 Welcome to the fifteenth lecture of cryogenic engineering under the NPTEL program. In the earlier lectures on gas liquefaction, we have talked about various cycles and I will just go in brief about those cycles. We have seen an ideal thermodynamic cycle in which all the gas that is getting compressed is getting liquefied. So, whatever gas ideally in the ideal cycle, whatever gas gets compressed, it gets liquefied 100 percent. Then we came from this ideal cycle to a little practical cycle which was a Lindy-Hamson system. So, in a Lindy-Hamson system, a heat exchanger is added to the thermodynamic cycle and so whatever cooling effect is produced, the cold is conserved. So, a heat exchanger is used to conserve the cold and only a part of the gas that is compressed is liquefied. So, in first case all the 100 percent gas which was compressed got liquefied. Here only a part of the gas that was compressed got liquefied. Then we went for the second modification which is the pre cooled Lindy-Hamson system. It is an independent refrigerating system is used to pre cool the gas before it enters the heat exchanger. The mass ratio of this refrigerant cycle corresponding to the maximum yield is called limiting value meaning which that one cannot go on increasing the refrigerant flow in the auxiliary cycle because after a particular R value, addition of that R value there is no significant effect or there is no effect on the yield and therefore, this is limiting value of the R or the amount of refrigerant that is flowing through the auxiliary refrigerant system. We have seen this system in detail in the earlier lecture. Then we went for a Lindy-Dual pressure system wherein we used two compressors. The Lindy-Dual pressure system is a modification of simple Lindy-Hamson system in order to reduce the work requirement. In this system, the work requirement per a mass of gas liquefied decreases when the compression of the fluid is done in two stages and for different mass flow rates. I think this is very straightforward and this is what we did during the last lecture. So, we had two compressors over there and both the compressors had different flow rates as part of the gas which was compressed in the first was the second stream joins there and the total amount M is getting compressed only in the second compressor while the first compressor had less amount to compress. Now going ahead from there, we come to a different cycle now and the outline of this lecture which I am going to talk about is again continuing of the same topic gas liquefaction and refrigerant system. What I am going to talk today about is Claude cycle or sometimes called as cloudy cycle. I will call this Claude cycle. Now Claude was in 1920 Claude developed an air liquefaction system which is famously known as air liquefaction today a French company. And during my first lecture when I talk about the chronological event that in the in the field of cryogenic engineering Claude figured in that list also this is a very big company as of now which makes Helium liquefier nitrogen liquefiers. Under this Claude system I am going to talk about the expressions for liquid yield, the work requirement and also the parametric study which is a very important constituent of the Claude cycle. So introduction to the Claude cycle in order to achieve a better performance and to approach ideality the expansion process should be a reversible process till now what you have was a Jules-Thompson expansion which is a irreversible process. In the earlier lecture we have seen that a JT expansion or a Jules-Thompson expansion is an irreversible isenthalpic expansion and expansion using an expansion inging is an reversible isentropy process. We have talked about this earlier in one of the earlier lectures that expansion could be assured by an isenthalpic expansion or an isentropic expansion. In a domestic refrigerator for example what we have is a Jules-Thompson expansion achieved using a capillary tube till now whenever we have expanded the gas in the earlier cycle we have found that a JT expansion is used which is an isenthalpic expansion and it is a irreversible process and therefore we have to worry about the inversion temperature of this gas below which only when the JT expansion is carried out would result in cooling. Now in this Claude cycle we will talk about an expansion engine where you got a piston and cylinder kind of an arrangement where the work also is done by the system in an expansion engine and in this case when the expansion is carried out using such an engine the process is a reversible isentropic process ideally it will be reversible isentropic process and this would always result in cooling and one would not have to worry about the inversion temperature of the gas. So, for any gas an isentropic expansion results in lower temperature irrespective of its inversion temperature that is T i and V. So, this is clear now that isentropic expansion is always preferred and is always basically result in lowering of temperature which is not so in case of a JT expansion. So, this is a schematic of a Claude cycle and what you can see from here it has got a compressor over here and it has got an expander over here and we got a heat exchangers 1, 2 and 3 and there is the JT wall at the end where the liquid is yielded in a container at this point here. So, the schematic the Claude cycle is as shown here it consists of a compressor which is this and a 2 and a 3 2 fluid heat exchangers. So, there are 3 heat exchangers 1, 2 and 3 and there are 2 fluids as in earlier cases we have found that there were 3 fluid heat exchangers also and a JT expansion device and a make up gas connection. So, after getting the liquid at this point the gas would go back and whatever amount m dot f as you get this point in time point here the make up gas is added equal to m dot f at this point. The system has an expansion engine operating across the second heat exchanger as shown over here. So, what you can see from here is when the gas is getting compressed is goes to the heat exchanger and a part of the gas that is ME is taken off from here the gas is expanded to the expansion engine over here which produces cold and this gas the cold gas at the end of the expansion joins the return stream. The return stream is coming basically from the gas which has not got liquefied at this point. The whatever amount of gas is coming from 6 of which only m dot f gets liquefied and the return stream will have the remaining gas this gets joined to m dot e which is coming from expansion engine and the whole gas goes back and gets compressed with a make up gas connection at this point. So, in this system the energy content of the gas is removed by allowing it to do some work in an expansion engine. So, what you get at the end of the expansion over here is WE which is expansion work that is the work done by the expansion engine. So, here on the compressor we are doing the work while at this point here during expansion we get WE as the work done by the system. As shown in the figure a part of the main stream of gas is expanded from 3 to e. So, a part of the gas which is ME is taken from this m gas which is compressed and this expands from 0.3 to 0.e where it joins the return stream and this is the stream which is going to produce cold so that you go near the dome and then we have a JT expansion resulting in getting liquid at this point. This process of expansion is a reversible adiabatic expansion that is what we are talking about ideal expansion in this case. However, it will have its own efficiency and therefore, it will not be a clearly 100 percent isentropic expansion but as of now we will take it as 100 percent expansion and we will worry about the irreversibilities later. So, if I plot this cycle the Claude cycle on a temperature entropy diagram let us try to understand this the 1 to 2 process is a compression which is what you can see here 1 to 2 process is the isothermal compression process. In the heat exchanger number 1 this m dot is cooled by the return stream from 0.2 to 0.3 and you can see from 0.2 to 0.3 and this temperature matches with the return stream at this point which is 9 over here assuming that this heat exchanger is a 100 percent efficient heat exchanger. Now, at this point now from where the gas is expanded is a very important question in case of Claude cycle. So, why should I expand the gas from only 3 why not it should be at still lower temperature and things like that that has to be worried and the effect of this 0.3 we can understand in the subsequent slides. So, the gas is taken off from this main circuit at 0.3 and it is expanded isentropically from 3 to E and this you can see on the T S diagram that the gas is taken from high pressure line at 0.3 and it is expanded from 0.3 to 0.E. So, what you can see being an isentropic expansion you can have a vertical diagram vertical line over here on a T S diagram. So, one has to locate this point on a T S diagram when this isentropic expansion what you have to do is to draw a vertical line from 0.3 and see where it intersects on the low pressure return line and that is a point where the point E would lie because this is a completely 100 percent isentropic expansion. So, the gas expands from 0.3 and joins the return stream at 0.E where it gets mixed up with the return stream where it joins the return stream coming to the high pressure line again at 0.3 the gas which is going to be here at this point is M minus M E because M E has been diverted through the expansion engine. So, what is going ahead from after the 0.3 is M minus M E this gas further getting pre cooled from 3 to 4 and this is what you can see on the T S diagram it gets pre cooled from 3 to 4 and this 4 would match the outlet temperature of this heat exchanger which is 7. Being a 100 percent heat exchanger effectiveness we can see that this point matches over here. Now this gas will be getting further pre cooled from 4 to 5 this gas amounting to M minus M E is getting further pre cooled from 4 to 5 by the return stream which is going after liquefaction. So, what is the return stream will have the M minus M E is the gas which is getting expanded from 5 to 6 it will fall in a dome and corresponding to length of the 6 to G you will get liquefaction depending on the point where it falls. Here the M dot F is collected from the container and the return stream would therefore have M minus M E minus M F ok. So, out of the gas which is M is compressed M E is going to be expanded through the expansion engine and it joins here at point E what is going to come ahead beyond this point is M minus M E out of which M dot F is collected. So, what is remaining and what is going back in the return stream is M minus M F minus M E this is very important because we have to understand about this in subsequent slides ok. So, the gas which is going back is M minus M F minus M E which is at point G which is a gas over here at this point the gas comes out at 7. So, you can see from here it comes to 7 which is a warmer which is on a higher temperature side because it matches with the temperature 4. However, the cold gas which is expanded from 3 to E comes at a much lower temperature as you can see it will depend on what is the point number 3 where the point number 3 is ok. In a TS diagram where the point 3 is depending on that the point E would be decided and this stream which is coming at point 7 will combine with the M E flow rate and coming out at temperature at point E. So, as a result of which as you can see the heat exchanger output is actually at T 7 while the expansion engine output as T E combining both these the point at the entry of this second heat exchanger is coming at point 8. This is because of the cold which is going to be generated from during the expansion 3 to E. I hope this is clear depending on the expansion point E and depending on the point 7 which comes out out of this heat exchanger this two streams will get combined and they will getting combined by enthalpy rule that means they have to multiplied by their respective flow rates that is M minus M F minus M E into H 7 plus M E into T H enthalpy at this point E which would result in a some enthalpy which will corresponding to point number 8 the temperature corresponding to point number 8. Now, the gas enters at point 8 and it would come out at point 9 and this now gas will be M minus M F over here because M E has joined return stream the stream going to the second heat exchanger now will have M minus M F as the mass flow rate. So, this get this gas entering at 8 will get warm up to point 9 which matches with point 3 and further this gas point 9 enters through after the first heat exchanger comes to point 1 with the make up gas would join amounting to M dot F and the cycle continues. I hope this T S diagram which is a very important diagram to understand has been understood by you what is important to understand is what is that M E has been drawn from the main circuit at what T 3 temperature it has been drawn of and how much amount of M dot F or what is your return stream gas is left. What you can see is all these heat exchangers are basically having different flow rates for example, in this heat exchanger you got M dot flow rate over here while it is the return stream have M minus M F this has M minus M F and this has M minus M E well this has M minus M F and the third heat exchanger will have M minus M E over here while it will have M minus M F minus M E over here. So, these heat exchanger are kind of imbalance heat exchanger means that the flow rates in both the directions the hot stream and the cold stream flow rates are different. If you understood this diagram we can go ahead now in order to compute the expression for the yield. So, if we take this as a working volume. So, consider a control volume as shown in the figure and applying first law what we have is this we have done plenty number of times earlier for various cycles. So, I will not go into details of this calculations, but I will just show you the final expression which comes out in 1 or 2 steps. So, what is going in has to come out that is what your first law says. So, what is the entering at this point is M dot H 2 and what is coming out of this control volume is M dot E. So, M dot H 2 is equal to W E plus the return stream which is coming to come at this point which is M minus M F into H 1 plus M F H F which is going to come at this point. The expander work output that is W E is given by what is entering the expansion engine is M E into H 3 while what is leaving is M E into H E. So, W E is equal to M E H 3 minus M E H E. So, basically the flow rates are same it depends on the enthalpy difference between H 3 minus H E or T 3 minus T E on a T S diagram. So, if you put this value of M E over here what you get is M dot H 2 into M minus M F H 1 plus M F H F plus M E H 3 minus M E H E by reorganizing this or rearranging this what we get ultimately is this y which is nothing but M dot F upon M dot is equal to H 1 minus H 2 upon H 1 minus H F which is what we are familiar with this expression now plus x time H 3 minus H E upon H 1 minus H F and what is x is nothing but the mass flow fraction of the stream which gets expanded. So, M E M dot E divided by M dot. So, depending on how much M dot E how much mass fraction of the total mass flow rate has been taken off for expansion this value of x will be determined. So, basically the amount of gas which goes through this if the more gas goes through it more and more cold will get generated. However, if you take lot of gas from here as you can understand less gas is going to in front. So, M minus M E which is going front is basically some gas out of that stream is going to get liquefied when has to be very careful about how much gas is taken and at what temperature is taken and this is what we going to see in subsequent slides. So, where the expander mass flow rate is denoted by x over here. So, what you can see is H 1 minus H 2 upon H 1 minus H F is a very familiar expression for you and the Claude cycle brings this additional component of the next bracket because of which y gets increased. So, the first term is the yield of a simple Lindy-Hamson cycle which we have seen that it depends only on the point 1 and 2 the compression process 1 and 2. The second term which is an addition to the simple Lindy-Hamson cycle is basically coming because of the expansion engine over here. So, second term is a change in the yield occurring due to the expansion engine in the cycle. For a given initial and final conditions of P that means if the point 1 and 2 are fixed the yield y depends on H 3. So, if you see this expression the y value depends on 1 and 2 and 3 because as soon as your 3 is decided the point e is also decided immediately. So, 3 will be decided on the enthalpy 3 will depend on what is the corresponding temperature at which the expansion starts alright. So, y would depend if you are 1 and 2 points that means the pressures are fixed it will depend on the value of x and the temperature T 3 from where the expansion takes place alright. And if we say that if the T 3 also is held constant then the yield depends on only x. So, this is constant if your point 1 and 2 are fixed and on your pressure number 2 on your pH value your temperature is also fixed because of which your H 3 minus H e also gets fixed this becomes a linear function of x now because this is a constant this also becomes constant and the value of x however could be variable. So, in this case y will be a linear function of x when the temperature T 3 is fixed, but for a case is equal to x is equal to 1. Now, can I do this can I fix the value of x I can go on increase in the value of x I am sure I will get y increased. So, I can theoretically I can go on increase in the value of x, but take a limiting case that if I make x is equal to 1 that means whatever gas is coming from here is expanded that means no gas is going to go in front. In that case y should be equal to 0 and if you get look at this expression I will not get the value of when x is equal to 1 I will not get the value of y to be equal to 0 why because that particular case has not been taken into account over here in this expression alright. So, for x is equal to 1 what happens when x is equal to 1 the gas in the return stream which is m minus m f minus m e is equal to 0 actually actually the gas which comes over here also equal to 0 basically. So, if we want that the heat exchanger has to function that means there should be return stream you got a onwards stream and you should have a return stream also and in order that return stream is finite value your m minus m f minus m e should not be equal to 0 alright. So, it means that in order to have a finite yield that means in order to have a finite value of y at this point your m minus m f minus m e which is nothing but the return stream at point g has to be always more than 0 that means it should be some finite value. So, that the heat exchangers function so if we divide the whole thing by m what we get is 1 minus y minus x which is more than 0 and if I rearrange that what I should get is x plus y should always be less than 1 it means that in order that heat exchangers function in order that I should get some finite return stream I should see to it that x plus y is always less than 1. If your x plus y is less than 1 all these things possible and this expression of y is equal to this bracket plus x times this bracket is valid only when your x plus y is less than 1. When x plus y is equal to 1 that means there is no again there is no return stream basically x plus y is this thing whatever x has been directed and remaining y is equal to 1 then I will not get any return stream at this point. So, one should ensure that x plus y is less than 1 always less than 1. However, from this expression I can go on increasing the value of x and I can get y also has increased value. So, it has no meaning. So, therefore, as soon as this particular expression start giving x plus y as more than 1 we should not use this expression this expression is valid for x plus y less than 1 only. Therefore, the above equation is valid only when x plus y is less than 1 the yield y of the system increases with the increase in the value of x for a constant value of t 3 that is what we have seen that if we start increase the value of x from let us say 0.1 at a particular value of t 3 it will start increasing with the value of x in a linear fashion based on y calculated from the above expression when the sum is x plus y becomes more than 1 that is what I just talked about. When utilizing this equation when we reach a condition that x plus y is more than 1 we know that a limiting value of y must be calculated now and what we say as we say that x plus y has to be less than 1 let us compute a limiting value that is x plus y is equal to 0.99. In that case I will calculate my y when my x plus y happens to be more than 1 in this case I will calculate the y is equal to 0.99 minus x this could be 0.999 also minus x, but I am taking only 2 digits. So, y is equal to 0.99 minus x when we reach a condition that x plus y is more than 1. In summary how do I calculate y? y is calculated using the above equation until x plus y is less than 1 or 0.99. So, this expression is valid when x plus y is equal to 0.99 after which a limiting value of y can be calculated by y is equal to 0.99 minus x and this is what a general principle we are going to follow to calculate the yield choosing Claude cycle I hope this is clear to you. This value is the maximum y value because I am taking 0.99 this is the maximum y value that is possible, but the actual value could be less than this value the actual value could be less than this value, but this is a limiting value that means the y cannot exceed this value in any case it is clear that the work interaction of the system with the surrounding is due to. Now, let us calculate the work requirement for this system and where the work comes into picture is compressor and expander. Compressor we do the work on the system and expander we get work output that means work is done by the system. So, the net work requirement if the expander work is used in compression process ideally whatever work is done by the system it could be you know transferred to the compressor basically. So, that it decreases the net work which has to be done by the on the compressor. So, minus w net is equal to minus w c minus w e the negative minus w c is the work done on the system. So, work done on the system is negative and work done by the system is positive. So, it is a net difference between what is done work on the done on the system minus work done by the system. As we have stated earlier using the control volume first and second law for compressor this is a very standard equation which we have used almost in all the cycles that work done on the compression is equal to m dot into t 1 into delta s across 1 and 2 minus delta h across 1 and 2 very standard equation which has been used. Similarly, the control volume for expansion engine also gives us what is entering is m e at point 3 what is leaving is m e at point e at this point and also what we are getting at output as output is w e meaning which w e is equal to m e into h 3 minus h e we have talked about this earlier also. So, if I put them together I will get dividing by m dot to the expression I get w net upon m dot that means, net work done per unit mass of gas which is compressed is equal to t 1 into s 1 minus h 2 minus h 1 minus h 2 minus x into h 3 minus h e where x is equal to m dot e upon m dot. So, basically the first term again x is the expansion engine flow rate. So, what do you understand from this that this is the work done on the compressor the first term is the work requirement for simple Linde Hampson cycle where the only compressor comes into picture while the second work is the work done by the system and it is coming only because the expansion engine is in place in the cycle now alright. So, this is work done by the system amount to w e the second term is the reduction in the work requirement occurring due to the modification. But you may understand that the work output what you get at this point which is expansion engine is much smaller as compared to work done on the system. So, if they are talking about kilowatts over here what you get here would be 50 60 watts only over here. However, that is also we have considered that sometimes we may consider sometime we do not consider also, but theoretically we can consider that whatever net is available is work done on the system compressor minus work you get from expansion engine although it could be possible that this work output you get from the expansion engine is very very small as compared to the work done on the compressor. Based on the what we have learnt in the Claude cycle it will be very useful now we take a small tutorial and we do the parametric studies using this tutorial where we can understand the effect of values of the gas x the concept x or the gas amounting to m dot e which is diverted to the expansion engine. And also we will try to worry about what happens if the temperature T 3 from where the expansion happens what is the role of T 3 in this calculations. So, from that point of view this tutorial has been framed it reads as it has got a part A and part B the part A is determined W by M F that is work up compression per mass of gas liquefied for a Claude cycle with nitrogen as a working fluid. So, one can have air oxygen nitrogen etcetera with nitrogen as a working fluid in this case the system operates between 1 atmosphere and 40 atmosphere or 1.013 bar and 40.52 bar the expander inlet T 3 value is 225 Kelvin that means it is below ambient the expander flow ratio is varied between 0.1 and 0.9. That means you can see the value of x is varied between 0.1 to 0.9. And what we want to see is what is the effect of this variation of x on the value of W by M F now in this problem what we have done value of T 3 has been fixed at 225 Kelvin. In the part B we want to study repeat the above calculations for T 3 is equal to 300 k 275 k 250 k and 225 k. So, here in we can see that what is the role of this T 3 in the entire cycle if we reduce the value of from 300 275 250 225 etcetera how does it change the value of y and W by M F. So, what the problem says is plot the data that is y and W by M F versus x graphically and comment on the results. So, basically here we want to study the effect of the value of T 3 and the value of x which is varied from 300 point 225 k and 0.1 to 0.9 Kelvin respectively. And we want to study how do these parameters affect y and W by M F versus x when we want graphically plot we can the nature would be clear. So, as we go in the every circuits every cycle we see that what is the problem the problem is we got a cloud cycle with a cloud system. The working pressure is 1 atmosphere to 40 atmosphere here the working fluid is nitrogen the value of T 3 variations what we want in total is from 300 Kelvin to 225 Kelvin. So, 300 275 250 and 225 that is in steps of 25 Kelvin the mass flow fraction that is x is varied from 0.1 to 0.9 as I said earlier x cannot be equal to 1 that means all the gas which comes cannot be diverted through the expansion engine alright. So, it has to be less than 1 for the above system what is asked from you is to calculate the work per unit mass of gas liquefied. And for which you have to do the calculations of y one has to calculate y first and this W by M divided by y will give you W by M F. And this study is as you can see is we are going to have 1 2 3 and 4 for 0.3 at 375 250 and 225 Kelvin and this is what the overall problem is. So, in the part a the expander inlet condition under study is 225 Kelvin and at 40.52 bar the expander mass flow ratio varies between 0.1 and 0.9 what we are going to do now I am not going to solve all the problems all the for all the values of x from 0.1 to 0.9 what we are going to do in detail is to understand in this tutorial how y and W M F are calculated for x is equal to 0.2 for example, and T 3 is equal to 225 Kelvin for example, as an inlet condition what we expect you to do is now completed table we will give you a complete table and what we expect you that you will solve it out yourself. So, that you can see that you have understood the problem. So, now let us see the Claude cycle and here is the Claude cycle having a compressor and expansion engine and what we have is a 0.1234 E and F. So, one can see that 0.1 is over here let us try to look at the points 0.1 is at 1 bar 300 Kelvin corresponding enthalpy entropies are given at 0.1 0.2 is the after the compression and again at this 0.2 we got the pressure temperature conditions given over here with the corresponding enthalpy entropy values. The 0.3 happens to be what is given in the problem which is what we have taken as 225 Kelvin and the pressure is same as what it is at 0.2 corresponding enthalpy are again calculated out here. As soon as the 0.3 is fixed on a T s diagram at 40 and 225 Kelvin we get the point E which is over here and the temperature at this point E happens to be 80 Kelvin. So, what we do basically we have got a T s diagram we draw a vertical from 225 Kelvin on 40 bar line vertically down where it intersects the temperature on 1 bar isobaric line and wherever it intersects the corresponding temperature happens to be 80 Kelvin. So, the point E is located on p is equal to 1 bar isobar line by drawing a vertical line from 0.3 and this has to be graphically or it from a T s diagram it has to be taken and the pressure at this point is going to be low pressure line because this is a return string. The point F happens to be the boiling point of nitrogen which is 77 Kelvin. So, here F is where from you collect the liquid from where you get the yield value. So, different properties in terms of entropy and enthalpy this is the most important part if you commit any mistake on this then the whole problem is going to go wrong. So, one always should have a habit of locating this point 1 2 3 E and F and correspondingly one should correctly write the value of enthalpy and entropy this is a very common mistake students commit because they read the T s diagram very fast the current points are not located and sometimes what you get is your yield also happens to be negative and thing like that. So, the enthalpy entropy value have to be correctly taken from the T s diagram. One can see the corresponding T s diagram here 1 2 is a compression 2 2 3 is a cooling 3 2 E is where you get the point E that is what I just said that we draw a vertical from high pressure line at 0.3 to the low pressure line which intersect this point at E and corresponding to this what you get is 80 Kelvin in this step in this particular case temperature at 0.3 is 225 Kelvin. So, the expander inlet condition and its mass flow rate at R 225 Kelvin and how much gas is going in this particular problem is we are saying x is equal to 0.2. So, 0.2 times m dot is the gas which is sent to this expansion engine. So, fraction which is going through the expansion engine is only 0.2 that means what is coming out at this point is 0.8, 0.8 fraction of the total mass flow rate over here. So, let us start doing simple calculations and calculate the yield by the formula y is equal to h 1 minus h 2 upon h 1 minus h f plus x time h 3 minus h e upon h 1 minus h f where x is equal to 0.2 in this case T 3 is equal to 225 Kelvin and 40 atmosphere pressure. So, taking the respective points from this table from where enthalpy entropy have to be read let us calculate the value of y is equal to h 1 minus h 2 which is 462 minus 453 upon h 1 minus h f 462 minus 29 plus x time that is 0.2 times h 3 minus h e which is nothing but the difference between these two values divided by h 1 minus h f and this will give you 0.086. So, y that is m dot f upon m dot is only 0.086. So, this much fraction of gas gets liquefied when 0.2 fraction is diverted through the expansion engine and this is what we have calculated. Now, let us calculate even w by m f by the same technique. So, if I want to calculate work per unit mass of gas compressed the formula is w c by m dot which is nothing but T 1 into h 1 minus h 2 minus h 1 minus h 2 minus x times h 3 minus h e again taking the different entropy values and enthalpy values as given over here. This is the calculation you do and what you get at the end is 299 joule per grams. So, this is the net work done on the system sorry this is the work done on the compressor not net work done it is the work done on the compressor and work done per unit mass of gas liquefied is w c by m dot is equal to 299 y is equal to points 86.086 and w c by m dot f is 299 upon 0.086 which is 3476.7 joule per gram all right. So, what you get from here is the work done on compression per unit mass of gas which is liquefied we have not taken the value of w e over here we have not taken the value because that value happens to be a very small quantity extending this calculation for all the values of x and tabulating the result. So, what we have done I have just calculated the value of y and w by m f for x is equal to 0.2 for the value of t 3 to be when it is equal to 225 Kelvin. Now, what I will do I will give a sweep to the value of x from 0.1 to 0.9 remember all the discussion what we had earlier regarding where do we use this expression and after limiting value what expression do we use to calculate the value of y. So, extending this completely I am not showing all the details what I show is a table which gives when t 3 is equal to 225 Kelvin when you vary the value of x from 0.1 to 0.9 the corresponding y values are given over here and corresponding w by m f values are also given over here. So, what you can see here as we had seen that as the value of x 1 is increasing as the value of x is increasing y increases linearly up to the point when x plus y is limiting value in this case what we have done it has to be less than 1 or in this case is 0.99. So, you can see that when the x is increased from 0.1 to 0.73 at x is equal to 0.73 y goes on increasing and meets a limiting value of 0.26 where after that if x goes on increasing I cannot use the earlier expression because x plus y happens to be more than 1. In that case we have used the limiting value of y which is 0.99 we have did the limiting value of x plus y is equal to 0.99 and therefore, y has been calculated as 0.99 minus x. So, you can see the sum here happens to be 0.99 all the time this is a limiting value of y that means actual y could be less than this y, but definitely not more than this. Having calculated y in this fashion we have calculated w by m f also and that also uses this y values and what you can see the results over here. What does the result show you? The result show you that the as the x increases the y value increases and goes through a maximum of 0.26 and after that it meets a limiting value and it starts coming down. So, the value has start coming down it has reached maximum. Similarly, what you can see on w by m f that as the value of x increases the w by m f values or the work done per unit mass of gas liquefied decreases and that is basically where the principle of using expansion engine happens that expansion engine the amount of gas which is transferred or which goes to expansion engine should be such that the w by m f has to be minimum and y has to be maximum for that particular temperature T 3. So, w by m f goes on getting the value goes on reducing and it meets a minimum value at which y happens to be maximum because w by m f nothing, but w by m upon y. So, wherever y get minimized the value of w by m f will be maximum over there. So, the value of w by m f reaches the minimum value over and again it start increasing. So, this y goes through a maximum and w by m f goes through a minimum that is what we understand from this. So, if I plot these things that is y versus x what we have just calculated and I just take four representative values or five representative values over here and if we calculate case number 1, 2, 3, 4 and 5 and what if I join what you say is this is a x when the x is varied from 0.1 to 0.9 y goes through a maxima and the maximum value happens to be around 0.25 or 0.26 for T 3 is equal to 225 Kelvin temperature. If I do this from the plot it is clear that y crosses a maxima when the value of x is increased from 0.1 to 0.9 and the optimum value of x at which the y is maximum happens to be around 0.73. Beyond this maxima with up to this point we had used the expression which was given as y is equal to h 1 minus h 2 etcetera what we have calculated earlier. Beyond this value what we have used y is equal to 0.99 minus x it is a limiting case of y over here. Similarly, if I compute the value of w by mf versus x I see that the value of w by mf as I said earlier it starts coming down it hits the minimum value and then it starts going up and if you plot this 1, 2, 3, 4, 5 cases over here what you see is for T 3 is equal to 225 Kelvin I get minimum at this point again around 0.73 and this condition is for T 3 is equal to 225 Kelvin temperature. The trend shows that w by mf crosses a minimum with an increasing value of x. Beyond this minima the w by mf is estimated based on the limiting value of y it is clear. Now, all the calculations are pertaining to the part b which is what we have done till now is part a and it part b now what we are going to change is the value of T 3 and we will give a sweep to the value of T 3 also. That means, we will go for T 3 is equal to 300, 275, 250, 225 etcetera and for every value of T 3 we will have a sweep of even x value from 0.1 to 0.9. It means there is a lot of calculations involved and what we have done is written a small program and from where I will now show the table the results. The results are calculated and shown here. So, what you can see now here is a the value of T 3 is equal to 300 Kelvin, 275 Kelvin, 250 Kelvin. I have already done the value calculation for 225 as T 3 value. So, this is the value of for T 3 is equal to 300 Kelvin this is actually room temperature basically the expansion happens at room temperature only. If you see from here I am going to see from 0.1 to 0.9 and correspondingly the value of y are calculated and here you can see that the maximum value of y happens to be for x is equal to 0.67 and the maximum value of y is 0.32. If I reduce the value of T 3 to 275 Kelvin the maximum value of y happens to be 0.3 which is little less than this and the corresponding x happens to be 0.69 that means little higher than this. If I come further down from here T 3 is equal to 250 Kelvin the maximum value of y is 0.27 which is less than this and the corresponding x value is now 0.72 and if you remember the 225 Kelvin it was 0.73 and around 0.26 for T 3 is equal to 225 Kelvin that means as you go on reducing the temperature over here the y value goes on optimum y value or the maximum y value goes on decreasing and correspondingly the amount of gas which can get transferred to the expansion engine has shown an increase over here 0.67, 0.69, 0.72 and 0.73 for expansion engine when T 3 is equal to 225 Kelvin. However please understand that all these calculations have been done for heat exchanger effectiveness 100 percent and expander efficiency 100 percent. As soon as the inefficiency parameter starts coming in these values can change alright the optimum can shift accordingly. So, all the calculations pertaining to part B are again shown for W by M F here. So, again the same thing is done for T 3 is equal to 300 Kelvin 275 Kelvin 250 Kelvin and again what you see is a minimum is at 0.67, 0.69, 0.72 while W by M F shows an increase value when the T 3 value is decreased. This is very difficult to follow a table if I plot this graphically it will make more sense and therefore all these calculations are now are shown graphically where some conclusions can be directly drawn. So, if I draw this liquid yield y versus x what you can see is y versus x for different T 3 values and this is for 225 Kelvin this second curve is for T 3 is equal to 250 Kelvin the third curve is for 275 Kelvin now and the fourth Kelvin what you can see is for 300 Kelvin. So, what you can see is from here is as the value of T 3 is increased that means 225 to 275 to 300 Kelvin my y or the yield has increased. So, theoretically I can understand that I can carry out the expansion from as high temperature as possible which is 300 Kelvin in this is going to be beneficial. As mentioned earlier y crosses a maxima with increase in the value of x for each T 3. So, every T 3 what you see it has gone through a maximum and what you can see from here the position of this maxima shift to the right when decrease in temperature value of T 3. That means if I come down from 300 to 225 Kelvin the optimum x value has changed it is a marginal change however from 0.67 to 0.73 or so that means, but because the temperature has decreased down that means when the temperature decreases down I can send more gas through the expansion engine. I can divert more gas through the expansion engine to get an optimum. The optimum is optimum requires more gas to be diverted through the expansion engine why this occurs because the expander work h 3 minus h e decreases with the decrease in temperature T 3. So, I go down the temperature from in this digression in this dimension what you can see is the y in the formula is a function of x into h 3 minus h e. If you remember the formula y is equal to h 1 minus h 2 upon h 1 minus h f plus x time h 3 minus h e upon h 1 minus h f. So, the product of x into h 3 minus h e I would like to maximum I would like to get maximized because I want to see the optimum associated with those values. So, as soon as I go on lowering the temperature the value of h 3 minus h e is going to get less and less depending on a T s diagram for a particular gas. So, because h 3 minus h e gets reduced I can divert more and more gas so that I get a optimum value alright and that is the reason that I can divert more and more gas when I come down lower in temperature. Also at the lower values of T 3 more amount of gas can be diverted to the expansion engine that is what I just talked about this is because the product x into h 3 minus h e is to be maximized I hope I am clear on this. However, the T 3 is limited by the position of the point E one cannot really come down below a particular temperature because the isentropic expansion from that point 3 might fall in the point might fall in the dome which is undesirable because expansion engine would not like to see the liquid in the expansion engine it wants to be in gaseous stage and therefore, the how much to be lowered the temperature how much T 3 how much it has to be lowered is limited by the value of isentropic expansion from point T 3. If I do the similar thing now to understand W by M f versus x this is the curve for T is equal to 225 Kelvin and if I draw for 250 Kelvin followed by 275 Kelvin and followed by 300 Kelvin exactly in the similar line what I talked about at 300 Kelvin because I get maximum y value there I get minimum W by M f over here while if I lower the temperature my W by M f requirement increases as stated earlier the W by M f crosses a minima with the increase in the value of x for each of the T 3 values also the position of this minima shifts to the right. So, my optimum value shift to the right at with the decrease in the value of T 3. So, as I decrease the value of T 3 the optimum value shift to the right because my optimum y happens to be for those respect to x values this can be well understood from what we have studied for the y values y versus x behavior is absolutely similar to what we are seeing in this particular case. The minima shift to the right because expander work h 3 minus h t decreases with the decrease in the value of T 3 also at the lower values of T 3 more amount of the gas can be diverted so that the product x into h 3 minus h t is maximized this is what we talked about earlier for the y versus x case and this is what has to be well understood that why it goes to a minimum and why it shift to right in this particular case with this if I summarize whatever we have learned today the J T expansion is an irreversible isenthalpic expansion and an expansion by an expansion engine is an reversible isentropic process is clear in a Claude system the energy content of the gas is removed by allowing it to undergo an isentropic expansion alright. So, the energy is basically because the work done by the system due to the isentropic expansion the yield and the work requirement of the systems are y is equal to this formula which I am sure now you are conversant with and W net by m dot is equal to this formula of which this is the work done on the compressor this is work done by the expansion engine. If T 1, T 2 and T 3 of the system are held constant the yield y of the system is a linear function of the expander mass flow ratio x again this is clear from the yield expression for y the equation for y is valid only when x plus y is less than 1 beyond a certain value of x a limiting value of y is estimated as y is less than 1 minus x I am sure this clear why are we doing this because we want the return stream to be having a finite mass flow rate this m minus m e minus m f has to be more than 0 for a given temperature of T 3 the yield y crosses a maxima with a increase in the value of x. So, if you give a sweep to x the value of y goes through a maximum and this is what the value of y we want to attain also the maximum shift to the right with the decrease in the value of T 3 we have talked about this for a given value of T 3 the W by m f of the system goes through a minima with an increase in the value of x and this is again the optimum value of x we would like to attain or use for design purpose for cycle optimization also the position of this minima shift to the left with the decrease in the value of T 3 with this background and with this tutorial an assignment is given to you on the similar line need nitrogen as a working fluid, but we have changed the pressure to 50 atmosphere and again the T 3 has to be a 250 Kelvin expander flow rate to be varied from 0.1 to 0.9. And again I would like you to repeat the cases for temperature because what happens to the optimum we have done a case for 40 bar or 40 atmosphere pressure as soon as you go to 50 bar what happens to optimum x value of what happens to optimum y value and what happens to the optimum W by m f value has to be well understood alright. So, this is these are the answers for your questions and I am sure you will take care of y versus x and try to tally your answers. Similarly, W by m f upon x variation also has been given for different T 3 values please carry out this exercise and try to see that your results are matching with these results. Thank you very much.