 Welcome to this lecture number 9, which is in which we will continue with our previous class, previous lecture as for that is on heterogeneity, we will take a numerical problem in this heterogeneity and anisotropy in the aquifers. And let us consider say for example, a stratified aquifer, a numerical problem of a stratified aquifer of say thickness 12 meters. So, this total thickness is say 12 meters and it consists of 3 layers as shown here and this is the flow direction, the flow is along the layers or stratifications or strata and the bottom layer has say this is the bottom layer, it has k value of 30 meter per day and it has a thickness of say 5 meters and the middle layer it has middle and top layers have permeability of say 20 meter per day, this top layer also has a permeability of 20 meter per day. Now, I am sorry top has let us consider take it as say 45 meter per day and are of equal thickness. So, obviously, here if you direct this 5 meter thickness of the bottom layer from the total thickness of 12 meter. So, we are left with 7 meters and that 7 meters is equally distributed between the top layer and the bottom layer, top layer and the middle layer. So, that is 3.5 meter is the middle layer and so is the thickness of top layer which is 3.5 meter. So, the now we know the and the flow is along the stratifications or layers or strata and for this let us say find out what will be the equivalent permeability as well as what will be the transmissivity So, now let us come to the equivalent permeability. So, K e in this case is given by summation K i B i from 1 to n divided by summation B i. So, in this case we have to sum there are 3 layers. So, there will be 3 terms in the numerator. So, that is 45 into 3.5 plus 20 into 3.5 plus 30 into 5. So, this is in the numerator and in the denominator the total thickness which is say 12 meters. So, this will be the equivalent permeability and in this case. So, this equivalent permeability K e will be simply that is 45, 65 into 3.5. So, that is here I am computing here. So, this is 65 into 3 into 3.5 plus 150 which is 377.5 divided by 12 which is 31.46 meter per day. So, this is the equivalent permeability and let us also determine the estimate the transmissivity or it is also known as transmissibility. So, this is simply equal to K e into sigma B i or which is simply equal to this is summation 1 to n K i into B i. So, this is as will be the numerator of the 1. So, this is 377.5 meter square per day. So, this is the value of transmissibility. So, here I was mentioning here. So, there were 3 layers the top layer and middle layer having 3.5 meter thickness each and bottom layer having 5 meter thickness each with the hydraulic conductivity or coefficient of permeability of 45 meter per day, 20 meter per day and 30 meter per day respectively. So, in this case and the flow is along the stratifications. So, in this case it is simple we just apply the formula and get the equivalent permeability which is 31.46 meter per day and the transmissivity for this entire aquifer consisting of 3 strata is 377.5 meter square per day. So, this is just an illustration of one simple numerical example consisting of an anisotropic aquifer confined aquifer with 3 layers or 3 stratifications and flow along the. So, now let us come to the ground water flow rates flow directions of course. So, this ground water we have we have been abbreviating as G w. So, we know that with by Darcy's expression the ground water flow velocity is given by the hydraulic conductivity K into I and here and the ground water flow rate obviously is given by the velocity into the area flow which is K into I into A. So, accordingly the we can if we can estimate all these 3 parameters the hydraulic conductivity K, the hydraulic gradient I and the area of cross sectional area of flow for the ground water. So, if we estimate each of these parameters then the product of these 3 estimated parameters will give us the estimated ground water flow rate. Now let us consider say for example, a stream or a river which is flowing above an alluvium something like this and here let us say this is the water table this is the stream. Let us say this. So, this is the alluvium which is essentially an aquifer and say let us say the width of this alluvium is something of the order of say 1000 meters and the average depth of this alluvium. So, this is the average width and the average depth say that is say of the order of say something like say 50 meters. Then in that case our cross sectional area A will be 1000 into 50 which is 50000 meter square and if we multiply this by the ground water flow velocity given by the Darcy's law which is K into I and suppose the hydraulic gradient is say the hydraulic gradient is say 10 meters or say maybe 5 meters say let us take this to be 8 meter in 1000 meter length. So, this will be 8 divided by 1000 which is 0.008 and obviously that is hydraulic gradient and then the coefficient of permeability if we take this as say something like say 100 meter per day then in that case simply these are the estimations permeability or hydraulic conductivity is estimated as 100 meter per day and the hydraulic gradient is estimated as 0.008 in the area again this is also an estimation only that is why I am using this approximately equal to sign because this is based on average width and then average depth. So, in that case so now we can write down this Q is equal to 100 into 0.008 into 50,000. So, this will be 50,000 into 0.8 so this is of the order of say 40000 meter cube per day. So, this is our estimated value of ground water flow this is just to give an idea. So, basically here we need to estimate each of the 3 parameters the first one is this the area of cross section of flow area of flow cross section for the ground water the second one is the hydraulic conductivity K and the third one is the hydraulic gradient I. So, these estimates should be as realistic as possible and based on that once we estimate each of these 3 parameters then simply take the product of this. So, that will give us the estimation. So, that is why let me use the approximate equal to sign here. So, this alluvium with say these dimensions average depth of 50 meters and average width of 1000 meters which is essentially the aquifer below the stream. So, we will yield a ground water of approximately 40,000 meter cube per day when the estimated hydraulic conductivity is 100 meter per day through this the material of this aquifer or alluvium constituting this aquifer or alluvium as well as the hydraulic gradient which is estimated as 8 meter and 1000 meters length along the flow. So, this is just to give one idea of is estimating the ground water flow rates now let us come to this ground water flow directions. So, here so we need to go by what is known as the flow net which is essentially a network consisting of 2 sets of orthogonal lines the stream lines shown by the green color one and the equipotential lines let me show them by the. So, these are the equipotential lines and these are the stream lines or the flow lines also known as flow lines now let us consider the stream wise let us partition this flow net into square grades and say let us say the total flow between 2 neighboring stream lines the total flow between 2 neighboring stream lines this be d cube and here also this is the flow is d cube between the bottom and the middle stream line as well as the middle and the top stream line and let us write down the head values or the total hydraulic head value along each of the equipotential line. So, let this be h here and obviously due to friction and other losses there will be a continuous head loss. So, for the next along the next equipotential line let the head be h minus d h and in this case it can be h minus 2 d h. So, now and also let us consider the flow this stream wise length as d s and let us consider the length that is the d s is the distance between 2 adjacent equipotential lines which is by along the stream wise that is why it is denoted by d s and this the perpendicular of the meridional. So, which is which represents the distance between 2 neighboring stream lines let this be d m. Now, we know that by the definition of this hydraulic gradient. So, this is d h by d s basically the change in the head per unit displacement along the stream lines. Therefore, here if we consider unit width perpendicular to the paper. So, therefore, this total discharge q is simply given by the hydraulic conductivity k into the hydraulic gradient d h by d s. So, this is the velocity multiplied by the area. So, this is simply given by 1 into d m. So, this is the area. So, therefore, this the ground water flow rate and here because since the grid is almost a square grid. So, we can say this d s is approximately equal to d m that is a stream wise distance is approximately equal to the perpendicular distance along the flow net. So, therefore, we can write down the expression for q is k into d h and suppose this d h the total head h which is there in the along the upstream most equipotential line. If this is and along the downstream most point if the head be 0 it is the entire head is lost due to friction and other losses in say n along n such equipotential lines. So, in that case. So, this d h can be approximated as h by n where n is the number of segments of equipotential lines. Therefore, q which is the flow per unit width this is flow rate per unit width is given by k into h by n and the total flow rate for this we have to if the flow is divided into m channels m channels m number of channels in the stream wise direction by stream lines then total flow. So, that is q is equal to m into small q which is simply equal to m into k h by n. So, this is the expression for the total flow rate. So, here m is the number of channels along the stream wise direction and the total flow rate k is the hydraulic conductivity h is the total head in the upstream most along the upstream most equipotential line n is the number of segments in the along the of equipotential lines. So, in that case the total flow is simply given by this m k into h divided by n. So, this is how we can estimate the total flow and also we can estimate say using the direction. So, we can estimate the ground water flow direction also. So, like this we can estimate the flow rate as well as the flow direction. Now, let us consider an anisotropic case wherein say consider anisotropic aquifer. So, here which is which has in the x z plane with hydraulic conductivity is k x and k z along x and z directions and typically in case of the alluvium or aquifers. So, this k x will be greater than k z. So, the equivalent hydraulic conductivity. So, this is k e is can be approximated as the square root of the product that is k x k z and say in such case because of this anisotropy which results in different values of the parameters the ground water flow parameters such as hydraulic conductivity and other parameters like transmissivity and so on. There is aquifer thickness in the either the geometric parameters or kinematic parameters. So, what happens is we need to estimate the flow rate through such anisotropic aquifers by assuming by estimating the equivalent hydraulic conductivity and in such case the flow lines that is the stream lines will not be fully perpendicular to the equipotential lines because of anisotropy. And in such case if we transform this into anisotropic condition equivalent isotropic condition in that case only the flow lines will be perpendicular to equipotential lines. So, here let us say the flow net analysis of for an earth ram let us consider say for example, this is the and in this case suppose these are the flow lines and the equipotential lines because of anisotropy they will not be intersecting orthogonally and in this case say this is the total length L. So, here let me write this is the this is a true anisotropic section with say K x is equal to 9 K z. So, this is the downstream and this is the x z plane this is the x direction and then this is the z direction. So, this is figure A which represents the true anisotropic section with say K x approximately equal to 9 times K z then so, this can be transformed into an equivalent transform the isotropic section for transforming this into so, this is B is the transformed isotropic section. So, in this case so, this length the length L will be so, this is a L dash. So, this L dash will be equal to L into under square root K z by K x in this case this will be L into under square root K z by K x. So, this L dash is simply equal to L by 3. So, this L dash is L by 3 and in such case for this transformed isotropic section in this case. So, this distance will be m whereas, in this case this distance of a grid so, that will be 3 m. So, it is only in case of this because of this anisotropy the stream lines as well as equipotential lines they will not be perpendicular to each other they will have some one and only in case of a transformed isotropic section. So, it will have a transformed length which is given by that is L dash will be equal to L by 3 in this case. And similarly in this case since suppose we are since we are taking this K x is approximately equal to 9 times K z. So, therefore, here we can say this K e is approximately equal to 9 K z into K z which is 9 K z square under square root. So, this is 3 K z. So, this is the equivalent is equal to K e. Therefore, so for this anisotropic aquifer with varying hydraulic conductivity values along x and z directions. So, the equivalent hydraulic conductivity is given by 3 times the hydraulic conductivity along z direction. And the base width of the transformed isotropic section through which the flow ground water flow takes place will be given by L by 3. And so this is corresponding to the expression this L dash is equal to L into under square root K z by K x. So, like this the and say let us consider another case wherein there is a wear or a barrage with say some kind of upstream head. And of course, wear or a barrage. So, let us consider this to be almost impervious. And in this case suppose let us consider this as the channel. In this case what happens is the channel is pervious whereas, this is impervious dike or barrage or wear. So, in this case the stream lines will be and let us consider this also to be the impervious sheet pile. So, in such a case the stream lines will take the shape if the downstream water level. So, in this case the stream lines will be having stream lines of the flow lines. So, these are the flow lines and we can construct the. So, if this is the head difference between the upstream and downstream. So, if we divide this head difference into. So, in this case there are 1, 2, 3, 4, 5, 6, 7, 7 stream lines are there. And if we divide this into say 7 it is let us I am sorry that is a this is not the downstream. So, this is the downstream water level. And if we so this difference between the upstream water level and the downstream water level if we divide this into 7 divisions. So, here we get our. So, this will be the downstream most equipotential line. Then followed by the next equipotential line will be like this the third equipotential line like this and the fourth equipotential line and the fifth equipotential line the next equipotential line. So, to get the to know the flow direction at any particular location we need to find use the concept of flow net. And we need to construct the flow net. So, therefore, the stream line or the flow line at any particular location will give the ground water flow direction at that particular location. And so, this is regarding the now let us consider let us go to the general flow equations through porous media. So, here so let us consider a two dimensional flow initially and again later on let us extend it into a three dimensional flow domain. So, in this case this is the flow domain two dimensional flow domain. And then so these are the stream lines shown by this green color. And let considering the unit width let the flow across the this boundary of the square flow domain the q x i. So, i standing for inlet. Input inlet section and similarly the flow here per unit width let this be q x comma o and then similarly along the bottom boundary. So, let the flow be q y comma i and then so this is the the inlet and then so here let this be q y comma o. And let the dimensions let us since we are considering this to be a square grid. So, let us assume the dimensions of each side of this flow domain as w. And in this case we know that the Darcy's expression holds good that is the ground water flow velocity is given by minus k into d the partial derivative of h with respect to s. This h is the head and s represents the displacement along the flow line or stream line. Now, let us consider let this for this flow domain let t x comma t y be the transmissivities in x and y direction. So, therefore, this q x i we can write this as q x i can be written as minus t x which is a transmissivity or transmissivity. In the x direction into the width of the flow domain that is w and we are considering unit thickness. So, the area will be w into 1 into d h by d x in the i direction. Similarly, q x the flow the outflow through the right boundary which is denoted by q x o is given by minus t x which is transmissivity along x direction into w into d h by d x at this point. This outlet section now let us take down the change in the flow rate change in flow rate along x direction and y direction. So, similarly here we can also write down the expressions for q y i and q y o. Let me write here. So, this is q y i can be written as minus t y into w into d h by d y i and so q y o is equal to minus t y w d h by d y o. So, therefore, the total change in flow rate along x and x direction and y direction. So, this is basically this is essentially that is the flow rate. So, this is simply given by q x i minus q x o. So, this is the change in the flow rate along x direction plus q y i minus q y o and this must be equal to this can be expressed in terms of the storage coefficient s and multiplied by the area in this case this will be w square. This will be the cross sectional section and multiplied by the rate of change of h with time the partial derivative of h with time. So, therefore, so we can write down the minus t x into w into w into w into w d h by d x i minus d h by d x o. Let us divide the whole thing this is d x into w minus t y into w d h by d y i minus d h by d y o. This must be equal to minus d h by d y i minus s into w square into d h by d t. So, we can write down we can simplify this as t x into d h by d x i minus d h by d x o divided by w minus t y into d h by d y into w into w by i minus d h by d y o divided by w. So, this is equal to minus s into d h by d t. So, here in this when the when this w is extremely small or infinite extremely small in this case so this the term under the square bracket under the square brackets can be approximated as the second partial derivative of h with respect to x and second partial derivative of h with respect to y. So, therefore, so the terms in the square bracket can be approximated as second partial derivatives of h with respect to x and y. So, therefore, suppose if we call this as equation 1. So, this equation 1 becomes t x into d square h by d x square plus t y into d square h by d y square this is equal to the storativity s into d h by d t. And this t x and t y. So, we can can be simplified as k x into w k x into say b which is the b x you can say b into d square h by d x square plus this t y can be substituted as k y into b where b is the saturated aquifer thickness into d square h by d y square this is equal to s into d h by d t. Or the same term can be substituted as k y into b where b is the this. So, therefore, if this is equation 2 and then this is equation this is equation 2 and then this is equation 3. So, equation 2 can be simplified as d square h by d x square plus d square h by d y square which is equal to s by t into d h by d t. So, this equation 2 here this is for a for an isotropic aquifer. So, in this case this is t is equal to say t x is equal to t y. So, this is the basic ground water flow equation in the two dimensions and of course, if you add the third dimension also. So, in that case. So, this is the basic 2 D ground water flow equation. So, in the next lecture we will discuss we will extend this two dimensional ground water flow equation to three dimensional one and also we will see the its variation in when the flow is steady in this one. Thank you.