 In the previous lecture we were discussing about interfacial equilibrium from energy minimization considerations. So I will just briefly recapitulate what we are trying to do through this view graph that we considered droplet which is a part of a spherical shape and then we have written the total energy as the sum of the interfacial energy and then we have minimized the total energy by constraining the total volume of the droplet where lambda in this expression is the Lagrange multiplier. So we have derived these expressions of what is the volume of the droplet or what is the liquid vapor interfacial area solid liquid interfacial area. All these expressions we have derived and the expression for E is summarized in the view graph. For minimization of E you can set the partial derivative of E with respect to theta equal to 0 and with respect to r equal to 0. So let us try to see that what is the consequence of that. Let us get back to the board and look into the expressions. So the first expression partial derivative of E with respect to r equal to 0 inlet this particular expression. So we can just rearrange that and write this is equal to 3 lambda pi r square by 12 into 8 plus cos 3 theta minus 9 cos theta and this also we can rearrange to get this form. Now let us say this is equation number 1 and this is equation number 2. We can divide equation 1 by equation 2. If we divide that then lambda gets cancelled the Lagrange multiplier gets cancelled. So 1 divided by 2 what does it give? It gives sigma SL minus sigma SV you may divide both numerator and denominator by sigma LV into 2 pi r sin square theta plus 4 pi r into 1 minus cos theta divided by sigma SL minus sigma SV by sigma LV into 2 pi r square sin theta cos theta plus 2 pi r square sin theta is equal to 3 pi r square by 12 into 8 plus cos 3 theta minus 9 cos theta divided by pi r cube by 12 into 9 sin theta minus 3 sin 3 theta. I hope this expression is alright because we will proceed with that. So this expression sigma SL minus sigma SV by sigma LV let us say this is equal to k. So if you take 2 pi r common then k sin square theta we can write 1 minus cos square theta plus 2 into 1 minus cos theta divided by 2 pi r square sin theta into k cos theta plus 1 is equal to 8 plus cos 3 theta minus 9 cos theta divided by 3 sin theta minus sin 3 theta. Please check the algebra of this and let me know. See if we make a mistake here I mean the final expression will have no meaning okay. So we can cancel this 2 pi and then 1 by r again and 1 by r on this side that 1 by r gets cancelled with this. Now let us simplify the numerator and denominator of the right hand side numerator of RHS is equal to 8 plus cos 3 theta minus 9 cos theta. So this is 8 plus 4 cos cube theta minus 3 cos theta minus 9 cos theta. So that is equal to 4 cos cube theta minus 12 cos theta plus 8. So you can take 4 as common cos cube theta minus 3 cos theta plus 2. So 4 cos cube theta minus cos theta minus 2 cos theta plus 2. So 4 into cos square theta into cos theta minus 1 minus 2 minus if there is any mistake please let cos theta into cos square theta minus 1. So 4 into cos square theta minus 1 minus 2 into cos theta minus 1 alright. So 4 into cos theta cos square theta minus 1 we can write cos theta minus 1 into cos theta plus 1. So that cos theta minus 1 if we take as common then cos theta into cos theta plus 1 minus 2 right. So that is equal to 4 into cos theta minus 1 into cos square theta plus cos theta minus 2. So 4 into cos theta minus 1 into cos square theta plus 2 cos theta minus cos theta minus 2. So 4 into cos theta minus 1 into cos theta into cos theta plus 2 minus 1 into cos theta plus 2. So 4 into cos theta minus 1 whole square into cos theta plus 2. So this is as good as 4 into 1 minus cos theta whole square into cos theta plus 2. If this expression is alright I will erase the derivation and just keep the final expression because we have to simplify the other expressions in the board. So let me do one thing just in place of this keep a note that this is 4 into 1 minus cos theta whole square into cos theta plus 2 and then let me erase the derivation. Then the denominator of the RHS is equal to 3 sin theta minus sin 3 theta. Sin 3 theta is 3 sin theta minus 4 sin cube theta. So 3 sin theta minus 3 sin theta minus 4 sin cube theta. So this is 4 sin cube theta. So this we can write 4 sin theta into 1 minus cos square theta. That means this expression now becomes k into 1 minus sorry 1 minus cos theta if you take as common k into 1 plus cos theta plus 2 divided by sin theta into k cos theta plus 1 is equal to this 4 and this 4 get cancelled. So 1 minus cos theta whole square into cos theta plus 2 divided by sin theta into 1 minus cos theta into 1 plus cos theta that is 1 minus cos square theta. So let us see what are the things which get cancelled. So 1 of the 1 minus cos theta gets cancelled here theta is not equal to 0 in general. Then again this 1 minus cos theta gets cancelled and sin theta gets cancelled. So k into 1 plus cos theta whole square plus 2 into 1 plus cos theta. I hope there is no algebraic mistake in between because then we will get not the correct answer. Let us see cos theta plus 2 into k cos theta plus 1. So k plus k cos square theta plus 2 k cos theta plus 2 cos theta plus 2 is equal to k cos square theta plus cos theta plus 2 k cos theta plus 2. So k cos square theta gets cancelled 2 k cos theta gets cancelled 2 gets cancelled. So you get cos theta k is equal to minus cos theta. So k equal to minus cos theta that means sigma s l minus sigma s v by sigma l v is equal to minus cos theta or cos theta is equal to sigma s v minus sigma s l by sigma l v. This is the condition for equilibrium of the droplet and this is known as Young's equation. The previous equation that was having a similar name is Young's Laplace equation. This is Young's equation. Now let us substitute this expression for k in one of the expressions to get what is lambda. So if you do that I mean we have done enough of algebra I will not take more time in doing that but if you do that you will get. So how do you get all this? You simply substitute back the expression for k in one of the expressions daily del r equal to 0 or daily del theta equal to 0. So if you do that lambda will be the only unknown and this is 2 sigma l v by r. Now I mean let us summarize this what we have discussed so far that for minimization of E setting daily del theta equal to 0 and daily del r equal to 0 and eliminating lambda from the above expressions above 2 expressions. I mean sometimes you know that like very commonly in books it is told that by trivial calculation it follows. Sometimes trivial calculation can take pages. So every time I do not want to do those quote unquote trivial calculations but for just one example I did it reminded me of my school days when I started doing trigonometry so gives me some kind of feel good effect. So I did it but it is not that all the time we will be going through such trigonometry and algebra but like if you look into the expression. So cos theta equal to sigma s v minus sigma s l by sigma l v that is Young's equation and substituting the value of cos theta the Lagrange multiplier can be obtained as 2 sigma l v by r. This we can confidently say even without doing the algebra because it carries some physical concept associated with it. See in the expression for E actually we had lambda into the volume and if that lambda is delta p then that delta p into volume actually becomes physically the work done. So from physical intuition we can argue that delta p is nothing but or lambda is nothing but the delta p itself. Look into the beauty of mathematics even if you do not have this concept of that understanding from algebraic simplification it straight away comes that this lambda is equal to 2 sigma l v by r and from the Young Laplace equation we know that 2 sigma l v by r is delta p okay. So these are the 2 equations that we get and the Young's equation that we have got we can clearly get it from force balance considerations without going through the involved energy balance. So if you look at this diagram it will show you that in the 3 phase contact line you have the forces per unit length expressed sigma s l, sigma l v, sigma s v these notations are now clear. So along the horizontal direction if you resolve the forces so you see that let us say that the negative horizontal direction is the or this direction is actually the positive direction what we are considering. So the component of sigma l v in that direction is sigma l v cos theta. So sigma l v cos theta minus sigma s v plus sigma s l equal to 0 for horizontal force balance and that straight away gives the Young's equation okay. Now there is a very interesting question. This the interesting question is there are couple of interesting questions but let me first pose the first interesting question. We have discussed a condition of equilibrium or derived a condition of equilibrium by considering the horizontal force balance. What about the vertical force balance? Nowhere it is discussed. So if the vertical forces are not balanced there is an unbalanced force droplet might suddenly start accelerating on the in the vertical direction. We never see it to happen in experimental practice that means something intuitively takes care of the vertical force balance. This is not a true free body diagram because this shows an unbalanced vertical force. So actually what happens the force balance in the vertical direction is achieved by the normal reaction. So just to visualize it instead of the droplet think it like a rigid body. If it was a rigid body sitting on a substrate there will be a normal reaction and similar kind of normal reaction is acting on the droplet. The difference between the droplet and the rigid body is of course quite clear the droplet has its own deformability. But in the normal direction in the vertical direction the force balance is achieved by the normal reaction. So this is one important note. Now how to calculate the normal reactions and all those things I mean there are various ways to look into the forces over the small scales. Sometimes one has to go for molecular dynamic simulations to figure out what is the resultant force in the normal direction and resultant force in the tangential direction and all this. So let us not complicate those matters any further here. The Young Laplace and Young's equations are necessary conditions for equilibrium but not sufficient conditions. This is very very important that means even if these equations are satisfied we cannot say that the system is in equilibrium. Question is why? The answer is very straightforward. We have said the first derivative equal to 0 but that could be either a maximum condition or a minimum condition. So the first derivative equal to 0 does not necessarily imply that it is a minimum condition because the equilibrium condition is the minimum energy consideration. So it is only we can say from this that it is only a necessary condition for equilibrium but not sufficient condition because then we have to test for higher order derivatives and so on. That is not tested while deriving these equations. Now so we have so far discussed about the basics of like the pressure difference, the relationship between the pressure difference, surface tension, contact angle, the basics of that. Now we will try to use these concepts in the context of fluid mechanics and typically fluid mechanics through micro and nano channels. So to do that as we do in all discussions in fluid mechanics we will first start with statics and then we will go to dynamics. So to do that let us come to the board and we will explain. Let us say that we have a tube like this capillary tube immersed in a fluid. So when we dip the capillary tube in the fluid a common observation is either the fluid level goes up or the fluid level goes down. That observation is typically very common when the capillary tube is quite narrow. If it is instead of a narrow capillary tube if it is a wide pipe that observation is not very common and we will very soon see why that observation is not common if it is a large pipe instead of a narrow capillary. So let us say that there is a meniscus that is formed because of a capillary rise. This phenomenon that what we are calling as capillary rise can also be a capillary depression depending on the contact angle. So the pressure on this side is p atmosphere and on this side pressure is p liquid. Let us say that the average static displacement is h. When we say average it is some sort of average representation of the meniscus. Now which one is more p atmosphere or p liquid p atmosphere right because we have in the previous lecture shown that like that is like the inside and outside that is like the inside that we explained in the previous. So p atmosphere – p liquid what is that 2 sigma by if we assume that this is a hemispherical then this is 2 sigma by this r. But let us say that small r is the radius of the capillary. Small r is different from capital R. So how do you relate these 2? Look at this diagram carefully. Let us say this angle is theta which is nothing but the contact angle. So if that is theta then this small angle is also theta. Therefore capital R is equal to small r by cos theta. So p atmosphere – p l is equal to 2 sigma by small r cos theta. Now in this free surface if the pressure is p atmosphere here pressure here is also p atmosphere because this is exposed to the atmosphere. So what is p atmosphere – p liquid? It is the height of this liquid column into rho into g. So this is h rho means rho of liquid into g. So we can write h equal to 2 sigma cos theta by rho g r. What is this sigma? Sigma between which phases? Sigma lv. So see very loosely we write sigma but we have to keep in mind that this is actually sigma lv. So now you look at this expression carefully. First of all I commented that it can be capillary rise or depression depending on the contact angle. That is clearly seen from this. If cos theta is positive then h is positive. If cos theta is negative then h is negative. So it depends on the fluid surface combination whether what will be the contact angle. And the engineering of the surface can alter the contact angle. So just by telling that this is the surface, this is the material we cannot say what is the contact angle. You can engineer the surface to tune the contact angle that is what the micro and nano fabrication facility has provided now. The other important point is that how can you increase the h? See for a given contact angle and for a given liquid vapor interfacial tension the only way in which you can manipulate is I mean rho of the fluid is fixed given a particular fluid g you cannot play with here. So h is proportional to 1 by r. That is where the importance of this in the context of microfluidics comes or nano fluidics comes. You bring down r from 1 meter to 1 micron. You see that there is a 10 to the power 6 times rise in h. So many times people ask that we are discussing this in the context of micro and nano scale transport. Is it not there in a large scale pipe? Yes it is there but in that case this h is so small that it does not make any effect. But as you reduce the scales because h scales with 1 by r there is a dramatic influence. There is a dramatic augmentation in h. So that makes the surface tension such a powerful force in small scale systems. Now we can interpret this expression in a somewhat different angle. This is if this is the pressure difference then what is the force due to the pressure difference? If this is delta P then delta P into pi r square. So multiply both sides by pi r square. So this is 2 sigma by r cos theta into pi r square. So 2 pi r sigma cos theta. Of course it has to be like that. So 2 pi r sigma cos theta. So what is 2 pi? So what is the delta P into pi r square? This is 2 pi r sigma cos theta and then we can write if we write so the first we can interpret something from this. So we can write the surface tension force as 2 pi r. 2 pi r is what? The perimeter of the surface. The perimeter into sigma into cos theta. So if you look at the surface it is actually like this. So surface tension is force per unit length. What is the length on which it is acting over this dotted line? That is 2 pi r. So 2 pi r into sigma and that is making an angle theta with the vertical. So vertical component is 2 pi r sigma cos theta. Now what is delta P into pi r square? Is equal to h rho g pi r square. So you can write that as rho into pi r square h into g. What is this? This is the weight of this liquid column mg. So you can perceive the static equilibrium condition as a condition in which the surface tension force balances the weight of the liquid column. That is also a way in which it can be perceived. So these are just, see I am trying to give you different interpretations in whatever way possible to get the physics of the problem because I mean these are important concepts. This is simply static but we will get into the dynamics pretty quickly. So to summarize this discussion let us go to this view graph. So you can write this h. Only difference is that in the view graph we are writing it as small h and here I have used capital H. Other parameters are the same. So the height is proportional to the inverse radius. This is what we should remember and that gives the perspective in micro nanofluidics. Now we will move into the dynamics and the capillary filling dynamics is something which has been worked upon by researchers over the years actually. So there have been various levels of simplification with which these theories have been handled and we will begin with the most elementary consideration when we have, when the researchers have considered that there is no inertia force. That means the capillary front, the capillary front is displacing without inertial effects. We have to keep in mind that there are conditions under which there are constraints under which this assumption is justified because there are cases, there are scenarios when the inertial forces are important. But if you neglect the inertial forces it has been seen that remarkably it can represent the physics of capillary advancement or capillary retraction whatever because of one simple reason that in micro and nano scale systems where the capillary effects are very important. There inertial effects are not that predominant except over a small temporal regime except over a small span of time the inertial effect may not be important. So we will get into the corresponding analysis first without inertial effects and that was initially reported by several scientists and one such celebrated model is called as the Lucas Washburn model and we will enter into the description of the Lucas Washburn model. So we will come to the board to discuss about that. So let us say that again we are considering a vertical capillary where you have a meniscus form like this. Meniscus is advancing in this direction and the gravity is acting in this direction. Let us say that r is the radius of the capillary. We have made a small change in the notation instead of small r as the radius of the capillary we have considered as capital R. So now if inertial forces are not present then what are the other forces which are present? Now the difference between the previous case and this case is that the capillary is moving. See why the consideration of capillary is moving is important that the height that we derived in the previous calculations that is the equilibrium height but it should take a time to reach the equilibrium height and shorter that time taken more rapid is the transport and if that is done for example to extract blood or inject of medicine through a micro needle or a nano needle then that means there is a rapid blood extraction or drug delivery. So I mean there are significant consequences why we are doing that, why we are trying to understand the capillary dynamics. Not only that in human bodies there are micro capillaries through which blood is flowing. So I will first discuss about a Newtonian fluid and once our discussion is over about a Newtonian fluid I will discuss about some research issues on blood being transported through a capillary instead of a Newtonian fluid. So to do that, so first is let us write the various forces. So one of the ways in which this analysis is commonly done is like making a one dimensional model. Sometimes people make fun of one dimensional models because many times people think that those are outdated and nowadays anything and everything can be solved by 3 dimensional, unsteady whatever all sorts of complicated models. But we have to remember that in old days when lots of computational facilities were not available actually people had good physical insights. So in a simple way by getting a lot of physical insight they could solve a problem which nowadays everybody is trying to solve by brute force CFD. So one has to see what is the perspective. I will tell you where for these particular problems the detailed CFD is necessary but for all situations for getting all possible I mean interpretations it is not necessary to do a full scale CFD. Like if you have a small mosquito in this room I will not advise you to shoot a cannon to kill the mosquito. You could have other simple tricks of killing the mosquito. So if you now consider this as a simple one dimensional system with the fluid in the capillary as a lumped mass. So what this lumped approach this is called as reduced order model or lumped mass approach. Then actually the details of the variation along the cross sectional direction is lost. It is effectively like a lump moving along the x direction. The details of like the mass distribution and all those things are not important. So if you consider that then there are certain forces acting on this lump. So what are the forces surface tension plus gravity plus viscous I mean I am writing this as algebraic sum. So some of these will be positive which will be aiding and some of these will be negative that will be opposite. So that is equal to 0. So in the previous problem what we discussed that was a statics problem. So only these 2 forces are there. Now it is a dynamics problem. So this will come into the picture. So what is the surface tension force? Let us try to write the surface tension force 2 pi r let us say the contact angle is theta 2 pi r sigma cos theta. We have just derived this. What is the this sigma is sigma lv again I am reminding you all the time we are not going to write that. Gravity force first it is minus because it is surface tension is pulling the fluid in the positive upward direction and gravity is resisting it opposing it. So minus pi r square let us say so now this is the initial displacement. Let us say that the initial instantaneous not initial this is the instantaneous displacement. So let us say this is s as a function of t. The final equilibrium value of s is h but s changes with time as the capillary front advances. No, no, no neither lowest now the lowest highest all those things are lost because you do not resolve the other dimension that is the essence of the lump model that you have to understand carefully. So from a fundamental perspective it could be the average location of the meniscus somewhere but again that is not very important because once you are not resolving it whether it is minimum maximum average it makes no sense. So minus pi r square s rho g what is the gravity what is the viscous force? What is the viscous force? So first of all we can write it as tau wall into 2 pi r into s provided tau wall is constant throughout right. When is tau wall constant throughout? When the entire flow is fully developed hydrodynamically fully developed then the velocity profile is constant throughout and tau wall is like the mu into velocity gradient at the wall so that is also constant throughout. So this is the simplification that we are making and I will discuss in details about the consequences of this simplification later on. There are cases when there is a deviation from that but still this gives a good analytical way of solving the problem. So minus tau wall into 2 pi r into s what is tau wall? So what is the velocity profile? u by u average is equal to what? What kind of flow is this? It is a circular tube fully developed flow what is the name of this flow? Again Poiseuille flow. So u by u average is equal to 2 into 1 minus small r square by capital R square where small r is the radial coordinate. So what is del u del r is equal to 4 u minus 4 u average del u del r at small r is equal to capital R is equal to small minus 4 u average by r. Here when we have written tau wall is the magnitude. So what will be this? 4 mu u average by r. So this expression becomes 8 pi mu u average s right minus of course I have excluded the minus sign so minus is there. Very interestingly the viscous force has no relationship no scaling with r make r large small medium whatever this does not change. Now one important consideration a very important consideration for this model that in a one dimensional paradigm this u average we can simply write as ds dt the rate of change of s with respect to time. So we can write 8 pi mu s ds dt is equal to 2 pi r sigma cos theta minus pi r square s rho g and in place of 2 pi r sigma cos theta what we can write? This is as good as pi r square h rho g where this capital H is the static equilibrium height or the capillary rise. So now let us quickly solve this equation in some limits the limit one short time limit. So the right hand side is pi r square h minus s into rho g right in short time how can we simplify this? In short time s is much much much less than h. So short time means s is much much less than h. So the right hand side is approximately pi r square h rho g. So 8 pi mu s ds dt is equal to pi r square h rho g. So s ds dt is equal to r square h rho g by 8 mu. You might elusively infer that this does not depend on the surface tension contact angle etc. But all those are hidden in the expression for capital H. So it is very much dependent on contact angle surface tension coefficient and all this. This is just a short hand way of writing the expression. So now if you integrate this then you can write s square is equal to 2r square h rho g by 8 mu t plus c right. And if you take the limit that adds t tends to 0, s tends to 0 then we can clearly see that the c will be 0. Otherwise if at t equal to 0, s equal to s0 then this will involve s0. So this with an understanding that at t equal to 0 s is equal to 0 that may not be the case always. You may have a initial length up to which the capillary has already been filled. So this is just a special case. So from here you can write that s is equal to something into square root of t. So this square root of t dependence is a classical dependence and most of the flows through micro and nano fluidic systems follow this. Such a very simple derivation see how it captures the essential physics. Now this is the solution for short times and neglecting inertia. Neglecting inertia the solution can still change. The nature of the solution can still change when we go for large times. Large time means what? That means the capillary is almost on the verge of being filled. And what happens at that asymptote in the large timescale limit? We will take up that in the next lecture. For the time being thank you very much.