 Welcome to the 34th lecture of cryogenic engineering under the NPTEL program. In the earlier lecture, we were talking about insulation and what we had covered with the following points. We found that cryogenic vessels use insulation to minimize all the modes of heat transfer. Basically, they should be designed to minimize all the modes of heat transfer. The apparent thermal conductivity and we had defined this apparent thermal conductivity kA and it is calculated based on all the possible modes of heat transfer. It takes into account the conduction, convection, radiation, etcetera and based on all these modes of heat transfer, value of kA is normally calculated. We found out that expanded foam is a low density cellular structure. A gas filled powder or a fibrous insulation reduces the gas convection due to the small size of voids. We also found that radiation heat transfer is reduced by using radiation shields and I had shown with this with an example that if I put many shields, the radiation heat transfer gets reduced. Extending this further, the topic is cryogenic insulation and what I am going to talk in this today's lecture is vacuum evacuated powders, opacified powder and we will have a small tutorial in order to compare the performances of these different insulations. Just to go back and just to let you know how many types of insulations we are studying and we went through expanded foam which is a mass type of insulation. Similarly, gas filled powders and fibrous materials and both these things were covered during the last lecture. And now we are going to covered vacuum alone, evacuated powder, opacified powder and then in the next lecture we will have multilayer insulations. So what is going to be covered in this lecture are all these three possible insulations. So as seen earlier, the different modes of heat transfer are conduction, convection and radiation. If the physical matter between the hot and the cold surfaces is removed that is by maintaining a perfect vacuum conduction and convection are eliminated. So if I have a vacuum between two hot surfaces or two different surfaces of different temperatures, the medium air or any gas for that matter, which would cause gas conduction or any other type of conduction and convection because of the movement of the gas, this will be completely eliminated. So what this can be assured by having a perfect vacuum, however what will remain now will be radiation. So however, radiation heat transfer does not require any medium and in such cases it is the only mode of heat transfer. So if I have got a vacuum between two surfaces conduction and convection modes of heat transfer will be eliminated, while what will remain now is going to be only radiation heat transfer which depends on the temperatures of the two surfaces. So let us come to the vacuum first and as we just found that even if the vacuum is there the radiation heat transfer will be there. It is important to note that even in vacuum there is some residual gas. Now this is an additional aspect I had not talked about. So depending on the kind of vacuum, depending on the level of vacuum there will be some gas and this gas also would cause some gas conduction. These gas molecules contribute to the heat transfer by gaseous conduction. In comparison, this gaseous conduction, the quantity is going to be very very less. However, depending on the quality of the vacuum, depending on the degree of vacuum these gas molecules will contribute towards heat transfer by the mechanism called gaseous conduction which will depend on the conductivity of the gas. As the vacuum improves, this gas conduction will decrease. Naturally, if the vacuum is more and more is better and better, the amount of gas between the two surfaces will not be less and therefore the gas conduction will be less. So we can find that the gas conduction quantity will actually be related to the kind of vacuum we will have between the two surfaces. Now in an ordinary conduction, when we are talking about general conduction a linear temperature gradient is built up. We know that a surface is at T1, surface is at T2 and if I have got some medium having a conductivity k which is constant then we will have a linear temperature gradient across the two surfaces. The molecules we are talking about having gas between the two surfaces, the molecules exchange heat with each other and as well as with the surface. So these molecules are basically responsible to conduct heat from surface at T1 to the surface at T2. Each molecule will conduct the heat to other molecule and in this way conduction would happen depending on the gas conductivity and ultimately these molecules will actually touch the walls, touch the surfaces and therefore the heat will be transferred from surface number 1 to surface number 2. But in vacuum, if I am going to talk about a very high vacuum, very high order vacuum then it basically is represented by something called as mean pre-path lambda. In vacuum, the mean free path lambda of the molecules is more than the distance between the surfaces. The molecules rarely collide with each other, this is very important to understand. If I got a very high vacuum or very good vacuum, the number of molecules between these two surfaces will be very very minimum, I mean they could be of the order of 4 or 5 or some number basically. In that case, the molecule to molecule conduction will be completely eliminated or is assumption that the molecules will never emit the molecules, it will always emit the wall directly. That means the collision between the molecules will be avoided, will not be there. If the mean free path that is the distance between the two collisions, the time between the two collision, the length between the two collisions is going to be comparable with the length of the two surfaces or the distance between the two surfaces. In that case, the molecules will rarely collide with each other. Now, the energy is exchanged only between the surfaces and colliding molecules. The molecules will not collide with molecules, the molecules will directly collide on the surfaces and therefore whatever heat transfer we had said that it occurs because of molecule to molecule conduction will not occur over here. It will be molecule, one molecule will take heat from surface 1 and it will directly bring this heat to surface number 2 and this is what we are going to talk about. This type of heat transfer is called as free molecular conduction or residual gas conduction. Please understand that the free molecular conduction or residual gas conduction whether it occurs or not is going to be completely dependent on the vacuum level because this vacuum level will determine what is the mean free path of the molecules. The mean free path is very, very large as compared to the distance between the two surfaces we will have free molecular conduction or residual gas conduction and this is one of the losses that will be encountered in vacuum when vacuum is used as insulation. This exists only at very low pressures or very high vacuum or at very good vacuum that is what we are talking about. Now, let us understand the mechanism of this free molecular conduction. For the sake of understanding consider two plates with temperature T1 and T2 as shown in the figure. So, here we have got a warm plate which is at temperature T2 and here we have got a cold plate which is at temperature T1 and we say that lambda which is a mean free path is more than the distance L. That means this molecules in between these two surfaces will never collide with each other. The molecules coming from this surface from the warm plate will straight away meet the cold plate without having any molecules in between for collision alright. So, mean free path for these molecules is more than the characteristic length that is L which is between the two surfaces and we found that T2 is more than T1. So, conduction will occur and the heat will be brought from this warm plate by the molecules when the molecule hits to the cold plate it will bring this heat and dump it on the cold plate then again from cold it will go to warm plate and this will continue and let us analyze the motion of a simple molecule in this case. The gas pressure is very low in order to ensure that the mean free path lambda of the molecule is greater than L. Therefore, lambda is more than L which means that it is in free molecule conduction zone. In such situation the gas molecules collide only with the surfaces and exchange energy alright. So, consider a molecule colliding with the bottom plate and leaving towards the upper plate. So, let us see a molecule which is shown by this red spot the molecule has just hit the cold plate and this cold plate is at temperature T1 and what is you understand from here is this. The gas molecule collides with this surface at T1 and it transfers some energy to the surface alright. So, this gas molecule which is coming from warm plate it will bring some energy and it will dump this energy on the cold plate which is maintained to be at T1 temperature. Now this gas molecule then having dumped the heat on the cold plate it will leave this plate at a different temperature than the cold plate temperature. So, it leaves the cold surface with a kinetic energy corresponding to a temperature T1 dash which is higher than T1. So, why so? It is not leaving at T1 because the heat transfer is not perfect because they could not get time to have a thermal equilibrium between the molecule and this cold plate. So, naturally the gas would leave at an higher temperature than this cold plate. It will not be able to attain the lowest temperature of T1 because the time spent by this molecule on the wall is going to be very very less. So, when it leaves the surface it will leave at a temperature T1 dash and not T1. So, this was this is what we just analyzed for the cold plate. Now, let us see what happens at the warm plate. So, again consider a molecule colliding with the upper plate and leaving towards the bottom plate. So, here a molecule comes and hits the warm plate and now the molecule color we have shown a different one blue one. This gas molecule collides with the surface at T2, this surface temperature is maintained at T2. The molecule now will leave this plate at a temperature which is T2 dash which is lower than T2. Again with the same reasoning what we gave for the cold plate they will not be in thermal equilibrium or the time that is given for this molecule to have heat transfer over here will not be sufficient enough and its temperature will not reach to T2, but it will leave this plate at a temperature which is lower than the warm plate temperature T2 and therefore, it would leave at T2 dash. So, what we found that at cold plate it leaves at T1 dash, at warm plate it leaves at T2 dash. T1 dash is going to be more than T1 while T2 dash is going to be less than T2. If there is a perfect thermal equilibrium this T1 dash should have been equal to T1, this T2 dash should have been equal to T2. It is clear that in both these impacts thermal equilibrium is not attained, the process is repeated and this contributes to free molecule conduction. So, this molecule again will come from T2 dash it will come to the cold plate again would leave at T1 dash and this process will go on happening forever alright. As long as the temperature is maintained at T2 the warm plate temperature as long as the temperature of the cold plate is maintained at T1 this process will get repeated the molecule will leave at T1 dash cold plate the molecule will leave the warm plate at T2 dash, but the heat transfer would happen in this manner. What is important to understand is why does this happen and who is responsible for that. In order to measure the degree of thermal equilibrium between the molecule and the surface we define a parameter called accommodation coefficient given as A. This accommodation coefficient will tell you what is the difference between T2 dash and T2, what is the difference between T1 dash and T2, T1 dash and T1. So, this will be defined by this accommodation coefficient and it will depend on what kind of gas you are talking about at what temperatures you are going to talk about and things like that will dominate this parameter called accommodation coefficient. Let us see what it is. So, accommodation coefficient is a ratio of actual energy transfer to the maximum possible energy transfer like a heat exchanger effectiveness when the molecules of a different temperature let us say T1 hits the surface at T1 it will leave at T1 dash. So, actual energy transfer is going to be less than maximum possible energy transfer and this ratio is going to be defined as accommodation coefficient. So, mathematically we can write A is equal to actual heat transfer divided by maximum heat transfer. Its value depends on gas surface interaction and of course, the temperature of the gas. So, it will depend on actually type of gas, the gas surface interaction and the temperature of the surface. So, again looking at the same plate now we see that the molecule is living at T2 dash hitting cold surface which is maintained at T1, leaves this surface at T1 dash, then hits the warm plate, leaves at T2 dash again it will come here and leaves at T1 dash and this process will continue. So, from the figure for the cold surface if I talk about the cold surface the actual temperature change is T2 dash minus T1 dash. For the cold surface the gas is coming at T2 dash while it is living at T1 dash. So, the change in temperature of the molecule is going to be T2 dash minus T1 dash while the maximum possible temperature change could have been if the gas is coming at T2 dash then it would have actually attained if there was a perfect heat transfer it could have attained the temperature of T1. So, the maximum possible heat transfer would have been T2 dash minus T1 alright. So, actual temperature change was T2 dash minus T1 dash and maximum possible change was T2 dash minus T1. So, by definition now the accommodation coefficient for the cold plate is going to be A1 let us have a cold plate A1 T2 dash minus T1 dash divided by T2 dash minus T1. If we find that the mass of the molecule remain the same the Cp of the molecule remain the same therefore, it will come only in the ratio of the temperature differences. The numerator shows actual temperature change divided by denominator shows maximum possible temperature change. Now, we can repeat the same thing for the warm plate also. So, this is what is going to be defining the A1 at a cold plate temperature accommodation coefficient at a temperature T1. Now, let us see what is the accommodation coefficient at the warm plate which is maintained at temperature T2. Similarly, for the hot surface the actual temperature change again is going to be T2 dash minus T1 dash. So, actual temperature change for both the surfaces is going to be the same while what changes is the denominator which is the maximum possible temperature change. But the maximum possible temperature change is going to be in this case now the gas is leaving this at T2 dash. So, therefore, the maximum possible when the gas is coming at T1 dash the T1 dash the molecule should have attained the maximum temperature up to T2 alright while it leaves at T2 dash because of the imperfect heat transfer. So, the maximum possible temperature change could have been T2 minus T1 dash. So, this is what we call is denominator. Therefore, the accommodation coefficient for the hot surface is given by A2. So, A2 is for the warm plate which is maintained at T2 temperature is going to be T2 dash minus T1 dash divided by T2 minus T1 dash. So, we have defined now accommodation coefficient for plate 1 and accommodation coefficient for plate 2 which are maintained at T1 and T2 temperatures. From the earliest slides the accommodation coefficients are A1 as T2 dash minus T1 dash divided by T2 dash minus T1 while A2 is going to be T2 dash minus T1 dash divided by T2 minus T1 dash. So, rearing is above terms what we get is now if I write in terms of T1 I will get T1 is equal to T2 dash minus T2 dash minus T1 dash divided by A1 if I look from A1 definition and I can define T2 from here from definition of A2 which is T2 equal to T2 dash minus T1 dash upon A2 plus T1 dash. So, I am just rearing this terms to get expression for T1 from A1 and T2 from A2. From this now again I can rearrange terms because I can see that T2 dash minus T1 dash terms are common over here and therefore, if I got T2 minus T1 if I say difference of temperature which is T2 minus T1 then I can take T2 dash minus T1 dash as common which will give me 1 upon A2 plus 1 upon A1 minus 1 alright. So, from here I can define that T2 minus T1 I can I am relating basically T2 minus T1 to T2 dash minus T1 dash and there is one more parameters coming which is 1 upon A1 plus 1 upon A2 minus 1. So, again if I rearrange these what I get is a parameter which is basically like emissivity factor similar to an emissivity factor if you got two surfaces which have got emissivity of E1 and E2 then we say that the net emissivity Fe will be 1 upon E1 plus 1 upon E2 minus 1 similarly here similar to an emissivity factor we define a term accommodation factor FA will be which is given by 1 upon A1 plus 1 upon A2 minus 1. So, 1 upon FA is equal to 1 upon A1 plus 1 upon A2 minus 1. See if I know accommodation coefficients for both the surfaces I can now calculate accommodation factor which is FA A1 and A2 are accommodation coefficients while FA is the accommodation factor. So, how do I write this expression as now incorporating this accommodation factor. So, T2 minus T1 is equal to T2 dash minus T1 dash into 1 by FA. So, FA is equal to which is accommodation factor T2 dash minus T1 dash divided by T2 minus T1. So, now this is a general expression which could be used to calculate FA or I calculate FA from A1 and A2 and it could be realized it could be basically used to calculate the actual heat transfer that is going to occur because of free molecular conduction. The approximate accommodation coefficients for concentric spheres and concentric cylinder geometries are as tabulated below. So, what you can see if I got a cylinder A1 and A2 maintained at temperature T1 and T2 and you can see that I got different temperatures 378 20 for A1 and A2 and then we have got different gases helium, hydrogen, neon, air and what you can see that as at a particular temperature if the gas changes or if the gas becomes heavier and heavier the accommodation coefficients at this temperature has increased while if the temperature decreases for a particular gas the accommodation coefficient increases. When there is an increase in accommodation coefficient it is basically because of better and better heat transfer between the molecule and the surface. So, actual heat transfer is getting better and better as compared to maximum possible heat transfer alright. So, this is what the subscripts one denote the enclosed surface where subscript to denote the enclosure. So, from this table what we see at a given temperature the accommodation coefficient increases with the increase in the molecular weight of the gas. For a given gas if I see a given gas the accommodation coefficient increases with the decrease in the temperature due to better heat transfer at lower temperatures. So, this is what we just saw from the table. Now, let us come to further calculations from the kinetic theory of gases the total energy of a molecule is the sum of internal energy and kinetic energy by neglecting other energies we can say that mathematically E is equal to U plus K E this is what is given by kinetic energy theory of the gases. Now, this E is equal to U is equal to C V into T we know that. So, I can write this as C V plus R by 2 T kinetic energy is half R T while internal energy U is C V T alright. So, I am just writing those values and therefore, delta E is equal to C V plus R by 2 into delta T where R is specific gas constant C V is equal to R upon gamma minus 1 which is gamma is nothing, but C P by C V and delta T is nothing, but actual delta T which is T 2 dash minus T 1 dash this is why we are doing that thing we are basically calculating the now energy transfer happening because of this free molecular conduction alright. So, taking this further delta E is equal to C V plus R by 2 delta T the definitions of C V and F A are given as below C V is equal to R upon gamma minus 1 again this is known from kinetic theory of gases while we know the F A is defined as T 2 dash minus T 1 dash is equal to F A into T 2 minus T 1 putting these values over here we get delta E expression I will put C V by R upon gamma minus 1 and I will put the value of F A over here alright. So, I have got now delta T over here so, I am replacing the delta T by F A into T 2 minus T 1. So, rearranging these terms I will get delta E is equal to F A R I am just taking R common from here and evaluating this bracket F A R upon 2 T 2 minus T 1 into gamma plus 1 divided by gamma minus 1. Now, if I want to calculate now mass flux per unit time is going to be m dot upon A which is nothing, but rho into velocity of the gas. So, rho is the density V is the average velocity of the gas and this will give you basically the mass flux alright. So, from kinetic theory of the gas the average velocity the value of V bar average velocity is given by 8 R T upon pi to the power 0.5 or square root of 8 R T by pi putting this value over there combining the above expressions we can find out what is m dot by A. So, m dot by A is equal to replacing this rho by P upon R T alright. So, by assuming the ideal gas equation we have got rho is equal to P upon R T 1 by 4 and putting the value of V bar I get this expression and again rearranging I get R T and R T gets cancelled m dot A is equal to P into 1 upon 2 pi R T to the power 0.5. Once I get the value of m dot E the total energy transfer per unit area o into molecular conduction is given by q by A is equal to m dot A into delta E I am basically interested in calculate this q that amount of heat transfer that occurring that is occurring because of the molecular conduction. So, what is the m dot A we have just found out what is the delta E we have calculated putting those values m dot A is equal to this delta E is equal to this putting this value over here replace this m dot A by this replace the delta E by this parameter I will get an expression for q by A alright. Just rearranging that I will get q by A is equal to this expression you got a big bracket over here multiplied by P which is the pressure into T 2 minus T 1 T is the temperature this T is the temperature of the pressure gauge measuring the gas pressure. So, wherever I am measuring this P this T is going to be corresponding temperature at where the pressure is measured. So, rearranging this terms again I will call the entire bracket as G because in this bracket you can see there lot of constants gamma is a constant R is constant the specific gas constant gamma is specific to the gas again temperature is constant because I am measuring pressure at that particular point while F A is also going to be constant if I know the temperatures of the surfaces and if I know the gas by knowing accommodation coefficient A 1 and A 2 I can calculate F A then. So, this bracket is more or less constant I can call this bracket as G. So, my q is equal to now G into P into A into T 2 minus T 1 and this is what is going to be my heat conducted because of the free molecular conduction alright. So, q is valid only when the distance the point to be noted is that this equation is valid for molecular conduction region. So, q is valid only when the distance L between the plates is less than the mean free path lambda mathematically lambda is going to be given by this and we say that L is less than lambda lambda will depend on the gas properties and the temperature and the pressure of course. So, we first find out lambda compare that lambda value with the characteristic length L and decide if we lie in a free molecular conduction region. If lambda is going to be more than L if the mean free path of the gas is going to be more than L that means there is no gas molecule to molecule collision the molecule is going to heat to the surface straight then I am sure that I am in this region I am lambda is more than L having ensured that then I can apply this equation to calculate heat conduction by residual gas or I can calculate residual gas conduction alright. So, q is equal to G into P into A into T 2 minus T 1. What is this G? G talks about the perfect gas and takes care of the accommodation coefficient P is a pressure, A is the area and T 2 minus T 1 is the temperature difference of the two plates. Now, I am not bothered about what is T 2 dash and T 1 dash because that has already been taken into account by this F A which has been enclosed in the value of G and the lambda value is given by this expression. From the above two equations it is clear that the free molecular regime can be achieved by achieving very good vacuum. As the pressure goes on reducing that means as the vacuum gets better and better the value of lambda is going to increase and when lambda is going to be more than L we are going to be in the free molecular regime and therefore, then I can use this expression to calculate residual gas conduction. Also one can understand from this expression that the free molecular conduction heat transfer can be made negligible compared to other modes by lowering the pressure. So, if my P value is directly proportional to P if the P value is much lower that means you got a perfect and perfect vacuum you ensure that there are hardly any molecules then which will cause the free molecular conduction. So, if I reduce the pressure my Q is going to be much less. If I reduce the T 2 minus T 1 my Q is going to be perfectly less. So, we can understand that the free molecular conduction heat transfer can be made negligible compared to other modes by lowering the pressure by decreasing the value of F A or by decreasing T 2 minus T 1 alright and this is what the formula we will use to calculate free molecular conduction. So, now let us come to evacuated powder we have seen earlier what happens when we use pearlite powder we have also seen what happens when we can use vacuum can we have this two together alright. So, basically the idea is now to have powder, but then evacuated powder is what makes more sense in order to reduce radiation also in order to reduce conduction. So, let us see what is this evacuated powder. So, gas conduction is the primary and the dominant mode of heat transfer in a gas filled powder and the fibrous insulations and this is what we have seen earlier that when we got a pearlite powder which is a gas filled pearlite powder we got a gas conduction as a one of the dominant modes of heat transfer and of course the solid conduction also one of the obvious ways to reduce this heat transfer is to evacuated. So, if I want to get rid of this gas conduction and as I have just seen that if I have a perfect vacuum I got a very good vacuum then the gas conduction will be minimized in that case. So, one of the obvious ways to reduce this heat transfer is to evacuate the powder and the fibrous insulation. So, just remove whatever gas is there, but for that you have to have a perfect vacuum now. So, one can have a perfect vacuum now. So, what remains is only powder in that case and usually the vacuum that is commonly maintained in these insulations is in the range of 10 to the power 3 to 10 to the power upon minus 5 torr. So, you can see that this is not a vacuum actually what you can see is that 10 to the power minus 5 is a good vacuum and it is not a good vacuum. This is what the range could be normally in such insulations and what is this torr? We will talk about this torr more when we deal with vacuum, but for the benefit just understand that 1 torr is equal to 1 millimeter of mercury. You know that 760 millimeter makes 1 atmosphere and we are just talking about 1 millimeter of mercury now which is equivalent to 1 torr. So, this is what if I plot the apparent thermal conductivity with residual gas pressure in torr this is what the conductivity curve would look like. What is this? The adjacent figure shows the variation of Ka apparent thermal conductivity with the residual gas pressure inside an evacuated powder insulation. So, I have put as insulation an evacuated powder and if I vary the pressure if I evacuate this powder the pressure in the powder will go on reducing and according to this reduction in pressure the apparent thermal conductivity of this evacuated powder will go on changing and it will be taking such a S kind of a shape alright. So, let us see how it behaves. So, if I talk about this region which is from atmospheric pressure to around 15 torr 15 torr means 15 millimeter of mercury column and atmosphere is around 760 millimeter of mercury column. So, you can see from 760 to 15 torr Ka is hardly changing the value of conductivity is very high this is in milli watt per meter Kelvin. The conductivity value is around 50's kind of a thing and we found that Ka is independent of residual gas pressure between the temperature pressure range of atmospheric to 15 torr alright. If you reduce the pressure further from 15 torr to let us say 10 to the power minus 3 torr we can see that the apparent thermal conductivity is linearly changing with the lowering of pressure from 15 torr to 10 to the power minus 3 torr Ka becomes directly proportional to the pressure. You got a almost a straight line and therefore it is showing a linear variation or linear decrease if I decrease the pressure from 15 torr to 10 to the power minus 3 torr and this is a very important thing having this lowest value. So, it varies almost linearly on the logarithmic chart as shown over here. Here in this case now the modes of heat transfer are due to radiation solid conduction and free molecular conduction. So, depending on the kind of vacuum you are having will have solid we are not talking about this region because it is a gas filled order actually is not it because it is a gas pressure. Well now we are evaluating this pearlite powder if about any powder basically and now the modes of heat transfer will be due to radiation solid conduction and free molecular conduction. But in all these things the free molecular conduction is going to be dominant here because the vacuum is not very good from 15 torr to 10 to the power minus 3 and we have just found that you are not possibly gone into free molecular conduction region. Here you could be causing lot of heat because of the molecular to molecule collisions in this case because the vacuum is not very very good. So, in this region because there are you know powder radiation will be minimum because entire area is occupied by the powder they will be solid conduction. But may be the free molecular conduction in this region depending on the kind of vacuum we are talking about is going to be dominant mode of heat transfer. Now if we decrease further pillow 10 to the power minus 3 torr and let us go up to 10 to the power minus 5 torr with the further lowering of pressure below 10 to the power minus 3 torr the variation of k is almost the minimum. So, here I have reached the minimum value of k a apparent thermal conductivity and bringing the vacuum further down will not help towards sort of improving or decreasing the value of apparent thermal conductivity. So, normally the vacuum in this using a powder will be in this region 10 to the power minus 3 to the power minus 5. So, here I would have taken the the molecular conduction will be minimum in this case now because we got a very high order of vacuum. The modes of heat transfer is primary due to now solid conduction because now there is a powder and there is a solid conduction because of the material of powder and of course radiation because you got to two temperatures, two different surfaces at two different temperatures. But what has been taken care of is the free molecular conduction it will be minimum free molecular conduction in this region. So, if you at all you want to have a good vacuum at all you want to take care of the molecular conduction you should be in this region the vacuum should be below 10 to the power minus 3 alright. So, evacuated powder are superior in performance than a vacuum alone. If you have got a vacuum alone then you got a radiation heat transfer which is proportional to 300 to the power 4 minus 77 to the power 4. However, in this case because there is a powder the radiation parameter is going to be minimum and the molecular conduction has also been taken care of you got a apparent thermal conductivity of around 1 or 2 or 1.1 or 1.2 like that as compared to that vacuum which could be higher in the same temperature region of 300 to 77 because in 300 to 77 Kelvin the vacuum the radiation heat transfer is going to be maximum for a vacuum alone. Well in evacuated powder the radiation heat transfer is going to be comparatively less alright. So, evacuated powders are superior in performance than a vacuum alone in 300 to 77 Kelvin as the radiation heat transfer is comparatively less in this case. So, if I want to have liquid nitrogen and if I want to have a container or liquid nitrogen around which I would prefer to have evacuated powder rather than having only vacuum around it as insulation. I will have a powder which is evacuated powder and evacuated powder will take care of radiation heat transfer from 300 to 77 Kelvin in that case. Please understand these points. However, if I go at very low pressures and very low temperatures now using this evacuated powder the solid conduction in evacuated powder will dominate the radiant heat transfer alright. Radiation heat transfer will be there, but now if I go for lower temperatures the solid conduction with the powder material is going to be dominant and therefore, in this temperature range when I am going very I am going to go for lower temperatures the evacuated powder may not help, but they are definitely better up to 77 Kelvin or liquid nitrogen temperature temperature range. Hence it is more advantageous to use vacuum alone in 77 to 4 Kelvin. So, if I have a 2 surfaces at 77 and 4 I may have to go for a vacuum alone as compared to evacuated powder in this temperature range alright. Now, from the Fourier's law we have q is equal to k into a into dT by dx. So, we got a instead of k we got a apparent thermal conductivity and value of am is going to be decided whether we are going to have a cylinder surface or we are going to have spherical surfaces. So, q is equal to k into a into dT by dx where k a is the apparent thermal conductivity, T h and T c are the temperature differences, delta x is the difference in the 2 temperatures surfaces having T h and T c as temperatures. While am is a mean area of insulation and am for concentric cylinder and concentric spheres as given below. So, for cylinders it is a 2 minus a 1 divided by log a 2 by a 1 and for sphere is going to be under root of a 1 into a 2 or a 1 into into a 2 to the power half. I will put this area in this equation I will get the value of apparent thermal conductivity and I will calculate the conduction the loss due to this conduction. The apparent thermal conductivity and density of few commonly materials are given over here. So, you can see now for 377 I got pearlite coarse pearlite lamblack fiberglass and I got a conductivity value of 0.951.9 1.2 between 1 to 2 kind of thing for evacuated powder alright. Let us go to opacified powder insulation. The radiation heat transfer still contributes to the heat in leak in 377 k temperature range in case of evacuated powder we have just seen that thing. Radiation heat transfer still exist and over there. If I want to get rid of that comes opacified powder. In the year 1960 Reed and Wang Hunter et al minimized this radiant heat transfer by addition of reflective flex made of aluminum and copper to the evacuated powder. So, if you put the flex of aluminum copper it will stop radiations reaching the internal surfaces. This flex act like radiant shields in the tiny heat transfer pass and are formed in the interstices of the evacuated powder and therefore they will prevent this radiations also further. So, I can I am reducing now all the modes of heat transfer using powder. So, I got a opacified powder which will now block even the radiations reaching the lower surfaces. So, these are the apparent thermal conductivity for opacified powder in the relation with how much opacifier by weight has been used. How many aluminum flex how many copper flex have been used as a percentage of the total insulation material. So, figure shows the variation of percentage of opacifier with thermal conductivity for copper or santocel is called as commercial name and aluminum santocel. So, you can see that there exists an optimum operation point for each of these insulations alright. So, if I use 40 60 case the opacifier ways 40 percent out of 100 percent my apparent thermal conductivity has come to 0.4 which was 1.5 or 2 earlier that means I have reduced my apparent thermal conductivity by almost 5 times. Similarly, if you have copper santocel I can go for 60 percent opacifier by weight and again I can reduce the apparent thermal conductivity to 0.4. It shows that by taking care of this radiation heat transfer board in the evacuated powder by adding some aluminum flex or copper flex I can reduce my apparent thermal conductivity further down alright to around very low value of 0.4 and this is what we are talking about optimum value. It has been observed that with these additions k a can be reduced by 5 times. If you go on adding this opacifier further then you will have a more and more conduction also because number of opacifiers have increased and therefore there will be minima somewhere in between. You cannot just go on adding that thing because as the opacifier weight goes on increasing that means there are more and more flex now and they might start touching each other because of which you will have a solid conduction path increase in that case. Copper flex are more preferred as compared to aluminum flex. The aluminum flex have large heat of combustion and that is why copper flex are preferable. These together with oxygen can lead to accidents when used in locks condition therefore aluminum should not be used or normally is not used. Now the disadvantage one is we talked about copper is preferable. The another disadvantage of using opacified powder is the vibrations tend to pack up the flex together. If you are carrying let us say some cryogen in a moving container or in a mobile container over a period of time this flex can cup together. The vibrations can tend to pack the flex together and therefore you will have certain zones where flex are plenty while certain zones are there where flex are not existent and therefore radiations in that case can reach directly to the low temperature surface which is not advisable while only in one area radiations might get blocked. This not only increases the thermal conductivity but also short circuits the conduction heat transfer alright. So, lot of flex together means you got a more conduction happening out there. So, one has to ensure that this flex are evenly distributed first thing and also ensure that there are no vibrations given to this particular vessel which is having insulation of opacified powder. But if you got a vibrations then these problems can be faced over a period of time. The apparent thermal conductivity and density of few commonly used opacified powder insulations are as shown. The residual gas pressure is less than minus 3 and we are talking about temperatures of 377 Kelvin and you can see here. You got a copper santo cell, you got a aluminum santo cell, you got a brand santo cell and you got a silica carbon combinations for different densities and you can see the apparent thermal conduct is 0.33, 0.35, 0.58, 0.48. So, what was between 1 to 2 for evacuated powder got reduced to now 0.3 to 0.6 in this conductivity range alright milliwatts per meter Kelvin. So, we could show from here that having this opacifiers in the evacuated powder the apparent thermal conductivity has decreased by almost 5 times. And therefore, opacifiers are preferred but then we have got conditions like you know not having vibrations having it you know evenly distributed and things like that. With this background now having studied vacuum, vacuum plus radiations in earlier case per light powder, evacuated powder, opacified powder. I would like to have a small tutorial to understand the comparative effects of all these insulations for which I have taken a little practical problem. And let us see this practical problems and apply all these equations which we have used earlier or which we had derived earlier to understand the losses using different insulations ok. So, the problem statement is a spherical liquid nitrogen vessel and this vessel have got emissivity of 0.8 for the both surfaces. Emissivity of 0.8 is as shown over here the inner and outer red eye are as 1.2 meter and 1.6 meter. So, it is a pretty big vessel compare and comment on the heat in leak for the following cases. So, we got a pearlite powder which has got 26 milliwatts per meter Kelvin as apparent thermal conductivity. Then we got a case whether the vacuum is maintained between these two surfaces pearlite is removed and we got a vacuum and the vacuum is of the order of 1.5 MPa. Then we got a vacuum alone this is also vacuum alone, but the vacuum level is 1.5 MPa while here is a good vacuum or a perfect vacuum in this case. Then we got a vacuum plus 10 shields and the shield emissivity is as low shield as low emissivity 0.05. Then we got evacuated fine pearlite which has got 0.95 milliwatts per meter Kelvin as apparent thermal conductivity and we got a copper centocell 5050 where the apparent thermal conductivity is 0.33 milliwatts per meter Kelvin. So, I would like to put these values over here get apparent thermal conductivity for different insulations and compare the heat that is being conducted or heat loss that will occur using this insulation and come out with a perfect insulation that one can recommend from this study. So, what is a tutorial the given is spherical vessel which has got a emissivity of 0.8 working fluid is liquid nitrogen that is why temperature is 77 Kelvin of the inner vessel while 300 Kelvin which is ambient temperature for the outside vessel and we have to calculate the heat in leak for this six different insulations for which the data regarding apparent thermal conductivity has been given. The shape factor between the two containers is assumed to be 1 fine. So, let us have first pearlite for which apparent thermal conductivity is 26 milliwatts per meter Kelvin. So, we have got a sphere here which has got a R 1 of 1.6 and R 2 of 1.2 R 1 is for outside R 2 is for the inside vessel apparent thermal conductivity has been given delta t is 300 minus 77 which is 223 find out the Q by this formula 4 pi k R 1 R 2 delta t upon R 2 minus R 1. So, this is what k into a into d t by dx this is what the formula to be used put those values 4 pi into 26 milliwatts. So, you know take care of all the units in such problems let us have meters everywhere R 1 and R 2 delta t R 2 minus R 1 over here what we can see that Q is equal to 349.7 watts which is fantastic which is very high heat in leak in this case. So, if we just have pearlite powder between these two vessels which is at 300 to 77 Kelvin will have 349.7 watts of heat in leak happening R 1 and R 2 across two surfaces. Now, if you got a less vacuum of 1.5 MPa only again we can apply, but now in this case we will have free molecular conduction, but before that we have to worry about are we in free molecular conduction region for which we have to calculate lambda. The net heat transfer is due to both radiation and residual gas conduction and if I want to calculate residual gas conduction if I want to calculate radiation first I will calculate what is my emissivity factor which is 1 upon E 1 plus A by A 2 1 upon E 2 minus 1 put in the value because both emissivity E 1 and E 2 are 0.8 I can get the value of Fe is equal to 0.72. So, less vacuum and if I want to calculate radiation now I will have this formula Q is equal to Fe into chef factor sigma which is stiffen Boltzmann constant into A 1 T 2 to the power 4 minus T 1 to the power 4 Fe I have just calculated to be 0.72 put those values over here I will get Q radiation is 2648 watt 2648 watt which is much higher than the pearlite powder over here. So, we can see that vacuum alone is not going to be helpful at all in this region only pearlite powder is much better than having only vacuum in this region. In addition to this radiation because of the vacuum we will have even the residual gas conduction. So, for that let us establish whether we are in free molecular conduction for which we will calculate lambda putting the value of different properties and temperatures I get lambda to be 4.53 meter which is much higher than R 2 minus R 1 all right. So, it is clear that mean free path lambda is greater than the distance between the surfaces which is 0.4 meter and therefore, we establish that we are in free molecular conduction region and therefore, I can apply the formula which we have derived. Let us have accommodation coefficient for two temperatures. So, 300 k I can have a accommodation coefficient 0.8 or 0.9 and around 78 Kelvin we got accommodation coefficient of 1 for air which is what we say lies between the two. I can calculate now accommodation factor which is given by 1 upon alpha 1 plus a 1 by a 2 1 upon alpha 2 minus 1 inverse put the value of alpha 1 as 1 alpha 2 as 0.85 I will get accommodation factor as 0.91. So, now I can calculate my Q because of free molecular conduction put all the values over here. I know all the values put the value of F A put the value of pressure which is 1.5 MPa put the value of area and I get T 2 minus T 1 over here. So, by getting all these parameters I will get Q is equal to 0.356 watts over here all right. So, which is much less than even the radiation heat transfer when we have vacuum alone. So, total heat transfer will be what happens in radiation plus gas conduction or residual gas conduction. If I got vacuum alone then I will have the same formula and I will get only 32648 watts, but if I add now vacuum plus 10 shields then I can apply the formula using shield emissivity of 0.05 and my F E now will get reduced from 0.72 to 0.003. So, I can if I can allow to have 10 shields in between my emissivity factor comes down to 0.003 and therefore, my Q will reduce from 2648 to 11 watts. So, you can see the effect of having 10 shields and the problem is how practical it is to have 10 shields between these 2 in this sphere in this R 1 and R 2 because that will going to increase lot of weight and that may add some conflicting problem in the manufacturing. So, this is the problem. Now, let us go to evacuated fine pearlite where k a is equal to 0.95 and put the formula in this again by using I get Q is equal to only 12.7 watts in this case. So, evacuated fine pearlite is much better than having vacuum alone that is what we have said earlier and we just proven it here and if I go for a copper centosil now, if I go for opacified powder now, my k is going to be further reduced 0.33 and if I apply the same formula my Q is 4.41 watts only which shows that copper centosil or opacified powder is going to be much better for this. So, with this you get a feel of what the values of heat in liquid will be for all these different insulation. So, if I want to compare all these values you can see it here. If I were to go for only pearlite which is gas filled pearlite it is going to be 349.7 watts. If I got to have vacuum which is going to 2648 plus some gas conduction which is absolutely negligible in front of this radiation heat losses then vacuum alone is going to be same, but vacuum plus 10 shields is going to reducing the entire thing to only 11 watts. But then the problem was having how to have this 10 shields over there in place. Then evacuated fine pearlite powder is much better 12.7 watt and what is most preferred will be opacified powder which is copper centosil giving only 4.41 watts of heat in liquid. So, this gives an idea that evacuated fine pearlite and opacified powder is going to be quite good. We cannot practically have really 10 shields in vacuum over there going to be weight increase also the mass increase over there and going to have a lot of manufacturing problems over there. So, one can get an idea of different you know heat inlix amount for different insulations using this tutorial. To summarize all the insulation which we have studied we found that in vacuum the radiation is the dominant mode of heat transfer. Evacuated powders are superior in performance than a vacuum alone in 377 Kelvin as the radiation heat transfer is completely comparatively less which is what we have just proved with the tutorial also. At low pressure and temperature the solid conduction in evacuated powder dominates the radiation heat transfer and therefore, for lower and lower temperatures up to helium temperature evacuated powder may not be used and this is what we will see in the next lecture also. Also we found that in an opacified powder the radiation heat transfer is minimized by addition of reflect to flex and that we found that actually this was much better actually, but understood there are some problems associated with this opacified powder because the flex try to come down to a one point due to the vibration for this container and therefore, they may not be preferred at many places, but what is important is evacuated powder is always preferred solution for up to liquid nitrogen temperature and this is more kind of a practical solution to take care of various problems. Thank you.