 11 in which we will continue with the 1D ground water flow in an unconfined aquifer. And firstly let us consider the case of a 1D ground water unconfined ground water flow between two water bodies having a discharge with a discharge in the vertically downward direction and here we know that when there is a 1D ground water flow the obviously the flow is in the x direction and in this case the governing equation will be d square H by dx square and since the flow is only in the x direction. So, therefore, the second order partial derivative terms in the y and z direction they will not be there and this will be equal to minus 2 R by k where R is the rate of recharge and this is in the vertically downward direction and k is the aquifer permeability or the hydraulic conductivity. So, this is the governing equation and so, this equation if we I am sorry so, I made a small one. So, this is a second partial derivative of H square. So, this is d square by dx square of H square is equal to minus 2 R by k. So, now, let us integrate this governing equation twice. So, we get H square will be equal to minus R by k x square plus c 1 x plus c 2. So, here the c 1 and c 2 are the constants of integration. Now, let us supply the boundary conditions. So, the boundary conditions so, at the upstream boundary upstream boundaries x is equal to 0, H is equal to H 0. Therefore, H 0 is equal to c 2. So, this is the first boundary condition. The second boundary condition that is the downstream boundary at the downstream boundary we have x is equal to L and at that location the head H is equal to H 1 which is the head the ground water head in the downstream water body. So, now so, therefore, here so, we get this H 1 square will be equal to minus R by k L square plus c 1 into L and c 2 is equal to H 0. So, therefore, here you can write down. So, this c 1 is equal to minus H 0 square minus H 1 square minus R L square by k. So, this whole thing divided by L. So, therefore, if we call this equation 1 therefore, equation 1 becomes H square is equal to minus R x square by 2 or R by k x square minus of H 0 square minus H 1 square minus R by k into L square this divided by L this into x plus H 0 square. So, this is the expression for the square of the head the ground water head in an unconfined aquifer having one D flow between two water bodies the upstream water body having a head of H 0 and the downstream water body having a head of H 1 and there is also a vertically downward recharge in the vertically downward direction with an intensity of R and the aquifer is a homogeneous isotropic with a hydraulic conductivity of k the distance between the upstream and downstream aquifers is L. Now, I am sorry the upstream and downstream water bodies is L. So, here as you can see from the figure due to this constant or uniform rate of recharge over in the vertically downward direction the water table shows a hump somewhere in the middle and where the this H value is a maximum equal to H max and so from there the water table dips slightly towards upstream water body to the left and then more towards the downstream water body to the right of this and this section is known as the water divide. So, this is known as the water divide wherein this H is the maximum the ground water head is maximum and the to the left of that the water will be flowing towards the upstream water body and to the right of that the ground water will be flowing towards the downstream water body and now let us and this the water divide. So, let us denote the distance of this water divide at x is equal to a from the upstream water body. Now, let us find out an expression for this the distance a of the water divide from the upstream water body and we know that the expression is. So, this is the expression and if we take the derivative of H with respect to x and equate that derivative to 0. So, that will give the condition for maxima that is H is equal to H max. So, we will get the expression for. So, we know that at x is equal to a H is equal to H max and d H by d x is equal to 0. So, this is the condition for maxima. So, therefore, let us differentiate this expression. So, that is 2 H into d H by d x this is equal to 0. So, this and here let us substitute because this is d H by d x which is 0. So, therefore, we are equating it to 0 and in the right hand side and the in place of x we have to substitute the value of a. So, therefore, here on the right hand side we get minus R by k into a square minus this is H 0 square minus H 1 square minus R by k into L square this divided by L into a plus H 0 square. So, therefore, so we get. So, this is R by k a square which is equal to H 0 square minus H 0 square minus H 1 square minus R by k L square into divided by L into a. So, if you substitute this if we simplify this after simplifying. So, here we will get the we get and obviously, so here we can cancel out one a here and so when we are cancelling out this a. So, obviously, here there will be a here. So, after simplifying we get this a is equal to L by 2 minus k by R into H 0 square minus H 1 square divided by 2 L. So, this is the expression for expression for the distance of water divide from the upstream water body. So, now, we will get the what we will do is let us find out the expressions for this the flow from the water divide to the left of the water divide as well as to the right of the water divide. And obviously, at the water divide there is no flow. Now, let us write down. So, this is the flow per unit width. So, flow per unit width that is q x is given by minus k into H into d H by d x and in this case. So, this is minus k and so H into d H by d x is given by that is minus R x by k minus. So, here we get this is a H 0 square minus H 1 square minus R L square by k this whole thing divided by 2 L. So, this is the expression for q x and here in this case we are taking the here is one the substituting the value of H into d H by d x and multiplying it with minus k. So, we get this expression. So, now, if we simplify this. So, we get this q x will be equal to R into x minus L by 2 plus k divided by 2 L into H 0 square minus H 1 square. So, this is the expression for q x. Now, to get the upstream and the downstream the flow per unit width into the upstream water body and the downstream water body, we need to substitute x is equal to 0 to get the flow per unit width into the upstream water body. And we need to get x is equal to L we need to substitute x is equal to L to get the expression for the flow per unit width into the downstream water body. So, here you can write down this is the expression for flow per unit width in 1 day unconfined flow, unconfined steady flow steady ground water flow between two water bodies. So now, for the upstream water body q is equal to q 0 in which is simply given by substituting x is equal to 0 in this expression. So, therefore, this q 0 is given by minus R L by 2 plus k by 2 L into h 0 square minus h 1 square. Similarly, for the downstream water body, so for the upstream water body x is equal to 0 and for the downstream water body x is equal to L. So, therefore, this q x in this case q x is equal to q L which is equal to this is R into L by 2 here x in place of x we are substituting L R into L by 2 plus k by 2 L into h 0 square minus h 1 square. So, therefore, this can be expressed as this q L is equal to q 0 plus R L. That means, the flow into the downstream water body obviously, it has to be more because there is more gradient between the water divide and the downstream water body. So, that is obtained by adding this the term the product of R into L the recharge intensity as well as the length the distance between the two water body upstream and downstream water bodies to the flow per unit width into the upstream water body. So, therefore, so here we get this is the so this q 0. So, this is the expression for expression for flow per unit width into upstream water body and obviously, this is for say 1 D confined steady groundwater flow and this is the expression for flow per unit width into the downstream water body and obviously, this is also for 1 D unconfined steady groundwater flow. So, this is the expression for so now, we have discussed this 1 D groundwater flow without a with recharge. Now, let us come to this another case that is the 1 D unconfined steady groundwater flow without recharge. So, here we know that so in this case. So, let me draw the figure here in this case, this is the bottom confining layer for the unconfined aquifer and then there is an upstream water body and then this is the general ground here and then there is a downstream water body. So, this upstream water body has a head of H 0. So, this is the upstream water body and then this is the downstream water body has a head of H 1. So, this is the downstream water body and here this is the ground level and obviously, since there is no recharge. So, the water table will simply will assume a curve curved shape and here x is equal to 0 and at the downstream and x is equal to L. So, this is the distance L between the upstream and downstream water bodies and here so at any general point. So, this is H and obviously, let us say at this general section x, the discharge per unit to this cube and this aquifer permeability. So, permeability aquifer permeability is equal to k. So, now here in this case the expression that is in this case the governing equation will be simply that is d square H square by dx square will be equal to 0. Therefore, if you differentiate this one twice or I am sorry if you integrate this twice. So, after integrating we get H square is equal to c 1 x plus c 2 and again the conditions are there the boundary conditions are at x is equal to 0, H is equal to H 0 and at x is equal to L, H is equal to H 1. So, here so therefore, so this is H 0 square is equal to c 2 and H 1 square is equal to c 1 L plus c 2 c 2 in this case is H 0 square. Therefore, c 1 is equal to H 1 square minus H 0 square divided by L which you can write this as minus H 0 square minus H 1 square divided by L. So, therefore, so this equation if I call this as if I call this as equation say 3. So, this equation 3 becomes so H square is equal to minus of H 0 square minus H 1 square divided by L into x plus H 0 square or in other words H square minus H 0 square is equal to H 1 square minus H 0 square divided by L into x. So, this is the so this is the expression for hydraulic grade line for 1 D unconfined steady ground water flow between 2 water bodies without recharge. So, it gets simplified. So, therefore, and we can also write if you differentiate this so the we get this 2 H into D H by D x and so that can be half of that term can be used can be multiplied with k. So, we will get the discharge per unit length unit width. So, the discharge per unit width so that is q. So, this is equal to minus k H D H by D x. So, this is equal to minus k and H into D H by D x is simply given by it is H 1 square minus H 0 square divided by 2 L into so that is the expression for this H into D H by D x. So, therefore, the discharge per unit width so this is the expression for this is the expression for the discharge per unit width. So, for 1 D unconfined steady ground water flow between 2 water bodies without recharge. So, therefore, so thus we have come to the end of the second chapter that is an occurrence and the movement of ground water. Now, we will go to the third chapter that is the advanced well hydraulics and in this case specifically we will discuss with the initially we will start with the flow through the wells and that to initially the steady flow through the wells. So, we are entering the third chapter that is the well advanced well hydraulics. So, this is chapter 3 of this NPTEL video course on ground water hydrology. So, this chapter 3 is on advanced well hydraulics and specifically we will start with the steady flow through wells steady flow into the wells. So, firstly we will consider the confined aquifer wells in the confined aquifer and then we will go for the wells in the unconfined aquifer and firstly let me start with the basic figure of the diagrammatic representation of a well which is used for extracting ground water. So, this is the well and let us consider this well to be fully penetrating an aquifer. So, this aquifer may be a confined aquifer or an unconfined aquifer and in this case this is the ground level and here what happens is. So, this is the water table the original water table. So, this is the original water table which is horizontal in this case and then because of the pumping. So, this water table will show a depression like this and this is known as the cone of depression. So, this is the cone of depression. So, this is the steady rate of pumping from this well and obviously the flow is radially there is a radially symmetrical flow into this well and the diameter of this well. Let us consider the diameter to be 2 R w and here. So, this is the water surface and this curve is known as the drawdown curve. Obviously, there is no pumping. So, the original or the static water table is a horizontal surface with no drawdown and no cone of depression and in this case suppose we have an observation well. Let us say this is one observation well and this is another observation well which were. So, in this case the drawdown in this observation well that is basically the head difference between the original water table and the depressed water table. So, this S is known as the drawdown and the maximum radius up to which this drawdown is felt. So, this is from the axis of the well is known as the radius of influence. In this case here say we may obviously. So, these are the strainers basically well casing with perforations. So, through which the ground water enters radially in a radially involved flow and obviously. So, this is axisymmetric flow. So, it is symmetrical in all the directions because we are assuming the aquifer table to be homogeneous and isotropic and in this case of course I have considered this as an unconfined aquifer and same thing can also be shown as a confined aquifer. So, this is the original this is the water table that is W t and so these are the terminology here R W the well radius and this radius of influence. So, this is generally denoted by R and this is the impervious boundary. In this case the bottom impervious boundary of the aquifer and then this is the drawdown. This is the drawdown curve and then this is the cone of depression and this is the pumping well with a steady pumping rate of Q and these two are the observation wells. So, these are the observation wells and this is the pumping well it is also known as a discharge well. So, this is the typical sketch indicating all the terminology and let us consider first the steady flow into a well fully penetrating a confined aquifer. Let us consider steady confined ground water flow into a fully penetrating well. In this case the well is penetrating for the entire depth of the confined aquifer. So, let us draw the sketch here. So, this is the bottom impervious layer of the confined aquifer and here so this is the well. In this case let us say this is the top confining layer of this confined aquifer and obviously, so here there are strainers basically well casing with perforations and so this is the confined aquifer and here the total let us consider this is the ground level and here let us consider the so this is the water table. So, this is the water table with the drawdown and the total height of the water table with respect to the bottom confining layer of this aquifer is H and here the flow is radially inward flow and by Dupit's assumptions we are assuming the stream lines to be horizontal and this well which is fully penetrating the confined aquifer has a diameter of 2rw and this is the radius of the influence is r and at any two general points so the drawdowns are s1 and this is h1 and then say for the second so this is s2 and this is h2 and obviously, at the well so this is hw and this drawdown here is sw and at any general point the drawdown is s and the variable head is denoted by small h and this is the pumping well with discharge that is q. So, this is a fully penetrating well through a confined aquifer and then there is a steady ground water flow into this confined aquifer and obviously, here this s1 so this is at a radius of r1 and similarly, this s2 the drawdown s2 is observed at an observation well which is at a radial distance of r2 and let the thickness of this confined aquifer be b and obviously, the this one that is so the radial direction is the axis is one is radially there is a radially inward flow throughout and so these are the so this is the drawdown curve which represents a piezometric surface. So, here this is the drawdown curve that is piezometric surface under pumping so now, for such a well which is having which is receiving a steady flow which is radially inward flow and then the flow is axis symmetric so the flow is symmetrical about the vertical axis of this fully penetrating well. So, now, let us write down the expression for the radial velocity vr and this is simply given by the Darcy's expression by Darcy's law radial flow velocity that is vr is simply given by k into dh by dr k is the aquifer permeability and then dh by dr is the hydraulic gradient. And so this is the expression for the velocity and if you multiply this by the area of flow so in this we will get the expression for the steady rate of pumping so this q is simply given by 2 pi r into b so this is the area of flow for the radially inward flow into k into dh by dr so this is the area multiplied by velocity that will give the expression for the steady flow rate into this well. So, therefore, let us rewrite these terms so this is q divided by 2 pi kb into dr by r this is equal to dh and this we need to integrate between the appropriate limits so in this case the limits are between say r 1 and r 2 the lower limit is r 1 the upper limit is r 2 and then similarly for the right hand side the lower limit is h 1 and the upper limit is h 2. So, in this case we get so this is q divided by 2 pi kb into natural log of r 2 by r 1 so this is equal to h 2 minus h 1 so if we rewrite this one so this the expression for the steady flow rate is given by this is 2 pi kb into h 2 minus h 1 divided by natural log of r 2 by r 1 so this is the expression for steady flow through a fully penetrating steady fully penetrating well in a confined aquifer and this is generally known as the Themes equation after the hydraulic hydrolysian theme who initially proposed this equation. So, and obviously so this k into this b so this can be replaced by this transmissivity T. So, if we know the transmissivity and then the depths of the water table above the lower confining layer in two observation wells and also the radial distance from the center of the vertical well access. So, then we can determine the expression for this steady flow through the fully penetrating well this will also be equal to the steady discharge which is given out by this which this well is giving out so in that equilibrium condition the amount of the inflow which is through the radially invert direction that is equal to the pumpage the rate of pumping through this well. And so in this case so this h 1 the head at the observation well the piezometric head at the observation well is simply given by h minus s 1 and h 2 is equal to h minus s 2. So, therefore, and this k into b is equal to T. So, therefore, we can write down this q is equal to 2 pie into the transmissivity or transmissibility multiplied by so this is the s 1 minus s 2 divided by natural log of r 2 by r 1. So, this is another expression for the steady flow into a fully penetrating well through a confined aquifer. And suppose we consider the extreme points that means so we consider the downstream extreme point which is the radial surface the cylindrical curved cylindrical surface of the well and the upstream extreme point which represents the point on the circle of influence which is at a radial distance of r the radius of influence. So, in that case so we get at the edge at the extreme points this the drawdown s is equal to 0 and r 2 is equal to r which is the radius of influence and h 2 is equal to h at the cylindrical wall of the well this s is equal to s w r 1 is equal to r w and h 1 is equal to h w. So, therefore, we can write down this q is equal to 2 pie into t into s w divided by natural log of r divided by r w. So, this is another expression wherein if we know the drawdown in the well the radius of influence as well as the well radius. So, then we can and of course the transmissivity we can get the expression for the discharge through the steady flow rate through this fully penetrating well in a confined aquifer which is the same as the rate at which the water is pumped out through this well because the flow is steady obviously in flow is equal to outflow. So, in the next class we will continue with the steady flow through an unconfined through a fully penetrating well in an unconfined aquifer. Thank you.