 This is the seventh lecture on cryogenic engineering under the NPTEL program. What we are studying right now is the properties of materials at low temperature. This topic we have been studying since last two lectures and this is the third and the last lecture covering this topic. We have been covering various properties under different subheadings, mechanical properties at low temperature, thermal properties of low temperature, the electrical properties at low temperature and finally the magnetic properties at low temperature. The first two properties, the mechanical and the thermal properties are very very important for you, when I say you means the students of mechanical engineering. At the same time electrical properties are also important, but these properties, the electrical properties and the magnetic properties I am going to cover under the section of superconductivity because superconductivity is a very important aspect of cryogenics. In fact, cryogenics is the cause and the superconductivity is effect and here I will cover some basics of superconductivity under the section of electrical and magnetic properties at low temperature for any material. In the last lecture, I emphasis more on the thermal properties and before that lecture I had talked about mechanical properties. So, during the last lecture, I introduced you to various material properties with reference more to the thermal properties and what we covered under thermal properties were thermal expansion and contraction at low temperature, the specific heats of solids and we covered Debye theory, we talked about Debye function, we talked about thermal conductivity of solids and how the thermal conductivity varies at low temperature. In addition to that, we talked about integral K D T, that is K D T integrals which basically takes into consideration the thermal conductivity variations with temperature especially at low temperature and also we talked about electrical resistivity of solids at low temperature. This was what we covered during the last lecture. In the present lecture, I am continuing with the material properties at low temperature but my focus here is basically going to be superconductivity which is a phenomena which one comes across only at low temperature and what it is, how it happens, what are the effects of all these things we can see and also I will show some video which show some very peculiar effects at low temperature. Now, based on all the lectures what we have had under this topic, I am going to solve some problems under this tutorial section, I am going to solve around 3 to 4 problems on different properties, how to calculate those properties at low temperature and finally, I would like to give you some assignments and I will expect that all of you spend time and solve these assignments and broadly whatever I have covered under the topic of properties of material at low temperature, I will draw some conclusions covering all my total 3 lectures under this topic. As you all know, we have seen a video which shows that the properties of material change and change drastically when cooled to cryogenic temperatures. If you recall the video of liquid nitrogen and we had put some materials like rubber, like potato, we had put because those materials were available to us, we showed what happened to those materials at low temperature. The electrical resistance of a conductor decreases as the temperature decreases. The resistance decreases and the material becomes more and more conductive, electrically conductive and finally, it becomes superconductive. For example, the wires made of materials like niobium, titanium, this is alloy, niobium and titanium, it exhibits zero resistance when subjected to liquid helium temperatures. So, a material like this, if it is subjected to very low temperature as well as liquid helium temperature that means 4.2 Kelvin, it becomes superconducting. So, let us see a behavior of how the electrical resistance changes at low temperature. So, this particular graph shows the electrical resistance of metals decrease with the decrease in the temperature. You can see a general behavior of these metals. Some of the metals become superconducting and suddenly the resistance become equal to 0 at critical temperature. Now, some of the materials become superconducting, alright. So, few of the materials when cooled to lower temperatures, the resistance suddenly drops to 0 at a particular temperature. This is what is shown by this particular graph. So, here you can see that as soon as the material starts getting cooled at a particular temperature, see resistance decrease gradually, but at this particular temperature, the resistance suddenly become equal to 0 and this particular temperature what we call as critical temperature. Suddenly the resistance become 0 and the material become superconducting at this point, alright. So, this is a phenomena or this is a behavior which is shown by a superconducting material that as soon as it hits the value of T c, as soon as the temperature comes to T c, the material has become superconducting. That means, its electrical resistance has become equal to 0. Now, in 1911, Camerling Onas from Holland discovered the phenomena of superconductivity and what he did is all shown here. He used basically mercury then and when mercury's temperature was decreased, at a particular temperature let us say equal to 4.2 Kelvin temperature, its resistance suddenly become equal to 0. It suddenly kept down from this value to 0 and this is where he understood that some of the properties have drastically changed as far as mercury was considered and then he said that mercury has become superconducting as its resistance become equal to 0 at 4.2 Kelvin. During his investigation on mercury, he observed that the resistance drop to 0 at 4.2 K. So, T c of mercury can be called as 4.2 Kelvin. So, what is superconductivity? This particular graph in three dimension shows that is a three axis. One is a temperature axis, one is a current density axis and one is a magnetic field axis. We say that the state of a superconducting material is governed by three parameters. What are those three parameters? They are temperature, current density and magnetic field as shown in this figure. So, the blue section what we show is a superconductivity region. As long as the material is in this blue section in terms of all these three axis parameter that is temperature, current density J c and the magnetic field H c, if they are in this region the material always remains in superconducting state. However, if any of these parameters exceeds for example, the temperature goes beyond this T c or the current density exceeds this J c value or similarly, the magnetic field exceeds this H c value. They are all the critical values. If any material exceeds this value, the material ceases to be a superconducting material. It will not show the superconductivity phenomena anymore and therefore, superconductivity phenomena is shown as far as the temperature is less than T c. The current density is less than J c and the magnetic field is less than H c which are all the three critical parameters so that the material remains in superconducting state. So, here the blue region in the figure is enclosed by T c, J c and H c. If I say that J c is equal to 0, H c is equal to 0 that means the material temperature has to be less than T c in order that it remains in a superconducting state. So, one has to go on lowering the temperature so that the material becomes a superconducting material. So, this is a very important parameters that if a material is superconductivity, the current density cannot exceed J c. As soon as one of the parameters is violated, the material ceases to be a superconducting material. Now, why does this happen? Let us understand what are the reasons that material becomes superconductive in brief. The electrical resistance is due to the scattering of electron motion as you know every material has got free electrons. The resistance to the flow of electrons is basically due to scattering that is inside the material and that scattering is happening through the lattice imperfection because of the existence of lattice imperfections like presence of impurity and dislocations. Electrons move in a material, but it will come across presence of impurity, it will come across dislocation and the electrons may get scattered alright and this is what offers the resistance to the flow of electrons. The first imperfection that the presence of impurity is a temperature independent parameter. So, whatever impurity is present in the material presents all through irrespective of the temperature. While the scattering phenomena occurring due to the lattice imperfection decreases with the decrease in temperature what does it mean? At lower and lower temperature the scattering phenomena is going to get less and less alright and therefore, the resistance to the flow of the electrons is going to be less and less alright. This is what we the analogy shows that at lower and lower temperature because the scattering phenomena is going to be decreasing the resistance to the flow of the electrons is going to be decreasing. As a result the electrical conductivity when the resistance is decreasing the electrical conductivity is more at low temperature. simple argument to show that at lower and lower temperature, electrical conductivity of the material increases. Now, why does the material become superconductivity? We understood the fact that conductivity increases, electrical conductivity increases, but Bardin, Cooper and Schriefer put forth a theory that not only that the electrical conductivity increases, but some of the materials become superconducting at low temperature and this is called as BCS theory named after Bardin, Cooper and Schriefer. So, electrons being negatively charged particles, they move easily through the space between the adjacent rows of positively charged ions. As in every material, we have got positively charged ions and the electrons flows through the positively charged ions. This motion is assisted by electrostatic force which push the electron forward because of the attractive force between the positively charged electron ions and the electrons which are basically forming an electron cloud. The electrons move by the electrostatic force which push the electron inward. However, in the superconducting state, the electrons interact with each other and they form a pair. This is what happens at very very low temperature, below the critical temperature of any particular material, the two electrons come together and they form a pair. In fact, one electron pulls the other electron and that is why we call it, it becomes a pair and this pairing is a low energy process basically and it is a stable process. Every low energy process is basically a stabilized process and therefore, formation of electron pair is a low energy process and this is called as phonon interaction. This is what put forth by the BCS theory. So as soon as the electrons form a pair, the motion of the electrons is favored. The motion of the electrons is now more and more conducting. The electron pair so form, they move easily and the second electron follows the first electron during the motion. As I said, one of the electrons will pull the other electron and this results in increased motion of electrons. As a result, this electron pair traveling together encounters less resistance. This is clear from this. This electron pair is called as Cooper pair. This is one of the highlight of the BCS theory and this is what makes the material superconducting and therefore, it can carry more and more current in a superconducting state. This theory was first explained by Bardeen Cooper Schrieffer BCS in BCS theory in the year 1957 and for which they are awarded Nobel Prize in the year 1972 for this theory. This is some of the basics of the superconductivity and the BCS theory. We are not going to go in the details of this theory. Now, here you understand how the material behaves at low temperature and how the relation for different material with respect to the magnetic field and the temperature as shown in this figure. So, what you can see that on the y axis, what you have got is the magnetic field strength. On x axis, what you have got a temperature for different materials like tantalum, lead, tin and aluminium. If you see it for lead for example, at no magnetic field, the T c is around 6.5. Similarly, for aluminium, the T c is around 1, but as soon as you got a magnetic field, the behavior is like this. That means, at this particular magnetic field, the corresponding T c value is less than what it was earlier at 0 magnetic field. So, this basically gives a reason how the magnetic field strength increases at lowering of temperature. At a particular T c value, it has strength of only 0 magnetic field, but as soon as the temperature decreases below the T c value, magnetic field strength increases and the maximum value it at 0 Kelvin. So, for every material, the magnetic field strength is maximum at 0 Kelvin. So, what is the relation between this threshold field H t and its temperature? This is the relationship. H t is equal to S 0 in the bracket 1 minus T upon T 0 square. So, H t is a field strength at any temperature T, while S 0, we can see, S 0 is the critical field at 0 Kelvin. So, S 0 for different material is what it is on y axis. At 0 Kelvin, whatever field is there, it is called as S 0 and T 0 is the critical temperature at 0. That means, the critical temperature T 0 is these values or what we call earlier at T c or critical temperature. So, if I want to see what is H t value at t is equal to T 0. As you know, this is the T 0 value. So, if I put at H t at t is equal to T 0, if I put T 0 value here, then 1 minus 1 and the value of H t at t is equal to T 0 is nothing but 0, which is what is shown in this figure alright. So, basically at H t is equal to 0, the value of T is equal to T 0, which is what comes from this formula. So, if I want to find out what is the value of H t at any temperature let us say 5 Kelvin, I just put this value t is equal to 5 over here. Correspondent T o value is around 6.5 and I know the value of H o, which is around 850 for example, I can calculate what is the field strength at temperature equal to 5 Kelvin in that case. Now, there are different materials like high T c and low T c material. Some material show high critical temperature, while some material show low critical temperature. So, superconducting materials are distinguished depending on the critical temperature of the exhibit. Earlier these materials having transition temperature above 30 Kelvin were called as high T c material. Nowadays mostly the materials which use liquid nitrogen for their superconductivity are termed as H t s materials, while below that particular temperature other materials are called as low T c material. So, what are these materials? Here you can see a graph again on the x axis you got lot of material with the ear, with the discoveries and on the axis what you see is their T c. So, you can see that H g was discovered, mercury was discovered as superconducting in 1911 by Caroline Onnes. Niobium Titanium has a T c of around say 1 or 8 Kelvin, which was discovered somewhere in 1930s. Niobium tin was discovered later around 1950s and these are all called as L T s that means low T c material, low T c superconducting material, while what you have above those is high T c material. So, previously as we said previously this division line was at 30 Kelvin, but nowadays it has become 77 Kelvin. So, all the materials above these are normally called as H t s material, all the materials here are called as low T c material or L t s material. The material to be talked about more at H t s are normally the copper oxide conductors, while these are normally the metallic conductors alright. And the materials to be called as bismuth based, yttrium based, these are called as bisco and they are called y b c o. These are the two highly used materials, high T c material in manufacturing of current leads or something like that or the windings if you want to have for high T c superconducting magnetic materials alright. So, these are basically high T c material or H t s material or these are L t s material. Now, we will see the phenomena of Meissner effect, this phenomena is shown by a superconducting material. So, what you see in this particular effect is at room temperature, if a material is subjected to magnetic flux. So, this is basically a material at room temperature and if it is subjected to magnetic flux, the flux lines of force penetrated through the material as shown in this figure. However, as soon as this material becomes superconducting that means, it is subjected to low temperature and the material becomes superconducting, it repels the magnetic flux line. So, what happens to it? It behaves in a different way and it behaves like this. All the magnetic flux line which were passing through the material earlier are no more passing through the materials and they are going outside the material and because they are repelled by the because of the superconducting material. Now, this phenomena or this effect is what is called as Meissner effect and was discovered first discovered by Meissner and Robert in the year 1933. So, the phenomena in which the magnetic flux is repelled as soon as the material becomes superconducting is now used for various end applications. What I am going to show you right now is a small video to show this effect so that you understand this effect in a better way. In this video now, what you are seeing is that are two parts. One is a magnet which is ND FEB, neodymium iron boron, this standard magnet and what you see here is basically a superconducting material. So, the base of this what you are seeing here is a YBCO material having a TCO of around 95 Kelvin and what you see around that is a plastic container. Basically, it is a plastic container and the superconducting material forms the base of this material. So, what I am going to do is put this YBCO container on the magnet and as you can see it is sitting perfectly alright. This is at room temperature so you can understand that YBCO is not in a superconducting state right now. Now, what we are going to do is we are going to see basically how it behaves as soon as it becomes superconducting. So, what should I do? In order to make it superconducting I can pour liquid nitrogen inside this. The liquid nitrogen will bring its temperature down to 77 Kelvin and as I said the TCO of this particular material around 95 Kelvin that means definitely at 77 Kelvin it will become a superconducting material. Now, let us see how it behaves. It shows the Meissner effect and the material will start repelling the magnetic flux line. So, now let us bring it nearer to this and now what we can see suddenly is a whatever flux lines were passing through the YBCO material it is not letting it pass now through it and because now I am putting a pressure on it, it is creating a back force and the magnet is trying to go away because YBCO is trying to repel this magnetic flux line to pass through it alright. So, this effect basically shows that as soon as the material becomes superconducting it repels the magnetic flux. Now, this leads to the application of Meissner effect which is a magnetical levitation train or maglev. So, maglev train runs on the principle of magnetic levitation. When YBCO is cool to temperatures less than 90 Kelvin it turn diamagnetic alright. The proper balancing of the vessel as shown in the video levitate it is from the magnetic track. Now, this is what we will see using the same principle maglev train gets levitated from the guide way. This results in no contact motion and therefore, no friction. Now, the same principle what we saw in Meissner effect I will show in a application in the terms of maglev right. So, in this video now what you see is a the ND FEB magnets basically are laid down as track. So, you got magnetic track you can say that the train should move on this you must have seen this in some kind of toys earlier. Now, what I am going to show here is how the levitation takes place. So, the YBCO magnet container what we saw earlier is kept here and it is now the liquid nitrogen is put in that and it is already kept over there on the track with some kind of material in between right. So, let us say the train is already standing on the track over here alright. As soon as now the material become superconducting it will start floating because the magnetic flux lines are now repelling and which kind of produces a force and therefore, the material starts floating on the trains sorry floating on the rails over here. So, what are you getting from here you are getting a lift thus this particular train or material is floating or levitating on the rails which means there is a lift with the train obtained and if you give a push to this now it is a kind of a propulsive force because of will the train will start moving. So, basically now we can see that the propulsive force is given by some kind of engine well let us not talk about that, but the levitation basically makes the train float on the train on the rails. Now, it will cease to be a superconducting material with the passage of time because nitrogen will get evaporated and as soon as the material ceases to be a superconducting material it will become a normal material and the mice nary field will come to stand seal and this is what happens right now alright. As soon as the temperature of YBC is more than 90 or 95 Kelvin the material is no more in a superconducting state and now it has sat down. So, this particular kind of motion does not result in any contact and therefore, the frictions will be absolutely minimum and this is the principle of magnetically levitated train. Now, the similar applications are there for superconductivity and possibly we have discussed that in the first lecture and one of them is a NMR the nuclear magnetic resonance NMR apparatus which is a magnet over here and it is kept dipped in liquid helium and liquid helium is surrounded by liquid nitrogen. So, this is what it becomes superconducting magnet at low temperature it results in nuclear magnetic resonance one can do NMR of any sample which is kept at the center of this magnet and similarly, what you have is a MRI, MRI magnet also is kept dipped in liquid helium surrounded by liquid nitrogen or a cryocooler cooled shields so that the body scanning could be done as one can see the internal details of body here in this case and one can see the internal material details in the case of NMR. These are very well known usages of superconducting magnet as we saw earlier alright all the winding in this case are made of niobium titanium and niobium tin depending on the magnetic field strength. Now, what I am going to do in this remaining time is to solve some problems under this tutorial section. We have seen various properties still now and each property has got some way to calculate the property variations at low temperature. Now, in this tutorials I want to solve some of the problems that we can calculate these properties at low temperature. I am going to have 4 tutorial problems one is on a thermal expansion and contraction. So, one tutorial on that then we have got an estimation of specific capacity using Debye theory we got 2 tutorials on that and finally, we got thermal conductivity of materials at low temperature we can calculate the value of k or the loss due to k at low temperature and we got one tutorial on that. So, tutorial number 1 the problem statement is calculate the overlap length of a brazed lap joint formed by 2 materials one is SS 304 and the other one is copper the SS length of L naught or length at 0 Kelvin is given as 1 meter while copper the length is 0.5 millimeter. Now, it is desired minimum overlap should be greater than 5 millimeter the joint is subjected to low temperature of 80 Kelvin and following data could be used. The problem statement says that you make a joint and the joint is made at naturally at room temperature which is at 300 Kelvin. However, this joint is going to be subjected during operation to a temperature of 80 Kelvin and therefore, what you know is a shrinkage will happen the statement also says that the overlap should be of this 2 material should be greater than 5 millimeter at low temperature because the shrinkage would happen at low temperature and overlap will decrease at low temperature and therefore, what should be the overlap at room temperature in effect that is basically the question. So, what you can see here is a problem statement that at 80 Kelvin condition this is a SS material this is a copper material both the lengths are given as 1 meter and 0.5 meter respectively. What is the problem? The problem is that at 80 Kelvin this overlap of material should be more than equal to 5 millimeter. So, it should basically be more than 5 millimeter. So, that it has got good strength the data is given at 300 Kelvin and 80 Kelvin. So, del L by L 0 we have seen this behavior of mean linear thermal expansion the values for SS and the value of copper are given as this all right at 300 Kelvin it is 304 at 80 Kelvin it is 13 and at copper for copper it is 337 and 26. So, what we are going to do basically is calculate shrinkage of SS at 80 Kelvin calculate shrinkage of copper at 80 Kelvin and then based on that based on the relative shrinkages we should calculate what should be the overlap at room temperature. Let us divide into two parts one for SS 304 and one for the copper getting the values from the data it is a mean linear expansion in SS 304 is del L by L 0 is L T 1 by L 0 minus L T 2 by L 0 to 10 to the power minus 5 this is what we get this is what we have seen. See, if I put the values of L T 1 by L 0 at T 1 temperature which is in this case is 300 Kelvin and T 2 temperature which is in this case is 80 Kelvin. See, if I put those values what I get del L SS by L 0 is equal to 304 minus 13 304 is the value at 300 Kelvin 13 is the value at 80 Kelvin. If I take L 0 as 1 meter which is what the problem says delta L SS will be 2.91 millimeter the material is going to be subjected from 300 Kelvin to 80 Kelvin for a length of 1 meter the shrinkage would be 2.91 millimeter as far as SS 304 is considered. For copper if you do the similar calculation we have got del L Cu by L 0 is equal to L T 1 by L 0 minus L T 2 by L 0 L T 1 by L 0 for copper is 337 at 300 Kelvin L T 2 by L 0 of copper is 26 at 80 Kelvin. But this values were meant for L 0 equal to 1 meter. So, at L 0 equal to 1 meter del L Cu is going to be 3.11 millimeter while if the length is only 0.5 meter it will be half of that and for del L Cu will be 1.55 millimeter. What does it mean? The shrinkage of copper will be 1.55 millimeter while the shrinkage of SS will be 2.91 millimeter when the joint is subjected to 80 Kelvin. Now, the problem says that the overlap should be more than or equal to 5 millimeter. So, I have to take the worst of the shrinkages which is 2.91. If I take that worst of the shrinkages I should be able to calculate what should be never overlap at room temperature B. So, the greatest of the two expansions is shrinkage of stainless steel. So, the safe lap joint should be more than del SS which is shrinkage of SS plus 5 millimeter that is 7.91 millimeter. If I take this as 8.1 for example, which is more than 7.91 then the lap width in copper after shrinkage will be 6.55 millimeter. Similarly, lap width in SS after that will be 8.1 minus 2.91 I think as 5.1 millimeter which means that the overlap is going to be more than 5 millimeter and it can stand the required strength at 80 Kelvin. So, basically the problem is to calculate the shrinkages of these two materials and see that at 80 Kelvin the joint does not fail. That means the overlap is basically more than the minimum value of 5 millimeter alright. This is a very practical problem. I am sure you will understand from here what are different parameters that have to be considered while designing such a joint at low temperature. Hence, the overlap being more than 5 millimeter is a good design. Second problem is Debye theory. We know in Debye theory we got an expression for the value of specific heat capacity. So, C v is equal to 3 R T by theta d cube into a Debye function represented by D of T by theta d. The theta d is basically Debye characteristic temperature which is known and it is a characteristic temperature for every material. Also we know from earlier lecture that at the temperature which is more than 2 times theta d, C v approaches 3 R value which is this coefficient over here. This is called as Dolong and Pettit value alright and very low temperature that is T less than theta d by 12. C v is given by following equation it reaches this value. Now, we see the problems based on both these extreme cases. At D T is equal to 0 the constant value is 4 pi to the power 4 by 5. So, this particular curve shows the C v by R variation against T by theta d values alright. So, this variation of C v by R versus T theta d is a cubic at low temperature and it constant at high temperature as we have seen earlier. We know the theta d values for various materials and they are as given over here in the table. Now, based on this we will solve the second tutorial. Determine the lattice specific heat copper at 100 Kelvin. So, calculate the value of C v at 100 Kelvin for copper given that the molecular weight of copper is 63.54 gram per mole. So, the step one for this is basically to get the value of T by theta d. So, T is known to us which is 100 Kelvin and theta d is a characteristic day by temperature for copper which from this table you can see that the theta d value for copper is 310. So, T by theta d for copper is 100 upon 310 which is 0.3225. This value of T by theta d is greater than 1 by 12 of T by theta d and therefore, what we can say now we can basically refer to this particular graph. So, for T by theta d to be 0.3225 I have to locate the value of in here and get correspondingly the value of C v by R. So, from the graph what you see is a corresponding to 0.3225 of T by theta d I get the value of C v by R which is equal to 1.93. Once I know C v by R I just multiply it by R and to get R what I do universal gas constant divided by the weight which is given which is 63.54 grams per mole and which gives me the value of R to be 130.85 which is a specific gas constant. So, the value of C v therefore, equal to 1.93 into 130.85 and what you get ultimately is a 252.534 joule per kg Kelvin. So, the C v of copper at T is equal to 100 Kelvin is 252 joule per kg Kelvin. Naturally the C v of copper at 300 Kelvin is much higher and as the temperature is reducing the value of C v is decreasing. Extending the same I go through tutorial 3 where the problem statement is determine the lattice specific heat of aluminum at 25 Kelvin given that the molecular weight is 27 grams per mole. So, here I am talking about aluminum and I want to calculate the value of specific heat at 25 Kelvin is a very low temperature. The data is given to me in terms of its molecular weight which is a standard property again going by step 1 I am going to calculate the T by theta d. So, what is T by theta d ratio for which I have to know what is the theta d value of the Debye temperature for aluminum which is 390 the Debye temperature for aluminum is 390 and therefore, T by theta d is going to be 25 upon 390 which is 0.0641. Now, this T by theta d is less than 1 by 12 of theta d. So, 1 by 12 of theta d is basically 0.0833. So, T by theta d in this case of aluminum is 0.0641 and if this T by theta d is less than 1 by 12 that is 0.0833. So, this 0.0641 is less than 1 by 12 that means, we can apply a simple formula to calculate the value of C v in this case. Since the T by theta d ratio is less than 1 by 12 the equation to calculate the specific heat is as given below. Now, with this equation I can calculate the value of C v which is completely T by theta d dependent. We know the temperature T is 25 Kelvin in this case and T by theta d value is known to us if we put those values for corresponding R is universal gas 8.314 divided by the grams per mole for aluminum. What you get is a specific gas constant which is 307.9 for aluminum. Putting those values over here what you get ultimately is a value of C v which is 18.958 joule per kg Kelvin right. So, what you see here is the value of C v is drastically less at 25 Kelvin in this case of aluminum. In the earlier problem what you saw first for copper which was a correspondingly very high value and at 25 Kelvin aluminum shows a very low value. We have studied two examples to calculate the specific heat at low temperatures for one for copper one for aluminum. In one case we refer to the graph in the other case we have calculated based on a available formula at low temperature. The next tutorial will be on thermal conductivity integrals. Just to revise that the Fourier law of heat conduction is as you know q is equal to minus k into a into dT by dx while k is a strong function of temperature in this case at cryogenic temperature. To make calculations less difficult and to account for the variation of kT with temperature q is expressed as q is equal to minus g into theta 2 minus theta 1 alright. We have seen this earlier and here k dT is taken as an integral called thermal conductivity integral which takes into account the variation of k with temperature and if I am talking about temperatures I will talk about a range of temperature and the theta 1 and theta 2 are basically 41 and t2 respectively both of them should have the same base of td here. So, theta 1 is nothing but integral k dT from some datum temperature td up to t1 and this datum temperature could be from standard books it could be 0 Kelvin or 4.2 Kelvin. Here the ACS is constant the cross section area is constant and g is defined by ACS by length. So, what you see is a graphical variation of integral k dT in a graphical form with temperature on the x axis for different material stainless steel aluminum and phosphor bronze and the variation of k dT for few of the commonly material shown over here in the calculations the actual temperature distribution is not required. So, if I know my t1 is going to vary from 100 Kelvin to 20 Kelvin 100 Kelvin to 10 Kelvin I have to just see the end values integral k dT at 100 Kelvin integral k dT at 10 Kelvin and the difference of this 2 will give me integral k dT between 100 and 10 Kelvin alright. So, the formula is if I want to calculate integral k dT between 10 Kelvin and 100 Kelvin I should get the value of integral k dT at 100 Kelvin which is basically taking 0 as a base minus integral k dT at 10 Kelvin again the base remaining same at 0 Kelvin. So, if I get a difference of these 2 values it is nothing but integral k dT between 10 to 100 Kelvin this is what I want to use if I want to calculate the loss due to thermal conductivity or loss due to conduction or heat transfer due to conduction if the material is subjected at one end from 100 Kelvin and the temperature at the other end is around 10 Kelvin. So, my next tutorial is based on this formation the problem statement is determine the heat transferred in an copper slab of uniform crash section of area 1 centimeter square and length of 0.1 meter. So, the L is equal to 0.1 meter while A is equal to 1 centimeter square when the end phases are maintained at 300 Kelvin and 80 Kelvin. So, one end of the copper slab it has 300 Kelvin the other end is subjected to 80 Kelvin respectively compare the heat transferred by k average and k dT methods. So, k average is basically taking average of the k value at 300 Kelvin and the k value at 100 Kelvin while k dT takes into consideration the k variations between 300 Kelvin and 80 Kelvin at all the points. So, let us compare these two heat transfers obtained using k average method and an integral k dT method. So, what is given to us is a copper slab one end of the copper slab it at 300 Kelvin the other end is at 80 Kelvin naturally there will be heat transfer from a high temperature to the low temperature in this case which is equal to q. What is given to us is area of cross section A is 10 to the power minus 4 meter square which is nothing but 1 centimeter square and the length of specimen is 0.1 meter T 1 is equal to 300 Kelvin while T 2 is equal to 80 Kelvin. So, let us first go by k average method in which I will take the average value of thermal conductivity between 300 Kelvin and 80 Kelvin. So, what is k 300 is 78.5 as first copper is considered 78.5 at 300 Kelvin and k at 80 Kelvin is 37 watt per meter Kelvin. If we take the average of this two the k average value is going to be 57.75 watts per meter Kelvin. How do I calculate q in this case q is equal to minus k into A into dT by dx the value of k is taken as k average in this case. I put the values over here what I get ultimately is 57.75 into 10 to the power minus 4 which is A and this is k T 1 minus T 2 which is nothing but dT 300 minus 80 divided by the length which is 0.1 meter and the q calculated in this way based on k average is 18.958 watts. This is based on a k average method. If I want to now calculate the same thing by using integral k dT method or a k dT method what I should know now q is equal to minus g into theta 2 minus theta 1. Now, the base line here is in case 4.2 4.2 to 300 k T dT is 15000 for copper while theta 2 at 80 Kelvin taking base at 4.2 Kelvin is nothing but 1600. G is equal to ACS upon L which is 10 to the power of minus 4 upon 0.1 and the q if I put all those values over here I get the value of q to be equal to 13.4 watts which means that if I compare the two values based on k average I got 18.98 watts while based on k dT method I got 13.4 watts. That means my calculations based on k average is quite higher as considered to what it is which is a realistic picture taking all the property variations in this temperature region of 300 to 80 Kelvin region which is just 13.4 watts and this is what I should use for actual calculations. So, k average is basically predicting more heat transfer as compared to the realistic heat transfer predicted by integral k dT method. So, we just saw several tutorials based on those I am going to give assignments which you can see calculation of specific heat, calculation of specific gas constant is given determine the energy required to warm calculation of q over here then specific heat of aluminum has been given at 60 Kelvin minus to calculate specific heat over here. So, based on whatever problem we have solved in the tutorials you have to solve these assignments and this is calculation of the heat transfer which we just saw and here again based on what we did calculate of shrinkages. So, that the overlap length is always maintained. So, please go through the assignments and solve these problems and for all these problems use the standard data that is available both in the literature and as given in my earlier lectures. Just to conclude what we have done till now we know that the properties of material change when cool to low temperatures or cryogenic temperature. We have found that stainless steel is the best material for the cryogenic application from strength point of view. So, if the strength is the requirement one should go for stainless steel only. Carbons steel cannot be used at low temperature as it undergoes a d B T transition or ductile to brittle transition. Heat strength and yield strength fatigue strength of material increase at low temperature while impact strength ductility decrease at low temperature. PTFE or Teflon which is a non metal can be deformed plastically at 4 Kelvin as compared to other materials and therefore, Teflon is preferred in cryogenics. The coefficient of thermal expression lambda decreases with that decrease in temperature. For pure metals K T remains constant thermal conductivity remains constant above l n 2 temperature which is 77 Kelvin. Below this temperature it reaches a maxima and then decreases steadily. This is what we saw in the last lecture. For impure metals K T decreases with decrease in temperature while as we just saw in a problem integral K D T is used to calculate the heat transfer at low temperature. We also saw that the electrical conductivity of metallic conductors increases at low temperature. That is what leads to superconductivity. Also we saw that K E electrical conductivity and thermal conductivity are correlated by Wedemann-Franc's law. Basically K T by K E is just a function of temperature. The sudden drop in the resistance when a particular material is cool to lower temperatures is called as superconductivity. This is what the phenomena we studied in this lecture. This state is governed by three parameters namely temperature, current density and magnetic field. The superconducting materials are distinguished into high TC material and low TC material depending on the critical temperatures they exhibit. We also saw that Meissner effect, Maglev superconducting magnets are some of the applications of the superconductivity. Thank you very much.