 So, welcome to the 35th lecture of cryogenic engineering under the NPTEL program. In the earlier lecture, I had covered the part regarding insulation and we are continuing with the earlier lecture only regarding insulation in cryogenic engineering. So, in the earlier lecture, we had seen that radiation is dominant mode of heat transfer in vacuum. We found that evacuated powders are superior in performance than vacuum alone in the range of temperature which is 300 Kelvin to 77 Kelvin that means, still liquid nitrogen temperature and later we found that radiation dominates. So, in this range as radiation heat transfer is comparatively less, but below 77 we found that the radiation takes over and therefore, we will have to look for other ways of insulation in those cases. Also, we found that in an opacified powder, radiation heat transfer is minimized by addition of reflective flex because of the presence of this reflective flex, the radiation heat transfer is minimized and therefore, this we found was beneficial as compared to the evacuated powder. However, we talked about various disadvantages of having flex also and then we had taken a tutorial problem in order to understand the effect of various insulations we had understood till then. So, a tutorial problem is solved to compare different types of insulation so far discussed and what I want to do now is basically continue with this problem and then you know take away from that problem and basically bring about the importance for new insulations that could be used for lower and lower temperatures. So, in this particular lecture we will talk about multilayer insulation which is a very important insulation in cryogenic engineering and then I will take a tutorial to cover or understand what this multilayer insulation's calculations are all about and we can take some cases or some studies regarding those. And finally, we will conclude the this particular topic of cryogenic insulation all right. So, in the earlier lecture we have solved the following tutorial and I will now go through this tutorial we will find what results we had got and then I would extend that particular tutorial to understand the need of multilayer insulation all right. So, let us look at this problem so this was the problem basically where we had a 77 Kelvin or liquid nitrogen bath or liquid nitrogen cryogenic inside and this is 300 K outside and we had a small cryostat spherical in nature and we had this dimension 1.6 meter outside diameter 1.2 inside diameter and this is just for an academic purpose basically and we are not bothering to calculate the neck conduction we just want to calculate the heat in link that is happening because of the insulation material. So, a spherical l n 2 vessel emissivity given as 0.8 is as shown over here the inner and outer radii are 1.2 meter and 1.6 meter respectively compare and comment on the heat in link for the following cases and the following cases were basically the insulation cases when we had pearlite as insulation between this outside 1.6 meter and 1.2 meter radii. We had a less vacuum case we had only 1.5 MPa vacuum case then we had a good vacuum or vacuum alone perfect vacuum we can say then we had a vacuum plus 10 shields we had evacuated fine pearlite we had a 50 by 50 coppers and 2 cell as opacified powder and then we had computed the heat in link for all this possible insulations and result was as given over here. So, we had pearlite powder we had 349.7 watt as heat in link we had less vacuum then we had lot of radiation heat transfer happening because there was no medium inside there was basically only vacuum. So, 2648 watts where the loss due to radiation and there we had a free molecular conduction also which is just 0.356 watts which is actually can be neglected because it is very small amount as compute the value of q r when we had a vacuum alone was 2648 watts then we had vacuum plus 10 shield we had only 11.02 watts as heat in link, but then we talked about that it is not very simple to have 10 shields in a cryo strut because it will increase the weights and thing like that then we had evacuated fine pearlite which is giving you only 12 watts as heat in link and when we had a opacified powder with copper flex or what is we call as a trade name of sentocell we had only 4.41 watts as heat in link in this cases. So, it is clear that opacified powder is the best solution to this. So, we say that now we could minimize the heat in link for 77 Kelvin or liquid nitrogen Kelvin case and we found that opacified powder was the best candidate followed by fine pearlite assuming that vacuum plus 10 shield is not a very practical solution. So, we can go for one of these two cases as insulation for a liquid nitrogen cryo strut or a container. So, a heat in link of 4.41 watts to ln 2 to liquid nitrogen would vaporize around 2.36 liters per day as shown in the next slide. So, it is not going to evaporate lot of liquid nitrogen alright 2.36 liter per day is a reasonably conservative estimate if 4.41 watts of heat in link is incident it is coming on the liquid nitrogen cryo strut. How did we calculate this we can do some algebraic computation to arrive at this rate of evaporation of liquid nitrogen it is very important to see how we calculated this because this will be useful tool for various calculations further and therefore, I will show the simple calculation. So, if you want to calculate the amount of heat is evaporated because of the incident heat in link we know the latent heat of liquid nitrogen is around 200 kilo joule per kg while density of liquid nitrogen is 807 kg per meter cube we know that 1 meter cube is equal to 1000 liter and 1 day has 24 hours. So, if I want to calculate the boil off required for 1 liter per hour if I have got a boil off of 1 liter per hour of liquid nitrogen let us calculate how much watts of heat in link should be there how many watts should be coming on the liquid nitrogen bath. So, that the resultant boil off is around 1 liter per hour. So, if you have got a 1 liter per hour boil off of liquid nitrogen it is equivalent to let us convert this 1 liter per hour to 1 kg per hour alright. So, that we can rope in the density of liquid nitrogen also and this amounts to this. So, I got a 1 liter per hour that means, 1 by 1000 meter cube here if I want to 1 meter cube is equal to 1000 liter that means, 1 liter equal to 1 upon 1000 meter cube which is this 10 to the power 3 multiplied by its density which is 807 kg per meter cube alright. That means, we will get so many kg now in 1 liter multiplied by so much of kilo joule per kg is the latent heat. So, in order to evaporate 1 liter we will have to have 200 kg kilo joule per kg as a latent heat part alright and then we got a per hour. So, we want to basically convert this per hour 3600. So, we get now per second because we want to get watts and this is kilo joule and that is why we are taking the power 3 here. So, this will amount to so many of watts which are responsible to cause a boil off of 1 liter per hour alright and this amounts to 44.83 watts. So, what does what does it mean if you got a heat in look of 44.83 watts it will amount to 1 liter per hour boil off of liquid nitrogen. So, if we have a boil off of 4.41 watts which is what we calculated earlier it would result in a boil off of 0.098 liters per hour which is very small and which is very much acceptable. So, in 1 hour we will have 0.098 or 98 cc per hour as boil off of liquid nitrogen. Therefore, the total boil off 1 day if you multiply by 24 hours will be 2.36 liters per day which is again acceptable solution taking it ahead. Different types of insulation discussed in the earlier lectures are applicable only to 300 K 277 Kelvin. We found that all these insulations calculations what we had done were up to 300 K 277 Kelvin. However, if I want to now understand what is my insulation for liquid helium let us see the gravity if I got a same amount of heat in leak coming for liquid helium how much boil off will occur for liquid helium now. So, given that latent heat of helium is only 20.2 kilo joule as I guess 200 for liquid nitrogen. So, you can understand that latent heat is very very small and therefore, smallest amount of heat in liquid will cause lot of helium to boil off and its density is 124.8 kg per meter cube. So, if I want to do the same calculations now I can repeat those calculations for liquid helium and 1 liter per hour boil off now for liquid helium will be equivalent to only 0.7 watt. So, understand this figure for nitrogen it was 44.7 watts would result in a boil off of 1 liter per hour while for helium 0.7 watts of heat in liquid result in boil off of 1 liter per hour and that is because we are talking about the latent heat is almost 10 times less as compared to that of liquid nitrogen. At the same time we got a density which is much smaller for liquid helium as compared to that of nitrogen and because of this will have only 0.7 watts required or 700 milli watts requirement to cause a boil off of 1 liter per hour of liquid helium that means what I want to show basically here is the smallest heat in leak would cause liquid helium to boil off. Now, if we take the same amount of heat in leak as that was present for liquid nitrogen case the same amount of heat in leak that is 4.41 watt if that is incident now for liquid helium it would vaporize 151 of liters of liquid helium in one day alright. So, what was 2 liters only per day now suddenly becomes 151.1 liter of liquid helium in one day which is not at all acceptable looking at the cost of liquid helium looking at the cost associated with the liquefaction of liquid helium having a boil off of around 151.1 liter per day is not acceptable at all and therefore, what is understandable from this that opacified powder pearlite powder may not be good insulation for liquid helium and therefore, we will have to go for something else for liquid helium what was good for liquid nitrogen is not good enough for liquid helium that is what we want to basically learn from this if I have got a heat in leak of 4.41 watt at 4.2 Kelvin it would cause such a big boil off which is not acceptable at all in cryogenic engineering. Therefore, there is a need to develop better insulations for now a case let us say a 77 Kelvin to 4 Kelvin or at 4 Kelvin temperature levels that is why we are going to now multilayer insulation. So, these were the insulations which we had talked about of which we have studied all these first 5 cases what we are going to learn now is about multilayer insulation which basically have vacuum which works in a vacuum and also which has got a reflective medium also. So, here we talk about multilayer insulation now in this particular lecture. So, let us see what is this multilayer insulation is all about. So, multilayer insulation was first developed by Peterson of Sweden in year 1951 and as I have shown the figure over here it consists of alternate layers of high reflective shields of foil here you can see here with the high reflective shield or foils and we have got a spacer material it is separated by low conductivity spacer. So, multilayer insulations are different layers put together and each layer would consist of high reflecting shield or foils which will take care of radiation while the next layer in between the two layers what we have is a kind of spacer which is separated by low conductivity spacer. So, I just would like to show you I got a sample to show you here. So, this is what a multilayer insulation would look like you got a highly reflective multilayer insulation or a shield over here and if I open this you can see that it is separated by some kind of a nylon net which is followed by again multilayer. So, you can see so many layers separated by spacer and this spacer is basically a non-conductive spacer. So, you got a various layer separated by spacer and they are stacked the entire thing will be stacked around the cryostat or liquid helium container or liquid nitrogen container. So, you can see several layers coming together each of them is separated by a spacer followed by reflective material and this is what constitute a multilayer insulation. So, as I just shown you it has it consists of a alternate layers of highly reflecting shields or foils separated by low conductivity spacer and a very good vacuum. Now, this is applicable multilayer insulation can work only when you got a very good vacuum that means we do not have any gaseous present any gas presence over there no molecular conduction no free molecular conduction no convection if we have ensured that no conduction and convection then only multilayer insulation will be effective otherwise they will not be. The high reflecting shields are generally made up of aluminum, copper or aluminized mylar. The aluminum sheet of 6 micrometer thickness is commonly used at low temperature. So, you can see the thickness is very very small. In order to improve mechanical strength and age of application plastic materials like mylar and capton are coated with aluminum. So, aluminum is coated with some kind of a mylar material in order to get strength. So, that they are put vertically down in the cryostat or containers. Low conductivity spacers are made of coarse silk or nylon net. I just showed you this spacers material. Very often substances like glass fiber, silica fiber, low density foam or fiberglass mat are also used. The spacer could be of any types alright. Most common material among fibers are dexiglass and tissue glass. These are basically the trade names for the spacer. Normally, dexiglass could be used for such multilayer insulation or nylon net could be used for such multilayer insulations. One layer of multilayer insulation normally short form as MLI is defined as one sheet of reflective shield plus one sheet of spacer material and this is what I just showed to you. Each component of this insulation is designed for a particular function alright. So, what are different functions? We got the radiation shields. Radiation shield or foils with high reflectivity it will reduce the radiation heat transfer alright. So, the highly reflective material we found that highly polished material is basically to take care of radiation heat transfer. The spacer which could be nylon spacer with very low thermal conductivity it reduces conduction. So, whatever solid conduction may have because we have got this layers put in the form of layers and these layers do not the radiation shield do not touch each other. Basically, they are you know spaced across a spacer and this spacer is of low thermal conductivity and this will take care of the solid conduction. And therefore, these are very important component of multilayer insulation and we have got a vacuum because I have said that multilayer insulation works only in the presence of vacuum and good vacuum and this vacuum will take care of residual gas conduction convection is going to be minimized because of this vacuum. This is a very important three aspects of multilayer insulation. So, what are different types of multilayer insulation? Let us have a look at this point. So, multilayer insulation are classified according to the type of spacers used. One can have different classification what I am going to show is basically the classification made depending on the spacers that are used in the multilayer insulation insulation. So, you can see the multilayer insulation is first type is multiple resistance spacers. So, you got a different types of spacers and they got a multiple resistance aspect associated with it. What is it? It is the fibers are arranged in a parallel fashion to minimize contact area as shown in this figure. I can show the same thing to you. So, you can see the same thing over here. So, you can have a spacer basically which is having the spacers are arranged in a kind of a parallel and then you got a reflective surface and then again followed by the spacer and again followed by the things. So, this is what the first type. So, multiple resistance spacer I just show you fibers are arranged in a parallel fashion to minimize contact area. Then we got a point contact spacer. I do not have that sample. A grid of nylon spheres is used to separate adjacent radiation shields. So, this is a different type of spacer. Continuing further what we have is a single component MLI. The reflective shields are crinkled or embossed to minimize contact area. Instead of having a plain reflective surface you got a crinkled that means point touches in order to basically minimize the solid conduction. We got a crinkled surface of multilayer insulation and in this case we do not use any spacer material. So, I just want to show it here also. So, this is the type. You got a crinkling done at various point and the multilayer insulation touches each other at only those points. This will basically minimize the solid conduction part and you can see that we do not have any spacer in this. So, basically they have a crinkled surface alright. There is no spacer used in this case. The multilayer insulation will touch each other only at those spots where it is crinkled. It will not have a continuous contact. It will not have a area contact. It will have only point contact which is what is responsible for minimizing the solid conduction which otherwise was taken care in the earlier case was by the net or by the spacer. So, this is called crinkled multilayer insulation or single component multilayer insulation. This is also shown in the figure over here and then we got a composite spacers. In this case we can have different materials. So, composite spacers few spacers consist of two or more materials. Each material has specific function to perform. So, we can have different spacer material also which will have its requirement as per in the applications alright, but I have not come across these spacers being used normally. It is a very special case where some basic expectations are there in terms of let us say conductivity or some other function. We can use different material spacer at different points. So, these are different types of multilayer insulations. Typically thickness of each layer is around 6 micrometer. So, this is what we earlier also talked about. Then the residual gas conduction inside an insulation depends on residual pressure of the gas. So, we are saying that we are in a much better vacuum now and therefore, the residual gas conduction is absolutely minimum which could be which could be considered 0 in this cases. So, multilayer insulation works only we got when we got perfect vacuum alright. For an optimum performance the usual level of vacuum that is maintained around an MLI are in the range of 7.5 into 10 to the power minus 5 torr. So, this is the vacuum level which is acceptable for multilayer insulation. We should not have minus 3 or minus 2 torr vacuum level. You should have minus 5 and below as acceptable vacuum for multilayer insulation to be effective. So, if I want to apply multilayer insulation now for an optimum performance multilayer insulation is placed perpendicular to direction of the heat flow. So, you got a direction of heat flow like this multilayers will be kept across it perpendicular to the direction of the flow in this fashion T 1 and T 2. T 1 is a high temperature T 2 is a lower temperature heat would rush from T 1 to T 2 and there will be several multilayer insulation which would take care of this which will not let this Q go inside. So, it is placed perpendicular to the direction of heat flow. The insulation performance is a function of following parameters. The insulation will vary the performance would be dependent on various parameters and what are these parameters? Applied compressive load if the multilayer insulations are compressed too much then the conductance parameter will change and more compressive load would result in heat loss. So, you will have more Q going inside if you have got a more compressive load. So, we should keep minimum compressive load just to make sure that this multilayer insulations stack you know stands vertical and therefore, for that you will have to have some compressive load, but that should be kept as minimum as possible. Number of shields now it is a very important component and I will touch upon this component later just to tell you the point that number of shield is a very important parameter and it will be optimum number of shields for a particular application for a particular size we will have an optimum number of shields that has to be taken into account and only so many shields have to be used normally it could be 20, 30s or 40s 40 layers per inch or in centimeter that could be given as 40 layers per centimeter 50 layers per centimeter that is what could be optimum number of insulations shields for a given application. Then gas type of and its pressure we just talked about what kind of gas could be there and what kind of pressures we are talking about we should have a perfect vacuum and therefore, we should not have any gas if at all gas is there its conductivity should be as minimum as possible. We have said that depending on the gas type it should be have no free molecular conduction and therefore, we should have perfect vacuum maintained in multilayer insulation. The size and number of perforations if at all number of perforations are maintained on the in the insulation it should ensure that a good vacuum is present between two layers of insulation also. So, having insulation but what will be the vacuum between the two layers in order to have a good vacuum between the two layers of insulation we should have some perforations on the insulation because of which the gas there could be sucked out by to maintain a good vacuum and therefore, number of perforations also will make a difference and of course, the operating temperature because operating temperature will determine how much Q is incident due to radiation and how much Q will be there because of which the solid conduction also would take place. So, less delta t that is what would be preferred over here. So, just to show how it is applied I have got a photographs courtesy TIF for Mumbai you can see a liquid helium or liquid nitrogen whatever they were we are talking about this is inner vessel and this is outer vessel and you can see the gap in between is filled with the multilayer insulation put from top to bottom on the top as well as on the bottom side. So, you can see depth various number of so many layers of multilayer insulations are kept in the gap from outer vessel to the inner vessel from 300 Kelvin to 77 Kelvin or 4.2 Kelvin whatever this content is made for. So, you can see very nicely how they are stacked. Now, stacking of multilayer insulation also is an art how do you keep how do you put them over the inner vessel all right. So, it is a very important thing to arrange this multilayer insulation nicely over the inner vessel this is very important to understand how the multilayers are applied on the inner vessel all right. So, multilayer insulation now the adjacent figure shows the variation of apparent thermal conductivity for multilayer insulation with residual gas pressure. So, basically this will tell you what is the effect of vacuum on the apparent thermal conductivity for a typical multilayer insulation. So, you can see that insulation layer density is 24 layers per centimeter in this case and the temperatures are 300 k and 90 k for which the values are given over here. It is clear that k a is independent of residual gas pressure between atmospheric and 15 TOR. So, you can see for this atmospheric condition to 15 TOR k a is very high and therefore, multilayer insulation cannot be used in this in this pressure range because k a we want to basically minimize all right. So, in this case we cannot use multilayer insulation at all or if you use the apparent thermal conductivity is going to be very very high with the lowering of pressure. Now, if you reduce the pressure from let us say from 15 TOR to 10 to the power minus 3 TOR k a linearly decreases that is what we had seen earlier also and we can see it right now also with lowering of pressure k is directly proportional to residual gas pressure. So, if we come to 10 to the power minus 3 from 10 to the power 15 TOR k a will linearly get reduced from very high value of around 20 or 30 to around 0.03 to 0.04 that is what we can see from here. The variation is almost linear on the logarithmic chart as shown here. Now, suddenly as you reduce the pressure from 15 TOR to 10 to the power minus 3 TOR value of k a will come down from 30 milliwatt per meter Kelvin to around 0.04 milliwatt per meter Kelvin. So, such a drastic reduction will happen as soon as you have a good vacuum now and that shows that will basically make us realize the importers of vacuum to be maintained for multilayer insulation. If you go further down now the mode of heat transfer in this region will be made mainly due to free molecular conduction or residual gas conduction alright. Redition will be taken care of because of multilayer insulation, but we will have a gas conduction now which is very very important to be minimized by lowering this vacuum. With the further lowering of pressure now if you come below 10 to the power minus 3 now the k a value gets minimized which is around 0.03 milliwatt per meter Kelvin. k a remains fairly constant that means we will have to have multilayer insulation vacuum to be maintained below 10 to the power minus 3. So, we say normally minus 5 is the most acceptable vacuum pressure for application of multilayer insulation alright. So, this shows that if we are below minus 3 the multilayer insulation are base suited over here while not in this range and not in this range definitely not in this range. Now, multilayer insulation bulk density rho a is an important parameter of the insulation and basically now we are calculate how many multilayer insulation should be kept what is this optimum multilayer insulation business. So, the bulk density rho a is an important parameter and it will depend on various parameters like what is the thickness of each reflective shield. Suppose it is T r the density of each reflective shield is going to be rho r that is the material of multilayer insulation coming into picture then mass per unit area that means kg per meter square of the spacer material which is SS alright this is spacer material and these two are the reflective material and the layer density which is very important concept. So, layer density per unit thickness that means so many layers per centimeter per inch that is what normally this is referred to which is equal to n upon delta x delta x is the distance from outer to inner multilayer insulation. So, how many number of layers will be there per centimeter or per inch which is what we call as layer density and this is very important concept as I just told you. So, total mass per unit area therefore will be given as SS plus rho r T r. So, total mass per unit area will be SS plus rho r T r that will be total mass per unit area. Density being mass per unit volume for n layers rho a will be given as rho a is equal to SS plus rho r T r into n upon delta x for n layers it will be n times this divided by delta x. So, this will be so many kg per meter cube will be what we call as bulk density or m li bulk density. So, basically we will talk about n delta x n by delta x which is layer density and corresponding components are coming because of the spacer material as well as with the reflective materials. Now, the apparent thermal conductivity which could be in microwatts per meter Kelvin for multilayer insulation and the layer density referred to as layers per centimeter a few commonly used m li r as shown over here. The residual gas for this is 10 to the power minus 5 torr and n temperatures are 377 Kelvin as just shown over here. So, you can see now if I got a 0.006 millimeter of aluminum foil and a spacer of 0.15 millimeter fiberglass spacer and a layer density of 20 layers per centimeter will have K a apparent thermal conductivity as 37 microwatts 37 to the power minus 6 watt per meter Kelvin such a low value here as compared to any other earlier values if you see. Similarly, if you got a spacer of rayon net will have different layer density for this material now and we can have K a value of 78 microwatts. Similarly, we can have NRC 2 crinkled aluminized miler film we can have around 42 microwatts meter Kelvin. So, you can see that we are talking about in microwatts 2 digit that is my apparent thermal conductivity for a multilayer insulation. So, apparent thermal conductivity if I want to calculate for multilayer insulation now for an evacuated m li heat is transferred by radiation and by solid conduction assuming that conduction and convection are taken care of the gas convection and gas conduction has been taken care of the only modes of heat transfer that could be present here is a radiation and solid conduction. The solid conduction will be more and more if more and more multilayer insulation are put together in a given space of delta x. So, for one layer let us calculate net heat transferred as q net is given as q net is equal to q radiation plus q solid conduction this is for one layer we are talking about. So, we can put formula for it. So, q net is going to be the formula for radiation now similarly the formula for solid conduction alright. So, K c into a T h minus T x T c upon delta c please refer that K c will depend on the number of layers also this is going to be now effective thermal conductivity of the spacer material. So, F e is a effective emissivity of the shield over here which is a radiation parameter F 1 to the shape factor which normally we take as 1 while a and delta x are the contact area and the width here and K c is a effective thermal conductivity of the spacer material or with the what is coming as thermal conductivity of the m li basically. So, what we call referred to as solid conduction also here, but that will be function of how many layers are there and therefore K c is a parameter which will vary with number of shields number of spacer materials also extending this further we can put the value of 1 upon F e as 2 minus e upon e F 1 to as 1 and also referring H c taking K c upon delta x as S c which what we call as solid conductance. So, putting this values over here we get combining this equations I can put this my q net is equal to this parameter H c into a into T h minus T c where e is the emissivity of the shield and e is H c the thermal conductance per unit area. So, I am just replacing H c as K c upon delta x we found that K c also would vary with the number of layers or layered density. So, H c is thermal conductance per unit area. Let us now find out what is apparent thermal conductivity for multilayer insulation. Let K a be the apparent thermal conductivity of the multilayer insulation and therefore q net in this case is equal to K a into a into T h minus T c upon delta x equating these two parts I will say K a into a into T h minus T c upon delta x is equal to what q net we got over here alright. So, here we got to calculate now what is my apparent thermal conductivity for multilayer insulation. So, I would now like to calculate here multilayer insulation and if we can see that we can get rid of this a into T h minus T c a into T h minus T c in both the sides. So, getting rid of a into T h minus T c and getting the value of K a now delta x would come on this side. So, apparent thermal conductivity now is equal to K a is equal to delta x into the bracket and this bracket has a component due to radiation and due to solid conductance H c. So, for n layers now we will have K a is equal to delta x upon n sigma radiation component plus H c which is going to be solid conductance. So, you can see now entire thing divided by layer density we can say n by delta x which is my layer density which will determine now the value of K a where T h and T c as the boundary temperature n by delta x n by delta x is the layer density. Now, let us see what happens to apparent thermal conductivity as a parameter of n by delta x and we can see that this curve is there. So, layers per centimeters given here which is nothing but n by delta x against which we found a apparent thermal conductivity and you can see that this kind of variation is shown and we can see that at a particular number of layers per centimeter we can have minimum K a value in milli watts per meter Kelvin why does it happen? So, we know that q net is equal to q radiation plus q solid conduction for multiracial insulation. If I go on increasing the number of layers if I go on increasing the number of layers here with the initial increase in the layer density the decrease in the radiation heat transfer is more than the increase in solid conduction. We know that if we go on increase the number of layers in a given centimeter we have 100 layers per centimeter for example, we will have so many layers that they will start touching each other and therefore, the solid conduction would start increasing. So, as you go on increasing the number of layers the radiation losses will reduce, but at the same time the solid conduction will start increasing. So, if you go on increasing n by delta x initially your q r will come down, but q solid conduction would start increasing, but the decrease in q r is going to be much more pronounced than the increase in q solid conduction and therefore, initially we will have a decrease in the value of K a as we go on increase in the layer density. So, hence K a of the insulation decreases in this range, but if the layer density goes beyond a particular value with the further increase in the layer density K a increases due to increase in solid conduction because after some time there will be so many layers which will be touching each other now that the solid conduction h c which depends on n by delta x will start increasing alright. So, in this case after some time after some number of some optimum number of layers per centimeter this will start increasing which means that initially it was decreasing later on it is increasing that means, it will go through a minimum at some point in time at some layer density. So, therefore, K a goes through a minimum at and then it rises as shown in this figure. So, our layer density should be kept as this minimum layer density. So, whatever is layer density we should have minimum value of K a associated with that and that should be used for our application. So, it is very important to understand that there is a concept of optimum layer density for multilayer insulation in the initial case radiation gets reduced, but solid conduction starts increasing. However, the radiation decrease is more than the increase due to solid conduction, but beyond this optimum point the solid conduction increases much more than the decrease in radiation heat transfer. And therefore, it goes through an optimum value of or the minimum value of K a corresponding to which you have got a optimum layer density and for every application we should find out what is this optimum layer density and that is what we should use for calculation purpose that is what we should use for application also. So, with this concept of optimum number of layers to minimize the thermal apparent thermal conductivity of multilayer insulation we have now understood what is multilayer insulation, how does it work, how do we calculate the conductivity apparent thermal conductivity of multilayer insulation, why does it have this optimum concept of number of layer density and also we found that what are different types of multilayer insulation and I just showed you different types of multilayer insulation also. In addition to that I have got two numbers of multilayer insulation which I would like to show again before I go and take a problem. So, you have got a you can use only aluminum sheet also. So, some foils could be used alright this is a thicker one and that also can be used as multilayer insulation which could be sometimes separated by some kind of a glass fiber or, but this is thicker one this is also a different type and also I got a different type which is normally plain and you have got a different kind of spacer which you can see very special spacer which thin digsiglass kind of a spacer also has been used over here. So, I showed you various multilayer insulation and we got a different spacer also that could be possible because there are lot of manufacturers of this multilayer insulation and depending on availability and the cost we can have different combinations of spacer as well as reflective surfaces alright. So, we have seen various combinations of reflective material as well as for spacer and you can have different manufacturers of this multilayer insulation. However, what is most important is how do you put or how do you apply this multilayer insulation on a cryo start or a cryo container it is an art. And therefore, normally an experienced person will would put this multilayer insulation and this basically depends on how you cut it how you put it minimum number of minimum layers minimum compressive load etcetera the parameters which affect the performance of multilayer insulation. So, with this background now I will take you to a tutorial where you can understand the effect of multilayer insulation and how do we calculate the boil of that occurs because of multilayer insulation. So, let us take a tutorial now and then I will end this particular topic later. So, I will go back to the tutorial which I had taken last time, but now here I am now having a typical application of multilayer insulation. We got a outside temperature let us say at 294 Kelvin then we got a multilayer insulation put from outside till we got a bath of liquid nitrogen put over here. Then again we have got a vacuum and we got a multilayer insulation put around here and we are storing liquid helium at 4.2 Kelvin I have written 4 Kelvin because that is just a way of designation of temperature, but this is liquid helium here this is liquid nitrogen here. Why do we have a liquid nitrogen here? So, that the radiation from 294 straight away do not go to 4 Kelvin or liquid helium which will cause a huge boil of. So, instead of that we got a buffer temperature of 77 Kelvin and the radiation now go only from 77 Kelvin to 4.2 Kelvin to liquid helium alright. So, we will have two boil offs one is the boil off of liquid nitrogen from here and one is a boil off of liquid helium from here. I have not shown in this case, but there should be way of taking out this boil off because continuous boil off would happen and therefore, liquid nitrogen will come out continuously liquid helium will come out continuously the boil off should not be kept inside otherwise the pressure inside will increase. Also we can neglect the conduction occurring across the next because we just want to study here the calculations due to insulation or heat leak that is going to happen due to this insulation alright. So, let us read the problem consider a spherical liquid helium vessel shielded with liquid nitrogen bath. The red eye of spherical shells are as shown in the figure the MLI are around 24 layers per centimeter is applied at each stage because of course, we have got a vacuum at this blue whatever color has been shown over here is a basically vacuum plus multilayer insulation and also you got a vacuum plus multilayer insulation. In both the cases we have got a 24 layers per centimeter as the layer density and this is applied at each stage that means, between 294 to 77 Kelvin we got a multilayer insulation of 24 layers per centimeter and from 77 Kelvin to 4.2 Kelvin we got a multilayer insulation of 24 layers per centimeter. In both the cases we have kept perfect vacuum of the order of 10 to the power minus 5 torr. Now, what we have to do is to calculate the boil off per day for liquid nitrogen as well as for liquid helium. What is given now given that the emissivity of shield is 0.05 the solid conductance of spacer is 0.0851 and we are going to neglect the neck conduction. Neck conduction also is very important way of having heat in leak, but at this point we just consider the losses due to insulation and corresponding to those heat in leaks we will calculate what is the boil off for nitrogen and what is the boil off for helium. This is my problem. So, what is the data which is given to us is multilayer insulation operating l into boil off. What is the temperature for this operating temperatures 294 Kelvin to 77 Kelvin and operating temperature for liquid helium boil off will be from 77 Kelvin to 4 Kelvin. So, operating temperature for liquid nitrogen and liquid helium are the heat in leak is going to happen from 294 to 77 Kelvin for liquid nitrogen case and for boil off calculation for liquid helium we got a temperatures of 77 Kelvin to 4.24 Kelvin or 4.2 Kelvin. The emissivity of shield is given to as 0.05 the number of layers are 24 layers per centimeter and solid conductance is 0.0851 watts per meter square Kelvin. The layer density also has been given. What we have to calculate is boil off of liquid nitrogen and liquid helium on per day basis. So, so many liters per hour we have to calculate and convert it to so many liters per day. So, what we have to do we have to first calculate what is the apparent thermal conductivity for the temperature range of 294 Kelvin to 77 Kelvin which will be important to calculate or we will responsible to calculate the boil off of liquid nitrogen. So, delta x by n is going to be 1 upon 2400. So, converting into millimeter will have 1 upon 2400 then H c is given as 0.0851 and then we will have E is equal to 0.05 as given and we got T h and T c specified as 294 and 77 respectively. So, I got a formula for application to calculate what is apparent thermal conductivity which is equal to delta x upon H c which is solid conductance this has been given to us as 0.0851 plus a component which is coming because of the radiation. So, if I start putting these values now I will get K a is equal to 1 upon 2400 0.0851 5.669 into 10 to power minus 8 which is sigma value then E is given as 0.005 divided by 2 minus 0.05 then we got a temperature T square plus T c square which is given by this bracket and then we got a T h plus T c as given as 371 which is 294 plus 77 Kelvin. Calculating this we get K a is equal to 56.2 micro watt per meter Kelvin. So, it is very important calculations and we get apparent thermal conductivity for whatever assumption we have got for layer density and H c etc. We get a K a of 56.2 for the temperature range of 294 to 77 Kelvin this K is responsible to cause the boil off of liquid nitrogen. So, can we calculate now therefore heat in leak for liquid nitrogen is going to be given by for this parameters we got a 56.2 micro watt meter Kelvin as apparent thermal conductivity R 1 is equal to 2.4 R 2 is equal to 2 this is what has been given delta T is equal to 217 now here and therefore Q is equal to 4 pi because this is spherical in construction we got a 4 pi K a R 1 R 2 delta T divided by R 2 minus R 1 putting the values 4 pi into K a value as 56.2 into 10 to power minus 6 then R 1 R 2 2.4 and 2 delta T of 217 divided by R 2 minus R 1 which is 2.4 minus 2 and if we calculate this Q is equal to 1.84 watts. So, heat in leak from ambient to 77 Kelvin across this vacuum plus multilayer insulation is going to be 1.84 watt and if I want to calculate the boil off of liquid nitrogen. So, boil off of liquid nitrogen for the temperature range of 294 K 277 Kelvin is going to be depending on the latent heat of liquid nitrogen 200 kilo joule per kg density of liquid nitrogen is 807 and we have shown this calculation earlier and we know that 1 litre per hour of liquid nitrogen boil off is equivalent to 44.83 watts of heat in leak. In the current case we got only 1.84 watt of heat in leak and this would be amounting to 0.041 litre per hour boil off which is a very small. So, we will have only 0.041 litre per hour boil off and per day therefore we will have 0.985 or around let us say approximately 1 litre per day as boil off all right. So, this multilayer insulation it put from 294 to 77 Kelvin would cause an approximate 1 litre per day as liquid nitrogen boil off which is pretty ok which is acceptable. Now, let us go to 77 K 24 K to calculate apparent thermal conductivity for responsible for boil off of liquid helium. So, here we have got again the same layered density as given same solid conductance as given is equal to 0.05 and T H and T C as 77 Kelvin and 4 Kelvin respectively. So, I put those values over here and calculate the value of K A if I put those respective values the T E now is going to be T H plus T C is going to be 77 K plus 4 Kelvin 81 Kelvin and again these two values all also will be different in this case. The apparent thermal conductivity in this case is going to be less now 35.7 micro watt per meter Kelvin. So, in this case now we have calculated the for 77 Kelvin to 4 Kelvin the apparent thermal conductivity is 35.7 put those values and get the heat in leak for liquid helium now. So, heat in leak calculations again would be done delta T is going to be 77 minus 4 which is 73 Kelvin R 1 and R 2 are going to be different R 1 is 1.6 meter R 2 is 0.6 meter if I put these values in the same formula now I will get Q is equal to 0.031 watts that means only 31 milli watts is going to be my heat in leak from 77 Kelvin to 4 Kelvin. If I want to calculate now the boil off of liquid helium because of this I will take into consideration the latent heat of liquid helium which is only 20.2 kilo joule per kg density of liquid helium is 124.8 kg per meter cube and we know from the earlier calculation that one liter per hour of liquid helium boil off is equivalent to heat in leak of 0.7 watts. In this case we have got only 0.031 watts of heat in leak and therefore it will vaporize only 0.044 liter per hour which means around 44 cc per hour of liquid helium boil off. So, therefore the total boil off per day for one day is going to be around 1.062 liter. So, boil off of liquid helium actually per day is going to be less than is going to be more than that for liquid nitrogen may not be sometimes acceptable. So, but for the given dimensions we got a boil off of liquid helium also close to 1 liter per day and boil off of liquid nitrogen also close to approximately close to 1 liter per day. And this is what the calculation show and this example would show you how to calculate the boil offs and how to calculate the heat in leak due to multilayer insulation alright. So, this is the way multilayer insulations work this is the way we have to calculate the boil off helium or nitrogen for multilayer insulation. So, to summarize the results for this problem we will get for liquid nitrogen boil off temperatures were 294 to 77 Kelvin the apparent thermal conductivity calculated was 56.2 micro watt per meter Kelvin Q was calculated at 1.84 this would basically resulted into 0.985 liter per day boil off while for liquid helium boil off we had a temperatures of 77 Kelvin to 4 Kelvin apparent thermal conductivity was 30.7 micro watt per meter Kelvin Q is only 0.03 you can see that Q is so less as compared to what it was for nitrogen, but because of the less latent heat of liquid helium and its properties we had a boil off around 1 liter per day for liquid helium. So, this is just to show you comparison that how much boil off for nitrogen and helium would occur if multilayer insulation is used with this dimensions in the present container. So, conclude to conclude multilayer insulations or to conclude basically the insulations we can see different points whatever we have studied under this topic. So, we have understood that cryogenic vessels need insulations to minimize all modes of heat transfer this is what we have studied throughout K a or apparent thermal conductivity is calculated based on all the possible modes of heat transfer all right. It takes into consideration conduction convection radiation and therefore, we can compare different insulations by having their apparent thermal conductivity for calculation purposes. In an expanded form heat is transferred only by solid conduction with decrease in main cell diameter we have seen that K a decreases with an increase in bulk density we have seen that K a increases. In a gas filled powder or fibrous insulation heat is transferred by gas and solid conduction all right. So, we have to take care of these two aspects or two modes of heat transfer in a gas filled powder or fibrous insulation. In vacuum we have seen that radiation is a dominant mode of heat transfer and it is minimize it can be minimize using radiation shield, but we know that having radiation shields will not be a practical solution for various cryo containers. In an evacuated powder heat is transferred by free molecular conduction solid conduction and radiation at low pressures and temperature solid conduction will dominate the so it will dominate the mode of radiation heat transfer. In an oppressive hot powder we have seen that radiation heat transfer is minimized by addition of reflective flex. We have seen that all these insulations can be used up to liquid nitrogen temperature while at lower temperature we will have to go for a multilayer insulation. So, multilayer insulation consists of alternative layers of high reflective shield and low conducting spacers. Multilayer insulations are more effective in 77 k to 4 k also from 300 k to 4 k temperature levels. So, if you come to very low temperature levels we will have to think about having multilayer insulation and that is basically for liquid helium containers due to its properties when provided with a good vacuum. So, multilayer insulation has to be always be with good vacuum. Multilayer insulation without having good vacuum is of no use because gas conduction and convection now will cause lot of heat in leaks. And we also seen that there is an optimum layer density at which k a of multilayer insulation is minimum. So, we got to have this optimum layer density for our applications depending on that we should apply particular layer density for minimizing the heat loss minimizing the heat in leak. So, we can see a comparison at the end of all the insulations which we have seen till now. So, the following table shows the apparent thermal conductivity for different insulations so far discussed. The operating temperatures here taken are 77 k to 300 k and just to give indicative apparent thermal conductivity values for let us say perlite the apparent thermal conductivity is 26 milliwatt per meter Kelvin. For evacuated fine perlite this gets reduced down to 0.95 milliwatt per meter Kelvin. Then we got a opacified powder 50-50 copper and this isantocell and again it gets reduced to 0.33 milliwatt per meter Kelvin. And then when you come to MLI just to see the comparison of powder evacuated powder opacified powder and then we got a MLI having 25 layers per centimeter and we can see that the apparent thermal conductivity has come down to 56.2 micro watt per meter Kelvin. And here what you can see is all the values for comparison of different insulations. Thank you very much.