 So, welcome to the 32nd lecture on cryogenic engineering under the NPTEL program. Just to see what we had done in the last lecture, we had seen that the phasor analysis for the OPTC, we had also seen for the DI-PTC the phasor analysis and we know that there exists a phase angle between the mass flow rate at the cold end and the pressure vector which is the cos theta is responsible for increasing the cooling effect and we had said that this cos theta has to be maximum in order that cooling effect is maximum and in order to make this cos theta maximum that means theta to be smaller, we employ various phase shift mechanisms for the pulse tube cryocooler. So, various phase shifting mechanisms have been developed in order to optimize this phase angle and so we got a basic pulse tube, we got OPTC, we got DI-PTC and now I am going to talk more about inheritance tube pulse tube cryocooler which I had not talked during the last lecture and I had told that I will talk about the inheritance pulse tube cryocooler in a separate lecture. The phasor diagram for basic orifice and double inlet PTC also we had seen, we had seen that the heat lifted at the cold end that is the cooling effect depends on the mass flow rate at the cold end, the P1 by Po ratio that is amplitude of P1 and that is actually indirectly reflect the pressure ratio generated in a system, the cold end temperature TC at which cooling effect is obtained and the phase angle and all these are connected by this formula. So, q dot c is equal to RTc P1 by P0 m dot c into cos theta divided by 2 and we had seen the phasor diagram and in the phasor diagram in the phasor analysis an adiabatic process is assumed in the pulse tube and an isothermal process is assumed in other elements like connect tube after cooler, cold end heat exchanger and hot end heat exchanger. Depending on these assumptions we had ultimately derived this relationship between the cooling effect and various parameters. In the phasor diagram the relative lengths of the vectors indicate the mass flow rate in those parts. We had seen the mass flow rate for the entire pulse tube cryocooler and I will just show that again in order that you understand what we had done in the last lecture. So, in this lecture now taking ahead from what where we left during the last lecture I would like to continue with the phasor analysis and I would like to give a electrical analogy to understand what is this inerton tube business basically. So, normally we understand the RLC circuit the resistance the inductance and the capacitance circuit and we know the relationship between the voltage and the current and therefore I would like to explore the fact regarding RLC circuit in order to understand the phasor diagram. I will take a small tutorial based on this phasor diagram so that you understand the concepts of related to calculations of the pulse tube cryocooler. And then I will spend the second half of this lecture on the pulse tube cryocooler research at IIT Bombay and I would like to show various parts which we have developed at IIT Bombay various pulse tube cryocooler which we have developed at IIT Bombay and the results related to those pulse tube cryocoolers. Finally, I will conclude the entire chapter on the cryocoolers because this is the last lecture on the cryocoolers. So, let us see phasor diagram as seen in the earlier lecture the phasor diagram of an OPTC is as shown in the figure and this was our m dot h and on this the pressure axis we got m dot h which is in phase for an OPTC because it is a orifice only and on which we calculate T h by T c into m dot h this is this vector then we got a length associated with heat exchanger hot end heat exchanger of pulse tube this is a hot end of the pulse tube. So, we got a omega 1 omega p 1 V h s divided by p m into T h which is a volume of the hot end heat exchanger divided by corresponding temperature T h and then I get mass for it at the inlet of the heat exchanger which is let us say pulse tube hot end and they got I got a pulse tube corresponding to pulse tube I got a V p t upon T c and what I got here m p t h and what I got here m m dot c this is the mass flow at the cold end and this is which is very important because it generates cooling effect then at the other end of cold end heat exchanger what I have is a regenerator at the cold end. So, depending on the volume of the cold end heat exchanger divided by its temperature I got m dot r c which is a mass flow rate of the regenerator at the colder side and then I got the regenerator. So, depending on what is the volume of regenerator divided by its temperature there I will get now m dot r h that is mass flow rate at the hot end of the regenerator alright. So, understand the fact that I am basically adding each volume divided by its temperature to give the mass flow rate in that part alright. So, depending on the volume of that particular part divided by its temperature and depending on whether it is a adiabatic process or isothermal process for example, p t is adiabatic process. So, you got a gamma over here while we are assume that these portions are basically the isothermal and therefore, there is no gamma below in the denominator for these mass flow rates. Then we got a after cooler and associated with the volume of the after cooler we have got a vector again which is basically coming from the compressor and then we got a dead volume associated with the compressor. So, I got a v c p d and this is the dead volume associated with the compressor and the connecting tube if any this is m dot c p this is the mass flow rate from the compressor to be delivered to the entire cryocooler. Now understand that and beyond this we can have now the step volume of compressor if one wants to. So, I need to compress so much of gas to get m dot c at the cold end of the pulse tube to generate cooling effect and this is my net theta the theta being angle between mass flow rate at the cold end and the pressure vector alright. So, in order to get this theta between m dot c and pressure I have to compress so much of gas at the compressor. The compressor power will definitely increase depending on the dead volumes of the system if my dead volume is more here my compressor volume will be still more and more and therefore, by this vector of m dot c p the length is going to be much higher as compared to m dot c. So, actually indirectly talks about the C O P of the system how much gas has to be compressed that means how much power input has to be there and how much cooling effect do I get basically when I compress so much of gas. So, cooling effect divided by power will give me the C O P and therefore, the phasor diagram is a simple indication it is just an indication we have have taken various assumption of you know having isothermal processes and adiabatic processes, but it is an indication it also shows you the relationship between m dot c and the pressure, but it also talks about how much power input goes in order to get the corresponding cooling effect. So, in the phasor diagram the relative lengths of the vector you can see the relative lengths of the vector in a vertical direction as well as in the inclined directions they give basically they indicate the mass for it in those parts. So, this is mass for it in the pulse tube this vector gives you mass for it at the outlet of the coalescent which is m dot c then mass for it at the outlet of the regenerator which is m dot r c then mass for it at the inlet to the regenerator which is m dot r h and things like that. So, it is a very important diagram and we have done the entire derivation during the last lecture. Going ahead from there coming back to the pulse tube classification which I had shown you earlier we got a basic pulse tube, we got a orifice pulse tube, we got a double inlet pulse tube and what we also have is inner returns tube pulse tube and about which we had not talked during the last lecture and I had said that I will talk about inner returns tube in a separate lecture. So, this is how this is something I would like to talk about and to understand this concept I would like to have a electrical analogy. So, that you understand how do you compare how do you see a performance of inner returns tube as compared to the for example, the orifice pulse tube normally we do all comparison with respect to OPTC. So, let us see in OPTC and let us in the relationship to that let us see ITPTC that is inner returns tube pulse tube cryocooler. So, this is my inner returns tube pulse tube cryocooler, we got a after cooler regenerator cold end heat exchanger pulse tube hot end heat exchanger and this hot end heat exchanger in OPTC will be connected through an orifice to a reservoir. Now, I am getting rid of orifice and I will have a inner returns tube which is a small diameter kind of a capillary tube running into length of let us say 1 meter to 2 meter depending on the diameter and it is getting connected to a reservoir. So, the figure shows the schematic of an inner returns tube pulse tube cryocooler ITPTC and orifice here for a OPTC gets replaced by a very long tube of a small diameter and this tube is called as inner returns tube or IT in short the word inner returns comes from the word inertia and inductance. So, we know the electrical circuit in electrical circuit we have got a resistance then inductance and capacitance. So, in fluid flow now we have got a inductance replaced by a term called inner returns which basically talks about inertia, inertia nothing but the mass mass flow rates. So, in fluid mechanics I will have inertia replaced by mass flow rate and inductance and if you combine these two what you get is a basically the term called inner returns. So, inner returns is nothing but inertia plus inductance which shows that mass flow rate through a inductive circuit or through a tube which is running in length it has got a diameter it offers impedance to the flow rate. It offers inductive impedance to the flow rate and that is why we call it take inert from here and tons from here and this a word called inner turns has been generated in fluid mechanics. For the sake of understanding let us study now RLC circuit and thereby the electrical analogy between RLC circuit and PTC I would like to deviate from here give some glimpses of basics of resistive inductive and capacitive circuits and we will come back to the pulse tube cryocooler so that we understand the inertous tube business in detail. Let F be the frequency and J represent the imaginary part you know the real part and imaginary part any vector can be replaced can be written in a real plus imaginary part and that is what I would like to use in this case. So, let us see a simple RLC circuit which is resistive plus capacitance circuit and you got a voltage across is and therefore a current will flow through it alright and this current will have impedance in the form of resistance plus capacitance there is no inductance in this case and voltage across the resistance and C capacitance is called V. So, schematic of a series RLC circuit is shown here the impedance is given by Z and this Z is called as V upon I in a pure resistive circuit this Z will be equal to R alright but in a inductive circuit this Z is equal to R minus J into 1 by omega C I will not go into the very basics but most of you know this definition how the impedance of a RLC circuit RLC circuit and RLC circuit is written as. So, this Z is equal to V upon I which is equal to R minus J into 1 by omega C alright this is a vectorial addition basically the angle between the current and the voltage will be given by tan theta which is nothing but tan inverse of 1 upon omega C divided by R that is 1 upon omega R C and you can see a negative sign over here that means this theta is going to be negative in this case alright. So, in that case my Z because Z being negative now if I add a vectorially this thing Z will always be in the fourth quadrant if I plot this Z my depending on R because R is going to be positive and this term is going to be negative. So, if I want to have vectorial addition R minus J upon omega C I will have R as positive direction towards x axis and in minus 1 upon omega C which is coming to fourth quadrant and therefore Z will come in the fourth quadrant as is obvious from here because tan theta is negative. So, theta being negative in this case this is my plotting of Z in this case you got a real part on y axis you got imaginary part. So, Z therefore will have R minus 1 upon omega C J time minus J 1 upon omega C and therefore Z will always be in the fourth quadrant as shown here. From the above equation it is clear that the angle is always negative now in such a situation in such a situation my current will always lead the voltage. So, you know that in a R C circuit in a resistive circuit current and voltage are in phase in a R C circuit the current will always lead the phase that means current will be ahead. So, let us say this is current and this is going to be voltage in that case all right direction this is actually a Z vector basically, but in this case if I take this as I this I will lead the voltage vector and therefore current will always lead the voltage this is basically to understand from a R C circuit and magnitude of the impedance is under root of R square plus 1 upon omega C whole square this is standard basically for any imaginary any any A plus I B vector. Now, let us go to R L circuit the schematic of a R L circuit is shown over here the impedance is given by V upon I R plus J omega L if you see earlier equation we have got now R plus J omega L that means this is positive and this is also positive. In this case now voltage will lead the current that means current will lag all right if you add because R is going to be positive towards x direction while omega L is going to be at a angle over there and therefore V will now lead the current in this case. The angle between the current and the voltage therefore will be a positive angle tan inverse of omega L upon R. So, this is my theta and in this case now Z will be in the first quadrant earlier case Z was in the fourth quadrant. So, in a in a resistive plus inductive circuit now the theta is going to be positive and because the theta is positive from the above equation it is clear that angle is always positive and therefore in this case current always lags the voltage or voltage leads the current. In earlier case current was leading the voltage in this case current always lags the voltage. So, R L circuit current always lags the voltage and magnitude of the impedance is R square under root of R square plus omega L square. Now, let us come to the R L C circuit where resistance inductance and capacitance are in series basically. So, schematic is shown over here it is a series combination of R L and C and a sinusoidal voltage V. In this circuit both L and C have a collective effect on the performance of the circuit as you know this thing and the impedance will be given by or Z will be given by R plus j time omega L minus 1 upon omega C alright and now theta will be given by tan inverse of this component divided by R that is omega L minus omega C inverse 1 upon omega C divided by R. Now, the value of theta will depend on what is the net result of this what is my omega what is my L and what is my C. So, omega L minus omega C will determine where theta is going to be positive or theta is going to be negative and this is very important to understand alright. This is a very simple state mathematics let us please understand that theta is going to be determined by this numerator because R is always positive. So, theta is going to be tan inverse of omega L minus omega C inverse divided by R and depending on the value of this I can have a theta from positive to negative that means I can have z vector in the first quadrant and also in the fourth quadrant I got all the possibilities depending on the value of the inductance the omega and the capacitors. So, it is clear that depending upon the value of L C and F, F is basically given the omega 2 pi F is nothing but equal to omega the angle can either be positive negative or 0. So, it can have any variation possible from being in the first quadrant being on the z axis being on the x axis as well as being on the fourth axis. So, all the variations are possible positive 0 or negative depending on the value of omega L and omega C. And therefore, in RLC circuit now I got lot of leverage to play with the z vector and that is what I want to come back to basically alright. So, in addition to L and C frequency plays a very important tool. Now, if frequency is very high then only this circuit become very important and this becomes a frequency dominant circuit. If omega is very small the inductance will not play any role in that case it becomes purely RC circuit, but omega is being high the inductance comes into picture and therefore, inductive circuits are very very important with during high frequency for high frequency operation. Now, coming from electrical analogy to pulse tube cryocooler. Now, I got comparative parameters to the RLC circuit in the fluid mechanics and fluid mechanics I am talking about the pulse tube cryocooler. For example, oscillating pressure will give me the voltage alright. When I got a delta P across when I got the pressure of variations current passes. So, because of the voltage current passes between because of the delta P because of the pressure variation mass flow rate is there. So, mass flow rate is nothing but analogous to the current because of voltage current is there because of pressure variations mass flow rate is there. So, this is I this is V then we got the reservoir alright we got a capacitance therefore, which is called as compliance sometimes in fluid mechanics. Therefore, we got a voltage current capacitance we got orifice as you know orifice flow flow through orifice will be always be proportional to delta P across it and therefore, a pure orifice normally is called as resistance to the flow alright. So, this is always going to be kind of resistance in the circuit and inner turns to be nothing but inductance which is having mass flow rate. So, we have got inertia plus inductance. So, inner turns to basically is nothing but a tube which offering inductive impedance through with the mass flow rate flows. So, the mass flows alright. So, we got a inner turns tube also here and the inductance of an inner turns tube will be given by rho into L by a rho is the density of the fluid L by a. The reservoir compliance is given by V volume of the reservoir divided by gamma which gas divided by P m again these are standard parameters. Now, let us have a orifice. Let us see this electrical analogy to OPTC and ITPTC. So, an orifice in an OPTC analogous to resistance. So, we got theta is equal to tan inverse of minus omega RC this is what we had seen and here now we know that the current the impedance lies in the fourth quadrant we had seen this and in this case now we know that mass flow rate leads to the phase pressure vector. So, if you remember the phasor diagram we had m dot c is leading the pressure vector m dot c was having an angle of theta to the pressure vector and this is what is getting proved in case of a with the RLC circuit in OPTC I have got only resistive and capacitance I got a orifice and the reservoir. So, this is a resistance and this is the capacitance and in this case my impedance vector is going to be in fourth quadrant and my mass flow rate which is m dot c leads the pressure vector this is what we had seen earlier and for a inertance tube now I want to prove that an orifice together with the inertance tube now is analogous to resistance and inductance it has got small opening. So, you got a inductance capacitance alright. So, you got a RLC circuit over here now and in this case my tan theta will be omega L minus omega C inverse divided by R depending upon the value of L C and F the angle can be changing through a large value now my angle now we will have we have seen earlier my impedance vector can be positive in the first quadrant or in negative in the fourth quadrant. So, now I can have a large variation that is possible of theta for a RLC circuit and therefore, when you have inertance tube you have to work with high frequency because inertance is always omega into L and now in this case depending on the diameter of this tube and the length of the tube I can have theta whatever I want to. So, my impedance now could be in the first quadrant or in the fourth quadrant and I can play with the theta angle. So, here in a inertance to PTR PTC I have got a large possibility of changing my theta and therefore, always for a high frequency PTC I will have a inertance tube associated with it I will not have only orifice or I will not have double inlet PTR, but I will have inertance tube and maybe I can add double inlet to it also, but what is most important for high frequency is the inertance tube and because of now diameter and the length of the inertance tube I can have theta whatever I want alright. So, this is the most important thing related to ITPTC. So, what I have understood what we have understood from here I just want to show you that from the electrical energy following things have been well understood in an O P T C the mass flow rate always leads the pressure pulse this is what we are saying. So, always the theta angles theta angle remains in O P T C because it is a resistive plus capacitance circuit while in a high frequency ITPTC depending on the inductance of the inertance tube the phase angle can be changed through a large value and therefore, ITPTC I am able to make minimize this theta by optimizing the IT inertance tube length and diameter. So, now I have got a very big possibility to change the inertance tube inert inductance of the circuit. So, that my theta gets minimized because now my z vector can lie in the first quadrant or in the fourth quadrant, but I have to see that correct diameter and lengths have been properly chosen. This is the most important thing and therefore, ITPTC I can now make my theta very very close to 0 or even negative if I want to. So, inertance tube is significant only for high frequency pulse tip cryocooler I have just said that because it is omega into L while for low frequency P T C normally double inlet phase mechanism is used not the inertance tube will not be used for low frequency machine that is the GM type pulse tip cooler for a GM type pulse tip cooler you will hardly find inertance tube being used, but for sterling type pulse tip cooler where inertance tube is machine works at a very high frequency will always use inertance tube for that and this is what I wanted to tell you choosing electrical analogy for the ITPTC. I hope you understood this please understand your basics principles of RLC circuit and try to compare or find an analogy with the ITPTC OPTC etcetera using what I have told you in this slides. Going ahead from here I will just like to take a small tutorial. So, that you understand the phase diagram. So, the tutorial is basically aiming at development of a small phasor diagram. So, that you understand the principle of phasor diagram. So, draw a phasor diagram for a 50 k OPTC with helium as a working fluid the other operating parameters are as given below. So, I got a frequency of 30 hertz charge pressure is 20 bar dynamic pressure as 4 bar the positive volume is 8 cc the regenerator volume is 20 cc the compressor data volume is 20 cc which is very high I can have even compressor swept volume added to it, but I have not added this heat exchanger volume is 2 cc. Temperature is 300 Kelvin on the ambient side well the orifice the mass flow rate through orifice that is m dot o is given as 2 gram per second. The idea is now draw a phasor diagram for this OPTC this is orifice pulse tube cryopolar. So, these are the parameters which are given VPT V region VHX VCP dead volume m dot o then F 30 hertz P o as 20 bar P 1 as 4 bar T o as 300 Kelvin and T c as 50 Kelvin. What we want to find out is phasor diagram that is phase angle m dot c between the m dot c and the pressure vector and also phase angle between the mass flow rate and the compressor and the pressure vector. So, begin with I know m dot o is equal to m dot h for OPTC which is given as 2 gram per second which is 0.002 kg per second then we know the pressure vector is going to be horizontal and on the pressure vector we have got a m dot h which could be plotted and then what we have is m dot h into T h by T c all right. So, I will have now T h by T c into m dot h which is 0.002 into 300 by 50 because my cold end temperature is 50 and this gives you 0.012. So, on this now I will build up a vector which has got a length of 0.012 this you can take any scale to begin with and draw all these diagrams. Now, depending on the volumes of hot end heat exchanger pulse tube regenerator I am going to find the mass flow rate through these different parts. We have already given the equations to this. So, for example, I want to find out mass flow rate through the hot end heat exchanger where I will use therefore, omega into P 1 V h x which is volume of the hot end heat exchanger divided by T h and divided by R. So, omega is 2 pi f P 1 I have given as 4 bar convert it to the appropriate unit V h x is given as 2 c c convert it to appropriate units of meter cube R into T h. So, having done this I get this value this is kg per second and I will add this vector at 90 degree to earlier vector I will get now mass flow rate at the inlet to the heat exchanger. This is mass rate out of the hot end heat exchanger or through orifice and this is my mass flow rate to the inlet to the hot end heat exchanger. So, depending on this now I will have a mass flow rate through the pulse tube like this and now I will add on this vector. So, now at the cold end of the pulse tube I will have m dot c this is the volume through the pulse tube divided by it temperature as given by mass flow rate in the pulse tube alright. Then mass flow rate at the cold end heat exchanger depending on the volume of the cold end heat exchanger and temperature there I will have vectors. So, I will complete this diagram and I will not go into this state. So, I want to have respect to temperatures in the denominator. So, I will normally we will take logarithmic mean temperature for regenerator. So, this is my temperature of logarithmic and I will have a vector which gives me the mass rate through the regenerator as 5.2 gram per second or 5.2 into 10 to power minus 3 kg per second and I will draw according to the length I got m dot r h. This is mass flow rate at the regenerator entrance for the hot end basically alright and then I got after cooler and now I will get mass rate at the compressor outlet which is depending on the whatever data volume I have I have got a clearance volume of compressor I have got connecting tube of compressor and correspondingly I will get now. So, I should get mass flow rate corresponding to this length to be compressed in the compressor it should leave the compressor which has got this much flow rate in order to get m dot c equal to this alright this is what we learn from this and what is the phase angle this is the phase angle theta the phase angle between the mass flow rate at the cold end and the pressure vector is the m dot c is 13.1 gram per second and theta is 23.3 degrees the phase angle between mass flow rate at the compressor in the compressor and the pressure vector is this is 16 gram per second here to be compressed and correspondingly I got alpha is equal to 47.5. So, alpha is equal to 47.4 while theta is equal to 23.3 which is a good job done by this OPTC of given dimensions. So, this is my phasor diagram it shows me that the compressor flow at the outlet should be of this length corresponding to this length approximately in order to get m dot c at an angle of theta of this length alright. So, this is where I understand that if my dead volumes are more and more my compressor gas to be compressed going to be very very high. So, understand that these dead volumes in the regenerator the heat exchanger play a very important role and therefore, depending on dead volumes I will have to compress more and more gas in the compressor which means my power input to the compressor will go on increasing accordingly alright. This is a very important thing. So, this is my final diagram I have drawn on excel sheet what it will look like to show you exactly on the pressure axis you got a m dot c m dot hx and this is drawn to the scale. So, you can see now how does this diagram look basically it is the same thing as what we have drawn but this is with correct dimensions given over here alright. So, with this background on the pulse tip curricular I will now show you the pulse tube research at IIT Bombay. So, let us see what we have done and so, you can see what kind of research has been happening at IIT Bombay and you can see I would like to show you different developed parts of pulse tube curricular at IIT Bombay. So, let us see the pulse tube research at IIT Bombay is in two parts one is the theoretical research as well as the experimental results. So, we got both the works going on theoretical front as well as on the experimental front. If I see theoretical work is normally the modeling and simulation kind of thing where we have developed isothermal model then we have got a phasor analysis we are working on CFD analysis of the pulse tube curricular as well as as you know a lot of heat exchangers involved the regenerator the hot end heat exchanger the cold end heat exchanger etc. So, we are doing a lot of work on the heat exchanger analysis also as far as theory of the pulse tube curricular is considered. Now, under the experimental work we have developed various curriculars taking various projects from different agencies the funding has to come for experimental work and we have developed various single stage pulse tube curricular we have developed two stage curriculars as well as we are developing we are in process of development of the three stage pulse tube curricular also and in both cases both for theory and experiments we want to understand or we are understanding different geometries of the pulse tube curricular which are basically in line U type and coaxial. I have talked about these during earlier presentation so in line is the one with the regenerator and the pulse tube are in line in the U type the regenerator and the pulse tube are in this U direction are parallel to each other and coaxial the regenerator and the pulse tube are actually the regenerator is outside and the pulse tube is inside so the regenerator kind of annular to the pulse tube. So, let us see all these parts I would like to show the regenerator mesh material what we use and things like that try to understand I may be going very fast to show in the limited time but please you can see those parts fabricated also. So, the pulse tube research at IIT Bombay is development of theoretical models development of single stage inline pulse tube curricular single stage U type pulse tube curricular and single stage coaxial pulse tube and this is I am getting my lecture limited to only single stage units because I do not have much of time to show even two stage units right now. The compressor I use or a pressure wave generator which I use is a linear compressor the linear compressor work on the principle of linear motor and this is a compressor which is developed at IIT Bombay based on a post piston linear motors you can see there are two pistons driven a post side and from here the pressurized gas comes out in the form of pulse at this point. So, you got a one linear motor here we got a one linear motor over here and both pistons are moving towards each other which what we call as a post piston compressor developed at IIT Bombay and similarly I have got one imported piece from CFIC this is a clever fellow incorporation in USA and because of some funding available we have bought this compressor from outside party which is also a post piston compressor this was a moving coil compressor this is a moving magnet compressor, but this is also is basically based on the same principle of having linear motors at two ends. So, this is a pressure wave generator or this is a compressor which we use to drive the pulse tube curricular. Now, I would like to show different parts of the fabricated parts at IIT Bombay. So, this is the after cooler and this is my fabricated part I will just like to show you the fabricated part over here. So, you can have a look here this is the after cooler through which water flows outside these are fins cut and this is the inside part and you can see the fins cut on this part. So, the compressed gas comes through these fins which is kept inside this. So, the compressed gas will come through these fins this is not meant for this cooler I have taken different parts, but this will be tight fit this will be inside fitting on this. So, the gas will come through these fins across and this gas while traveling through fins will be cooled by the water which is flowing outside. So, I got this circular fins on the outside through with the water will flow and this during this flow of this gas from one end it will enter from here and it will go out and during this travel it will cool the gas and it will take the heat of compression. So, this is basically the after cooler. Now, let us see slide again. So, second part is the regenerator. I will show a regenerator is going to be a thin wall tube and I will show a small thin wall tube. So, here you can see a small thin wall tube this could be any thin wall tube, but you can see this is a thin wall tube which has got a thickness of let us say 0.15 millimeter all right. Why do I want such a thin tube? Because I got a hot temperature on one side I got a cold temperature on this side and in order that my conduction is minimized my Q should not get lost I would like to have the cross section area as minimized possible. So, these are thin wall tube thick wall tubes are different and thin wall tubes are very very important in cryogenics because I will not have this thin wall SS tube thin wall SS 304 and the thickness as I said is 0.15 millimeter specially fabricated for the regenerator and the pulse tube. Next component is cold end heat exchanger. I will show a small cold end heat exchanger which is very important again because the gas gets cool. So, you can see this part again you have got fins on this side. So, the gas at the exit of the regenerator comes through this and passes through fins and this presses across outside and the outside is the cold end heat exchanger and this cold end heat exchanger is the one which transfers the cold generated in the regenerator on the pulse tube to outside object to be cooled all right. So, this is a very important component one can use mesh also for this. So, this is my cold end heat exchanger again you got a pulse tube which is a thin wall tube like the way I have shown it as regenerator. So, I will not go into showing the pulse tube but as I have shown earlier it is exactly again 0.15 millimeter thickness of that tube also. This is a hot end heat exchanger which also looks like the after cooler. So, exactly in a similar way what I have shown you cooled by water from outside in the hot end heat exchanger and the after cooler is basically they both are cooled by water from outside and you got a inside piece which is in tight fit with this through which we can always have mesh also here through with the gas goes. So, you have to have porosity through with the gas to come and it should be good contact with this. So, outside water will cool the gas coming or flowing through this fins. So, these are my two fabricated parts which are which really sometimes could be wire cut or sometimes can be done by milling machine also and this fabrication is very important. So, the water will come from one direction and leave from other direction cooling the gas which is passing through this tube and then I will have a vacuum jacket which is very important I cannot show you vacuum jacket because it is a huge thing alright. So, vacuum jacket also is a very important entire assembly will be kept in the vacuum jacket. Now, I would like to show a typical U type pulse tube cooler how will it look. So, this is my engineering drawing assembly. So, the gas comes from here this is my after cooler which I just shown you outside fins you can see and inside fins part will be over here through with the gas will come and then it will enter the regenerator. The part I had shown was cold and heat exchanger is kept at the other end of the regenerator which is like this and then I got a U tube connecting to a pulse tube and this pulse tube has a after cooler at this point and entire thing is kept in a vacuum jacket and I have shown all this part to you I will again show you a photograph of this. So, this is my top flange to which my cold and heat exchanger is attached or a sorry the after cooler is kept over here the gas will come from compressor and it will enter the regenerator and this is my regenerator and at this point what I will have is a cold and heat exchanger alright and then I got a collecting tube which is why it is called as U type pulse tube cooler and you can see the scale also here which gives you idea of the different dimensions. This is my pulse tube this is my cold end of the pulse tube this heat exchanger can be regenerator cold end side or at the pulse tube cold end side also alright because they going to be at same temperature because of the copper connectivity more normally the temperature at the cold and heat exchanger at the regenerator cold end and the cold end pulse tube will be same and because the diameter of the regenerator is little higher you have got more heat transfer area available at this point and therefore, we put this cold heat exchanger at the regenerator colder end and now you can see the hot end of the pulse tube and this is the heat exchanger at the hot end of the pulse tube both of these heat exchangers are cooled by water. So, you can have water coming and then divide into two parts one part is cooling the gas at the compressor and one part is cooling the gas at the hot end of the pulse tube alright. So, you can see all this fabricated parts over here this is my top flange this top flange sits on this vacuum jacket. So, anybody wants to work on this research you can see these parts and you can have a you know first hand information about different fabricated parts at IIT Bombay and anybody wants to work on this can always send me an email and we can always talk about how these parts are getting fabricated who are the fabricators because we have developed various fabricators for example wire cutting the fin cutting and thing like that. So, let us see the compressor and the pulse tube heat exchanger and the pulse tube cryocooler and you can see here inline pulse tube cryocooler mounted on the heat compressor developed at IIT Bombay. So, this is the compressor on which I have kept inline pulse tube cryocooler this is developed by one of our PhD students at IIT Bombay and this is inline pulse tube cooler and integral unit because the compressor and a pulse tube cooler are attached to each other directly there is no gap there is no connecting tube between the compressor and the pulse tube cryocooler while on this unit you can see that this is the compressor and this is connected through a tube and therefore, this is a split unit compressor and the expander are away from each other they are connected through a tube and inside this what you have is a u 2 pulse tube cryocooler which you cannot see right now, but you have seen earlier and you can see the water coming over here and cooling from this side and leaving from this side. Similarly, in this case also you can see the hot end hot end is going to be coming at the top while the after cooler is going to be coming over here. In u type the hot end and the after cooler come in the same side because this is the u connection alright and therefore, water can cool both after cooler as well as the hot end heat exchanger in the same direction, but here it cannot do that thing the after cooler is here while the hot end it at the other end and at the top end comes the inheritance tube. So, I have not shown you cannot see the inheritance tube over here similarly, I will have inheritance tube coming from the hot end of the pulse tube cryocooler in this case. Now, this is a complete experimental set of you can see and here I have got a compressor, I have got a expander or a vacuum jacket and you can see the inheritance tube opening into a reservoir at this point. So, I have got a linear compressor here from a CFIC company I have given the model it can give you maximum power of 350 watt. Then I have got its power supply, then I have got other units I have got a double inlet wall used over here, I have got the pressure measurement sensors good end wave co pressure sensors have been used at this point which is PZ resistive sensor and I have got a temperature measurement you can see this is why lecture temperature measurement device and what sensor we have used is silicon diode to measure the cold temperature. So, this is my power supply to the compressor I can change the frequency of this power supply also this is the expander a vacuum jacket which houses the pulse tube cooler this is the compressor inheritance tube opening into reservoir while this is my lecture device to measure the temperature on the using the silicon diode and these are also a temperature measurement if I am using P T handed as a temperature sensing device. This is entire experimental set of would look like on a pulse tube cryocooler. So, here I got a small video of the pulse tube refrigerator that has been developed in our laboratory our past research student and professor Narayan Kedkar also has worked towards making this pulse tube refrigerator at IIT Bombay. So, this is my pulse tube refrigerator and this is the linear compressor alright. Now, I am charging some gas from the gas cylinder to this pulse tube refrigerator and I have used a turbo molecular pump to vacuum out the vacuum jacket which is across the pulse tube refrigerator in which the pulse tube refrigerator. So, this is my pulse tube refrigerator and this is the vacuum. So, you can see I have used a K F coupling to connect this vacuum hose in this vacuum jacket this is my linear compressor in which there is a single stage in lined pulse tube refrigerator. This is a hot end of the pulse tube and this is the after cooler which is what we where the again the water is circulated to remove the heat of compression alright. So, you can see this vacuum connection and this is my temperature measurement. This is the power supply I am giving some 100 watt power supply to begin with and this is the oscilloscope here in I see the pressure fluctuations. You can see the pressure fluctuations now here as soon as the power is given is around 102 watts are fade actually and you can see that the pulse tube refrigerator has started functioning the temperature is coming down to 251 Kelvin and power is just 100 watts and you can see a pressure pulse monitored at this point at the hot end of the pulse tube. So, you got various instruments here this instrument monitors the charging pressure of the filling pressure over here you got a turbo molecular pump here used for vacuuming the vacuum jacket and this is the inner tons tube of the pulse tube alright. You can see this is the power supply which gives you around 100 watts at 50 hertz alright. So, I am giving power supply at 50 hertz and now temperature around 122, 182 Kelvin maybe after 5 to 10 minutes and now it has come to 100 Kelvin 91 Kelvin alright and now the power was increased to 160 watts and now increase the power to higher values and correspondingly the pressure ratio will increase and the temperature will start coming down it has come to 82 Kelvin now for 205 watts power input it has come to 78 Kelvin now. So, we increase to from 100 to 200 to 250 and now the power is 300 watts and temperature of 48 Kelvin has reached this is the lowest temperature that it could be achieved by this single stage inline pulse tube refrigerator developed in our laboratory at IIT Bombay. Now, I would like to show different fabricated parts of the single stage pulse tube cooler for example, this is a inline pulse tube cryocooler alright. So, you got a inline pulse tube cryocooler the regenerator and the pulse tube cooler are inline which is each other this is the hot end after cooler this is the hot end of the pulse tube. Let us see this fabricated part here in this camera. So, you can see this part over here I have just shown you the picture of this. So, the gas will enter from here and I will have lot of meshes mounted over here which basically from the regenerator the gas will go through this regenerator which has got mesh material put over here which acts as a matrix material at this point I will also have a cold end heat exchanger sitting over here which is in direct contact with this. So, whatever I want to cool I will cool with this. So, this is inline unit in which you know that the cold end heat exchanger is at the centre leading into pulse tube cooler here. So, leading into pulse tube and the pulse tube opens at the hot end of the pulse tube. So, I will have a hot end heat exchanger on other side followed by inner turns tube coming on this direction. So, this is a simple pulse tube inline where the regenerator and the pulse tube are inline the cold end heat exchanger is at the centre. Also I would like to show right here what is this regenerator. So, in this now coming back to the same system I got a regenerator meshes over here. So, you can see this regenerator mesh if you could focus on this this is my regenerator mesh and this is a 400 size that means I have got 400 meshes in 1 inch. I have cut it from a cloth of stainless steel and I have made it I have punched them in this size you have to do dye and punching for this and all these things are you know thousands of these put together they have to be inserted in this through with the gas flows and the gas gives this heat to the mattress or these meshes basically store the heat. So, it is a very important how the meshes are cut. This is the most important component because if your meshes are not good if they are having dust they will not be good flow of the gas and this mesh is connected from a cloth and this is a cloth a simple cloth. So, you can buy meshes from the market and it is like this I can put them together and I can cut meshes from here I can fold this cloth and then I can have a punch like this and then from which I can generated the punched meshes like this for whatever diameter you want so that it fits in the regenerator and this is the most important operation you can see that this is a very fine mesh 400 meshes in 1 inch. So, you can see that a dust sometimes can get occupied it can occupy the space over here and therefore, it will not allow the gas to flow. So, it has to be absolutely properly cleaned normally this operation in a commercial organization will be done in a clean room. So, that dust will not be there alright. So, this is my inline pulse tip cooler and now I will show you U type pulse tip cooler. So, this is my U type pulse tip cooler and with this cooler we have got inline color we have got 50 Kelvin temperature U 2 pulse tip cooler we have got a 54 Kelvin temperature and I have got a coaxial pulse tip cooler also which we have got around 60 Kelvin temperature alright. So, I will show you all these three parts. So, let us come back to this camera again. So, this is my inline pulse tip cooler alright then I have got a U type pulse tip cooler. So, you can see that the gas will enter to this regenerator this is the gas to enter. So, compressor then it comes like that and with a U connection it enters the pulse tip cooler. So, you can see that in a U type pulse tip cooler the after cooler will be here and the hot end of the pulse tip cooler cooler will be here while it will not be so in this. So, you can see and compare both and this is my cold end heat exchanger whatever I want to cool I can connect you. So, this I had done in order to connect something to cool something here. So, you can see that how the cold generated is going to be transferred and whatever object I want to cool I will keep connected to this. So, this is my inline connection inline pulse tip cooler this is my U type pulse tip cooler and now I have got one more which is a coaxial pulse tip. So, you can see now a coaxial pulse tip can you see this. So, the annulus part you can see is a regenerator in which I will fill the mesh. So, the gas will come and go through the annular part and it will come through regenerator out now. So, at the bottom end where my heat exchanger is going to be there the gas will turn like this. So, you can see how compact this unit is. If you remember my earlier lecture I had said that the user will always love to have a very compact unit and it offers a lot of heat transfer area at the bottom to be cooled this is my cold end alright. But construction wise is a little problem the fabrication is always a problem and also the pressure drop across is a very large pressure drop. So, the although I prefer to have a coaxial pulse tip cooler the fabrication wise is a problem and now you can see all the three units over here. You can see inline unit the heat exchanger at the center the cold end heat exchanger at the center the u type and the coaxial unit alright. So, I have shown you all the parts which are fabricated and now let us come back to our slides to see the results. So, now let us see the results for the inline the u type and the coaxial type. Here you can see I have plotted temperature versus cool down curve versus time and you can see that with inline temperature I could reach 50 Kelvin and the temperature reaches to lower value within half an hour you can see or 40 minutes. While a u type unit I have got 54 Kelvin and you know the fact as I said earlier the gas thermodynamically always prefers you know it travels in the same direction therefore, the pressure drop losses are minimum while you have to have gas has to take 180 degree turn and therefore, pressure drop losses are more and therefore, the temperature in this case has not reached to as low as what we got in a inline unit. So, we got a 14 bar charging pressure in this case and operation is done at 50 hertz what we have reached is 50 Kelvin for inline and 54 Kelvin for u tube and this is what is called as cool down curve. Now, I want to see the cooling effect also. So, what I do normally I will put a electrical heater at the other end or the cold end and I will go on increase in the wattage which will actually acting as a dummy load in a system and this will give you basically what is the wattage or a cooling load delivered from a system. For example, at 80 Kelvin I can see that I got around 9.4 watts at 80 Kelvin and this is what we call as a cooling load. The cooling load will always be associated with some temperature while u tube pulse tube cryocooler I got 6.5 watt at 80 Kelvin again showing that some cooling effect is lost because the pressure drop and increased dead volume at the cold end. So, I get around let us say 10 watt at 80 Kelvin for 14 bar charging pressure and 50 hertz operation and now I have got a coaxial unit again here I reach around 61 Kelvin temperature, but you can see how fast I have reached here basically in almost 20 minutes. So, I got a operation at 46 hertz, but I got a 17 bar higher pressure little bit as compared to earlier case and power input of 250 watts. So, these are my different results I just wanted to show you various results how are they plotted and we can compare the in line versus u versus coaxial. As I earlier told you in line is always beneficial from thermodynamic point of view, but it has got the cold end heat exchanger at the center u type and the coaxial will be preferred, but in coaxial unit the fabrication is a problem, but unit is always very compact. So, normally the u type pulse tube cryocooler is normally manufactured by most of the fabricators. Having seen all these parts having seen all the theory behind the pulse tube cryocooler the sterling cryocoolers the GM cryocooler the sterling type machines sterling type pulse tube cryocooler the GM type pulse tube cryocooler. I would now ultimately would like to take you through the summary of the all the lectures of this cryocoolers. So, summary is let us go step by step. A cryocooler is a mechanical device which generates low temperature due to compression and expansion of gas this is what we have understood. The example of recuperative cryocoolers are JT cryocoolers, Brighton cryocoolers, Claudic cryocoolers. They use recuperative heat exchanger and not regenerative heat exchangers. The examples of regenerative cryocoolers are Sterling cryocoolers, Gaffer-Mackmann cryocoolers and pulse tube cryocoolers. A sterling cryocooler was first conceived by Robert Sterling in 1815. Depending upon the relative arrangements of piston, displacer piston we had alpha, beta, gamma combinations possible and depending on the geometry of this arrangement there are different types of sterling cryocoolers. For an optimum pulse tube cryocooler, for an optimum design of a cryocooler a compromise between operating and design parameter may be sought. So, operations may demand some different parameters, the design may demand some different parameters and you have to do optimizations based on this. A combined effect of parameters on performance of a system as a whole is given in Wokker's Optimization Chart. So, we had used Wokker's Optimization Chart to design a sterling cryocooler and you can go back to those lectures to understand what was this optimization all about. Now, in a GM system for an optimum performance the relation between the pressure pulse generated by wall mechanism and expander motion is vital. So, we know that in a GM system you got a wall in between the compressor and expander and therefore, the relationship between the pressure pulse generated by the wall mechanism and expander motion. You got a displacer moving over there and this relation is very important. While in sterling system the relation between the piston and displacer is what is most important. A GM system can reach much lower temperatures as compared to sterling system. Normally in a single stage or in two stage unit we can reach much lower temperature because it operates at low frequency. However, the power input required for a GM systems is much higher than the sterling machines. The single stage normally will reach around 30 Kelvin using single stage using the stainless steel mesh for a typical GM cycle. While a two stage machine can fetch you 10 Kelvin temperature while which can use on the first stage stainless steel mesh and on the second stage it will use lead balls as regenerator matrix. Also we can get two stage reaching 4.2 Kelvin temperature in which case we will use first stage stainless steel mesh plus lead balls if possible and in the second stage we can have a lead balls plus erbium 3 nickel. This is a magnetic materials to be used in the second stage of the material where the Cp of this material increases at a very low temperature. It could be hybrid regenerator having lead balls plus erbium 3 nickel or it could be 100% erbium 3 nickel also. In a pulse tube cryocooler the mechanical displacer is removed and an oscillating gas flow in the thin wall tube produce cooling. This is what we talked about pulse tube cryocooler. The pulse tube systems can be classified based on sterling type or GM type. The geometry and operating frequency we know that in line U go axial having high frequency low frequency kind of machines. We can have phase ship mechanism so we can have OPTC, BPTC, ITPTC, double inlet and all these things which is what we have studied from phase diagram. There exists a phase angle between the mass flow rate at the cold end and the pressure vector in the pulse tube cryocooler which we try to minimize for which we employ various phase ship mechanisms. Heat lifted at the cold end depends on m dot c P1 by P0, Tc and phase angle for a pulse tube cryocooler and orifice in the OPTC analogous to resistance and orifice together with inner tons tube is analogous to resistance plus inductance connected in series and this is what we have studied today using the RLC circuit and we have seen that various units have been fabricated at IIT Bombay. We are working on two stage and as well as three stage units. In fact, we have reached around 20 Kelvin now using a two stage PTR. We have seen that we have read around 50 Kelvin using a single stage PTC. We are already started working on the three stage pulse tube cryocoolers also at IIT Bombay. So, we have seen that fabricated components at IIT Bombay. We saw I will I have shown you various components of you know after cooler, hot end heat exchanger you also seen the mesh material which is used for regenerator alright. So, with this background if you want to work on the pulse tube cryocoolers, sterling coolers, GM you are most welcome and with this lectures I feel that you can start working in the cryocooler research. Now, various publications can be seen. So, we have generated lot of publications and this is what I would like you to go through. So, there are publications in the conferences and journals, various journals. I have given all the details I cannot go through all these things, but there are various lectures given in the national conferences as well as in the internal conferences and they are all given on the website. So, this is for your data. Please go through those papers, get access to these journals and you can see all those publications which will talk about what I have developed, what I have shown you here in this lecture. Thank you very much.