 In the previous lecture, we had looked at the ground water which is the water in the subsurface and we saw that if we want to utilize the ground water, we have to know how much water is present, how much is available for withdrawal and how fast we can withdraw it and the other thing which we should look at is how deep is the water table because that will affect the cost of pumping. So, the ground water table has let us say horizontal level initially, but when we start pumping it, it will start to go down with time and we should know how this water table will change with time. So, that we know how we have to lower the pump if the water level goes down very much and to look at this, we had looked at a mass balance or continuity equation in which we take an element of the aquifer and look at how much is the mass of water flowing into this element and what is the change of mass within this element and the continuity equation tells us that these should be balanced and looking at that mass balance will allow us to determine how the water level or the pressure in the aquifer is changing at that point. So, in order to look at the mass balance, we should know what is the amount coming in and what is the amount which is going out. If we take an elementary volume like this with we have delta x, delta y and delta z, then we can find out what is the storage of water within this volume and how it is changing with time. So, we need to know two things, one is how much water is coming in, how much is going out, this will depend on the velocity which can be obtained from the Darcy's law and therefore, it depends on the hydraulic conductivity k and the hydraulic gradient i and q also depends on the area of cross section which in this case will be delta x, delta y or delta y, delta z depending on which phase we are considering. So, combining Darcy's law and the area velocity term, this will give us an idea about q. The other thing which we need to look at is the storage within this volume and how much water can come out of this storage for a unit drop of head. So, we will look at these things today. Let us first look at the hydraulic conductivity k, we had discussed briefly that this k is a function of both the fluid property and the medium property. If we take the analogy with flow in circular pipes in which let us say we have a drop of head equal to delta h in a length of l, then for laminar flow we know the well known equation for the head loss in which mu is the viscosity, dynamic viscosity of the fluid, v is an average velocity, gamma is the specific weight and d is the diameter of pipe. If we want to compare this with flow through porous media, we can think of the d as being the diameter of the particles because that will be related to the diameter of the pores. So, in the porous media the equation which we have is the Darcy law which says that v is equal to k delta h over l. So, if we compare these equations, the k will come out to be dependent on the medium property, the diameter of the particle and it will also depend on the fluid property, the specific weight and the dynamic viscosity as gamma over mu. This is fluid property and d square denotes the porous medium property. So, typically gamma over mu is taken out of this equation and we can write the hydraulic conductivity k as gamma over mu times some intrinsic permeability or a specific permeability which is only a property of the medium and typically this k is some constant into the grain size square based on the pipe flow analogy. For different porous medium for example, for sand and clay the value of c may be different and there are lot of empirical equations which are used to correlate the intrinsic permeability with the grain size. Gamma over mu varies with the fluid. So, if we have flow of water this value will be different, oil it will be different, but in general for ground water flow typically we have flow of water even for water the value of gamma and mu may be different at different temperatures. In general in ground water flow we will assume that the temperature variation is very small and water is the flowing fluid therefore, k is commonly used not the intrinsic permeability. So, commonly used term is the hydraulic conductivity since water is the flowing fluid and temperature can be assumed constant. If not then it is better to use the intrinsic permeability because that will not change with fluid properties. The second thing which we want to see is how much water can come out of the aquifer for a drop of head. So, for a given drop of head suppose there is this volume which has some solid particles and the rest filled with water. We have already discussed that we will only consider saturated flow conditions. So, there is no air inside this. Now, this volume is subjected to some pressure and we will start with the confined aquifer case because that is easier to derive. If we put a tube here a piezometer here water will rise up to a certain level which may be below the ground level. It may even be above the ground level in which case it is called a flowing aquifer. Now, suppose the piezometric head is here. Now, what we want to see is how much water will come out if we lower the piezometric head by certain amount delta H. This lowering of the head may be because of pumping or it may be because of head drop in some other area where this aquifer is connected to. But let us take this delta H is the head drop and if we take an element of the aquifer here. Let us say we take some element here. We want to see how much water can come out of this aquifer for a head drop of delta H. Now, let us look at the mechanism of how this water is coming out. For unconfined aquifer it is quite straight forward. This is the ground water table bedrock. Now, if we lower the ground water table by some amount delta H, water will come out because of drainage of this volume. But in unconfined aquifers there is no change or not much change in volume. There is little change which we will see. But there is not much change in the volume. Therefore, the water which comes out is because of the compressibility. When we talk about compressibility it would be compressibility of water as well as the compressibility of the medium. So, we can call that aquifer or formation compressibility. When we lower the piezometric head the water from this element, suppose it comes out of this element. Now, this water is under some pressure which has not been which has been kept constant, but now it has been lowered because of this lowering delta H. So, when we reduce the pressure water will expand and therefore, it will come out of this element. The other thing which happens is because of this lowering of pressure in the water since the total stress remains same, the overburden pressure remains same. Lowering of water pressure means increase on the grain pressure. So, if there is some grain pressure let us call that sigma which is acting on the grains that pressure will increase if we lower the value of. So, the pressure in the water is let us say p. The pressure on the grains is sigma. Now, sigma plus p will be the total pressure which remains constant. So, by lowering the piezometric head we are lowering p and therefore, sigma will increase and due to this the aquifer or the formation will be compressed because of this increase in stress and that compression will lead to further release of water from this element. So, we will look at these how to derive these terms which represent release of water due to matrix compressibility or the formation compressibility, release of water due to water compressibility. So, let us define a term which is known as the storage coefficient and typically denoted by S. S is the volume of water released from the formation from a prism of unit cross section area with unit drop of piezometric head. So, there are three things here that S is the volume and it is from a prism of unit cross section area for a unit drop of piezometric head. So, if we take this confined aquifer of height b and consider a unit area of aquifer. So, this area is 1 and this has some piezometric head which we will call h. The datum can be anywhere typically the datum is taken at the base of the aquifer that is at the impermeable rock. Now, this is some value of piezometric head and now if we lower this by a unit amount how much water will come out of storage. So, that term is denoted by the storage coefficient S. Now, the mass of water which is present inside this prism m can be related to the volume and the density. So, rho is the density of water into the volume which will be since the area of cross section we have taken as unity the volume will be equal to b into porosity. So, the mass of water contained within this confined aquifer the prism of base area 1 can be written as rho p eta which is the porosity and therefore, if this mass is changing it will change because of change in all these parameters. Suppose we have some change here delta b some change here delta rho. So, when we are lowering the piezometric head the aquifer thickness b will also change as we have seen that by lowering p we are increasing the effective pressure on the grains and therefore, the aquifer will be compressed. So, b will reduce the porosity similarly will also change. So, all these three parameters they change and their net effect will cause some change in mass delta m. So, our aim is to find out this delta m because then we can relate the storage coefficient S with delta m. So, let us look at how to find out this delta m for certain change in pressure or the piezometric head H. So, we start with our original mass since all these three can change we can write this as. So, change in porosity will cause some change in mass change in formation thickness. So, this is the effect of change in porosity formation thickness and density. So, the total change in mass occurs because of change in porosity change in the formation thickness and change in the density of water. So, we need to evaluate these changes and for that we introduce a property which is known as compressibility which is inverse of the modulus of elasticity and is defined as change in volume per unit volume divided by the change in pressure which is basically strain over stress. For water we can define the compressibility which we can say represented by beta. In general a positive change in pressure will cause a negative change in volume and therefore, we put a negative sign here where V w is the volume of water and delta V w is the change in this volume. Delta P of course, is the change in pressure as we know the piezometric head H is P over gamma plus z where P is the pressure and z is the elevation. Therefore, a change in pressure delta P can be written as gamma into change in piezometric head delta H. So, we will be using this relation later on to find out the storage coefficient. So, let us look at the water compressibility equation beta. Now, V w is the volume of water and therefore, if we write rho V w this will be the mass of water V w will be the volume within the aquifer. Now, if we look at this mass of water rho V w continuity tells us that matter cannot be created or destroyed therefore, this mass should remain constant. So, we can relate a change in volume of water with a change in density. So, if this is not changing then the sum of these two changes should be equal to 0 and from here we can find out the term minus delta V w over V w will be equal to delta rho over rho. So, we can write this as delta rho over rho delta P and therefore, the change in density of water can be written in terms of its compressibility as delta rho beta. So, using this equation we have related the change in the density of water with compressibility of water its density and the change in the pressure. The next thing which we look at is compressibility of the formation or the aquifer material. So, we define the formation compressibility or aquifer compressibility and use a symbol alpha. In a same way as we did for water we have where V now denotes the volume of the formation or the aquifer. So, in this case since we are taking unit area we can say that V will be equal to B into 1 and delta V is the change in V delta P is the pressure but since we are using P for as a symbol for the pressure in the water we can use a different symbol here and we can write P f as delta P in the formation or typically we can denote this term also by sigma. So, we can use delta P f or delta sigma for this term. So, we can find out delta B from here delta rho from here if we look at the equation for delta M we had delta rho delta B and delta eta. So, we need these three terms to find out the change in mass using this equation delta rho we have seen from here can be related with the water compressibility delta B can be obtained from here over B delta sigma we have also seen earlier that sigma plus P is constant. Therefore, delta sigma will be equal to minus delta P and therefore, we can also write delta B from here as alpha B delta P. Now, the third term which we need is how the porosity is changing with change in pressure and for that we can assume which is generally a very good assumption that the volume of solids is constant. So, if we have an aquifer material here no matter how the pressure is changing in this aquifer water pressure and the grain pressure the amount of solid material present in this area volume of solids which we typically says V s it remains same because the grains are assumed to be incompressible. Now, V s we can write as 1 minus eta if the thickness as we have taken as B and cross section area is 1 then the amount of solid volume present inside this unit area and height B will be 1 minus eta times B because total volume is B out of that eta B is the volume of the liquid water in this case therefore, total volume of solids V s will be equal to 1 minus eta B and if this is a constant then we can use the same formulation as we used before that delta V s will be equal to 0 and this would imply that B times minus. So, from here we can obtain delta eta which is the change in porosity we can relate it with change in the formation thickness B. So, now we have all three parts of the equation formulated delta eta from here which is a function of delta B delta B we have from here and delta rho we have from this equation. So, putting all these three in this equation which tells us total change in mass we can obtain delta M in terms of the compressibility of the water alpha and the compressibility of the aquifer formation beta. So, we write delta M in terms of the aquifer thickness B into alpha plus the porosity into beta multiplied by the specific volume of the liquid water weight gamma. Now, this we will need because we have said that we want to find out a storage coefficient S which is the volume released per unit change in H. Now, this H is the piezometric head and is given by P by gamma plus z. We have related all the changes with delta P. So, delta P we can replace by gamma delta H. So, we would write in terms of delta P and then write that delta P as gamma delta H which is nothing but delta P and then to find out a storage coefficient we know that this is volume and we have written our equation in terms of mass. So, S will be equal to the mass released divide by rho and then per unit change in the piezometric head and therefore, S will turn out to be gamma alpha plus eta beta. This equation gives us the storage coefficient and it represents the volume of water released for a prism of unit cross section area over the entire thickness B. Similar to the storage coefficient S, we have a term which is known as specific storage S S and this is the volume of water released from unit volume. So, if we compare it with the definition of the storage coefficient if we divide S by B, we would get S S because S was the volume released from the entire thickness B and therefore, if we take unit volume the amount of water released will be given by S over B and then we can write this also in terms of the compressibilities and the porosity as gamma alpha plus eta beta. S S will be used when we derive our continuity equation for confined aquifer case. For unconfined aquifer the storage coefficient S is equal to the specific yield although there is a small component which comes from the compressibility. Typically, it can be ignored in with respect to S y. So, compressibility terms are generally negligible and therefore, for unconfined aquifer the storage coefficient S is taken as S y. Using these storage coefficient S, the specific storage S S we can derive the continuity equation and that gives us a relation between the head the piezometric head in the aquifer and it relates with the how much water we are pumping out of the aquifer. So, as we have seen the storage coefficient can be defined for unconfined and confined aquifers. The specific storage for confined aquifers can also be defined and we will use that now to derive the equation of motion for a confined aquifer. Ground level then there is confining layer and at the bottom we have this bedrock. So, in this confined aquifer we can look at an elementary volume and see how the head change effects the amount of storage and the amount of water which is coming in and then by balancing these two we can derive an equation which will tell us how the head is changing with time and space. So, the spatial and temporal variation of the head can be obtained by the continuity equation. So, if we take this element let us enlarge it. Typically, we will be using the collision coordinate system, but sometimes it is better to use a radial coordinate system. For example, if you are pumping from a well then the piezometric head will be symmetric about the well in the radial direction and there we may be better of using a radial coordinate system, but let us start with the collision coordinate system. So, we take this elementary volume which has size delta x delta y and delta z. Of course, our x y and z directions are like this. Let us consider the flow in the x direction. Same thing can be done in y and z direction also, but let us start with the x direction and say that there is some velocity q x which is the Darcy velocity in this case not the seepage velocity. Because of this velocity q x the mass which is coming in the mass flux will be rho into q x. What we want to see is how much mass net mass is flowing into this control volume. The mass which is going out can be written by using Taylor series as rho q x which is the mass coming in from the left face plus what is the rate of change multiplied by the distance delta x. Sometimes we assume that this whole term rho is inside the differential here. So, when we take Darby Dirac's we put rho inside the derivative, but typically the variation of rho is very small with respect to the other term which is del q x by. So, if we write rho delta x del q x by del x and write q x, then it compare to this term it can be neglected. And therefore, generally we write this term only as rho del q x by del x. So, we will be making this assumption that the change in density can be ignored and therefore, it can be taken out of the derivative and we can write as rho del q x by del x. Therefore, the net inflow would be rho q x which is the inflow from the left face minus the outflow from the right face which is rho q x plus and using our assumption we have taken rho out of the derivative there by the x of q x into delta x and this will result in minus rho. Now, this is the mass flux rate per unit area. So, we have to multiply it by the area. So, we can say net inflow per unit area is given by this and therefore, inflow in the x direction. Remember that we are talking only about the x direction now other directions we will do similarly. So, inflow in x direction can be written as minus rho multiply by the area which is in this case delta y delta z as you can see that this face has an area delta y delta z. So, this term gives us the mass coming in the element in the x direction we can now use Darcy law to relate it with the head. So, as we know q x now we will make another assumption here typically we have medium which is isotropic if we make that assumption then we do not need to use this subscript x for isotropic medium we can remove this x because k will be same in all direction. Sometimes the k is not same in all direction for example, if you have a layered formation then k will be different for flow along the layers and it will be different for flow perpendicular to the layers, but here we assume that the medium is isotropic and therefore, we will ignore this subscript x using Darcy law. We can write q x as minus k del h by del x and therefore, from this we can write the inflow in the x direction as rho k. So, this term represents the mass of water flowing in the x direction net inflow of mass into this control volume delta x delta y delta z and similar expression can be derived for y and z directions and all of them can be added together to get the total or the net mass inflow into the system as rho k delta x delta y delta z we can combine them as delta v this is the volume of the element into we have the contribution from the x direction del 2 h by del x square and similar terms for y. Now, the continuity equation says that this net mass inflow should be equal to the change of storage within the control volume. Now, if we use the storage coefficient s we have seen that s represents the volume released per unit cross section area using a prism, but if we want to use elements with area delta x delta y and delta z we would need to use a specific storage rather than a storage coefficient. So, we use s s which is the specific storage which represents the volume released from a unit volume. So, volume of water released from a unit aquifer volume for unit head drop. So, if we want to find out what is the volume of water released from this elemental volume for a head drop of delta h we can write an equation which tells us that the volume the mass released per unit time from this elementary volume. Can be written as rho s s delta v delta h and the way we write this expression is that s s is from a unit volume. Therefore, we have to multiply by the volume of the element s s is per unit head drop. Therefore, we have to multiply with delta h s s gives us the volume and therefore, we multiply with rho to get the mass. So, the mass change of mass per unit time in the elemental volume delta m by delta t can be related with the specific storage s s the density delta v delta h and delta t and in the limit as delta t tends to 0 we can write and the other limit which we will take is that when this element reduces to a point that delta v tends to 0. So, when we equate net mass inflow which is given by this expression with change of mass within the control volume. So, these two terms should be equal and therefore, delta v will cancel out rho will also cancel out and we should note that rho will cancel out because we have assumed that change of rho is very small negligible if we do not ignore that term then rho will also be included in the equations. So, if we equate these two we finally, get an equation which can be written as del square h this equation represents we have made some assumptions. So, I will write them here first assumption is that it is a confined aquifer second assumption which we have made is that the aquifer is isotropic that is why we have here only k otherwise we will have k x k y k z three different terms which will be included in this. The term del square can be expanded in Cartesian coordinates it has a very simple form and if you have some so situation where you have radial and axis symmetric flow. So, symmetric means that theta will not be coming into the equation then you get a term which there is del x by del theta is equal to 0. So, that term has not been considered there is a del 2 h by del theta square term which has been neglected because h is not a function of theta. So, using this equation we can obtain the variation of h for a given condition we can also write this equation in a little different form by noting that the storage coefficient s is a specific storage into the depth of aquifer and we have already defined the transmissivity as k into b. So, if we use this s and t we can also write this equation in terms of the storage coefficient and the transmissivity. So, either we can use a specific storage and hydraulic conductivity or we can use a storage coefficient and transmissivity and solve this equation for given conditions. Since, this is a second order equation in space first order in time we need one initial condition and two boundary condition to solve this equation. For example, if we consider the case of one dimensional flow. So, we have a confining layer ground level and suppose we are looking at a case where flow is occurring between two water bodies one of them here such that the height of water is h 1 and the height of water here is h 2. The piezometric head in the confined aquifer may be like this or like this or it may be a straight line like this. So, in order to find out which is the variation in the confined aquifer we need to solve the equation of motion which is again written here as. Now, if we make the assumption that the flow is steady then del h by del t because steady means the parameter is not changing with time. So, del h by del t will be equal to 0 and therefore, we will get a simple equation which is known as the Laplace equation. This equation has a form similar to the heat diffusion equation therefore, this is also called the diffusion equation. Now, let us say that this is a steady state flow condition therefore, we will need to solve this and let us also assume that this is a one dimensional flow. So, the flow is taking place only in x direction. So, we can write since the head is not changing in other directions this equation simplifies to del 2 h by del x square equal to 0. The solution of which is h equal to c 1 x plus c 2 c 1 and c 2 are constants which will be obtained by the boundary conditions. In this case considering the length of the aquifer to be l the boundary conditions are at x equal to 0 h equal to h 1 and at x equal to l h equal to h 2. So, using these two boundary conditions we can obtain the solution of this equation which tells us that h is linear. So, in this case if we have a flow between these two water bodies the piezometric head in the aquifer will vary linearly between h 1 and h 2 and the variation can be obtained directly from applying these two boundary conditions from which we can write h equal to h 1 minus. So, this is obtained by applying the these two boundary conditions in this equation. If we want to find out the flow rate we can use the Darcy's law which gives us q equal to k i. i in this case is del h by del x because we are considering one dimensional flow which is simply h 1 minus h 2 over l. Since the piezometric head is linear i is constant throughout the length and is given by the delta h which is h 1 minus h 2 divided by l and this will give us the apparent velocity or Darcy velocity. Using that we can find out q into area if we consider unit width then area will be equal to the thickness of the aquifer which is b. So, we can write this as k h 1 minus h 2 over l. So, we have seen today how to obtain the equation of motion for confined aquifer how to solve it for given boundary conditions and the same thing we can do for unconfined aquifer. The thing is that unconfined aquifer is little more complicated because the water level itself determines the thickness of the aquifer. So, thickness of the aquifer is not constant what is varying from place to place. So, we need to make some simplifying assumptions in order to derive that equation. Similarly, for unconfined for confined aquifers also we have solved the equation for one dimensional steady state flow conditions. If we go for unsteady flow it will be little more complicated or if we go for radial coordinate system the solution will be little more complicated. So, we will look at these solutions for confined aquifer and we will also look at how to derive the equation of motion for unconfined aquifer in the next lecture.