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Module 1: Flow Behaviour of Unsaturated Soils

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Video 1
Hello everyone in the previous lecture we discussed Transient flows through horizontal and vertical columns or the transient flows without the influence of elevation head and with the influence of elevation head we have considered and we got analytical solutions. These equations are called Green and Ampt equations. This is given by the two gentlemen Green and Ampt in 1911 very old literature.This particular equations this is for the horizontal flow the x by the square root of time is = square root of 2 ks times h o - hi by Θ o - Θ s. Here Θ o or Θ s. So, here x is the spatial distance and t is a time and ks is the saturated hydraulic conductivity of the soil h o is thepositive head you maintain it could be 0 or more than 0. And h i is initial head and Θ s is saturated volumetric water content and Θ i is initial volumetric water content. Similarly, for the vertical flows, we have this expression instead of x you will have the independent variables z here this expression is implicit and you can obtain for a giventime what is wetting front location in the vertical direction can be obtained using this particular equation. Here the z is assumed to be positive upward. So, when there is an infiltration against the gravity that is taking place then we have used positive value for zand this is an expression for this. Again once again I remind you that these expressions are derived in 1911. So, when the computation was nearly a very costly and during that time the analytical expressions for the transient flows were derived. This expressions can be used for the coarse-grained soils, but not applicable for fine-grained soils because sharp wetting front generally you do not find for most of the soils, but that approximation is can be born for coarse-grained soils; for some coarse-grained soils. Interestingly capillary rise in soils has been studied slight little earlier by Buckingham in his classical work on studies on the movement of soil moisture which is published in 1907 very old work which is a classical work in soil science literature. In this particular work here done to establish the rate of capillary rise, he found the capillary rise data with time for several soils. The experiment involves somewhat like this; this figure is not from this work this is setup from our lab at IITG. And you will have a soil column it could be a sand column in this case and you have a reservoir, water reservoir these two are connected through these pipes.So, you have a water table maintain at a particular level. So, this is the water table that is maintained which can be read on the scale. So, when the water table level in the reservoir is this at equilibrium the water table level in the soil also is expected to be here somewhere here. So, this is the water table and z is measured upward positive from here onwards and here you can see the capillary rise that is taking place and capillary fringe you can be seen the fringe the way the water moves in the sand which is called a capillary fringe you can be seen here. So, the rate at which the movement of water takes place within the soil mass wasobtained for different soils where are the porosity values for different and all compassion densities could be different, therefore, the porosity varies or the saturated hydraulic conductivity varies than at equilibrium what is a maximum a capillary height that was achieved, that was found which is indicated with h c capillary height. So, for these three different values, for example, the capillary height values for different soils were correlated this capillary height correlated with different soil pore size distribution parameters by several researchers afterwards also, for example, several works in later on were conducted by lane and wash burn. This is in 1946 he establishedrelationship between h c and detain of soil h c measured in millimetre and here - 990 log d 10 - 1540. This is one im-prickle equation he derived by conducting experiments on several soils and here detain is also substituted in mm. So, h c was predicted using such empirical equations. So, this is one such empirical equation, another empirical equation given by peck et al this is in 1974. So, here the h c is = c by e D 10. Here D 10 is substitute in mm and c s units of mm square and h c has units of mm, e is a void ratio and c generally it varies between 10 to 50 mm square. So, such empirical correlations are available for predicting the capillary rise or capillaryheight for different coarse-grained soils generally, because for fine-grained soils it takes enormous time to reach equilibrium and the height will be very significantly high. 
Video 2
in 1943 Terzaghi’s infamous book theoretical soil mechanics providesone solution for the rate of capillary rise. The rate at which the capillary rise occurs in soils a simple analytical solution is derived which has some interesting aspect which I will discuss after the derivation. So, which again assumes that the Darcy velocity value it is. So, he uses q = - ks*i take also - ks i and q we substitute the flux or the Darcy velocity as earlier we substitute as n times d z by d t. So, this is velocity and this is the porosity of the soil. So, this is = - ks heassumes that the saturated hydraulic conductivity is still valid for this particular soil because he also assumes that there is a sharp wetting front exist within the soil when the capillary rise takes place which assumption is similar to Green and Ampt equations. So here he assumes the gradient that governs for the flow is hc the capillary height - z which is a distance, for example, this is a water table within the soil mass and z varies upward and which is, now if this is the capillary height. So, this is the capillary height h c - z which is the head that controls and distance is z. So, this is assumed to be the gradient for the flow. So, when z = 0 there is this gradient becomes infinity that means, it wants to achieve this value and when z is = h c that means, it reaches the capillary height then gradient becomes 0.So, this is a hydraulic gradient he used, this expression is very one similar to the Green and Ampt equations which were given in 1911 and this is 1943.So, we can understand that rate of capillary rise equations are derived based on existing green and Ampt model for the transient flows through soils during that time. So, he further simplifies this expression. So, if we simplified. So, this you get this expression zdz by h c - z = - k s by n integral dt. So, after rearranging the terms all z terms are taken into one side and time terms aretaken into other side and we integrate on both sides then. So, this can be simplified further into if you put - here and add h c to it I will put d z somewhere here and - h c then this expression becomes - first two terms if you used this is - d z + h c times d z by h c - z = - k s by n this is time t + constant of integration. So, this is what results and this is - z + the integration of dx by a - x is - log a - x. So,therefore, so, this becomes - h c log h c - z there is = k s t by n + c. If we invoke the initial condition that is at time t equal to 0 z = 0 there is no flow there is no rise. So, then if we invoke that the c is = - h c log h c - z sorry h c. So, this is the value for c if we substitute we get - z. Then if you bring this value this side + h c - h sorry + h c log h c - h c log h c - z - k s t by n. So, this can be written as - z - ofh c. If you take common or + h c if you take common. So, this simply becomes log h c by h c - z = - k s t by n. So, therefore, the t is = n by k s. If I take common h c also then log see if I put k s + k s here then everything should be fine I think then this is simply. So, this is + k s then everything is fine then. So, t is = - (Refer Time: 14:33) k n t by k s k s times h c log h c by h c - z - z by h c. So, it assumes the validity of Darcy’s law general Darcy’s law that is we consider totalhead here. So, if we consider that this can be written as - ks the hydraulic gradient he considers is h c that is capillary height - z divided by z. This form is very much similar to the Green and Ampt equations which are derived or given in 1911. If you notice the time interval where the Tezarghis expression is given in 1943 for thecapillary rise rate and by then already the Green and Ampt equations are available for transient flows through unsaturated soil. So, therefore, it is very much clear that Terzaghi has considered the Green and Ampt equations for the derivation of capillary rise rate inunsaturated soils. So, here this expression q is = - ks time this one if is it written and here the q is similar to the green and Ampt their Θ s - Θi it is written. So, here initially the soil is assumed to be completely dry then Θ is = 0. Then that is Θ s which is nothing, but n times d z by dt there is a velocity term. So, then this is flux is = - ks again he assumes that the saturated hydraulic conductivityprovides because he assumes that there is a wetting front; sharp wetting front exists during the capillary rise which is similar to what shown here. If this is a water table location and above this the z the elevation is shown with z and if this is the capillary height h c then the gradient is considered to be h c - z that is head that governs the flow divided by the z at any given location for example, at z equal to 0. So, it has it becomes infinity that means.So, the gradient is with the full capacity it has infinite gradient or large gradients. So, when z is = h c that means when the flow reaches the estimated capillary height. So, because the capillary height is should be known in this particular equation if it reaches the capillary height then the gradient is 0 and the flow stops. So, that is the expression for when we simplify this here we need to observe one thing here as we are writing the flow is taking place from higher head to the lower headbecause this is the higher head and it is taking towards the lower head this should be a positive value and then you will get positive quantities and it is solvable. Now, when you simplify this particular expression by rearranging the terms on both sides you get z d z by h c - z and integrate on both sides. So, you get and if I take it on the other side ks by n and integrate the d t term. So, then if we integrate this we can simplify this a priori. So, you can take negative here and a put negative here and then you can d z I will write here then I can add c h c and deduct h c. So, then I can simplify this as d z - d z. If I write the first two terms then this is 1. So, therefore, - d z + h c integral d z by h c - z which is = ks by n t + an integration constant.So, this expression when you solve here this is - z + hc. So, the integration for this is 1 by a - x is - log a - x. So, therefore, the integration for this is you get negative here - h c log h c - z = k s t by n + c. If we invoke the initial conditions, the initial condition is at time t equal to 0 z is = 0 there is no flow, the capillary rise is 0.So, therefore, c becomes - h c log h c. So, if I substitute this constant here you get - z and take the c on other side + h c log h c - h c log h c - z = k s t by n and if I simplified this - z + h c log h c by h c - z = ks t by n. So, this can be further written as t is = n by k s. If I take the h c common this is log h c by h c - z - z by h c. So, this is the expression given by Terzaghi in 1943 in theoretical solve mechanics book and this expression is an implicit expression describing the rate of rise. So, there is some problem with this figure here. So, ignore this soil to figure it should be somewhat like this. So, there is some issue and then you can ignore this soil to data you can only assume the soil one data now let us see. So, the expression for time t is = n h c by k s times' log h c by h c - z - z by h c. Here for a given soil see you take soil one the porosity of the soil is known ks of the soil is known hc should be determined or hc should be known thecapillary rise at equilibrium should be known for a given soil then if such information is known one can predict the rate at which the capillary rise takes place through partially saturated soils can be determined. So, for example, if this data is given this particular data is given we substitute for time = at any given says space. Here it is an implicit form therefore, what is a time taken for reaching the capillary rise to given an elevation can be formed. So, for example, the time required to reach 0.1 m 0.01 m can be obtained. So, which is n = 0.31 times H c is 0.284 this is substituted in metres ks again substituted in metres metre per second this is 1.6 into 10 to the power - 4 the data is taken from here times log H c is 0.284 this should have been smaller letter small h c 0.284 divided by 0.284 - z I said 0.01metre - 0.01 metre divided by H c 0.284. So, this comes out to be a very small value of 0.39 seconds. So, to reach 0.01 metres ithardly takes 0.39 seconds for this particular soil having these properties. Similarly when z is =, here z is = 0.01 metre if z is = 0.01 metre, but the time required is 45 seconds.So, in this manner, we are using different data points at different heights or elevation thetime we have found and drawn in this manner. So, this is a data somehow there is an error in plotting. So, dimensions. So, this is somehow distorted, but the data should have been like this. So, as the hydraulic conductivity decreased nearly 2 4. So, this is for soil one right. So, this is for soil two this is for soil one. So, the colour also there is some error in this while copying from a spreadsheet there is an error in the error or the entire figure got distorted.As the capillary height is 0.82 for soil 2. So, this data for this and 0.284. So, the data somewhat like this it should come here 0.284.So, this is soil 1 and this is soil 2 for the soil 2 data the capillary height is 0.82 and the rate is very small because it is taking nearly 8 days to reach equilibrium. After all, the hydraulic conductivity is 10 power - 6 mitre per second. Here the equilibrium time is very less hardly 0.03 days are nothing, but a few seconds because the hydraulic conductivity is 10power - 4metre per second. So, with this one can use Tezarghis expression, it is a very simplest expression to find out the capillary rate. However, researchers have found that the Terzaghis expression over estimates capillary rate. This is because especially the limitations in the expression, one of the major limitation is he assumes that the saturated hydraulic conductivity for the soils he uses saturated hydraulic conductivity for k.
Video 3
So, which is the major limitation and in 2004 after in 1943 and Lu and Likos in 2004. They have given expression by modifying the hydraulic conductivity function. So, they use the same expression here k times h c - z by z. So, everything else is same instead of k = ks they use Gardner’s expression they use. the expression for k as ks time exponential of α h m. So, this is what the expression they useand they got a close form analytical solution which is a ks times t by n which is similar to what we got is = σ j = 1 to m α j divide by j! - σ s = 0 to j, h c power s into z power j + 1 - s divide by j + 1 - s + h c power j + 1log h c by h c - z. This expression will be same as Terzaghis expression if the series form is not considered if j = 0 to m equal to 0 if you substitute m = 0 this expression will converge to Terzaghis form. So, the Lu and Likos in his in the paper rate of capillary rise in soils published in the ac journal. They showed that this particular expression is a very good match for coarse-grained soils for predicting the rate of capillary rise. When compared to Terzaghis expression Terzaghi expression over estimates significantly. So, far we have seen several preliminary expressions for transient flows through unsaturated soils, but then 1931 researcher called a Richards had given an expression for transient flow through unsaturated soils, this expression is even now it is used.This is based on Darcy’s generalized Darcy’s law and combining with mass conservation expression which I discussed earlier also which involves considering unit volume, considering a representative volume when the flow takes place through this representative volume. So, Q z in the upward direction and Q z + Δ z because this quantity is Δ z. And this is a Δ x and another direction is a Δ y. So, the flux which is this is volumetric flux. So, the volumetric flux which is going into the element is volumetric flow rate Q z of the water which is entering into the element, the elemental volume is there is Darcy’s velocity Q z times cross-sectional area Δ x Δ y. Similarly, the amount which is coming out is Q z + there is some change in the flux δ Q by δ z times Δ z times Δ x and Δ y. So, this change exits there is what is the difference between these two the difference is δ q - δ q by δ z times Δ x Δ y Δ z. So, this difference exists if this is a Transient flow. So, in the Transient flows, this = what. So, this is Q has units of metre per second. So, this = this quantity = the change in volume of water with time because there is a change in the volume of water, when wateris water content increases in this particular element then the volumetric flow rate also changes. So, therefore, in Transient flows this = this if you look at the dimensions this has units of metre per second and this has metre and this is metre q. So, this has units of metre cube per second and here volume of water is metre cube and t equal to time. So, dimensionally this form is correct this equation is correct, now this can be written as δ θ by δ t times volume element because θ is defined as volume of water by total volume. So, this can be written as this. So, therefore, the expression which you get using the mass conversation is δ q by δ z = δ θ by δ t using mass conversation; if you use generalize Darcy’s law to express q then this becomes δ by δ z of - k s gradient. So, there is d or δ h by δ z this is a head the hydraulic potential varies in the vertical direction that is why the flow is taking place only in the vertical direction. So, that is why the gradient is δ h by δ z = δ θ by δ t. So, this expression δ by δ z of sorry, this should not be ks this is k. So, this is unsaturatedhydraulic conductivity times k times δ h by δ z = δ θ by δ t this expression is Richardsexpression Richards equation this is the very famous equation which is even now used for solving Transient flow problems. So, this original expression is in the mixed form it is called mixed form because the dependent variables on one side the right-hand side is θ volumetric water content and the dependent variable on the left hand side is h suction head or looks are present asdependent variables this form is called mixed form. This can be written in the other form as a δ by δ z of k δ h by δ z. By using chain rule you can write this as δ h by δ t times d Θ by d h. So, this quantity is a slope of your soil water (Refer Time: 34:49) of c Θ it is called its quantity is called specific moisture capacity. So, this is head form because when you have a smooth soilwater characteristic curve function smooth function of the soil-water characteristic curve the slope can be determined and then if you solve you will get h of z comma t using this expression so, this is called head form. Similarly, you can write this expression as δ by δ z of k this can be modified again using the chain rule. This can be written as d h by d θ times δ θ by δ z = δ θ by δ t. So, this particular quantity is called diffusivity, hydraulic diffusivity d θ which may be different from diffusion coefficient which is used in and diffusion this has units of metre square per second. Because it has units of metre per second and h has units of metre this as units of metre square per second this is called hydraulic diffusivity. So, this expression finally, is δ by δz of d θ times or I will write it here this expression is δ by δ z of d of θ δ Θ by δ z = δθby δ t this form if you solve you get θ of z t. So, this is called θform. Many researchers have studied all this by using several numerical techniques to solve these expressions and get the solutions, they found that the gives good solution because itdoes not approximate any mass conservation principles, but the head form was found to be not conserving the mass and which gives erroneous results. And this particular form we have utilized earlier for when we discussed the multi-step outflow technique this is again simplified further into this particular form. Δ by δ z of D of θ δ sorry δ θ by δ z = δ θ by δ t. This particular form when we simplified, we assumed that the variation of D with respect to z is ignorable then are we assume that the variation of Θ with respect to h is linear. Then we got an expression for δ square h by δ z square equal to δ h by δ t. And then Richards provided an analytical solution for this particular expression and we estimated d Θ and from knowing D Θ and solver characteristic we estimated k Θ earlier. So, this Richards expression is a very popular expression popular equation which is used for solving many problems. In this expression δ by δ z of k δ h by δ z = δ Θ by δ t is a mixed form we wrote, if the flow is taken place in vertical direction this will be transformed into k here h consist of both metric suction head and + elevation head. If we considered upward positive then if there is an infiltration taking place in an upward direction or evaporation taking place from initially saturated soil this expression becomes d h by d z + k = d δ Θ by δ t. If you considered this is negative then you get - k. Because when we substitute h = hm + z you get this expression. So, this is theee expression for vertical flows. Vertical flows or the influence of gravity is considered, this expression is valid for horizontal flows or the influence of gravity is ignored.
Video 4
For the expression which I just discussed that is Θ form that is δ by δ z of d Θ of δ Θ by δ z = δ Θ by δ t for this particular Θ form researchers derived analytical solution earlier so, which is called Boltzmann transformation. So, this particular derivation can be found in Lu Likos textbook or many other resources like Rumynin which is published in 2012 on subsurface solute transport models and casehistories with applications to radionuclide migration and many other researchers in much other literature also this particular formulation is provided. And all us most of the soil science literature we will have the derivation for this particular form of expression form of the equation using Boltzmann transformation. So, because this is a partial differential equation the analytical solution for this particularnon-linear form of an equation is not possible difficult. So, therefore, if these two dependent variables can be transformed into z and t forexample, this is a horizontal flow then it will be better if you write x that means it is a horizontal flow. So, the influence of elevation by gravity is not there. So, for this particular form, you have two independent variables like x and t. If these independent variables can be transformed into one independent variable then this can be this will become simply ordinary differential equation then solutions may be possible. So, this concept is utilized based on Boltzmann transformation. So, here the simple boundary conditions are considered, the boundary conditions are Θ x = 0 and time any given time this = Θ s and Θ at x = infinity at any given time th= 0.So, this means that you have an infinite column that means, very long column long soil column which is connected to you may have some porosity or a something here porous plate and which is connected to a reservoir. So, when the flow takes place through the soil mass you discard the experiment or you stop the experiment before the flow reaches the other boundary this x = infinity boundary and this x = 0, at time t = any given time. So, Θ = Θ s all the times. So, at this point this is a saturated water content this is the saturated water content at any given time and the flow does not reach the other boundary. So, now, the independent variables Θ of x t is an independent variable the independent variables x t if they can be transformed Θ of x t can be transformed to one independent variable that is λ send a where λ = x y square root of t this is called Boltzmann transformation. This is called Boltzmann transformation now how to convert this. So, the δ Θ by δ x terms. So, this can be written as δ Θ by δ λ times d λ d x using chain rule, but as the Θ is not dependent on only one single independent variable this will be d instead of δ. So, this is d Θ by d λ and this is solved δ this is δ δ λ by δ x. So, if λ = x + square root of t this is d Θ by d λ times d λ by dx is simply 1 by square root of t. Similarly δ Θ by δ λ = d δ t sorry, δ Θ by δ t δ Θ by δ t = d Θ by d λ times δ λ by δ t. So, this = d Θ by d λ times this one is nothing, but - x by t root t because the derivation of D by dx of 1 by square root of x is - 1 by 2 x root x and because this is a x power -half and the derivative is simply - half and x power - 3 by 2. This can be written as 1 by 2 x - 1by 2 x square root of x. So, similarly here x is t. So, therefore, d Θ by d λ into - x - there and you are derivatingpartial derivative only derivative for t is considered. So, therefore - x by t square root of t. So, if you substitute these into the main expressions then δ by δ x gives d by d λ of 1 by square root of time by t divide λ d Θ and δ Θ δ Θ by δ x is again d Θ by d λ and 1 bysquare root of time can be taken out and this expression which =.So, δ Θ by δ t is - x x by square root of t is again λ from the Boltzmann transformation. So, this is simply - λ by t d Θ by d λ if you get two here. So, as this t gets cancelled oneither side this expression simply becomes λ by 2 d Θ by d λ + d by d λ of d Θ d Θ by d λ = 0. So, this is ordinary differential equations this can be solved by integrating the boundary conditions will be transformed to Θ of x = 0 means λ = 0 this = Θ s. And similarly when x = reaches infinity λ also reaches infinity which approaches to infinity which is = Θ initial; initial value could be 0 or anything. So, then if you integrated then you get value for if you integrate twice you get d Θ. So, then d Θ can be written as - half d λ by d Θ at Θ Θ initial to Θ s λ of Θ times d Θ. So, this is the solution for the Θ form of Richards equations by considered the Boltzmann transformation. So, essentially when you conduct such an experiment when you obtain the variation of Θ with x and time if you get then this can be transformed to this can be written as Θ of λ if that information is known. So, how λ varies with Θ or Θ varies with λ that is known. So, because a λ is simply express square root of time. So, for different values of x and different values of time if you get the data of Θ then this can be compiled and Θ of λ or λ of Θ can be written, once the data is substituted here you can get d of Θ there is a diffusivity.How diffusivity changes with volumetric water content can be obtained using this particular expression. So, from the measurements of Θ of x t is measured from the data from this particular data which is converted to Θ of λ or λ of Θ. So, then from this using this expression d of Θ can be obtained the diffusivity changes with volumetric water content. Thank you.