value will be equal to I in let us say soil A multiplied by kA into area of cross section area of cross section in this case remains same of the sample. So, this is equal to Ib multiplied by kB into A. A can be eliminated all right. So, this is going to be now kA into h1 minus h2 upon La equal to kB into h2 over Lb. So, I have got another relationship between h1 and h2. The principle unknown is h because I do not know what is the variation of h along the flow path. So, from here I can get h1 as a function of h2 and if you solve this expression you will be getting this as h1 upon h2 will be equal to 1 plus kB into Lb La upon kA into Lb all right. So, I can use this expression, I can use these expressions which we have derived and we can solve h1 and h2. So, this is the application of the Laplace equation for analyzing the one dimensional flow. These type of problems we have done earlier also. I just wanted to cite an example of how to use the generalized seepage theory to obtain the solutions to the problems which are going to be more complicated. If I use the two terms as phi and psi, so the flow lines will be defined with or designated with psi and the flow lines will be designated with phi all right. So, this is the equipotential function and this is the flow function. The characteristics or the properties of these functions are that if I say del f del phi by del x, this will be equal to Vx all right. And this is what is equal to minus kx del h upon del x and del phi over del z will be equal to Vz the charge velocity minus kz del h upon del z. The minus sign indicates that as x increases the h decreases all right. So, this is what the interpretation is. I can show that del square phi is equal to 0 and del square psi equal to 0. Now what this indicates is that the functions phi and psi also follow Laplace equation. Now if I solve this expression this will give me phi equal to minus k h as a function of x and z plus c. Now this could be an equation of a curve or this could be a straight line also. So, what this indicates is that the equipotential function or equipotential line could be either a curve or a straight line. Now suppose if I take phi as constant clear. So, h becomes constant and c also becomes constant. If this is a situation I can say that del of phi will be equal to 0 and del of phi can be written as del phi by del x into dx plus del phi by del z into del z, dz sorry. Now just now we have defined del phi by del x as Vx plus del phi by del z as Vz. If you solve this expression what you will be getting is dz upon dx will be equal to minus Vx upon Vz. So, this becomes your function number A. Now the same thing I can do for the flow function also. If I define del psi by del x equal to minus Vz and del psi by del z equal to Vx. So, this function can be written as minus k x into del h upon del z and this can be written as minus k z into del h upon del x. I can use this functions again to show that if psi is a constant d psi will be 0 and this will yield dz by dx as Vz upon Vx. Now if I designate this as B, you can make out that A is perpendicular to B. So, A is perpendicular to B. Is this part okay? Now A happens to be a equipotential line. So, equipotential line is always perpendicular to the flow line. This is what we have derived. So, when we talk about the seepage theory where all this is going to be used, this whole thing can be applied to solve the or to establish the seepage regime. We call this as seepage regime. Seepage regime is defined by phi and psi functions and this is also known as the flow net. Clear? That means if I maintain this condition and if I define this as a psi function which is the flow function, this could be phi plus delta phi psi. This is an equipotential function which is phi and another function here I can take as delta phi as long as the perpendicular or the perpendicular you know what do you call it as a normality criteria is established. That means phi is perpendicular to psi function. This is a flow net. So the first requirement of the flow nets is that phi function should be perpendicular to the psi function. So what we have done until now is we started from a three-dimensional flow which is entering into a control volume of the soil. We are assuming that because of the seepage the volume of the control volume does not change. Soil is incompressible. Flow takes place, goes inside, it comes out. We have used the continuity equation for flow and then we have derived the functions for equipotential lines and the flow lines and we have shown that these two functions are always perpendicular to each other and this type of arrangement of phi and psi function is known as flow nets. Now I will discuss about how to utilize flow nets for solving different problems. So let us play with this flow nets a bit more. Let us generalize this. So suppose if I assume that this is the psi plus delta psi and there is another flow line which is depicted as psi. The perpendicular to this would be the equipotential functions depicted as phi phi plus del phi. In such a manner that this is delta s distance and this is delta n. So there is a specific reason of assuming this as delta s and delta n. This is the direction of the flow. So that means the velocity is in the discharge velocity is in the s direction. Now if I take the components of this vector in such a manner, this is x axis and this is z axis what is known as rotation of the plane. So I am just rotating the axis from s n to x z with this as alpha. I can define v x as v s cos alpha and v z as v s sin alpha. I can also define the term cos alpha as if I take the slope of this line and if I define this as delta x delta z. So cos of alpha will be equal to del s del x upon del s del x upon del s. This can also be written as d x upon d s and sin of alpha will be equal to delta z upon delta s which will be equal to d z upon d s. Now suppose if I say what is del phi by del s. Incidentally del phi by del s is nothing but rate of change of eq potential in the s direction. So how phi is changing along the direction of the flow is what is being depicted as del phi by del s. So this will be equal to del phi by del x into del x upon del s plus del phi by del z into del z upon del s. So this is nothing but what is del phi by del x. Del phi by del x is v x. So v s is v x into cos alpha and d x by d s is cos alpha plus this will be v s sin alpha into sin alpha. So this is equal to v s cos square alpha plus v s sin square alpha. This will be equal to v s. Similarly I can also show that del psi by del n will also be equal to v s. Now what this indicates is that del phi by del s equal to del psi by del n which can be written as delta phi over delta s equal to delta psi over delta n. And that is what I had said that there is an intention of choosing this delta s and delta n as the steps or the spacing between the eq potential lines as delta s and delta n as these spacing between the flow lines. Now if I put a condition that delta s is equal to delta n and suppose to simplify things if I make it equal to unity. So this is the typical property of a square. That means the flow net are going to be geometrically square units either they will be curves or they will be linear. So there is no harm if I assume delta s equal to delta n equal to 1. What I am saying is delta phi is equal to delta psi. So the flow net by definition is a square entity made up of linear system or this could be non-linear system also provided the delta s. So this becomes the delta n the curved path and this becomes the delta s. The condition of normality cannot be sacrificed. So this is 90 degree this is 90 degree this is 90 degree and this is 90 degree. So this is a typical unit of the flow net. Suppose if I start from this function that is delta phi is equal to delta psi. This is nothing but delta q and this will be delta h. Now if I introduce two terms that is the total head you remember the h is equal to h 1 minus h 2 which is also equal to delta h. So this head is causing the flow to occur. A quick review of the discussion which we had earlier if this is the body of the dam this is h 2 this is h 1 head downstream upstream if this is the datum at the tail water or downstream water this is delta h which is equal to h and this head is causing the flow to occur. So if I introduce a term delta h as number of drops this is going to be equal to delta h and what about the q? The q is taking place through number of channels. So what will happen to a number of channels? It gets multiplied by number of channels is it not? So what will be that term? This will be q the total discharge. So the delta h is the total head is getting divided by number of drops the total discharge is getting multiplied by total number of flow lines. Now this is what we are going to utilize in further analysis say q l and sin we will be using these terms. So if I say that the total discharge is equal to k into i into a normally what we do is we consider this as 1 into 1 which we have done and area of cross section is defined always as per unit third dimension of the element as if the flow is entering into the system. So that means a will become 1 and then this q is nothing but q into n f and then k into i h upon n d is it okay? Now I was just trying to show the equivalence over here. So let it be this is the equivalence. So basically what I am trying to show is when we talk about the total q value if I am dealing with a delta q function. So this function will get multiplied by n f. So that means the total q will be equal to this multiplied by n f is this okay? Because there are so many flow channels which are contributing to the discharge. So this concept I have used over let us say I think I will eliminate this step to not complicate things then it is alright? No no my idea was that delta phi is nothing but delta h function and delta psi is delta q function this is what I am trying to show. So basically if you use this function q which is the total discharge taking place through a system what I have to do is I have to plot the flow nets I have to compute what is h which is causing flow to occur. I should be knowing what are the number of drops of the potential which are occurring and what are the number of flow channels. This is what is going to give me the discharge per unit length per unit thickness of the element of the porous media. So remember the discharge units would be meter cube per second area of projection we have taken here as 1 into 1. So this is your discharge will be meter cube per second per unit length in the third dimension. Now if you have anisotropic soils this was for the isotropic situation you know whatever we had done. So here we are assuming that kx is not equal to kz is not equal to ky alright. But suppose if I still consider the two dimensional flow I can use the Darcy's law and I can say vx equal to minus kx yeah this is already written there. So I can use this function I can say vs vs equal to minus ks into del h upon del s. Now if I solve this function del h upon del s this will be equal to del h upon del x del x upon del s plus del h upon del z into del z upon del s. This can be shown to be equal to 1 upon ks if you substitute vs term vx term and vz term you can show this to be equal to cos square alpha upon kx plus sin square alpha upon kz. I hope you can realize that this is the equation of the ellipse alright. And if I ask you to draw the ellipse this is the x direction, z direction, this is the alpha term, this is under root of ks, this is under root of kz and this is under root of kx. This can be also written as s square upon ks equal to x square upon kx plus z square upon kz. Now there is an interesting way of depicting the anisotropy by using the Laplace equation which we had earlier assumed to be one dimensional flow and for isotropic conditions. So if I say that kx del square h upon del x square plus kz del square h upon del z square equal to 0 this is the equation for anisotropic condition. I can create isotropy by assuming that z is equivalent to z let us say t alright. In that case I can write del square h upon this plus if I say this is del square h upon del x square and this kz comes over here divided by kx is this ok. So this is equal to 0. If I assume that x t is equal to kz over kx into x can I replace this term with del square h upon del x t square plus del square h upon del z t square. So what I have done is anisotropic situation has been converted into isotropic situation by transformation. So the flow net which we have been talking about on x z plane suppose which was a non mean sorry which was a rectangle why because we have kz and kx this can be transformed to a scale x and z t. Now the moment I transform it over here what is going to happen this is going to be a square this is fine. Now this type of transformation is used normally to deal with the anisotropic situation. Here see what we have done here we have manipulated with the permeability coefficient. So kz upon kx is treated as del x t upon del x equal to kz upon kx. So this is the anisotropic coefficient I can also write x t upon so there is a mistake I have done. So this part would be under root because when you are doing a square term over here this should be the correct function is it not it is only this will be outside the under root because this whole thing is x square. So if you say del x t will be equal to under root of kz upon kx into del x I can define another term as k equivalent of the anisotropic system and that would be equal to suppose if I write that k equivalent would be under root kx into kz can you prove this again you go back to the basics. So vx is equal to minus kx into del x del h upon del x. Now if I replace this term by k effective or equivalent this will become del h upon del t del x t. Now this will be equal to minus k equivalent what will be del h upon del x t this will be under root kz upon kx into del x that means kx will be equal to k equivalent divided by under root kz by kx and this would give you the expression that k equivalent would be equal to under root kx into kz.