 I would like to welcome all of you to the 12th lecture of the course. What I will do is I will just start working out some of the problems and things like that in the second half of today's lecture. I want to just wrap up a few things which I think is necessary so that there is some clarity. The first thing I want to do is talk about the problem of the cylindrical surface and actually evaluate n dot t dot n so that we have a feel for how it looks okay. So what we want to do is we have the cylindrical surface like we had last time which is given by r equals f of z okay. This is the cylindrical surface and like you saw last time what we do is we write this thing in an implicit form equal to 0 and the normal vector happens to be the gradient of f divided by the mod of the gradient of f which in this case will be er minus f dash e z divided by square root of 1 plus f prime squared okay. This we saw last time. What I want to do is see when you are actually working out a problem you need to evaluate the quantity n dot t dot n to get the normal stress okay. I just wrote it to you as that is the formula but then when you are working on a cylindrical coordinates or when you are working in a problem in rectangular coordination coordinates you need to be able to evaluate it. So let us just do one case where we are actually doing the evaluation and since we have this pet problem of cylindrical surface we will just use that to illustrate okay. So now our job is to evaluate n dot t dot n find n dot t dot n for this problem okay. So n is we have already found out what is n and in order to since we are talking about a very specific problem I am not going to talk in terms of indices i and j. I am going to talk in terms of r and z okay and t remember will be just a 2 dimensional matrix. I can write it as tau r z tau r r tau r z tau z r tau z z. So this would be the vectorial form by which you normally write represent your stress tensor. We have assumed theta symmetry so that all the theta dependence is gone and before I proceed further I just want you to know that tau r r can be written as 2 mu times the derivative of the velocity component in r direction by r for a Newtonian fluid and similarly you can express each of the stress components in terms of some velocity gradients or strains okay and if you have a different rheological property this particular relationship will change. Idea is that I have everything in terms of velocity now and we will be and my differential equation also has velocity my boundary condition also is going to be in the form of velocity so everything is consistent okay. Now I want to evaluate this right. So now n is how do I go about the 2 ways by which you can evaluate both of them are equivalent. So n dot t dot n is going to be given by my unit vector which is E r minus f prime of E z divided by square root of 1 plus f prime squared dotted with my stress tensor. So what I am going to do now is I am going to write my stress tensor just expand it and each of this is going to be associated with a particular unit vector one representing the outward normal one representing the direction. So tau r r for example is going to be associated with E r E r okay is a normal component tau r z is going to be associated with E r E z tau z r is associated with E z E r and I do not like this tau z z is associated with E z E z again it is dotted with E r minus f prime E z divided by square root of 1 plus f prime squared. Now this is what I am saying when you are solving a problem you have to be able to write the n dot t dot n in terms of variables which are of interest to you which could be f prime and the velocity gradients okay. How to evaluate this? I am going to take this dot product first I am going to take this dot product first so this E r dotted with this term when I am doing this I need to just look at the unit vectors adjacent to the dot so I need to take this E r with this E r that is going to be unity okay and I have tau r r so evaluating the dot on the left what do I have tau r r E r okay I have tau r z E z because E r dotted with this E r is going to be unity and I am left with this E z tau r z stays as it is tau r z E z the dot of this E r with the other 2 terms is going to be 0 because E r dotted with E z which is 0 E r dotted with E z so these 2 terms do not contribute okay I am left with this term now this term minus f prime E z dotted with this term is going to be 0 E z dotted with this term is 0 but E z dotted with this term is going to contribute and I will have minus f prime tau z r E r okay and again I have minus f prime tau z z E z okay that is just the dot on the left the dot on the right is still to be evaluated so I am just going to write that as it is square and I must remember that is this denominator which is still there okay now is straight forward for you to evaluate this this E r dotted with this E r is going to give me unity so I am left with tau r r this simplifies to tau r r and then this E r with this E r is going to give me a minus f prime tau r z okay clearly this E z with this E r is 0 this E z with this E r is 0 okay so I got 2 terms here and now I need to worry about tau r z E z with this I have minus f prime what did I do just now tau r r E r I did these 2 right tau r z with f prime so I need to do tau z r with this I have minus f prime tau z r and then I have plus f prime squared tau z z divided by 1 plus f prime squared the square root and the square root together will combine and give me this. So basically this is the normal stress balance or a normal stress component so when you are actually going to be solving an actual problem that n dot t dot n is actually useless you need to get this form this form is what you are going to be using when you are solving a problem whether it is in cylindrical coordinates or in Cartesian coordinates or spherical coordinates so while n dot t dot n is nice to put it in a vectorial form when it comes to actually solving this is the guy who is going to be used. So this is one component supposing you have a problem where you just say that the n dot t dot n from one phase is equal to the n dot t dot n from the other phase all you have to do is evaluate this in both the phases on the surface and equate them and that gives you your boundary condition okay. So if the normal stresses across the interface are equal for instance then evaluate the above on both sides and equate this is the boundary condition what is going to be different on one side of the fluid you will use viscosity of the first fluid mu 1 on the other side you are going to use viscosity of the second fluid mu 2 okay and then you will go back to using this constitutive equation here you will use mu 1 mu 2 and then use it okay. So that is basically important when you are actually solving problems because it is this form of the thing which you need and what you would do is you would write the tau's in terms of the velocity gradient remember your Navier-Stokes equations are also in terms of velocities. So you have your unknown velocities your boundary conditions on velocities everything is fine okay and it is important to do this because the boundary condition is usually an expression of a physical condition balancing of normal stresses balancing of tangential stresses okay. So that is the reason why you need to find the normal component of the stress there is another way I will just mention what it is and then you can check for yourself that they are indeed equivalent. See n dot t dot n can also be evaluated as a using a matrix multiplication idea approach at the end of the day I want to get a scalar right I want to get the component in the normal stress. So what is n? n dot t dot n if I have this as a 1 by 2 vector and I am going I am referring to my specific example which I saw just now t is a 2 by 2 matrix which you saw already okay and n can be written as a transpose and then if I write this n as a 2 by 1 okay. You have already calculated n it had an er and ez components so er is the first component, ez is the second component. So if you write this as a vector the 2 components and the matrix you already know in terms of the tau rr tau rz n you already know again but now write it instead of writing it as a row you write it as a column so you get a 2 by 1 okay and if you would evaluate this you get a 1 by 1 scalar okay. Now what I want you to do is whichever way you are comfortable with you are going to use when it comes to actually solving a problem either a matrix multiplication approach or that approach it does not matter you will get the same result. So you guys can do this and check if you are indeed getting the same result I am not going to do this. So what I am saying is how would you write the n as er the er component is 1 the ez component is – f prime okay this is my 1 by 2 and the t is tau rr tau rz tau zr tau zz. Now I am going to write the n as a 2 by 1 which is basically 1 and – f prime okay. I want you to understand that this tau rr is actually sigma because this is a normal component and remember what I did last time was I actually when you did a force balance you write this as sigma. So I should have been careful right at the beginning but since I was it I am just mentioning it now okay. So remember the tau rr is actually got the pressure as well as the hydrodynamic contribution okay. So if you remember this is normally written in the form of – p plus tau rr – p plus tau zz okay and that is what you do is you separate out the p and you get your gradient of p term would say. Yes I have forgotten that you are right. So what I need to do is I must remember what he is saying is I have forgotten the normalization factor square root of 1 plus f prime square and I must remember that there is a scalar which is 1 plus f prime square here yeah thank you okay. This is coming from the normalization condition okay I kept telling myself I should do sigma but I did not just want to do the boundary condition and talk a little bit about the generalization of the boundary condition okay and then we will move on. So you have already seen something like this earlier where we use the Young Laplace law okay and I believe it was done for a cylindrical geometry okay so it has been done for a cylindrical geometry the pressure here let us say is p1 inside the thing as well as an outside it is p2 okay. Now what you have because of the curvature is there is going to be a pressure difference okay and you have p1 minus p2 is given by sigma divided by r where r is the radius in fact what I should do is I should write this as sigma divided by capital R and because this small r actually represents my radial coordinate okay. Now I just wanted to when I wrote down the boundary condition in the last class what I did was I just told you that n dot t dot n in the first fluid or was it the second fluid minus n dot t dot n in the first fluid equals sigma times del dot n this is the most general form of the boundary condition. I want to basically show that this particular condition that was derived last time using a work energy principle in terms of you know how much of energy stored in the on the interface when you change it by dr what is the work done and by equating that that is how you got this relationship. So rather than do a formal derivation of this which we will do later on in the course I just want to show that this is basically a generalization of this and the way I am going to do this is evaluate each of these terms and show that it collapses to this and remember the way that n is defined as outward normal from 1 to 2 that is the direction. So what is the direction of n? This is n, n happens to be equal to er because it is the radial direction okay n is er and our job now is to calculate del dot n for this interface. So what is the gradient operator? The gradient operator in cylindrical coordinates is er plus er by r plus d by dz of ez all I have to do is get del dot n which means I must do the dot product of this with er, er dotted with er is unity when I differentiate unity I get 0. So this guy is not going to contribute, er dotted with er is unity this r remains so I get a 1 by r term here and I am going to be evaluating it on the boundary remember so therefore it is going to be 1 by capital R and ez dotted with er is going to be 0 okay. So all I am trying to show here is that del dot n reduces to 1 by r for this problem and if you had a more complicated shape rather than you know worry about the interface you would just do del dot n and possibly in your calculus courses you already see in the del dot n indeed represents the curvature. So that is the idea here that del dot n, del dot n is nothing but del dot er and that is equal to 1 by r and on the interface we have 1 by capital R because we are evaluating the boundary condition on the surface. So r is equal to capital R okay what about the guys on the left n dot t dot n now t now I am going to be careful how do I write t I am going to make sure I do not make this mistake I made last time sigma rr tau rz tau zr and sigma zz okay just to tell you that in the normal direction we are using sigma to represent the total component and now I am going to break this up in terms of the pressure and that due to the flow okay and so now I am going to write this as minus p plus tau rr and this is what I should have done last time minus p plus tau zz. So since we are talking about our objective remember is to derive the Young Laplace law from this generalized boundary condition this generalized boundary condition basically includes the effect of the flow as well now in the case of a static situation the only thing that is going to contribute is going to be the pressure terms in the diagonal element okay these guys are going to be 0 because there is no velocity and if you were to now evaluate n dot t dot n using whatever we did earlier you will find that this is going to basically boil down to just the pressure term yeah. So whenever if I were to substitute this particular form for the tensor t here and what evaluate n dot t dot n using the method that I just discussed earlier what you will get is minus of p2 coming from the first term and you will get minus of minus p1 plus p1 coming and that is something which I want you to do okay I want you to find out that this basically boils down to sigma divided by capital R okay on substituting all this on substituting that is what we get this is for the case when the fluids rest. So what I want you to do clearly is do the spherical analog the spherical analog use the generalized formulation of this boundary condition and do it for a sphere and see what you get okay and the relationship of course is classical but I just want you to derive it this way just so that you get comfortable with this whole business of evaluating the boundary conditions starting from this point okay yeah. Now if you have a complicated surface wherein we define it by f of capital F of r, z then we will be we should be careful with del f right we can express it in the other way also the negative of that. Yeah that is the reason I wanted to be explicit if you want to define it from 2 to 1 then you will flip the things on the left hand side you would you have to make sure of that while looking at the way you are defining the coordinate system. So I think the question is how do you make sure that the outward normal is pointing from 1 to 2 okay the definition of the outward normal is depending upon the way we have defined the surface in the sense I am going to write it as r is equal to f of z or z is equal to f of x. So you are looking at when you write it as r is equal to f of z you are going in the radial direction so the as you are going in the direction of increasing r you are going from 1 to 2 that is the idea okay. So I think if you are going in the other direction it would be in the direction of minus n because n was equal to the way I defined it n was in the r direction as I was going in the direction of n I went from 1 to 2 this is going in the direction of the radius. If I were to go come in the negative r direction then n would be automatically minus of er okay and then you are coming from 2 to 1. So I think the best thing to do is to just make sure that you are going in one direction and calculate the normal and look at the sign okay. I think when we do a couple of problems it will become clear okay. So I think what we have done is we have actually established the framework in the sense we got the boundary conditions we got the differential equations and since this course is basically on analytical methods so we will now talk about using some analytical methods and the basic concept is using a perturbation theory approach. Now here what is important is sometime back you had some couple of lectures on scaling right you transformed your equations in terms of dimensionless variables you had Reynolds numbers you had different kinds of dimensionless groups which came up and under some conditions these dimensionless numbers can have different orders of magnitude. So for example if your flow is highly viscous okay then the Reynolds number is going to be very low and if the Reynolds number is very low you can drop the inertial terms because the inertial forces are low you can make a simplification. So in the limit of Reynolds number equal to 0 you can possibly get a solution. Now the question that arises is supposing Reynolds number is not 0 but it is small but finite you want to get a better approximation to the solution how would you go about doing that okay. So what we want to do is we want to be able to get some insight about behavior of a system which in this case happens to be a flow problem by looking at regimes where some parameters could be small. So can I exploit the fact that some parameters are small to get some analytical solution okay. I am going to illustrate this idea with a very simple problem now and then later on we will solve this thing with an actual flow problem. So let us take so idea is we do a scaling of the equations and maybe some parameters small for instance the viscosity is high then Reynolds number tends to 0 okay Reynolds number is low and can I actually use the fact that Reynolds number is low to get a solution. So one thing you can do is we can put Reynolds number equal to 0 and find a solution why do I say that because when I put Reynolds number equal to 0 the left hand side basically contains the non-linear terms okay all I have on the right hand side are my pressure gradient and my viscous forces. So that is a linear system and I should be able to solve it. But the question is is the solution valid for finite Re. Can I use the information that I have as a solution for Re equal to 0 and how can I use this information to make a correction and find the solution for a finite value of Reynolds number which is small okay that is the idea okay. So this for low Re can we make a correction okay and get improved estimate of the solution okay because maybe and if I can get an improved estimate of the solution then I am more confident rather than use the solution Re equal to 0 for Re equal to 10 or Re equal to 100 I would rather use this improved estimate to find out what my flow is when Re is 10 or 100 okay. So how do you go about doing that? So we will take a very simple problem first to illustrate the idea and then we will go back to solving fluid flow problems okay. So remember one of the important things using perturbation theory is you have to do the scaling you have to make things dimensionless and you had a couple of lectures earlier on how to make things dimensionless. So depending upon the values of those parameters some parameter may be large some parameter may be small. If a parameter is large you can treat the reciprocal of that as a small parameter. If a parameter is low then you can use it as it is okay. So let us just look at a very simple problem just an algebraic equation. So consider the algebraic equation x squared-1 equals 0 okay. So everybody knows how to solve this you may possibly did this in high school and you know the solutions are x equals plus or minus 1 okay. Consider now a modified problem consider a problem x squared-epsilon x-1 equals 0. What I have done this is of course a fictitious mathematical problem okay. I want to just illustrate the ideas on a fictitious mathematical problem then you can go back and do your fluid flow problems but you can apply things. This epsilon is a small parameter okay and here again so epsilon is a small parameter and do you know the solution to this equation of course you know the solution to this equation also this x the 2 roots the solutions to this equation is given by-b plus or minus square root of b squared. So in this case of course although you had a small problem is this okay. In this case of course since the problem was simple you already know the solution but supposing you do not know a formula supposing this equation has been a cubic or a fourth order equation which contains this maybe you do not have an explicit relationship okay. In fact if you have up till fourth order fifth order you do have explicit relationships but we do not worry about this but now the question that I want to ask here is for the case where epsilon is 0 I know what my solutions are okay and is it possible for me to find out the solutions for this equation as because for small values of epsilon I expect that my roots are going to be only slightly different from the case where epsilon is 0. So when epsilon equal to 0 I have 2 roots plus or minus 1 when epsilon is small maybe 10 power minus 5 10 to the negative 6 okay I do not expect that to be a very significant change there would be a change but maybe not a very significant change. So can I look at the variable x as a function of epsilon and look at it as a do a Taylor series expansion or do a power series expansion and seek the solution of x in terms of epsilon as a power series okay. So that is the idea. So for small values of epsilon small epsilon we expect the solutions to be close to plus minus 1. So small changes in epsilon give me small changes in x okay give small changes in x the solution. So can we do a Taylor series expansion or a power series expansion. So we seek x as a power series in epsilon. So clearly you all understand that x depends upon the value of epsilon okay we do not know what the function is but what we will do is instead of writing it as f of x we normally write it as a power series expansion I am going to write the power series expansion about a point which I already know when epsilon is 0 x is plus or minus 1 okay. So that is the idea when we are doing this perturbation series. So we seek x as x0 plus x1 epsilon plus x2 epsilon squared etc okay. So the idea is x is a function of epsilon okay this is an approximation to x which is a function of epsilon. Clearly different values of epsilon give you different values of x. So x depends on epsilon and the functional dependency is written in this form. What is x0 x1 x2 these are going to be numbers these are going to be constants and if you can actually calculate what x0 x1 x2 are you can actually calculate what x is as a function of epsilon okay. So x0 in this particular problem it is very simple. So x0 x1 x2 are numbers or scalars okay and if we can find these we know x as a function of epsilon okay. So whatever the arbitrary function is given right there in terms of square root sign I am just going to write in terms of a Taylor series or a power series. So how do you go about finding x0 x1 x2? So if x is going to be in this form clearly x must satisfy my algebraic equation which I had. So I am going to substitute this particular form of x in the original equation and I am going to invoke the fact that this particular thing has to be satisfied for every epsilon okay for any arbitrary epsilon. What that means is I would get something like a power series and we will be equating terms of the same order of epsilon all the terms that of order epsilon to the power 0 I group epsilon to the power 1 I group epsilon squared I group okay and that is the general approach. So what we do is we substitute this in that particular form that x is squared minus epsilon x. So substitute this in the quadratic what do you get whole squared minus epsilon multiplied by x2 minus 1 equals 0 okay. I need x to satisfy the equation. So this particular form has to satisfy the equation and we are substituting it here. And now you just need to expand all this and see what you get. So to expand the quadratic term the square term I get x0 squared plus 2 epsilon x0 x1 okay plus a squared. I do not all other terms will have this thing plus my epsilon squared term which is going to arise from 2 x2 x0 plus x1 squared plus higher order terms which I am going to neglect. What I have done is I am just writing terms up till order epsilon squared okay. This is going to be coming from my x0 squared plus 2 this and then some of you must tell me if this is right or not okay. Then I have the other term which is minus epsilon times the same thing x2 minus 1 equals 0. I am going to now I want this equation these terms to be valid for any epsilon for any choice of epsilon I want this to be valid okay. So what I am going to do is I am going to equate terms since we want this to be valid for any epsilon we equate powers of the order of epsilon to the power n okay. So I am going to group the terms that are order epsilon to the power 0. So which are the terms which are independent of epsilon it is just x0 squared and minus 1 this must be equal to 0 okay. That is what it gives me and it gives me x0 is plus or minus 1 what about order epsilon to the power 1 I have a term here x0 x1 okay. All these are higher order terms I have one here minus x0 equal to 0 okay. This gives me x0 is already found out so this implies x1 is half okay and you can similarly calculate x2 by looking at the term which is our order epsilon squared. So now we look at the next term the second order term and at order epsilon squared the equation which has to be satisfied is 2 times x2 x0 plus x1 squared minus x1 equals 0 and we already know x0 and x1 and so what I can do is I can use this to find out x2. In fact we know that x0 is plus or minus 1 and x1 is half. So what this gives me is when I substitute these values I get 1 by 4th here. So I get 2 times x2 times x0 plus 1 4th minus half equals 0 which implies that x2 is equal to half minus 1 4th divided by 2 1 8 1 8 of x0. Remember x0 can take plus or minus 1 as 2 values. So when x0 equals plus 1 I have x2 equals 1 by 8 and when x0 for x0 equals minus 1 we have x2 equals minus 1 by 8 and now since I know x0 x1 and x2 I can substitute the values in my power series and this gives me the 2 expressions x as being plus 1 plus half of epsilon where x1 is plus half and x2 is plus 1 by 8 of epsilon squared plus etc. And the other root gives me minus 1 plus half of epsilon minus epsilon squared divided by 8 plus etc. So these are the power series expressions for x in terms of epsilon which we have obtained using the perturbation series method and depending on the level of accuracy that you want you would take you know higher order terms should you be more interested in getting more accurate values. So when epsilon is 0 I get back plus or minus 1 which are my roots. My first order correction is the same for both but my second order correction has a different sign. So x was given by epsilon plus or minus square root of epsilon squared plus 4 divided by 2. Now this can be expanded in the form of a binomial series right and you know how it is epsilon squared plus 4 to the power half you can factor things out and you can do this. I want you to do that and see if it boils down to this expression that we have and then you will get a better feel for what exactly is going on. So one way to do an approximation to this particular exact solution is to do a binomial series expansion take a few terms in epsilon. The other way is to do what we did which is just assume a power series expansion and you get the solution and you know when you see that both of them are equal which is how they should be then you are convinced that things are working fine. There are of course some limitations to this there are times when this method is not going to work okay and those are things we will see as we go along. One particular book which deals extensively with perturbation series expansion is authored by Nefe and I would you know recommend those of you are interested in understanding this a bit more deeply. This is a good book for perturbation expansions. So what I have done today is just give you an idea about how this perturbation method works and what you will do is when any problem where you have a small parameter you can actually go about exploiting the existence of the small parameter and finding solutions.