 So, welcome to the 11th lecture of multi phase flows. What we will do today is just continue from where we left off in the last class, okay and the idea is to try and understand how boundary conditions can be formulated. So, if you remember what I said was that if you have a surface whose outward normal is given by say N, okay then stress component that surface acting the t direction is given by dot t dot t. So, the reason why I have used this kind of a notation is this tells you that you have a surface on which you are trying to find the stress components, okay. This is the direction of the normal to the surface because that is what you are interested in, okay and on that surface remember you can have different directions just like and this is the second index that we were talking about when we talked about the stress tensor. So, if you want to find out the component in the direction t this is the way you would go about calculating it. Now I am going to explain to you how this is actually evaluated, okay. So, remember the N is going to be given by I can write this as NiEi. So, this are the unit vectors. So, let us for the sake of simplicity assume it is a Cartesian coordinate system. Though I am not specifically saying x, y, z I am just going to keep it in terms of i, j, k but these are the unit vectors in the three classical directions x, y, z or r, theta, z depending upon cylindrical or Cartesian or xy versa, okay. Similarly, t is a vector and that I am going to write as T subscript l E subscript l. So, again l is i, l go from 1 to 3, okay. So, i, l go from 1, 2 and 3. So, these are the components of a vector. This is a unit vector. So, E1, T1, E1, T2, E2, T3, E3, okay plus. So, because we are using the summation notation and what about the stress tensor? This we will write as tau i, j, i, j, k, Ej, Ek, okay. The stress tensor is written in this form. You have two subscripts. One telling you the direction of the normal. The first one tells you the direction of the normal of the surface. The second one tells you the direction of the, in which the component is acting, okay. So, tau xy tells you that it is acting on a surface whose outward normal is x and in the y direction. So, that is basically what it is. What I want to do is explain to you how this particular term is evaluated, okay. So, now n dot t dot t is going to be written as niei dotted with tau jk, Ej, Ek dotted with Tl, El. There is a specific reason why I have chosen these indices differently because I want to make sure that the indices if I have chosen to be the same, then you would not know how to take the dot product because Ei dotted Ei will always be 1, okay. We want to make sure that the contribution of Ei dotted Ej is only when i is equal to j, okay. So, how do I go about doing this evaluation? You see a dot here and when you want to take this dot product, you are going to look at the unit vectors adjacent to the dot product. The unit vectors adjacent to the dot are the Ei and the Ej. So, that means I am going to do Ei dot Ej. I look at this dot, the unit vectors adjacent to this are the Ek and the El, okay and therefore going to do a dot of Ek with El. So, when I do the dot of this, I would get a scalar, I would do a dot of this, I get a scalar and the end of it I get a component acting in the normal direction, okay. So, what I am going to get at the end of doing this operation is a scalar. So, let us do this guy first. This is going to contribute only when i is equal to j, okay. When i is not equal to j, Ei dot Ej because they are perpendicular is going to be 0, okay. So, I am going to write this as Ni, I am going to write this as tau ik because I am just putting i equal to j. Ei dot Ej will be delta ij which is 1. So, that goes off. I am left with Ek dotted with Tl El. Now, I have just taken care of the first dot. I need to take care of the second dot and that is going to be contributing only when Ek is equal to El, okay and this gives me Ni tau ik Tk because El must be equal to k only then this guy will contribute, okay. So, I have Ni tau ik Tk. This is the expression for n dot t dot t. What is Tk? Tk is nothing but the components of the tangent vector. What is Ni? The components of the normal vector, okay and tau ik you know how to evaluate because depending upon the axis. So, what I have done is given a particular axis x, y, z. I know the components of n in terms of x, y, z. I know the components of t in terms of x, y, z, okay. And once you know that you know what the stress-stress components are in terms of x, y, z, tau x, y, tau x, z, etc. And you can just evaluate this. Remember this is being summed over both i and k, okay. There is a double summation because i is being repeated, k is being repeated, okay. So, this is a double summation since both i and k are repeated, okay. And if you want me to be explicit, I will just let i equal to 1 first. I have i tau 11 T1 plus n1 tau 12 T2. This is being summed over k, the second index for i equal to 1. And then I do for i equal to 2 plus n2 tau 21 T1 plus n2 tau 22 T2. These are the 4 terms that I get, okay. i is 2 and k is 2, okay. So, this is how you would go about evaluating the stress component. What we are going to do next is see, go back to the example of the triangle which we took. We are going to evaluate a stress component using the physical argument and using the mathematical formula to just convince ourselves that they are both giving you the same expression, okay. So, is this clear? We will go back to this very easy to draw triangle, triangle theta. And this remember was my x direction. This remember is the y direction. So, this is delta y and I have sigma xx this way. I have y in this direction and this direction is tau yx, okay. And this is the normal direction. Is this correct? This is theta. This is also theta, okay. What I am going to ask is what is the stress component acting on this inclined along the hypotenuse, on the hypotenuse but in the x direction. That is the question, okay. The stress component on the hypotenuse in the x direction. So, that is going to be given by tau nx because this is the one is acting on the hypotenuse. The first subscript tells you that the perpendicular to on that surface is the n direction and the direction of the stress is actually in x direction, okay. So, tau nx tells you what this stress component is that what we are interested in is finding out tau nx. So, let us do it the physical way. We keep in mind that those acceleration terms, the body force terms are all of higher order. So, they disappear and what we have is only the forces which are acting on the surface, okay. Retaining only the surface terms, surface force terms which are of order epsilon, what do you get? You should get sigma xx multiplied by delta y plus tau yx multiplied by delta x, okay plus tau nx multiplied by delta l equal to 0. This is the force component balance in the x direction, okay. This is the force balance in the x direction and I am going to rearrange things a little bit here and tell you that tau nx is going to be equal to minus sigma xx delta y divided by delta l minus tau yx delta x divided by delta l and clearly from the figure delta y by delta l, this is delta l is the hypotenuse for you. It is nothing but sin theta, is that right? So, this is minus sigma xx sin theta and this is tau yx cos theta. So, the point I am trying to make here is that on this surface, on this surface the stress component acting in the x direction is given in terms of my classical stress tensor components sigma xx and tau yx but then there is a sin theta and a cosine theta which I have to factor in, okay. So, this I have got from my physical argument. Now, what I want to do is I want to redo the same thing using my formula, okay because that is the whole idea. We had a physical argument, we had a mathematical argument. So, now we are going to evaluate tau nx, okay using the mathematical formulation. So, how would you define tau nx? Tau nx using the mathematical formulation is going to be given as n dot t dot, I want it in the x direction, okay. I want it in the x direction. So, I am just going to write this as Ex, okay. Now, just for the sake of illustration rather than talk about 1 and 2, e1, e2 etc. that is what I really should be doing. I am going to talk in terms of x and y, okay because we are doing things in Cartesian coordinates, okay. So, now I could have put this as e1. We can do that also right now in fact. We can do this right now. Why don't we just do that? e1 where 1 is the x direction and 2 is the y direction. I just change my mind. But in order for me to evaluate this, I need to know what is n, the direction of the normal, okay. Now, if you look at the figure over there, the direction of the normal is such that it has 2 components. It has both the x component and the y component. The y component is going to be given by cosine theta, negative cosine theta and the x component is going to be given by negative sin theta, okay. So, n is x component is minus sin theta. Instead of ex, I am going to put e1 and it is minus cosine theta e2, okay. That is n. And this of course is simple for me. I have only e1 here. And t remember is ij, ei, ej because the others are fixed e1, 2. I am just going to use ij here. So, let us evaluate this quantity now. What is n? Minus sin theta. So, n dot t dot e1 gives me minus sin theta e1 minus cos theta e2 dotted with tau ij ei ej, okay, dotted with e1. Now, when you are doing this dot product, I am going to look at this term. This term dotted with this will contribute only when i is equal to 1, okay. So, I have minus sin theta. When I take this term with this, I will get tau 1j, okay. And when I do, take the second term here and I take the dot product here. Remember, I need to take the dot product with the adjacent vectors and I need i equals 2 here. Only then this is going to contribute. So, I am going to have minus cosine theta tau 2j, okay. E2 dotted with e2 will be 1. I get cosine, yeah. i will be equal to 2. So, it is 2 and j remains as it is, okay. And this guy, the ej remains as it is dotted with e1. Now, this is going to contribute only when j is equal to 1. This will have a contribution only when j is 1. If j is equal to 2, e2 dotted e1 is 0, okay. So, if j is equal to 1, I have minus sin theta tau 11 minus cosine theta tau 21. Remember, y was equal to x and x is equal to 1. 1 is equal to x and 2 is equal to y, sorry, okay. So, if I go back and I write this, minus sin theta, I have tau xx minus cosine theta tau yx. And this should be the same as what I had earlier. Only thing is, I used for the normal thing there, I used sigma as my, this thing. So, remember, here the tau represents actually the complete normal component, which means the sigma and the p are included, sorry. This is the way I have written it. I have written it as tau but this is actually sigma. This is the complete normal stress component, okay. So, I just want to tell you that this is the same as what we had there. So, this is just to illustrate to you that physically you could have gotten the component acting in the direction x on the surface whose normal is n doing those force balances. But I mean, you cannot keep drawing those surfaces again and again. So, tomorrow if you are given a surface, you should be in a position to directly calculate what the stress component is, okay. And then I would just use this formula. If you want to get the component in the y direction and that is something for you to do. You can, I am not going to do this but what I want you to do is, I have done this in the x direction. I want you to do the same thing in the y direction. Do it physically. Do it using the formula n dot t dot e2 in this case and see if you are getting the same thing, okay. That is for you to just practice. So basically as we wanted, the mathematical and the physical way, the mathematical and the physical approach give the same result. In fact, if they did not, we would be in trouble, okay. So, that is just a justification. I did not really do any proof of how that thing came n dot t dot t but this is just more of an illustration of how that actually gives you the component that we are interested in, okay. So, homework problem establish the similarity in the y direction, okay. So, now the question comes as to how do we go about using it for an actual problem, right. Let us go back because we start off with this beautiful tapering jet, right. I do not know if you think it is beautiful but anyway it is a tapering jet or and the way the surface is going to be defined here the surface of the tapering jet is given as r is equal to function of z, clearly, okay. It can also be a function of time and for sake of simplicity just to tell you how the calculation of n is done. So, what are we going to do now? We are going to talk about for simplicity, we assume theta symmetry steady state, okay. So, there is no time dependency and no theta dependency. Theta symmetry means irrespective of what the theta position is it is the same it is independent of theta. Now, so that I can you know do things in a very simple way. I can give the more complicated problems as homework for you guys to do, right. I can write the same surface equation. We can also describe the surface as f of r comma z equals r minus f of z equals 0, okay. This f is an alternative shall I say implicit representation of the surface. r equals f of z is an explicit representation of the surface, okay. So, now if you have a surface which is given by r is equal to f of z explicit. I can always rewrite it as r minus f of z equal to 0 and that is your implicit representation f is a function both r and z. Clearly, now if you remember your calculus, what is the normal to the surface called? Because that is what I interested in on the surface how do you go about finding the normal? Normal is going to be given by the gradient of f, okay and since we like to deal with unit vectors the unit normal vector I am going to divide the vector by the absolute value or the magnitude of this. So, this gives me the unit normal vector, okay. How is the gradient going to be calculated? You just calculate the gradient operator is nothing but d by dr of er, f is a scalar of course plus d by dz of ez. I am forgetting about the theta direction so I am not writing the theta component, okay. So, gradient of f is what? I am going to have to differentiate this thing with respect to r that gives me unity, okay. So, the component in the r direction is just er because this is independent of r and the component in the z direction I just differentiate this with respect to z. I get minus f prime of z ez that is my numerator that is my gradient. I want to normalize it so how do I normalize it? I just divided by the magnitude of this guy so I get 1 plus f prime squared. So, that is what have I done? I have just told you for a given surface how to calculate the normal. So, supposing you have a surface in a problem which is given by z is equal to f of r or z is equal to f of x or y or t whatever it is you know how to go about calculating the gradient, okay. Once you know about calculating the gradient and you get the normal then you are in a position to at least get the n dot t and supposing you are interested in the normal stress balance you need n dot t dot n. So, you already got the stress component that you are interested in, okay in terms of the unknown surface f, okay. Now, let us now that the n is given so this is my n, okay. So, wait, wait, wait, I need to this is gradient I think I need to do this thing step by step that is just the gradient and I need to do this as gradient of f I am doing too many things in one shot and n is this, okay. So, the gradient operator defined as this gradient of f is given by that and the unit normal is given by this, okay. What about the tangent because those are the 2 things we are normally interested in because when I am trying to impose boundary conditions I am normally doing a balance of the forces on an arbitrary surface in the normal direction and in the tangential direction. So, I need to know the normal direction and the tangential directions, okay. So, the tangential direction how is that going to be given clearly the tangential direction has to be perpendicular to the normal direction, okay and the tangential direction is going to be given by f prime of z, E r plus E z divided by square root of 1 plus f prime squared. How would I get this? I just I am making sure that the dot product of n and t is 0, okay. I just the denominator does not bother me it just comes there because it is normalizing. The f prime has been moved to E r I have changed the sign here and now this has a component of unity. So, now I have this follows from t dot n equal to 0 just want to mention a couple of small things. What we are talking about here is f prime of z, f prime of z remember is nothing but df by dz, okay and the way we have written it f of z is nothing but r, small f. So, this is the same as dr by dz because usually there is some confusion people have regarding this implicit and explicit dependency. So, I just want to do one thing here to show you how these are related. It is pretty much simple calculus. We also have f of r comma z equals 0 on the surface. So, everywhere on the surface the capital F is 0, correct. So, that means what? The changes in f will also be 0. This means df equals 0 on the surface but what is df? When I am moving along the surface there is change both in r as well as in z. So, df is nothing but the partial derivative of f with respect to r times dr plus partial derivative of f with respect to z times dz this must be equal to 0, okay. So, what I have done is I am trying to show the relationship between the total derivative of the small f with z with the partial derivatives of capital F with r and z. Clearly you can just move things around a little bit and you can get dr by dz as being equal to minus df by dz divided by df by dr and this is equal to f prime of z. So, if you had an explicit relationship I would use f prime. If you had an implicit relationship I would calculate f prime using this, okay and then I will use it in the formula. So, you have to be careful about whether you are using an implicit relationship or an explicit relationship when you are proceeding, okay. So, I just wanted to show you this equivalence of these 2 because sometimes people do the wrong thing, okay. So, we have learned how to calculate the normal direction, we have learned how to calculate the tangential direction and given a particular surface f. There are 2 things we need to do now. One is to learn how to use the boundary conditions in the normal direction in the tangential direction and one more thing which is when the surface itself is changing with time how do you track the surface, okay. So, those are the things which are remaining. So, now the 3 questions are tracking the interface, applying the normal stress boundary condition and 3 the tangential stress boundary condition. Since we are talking about tracking the interface we clearly have a dynamic problem things have to change with time, okay. So, now I am going to go to the first question or rather the answer to the first question a1. Since the interface changes with time we have a dynamic situation which means I function the interface let us say we will keep things simple z is now going to be a function of x, y and t. I have just pulled a fast one I have just gone back to the Cartesian coordinates, okay. So, now we are just doing this in Cartesian coordinates. So, what does this mean? Let us say you have an interface I mean you know this interface looks very smooth. So, things are changing this is my z direction this is my x direction and that is my y direction. So, remember you are on the beach you see these waves on the surface. So, the interface is actually the going to be a function both of the distance into the beach that is the y direction along the coast that is the x direction and z is the height of the interface. So, clearly things are changing with time at every point, okay and what I am doing is I am just writing this dependency. So, this is my f or in this case this is height maybe I should use h, okay. Let us just not confuse the issue here but I am just telling you that the interface is going to be given by z of f of x, y, t that is my interface. So, instead of only one direction now I have just generalized it to 2, okay and since I am working with Cartesian you know how do you calculate the gradient operator and all that very elegantly and it is dynamic. I am going to go back to what we did earlier which is go through the implicit representation of the function of the surface. So, I am going to write f of x, y, z, t as being equal to z-f of x, y, t equal to 0. This is the implicit representation of the surface, okay. Now clearly at any time any point on the surface capital F has to be 0, okay. So, what does this mean? If you want to look at the material derivative how the particles on the surface are actually moving that is going to be characterized by the material derivative being equal to 0, okay. So, what I am saying is f, just like the argument last time f was 0 whatever df was 0 along the surface. So, f equal to 0 on the surface. So, here we have for the particles on the surface df by dt equal to 0, okay. Now you know how to calculate this material derivative df by dt is nothing but the partial derivative of f with respect to t plus v dot del f that is what we did sometime back when we were talking about the Euler's acceleration formula. So, I am just going to use that now. So, we have df by dt equals partial derivative f with respect to t plus v dot del f, correct? v dot del f. This is from what we saw a few lectures back and this must be equal to 0. Now when I look at the partial derivative of f with respect to t, capital f with respect to t is the same as the negative of the small f with respect to t. Remember x, y, z are actually independent now in my explicit formulation. So, this is nothing but minus df by dt plus I am going to stick to the velocity vector as it is and the gradient vector is nothing but the partial derivative of f with respect to x times Ex, okay, minus of partial derivative of f with respect to x times Ex minus the partial derivative of f with respect to y times Ey. When I differentiate this with respect to z, I get unity plus Ez. This is my gradient of capital F in terms of small f. All I have done is just taken the gradient of that, this dy dx of that and I get df by dx, df by dy and that, okay. So, and this of course equals 0. So, this implies, let me just say that this implies 0 equals that and I am going to do something very simple which is write this in terms of the components, take the dot products and move these guys which are negative to the left hand side. So, but maybe I have learnt my lesson, I should do this one step at a time. So, I am just going to move this dx by dt here. I am going to write this as Vx Ex plus Vy Ey plus Vz Ez dotted with minus df dx Ex minus df dy Ey plus Ez and do the dot product. This gives me minus Vx df dx minus Vy df dy plus Vz, okay and this basically means df by dt plus Vx df dx plus Vy df dy equals Vz. This particular equation it tells me how f is changing with respect to time. How does the interface evolve with time, okay and this is called the kinematic boundary condition. So, if you have an unsteady set problem and this we are going to see later on. Later on when we talk about stability of multi phase flow systems, we will have the interface which is actually changing with respect to time. So, when the interface is changing with respect to time, you need to be able to track the interface and the tracking of the interface is actually done using this, okay and so then I would use this to find out how my interface position changes. If you have a steady state situation, of course this particular df by dt is not going to be present, okay. So, when we solve some perturbation problems later on, we may be neglecting the time derivative term. What I want you to emphasize see here is supposing the interface is flat. Supposing the interface is flat, it means that f is independent of x and y. It means the partial derivative of f with respect to x and the partial derivative of f with respect to y will be 0 which means that df by dt will be equal to Vz. That means the rate at which the height is changing will be given by the vertical component of the velocity. So that is basically consistent. So, I think whenever you derive an equation, I expect you to, you know, sit down and see even some limiting cases, it boils down to something which is consistent with what you expect or is there an inconsistency. So, that is basically what I am trying to show you here. If you have a flat interface, if the interface is flat, then f is independent of x and y, okay because the same is, I am telling you it is flat. Do not ask me how it is being flat, it is flat, okay. So, now z is a function of t alone and what are the kinematic condition give me? I am going to write this as the kinematic boundary condition implies df by dt equals Vz. So, the rate at which the f, the height is increasing with time must be the same as the velocity, okay and that is possibly common sense, right. So, basically what I am saying is this is consistent with physical intuition. So, I will just write here that this boundary condition is useful in determining stability of multi-phase flows where the interface deforms. Now, in order to keep life a bit simple, what I will do is I will just write the answer to question number 3 which is A3. I think that was the tangential stress condition, right, condition. The tangential stress condition basically tells you that on an interface with outward normal n dot t dot t in the first liquid equals n dot t dot t of the second liquid. I want to qualify this a little bit now. This tells you that the tangential stress exerted by one liquid on the other is equal to that exerted by the second liquid on the first. This is true only as long as there is no variation of surface tension along the interface, okay. This is an implicit assumption. Later on in the course we will relax this assumption, okay. So, this is we have as long as the surface tension does not change along the interface. And there is a small derivation to take into account the variation of the surface tension. I am going to do it later on in the course. I do not want to make it too mathematical because right now I think we need to establish some framework. We need to start solving some problems. So, that will possibly keep the interest alive otherwise it becomes too mathematical and people go crazy, okay. So, what we are doing is this particular thing I am tacitly assuming that there is no surface tension variation along the course along the interface. So, now but that is a problem which is basically called a Marangoni convection problem. And in the Marangoni convection, some of the people read Marangoni convection you will see that that is the tangential stress has an extra term which incorporates the gradient in the surface tension. So, if we have a surface tension gradient exists, then this condition is modified. An example of this problem is the Marangoni convection problem. And I guarantee you we will see this later on in the course, okay. Just to build up some suspense here. I will answer A2 now. I am not sure if this has already been done in the class before. So, I will talk to them and I will possibly derive it tomorrow if it has not been done. The normal stress boundary condition supposing we have an interface, okay. And this is fluid 1 and this is fluid 2 and this is the direction of the normal. So, the normal is pointing from fluid 1 to fluid 2. The normal stress boundary condition and in the most general form is going to be given by n dot t dot n in fluid 2 minus n dot t dot n in fluid 1 has been equal to sigma the surface tension times del dotted with n, okay. Now, if this has not been derived, I may not formally derive it, I will derive this equation and that other equation formally later. But I will give a hand waving derivation later on, I mean maybe tomorrow. What is going on here? This sigma is the surface tension and what does this term represent? del dot n, it represents the curvature, okay. So, you have actually seen this formulation in your courses in hydrostatic when you had a meniscus or a bubble, you talk about pressure terms. Things are not moving there, okay. You do not have liquids in motion, fluids in motion but everything is static. So, the only contribution to the stress tensor terms are going to be the diagonal elements which are the pressure elements, okay and you have p1 minus p2 is equal to sigma divided by r something like that, okay or 2 sigma divided by r. So, that the 1 by r or the 2 by r that you have seen earlier is coming from the curvature term. What I have done here is instead of writing in terms of 1 by r, I have just generalized the formulation in terms of del dot n because in general you will have an arbitrary surface where you have a normal. So, given the normal, if you were to find the divergence of the normal, that tells you what the curvature is and that curvature should simplify. So, this formula should simplify to your 1 by r or 2 by r for a cylinder or a sphere, okay and that we will check. So, this is basically your sigma by r and this would be p1 minus p2 equal to 2 gamma by r or something like that, okay. So, what I have done is this is just a generalized boundary condition in the normal direction, okay. So, I think basically what we have is we have established all that we need. We have the differential equations, the equation of continuity, the equation of momentum, the Navier-Stokes equation, we have the boundary conditions, the kinematic boundary condition. So, we will, we are all set to solve problems, okay.