 we will move on to radiation 2. So we are going to introduce concept of black body. So in the black body what is black body? What is black body? The black body is what is black body? We have understood that any body which is at a finite temperature is going to emit radiation. Now the radiation is dependent on various other things that is material of the body whether the body is surface, the surface is rough or smooth because that is going to decide the reflectivity and the absorptivity. And the amount of the radiation emitted by a surface depends on its temperature, material and the roughness of the surface. We will worry about these parameters little later on. But first thing is what we have to consider the ideal case always I take this example in the class in thermodynamic cycles whenever we are teaching we always refer back to Carnot cycle. Why Carnot cycle? Because the efficiency of the Carnot cycle is highest. I cannot get anything more than that that is the ideal situation. Similarly here I have to define an ideal situation or an ideal body which cannot emit any other body should not be able to emit more than this or any other body should not be able to observe more than this that is the ideal situation that is the black body. Black body is serving as a standard. Black body serves as a standard against which radiative properties of real surfaces can be compared. Why because black body? Now let us see what is black body? What is so ideal about a black body? Black body absorbs all the incident radiation which falls onto it regardless of the wavelength whether it is smaller wavelength or larger wavelength all the incident radiation is absorbed by a black body. The thing reflected that is the point what we understand from this and for a prescribed temperature and wavelength no surface can emit more energy than a black body. So although the radiation emitted by a black body is a function of wavelength and temperature it is independent of direction that is black body is independent of direction. There is no directional sensitivity so black body is a diffuse emitter three points that is it absorbs all the incident radiation second point it no body can emit more than black body for a particular wavelength and temperature that is very important and it is not having any directional preference or dependence it is directionally independent. Black body is a diffuse emitter these are the three points which we need to note down for black body. Now the question is do we come across any black body in real life? Yes, yes what is that? That is I think the cavity. Cavity can be I will cover this and then go back to spectral distribution of Planck's distribution that is if I take a large cavity and now let us say there is a small hole in this cavity what is happening whatever is entering into this first let us see how will become a perfect absorber whatever is entering into this whatever radiation is entering into this cavity what should happen? What should it happen? It has to be absorbed if it has to be a perfect absorber that is the first definition of our black body what will happen if it comes into the cavity it undergoes multiple reflections and it gets absorbed completely so nothing is lost so nothing so the cavity can be visualized as a perfect black body. Now how can it become a perfect emitter? Now whatever is coming out in the same same way with the way we argued for absorption same way whatever is getting multiply reflected can also get reflected and get out through this hole so that is why it is called as a perfect I mean it can act as a perfect emitter that is I will rephrase this radiation coming in through the opening area undergoes multiple reflection and thus it has several chances to get absorbed by the interior surfaces of the cavity before any part of it can possibly escape. If the surface of the cavity is isothermal the radiation emitted by the interior surfaces streams through opening undergoing multiple reflections that is why it is diffused in nature number one and it is a perfect emitter number two it has no preference in direction that is why cavity acts as a perfect absorber or emitter and I also take this example always if I have to calibrate heat flux sensors radiative heat flux sensors what do we do? We create a cavity we create a cavity and that is a thin metal foil I try to keep it spherical I try to keep it spherical and apply and try to maintain that at constant temperature and put my heat flux sensor somewhere here this is my sensor. So whatever this sensor is seeing is essentially because of radiation of course this chamber is now evacuated why do I say evacuated because I need to exclude natural convection if there is no medium inside there is no natural convection then this cavity can be considered as a perfect black body against which why I say perfect black body because I can get whatever voltage into current upon the surface area I have this is my net heat flux which is falling on to my body so that is how I can calibrate my heat flux sensor against this black body against this black body so that is one of the examples where in which cavity can be used for calibrating heat flux sensors and cavity can be used as a perfect black body. So with this we will move on and try to understand what is called as Stefan Boltzmann law. So what does Stefan Boltzmann law say that Stefan Boltzmann law says that radiation emitted by a black body per unit time per unit surface area is that is given by sigma t to the power of 4 that is eb equal to sigma t to the power of 4 what is sigma sigma is 5.67 into 10 to the power of minus 8 what is the unit watts per meter square Kelvin to the power of 4 watts per meter square Kelvin to the power of 4. So little later we will understand how did we come across this 5.67 for now we can just take it as a relation which was given by Stefan Boltzmann okay actually this was first found empirically that is through experiments that was found by Stefan and Boltzmann found out this through theory that indeed the sigma is 5.67 into 10 to the power of minus 8. So it came through experiments first and it was explained through Boltzmann statistics through later on by Boltzmann himself and Boltzmann happens to be student of Stefan and Stefan is his teacher. So I just want to stop here for a minute and tell who is Stefan and who is Boltzmann Stefan is an Austrian physicist and of course Boltzmann is also an Austrian physicist and Boltzmann worked with Stefan. Stefan took all the Sun's data whatever was available at that time and did the simple Karoo fitting and got that eb equal to sigma t to the power of 4 but why others could not do this no they could not do this why because in all the experimental data there was conduction losses which was involved he was the first person to measure the thermal conductivity closest to what we have thermal conductivity of air closest to what we have today that is k equal to 0.02635 he got 0.025 because he could measure the thermal conductivity of air he could rule out the conduction losses which were there in the experimental measurement he removed those conduction losses convection losses and got the radiation alone and did the Karoo fitting and he could get that sigma as close to not exactly 5.65 little lower than 5.65 5.67 and Boltzmann derived this as 5.67 and Boltzmann was the student of Stefan so through statistical mechanics or Boltzmann statistics he could arrive with this to this number sigma of 5.67 into 10 to the power of minus 8. So with this we will move on what is the distinction between idealized black body and an ordinary black body so point here is in this transference I will not read this the point here is all black bodies that is black painted bodies need not be black and all white surfaces visible through visible eye whatever white surfaces are visible whatever white surface we see or perceive through eye or not we cannot consider them as non black bodies no snow is a perfect black snow is emissivity is one that is it can act as a perfect emitter so it cannot be thought of as non non black body so whatever we see the color of the surface does not mean anything about idealized black body or all black bodies need not be ideal black body that is the point what I want to tell through this transference okay and we have seen this perfect cavity now what is spectral distribution here we need to go little slow I want to write this relation that is I want all of you to write along with me so I lambda comma B that is I lambda comma B equal to I lambda comma B equal to 2 h c naught squared 2 h c naught squared upon lambda to the power of 5 lambda to the power of 5 upon lambda to the power of 5 lambda to the power of 5 exponential of h c naught by exponential of h c naught by lambda k t minus 1 okay this was given by Planck's distribution that is why it is called as Planck's distribution what is h this is universal Planck's constant I have presented this in the morning 6.6256 into 10 to the power of minus 34 joule second and k k is not thermal conductivity 1.3805 into 10 to the power of minus 23 joule per Kelvin this is universal Boltzmann constant c naught is velocity of electromagnetic wave in perfect vacuum that is 2.998 into 10 to the power of 8 meters per second t is in absolute temperature in radiation one important thing we need to remember is that temperature is always in Kelvin temperature is always in Kelvin if we take degree Celsius we are gone so we are going to give way off from our actual result so with these constants what is that we see from this intensity of radiation coming out from a black body is a function of only wavelength and temperature the function is looking something like this and of course wavelength is sitting here also it is a non-linear function now this was given by Planck now what do we do with this let us plot this so who is Planck of course in radiation there are no many noble prices okay everyone has got a noble price Planck has also got noble price for giving the Planck's distribution so anyway I will not spend too much of time because of velocity of time you but I usually give lot of time for historical perspective in the class but I would request you also to give lot of historical perspective because the students understand only when we give the historical perspective so he got noble price in 1988 for this Planck's distribution now what did he do what did Boltzmann do so e lambda, b lambda, t equal to pi i lambda comma b lambda comma t what does this lambda comma b represent lambda means that intensity is a function of wavelength and what are whose we are talking about we are talking about black bodies and it is a function of wavelength and temperature that is given by just now whatever I had written that is h c naught squared upon lambda 2 h c naught squared upon lambda to the power of 5 exponential of blah blah blah but where did this pi come from pi i lambda comma b we said that black body is a diffuse emitter so emissive power in the morning professor Arun has derived it for us e equal to pi i if it is a diffuse emitter so black body is a diffuse emitter so I can take emissive power equal to pi i lambda comma b and that pi is sitting here so 2 h c naught squared I am calling that as c 1 and h c naught by k I am calling it as c 2 so that it looks little easily it looks amenable little bit so c 1 upon lambda to the power of 5 exponential of c 2 upon lambda t minus 1 so that means what all these constants I know that is c 1 and c 2 I know and they are represented here all that I get is emissive power is a function of wavelength and temperature now let us plot this as a function of wavelength and temperature that these plants distribution so what is this y axis the y axis is emissive power that is wad per meter spectral emissive power sorry it is spectral emissive power e lambda comma b that is wad per meter squared micrometer and x axis is wavelength you see and I am plotting it for different temperatures I said that emissive power is a function of both wavelength and temperature so for different temperatures if you see any temperature one of the things which we see here at any given wavelength emissive power is increasing with the increase of temperature number one and for a given temperature emissive power increases initially there is a maxima and subsequently decreases all these maxima if I connect them then I if I that is a linear line then if I curve fit that I get lambda t that is lambda maximum that is the wavelength that is it is maximum lambda maximum into t is a constant that is what is called as Wien's displacement law lambda maximum t equal to 2897.8 micrometer Kelvin now here I want to stop and I am answering a question perhaps you have this question here when we said in electromagnetic spectrum we said that only infrared waves and visible light and a little bit of ultraviolet light ultraviolet light is responsible for thermal energy that is what we said students usually ask the question why only this why not x-rays no you see here they cannot other than these waves they do not carry any thermal energy there is no emissive power contribution after this wavelength what is the emissive power 0 after this wavelength what is the emissive power 0 that is the reason okay so here we have been told to just do some upload a different file and get back to this file and what I am doing is I am opening radiation one it is for good it is for good in radiation one I want you to show the spectrum that is you see in the spectrum what did we say the thermal radiation blessing in disguise what you said is right I was lazy to go come here actually so thermal radiation is contributed by ultraviolet infrared so this only why these two are contributing for thermal radiation I said I will come back to this question later on this can be answered through Planck's distribution in Planck's distribution in Planck's distribution only within a particular wavelength in a particular wavelength you see that is for all wavelengths lower than around 0.0 what is that 0.08 I is that 0.008 yeah yeah 0.08 and may be about 1000 micrometer the contributions for thermal radiation is negligibly small that is why we do not say that that is why we say that only these waves contribute for thermal radiation this can be explained only through Planck's distribution otherwise it is quite difficult to explain why only these waves are contributing for thermal radiation no matter whatever is the temperature you take an x-ray there is x-rays there is no contribution for thermal energy that is the point you can understand from this Planck's distribution with this with these insights let us move on to of course again Wien also has got Nobel Prize so no so many Nobel Prizes in radiation okay he has also got Nobel Prize in 1911 actually Wien's displacement law was found before then before the Planck's distribution so always student asked why Wien has got although it is a very trivial thing compares to Planck's but he had done that first compared to Planck's distribution that's why Wien has got Nobel Prize okay now if we take this if we take the bull this Planck's distribution and integrate this if you are lost what did we do this is what the emissive power we got by putting the intensity of radiation through Planck's distribution if I integrate this for all wave length what wave length the wave lengths which are contributing for the thermal radiation let me let me go there and get that wave length because I am not able to remember that wavelength but I just want to go and do that otherwise I will be just hand waving that that is 0.3 micrometer to 3 micrometer that is from 0.3 micrometer to 3 micrometer that is if I have to be very clear up to 10 to the power of minus 5 micrometers to oh there is a problem in this transparency okay so everywhere it is 10 to the power of minus 5 so it is between 0.3 to not 3 15 micrometers it is 0.32 I think I can answer this the previous transparency yeah that's right so that is from 0.3 micrometer to around 10 to the power of minus 8 micrometer is the wavelength range in which thermal radiation contribution is there okay so with this we will come back to radiation 2 now what do I do is I will take the spectral distribution and integrate over complete wavelength if I integrate that I am going to get sigma t to the power of 4 that's what I am going to get I think that you can give as an exercise for the students we have not put that you can give that as an integral that integration can be given as exercise that is you take although it is 0 to infinity what my point is it is not 0 to infinity even if you take 0 to infinity you will be getting automatically but it is between only that thermal radiation range band that is 0.3 to 10 to the power of minus 8 micrometer so now if you get that 10 to the power of minus 8 meters not micrometer so if you integrate that you are going to get eb equal to sigma t to the power of 4 that is ib equal to eb by pi this is what is what we use as Stefan Boltzmann now you see why two names are there Stefan found it experimentally Boltzmann did this integration and he could find that this is indeed sigma equal to 5.67 into 10 to the power of minus 8 so this is how we can I think this is the equation which is used all over the radiation okay so with this so professor Arun wants to add some point here see this area eb lambda when you integrate with respect to lambda what do you get you get eb so this area under the curve under each of these curve represents the emissive power of the black body at that temperature so higher the temperature greater is the emissive power and that is logical right body black body or any body black emissive power of black body is easy to understand 500 degree Kelvin it has a certain amount of emission 1000 Kelvin it has a greater emission and it is obvious from this diagram because at 1000 Kelvin it encompasses a larger spectrum of wavelengths associated okay okay so what I would request all the participants is that we will be visiting several centers to the extent possible but limit your questions only to the radiation we are not going to take any questions on convection if you have we will take the questions on convection during tutorial session please restrict your questions on radiation number one because professor Arun is not going to be there in the afternoon number two because subsequently if your questions are clear you will be able to understand all whatever we are going to teach radiation downstream you will be able to understand very easily if you have understood the definition so over to participants for questions sir the so the assumption of diffuse surfaces when is it applicable sir or when does it fail what kind of surfaces will the diffuse assumption fail okay so the question is when does a surface is diffuse or when does a surface is not diffuse that is the question so that surface which is a black body is a diffuse surface all surfaces which are closer to black body are diffuse surfaces see in the definition of the black body itself we said three constraints three conditions one is perfect emitter perfect absorber and diffuse when will it be diffuse I will tell you in a very what to say in a very layman terms if you take how do you make a black surface in your lab let me ask that question you take a plate and typically you make a surface black by making by putting lamb black that is by taking a candle and putting the plate below the candle so that all the suit gets deposited on my plate okay so now after getting depositing after depositing if you just see the surface it looks diffuse okay I shouldn't be using the word diffuse it looks very what to say not shining it is not shining it is not at all shining it looks it it's not it is dull it is dull it is dull why the dullness gives it diffuseness that is it is not having any preference if it is shining then it becomes it is not diffuse so to answer your question all shining surfaces are not diffuse surfaces mostly dull surfaces are diffuse surfaces as long as you paint them with black black paint and there are there are certain black paint for example there is a paint called pyromark if you paint that if you paint that paint if you paint any surface with that paint generally emissivity and absorptivity are equal to closer to one not just one it's roughly around 0.9495 and even with lamb black you are going to get an emissivity of the absorptivity of 0.9596 I am not going to answer you the question emissivity and absorptivity how do I measure when I touch upon emissivity and absorptivity I am going to tell you how do we measure emissivity and absorptivity I have answered you just when does a surface is a diffuse okay sir is it just a surface property or yeah we wake we will we wake we will understand this when we when we get to emissivity and absorptivity emissivity is a function of surface absorptivity is a function of source we will understand this when we go to emissivity and absorptivity yeah hello sir I have question regarding comparison between radiation and conduction suppose I am taking two body at a finite temperature difference A and B now I am taking other losses 0 now if I want to thermal equilibrium in which case in heat transfer by either conduction or heat transfer by other radiation I will get earlier thermal equilibrium condition the question is if I have two surfaces if I have two surfaces and now between the two surfaces in order to reach thermal equilibrium whether conduction makes me reach faster or radiation reaches me faster I cannot answer this in terms of time but all that I can say is that conduction takes place conduction is very good in solid and radiation is very good in vacuum for conduction I said there is a medium required if there is no medium there is no conduction they one molecule has to get in touch with the other molecule if there is no molecule itself then there is no radiation sorry then there is no conduction so for radiation I do not need any molecule for conduction I need a molecule so I cannot compare the way you asked in terms of time from one surface to another surface because from one surface to another surface something has to be filled for the conduction to take place and nothing should be filled for radiation if it is filled there is no radiation or minimal radiation between the two surfaces if there is a media there is a medium then what happens radiation gets stunted because the medium which is there starts participating and conduction overtakes so that is all I can answer in a case of conduction I assume that both the body are in a contact that is precisely what we are saying if the both the bodies are in contact then there is no question of radiation to take place if when they see the question separate case where temperature difference is same the question is between the two plates if I if I have two plates if I have two plates and both are in contact with each other and one plate is higher temperature than the other one is lower temperature which is the mode of the heat transfer which is taking place it is conduction what his question is is that if I have two plates at two different temperatures will bringing them together and keeping them next to each other make them reach thermal equilibrium faster or will radiative heat transfer give the thermal equilibrium faster radiative heat transfer between surfaces depends on several things geometry orientation how far how close are you from each other all these things are parameters if I keep them one meter apart it is going to be much slower than if I keep them one feet apart okay so these things are all very variables actually I will pitch in now actually we will be able to appreciate how much closer or farther when we when we get to resistance concept and when we study view factors will you please hold on until then sir in radius we have defined the term radio city so in that radio city we have taken the emission radiation which is emitted plus the reflected rays so actually in case of a body suppose a suppose a ray is incident on a body so it will be some part of it will be absorbed some will be reflected and some will be transmitted so in case of radio city why do we take only the emission and the reflected rays why don't we take the transmitted rays transmitted has gone away no the question asked is in case of the definition of radiosity for the case of definition of radiosity what we are taking is only the emission plus reflection we are not taking transmission why is it so that we don't account transmission into account where is the transmission going the answer is where is the transmission going transmission is going downwards the surface I am not interested what is going downwards the surface I am interested what is coming out from the surface what is coming out from the surface is only emission and reflection nothing else so transmission is what goes down not what is coming out it is what it is just going inside what is what is transmitted passing through. Passing through the plate I am not interested in that what is getting out of the plate again back is what is I am interested. Because I am interested heat transfer between that surface and another surface. So, that surface is giving out energy and also reflecting something correct. So, this part is what is interacting with the other surface. That is what we need to account for. In fact, we this radiosity is going to be the major potential when we study the furnace problems or the enclosure problems. We are just going to keep the radiosities into account. Those are the potentials for us in radiation like temperature was the potential. Any other questions quickly? Sir, how the radiation energy emitted by a surface is changing with respect to dry and moist surface? Is it place any road for a dry and moist surfaces? How the energy emitted will be? The question, it is little the question asked is if I have a surface dry surface and another surface which is moist surface. How do these two surfaces who are radiation intensity from these surfaces going to get affected? Does it get affected at all or not? It is little premature, but still I will go ahead and answer your question. The emissivity of the moisture is usually closer to 1. That is emissivity of water is 1. In fact, if you take the emissivity of the human body, emissivity of the human body is 1. Why? Because we are all made up of water. So moisture, the plate which is having moisture on it will have higher emissivity compared to that of the emissivity of the dry plate. That is all the answer for your question. Hello sir, good afternoon sir. When we have derived the expression for differential solid angle in the morning, we have taken that area in the hemisphere and drawn the right angle triangle. Whether it is approximation or it is exact right angle, one figure we have drawn on the white board sir. Correct. So the angle subtended at point A, whether it is a exact right angle or it is approximation because that area can be anywhere in the hemisphere sir. This figure, this figure. Here the question is actually the triangle, this is perpendicular sign is missing here. Let me put the pointer. The perpendicular sign is missing here. That is not an approximation. It is indeed right angle triangle. That is not an approximation. The question asked is, is this triangle which we have taken, it is a right angle triangle which is taken, is it an approximation? It is not an approximation. It is indeed a right angle to triangle. I think your question is, because we have drawn it on the surface you are asking. What if the point is on the inside, maybe your, if that is the doubt, then we do not consider this outside hemisphere. We will consider a hemisphere on which this point lies. So still always that will be a right angle. So it will be a smaller hemisphere if you are inside. If you think of the fruit, if you are inside the fruit you will consider the smaller sphere. That is all it is. So the angle is always right angle. The question is, refrigerator backside of the refrigerator is always painted black and the condenser and compressor is always painted black and the air conditioning is not having the same criteria. So is it something related to radiation or it is just the normal convention what we are following? Professor, the question asked is for the always the refrigerator backside is painted black and for evaporators and condensers in air conditioning it is not done so. This is an application oriented question. I have no answer. Please put up this in the model. We will think and get back. I want a question pertaining to basics which are taught since morning. The question is if I take a cavity for defining a black body I took isothermal. That is a very good question. Isothermal body, why do we take it as an isothermal body? Let us say the radiation we said that it is a function of temperature. Now if I take a cavity and in the cavity half of the cavity is half one temperature, another cavity is another temperature let us say 1000 and 300 then there is radiation between one wall to the half of the cavity to the other cavity. No it is not so. We have to take complete cavity as one temperature so that the emission and absorption is same. Otherwise I have involved one more parameter that is the temperature. I do not want to take the temperature as a parameter when I am taking the black body. Another thing is if different parts of the cavity have different temperature interaction we will bring them to a thermal equilibrium eventually. So, it will come and settle down at 750 degree Kelvin or whatever, but it will have to come and settle to a unique temperature isothermally completely then only this thing will be valid otherwise it is no longer correct. One last question from for this session. Question is regarding, yes sir basic question as you told us while answering one of the questions that air is medium is not participating in the radiation effect. As I know that polar gas is like CO2 and NO2 is taking part in the radiation effect. As the air contains such in polar gases why miss air is not a participating is not participating. I think the question asked is the air is combination of carbon monoxide, carbon dioxide, oxygen, nitrogen and all etcetera, but why is it that air is not participating. You see usually the inert gases the air is combined is a combination of nitrogen, oxygen and then carbon monoxide and carbon dioxide. Carbon monoxide and carbon dioxide are participating, but oxygen and nitrogen are thoroughly non participating. So, that is the reason why we can consider that as air as a non participating medium. Major portion that is almost 90 percent of the air is consisting of nitrogen and oxygen. Left out is only carbon monoxide and carbon dioxide the extraneous gases. While quoting that examination as you told that air will get heated up certain temperature. That is because of carbon, air is getting slightly getting heated up because air is having slight amount of carbon monoxide and carbon dioxide because of which it is getting heated up that is all. Otherwise because of oxygen and nitrogen it is not going to get heated. For all practical purposes we can take that air is a non participating medium. That is what we do while designing furnaces for that matter. That is non participating, but it is taking participation in the very minimum. I think we will all break for lunch and come back.