 Next we shall emphasis I think I have taken little longer than what I should have taken by digressing here and there because it was an introductory lecture. Let us now focus on the conservation of energy. So I would like to write because instead of doing this we will not appreciate. So I would like you also to write along with me whatever little I am writing that is E dot in minus E dot out plus E dot g is equal to E dot st. This is the main equation if I understand how to apply this equation I think I have understood the heat transfer. I can formulate a problem much easier ok. In real life what I understood is that after reading and all if I have to if someone comes and defines a problem how to formulate a problem is more difficult than solving textual problems. You can always solve textual problems but real life problems you cannot crack until we understand this basic laws how to apply those laws to real life problems ok. So what is E dot in? This is basically I should be choosing a control volume. I should be choosing a control volume. I should be choosing the control volume. Saket you are able to see what I am writing. So once I choose the control volume then things become easy that is E dot in is the what is E dot in energy transferred into my control volume energy transferred into control volume please write along with me otherwise it gets boring please write along with me into control volume and I would request you also to write. I know you know everything but let us write otherwise it becomes boring for you also. E dot out energy transferred out of the control volume E dot g is energy generated this need not be always positive what can be the word opposite to generated yeah but energy dissipated maybe or energy getting out of the control energy opposite to generated that is all I am I am not getting the right word energy absorbed maybe energy absorbed maybe the right word ok. So is that energy storage this is always tricky between probably ok. So energy generate see why I want to emphasize here because generated gives a connotation in our minds that it is always positive it need not be positive all that I want to say it can be either positive or negative E dot s t is the transient term that is energy stored energy stored within the control volume I am using c v for control volume I do not know how legible it is because I am used to writing on the board first time I have been forced to write on paper fine. So this is where I think we need to apply this is all it is it is very simple but we need to apply I think what I will do is we will take up immediately a problem on this and then solve it because then only we will be able to understand how to apply this. So there are various terms let us not get into this it can be conduction inflow outflow can be either conduction convection radiation so volumetric phenomena it can be nuclear we always give the example for energy generation nuclear rod so nuclear electrical chemical it can be anything it can be anything so let us not get into all this but let us try to solve a problem ok. So this is I have just put for the sake of completion I do not think I need to worry so much Professor Sukhatme already has told in day to day world we cannot get detached ourselves from heat transfer it is there everywhere so I do not want to get into this ok let us solve this problem and another thing it is not about solving a problem whether any textbook if you see especially incorpora and David and chungal whichever problem they solve there are five headings they cover all this one is known find next important thing is assumptions whenever we derive we need to emphasis more on assumptions in fact students usually go by derivation problem derivation problem but for them under what circumstances this relation is valid up to is to be emphasized more than the relation itself. So for every problem assumptions known find one can write and of course there should be a schematic we all think through pictures ok if you after ten years if you come you may not remember my name but you will definitely remember my face wherever you see why because we think through pictures images. So a schematic is very important because the clarity of mind improves if I have a picture if I have a figure assumptions so known find assumptions analysis and comments always students this time also the feedback in our model is what should I write in comment I have nothing to write in comment or my comments will be one page is that ok this is the feedback of from our VDAC always because why comment is important because comment is sometimes it design we write diameter of the shaft we will get point 0 0 0 0 4 5 mm we will write it like that but do I really think can I fabricate a shaft of that is there anything wrong so that is what thermal conductivity maybe I am trying to compute for a solid I end up getting point one for a metal and I end up getting point one is that right is there something wrong that stop and think is possible only when I insist with my students to comment. So it is not just problem solving here for our quizzes and mid-sense and n-sense also we insist that each heading has to be there we are copying this from incorpora and David and chungal but it is ok this methodology is very important otherwise you will solve the problem but to what extent it is not about just getting that answer but how am I following the procedure is more important than the answer. So we need to insist our student I do not know in the university how we can inculcate this I do not know if time we can perhaps change this but because I am also done in university I know very well how someone teaches someone corrects someone sets the question paper but somewhere we should try to bring this change that not only for this subject for any subject for that matter we should have known find assumptions analysis is what we solve and comment most of the times we just write analysis maybe we will write what is known and find we will write and just do the analysis we will never comment and we will not write assumption. So coming to this problem I would request you to solve this problem take a while solve this problem yourself there is a two meter long 0.3 centimeter diameter that is 3 mm 3 mm it is just very small 3 mm diameter electrical wire across a room extends across a room at 15 degree Celsius as shown in the figure your figure has been given heat is generated in the wire as a result of resistant heat resistance heating that is I squared R heating there is no induction heating or some other heating method and surface temperature of the wire is measured to be 152 degree Celsius in steady state operation this word steady state it is a very important word and I every day make sure that I use this word in the class have we reached a steady state because before starting the class always students are making noise that is natural thing for you you always hit your duster or you have your own methods to quell their noise. So I just keep silent but the point is I always before starting the class I always make sure that I use this word have we reached a steady state intentionally I keep using this at the point of getting bored why because the word is steady will get ingrained steady means things do not change with these two words are very important word uniform and uniform is that thing which does not change with steady is something which does not change with opposite to steady transient and opposite to steady can be unsteady also opposite to uniform is non-uniform these words have to be carefully used and they should be properly told what they mean to students for as it may be very obvious because we are using day in and day out but for a starting be student it takes a while before he absorbs these words that is the reason I suggested that you use the colloquial words in sorry technical words in your language regularly. So come on go ahead go ahead and solve this problem so you have a voltage of 60 watts in a current of 1.5 amperes you see this is a very important disregarding any heat transfer by radiation can I really disregard heat transfer by radiation here no 152 plus 273 is a quite sigma t to the power of 4 if you take epsilon of wire I think I can assume it is around 0.3 0.4 it is not it is not small I cannot disregard but I am assuming I am assuming that it is not involved here I am putting off radiation disregarding radiation determine the convective heat transfer coefficient for the heat transfer between the outer surface of the wire and air in the room. So what is known wire dimensions are known diameter length room temperature is known surface temperature is known and voltage drop of 60 volts and current of 1.5 find of course we need to find H schematic is already there what are the assumptions till now we have not taught anything transient or anything so and we have told that in steady operation and of course I have to assume that it is of course in the assumptions I know do not have to state but still it is a good idea steady state operations I am assuming and radiation heat transfer is a negative that is what I am assuming okay and then analysis how did I apply E dot in what is there here E dot in E dot out E dot g E dot st what are all there here Vi can be taken as E dot g because that is what is being generated and what is happening here in this wire I am heating this wire I am heating this wire and the wire is losing the heat by which mode through the outer surface by connection naturally here which mode of convection we do not know let us see it depends on the value of the H because nothing is being told in this problem whether fan is there or not so E dot g is 90 watts and what is this HATS minus T infinity that is that is E dot out E dot out okay here why are we using dots rates here we intend that these are rates so E dot g minus E dot minus E dot E dot in is 0 and E dot st is 0 so you have E dot g equal to E dot out of course I have not written that in so many terms there but that is what we need to emphasize for the student H equal to Q dot we have formed and you get and what is that you think this H would be it can be mixed convection but okay sort of yes but it is more leaning towards natural convection because typically we take 10 to 15 as natural convection and maybe around 60s and 70s as forced convection that is why professor said it is mixed convection so that is what we need to comment that is what we need to comment here but here now in the problem we are not told anything whether fan is there or not there but that can be yes it is more as professor said I would agree for mixed convection because H otherwise maybe Reynolds number Reynolds number is very less yeah but one cannot conclude just like that because it is air one cannot conclude just like that it depends yes so how do I decide okay yeah these are all quality typical so I can only make a statement I can only make a statement when I compare my Grashof number with Reynolds number so we will see when we handle g r by r e squared is what I keep I need to keep a tab off but I guess you can calculate do I have everything for calculating here perhaps yes re is not there because there is no so to that we do not have the information okay so there is another problem so at the end in fact I intentionally or consciously we are not giving notes up reary CD everything is ready and hard copies also I am not giving so we will give you CD on the day of leaving in fact it is occurring to me now why should we give CD also because if it is there on the model you can as well download it okay but anyway you will get the CD because model site is up by the time you leave if everything is there on the model website then you can download everything okay so I think I will stop here and Professor Arun will take over from here on steady state conduction yes please go ahead. Or can we take heat flow rate also for a definition that is we always taking q dot q double dot no not really not really you can take q dot q dot but especially heat flux is very convenient thing to handle why because if you are trying to measure any system let us say boiling heat transfer or radiation or anything it is good to always quantify things q double dash and delta t if someone wants to compute heat transfer coefficient let us say he can always compute it is not now when you say q dot then you will have to give the area dimension so that is why it becomes little generic all the not non-dimensional just to add if I am dealing with antigen system plane wall or just kind of thing the area is not a function of the direction of heat transfer in such cases q or q double prime heat flux should not matter correct for analysis but if I am dealing with radial system cylindrical or spherical it is always always always because heat flux is what is going to change because q is constant heat transfer rate is constant q by a is going to change radially q by q by a is going to change so heat flux is going to change so heat transfer rate is a good thing to deal with if you are dealing with radial systems heat flux if you are dealing with radial system you have to be careful what area you are using so this something is you have to keep in mind. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . that adaptation we have. So, we are hair splitting now, we are hair splitting, but I think in the back of the mind what I am doing if I know how I write I do not think it should matter. But yes, what you said is right. Good morning, just couple of minutes on this equation because after hearing his question why do we take E dot G. So, I wanted to ask a question about this equation to you when he was teaching, but I said anyway I am going to take over in a few minutes. How do you explain this concept to your students? First time, how do you explain this to your students? What kind of examples do you give? E dot G. No, this equation you are talking of an energy equation. How do you relate this equation to something in general in real life or something like that? Energy, forget energy. This equation is valid for any quantity. How do you teach this concept to the person? Balancing, account, accountancy you will balance your budget books. So, how is this equation taught to the student? That is very important because we were not taught like this. I can tell you. You take a turbine or compressor? Turbine or compressor would require lot of other information. So, the base way is internal combustion engine because. So, internal combustion, any other examples? One by one, one by one, one by one. Energy consumed by the body, human body, energy in and energy out, human body consuming energy and consuming like. So, will we be able to get some numbers for it? No. No, fine. Any other example? Nuclear fuel. Which one? Nuclear fuel. Nuclear fuel, any other examples? Boiler. Boiler, drying of the wall. So, everybody has thought in terms of engineering. Sir, we can quote an example of bank account. Which is what I do all the time. It is a classic, simple, easy to understand. Your grandmother will understand this. E in your salary is your energy in. E out is expenditure. What you withdraw for your monthly expenses, what you, online transaction, what you do, online billing payments that you do, that is all E out. What interest that you generate per day, that is your E generated. Now, that question is answered. E generated is something which can be quantified in such cases. E generated can, will be 0 if you are having a current account where there is no interest. But if you are putting in a savings bank account, there is a 3.5 percent interest. So, that is your generation. You have put 100 rupees. That is some money which the bank is paying you, which is generating, which is a generated term. And at the end of the month, when you balance, if the sum of the left hand side equal to 0, then whatever you have earned plus interest is equal to what you have spent, you are at steady state. So, no gain, no loss. E B can be minus. E B can be minus. In kind of, exactly. So, if you have overdraft, no correct, this is a very, I was going to say this, overdraft, penalty. If you withdraw more than 5 times in a non-bank account, you have to pay at 10 rupees thing. So, those are all E generated, which is a negative thing. So, this concept of balance is there in our life everywhere. So, it is there right from your bank. It is there in, as he said, body. So, your body is generating heat by means of what you eat, calories. Calories is another form of, another unit for heat. So, this energy generated, if you do not do any exercises, you are just sitting in front of the computer doing nothing. E out is very, very less. E s t is going to increase tremendously because E generated by the food intake is going to be large. So, this kind of a balance is what we need to appreciate. We need the students to appreciate. This rate equation is probably the mother of all equations. If you replace E with any other thing, still this equation is valid because the budget always has to be satisfied. How do you explain steady state to a student? I will give an example of a steady frame for a long period of time. If you have the same frame, you can state it here. If you are defining, explaining a term, you should not use that term. So, that is the general rule. What now I will try to explain? Time passes when you are having the same frame. Same? Same frame. If you do not use over a period of time, you have the same frame. Okay. Suppose for 10 years. Okay. Same, steady friendship. Okay. So, you can state that you have a steady frame. Okay. That is qualitative. Any other? Okay. Eating. Food item, for example. Initially, as your stomach gets full, you start eating less and less. Okay. So, steady concept again can be explained very easily by traffic. Okay. So, if your roads are choked, that means traffic is accumulating. Four roads are meeting at a junction. Typically, in all cities, we have this problem. We have a traffic jam. What is traffic jam? Essentially, that each road is increasing. At that intersection, there is a bunch of cars, vehicles, all those things which are creating confusion. If number of vehicles coming in one direction is what is going out, then you are at steady state. So, all these frames of reference which you have studied, fluid mechanics, Lagrangian, Eulerian, etc., forget all that. Just what is coming in is what is going out. If I take a snapshot at this time and at some other time, if I take a picture or stand there at the road at the gate of IIT and count the number of vehicles passing through from right to left for 10 minutes at 12 o'clock, at 1 o'clock, if I count if the number is the same and all through it has been the same in the situation at steady state. So, simple such examples will always, a student will remember this and conceptually, he might forget the equation. E hack, E dot here, all those things he will not remember. But at the end of the day, he will know it is a budget equation. So, something is coming in, something is going out, something is generated. Balance is something which is E stored. And the right hand side, if you write it in another way, thermodynamics, we also write this as, we have this in thermodynamics. De Cv by dt, first law actually says, De Cv by dt equal to q dot minus W dot plus m dot h i plus v i square by 2 plus g z minus m dot e h e plus v square by 2 plus g z e. So, this De Cv by dt in thermodynamics, we have seen variably cancelled. And q minus W, all those things were all black boxes, we just give numbers for that. Now, we are finding how to get that q, how to get those energy interactions by heat transfer. So, I think the connection, the perspective to a layman is also established. Always a rule of thumb, whether you are clear is, you should explain your stuff to a non-technical person first. If that person understand, your student will definitely understand. So, quickly we will cover the steady state conduction issues. So, as the professor has told, heat transfer, our main aim in heat transfer is to get a temperature distribution. Why? Because the temperature distribution, if I know, I know the temperature at every location and it will give me the heat transfer rate on q dot at for that process or for that medium. So, in general we can write T as a function of x, y, z and time. So, that is what is written here, T as a function of x, y, z and t. And x, y, z and t can be replaced by r theta z and time for cylindrical r theta phi and time for spherical coordinates. But in general what we want to say is three-dimensional transient situation. And by definition we say steady means temperature does not vary with time and steady means temperature varies with time. So, we can have a situation where we have a transient three-dimensional conduction problem in general or as a first step we will go with steady state one-dimensional conduction. That means T is just a function of one variable. We will call it x. If you are talking of radial system we will call it r. So, that is the first step. So, in general we have multiple dimensions, but as a first starting step we will deal with one dimension. So, conduction rate equation, again Fourier's law you should have, you would have been explained by professor. Fourier's law essentially tells you what it is a law which has been formulated by observation. The q is equal to minus k A dT by d x. So, it tells me that there is a material property which governs the heat transfer rate. The area which area, the area normal to the direction of heat transfer becomes important. Smaller the area which is available, smaller is the heat transfer rate. Not only delta T it is delta T by delta x which is important. I can have a 100 degree temperature gradient between the inside wall and the outside wall, but if it is over 1 kilometer that is not what it is going to be a very small number. But if this 100 degree temperature gradient is across 10 millimeters or 1 centimeter then that is a larger heat transfer gradient. So, dT by d x becomes important not just delta T. So, when we say, when we say q is directly proportional to delta T, it is proportional to delta x. All these things we say for the sake of thing, but we have to look at dT by d x as one quantity and not just dT and d x. That is also something which the student has to be able to appreciate. So, and classic professor Sukathmi had mentioned, classic example is this pipe insulation problem. So, this is told here. And remember one thing, there are two areas. One is the flow area. If you are talking about cylindrical coordinates, cross sectional flow area and surface area for heat transfer. Heat transfer is normal to the surface area. That is always the case. Plain wall does not matter. It is always going to be perpendicular for a radial system. The area is going to change because surface area is 2 pi R L for a cylinder 4 pi R square for a sphere and that is changing with every radial location. So, this is the general expressions which are there. We want to spend time. These are straightforward stuff. So, what is written here? The last equation 2.4 basically tells you heat flux in a vectorial notation is minus k del T. The operator del is nothing but i partial derivative with respect to x plus j cap partial derivative with respect to y plus k cap partial derivative with respect to z operator on the variable temperature. So, vectorially I can say minus k dT by d x represents physically the heat flux in the x direction, minus k dT by d y represents the heat flux in the y direction, minus k dT by d z represents the heat flux in the z direction and the combination of these three will be the total heat transfer rate. Now in all real life situations, whatever we give the situation, we will have a multidimensional heat transfer, whatever be the situation. So, what is this 1 d? What are we doing? What we are saying is one dimensional means the gradients of temperature, just listen carefully. Gradients associated with temperature are stronger or larger compared to the gradients in the other direction. For example, if I think of this fuel rod which is here I should keep. This is the radius R and this is the length L and L is say about 3 meters hypothetically. R is about 12 millimetre. So, this is about 3000 meters versus 12 millimetre. This is a nuclear fuel rod, typical fuel rod is of this dimension. One rod is extending to about 3 meters in length. Inside this there are several fuel pellets which are there. We are not concerned with all that. We are trying to explain where 1 d versus 3 d is going to be used. So, what we are saying is, if the gradient of temperature evaluated with respect to R and the gradient of temperature evaluated with respect to Z are calculated, it will be seen because R is much smaller than Z, length scale associated with R is very, very small compared to Z. All of you are familiar with boundary layer theory and fluid mechanics. In the boundary layer we will say this is delta. Delta is very much smaller than L which is the length scale associated with the x coordinate direction. I am using the same logic here. R is much smaller than the length in the axial direction. The gradient associated with temperature in the R direction is much larger than the gradient associated with the temperature in the Z direction and I assume azimuthal symmetry. Azimuthal symmetry means temperature is not a function of, in such cases, though two-dimensional heat transfer is going to be there, I can neglect for simplicity the heat transfer in the second dimension because of this factor. So, calculate K dT by dr and K dT by dz. K dT by dr is going to be at least two orders of magnitude larger than K dT by dz. So, I can neglect one in comparison to the other. If I am going to use a sophisticated computer code, then it might have a three-dimensional formulation, it is okay. But if I am doing a regular analysis back of the envelope classroom type analysis where no great design is going to take place because of our calculation, 1D is very, very useful in such cases. So, as an engineer, we have to inculcate when a student can use one-dimensional, when he or she can use two-dimensional. So, accuracy is not lost in this case if I use 1D. Whereas, this length is 30 millimeters compared to 3000 millimeter, then I am in trouble if I use one dimension because it is a short cylinder whose height is comparable to the radius. So, in that case, the gradient in the z direction will also become important and 1D approximation will give me a much less realistic result as opposed to a 2D analysis. So, if I calculate the temperature distribution for the long rod using 1D and 2D analysis, the results will almost be the same. But whereas, for a short cylinder, it is going to be largely different and that is what we have to tell them. Look, assumptions are there. You have a lot of time, you have a computer, you know to write a program, do what you want. But where to use 1D, where to use 2D, that field should be inculcated. So, these are the heat fluxes. In previous slide, sir, I have a problem over here that q double bar is equal to minus k i bracket i del t by del x. I think it is written only for the limit of water meter square. Because here we have taken A x. 2 double dash is already mentioned. Sir, that is ok. But it is also vector and if you are taking care of i j m k, it should be A x i e y z e z k. So, it is, I think it has taken that in x, y and z direction it is unity, that is why it probably you have taken q bar. Otherwise, it should have been written like this, minus k A x i dot del t by del x i. So, it is a product of cross sectional edge. Sir, the area in that radium, that t by del x. Because area is changing in x, y and z direction. The normal to the x direction area, normal to the y direction area, you are correct. It should be written like water meter square, area is unit. Point well taken. Yeah. In here, water meter square. At that point, it should be mentioned that A x equal to where E y equal to A z. So, you can say that it is per unit area in all direction, in all direction. It should be. We can put that in unit area in each of these directions. Otherwise, you can write it like q double bar is equal to minus k A dot diver. Got it, got it, area in all direction. So, that is the general schematic, where that is generally to emphasize only the. So, we are emphasizing a delta diversion of t and area both are a vector and the product of this will be the scalar. Correct. So, by this way, we can bring to the knowledge of the students also. Correct. That which area probably we would take care when we are writing down the gradient. Really. Yes. Yes. Okay. So, we have assumed. This is right. We have assumed that conductivity is independent of the coordinate direction. So, k has been taken as constant. Otherwise, that also has to be taken as a variable. So, these are just some slides to give you introduction on thermal conductivity. We are not going to spend too much time on it. As we said, it is a material property which is essentially, if you want to put a definition to it, thermal conductivity is defined as the rate of heat transfer through a unit thickness of a material per unit area per unit temperature difference. So, it is like a specific heat definition if you want to put in a definition for it. It is a material property. It is dependent on the molecular activity associated with that particular material. And all of us appreciate that larger the thermal conductivity, larger is the, for metals typically larger is the thermal conductivity and heat transfer rate will be higher. And non-metals or insulators will have low values of thermal conductivity. And all of us know the examples, fiberglass, sawdust, all these things are insulators. Metals typically are conductors, good conductors of heat. So, these are essentially the ranges of thermal conductivity. And metals are around the range here, 10 to 1000. Non-metallic solids are here and insulating systems are somewhere in this region. Liquids and gases, this is something which like to keep in mind, gases are much lower in thermal conductivity than liquids, one order of magnitude at least. And liquids are also not great, maximum of to about 10. And especially in design of heat exchange equipment in fins, etcetera. So, this thermal conductivity associated with the gas or the fluid which is flowing is going to dictate the, how much heat is going to be carried out. So, this is general stuff related to thermal conductivity. But actually there was a discussion. Which one? K e, K l, there was a discussion. Okay. Okay. So, actually we were discussing that the thermal conductivity is having two components. One is the thermal conductivity because of the vibration of our molecules and the other one is because of the lattice arrangement. So, usually in the metals the thermal conductivity in general is combination of K e and K l. K e is because of the vibration of these electrons or the molecules and K l is the lattice arrangement. So, lattice vibrational waves. The other one is K e is because of the migration of the free electrons. So, for metals K e is quite high compared to that of K l. But for example, quads. Quads is having, I think that is mentioned here. This is only for pure metals. Pure metals. Pure metals. For alloys yes, you are right. For alloys again K e and K l becomes more or less significant. I mean same order. Yeah, yeah, yeah. When I say the order I guess we are reaching you and we say same order. We use this word very easily. What we mean here is order means we say let us say tens of rupees or hundreds of rupees. Tens of rupees we do not say 20 rupees is equal to 30 rupees. 20 and 30 are of same order. But 1000 rupees is definitely two orders of magnitude greater than 10 rupees. The professor said r is significantly lower than L that is order of magnitude. Whether we are comparing in terms of mm or meter. So, we have to feel this order of magnitude. Here also when we say for alloys K e and K l are of the same order if not equal. So, similarly for crystalline structure if it is a it is a very perfect arrangement for example, quads that is what is coated here. It will have very high thermal conductivity but the contribution is because of K l. Of course, there is a book itself for heat transfer on molecular level whether it is conduction or convection or radiation. Stefan Boltzmann law can be derived from molecular level. So, that is there is a book by K V Yani. But that we are not intending there are lots of theories on predicting K itself. That is what Sandeep's PhD is all about. He is going to model for nanoparticles. So, we are not getting into that but this is vaguely what we are saying. We are not really answered any of the questions. But all that we are saying is thermal conductivity is because of free electrons or particular lattice arrangement that is all we are saying. So, that is why it is called electronics conditioning pure metals. Yeah, maybe yes. So, the key is very important. Yes, yes, yes, yes, in metals, yes, K e is significantly larger. And many of the K s, many measurements over the K e. At the point of time we should not forget that or what type of material we have. Definitely. It is laminated or something like that because it in each direction it has some different values. Correct. The concept of isotropic comes into the picture. Correct. Maybe here it is not appropriate isotropy. Maybe we can discuss now itself. There are materials. Yes. There are. Even with that first equation what we discussed over A ix and the other. At that point of time we have forgotten that K is the isotropic material. K can have preference to direction. K can be dependent on direction also. I can have K x, K y, K z. But for engineering purposes it becomes too difficult. But yes, we go ahead and assume that my surface is in fact we had put, I do not know whether we have put in our problem or most of the problems we assume that properties are constant and homogeneous. We use the word homogeneous. That is it is same everywhere. Uniform everywhere. It is same. It has no preference to direction. Composite material it has to be. Yes. There are, there are materials which they generate in which K is only in one direction but not in other direction. But we are not talking about them. Yes. In case of nanoparticles, it was not similar to the key what you are saying here. It is because of free electrons. Ok. There it was called as effective thermal conductivity. I do not know. I do not know about nanoparticles at all. Ok. Please go ahead. It is called as effective thermal conductivity which is going to take care of conduction part as well as convection surrounding this because of the particles. I am not aware of it at all. There is a long theory because it is taking care of any sleeping mechanism and everything. Sandeep, can you elaborate? There it is thermal conductivity of the fluid plus thermal conductivity of the material put together. Ok. Yes. Do we really have any defective thermal conductivity? Good question. Good question. Good question. Good question. I will tell you. Yes. The thermal conductivity per se is not so much dependent on pressure. It is dependent on even for solids, even for solids. It is not such a strong, I am not saying that it is not at all a function of pressure. It is, but it is not as strong as temperature. Suppose we are working with very high pressure in the fluid and at that point of time if we are bothering the effect of pressure on conduction. But it depends whether my pressure is changing from location to location or not. Here what is the question we are asking? The value at 10 bar and 100 bar will be different. But if the system pressure itself is varying from top to bottom, then it is a variable thermal conductivity. But at 100 bar if the system is homogeneous in pressure, that means it has reached equilibrium. Thermodynamic equilibrium is there. Then whatever is the thermal conductivity associated with that pressure is what it will give. To my mind it is because of the temperature variation we are taking care. We are clubbing to this one number and that is not right. Yeah, but see if there is a temperature. Let us not digress. So, parental temperature. No, temperature variation implies non-homogeneous, non-equilibrium. So, that means you are not reached thermodynamic equilibrium. So, steady state is not there then. If temperature at one point in the system and the other point is changing, then there is going to be heat transfer between those two molecules, right? We are mixing up too many things. We are mixing up. See question asked is I guess if it is homogeneous, k is same everywhere. k is not such a strong function of pressure as much as it is temperature. Yes, it is a function of pressure, but not as strong as temperature. It is like viscosity. Prandtl number is not only related with thermal part. It is also related with velocity also. Let us not mix up. Prandtl number is our Prandtl. Let us not mix up. We do not know Prandtl number. At this point of time let us not take Prandtl. We will handle when we reach Prandtl number. One more question. Yeah, yeah. It is obvious that for thermal conductivity of solid is higher than thermal conductivity of liquids. Yes. But then we are taking thermal conductivity of mercury as 8 watt per meter Kelvin. So, this 8 value, is it based on the liquid part of the mercury or of solid part? Solid. Solid. Solid. Thermal conductivity of liquid metals. Incidentally, we are working on melting of solids. It is solid. The number what you are telling is solid. When you take liquid thermal conductivity, it will be small for melting. See, to answer your question. So, if I have a container which is filled with lead, which is what the problem we are solving. So, thermal, now it starts melting. So, there is an interface. Now, certain portion is liquid and certain portion is solid. For the solid portion, I should be taking solid thermal conductivity. For liquid portion, I should be taking liquid thermal conductivity. Let me not get into, there is, in between there is mushy zone, where I will have to take an intermediate of these two thermal conductivity. To make your, to answer question in a very simple form, if it is in the liquid form, thermal conductivity of liquid metal will be lower than the thermal conductivity of the solid, of the liquid metal in the solid form. Yes, sir. Similarly, the thermal conductivity of the alloy metal. For example, thermal conductivity of copper is around let us say 300. 400, 300 and 80 per meter square kilometer. And then for brass, let us say 210. Yeah. If I mix these two together, copper and brass. So, what will be the thermal conductivity of that alloy? Will it be in average or will it be less than brass? I do not know really. I cannot. Probably it will be average because. No, it depends. It depends when you make, alloys, when I get into alloys, it depends which one is dominating over K and KL. Brass itself is an alloy. But in one of the textbooks, I read that the thermal conductivity of that material will be less than the least of two materials. I see. Brass itself is an alloy, right? Yeah. So, thermal conductivity of that alloy will be less than the thermal conductivity of brass. Okay. So, there is a combination of 400 and 210. So, there is a combination of 400 and 210. I will write down this question. It will be less than 210. The point I understood, the question what I understood is that if I mix an alloy and a pure metal, the thermal conductivity of the mixture will be lower than the thermal conductivity of the alloy. Because pure metal, least metal, pure metal will have higher thermal conductivity compared to that of the alloy. Is that right? Okay. I will think, we will think over this and get back. Okay. But off the hook, I do not have an answer. I do not know. Yeah. Mangesh, yes, please go ahead. That means nowadays the students are very enthusiastic to know that how to find the thermal conductivity of a composite material. Because we are conducting the practical of insulating powder and the composite slab and all that. But they are very eager to know about this thing also that, sir, if there is a mixture of two things, things are composite material there. So, how we can find the things? I think this question is not related. No, that is fine. That is fine, Navi. It is not that we should... I said, no, we cannot put under the carpet the questions just because that we do not know the answer. It is fine. Yeah, Sandeep. We have developed a measuring nanoparticle of this alloy in our lab. So, tomorrow we have a lab session. So, we will discuss that. Yeah. In fact, he is Sandeep Sonwane. He is doing his PhD on nanoparticles. He is putting nanoparticles in a fluid, 80 of fluid. Is that right? 80 of fluid to increase the... Thermal. Hopefully, he... Of course, thermal connectivity will increase. This is what we are talking about. And in that pursuit, hopefully heat transfer coefficient will increase. That is what his PhD about. That is what I request to put that point in the presentation of that 10 days workshop. Okay. That to find... Deal with... So, you are saying add a little more in detail. Little more, yes. Molecular level interpretation of thermal conductivity. Is that right? Right. Okay. And you can put also of mixtures. Of mixtures. Yeah. I know for sure it is there in KVANI. I have downloaded this book. I have it. Okay. In fact, I attended Professor Bandarkar's lecture. He derives in thermodynamics from molecular level to Stefan Boltzmann law. So, it is possible. So, I will try to read. I mean, we will try to add this molecular level of interpretation for thermal conductivity of not only pure metals, but also for composite material. Composite. Okay. Directionality basically. Thank you sir. Yeah. Okay. So, with lot of time spent on thermal conductivity, obvious next thing is thermal diffusivity. So, again thermal diffusivity is something which is hard to feel many times. Conductivity we can feel. But diffusivity is something related to storage of energy. And storage of energy is something which we cannot, we do not have a feel for it. So, this is another thing which students get stumped a little bit. What is the difference between conductivity and diffusivity? So, diffusivity is essentially related to the rho Cp associated with that material. Rho Cp everybody is m Cp delta t. From high school you have been studying m Cp delta t. M is nothing but rho V Cp delta t. That is what it is. Rho Cp is a material property. V is incidentally the volume. And if it is 1 meter cube, we can throw away that. So, rho Cp delta t. So, the amount of energy that if I put in so much energy, how much will go across and how much is going to be retained by virtue of its material property which is the product of density and specific heat. So, ratio of what is being conducted to what is being stored. Now, that occurs very frequently in our heat transfer analysis and we have to give that a specific name and that is called as thermal diffusivity. So, k by rho Cp this all of us are familiar with and larger the alpha value we call that by symbol alpha. And it has the dimensions of meter square per second which is similar to what kinematic dynamic viscosity, kinematic viscosity influence mechanics has. So, we will come to that later. Large alpha will indicate a larger thermal conductivity that means larger conduction heat transfer of post storage and smaller alpha would be vice versa. Now, this part I will just ask a quick question. How many of you derive the heat transfer equation in class? Full three dimensional equation everybody? So, there is nobody who is not so everybody is deriving. So, we will quickly go through this. So, in Cartesian coordinates typically all of us will derive and leave the cylindrical and spherical coordinates for students to break their heads with. So, here also we will do the same thing conduction analysis idea is to get the temperature distribution. So, my dependent variable is T independent variable that is x, y, z and time r theta z and time r theta phi and time. So, these are four independent variables and one dependent variable T. So, our aim here is to get a differential equation which gives me the temperature distribution in this form where T is the function of x, y, z and T. And we are going to make couple of simple assumptions that it is a homogeneous medium at least k is and no there is no bulk motion of the material. So, the logical way is to always start with an infinitesimal control volume. Students sometime you know the I think when heat transfer is taught they are in the third year. So, they are I think they have designed themselves to dealing with the infinitesimal control volume. But in second year et cetera the infinitesimal control volume is something which puts students off. So, essentially what we do is we take a small region, apply this concept of energy balance to that and then say if this is valid for this we expand it to the entire solid that is the concept of differential control volume analysis. So, what we do here is take a cube and this is the coordinate direction x, y and z into the plane of the board heat transfer from left to right in the x direction qx, qx plus dx, qy, qy plus dy, qz and qz plus dz from the face on the board and the face inside the board. E dot g and E stored, E g and E stored are the total energy generated by that unit volume or by that differential volume of dimensions dx, dy and dz, E stored is the energy storage. So, it is like a bucket, water is coming in, water is going out and somebody is there is ice which is melting which is also generating water and E stored is the change of water content. Similar thing we are dealing with energy. So, how do I relate? This students somehow do not like, but qx and qx plus dx. This is where mathematics comes in and we deal with Taylor series. We say qx plus dx or f of x plus h is equal to f of x plus derivative, first derivative and higher order terms. So, qx by dx times delta x, the change in the x coordinate here we call it dx. Same thing for dy and dz and then E stored, E dot stored is volumetric heat generation given by q dot times dv that is dx, dy, dz. E stored, E dot stored is rho Cp, this is dv, dx, dy, dz is dv. So, m is rho times dv that is delta m associated with that control volume delta m Cp delta t would be the energy change m Cp delta t, m Cp delta t by delta time will be the rate of energy change. So, that is what is this dp here. So, rho Cp dv times delta temperature by delta, this is energy stored. So, now we go to the familiar most fundamental equation E dot in minus E dot out plus E dot generated equal to E stored. E in is heat transfer in all the three directions that is x, y and z. E dot out is qx plus dx, qy plus dy plus and qz plus dz. I substitute everything in and cancel off the first term in this derivative expansion. So, I will get minus dq by dx delta x so on so on and so forth and q dot dx dy dz rho Cp dx dy dz dt by dt. What is wrong with this or why cannot we stop here? You cannot stop here because I require a temperature distribution not the heat transfer rate. Only when I get the temperature distribution can I get the heat transfer rate. So, the fundamental quantity is t and not q. So, therefore, I will say for conduction q is related to the temperature by Fourier's law and I say qx is minus k area for heat transfer in the x direction times dt by dx. Area for heat transfer in the x direction is this yellow red area that is given by dz times dy because heat is flowing across this phase. So, this is the area normal to the direction of heat transfer the area is dy dz and so if I incorporate that minus k a dt by dx is for heat transfer in the x direction k a heat transfer area in the y direction is dx dz dt by dy and corresponding term for the z direction. This is where students make a lot of mistakes because many of them deal with heat fluxes. So, heat flux when you use Cartesian you can get away somehow it will come correct. Now, what will happen is that heat transfer area in the radial direction is 2 pi r dr that that r is going to be staying inside the d by dr term and that they will miss out. Heat transfer rate all these are heat transfer rate we will we stop putting the dots because it becomes too cumbersome ok. So, Cartesian coordinates because heat flux and heat transfer rate are the same meaning it does not change with area we are in dealing with this way or even if the student makes a mistake by taking it does not matter, but in cylindrical he will be in for serious trouble. So, if I substitute this back into equation 2.16 here minus sign gets absorbed because of this d q x by dx will be replaced by k dy dz dt by dx into the bracket here dx is outside dy dz is inside the bracket that can also come out because that is not a function of x. So, everywhere in all the five terms I will have dx dy dz dx dy dz etcetera. I can cancel that off because volume is infinity small not equal to 0. So, I can divide through by that and I will get this differential form of the heat transfer conduction equation heat diffusion equation many of us have studied this even mathematics actually ok. This represents three dimensional transient conduction equation with variable thermal conductivity because k is a function of x y and z and uniform volumetric heat generation. What do I mean by uniform volumetric heat generation q is not location specific it is constant whatever it is it is a uniform value given by some watt per meter cube ok. So, this is the parent equation now to solve this or to be able to make any progress in understanding or getting a temperature distribution. I have t as a function of x y z and time second order in x second order in y second order in z first order in time. So, how many conditions do I need to solve this equation 2 in x 2 in y 2 in z and 1 in t 7 conditions have to be specified for me to even get a starting point for the solution. In real life it is not going to be possible at least in class to be able to do something and that. So, as I said most engineering applications at least back of the envelope calculations we can do with 1D analysis or maximum 2D analysis. So, what we can say is we can make several assumptions in the analysis in the first and the most obvious one that we will make the course is transient conduction is not there that means you are talking of steady state. So, this d t by d t will go to 0 we can also say volumetric heat generation sometime is not there. So, q dot will go to 0 constant thermal conductivity. So, k can come out of the differential sign if the right hand side is 0 then k will also vanish because I can divide through by k 1D or 2D I can cancel of the appropriate d t by d whatever x or y whichever I am saying it is not important and have a one dimension or two dimensional equation that can be solved quite easily and in fact some forms are given here. If thermal conductivity is constant I can pull k out divide through by k rho C p by k will be 1 over alpha. So, this is a 3D transient conduction equation with uniform volumetric heat generation and constant thermal conductivity. So, one assumption I mean k is constant. Now, if I have steady state right hand side is 0 2 assumptions k is constant and steady state if I say q dot is 0 then this goes off 1D means d t by d y d t by d z goes off I am left with a form. So, whatever be the assumption the student makes accordingly the equation is going to get simplified. In this equation everybody can solve second order ordinary differential equation as opposed to a second order PDE in three different variables it is only it is only dependent on x variable you need only two boundary conditions which we can specify very easily okay 2.10 which is 10 e dot g that is EGEL xi good EGEL xi okay. So, this equation what does it tell me yeah one dimensional in x that is obvious what else what is the implication first derivative if I integrate this once what will I get d t by dx is equal to constant because k if I take it as a constant d t by dx is equal to constant what does that mean constant temperature gradient what can you say it with respect to heat flux k d t by dx is also constant that means heat flux is a constant d t by dx is a constant k is a constant k d t by dx which is nothing, but heat flux in that direction is also a constant very, very important useful piece of information just by looking at the equation or first integration you will be able to get and for therefore, for a plane wall for Cartesian systems this is a very useful piece of information when you do not have volumetric heat generation heat flux is a constant that is why in students make a mistake of heat flux or heat transfer rate does not matter when I have q dot present then that mistake will not be pardonable because because of heat generation locally heat flux is going to be we will see that. So, I will just I will skip the problem I just want to introduce the boundary conditions and initial conditions of time. So, we have k d square t by dx square if k is a constant. So, essentially d square t by dx square is equal to 0 this I say this is my governing equation for this I need two boundary conditions what are the possible in fact, I will integrate this right away and then we can see this first integration will give me d t by dx is equal to c 1 and t of x is equal to c 1 x plus c 2 I think all of us will be able to tell this even in a sleep what does this tell me if this is a plane wall if this is the x direction t is a linear function this is the temperature distribution t of x and heat transfer rate is in this direction and q double prime is also a constant which is given by c 1 times k minus c 1 times k. So, q double prime x is equal to minus k times c 1 correct how do I get c 1 and c 2 I need two independent boundary conditions. So, let us say I have x is equal to 0 and x is equal to l. So, what are the possible boundary conditions all are teachers. So, one is obvious specifier temperature. So, first case is temperature is equal to T s at some surface second one incident heating some form of heating. So, specify a heat flux third thing is specify a convective boundary condition h comma T infinity and as a special case here heat flux is equal to 0 or non 0 if heat flux is equal to 0 I will call this as an adiabatic or insulated boundary condition that means if the wall is insulated there is no heat transfer through that wall. So, these are the family of boundary conditions which I can specify. So, here is a list because of the equation is in second order in special coordinates we need two boundary condition. If you are talking of a transient equation we need one condition in time which will typically be the initial condition we will go to transient thing when we come to transient conduction. So, here if I specify T at x is equal to 0 at this wall temperature at this wall at all times is a constant given by T s. So, I can anchor I can fix that temperature to a value the wall is always maintained at 100 degree centigrade. If you have steam condensing on the outside of a pipe steam at atmospheric pressure will condense at 100 degree then if you say the temperature remains pressure remains constant the condensation temperature remains constant wall is also at roughly the same temperature. So, constant wall temperature condition very classic boundary condition other thing is constant wall heat flux this is something which we can provide experimentally most of the times electrical heating or wrapping a wire around the tube or whatever we can get a constant wall or close to a approximately constant wall heat flux condition which tells me that minus k dT by dx at x equal to 0 at this location is equal to q double prime s. So, q double prime s is known dT by dx therefore, can be computed therefore, I can get the temperature distribution special case of constant wall heat flux is when q double prime is equal to 0 that means, dT by dx is equal to 0 that is a very very special case of constant wall heat flux and last and the most difficult or whatever you want to call it is convection at the surface minus k dT by dx at x is equal to 0 is equal to h times we have to be a little bit careful here in writing this heat is going to flow from left to right correct because the temperature is shown to decrease by this red line like this. So, heat has to flow this way. So, the fluid has to be at a higher temperature than the solid wall. So, it has to be h times T infinity the fluid hot fluid is giving the heat to the colder wall. So, it is h times T infinity minus T at this location T at the surface that is the heat flux which is given by minus k dT by dx.