 Today, we will start with a new topic that is called as dispersion. We have discussed about several aspects of species transport in this particular course and dispersion is very much linked with species transport in a fluidic system. So, before introducing what is dispersion and why it is important, I think it is useful if we revisit the concept of diffusion. So, diffusion refers to the transport of species in a solvent from a region of high to low species concentration. So, it is driven by concentration gradient. Pure diffusion refers to the case when velocity of the solvent is 0. For example, add sugar to water and let it sit, observe that after a sufficient time the sugar dissolves completely, but diffusion itself is a slow process. So, it takes quite a bit of time for the sugar to dissolve completely in the solution if it is not stirred. In presence of a non-zero solute velocity, the transport of species is partly diffusive and partly advective and sometimes it is also referred to as convective, but I mean a better terminology for transport due to fluid flow is advection and not convection. So, add sugar to water and start stirring, observe that after a relatively small time the sugar dissolves completely. So, advection helps a lot in augmenting the rate of species transport. So, now diffusion is generally described by a suitable constitutive law known as the Fick's law. Fick's law of diffusion states that the mass flux due to diffusion is proportional to negative of the concentration gradient which means that you can write j is equal to minus a constant of proportionality times the concentration gradient. The constant of proportionality is called as diffusion coefficient. So, now if you add the concept of the Fick's law to the conservation equation for the mass of the species under consideration, then you can write this equation. We have derived this equation earlier that is why I have not tried to be too formal in deriving this at this stage because we want to revisit the essential meaning in a very simple way and we will then relate that to dispersion. So, you have del C del t is equal to minus del dot j. So, this is like this has been obtained from the balance of mass of a species in absence of any chemical reaction. So, you can write j is equal to minus d grad C. Remember that when we are writing j is equal to minus d grad C, we are not including the electro migration flux. That means we are considering this as an uncharged species even if it is a charged species but not subjected to an electric field, then it is as good as an uncharged species. A charged species subjected to an electric field will be subjected to an additional flux known as electro migration flux but we have just considered here the diffusion flux no advection no electro migration. So, this is pure diffusion. So, you can get this equation del C del t is equal to del dot d grad C. So, if d is a constant then del C del t is equal to d del square C. So, this is a very simple picture of simple mathematical picture of diffusion. Now the question is that how do you obtain the diffusion coefficient? So, it must be born in mind that diffusion is an inherently molecular process. For typical dilute solutions, the diffusion coefficient is estimated by the Stokes Einstein relationship. Again we discussed about the Stokes Einstein relationship in the context of electrokinetics. So, the same thing is written here. So, diffusion coefficient is written in terms of the Boltzmann constant absolute temperature and the ionic radius of the solute. So, like if you have a particular ionic species for example, so you can write like several examples of various cations and anions are there. So, it is the ionic radius if you are bothered about the ionic species being diffused. So, this is an example where we are talking about ionic species but I mean this is just to give you a feel of the values of the ranges of common values of the diffusion coefficient. So, if you look into the table you will find that the diffusion coefficients no matter what ion and what is the ionic species and all these are typically very small. So, like mostly of the order of 10 to the power minus 9 meter square per second. So, 10 to the power minus 9 meter square per second is reasonably low value of the diffusion coefficient but this reasonably low value is what is commonly encountered in practical scenario. So, now we will discuss about various scales of diffusion and one or two fundamental problems in diffusion before we move on to dispersion and we relate dispersion with diffusion. So, let us think about this problem which is described here. So, let us say that we have a one dimensional problem. So, the del square becomes del square C del x square. So, del C del t is equal to D del square C del x square. Now, you can see that like at time t but at infinity C is equal to 0 and at time t at x equal to minus infinity C is equal to C 0. So, let us say this is the condition to which this equation is subjected. Now, let us try to make an order of magnitude analysis of the equation. So, we come to the board del C del t. So, this is of the order of C 0 by T c and this is of the order of D C 0 by x c square or L c square whatever. Taking these two orders, we can say that the diffusion length scale is related to the diffusion time scale in this way. So, the first important understanding is that for such a problem, the length scales and time scales may be related to each other and this is how these scales are related. Now, with this little bit of background, we will derive a fundamental solution for a diffusion problem. Let us address a fundamental problem in diffusion. There are some mathematical steps involved. The steps are summarized in the slides but I will work out the details of the steps to the extent possible in the board but let us go through the basic problem definition using the slides. Consider a point source of species located at x at time t equal to 0. So, it is like a mass source. We are required to determine how the concentration varies with space and time. As time progresses, what will happen? The total amount of species which was originally present in the solution will simply get distributed owing to diffusion. So, the mass will be conserved. So, typically at a particular location, if you inject a species, then that species will diffuse to the sample but the total mass will remain constant, total mass of the species will remain constant. Only the concentration wise, it will be distributed differently because of diffusion. So, our aim is to solve a governing equation of this form. u t, you can understand these subscripts like u subscript t is del u del t, u is like concentration. This is just a generic mathematical form is equal to d u xx, u xx is del square u del x square. This is for t greater than 0 and minus infinity less than x less than infinity. At t equal to 0, you have a species source. So, that is represented by this concentration times a function. This is the typical representative of a point source c0 times the direct delta function. We have already come across this direct delta function earlier in some other context. So, the delta function as the property that delta 0x is equal to 1 at x equal to 0 is equal to 0 at x not equal to 0. So, this is one important aspect of the delta function. The other important aspect is like integral of delta 0x from minus infinity to infinity is equal to 1. This is one important property of the direct delta function. Now, if you are interested to solve this equation, then the solution to the above equation is not a function of the form of function of t into a function of x, not like the standard method of separation of variables. So, typically because there is a similarity variable here, one can use a similarity transformation. But for a point source of this kind where you have a delta function to work with, it is better to use transform calculus instead of using a simple similarity transformation based technique. So, the transform calculus based method will be based on first taking a Fourier transform and then taking a Laplace transform, then taking inverse Laplace transform and inverse Fourier transform. So, we will go through the steps. So, the governing differential equation ut is equal to du xx plus c0 into delta xd. The delta function signifies that this delta is the delta 0 basically. So, at x equal to 0, there is a peak. So, concentration wise basically you introduce the solute at a given x, let us say x equal to 0 and then it will spread in the solution as you allow time to grow, that is what is physically happening. So, the definition of the Laplace and Fourier transform, in case you these have gone out of your mind, so I mean I do not think I need to like discuss about these definitions, but I mean essentially these transforms you can use for transforming from time domain to frequency domain etcetera. So, you can see for example, this is the definition of the Laplace transform e to the power st phi dt and the Fourier transform e to the power ikx phi dx and importantly the in the Laplace and the Fourier transform, the domains of integration are different. I mean in the Fourier transform, you are integrating in the space domain from minus infinity to infinity and in the Laplace transform, you are integrating over the time domain from 0 to infinity for obvious reasons because time domain in the range of minus infinity to 0 has no physical meaning. So, the inverse Fourier transform is obtained by this formula, where you integrate over the frequency and we will use these results for making the Laplace and the Fourier transform. So, let us work out a few of these steps in the board that will make you feel a little bit easier. So, we will start with the differential equation del u del t which is a function of x and t is equal to d del square u del x square which is a function of x t and this is our differential equation. So, we will first take a Fourier transform, so Fourier transform, so ut e to the power i kx dx from minus infinity to infinity, so what will be this? Just by the notation it will write it as ut hat, ut hat is the Fourier transform of ut, just a shorthand way of writing this. It is a function of what? It is no more a function of space because it has been integrated over space, so it is a function of k and t, then uxx e to the power ikx. So, this we can integrate by parts to reduce the order of the derivative of u. So, then in that case we will take this as the first function and we will take this as the second function. So, then first function into integral of the second minus integral of derivative of first into integral of the second. So, at now at both minus infinity and infinity del u del x is equal to 0. So, that we will make this term as 0. The next term again we can integrate it by parts with this as the first function and this as the second function. Then minus ik first function into integral of the second minus integral of derivative of first into integral of the second. The concentration is 0 at both minus infinity and plus infinity. So, this term is 0. So, you are left with i square k square. So, that means minus k square because i square is minus 1 into what? u hat kt. This is Fourier transform of u. So, u hat ok. Now, so we had taken Fourier transform of this term this term and only the delta function is left. So, minus infinity to infinity delta into e to the power ikx dx. This is delta of t by definition of the delta function. This is by definition of the delta function. So, our governing equation if you now combine all these 3 terms. So, if we have taken the Fourier transform then you will be getting u hat t kt is equal to d minus dk square u hat kt plus c0 into delta t taking the Fourier transform. Now, we will take a Laplace transform. So, Laplace transform integral of ut kt into e to the power minus st dt from 0 to infinity. So, again we will take e to the power minus st as the first function and this as the second function. So, first function into integral of the second minus integral of derivative of first into integral of the second. So, this term will be 0 because at time tends to infinity this will tend to 0 and at time equal to 0 u hat equal to 0. So, this term will go away and you will get this as s. Now, what is this? This is basically 0 to infinity let us write. This is the Laplace transform of u hat. So, u hat it is a function of what k and s. So, if you take the Laplace transform now of this equation this will become s u hat ks is equal to minus dk square Laplace transform of u hat kt will become u hat ks Laplace transform of delta will become 1 that is that unit step function. So, this see eventually after these two transforms you get a simple algebraic equation. So, you will get u bar ks sorry u hat ks is equal to c0 by s plus dk square. So, with this background. So, now the next steps are quite obvious you take an inverse Laplace transform and inverse Laplace transform of this is of the form of e to the power at because e to the power at Laplace transform related to s minus a will give rise to a form of e to the power at. So, if you take inverse Laplace transform of this you will get back e to the power at where a is basically like this dk plus dk square you can treat as minus a. So, let us summarize as we go to the board. So, taking an inverse Laplace transform. So, you will get u kt using the fact that l is Laplace transform 1 by s minus a equal to e to the power at. So, you will get back this then you take an inverse Fourier transform of this by taking the by making the integral that we have defined for the inverse Fourier transform and then you will get this form this particular form. So, this is a very classical solution and this is a this classical solution for point source we will be using in the context of something called as dispersion that we will be learning. So, look at the solution look at the graphical nature of the solution. So, this is just it is like a step function but we have smoothened the step function a little bit just for your clarity in understanding. Let us say this blue line represents the species concentration at time equal to 0. As time evolves the different species concentrations will be determined by this other dotted lines this green line the red line all this at different instance of time which are mentioned in the legend of the figure. So, you can see that the initial sharp peak which was at t equal to 0 it gradually flattens down. So, this blue line is not at t equal to 0 this is t greater than 0 but slightly greater than 0. So, that it has almost preserved the shape of a delta function but after that you will see that it will be flattened the species concentration from assuming a peak it will come down to a flattened nature but the area under the curve will be preserved that is mass conservation. So, you can see here that when it is flattened it is spread more when it was not flattened it was spread up to only this. But when it is more and more flattened it gets more and more spread because area under the curve has to be conserved that is the graphical interpretation of the mathematical expression that we have derived. Now constant source planar diffusion it is something which again is a matter of interest but we will not work out it completely because it can be shown that this solution is just like the stokes first problem that we discussed in the context of fluid mechanics. So, I will show you that where is the similarity I mean this is the interesting aspect of mathematical physics that no matter what is the context the context may be totally different but if there are mathematical similarities then the solution technique becomes quite similar despite the physical disparity. So, the definition of the problem is ut is equal to du xx now there is no delta function because it is not a point source and u at x equal to 0 at time t is equal to u 0 that is you specify the concentration and u at x for t equal to 0 is equal to 0 this is the initial condition. This is initial condition this is boundary condition okay. Now we have seen that the time scale and the length scales are related in this particular problem. So, if you have these 2 terms only so you have u 0 by t is of the order of du 0 by x square that means x square by t is a similarity variable for this problem this is how the length scales and time scales are related. So, we had worked out a problem I mean where is the similarity between these and the stokes first problem this is the governing differential equation for the stokes first problem with this is replaced with this being replaced by the kinematic viscosity. So, what is the stokes first problem if you recall there is a plate which is impulsively moved at time equal to 0 and you want to study the velocity field because of the movement of the plate. So, you can see here that just if you replace d with nu the problem is essentially the same and the boundary condition at x equal to 0 at time t you impulsively move the plate instead of with the velocity u 0 instead of that here is the concentration c 0, but I mean the mathematical analogy is pretty clear. So, earlier we had taken a similarity variable eta is equal to x by square root of t with some additional adjusting terms like may be square root of nu t or square 2 into square root of nu t whatever. So, now just for just by looking into this you can say select a variable x square by t this is not a dimensionless similarity variable, but this is a similarity variable with that with the dimension involved with it. So, you can write del u del t is equal to du d eta into del eta del t del eta del t is del eta del x 2 x by t. So, del 2 u del x 2. So, this is a basically what we have done this is a product of 2 functions. So, we have used the product rule for differentiation. So, you have del u del t that is minus x square by t square du d eta this is du d eta right. We have differentiated this with respect to x and sorry sorry sorry this was yes. So, we have differentiated this with respect to x. So, this has become 2 x by t du d eta when you differentiate it again with respect to x this becomes 2 by t du d eta then this differentiated with respect to x right. So, minus x square by t t square du d eta is equal to 2 by t du d eta plus 2 x by 4 x square by this is d. So, you can take x square by t square on one side and write 4 d d 2 u d eta 2 plus du d eta plus 2 d by t du d eta is equal to 0 right. Now, you can cancel 1 t and you get back x square by t which you can write as a similarity variable. If you could not express the entire equation in terms of the similarity variable then your similarity transformation is not successful. That means then the equation is not amenable to similarity transformation and the similarity transformation what it essentially does is it converts the PDE to a ODE. So, now x square by t you can write as eta. So, this entire equation is now in terms of eta and du d eta okay. So, we will get back to the slides to complete this because the remaining steps are very trivial you get this differential equation. So, I have shown how do you convert the PDE to this ODE and then the analogy with the Stokes first problem this I have already discussed. So, the Stokes first problem is an impulsively started plate the fluid is initially at rest and the plate is suddenly moved with a velocity u parallel to itself 40 greater than 0. This is the governing equation and this particular problem of diffusion is that it is suddenly exposed to a species source 0 concentration everywhere the region exposed to concentration u 0 for t greater than 0. So, very same equation just nu and d are different coefficients but same equation with same boundary condition. So, the solution for this is exactly the same as the solution for the Stokes first problem. So, what is the solution the ODE that I have just shown you integrate it once. So, you see that it is that is of the form d 2 u d eta 2 plus something into du d eta. So, if you take du d eta as a variable then integrating that you will get this equation and this may be integrated once more to get you as a function of eta. So, what are the boundary conditions that you need to use at x equal to 0 u equal to c 0 right that is the concentration specified concentration boundary condition for fluid mechanics Stokes first problem that is u equal to u 0 the impulsively started plate velocity and at x tends to infinity u equal to 0 that at far away from the plate you had subjected the plate to a species source. So, far away from the plate located at infinity it does not feel the effect of the species source. So, at x tends to infinity u equal to 0 and that will give the solution u equal to c 0 into complementary error function of x by 2 root d t the equivalent solution of the Stokes first problem was u by u 0 u equal to u 0 into complementary error function complementary error function is 1 minus the error function is equal to complementary error function of x by 2 root nu t the kinematic viscosity. Now, we will extend the diffusion concept to the concept of dispersion. Now, consider an axial plug flow in a look at the schematic you will understand the thing very nicely if you look into the schematic consider an axial plug flow in a cylindrical channel. So, this is a initial band imagine a concentration solute plug release at the inlet of the channel as shown in the figure the solute plug is advected with the velocity field right. Because of the flow velocity the solute plug will move at the same time the solute will spread because of diffusion. So, the had there been no diffusion the solute plug would have come from here to here without spreading and from here to here without spreading and from here to here without spreading, but you can see the width of the plug gets broadened and broadened and this broadening of the plug is because of diffusion. So, in totality you are having a phenomenon which is governed by combination of advection and diffusion of the species. Now this combined phenomenon is called as dispersion. Now dispersion is unfavorable in chemical or solute separation and fractionation process because let us say you want to separate solutes that means there are different plugs of solutes that you inject at the inlet of the sample and now what will happen is that the solutes will reach the end of the channel at different instance of time may be different instance of time while why it will reach may be you have applied an electric field and different solutes have different electrophoretic mobility. So, based on different electric field based on an electric field the electrophoretic mobility may be varying from one species to the other or it may depend on the size of the species. So, different species will collect at the exit of the channel at different instance of time. So, you can separate different species from a collection of species, but this separation becomes difficult if the band is wide because then you may not be able to distinguish one species from the other by the time one species has left the channel the other species has already come and overlap with the band it becomes difficult for you to separate. However, dispersion may be favorably employed in achieving homogenized volume of the solute, but if you want a uniformity in the solute concentration then diffusion helps in mixing the dispersion helps in mixing not only that you also have to consider transverse diffusion and not just diffusion of species, but diffusion of momentum in case of a pressure driven flow the velocity varies along with the radial direction. The solutes near the center line are advected faster than those near the wall leading to the plug deformation. So, whatever was the plug the plug will deform because of the parabolic velocity profile along the center line the fluid will tend to move with the faster velocity and at the wall the fluid will tend to move with the slow velocity. It leads to longitudinal diffusion along with the transverse diffusion and the transverse diffusion may be reduced by having a plug type of velocity profile. So, if you have a electro osmotic velocity profile with a uniform velocity that means there is no distortion in the electro osmotic velocity profile then that kind of velocity profile is very much suited to avoid the transverse dispersion. So, just to summarize the physical picture of dispersion if there is no diffusion you can see this blue concentration profile will just be moved with just like a rigid body to the red concentration profile at over the time delta t and the displacement is just u into delta t. There is no distortion in the concentration profile although this is true for ideally for 0 diffusion but actually even with very low diffusion this is approximately valid. In case of 0 fluid flow in case of 0 advection the solution is the problem the solution is the solution to the problem of diffusion fundamental problem of diffusion that we are discussed with the solution with the Laplace and the Fourier transform. So, this is a special case when the advection is 0. So, then you can see that in now when can you transform the physical scenario to this case let us say that you are watching the phenomenon with a reference frame that is moving with the fluid. If the reference frame is moving with the fluid then by sitting on that reference frame you will study a diffusion like phenomenon because the advection is absorbed within the reference frame itself the reference frame is also translating but still there is a net advective transport because of what? Because of the fact that the reference frame may translate with an average velocity but along the transverse direction there is a deviation from the average fluid velocity because of the velocity profile. So, even if you are moving with a reference frame that is translating with an average flow velocity with respect to that also you will see some dispersion and that is possible because you have a transverse velocity profile. But broadly with respect to a moving reference frame you mainly study you in the reference frame you absorb the advection effect and you mainly study the diffusion phenomena. So, in the reference frame the molecules undergo random thermal motion causing the solute to diffuse from the band to the outer region of the band because of concentration gradient. So, that is what you see from the blue concentration profile you will see the emergence of the red concentration profile at t equal to delta t. In reality the combination is what is there. So, you have the effect of flow plus effect of diffusion relative to the flow. So, these 2 effects combine together give rise to a translation as well as broadening of the solutel band and we want to get a mathematical picture of that and there are many ways to address this mathematically we will for this particular course we will consider a theory which is known as Taylor's dispersion theory. So, let us work out this theory I will work out the conceptual part and the algebraic parts are presented in the slide the algebra is very trivial numerical algebra I mean there is no complexity with the algebra. So, I will work out. So, now you assume that there is a coordinate transformation x dash before the coordinate transformation we will do a scaling analysis. So, let us say that you have a cylindrical tube and in the cylindrical tube there is a fully developed flow. So, that u by u average is equal to 2 into 1 minus r square by say a square where a is the radius of the tube v0 is v average this is the Heggen Poiseuille velocity profile all of you can remember this. So, the governing differential equation del c del t plus u del c del x is equal to d 1 by r del del r of r del c del r let us find an order of magnitude of this. So, what is the order of magnitude of this say v0 c reference by x reference this also if you use the advection timescale it will come out to be the same if you use the advection timescale. What is this order of magnitude d c0 by a square this is of the order of d c0 by l square right because the tube radius is much much less than the axial length scale we will usually neglect this contribution ok. Now this c0 is c reference c reference is c0. So, advection by diffusion this you can write v0 a by d into a by x reference. So, this parameter v0 a by d this is called as the piclain number ok. So, we can write that this is the piclain number into a by x reference and the piclain number the physical meaning is that it represents the advection ratio of the advection strength to the diffusion strength. Now let us say that we want to study this phenomenon in a reference frame that moves with the average flow velocity. So, we make a coordinate transformation x dash is equal to x minus v0 into t. What is the timescale over which we are observing the phenomenon we are observing the timescale which is greater than the diffusion timescale. What is the diffusion timescale typically a square by d and less than the advection timescale which is l by v0. So, the timescale under which we are observing the phenomenon is typically greater than the diffusion timescale, but less than the advection timescale that is what is our regime. So, del c del t you can write as del c del x dash into del x dash del t. What is del x dash del t minus v0. So, del c del t plus u del c del x what is this minus v0 del c del x dash plus u what is this del c del x dash what is del c del x del c del x and del c del x dash are the same. So, this is u minus v0 del c del x dash what is u minus v0 this is u dash see one very interesting thing has happened in the process this had an unsteady term in the moving reference frame with this transformation the unsteady term has now vanished. This is because of a transformation which is this one it is called as Galilean transformation. So, by Galilean transformation what we have done is we have been successful in converting an unsteady problem to a steady problem in the moving reference frame. So, with this background we will come to the slide and complete the description of the problem. So, you can see that the left hand side will become this is u dash into del c del x dash and u minus v0 remember u is u by v0 is equal to 2 into 1 minus small r square by a square that is r bar square. So, u minus v0 is v0 into 1 minus 2 r bar square del c del x dash right. So, the governing differential equation the right hand side we have neglected the axial diffusion as compared to the transverse diffusion. So, that is the governing differential equation and we will integrate this equation. So, to integrate the equation one very important assumption that we will make is that del c del x dash is a constant that means r independent then the integration becomes very easy. So, then you integrate it with respect to r the right hand side is 1 by r del del r r del c del r. So, if you integrate with respect to r basically you will get this with the constant of integration c1. Now, we can apply the boundary condition that at the center line of the tube del c del r is equal to 0 because of symmetry. So, that will give you this c1 constant of integration equal to 0. So, you will get this equation. Now, we will integrate it once more with respect to r to get c as a function of r and x. So, we integrate it with respect to r to get another constant of integration c2. We will just express c2 in terms of the cross sectional average concentration. So, what we will do? See the cross sectional average concentration is defined in this way 2 pi r c dr by pi integral 2 pi r c dr by pi s just the cross sectional average. So, if you take the cross sectional average of this equation then you will be able to express c2 in terms of c average. So, then your equation in terms of c average becomes this one. Now, you can say that r independent axial diffusion is possible if this term is small. That means v0 a square by 4 dl is much much less than 1 or Peclet number much much less than 4 into l0 by a because Peclet number definition is v0 a by d. So, this is one important assumption conclusion. Then you can write this c as a function of c average and then a very important consideration the average flux density through the cross section. What is the average flux density? So, what is happening is that the species is in the moving reference frame with respect to the moving reference frame what is the velocity at which the species is moving u dash. So, the species transport is 2 pi r c u dash dr by pi s square this integral of that that is the species flux. It is just like an advection flux of the species. So, then basically you substitute the values of c in terms of c average and substitute the velocity u average in terms of u dash in terms of the average velocity and do the integration. In the integration this term becomes 0 and I have raised a question why this term is 0? That term with the average species the average species concentration the c average that term with the c average which will mathematically come out to be 0. Think about it that why should it physically come out to be 0? That this you can follow this like the c will contain a c average term. The c average term when integrated with respect to the relative velocity u dash will become 0. So, then the remaining term when it is integrated it will this flux will become v 0 s square by 48 d del c del x dash ok. So, this j now had it been a pure diffusion you can write the diffusion flux as effective coefficient times the concentration gradient right. So, this term v 0 square a square by 48 d this becomes an effective coefficient which is called as dispersion coefficient. It takes the combined effect of advection and diffusion into account to give an equivalent picture based on diffusion in a moving reference frame. So, d effective so this j this is just relate the fix law j is equal to minus d effective into the concentration gradient. So, here you have d effective is equal to v 0 a square by 48 d which is piclet number square d by 48. Now, when you have this equation you can get the solution for the point source. So, this is the concentration solution as a function of time ok. So, time and position. So, that means we have given a described a picture by which you can make a framework of finding out the species concentration as a function of position and time using Galilean transformation and using the Taylor's dispersion based analysis that we have just demonstrated. You can use of course other techniques for getting the dispersion coefficient, but we will not discuss those techniques for this particular introductory course. So, for the time being thank you very much.