 Sir, my question is, can temperature of the coil flowed, cold flowed rise above the inlet temperature of the hot fluid at any location in the heat exchanger? Yeah, actually the question is, in heat exchanger, can anywhere the cold fluid temperature exceed the hot fluid temperature? Yes, I remember, I forget exactly what word that is called as, but yes it can occur, why because it is going in cross flow and locally somewhere, locally somewhere the fluid temperature can exceed, but that is locally only, that is locally just because that the heat transfer coefficient there has become very less because of which the wall temperature might have gone very high. So, or the fluid or other way around, the heat transfer coefficient somewhere has become very less, very high because of which the fluid temperature has gone very high. I am not getting the exact word, there is a word for this, I am not getting, I have taught this in the heat exchanger course. Please put this question, this is called as some reverse factor of the inverseness, I do not remember exactly, yes it can occur in few locations, but not in tube in tube parallel and counter flow heat exchangers, it can occur in cross flow or shell and tube heat exchangers, where in which all sorts of, in all sorts of directions the flow is taking place. Okay sir, my second question, the effectiveness of the heat exchanger can be greater than 1? No, effectiveness, can the, effectiveness of the heat exchanger greater than 1, never, effectiveness of the heat exchanger can never be greater than 1, because I have already taken the maximum temperature difference, how can I get more temperature difference than what I have taken, I can never get. Okay, Ruba college indoor, any questions please? I have got one query, what is the difference between the effectiveness of heat exchanger and efficiency of heat exchanger? As such, effectiveness is the ratio of actual heat transfer to the maximum heat transfer, how one can define the efficiency? Can it be like that, it is, it can be the ratio of actual heat transfer to the heat transfer when we take the average temperature difference? Okay, question is why do not we define efficiency, why do we handle only effectiveness? How can I take average temperature difference? Does the average temperature difference have any meaning? No, see whatever temperature difference I take should have a meaning, it is LMTD if I take, it is varying, that is if I take the temperature difference from hot side to the cold side, continuously there is a temperature difference. If I take average temperature difference, we said in convective heat transfer, if I take average temperature difference, I can be off by 15 to 20 percent or even more. That is the reason why we have introduced the concept of delta TLMTD, question now answer is first question, first part of the question answer is we cannot take average temperature number one and where is the question of efficiency definition, where is the question of efficiency we have told already that in terms of effectiveness, what is the ratio of the maximum temperature difference, why we have defined effectiveness, why not efficiency in terms of actual heat load to the maximum heat load, why do not we do that because the temperature gradient is the one which drives it. So that is the reason why we have defined effectiveness but not efficiency. All this while we have been emphasizing all the time that whether it is constant flow heat exchanger and constant wall temperature case or even counter flow or parallel flow heat exchanger, what drives is the temperature difference, that is the reason we have defined effectiveness as temperature difference, we have not defined efficiency as actual heat transfer to the maximum heat transfer, this is because the driving potential is the temperature difference, that is the answer. Okay, MES Pillai Panvel, any questions I hope your audio is fixed. Sir, one example you have taken in that example you have incorporated the heater at the bottom of the tank, suppose I insert the heater at the top of the tank, so the upper portion of the water will get heated and the remaining water which is present at the bottom of the tank that will be heated by the conduction only after some time. This concept is clear or not sir? So what you are saying is that we are having a geyser, in the geyser I said that in the morning we are having a heater at the bottom, so that is I have a heater at the bottom and because of which the hot air goes up and the cold air comes down and the natural convection current is used for maintaining the uniform temperature distribution, what if I have the heater at the top, what will happen the water which is there here will get heated up and the only way the heat can transfer itself to the cold fluid which is at the bottom is through simple conduction. So it will take enormous time for the bottom water to get heated up number one, number two it will perhaps not get heated up as much as this gets heated up, okay. So uniform ultimately why are we heating, we want to get the uniform temperature distribution of the water, but that temperature uniform temperature distribution is managed or exploited or we are able to generate that here in this situation, but not in this situation. Why because conduction is not as effective as natural convection, okay, that is the reason. Is that okay? Pallapur, will you please suggest books on experimental heat transfer sir? Yes, that is what I am doing, I have that is why every day I am carrying but I forgot to mention that please write down this is the best book what I have got and this experiment what I have given also is from this book, please write this down and order this book for your library or if you can purchase nothing like it. Experiments in heat transfer, experiments in heat transfer and thermodynamics very difficult to separate out thermodynamics and heat transfer although we teach them separately, okay, thermodynamics by this is given by several authors, the experiment what I gave was by Professor Bajan whom I like very much, why because that scale analysis concept is a strong proponent of scale analysis concept, I like him very much, I am his great fan. So, this book is edited by Robert A. Granger, Robert A. Granger, let me rewrite this Robert A. Granger, Cambridge University Press, University Press, let me repeat experiments in heat transfer and thermodynamics by edited, this is edited not written, there are around 31 experiments listed here, there are 31 experiments not all are very easy starting from single phase to two phase flow, all experiments interferometry, all experiments are listed out but contributed each experiment is contributed by an expert himself, okay, so this is the best book according to me where in which you can build setups on your own, in fact that was the reason good that you asked me this question, I wanted to tell every day but I was forgetting because of so many things, so I used to carry this book, I do not have the original with me, I have a Xerox copy of this but original is in our library but I think we have to purchase this book, okay, so I recommend everyone to purchase this book and add this to your library and give this as projects, BTEC projects for your students, each experiment can be given as a BTEC project for your students or part of the heat transfer lab you can give that as a project lab, okay, so if you give that it will be very good, little higher end book if you want, there is another book by Goldstein, this is again experimental heat transfer by Goldstein, not too sure about the heading of the title of this, I think it is experimental heat transfer, I am just searching Google book search so that I will not make a mistake, fluid mechanics measurements, it is experimental heat transfer, correct, it is experimental heat transfer, correct, I am not wrong, experimental heat transfer by Goldstein, okay, so these two books should be very good study material or it can be good beginning for and of course there are plenty of ideas in various journal papers that is very diffused way of telling but most of the ideas we get it from text, from papers only, journal papers only, anything else professor? Sir as a student, will you suggest the must books which we read on heat transfer sir, okay, so as a student, as a beginner, what are the books we should be reading, I would suggest the following, if he is a UG student, if he is a UG student, if he is a UG student that is undergraduate student that is for DE or BTEC, the best book what I like or the best books what I like or I refer myself and I think as a student it is a very good book is Jungle, C E N G E L, actually you may feel that I should be pronouncing it as Sengal but actually this is an Israeli name, there is a small tilde here sitting in the denominator, so that is what makes it called as cha, chengal, it is chengal heat transfer by chengal and recently I came to know that chengal along with chengal in the fourth edition there is a co-author, I do not remember because I forgot to bring that book with me today, the publisher has just given me the sample copy just on the, it was a coincidence that just on the day of starting workshop I got that book for fourth edition but chengal is a very good book, why I say very good book because I call this as armchair reading, armchair reading means I do not have to take special effort in understanding from professor chengal, he writes it so well, he writes it so well that every concept from one sentence to another sentence there is a flow, there is a logic flowing and he tells the same thing in two, twice in two different ways so that I cannot miss out the concept, another thing which is good about chengal which is good for a student is that there is left hand side, there is there is the summary of the whole page in form of figures or in form of equations, after I read the page on the left hand side if I close my eyes and that left hand side whatever is there if it comes on to my mind that means I have understood that page, that is why I that means this book is like a notes it is not like a text it is a notebook that is about chengal, now another good book for problems is definitely all of us know that is incropera and David, incropera and David it is not armchair reading book you will have to make for example fins if you are reading infinite fin length you will just give in one sentence I did not understand myself first time when I read that it took lot of time for me to understand that is why the tan hyperbolic ml if you remember we have discussed that so much in length but that thing is made understood by chengal in one shot because he has given one figure there where in which there is a temperature variation from base to the tip. So each book I say every book is a good book but there are few books which are not good which we should not read I will not take their names but chengal and incropera are the best books for UG this is what I suggest to my students incropera and David if someone does all exercise problems as a teacher I would think I should be doing all exercise problems and also as a student if one does all exercise problems of incropera and David is master of heat transfer why because incropera and David all problems all numbers are real life numbers they are not just cooked numbers are cooked problems they are all real life problems and apart from this there are n number of books one is bhejan heat transfer by bhejan very expensive book but I like this book very much this book is heat transfer he is having two books heat transfer and also convective heat transfer as a teacher I would think bhejan for convective heat transfer has to be read I know my writing is not coming proper because there is some problem with the pen but I am telling please write down this is convective heat transfer by bhejan and of course apart from this there are n number of books I think I will stop here there is what is ojisik is there then frankreeth is there frankreeth and bonn frankreeth and bonn frankreeth and bonn is also very very good book but apart from all these books main the stand all stand apart are the best books I would think are changal incropera and David. Of course there are plenty of indian books but I will not take them ok so apart from these books you read any book you will it will be a value addition only there is nothing you are going to lose after you read changal and incropera and David you read any book it is fine with me ok but first time if someone is reading I would think he should start off with changal ok you have a question sir yesterday I have seen one slide in radiation that is the roof top of the house wherein the radiation is coming from the sun and one is coming from the trees ok so in radiation when it is coming from the sun if the radiation passes through some other medium and then falls in the roof then whether it affects the absorptivity see the question is the roof top problem roof top since yesterday we have been taking see in the morning also I solved see what did I say morning how do I measure absorptivity absorptivity or emissivity how do I get this reflectivity transmissivity absorptivity emissivity how am I getting what we are saying is that it does not mean that surface properties are not at all important absorptivity comes from emissivity emissivity is the one which can be measured emissivity reflectivity transmissivity can be measured when I measure this emissivity reflectivity and transmissivity they become the properties of the surface ok now what we are saying is that for the absorptivity absorptivity of that material will depend on the source temperature that is absorptivity of absorptivity of a plate or a given material at source temperature will be equal to emissivity of the source temperature but emissivity at the source temperature is again dependent on the material properties that is how material properties are accounted and also the temperature ok. Sir if it is passed through a glass and then it is falling in the roof top if it passes through some other medium before falling it on the roof top then whether the if it passes through the roof top if the if it passes through the if it passes through the roof top what happens to the absorptivity of my material then where is it coming from it is not coming from the sun it is now coming from the roof top glass. So, sorry between the roof top and the sun there is a glass now my source has become glass ok. So, now you have complicated the problem now you have complicated the problem. So, because now I really do not know I think it will be still dependent on the temperature of the source I still insist that but how much it will come that is the wavelength dependency that is what wavelength is passing through this glass is dependent on the transmissivity of the glass but still the absorptivity will be dependent on the source temperature ok. So, I think we will get started with tutorial today we are lagging behind let me get started with the first problem yes there is in the chat there is a question that you will give me a download historical perspective it is not a link it is a journal paper I will give you the reference of that paper ok just bear with me I will today send it across ok I was still yesterday night I was working out with that emissivity concept I had to spend time I have not checked even a single email till today ok. So, I will do that before I go today I will send it across ok. Now, let us get started with see let us take up problem number 47 let us take up problem number 47. Now, there are two parallel plates there are two parallel plates one is surface one another one is surface two ok and the parallel plate size is given by 1 meter by 0.5 meter both the surfaces are of same size and the distance between the two plates is 0.5 meter apart ok. So, now the emissivities of the plates are epsilon 1 equal to 0.2 epsilon 2 equal to 0.5 then these two are exposed to and one plate is maintained at 1000 degree Celsius and the another one is maintained at 500 degree Celsius and both are exposed to ambient which is infinity equal to 27 degree Celsius. The plates exchange heat with each other and the room, but only the plate surfaces facing each other need to be considered in the analysis find the net transfer to each plate and to the room that is the question ok. So, now what are we saying the same whatever circuit I had drawn will be valid, but with an assumption or with a truncation what is that? Let us call this as surface one this is already 1 2 let ambient be 3 ok. So, same circuit let me draw this is e b 1 j 1 what will be this resistance what is this is it space resistance or surface resistance surface resistance 1 minus epsilon 1 a 1 epsilon 1 then you have j 2 that is 1 upon a 1 f 1 2 ok. Next is this is j 2 next is e b 2 and what is this resistance 1 minus epsilon 2 upon a 2 epsilon 2 ok and then I have j 3 and j 1 j 3 are connected j 2 j 3 are connected j 1 j 3 what is the resistance between j 1 j 3 1 upon a 1 f 1 3 and this is 1 upon a 2 f 2 3 ok. Now what about j 3 what about e b 3 here actually we had resistance e b 3 and we had 1 minus epsilon 3 upon a 3 e b 3 what was that that was the what was that resistance resistance of the surface is my air participating is it having in emissivity no it is having in emissivity that means 1 minus epsilon becomes 0. So, that is why I do not have resistance if I do not have resistance what will be j 3 j 3 will become equal to e b 3 j 3 will become equal to e b 3 that is all the catch in this problem ok. If we understand this j 3 equal to e b 3 rest all is straight forward nevertheless I will go ahead and solve this problem other way of looking at it is instead of looking at it from epsilon alone epsilon 3 a 3 because you might be thinking 1 by 0 is coming what should I do my a 3 is infinite you think like that if a 3 is infinite what will happen to my resistance as a 3 tends to infinity this resistance tends to 0 that is what is happening here that is what makes it j 3 equal to e b 3 is that ok. So, now let us do the problem for the sake of completion. So, that I do not miss out anything. So, if I set up the equation what are the equations e b 1 some you will have to help me e b 1 minus j 1 upon 1 minus epsilon 1 upon a 1 epsilon 1 plus j 2 minus j 1 upon 1 upon a 1 f 1 2 plus plus j 3 or e b 3 minus j 1 upon a 1 f 1 2 upon 1 upon a 1 f 1 3 equal to 0. Similarly, for node 2 what does let me take yeah let me take node 2 what do I get e b 3 minus minus j 2 upon 1 upon a 2 f 2 3 plus plus plus e b 2 minus correct e b 2 minus j 2 upon 1 minus epsilon 2 a 2 epsilon 2 plus j 1 minus j 2 upon 1 upon a 1 f 1 2 equal to 0. What is e b 1? e b 1 equal to sigma 1 upon a 1 t 1 to the power of 4 that is sigma t 1 to the power of 4 mean 148.87 kilo watts per meter square. What is t 1? 1000 degree Celsius t 1 is 1000 degree plus 273 you get 1273 to the power of 4 then e b 2 what is e b 2? Sigma t 2 to the power of 4 t 2 is 500 that is 773 Kelvin if I substitute that I get 20.24 kilo watts per meter square. Similarly, e b 3 equal to sigma t 3 to the power of 4 t 3 is 300 Kelvin t 3 is 300 Kelvin I get e b 3 of 0.4592 kilo watts per meter square. I got e b 1 e b 2 e b 3 in these two equations what are the unknowns? J 1 and J 2 what are not known is J 1 is not known and J 2 is known do I know epsilon 1? What is epsilon 1? 0.2 epsilon 1 is 0.2 what is a 1 equal to a 1 equal to 0.5 into 1 that is 0.5 meter square is that right? Now, epsilon 1 I know do I know f 1 2 and f 1 3? No I do not know I have to calculate. So, I will calculate f 1 2 and f 1 3 if I calculate f 1 2 and f 1 3 and I substitute that and solve these two equations I will get J 1 and J 2. Let us do that that is what is f 1 2? How will I get f 1 2? f 1 2 it is very difficult it is not impossible, but difficult to get from the equation. I have just put I am just putting in a value which you can get from the chart that is taking the between two parallel plates you take two parallel plates you get I am writing directly f 1 2 as 0.285. Now, what is f 1 1 plus summation rule f 1 2 plus f 1 3? 1 1 to itself 1 2 is 1 plate to the second plate 1 3 is plate to plate to surrounding that is atmosphere. So, f 1 1 what is f 1 1? 0. So, f 1 2 I have got point I have made a mistake no one has corrected me this is equal to 1. So, 0.285 plus f 1 3 equal to 1. So, f 1 3 equal to 0.715. Is that enough or anything else I need? I need in my thing f 2 3 sorry yeah f 1 3 I have got f 1 2 I have got I need f 2 3. How will I get f 2 3? f 1 3 is equal to f 2 3 by symmetry because both the plates are there if you see here both the plate are seen the atmosphere they are symmetric with respect to atmosphere. So, f 1 2 is equal to f 1 3 this is equal to f sorry f 2 3 I said f 1 2 equal to f 1 3 not that f 1 3 equal to f 2 3. Now, if I substitute all these numbers in the previous equation in the previous equation what do I get? I am going to get what do I get? J 1 and J 2 you will find that J 1 is 33.43 kilo watts per meter square and J 2 equal to 14.4884 kilo watts per meter square. What is the question asked? What is that we are looking for? What is that we are looking for? We are looking for net heat transfer to each plate and to the room. What is the heat lost? What is the heat transfer by the plate? Let me go ahead and write heat lost by the plate because I know that heat lost by the plate because why it is heat lost by the plate I am saying so confidently because it is sitting at a higher temperature plate 1 it is sitting at 1000 degree Celsius. What is that given by q 1 equal to e b 1 minus J 1 upon upon 1 minus epsilon 1 upon a 1 epsilon 1. What do I get from this? I get 14 minus 4 3 kilo watts per meter square. So, similarly heat lost by plate 2 q 2 would be equal to what? e b 2 minus J 2 upon 1 minus epsilon 2 upon a 2 epsilon 2 is equal to what do I get? If I substitute I get 2.6785 kilo watts per meter square. Now, what is the heat received by the room? Heat is received by the room. What is that heat received by the room? Next is heat received by the room that is equal to that is equal to what is that equal to? Heat received by the room there are two resistances. You see you go back to the circuit I am just seeing the heat flux that is what is the current? See why did I write for plate 1 e b 1 minus J 1 upon 1 minus epsilon 1 by a 1 epsilon 1 that is the heat flux that is the heat lost by the plate and for what is the current which is go to the circuit here you will understand. How am I rating? I am not rating from the air this is the heat flux which is heat lost by the plate. This is the heat flux which is heat lost by the plate 2. What is the heat gained by room is J 1 minus J 3 upon this plus J 2 minus J 3 upon this current is heat flux is it not? So, J 1 minus J 3 upon 1 upon a 1 a f 1 3 plus J 3 J 2 minus J 3 upon 1 upon a 1 a f 1 3 plus J 3 J 2 minus J 3 upon 1 upon a 2 f 2 3. What do I get? I get heat received I get it as 16.945 actually I will get them as negative. So, net energy from plates 1 and 2 should be absorbed by the room is it not? Whatever is gaining from the plates has to be absorbed. So, what is that we got for plate 1 and 2? Q 1 plus this is let us call this as Q 3 Q 1 plus Q 2 if you do that is 14.43 plus 2.68 if you do you are going to get it as 17.1085 these two are just that is the net. See at the end of the day I as professor Arun was saying you do whatever circus energy balance has to occur. So, this is the energy balance any radiation problem for that matter why any problem we need transfer check for energy balance. So, we will take questions we will go to Amrita Koyambattur. Sir, we have something like 1, 1 parallel flow shell and tube bed exchanger and 1, 1 counter flow shell and tube bed exchanger sir. It is a 1 time flow on shell and 1 time flow on tube side. Okay, okay, okay 1 pass you mean okay. Yeah, 1 pass 1 pass in shell side and 1 pass in tube side. So, we have something like 1, 1 parallel flow shell and tube bed exchanger and 1, 1 counter flow shell and tube bed exchanger even whereas in the case of while the number of pass increases in tube side while the number of pass increases in tube side like twice or four times right only then we are unable to classify the exchanger as a parallel flow and a counter flow. And this has been quoted in a book see even sir. Which book you said? Okay, what is the issue now? What is the issue now? What is the issue? No, the issue was say there is not an issue sir, but we have something like 1, 1 parallel flow even say in epsilon tube bed exchanger even you have something like parallel flow and a counter flow. If it is 1 pass single pass. No, I do not agree sorry I do not agree because even if it is 1, 1 shell on the shell side it is coming normally the flow is coming normally and see the question asked by one of the participants is that even a 1, 1 shell and tube bed exchanger I am having a shell like this and I am having only a single tube and now the claim is that and the claim is that now if it is a single tube it is going to be a perfect parallel flow heat exchanger. I do not agree professor with you why because the flow is coming normally now it has to turn around and then go may be the argument is that may be for the first initial phase where in which it is taking a turn there it is sort of there is some variation in the heat transfer the heat transfer coefficient will be very high subsequently then it has to become flow is in the developing region and if I have baffle plates if I have baffle plates then everything is mixed maybe you are talking when there are no baffle plates even if there are no baffle plates I do not agree with you that it is going to be a perfect parallel flow heat exchanger. You have a shell and tube heat exchanger with 1 tube and 1 pass then why have a shell and tube heat exchanger you can have a double pipe heat exchanger. That is a tube and tube no. Then it is a tube and tube if you call 1 shell 1 pass then it is going to be tube and tube then it is no longer shell and tube heat exchanger. So, perhaps you are meaning tube and tube heat exchanger and then calling it a shell and tube heat exchanger is that right? Tube exchanger but that may number of tubes in it and the pass goes like 1, 2. Professor I think this is getting very specific you please put up this in moodle we will get back to this this is just a matter of nitpicking I do not think there is any conceptual issue any other questions from the center please. That is it sir. Thank you sir. Can I take any questions please. Sir one question sir. So, boiling during nucleate boiling the bubble is formed at the surface and it is moving upwards it is going against the gravity when you are going up against the gravity we have to go velocity must be 11.2 kilometer per second escape velocity whether it is due to the buoyancy effect or any other reason for this sir please comment on this. Yeah, yeah, yeah okay see very interesting question actually the question is the bubble is formed and the bubble is trying to go up and it has to go against the gravity and the conception is that the thinking here is that it has to go at escape velocity see here we are not trying to escape from earth's atmosphere we are not transporting the bubble to the atmosphere to the space here we want our bubble to get into the from the bottom surface to the top surface of my boiling boiling vessel is not it. So, all that I need is density gradient the density gradient creates the velocity again natural convection sort of thing only thing is that while the bubble is going up it gets condenses and it may not all the time reach the top wall that is what professor was saying earlier so all the time it may not reach the top wall before it reaches the upper surface I mean top surface means free surface of the liquid it may get condensed okay so it is first thing is it is not escape velocity okay buoyancy force is still there buoyancy force is still there that is what is density gradient only is driving that and that is creating the velocity is that okay so please do not confuse with escape velocity with the velocity with which the bubble is moving very good thought process no problem but escape velocity is that velocity for my rocket to zip into the space but here bubble is not zipping into space it is just coming from the bottom of the base to the free surface of the liquid that is all it is okay. No sir again it is there is effect of gravity no sir that point only I am asking yes the gravity will pull that bubble downwards but it is moving against the gravity so. Yeah understood see the question yeah understood your thought process is it is going against the gravity how come it is able to go against the gravity now you see it has become vapor if it has become vapor means what has happened to its density what has happened to its density what is the density of the vapor it is almost nearing to one okay but what is the density of water thousand so now it has become thousand times lighter if it has become thousand times lighter it has no option but to float it has to move up but while moving up it may lose its this lower density because the vapor may get condensed and become liquid in that process density difference is lost and drop has collapsed so it is the density difference yes you are very right gravity is required if there is no gravity it could not have floated if this experiment I cannot do it in space you are very right that is what perhaps you mean by saying that gravity is required as we said in natural convection you need not only density difference but also the gravity that is what perhaps you are emphasizing is that okay professor. Sir thank you sir. Over and out. GPCOE Baramathi quickly a question. Hello sir much question is about the typical boiling curve. You know practically in our lab can you achieve the curve C2D curve and whether the point E or C we are getting that is the melting point of the wire. The question asked is when we do the experiment will we be able to capture the point D see when we do the experiment under it depends on the boundary condition see we have to thank professor New Kiyama for this for generating this equation. In fact this equation should be called this curve should be called as New Kiyama curve it is professor from Japan who has considered as expert in boiling long time back okay just I will give a small history because I like to give this history. Professor New Kiyama was such a good professor in boilers in building boilers he was invited by Americans and he was called to US Embassy in Japan what is the capital of Japan Tokyo he was called to Tokyo from one of the villages so he went and he did not take TADA from their embassy he was so nationalistic because it was just immediately after second world war and he was so much against America because they had bombed them he said that I am going to do nothing for this Americans and I am going to build boilers only for my Japan that was so nationalistic he was the full credit of this boiling curve goes to professor New Kiyama in fact we have to call this as New Kiyama curve if you see this paper how this curve has been generated it is published in IJHMT that is international journal of heat and mass transfer it is actually first published in 1952 but on invitation it was republished in 1972 such very rarely we get papers invited again because we usually send papers and they get accepted or rejected this paper is not like that it was specially invited what is so great about this paper is that experiments have been done so carefully they have done once with nichrome wire they have done with stainless steel wire they have done with various material repeatability tests have done so carefully so meticulously the word meticulous only can fill in this carefulness here what is mentioned in this paper such a care has been taken in doing this experiment now coming back to the issue it is the issue is I can capture C2 usually when we do the experiment under constant heat flux boundary condition I am moving from C to E directly C to E directly only when I do under constant temperature boundary condition I will be traversing through to C D and E and there is hysteresis if you see there are two dotted black lines here from E D and A so even if I come back on constant heat flux boundary condition starting from E you will come from E to D and then from E to nearing about a point little above A point is cool boiling curve is having hysteresis it is not having in both the directions it is not going to follow the same curve so it depends on the boundary condition new kiyama when he first did the experiment he only found C and he went into D he understood arrow I do not need arrow actually I they know what it is so here the point is he predicted what would happen between C D and E again it took 10 years I do not remember there was one more professor who did the experiment under constant wall temperature and built this and all this data has been amalgamated and given to us as pool boiling curve actually this should be called as new kiyamas curve okay hello sir one more question in yesterday's discussion about emissivities we have observed that with increase in absolute temperature emissivity of all the bodies was increasing not all the whereas for aluminum oxide it was decreasing please aluminum oxide was having lesser emissivity with increase in temperature sir yeah okay question make this point clear sir yeah yeah yeah question is the question by one of the teachers is that we have emissivity for most of the metals increasing with increase in the temperature but for aluminum it is decreasing I cannot answer directly this because in the morning session I have given epsilon as a function directly related to temperature which is sitting in the numerator and there is another parameter called sigma e which is electrical conductivity which is sitting in the denominator how this emissivity increases with the increase in the temperature and decreases with the increase of electrical conductivity decides my nature of the emissivity variation with temperature that is all I can say you will have to sit down and calculate for aluminum oxide whether the sigma e and t how does it vary but still it is very simplistic explanation very difficult to explain from electromagnetic wave theory all these things very specifically thank you sir one more question the question is about the compact heat exchangers air density that is a meter square per meter cube for human lungs is a twenty thousand sir so can you elaborate its structure of the particular that part yeah yeah see the compact density that is the we say that I did not mention that because of lack of time it was there very much beta we define what is that area volume per area per volume or volume per area I do not remember let me open that the point is whatever we do we can never mimic nature okay so it is very difficult to mimic nature that is what I wanted to represent by that example and it is given in professor Changel's textbook that is why I put that lungs beta is beta of lung so beta of car radiator is thousand what is beta heat transfer surface area upon volume we need we want in order to make it compact for a given volume I need to get more surface area why more surface area because I can get more load because for a given u1 delta t uas delta t decides my heat load so but for human lungs I do not know the design of the human lungs okay I do not know that but all that I know is is human lungs is highly dense highly compact that is it is having twenty thousand meters squared per meter cube I have given that number why because even if we design whatever circus we do I do not think we can exceed that number twenty thousand it is like it is like the ideal number what I should have it is very difficult to mimic nature it is very difficult to mimic we are talking about micro air vehicles and many things how difficult it is for the how difficult for us to take off the way pigeon takes off it is not at all easy we can make videos and see how pigeon takes off and how it modulates it modulates its wing shape while taking off and while what is that while touching down while landing it is not that easy if you are doing with with aircraft it is not easy point is by taking all these examples is that we can never mimic nature or we cannot be performing any or the any machine cannot perform as good as a human being that is why that beta of twenty thousand is given so if we can make a compact heat exchanger which is having beta greater than twenty thousand on that day we can say that we have beaten human design that is nature's design okay so I think we will take the next problem and we will come back to questions little later again so for a change we will do a problem on view factor problem number forty five a truncated cone as a top and bottom diameter as ten centimeters and twenty centimeters respectively the height of the truncated cone that is the top one is ten centimeters and bottom is twenty centimeters and truncated cone length is having ten centimeters height find the shape factor between the top surface side surface and this is another question everyone ask what is the difference between side shape factor configuration factor and view factor actually they have all three are different however in our course as long as we are handling non participating media all three mean the same configuration factor shape factor view factor all three mean the same so we should not be puzzled about seeing any three of this and another thing what I forgot to tell in the morning is that emissivity absorptivity transmissivity that is evt is referred usually for smooth pure surfaces emissivity transmissivity and absorptivity is used for smooth surfaces emittance absorptance transmittance is used for rough and contaminated surfaces this is according to who has told us this is national standards n i s t national institute standards American American national institute of standards follow follows this that is what in fact whatever material I taught in the morning it is all from professor modest textbook radiative heat transfer from modest if someone wants to dig and understand more about this reflectivity and all it is professor M F modest and of course professor Arun has studied with professor modest itself professor modest has taught him conduction okay so now let us get back to problem number 45 that is we said that there are three surfaces now let us find the shape factor between top surface on the side and side surface and itself okay that is F 23 and F 33 okay so now r i r j r i equal to capital r i equal to r small r i by L and r j equal to that is r i by L L is the height what is r i now r i is 10 by 20 by no r i I have taken 10 10 by 10 I am calling this is i and this is j another thing okay so r i equal to r i by L equal to 1 yeah r j equal to r j by L equal to 0.5 small r j by L equal to 0.5 you are right now r i by L is 1 r j is 0.5 now r j is 0.5 okay s equal to 1 plus 1 plus r j squared no again 1 plus r j plus 1 within the bracket upon r i square equal to 2.25 now F i j equal to 0.5 into bracket open s minus s squared 4 into r j by r i whole squared this bracket there is a bracket between here and there okay sorry to the power of 1 by 2 again 1 power bracket is important okay that is equal to 0.117 that is for F 12 we got now F 12 as 0.117 why we have used i j notation because that is in line with the relations what are typed in our notes that is why I have gone back to i j notation okay now reciprocity rule a 1 F 12 equal to a 2 F F 21 a 1 F 12 equal to a 2 F 21 so I get F 21 equal to 0.469 now summation rule tells me F 11 plus F 12 plus F 13 equal to 1 F 11 is 0 F 12 little while ago I have got 0.117 and F 13 will become 0.883 now F 23 will become equal to 1 minus F 21 because F 22 is 0 okay that is equal to 0.531 now F 31 equal to a 1 F 13 upon a 3 okay and similarly F 32 equal to a 2 F 23 upon a 3 now if I substitute the areas I get F 31 as 0.5265 0.5265 and F 23 as 0.079 so F 33 will become equal to 1 minus of F 32 and F 31 that I get 0.3945 that is how note this F 33 is not equal to not equal to 0 okay so it is it is all by geometry see view factor is nothing but finding areas how much one surface is the other surface that is all it is that is all it is actually it is geometry okay so with this we will take maybe I think we will take next problem and then we will start taking questions so we will quickly solve one problem let us take the last problem problem number 53 so we have saturated steam yeah the data is correct so we will get started with the solution so that is what some confusion was going on okay q equal to m dot h f g that is the load m dot is given to be what is the problem given m dot is given to be 2.3 kg per second so 2.3 kg into h fg is 2257 kilo joule per kg I get 5.191 into 10 to the power of 6 watts okay so this heat should be equal to the heat gained by the water so m dot ccpc into delta tc that is tc0 minus tci tci is given to be 15 degree Celsius and m dot do I know m dot m dot is not known velocity yeah so we will we will come to this little later on because here number of tubes will come into n into m dot cp into n assuming that m is per tube okay now let us calculate Reynolds number Reynolds number equal to umd by nu um is 3.5 diameter is 14 mm 14 into 10 to the power of minus 3 we have goofed up okay upon 959 into 10 to the power of minus 6 that is right now it is right 959 into 10 to the power of minus 6 so equal to 50,993 almost same problem so this is comedy of errors okay Nusselt number I get 0.023 that is I get 285.8114 using Dittus-Bolter correlation so I get hd hd as 12371.55 yeah now it makes sense there is no thickness given so whatever we wrote earlier that is 1 upon hi plus 1 upon ho equal to 1 upon u so hi is 1237.55 and ho is 21,800 is that right 21,800 is ho and 12371.55 I get overall heat transfer coefficient as 7892.53 okay so both are of the same order so that is why my overall heat transfer but still my overall heat transfer coefficient is lower than the two terms because of averaging okay so now effectiveness we are trying what are we finding what is the question asked we need to find the number of the tubes okay we are trying to find the number of tubes so how will I get number of tubes I have to find NTU I have to find NTU for finding NTU what should I know I should be knowing the effectiveness and CR that is the ratio of the specific things now first let us find effectiveness equal to effectiveness equal to what is effectiveness equal to here q upon q upon q maximum that is equal to I get q is 5.191 into 10 to the power of 6 we just now found 5.191 into 10 to the power of 6 upon m dot c CPC into th i minus tci that is the maximum temperature difference this is by definition so that should be equal to yeah which is having all number of tubes in that it is embedded in that okay so now Cc equal to capital Cc m dot c CPC equal to m dot c is rho AC um into n rho AC um into n into CPC which is CPC so rho is we are just now found mass flow rate didn't we know rho is what is the rho 998 area is 5 by 4 into 14 into 10 to the power of minus 3 whole square into into 3.5 okay into 41 81 CPC so I get 2249 n okay now epsilon n equal to where is my epsilon okay so from this I get epsilon n equal to 27.154 that is epsilon I need NTU and then into n still we have two unknowns NTU equal to NTU equal to UA upon quite a lengthy problem heat exchanger problems are going to be like this UA upon C minimum UA is 7892.53 into PIDL PI into D is 0.014 into L is 0.5 so that is 222249 n so n n gets you are right n n gets cancelled out so this is we have to take the net surface area this is the overall heat transfer coefficient and in the denominator we have C minimum which is accounting for overall so that is the reason n n gets cancelled out and I get 0.1543 so effectiveness equal to 1 minus of e to the power of here in the numerator I have to multiply by 2 because it is 2 to pass so yeah okay so epsilon equal to 1 minus of e to the power of minus NTU I get 0.143 so I have already got epsilon n equal to 27.154 from that I will get n equal to 189.88 which has to be 190 tubes outer wall temperature one can calculate by simple energy balance let us do that outer water temperature TCO equal to 15 plus in the numerator 5.191 into 10 to the power of 6 upon 2249 into 190 what is this this is 9 to 190 tubes and 2249 was per tube equal to you get 27.15 it has not increased much okay and condensation maximum condensation rate is when q equal to q maximum what is q maximum 5.191 into 10 to the power of 6 that upon m dot sorry hfg 2257 upon is equal to 16 kg per second effectiveness is 0.143 so we get 16 kg per second so much about the tutorial for today we will take some 5 to 6 questions here and there before we sign off indoor SGSITS any questions there is a question related to NTU number of transfer unit yes I know that number of transfer unit is the measure of the size of heat exchanger but I want to know more about it so my first question is what is the physical significance of NTU and next question is what we can predict about heat exchanger if we have calculated the NTU over to you okay question is what is the physical significance of NTU over and above understanding that it is the size of the heat exchanger that is AS and apart from that what you can predict about the okay what one can predict what one can predict about the heat exchanger with NTU okay NTU only means we keep saying this area this also we have to keep in the back of our mind whenever we say NTU is a measure of the area for the same overall heat transfer coefficient and C minimum not always so if the overall heat transfer coefficient changes again area changes but when I change the design configuration many a times my overall heat transfer coefficient also changes then my NTU changes for a given you and for a given C minimum that means for a given mass flow rate and for given two fluids for a given heat transfer coefficients that means for a given mass flow rates heat transfer coefficients get fixed if my dimensions get fixed that is diameters what is not fixed is length so this AS only means for a given you and C minimum means actually my diameters are fixed what is not fixed this length it only gives us the idea of length when I say that I say that because I am keeping my you and C minimum constant or same but if I change that then area also increase or decrease has no meaning to say now professor wants to add something else he has written something else over to professor so what professor is saying is that it is a measure of NTU is NTU or delta T LMTD more or less is turning out to be the same if you see you AS I can write it as Q upon delta T LMTD and C minimum I can write it as Q upon TC not minus TCM what is NTU NTU is nothing but delta T LMTD so higher the NTU means higher delta T LMTD that is all lower the delta T LMTD so NTU is also giving me a measure of delta T LMTD if lower delta T LMTD is there means I will have higher NTU which is correct now if I have lower delta T LMTD I will definite that means temperature differences are less then my surface area required is or the size of the heat exchanger has to go up I think other than this I do not think I can give any other answer and the second part NTU does not tell anything about the heat exchanger Thai for anything so second part I do not think I can we can give any conclusive answer with just make with just NTU why C college not poor Hello sir my question is two fluids with different properties flow with equal stream velocities parallel to a flat plate sir what property of the fluid determines whether the velocity boundary layer of one is thicker than the other anyway the question is there is a flat plate and on a flat plate two fluids are flowing two fluids are flowing and nothing has been specified but we are assuming that these two fluids are having different viscosities who will decide the boundary layers and which one is thicker than the other it depends on the viscosity it depends on the viscosity definitely the denser one will flow below and the lighter one will flow above definitely but that does not decide the boundary layer thickness but the boundary layer thickness is decided by my viscosity if you see we derived by scale analysis delta is of the order of L REL to the power of half that is is that right L REL to the power of half correct so as Reynolds number is increasing that means as I am marching from the tip of the flat plate the boundary layer is increasing so this is going to decide which one is going to be thicker whether fluid one or fluid two because in Reynolds number both density and viscosity together are embedded and more over both the fluids need not have to move at the same velocity there will be relative velocity also ok. I shall my question is value of heat transfer coefficient does not depend upon temperature difference and rate of heat transfer so does this concept hold good for boiling and condensation also? This is a very good question the question is we were telling that in the forced convection heat transfer coefficient is not going to be dependent on temperature difference and heat flux is this true in case of boiling and condensation very good question we have missed at this point you have asked a very good question yes I cannot make this statement heat in fact if you take pool boiling curve no one plots heat transfer coefficient everyone plots q double dash versus T wall minus T sat y let us go to the definition of heat transfer coefficient how have we defined heat transfer coefficient in two phase flow h equal to q double dash upon T wall minus T sat is this T sat right reference temperature not y professor told us there is sub cooled boiling and also saturated boiling in case of sub cooled boiling my reference temperature is not going to be saturated temperature my temperature of the bulk fluid will be lower than the saturated temperature. So reference temperature in the definition of heat transfer coefficient itself is under problem that is h equal to q double dash upon T wall minus T sat T sat it is arbitrarily chosen because it is a fixed number for two phase flow. So we should not plot h as a function of I mean h we should not plot we should plot always q double dash and T wall minus T sat separately to give one line answer for your question h is not independent of q double dash and T wall minus T sat for the simple reason that T sat taken as a reference temperature is not right okay. Sir this h also independent of temperature in case of natural convection yeah so coming to natural convection whatever this concept is h is independent of temperature difference correct correct I had answered this question in the morning again yeah yeah the question is in natural convection is heat transfer coefficient dependent or independent of heat flux and temperature difference let us get to heat transfer coefficient in natural convection how is it what is it dependent on Nusselt number is dependent on Grashof and Prentel number what is Grashof number G beta delta T l cube divided by alpha nu so delta T is sitting that means heat transfer coefficient is a function of delta T if I take constant heat flux boundary condition I can define Grashof number on the basis of constant heat flux. So the crux of the matter is for free convection heat transfer coefficient is going to be a function of temperature difference and heat flux let me summarize because so many questions heat transfer coefficient is independent of heat flux and temperature difference only in forced conduction forced convection not in natural convection and two phase flow heat transfer no matter whether it is boiling or condensation in natural convection looking at the equation it is very clear but think of it physically only when you have a temperature difference you have a buoyancy effect if I have a larger temperature difference larger is the density difference larger is the buoyancy effect larger is the heat transfer rate. So there itself I am able to say that you know your coffee cooling that is just think of real life example larger the temperature difference the cooling is faster. That is the driving force that is the pump in a natural convection temperature difference is the pump there is no other pump that is the driving force if there is no temperature difference there is no driving force no natural convection that is what professor is trying to emphasize. My question is that if we consider heat exchangers then in heat exchangers is it really necessary to consider LMTD as considering AMTD also we are going to get the same value for approximately same value for AMTD and LMTD. Question is whether I should take arithmetic mean temperature difference or logarithmic mean temperature difference for the design of the heat exchanger you see professor Arun emphasized while he taught NTU sorry LMTD in convective heat transfer he said that there was one transparency where in which it was emphasized that if one takes average mean temperature difference one can make an error of as high as 15 to 20% only if the temperature differences are not very large we can get away with arithmetic mean temperature difference. So the safest route is to always take logarithmic mean temperature difference so that we do not commit mistake the summary is we cannot afford to take arithmetic mean temperature difference better we stick to logarithmic mean temperature difference that is the answer for your question. I think we will stop for the day.