 So, applications of course, everything actually to say everything is based on this conservation of energy, it is a different more general form of the first law of thermodynamics, it is the law of conservation of energy, all our problems will have this thing as the undercurrent. Maybe we may not write this explicitly, but this is what we are going to budget all the time. So, in as if you have a class test or a quiz, please ensure that this form of the energy balance is tested, ask them to write this equation in this form in the mathematical form and also ask them to write the word form of this equation. So, that this is ingrained in a student and as a heat transfer teacher, I would say that there are 3 or 4 things which a student has to take away from a course in any course for that matter and for this course, this equation definitely he or she has to take hold and even in the sleep he or she has to be able to write this. So, these are a bunch of examples which are there, where heat transfer is important, we have heat transfer in almost all spheres of everyday life, air conditioning system, refrigerator, water heater, so on and so forth computers, solar collectors, power plants, spacecraft everything. So, from a microscopic device to a large application like a power plant or a spacecraft, practical things is like you are insulating your house that is again a heat transfer problem, how much time it is going to take to cool, the person making sweets in a shop definitely has to know some heat transfer, he does not know heat transfer, but if he is making a sweet and has to be cooled, so that he can put it into appropriate shapes, it has to cool. So, larger the volume and it is going to take longer time, he knows that, but probably he does not know the mathematical implications of it, but heat transfer is there even in that everyday application. Then water for a cup of tea, has heat transfer implications into it, so it is a very, very vibrant useful subject and with this I would open the forum for any questions on this aspect, before we proceed with the further example problems. Next five minutes, we are planning to take about five questions at most in each of these sessions, during the tutorial of course, as you are doing the problems we will have, we will be free, so you can ask us questions appropriately at those times, but in view of continuity we would like to take five minutes of questions between every session. We have NIT Tirchira Palli, are you able to hear us? What is the question please, over to you. The question is as follows, K is a material, it is a material property, is it correct? K is a question is, K is a material property, is that correct? Over to you. Sir, I just want to know whether, I just want to know whether it is, material property is nothing to do with the other factors, no, over to you, please. Yeah, it is purely a material property, of course this material property will depend on the temperature, actually K is a material property, yes it is only material property, but of course this material property depends also on the temperature, we will know much, many more details about K in the just next one hour lecture, so I think we will take up this, we will, we will wind this question. We are with JN to you Hyderabad. Yeah, please go ahead. Plus advection, so diffusion is similar to conduction or something different? Yeah, diffusion is nothing but conduction, diffusion is nothing but conduction, everything in fact, the knowledge is getting diffused from here to you, isn't it? So, that is how even the conductive or the heat transfer is getting conducted through molecules within the boundary layer in convection, that is all we are saying, that is all we are saying. But anyway, we will take up this in much more detail when we go to convection, but it would suffice to say that convection is nothing but conduction within the thermal boundary layer. Okay, fine, I think we will get back to again teaching mode, we will stop the questions. Okay, we will get back to questions, I got again two more questions, yeah please go ahead. What is the significance of heat transfer coefficient and utility of the H to solving the problems? Yeah, that's it. What is the significance of the heat transfer coefficient? Yeah, yeah, that's a very good question actually, that's a very good question. To answer that question, the question is for everyone, the question is what is the significance of heat transfer coefficient in computing the heat duty? Actually, the Newton's law of cooling, we will discuss in detail, but for now I would just say that the Newton's law of cooling is not about that equation q double dash equal to H into T wall minus T infinity, it is not that actually. H definition comes back to temperature gradient within the thermal boundary layer, but to answer this question, H is the one which is going to define the amount of the heat transfer in any given problem. See for example, if we take jet impingement or we take natural convection. In the jet impingement, the heat transfer coefficient is significantly large compared to that of the natural convection cooling, so that is that is signified or quantified on the basis of heat transfer coefficient only. But, H is not computed from that equation, H is computed on the basis of the temperature gradient within the thermal boundary layer, for now this much would suffice. I will come back to this definition of H much more later, but H is the one which is required for computing the heat duty, not q double dash equal to H into T wall minus T infinity. Is it a material property? No, no, actually now questions are multiplying, I can understand. It is not the material property, it depends on the thermal boundary layer, it depends, it does not depend whether my fluid is flowing in a copper pipe or stainless steel pipe, but it depends on the fluid that is whether my fluid is water or air or some other fluid or oil. I think now, I think now I will stop my discussions and we will get back to teaching that. We will get started with the problems, but on a quick note I will just take a recap of what professor Arun has taught us little while ago. That is this is the summary, that is we said that there are three modes of heat transfer, one is conduction, convection, radiation, conduction and convection require medium, radiation enjoys or releases no medium, that is radiation is perfect when it is in vacuum and conduction and convection do require medium, without medium they cannot happen and the general notion is that conduction usually occurs in solids, it is not true, conduction does take place also in liquids and gases, but it is not so effective, why we will come to that little later because of thermal conductivity, convection of course requires medium. That was the basic definition or three modes of heat transfer introduction what professor Arun gave and he gave us he introduced as the energy budgeting that is E dot in minus E dot out plus E dot g equal to E dot st. Now, let us just apply one or two problems for this energy budgeting, what does this problem say, this problem says that there is a long wire we all use this wires for heating there is a 2 meter long wire and 0.3 centimeter diameter, 0.3 centimeter means it is hardly 3 mm, 3 mm wire which is electrical wire means generally it is made of nichrome, generally made of nichrome wire which extends across a room at 15 degree Celsius that is my room is sitting at 15 degree Celsius and heat is generated in the wire as a result of resistance heating that means I squared R heating, this I squared R heating we use in IM boxes as well and the surface temperature of the wire is measured to be 152 degree Celsius you see the temperature is quite high it is 152 degree Celsius in steady state operation that means I have put on my iron box or put on the wire put on the current and my temperatures of the wire have reached steady state whenever I see steady state I mean my temperatures are not varying that means my wire temperature has reached 152 and it is going to stay there and it is staying there also the voltage drop and electrical current through the wire are measured to be 60 volts and 1.5 amp this gives me how much power I am electrical power I am dumping into my electrical wire this regarding any heat transfer by radiation the temperatures are quite high it is not proper to neglect radiation however because we are not studied yet radiation and convection we will just go ahead and apply we will neglect radiation and we are going to determine or the question asked is determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room so what is known here we know the wire dimension we know the room temperature we know the surface temperature and we know the voltage drop and the electrical current as professor told we need to first list out known find assumptions and then draw a diagram and then only go for solution so we have listed down what is known what is that we need to find convective heat transfer coefficient and what are the assumptions involved as I said it is steady state and we are neglecting radiation we are only taking convection into account so here of course we are not written that here e dot in minus e dot out plus e dot g equal to e dot st so here it is being steady state e dot st nothing is changing with time so e dot st will turn out to be 0 and energy is being generated that is e dot g is equal to voltage into current that is 60 volts into 1.5 amp so we have 90 watts so what is e dot out the heat is lost the heat is lost from this wire it is being generated by I squared r heating and it is lost it is lost by convection so convection is accounted by Newton's law considering that is h a into t s minus t infinity whatever is generated is lost by convection so that is e dot out to put it e dot g minus e dot out equal to 0 that is e dot g equal to e dot out so that is q equal to h a into t s minus t infinity a is the bathing area that is surface area that is pi into diameter into the total length so that is pi d l is my surface area so we have q dot equal to h into surface area is 0.01885 meter square into t s minus t infinity t s is 152 degree Celsius and t infinity is 15 degree Celsius so now I said q dot g equal to e dot g equal to e dot out so I have e dot g as 90 watts and that is equal to e dot out which is equal to h a into t s minus t infinity if I plug in area into t s minus t infinity I am going to get 34.9 watts per meter square degree Celsius this is quite low so this suggests that the mode of the heat transfer is generally natural convection for this problem at least so the heat transfer coefficient in this problem is significantly low so this is one of the problems where in which we have demonstrated how we can elegantly apply the energy budgeting okay so actually we had one question from one of the participants that one of the teachers that is the h being computed from q double dash equal to h into t wall minus t infinity here in this problem we can say h is being measured so it is so h cannot be directly computed using this relation it can only be measured we can be by applying the boundary layer concepts we can compute this but that we will analyze that or handle that when we come to convection so there is another problem so I think we will quickly try to complete this problem before we move on to one dimensional conduction so that is the problem says it is a simple combustion chamber that is there is a furnace the hot combustion gases of a furnace are separated from ambient air and its surroundings which are at 25 degree Celsius by a brick wall of 0.15 meter thick that means furnace means we can all visualize that as a chamber an enclosure so within the enclosure there are all hot combustion gases so may be some natural gas or some methane is being burnt and this furnace is separated from the ambient through a brick wall which is of 0.15 meter thick 0.15 meter means 15 centimeter that is half a scale 15 centimeters half a scale it is and we have an ambient which is sitting at 25 degree Celsius and which is sitting at 25 degree Celsius that is what we understood and the brick has a thermal conductivity of 1.2 watt per meter Kelvin the brick material which is usually an insulating material is having thermal conductivity of 1.2 watt per meter Kelvin under steady conditions you see here the word is steady state conditions that means the temperatures are not going to change and outer surface temperature is 100 degree Celsius. So here in this figure T2 that is the temperature of the wall which is facing the ambience that is the atmospheric air is sitting at 100 degree Celsius and the free convection heat transfer coefficient or the free heat transfer to the air adjoining the surface is characterized by convection coefficient of 20 watts per meter squared Kelvin. So here 20 itself suggests that it is natural convection so h equal to 20 watts per meter squared Kelvin so what is asked is what is the brick inner surface temperature that is T1. So here to take a recap we have a brick wall which is 15 centimeter thick and 1 wall temperature is known that is outer wall temperature is 100 degree Celsius and it is exposed to ambient which is undergoing natural convection so we have natural convective heat transfer coefficient at this wall as 20 watts per meter squared Kelvin and the ambient temperature is 25 degree Celsius. As we can see there are three modes which are occurring here also we can say that we are neglecting radiation. So known is told what we need to find is the inner wall temperature so what are the assumptions the first assumption is steady state condition and we are assuming that it is one dimensional heat transfer and radiation heat transfer is being neglected. So as professor has taught us e dot in minus e dot out plus e dot g equal to e dot st e dot st is 0 because it is steady state e dot generated nothing is being generated here in our brick wall nothing is being generated so it is e dot in minus e dot out equal to 0. So that is e dot out is convection and e dot in is conduction which is taking place in my wall. So I have q conduction equal to q convection that is e dot in equal to e dot out so q conduction is operated by Fourier's law so k into t1 minus t2 upon l equal to h into t2 minus t infinity note here I have taken per unit area double dash I have taken q conduction double dash minus q convection double dash so area per unit area of the wall. So I do not have to budget that area even if I take that area it is going to get cancelled in fact in our problem area is not given at all so we do not need to know that because we have taken it per unit area so I have k is 1.2 t1 is what is to be found t2 is 100 upon 0.15 equal to 20 into 100 minus 25 so I get t1 is 287.5 definitely this temperature has to be higher than the outer wall because this is the wall which is facing the combustion gases so this temperature has to be higher than 100 which is what is happening here. Yeah here actually if you see here what Professor Arun wants me to emphasize is that actually if you write Fourier's law of conduction actually it is minus k into t2 minus t1 upon l but that minus has been already absorbed so k into t1 minus t2 by l so as such if I write the Fourier's law it will be minus k into t2 minus t1 by l that minus has been absorbed that is why I have k into t1 minus t2 by l so with this I guess we are through with the conservation of energy that is we have energy budgeting we have taken care so now this is the conservation of energy applied to a control volume we have taken care of course the problems which we have solved is both for steady state we have taken for steady state and no energy generation we took for the second problem at least. So now let us move on to the steady state conduction and of course the order in which we will go is first we will take steady state and then we will take 1d and of course we are not going to touch upon 2d and then we will go to transient that is studying with temperature variation with time so let us what are the objectives of this lecture so what we are going to focus is we are going to focus that we are going to emphasize what is steady state and what is unsteady and what is one dimensional two dimensional and three dimensional so if we see here temperature we have written as temperature as a function of x y z and that. So if we say that the heat transfer is not going to vary or the temperature is not going to vary with time then we can say that temperature is going to be a function of only x y and z that means the temperature if it does not vary with time then we say that the heat conduction is in steady state that is it is steady state heat transfer but if the heat if the temperature is going to vary with time then it is going to be unsteady for example let us say I take I put on the iron box I put on the iron box initially my iron box is going to go through unsteady that is my wall of the iron box is going to the temperature is going to continuously increase but after some time it will reach constant then I would say that my iron box is reached steady state there is a reset there is an automatic on off button there which takes care of this steadiness if my temperature of the iron box comes down again my heater goes on. So that is what we mean here that is if the temperature does not vary with time it is steady if the temperature varies with time it is unsteady or transient the word unsteady and transient we use replacing. So then if temperature is going to vary only with x then it is 1 d either it can be only with x or only with y or only with z then it is going to be 1 d let us say if the temperature is going to vary with either x or y or y or z or x or z that is if temperature is going to vary in two dimensions then it is going to be two dimensional if temperature is going to vary with x y and z then it is going to be three dimensional. So as is the case always we go from simple to complex we first take one dimensional heat transfer. So before we move on what is the plan of this lecture is actually in this lecture what we are going to do is we are going to derive what is called as heat diffusion equation. So let us see we before we move on how do we do that. So as I have whatever I told in the last transparency is retold here at the cost of being repetitive let me tell that again because it is told here nicely. One dimensional is temperature can be either a function of x or y or z 2D means temperature is a function of x or y or y and z and x and z three dimensional means it is a function of x y and z. Now as professor had told it is conduction rate equation. So Fourier had done this we know Fourier very well we know Fourier through Fourier transforms actually he has contributed in a great deal in Fourier transforms it is the same Fourier who has given us this Fourier's law of conduction. So this is told that this is not derived this is not derived this is coming from empiricism professor Arun wrote very nicely q double dash or q is proportional to area q is proportional to delta t and q is inversely proportional to thickness of the plate he had written this came from experiments when I say empiricism that means empirical empiricism comes from only experiments that means I will keep delta x constant vary delta t then find that q double dash is proportional to delta t and then I will vary delta x keep the delta t same then I will find that it is inversely proportional that is how this relation Fourier's law of conduction has been originated it is purely empirical that means it is coming from purely experiments. So then they found k is the constant that is it is a property of the material one of the questions was during interaction session k is a property of the material yes k is a property of the material alone. So q double q equal to k a delta t by delta x sorry here there is a mistake this is not q x double dash this is q x q x equal to k a delta t by delta x when I use double dash I should not be writing a here q x equal to k into delta t by delta x I mean here x means heat transfer in the x direction. So this is the same thing telling in different ways professor has already told us minus sign is necessarily telling us that heat is being transferred in the direction of decreasing temperature so to make it positive we are putting it as negative sign. So if I take heat transfer in all directions here you see q double dash we are not putting x or y or z it is taking place in all directions so this is the victorial notation that is q double dash equal to minus k into I delta t by delta x I is the unit vector that is in the normal to the x direction this is y direction this is z direction delta t by delta x delta t by delta y delta t by delta z if I have to represent all three dimensions this is q double dash equal to minus k delta t by delta x plus delta t by delta y plus delta t by delta z. So this is the same thing q double dash I q x double dash plus j j q y double dash plus k q z double dash and each one being represented in different key here del t del x del t del y del t del z and of course we have taken k is independent of direction that means medium that conduction occurs is isotropic iso means the same tropic is direction it is same in all directions and thermal conductivity is independent of the coordinate direction it is the same that is it is same in all directions. So that is about the Fourier's law of the conduction retelling so here we will before we go to heat diffusion equation we will just touch upon what is thermal conductivity we will touch upon two things here that is what is thermal conductivity and thermal diffusivity so thermal conductivity is again defined on the basis of Fourier's law of conduction only that is I think thermal that is again here you just remember yourself that q x equal to k into a into del t by del x that is what it is told here rate of heat transfer through a unit thickness of the material per unit area per unit temperature difference that is watts per meter Kelvin that is here meter here and meter squared they will get cancelled out so I end up with watt per meter Kelvin watt per meter degree Kelvin sorry not degree I should not be using degree Kelvin watt per meter Kelvin. So this thermal conductivity is a very important material property see if the thermal conductivity is high it means that it is a good conductor if the thermal conductivity is less it is less the conductivity or the conduction is less in the if you remember in the previous problem the thermal conductivity of the wall was the insulating material was 1.2 the furnace wall was made of a insulating material wall whose thermal conductivity was very less if you take thermal conductivity of copper it will be very high 400 so that is what is being told here so that is you see the thermal conductivity of pure metals is of the order of hundreds the thermal conductivity of liquids let us say water it is 0.6 so it is the order of 1 the thermal conductivity of gas let us say air it is 0.02635 watt per meter Kelvin so solids are very good thermal conductors next come in line is liquids and gases are not so good thermal conductors so even air can be used as an insulator if you have two walls if you want to insulate the heat transfer from one wall to the another you separate that with air and then that itself becomes a very good thermal conductor so thermal conductivity of solids is higher than that of liquids which is in turn higher than that of gases that is what this transferency tells us. So this is essentially why is this happening because the molecules are densely spaced in case of metals and lesser densely spaced in case of liquids and very the molecules are very farther away in case of gases that is all that is the reason why pure metals have very high thermal conductivity. So this thermal conductivity is because of it is going to material properties as we said so material properties means why who is contributing with for this thermal conductivity so that is thermal conductivity is because of two components one is K e that is because of migration of free electrons that is and the other one is K l that is lattice vibration the subscript e itself represents free electrons L represents lattice vibration so most of the pure metals is because of K e K e is much larger than K l but for alloys with the thermal conductivity is usually because of lattice vibration the contribution of K e is less. So for alloys K l becomes quite significant compared to that of K e but for pure metals K e is significantly large but for non metals for example quads the K l is significantly large K e is contribution will be almost negligible so that is what is being told here quads is very very crystalline structure that is there is a perfect lattice arrangement we have studied in material science there is a perfect lattice arrangement then the K l contribution will be significantly large and for insulation systems for insulation systems the thermal conductivity is very less. So just to take a recap of what I said in the last transparency for solid state the thermal conductivity is because of K e and K l for pure metals K e is significantly large it is basically because of the free electrons K l is the lattice arrangement that is how the molecules are arranged whether they are in FCC or BCC so for quads that is for non metals K l is larger than that of K e for metals again for alloys K L takes over or K e and K l will be of the same order. So now thermal conductivity of insulation systems of course here thermal conductivity depends on the how for example if you have thermal conductivity of a thermo wool how densely you place the thermo wool also affects the thermal conductivity that is what is being said here as volumetric fraction or wide space. So if I have no metal that is a no material let us say if two metals are separated by vacuum then the thermal conductivity will be almost negligible almost negligible but the radiation takes over there radiation takes over there. So there is another important material property which is thermal diffusivity that is thermal diffusivity is given by density specific heat and thermal conductivity thermal conductivity we just got acclimatized little while ago. So thermal diffusivity is given by K by rho C p that is alpha equal to alpha is thermal conductivity and sorry diffusivity alpha is thermal diffusivity is given by thermal conductivity K by density rho C p is specific heat of the material. So if alpha is large that means K is large rho C p is less see rho C p is like capacitance it is like inertia if rho C p if the specific heat is large means then it has more inertia so if K is large means thermal conductivity is very high so if alpha is very high means inertia is less thermal conduction is very high that means if alpha is large means it will respond very fast. So we will understand this alpha thing when we come to transient conduction through Fourier number for now it would suffice it to say that if alpha is large means my material will respond very fast to the temperature changes wherever I am putting that means if I take a copper ball and put it in hot water it will respond fast compared to taking an insulating material ball that is perspex ball and if I put that it will respond to hot water very slow because thermal diffusivity of the perspex ball is much lower than the thermal diffusivity of the copper ball. So this is what is the importance of thermal diffusivity just to take a recap we have introduced you two material properties very important material properties that is thermal conductivity I would just like to write that so that we do not rush through so that is I just want to write thermal conductivity I always remember thermal conductivity unit like I do not remember thermal conductivity unit of course now I got to remember that is Watt per meter Kelvin we do not have to remember this it comes from Fourier's law of conduction that is Q equal to minus K A delta T by delta X so little bigger right yes Watt Q is Watt and I do not know the unit of K area is meter squared delta T is degree Celsius or Kelvin delta X is meter so the unit of thermal conductivity would be Watt per meter degree Celsius so this is how this is how I remember the unit of thermal conductivity but now with the important unit of what is the unit of thermal diffusivity let us just write alpha what did we see alpha as K upon rho C P K upon rho C P what is the unit of K we just found that it is Watt per meter Kelvin rho is kg per meter cube C P is joule per kg Kelvin so now but we know that Watt is also joule per second so bigger you want yeah that is let me rewrite that is joule per second meter Kelvin kg per meter cube joule per second meter Kelvin Kelvin let us see what all I can cancel out so joule joule gets cancels out and kg kg gets cancels out Kelvin Kelvin gets cancels out so meter and meter cube becomes meter squared so unit of alpha becomes meter squared per second very interesting unit it is why I say interesting because we have come across this unit in fluid mechanics also so if you remember here from this transparency this is thermal diffusivity let us take up what is called as momentum diffusivity if I say momentum diffusivity I am sure you will not get me but you will realize within a few seconds what is that I am talking about see we know we know kinematic viscosity nu equal to dynamic viscosity upon density we know this from fluid mechanics so dynamic viscosity what is viscosity viscosity just gives us the fluidity that is if you see the oil it is quite thick but if you see water it is quite thin so viscosity just tells us the fluidity how fluid is my fluid so how thick is my fluid or how thin is my fluid is given by my viscosity and of course density we all know it tells me the heaviness so if it is dense it is quite to be it is got to be heavy if it is if it is not so dense it is got to be light so the unit of viscosity what is the unit of viscosity we know that it is or where does this come from let us let me go back let us go back I have to go back because we have not touched upon fluid mechanics so I need to before writing the unit of viscosity directly let me write it is coming yeah so tau equal to minus mu del u by del y this is shear stress we know stress means we can get attached quite quickly that is the greatness of Newton again we forgot Newton while studying Newton is the same Newton we gave who gave us the classical mechanics is one who is responsible for convective heat transfer quite interesting it is so del u by del u is the del u by del y is the shear strain so this is if we just do this what is that I get the unit of shear stress is Newton per meter square mu is del u is meters per second and unit of y is meter meter meter gets cancelled so unit of viscosity would be Newton second per meter square let us put those units back in my kinematic viscosity Newton second per meter square and unit of density is kg per meter so again okay we will have to get back to Newton again what is Newton its force that is mass into acceleration that is mass is kg acceleration is meter per second squared and I have in the numerator second and I have in the denominator meter squared kg per meter cube I cannot write bigger I guess okay so kg gets cancelled out so meter squared and here I am left out with one meter you see the unit turns out to be meter squared per second so what is this this is correct kinematic viscosity see as I said this kinematic viscosity I was telling this as actually let us write we wrote thermal diffusivity is equal to k by rho c p this is having the units meter squared per second this is thermal diffusivity that means temperature is getting diffused mu equal to mu by rho this is also having the unit meter squared per second this is momentum diffusivity in fact these basics are required when we go to convection so this is momentum diffusivity this is thermal diffusivity both have the unit meter squared per second so in one case in heat transfer in conduction the temperature is getting diffused by virtue of alpha and momentum in fluid mechanics momentum is getting diffused by virtue of nu that is kinematic viscosity we should not forget this these two are very very important properties that is the reason why we are emphasizing these two properties quite heavily so I now I guess with these basics we will get back to our transference is back where in which we introduced alpha and k so now let us get to the next important part that is the heat diffusion equation so here this is the mother of all equations of course this is next to energy conservation conservation of energy so the heat diffusion equation what is that I will just take a look at it and then come back what is that we are doing here is we are trying to write the governing equation write the governing equation for conduction alone and then how to get this governing equation is what we are trying to derive and that governing equation is the heat diffusion equation so as usual what we do here is I will what do I do is I will multiplex between or I will parallely use both the transparencies and the whiteboard so but I want all the participants across I want all of you to derive along with me so let us get to whiteboard let me see how much I can do so that is I have let me take a cube let me take a cube the dimensions of this cube are delta x delta y delta z okay. And of course we know now that defines x y and z directions and this is the heat transfer in the x direction that is q x which is entering at x equal to 0 q x this is in the x direction and which is getting out let me call that as q x plus dx q x plus dx and in the y direction it would be before we go to the y direction let us understand what is q x in the x direction would be we know from Fourier's law of conduction minus k a dt by dx what is the area here this is the area through which the heat transfer is taking place so that is let me get back to document mode and show you this again elaborately through this figure we will keep multiplexing this you see here what is that we are doing this is q x in the x direction this is getting out that is q x plus dx and this is dx dy and dz and we are seeing heat transfer in the x direction is q x similarly we will have q y that is this is q y and this is q y plus dy this is small mistake I will get it corrected this is supposed to come here and there is a subscript q y plus dy similarly you have q z and q z plus dz so here what we have here doing is we are assuming that there are two assumptions which are very important for any derivation as professor Arun told also earlier assumptions have to be emphasized over emphasized and told again and again that is here what are the assumptions in this derivation of heat diffusion equation what we are saying is it is homogeneous medium it is homogeneous medium and there is no bulk motion that is the fluid is not moving at all it is still fluid is not moving that is why do I make this assumption this is to ensure that there is no convection there is only conduction and homogeneous because all my material properties are uniform throughout that is what I mean and I am saying that the temperature is varying in all directions it is three dimensional that is x y and z so if you remember we had written we will get back to whiteboard we will use before writing q x and all let me rewrite again e dot in minus e dot out plus e dot g yeah is equal to e dot s t okay e dot in minus e dot in minus e dot out I know I have not written so great but you please remember e dot in minus e dot out plus e dot g equal to e dot s t so here this is what exactly what we are going to do in this derivation e dot in is going to be q x q y and q z q z and e dot out is going to be q x plus d x q y plus d y and q z plus d z okay so what is q x q x equal to I will get back okay q x equal to minus k what is the area area is delta y delta z into into delta y delta z delta t by del x that is my q x so I am now switching back to document so with that let us now I guess we can we can use this transparency so I have q x which I wrote as minus k del del y delta y delta z delta t by del x and q x plus d x here I have written q x plus del q x by del x into d x how did I get this how did I get this so that is I will use next transparency that is what am I trying to say is see we all know Taylor's expansion series that is if I have an expansion f of x plus delta x how do I write this yeah that is f of x plus delta x equal to by Taylor's expansion series I can write this as f of x plus del f by del x into delta x plus del squared f by del x squared into delta x whole squared and this goes on and on this is 1 upon 2 factorial 2 factorial happens to be 2 into 1 so if I write the same way q x plus d x or delta x that may write x plus delta x equal to what will it happen so it is going to be q x plus del q x by del x into delta x this is how I have written in that transparency but plus all higher order terms I am neglecting how can I afford to do that because my delta x is small my delta x is small that is the reason why I can afford to neglect because delta x whole squared becomes very small so I have q x plus del q x by del x into delta x that is what I have written here in my transparency so q x plus delta x equal to q x plus del q x by del x into d x again similarly I can write for q y plus d y equal to q y plus del q y by del y into d y and again q z plus delta z equal to q z plus del q z by del del z into d z this is if you remember all these three terms are going to be e dot out in our previous these are all getting out q x plus d x q y plus d y and q z plus d z these three are getting out getting in that is e dot in or q x q y and q z q y and q z so next is e dot s t. So what is e dot s t? e dot s t is energy generation per unit volume into d x into d y into d z so I should be knowing this q dot as a as professor said for nuclear reactor application we will perhaps know the energy generation or the heat generator so now that is this q dot d x d y d z next is so we finished e dot in e dot out e dot g e dot this is wrong this is e dot g here what is written is wrong this is e dot g this is not e dot s t this is e dot g so in the next term we will be having e dot s t that is what is e dot s t sorry here while writing also I have written wrongly this is e dot g please correct yourself this is e dot g that is I am keeping track of e dot in minus e dot out plus e dot g is equal to e dot s t at least you can correct me if I am wrong making wrong you realized it you can correct me so e dot s t so this is q x q y q z q x plus d x q y plus d y q z plus d z and this is q dot d x d y d z e dot s t what is this e dot s t equal to rho c p into volume that is d x d y d z into del t by del t. This is the term I usually you remember I usually use to take this for steady state this is 0 there is no temperature variation with time for steady state this term will become 0 so this is the energy stored term so we have written all terms now we have to just book keep all these terms so we have written e dot in is q x q y q z e dot out is q x plus d x q y plus d y q z plus d z and e dot g is q dot into q dot into d x d y d z and e dot s t equal to rho v c p that is rho c p into volume d x d y d z into del t by del t. Next is if I put all substitute all of this here q x and if I expand this q x plus d y q z plus d z plus d x if you remember q x plus d x is nothing but q x plus del q x by del x into that is if you remember this is the term we had q x plus del q x by del x into del t x so that is what I am using here if I put that q x q x get cancelled out I am left out with minus del q x by del x into d x minus del q y by del y into d y minus del q z by d z del z into d z plus q dot d x d y d z equal to rho c p del t by del t d x d y d z so this is of course this I had written already. So if I rewrite this what did I do between this step and that step I divided throughout by d x d y d z number one and I have substituted for q x as minus k d y d z into del t by del x del t by del x from Fourier's law of conduction in x direction and this is Fourier's law of conduction in y direction minus k d x d z del t by del y and q z is minus k d x d y del t by del z. So if I substitute this this reduces to del by del x into k del t by del x plus del by del y k del t by del y plus del by del z k del t by del z plus q dot is equal to rho c p del t by del t. This is nothing but the heat diffusion equation this is nothing but heat diffusion equation see if I take if I assume that my material properties are homogeneous and isotropic the k can be taken out that means k is independent of direction that is isotropic if I take then I can pull out this k and if I pull out this k and divide this equation throughout by k my equation is going to be del square t by del x square plus del square t by del y square plus del square t by del z square plus q dot by k equal to 1 upon alpha del t by del t del t by del t. So remember here alpha how did it come by because alpha equal to k by rho c p rho c p was sitting here and I divided it by k. So I got k by rho c p that is 1 upon alpha del t by del t I think I will stop here I think maybe I will take one question if anyone is having a question maybe I will just take one question here. So I see a question here with VJTI Mumbai I am going to VJTI Mumbai yeah what was your question. Good afternoon sir myself P K M away attending your program from VJTI sir one basic question which I am having is the Fourier's law assumptions they state that the material should be isotropic and homogeneous. So when we say homogeneous material is it applicable to can you extend it to a cube or a rectangular parallel or a pippoid made up of an alloy. Alloy it becomes difficult see for alloy it becomes difficult whether the k is going to be uniform everywhere or not but still yes even for an alloy we can assume that k is k is going to be independent of direction but when when does this k become not because we can generate materials where k can be let us say I want heat transfer only in one direction but I do not want heat transfer in other two directions let us say y and z I want heat transfer only in x direction I can generate a material if you ask me how to generate that material I do not know but that metallurgists will generate that material if they generate a material which will which is going to have k only in one direction then I cannot assume that the k is independent of direction then I am going to have k dependency only in one direction but for an alloy why for an alloy I can still go ahead and assume that it is going to be independent of direction but not on a molecular level but on a continuum level on a continuum level that means when I say continuum level I am going to take an infinite for example in this cuboid for an infinite decimal small volume I am going to have thousands of molecules that is they put it for 1 mm by 1 mm by 1 mm typical number of molecules are going to be 22.5 into 10 to the power of 6 number of molecules so we are not talking about molecular level but for infinite decimal level we are saying that it is independent of direction even for an alloy even for an alloy over to you specific value of k for the alloys that is the value of k for alloys that is also specified like you have shown also in the transparencies that k the range of value of k for alloys that is also fixed so the question is how is the k for an alloy is going to be unique over to you property wise what I can conclude from your explanation is that properties wise we can take it as a homogeneous but material wise it is not homogeneous is it right am I right no no no no see when I say property wise homogeneous material wise homogeneous both should mean the same no when the material is homogeneous then the properties are also going to be homogeneous so what we mean here is that the thermal conductivity is independent of direction it is going to be same even for an alloy even for an alloy when I say it is not having any preference of direction so otherwise in my heat diffusion equation what will happen I have to take separate k in x direction separate k in y direction separate k in z direction I am not doing I need not have to do that even for an alloy even for an alloy kx equal to ky equal to kz this can happen when my material is homogeneous and isotropic yeah you have that okay thank you sir thank you very much okay that is it I think we let us wind up this session let us wind up this session and we will assemble back again at 2 o'clock sharp 2 o'clock bye bye.