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Module 1: Flow Behaviour of Unsaturated Soils

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Video 1
Hello everyone, we have been discussing the transient flows through Unsaturated Soils. Extend where we can solve the exact formulation we have an exact formulation for the transient flows. So, the using the Boltzmann transformation, we can solve the θ form of Richards partial differential equation and we get an ordinary differential equation then we integrate it we get the equation for d of θ. Diffusivity; here the diffusivity = - half times d λ by d θ at θ x earlier I would have written θ i that is at given point into integral θ initial value that could be 0 or something θo θ o maybe 2 θ x that is it a given point and λ of x dx this if you solve this here x is anintegrational variable. So, if you solve this we get D θ D variation diffusivity variation with θ d is a hydraulic diffusivity which has units of metre square per second we discussed earlier how to solve this particular question is a question. Now we need to conduct an experiment for obtaining θ of x and t. So, this can be done by compacting a soil sample into a columnand connecting this column, this is how you (Refer Time: 02:24) or something and then which is connected to a reservoir. So, if you maintain a constant head of water and you take a sufficiently long column so, that on infinite boundary condition can be booked. So, then this is a particular head that is maintained, positive head that is maintained there is ho this is the initial condition. And in terms of θ initial condition is at this particular point that is at x =0 at any given time, this = the θ s volumetric water content. Here within the soil sample, the initial condition is at any given x greater than 0 when time t =0, this =the initial value or you can write is θo or whatever. This is the initial condition and this is one boundary condition inlet boundary condition and other boundary condition that is at θ at x = infinity at any given time this is invariant; that means, d θ by is =0 or with respect to is x is 0. So, there is no flux here. So, with respect to x, this is 0. This is the boundary condition we invoked or simply this is θ at x =infinity at any given time this is simply 0 this = θ initial. So, this is the constant value therefore, if you differentiate it to become. So, 0 it is 0 values now such an experiment can be conducted where we have a long column. So, that this boundary condition is satisfied and always a constant head is maintained, then you can maintain than the θ also becomes θ =θ s then after a certain time, we can measure the water content at different points within the soil sample. Either by the placing sensors at different locations or by destructive technique, we can establish θ at any given x at one particular time interval this can be estimated. So, the θ variation with respect to x at given any particular time t one can be estimated from the measurements. So, this is the measurement data. So, knowing spatial location; so, this how does it varies? So, this varies in this particular manner. So, the grammatical water content or θ volumetric water content. if you plot with respect to x. So, then it varies in this particular manner. So, this is =θ initial or θ o and this is θ s. So, beyond this point, the water has not reached. So, wetting front somewhere here, but there is water distribution within this entire space from θ s to θ initial value. So, the sharp wetting front cannot be seen, you know that is only hypothetical and mostly that can be invoked in coarse-grained soils without much loss of accuracy. So, this is volumetric water content versus x this is an observation you get.So, this is the at one particular time t = t1, that is observation. As we know how θ varies with x and t, we can obtain λ which is x by square root of t. So, we can also plot how θ varies with λ because this variation is also similar though even though the x-axis scaling would be different. So, this may be similar because is simply x by square root of t, t remains constant because at one particular time interval for different x value you are obtaining it will be scaled and you will get the same qualitative similar data. So, this is the data we require for this particular analysis, I will draw this one here. So, this is λ Boltzmann variable and this is θ. So, now, this is θ and. So, this varies in this particular manner. So, this is theobservation. So, this is θ initial value. Now, this integration means at any given point that is θ x at any given point x, x square root of t.So, this area is simply integral θ initial that is this point to θ x and λ of x dx. So, this can be simply obtained by numerical integration you can take the discrete point because the data will be available. So, we can take discrete points and we can calculate the area under this particular curve, you know this hashed portion can be obtained for any given data and if this is obtained and at this particular point the slope can be obtained that is d λ by d θ. So, this is in fact, is the slope is - d θ by d λ at θ x that is what you are getting here; because the slope of this one is d θ by d e λ. So, if you inverse it, this value can be substituted into this and this portion area is coming directly can be substituted here. So, essentially by multiplying with - half and the slope of this point at on the curve at this particular point θ x, and integration that is an area of the hashed portion should give you d θ at x at this particular point. Similarly, discrete data of d at different θ x values can be obtained. So, if you plot that so, that appears to be. So, on the y-axis, you can plot D θ or simply D and x-axis you can plot θ, and θ varies from θ initial to θ s. As θ increases the D θ valueincreases like this may be several orders of magnitude changes from here to here when θ changes as over a small value like say the initial value of θ could be 0.1 or something and are 0.01 θ s could say 0.4 or something. So, then it can change in this particular manner.So, this is how the hydraulic diffusivity can be obtained using a graphical technique it can be easily obtained. So, this is a very popular technique in soil science and this particular equation is called Bruce and Klute equation because of their contribution and the derivation of this particular expression, and a graphical solution is also commonly used for obtaining the diffusivity. So, the diffusivity helps in understanding how fast the water diffuses in the soil.
Video 2
Often we may require to solve the original partial differential equation of either mixed form or θ form or head form depending on the situation and the Richards equation that is a that equation is θ form is ∂ θ by ∂ t = ∂ by ∂ x of D ∂ θ by ∂ x. So, this is the θ form of Richards equation and we earlier simplified this to ∂ square θ by∂ x square times D. Assuming that the d variation with respect to x is negligible. So, this we applied in case of a small sample which is trusted in axis translation technique or plate operators using the method of multistep outflow technique for the estimation of hydraulic conductivity function. So, in that we made a similar approximation; however, the representation was in the head form or the suction was used as a dependent variable. Here is the dependent variable is θ volumetric water content, and this form there is a ∂ θ by ∂t equals to D times ∂ 2θ by ∂ x 2 exactly replicates the fixed diffusion equation. Fixed diffusion equation for ionmovement within any media porous media or any other media; so, this is exactly similar to that either this can be approximated like this or the whole equation needs to be solved or mixed form needs to be solved very often. So, that can be achieved by considering numerical approaches. So, far we are trying to understand how analytically the problem can be addressed, but for complex problems and complex boundary conditions, it is not possible to obtain analytical solutions. Moreover, in the field, we do not expect simplified boundaryconditions and 1Dimensional flows very often. So, in that particular case, the numerical solution can be advantageously used to solve these problems and address those field issues. So, I will demonstrate how one particular type of finite difference approximation can beused to solve the partial differential equation, and this is the demonstration purpose only and extensive understanding on this particular finite-difference or computational fluid dynamics requires specialized training. So, here only its demonstration will be done. So,I take the simplest form of the expression that is ∂ θ by ∂ t is =D ∂ square θ by ∂ x square. This is a linear form of the partial differential equation and the parabolic partial differential equation. So, I will demonstrate how a numerical solution for this particular case can be obtained. Generally in computational fluid dynamics or finite difference techniques, we invoke Taylor’s series expansion, for example, if you have a function f which is varying with x if there is a function that exists like this, then at any given point x I can expand the function around this particular xi. So, this difference would be Δ x. So, around this particular region of Δ x, I can expand this particular function. So, if I expand that I get f of x + Δ x. So, therefore, I can invoke Taylor’s series expansion that is for f of x Δ x th= f of x + Δ x times f dash of x + Δ x square by 2! f`` of x + so, on and so, forth.Similarly, for f x - Δ x; so, that is for this particular point. So, this was expression becomes an f (x) -Δ x f dash of x + Δ x square by 2 factorial f double dash of x - so, on and so, forth. Using these expressions, we can obtain first-order difference expressions this is the differential equation will be expressed in the form of difference equation so, that we can generate an algebraic equation to solve these equations and finally, solve.So, as we are saying there is a small increment of Δ x considered and around that we can expand this function f of x in this particular manner. So, therefore, what we could is, the θ variation with x and t we want and would not know how it various here 2 independent variables is time and spatial variable x. So, now, what we assume is that we consider a grid a domain; so, a θ variation over a column length of say x centimetres. So, this is thedomain that is shown, we divide this domain into small cells are grid points. So, now we have said, this particular thing is i, this is i + 1, this is i - 1. So, starting from 1 etc 2 etc up to say n number of points I can consider on this particular domain. So, I intend to obtain this is a spatial x and this is temporal. So, how the θ variation happens along this spatial length at any given time? So, there is t equals to 0, I want to understand. Here as it does not start from 0, this is 1, this is 2 and this is three and so, on and so, forth. So, here at x =0, there is first point grid point and x =n that is the large grid point on this domain. Similarly, in the time domain time, t =0 that is nth time step or n =0 one-time step. So; that means, n =one-time step, this is I equals to one-time step. So, again we divide this into several time intervals; so, n =2 etcetera. So, this is n this is n + 1 and this is n - 1so, on and so, forth. We are now getting the volumetric water content variation how it varies in the spatial domain of the problem how it varies at discrete points we can obtain similarly with time, how it varies at discrete time intervals what is a value we can obtain. So, essentially we are changing in a continues to form into a discrete form. So, that is what is called difference form we are getting from the differential equations. So, this is n =2 and this is n =3 or some other time interval. So, here what we are trying to do is, we express the θ to obtain this different form. Now, we can simplify the ∂ θ by ∂ t term there is a differentiation of this particular equation can be made in different form by considering ∂ θ by ∂ t as the first derivative of θ with respect to time can be obtained by considering this particular equation itself. So, from this particular equation f dash of x can be written as f of x + Δ x - f of x divided by Δ x + some higher-order terms.So, if I consider Δ x terms which is very small, Δ x approaches 0 very small Δ x then this higher-order terms can be ignored. So, when we ignore essentially the slope or thederivative of the function can be approximated into different form like this. So, now this particular ∂ θ by ∂ t that is the first order differential form can be expressed as θ x + Δ x I can write or I can simply write that this is( i + 1)th value because x + Δ x if I consider at ith point then this is i + oneth point - θ at ith point divided by Δ t.So, this is in the time domain. So, now, I can solve this particular form in the nth level itself n could be 1 and i could be 1 in this particular case likewise or i could be 2 in this particular case because i =1 because of boundary condition. So, i =1 becomes the boundary condition what happens at the boundaries we already know the information θ at x =0 that = θ s that information is known. So, this can be approximated to this now we have second order differential equation on the right-hand side ∂2θ by ∂ x2how to obtain difference form for this particular equation. So, we can use these 2 equations, when you add these 2 expressions what you get is f of x + Δ x + f of x - Δ x is =2 f of x, these 2 terms get cancelled +. So, this is simply Δ x square f double dash of x + some higher-order terms. . So, now, if I write the expression for f``( x), this is simply f of x + Δ x + f of x - Δ x - 2 f of x divided by Δ x square. This is called a central difference and this is called forward differencing etcetera, but I am not discussing the details of what is central difference etcetera central difference higher-order terms. This is the second-order approximation or these things need not be discussed here. So, the ∂2θ by ∂ x2 term can be approximated as based on this θ, x + Δ x is i + 1 - 2 θ i f of x is θ i + θ x - Δ x is i - 1 divided by Δ x square. So, now, if I substitute this one when this one in the differential equation, what we get is θ i + 1 n - θ i n divide by Δ t = d θ i + 1 - 2 θ i + θ i - 1 by Δ x square. If I simplify this is an explicit form of final difference approximation. So, here I can write for θ i + 1 sorry this one is not θ i + 1. So, this is at essentialist at same i th level. So, this is at i and n + 1 only time step is varying because we are actually doing the difference operator on the time steps. So, thisis n + 1. So, here sorry for the mistake; so, this is n + 1. So, therefore, this is n + 1 can be written as θ i n + D Δ t by Δ x square times θ i + 1 these are all at n th level this is my approximation I am considering at n th level these values. So, - 2 θ i n + θ i - 1 n I choose to consider them at nth level. So, now, if you closely observe θ i at n + 1 for example, if n = 1. So, at the second interval i th level this point is calculated using θ i at n th level. So, that is this particular point and θ i + 1 at n th level. So, that is this point and θ i - 1 at nth level so, this is this point. So, using these three different points this point is calculated. So, as we have an initial condition in this particular case is, θ at any given x greater than0 when time t =0 is =θ initial. So, therefore, we have the knowledge of θ along this grid. So, knowing these three terms, we can obtain θ in the next time step. So, this is Δ t and this is grid point between 2 different points is Δ x. So, essentially for any small Δ time increment, what happens to θ how θ distributes along the spatial domain can be obtained spatial distance can be obtained. So, the next time level again. So, similarly for this particular point utilized these three points, for this particular point θ i + 1 θ i + 2 utilized these three points known points. Similarly, you can obtain all points on this particular n th n =2 th level. Then once you obtained all these values, you increase the next level increment again this point can be obtained by utilizing these three known points. Similarly, you march in timefrom n =1 th level to 2 th level to third level etcetera you march in time. So, this, therefore, this is the final difference techniques are called time marching techniques. So, we march in time and step by step we estimate how θ value changes which spatial distance for different time intervals and now any given theoretical time what is the θ value can be obtained along the spatial length. So, this is a simply the explicit finite difference scheme we call. So, we can one can obtain the stability and convergence of these particular expressions or equations and based on how this d Δ t by Δ x square changes, one can obtain the convergence and stability criteria can be developed; how error propagates if you change the Δ x or Δ tetcetera can be studied. And this is one type of scheme in the finite difference, and there are other schemes like implicit scheme where if you put n + 1 th values here.So, then so, if you put n + 1 th values here, then this becomes implicit scheme. So, because we have n + 1 on either side of the equations only n th term only one particulardata point is there then it needs to be solved by numerically it will be more rigours because it is an implicit scheme. So, by Gauss elimination or such kind of techniques we utilized to solve theseexpressions. To theoretically obtain θ of x and time, so, in inverse problems essentially you may want to estimate what is a D value knowing θ of x and t experimentally. So, then in a particular case we utilizing optimisation techniques. Because earlier I have shown that if you have θ measurements x t and θ theoretical then the difference between these 2 data points can be taken, and error can be computed and this error can be minimized to by utilizing several optimisation techniques. So, now, I will demonstrate one way of estimating the θ of x t for a known data of hydraulic conductivity functions and soil-water characteristic curve. Because a once the d diffusivity is known are knowing actual equation is that this is a ∂ θby ∂ t = ∂ by ∂ x of K ∂ h by ∂ x. So, here h of θ if this function and k function are known. So, then you can obtain θ of xt by numerically like what I have shown here. So, such a methodology is utilized in one of the freely available software called HYDRUS 1D; which is developed by the same authors who developed ret c r etc 
Video 3
So, this is a HYDRUS one Dimension 1D and HYDRUS 2 D and 3D isalso available, but they are paid software they are commercial and here you can choose one particular name and you can write some description. So, then you have this free processing window you can select the main process in that the water flow that is what we want to solve through unsaturated soil and vapour flow we have not discussed and we have solute transport. Solute transport and many other options we transport and root water taken many other options. Here we will concentrate on water flow alone and this inverse solution is nothing, but what I discussed earlier that if youwant to estimate the hydraulic conductivity function soil-water characteristic curves, knowing the volumetric water content distribution with time and space, you can estimate those functions by inverse solutions. So, the inverse solution means that. Then you select next. So, then length units can be selected as in centimetres metres or millimetre and number of soil materials. So, it can handle the layering system or nonhomogeneity. So, you can select several layers and each layer you can give what is the difference soil material that is available; and whether the flow is taking place verticallyand horizontally that can be seen here that can be understood because here it mentioned that decline from vertical axis if you put one; that means, it takes vertical flow and if it is 0 then it simulates horizontal condition. So, here I was saying the number of soil materials can be considered for nonhomogeneity. So, the depth of the soil profile by default is 100 centimetres and here you can change if you have any other values then if you select next. So, here the time information you have units per time seconds minutes hours.days and years and the default values days and I keep the days initial time is 0 and the final time is 100 days. So, you may choose may be seconds or minutes and then this is the initial time and final time and you can consider time-variable boundary conditions also. So, for every one time step, it gives the output print options and prints regularly you can play with these options available. So, several iterations you can select maybe 100 you can select or you can give the. So, you can select maximum number iterations to be 100and these are tolerance etcetera and if I go to next. So, here the choice for hydraulic models is given, here if you consider single porosity model your van Genuchten Mualem this is what we discussed in our one of the lectures. And with air entry value if you enforce - 2 centimetres then the stability will be satisfied. So, it will be smooth it seems as it mentioned in the manual and you have modified Van Genuchten and Brooks Corey and Kasogi models, and you have other dual-porosity models also and you can consider hysteresis you are considering of cyclic effect of drying and wetting.So, if you consider a simple problem and here θ r are residual water content, and the q r means θ r residual water content is 0.078 there is a default value, and θ s is 0.43 this is the porosity or saturated volumetric water content is 6.43, alpha is 0.036 and n is 1.56and ks is this value 0.017 centimetres per minute which is considered.So, these are the values for loam soil and you can also select difference soils from the menu. Here you can select boundary conditions for these particular cases you have constant pressure head constant flux and atmospheric boundary condition with surfacelayer etcetera. These are under pressure boundary conditions. So, you can select either pressure boundary conditions because is essentially you have a mixed form to solve and you can input the boundary conditions in terms of head or water content. So, if you change into water content boundary conditions then you have constant water content at the upper boundary, our boundary this is θ s you know constant water content θ s saturated water content or constant flux etcetera and you can go and lower boundary constant water content constant flux or free drainage etcetera we can select.So, we give in water contents only constant water content we give, and next do you want to run profile application save data yes. So, this is the, your mesh and you can consider. So, the initial input so, far it is not taken anything. So, this is the initial input and if youclose this. So, this is how the grid is considered; here the θ variation is not considered at all and here we need to give input. So, here θ is given as. So, this is not θ this is a head. So, it is wrongly considered we go to previous. So, here we go and go for options. So, here we go for the pressure head conditions itself because it is not the taking water contents now at present. So, pressure head consideration if you take and then if you consider the initial conditions then, there is the initial conditions is selected here. So, then we can proceed and initial condition is that at x is =0 or z =0 the head is 0. And allother points this is - 100 centimetres. So, this is an initial condition next ok. So, continue and then profile information. So, here it is not able to generate anything. So, this is the pressure head profile and this is a water content information at one particular time interval. And more information can be obtained if you increase time andif you run the software for different time intervals and you get the data, and you can analyse how the water content changes depth it can be seen. And here if you put realistic values, so, the variation in the water content can be very clearly seen. So, in the particular case what we cloud see that when soil column and you maintain some certain pressure head. Say you can have your porous plate here andconnect to a reservoir, water reservoir you can maintain certain head it could be a constant head. So, then the water flow takes place into the soil in this particular manner. So, this is how the waterfront advances within the soil due to head gradient as well as elevation or the gravity effect, this is how it advances in this particular direction. So, the distribution of water with depth; so, for examples, this is depth Z in centimetres and the volumetric water content if you plot.So, here the initial condition is that θ s here and θ i or θ initial value here everywhere else. So, you have. So, here it is θ s and this is how it is slowly advancing with time and slowly it reaches like this is how the profile advances. So, volumetric water content increases in this particular  this is the time t1, this is t2, and this is t3 and this is t4 here t4 is more than t3 more than t2 more than t1. So, this is how the volumetric water content increases within the soil with time such kind of a profile can be obtained using such free software like HYDRUS 1D. So, here our discussion on steadystate flows and transient flows is complete for our course.So, far I have discussed only the water flow or water movement through unsaturatedsoils under steady-state and transient conditions. However, in unsaturated soils, there could be vapour flow that can take place because of the thermal gradients are any other humidity variations or density of the air variation in air density across the soil length andthere could be vapour flow that takes place. And similarly, in unsaturated soil, the air also can get dissolved in water and this air can diffuse through water. So, those 2 cases also prevail in unsaturated soils.
Video 4
In unsaturated soils the vapour flow can take place the steady vapour flow forunsaturated soils can be described using Fick’s law.So, Fick's law for steady vapour flow. So, here this is a vapour flux which has units of kg or metre square second. So, this is adiffusivity which has units of meter square per second and this is a density of vapour,which has units of kg per meter cube. So, the expression is q = - Dv del ∂ v. So, thismetre square per second in 2 kg per metre cube. So, the metre square and here ∂ /∂ xcome because of the from del operator will get ∂ by ∂ x. So, this is 1 by metre also.So, therefore, the units are satisfied on both sides. So, this particular expression the rho vwe have earlier seen the density of water which can be written as the molar mass ofwater is moist air times Uv vapour pressure divided by RT gas constant and temperature.So, the variation in density of vapour is essentially due to change in RH and change intemperatures which can be expressed as M w Uv sat times RH you get.So, therefore, divided by RT times change in RH - M w times RH divided by R times Uvsat del T by T square - del Uv sat by T. So, this is an expression for δ ρ v once yousimplify this and substitute in this expression. So, the expression for steady vapour flowhas given in Lu Likos is q v = - D v ρvsat times δRH by RH - δ T by T + λ Mw δT by RTsquare. So, this is an expression from Lu Likos textbook for study vapour flux is this,here the variation of relative humidity and variation of temperature could be consideredthis λ is the latent heat of vaporization of water.So, especially these steady vapour flux conditions are encountered in the nuclear wasterepository which I mentioned at the beginning of the course that you have a coppercanister this is a copper canister, which contains the nuclear waste these are all coppercanisters which are place at different spatial distances and this is a compacted back in apit this is placed and which is backfilled with the bentonite. tunnel and you makepit here and the disposal pit you place the canister and then backfill it with bentonitesthis is called buffer bentonite buffer.