 Today, we will continue with our discussions of ground water. We have already seen the Darcy's law which relates the hydraulic gradient and the velocity with the constant of proportionality, which is known as the permeability. And we have also seen how when we pump a well the water level or the piezometric level goes down. So, let us see what happens when we pump a well. So, this is the ground level and let us say we have a confined aquifer being pumped through a well. As we have seen the piezometric surface will go down as we start pumping and the difference is known as the draw down typically denoted by s. So, this s as we have seen is proportional to q because we have assumed that the flow is laminar. This is generally true, but when we are pumping from a well typically the velocity is very large near the well and therefore, near the well the flow will be turbulent. Similarly, near the well because of the casings when the flow passes through the casing there will be additional head loss. We combine these and call them the well loss and we write the total draw down at the well as summation of the laminar loss, turbulent loss and the casing loss. So, this will be total draw down at the well and if we draw the same figure again. This is the initial piezometric level before the pumping is started and then a final piezometric level after the steady state has been reached and this is the draw down at the well. So, if we say that there is only laminar flow then there will be head loss or a draw down because of the laminar flow. This is because of turbulent flow near the well and this is the loss due to casing well casing. As we have seen the laminar loss generally is proportional to the discharge. So, swl will be proportional to q. The turbulent loss is normally q to the power something and typically we can take it as q square. So, we can write swt is proportional to q square. The casing loss has also been seen to be proportional to q square. So, both of them swt as well as swc both of them are proportional to q square. These are known as the well losses and since both of them are proportional to q square we can write that the well loss is proportional to q square or we can write as w, some constant into q plus another constant into q square, where this c 2 q square represents the well loss. And we want the well loss to be as small as possible, so that the well is considered more productive. To find out these constants c 1 and c 2, we can pump the well at different rates and note down the draw down at the well. And we can write this also as which indicates that if we plot sw over q versus q, we would get a straight line and the intercept will give us c 1, the slope will give us c 2. So, if we pump the well at different rates note down sw for different q values, we can compute sw over q and plot it against q and get the values of c 1 and c 2. We want c 2 to be small, so that the losses are minimum. If well losses are large, for example, if c 2 is large, large c 2 will indicate in efficient well. And if we plot sw versus q, you would get a straight line with the coefficient c 1 and the slope as c 2. So, this method gives us a way of finding c 1, c 2 and we see that if c 2 is large, the well is inefficient. So, we want to have a small c 2 for the well. If the c 2 is very large, then the well may be inefficient and we may have to look for replacements. The other concept which is related to the well loss is known as the specific capacity and specific capacity is defined as the discharge per unit drawdown. So, naturally we want large discharge and small drawdown. So, specific capacity of well should be high, so that it is efficient. If we look at the equations, we have already seen that we can express the drawdown as q over 2 pi t log of r over r w, where r is the radius of influence, r w is the radius of the well, t is the transmissivity. So, we can see that a specific capacity is proportional to the transmissivity from this equation. If the flow is transient, then we have the equation using the Thais equation and the Cooper Jacob approximation and including the well loss. If you look at the previous equation, there was no well loss term included here. So, we have assumed ideal conditions here without any well loss, steady state conditions, but if the flow is transient and there is well loss, then we have to include this c 2 q term also and this is using the Jacob approximation for well function. So, on the transient conditions, the specific capacity which is q over s w will be given by this equation and we can see that a specific capacity decreases with increase in q and t, because the increase in q, this term will increase, will increase in t, this term will increase and therefore, specific capacity will decrease. So, any well, if we pump at a higher rate, specific capacity will be smaller, if we pump it for a larger time, specific capacity will decrease. Now, let us look at some of the conditions which occur in the field. For example, till now we have discussed most of the things assuming the aquifer to be homogeneous. That means, there is no variation of conductivity with distance, but most of the times when we see natural porous medium, they will be layering. For example, deposition of if this is the bedrock, then there may be a layer of material here. Depending on how the porous medium is formed, there would be layers of different materials which will have different conductivities. So, for example, this may be a thickness b 1 and conductivity k 1. Similarly, there will be another layer which may be of thickness b 2 and conductivity k 2 and so on. So, under these conditions, we want to look at how to obtain an effective conductivity for this area. For example, we can assume or we can replace this by an equivalent porous medium. Let us say that there is an impermeable layer here also. Then, what we would like to do is, we would like to replace this porous medium by an equivalent homogeneous porous medium of height b. So, to find out this k equivalent it depends on whether the flow is taking place along the layers or perpendicular to the layers. For example, there may be some recharge from the top, then the flow will be perpendicular to the layers and if there is natural groundwater flow horizontal direction, it will be along the layers. Under both these conditions, the equivalent conductivity are different. For example, in this case, the discharge q 1, q 2, q n, they would be added up to get the total discharge, but the head loss from one point to the other point would be same for all the layers. In the case of flow being perpendicular to the layers, the discharge is same through all the layers q, but the head loss in each layer will be different and has to be added to get the total head loss. So, doing these computations, we can obtain the equivalent conductivity for this case. For example, in this case, the equivalent conductivity can be given by the arithmetic mean and then the harmonic mean. So, these two can be used to obtain the equivalent conductivity and then we can treat the porous medium as homogeneous with that equivalent conductivity and the thickness b, which is b is nothing but sigma b i and i will go from 1 to n, where n is the number of layers. So, after looking at these flow situations, where the flow may be steady, unsteady, we may be having a one dimensional flow or radial flow towards the well, we may have layering and non-homogeneous conductivity. So, we have looked at all these flow situations. Now, we will take some examples to explain some of these concepts. So, we will take the example of a confined aquifer and unsteady flow through the confined aquifer. The data, which is assumed is that there is this charge of 2 meter cube per minute. So, we have a confined aquifer being pumped at this rate q. The transmissivity of the aquifer is given as 1 meter square per minute. Storativity as 0.001. Now, there is an observation well, which is 50 meters from the pumping well and the water level in this observation well is being monitored. So, r is 50 meters. Using this data, we want to find out how the water level in this piezometer. So, initially there is some water level. So, the piezometer will show the same water level. When we start pumping with time, this water level will be going down. So, we want to see how this water level goes down and that we can do using the Theis method. For example, suppose we want to find out what will be the draw down at let us say 1 minute after the pumping starts. So, if we take t equal to 1 minute, we know that the equation for the draw down is the draw down q over 4 pi t times the w u well function and u is r square s over 4 t. So, in this case, suppose we take r equal to 50 meter that is what we want. So, r square s will be 2500, s is 10 to the power minus 4, 4 t is 1 and r square then t is also 1 minute. Now, we could have used second also, but typically in the draw down conditions, minutes are preferred. So, we will use the minutes here and for discharge meter cube for transmissivity meter square, but the time units we will be using as minutes. So, if you do this, u which is a dimensionless quantity will come out to be 0.0625 and corresponding to this u, we can look at the well function tables or there are some formula which give us the well function for any given value of u and using that we can see that w u is equal to 2.26. So, putting the value of w u here and the value of q s 2 t as 1, we can get s equal to 0.36 meter which is what is shown here. So, what we can do is we can draw down. So, for 1 minute we have 0.36 meter of draw down. Similarly, you can see that at the end of the measurement which is 240 minutes or 4 hours there is a 1.22 meter draw down in the observation well which is 50 meters away from the pumping well. This data is shown in the plot here. So, we have the time here in minutes and s is in meters. So, this shows the time was a draw down curve at the observation well which is 50 meters away from the pumping well. And we can see that initially the draw down is increasing very fast, but then as time progresses the rate of increase becomes slow. It will not reach a steady state because the Thais equation assumes that the aquifer is infinite. So, slowly the radius of influence will increase and there is no limit to it the increase of the radius of influence. Therefore, this will not achieve a steady state it will continue to increase, but at a very slow rate. So, this data which we have computed based on the Thais curve most of the times what we will not have is t and s. We will not be knowing the values of t and s, but what we will have in the field is the time versus draw down data or the time versus draw down curve which is this curve. So, the second question which comes to the mind is if this curve is given to us given s versus t curve how to find t and s. So, this is an inverse problem or a parameter estimation problem in which from a given draw down data we want to estimate the transmissivity and the storage coefficient. One of the methods which we have seen is known as a Jacob's method or Cooper Jacob method in which we say that for very large value of time. So, large t the well function can be approximated by the first two terms and then what we do is we fit on a log plot we fit a straight line through the data corresponding to large time and extend that straight line up to the 0 point the draw down equal to 0 point and we call this t 0. In this case we see that t 0 comes out to be about 0.11. The other thing which we need in Jacob's method is what is a draw down per unit log cycle of time. So, we can look at 10 and 100 we can take any log cycle, but let us take this and denote this by delta s. Delta s it turns out to be is about 0.36 meters. So, that is the difference between this point and this point which is one cycle of log here. We could have taken 1 and 10 also and get the same value here or here. So, using the data on delta s and t 0 the Cooper Jacob method has the equations 4 pi delta s and s. So, as we have seen q is 2 delta s is 0.36 and this gives us a transmissivity of 1.02 meter square per minute which is very close to the transmissivity 1 meter square per minute which we had assumed to generate this data. So, this method works quite well for estimating the value of t. Similarly, once the t is known we can put that t here t 0 is obtained as 0.11 minute and r is 50. So, s will come out to be 1.01 again very close to the value 1 into 10 to the power minus 4 which we had taken to generate the data. So, the Jacob method works very well for cases for cases where we have a straight line portion available to us or where the measurements have been carried for large time. If we do not have this large time data available suppose we had only this data available to us then we would not be able to fit a line here or even if we fit the line it will not be the correct straight line. So, Jacob's method will not work if the data is or the experiment is not carried out for a long time and data is not available for very large time. So, this straight line portion has to be available in order for Jacob's method to be applicable. Therefore, what we do mostly is use the Thais curve fitting method or the type curve matching method in which we prepare a generic plot. So, this is a generic plot, generic curve or the type curve which plots 1 over u versus w u. Now, 1 over u as we have seen 1 over u is equivalent to t and w u as we have also seen is equivalent to s. So, on a log log scale if we plot 1 over u versus w u it should be similar to the behavior of the drawdown curve. So, this should be similar to s versus t. So, the idea of Thais curve matching as we have discussed is try to match this curve, the type curve with the actual data. The actual data which we have plotted on log log scale. So, this is the same data which we had earlier, but earlier we had plotted on simple scale now we plotted on log log scale. So, this is s in again in meters t is in minutes and now this curve should match with this curve. As you can see both curves it will not match as we put it as it is. So, we have to shift the axis in order to make a match and that is what is done here. This is the drawdown curve the symbols and the line shows the type curve. So, we shift the drawdown curve or the type curve the axis should be parallel. We shift it in such a way that all the drawdown curve data matches with the type curve and by this shifting when we get the perfect match then we note the ordinates which match for example, for some w u we find out corresponding s and for some value of 1 over u we find corresponding t. And then we use these values to estimate the transmissivity and the storage coefficient. So, the if we match it suppose we take 1 by u as 10. The corresponding t star which is the point which corresponds to 1 over u equal to 10 is about 0.6. Similarly, we can match w u equal to 1 and 1. So, w star equal to 1 and corresponding value of s comes out to be 0.16 from this figure. So, this is s star this is t star this is 1 over u star and this is w star. So, using these 4 values of the starred variables we can use the thies curve matching equations to obtain the transmissivity. So, the equations which are given t equal to q over 4 pi w star over s star and s equal to 4 pi t star u star over r square. So, w star here is 1 s star is 0.16 putting these values we get t equal to 0.995 again this value is very small very close to 1 and s comes out to be 9.6. So, this is also very close to 1004 minus 4 which we had obtained or which we had used to generate this data. So, type curve matching method involves some subjectivity in this case you can see that the fit is very good here that data points all lie exactly on the type curve because the data which we have generated is synthetic there are no errors. In general the data would not lie exactly on the line it may have some values like this. So, we have to adjust what is the best fit once we get the best fit w star s star u star and t star can be obtained and they will give us the value of the transmissivity and the storage coefficient using this method. Now, let us look at the first slide which we had showed us time versus draw down for values of q equal to 2 t equal to 1 s equal to 0.001 and r equal to 50. Now, we can see what will be the effect of changing t and s. So, we can generate a new data set which includes t value which is one-tenth and s value which is 10 times. So, the value which we use for t and s now are t equal to 0.1 and s equal to 0.001. So, using these values we can regenerate the data using the thys curve and you can see that this figure shows time versus draw down. Earlier for a time of 1 minute we had a draw down of 0.36, but here you can see that there is almost no draw down till 1 minute, but at the end we had a draw down of about 1.22 meter here we have a draw down about 4.93 meters. So, changing t and s will affect the rate of draw down or the shape of the draw down curve and if we plot the draw down curve it looks like this. So, we can see that earlier one had a shape like this, this one has a shape which is much more steep. So, again we can use the thys curve matching technique to match this observed draw down curve with the type curve and of course, in this case the shifting will be different than the previous case where the curve was like this. So, here you can see again that the data now plots in this portion again a very good match is being observed. Now, we can again obtain the same for example, this is 1 over u equal to 1 and it corresponds to t of t star of 7. Similarly, w also we can take as 1 and it will correspond to s star of. So, using this data we can again obtain t in this case t is obtained as 0.094 meter square per minute compared to the value which we have used as 0.1 and s is obtained as 1.05 minus 3 compared to the 1 into 10 to the power minus 3 which we had used to generate this data. So, different values of t and s can be obtained using the type curve matching only thing is that the axis will have to be shifted literally different for different kind of data and the match it depends on our judgment a little bit where to match this. Similarly, since it is graphical method the reading of t star and s star would also be a little approximate for example, the value here and the value here there may be some approximation involved in reading these values which may cause some deviation of the values. For example, this shows about 6 percent error this shows about 5 percent error, but for most groundwater practical cases 5 percent 6 percent error is tolerable. So, this method can be used these days there are lot of other methods which are proposed for example, we can use computational methods to try different values of s and t and generate the low down data try to match it with the observed low down data and then try to minimize the error by choosing optimum values of s and t. So, computational methods are becoming more popular these days, but graphical methods are still very good for a first approximate guess. Now, let us look at another example for example, this data we had generated for r equal to 50 meters or the first slide showed data which had t equal to 1 s equal to 0.001, but again at r equal to 50 meters. Now, what happens if we have two different wells. So, suppose we have the confined aquifer being pumped and then we take the observations in two different observation wells one may be let us say r equal to 50 meters the other may be let us say 100 meters. So, naturally the 50 meter well will show a larger draw down 100 meter well will show a smaller draw down. So, this curve shows t again in minutes and s in meters the draw down corresponds to 100 meter well and 50 meter well. We can see from this figure that these open circles which represents the 50 meter draw down they are higher than 100 meter. So, the draw down at the 50 meter well would be higher than that at 100 meter well and these two are shown here. Now, since these draw down curves are different we can estimate the t and s values either separately for both of them or there is another method which uses combined data and if you look at the definition of u then we can see that this r square over t term occurs in the definition of u. So, instead of plotting t versus s if we plot t over r square versus s then both of these curves may come at the same location. So, that is what is shown here again the filled circles are 100 meter draw down and the open circles are 50 meter draw downs. We can see also that at the end of 4 hours which is 240 minutes the draw down in the well at 100 meter is about 1 meter and the draw down in the well at 50 meters is about 1.2 meters. So, the combined curve shows the same thing the data for t over r square for 100 meter well stops here which is 1 meter draw down the data for the 50 meter curve goes up to 1.2 meter draw down. So, t over r square what we are doing is kind of non dimensionalizing the time with respect to the distance. So, t over r square versus s curve t again s is meters t over r square here is minute per meter square. So, the curves the two curves which are quite different here. So, these two are quite different here, but may be make it non dimensional respect to r square then we get the same curve and now we can use this single curve for curve matching or the type curve fitting and we can see that the data fits again very well. So, here these again open circles these are the 50 meter well the filled circle there represent the 100 meter well this is the type curve and this match again shows a very good matching between the data and the type curve. The procedure is exactly the same. So, let us say that we take 1 over u star equal to 1 which is this line and it gives us a t by r square equal to 2.4 10 to minus 5 which is corresponding to this point the scales are shown on this line. So, I have taken it here similarly I will write this as also star t over r square star similarly for w star we can take n 1 which is this and it corresponds to s star of 0.16. So, this is 0.1 this is 1. So, this point is 0.16 and using these now we have t 0.995 and s. Again these two are very close to the values which we had used to generate the curve for t over r square versus s. This was used as 1 again this was used as 1 into 10 to power minus 4. So, this has 2 percent error this has 0.5 percent error which is quite good. The only thing which needs to be mentioned here is that s is given as 4 pi t star u star over r square. So, now instead of t star we will have t star over r square directly substituted from this value. So, using the thious curve we can estimate the value of s and t for any combined aquifer. Most of the times we will have data for one well or sometimes more than one well and we can combine them using the t over r square to get all the data on a single curve and then match with the type curve. Cooper Jacobs method can be used if the data is available for a longer time or as the dimensionless parameter u should be a small. So, if it is not available then we will have to look for other methods. For example, the graphical method of curve matching or numerical methods of parameter estimation. There are some other advanced methods also for example, slope matching or derivative matching methods, but we will not discuss them here. So, this method tells us how to estimate the quiver parameters if we have data available for a well. Sometimes data from wells may not be available or there may not be a pumping well. Sometimes we may have just a few observation wells in an aquifer. So, we can look at a technique to estimate the gradient and then knowing the conductivity how to estimate the ground water velocity or if we know the ground water velocity we can estimate the hydraulic conductivity. So, suppose we have the ground level impermeable bed and then there is this water table which is at certain gradient. So, that the flow occurs because of this gradient. If we have observation wells here we can note down what is the ground level here and what is the ground water table at this point. So, this will give us an idea about what is the ground water level at this point. Typically what we will have is elevation at the ground level this data is easily available and what we will have the other measurement will be depth to ground water. If we have another well here and we also have the same data available here ground level as well as depth water table this will give us some idea about the gradient here. If we know that the flow is in this direction, but a number of times it is not possible. For example, if we look at the top view there may be some area here where there are wells here here and here and we do not have any idea about what is the direction of flow of the ground water. For example, it may be flowing like this it may be flowing like this or it may be flowing like this. Some idea about this movement may be obtained by noting down the ground water elevation here. So, if this is higher elevation and these are lower then we know that the flow will be towards this side, but exact direction of flow will not be known. So, for this case if we have three different observation wells and we have the measurement of ground water elevation and depth of ground water at the three wells. We can estimate roughly what is the direction of flow and if we know the conductivity hydraulic conductivity k. We can also find out or estimate what is the velocity of ground water movement or if we somehow measure the ground water velocity then we can estimate the hydraulic conductivity. So, in this case what we do is let us say that we have one well here, another well here, another well here and let us say we have measured at the wells A, B and C the ground water elevation by measurement of the ground level and the depth of water. So, let us say that at A ground level is let us call it 105 meters and depth to ground water let us call it 4.6 meters. So, if we have these two data available to us we know that the ground water elevation ground water level at A would be nothing but the difference of these two and it will be 100.4 meter. So, at A we have ground water elevation which is of 100.4 meter and similarly, let us say at B we have all the data available at C also and let us say that at B we have an elevation of 101 meter this is all ground water elevation. So, these are all ground water levels and at C let us say that the elevation is 100 meters. So, if we look at these three elevations 101 here, 100 here, 101.4 here we know that ground water flow will be from the higher head to the lower head, but it may be in this direction it may be in this direction it may be in this direction. So, in order to find out or estimate roughly what is the direction we can do a linear interpolation for example, this is from 100 to 101. So, we can estimate where on this line would be 100.4. So, if we divide this into five parts then this would roughly be 100.4 and that means the line joining A with this point will roughly be a line where the ground water elevation is 100.4 and then we can draw lines parallel to this from all the points they will represent ground water elevation of 100, 100.2, 100.4, 100.6, 100.8. So, knowing these ground water contours we can say that the ground water flow direction would be perpendicular to these and we can make a rough estimate of the gradient by measuring this length and saying that the head loss delta H is 0.2 meter in a length of L. So, delta H over L will be 0.2 over L. Knowing the conductivity we can say the velocity will be k i and knowing the conductivity and knowing the i we can estimate what will be the velocity of ground water or if somehow we have obtained the velocity of the ground water if we know this we can estimate what is the hydraulic conductivity. So, this will give us an estimate of the conductivity if we do not have a pumping well and we have some three observation wells which measure the ground water elevation. So, in the ground water chapter we have looked at various flow situations we have looked at the Darcy's law which relates the conductivity gradient and the velocity. We have also seen what happens when flow occurs between two parallel bodies of water one dimensional flow or sometimes we have a pumping well which will cause radial flow. We have looked at confined aquifers and uncomfined aquifers in which the behavior of or the distribution of the head is very different from each other. In the confined aquifer it is the piezometric head with changes, but the area of flow remains constant. In the unconfined aquifer the area of flow also depends on the head in the aquifer and therefore, it becomes more complicated. We have looked at unsteady state flow towards a well in confined aquifers and we have looked at a steady state flow in both confined and unconfined aquifers. We have seen some methods which are used to predict the drawdown for known aquifer properties and pumping conditions and we have also looked at the inverse problem in which we estimate the aquifer parameters from the measure data of drawdown versus distance or time and for a given pumping rate. So, for any aquifer we can design an experiment with different pumping rates to find out the aquifer properties. Similarly, we have looked at the well loss, how to estimate the specific capacity of a well, how to estimate the C 1 and C 2 constants in the well loss term and how to obtain or how to estimate whether the well is efficient or not and when it should be replaced.