 In the previous lecture, we have seen some aspects of flow through confined aquifers. We have seen how to use the mass balance along with the Darcy's law to derive an equation of motion which relates the variation of head for different times and distances. The equation which we had derived was del 2 h. So, this equation tells us that the Laplacian of h is equal to a storage coefficient divided by transmissivity times the time derivative of h. This term represents the change in water volume inside the aquifer and this term represents with multiplication by T the net inflow into the aquifer from all sides. In the confined aquifer case, we have seen that the thickness remains constant, but for unconfined aquifers the thickness varies and it is real equal to h. So, it is more difficult to derive equations for unconfined aquifers. So, that is why we have first started with the confined aquifer case. We have already seen how to solve this equation for a steady state. So, if we have a steady state flow that means the parameters are not changing with time the head remains constant with time only changes with space and therefore, del h by del t term can be ignored as 0. The equation which we then get is the Laplace equation. If we assume that the flow is one dimensional we have a very simple equation. So, for one dimensional flow in the x direction we can write that the second derivative of h with respect to x will be equal to 0 which gives us a linear head profile as h equal to c 0 plus c 1 x. The values of c 0 and c 1 would be obtained based on the boundary conditions as we saw in the previous lecture. If the flow is occurring between two water bodies which have elevations of h 1 and h 2 the head variation within the aquifer is linear and therefore, the piezometric level would be a linear profile joining h 1 and h 2. That means that if we put any piezometer we install the piezometer here the water will rise up to this level in that piezometer. Similarly, if we have a piezometer here water will rise up to this level. Now, this one dimensional flow it is easy to solve, but it is not very practical in most cases we are interested in flow through or towards a well. So, that is why the next case which we will discuss is the flow of water in confined aquifer towards a well. So, we have an impermeable bottom here which is typically rock then we have a confining layer which is another impervious layer the ground level and this confined aquifer has a thickness of b. Now, when we install a well we will assume that the well is a screened throughout the aquifer thickness. So, water can come in this well throughout this thickness and there is no water entry from above the aquifer. We have some pumping rate let us call it q and the conditions are typically when we start the pumping we can assume that the piezometric surface is horizontal. So, before start of the pumping the piezometer water level in any piezometer if we put here we assume that it is horizontal. Once we start pumping then this piezometric head will go down and with time it will slowly. So, at certain time it may be like this after some more time of pumping it may go like this. So, with time this piezometer level is going down and typically if you look at this it is in form of a cone. So, this is called a cone of depression which will go deep as we keep on pumping. Now, there is some recharge into the aquifer and ultimately this cone of depression will reach a stage where this q which is being pumped out will be balanced by the q which is coming in and we will say that it has reached a steady state. Now, this cone of depression is important for us because we know that how much is the head at any location h. So, let us draw it here again what we are interested in is finding out the location of the cone of depression at any distance r from the center of the level. Let us say that the head is h this is the initial piezometer level. The difference between the initial level and the cone of depression at any time is called the drawdown typically denoted by s. So, as you can see with time s will increase when we start s is 0 because the piezometer level is horizontal and as we keep on pumping s will increase and s will typically be a function of time and the distance from the well. Once it reaches a steady state then that profile s will only be a function of r. If we go very far from the well we will reach a point beyond which the drawdown is 0. This distance r may be thought of as radius of influence. So, this is the distance up to which the well is influencing the water level and therefore, we call is the radius of in other words there is no effect of the well felt beyond a distance of r from the well. If we stop pumping then slowly this cone of depression will be going up because once we start stop pumping there is still some water coming in and that will cause the piezometer head to rise and that portion or this phenomena is known as recovery. So, we have a drawdown in which by pumping we are lowering the piezometric head and then once we stop pumping the piezometric head will try to go back to its initial level which is called the recovery. So, all these are important in the sense that if we measure the drawdown it depends on the aquifer properties s and t. If we measure the drawdown we can estimate the parameters s and t from the observed values of s small s which is the drawdown. So, we will see first how h and therefore, s because s is nothing but initial h minus h where capital H is height of the initial piezometer level. So, if we find out how h is changing with time we can derive an equation for the drawdown and that will help us in finding out the storage coefficient and transmissivity of the aquifer. So, we will first look at steady state condition because that is easier to derive there is no time factor involved and s will only be a function of r. So, let us start with a steady state flow the assumptions we are making is that the thickness of the aquifer b is constant. Initially we have a horizontal piezometric head level as capital H the pumping rate q is assumed to be constant. So, we are pumping the aquifer at a constant rate and then we also assume that we have reached a steady state conditions. So, the drawdown is not changing with time that is the cone of depression remains constant with time if we use the equation the Laplace equation del 2 h equal to 0. Now, since this is a radially symmetric flow condition we would like to use radial coordinates here and therefore, the equation becomes this. The Laplacian operator in radial coordinates also involves partial derivative with respect to theta, but here because of the radial symmetry H will not be a function of theta. In other words any distance r from the well the value of H will be same irrespective of what is the angular direction in which we are measuring it. So, due to radial symmetry del 2 h by del theta square term is ignored from here. So, this is the equation of motion for steady state flow in a confined aquifer towards a well which is screened throughout the aquifer thickness and which is pumping at a constant rate of q. To solve this equation we can write this as 1 over r del by del r of r 2 h by del r and this implies that partial of h with respect to r will be equal to some constant over r because from here we can say that this will be a constant because it is derivative with respect to r is 0. So, this constant if we assume to be c 1 then we can write partial of h with respect to r will be equal to c 1 over r. So, this gives us the derivative or the gradient of head at any location r and you can see that the gradient is inversely proportional to r. So, as you go away from the well the gradient will decrease which is also clear from the cone of depression because the slope will be high in the early portion and then as you move away it becomes flat till the slope becomes 0 very far away from the well. The radius of the well is typically very small and denoted by r w. So, r w is the radius of well and r capital R will be the radius of influence thus we have already discussed. So, to solve this equation we would need some boundary conditions and for example, in this case we can use the conditions that r is equal to r w s equal to s w where s w is the draw down in the well. So, we can show it here s w is the draw down in the well. So, one condition may be that r equal to r w s equal to s w, but typically the draw down in the well is affected by some losses in the well the flow is also not laminar it is turbulent. Therefore, s w is typically not used and we take some other location let us say r equal to r 1 and find out the draw down there as s 1. So, these boundary conditions need to be used in order to be able to solve this equation, but this equation tells this is the governing equation for steady state flow in a confined aquifer. If we solve this equation we can write h plus c 2 and as you can see there are 2 constants here c 1 and c 2 which need to be evaluated which means that we need to have boundary conditions at 2 different distances for example, r 1 and r 2. So, if we assume that at r equal to r 1 h is h 1 h is h 2 now these r 1 and r 2 may be any 2 radius for example, r 1 may be r w and r 2 may be r capital R. So, if we take this as r w then this would be s w and if we take this as capital R then this would be h because at the radius of influence the head remains constant at the initial value, but we will not assign these values we will just assume that they are 2 different r values r 1 and r 2 and 2 different h values h 1 and h 2 which will allow us to obtain the value of c 1 as h 2 minus h 1 over natural log of r 2 over r 1 and if you look at this equation del h by del r is c 1 over r and c 1 is obtained from here. So, what it tells us is that by measuring the drawdown at 2 different locations r 1 and r 2. So, h 1 and we can get the gradient del h by del r and we can also from there write the discharge as k i at any distance r the area would be equal to 2 pi r into the thickness of the aquifer b and i is nothing, but del h by del r which we have already seen from here is c 1 by r and c 1 is given by this. So, h 2 minus h 1. So, if we put these values of i and a in this equation then q can be written as k h 2 minus h 1 and r 2 over r 1 2 pi b because this r and 1 over r there cancel out. Now, k b as we know k and b can be combined as the transmissivity. So, we get a simple expression for q in practice it is difficult to measure h because the impermeable layer here would be quite deep we may not even know how deep it is, but if we measure s which is the drawdown we can see that h 2 minus h 1 would be equal to s 1 minus s 2 because h plus s is constant which is equal to h. So, h 2 plus s 2 and s 1 plus h 1 will be same and therefore, h 2 minus s 1 h 1 will be equal to s 1 minus s 2 and therefore, typically we replace this term by s 1 minus s 2 s 1 and s 2 they are easy to measure because we know the initial water level or the piezometric head level in this case and the final piezometric head level the difference between those two will be the drawdown s. So, this gives us a method of estimating t if we measure q the drawdown and distances of course, we know where we have put the piezometers. So, knowing all these three terms we can obtain the value of transmissivity. So, this test is very useful in estimating the transmissivity or hydraulic conductivity because the thickness b will be known and therefore, we can obtain hydraulic conductivity also. This equation is commonly known as the theme equation based on the first derivation by him. So, the theme equation is valid for a steady state flow towards a well in a confined aquifer assuming that the well is fully screened. So, it is drawing water throughout the aquifer width and all the assumptions which we have earlier made in deriving the equation would still hold good. For example, we had assumed that the aquifer is isotropic. So, k x k y k z they are all same. We have also assumed that the aquifer is homogeneous that means k is not changing from place to place. We have assumed that the thickness of the aquifer remains constant. So, this b is a constant value. So, if all these assumptions are satisfied then for a steady state flow towards radial well themes equation can be used. The steady state equation although it is useful, but typically it will take a very long time to reach the steady state and therefore, it is better to derive some equation which accounts for the transient behavior and which tells us how the cone of depression is decreasing. So, initially we have some piezometer level then we start pumping with time we have this level going down. The radius of influence would be a distance beyond which there is no influence and at different times the radius of influence will also be increasing. So, as we go on pumping the cone of depression will expand and the radius of influence will be increasing. So, if we are able to derive an equation which will tell us how the h or s varies with radius as well as time it will be very useful to know how is the cone of depression expanding, how is the radius of influence increasing and it will also help us in deriving a technique to estimate s and t. As we have seen in a steady state case we would be able to estimate the transmissivity, but not the storage coefficient because of the nature of the equation. The term s affects the transient behavior and if we assume a steady state then there is really no effect of the storage coefficient because things are not changing with time. So, storage coefficient does not enter into the equation, but if we are using a transient flow condition then we get an equation where s will also be used in the equation and therefore, we can derive the values of both s and t from that equation. So, we will now look at the transient flow radial again towards a well all the other assumptions will still be valid. So, we have a fully screened well constant thickness of b. The only assumption which we have now relaxed is that instead of being a steady state flow the flow now is transient and therefore, the draw down and the h will both depend on time also and therefore, the equation governing will include the transient term del h over del t. Now, partial of h with respect to t will not be 0 in this case while in the steady state case we have taken it as 0. The solution of this equation is a little complicated it is not as easy as it was for either one dimensional flow or for a steady state flow and therefore, the person who derive this solution first was Thais. In fact, Thais and Lubin both contributed to this solution, but it is commonly known as the Thais equation which relates s because as we know h is h minus s h is a constant. So, the same equation can be written in terms of the draw down. So, del for example, if you take this term it can be written as minus because h is capital H minus s and capital H is a constant it is independent of r. Therefore, the second derivative of h with respect to r would be equal to negative of the second derivative of s with respect to r. So, similarly we can put these terms also in terms of s and the Thais solution gives the draw down s at any radius r and at any time t. The q is the pumping rate as before we assume that q is a constant and Thais introduced a function which is called the well function. So, we will now look at what is u and what is w u. So, this is the Thais equation where w is called the well function, u is a function of time and distance. So, you can see that u increases as t decreases and u increases as r increases. w u is an integral, this integral is not very easy to obtain and therefore, tabular values are given in some books and you can refer to any ground motor hydrology book it will have a table of u versus w u values. So, knowing value of u r square is known to us, suppose s and t are known. So, it is not estimation of parameter problem, suppose it is a problem in which s and t are given and we want to estimate the draw down at any observation well at any given time. In that case we will be able to obtain the value of u for any time then from this table for a given value of u we may need to interpolate the values here or you can directly get the value from this table. Sometimes we prepare a curve also and from that curve we can obtain the value of w u. Once we obtain the value of w u, getting the draw down is straight forward because q and t are assumed to be known. If you look at this expansion of the exponential function we can expand w u in terms of a series infinite series and that series is written as 0.5772. So, if we do term by term so minus u square over 2 factorial 2 plus u cube over 3 factorial 3 and so on. If u is a small we can ignore the higher order terms and there is an approximation which says that if u is less than 0.01 only the first two terms can be used. So, that approximation we will see a little later, but w u can be obtained using this series and once either from table or from figure or from this equation w u is known the draw down can be easily obtained. The approximation which is known as Jacobs approximation it says that if u is less than about 0.01 then only the first two terms of the series can be used and this approximation is very useful in deriving the or estimating the parameters. Parameter estimation means estimating the values of storage coefficient s and transmissivity t. So, since the transient method involves measurement for smaller times it is preferred over the steady state method. In transient conditions what we can do is we can select an observation well. So, the pumping well is here we can select an observation well at distance of r and note down how the water level in this observation well is decreasing. So, with time the draw down s will increase and using this t and s value we want to estimate the parameters s and t. So, that is what we will look at next how to estimate or compute the values of storage coefficient. And transmissivity from a given observed set of data for draw down at well which is located at distance of r sometimes we may choose wells at different radial distances. So, we may have another piezometer and you can measure the water level in this with time may be some distance r 2 and r 1. So, if you have two different wells we can use that data to estimate the value of s and t using both of the wells and if you look at the definition of u is r square over t s over 4 t. So, individual values of r and t are not important this r square over t factor is what will affect the draw down. So, if you have two different wells at distance of r 1 and r 2 if we plot t versus s they would be on different lines but if you plot r square over t or t over r square versus draw down both of them will fall on the same curve and that curve can be used to estimate the value of s and t. If you look at the equation for s and u one thing which is clear is that if s and t are known and we want to find out draw down there is no problem because for any given s and t and knowing the distance of the observation u can be computed. Once u is computed w u can be easily obtained from the tables graphs or some equations and therefore, we will be known. So, knowing u we get w u knowing w u we get s, but the other problem where s is known or measured and find s and t. This is an inverse problem in which we are estimating the parameters and this is not directly solvable because if s and t are not known then u cannot be computed and if u cannot be computed then you cannot obtain w u because this also involves the unknown t. So, we will look at some of the methods which try to estimate value of s and t from measure data of s and there are some methods which are graphical some which are numerical and we will look at all these methods. The simplest method to estimate the values of s and t is based on the Jacob's approximation. So, in the Jacob's method as we have discussed we use the approximation that w u and since u is r square s over 4 t t. It gives us the value of s and t. So, the drawdown s as q over 4 pi t w u which is approximated by this and it can be written as natural log of 2.25 where this 2.25 comes from this factor of 0.5772 using this approximation and from the data of s versus t. So, what we have in the field is measurement of drawdown at different times. The t value typically is taken in minutes, but any other unit can also be used s generally will be in meters and the plot would look like this. The rate of increase will go down as we continue the bumping. So, initially the drawdown will be increasing at a fast rate and then it becomes almost flat as we move away. So, this one the data available to us is this and what we want is estimate t and s. We assume that Jacob's approximation is valid, but we know that it is valid only for u less than 0.01. So, that is an implicit assumption when we use Jacob's method is that whatever data we are using has values of u less than 0.01 and if you look at the definition of u, you will notice that t has to be large which means that some of the early data will typically not follow the Jacob's approximation. So, what we do using this approximate equation, we see that s and log of t natural log of t should fall on a straight line. So, that is what we will plot we prepare a curve in which t is on the log scale. So, we may start with 1, 10, 100,000 and s is on simple arithmetic scale. So, s may be 0, 1, 2 the data once it is plotted on this let us say the data points lie like this in which as you can see there is a straight line which fits the data, but some of the data which are circled here will not fall on the straight line. This is because the time is small and therefore, u value is larger than 0.01, but once we see the data point which is falling on a straight line, we know that this data follows Jacob's approximation and therefore, we can use this straight line to obtain the parameter values s and t. So, let us write it again the value of s in terms of now if we take the slope of this line this is a straight line the slope of this straight line which is d s over d of l and t would give us q over 4 pi t, we are assuming that r is constant. So, we are using data from a single well otherwise instead of t we have to use t over r square, but for this analysis let us assume that r is a constant. So, we have a single observation well at which we are measuring the draw down if we assume that r is a constant then we can use here l and t and keep this r square within this constant value here. So, the slope of the draw down versus log time curve will give us q over 4 pi t and from that slope t is the only unknown once we find the slope we can obtain the value of t. The slope d s by d l and t can be easily obtained from this plot. So, let me draw the plot here a little more detailed. So, these are the data points which follow a straight line and then there are some data points which are away from the line and this is the line which represents the Jacob's approximation. The t value is let us say 1, 10, 100 and then s value may be 1, 2, 3. In order to find out the slope what we can do is we take one log cycle of t and we can find out what is the change in draw down in one log cycle of t. That will give us the slope of s versus log t on base 10 and then we can convert it into natural log and we get the equation for estimating t 2.3 q over 4 pi delta s where delta s represents the change in s in one log cycle of t and 2.3 comes because of conversion of the base from 10 to natural log e. So, we can see that the slope of this line gives us transmissivity directly. Now, this line intersects the s equal to 0 line at some time which we will call t naught. So, this means that at t equal to t naught s is equal to 0 and if we use that condition we can see from this expression s equal to 0, t equal to t naught which means that this term should be equal to 1 because natural log of 1 will be equal to 0 and from there we can obtain the value of s and t is already known from here in 2 5 t 0 t over r square. So, the procedure is first we have to plot the data on a semi log plot where t will be plotted on log axis s will be plotted on arithmetic simple axis. Once we plot the data we look at the data and see what portion can be fitted by a straight line, early portion will not be on the straight line, but the late portion will be on the straight line. We extend the straight line to intersect the s axis at time t equal to t 0, we also note down the change in draw down in 1 log cycle of t and we call that data s. Using the value of data s we can estimate t using the equation 2.3 q 4 pi data s q of course, is known to us the pumping rate constant value, data s we obtain from this line. Once we get the value of t we use this intercept t naught, we say that s will be 0 at t equal to t naught from the j cups approximation for s to be 0. The term here for the argument of natural log should be equal to 1 and therefore, s will be obtained as 2.25 t naught t over r square where t already we have obtained from here. So, in this way both t and s can be estimated from the data from an observation will draw down data from an observation will, if we have more than 1 wells then instead of t we will be using t over r square. The only problem with this method is that we need data for very long time, sometimes when we plot t versus draw down. The data may not show straight line behavior even for very long time. So, this may be 1000 minutes or sometimes even more than that it is possible that depending on the storage coefficient and transmissivity values, the value of u corresponding to this time the largest time may be more than 0.01. So, in that case we need to continue the test for a very long time, which may not be economic. Therefore, we should look at some method which uses the entire data set rather than depending on j cups approximation for fitting a straight line. That method is called one of these methods is called the Thais type curve method and in this method we use the entire curve. The type curve tries to compare two different curves in one of them we plot the well function variation with respect to u or 1 over u and in the other curve we plot t over r square versus the draw down s. The idea is that these two curves should be similar, because if you look at the equations. So, if you compare these two equations the t over r square let me write it here in the form of t over r square. Sometimes we use w u and u here and then s versus r square over t, but most of the times it is preferred w u versus 1 over u and s over t over r square. So, if you look at these two equations we can see that relation between s and t over r square and relation between w u and 1 over u would be similar only a shift of the axis would be there. For example, if you take the log of this and similarly if you take the log of the second equation. So, this tells us that there is only a shift which is given by these two parameters in the curves s versus t over r square on log scale and w u versus 1 over u again on log scale. w u versus 1 over u does not depend on the data points this is a fixed function w of 1 over u and the plot would look like this. So, this is the type curve with which we have to match our plotted data s versus t over r square and that data may be like this or it may be like this depending on the values of s and t you may have a curve like this the idea is that we match these two curves. So, we plot the values in this curve move this curve here in such a way that the data lies exactly on the type curve or in some cases if the data is like this the curve may have to be plotted somewhere here or the axis has to be moved somewhere here to match with the type curve. So, depending on the s versus t over r square data the shifting of the axis will be different this is known as the type curve matching and this gives us a method of estimating the values of s and t. So, let us say that we have matched the type curve this is 1 over u versus w u which is a fixed curve and let us also say that we have matched the data by shifting the axis here and now we want to find out the values of s and t. We can choose any point on this graph suppose 1 over u values are 0.1, 1, 10, 100 w u values may be 0.001, 0.01, 0.11. Similarly, this t over r square values may be 0.11, 10, like this and s value again may also be 0.1, 1, 10. So, the scale of course has to be same the log scale here and log scale here they would be the same. Now, we can choose any point on this let us choose typically we choose point which gives us easier computations but we can choose any point we can choose any point on this curve. After matching we can choose any point on this graph and we note down the coordinates for 1 over u for w u for t over r square and for s. So, let us call these values as w star s star t star over r square and u star. We are measuring 1 over u but let us call that value as 1 over u star. So, this will be 1 over u star this will be w star this value on t over r square scale will be t star over r square and this value will be on the s scale s star and using these we can easily obtained the values of transmissivity as q over 4 pi and then the storage coefficient s this is using the two equations which we had derived earlier. So, once w star s star t star and u star are known then t and s can be estimated the only thing is that it is a graphical method. So, there is some subjectivity involved in choosing or matching the type curve but once it is matched the computations are very straight forward and as you can see if we take w star as 1 and u star as 1 the computations become a little simpler that is why typically we choose a point which gives us. So, this point will give us w star of 1 and u star also of 1 for that we can find out s star and t star and use this to get t and s. So, we have seen some methods of estimating the parameters s and t for example, Jacob's method which require data for very long time then we have theis type curve which can use all the data set and compute the values of s and t. We have also discussed all these methods only for confined aquifers unconfined aquifers are little more complicated because the thickness is not constant it depends on the head edge and we will have to make some simplifying assumptions to obtain the equation of motion that we will do in the next lecture.