 Let us continue with the topic of modeling and simulation of distillation systems and this is the concept of a very generalized stage which we had seen earlier. Let me just go through the important aspects once again. This is stage J within the column and what we have shown here is that the stage can receive feed which is characterized by these variables and it can be at a pressure equal or higher than the pressure existing at stage J. Provision has been made to take care of the static head which will be there for the liquid which is on the tray. Provision is also made for the resistance which the vapor will encounter when it bubbles through and there will be a pressure loss. Provision has also been made to allow heat transfer on this tray in either direction. Then provision has been made to withdraw a vapor stream as one of the products, side products have a liquid stream and another side product. I had also mentioned that the pressures will remain as specified variables because in the model which we are going to see now, we are not going to bring in the momentum balance. So, we are going to write mass balances, energy balance, equilibrium relations and other relations if required, but momentum balance certainly is not in picture and therefore, pressure has to be a pre-specified variable. I also mentioned that we are interested in finding the vapor flow rates leaving every tray, the liquid flow rates leaving every tray, the compositions of vapor streams leaving every tray, the compositions of liquid streams leaving every tray of course, they are enthalpies which will be known once the temperatures are also known. So, we are interested in a vapor profile, a liquid profile, a liquid composition profile, a vapor composition profile and the temperature profile. So, these are the profiles which we need to generate through the regress model. So, we start with modeling and we write down what are popularly known as mesh equations and the first component of mesh equation which is designated as M and this is written in a residual form is the material balance for each component. So, these are component material balances. So, when I say M ij, it simply means that I am working on the jth tray and the component balance is written for ith component. I will go from 1 to number of components which are there in the chemical system which you are handling through distillation. Now, this is very easy to work with, lj minus 1 multiplied by x ij minus 1 represents the moles of component i which are received on tray j from the tray above because the numbering is from the top. Similarly, vj plus 1 multiplied by y ij plus 1 is moles of ith component which are received on tray number j from the tray below. This is coming from the vapor. So, this is the sum of the two. So, this is component i which is received through liquid and vapor which are internal to the column. In addition, we have made provision for feed to the column. So, moles of i may come in through the feed to the column which is on the jth stage. So, first three terms represent all the inputs. What leaves the column is written here? Now, we can combine lj and uj together because the side draw which is uj has exactly the same composition of that of lj. So, lj times xij is what is coming to the next tray and this tray is j plus 1 and uj xij represents what has been withdrawn as a side product. Similarly, this is the vapor which is going to the j minus 1th stage and its composition being y. So, ith component leaving the tray through the vapor stream to the next tray next stage and w multiplied by y, this represents the ith component which has left the column through the side draw. So, in total in minus total out and that should be 0 if we are at the solution. So, mij will be called the residual and we would like these residuals to go to 0. If you are not at solution, if you do not have the right values of the profiles or the compositions, then mij will be either a positive or a negative number. It will not be 0 which means balance is not satisfied. So, on the jth stage, how many such equations do you have? Well, it is written here c equations for each stage. If c capital C represents the number of components in the system, so if I am working with a 10 component system, then I have 10 such equations. So, these are known as m equations. Easy to remember because m stands for material balance, but we have to remember that it is component material balance. The next set of equations is what are known as e equations and e here stands for equilibrium. So, we are working with stages theoretical stages and again these equations are written in the residual form. So, yij for every component now we are talking about. So, yij is the mole fraction of ith component on the jth stage and this is leaving, this is leaving the vapor composition that is that should be equal to the kij the equilibrium constant for that particular component on the jth stage multiplied by the liquid composition and this is the leaving liquid. So, xij or y minus kx equal to residual. So, eij again is residual and for every component this equation can be written on the jth stage and hence for the jth stage we again have c equations. For every stage we have c equations. So, we have c of m equations, c of e equations. So, total so far we have 2 c equations. Now, we have taken it for granted that these x's and y's on the jth stage they have a physical significance and that of mole fractions because we use them as mole fractions in the component material balances. If they were to be mole fraction then a required condition is that on every tray on every jth stage all the x's they should sum to 1 because the numerical solver does not know that these have the physical significance of mole fraction. For numerical solver these are just variables and therefore, for every stage we need to put two more conditions which are known as summation equations or s equations. So, mole fraction summations where we have to say that sigma y the summation is over i we are sitting on the jth tray the summation is over i. So, sigma y i should be equal to 1 or in residual form sigma y minus 1 equal to 0. Similarly, sigma x i minus 1 equal to 0. So, again this is the residual of sy this is the residual of sx. So, we have now 2 c plus 2 equations 2 c plus 2 equations. Then the last element is the h equations h equations are representing the energy balance normally when we talk about energy balance we should have taken into account all types of energy which are possible, but it turns out that even for tall columns if you try calculating the potential energy contribution of the species sitting on the top tray let us say and compared that with the enthalpy at the temperature and pressure conditions which that element may have the species may have the potential energy is negligible. Kinetic energy terms are anyway very small because you do not have that large velocities and they are negligible as compared to the enthalpies and therefore, the energy balance in this particular case can be very nicely approximated by an enthalpy balance. So, we are saying that kinetic and potential energy terms are ignored. So, write down the enthalpy balance now enthalpy balance each tray has just one balance. So, what we have to say here this is the let us say the sensible heat content which is coming through the liquid which is falling on the jth tray because this is coming from J minus 1. So, the liquid flow multiplied by the molar enthalpy. So, this is the energy which is coming from J minus 1th stage, this is the energy which is coming from the tray below through the vapour. So, J plus 1th stage this is the flow of enthalpy which is coming from the fresh feed. So, this is the total amount of energy which is received on the jth tray. What leaves the tray in terms of enthalpy content is simply this is the sum of the liquid leaving and the side draw multiplied by the liquid enthalpy, this is the sum of the vapour leaving and the side draw multiplied by the vapour enthalpy and we have made provision for the heat transfer of course, this sign decides whether it is leaving or being added. So, that is where the convention of plus minus comes into picture. So, this is your energy balance, I really has no meaning because there is only one balance and the balance is on jth stage. So, this I index you should ignore it should be H j. All right. So, how many equations do we have now? We have 2 C plus 3 2 C plus 3 C we got from component material balances, another C we got from equilibrium relations, additional 2 we got from the summation equations and one equation we got from the energy balance. So, 2 C plus 3 equations are there and they are all written in residual form. Before we go further let us try to understand that whether it is enthalpy in this particular case let us say if I focus attention on the liquid enthalpy, we know that the liquid enthalpy ultimately is going to be a function of the temperature which prevails on the tray, the pressure which is there on the tray and the liquid composition which is on the tray. Similarly, the vapor enthalpy will be a function of the temperature, the pressure and the vapor composition. So, temperature is one of our desired variables which we would like to see in our profile, it is a variable to be computed. Then the liquid composition here or the vapor composition here, they are also our desired variables that is right. If there is no feed on that particular tray, then this F j will be 0 and if this F j is 0, there is no contribution from this term. Q j is the amount of heat transfer done on that particular tray. If you have a coil, you may remove energy or you may add energy through a heating or cooling coil. Rebooters need you minus, if it is a condensed side it will pass. Yeah, I had given the convention earlier. So, we are going by that convention if you see, in the model we had written this down. Here plus if from stage and minus if to stage, it is just a convention. If you. Yes, in the conventional column, only tray number 1 and the tray number n will have heat transfer in the conventional column. But if I have a complex column, I had shown you some configurations of complex columns, I can have heat removal on other trays, I can have heat addition also other trays. So, I am making provision for heat transfer on every tray. It is just a convention, I mean you can reverse the. Out. Out. Yeah. Is there removing heat energy? Right. So, those things are negative. You mean it has been added on the wrong side? We can check that. I am saying yes. Yeah. All out. Right. They have been shown as negative in this enthalpy balance. Right. You can do that. It is just a convention as I said. I have no objection to that. But the numbers will change. No, no. The only thing is depending upon this convention, depending upon this convention, then you have to properly put the sign here. This Q j, if you change the convention, it will get added here rather than getting subtracted here. That is the only difference. Numerically, there is no difference except that the sign has to be properly accounted for. Net change in energy in the particular plane? H j. H j is the total residual of the enthalpies. It is the residual. It is the error in the kilocalories which exist on that trace. So, there are certain kilocalories which are flowing as sensible. Accumulation. Pardon? Accumulation. You can say that yes. You can say accumulation. So, I have certain kilocalories flowing into the tray per hour. I have certain kilocalories flowing out of tray and to that I have to add the energy which is added externally or removed externally. So, the unit of Q j will be again kilocalorie per hour. I agree with you that the sign convention can be reversed. Absolutely no problem. In fact, many a times in chemical engineering, we say that heat added to the system is positive and heat taken out of the system is negative. So, that is right. They are adding into the tray. That is right. So, what we could do is that minus means if it is two stage. So, rather than me writing this particular term towards the end, I can write this three terms before. That is all. Do you understand? What I am saying here is it is negative if it is two stage. Two stage means it is added to the stage. So, when Q is negative, it is added to the stage. That is the convention. Now, if under that convention if I operate, the only thing I need to do is I need to pull this minus Q j from here and put it here. If Q is only the magnitude of the energy, the sign will make the difference. If you are careful about the sign itself, the convention, then you just put it at the appropriate place. So, what I am simply saying? If energy is added, you add to these three terms. So, fourth term will come. If energy is removed, you add to these terms and another term will come. That is all. Alright. So, we have two C plus three equations written down. And these equations, once they are solved, they will guarantee that the x's and y's, they will sum to one because two summation equations are part of this equation set. All other variables, all other variables for example, L, V or U, U of course, will be specified. But we are solving for L, V and intermediate variables like enthalpies, etc. They will change numerically. And we hope that variables such as L and V, they will turn out to be positive variables. They should be positive variables. They should be greater than 0. Either L can be 0 or a positive number. We do not expect negative L's. We do not expect negative V's. Alright. So, that kind of provision is not written in the model so far. And therefore, the solver has to be very carefully chosen so that we maintain the physics of the process. If the solver takes the solution in the direction where L's and V's are becoming negative, the physics of the process will be lost. Numerically L and V can be negative. But the movement we say L represents liquid flow rate, V represents vapor flow rate, then they cannot be negative. They have to be positive. So, the moment you attach physical significance to the variables, we have to be very careful. Alright. Now, stage number one in the model I had said is partial condenser because vapor was taken as a product. And stage number N represents the reboiler. So, I have total N stages, capital N stages. And I have 2 C plus 3 equations on every stage. And therefore, total number of equations which I have are N times 2 C plus 3. 2 C plus 3 within parenthesis multiplied by capital N. What I was trying to explain before the questions came that H is a function of temperature, pressure and liquid composition. In this particular case HL, liquid composition is one of the computed variables. We have to compute those variables. They are unknowns. They are dependent variables. And so is the case for the liquid flow rate. So, I have product of two unknowns. I have implicit enthalpy sitting here. So, I have a variable which depends upon composition which also appears in the same term in which there is another unknown which is sitting. And I can go on giving you examples like this if I travel from this end of the equation to the other end of the equation. The point I am trying to make is that in terms of unknowns which need to be solved for this particular system, the system of equations is a non-linear system. Necessarily a non-linear system because you have products of unknown sitting there L multiplied by x or V multiplied by y. And both are unknowns. So, it is a non-linear system. So, what I can say that this n times 2 c plus 3 equations which I got, it is a non-linear algebraic system. It is a non-linear algebraic system. And I need a good solver to solve this system. Of course, we have written balances for steady state and therefore, these equations are all algebraic equations. There are no derivatives here. These are algebraic equations. So, what this now represents is the cascade, the concept of the cascade. We have written equations for let us say jth stage which was this stage and then we extend this all the way from stage number 1 to stage number n. Now, the advantage of doing this is that once I have a solution methodology for this particular cascade which has given me n times 2 c plus 3 non-linear algebraic equations, I can configure any column. Configuration does not remain a constraint. If I were to configure the conventional column, the simple column with a partial condenser, the only thing I need to say is that all f's are 0 except one of the f's somewhere in between. Let us say if I had 20 trays and if the feed was coming on tray number 10, then stage 1 to stage 20, n is capital N is 20, this fj then will become j is equal to 10. And all other f's will be 0, f1 to f9 is 0, f11 to f20 is 0. Similarly, q1 will remain non-zero because there is a condenser, qn will remain non-zero which is a reboiler, but all other q's q2 all the way up to q19 they will be 0. There are no liquid side draws. So, all these side draws which we had written in terms of u's, they will be 0 and all the side draws for vapor they will also this is u and this is w. So, all the u's will be 0, all the w's will be 0. So, if you did that, the same model will reduce to a conventional column. The number of equations remain the same, only variables we are dropping specified. u1 will be this one. Pardon? u1 will be this one. Yeah, this one, yeah. Then u1 will be this one. Yeah, on the top tray there is nothing like a side draw. The term itself you know you have to interpret the term, the term says side draw. So, side draw starts from here, side draw is not on the top. And this is v2, the one which is coming here is v2. So, if you talk about u and w, so w will have w2. Similarly, this is l1. So, l1 corresponding is u1. You are not withdrawing anything, excuse me. Because inside whatever is getting condensed and returned from the vapor here that takes care of the reflux. Reflux is internal to the system. So, this is the strict side draw. Now, if you had a situation that you have a vapor product, vapor distillate and you also have a liquid distillate, then this anyway is there because vapor distillate is there. This will come as the liquid distillate. So, liquid distillate we are going to call as side draw taken from tray number 1. Is this clear? Alright. So, let us now see how we can configure various columns. Same model we will keep on applying and we should understand this process. Now, there are various operations which exist in industry and they all can be viewed through the window of distillation once you have this mesh model available. So, what is this I have shown here? I mean if you pardon this slight mistake here, you know like this particular liquid it should have been on the top tray or you can assume that this is the top tray and this is coming just below the bottom tray. So, what is this I have shown here? When do you encounter this scenario? There is no condenser, there is no reboiler. So, I may have a feed which is in liquid form which I am putting on the top of the column and I have a vapor which I am putting at the bottom and I am calling this vapor as the mass separating agent. The mass separating agent. So, yesterday I had shown you a diagram and we said mass separating agents. Steam could be one of the mass separating agents. So, now think about a problem where you have liquid coming in here and you have steam as a mass separating agent. It need not be steam, it can be any other compound also. So, what does this represent? Certainly this represents a steam stripper. This could be a hydrocarbon liquid and you are trying to strip out the lights out of this and the heavy will come and join this stream. So, this is a stripper. Typically this is what happens in a stripper. Is that right? Does this fit into our mesh system? The answer is yes because there is a feed on a tray number 1, there is a feed on tray number n and every tray allows a feed. All other intermediate streams and intermediate trays will not have any feed. There is a top vapor product, there is a bottom liquid product which we had. We have no side draws, no side draw liquid, no side draw vapor. We have no provision for any heat transfer on any of the trays. All Q's are identically 0. So, if I define that then the same mesh system will solve for stripper calculations. So, I do not need any other model. The same model will work for stripper calculations. Is this clear? The only thing one has to keep in mind is that thermodynamics may have to play a very different role, but then we have not focused on thermodynamics in this particular model formulation. We have focused only on the distillation configuration. What is the difference in thermodynamics we are going to have? Well, in a situation like this you may have supercritical compounds, light ends may be there and therefore, in addition to vapor liquid equilibrium you may also have the gas liquid equilibrium, but then there are models like we had discussed equation of state. So, we are at Leuphang or Peng Robinson. They handle VLE as well as GLE. So, absolutely no problem. If GLE feature is not there may be Henry's law can be brought in and the supercritical compounds can be taken care of. Small amount of non-condensibles can be handled if you bring in the Henry's law. So, it is the thermo which can become a critical component of this calculation, but as far as the structure of distillation is concerned the mesh system will solve it. Let us look at some other operations. So, the same column I have a feed vapor and I bring in a liquid mass separating agent and then the same unit is called absorber. I may have let us say air containing some useful hydrocarbon acetone or may be ammonia and I use water as a solvent to absorb it. So, then the solvent becomes the mass separating agent and it is in liquid form and this is your feed vapor. Why we are calling this as feed because the desired component resides in this that is what we are trying to remove. So, this is your absorber. Now, if we could model stripper using mesh system we can model absorber also using the mesh system, absolutely no problem. Identical configurations. Now, we go one step further. This is a piece of equipment it is a variant of a distillation column which is normally called a reboiled stripper. In stripper we were putting a mass separating agent from the bottom we said it could be steam and that was the source for vapor. Here the source for vapor is a reboiler. So, you are putting energy into the reboiler and you are generating vapor and liquid product you are withdrawing from the reboiler. So, it is a reboiled stripper rest of the concept is the same the liquid feed is coming on the top of the column and the vapor product is leaving from the top. Can I model this? The answer is yes. Again nothing has changed only there is one feed one top product one bottom product Q n is non-zero Q n is non-zero all other Q's are identically 0. There are no liquid side draws there are no vapor side draws and therefore, this also fits very nicely into the mesh system. Similarly, I can have another configuration where the feed is at the bottom this should have been at the bottom all these arrows mean that it is at the bottom. So, the source of vapor is the feed itself and the source of liquid is generated by a reflux which is coming from a condenser here and therefore, this is called a refluxed rectifier because essentially this is the top portion of a conventional distillation column which is the rectification zone this is the bottom portion of a conventional distillation column which is the stripping zone. So, this is a reboiled stripper and this is a refluxed rectifier. In refineries if you really pay attention to the crude column and for the time being if you ignore the side draws which are taken out a crude column resembles a reflux rectifier except that at the bottom because this hydrocarbon stream which comes at the bottom it has some useful light hydrocarbons which are to be knocked out. So, stripper is used steam is used and those light ends are or light hydrocarbons are taken back into the system. So, if you ignore that stripper here it resembles a reflux rectifier. I can go on and on now this is another scenario where I have a reboiled absorber, but this is not only it is not only that the separation is being done by thermal energy which is put into the reboiler as was the case earlier. I also have a liquid mass separating agent which is being put from the top. So, this is a reboiled absorber you have a refluxed stripper where the mass separating agent is at the bottom. If the feed is not necessarily at the bottom and reboiler is the source for the vapor at the bottom of the column and I am using a mass separating agent which is heavy and is expected to come out here and its presence affects the vapor liquid equilibrium of the constituents which are there in the feed or the species which are there in the feed this is the act of extractive distillation. Now again where is the difference? The difference is only in terms of characterization of the chemical system. Now if chemical system characterization we are worried about then the K values have to be properly calculated, their enthalpies have to be properly calculated. The mesh system remains the same. So, even this will fit very nicely into the mesh system. Same number of equations we are dealing with n times 2 C plus 3. Another variant is your azeotropic distillation. Of course, in azeotropic distillation there is a slight modification one has to make and that is that typically the reflux drum here after the condenser it will be a heterogeneous system. It is normally intended that way. So, the entrainer which you add is supposed to form another liquid layer. So, that you can easily decant or you can easily separate it out. That provision we have not made in the model which we have written, but again that is a question of thermodynamics. So, I can do that. I can take rest of the model from the mesh system and make this provision for decantation or rigorous thermodynamics to take the second liquid layer out and I should be able to simulate the azeotropic distillation. In azeotropic distillation the entrainer normally is light at cold temperatures. It is supposed to separate out from rest of the material. It should become a heterogeneous azeotrope and also we do not want this mass separating agent or entrainer to appear in the bottom. Whereas, in extractive distillation the mass separating agent is normally heavy and it is expected to have very little or very small vapor pressure and most of it is expected to come out at the bottom. Here it is a coupled effect of extraction and distillation. Here you are breaking the azeotrope and separating the entrainer in the condenser, but it is not done in the condenser, out of the condenser when it comes to the reflux drum that is where a three phase separation has to occur. So, when we talk about mesh system, the mesh system is applicable to these configurations also. This is the complex column which I had shown you and now you can satisfy yourself that the mesh system will be applicable to this system also, but there is a slight modification we have to worry about. What is that modification? The modification is that we have shown what are typically called pump arounds. I mentioned to you earlier that the main purpose of pump around is to balance the internal traffic of the distillation column and what is the secondary purpose? The secondary purpose is because the energy is removed at fairly or relatively larger temperatures as compared to the temperature at which you are condensing the mixture at the top, this energy can be utilized within the process to cause preheating. So, it is a useful energy. Normally what you discard in the condenser is not a useful energy. Most of the condensers they operate with cooling water and this cooling water goes to the cooling towers. So, all that millions of kilocalories they are dumped to the environment, they are lost whereas this energy being at higher temperatures can be used. So, essentially there are two main advantages of having pump arounds. One is that they become a good tool for balancing the internal traffic. So, that the column diameters remain uniform and the second that they become sources for energy integration. Now, we have not made this provision in our mesh system. We have made provision in mesh system to remove energy from a given tray, but pump arounds do not work this way. Pump arounds they withdraw the stream let us say on tray number 7, remove it and then return it either on tray number 6 or tray number 5. One or two trays above. So, basically you are causing disturbance in the neighborhood not on the tray. Is this point clear? And therefore, we need to do something with this model. So, typically when mesh model is used what is done is that it is assumed that this is returned on the tray just next to it. We do not allow the gap. If there are gaps in between we call them as heat transfer trays because on those trays we do not expect any mass transfer to occur. And that is true also because when this is returned this will be a sub cooled liquid because you have removed the energy. When you withdraw the product it will be at its bubble point because inside the column all liquids are at their saturation condition. So, this will be at its bubble point. This bubble point liquid is now losing energy. So, this will give you a sub cooled liquid. This sub cooled liquid when it falls into the column it is going to absorb energy to come to its bubble point. At the same time that energy is not available anywhere other than in the vapor stream. Vapor is warmer. So, vapor condensation will occur inside. So, there are internal refluxes which are generated inside. Vapor rising here comes in contact with this liquid and vapor starts condensing this liquid absorbs energy and it goes to its bubble point. So, we generate more liquid inside and reduce the vapor traffic and that is what does the balancing of the internal traffic. Whenever vapor volumes are large a presence of a pump around will help. You can condense part of the vapor. So, with this kind of approximation even a refinery column this is nothing but a refinery column you see the fresh feed is coming here and we have shown a steam feed which is for stripping. So, if I ignore this part if I ignore this part then this rest of the column is like a reflux rectifier. It is like a reflux rectifier and this particular case it is a reflux rectifier plus two side exchangers. All the heat is not being discarded in the condenser part of the energy is discarded in the side heat transfer equipment. We would not call them as condensers it is not a side condenser it works like a side condenser it is a pump around. Had it been condenser the provision should have been made to condense the vapor right on that tray. I should not take it further and bring it from the top. Here is a vacuum tower schematic of a vacuum tower again this is something which is done in industry. So, if we do if we do not focus attention on the furnace here and only focus attention on this part of the column everything here like you have a packing here you have a packing here you have a packing here and you have these pump arounds you have withdrawal this again can be simulated using the mesh system. This is a light vacuum gas oil heavy vacuum gas oil this is the vacuum residue and this what the feed to this system actually is the bottom product from the crude column the long residue as it is called this is called short residue. Here we have packings and therefore we have to do something about either representing these packings in terms of equilibrium stages through HETP concept HETS some people call it height equivalent to a theoretical plate or a theoretical stage or there are other methods available where mesh equations can be further modified and use for the packed columns and I will come to that I will show you that. So, this is the system now before I go further let me tell you that there are various ways people have tried solving this mesh system I am not getting into the mathematics of solution here I will give you some popular algorithms how they have been developed over the years I will just name them and give some background about the development. There are two types of systems which are normally encountered by enlarge in industry one set of systems we call as close boiling systems close boiling systems means that multi component distillation when you are doing we do not see too much of a temperature profile or temperature drop from top of the column to the bottom of the column. Now, separations such as propane propylene separation focus may be on propane and propylene, but there may be small amounts of ethane, ethylene, butane all these components may be present. So, it will be a multi component system. So, propane propylene system will be put in this category a close boiling system ethane ethylene separation again is a close boiling system ethyl benzene styrene separation also will be termed as a close boiling system. Because the variation in temperature is not too large. So, close boiling systems are those systems where alpha stays close to 1, 1.1, 1.2, 1.15 it has been observed that most of these systems are encountered in hydrocarbon industry hydrocarbon systems are not highly non ideal systems, they are mildly non ideal systems. We had mentioned that even regular solution theory predicts the behavior very well. So, you do not you do not really require very deep characterization of non-ideality in in these cases. So, equation of states work very well and systems are well behaved from thermodynamics view point. So, what it means that the k values the enthalpies the liquid and the vapor enthalpies can be very nicely calculated by an equation of state. Now, when the systems are close to ideal and if we go back to the thermodynamics which we had discussed, do you recall what I had said on the property changes between like the volume change and the enthalpy change for the ideal system or close to ideal system. I had said that the volume change of mixing is approximately 0, the enthalpy change of mixing is also approximately 0 alright. And therefore, the vapor and liquid profiles the vapor and liquid profiles they are strong functions of the k values and weak functions of the temperatures. What it means that out of mesh system M and E they need to be solved simultaneously they should be combined of course, S also is required S is a mathematical requirement, but the H component the enthalpy need not be solved simultaneously because the k value dependence on composition is not very strong. And therefore, you can keep the compositions on one side temperatures on the other side. And this is what used to be the case earlier for hydro carbon systems. There are algorithms where the researchers tried to solve the M and E equations together for a given temperature profile. Once you have solution for the M and E equations then you can come back and update the temperature profile. So, rather than saying that temperature will be simultaneously solved you can say that I will load a temperature profile just like we have loaded a pressure profile, solve for the composition profiles and the vapor and liquid profile. Once you have that you apply the condition that on every tray the liquid leaving is at its bubble point. So, you can calculate the bubble point and that can give you a temperature backwards. So, you back calculate the new temperature profile and you update the temperature profile. So, there are methods where the M E or M E S you can call set of equations have been D linked from the energy balance. So, mass balance is D linked from the energy balance and energy balance forms the outer loop on the calculation the material and energy balance form the inner loop in the calculation. So, if you have seen methods such as Wang-Henke, are you familiar with Wang-Henke any of you? Wang-Henke method, it works that way the temperature profile is kept separate in the outer loop the material balance is solved inside.