 So, for the next 10 minutes or so, we will introduce one of the performance evaluation parameters for a fin. Anything that we do, we want to see how good it is. So, as engineers, you have put a fin, you want to see whether it is meeting the requirement well. So, there are two definitions or two parameters by which you study or you quantitatively say how good a fin is. So, these two quantities, one is called as fin efficiency, other is called as fin effectiveness. Now, both these quantities, in any heat transfer application, you are not going to put one fin and be done with it. You are going to put several fins or what we call as an array of fins. So, this definition of fin effectiveness and fin efficiency are for an individual fin. So, keep that in mind, we are talking of an individual fin and for that we are going to define two performance parameters, one is called as a fin efficiency, other is the fin effectiveness. So, what is the fin efficiency? Fin efficiency, I will go to the definition and then I will come back to this. Fin efficiency is defined as the actual heat transfer rate from the fin divided by the ideal heat transfer rate, if the entire fin was at the base temperature. That means, if I have a material of infinite thermal conductivity, for that material, there will be no temperature drop along the length, the entire fin would be at what we call as the base temperature T b, that for that fin, q will be h a T b minus T infinity, T b minus T infinity, because the entire fin would be at the base temperature. In reality, the actual fin would have h a T minus T infinity, where T is the local temperature at any given location and we have expressions for heat transfer through the fin for any given boundary condition. So, actual heat transfer rate divided by ideal, that is the best possible heat transfer rate. When is the best possible heat transfer rate achieved? The best possible heat transfer rate is achieved when the entire fin is at the base temperature. So, that is what is implied by the best possible heat transfer rate. Now, let me just go back to geometry and this is essentially what I had written in the beginning. q is h a T s minus T infinity. If I take a base area, remember when we started fins, we said this. I have this base area, which is on which I am putting a fin. So, this area is lost as far as heat transfer is concerned. This orange area is still there, but this red shaded area is lost. It is replaced by a extension or an extended surface, which has an additional area on the top and the bottom in addition to this area a fin or a base w times T. We have the top, bottom and neglecting the sides, we are going to have top and the bottom and this portion, the tip portion. We are saying heat will flow from the surface, base surface through the fin by conduction and by convection from the fin tip, from the surface here and this by convection to the free stream. The temperature of the fin would be T b at the base and will gradually decrease towards the fin tip and we do not have to go any further. We just have to look at this problem and we have this established already a few minutes ago that we are going to have a temperature drop, no matter how high thermal conductivity material we are going to choose. Even when we had copper of 400 watt per meter Kelvin, we had a temperature drop. So, irrespective of the material, we are going to have a temperature drop. So, that is what is implied here. However, cross sectional area of fin is usually small, the temperature across any cross section can be considered uniform. You might have a change in the temperature at a given cross section, but since these are usually thin, we say that entire at a given cross section, entire location has uniform temperature and fin tip can be assumed as insulated just for convenience for understanding this definition. So, in the limiting case of zero thermal resistance or infinite thermal conductivity, the temperature of the fin will be uniform and its value will be equal to T b. So, if I go back and write this, I will say q fin is h A T minus T infinity. We all we said T b minus T infinity is always greater than T minus T infinity. I think this everybody has understood, T b minus T infinity is always greater than T minus T infinity, when in the limiting condition, when k of the fin tends to infinity, T is equal to T b everywhere. I think this is what we are going to use. So, when k tends to infinity limiting condition, this would be the q fin. So, when k tends to infinity, when k tends to infinity, q fin maximum, I will write this as the maximum or the best possible situation is h A T b minus T infinity. Do I get this in practice? I will never get this in practice, because we saw even for copper with 400 watt per meter Kelvin thermal conductivity, we had a drop, but we are saying this is the maximum or the best possible performance and using this thing, we now define efficiency of the fin. So, efficiency of the fin therefore, is q actual divided by q max and this max is when the entire fin is at the base temperature, I do not have to keep writing that, it is implied that the entire fin is at the base temperature and if I put a mathematical representation, this is h A T minus T infinity divided by h A T b minus T infinity. So, we are assuming here that you do not have any variation in the heat transfer coefficient and for this, we already have an expression, why cannot we calculate this directly? This part, we cannot calculate directly, because this is locally varying T minus T infinity is locally varying therefore, we have four different expressions are there, we have derived those expressions. So, essentially we will say eta f or fin efficiency is q fin, u is one of those four expressions divided by h A fin times T b minus T infinity, yes q f in itself if you go back will have T b minus T infinity as one of the terms, if I go back here look here q f all of these you see here it has a theta b. So, it has T b minus T infinity, but instead of a plane h A T theta b, it is theta b times some parameter which has all the effects of the boundary condition that we have seen. And now we are saying with this definition let us just use it for a case where we have an insulated tip, why do we use it? Because that is the easiest case to show mathematically here. So, this relation enables us to determine the heat transfer from the fin when the efficiency is known, yes if I know the efficiency I can get heat transfer from the fin vice versa. For constant cross sectional area of very long fins with insulated tip the fin efficiency basically is obtained from q fin divided by q fin max and q fin comes from this formula h p k A under root tan hyperbolic m l times theta b that divided by h A theta b h A theta b. So, theta b theta b cancels and you have one by l times under root h p k A which is nothing but 1 by m l. So, this tells me that for an insulated boundary condition I will have tan hyperbolic m l by m l for an infinitely long fin I will have 1 over m l. We do not have to remember any of these expressions please again let us not ask our students to say for each of these cases please remember these expressions and come for the exam that is not the aim of this exercise. Here we are just trying to tell you that efficiency of the so called long fin is inversely proportional to m times l and now let us try to correlate this with what we saw m times l as it increases here. This is what we are going to say as m l increases you are going to get the effectiveness or sorry the efficiency of the fin is going to drop meaning larger and larger length if I keep adding the farther and farther away from the base temperature my fin material is going to be and therefore my efficiency or the measure of performance when the fin was at the actual condition versus the ideal condition is going to deviate more and more. So, I am going to have a poorer and poorer performance that is what we want to interpret from this rather than derive a formula and sit and do a memorize the same. So, since a fin is nothing but perimeter times length for fins with constant cross section area this expression for an insulated fin can be used for various other geometry if l is replaced by lc anyway. So, fin efficiency relations are developed for fins of various profiles and plotted and these are known to us from these kind of charts are there already available in textbooks this is what a student typically uses where a certain set of parameters are evaluated based on the geometry and this efficiency of the fin is read out based on the parameters that are calculated. So, this the next definition would be effectiveness. So, I do not want to start that now we will begin that at 2 o clock. So, keeping this in mind we will deal with effectiveness then merge these two efficiency and effectiveness we will try to combine together and keep in mind that these two are valid for an individual fin and not for an array of fins. For an array of fin there is something called as overall performance, but for individual fin if there is one fin how is that fin performance as opposed to the surface without a fin. So, that is what is quantified by these two definitions on chat there is been a message that equation 3.37 there is a typographical error q fin by q fin max has been written as 1 by m l that 1 by m l is correct, but the term before that it is 1 by l times it should be under root k a c by h p it has been typed as under root h p by k a c. So, please make a change equation 3.37 1 over m l is correct, but the term before that is 1 over l times under root k a c by h p. Fin's in the morning professor Arun has extensively taught us how we analyze a fin and he has introduce the concept of efficiency. So, that is what I will just brush through why because I need this for defining what is called as effectiveness. So, fin efficiency how did we define is that we define what is called fin maximum that is we realize that in a real life case the fin temperature is not going to be uniform is going to decrease with the increase of the length from base. However, in an ideal situation the complete temperature of the fin at any location is going to be same. So, that is that of same as that of base temperature that is T b. So, that can create maximum heat transfer that can transfer maximum heat transfer rate or heat transfer rate that maximum heat transfer is h into area of the fin into T b minus T infinity. So, the actual fin heat transfer rate to the maximum heat transfer rate or the ideal heat transfer rate was what was defined as efficiency. Now, let us see how do we define what is called as effectiveness. So, for fin efficiency the reference was the maximum heat transfer, but for fin effectiveness the reference is no fin case that is what is the heat transfer rate when there is fin and what is heat transfer rate when there is no fin. So, definitely effectiveness has to be greater than 1 as you can see. So, instead of just rushing through let me just write this fin effectiveness that is effectiveness equal to I would strongly encourage you to write along with me this simple definition q fin upon h a b into T b minus T infinity. This actually the denominator is basically no fin situation that is at the base there is no fin placed. So, the definitely the heat loss would be lesser in case of no fin compared to that of the fin if my fin is effective that is how the word effectiveness comes. So, if I take the effectiveness see the effectiveness is 1 if the effectiveness 1 means what if effectiveness is 1 means the no fin case heat transfer rate is equal to fin heat transfer rate. So, it is not doing any job why are we putting fin after all we are putting fin because we want to increase the heat transfer rate. If effectiveness is greater than 1 this is what is the ideal situation or not ideal this is what is the situation which we are looking for or aiming at this is what we are aiming at that is the heat transfer because of the fin is more than the heat transfer without the fin. Now, if effectiveness is less than 1 that means what that means what instead of actually helping us it is actually creating negative effect that means it is decreasing the heat transfer rate that is possible only when I use it is acting like an insulator it is acting like an insulator this is not at all required and of course this is also not required we are always looking for effectiveness greater than 1. So, that is the whole idea of this effectiveness so that is what is the summary of this whole transparency. Now, let us go to how to relate this fin efficiency. So, how to relate fin effectiveness with fin efficiency if you know 1 you can get the other provided you know the areas that is all it is. So, what does that tell or how do we get that so that is effectiveness of a fin equal to effectiveness of a fin is equal to q fin upon q no fin that is the definition. So, q fin upon q no fin is h a b into t b minus t infinity. So, what is q fin we know already q fin has been defined as efficiency of the fin efficiency of the fin into h into area of the fin area of the fin into t b minus t infinity upon h a b into t b minus t infinity if you see here h is same t b minus t infinity is same. So, I end up getting efficiency of the fin equal to sorry effectiveness of the fin equal to efficiency of the fin into area of the fin divided by area of the base. That means you know efficiency you can get effectiveness provided you know the area of the fin and the base area. So, it is just the areas which are going to connect the effectiveness with the efficiency. So, that is what is told here effectiveness of the fin equal to efficiency of the fin into area of the fin upon area of the base. So, I have just taken again in long fin if I take the long fin we know that it is q fin equal to t b minus t infinity into square root of h p k ac upon h a b into t b minus t infinity a b is of course equal to a c. So, I am going to get effectiveness equal to square root of h p by k a c sorry. So, that is sorry it is k p by h a c k p by not it is not in terms of m it is square root of k p by h a c. Please note this carefully for a minute I also got confused, but it is k into p upon h into a c because here we are taking a c equal to a b why I can do that a c equal to a b because I have put it a b and my fin is along a b. So, a b is equal to a c. So, now what does this say this little fellow that is effectiveness of the fin equal to square root of k p by h a c tells that my thermal conductivity has to be very high. That means I have to choose a material whose thermal conductivity is high and p by a c definitely the perimeter has to be greater than the cross sectional area. I think that is very easy to see why I say very easy to see because the bathing area the bathing area is the one which is taking away the heat. If you remember our derivation which is taking away the heat conducted who is taking away the convective heat that is h a s into t minus t infinity who is having a a s a s is having the perimeter. So, perimeter is to be larger compared to cross sectional area that why because the surface area or the bathing area will increase. So, p by a c has to be very large and thermal conductivity should be very large. This equation tells us that h should be low do we have control on h no this only tells when not to use fins. See let us say if I take a heat exchanger what we are saying is h should be low means if you have let us say we will take a situation. The question what I am trying to answer is the equation effectiveness is saying that effectiveness is inversely proportional to square root of h. It does not mean that I should decrease h it only tells in which situations I should be using h. Now let us say I have a pipe and in which I am trying to put fins now outside there are fins, but now in this situation there are two situations in one of the situations there is no flow around these fins and in another situation there is flow around these fins. If there is no flow that means h of no flow here would be lower than the h of flow situation. So, that means when do you go for fins that means one would go for fins only when there is no flow situation why because h in no flow situation is lower than the h of flow situation that is what this equation is telling it does not tell that I have to decrease h no it does not tell that because I do not have control on decreasing h it only tells in whichever situation h is lower that is usually where natural connection is taking place you recommend to use fins that is the usual thumb rule one takes for taking the fin. So, that is what this equation is telling quite interesting or quite far reaching explanations we can give with this simple harmless equation. So, now this is book keeping this is the equation what is it telling is when I am using fins I am just not using only on the base see if you see here in this figure what is there for this cross sectional area there is a fin, but this surface which is between two fins is also contributing for the heat transfer it is not that it is not contributing it will be less in the next problem we will see how much less, but we are calling the area where there are no fins as area un-fin and where the fin is there we are calling that as area fin ok. So, in the area un-fin also there is some heat transfer rate that we have to book keep so that is what is being done here total heat transfer is because of the q of the fin and q of the area over which there is no fin that is area un-fin which we which we have called that is h into area of the un-fin that is when there is no fin that temperature is corresponding to T b that is the base temperature so that is h into area of the un-fin into T b minus T infinity plus q fin we know already has been defined in terms of or expressed in terms of efficiency of the fin that is efficiency of the fin into area of the fin into T b minus T infinity I am just doing taking common h into T b minus T infinity if I take common I get area of un-fin plus efficiency of the fin into area of the fin. So, this is the total heat transfer rate taking into account both the areas of the fin and areas of the base where in which the fin is not covered that is what is called as area un-fin so here effectiveness of the fin equal to q fin upon q no fin q fin is what we derived just a little while ago that is this if you substitute here this is what we get for effectiveness of the fin. So, with this I think we have the only summary what I want to express or tell is that proper length of the fin has already been covered by professor Arun all that I want to emphasize is that effectiveness or efficiency both the parameters can be used for expressing but one usually ask the question when do you use effectiveness and when do you use efficiency I think both can be used all the times it is a matter of convenience when do you go for effectiveness and efficiency. So, one would use either effectiveness or efficiency of the fin why effectiveness is little interesting because you want to check first in the first place whether your fin is doing any good any good or not first of all is it better compared to no fin case if you are comparing let us say 2 to 3 configurations of fins then perhaps you can use efficiency but still you can still use effectiveness whichever is having maximum effectiveness still can be used to compare different fin configuration. So, it is not that I will be using effectiveness or efficiency for a particular situations they can be used in all the situations so with this we will just move on to what is a simple problem. So we have a very interesting simple problem how to use this calculations of and fin fin all this several terms we have introduced let us see how do we use this situation use all fundamental for designing or calculating the heat transfer rate. So, now if I take this problem the problem what it says is my fins are as you can see they are circular fins so I have just put a circular rings around a circular tube. So, unfin is this portion where there is no fin or when before putting fin it is completely simple circular pipe in which steam is flowing what is happening let us read the problem once carefully steam in a heating system flows through tubes whose outer diameter is 3 centimeters that is outer diameter is 3 centimeters that is d 1 is 3 centimeters that is 3 into 10 to the power of minus 2 meters or r 1 is 1.5 into 10 to the power of minus 2 meters at what temperature it is being being maintained the temperature of the walls is being maintained at 125 degree Celsius what does that mean that means that T b is 125 that is T b is equal to 1.5 into 1.5 into 25 degree Celsius. Circular aluminum fins I was looking for thermal conductivity of aluminum here it is thermal conductivity of aluminum is 180 watt per meter degree Celsius but for copper it is 400 why do we go for aluminum although copper thermal conductivity is high as I said in the morning it is because of economics because aluminum is cheaper than copper. So thermal conductivity of aluminum fin is 180 so K is 180 watt per meter degree Celsius okay and outer diameter d 2 is 6 centimeters and a constant thickness of 2 mm are attached to the tube that is d 2 that is the ring diameter the ring diameter is 6 centimeters the tube diameter is 3 but the ring diameter is 6 centimeters. So that is d 2 is 6 into 10 to the power of minus 2 meters that is r 2 equal to 6 into 10 to the power of minus 2. Now what is the thickness of the fin and a constant thickness of 2 mm the thickness is 2 mm let us write that thickness is 2 mm if I convert it into meters 2 into 10 to the power of minus 3 meters r 2 is 3 that is right I made a mistake here r 2 is 3 into 10 to the power of minus 2 thank you. So now the space between the fins is 3 mm that is the pitch between one fin to another fin is 3 mm that we are calling as S. So S equal to 3 mm okay so now how many fins are there like that there are 200 fins along 1 meter per meter length of the tube so that means what is the tube length we are considering length we are considering 1 meter number of the fins 200 okay. So now heat is transferred to the surrounding air at T infinity equal to 27 and combined heat transfer coefficient of 60. So 60 is quite high let me come to that little later that is T infinity equal to what was that 20 so 27 degree Celsius 27 degree Celsius and heat is 60 watts per meter square degree Celsius okay. So you might be wondering why is this 60 so high natural convective heat transfer coefficients generally are between 15 and 20 or maximum a minimum of 10 we can take but why is it 60 because he has combined both radiation and convection that is why in the write up he writes very carefully with a combined heat transfer coefficient there is radiative heat transfer also because you see the wall temperature is at 125 we cannot neglect radiation he has combined both convection and radiation and given as a combined heat transfer coefficient as 60 okay. So that is why the number is significantly large okay. So now what is the question asked so determine the increase in heat transfer from the tube per meter of its length as a result of adding fins. So actually this tube runs into few meters but we are not going to do for that many meters we are going to do only for one meter. So how do I go about this problem the first thing is we have listed out all the known things okay. The second thing what I should do is as we have been telling known is what properties of the fin ambient conditions heat transfer coefficient dimensions of the fin. So what is that I need to know I need to find out essentially is asking effectiveness the question asked is effectiveness so what are the assumptions what all assumptions we made for deriving the relations for fins are valid here also that means steady operating conditions. So we are assuming that nothing is changing with time that is nothing when I say nothing here I mean temperature temperature is not changing with time and the heat transfer coefficient is uniform over the entire fin surface this is a question which we need to this is a real assumption. So if you really see a natural convection will not take care that h is throughout uniform no there will be convection cells convection currents will be there h might be more at the center h might be less near the fins but we are going ahead and making the assumption that h is uniform when I say 60 it is same throughout here h is same on these walls and on these walls okay. So now next is heat transfer by radiation is negligible we are neglecting the heat transfer by radiation now first let us calculate area of no fin I am giving you two minutes time I want all of you to solve this problem along with me I hope you have calculators if you are not brought I hope why I should say I hope you have brought calculators for the tutorials you anyway need calculators you should have calculators when the class is going on. So what is what is being done here is what are we going to do is area of no fin that is when there is no fin what is the area what is the area of no fin pi into d 1 into L pi into d 1 into L that is pi into 3 into 10 to the power of minus 2 into L is 1 meter. So I get 0.0942 meter square q dot no fin is equal to h area no fin into t b minus t infinity that is h is 60 area of no fin we have just figured out it is 0.0942 and t b minus t infinity is 125 minus 27 I get what is the value I get 554 watts not a small number not a small number but still let us see whether putting fins is it going to take any higher compared to 554 watts anything wrong with the number some it fine. So now let us take up this is what I have put here you see this is what I just calculated area of the no fin comes to 0.0942 area q dot no fin equal to h area no fin t b minus t infinity if I plug in that I get 554 now how do I get the efficiency of the fin what is that we need to find out is how do I go about this problem is what we are saying is let us go to the definitions of the efficiency of the fin what did we say for efficiency of the fin we said efficiency of the fin equal to q fin upon q fin maximum q fin maximum is what we have calculated now is it not h no this is not h area of the no fin into t b minus t infinity is what we have calculated. So we need to calculate q fin maximum and also efficiency then that is efficiency of the fin equal to q fin upon q fin maximum. So if I calculate q fin maximum and efficiency of the fin I am going to get q fin so let me see how do I get q fin maximum and efficiency of the fin just if you are getting lost let me write this for the sake of completion that is efficiency of the fin equal to q fin upon q dot fin maximum. So now let me figure out if I get efficiency of the fin and if I calculate q dot fin maximum I can get the q dot fin. So efficiency how do I get efficiency of the fin so if you see we have drawn for this configuration efficiency of the fin see this figure what is it told what is it written here efficiency of the fin is a function of zeta that is which is involving length thickness of the fin heat transfer coefficient thermal conductivity and thickness again and area of the fin is 2 pi r 2 squared minus r 1 squared plus 2 pi r 2 t. So that is for the benefit of who are not able to see this let me write this that is it is a function of r 2 plus t by 2 upon r 1 and l plus t by 2 into square root of h by k t. What is r 2 we just now saw know r 2 is 3 into 10 to the power of minus 2 what is the thickness? Thickness is 2 mm that is 0.002 upon 2 upon r 1 is what what is my r 1? 0.015. So if I substitute this I am going to get this ratio as 2.07 okay if you check with the calculators you are going to get the same number. So what is here l what is here l? l is not the length actually l is if you see this figure if you see this figure you see this what is the additional length of the fin that is additional radius that means what here it would be r 2 minus l here is r 2 minus l here is r 2 minus r 1 that is 3 into 10 to the power of minus 2 in minus of 0.015 that means you are going to get 0.015. So if I substitute that 0.015 plus 0.002 upon 2 into square root of what is h? h is given to be 60 h is given to be 60 60 upon thermal conductivity is given to be 180 into thickness is again 2 into 10 to the power of minus 3. So you get this ratio as 0.21 remember these two numbers 1 is r 2 plus t by 2 upon r 1 which happens to be legend in our graph and the x axis is l plus t by 2 square root of h by k t is 0.21. So let me flash that graph you see 0.21 this is on the x axis is 0.21 and this graph is this graph this ratio is 2.07 I will just take something nearer to 2 and 0.27. So you will get somewhere around 0.95 0.95. So if you get back so the efficiency of the fin efficiency of this fin is equal to 0.95 but what is left out now we have got this efficiency of the fin now I need to calculate q dot fin maximum let us see how do we do that q dot fin maximum equal to efficient sorry h into area of the fin into t b minus t infinity isn't it. If I replace this fin with area of the fin with area of the base my q becomes q no fin case if I replace that area of the base with area of the fin it becomes q fin maximum. So now let me calculate area of the fin area of the fin is equal to 2 pi into r 2 squared minus r 1 square plus 2 pi r 2 into t I know you are not with me for this I need to go to that figure let me go to that figure yeah. So what is this 2 pi that is this area this ring area is 2 pi r 2 squared minus r 1 square but my fin is having a thickness there is a perimeter there is a surface area around the thickness. So that is that is this that is 2 pi r 2 into thickness t. So if I substitute these numbers if I substitute these numbers 2 pi into r 2 squared is 0.03 square minus 0.015 square plus 2 pi into 0.03 into 2 into 10 to the power of minus 3. If I substitute that I am going to get area of the fin as 0.00462. So q dot fin maximum is equal to h is 60 into 0.00462 into T b is what was T b 125 and T infinity is 27 we are going to get maximum heat transfer rate is 25.8 watts do not be puzzled earlier without fin we got 554 we are not taken all fins into account we have taken only one fin. So we have how many fins we have 200 fins so we do not have to get disaltern. So if we take all fins into account so what will what will I get q dot all fins so before we go to all fins we need to take into account see we just calculated what is the heat transfer caused by the fin portion. But there is some portion where there is no fin so that is unfin which we called so what will be that area unfin what will be that area unfin area unfin is equal to pi into pi into d1 into s. So pi into d1 is how much what is d1 0.03 into s is 3 mm so that is 0.003 so that area unfin turns out to be 0.000283 meter square. So q dot unfin portion equal to h into area unfin into T b minus T infinity h is 60 area unfin is 0.000283 into T b is T b is 125 T infinity is 27. So I am going to get a value of just 1.66 watts you see area of the heat transfer because of the fin one fin is 25.8 and one left out space where there is no fin is 1.66. So total heat transfer rate should be what total heat transfer rate q dot total because of fins is number of fins into q dot fin and q dot unfin is that right. So number of fins are 200 q dot fin is how much 25.8 and q dot unfin is 1.66. So I am going to get q dot total of fin is 5492 watts it is quite high if I have to calculate effectiveness what is the effectiveness q dot fin upon q dot no fin q dot fin is 5492 q dot no fin is what was that we found 554. So what is the number I get 9.91. So effectiveness in this case indeed is much larger than 1 and indeed putting fins is increasing the heat transfer rate almost 10 times almost 10 times. So this is how one evaluates how to put fins. So you can go on you can ask your students to decrease this spacing you decrease this spacing you increase the number of the fins you can go on optimizing this. So but then you do not have to really do that why I say that because you can figure out that from this figure itself you see we have already chosen properly why I say properly our effective efficiency of the fin is already hovering around 1.95. So we have already chosen spacing properly is that okay for R2 and R1 and thickness is fixed the spacing sorry the R1 and R2 is fixed here it is not going to tell the spacing the spacing if I decrease what will happen what will happen if I decrease the spacing yes number of fins will increase but can I increase the spacing in this problem no it is already 3 mm 3 mm is very small fabrication will become difficult if I decrease the spacing still further okay. So I think I am already running my fin at the best case in this problem. So that is how one can go ahead and solve any configuration of the fin we have just demonstrated for the fins the way it has it is handled generally. We will take questions for next 10 minutes and then maybe we will take introduction for transient conduction little later Hyderabad Mufakkam job college you have given effectiveness of the fin is proportional to 1 root h that is inversely proportional to 1 by root h can you explain me with respect to a 2 wheeler engine with fins when the vehicle is moving yeah let me rephrase your question if I understand what you are asking is the effectiveness of the fin we just now derived as square root of k p by h ac so in a moving engine I am having a fins compared to stationary engine and a moving engine so what will happen to the effectiveness definitely in if it is a moving engine so the h will go up compared to the stationary engine. So h will go up that is it is almost like forced convection so the effectiveness in a moving engine of the fins effectiveness of the fins in a moving engine is going to be lower than the effectiveness of the fins in a stationary engine because the h in a stationary engine is natural convective heat transfer coefficient which is less but while the engine is moving it is forced convection the convective heat transfer coefficient is larger did I answer your question so but we are providing these fins to increase the rate of heat transfer so here effectiveness is once again proportional to efficiency. I understand your question but we are using fins in a moving engine yes to answer your question I said that the effectiveness of the fin in a stationary engine is more effective compared to moving engine but when I when I am putting the fin I am putting the fin for moving engine only I will design my fin that is I will choose my perimeter cross sectional area and the thermal conductivity that is of course I usually use aluminum fins only having chosen aluminum fins I will play with my perimeter and cross sectional area such that the moving convective heat transfer coefficient is chosen and I will design my fins such that the effectiveness of the fin is effective is high for a moving engine so fins are going to be effective even in moving engine but definitely effectiveness or efficiency of a moving engine fins will be lower than that of the stationary engine no doubt about that okay I think I have answered your question over to saint joseph college kerala. Sir my question is that sometimes fins may not be of no use because in the case of some things the velocity of flow is very high so in that case what is the better option for effective heat transfer yeah fins are not always useful see as I said in the initial case itself if I if I took the example if I took the example earlier itself and I told that not always the fins are going to be useful so in fact if you see this I have explained here yeah so the question asked is if the velocity of the flow is very high fins are not effective yes it is true I have already mentioned that effectiveness is inversely proportional to square root of H and in the no flow case and the flow case if I take H of the no flow is going to be lower than H of flow so the fins would not be as effective in case of the flow situation so yes but if I take flow there again flow is of two types if I have a flow in air and flow in water let us say so H of flow in water to H of flow in air they would be different so H of flow in water will be higher than H of flow in air that is putting fin in a flowing medium with air it would be more effective compared to that of water why because heat transfer coefficient in case of water will be higher than that of in air so even in a flow situation we can use fins provided the heat transfer coefficient is less that would be the answer for your question so one more question so in the case of triangular fins is there any effect of this angle angle of the surface in the case of heat transfer often readily I cannot answer this question because it depends on the the question asked is is there any influence of the angle in a triangular fin that is the question asked by one of the participants the answer is I cannot answer it directly I am sure it will be having influence how it will have the influence the angle it will have an influence such that because I have to keep my what is the thing what I have to keep perimeter by cross sectional area should be high I have to choose my angle such that my perimeter to area is larger whichever configuration is having P by AC larger is the one which is going to give me the higher effect of course we have to sit down and calculate but by and large all that I can say is that I have to choose a case in which the perimeter is larger than the cross sectional area thank you Jaipur college kukas if you have any question please shoot them what is the effect of surface smoothness on the fin effectiveness surface smoothness on the fin effectiveness okay see this way we can look at it see again I will come back to this so what will happen if there is surface roughness on this smooth surface where there is the question asked by one of the participants is that what is the influence of the surface roughness of this smooth surface where there is no fin so I would answer it this way if there is no fin in this portion if it is rough let us say for natural convection perhaps this H is not getting influenced by the surface roughness but if it is forced convection the H in the flow if it is rough would be higher than otherwise that is the smooth surface so that way effectiveness would come down if it is flow if the flow is there and the surface is rough so we have created four situations now flow is there and rough surface H increases effectiveness decreases compared to smooth surface but if I have a rough surface but there is no flow it is simply natural convection then H is natural convective heat transfer coefficient generally is not so influenced by the surface roughness of course there are some special cases where there might be effect but generally the heat transfer coefficient in natural convection is not influenced by the surface roughness in that case the surface roughness has no influence on the effectiveness because it has no influence on the H but on the area of the unfin area of the unfin increases if my surface area if my surface is roughened because there are lot of serrations all over the place so but I do not know how much the surface area is going to increase that one has to sit down and calculate but the gut feeling is that it is not going to increase significantly it is like putting a sandpaper I do not think sandpaper is going to increase the surface area significantly higher as opposed to no sandpaper so that is the answer for your question thank you.