 Welcome to this lecture number 13. This will solve one or two numerical problems in the steady flow into wells and then we will move on to unsteady flow into wells. First, let us consider a numerical problem of steady flow fully penetrating well in an unconfined aquifer. So, the data is well diameter that is 2RW that is a 45 centimeter. So, this is the given data in thickness of unconfined aquifer which is a saturated thickness that is h is 30 meters I am sorry it is 25 meters and the steady discharge that is Q is 2100 liters per minute observation well radial distances that is R1 and R2 R1, R2 this is 30 meter, 90 meter drawdowns in observation wells S1, S2 that is 5 meter, 4 meter and to estimate so based on this data that is well diameter saturated unconfined aquifer thickness steady discharge observation well radial distances obviously this is from center of the well that is R1 and R2 and the drawdowns in these observation wells in these two observation wells. So, these are the data given so to estimate the coefficient of permeability 1 coefficient of permeability K that is transmissivity that is T. So, now we know that for steady flow through an unconfined aquifer we have the relationship for the discharge in terms of the two drawdowns or the one is given by pi K h2 square minus h1 square divided by ln of R2 by R1 and here so this h2 is equal to h minus S2 and in this case this is h is given which is say 25 minus 4 which is 21 meters and h1 is equal to h minus S1 which is 25 minus 5 which is 20 meters. So, let us substitute the values therefore this Q which is given in terms of liters per minute. So, that is 2100 and so let us convert this into meter cube so divided by 1000 so this will be meter cube per minute this is equal to pi into K into h2 minus h1 square that is 21 square minus 20 square divided by natural log of that is R2 by R1 and in this case this R2 by R1 is R2 is 90 and R1 is 30. So, therefore K will be given by 2.1 into natural log of 3 90 by 30 is 3 divided by pi into so this is 441 minus 400. So, this is 21 square is 441 and 20 square is 400 and so this will directly give the value of K in terms of meter cube per minute. Let me use here the calculator it is 2.1 into ln of 3 divided by into 41. So, this will directly give the answer of coefficient of permeability K as 0.0179 so this is meter per minute and this the transmissivity T is simply given by K times h. So, this into h is 25 so this will be 0.448 meter square per minute. So, these are the answers the K value is 0.0179 and then T is 0.448 meter square per minute. So, this is how we can use this the data from the well to determine the to estimate the formation constants of either formation constants of the aquifer or if they are given the steady rate of discharge. Now, let us come to the unsteady flow into wells here. So, the flow is varying with time and firstly let us consider the unsteady flow through a confined aquifer. Consider a well a confined aquifer and that to a fully penetrating well. So, wherein the there is an unsteady flow. So, let us draw the diagram. So, this is the ground level this is the confined aquifer and this is the well this is the lower impermeable boundary of the confined aquifer and this is the upper impermeable boundary of the confined aquifer. And for simplicity let us consider the thickness as b and the flow is varying here. So, in this case this is the unsteady flow through a confined aquifer. Obviously, so here so there is a radial flow. So, consider a fully penetrating well through a confined aquifer receiving radial inflow. So, in such case the equation is given by the equation for the ground water flow in the radial coordinates and here obviously. So, let me also show the water table the original water table which is horizontal. So, this is the original W t and then so this is the draw down and obviously. So, here at a distance at a for a general section the height of this one above the this is s and the draw down is s and the height is h. So, now in such case we can write down the governing of ground water flow equation radial coordinates in say cylindrical coordinates and here you can say this is the homogeneous comma isotropic confined aquifer is d square h by dr square plus 1 by r. So, this is the radial distance r where the head is h and the draw down is s. So, this is 1 by r into dh by dr. So, this is the second partial derivative of h the variable head h in the unconfined flow. I am sorry in the unsteady flow through a confined aquifer through a fully penetrating well in a confined aquifer. So, this must be equal to s by t storativity by transmissivity into dh by dt and since this is an unsteady flow. So, this term so this dh by dt is a nonzero value. So, this is the governing equation. Now, in this case and here you can write down this the total head the undisturbed head or the piezometric head is capital H and here you can write down. So, this obviously we can write down the expression the draw down small s is equal to the piezometric head capital H minus the variable head in the well along the draw down curve. So, this is the draw down curve this is the unsteady draw down curve. So, this draw down s is given by this capital H minus small h. So, for this case if we so it is Thase in 1935 obtained the solution using electrical analogy as follows. So, that is why this equation is known as Thase equation. So, the equation is for the draw down s the variable draw down s which is equal to capital H minus small h the piezometric head the constant piezometric head minus the variable head of the draw down curve H measured with respect to the bottom impervious layer of the confined aquifer. So, this is equal to q divided by 4 pie into t into integration that is u to infinity e to the power minus u divided by u into d u. So, here so this term u is known as the well function parameter. So, this is known as the well function parameter this is equal to r square into s the storativity divided by 4 t into t. So, this is the transmissivity t and then this is the time since pumping since beginning of pumping and this r is a radial distance of the where the draw down is s. So, in this case so this r square s by 4 t t is denoted as a well function parameter and its integration that is e to the power minus u divided by u into d u between the limits u to infinity and this whole thing is known as well function denoted by w u. So, therefore, the equation for this the solution for the governing equation is the draw down the variable draw down s is simply given by q divided by 4 pie into t into the well function w u. So, this well function where this well function w u is equal to integration u to infinity e to the power minus u divided by u into d u. So, this has been evaluated as a minus 0.5772 minus natural log of u plus u minus u square divided by 2 into 2 factorial plus u cube divided by 3 into 3 factorial minus and so on. So, this is the expansion of this integral. So, therefore, we can write down the and this is known as the this is known as Theis equation for unsteady draw down through a fully penetrating well and it is in an unconfined aquifer. So, this fully penetrating well in I am sorry confined aquifer. So, this is the Theis equation and now let us write down say that is and here obviously. So, there are the various tables are given for well function value for different well function parameter and so this Theis equation can be solved by these graphical methods. So, solution for Theis equation by let us say this is the graphical methods. So, in this the first one let us consider it as the type curves one is that is by type curves two by Cooper Jacob approximation. Now, let us consider the solution of Theis equation by the first method which is by type curves the graphical method by type curves. So, here what is done is the expression for the unsteady draw down is given by q divided by 4 pi t into the well function W and we know that. So, this is the well function parameter u is given by r square s divided by 4 t into t. So, that is we can write this down as t is equal to r square s divided by 4 t into 1 by u. Let us denote this as equation 1 and this as equation 2. So, this equation 1 and equation 2 can be written can be written as log of s this is the unsteady draw down s. Now, basically here this q is the steady rate of pumping. So, here it has not reached an equilibrium. So, therefore, even though this q is steady the draw down s is unsteady it is varying with time. So, this is equal to log of q divided by 4 pi t plus log of the well function W and similarly. So, for the second equation. So, we can write this as if you take the logarithm. So, this log of t is equal to log of r square into s divided by 4 t plus log of 1 by u. So, here so these are the two equations the first. So, let us denote this as equation 3 which is simply the logarithm of equation 1 and this is equation 4 which is the logarithm of equation 2. So, here so this comparison of 3 and 4 will tell us that if we plot s versus t that is the draw down versus the time since the pump since the beginning of pumping on a log log scale and compare that with the plot of this log of W u versus log of 1 by u and match them. So, that the both the coordinate axis are maintained parallel. So, in that case we get what is known as the first let us say this is the plot of log s versus log t and we get the second plot of log W u versus log u log of 1 by u. So, in this case so this is log s versus log t is based on the pumping test data. Obviously, as t increases so this s will increase. So, this the number of data points the represent the number of readings in the pumping test taken and then this log of W u versus log of 1 by u. So, from the tables so this is the type of curve. So, here in this type curve so what is done is so these 2 plots are matched that means on this. So, on the plot of this log s versus log t this log of well function W u and log of reciprocal of the well function parameter 1 by u. So, is superimposed such that both the pairs of axis are parallel in such a way that. So, there are the curves more or less match in such case what is done is. So, this is by trial and error this is done and then corresponding to any matching point on the curve. So, here say suppose this is the matching point. So, this is the and corresponding to this we have the log s value given along the log s axis. The log of W u value given along the log of W u axis and then this will be log of 1 by u and this will be log of t. So, if you substitute these values then and of course, we have the data that is log s value is there and then log t value is there and then. So, we get this then we can find out the that is both this log of q by 4 pi t as well as log of r square by 4 pi t. So, then we can solve for this and get the values of this is. So, after get the values of the aquifer parameters. So, after obtaining the matching point ensuring the matching of both the type curves. So, this is type curve 1 which is basically the plot of drawdown versus time on a log log scale. In type curve 2 is a plot of log of well function W u versus log of the reciprocal of the well function parameter 1 by u. So, maintaining the parallel nature of both pairs of axis. So, we get the matching point and corresponding to any matching point we can get this log of s log of W u log of 1 by u and then log of t. So, once we get all these. So, simply substitute the values. So, then we can solve for. So, substitute the values of log s log t log W u log 1 by u and estimate s comma t by equation 3 and 4. So, this is the first graphical method for determining the aquifer parameters using the Theis equation. Now, let us go for the second graphical method which is the aquifer parameter that is. So, this is the graphical solution of Theis equation by Cooper Jacob approximation. So, in this case. So, obviously here and of course, this has some limitation for u the well function parameter less than or equal to 0.01. So, in such cases only this approximation can be used. So, in such cases the well function. So, here say for u less than or equal to 0.01 the well function that is W u is given by just the two parameters. Just the first two terms well function is approximately equal to minus 0.5772 minus natural log of that is the well function parameter which is r square s divided by 4 t into t. So, therefore the expression for the drawdown s will be given by q divided by 4 pi into transmissivity t multiplied by minus 0.5772 minus natural log of r square into storativity s divided by 4 t into t. So, this is the transmissivity and this is the time since pumping and this is the drawdown. So, in this case. So, obviously so this is the approximation or so for this the value the well function has to be less than say 0.01 less than or equal to 0.01. So, in such case so this is this s can be approximated as q divided by 4 pi into t multiplied by the natural log of that is 2.25 t t divided by r square into s. So, this is the approximation and in this case. So, what is done is so here a semi log plot of drawdown versus time the drawdown s the unsteady drawdown versus the time. So, that is it is plotted on a log scale log of t this case. So, obviously it will have to so this is here it starts with it has to start with this one. So, here we get a number of this data points and this one. So, these data points the data points having the higher values of time. So, they for a straight line and if we extend this. So, this is log of t 0. So, in this case so basically so when s is the drawdown the unsteady drawdown s is equal to 0. Then t is equal to t 0 or in other words in this case and here we can say that so that is the in this case the argument of this logarithm has to be 1. So, that natural log of 1 is equal to 0. So, in that case s is 0. So, that is that is 2.25 t t 0 divided by r square into s this must be equal to 1. So, therefore in the Cooper Jacob approximation what is done is a semi log plot of drawdown on the linear scale and then this time t on the logarithmic scale. So, that is plotted and then so the data points having higher time values. So, they will be forming they will be having a linear relationship. So, this linear relationship is extended backward. So, to intersect the logarithmic scale of t and wherever it intersects. So, that intercept will give the value corresponding to the time that is t 0 where in the drawdown is 0. So, therefore here what is done is in the Cooper Jacob approximation. So, this t 0 is determined and once this t 0 is determined. So, then we can say that. So, the storativity s is simply given by 2.25 t into t 0 divided by r square and before this we need to know this value of t and we know that. So, the suppose denote this as equation 5 and so based on this equation 5 equation 5 any 2 data points any 2 pumping test data points having drawdowns s 1 comma s 2 at times t 1 comma t 2. We have s 2 minus s 1 is equal to q divided by 4 pi into t into natural log of t 2 by t 1. So, here this is equation 6 and from this equation 6 you can write down the expression for this transmissivity t. So, this t is equal to q divided by 4 pi s 2 minus s 1 into natural log of t 2 by t 1. So, this is the equation 7 and by this and here obviously this s 2 s 1 t 2 t 1 they are obtained by the pumping test and obviously q is also obtained by the pumping test. So, by the pumping test data where the discharge is constant only the drawdowns are variables are unsteady. So, therefore, we can estimate the value of transmissivity t and once we estimate the value of transmissivity. So, then substitute this value of transmissivity and the value t 0 obtained by the extending backwards the linear portion of the semi log plot of drawdowns versus the time t. So, we get we can estimate the storativity. So, this is how we can estimate the parameters the aquifer parameters the storativity s of the aquifer as well as transmissivity t of the aquifer. So, this is this we can get the this unsteady this one and here we should know the approximations. So, it is the assumptions on which this Thay's equation is based on. So, assumptions for Thay's equation for unsteady flow fully penetrating well in a confined aquifer. So, let us list these assumptions. So, the first assumption is aquifer is homogeneous isotropic of uniform thickness and of infinite radial extent I am sorry, aerial extent. The second assumption is before pumping the piezometric surface is horizontal. The third assumption is the well is pumped at a steady discharge rate. The fourth assumption is the pumped well is fully penetrating the confined aquifer the streamlines are horizontal everywhere. The fifth assumption is the well diameter is small. So, that there is negligible storage and the last assumption is water removed from storage is discharged instantaneously with the decline of head. So, these are the six assumptions on which the Thay's equation for unsteady flow is based. And so, in the next lecture we will discuss the unsteady flow through wells in unconfined aquifers as well as leaky aquifers and here we should note that. So, of course, I have already made it amply clear. So, the well is pumped at a steady discharge. And so, in the next class we will discuss the unsteady flow through the well in unconfined aquifers. Thank you.