 We will get started with interaction between regular surfaces that is non black surfaces first. So, before going for lunch we looked at interaction between one I mean two black surfaces or multiple black surfaces in an enclosure. Now, what we are saying is let us look at a diffuse gray surface what we mean by a diffuse gray surface we already seen when we looked at Kirchhoff's law in detail that is a surface where things are not dependent on the direction and the wavelength. So, if I have any regular surface I am just going to go slow at this part afterwards once we go to the concept of space and space resistance and surface resistance I am going to go fast. So, what this is we will see any surface you have G is the incident radiation that is coming in and rho times G is the reflected component of that incident radiation epsilon times E b would be the emitted radiation. So, we are talking of a regular surface. So, epsilon is going to be there E b would be the emissive power of the black body at that same temperature the surface of black epsilon would be equal to 1 otherwise it is epsilon E b. So, radiosity therefore, we define as rho G plus epsilon E b. So, epsilon E b i plus rho i G i is equal to J i. So, this is going to have dimensions what per meter square for a diffuse gray surface what did we say that epsilon equal to alpha that is one thing. If I if I make an assumption of an opaque surface. So, please make a note here non transparent and diffuse both are going to go together. So, these are important things only then can I get some relationship which I am going to direct. So, non transparent means opaque tau is equal to 0. So, alpha and rho are related by alpha plus rho equal to 1 and then if I say diffuse gray surface this is rho plus epsilon equal to 1 because of Kirchhoff's law. So, Kirchhoff's law tells me Kirchhoff's law tells me epsilon equal to alpha for diffuse gray surface. What have I said if I have a regular surface i G is coming in rho times G is going out G i is coming in epsilon i E b i is what is emitted, emitted plus reflected component I call as radiosity. So, J i is equal to rho i G i plus epsilon i E b i. Now, I am saying alpha i plus rho i plus tau i equal to 1 tau i is 0 because it is opaque. So, rho i therefore, is equal to 1 minus alpha i and with Kirchhoff's law I get this as 1 minus epsilon i. So, I can recast this as 1 minus epsilon i G i plus epsilon i E b i what is the use of this where what am I trying to do? I am trying to do something which is going to help me to get temperature distribution or heat transfer effect from one surface to another surface interacting regular gray surface is what I am trying to do for that what do I need? J is referring to radiosity associated with that particular surface E b is something which is because of its finite temperature epsilon i is a material property associated with that surface. Now, this G this is not going to be dependent on the receiving surface it is going to be something which is coming from outside. So, this is the incident radiation. So, that depends on the other surface. Now, what I am saying is if I write this like this for one surface I write it for another surface third surface etcetera I will have a similar set of equation 1 3 surface let us say J 1 epsilon 1 E b 1 1 minus epsilon 1 G 2 plus 3 it could be some G let us say only 2 surfaces this would be G coming from 2 to 1. Similarly, for surface 2 I will have energy going from 1 which is going to 2. So, I will have a family of equations of this can which can be solved. So, right now we are looking how to formulate things for one surface. So, we will complete that and then go to interaction. So, this J is there this J depends on something from outside and a material property epsilon and its temperature. So, if I have a black body this is epsilon is E b i is sigma t to the power 4 epsilon is equal to 1 and then for a black body this is J equal to E b is equal to sigma t to the power 4 how do I get that because epsilon is 1 this goes to 0 we know that now I go to equation it absorbs all radiation. So, reflected component is going to be 0. So, G is what is coming in everything gets absorbed that what is going out is nothing but E b epsilon is 1 E b is equal to J for a black body it is an important relationship, but again there is no need to memorize it it just comes from fundamentally you draw this diagram what I have drawn you will get when you substitute epsilon equal to 1 rho will become equal to 0 you will get what I am saying. So, for a surface that can be approximated as a black body epsilon equal to 1 it is a very very useful relationship radiosity therefore, can be found out as nothing but emissive power of the black body sigma t to the power 4. Thus radiosity of a black body is equal to emissive power where is this useful you will see when we solve problems it looks very innocuous, but it is a very powerful equation this equality makes life very easy for us when we are trying to solve a problem if the surface becomes black. So, having come this far we will now say I know what is happening for one surface let me deal with two surfaces which are diffusion grade. So, what do I do during radiation interaction we are going to see radiation leaving the surface and what is the incident on a surface that is the net energy transfer always we remember this one thing net energy interaction is related to what is coming in also. So, what is going out minus what is coming in is what is going to be the net energy transfer remember that cubical enclosure that we saw some 300 and something watts was kilo watts was coming in I mean sorry 1000 something watts was coming in and only 300 was going out there was net energy coming in. So, here also if I write q i is nothing, but where did this come from q i is a net interaction is nothing, but j i a i surface you take surface i j is what is going out this encompasses e b i epsilon i plus rho i g i does not matter it is j. So, much is going out per unit area this is the total watts that is going out what is coming in g i. So, e in minus e out is what I am doing I am taking this control volume here control surface e in minus e out plus e stored equal to e dot s t e dot s t is 0 steady state we have reached. So, every every object is at a prescribed given temperature what is the interaction. So, e out is this e in is this of this g what is participating what is giving what is important in the net heat transfer what is important in the net heat transfer the full thing because the reflected part we have already reflected part we have already taken care here when I am doing a budget of what is coming in I am not bothered about what happens to it after it comes in. So, this surface c is g times i which is coming in. So, we have to be careful we should not introduce rho or anything here that rho part has already come here because this involves epsilon i e b i plus rho i g i times a. So, this we have to write carefully. So, this is the total energy coming in e dot in this is e dot out going out I have put this is been flip the other way does not matter it is the same energy balance. So, if I know to write this now it is just algebra. So, j is this therefore, I can write g is equal to what am I trying to do and why am I trying to do what I am trying to do g we have taken this is coming from another surface another surface multiple surfaces have to be budgeted before I calculate a g. So, you are sitting in a room there are in my room there are some 25 tube lights I have to take energy coming from each of the tube lights into me that budget becomes very difficult for me at for every time. So, if I can recast this g in terms of j and epsilon by some algebra like this. So, what I am doing I am just rearranging j is this is definition of j. So, I write for g i in terms of j and e b i just using the definition and then what I do let me write me do it on the white board it will be easy for you. So, j i is equal to epsilon i e b i plus rho i or 1 minus epsilon i because of the usual assumptions of diffuse grey and opaque. So, I am using this relationship directly. So, 1 minus epsilon i g i I do not want this I do not like this I like this and this. So, I will say now I will write g i is equal to j i minus epsilon i e b i divided by 1 minus epsilon i what I do next I substitute for this here because why you will see that in a minute. Therefore, q 1 is equal to j i j i a i minus substitute for g in this part directly I will get a i j i minus e epsilon i e b i divided by 1 minus epsilon i I have just substituted nothing else I have done. Everybody I hope is with me please do this now I multiply this by 1 minus epsilon i I will get j i a i there is a plus j i a i minus j i a i which will cancel I am left with minus epsilon i j i a i plus epsilon i e b i a i divided by 1 minus epsilon i what have I done I have just multiplied this and cancelled of a i j i term that is all I have done. So, when I do this I get some one very useful very important result actually this will give me rearrangement q i therefore, is equal to a i I will keep epsilon i I will keep e b i minus j i divided by 1 minus epsilon i implies e b i minus j i divided by 1 minus epsilon i divided by epsilon i a i this result as far reaching implications let us just stop and look at what this is what is this is talking about a surface let us take a surface i this has a this is a gray surface. So, it has a property epsilon i associated it is at a temperature t it is m s a o power is given by this one it is reflected part is given by rho i g i and this is the g i coming in this we together called as j i what am I saying I did this energy budget as j i minus g i times a I did that, but now since I do not like g i because I have to account for 25 different surfaces which I am interacting. So, instead I have casted in a form of e b i minus j i I have not got a solution or anything I am just writing it in a convenient form because this quantity I do not know this quantity has all the g embedded in it. So, it is not something which I can get from a text book e b I will know because of a temperature, but j has all the g embedded in it from source 1 from 2 3 4 all the budget has been taken care here, but this result tells me something much more important though it has been done from a convenience point of view what does it tell me we know from conduction t 1 t 2 it flows like this l over 3. What was this when there was a temperature difference this was the resistance associated with conduction heat transfer this was the heat flow from high temperature to low temperature what am I saying here when let us listen carefully what is e b I e b I let us go back one step to the definition of j i j i is epsilon i e b I plus 1 minus epsilon i g i correct this was what I got from energy balance if my surface were a black body what did I say j i is equal to e b I we proved that by putting epsilon equal to 1 that means radiosity of a black body is same as MSU power. Now, when my surface is a grey surface what am I saying because of the fact that my surface is not a black body anymore there is a difference between what is e b I and what is j I how are they related they are related by this expression. So, by virtue of being a black body the heat transfer from it the net energy interaction from it would have been e b I equal to j I which is equal to 0 because black surface is this e b I equal to j I it absorbs everything MSU power is the same as radiosity. Now, for a real surface by virtue of it not being black there is a difference associated because of it surface material characteristics epsilon emissivity that it is not able to behave like a black body that difference in that behavior that potential difference which is responsible for this heat transfer from that surface because it is not a black body because it is not a black body what it has to do it has to throw out something it has to reflect something it has to do something more than what a black body would do that is this potential e b I minus j I e b I would have been the MSU power of the black body at the same temperature. If this body behaved exactly like a black body this would have been equal to 0 numerator now because it is a real surface this j is going to be some quantity which is different from e b I and what does that some quantity which is different from e b I depend on that depends on whatever is coming in and also it is own temperature. So, this concept is exactly like what we had in conduction except that we are not dealing with temperature differences we are saying this potential this driving potential exists for heat transfer because of its difference of from being a black body and this quantity we call as a surface resistance. So, this surface resistance is not there when I have a surface behaving like a black body because epsilon is 1 this would go to this will not be there j equal to e b I. So, there is whatever the surface emits that is the radiosity that is the black body, but in real life there is a potential which exists which is different from the black body potential and that is what we call as a we quantify that difference by what we call as a surface resistance. So, e b I minus j I by 1 minus epsilon by a times epsilon is the surface resistance. Therefore, e surface which is diffuse gray which is at a temperature will not emit energy equal to e b I that is going to be quantified by this. Now, let us quickly go. So, direction of the net radiation heat transfer depends on relative magnitudes of course, we know that if e b I is greater than j then it will be from the surface to the outside and to the surface if j is greater than e b I we saw that in that cube problem a negative value means it is coming in going out so on and so forth. So, now we will come back to this radiating surface little later let us quickly go to another concept this is also an important concept which is very useful in radiation that is called as space resistance. What is this? We said if I have two surfaces interacting with one another two surfaces are interacting they are arbitrarily positioned energy net energy transfer between i to j is nothing but what q going from here to here minus q coming from there to here. So, q 1 to 2 minus 2 to 1 is the net energy transfer. So, what is that? Let me just write quickly this q i to j two arbitrarily space surfaces 1 and 2 that is q 1 to 2 is nothing but q 1 minus q 2 what is that energy leaving 1 j 1 a 1 is the total energy of that what is going from 1 to 2 is only a fraction of it. So, f 1 to 2 times this quantity is what is going from 1 to 2 1 is in all direction only 2 is seeing only this much minus f 2 to 1 this is 1 this quantity f 2 to 1 j 2 a 2 is this quantity. So, the net q 1 to 2 is nothing but let us use view factor relation a 1 f 1 to 2 is a 2 f 2 to 1. So, I can pull that out a 1 f 1 to 2 j 1 minus j 2. So, just only when there is a temperature difference between two surfaces can I have heat transfer only when I have potential difference in the form of radiosity difference can I have energy interaction radiosity is not just emissive power please remember it is emissive power contribution plus the reflected portion. So, by chance if j 1 and j 2 come out to be the same even though there are different temperatures we will still have no interaction because of this. So, this quantity q 1 to 2 is j 1 minus j 2 divided by 1 minus a 1 f 1 to 2 what is this quantity. So, this quantity which is there because of the relative position of the two how did this come this view factors are embedded in this relationship this heat I have not derived anything from the air. I have just written energy balance I am saying heat is going from 1 to 2 2 to 1 what is how is it related it is related like this. How much heat will go from the net heat transfer how much will it be it depends on what is j what is j 2 what is f 1 to 2 and what is a 1. So, this quantity essentially is dependent on the geometry the related what is view factor definition let us go back it depends on 1 by a 1 double integral a 1 a 2 right 1 by a 1 double integral a 1 a 2 cos theta 1 cos theta 2 d a 1 d a 2 by pi r square. So, it depends on theta 1 theta 2 d a 1 d a 2 and r. So, relative sizes relative positions orientation whatever you want to call. So, this quantity therefore we will say is now called as the space resistance why is it called a resistance when two objects are placed very close to each other that is you put your hand just above the gas stove it will start to burn that is because the space resistance the view factor is almost equal to 1 use whatever is coming out is being received by you. If I take it away I have moved away it is the same hand the same gas stove that f i j has changed because of r now I think it is easy to imagine because this f i j has changed the net heat transfer between 1 and 2 has changed that though nothing has happened to what is coming out from 1 nothing has happened to this relation it is just that f i 1 to 2 has changed. So, this is a thing which is related to the space aspect of the of the two geometries and j i minus j j is the potential difference if that goes to 0 there is no heat transfer positive means heat is going from i to j negative from j to 1. Now, if I have multiple surfaces I am not going to go into the details of this if I have multiple surfaces interacting I will just write it down and show it to you all of us are familiar with this. So, surface i is interacting with 1 2 3 surfaces there exists a surface resistance of i because of it being not a black surface a space resistance between 1 and 2 between 1 and 3 and between 1 and 4 and for this surface 4 this surface 3 and 2 inherently because they are diffuse gray surface I will I will this is surface 2 this is surface 3. So, I will make a better diagram here surface 1 2 3 4 let us say we have like this this surface by virtue of being a gray surface will have a surface resistance e b i j i then between surface 1 and 2 there is a space resistance between 1 and 3 there is another space resistance between 1 and 4 there is another space resistance what are these 1 by a 1 f 1 2 1 by a 1 f 1 3 1 by a 1 f 1 4 this is there added to that each surface by virtue of it not being a black body as a surface resistance which is given by e b 3 and j 3 and e b 4 and j 4 this diagram is meaningless if we do not put what are the driving potentials e b i e b 1 j 1 j 1 to j 2 this is the resistance j j 1 to j 3 this is the space resistance j 1 to j 4 this is the space resistance and between these e b's and j's these are my surface resistances associated. So, such a diagram we can draw for any set of geometry and that is what is written here we have shown here. So, this surface and summation of all space resistances I have done this for one surface now we are we are talked of surface 1 talking to 2 3 and 4 surface 2 similarly can talk to 3 surface 2 can talk to 4 surface 2 is already talking to 1. So, I will write what do I get what do I get for surface 1 if I want to write this I will get I will get a set of equations which are going to be identical in form for each of the surfaces for surface 1 I will get an equation which relates e b 1 j 1 and e b 2 j 2 e b 3 j 3 e b 4 j 4 this I have done for surface 2 if I write I will I will relate whatever is happening at this surface to what is interacting between 2 3 2 4 1 1 2 like that I will get a family of relationships with equations which will have to be solved what is known in doing this we have forgotten the picture temperatures if they are known radiosities are unknown what is the net energy transfer that I do not know. So, this might be known these will have to be calculated how many radiosities j 1 j 2 j 3 b 4 4 equations should be there if I write for all 4 surfaces I will get 4 equations 4 unknowns technically I should be able to. So, how you solve it go to a computer mat lab math card matrix inversion software whatever it is you will be able to solve it. So, that is what is explained here. So, combining this what am I writing surface and space resistance whatever it is coming at this node see you do any complicated circuitry, but energy balance is again valid 100 kilo watt is coming from here to here e b i minus j i divided by r is q dot i that has to be split that is all we are saying. So, how it is split how much goes here how much goes here depends on the individual space resistance, but what I know for a fact that e b i minus j i by r 1 should be equal to summation of all the energy transfer and that is what is put in this line q dot i net radiation flow from surface through its surface resistance equal to the sum of the radiation flow from that surface to all other surfaces including itself one thing we have forgotten here in this diagram I have not forgotten if this was a curved surface like this there would be a a 1 f 1 to 1 also. So, that also has to be there. So, a space resistance associated with all the interacting surfaces. So, how to solve I am not going to do you will have a set of simultaneous equations there are two things surface with heat transfer rate specified that is ok surface with specified temperature all the j's would be unknown you will get a set of simultaneous equation and you can solve for insulated surface pre radiating surface q dot i equal to 0 what do you mean by insulated surface no heat transfer nothing goes here. So, q dot i has to be equal to 0 and q dot i equal to 0 would give me j is equal to e b. So, please bear in mind one thing black body boundary condition black body condition also gives me j equal to e b correct that comes because epsilon equal to 1 that is because of the nature of the surface on the other hand nature of the boundary condition insulated boundary condition makes what 0 epsilon does not become 0 this is therefore, j will be equal to e b it is not because the surface is black though the end result is the same the cause is totally different one is because it is a black body other is because of a boundary condition that has been imposed this we have to be very careful in explaining and this surface such a surface is called as a re-radiating surface. So, that is what is written here n algebraic equation j 1 j 2 j 3 are the unknown heat transfer rates can be calculated once the matrix is solved for j 1 j 2 j 3 and j n and what is the heat transfer rate I do not have to do anything I just if I if I draw this diagram I will know. So, what if I know this and from here to here this is the heat transfer the net heat transfer rate we can find out like that. So, we will just quickly go to now two surface enclosures all of us have just finish this and you can go q 1 to 2 two surfaces if we are talking of black bodies then sorry if you are talking of regular surfaces we are dealing with regular surfaces q 1 to 2 is q 1 that is q dot 1 equal to minus q dot 2 what we are saying is net interaction is such that if so many watts is going out this surface is going to receive the same amount of energy. So, how are we going to write that that is written essentially by this network in this network resistance network that we have drawn there is a surface resistance associated with one surface resistance associated with two and there is a space resistance between one and two I will just write for one geometry because we are doing it for the first time two two surfaces one and two there is a surface resistance e b 1 j 1 between j 1 and j 2 we have a space resistance a 1 f 1 2 this is 1 minus epsilon 1 by epsilon 1 a 1 j 2 there is another surface resistance 1 minus epsilon 2 by epsilon 2 a 2 this diagram should have all these ingredients. So, direction of heat transfer the relative respective potentials differences which will be there and the associated resistances and the net q therefore, is e b 1 minus e b 2 divided by summation of all these plus 1 minus epsilon 2 by epsilon 2 a 2. So, I have all this so once I write the resistances I know the temperatures I can calculate the net heat transfer e b 1 e b 2 put sigma t to the power 4 you are done. So, small object in a large cavity all the subsequent cases that I am going to show are special cases of this formula. So, if I am talking of a small object in a large cavity f 1 to 2 1 refers to the to the enclosure 2 refers sorry 2 refers to the enclosure 1 refers to the body. So, net a 1 by a 2 is approximately equal to 0 f 1 to 2 I said the table tennis ball in this large room whatever energy is coming out of the ball is going to be received by this large room. So, f 1 to 2 is equal to 1 based on that you can simplify this. So, you can divide I mean a 1 by a 2 is approximately equal to 0. So, you divide appropriately you will get terms which involve a 1 by a 2 you can neglect that the other terms will have u factors f 1 to 2 this will be 1 and you will be left with a 1 sigma 1 epsilon t 1 raise to 4 minus t 2 raise to 4 this is what we have used actually brute force we have used this in various problems before dealing with radiation inherent assumption was that whatever is leaving the surface in this problem is going to interact with all other surface it does not come back to it and no other surface is there that is what is the inherent part. So, sigma a epsilon t 1 raise to 4 minus t 2 raise to 4 infinitely large plates a 1 equal to a 2. So, this a 1 can a 1 and a 2 will come up it will be just a simple a if you are talking of f 1 to 2 between these 2 the view factor is equal to 1 what will happen I substitute view factor is equal to 1 take a is up it is just 1 minus epsilon 1 divided by epsilon 1 that will be 1 by epsilon 1 minus 1 plus 1 plus 1 by epsilon 2 minus 1 minus 1 plus 1 will get cancelled. So, I will be left with these 2 terms a 1 equal to a 2 you will have a a here up that a is missing here a is missing here if you want to write it in terms of heat flux then this is the appropriate expression and infinitely long cylinder I am not going to go into all these view factors again area is written in terms of the radius surface area. So, 2 pi r l or pi d l concentric spheres 4 pi r square. So, area ratio is r 1 by r 2 square inside to outside view factor is 1. So, I can do it appropriately this is a simple problem which illustrates the use of this formula direct application. So, I have 2 surfaces 800 and 500 Kelvin determine the net radiation heat transfer between 2 large parallel plates maintain at the same uniform temperature 800 and 500 emissivities are given. So, 1 surface resistance another surface resistance space resistance you put all that you will get this formula directly and I mean you will get this formula and substitute the numbers you will get 3625 obvious this is at higher temperature this is at a lower temperature heat is going to flow from top to bottom. So, this number has to be a positive quantity. So, most of these things are common sensical actually. So, with this I will stop will give it to professor Prabhu he is going to do the last part involving radiation which is multiple surface enclosure re radiating surface business.