 We will get started back again. So, one of the questions over coffee was that, is this tan hyperbolic ML very sacred? Is it true for all cases? Is it true for all cases? Tan hyperbolic ML, we said that it is going to represent the, yes, infinitely long field is it, is it sacred number ML equal to 2.5? What is the boundary condition I have used while putting this? Q fin of finite length upon Q fin of an infinite length. I have taken in the top for finite length which boundary condition I have taken? Edyabatic. If my boundary condition changes my ML also should change. This ML of 2.5 is pertinent only to Edyabatic boundary condition. If my boundary condition changes my ML also is going to change accordingly. This is just to show that how do we arrive at infinite length, infinite way that is all. So, we I guess we had broached up efficiency, now effectiveness. What is effectiveness? Effectiveness is we took for efficiency reference what was the reference? Q fin upon Q ideal, Q fin maximum Q ideal. Effectiveness is Q fin upon Q no fin best case and the worst case we have taken. Best case if I compare with it is efficiency of the fin. Worst case what is the worst case? Having no fin. So, that is effectiveness. So, effectiveness you would like to have which effectiveness is the best effectiveness? If this is the ratio of the effectiveness definition that is the Q fin upon hab into tb minus t infinity what is denominator representing that? What is denominator representing here? No fin case. So, you want this effectiveness of the fin to be what? Greater than 1. If it is equal to 1 what does it mean? There is no use of putting a fin. Is less than 1 good or bad? Bad because you do not want your heat transfer rate to be lesser than no fin. Yes, access in exactly that is what is there here. It exactly access in insulator which is not intended when we use a fin. We want the heat transfer rate high. So, you can now you know what it is. So, you can relate effectiveness of the fin with the efficiency of the fin with area of the fin upon base. So, I think with this you can generate whole lot of equations for effectiveness of the fin and your total heat transfer rate of the fin is Q unfin plus Q fin. What is unfin is this portion where there is no fin is the base and the area of the unfin to the area of the fin. So, that is how we do the bookkeeping. That is it. So, one thing what professor Hemant Kumar Mehta has just suggested which is right. In fact, just two days back we realized that in today's mid-sem question paper sort of we have given that we have only. So, we have to define what is called as efficiency overall efficiency of the fin when there are multiple fins. How to handle that in the form of circuit diagram where we can add that it is there in incorpora and David. In fact, why I say if you see incorpora and David he will not deal so much about this tan hyperbolic energy the way or he will not even draw this figure. In fact, I myself understood much better when I saw this figure what is efficiency of the fin. This figure is there in chamber. I mean what I mean is images get retained in our minds without any difficulty. This figure is not there in incorpora. He tells that in one sentence, but that sentence has to get into my mind. So, that is why I say chungal is better. This figure is there in chungal. Anyway, I like chungal very much. So, maybe I am praising too much that is fine. So, so much about fins. So, now you have a problem. What is the thin fin and thick fin? Effectiveness. You have it. Effectiveness you may have. It looks like that is a longer length of few number of fins of shorter length of fins. That is there. Here it is there. Yes, it is there. That is there. Now the concepts are there whether I can compute now whether it is a thick fin or thin fin. No, I may be wrong. You just correct me. So, if I understand your question correctly what I understand is that see we are trying to optimize two things. One is spacing of the fin. Another thing is thickness of the fin for a given material of the fin chosen. Is that right? So, your question is how thick should be the fin and how it should be optimally spaced? Is that right? One can compute for once I decide on a configuration. Once I decide on a configuration, when I say configuration, what do I mean? Fin can be of different types, right? Let us say I decide on in this case for example, I decide on annular fin case. So, now the question is thickness and the fin spacing. I have to play with those with this chart. I can get those numbers. I can then decide which one is the best. Does that answer your question or not? I am not sure. Yes. Correct. Yes. Definitely, it will affect but then I will have to sit down and do the calculations myself. But are you asking for general derivation? Yes, I vaguely remember long, long time back something like the thick and thin fins I had read somewhere. Maybe that is what you are broaching up. If I vaguely remember that is there in the Hong Kong War. During the last test time I remembered. What is the maximum weight of the thin fins? Those things cannot be given as numbers. So, in general, thin, grossly spaced fins is better as opposed to thick for… No, actually now I remember. I think I can get back. I do not know whether I have the Hong Kong War or not. No, actually in few textbooks yes, thick and thin fins are separately dealt with. I can understand that. I do not recollect what exactly is a thick fin and what exactly is a thin fin. We will write down. P by A. Yeah, no. But there are standard relations derived. So, we will write down. I will write down. But all that actually is covered in post-graduates. But anyway nevertheless let me put thin and thick fins and then let us see whether we can calibrate accordingly. Is that okay? So now, but by and large I guess we have covered the basics. Using this I guess one can cover several other things. So we will move on to transient conduction because the idea is we will spend half an hour now on transient conduction half an hour 40 minutes. And then whatever wherever we are we will stop there and we will start taking feedback from you whatever time it takes. That is the reason why we have put at 430. So that there is no imposition and over gels on the time to leave. So, the first thing here is here so far we were dealing only with steady state conduction. We never bothered for temperatures being varied with time. But now we will take up a simple case. We always go from simple to complex. So in that pursuit we will take up temperature as a function of time alone but not pace. So that body in which the temperature is a function of time but uniform all over is called lumped ball because it is lumped everywhere. It is lumped so it does not matter where the temperature I am taking. So temperature is same everywhere but it is changing with time. That is lumped ball. So let us see how do we handle this. So this is what I have just taken copper ball and potato example but it is fine. Copper ball because thermal conductivity is quite high. Potato thermal conductivity is not so high. Potatoes thermal conductivity would be copper thermal conductivity. I think we should be having the feel of these thermal conductivity as a teacher. What is the thermal conductivity of copper? 386. 386. At least if I know it is of the order of 400, fair enough. Thermal conductivity of water? 0.6. Thermal conductivity of air? 0.02635 at room temperature. It tells you see 1 is 0.02, 0.6, 400. Morning we were showing order of magnitude. That tells you how much is the thermal conductivity different. These numbers should be there in our mind. Should I buy hard or should it come naturally it is up to us but we have to know those numbers. I do not know. We need to know those numbers otherwise we cannot convey it to our students properly. So here also I can answer this straight away through thermal conductivity why because so I have thermal conductivity of copper very high. As opposed to thermal conductivity of potato why did I ask thermal conductivity of water because potato thermal conductivity can be taken as water because potato is full of water. So thermal conductivity of potato is around 0.6 but copper is 400. So copper is logical that temperature gradient should be k dT dx. So dT dx will be less if k is high for copper as opposed to potato. So now let us take this lumped body. I have taken a hot body and thrown it into cold water actually this I am going to I know all of you would have done tens of times. Nevertheless I want all of you to derive along with me after the coffee. I know you are all little bit rejuvenated but still let us derive this why because I am going to give experimental implications of this equation. We can design at least couple of experiments based on this simple derivation. It is innocuously simple but it is having very deep understanding. Let us see what is that. So now I have taken a hot body and immersed in a fluid and the fluid temperature we are assuming that the fluid temperature is not going to vary with time. My fluid sink is sufficiently large that my T infinity of my fluid is not going to change with time. So if I do that let me write my first energy balance equation which is what we do always e dot in minus e dot out plus e dot g is equal to e dot s t. So I have a body this is my control volume. What is e dot g here 0 there is no energy generated. Now what is e dot s t? rho v c p d t d t rho is here density of my body not fluid v is the volume of my lumped body c p is specific heat of my lumped body d t d t is the temperature gradient with time. Now what is there here e dot in is there or e dot out is there. It is a hot body out ok. So e dot in is so what is e dot out what is the mode of the heat transfer for the e dot out to occur? Conversion. Conversion. Conversion. So that is minus h A s into into t minus t infinity is that right. So if I write this equation so what do I get if I write come on do it for me take t minus t infinity as theta ok. So if I push this h this negative sign inside what do I get h A s t infinity minus t equal to rho v c p d t d t ok and what do I get for t infinity minus t minus theta ok. And what is this h A s by rho v c p or rho v c p by h A s what is the unit of rho v c p by h A s per second or second. Second it is second isn't it kg per meter cube meter cube joule per kg h is what is the unit of h watts per meter square Kelvin that I can write as joule per second meter square Kelvin meter square everything gets cancelled out you end up with seconds rho v c p by h A s. So what do I get here what do I get h A sorry minus theta equal to yes yeah joule per kg Kelvin you are right otherwise that will not get cancelled you are right here minus theta equal to what do I get come on you are not helping me I am not supposed to do myself yes d t is not going to be there because t I am writing in theta d theta by d t on the right hand side rho v c p by h A s let me call that as tau ok. So if I separation of variables if I do minus d theta by theta equal to equal to d t by tau. So if I integrate this between theta i to theta and this I can write 1 by tau integral of d t 0 to t what do I get and push this minus side this side can I push this minus side this side so what do I get theta by theta i equal to sorry log of theta by theta i log of theta by theta i is equal to minus t by tau. So this implies theta by theta i equal to e to the power of minus t by tau or t minus t infinity upon t i minus t infinity is equal to e to the power of minus t by tau ok. So now what is what does this tau mean I am now flipping back are we through with the derivation yes so what does this tau mean actually we call this tau time constant time constant ok. So what is this time constant here define first order system this is actually a first order system in fact the same equation you have seen in first order first order systems definition in measurements is that right everything is same even in simple r c circuit it is the first order system ok. So here time constant if you see it is resistance into capacitance it is resistance into capacitance now how can I utilize this before I want to e transfer rate and all that how can I use this equation for various measurement applications for various measurement applications before we go to measurement let me go to this problem I am going haphazard I know I am consciously doing that. So I have taken a small thermocouple I have taken a small thermocouple and here this thermocouple can be approximated as a sphere and the convective e transfer coefficient between the junction and the gas is taken as h equal to 400 watts per meter square Kelvin quite a high value and junction physical thermo thermo physical properties are k equal to 20 and c p equal to 400 and density is given to be 8500. So determine the junction diameter needed for the thermocouple to have a time constant of 1 second if the junction is at 25 degree Celsius I want all of you to solve this problem if the junction is at 25 degree Celsius and is placed in a gas stream at which is at 200 degree Celsius how long it will take the junction to reach 199 degree Celsius come on solve it for me yes so I have k equal to 20 c p equal to 400 rho equal to 8600 junction temperature is thermocouple initial temperature is 27 degree Celsius okay tau has to be formed tau is given tau is given to be 1 second but what is question is diameter so tau is given to be rho v c p upon h a s is that right tau is equal to rho v c p upon h a s so you see here tau is equal to rho v c p upon h a s. So tau is known to be 1 second h is given a here is pi d squared so into rho pi d cube that is the volume pi d cube by 6 is the volume c p so you get diameter equal to 6 h tau t into rho c p if you substitute what is the diameter you are getting does that make any sense what is the diameter you are getting into 1000 if I make I get 7 into 10 to the power of minus 1 that is 0.7 mm it should be 0.7 mm what does this mean it means a lot it means a lot for planning my experiments it means a lot let us say I am measuring a flame temperature let us say I am measuring a flame temperature we all know any fire or flame if I am measuring a flame temperature and my flame for some reason is going to vary with time I want to capture those variations with time let us say and those variations of flame temperature are less than or of the order of 10 seconds let us say they are varying of the order of 10 seconds then my thermocouple time constant should be faster than that or slower than that faster than that so that is why I have taken 1 second the problem we are given is 1 second okay it is very nice problem and here h is also very high which is a quite high okay 400 400 so if you are the time constant of a thermocouple is not dependent on the properties of the thermocouple they are dependent on the environment in which you are going to measure for example here h if the h is high my time constant is going to be time constant go back rho v cp by h as that is why I made you write h is high my time constant is going to be less if I am measuring temperature of a water whether it is natural convection or forced convection its time constant will be much faster than what I am measuring in because we have ingrained somehow that thermocouple material decides the thermocouple time constant it is not sure where we measure is what is important in the morning I was telling in the rarefied gas dynamic it is very difficult to measure thermocouple temperature because h is there of the order of 0.01 now how much time it will take to reach steady state it will not reach steady state at all that is the meaning that is why I said this sounds very innocuous but there is more depth in understanding this there is more depth and you see 0.7 mm and flame temperatures we usually for measuring flame temperatures the diameter of the thermocouple is of the order of 25 micrometers means how much mm 25 20 to the power of when you can anyone help because I cannot think 0.025 mm what is my air diameter air diameter is 75 micrometer that is how I think 25 micrometers means 3 times lesser than my air diameter I should have the thermocouple bead size not the wire size then only I can capture the flame temperature properly and not just the flame temperature here we are not just talking about the steady state temperature, temperature fluctuations of the flame because burner is going to burn and there are going to be temperature fluctuations of the flame if I have to capture that my thermocouple has to be faster than the temperature variation of the phenomena what I am capturing that is very important that is that is that is what this little fellow tau equal to rho V C p upon h a s is going to tell us this is going to aid us in designing of our experiment designing of our experiment because we always think that we put a thermocouple anywhere and we can get anything and everything that is not true always that is not true always okay so much about this problem yeah so what professor is saying is that what will be the time constant how will the time constant compare with forced convection and natural convection for a given fluid let us say air this room is filled and now fans are rotating that is forced convection same air temperature I want to measure natural convection mode fans are off which one in which case thermocouple will respond faster so tau depends the thermocouple response time depends on the environment in which we are measuring the temperature okay that is very important another thing now let us get back to the same equation let us get back to same equation that is rho V C p now I do not want to see this equation in this form let me get back and try to see this equation in a little different form that is let me write back my equation as rho V C p dt dt equal to h a s into t infinity minus t I am getting back to the same equation you might be wondering why am I revolving around that you will understand little while from now now let us say t infinity I just want to measure the heat transfer coefficient heat transfer I have now let us say I have a flame I am getting taking back again the example flame now in the flame I insert a body let us say that body is a small body why I am taking small body why I am taking small body because I want to make it lumped because this equation is valid only for lumped if I take a huge body as huge as this then it may not be lumped so to make it lumped I have taken a highly thermal conductive material either copper or brass or bronze or stainless steel or aluminum I made a small ball spherical ball now I want to measure the heat transfer coefficient around that ball that is I have a flame and I put a ball now I want to measure the heat transfer coefficient around this can anyone suggest with this simple equation how to measure that heat transfer coefficient yes it is possible now this ball inside that ball let me embed a thermocouple let me embed a thermocouple and I measure temperature versus time plot I measure using a data logger I measure temperature versus time it picks up some way it is going to pick up now now yes so now I have temperature as a function of time because I have measured it so I get from temperature versus time I get dt dt okay and I know rho v ct I know as and t infinity I know and t I have I have I have measured and I have measured so I can get I can get people do this people do this now let us say this I just took for external flow I can do the same trick for internal flow now let me take a pipe let me take a pipe simple pipe okay I will take questions I will take a pipe and in pipe you would have seen in most of the papers you would have seen they use copper bodies small copper blocks for mungation another day was using it but he was doing steady steady experiment so there are small copper blocks if I have a test section like this you imagine square test section and in this test section I am putting plenty of copper blocks and each copper block I am separating it out by wooden piece why am I separating it by one wooden piece in between two copper blocks so that one copper block does not thermally talk with the other okay now each copper block I put a thermocouple okay so now why I have those copper blocks will be very small very small means thickness will be 3 mm maybe width is 25 mm and length is also around 10 mm why am I taking these small little dimensions so that they are lumped so that they are lumped so if they are lumped now all of a sudden I will draw either hot water or hot air directly into this directly into this what will I measure now I will measure I will measure one is fluid temperature I will measure another one is what is that I would call lumped temperature I would measure okay so fluid temperature because I have generated some more hot air all of a sudden I have asked the hot air to enter into my test section which is this so what will happen my fluid temperature will go up like this after a while because there was a cold air it is picking up picking up and then it will become constant now what will happen to my lumped body how will my lumped body temperature respond that will also pick up something okay now I will choose this portion why do I choose this portion because my T infinity now has become constant otherwise my T infinity is also function of time which is not what case I have derived this one okay so now I have to do some little rearrangement to get my H what am I trying to do what am I trying to do in all these examples I am trying to devise methods to measure H in different class of problems I showed one for external flow and I am now trying to show for internal flow okay so what do we do rho vcp same equation rho vcp dt dt if I take I will take I will call this as T initial T final okay T initial and T final this is T initial and this is T final so now let me integrate this because I am looking for H so H equal to rho vcp upon As into what do I get T infinity minus T dt upon dt reverse okay let me rewrite this so that is H is equal to H is equal to rho vcp upon As right dt upon T infinity minus T dt now I have to integrate this numerator will get integrated between T initial to T final okay and denominator will get integrated between small T initial to small T final so that is small T initial to small T final what is this below denominator what is this what is this if I have to show in the figure that is this area under this curve is that right so what do I understand from this equation can I get this or not now H I have measured I have measured all the temperatures so I know all the temperatures so if I find the integral T infinity minus T this I can find out either with since one third rule or trapezoidal rule if I have temperature as a function of so I can find heat transfer coefficient even in internal convective flow but only condition is that what I will care I should take care what all precautions I should take care I should devise my body is such that it is lumped that is why I have put wooden pieces in between so that one body is not talking to another body if I make full test section full of copper then it is not going to become lumped and another beauty of making small small pieces is that I can capture local heat transfer coefficient that is the beauty of this so what I wanted to tell by taking these two examples is that what we read as rho V C P D T D T H A S T infinity minus T V which we routinely derive in the class has tremendous experimental implications experimental implications not only this you can you can go on devising your method to be honest with you my first my first PhD student on flames on premixed flames he is indeed going to measure heat flux this heat flux and heat transfer coefficient using this while teaching only I understood what I am trying to say is that if you teach more involvedly it will give you ideas for your research okay fundamentals definitely help us in research they are not two disconnected things they are both connected the more we get involved in teaching in teaching of fundamentals the more it will help us in doing that is the reason why I took examples I can go for measurement of here H on and on at least two or four three more examples but I will stop here