 So, azeotropic distillation, what is the difference in extractive distillation and azeotropic distillation? A basic difference, of course, there are many definitions, you see one book, we will have one definition and all that. But the simplest definition that we use here is when you have mixture AB which is difficult to separate. When I say it is difficult to separate means either there is a formation of azeotrop between A and B or they are very close boiling, like acetic acid water, very close boiling. In that case, I add the third component C is external compound, in extractive distillation I will do the same thing. But there this C does not form azeotrop with A or B, remember there are same, there is no azeotrop between C and A, there is no azeotrop between C and B. In this case, the difference is this C is an azeotropic agent, which forms azeotrop with either A or B or both and makes the separation easier, so that is the difference. In both the cases I am adding external component, azeotropic distillation, this external component forms an azeotrop with either A or B or both A and B. For example, ethanol water, if I add benzene, you have seen that, you add benzene, benzene forms azeotrop with ethanol, benzene forms azeotrop with water, benzene forms azeotrop with both of them, that means there is a ternary azeotrop as well and we are going to see that case now, toluene same thing, cyclohexane same thing, right. If we add ethylene glycol, it does not form azeotrop, so the same mixture ethanol water, I can separate it by extractive distillation, I can separate it by azeotropic distillation. When I use ethylene glycol as external component, which does not form azeotrop, I call it as extractive distillation and if I add benzene, cyclohexane or toluene as external component and just because these compounds form azeotrop with ethanol, benzene and both rather, the three azeotropes form, right, we call it as azeotropic distillation. Now, this is very important, in most of the cases, the azeotrop is heterogeneous, ok. Now, we are going to talk about heterogeneous azeotrop, ok. Now, what is heterogeneous azeotrop? Phase splitting, ok. You have phase splitting taking place, y is equal to x anyway, the vapours when they condense at that composition, they will form liquid. In homogeneous azeotrop, you have single liquid phase, in heterogeneous azeotrop, you have two liquid phases, right. Now, when I say y is equal to x, that means vapour composition is equal to liquid composition. In the case, when you have two liquid phases, this liquid composition is the overall liquid composition and not the composition of any of the phase, any individual phase, ok. Always remember that, overall liquid composition, ok. So, y is equal to x. So, that is your heterogeneous azeotrop and in azeotropic distillation, right, when I add x-ray component, it forms azeotrop and that azeotrop has to be heterogeneous. And y, of course, the answer is given here. See, a forming azeotrop, the azeotrop has to be broken later, right. If benzene is forming azeotrop with ethanol or water, what will happen? I can separate that azeotrop and make the separation of ethanol and water easier, right. See, A and B, ethanol and water, they are forming azeotrop. I want to break this azeotrop by adding another component which forms azeotrop, ok, right, ok. So, fine. I will break this azeotrop and I will separate the azeotrop formed by this component. Now, this separated azeotrop again has to be broken, right. Otherwise, how will I recycle the external component or normally it is called as entrainer in azeotropic distillation. I need to recycle that. So, that azeotrop, I cannot break it otherwise. So, I am solving one problem, but getting into another problem, right, ok. If it is homogeneous azeotrop, but if it is heterogeneous azeotrop, ok, separation is not a problem, because you have phase splitting taking place. I can separate one layer from another layer. Now, X is your overall composition, but when I separate these layers, the compositions are different, ok. And I can exploit this behavior of heterogeneous azeotrop, ok, to get rid of this problem or get rid of this limitation, right. So, most of the times, in fact, in every case that is practicing industry of azeotropic distillation, you have the, you have the azeotrop heterogeneous and not homogeneous, ok. Because practically it is always easy to break this azeotrop by just phase splitting, ok, right. Heterogeneous azeotrop, if you separate the layers, two layers, you have broken the azeotrop, right. That is right, that is right, yeah. Now, can we use azeotropic distillation for binary systems, ok. What is the purpose? Now, for example, butanol water, ok, or butyl acidate water. It is a binary system, forms azeotrop, heterogeneous azeotrop, ok. Now, if you see Y versus X diagram, binary system. So, the capital again Y versus X. What is this diagram? Heterogeneous azeotrop, heterogeneous azeotrop. What is the difference between this and homogeneous? See, homogeneous is this. See the difference? This is straight line there. What does it mean? Suppose, you are in this range, ok, phase splitting takes place, ok. And I get two layers with these two compositions, right. I get two layers with these two compositions. The overall composition is here. That is azeotropic composition, ok, right. Now, I have a field X F here, butanol water. I am plotting it for water, which is more volatile, right. I have this field. Now, I am in the homogeneous region. There is no phase splitting. Phase splitting is here in this region, right. So, I have homogeneous single phase mixture of butanol and water. I want to separate butanol from water. It may be less, say 5 percent, 2 percent, ok, but I want to separate it. I want to get both these compounds in pure form, ok. So, I can exploit this behavior. I just give this to a distillation column, right. This mixture, X F, what will you get at a bottom? I will just look for a stable point, right. What do you get? Pure butanol, right. What will you get at the top? Water? Azeotrope. Azeotrope. So, you are somewhere here in this region, right. If you have sufficient number of stages, you can go up to this point. When the vapors they condense, you have 2 liquid phase, you have decanter there and a 2 layers, ok. You have 2 layers with compositions given by, compositions given by these 2 points, right. So, even if there is azeotrope, now even if there is azeotrope, I am able to cross. I have a separate layer now, ok. I have 2 layers. If I just take the aqueous layer out, if I take the aqueous layer out, this is my aqueous layer, what is the composition? This point, right. So, this A and this is A, right. So, I have gone from this region to this region now. So, phase splitting helps you to cross the boundary, which was otherwise not possible for homogeneous systems. I hope it is clear. See, I was here by doing distillation, by doing distillation, I go up to this point, right, providing number of sufficient number of stages. I go up to this point, right. And in this region, since I am in this particular region, phase splitting takes place and I have 2 strains with compositions given by this point and this point. So, I have phase splitting taking place, right. I have this composition A, this is the one, and this composition is, I will call it as B, B, right, ok. So, I am able to cross the boundary, which was otherwise not possible when you had homogeneous azeotrope. Homogeneous azeotrope or in this case, you get stuck up here. There is no phase splitting, right. So, this phase splitting, it comes because of non-ideality, ok. But I am exploiting this non-ideality to get a separation. Now, what happens later? Now, this A, this stream, now I have crossed the region. I am interested in pure water now, already got pure butanol, right. So, I put it to another column. Now, you are here, this goes to the feed, right. This goes to the feed. You will get this point as top or bottom, bottom, yeah, because it is below the diagonal. So, you will get this point but what is this water? What is the top composition? Somewhere here, azeotrope, ok. So, I can just put it to the same decanter, right. So, effectively what am I doing? I have feed going in, I am separating butanol and water, and this can go as a reflux, very interesting system, ok. I need reflux for both the columns. So, I can reflux the organic phase to the column. Oh, A means A is this point actually, ok. That means, it is a binary mixture. I am just denoting this because we should be able to correlate. I am not saying A is the pure component, A is the name of that point, ok, the composition, ok. So, it is a binary mixture. Similarly, B is also a binary mixture, ok. So, this is a binary mixture. So, it has water in it, it has butanol in it. And this goes to the column and you remove water from the bottom and again some azeotrope, ok, a close to azeotropic mixture comes out. I will just put it again to the same decanter, ok. So, I have a complete system where I am giving feed and is removing butanol and water in pure form. So, I am using, it is a very simple system actually, but I am using the heterogeneous azeotrope, ok, right, to get this separation. If there is no heterogeneous azeotrope, separation was not possible at all, right, because I get stuck up here, homogeneous azeotrope. See the difference? You have butanol water system. This is formation of heterogeneous azeotrope. If the azeotrope was not heterogeneous, first column itself, though I would be able to separate butanol, at the top you would get this azeotrope, right. So, that separation you get stuck up there. But just because there is a phase splitting taking place, I exploit that phase splitting to go from one region to another region by just doing decanting there, ok, right, and I can separate pure water because I have crossed the region now, ok. So, this is a simple system azeotrope. Now, we are going to see a very popular and at the same time a complex system, ok, again the same thing ethanol water using benzene as entrainer. The point that I wanted to make as far as this system was concerned that binary system was concerned is that the phase splitting helps you to cross the region, ok, or you cross the boundary, ok, go from one region to another region. So, that is what I wanted to tell you from that example. A same principle is used in more complicated systems like terminal systems, ethanol water benzene system to cross the region and we will see that. Before we go ahead, let us see what happens to the RCM when we have two liquid phases formed, ok. Now, this is your triangular diagram, ok. This is your triangular diagram. Now, in this case phase splitting is taking place. Now, look at this red envelope. You must have seen such diagrams in your liquid-liquid extraction, ok, the phase splitting. What does it mean? It means that if you are in this particular region, red part, ok, in this region, there is phase splitting taking place and you have this tie lines, right, tie lines giving the compositions of individual phases, right. This is where your phase splitting taking place. Suppose you are in this region, you have homogeneous system, right. Your single liquid phase here, ok, and you have two liquid phases here. This is the immiscible region, rather, ok. Now, example is again benzene here, benzene, ethanol and water, ok. Now, benzene and water, benzene and water are immiscible. This is very small solubility, ok, very small solubility of benzene in water and water in benzene or other way around, rather, ok, water in benzene and benzene in water, right. But in this region, there is a phase splitting taking place. Now, residue curve map, what will happen to the residue curve map? If you are in the homogeneous region, it is one and the same, like, it is like what we had before. But if you are here, you can derive. Now, we are not going to spend time in derivation, but you are going to see that instead of x, now I have x overall, ok, x overall. That means, the overall liquid phase composition, right. Earlier you had dx by zeta or yeah, it should be d zeta here, is equal to x minus y, right. Now, I have dx o by d zeta equal to x minus x 0 or x o minus y, ok. So, it behaves in the same way. I do not have to worry about two liquid, presence of two liquid phases when I am in this particular region. Here, it is overall composition. I should treat it as overall composition, whereas, anyway in homogeneous, it is going to be overall composition. So, you have continuity in residue curves, ok. If there is a phase split, does not mean that there is split in residue curve. The moment I go into phase region, x becomes x overall, ok. You do not have to worry about the presence of two liquid phases. Though, individual phases are present, ok, we do not need to consider that. You can derive that, ok. Now, you have two liquid phases, you boil it, right, component balance for both the phases, ok, and you will get this particular equation, ok. So, I do not have to worry about whatever rules we have developed or rather come out with based on the residue curve maps are going to be same here as well. So, boundary will get extended in the two phase region and off, ok, right. So, just read this note. The comparison with homogeneous systems indicates that the curves behave in the same manner as that of homogeneous system, because the separation is same, ok. Only difference is instead of x, you have x 0, right. Only difference is that in the two liquid phase region, one should plot overall composition instead of the composition of individual phases, right. So, I am not looking at individual phase. Suppose I am here, actual individual phase composition would be given by the tie line, right, but I am not plotting that as per as residue curve is concerned. So, again the problem I am defining here, separation of ethanol, water, azeotrop. Now, I have ethanol, water, azeotrop, ok. We have seen that experiment as well, ok. Typically fermentation growth would give you 10 percent solution of ethanol, dilute solution. I get I remove water from it and I get azeotrop from that. Now, this azeotrop is difficult to break, ok. I am going to use azeotropic distillation to break this azeotrop. I am going to use benzene or cyclohexane as all toluene as entrainer, ok. So, all of them from azeotrop, ok, which is heterogeneous binary azeotrop with water, ok. All of them means benzene, cyclohexane and toluene, ok. And a ternary azeotrop is very important, ok, which is again heterogeneous. And these both are minimum boiling azeotrop. Now, let us draw a residue curve map for this system, ok. This is the data given to you, ok. Benzene, boiling point, water 100, ethanol 78 near this binary azeotrop form, ok. Benzene water 70, then ethanol benzene 67, ethanol water 78 and is a ternary azeotrop, ok, right. So, how many nodes you have? How many nodes you have? In this case, how many nodes? 7. 7, right, because you have 4 azeotrops and 3 pure components. Now, I have to know which of these are stable points, which of these are unstable points and which of these are saddles, right. And water benzene binary is heterogeneous and even this ternary is also heterogeneous, ok. So, can you draw RCM? Benzene, alcohol and water, there is one azeotrop here, there is one azeotrop here, there is one azeotrop here, there is one azeotrop here. How many total nodes? Total number of nodes is 4 plus 3, 7, ok. Let us write down the boiling points, can you tell me the boiling points? Benzene 80, yeah, then alcohol 78, yeah, I am just changing my convention now, because normally I write the most volatile component here, but does not matter, ok, like, now water 100, right. Now, these azeotrops, alcohol benzene, alcohol benzene AB or BA? 67. 67, yeah, water alcohol 78 point something, right, should be less than this, less than alcohol, alcohol is what, 78 point, pure alcohol? 78 point 3. And this is 78, right? 78 point 06. Oh, I see, very close. Then, what are benzene? 70 point 5 0. 70 point 5 0 and ternary is 66, right? 66. Now, as I said, these two are heterogeneous, these two are heterogeneous, that means the liquid envelope will be, right, because these two points would come in the phase splitting zone, right, zone in which you have two liquid phases, these two points, right. Can you draw RCM? Do not worry about phase splitting. As I said before, RCM just ignore phase splitting thing and you draw RCM. First step, draw arrows, ok, binary edges, all are minimum boiling. So, arrows will go away from those points, ok. Now, I will join these 66, 67, so arrow will be in this direction. 66, 78, arrow will be in this direction. 66, 70, arrow will be in this direction, ok. How many pairs? Unstable nodes, how many unstable nodes? 1 is only 1 and unstable node. That is at ternary as your draw, ok. How many stable? 3. 3. So, how many pairs? 4. 3. 3 only, yeah. So, 3 pairs and 3 zones. So, I can see these 3 zones. This is 1, this is 2 and this is 3. So, 1, 2 and 3. 3 zones or 3 regions and these are the boundaries, ok. Now, I need to use this RCM to separate or break this azeotrop. This is my feed, ethanol water azeotrop. That is my feed, ok. Can I draw RCMs now? I will just follow this, ok. Just look at what I am doing. Just follow this arrow. You go up to this point, then here, here, here. I am ignoring the presence of liquid-liquid split. This is my RCM, ok. Now, before I go for a sequence, see I have this as a feed. I have this as my feed and I want to separate alcohol and water in pure form, right. Alcohol and water in pure form. So, in this system, you have this as a feed, right. And I am interested in alcohol and water. And alcohol and water, both are stable loads. So, whatever sequence that I am going to come with, ok, my both products are going to come as bottom products, ok. So, this is something that you should learn just by looking at RCM, ok. You do not need to do the complete exercise of synthesizing the sequence and then realize it, ok. These are some tips, ok. This is how you read RCM, right. So, you are going to get both ethanol and water in pure form as bottom products, ok. Now, this is your feed. Now, I am going to design or synthesize a column sequence, right, that would give me pure ethanol and pure water, ok. I am going to add benzene. The benzene is going to go into the system, ok. And it will be recycled inside because you know external addition of benzene if there are no losses, right. So, I have a feed going in ethanol water. I have a column. See, you have a choice, ok. Of course, you have a prefractionator first where you have 10 percent ethanol. I think I have in the nodes I have shown you, ok. You have one column prefractionator, ok. I am not showing that here. I am directly starting with this as a feed, ok. That is your isotropic mixture, ok, right. So, with this as a feed, now I want to take bottom product in the first column, right. You have a choice. You can either go for ethanol or water, ok, right. In this case, so was that it is ethanol, ok. You remove ethanol from the bottom, right. Ethanol is a stable point, ok. Water is a stable point. In the first column, I remove ethanol from the bottom, right. So, the top composition will be what? This is coming as the bottom product. Now, you are in this region, ok. Your top composition, the vapor composition is not going to be here, not going to be here. It will be in that region only, right, because I am not able to cross the boundary, right. I am not able to cross the boundary. So, the top composition here is going to be in this region, ok, in the upper region, right. Now, you should design a column. The number of stages should be such that my top composition should be here or here, right, below, here, here, why? See, if you are here, then you are not exploiting phase split. I want to cross the boundary later. See, your water is sitting in one region, ethanol is sitting in another region, right. I want to cross the boundary, ok. So, I should exploit the liquid-liquid split. So, the column number of stages should be such that I am in this region. The top composition should be in this region, right. So, this is your top composition. This is your vapor composition here, right. Then, what do you have? You have a decanter, right. You have a decanter. So, suppose this is your vapor composition. The decanter would be giving you the phase split and the two layers with these two compositions, right, ok. One layer would be benzene rich, this is benzene. One layer would be water rich, right. So, this is your water rich phase and this is your benzene rich phase. So, benzene rich phase, I should recycle it back because benzene is added as an entrainer, ok. So, I will put it back to the column. Now, you have got this bottom layer. This is your bottom layer, ok. Benzene rich, sorry, water rich. I will take it to another column and here, ok. What is the stable load in this zone? Water. So, I will take water out from the bottom, right. It is not yet complete, ok. What will you get at the top? Will you get at the top? See, that is why I told you, do not forget that lever rule. I have identified two compositions, ok, water. This is my feed. The top composition would be on the line joining these two, ok. I will not be able to cross this. I will not go here. So, my top composition would be somewhere here on this line. Let us say this, ok, right. So, this is my top composition. This is the composition I have here, right. I should connect that stream somewhere, you know. I just cannot leave it like that. It should be recycled, ok. See, I have got what I wanted, right. I have got ethanol, pure ethanol. I have got pure water, but I have just leave it like this. This column also needs reflux, right. At the same time, if it just goes out, then that is loss. I will get one stream with this composition, with all the three components present. What will I do with that? It should be recycled. So, I can give it back to either decanter or I can just give it to here, because decanter is no meaning because it is not going to split, because this is homogeneous, right. I will just give it back to the column, ok. And then you can check the material balance lines, ok. There are many material balances that you can perform in this, ok. For example, is this clear? See, how I have used the phase split to go from one region to another region? Even if benzene forms as you have took with this and you have so many regions possible, just because of phase split, I was able to separate water and ethanol in pure form, right. You can do it overall material balance. Now, for this particular distillation column, say column D2 and this is D1. For D2, this is my feed. This is my top product. Of course, you may have reflux here for this. This is my top product. This is my feed and this is my bottom, right. Are they on the straight line? Water, feed and a top product. They are on the straight line, fine. The limit of column 1 which is slightly more complicated, ok, right. What is the resultant feed to the column? Now, when I do a material balance, I have to identify the boundary, ok. What is that boundary? See, this is that boundary. Which are the streams going in? The two streams going in, ok. The two streams coming out. There is no other stream crossing that boundary. There are four streams crossing that boundary, two going in, two going out, ok, right. Now, I want to see whether the material balance is satisfied or not. Leave a rule. Resultant feed. What is the resultant feed? Which will be the combination of these two. Now, which are these two? Ethanol, water is this, right and this particular stream, this, right. So, your resultant would be on the line joining these two points, right. That is your resultant feed to column 1, ok, D1. That is the resultant feed going in to the boundary and these two are the products. Pure ethanol is sitting here. This is my feed, right. This is my resultant feed. Pure ethanol, sorry, pure ethanol and this stream. Where is this composition? Where is it? Here, right, ok. So, this point, this point and resultant feed should be on the same line, ok. Resultant feed and top and bottom compositions should be on the same line. Are these on the same line? Ethanol, this particular stream and the resultant feed. Ethanol, this particular stream and the resultant feed. Azeotropic distillation. Now, instead of benzene, if I take cyclohexane, it will have a similar nature. The RCM will be quite similar, ok. So, I can say, I can use cyclohexane. If I use some others, I want and give some different RCM, then I have to do this exercise again. It is not that this is the only RCM that is used for azeotropic distillation. There are many other possibilities, ok, alright. So, this is what we have seen, ok. The three regions form and this is the configuration, ok. Of course, I have shown you three columns here. First column is a simple pre-fractionator, ok. And from that, I get a azeotrope. This is something that I do as azeotropic distillation, sorry, yeah, it will contain benzene. So, it is this particular stream, which is used as a recycle, ok. And P is this stream. So, both are given back to the column. So, benzene is going back to the column. So, in this, there is no fresh benzene required if there is no loss of benzene, right. It is totally recycled through these two streams, P and Q, right. So, in a continuous system at steady state, I do not need extra benzene if there is no loss of benzene through these. If there is a loss of benzene through ethanol and water, then I need some make up benzene, ok. Now, as far as phase plate is concerned, do you have a phase plate in the column? In this case, look at the profiles. I am choosing this point in such a way that this part of the phase plate or two phase region is very small. So, at the most, you will have phase plate in the upper portion of the column, first two stages, right. And this is very important. Normally, operational point of view, column is always good to have single liquid phase in the column. If there are two liquid phases, there is so many problems handling or operating the column, ok. So, that is why I choose this point in such a way that it is very close to the boundary. When I am saying boundary, it is a liquid-liquid boundary, not our RCM boundary. It is very close to it. So, that at the most, we will have phase plating in top one or two stages. If you can avoid that also, it is better. So, the column is operated in homogeneous mode, that means in single liquid phase, right. Whereas, phase plating takes place outside the column. A very wise or rather judicious way of operating the column, ok. I am using phase plate, I am exploiting the phase plate, but not allowing that to happen inside a column, right. That is a suggestion, ok. See, Z should be chosen such that there is no phase plating inside a column. Why? Because it is difficult to operate and control a column with phase plating taking place on stages, ok. It has some practical issues as well. Now, this is another system that I have shown here, which is a very specific example. I would like you to rather just have a look at it and try it out yourself, instead of we spending time on this. It is just additional system, ok. I will just explain what exactly is we are doing here. The purpose is to separate acetic acid water mixture by azeotropic distillation. It is a very famous example, popular example, commercial example. We will do that in industry. Is there any azeotrope between these two? There is no azeotrope. Unlike ethanol water or unlike ethanol water, ok, there is no azeotrope between acetic acid and water. Then why do you need azeotropic distillation? You need azeotropic distillation because it is a close boiling mixture, ok. It is a close boiling mixture, right. If you do normal distillation for acetic acid water, distillation will take place. I can get pure acetic acid, I can get pure water, but you will need large number of stages, you will need large reflux ratio. It is an expensive affair, right. Whereas, if you use some entrainer, use some entrainer, let us say butyl acetate, ok. Then it will form azeotrope with water, which is heterogeneous azeotrope and makes the separation easier, makes the separation easier. Then need less number of stages. Energy cost will go up because reflux ratio comes down, right, ok. And that is a purpose. So, if you see on the RCM, your feed is here, but it is not azeotrope. It is not azeotrope. It is dilute acetic acid mixture, ok. This is not azeotrope. So, if you just look at this binary, I can separate acetic acid in water in pure form. It is not a problem, ok. But I do not want to do that because it is expensive. So, I add butyl acetate. Butyl acetate forms a phase split region with water and there is azeotrope with water as well. So, this is one azeotrope sitting here, ok. There are no other azeotropes in this system, right. Can we use this RCM to generate or synthesize a column sequence? So, I have shown it here clearly, ok. You just go through that and if you have any questions, you can ask me today or tomorrow. So, I have shown it all the points here, the Q corresponds to this, T corresponds to this. And I have used all the rules that we have looked at so far, RCM, ok. We should not cause the boundary, right. So, once we are very clear about this methodology or feasibility issues, then there is no problem. Once you have VLE, it is easy to rather design sequence or synthesize a sequence for azeotropic distillation or extractive distillation, ok. Of course, whatever we have learned is just the beginning, like you will have to really spend more time, read it on your own and do some exercises given in the book and probably whatever I am telling you here, ok. To get mastery over this particular technique, ok, then only you would be able to extend it to the systems of your interest, ok. What is the purpose doing all this? One can say always that, ok, I will just do experimental laboratory, I will do some simulations and find out, ok. Why should I learn this residue curve maps and all? Of course, for teachers, it is good, ok. Academic purpose, like students will also enjoy drawing residue curve maps and all that. But then in industry, is it really useful, ok. See, if you are dealing with ideal systems, ok, it is not at all useful, ok. If you are dealing with ideals, like example FCC product or even refinery, refinery know where this is useful, ok. Take it from me, no need to learn RCM at all, right. Whenever you are dealing with ideal system or non-azeotropic systems, RCM is not required. This is for azeotropic systems, ok. So, most of the time in fine chemical industry, ok, right or even for that matter chemical industry where you are talking about azeotro formation, close boiling mixtures, ok, where you are doing some chemical reactions, ok. You have polar non-polar components mixture and all and you are talking about separation will come across such systems, ok. And there you will need, at least you need to know the thermodynamic behavior of the system. And there is no other way to visualize this behavior other than the residue curve map technique for multi-component systems. So, whatever we have learned, ok, for binary system macabre thing, ok, it improves our knowledge, ok, it gives us insight into distillation, but that is restricted to binary system, ok. And if you want to extend it to multi-component system, this is the way, ok. Now, what we have done here is just looked at thermo systems, ok. Now, we may have quaternary systems, we may have five-component systems, right. It is difficult to visualize, ok, such systems, ok. You have mixtures with many components and all that. But what people have done, ok, those who are involved in this research, ok, they have as I told you in the very first lecture on conceptual design, they have come up with certain rules, some mathematical tools, ok, which are based on thermo systems visualization, ok. And they have extended it to even the multi-component system. And they have made programs and their software is available, ok, to come up with the sequences like what we have done, ok. So, these softwares, they make use of residue curve maps, ok. And then even for multi-component systems, you have a residue curve map, ok. Even if I cannot visualize it, yeah, I have it in mathematical form, ok, a residue curve, ok. How many stable points? You may have 15 stable points, you may have 10 saddles, possible depending on how many components you have, ok. So, there is a mathematical representation of residue curve map. This is a graphical representation, right. So, there are softwares, ok, which are based on such technique, ok. And aspen split, ok, aspen has a software called or a module called aspen split, which makes use of this particular technique, ok. For ternary and cortanyl systems, it would be very easy to visualize, ok. And you can come across such systems many times in industry. And you can just look at a residue curve and see whether the separation is possible or not. And you can take a decision there itself, ok. So, you can draw a residue curve map in aspen, ok. You do not have to solve those equations yourself. There is a software called distil, of course, aspen itself has something called as distil, but that distil is different, ok. I am talking about another software, a separate software called distil, ok, which makes use of the conceptual design concepts, ok. And of course, this is not the end of it, because this gives input to your simulation, right. So, whatever reserves I am getting from this, ok, I have synthesized the column sequence. I have got number of stages for every column. I have got reflux ratio for every column, right. But then that is approximate based on what the constant molar overflow assumption, right. And that is going to go as input to your simulator. That is where you will do some fine tuning and after that you will do optimization, ok. So, the complete exercise of designing, overall design, I am not talking about conceptual design, complete overall design would include all these steps. So, this is just one step, ok, that goes as of course, the outcome of this goes as input to the simulator, ok. Otherwise, there are ways to deal with the simulator. People go by trial and error, give some number of stages, give some feed flow rate, feed composition, give reflux ratio and see whether what you are getting at top and bottom is matching with your requirement or what is desired, ok. Sometimes it may match, ok, and you are lucky in that case. But of course, in that case, since you do not have understanding, simulator is like a black box, it may be, may not be an optimal design, ok. You can still improve that design, ok. For that, this understanding is necessary, ok. So, if you are working in refinery, if you are working in ideal systems, forget this, but then if you are working with non-ideal systems, ok, this is very important, ok, especially the azeotropic systems. And of course, when you have tangent pinches and all, we have not covered that part, ok, that is also very important. Sometimes you may have tangent pinches, ok. Multicomponent system also you may realize, ok, how to deal with them and all. Again, one more exercise and if you go through the book, they have talked about it, ok.