 We will get started with tutorial, problem number 39. What does problem number 39 tell? Problem number 39 tells I have an aluminum plate heated to a uniform temperature of 227 degree Celsius and is allowed to cool while being vertically suspended. This is natural So, I think natural convection problem it is we have solved problem number 39. So, I do not think we should be solving again. So, even that 40 is also natural convection. So, we have solved forced convection all that we are saying is we have to compute the Rayleigh number and compute that I think this is a plug in problem there is nothing difficult in that. So, what we shall solve now is we shall take up 41 a furnace with an aperture of 20 mm diameter a furnace of 20 mm diameter and an emissive power of 3.72 into 10 to the power of 5 watts per meter square is used to calibrate a heat flux gauge having a sensitive area of 1.6 into 10 to the power of minus meter square area is 1.6 into 10 to the power of minus 5 meter square at what distance measured along a normal from the aperture should the gauge be positioned to receive irradiation of 1000 watts per meter square irradiation is 1000 watts per meter square. The question asked is what should be the distance between the gauge and the aperture of the furnace. So, now you have taken it. So, G equal to this is R G equal to Q F D Q F D that is furnace to gauge upon A D is equal to I E intensity D is the diameter for gauge the aperture correct I E A F I E A F area of the aperture into cos theta F into omega D F that is the solid angle D F omega D F upon A D. So, this is equal to what is intensity I E A F area of the aperture into cos theta F into omega D F that is the solid angle D F omega D F upon A D. So, this is equal to what is intensity I E what is intensity here this is we can be this is coming out from a furnace. So, we can consider this as a like a black body. So, I E equal to E B by pi E B by pi into we will substitute everything A F I E equal to E B by pi we got G equal to G equal to that is 1000 equal to G equal to 1000 equal to G equal to 1000 equal to E B by pi is 3.72 E B is what E B is what is the irradiation given E B by pi 3.72 times 10 to the power of 5. How did I get that 150 it is given to us. Yeah emissive power has been given to us that is 3.72 into 10 to the power of 5 divided by pi divided by pi into A F what is A F that is pi by 4 into 20 into 10 to the power of minus 3 whole square that is A F that corresponds to we will write A F equal to pi A F equal to pi into 20 into 10 to the power of minus 3 whole square by 4 that is that area turns out to be 1.6 into 10 to the power of minus 5 into cos into cos theta 0 cos theta F it is normal. So, 0 into A D cos theta D upon L square remember whenever I said there are two areas they should be normal to the irradiation. So, now here normal again A D is given to be what is the area given for A D that is pi by 4 D square that is coming out to be correct only 1.6 times 10 to the power 5 this is pi by 4 D square that is one area this is given as the other area see we have. Yeah right right. So, into does not matter which we are putting into divided by R square divided by R square that is it. So, if I get I will get R equal to point. So, let me just write here omega D F is A D cos theta D cos theta D divided upon L square R square or L square. So, A D is what area of the aperture correct that is fine. So, we get L equal to 0.193 meters 0.193 meters okay. So, if the gauge is tilted if the gauge is tilted of normal by 20 degrees what will be its irradiation. So, it will be 1000 into cos 20 G at 20 degrees G at 20 degrees would be G G G at 20 degrees is 1000 into cos 20. So, that turns out to be 940 watts per meter square. So, a glass plate let us take problem number 42 that is a glass plate 30 centimeter square is used to view radiation from a furnace that is the glass plate is 30 centimeter square 30 centimeter square glass plate that is 30 into 30 into 30 centimeter square okay. The transmissivity of the glass is 0.5 from 0.2 to 3.5 micrometer that is lambda 1 equal to 0.2 micrometer and lambda 2 equal to lambda 2 equal to 3.5 micrometer. That means rest all in rest all other wavelength the transmissivity is 0 that is what we need to assume okay. So, all that we are saying is there is a transmissivity of the glass only within a wavelength of 0.2 to 3.5 micrometer anything less than 0.2 micrometer it is having transmissivity is 0 anything above 3.5 micrometers it is having 0. Now, the emissivity of the glass yeah. So, next sentence the emissivity of the glass may be assumed to be 0.3 up to 3.5 micrometers and 0.9 above that. So, if you plot that emissivity versus lambda 0.3 up to 0.3 3.5 micrometer and now at 3.5 micrometer there is a sudden jump and that will take to 0.9 and there from there on till infinity it is going to be infinity means it is not infinity that is until thermal radiation is contributed we will have an emissivity of 0.9 and the transmissivity of the glass is 0 except in the range of 0.2 to 3.5 micrometers I stated that already assuming that the furnace is a black body at 2000 degree Celsius the black body that is this glass plate is see in a furnace typically we put a glass. So, that we can see what is happening inside a furnace. So, that is what we are trying to do the energy balance for this glass assuming that the furnace is a black body and sitting at 2000 degree Celsius what is the energy absorbed by the glass and how much energy is transmitted through the glass outside that is the question which is asked. So, we have studied fraction of energy transmitted yesterday so if we take that concept we if you remember we can go to the table lambda 1 t table is there that is lambda we have listed the fraction as a function of lambda t is it so let us first calculate lambda 1 t and lambda 2 t and then we will go to that table so what is lambda 1 t lambda 1 is lambda 1 t that is 0.2 micrometer 0.2 into t is 2273 why did I get where did I get this 2273 from 2000 plus 273 remember in radiation always we have to keep temperature in Kelvin 2273 I get 454.6 micrometer Kelvin similarly lambda 2 t I get 3.5 into 2273 7955.5 micrometer Kelvin. So, now let us go to the table and see the radiation so for lambda 1 t 454.6 that means it is somewhere between 400 and 600 that is 0. So, f no f 0 lambda 1 0 to lambda 1 is 0 now for lambda 2 t is 7955 let us go back to the table again that is let us see for 8000 7955 we would not get if you see 8000 it is around 0.856. So, we will get f 0 lambda 2 as 0.85 this is the value this is the value 0.856288 f 0 lambda 2 is 0.85443. So, f of 0 lambda is equal to we will not do that I will not let us not write that let us just sigma t to the power of 4 the total incident radiation if we compute we are what are we supposed to find we are supposed to find the calculate calculate the energy absorbed in the gas and the energy transmitted. So, we are supposed to find the energy transmitted now let us find what is the energy absorbed what is the energy absorbed. If this is the glass plate g is coming from furnace and what is getting transmitted what is it being falling sigma t to the power of 4 is falling that is because it is black 0.3 square is the area. So, transmitted radiation is equal to this is sigma t to the power of 4 and area is 0.3 into 0.3. So, total incident radiation equal to first of all let us find out total incident radiation equal to total incident radiation equal to sigma t to the power of 4 total incident radiation equal to sigma t to the power of 4 into area into f of actually it is lambda 1 to lambda 2, but what is this f lambda 1 to lambda 2 equal to f of 0 to lambda 2 minus f of 0 to lambda 1. So, let us substitute all this I get yeah. So, sigma is 5.67 into 10 to the power of minus 8 2273 to the power of 4 0.85443 into 0.3 square I get 11,673 to the power of 4. So, this is the incident radiation that is coming from the furnace, but how much is getting transmitted if we see the graph of the transmissivity see it is getting transmitted only between 0.2 to 3.5 that is why we also calculated what is the incident radiation within that bandwidth that is 0.2 micrometer to 3.5 micrometer. So, what will we get total radiation transmitted what is the transmissivity within this bandwidth we have 0.5 how much is falling we have the G. So, 0.5 into G is the G transmitted G transmitted how do we show the transparency no 1 minute I just want to say why we did what we did here just 1 minute you might understand why we did this why we took from 0.2 to 3.5 is because we could have taken from 0 to infinity, but the thing is even if you have energy incident over all wavelengths the energy that is useful for this case is only what is coming between 0.2 to 3.5 micrometer. So, there itself we have used this fraction nothing else is going to go through that otherwise it might be falling on to it, but nothing is going to go through it. So, total radiation transmitted equal to 0.5 into 116381.7 which turns out to be 58190.85. Now, would I just like to add one thing here in this problem see actually we have this tau which is a constant in this range. So, we just took it as 0.5 if in this range also tau had some variation then we would have had to go through an integration to get the actual value of tau we will not do it for this problem there are couple of other problems where we will have to do an integrated value for tau or epsilon or alpha whatever and the procedure will be similar. So, we are leaving the other portion for yourself to work out that is the energy absorbed in the gas you please work out yourself because they have to do same thing. So, energy absorbed in the glass you will be working out yourself because we have emissivity of the gas given. So, from which you will have to work out the energy absorbed. So, let us move on glass gas ok. So, now I think if we have any questions here and there we will take. Sir my question is that sir in case of suppose you are we are considering a flame yesterday you told that suppose you are considering a blue flame it is having an emissivity of 0.2 to 0.3 something 0.2 emissivity it must be having some absorptivity term or transmissivity term along with that now how is the flame going to absorb something. Now my question is how is the flame going to absorb something. So, flame is something which is does not have any mass it is just a radiation. So, how is that absorbing any other radiation or it is absorbing a radiation or it is transmitting anything. See one of the question is yesterday we broached up that there is a flame because one of the questions was how does the blue flame how does the blue flame affect the heat transfer on my pan. So, what did we say if I have a blue flame where did this blue flame come from the question the there is no mass in the flame is the not right thing there is mass there is mass what is this mass consisting of it is what is coming inside. First is methane let me take simple in a LPG of course there will be butane and propane also I am just taking methane gas or a hydrocarbon. Let me put it very safe way answering is hydrocarbon you have an hydrocarbon and you are taking oxygen plus nitrogen that is air oxygen and nitrogen other extraneous things I am excluding. So, if you take this what is that which is there in the flame if the complete combustion takes place. So, you get C H 4 plus O 2 you get what C O 2 plus of course nitrogen is sitting here and you will have nitrogen and you will have H 2 O also. So, do not say that there is nothing inside the flame there is mass that is this carbon dioxide nitrogen and water vapor. Now that we are talking about emissivity absorptivity of these components carbon dioxide nitrogen and water vapor what we are saying is that because it is complete combustion it is blue flame number 1, number 2 it is emissivity not 0.2 it is almost 0.03 if the complete combustion is taking place it is almost 0.02 to 0.03. So, if it is 0.03 what does that mean what does that mean emissivity equal to absorptivity absorptivity is also equal to 0.03 assuming that there is no reflectivity which is true for the flame there is no reflectivity what is the transmissivity 1 minus 0.03 that is 0.97 0.97 that means what everything is getting transmitted through it it is a non participating gas that is why I said there is no radiation from the flame to the human body or to the plate whatever heat transfer is taking place between the blue flame and the plate is completely or majorly dominated by convective heat transfer convective heat transfer the mode of the heat transfer between the blue flame and the plate what I am putting on my gas stove is convective heat transfer majorly. If we need pick may be 3 to 5 percent it would be radiation because that radiation contribution is because of this small emissivity and of course there are lot of chemical reactions and all are there there will be a thermo chemical heat release also what is called as TCHR let us not get into all of that majorly the heat transfer is convection. So, please do not say that there is no mass in my flame there is mass what is that mass that is consisting of carbon dioxide nitrogen and water is this an assumption no this is experimentally measured value this is experimentally measured value we ourselves have measured this actually. So, we put a thermocouple and put a thermal camera the way I explained in the morning and we have measured the emissivity of the LPG flame. So, it is coming around 0.03 even lesser than that sir in another one I have another question. So, in temperature measurement by using thermometer you said that we have to consider the temperature of the wall it will affect the thermometer reading sorry temperature reading in a thermometer. So, suppose in case of a fluid which is having a very high temperature do you want to take the radiation effect of the fluid will it affect the temperature measurement by using thermometer. The question asked is in the radiation error in the radiation deviation what did we say yeah what did we say T fluid is equal to T thermocouple plus sigma epsilon thermocouple into T thermocouple minus T wall to the power 4 T thermocouple to the power of 4 T wall to the power of 4 upon H. But now what you are saying is that if there is a fluid and that fluid is moving and let us say water but water is not participating no water is not participating and even if you take air for example that is not participating as long as there is if we take participating medium it becomes complicated it is much difficult than what we have told. If it is non-participating then I do not think there is any problem in calculations of this. Did I answer your question. Suppose it is having a very high temperature fluid suppose a 700 degree or 800 Kelvin fluid is used will it affect now it will be also having mass which is emitting radiation no. See the question is if the fluid which is flowing is also sitting at high temperature will it affect why that is what it is not a function of temperature it is a function of participation. So, this in fact this T fluid here could be 700 800 how does it matter. Yeah here in the problem what we took itself it was it was quite high temperature actually. So, fluid temperature is not the issue the issue is whether my fluid is participating or not that is all the issue is. Instead of water if in the same place there is mixture of air I mean so instead of air if it is air plus some steam then it is a different problem it will be difficult. So, then it is air at that high temperature it is not going to be participating even if there is some amount of water vapor it will start to become participating that is what will matter that temperature might be the same for the two fluids, but the nature of the interaction is different that is what is more. So, what is the interaction we are looking at even if I have a participating medium what will happen. So, what will happen my thermocouple whatever temperature is there it cannot it cannot the fluid cannot transmit or radiate the heat to the wall that is being stunted by the medium which is there. So, in fact the participation the radiation from the fluid to the wall sorry from the fluid to the wall of course there is no radiation, but from the thermocouple to the wall is minimized in fact it is good no my radiation error is going to come down actually is that ok. How do we identify the participation is it depend upon viscosity or something like a degree of participation how do we identify a medium is participating or not. How do we identify. Does it depend upon the emissivity of the medium or something like that. The question is how do we how does one make out whether a medium is participating or non participating I think let us wait till tomorrow morning I am I have planned that I will be giving that participating mediums transmissivity emissivity and the of course again as you said whether they are participating or not will be dependent on emissivity and absorptivity. If their emissivity and absorptivity are equal to 0 then transmissivity is 1. We say that in the morning itself we said if I have a gas whose absorptivity is almost equal to 0 that is which is the case of optically thin medium then its transmissivity is equal to 1. As long as any medium has having a transmissivity of 1 everything which will come through it will pass through it. So any medium which is having transmissivity equal to 1 is a non participating medium ok. So if I have to imagine in terms of emissivity absorptivity reflectivity transmissivity is if it is a gas there is no reflectivity so it is only transmissivity it should be having 100 percent transmissivity. If you have 100 percent transmissivity then my medium is non participating ok ok. We will come out from this college now. In natural connection experiment if the geometry is changed from rectangular block to cylindrical block so we will be considering all entire surfaces for the calculating nozzle number no. So what will be the difference? In the natural connection experiment if I take a cylinder instead of a flat plate can I go ahead and do the natural connection in fact it will be good I think that is a good idea actually. So instead of taking a flat plate one can even take a cylinder but only thing is that yeah we can take a cylinder and do the experiment there is no problem you can go ahead and do this may be making a cylinder block is much easier than making a flat plate one can try this this is a very good idea really good idea we can go ahead and that way actually you have given idea that any shape we can do as long as I can get that shape in ice and with ice getting any shape is easy because template is in our hands. So question is related to absorptivity yeah very much. So in your slide there is a tree source temperature 300 Kelvin and the sun temperature around 5000 Kelvin and absorptivity related to the source temperature in that case at a lower temperature source have a high absorptivity while it is cold. The question is that in that slide with a house roof sun is at 5800 Kelvin absorptivity is 0.3 a tree is at 300 Kelvin absorptivity for that is 0.6 or something. So is it that for the roof material absorptivity decreases with let us increase in temperature no this is not that absorptivity decreases with what is it no temperature yeah absorptivity is this question actually we had in our class also okay I am just trying to open please bear with me see the point here is the absorptivity is a function of source temperature. So ha your question is this source is sitting at one temperature another source is sitting at temperature is that because of that the absorptivity is less no I cannot make a conclusion like that I cannot make a conclusion that absorptivity of the roof decreases with the increase in temperature no no no so because the each source is different is the source of each one is different it is a function of absorptivity is a function of source material and source temperature and source material properties like whether it is rough or smooth and things like that okay actually if these are the separate sources then absorptivity is 0.9 high value for the separate source as a sun its absorptivity is 0.6 that temperature is high that that time absorptivity is less compared to 300 Kelvin source that is my question yeah yeah the question we have understood your question is tree temperature is 300 Kelvin that is why my absorptivity is 0.9 and sun's temperature is 5780 Kelvin my absorptivity is 0.6 so absorptivity decreases with the increase of temperature no that is what we are saying we have got your question the conclusion is wrong just because this is an example which just shows that the absorptivity of the building material is a function of the is a function of which source it is coming from I am not told that this temperature is less that is why it is less see sun is going to energy coming from the sun is going to have this spectrum the last one 5800 for that 5800 Kelvin spectrum which is there absorptivity of this roof material is 0.3 now tree is not going to emit at this spectrum no it is going to be at a different temperature its characteristic of emission is going to be different so it will have a spectrum which is different for that spectrum the absorptivity is 0.9 now if that incidentally the temperatures are like that there that is one is less than the other one so do not look at temperature in isolation look at the source it is a different source all together two different sources at same temperature absorptivity of from that let us say I have source 1 which is a 500 I mean 500 Kelvin another source 2 at 500 Kelvin absorptivity of that roof for source 1 will be different from absorptivity at source 2 it is not related to temperature alone by chance it may come out to be the same that is that is incidental it is not necessarily a function of temperature alone it is a nature of the emission or nature of irradiation coming out from that surface to the surface. So, we will take up I hope you have pointed both today when is it problem 46 it is a slightly involved problem a small work piece is placed in a large oven having isothermal walls at T F equal to 1000 Kelvin this is F we are calling because furnace furnace so subscript F is for that with an emissivity of 0.5 that is furnace wall is having emissivity of 0.5 the work piece experiences convection with moving air at 600 Kelvin T infinity is 600 Kelvin and convective heat transfer coefficient is 60 watts per meter square Kelvin the surface of the work piece has a spectrally selective coating where whose spectral emissivity varies as epsilon lambda equal to 0.2 for lambda 1 equal to 0 to 5 micrometers and from 5 micrometers to infinity we have 0.8 we have 0.8 these are the inputs what is the question as perform an energy balance on the work piece and determine the steady state temperature T S let us try to solve this problem it is little lengthy we prepared and it is a little bit involved you might have questions please wait till the end I have drawn a furnace or the oven here this rectangle it is at temperature T F having emissivity 0.5 this is that work piece it is much smaller than the enclosure. So, I have arbitrarily drawn it as a circle does not mean anything it could be any arbitrary shape I am taking a control volume around this one. So, I have to do energy balance some energy is coming in some energy is going out. So, what is the mode of energy which is coming in very carefully we have to write this. So, e dot in minus e dot out equal to 0 because there is no e dot g and e dot s t term. So, what is e dot in e dot in should be the irradiation irradiation means it should be coming from the furnace wall. So, e dot in equal to epsilon e dot in equal to epsilon F sigma T F to the power of 4 why I am telling this only this much means what is this epsilon F sigma T F to the power of 4 what is this epsilon F sigma T F to the power of 4 emissive power of what of the furnace emissive power of the furnace is that equal to what my body has taken that is the irradiation how much my body has taken is dependent on what yes absorptivity that is this has to be multiplied by the absorptivity of my body very important this is where this is where we will go wrong alpha not of furnace, but of source my source is emitting epsilon F sigma T F to the power of 4 that is alpha of the body not work piece my work piece can take only as much as my absorptivity allows it to take it cannot take more than that. So, this is e dot in now let us write e dot out e dot out can be of two modes one is convection another one is radiation H into T S minus T infinity plus epsilon S e B at T S that is sigma T S to the power of 4. So, this is equation 1. So, these two are equated. So, I will get alpha S epsilon F sigma T F to the power 4 minus sigma S sorry epsilon S sigma T S to the power 4 minus H T S minus T infinity equal to 0. So, this is my governing equation in this quantity is for the work piece this quantity is for the work piece this is for the work piece we have to find this quantity it occurs here as we see it is a non-linear equation first of all. Now, alpha S also is not given to us directly sigma S also is obviously not given to us directly. So, we have to get these two somehow to get temperature that is what we are going to do. So, first let us find out alpha S alpha S alpha equal to we know we have written already the relation alpha equal to in the numerator 0 to infinity alpha lambda g lambda d lambda upon integral g lambda d lambda 0 to infinity. But alpha lambda equal to epsilon lambda 0 to infinity. So, alpha equal to epsilon equal to 0 to infinity in the numerator 0 to infinity epsilon lambda g lambda can be visualized as e b lambda at this is you see it is irradiating from the furnace. So, furnace we can take it as a blackboard. So, e b lambda T F upon e b of T F. Yeah sorry I made a wrong statement we are not taking furnace as a black body I am multiplying that with epsilon lambda it cannot be black body otherwise if it is I am a black body then its emissivity will be equal to 1 it is not there. So, I am multiplying epsilon lambda into e b lambda T F. So, now I have to take fraction what is the fraction which it allows we will have to see the you will have to see what we have plotted just now professor has plotted. Just one second what will what will the confusion be in fact when I write this each time I get confused see this alpha and epsilon are for the work piece. This epsilon I mean this e b is the furnace emissive power that is obvious what we have assumed is that this distribution spectral emissivity distribution for the work piece is the same as spectral absorptivity absorptivity work piece. So, Kirchoff's law is valid this way epsilon lambda equal to alpha lambda epsilon is not equal to alpha epsilon lambda equal to alpha lambda for the work piece that is what that and bear in mind this thing is this is going to be dealt with at 1000 Kelvin because energy is coming in at 1000 Kelvin. So, this is this is the as professor said this is where most of us will get confused ok. So, now what is lambda 1 and lambda 2 lambda 1 T F is equal to 5 5 times lambda 2 lambda 3 T 5 into 1000. So, you get 5000 micrometer Kelvin and lambda 2 into T F is 5000 lambda T you get 0.6337 0.6337. So, for this I will get F 0 lambda 2 lambda 1 as 0.6337 that is enough actually that is enough. So, epsilon equal to therefore, epsilon of what epsilon of the work piece at what temperature based on T F T F equal to equal to alpha is equal to 0.2 into 0.6337. Where did this 0.2 come from? 0.2 has come from this distribution here this is 0.2 into 0 to 5 micrometer that is that fraction that fraction that is 0.2 into 0.6337 plus 0.8 into 1 minus 0.6. Because the emissivity is 0.8 for all other wave lengths greater than 5 micrometer up to infinity. So, this turns out to be there itself you write as of 0.419 0.419 see that is what professor has emphasized this alpha s is based on T F. Now, using this alpha where will I use this in my previous equation? This governing equation this one has we solved. So, I got alpha s 0.419, but I do not know epsilon s and the most common mistake that we which most of us would have made even we would have made probably is that we would have taken alpha s equal to epsilon s that is wrong. Why is it wrong? Because this alpha was computed based on furnace temperature which is 1000 Kelvin. This is married to T subscript s because that is based on the temperature of the work piece. So, I do not know T s nor do I know epsilon. Let us write the equation in terms of epsilon s and T s. So, I will rewrite this equation now I will tell that equation again for you epsilon f epsilon f epsilon f epsilon f alpha s sigma T f to the power of 4 T f to the power of 2 or minus equal to h into T s minus T infinity h into T s minus T infinity plus epsilon s sigma T s to the power of 4. By the way this was our end same question final exam final exam end semester examination So, this is known 0.5 given to us very nicely this we have calculated 0.419 sigma we know this is 1000 Kelvin. So, left hand side we know. So, this quantity we know h is 60 this is also known T infinity is known sigma infinity 600. So, I have epsilon s and T s as unknown what do I do how do I solve this. So, I know by intuition by observation that T infinity is going to be 600 given T f is 1000 given. So, it has to be between these two. So, logically the first thing I can take is assume T s to be equal to 800 Kelvin. So, for 800 Kelvin I have my epsilon distribution given already this one 0.2, 0.8 this 5 micrometer this was already drawn earlier that one is there. So, knowing temperature T s I will find lambda T s assume get a value get a fraction associated with this one and then evaluate epsilon using epsilon 1 into f of 0 lambda 1 T s plus epsilon 2 into 1 minus f 0 to lambda 1 T s. So, this fraction I will get based on the assume temperature 800 Kelvin. So, for 800 Kelvin I will go to that table get this fraction I have the distribution I will get a new epsilon s calculated value of epsilon s substitute this epsilon s and T s on this equation right hand side left hand side is a number which I know right hand side I will substitute. Obviously, it is not going to match. So, then I will go to a next value if the right hand side is more that means energy out is coming out to be more. So, I should make a judgment based on that. So, I should take a lower value of or I will substitute this epsilon and get T s and take that T s to compute again new epsilon I will go on doing this until my convergence is reached left hand side equal to right hand side. So, after doing that we get an epsilon of 0.5. So, after substituting 2 3 times trial and error you get epsilon equal to 0.57 and T s equal to 680 Kelvin note please note what was the alpha s 419 at furnace temperature of 1000 Kelvin our epsilon s is different why because my furnace temperature is at 1000 but my surface is at 680 Kelvin. So, see although Kirchhoff's law is there always epsilon is not equal to alpha please remember that we always abuse Kirchhoff's law on this note. So, we have to say the temperatures have to be the same source is not going to be at the same temperature as my body if source is at the same temperature of the body the problem is gone there is no radiation. So, this is this is some even we are laughing but even probably by if you are student we would have made a mistake. We would have made a mistake when so that is what that is another thing there is there is no question which is trivial in this world as long as I understand that. So, I think we should as teachers we should make a conscious effort to not to use these words I would I at least make a conscious effort not to use these words trivial and obvious actually I have developed a habit to hate these words why because these words put off the audience actually they think that this is this is too trivial and I am asking something very trivial nothing is trivial until I understand. So, I think if we get into that comfort zone that is if the teacher and the student gets into that comfort zone that I can ask this teacher any nonsensical thing and I can get an meaningful answer that is the ambience I need to create in the class that is why in fact if you if you have watched me I have gone to every center and pastored you for question why because we need to inculcate our students also to ask questions not put the questions under the carpet and go home. So, this is this is what we need to understand we will take quickly one or two questions if they are there anywhere the question is what is the influence of distance between the thermal camera and my object on the measurement of temperature. Now see basic question how is my thermal camera working it is capturing the thermal radiation it has to transmit through with the air. So, how does that matter where do where do I keep my camera it is as long as my in fact in all of this air is non-participating that is not an assumption that is indeed true. So, because air is non-participating medium that is why my camera is able to capture if air were to be non what to be participating my camera will be seeing something less it cannot see complete heat flux the radiative heat flux which is incident on it. So, it is the answer is the thermal camera is independent of the distance between the source and the thermal camera, but if you take very far away distance what will happen your number of pixels are fixed you know I said typically 320 by 240 pixels. So, if you go far away your resolution of the object what you are seeing is going to get decreased that is all because you are going to see your full object in may be few pixels if you go closer your complete 320 by 240 pixels you would have used that is all. Will the absorptivity depend only on the source temperature? Will absorptivity depend only on the source temperature will it not depend on the receivers temperature. Oh yeah will the absorptivity dependent only on the source temperature, but not the source sink temperature that is the receivers temperature. Yes, it is not dependent on the receiver's temperature, it is dependent only on the source temperature. But energy transfer, heat transfer interaction, net heat flow rate will will change depending on the source temperature. That is that is the energy value. But alpha will not change. Yes. What about receiver's properties? Receiver's properties. That is what we are saying. No, it is not dependent on the receiver's properties. It is dependent on the source only. See this is the general perception. This is what we have been telling again and again. Absorptivity is not dependent on the receiver's properties or temperature or its surface roughness. The absorptivity of any material is dependent only on the incident that is the source temperature. Sir, another question is on, while defining solid angle, we took the sphere. But while defining emissivity or radiocity or irradiation, we took only hemisphere only. What could be the reason? What you were there yesterday morning? Because we went through this, I will repeat for the sake of being repetitive. See in our normally radiation is going in all, will go in all directions. So, you will get a sphere. But in heat transfer applications that we are dealing with, we are dealing with interaction from one surface to another. So, if this energy is coming on from a from a source like a tube light, you are looking at what is coming down to us. So, we are looking at only the bottom part or the top part if things are going from here to upper part of the room. So, if I have a source of light in this room, I am looking primarily from the it is as if I am covering the bottom part by a cardboard and just looking at the top part so that I get a hemisphere. If there is let us say, if both the sides of the plate are emitting something, let us say. So, I will have two hemispheres, this hemisphere and again this hemisphere. See even in natural convection, we said that one side is insulated, we are taking only the other side. But we have solved problems where a plate is suspended and we have coolly multiplied everything by a factor of 2, exactly this. Once we know one half, the other half can always be built upon. What are the different methods to measure the heat transfer coefficients? Of course, you told using the lumped capacity. Apart from that, is there any other methods to find the heat transfer coefficient practically? Actually it is a very big, the question asked is what are the methods to measure the heat transfer coefficient? There are plenty of methods to measure the heat transfer coefficient. One of the methods is what we told yesterday that is the lumped capacitance. So, but it is in case of internal pipe, let us say if you want to measure the internal pipe, what should I do? I should take a pipe and wind a wire, wind wire that is wire should be as close as possible, there should not be any gap. So, why I am winding the wire, this is the nichrome wire which is having a thin insulation around it, so that I can maintain constant heat flux boundary condition. That is I will supply voltage into current upon pi into d into l, that is d is the diameter of the pipe and l is the length of the pipe. So, I get the heat flux. Now, if I measure, if I put thermocouples on the plate sorry on the pipe at various locations, so I get wall temperature, but how do I get the bulk temperature? For the bulk temperature, we said that bulk temperature is going to for constant heat flux boundary condition, temperature versus x, professor has taught us that the surface temperature and the bulk temperature are going to be parallel to each other. I can measure only the bulk temperature at the inlet and bulk temperature at the exit. So, in between the two I will assume linear. So, wherever I have measured the wall temperature, I will use this linear relationship and get the bulk temperature. So, that is how I have the bulk temperature at every location and wall temperature I have measured at every location possibly where I can put the thermocouple. Now, q double dash equal to h into T wall minus T bulk. So, h equal to q double dash upon pi. So, on T wall minus T bulk, this is assuming fully developed flow. If it is not fully developed flow, bulk fluid temperature is difficult to, now bulk fluid temperature will still be linear. Only the delta T will be linear. Yeah, even in developing flow I can measure this, there is no problem, because even if it is developing flow, it will be like this, this would not be there, but the bulk fluid temperature is going to vary linearly that cannot change. So, this is how one can measure the heat transfer coefficient. If I want to get the local, what should I do? I should not use this pipe as a copper pipe. I should use SS pipe and that too thin pipe I should do. I should use, why? Because there is no lateral conduction within my pipe. It is only going to what is that is outer wall temperature. I am measuring two things that is I am measuring outer wall temperature, but actually I need inner wall temperature. If I have thin the conduction within this is going to be minimal and I am going to get the same temperature, same temperature. So, the conductivity has to be as low as possible so that my lateral conduction is avoided and thickness has to be so small, because the inner surface temperature should be equal to the outer surface temperature. That is the differentiation. I have told you only one method. Actually there are several methods. So, please put this question on the model. I will put it across. I will answer, but I think I will tell you one more method just for the heck of it. Let us take, because I will tell with reference to thermal camera. So, now let us say I have a flat plate. As I said in the morning, I have taken a flat plate. Let me take a natural convection situation why every time force it. I have told for external flows. So, now I am going to tell for natural convection. Let me take a thin metal foil. As I said in the morning I have explained why thin 0.05 mm thickness which is the paper thickness if I take that and give direct DC power supply. In the morning I have explained you why DC power supply. If I give DC power supply, because if it is a thin metal foil resistance equal to rho L by A cross sectional area is less. So, resistance is we want to increase, but still if I take a nichrome foil or a stainless steel foil resistance will be very less. So, current will be very high. Voltage will be very less. It is a safe way of doing experiment. If you and I touch that foil nothing will happen. So, if I give the DC power supply for this. Now, what will happen? It will measure the I am giving constant heat flux boundary condition. Now, I will get the wall temperature everywhere using the thermal camera provided I have painted this with lamp black let us say because emissivity is known. So, in from the thermal camera I get temperature temperature at every pixel at every pixel you are going to get heat flux equal to h into T wall minus T infinity ambient I know wall temperature at every pixel I have measured and q double dash is nothing, but voltage into current divided by the area of my plate that will give my heat transfer coefficient. In all of this in all of these methods what is very important is to characterize the losses like in ice experiment there was radiation losses here also I have to deduct the radiative. There is radiative heat transfer also while there is natural convection from this plate I have to deduct that from my voltage into current that is sigma into epsilon into T s to the power of 4 minus T infinity to the power of 4 at every pixel I have to deduct. If I do that I will be getting the natural heat transfer coefficient these are not that it is a comprehensive we have covered everything there are tens of methods of measurement laser interferometry is there. There are lots of methods there are lots of methods, but everyone uses constant heat flux with a heater wire and a thermocouple because it is very easy to do in the lamp and you do not need much money to do that. I have one question, why lamp black when this is going to be more? No, why lamp black the question asked is why lamp black the answer is if I have lamp black the emissivity of that lamp black is known which is the input which I have to give to thermal camera then only I will get the temperature. So, I think yeah there are some questions Sir, we have some confusion regarding the value of alpha for the black body is it depend on temperature for the black body also? Yeah, it depends on the temperature for the black body also, but for the source temperature for black body see, am I what the what is the question I am asking let us go to the spectral what is that in this it is that So, let us see this what is this this is the emissive power emissive power of a black body what is the absorptivity of a black body once you say it is a black body what is the emissive what is the absorptivity of black body it is 1 and that is all it is. Now, if we say is it dependant on temperature no it is not dependant on temperature because it is a black body once it is a black body absorptivity is 1, but emissive power is continuing be going to a function of temperature that is what we have plotted as Planck's distribution. Is that ok? May be you may be. I do not know may be this is not your question. May be this is not your question. You have you are having some other question but you are not phrasing it properly. Yeah. Is that your question? Ok. Second question is regarding to the thermal penetration depth for radiation. Like how it is depending on the radiation that. Ok. Thermal penetration the question asked is what is the thermal penetration depth. So, what we have been saying is that this is for participating medium. What we are saying is that absorptivity equal to 1 minus of e to the power of minus k beta l actually that is that is the thing or usually this k beta is given as minus tau s l that is what they say. This is what is called as tau l together is called as optical thickness. So, if my tau s l is very less what will happen if tau s l is very less is let us take it as 0. If it is 0 this becomes 1. 1. e to the power of 0 is 1. e to the power of minus 0. 1. No no 1 upon e to the power of 0. 0 is 1. 1 by 1. 1 by 1. So, that is right that is right 1 minus 1 is 0 absorptivity is 0 absorptivity is 0 means what transmissivity is 1 that means this is non participating that means it is not affecting my non participating medium, but now on the other hand let me take tau s l very large how large I do not have to take very large. If I take that as 5 as I said in the morning optically thick e to the power of minus 5 you will get 0.99 you will get 0.99 something you will get tau alpha equal to 0.99. What does this mean what is transmissivity? Transmissivity is almost 0.01 that means it is highly participating it is highly participating medium. So, how much it is participating it depends on the thickness or the length or the distance between the one plate to the another plate is that ok. So, for each measurement for each participating medium these tau s have to be measured in fact these tau s are measured in terms of transmissivities and they are done through FTIR spectrometer Fourier transform infrared spectrometer. If you still want to know more about FTIR spectrometer put this on the model where you will answer it and each FTIR spectrometer will cost around 1.5 crores. Sir my question is what is the absorptivity of ice? Ok what is the emissivity we what is the absorptivity of the ice is the question what is the emissivity of the ice I took yesterday for my natural convection experiment you have to answer me emissivity of ice we took it as 1. So, we said that ice acts as a black body. So, emissivity is 1 means absorptivity is also equal to 1 come on here I am saying ice is black body once I say absorptivity why it is black body where is the question of temperature coming in the picture come on here I am saying black body again you are asking me the same question. So, ice is a black body. So, there is no question of functionality of temperature moment you are characterizing something like a black body then you have to forget temperature that is it that is it absorptivity and emissivity of a black body are not functions of temperature that is all. I understand what you are saying suppose emissivity of a body is 0.5 till some wavelength and then 1 after that it is not then ok. And here in chat we see lot of questions balance experiments for paucity of time we could not take those questions because tutorial also we have to solve two three problems. Tomorrow we will try to fill in these experiments sometime we will keep that in the back of our mind may be immediately after heat exchangers we will do that for today we are closing the session and we will meet you tomorrow morning at 9 o clock bye bye.