 After having done black surface, why do we do black surface analysis first? Easy, easy to understand because we do not have the epsilon coming in. So it is a good emitter, it is the best emitter, it is good in all respects. So now we say okay having understood black body enclosure or multi control area. So we are now looking at diffuse gray surface. We are going to reiterate what is diffuse surface and gray independent of wavelength. Okay, everybody is having a go. We spend so much time, we spend so much time talking this wavelength and direction, but when it comes to real radiation problems, we throw that out actually in engineering. We never are bothered about wavelength. That is why engineers do not get Nobel Prize, so probably that is why. We make life simple. Okay, so if we assume surfaces in an enclosure to be opaque, what is opaque? What does it mean in terms of radiation? And nothing, nothing is, nothing, either it is absorbed or reflected. Okay, so if we assume, so they are non-transparent, diffuse emitter and reflectors and radiation properties are independent of wavelength. So everything is nice. So I can use formula. Okay, each surface of the enclosure is isothermal. Okay, both incoming and outgoing radiation are uniform. So life has been made so simple for us. Everything good. Now radiosity is of course again definition is emission plus reflected component of incident irradiation. So J i, if a surface i is there, epsilon i E B i represents the emissive power. That is heat flux coming out of that surface by virtue of its finite temperature. The epsilon represents that it is a gray surface plus reflectivity times the incident radiation or irradiation G i. That fraction plus this one is J i. We have assumed transmittivity is 0. So this goes to 0. So alpha plus rho is equal to 1. So rho i is equal to 1 minus alpha i and this one. What is the inherent assumption in using alpha i equal to epsilon i? Kirchhoff's law. What is the inherent assumption? Temperature is the same. Temperature of which is the same? Both surfaces is the same. So J i therefore becomes, why do I want to write it in terms of epsilon i? What happens if I keep this alpha i there? Alpha i becomes what? See, why can't we sit with this reflectivity? I don't know the answer for this. I am asking why? No, no, no. Ultimately we want to reduce the number of unknowns. That is probably the simplest. So we want to and easiest thing to measure is, reasonably is emissivity. Emissivity. So that is the, it is a, I think it is an expediency. It is an engineering expediency or engineering necessity, which again drives us to throw out all the things and make all sorts of assumptions and end up with only one unknown emissivity. Measure that and keep using it. So therefore J will be equal to epsilon E B i plus, this is 1 minus alpha which becomes 1 minus epsilon G i. Next, this E B i will be sigma T i to the power 4, not a problem. G i is the radiation energy incident on surface per unit time per unit area, watt per meter square J i equal to E B i equal to sigma T i raised to 4 for a black body. That is also logical because for a black body all these things will not be there. Now what happens? This is J I have calculated. I am coming there. For one surface I have written this equation, J. If I have multiple surfaces which are interacting with each other, get radiation heat transfer. So we wrote for a black surface very easily, sigma A 1, F 1 to 2, T 1 raised to 4, minus T 2 raised to 4, black body. That was so nice because we did not have to deal with J. We just have to deal with E B. Now I have J. So there is a J from this surface. There is a J from the other surface also which is coming in. So in an interaction what we have to do? Q i is nothing but radiation leaving entire surface i minus radiation incident on the entire surface. So A i J i represents radiation leaving. What is J? Again let us not lose sight. J is epsilon E B i plus rho G i. Again there is a G here minus radiation incident on the entire surface. So A i G i. So this aspect has to be clarified. So this J I will substitute here from the previous slide as this quantity epsilon i E B i plus 1 minus epsilon G i. Shall I write this or is this okay? So I will also make it, I will also work it out because. Q i A i J i minus G i. From the previous slide I will write J i is equal to epsilon i E B i plus I do not want to memorize. What was it? Rho i G i. Rho I do not like. So I will write 1 minus alpha. Again I do not like that. So I will write this as 1 minus epsilon i G i. See because I do not like so many things I throw them away. That is all. So what do I get? G. Again here see look at the first expression. I have G. What does G depend on? All other surfaces. So I have to keep budget. This much is coming from here, this much is coming from here, this much. So I have to become an accountant. I do not have the time as an engineer to account for this much energy from here, here, here. So again I will eliminate that G into something which I like, which I know because my field of reference, frame of reference is the surface i. So everything from J, I mean 2, 3, 4, etc. I will write in terms of surface i only. So I will write this G i which is energy coming from all other surfaces to surface i as J i minus epsilon i E b i divided by 1 minus epsilon i. So what do I do? I want to eliminate this. So I will substitute it there. Therefore Q i is equal to A i times J i minus J i minus epsilon i E b i divided by 1 minus epsilon i E b i what does this give me? A i, let us do the maths here. Take it up J i minus epsilon i J i minus J i plus epsilon i E b i divided by 1 minus epsilon i. J is cancelled, J i cancels off, I get epsilon i A i divided by 1 minus epsilon i E b i minus epsilon i E b i minus epsilon i J i, not done as yet. What is this E b i minus J i telling me? Correct, finished, finished, actual, great. So that difference is a function of what surface characteristics of the body, that is why the epsilon is coming in there. Is that correct? What did she say? Madam, would you like to repeat for the benefit of this numerator, this quantity, E b i minus J i, what is it? The variation of the physical surface of the black body, difference in the variation, if it was a black body and if it was a natural surface. The real surface, the real surface, real surface. Is that understood? What she said was, E b i minus J i, we are writing this for a given surface, we are not talking of surface 1 to 2, we are writing the same surface only. If this surface were a black body, it would have emitted energy E b 1, E b i, but now that has become less, something is lessening that energy, it has become less because it is a diffuse gray surface and that difference, that change is given by E b i minus J i and beautifully it will come out Q i therefore, will be E b i minus J i divided by 1 minus epsilon i by epsilon i a i. This quantity in the denominator is called as surface resistance, can, space resistance, that was in the space between the two surfaces. This is physically you are not able to put a picture to it, like conduction or something, but this thing strongly deals with the surface properties. Poorer the emissivity, what happens? This is value, epsilon is, epsilon is 1, what happens to this? What does it mean physically? The black body, correct? No surface resistance. So whatever I have drawn here in this diagram, very nicely drawn, see here. This E b i will collapse to J i or sorry, J i will collapse to E b i, if the surface were a black surface, correct? I mean this resistance would be equal to 0. When I put it here, I will, it will look as something by 0, that is not the way you should look at it. You should look at it from the physical point of view, E b i and J i will become equal to 0 themselves, will be the same. So the numerator itself will become 0. So there is no difference by virtue of its surface characteristic because it is a black surface, black body. So this quantity we will call as the surface resistance and this is the potential difference or the difference between the characteristics because of it being a gray surface compared to a black surface. So E b i minus J i and Q i represents a net radiation heat transfer corresponding to whatever we have in the electrical analysis. So if I have a general geometry like this, a surface which is gray, another surface which is also diffuse gray, what is the net radiation heat transfer given by? So black body always I can remember very easily sigma A 1, F 1, 2, T 1 raise to 4 minus T 2 raise to 4. Here I cannot remember like that because in addition to the space resistance which was there even for black body, I have two additional resistances which are there because of the surfaces being non-black or gray and that is what is manifesting itself in terms of these three resistances. What is it doing? What is this additional resistance doing to the heat transfer? Reducing, isn't that logical? Black surface heat transfer is maximized. Here my heat transfer is reduced because I am physically having more obstructions to the path. One of the obstruction even if the two bodies are the same, if two parallel plates are there, black same geometry, two parallel gray plates are there, A 1, F ij would be the same. So space resistance would have been the same. I am impeding energy transfer from 1 to 2 because of the surface. So these two are always going to be non-zero even if these were the same. So what I am saying is Q ij is A ij i F ij minus I will also write that is easy and the problem is it looks obvious when we read it. Honestly it looks obvious when we read it. Only when we write it we know why is this? Why is it j? Why not e? So Q ij is nothing but so I have two surfaces one is i and another is j. This is at t1, this is at t2, this is eb1 and eb2, ebi and ebj and I have three resistances 1, 2 and 3. This is r1, I will write r2, r3 just for simplicity and heat is flowing this way, Q ij. So Q ij therefore is equal to, help me please, why ebi? It is a diffuse gray surface, diffuse gray. J i, that is a flux, J i times A i, what so we are then F ij, very good. This is energy going from one reaching two minus, madam seems okay, Savarkar is having his hand on his head. Savarkar, okay, not okay. So he had his hand like this. So jj, aj, what is again F ji, okay. Anybody has difficulty in writing this? Hopefully no. Now I will use you factor A algebra, A i, F ij and this one are the same. I like to deal with surface i. So I will keep this form, throw away this form. So this is going to be A i F ij, J i minus J j. Therefore Q ij therefore is J i minus J j divided by 1 over A i F ij, okay. So j is, who gives me this j values? I do not know, correct. So we have to come back finally to this one. This quantity is of course our same concept what we have for black surface, space resistance. Only thing what was changing is the numerator. Blackbody if you remember, what was it? Sigma T 1 raise to 4 minus sigma T 2 raise to 4 divided by 1 by A 1 F 1 2 or replace i by 1 and 2 by j, this is the same. Nothing has changed. Only thing which has changed is the numerator. So conceptually both are the same. This is the potential difference. Here is the potential difference for heat transfer or for the diffuse gray surface. This is the potential difference for a blackbody. That is it. So Q ij is that. Now I have to deal with the surface temperatures between the two. So this is what? J 1, J 2 or J i, J j. So if I make the diagram a little bit better, three resistances were there. This was surface 1 E b i J i. This is actually inside the surface you can say. This is the resistance which is there because of the surface property. This is the resistance which is there because of geometry 1 by A i F ij and there is a resistance which is there. This is 1 minus epsilon i epsilon i A i 1 minus epsilon j epsilon j A j. So what this resistance tells me is the reduction or the drop in the driving in the heat transfer or the energy content because of it being a gray surface. So therefore my net energy transfer would be, it is just summation of the resistances between these surfaces and I can write all the resistances together. So if I have multiple surfaces, so for I do not have to go through this here. I will just show the slides here. If I have an n surface enclosure, conservation of energy tells me Q i is nothing but summation of this. So it will interact with surface 1, it will interact with 2, 3, 4. You can take 3 surfaces and deal with it. It is easy. And then I will write this as this form and putting it like this, I will have, this is surface 1. This is surface 1. This whole thing is related to surface 1. Though we are drawing, this is the surface 1, this resistance is associated with that surface. This is nothing but 1 minus epsilon 1 divided by epsilon 1 A 1, this quantity. That is this R i. So I have to draw this point J i will essentially be what is there beyond surface 1. This is interacting with 1 to n minus 1 to n. Several such surfaces are there. And each of these J 1, J 2, J n minus 1 and J n will be linked by a resistance to the corresponding E b 1, E b 2, E b n minus 1 and E b n. Through which resistance? Same surface resistance. Only thing different will be, it will be 1 minus epsilon 1 by A 1 epsilon 1, 1 minus epsilon 2 by A 2 epsilon 2, 1 minus epsilon n minus 1 by A n minus 1, epsilon n minus 1, so on and so forth. I think everybody has got it. So if there is a chorus answer and all, there is no discontinuity, this correct that means. So the net radiation therefore we can calculate and I think all of us know to do this. So if I have to solve these things, matrix method will always be there. So I will just write, I think solution methodology we can see. So if you are dealing with two surface enclosure, 1 to 2, 2 to 1, only two surfaces are there in the enclosure. I can write Q 1 is equal to minus Q 2 and this is the surface characteristic associated with this surface and this is the surface 2. So I will write this in general as E b 1 minus E b 2 divided by these resistances. What is it actually? It is essentially this. E b 1, J 1, J 2, E b 2. What is known? What are known quantities here? E b 1 is known, you do not like this one. E b 1, what is E b 1? Sigma t 1 raised to 4. E b 2 is known, sigma t 2 raised to 4. J 1 and J 2 are known. J 1 is not known. J 2 is not known. Epsilon 1 known, A 1 known. So this one is also known. Similarly this one is also known. A 1, F 1 to 2, is that known? Yeah, if I know the geometry view factor is known. So this one is also known. Do I need these two to solve for the heat transfer rate? No, I do not need that. So what I will do is Q therefore is equal to E b 1 minus E b 2 divided by summation of r. I know each of the r's get the value of Q. Then I will write this Q is equal to E b 1 minus J 1 divided by 1 minus epsilon 1, epsilon 1, A 1. Everything is known. J 1 is obtained. J 2 is also obtained. What does this mean? J 1 and J 2. What am I calculating? Why am I calculating J 1, J 2? E b 1 and J 1 are different, right? E b 1 is because of its temperature as it was a blackbody, E b 1. What is J 1? It is because of the emission, because it is a non-blackbody and also a reflected component of what is coming in. So we can solve for that and that is given here to you guys. So Q 1, 2 can be obtained based on these values and once I know this other things can be obtained. Small object in a large cavity, very, very straightforward. A 1 by A 2 is approximately equal to 0. View factor F 1 to 2 is almost equal to 1. Therefore, we can deal with it exactly like a blackbody. What is this you will tell me? My question is, can we put any example for this? Yes, H B ball, you get H B ball moving in or out. It is to, how many as long? Thermal couple, I have put inside for material. Furnace. Furnace inside. So usually for furnace, we will see directly that equation. So we keep on doing. So this is how. How did it come? Come. So this is how we understood. So even though it is a non-gray surface, a black. The same thing is that it is dependent on the emissivity of the thermocouple. It is not dependent on the emissivity of the furnace. Why? Why? Who is doing that? Because of? It is very, very small area of lung. Area of lung. A 1 is so small compared to A 2. So what actually what is Q 1 to 2? We transfer from 2 to 1 and 1 to 2. Both are there. One of them is so small compared to the other. Right? Because the area ratio, I mean area is so small. Am I right? Okay. Okay. Infinitely large parallel plates again. These are all classic problems. I am sure all of you have assigned these and. But usually two guys are teaching. When I was studying, we used to just remember them without understanding. Yeah. That was the problem. That is why although it may sound very trivial to you, but I am trying to. See one of the things that we want to emphasize and hopefully this will change with time you know next 5 to 10 years. Hopefully this culture of not memorizing and trying to get the formula on the spot if needed. That culture should come. Otherwise for parallel plates it is like this. For triangles it is like this. For circle it is like this. You know thermodynamics when we studied, we had problems constant pressure, constant volume work done. So I used to learn constant pressure. P will come out V1 – V2. Constant volume work is 0. Esothermal there will be a formula. Adiabatic there will be another. Why so many formulas? Everything comes from P v raise to n is equal to constant. If you can derive P v raise to n is constant. And then P v raise to n equal to constant work done will be a big derivation. Oh my God, such a big thing. Ultimately it is all the same thing. So that conceptual understanding was never there and that is unfortunate part. So we just give so much information. Take this, take this, take this. No. All I remember is one black surface. So if I can always come back to the black surface by cancelling terms my derivation is also correct. So one very good thing in my PhD guide used to say any complicated problem anything always have a baseline case. If these, these, these go to 0 are very small order of magnitude. Can you get a known solution from this? If the remaining term is like your known solution for an existing problem then you are correct. I think that thought has to be emphasized. I do not know how much more to tell this. Okay, anyway. All of this analysis we have said in no, we do not have any confusion that no medium is possible. It is like perfect right here. Once the participating medium comes then things get. This is only for right here. This is only for right here. But in furnaces for example, otherwise what is furnaces filled up with generally? It may not be such a bad idea but yes if there are gases inside yes I should be worried about. I am not teaching, I have the material, we have the material for participating medium. For that matter actually the best book for this part of radiation is Hallman, not Modest Hallman. Actually this network and even participating medium patterns charts are there. Actually emissivities get decreased. How much they get decreased are again functions of pressures and how much more fractions are there which we represent in terms of partial pressures to answer your question. But for usually we never teach participation. And same thing what is there in Hallman? The most enormous same Hallman. Hallman is this portion I like in Hallman especially in network method what we did and in fact he goes much more than what we are going to do. In fact I would recommend that when you go back home please open because I see in all of your reference books Hallman is one. So Hallman there are n number of problems even with enclosures and enclosures open lot of solved problems. Once if I ask the students to write the set of equations they will not solve them. But at least they know how to In fact. Formulate a problem. So that is there in Hallman. Where did we learn question was participating medium. But in all of this we are not considering to participate. In fact one is again digression but just because of the nature of discussion. I had the good or bad fortune. Bad fortune because I was not ready for it. Good fortune because the person is really great. Modest teach conduction heat transfer for me. Very good teacher but very unsympathetic towards students. Anyway so his questions would be please formulate the problem. Do not attempt to solve it. So formulation of the problem is one of the most important things. Hardly we have solved problems. All you used to say is you can solve. You can write a computer program. Do this do that. We have solved problems but formulating a situation into a mathematical form is the most difficult thing because you have to understand. You have to assume. You have to put it in words and go to the right appropriate equation and make simplification. So that aspect unfortunately our UG curriculum does not emphasize. It says given this, this, this what can I get. 5 are unknown, 6 is unknown. I mean 5 known quantities get the 6th one. Permanentation combination of that is one. We should avoid a plug-in problem. What is called as plug-in problem. That is the mundane here. Anybody will do. And you know one thing is not known and you will anyway get. We should not give any question which is plug-in problem. If a problem is plug-in type, then that question is not worth doing a question problem. See that can be given for assignments where the student is exposed to it for the first time. When he is doing it for the first time, he knows how to use the formula, which table to go to get this value etc. But exam, you do not have to get. Because none of the real life problems are plug-in. And you are saying that they have all the relations. So we can give them all thinking questions. So parallel plates, the areas will be the same. So what will happen to this? I am on the correct page. What will happen to this equation? Areas will come out. A1, A2, everything is the same. It will go in the numerator. So I will get A, sigma, T1 raise to 4 minus T2 raise to 4. And this epsilon 1, epsilon 2 etc., this one will basically become equal to 1. And what happens is 1 by epsilon 1 minus 1 plus 1 plus 1 by epsilon 2 minus 1. One of these plus 1 and minus 1 cancels. That is how you are left with this equation. So not rocket science. So spheres, cylinder etc. I can do. No big deal. Excessive problem is also there. So conceptually what have we learnt in this? So much of discussion etc. Conceptually what have we learnt? Two minutes. What have we learnt? We are not caught, right? Anil Agarwal, Shibilman or Vijay Tia we have caught several times. Nirmah? Anil Agarwal. Anil Agarwal. Anil Agarwal is going to summarize what we learnt. What have we learnt so far? That is all we need to know. You can look what you have written. No problem. Not my very test. You want to? Yes, please. Given the resistances is offered by radiative heat transfer. And from basic principles we have formulated that problem. So we have formulated that equation in the form of T B 3 4 2 N i B divided by something like that. So that in any serious problem in the situation of the basic principles we can form an unsolvable formative and within the unknown. And both things are temperature, emissivity, area, we divided the parameters. So from unknown from un-formulated to formulated. By putting these known things you can hide that. So only difference between conduction, resistive analogy, radiative analogy to radiation. Temperature is a potential but here we are taking this power as the difference. Power or radiosity, whatever. Something which is related to energy content. That is what we are going to take. Why do we always try to reduce in terms of resistances and capacitances? Why do we do that? That is a good question. It will take us into different world but that is ok. Even for transient conduction for example. One upon H A is resistance and rho V C P A is capacitance. So I can do an experiment not taking by a thermocouple and a hot water but I can take RC circuit and demonstrate the same thing with RC circuit what I learnt in heat transfer. In fact people earlier there are papers. People have done experiments with electrical resistances and capacitances. All this what we are network analogy is there. People have done experiments with resistances. But you have to keep that resistance. Each resistance has to be kept as which are representative of 1 minus epsilon i upon A epsilon i and 1 upon A 1 F 1 2. Whatever potentials I get they are supposed to be the answers. So now you make as complicated as possible your network is. If you cannot solve you can do an experiment. Of course today's world whatever be the complicated you can solve it but earlier people have done that. In fact even in vibration we have studied. In vibration what did we study? Mx double dot plus Cx dot plus Kx equal to forcing function. That also can be written in the form of resistance, capacitance and damping. Everything what we want to say is whether it is electrical circuit, whether it is vibration, whether it is heat transfer you understand one same thing you can understand. It can aid in understanding others. Why did we like Reynolds analogy? That is there. Strength is there. The question answer is there. What did I say about Reynolds analogy? Standard number equal to CFx by 2. Why was I so excited about it? If I measure friction factor I get Nusselt number. I do not have to do the experiments for heat transfer. We all know that doing experiments for fluid mechanics is much easier than heat transfer. This is short cut. To put it in layman term this is a short cut. Analogical approach is always a short cut. Yes it is going to give because the equations are same. Why will it not give? Oh yes you can do it and go on. No two ways to do it. That is the strength of analogy. Why we always try to put electrical resistances because we have studied from high school days the resistance approach and we can feel it very easily. That is the reason this electrical resistance approach is always appreciated. That is what we are going to study. That is what we are going to study. In fact Holman is taking more than three. He formulates for four, five, six. Enclosure he is taking. Six when I say he is taking a complete enclosure. So when you go back home please open your Holmans and solve the solved problems. Solved problems are plenty. Unsolved of course plenty are there. So solved problems itself you can solve. Anyway I think we will break for tea.