 we will take over from where we had left that is so far till this problem we have considered effect of we have considered lumped capacitance. That means the entire object entire body is that uniform temperature and we developed some kind of a criteria for this lumpness so called lumpness that is typically shown by a diagram like this where entire object specially is at uniform temperature. But so what does it mean situations where lumped capacitance is not valid we do not solve the problem now that is not the case. So, what we do is situations where lumped capacitance is not valid or bio bio number is less greater than 0.1 we have to take into account the special variation in the temperature which means we have to take into account the fact that temperature is going to change with respect to the coordinate position and apply not so simplified lumped capacitance approach, but a slightly more elaborate approach to try to solve the problem. I am not sure how many colleges how many universities cover topics beyond lumped capacitance and transient conduction for those of you who cover it definitely is going to be very useful for others who are not covering this topic as a part of their undergraduate curriculum. Still I am sure this is going to be useful because we are going to cover certain mathematical formula certain charts and graphs which we can use directly for getting the solution it is not complicated it is just initially it looks little bit cumbersome, but it is not so. So, what are we doing here spatial effects means temperature change with respect to space with respect to the coordinate position that is what is meant by spatial effect. So, if I have if you look at this diagram here you have a plane wall which is a large plane wall you have a long cylinder you have a sphere now why is this a large plane wall why is the word large important why is the word long cylinder important that we will we will see when we come in and deal with a plane wall of you know finite dimension or a short cylinder at that time we will see because in case the dimension in the other direction let us take a cylinder this radial direction is here and the length is comparable to the radius then my two multidimensional or two dimensional conduction effects come into place. So, I will have a transient two dimensional conduction problem if I have a l by d ratio of the order of 1. So, that is why we are not dealing with short cylinders or plane wall of a finite dimension when we say a plane wall large plane wall it means it is of a finite thickness and extends in both directions to whatever length which we are not concerned with. Similarly, long cylinder would refer to a cylinder whose length is much greater than the characteristic dimension which is the diameter and of course sphere there is no such issue. So, what are we doing here we are saying we want to take care of variation of temperature with respect to space along with variation with respect to time what did we do in lumped capacitance in lumped capacitance we said well temperature is uniform. So, if I if I go to the white board and try to draw this for you in case of lumped capacitance method we said temperature at all x y z is equal to t at any given instant we did not have different numbers. Now, what we are saying is spatial effects when I am talking of spatial variation in temperature I am going to say that t at is a function of x probably a function of y and function of z, but we are restricting ourselves to only one dimensional because we are considering a long cylinder or a large plane wall with a finite characteristic dimension. And we want to take the effects of this variation in the heat transfer direction it is not some random direction it is the heat transfer direction. So, what it means is let me just go to the next figure that is a little bit this figure is actually very nice maybe I will draw this and explain that is better. Suppose I have a plane wall and center line is shown here let us say initially it is at temperature t i. So, at time t equal to 0 temperature is equal to t i and all through this surface I mean all through the plane wall my temperature is uniform because you are not started any change in the boundary condition. Now, let us apply a convective boundary condition let me say sorry this is t i is here let us say t i is here it is high. So, we will not look at this we will look at t i being high the body is at a high temperature initially. And now I am going to establish cooling at t greater than 0 at the boundary surfaces both surfaces I am going to have convective cooling given by h comma t infinity h comma t infinity at the this let me call x is equal to 0 and this is plus l this is minus l at the surfaces x is equal to plus minus l we have h comma t infinity as the boundary condition. So, at a given instant. So, this is a common thing which you which we will have in case of cooling quenching of hot metals metallurgical various metallurgical processes we are going to have this kind of a phenomena. So, it is not something which is out of ordinary we are going to see it every time. So, one hot body is suddenly subjected to a cold environment and we want to see what happens to the temperature what will happen I do not know any heat transfer I just know logic. So, let us use logic and try to find out what is going to happen. First thing I have applied a convective boundary condition at the free surface the complete layout is here. So, I am going to now say this temperature was the this temperature was the initial temperature I want to see let us say let us even take a number say 600 degree centigrade and h t infinity t infinity is say 30 degree centigrade. So, when I apply this boundary condition what happens logic tells me that first the two phases are going to reach the fluid temperature. So, it is not going to happen instantaneously. So, I have applied h comma t infinity what will happen is just at t greater than 0 I will have the boundary coming to t this is t infinity let us say this is t infinity here. This portion alone x is equal to l and minus l this will come to t infinity and everywhere else the temperature is going to be high. So, from the surface to the inside the effect of the change in the boundary condition is going to propagate that means at any distance at a small distance inside the wall inside the solid I would have a temperature slightly greater than 30 further inside I am going to have a temperature slightly greater than 30 further much more greater than 30 and at some distance you will see it is almost equal to the initial temperature of the solid. Why go to all these examples those of us who have cooked will know if you have boiled a full potato in the pressure cooker and you are trying to peel it common sense tells me you put it under tap water. So, outside surface is cool you try to peel it and then you try to you want to break it into small pieces. So, you use your thumb to press it and you will see the inside is still very very hot that is exactly what is happening here. You have t infinity the effect of the cold boundary condition which is propagated not all the way through in the in just because you have put cold water on the potato you have your surfaces become relatively cooler or maybe it is become much more easy to handle. So, you peel it off the surface is cool tap water is running you feel nice. So, it is become cold I will now be able to crush it the moment you go a little deeper you feel the hotness still there that is precisely because of this reason anyway. So, this is for let me call this as curve 2 at t slightly greater than 0 when the after boundary condition is imposed. Let us say t equal to t 1 we get curve 2 if I go further in time that means if I am going to say now t is equal to t 2 which is greater than t 1 the effect of this change in the boundary condition is going to propagate further. And I will see a curve which is generated like this that means further inside from the boundary the effect of the coldness associated with the boundary condition is felt. So, this curve 3 can be a situation where you have still some portion at t i depending on your time and depending on your boundary condition it can be a situation where even the center line portion where the hottest portion was there is has started to come below t i. Now, let us just go and look and see whether we are doing it correctly yes we are doing it correctly t equal to 0 t i is the horizontal line at t greater than 0 I see the first curve which is going like this and coming. Now, I come back at t equal to t 3 which is greater than t 2 I am going to draw curve 4 which it says that something like this this is curve 4 that means the effect of the cold boundary condition has propagated further inwards and has caused the center line temperature also to drop from t i value further increase in time if I wait for some more time this effect would be felt further you get a curve which is resembling curve 5 that center line temperature has dropped what am I keeping constant here this is anchored this is t infinity this is anchored here that does not change because that is the boundary condition t infinity is fixed t i is coming down. So, this curves come in as t increases. So, with increase in t my nature of the temperature distribution I do not know any heat transfer I am just using common sense it is going to come in further and if I wait for time t equal to infinity that is very long time I will achieve a temperature distribution which is almost uniform equal to t infinity whether we want that or know that we will see later, but qualitatively temperature changes in this manner in case of a situation where spatial variations are important that means for cases where bio number is large is greater than 0.1 0.1 was the criteria we have chosen here we are having bio number greater than 0.1 for bio number greater than 0.1 this is the nature of the variation. So, you start with almost uniform temperature further increase in time after application of the boundary condition you get curve 2 next you get curve 3 4 5 6 7 so on and so forth I have just drawn few representative curves but the point I am trying to make is the temperature trend if I close this part and have a line of symmetry here we notice that the curve is symmetric about the center line why is it symmetric it is symmetric because our impose boundary conditions are symmetric h comma t infinity is same on both sides of the plane wall if it were different then this half would behave slightly differently this half would behave slightly differently you would have a non-symmetry associated with the temperature distribution, but the overall behavior is going to be the same. So, the maximum temperature in that case would in a non-symmetric boundary condition would occur on the side of the one with the lower heat transfer coefficient. So, if the right hand side h is low if this was low then the peak would be on the right hand side if this was if the left hand side was low the peak would be on the left hand side nevertheless you would have a temperature distribution which looked like this in this case we are doing the symmetry boundary condition so we get a nice set of curves. So, that is what is shown in this graph and as t tends to infinity you come up with a horizontal line. So, all that is given in words here when the wall is first exposed to a surrounding medium t infinity greater than t i at t equal to 0 the entire wall is at initial temperature once the boundary condition is established flow is established wall temperature at and near the surface starts to drop because of increased heat transfer creates a temperature gradient what can we say about the temperature gradient what is the temperature gradient temperature gradient is nothing, but the slope the steepness of this curve very high temperature gradient initially temperature gradient is going to decrease as time progresses can that tell me something about the heat transfer rate yeah because of temperature gradient d t by d x is directly related to the heat transfer. So, larger the temperature gradient larger is the heat transfer that is why when you have this potato and you have put it under cold water you feel things are very nice because lot of heat has been taken off. So, the surface has gotten cool very nicely and the moment you stick your hand in and try to open it you see the effect of the special non uniformity that is why you have to spend if you want the entire potato to become cool you will realize that you have to spend lot of time holding it under the cold water rather if you just want the surface to be cool to get a reasonable temperature it requires a very less amount of time. So, heat transfer rate is also non uniform because the temperature gradient is non uniform heat conduction from the inner parts towards the outer surface note that temperature at the center point remains at T i until some finite time that I have already told. So, depending on the time at which you are plotting you will have a large portion of the near the center still at the initial temperature temperature profile gets flatter as time progresses and results in and as a result of heat transfer eventually becomes equal to T infinity which is the horizontal line wall reaches thermal equilibrium with its surrounding that typically happens after a very very long time that the wall reaches thermal equilibrium at this point there will be 0 heat transfer because there is no driving temperature difference between the fluid and the solid. So, similar discussion of course we can have for other geometries now what we are trying to say because of all this is all this I can write a story write a book how do I solve this mathematical for mathematics we still have the same heat diffusion equation this is something which those universities those colleges which cover this transient conduction spatial effects etcetera. You can have questions of this type where you ask for qualitative variation sketch this draw this please explain like this rather than asking them you know to write just instead of just writing graphically explain what happens when this kind of boundary conditions are imposed if bio number is greater than 0.1 give them a non symmetric boundary condition and see how they are able to do it in terms of the explanation. So, that you can test them on this concepts also so the governing equation for this case of course is the same whether it was lumped or non lumped it is the same equation, but now we have initial and boundary condition which are given like this initial condition at T at x is equal to 0 and sorry T at all x and time equal to 0 temperature initially time T equal to 0 and all x everywhere is equal to T i second one boundary condition this is an initial condition initial condition is required for all problems involving transient behavior. So, we want to know where the problem has started from at what temperature was the initial was the surface initially. So, that is why we have to give a initial condition boundary conditions we need to because this is second order in space first one of course is a very nice one easy symmetry boundary condition dT by dx is equal to 0 at x is equal to 0 this is true for all times. So, at all times so I will write this as dT by dx at 0 comma T equal to 0 I will write this in this fashion. So, initial condition T at all x and time T equal to 0 is T i boundary condition first one dT by dx at x is equal to 0 and at all times is equal to 0 this boundary condition is possible only only only if I have symmetry on for the size if I have the same boundary condition imposed on the exposed surface of the plane wall remember we had the plane wall x is equal to minus l n plus l was exposed to the same h and same T infinity only then I will see this symmetry here. If I have h 1 T infinity 1 and h 2 T infinity 2 I will not be able to write this boundary condition like this in that case I will have to know the line of maxima or minimum maximum temperature based on some kind of a analytical solution. So, I have to have a line of symmetry x is equal to some value depending on the boundary condition. So, that becomes a little bit more complicated right now let us just stick with the symmetrical boundary condition therefore, this is the boundary condition as a result of that symmetry the second boundary condition in space is minus k dT by dx at x is equal to l or minus l does not matter whichever you say is a convective boundary condition h times T at l special location is l at all times that is why the general time is used minus T infinity what are the assumptions here h on both sides is the same h is uniform throughout T infinity remains fixed of course properties are all constant. So, this body could be at 800 degree centigrade initially is coming down to 30 degree or 100 degree centigrade after a long time still we are saying this rho v C p all these properties remain the same with respect to this problem. So, that is a big assumption in reality it would not be the case we can account for this of course, once we get if you are going to solve this numerically or anything you can have a next temperature at the next time next time so on and so forth and once you can you get that temperature you can iterate you can get the new set of properties and apply that whatever be the case this is the boundary condition that would be there and we have to solve this how do I solve this. So, the differential equation and the boundary conditions are given by this bio number is greater than 0.1 spatial variation in temperatures do exist and this is the nature of the distribution now to solve this problem we have to take recourse to some kind of a series solution. So, that is what is given in the next few pages and many universities where post graduate heat transfer is there definitely cover this topic. So, for them definitely it would be useful others if you are not seen this I am sure there will be lot of questions. So, please make a note and ask them after the discussion is over on this topic. So, now let us get to the solution there are two methodologies one is the analytical solution which is obtained by using paper and pen the other is a graphical solution both are not two different solutions it is the graphical solution is just a graphical representation of the analytical solution. So, that you can if you are hard pressed for time you can just use the graph and get a approximate answer. So, what we do first thing we do is we have to non dimensionalize this equation. So, the formulation of the problem for determination of 1 D transient temperature in a wall results in a PDE which can be solved using advanced mathematical techniques. The solution however involves infinite series. So, those of you have mathematics you would have seen solution of this would have infinite series which are inconvenient and time consuming to evaluate. So, what do we do we do not do it that is not the thing we say that we can do it with restricting ourselves to 1 or 2 terms in the solution or go to the graphical approximation etcetera. So, what are the variables that are involved parameters are x independent variable time another independent variable L which is the dimension characteristic dimension k alpha which are material properties T i T infinity initial temperature boundary temperature H which is the heat transfer coefficient which are far too many we have 1 2 3 4 5 6 7 8 variables to handle. So, the obvious logical thing to do is non dimensionalize the governing equation. So, what do we do we say let us take a non dimensional temperature generally when we non dimensionalize any quantity we try to take the two extremities associated with that quantity. So, non dimensional parameter typically let us take for example, length scale x divided by L. So, when x is equal to 0 when you are at the center line this non dimensional capital X would become equal to 0 when small x is equal to capital L the non dimensional capital capital X would become equal to 1 or minus 1 depending on which side you are. So, that is the way non dimensionalization is typically done temperature the largest temperature difference you have is T i minus T infinity whether it is heating or cooling this is the largest temperature difference and local or T at any location at any given time minus T infinity divided by T i minus T infinity is your non dimensional temperature. Bio number is a non dimensional heat transfer coefficient do not take this like the you will see Nusselt number later on. So, do not get confused with this this is only for this lumped capacitance approach H L c by k of the solid and last thing is the non dimensional time which is called as the Fourier number alpha T by L c square. So, we have 8 variables combined together as 4 theta capital X bio number and tau. So, what do we want in doing all this once we start mathematics most of us forget why we are doing what we are doing we want let me write down what we want we want T as a function of X comma T this is what I want let us not lose focus of this. That means in a non dimensional thing I want theta so, if I want theta as the dependent variable. So, it has to therefore, depend on the all other parameters we know Buckingham pi theorem says many pi groups are there you can relate one with respect to the other same thing I am doing here theta is a function of what are the other non dimensional variables capital X this involves small x and L bio number it involves H L c by k what was the other number non dimensional time which is Fourier number which is alpha T by L c square. So, this is T minus T infinity local divided by T i minus T infinity is therefore, of function of all these 3 parameters and that is what we have written non dimensionalization enables us to give a temperature in terms of only 3 parameters capital X bio number and non dimensional time. So, what I do I have non dimensional I want to cast this let us actually look at what has happened. So, a partial differential equation with two independent variables X and T and one dependent variable capital T you will see now it is going to get transformed into something else. So, these were the boundary conditions T of X comma 0 is T i d T by d X is equal to 0 minus k d T by d X at X is equal to L is this these are the boundary conditions we wrote. Now, I am going to transform this equation by using the non dimensionalization which I have given and let me start here itself. So, theta is given like this. So, d T by d T has to be transformed to d theta by d tau because we do not have T anymore we do not have small t anymore we have replace this by theta we have to replace this by non dimensional time. So, let me take a look d T by d tau d time therefore, can be written as question mark for this we need to go to the definition of these quantities theta is equal to T minus T infinity by T i minus T infinity all of you please do this with me. So, I can write I am going to do this step by step later on when we do non dimensional analysis for convection we are going to do it fast because this is probably the first time we are doing this T i minus T infinity times theta plus T infinity is equal to T. Therefore, also we have alpha T by L square is equal to tau implies T is equal to L square tau by alpha. So, d T by d T is equal to T infinity therefore, can be written as if I take derivative of this this is my temperature expression in terms of theta derivative of T infinity is 0 whatever be the independent variable because it is a constant T i minus T infinity is the constant it is a number initial temperature minus the free stream temperature. So, I can write this as T i minus T infinity d T would be replaced by d theta now derivative with respect to time we have to take now. So, I am going to transform time in terms of this one. So, this d T would be equivalent to L square by alpha. So, this is going to be L square by alpha theta tau. Therefore, I would get alpha T i minus T infinity by L square d theta by T tau this is my transient term in the governing equation which I have non dimensional. Let me just go back and see if I have done it correct we do not have the intermediate steps, but let us proceed. Second thing is the left hand side which is d square theta by d x square I want to non dimensional as this. So, d square T by d x square is nothing but d by d x of d T by d x and this I can write what was x how was x related let us go back to the non dimensionalization x was directly related to the length scale like this. So, capital x is equal to small x by L. So, capital x is equal to small x by L therefore, d capital x is equal to d small x by L I take the L on the left hand side I will have this. So, now I will write. So, this is going to be d by d x small x I will keep it like this d T by d x I have temperature here the derivative with respect to x or whatever is going to be just T i minus T infinity d theta d by d x would become L d by d capital x this quantity can just come out. So, I will write this as T i minus T infinity by L d by d x here can therefore, be replaced as this way and when I just go to the next step I will write this as T i minus T infinity by capital L square d square theta by d capital x square equal to d square T by d x square small x square. So, this was the left hand side of the governing equation right hand side we already wrote d T by d T is equal to I have already written that as alpha T i minus T infinity by L square d theta by d tau. So, what was my fundamental governing equation basic governing equation is this one. So, I am going to now ask this in a nice form put this non dimensional things together I will say d square T by d x square is the left hand side that is nothing but T i minus T infinity by L square d square theta by d capital x square is equal to 1 over alpha d T by d T I will substitute as alpha T i minus T infinity by L square d theta by d tau alpha cancels T i minus T infinity by L square occurs on both sides that also can be cancelled off and my governing equation therefore, looks like this. So, what have I done one p d e to another p d e that is all why why did I do so much why did I talk so much about this. Well this equation if you look though it is a partial differential equation with capital x and tau as the independent variables and theta as the dependent variable it does not have the effect of all other parameters which were sitting k h so on and so forth that is why this is a far more convenient equation the boundary conditions also have to be transformed. So, we transform the governing equation we have not transformed the boundary condition let us go back to the boundary condition T at x is equal to 0 all time is equal to T i that is very easy to transform go back to the substitution T i minus T infinity by T i minus T infinity theta would be equal to 1 that is what we have here theta equal to 1 at any small x it is equivalent to any capital x time T equal to 0 time T equal to 0 same thing except that instead of T i it is becoming equal to 1 d T by d x at x is equal to 0 we have taken the first derivative of temperature with respect to the spatial coordinate let me just go back first derivative of temperature with respect to spatial coordinate is this one d T by d x has become T i minus T infinity by L d theta by d capital x when I apply this to the center line small x is equal to 0 it is equivalent of saying at capital x equal to 0. So, the boundary condition would have been second boundary condition or rather the first boundary condition was we have written it as d T by d x at x is equal to 0 is 0. So, transforming this would be d theta by d capital x 1 over L at capital x is equal to 0 is 0. So, because x capital x is equal to small x by L when small x is 0 capital x is also 0 this 1 by L is there take it to the other side. So, the boundary condition remains almost identical the second boundary condition is what minus k d T by d x at x is equal to L is equal to h T minus T infinity that has to be transformed minus k d T by d x at x is equal to L is given by h times T at the boundary L at all times T infinity this is what was the boundary condition this needs to be transformed. So, this left hand side would be minus k d theta by d T by d x we wrote here the previous step here also we have written d T by d x is 1 by L. So, that k by L this k is from here already minus sign is there already k by L d theta by d capital x when x is equal to small L capital x is equal to 1 because capital x is equal to small x by L L over L becomes 1 right hand side would be h fine T L comma T minus T infinity that is nothing but theta L pi why is that. So, theta is equal to local temperature x comma T minus T infinity divided by T i minus T infinity. So, when x is equal to L I get T of L comma T minus T infinity divided by T i minus T infinity. So, I have theta L times T i minus T infinity. So, this T i minus T infinity. So, this temperature is nothing but T i minus T infinity will be there here. So, boundary condition would become not it completed it let me just go back to it. So, T i minus T infinity will be there there is a minus sign x is equal to 1 theta h L c by k T i minus T infinity is there x is equal to 1 L comma T T i minus T infinity. So, the boundary condition therefore, would become minus k by L d theta by d x at x is equal to 1 is equal to h this is nothing but T i minus T infinity. So, this is the T i minus T infinity times theta L and one of you did not catch me here when I did a small error d T by d x when I transform to d theta by d x I had a T i minus T infinity here also. So, there is a minus sign there is a T i minus T infinity which is sitting here. So, let me just rewrite it instead of causing confusion T i minus T infinity minus k by L d theta by d x at x is equal to 1 is equal to h T i minus T infinity at theta at I should not retain L I will retain make this as 1 because x is equal to L implies capital X is equal to 1 T is equal to general T implies tau is equal to general tau. So, it will be like this where did this come from this I had forgotten in the previous slide here and I had missed out T i minus T infinity here also I had missed out T i minus T infinity let me just you please write down this here I will show it to you here where I had made a mistake yeah d T by d x is not just 1 by L d theta by d x it is T i minus T infinity by L d theta by d x I had forgotten to carry this T i minus T infinity in the next page here. So, here it should be T i minus T infinity by L in this portion T i minus T infinity by L I am circling this times d theta by d x it does not matter in the final equation because it is a constant you have the right hand side equal to 0. So, this remains the same here I am going to have T i minus T infinity which is sticking here I have written that in the next part. So, T i minus T infinity times d theta by d x divided by L now things look correct this will cancel off this theta because we are doing it at x is equal to L it corresponds to capital X is equal to 1 T equal to general T corresponds to tau equal to a general tau and therefore, this boundary condition becomes d theta by d capital X at capital X is equal to 1 equal to h L by k that is nothing but bio number there is a minus sign. So, minus bio number times theta at 1 comma tau. So, this is my second boundary condition first boundary condition was very straight forward we could write it as d theta by d capital X at X is equal to 0 is equal to 0 this was the first boundary condition this was the second boundary condition. And your temperature related condition we have already written there let us just go back d theta by T T condition we have already written this to the straight forward time dependent condition initial condition was already written. So, the transformed governing equation looks like this with the transformed boundary condition once this is done correctly. And how do we know that actually how did I know I made a mistake here meaning how did I know I forgot to have this T i minus T infinity is because on the right hand side I was getting this T i minus T infinity it should not occur when I have a non dimensional equation everything has to be non dimensionalized h L by k I know it looks like bio number. So, that is a non dimensional quantity theta is non dimensional capital X is non dimensional this theta is non dimensional. So, this T i minus T infinity cannot occur here alone it cannot stand alone therefore, it has to get cancelled somewhere that is how I figured that there was a mistake in this path. So, this T i minus T infinity has to come and this is your transformed boundary conditions. Now, once this is done we are just going to present the solution to you. So, theta therefore, function of capital X tau and bio number and the exact solution of this partial differential equation looks like this it is theta at X comma T is nothing, but this series solution summation of T i minus n is equal to 1 to infinity a n e to the minus lambda n squared tau cos of X n time sorry X by L times lambda n where a n the coefficient is given by 4 sin lambda n by 2 lambda n plus sin 2 lambda n. I hope all of you can see this part clearly and what is this so called lambda n the discrete values of lambda n are positive roots of this transcendental equation given by lambda n tan lambda n equal to bio number. So, what is to be done I have this theta it is given by a series solution what is this a n what is this lambda n all these things we are trying to explain a n is given by this equation and a n requires lambda n which is nothing, but a solution of this equation X times tan X is equal to bio number something of that form. So, I have to have a solution for X or in this case solution for lambda n and we do not have to worry about this solution these are all given to us in mathematical tables. So, what do I do theta I have to calculate I need suppose let us take first two terms I will have n equal to 1 it will be a 1 let me write it for theta of minus t divided by t i minus t infinity this is nothing, but the definition of theta. So, this is nothing new now I want to write the series solution the series solution you had was this one. So, we want to write one or two terms associated we want to write the first term associated with this expansion. So, the first term therefore, would be a 1 e to the power minus lambda 1 squared tau cos of lambda 1 x divided by l plus a 2 e to the power minus lambda 2 squared tau cos of lambda 2 x by l plus. So, on and so forth what is this a 1 a 2 we have a 1 is equal to 4 sin lambda 1 divided by 2 e to the power minus lambda 2 squared tau cos of lambda 2 x by l plus. So, on and so forth what is this a 1 a 2 we have a 1 is equal to 4 sin lambda 1 divided by 2 e to the power minus 2 lambda 1 plus sin of 2 lambda 1. Now, what is this lambda 1 lambda 1 is obtained by solution of this equation lambda 1 times tan lambda 1 is equal to biot number. So, the solution of this equation will be given will be used here and that is going to be used here. I have written for a 1 we can write similar expression for a 2 a 2 would be 4 sin lambda 2 divided by 2 lambda 2 plus sin 2 lambda 2 the transcendental equation for lambda 2 would be lambda 2 tan lambda 2 equal to biot number. So, what we are saying is I have the solution I have n terms n going from 1 to infinity how many terms do I have to take 1 2 4 14 50 well logic tells me more the number of terms better it is for me, but in this case what we are saying is many times it is given here very nicely and just go back to the power point we will come to this. This is what I have written first this is only the first term a 1 e to the minus lambda squared tau cos lambda 1 x by l we are doing it for plane wall. So, this is the solution for the plane wall I have written this term on the white board first term and the second term I have written this slide here shows the first term alone and what is being told is that it can be shown that values for values of tau greater than 0.2 when tau is greater than 0.2 this series solution can be approximated by only the first term in the series. So, I do not have to take care of second third term also only the first term is almost equal to the correct solution in case tau equal to tau is greater than 0.2 what is tau let us remember what is tau tau is non dimensional time alpha t by l squared. So, for a given problem alpha is the material property we know the characteristic dimension l at what time I want the temperature distribution that is going to be specified you have seen problems 200 seconds after the process has begun 50 seconds. So, that tau is known. So, I can calculate tau because alpha t and l squared are known and once we know tau is greater than 0.2 what we are saying is that in such situations this big series solution though it has infinite terms only the first term is sufficient to give a reasonably correct solution. So, that is what is given here for a plane wall we have given this solution first term approximation this is called as the first term approximation where constants a 1 and lambda 1 are functions of bionumber only how do I know this because of this a 1 I put instead of n n equal to 1 where lambda 1 lambda 1 lambda 1 will be there what is lambda 1 lambda 1 is of this form lambda 1 tan lambda 1 is equal to bionumber. This equation will have multiple solutions because tan is a tan function is a periodic function sorry it is not it is going to have various values from 0 to infinity. Therefore, I am going to have different values of lambda 1 and these values are tabulated for us in this table. So, most of the text books would have this table where the first column for those of you who cannot read this first column is the bionumber going from 0.001 to infinity why are we using 0.001 this whole exercise started because we are saying bionumber was greater than 0.1 lumped capacitance is not valid does this mean that this mathematical solution is invalid for small bionumber no we are saying this is the general solution lumped capacitance is a special case where I do not have to deal with these kind of big mathematics I can just use uniform temperature. So, we do not normally care about these tables for lower bionumber less than 0.1 does not mean that solutions do not exist. So, bionumber ranging from 0 to infinity we have these values and this table gives this is psi right psi psi 1 psi 2 psi 3 psi 4 which is nothing but lambda 1 lambda 2 lambda 3 lambda 4 this is zeta and this is zeta zeta 1 zeta 2 zeta 3 zeta 4 is given in your slides it is nothing but lambda 1 lambda 2 lambda 3 lambda 4. So, the first 4 roots of this equation are given in this table. So, you know the bionumber you have calculated tau. So, you know whether first term approximation is once you think first term approximation is just go to the bionumber row let us say bionumber is 1 lambda 1 for that bionumber is 0.8603 that is what we need. So, once lambda 1 is known a 1 can be given obtained. So, it will be 4 times sin 0.8603 divided by 2 times 0.8603 plus sin 2 times 0.8603 remember these sin functions are in radiance. So, you can have sin greater than 1. So, you will get a 1 once a 1 is known very nicely I have written everything here a 1 is known lambda 1 0.8603 square tau we have calculated was greater than 0.2 cos lambda 1 I have calculated from the table x what location I want the temperature is it mid plane is it 0.25 from the mid plane is it at the surface whatever. So, x by l is already there. So, the ratio of x by l is there everything here is known I can get this right hand side only first term I need in case I need the second term I do the same thing from this table I go get for bio number is equal to 1 get the second lambda 2 3.4256 if I need the third term I have this one 6.4373 so on and so forth. So, the point I am trying to make is though things look very complicated on face value it is not. So, and it is just good book keeping that is going to give us a proper solution. So, once all these are known theta is obtained we have an example cooked up example problem which we will do after the t break let me just go here I will just show you this cooked up example problem is there we are going to do the analytical part from 1130 onwards we will have t n assemble here by 1115 or 1120. So, we will sign off for now thank you.