 Good afternoon everybody. I will come back to the post lunch session of the first day. Before breaking for lunch, we had derived the heat diffusion equation and one of the things which we wanted to emphasize again was what is this heat diffusion equation? What are we trying to find out? Essentially, our aim in any energy equation is to get the heat transfer rate. And heat transfer is directly related as we saw in the first lecture, heat transfer is directly related to the driving temperature difference. So, our aim through in any of these issues actually whether it is conduction or convection or even radiation for that matter, we are trying to get the temperature distribution. It means how does temperature vary in the given problem in the given domain of interest. So, whether if it is a conduction problem, how does temperature vary with respect to that space variable which is x, y or z. And if we are able to get a solution to an equation which gives me the final answer in the form t is a function of x, y, z, then I can calculate q by using Fourier's law by getting the differentiation of the temperature distribution. I can calculate heat flux etcetera in all possible direction. So, the bottom line or the main main objective which we are interested in when we derive the heat diffusion equation is to obtain the temperature distribution or the temperature field as we call it. So, we our aim is to not get q directly, our aim is to get t as a function of x, y, z or time. So, this is the time. So, if you are able to get this distribution that means I want t as a function of x, y, z and t which means my differential equation should not be in terms of q which is the dependent variable. What are my independent variables x, y, z, t a combination of these or all of these, these are the independent variables. By independent variables I mean if I have a differential equation d y by d x is equal to 2 hypothetically, x is called as the independent variable, y is called as the dependent variable. This is the dependent variable, this is the independent variable that means what I control and what is dependent on what I control. So, x, y, z and t are the independent variables here and temperature is the dependent variable. So, I want therefore, a differential equation not in terms of q, but actually in terms of t that is the aim. In fact, you start off with this box with this rectangular parallelepiped etcetera where you write everything in terms of q x, q y, q z so on and so forth. This was what you did and when we did that we did not stop with the final form of the equation as d q x by d x plus d q y by d y plus d q z by d z plus volumetric generation term is equal to rho v c p d t by d d time. We did not stop with this form. In fact, I will just go back to the document see here. We came up with this form and right here 2.16 this equation this was not the differential form that we stopped with. This is also ok, but this relates temperature on one side and heat transfer rates on the other side. I do not want heat transfer rate, I want the temperature distribution. Therefore, we substitute q in terms of the temperatures by using Fourier's law and substitute for the heat transfer rates here and then once I finish this mathematics, I come up with what I call as a complete general. This is the most general form of the heat diffusion equation in the Cartesian coordinates. Most general y because when we derive this, we did not make any assumptions on the property k. We took temperature to be a function of space x, y, z and time only we made a statement that the volumetric heat generation rate is uniform. That means it is not that the volumetric heat generation rate is more in one part and less in other part. It is not more here and less here. We said q dot is a constant for the given control volume. So, volumetric heat generation rate is constant. Apart from that, we said that rho C p of course are constant. Beyond that, we did not make any assumption. Therefore, this is the most general form of the heat diffusion equation. This I think all undergraduate institutions teach this. It is one of the favorite questions for teachers to ask, students to derive this etcetera, but when we are deriving it more often than not, we derive it just as a routine exercise without understanding why what is done. For example, this one Taylor series which Professor Prabhu went over that Taylor series expansion is a logical obvious step, but people, students most of the time just what we say memorize this step without understanding what this is and how this is come. This is not come from thin air. This is just a mathematical function f of x plus h is equal to f of x plus d f by d x times h so on and so forth. One of the first things that we study when we do calculus definition of limits, this comes from there. So, this is the most general form of the equation and from this solution of this equation, we can get temperature distribution as a function of x, y, z and time as I mentioned. So, you would get, I would get when I solve this differential equation, I would get t as a function of x, y, z and time. So, I would get t in general as x, y, z and t and if I want this to be a two dimensional conduction, then I would have to eliminate these two. If I want it to be one dimensional, it can just be function of x, y, z etcetera. It need not be a function of t at least for the first part of this conduction course, we would have temperature or a steady state distribution that we are going to talk about. Now, various simplifications, well what kind of an equation is this? This is a second order, why do I know it is second order? Because it is d square t by dx square form. So, it is a second order, obviously it is a non-homogeneous, non-linear equation. So, it is quite a complicated equation in terms of its appearance actually and almost all of us will say I cannot solve this. So, we can solve this numerically, but to obtain analytical solutions to obtain solutions using paper and pen, it is very difficult to do using this equation therefore, we go in for certain simplifications or approximations. So, in fact that is why I said in the morning we need to list down the assumptions that we make when we solve a problem. Are we taking 2 d? Are we taking 1 d? Actually probably in real life the problem might be a two dimensional problem, but if we feel that one dimensional solution is to get an idea of what is the range of this temperature, may be it is off by a factor of 0.2 or something, but that is ok. I will at least get an idea of the range of operating temperatures then may be one dimensional conduction analysis is also ok. So, looking at the problem, looking at the situation, looking at the accuracy needed we can go in for 1 d, 2 d or 3 d or transient mode of solution. So, first approximation or assumption is when k is constant this comes out, k comes out of all the differential sign I get q dot by k here and rho C p by k was defined as 1 over alpha. If I say the steady state problem temperature does not change with respect to time. What do I mean by steady state? If I if I measure the temperature now and measure it say after 15 minutes after 2 hours, tomorrow day after tomorrow whatever it is things do not change with respect to time. That means it is invariant with respect to time right hand side drops off I am left with only left hand side terms right hand side goes to 0 automatically that is the case this is my general form with where we are dealing with steady state 3 dimensional conduction with variable thermal conductivity. Why is it variable? Because k is inside the bracket it is not taken out therefore it is a variable thermal conductivity q dot with uniform volumetric heat generation. So, equation 2.20 represents 3 d steady conduction problem conduction equation with uniform volumetric heat generation and variable thermal conductivity. Now in this I put q dot equal to 0 this goes off I put 1 d in x direction this term that is d by d by and d by d z go off ultimately I am left with d by d x of k d t by d x equal to 0 1 dimensional steady state 1 dimensional x steady state no volumetric heat generation this partial derivatives therefore become regular derivatives because we are talking of 1 d problem and if k is a constant k comes out you can divide through by k I get a very very nice equation that is d square t by d x square equal to 0. Let me just go to the next right here page 2 see we had the most general form d by d x of k d t by d x plus d by d y of k d t by d y I want all of you to write this with me d by d z of k d t by d z plus q dot equal to rho C p d t by d t. So, this is the basic general 3 d unsteady equation with uniform volumetric heat generation and variable thermal conductivity. So, this is the most general equation now I said let us get 1 d form for this. So, if I say steady state. So, when we list the assumptions in a problem we will put steady state 1 this implies d by d t of everything is equal to 0. So, I will cancel this and put this reason as number 1 assumption 1 cancels it this no volumetric heat generation implies q dot equal to 0. So, this goes to 0 this is approximation 2 assumption 3 is 1 dimensional in x implies what d by d x. d y of everything is 0 and d by d z of everything is 0. So, this and this go off because of assumption or approximation number 3 and I will therefore get the differential equation as d by d x of k d t by d x equal to 0. Now, fourth assumption was k is a constant this implies d square t by d x square equal to 0. So, k is a constant this gives me a very easy to solve. Simple what kind of an equation is this second order equation partial or ordinary ordinary differential equations. So, this is second order ODE constant coefficient basically there is one in front of it right hand side is 0 it is a very easy equation for which I know the analytical solution. I know the analytical solution means I can derive this very easily. So, second order equation means I need two boundary conditions. So, at x is equal to x 1 I should specify t at x is equal to x 2 I should specify another t. So, two boundary conditions that means when I integrate this equation when I integrate this equation I will get let us do it first step will give me d t by d x is equal to c 1 second integration will give me t of x is equal to c 1 x plus c 2. So, in order to get c 1 and c 2 I will apply this boundary conditions x is equal to x 1 I will get t 1. So, I will get t 1 is equal to c 1 x 1 plus c 2 t 2 equal to c 2 x 2 plus c 2 you subtract one from the other you can eliminate c 1 you can eliminate c 2 get c 1 so on and so forth it is not a big deal. So, all of us know how to solve this equation. Now, let us just go and do this here it is already done for us I will come back to this problem in a minute. This governing equation we said was second order ordinary differential equation and this one has k inside we have taken k out in this equation. So, 2.220 a is almost identical to the one we wrote on the white board except that the k has been pulled out. What are the possible boundary conditions? I said we need two boundary conditions here and the possible boundary and initial conditions initial conditions are important when we are dealing with a transient or time varying problem means what? If I have temperature as a function of time that means I have where I am saying temperature varies with respect to time then it is only logical that I give a temperature initial temperature. For example, if a metal is coming out from a casting or something it has to be cooled so it is coming out at say 800 degree centigrade that has to be given it cannot come out at 500 and expect the same behavior as you would expect one hour later if it came at 800 degree centigrade. So, we always have to given initial condition in case of a transient problem. So, that is why we say boundary and initial conditions what are the boundary conditions? What are the forms of boundary conditions we can have? One I already told you guys this is this one. So, I have a prescribed temperature at two different boundaries x is equal to x 1, t is equal to t 1, x is equal to x 2, t equal to t 2. What are the other forms? So, let us just go to that because heat conduction equation is second order in the spatial coordinates two boundary conditions must be expressed for each coordinate to describe the system. Suppose I had a two second order differential equation this one I had shown was p square t by dx square. Suppose I had governing equation for example like this I had a plate this is a hypothetical thing, but in 2D conduction we always give this kind of problem this is the coordinate axis t 1, t 1, t 0, t 0. So, obviously this is a two dimensional problem 2D heat conduction problem we know logically that if these two are at higher temperature let us say t naught greater than t 1. So, heat is coming from here and here higher temperature heat flow is going to be like this. So, I have to have d square t by dx square plus d square t by dy square equal to 0 with k is equal to constant. So, this is a second order PDE in x and y. So, for this equation I need four boundary conditions y I need two boundary conditions in x is equal to x 1 x is equal to x 2 y is equal to y 1 y equal to y 2 this y 1 y 2 may be 0 etcetera that I do not care, but definitely let us say if this is a plate of length l and height h I can specify four boundary conditions very easily. So, at this is the x axis at for a y axis we say x is equal to 0. So, x is equal to 0 t equal to t naught y equal to 0 t equal to t naught we got two boundary conditions let us write it down. x is equal to l on this phase x is equal to l t is equal to t 1 y is equal to h t is equal to t 1 y is equal to h represents this phase distance h from here. So, I have four boundary conditions which have been specified for this problem. So, all these boundary conditions incidentally have been constant temperature prescribed boundary conditions, but in real life we can have need not be constant temperature it can be constant heat flux it can be insulated it can be free convection or force convection boundary conditions by that what I mean let us say let us take the same rectangle and I say this one is t naught this is insulated what do you mean by insulated no heat is flowing through the surface. So, instead of t naught I specify this as an insulated boundary condition which means I mean which means I say q is equal to 0 through the surface. So, that gives me an insulated boundary condition. So, we are going to see how to mathematically represent these conditions the commonly listed conditions are given here in this table the first one is the first one is prescribed constant temperature at x is equal to 0 and at all times t temperature is fixed at t s this one prescribed surface temperature prescribed or constant heat flux by constant heat flux condition I mean that let me just go back here constant heat flux condition heat flux condition by that I mean q double prime is equal to constant what does that mean how do I relate temperature and heat flux how do I relate temperature and heat flux temperature and heat flux are related by Fourier's law. So, if q double prime is constant I can basically say what let us just go back here minus k d t by d x at the wall is constant what is that q double prime is what minus k d t by d x by Fourier's law how did we get that q is equal to minus k a d t by d x I have divided by a I got this one. So, this is a constant. So, why have I converted this heat flux boundary condition to a temperature boundary condition because my differential equation was d square t by d x square form we have to remember that the dependent variable is not q, but t. So, though I have a constant heat flux boundary condition let us say solar radiation I have a flat panel this is the collector I think many of you are familiar with solar radiation business. So, I have radiation is incident at the rate of g watt per meter square this is a constant heat flux boundary conditions g is specified and this remains constant at the surface. So, if I put the coordinates like this at x is equal to 0 q is q double prime is equal to g that is not useful for me to solve this equation because it has independent dependent variable as t what do I do therefore is to write this in terms of d t by d x at x is equal to 0 at the wall at x is equal to 0 q double prime is fixed. Therefore, I can relate temperature or its derivative to a heat flux. So, with this in mind again why am I emphasizing this is that you can have heat flux boundary condition, but we have to be able to convert it to a temperature function. Now, if the surface is insulated that means there is no heat transfer to the through the surface insulated or adiabatic what is adiabatic thermodynamics tells me adiabatic means something where there is no heat transfer. So, in adiabatic or insulated surface implies q double prime of s equal to 0 very nice why is it very nice because this tells me d t by d x d t by d x is equal to 0. So, very powerful equation I have this q double prime is minus k d t by d x at x is equal to 0 I have this equation I am just setting the left hand side q double prime to 0 k you can divide through by k I will get d t by d x at the wall is equal to 0 what is this d t by d x at the wall temperature gradient at the wall is equal to 0. Remember fluid mechanics we had d u by d y d u by d y was related directly to the shear stress here also d t by d x we are calling we can call that y basically gradient gradient of temperature with respect to the coordinate direction that is directly related to the heat transfer rate we will see that later. So, d t by d x equal to 0 is a very very useful boundary condition not anywhere it is at the wall this is for insulated boundary condition. And it essentially tells me that at the surface and just before that there is no temperature gradient last of the this is called boundary condition of the first kind this is called the boundary condition of the second kind this is a special case of 2 2 is constant wall heat flux case finite heat flux. And when the finite value is turned to 0 you get this one d t by d x is equal to 0 and you can see very nicely it is shown these are all taken representations from various text books only they also reproduce them this represents the slope d t by the blue line here represents the slope d t by d x essentially is a slope of the temperature distribution where is it is not at some internal location it is at the wall location. So, d t by d x at x is equal to 0 is nothing but q double prime s divided by k with a negative sign the slope is related to the heat flux directly. And when the slope is equal to 0 it means that look at the curve here it is asymptotically it is basically the blue line is at tangent to the temperature distribution. Last boundary condition is again a specific form of this heat flux if this heat flux essentially is because of convection I can relate the left hand side minus k d t by d x that is x is equal to 0 to the convective heat transfer coefficient that is h times t infinity minus t of 0 comma t what is it looks complicated it is nothing it says conventionally of course, this arrows here show that there is a flow of a fluid with temperature t infinity and it is giving you a heat transfer coefficient h associated with it. Now, in conduction heat transfer when we are doing conduction now this h is given to us a priory that means it is already known we do not have to compute this. So, 25 watt per meter square Kelvin these are all known quantities once we go to convective heat transfer this same conduction problem where we are we were given flow over as a reactor fuel rod was cooled with a convective heat transfer coefficient of 100 watt per meter square Kelvin. The same problem you can have where h has to be calculated and by an appropriate correlation you will get h and then we solve the problem for now in conduction this is specified h t infinity is specified and we are looking at heat flow from left to right why because temperature is decreasing in the positive x direction. So, h times t infinity that means this fluid temperature has to be larger than the surface temperature here. So, t infinity minus t of 0 comma t why is this 0 because it represents x is equal to 0 and t is for at all times this boundary surface temperature. So, convective heat transfer coefficient times t infinity minus the surface temperature at all times that is equal to minus k d t by x equal to at x equal to 0. So, these are given special names first condition is called boundary condition of the first kind which is called Dirichlet's condition. Second condition where constant heat flux is specified it is called as the Neumann condition, but these are all information that is given to you special case of the constant heat flux is adiabatic or perfectly insulated case and convective heat transfer condition is the most general and obviously as you can see it is a slightly more involved boundary condition why is it involved we will just see convective heat transfer condition is minus k d t by d x at x is equal to 0 is equal to h times t infinity minus t at 0 comma t both these represent surface temperature at the wall x is equal to 0 this is the solid here this is the fluid here. So, I have temperature at this wall which is to go here the gradient of the temperature at the wall actually I do not know that I have to calculate that. So, as you see here this quantity and this quantity is related both these are directly related when the right hand side was 0 it was very easy I could write d t by d x equal to 0 I can relate this temperature at this location to a temperature at another point slightly away from it by d t by d x equal to 0, but here I cannot do that very easily because this derivative also has this unknown temperature in it. So, it is a slightly bit more involved boundary condition nevertheless it is very useful it is used most of the time in fact we will write recast this at x is equal to 0 is equal to minus h over k t infinity minus t at x is equal to 0 comma time. So, this will be recast like this and you can apply it to the solution of the equation. Now, we go back to the problem that we the problem that we had left the this problem we had intentionally left out because I did not talk about the boundary conditions then. So, this is a temperature distribution across a wall of one across a wall which is one meter thick at a certain instant of time. It is actually most of these problems you know are very very straight forward and it turns out that these are exercises in understanding English. So, if we are careful in interpreting each and every word I think all of us will get full marks in any course problem is we are unable to interpret things correctly. So, temperature distribution across a wall what is given a one meter thick wall is given here shown in blue here at a certain instant of time. So, at some instant of time it is given as t of x is equal to a plus v x plus c x square now a question will rise. So, at another instant of time is it different yes the answer is yes because what has been done is at a given time when these conditions were there this is the expression or this is the curve fit to the temperature distribution that has been obtained and given to us a plus v x plus c x square where a is 800 b is minus 350 and c is minus 60 and temperature is in degree centigrade x is in meters. So, it is a polynomial curve fit which has been given at a given instant of time notice that this one does not have a time dependency why because time dependency has already been observed when I say that temperature distribution at a given instant of time when I say at a given instant of time t equal to 15 seconds after start of the process t is equal to 500 seconds after start of the process. So, I have specified a time and then given you this distribution whereas, if I say the general temperature distribution of for this surface with respect to time is then this equation is not valid there has to be a time factor in the general expression. I hope I am clear in this I am sure there will be some questions related to this problem keep it waiting we will take it later. So, somebody has done a curve fit for this problem where t of x has been given to us a uniform heat generation of 1000 watt per meter cube is present very nice. So, e dot g is given to us in the wall area 10 meter square. So, this is some furnace wall or something whatever it is there is a wall whose area into the plane of the board into the plane of the paper is 10 meter square. The properties of this material are k is equal to 40 watt per meter Kelvin constant density there is a row missing here. So, 1600 kg per meter cube and c p is 4 kilo joule per kilogram Kelvin the figure c p units are incorrect 4 kilo joule per kilogram Kelvin. So, there is a kg missing in the figure what are we supposed to find determine the rate of heat transfer entering the wall at x is equal to 0 and leaving the wall at x is equal to 1 meter. What is this they say please find rate of heat transfer that means wattage what is the watts for heat coming in at this boundary and what is going out at this boundary. We will come back to this question second one the rate of change of energy storage in the wall the third is time rate of temperature change at x is equal to 0.25 somewhere here and x is equal to 0.5. So, very interesting problem actually. So, what is asked first have you understood what is given to us. I reemphasize we have been given a temperature distribution which appears steady state that means it is at a given instant of time very nice. But we are asked to find rate of heat transfer entering the wall at x is equal to 0 and leaving the wall at x is equal to L will these be different. On first thoughts that means can I just look at the problem and just want to blindly solve it the answer would be yes these both are equal because we are talking of steady state at that instant of time. But we have to remember there is this E generation term which is sitting here that means some heat is coming in some more is generated. So, it has to be this what is going out has to be different it obviously is not going to be the same as what is coming in. So, if you are able to understand and approach the problem we will not mean for surprises when we get mathematically that these two numbers are different. So, if you are fixing in our mind that these two have to be same then we are going to look for mistakes that we have not made and try to make these two equal. So, with this background let us see assumptions 1D conduction in x direction yes we are not concerned with 2D homogeneous medium with constant properties uniform volumetric heat generation constant properties means k rho C p everything has been taken as constant uniform heat generation that means this q dot which is given is uniform everywhere. And this temperature distribution is coefficients are all known to us. So, first part is very straight forward rate of heat transfer entering the wall and leaving the wall. So, entering means what coming from left to right this way I have a temperature distribution for the solid. So, Fourier's law gives me the heat transfer rate q dot sorry q is equal to minus k A d t by d x. So, if I have t as a function of x which I have here d t by d x if I calculate I should just apply the formula and get q let us see what we have done. So, d t by d x has been evaluated first so I will just do it here in a minute. Temperature distribution temperature distribution is A plus B x plus C x square d t by d x why do I write a partial derivative here? So, remember though the temperature distribution looks to be a steady state distribution we when we wrote the when we read the problem we said at a given instant of time has been given that means at another time the functional form is not going to be this. So, therefore t is essentially actually it is a function of x and t actually it is this, but when we fix this t becomes only a function of when we fix a value of t when this is given t becomes only a function of x. So, it appears as a steady state solution, but actually it is a transient solution at a given instant of time. So, derivative will give me B plus 2 C x. So, heat transfer rate at x is equal to 0 is minus k A d t by d x at x is equal to 0 I have k I have A I have d t by d x when I substitute d t by d x at x is equal to 0 I will get B plus 2 C times 0. So, this term will vanish off and I will get minus k A times B. So, Q at x is equal to 0 is minus k times A times coefficient B same expression I can use for x is equal to L Q is equal to minus k A. So, B plus 2 C L so, beyond this there is nothing to write I have this equation minus B k A put in the numbers 140 kilo watt is coming in from the left hand side outgoing is 188 kilo is expected coming in is 140 outgoing is 188 why is it so? It is obviously going to be larger because 140 has come in some 100 watts has been generated per unit volume I do not know that is 100 let me look at the number it is 1000 watt per meter cube there is an error in the figure this is watt per meter cube here 1000 watt per meter cube has been generated and that is going to some of it is going to go out therefore, I am going to have a Q out which is larger than Q in. Now, this itself will tell me that we are going to have a temperature change with respect to time because 1000 watt per meter cube is generated and 140 came in 188 is going out. So, there has to be some kind of storage so, we are going to have a temperature change. So, that is what is calculated in part 2 calculate the rate of change of energy storage in the wall right hand side of the energy balance E dot in minus E dot out plus E dot generated equal to E dot stored. So, we are asked to find rate of energy storage or change in energy of the wall. So, E dot stored therefore, is this E dot in I have calculated 140 E dot out is 188 E dot generated Q dot is watt per meter cube times the volume associated with this geometry area is 10 meter squared times 1 which is 10 times 1 10 meter cube is the volume. So, 1000 times 10 meter cube is here that is what is multiplied here. So, E stored comes out to be minus 38 kilo watt very interesting because see energy coming in 140 energy going out 188. We said there is a heat generation and is it contributing to a rise in temperature or a drop in temperature that is clear when you look at this term. So, just by looking at these 240 and 188 we will not be able to say whether there is a rise in temperature or a drop in temperature that will become clear only when I put in this balance equation. So, it comes out that E dot stored is negative which means rho V C p d t by d t is negative all the terms except d t by d t are positive. So, d t by d t comes out to be negative it means temperature is going to drop with respect to time. So, last part yes we got an answer for that temperature drops with respect to time we are asked to find time rate of change 2 changes are here time rate of change of the temperature at any point in the medium may be determined by the governing equation this one just 1 d transient equation has been recast what is it rho C p by k d t by d t equal to d square t by d x square plus q dot that is the governing equation I have I want d t by d t. So, everything else comes to the right hand side and therefore, I can write d d square t by d x square I will write it in a form which is convenient to me second derivative of the temperature I have temperature distribution given to us already first derivative was calculated first derivative was calculated here B plus 2 c x second derivative will come out to be B derivative of B is 0. So, I will have 2 c is the second derivative this is going to be used directly there 2 c. So, d square t by d x square will come out to be 2 c substitute that here q dot is known 1000 watt per meter cube put in all the numbers we get d t by d t is minus 5.94 times 10 to the minus 4 degree centigrade with respect to per unit time or per second means there is a drop of about 0.006 degree centigrade per second it does not seem much, but if you are going to have this process go on for an extended period of time say per hour that means you are having 3600 seconds then this could start becoming a significant quantity. And what this problem also tells me that merely looking at these 2 numbers I cannot get an idea of whether it is heating or cooling I have to look at this complete form of energy balance minus sign here e dot stored indicates that temperature is going to drop and mathematically if I made a mess here and I got positive 5.94 we should wake up and say whether this is what is correct because this involves energy balance and this there is some mistakes we have to go back and look at these things. So, double multiple ways of checking the same thing conceptually Fourier's law can always be used to compute and it is evident that temperature at every point of the wall is decreasing with respect to time. So, I can take couple of questions if there are I see 3 questions here let us quickly go to Amal Jyoti Kerala, yeah please go ahead. So, my question is that sir in the derivation of the heat conduction equation. So, we have taken the amount of energy which is stored as rho Cp dot t by dot t one term is the final which is on the right hand side rho Cp dot t by dot t d x d by d z. So, in that term it is given as the it is the internal energy which is stored in the material. Now, internal energy is defined as it is m Cp d t sorry m Cv d t. So, why do you take as Cp instead of Cv? Suppose you are having a material gas which is a stationary gas condenser we can take it as Cv you know rather to take it as Cp over. See rho is m rho times v is clear rho times d x d y d z that is m. Now, if you are conduction if you are talking of inner solid Cp Cv is all the same if you are talking of a gas then Cp and Cv will be different, but Cp and Cv we are saying is energy change with respect to temperature. If you are talking of gas where you are dealing with conduction and there is no you know what you say precisely only conduction then we have essentially to if you are dealing with internal energy it has to be Cv. If there is no flow Cp and Cv will still become the same thermodynamics if you see what is H? H is Cp times t u is Cv times t, but what is H and u related? H is equal to u plus Pv and that Pv essentially has come because of the flow aspect of the fluid. So, if I am not going to have a fluid if you have a flow you can actually take Cp Cv all coming out to be the same. So, Cp Cv let us not get into the integrities of it it is the change of energy with respect to time. The moment I have a flow then I cannot use Cv I have to use H plus therefore, in fact we do not use this form of the equation at all we will have to go to the complete derivation of the actual equation for conduction. Conduction if you are talking it is next question and I take a start in fraction. Hyderabad. Hello sir this general equation, can I make use of it for a molten metal? Molten metal means there will be flow right. Yes, yes sir. Hello actually to answer your question for molten metal I cannot take directly you see why because in a molten metal that is what standard Stefan problem is told for a Stefan problem you have both the molten metal that is the you have both solid and liquid as the interface is moving see if you have both solid and liquid you have half of the portion solid and half of the portion liquid. See let us just take standard Stefan problem which is called let us say this is my plane wall this is a container in which the conduction is taking place only in this direction. And this is let us say I have applied either constant wall temperature or constant heat flux boundary condition earlier it was solid. Now after certain time it just melts that is this complete thing has become now liquid and this is solid. Now to answer your question within liquid there is conduction and convection because there is natural convection but within solid yes I can apply Fourier's law of conduction within liquid yes I can apply Fourier's law of conduction but in addition to conduction there is natural convection. So, it is very hard to answer your question directly there is no one way of answering it. So, it depends how we are handling or which type of problem we are solving. So, we cannot to put it very simplistically we cannot use it directly because there is convection. For the entire volume. For entire volume it is not possible to use Fourier's law of conduction or the heat diffusion equation. How about heat transfer between this molten material and the solid material. They are also there also you can only apply the boundary condition but you cannot apply the heat diffusion equation. For liquid zone for liquid zone how can I apply heat diffusion equation because natural convection is taking place. See in the liquid zone I think it is not possible for us to apply heat diffusion equation only for the solid zone only for the solid zone we can apply the heat diffusion equation but not for the liquid zone why because there is natural convection there are fluid velocities which are involved. So, we cannot apply heat diffusion equation for liquid but we can definitely apply for solid but the solid is moving I mean the solid portion is decreasing. But as long as we know which is the solid portion I can always apply the heat diffusion equation. At the interface I can only apply the boundary condition not the heat diffusion equation. So, we will have minus k d t by d x on the left hand side should be equal to minus k d t by d x on the right hand side whatever be the appropriate values of the thermal So, precisely what is called as Stefan problem it is a moving boundary problem ok. Thank you. Yeah, VIT Pune please. Hello, my question is there will be problem 3.1 which we have solved. In the third parameter that is we in that in that in that we calculate that is time rate of change of temperature but in this equation while calculating daba 2 by daba 2 t by daba x square we have a x that is and which is which is basically for the I mean the steady state temperature dispatcher how can we use this particular temperature distribution for calculating the unsteady state equation. The question is that in this problem which we just solved problem 2.1 part 3 this question is though we are saying it is a steady state distribution I am I am just phrasing what you have told it is though it is a steady state temperature distribution how am I using the unsteady form of the governing equation the slide which which has part 3 how can I use dT by dT equal to ok. The answer to that is actually that again I have told you this part this problem it is a local steady state that means dT dT by dT is going to represent a change in temperature with respect to time. Now at this instant of time I have this particular temperature distribution that is all I am getting by this A plus B x plus C x square at another instant of time because of this kind of energy transfer which has happened and e dot stored which has become minus 38 kilowatt I am going to have a different temperature distribution. For example, if I want to draw some kind of arbitrary temperature distribution this need not be representative of this problem. Let us say I have some some wall like this and at a given instant of time T equal to T 1 I have this kind of a temperature profile all I am saying this this has been given to us T of x is equal to A plus B x plus C x square this is given at this time. Now what we are saying is this is a steady state distribution at T equal to T 1. Now dT by dT is contributing to the shift or to the change in the temperature distribution or the temperature profile because of the fact that the problem is unsteady because we have e dot stored is equal to minus 38 kilowatt what is this going to contribute this is going to contribute to a net drop in the temperature. So, how is that drop in the temperature manifested that drop in the temperature is going to get I cannot have a same curve which is going to say that it represents a new temperature distribution at a new time no it has to shift and probably it will shift something like this or I do not know how what is the nature, but there will be a shift for time T equal to T 2 where new constants will have to be evaluated it may not even have the same form A plus B x plus C x square, but this thing is these two are related these two curves are related by energy diffusion equation definitely because that is the unsteadyness associated with dT by dT rho C p dT by dT equal to d square T by d x square plus q dot by k and here also there is a k. So, time T equal to T 1 and time T equal to T 2 I am relating when I am going to say what is the rate of decrease of temperature with respect to time because that is what T 2 minus T 1 divided by delta T I am bringing in time here. So, I have to use the unsteady nature of the governing equation that is why I have to use this form.