 V N I T Nagpur, professor you have any question? Sir, how we calculate solid angle for hemisphere? The question by one of the participants is, how to calculate solid angle for an hemisphere we have done that. For a sphere we have done, now you can go ahead and integrate with respect to what is that? That is d A N by r square solid angle therefore for a full sphere is going to be 4 pi d A N by r square would be 4 pi instead of integrating from this way I take the appropriate limits. So this what is it? This angle goes from 0 to pi by 2 and that is 0 to 2 pi that is theta is 0 to pi by 2 and phi is 0 to 2 pi you will essentially end up with getting A N equal to 2 pi r squared instead of 4 pi r square. And then put it here you will get that is that okay professor? My second question is related to convection sir, in a convection volumetric bit is equal to 1 by T that is for a perfect gas but for a liquid metals can we use this equation beta equal to 1 by T? Okay, so no we cannot use this we cannot use that you are very right the question asked is for perfect gases we can take beta equal to 1 by T but for liquid metals on all can I use this? No we cannot use this this is applicable only for perfect gases that is why in the class yesterday I told that for that problem which we solved specifically that was air that is why we took beta equal to 1 by T infinity but generally we should not take beta equal to 1 by T infinity we should give the properties if we see the properties generally they are given directly as G beta by alpha nu if you see professor Bejan's book he directly gives G beta by alpha nu but the point is beta has to be obtained it is not going to be 1 by T infinity is that okay? How we find the beta for liquid metals from the tables or which table? Yeah how do we get the beta for liquid metals from the tables yes the answer is we will get it from the tables someone has measured it someone has measured it and tabulated it as a function of temperature so beta we will be taking it from the table. Jayantiyo Hyderabad? Jayantiyo Hyderabad any questions professor why cell phone radiation is harmful to human beings honestly I do not know the answer but the I think that again cell phone radiation is again electromagnetic waves but I do not know their frequency see I think the frequency of the cell phone waves is much lower than what we are talking as thermal radiation that is why they will carry large energy as we saw equal to H nu so they are I think low frequency waves but we have to check out that is why they will probably hurt our biological tissues that is the theory but this is not actually proven theory that indeed the cell phone waves are going to affect there are two schools of thought one school of thought is that these cell phone waves whatever electromagnetic waves are emanated from because of the cell phone electromagnetic waves they are going to hurt or harm the biological tissues another school of thought says nothing doing they are not going to hurt so there is no clear evidence that it is indeed going to hurt or not going to hurt there is no correlation directly available but nevertheless what we can go back and check is that what is the energy pertaining to cell phone electromagnetic waves we can check out that is what is the frequency of this waves which are emanating from cell phones we can check out in fact there is a theory also that sparrows have vanished because of cell phones if not cell phones at least that cell phone what is that tower cell phone towers but these are all theories only there is no direct correlation anything else you want to add over this question how does medium affect radiation ok the question asked is how does medium affect radiation see if we see in the first transparency in the very first transparency what we showed in the morning what did we say now if we see here the first thing is said we said that the wave length equal to speed of light speed of not light speed of electromagnetic wave in that medium that is sorry lambda equal to c upon nu where c is the velocity of wave in vacuum and the velocity of the electromagnetic wave in that medium is given by the speed of light in vacuum upon n so that is n is the refractive index there are two things here n is the refractive if it is a perfect vacuum or air if it is non participating that is if it is a vacuum or air is also generally non participating in that case the refractive index is one that means my wave length is not getting distorted number one number two if it is non participating means we will see later on in the today's lecture we will not have absorptivity transmissivity reflectivity involved with any of because of this medium that is I took in the I took an example in the morning that is if I take an enclosure if I take an enclosure that is this room and it is filled with air and this is non participating that means one walls heat will interact with the other wall directly the air is not participating but now let us say this room is filled with suit suit is having lot of absorptivity suit is having lot suit means what you have burnt some wood in the middle of the room let us say suit is filled all over the room the suit is going to absorb some of the thermal radiation which is coming out from one wall and trying to reach the other wall so it is starting participating so participation is usually quantified in terms of refractive index of the medium and in fact the absorptivity transmissivity reflectivity of the medium are represented in terms of refractive index in fact if I have to go overboard and say this refractive index will have what is that imaginary component and the real component usually the real component is the absorptivity so anyway let us let us not get into that much detail but the point is the participation is usually quantified by the refractive index of the medium one other way of looking at it not from a radiation point of view but from if you are thinking from a light point of view in winter season you see lot of use about fog delaying trains etcetera what it means is that your visibility is reduced completely or to a great extent so if you are shining a light you would have seen pictures on television the train is coming the light is not even seen by the cameraman so because there is so much of moisture present in the air you are not able to see the light if the moment the sun comes out at around 12 o clock or something in the afternoon the same train the same cameraman same light were able to see much better that is because of the what you say the amount of moisture in the given space has reduced so this probably is a little bit easier to understand for participating that means the particles of water vapor which are there are actually taking away whatever you can see it is forming like an opaque medium in between the light and the observer so if the medium is going to act along if it is a passive if you are a participant in a classroom and you are sitting in one corner and not bothered about what is happening you are passive non-participating but if you are actively interacting you are going to be interacting or you are going to be participating so the medium is not participating it does not care what is happening between the two surfaces it is there it is not there it does not make any difference but when it affects the interaction that is called as participating medium. This effect is only with reference to thermal radiation that is heat transfer. Sir my question is in convection heat transfer how does Prandtl number and Reynolds number varies in capillary tube for two phase fluids I mean liquid and vapor phase. The question is in capillary tube how does Reynolds number and Prandtl number vary especially when there are two phase flows why capillary tube okay the answer let me elaborate the way I have understood it capillary tube is almost like a micro channel in fact one of our colleague here started off with micro channel measurements with capillary tube so what I understand capillary tube is whatever tube you get in the inside the thermometer is the capillary tube so he just broke the thermometer and did the experiments. So now what is capillary tube essentially the diameter is very small if the diameter is very small means my pumping power required is very high you have seen yesterday in our derivation delta P in a channel is inversely proportional to diameter that is delta P is proportional to 1 upon D to the power of 5 so as the diameter comes down the pressure drop is very high so so much pumping power goes up. So in the capillary tube my point is in the capillary tube essentially the flow is going to be laminar it is not going to be turbulent most of the times 99% of the time I can say that the flow is going to be laminar it is very difficult to get turbulent flow in a micro channel that is in a capillary tube because you are pumping power even if you put 10 bar pressure you are going to get 10 bar pressure drop also you are going to get a velocity you are going to get a very low velocity as long as you are taking a liquid if it is air perhaps you can still manage to get turbulent flow generally in a capillary flow it is laminar. Now I am answering with reference to single phase flow so then if it is laminar we have seen in the basics yesterday that the friction factor is dependent only on Reynolds number and Nusselt number is not dependent on either Reynolds number nor Prandtl number so as far as the heat transfer is concerned it is only going to be either 4.36 if it is constant heat flux or 3.66 if it is going to be constant wall temperature for pressure drop it is going to be 64 by Re. Now for 2 phase flow if you ask me it is a 2 phase flow in a micro channel so it is quite difficult yes but I can answer in capillary flow also little bit better see generally in capillary tubes yeah very good in capillary tubes what happens is the nature of the flow the flow pattern if I imagine it is water and vapor if I imagine generally it is bubbly flow generally it is bubbly flow but not always not always generally when I say bubbly flow I can imagine that fluid as a homogeneous fluid yes what I mean by homogeneous fluid is there are 2 phases that is one is liquid and another one is vapor but I will imagine a third fluid which is having a properties which are equivalent to the combination of liquid and vapor and I will define equivalent viscosity equivalent density and so I will go ahead and define equivalent Reynolds number but this is too simplistic an answer for 2 phase flow I have answered for your question only for single phase flow I have not answered for 2 phase flow 2 phase flow what you have asked is actually 2 phase flow in a micro channel over to you sir one more question sir how would be the thermal boundary layer and hydrodynamic boundary layer for say 2 phase fluids whether it is condenser or evaporator in a refrigeration system okay see in a condenser or evaporator how do how does thermal boundary layer and hydrodynamic boundary layer develop see it develops definitely both hydrodynamic boundary layer and thermal boundary layer develop but the way they develop depends on the flow pattern let us say if I take stratified flow when I say stratified flow what it means is that yeah so what pressure is drawing is a stratified flow that is the denser fluid will flow at the bottom and lighter fluid will flow at the top then what is happening in fact we have answered this question yesterday also so what will happen boundary layer will grow on the top also on the vapor side and boundary layer will grow on the bottom also on the liquid side and in addition there is interfacial interfacial shear stress there will be boundary layer between the two also because why we said boundary layer will be there as long as there is a relative velocity now there is a relative velocity between liquid and vapor vapor tends to go fast liquid tends to go slow you would have seen liquid velocity should never exceed 0.3.4 meters per second but the vapor velocity can be as high as 4 meters or 10 meters per second so there is a boundary layer growth at the interface also and also on the let us say if it is flow between two parallel plate on the top wall and the bottom wall also so that is how we have to handle we have answered only for stratified flow if you now say it is going to be let us say slug flow or what other flow pattern angular flow it is all going to be difficult to answer but definitely boundary layers are going to be there even in two phases see one other thing with respect to condenser and evaporator what will happen is that see condensation will happen in such a way that you will have if it is like a vertical pipe condensation will form something like this. So, vapor will be here liquid will be here this is called as an annular flow the bottom line is depending on the distribution of the liquid and vapor phase in that geometry which we call as flow pattern depending on that you will have the same fundamentals associated to single phase fluid mechanics applied only thing that you have to worry about is the what is the nature of flow pattern if the condensation rate is very very small the nature of flow pattern will be very different if the condensation rate is very high it is going to be different see one point I want to add why our basics are going to be so good so important for example you are asking condensation in case of condensation in fact if you see correlation which is the most used correlation for condensation condenser for design of condenser is again non dimensionalized using dittus bolter only this nu naught is the nusselt number which is used for single phase flow dittus bolter correlation which is used in which all the properties are calculated at saturation temperature that is nu naught, but what is there on the right hand side it is a function of thermodynamic quality that is all otherwise for condenser also nusselt number is again a function of reynolds number Prandtl number in addition to reynolds and Prandtl you have what is called as thermodynamic quality this is what we do for condensers if you want many more details on this this correlation what I am talking about is what is called as Shah's correlation which is the most extensively used correlation for design of condensers if you want more details on this you please moodle we will send you the notes what we teach for heat exchangers that is for condenser design we will send upload that is that okay professor. Thermic fluids as far as I know it is a the thermic fluid which we had used in our lab is ethylene glycol why do we go for ethylene glycol is that the Prandtl number I remember what do I remember do I remember or do I do not remember just give me a minute for high temperature applications we use and its viscosity is is equivalent to water see we this thermic fluids are generally used for thermal storage applications and for chiller applications why because its viscosity is such that it is it is almost equivalent to that of water but the Prandtl number is high because its specific heat is high its specific heat is high but the viscosity is as much as that of water but not as that of oil because for thermal storage the obvious choice would be high Prandtl number that is high CP fluid but when you try to go for high CP fluid which is usually oil you the negative point of the oil is that its viscosity is high but if viscosity is high your pumping power goes up in order to keep your pumping power low but still try to get the advantage of high specific heat you use this thermic fluid which is having high CP but low viscosity low viscosity is the viscosity not as much as that of oil but as much as that of water so that is what is called as thermic fluid we have used this in chiller but in thermal storage that is for solar power plants and all here in one of the applications in solar power plants thermic fluids are used so these are essentially what I said high CP low viscosity fluids that is the summary of the question. Sir my question is regarding the basics of radiation you mention about electromagnetic waves that it is a vibration of molecule which we take as a sinusoidal wave but when you go to basics of radiation we mention it travels through vacuum where there is no molecules so can you please elaborate on this. Mechanism of emission we said that the question is mechanism of emission not transport see this is the question asked is in the emission while emitting there is we said that the electrons are oscillating because of which we can imagine that this oscillation as a sinusoidal wave that is what we have studied in plus 2 also for a simple pendulum there is an oscillation so but then in a perfect vacuum how can it get transmitted that is the question see there are these 2 are 2 different things when we said this with reference to this transparency it is emission where is the emission happening emission is happening on this surface and we also said that radiation we consider the emission only at the tip of the surface so we are not taking it as a volumetric phenomena we are taking it as a surface phenomena if it has to emit the electron on the top surface has to vibrate and generate a wave wave form so that is all we are saying energy this is released because of the oscillations of the electrons at the surface are emitting electromagnetic waves that is all that is all the answer so we should not get confused with emission and propagation so what you are telling is propagation what I have told is emission one more question yeah we have studied Reynolds analogy okay so can you please explain with a practical example how it can be used in practice one of our professor uses the question asked is what are the practical applications of Reynolds analogy see if we go back and use if we just say that we all know that we all know that doing experiments with heat transfer is sorry fluid mechanics is much easier than doing measurements with heat transfer now let us say I said in the Reynolds analogy standard number equal to CFX by 2 now if let us say for flow over a flat plate as I said doing pressure measurements are easy and the measurements with pressure or fluid dynamic measurements have lesser uncertainty I have also told in the yesterday's class that typical uncertainties in fluid mechanics are 6 percent and typical uncertainties in heat transfers are more than 10 to 15 percent so if I do pressure measurement and estimate CFX and let us say this is for air why I am taking air because my Prandtl number is sorry my Prandtl number is equal to 0.71 which is which is closer to 1 which is closer to 1 it is not equal to 1 but it is closer to 1 then in that case I can assume that Prandtl number is equal to 1 then my Reynolds analogy one of the assumptions sorry one of the conditions for the Reynolds analogy to be valid is that Prandtl number is should be equal to 1 so for air Prandtl number is closer to 1 that is the closest one can get so for air if I have done pressure measurement and measured CFX by 2 without measuring I can now get Nusselt number because N u by R E P R equal to CFX by 2 so let me write it slowly N u upon R E P R equal to CFX by 2 so R E I know because at that Reynolds number only I have measured the CFX now and P R I know because it is 0.71 now I can get Nusselt number from Nusselt number I can infer heat transfer coefficient the point is if you have done fluid dynamic measurement you can get the heat transfer coefficient this is what I used to think earlier but one of the professors in our aerodynamics lab that is professor Menanges he uses reverse way in supersonic flow interestingly I was thinking that for supersonic flow Reynolds analogy is not valid interestingly for flow over a flat plate in supersonic flow he measures actually Nusselt number and infers the CFX very interesting it is and he says that measurements have demonstrated that even for compressible flow Reynolds analogy works. So of course there are assumptions in fact when we did all this we have done for incompressible flow most of the times when you do measurement you come upon you come up with this conclusion that Nusselt number this Reynolds analogy is working so the point is if you do one measurement you can do away with the other measurement if you apply Reynolds analogy that is the beauty of Reynolds analogy actually in fact people say that even in sub cooled boiling people have seen that Reynolds analogy is working but it is little difficult to understand because in a pipe flow there is a pressure gradient that is the another condition for the Reynolds analogy to work we made two conditions to be valid that is one should be parental number should be one and there should not be pressure gradient otherwise in my momentum equation I have dp by dx and in my energy equation I do not have this pressure drop term so but people have observed that even in two phase that is sub cooled boiling Reynolds analogy works so in spite of having these conditions not being made here and there they do Reynolds analogy works okay in the morning in the second or third transparency you told us that by giving the example of that campfire and the person that you told us that whatever the temperature of the air remains constant in between that person and the campfire so I cannot understand how that temperature remains constant for the okay there will be no convection so the question is actually I have a campfire and I have a human being and what is not able what we what we are not able to understand is how does this air which is sitting in between is not going to get heated up at all so that is the that is the question so let us look at it in a different way so we have a campfire what is the temperature of the campfire 900 degree Celsius what is the human body temperature 35 degree Celsius so 37.5 degree Celsius now because of radiation because the electromagnetic waves are traveling from fire they are able to reach the human body now let us say instead of this air if air was work to be highly participating that is if they are going to absorb all the radiation which is coming out from the fire then I would not have received any heat now if I put if I have to insulate myself if I have to not receive any radiation what should I do I should put a asbestos asbestos sheet in between the fire and myself so what is doing what is this asbestos sheet doing sheet doing in between why am I not feeling the heat now asbestos sheet is absorbing almost all of the heat why because its absorptivity is very high reflectivity is very minimal and transmissivity is almost zero for a solid body especially for asbestos sheet that is what we do know that is what the firefighters do when they go near fire they keep asbestos sheet I do not know whether you have watched or not the material whatever they keep the they are all about asbestos sheet now if I remove the asbestos sheet I am filled I have what is the medium which is in between the fire and myself air but air is having almost nil absorb absorptivity so it is not absorbing any of the thermal radiation which is coming from the fire to myself so that is why I am absorbing almost all of the heat or I am able to receive all the heat so of course air also gets heated up slightly when I say this I am I am saying with the pinch of a salt it is getting heated up slightly but not to a greater extent why because it is reasonably non-participating medium in fact this is the example quoted this example I have taken from chungal this example is given by chungal so I think this is a very good example I hope you have understood now the crux is that whether my medium in between the source and the sink is it participating or not if it is not participating everything reaches the sink if it is participating some of it only reaches the sink as much as my participating medium allows it through it allows the radiation through it that is the answer for your question so I have got a q question regarding boundary layer theory the basics so my question is in fluid mechanics we have the local velocity and the mainstream velocity ratio u upon u infinity given as 0.99 and similarly when we come to thermal boundary layer we have we take t minus ts upon t infinity minus ts is equal to 0.99 instead of taking t infinity upon ts is equal to 0.9 or ts upon t infinity is equal to 0.9 why is that yeah see this is actually actually we should have told yesterday this is a common question for velocity boundary layer we take u by u infinity equal to 0.99 but for temperature infinity temperature boundary layer why do not I take t by t infinity equal to 0.99 yeah professor had told this yesterday why because you see heat transfer is all because of temperature gradient what is that which is changing what while I am moving up it reference to the temperature gradient what is the maximum temperature gradient I have ts minus t infinity so as I come down this temperature gradient has to be decreasing so in heat transfer we are worried about temperature gradient and in velocity in the velocity boundary layer we are worried about the velocity gradient and interestingly velocity gradient is at y equal to 0 it is u equal to 0 but at y equal to 0 t is not equal to 0 it is equal to ts so I have to take ts minus t infinity for temperature boundary layer and I have to take u by u infinity equal to 0.99 for velocity boundary layer that is the answer for your question one more question sir it is regarding radiation can we consider a hot tea in a thermos flask as a example the tea cools after certain amount of time right so why that time is not a variable here yeah okay okay okay the question is the question is if in a thermos flask whether why does the tea does not get cooled initially but with the passage of time why does it get cooled so what is that we are doing in a thermos flask as I understand what is that we are doing in a thermos flask this is my container in which where I have filled my tea and outside that there is a chamber and in this chamber I have filled with vacuum I have filled with vacuum so what is that yes we can give this as an example for radiation what is the mode of the heat transfer with reference to atmosphere it can interact with the atmosphere only through radiation in fact if you have closely watched the thermos flask they are highly reflective surfaces reflectivity of the surface is very high because they are not supposed to absorb any heat so the reflectivity is very high so what is the essential mode of the heat transfer which is taking place between these two plates where the vacuum is there is radiation but eventually tea has to get cooled why again there is a radiative heat transfer there is no conduction and convection but there is radiative heat transfer because of which there is the tea will eventually cool and also this vacuum cannot be perfect vacuum correct see what professor Puranic who is sitting by my side is saying is that if I take tea the temperature of the tea definitely is varying with time no doubt about that because there is there are other modes of the heat transfer but for radiation it is only the boundary condition so the point is that the tea has to eventually cool with passage of time I cannot make any system thoroughly insulated thoroughly insulated so this is essentially what we have done is an insulator made by what is that by maintaining vacuum but if I that is what is super insulator which we were anyway going to touch in radiation if I have two surfaces and if I evacuate that is going there is no conduction there is no convection but then radiation takes place that is why they put lot of shields if you put shields then what will happen highly reflective shields then the heat transfer from this surface to the next the surface is impeded that is what is super insulator which is used in cryogenic applications so and another point I want to emphasize is that here in this vacuum there is nothing like perfect vacuum no matter which pump you use for creating vacuum there will be small amount of air and that amount of air will create small not create will make a natural convection take place so some and substance of this question is that yes thermos flask can be given as example for radiation but T will eventually cool because there is no perfect insulator in this world okay so now I think we will stop the question answer session and move ahead with the regular lecture we looked at this definitions for related to intensity related to emission irradiation and radiosity and before signing off I told you that what we define let me just go back here what we defined yeah we defined for emission e we defined g we defined j we did not have any subscript of or brackets in them those were total hemispherical quantities total hemispherical emissive power total hemispherical irradiation that is capital G total hemispherical radiosity capital J that means it encompasses all wavelengths does not care about wavelength does not care about direction it relates to the every it relates to all wavelengths all possible directions okay so this was the whole picture now let us break it and go into impose one constraint after another so that constraint now we will say is spectral quantity when I say spectrum or spectral quantity refers to wavelength dependency so I had told this already now you have put this in a logical fashion so total radiation quantities which have been integrated over all the wavelengths have been considered so far e g j were total spectral quantities variation of radiation with wavelength as well as direction both will be considered in general right now we are saying we will do both together and then we will eliminate one of the dependencies and get the other one so spectral directional quantities refer to situations where variation of radiation with wavelength as well as direction are considered and to express these quantities at a certain wavelength or per unit wavelength interval about lambda so this is jargon now let us read the definition and again understand English point of view spectral radiation intensity for I mean what it what does this tell me I lambda e theta phi okay let me write it without being lazy I lambda e theta phi I is for intensity e is for related to emission okay then lambda in front here refers to spectral okay so this subscript comes because we are talking with a spectral quantity this refers to emission related to emission when I associate intensity related to emission I will have the subscript this is our dear intensity what is intensity again we have already studied something which is related to the normal okay so this I will now write is equal to dq divided by I am flashing this and I will write is defined as rate same definition exactly the same we have just added one extra term related to wavelength rate at which that means what joule per second watts rate at which radiation energy dq is emitted that means by virtue of its temperature is emitted that is why this emission is here at a wavelength lambda in the theta phi direction per unit area normal to this direction per unit solid angle about this direction let me just go back to the earlier slide rate at which radiation energy dq is emitted in the theta phi direction per unit area normal to the direction per unit solid angle what is different between this and that definition which I have flashed is that there I have a d lambda because that is wavelength specific I told you to imagine red color light how much of intensity of red color light at a wavelength given by lambda of red light that is what is given by that definition so if I go back quickly it is nothing but rate at which radiation energy is emitted in the theta phi direction was there I have introduced an additional word at wavelength lambda that is all there is and here therefore dq by d lambda will come integrated over the differential wavelength d lambda so the units essentially will be watt per meter square steradian for solid angle micrometer and I can write this d omega as sin theta d theta d phi so I will have essentially the same units that we had before dq divided by dA1 cos theta sin theta d theta d phi this is what we had this was I this was I e theta phi this was what we defined at 930 today this quantity divided by that is what we write I lambda e lambda theta phi is equal to dq divided by dA1 cos theta sin theta d theta d phi d lambda so right now if I want to get the total energy total watts I have to integrate it with respect to theta I have to integrate it with respect to 5 and also over the wavelength of interest that was done what was given in the earlier transparency this quantity was essentially one integration and already been performed integration with respect to lambda was already done this is one more step this is like the first we have taken the wavelength dependency out here now we are saying if it was wavelength dependency was there this is my definition of intensity watt per meter square steradian micrometer that is what is given here so if I understand this for emission the same thing will be valid for irradiation and radiosity so the spectral emissive power let us just do couple of integrations here spectral emissive power so e subscript lambda that means emissive power relates to all directions so I do not care about direction anymore I am going to integrate with respect to both theta and 5 spectral hemispherical emissive power correctly if I have to write strictly it will be spectral hemispherical because the integration is like this theta is equal to 0 to pi by 2 phi equal to 0 to 2 pi it forms a hemisphere so I lambda e theta 5 this quantity cos theta sin theta d theta d 5 this I have not integrated with respect to wavelength so the subscript lambda will remain the units for this emissive power would be watts per meter square was the original units flux this will be watts per meter square micrometer so this we have to be careful this is at that wavelength but when we say spectral emissive power we are saying it is like flux only so watt per meter square but we use the word spectral emissive power so do not get confused this subscript takes care of the fact that you are using this spectrum so I am not concerned with other wavelengths at this wavelength how much is the emission occurring what is the flux what is the intensity what is the emissive power at that wavelength so I do not have to say red light how much is the emission so I do not have to use per micrometer there it is if I am talking of emissive power associated with red light then that will be watts per meter square associated with that wavelength so when the variation of spectral intensity I lambda varies like this with lambda I can just integrate it the total radiation intensity for emitted incident and reflected part why reflected because we need radiosity we need the reflected part of incident radiation we can get this by integration over the entire wavelength spectrum so total intensity I e is spectral intensity that is a spectral variation something like this integrated over all wavelengths going from 0 to infinity so these things will become clear when we do problems so right now these might look like what double integral triple integral so on and so forth but when the problems are done it will be very clear so actually this is the only definition which I have to know e total emissive power is nothing but a total quantity because it is a wavelength independent now all wavelengths have been taken care because it is a triple integral one of the integrating limits is lambda 2 lambda is 0 to infinity it is a hemispherical quantity because it has been integrated to form a hemisphere so I e lambda theta phi cos theta sin theta d theta d phi was there already d lambda if I put here d lambda is missing here d lambda here is missing so this triple integral should have d theta d theta d lambda so d lambda is missing in all these three so this is the effect of cut and paste the total e therefore is obtained by integration of this who gives me this function if I know this function I can integrate it now these functions will be known to us we will see in the problem some kind of in real life functions will be arbitrary in terms of it will be fluctuating but we can approximate for calculation purposes functions to be straight line triangular profile trapezoidal profile so on and so forth so piecewise linear or some geometric representation can be made so similarly total irradiation is intensity I lambda spectral subscript is I because it is referring to intensity coming in incident that is why this I exactly identical I am not going to even remember cos theta sin theta d theta d phi d lambda same things here everything looks same except the subscripts I and lambda has been there theta phi is there therefore it is a triple integral for total j will be e plus r so we can do all these things very easily and therefore if everything is diffuse as we saw e is going to be pi of that intensity whether it was spectral quantity or total quantity this is valid because this is been obtained by integration over all solid and this so with this I will stop there are problems which we will do now where these definitions are involved and once these definitions are over it is only black body and absorptivity transmittability. So now I think we have spent ample time although I was getting restless that we are losing time but I think professor Arun has spent enough time teaching us the definitions. So now with these definitions that is emissive power and basics with these basics I think we can move quite fast there is one simple problem which has been given irradiation there is a spectral distribution given that is from between wavelengths 0 to 5 you have an irradiation of 1000 watts per this is meter squared this is watts per meter squared micrometer so and from 5 to 20 my irradiation is constant and again from 20 to 25 micrometer it drops so how do I integrate how do I get the total irradiation total irradiation is integrating over all wavelengths so g lambda d lambda between 0 to infinity. So I get g lambda d lambda 0 to 5 because that nature linear nature I know and g lambda d lambda between 5 to 20 micrometer it is constant and again 20 to 25 micrometer there is particular wavelength particular irradiation and 25 to infinity it is 0 because there is no distribution if I substitute that the same thing that is this is area under the curve area under the curve is area of the triangle that is half into base into height that is half into 5 into height is 1000 if I do that I get the area under this curve that is this integral can be obtained as half into 5 into 1000 plus 20 minus 5 into 1000 that is again area under the curve 20 minus 5 15 into 1000 gives me the area of this rectangle and again area of the triangle 25 minus 20 that is half of 5 into 1000 so if you add up all of this you are going to get 20,000 watt per meter square so this is how we typically have the spectral irradiation and this is just a given a simple example but you will have all sorts of variation where in which you will break that into triangles rectangles trapezoids and squares so this is how okay there is another problem this is a very important problem why I say very important because you see I keep saying this always I take these hands only in my in my class also I take the hands only into account can you just yeah these are the two hands my left hand is let us say at sitting at one temperature and the right hand is sitting at another if both are normal let us say this is dA1 and this is dAn so if both are normal there is no issue now let us say if my the left hand is not normal it is inclined what should I do I have to make it normal to make it normal this has to become dA1 cos theta okay it has to be made normal similarly for this also if it is not normal I have to make it normal so to make it normal whatever angle is there that area into that cos theta so that is why always we say normal area to the direction of intensity that is to the direction of intensity both the emitter and the receiver are supposed to be normal and in the solid angle I am having who is receiving and in the other area in the other area we worry about the area which is emitting that is if you see here q1j that is the have I yeah see let me reread the question see I have area A1, A2, A3, A4 all are of 10 to the power of minus 3 meter square and the intensity is 7000 Watt per meter square steridium that is per unit solid angle and the question and the distances are all 0.5 meter but their orientations are not same you see now A1 and A3 are normal the way I showed my hand A1 and A3 are normal so let us first find out that solid angle for omega 3 1 is A3 by r squared I am taking A3 only why because it is normal similarly for 4 also 4 also is normal to the direction of intensity but for A2 it is not normal so I have to take A2 cos theta that is I get A2 cos theta 2 to make it normal by geometry you can just work out I am not going to do that the angle between if I tilt this normal to this normal to this I am going to get that as 30 so A2 cos 30 is what is the normal so that is how I get A2 cos 30 that is solid angle omega 2 1 is A2 cos 30 upon 0.5 squared that is how I get this one so point is they have to be normal now q1j that is what is the net heat transfer leaving the surface 1 intercepted by surface 2 surface 3 surface 4 that is given by intensity into the area which is emitting this intensity of radiation that is A1 but that A1 has to be normal to the direction of intensity that is A1 cos theta for 3 it is cos theta is cos 0 1 that is it is in the it is normal to the direction of intensity but that is not so for 2 and 4 so for 2 and 4 I have to tilt this tilt A1 to make it normal to the direction of intensity tilt A1 again to make normal to the direction of intensity so that tilt angle is cos theta that is all here it is cos 45 here it is going to be cos 60 so that is what I have put that as cos 45 and cos 60 and if you substitute that and if you substitute the solid angle you are going to get the you are going to get the net heat flux you see there is a there is much you can understand by these numbers you see q 1 3 is higher why because they are they are just normal directly it is they are seeing directly each other that is why q 1 3 is having higher q 1 4 actually is next higher why q 1 4 because at least 4 is normal but q 1 2 both are not normal to the direction of intensity I have to tilt that is why the net heat flux is lowest in case of q 1 4 I mean this is how you can understand how much heat flux is emitted or received by whatever is emitted is received by the other surface that is the beauty this example shows us how we can understand the solid angle I think with these examples we have understood these definitions we will move on to radiation 2.