 So, welcome to the 10th lecture of cryogenic engineering under the NPTEL program. Just to take a overview of what we learnt in the last lecture, we found that for the liquefaction for real gas, JT coefficient that is del T by del P at constant enthalpy depends on inversion temperature. So, we learnt the concept of inversion temperature which is nothing but del T upon del P at constant enthalpy and we found that in order to get JT cooling effect for a real gas the temperature or the initial state of the gas should be less than T inversion temperature. Then we went away from isentropic process to isentropic process using reciprocating expansion engine or turbo expanders and we understood that the isentropic expansion of a gas always results in cooling irrespective of its initial state. So, as it was in a JT expansion which depends on the initial state, the isentropic expansion does not depend on the initial state. The JT expansion is normally used where phase changes are required, this is a very important thing during JT expansion normally there is a phase change while in an isentropic expansion it is used for a single phase fluid. So, the inlet is gas, the outlet also should be gas while it is not true with JT expansion in JT expansion the inlet could be gas or a two phase mixture and outlet will be liquid or a two phase mixture. The isentropic expansion coefficient is nothing but delta T by delta P at constant entropy. So, delta T by delta P at constant S this was delta T by delta P at constant H for JT expansion. We also studied that the gases like air nitrogen they show JT cooling when expanded at room temperature while other gases like helium hydrogen neon are required to be pre cooled to result in JT cooling. This is owing to the fact that the inversion temperature of helium hydrogen and neon is below ambient temperature and therefore, if you want to have JT cooling for these gases the initial state of the gas should be brought down below its inversion temperature and therefore, they need to be pre cooled and then only they can result in JT cooling. We studied thermodynamic ideal system for gas liquefaction where in a thermodynamic ideal system all the gas that is compressed gets liquefied. Using the ideal thermodynamic cycle one can calculate ideal work requirement for liquefaction of a unit mass of a given gas. So, ideal thermodynamic cycle basically teaches or basically gives you the ideal work requirement to liquefy a unit mass of a given gas. This ideal work of requirement depends on the initial condition of the gas which is nothing but initial pressure and temperature condition for the given gas right. Using this whatever we have learnt in the earlier lecture the outline of today's lecture again holds around the topic gas liquefaction and refrigeration systems and what we study under these today is the parameters of gas liquefaction systems and then we will come to the first practical system which is Linde-Hampson system where we will talk about liquid yield, work requirement and optimization of liquid yield all these are calculation based on family based on their derivation we will try to understand how Linde-Hampson system behaves. This is the ideal system as seen earlier the schematic of an ideal system and its T s diagram as given here. This is the ideal system wherein you got a gas input, you got a compressor, you got a expander and corresponding T s diagram is given here where 1 to 2 is a compression process and 2 to f is a isentropic expansion process. This is the ideal thermodynamic cycle the process of compression and expansions are from 1 to 2 and 2 to f respectively. So, 1 to 2 is a compression 2 to f is a expansion process. The initial condition of gas at point 1, point 1 is initial condition it could be ambient 300 Kelvin and 1 bar that could be standard operating condition as a initial condition and this determines the position of f as you can see that the point 1 determines the location of point f because this is a 1 bar line and this is corresponding to the boiling point of the gas at 1 bar for nitrogen it will be 77 Kelvin. So, as soon as your point 1 get determined your point f is always known and therefore, we say the initial condition 1 of the gas determines the position of point f the point 2 determines the final state of the gas after the compression process. So, point 2 comes over here which is nothing but a point which is perpendicular to this line. So, this is isentropic expansion entropy remains constant. So, if you extend this line further up and if you draw a horizontal line temperature remaining constant the intersection of this 2 lines will be nothing but the point 2 which is the final state after compression process 1 to 2. Let us take an example of nitrogen and initial condition at point 1 which is 1 bar and 300 Kelvin. We can find out now as soon as your 1 bar and 300 Kelvin is determined we know the point f and if we draw a vertical here and if I draw a horizontal at this point the intersection at this point will give final condition after the compression process 1 to 2 and here we can understand the required pressure at point 2 to fall an ideal cycle is more than 70,000 bar. This is a very very high pressure absolutely impractical to attain in practice. So, what we want to conclude from here is such high pressures are impractical and hence there is a need to modify the system to lower the maximum pressure. What we understand from here is if you want to have ideal thermodynamic cycle for gas liquefaction you have to liquefy all the gas that is getting compressed and if you were to do this the point 2 lies at a very high pressure and it is as high as 70,000 bar which is absolutely impractical to attain. And therefore, what we say is ideal thermodynamic cycle for liquefaction is impractical and hence we have to modify this system so that we lower the maximum pressures which are normally attainable using the compressors available. In the first lecture I had talked about devices like heat exchangers, Joule-Thompson wall, turbo expanders all these systems could be used to modify the system. And now that we talked about changing the ideal thermodynamic cycle to get a relatively practical cycle we will have to use all these components. It is a two way heat exchanger one gas coming in and one gas coming in from this direction. These are the turbo expanders and this is JT wall. The heat exchangers are used to conserve cold and JT devices are used to achieve lower temperatures. This is all you know now. The following slides the lectures after these will explain various cycles that are used for gas liquefaction. So now we will come to the important cycles before we go into various liquefaction cycle let us find out what are different gas liquefaction parameters. So we have got different cycles and if I were to compare these different cycles I will have to compare different parameters. What are these parameters? In the refrigeration system for example the Carnot COP is often used as a benchmark to compare the performances alright. On the similar line there is a need to compare different liquefaction systems. So there are different cycles and therefore we will have some parameters so that we compare the performance of cycle A to cycle B to cycle C. In liquefaction systems an ideal cycle is used as a benchmark to compare the performances. So one way of comparison is to compare its cycles performance with an ideal cycle. Different ratios and functions are defined to give a qualitative and quantitative information of different liquefaction cycles. So what are these different ratios and functions? So if you take a typical example of this cycle and you got a compressor and expander and the gas m dot is compressed at this point out of which m dot f gets liquefied and therefore the gas which returns to the cycle is m minus mf dot and this gas goes in here and a replenishment is done at this point which is m dot f. So whatever mass goes out it is replenished over here and the cycle continues. So from here m dot gets compressed and m dot f gets liquefied. So what are different parameters? Now at the same time what you can see here the power input to the compressor is wc and the heat rejected to the atmosphere is qr or the heat of compression. So first parameter is work required per unit mass of gas compressed. So work required is wc and the gas which is compressed is m1 dot. So minus wc upon m1 dot is the work per unit mass of gas compressed. Work per unit mass of gas liquefied so the gas liquefied is m dot f and therefore the work required is minus w upon m dot f. The minus sign is given for work done on the system which is in this case the work done on the compressor and therefore we are calculating work done per unit mass of gas liquefied work done per unit mass of gas compressed. The compressor isothermal efficiency we are assuming that the compressor is isothermal however in actual case the compressor may not show isothermal compression and therefore we have a efficiency which is called compressor isothermal efficiency. Similarly we have got compressor mechanical efficiency which amounts to all the irreversibility or the frictional losses in a system and it is called as eta c mechanical for efficiency mechanical. Then we define one more parameter called figure of merit or FOM and this is ratio of ideal work input based on the ideal thermodynamic cycle divided by actual work input to the system. So naturally the ideal work input is going to be less than the actual work input to the system. So this ratio is going to be less than 1 and this is called as figure of merit of any cycle and fraction of total gas liquefied. So you can see here y represents what is the ratio of m dot f divided by m dot. When the m dot gas is getting compressed the amount of gas which is liquefied is only m dot f. So m dot f upon m dot is nothing but fraction of the total gas liquefied. With these parameters now let us see some fundamentals which would help us to analyze this liquefaction cycle. The sign convention I just talked about that the work done by the system is taken as positive. Similarly the work done on the system will be taken as negative. The heat transferred to the system is taken as positive. The heat leaving the system will be taken as negative and in these all these cases the pressures we are using bar or Pascal. The conversion table is 1 Pascal is equal to 1 Newton upon meter square, 1 bar is equal to 10 to the power 5 Pascal's and 1 atmosphere is nothing but 1.013 bar. In actual cases I will assume that 1 atmosphere is equal to 1 bar as you can see from this approximate expression. Now with this background let us come to Lindy-Hamson cycle which is the first sort of practical cycle taking us away from the ideal thermodynamic cycle. So this is the cycle this is the schematic of the Lindy-Hamson cycle and what it has you can see over here. It has got a compressor, it has got a heat exchanger, it has got a JT valve, it has got a container, the gas goes back and the gas is replenished. So in this sense if you see the thermodynamically ideal cycle to this cycle what we have added is a heat exchanger. In the ideal cycle all the gas which was getting compressed had got liquefied while here no that is not happening only some of the gases is getting liquefied while the remaining gas which is m minus m f dot is coming back to the compressor and m dot f gas is replenished or make up gas is added here. With this background let us see what is the salient features of this system. Lindy-Hamson cycle consists of a compressor, heat exchanger and a JT expansion valve this is what we have seen only a part of the gas that is compressed gets liquefied only m dot f is getting liquefied out of m dot which is compressed. Being an open cycle the mass deficit occurring is replenished by a make up gas connection we have just seen that. All the processes are assumed to be ideal in nature and there are no irreversible pressure drops in the system. This is the assumption which will be used for analyzing this system. We assume that the compression process is isothermal while the JT expansion is isenthalpic and also we assume that the the system incorporates a two fluid heat exchanger which is assumed to be 100 percent effective alright. So it is a perfectly ideal system however the heat exchanger here also is assumed to be 100 percent effective. The heat exchange process is an isobaric process and it is used to conserve cold in the system that means when both the fluids are travelling through the heat exchanger there is no pressure drop across the heat exchanger. That is the stream 2 to 3 2 to 3 and G to 1 this is what the points the direction of the fluids flowing through here the stream of the gas 2 to 3 is cooled by the stream of the gas going back. The JT expansion device is used for phase change of gas stream to liquid stream by lowering the temperature. So once you get gas at this point what you get at this point is a two phase or the liquid gets collected and the gas goes back at this point. Now this is the most important thing and what you have to learn from here is how to draw a corresponding temperature entropy diagram or a TS diagram from a given cycle. So what you can see from here is 1 to 2 process is a isothermal compression 2 to 3 is a isobaric heat exchanger process right. The pressure remains constant temperature gets lowered because of the return gas stream which is from G to 1. So G to 1 because of this low temperature gas which is going back to the compressor the temperatures gets down from 2 to 3. At point 3 the gas gets expanded from 3 to 4 and as soon as the expansion happens state of the gas lies in the dome and this is what where we say that now the state of the gas at this point 4 is two phase gas that means you got some liquid and some gas. The liquid is collected at point F of which some of the liquid is drawn out or the remaining gas will go back and this will pre cool the incoming gas from 2 to 3 alright. So at point G the gas would go back from G to 1 pre cooling the stream from 2 to 3. So these two process 2 to 3 and G to 1 are basically heat exchange process in the heat exchanger while process of 1 to 2 is a compression process and 3 to 4 is an expansion process alright. And 4 to G is a place where we get vapor at this point or gas at this point and the liquid is connected depending on where the point 4 is. Depending on where the point 4 is amount of liquid will be decided amount of gas which has got converted to liquid will be decided and this is what we will calculate now. This should be very very clear to all of you how to draw the T S diagram for various cycles. Now having done this let us have a control volume over here in this region. So we want to basically now analyze this cycle in order to understand what is the work input to the system or what is the fraction of the gas that has got into liquid region as a fraction of m dot how much m f has come into in a liquid form. So consider a control volume for this system as shown in the figure it encloses the heat exchangers the control volume is basically enclosing heat exchanger J T wall and liquid container as you can see from this. And let us apply the first law of thermodynamics to this system the changes in the velocities and the datum levels are assumed to be negligible as we had done during the last lecture also for ideal cycle. The quantities entering and leaving the control volume are as given below. What is entering is only the gas which is coming at m dot 2 right mass of gas which is coming at point 2 while what is leaving the system is m minus m f dot at this point at the same time what is leaving the system is m dot f here ok. So using first law now energy is equal to energy out and therefore what you get is m dot at this point multiplied by enthalpy at this point. So m dot h 2 is equal to m dot minus m f dot and corresponding to this point this point is nothing but point 1. So m dot minus m f dot into h 1 which is enthalpy at point 1 plus whatever is leaving this control volume is m f dot into h f alright with this background if I get a ratio which is m dot f upon m dot is equal to h 1 minus h 2 divided by h 1 minus h f. So what does h 1 minus h 2 gives you enthalpy difference across the compressor divided by h 1 minus h f here. The fraction of gas liquefied or liquid yield is defined as m dot f divided by m dot is equal to y this is nothing but y is equal to h 1 minus h 2 divided by h 1 minus h f. What does it mean? Y depends on the initial condition and the compression pressure. So h 1 h 1 and h f depend on the initial condition while the point h 2 depends on the compressor pressure. So here we can see that the values h 1 and h f are governed by the initial conditions which are often ambient conditions. So h 1 is corresponding to 1 bar and 300 k corresponding to 1 bar h f dot will be at 77 Kelvin which is the boiling point of liquid nitrogen at 1 bar pressure. Now in order to maximize y the value of h 2 should be as small as possible or h 1 minus h 2 should be as high as possible. So if you want to get maximum yield or m dot f upon m dot the numerator should be as high as possible while denominator is fixed because h 1 minus h f is immediately fixed once your initial conditions are fixed. In order to maximize y h 1 minus h 2 should be as high as possible or h 2 should be as small as possible. So h 1 is h 2 enthalpy at a point after the compressor. So it depends on what is the pressure of the gas after the compression and corresponding enthalpy at that particular point. So to have minimum h 2 the change in enthalpy for even change in pressure should be 0 at temperature T 1. h 1 minus h 2 is a compression process or process 1 to 2 is a compression process it is happening at constant temperature. So it is a isothermal compression process. Now in order that the value of h 2 should be minimum the change in enthalpy or del h 2 by del p 2 change in enthalpy for a given change in pressure should be 0 at a given temperature T 1 this can be better understood later. So what I want is del h 2 by del p 2 if I go on changing the pressure p 2 at certain point or at certain pressure the value of h 2 is going to be minimum it also has to take into consideration that during this process the temperature remains constant. So mathematically I will say that del h by del p at temperature T 1 is equal to T 2 at point 1 is equal to point 2 this should be equal to 0. So that I get minimum h 2 value here or I get maximum h 1 minus h 2 in that case. Using calculus h p and T we have seen earlier that del h by del p at constant temperature into del p by del t at constant enthalpy into del t by del h at constant pressure is equal to minus 1. So what is this del t by del p is nothing but 1 upon mu j t and del t by del h is nothing but 1 upon c p if I take this points on this side what is this point del h by del p is equal to 0 it means that mu j t into c p is equal to 0. If I say mu j t into c p is equal to 0 quantity c p cannot be 0 because c p is always a finite quantity and therefore what you learn from this is mu j t is equal to 0. So point 2 is in such a way that mu j t should be equal to 0 in that case for that particular temperature it implies that in order to maximize y the state 2 should lie on inversion curve as you remember that on inversion curve mu j t is equal to 0. So point 2 should lie in such a way that the point 2 lies on inversion curve for a particular gas at a temperature of compressions the temperature remains constant and if you were to get h 2 as minimum value in that case point should lie on inversion curve this is the most important requirement one can understand from in order to maximize the value of y or in order to minimize the value of h 2 what do you understand from this if you see a disc curve which is a temperature pressure diagram and this is your inversion curve and there are different enthalpy lines. If I want to compress the gas from point 1 to point 2 isothermally that means the temperature remaining constant you can understand from here that I can have various infinite values of different enthalpies but my h 1 minus h 2 is going to be maximum only when the point 2 lies on the inversion curve all other points in between will have enthalpy more than point h 2 and therefore h 1 minus h 2 will not be maximum in that case h 1 minus h 2 is maximum only when point 2 lies on inversion curve one can easily understand from this curve. So, consider three constant enthalpy lines here as h 1 a intermediate values over here and h 2 and what you can see from here is h 1 is more than any intermediate values over here and more than h 2 intermediate values lies between h 1 and h 2 from the figure it is clear that h 1 minus h 2 is maximum when the point 2 lies on the inversion curve. So, that y is maximum this is very clear from this figure that if you want to have y as maximum h 1 minus h 2 has to be maximum and then h 2 has to lie or the point 2 has to lie on the inversion curve. Now, let us find the work requirement for this the work requirement for a Lindy-Humson cycle can be derived by considering a control volume enclosing the compressor. So, let us have a compressor over here and again do the energy balance at this point the quantities entering and leaving this control volume are as given below. So, what is entering is mass at point 1 and what is also entering is W c or the work done on the system which is written as negative minus W c what is leaving the system is mass at m 2 and also the q r q r because it is leaving the system is given as minus q r. Using the first law for the following tables whatever coming in is equal to whatever going out as e in is equal to e out we can find out that therefore, m dot h 1 minus W c is equal to m dot h 2 minus q r. If we rearrange this terms what we get is q r minus W c is equal to m dot into h 2 minus h 1. So, q r minus W c we are taking a energy balance across the compressor what you get q r minus W c is equal to m dot into h 2 minus h 1. The expression for q r can be obtained from a second law for an isothermal compression and it is given by q r into m dot into t 1 s 2 minus s t d s what you know as q r is equal to m dot into t 1 into change of entropy across the process which is s 2 minus s 1. If you put this value over here and rearrange the terms combining the above equation the work requirement per unit mass of gas compress is minus W c upon m dot is equal to t 1 into s 1 minus s 2 minus h 1 minus s 2. This is the work of compression per mass of gas which is compressed m dot is nothing, but per mass of gas which is compressed and this is the expression for that which one has to calculate. If you want to calculate the work of compression if the gas is compressed from 0.1 to 0.2 isothermally at temperature t 1. The liquid yield y is given by already calculated m dot by m is equal to h 1 minus h 2 upon h 1 minus h f. Combining the above equations the work required for a unit mass of gas liquefied if I want to calculate now W c by m dot f. So, W c by m dot f is nothing, but W c upon m dot divided by y which is nothing, but W c upon y m dot. So, if I know what is my W c upon m dot just divided by y and what you get therefore, is W c upon m f or work of compression per unit mass of gas which is liquefied. So, one has to be very sure about what is the question being asked is it the work done per unit mass of gas compressed in which case we have to use this expression. If the question is what is the work done per mass of gas which is liquefied then whatever you have calculated here should be divided by y and therefore, what you get is a work of compression for unit mass of gas which is liquefied. With this background now I would like to solve a few problems. So, that whatever we have learnt till now could be first used and you will understand how to do this calculations and importantly how to interpret the results. So, let us have a small tutorial of one or two lectures and the first tutorial is now it is here. The problem statement is determine the liquid yield which is nothing but y, the work per unit mass of gas compressed which is nothing but W c upon m dot and the work for unit mass of gas liquefied which is nothing but W upon m dot f for a Lindy-Hamson cycle with air as working fluid. Now this is very important to understand what has been asked in the problem. Sometimes the problem statements here are very very large and you have to really split those into small sub problems and then solve the problem. So, here you have to understand first what has been asked. So, what is asked is y which is m dot f upon m dot work per unit mass of gas compressed which is W upon m dot and work per unit mass of gas which is liquefied which is W upon m dot f for a Lindy-Hamson cycle and your working fluid is air therefore, you should have is the T s diagram for air and that is the basic requirement that you should have a temperature entropy diagram of air. This can be obtained from NIST website or various books. Now the system is operated between 1.013 bar which is nothing but 1 atmosphere and this is the 0.1 and 200 atmosphere or 202.6 bar at 300 Kelvin. So, your 0.1 is going to be 1 atmosphere 300 Kelvin pressure and temperature respectively and the 0.2 is equal to 200 atmosphere and 300 K respectively. So, what is the step one? Step one is basically the T s diagram for a Lindy-Hamson cycle is as shown. So, first plot T s diagram and get the different enthalpy values entropy values associated with those point from the chart which is available with you. So, here I am drawing small schematic and also you should draw a small schematic showing the cycle 1, 2, 3, 4 of the Lindy-Hamson cycle as given over here. So, the 0.1 is at 300 Kelvin and 1 bar. So, locate the pressure line at 1 bar, locate the temperature at 300 Kelvin and look at the 0.1. Process 1 to 2 is a isothermal compression and the 0.2 should be located in such a way that this pressure is 200 atmosphere or 200 bar at this point temperature remaining constant 1 to 2 because this is a isothermal compression process. Now, as soon as your 0.1 is fixed we know where the point f is, what we need to know is state 1 and state 2. You may not know state 3 immediately right now which may not be required for solving this problem for the data which has to be calculated over here. So, what you understand from here is the state properties at different points are as shown over here. What are the state points? These properties are taken from NIST. I have talked about NIST in the first lecture, National Institute for Standards and Technology. At this site if you go we will have all the TS diagram for various gases which could be downloaded, which could be used or there are other books on cryogenics which always give this TS diagrams. So, if you see the formula which we want to use now, what properties it requires the property at 0.1, the properties at 0.2 and the properties at point f, this is what is required. So, we have got a small table and I will request it also while solving such problems is to make a small table of all the points and get all the property data like pressure, temperature, enthalpy and entropy at those points. Why this only 3 points? Because only 3 point data is required to solve all the problems associated with whatever questions have been asked in the problem. So, we got a 0.1 which is this, the pressure is around 1 bar, temperature 300 Kelvin, the corresponding enthalpy and entropy are this. If I go to 0.2, the pressure is 202 bar, temperature remains constant which is 300 Kelvin, corresponding enthalpy and entropy values are taken over here. Similarly, at point f which is this, pressure is 1 bar or atmospheric pressure, the temperature at this point is the boiling point of air associated with this pressure. So, what is the boiling point of air at 1 bar which is 78.8 Kelvin, corresponding enthalpy and entropy are the values over here and this is the most important thing. If you make mistakes in finding out these properties, it will go wrong, you will get very absurd values which will be very difficult for you to explain or interpret later and therefore, you should get this point 2 or 3 times. Some error is always allowed because you are reading the chart and therefore, some mistakes can happen. However, you should see that 28.47 should not become 35 or 40, it can be 30 or 26, 27 on this side, always 10 percent error could be allowed in that case. Now, the first problem is to solve the liquid yield, the formula for this is y is equal to h 1 minus h 2 upon h 1 minus h f. So, what you understand from here, the properties are required from point 1, point 2 and f. We know all the enthalpy values which have been given in the earlier table. So, this is the table, I have just reproduced it for you, you can read now what is h 1, what is h 2 and what is h f. This is the third row of this table, y is equal to h 1 minus h 2 upon h 1 minus h f. If I put these values 28.47 minus of 8.37. So, I will get plus 8.37 over here. Similarly, h 1 minus h f again minus of 406 which is going to be kept over here. If you solve this, what you get is 36.84 divided by 434.47 which is equal to 0.085. What does it mean that y is equal to 0.085? What does it mean that m dot f upon m dot is equal to 0.085 which also means that m dot f is only 8.5 percent of whatever m dot is compressed. This is a very important thing that whatever amount of gas which is getting compressed in the compressor at point 1, only 8.5 percent of that gas gets liquefied and that is what the product is. The liquefaction product is only 8.5 percent of whatever is getting compressed at point 1. Now, let us calculate the work per unit mass of gas which is compressed. Again understand that this is w upon m dot. So, apply the formula which is minus w c upon m dot is equal to t 1 into s 1 minus s 2 minus h 1 minus h 2. Again take the property data from this table put the value of s 1 and s 2 which is this s 1 and s 2 and h 1 minus h 2. So, if I put those values w c by m dot is equal to 300 which is nothing but t 1 into s 1 minus s 2 which is this 0.1 minus of minus 1.5 minus h 1 minus h 2. So, h 1 minus h 2 over here which is nothing but 443.16 joule per gram. So, the amount of compressor power required per unit mass of gas compressed is 443.16 joule per gram. Now, the third requirement of the problem is calculate the work required per unit mass of gas which is liquefied. What we calculated earlier is work per unit mass of gas which is compressed. So, what we have earlier found is minus w c upon m dot is equal to 443.16 joule per gram and what we know is y is equal to 0.085. So, using this we can find out what is work done per unit mass of gas which is liquefied which is nothing but w c upon m dot f which is nothing but equal to minus w c upon m dot divided by y or minus w c upon y into m dot. So, what is w c upon m dot is 443 if you divide that by y what you get is w c upon m dot f. So, work required per unit mass of gas which is liquefied is 5213.64 joule per gram. This is way high as compared to what is the work required per unit mass of gas which is compressed. This is quite understandable because m dot f value is very very small and therefore, work done per unit of mass which is liquefied is going to be very very high. It is just a way of representing the values. This is work done per unit mass of gas which is compressed. This is work done per unit mass of gas which is liquefied and therefore, m dot f being very small quantity. This quantity is always going to be much higher as compared to work done per unit mass of gas which is compressed. Now, this was the problem wherein we have got all the three values. So, we have understood now how to calculate these values. Now, I will take you to the next tutorial where we will understand the effect of pressure of compression and also the temperature at which the gas compression happens. So, let us see the next tutorial again as I said it is very important to understand what is the problem because the problems are normally statements are very very bigger and therefore, you have to understand them in parts. In this case now, determine the liquid yield for a Linde-Hamsen cycle with nitrogen as a working fluid for the following operating conditions and ultimately what is important is comment on the results or interpret the results. So, again Linde-Hamsen cycle but the working fluid in this case is nitrogen and not air. So, this is a schematic of the Linde-Hamsen cycle and these are different operating conditions. What do you understand from this? There are 4 cases case 1, case 2, case 3 and case 4. In the first case the 0.1 and 0.2 are at same temperature but the temperature is 300 Kelvin and the pressures at 0.1 is 1 bar. So, 1 bar 300 Kelvin is the condition at 0.1 over here while the 0.2 condition is 300 Kelvin and 50 bar. So, the gas is getting compressed from 1 bar to 50 bar. In the case 2 the temperature is not now 300 Kelvin but it is at 200 Kelvin alright. So, now here the compression is happening at 200 Kelvin which is way below the ambient temperature while the pressure is same as 1 bar compressed to 50 bar pressure after the compression is 50 bar. In the case 3 here is the pressure now is changed from 50 bar to 100 bar. Again the cases are taken at 300 Kelvin and 200 Kelvin respectively. In both the cases the compression is happening from 1 bar to 100 bar. The objective of this giving this tutorial is to understand what happens if you have a compression at 300 Kelvin or at 200 Kelvin. This is the isothermal compression from 0.1 to 0.2 here. This can happen at 300 Kelvin which is room temperature or we can have the compression as reduced temperature which is at 200 Kelvin and we would like to understand how does it affect the yield or work done etcetera. Similarly now instead of compressing the gas up to 50 bar if I went ahead and if I compress it up to 100 bar what is the corresponding changes and we will like to understand these changes even when the temperature of compression is reduced from 300 to 200 Kelvin. So, this is a very important and interesting problem where you understand the effect of compression pressure as well as you understand the effect of compression temperature isothermal compression temperature. And let us study the effect of this parameter in this problem. So, step one as I said the assumption is that the heat exchanger is effective 100 percent effectiveness. The T s diagram from Lindy Hampton cycle is as shown. So, what is important what I want to say was first of all get hold of the T s diagram of nitrogen again it could be downloaded from various places. In this tutorial we will assume 1 atmosphere is equal to 1 bar I mean I do not want to write every time that 1 atmosphere is 1.013 bar we can fairly assume that 1 atmosphere is 1 bar and take the values accordingly. So, the point 1 the first case temperature at 300 Kelvin and the compression is happening from 1 bar to 50 bar the T s diagram is here 1 bar to 50 bar 1 to 2 is a compression pressure is happening at 300 Kelvin. The properties at 1 2 and f are as given over here as soon as I say 1 bar nitrogen the point f will be at 77 Kelvin. So, the point 1 is 1 bar 300 Kelvin corresponding enthalpy and entropy given over here the point 2 is 50 bar 300 Kelvin corresponding enthalpy entropy given here the point f which is again at 1 bar 77 Kelvin and enthalpy entropy values are given over here. So, I can straight away start calculations now which is y for the first case is equal to h 1 minus h 2 upon h 1 minus h f put all the values over here what you get ultimately is 0.023. So, for the first case I am getting yield as 0.023. Now, I go to the second case which is the compression process happening at 200 Kelvin the pressure remain in the same that is 1 bar to 50 bar compression. So, what I am doing is now instead of 300 Kelvin I am carrying out the process of compression at 200 Kelvin this is schematically shown 1 to 50 bar at 200 Kelvin now alright. Again I go up to T s diagram and I get values the point 1 is at 1 bar 200 Kelvin corresponding enthalpy entropy here the point 2 is 50 bar 200 Kelvin corresponding enthalpy entropy over here and the point f is at 1 bar and 77 Kelvin this will not change. If I calculate now y for case number 2 same formula putting the values of enthalpy at this point I get 0.076. So, as compared to 0.023 what we got in the case 1 I am now getting 0.076. What is the reason? The reason is now the process of compression is carried out at lower temperature at 1 to 2 is happening at 200 Kelvin while the earlier process was happening at 300 Kelvin and you see correspondingly I got an increase in the value of y in this case. Now, let us go to case number 3 which is now which is happening at 300 K the process of compression is happening at 300 K only, but the pressure at the end of compression is now is 100 bar and not 50 bar anymore alright. So, corresponding now I have got a point 1 at 1 bar point 2 is at 100 bar and 300 Kelvin. So, again go back to T s diagram and get the properties the point 1 is now 1 bar at 300 Kelvin the point 2 now is 100 bar and 300 Kelvin the point f remains at the same condition. If I put these values y now for case 3 is going to be putting all the respective value of the enthalpy in this formula what you get here is 0.044 alright. So, if you again compare it with the first value which was at 50 bar this value is going to be higher as compared to what it was for 50 bar 300 Kelvin compression temperature. So, as soon as your pressure after the compression at this point has increased what you get is a higher yield or higher y value and now I got a case 4 where the same compression instead of doing at 300 Kelvin now from 1 bar to 200 bar I am doing it at 200 Kelvin again I will go back to T s diagram again I will get points 1 2 f for the 1 2 f respective points from this chart. So, here my point 1 is at 1 bar 200 Kelvin point 2 is at 100 bar and 200 Kelvin. So, process of compression is happening at lower temperature we can say it is a pre cooled cycle which is happening at 200 Kelvin and not at 300 Kelvin again I will take enthalpy entropy values from the T s diagram and calculate y. So, y for the fourth case is by this formula is going to be 0.14 which is very very high as compared to what it was for earlier cases. What does it mean? It means that if you carry out the compression at lower temperature you get higher yield at the same time if you compress the gas to higher pressure you get higher yield this is what we understood from this. So, now let us summarize the result. So, here are the 4 cases case number 1 2 3 4 the case number 1 was compressed from 1 to 50 bar at 300 Kelvin gave us 0.023 y while if I carry out this compression not at 300 Kelvin but at 200 Kelvin the yield increased from 2.023 to 0.076. If I do the compression not up to 50 bar, but up to 100 bar yield increase from here to here again 0.023 to 0.044 at 300 Kelvin compression temperature while if I reduce the temperature of compression from 1 to 100 bar at 300 Kelvin to 200 Kelvin yield increase drastically from 0.044 to 0.14. This can be now graphically represented on the slides over here. So, understand from here that as the compression process increases the liquid yield increases at a given compression temperature and as the compression temperature decreases the liquid yield y increases at a given compression pressure this is what we learnt from this. So, if I want to plot this now on a y as a function of temperature T 2 that means the compression temperature or the isothermal compression temperature what you can see is point number 1 is this 0.2 if I plot at 200 Kelvin and 50 bar pressure is this and if I join this point and extend this point further this is the line of 50 atmosphere and you can see now as you go on lowering the temperature of the compression the yield is going to increase in this fashion. If I take into consideration of 0.3 and 0.4 or case 3 and case 4 this is the case 3 which is 0.044 which will come at this point at 300 k and 100 bar and this is 200 k and 100 bar and now if I join this 2 points this is the 100 bar case and you can see now. So, you can see how much yield has increased if we went from 50 bar to 100 bar or you can also see how much yield has increased if the process of compression instead of carrying out 300 Kelvin I carried it at 200 Kelvin the yield increased and if I now extend this further for different pressures what you get is this is at 200 atmosphere and this is at 300 atmosphere. So, it is fairly easy to say that I should go for as high pressure as possible and as low temperature as possible for the compression process summarizing we have as the compression pressure increases the liquid yield y increases at a given compression temperature as the compression temperature decreases the liquid yield y increases at a given compression pressure this is what has been summarized from this. It is very important for to understand what happens when the pressure changes what happens when the temperature changes based on this we have got different assignments please calculate the liquid yield work done compression work done per unit amount of gas liquefied and for the above parameters again as air as a working fluid nitrogen as a working fluid please do these assignments and similar to what we have learnt I have got a different assignment for you please carry out these assignments. To summarize today's lecture the ideal cycle demands very high pressure which is impractical and hence modified cycles are proposed to lower the maximum pressure. An ideal cycle is used as a benchmark to compare the performances of different liquefaction cycles. In a Lindy Hamson system only a part of the gas that is compressed gets liquefied that is why you got a m dot or m dot f which is much less than m dot and therefore a fraction come which is called y. A heat exchanger is used in Lindy Hamson system to conserve cold in the system this process is an isoboric process and is assumed to be 100 percent effective. In order to maximize y or m dot f upon m dot for a Lindy Hamson system the state 2 should lie on the inversion curve at the temperature of a compression process. This is made fairly clear in this lecture the work required for a unit mass of gas compressed for a Lindy Hamson cycle is given by this formula. If we have seen the derivation of this formula also for a Lindy Hamson cycle falling holds true as the compression pressure increases the liquid yield y increases at a given compression temperature. This is fairly clear from the examples also which we have taken at the same time as the compression temperature decreases the liquid yield y increases at a given compression pressure. This is what we have understood from the last problem. Thank you very much.