 Good morning everybody. So, let us start off with the heat transfer. The overall plan is that I will just take a recap for the next 5 minutes, whatever we taught you in the yesterday's class. So, what we said is that there are 3 modes of heat transfer. One is conduction, convection and radiation and conduction is essentially quantified by 4 years law of conduction and we have T 1, T 2 and Q double dash and T 1 and T 2 should be visualized as electrical potential and heat transfer rate as the current and convection is quantified by Newton's law of cooling that is heat flux equal to heat transfer coefficient into T infinity minus T s. Here at least it would be important to recollect that h is not dependent on the plate material, but it is dependent on the fluid properties and the velocity and the radiation does not require any medium and the radiation heat transfer is quantified by Stefan Boltzmann law. So, conduction essentially takes place even when there is no bulk motion that is if the gases are there even if it is still and if the liquid is there it is still and solids of course are still all the time. So, the conduction is essentially an atomic or the molecular activity when there is no bulk motion and this is quantified by Fourier's law of conduction and there is a minus sign. So, that we tell that the heat flux is transferring from in the positive direction. So, this negative sign is to make the heat flux positive in the x direction and we did solve a problem and we said the convection is essentially because of the conduction within the thermal boundary layer and fluid mechanics is very much essential and we are going to spend indeed around 5 to 6 classes in taking the recapitulation of or brushing up our memories on fluid mechanics itself. There are various types of convection, forced convection, natural convection, boiling and condensation also falls under the classification of convection only. And then this is the Newton's law of cooling and these are the typical values of h as you can see natural convective heat transfer coefficients are lower compared to that of the forced convection, but for liquids generally whether it is forced or free the heat transfer coefficients are high compared to that of gases. Essentially this is coming out because of fluid properties that is thermo physical properties that is Prandtl number, what is Prandtl number we will come to that when we come to convection. Then we have Stefan Boltzmann law which is given by MSU power or you can take it as heat flux for now equal to sigma T s to the power of 4 where sigma is the Stefan Boltzmann constant given by 5.67 into 10 to the power of minus 8 Watt per meter square Kelvin to the power of 4 quite easy to remember and of course we have real surface which introduces emissivity. So we went ahead and this is the summary which tells the mechanism and the rate equation and the important concept professor Arun taught us is E dot in minus E dot out plus E dot g is equal to E dot s t and E dot g will not be there when there is no volumetric heat generation or volumetric heat depletion if you want to call and E dot s t is if there is no transient thing that is if my things are all steady then E dot s t would not be there then it would be only E dot in equal to E dot out. So this is budgeting of thermal energy and we specified where are what are the applications of heat transfer there are plethora or plenty of applications of heat transfer all over all around us every in day to day's life. So we solved couple of problems and we reminded ourselves that if it is 1 D conduction temperature is either a function of x or y or z and if it is 2 D it will be function of x y or y z or x z and 3 D means it will be function of x y z and transient means of course it will be function of temperature also and then we took a recap of conduction rate equation in a vectorial form and there was lot of discussion about this isotropy. Isotropy only means that the properties in this case let us say the thermal conductivity is independent of direction iso means same tropic means sort of direction so it is same in all directions and of course we studied what is thermal conductivity and we said that the thermal conductivity of a conductor is quite large compared to that of non conductor that is for example insulating material. So for solids it is quite high and for liquids it is of the order of 1 and gases it is of the order of 0.1 so then we saw that for the alloys generally the thermal conductivity is lower than the individual thermal conductivity of the material. So we have not answered the question yet why it is so but one of the one of my students sitting here has offered an answer that is the resistance is essentially going to be that is what he says is that if I have thermal conductivity of material k a let us say in this case copper and thermal conductivity k b let us say it is nickel the thermal conductivity of copper is around 400 and nickel is around 100 so copper nickel combination would be less than 100 how he explains that it is because the net resistance that is 1 by k equal to 1 upon k a plus 1 upon k b that is k equal to k equal to k a k b upon k a plus k b which is going to be lower than the lower than the independent thermal conductivity this is one way of explaining why k l i is going to be less than k a r k b sounds plausible but still we will check up check out the validity of this explanation as we go along and we said that the thermal conductivity is two components because of free electrons and another one is the lattice arrangement. So and then we define what is called as thermal diffusivity which is k upon rho Cp so then we define derived heat diffusion equation so the heat diffusion equation states that I will take the final equation that is this is the conduction in the x direction conduction in y direction conduction in z direction this is the energy generation or energy depletion this is the energy stored term that is rho Cp dt dt okay. So with this we will move on to today's thing I will not take recap of 1D conduction so what I will do is on 1D conduction we had solved a problem so I will just take recap of 1D conduction in all coordinates we will solve this problem now before we solve that so we said that this is the 1D conduction equation for plane wall cylinder and spherical and sphere this is plane wall cylindrical wall spherical so this is linear temperature distribution this is logarithmic and this is non-linear and this is the heat flux so here we should realize that the heat transfer heat flux is not independent of r in case of cylindrical wall and spherical wall but here in heat flux at any location it is going to be that is it is independent of x so the resistances are l by k a for plane wall and for cylinder it is log r2 by r1 upon 2 pi lk and spherical wall it is 1 by r1 minus 1 by r2 upon 4 pi k so let us just quickly solve this problem which I had skipped intentionally yesterday that is this is about a furnace we have a furnace that is a leading manufacturer of household appliances is proposing a self-cleaning oven design that involves a composite window separating the oven cavity from the room air that is you have the oven and this is the window what is made of two materials A and B and the two materials are having each thickness L A and L B and it so happens that the thickness of L A is twice that of L B and thermal conductivity of A and B are 0.15 and 0.08 you can note that these values are very less because we do not want the heat to be lost from oven to the outside atmosphere and the oven wall is maintained at 400 degree Celsius and ambient air is at 25 degree Celsius so now interesting thing is that he has given that the convective heat transfer coefficient and radiation heat transfer coefficient and the outside heat transfer coefficient that is convective heat transfer coefficient inside because of perhaps because of natural convection and there is some radiation but instead of accounting radiation he has given an equivalent heat transfer coefficient this is little hard to understand but instead of writing sigma epsilon Ts to the power of four minus T infinity to the power of four we we are writing that equal to h into T s minus T infinity. So, this is just artificial way of looking at it there is no physical meaning as such. So, h r is also same as 25 and h i is 25 and h o that is the outside wall to the atmosphere is also 25. And the minimum window thickness is to be found out to ensure that the outer surface of the window is 50 degree Celsius that is inner wall it is T s i is this temperature T s o should not exceed 50 degree Celsius. If I draw the resistance circuit for this remember both convection and radiation are simultaneously occurring. When I say simultaneously you can imagine that convection and radiation and see that the temperature potentials are same when I say temperature potentials you see here it is told that the wall temperature and the air temperature is given to be 25 degree Celsius. You see here sorry wall temperature and the air temperature not at 25 400 degree Celsius. The air temperature the air inside the oven and the wall temperatures are given to be 400 degree Celsius that means potential is same and only the resistances are that is the heat transfer by convection and heat transfer by radiation are simultaneously occurring that means we can imagine that the resistances are parallel to each other because potentials are same they cannot be in series if the potentials are same. So, that is essentially is what is occurring here. If you understand this the rest all is straight forward this is the resistance because of the material A of the thermal window and this is the resistance of the material B of the thermal window and this is the outer wall convective resistance. So, this is radiative resistance this is inner convective resistance. So, if I just plug in what all I know that is E dot in minus E dot out before we go to that let us put the overall resistance that is these two resistances are in parallel that is what is this term and this resistance and this resistances are series. So, that is what is L by L A by K A and L B by K B and these two are in series with the net resistance because of convective and radiative resistance inside the oven. So, this is the net resistance. So, if I put that T A that is the air temperature minus T S naught that is the outer wall temperature which is not to exceed 50 degree Celsius is the net resistance equal to outer wall that is H naught into T S naught minus T infinity. What is T S naught? T S naught is this temperature that is this outer wall temperature here essentially convection is taking place. So, if you put that and plug in everything and what is required I need to find L A and I am substituting for L B I have substituted already for L B L B equal to L A by 2. If you see here we have already mentioned L A equal to 2 L B. So, L B equal to L A by 2. So, if I substitute that L A by 2 here if I substitute that and put in all the convective heat transfer coefficients and the radiative heat transfer coefficients and the thermal conductivity I am going to get the thicknesses 41 mm. So, and L A plus L B is going to be 62.7 mm, but there is a one major assumption what we have made here is that ambient air temperature is equal to wall temperature which is not going to happen in real life wall temperature is going to be different from ambient air temperature. So, if that is the case how do I handle this problem? So, that is what essentially is put here. So, if I put for the kitchen wall that is sigma epsilon T wall minus T s I to the power of 4 plus H I. So, you see now I am taking radiation and convection in series equal to this is the radiation inside between the wall and the inner wall temperature that is oven wall temperature and the thermal window temperature T wall to the power of 4 minus T s I to the power of 4 and the convection that is T air to T s I air temperature is going to be high, but definitely air temperature is going to be lesser than T wall that is the oven wall temperature that should be equated to the conductive resistance of the wall and again this conductive resistance we are again equating to the outer wall. Here again we are taking the radiation between the outer wall that is T s O minus T wall O that is T wall O is nothing, but the T ambient that is T infinity this is not T wall O this is T infinity to the power of 4 actually that has this has got shifted here you can see here T wall O equal to T infinity. So, that is T infinity to the power of 4. So, if you simultaneously solve these two equations you will get both T s I and T s O. So, this is how we need to handle just to make our life easy we are assumed that the oven wall temperature and the air temperature same. So, much about this problem. So, we will now move on to what is called as heat generation. So, let us just rewrite I want all of you to write all participants to write the heat diffusion equation in Cartesian coordinates. So, let us write that that is del square T by if assume k is constant del square T by del x squared plus del square T by del y squared plus del square T by del z squared plus q dot upon k equal to rho C p by k that is 1 upon alpha del T by del. So, what we are saying is that we are having 1 D conduction that is temperature is a function of x only and steady state steady state when I say steady state this term is going to vanish and when I say temperature is a function of x only temperature variation with y and temperature variation with z will vanish. So, I am essentially getting notice that I am converting p d e to o d e d square T by d x square plus q dot by k equal to 0 this is again second order differential equation. So, I need two boundary conditions, but the only difference between yesterday's 1 D conduction and today's 1 D conduction is that I have taken heat generation term that is volumetric heat generation I have allowed it to happen. So, let us take a plain slab wall and have heat generation and see how does the temperature distribution vary in this case. So, that is what we are going to do in next few minutes. So, this is the plain slab wall this is the plain slab wall and I am having convective boundary conditions on both the sides that essentially means that the my wall is maintained at T s 1 and T s 2. So, in one case T s 1 is different from T s 2 that means the here the it is hot side and this is the cold side, but I can take both the water temperature both the fluids of the same temperature and same heat transfer coefficient then in that case the wall temperatures on both the sides would be T s. Notice that I have taken origin here at the center x equal to 0 and this side it is going to be minus L and this side it is going to be plus L. I want all the participants to draw along with me this plain wall and then we will derive this equation. So, let us do that that is we have a plain wall this is the center to start with let me take this as T s 1 and let me take this as T s 2. So, we found that in the last case the symmetric case it was parabolic, but we have not a derived y how it has come out to be parabolic that is what we are going to see now and this is 0 x equal to 0 and this is x equal to minus L and this is x equal to plus L. So, now what is the equation we wrote d square T by dx square plus q dot by k equal to 0 and what are the boundary conditions when x equal to minus L T equal to T s 1 and when x equal to plus L I have T equal to T s 2. So, if what should I do to solve this problem I need to integrate this if I integrate that I am not going to integrate it step by step I will write the equation directly or even we can do that if I integrate that d T by dx equal to q dot upon k into x plus c 1 again if I what is that yeah there is a minus sign please correct me like that. So, T equal to minus q dot by k into x square by 2 plus c 1 x plus c 2. So, I need to utilize these two boundary conditions if I substitute x equal to minus L and then my left hand side of this equation becomes T s 1 and if I substitute x equal to plus L my here it would become T s 2 then I have two equations and two unknowns what are the unknowns c 1 and c 2. So, that is what we have done here. So, if I solve that I get c 1 equal to T s 2 minus T s 1 upon 2 L and c 2 equal to q dot L square by 2 k plus T s 1 plus T s 2 by 2. Let us write that so that you all are with me so that is c 1 equal to T s 2 minus T s 2 minus T s 1 upon 2 L and c 2 equal to q dot L square upon 2 k plus T s 1 plus T s 2 by 2. So, if I substitute this c 1 and c 2 my equation for temperature would become T of x equal to q dot L square by we can even substitute and c q dot by k x square by 2 plus c 1 is T s 2 minus T s 1 upon 2 L plus c 2 is q dot L square by 2 k plus T s 1 plus T s 2 by 2 something is missing here x is missing. So, now if I rearrange this so that it looks little elegant q dot L square by 2 k if I take it as common between these two terms. So, I would get 1 minus of x by L whole square or x square by L square plus T s 2 minus T s 1 by 2 L into x plus T s 1 plus T s 2 by 2 is it. So, this is the temperature distribution. So, now we will get back and see is that what we got yes. So, this is what we got as the temperature distribution. Now, we have taken a complicated not a complicated a case in which the temperatures on both the walls were different that is T s 1 and T s 2 were different. If T s 1 is equal to T s 2 equal to T s this equation this term will vanish and this will reduce to T s plus T s by 2 that is 2 T s by 2 that is T s. So, T s plus q dot L square by 2 k into 1 minus x square by L square and what is the temperature at the center and that is T of 0 equal to that is q dot L square by 2 k plus T s no to this what is this distribution this is this is parabolic this is parabolic. So, this is precisely what we have plotted earlier when we plotted we did not know that it was parabolic but now we have indeed shown that the temperature distribution in the presence of volumetric heat generation with one dimensional heat transfer the temperature distribution is going to be parabolic is going to be parabolic. So, that is essentially what we have done now if someone asks us how do I relate or how do I calculate the T s from this convective boundary condition that is h and T s how do we do that. So, we can do that now how do we do that. So, that is what should I do I should basically if you see here in this transparency what is that you are doing you are essentially applying energy balance on the outer wall or one of the walls that is let us take this was our wall let me take symmetric boundary condition for the sake of simplicity that is this is T s T s h comma T infinity h comma T infinity if I apply energy balance here what is that minus k d t by d x at x equal to l is equal to at x equal to l is equal to h into T s minus T infinity T s minus T infinity what do I get what do I get what is my temperature distribution if I differentiate my temperature distribution what was my temperature distribution let us go back and see so that we do not remember see this was my temperature distribution. So, if I differentiate that this equation what do I get. So, I get 2 dot l square by 2 k into 1 minus x square by l square T s constant. So, I get what do I get. So, I get q dot l square by minus k into q dot l square by 2 k into minus k into q dot l square by 2 k into minus 2 x at x equal to l square equal to h into T s minus T infinity essentially what I am trying to say is that if you differentiate that temperature distribution and substitute x equal to l you are going to get T s equal to T infinity plus q dot l by h. So, this is this is basically because yesterday professor had taught us that this is the convective boundary condition. So, that convective boundary condition by applying that we have got T s equal to T infinity plus q dot l by h for those guys you are not understood what I have done what I have done is that this d T by d x I have obtained by differentiating this equation and substituting if I differentiate what do I get q dot l square by 2 k into minus 2 x by l square this l square this l square will get cancelled out and minus 2 x becomes minus 2 l 2 2 gets cancelled out. So, I have got minus q dot l by k. So, if I substitute that minus q dot l by k here minus into minus becomes plus k k gets cancelled out. So, I have q dot l. So, q dot l upon h plus T infinity gives me T s that is essentially what I have tried to do in the last few days. So, next is on the same lines we can do the radial system before I do radial system I would like to solve this problem which is there for plane wall then we will approach towards radial system. So, it is a very interesting problem and it is going to tell us something about when to apply resistances and when not to apply resistance analogy we cannot blindly apply resistance analogy anywhere we like. So, that is what this problem is going to tell us. So, the problem is a plane wall is a composite of two materials A and B. So, we have two materials A and B material A is 50 mm this is a typographic mistake this is not 50 meters it is 50 mm and L B is 20 mm and here the wall of material A has volumetric heat generation 1.5 into 10 to the power of 6 watts per meter. So, that means in the material A we have 1.5 into 10 to the power of 6 watts per meter and remember this volumetric heat generation is only in material A not in material B this is the major difference in this problem and thermal conductivity of the material A is 75 seeing that number itself you can say that it is some something like a metal. So, 75 watts per meter Kelvin and L A is 50 mm and material B has no generation and thermal conductivity of material B is 150. Please note this number the thermal conductivity of material B is very high as we go along we will see that this resistance is very less because of high thermal conductivity and the L A sorry this is L B typographic mistake L B is 20 mm and the inner wall surface temperature the inner surface of the material A which is black in color here that is insulated that is this side is insulated. But on the other side of the material that is the one side of the material A is insulated other side of the material A we have material B and for material B to the left of it we have material A and to the right of it it is actually convective boundary condition that is while the outer surface of material B is cooled by water stream T infinity equal to 30 degree Celsius and H equal to 1000 watts per meter squared Kelvin. So, you H you see it is significantly high why it is high because it is water and it is flowing so 1000 it is. So, typically for water H is very high and it is flowing as well so naturally H should be very high 1000 watts per meter meter square meter squared Kelvin this is typo it is not meter Kelvin squared it is meter squared Kelvin and T infinity that is the water temperature is 30 Kelvin. So, here we should not be blindly applying the resistance circuits. So, we need to carefully solve this problem so what is known we know the properties of material A and B and there is volumetric heat generation in A and there is no volumetric heat generation in B and this is insulated and this is convective. Before we solve this problem we need to appreciate one of the major point which we studied just a while ago. If you just see here we have let us take symmetric boundary condition let us forget about asymmetrical boundary condition. If I take symmetrical boundary condition you see what is happening I have convective boundary condition and I have at the origin at the origin what is the thermal gradient what is dT by dx at x equal to 0 even if you solve this you will figure out that because of symmetry itself you do not have to really solve because of symmetry itself you can say that dT by dx is 0 that is the mathematical implication. But what is the physical implication what is the physical implication let us say I have a situation that is there here. Let us say I have a situation where in which this is insulated and this is convective boundary condition still whatever solution we got for this situation is valid isn't it that is even if I have an insulated boundary condition physically and a convective boundary condition on the right of my wall the same temperature distribution is the same temperature distribution is valid is that okay and the center line temperature is also going to be same this is what this is just we have played with the mathematical implication and the physical implication the mathematical implication is that it is symmetric but the physical implication that dT by dx at x equal to 0 is 0 means it is physically insulated okay so that is precisely what we are doing here. So in fact that is what is being written here it is important to note that the plane of the symmetry the temperature gradient is 0 if you differentiate this and put x equal to 0 you are going to get that as 0 or you can easily see that it is because of symmetry you can say that it is 0. So one of the implication is that this solution also applies to plane wall that are perfectly insulated at x equal to 0 and maintained at a constant temperature on the other side at x equal to L that is the physical implication this is what we are going to capitalize in our problem now if you see here in this problem what is that we have done one side is insulated and the other side is constant wall P1 okay. So if I apply that relation here if I apply that so that is how you can see here there is a parabolic temperature distribution in material A linear temperature distribution in material B note this parabolic is because in material A there is heat generation in material B there is no heat generation that is why it is linear and this is the temperature drop in the thermal boundary layer okay. So that is essentially what we are doing here now if I see the if I see what is that I should be doing if I apply what is the energy balance or the what is the heat flux if I see if I make the energy balance at this interface at this interface E dot in to E dot out so what is the what is the heat leaving here what is the heat leaving here it is because of the volumetric heat generation Q dot into heat flux if I have to calculate that would be Q dot into L A Q dot into L A has to be equal to Q dot into L A has to be equal to heat flux please note here one important thing here I cannot apply resistance circuit but here I can apply why I cannot apply heat resistance circuit here because there is volumetric heat generation but here there is no volumetric heat generation so I can apply resistance. So if I apply resistance I will get L B by K B plus the convective resistance that is one upon H should be equal to the heat flux so what is the heat flux Q dot L A equal to heat flux so that should be equal to that is the same heat flux which is going which is going by conduction and convection so the heat flux equal to H into T infinity minus T 2 I do not know T 2 that is what I can compute that is Q dot L A equal to H into T infinity minus T 2 so that is sorry T 2 minus T infinity so that reduces to T 2 equal to T infinity plus Q dot L A by H so I know T infinity 30 Q dot I know L A is 50 mm and H is 1000 you get T 2 as 105 so same heat flux if I apply that is T naught minus T 1 or T 1 minus T 2 equal to we can solve it in multiple ways but we are applying this equation this in a volumetric heat generation case we had derived this that is T naught equal to T 1 plus Q dot L A square by 2 K A we had just derived little while ago so I will compute T 1 from the thermal circuit so T 1 needs to be found I do not know T 1 as of now I know only T 2 how will I get T 1 T 1 will come from net resistance that is Q double dash equal to L B by K B plus 1 by H if I do that and I know the heat flux that is 1.5 into 10 to the power of 6 into 15 into 10 to the power of minus 3 I get T 1 if I put this T 1 I will get my T naught if I put this T 1 and I know all other parameters I will get T naught please note here let us take stock of the numbers what we have got T 2 we have got T naught 5 T 1 we have got 115 see here as you can see that the conductive resistance is much larger than convective resistance if you see here conductive resistance is 150 by 20 mm but convective resistance is 1 by H is 1000 so convective resistance is very less so T 1 to T 2 temperature drop is significantly less compared to that of T 2 to T infinity so what I am trying to say is that you see T 1 is 115 T 2 is 105 in the wall there is a temperature drop of only 10 degree Celsius that is material B wall there is a temperature drop of only 10 degree Celsius but between the outer wall and the water there is a temperature drop of 105 to 30 why because the convective resistance is much lower than the conductive resistance this is the significance that is how the resistances help us in feeling the numbers feeling the numbers so the important point the message what we are going to carry from this problem is that the electrical resistance analogy we started applying only from material B not in material A because in material A there was there was volumetric heat generation so this is the most that is why this problem has been cookable. So having understood how to handle volumetric heat generation on the same lines one can handle radial systems so we have just now till now we have handled volumetric heat generation in plane systems that is the plane wall so similarly we will touch upon radial systems in a radial system if you see that is the cylinder if I take so this is the equation which we have taken yesterday this was the equation we equated that to 0 why because we had neglected the volumetric heat generation now we take volumetric heat generation that means essentially we are saying that the temperature is there in the radial direction and there is volumetric heat generation so that is this cylinder and we have volumetric heat generation and I can either have constant wall maintained or I can take convective boundary condition both mean the same why I say that because I can apply or I can I can balance the heat flux on the wall between convection and conduction and get relation between T s and T infinity through H. So if I differentiate this equation like we did for the last case I am not going to do step by step on the paper I am just going to take you through the steps because I have done that for plane wall so if I differentiate that if I differentiate this sorry integrate this I get r dt dr minus q dot r squared r squared why it has come because I have multiplied this equation throughout by r if I integrate this first term I will get r dt dr and I have r q dot by k that is q dot by k into r squared by 2 that is what is this plus c 1 minus is because I have pushed this from left hand side to the right hand side. Now I will divide this by r dt by dr equal to minus q dot by 2 k into r plus c 1 by r now if I integrate that I get t of r equal to minus q dot by r squared by 2 k already 2 is there so I get 4 k plus c 1 by r that is c 1 log r plus c 2 what are the boundary conditions I am what are the boundary conditions here you can expect that at the center the temperature is going to be symmetric. So that is the symmetric boundary condition dt by dr that is the gradient of the temperature at r equal to 0 is 0 that is at the center and at outer wall that is t at r equal to r naught that is r naught we are taking it as equal to t s that is t s so if I substitute that I am going to get if I substitute this boundary condition in this we can directly see that dt by dr equal to 0 and r equal to 0 so c 1 will tend to become 0 so c 1 is 0 so all that I need to find is c 2 that is for finding c 2 I have to substitute for r as r naught and t at r equal to r naught is t s so c 2 will be equal to t s minus q dot r naught squared upon 4 k so that is what is substituted here. So if I substitute that I will get this c 2 if I substitute that in this equation there is no c 1 anymore so I get t of r equal to t s plus q dot r naught squared by 4 k into 1 minus r squared upon r naught squared again so this is again parabolic you can see that this is again parabolic I am just rearranging this t of r minus t s upon if I find again central line temperature t naught and divide it by t naught minus t s I will get 1 minus r by r naught squared you do not have to worry about this it is just an algebra if you just put r equal to r naught you are going to get that as t s plus q r naught squared by 4 k so if I again divide it by t naught minus t s this is going to vanish that is all we are saying. So if I have to relate t s with t infinity and h if I do the energy balance so I get q dot into volume what is my volume pi d squared by 4 or pi r naught squared into l equal to h into bathing area that is complete cylinder bathing area 2 pi r naught l into t s minus t infinity if I equate this pi pi gets cancelled one r naught vanishes l l gets cancelled out so I am going to end up with t s equal to t infinity plus q dot r by 2 h so basically this is what we have understood in case of volumetric heat generation if I just take a recap of what we have done so the main point what we need to understand is that in this problem wherever volumetric heat generation is there you cannot apply resistance circuit I am telling this again and again because while solving problem usually students tend to make this mistake they think that resistance analogy can be electrical analogy can be applied anywhere and everywhere no everything comes with restrictions so this is the restriction. So we need to appreciate this so now after doing this let us just take couple of problems one more problem I would like to take this is a formulation problem I would not like to take this yeah this problem let us see it is a very interesting problem because there were various questions yesterday on thermal conductivity variation with temperature and everyone wanted a unique answer how does temperature vary so generally we found when we went back and saw the equation when we went back and saw the variations of temperature for various metals generally for metals we are seeing that the thermal conductivity is decreasing with the increase of temperature so but it need not be same for quads we saw that the temperature was increasing so why we do not know that we have to answer out answer little later but now the question is is the temperature distribution in a plane wall always going to be linear may not be so what is that I am trying to do is I am asking a question myself yesterday's till now I have been always saying K is constant now let me relax this assumption and take K equal to K naught plus A t that is K naught is varying with temperature that is that is that is that is the formulation so that is K equal to K naught plus A t in any textbook usually you would see this is a standard problem usually this is the question which is asked so far what we have solved or we have found that it is linear is that if I have a plane wall I have taken linear in which case it would be linear where K equal to K naught that is when A equal to 0 but now let me go back and take a case in which A is positive let me take a case in which A is positive so you will see that for A is positive my temperature distribution is going to be is going to be something like this that is convex and for A negative it is going to be K so why so it is coming from Fourier's law of conduction so the argument is there here so that is when A equal to 0 there is no problem so the question what we are posing is when A is positive it is going to be something like this A is positive A is positive why it is so so if A is positive what will happen K equal to K naught plus A t so what is happening if I differentiate this D K we all know that in a plane wall when heat transfer is taking place from left wall to right wall what is happening to the temperature temperature is decreasing how it is decreasing we do not know how means whether it is decreasing linearly parabolically we do not know that but definitely temperature is decreasing so if I differentiate this K with respect to temperature D K by D t if A is positive this is a mistake D K by D t is not alpha it is A so D K by D t is positive then A is positive so that means what K is what does the D K by D t positive means K is increasing with the increase of temperature or K is decreasing with the decrease of temperature so what will happen K increases with the increase of temperature what does my Fourier law say Q equal to minus K into D t by D x so as K increases this is capital T temperature is decreasing as K x increases that is as I move from left to right of my wall what is happening to my temperature is decreasing so is K because D K by D t is positive so with the decrease in the temperature K is also decreasing but what is constant for a plane wall Q by A has to be constant the heat flux is constant so in order to maintain the same heat flux what should happen might slope that is D K by D t should increase so that is what is happening here so if you see back so D K by D t is increasing D K by D t is sorry D K by D t is decreasing because of the temperature decrease but my Q by A has to be remaining constant in order to keep it constant the only way to compensate it is D t by D x should the slope should increase that is why compared to the slope of the straight line the slope is high here so that is why my nature of the curve is going to be like this for A equal to positive so conversely you can argue out for A equal to negative is that okay so that is how we can understand this is a very important concept so this is just book keeping the slope so if you just take a recap what is happening temperature is decreasing so but the K variation with T is always positive so D K by D t is going to be decreasing so but in order to compensate for this decrease D t by D x has to increase that is the slope of or the variation of the temperature with x has to increase in order to keep my Q by A as constant that is all it is just book keeping okay so this is actually in fact I would think that this is this why I am harping so much on this is because this is usually our PhD interview question so we always ask this question and most of the times people find it very difficult to answer this question so why because we are not used to think this way so I would request all of you all the coordinators in the class spend considerable time in explaining this concept okay so I think with this problem this problem I am skipping because this is there in the tutorial today afternoon we are solving this problem okay so so much about volumetric heat generation so I think we are at the end of the session for next 5 minutes I will just take 3 to 4 questions first question is VNIT Nagpur any question in the plane for a plane P is equal to A plus B x plus C x square the plane for a plane P is equal to A plus B x plus C x square is the math cad method math cad method applied and what is the peculiarity of that math cad what is math cad as far as I know the math cad is just like excel it is just an off software what is this math cad method sir math cad applied for determining the temperature mean or at a particular length of that plane if you want to determine the temperature so in math cad there are certain limitations so I need to know about math cad math cad is something like an excel sheet it is just a software which calculates temperature based on the values what you give A B and C there are no limitations of math cad math cad is limited by our mind so it is just how we give the temperature distribution as A B and C it is going to compute that is all it is okay sir thank you over to you M.A.S. Pillai on well good morning any questions see one of the participants wants to know wants me to re-explain this k variation with temperature so let me just go slow as slow as I can see what we are saying in a linear case k is constant and what does Fourier's law say q equal to minus k A dT dx so there is a linear temperature gradient that is what we found yesterday's derivation and that is this linear temperature variation now what are we saying so and we know that when there is a heat transfer from left wall to the right wall temperature is going to continuously decrease that means T is going to decrease with the increase of x so what will happen to my dk by dt so if A is positive dk by dt equal to A so but my temperature is decreasing so dk by dt is what will happen my dk by dt is positive so as my temperature is decreasing my k is also going to my k is also going to dk by dt is positive as my temperature is decreasing my k is also going to decrease but I have to keep my heat loss rate constant that is q has to be constant because it is given by minus k A dT by dx the only way to compensate this decrease in thermal conductivity with the increase of length is by dt by dx so dt by dx has to go up compare to the case in which k was constant so that is why my distribution temperature distribution is going to be convex for A equal to positive that is upward I hope I have made I think I have made you understand okay I think we are stopping the question and answer session no professor Arun will take over.