 So, welcome to the 14th lecture on cryogenic engineering and what we have been dealing is related to gas liquefaction and refrigeration. In the earlier lectures, we have covered several topics related to this particular gas liquefaction topic. In the earlier lecture, we talked about ideal thermodynamic cycle in which what we know is whatever gas is compressed gets liquefied. So, all the gas which gets compressed gets liquefied and that is why it is called ideal thermodynamic cycle which normally cannot be realized in practice. The first modification of that happened in a cycle called Lindy-Hampson system. It has got a heat exchanger in it and this heat exchanger is used to conserve the cold. In this system, only a part of the gas that is compressed is liquefied. So, if we compress x gas, only 5 to 10 percent of that gas will get liquefied and the remaining gas in the cold condition will go back to this heat exchanger and it gives the cold to the incoming compressed gas and thus the cycle continues. So, the Lindy-Hampson cycle or system was the first modification from the ideal thermodynamic cycle and then came several modifications of the Lindy-Hampson system itself and one of which is pre-cooled Lindy-Hampson system in which the liquid yield and figure of merit gets improved by pre-cooling the working fluid using an independent refrigerating system. So, this pre-cooled Lindy-Hampson system, what happens? We have got one more auxiliary system to the major system and this auxiliary system runs a refrigeration cycle which pre-cools the gas when it enters the heat exchanger. So, basically what is done is pre-cooling and thereby the liquid yield and the figure of merit gets improved. This was studied during the last lecture and the highlights of this particular lecture are in a pre-cooled Lindy-Hampson system, the liquid yield y what we call as and the figure of merit are dependent on the refrigerant flow rate m dot r is what is the refrigerant flow rate we call as in the auxiliary circuit and the liquid yield y and figure of merit depend on this m dot r also the compression pressure and the pre-cooling temperature. So, there are three major parameters which is m dot r compression temperature and the pre-cooling temperature. The yield of the system increases with the increased in the refrigerant flow rate and the compression pressure. So, this is a natural output which we understood from the earlier lecture that the y or the yield increases with the increase in the refrigeration flow rate naturally increase refrigerant flow rate means increased cooling effect and the compression pressure if we increase the compression pressure also in that case also the yield increases. So, one can understand that as the flow rate increases we get more benefit in terms of having y. In this system however the mass ratio r corresponding to maximum yield is called as limiting value that means one cannot go on increasing the refrigerant flow rate m dot r it has got some limiting value and we have talked about in detail what this limiting value is all about. So, this limiting value is a function of compression pressure also. So, if we increase the compression pressure in the major circuit in the major system the value of limiting ratio also increases that means you can admit you can have more and more mass flow rate in the auxiliary circuit or in the pre-cooling circuit. So, the relationship between the value of r and P 2 or the compression pressure was shown during the last lecture it was done using a problem or a tutorial and we could draw various curves we could understand what is this limiting value of r as a function of this compression pressure. From this we understood that work per unit mass of gas compressed increases with the increase in refrigerant flow rate and compression pressure it is very directly one can understand from this as the flow rate increases we have got two compresses now one is the major system one is the pre-cooling circuit. So, as the refrigerant flow rate increases the work per unit mass of gas will increase also if the compression pressure increases work per unit mass of gas components will increase, but the work per unit mass of gas liquefied decreases with the increase in the refrigerant flow rate and the compression pressure this is the most important thing why are we doing all this thing because we want to decrease the value of w upon m dot f that is work per unit mass of gas liquefied and that is the whole objective of this pre-cooling circuit also the figure of merit increases with the increase in the refrigerant flow rate and the compression pressure. So, whole exercise was basically done in order to decrease the value of work per unit mass of gas liquefied or increase the figure of merit with the increase in the refrigerant flow rate and the compression pressure with this background what we have understood what we have done during the last lectures the outline of this present lecture is again on the same topic of gas liquefaction and refrigerant system in this system in this particular lecture what we are going to talk about is linde dual pressure system as the name suggest it has got linde two pressure systems or we got two devices or two compressors and two expander to be specific in this particular cycle it is also again is a modification of the linde hamson cycle with dual units existing in the cycle in the system what we are going to study about in this particular lecture is what is the liquid yield in this particular case what is the work requirement in this particular case there should be some advantages and we have to study all these advantages and lastly there are various parameters and we will do the parametric studies, but this parametric studying I am going to carry out again as we have done previously using the tutorial because here we understand with the values how to solve the problem at the same time we can draw various curves ourselves. So, this is what is going to be studied in this particular lecture mathematically the work requirement for ideal isothermal compression process is given by w dot is equal to m dot r t 1 log p 2 by p 1 this is the work of compression for a isothermal process what does this mean the work requirement decreases either with the decrease in mass flow rate or with the decrease in compression ratio. So, in this equation you can see there are three parameters m dot t 1 and p 2 by p 1 suppose the compression process is being carried out at a given temperature or at room temperature only will not consider t 1 option open to us because we want to carry out the process of compression always at room temperature. In this case if I want to decrease the value of w that is the work of compression I can touch upon two parameters m dot and p 2 by p 1. So, the work requirement decreases either with the decrease in mass flow rate that is m dot or with the decrease in the compression ratio actually this is a well known thing for thermodynamic students this is what exactly is being exploited in a lindade dual pressure system. In a lindade dual pressure system the work requirement decreases when the compression of a fluid is done in two stages and for different mass flow rates. So, if I want to go for a very high pressure I will not go to that high pressure in one shot, but I will go in two stages and this is again known as a compounding of compressor or having two or three different stages of compression and also in each stage we will control the value of m dot. So, we want to minimize the work input to the compressor in such a way or by playing with this parameters m dot and p 2 by p 1 and this is a real basics of this lindade dual pressure system. Now, how it is done we will see in the coming slides. So, here is a slide which shows you how a lindade dual pressure system works. So, as I said dual means two and we have got two compressors this is what you can see from this particular schematic and this is a lindade dual pressure system schematic there are two compressors and two expanders. So, we have got two compression processes and we have got two expansions over here at the same time we have got two containers also. So, one can see a lindade hamson system having one heat exchanger and a compressor process and an expansion process. However, what we have is a two compressors two expanders and two containers and this is what a highlight of a lindade dual pressure system is. So, what does lindade dual pressure system has the system consists of two compressors a three fluid heat exchanger and you can see in this heat exchanger you got one flow which is going down and one stream is going up and one stream is going up and this is a very critical heat exchanger in this case and this is called as three fluid heat exchanger. So, you got two compressors three fluid heat exchanger two Joule-Thomson expansion devices. So, you got a one device over here and we got a one device over here making it a dual expansion kind of a system and again two liquid containers and a makeup gas connection. So, you got a two connections two containers one over here one over here and here you got a makeup gas connection m dot f at this point. So, this is what a lindade dual pressure system is all about. In this system the entire mass fluid of the gas is not compressed to the required high pressure as was the case in a previous systems and that is what a highlight of having a two stage compressor. We do not compress the gas to the final pressure in one shot, but we do it with two compressors. In each of this compressor we are playing with the mass flow rates also as I said in the earlier slide we can vary the mass flow rate as well as the pressure ratio across this compressor. So, you got a compressor number one and compressor number two over here. Now, as you see I am talking about now the mass flow rates in each of these compressors only a part of the mass flow rate that is m minus m dot i which is coming from this side is being compressed in a compression number one is compressed to an intermediate pressure here P i. So, we have got two compressors and at this point I get a final pressure that is what is in my mind. I compress the gas equivalent to m minus m dot i at this point and compress it to some intermediate pressure P i and the next compressor compresses this gas at P i to the final pressure at this point. So, this is what a compression process takes place there after that means at point two the mass flow rate m i is added to the above stream and is taken and is then compressed from two to. So, here you can see I got one more stream coming from here after the first expansion it joins the stream or the gas flow at this point the gas flow which is compressed at this point is m minus m i having m i joining this stream what you have is a m dot over here. So, I am compressing full mass flow rate in the second compressor from point two to point three or from intermediate pressure P i to final pressure or P 3 at this point. So, this is what is to be understood that mass flow rate here is m minus m i and mass flow rate here in this second compressor m dot the compression is from P 1 to P i or P 2 and from P 2 or P i to P 3 in the second compressor this is the basics of lid day dual pressure system. This arrangement not only compresses the gas in two stages, but also reduces the work requirement and this is why we are doing all this thing basically we want to reduce the work done per unit mass of gas which is compressed. The stream m dot i along with the written stream from the container number two here is used to pre cool the gas at point three in a three fluid heat exchangers. Now, the circuit is the gas when compressed up to the maximum value P 3 gets pre cooled by the written stream over here and by written stream over here. What is this written stream? This gas is going to pass or m dot which is compressed to the full pressure is pre cooled first in this heat exchanger and it goes undergoes the first expansion and this expansion happens from value of P 3 up to P i and this is happening the part of the gas or m dot i goes back. So, the written stream having m dot i goes back and it pre cools the gas coming over here and from this gas at this intermediate temperature and pressure next expansion happen from P i to P f or the lowest pressure which is P 1 and here what you get is a the m dot f that is the fluid which is getting liquefied the gas which is getting low liquefied. So, this written stream will have m minus m dot f minus m dot i because m dot i has already gone back m dot f has already been taken off the written stream will have m minus m f minus m i and this is going to get compressed in the first heat exchanger when m dot is m dot f is added over here as a make up gas and this is how the whole cycle works. So, the pre cooling heat exchanger is basically 3 fluid heat exchanger it is getting pre cooled by this written stream at intermediate pressure and this written stream at the lowest pressure at P 1. It is important to note that 3 fluid heat exchanger has 3 streams with the 3 different flow rates they are as I just talked about m dot which is coming from here m dot i which is going back and m minus m f minus m i which is going through this point. Now, most important thing if I want to plot the whole cycle here this is my T s diagram or temperature entropy diagram the gas is compressed from 1 to 2 intermediate pressure from 2 to 3 to the final destination pressure then gas is getting pre cooled up to 0.4 which is over here at this point the gas gets expanded from P 4 to P i or P 2 value here and you can see the expansion process with isenthalpic. The written gas m dot i goes back and joins the main stream over here the remaining gas which is m minus m i gets expanded further from 0.6 to 7 over here here we get m dot f and the written stream has m dot minus m dot f minus m dot i which goes back and the cycle continues. Now, this is the most important thing that we have got 2 compression process 1 to 2 and 2 to 3 and we got 2 expansion process from 4 to 5 and 6 to 7 ok. We got a written stream which is going at m dot i at 0.8 and written stream at m dot minus m dot f minus m dot i at point g and this is how the whole TS diagram would look like. This most important thing is to plot this TS diagram correctly. Now, we will come to our routine job to sort of get expressions for the work input and a value of y. So, consider a control volume as shown over here and see what is going in and what is going out of this system ok. So, what is going in here is m dot at 0.3 what is going out is m dot i at point 2 m minus m f minus m i at point 1 and m dot f at f over here. Applying the first law we know all these things getting the respective enthalpy values at those point we write this expression ok. I will not go through this all the enthalpy values are associated with all the mass flow rate over here and if we reorganize the whole thing what we get over here is m dot f upon m dot which is nothing but y is equal to h 1 minus h 3 upon h 1 minus h f minus m dot i upon m dot into h 1 minus h 2 upon h 1 minus h f. So, it is a very important expression and you can see that there are two terms. Again if you could denote m dot i upon m dot as intermediate mass ratio m dot i upon m dot as i. So, it could be called as i which is intermediate mass ratio. We have the liquid yield as m dot f upon m dot is nothing but y, y is equal to h 1 minus h 3 upon h 1 minus h f minus i times h 1 minus h 2 upon h 1 minus h f. That means that y is equal to this expression and what you have is first expression which is the first term is the yield for a simple Lindy-Hamson system considering that entire mass of the gas is compressed from 1 to 3. So, if you look at this first term it is nothing but this that is a simple Lindy-Hamson system where the gas is compressed using one compressor from 0.1 to 0.3 and the yield in that case would have been this. However, it is not a Lindy-Hamson system which is compressed in the gas from 1 to 3, but we have got 1 to 2 and 2 to 3 and we have got m dot i joining at an intermediate point at an intermediate pressure and therefore, we got a term called i. So, this second term takes into consideration the Lindy-Duell pressure system and it differentiates otherwise from Lindy-Hamson system in which the compression will happen from 1 to 3 process. So, the second term is a reduction in the liquid yield occurring due to modification of the dual pressure system. So, if one wants to understand the first term indicates what would be the yield if the gas is compressed from 0.1 to 0.3 directly and the second term actually brings it to the actual case of Lindy-Duell pressure system. It deducts the value of y because it is a dual pressure system. So, you will get less yield as compared to what it would otherwise have got in the Lindy-Hamson system compressing the gas using one compressor from 0.1 to 0.3 directly. For the work requirement now, what we have done till now is a finding an expression of y, but what is most important is what is my WC1 and WC2 which is the work input to the compressor number 1 and work input to the compressor number 2 and this is what will basically bring about the advantages we get in a Lindy-Duell pressure system. So, we should first quite find out what is my WC1 and WC2. What you understand from here is for WC1 what you have is a mass flow rate of m minus m dot i and what you have corresponding to that is a QR value which is minus QR1. In both the case system what is going in a system is m minus m i what is coming out from that is m minus m i at 0.2. What is going in is minus WC1 what is coming out is minus QR1 this is according to the direction and a science convention we are going to follow with WC1 and QR1 as you all know. So, using first law what we have is a energy in is equal to energy out. So, what is going in is m minus m dot i h1 minus WC1 is equal to what is coming out m dot minus m dot i h2 into minus QR1 which is going to come out. Rearranging this terms what we have is a QR1 minus WC1 is equal to m dot minus m dot i h2 minus h1. By second law we got a value of QR1 and we know that QR1 is equal to m minus m i t1 into s2 minus x1 tds basically multiplied by the corresponding mass flow rate. If we put this expression of QR1 in this what we get is WC1 which is equal to m dot minus m dot i into t1 s1 minus s2 minus h1 minus h2. So, you can see that what is the mass flow rate in the first heat first compressor m dot minus m dot i multiplied by temperature into delta s tds minus h1 minus h2 this expression is similar to what we have had earlier for a compression process. If I go to the second compressor now what is the flow rate through that is m dot. The second compression has full mass flow rate of m dot and corresponding value of entropy and enthalpy have to be taken into account. The mass flow rate across the second compressor is m dot following the similar procedure for the work requirement for the compressor 2 what we get is minus WC is equal to m dot into t1 into s2 minus s3 minus h2 minus h3 because the process WC2 compression process is happening across 0.2 and 0.3 and we are taking the entropies and enthalpies related to those points. So, this will form the work of compression for the second compressor. Now the total work requirement is given by WC is equal to WC1 plus WC2. So, the total work is not now going to be determined by what is the value of WC1 and what is the value of WC2. WC1 depends on what is the gas what is the amount of gas which is getting compressed in the compressor number 1 and what is the pressure ratio across 0.1 and 2 alright P2 by P1. Similarly, WC2 depends on what is the mass total m dot going into it and what is the pressure ratio across it 2 and 3. It is this WC1 and WC2 which makes a difference to the value of WC or the total work of compression and this is the most important to understand because by doing this during the dual pressure system we are basically trying to minimize the work of compression in this case as compared to a simple Linde-Hampson cycle. Denoting the ratio now if we add both the expressions together and if we put the value of I in that expression we get work per unit mass of gas compressed that is WC by m dot as this expression that is T1 into S1 minus S2 minus H1 minus H3 as if the whole thing is getting compressed from 1 to 3 part which is what we saw in expression for Y. In the similar line again you have got expression for WC by m dot first term denotes as if the whole thing is compressed from 1 to 3 and the second term denotes depending on the value of I what is to be deducted from it that is I into T1 into S1 minus S2 minus H1 minus H2 this much of work will be done less as compared to the process of compression carried from 1 to 3 process. So same thing is done the first term is the work requirement for a simple system considering that the entire mass of the gas is compressed from 1 to 3. The second term is the reduction in the work done work requirement occurring due to the modification and this is what we basically are going to exploit in order to reduce the work done per unit mass of gas which is compressed. With this background on linde dual pressure system linde dual system as I said we have just derived expressions for Y and the power input compression power basically I want to now take you to the to understand what is the effect of various parameters on these parameters. So what I am going to do is I am basically giving you a tutorial and will solve this tutorial and understand from this how this parameters play a very important role in order to get expressions for what is the work done what is the work of compressor per unit mass of gas which is compressed and thing like that. So this tutorial is the most important tutorial. So what is the problem to understand is determine W by m dot f and figure of merit for a linde dual pressure system with argon as a working fluid for the following intermediate pressures. The system operates between 1 atmosphere and 120 atmosphere and the intermediate mass ratio I is 0.6. There are various intermediate pressure points that is 0.2 and in this problem we want to understand what is the value of W by m dot f and figure of merit for these intermediate pressures. So I am going to compress the gas first from 1 bar up to 4.05 bar and solve all the values to get what is the W by m dot f and figure of merit for that case. Then I will get the second value of pressure 20.3, 75.9, 101.3 and in such a way that my final pressure always remains at 120 atmosphere alright. So here I want to understand what is the effect of this intermediate pressure. So first of all we should calculate how much worked per unit mass of gas which is liquefied when gets at different intermediate pressure. The intermediate mass for ratio I is in this case 4.6. Then my tutorial also want to understand what will happen if I go from I value of 0.6 to 0.7. So repeat the above problem for I is equal to 0.7 and ultimately plot the data graphically and comment on the nature of Y W by m dot f figure of merit versus I alright. So it is basically I am trying to understand what is the effect of intermediate pressure on W by m dot f and figure of merit and what is the effect of value of I or the mass ratio 0.6 and 0.7 on all these studies. So there are intermediate two problems in order to understand the effect of intermediate pressure and effect of I on these important parameters alright. This is my problem. So what is given in this particular problem is we want to work on a linear dual pressure system. The working pressures are from 1 atmosphere to 120 atmosphere. The working fluid is argon. Now this is very important that one should have all the properties of argon. The temperature is 300 Kelvin that means the process of compression is carried out at 300 Kelvin. The intermediate mass ratio is 0.6 and 0.7. For above system calculate work per unit mass of gas liquefied is W by m dot f and figure of merit and the intermediate pressures are 4, 20, 75, 101 atmospheres alright. So this is my problem. What is the methodology? The methodology is for solving in this particular class is the two mass ratios I condition under study are 0.6 and 0.7. In this particular tutorial I am not going to repeat the calculations for each of these values. I am going to demonstrate to you how to calculate these values for one pressure and for one I value alright. What I expect you to do is then carry out this exercise further for all the pressures and for both the values of I is equal to 0.6 and 0.7. So in this tutorial the liquid yield Y and work plus unit mass of gas liquefied W by m dot f are calculated only for I is equal to 0.6 and 4.05 bar or 4 atmosphere as intermediate pressure conditions. So I am just solving one problem for I is equal to 0.6 and P2 or Pi is equal to 4 atmosphere while what I am leaving it to you is to calculate I is equal to 0.6 and all the pressure and then I is equal to 0.7 and all the pressures for yourselves. However, in the final destination or in the final slides I have got values for all these calculations. So we have done all this calculation ourselves and what we want you to do is to check your calculations so that you also get those values saying that your calculations are correct. It is basically an exercise for you all other calculations pertaining to value of I is equal to 0.6 and 0.7 and for all other intermediate pressures conditions are left for you as an exercise the students alright. So, let us go for I is equal to 0.6 and P2 or Pi is equal to 4 atmosphere. So, before as you all know with earlier problem first we want to do ideal cycle analysis and therefore, we got a ideal thermodynamic cycle from which we will get ideal work requirement and this is very important to calculate the figure of merit. For ideal work requirement this is the formula W i by m dot t 1 into s 1 minus s f h 1 minus h f and here as you know whatever is compressed is getting liquefied. So, which one we know one we know the point f also. So, pressure is 1.013 bar temperature is 387.3 which is the boiling point of argon at 1 atmosphere. The enthalpy values of given at 0.1 and f the entropy values are given at 0.1 and f. If you put these values in this equation what you get is W c by m dot work of compressor per unit mass of gas liquefied is equal to 461 joules per gram. So, this is my first basic calculations which we do almost for all the problems. Now, coming to our problem that is linday dual pressure system and this is our schematic in this schematic the enthalpies and entropies one should get from the T s charts for argon alright. So, get those charts for argon and get the enthalpy values. So, here you can see 0.123 and f these are the values which are required for our formula where we require enthalpy values and entropy values. Now, where are these points the point 1 is at 1 bar 300 Kelvin at this where m minus m dot i will be getting 2 compressor number 1, point 2 is the intermediate pressure or after the first compression which is this point. The point 3 is over here and the corresponding properties are given. Now, point 2 is going to be at 4.05 bar while point 3 is going to be at maximum pressure that is 120 atmosphere or 121.5 bar and again point f is going to be the last point at this which is going to be at again 1 bar. So, one should check all these values at 1 bar, 4 bar and 3 bar 300 Kelvin, 300 Kelvin, 300 Kelvin here while at point f it is 87.3 Kelvin corresponding value of enthalpies 349, 348 and 326. So, one can understand that as the pressure increases the enthalpy at the same temperature at 300 Kelvin temperatures are same over here the enthalpy decreases. Corresponding to again similar thing happens with the entropy as the pressure increases the enthalpy and entropy values are decreasing over here. So, these are the 4 different points and now what we want to see is take the enthalpies and entropies of this value and put in the equation. So, corresponding temperature entropy diagram is 1, 2, 3, 4. So, one compression from 1 to point 2, 1 bar 2, 1 atmosphere to 4 atmosphere which is intermediate pressure, 4 atmosphere to 120 atmosphere which is the maximum pressure over here. The T S diagram for linear dual pressure system is shown here. The compression process is from 1 atmosphere to 4 atmosphere and then from 4 atmosphere to 120 atmosphere. So, you got a W C 1 over here from 1 to 2 and W C 2 here from 2 to 3 in this case and respective points are given over here in this T S diagram. Now, let us first calculate the liquid yield which is this expression as you know we got a first term and a second term. What we want to know is now enthalpy at point 1, 2, 3 and f and we have already made those tables we will get the points enthalpies at h 1, h 2 and h f and of course, h 3 also. So, the first is Aragon i is equal to 0.6 at this point which is the intermediate pressure of 4.05 bar and the corresponding values are given over here 1, 2, 3, 4 all the temperatures 300, 300, 300, 87.3 corresponding enthalpy values corresponding entropy values. If you put those values in this formula what you get is 0.0817. So, y for i is equal to 0.6 and intermediate pressure is equal to 4 bar or 4 atmosphere we get the value of y to be equal to 0.0817 that is my calculation as I said I am going to do this calculation for only 1 i and 1 pressure which are these 2 values over here. Now, let us complete all the calculations for this particular value. So, now let us calculate work per unit mass of argon compressed for i is equal to 0.6 and this is the expression again we have got t 1 into s 1 minus s 3 minus h 1 minus h 3 minus i times t 1 into s 1 minus s 2 minus h 1 minus h 2 putting this values from this table if I put this value I get w c upon m dot is equal to 235.6 joule per gram this is work done per unit mass of gas argon compressed. Now, I want to calculate work done per unit mass of argon liquefied for which what I want to know is w c upon m dot and I want to know what is the value of y which is 0.0817 which is what we have just calculated. What we know is w c upon m dot f is equal to w c upon m dot upon y. So, what we have is both the values this upon this will give you w c upon m dot f. So, I will do 235.6 upon 0.0817 which is equal to 2883.7 this is my work done per unit mass of gas argon liquefied which is rather high as compared to the ideal work input which we had seen earlier in the ideal thermodynamics cycle. The figure of merit is equal to w i upon m dot f ideal w i upon m dot f divided by actual w c upon m dot f which is what w i upon m dot f is 461 from ideal thermodynamics cycle and if I put those values what you get is figure of merit is 0.1598. So, here we have got all the values calculated so far that is y and figure of merit y and w c y and total work done and of course, the figure of merit. Now, I would like to calculate all these values for all the intermediate pressures and without giving the details of those calculations I will give you a table which will have all these values. So, tabulating the results for i is equal to 0.6 we have the following comparison for the various values of intermediate pressure here. So, intermediate pressure is 4.05 bar. So, case number 1, 2, 3 and 4 this table gives for all the intermediate pressures that is 4 atmosphere 20, 75 and 101 atmosphere pressure as the value of p 2. The final value of the p 3 is always 120 atmosphere and you got a value of y w by m dot work done per unit mass of gas compressed w by m dot f work done per unit mass of gas liquefied and figure of merit. What you can see from this table is I will talk about each of this vertical column in detail in the next slides, but what you can see is as the intermediate pressure increases if the entry pressure and exit pressure remain the same and if the value of intermediate pressure if it is increased the value of y has started decreasing. So, 0.08, 0.07, 0.05 and 0.04 are the values of y that means the value of m upon m dot f upon m dot has decreased the yield has decreased if we increase the intermediate pressure, but the value of w upon m dot m dot has decreased 273 minus 172 235 172 118 111. So, this value has decreased what is interesting to see is the value of w upon m dot f that is work done per unit mass of gas liquefied what you can see this has come down from 2883 to 295 and again started increasing. So, it is decrease over here, but then there is a increase over here it means that it has gone through a minima and similarly to that what we get FOM figure of merit we have got 0.15, 0.20, 0.8 and 0.2 and 0.17 that means FOM goes through a maxima and this is what we will see in the next slide. So, what you understand from here is as the intermediate pressure increases y decreases w by m dot decreases, but w upon m dot f goes through a minima and figure of merit goes through a maxima this is what is the output of all this value for i is equal to 0.6. Now, the same study is done at 0.7 also when i is equal to made equal to 0.7 we have carried out the same studies and the results for i is equal to 0.7 for the various intermediate values is this. So, again you can see now as the intermediate pressure increases from 4 atmosphere to 20 to 101, y again decrease from 0.08, 7, 4 and 3, w upon m dot that is work per unit mass of gas compressed decreased from 228 to 154 to 91 to 83, but w upon m dot f work per unit mass of gas liquefied again went through a minima the minima is happening possibly between 0.2 and 0.3, case 2 and case 3 and it is maximum over here goes down reaches a minima and starts going up over here the figure of merit again goes through a maximum somewhere and between the case 2 and case 3. So, again it follows absolutely the same train as what it was for i is equal to 0.6. Now, here as I said we have carried out only one exercise for 0.6 and 4.05 bar what I expect all the students to do is get this table calculated by yourself get the earlier table also calculated by yourself and then plot it and then you start understanding what exactly happens and why it happens. So, all these things has been laid for you as an exercise while I have just got all these value calculated I am putting in front of you. So, this may serve as the answers for you while I expect you to solve this as a problem or as an assignments for you to see that you can read the values of enthalpy and entropy properly. Now, what is important is to understand why it happens I mean plotting this or calculating this values not a big deal, but what I want you to understand is what happens when I go from 0.6 to 0.7 as i values and for a 1 i why if the intermediate pressure goes on increasing why the value of y decreases why the value of w by m dot decreases why the value of w by m dot goes through a minima and why the value of f by m goes through a maxima all right. So, this is the most important thing for the engineers to understand. So, let us now therefore whatever table I have just shown you let us try to plot this table graphically and therefore the understanding will be much much better. So, first table we will calculate is liquid yield y versus i. So, what I am showing here you are the 2 axis on the left side you got w by m dot and the right side you got a y in the first this thing right now you understand what is the relationship and on the x axis what you have is the intermediate pressures which are varying from 2 bar to 102 bar atmosphere and let us compare the as you increase this intermediate pressure what happens to the value of y which is on the right axis right y axis do not see at the left y axis. So, see whatever points we have got let us plot those y values versus the intermediate pressures. So, this is my table for i is equal to 0.6. So, if I plot this points the first point 4.05 0.08. So, if you see on the right axis 0.08 value and 4 bar the second value third value and the fourth value alright. So, if you join them together you can see kind of linear variations when you plot y with respect to the value of intermediate pressure and this is the value for i is equal to 0.6 we got a second table for i is equal to 0.7 again get those values of case number 1 over here case number 2, case number 3 and case number 4. So, if you could plot these values and connect them you can see this. So, this is to be compared with the value. So, what do we understand from this particular graphical graph that as you go on increase in the intermediate pressure the value of y decreases this is exactly what we saw in the earlier table. So, for a given value of mass flow ratio i the yield of the system decreases with the increase in intermediate pressure why does it happen. Now, if you recollect the expression for y we got the first term and the second term. What you understand second term has a negative value which is h 1 minus h 2 which has numerator as h 1 minus h 2 and it has a negative sign as you understand that as the value of p 2 or the p i increases this second bracket starts increasing and therefore, the value of y decreases all right. If the intermediate pressure increases it means that mathematically the expression for y whatever we had the second part which was i times 1 bracket that bracket has a numerator which was h 1 minus h 2 and this value starts increasing all right. And therefore, the negative part start increasing and therefore, the value of y decreases as you go on increase the intermediate pressure. The second aspect to understand for this is as the intermediate pressure increases the intermediate temperature in the T s diagram also will increase the intermediate temperature in the dome also increases all right. If you remember the dome and we got a point intermediate pressure temperature corresponding to the intermediate temperature also and this intermediate temperature starts moving to the critical temperature value or it is starts going up the dome and therefore, when the second expansion happens it will happen from this intermediate temperature. And again at this point the point after the expansion will fall towards the gas region and therefore, if you increase the intermediate pressure the intermediate temperature from which the second expansion happens that also gets lifted up corresponding to which the point after the second expansion falls near the gas side. Therefore, again you will get the less less y value and this explains why if you increase the intermediate pressures you will get less y in those cases. Also as the mass ratio i increases that means if I go from 0.6 to 0.7 the yield of the system decreases. Now, understand again if my intermediate gas ratio increases that means 70 percent for example, i is equal to 0.7 70 percent of the mass flow is going back. It means that only 30 percent of the gas is going for the second expansion all right and the liquid yield is going to come from this 30 percent only. It means that more the gas goes back more the value of i is the m dot f is always going to be less because the second expansion has only 30 percent of the gas which is compressed all right. So, as the value of i increases less and less gas is going to go for the second expansion which means less and less yield all right. So, corresponding to that however, the work per unit mass of gas which is going to be compressed is going to be becoming less and less, but the value of y effectively is going on reducing the value of i increases. So, as the mass ratio i increases the yield of the system decreases because the mass of the gas actually expanded in J T device decreases and this is the reason that as i increases the y value will decrease. Now, let us plot the second parameter which is w by m dot upon versus i all right. So, the plot between w by m dot versus i for this value is over here. So, you got a various values and you can see that as you go on increasing the value of intermediate pressure the w by m dot decreases. What is important note is the slope of this is much higher initially while it is less and less as you go for higher value of intermediate pressure the maximum value of p 3 is always 120 atmosphere. This was the case for i is equal to 0.6 as you go for i is equal to 0.7 again similar trends you can see, but here w by m dot is still less. That means, if you increase the intermediate ratio 0.6 to 0.7 the w by m dot further decreases, but again you can see that the slope here is very high initially and it flattens down as you go for higher intermediate pressures all right. So, this can be definitely understood that as you go on increasing the intermediate pressure your w c 1 and w c 2 will change. The w c 2 will have only m minus m dot i which is compressed while w c 2 will have all the m dot which is going to be compressed. So, as you go on increasing the p i the pressure ratio for the second compressor is less and less while the pressure ratio for the first compressor increases corresponding to that the value of w c 1 and w c 2 change, but if you go on increasing the value of the intermediate pressure all right. The whole mass which is getting compressed actually is in the compressor number 2. So, the second compressor is now subjected to less pressure ratio and therefore, the value of w c 2 in this case gets less and less all right. So, as you go on increasing the intermediate pressure ratio the value of w c 2 decreases and therefore, if you add both of them together w c 1 plus w c 2 the value of work done per unit mass of gas compressed goes on decreasing if you go on increasing the intermediate pressure. This we will have to understand with all the logistics which I am just talking about what happens to w c 1 what happens to w c 2. As you go on increasing the intermediate pressure ratio though the pressure ratio for the first compressor increases, but it is subjected to less mass flow rate. The while the pressure ratio for the second compressors gets reduced, but it is subjected to maximum mass flow rate. So, it is a net addition of w c 1 and w c 2 which will determine what is the value of w by m dot in this case. Now, for a given value of mass i and w dot the system decreases with the increasing intermediate pressure as the mass ratio of i increases the w by m dot decreases because the more of the mass flow rate is bypassed from compressor number 1. So, if we increase i now your work w c 1 now in this case will decrease while w c 2 will remain the same, but w c 1 will decrease because the less mass is going to the compressor number 1 in this case. So, you can see the relative curve in both the cases that is y curve and w by m dot. It is important to note that initially the slope of w by m dot is much steeper than that of y. Later on as the intermediate pressure increases the slope of y is steeper while the slope of w by m dot decreases in this case. What is important now is to understand w by m dot f versus i which depends on both this waiver of y and w by m dot and this is the values. So, if I plot these values 1, 2, 3 and 4 and if I join them what you can see is the clear cut minima between this point over in this two case and again for i is equal to 0.7 if I put those values alright. So, this is the whole exercise is carried out in order to choose a point of intermediate pressure for which w by m dot is going to be minimum work of compression per unit mass of gas liquefied alright. So, this is a 0.6 case and this is a 0.7 case alright. So, this is the most important curve which we get from the values of division of w by m dot divided by y because the plot of the w by m dot f value is determined by w by m dot divided by y. So, mathematically as you understand w by m dot f is nothing but w by m dot upon y, w by m dot f being the ratio of w by m dot and y the relative decrease in the numerator and denominator determines the slope of the value of w by m dot f and that is why the ratio is the curve nature is like this. For a mass ratio of i w by m dot f decreases with the increase in the intermediate pressure alright. So, we got a it goes through a minima and it decreases first goes through a minima and again starts increase the work falls to a minima and then increases with the increase in the intermediate pressure and you can find that for this case 0.6 one should operate at a pressure somewhere around 50 bar 50 atmosphere while somewhere around 62 is a corresponding optimum pressure for i is equal to 0.7. The working point is compromised value of y and m dot. So, as you know that as you go on increase the intermediate pressure the value of y falls down while the w by m dot f is touching minimum at these two values. Therefore, what value of p i one should select is actually going to be compromised between what y you want corresponding to that what w by m dot you might get. So, what is your liquefaction requirement and what should be your w by m dot f? This is a compromise value one has to select in this particular case alright. So, depending on your liquefaction requirement corresponding w by m dot minimum we should be selected. The figure of merit is also plotted from the same thing and exactly reverse will happen the wherever there is a w by m dot f minimum you will get a maximum by the definition that definition of figure of merit you will get a maximum w by m dot f here and for 0.7 case i is equal to 0.7 case you will get a value of figure of merit as maximum as around 62 bar somewhere here and as I said around 50 bar somewhere at 0.6 and this is the value one should select if one has choice or one has to compromise and on this top portion and corresponding to those values the value of p i should be chosen alright. So, for a mass ratio of i the figure of merit increases with the increasing intermediate pressure with the further increase in the intermediate pressure the figure of merit reaches a maximum value and thereby it decreases. It is important to note that the figure of merit reaches a maximum value at the same intermediate pressure at which w by m dot f reaches a minimum for a given value of mass ratio i this is what I just talked about with this background with all this the way the tutorial has been solved I got assignment for you and assignment solving only this will make you perfect. So, what I want you to understand solve the problem again for Aragon for different pressures of 1 bar and 200 bar now intermediate mass ratio is taken as 0.6 and we are given the answers also for this case for y w by m dot w by m dot f and figure of merit. So, please do this exercise and check your answers check your calculations to get these answers. In summary of this particular lecture the linde dual pressure system is a modification of a simple linde hamson system in order to reduce the work requirement alright. The linde dual pressure system is never to increase the yield alright. In fact, you know that the yield decreases with the intermediate pressure increasing the linde dual pressure system is basically aimed at reducing the work requirement per unit mass of gas compressed or per unit mass of gas liquefied. In this system the entire mass flow rate of the gas is not compressed to the required high pressure as was the case in the previous systems alright. In this system the work requirement decreases when the compression of fluid is done in two stages and for different mass flow rates this is the most important thing. The pressure ratios for the compressor number 1 and compressor number 2 are going to be different at the same time the mass flow rates in the compressor number 1 and 2 are going to be different. The yield of the system is given by the following equation which is this as you know y is equal to h1 minus 3 upon h1 minus hf minus i times h1 minus h2 upon h1 minus hf. The work requirement also is given by this expression as if we are going to compress from 0.1 to 0.3 and then reducing the work in the ratio of i from 1 to 2. In the above expressions of the equation the first term correspond to the Linde-Hamson system and the second term is the reduction occurring due to the modification. So basically the first term in both the cases is as if we have got the Linde-Hamson system from 0.1 to 0.3 which is normally not possible. One cannot compress from 0.1 to 0.3 directly and therefore what we are doing is reduction which is happening in value of y and Wc by m dot because of the modifications of the Linde dual pressure system. For a given value of mass ratio i the y and W by m dot of the system decreases with the increase in the intermediate pressure. This is just what we saw in the tutorial answer sheet whereas you go on increasing the intermediate pressure the value of y and W by m dot decreases. For a value of mass ratio of i W by m dot f passes through a minima as the intermediate pressure increases. So this is the minima which we want to attack basically at what pressure at what intermediate pressure one gets minimum work done per unit mass of gas which is liquefied. We want to attack that minima which will ensure that you are doing minimum work of compressor to get the work of liquefaction per unit mass of gas which is going to be liquefied. On the other hand for a mass ratio of i the figure of merit passes through a maxima with the increase in the intermediate pressure. So when you attain minima over here at the same particular pressure we attain maxima for the figure of merit. The operating point of the system is a compromise value between y and W upon m dot f. As you know as you go on increasing the intermediate pressure the value of y decreases but the value of W by m dot f will fall through a minima. So one has to take an optimum value between two between these two. In fact one should design to get W by m dot f minimum corresponding to that your desired y value should match. So whatever my desired value of y is I should get W by m dot f minimum. It is important to note that the figure of merit reaches a maxima value at the same intermediate pressure at which W by m dot f reaches a minima for a given value of y. Thank you very much.