 the previous lecture we were discussing about the lubrication theory and we eventually derived the governing equations for the leading order velocity component along x along with the boundary conditions. So before proceeding further I will try to summarize what we have discussed so far so that you can recapitulate and then we will move forward. So if you look into the view graph which is there then basically you have a narrow passage where the length scale along y is much smaller than the length scale along x. So the epsilon which is the length scale along y by the length scale along x is considered to be much much less than 1. So we have used various non-dimensional parameters I will go through these slides pretty quickly because this is just to sum up not to give you any new concept and we have obtained the velocity scale along y in terms of the velocity scale along x using the continuity equation. Then important notes for non-dimensionalization we have used the characteristic values of the variables to non-dimensionalize the pertinent variables. This is a very important step this ensures that all non-dimensional variables and their derivatives have order of magnitude of 1 that is variables are of the order of 1. This means that the contribution of a variable or its derivative in a given equation solely depends on its coefficients because the derivative is of the order of 1. This process thus enables us to neglect certain terms in the corresponding equations based on only the coefficients. The concept is of utmost importance in fluid mechanics as a number of methods employed in the subject comes straight from this concept. For the sake of completeness we will assume that the body forces fx and fy act along x and y directions. These forces might be gravity, magnetic field, electromagnetic effect whatever and the characteristic scales are fcx and fcy. So I will not go through all the details of the governing equations because these have already been worked out. So the analysis of the x momentum equation gives a choice of the pressure scale with scales with 1 by epsilon square. Then we can rearrange the x momentum equation and the various terms are shown here and similarly we can do analogous stuff for the y momentum equation and then we can use an asymptotic expansion. So this asymptotic expansion we can do provided that all variables of different orders must be of the order of 1 which is ensured by non-dimensionalization. We cannot guarantee these always to be satisfied because I mean why we cannot guarantee these always to be satisfied we are unsure about certain scales. We have selected u as the dimensional u by uc but we do not know what is uc. So how can we ensure that u by uc is of the order of 1 that is our hope. So if that hope does not work then it may so happen that the individual dimensionless parameters are not of the order of unity and then this method will not work. So choice of the scale is something which is very important. However if the suitable characteristic values can be found then non-dimensionalizing with the characteristic values almost always make sure that of the order of criteria is satisfied. So with the asymptotic expansion whatever we could not write on the board here you can see on the slide with all the different terms with expanded in terms of epsilon, epsilon square you get big expression but by looking into the equation you itself you can understand that which terms will be important. So you can have the asymptotic expansion along x leading to this equation and one of the important considerations is that in the process of equating powers of epsilon we have assumed that epsilon square Reynolds number tends to 0 as epsilon tends to 0 and that is true if Reynolds number is small otherwise you cannot ensure that. Similarly the y momentum equation and the continuity equation we have not yet used the continuity equation. So these equations which we have obtained of the order of epsilon to the power 0 are also known as the leading order equations in asymptotic analysis terminologies. So these are certain terminologies in mathematics and the reason is self-explanatory because these are the leading order representatives of the leading order effects in the physical system. The particular form this is some note on the asymptotic analysis the particular form of the asymptotic analysis that we have applied here is called as regular perturbation. There are different types of perturbations which are possible this is called as regular perturbation. It is regular because a uniform expansion like the ones in the previous equation equation number 14 is this kind of expansion. This kind of expansion is valid throughout the domain. It cannot be ensured I will give you some example where it cannot be ensured that this kind of expansion is applicable throughout the domain. So some further discussions on the asymptotic analysis this is for those of you who are mathematically minded but it I mean even if we skip this discussion that will not I mean hamper much of our discussion so far as what we are discussing right now but just for the sake of your interest regular perturbation is probably the easiest kind of perturbation. Apart from regular perturbation there exist many other forms of asymptotic analysis some of which are singular perturbation domain perturbation method of multiple scales the WKBJ theory method of dominant balance these are several perturbation methods. I will talk about the singular perturbation a little bit because singular perturbation is something which is also very important in fluid mechanics. So the reason I mean I will talk about the physics where the singular perturbation is required. So if you are considering a boundary layer let us just go to the board to little bit to draw a sketch and understand. Let us say that you have a high Reynolds number flow so you have a thin boundary layer and the remaining fluid is behaving like a inviscid flow. Now the viscous effects are important in this region which is very thin as compared to the total dimension of the problem. Question is can we neglect this region altogether? If we neglect this region altogether the problem is that we neglect the viscous effects altogether and mathematically we bring down a second order PDE to a first order PDE because when viscous effects are not considered the Navier stokes becomes the first order PDE second order effect is not there. So this particular region has to be considered but the problem is this region is small. So if this region is considered the resolution of this region may be lost until and unless you blow up or zoom up this region and apply a different asymptotic expansion in this region. So applying a particular asymptotic expansion for this region and applying a different asymptotic expansion for the remaining bulk is something which is of the order of the day for a case where a thin boundary layer exist because you have to resolve the boundary layer. So that kind of perturbation where you treat this differently is called as a singular perturbation and the whole idea is that like physically try to understand that this is a small region. So to make it prominent you try to apply a transformation so that this small region is magnified as if you have a magnifying glass. By putting a magnifying glass you have magnified or stretched the boundary layer region and have applied a transformation to achieve that stretching this is called as stretching transformation okay. So this is something this singular perturbation theory is commonly used to analyze the boundary layer type of flows. So singular perturbation is also important. So let us look into the summary. So perhaps you have already encountered singular perturbation in your fluid mechanics course. The other widely used name of singular perturbation boundary layer theory theory will perhaps sound more familiar to you but I mean singular perturbation is a general mathematical treatment and boundary layer theory is a special case of application of that. A perturbation series usually involves expansions in terms of small valued parameters which is epsilon is the present problem. A general perturbation series has the form you can see that y is equal to summation of fk epsilon into ykx where fk's are known as gauge functions. Here the gauge functions are epsilon, epsilon square, epsilon cube like that. So you can see the lubrication equations summarized I mean in the special case that I was working on the board eventually I neglected the body forces but if you keep the body forces this is what the equations will look like. And then the transformation we attach our reference frame to the bottom wall as shown in the schematic. It moves along the x axis with a particular velocity. Whatever I mentioned as u star in the board that is given as utilde in this particular slide. So this is u star. So the dimensional boundary conditions at y is equal to 0 u is equal to minus u star. The boundary condition at y equal to h again u equal to 0. The u boundary condition is good enough to calculate u0 but that is not good enough to solve the pressure distribution because you have to link it through v. So you require a boundary condition for v and that boundary condition is v is nothing but del h del t right. The rate of change of the height with respect to time is v. So this is like the this is the kinematics of the problem that is deciding. So it is just a kinematic parameter. These kinds of boundary conditions are also called as kinematic boundary conditions. So in the non-dimensional form the same equations in the non-dimensional form. Now the choice of scale very important. If the x motion of the upper plate is dominant so you have to see that what is the physics. As I told you that the choice of the scale depends on the physics of the problem. So if the x motion of the upper plate is dominating then the velocity scale is nothing but the velocity of the top plate. If the y motion of the upper plate is dominating then the dominant velocity scale along y is your v tilde and the dominant scale along x is that divided by epsilon because vc by uc is epsilon. If the x body force is dominant then we equate the coefficient of the x body force as equal to 1 right. This is of the order of 1. So that gives a choice of uc. Similarly if the y body force is dominant then we equate the coefficient of the y dimensionless y body force equal to 1 that will give you a choice for uc. So whichever is dominant that you have to decide from the physics of the problem. See mathematics cannot decide each and everything. You have to give your physical judgment to figure out that what is dominant and what is not. Accordingly you have to select the suitable scale and why the selection of suitable velocity scale is important. Again I am repeating it is important because that will ensure that all the derivative terms or all the velocity individual u0, u1, u2 all these terms these are constrained to be of the order of 1 and these are possible in the dimensionless form only if you have non-dimensionalized with a suitable scale okay. So we will do one further simplification. We will assume that there are no body forces with this we shall proceed toward the solution of the problems. So these are the final equations. We have derived these equations but just to summarize and general steps for the solution. I will go through the general step later on because let us work out the steps and then you will yourself appreciate that what is the general steps so that we can write it in the form of an algorithm. So that for any problem using lubrication theory you can use the same algorithm. So with that let us come to the board and try to solve the equation. So let us write the x momentum equation the leading order. So x momentum equation right this was the equation. So let us integrate it del u0 del y. So this is the first step in the algorithm you integrate the x momentum equation to get u0. So del u0 del y what is that u0 is equal to this is function of x again. How will you get c1 and c2 you apply the boundary conditions at y is equal to 0 u is equal to minus u star right u0 is equal to minus u star. So minus u star is equal to c2. Boundary condition number 2 at y is equal to h, h remember is a function of x and t u0 is equal to so 0 is equal to dp0 dx into h square by 2 plus c1 h c2 you can write as minus u star. That means c1 is equal to u star by h minus dp0 dx into h square by 2 h by 2 right h by 2. So what is your u0 let us write it here u0 is equal to 1 by 2 dp0 dx into y square minus yh plus u star y by h minus 1 right. So just check I mean this is always a nice thing to do when you solve a problem after you solve the differential equation you check at least that the boundary condition is satisfied by the solution. So at y is equal to 0 u0 is equal to minus u star and at y equal to h both the terms are 0. So u0 equal to 0. So now how do you get v0 you have to relate this with the continuity equation. So continuity equation is del u0 del x plus del v0 del y equal to 0 in the leading order. So del u0 del x because h is the function of both x and t in general so I have written the partial derivative this is the first term right. Then minus u star by h square y del h when you differentiate this with respect to x you have to remember that p0 is a function of x and h is a function of x okay. So this is equal to minus del v0 del y continuity equation. So v0 is equal to you have to integrate this with respect to y minus half this is y cube by 3 minus y square by 2h plus dp0 dx minus y square by 2 del x. We have taken a minus sign outside then plus u star by h square y square by 2 del h del x plus some c3x what is the boundary condition at y is equal to 0 v0 is equal to 0 right. This is no penetration boundary condition so that means all the terms have y coefficient that means c3x equal to 0. Now our objective is to get the pressure distribution. So to get the pressure distribution we can use one condition which we have not yet utilized that is the kinematic boundary condition at y equal to h. So at y is equal to h v is equal to del h del t v0 basically. So del h del t is equal to minus half d2p0 dx2 y equal to h h cube by 3 minus h cube by 2 so minus h cube by 6 dp0 dx h square by 2 del h del x plus u star by 2 del h del x right. So we can simplify this and write so this term will become plus this term will also become plus. So we can write this as del del x of p0 into del p0 del x into h cube by 12. See when we consider when we write this the first term is this one. Second term you see del p0 del x into del del x of h cube by 12 that is 3h square by 12 into del h del x that is h square by 4 into del h del x. So these 2 terms are covered by this plus u star by 2. Now that because we have taken common so I mean this is more general mix okay. This equation is known as Reynolds equation in lubrication theory. So this gives a pressure distribution. If you now integrate this equation for a given scenario then that will give the distribution of pressure and when you integrate the pressure over a surface that will give you the total force which is of engineering interest. So now let us go through the general steps of the solution and we can now appreciate that how the general steps of the solution are have been applied. So you see the y momentum equation shows that pressure is not a function of y. So this means we can integrate the x momentum equation with respect to y to obtain the solution for u0. Enforce the boundary conditions for u to evaluate the full solution. Integrate the continuity equation to calculate v. Since continuity equation is a first order equation we need only one of the two available boundary conditions for the v to completely specify v and use the remaining boundary condition for v to deduce an equation for pressure. Now you can understand it much better because I thought that instead of telling the algorithm first let me work it out first and then you can generalize it for any scenario. So let us so this is the step up to which equation number 32 in the slide. This is the step up to which we have come. So this is called as Reynolds equation. Now we will work out two problems to demonstrate the use of the Reynolds equation. And all the problems are worked out in details in the slide. So I will only discuss about the concept and not necessarily go through all the integration differentiation and all which is mostly a trivial matter. But we will discuss more on the concepts of the problem. Before discussing on the problem we will summarize some of the important observations which we made in the previous lecture. First observation the scaling of pressure is one of the most important part of lubrication theory. We have noted that the pressure scales as 1 by epsilon square therefore the force on any one of the plates also scales as 1 by epsilon square. The viscous stress on the other hand scales as 1 by epsilon. Hence the force due to viscous stresses on the plates is also of the order of 1 by epsilon. I have already discussed but just to summarize because these are very very important conceptually. Therefore the viscous stresses are simply asymptotically smaller as compared to pressure. Thus when we calculate the force on the walls the contribution from the pressure dominates. In fact in the leading order of epsilon it is the only contribution that we will take. So when we calculate the force do not ask me that why are we not considering the viscous stress because the reason we have already clarified. So that is why pressure finding the pressure distribution is one of the big things in lubrication theory because that will give the force on the boundaries. Since the pressure scales as 1 by epsilon square it is usually very high therefore the forces resulting from it are also of very high magnitude. If the pressure is positive so I will show you examples to show that the pressure itself can come out as negative from the mathematical calculations and then one of the problems will work out like that and then what is the consequence. If the pressure is positive that means what? So you have two plates if the pressure is positive so let us try to draw this in the board and try to understand. So pressure is positive. This means positive pressure normal force onto the plate inward normal force to be precise. So it will require a huge amount of force to squeeze the two plates together because there is a force that is trying to separate the two plates. On the other hand if the pressure is negative if the pressure is acting like this then it will require then what will be the tendency of this plate. The tendency of this plate will be to come down relative to the bottom plate and there will be a high amount of force necessary to pull the two plates apart. So the fluid is acting like an adhesive. This is how you can engineer fluids to make work like adhesives. So in microfluidic technology or in nanotechnology these kinds of things are very commonly used. So if you can generate a negative pressure then you can make this fluid acting like an adhesive. Now let us get back to the slides. So the first problem a very classical problem in mechanical engineering a slider block problem. So the description of the problem is given in the slide but I will read it out for clarity so that you can understand the definition of the problem. We will work out a few steps but I will not work out the integration differentiation part but we will cover out the concepts. So consider the schematic shown in the figure. The bottom wall is fixed while the top surface moves in the x direction with a velocity u okay. So the top surface has no y component of velocity and only x component of velocity which is capital U. So what is an appropriate choice of velocity scale in this case? uc is capital U because the motion of the top plate is dictating the physics of the problem in this case. Had the top plate not been moving there I mean the physics of the problem would not have been described in this way. So the motion of the top plate is dictating the physics of the problem. The top block or surface is also known as the slider block. These are certain engineering terminologies but it is important to know about these engineering terminologies. The height of the space between the 2 blocks varies linearly. So h as a function of x linearly varies from a to b. So you can write h as a function of x. The length of the block is l in the horizontal direction. The axis system has also been depicted with like x and y axis as shown in the figure. We wish to find out how much force is required to pull the block forward using the lubrication theory. This problem is classically known as the slider block problem. This is a very classical problem. So let us get into the solution. So let us discuss a little bit. See every problem has certain aspects. One aspect is mathematics doing integration and differentiation. That is important because without that you cannot solve the problem. But there are other aspects which are physical aspects and let us try to understand some of the physical aspects some of which we have already discussed. We have to find the force exerted by the fluid on the top block as it is being pulled. The force required will be equal and opposite to that. So this is basically Newton's third law type of thing. So basically we have to find out the force distribution on the top plate. Force exerted by fluid on the plate. We have already discussed that the dominant contribution in the force comes from pressure in lubrication problems. Until and unless you have additional body force that comes into the picture. But here we have not considered any other body force. Of course if we are to apply lubrication approximation. See the statement of the problem does not give you a guarantee that lubrication theory can be applied. This is where many people make mistakes. That is given a problem because you have learned lubrication theory you tend to apply lubrication theory for anything and everything. But where is the guarantee that A and B are dimensions much smaller than L. Until and unless A and B are much smaller than L then you do not have the small parameter epsilon dictated by the ratio of either A by L or B by L whatever you take. Number one. Number two is the variation from A to B also has to be a slow variation. Otherwise if A is small you cannot ensure that B is also small. If there is a slow variation from A to B then if A is small B is also small. So these are certain things certain assumptions which need to be justified from the physical dimensions that you are considering. Choice of scales. The choices of the scales are quite obvious here. The obvious choice for uc is equal to capital U. The obvious choice for length scale in x direction is capital L. The obvious choice for length scale in y direction it could be either A or B. I mean whatever you choose because it is a slow variation it does not matter and because it is a slow variation B by A is of the order of 1. And here the V which is del h del t with that is equal to 0 because there is no y movement of the top plate. And the dimensionless u at the boundary that is u by uc that is equal to u by u that is equal to 1 because your uc is itself capital U. Now one interesting thing which we have not shown but you can show it easily that the dimensional form of the Reynolds equation and dimensionless form of the Reynolds equation look exactly the same. So if you revert back the dimensional terms from the dimensionless terms they will look exactly the same. So now let us solve the problem. So we will see the key steps here. The Reynolds equation now takes this form the left hand side that is del h del t is equal to 0. So now you separate variables. So h cube by 12 del p0 del x plus h by 2 the del del x of that equal to 0 that means if you integrate that h cube by 12 del p0 del x plus half h is equal to c1 or h cube dp0 dx is equal to so del p0 del x you can write dp0 dx for the reason we have already explained that p0 is not a function of y. So is equal to c1 minus 6h. Now you can integrate this to get p0 as a function of x once you substitute h as a function of x. So how do you get h as a function of x? Let us quickly go to the board and see. So this is a this is b let us say this is h at x this is l. So if this angle is theta you can write tan theta is equal to h minus a by x is equal to b minus a by l right. One from the small triangle another from the large triangle. So h is equal to b minus a into x by l plus a. So a plus b by a minus 1 into x by l. So this x is x dash actually the dimensional x b and a are physical dimension. So then if you sorry this is 1 h by a equal to I have written 2 steps together dividing by a. So this is dimensionless h if you are considering h0 equal to a dimensionless h is equal to 1 plus let us say b by a is equal to r. So 1 plus r minus 1 into dimensional x x dash by l. So we can write h is equal to in a dimensionless paradigm 1 plus r minus 1 into x all are dimensionless and r minus 1 you can give it a new name r bar okay. So this gives h as a function of x. So let us get back to the slides. Now so h as a function of x. So p as a function of p0 as a function of x. So you have these are the coefficients. So in place of h you can write 1 plus r bar x do the integration. This integration is simple enough I do not want to spend time on this. So once you integrate this you will get p0 as a function of x with 2 boundary with 2 constants c1 and c2. How will you get c1 and c2? The boundary conditions at x equal to 0 and x equal to l x dash equal to 0 and x dash equal to l. So the boundary conditions for the pressure at x equal to 0 is atmospheric pressure equal to 0 gauge pressure and at x equal to 1 also atmospheric pressure which is 0 gauge pressure okay. So that will give you the 2 constants c1 and c2. So eventually you will get this pressure as a function of x dimensionless. As we see that the Reynolds equation has indeed given us the pressure distribution in the channel. All it remains now is to calculate the force along the x direction originating from the pressure on the top plate okay. So force you can see that the force we have kept the contribution of pressure and the kept the contribution of the viscous stress okay. But we have to keep in mind that the viscous stress is one order less than the contribution of pressure. So how do you calculate the force due to pressure? Let us come to the board. So this is the plate, this is the x direction so at a given x. So we use that at a given small l we take a strip of length dl. If this angle is theta then dl is dx by cos theta right. So what is the normal force on this p into dl into the width? Let us say width is equal to 1. So p into dx by cos theta the force in the axial direction is the x component of that. So that into sin theta right plus or minus I am not bothering about so much depends on whether you are finding out force on the fluid or force on the plate and that is there. So this is tan theta and tan theta is you know b-a by l. So you can express it in terms of r capital R. So with this mechanism so basically you have to calculate integral p dx with an adjustment parameter which is r-1. So you look into this slide that is what it gives equation 42. The force required to pull the blocks is of equivalent magnitude and opposite sign to this evaluated force the answer for the force is this. This I leave on you as a homework. So you calculate the force which is coming from the previous expression do all the integration and this is the final answer to the problem. I mean this is just integration work it has I mean does not involve a lot of fluid mechanics. So this is a graphical representation of p and dp dx as a function of x for capital R equal to 2 as an example. Example 2 which is the last example that we consider in this chapter adhesive problem. Consider the schematic shown in the figure the bottom wall is fixed while the top surface is being pulled in the vertical direction at a fixed velocity. Now there is a vertical motion of the top plate. We have not yet mentioned that whether the vertical motion is in the positive y direction or negative y direction. So both are possible then I will discuss about both possibilities. So the description of the problem is pretty straight forward. Now the upper plate is having a vertical motion instead of a horizontal motion. So now what is the appropriate choice of the velocity scale appropriate choice of the vc is the velocity of the upper plate and therefore the appropriate choice of uc is that divided by epsilon okay. So here the vertical motion of the top plate is governing the choice of the scale. So now you look into the Reynolds equation as I mentioned the dimensional and dimensionless versions are the same for the Reynolds equation. So I mean here we let us deal with the dimensional version. So we have used all that dash and here u dash u tilde which is u star is equal to 0 and h dash is equal to a function of time. The top plate is moving with respect to the bottom plate as a function of time. So del h dash del t dash is equal to v dash. So that has given rise to this form of the Reynolds equation. Assuming h is equal to h0 at t equal to 0 we can write h as a function of t right. dh dt is equal to the velocity of the top plate. If you integrate that as a function of time with at t equal to 0 h is equal to h0 you will get h equal to h0 plus v tilde into t right. So that h you can substitute in this equation and integrate this twice. See that integration of that twice will not it is so straight forward because this is a function of time this is not a function of x. So you just integrate it just del square p del x square just twice for that this term in the right hand side will appear to be like a constant because it is not a function of x. So we can integrate it twice and use the atmospheric pressure boundary condition to obtain the c1 and c2 and you will get the pressure distribution like this. And the force now I mean the plate is not inclined so simply integral p dx so that if you calculate this is the final expression for the force okay. Now you can use the force required as the same magnitude as that of this but with opposite sign. Now look at the magnitude of the force if v tilde is positive if v tilde is positive then as time progresses the numerator becomes at the denominator becomes larger and larger like because h0 plus v tilde into t as if v tilde is positive then as t increases the denominator also increases. So that means the height as a function of time will increase more and more and the force will decrease like that. So at a very large height you will require basically no force to separate but is it within the purview of lubrication theory? The answer is no because when h becomes larger and large at infinite time it can become comparable and even may be larger than the axial length scale l. So then lubrication theory will break down you cannot use the lubrication theory. The other interesting part is when v tilde is negative then h with time will decrease right the gap between the 2 plates will decrease that is the top plate is coming down towards the bottom plate. So then the force will be asymptotically increasing and as h tends to 0 the force will like tend to infinity. So it is extremely difficult to like bring the 2 plates to such a situation when the gap between the 2 plates is tending to 0 that is the physics that we get from the solution. So since v tilde is positive the force exerted is negative as shown in equation 54 and hence the force required to pull the plates apart is positive. So we observe that this indeed scales as 1 by epsilon square if you look into the expression you will see. The magnitude of the force required decreases as the plates are pulled further apart since the denominator grows with time but after the critical time the magnitude will grow so much that the assumptions of the lubrication theory will break down. Now if the v tilde had been negative then the force required would be negative not only that it will grow with magnitude as time progresses because the height will become smaller and smaller and height is coming in the denominator. After a sufficient amount of time this will actually blow up to become infinity and that time is h0 by v tilde then the denominator will become 0. Physically this means that as we squeeze the plates more and more higher and higher amount of force is required to keep up squeezing. So very difficult to really squeeze the 2 plates. So with these 2 examples I will try to conclude with the discussion on the lubrication theory. Now I mean there are many more issues and if you want to look into the various issues the mathematical as well as the physical issues you can look into these books. Typically I would like to give a pointer to the reference number 1 which is a text book Advanced Transport Phenomena authored by Professor Gary Lille and very nice book published by the Cambridge University Press and I mean if you want more advanced material on lubrication theory you can look into that text as well as the other references which are given in this slide. So we stop here today. Thank you very much and we will start with a different topic in the next lecture.