Extended Surface Heat Transfer Example - Part 1
 
Hello everyone again we’re starting a new topic of heat exchangers. And if you can recall that we have started discussing different augmentation techniques and then we have told that extended surfaces and fins they are very versatile augmentation technique. their plastic picnic not need any of the power but they can be used in various conditions for various modes of heat transfer on various geometry and buy some innovative design we can help both increasing surface area whenever we are using any kind of extended services primarily with increasing surface area but at the same time by our innovative design we can also enhance the heat transfer coefficient convective heat transfer coefficient so there is a double gain in many cases and we can have a good amount of heat transfer enhancement by providing extensive surface surface and extended services all things are very extensively used for gas side heat transfer augmentation do they use for liquid side transfer recommendation or transfer augmentation for free state cases but mainly they are used for get side heat transfer fermentation and what we want to do today we like to take several examples because things are extended successive transfer devices are very important we want to take several examples particular example how the calculation or estimation can be done for extended service transfer this is kind of it is not truly speaking a heat exchanger problem but basically transfer problems but we will have some idea that her calculation can be done very simple what we have done is like this but we had taken the Field on a plane surface and this thing is it very narrow cylinder this type of things are called pain feels so only one thing is there whose diameter is d and link this earth and all the information given here it is said that the an aluminium fin fin of uniform diameter 2 mm LinkedIn millimetre is attached to a hot surface with a temperature of 200 degree celsius in his own bypassing air across it at 15 degree celsius resulting in a heat transfer coefficient of 100 watt per metre square calculate the rate of heat transfer from the effectiveness and efficiency for doing this we need the thermal conductivity and thermal conductivity of aluminium is 250 watt per metre KB many of you have done this kind of problems why does Daddy hit transfer but for the benefit of everyone and for the sake of generators have a very simple problem so this thing that is the diameter of the penis 2 mm length is 10 mm you see from hearing we get some idea the length is at least five times more than the diameter with a heat transfer coefficient that is also given 100 watt per metre square in and thermal conductivity of the material that is 250 watt per kilometre reset CSS is 15 degrees Celsius to be that is the best temperature of the field and that is 200° Celsius that is the difference between the base and the fluid that is 15 degree celsius cross-sectional area of the field through which transfer takes place that is 5 by 40 square and we get this and then perimeter perimeter of the field that is a link to perimeter of the field is equal to pi li so we have calculated if we go to the next light now if it is under root HP by KFC where it is the heat transfer coefficient he is that is if it is under root HP but so we have calculated if we go to the next light now first we have to calculate geometric parameterHP buy KFC where it is the heat transfer coefficient be is the perimeter of the figure is the thermal conductivity and ac is the cross-sectional area putting all these values we will get to 8.27 park means metre to the power one that is what the unit of stain is so in the event of age is metre to the power minus one insulated transfer is given by the formulatransferred from the field or picking a given temperature and before assuming that I finished infinitely long fruits ultimately it achieves the atmospheric temperature temperature of the surrounding these things and material of heat transfer any standard transfer but I’m almost there be insulated feel team is most commonly used there are many reasons for that first thing is that in many cases the heat exchanger find that deflated that means there is no transfer fee we do not provide any information on there but the design is such that from which there is no picture so that is one point below so she is very small from the fieldpod point is that in many cases what we can do we can do an approximation we can correct the length of the fee taking into account that there is some amount of heat transferred from the field and if we use the formula for feeling with insulated because that formula is easier to use compared to using a formula for convicted applause from the fee so that is what we will do in the entire exercise 41 examples insulated what we have done shoppingfind a single free number of feet and in that case we have to calculate what is the heat transfer from all this is not only that there will be spaces in between also want to know that what will be the heat transferred from the internal spaces then we will be able to find out what is the total is transferred from the Steam surface now find out what is the transfer and effectiveness by giving the formula by using the formula and also I have written so it is coming 19.4 in simple words what does it mean it means that if we do not have a fee and we propose to use afrin by using a single feel we can have this much amount of heat transfer enhancement ok very large amount of enhancement we can have this is what is same in given by this particular calculation so this really justify the use of the fee then so fast it justifies the use of the field but then later see how good is the fee we are using it is finish the heat transfer and the maximum be transferred that could be possible while you’ve been absent geometry and it is point 97004 23560 suppose we decide to use this geometry from some other consideration then our ceiling design for this case means the length of the field the material of the fee etc we have selected in such a way so that we are getting 97% capability of an ideal really want this design is good time to time we have to calculate the effectiveness and efficiency of the field to check whether we have done a good job or notthe problem is from the book of Esther and circular cone heat exchanger basically do it is associated with a heat exchanger but the problem is one can take is as a simple heat transfer problem any problem let us read the problem AA plate-fin heat exchanger has 24 mm high point 12 mm thick rectangular face with its intensity of £600 per metre the heat transfer coefficient for airflow over the fence is 170 watt per metre square km in determine the phanaticmc beliefs are made from copper with the thermal conductivity of 401 what form it will be used 2.06 what happens to me and associated how do you change the intensity to bring the same level of heat transfer vinyl fence if you realise that there is no change in a transfer coefficient with the change in phenix city no change in material from copper to aluminium with the thermal conductivity as 237 what per metre square shelving discuss the implication of inefficiency and transfer of changing the material from copper to aluminium the mass density of these materials are given for aluminium obviously it is much lighter and copper it is having a I mean the density of copper is higherthen conductivity of the field that is in density of fence per metre that is given and then confirm we are considering two cases one is .12 mm and another is .06 mm thickness show all the data archive our school to the next diagramin the hole inside there in a plate fit extended that is so hopefully next year if you notice careful if this will be explain when we will like to discuss play winner takes it back here in connection with fins latest have some idea so you said to place in between their offence like this so you see that very thing and this one acts are in a kilometre in an accident and you see if we assume that the plates are at same temperature then one can think of that this is the base of the feet and here and teammate play we will have the same teeth at insulated condition this is very important to appreciate that most of the things we have seen that it is attached to the base and it goes out of the primary surface and the team is not connected to any solid surface but here it is kind of a  which is connected between other two surfaces which are at 90 degree 90-degree to the field where there is no free and no one can assume as if it is made up of two Fields one thing is attached to the office yet another thing is attached to the lowest rate so this is the best condition of one feels this is the best conditioner for the field and I did meet play the Finn is insulator insulator similar things give and the thermal conductivity is it is then what we can do we can calculate this thing a man we can calculate and as a function of email we can find out what is the efficiency of the suspect then we know there are a number of Cars 1 2 3 4 5 cars at the earth below the top card should therefore radial feel or a new lock screen or sometimes they’re called succulents that means being around a circular to share what we need we have to calculate him into minus r r y and are these two things that give ok so if the fig tree is known as all should be known I should be known we can calculate this and he is thermoelectric pyrometer how to calculate if we have discussed andestimate the efficiency of the circular thing without going into some sort of interest calculation which involves making sure you know that in the estimation of at c&c for this one straight feel we need to have the the the hyperbolic trigonometric function and the estimation of the stuff we need to have the basic function now if we want to avoid this thing that can be done for ready calculation for quick calculation and then this card will be very hot this graph will be useful these are kind of data given for the solution of the problem should now let us move to this solution so we have to determine the change in the change from .1 to millimetre 2.06 millimetre next to do the change in density for the same heat transfer in reducing the thickness from .1 to 2.06 minute and how is it different material is made from copper to aluminium keeping the same inventory what are the other design invitation what we assume we are showing the heat transfer coefficient does not change if you change the field density heat transfer coefficient we are what we are doing we are changing the same thickness accordingly sometimes we are changing the field density how many films will be there on a particular Lane but our heat transfer coefficient is not this is one assumption this assumption is not to connect but just to get an idea that what is the effect of intensity what is the effect of the material on this simplified assumption we are working first let us try to find out the efficiency of the Copperfield we have calculated a&m you see him how it is calculated giving all the values etc and then what we have done we have putting all the value of 4.057 ^ 18 is your hair so we have got email and with this formula we have got beef in efficiency is equal to .759 so what is the point of having that car so we can go to that car I have shown in the previously previously and let’s see if we can see that what we callnot want to calculate we want to get from the car show email when we have gone and from the evil value we can get directly from that is one way of doing it I am requesting you to check what my calculation we have got that from the start we are getting the same one another way that we calculate the hyperbolic trigonometric function 10 hyperbole and then we calculator Philippines collect76% in the second case it is 62 but we get if we know that there is 80% reduction so what we can do if we increase the density number of sparkling then we can get the same amount now what we can do and then all the calculation andaluminium and keeping everything constant then there will be a falling everything we got the two efficiencies we got that there is a 18% reduction in his duty if we moved from a 12-11 many metres are in 12.12 mm thickness 2.06 mm thickness and then we are considering the change in material from copper to aluminium we have kept everything alright uniform point to 12 mm thickness of the fence and with that we get and efficiency of 1659 that means 66 percentage since we’re getting which is less than our efficiency of copper fin and it is expected then what we like to do that what is a costly material buy aluminium nowaluminium that means the minimum thickness of copper which we can use in our pen design there is a minimum thickness we cannot be used for aluminium design so here it is very important we can see that changing the material from copper to aluminium reduces efficiency from 76% to 66.12% in Fenwick .06 mm copper then v efficiency is about the same that means let’s say we have seen earlier two cases in one case we have used .06 mm thickness of thing and if we take that is one design option and on another side we have got another design option that is .12 mm aluminium be there in the efficiencies are competitivenow we have compared the material used for this to feel that point to 12 mm of aluminium and .06 mm subcompact and then since the mass density of aluminium is much lower compared to the mass density of copper we can make and we can make a comparison like this aluminium sheet material and that is the film it really this is the density this is coupled filmed material and this is the density and you see by comparing them or by taking a resume pt6 that means for same kind of the transfer if you use copper we can see if we need to use only 60% else so it makes the material used much was it makes the weight of equipment that could be that could have some relevance relevance or equipment which is not stationery which are moving because the weight of the movie moving equipment is larger then the traction power is more so by providing aluminium aluminium parts exercises part so what we can do we can reduce the we can reduce destructive power without sacrificing. With this one point I would like to keep in mind and with this I like to end today's lecture.  Thank you