Mega May PDF Sale - NOW ON! 25% Off Digital Certs & Diplomas Ends in : : : Study Reminders Support
Text Version

We will email you at these times to remind you to study.
• Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Sunday

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