 In the previous lecture, we were discussing about surface tension driven flows and their simple one dimensional mathematical representations. In the mathematical representations, we started with one particular scenario in which we neglect the inertial forces or inertial effects and then we came up with an equation. You can go to the view graph to recapitulate this that this is the viscous force. This is the actually the surface tension driven surface tension force but expressed in terms of the static height capital H and this is the weight or the gravity force that is opposing the motion of the capillary. In the small time limit when S is much much less than h, then the forces which are balancing each other are the viscous forces and the surface tension forces. So this is the viscous force. We assume the fully developed flow to describe the viscous force and that is balancing the surface tension force and then we got relationship between the capillary displacement S and the time t where S varies as square root of the time. Now that is about the small time limit. What about the large time limit? So by large time limit what we mean? Let us come to the board to explain this. So let us say that this is the capillary and this is the final static height. So small time limit means temporal regimes where the height is much much less than the static height. Large time limit means the temporal regimes when the capillary front is almost near the static height. So to express the corresponding relationship what we will do is we will write, so we will be doing now large time limit. So 8 pi mu S ds dt is equal to pi r square h rho g – pi r square s rho g. Now let us say that what is the hallmark of a large time limit? The hallmark of a large time limit means the difference between h and s is very small. So let us say that h – s is equal to delta s. So 8 pi mu in place of s we can write delta s plus h, sorry h – delta s, h – delta s into d dt of h – delta s is equal to pi r square rho g into h – s is delta s. So now one can work out this that is h dh dt – h d delta s dt – delta s dh dt plus delta s d delta s dt. Clearly capital H is not a function of time, it is a static, the final static height. So this is 0, this is also 0, this being involving 2 small numbers will be less than the order of the other. So eventually what you get is 8 pi mu h d delta s dt is equal to – pi r square rho g delta s. So we can write d delta s by delta s is equal to – pi r square rho g by 8 pi mu h dt. So ln delta s is equal to – r square rho g by 8 mu h dt plus ln c1 if you integrate it. So delta s is equal to c1 into e to the power – r square rho g t by 8 mu h. So that means we can get an expression because once you know delta s, you can find out what is h – delta s. So that is equal to s. So we will summarize this discussion on the last time limit through the slides. So let us go to the slides there. So in the slides I mean maybe there is a different notation instead of c1 it is c2 that is used, k2 whatever. But the form is pretty clear. So ln delta s is equal to ln c2 – k2, k2 is that the combined term r square rho g by 8 mu h. So that in the shorthand notation is written as k2. So I mean in board whatever we mentioned as c1 that is mentioned as c2 here. So you can get s as a function of the constant c1, c2 whatever name you give. As time tends to 0, s tends to 0. So time tends to 0 has to be analyzed very carefully. In the short time limit whatever is time tends to 0 is different from what is time tends to 0 in the large time limit. In the long time limit the time tends to 0 means the time under consideration becomes small as compared to the notion of large time solution. So it is not literally the beginning but it is a condition when you start thinking about the large time solution. So it can be possibly like when you are thinking of the large time solution your initial condition may be the end of the small time solution. So at the end of the small time solution your height of the liquid column becomes small within the paradigm of the large time solution. It is not actually 0. See these things have to be notionally understood because this we are considering the large time limit. In the large time limit whatever is applicable that is what we are considering. So this is not technically addressing the situation from the beginning of the capillary filling. Because beginning of the capillary filling is not actually within the purview of the large time solution. Beginning of the capillary filling is within the purview of the small time solution. So these things when we are putting these as tending to 0 these are not referring to the beginning of the small time solution or the beginning of the capillary rise. But these are referring to temporal instances when the height is much smaller as compared to what you get in the large time solution. So that is as good as tending to 0. So with that you get s is equal to h into 1- e to the power – k2 into t where k2 is that rho g r square by 8 mu h. So you can see that it is an exponential variation with respect to time towards the end of the capillary rise. Now we will proceed a little bit further forward and we will try to assess the situation when inertial effects are important. So far we have neglected the inertial effects but we will now consider the inertial effects for simplicity and for its relevance in micro and nano scale we will neglect the gravity effect. That means we will neglect that gravity as a force as compared to the surface tension as a force. The reason is that in small scale the surface forces are overwhelming as compared to volumetric effects. So that means that if you reduce the length scale of your capillary or if you make the capillary narrower and narrower the gravity effects will be less and less important as compared to the surface tension effects. So with that understanding so once we say that is not very important then it really does not matter whether the capillary is vertical horizontal inclined or whatever we will neglect gravity as a body force. So we will consider reduced order model with inertial effects. Let us say that we have a capillary of radius r. Now the relative orientation of the capillary is not important because we are not going to consider the gravity forces anymore. So we can write if you write the inertia forces if we are considering this lumped mass this v could be the velocity of the center of mass for example and that is the resultant force acting on it. Let us assume that the contact angle here is theta. So what is DDT of MV? So well before that let us write what is this F? F surface tension plus F viscous plus means algebraic plus it will actually be a resistance force. So if S be the average displacement then DDT what is M? What is the mass in the shaded region? Rho into pi r square into S and v is DHDT that we discussed in the previous lecture is equal to F surface tension. So F surface tension is 2 pi r into sigma cos theta and what is the viscous force? – 8 pi mu s ds dT. This sigma is sigma lv. Now a very interesting thing is that if you could write this in terms of the static height h then that will that may give you an illusion that as if gravity effect is considered because that they will then consider rho g and all this but that is not true because the static height is just like a parameter which can be expressed in terms of the contact angle. So if this could have been dipped vertically in a fluid then the amount of rise in the fluid due to surface tension or fall in the surface tension is characterized by the static height. That does not mean that in a horizontal or inclined or whatever situation equivalent rho g or gravity effect is actually coming into the picture in all configurations. What it basically means is that if you dip it vertically then whatever height it could have attained is the static height that could be expressed in terms of the contact angle. So that this is not simply this is not at all body force due to gravity. It can equivalently be expressed in terms of a static height that is all. Now we will try to solve this equation but before solving this equation we will try to non-dimensionalize the equation so that we will get various scales. What is the order of this? Pi r square s ref square by t ref and what is the order of this? Actually putting pi and all those things in the order has no relevance. I mean you might as well drop it but just we are writing it for the sake of writing. 8 pi mu s ref square by t ref. Without the inertial limit the surface tension force always be very important see in the capillarity surface tension force must be important. Whatever is not important is secondary but whatever is important must be focused. So surface tension force is always under focus in capillary dynamics. Now the surface tension force may eternally drive it if it is a driving force and the fluid will not stop. So that will create a sort of a perpetual type of motion but that will not be possible because the viscous force will come into the play as the fluid is moving. So that means the viscous force will also be important and when inertial effects are important then only this term is important. Therefore inertial term also has to be comparable with either of this or of this in a competing environment. That is why when we are trying to establish a proper timescale of the problem we can equate this with this to get the timescale where the length scale gets cancelled out. So rho pi r square again I am repeating that the pi and 8 and all those things are not important for the scale but for the sake of non-dimensionalization this will help this will help us to clean up the coefficients of the equation. That is why we are retaining this by t ref square yes t ref square this is t ref square s ref square by t ref square right. So it is of the order of 8 pi mu s ref square by t ref. So t ref is of the order of rho r square by 8 mu this is the characteristic timescale of the problem. See this characteristic timescale of the problem we have determined from a competition between the inertial force and the viscous force. That is why surface tension has not come into the picture but we could have as well calculated the characteristic timescale by comparing by considering a competition between the inertial force and the surface tension force that could have given rise to a different t ref. So that has to be understood very carefully. Here we are not bothered so much about the scaling of different forces considering the inertial effects what we will simply use this is we will do this is we will use this for non-dimensionalizing the governing equations. And then I mean it will not notionally differ so much whether we do it on the basis of inertia versus viscous balance or inertia versus surface tension balance. But to understand the proper timescales governed by inertia versus surface tension balance and inertia versus viscous force balance then those different timescales will conceptually matter. So far as understanding the scales is concerned remember here our objective is more to solve the differential equation with non-dimensionalization. So then I mean this is just fine but this also physically gives a timescale over which you have a balance of the inertial and the viscous effects. Now you can also calculate the s ref the s ref got cancelled as we will consider these 2 terms and in fact that is why we consider these 2 terms instead of this with the surface tension term because then that will have both s ref and t ref. So it will not be possible to get explicitly s ref or t ref. But here these 2 terms are fortunately such that s ref got cancelled and we could in one step one short get t ref. Now to get s ref we can now use a scaling of balance between viscous and surface tension force or competition between viscous and surface tension force. So to do that what we will do is we will use a little bit of force site. See this is a differential equation on this term you have s ds dt, s ds dt is d dt of s square by 2 right. Here also you have s ds dt so d dt of s square by 2. So if you take a half with this term also these 2 will become 4 then half will get cancelled from all the terms. So with this little bit of force site we will take this term as 4 pi r sigma lv into cos theta by 2. So the surface tension term 4 pi r sigma lv we will take as the coefficient of this term because we will keep cos theta by 2 then this half will cancel when we integrate the expression. So that we will again I am repeating over and again this 4 pi and all those things are actually not important for scaling but here we are trying to do it not merely for the sake of scaling but for the sake of non-dimensionalization with cleaning up of the coefficients and that is why we are accommodating the coefficients in this expression. So for a scaling is concerned it is simply r sigma lv okay. So now this is of the order of 8 pi mu s ref square by t ref. So s ref square so this 4 pi 8 pi will become 2 s ref square will become r sigma lv t ref by 2 mu. So that means this is t ref is rho r square by 8 mu so rho r cube sigma lv by 16 mu square right. So s ref is of the order of r by 4 mu into square root of rho r sigma lv. So now we have considered non-dimensionalizing parameters which consider competition between 3 terms simultaneously because the time scale was given by a competition between these 2 but using that same time scale and using the competition between these 2 terms we have considered the s ref. Therefore in effect we have considered the competition between the 3 terms simultaneously. So with this consideration now if you non-dimensionalize s bar is equal to or s star is equal to s by s ref and t star is equal to t by t ref then what will this differential equation give? Clearly as we have chosen the scales it will be d d t star of s star d s star d t star is equal to cos theta by 2 minus s star d s star d t star right. That is how we have chosen the reference things other things we have taken in the non-dimensionalizing coefficient. So they will get cancelled okay. Now s star d s star d t star is d d t of s star square half and let d s star square d t is equal to y t means t star basically. So half d y d t star is equal to cos theta by 2 plus half y. So d y d t star sorry minus half y d y d t star plus y is equal to cos theta. So this is a differential equation of the form d y d x plus p y is equal to q. This is the form of the differential equation. So for solving this differential equation the technique is you multiplied by an integrating factor e to the power integral p dx okay. So you multiply this with an integrating factor here x is t star. So e to the power integral p is 1. So e to the power integral d t star. So that means e to the power t star. So we multiply both sides by e to the power t star. So the left hand side becomes d d t star of e to the power t star y is equal to e to the power t star cos theta. So if you integrate this e to the power t star y is equal to integral of e to the power t star is e to the power t star e to the power t star cos theta plus C 1. So y is equal to cos theta plus c1 e to the power minus t star, right and y is ds star square dt, dt star. So s star square is equal to t into cos theta, t star into cos theta minus c1 e to the power minus t star plus c2. Now we require c, we require to evaluate c1 and c2. So for that purpose we require 2 initial conditions at t equal to 0, let us say s star is equal to s0 star. It could be 0 also but in general we may start with an initial length. I will show you that see I have purposefully avoided 0 with something in mind and I will show you that what problem can arise if you take it as 0, we will come into that. So however to generalize we can take it something non-zero but if somebody takes it as a 0 as a special case what will happen then? We will talk about that. So s star is equal to s0 star and let us say ds star dt star equal to 0, initially the capillary front advances with a 0 velocity. So if you do that then you will get what you will get from this cos theta plus c1 equal to 0, right. So c1 is equal to minus cos theta and from here s0 star square is equal to t star cos theta that term will be 0, c1 will be minus cos theta. So cos theta into e to the power minus t star, t star is 0 plus c2. So c2 is s0 star square minus cos theta. So s star square is equal to t star cos theta plus e to the power minus t star cos theta minus cos theta plus s0 star square. We can of course rearrange it and write it nicely I am not going to that step, okay. So similar considerations can be applied for channels of other cross sections. So far we have done for circular channels but circular channels are basically called as circular capillaries in when the channels are small and capillary effects are important these are also called as circular capillaries but many of the micro channels by virtue of the micro fabrication process are not actually of circular shape but they may have other cross sections like approximately rectangular or trapezoidal or several other cross sections may be possible. And I will show you in few lectures from now when we discuss about micro fabrication I will show you that how the fabric by the fabrication process you can make actually angles in the cross section of the micro fluid channel that you are going to manufacture. So now we will come to the board to summarize the reduced order model accommodating the inertial effects. So we have neglected the gravity force that we have already told. So then ddt of mv for circular micro channel of radius r this is this for rectangular micro channel instead of pi r square it is the width into height right that is the only difference. Surface tension force for circular micro channel 2 pi r sigma lv cos theta so that is basically perimeter into sigma lv into cos theta. So for rectangular micro channel the perimeter is 2 into width plus height. So see the similarity the viscous force for a rectangular micro channel we have solved the problem of flow through a rectangular micro channel the exact analytical solution. So from that you can calculate the viscous force it will come out to be in the form of this tan hyperbolic form which is given in this slide. Parallel plate channel is a special case see we are here we are talking about the rectangular channel where the width and height none of them is negligible as compared to the other or none of them is too large or too small as compared to the other. So but if one of the dimensions is much smaller as compared to the other it effectively becomes a parallel plate channel but here the expression that is written in this slide is for a rectangular micro channel. So how do you get this expression it is very straight forward we have derived the velocity profile for rectangular micro channel I mean quite some time back when we are discussing about low Reynolds number flow and in that expression you from that expression you can find out what is the wall shear stress and from the wall shear stress you can get the force. So for a circular capillary this is what we have done so this is just a brief recapitulation I will not go through the details of these steps because we have already gone through these in the board. So this the final solution the last expression if you whatever we have derived in the board if you write it nicely in a compact form this is what it will this is the form that it will assume. Now limitations of the simple model see this is what is very important in science and engineering many times we are doing with certain simplifications there is nothing wrong with doing with the simplifications provided we know what are the limitations right I mean this is also philosophically very appropriate in all aspects of life right I mean all of us are born with certain limitations right I mean none of us are perfect and none of us are I mean very very proficient in all and every aspects of activities but if we know our limitations I think that is where we can actually excel that is it is important to know our strengths but it is also equally important to know our limitations. So for a mathematical model see a mathematical model cannot be a true representation of a physical reality no matter whatever you put in the model because in the physical reality there will be certain factors which will not be actually taken into account by the mathematical model. So whenever we are saying that say let us say there is a pipe flow and fluid is entering with an average velocity of 2 meter per second let us say now how will you know that the fluid is entering with an average velocity of 2 meter per second then base you can well argue that I have measured the flow rate and I have measured the flow rate very accurately I have measured the cross sectional area very accurately but there are uncertainties in the measuring the flow rate because there are uncertainties in the measuring devices and there are uncertainties in measuring the cross section in micro channel there are lots of uncertainties in measuring the cross section see I mean this is where see sometimes in experimental result we plot the friction factor versus Reynolds number and different research groups come up with different types of plots under very similar experiments condition is I mean the problem is that standardizing the Reynolds number the hydraulic diameter itself is a big question because if you look into a micro channel surface if you come to research in the area of microfluidics you will definitely get a chance to have a look into how the micro fabricated surface looks if you look maybe through an atomic force microscope and all these things and you will realize that I mean forget about the paradigm of an ideally I mean rectangular channel and all those things. So the surfaces are undulated in a very peculiar way where the notion of an average height may become actually questionable I mean an average height may not represent the physical reality. So there are uncertainties in the height there are uncertainties in the flow rate in the flow rate measurement so all these uncertainties are actually carried over in the boundary condition that itself you are putting for your numerical simulation. So no matter how good your equations are you may not be able to solve the problem properly until and unless you have you give account of uncertainties in your CFD simulation. In fact that is one of the branches of modern CFD study that is how to take how to give account of uncertainties in numerical simulations I mean uncertainties in boundary conditions uncertainties in physical parameters and all these things. I am little bit deviating from the main topic of discussion but this is something which is conceptual or philosophical rather than very specific to what we are discussing here. So my idea of telling this is that even for a very rigorous CFD model you can see that there are so many limitations. So for a simple one dimensional model you can clearly see that limitations are limitations will be even more but despite that limitations we can get some answers but we cannot get certain features and what we cannot get that we will discuss through limitations of this simple model. So this graph this kind of graph we can get the S star versus T for different theta S star versus T star this is basically a plot of the analytical expression that we just now derived. So you can see I mean this is quite intuitive that is how S star varies with T. A lumped system analysis does not capture point to point variations right this is just a simple lumped system means what the entire liquid mass within the capillary. So here we have considered a liquid gas or liquid vapor type of system. So the entire liquid mass is idealized or modeled as a point mass. If you do that then point to point variation is lost you are not able to capture point to point variations if somebody is interested to know what is happening in all details within the liquid then that cannot be captured by here by this model. The viscous resistances of different flow regimes are not considered this is very very important and we will discuss about this we have assumed that the flow is fully developed but we will discuss that what are the pitfalls of that when you are considering a capillary advancement or capillary retraction whatever. So the consideration of fully developed flow it might appear that well in a low Reynolds number regime you know the flow is expected to be virtually fully developed. But we will see that when there is a capillary meniscus that is formed that paradigm has to be little bit adjusted and we will see that how it needs to be adjusted. The third point is a very important point and I want to discuss I want to use the use the board to discuss about that not that we will be writing some big equations but just to have a distraction from the slides. So d dt of mv is equal to f that is what essentially we are writing see actually you know in approximation you do well once you know that well the first step that I am writing itself may not be correct this first step that we have written actually may not be correct why can you say this is true for a control mass system that a system of fixed mass but the liquid that is being pushed into the capillary this mass is a variable mass. So somewhere that conceptual error should reflect where does it reflect it reflects under that condition when m tends to 0 when is m tending to 0 the liquid is just entering the capillary that is at t equal to 0 at time equal to 0 if s 0 equal to 0 now we are discussing about the consequence of s star equal to 0. So at time equal to 0 if you take s 0 equal to 0 then what it basically means m equal to 0. So this is nothing but mass into acceleration this force being finite and the limit as time tends to 0 then if m tends to 0 then a mass tend to infinity so but you know in a physical system you cannot have an infinitely large acceleration. So this infinite initial acceleration is a defect of this model and I will show you how this defect can be corrected but this is the defect that we must appreciate. Then let us go to the slides the next point there is no explicit description of the contact line see because it is a lumped model it does not capture the meniscus it does not capture the meniscus shape. So you are just representing it in terms of the contact angle but the topography of the meniscus is not captured right just the contact angle is coming as a parameter it is not that you are resolving it and putting it it is just coming as a parameter. Not only that we have assumed that the contact angle is a constant but as the meniscus progresses along the channel the contact angle changes dynamically this is known as dynamic contact angle the contact angle does not remain at its static value. So whatever value of contact angle that we are using that is actually called as a static contact angle but the static contact angle may not remain as it is and it will evolve in the form of a dynamic contact angle as the capillary front progresses. So these are certain considerations which need to be put into the mathematical model some of these considerations remember are big research issues which have yet not been very convincingly resolved see this is like this is always gives us a little bit of difficulty in teaching a course like this like on one side we have fundamentals which have been existing over the years and it is I mean it is a duty for us to like transmit that knowledge to you in terms of the fundamental advancements in the subject. But microfluidics is such a subject where if you go on discussing starting from the fundamentals you very soon come up to research topics and as a researcher in the field I do not mind in sharing with you those research topics but at the same time we have to keep a break because I cannot go on discussing open-ended questions in a structured lecture framework and it is better that I will leave these topics at some point of time from where if you are interested you can take up reading of further research materials research papers like that. So that is that is where I believe that this course I mean is what I have ideally designed it as like a bridge between a sort of a theoretical course and a research level advanced theoretical course a basic theoretical course and a advanced research level theoretical course. But remember so I will not be able to satisfy all your inquisitiveness in this particular course because we will leave it at somewhere all expressions we will not be able to derive from the fundamentals because you know I mean that will take huge amount of effort like I mean I will show you one or two equations I mean their original derivation if you look into that paper that is a 50 page paper. So it is impossible to cover that in the framework of this lecture but I just want to give you the concept give you the idea and leave you there so that you know if you are interested with that you can proceed further. The first concept is that like the first point which we would like to address is that like the infinite initial acceleration issue how that can be corrected. So that can be resolved by referring to the equation of continuity look at this description the onset of the capillary motion the liquid present in the dip portion of the capillary or the bulk reservoir to which it is connected also starts moving at the same time right. Here we have considered what here we have considered that like if you go to the board this is a capillary this capillary may be dipped in some reservoir it is I mean capillary is always going to suck I mean this channel is always going to suck fluid from somewhere right I mean there must be a source of fluid so that is there in the reservoir we are not showing what is there in the reservoir. So ideally whatever is there in the capillary plus like it forms a connected system there is an additional mass which also when this fluid is moving this is also sucked up and that additional mass together with this mass in the capillary in a system environment actually forms a control mass system not just the mass within the capillary. So if you continue with it this additional mass of fluid inducted into the motion initially is known as the added mass or sometimes it is called as virtual mass this is called as added mass or virtual mass for the reason that actually this is not the mass seen within the capillary but it is a mass that is integrated actually with the motion that is taking place within the capillary. Now this added mass how can you estimate this added mass I mean this is actually like the expression that I have written here I mean these are apparently simple expressions but you require or rigorous potential flow analysis to come up with these expressions and I mean there are research papers which actually derive these expressions I mean there is not enough scope of deriving these expressions but for solving a problem you can use these expressions these expressions are derived by applying a potential flow analysis for an incoming spherical liquid element that moves with the velocity ds dt why potential flow analysis because it is not already within the channel so viscous effects are neglected okay. So that makes inviscid flow type of analysis you know potential flow is inviscid irrotational flow so this yields the following added mass the following expressions for the added mass one for a cylindrical tube and one for a rectangular channel of height h and width 1 so that is basically parallel plate channel so for these two types of channel these are the expressions so when you write d dt of mv if you now so let us go to the board and see instead of this m you write m plus m0 then at time tends to 0 even if this is 0 because of this m0 not equal to 0 you will not get infinite acceleration and this discrepancy may be corrected. So this is one of the discrepancies in the basic model and we have seen how it is it can be corrected there are other discrepancies in the model that we have discussed and we will see how those can be corrected and then we will see that how this model can be extended to a simple non-Newtonian model for example to capture the behavior of blood in a micro capillary or blood flow in a micro needle and that we will take up in the next lecture thank you very much.