 Good afternoon, we will just get started. Just before lunch, we had looked at the full equation for heat diffusion and all the three boundary conditions were also presented to you and we had come up with the solution for the differential equation for a plane wall, audible at the end circuit. So, this was the ODE, the PDE gets converted to a ordinary differential equation and again just for iteration, this is based on the assumption that you are dealing with a steady state problem, no volumetric heat generation, conduction in only x direction and in this case, we will make an assumption about the k a little later. So, this differential equation 2.22 is for 1 d steady state conduction with no volumetric heat generation. So, if I pull k out and take it to the other side, I will get d square t by dx square. In fact, this differential should be a regular differential, the solution will be t of x is equal to c 1 x plus c 2, we saw that and it tells me that the temperature distribution is linear in x. That means, it is a straight line distribution because there is no volumetric heat generation and two boundary conditions are needed and all of us have seen this as teachers also. Two boundary conditions easiest to deal with is temperature because it does not involve any derivatives. So, at the boundary x is equal to 0, I specify constant surface temperature given by T s 1 at the boundary x is equal to L, another constant temperature T s 2. The linear variation is shown by a straight line here. So, T of x is equal to 0 is T s 1, T of L is equal to T s 2. When I solve, I get the temperature distribution in this form equation 2.24 just linear with a constant T s 1 here and tells me that when x is equal to 0, this term in the bracket goes to 0, this first term goes to 0. So, temperature at x is equal to 0 is T s 1, when x is equal to L, this becomes 1, T s 1 cancels I am left with T s 2 as the temperature at x is equal to L. So, it satisfies the boundary condition also correctly. So, the mathematics is correct and the first derivative would give me d T by d x is equal to c 1. So, that was my functional form, c 1 is basically what will come in the heat flux term. So, heat flux term is nothing but q double prime is minus k d T by d x and that comes out to be this one, equation 2.26. So, these are all things which we have studied and nothing conceptually new here. Point to remember is two boundary conditions are needed, a straight forward solution is obtained when I have constant temperature boundary condition, other situation we will have to have a little bit of involved mathematics that is it. And look at this equation 2.26, we say q double prime x is equal to k times delta T by a, what is this delta T, T s 1 minus T s 2, what is that? If I go back to the figure, you can answer, what is the what is this delta T, T s 1 minus T s 2 represents the maximum driving temperature difference. That means, if this is say 100 and this is 20, 80 degree temperature drop maximum is possible. At any point inside this solid, we are going to have a temperature which lies between T s 1 and T s 2. So, the maximum temperature drop that is a constant, k is a constant, we have assumed that area is a constant for a plane wall. So, q double prime x is also a constant. So, for a plane wall, whether I am dealing with q double prime x or q, things are not that complicated, you can keep dealing with either or because the area term is going to be a constant for a plane wall. Now, if I have multiple. Where? 2.2, yeah, l is a thickness which is constant. So, the a from here comes to the denominator. So it is independent of x. So, indicate that both heat transfer rate and heat flux are constant and are independent of x. One question somebody asked just before we dispersed for lunch was the following. You say heat flux is a constant. What is in this situation? I am giving you a situation. You tell me how heat flux is constant. I think all of you will be able to identify with that. This is a plane wall, part of a plane wall and this is a composite structure. Material a, material b and material c and heat is flowing from left to right x direction. So, whether in this is say l by 2, this is l by 2 and this is w whatever. So, the question was you said heat flux is a constant. What about heat flux through this part and heat flux through this part? Are they the same? What we said was this surface and this surface, the total heat transfer is the same because we are dealing with a steady state problem. The sum of the heats going through this path and through this path should add up to the total heat transfer rate. What is the split? 50, 50, 40, 60, 60, 40 does not matter. That depends on the individual properties or the we will call this as the resistance of the solid to heat transfer. So, that comes a little later, but since the question was raised we just brought it up. If you are talking of a single material especially and non-parallel kind of circuit then the heat flux everything is a constant independent of x direction. Now, this brings us to the concept of thermal resistance. Again, one of the most user friendly, favorite for teachers, for students because easy to understand and analogy is also very nicely put. I think he had also explained about pressure drop and volumetric flow rate. X books usually give only resistance analogy, but for a mechanical engineer in fact delta P versus Q is a very good analogy for this one. So, what we are trying to say is driving temperature difference is what governs the heat transfer rate. So, larger the delta T across which heat is going to flow larger is the heat transfer rate. So, if I am able to cast my equation, so if I am going able to cast this thing as Q is equal to delta T over I like to call this just denominator that is all. Then this so called denominator takes care of everything K A everything associated with the geometry and the material and this I can call it as the thermal resistance to heat transfer. This can be convection, conduction I do not care at this point. All I am saying is this delta T is the driving temperature difference for the heat transfer. This is like your driving potential. In case of fluid mechanics this will be your driving pressure gradient. This Q which is your heat transfer rate will correspond to an electrical current in fluid mechanics volumetric flow rate and this resistance would be your electrical resistance and in fluid mechanics what will it be? What are the resistance friction? So, if I am able to cast my any heat transfer equation Q is equal to some delta T, driving delta T over everything else. In fact that everything else becomes a mess when we are dealing with cylinder or sphere that is why I just call it as a denominator term. This denominator term we will call as R thermal resistance and larger the resistance smaller is the heat transfer. So, that is also logical. So, larger is the obstruction to the flow smaller is the heat transfer. So, choked drains if you are talking of choked pipes you have a larger resistance to the flow therefore volumetric flow rate is small. No matter how much you try to push in water there is a large obstruction it is not going to flow. So, delta P would be resistance would be large to the flow same way here you are larger the resistance you have a smaller heat transfer and that is put in the slide here. So, for a plane wall just going back to this equation this is L this equation 2.25 I will cast this as delta T divided by L over k A in the denominator and that L over k A I will call this as the conduction resistance for a plane wall. L by k A and this is analogous to what you see in electrical resistance concept all of us are familiar with this. So, I do not want to spend more time. So, convective heat transfer similarly the same heat is going to flow if I am having a boundary condition which is convection here. So, delta T which drives the convective heat transfer Q whatever comes in by conduction is what is going by convection. So, the same Q is equal to h A T s instead of T s 2 we are just calling it T s minus T infinity and I am I have to keep it in terms of delta T divided by everything else. So, I have this in the denominator 1 over h A is in the in the denominator and this 1 over h A is called as a convective resistance to heat transfer. So, conduction resistance for a plane wall L by k A convective resistance is 1 over h A units you can work it out Kelvin per watts. So, this is a typical simple first exercise problem which most U G students are subjected to draw the thermal circuit diagram for a geometry shown like this. So, all of us all teachers will give this problem and if the student is not able to do even this then the student does not deserve to pass this is what we say. And the thing is very very easy and what are the common mistakes which students make that also we will see the first thing for this diagram to carry any meaning the direction of heat transfer has to be important. So, the Q has to be shown clearly the driving temperature differences or the nodes should have the temperature values associated with it the resistance should be clearly put. So, it can be put here as 1 by h or r convection 1 or something like that and correspondingly the Q has to be put correctly meaning the direction of heat transfer the temperatures and the resistance values have to be put correctly for the figure to carry any meaning. So, for this plane wall subjected to convective boundary condition hot fluid on the left hand side cold fluid on the right hand side heat is going from left to right. So, T infinity 1 is larger than T s 1 which is larger than T s 2 which is larger than T infinity 2. So, h a what area I am talking about here cross sectional or surface what is the surface area into the plane of the paper. So, h a T infinity 1 minus T s 1 why why not the other way I think 2 o clock post lunch cannot be monologue unfortunately becomes a monologue I am saying it should be Q is equal to h a T infinity 1 minus T s 1 which of these is correct which one is correct anybody for second one. So, this is wrong majority wins. So, this is correct why is this correct what is the direction of heat transfer assumed direction is from left to right because I have shown that the temperature T infinity 1 greater than T s 1 greater than T s 2 greater than T infinity 2. So, Q is in positive direction shown like that. So, for the plane wall my with this assumption of T infinity 1 larger than the others T infinity 1 minus T s 1 will be Q which is the same as T s 1 minus T s 2 divided by L by k a is equal to h a T s 2 minus T infinity. So, I have put those 3 together cast it in a form which I like delta T divided by everything else that is the resistance and if I just put these separately and add that step is not there I just put that here. So, Q x times 1 over h 1 a is equal to T infinity 1 minus T s 1 Q x times L by k a is equal to T infinity 1 minus T s 2 and Q x times 1 over h 2 a is equal to T s 2 minus T infinity 2 these are my 3 equations convection conduction and convection. So, if I eliminate the temperatures by adding these 3. So, this one cancels this one cancels of I am left with T infinity 1 minus T infinity 2 is equal to Q x times summation of all the 3 resistances I would like to bring that on the right hand side on the other side this will be called as summation of r thermal which includes 1 over h a h 1 a plus L by k a plus 1 over h 2 a summation of all the 3 resistances will be there. Now, why did I use T infinity 1 minus T infinity 2 and why why did I not add only these 2 equations I can do that also. That means, I will have only 2 resistances to deal with provided I nu T infinity 1 and T s 2 because T s 2 will be left there in the equation. When I add all the 3 that means, I know the maximum delta T which is there T infinity 1 minus T infinity 2. So, many times students have how what to add you have added 3 suppose T s 2 is known many times that is a constraint like the surface of the oven should not exceed 40 degree centigrade. So, that T s 2 is specified it is anchored it is fixed. So, I do not care about T infinity what is the temperature of the air inside the room you have a oven in which you are baking something inside is very hot there is a insulating glass and for safety reasons the outside temperature of the glass should not exceed 40 degrees that means, that T s 2 is anchored at 40 T infinity 2 is some 200 degree centigrade. So, you deal with the resistances associated with convection in the inside of the oven conduction in the window in the oven glass you do not deal with the third equation. So, depending on the situation you have to use the appropriate summation and the appropriate resistances that is all there is not rocket science, but at a third year level when students are first exposed to this invariably a confusion because textbooks will give solve problem where all 3 are added and then if you have a twist and say this is the condition they will not know what to do, but very straightforward conceptually nothing new. So, this is clear equivalent circuits concept are can be used for much more complex systems such as composite walls any number of series and parallel combinations and one such example is given here you have 3 different materials 3 different thicknesses 3 different thermal conductivities and I am writing the same equation is similar for all of us know this. So, I do not want to spend time. So, this will give you q x is equal to this quantity. So, now one very very important thing which we will emphasize during heat exchangers, but it is a good idea to bring this here is every time to deal with R thermal whatever that somehow we do not like we always like to deal with equation of the form h h a delta t it is very nice to see nice to interpret also. So, what is written this q is equal to u a delta t many times this is introduced this is nothing but delta t over R thermal you have seen this. So, what I am saying is delta t divided by 1 over u a correct this 1 over u a therefore, is nothing but summation of the thermal resistances and if you go back to the single plane wall it is nothing but 1 over h 1 a plus l by k a plus 1 over h 2 a. So, it is of the same form as a thermal resistance only thing every time to carry these three terms together I do not like I call this by a u a term and this u times a therefore, is nothing but 1 over 1 pitfall students what they do this is correct till here they will do correctly instead of putting it like this this is the correct thing reciprocal of thermal resistances many of us even as students we would have done this we will just h 1 a plus k a by l plus h 2 a this is happened it will still happen. So, these are some things which students make a mistake the units probably will come correct because it is just a reciprocal, but conceptually it is not correct because this is nothing but reciprocal of the thermal resistances and why is it introduced this is introduced primarily to deal with heat exchangers especially where u a l m t d all of us are familiar this u a is called as the mean u is called as the overall heat transfer coefficient it is a artificially created quantity especially in this context where all the resistances are put together in this form and I am writing q is equal to u a delta t which is similar to a form h a delta t this delta t u a delta t in this equation this delta t is the driving temperature difference this u is called as the overall heat transfer coefficient it has the same dimensions as h point to note is u times a is nothing but 1 over r thermal. So, for a given system r thermal is a constant. So, this is a constant say if I have a pipe with an insulation hot fluid inside the pipe cold air on the outside it is a given dimension system u a is a constant because the resistances are all constant that means the product u times a is a constant is u a constant no let us let us understand this this is a pipe hot fluid is flowing inside h i is the heat transfer coefficient this pipe has a finite thickness. So, this is a wall and there is a insulation on the outside I am not drawing the full circle in view of the quality of the diagram and h 2 is the heat transfer coefficient on the outside we are going little beyond what we have done so far. But the point I am trying to make is this is the convection resistance on the inside t infinity 1 this is the conduction resistance t 1 I will call t 2 this is the wall conduction resistance this is the insulation conduction resistance and this is the convection on the outside ok. My question is is u a constant in this case or u a a constant u a is constant because these resistances are all fixed is u a constant why? Because it depends upon u is not a constant u is not a constant for a plane wall geometry because area is independent of the location does not matter u is also a constant. But for a radial system sphere or cylindrical geometry area is going to change if I take this plane it has a different surface area if I take this plane it has a different surface area if I take this inside wall of the pipe it has a different surface area. So, product u a is a constant therefore, many times in heat exchangers you will write u i a i or u out a out some such thing will be there. So, always the heat transfer overall heat transfer coefficient is told in conjunction with the associated area it has no special identity it is always a pair u times ok. So, this is something which again is a pitfall for many of the students because they realize it later on because u a is introduced here, but its application or its logistical use is put in only heat exchangers. So, at this point I think it is a good idea to say u is not a constant except for a plane wall geometry in general u is always associated with the area of heat transfer. So, this was the concept of overall heat transfer coefficient and extension of this series series network is now done for a parallel system when will resistances be in parallel? We have drawn a plane wall and we have put a resistances in parallel, but when will a resistance be in parallel? Why should a resistance be in parallel? Let us ask question little bit different why should resistance be in parallel in a given problem? High metallic strip. High metallic strip ok. You are giving the examples my thing is when under what situation can heat transfer resistance be in parallel? What is the heat transfer resistance representing? So, for a given amount of energy to flow say 100 watts if this split if there are multiple paths for the heat to flow and outside it has to be the same again then in that case the resistance have to be in parallel. For example, in this room so if this is a hot wall 200 degree centigrade there is convection air is at 25 all the other walls are also at 25 degree centigrade. So, heat is going to flow by radiation and convection across the same temperature difference ok that means I will have to say q is equal to q convection plus q radiation. This is H A H for convection C is for convection A is a surface area T surface minus T infinity plus this would be sigma epsilon A. T s raise to 4 T surroundings I am also having an approximation that it is the same as T infinity because the driving temperature difference is the same and this I will call as a pseudo radiation heat transfer coefficient times T s minus T infinity where H r is nothing but sigma epsilon T s square plus T infinity square times T s plus T infinity. Expanding this A square minus V square A square plus V square etcetera. So, this q radiation therefore will be H rad A T s minus T infinity. Now, let us just operate this carefully T s minus T infinity common H convection plus H radiation ok. So, for the same heat to flow total heat there are two resistances which are going to be in position preventing the heat transfer and that if I say it will be T s minus T infinity divided by 1 over ok. So, multiple parts for the same heat to flow across the same delta T in that case resistance will be in parallel. Now, we will just go back to the screen wall. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Qx is flowing from left to right thermal resistance associated with this wall is Le by KeA for this wall is Lh by Ka and in this Fng it is just split according to the associated heat transfer area. And the heat is divided in such a way that Qx is coming at this junction at this node coming back at this node it is split into two parts just as you had in your electrical analogy. In a serious circuit what remains constant? Current or voltage electrical circuit current in a parallel circuit same logic ok. Another again concept which is there but most of the time that is neglected or pushed under the carpet is this concept of contact resistance. Contract resistance essentially is because of layers which are going to be in touch with each other. So when I have two different material in touch with even same material why different material in touch with each other definitely however nicely it is machined there are going to be some microscopic gaps which are going to be present. And these gaps will have some entrapped air which are going to cause a reduction in heat transfer why? Because of the poor thermal contact thermal conductivity associated with the air. So what happens is if it is just placed like this versus if it is pressed nicely I will have a different heat transfer rate. That is why many many times we have heard this thermal grease or all these kind of things. It is basically something which is put between two surfaces to ensure good thermal contact ok. So that is a function obviously as I am even in a layman's language if I press two surfaces well heat transfer is better. That means it depends on the materials depends on the applied pressure also. So what it does what this contact resistance this contact resistance is nothing but the resistance offered to heat transfer across these two surfaces which are in contact. So there is a temperature drop associated with this interface. Normally we are taking between two materials T1 is equal to T2 at that interface. But in real life there is a significant drop T1 is the temperature on the edge of the left layer T2 is the temperature on the edge here of the right layer and T1 minus T2 is the delta T. Remember Q is a constant Q is equal to delta T by everything else. This delta T is now T1 minus T2. It is flowing across this temperature difference that everything else again is your contact resistance. This contact resistance is something which is given as a function of pressure and materials etc. So what it tells me is for this T1 and T2 to come closer and closer to each other. What should be the contact resistance ideally? Contact resistance should be 0 or infinity. So this is essentially T1 minus T2 by Q. Q is finite. This has to be 0 that means R should be equal to 0. I am just going to skip all these problems associated with this time. And quickly let us just go to this radial systems. If you know to derive Cartesian geometry, good bookkeeping will help you derive the cylindrical geometry equations and the spherical geometry little better bookkeeping is needed. Thing which you have to keep in mind is the area term associated. So for the curved surface, theta has no dimensions. R d theta has to be taken that is the common thing. Of course sphere is little bit more complicated. So you get a differential equation which is again third order PDE in space and time 1D with no heat generation and steady state I get this as the governing differential equation. So A is nothing, if you write it in terms of Fourier's law minus kA dt by dr, we are assuming one-dimensional heat transfer in the radial direction of the cylinder and A essentially is 2 pi R L where this R is the local radius associated with that geometry. So if I have a wall of thickness delta R, R i being the inner radius, R o being the outer radius, the area would be 2 pi R i L for the inner wall, 2 pi R not L for the outer wall. Any intermediate location will have a corresponding radius. So that has to be kept in mind. So all tells me prescribes the quantity kR dt by dr is independent of R, it follows from this equation that conduction heat transfer rate qR and not heat flux is constant. This we have been emphasizing time and again heat transfer rate is constant. So let us always keep that as a sacred truth. Heat transfer rate is constant, heat flux, plane wall it is a special case. So let us not keep plane wall as the general case and these are the exceptions. Exception is only a plane wall. In reality heat transfer rate watts is constant, watt per meter square is a constant only in case of the plane wall. That I think we should clarify all the time. In the radial direction q double prime R is not a constant because area is strong function of the radius. So this differential equation if I solve first integration will give me kR dt by dr is equal to C 1, C 1 by kR. So it will give me a log term. So first integration will give me C 1 by R and second integration will give me C 1 log R plus C 2. Two boundary conditions are needed and we have two surface temperatures being specified. Temperature at inner radius is Ts 1, temperature at the outer radius is Ts 2. Ts 1 is greater than Ts 2, heat is flowing from inside to outside. So if I use those boundary condition I will get C 1, I will get C 2. Substituting that back I will get temperature distribution given by these equations which are there on the in the red box. Both are the same equations except that one has Ts 1 and Ts 2 and the corresponding radii are taken care here. So answer will come the same whichever you use, it is just the functional form. So obviously it is not a very nice looking equation, you can see some log etcetera is it. So it is not very user friendly. But so what is the nature of the temperature distribution? Non-linear. Non-linear, so you can draw a straight line. So it is some logarithmic distribution that is shown to us like this. Another thing, T is essentially this one going back to this equation, this form. Ts 1 minus Ts 2 if I go into this form and look right for the temperature heat transfer across this finite temperature difference. I will get Tr is equal to this way and heat transfer rate given by this equation K A dT by dr, K times 2 pi R L, dT by dr is nothing but your C 1 by R, C 1 is here divided by local radius. So I substitute that here that 2 pi R L that R gets cancelled with the R in the denominator. I get qR heat transfer rate is given by delta T Ts 1 minus Ts 2 divided by everything else that is coming out to be log R 2 by R 1. This 2 pi L K will come in the denominator. So this whole thing is called as the conduction resistance for a cylindrical geometry. So log R 2 by R 1 divided by 2 pi L K. Again all of us are familiar with this composite wall same thing, multiple cylindrical layers just add them in series. Appropriate areas have to be taken for the convective heat transfer 2 pi R 1 L here, 2 pi R 4 L here for H 4, 2 pi K A R 2 by R 1, R 3 by R 2, K B, R 4 by R 3, K C, ultimately good bookkeeping. If you are sloppy in your writing you are going to make mistakes that is it. And again we come back to our friend overall heat transfer coefficient expressed in terms of A 1. So is this a constant? Is this U a constant? U constant? Is it a constant or not a constant in the area? U times A is a constant, U is not a constant. So this is just based on the A 1 area. Composite wall again resistance diagram is drawn. Sphere quickly I am just giving you this equation 1 by R square d square R dT by dr and solution will give me P of R is equal to derivation is there, it is just same form. Integrate once you get R square dT by dr is equal to C 1, dT by dr is like this. Second integration will give me C 1 by R plus C 2, two boundary conditions at the two radii I will get C 1 and C 2 and the temperature distribution. 1 by R square, the first is R square, this square is missing here. The governing equation is written correctly here, this is by transferring when R square is gone. So heat transfer rate is minus K A dT by dr having gotten the temperature distribution, always see the logical flow is we get the temperature distribution, then the heat transfer rate, then the heat flux and then the resistances. So K A dT by dr local area 4 pi R square dT by dr was here given to you C 1 by R, C 1 was obtained. So I get dr like this and q R is given by this form. I like delta T by everything else. So I just keep the delta T in the numerator, bring this 4 pi K down, 1 by 4 pi K, 1 by R 1 minus 1 by R 2, this whole thing is your conduction resistance through a hollow sphere of inner radius R 1, outer radius R 2. So these three resistances I think the students remember by hook or by crook, but something is unavoidable but they should know how to derive this. So this is just a simple energy balance thing. Another I think favorite topic for testing in exam is this critical radius of insulation. So what is this radius of insulation? So it is not, everybody knows when I add an insulation I am going to reduce the heat transfer, that is what logic tells me. So for how much should I add? Should I keep on adding or should I stop somewhere? So that depends on the economics. So if you are outside temperature is minus 30 degrees and inside is 25 degrees, you have to have a finite insulation will give you a finite heat loss and accordingly your electricity bill to keep the place warm at 25 is going to increase. If I have a larger insulation lesser the electricity bill you are going to have. So insulation from that point what you are saying is right, larger the insulation thickness better is for me from heat transfer point of view that is only partially true. So if I am thinking of an insulation here for a plain wall, so this blue one is the wall, this violet one is the insulation, T1 is the temperature at the interface, HT infinity are the temperatures of the fluid and the convective heat transfer coefficient, K is the thermal conductivity of the insulation. So Q is nothing but T1 minus T infinity divided by L by KA plus 1 by HA, this is T1 minus T infinity divided by R conduction plus conduction, this I think is very clear. So what does this tell me? It tells me that irrespective of how much ever thickness I add, L if it increases, L becomes large, denominator becomes large, Q is going to drop. That means heat transfer rate decreases, larger and larger the thickness good for me. Plain wall again exception, not a rule. Let us go to radial system. So you have a cylinder, T is the temperature at this interface T1, insulation is here, convective boundary condition is on the outside. So for the insulation I will write T1 minus T surface divided by log R2 by R1 by 2 pi LK that is the heat transfer rate. For the convection I will write T surface minus T infinity divided by 1 over HA and when I add composite resistance it will give me T1 minus T infinity divided by R conduction plus R conduction. What is happening here is what makes it interesting. R2 appears here as well as here. So if you look at this diagram, outside surface area which is associated with convective heat transfer is going to increase as I increase the thickness of the insulation. So logically thicker the insulation, more resistance to conduction, better it is for me. So R2 becoming large, conduction resistance increases, R2 becoming large, 1 by H this one this term is going to decrease. So conductive resistance increases, convective resistance decreases and there comes a stage when one of them dominates over the other and that thing is what is important in radial systems. So if this is the expression for heat transfer what we say, value of that critical radius of insulation where heat transfer rate reaches the maximum is determined by the requirement that dQ R by dr is set to 0, maximum heat transfer is given by dQ R by dr equal to 0. Doing the mathematics the numerator is constant, this is what needs to be taken care of. So if I differentiate I will get K by H, K for the insulation divided by H on the outside that gives me the critical radius of insulation for the cylinder and what it tells me is for radii which are smaller than the critical radius of insulation and for radii which are larger than the critical radius of insulation the trend in heat transfer is going to be different. So rate of heat transfer from the cylinder increases with the addition of insulation for R2 smaller than R critical and reaches the maximum value after which it starts to decrease. So if I want to essentially if this is the bare pipe heat transfer Q bare, if I want to have a reduction in heat transfer I should not stop with a small thickness which is coming in this in this band. In fact I should cross this and go beyond this point corresponding point for the heat transfer to decrease compared to the bare value. So this is the concept of critical radius, similar thing can be derived for a sphere, it will come 2 K by H and let us come here, we want to spend time on these things. This is a brief summary the one dimensional steady state equation for a plane wall d square T by T dx square constant properties cylinder sphere temperature distribution is linear temperature distribution is logarithmic heat flux, heat transfer rate and heat flux, heat transfer rate constant constant constant heat flux constant variable variable. Thermal resistance conduction thermal resistance for cylinder thermal resistance for sphere convective resistance is 1 by h a depends on the area you will get the appropriate convective resistance. So now let us move ahead now I will be going fast compared to Professor Arun I will be going quite fast. So let us see how fast I can go because basically now we have gotten into we have so far we have not taken we had made major assumption that is steady state conduction that is there is no transient term E dot ST is not there and then we said that there is no energy generation and it was one dimensional. Now we will add one more complication that is it continues to be one dimensional it continues to be steady state but only thing is that we are going to say that now energy generation term also we are going to add that is all we are going to do. So with the energy generation term added it can be again generation of energy can be any form here in this case let us say we have taken I squared R. So if I take plane wall same if I if you remember our heat diffusion equation you had d squared t by dx squared d squared let me write that so that we do not keep we do not lose side of what we are doing that is we have heat diffusion equation as for constant thermal conductivity d squared t by dx sorry del squared t by lx squared plus del squared t by del y squared I would request all of you to write along with me because I see everyone sleeping or at least in slumber mode I want everyone to write along with me that is precisely why I am also writing. So let us not deceive each other so del squared t by del z squared plus what is the next term q dot upon k is equal to 1 upon alpha del t by del t right. So what we are saying is that our situation is steady state steady state means which term will vanish here right hand side term this will vanish because it is steady state and now I am saying it is one dimensional conduction that is conduction is there only in the x direction Mangesh some problem with this equation del z squared yeah that is wrong you should correct me sometimes I make intentionally sometimes unintentionally. So now heat transfer is only in the x direction so this is not there and this is not so my equation reduces to del now I can write ordinary differential so d squared t by dx squared plus q dot by k equal to 0 okay some problem so this is my equation how did we get this equation I have just shown so that okay so that is what is this equation so d squared t by dx now rest is if we just integrate it twice I get this equation I am not going to do that okay so this is my equation so I have here two constants C1 and C2 so to get these two constants I need appropriate boundary condition so I can be constant wall temperature or convective heat flux boundary condition if I apply convective heat transfer convective heat transfer boundary conditions here this is H1 and H2 with two different fluid temperatures I may get asymmetric temperature distributions but if I have the same convective boundary conditions on both the sides my temperature distribution is going to be symmetric okay so let me skip so if I just take constant wall temperature boundary conditions on the left and the right hand side that is TS1 and TS2 that is this case in which I have taken H1 and H2 different with two different fluid temperatures so I am going to end up with two different constants C1 and C2 okay I am not going to differentiate and show you that so if I just plug in that I am going to get this temperature distribution okay so that is how we can do this we can go ahead and solve this temperature distribution for various boundary conditions incidentally what happens is that here in case H and T infinity is same for both the cases let us say H and T infinity is same so what is the boundary condition at the center at x equal to 0 what is equal to 0 so what does that incidentally we have solved another problem in which physically there is a insulation on one side and convective boundary condition on the other side that is this okay so here we decided that there is no heat flux at the center because of the symmetric temperature boundary temperature distribution okay so that is I think that is about the plane wall I am not going to deal with radial systems same theory can be adopted for radial systems and spherical systems and there are two problems I am not going to solve that but I am going to leave you with a question for thinking maybe I may get the answer right away let me say that instead of giving convective boundary conditions on both sides if I give constant heat flux that that was the question which we had given constant heat flux boundary conditions on both the sides can I get the temperature distribution let me repeat my question I have plane wall that is this I can draw it here yes this is again to get you out of slumber that is all just to make you think so this is my plane wall let us say and this is my center and this is x equal to 0 x equal to minus l I cannot write as nice as professor Arun he writes very nicely x equal to plus l here I am applying yeah I think I think I will force you to solve this problem and physically interpret no right away physically let them interpret q s 1 double dash is there on this side and q s 2 double dash is there on this side okay and there is q dot all that I want you to do is now within next two minutes apply energy balance and physically interpret this problem to me before I ask for solution I am not asking temperature distribution to start with just apply energy balance and tell me what is the physical interpretation whatever drawn is right or wrong or what please tell me this is just to get out of slumber that is why I just thought that I will share this problem apply e dot in minus e dot out plus e dot g equal to e dot st e dot st is not there because it is steady state come on yeah I am not going to take any of your questions for a while until you come up with the solution two more minutes where is e dot out and all you have to answer me whether there is e dot out or not if at all if it is there what it is that is that is precise it can never reach steady state the way it has been drawn because I am giving heat flux on both the side into the energy is also getting generated so that is why whether why why I took up this problem because we should not forget our equation e dot in minus e dot out energy balance equation should not be forgotten otherwise I might be telling it is reaching steady state but I may not be right when I tell that it is reaching steady state but when can it reach steady state no I am saying it is one dimension yes so see if I put energy balance what is that I get q s 1 double dash plus q s 2 double dash plus 2 q dot equal to yeah I am removing area is equal to 0 one of the three has to be made negative to make it reach steady either my q dot can be negative or I have to flip the directions of q s 1 or okay so I mean why I took up this example and still the question is I will postpone this question tomorrow morning we can discuss this can I get the temperature distribution for this case okay please go back and work out if you can get this okay so that is about the what is that energy generation concept the radial systems you can go ahead in the same direction so I am not going to broach up broach up on that issue at all I think with this there are couple of problems which you can solve it yourself so I guess with this I would say that we will we have concluded on conduction 2 that is energy generation.