 Welcome to this lecture number 12. So, in which we will continue with our previous lecture article on steady flow through wells. So, in lecture number 11, we discussed the steady flow through a confined into a well through a confined aquifer. So, in which the confined aquifer thickness was B and the well was fully penetrating the entire thickness of the confined aquifer and the water table was horizontal before the pumping began which resulted in a steady flow into the aquifer into the well and the which also resulted in a steady discharge of Q from the well and we have got the expression for the discharge which is popularly known as the, yeah this is the expression and yeah the original expression in terms of the hydraulic conductivity is Q is equal to 2 pi K into B H 2 minus H 1 there by natural log of R 2 by R 1. So, this is popularly known as the Thames equation and this when we simplify as you applying the boundary conditions at the extreme points that is the extreme point wherein the there is no drawdown and the radius is equal to the radius of influence R capital R and H the head is equal to the total this head or the piezometric head and that the so this is the upstream most extreme and the downstream most extreme is the cylindrical wall where this S the drawdown is equal to the drawdown in the well equal to SW R 1 is equal to the radius of the well that is R W and H 1 the head is equal to the depth of the water in the well in the confined aquifer. So, we get this expression and these expressions can be used for with the in case of pumping test to determine the hydraulic the aquifer parameters such as the hydraulic conductivity and transmissivity. So, therefore, so these expressions can be used in steady flow pumping test for fully penetrating wells in confined aquifer for determining the hydraulic conductivity K and the transmissibility or transmissibility T of the aquifer. So, here we get an expression for the hydraulic conductivity K for the steady pumping test through the confined aquifer having a fully penetrating well as Q divided by 2 pi B into H 2 minus H 1 into natural log of R 2 by R 1 the same thing can also be written as Q divided by 2 pi B and in this case that is SW natural log of R divided by R W. So, here so this is the expression for K steady flow pumping test a fully penetrating well in a confined aquifer. Similarly, the same expression using these equations you can write down the expression for T as Q divided by 2 pi H 2 minus H 1 into natural log of R 2 by R 1 this can also be written as Q divided by 2 pi into SW into natural log of R the radius of influence divided by R W the weld radius. So, this is the expression for K, T. So, this is the expression for K and this is the expression for T. So, in a steady flow pumping test through a fully penetrating well in a confined aquifer. So, we need to know to determine this hydraulic conductivity K and transversivity T. We need to know the steady flow rate from this fully penetrating well in a confined aquifer. We need to know the thickness of the confined aquifer. We need to know either the depths between two the observation wells the depths of water in two observation wells that is H 1 and H 2 and their radii with respect to the radial distance with respect to the well axis the vertical well axis R 1 and R 2. And either we need to know these H 1, H 2, R 1, R 2 or we need to know the weld radius R W the drawdown in the weld SW and the radius of influence that is capital R. And so, in this case, so we can easily determine easily estimate the value of the two of the three aquifer parameters formation constants that is the hydraulic conductivity K and transversivity or transversibility T. Now, let us go to the case of steady flow a fully penetrating well in an unconfined aquifer. So far we have seen the steady flow through a fully penetrating well in a confined aquifer. And now, let us go to this steady flow through a fully penetrating well in an unconfined aquifer. So, for this the diagram would be a well which is fully penetrating through an unconfined aquifer. And in this case, this is the water surface that is the water table before the pumping started and after the pumping started. So, this water table will show a steady drawdown. So, this is the well axis and this is the steady discharge from this well in the unconfined aquifer. This is the lower impervious strata for the unconfined aquifer and this is the radius of the well is R W and the diameter is 2 R W. This is the radius of the circle influence or radius of influence radius of the circle of influence is R and the head there measured above the bottom impervious layer as capital H. And at any two points and here what happens is so this is the we need to apply the Dupitz assumptions and so this is the water table by Dupitz assumptions and this is the true water table. And here this is H 1 and the drawdown there is S 1. And similarly for the second observation well the head is H 2 and the drawdown there is S 2. And obviously here the Dupitz assumptions which we discussed in the previous lecture considered that is applied. So, the Dupitz assumptions accordingly that is say for small inclinations of free surface the streamlines can be approximated as approximated or assumed as horizontal. And second assumption is the hydraulic gradient is equal to slope of free surface is independent let me write here independent of depth. So, here the first assumption that is the streamlines are assumed to be horizontal. So, this is more or less true except close to the well where the streamlines are significantly deviating from this one but still with these assumptions that means the error in estimating the discharge is quite less. So therefore, because if we consider the inclination of the streamlines near to the well. So, then we have to modify our expression and for the discharge and it will make it more complicated it will make the expression more complicated. So, therefore, we are making the two Dupitz assumptions and so here the so let me show you the flow that is radially into the well and here this is hw and this is sw and at any general point this is h at a distance of r from the well axis the drawdown is s. Now, we will derive the expression for this steady flow. So, we know that the radially inward velocity vr is given by k into dh by dr this is by Darcy's law and this is by Darcy's law. So, therefore, the steady flow rate through the well is given by this vr into the area which contributes and that area which is contributing in this case will be equal to 2 pi r which is the circumference of the circle at that radial distance r multiplied by h which is the depth of water measured with respect to the horizontal impervious flow multiplied by this is the area multiplied by the velocity that is vr and this is equal to 2 pi r h into k into dh by dr. So, therefore, we can write this as q divided by 2 pi into k into dr by r this is equal to h dh. So, this expression needs to be integrated that is between the limits r 1 the lower limit r 1 and the upper limit r 2 and for the h the lower limit is h 1 the upper limit is h 2. So, therefore, we get q is equal to pi k h 2 square minus h 1 square divided by natural log of r 2 by r 1 and in this case. So, this is the expression for the flow rate. So, at the extreme points. So, here this case there is h 2 is equal to h r 2 is equal to r and h 1 is equal to h w r 1 is equal to r w. So, therefore, this expression can be rewritten as q is equal to pi into k into h square minus h w square divided by natural log of r divided by r w. So, these are the expressions for the steady flow rate in terms of the hydraulic conductivity k h which is the depth of water above the impervious layer in the unconfined aquifer h w is the depth of water in the well and r is the radius of influence r w is the well radius. And of course, if you express the same thing for two observation wells in that case. So, this is the expression two observation wells at a distance of r 1 and r 2 at a radial distance of r 1 and r 2 where the depth of water is h 1 and h 2. So, these are the this is the expression. Now, let us so in this case it should be noted that it is to be noted that time required for achieving a steady state is longer for unconfined aquifer. And so these two expressions let us write down the approximate forms approximate forms of equations for the steady flow rate q. So, when s w that is the draw down in the well is very small. So, in that case that is s w is equal to h minus h w is small relative to h. Therefore, h h square minus h w square you can write this as h plus h w into h minus h w. So, this h plus h w is approximately equal to 2 h and h minus h w is approximately equal to s w. So, therefore, the approximate expression for the steady flow rate is equal to 2 pi that is pi into k into 2 h s w divided by natural log of r divided by r w. And this can be written as 2 pi into t into s w divided by natural log of r divided by r w. So, these are the this is one approximate expression for q when the s w is very small. Similarly, this we can also write this q is approximately equal to 2 pi into t into s 1 minus s 2 divided by natural log of r 2 by r 1. So, in this case that is. So, here that is h 2 square minus h 1 square. So, here that is assuming h 2 square minus h 1 square is equal to h 2 plus h 1 into h 2 minus h 1 and this h 2 plus h 1 we can take this to be approximately equal to that is 2 t that is h into and this is s 2. So, this is s 2 h 2 minus h 1 is equal to s 2 minus 1. So, therefore, we get this expression that is. So, these 2 expressions will give the approximate expressions for the steady flow rate through this fully penetrating well in an unconfined aquifer. And obviously, they are almost the same as the expressions for steady flow through a fully penetrating well in a confined aquifer. So, now we have discussed the steady flow through a fully penetrating well in an unconfined aquifer as well as a fully penetrating well in a confined aquifer. And now in this case. So, here we can also write down. So, therefore, this t is approximately approximated as k into h 1 plus h 2 by 2. So, therefore, this k into h 1 plus h 2 is equal to 2 t. So, this is with this assumption. So, we get this approximate expression for this q. Now, let us consider the unconfined aquifer. It is unconfined aquifer with uniform recharge. So, in this previous 2 cases, we did not consider any recharge into the well. So, now let us consider the unconfined aquifer with uniform recharge. So, in this case that is the ground level and this is the well in the unconfined aquifer and this is the water table. And of course, so here we are getting the uniform recharge. And let us take this as W, recharge rate W. And this is the steady flow through this well is q w. And here, so this is the lower impermeable strata. Let us consider where in to the inflow is q and outflow is q plus dq. So, this dq is the one which gets contribution from this recharge, the constant recharge. In this case, let us consider the water table. Let us consider the steady height as say h 0 and the radius of influence as r 0. And here, so this radius is r and then for the elemental strip of radial distance dr. So, now we can write down for this case, we can write down the expression for the discharge dq is given by the area that is minus 2 pi r into dr into the uniform rate of recharge. So, if we integrate this one, we get that is q is equal to minus pi into r square into W plus the constant of integration. And the boundary conditions are at the well r tends to 0 and q tends to q w. Therefore, q w is equal to c. Therefore, we can write down this q is equal to minus pi r square into W plus q w. And if we substitute the expression for q which is minus 2 pi r into k into h dh by dr. So, here this k into dh by dr is the velocity as per Darcy's law and 2 pi r into h is the area which contributes flow. So, this is equal to minus pi r square into W plus q w. So, if you integrate this expression integrating applying the boundary condition, it is h is equal to h 0 the upstream most boundary condition and at r is equal to r 0. So, therefore, we can write down after integration, we will get an expression for the drawdown curve h 0 square minus h square will be equal to this is W divided by 2 k into r square minus r 0 square plus q w divided by pi k into natural log of r 0 by r. So, this is the expression for this drawdown curve in an unconfined aquifer uniform recharge. So, if we know the value of the depth of water above the lower impervious layer, the uniform rate of recharge hydraulic conductivity and the radius of influence r 0 q w which is the steady flow through this well fully penetrating the unconfined aquifer. So, in that case, so this will be the expression for the drawdown and we know that when r is equal to r 0 q is equal to 0. That means beyond this the cone of influence, there is no flow contribution. Therefore, substituting these values, we get q w is equal to pi into r 0 square into W. In this case, so that is so when r is equal to r 0, in that case, so this q is equal to 0. Therefore, if we apply this expression, so in that case, this q w is equal to pi into r 0 square into W. Therefore, so this is the expression for the steady flow through this well in an unconfined aquifer having a uniform recharge and this is the expression for steady flow a fully penetrating well in a confined aquifer, in an unconfined aquifer with uniform recharge. So, it is fairly simple. So, simply the area of the circle of influence multiplied by the uniform rate of recharge. So, that will be given by, that will give the expression for the steady flow through the well. And now, let us consider a well in a uniform flow in an unconfined aquifer. So, again this is a fully penetrating well in a uniform flow through. So far we have seen, we have considered the water table to be horizontal, so that only after the pumping starts, the well starts getting contribution from the aquifer. Now, instead of that, suppose we consider a well, which is drilled through an unconfined aquifer having a sloping water table below the sloping ground. In this case, let us consider, so this is the sloping water table, so with slope of I and let me denote the well axis here, the uniform discharge through this well. And here what happens is, after the pumping starts, then the water table will show a drawdown curve like this, so this is the drawdown curve. And in this case, obviously here, so up to this point, only there will be contribution to the well. And obviously, on this right side, it is fully flowing this one. And if you draw the, so in this case, let us, I am sorry, so this is let us say, this is a confined aquifer. And here, this is the top confining layer and the confined aquifer thickness is B. Now, in let me, so this represents the stagnation, this vertical line represents stagnation. And let us also draw the top view for this, if this is the, let me draw the top view. In this case, this is the well, so here, so this represents the stagnation surface, within this, so these are the stream lines. And here, at any, let us consider this as plus y l, let us consider this as, I am sorry, so this plus y l goes all the way up to here, this is minus y l. And this is, this distance is say, minus x l and obviously, so this is the x and y coordinate. And beyond this, the flow lines will be like this. So, now, for this case, so there is stagnation surface, because of the original slope, that is the original hydraulic gradient due to an inclined water table. So, in this case, we can write down the expression for the hydraulic conductivity as 2 q, k is equal to 2 q divided by pi into r into h u plus h d into i u plus i d. And here, this h u is the upstream saturated thickness, h d is the downstream saturated thickness and this i u is the upstream water table slope and i d, this is the downstream water table slope. By superimposing radial flow and 1 d flow, we get, we can write down the expression for this stagnation surface, that is y by minus y by x is equal to tan of 2 pi k B i divided by q into y. So, this is the expression. So, this B is the saturated, that is the aquifer thickness and q is a steady discharge and of course, I am not going into the variation of this. So, therefore, and this y l is given by plus or minus q divided by 2 k B i. So, this is the distance of the stagnation surface at a distance and x l is given by minus q divided by 2 pi k B i. So, we will stop here and we will continue in the next lecture. Thank you. Bye.