 Any questions? My question is about solid angle definition. It is written like d a n by r square. My question is why it is r square? Or to you? One of the questions by one of the participants is for solid angle y d a. It is d a n by r square. Why is it r square? First of all, definition cannot be reasoned out. Definition is a definition. But still, let us attempt how do we do this? See, perhaps I would think that the solid angle with the increase of the radius, what is happening with the solid angle actually? Increasing. It is no, with the increase of the radius, solid angle should decrease. For a given d n, for a given d n with the increase of the radius, what is happening? The solid angle is decreasing. So, that is all we can offer. See, point is this is a definition. See, why are we defining solid angle? Let me ask a question like this. Professor Arun took as a very good example. That is he took, what is this solid angle? It is all ice cones. I have to fill all the hemisphere with the ice cones. So, how many ice cones I need to fill is defined by, is decided by this solid angle. So, that is how the definition comes through. I think that is all the answer I can offer. Probably one other way of looking at it is that if you look at this figure, see this, this one which is shown in black, let us say that is one area A1 and this one which is in red is another area A2. See, the area is going to be directly dependent on this radius. But whether it is this area or this area, the solid angle subtended at the center is the same. So, what we are trying to say, hopefully is, what we are, what we are, yeah, what I am trying to say is if I look at this black circle and this red circle, this black circle is at radius much smaller than this radius. But the solid angle subtended by both these areas is the same. So, area is proportional to, by definition, any area is proportional to the characteristic length square. So, in this case, the dimension of importance is obviously the radius. So, that is why probably the solid angle, even though for this two areas it is the same, the, even though the solid angle for these two areas is the same, the area is different, the radius is different and that is taken care hopefully by this definition. I think that is probably what I can say. And another thing is, the solid angle is related to the area means the circumference is also changing. So, circumference times, but that is not a correct way of looking at it. I think the area is the correct way of looking at it, yeah. Do you want to add something on to this? No, sir. My second question is about Kirchhoff's law. The problem we have solved is, T brick is about 500 degree Celsius, 500 degree Kelvin and T coal is about 2000 degree Kelvin. Now, there is considerable difference between source temperature and the wall temperature. Is it good approximation to treat Kirchhoff's law that is sigma epsilon is equal to alpha? No, only. Let us, the question asked is, in the problem, in one of the problems, yeah, in this problem only, we are saying here, this is at 2000 Kelvin, this is at 500 Kelvin. Your question is, is it a good approximation to treat alpha, is the alpha lambda equal to epsilon lambda? No, you are saying, although temperature, sir, the question is, although the temperatures are different, how is that I am taking absorptivity equal to emissivity? No, we are not doing that. We are doing, if I am taking the emissivity of, if I am taking the emissivity of this brick wall, I think what I am calculating, emissivity of brick wall, brick wall is sitting at what temperature, 500 Kelvin. Whatever I have done for emissivity calculation, I have taken that, whatever emissivity calculations I have done, I have taken that at temperature 500 Kelvin. Whenever I do this, I am here, I am freely to take emissivity equal to absorptivity provided I have taken both of them at the same temperature. Now, when I go to coal, again, coal is different. Coal, I have taken at what temperature in my calculation, I have taken calculation at 2000 Kelvin. So, point here is that, whenever I make a, whenever I equate absorptivity equal to emissivity, which is what Kirchhoff's law is saying, the back of the mind, the assumption which is involved or the validity of the Kirchhoff's law which is involved is at the same temperature. So, as long as I take it at the same temperature, I will take it at the same temperature. Another thing which I want to tell you is professor, that if I do the same calculation of emissivity, the same calculation for 2000 Kelvin, I will get a different epsilon and that different epsilon has nothing to do with this epsilon, it has nothing to do with this alpha. That you, you can try that at home, actually it is a good question. Please try this at leisure and you will see, this we are taking epsilon lambda at 2000 Kelvin, which is referring to the radiation coming from here. I am saying the absorptivity for this temperature radiation is given by alpha lambda and at 2000 Kelvin, alpha lambda equal to epsilon lambda. I am telling that, I am not saying that it is at the same temperature of 500 Kelvin, that is wrong. So, for the same brick wall, if you do it at 2000 Kelvin, it will be a different epsilon and that will not be equal to this alpha, that also. In case of solid angle, I tried myself, I tried myself in a different way. Suppose, a sector d l 1, subtends an angle d theta 1 at the center of a circle at radius r, then d l 1 is equal to r d theta 1 and suppose d l 2 will take r d theta 2, suppose d l 1 d l 1 2 become the sides of a small rectangle, then the area would be d l 1 into d l 2, that is equal to r squared into d theta 1 into d theta 2. Probably the radiation, the product of this d theta 1 d theta 2 can be called as a solid angle. Is it right sir? The question is, one participant said, for defining solid angle, he took a small rectangle whose sides are d l 1 and d l 2. Let us say d l 1, I think this we have done, d l 1 subtends an angle d theta 1 at the center, that is an arc, d l 2 subtends another angle d theta 2 at the center. So, what he is saying is the area of this rectangle is d l 1 times d l 2. So, we are trying to draw what you have said, what I think is your question is this one, this is d l 1 and this is d theta 1 and I have another arc which is like this, this is d theta 2 and this side is d l 2. So, I think the question is, is the solid angle defined as the product of these two angles or something like that? This is my area d A n which is d l 1 times d l 2, no, that is not the case. This angle d theta 1 is a planar angle. This is a planar angle, this is another planar angle. You can try this at home actually. You take two strips, two triangles and keep them next to one another and you have, I mean two equal triangles and two different other size triangles and you try to make this geometry. d theta 1 will be the central angle of one triangle, d theta 2 will be another central angle of another triangle, but the solid angle definitely is not going to be that. So, I do not think you can connect this planar angle to a solid angle. Because solid angle is a three dimensional concept. It is not two dimensional concept. The very definition of solid angle we have gone to cover the 3D. It is not going to cover, if we can manage with 2D then why do we define in 3D? And the same thing, if I keep this, if I make the, what do you say, the planar angle is different, I can have the same solid angle still. I mean I can have a different solid angle if I diverge it. It is like a petal of a flower, when it is like a bud, the petal has the same solid angle, but when it expands and becomes a flower, the central angle has changed and the solid angle has changed. So, next we will go to Anna University. So, for the black body, emissivity is one. In that case, same time the black body is an isothermal body. So, thermal conductivity is very high. So, in one of the graph shown that for a conductive material, the emissivity is less compared to non-conductive material. Why it is so? I told. See, the question asked by one of the participants is that why is for emissivity, if non-conductors is higher than the emissivity of the conductor? See, I told that this is related to electrical permittivity and magnetic permeability. And I also said that in the afternoon class, I will try to relate this, but I cannot give complete solutions because it requires complete electromagnetic wave theory. You will have to wait till afternoon. Okay, I think we will stop the question and answer session and we will move on with the view factor session which Professor Arun is going to teach us. Yeah, we will go on a very fast note. So, we want you to be very alert while this class is going on because it is essentially geometric. Okay, this aspect of view factor, I think all of us understand, teach our students also quite well. So, I am going to go very fast on this. What is the view factor? View factor essentially, it goes by different name, configuration, factor shape, factor so on and so forth. It is essentially giving you an idea of how much energy from a given source is being received. I am not using the word absorbed. It is not absorption. It is being received by another surface. That another surface could be very close, far oriented differently so on and so forth. Depending on that, the fraction of energy is going to change. A good example always is the tubelite. If you are standing under the tubelite, the tubelite is the same. You are the same person. You will get more energy from the tubelite if you are directly below it. You move away. It is going to be less. Same thing, you are making chapatti or natava. You keep your hand just above it, very close to it. You will feel the heat. If you take it little further away, it will be little less heat. You move sideways away from the pan. It is going to be much cooler. The logic is common sense actually. Now, we want to put it in mathematical form. The physical distance between the emitting surface and the receiving surface, the relative position of the receiving surface with respect to the sending surface, all and the size of course. If I take a spoon and hold it above a pan versus if I take a lid, big circular lid of the same dimension of the pan and hold it, obviously it is going to receive different amount of energy and send different amount of energy. So, it depends on the shape, size, relative positions and the distances of the object. I am not including temperature because we are talking of a fraction. If 100 watts is coming out, 10 watts is being received. If 50 watts is coming out, 5 watts will be received by that object. If it is at the same relative position and if the same shape and sizes are maintained. So, we are talking of fractions not absolute quantities. So, it does not depend on the temperature. So, let us quickly go over these definitions. If you stand in front of the fire, you are going to face maximum radiation. Your back of the body is going to face less amount of radiation. This shape factor, view factor or geometry factor or angle factor is given by F subscript i j. i refers to the sending surface, j refers to the receiving surface. So, subscript i is for the source, j is for the receiver. So, fraction of radiation leaving surface i that is intercepted or that strikes surface j directly, very important directly. It is not reflected from something else in coming. I do not care about that. That I will take as how much of that reflected component is being received by me. So, that is not directly from the source. So, from a source, how much is directly received? F 1 to 2 is referring from surface 1 to 2. F 2 to 1 refers to the fraction from surface 2 being received by surface 1. So, let us take two arbitrarily shaped geometries A 1 and A 2 and this D A 1 refers to a small area in A 1, D A 2 refers to a small area in A 2. They are separated by a distance small r and the solid angle subtended by 2 when I look at it from 1. So, I am sitting at 1. My i is here. I am looking at surface 2. So, what do I see? I see a solid angle d omega. So, I am sitting here. I am having another surface. What is the interaction between these two? You come in front of me. I can see whether you are smiling, laughing, sleeping, everything. But if you are in the farthest corner of the room, I probably cannot see too much. So, that is what is a view factor. So, the outward normal is N 1. Outward normal is N 2. Angle theta 1, angle theta 2. Solid angle is d omega subscript 2 1 of 2 seen from 1. That is why it is 2 1. So, now we are going to just use definitions. Consider two differential surfaces, D A 1, D A 2 on two arbitrarily oriented areas. Distances so on and so forth. Surface 1 emits energy and also reflects radiation diffusely. What is diffused surface? Something nothing is dependent on the direction. Constant intensity i is maintained. So, we are talking of constant intensity. So, solid angle d omega 2 1 that means, subtended by, I am sitting at the vertex of the ice cream cone I am looking out. That is the solid angle. Subtended by this is D A 2, this area. But what direction? Normal. So, if it is, if I am seeing like this and this is oriented like this, I have to make the projection and get the normal area. So, the normal direction of the area is what we talk about. D A 2 cos theta 2 divided by R square. Why? Because this is going to form a cone or whatever you want to call it of radius R. Now, we are going to, first time we are going to start talking of energy transfer, interaction. So, q dot from D A 1 to D A 2. So, what is the energy going from 1 which is going to 2? 1 is going to emit in all direction. You take a gas stove, you turn it on. It is going to show give heat and light to all directions 360 degrees, atmosphere. But what is going in a particular direction coming towards me? So, in that direction which is having area D A 2 oriented at distance R 2, I mean distance R, subtending a solid angle D omega 2 1. So, let us look at it now. q therefore, is nothing but I 1 cos theta 1. Who gives me this? I am not writing it out of magic. Remember my definition of intensity. I, I will just go back. So, even though I will spend 2 seconds going back to the slide, it is important. Intensity is given by D q divided by D A cos theta D omega. Let me see. It is here cos theta D omega, D q, D A cos theta energy leaving the surface q is intensity associated. We have said diffuse. So, this theta and phi business is not there, but essentially the same thing. D q is D A 1 cos theta, sorry pointer D A 1 cos theta D omega into I. That is what I have used. So, nothing new I have used. So, let us just go back to. So, this is I cos theta 1 D A 1. So, this is instead of I theta phi you just have I, I 1 because it is coming out from surface 1 cos theta 1 D A 1 refers to the normal component of that area. D omega 2 1 refers to the solid angle subtended by 2 from 1. That is all. So, now I am going to substitute for solid angle. It will be D A 2 cos theta 2 R square. There you see area 1 has come, area 2 has come. Relative orientations also have come cos theta 1 cos theta 2. So, we are now going to see slowly how mathematically we are going to implement whatever we have told. Dimension relative distance R, relative positions theta 1 theta 2 everything has come. Now, we have to cast it in a nice form. That is all. So, this is Q total rate at which radiation leaves. We do not care whether it is emitted or radiation. I said leaving surface 1. So, actually what is leaving E plus row times G? G is from all other surfaces. I do not care what it is. Now, I am doing thermodynamics like thing. Whatever is going out, whatever is going out is what? J times D A 1 infinitesimal area D A 1 emissive power E plus row times G times D A 1 is the reflected part. These two things are going out. So, that is J 1 D A 1. Now, what I said? Surface is diffuse emitter and a diffuse reflector. Inherent assumption in this whole definition, it diffuse means there is no directional preference. So, this quantity becomes pi times I 1, intensity times pi. How did we get this? Go back to yesterday's notes. The cos theta sin theta D theta D phi when it is integrated from 0 to pi and 0 to 2 pi, 0 to pi by 2 and 0 to 2 pi. You will get this pi definition. That is it. I have now put Q. See, I have got the Q D A 1 to D A 2. Now, I want to say this is what is received. How much is the surface sent out? The surface is sent out this. Division of that by this, this is what is coming out from 1 going to 2 divided by what is being sent out by 1. Is my view factor by definition? Fraction of energy going from 1 to 2 received by 2 from surface 1. So, that is F D F I use because we are talking of infinitesimal areas. That is nothing but the quantity written here divided by this quantity. So, I just substitute all that. I get cos theta 1 cos theta 2 D A 2 by pi r square. This view factor from a differential area D A 1 to a finite area D A 2, I started off with what a small infinitesimal area to another small infinitesimal area. If I integrated this for area A 2, I will get what is coming out from area D A 1 to this whole area D A 2. Then, if I integrate from this over this whole area, I will get a double integral essentially integrated over A 1 and A 2, which will give me the total view factor. Right now, we have got a differential view factor from D A 1 to D A 2. D A 1 to total A 2 will be integration over A 2. D A 1 to D A 2 will be integration over total area. What is the energy coming out from D A 1 which is being received by whole area D A 2. Now, I am going to say we talked of D A 1. Now, let us talk of full area A 1. What is the energy going out from the full area A 1? J 1 A 1. I do not I just throw away the D it is J 1 A 1 which is pi I A 1. From A 1, this whole thing this whole area how much is being received by this area. I can do the same logic logical steps and write Q A to D A 2 is nothing but Q D A 1 to D A 2 which is from here Q D A 1 to D A 2 from this part this this equation. This is for infinitesimal both infinitesimal. I am now writing this this was written for this was written for infinitesimal source to a finite receiving surface. I am now writing this for finite sending surface to infinitesimal receiving surface just because I want to do some algebra. So, this is integrated over A 1 I get this quantity. Now, all I am doing is integrating it again from A 1 to A 2. I therefore, get pi I 1 that is this one fraction I have to put divided by. So, Q Q D A 1 to D A 2 will give me energy coming out from D A 1 received by going to D A 2. Q A 1 to D A 2 will give me energy coming out from total area A 1 going to D A 2. This quantity when I integrate with area A 2 will give me the total energy coming out from A going to A 2. So, that is why this is one integral this quantity I will integrate again this one I put this expression here I will get a double integral A 1 A 2 I 1 cos theta 1 cos theta 2 by R square D A 1 D A 2 I 1 came from this one. So, I 1 cos theta 1 cos theta 2 by R square D A 2 D A 2 who gave me this this is come from fundamental definition of intensity we saw that. So, this was from here. So, if I do the magic fraction F 1 to 2 is nothing, but this quantity Q A 1 to 2 is energy going from 1 received by 2 divided by energy leaving this quantity leaving 1 that is this quantity. So, this double integral divided by pi 1 I 1 A 1 is view factor F 1 to 2 right why because this term refers to quantity heat energy going from 1 received by 2. So, if 1 is sending 100 watts only 20 watts is received that 20 watts is this quantity 100 watts is this quantity what is being thrown out emitted plus reflected by 1. So, view factor therefore, is this one and when I substitute double integral I cos theta 1 cos theta 2 D A 1 D A 2 by R square divided by pi I 1 A 1 pi I 1 A 1 you not yet over. So, F 1 to 2 therefore, is this quantity I is will cancel off I will get 1 by A 1 double integral A 1 A 2 cos theta 1 cos theta 2 pi R square D A 1 D A 2 similarly, I did this I did this for 1 to 2 I can do the same thing for 2 to 1. So, I will write Q A 2 which is J 2 A 2 which is pi I 2 A 2 everything I will write the same way the Q A 2 to A 1 will be double integral A 1 A 2 I 2 cos theta 1 cos theta 2 R square D A 1 D A 2 this when I divide by pi I 2 A 2 I 2 will get cancelled off I will get 1 by A 2 pi R square that will be inside the integral cos theta 1 cos theta 2 what does this tell me this whole thing is the same for both. So, A 1 F 1 2 is equal to this double integral A 2 F 2 1 is the same double integral which means A 1 F 1 2 is A 2 F 2 1 all of us know this reciprocity rule. So, A 1 area of the sending surface times the view factor of sending to the receiving surface is equal to A 2 F 2 1 very useful very powerful rule. So, this is what all students also uses very nicely very easily they use, but when we use this we have forgotten one thing what is that we have made one assumption which we do not even bother to state what is that assumption constant intensity which we can say is related directly to constant temperature. So, if the object is having a constant temperature throughout this A D A 1 obviously, because if I want to pull that eyes out come out of the integral that is an inherent assumption. Now, in real life do I have surfaces where this is constant you can approximate many times, but where you know for sure that surface temperature variation is there in those cases directly using this formula directly the fundamental way of dealing with it is the same. So, that will be different. So, what you have to do is for example, if you take if I have one surface where I have 3 or 4 different regions 1 2 3 4 5 and these are at 5 different temperatures and I want to calculate the view factor between this surface 1 and this surface 2 or A to B I want to do this. So, A to B I want to calculate F A to B I want to calculate and let us say this is a pipe inside another pipe for example. So, if I do this I will get this is equal to 1 F A to B is equal to 1 is what I am going to understand, but if I want to take energy interaction view factor is ok, but real life if I want to take how much energy from this part is going here how much is going to this part how much is going to this part how much is going to this part from here how much is going to each part those will be non uniform quantities not the same quantities because inherently these temperatures are different. So, for such cases I will have to take some kind of a strip element or some kind of a differential area and this interaction with this surface A B C D E. So, 1 with A 1 with B 1 with C 1 with B 1 with P 2 with A 2 with B 2 with C 2 with these one and so on, but summation of all these 25 things should add up to what is going out from here that is a check we have to do and some probably will go out we account for that whatever, but because that surface temperatures are non uniform we have to be a little bit more careful that is all I want to tell. Our U G heat transfer problems we do not deal with such kind of situations. So, it is ok, but when we are dealing with real life situations we have to keep in mind that this non uniform temperature can lead to heat transfer related issues. I am U factor of this surface with this if this is a inside tube U factor with respect to an outside tube A to B is 1 that we are agreeing we are not denying that, but energy transfer I cannot use T 1 and T 2 T A and T B as constant values. So, in that case I will have to take fractions. So, I will have to take energy coming out from here by virtue of T 1 how much is intercepted by each of these surfaces. So, that care I have to take this is reciprocity rule plane surface we know it is not going to see itself. So, view fact everything F 1 to 1 is equal to 0 surface to itself is 0 convex surface ball everything what is coming out goes out nothing comes back a dish saucer it will have a view factor with respect to itself concave surface one cylinder inside another inside to outside will be 1 outside to inside will not be 1 because this surface is going to see itself some part from here is going to be received by this one. So, therefore, that is not going to be equal to all of us know this. So, I am not spending time on this so much then for now if you have geometries various geometries that you come across in literature people have evaluated this how they have done this by formulae by this double integral which they have done and you have this formulae in view factor catalog handbook text books whatever. So, parallel rectangle coaxial parallel same size different size one and so forth. So, perpendicular rectangles with one common edge. So, it is just good bookkeeping if I define these things and put everything properly carefully I will get the answers for the view factors one parallel plate which is large with another parallel plate inclined one plate inclined with another. So, it is like this mirror two mirrors kept next to each other at some angle then right angle geometry three sided enclosure infinitely long 2 D geometry. So, does not matter whatever be the geometry you can get view factor either from catalog if it is not there you have the tools to derive this. So, that is what we are trying to say and now let us quickly come to the relationship first one is the reciprocity theorem or reciprocity relation a i f i j is equal to a j f j i this is surface i to j in general we have made this. So, we use this quite often and with good accuracy with correctness summation rule conservation of energy this is nothing but conservation of energy. So, if fractions if q is the energy coming out from surface i fraction f 1 goes to surface j f 2 goes to surface k f 3 goes to surface l so on and so forth summation of all the fraction should add up to 1 that is what this is. So, view factor summation from surface i to j where j goes from 1 to n equal to 1. So, this could be also with to itself. So, if I have let me not rush rush. So, if I have 1 cylinder inside another 1 is this 2 is this f 1 to 1 is 0 f 2 2 to 2 is not equal to 1 because this is going to see some part of it. So, I can get f 2 to 1 a 2 times f 2 to 1 is a 1 times f 1 to 2 and f 1 to 2 is 1 by summation rule because a 1 1 sorry f 1 1 plus f 1 2 is equal to 1 this is 0. So, f 1 2 is 1 I substitute I get this in terms of the areas which is nothing but diameter ratio. So, I can get f 2 to 1 like this now what we are saying here is summation rule multiple surfaces are there 1 2 3 from the base of a cylinder to the side walls of the cylinder. So, f 1 to 2 plus f 1 to 1 plus f 1 to 3 equal to 0 1 is this 2 is the top surface 3 is the curved surface of the cylinder we know f 1 to 1 is 0 f 1 to 2 plus f 1 to 3 is equal to 1 summation of the energy going from here to here everything should add up to what is being thrown out. So, the fraction should add up to 1. So, that is what is explained by this if I make this a curved surface for example, if I make this a curved surface this 1 2 and 3 3 is the curved surface then from here I will have 1 to 3 1 to 2 and 1 to 1 also all those 3 have to be taken into account in this expression. So, we are doing just careful that is all not rocket science this just tells you that if I have n surfaces in an enclosure we have n square factors. So, this is number of u factors for 3 surfaces is going to be 3 and some of them would become 0 and 1 by inspection. So, I am not going to do this problem it is very straight forward. So, we have done this already for 1 cylinder inside another 1 sphere inside other also you can do like that other thing is this so called super position rule. This super position rule is nothing but splitting of a particular surface into 2 surfaces this 1 has to be a little bit careful basically 1 to 2 comma 3 if I have this split into 2 parts then I can say this u factor from 1 to 2 plus u factor from 1 to 3. So, if I want for example, u factor from this surface to something which is here. So, what I do is I artificially connect this and call this surface 2 plus 3 as 4 f 1 to 4 this I know from a formula f 1 to 2 I know from the same formula with a different dimensions. So, f 1 to 4 is equal to f 1 to 2 plus f 1 to 3. So, 1 to 2 I calculated 1 to 4 I calculated the remaining is f 1 to 3. So, you can do it from fundamentals you will get the same thing the view factor from a surface i to a surface j is equal to the sum of the view factors from that i to different parts of surface j. I cannot do the other way I cannot say view factor of 2 plus 3 to 1 is equal to view factor of 3 to 1 plus view factor of 2 to 1 that is not correct. Why because the fractions are not going to adjust itself like that this is just the definition of you are saying 1 is throwing out energy 1 throws out energy how much goes to 3 divided. So, q 1 to 3 divided by q 1 is equal to q 1 to this whole divided by q 1 minus q going from 1 to 2 divided by q 1 that is the fraction directly. So, that is why this is correct the other thing is not correct. So, please understand this this is not possible. Symmetry rule most of the time by inspection we can use this if the geometry is symmetrical if you have a pyramid base of the pyramid to all the four sides if you have a square pyramid or a triangular pyramid if it is a triangular pyramid the base is going to see all the three sides of the triangle equally. So, I will have 1 by 3 as the view factor. So, here from I to 1 to 2 is same as 1 to 3 because it is identical in geometry that is what is shown here 1 2 3 4 5 this is a square pyramid 1 to 5 is same as 1 to 4 is same as 1 to 3 is same as 1 to 2. So, if I use summation rule all this equal to 1 each of them are equal. So, each of the view factor is 0.25 that is all. So, what is given here? So, we now determine the view factor from any one side to another side of an infinitely long triangular duct. So, again that is straight forward this side l 1 l 2 l 3 if it is an equilateral triangle then this l 1 is going to see 2 and 3 equally you are going to get 50 percent to a non equilateral triangle then you have to take care of the geometry aspects that is it. So, this is given as a general rule this is a general formulation there is one other thing which is called as cross string method which is used for infinite geometries. These many problems encountered involve geometry of constant cross section such as channels etcetera which are very very long in one direction. Such geometries we can consider as a two dimensional since the radiation interaction the end is negligible we do not care about the end effects are negligible. So, this hotel has given as this cross string method the surface of the geometry need not be flat can be convex irregular shape and what it tells me is something very easy f 1 to 2 is nothing, but if I connect these by strings cross strings basically from here to here and here to here and then the straight strings l 5 plus l 6 l 5 is this cross string l 6 is this cross string minus l 3 plus l 4 divided by 2 l 1 is going to give you the view factor. So, summation of the lengths of the cross string between two surfaces minus length of the uncross string divided by two times string on surface I. So, even when the two surfaces share a common edge you have a you can use this by treating this as a cross string of 0 length anyway that I think all of us have understood these concepts also quite easily view factor essentially is a widely used tool. So, that is why I am going so fast all school all colleges all teachers I think teach these things very nicely. So, there is a small problem on how to do this I am not going to spend time on that. Now, we come to after having decided this view factors somebody had asked this question in the morning if I have one surface at emissivity one other that emissivity two and they are at same temperature what happens to the extra energy that goes out of something like that till now till this point from yesterday morning we have just defined quantities related to a given surface surface related things emission radiation irradiation radiosity only now after T we started this interaction between one another surface and we gave what we call as view factors. Now, what we are saying let us now talk of interaction between one surface in another. So, if I have surfaces interacting with one another how do I get the net heat transfer between them that is all is the aim. So, what we are going to do now between me now till end of radiation we will take two surfaces in general couple we will do for two surfaces then we can extend it to three four n surface easiest to deal with is a black body. So, if two black surfaces are there how are they going to interact then we relax this assumption of black make it diffuse gray surface. So, these are two surfaces then we say let us have multiple surfaces and then last concept is this so called enclosure. Enclosure means it is a closed surface and then in enclosure we will have multiple modes convection plus radiation and you can have enclosure problem where one of the size is insulated. So, these from now on will be direct application of the concepts. So, hardly any new definition that we are going to study all the concepts all the difficult part of radiation is over now is the easy part. So, let us look at interaction between two black surfaces two black surfaces any general dimensions A 1 and A 2 temperature T 1 and T 2. So, first thing I have to do when I talk of interaction between any surfaces is to evaluate the view factor. So, that I have to do by geometry by formulae by catalog whatever I do not care I should get this f i j and f j i. So, first step I will do that that does not require knowledge of temperature it requires only knowledge of the geometry relative position between the two orientation so on and so forth. So, assume that I have calculated f 1 2 and f 2 1. Now, we are going to write something which we have written from day 1 net heat transfer between 1 to 2 is equal to radiation leaving surface 1 that strikes 2. I am not concerned with what is going from 1 to this gap between the two going out. What I am saying is net energy transfer from 1 to 2 that is it is what is going from 1 which is received by 2 minus this is net budget 20 rupees has come in you have spent 15 rupees you have 5 rupees so it is a net budget. So, this is going from 1 received by 2 minus what is leaving from 2 received by 1 that is what is going to trans give the net. So, q 1 to 2 we are talking of a black body so q 1 to 2 is nothing but energy leaving surface 1 received by 2 minus energy leaving surface 2 received by 1. So, let me make a example I have say surface 1 surface 2 t 1 t 2 q 1 to 2 is nothing but what is the energy leaving surface 1 which is received by 2. Let us write what is the energy leaving surface 1 a 1 times e b 1 this is the energy leaving surface 1 meter squared watt per meter squared. Of this energy this is going here here here here everywhere of this energy how much is coming to surface 2 f 1 to 2 of this energy is coming to surface 2 one way path you go to your grandparents house you buy something and go your grandparents give you more gifts that is the net. So, after you make a trip your net worth is more either in terms of cash or in terms of gifts that is what is happening surface 1 has given energy surface 2 says no no I will give you something back. So, that something back will be a 2 e b 2 it is giving out, but it is you know it does not know where 1 is it will give it an all possible direction by virtue of it is temperature of that what is coming to surface 2 for surface 1 is f. So, 100 watts is coming out from 1 50 has gone to 2 80 is coming out from 2 only 20 is coming back to 1 does not matter, but something is coming back this difference is going to be the net heat transfer. So, I will get f 1 to 2 a 1 e b 1 minus f 2 to 1 a 2 e b 2 is equal to q 1 to 2 now these two are equal. So, I can write this as what a 1 f 1 2 e b 1 minus e b 2 correct a 1 f 1 2 is same as a 2 f 2 1 that is q 1 to 2 is this and we have used this 100 times we have used this so far e b 1 e b 2 sigma t 1 raise to 4 sigma t 2 raise to 4 have we not used this we have used this except this was not there that is why remember we said all our conduction convection problems radiation we blindly used the formula epsilon a or forget even epsilon sigma a t 1 raise to 4 minus sigma a t 2 raise to 4 what was the inherent assumption this is 1 it means I have some pipe which is going inside this room view factor of this pipe with this room is 1 whatever energy is thrown out by this pipe everything is going into this value was made to 1 force to 1 that is understood we did not bother to write it but the parent formula is this parent formula is in fact the more general formula when it is the gray surface where epsilon are there, but for a black body this is the formula. So, this we have used without knowing what we are using this is where it is coming from. So, a 1 f 1 to 2 sigma t 1 raise to 4 minus e 2 raise to 4 and now you tell me logically if 1 is at 1000 and 2 is at 800 Kelvin what is the net direction of heat 1 to 2 because 1 is going to throw away more energy 2 is going to throw away less amount of energy. So, net heat direction would be 1 to 2 this would be a larger number this would be a smaller number and eventually at steady state 1's temperature would go down 2's temperature will increase correct all of us know this actually you know the problem with why radiation is felt or why any subject is difficult is because we fail to get in touch with ground realities logic which all of us have applied for most of our lives common sensical examples are forgotten that is the problem. So, now if I do this with n black surface why only to have n surfaces I am going to have this thing summed up correct I will write it for benefit of those who are finding it with a little loss 2 minutes is not going to matter too much here. So, I will write I have written next page. So, surface 1 to 2 I wrote which is a 1 f 1 to 2 sigma t 1 raise to 4 minus t 2 raise to 4 if I have 1 surface 2 3 4 I will do the same thing for each I am concerned with 1 2 is going to interact with 4 I am not concerned about that at this stage what is the net heat transfer to 1 when it interacts with this this this. So, you go to your go to a wedding function that person gives you a gift your grandparents give you a gift some aunt gives you a gift everything that is interactive between there you do not care about their transaction same thing here 1 to 2 1 to 3 1 to 4. So, let me write 1 to 2 I have written 1 to 3 I can write similarly. So, I will make this a 1 f 1 to 3 sigma t 1 raise to 4 minus t 3 raise to 4 plus a 1 f 1 to 4 sigma t 1 raise to 4 minus t 4 raise to 4 exactly the same way and if I sum this up this is going to be summation of a i f i to j t i raise to 4 minus t j raise to 4 j is equal to 1 to q 1. Let us say this is at 800 Kelvin this is at 300 Kelvin this is at 600 Kelvin this is at 1200 Kelvin I cannot now intuitively predict whether this temperature is going to rise or come down right because this is losing heat here net net will be in this direction between 2 and 3 there will be a net interaction like this between 4 and 1 logically this is going to give more energy, but we have not looked at interaction between these things. So, I have to do a complete analysis right now with just 4 surfaces in an enclosure I could write like this net heat transfer. So, let me go back and that is what is written here I do not think there should be any questions regarding this. So, in an enclosure it is determined by adding the net radiative transfer from the surface i to each of the surfaces in the enclosure negative value indicates that this is net received is more positive value indicate energy going out from that is more. If it is 0 whatever is coming in equal to going out if both objects are at same temperature there is no heat transfer we know that. So, that is what it is mathematically sometimes you can get 0 here if the temperatures are adjusting themselves like that. So, I have now writing this for a 3 surface enclosure which are black. So, q dot 1 is nothing but summation of energy transfers from 1 to 2 1 to 3 1 to 1. So, I will write this I mean I will go through this slowly because this is the first time you are going to formulate something like this. So, q dot 1 energy going from 1 to 2 1 to 3 1 to itself this is a flat surface 1 to itself is 0 2 to itself is 0 3 to itself is 0, but if this was a curved surface 1 will also see itself. So, this is now going whatever I have written on the white board see here I have written 3 terms. So, this is 1 to 2 this is 1 to 3 and this is 1 to 4 I will recast this now I do not like it in this form I will recast this this q is nothing but this is the net energy interaction from 1. So, q 1 q dot 1 what is going out from 1 I will write t 1 raise to 4 minus t 2 raise to 4 sigma divided by 1 by a 1 f 1 2 plus sigma t 1 raise to 4 minus t 3 raise to 4 divided by 1 by a 1 f 1 3 plus sigma t 1 raise to 4 minus t 4 raise to 4 divided by 1 by a 1 f 1 4. Why am I writing this like this you will see a little later today. Similarly, I can write for q dot 2 and q dot 3 in that triangular enclosure 1 2 3 3 sided black surface I can write all these things. So, if I if I write all this I will get some expression where some things will go to 0 because view factor of 1 to 1 will be 0 I will get interacting with 2 and 3 2 will interact with 1 and 3 3 will interact with 2 and 1. So, I have a set of equation which I can solve to set whatever I want. So, let me do this problem because this is the first time we are doing interaction between multiple surfaces you have a 5 meter by 5 meter 5 meter cubical surface whose surfaces are approximated at black bodies base top and side surfaces are at 800 1500 and 500 Kelvin base is at 800 top is at 1500 and sides are at 500 Kelvin 1 2 1 is base 800 Kelvin 2 is top 1500 Kelvin and 3 4 5 and 6 are all side surfaces which are at 500 Kelvin. So, 3 4 5 and 6 all are black. So, this is black body all black epsilon is 1 for all surfaces. Then determine the net rate of radiation between the base and the side between the base and the top and net radiation heat transfer from the base surface. What does this mean? What are the three questions? Actually you know problems we solve, but many times we do not understand what we are solving for net heat transfer between base and the side. So, when I am when the base is interacting with the side how much is the net heat transfer from the base to the side that is all. In size they are not there what is the what is the talk between base and the top. Third part is overall complete net interaction between what is the net heat transfer from the base surface. So, it involves heat transfer from the base to the sides plus the top. Assumption isothermal surface because every point on that surface is that same temperature given to us surfaces are black there is no convection between any plates. So, this cube is evacuated inside. So, there is no convective heat transfer that is important. So, there is no convective heat transfer inside the cube and all surfaces are at uniform temperature. So, if I do this thing I have to calculate first view factors that is no problem absolutely not a problem. So, first step in all these interaction problem whether you please tell our students you know if they leave a question blank the only option we get is to give them 0. So, at least if they calculate view factors in these radiation problems they will get some marks blank is 0 anything they write will at least give them some mark greater than 0 that does not mean copying the question in their own words it means something more useful if they write. So, view factors I think nobody has to remember catalogs are there formula is there data book is there. So, with the geometry given at least the view factors the guy should be able to calculate. So, what are the view factors associated let us take this surface 1 1 to 1 is 0 because it is a plane surface I can also write 1 to 3 is same as 1 to 4 is equal to 1 to 5 is equal to 1 to 6 can I write that because it is symmetry also I can write summation of f 1 to j is equal to 1. So, view factor from 1 to all surfaces when I add is going to be 1 f 1 to 1 plus f 1 to 2 plus f 1 to 3 1 to 4 1 to 5 1 to 6 I can call that as 4 f 1 to 3 that is equal to this is already 0. So, I get f 1 to 2 plus 4 times f 1 to 3 is equal to 1. So, f 1 to 2 if I determine from a catalog formula I can use this equation and get f 1 to 3. Conversely if I go to a catalog and determine view factor between 2 surfaces which are like that I substitute it here I will get f 1 to do whatever you like, but you can get the view factors easily and those view factors are computed for as already this problem. So, f 1 to 3 f 1 to 1 is 0.2 and all sorry I did not get you f 1 to 2 is computed as 0.2 and all the view factors I will get whatever I need from this formula or there are graphs most books carry some graphs which are there. So, you can do the up or eyeballing from the graph and get the form get the view factor does not matter whatever it is the view factors are obtained. So, let us see what they call this 3 is the side surface all the side surfaces we are treating as 1 surface that is also that is also something we can do why which rule tells me that we can do it is that splitting superposition which we call. So, f 1 to 3 plus f 1 to 4 plus f 1 to 5 plus f 1 to 6 can be treated as f 1 to 1 full surface which has a view factor which is given by 4 times view factor associated with that surface because it is symmetrical. Now, knowing that this view factor from base to top is 0.2 view factor from base to the all the sides put together is 0.8. It so happens in this case that all the view factors come out to be 0.2 individually also, but if the cube is not a cube and it is a long structure where the distance between the base and the height is longer than the size then appropriate geometry will give you that these two are not going to be equal 0.2 it will not be the case anyway. So, view factor is immaterial now it is it is trivial. So, I calculate energy interaction between 1 and the side surface all sides have been told. So, that area a 1 where do I get this from I get this from this simple formula and again please do not ask students to memorize this black surface one formula gray surface another formula green surface and no that is not the way they should be asked to do problems. There is no need to memorize anything I did not memorize anything I just wrote this one this all what I wrote. So, if I have two surfaces I write another term that is all. So, what I am telling them is from fundamentals we should emphasize again and again do not mug up anything and do not if you can if you are able to do this two steps I think you have learnt heat transfer calculation anybody can do as I said even a technician will do calculations. So, this q 1 to 3 is given by area view factor of 1 to 3 sigma t 1 raise to 4 minus t 3 raise to 4 it comes out to be 394 watts from 1 to 2 is area 1 f 1 to 2 sigma t 1 raise to 4 minus t 2 raise to 4. So, 394 kilo watts is going from the base to all the 4 sides put together, but from base to the top energy is not going and that is logical because temperature is associated with top surfaces 1500 Kelvin temperature associated with the base 800 net heat flow is from top to bottom and that is exemplified by this minus sign here minus 1319 kilo watt energy is coming in 394 kilo watt is going out from base to all the sides. So, this interpretation a student should be able to give when we said you know in the first thing we should ask we should write comments what comments will 394 watt is the heat transfer no this sign is something which we have to notice. If this may be made a mistake and got 1319 we should open our eyes and say oh this is wrong because I expect heat to flow from high to low temperature. So, this should have been negative I should go back and check the sign if we close our eyes and just punch numbers in calculated it is whether it is 1319 or minus 1319 it is not going to matter. Magnitude is correct but the direction would be wrong. So, now finally the net rate of heat transfer net radiation from 1 is what what is going from 1 to 1 plus what is going from 1 to 2 plus what is going from 1 to 3 1 to 1 is 0 1 to 2 is 1 to 2 is this 1 to 3 is this I add minus 925 kilo watts what does this tell me that heat is coming even though this is throwing out 394 watts to all the sides it is getting so much more that with time per every second I am getting 925 kilo joules of energy from the top surface. I am getting minus 1319 actually from the top surface but I am throwing out something. So, my net thing is minus 925 kilo watt or 925 kilo watt is coming in what does this translate to this will translate to rho ct dt by dt later on it will translate to a rise in temperature and thermal equilibrium will be established after that. So, these the problems are trivial as you see you know it is nothing you factor anybody will evaluate there is some formula 4 number there you can substitute and get that is not the issue issue is this interpretation. So, our students should be able to understand what these means and the question is worded such that you have to just interpret it carefully. So, we have done for black surface the black ok. So, we will go to NIT Tricci now if you have any questions please shoot. Sir this is regarding the view factor two surfaces that possess symmetry amount a third surface will have identical view factors from that surface. Is it true for those two surfaces which are made of different material go over to you. If question is if two surfaces are oriented symmetrically with a third surface we say that the view factors are identical. Now is this statement true if those two symmetrically oriented surfaces are of different materials to let us go back to the definition of view factor what is view factor it just is a fraction of energy coming out from one surface which is intercepted by another surface. If the material is different it is going to absorb reflect transmit differently, but it is not going to see it different ok. So, let me just say I will just explain this by simple drawing forget view factors or radiation if I just say here I am drawing a house with a roof. This is made of tiles and this is made of you know coconut husk whatever you know the olden type roof you sit here and you see the floor you sit here and you see the floor it is going to be the same. If I sit here and see here and here it is going to be the same it is all the perception. Yes what energy is going to come from here all those things the fraction if 100 watts is coming from here in and only 50 watts is coming from here the fraction that is absorbed which is received by this will be different, but I mean the magnitude of energy will be different, but the fraction would be the same that is the essence of view factor. Ok quickly any other question here. Sir in the last problem what we discussed so q dot 1 3 we have calculated 394 kilowatt so would we multiply that by 4 because 4 side surfaces. Ok question is for q 1 3 we have 394 kilowatts should we multiply by 4 see individual view factors are 0.2 already we have multiplied by 4 so we have already multiplied that by 4 and 0.2 into 4 only I have got it as 0.8 so again I need not have to multiply by 4 I should not be multiplying by 4. Ok next question please. The Stefan Boltzmann law is applicable only for thermal radiation or it can be used for nuclear radiation also. The question is Stefan Boltzmann law can it be used for nuclear radiation in addition to thermal radiation. We said that Stefan Boltzmann law what is Stefan Boltzmann law saying what is this emissive power we are saying emissive power is a function of temperature. If there is no temperature you put temperature equal to 0 no matter what wavelength you take in that Stefan Boltzmann law if you put the Planck's distribution what will we get we will get 0. So why because this accounting is only for thermal radiation for not for nuclear radiation so this Planck's distribution is only for thermal radiation that is all the answer. Next question please. Sir regarding the absorptivity whether it the surface property is dependent will affect the absorptivity of the material. Ok see absorptivity whether surface property will depend see actually question itself is little open ended I think perhaps you mean see there are two surfaces I have absorptivity of the surface that is surface material itself and I have a source source also can have a material if it is sun it is a different issue but source also can be another hot plate so but what we are saying is that the absorptivity of the surface on to which it is incident upon it is dependent only on the material property of the source and the temperature of the source that is all it is it is not dependent on the sink receiver. Now a question is whether if the receiver is a polished surface and the surface is with rough very rough then there will be a difference or it will be there will not be any difference at all. Question is if the receiver surface is rough or if the receiver surface is polished whether they would be there would be any change in principle there should not be any change absorptivity is dependent on the source temperature only largely nothing will change but what will change is emissivity will change so you have given a very good idea for getting high absorptivity but for low emissivity you take a you take a polished surface absorptivity of that polished surface can be very high but emissivity can be very low because I have made it polished so you have given through your question you have given a very good solution for a surface which can which can have high absorptivity but low emissivity. This is regarding the radiation heat exchange between two black surfaces it is mentioned in the text that it should be separated by a non-absorbing medium for a textbook it is mentioned it is non-absorbing medium non-absorbing medium here does it mean vacuum or a medium having very low absorptivity and if it is not a it is not a non-absorbing medium there is some absorption then what will be the what will be the expression or what changes are incorporated. The question is in an enclosure if I have a medium which is absorbing medium what will happen if it is not absorbing or if it is absorbing what will happen to my temperatures of the body temperatures of the two walls question number first question the answer for this question is what you are essentially asking is if it is non-absorbing means loosely it is being used there as non-participating medium. So the medium in addition to being absorbed it can also transmit it can also reflect but here absorptivity is the one which is going to affect more but nevertheless reflectivity and transmissivity also are going to affect. What we are saying is in this enclosure if it is non-participating medium means whatever is being emitted is not going to get hindered by the medium which is there in between these two surfaces it can be absorptivity that is nothing is absorbed it can be transmissivity that means everything is transmitted and if it is reflectivity nothing is reflected back. So it can there is that is why transmissivity is one so nothing is reflecting that is why people are saying it is that that as non-absorbing medium. Presence or absence of the medium does not matter that does not matter. So whatever is being emitted by the surface without knowing that there is a medium there it will reach the other surface that is what we mean by non-participating medium but usually for non-participating medium people use it as non-absorbing because if there is no absorptivity and if transmissivity is 100 percent and no reflectivity is there it will reach the other surface and now if that medium becomes participating medium how much it will reach definitely everything cannot reach something is getting absorbed and something is getting scattered. So it becomes very difficult it becomes a complex problem precisely for this reason in the radiation whatever we teach for the under graduates or post graduates in radiation we usually take non-participating medium. We will we are closing the session now.