 So, now I will start a topic which is called as special topics in conduction heat transfer. Here, I will take two types of problem, first which I called as multi-solid heat conduction problem and the second which I called as non-linear heat conduction problem. So, with this the idea is to introduce to you like in fluid we have multi-phase flow. So, what happens in what is multi-phase flow? Two different fluid flowing like you can have boiling condensation, you can have two fluid stratified flow, let us say oil water flowing in a pipe together where they are separated. So, in multi-phase flow you have interfaces. So, analogously one can teach multi-phase flow simulation as an advanced course on CFD, but to introduce you the concept of interface, in conduction I thought that let us have multi-solid heat conduction, but the difference between the multi-solid and multi-fluid is that in multi-fluid the interface moves with respect to time, whereas in this because of multi-solid heat transfer in if there is no melting the interface between the two solid will be fixed. So, the interface is fixed here whereas in multi-phase flow the interface changes with respect to time. So, when you solve a multi-phase flow problem computationally you need an additional equation for moving the interface which is not needed here and because the interface is fixed no melting. So, to give idea about the when we have two different solids how you take care of the conductivity because at the interface the question comes is that what should be the conductivity? So, those questions will be answered here. The second problem non-linear heat conduction, non-linearity mainly you see in fluid mechanics, but there is one case at least where you get a non-linearity heat conduction when conductivity becomes function of temperature. I think someone asked you that asked me in the yesterday's lecture that. So, I am anyway we teach this in our undergraduate CFD course. So, I am taking two topics into that, this I call as special topics, one may find nothing special about it, but just for a beginner I think it is special. So, let us take a simple problem, let us take a square plate made up of four materials of equal size copper, iron, stainless steel and aluminum and let us suppose as a as you know by this time you will be knowing that I use when I take example problem I use two types of boundary condition, one all Dirichlet, second all possible boundary condition. So, let us suppose left wall is 100, bottom wall is 200, right wall 300, top wall 400 degree centigrade. Now, here for a beginner in CFD this is the simplest grid generation equi-spaced horizontal and vertical lines I would like to. So, these are the grid points which I have shown earlier also I would like to draw your attention into this, what is this? Interface between the two different materials, I am showing it by red lines, because this is where we have to focus mainly, otherwise the problem is easy. The only difference which lies in this is this interface, we are considering the heat transfer that there is no melting. So, interface will not change with respect to time, the computation stencil remains like this. So, if I take a stencil which is border as far as this material is concerned, then when you apply as you know by this time that the way we solve infinite volume method is that we have to calculate heat fluxes on the different phases. So, when you want to calculate heat flux at this phase, what will be the conductivity you use? You know that the to calculate heat flux unit conductivity by Fourier Lafette conduction and note that I have generated the grid in such way that the solid two solid interface lies at the phase of the control volume, not at the cell center of the control volume, correct? Even the problems which we will be giving you in the lab session today, we will tell you that enter the even number of grid points. So, that at least we ensure that the interface is coinciding with the phase of the control volume. So, the question, the only question I would say in this case is that how to calculate k at the interface, these are the boundary border and interior control volume that I had explained it. Here we use, so the conductivity is calculated using simple idea which you teach in let us say under graduate fluid mechanics course. So, the idea is this that let us suppose I have plate, here we use one dimensional approximation. So, let us suppose you have plate made of two different materials and the thickness of one plate is delta x by 2, the second thickness of the second material is again delta x by 2, okay? Then what you do? You use resistance concept. What is the resistance concept? The difference, so I am using, you are using one dimensional conduction between the two grid points only. So, the temperature difference divided by the thermal resistance. What is the temperature difference? T capital E minus T capital P divided by, so you use the resistance concept and what is, how do you calculate resistance concept? Resistance, thermal resistance in a material L by k A. So, what is the length here? Delta x E by 2 and the conductivity is k E 2 and here it is k E 1. So, these are the thermal resistance and they are in series, okay? So, using this resistance concept, we can equate this to an equivalent conductivity also and then you can get a relationship between the equivalent conductivity and k E 1 and k E 2, the conductivity of the two different materials. Is that clear? So the point is that, let us suppose if you take one material, let us say this copper, if so in this material you will generate some control volume and in those control volume there will be face center, certain number of face center, horizontal face center, vertical face center, those face center which were, which are lying at the boundary. For those you need to, so the face center which is lying at the boundary, let us say here, you need to use a different k. You cannot use k of this or k of this and that harmonic mean of k which you will calculate, which will be used as q x for the east face of this and west face of this. So, there will be continuity of heat flux, note that. There is no jump of heat fluxes, there is no, what we call as contact thermal resistance. Contact thermal resistance is taken as 0 in this case. So, with that using, so that is the difference which you have to, the way the, we will give you the code, the code which has been generated in such a way that everywhere it calculates the harmonic mean, but the thing is that if there is a change it calculates the correct ones, but it unnecessarily correct calculates for the interior grade points also. Is this correct? So, for simplicity you will see that in our code generation, we are doing some lot of unnecessary things you may feel, but this is the simplest way of thinking and doing. So, for a beginner in CFD code development, we felt that this, the way I defined that q x, q y grid point also, that is also not necessary, that is not an efficient way of code development, but for a beginner I think it is okay to get a feel of it while you are developing a code. Now, let us take this problem. So, the solution algorithm remains more or less same what I thought. So, I am not discussing the implementation detail solution algorithm, those are same what I thought in the previous lecture. So, using all those, if you do this modification in the code and if you run the code for this, then you get a movie of the temperature contours which is as like this. Left wall 100, water wall 200, right wall 300, top wall 400 degree centigrade. This is the unsteady simulation, you have a series of picture of oscillotherms at different time instants and it reaches to a final year study state. I had shown this problem earlier also. What is the difference between the isotherm here and the earlier case? Yeah, that is a discontinuity, there is a discontinuity of the interface at the interface between the two different solids. So, there is a change in the, let us say temperature gradient across the interface and the change in temperature gradient across the interface is inversely proportional to the conductivity issues. Because basically, Fourier law of heat conduction has to apply at the interface. So, there is a continuity of heat flux. So, there will be the, as the conductivity is different. So, if you apply Fourier law of conduction, it says that there should be change in the slope of the isotherms, that is what you are getting here. Now, let us take the second problem. Second problem will be left wall 100, bottom wall insulated, right wall constant heat flux, top wall convective boundary conditions. For this case, if you generate the animation, it will look like this. Bottom wall is insulated, you always see that each time step, each isotherm is hitting the bottom wall perpendicular. Right wall, constant heat flux, the isotherms are hitting at a constant angle. But that constant angle is different in the top portion and the bottom portion because you have two different materials, note that. Just see it is hitting normally. Here it is hitting at a constant angle on the top part, a different constant angle in the bottom part because you have two different materials. So, this is the steady state temperature distribution. So, before I move into next topic, if you have any question, I will be happy to answer. So, if there is no question, let us move to the next topic. Next topic is non-linear heat conduction. As I mentioned that in heat conduction, there are very few cases where you get non-linearity and one of the cases that if conductivity becomes function of temperature. So, let us suppose conductivity varies in a linear fashion k 0 1 plus b t where k naught and b are constants and this b can be positive or it could be negative. And there are certain materials for which we know k naught and b are given. I will show you in the next slide and the good thing is that for this case, we can obtain analytical solution also. So, this is you can obtain exact solution by just integrating and this I am telling you can obtain analytical solution for the steady state situation. Note that not for the steady state. For the steady state situation, this will be 0 and then you just integrate it twice using this k expression and you will get this exact solution. So, the objective of this will take a problem and we will I will show you how the what is the solution algorithm and how code is developed. Once you develop a code, you can test it with exact solution. So, you can can be used for validation code validation. The finite volume discretization if you do you will get unsteady term diffusion term as this. So, thermal conductivity just for reference I am showing you that for stainless steel what is the value of k naught and b and for pure iron at less than 1000 Kelvin what is the value of k and b. Note that b is positive for stainless steel and negative for iron although the value of b is very small. So, here the solution algorithm procedure is same what I discussed earlier. The only difference which we need is that here you find that after each time step as at each grid point temperature is changing and your conductivity is a function of grid point temperature. So, k will change with respect to time. So, k is changing with respect to time and temperature calculation depend on k. So, it is a coupled equations. What in CFD we do is that we solve one equation at a time. So, let us suppose I solve k using the temperature of previous time level. Once I get updated value of k, I solve the heat conduction equation. It can be vice versa also. So, I am just suggesting that let us calculate conductivity using the old time level and then use the updated k to calculate temperature at the new time level. So, in this case calculation of k is lagged by one time step. I will take an example problem. If you take that stainless steel iron b, you do not get much change in the curve. Steady state linear curve is one dimension, it is a linear curve. But if you take that b of iron and stainless steel, there is hardly any change. So, I wanted to show you some appreciable change. I have taken some hypothetical problem. So, let us say k naught is 1 watt per meter Kelvin. It is a hypothetical problem and from exact solution we have found that the value of b had taken three different values 1, 0 and minus 0.01. And k in this case conductivity, this was capital K. So, I was confused. It should be small k. The conductivity from if you take the variation from 0 degrees because the boundary condition is left wall it is 0, right wall sorry left wall it is 0 and right wall it is 100 degree centigrade. So, for b is equals to 1, the conductivity varies from 1 to 101. This is a steady state conduction problem with no volumetric heat generation. So, temperature has to be within limits boundary limits 0 and 100. So, the k with when b is 1, it varies from a very large extent 1 to 101. When it is minus 0 1, it varies from 1 to 0. But in this case k is increasing with temperature, here it is k is decreasing with temperature. So, at the center point length of the plate which we have taken is 1. So, at times equals to 0.5 when you have 0 which means linear variation you get 0 plus 100 divided by 2. When b is positive at the center you get 70.4. When b is negative you get 29.3. So, a code is developed and what we did is that we compared our results with the exact solution and we find that the results matches exactly. So, you can see that b is equals to 0 is the linear variation which you anyway know. But when b is positive you get a curve which is convex upward and when b is negative you get a curve which is concave upward. Actually, this is one of the question we typically asked in our MTech interviews. May be even in PhD interviews we ask and we always find that the student struggle and we wonder that why they struggle. I am say in undergraduate teaching what is there any issue because you may be also asking this type of question is that what is the temperature distribution for one dimensional steady state conduction and then you make it more complicated saying that conductivity is increasing with temperature then how it will vary if it is decreasing then how it varies. I hope that this problem you should at least discuss in the class. So, that they have an idea of application of the Fourier Lafitte conduction in this case. So, to show that actually this is 70.4 and this is 50 and this is 29.3 because this is what you get from the exact solution I had zoomed on this figure and you can see that it is close to 70.4 this is 50 and this is indeed close to 29.3. This has been obtained numerically and this is an exact solution. So, I think I had come to the end of this lecture but if we have any question I will be happy to answer. Any question? We have taken two problems which I called as special topics in heat conduction, multi-solid heat conduction and non-linear heat conduction. Any question? Please raise your hand. How this non-iliarity thermal conductivity take and care in the finite volume? Okay, maybe things are not clear I will go back to the domain and then I will explain it. This question is how I incorporate non-linearity sorry non-linearity in the code you are saying. There what we do is that the idea is something like this you okay let me start with the situation. Let us suppose you have a plate which is at you are taken from a furnace okay. So, everywhere let us suppose temperature in the plate is 250 degree centigrade but suddenly let us suppose we do something and subject the left wall at 100, bottom wall at 200, right wall at 300, top wall at 400 degree centigrade okay. For that material somebody has done an experiment and he has given some let us say table or some expression that k varies with temperature like this. Then what you can what you do is that from the initial condition you know that all the yellow circle which are interior point will have 250 degree centigrade and the boundary point blue points on the left wall 100, bottom wall 200, right wall 300, top wall 400 correct. So, using this what you can do is that you need conductivity at the phases. So, one way of doing it is that you know the temperature at the cell center. So, you calculate conductivity at the cell center and take linear interpolation of that you know you will have an expression for k as a function of temperature. So, if you know the temperature you can calculate k using that expression but this k you will be calculating at cell centers yellow circles and blue circles. So, you do linear interpolation because conductivity you need to calculate at the phase centers because heat flux comes at the phases not at the cell centers. So, you do linear interpolation of that conductivity or you take the linear interpolation of temperature and calculate conductivity. So, either you do first averaging and then do then use the expression for k or use calculate k at the cell center and then do average either of the two is fine okay. So, this way you get a matrix for k at all the phase centers. Once you know k at all the phase centers when you are developing code you follow the same procedure which I had discussed earlier. But the only difference is that now k will not be a constant value k will be coming from the matrix. So, for a particular cell you will have that value at the phase center when you calculate q x q y at only yesterday where do you calculate horizontal phase center vertical phase center for this that goes phase center earlier k was constant. In this case k will be calculated after each time and it will be updated in a matrix solving solution procedure is there any we have given you a code for that. So, no no no for a generating code for this type of problem nonlinear problem. What I suggest is that if you understand how to what I mean is that you look into the codes which we had given to you and I had in my earlier lecture this slides will also be given to you there is an implementation detail there is a solution algorithm there is a code also. So, we hope that but at least you have to make some effort I am saying everything will not be 100% clear through this lecture. But at least we are feeling that you have lot of resistance in teaching this course. So, we are trying to help you we are ready to help you now also we are ready to help you later also by email. But you have to make some effort that is very important. Sir that different material problem I did with putting additional points at the interface. So, whenever it is on the left side we can use thermal conductivity of say material 1. And right side we can use conductivity of material 2 it is like a boundary point. And it works and it also gives the interface temperature which we can compare with analytical solution. So, what you are saying is that you find you have found some different method of solution correct. And you feel that it is working properly you have not use harmonic mean that is what you are saying. Yeah that harmonic mean is given in that Patankar's book. Yeah. No, no but what you are saying is that you are not using this procedure and still you are getting the correct solution. So, you have found an alternative methods you are saying correct. Yeah, wherever you have shown that red lines. Yeah. I have put additional point just like your blue points on the boundary. So, zero control volume. Okay. So, as long as it is on left side we can use conductivity of material 1. But what is the boundary condition which we will be using at that if you say what is saying is that in this red line it puts blue circles correct. Yeah. But you are saying. Yeah. But if you put the blue circles here. Yeah. You need boundary condition to solve it. So, there we have to ensure that flux is same. So, k dT by dx on left equal to k dT by dx on the. So, actually you are doing in a slightly different way but I think idea is same. Yeah. Yeah. But we can get interface temperature. Yeah. It is good that you have found some. Which we can compare with our analytical solution. Okay. No, it is good that you have found some alternative method. But the idea remains same. So, we are presenting some method but we will be more than happy if you come up with some alternate method. But make sure that you get the same result because finally the physical solution will be one solution. So, your numerical formulation should not change the solution. Sir. Yeah. I want to ask more physical situation practical case. Yeah. There will be some contact resistance. Yes. How do you solve that problem? For that you have to based on the contact resistance there will be discontinuity in the heat flux. So, from the contact resistance you calculate what will be the discontinuity and that can be incorporated in the code. Then the boundary condition will be jump of heat flux. It is not continuity of heat flux. Then I think the method which he has proposed will be a better one.