 So, today's topic is yesterday topic was computational heat conduction. Now, here I have taken what I call as special topics in computational heat conduction. It depends upon the word special you may think that what is so special about let us say competition heat conduction in cylindrical coordinate. It is just that Cartian coordinate the formulation is quite straight forward, but when you move to cylindrical or spherical coordinates the geometrical information which are involved get more complicated and then I had taken multi solid heat conduction problem and finally, the non-linear heat conduction. So, for first course in CFD you can consider this as special topics in computational fluid dynamics. So, let me first start with computational heat conduction in cylindrical coordinates. So, as I try to always connect you to the your undergraduate courses of let us say here I am doing it for heat transfer the way you derive the partial differential equation governing partial differential equation in a cylindrical polar coordinate system. Here we are taking a two dimensional control volume in r phi direction where this is the radial direction coordinate and this is the phi coordinate angular coordinate. So, what you do in a control volume now it is not a Cartian control volume. So, you do not have x plus dx y y plus dy. So, here you have q phi q phi plus delta phi q r q r plus delta r. Now, the area here is delta r on and here the area is this is r and this is r plus delta r. So, the area here is r delta phi here the area is r plus delta r delta phi. So, you what I am showing is q conduction is basically total heat gain by conduction per unit volume. So, if you apply the limits do a balance divide by the volume apply the limits you end up with a partial differential equation in heat conduction in cylindrical coordinate r phi directions. Now, we apply the same conservation law the same control volume, but instead of writing this q phi q phi plus d phi you will see in the next slide we do what we call as first level of approximation where we do surface averaging of heat fluxes. The heat flux on this surface if you take any phase it varies by point to point. So, we express the average value on that surface by the value at the centroid. So, this is what I call as surface averaging of to calculate heat flux at control volume phases. So, we express this let us say q phi as q small w here q phi plus d phi here as q small e here. So, that way we represent q s here and q n here and then we do a balance and obtain the total heat gained by conduction. Then we go to the second level of approximation where we do the discrete representation of Fourier law of heat conduction which is shown here. This q e and q w are the heat fluxes in the angular direction and the Fourier law of heat conduction for heat flux in the angular direction is minus k del t by r del phi. So, this is the discrete form of that q n and q r q s r in the radial direction and the Fourier law of heat conduction in that direction is minus k del t by del r. So, this is the discrete representation of that. So, we substitute this discrete form of Fourier law of heat conduction in cylindrical coordinate system to this equation and we end up in getting an equation in getting a linear algebraic equation which is shown here. Now, here the total heat gained by conduction is calculated where temperature you can see the superscript is n which is from the previous time level. So, this is an explicit method. So, this is finite volume equation linear algebraic equation using explicit method in cylindrical polar coordinate system. This is the finite volume method in cylindrical coordinate system. Now, here I am showing you a simplest form of grid generation in a Cartesian coordinate system we had drawn equispaced horizontal and vertical lines. Here we will not draw horizontal and vertical instead what we do is that we draw always you will see that there are 2 family of lines there it was horizontal vertical here it will be. So, horizontal vertical means x equals to constant y is equals to constant lines here you will have r is equals to constant theta is equals to constant lines what is r is equals to constant line circle. So, you have concentric circles in this case equispaced concentric circles and you have equispaced radial lines and when you draw this equispaced concentric circles and equispaced radial lines you get certain control volume and the yellow circles which I am showing you is positioned at the centroid of the control volume. So, when you want to crawl through the different because in basically what happens in a finite volume method is that we solve control volume by control volume. So, we have to solve we have to scroll through all the control volumes in the domain then only we will get the solutions for all the grid points in the domain at each time step. So, when you want to move in let us say in this direction this direction this phi direction I am denoting as running indices i the radial direction I am denoting as running indices j and when you want to scroll through let us say in the i direction you then you start from let us say from this control volume solve the linear equation for this control volume then you move to the next control volume then to next. So, this way when you keep moving in the angular direction what you see is that you reach to the last control volume, but this two control volumes are connected in a Cartesian coordinate system they were separated. And another thing to note is that when you want to solve for let us say this control volume you know that it needs its neighbor. So, the neighbors which are involved is let us say this neighbor this neighbor and this neighbor. So, to solve this problem we convert this into what we call as we do a branch cut. So, what the green line which I am showing you is the idea something like this that let us break this domain from here and then open it out. So, in the next slide I will show you the domain which seems to be a Cartesian, but i direction represents the angular direction j direction represents the theta direction sorry yeah radial direction and the note that this domain which I am showing you is something like this that if there is a square plate with a circular hole and if it is a heat transfer problem. So, if there is a square plate with a circular hole. So, the solid region is the domain and this is the circular hole. So, the circular hole there are two boundaries inner boundary inner circle outer boundary which is outer circle. So, you have competition in grid points boundary grid points on this inner boundary as well as you have a boundary grid points on the outer boundary where you have grid points are shown by blue circles. Now, as I said that to solve this problem when you want to scroll through the different grid points you have to do a branch cut. So, if you cut from here then your i this will be i is equals to 1 this is 2 this is 3 4 and so on and then it goes to this the way we are defining is that we are using the same convention which we had used in the earlier for Cartesian coordinate system. So, I will clarify here that i is equals to 1 and i max is the same although we do not have a boundary grid point any grid point here, but 1 and i max are the same here this is 2 3 and so on I will show in the next slide. So, this is i is equals to 1 this is 2 this is 3 4 5 6 7 8 9 and right now we have a 10 although we do not have a boundary grid point in the angular direction. Right now we have note that boundary grid points only in the radial direction this is on the inner radius and this is on the outer radius. So, this is let us suppose your j it is starting from inner radius j is equals to 1 is the on the inner radius and j is equals to j max is on the outer radius. So, the bottom surface of this domain is the inner boundary the top surface of this domain is the outer boundary I would like to point out that this is what is called as physical domain. Now, for computation we do this branch cut and then we get this what this is called as computational domain. Now, we solve in this computational domain where the coordinate note that it is still r theta. So, this is bottom boundary it is r is equals to r 1 let us say whole radius the top boundary is r is equals to r 2 let us say outer radius of that plate and this is the branch cut which we had done which had. So, actually in this boundary and this boundary in the physical domain are same and on this boundary i is equals to 1 as well as i max. So, in the next slide here I have taken i is equals to 1 and i is equals to i max and here what I am showing you are the again I am using the convention here in Cartesian coordinate system earlier you had q x q y here you will have heat flux in the theta phi direction q t and in the radial direction q r. Now, the convention which is being followed let me go back. So, the in this if you take any control volume let us say if you take this control volume. So, the in the i direction which heat flux you have i direction represents the angular direction. So, the i direction on the this face center this face center it will be q phi and q phi plus d phi and on the radial direction you will have q r and q r plus d r. So, what I am showing you here this is the heat transfer i direction is which direction angular direction. So, this green squares which I am showing you is the grid point to calculate the heat flux in the angular direction. Now, vertical line now the red inverted triangles represents the heat transfer in the radial directions and the convention which we are following is that as we had used in the Cartesian coordinate system that the grid point for a particular control volume may be I can clarify this again in white board. I am using certain green squares and red inverted triangles and the idea behind this is something like this that let us suppose I have a control volume which is a representative control volume p. There is another control volume which is capital E. Now, what you may do is that you can to calculate heat fluxes let us suppose if you take this face center this face center if I draw the diagram separately this is a east face for this control volume and this is a west face of this control volume, but the heat flux heat transfer which is occurring here which is q E for this and which is q W for this. The expression which is used to calculate q E here is T capital E minus T capital P divided by delta S E this is basically multiplied by minus k and similarly here q W is calculated as minus k T capital E minus T capital P divided by delta S E. So, the point here I want to emphasize is that if I go to the control volume and calculate q E and then I go to this control volume and calculate q W I am using the same expression two times. So, I am doing unnecessary calculations or computations. So, the idea is rather than doing that way let us define and another thing is that whether if I say that there are three type of grid points this is the grid point for temperature this is the grid point for let us say q X and let us say this is the grid point for q Y. Now, if I use a convention let us suppose for temperature I have I 1 J 1 for q X I have another running indices I 2 J 2 if I have another running indices for q Y I 3 J 3. So, it is not a good idea to use three different running indices specially if you can come up with a convention where there is a relationship between I 1 I 2 I 3 J 1 J 2 J 3. So, we are coming up with a relationship between this and we are only. So, rather than using three different I comma J. So, we are not using I 1 J 1 I 2 J 2 I 3 J 3, but instead we want to use only one convention which is I comma J. But if you want to use one convention for three different grid points you have to come up with some proposition. So, the proposition which is being said that any control volume what is the east face of this control volume this is the east face of this control volume. What is the north face of this this is a east or positive face this is also a north is also a positive face. Now, q X is defined here q Y is defined here. Now, what is the we have to just see what is the I comma J corresponding to this. Let us suppose if this is I comma J here. So, this will be I comma J. Now, for a neighboring control volume let us suppose if you go to the west control volume what is the running indices here it is at X minus delta X. So, it will be I minus 1 comma J. Now, what is the east face of this control volume this is the west sorry east face of this control volume. So, the q X here will be I minus 1 comma J. If you go to this control volume this will be q X I minus 2 comma J. So, the convention which we are following here is that for any control volume the there is a particular running indices at the centroid of the control volume. Now, to know the running indices of q X because q X for a control volume is defined at two different positions such as q X is defined on vertical face center there are two vertical faces in a control volume. So, you will have q X on the west face on the east face and there is only one running indices which is for the centroid of the control volume. So, we have to come up with a proposition or a convention that here is what we are using is the positive face will have the same running indices which is at the centroid when you go. So, what will be the that for the next this face this face running indices this is the west face for this control volume, but it becomes an east face for the west control volume and the running indices for the west is I minus 1 comma J. So, the q X here will be I minus 1 comma J which is shown here. Similarly, the running indices for this year will be q X I minus 2 comma J. So, this way we these are some of the details which you have to work on before you start developing the codes. So, this is just for that thing. So, here in this slide I am showing you the running indices for. So, instead of q X here you have q in the angular direction is the heat flux in the radial direction and in the next slide I am showing you the pseudo code for the as the procedure which we are adopting here as I had shown you earlier. So, let me go to the whiteboard and show you in a finite volume method. The way we are working here actually the procedure which we are using to solve is that we are not solving the direct linear and laboric equations, but instead we are solving it or we are calculating or we are programming in two steps. First step we calculate this heat fluxes where we calculate this heat fluxes at the face center. Like this q e q w is basically the heat flux in the x direction. This is the heat flux in the y direction. So, note that in the finite volume formulation which is being proposed here and the programming programs which are given to you which you which have you have used yesterday as well you will be using today is based on a formulation as well as procedure where in the first step we calculate this heat fluxes and in the second step we use this equation and because I believe that this is the way the conservation law is obeyed. So, first we calculate the heat fluxes then calculate the total heat gain by conduction and then using the total heat gain by conduction in an unsteady problem you can equate it to the unsteady term and obtain the temperature for the new time step. That way here first step is we calculate the heat fluxes at the face center using this convention using the discrete form of the Fourier law of heat conduction. So, after that first step this so this is the first step. So, in this step what you get is you get the heat fluxes on the faces of the control volume and then you calculate in this step total heat gain by conduction and then you use this expression to calculate the temperature of the new time. If you program this way you also get a feel that it is not mathematical but it is more physical in sense. At each when you are programming each step you understand that because many times when you program if you are using a coefficient based formulation you get certain coefficients and many times you do not get feel of what the coefficient has a physical. So, at the another the only point which I would like to emphasize is what I am shown here by underline. This underline which I am showing you you can see that the q heat flux in the theta direction there is a periodicity. So, there is a procedure which I am using and I have given a code for this I will due to the shortage of time I will not go into the detail of this code. But if you have any question I will be happy to answer this is just I would like to emphasize that when you want to calculate heat flux at let us say this point this is equal to this value temperature here minus temperature at this point divided by the distance between this point and this point actually this boundary and this boundary physically they are same. So, the distance if I show it here the distance between this two point I am talking of it is basically r radius here into delta theta delta phi. So, the heat flux on the left boundary is calculated using this temperature minus this temperature divided by the distance between the two and this heat flux is same as this heat flux which is shown in the here. So, q at i is equals to 1 is q heat flux in the angular direction at i is equals to 1 is calculated in the same value is substituted at i max minus 1. Now, let us go to the next topic computational multi solid heat conduction nowadays in fluid flow analogously there is a multi fluid flow problem which we call as multi phase flow problem worldwide this area is getting a lot of importance. In fact, a new subject is also coming which is also called as computational multi fluid dynamics CMFD because multi phase problem I would like to point out we are doing research in this. So, we have an idea about it that although we say that CFD can solve any problem, but I would like to point out that CFD can solve those problems only those fluid flow problems for which the mathematical models are well established. Like in multi phase flow problem just to give an example let us suppose you do someone simple experiment you take a pan filled with water and put it in a gas stove and you see how the boiling occurs you do this in summer you do this in rainy season you will see the difference you will see that the site at which the nucleation occurs you will see certain at the bottom of the pan certain bubbles will be formed vapor bubbles. Now, the site of the nucleation or that vapor bubble formation depends on many things it depends upon the surface roughness of your pan it depends on the moisture content. So, the mathematical modeling of that is still a challenge. So, in multi phase flow problem if you see the methodology numerical methodology they are of different types there is not only one methodology there is let us say for separated flow there is another one methodology for dense flow there is other methodology if you have a solid gas flow you have another methodology langrangian methods. So, there are multi phase flow problems is one of the challenging research areas as far as CFD is concerned. Now, analogously a simple problem in heat conduction I am proposing here it is a multi solid heat conduction problem. So, the idea which is being proposed here is let us suppose you have a plate which is made up of 4 different materials stainless steel aluminum copper and iron. So, we are talking of heat transfer in such a plate you can also called as a composite plate. Now, for this how to solve this problem the you use the same finite volume method I will show you through an animation. So, let us go to this window I will show in animation of grid generation and the grid point distribution. So, you can increase the size of this window where animation will play ok. So, this is the plate now let us what will be the grid generation we the size of the plate is l 1 in the x direction l 2 in the y direction we draw equispaced vertical lines equispaced horizontal lines and we get this are look this are the grid points located at the centroid of the control volume. These are the grid points located at the left boundary bottom boundary right boundary top boundary. So, with this how many grid points are there in the 4 materials how many yellow circles are there in aluminum stainless steel copper iron 9, 9 control volumes 9 grid points and let us suppose this is aluminum it is boundary is on the left boundary and the right boundary and what is there in the sorry left boundary and bottom boundary what is there in the right boundary this is the running indices in the x and y direction just see this what is this this is the interface between the two material. So, the question is how to handle this interface because note that we have generated the grid in such a way that the at the interface between the two different materials the interface between the two material which is shown in this slide by red line is coinciding with the intersection of the two material. The face of the control volume note that not the sale of the control volume we have drawn the control volumes in such a way that the face of the control volume is coinciding with the intersection of this two materials ok. So, at the joining between the two material you have certain face centers and what you do at those face center you calculate heat fluxes like on this face center you want Qx on this face center you want Qy. Now, when you want to calculate heat flux you need conductivity. So, the biggest question in this problem is if I want to calculate conductivity I have to take conductivity of let us say this material or this material or what should be the expression whatever has been discussed in the previous lecture for Cartesian coordinate system that same finite volume method everything is same, but here the only questions which come you have to two things which you have to make sure is that you have to make sure that you use appropriate conductivity when you are calculating the fluxes. Whether you have to take conductivity of aluminum stainless steel copper or iron you have to see whether the face center is lying in which material. So, whichever material it is lying you use appropriate conductivity this is one thing. Second thing is that at the intersection between the two materials two plates two different plates you have a face where you want to calculate heat fluxes. Now, the question is what should be the conductivity. So, and this is a typical computational stencil and the biggest challenge is how to what conductivity have to use. So, let us come back to this is the boundary control volume this is the border control volume with a thick black at the circumference and this is the interior control volume. So, now let us come back to the slide. So, this I had already shown through an animation and I mentioned that the biggest challenge is how to calculate the conductivity. So, this material is stainless steel this is aluminum. So, let us suppose this is I this is this is a face which is common to two different material. Let us suppose left hand side you have a stainless steel on the right hand side you have an aluminum. So, what should be the conductivity here? Now, this we calculate using simple idea which we use in our earlier lectures of let us say under graduate heat transfer course. The idea which we use is what is the thermal resistance. So, Q e on this face center is equals to some equivalent conductivity multiplied by temperature at this point minus temperature at this point divided by the distance between these two point which is shown as delta x e. Now, using a resistance concept you can express this as the potential here is the temperature difference divided by the thermal resistance. Thermal resistance for heat flux it comes out to be L by k. So, what is the length of let us say stainless steel here delta x by 2. What is the length of aluminum here delta x by 2. So, it will be delta x e divided by 2 k e 1 let us suppose this k e 1 is that of stainless steel plus delta x e divided by 2 k e 2 let us suppose this is conductivity of aluminum. So, this equivalent conductivity using this expression we can obtain a relationship 2 by k e is equals to 1 by k e 1 plus 1 by k e 2. So, you calculate the equivalent conductivity by this expression which is the harmonic mean of the properties of the two materials at interface. Note that today afternoon you have a lab session where we have given you a problem corresponding to this. Actually whatever example problems or animations I show you for the example problem we had given you programs. So, you can run the program and you can get similar results. So, the problems which example problems which are being discussed here we have already given the setup of that problem and given you the code. So, that you can see that the results which have been obtained in this example problems you are also able to obtain. Now, let us take the first problem. So, the problem is which I have shown here left wall T b 1 bottom wall T b 2. So, the problem which we are taking is let us suppose this left wall is maintained at 100 degree centigrade, bottom wall at 200, right wall at 300, top wall at 400 degree centigrade. So, this is the definition of the problem which is shown here. So, for this I will show you a movie of temperature distribution temperature contours. So, I am showing you unsteady state temperature contours. Now, temperature contour what I am showing you here are two types line contour where you see certain lines like on this line temperature is 150 on this line temperature is 200 on this line it is 250 here it is 300 and so on as well as you see some color. So, this red color represent 300 degree centigrade this green color represent 200 blue color represent 100 and there is a interpolations of this colors and similarly the temperature values. So, there is a flooded contour flooded color contour and line contours of the temperature and this is an animation for that. Note that at the interface between the two material note the continuity of the stream the temperatures isotherms see the slope of the line when it is crossing the interface of the two material I had taken the same problem for single material also yesterday I had shown you an animation for single material. Now, I am showing you an animation for four different materials just see the smoothness of the line at the interface between the two material there is a jump in the slope of the isotherms at the interface between the two material why because there has to be we are assuming that the thermal resistance is 0. So, if the thermal resistance at the junction between the two material is 0 then there should be continuity of heat fluxes. So, heat flux let us say coming coming from the stainless steel side should enter into the aluminum side. So, if you equate the heat fluxes the ratio of the maybe I can show you in the white board what I pointed out is that at the interface between the two materials there is a change in the slope of the isotherm. And the reason is what I said is at this let us suppose this is stainless steel and this is copper sorry aluminum this is stainless steel this is stainless steel and this is aluminum. So, what will happen at the interface let us suppose there is a heat flux which is coming from the stainless side and it is entering into the aluminum side we are assuming that the thermal resistance at the junction is 0. So, with that assumption I will show you. So, that as I said that this is a junction on this side you have stainless steel here you have aluminum let us say let us assume that there is certain heat flux which is coming from the stainless steel side and which is entering into aluminum side. So, this is an interface between the two materials. So, or there is an interfacial boundary condition like in fluid flow there is a continuity of shear stress here analogously if there is no thermal resistance note that there is a no thermal resistance or contact resistance with this assumption q from the stainless steel side should be equal to q from entering into the aluminum side. So, it will become minus k del T by del x on the stainless steel side is equals to this is k of stainless steel minus k of aluminum del T by del x on the of aluminum. So, the ratio of the slope of let us say del T by del x aluminum divided by del T by del x stainless steel will be equal to k of stainless steel divided by k of aluminum. So, the ratio of the slope is inversely proportional to the ratio of conductivities. This is the reason if there is a single material then this becomes one. So, there is no discontinuity that is the reason for discontinuity in the slope at the contact between the two material. I would like to point out taking an analogy to the multi-phase flow problem that in this case the interface the contact between the two material also I can consider it as an interface. Here in this case if there is no melting then this interface does not change with respect to time. However, if there is a melting then there will be a fluid flow and then this interface does changes with respect to time. Then in that situation we have to come up with an equation and a procedure to capture a movie for the interface with respect to time. So, although I am taking a multi-solid heat conduction, but I am trying to give you a feel that in multi-phase flow problem we not only create movies for let us say flow, but we have to also create a movie for interface. We also have to use let us say conservation law or equations for representing and capturing interfaces with respect to time. So, multi-solid conduction is much easier as compared to multi-fluid flow because in multi-fluid flow the interface changes with respect to time. It is just that in this case at the contact we have to calculate the conductivity otherwise the procedure is quite straight forward. Let us take a second problem where left wall is maintained at 100 degree centigrade bottom wall is insulated right wall there is a constant heat flux boundary condition and top wall it is a convective boundary condition and let us see how animation looks. Note that the isotherms the lines are hitting perpendicular on the bottom wall why because bottom wall is insulated at each time step each picture note that the line is hitting vertical. Note that on the right surface the lines are hitting at constant angle although that constant angle is different in aluminum and iron why because on the right wall you have a constant heat flux boundary condition. So, slope is equals to minus q w by k. So, this is an animation for that will stop this animation and come back to the slide. So, this is the steady state temperature distribution. So, you can so that movie goes on and finally it reaches to a steady state and it does not change with respect to time. So, this is that steady state temperature distribution. So, I had shown you two problem first in which all the boundary conditions temperature were prescribed Dirichlet boundary condition second where both Dirichlet as well as non Dirichlet boundary condition is prescribed. These are the two class of problems which I take in heat conduction in the topic of heat conduction. Now, non linear heat conduction this is another special thing as far as heat transfer is concerned. So, we started with cylindrical solution in cylindrical coordinates moved on to multi solid heat conduction and now we are talking of non linear heat conduction. You know fluid flow equations are non linear you know convective heat transfer equations are non linear, but in conduction non linearity can come due to only one reason and what is that when the conduction equation becomes non linear when conductivity becomes function of temperature. So, we are taking a situation where let us suppose conductivity is a function of temperature. So, when conductivity is a function of temperature this is function of temperature you have it is multiplied by a temperature gradient. So, there is a multiplication then that is why it is become non linear. Now, this equation so, what I am trying to tell you is that you cannot write k del square t by del x square when k is a function of temperature. The governing equation in that case it is del by del x k del t by del x because k is at different x location temperature is changing and k is changing. So, in this case if you take the steady state equation here I am showing you the exact solution, but this exact solution is obtained where this left hand side is 0. So, for steady state conduction situation you can take only the right hand side and you can integrate it twice use this k also as k naught 1 plus b t use this boundary condition at x is equals to 0 t is equals to t 1 at x equals to l t is equals to t 2 and this is the exact solution you will obtain. Now, here either we can solve the steady state equation by steady state formulation or right now what I am showing you is an unsteady state formulation. So, you can take this unsteady state equation right now I am taking a one dimensional situation. So, one d unsteady state equation and we do a finite volume discretization which is shown here on the east phase the fluid is heat is going out and the on the west phase heat is coming in. So, after the finite volume discretization I am just showing you that just to give you an idea that in this expression k is equals to k naught 1 plus b t what are the what is the value of k naught and b for stainless steel k naught is shown k naught and b values are shown here for pure iron at less than 1000 Kelvin the value of k naught and b is shown here. Note that although the value of b is very small in most of the cases just see that when b becomes very small the dependence of k is weak on temperature. So, for most of the material there is a weak dependence. So, in most of the heat transfer problems are not long linear, but just to give you an idea I am showing you the values and there is a positive values and there is a negative value of b. And what is the meaning of positive value of b and negative value of b when b is positive k naught is always positive. So, when b is positive k increases with this temperature and when b is negative k decreases with temperature and the procedure which we use to solve this finite volume equation is whenever we are calculating the procedure which is being proposed here first we calculate the heat fluxes at the phase center vertical phase center to calculate the qx horizontal phase center to calculate the qy this is a one dimensional problem. So, you will calculate only qx when you are calculating qx there will be conductivity at the phase center. Now, let us suppose you want to calculate conductivity here now to calculate conductivity here k is a function of temperature. So, you will need a temperature here. So, to calculate a temperature here what you do is that you do linear interpolation. Let us suppose temperature here and temperature here and you do a linear interpolation such as if you slang exactly in the middle it will be t capital E plus t capital P divided by 2. So, that we calculate the temperature using that temperature you can calculate k and then calculate q here as t capital E minus t capital P divided by delta se into minus k k which you have calculated from here. Now, the question is that when you are doing unsteady state simulation and when it is a non-linear in fact in computational fluid dynamics whenever you encounter non-linear system of equations you solve one equation at a time such as in this case there are two options because you have to solve for k as well as you have to solve for heat flux. So, if you let us suppose you are first solving for k. So, what you can do is that when you are solving for k conductivity you have temperature of the previous time level. So, using the temperature of the previous time level first you calculate k then use this k to calculate the heat flux and then once you have calculated the heat flux calculate the temperature. So, what does that mean is calculate thermal conductivity using the previous or old time level value of temperature then use the updated value of k to calculate first heat flux then do the balance of heat flux and finally, obtain the temperature for the new time step. Thus in this calculation k is lagged by one time step. I have taken up a problem where I have taken values of B not from any materials but I have just taken some values so that I can show you that there is a large change in temperature. As I said that for most of the commonly used material the value of the B is very small of not minus it is minus 0.01 I have taken but if you look into this expression it is minus of 0.0007. If I use this values you will not see much difference in this curve. So, to show the difference I have taken larger values of B I have taken in the positive side 1 or for negative I have taken minus 0.01 and just to show you that if we and k naught I have taken as 1 watt per meter Kelvin. Now, if you use this equation k is equals to k naught 1 plus B T k naught is equals to 1 watt per meter Kelvin. If you take B as 1 and if you use the boundary condition temperature is equals to 0. So, when you substitute temperature is equals to 0 B is equals to 1 k naught is equals to 1. So, k comes out to be 1. So, from the limits 0 to 100 is a limit of this is without volumetric heat generation so that temperature will vary between 0 and 100. So, if you substitute 0 you get k is equals to 1 if you substitute 100 degree centigrade you get k is equals to 101. So, this the k the variation of k is from 1 to 101. So, there is a large difference of k in this range of temperature when the value of k naught and B is this. So, for this case actually there is a code which has been given to you and in the afternoon lab session between 2 to 3.30 this problem is given to you and you will get this type of figures also this results. So, in that you can see that the whatever is been discussed here the formulation the methodology is been used to develop that code and you get for this case you get the temperature variation as shown in this figure green line. So, the red line corresponds to B is equals to 0 which is when k is constant you know that in steady state conduction you get a linear variation. So, the red line corresponds to that where it is a linear heat conduction. Now, when B is 1 positive value and then you get a curve the point to note here is that what is happening to the slope in this curve what is happening to dT by dx in this curve it is increasing or decreasing dT by dx is reducing in this case why because k is increasing with the increase in temperature. What is happening to temperature when as x is increasing as x is increasing temperature is increasing when temperature is increasing k is increasing. Now, this is steady state conduction heat flux has to be constant. So, k dT by dx has to be constant if k is increasing dT by dx has to reduce then only heat flux will be constant this is the common questions we ask in the interviews for non-linear heat conduction how what will be the curve when this B is positive. So, the answer is when B is positive as x is increasing in this case temperature is increasing. So, value of k will increase, but q has to remain constant. So, dT by dx has to reduce. So, this is the curve corresponding to that whereas, if B is negative what is happening to the temperature with increase in x it is reduced temperature is increasing, but when B is negative k will reduce. So, when k is reducing dT by dx has to increase. So, the slope of this curve you can see as x is increasing it is increasing. So, this is just as an example problem for non-linear heat conduction with this I had come to the end of this topic special topic on heat conduction, but we have still have 10 minutes and I would like to take few questions before I move to the next topic. So, I will start with PVPS Vijayawada. I am talking about the first problem the different materials heat transfer. Now, that is for the materials of isotropic nature where all the four materials will have independent values of k. So, in case of composite material where you have kx and ky components then how the finite volume method can take into the formulation about this where you required to include kx, ky as well as the cross component in the matrix kx, ky over ok. So, the question is on the computational heat conduction and the question is right now we have taken materials where this material individual material are isotropic in nature where conductivity is not function of x and y. So, the question is if conductivity is a function of x and y and he said and then there is a cross component also kx, y. So, whatever I understand I feel that although I had not done that because I had not solved those type of composite material heat transfer problem, but if there is a kx value and ky value then what I believe is that you can calculate qx using the kx value and you can calculate qy using ky values to calculate the heat fluxes. So, the green squares which are shown which was for qx at those point you calculate using the kx value and for the red inverted triangles you use ky values. Thank you. Srinati Suresh. My question is regarding the composite plates subjected to different boundary conditions. Here if we are having the four and more solid materials are there, how do we take care of the interfaces, interfaces nodes? Is it to generalize the equation first and then we define the nodes near the interfaces in terms of arrays? We can have the same running indices of i and j because as per the figure you have shown over here you are using the simple running indices i and j and you are make it generalize. Okay, so the question is about when the programming details, when implementation details although I had not shown in this problem. This question is that I am using a single running indices. So, how do I take care of while programming? I would say that the single running indices is good enough to take care of the application of law of conservation of energy in the different control volumes and we have given you a code. I would suggest if you want to look into the detail the single indices is good enough and we do not need to use different running indices. However, one problem comes that let us suppose I want to know that particular grid points whether it is in aluminum or stainless steel or copper or iron. So, for that you have to see that you have to define certain tags for the running indices, border running indices to take care of so that you filter out that when you want to use k, which k value which you are taking. But the single indices is good enough. However, you have to take a tag near to that for the running indices at the interface. I have three doubts sir. First one is how to use dT by dx equal to 0 for insulated type of boundary condition in discretization. That is first doubt and second doubt is for solving two dimensional steady state heat conduction problem any algorithm is there other than iteration method. And third doubt is in a topic 1 slide number 9 check for convergence is there actually we have to check for convergence or stability. Convergence is the nearest value of nearest value without error and stability is whether it is going properly or not. So, we have to check for convergence or stability that is my doubt sir. Thank you. First let me answer the last question. He has asked that convergence told that this epsilon it is a convergence criteria whether it is a convergence criteria or it is a stability criteria. It is ultimately whenever you are following a numerical method whether if you are following a steady state formulation the solution proceed iteration by iteration. And if you are solving an unsteady state problem it proceeds time step by time step. But in both the cases what you do is that you take the difference of the value between two consecutive iterations if it is a steady state and between two consecutive time step if it is an unsteady state problem. So, in both the cases you take the difference and you try to check that whether that difference is reaching towards 0 and that is what we call as convergence criteria. So, this is known as a convergence criteria not as a stability criteria. Stability criteria is in CFD it is an expression which is used to calculate the time step restriction. So, this is a convergence not stability criteria. Second question is that you have asked for the discrete representation of dT by dx at the boundary insulated boundary condition. So, I will show that in a whiteboard the discrete representation of let us suppose you have a vertical wall which is insulated. So, Qx is equals to 0 in this case. So, Qx discrete the differential representation is minus k dT by dx is equals to 0. So, here dT by dx will come out to be 0. Now the way we do is that there will be a control volume here there will be a grid point here there is a boundary grid point here. So, let us suppose you have a plates like this then this becomes i is equals to 1 this becomes i is equals to 2. So, this is your i is equals to 2 this grid point corresponds to i is equals to 1. So, this we do forward differencing T2 minus T1 divided by delta x is equals to 0 and then it reduces to T2 is equals to T1 and what was your third question I am not or the your second question is for the solution of study state problems is there methods other than the iterative methods your second question is to solve a study state problem whether there are methods other than the iterative methods there are methods definitely non iterative methods, but they are not used commonly in computational fluid dynamics. You can take the inverse of the matrix there are various matrix inversion techniques which you can use, but those are not used the only reason being the coefficient matrix in computational fluid dynamics are sparse in nature. So, we use iterative technique like Gauss-Seidel, Jacobi there are more advanced method like conjugate gradient method, multi-grade methods all are iterative in nature. Thank you. So, we will break for T.