 So, we will get started where professor Arun left us that is to just to take a recap what did we do is we realize that we need to put fins there is no other option why because there are three ways of increasing the heat transfer rate as is obvious from this equation that is either I can increase h which as we discussed earlier increasing h always involves improving the pumping power or increasing the pumping power. So, it comes at a cost and decreasing t infinity that also as to involve pumping power the only innocuous or harmless way of increasing the heat transfer rate is increasing the area that is essentially what we are doing in by putting the fins now. So, there are multiple ways of putting fins how exactly we should be putting that is what is the question what we are answering. So, now we saw various applications and we saw a general conduction analysis we took variable area and we applied the energy balance Q x is the Q in E in and E out is Q x plus dx and d Q convection. So, if I put those terms here and substitute for each one of them from four years law of conduction I am going to get an equation something like this where in which if I assume that the AC is constant that is the cross sectional area of my fin is constant and this term is going to vanish that is 1 upon AC d AC by dx is going to vanish. So, I am going to left out with d square t by dx square minus of 1 by AC h by k d as by dx into t minus t infinity. So, now to make it look elegant the fin can be any cross section as he shown here we have just taken a plane surface or a plane plate or it can be pin fin this is usually called as pin fin because it is like a pin. So, this can be rectangular fin or a pin fin. So, essentially in these two cases the cross sectional area is constant or any other shape which you can conceive in however the cross sectional area has to be constant that is what this derivation is relevant for from now on. So, the equation is d square t by dx square minus h p by k AC t minus t infinity equal to 0. So, we said that we will write this little more elegant that is t minus t infinity equal to theta and m square equal to h p by k AC what is that we are looking for from this equation let us stop and ask ourselves the question why did we frame this equation what for the main reason for framing this equation is we are interested in the temperature distribution of the fin why do I need temperature distribution from the temperature distribution I can compute what is the heat loss or what is the heat transferred from my fin to the atmosphere which is what I am trying to maximize by putting the fins. So, it is always a good idea yesterday since yesterday we did not tell neither myself nor professor Arun told in any derivation in the middle of the derivation it is we need to emphasize two things one is assumptions. In fact, professor Arun you would have noticed in the morning he spent almost half an hour in explaining assumptions equations we went fast the main reason for doing is that we need to know to what conditions this derivation is relevant for that is to that extent the assumptions are important. Second thing is that in the middle of the derivation every time we need to stop ourselves and ask where are we and what are we so what are we why are we doing this derivation. So that is that is always an important thing as a teacher to stop and ask the students so that they remind themselves what we are looking for and we will be able to get to the or we will be able to reach all the students even if they are lost in between. So the equation reduces to d squared theta by dx squared minus m squared theta equal to 0 and m squared is h p k ac what is h it is the heat transfer coefficient between the fin surface and the atmosphere p is the perimeter k is thermal conductivity of the fin material ac is the cross sectional area of the fin. So as professor told the temperature distribution or the solution of this equation is this is the second order linear differential equation so the solution of that is C1 e to the power of mx plus C2 e to the power of minus mx. So if I need to get C1 and C2 I need to know the boundary conditions so what are the boundary conditions one of the boundary conditions is imposed on us that is the base temperature so if I take any fin yeah if I take this fin the base temperature is fixed but the other boundary condition can be changing in this case what I am showing here is a convective boundary condition the boundary conditions what we are going to choose are four that is convective heat transfer boundary condition from the fin tip adiabatic that is insulated prescribed the temperature maintained at the fin tip infinite fin as professor told the closest would be among all these boundary conditions for the real life would be something like convective heat transfer from the fin tip that would be the closest most of the times adiabatic I do not think we will be coming across most of the times because insulating also is quite difficult for fin and why do you insulate when you want to have more heat loss and of course infinite fin is a theoretical concept to arrive at what is the maximum length because this is always infinite is not infinite for engineers I keep telling this in all my classes for engineers it is not infinite infinite is also finite but how big or how large is that infinite is the question we can answer when we take up this infinite fin now we are not going to solve all the four cases I am just going to solve one of the cases that is the adiabatic condition and each and every step is there in the transparency I am not going to write on the white board I am going to follow only the transparencies please come along with me while I am taking this is pure algebra there is no heat transfer involved in this so two boundary conditions let us remind ourselves that is at x equal to 0 that is if this is my fin and this is x equal to 0 and this is x equal to l what we are saying is that at x equal to 0 t is equal to tb and at x equal to l we are taking insulated boundary condition by now we know insulated dt by dx equal to 0 at x equal to l so these are the two boundary conditions which we are going to put for our equation which we derived in terms of theta so what is theta theta equal to c1 e to the power of mx plus c2 e to the power of minus mx for those people who have lost what is theta theta is theta equal to t minus t infinity theta equal to t minus t infinity we are interested in getting this so theta equal to so now coming to the boundary conditions so I have one of the boundary conditions d theta by dx at x equal to l equal to 0 and theta equal to c1 e to the power of mx plus c2 e to the power of minus mx so theta equal to theta b because what is theta b at x equal to 0 t equal to tb tb minus t infinity we are calling it as theta b that is at x equal to 0 we have t equal to tb and theta of 0 equal to tb minus t infinity equal to theta b now what do I get so I get theta b equal to that is at x equal to 0 e to the power of 0 is 1 e to the power of minus 0 is also 1 so I get theta b equal to c1 plus c2 so that is one of the boundary condition what is the other boundary condition that is d theta by dx at x equal to l equal to 0 so if I differentiate this equation m c1 e to the power of mx but at x equal to l so I have e to the power of ml minus of y minus when I differentiate this I get minus mx minus m so that is what I got minus m c2 e to the power of minus mx at x equal to l so this becomes minus ml which is equal to 0 because d theta by dx equal to 0 at x equal to l if I equate this e to the power of ml if I push this e to the power of ml to the right hand side so I get sorry the other way round that is e to the power of minus ml if I push to the other side so I get c2 equal to c1 e to the power of 2 ml on the other hand I know theta b equal to c1 plus c2 so if I substitute for c2 I get c1 plus c1 e to the power of 2 ml that means c1 equal to theta b upon 1 plus e to the power of 2 ml so I have got c1 so if I substitute this c1 here I will get c2 so c2 would be equal to theta b into e to the power of 2 ml upon 1 plus 2 ml e to the power of 2 ml that is what is done here. So you get c1 and c2 so if I am going fast let me write this for the sake of completion so what did we get so we got c1 as we got c1 as we got c1 as theta b upon 1 plus e to the power of 2 ml is that right okay so then c2 equal to c1 e to the power of 2 ml so all that I need to do is substitute this c1 here I am going to get c2 so once I get this c1 and c2 I have the temperature distribution written as theta equal to c1 e to the power of mx plus c2 e to the power of minus mx so I need to substitute now this c1 and this c2 here and get my theta that is essentially what I am trying to do in this derivation. So when I substitute that c1 and c2 so I get theta by theta b which is theta b is common for both the sides so I get e to the power of mx upon 1 plus 2 ml plus e to the power of 2 ml into e to the power of minus mx that is this is c1 and this is c2 so this is c1 e to the power of mx c1 is 1 upon 1 plus 2 to the power of e to the power of 2 ml and c2 is e to the power of 2 ml upon 1 plus e to the power of 2 ml so you have theta b common so you get theta by theta b equal to whole of this equation so if I go ahead and do the algebra theta by theta b equal to e to the power of mx upon 1 plus 2 e to the power of 2 ml and e to the power of minus mx I am pushing this e to the power of 2 ml to the denominator so I get 1 upon e to the power of 2 ml plus 1 this 1 upon e to the power of 2 ml can be written as e to the power of minus 2 ml that is what essentially I am doing in this step so theta by theta b equal to whole lot of this equation now I do little bit of circus so that it I can write this in the form of hyperbolic functions so I multiply throughout for this term by e to the power of minus ml multiply both the numerator and denominator by e to the power of minus ml so that is e to the power of mx into e to the power of minus ml upon e to the power of minus ml plus 2 ml minus ml gives me e to the power of ml so similarly here I will multiply and divide by e to the power of ml so e to the power of minus mx into e to the power of ml upon e to the power of ml plus e to the power of minus ml that is actually this is e to the power of minus 2 ml minus 2 ml into e to the power of ml I get e to the power of minus ml and this term becomes e to the power of ml so now if If you see this I can take common here in the powers m into x minus l and here m into x minus l. If I rearrange this the denominator is common so I can write e to the power of minus e to the power of m into x minus l plus e to the power of minus m into x minus l. So this is of the form e to the power of x plus e to the power of minus x by 2 if I just take only the numerator that is if I divide numerator by 2 and denominator by 2 I can write the numerator as cos hyperbolic m x minus l upon cos hyperbolic m l. So remember remind yourself what is m? m is square root of hp by ka so if I fix the h if I know the material of my fin I should be getting m. I know m, I know the length of my fin I know the base temperature this is Tb minus T infinity I know the base temperature so I this theta is T minus T infinity I should be getting the temperature of my fin at any given location x throughout the length of the fin. So what have we achieved? We have achieved the temperature distribution what is next left out we have got the voltage now we need to get the current so that is the heat flux that are the heat transfer rate. So that is what we are trying to do here q equal to minus ka c dt by dx at x equal to 0. So if you do that I know I need to I have the temperature distribution in terms of theta so let me rearrange this dt by dx in terms of theta. So that is dt by dx is also equal to dt by dx why because theta equal to theta is equal to T of x minus T infinity. So T infinity is constant so dt by dx will be equal to dt by dx minus 0 so that is what is coming out that is why I am able to write dt by dx as dt by dx that is what is happening here. Minus ka c dt by dx at x equal to 0 but what is my theta just little while ago we derived this but differentiating this becomes little difficult so I am taking one of the intermediate steps from here and putting that that is e to the power of mx upon e to the power of 2 ml plus e to the power of minus mx plus upon 1 plus e to the power of minus 2 ml. I think you can very easily see why am I doing this because denominators are constant if I differentiate I need to differentiate only the numerator. So this is just mathematical ease for mathematical ease I am taking one of the intermediate steps if you want to see that you have that step here this is the step this is the step I am taking this is the step I am taking. So if I differentiate this that is d theta by dx that is this if I differentiate d theta by dx at x equal to 0 theta b is constant so theta b will stay as it is into all of the bracket denominator is a constant so we are going to retain as it is e if I differentiate e to the power of mx I am going to get m e to the power of mx but x equal to 0 so e to the power of 0 is 1 so I am left out with m similarly if I differentiate e to the power of minus mx I am going to be left out with minus m because e to the power of minus 0 is 1 so minus m upon 1 plus e to the power of minus 2 ml I get. So now if I pull out this m outside because it is common so if I pull that out and do the algebra so that is m theta b into 1 upon 1 plus e to the power of 2 ml minus 1 upon 1 plus e to the power of minus 2 ml this if I substitute in my heat loss what am I up to using the temperature distribution I want to find the heat transfer rate which is again given by Fourier's law minus k ac d theta by dx at x equal to 0 I am always asked this question in the class how can I use Fourier's law now we are applying fin and there is convection remember Fourier's law can be applied as long as my heat transfer is one dimensional and there is no energy generation there is no energy generation in my fin okay so and it is one dimensional I have assumed that in my fin it is one dimensional you can ask me how correct is this one dimensional heat transfer yes it is not always one dimensional but usually the fins are quite slender that means if they are quite long compared to their thickness so that is why it is a it is a reasonably good assumption to take the fin as one day of course with the present day computers and present day software we can use the software's CFD analysis for numerical methods for solving complete 2D or 3D but still these derivations are these results are going to tell us the influence of various parameters very elegantly so coming back so I use this d theta by dx at x equal to 0 all of this and substitute this here so I get minus k ac m theta b into whole of this this is a reminder for m m is equal to square root of hp by k ac this is ac cross sectional area so minus k ac into square root of hp by k ac theta b into what am I doing here again I am multiplying this with e to the power of minus ml numerator and denominator here I am multiplying e to the power of ml in the numerator and denominator so if I do this what can you see here so k ac so k ac one k ac will get cancelled out with another k ac so I am left out with square root of k ac I get square root of hp k ac theta b remember theta b is not inside the square root it is outside the square root so now this minus I have pushed inside so I get first e to the power of ml why am I doing this because numerators are same so e to the power of ml minus e to the power of minus ml upon e to the power of minus ml plus e to the power of ml so this is sin hyperbolic ml this is cos hyperbolic ml if I divide both sides by 2 numerator and denominator so I get tan hyperbolic ml so heat transfer rate is given by square root of hp k ac theta b into tan hyperbolic ml very this equation is going to have very very far reaching implications in the sense that you can see very easily very easily that if I use a highly thermally conductive material my heat transfer rate is quite high in the next problem we are going to see that the copper is going to be much much better than stainless steel or aluminum why because k is high and heat transfer rate is directly seeing that is it is directly proportional not directly proportional to k but it is proportional to square root of k so definitely if I increase k my heat transfer rate is high so that is one of the primary reasons why you see always all fins made of copper or aluminum see copper we cannot use because it is quite expensive okay so we tend to use aluminum copper thermal conductivity is 400 aluminum thermal conductivity is around if I am right 100 if it is around 100 so it is 3 times not so good but still because of its cheapness we go for aluminum just check for thermal conductivity of aluminum okay so just to take recap what did we do we took one of the boundary conditions that is adiabatic boundary condition of the fin tip and we took the boundary conditions d theta by dx at x equal to l equal to 0 and the base temperature is theta b that is t equal to tb we solved this equation found the boundary condition found the constants c1 and c2 and found the temperature distribution and using this temperature distribution we found the heat loss or the heat transfer rate heat loss is not a good word heat transfer rate from the fins whenever I use the word loss I would like to minimize it but here in the fins heat transfer rate is appropriate because I want to maximize the heat transfer so likewise all other boundary conditions we can work out for adiabatic prescribed temperature and infinite fin so by and large we have given in the notes salient steps if not all steps so this is the convective boundary condition that is hac t at l minus t infinity equal to minus kac dt by dx into at x of l so if I transform this in terms of theta it reduces to something like this if I apply this boundary condition and get we will be able to get there are certain steps which are skipped here we would strongly encourage you to go back and sit down and do those steps yourself and we will get theta by theta b essentially as a function of again m and l so again it is it will look elegant if you put in hyperbolic functions that is why we have to do some mathematical manipulation all the time and the heat transfer rate also we will be able to get using minus kac dt by dx at x equal to 0 only because once I get the temperature distribution it is quite easy to get the heat transfer rate that is what we are getting here similarly for adiabatic boundary condition also one can work out and prescribed temperature maintained at the fin tip also can be obtained similarly and infinite fin also one can come from finite case that is from kc that is prescribed temperature maintained at the fin tip only we are going to use that is this equation we are going to use and extend x to infinity so that is what l tending to infinity theta l tends to 0 so that is there is no temperature gradient that is what we are going to substitute and we get theta by theta b as e to the power of minus mx and qf equal to theta b into square root of hp kac so this is easy to remember of course we do not advocate you to remember anything but this is the essential summary for all boundary conditions this is the tip condition convection adiabatic prescribed temperature infinite fin this is the temperature distribution this is the fin heat transfer rate so these are all essentially functions of m and l so except for infinite fin where it is e to the power of minus mx and here also e transfer rate for infinite fin it is theta b into square root of hp kac so I think with this we are ready for a problem now professor Arun will take this problem and I am giving this problem over to professor Arun. So it is a very interesting problem actually if you take a look at the problem statement lot of things actually can be learnt from this problem so that is why we will spend time on this a very long rod 5 millimeter in diameter has one of the ends maintained at 100 degree centigrade surface of the rod is exposed to ambient air at 25 degree centigrade with a convective heat transfer coefficient of 100 watt per meter square Kelvin. So you have basically a fin so this is the rod the red color one so the diameter is 5 mm and 1 n that is the base portion is maintained at 100 degree centigrade and the ambient condition that is h is 100 watt per meter square Kelvin T infinity is 25 degree centigrade. What are you supposed to do determine the temperature distribution along the rod constructed from pure copper 2024 aluminum alloy and AISI 316 stainless steel what are the corresponding heat losses or heat transfer rates from these different material rods. The next part of course is the more interesting one estimate how long the rods must be for the assumption of infinite length to yield an accurate estimate of the heat loss we will come back to this in detail this requires reasonable amount of discussion so we will first state the problem and try to come up with the solution for this. So if you are with me this diagram is drawn so there is a base temperature specified there is h and T infinity also specified the material is known so thermal conductivity is the thermal conductivity is also specified for these materials. Now assumptions typically similar assumptions to what we had done during the derivation steady state condition one dimensional conduction along the rod the whole derivation was based on this assumption constant known properties thermal conductivity etcetera is constant negligible radiation exchange because that h what we have is primarily convective heat transfer coefficient and this heat transfer coefficient is uniform so we do not we are going to consider it as a constant value throughout properties again this is something which we tell our students in real life properties will not be given you have to go to a data book or back of the text book or the net to find out the properties. So properties are going to vary with temperature as most of us know so what is the temperature at which you want to take this properties typically if you do not have an idea of what the actual temperature would be which is proper in this case we do not know what the temperature of the fin is going to be from here to here all we know is that the base is at 100 degree and the air is at 25. So roughly at the average of the 200 plus 25 by 2 that is 62 and a half degree centigrade we take the properties and we say that things are reasonably constant over the working range of temperatures. Now if the temperature range is very high so if one one if the base temperature is 500 and the ambient is 25 or 30 then still you will have to take the average value but in that case the average value would be significantly different from the temperatures at from the properties at one of the ends of the other. So 225 if it is the average and the base is at 500 and the tip is at 25 then you are talking of a difference of 200 degrees there itself so in such cases you might have to go especially if you are writing a computer program where you are going to do something by iterative method you can actually feed in the properties as a function of temperature but for a classroom exercise where you are trying to just solve and get a back of the envelope solution this average temperature or mean temperature properties are fine. So at this temperature copper thermal conductivity is about 498 aluminum is 180 and stainless steel is 14. So 398, 180 and 14 you see this is about you know 14 times so 13 or 14 times stainless steel and this is more than double copper is more than double of aluminum so this is about 28 to 30 times the properties. So thermal conductivity you will see plays a very very very important role in the heat transfer rate through fins. Now we are going to assume an infinitely long fin why do we do that wherein the problem are we given that it is an infinitely long fin are we given the length we are not given the length we are just given the diameter. So we say let us assume an infinitely long fin and of course the next part is the giveaway estimate how long the rod must be in order that the assumption of infinite length is accurate. So this also tells me that we can start with the assumption of infinitely long fin. So we start with that and Sirus taught us the solution methodology and we have these solutions. Now in our engineering colleges exams etcetera unfortunately at least when we were studying when I was studying we had to memorize these formulas to come for an exam. I hope that trend is not there and people have access to these formulae because in the real life you have everything at your fingertips either in the form of a book or the internet or some kind of a program or notes whatever. So formulae memorization should be kept minimum because whole time the students energy goes in remembering whether it is a plus or a minus sign or whether it is this value or that value. So let us not emphasize in fact the in universities which are non autonomous it is probably very difficult to change to effect an overnight change but where colleges which are autonomous are where the teachers have the power to set appropriate questions and test students appropriately. Sincerely urge you not to have anything where formulae have to be memorized. Of course you have to say Reynolds number we have to know Prandtl number we need to know what it is but I do not I should not expect a student to memorize any of these formulae. It is criminal because we ourselves do not remember. So it is unfair to expect students to remember these and let us hope that wherever we have the ability to effect a change we should not ask questions where these have to be remembered by the student. Give the table how does it matter even in the university system if you have to give a problem involving fins give the table let the student choose the appropriate formula because in the real life he has access to all these and he should know how to use he should know what is this m what is this k where to get this h from because in a real life situation h will have to be calculated. So those are the things that you are going to test not a brute memorization of these formulae. So this is something which we are harping on we need to effect these kind of changes otherwise we are just going to produce good people who know to memorize things well and students who memorize well will obviously get good marks whether they are good engineers or know that is left for everyone to debate. So let us try to change things as much as we can so that these things are avoided anyway coming back to the problem this formulae is available where did this come from it came from here. So for infinite fin temperature distribution theta by theta b is equal to e to the minus m x. So theta by theta b I write it as t minus t infinity that is theta divided by t b minus t infinity that is theta b that is equal to e to the minus m x. So I have rearranged this to give you t because t is the unknown for us what are we supposed to find we are supposed to find that temperature distribution and the heat transfer rate heat loss or heat transfer rate when each of these materials are used. So I need the temperature distribution so I know the base temperature I need to find what is the temperature distribution. So m of course is known to us h b by k under root. So everything is known I can get the temperature distribution that is perimeter is pi times d for a circle I will I will write this I do not want to speak in air. So for this rod m is equal to under root h p by k a c h times pi d divided by k times pi by 4 d square this is the cross sectional area this is going to give me 4 h by k d under root that is what I have here. So 4 h by k d square root so I have theta over theta b theta over theta b is equal to t minus t infinity divided by t base minus t infinity is equal to e to the minus m x. So I substitute for this m so t minus t infinity is equal to t base minus t infinity e to the minus 4 h by k d square root times x therefore t of x local temperature at any location is given by this form e to the minus 4 h by k d square root times x. So if I know this t base is known t infinity is known h is known k is known h is known k is known diameter is known everything is known. So all I have to do is substitute various values of x to get a temperature distribution this can be put in a spreadsheet or substitute various values of x point 1 point 2 so on and so forth you will get the temperature distribution and knowing the thermal conductivity of aluminum copper and stainless steel that is what is going to change. So for each of these three materials I can get the temperature distribution so I get t of x and that is what is plotted here. So if I substitute for h and d and thermal conductivities please make a note of this value of m that is under root h p by k c square root comes out to be 14.12 for copper 21.2 for aluminum 75.6 for stainless steel this m is what is going to control because that is what is going to have the k embedded in it. So you see when I make a plot of this temperature distribution on the same graph again when in our assignments now most students have access to some kind of computer. So we can give assignments of the form where they try to do these things plotting interpretations by looking at graphs etcetera because merely solving a problem you know if I give numbers what is the temperature at the tip what is the temperature at this location I am going to get one unique answer. Whereas if I give a problem of this nature all students almost will be able to come up with this equation and even if they make 10 data points x is equal to 0.1, 0.2, 0.3 so on and so forth and I do this for three different materials I can get some kind of a plot ask them to plot it on the same graph they will at least get to see you know this kind of a trend and believe me this kind of a trend I think they will learn a lot more than giving 20 problems of the same type even this one problem is quite a powerful tool. So your assignments because you cannot ask like this in an exam. So we cannot do this in an exam but definitely in a class assignment or in homework assignment which students take for a week or something to do you can definitely ask them this kind of an exercise. So you can give for one group of 4 or 5 people stainless steel another group copper aluminum so on and so forth and actually have a discussion on what are these. So you can discuss this problem and that way fins are so much well understood and the person who has done stainless steel will appreciate that oh yes this is a sluggish distribution whereas in copper it is a very sharp distribution I mean the pink one is copper it is a very nice distribution for copper whereas for stainless steel it is a very very sharp decrease in temperature and then you come to almost a constant value which is equal to about 40 degrees or 35 degrees centigrade. So what when we give assignments also let it not be run of the mill kind of problems anyway. So this plot tells me lot of things first the pink line represents the temperature distribution in case of copper the red line represents aluminum and the blue line represents stainless steel. Now remember several I mean about 2 hours ago when we started fins I said heat transfer through the fin is equal to Q H A T minus T infinity this was the definition we used and we said yes very nice this is increased very nicely the heat transfer rate has to increase. But we also said this product A times T minus T infinity is what is going to control because in case of a bare surface the temperature difference is T base minus T infinity in case of a finned surface this temperature difference is T minus T infinity and if I take the length this difference progressively decreases with length. So this is temperature or delta T I might say T minus T infinity it is going to decrease with length. So section 1, 2, 3, 4, 5 if I divide I am going to see a driving temperature difference reducing with length which means the heat transfer if I take the heat transfer in this part it is H A this delta T next part H A this delta T so on and so forth and that is precisely what you are seeing here the driving temperature difference in case of stainless steel is almost equal to 0 what is your ambient temperature your ambient temperature T infinity is 25 degree centigrade the plot comes and settles to about 25 or 25 degree centigrade this middle mark is 20 and this is slightly above that so it is about 25. So in roughly about 40 mm length U that stainless steel fins temperature drops considerably to reach almost 25 degree centigrade. So if I have a fin of stainless steel which is going anything beyond 50 mm it is actually useless because you see this blue line it is almost horizontal delta T which is there for that part is almost equal to 0. So this portion of the fin the horizontal portion of the fin is going to offer nothing meaning no heat transfer or negligible heat transfer but you are going to increase the cost you are going to increase the weight you are going to add to the pressure drop associated with the flow you are going to have all sorts of negative implications rather than anything rather than getting anything useful from this extended surface beyond 50 mm. Also you our equation we had let us go back to the whiteboard T minus T infinity one of the first criteria for in the fin lecture we said was the temperature should be as close to the base temperature as possible. And we also said only when the thermal conductivity is infinite can I have uniform temperature throughout I showed this plot also I remember showing this plot when I have infinite thermal conductivity this is going to be the case we always will have a drop in temperature. But how close this is to the base temperature is what I am looking for I know this is the more realistic case a second one this will never be there but how close is it is this temperature coming to 58 or can I somehow have a material which makes it 68 or 70 as compared to 40. So, I want to look for a material whose temperature distribution is almost a straight line with respect to the base temperature. So, that is not possible but I want to slow down the drop in stainless steel the drop is very rapid whereas in copper if you see it takes almost about 250 to 300 millimeter length for the fin temperature to reach the base temperature sorry to reach the free stream temperature of 25 degree centigrade. So, even at about 200 millimeter length there is at least a 10 degree delta T available for heat transfer of course, the good heat transfer happens in the first 100 meters but nevertheless unlike stainless steel here you still have utility up to about 200 to 250 millimeter length in case of aluminum you have some kind of utility at least up to 100, 150 millimeter length. So, it is a very very important observation that we make from this problem. So, this k is what is controlling and you see this is 14 times roughly 14 times higher than this one and therefore, the length you see is also much shorter what is desirable this is absolutely not desirable this is what is desirable. The copper one is what is desirable that is why fins are made of typically materials like copper aluminum bronze etcetera which have much higher thermal conductivity. So, from these distributions it is evident that whatever I have told is being summarized little additional that is any additional delta T or any additional length that you introduce is going to contribute to negligible increase in heat transfer. So, 50, 200 and 300 mm they have written. So, beyond 50 mm for stainless steel beyond 200 mm for aluminum and beyond 300 mm for copper there is no use of adding any more fin material. And next we are asked to find the heat transfer rate there is a formula directly for it theta B times under root HP k A everything is known HP k A everything is known theta B is also known. So, I will be able to calculate this for all the materials 8.3 watts for copper for aluminum it is 5.6 watts for aluminum quartz and for stainless steel it is 1.6 watts. Remember this formula this heat transfer rate does not take into account any length. So, we are not considering any length issue here at all we are calculating the heat transfer rate based on the temperature distribution. It only involves H perimeter which is pi times d k and cross sectional area pi by 4 d square theta B which is the base temperature minus the free stream temperature. So, this implication I mean this graph is not anywhere used, but this graph is essentially what is going to give you the heat transfer. So, in this case you have very low heat transfer medium and highest heat transfer in case of copper fins that is why we in fact most fins are made of aluminum because it is light in weight and also less expensive compared to copper. So, we would rather go for aluminum than for copper and sacrifice little bit of performance, but stainless steel definitely is not an option as we see from this problem. I am sure there will be lot of questions we will complete the problem and then take questions. So, please hold on please write down do not forget to note down the questions. Second part this is actually the more interesting part we have learnt one thing from the problem about the length. The second part is how long must the rods be to assume infinite length. Remember when I close the discussion on the boundary conditions I told you infinitely long fin, infinitely long fin does not mean the fin is a few kilometers long that is not correct, infinitely long fin refers to something else what that is what we are going to see. So, what it means an infinitely long fin means some if the fin is very long whatever be the length there will be a situation beyond which there is adding more length you are not going to get any more enhance any more heat transfer or you will not be able to carry away any more heat. As we saw here in this case beyond this point any length is useless in this case beyond 200 in this case beyond 300. So, what we are going to say we are going to use this concept to define an infinitely long fin. Since there is no heat loss from the tip of an infinitely long fin why is there no heat loss from the tip because there is no driving temperature difference because the if the fin was extending to infinity somewhere along the length at the tip the temperature of the tip is going to be equal to t infinity that is asymptotically it will approach t infinity never will it be equal asymptotically it will approach t infinity therefore, you will have no heat transfer from that. So, an estimate of the validity of this approximation may be made by comparing equations 3.25 and 3.34 what are they let us just go back what we are saying is 25 and 34 25 is this. Q f is h p k a theta b times tan hyperbolic m l this is the heat transfer rate through an infinitely long fin. So, this is the equation expression this is made equivalent or equal to sorry sorry sorry I think I 3.20 is yeah this is case b adiabatic condition at the tip. So, when there is no heat transfer at the tip as I said if the length becomes infinity or infinitely long because there is no heat transfer we can call that tip as an adiabatic tip. So, this expression that we have is essential expression for adiabatic tip condition heat transfer rate that is equated to the infinitely long fin case and we will see why we are doing that this is the heat transfer rate in case of an infinitely long fin. Because the logic we are using is that when the fin is sufficiently long at the tip of the so called infinitely long fin I am repeating this again because this is where lot of questions and confusion is there when the fin is infinitely long at the tip of the fin the driving temperature difference becomes almost equal to 0. Therefore, that tip of the so called infinitely long fin represents a situation where there is no heat transfer. So, I can call that as an adiabatic condition. So, I am equating the heat transfer rate obtained by this so called infinitely long fin to the adiabatic tip condition heat transfer rate equation. Let me just do that for you here. So, I have I will come back to this instead of going to the power point and this I will use the nodes which are here that way it is easy. So, I have adiabatic tip q is equal to h p k a c under under root theta b tan hyperbolic m l. So, this is for this adiabatic tip then I have an expression for for the infinitely long fin. As you see I I do not remember I hope none of you also remember any of these formulae. So, if teachers do not remember then we cannot expect students to remember. So, we should not be asking them these questions. So, under root h p k a c theta b what I am going to do is this is for the l infinity case. So, these two heat transfer rates are going to be equated and when I equate this that is where I am going to see something nice happening. So, I will write tan if I equate this basically I divide one by the other I just get tan hyperbolic m l h p k a c theta b cancels everywhere is both cancel off equating the q I get tan hyperbolic m l comes out to be this gives me those of you who have calculators can we see that you will get m l equal to 2.65 there is a typographical error I thought this was corrected, but unfortunately it is not been corrected sorry about that this one m l is 2.65 equal to sin is missing this is not tan hyperbolic m l oh sorry sorry sorry sorry sorry yeah here point number two fourth line it says tan hyperbolic m l is 2.65 it is not tan hyperbolic m l those of you who have calculators can do it and see when tan hyperbolic m l is 1 m l will come out to be 2.65 anybody wants to check for me as somebody doing it yeah. So, m l comes out to be 2.65. So, in the slides please make a change page 33 point number two fourth line the tan hyperbolic should not be there it is m l equal to 2.65. So, what I get from this is that when these two are made equal I they are equal under this condition. So, in reality what is the what is happening in reality the infinitely long fin can be of whatever length, but when it reaches this condition product of m times l is equal to 2.65 beyond that point any length that you have is going to contribute to negligible heat transfer and addition. So, I can say when l is equal to 2.65 by m if any l that is greater than the so called l infinity the subscript infinity is used to indicate this infinite so called infinite length physically there is nothing like an infinite length. We are saying beyond this length value what is that length value we are trying to calculate beyond this so called length value after which any additional length is useless from heat transfer point of view that length I want to calculate what is that length given by that length is given by l is equal to 2.65 by m. So, that is 2.65 by m is here and therefore, I get l infinity or the so called infinitely long fin is one where l infinity comes out to be 2.65 divided by under root h p k a which is nothing but 2.65 times under root k a by h p and for copper this so called infinite length comes out to be 0.19 meters for aluminum it comes out to be 0.13 meters and for stainless steel it comes out to be 0.04 meters. So, 19 centimeter or 190 mm, 130 mm and 40 mm and you look at the graph here 190 mm is roughly 200 you see the utility is pink line if you see the utility is so much less beyond this point here 130 is roughly here again it is almost reached the infinity here about 40 you have reached almost the infinity. So, what it tells me just without any sense I should not have a fin which is of several meters or several centimeters long we have to see what length works out as the so called infinitely long fin. Yes infinite long fin is good, but it comes with a big penalty that there is a temperature drop very quickly and this temperature drop very quickly depending on the thermal conductivity of the material will have different implications on the heat transfer rate. Yeah very very small 40 mm of stainless steel fin small nice compact, but we are having a heat transfer rate which is very very low whereas copper about 190 mm if I put I have very high heat transfer rate. If I put 350 mm of copper fin there is hardly any improvement which is contributed by this length beyond 190 mm. So, that is what is the implication of this problem and we have not completed it there is some more discussion on it I will come back to a table just give me a minute. So, I will read the comments here and essentially what I have told the above results suggest that the fin heat transfer rate may be accurately predicted from an infinite fin approximation if M L is greater than or equal to 2.65. However, if the infinite fin is to be approximation is to be is to accurately predict the temperature distribution a larger value of M L would be required why why is that we will answer that this value may be inferred from equation 3.32 and the requirement that the tip temperature may be very close to the fluid temperature hence if we require this one equation 3.32 which is here because there is a temperature distribution associated with this M L greater than 4.6 is what is going to give you the so called infinite approximately infinitely long fin approximation. The first one was done using heat transfer being equivalent adiabatic as well as this the second one is actually a more correct one because we are going to take the temperature distribution and say that these two are going to be made equal. So, you can take a look at this 3.32 theta by theta b equal to e to the minus M x is going to be used along with the temperature distribution equation. You will get so called infinitely long fin would come out to be M L greater than 4.6 there is a before we go to fin efficiency etcetera there is this table which I want to show which is to be done along with this problem see here the variation of heat transfer from a fin relative to that of an infinitely long fin. We said that M L about 2.65 was what we came up. So, M L is 0.1 tan hyperbolic M L this is a table of M L and tan hyperbolic M L and if you see the values M L is 0.1 tan hyperbolic M L is 0.1 M L is 1 tan hyperbolic M L is 0.6762 at 2 it is about 0.96 or 96.96 2.5 is 0.987 3.995 and when I come to M L equal to 5 tan hyperbolic M L is equal to 1 basically M is has nothing to do with the length it is H p k a H p by k under root. So, that is just material property dimensions and H L occurs only in this term explicitly as L. So, when I go from M L is 2.5 to 5 I have increased the length by a factor of 2 tan hyperbolic M L which is a representation or a measure of the heat transfer is gone up by 0.013 percent 0.987 here and 1 here the difference between that is 0.013. So, 1.3 percent improvement has happened when I increase the length by a factor of 2, but it comes with so much greater penalty of pressure draw of weight cost etcetera size implications that we would not want to do it. So, therefore, we observe from the table that heat transfer from a fin increases with M L almost linearly, but the curve reaches a plateau later and reaches a value for a infinitely long fin at M L equal to 5. So, therefore, for a fin whose length is L equal to M by 5 can be considered as an infinitely long fin. We observe that reducing the fin length by half that is from M L equal to 5 to 2.5 causes a drop by just 1 percent I said 1.3 percent it is about 1 percent only. We certainly would not hesitate to sacrificing 1 percent of heat transfer performance in return for 50 percent reduction in size and the cost pressure drop all the other problems. In practice the fin length that corresponds to about M L equal to 1 which will transfer 76.2 percent of the heat that can be transferred from the infinitely long fin we can use and that should offer a good compromise between fin performance and the size. So, just go back here 1 M L equal to 1 gives me tan hyperbolic M L is 0.762 it is not a sacred thing it is just a suggestion that this is ok. So, this table actually should be considered along with the problem. So, that this goes hand in hand with that and with this actually you see here this temperature distribution as you saw in the problem is here higher delta T in the beginning lower delta T as we move away and the heat transfer characteristics is going to be just like this. The just the mirror image of this you would have high heat transfer rates in the beginning because of a large temperature difference and then it is going to drop significantly because of the lower delta T. Yeah, we would like to take some questions I am because I see a bunch of questions question marks here when we started after we did this problem. So, between now and some 15 minutes we will spend on questions and then probably go back to additional topics. Yeah, PSG. PSG college Coimbatore. Sir, can you or will you be dealing with rectangular fin sir, sorry triangular fin. Once we have dealt with rectangular and circular fin the same concept can be extended to any cross sectional shape. So, I do not think particularly we will be dealing with triangular fin, but we believe that with these basics any cross sectional area can be approached. And there are textbooks which deal with the solution for these non uniform geometries. So, you can get the analytical solutions available of course, otherwise you have these in the form of tables in textbooks also. Extended surface by cross there is a textbook by name extended surfaces heat transfer in extended surfaces by cross all sorts of configurations one can think of have been derived and kept there k r a u s cross. P. I. T. Pune. In a graph it shows copper aluminium and stainless steel. So, for infinite length we are looking for finite difference before that we can consider the length which to be considered for application. So, from the graph it is seen that stainless steel may be just matter of 10 mm it can give a temperature drop required corresponding to copper and aluminium. So, I think we need to plot heat transfer versus x to get a proper idea of the final things to be chosen in terms of copper aluminium and stainless steel. You are right we do agree with you we should be plotting heat transfer rate, but as we have said already in the problem the heat transfer rate in each of the cases for example, the copper is 8.3 watts for aluminium it is 5.6 watts and stainless steel it is 1.6 watts. These three numbers suggest that the copper is the better one yes, but then what is there to plot the heat transfer rate there is only one value. No, may be what what you are are you saying that I should plot the differential heat transfer in every element. So, as the as a function of length, so heat transfer rate is one unique number I do not think there is anything to plot for the heat transfer rate the overall heat transfer rate we have got it as for copper it is quite significantly high. So, that answers the your question over to Amritha if you have any questions please come back. Sir in the comments we have seen a problem in the comments you said ML is equal to 5 and you gave L is equal to m by 5 how is it possible sir. What is the issue I do not understand. See the question asked is ML equal to 5 and m equal to 5 by sorry L equal to 5 by m. So, I we do not see any problem in this why because we are saying that product of that table. So, in that table if we see tan hyperbolic ML and ML where we have plotted in that table what we are saying here is that ML equal to 5 is equal to tan hyperbolic ML that means maximum heat transfer we cannot get anything better than that. So, if you want maximum heat transfer for that happens only for ML equal to 5. So, that means my length has to be m is fixed why m is fixed because heat transfer cross sectional area is fixed means. For a given coefficient. And given material that is the thermal conductivity if I choose these three things the only variable is length, but if I have to achieve complete heat transfer my length has to be 5 upon m type. I do not see any problem in expressing L equal to 5 by m nothing wrong. But in the slide it is L is equal to m by 5. So, only I asked. Then it is a typographical mistake 5 by m. We are wrong you are right we are wrong it is L equal to 5 by m not m by m. It is a typographic mistake over to Baramati. Sir that ML equal to 2.65 is it a compromise or because tan H ML function will never approach to work will become 0.999 etcetera. So, the value of ML depends upon how many decimal places we take in the calculate. So, whether it is a compromise to take 2.65. The question is ML tan hyperbolic ML will never approach 1 it is going to be 0.999. So, what ML I should be taking? So, the answer is engineer will never look at 0.999 and all. So, already we said ML we will never take 5 actually we as it said one of the suggestions was ML equal to 2.5. So, I would not take 2.5 if I want to maximize my heat transfer I will take an ML of let us say 0.964 or 0.987 somewhere between 2 to 2.5 of an ML I would take. So, I do not think there is any concern about tan hyperbolic ML reaching 1 for an engineer is never going to depend on this number. So, much he goes by practicality ok. Thank you over to Ashwant Rao Chawan college Nagpur. Yes derivation in the derivation on equation we derived Q value in partial derivative, but in the fin analysis we taken the same, but we taken the ordinary differential equation. Will you please elaborate the reason behind this? Over to you sir. The question is why are we taking only the total derivative? Why are we not taking the partial derivative in whole of this analysis? But let me answer this question I think it is we do not have to read any equation. So, what was our equation to start off with? It is actually it is a one dimensional conduction number 1 and number 2 it is having no generation. So, it is one dimensional that means temperature is a function of x only. So, when temperature is a function of x only then it has to be ordinary derivative it cannot be partial partial comes into picture only when temperature is function of 2 parameters not 1 that is it can be x y or x z or x t that is space and time. But in this case temperature of a fin is a function of x only that means it is going to vary only with length. So, it has to be ordinary derivative. Thank you. K K what? Thickness. How do we decide on the thickness of the fin? Thickness means how do you define the thickness? See thickness when do you how does thickness come into picture here? Thickness comes into picture. First let me rephrase the question. Let us go back to that figure. The question asked is how does one fix the thickness of the? For example, in this figure how does one fix the thickness? See how does thickness come into picture here? Perimeter is 2 w plus 2 t area is w into 2 t. What are we saying w into t? What are we saying m is equal to square root of h p by k a. So, what is that we need to look for? p by a has to be as large as possible that means perimeter to cross sectional area has to be as large as possible that only means that what the either I can increase the perimeter by either increasing the width also by increasing the thickness. So, there is no one unique answer for this question. All that we can say is h p by k a has to be large. So, that means p by a has to be large accordingly you choose the thickness. But many times the width of the fin the w is probably fixed by the base surface geometry ok. So, the space. So, we will have control only over the t. So, that is probably why you can say if I choose w t or 2 w plus t divided by w t p by a you can optimize it with the fact that you have to have thin fins as opposed to thick fins. Yes definitely if you have to qualitatively answer thin fins are better than thick fins why because we can stack more fins because w is going to increase. Thank you.