Welcome to the 16th lecture of cryogenic engineering under the NPTEL program from IIT Bombay. Before I go further in this lecture, I just wanted to take or take you ahead from where we left during the last lecture and that is the Claude cycle. We were talking about Claude cycle and in the earlier lecture, we have seen a Claude system in with the energy content in the gas is removed by along it to do some work in an expansion device and we talked about having y as yield and w by m as the work per unit mass of gas which is compressed and they are given by this formula. Just to give a recap of the cycle, we have got a compressor, we have got 3 heat exchangers 1, 2 and 3 and we have got an expander over here from where you get a work output. The gas gets compressed, it gets pre cooled and a part of the gas amounting to m dot e is taken off, it gets expanded to low temperature from 0.3 and meets the return stream at point e. The remaining gas that is m minus m e goes ahead, it gets expanded in a isenthalpic pattern from 5 to 6 and what you get here is liquid.  When it goes back and gets warm up to 0.7, the cold stream comes from here after expansion joins the point at e and the gas then gets into at the point 8 as a result of this mixture. Then this gas gets warmed up up to 0.1, where the make up gas comes and the cycle repeats. This we have talked about in great details in the earlier lecture. The yield given by the Claude cycle depends on this formula, which is h 1 minus h 2 upon h 1 minus h f, which is a parameter comes based on this point 1 and 2 plus addition to this is coming because of the x fraction, which is nothing but m e upon m dot which is diverted to the expansion device and therefore, x into delta h or h 3 minus h e, which happens across the expansion device is an additional yield what you get in the Claude cycle. Now, as you get work output or as you get work done by the system by the expansion device, the W net by m dot that is work of compression per mass of gas which is compressed is given by this formula. This is a standard formula minus x into delta h basically this is nothing but the work done by the system, because of which net work input becomes less. This is what we had done last time and we also studied the parametric effect of variation of x and its effect on the value of y and this we studied for a different t 3 values that means the temperature from where the gas expands. So, what we have studied was in a reversible Claude cycle if t 1 and t 2 and t 3 are held constant t 1 and t 2 basically are the compression pressure ratio dependent, one is a ambient pressure ambient temperature point t 2 is a temperature isothermal compression depending on the value of the point 2 and t 3 is nothing but the point or temperature from where the gas expands. We had seen from this curve that the yield y which is plotted on the y axis and on the x axis what we have plotted is x which is nothing but the fraction of the gas which is diverted to the expansion device. The yield y goes through a maximum with the increase in the value of x this is obvious from this particular figure. At the same time what you can see is this maxima shifts to right and the maximum value decreases with the decrease in the value of t 3 which we can see from different t 3 values as 300 k 275 250 and 225 k. What you can see from here is as the value of t 3 decreases from t 300 to 25 the maximum value of y also decreases from let us say about 0.32 to around 0.25 as the temperature or the t 3 point decreases. At the same time what you see is the optimum value of x which in this case is just below 0.7 increases marginally as you go on reducing the value of t 3 and it goes beyond the value of 0.7. So, one can see that the maxima shift to the right and the maximum value decreases with the decrease in the value of t 3 this is just to recap what exactly we did during the last lecture. Similarly, what we can see from this curve is how is the variation of w by m f that is work of comparison per unit mass of gas liquefied how it changes as a function of x. So, again we can see that in a reversible cloud system if t 1 t 2 t 3 are held constant the w by m f value of the system goes through a minima with an increase in the value of x this is what you can see for every temperature t 3 we have got a minima that means the work done per unit mass of gas liquefied is going to be minimum normally aim at this particular point and this point essentially matches with the value where the y becomes maximum. So, at this point your work input is minimum and yield is maximum also what you can see from here is as the value of t 3 decreases in this direction for example, this green curve is for 300 Kelvin value of t 3 and as you go up what you can see the value of t 3 is decreasing what you learn from this curve is the position of the minima shift to the right. So, you can see that the minima again was at a value of x which is just less than 0.7 and it shifted to the right as we went on reducing the value of t 3 from 300 to 220 Kelvin. At the same time the value increases w by m f increases with the decrease in the value of t 3. So, as you go on reducing the value of t 3 the w by m f or the net work increased here as shown in this particular figure. This we had seen during the last lecture the current lecture would aim at same topic, but what we want to do is now study the cloud system with irreversibilities in compressor and expander. So, what we had done was we had assumed 100 percent compression process, 100 percent expansion process basically and what we can see now the different efficiencies associated with these two devices. Then we will also study the Capitza system, the Helene system and the Colleen system which are actually in a ways some kind of modification of the cloud system and that is why I have clubbed this clubbed together in the lecture regarding cloud cycle. So, first let us see how cloud system behaves for different irreversibilities. Under Colleen cycle we will see the value of y and again work requirement. As you know the compression and expansion process in actual cloud cycle are irreversible. Also the heat exchange process also is going to be irreversible during this lecture I am going to talk only about the compression and expansion process. These irreversibilities cause inefficiencies in a system and it deteriorates the performance of the system. This is obvious. To study the effect of this irreversible inefficiencies or irreversibilities a two-terror problem is taken up here. The results are graphically plotted and compared with the reversible system. So, let us see the effect graphically as to what exactly happens if these inefficiencies are taken into account. So, this is the cloud cycle. You can see here the gas is expanded from 3 to E and 1 to is a compression process, 2 3 is a heat exchange process, 3 to 4 and 4 to 5 also are heat exchange process, 5 to 6 is isenthalpy process and this is a return stream. The expanded gas joins the return stream at point E. This is what we show on a T S diagram. Now, here what you see is a 1 to 2 is a isothermal compression process which is given by 1 to 2, but in actual system this will not be a 1 to 2 process. It will have its own efficiencies and therefore, 1 to 2 process would behave like a 1 to 2 dash process. So, actual work done will be more than what is ideally assumed 1 to 2 process. Similarly, the expansion process which happens from 3 to E when it is 100 percent isentropic or adiabatic in actual case it will be 3 to E dash depending on the inefficiencies involved and therefore, the process of 3 E will become 3 E dash and these are the inefficiencies of compressor and expander which we would like to take into account and solve in a tutorial problem. The compressor inefficiency is due to both frictional losses calling as mechanical efficiency of compressor that is eta mechanical and the non isothermal process that is eta isothermal compressor. So, basically these are 2 efficiencies eta mechanical and eta isothermal and the net irreversibility is given by eta overall which is nothing but eta mechanical into eta isothermal and this is clear in most of the thermodynamic course this is taught. Similarly, the expander inefficiencies due to both frictional losses which is eta mechanical expander again and non isentropic nature of the process which is given by eta adiabatic for the expander process. So, again we will have overall efficiency for the expander which is nothing but eta mechanical into eta adiabatic for expander. So, we have got 2 efficiencies one for compressor and one for expander and these have to be taken into account in order to get actual work input and in order to get actual yield from this system. So, with these efficiencies taken into account the yield of the system decreases and the work requirement increases. This is clear that because of the irreversibility is in a system the yield that is y value will decrease while the work input will increase. How do we give this formula for the mathematical formula for this yield and work requirement they are given by this. So, over formula becomes now y is equal to h 1 minus h 2 upon h 1 minus h f plus x time eta adiabatic because actual enthalpy drop will not be now h 3 minus h e, but it will be eta adiabatic into h 3 minus h e. So, these values have been replaced by multiplying the parameter x by efficiency of the expander everything else remains the same and work input the net work input now will increase depending on what is the overall efficiency of the compressor. So, instead of having the work input as t 1 into s 1 minus h 2 minus h 1 minus h 2 for normal process will have t 1 into s 1 minus h 2 minus h 1 minus h 2 divided by the overall compressor efficiency. Also from where we subtract the work done by the system and this work done by the system also will reduce depending on the overall efficiency of the expander. So, the formula is going to change in this manner for net work input and net yield that one gets from a inefficient cloud system. Now, in order to understand the effect quantitatively what I am going to do is basically solve a tutorial and therefore, we can have some quantitative field of as to exactly what happens to these values and how much corresponding work input increases and how much correspondingly the y or the yield decreases. So, let us do the tutorial which takes into account all the inefficiencies involved in a cloud system. So, now the problem definition is determined W by M F for a cloud cycle with nitrogen as a working fluid. The system operates between 1 atmosphere and 50 atmosphere the expander inlet assumed to be t 3 which is 250 Kelvin. The expander flow ratio is varied from 0.1 to 0.9 that is the value of x the efficiencies are given as below. The compressor efficiency that is overall efficiency of the compressor is 75 percent or 0.75 and the expander has mechanical efficiency of 0.86 and adiabatic efficiency of 0.86. So, overall efficiency of expander will be 0.86 into 0.86. Also we would like to understand the effect of this inefficiency on the value of t 3. So, repeat the above problem with t 3 is equal to 300 275 250 Kelvin plotted data of y versus W by M F y and W by M F versus x graphically and comment on the result. So, you can realize that this problem essentially is similar to what we have done during the last lecture during in the tutorial we conducted during the last lecture. What I am going to do now is basically add on the inefficiencies in the Claude cycle and see how do they affect the one the values which we got assuming that all the efficiencies were 100 percent. So, the given data is the Claude system 1 atmosphere to 50 atmosphere working pressure for nitrogen fluid t 3 value 300 275 250 mass flow ratio is the mass flow ratio that is goes to the expander 0.1 to 0.9 efficiencies are given over here. What we have to do is work per unit mass of gas liquefied to be found for this 3 temperatures of the value of t 3. So, methodology which we are going to follow here is I am not going to go into the details of solution of this problem, but I would come to the result straight away by following this methodology because we have solved this problem in detail in the last lecture. So, in the earlier lecture an assignment problem on a reversible Claude cycle with an answers was given as stated earlier the same problem is taken up and the effects of inefficiencies of the compressor and expander are studied here. All the calculations detailed calculations are left for you as an exercise for the students and the final results are graphically plotted. So, what I expect you to do is to carry out all this calculation by yourself and check your answers with the graphical plotting which we have done. So, these are the results actually. So, what do the results tell you these results are now variation of y versus x. For a system the continuous line basically shows 100 percent efficiency which are given over here while the dotted line gives for the inefficiency system for which the efficiencies are given over here and these two plots are done for 300 k and 275 Kelvin for 1 to 50 atmosphere. So, what you can see from here the y again goes through maximum when y is plotted versus x and what you can see here is as the system become inefficient the y value reduces from this point to this point this is the dotted line while the violet line is for 275 Kelvin value of T 3 and here also the y reduces from this value and it comes down to this value which is around 0.27. So, the reduction of y value and for both the cases for 300 Kelvin as well as for 275 Kelvin alright. So, the plot for y versus x for T 3 is equal to 300 Kelvin and 275 Kelvin is shown over here it is clear that maximum yield of the system decreases due to irreversibility. So, because of the inefficient system which is dotted line showed here in this plot the y value decrease from this point to this point and therefore, what we can see here is decrease due to the irreversibility or the inefficiency in the system also what you can see the percentage decrease in the y max is around 10 percent and 9 percent for T 3 value of 300 to 275 Kelvin. That means, as you went on reducing the temperature of T 3 the percentage decrease is less alright. So, for a high T 3 value the percentage decrease is higher for an inefficient system alright. I hope this is clear to you now same thing we can study it with W by M F plotting W by M F versus x again you can see that the continuous lines are meant for the efficient system as shown in this table here. While the dotted lines show the inefficient system the values for inefficiencies are given in this rectangle as value of 0.7586 and 0.86. So, again you can see that the value of W by M F or the work done per unit mass of gas liquefied gets lifted up that means, the values increase as the system become inefficient and here you can see that the increase in the value is tremendous. If you see the values on the y axis you can see that this increase is tremendous and therefore, the figure of merit also is going to be changing substantially alright. So, the plot shows for 300 Kelvin and 275 Kelvin values of T 3 it is clear that the minimum work requirement of the system increases due to the irreversibility alright. So, here one can see that the minimum work input has increased dramatically this was close to around 500 here while it has gone about 1250 over here. So, one can see that the increase has been almost of the order of 90 to 100 percent. The percentage increase in the value of W by M F minimum is 89 percent and 87 percent for 300 and 275 Kelvin respectively. Again the increase is more for higher value of T 3 that is 300 Kelvin as compared to that the increase is little less marginally less for 275 Kelvin, but the point to be noted here as soon as the inefficiencies came into picture the work input the net work input increase substantially. Mind well that we have not included the inefficiencies of the heat exchanger as soon as that if inefficiency also come into picture will have lot of problems with this curve and increase in the work input or decrease in the y value the yield value will be substantial alright. So, this was basically an indicative exercise to let you know what is the effect of inefficiency of a system in relation with compressor and expander only on the work input to the system as well as on the yield value. So, one has to really understand the importance of these devices the expanders and the compressors and the effect of their efficiencies on the yield and the W by M F values. This makes this clear that how the inefficient system increases the work input or decreases the value of y. This actually concludes the Claude cycle from here I would like to now take you to the next cycle which are basically nothing but the modified Claude cycle and they are now named after the people who invented those cycle as usual they are called capita cycle and heland cycle alright. So, that why do these cycles are required because now we are moving towards more practical cycles these are the cycles which normally exist and they are being used in actual liquefaction of the gases. So, you know that the transportation of gases across the world is done in liquid state by storing them at cryogenic temperature this is what we had talked about right. Also we talked about the air liquefaction is of primary importance because nitrogen and oxygen or liquid nitrogen and liquid oxygen are separated from liquid air. So, air liquefaction industry is a very big industry because from this air liquefaction only by cryogenic distillation what you get is a liquid nitrogen and liquid oxygen of a desired purity alright. So, therefore air liquefaction plant utilize these cycles and therefore these cycles are of very high importance in cryogenic engineering. Capita and heland systems are the two different modifications of the Claude system which are generally used in the air liquefaction these are two very commonly utilized cycles or systems and which are nothing but modification of the Claude system we will see how these modifications are brought about. And finally, we will talk about the Collins system which is nothing but the modification of Claude cycle again, but it is for now liquefaction of helium till now we are talking about around 80 Kelvin region that is oxygen nitrogen air etcetera, but then comes the next cycle which is Collins cycle which is now we are talking about helium temperature that is 4.2 Kelvin. So, for a very low temperature gases Collins cycle comes into picture. Now we will talk about the capita and heland cycles. So, this is what basically we are talking about the capita system a capita system is a low pressure system which is used in air liquefaction this is not a capita right now this is as you know is a standard Claude cycle. What I want to show now here what are the modifications are done in a standard Claude cycle in order to get from Claude to capita system. So, a capita system is a low pressure system which is used for air liquefaction it was invented in 1939 by Poitre capita in which the first heat exchanger is replaced by a set of wall regenerator. So, this heat exchanger which is the first heat exchanger now is being replaced by another heat exchanger called as regenerative heat exchanger the regenerator alright. So, you can see that this first system is of and a second heat exchanger comes into place and this heat exchanger is called as regenerator and please see that there is a arrow I am showing here means that this is a kind of a rotary heat exchanger which rotates like this. Also in addition to this we will talk about the regenerator and this rotary business in the coming slides, but what other changes happen from the Claude cycle when we go from Claude cycle to capita cycle is third heat exchanger is eliminated in the Claude cycle and therefore, I will just remove that and the cycle gets reduced to now only two heat exchangers and one expansion device alright. So, it is become very simple now to understand that I got the first heat exchanger removed and replaced by something called as regenerator and I got a sign which tells me that this rotates and then the third heat exchanger has been removed and therefore, in effect capita cycle or capita system has only two heat exchangers and one expansion device. The regenerator or the heat exchanger performs two different operations this regenerator the first regenerator performs two operations it cools the gas when it goes from 0.2 to 0.3 and also it warms the gas when it goes from 9 to 0.1 alright this is what a heat exchanger is expected to do. In addition to that what it does is it purifies the gas a practical problem which comes in the actual cycle is the gas which gets compressed comes with contamination the contaminants could be moisture or other gases and the gas which goes beyond this heat exchanger has to be devoid of this all the contaminants like CO2 or moisture or thing like that. In order to remove those contaminants this first heat exchanger is of rotary type and in addition to cooling and warming the gas it purifies this gas also. How does it do? How does it purify the gas? During one cycle one unit purifies by freezing the impurities and cools the incoming hot gas. So, when the gas goes here and let us say this is impure gas this impure gas let us say this whole heat exchanger it divided into two part and this high pressure gas goes through one part to begin with because of the cold gas which goes back the impurities get frozen over here and the impurities will not allow later once the impurity get frozen it will not allow the gas to go through it because it blocks and thereby what we do now here the whole heat exchanger rotates and this block heat exchanger comes this side while this part will come on this side. When this block heat exchanger with impurities come on this side there is a counter blow alright. So, the gas come from the this side for warming and this basically because of the warm gas this basically evaporates the impurities which had block the regenerator or this part. There by releasing all the impurities and therefore the gas can get warm at the same time once this part of the heat exchanger gets devoid of the impurities this can come on this side and therefore the heat exchanger basically rotates it freezes down the impurities while the other side gets removes the impurities and therefore the improved part will come on this side the frozen part will go on this side and this continues. In the rotary manner this heat exchanger one side will get blocked and then this blocked side will go in other side where the counter blow gets into picture which evaporates all the frozen impurities during which time the after this the impurity driven part will come in the path here on this side and the cycle continues and this has to be achieved with lot of wall mechanisms alright. So, during one cycle one unit purifies by freezing the impurities and cools the incoming hot gas while the other unit warms the outgoing gas and simultaneously removes the frozen impurities by evaporation. So, two actions are happening simultaneously one side is freezing down the impurities the other one is removing the impurities once this gets blocked and once this is cleared the heat exchanger rotates and the cleared path will now allow the gas high pressure gas to go in while the impurity frozen part will come on this side it will allow the low pressure gas to warm up and at the same time the impurities are driven out. I hope it is clear to you and this is what exactly done you know this regenerative purifier or regenerative heat exchanger. The wall mechanism is used to periodically change over from one unit to another this walls are not shown in the picture all, but you have to see that the walls open at correct time and for correct amount of time the high pressure gas passes through it and for the correct amount type the low pressure gas removes this frozen impurities this is the most important thing. This periodic alternation of units along with the counter blow arrangement ensures continuous performance. This system was the first one also in addition to this changes what was introduced by Capitza is a use of turbo expander till now what expander we are talking about was of reciprocating type Capitza was the first one who introduced a turbo expander instead of a reciprocating expander. So, the mechanical engineering development in terms of having a high speed turbine came into existence in a air liquefaction cycle and now it can take higher flow rates, but you have to worry about that turbine runs with a very high frequency it has to have all the correct bearings shafts and all the mechanism which comes with a rotary expander that has to be taken into account. So, all those turbo expander machinery comes into picture and Capitza introduced for the first time in the Capitza system alright. So, the modifications were plenty as compared to Claude cycle the rotary expander came into picture the first heat exchanger was changed to regenerative wall kind of heat exchanger and the third heat exchanger was removed and these are basically the changes that happened when we modified Claude cycle for Capitza cycle. This modification allowed the elimination of third heat exchanger in Claude cycle basically now that turbo expander could handle more and more flow rates and therefore, it allowed to removal of this last heat exchanger alright. The yield and the work requirement of the system are given by the following equation. If you see that nothing has changed much from mathematical perspective and therefore, the equation actually remains the same. So, the equation for a yield y is given by h 1 minus h 2 upon h 1 minus h f plus x time h 3 minus h e divided by h 1 minus h f. So, equation is similar to what we had earlier and the W net also is a same formula this is the work of compression on compressor over here while this is nothing but the work done by the expander which is nothing but x into the enthalpy drop across the expander. So, work decreases by this amount and here x nothing but the mass flow fraction that is driven to the turbo expander here and this is what the Capitza system is for air liquefaction. Now, let us study what modification happened to the Claude cycle when we went from Claude cycle to Heland cycle. So, Heland cycle is basically high pressure system which is used in air liquefaction. The typical operating pressure for Heland cycle around 200 atmosphere in 1949 Heland observed that when a Claude cycle is operated with air with 200 atmosphere and x is equal to 0.6 x is nothing but the m e upon m dot the optimum value of t 3 before the expansion engine is close to ambient and this was the basic reason how Heland got idea he found out that if you have the pressure as high as 200 atmosphere and the value of x is equal to 0.6 that means 60 percent gas is diverted to the expander. Then the yield is maximum when point t 3 or the temperature at point 3 is at ambient that means here at this point as we have seen in our tutorial when the value of t 3 is at ambient temperature the yield is maximum. This gave him an idea that what is the need of this first heat exchanger why not have the gas expanded first without going into the heat exchanger and this is what basically draw Heland come up with this Heland system. So, what Heland did was he eliminated the first heat exchanger and therefore the system become like this. What does it mean? That means the expansion started happening that means m dot e is drawn from this cycle right at the ambient temperature and the gas is taken at right at the ambient temperature that is 300 Kelvin and expanded and joins the return stream at this point. If you remember in Capitza system we had removed the last heat exchanger, but in Heland system now you can see the first heat exchanger is removed and that is the only difference in addition to other differences like the regenerative heat exchanger introduced on the top plus the turbo expander as Capitza had introduced. Basic difference however he the Heland removed first heat exchanger and Capitza removed the last heat exchanger. This modified system is called the Heland system. In this system the inlet to the expander is at ambient. Now this solves lot of practical problems and what are those practical problems? The lubrication on the higher pressure side of the turbo expander or any expander is now at ambient temperature alright. So, the approach one can easily approach to the higher pressure side of the expander and the lubrication at least at the higher pressure side can be taken care of and the operation of the expander are greatly simplified. So, lubrication problem and also the operation of the expander got very much simplified in a Heland cycle and this is what I say that the Claude system was modified from practical perspective in Capitza and Heland systems. The yield and work requirement of the system are given by following equations again actually they remain the same. The formula mathematical formula remain the same as for Claude and Capitza. Y value is given by this equation and W net by M dot value also is given by this equation and essentially this points out that the expression for Y and W net by M dot are essentially same for Claude system, Capitza system and Heland system, but the cycle has changed in terms of all the changes we have just talked about and these are the three important cycles for mostly used for air recruff action or for around 80 k operation. So, with this background of the Claude cycle, Capitza cycle and Heland cycle. Now, I will talk about the low temperature cycle and the most important cycle and I will take only one cycle which is nothing but called as Colline system or Colline cycle. How does this cycle look? Now, this looks more complicated and now as I said the basic objective of this cycle is to reach lower and lower temperature and the Colline cycle is a very important cycle which is used for helium refaction or for those gases which have got very low temperature boiling point. So, we can see how complicated it looks. It looks a very big cycle over here. What does it comprise of? And now I am talking about reduction of temperature from 300 Kelvin to let us say 4.2 Kelvin if I am talking about helium as the working fluid. So, as soon as you see that the delta T span is from 300 Kelvin to 4.2 Kelvin, essentially it would have more heat exchangers because we would like to have conservation of this cold. At the same time, because the inversion temperature of helium is 45 Kelvin, I have to somehow come down much below 45 Kelvin and therefore, I need expansion devices to produce more and more cold. So, one device may not be enough as it was in Claude cycle, but now I may need two devices or more than two devices so that the temperature before the JT expansion let us say at 0.7 here is much less than 45 Kelvin for helium as a working fluid. So, a Colline system basically comprises of heat exchangers and expanders. It was invented in the year 1946 by Samuel Collins at MIT USA. Samuel Collins is a very well known in cryogenic engineering and it is actually an award given on Collins name. Conferences that are held every alternate year Collins award is a very prestigious award for doing some applied research work in the field of cryogenic engineering. He was from MIT USA. This system is considered as one of the biggest milestones in cryogenic engineering and lot of variations or modifications happened over it over a period of time. This was the first system or first practical system that came into existence for helium liquefaction. So, this system basically as you can see is an extension of Claude cycle or modification of Claude cycle. The system has a compressor which is what you see over here, a JT expansion device which is what you see here. So, every system will have compressor JT expansion and heat exchangers. It also has a make up gas connection which is coming at this point. There are five two fluid heat exchangers we can see one two three four five and there are all two fluid heat exchanger and not three as we had seen in couple of systems studied earlier and also it has two turbo expanders. This could be turbo expanders, this could be reciprocating expander also. So, this could be turbo expander as well as reciprocating expander, but it has got two in numbers alright. So, what you can see from here is the gas gets compressed from 1 to 2, it gets pre cooled from 2 to 3 and some part of the gas amounting to M E 1 is taken off from the main circuit and is expanded from M E 1 to E 1 and join the return stream at this point. The remaining gas which is M minus M E 1 goes ahead and again at 0.5 M E 2 gas is taken off again it is expanded over here and it joins the return stream. So, these basically two expander produce cold and join the return stream. The remaining gas which is nothing but M minus M E 1 minus M E 2 comes at 0.7 gets expanded out of which M dot F goes out. The gas which comes at point G is going to be M minus M E 1 minus M E 2 minus M F alright. So, this gas is going to come over here and at appropriate points the gas after expansion M E 2 and M E 1 join and the return stream will ultimately have M minus M F at this point M F is a make up gas and the cycle continues and this is the way this cycle operates. Depending on the helium inlet pressure we can have 2 to 6 expansion devices. So, basically if the flow rates are very high you may have to worry about having number of expander more alright you may have to have more expander in that case. But normally the number of expander are even number because the vibrations have to get cancelled if you go for a reciprocating expander. The expansion engines are used to remove the heat from the gas and thereby to reach lower and lower temperatures alright. So, ultimately these are the two expander which basically produce cold and once the temperature at 0.6 and 0.7 goes much below 45 Kelvin the gas can get expanded producing liquid till that time there is no meaning to send in the gas at this point because the isenthalpic expansion would actually result in warming of the gas. The inversion temperature of helium is around 45 Kelvin and in order to have a yield the T 7 point has to be less than 7.5 Kelvin. It can be seen from a T S diagram that the temperature at 0.7 is around 7.5 Kelvin then only after isenthalpic expansion the 0.8 would lie in the dome. So, if you want to have liquid helium the temperature at 0.7 should be less than 0.75 that is let us say 7 Kelvin or 6.5 Kelvin or around those values then only when you expand the gas isenthalpically from 7 to 8 the 0.8 would lie in the dome and therefore, you can imagine that we have to first decrease the temperature of 0.7 at around 7.5 Kelvin below then only you will get liquid helium the helium is a working fluid depending upon the mass for its as I said 2 to 6 expander are used. So, 2 expander may not be enough sometimes it could be 4 expander or sometimes it could be 6 expander also. Now, this is the corresponding T S diagram for the Collins system and what you can see here 1 to 2 is a compression process 2 to 3 is a heat exchange process and at 0.3 you got a isentropic expansion in the expander number 1. The remaining gas comes over and at 5 again at this point you again got a second expansion from 5 to this. Also you have to ensure that this point E or point E 2 should not lie in a dome because in that case it will be weight expander and that has to be designed basically that is a completely new technology. So, we have to ensure that the isentropic expansion results in gas only alright and the reciprocating expansion devices would not like to have liquid there because of the compressibility effects coming into picture and therefore, you have to select the temperatures of 0.5 and 0.3 correctly also what is important is what is the value of x 1 and x 2 how much gas to be diverted through the first expansion device and how much gas should be diverted second expansion device. Also what is important as I said earlier the 0.7 has to be correctly the temperature at 0.7 has to be less than 7.5 then only the isentropic expansion would come in the dome. The temperature at 0.7 should be less than 7.5 Kelvin. If it is more than 7.5 Kelvin the isentropic expansion would come at somewhere this point and not in the dome and therefore, if you want to have liquid helium you have to see that the 0.7 is below 7.5 Kelvin. This can be seen clearly from the T s diagram of helium for a corresponding pressure of around 10 to 15 bar. The pressure at 2 will be around 10 to 15 bar. Now, let us see what are the expressions for y or the yield and the power input for the calling system. So, if we take this as the control volume and see the how much is going in and what is coming out. So, what is going in is basically m at 0.2.  And if we apply the first law and if you do the energy balance what you see m dot h 2 is equal to we 1 plus we 2 into m minus m f at h 1 plus m f h f. I think this is a very standard formula and we have derived it now many number of times. So, one should be able to put in this control volume and derive such expressions. If the work done by each of the expander is going to be w e 1 and this w e 1 is nothing but m dot e 1 into the enthalpy drop across the heat exchanger which is h 3 minus h e 1. And for the second expander w e 2 is nothing but m dot e 2 into delta h 2 which is nothing but h 5 into h e 2.     As you remember in Claude cycle we have seen that x plus y cannot be more than 1. Similarly here we have to see that when the gas returns the quantity of the gas at this should be finite alright.  So, one should be ensure that this particular boundary condition is satisfied in order to calculate y by this formula where x 1 is equal to m e 1 by m dot x 2 is equal to m e 2 by m dot. Simple terms what you can see the first term here basically nothing but the yield due to Lindemansson and additional terms which is coming because of x 1 and x 2 is a addition to y value because of this diverting this gases m e 1 and m e 2 through the expanders 1 and 2 respectively. This is basically give the additional yield or y value. So, this is the net equation what you can see for y for a given initial and final condition of p the yield y depends on h 3 that is temperature t 3 h 5 that is temperature t 5 and x 1 and x 2. So, you can see that if the initial and final values are fixed your h 1 and h 2 are immediately fixed assuming that your compression process at ambient temperature. So, what does your y depend on the y depends on the value of x 1 and x 2 correspondingly the value of t 3 which determine h 3 and t 5 which is h 5. So, basically t 3 t 5 x 1 and x 2 are the critical parameters for a given pressure conditions of 1 and 2 and this has to be optimized in order to get maximum y or minimum work done and this is a very important optimization exercise that has to be decided that has to be calculated that has to be solved in order to get optimum value of t 3 and t 5 and corresponding to this value of t 3 and t 5 when you should get optimized value of x 1 and x 2 this is a very important exercise. I am not going to go in the details of those we have done that for cloud cycle may be if you are interested you should go into the optimization of the cloud cycle. Like the cloud system the values of t 3 t 5 x 1 and x 2 have to be optimized to obtain maximum yield or a minimum work done for getting gas liquefaction as stated earlier using a control volume first and second order loss for a compressor what we get is w c is equal to m dot t 1 s 1 minus h 2 minus h 1 minus h 2. Similarly, the control volume for expansion also one gets w e 1 is equal to m dot e 1 into delta h 1 w e 2 into m dot is equal to m dot e 2 into delta h 2. So, net work done is basically going to be this minus this minus this because this is work done by the system. So, the net work done is given by minus w net upon m dot is equal to minus w c upon m dot which is work done on the system minus w 1 m on m dot minus w 2 upon m dot and these are nothing but the decrease in the net work done due to the work obtained in the expansion devices. So, if you put those values your w net by m dot that is net work done per unit mass of gas which is compressed is going to be t 1 into s 1 minus h 2 minus h 1 minus h 2 minus x 1 into delta h 1 minus x 2 into delta h 2 where x 1 is this and x 2 is this. The first term here again is basically the simple Lindy-Hampson work and the second term is the reduction in the work done due to the expansion devices 1 and 2 respectively. Now, we are here we are determining the tutorial is going to be solved to understand why w by m f and figure of merit for a colline cycle with helium as a working fluid for 1 atmosphere and 15 atmosphere. The expander flow rates are 0.6 and 0.2 in the expansion 1 and 2 respectively the expander inlet conditions are given here 60 Kelvin and 15 Kelvin from 15 atmosphere. So, these are my input conditions find the work per unit mass of gas liquefied and figure of merit. This is the colline cycle these are the different values for 1, 2 and 3 and 5 and then e 1 and e 2 are nothing but the low pressure things and e 1 and e 2 are basically located by drawing verticals from 3 and 5 on low pressure line. What you have to do is basically now apply the formula and this is what a T S diagram is 1 to 2 is a compression from 1 bar to 15 bar and 60 Kelvin temperature is at 0.3 and 15 Kelvin temperature is at 0.5 where the expansion happens this is x 1 and x 2 0.6 and 0.2 respectively. So, liquid yield given by this formula these are my enthalpy values put up those values over here what you get y is equal to 0.066 all right this is straight formula application. If I want to come calculate the work per unit mass of gas compressed is again applying the same formula get the enthalpy values and this is what W net by m dot what I want is W net by unit gas of mass liquefied. So, I divide this by y and what you get is 25230.3 if I want to calculate figure of merit it is going to be W net upon m dot f divided by the ideal work done which is just 6 at 37. So, you can see 25230 and 6 at 37 if 1 divides this by this the figure of merit happens to be 0.271 and this is what a simple tutorial of Collins cycle just to summarize what we have done the compression and expansion processes in an actual Claude cycle are reversible these cause inefficiencies and deteriorate the performance of the system. Capitza and Heland systems are the two modifications of the Claude system in a Capitza cycle the regenerator or the heat exchanger performs both gas cooling and the warming and gas purification. This is the first heat exchanger of the Claude cycle which works as a gas purifier also and actually it is a rotary regenerator or a rotary heat exchanger why is called regenerator because it regenerates it gets purified of its own because of the rotation of the heat exchanger also it was the first system to use turbo expander a rotary type expander instead of a reciprocating expander. So, lot of changes took place in a Capitza cycle. So, as to bring it to a commercial level Heland system is a high pressure system which is used for air liquefaction and normal pressure is around 200 atmosphere. In this system the inlet to the expander is ambient and hence the lubrication on the high pressure side and the operation of the expander is greatly simplified alright. So, Heland system also has got its relevance because the expander is working on the high pressure side at ambient temperature. The Collins system is an extension of the Claude system basically it is called as modified Claude cycle also and depending on the Helium inlet pressure 2 to 6 expansion devices are used. The Collins cycle is normally used to reach down lower in temperature of let us say around 4.2 Kelvin temperature and therefore, the number of heat exchangers increase the number of expander also increase depending on the flow rates which we are talking about. The yield and the work requirements are given by for Collins cycle as this formula. So, you can see that first term is basically simple indiamson and the yield increases amounting to whatever x 1 into delta H that is happening across the first expander plus x 2 into delta H that is happening around the second expander. So, effectively there is increase in the yield amounting to the gas which is diverted in x 1 and x 2, but as I said that you have to consider that the return stream also has to have a finite mass flow rate during calculations. You cannot go on increase in the value of x 1 and x 2 like that they have to be optimized with corresponding the value of T 3 and T 5. The network input also gets reduced because of the work done by the system given by x 1 to delta H minus x 2 into delta H. There is a reduction in the work done as compared that of Lindenhamson cycle. A self assessment exercise is given after this slide kindly assess yourself for this lecture and there are various questions asked over here and I would like you to go through self assessment, try to answer those questions and rate yourself across and see how much you have gathered from this lecture. There are around 12 questions and this is just a quick response for you basically to know where you stand as far as this lecture is considered. There are answers given over here. Thank you very much.