 Now, we will go to probably the last part related to the fundamentals of radiation. Fundamentals I am saying because these are concepts which are going to be used in understanding exchange between two surfaces which we will study later on. Hopefully, after 11 o'clock we will begin interaction between two surfaces. So, what did we do? We started with basic definitions intensity, then we said there is something called emissive power, irradiation, radiosity and then we went to more specific things. Overall picture we understood by looking at the broad definitions, emissive power related to emission, irradiation related to energy coming in from all other surfaces on to the surface and radiosity related to sum of emission plus reflected component of all the radiation. After doing this basic definition we said now overall quantity is what we call as total hemispherical, emissive power or irradiation or radiosity. If I say I have it over a particular wavelength I will call it as a spectral quantity that means at that wavelength what is the emission overall directions. If I have a directional dependency and wavelength is not there I will say in this particular direction theta phi that I will call as directional emissive power or directional irradiation or directional radiosity over all wavelengths that is taking all wavelengths into account. The most general one would be spectral directional emissive power spectral directional irradiation spectral directional radiosity which would be a particular wavelength in a particular theta phi direction what is the amount of energy leaving the surface emitted by the surface or coming into that surface. So, these were broad definitions that we used and after doing this we went and studied various other things planks distribution means displacement law, absorptivity, reflectivity, transmittivity and in all these things why why did we study transmittivity absorptivity reflectivity emissivity these four definitions we studied and all these we again had directional spectral total hemispherical all these three categories because the nature of emission nature of irradiation nature of reflection nature of absorption are all dependent on e and g e and g are what dependent on direction in general they are dependent on direction they are dependent on wavelength or they could be total quantity. So, because the source that is emissive power or irradiation is a directional and a spectral quantity in general these definitions for alpha, rho and tau would be as general as possible that is they could be directional spectral or the total values what you are given in problems till now till radiation we were given emissivity is 0.9 when you say 0.9 emissivity that is the total hemispherical emissivity it has it has it has absorbed all the directional effects it has absorbed all the spectral or wavelength effect. So, till now till radiation was studied we had this we did not even know that there was a directional or a wavelength dependency now having heard having seen that we are going to say that whatever we should understand when we did not know convection we took H as a constant value and when we learn conduction convection we realize oh what we are given as an average value there is a local variation similarly now emissivity all these things which we have used so frequently we have to appreciate that these could be spectral directional quantities what you have is an integrated value it is like an average value over all wavelengths over all possible directions. Now, having come to this stage in studying radiation we have to be able to distinguish between epsilon lambda epsilon total that is just epsilon epsilon theta phi epsilon lambda theta phi that means emissivity in a direction and for a given wavelength. So, that distinction I have to appreciate and that distinction I will appreciate if I appreciate the inherent definitions of E G and J. So, with all the story let me go ahead what is Kirchhoff's law all of us teach Kirchhoff's law all of us can even recite in our sleep what Kirchhoff's law is we will say absorptivity is equal to emissivity what is that. So, that means why do we study a second definition for absorptivity emissivity itself is enough right. So, we could have just studied emissivity reflectivity and transmissivity why give one new term called absorptivity that is what is Kirchhoff's law's limitation and unfortunately we forget the limitation when we apply Kirchhoff's law we just say Kirchhoff's law is valid alpha I know therefore, epsilon I know or epsilon I know therefore, alpha I know that is far from the truth Bernoulli's equation fluid mechanics p by rho g plus v square by 2 g plus z equal to constant wherever whichever problem first reaction for fluid mechanics apply that whether there is a pump whether there is a blower whatever it is you will apply that and realize that you are made a mistake Kirchhoff's law also we have to understand the limitations our students do not understand they do not they probably do not even know that such a limitation exists. So, now we are going to see what is Kirchhoff's law and under what condition is this alpha so called alpha is equal to epsilon because it does not make sense to define two different quantities if they are equal right. So, all of us know this whole procedure to derive Kirchhoff's law there is no derivation per se just logical arguments. So, consider a large isothermal enclosure of constant surface T sub s within which several small bodies are confined now all these are important isothermal enclosure why yesterday somebody asked black body why we took isothermal why not half side of a body is one temperature another half another then that will not be uniform now because the small object which is placed inside will interact differently with one half will interact differently with other half which has a different temperature. So, and those two would interact with each other because of the difference in temperature until thermodynamic or until thermal equilibrium is achieved. So, isothermal is a necessary condition then several small bodies are confined why should it be a small body why should it not be a large body small body means what your size dimensions a of the surface divided by a of the enclosure surface area should be very very small we will appreciate this a little later when we do view factor what it means all the energy from the enclosure comes on to these surfaces that is it is view factor is one means that the surface is going to see surface a 1 I am I am a table tennis ball in this large room this table tennis ball is such a small surface area. So, the view factor or the fraction of energy I have not defined view factor that is why I should not be using this word, but since all of us are teachers we understand this whole surface which is a table tennis ball anything going out will be taken from the enclosure taken by the surrounding enclosure. So, that is why the small thing is very important then since these bodies are small relative to the enclosure they have negligible influence on the radiation field which is due to the cumulative effect of emission and reflection. So, there is emission from this isothermal surface there is a reflection of any incident radiation. So, regardless of radiative properties such a surface obviously, we know forms a black body. So, inside a black body inside a black cavity we are having very very small surfaces and accordingly because we have now proved that this large cavity is a black body we say regardless of the orientation. So, whichever position these small balls are table tennis ball wherever they are because the surface of the enclosure behaves like a black body there is no directional preference. So, the body is cavity is a diffuse emitter because that is the inherent characteristic of a black body. So, there is no directional preference. So, the ball in that part of the room or this part of the room is going to be treated the same way by the enclosure. So, regardless of the orientation irradiation experience by the cavity is diffuse and is equal to emission from the black body at that temperature irradiation means what what is coming to me what is coming to me is what is given out by this cavity ball. What is given out by the cavity ball is because the cavity ball is behaving like a black body it is giving out e b at what temperature p subscript s which is that surface temperature is this clear. So, now now it is just useful things. So, under steady state conditions why this again steady state condition if it is unsteady then what will happen there will be interaction between this part of the cavity to that part between this ball and that ball everything else. So, under steady state these balls are there for a sufficiently long period of time and they have reached an equilibrium temperature p subscript s under steady state condition this is very very very important this is probably the biggest limitation of Kirchhoff's law. If the body is at 300 Kelvin and the room is at 1000 Kelvin this alpha equal to epsilon is not valid we will see why it is not valid in this in this slide itself we will understand. So, under steady state condition that means thermodynamic I mean thermal equilibrium has been established thermal equilibrium must exist between the bodies and the enclosures and between the bodies themselves. So, everything is at some uniform temperature you put it at 300 the enclosure was at 1000 there you left it there for about 1 hour everything has become 800 does not matter, but it is all at the same temperature. Hence T 1 equal to T 2 equal to T 3 equal to T s and the net rate of energy transfer to each surface must be equal to 0 that is what is thermal equilibrium only when there is a delta T there is heat transfer now everything is at same temperature does not matter what whether it is a big object or small object there is no energy transfer between one another. So, now I will take this control volume which is around this ball or surface A I will write for surface A and all of you please write with me energy balance energy balance means what E in minus E out I am going to do that only. So, this is a large cavity which is at some surface temperature here let me take one object this is reach thermal equilibrium. So, this is also at T s if I take a control volume around this object E dot in minus E dot out plus E dot generated do not skip any steps we are going to teach our students we should be doubly sure what we are writing this is valid radiation conduction convection does not matter this is valid this control volume steady state has been reached. So, there is no e stored energy generation is 0 because nothing is being generated. So, E dot in equal to E dot out. So, everybody is with me E dot in minus E dot out for this surface 1 equal to 0 let me put another surface 2 I will put another surface 3 also whatever I am doing for surface 1 I am doing for surface 2 and why am I not considering interaction between 1 and 2 1 and 3 because they are. So, small that they have negligible influence on each other and even if they have everything is at same temperature. So, we do not care about emission and absorption from all those things all those things get included in what is coming from the cavity we are just talking about what is the energy coming in to this ball g everybody agrees irradiation g how much is going to be absorbed. So, g is coming in at what temperature is this g it is coming in at the same surface temperature T s is entire g taken by the by the object no alpha times g is what is coming in am I right alpha times g is what is coming in we go back alpha times g is the energy coming in per unit area I will use subscript 1 because we are doing it for surface 1 can I multiply it by the surface area a 1 that will be the total watts that is coming in what is going out steady state no rise in temperature. If it were a regular surface it will have an emissivity epsilon 1 it has a same temperature T s. So, a black body at that temperature will have e b emissive power times a 1 times we put the minus sign here I will put here a 1 a 1 epsilon 1 e b 1 and that is equal to 0 from here I have not done anything see 1 1 1 major problem with this is apart from radiation this I have noticed when I was a student I learnt it the hard way after lot of struggle actually after getting 0s on several problems in heat transfer I have learnt this energy balance is probably the most important thing in heat transfer. If you are able to quantify e in and e out and e stored and e generated for any problem half the battle is 1 after that it is just tools some formula something something you are going to apply concept, but if you are unable to do this budget for any problem you cannot even think of getting a correct solution. Whether it is combined conduction convection radiation convection alone radiation alone convection plus it does not matter any real life problem if I can you take a IC engine if I can put e dot in minus e dot out plus e dot generated equal to e dot stored and write equations write terms may be let us first write in words if we cannot write the equations let us write in what is e in for IC engine combustion of the energy coming out from whatever fuel will be e generated. So, e generated will be mass times calorific value so on and so forth. So, I should be able to put each of these things in words and put this equation if I write this equation and at least budget all the terms whether I know how to solve or not solve is the next step if I do not even know how to write this for a given problem I cannot do anything. So, our emphasis unfortunately in undergraduate courses in heat transfer is brute application of formulae I am not demeaning any college or university I have also come through colleges where we have been taught like that that is we have to change. So, do not give a simple k is given a is given t 1 is given t 2 is given l is given fine q that does not mean anything even a technician who is given a formulae and a calculator can solve it that is not a engineer does not you do not require an engineer to solve Fourier's law of heat conduction or h a delta t problem you do not require an engineer two times you teach somebody even a ten standard fellow will do. So, concepts are what have to be tested please test energy balance everywhere possible even though it comes in the beginning of the course every stage of the course test energy balance whether the student is able to write this and do it for this problem. In fact, one of the tutorial problems of yesterday or today yesterday's natural convection is there I do not know whether it was solved I was not here a rate of cooling of this aluminum slab or something was there first problem in natural convection there was natural convection and radiation problem in that you had to find d t by d t for cooling of the slab. So, we had to go through this step. So, this is cannot be over emphasis this is probably the most important equation. Now, after having written this energy budget a 1 cancels off. So, I will get alpha g t s is equal to epsilon e b t s. So, what is the big deal now is the big deal because g is nothing but e b t s what is this g what is this g this is irradiation I am writing in words just for being clear irradiation from where cavity walls what temperature is the cavity wall t s we have just argued that the cavity behaves like a black body therefore, this should also be equal to e b t s. So, I am done with the derivation alpha 1 e b t s minus epsilon 1 e b t s equal to 0. Therefore, I get alpha by epsilon equal to 1 or alpha equal to this is your Kirchhoff's law we use this blindly without even realizing why we are using it now you are now you see this what I have written when can I cancel when can I write e b equal to g at t s only if and only if see this is written from cavity to object 1 this is from 1 to outside right first term is from cavity to the body only when these two are a of same temperature are at the same temperature can this g be replaced by e b of t s if this is at 1000 Kelvin this is also at 1000 Kelvin if this is at 300 and this is at 1000 we know g is not going to be equal to e b at 1000 obviously right. So, most important overlooked assumption is this business of steady state meaning the objects have to be at the same temperature. So, I do not know how much more to over emphasize this only then can I write epsilon equal to alpha in real life. So, I have to wait till steady state is reached to solve a problem when these are almost close to each other temperatures you will get reasonably ok solution if you use Kirchhoff's law. So, that is the hand waving we do to be able to apply Kirchhoff's law. So, we are just going to read now things from here this relation is called as Kirchhoff's law no real surface can have emissive power exceeding that of a black surface at the same temperature. See here we have written e 1 at t s is nothing but e b times epsilon 1 we have written in our white board already. So, e 1 divided by e b at t s is nothing but alpha which is nothing but epsilon what does it tell me e b is always greater than e 1 we know that and this is also showing that notion of black body as an ideal emitter is confirmed total hemispherical emissivity and absorptivity this is relating the total quantities why total because I do not have any subscripts of lambda theta phi nothing I have not taken any directional dependency I have not taken any wavelength dependency this is the total energy. Therefore, it is total hemispherical emissivity is equal to total hemispherical absorptivity for a surface in an enclosure universe is also an enclosure for a surface in an enclosure, but the restrictive condition in this derivation are surface radiation has been assumed to correspond to emission from a black body at the same temperature as the surface this thing was converted to e b because of this assumption then only I get this. So, please note this please emphasize this to our student everybody can say epsilon equal to alpha, but under what conditions it is valid is what we need to transmit to our students even I think from J level or you know 12 standard also nowadays I think this they teach Kirchhoff's law that alpha is equal to epsilon, but that is actually a wrong way of doing because it gets so ingrained in our system thermodynamics all these things when they are taught at 12 standard level and only you are only giving the equations without telling why how under what conditions what happens is by the time they come to our third year or fourth year whenever those specific courses are there you have ingrained something which is next almost you know it is in the blood a b c d it is become so you have become so conversant with that concept that you fail to see the inherent limitations of that 12 standard J e or C e t you introduce the I think Bernoulli equation is also there probably I do not know, but Bernoulli equation is one of the equation instead of introducing Bernoulli equation if they introduce Navier-Stokes equation I will be very happy because that is the mother of equation then Bernoulli is a simplified form, but anyway we have to put this in our students mind that this assumption of temperature being the same is very important. Now, I have done this for total hemispherical let it be directional let it be spectral I do not care as long as the temperatures are the same I can do the same derivation get epsilon lambda is equal to alpha lambda if I want it in theta phi direction and lambda this is the most general thing spectral directional emissivity is equal to spectral directional absorptivity please remember epsilon is related to my surface which is emitting alpha is related to the surface which is sending. So, that is why this alpha and this epsilon I do not belong to the same surface we say that this alpha is related to what absorptivity of the surface receiving only, but this temperature is what is playing a role in the value of alpha. So, Prabhu had told this sun is going to be at a different temperature your roof is going to be at a different temperature tree is going to be at some temperature you have let us go back to the slide it is actually very important slide actually it tells something which most of us tend to ignore this one this alpha for this thing is more than the alpha for the sun though it is at a higher temperature it is this it is the same roof which is seeing different absorptivity to different kind of radiation. So, that concept we cannot under emphasize ever we have to keep in mind this alpha is for my surface only, but it is dependent on what is sending ok. So, same surface will have different absorptivity for a different type of radiation. So, let us go back. So, Kirchhoff law I think we can now teach very nicely spectral directional all these restrictive assumptions also we will have this statement I will read and it is very tempting to use Kirchhoff law in analysis since epsilon equal to alpha together with rho equal to 1 minus alpha for a opaque surface transmittivity is equal to 0. So, alpha plus rho plus tau is 1 will give me rho is 1 minus alpha rho enables us to determine all 3 properties of the opaque surface from a knowledge of only one property if I know rho I will know alpha if I know alpha sorry if I know rho I will know alpha then I know epsilon this is very tempting logical you know I think so what what is the big deal, but this epsilon equal to alpha gives us acceptable results in most cases. However, care should be exercise when there is considerable difference between the surface temperature and the temperature of the source of incident radiation. So, this is very important this is Gustav Kirchhoff see most of these things work you know you look at fluid mechanics heat transfer any of these things that we are studying now most of this has happened in the late 1800 early part of 1900 and if you look all this has happened in Eastern Europe Germany, Russia Austria those kind of places. So, something is there you know in olden days it is a partly putra or all these places were seats of learning you know same thing probably these were all fertile grounds for all these mines to be there and no wonder all our radiation convection and fluid mechanics prantle all these things that we are studying have come from this part of the world anyway this green house etcetera I leave it to you what is the grey surface this we will use in problems what is the grey surface grey surface is one for which properties are independent of wave length diffuse surface is one where properties are independent of direction. So, if I say a surface is diffuse and grey that is the easiest thing to do nothing varies with direction nothing varies with wave length diffuse and grey surface. Now, a surface can be diffuse and grey for emission, but need not be diffuse and grey for absorption it can emit without any directional preference, but it may not like certain wave lengths of light coming on to it. So, it might filter out something. So, just that is what I am saying epsilon equal to alpha is not secret we have to keep in mind epsilon equal to alpha is restrictive based on certain assumptions I might be diffuse emitter and grey surface for emission, but I need not be a diffuse or grey surface for absorption. So, I might like light coming only in a particular direction I might specifically like only one colour light white light only I might like. So, that becomes very important. So, emissivity of a diffuse grey surface when we say it is the total hemispherical emissivity because it is independent of wave length and direction. So, we leave this total hemispherical thing we leave it now and we just say emissivity means that value absorptivity means that value for a real surface this is now real surface in reality epsilon of theta is not a constant epsilon of lambda is not a constant diffuse surface epsilon of theta is a constant what is diffuse independent of direction. So, I get epsilon of theta as a constant grey surface epsilon of lambda is a constant diffuse grey approximation gives me all these things are equal and we invariably end up using this now commentary on Kirchhoff's law there is some this is just playing with the definitions so that you understand. Now, when spectral hemispherical emissivity how is the definition for that yesterday I think professor Prabhu has taught us this epsilon lambda theta that means it is a wave length dependent directional dependent emissivity in bracket I have lambda theta phi and t y t any of you here y t why this t is there when we remember when we wrote emissive power we had a similar definition but we did not have t t here emissivity is a function of temperature right. So, this is also important so I mean this is spectral directional at that temperature. So, that quantity if I am integrating over all directions will give me a quantity which is dependent only on the wave length if now this quantity I put a third integral and integrate over all wave lengths I will get epsilon at t I will not have this lambda anymore. So, this is the definition of epsilon lambda t spectral hemispherical absorptivity similar fashion please look at these equations together denominator has the same cos theta sin theta d theta d phi numerator also has what is here is epsilon lambda theta here absorptivity is related to incident radiation. So, alpha lambda of theta at lambda theta phi directional spectral absorptivity times intensity of incident radiation coming in that direction and that wave length. So, lambda theta phi divided by I lambda I subscript I is for incident these two when will they be equal this is the definition I have not written anything new Kirchhoff's law says these two are equal when are these two pertinent question is are these two equal I just copied the equation subscript the font is small I just copied the equation put an equal to sign and put a question mark if this is equal when will this be equal when epsilon lambda theta equal to alpha lambda theta it follows by inspection that when the spectral directional quantities are equal alpha and theta when they are equal when alpha and epsilon are equal spectral and directional then I can say this is also equal when irradiation is diffused what about this. Diffuse means let us go back diffused means direction independent. So, if in irradiation is direction independent this quantity is going to be a constant it will come out of the integral sign if this is direction independent it is only wave length independent. So, I am not integrating with respect to wave length. So, this can be treated as a constant denominator also this can be treated as a constant because it is no longer dependent on direction I will cancel these two. So, when the incident radiation is diffused then these two integrals I can cancel of that I the surface is diffused epsilon theta lambda is alpha theta lambda are independent of theta n phi only then is this value means the first condition is a reasonable assumption that incident radiation is direction independent. Let us assume second condition is reasonable for many surfaces especially for electrical non conducting surfaces. So, why this is been given is just to make you understand that this equality is not going to be achieved just like that I have to satisfy a certain criteria which has to be met by the definition of I lambda I or epsilon theta epsilon lambda. Only if this I is not a function of theta can this come out and get cancelled only when epsilon theta lambda and alpha theta lambda are equal can these two be cancelled if they do not have a same functional form then there is no way that I can equate these two. So, that is the necessary condition I hope I have made this clear I am sure there will be enough questions, but we will take them when they come. So, this is now extending this for all wavelengths I have integrated this suppose there was no directional dependency and only wavelength dependency in epsilon lambda alone is there can this be equal to alpha lambda. Remember for epsilon we are having this t always. So, emissivity total if I have taken the directionality away and dealing only with the spectral quantity emissivity is epsilon lambda of lambda t. This subscript lambda reminds me that it is a spectral emissivity this function multiplied by emissive power of a black body at that temperature and at that wavelength divided by emissive power total of the black body at that temperature this is just like averaging. Now, average velocity is total area rho total area times average velocity is equal to integrated rho velocity distribution and area distribution that is what essentially the same thing. When is this equal to total alpha? So, alpha definition it is dependent on g there is no t involved anyway. So, alpha lambda times g lambda divided by integral 0 to infinity g lambda for this to be equal what is the condition that needs to be satisfied epsilon lambda equal to e lambda if that is the case this will be equal provided radiation corresponds to emission from a black body at that temperature in which case g lambda of lambda is e b lambda. That means this functional form should be the same as this functional form then only I can do that. That means this integration would essentially be e b of t only then if I have two functions and they have to be equal every term in the function has to be equal that is all if it is especially if it is an integral function 8 can be written as 4 times 2. But, if it is an integrated function the independent variable dependent variable u v integral d x if I have integral u v d x I have to write and if I write it is integral f g d x u and f have to be equal h n v have to be equal then only the integration is going to be successful same thing here the surface is gray epsilon lambda equal to alpha lambda. So, only when these two are met we get epsilon equal to alpha. So, when I say epsilon equal to alpha temperatures are equal surface is gray and it is diffuse both gray because this one diffuse because inherently I have thrown away a directional dependency diffuse gray surface both are. Now, this figure just tells me how much the I could go wrong if I blindly assume alpha equal to epsilon why we will just see total absorptivity of a surface depends on what spectral distribution of irradiation it depends on what is coming in emissivity depends on what is leaving the surface by virtue of its temperature. Now, if for example, a particular surface may be highly absorbing to radiation in one spectral region and virtually non absorbing in another spectral region probable. So, up till lambda 1 it will have high absorptivity for these wavelengths beyond lambda 1 it could have 0 absorptivity. If this is my spectrum of g in this part till lambda 1 g 1 is having alpha is equal to 1. So, how will I calculate alpha just go back here this is the definition in till lambda 1 alpha is almost equal to 1 times g 1 when I go beyond lambda 1 alpha has gone to 0 that is what is written here alpha is 1 therefore, rho equal to 0 alpha equal to 1 rho equal to 1 alpha equal to 0 rho equal to 1. So, this surf for this irradiation the surface behaves like a reflector whereas, for this it behaves like a full absorber. So, I cannot put epsilon equal to alpha for all ranges that is what we are trying to say. So, it is a very very important thing subtle thing, but very important value of epsilon is independent of irradiation it does not know what is coming in it depends only on that temperature of the surface that surface characteristic. So, there is no basis for stating alpha equal to epsilon for this surface we have we are showing that we are wrong here. We are wrong because this surface has a preferential treatment it likes wavelengths from 0 to lambda 1 only it will absorb afterwards it is going to be completely reflecting why several coatings are there which are there in application several coatings are there which we will use. So, that it reflects energy of particular kind correct. So, you paint a surface black you paint a surface white so on and so forth why that particular color because you want a certain wavelength only to come in other has to go out. So, in that case I cannot take total epsilon equal to total alpha. Having understood all this we will go to two problems which are there I will do one of them step by step other one I will leave it to you it is solved completely I will leave that to you to solve. So, first problem I will do step by step it is an involved problem because you are doing such a thing for the first time that is why I will do it step by step. A diffuse fire brick wall of temperature 500 Kelvin is shown and is exposed to a bed of coal at 2000 Kelvin. I have a fire brick wall and it is at 500 Kelvin the spectral emissivity for that brick wall is shown is given by this graph. So, this is the coal this is a brick wall it could be a combustor or whatever does not matter. So, this is the coal brick wall coal is at 2000 wall is at 500 determine total hemispherical emissivity. So, I have what epsilon lambda versus lambda the moment I have this I should know by now that it is not total emissivity it is a spectral emissivity. So, what is emissivity as a function of the wave length I have been given that function. So, what are you asked to find you are asked to find total emissivity that means integrated over all wavelengths. So, I have to go to the fundamental definition of emissivity fundamental definition of emissivity is what epsilon at t is equal to integral epsilon lambda times e lambda at t divided by e b at t this will be integrated from 0 to infinity I will write that anyway do not worry. Then you are asked to find total absorptivity of the wall to the irradiation resulting from emission by the coal. So, what is the absorptivity of the brick wall to this temperature energy coming out absorptivity is dependent on the source the source of heat is this one source of energy is this. So, that has to be given if this temperature is not given it is difficult to find absorptivity. So, what are the assumptions what is known a brick wall t surface epsilon lambda is known cold temperature 2000 Kelvin you are asked to find total hemispherical emissivity if I write this is not written in the slide please note this will be epsilon without any bracket or subscript or anything epsilon at that temperature total emissive power of the brick wall e of the brick wall e capital E absorptivity of the wall due to the irradiation will be alpha. There is no spectral direction everything is total quantity you will see this epsilon is not equal to alpha both are for the brick wall only, but please bear in mind they are not going to be if Kirchhoff law if I use blindly having calculated emissivity I will have a tendency to write that equal to absorptivity that is not correct. So, let us go let us assume brick wall is opaque and diffuse why opaque opaque means tau is 0. So, reflectivity and absorptivity sum to 1. So, if I know reflectivity and absorptivity that should be equal to 1, but I cannot use epsilon equal to alpha second spectral distribution of irradiation at the brick wall approximates to emission from a black body at 2000 Kelvin. So, this energy that is coming from here to here is approximately equal to black body emission at 2000 Kelvin. Now, I go to the fundamental definition epsilon of t is nothing, but why am I writing this here I will write this from this slide epsilon of t is nothing, but 0 to infinity epsilon of t is nothing, but 0 to infinity integrated over all wavelengths epsilon lambda of t e b lambda note I did not I did not memorize this equation honestly when when I started writing I did not know whether I would write it correctly, but how did I write this I think this is correct let me cross check it is correct how did I write this correctly is simply by understanding this is a function of temperature. So, all these terms have to have a temperature denominator has to be e b it should not be e b lambda because we are talking of total quantity. So, this is total emissive power of the black body at this temperature. So, development has come easily this has to be integrated over all wavelengths. So, this integral has to go from 0 to infinity wavelength therefore, this has to be d lambda we are talking of the product of emissivity and black body emissive power this makes it the real surface the product is what makes it a real surface. So, if e epsilon lambda of t gives me emissivity of this surface at as a function of wavelength times emissive power of the black body as a function of wavelength at that temperature what is this related to this is related to my Planck's distribution correct. So, who gives me this this thing has come to me from Planck's distribution e b lambda as a function of lambda at various t remember this is Planck's distribution. So, that is how I remember this and this has to have a lambda because there is a lambda sticking here this is a function of temperature because black body radiation is a function of temperature that is how I wrote this luckily it was correct, but I use some logic when I wrote this. Now, this has to be split from 0 to infinity does not make sense because the definition here is given for is given for 0 to 10 and then sorry 0 to I think this is lambda 1 it is 1.5. So, this is not clear here this is 1.5. So, this curve this lambda 1 is 1.5 this is 10 the first step occurs is lambda is 1.5 second step is a lambda is 10 and then after that it goes on to it goes on and is a value of 0.8 all the way up to infinity. So, epsilon 1 when I go back to my white board what is this is nothing but sigma t surface to the power 4. So, I will write epsilon of t is equal to 0 to 1.5 micrometer epsilon lambda t e b lambda lambda t d lambda plus 1.5 to 10 the same thing plus 10 to infinity the same thing divided by sigma t to the power 4 what temperature is this whose emissivity I am trying to find coal or the brick wall brick wall. So, this has to be the brick wall temperature. So, sorry sorry sorry sorry brick wall temperature. So, now I will just do 1 term you can do the other terms on your own. So, what is this function epsilon 1 e b lambda d lambda by sigma t to the power 4 integrated from 0 to lambda 1 this thing is given to us by band emission fraction. So, f of 0 to lambda 1 is what is going to be this quantity how do I get this let me just write this separately next page epsilon of t is nothing but epsilon lambda e b lambda term 2 and term 3 I am not writing if I know to do this I can do that who gives me this fraction for that I have to take lambda 1 times t that is 1.5 micrometer times temperature which is 500 Kelvin 500 Kelvin is equal to 750 micrometer Kelvin for this lambda t product I get a fraction correct I get a fraction this thing is nothing but fraction which is obtained from that table which is known to us. So, this fraction comes out to be 0. So, this fraction comes out to be 0. So, epsilon 1 times 0 luckily it is coming out to be 0 does not matter if it comes out to be some other number use that fraction that is it. So, you will get the fraction. So, epsilon 1 f of 0 to lambda 1 second thing is going from lambda 1 to lambda 2 I think you have solved a problem yesterday of band emission. So, I am not going into the details. So, I will just write it for second thing from 1.5 micrometer to 10 micrometer I have epsilon lambda e b lambda of lambda and t divided by sigma t to the power 4 this means I can write this as integral a to b can be written as integral 0 to b minus integral 0 to a. So, this will be fraction from 0 to 10 micrometer minus fraction from 0 to b. Where do I get this fraction from 10 times t 1.5 times t I go to that particular chart lambda 2 t lambda 1 t read out the fraction. So, this is going to be epsilon times f of 0 to 10 minus f of 0 to 1.5 what is this epsilon this epsilon is this function which is already known to me between 1.5 and 10 emissivity spectral is 0.5. So, 0.5 times the difference between the 2 fractions 0.634 minus 0 to lambda 1 this there is a typographical error here this should be 0 it is typed as 1 it is 0 calculation is right the value is typed as wrong and the last one infinity if you are going it is basically 1 minus the fraction associated with that way you get epsilon of t purposely I went slow. Next one you are asked to find total emissive power of the brick wall black body at temperature 500 sigma t b t to the power 4 times this emissivity which I have calculated 0.61 that is what I get nothing great about this. Now is the nice part total absorptivity of the wall to what to a to a radiation coming from the coal that is important. So, this is my definition of total absorptivity. So, I do not have any subscripts or any parenthesis with it total absorptivity is defined as alpha lambda g lambda write it to you write it for you here alpha total is equal to alpha lambda g lambda g lambda lambda d lambda there is a d lambda missing here and here also there is a d lambda missing. So, please make a note of that d lambda is missing in this integral d lambda is missing in this integral. So, wide wide board this is lambda again an average versus that concept only. So, this quantity who gives me all this I do not I have not even talked about alpha what do I have I have epsilon of lambda which is given like this some such function is given. Now what I am going to make an assumption is the following opaque and. So, since the surface is diffuse this is the equality. So, total is not equal, but because it is a diffuse surface alpha lambda equal to epsilon lambda what is the diffuse surface for those of you who have forgotten even I keep forgetting that is diffuse surface is basically a surface where properties are independent of direction only dependent on the wavelength. So, because the surface is diffuse I have put here epsilon theta is a constant therefore, that will be equal to alpha theta. So, this one only I am saying. So, I will get alpha lambda is can be replaced by epsilon lambda I can use the same distribution from there more over since the spectral distribution of this radiation I made this assumption this coal behaves like a back body at 2000 Kelvin. So, the g that is coming out is equivalent to e b at 2000 Kelvin. So, that is what I am replacing this it comes out very nicely because this is now e lambda b at t of coal not 500 Kelvin we are doing for the same surface brick wall only except please note please note this very very important this calculation and this calculation is identical in form except that this gives me epsilon this gives me I am. So, this gives me epsilon this gives me alpha this epsilon is a function of t. So, I am having t in bracket this is not a function of t more importantly this e b lambda has come at what temperature 500 Kelvin e b lambda has come at 500 Kelvin this fraction was done for 500 Kelvin whereas, now this is going to be done for 2000 Kelvin. So, my lambdas are the same, but the t's are different. So, product lambda t is different. So, my fractions are different there will be a tendency to write the same fraction no that is changing because the temperature is changing therefore, I will get these fractions from there and therefore, alpha is 0.395 very important lot of information has come in this problem. The emissivity depends on surface temperature well absorptivity depends on irradiation source temperature. So, what I said this is t 500 Kelvin this is 1200 Kelvin which is the temperature associated 2000 Kelvin which is the temperature associated with the coal. So, that is one point second thing we see now by numbers we are seeing epsilon is 0.61 alpha is not 0.61 it is in fact, 0.395 alpha is 0.395 this is expected since emission is associated with surface of 500 Kelvin maximum spectral emission is approximate less as 6 lambda 6 micrometer this comes from where plant distribution in contrast since irradiation associated with emission from a source at 2000 Kelvin that is the coal coal is come is at 2000 Kelvin it is spectral maximum occurs at 1.5 micrometer that means, lambda max is roughly occurring at 1.5 now you go back to this 1.5 is that. So, for until 1.5 is that. So, 1.5 that energy only a small fraction is being absorbed because only 0.1 we are using here if you look here we have used the same spectral distribution. So, the even though the maximum wavelength is I mean the lambda max is 1.5 the energy maximum is coming at that wavelength only a small part is coming getting into the surface absorptivity is therefore, like that even though E lambda is equal to alpha lambda epsilon and lambda decrease with increasing T s and T c it is only for T s equal to T c that epsilon equal to lambda. So, think enough time we have spent on this concept next problem I will leave it to you please solve this during the break whenever lunch or tea whenever you are free it is solution is there it is hardly two sheets you do not have to do the numbers we want you to understand we want you actually to write this and set the problem up putting like e b means that what temperature g is that what temperature so on and so forth numbers you all can do I know that. So, put this and see whether you have functional form of the solution is the same if you have any questions you want us to discuss this we will discuss this later.