 Regarding insulation, I think this is also one of the favorite topics as far as undergraduate heat transfer courses are concerned, critical radius of insulation and what we do with that. Insulation essentially we have seen or from intuition we know that it is used to reduce the heat transfer. So, by adding more and more insulation, we have almost always thought that the temperature I mean the heat transfer rate is going to decrease. Therefore, how much ever you had it is not going to matter in the overall scheme of things, but that is not the truth because if you look at I will just quickly in couple of minutes draw these and go ahead. So, if I take a plain wall, this is the wall of thermal conductivity k and if I have an insulation with thermal conductivity k i n s and this is this temperature is T 1 and this is T infinity here. I have Q according to Fourier's law is T 1 minus T infinity divided by summation of R thermal, where R thermal is essentially L wall L insulation. So, L wall by k a plus L insulation by k a plus 1 over h a. This 1 over h a and L wall by k a are going to be there irrespective of whether we put in this insulation or no. So, these two are constants. This quantity L insulation by k insulation a is going to increase increase in thickness. So, as L insulation increases logically we are going to see that the overall total thermal resistance increases and therefore, Q is going to drop nothing great about it it is not a problem. And as we saw in from the morning for a plain wall the area associated with heat transfer is constant whether I take this surface or this surface or this surface the normal area to heat transfer is the same a. So, because for a plain wall a is constant along x direction Q is constant as well as Q double prime or heat flux is also a constant meaning it is invariant with respect to x. This is something which we have to understand and appreciate very clearly. In fact, students should be tested on this concept simply because we use heat flux and heat transfer rate interchangeably in numerical problems, but for a plain wall it does not matter whether we use heat flux or heat transfer rate we are going to have there is no effect of the area as a function of the cross sectional or as a function of the axial location, but if I look at a radial system like a sphere or a cylinder what is going to happen is the following. So, let us take this is a cylinder and whatever I do for a cylinder is valid for a sphere. So, this is t 1 and this is h comma t infinity and this is just a wall a pipe which is carrying steam is going to lose heat from here to here. So, this is the direction of heat transfer. So, I can write Q is equal to t 1 minus t infinity please note I have this resistance inside for steam flow is also constant. So, I am not taking that I am writing this temperature difference. So, I am writing the temperature difference t 1 minus t infinity. So, I will have the resistances associated with conduction through the wall and convection. So, this will be log R 2 by R 1 by 2 pi L k wall plus 1 over h a surface. And this 1 over h a surface is what is nothing but 1 over h times 2 pi R 2 L if this is R 1 and this is R 2 this is 1 over 2 pi R 2 L. So far so good no problems now I am going to put an insulation material to this pipe. So, R 1 R 2 and R 3 and we have the same h comma t infinity this surface is maintained at t 1. I can clearly say that I am going to have an additional resistance this is the conduction resistance for the wall this is the conduction resistance for the insulation and this is a convective resistance for the heat flow from inside to outside the direction of heat flow is fixed. We have an additional resistance in the path. So, I will write Q is equal to t 1 minus t infinity log R 2 by R 1 2 pi L k wall which is also constant plus these are the two important terms log R 3 by R 2 2 pi L k insulation plus 1 over h a 3. These two terms have come in instead of this one term what is what are these two terms log R 3 by R 2 divided by 2 pi L k insulation plus 1 over h 2 pi R 3 L. This is with insulation and without insulation it is going to be 1 over h 2 pi R 2 L without insulation. Every other resistance in the problem is constant if there is a convection on the inside of the pipe any other material associated with the pipe wall we do not care everything else is a constant what is changed is these two quantities. So, t 1 minus t infinity divided by something plus constant value plus 1 term in case 1 which is equal to in second case with insulation t 1 minus t infinity constant plus term 1 plus term 2 two additional terms which are given by this. Now, I cannot say what is the effect of keeping on adding insulating material why if I keep adding insulating material this R 3 is going to increase this conduction component of the resistance this conduction component. This is R conduction of insulation this is R convection on the outside of the insulating surface surface area for a larger cylinder is larger. So, when the radius was R 2 2 pi R 2 L R 2 is smaller than R 3. So, this area is going to be smaller. So, the resistance here is larger whereas, here this resistance is going to become smaller because R 2 is smaller than R 3 1 over h 2 pi R 2 L is going to be larger than 1 over h 2 pi R 3 L obviously. So, what is happening the bottom line is I am adding more and more insulation the thickness is increasing conduction resistance is increasing because this term is getting bigger and bigger. But I am going to see a drop in this quantity with additional resistance being added and when I keep doing this adding of insulation I will come up with a situation where I might actually see a increase in the heat transfer as opposed to a decrease in the heat transfer. So, if I plot these two resistances this is the x axis is resistance this is the thickness of insulation or thickness of insulation. So, or R 3 if you want to say as R 3 increases R convection decreases R conduction increases and you will come up with a summation of resistances which look something like this and all this has been illustrated here most of us are familiar also. So, additional insulation increases the conduction resistance of an insulating layer, but decreases the convective resistance of the surface because of the increase in the surface area. And heat transfer of the pipe therefore may increase or decrease and this happens because of the fact that in radial systems though q is equal to constant that is sacred we cannot change that. So, q at any cross section is the same how many watts is coming out from inside is what is going to go out, but heat flux is not a constant that means heat flux as R increases surface area increases heat flux progressively decreases as R increases means a is equal to 2 pi R L or a equal to 4 pi R square does not matter you go outer and go to the outside surface of the cylinder progressively from the inside the associated area for heat transfer is more. Therefore, heat flux is going to be progressively decreasing. So, that is the reason for this concept. So, all this is illustrated here in case of a cylindrical pipe this is the additional resistance which is there and as I told you all other resistances are constant these are the two resistances which are going to play a role conduction of the insulation convective resistance on the outside of the insulation. So, if I want to optimize if I want to have q R as optimized value then I should have an optimal thickness of insulation which can be obtained by differentiating this expression d q basically differentiate this one when this is minimized q R is going to be maximized so on and so forth. So, all of us have done this mathematically where the independent variable is now going to be R 2. So, d by the R 2 of this denominator expression differentiate set it equal to 0 you will get critical radius of insulation of cylinder is k over h and for a sphere it will come to k over h. So, what it tells me bottom line is the rate of heat transfer from the cylinder increases with the addition of insulation until where until the point where you reach critical radius of insulation. So, from a bare cylinder you add a little bit of insulation the heat transfer is going to increase as long as the value of the outer radius is smaller than the critical radius calculated this way it reaches a maximum when R 2 equal to R critical starts to decrease beyond that point that does insulating a pipe may actually increase the heat transfer rate instead of decreasing when R 2 is smaller than R critical. I told you here if you go back to this graph on the white board this is the summation of the resistances here R total. So, as the resistance progressively summation of the resistance progressively decreases your heat transfer rate is going to increase. In fact, that is what is plotted there the reciprocal is what is plotted the heat transfer rate increases between the bare surface to the location of R equal to R critical and then you see a drop in the heat transfer. So, this is something which all of us know critical radius of insulation people derive ask questions on that, but we need to keep in mind that this portion involves increase and where do we use this do we need to ignore this or do we have to be so careful in doing this we should put insulation little beyond R critical how much you have to put we have to be very careful. So, yeah the next transparency essentially will give you a feel of how important this aspect of critical radius is. So, value of critical radius will be largest when k is large and h is small as seen by these expressions k over h or 2 k over h here. So, lowest value of h typically we see in natural convection situation which is about 5 order of magnitude we are not talking of actual numbers around 5 in gases natural convection gases and insulate common insulating material let us say it is 0.05 watt per meter square Kelvin we are talking of a critical radius of insulation of about 1 centimeter or 10 mm. So, if you have radiation effects this value is going to be even smaller. So, what we are saying is though there is a curve which is shown like this this is the bare surface this is the critical radius essentially this portion is hardly 1 centimeter. So, even when we apply insulation we are almost always beyond this portion. So, that is something which we need to appreciate that is why in most general situations we do not care about critical radius of insulation that much when we are doing things practical. So, in case of force convection this is still lesser critical radius would be much less than 1 mm because of the larger h value associated k over h therefore, we can insulate hot water or steam pipes freely without worrying about the possibility of the increasing heat transfer by insulation of this pipe. So, the logic that we use for plain wall just more the insulation merrier is probably valid here electrical conducting wires where you want heat dissipation to be more you would want it to be below the critical radius of insulation. Therefore, plastic electrical insulation may actually be used to enhance the heat transfer from electrical wire and thus keep the steady operating temperature lower and at safer levels. So, this is with regard critical radius there is a small example problem which will solve this is a 3 millimeter diameter 6 meter long electric wire is tightly wrapped with a 2 millimeter thick plastic cover whose thermal conductivity k is 15.15 watt per meter Kelvin. Electrical measurements indicate that a current of 10 amperes passes through the wire there is a voltage drop of 8 volts along the wire. So, when there is a current of 10 amperes there is a voltage drop of 8 volts. If the insulated wire is exposed to a medium that is offering heat transfer coefficient of 12 which is typical natural convection value and temperature is 27 degrees centigrade determine the temperature at the interface of the wire and the plastic cover in during steady state operation also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature. So, it is a straight forward problem on insulation multiple resistances and we have to also deal with critical radius. So, if you look at the diagram this red portion represents the current carrying conductor whose diameter is 3 millimeter radius 1.5 millimeter length is 6 meters here then there is a plastic cover this blue one is a plastic cover material of thermal conductivity 0.15 this insulating material. And when you pass the current of 10 amperes voltage drop is 8 what it means is that electrical heating kind of situation is given in terms of voltage and currents. So, power is equal to V times I V and I are given. So, power or wattage can be obtained this heat has to be dissipated out. So, if the insulated wire is exposed. So, the heat is being carried away by a fluid which is basically air probably which is giving a heat transfer coefficient of 12 and a temperature around of the medium is 27 degrees. So, the resistance diagram here why have we drawn only this part of the network we have drawn this part essentially because we are dealing with the insulating plastic material that is the blue portion that is air plastic and air convection represents the outside convective heat transfer or resistance. For this wire it is a heat generating problem we cannot apply we cannot draw a thermal resistance directly because this is having a situation where q is not a constant there is a volumetric heat generation whatever heat is coming out that is what is generated that is going through this insulation and the conducting medium. Therefore, this q what I what we have shown here is V times I, but that resistances associated are this plastic and the convective part T 1 represents the temperature at the interface of the conducting wire and the plastic. This is what we have to find for this q T 1 minus T infinity T infinity is known if I calculate these two resistances I can apply my basic equation which gives me the appropriate answer, but again we follow a certain methodology. So, we have a list of assumptions that we have to write first thing is we have thermal conductivities which are known and constant there is negligible contact resistance at the interface between the wire and the plastic material. There is negligible radiation and even if there is any all the heat transfer coefficient value takes care of it one dimensional radial conduction. So, that is very important because we are studying only radial conduction at this point and it is steady state environment steady state system because there is no change with respect to time. So, after all this we get q is equal to 80 watts and a 2 which is the area associated with this surface that we have calculated 2 pi r 2 l. So, r convection can be calculated r plastic is conduction log r 2 by r 1 calculate the value at these 2.78 degree centigrade per watt. Therefore, the interface temperature is unknown T 1 I can calculate this as 89.4 degree centigrade. So, next part of the question was if the thickness was doubled what would happen that is also determined whether doubling the thickness of the plastic cover would increase or decrease this interface temperature. For doing that we first have to calculate the critical radius the critical radius comes to be 12.5 mm. This critical radius is definitely is larger than the radius of the plastic cover. Therefore, what is the radius of the plastic cover radius of the plastic cover is you have seen this here 2 pi r 2 l 0.0035 that is essentially 3 mm plus thickness which is been given to you already 3 mm of the diameter of the wire and 2 mm plastic. So, half of it would be the radius. So, 0.001 1.5 plus 2 1.5 is the radius 2 is the thickness. So, 3.5 mm. So, 3.5 mm instead of that if I make this thickness double I would be adding another 2 mm. So, 4 plus 1.5 which will be 5.5 mm. So, this which is larger than the radius of the plastic cover. Therefore, increasing the thickness would enhance the heat transfer until the outer radius reaches 12.5 mm. That means, we are in this part of the curve let me go back here we are somewhere here I have increased this is this number is 12.5. We are here at 3.5 if I make it 5.5 also we are in this part of the curve therefore, increasing the radius increasing the thickness of this plastic sheet will actually increase the heat transfer in this case rather than decreasing it. So, that is something which we have to keep in mind. And as a result the rate of heat transfer will increase when the interface temperature is held constant or T 1 will decrease when power or voltage times current the word power is missing T 1 will decrease when power is held constant which is the case here V times I is held constant. So, you can calculate whatever if the power is held constant you can calculate the temperature and I think it comes out to be 71.14 or 77.54 whatever. So, this is a plot which has been made to illustrate the temperature at that interface and the radius of the insulation. So, you have this plot which is essentially obtained from this equation which we have used. This summary we have already covered plane wall V square T by dx square just for even though we are repeating we do not mind doing this simply because we have to understand take home as we say at the end of this student what he what he or she takes home is this finally, what is going to be asked in the exam I think this is what most students will ask essentially if you remember all this we are in good shape and we do not have to memorize everything. Heat flux remains constant heat transfer rate remains constant for a plane wall heat flux not a constant for cylinder not a constant for spherical wall heat transfer rate is constant irrespective of the radius for both these cases thermal resistance conduction L by k a this one for cylinder this one for sphere convective resistance 1 over h a temperature distribution. We do not have to memorize it but point to be noted that this is the linear temperature distribution when you have no heat generation it is a straight line distribution whereas, this is logarithmic and this is some other functional form for spherical wall. So, we have left out one problem in the middle which we remember we will take it up tomorrow from now on till about 530 we are open for any questions across all the institutes preference will be given to people who have not asked any questions so far but we will try to accommodate everybody. So, S V N I T Surat yeah your question please. Sir, in case of the insulation thickness what would be the better 10 meter in case of the insulation thickness of the wall what would be the better heat transfer case 10 meter 1 wall or the each wall of the 1 meter 10 wall of each meter. Please rephrase your question. The thickness of the insulation the 10 meter and the second case is the 10 wall of the 1 meter so which one would be the give heat based heat transfer. The question is this the question asked by one of the participants is that if I have an insulating material is it good idea to take 1 meter whole as 1 insulation or I will break them into 10 pieces and then add one after another. So, that is the question. I think the answer is if you have 10 pieces added will add the interfacial resistance so that would be better but then I do not think it is fabrication wise or putting insulation wise putting 10 times is easier. So, we usually put whatever is available in the market we directly put so typical thicknesses whatever are available we directly put them but we cannot fragment them for 1 mm 1 mm 1 mm and then put them several times. Yes theoretically if you put several times interfacial resistances are increasing so definitely putting multiple layers is better than one layer no doubt about that but practically I do not think it is always possible if you are imagining a complete power plant and we have to insulate the complete power plant one of the heat exchangers of a power plant it is not going to be easy. Yes in a lab scale if you want to do something we can definitely do that. Gopal yeah please go ahead. Yes sir question is the regarding energy generation because this is the equation of energy conservation here why we are writing it yeah energy generation because there will not be any generation of consumption by a material if there is a reaction take place a chemical reaction in which generation of energy is taking place they are supposed to be a consumption of energy also will be that means energy generation may be a positive or negative or it will be a always positive. To rephrase this question there is a question on energy generation the thing is why is there energy generation considered in the equation I think the he is saying that there will be only energy in the reaction there will be conversion from one form to another or something like that is the question. But what we are saying is the heat diffusion or conduction equation is a very general form of equation and mind you chemical energy a chemical reaction we are not concerned with chemical energy or any other form of energy we are concerned with heat energy only. So, this is a heat balance equation if if the reaction is not generating any heat and it is proceeding like that H 2 H 2 plus O 2 is giving you two molecules of water we are not and if it is a non exothermic or non endothermic reaction that is not going to feature in the energy budget. So, when a reaction gives out or takes in heat energy in the form of heat that is what we are going to take in the energy generation term. So, we are chemical reaction nuclear reactor all these things where thermal energy is associated we are dealing with thermal energy only. So, actually to add on we can imagine like this in this room in your room whichever room you are sitting. So, if there is an electrical heater at the center of the room that is the energy generation term or otherwise if someone is spraying water on to us then that is the energy generation in this case will be negative. So, energy generation does not mean always it is positive the second part of your question. So, energy generation can be positive energy generation can be negative it depends on the process or it depends on the case by case as I said there are two examples in one of the examples electrical heater energy generation is positive if you are spray cooling if it will be energy generation will be negative because you are cooling. So, energy generation can be either positive or negative depending on the situation. So, any energy generation typically would involve conversion from one form to thermal energy if it is conversion from chemical to mechanical energy we are not concerned with that we are concerned with transformation we do not even care what is the source we just want to know whether in a given problem there is going to be thermal energy coming generated or thermal energy lost because of a process. Very good question this is only about thermal budgeting this is only about thermal budgeting. Not any other form that is why it is called heat diffusion equation not energy balance general energy balance equation common call it Salem yes please go ahead. As I said thermal conductivity is the property of material please elaborate the relation between thermal conductivity of material and its temperature. Thermal conductivity varies with temperature now whether it is going to vary linearly non linearly whether it is going to increase decrease is all dependent on the molecular structure associated with the material. So, there is no general rule or general relationship for any unique answer for this question actually. So, for a given material what is the variation you can find out next I think there is one more question quickly. The variation of heat transfer rate with radiation. It is what is the amount of heat that is here sir. Sorry your audio is really bad we are unable to hear anything there is lot of disturbance something related to 2.16 ok. Yeah yeah. Ok. This figure is common for both cylinder and radius. Yes yes. Qualitatively. Qualitatively this is not mathematical function it is a qualitative representation ok nature of the curve is like that increasing decreasing curve with a maxima. But the critical radius is going to be different for cylinder and sphere for cylinder it is going to be k by h and for sphere it is going to be 2 k by h. But the nature of the curve initially increase and later on decreasing and a maxima existing is going to be same for both cylinder and sphere ok. Ok. Thank you sir. Thank you very much. Sheerpur college over to you. Sir I have a small question in the equation of critical radius of insulation we got for the cylindrical system as k by r c is equal to k by h 0. And during your explanation through graphs you said that at this condition the heat transfer rate will be maximum. So how we can say that the heat transfer rate will be maximum. Will you please elaborate it to some mathematical expression over to you sir. For this we will have to go back to what we had done before on the white board. See this what will happen is as I keep adding insulation the surface area for convective heat transfer increases and this causes the convective resistance to decrease what is the convective resistance r convection is 1 over h a ok. This is going to drop as r increases that means as I add insulation conduction resistance is going to increase because you are adding more material. So greater the thickness conduction resistance is going to increase as r increases. So the summation of these two would give you a curve which looks qualitatively something like this r total and it comes to a minimum value somewhere which we essentially is what we call as the critical radius of insulation. How did we get this? It is the summation of these two curves and if you write q is equal to delta t divided by r r total and in this case all resistances except these two are constant. If I say d by d outer radius which is the variable of r conduction plus r convection is equal to 0 I come up with that so called critical radius of insulation and that is this point. This point represents the point of minimum resistance and therefore a point of maximum heat transfer rate. So for the application we have to put the insulation much much larger than this critical radius of insulation. So we cannot we are not supposed to put the insulation thickness equal to critical radius we have to put much larger. Our concern is by chance if we put our insulation much lower than the critical radius that is the concern. In the next transparency professor told us that typically for all applications it is around 10 mm. Typically any insulation you will not put anything less than 10 mm you will definitely put more than 10 mm. So most of the times this care is taken care of. However if one wants to do the calculations and show we have to ensure that our critical radius of insulation is much sorry our insulation thickness is much larger than the critical radius of insulation that is the concern. Yeah go ahead. My question is the same again on critical radius of insulation the graph you shown of q max versus R critical. If the situation is of adiabatic type then for which practical application how will we vary q with R? For adiabatic situation then for what critical radius of insulation we show. Let me ask you back the question how do you generate an adiabatic condition the question asked by one of the participants is how do I calculate or what is the usage of this critical radius of insulation under adiabatic situation. My question back to the participant is how do you generate adiabatic condition? Professor please tell me how do you generate adiabatic condition in any heat transfer application for example heat exchanger. How do you generate an adiabatic condition? No sir adiabatic it cannot be generated but we can provide insulation but I mean to say that how much by insulation we can go down for the heat transfer. Okay and for that what should be the critical radius that curve is going on decreasing up to what we can go. Okay see the question now I understood the question asked is to generate adiabatic condition what should be the radius of insulation. Okay to answer this question there is no way that you can make you can get purely adiabatic there is going to be heat loss no matter how much thickness we put there is going to be heat loss all that we are saying is that the temperature drop across the insulation is going to decrease with the thickness but there is going to be heat loss heat loss is bound to be there. See if you put a sweater if you put a sweater aren't we feel aren't we not feeling going to feel cold at all it's like asking how much thick sweater I should wear so that I am not going to feel cold at all. No matter how thick sweater I use I am going to use I am going to feel the cold but the temperature difference between the outer surface sweater temperature and my body temperature is going to decrease. There is going to be heat loss it is quite difficult or almost impossible to generate adiabatic condition just by thickening the insulating material. Good evening please ask the question we are able to hear you. So my important question is why suction lines are always insulated and whether it is to be insulated below critical radius or above critical radius. What lines? Suction lines of compressors in case of refrigeration systems. The question is why do we insulate only the suction lines of refrigeration equipment and if at all if it is insulated on the suction side whether that insulation should be above or below the critical radius. This is the question asked by one of the participants. So I don't think we should be insulating only the suction side. Why not on the pressure side? I think all the heat exchange equipment whether it is condenser or evaporator. Compressor. Compressor we will have whether on the suction side or on the pressure side whether it is evaporator or the condenser both have to be insulated and definitely whether it is condenser or evaporator the insulation thickness has to be more than the critical radius of insulation. He is talking about are you specifically asking about compressor? Suction line of suction line joining from the evaporator to compressor it is always insulated. I am interested to know why it is insulated and what is the criteria in case of refrigeration system it is always insulated and condenser is never insulated in case of the refrigeration system. My take on that would be the following. See evaporator typically operates at lower temperature. Compressor inlet has to be as close as possible to a liquid state because compressor power is directly proportional to this specific volume of the fluid. So, if there is a phase change and it becomes starts to become a vapor then you are going to talk of a larger power consumption because you are going to handle gas. Yes, you are going to handle some amount of gas but even the inlet therefore you do not want to increase the temperature as much as possible. That is probably the explanation I can think of at this moment. Yes. My question is you have spoken at length about the various forms of heat transfer the conduction, radiation and convection. Now I would like you to elaborate on heat transfer during orientation changes. For instance in equinoxes there is actually a tendency on people that there is an increase in radiation during equinoxes. So, what exactly is the reason? Is there an inclination of the range of what exactly? Equinox. Have I heard you right? Exactly. Second question? Yeah, the second question is you have given a very nice example, a very practical example of this various forms of heat transfer through a cup of coffee. So, give me more practical sort of an example while explaining the relationship between viscosity and temperature. Viscosity, we can easily give an example. Viscosity usually increases with the increase of temperature in case of gases and viscosity decreases with the increase of the temperature in case of liquids. This is general and we also explain that through cohesion and adhesive forces because in case of gases what happens is that the intermolecular interaction or the collisions increasing with the increase of the temperature. So, the viscosity or the resistance increases in case of gases. On the other hand in case of liquids it decreases because the cohesive forces the bond decreases because of the increase of the temperature. So, viscosity we can explain but I do not know your first question. First question I will take a short at it. Equinox. The question is whether it is equinox or whatever be the situation. Probably what I think happens in an equinox I have forgotten the actual phenomena. I know that it is the day when the when the day and the night are equal to 12 hours each. But I think it is to do with the relative position of the sun with respect to the earth. So, I think that is the time when probably from radiation point of view the distance is either minimum or minimum. So, that you are going to have a larger radiation component. Because the temperature of the sun is fixed the temperature of the earth is also roughly the same at a given location probably it is the relative distance which is going to matter. And we cannot we have to fix the location. So, if you are at a particular place probably what you are saying if it is true the only reason I can think of is the relative orientation of the two celestial bodies that is the earth and the sun. I do not have a better explanation than this. You know. Question number one, what is the influence of thermal diffusivity and kinematic viscosity on heat transfer when the heat transfer is taking place between the surrounding and spherical cell? Assume that there are some fluid inside the spherical tank. What are the influence of thermal diffusivity and kinematic viscosity of the fluid on heat transfer? Number one, question number two what are the factors which influence the overall heat transfer coefficient when the combination of convection and conduction is taking place? We will take the second question. There are two questions we will come back to first little later I will rephrase question number two. His question was what are the factors which influence the overall heat transfer coefficient which involves conduction and convection. The factors that influence overall heat transfer coefficient u are essentially the factors which influence the convective heat transfer coefficient and the conduction resistance. So, from definition we know u is equal to u a is equal to reciprocal of the thermal resistances. So, it depends on the nature of the fluid, it depends on the nature of the flow, it depends on the type of geometry that you are dealing with. So, exactly the same parameters which influence h the same parameters influence u, the same parameters which influence the resistance associated due to conduction that is what is going to influence u. So, it is a combination of these two which is going to influence u. The first question could you please elaborate again. The ambient is air at room temperature. What are the influence of thermal diffusivity and kinematic viscosity that is transport properties of a fluid on heat transfer. For example, if the more heat is transferred from the ambient to the inside the spherical shell for example, liquid nitrogen is available because of the transfer is there because of the change in properties transport properties. Then the evaporation rate will be higher, then there is a chance for the explosion of liquid nitrogen. Thank you. Actually we have not there is a question on influence of thermo physical properties that is alpha and nu on liquid nitrogen or spherical shell situation. We have not understood the question very well, but we will try to explain it with the problem that we have illustrated. So, see if we have to see that whether the leakage or the leakage loss is more all that we can say is that these resistances should be as less as possible. So, that means the heat loss can be high only when the resistances are less. So, I can see only when the resistances are less means the convective resistance can be made less only when the heat transfer coefficient is high. So, but then I cannot make inadvertently this heat transfer coefficient high. Yes, I think now perhaps I am understanding your question what you perhaps mean is yes, compared to natural convection if I have a forced convection let us say I put on a fan in front of my container then my convective resistance decreases and then my heat transfer rate increases and the leakage loss rate will increase. So, essentially it is boiling down to the resistances. So, to answer thermo physical properties that is the new coming in here is only in terms of Reynolds number if it is a forced convection. So, other than that I cannot see directly the kinematic viscosity effect coming in here and affecting the heat loss rate. If it is natural convection yes it is going to be Rayleigh number which is G beta delta T divided by alpha nu. So, as the nu increases my Rayleigh number decreases the natural convection effect will decrease. So, essentially whether it is natural convection or forced convection the thermo physical properties that is the kinematic viscosity is indeed having an effect, but ultimately this effect will be on H only. So, that is how I think we can understand this problem. There are some cases practically cases where the inside the spherical cell some fluid is there, liquid is there there may be a chance for turbulence while traveling. So, if the turbulence comes to a picture then the Reynolds number is also increasing. In that case what is the effect of thermo physical properties on heat transfer? See the question is the question asked is what if there is a turbulence inside the container of this problem inside the container of this problem. Yes professor had told that we have neglected the resistance because of the container because inside the container of the fluid. If there is turbulence inside ultimately what will happen there is no flow it is a container. If there is a turbulence then there can be some natural convection occurring inside. So, there is a resistance added on if there is resistance added on actually the heat loss rate should decrease. So, it is going to be a good situation if there is turbulence inside the loss will definitely decrease it is not going to make my life difficult. It is only going to aid my heat loss in terms of heat loss. How thermal conductivity varies with density of solids liquids and gases. See thermal conductivity is directly related with the density only. See why is for example if I ask myself a question why is thermal conductivity of solid I keep saying I have been telling this since morning twice or thrice thermal conductivity of copper is 400 thermal conductivity of water is 0.6 thermal conductivity of air is 0.02635 watt per meter Kelvin. So, if you see here solid is hundreds liquid is points one one decimal and gases is two decimals. So, why is this answer is in your question itself that is density. So, the density of solids is very high the density of solids is of the order of 8000 this density of liquid is of the order of 1000 that means density has decreased around 10 times. Now, if I go to the density of air it is almost 1000 times lesser. So, that is how even the thermal conductivity is also decreasing how closely my molecules are spaced so that the molecular diffusion or the molecular interaction is aided or abated decides my thermal conductivity. So, thermal conductivity and density are directly related. I am not comparing conductivity of metal solids liquid and gases, but in the same solids suppose we are taking more dense material then thermal conductivity will be different. So, within the same medium itself what is the effect of density? Same answer is again same if I have a denser material thermal conductivity of denser solid material will have higher thermal conductivity as opposed to again solid whose density is lower. Let us take an insulating material if you go on packing it densely the thermal conductivity of the insulating material will be very less but if I pack it loosely then the porosity has increased so thermal conductivity may not be very high. So, basically thermal conductivity of any material depends on how densely I pack it. Okay, thank you. Over to Vijay T.I. Yeah, my question is? Yeah, go ahead. The temperature varies with temperature. Question is how does the thermal conductivity of an alloy vary with temperature? Again, we cannot give an unique answer and say that thermal conductivity for all all alloys is going to increase with temperature or decrease with temperature. It depends on the material. It depends on the material but one question has been pending since morning is why the thermal conductivity of an alloy is lower than the independent materials? That question we have kept it as pending. We will look at that question and come back but there is no unique answer for this temperature variation of alloys. Back to you. Particularly for nonferrous alloys how the thermal conductivity varies with temperature? No, why only nonferrous alloys? For any alloy thermal conductivity is going to vary with temperature. It is going to be varying with temperature but I have no idea about nonferrous alloys separately or any other alloy for that matter. We have to look at, look into it. We will come back on the Moodle forum or during discussions in the morning tomorrow on this question. Yeah. My question is with critical thickness of insulation. Is there any relation between economic thickness and critical thickness of insulation? Okay. See, yeah, this is a question is is there any relation between economic thickness and critical radius of insulation? There is no unique answer. Definitely, yes. See, as we can see in the previous problem see if we see in the previous problem as let me answer this in a very generic fashion. First, first answer is that our insulation has to be definitely more than the critical radius of insulation so that my heat loss is less. But over that how much I should be putting is a question, open-ended question. So it depends on how much I can spend and how much heat loss I can afford to have. Definitely, as I said in one of the previous questions we cannot have zero heat loss. That is definitely not possible in any practical situation. We have to live with the heat loss but only thing is that we have to, we can minimize the heat loss with the increase of the insulation thickness but how much you can put that is decided by economics. If you take a power plant and even if you increase the thickness of the insulation by 1 mm, the heat exchangers are as big as one huge building. So if you increase the thickness by 1 mm the cost may turn into crores. So it is not always space. And also space, we need to have space to increase the insulation. So there is nothing like optimum economic thickness. It is all depends on how much money you have how much space you have how much you can afford to put. That is all. So there is nothing like optimal economic thickness. Only we have critical radius insulation. All that we need to take care is that the real insulation we put has to be more than the critical radius of insulation. Yeah, one more question I think. Yeah. Okay. Sir, I have a question. Is thermal conductivity of a material related with the valency of the atom of that material? Yeah, actually they might be related because it is related to free electrons. Free electrons which are there. They might be related directly if you ask me how directly they are related. I think number of free electrons dictate the thermal conductivity vaguely that is what I think. Number of free electrons, sorry the question asked is one of the by one of the participant is the valency is related to the thermal conductivity. The answer is yes I would think so. Why? Because the valency decides the number of the free electrons in any material. So, but how the valency and the number of the free electrons are related I have no idea we have to look into it. Okay. Because I perhaps feel that the answer of the answer of dependence of thermal conductivity with the alloying element I guess they are actually related. That's why I asked this question. If you have any idea please go ahead and answer that. Okay. Probably you are saying there is a decrease in the number of free electrons. It was just a guess I do not have. We also do not have clearly because in regular UG books I am not getting that. Probably true. It might be true. It might be true because We have to see that and Each parent material may have certain number of free electrons but when it is forming an alloy it might be lost completely or it might be reduced so much that the number of free electrons would be very less. Probably. Probably. Because we are not We are not seen. We have to see it and then corroborate. Yes. Yes. I. Yes sir. I also feel so. We will look in that direction. We will definitely look in that direction. It's a one clue for us. Yes sir. Thank you very much. Surat. Yes. So far we have been talking about the thermal conductivity and its variation and independency on the temperatures and the temperature. Yes sir. But my question is the material which is tested upon the Earth's surface which is at room which is at one atmospheric pressure if the same material is been capped in the space for which we are having very low pressure and temperature and suppose at the same time we are having the Mach condition of greater than 1 or greater than 5. At that time we are having the atmosphere which will be generating electromagnetic waves. So do we considering that kind of do we considering that kind of thermal conductivity variation for the space application? We have to. We have to definitely take care of there will be influence of pressure. See the question asked by one of the participants is he is the thermal conductivity going to be different he is going to be different if I measure in Earth and in space because the pressures where space and Earth are different so in space I am moving at a high Mach number that is the question asked definitely the thermal conductivity will be different dependent on pressure but how much depends on the amount of the pressure what we operate but there is no close form answer again for this question but all that I can say is that the thermal conductivity like it is dependent on temperature it is also dependent on pressure but it is not maybe as strong function as temperature. And any testing which is done has to be done at situations where it is going to be put in use so for space situation applications if you are if you are done calculations and obtain some value of thermal conductivity which is at room temperature definitely it is not going to be valid I think we are question is coming from Columbia accident where that there was a peel off because of the brick had to be peeled I mean fell off because unable to withstand the heat probably from that you are coming definitely will have to take care of the operating conditions which would be high speed pressure and all these also will translate to heat directly high speed will translate to heat I think directly so you will have to test it in those conditions in the lab simulate those conditions and that is what happened in even in the failure of challenger shuttle Dr. Feynman looked at the what was the reason for the failure of the challenger shuttle the gasket in the rockets which was being used was tested under ambient conditions but actually it was supposed to be used in cryogenic rockets which are operating at cryogenic temperatures and pressures different so definitely the conditions under which they are being used and under which they are being tested have to be same similar back to Hemant Sir my specific question is is there any is there any body force is effect on the solid material conductivity the last question you asked was is there any influence of the electromagnetic force on the thermal conductivity body force that is the body force electromagnetic force electromagnetic force that can be considered body force that can be considered as body force so the question is whether electromagnetic force which is a body force he is having an influence on the thermal conductivity on the face of it I cannot imagine any effect of the electromagnetic force on the thermal conductivity but honestly I do not know the answer if it causes rearrangement of the molecules then yes I do not know see it is a hand waving answer again so that is what professor is saying is that if structural rearrangement structure is going to change because of the application of the electromagnetic forces on my solid then only my thermal conductivity is going to change my molecular structure is not going to change why should my thermal conductivity change but then this is a vague answer we have no answer for this we have to look into it specifically Hemant why do not you come back with some answer yeah definitely sir I will get back to you later with the answer yeah okay yeah I think we are okay that is for the day