Yeah this is the fluke make, this is the fluke make I am just keeping it for the camera because when you get the video you should be able to rerun this and get this model number. If you can focus on the model number that would be great, no? If you just Google and ask for any Chinese supplier you will get n number of suppliers of thermal gun, they call this as thermal gun, so if you just put this is the temperature you see here I get the temperature directly here, 24.9 or something if I put to myself so I am getting a temperature of it is increasing, 29, 29.3 it is increasing. So because my emissivity as I said is 1 and I can change this emissivity also, I can change this emissivity okay, so how do I measure the emissivity of a body, I will take a plate, I will put a thermocouple again my mother is thermocouple I cannot forget thermocouple okay and in velocity pitot tube we cannot forget basics, so I take a plate put a thermocouple onto the plate I get independent temperature of my plate is that okay, now I put this gun onto the plate I adjust my emissivity here such that temperature shown by this gun and the temperature indicated by my thermometer or thermocouple not thermometer thermocouple matches, so that is how I measure the that is what we do in the lab we measure the emissivity not with the thermal gun we do with the camera okay, so but then this is quite this was this I guess costed us around 25 to 30,000 rupees but you will get now within 3 to 4000 rupees also okay, so you can stop using thermometer for measuring water temperature you can start using thermal gun okay, so my point was for your emissivity experiment you independently measure the emissivity by anyway you are getting the temperature there isn't it, so please add this tool to your lab at least that much care of this thermal gun business will go away, in fact Mumbai police is seriously thinking for surveillance purposes for surveillance purposes using thermal camera okay, now coming to thermal camera again thanks to professor vedula only he only gave me this tool in 2004 and asked me to play with it he just told me this is the camera which I have purchased from departmental money I have invested 30 lakh make sure that you use it left and right but that has been the eye for us since 10 years okay, so thanks to him otherwise I would have never learnt thermal camera what it was okay, so this is so this is the thermal camera okay whatever I want to do the beauty of thermal camera is it is not like photographic I have I need not have to be as smart as he is for normal photographic I can just zoom it any dumb guy like me also can measure okay, so that is the advantage of thermal camera okay point is missing luckily it is showing but I cannot get the colors so what is the what is that I said for thermal camera the advantage of this thermal camera is that I will get no I do not have I do not want to meddle with this, so yeah now you can see okay, so this is how pixel by pixel see I can get throughout in fact this thermal camera is being used for various applications various applications even in medical applications okay, so wherever if there is any tissue cancerous tissue apparently the temperatures there are going to be slightly higher, so this is a very non-intrusive tool to diagnose that okay people are working at it I came to know that one of the doctors in amdabad is working on that and so my request is if you can purchase not this expensive but nowadays cameras have come very cheap as cheap as around 1 to 1.5 lakhs if you can if you can afford that I would strongly suggest that you please purchase a camera any help regarding supplier or getting an appropriate quotation I will be of help no no no not that not that way because we have been working with these guys quite often so we can use this so that is the advantage of thermal camera imagine same thing measuring with so many thermocouple out if I have to measure my body temperature I have to put so many thermocouples all over the body instead of that I am getting pixel by pixel all over the place in one shot it is a magic only it is next to magic nothing else it is next to magic okay provided I know my it will give my temperatures provided I know my emissivity I am bugged throughout the institute oh you have a thermal camera I will measure temperature that is not true if a tool is there and if it is running tool and I want to measure the tool temperature how will I measure the emissivity is going to continuously vary with temperature and it is moving so answer is thermal camera is useful only if emissivity is known I make I make that unknown known by appropriately putting a paint but if you are handling some other surface on which whose temperature you are on whose surface coating you are no control you are gone to furnace and you just want to measure the temperature you cannot just like that measure because you do not know the emissivity okay so we will move on to lab I will be soft on you and I will only present the data you will not present the data so I am taking the responsibility myself but you can question me with reference to your results so first let me let me take the forced convection now in the forced convection which what all measurements we have taken Baba sorry yeah you are all teachers I should not be using the word Baba but I am I am used to use this word Baba from no from my teacher again back in convent so that is I cannot help it okay so what is that we are measuring what is that we are measuring here temperatures T1 T2 T3 T4 T5 T6 T7 voltage and current so what is the did we first to do energy balance e.in – e.out plus e.g equal to e.st did I do that I have a if I am not done that then I have not done what I am supposed to do what is the time supposed to do what is the time giving voltage into that is what I am dumping into my system so that is here in this case 100 into 1.05 in this case also perhaps he has not done in your cases let us take any of these cases correct let us read but first inputting I am inputting that much into my control value how much is your typical voltage into current let me ask anyone Mangesh what is your voltage into current 118 into now you calculate and tell me what is mcp delta T what was your m dot cp delta T 50 yes 42.9 so there is loss of 50% or even more than that so which of this heat I should take for calculating heat flux heat transfer coefficient or heat flux mcp delta T so if I take this mcp delta T now that delta T is of what order in your experiment 4 degrees is that a good number is that a good number why it is not a good number why delta T Professor Parshuram Chitrager says that delta T 4 is not good why 4 is not good 10 is good you are saying why 10 is good so actually we should not be using the word error actually there is nothing like error in this world why because error by definition is measured minus actual or actual minus measured I keep telling this all the time measured minus actual or actual actually you will never get in this world true value no one tells the truth in this world truth is always truth is always elusive so true value you are never going to get so error you can never estimate all that you can estimate is only uncertainty about this deviation okay so this is uncertainty in my temperature measurement is typically for a thermocouple whatever circus I do it will be of the order of 1 to 2 degrees everyone in papers write 0.5 it will be 1 degree Celsius at least so if it is 1 my deviation or uncertainty in my temperature measurement is 1 upon 4 25% is my uncertainty in my H directly so that is why my delta T's have to be okay so that is one of the points which we need to emphasis uncertainty in H can be as high as 100% when my delta T is 1 degree Celsius isn't it so people do quote uncertainties of H as high as 100% also because sometimes you cannot plan your experiment and try to make it delta T larger so then in that case you have to be you have to accept okay but here in this case at least we have a control why I am telling all these subtle details is that when you are doing I am sure all of you have this experiment in your you need to emphasize on these points to the students tell them up reary you maintain a delta T at least 10 degree Celsius precaution is wall temperature should not exceed certain number so that your heater does not burn out is that okay so now what was I telling yeah q double dash you will get q double dash now you get H now which temperature you have taken for calculating the heat transfer coefficient you have measured 5 temperatures I guess you have taken average of them and taken why did you average does that make sense or does that not make sense because we have studied yesterday a concept called developing flow and fully developed which of that thermocouple you imagine that it is going to be fully developed which of that right from the first thermocouple will it be fully developed no last one someone said who said that yes last one is what you can expect take only last one do not go by your manual manual is only for those guys who cannot think not for us okay it has no meaning because developing flow heat transfer coefficient is decreasing and then it is becoming constant taking something in the decreasing portion has no meaning is that right so I should be taking only the last value and checking if you check that if you have not done that you go back and check it will match with your detest voltage correlation within 20-25% not with standing so many errors in our so many uncertainties in our measurements. Water graph wall temperatures we know correct correct and where ever it is yes that is beauty what professor is saying is that you know the wall temperatures you plot the wall temperatures as a function of x now bulk fluid temperature you know at the inlet and exit you that we know for sure it is linear bulk fluid temperature is linear for what case constant this we can afford to do only for constant heat flux boundary condition check whether these two lines are parallel or not if they are parallel right from word go from the first thermocouple then it is otherwise I will get it deep to see you remember like this you saw and later on they became parallel so that is a that is a right way that is how the point is I keep telling always whenever you do a lab you should emphasize as professor keeps using the word marry we have to marry theory with practice practice within practice we have to keep coming back to our theory what we taught incidentally lab comes always after theory that is why it is that is how it is planned that is why we are planning so you go back and tell that this is what we learned that is why it is behaving for that your setup has to be strong that means you would have done you should have done your experiment reasonably well several times okay last thermocouple we have checked myself and we know last thermocouple matches with the voltage within 20-25% okay this is the this is the one of the experiments any other checks I can do ha in this another thing I can do I can estimate losses you got 50 watts how can I estimate losses can I estimate losses or maybe it is not possible yes Sandeep no that is I will get but can I estimate that loss can I estimate that loss because that has been encased in a container and that container of course I do not have a temperature now if I can measure on that container put a thermocouple now I can assume what are the modes of the heat transfer through which that container is going to lose heat to atmosphere convection and radiation assume emissivity of that surface whatever it is or if you have gun you can measure it so H you will have to assume for horizontal plate for horizontal cylinder or plate using natural convection calculate pressure of number and calculate the heat loss because of similarly sigma epsilon t wall to the power of 4 minus t surrounding to the power of 4 gives me the radiation if you compute within again 10% that computed value will come with what I should check now after calculating that vi minus mcp delta t should match with this all these checks see essentially what am I trying to do I am applying e.in minus e.out equal to 0 because there is no e.g yeah e.g is there because I am putting e.steady state e.st is not there make sure that they do the energy balance make sure that they do the energy balance that is what I want to emphasize in the lab and make sure that they do the uncertainty analysis how do I compute now the uncertainty in my H because I would take the experiment one experiment in detail other ones I will gloss over but for one whatever I am telling is applicable for everyone so H equal to q double dash upon t wall minus t bulk right what is the uncertainty of q double dash vi by area is vi by pi dl for a minute because reasonably we compute we measure voltage with right one but ours was very bad because our current meter has a least count because another thing we always have the habit that we take least count as uncertainty in the absence of anything else it is okay but always least count is equal to answer what is the least count of my watch point 5 seconds so any of our watches but need not be the uncertainty of my watch it might be running 5 seconds faster that is erroneous or it is having an uncertainty of 5 seconds when will I be knowing that it is deviating whenever I compare with my standard I compare with NDTV time or CNN IB and time because I am considering that okay so I should not take the least counts always but in the absence that can be one of the options okay so I should take the least counts of each of them diameter and length I am perhaps measuring with the vernier caliper it is going to have 0.01 mm calculate the uncertainties of each one of them and get del q by q double dash delta q by delta double dash is delta V by V whole squared plus delta I by I whole squared plus delta D by D whole squared plus delta L by L whole squared square root of this is RSS root sum squared root sum squared okay so they should calculate now what is the uncertainty in T wall I know 0.5 or 1 degree Celsius and I know the uncertainty in bulk so T wall minus T bulk I will get that much uncertainty so add up all that that much should be the uncertainty of my H typically my uncertainty whatever I calculate should be the deviation between the deters bolter and my calculated my measured value should be around that around the uncertainties if my uncertainty is coming around 30% the deviation between deters bolter and mine should be around 30% if it is not so if there is lot of deviation between the two there is some serious problem let it be problematic but we should make the students understand what is let answers not come the right way but we should offer explanations why they are not coming right way what might have probably gone wrong that is what we should be able to make them another thing I have I will and one of the one of the faculty members asked me do you have any book for experiments we have a book actually it is there in the library it is not coming out from my mind there is a book but in that book so nicely the experiments have been designed for example Bejan gives there design of experiments it starts with G that much I remember design of he transfer experiment he just gives very simple experiments you take ice plate you take ice plate put it in a container this is a flat plate put it in a container how much it melts it is a mass flow rate so that will from that we can compute the heat transfer coefficient of a flat plate you need anything more can I get a thermometer thermocouple or a thermometer and can I connect it in a container in a 100 ml flask what more I need for specs container I can make to me keep it insulated only thing I should have a fridge I should put a thread on that day of experiment and put that ice block and put it inside you do not need very high-five experimental setups okay so I will scan that book at the cost of being or maybe I will suggest that book to all of you you purchase that book through your institute okay so you can design experiments we do not have to take those experiments you can give them as projects I know very well in university one good thing is students if you tell to make they will happily make and they are very good at it and your workshops are quite strong okay you do not need workshops huge workshops you can simply cut with perspective you can give with I can assure you in that book you do not need for each setup more than 5000 rupees this is also on the highest side which you can draw as advances which you can manage very easy is that okay so before main workshop maybe I will come up with two three experiments at least so that we can demonstrate that okay I had given I had done this exercise in IIT Guwahati unfortunately our numbers are so high here we don't manage but we should be doing we should also do when I am preaching I should also practice so we should also attempt that giving that projects that way each year you give one project no one setup will get will get generated yes okay that's about that is about this next experiment was yeah we will see yeah any questions you have you please ask no you converted I have not taken mcp dt minus q radiation where did I take I don't I am not considering anything from the manual I am only thinking out of my out of the box so what is you ask me the doubt now you ask me the doubt okay okay okay okay actually people what they do is okay you have asked a good question people what they do is always people don't take mcp delta t how do you know that mcp delta t is right you are assuming that mcp delta t is correct but what people do is what people do is they take the container that is they take the pipe we have the heater anyway no flow there is no flow closed both sides close both sides now take give that much heat that is voltage and current such that the wall temperature would be almost same as what it would have been with the with the flow with the flow then if you do that experiment it will take enormous time to reach steady state because it is and you should stuff inside some insulating material so that you are avoiding natural convection if you do that what is that you get you get q double dash voltage into current upon area versus t wall minus t infinity I am taking what do I get typically you if you do this experiment if you do for four five values you get a linear line what is that what is the interpretation of this figure I can tell you this this is q loss there is no flow but this q loss embeds both convection and that is this q loss has to be deducted from my vi vi minus this q loss should be of the same value roughly within 5% with my mcp delta t these checks and balances I have to do every time I am doing energy balance only I am not doing anything else so that was a good question I had forgotten about this so this is this check we have to do before I lap up otherwise I do not know whether I might be making a mistake in measuring my bulk mean temperature okay so that is the another check I take care of yeah next what was the next next experiment natural convection have we done no no so what about heat exchangers what is the LMTD you are getting for parallel flow and for counterflow and for counterflow so you expect LMTD of counterflow to be larger than so what might be the problem now first question I would ask what are the delta t's on cold side and hot side all the time what is the delta t on the cold side and hot side 4 or 5 degrees that is the concept when we do our experiment we have to keep because you see there is not even measuring bulk fluid temperature what is the temperature I am measuring there no central line temperature I am measuring so we study after studying so much about bulk fluid temperature do you think that central line temperature is a representative of bulk fluid temperature okay it is a measure of bulk fluid temperature it is quite difficult to tell like u by u maximum equal to 0.8 you take what was the equation you take for velocity u by u maximum equal to we wrote power law or if it is laminar you take u by u maximum or u average by u maximum equal to 0.5 per laminar for turbulent you take u average by u maximum equal to 2n square divided by n plus 1 into 2n plus 1 where n is the power u by u average or u maximum upon y by r that is 1 minus r by r to the power of 1 by n that n it is so roughly if it is Reynolds number is of the order of 1 lakh I get this ratio as pointed but this type of simplification I cannot do for temperature can I do or can I not do if I know the velocity profile if I know the constant heat flux boundary condition can I get what is the question I am asking if I know the central line temperature can I get the bulk fluid temperature if I know the velocity profile yes what is the bulk fluid temperature definition let us get back TB equal to 1 upon something something within the integral what is there you should be you should be knowing velocity profile and you should also be knowing temperature have I measured temperature profile no how will I get my bulk fluid temperature I cannot get so central line temperature is a measure of something but in all heat exchangers that is what it is done you just insert central insert a thermocouple and believe that that is the temperature okay so delta T is my point is that whenever we do experiment for any heat transfer experiment we should take care that my potential differences that is the temperature differences are substantially large at least 15 at least 15 to 20 larger the better but in that pursuit if you go on doing my wall temperatures will go up I should be careful but in heat exchangers the tension is not there because you are heating and supplying so in heat exchangers you can easily afford to do you can easily afford to do so that is why we are getting the delta T LMTD now student will be puzzled or you are theory make with the product either to kuchavra so that theory never works with practical that is what he will he should not be under that impression we should show him that what we teach is what is happening in the real life or otherwise he will lose confidence on our theory that should not happen that should never happen okay we should take care of that if if at all it has happened the way it has happened we should be offering convincing explanations in this case at least I can get away saying that my delta T's are less and uncertainties are anyway going to be of the order of 2 degree Celsius 4 in 2 in 4 I am very much out that is what has happened that is what has happened in our heat exchanger problem okay second what is the third experiment yeah you have a question yeah only bend alone may not give see I have to take it to that is what Sandeep was saying we have to take it to a container I have to take it to a container if this is my flow direction I have to take it to a container and ask it to go through several serpentine passages and now put thermocouples as many as possible in this and then take the average of those thermocouples that is all right but water it is little easier if you take it to your flask and stir it or put some fan inside or even if you don't stir water more or less you are going to get because in the flask when it is falling anyway it is going out so anyway it is getting mixed so water is in fact that is how it is called as mixing cup it is it is getting mixed there okay that is that is that doubt always I used to have overall heat transfer coefficient of parallel flow heat exchanger how will it compare with the counter flow heat exchanger can I answer this question Sandeep you are not answering I am going to ask you at the end what is overall heat transfer coefficient because we know heat exchangers so I do not think I need to wait till tomorrow yeah you are welcome to give 1 by UA equal to 1 by HI AI plus 1 upon H naught A naught let me neglect conductive resistance now now tell me yeah theta m is not fiction no no it is not fiction no now I cannot argue like that because delta T LMTD of counter flow heat exchanger is is higher than delta T LMTD of HI and HO but HI and HO or am I keeping it same will what will HI and HO depend on mass flow rate mass flow rate am I keeping in both the cases same mass flow rate yes yes I should do the experiment such that my mass flow rates are same if the mass flow rates are same my HI and HO are going to be same your UA has to be same UA does not know whether it is operating in counter flow or parallel flow UA only knows the convective resistances and the conductive resistance so when you plan when you do the experiment make sure that the mass flow rate what you do for parallel are same for counter flow this is one another otherwise you cannot interpret the results it becomes difficult please keep the mass flow rates same for cold side and hot side both for parallel and always I used to wonder what is this earlier but later on with time I understood this is that okay any questions on heat exchangers before we move on to the next one delta T LMTD has gone up my heat load now you see q equal to what UA delta T LMTD if I keep the mass flow rate same what happened to my UA what has happened to my delta TLMTD because my length is fixed what has happened to my delta TLMTD counter flow is supposed to come more than the parallel flow so what has happened to the heat load q q dot larger in case of counter compared to larger load is coming out but what we tell in the class is that for a given heat load the length required for counter flow heat exchanger is less why because delta TLMTD is higher but here I have kept the I cannot overnight change the length so length has to be kept constant so that is why my all that I can show here is that load as my heat taking load of the heat exchanger of a counter flow heat exchanger is greater than that of parallel flow heat exchange that is a good question because we need to make them understand yes okay no problem so did you compute effectiveness okay no problem you calculate effectiveness because I do not recollect the relations myself but I remember that maximum temperature minus minimum temperature THI minus TCI is sitting in the denominator and numerator I remember whatever is the temperature difference so correct correct so we will take effectiveness tomorrow while we are teaching another question whether qh is equal to qc or they should be different theoretically same but are they going to be different yes they are going to be different why it is lost so then which one I should take average average that is what manual say why qh no that is the amount which I am dumping into my system but some amount is going to my cold fluid some amount is going to the surrounding so but outer wall can the cold side also can lose no heat whether whatever is going in or out there is going to be interaction with the atmosphere there is bound to be losses to the atmosphere let us say hot water is flowing inert in the annulus cold water cold water also eventually has to get heated up no from my hot water so it has to interact with atmosphere the hot or hot cold fluid if I have to put it that fluid if I have to put it that yes there is a confusion that is ok that is ok but all that you are giving is not going to cold fluid no what about the after I understood after going through this only it is going all of this 100 is not going to my cold fluid but because it is losing to the atmosphere as it is getting heated up it is getting my cold fluid as it is getting heated up it is losing its heat to the atmosphere so there is loss so you cannot argue once again you cannot argue that all that hot thing is going to the cold whatever circus we do there is going to be but my point is what are we trying to do in the heat exchanger one second one second what are we trying to do in the heat exchanger we want to transfer the heat from one fluid to the another but all the hot fluid thing whatever I am giving is not going to the cold fluid that is what I am saying and then you are talking like an engineer I am talking like a scientist what I mean is what I mean by saying this is that what you are saying is that ok I am going to spend on my hot water is this I do not care whether it is going to my cold fluid or losses eventually anyway I have to eat my cold fluid that is why I have to worry about 100 watts whatever he has to 100 watts but I am not saying that. Yes Sandeep that is true that is what that cold water will absorb that much amount of heat and then it will lose. I am saying simultaneously as it is gaining heat itself it has started talking to atmosphere it does not wait to get heated up and then start talking with atmosphere that cannot happen so parallelly it is happening both the things so I have to take 80 that is all it is so that is why I am saying that is why I said when I use the word I am talking like a research scientist and you are talking like an engineer as an engineer you should be worried like that only because you are spending 100 watts you should be worried about those 100 watts how much of that 100 watts is being utilized so I should not take average which all manuals tell I should take so please even your manuals also I am sure there will be lot of lot of things but you are interpreting let students follow those manuals let them allow them to commit mistakes we are not launching a rocket here that is what I keep saying we are not answerable for hundreds of crores everything can go wrong let it go wrong no problem but we need to explain them why things have gone wrong that is all it is that is all the aim of doing experiments okay so much about heat exchangers can we move on okay so what was the next experiment emissivity I do not think it requires any explanations I have given already augmentation may be Sandeep can take over from me and comment on forces convection what was the experiment flow over a cylinder what is the objective now let me why why you what is the objective then Sandeep can pitch in whenever he wants to determine the heat transfer coefficient around the cylinder so where all thermocouples I have put only one thermocouple so overall thermocouple so do I have any idea about at what all theta locations I have put no idea so I have couple of thermocouples and I am measuring the heat trans wall temperature what is the boundary condition I have put what is the boundary condition for the cylinder I have put constant heat flux so how did you calculate the heat transfer coefficient average again all temperatures h equal to q double dash upon T wall minus T infinity how did you get the T infinity here where did you take the T infinity at the entrance what was the what are the typical T infinity is for you 27 degree Celsius okay it is it is cold flow I mean maybe slightly heated because of blow air fine so now q double dash what did you take so can I take me into I complete me into I hear are there any losses but then which we have which you I should be taking here see I have a cylinder perhaps if I please correct me what I am guessing is that heater is wound on that or inside inside there is a heater embedded inside now you are heating this now whole of this VI is contributing for heating my cylinder now it is it can undergo two modes of heat transfer one is forced to convection another one is radiation yes conduction axial conduction how thick is the cylinder that is the question now how thick is my cylinder to answer whether it is axial conduction and what is the material maximum temperature is what one 173 173 plus 273 173 plus 273 37 means I should not be worried about radiation why 37 so why I am saying radiative losses will be very less okay 37 plus 273 310 to the power of 4 minus 300 to the power of 4 into sigma is 5.67 into 10 to the power of minus 8 into epsilon is 0.3 I will take okay so it is 19 watts it is coming to no 19 watts per meter square area is what pi DL into pi diameter is I want in mm please mm 15 mm is the diameter 25 is the diameter length mm again 0.22 watts I get 0.22 so I can it is inconsequential I can neglect the radiation so now I am convinced that radiative loss is less so what is the vi I have given total vi 85.4 watts okay so that is courtesy Sandeep current yes sir I can answer that this way 85 let us that is right whether that is right or not I can check it like this 85 divided by pi into D what am I trying to do is compute heat flux D is how much you said 25 mm length is 150 mm I get a heat flux of 7215 that is a reasonable heat flux I can achieve typically in the labs the heat fluxes what you can maintain easily without any damage is anywhere between 4000 to 10000 watts per meter square if it is above that then you usually tend to get into trouble with wall temperatures but this sounds okay for me but why do you say professor it is little high what was the current what was the voltage 12.2 12.2 and current is 12.2 divided by 7 I am getting a resistance of 1.74 ohms please remember this I am going to calculate now royal by a resistance let me check whether that is going to come out like that what is the material what is the material steel steel steel is 90 into 10 to the power of minus 8 ohm meter is the resistivity typically 90 to 110 I will go ahead and take 100 into 10 to the power are you with me I am calculating R equal to royal by a so 100 into 10 to the power of minus what did I say 8 rho L is yeah you are right your micro wire now I cannot calculate because how much wire is I have put you are right I cannot get because I am used to calculating directly heating stainless that is why okay so you are right I cannot calculate but for me with a micro wire 1.2 ohms seems okay 1.2 ohms 1.2 ohms you get now yes professor you please counter me it is not that always I should be right fine no problem that might be wrong now I would I have to concede and agree to you agree with you that or maybe earlier there was a different that is fine yesterday because he remembered that is what he said but that is okay earlier you might be using earlier different ammeter also see I am not at all worried I keep saying this those who never make mistakes never make anything at the cost of being repetitive I do not mind making mistakes at all it is perfectly alright but you should be worried yes I do not know because I have no confidence but now it has turned out to be 18 watts anyway let us not believe those numbers but my question is numbers part my question is how do I compute the heat transfer coefficient radiate all vi I can take okay so but if I were to know the location I can compute the local heat transfer coefficient perhaps there is a point at Professor Harp yesterday where should I expect slightly higher heat transfer coefficient you remember it was like this and again it went up and then came down so separation point perhaps you can get I do not know whether because those many thermocouples are there or not at least at the stagnation point if we know we can compute the stagnation point in a cell number but but I should take this with a pinch of a salt because if my thickness of my pipe is sufficiently large it is going to get my my heat transfer coefficient is going to get am I going to get local or average this is a very important point we need to drive home to students I kept I kept asking what is the thickness of the pipe why because larger the thickness more conduction within the pipe and the temperature becomes homogeneous if the temperature becomes uniform then my heat transfer coefficient is instead of me doing integration it has done the integration and given me the a challenge okay so in fact if you are doing cylinder experiment one experiment can be done with thin cylinder and one experiment can be done with the thick cylinder of course I think it is easier said than done but one can conceive of in the same set of with with one batch can do perhaps with the thick cylinder another batch can do with the thin cylinder and you can compare the heat transfer coefficient is that okay but my point is one general statement I would like to make is that please insist on uncertainty analysis they should know every number when I put H equal to 100 plus or minus what a tag has to be there plus or minus tag has to be there if the tag is not there my readings have no I think with this we will sign off quite a fruitful day yeah what is your point is his point is enclosure temperature I am taking only t to the power of 4 I should be taking t to the power of 4 minus t surroundings to the power of 4 that is one point another point is I do not know whether it is giving me the average temperature of the plate completely the plate is there and the thermocouple is put in so yes sir yes perhaps you put up the two wires of the thermocouple across the way but wherever my bead is there it will measure only there no whatever I do not at a single point you do not point through multiple junctions multiple junctions yeah yeah that also can be done that also can be done or you take copper plate and put take chromium wire automatically a junction is formed that is what people do for averaging but of course we are not doing anything but I am suggesting several things but anyway I am just suggesting because you guys are quite strong in experiment that is why I am suggesting all this is that okay so any more questions before we sign off okay so tomorrow also let us make it quite successful so we are starting at what time 815 sharp not just 815 okay.