 Good morning, any questions related to anything that has been covered so far, specifically again we are looking for fundamental questions of course, applied questions are also okay, but fundamental questions anything that you think we have, we should clarify or we want to answer a little bit better, please go ahead. I will go to Nirma Institute, Anagabad. Hello. Yeah. Good morning sir. Good morning. Hello sir, in case of derivation of today's equation that convective heat transfer coefficient for condensation, shall we consider mass transfer effect from vapor to liquid? No, that is we are see in the Nusselt condensation what are we saying we are not there is no mass transfer involved here. If there was the question asked is can we consider vapor getting mass transferred into water or water getting mass transferred into vapor, no that is not what we are talking about. Here we are talking about just film formed on the wall surface there is no mass transfer as such absolutely no mass transfer. So it is only purely heat transfer, okay professor. Okay, I will answer one more question. Yeah, please go ahead. As we neglect the inertia effect, so if we consider inertia effect what may be the change in the result? Yeah. Okay, the question asked is very good question I forgot to answer while presenting we neglected inertia as opposed to sinking effect and the what was the next term friction frictional force frictional effect sinking effect and the inertia. Professor Bejan in his book he again actually Chen professor Chen did those calculations taking the inertia effect also into account and he finds that there is a mild effect because of that and even if we have neglected inertia in fact we the results what we get with and without inertia they are not very different number one. Number two we have to relays another thing because here when Nusselt did he wanted to give the closed form solution you know because he wanted to give closed form solution we manage it just with integration and differentiation which was very simple which we could have done which we did only on piece of paper. So that is the greatest Chen used all similarity solutions approach all of that and then solve these differential equations because this differential equation becomes very involved. So Chen did using similarity solution approach the sum and substance of your question is that the effect of the inertia is not so great even if we take the inertia into account. So even if I neglect my Nusselt numbers are not way off. Sir what is the criteria the self side and the tube side of a heat exchange what is the criteria for selecting the number of pass? Side of pass okay what is the criteria for selecting number of process in a shell and tube heat exchanger there is absolutely there is nothing like criteria there are tens of hundreds of configurations which one can think of in shell and tube heat exchangers there is absolutely no criteria for multiple passes it is when do we go for multiple passes where I will go for multiple passes only when I do not have much space otherwise my heat exchanger length would have gone up instead of increasing the length I am increasing the diameter and increasing the number of the passes that is all what I am trying to do. So there is no unique answer for this question professor it is going by the economics and going by the size what I am imposed on and in fact if you see heat exchanger I forgot to tell heat exchanger length is decided by most of the tens is decided by the length of the pipe which a lorry can carry it may sound very phony but that is true okay how much a lorry can carry I think typically one pipe length that is 6 meter pipe length 6 meter more than 6 meter it is very difficult to carry by a pipe. So if I have to build a heat exchanger which is lengthier than 6 meter what should I do I should carry the 6 meter pipes wherever I am I am going to install my heat exchanger at the installation site itself weld my pipes such that my heat exchanger length is very high how many guys can afford this can be afforded only by power plants. So if I want to have the heat exchanger managed everything in my site where I am building it in my industry it cannot be more than 6 meters so the point is that there are plenty of constraints in a heat exchanger building so taking into account all those constraints these passes are decided there is no thermal answer to tell that okay you go for n equal to this many number you are going to get the best heat exchanger under the sun that is not one other thing is probably your question is if I have more passes then I will make the length less so is that acceptable probably that is not acceptable many times from other considerations like pressure drop and heat transfer characteristic see if I have same diameter pipe I have drawn here and the pitch is larger so I have a fewer number of passes in a given shell versus where the pitch is become very small then subsequent fluid the we have to remember fluid is flowing on the outside the shell side fluid is flowing on the outside of the tube. So there has to be some space for the fluid to flow some heat heat interaction has to happen if I if I keep if I stack it like you know stack of pencils if I hold it in my hand there is no space for any flow or heat transfer to occur so that also so every time you design you calculate the shell side heat transfer coefficient tube side heat transfer coefficient shell side pressure drop tube side pressure drop and then you decide okay any other questions from this centre quickly sir what is the difference between the term boiling vaporization and evaporation what is the difference between the question what is the difference between vaporization and boiling boiling see evaporation is a surface phenomenon okay meaning if I if I have a cup of tea which is left to cool in a room or a lake from which you know water is just converting itself to steam that is because of you studied in our high school also you know the molecules the inside molecules whatever is happening will get absorbed by the near my molecules whereas for the surface molecules it has it will have to because of surface tension it will have to escape off now evaporation is a surface phenomena there will be no bubble formation okay so there will be no bubble formation locally also whereas boiling involves nucleation that is birth or of small bubbles its growth to a larger bubble and then leaving the surface so even this nucleation we will call it as subcooled boiling and then when it flows and goes it is called as saturated boiling now I had told yesterday in case of vertical pipe flow or even horizontal pipe at when I go to the so-called annual flow regime nucleation gets suppressed and then evaporation takes over I have told that that is probably that the reason for that is a little bit advanced for this stage it is it has to do with the temperature distribution so on and so forth but what will happen is after a certain thermodynamic quality the concept of nucleation because let us understand it this way there is no there the fluid does not get enough time to form bubbles and cause them to grow and leave the surface because the bulk fluid has become has become hot okay so what is happening is this nucleation etc there has to be a certain time associated second thing the air which is the which is present in the cavities that would have gone by that time you know how much air can you have you are not pumping air from outside dissolved gases also would have escaped so there is no more entrapped or dissolved gases in the cavities or in the fluid itself so nucleation will get suppressed evaporation will take over okay the presence of non-condensable gases even with small volume that will that will affect greatly the condensation heat transfer coefficient so what are the is there any method by which one can tap the presence of non-condensable gases inside the condenser so that we can improve the performance of the condenser okay very involved question professor is asking a question what is the influence of the non-condensable gases or the condensable gases because of the because of whose presence how does the condensation take place see let me take up an example see to understand this I will quote this example see if we have a Fukushima nuclear reactor in the Fukushima nuclear reactor what is happening in the nuclear reactor we all saw that on that day people were trying to pump water in the Fukushima nuclear reactor so I have a reactor and I have a reactor and then what is happening it is full of steam okay now if I spray I spray water the water is the steam is going to condense on the on the water sheet now if let us say there is no steam and now I have only this spray nozzle in fact this spray nozzle characteristics itself is going to change spray nozzle characteristics itself is going to change okay so what is happening the condensation is the flow nozzle characteristics are changing means that those characteristics are going to decide my condensation so definitely presence or absence of non-condensable gases in steam are going to affect the are going to affect the condensation let me answer it little different way now let us say I have steam this much portion is steam and now I have air which is non-condensable gas which is air how much amount of air is there below the steam above the steam decides the condensation that means it will contain air and the surface tension between my sheet and this steam decides the heat transfer coefficient so the point is what you said is very right but people are still struggling to quantify it there are several papers on this if you are interested if you mail me in the moodle I can give those references but it is there is no answer for this saying that it is going to increase or decrease it is going to affect that's all one can say at this point of time okay one more question the condensation heat transfer coefficient depends on so many factors like the inside heat transfer coefficient and outside heat transfer coefficient and falling factor what how do you amend the condensation heat transfer coefficient in order to have better performance for condensation how does one increase the condensation heat transfer coefficient again again for augmenting the condensation there are microfin tubes are suggested why because you see how can one maintain augment the condensation number one we have to break the film condensation into drop wise condensation so what all things we can do by texturing I said Teflon plate I said they put Teflon coating that is one we can make the surface rough we can why because if I make the surf surface rough then I am having small small fillings that means it is tending towards drop wise condensation so whatever surface texturing we can do so that we can convert the film wise condensation to drop wise condensation that would increase the heat transfer coefficient so there are various heat transfer augmentation techniques with with which they are trying to convert film wise condensation to drop wise condensation nevertheless as I said in the beginning of the class whatever we do we are not sure about this drop wise condensation even if we think that our surface is textured we will continue to design our heat exchanger on the basis of film wise condensation only okay professor now we have to move on to next center sir in our lab we are doing that experiment but we are not able to get the newcomer the question is in the lab experiment we are doing but we are not getting new kiyama karo see how can I answer that why you are not getting I don't know the setup I don't know how under what conditions you are doing please put this question up in the mood with the details of your setup we will definitely come back y plus is what y utow by nu what is utow utow is tau wall by rho what is tau wall tau wall is shear stress if I am in the laminar sub layer I will get it based on the velocity gradient say if I measure the velocity profile using the hot wire animometer I will get the velocity gradient so that is that is laminar shear stress if I am in the buffer layer or in the turbulent layer then what is that I need minus rho u prime v prime bar so again I will use hot wire animometer measure the u prime and y prime I will get my minus rho u prime y prime that will give me tau wall that will give me square root of tau wall by rho that will give me u tau that will give me u u y u tau by nu that is y plus okay sir just give me one example to calculate y plus value give me an example to calculate y plus value okay roughly we can you are very insisting I like it okay so let me go ahead and calculate this let us open moody's chart let us do this calculation one of the participant's question is he is not happy with my answer about this y plus thing he wants to he wants me to calculate and show let me make let me just take what am I going to do I am going to take the shear stress from my moody's chart let me go to moody's chart that is so let us take this moody's chart in the moody's chart let me take a very rough surface that is epsilon by d equal to 0.06 that is epsilon by d equal to 0.06 now y plus what is what is the friction factor for epsilon by d 0.06 I am getting a friction factor of 0.07 okay that is friction factor equal to 0.07 okay now what is tau wall tau wall equal to f equal to 4 CF into tau wall upon or how can I get this yeah tau wall upon half row u average squared okay how I have to get to little more details that is diameter of the pipe let me take the diameter of the pipe as 25 mm and let me take a Reynolds number of because I have taken a very high Reynolds number that is let us say 10 to the power of 6 or 10 to the power of 5 it does not matter so 10 to the power of 5 if I take Reynolds number as 10 to the power of 5 what will be my velocity I am taking water 10 to the power of 5 equal to rho into velocity into 25 into 10 to the power of minus 3 divided by viscosity you take it as 0.001 I want velocity somewhere equal to this yes yeah 100 by 25 correct 4 meters per second now if I substitute this 4 here f is 0.07 by 4 equal to y by 4 because CF equal to tau wall by half row u infinity squared 0.07 4 into tau wall upon half into rho into u average is 4 square what do I get for shear stress 140 Pascal's now what is y plus y plus equal to y u tau by nu okay so now what what is u tau is square root of tau wall by rho so what is that I get for tau wall by rho so I get u tau equal to square root of tau wall by rho what is that 0.3742 okay let me take this as y plus equal to square root of y u tau by nu okay so now what is that I am getting y u tau by nu y is let me take this y as let me take no not I should calculate y sorry y plus let me take it as for breaking the laminar sub layer 5 okay 5 equal to y into u tau is 0.37 y into y into 0.5 equal to nu is 0.001 upon thousand 0.001 upon thousand equal to y into equal to square root of now what is the y I get I get a y of 6.68 into y equal to 6.68 into 10 to the power of minus 5 meters that means how much mm into 1000 multiply by 1000 0.00 0.0668 that means 0.07 mm should be the thickness so what was my epsilon by d in my case what I took epsilon by d is 0.06 epsilon equal to 0.06 into 25 mm what is that I get 25 into 0.6 1.5 mm this example states that this example tells that I have broken the laminar sub layer that is all I am telling is that okay through this problem you have but this is a rough estimate this is a rough estimate this gives you the feel that this is how one can break the laminar sub layer okay great why is he called as Nagpur good morning sir my question is what is the significance of boiling number okay boiling what is the significance of boiling number boiling number equal to q double dash upon ghsf let us open Roheson knows nodes we will open Roheson knows nodes just give us a minute yeah so what is boiling number you see here so boiling number directly has known relevance but it is q double dash upon hfg into rho actually boiling number is q double dash upon hfg into rho into what is that area into velocity will give you this is boiling number what am I telling yes this is boiling number q upon ghsf that is q double dash upon ghsf is the yeah q it is like this heat flux upon okay boiling number is let me define boiling number is q double dash upon mass flow rate per unit area into hsf what is mass flow rate into per unit area density into velocity so if you see here that is what we have done mass flow heat flux upon latent heat into rho f gives me velocity so the boiling number gives me a feel of the velocity which I have taken in my Reynolds number okay so this is the physical feel so in this tool boiling here heat flux is heat transfer rate upon plate area I hope I have answered your question okay relation related to the critical heat flux I think it is used for the modeling of critical heat flux this boiling number is used to okay your question is it is used for the micro channels and mini channels so sir please elaborate please elaborate the heat transfer mechanism in the when the flow is in micro channel and question is what will in this critical heat flux there are correlations for critical heat flux in terms of what is this boiling number so yes you are right and what is happening for the critical heat flux in case of micro channels and mini channels the answer is yes you are right critical heat flux is a function of boiling number correlations are there boiling number correlations for critical heat flux in fact is a function of I do not know how will this whiteboard work chf is a function of chf is a function of boiling number what else Reynolds number of course if it is a tube in this case it was pool boiling in micro channel it will be function of Reynolds number Prandtl number and Jacob number that is the amount of the sub cooling I have introduced to you cp delta t upon hfg that is Jacob number and then density ratio rho l by rho v and then viscosity ratio mu l by mu v and l by d that is the length to diameter ratio these are the various non dimensional parameters one can think of for chf ok chf is just not dependent on boiling number alone where in this case it is q double dash upon mass flux that is mass flow rate per unit area into hfg it is just not that in addition these are all the parameters coming to micro channels usually people claim that the boiling in micro channels is towards pool boiling why because usually this is a theory of course many people may agree and may not agree and there are journal papers which suggest that pool boiling and what is that the micro channel boiling more or less are closed why because the critical heat flux in the pool boiling and the critical heat flux in the micro channels they have found that they are more or less same that is the reason why they claim that the critical heat flux if you see here we could not teach this for the paucity of time you see here critical heat flux is a function of density variation rho f and rho g surface tension and hfg and g and 0.25 in micro channels what is happening flow velocity is very less why because it is laminar because it is micro channel because it is very less in addition to this what is the additional thing which has to come into picture Reynolds number because the velocity is very less people claim or people have found that in case of critical heat flux in case of pool boiling and critical heat flux in case of micro channels more or less they are same more or less more or less I am using in a big font in a bolder sense so but then if you still sit down and nitpick and go ahead and do experiments you may find 25, 30, 50 percent of variation ok. Looking at it from a correlation point of view and it said that it is pool boiling pool boiling like but what is happening in pool boiling is that there is no dirt of liquid that means liquid is still there in the pool available but it is just not able to reach the heated surface so what will happen is if I want to show it diagrammatically even in micro channels or in this one there will be vapor which is generated and which will just this is the liquid ok this is the liquid pool this is the vapor bubble which is formed ok. It is going to be so rapid the vapor generation is going to be so rapid that this going out and this coming in is not possible any longer and therefore you will have CHF which we are calling as in pool boiling in micro channels also typically of low qualities that means X thermodynamic quality is also low ok. It is not that liquid is gone when you have a pipe and pipe is very long and you are continuously heating there is no more liquid that situation does not happen here ok. Next question from VI-19 Akpur. Hello sir. The value of H in condensation and boiling is very high near about 5000 to 50,000 as compared to force and acetylmescin sir why. Ok because there is two phase flow is occurring the question asked is in case of generally in case of two phase flow the heat transfer coefficients are very high because of two phase flow what else to say let us see the point is the heat transfer is overwhelmed by the latent heat rather than the sensible heat. So, that is the reason the heat transfer coefficient for two phase flow is significantly higher compared to that of the single phase ok. So, my next question is the value of H equal to in a condensation 0.943 the relation is 0.943 plus equation in most of the book it is given 20 percent extra that is 0.943 into 20 percent that is the value comes H equal to 1.13 into this relation and HFG is liquid sub-pulling. So, that is effect. See I think you are right 0.943 is there and few textbooks takes 1.13 what perhaps they are doing is they are not taking HFG prime the way we have taken HFG prime HFG equal to HFG plus 0.68 into Cp into delta T we have taken instead of taking that there they have embedded that there outside and put that as 1.11 that is all the difference otherwise both are same ok professor. Any other questions from your center. Ok. Yeah we have to go to VIT Pune and I have to tell something about VIT Pune. Yeah please go ahead and explain please go ahead and explain how did you do this experiment chapati experiment. I have performed the experiment there before yesterday. Ok. The same volume of VIT dough was taken the dough was uniform both chapatis were rolled both chapatis were rolled to the same diameter both were roasted in the microwave oven to the same temperature that is around 78 degree centigrade. One chapati was wrapped in aluminum foil and one was wrapped in brown paper. Average temperature of chapati was measured every 5 minutes up to 20 minutes temperature of chapati in aluminum foil was higher and that in brown paper was lower. After 20 minutes both temperatures were around of course this was the preliminary experiment and no repetition was done. One first thing is. It can be said that for around 20 minutes. Yes ma'am I will take. Around 20 minutes hotness will be maintained and after it was same. Ok ma'am great I first want to congratulate you my salutations to you that you have done this experiment in such a quick manner taking the suggestion so seriously first my salutations to you for that. Now coming to the technical component it is a really good experiment I have one question to you did you do this experiment chapati open to atmosphere or you had put this in a tiffin box. Sir I kept it over and after roasting I have kept it on the table under natural convection. Yes ok see correct so usually when this is first of the ok ok see very good the point is why I ask that question is because see the major conclusion ma'am is drawing from this is very nice you see aluminum foil temperature chapati temperature is quite high this is a very simple experiment if you put data logger we will be getting still higher temperatures but still the point here is the aluminum foil temperature is higher but the brown paper temperatures are lower that means it has got cooled number one. Number two why it has sustained its temperature only at 20 after that it has become both same because the natural convection has taken over actually we keep this foil with the chapati complete thing in a tiffin box. So there radiative losses will be overtaking over the natural convection losses when it is open to atmosphere natural convection losses may be equal or I have not done that calculations but may be equal to radiation losses but when we put it in a tiffin box natural convection is also impeded so that is why maybe it would extend to an hour my chapati would be little warmer for one more hour but still whatever it is it is really a good experiment this suggests that we can do simple experiments and demonstrate for reaching conclusions very good we should all clap we should all clap for her for her effort and such a quickness and earnestness she has come back so I am at least clapping I want all of you to clap along with me. Sir I want to say few lines this workshop has definitely given us the motivation to perform small experiments in whatever resources we have and to learn and make our students learn heat transfer by simple experimentation and it is all because of both of you you have motivated us and our remote center. Okay ma'am because of you sir okay ma'am okay okay we will thank you sir okay ma'am thank you we will move on will you any other technical question from VIT Pune okay we are moving on sir in the ice block experiment natural convection as we have calculated H okay by definition H is equal to minus kf dt by dy divided by delta t so we know denominator we know left hand side H so is it possible to calculate kf back there no no so is it possible to calculate H equal to minus k del t by del y at y equal to 0 in case of ice melting problem what we have done it is not possible simply because I have not measured the temperature gradient I have not measured the temperature gradient within the boundary layer so how can I calculate minus del t by del y what del t by del y I have is t in 25 minus 0 but what am I looking for I am looking for the slope at y equal to 0 that slope at y equal to 0 I will be getting only when I get the temperature measured within the boundary layer that is what professor Arun was plotting all the time that he was plotting that I need to get the profile and in the profile I need to get the temperature gradient so I have to until I have this profile until I have this profile I cannot get the temperature gradient it is not this straight line correct correct correct what you are perhaps thinking is I will assume linear profile and do it is that profile will not be the profile what I get when I measure within the thermal boundary layer okay professor next question please sir what are metals to find then servo collective air or such gases air or gases in fact okay you please put up this in the moodle I will answer this question because there are various methods it becomes very difficult to explain all of them please put this up in the moodle I will get back next question please in this course you are not introduced the concept of conduction shape factor is it really important to two dimensional conduction shape factor question is conduction shape factor has not been introduced why conduction shape factor will become important only in case of two dimensional heat transfer just because that we are not covered two dimensional heat transfer that is why we are not covered that if we teach conduct two dimensional heat transfer shape factors come into picture next question please one and out sir okay thank you great next Amal Jyothi college Kerala sir this is the concept of evaporation okay how evaporation takes place at the room temperature and can we consider evaporation as a mass transfer phenomena or a phase change phenomena or both okay evaporation how does it happen evaporation you are right it is a mass transfer process it is the concentration which is varying okay if you put a chocolate we are not chocolate chalk piece we all are very comfortable with chalk piece because we use blackboard if you put the chalk piece in water what is happening chalk will slowly get diffused into water what is happening there that is mass transfer here also in evaporation also what is happening the water from the lake is getting into the air that is all the concentration of the water in the air is increasing so it is essentially mass transfer so that is it that is why clothes do not dry well when the air is humid that is it yeah next question please okay please put this in the moodle it is actually psychrometry mass transfer we need to get into fix law of diffusion and things like that it is little involved I cannot give a one line answer or one minute answer please put this up in the moodle I will definitely try to answer this question over to the next part participant for other question thank you sir okay over and out thank you I am Ritha Koyambathur okay the first question is from special effects so in special effects we found the temperature distribution we got the solution A1 plus something in that if you say the Fourier number is greater than 0.2 we are taking only the first part A1 itself if it is less than 0.2 we have to take A1 plus A2 and A3 up to how much we have to take and we have to calculate okay accuracy the question is for all tau greater than 0.2 only of first series solution is first term of the series solution is valid for lower tau how many how many terms in the series are to be taken there is a difficult question to answer tip I would think by gut feeling 4 to 5 terms should be sufficient but do not take my words you can do this calculation in the excel sheet put them in the excel sheet and calculate and see yourself how many terms are required next question you had second question the it is a simple question it is related to heat transfer or related to chemical species transport what is chemical species transport chemical species transport is mass transfer it is analogically we can study this but it is not within the scope of this course fundamentally so but still if you want to know more about mass transfer usually this is studied by chemical engineers and they study heat transfer analogically I think I will stop here please you want to know more about mass transfer please put up this in moodle we will answer this over and out MES fun well quickly any question what is the stability of heat exchanger tubes when it is subjected to cross flow please explain this okay. So what you are asking is if because of flow instability there will be vibrations and what are its influences on the heat exchanger okay that is an important aspect we do ever study that because actually vibrations are caused because of flow instabilities or heat exchangers are there in a power plant where motors and pumps are all running and vibration is definitely induced not only by fluid but also by machines but generally it is felt that the vibrations it is like pulsating flow it is going to create pulsating flow pulsations generally result in higher heat transfer coefficients than otherwise so incidentally it is going to be helpful for us if there is vibration next question please for and also great Anna University okay okay both the vapor and liquid even though they travel at the same velocity yeah consider to be two different liquids or two different in tau cannot be zero no in this case that is why now the question is why is that interfacial shear stress between the liquid and the vapor is being taken as zero if I in condensation at otherwise if I do not make this assumption I need to have the idea about the interfacial shear stress that is precisely then it becomes very difficult to solve my problem so that is the reason why we have assumed that the interface it is the experiment it is the expediency to solve this problem which has goaded us to make that assumption that interfacial shear stress is zero so but nevertheless now let us come back to the answer and see the heat transfer coefficient whatever we have computed is matching with the experimental results never the less in spite of making that assumption that interfacial shear stress is zero which is justified now okay next question please go to you okay NIT Tricci any questions so how can we select the orientation of the condenser this is a good question the question as this how can we select the orientation of condensation this is a good question yes condensation heat transfer coefficient is dependent on the orientation definitely that is what professor was yesterday emphasizing so much when I take a horizontal pipe when I take a vertical pipe what all things will happen definitely the heat transfer coefficient is dependent on orientation if you see here today also in our derivation we have said that H in case of orientation what will happen I said H into cos theta cos theta between 0 to 90 what will happen it is going to decrease in case of oriented plate my heat transfer coefficient is going to decrease that means orientation is going to be detrimental for my heat transfer or for my condensation so that is all the answer for this question so how can we select the boiling regime in boiler how can you select the boiling regime how can one select the boiling regime in boiler depending on the heat that you are going to supply how if you are if you if you have a little bit of water in a vessel and you keep the gas on very high it is going to boil very rapidly okay you can go to film boiling very quickly or CHF very quickly so same thing you have to control the heat accordingly to have nuclear boiling that's it and ensure that there is mass flow of liquid all the time okay over and out from this institute VJTM Mumbai yes sir this question is regarding compress incompressible flow at the time of teaching about incompressible flow you said that shear deformations cannot be included in incompressible flow I have a feeling that when incompressible flow takes place shear deformation may not lead to change in volume therefore please clarify on this point see it is just that what is that is that in energy equation you are telling or in momentum equation you are telling sir it is about the momentum equation okay see I didn't say that shear deformations will not take place in incompressible flow shear deforming angular deformation rotation linear translation linear deformation translation all will occur in case of both compressible and incompressible flows I did not say so I don't know in what context you are quoting me but the point is whether it is incompressible or compressible the fluid particle is going to undergo all of this for linear transfer what is that linear linear deformation angular deformation rotation and translation there is nothing specific with compressible and incompressible flow okay sir I have another question in Bajans experiment we have always shown the boundary layer from bottom to top is it going to be from top to bottom no boundary layer in case of Bajans experiment has to be from bottom to top because it is natural convection boundary layer there is nothing like from top to bottom condensation is for top to bottom no confusion here we are not confusing between condensation and natural convection this is natural convection so boundary layer has to be from bottom to top next. Please let me know the logic behind the assumption that while solving a problem we treated a furnace at 2000 degree centigrade as a black body no we didn't take it as a black body we take the emissive power see the point here is if I put a small body in a huge furnace all that we are saying is that the emissivity which decides the interaction is the emissivity of the small body that is all we are not saying that the furnace is a black body we didn't take the furnace as a black body. In fact in that example tutorial problem emissivity of the furnace was 0.5 we calculated emissivity of the object as some 0.57 so furnace is not a black body but for the interaction purposes we can treat it as a black body ok over and out. Sir in case of condensation whether this flow mass flow rate of condensate in vertical plate has to be considered or to be neglected no we cannot neglect it mass flow rate is there no in Reynolds number it is coming so we are indeed considering over and out. Good morning sir my question is the design of cell and multi two pass high multi two pass heat exchanger is carried by TEMA standards is the both the TEMA standards and bell delivery method is same bell. I don't know what is demo standard TEMA standard oh yeah ok ok see question is is TEMA standard same as bell deliver method in the design of the cell and tubular exchanger yes they are same TEMA has adopted bell deliver as the as it is standard it is right next question please. Next question is in condensation is always is always analyzed with respect to vertical plate if we if we consider horizontal plate how it will be laminar condensation over a horizontal plate no problem see question is we always take laminar condensation only for vertical plate why do not we take horizontal plate why why do you say that I have put a horizontal cylinder in my figure whatever analysis we have done today is valid even for the horizontal so it is is it not natural convection not taking place in horizontal plate it is taking place so it is ok even in horizontal plate it will condensation will be taking place but it will not be as effective as it is going to be in the yeah pressure is drawing for us so for horizontal cylinder also condensation is going to take place it is going to be something like this ok Next question please if we consider the horizontal plate condensation the condensation effectiveness of the condensation is less or more sir it will be less compared to that is of the vertical plate ok over and out MK triple S Pune any questions MK triple S Pune any questions please yes yes we talk about two bodies body A and B both at the same temperature and consider radiation we talking about electromagnetic waves between them would there be considerations of interference between these waves any two waves can interact right yes yes question is if there are two hot bodies which are sitting at two different temperatures which is going to have electromagnetic waves yes electromagnetic waves are going to interact that electromagnetic wave interference only we are on a macroscopic scale we are calling it as absorb tens absorptivity emissivity scattering and all of that that is all happening essentially because of the interaction or the interference between the two waves two multiple whatever number of waves it is handling whatever I think this this would be a question which we would have studied in light also right yeah similar phenomena only we can apply for this