 So, let us move on to the final phase of radiation that is radiation 5 that is okay. So here we have 3 wall enclosure, so if I go to 3 wall enclosure, so what are the, see one important thing, another important thing which we need to realize the resist electrical resistance analogy between radiation and induction. When our students in our NSM made a mistake in this that is especially we have to be careful when I have a conduction radiation combined problem remember in radiation the potential is emissive power and radiosity and in conduction the potential is temperature. So I cannot combine conduction and radiation problem both the resistances that resistance in conduction is having potentials of temperature and the current is heat flux no problem here also current is heat flux only but the potential is emissive power we should not forget this. This is the basic difference between the electrical resistance analogy in radiation and conduction. This is where we get into trouble usually when we have conduction radiation combined the problem we get into trouble we had unknowingly we had given a problem in the NSM and 50 percent of our class had goofed up this problem by combining the resistances that is why I am mentioning this okay. So we have now potentials what are the potentials one can conceive of eb1 j1 okay I will not construct this first eb1 j1 then eb2 j2 eb3 j3 why am I doing this because I have three surfaces I should be having three surface resistances. So those three surface resistances are 1 minus epsilon1 epsilon1 a1 1 minus epsilon3 a3 1 minus epsilon2 a2 epsilon2 so I have three surface resistances. Now these radiosities j1 j2 j3 are interacting with each other that is j1 is interacting with j2 and j2 is interacting with j3 j3 is interacting with j1 with individual space resistances. So j1 has to interact with j2 means what is the space resistance between 1 and 2 it has to be 1 upon a1 f12 or a2 f21 so I have this is 1 upon a1 f12 okay the space resistance between 1 and 3 is 1 upon a1 f13 and the space resistance between a2 f23 is 1 upon a2 f23 that is all. So it goes logically now what am I doing is I am applying Kirchhoff's law if I apply Kirchhoff's law this time Kirchhoff's law is not the Kirchhoff's law for radiation Kirchhoff's law for electricity okay for electrical circuits so that is V equal to voltage equal to current into resistance that is all I am applying. So what is that let me go through this that is I take three nodes j1 j2 j3 node at each node I say that the input current whatever is coming is the summation of the current is 0 the algebraic sum of the currents at each node must be 0 that is all current means what here it is heat flux so EB1 minus j1 upon R1 current has to flow in this direction so this has to be that is the direction assume it may be wrong if it is wrong it will come out as negative so I do not have to worry about that I think we have done this in electrical circuits I do not think I need to harp this again so EB1 minus j1 upon R1 plus j2 minus j1 why because current again has to go towards the node j1 and plus j3 minus j1 upon R13 so that is equal to 0 similarly if I do for j3 and j2 I am going to get three equations so total how many equations I have three equations typically if I know the temperature if I know the temperature of these three surfaces emissivity that is what I have taken I have taken that emissivity is known surface area is known and temperature of the body is known so I get if I know the temperature of the body what is EB1 sigma T1 to the power of 4 sigma T2 to the power of 4 sigma T3 to the power of 4 so I know EB1 EB2 EB3 do I know R1 R12 R13 yes because 1 minus epsilon 1 epsilon 1 A1 and 1 minus epsilon epsilon and A s are known means known means R1 R2 R3 I know that is surface resistances are known do I know F1 2 F1 3 F1 4 sorry F1 3 F1 F2 3 and F what is the other one F1 2 do I know these three I can compute from the geometry I can compute this this is not an issue so if I come that means I know all resistances I know all emissive powers what are unknowns then in these three equations j1 j2 j3 I have three equations three unknowns if I solve them I will get j1 j3 j3 okay so from there there is a problem here I do not think I need to spend time on this problem do you think this problem I need to solve because it is just yeah if time permits we will come back to this but actually in a you have a cylinder the top surface is one bottom surface is two and the circumferential surface is three so I have surface 1 surface 2 and surface 3 1 and 2 are having emissivities of 0.8 and 0.4 3 is having an emissivity of 1 it is black and temperatures are 1 is 700 and the bottom wall is 500 and the circumferential surface circular surface is 400 Kelvin so I can whatever I just derived the same circuit is valid okay so because there are three surfaces it need not be triangular only as long as there are three surfaces whichever shape you can think of that is applicable okay so if you put that so if you just take that and you have the equation that is eb1 – j1 eb1 you know eb2 you know eb3 you know okay and 1 – eb1 these three you can compute okay and now all that you need to do is you have to compute f12 f13 and f23 f11 is 0 okay because what is 1 the top surface 1 is f11 is 0 now you get f12 and f13 how do I get f12 that is top surface to bottom surface 1 to 2 it is a function of ri and rj that is the radius of the top surface and the radius of the bottom surface and the distance between the two surfaces so that is ri, rj and s if I plug in all this in this relation you are going to get 0.38 so you get 0.38 as f12 now you will get f13 as 0.62 but you need f31 so how do I get that is I will put reciprocity rule I get f31 equal to a1 f13 upon a3 you get 0.31 so now you got f12 f13 and f31 do I need anything else that is it that much is sufficient for me so now if I go ahead and substitute for and do I need yeah why have I not done f23 here f13 and f23 are symmetric by symmetric rule f13 and f23 are same that is why I have not set up or I have not completed again f23 separately is that okay so now if I substitute all these values I am going to get 3 equations and I am going to get j1 j2 j3 having got j1 j2 j3 this is important step yeah this problem has relevant in this step otherwise you may miss that that is q dot 1 equal to a1 into f12 j1 minus j2 plus f13 into j1 minus j3 how did I get this from this okay that is j1 minus this is the net heat flux see one is interacting with 2 and 3 so j1 minus j2 if I take j then I do not have to take the space resistance sorry surface resistance and if I take 1 and 3 I do not have to take if I take j1 and j3 I do not have to take the space resistance again for 1 and 3 so that is what I have done here so a1 j1 minus j2 upon 1 upon a1 f12 plus j1 minus j3 upon f13 plus j1 minus j3 upon 1 upon a1 f13 that gives me 27.6 kilowatts similarly for q2 I get minus 2.13 kilowatts and q3 I get minus 25.5 kilowatts so the net heat flux should be the algebraic sum of all this should be 0 the net heat transfer that is by conservation of energy so you get q1 dot 1 that means it is getting what is that what is happening q dot 1 see you see the temperatures you will be able to understand the temperature t1 is at higher temperature t2 is at lower temperature and t3 is till lower temperature so q12 that is q dot 1 what is happening q dot 1 q dot 1 is coming out to be positive that means it is it is radiating heat q2 is receiving heat and q3 are also receiving heat that is the interpretation of plus and minus side with this we will there is another problem here in which what we have done is another important concept here is insulated if it is insulated heat flux is 0 if heat flux is 0 eb3 equal to j3 so with that if you I think you are able to see that so with that I will move on to next concept called radiation shields so this I will be going really breezing through so let me see how fast I can go so here I have one surface I have another surface I have mentioned this already in case of super insulator in while building super insulator so these are eb1 and eb2 if there was no shield I would have had 1 minus eb1 eb1 a1 1 minus eb2 eb2 a2 but now I have a shield so the resistance between these two space resistances 1 upon a1 f13 here space resistance is 1 upon a3 f32 now what is this surface resistance on this side that is 1 minus eb3,1 the side which is facing 1 is called 3,1 the side which is facing 2 is called 3,2 1 minus eb3,1 upon eb3,1 a3 and 1 minus eb3,2 eb3,2 a1 so these are the two resistances which are added because of adding the shielding material and if I play with my epsilon properly I should be able to arrest the heat transfer from one surface 1 to surface 2 if I have to have this resistance what should be my epsilon low or high low if the epsilon is low that means my resistance will be high if epsilon has to be low means typically reflectivity has to be very high transmissivity is 0 so reflectivity has to be for shielding material you have to again use aluminum foil I am going back to my chapati example so that is why we use aluminum foil because we are using it as a radiation shield. So these are the resistances so I have taken if I take a1 equal to a2 equal to a3 and f12 equal to f32 equal to 1 the above equation reduces to this so if there is no shield I would have had this heat transfer if there is a shield this is the additional resistance which is coming to picture now if I have n shields if I have n shields these many n shields will be n resistances should be added up those resistances are this so q.12 with 1 shield equal to 1 upon n plus 1 into q.12 no shield under the condition that epsilon 1 equal to epsilon 2 equal to epsilon 3 comma 1 equal to epsilon 3 comma 2 so only says that if I use 1 shield the rate of the radiation heat transfer has become half if I use 2 it becomes one third if I become if I use 9 shields it becomes one tenth like that if you have infinite shield if I infinite number of shields there should be no heat transfer radiated to the other plate that is precisely what we are doing in super insulator. So with this radiation shield concept I know I have gone very fast but I request you to go through these steps there is nothing great in this it is just putting resistances so there is nothing great in terms of concept there so I request you to see and there is a problem again setup when there is no shield you have 3 6 2 5 watts per meter square when you have a shield you get 8 0 6 watts per meter square because emissivities are different now you can put in excel sheet and play with these emissivities and see what happens to heat flux. So now an important concept I want to touch upon before we move on to tutorials so that is radiation effect in temperature measurement why I am wanting to emphasize this more because we always believe that if I put a thermocouple or a thermometer anywhere it is going to give me a temperature not so under what conditions the thermocouple and thermometer is going to be representative of the temperature what I want to measure that is very important how will I get that so now let us just take a thermometer or a thermocouple which is inserted in a pipe okay which is inserted in a pipe what is that I am doing I am doing energy balance for this thermocouple bead or the thermometer bulb that is there are I am neglecting conduction so convection see what is happening here there is thermocouple which is going to show the temperature and fluid is having its own temperature which is what I want to measure through thermocouple and wall T wall is the wall temperature it is having wall temperature what is that it is happening I am assuming that the fluid temperature is very high fluid temperature is very high and wall temperature is definitely lower than the fluid temperature and thermocouple has to pick up the fluid temperature okay so now the heat transfer is from fluid to the thermocouple fluid to the thermocouple and there might be some radiative heat transfer between the wall and the thermocouple now if I set the energy balance between the thermocouple or the thermometer between the thermocouple and the thermometer and the fluid and the wall what is happening convective heat transfer has to be equated to the radiative heat transfer that means whatever the thermocouple is receiving heat by convection it has to be balanced by the radiative heat transfer between the thermocouple and the walls of the pipe how do I do that heat transfer convection heat transfer coefficient into Tf that is the fluid temperature minus T thermocouple T thermocouple that is thermocouple is not going to show the same temperature as the fluid there is a temperature difference why why because there is convective heat transfer coefficient that is H there is a heat transfer between the fluid and my thermocouple or the thermometer now that should be equal to epsilon thermometer that is you see here I had mentioned this when I inserted a small pressure around I had given as an example when I put a tennis ball in this room what matters is the emissivity of the tennis ball not the emissivity of the walls precisely that is what I am doing here that is here the radiative heat transfer I have taken the emissivity of the thermocouple only it does not matter what is the emissivity of the pipe wall assuming that here what is the assumption I have invoked is that my thermocouple bead or the thermometer bulb is very much smaller compared to the pipe size or the cavity in which it has been enclosed so that is epsilon of the thermometer or the thermocouple into sigma into T thermocouple to the power of 4 minus T wall to the power of 4 now let us rearrange this little bit little differently that is I get T f equal to T thermocouple plus epsilon thermocouple into sigma T thermocouple to the power of 4 minus T wall to the power of 4 upon H so you see ideally what is that I am expecting whenever I insert a thermocouple I expect that my thermocouple is going to show the fluid temperature not always so why because of this correction term when will fluid temperature be equal to thermocouple temperature or the other way around when will be the thermocouple temperature equal to the fluid temperature when the heat transfer coefficient is very high only for high speed flows high speed means not a very high speed what I mean is when the velocities are reasonably high for example air velocity is let us say around 3 to 4 meters per second the heat transfer coefficient between my thermocouple and the fluid fluid which is flowing over the thermocouple only when H is high that is for example if the water is flowing this radiation error radiation effect error will be very small but if it is air and the velocities are low H will be low then this error will be very high so or the deviation error may not be the right term the deviation would be high so if I have to keep the deviation low I have to keep this H high and the epsilon that is the emissivity of the thermocouple has to be as low as possible if I have to keep the emissivity low means what should I keep what should I what what can I control I can only control reflectivity all solid surfaces as I said transmissivity is 0 if it is highly flashing reflecting surface then reflectivity is high and emissivity is low but that is not always possible to maintain because once I insert it if there is a fluid it gets rusted then naturally that glaziness is lost and it becomes diffuse so you have this deviation we need to take note on I think I think we have 5 more minutes so we I will just quickly give you the feel of this deviation or the error not error is the right term although I am I am just using it just for the heck of it I will just give the feel of the deviation with the numbers I have a problem in which wall is maintained at 400 Kelvin and the temperature reading that is the thermocouple has is 650 degree Celsius who has given some thermocouple reading is 650 what is that I am looking for I am looking for fluid temperature and the emissivity of this bead is 0.6 Kelvin so now if I plug in this here and of course heat transfer coefficient has been given 80 watts per meter square degree Celsius it is significantly high it is not natural convection it is forced convection and I we can take this number as for air may be around 4 to 5 meters per second so if I plug in these numbers and with the thermocouple temperature as 650 Kelvin and wall temperature is 400 you see what is the deviation I get 65 degree Celsius if the H is low let us say I am taking a still room and H is low means we said that in under natural convection conditions H can be as low as 5 to 10 10 watts per meter square degree Celsius if I put in 10 if I put in 10 here what will happen 8 times my deviation will go up that means 65 into 8 so imagine the deviation between the fluid temperature and the thermocouple temperature so the point what I want to emphasize is that not always when you insert a thermocouple or a thermometer anywhere and everywhere it is going to measure the temperature what it is intended for that is the point what we want to drive home here if it is a low velocity flow or a natural convection most of the times measuring accurately the temperature is not an easy task so we need to keep this in mind because you may think that I will go ahead and measure I will try to go ahead and try to quantify this H and try to quantify this epsilon and make the correction most of the times it is just not possible simply because I will not be able to compute this H and epsilon thermo epsilon thermocouple that accurately and also it varies with I mean it changes with epsilon especially changes with time so there are various ways to circumvent this problem let us not get into that but for now what is important is there is going to be deviation where in which you have to control H and epsilon.