 In the previous lecture we had seen flow through confined aquifers in which there is an impervious layer at the bottom and also bounded by an impervious layer at the top. So, what it means is that the thickness of the aquifer remains constant or in other words it does not depend on the head. The piezometric head will vary if we take water out of the aquifer then it will typically decrease towards the well, but the thickness of the aquifer remains constant and therefore, the flow area remains constant. Compared to this the unconfined aquifer does not have an impervious confining layer at the top. So, if you look at an unconfined aquifer there is a layer at the bottom which is impervious and then the ground level the unconfined aquifer the water table will the ground water table would be open and therefore, the pressure on the top of this will be 0 or atmospheric. So, once we start pumping an unconfined aquifer with some discharge queue the water table will go down and it will keep on going down as we keep on pumping till it reaches some steady state conditions. And what we can notice from here is that the area of flow earlier when we did not have any pumping the area of flow is the whole thickness of the aquifer, but as the cone of depression increases the area of flow becomes smaller. So, the height h will represent the area of flow therefore, the governing equations will be slightly different compared to the case of the confined aquifer. So, let us look at the governing equation for an unconfined aquifer typically the let us say that this is the ground water table which is having certain slope that slope depends on what is the pumping rate higher the pumping rate more will be the slope. This is the height h and suppose we take an element which is delta x in the x direction and delta y in the y direction. There are some assumptions which we have to make in order to be able to solve this first of all these assumptions were suggested by Dupouth's and therefore, they are called Dupouth's assumptions which enable us to solve unconfined aquifer flow problems. The assumptions are two the first assumption is that the curvature of stream lines is small. So, if you look at the stream line pattern in the unconfined aquifer case they may look like this, but the first assumption says that the stream line curvature can be taken small and therefore, the flow will essentially be horizontal. The second assumption is that the hydraulic gradient can be taken as a slope of the free surface and the hydraulic gradient will remain constant over height. So, it will not be a function of z where z is the vertical coordinate. So, if we take an element like this element the hydraulic gradient for this whole depth would be given by the slope of the water table and it will not change with depth. So, using these two assumptions we can now write the mass balance for this element. So, we are drawing that element of the unconfined aquifer here which has certain velocity here q x the Darcy velocity which is coming in and therefore, we can write the mass which is coming in through this phase. So, let us first compare the x direction in flow and out flow the axis of course, are x y and z. So, in flow in the x direction would occur through this phase the Darcy velocity is q x multiplied by the area which will be h delta y. This gives us the amount of flow in terms of volume and since we are interested in the mass we will multiply it with the mass density rho. So, rho of q x h delta y will be the mass of flow coming in from the left hand phase. There is some mass which is going out in the x direction from the right hand phase. So, we can write that out flow as using the methodology as we used before plus change of this quantity multiplied by the length delta x. So, if we consider the x direction flow from both phases the net inflow of mass would be equal to minus if we look at this equation q x can be obtained from the Darcy's law. Rho we can assume to be constant delta y of course, is the element thickness which is constant. So, we can write this equation as rho delta x delta y q x let me write it here can be written as minus k using Darcy's law. We will again assume that the aquifer is homogeneous and isotropic. So, k does not change with location as well as direction and therefore, k can also be taken out of the differential. So, we can take k out of this and what we will be left with is del by del x of h partial h with x. We can further simplify this equation by noting that the h by del x term can be written as half partial of h square with respect to x. So, if we use this we can write the net inflow. So, we get the second derivative of h square. So, if we compare this with the unconfined aquifer unconfined aquifer if we compare we will see that in confined aquifer we had here the second derivative of h, but here we are getting second derivative of h square. Now, this was an x direction we can write similar expression for flow in the y direction. The mass inflow in the y direction would be in the z direction due to the proofs assumption there is no flow component velocity component in the z direction. Therefore, the only flow in the z direction for this element would occur if there is some infiltration from the ground surface. This is typically known as recharge. For example, there may be a rainfall event which will cause some recharge or there may be a river which might be contributing to the ground water. So, this recharge is generally expressed in terms of r rate of recharge which tells us about the volume per unit area per unit time. So, r has units of for example, rainfall intensity. So, it will be centimeter per hour or millimeter per hour or centimeter per second depending on what units we are using. So, this if there is rainfall then typically it will be the intensity it will depend on the intensity of rain I and depending on how much infiltration we are getting that r will be equal to that infiltration. So, let us say that there is some recharge rate r and the mass which will come in because of this recharge from the top again bottom we are assuming impervious. So, there is no flow from the bottom and therefore, the recharge which will occur would be per unit area we have a recharge rate of r. So, multiplied by the area of the element which is delta x delta y and multiplying by the mass density will give us the mass inflow in the z direction. So, after knowing these net inflow of mass into the element what we are left with is how to find out the change of mass within the element and for that we will be using the specific yield. So, how to find out the change of mass within this element specific yield as we have seen already a specific yield is the amount of water released from storage for a unit drop of head from a prism of unit base area. So, here since the base area is delta x delta y we will have to multiply the specific yield with delta x delta y. So, if we write now the total mass inflow combining all the x y and z direction. So, in the x direction this is the flow in the y direction this is the flow and then as we have seen here in the z direction rho this should be equal to the change of storage within the control volume which we are considering. So, mass released or we can write mass change would be equal to the specific yield s y area and then the change of head because specific yield is defined as the change per unit surface per unit area of cross section for a unit change in head. So, we have to multiply with the area and multiply with the change in head. So, this will give us the rate of change of mass within the control volume and continuity equation says that these two should be equal. So, we can write the equation as there is volume it has to be multiplied by the mass density. So, we can combine these and write it as k by 2 rho del x del y I will write here and then cancel it out these will cancel out and the final equation which we get plus 2 r over k would be equal to 2 s y over k. So, this is the equation which governs the change of head with time and space change with time is here and change with space will be given by these two terms. When it is subjected to some recharge r most cases which we will deal with we can assume that the recharge is 0 in confined aquifers all of course, since there is an impervious layer at the top unless the layer is slightly pervious there will be no recharge. So, therefore, typically we have not considered recharge for confined aquifer, but sometimes confined aquifers the over lying layer may not be completely impermeable it may be an aquitard which will allow some leakage into the confined aquifer in that case in confined aquifers also we should consider the recharge. So, this equation needs to be solved with different boundary conditions and initial conditions to get the variation of h in an unconfined aquifer. So, we will take some simple cases for example, if we take steady state flow then this term will be 0 steady state this term will be equal to 0. Similarly, if we say that there is no recharge then this term will drop off. So, in the absence of recharge and for steady state conditions the we will get an equation which is similar to the equation which we had for confined flow there we had Laplacian of h equal to 0. So, the only difference is that instead of h now we have Laplacian of h square as 0 which we can solve to obtain the value of h at different locations. So, let us take an example in which similar to the confined aquifer case which we had considered ground level and let us say there are two water bodies which have elevations of h 1 and let us say that there is a length l x is 0 here and then at certain x we want the location of the ground water table. So, in the confined aquifer case we have seen that the water table was a straight line because h was linear, but here it will not be a straight line. So, we should find out whether it is like this or it is like this or like this. So, this is what we are interested in finding out what is the value of h at any x and to solve this problem we can assume that there is steady state condition one dimensional flow. So, if we assume one dimensional flow then the equation becomes a straight forward what we are assuming is no recharge and steady state one dimensional flow. So, if we make these assumptions the governing equation will be reduced to this term will be 0, r is 0 this term will be 0 steady state and this term will be 0 because the flow is one dimensional. So, we are left with this very simple equation which we can solve and obtain that h square will be linear. Now, these constants can be obtained from the boundary condition and in this case the boundary conditions are x equal to 0, h is equal to h 1 and x equal to l. So, you can see that c naught will be equal to h 1 s square and c 1 will be equal to h 1 square minus h 2 square over l. So, using these two values of c 0 and c 1 we can obtain the value of h and since h square is linear the actual level the actual ground water table will not be linear and we can see the discharge q at the boundary condition. At any location x q can be given as k i a where k is the conductivity hydraulic gradient and the area of flow i in this case is del h by del x and this negative sign will be there because we are saying that the flow will be in the direction of decreasing head and the area of flow at any location where the height is h if we consider unit width perpendicular to this. So, the width perpendicular to the plane of the paper let us say delta y is equal to 1. So, per unit width the discharge can be given as minus k del h by del x into h minus k del h by del x into h and this can also be written as we have done earlier for deriving the equation we had seen that del h by del x h del h by del x could be written as half partial of h square with respect to x. So, using that we can write this in terms of partial of h square with respect to x square and we know that from this equation this is partial of h square with respect to x. So, we know that partial of h square with respect to x is equal to c 1 and therefore, it will be simply q will be given by minus k by 2 c 1. Now, if there is some recharge. So, here we had assumed that the recharge is equal to 0 if there is some recharge then we will have to add the recharge term and from this equation this 2 r by k term will also come. So, the equation derivation is similar but now we have to account for the recharge and in this case we will see that the ground motor table may rise above the level h 1 because there is recharge coming in here at the rate of r. So, because of this recharge ground motor table may rise and we can find out the equation of this line by solving this equation with the boundary condition which are same x equal to 0. So, using these 2 boundary conditions we can obtain the value of h if there is no recharge then of course, minus k by 2 c 1 and c 1 we have already obtained from here as h 1 square minus h 2 square over l. So, without recharge q will be k by 2 into c 1 k by 2 into h 1 square minus h 2 square over l and this there is a negative sign here c 1 is minus h minus square minus h 2 square over l and then this is the value of q. With recharge we have to add this 2 r by k term and then we can solve the equation in the similar form in which h square will be c naught plus c 1 x minus r by k x square. Now, this comes again by solving the differential equation which we had written here and applying the boundary conditions we can obtain the value of c naught. C naught will again be equal to h 1 square, but c 1 in this case will be different from the c 1 which we have derived earlier when r was 0. In this case the c 1 turns out to be minus h 1 square minus h 2 square using the boundary conditions x equal to 0 h equal to h 1 x equal to l. So, q again will be given by the same equation minus k by 2 c 1, but now c 1 will be a function of r. So, we have k then we have h 1 square while in this case it was only h 1 square minus h 2 square over l. Here it would be h 1 square minus h 2 square minus r over k l square over 2 l. So, if we draw the profile of the water surface this is the ground level. There may be a water body here in practice this may denote a river or a canal. So, if we have 2 rivers or 2 canals which are running parallel in the unconfined aquifer the water level in absence of recharge would look like this by plotting the equation and when we have recharge the water level. In order to find out this equation and by just by looking at this we can see that there is a point here where h will be maximum and if you look at this whatever flow is occurring on this side will move towards the water body on the left and similarly the ground water flow from this side will move towards the water body on the right. So, this portion can be called a ground water divide. So, because of the presence of the gradient in this direction here and the gradient in this direction here the water the ground water will flow towards one side. So, this side and towards this side. So, this line can be thought of as dividing the ground water such that one part flows into one water body and the other in the other water body. We can find out the location of this point also for example, the q value is known to us. Now, this may not happen in all cases it depends on what is the rate of recharge if the rate of recharge is small it may not happen it may be below h 1 all the time. So, it will depend on the rate of recharge and we can find out what is the critical rate of recharge at which the ground water divide will occur. So, for that we need to write the equation for q as we had done here and you can see that for at the ground water divide q will be equal to 0 because before that q will be in the negative direction after that q will be in the positive direction and just at the ground water divide q will be equal to 0 and this will give us a relation between the recharge rate and the location x. For example, when there is no recharge we know that q is k by 2 h minus square minus h 2 square over l which is not dependent on x. Now, if we have the recharge then q will depend on x also. So, there is a constant term q and there is which is the same as without recharge. So, if you look at this term this gives us the constant term and the second term will be with recharge. So, r by k l square and this q will be dependent on x also 2 r by k x. So, if we look at this point this line the q term has to be 0 on this line and therefore, what we get is 2 r by k let us call this point x naught. So, 2 r by k x naught will be the dependent space dependent or a spatially variable value of q this changes with space because of the recharge. So, there is some recharge coming in from the top and there is some flow let us say coming in from this water body or going into this water body. If we take any section here the amount of flow which is coming in because of recharge is r into whatever delta x we have considered. So, if you take q at any section here and q at a section here the difference between these two q's will be the amount of recharge coming into this from the top which is r delta x. So, when we look at this situation where we have some value of x naught minus 2 r by k x naught should be equal to minus c 1. So, x naught will be equal to k by 2 r c 1 and c 1 in this case we have already obtained is minus h minus square minus h 2 square minus r by k l square over l. So, we can write x naught as l by 2. So, this gives us the position of the ground motor divide and for ground motor divide to occur x naught must be positive and therefore, we can find out what is the recharge rate critical recharge rate which will cause a positive x naught. So, for x naught to be positive we can find out the critical recharge rate r should be greater than k h 1 square minus h 2 square. So, what we have seen here is that if recharge is greater than this value h 1, h 2, l and k they are the properties of the medium and the boundary conditions. If recharge rate is greater than this then we will have a ground motor divide otherwise we will not have a ground motor divide. So, for unconfined aquifer flow between two water bodies one dimensional steady state condition we have derived some expressions for example, this h square using this h square we have obtained as a constant value c 1 and a time space dependent value 2 r by k x. So, when we find out the q as minus k by 2 it will have a constant value and a time dependent value which is r x. This we have seen is the contribution of the recharge up to certain distance x. So, for example, if we have this recharge occurring r here and there is some q at a constant here I will show it in the positive direction that may be towards the water body. Then at any other x what it says is that q here would be q at 0 plus r x and this term r x therefore, represents the contribution from the recharge to the discharge. Then we have seen that there will be some point x naught at which the water table will reach a maximum value and therefore, it can be taken as the ground motor divide. The location of this point x naught can be obtained by putting q equal to 0 and since q is equal to minus k c 1 by 2 plus r x x naught will be given by k by 2 r c 1 which can be obtained as this. For a ground motor divide to occur x naught should be positive and therefore, r has to be greater than k h 1 square minus h 2 square over l. Typically, l will be large and h 1 square minus h 2 square and k all of them will be small. So, if r is equal to larger than this means if l is very large then this value will be very small and a very small value of r will be sufficient to cause the water divide. So, this q of course, you can find out q at any location in the portion before x 0 q will come out to be negative and in the portion after x 0 q will come out to be positive that means, it is going in this direction before x 0 it is going in this direction. So, this is a one dimensional flow situation which may not occur in practice very often what is the most common situation occurring in practice is flow towards a well. So, we will next look at this case of an aquifer unconfined aquifer being pumped at a rate of q and we will try to find out what is the steady state draw down cone. So, this cone of depression at any distance now we will be using r because now we want to use radial coordinates at any distance r what is the height of water table h or in other words we can also think of in terms of draw down s given that initial thickness of the aquifer is capital H. Now, the equation will remain the same we have let us assume that there is no recharge. So, r is equal to 0. So, no recharge and then we also assume that there is some water body supplying it enough water. So, that it has reached a steady state and there is no further depression of the cone. So, in that case we have our governing equation as Laplacian of h square e to 0 because there is no recharge steady state conditions and as we have seen earlier for radial coordinate system we can write the Laplacian of h square as this. We need two boundary conditions to solve this generally there will be two draw down values available to us or we must have two draw down values available to us let us say at r 1 h 1 and r 2 h 2. Now, these r 1 and h 1 and r 2 and h 2 can be taken anywhere, but a commonly used method is take r 1 as the radius of the well. So, if the well radius so on a very large scale if I show the well radius this is the ground level initial water table at depth of h and then is steady state draw down cone may look like from the center of the well we can write r w at the radius of well and we have already defined a radius of influence that is the distance beyond which there is no draw down and typically we denote it by capital R although we have used this r for recharge rate also, but there should not be any confusion. So, this r is the radius of influence. So, by measuring the draw down in the well s w or we can say the height of water in the well is h w. So, the equation which we are solving is the Laplacian subject to two boundary conditions r equal to r w h equal to h w and r equal to r h equal to capital H. The solution of this equation can be obtained as we have done for confined flow case where c 1 is some constant and therefore, it will give us identical to confined aquifer the only difference is that instead of h we now have h square this tells us how the head varies in the aquifer and we can also find using the boundary conditions what are the values of c 1 and c 2 c 1 is the one which is really important for us. So, we can write c 1 as h 2 square for any r 1 and r 2 we can write h 2 square minus h 1 square or if we take these r w and r values then it will simply be h 2 can be taken as capital H. So, if we take r w and r values then this will be equivalent to the discharge at any point q can be written as 2 pi r k h and I am not putting negative sign here because of this term. The charge is in the negative r direction therefore, this negative sign I am not putting now this h del h by del r this 2 pi r is the circumference at any distance r and h is the height of aquifer at that point. So, 2 pi r h is the area and k del h by del r is the velocity. Now, h del h by del r we can combine as pi r k. So, the term 2 h and del h by del r they can be combined to form this term. So, q becomes pi r k del h square by del r and as we have seen from here or here del h square by del r will be c 1 over r. So, it will become pi k c 1 which we have already derived in terms of any h 1 and h 2 or if we take the radius of the well and the radius of influence. Now, as we have seen for the confined aquifer case measurement of draw down is much easier therefore, we want to express this equation in terms of draw downs. The only problem here is that this h is a square and we know that h is capital H minus draw down s. So, when we do h square because of the presence of the square term we cannot say that the difference of these 2 squares is the same as difference of the draw downs square because it is not linear. Therefore, sometimes we approximate it and we write it we can write this equation in a little bit modified form and by using some approximation we can write this term h square minus h w square as h plus h w h. Once we separate them out then this h minus h w term is nothing but the draw down in the well s w which is very easy to measure in the field and the other term which we have is h plus h w and sometimes we have to make some approximation. For example, here we can say that h plus h w will be very nearly equal to 2 h if s w is a small for a small draw downs. So, the logic is that the draw down typically is very small compared to the thickness of the aquifer h. So, if s w is quite small then we can assume that h plus h w will be almost equal to 2 h and therefore, q can be written as 2 pi 2 k pi h and h minus h w is s w k into h is 2 pi 2 k pi h and h minus h w is s w k into h is the transmissivity of the aquifer as we have seen k into the thickness is the transmissivity. So, we can write this as 2 pi t s w. So, this gives us an equation which can be used to estimate the transmissivity of the aquifer this equation is exact, but it cannot be used to estimate the value of k typically because h w and h values may not be known very exactly, but the draw down in the well s w can be obtained easily r and r w also are easy to obtain. And therefore, for a known value of t we can find out known value of q we can find out transmissivity using this equation. So, this is useful for estimating the parameter values. So, after looking at the confined and unconfined aquifers in all cases we have assumed that Darcy's law is valid. Sometimes Darcy's law may not be valid the flow may not be laminar where we know that the head loss is proportional to when the flow is turbulent the head loss is typically proportional to some power of q. Generally we can take it to square sometimes it is 1.7 1.8, but we can say that head loss is proportional to q square. So, all the analysis which we have done till now does not account for variation of head loss as q square it only says that since we are using Darcy's law the head loss or we are saying that q equal to minus k delta h over l which says that delta h is proportional to q. So, we will look at some cases where what happens if the flow is not laminar, but turbulent especially near the well. So, here the velocity may be a small, but as we go near the well the velocity becomes very large and the flow may not be laminar it may become turbulent. And then the there will be an additional loss because of this proportionality q square. Similarly, near a well we have a screen and there will be some head loss when the flow passes through the screen. So, the total drawdown in the well will be a combination of these three terms laminar loss turbulent and the well screen loss. So, we will look at how to combine these to obtain the total well loss which will define some kind of efficiency of the well.