 So next thing is we will go ahead with heat transfer from extended surfaces, here also I am not going to go in great detail, I am not going to solve the way we solved for one dimensional conduction, I am going to go on a fast track for fins. So first how do you introduce fins in your class, when you start off fins, how do you introduce fins? So what are the common examples you give for extended surfaces, I see, I think I see engines seems to be very common or may be it goes or gels well with the students as well. Electrical motor fins we do, correct, so that is precisely what here also we have tried to do. What I usually try to introduce is again incorpora and David's methodology or Changel's method, I start off with Newton's law of cooling and say that I am interested in removal of heat. From this equation it is very apparent that there are three ways in which I can increase this heat, one is either increase my h or increase a or decrease T infinity, increasing h and decreasing T infinity definitely will not come free. It involves using a fan or a pump or a refrigerator, so some pumping power essentially we need some pumping power. We want to still increase the heat transfer but not use pumping power, let us see. The only option left out is to increase the area, so with that I get started saying that okay I need to have fins and this is the typical example in which, I mean typical case in which very perfect fin I am showing where in which there are extended surfaces, okay. So there is the same thing blah blah blah whatever I told in so many words has been put so many sentences it is all there, see this material we have generated for the sake that even if there is no teacher, if the student is reading he should be able to see the flow that is the reason it has been put that way, so there are various applications I think you are all teachers so I need not have to harp on that. So these are the various examples of various situations, so I just say okay this is what is called as straight fin of uniform cross section, this is straight fin of non-uniform and this is annular fin and this is pin fin of varying cross section. I can have a pin fin of uniform cross section so these are various examples which we said. So then here we come to general conduction analysis of a fin but here in this we have taken varying area, varying cross sectional area although we give closed form solution subsequently only for constant cross sectional area at least while deriving we will derive it for varying cross sectional area. So what do we do we just take a small control volume okay again if you see here I am going to apply again my E dot in minus E dot out plus E dot g is equal to 0 of course in this fins we are not having E dot g. So first thing is basic assumptions what are the assumptions main assumption is that we are going to assume that the heat transfer is going to be in one direction okay there is no heat transfer in other two directions and we are going to assume steady state and thermal conductivity is assumed to be constant is it going to be constant it depends on the material and the temperature gradients we are going to handle may not be true all the time may not be true all the time and then convict you the most important thing is that I and of course radiation is neglected radiation is neglected which may be not a bad assumption for fins because the temperatures at the fin tip is not usually very high but I think serious assumption here most of the times is convective heat transfer coefficient is constant is it really correct or is it really representative of real life may not be true if I take a fin if it is it need not be same throughout if it is if it is natural convection natural convection also is dependent on temperature gradient. So wherever temperature gradients are high the Rayleigh numbers of the Grashof numbers are high heat transfer coefficients are there are high subsequently when the temperature gradient is decreasing as I proceed along with the fin my temperature gradients are less my Grashof numbers are less my heat transfer coefficients are probably going to be less. So it is it is convective heat transfer coefficient need not be uniform in real life but we go ahead and make this assumption with these assumptions we get back to our equation we say that okay at any x the fellow which is entering in is qx and which is getting out is qx plus dx and out is also convectively it is getting out that is dq convection and here there are ac I am going to use two notations that is ac a subscript c is going to be cross sectional area a subscript s is going to be surface area in fact more than surface area professor Bajan uses bathing area I like bathing area rather than surface area that is that is the area which is getting bathed by the fluid okay. So that gives me the physical feel rather than the surface area so I would like to call it as bathing area rather than surface area although both are same but nevertheless the feel is that how much it is getting how much area is getting bathed by the outside fluid that is what it means the a s okay with this we are going to put again here e dot in is equal to e dot out because e dot g is not there and e dot st is not there and we have k ac dt dx and qx plus dx we have expanded this is not at all new to us now and I have substituted qx plus dx equal to k ac dt dx that is equal to minus of kd by dx ac dt by dx if I substitute that there these two terms this and this will get cancelled out and dq convection is h d as let us not forget this is d as t minus t infinity this t is going to vary with x so that is what we are interested in we are going to get temperature distribution what are we up to what are we up to another important thing while teaching what I practice is that whenever I am in the middle of the derivation I ask always myself the question what am I doing and what am I up to where am I why am I doing this what are we trying to do in this in this general conduction analysis if you ask what are we trying to do we are trying to write the governing equation by doing the energy balance why are we writing this energy why are we doing this energy balance ultimately to get the temperature distribution we are interested in temperature distribution why am I interested in temperature distributions because I want to find out whether by putting a fin whether my heat transfer rate is going to go up or come down I do not want my heat transfer rates to come down I want them to go up but how much they go up can be quantified only when I get my temperature distributions is that right so that is the reason why we are going to do this so it is a good practice to stop in between and all of a sudden ask the question why why are we doing and what are we doing what are we doing and why are we doing okay so if you just do this we end up with this and if the cross sectional area is constant means which of the term will vanish second term will vanish that is dac by dx so we go ahead with fins of uniform cross section we take various cross sections I am not going to get into details of this so we get this equation d square t by dx square and hkac dAs by dx because As is also not a function of x so we are going to get dAs if the cross section is constant so you get dAs by dx as perimeter P capital P I am using for perimeter so I end up with d square t by dx square minus hp by kac t minus t infinity we write we tend to write this in the form of theta because it looks elegant so I get d square theta by dx square minus m square theta equal to 0 as we can see general solution of this this is the second order differential equation so the general solution we have all studied in mathematics c1 e to the power of mx plus c2 e to the power of minus mx to get this c1 and c2 and m is squared of hp by kac h is e transfer coefficient P is perimeter k thermal conductivity ac is the cross sectional area so how do I get this c1 and c2 boundary conditions what are the typical boundary conditions I tend to take usually base temperature I take it as constant I take it as constant tb equal to constant and other tip I take different boundary conditions is that right so what we all do I do not know what how you do for all cases you solve one the another thing I wanted to ask you is how do you teach you use power point or you use blackboard see what we do is although we have power point we are not going to flash the power point in the class actually we take we vary both of them we take both of them together whatever derivations I can do the derivations we are skipping here because we are all teachers we know what we are doing so I am skipping I am not doing step by step but what we do is on the board we write each and every step but only for taking recap in the next class what I taught in the last class I take a recap for 2-3 minutes by taking the recourse of power point so I think the problem with power point we find that I cannot keep them engaged with me that is the problem so how you guys do I do not know yeah how many are blackboard excellent excellent so how many number of students you handled in a class we are better off yeah yeah 4th at 60 to 90 we are also not bad we are also around that actually we have 150 we divide the class into 75 and 75 and we share that fine so so fine why was I asking that because we derive on the board only for one case and subsequent ones I just leave it for themselves although the solution is there here of course I am not going to do for everyone I took here adiabatic boundary condition if I substitute adiabatic boundary condition is d theta by dx equal to 0 if I substitute that and do all the algebra which I am not going to do I get theta by theta b equal to cos hyperbolic k max minus l upon I think fins is a very favorite thing for everyone I do not think I need to well upon this I want to save the energy and time for transient conduction so if I do the same way heat transfer that is minus k ac d theta by dx or dt by dx at x equal to 0 I am going to get heat transfer okay so similarly one can do for all the cases which I am not going to show here which is summarized in this table so I have tip condition either convictive heat transfer adiabatic prescribed temperature infinite fin temperature distribution and finite heat transfer so that is what we do for fin so but then one thing we have to remember in all of this one thing is common what is the base temperature is assumed to be constant if I change this boundary condition to constant heat flux all are these equations value I have to sit down and derive for constant heat flux okay so but then if I have to incorporate radiation can I take up can I do that how but then I will end up with non-linear differential equation because I have t to the power of 4 so one way out is already professor Arun has introduced us of taking linearizing a non-linear equation into a linear form what did you do it for us sigma epsilon t to the power of 4 minus t infinity to the power of 4 can be rewritten as HR into AS into t to the power of 4 minus t infinity to the power of 4 so you can combine in H itself HC and H radiation so if you can combine it that way you can perhaps handle even with these equations for radiative boundary conditions as well okay so there are problems here so I think here I just usually I think I can keep this problem I think I can keep this problem do this problem yeah yeah so what is this fin is this problem is that maybe we will go slow maybe I am going too fast let me slow down myself yeah I will spend time on this okay so what are we doing here just just for a while you read this problem what is it so we have a fin of 5 mm diameter maintained as a base temperature at 100 degree Celsius and the ambient temperature is at 25 degree Celsius and the convective heat transfer coefficient 100 watts per meter square kilo the question is what should we the material chosen okay so here the question asked is okay go ahead and figure out the temperature distribution for pure copper aluminum alloy and stainless steel okay so that is one question and of course another one is estimate how long the rods must be for the assumption of infinite length yield and accurate estimate of heat loss we will come to that little later if I just go ahead and plug in the insulated boundary condition on the other side is that right I put the insulated boundary condition yes yes this equation is valid for insulated boundary condition let me go back and check sorry infinite the temperature distribution is e to the power of minus mx theta b into square root of hp kac okay so if I do that what do I get this this is the temperature distribution I get with the length okay so can you interpret this graph for me calculations and all we will be able to do it is just plugging in what am I trying to say is can I interpret this figure blue color is 316 s s that is this fellow and red is aluminum maybe this is indigo indigo is copper which material you would use stainless steel one answer is stainless steel another answer copper see gut feeling we always say even if I am not a heat transfer guy I would go for copper because copper is a very good conductor for some reason gut feeling is copper is very good just to give you the feel thermal conductivity of copper is around 380 aluminum I guess it will be around where have I written 180 stainless steel is around 14 or 15 that is true I have not I have not yet come to that yeah please go ahead please go ahead I should not be okay that is one interpretation that is one can anyone else interpret yeah no problem wherever temperature gradient is there that is all the place where the heat transfer is taking once the gradient is see when it is becoming concentrated is coming almost near to ambient after that it is of no use for is that right so without calculating the heat transfer rate just being the thing the temperature distribution we can guess that the heat transfer rate for copper will be more than that of aluminum will be in turn more than that of s s so it is very natural that we go for copper we go for copper so how much the heat transfer I have just summarized that yeah that is the impression in fact now so in fact yeah it's all it should be green color why is it transfer rate how do you calculate heat transfer rate don't worry yeah see I keep saying another thing I want to tell I keep saying this and I have written this in bold on my board also and you can my students for my words those who never make mistakes never make anything this is what I believe in okay if I don't commit mistake I am not throwing myself into the middle of the sea so I need to commit mistakes so please don't feel bad about getting it wrong okay yeah yeah so that is the reason he said he told me come on better stop yourself and explain it that's precisely why he told so what what is the heat transfer rate given by h a delta t where is that h a delta t coming from it is equal to what now that is my fin can give away what is k d t by dx for each of these three cases how why do you why are you keeping that as a fixed quantity you cannot fix it no you are you are dealing with the assumption that that is fixed number I am saying that is not a fixed quantity that is what we are interested no we are computing how much is the heat transfer rate taking this this gradient for stainless steel much steeper gradient for copper much shallower k times that next page he has put yeah I have put yeah you good so now you see the numbers you will get the field you see dt dx is you are right copper is slower in terms of not slower in terms of time in terms of length but degree Celsius per meter this gradient is quite less SS is very high but you see the wattage how much it is being convicted out so how much it is it can only come back one point but aluminum 5.8 point why are we why are we putting a fin I want to suck out more heat so that is why so point is I need to go for that material which is having high thermal conductivity that is what okay so that is that is one important thing which we get another thing is proper length of the fin it is there in that problem but still proper how what should be the length infinite means see I keep saying this in the class always infinite is not infinite for engineers may be infinite is infinite for mathematician infinite has to have some value for an engineer isn't it so how infinite is infinite for us how can I answer that see as I see the temperature the temperature gradient is going on decreasing as I go later portions but then here am I going to gain much in this if not if it is not much where should I stop that is the question so if I take Q of a fin of a finite length and Q fin of infinite length I get the ratio as tan hyperbolic ML you see tan hyperbolic ML here I think it is not here you see because here it is bolder what do you see see what does this mean Q fin of a finite length is equal to tan hyperbolic ML of Q fin of infinite length is that right that is what you get the ratio is this so this ratio you want it to be fin of a finite length should be reaching that of infinite length is that right so when we when is it reaching here in this table ML equal to 5 precisely it is reaching but perhaps I do not care even if it is 0.987 0.987 is 0.99 which is one for an in for an engineer it is perhaps okay because length increase from ML equal to 2.5 to 5 is that much costlier but it is law of diminishing returns in economics we have studied it is not going to commensurate for the input what I put in it may the output may not be commensurate to the input what I put in so it may not be viable for me to go above 2.5 so typically we stop for an ML of 2.5 that is for us infinite so infinite is not infinite for an engineer I keep saying this in the class all the time so when do I decide how do I decide this is how I do is that okay so that is that I have answered infinite length so if you go back and put the values of L you will see that L is 2 here I have taken 2.65 so roughly around 2.5 you can stop so you can get 0.19 meters of course this length will change for each one because case are for each material they are different yeah 0.33 0.23 and 0.07 okay so that is about proper length of the fin so then another important thing is I think we will take 5 more minutes and then we will move towards coffee so what is any question so far I have intentionally gone very fast you want to stop me and ask some questions for the portion covered so far do not mind even if you get questions in the middle of the night write down and come tomorrow we can discuss no issues okay so fin efficiency so how efficient is my actually we introduce in Fin's two concepts one is efficiency another one is effectiveness and we know that these two can be interrelated through areas just areas so if I know one I should be getting the other if I know the areas okay so fin efficiency is how do I define fin efficiency that is fin as one of you were telling when will be the fin be very efficient you are telling that when the fin is at the base temperature I have a figure for that yeah when fin is at completely at the base temperature it is going to transfer maximum but unfortunately real life is not going to be that way as we saw irrespective of the boundary conditions on the left and right that is the base and the fin tip we are bound to get temperature gradients so with the temperature gradients heat transfer rates are going to be lesser than the case in which the complete fin is being maintained at the same temperature so I take ideals the ideal situation is always we take and compare I take this example in the class also why do we study Carnot cycle although we know that Carnot cycle efficiency we can never reach why because they are ideal only when I know what is ideal I can live my life correctly why do we worry about ideals in life also so we should know what are the ideals then only we can live our life towards that may not be ideally but at least towards that okay so here also we are comparing the real life case with the ideal case so you get q fin that is the actual heat rays heat transfer rate from the fin to the ideal ideal heat transfer rate from the fin as if the entire fin was maintained at the base temperature so that is what is this so we know the q fin we have figured out all the relations for q fin now it is straightforward so q fin maximum is heat transfer coefficient into area of the fin into base temperature minus so that is how you one can figure out the efficiency and you can show that it is indeed efficiency equal to 1 by ml so for various cases we can show this is for insulated and this is for insulated tan hyperbolic ml by ml so the point here again same things I have put and these efficiencies have been figured out for various configurations for various configurations because in handbooks you might have seen various configurations fin efficiencies with various geometric parameters they are listed out so here for various fin configurations we have fin efficiencies okay so for various other configurations they are there I think I will stop here