 Okay, getting back to our radiation, so there are two aspects which I want to cover okay, so that is number one, we will get to three enclosures, it is an extension, so I can go pretty fast because all this while we were worried about only one and two surfaces, now I have added third surface, so and they are all grey surfaces, they are not black, so each surface is going to have a surface resistance and a space resistance between the other two. For example between one and two, there will be two surface resistances and one space resistance. Similarly between one and three, surface resistance is common and this is space resistance and again the surface resistance of three, so again between three and two, surface resistances of two and three and the space resistance between two and, so now I am going to use the Kirchhoff's law, that is the net current at each node should be zero, so that is why I get here Eb1 minus you see here, I will use this figure, you can relate with the below relation, Eb1 minus J1 upon 1 minus that is R1 plus J2 minus J1 because always the current is inwards, so J2 minus J1 upon R1 to plus J3 minus J1 upon R1, similar thing I can do for other two nodes also, so that is how I get, what is that I am getting, three equations, how many unknowns are there here, what are things known here for us, temperatures are known let us say, that means Eb1, Eb2, Eb3 are known, then how many unknowns I have, J1, J2, so once I get J1, J2, J3, what next, what am I up to, what am I trying to solve, what am I trying to solve, I am looking for Q dot 1, so I can Q dot 1 equal to Eb1 minus J1 upon the resistance, you will get the heat transfer, so there are problems, you please go through these problems, only one issue which we have intentionally skipped and now I will be broaching up through this is insulated surface, what will happen to an insulated surface, surface resistance is not there, so radiosity becomes equal to blackboard, is that okay, is that obvious or should I go back and spend time for that, you want me to spend time on that, no problem, yeah, so what is this, here you see it is there here, it is called as re-radiating surface, if I have Q i, Eb i minus J i equal to space, upon space resistance, sorry, surface resistance, 1 minus epsilon i upon ai epsilon i, so if Q i equal to 0, what am I saying, if it is insulated means Q i is equal to 0, so if that is equal to 0, what is this equation implying, Eb i equal to J, that is how I understand, why for constant, sorry for an insulated surface J i equal to 0, another thing I want to drive home, radiation shields, so let us spend, in fact I do not know how I forgot, how do I generate an insulating material, I want to generate a perfect insulating material, let us say, I will take your question little later, how will I generate a perfect insulator, I have two plates let us say, I want to make a perfect insulator between these two surfaces, how do I do that, I do not, yes, so conduction how can I reduce, no matter whatever insulating material I am going to use, there is going to be, there is going to be thermal conductivity, so the best thing is to have vacuum, but when I put vacuum, the other mode of the heat transfer which takes over is radiation, now how do I decrease the radiation by putting shield, so in why common sensically by putting shields my radiation should decrease, shields means what, I have two plates, shields means I am going to put shields, I am going to put too many plates in between, okay, so why they should decrease, no from resistances concept, we are introducing additional space, sorry surface resistances, so when I start from EB1 to EB2, it goes on decreasing, so the radiation interaction decreases, so that is how super insulators are fabricated, that is how super insulators are fabricated, okay, so now coming that is what is this radiation shield, this is simple bookkeeping, I think we are all very good in this, I am sure we are all asking questions in the exam, this I guess it is quite easy to see, so I do not want to spend too much time on this, because I am just putting a shield and you can see that two surface resistances are added and because of which you can show that with shield, without shield the heat transfer is going to decrease, how much it decreases, it depends on the number of the shields, all this I think you can figure it out, so I am not going to spend time on this, but what I want to spend time on is temperature measurement, because I had made a statement in the very beginning of the lectures that inserting a thermocouple does not ensure that it gives me the right temperature always, okay, so that is the point I want to emphasize now, so let us say I insert a thermocouple, thermometer, resistance temperature detector, whatever I want in a pipe, in a pipe, so in this pipe what is happening, now fluid what am I trying to do here, I am trying to measure, what is that I am trying to measure here, fluid temperature, when I insert my thermocouple and I want to measure the fluid temperature, whatever my thermocouple or thermometer or RTD is showing, I believe that it is going to be fluid temperature, but is it so or not is the question what we are asking, so what is this, I have a pipe which is maintained at a higher wall temperature, maybe I am doing a convection experiment, let us say I want to measure the bulk fluid temperature, so I have wall temperature T1 and TTH is the sensor temperature, I will call this as sensor temperature and TF is the fluid temperature, now there can be all modes of heat transfer, but for simplicity I am neglecting conduction, there will be conduction through this sensor, no matter whether it is thermocouple, thermometer it is less likely, in this case it is more nearing because thermometer is made of glass, so it is a good insulator, so less conduction effects, but nevertheless if it is a thermocouple conduction effects are there because thermocouple acts like a fin, so let me neglect conduction effects, if I neglect conduction effects I have convection and radiation, so what is the convection, convection between my fluid and the thermocouple, so let me just take H into heat transfer coefficient into TF minus TTH is equal to epsilon of my, why did I write this equation, I had reminded you earlier also that is the reason I had reminded you there, you remember it is like small thermocouple in a big large cavity, I had told that in that situation in my radiation case, my radiative heat transfer is dependent only on the emissivity of my thermocouple, not the wall emissivity I had reminded you, so that is what I am invoking here again, so that is why I get an equation epsilon thermometer sigma TTH to the power of 4 minus TW to the power of 4, so now what am I interested in, I want to infer fluid temperature from measured temperature very right, fluid temperature equal to thermocouple temperature plus blah blah blah, what is this telling me, what is the ideal, what is the ideal thing, what is the ideal thing it should be, I want my TF should be equal to TTH, so as you said this radiation correction term should be 0, but will it be 0 or when will it be 0, when H is very very high, numerator we cannot have control on the numerator, the only thing which is having on which you have control, do you have control on thermocouple temperature, no, do you have control on wall temperature, no, so do you have control on emissivity of your thermocouple, yes to a certain extent but not too much, you do whatever circus you want, you shine it, polish it, whatever you can bring it down to 0.2, 0.3 but nothing less than that, but you see TTH to the power of 4 minus TW to the power of 4 that is a huge number, to counteract that number only fellow I have control on is H, the only way H I have control means, do I have really control on my H, no it depends on my problem, if I am handling forced convection, if I am handling oil or high Prandtl number fluids incidentally my H happens to be high, so I am saved, but if it is laminar flow, if it is laminar flow your H is going to be nusselt equal to 3.66, very less, you have no control, there will be error, if it is natural convection let us say, your H is going to be still low air, in furnace this is a huge problem, this is a huge problem, to make you understand this let us just take the cooked problem, I am just taking numbers, I am just taking 650 Kelvin, epsilon 0.6, wall temperature 400 Kelvin, here the fluid temperature is higher than the wall temperature that is all, so if I plug in those numbers here of course H I have taken 80, so what is the difference you get, what is the fluid temperature you are getting, 715, what is the thermocouple temperature, 615, now you decrease this H to 10, H to 10 instead of 80 you make it 10, it goes 8 times your deviation increases, what was that now, 65 into 8, you see the magnitude, point what I want to strike or I have learnt it hard way is that, I used to always think put a thermocouple you will get the temperature, but that is not the case unfortunately, so we have to have a control on H, the only way people do is again play with the H only, but they do locally, they do locally that is you imagine a jacket, you imagine a jacket around this, can you imagine a jacket around that, now through that jacket you make the flow pass through locally that H can be kept as high as possible, you can decrease the radiation error although it should not be called as radiation error, it is a correction term or the error, deviation will come down, that is what is called as suction pyrometer, that is what is called as suction pyrometer. If you see in all furnaces or fire applications we are, people recommend suction pyrometer, temperatures you insert lot of thermocouples, but are they giving right temperatures are not God only knows, believe here another important thing conduction I have neglected, the conduction is going to create havoc much more actually, but without conduction itself I am having so much of problem, that means when I say conduction the length of the thermocouple you are using also is going to affect your deviation, there will be fin effect, there will be fin effect, my thermocouple is going to act as a like a fin, so fin means you know there is a locally there is a dip in the heat transfer, so the temperatures are going to be lesser than what actually they are going to be, okay physically. But I think I like this when I first read in chungal I did not understand this, I would like to share an experience in rarefied gas dynamics we are working, there H is going to be of the order of 0.01, in fact thanks to which professor I forget one of the editors of the journal he made me understand, when we sent our results he 3 was 3 over paper back 2 reviewers, 3 reviewers checked and we were adamant that our paper is right, because we were stupid enough not to understand things, so but then editor himself became reviewer and he said you put your thermocouple bend it and measure, put it straight and measure, why he is telling me to bend it and measure and put it straight and measure after a while I understood he is trying to see whether there is a fin effect or not, okay. So H is 0.01 let us say my H is 0.01 what will be my error, point is temperatures thermocouples cannot always give the right temperatures, it depends on the situation, natural convection that much more problem because H is going to be less, okay so we need to be cautious when we are reporting our temperatures, okay. So that is the advantage of sending a paper and we recently read an article where in which they said that doing research means realizing everyday how stupid you are, okay so that is what it is actually, yeah. So stagnation effect will be there only when my velocities are very high, see I said T not equal to T plus V squared divided by 2 Cp, if my V is 10 m per second let us say 10 squared upon 2 Cp is 2000, 100 upon 1000 it becomes 0.1, so static temperature and the stagnation temperature are nearly same, but stagnation temperature will be different from static temperature when I am handling high Mach numbers, so you take 300, so my numbers go up, so kinetic component that is V squared by 2 Cp which is contributing for my temperature will go up only when my velocities are high, so that is why that is a good point in the sense that I do not try to differentiate between static temperature and stagnation temperature when I am routinely doing the measurements because incidentally or coincidentally my velocities are less in my incompressible fluids, all that much more luckier we become when if we are using liquids. At the most I can imagine that it is going to vibrate because of vibrating it is not going to measure one temperature, it is going to measure different temperature, now I will take your question before I transit to natural convection, sorry because I had a flow I did not want to get resisted, yeah you had a question in between, yeah re-radiating surface you want me to go, this one, which exam sir, how do I get the insulating surface, practically I will insulate, so I am having insulating surface which is insulated over it, but here the gate we are taking care, it has certain emissivity, it has certain guidance, so how do I get my space is 0, all that I am saying is that there is no heat transfer, there is no heat transfer, there is no heat transfer from this node, there is no way that it can go out, no, no actually in this case telling surface thinking this in terms of surface resistance is little is not physical, is not physically right, how did I show that this is Q dot 1 equal to 0, why did I say that EB2 is equal to J2, did I make that is it because of surface resistance is 0, what is it making 0, there is no heat transfer, there is no heat transfer that is why my J2 is getting equal to EB2, it is not because of surface resistance becoming 0, because there is no heat transfer, there is no heat transfer, that is correct, yes, 1 minus 0 that is not 0, did I say that, if I had said it was wrong, what here we mean here is that Q dot there is no heat transfer, if there is no heat transfer there cannot be, there cannot be potential difference, that is what precisely we are saying that is how J2 is becoming equal to EB2, is that okay, is that convincing, still not convincing, okay, we will continue that, we will continue, we will discuss, I will discuss with professor around and we will continue that no issue, okay, so the last thing of today, so natural convection, so this is coming back to again convection, so we said that natural convection occurs because of density gradients which are created because of temperature variations and in the presence of gravity which is very essential, I am going to get the body force, that is what is going to create my natural convection currents, how I usually introduce natural convection in the class is that, I usually take this example or I will ask you, maybe you will all answer because you are all seasoned teachers, so how in a geyser, if I have earlier we used to make geyser at home, do not worry about the present day geysers, they come in all forms and shapes, that is not the geyser I am looking for, we used to make geysers at home itself, that is I used to just take a container and where do I put the heater here, bottom, why I put the heater at the bottom and sometimes in fact I do not know whether you have experienced this or not, at least as a PhD student whenever our geysers in our hostel were not working, we used to put in the bucket, we used to put the water and put the heater, immerse this heater and definitely it will not go till the base, it will go somewhere till one fourth and I connect it, so which one you recommend and why, the bottom one, why, so this is what used to happen, if here whenever I used to do this, I used to feel water is boiling in the top, oh funny, garam oh gaya, so I used to, I used to remove the geyser, go to bathroom, I have started taking bath till here my water is over somewhere and at the end I am ending up with cold water, this is at 4 in the morning, so then you are in trouble, so because you did not put the geyser heater properly at the right location, because you are not aiding natural convection, here very thoughtfully, very thoughtfully the heater has been put at the bottom, so that convection currents are aided. Olden, I do not know places, rural areas at least the concept of boiler is still there, right, old movies also if you see that, fuel will be at the bottom, huge brass vessel and there will be a tap, very nice arrangement actually, tea shop, yes, tea shops that is how they get hot water, so that used to be boiler for providing water for bathing also. And another thing we notice is that whenever I come from or whenever I am, how do I say that, yeah, whenever I enter into a room after a long time in summer, I go back to my room, I put on the fan, what is the first thing I experience in summer, especially if my floor is the topmost floor, okay, not for other floors, hot air comes, because it has got heated up and hot air has got settled in the top, these are the two examples I usually take to make them feel what is natural convection. Any other examples you have, maybe you can tell me so that we can reach them much faster and easier. Kapaktee, what will happen in Kapaktee? Yeah, but how will I feel it, it should be perceptible, what I am saying is in this example I should feel it. Please elaborate, you are right, I guess, but I need to understand. Why does it all of a sudden stop? Maybe there is a psychrometry, I am not too sure whether it is because of density difference. I cannot feel it, we are saying about heaviness in the air, that is related to the humidity connection. So, I do not know whether we can really associate natural convection to that, couple with natural convection, couple with hydroponics. Sometimes you can feel that very hardly, very hardly can feel that, it is not like this like that. You mean to say very sharp dip is there. How will my student feel that natural convection is there? Okay, so with these examples let us move on, so this is, let us move on, now the question is as I keep saying there is a density variation, there is a density variation because of the temperature at constant pressure. So that is taken care of by, before we do this, let us take up, how do I formulate the problem or write, we have to get back to our equations which we had thrown to wind, but now we have to get back to those equations that is again mass, momentum and energy. So what to do, we will have to live with them. So assumptions are, I am taking a vertical plate, this is a vertical plate and it is maintained at a constant temperature. So what are the major assumptions, let us here see the assumptions, we assume that natural convection flow is laminar, the flow is laminar, we know that the flow is going from bottom to top and that flow is laminar, it is not that we do, we cannot get turbulent natural convection, we do get, but for now we are assuming that this flow is laminar and flow is two dimensional, it is going to vary only with x and y, not perpendicular to paper or board and flow is steady, things are not changing with time and fluid is Newtonian and all properties are constant, you remember this, you will have to ask me a question later on because you will have to ask me why you said properties are constant but now you are making an exception, yes it is written already here. Although the properties are constant I will make an exception that I will assume that the density difference between the fluid inside and outside of the layer, only the density variation is there, that is I will consider only the density variation within and outside the boundary layer, rest all I am going to consider all the properties as constant, why do I do this assumption, why do I make this exception, yes very right, if I do not make this assumption, who is the driving force here, who has the driving force in my forced convection, no no in forced convection, fan, fan or a pump in natural convection as you rightly said it is by and see, if there is no density gradient then I cannot have any driving force itself is not there, then there is no question of natural convection, so I have to make this exception okay but for students I have always, always year after year I will have to spend 10 to 15 minutes convincing them why I make this exception okay, so I think students will be curious on this question okay, so del u by del x plus del v by del y equal to 0, what is this continuity equation and this is my x momentum equation and this is my y momentum equation, now which one of these two are important now, just a few minutes ago he said density variation is there, why are we not taking density, why are we taking the incompressible form of the continuity equation, density variation is there but we are taking the incompressible form of the continuity equation. No, later on we will take only the vertical direction only, we will not worry much about the horizontal direction, so how will you convince yourself, that is what professor is asking is I have taken rho here constant, here I have taken it out, otherwise I should have put it inside but I have taken it constant, in fact my continuity equation also I have written it for incompressible flow, otherwise I should have written d rho by dt plus rho del dot v equal to 0. You know what Heman said is also okay because it is not really causing any compressibility, it is not really causing any compressibility effects but then I have to take density variations in the body force term, no, it is not because of something you get to it, it is not because of cosmic approximation, sorry I am not joking, so now out of these two equations, which one is important and which one is not important, y momentum equation or x momentum equation, which one is, you are saying y is important or unimportant, y, velocity in the vertical direction is significantly higher compared to velocity in the x direction, converse of what we did for flow over a, there we threw y momentum equation, here we are throwing x momentum equation and now I have energy equation, energy equation, now I need to, now the requirement is this density, I cannot handle this density as it is, so I need to put this, now before I do that let me, okay this is what I have done, what have I done here, yeah y momentum equation I have thrown, sorry x momentum equation I have thrown and in the y momentum equation, what am I doing here, what am I doing here, what am I doing here, can you explain, can you explain, out of these two terms, out of these two terms which one is important here, same, see why I am asking questions, why I am not teaching because flow over a flat plate I have taught you, now you should be able to think yourself, this is my flat plate, this is my flat plate, which one of these two, there something is told but you will have to answer me why, all that I am asking, you cannot tell me like that, I have told you and convinced you in one particular way, order of magnitude analysis I have done yesterday, you will have to tell me like that only, what is the scale of x, what is the scale of y, in this case what is the scale of x, x is okay, to remind you x is this and this is y, okay, now what is the scale of x, what is the scale of y, so anything which is there x in the denominator, that x squared is there, that has to be larger than y squared, y squared is l squared, this is v upon delta squared, v upon l squared, so that is why this term is more important than the other term, now, okay, that is what I have done in the next step, this is kept, left hand side is kept as it is, right hand side, before I take up this, I am writing plus mu del squared v by del x squared minus rho g, what is this I am doing? I have spent ample time yesterday, now I have to take answers for it, very good, very good, so what is the, so what did we say, inviscid portion talks with the viscid portion through pressure, that is imposing pressure on it, what is the pressure, the inviscid portion imposing on my flat plate which is, that is right, dp infinity by dy, but what is the pressure which is acting, dp infinity by dy, what is the pressure which is acting, what is the pressure, it is still, see, it is still in the natural convection, it is still, atmosphere is what, is there any u infinity there, no, what is, it is outside the boundary layer now I am, outside the boundary layer now I am, outside the boundary layer my fluid is still, so if it is still means pressure variation how do I handle, hydrostatic pressure, rho g h I am supposed to use, is that right, so if you know one you will get everything, so that is why we spent so much time on flat plate yesterday, so now this is a mistake, dp infinity by dy it should be equal to minus rho infinity g, this is the hydrostatic pressure, this is what we have studied from hydrostatics, so now I can replace dp infinity by dy as minus rho infinity g, so if I do that rho u del v by del x plus v del v by del y equal to mu del squared v by del x squared plus rho infinity minus rho into g, now we will come to Heman's Bosnick approximation which he is always curious to get to, so now what is that I should look for, now if I recast my equations, if I recast my equations what is that equation looking, what are the three equations, I have del u by del x plus del v by del y, I have this equation again and again I have temperature, but I know that this density difference is created because of temperature, so I would like to see these density differences in terms of temperature, so in that pursuit I am going to make Bosnick approximation, okay, how do I relate, that is I am going to take the recourse of volumetric coefficient of expansion beta equal to 1 by v dv by dt at constant pressure, so this can be written as d rho by dt, so this can be written as rho infinity minus rho upon rho into t infinity minus rho and this can be recast and written as rho infinity minus rho into rho beta t minus t, so I am plugging in that, so I get, now I get the equations u del v by del x plus v del v by del y plus nu del square v by del x square plus g beta t minus t minus t infinity, if I recast and show these equations, what is the glaring difference between these equations and the flat plate equations which we wrote yesterday? In momentum equation, temperatures we never saw, you remember in our discussion also we said, Blasius who did PhD with whom, Prandtl, whatever flat plate case I showed yesterday I forgot to tell that was his PhD work, okay, so he first solved velocity profile and then he solved temperature profile, why, why could he afford to do that? Because they were uncoupled, but here it is not like that, they are coupled, so solving becomes that much more difficult, so with these boundary conditions that is I have taken constant wall temperature and no slip boundary condition and t infinity as x tending to infinity, what should be the next logical step, sorry I should not have flashed that, I should do dimensional similarity, why should I do dimensional similarity to identify the non-dimensional numbers which are going to govern this phenomena, so that is what I am going to do, so I do not have to spend too much time here, it is the same business what we did yesterday and what is that we are going to see here, what is this g beta delta t l cube divided by nu square, so that is Grashof upon Reynolds number square, so we are going to come across, REL we have been knowing, okay, but here Grashof number is the additional parameter, you see here I kept saying for natural convection h is not independent of delta t, it is because of Grashof number, it is because of Grashof number, so once I have this Grashof number and yesterday there was some discussion I think when to use whether it is going to be mixed convection or that can be decided by this ratio, that is Grashof is what does this physically mean, if Grashof is higher than Reynolds number means what, so that means which one should be dominating, forced convection or natural convection, so convection is forced convection, if it is of the order of 1 then both are dominant and I should consider mixed convection, okay, so now as we have been doing we will go in the lines of correlations, we are not going to derive anything, one can derive the same way what we derived with similarity approach, one can derive, but for UG I would think it is little difficult to absorb for because it is quite lengthier here for similarity approach, so I do not know at least we do not teach, I do not know about you guys, you guys teach, okay, so we state we only give the correlations, these are the correlations, so for vertical plate, for vertical plate with constant wall temperature various boundary conditions you have, so let us not get into that, so you have various situations, you get that, I think only one thing I want to, this is the same, this is what is this graph showing, Nusselt versus Nusselt by Rex to the power of half upon Grashof to the Grashof upon Rex squared, what is this showing, this is again indicating when does forced convection occur, when does natural convection occur and when does natural and forced convection occur, is that okay, so this is assisting flow, opposing flow, all these you can feel, I do not want to get into this, so this is the basic thing, but only one point I want to add which I am looking for which I am not getting is transitional Rayleigh number, see here I have not told one thing, what is Rayleigh number, Grashof into, Grashof into Rantel number, so what is the transitional Rayleigh number for laminar to turbulent is what I am looking for, yeah 10 to the power, around 10 to the power of 9, yeah it was there here, around 10 to the power of 9 it is till 10 to the power of 9 Grashof number it is above that it is going to be, so with this, yeah any questions we can take now, I think you were telling yesterday also same thing, yeah is it true, can anyone take his question, so we work coordinate is making a statement that for natural convection thermal boundary layer thickness at any given height is going to be same as hydrodynamic boundary layer thickness, is this right or wrong, some answer I need, I need otherwise I am going to hand pick someone and ask, we have dealt Prandtl number effect in external flows, in internal flows, now we are just coming to natural convection, no let us get the answer first before we convince others you should convince, that is what is my thinking, Prandtl number is 1 it will be equal, so for air only one can conceive or think of or approximate that delta is equal to delta, if my Prandtl number changes, let us do, let us say I start doing natural convection with water, what will happen, no you will have to answer me, I am not going to accept your answer, if Prandtl number increases what will happen, what did we study, what did we study, how does thermal boundary layer thickness vary with the hydrodynamic boundary layer thickness, what did we study, we have spent delta by delta t is equal to Prandtl to the power of n where n is positive number, this is what we have studied, in the background of this you will have to answer me, I want to answer from you only, having understood this basic okay and we have understood this also, this has been physically explained also by professor Arun which one goes faster and which one goes slower, now you will have to answer me, Prandtl number of water let me give you as 6, now my Prandtl number is 6, now what will be, will it be same now, do you say still it will be same? How will I believe you know, see you have to give me a logical reasoning to believe you know, see I, you see, density variation is causing, sorry temperature difference is causing buoyancy force okay, so this profile is coming and ending at P infinity okay, so density variation is there till that in this portion okay, so that is the temperature, now how is it feeding into the velocity, how are you bringing that into the velocity, I think now I am getting your question, I think you have a valid question, yeah you have a valid question, I need to think, let us just imagine and draw, I do not know whether we will be getting the answer or not, let us think, let us draw it, let us draw and see what will happen, see let us take Prandtl number is greater than 1 okay, let me do thought experiment okay, delta by delta t, I do not know whether I am going to get the answer or not but I am just trying to attempt, Prandtl to the power of n, so that means what delta is greater than delta t, I think t is possible, then if that is the case, so what we are saying is that delta is, this is going to my delta right and delta t is going to be okay, your question now okay, it cannot because if delta t, your point is that if the temperature differences are fixed here, so how will the velocity, I think it is a very valid question, what you are saying is right but still it is difficult for me to answer and conclude and say that Prandtl number I will have to think but this is a beautiful question, yeah. That is right, okay, why not, we did not say who is driving, we did not tell no in that, no I have to see myself and see before I answer, maybe he is right, I am wrong, that is delta by delta t equal to Prandtl to the power of n may not, he is not valid anymore but I need to confirm this, I need to confirm this but this is a very good question I never thought, I was always under the impression that this should continue to work, this now only I understood your question because when I saw this, because this is what the temperature difference is what is creating the velocity boundary, now I understood your question, sorry for taking so much time in understanding that, fine, I will think and get back. So this is another question, let us write this for record sake, I want to because in the morning we have written three questions, I want to add to this question, that is, is my boundary layer thickness, thicknesses delta and delta t going to be dependent on, okay, fine, yeah. The cases are available, Prandtl, natural conduction is possible in Prandtl, yeah, he has, P H D s are done, I use this word P H D s, P H D s because I mean when I say P H D s are done, so much effort is put, so much effort we do not put until and unless there is a relevancy for that, that is all I mean whenever I say I have inherited this from, I copy this from Prof. Dathir, that is all, okay, okay, that was a good question, you had in some things. Start from 0 and ending at 0 on the boundary layer, is that? Yeah. We will just start with this 0 and ending at 0. No, this is what we have shown, this is what we have shown in the textbook, but whether it is valid for all Prandtl. Can you not write that point and then you will feel it like this, which we are drawing in the previous lesson. Here the temperature, see I understand your point, what you are saying is that here the velocity is coming back to 0 and temperature also is coming back to t infinity, so temperature that way there also it gets back to 0. Let us, I do not want to answer now, you have made me think, I will think, that was a really good question, okay. Actually what is written, I just checked, delta by delta t is order of Prandtl number is to n, it is given in that first convection chapter, it does not mention anything about major of… Even in introduction to convection, but it is for a vertical play of… For a horizontal play only or something like that, we have to really look at it and it is back. Another thing I wanted to ask you, what is the boundary layer? Boundary layer. So, in this case, I have drawn this boundary layer, is there mass coming into the boundary layer? Yes. Inclignment from outside, is that happening there? Yes. So, in that case, this velocity really is 0 at the tip of the boundary layer, at the edge of the boundary layer. At the leading edge, where is the… No, not at the leading edge, what he is asking is at the interface, at the interface between the stagnant fluid and the boundary layer, where in which there is mild development. So, just now we have assumed that the outside this portion we are having… Correct. So, what are you… No, he is asking in real. In real, right. This is an assumption. Yeah. For the sake of solving, we have went ahead and made that assumption. All of us are… Burnt and insensitive, underburden. Yes. Right? So, it is a flow which is happening in three dimensions. So, it is going to go increasing bit as it goes upwards, right. Is there a cold air which it is making surrounding air, is it taking along with it or it is only the smoke which is there? It is diffused. It is diffused. It gets diffused. Yeah. It gets diffused and also in the process it is taking in the cold air. No, there will be actually there will be interface. There will be there will be a mild velocity here because it is pulling, it is sucking. It is getting entrained and another thing we need to realize is here REL I do not know, that is also an unknown. REL in our forced convection, we knew REL but here REL is unknown. It depends how much is entrained, is it not? Anyway, I think that was. The velocity near to the plate surface is higher or the velocity as shown in the velocity profile it is. Symmetrically. Symmetrically, velocity near the wall will it be higher. See, no slip boundary condition, see no slip boundary condition I have to satisfy near the wall and at the interface I am assuming that there is at the interface my fluid is stagnant, right. So, in between I have to have a maxima. Actually your question is whether my peak is situated near the wall or near the, definitely near the wall. This is just a representation maybe because you are seeing it close to the middle your thing. It is not here also it is not nearer the middle, it is leaning towards the wall only. This leaning towards the wall because temperature gradients are more nearer the wall, more nearer the wall. Mungage. Mungage. Yes, yes, please go ahead. This case surface temperature is higher than infinity. So, ultimately velocity of this air or the media which is moving very near to the surface need to have higher velocity than other levels towards right side. That is intuitively right. Let us see, I do not want to answer off the hook I do not want to answer I have to do my homework, okay then only I can answer. I have gotten into trouble now but that is okay. Can anyone take this question? Why is momentumic? No, no I did not take density as constant. I took the density variations within the boundary layer, within the boundary layer I have taken a but not in the if I have to be very precise I have not taken these density variations in my inertia force or in my viscous force. I am taking this density variation only in the body force because that is the one which is driving me. In compressible flows densities are varying by virtue of actually high speed and the variation is very very large also. There is no need of worrying about that. That is how I console myself. We are looking at d rho by rho you cannot look at d rho in isolation. Del rho by del t is compared to del rho by del p. Will you please allow it? Del rho by del p is change in density because of temperature is greater than change in density because of pressure. Is that what you are saying? Where is What is this? D rho by dt is greater than d rho by dp. Therefore we are taking it only as a No, I do not think so. No, no. Here I am assuming here pressure is reasonably constant. Right. That is what when I put beta itself the basic assumption there is that the definition of beta is having at constant pressure. It comes as coupling because of the buoyancy force term. Because rho g is there body force term is there. Let us not worry from equations perspective. Let us throw these equations out of our mind. Who is creating this natural convection? Who is creating density is an outcome but not the originator. Who is the originator? Who is the originator? Temperature difference. So naturally flow is being generated because of temperature difference. Let us say my temperature difference is not there. Will there be flow? No. So naturally in my momentum. Any kind of flow which is occurring is due to delta T. Delta. Therefore it is the in force connection the flow is appearing independent of whether there was a delta T. Because there will be u infinity. U infinity or u infinity. Correct, correct. Yes. Do I have a question? Question? Yeah. Any more question? What is that? This is applicable means which is applicable? Natural. No. Why? I can pull a 200 degree centigrade plate and 30 degree centigrade. Why? Why you only know delta T? No. No, no. Steam minus? Yes. There are two issues. You said natural convection will occur only for small damage. I am talking about those issues. You may want to consider. What is this approximation? You may want to consider this. Yes. Because here is sort of we are taking linear relationship. Actually I can go on. Yes. People do with various other. That is fine. I don't have a question on this thing. But I am having other question. Yeah. Please go ahead. There is always I am worried about that. Okay. In natural convection the convection currents we are talking about. Yes. Always we are dealing with the vertical plate. Yes. In the horizontal. Horizontal plate also can I have natural convection? Yes. Yes. Let us. But here the convection currents are showing into the horizontal plate. Then it is showing like this. It is coming on the from the center it is coming. Okay. Why? It is coming. It is so. Okay. Okay. Okay. So yeah, yeah. I know I understand what you are saying is if I take a plate like this. It comes something like this. Is that right? Yes. Which other way I can conceive otherwise? Let us think of because if I conceive other way is it physically possible or not? It is right. You are imagining that it has to just go like this. Going like this. Going like this. That is the one answer. But again it is not that much communicable that the air is coming like this and it is going up. No. So why it should form itself as a jet and then. But if we place this hot, the air will move on the plate. So it moves over against temperature. Then it has to be composed on the. But his question is why should this fluid. No. Fluid should know that it has to move in this direction. That is his question. Okay. So now, now, now okay. Professor has answered your question. How do you satisfy continuity equation in the second case? M dot in has to be equal to m dot out. So how do you satisfy in the second case? M dot in equal to m dot out. Where is the air coming? Somewhere it has to come and then get out. So if this is the case, where is it getting entrained from? That is how that interface is being. It has to suck to know. It has to suck to know. It will start sucking from sideways. Then the flow pattern is generated. Is that okay? Any more questions? Good to have so many questions. Although I went on a rocket speed. Yeah. Point is what? Fluid is entering. Second case? First case. First case? Case 1, case 2. Your explanation is? In first case. Okay. Fluid is entering. It attacks on temperature. Correct. So if density is decreased? Yes. So whatever the particle which is entering after that is going to go over density. Correct. In reverse of the case, in the reverse case what will happen? This is upward heating. Yes. This is upward heating. In the downward heating what is happening? In downward heating it will go like this. That means where is it coming from the bottom? So m dot in equal to m dot out. Okay. So I think with this, with one more question added to our kitty, we are signing off. We have completed convection and radiation. Quite a fruitful day, quite a fruitful day. So we will, with this we will come to lab sessions.