 Welcome to the 23rd lecture on cryogenic engineering under the NPTEL program at IIT Bombay. The earlier lecture we were talking about gas separation column or a distillation column. The earlier lecture we have studied temperature composition diagrams and enthalpy composition diagrams and we have understood the importance of these diagrams in gas separation. The separation of a mixture is more effective when the difference in the boiling points is more that is also we have understood. In this column the distillation column the low and the high boiling components are collected at top and the bottom respectively of the column. For example in a nitrogen oxygen you will get liquid nitrogen at the top and you will get oxygen at the bottom of the column. We also defined mercury efficiency which is a ratio of actual change in mole fraction to the maximum possible change that can occur in the mole fraction. In a way it actually points at the ineffectiveness of the column. This can help us to calculate the number of plates or number of processes of condensation and boiling that should occur in the column. In today's lecture I will again continue with the gas separation and what I would like to bring about is a beautiful animation to understand the functioning of rectification column or a distillation column and then from there just giving an animation so as to understand what all different processes that happen in a rectification column and then we will go ahead to understand the theoretical plate calculation or actually design of a plate column or a rectification column. So let us come to first the animation. I am showing you a distillation column. The animation has been done by Sneha Devrukkar and this animation has been done under the project OSCAR which is open source courseware animation repository at IIT Bombay. Please look at this and possibly you can come to understand different processes that happen when the feed enters the column. Now in this column I have shown the not the steady state operation but I have shown different processes that happen in this column. When the feed comes how the feed gets pre-cooled first before it enters the column and how the high boiling component and the low boiling component travel up and bottom respectively. So please see and each process has been highlighted and at the end what you see is a steady state operation of this column. So lot many ideas can get clear how exactly this process happened because the vapors go up and the liquid comes down. So you can see this is a column and the process gas and let us say air it is coming through this and this process air comes through this and it goes through something called as boiler you know about this. Now as I said I am just saying process by process and assuming that there is a high boiling component sitting over here for example there is a liquid oxygen already sitting over here and this process air will therefore get pre-cooled in this boiler. And there is this pre-cooling some of the vapors get evaporated and they start coming up and this process gas then gets expanded through a jetty wall and enters over here at the middle of the column at some location in the column and the liquid starts coming up over here and you can see that on this plate there are various plates the liquid starts getting accumulated and once the liquid levels goes to a particular level it starts coming down to the next plate which is below it. So you can see that as each and every plate gets filled up to the wear height the liquid starts coming down and from the bottom the vapour starts coming up because of this pre-cooling vaporization happen and the heat transfer happens as we have already seen between this vapour which is coming up and you can see this boiling happening over here. So the boiling and condensation that starts happening together the liquid starts coming down and as the liquid starts coming down it gets richer in high boiling component and slowly you can see that the vapours are going up. And this vapour reaches the condenser dome and you can see that the condenser needs to be cooled by some cooling effect which is supplied by an auxiliary circuit. Some cooling effect is offered over here some heat is taken out from here and whatever vapours come out it gets condensed over here and from here what you get is a low boiling component in liquid form not all the liquid is taken out some liquid starts coming down also in order to have a steady state you will have some liquid coming down and what you can see again is heat transfer between the upcoming vapour and the down going liquid and you can see that on each plate there are small holes or bubble cap through which vapours rises rise up and the liquid will go down through this wear. So the beautiful animation to understand how the vapour start coming up and how the liquid start going down and a steady state it ultimately reach as shown over here. So ultimately you will have some liquid at this point and you will have some liquid on the top of the column and this is where you can understand how this process happen. So we have seen the animation of a distillation column we have understood how the vapour rise up how the liquid start going down and what is most important is the heat transfer between this vapour which is going up and the liquid which is going down. This knowledge will help us now to understand how this different process of boiling and condensation happen that is the most important thing and based on this we do the distillation column design and in this case now we are going to talk about the number of plate calculations or how many plates will be required. Again coming back to some basics this was a column which we just saw what you get on the top is a liquid nitrogen or a low boiling component what you get from the down from the boiler over here is liquid oxygen or a high boiling component when we assume that the feed is air. So here you can see now the feed is coming at this point and as seen earlier in the rectification column the liquid moving down is enriched in high boiling component. So as the liquid is moving down the high boiling component means temperature will go up and as it goes down from here or as it comes down from here it will get richer and richer in high boiling component. At the same time as the vapour is moving up it will get enriched in low boiling component which is nitrogen. So as it goes up that means it will come down in this temperature composition diagram you can see here and as it comes down it will get richer and richer in low boiling component. This is the basic which we have studied in earlier lectures also what does it mean it means that forgetting 100 percent pure product that means if I want 100 percent nitrogen from this mixture or 100 percent oxygen to be separated from this mixture I would need to have infinite number of rectification processes that means I would like to have condensation and boiling which is happening on every plate I will need so many plates to be there and in principle I would require infinite plates because the heat transfer is never perfect it will always have some ineffectiveness associated with it. So in principle we need infinite plates but in reality the size and the cost of the column limit the number of rectification processes and therefore the purity. So I will have to sacrifice some purity that I need not have 100 percent pure nitrogen I need not have 100 percent pure oxygen but may be 98 percent may be 97 percent is good enough for me and therefore a column has to be designed for a given purity which is a input data that is very very important. So in order to calculate the number of plates that are required or in order to design this distillation column various researchers have developed various mathematical procedures which are very important to calculate the required number of rectification processes that means how many plates will be required which will indirectly determine the desired purity for nitrogen as well as for oxygen and this is what constitute a major design of a rectification column. Now these procedures require following data what are number of components so your feed should have so many components if the components are many the there will be difficulty in separation or even in calculation. So how many components are there can we have some reasonable assumption for example air we say that we have got only two components but if you have got four components then the data for all these four components in a mixture will be required and what is essential therefore is a phase diagram of the mixture. So more the number of component more important will be to have a phase diagram for this mixture. Also what is important is the property data of mixture at lower and lower temperature which is also very very difficult to get and ultimately what is most important is to know the heat transfer correlation because as you know boiling and condensation heat transfer occurring on every plate. So one needs to have all these correlations validated correlations for this data which is very very difficult. The method of calculation that are used for theoretical plates are so we have got several methods and we are just naming three methods which is called method of Ponchon and Savarit which is a first method. The other method is called method of Maccabay and Thill it is called Maccab Thill method and third is a numerical method. So these two methods are named after the scientist who developed these concepts or procedures. So we got a Ponchon and Savarit method we got Maccab and Thill method and ultimately what we have is a numerical method. Now let us just see what these methods constitute or what are the requirements basically in order to get the distillation column design. The Ponchon and Savarit method is an exact method for a plate calculation. It is applicable to any number of components and this method requires detailed data of enthalpy composition diagrams of the mixture. So I can have four component mixture, five component mixture but what I need is a correct detailed data of enthalpy composition which is very very important and which is very difficult to get also. While Maccab and Thill method was proposed by two scientists which is Maccab and Ernest Thill in 1925 and this method is less general and is simplest technique. This is a very simple technique less general however why it is less general because it is used widely used for binary mixture that means a mixture consisting of two components and at cryogenic temperature. So this method for binary mixture has been found a simple technique as well as it validates very well with the cryogenic conditions and cryogenic temperature also. The third is a numerical technique which as you know many of you know now with the mathematical tools we can have various equations for mass, energy, momentum etc. that we solved at every point as the vapor moves up and the liquid comes down. These methods as you know are tedious, time consuming and very very computer intensive. So for the sake of understanding in this lecture and in this course we are going to concentrate on Maccab Thill which as I just told you it is valid for a binary mixture and is relatively simpler and also validated at cryogenic temperature. So we will now see what is this Maccab Thill method comprise of alright. So of all these three techniques let us go for Maccab Thill and understand how do we calculate number of plates that are required to separate a given mixture for a given purity. That means if I want to have 98% purity, 95% purity etc are the requirement how do I design a distillation column. So this is the distillation column and now let us come to the Maccab Thill method. This method calculates liquid and vapor fraction of each component at every plate and also the number of plates. So what do we do basically? We march up or we march down from the feed this is the feed you go up or you go down and at every plate we calculate what is the liquid and the vapor fraction of each component and again we are talking about a binary mixture over here at every plate and also therefore we will calculate ultimately the number of plates required for a given purity. Now let us come to the mathematics involved in this right there are various mathematical steps I have tried to go in details but you will have to go through that many times to understand all the algebraic derivations involved but what is most important is to understand the physics behind the system. For the sake of understanding let the plates above the field which is called enriching section labeled as N similarly the plates below the field which is what we call as stripping section be subscripted as M alright. Let the total mole flow rate of top and the bottom products be D and B. So whatever moles we get at top as moles per hour moles per second whatever and what you get here is a low boiling component what you get at the bottom is a high boiling component. So let us say we receive D moles per hour and we receive B moles per hour at the bottom. So let the total mole flow rate of top and bottom be D and B respectively. Now consider small plate over here and you can see that it is important to understand the indexing pattern of the plate and its corresponding liquid and vapour. So as you know that for this plate we have shown just two plates and the vapour will leave this plate and it will also receive liquid from the top. So we have just shown two plates that the vapour will go up and liquid will come down. Let us call this as a jth plate any random plate in this and this is jth plate and this is a j plus 1th plate alright. So let jth and j plus 1th plate be any intermediate plate as shown in this figure. The liquid and the vapour leaving from the top of the jth plate are therefore called as lj and vj respectively. So consider any plate jth plate the liquid which is leaving this plate is coming down over this wear as you saw in the animation and let it be represented lj. So lj represents the liquid which is leaving this plate and vj which is going up the vapour which is going up the vapour which is leaving the jth plate be called as vj. So for any plate you got lj and vj as liquid leaving the plate and vapour leaving the plate alright. Similarly the liquid coming to the jth plate is from the j plus 1th plate. So this is the jth plate and it is receiving liquid from the top plate. So what is the liquid which is leaving the j plus 1th plate is lj plus 1. So therefore the liquid which is coming to the jth plate is lj plus 1 and liquid which is leaving the jth plate is lj. Please understand this nomenclature because it is very important. So jth plate receives the liquid lj plus 1 and the liquid which leaves the jth plate is called lj. Also the vapour coming to the jth plate is coming from the bottom. The liquid comes from the top the vapours come from the bottom. So what is that vapour which is coming from the bottom is the vapour which is leaving the j minus 1th plate alright. So vapour coming to the jth plate from bottom is vapour leaving the j minus 1th plate and therefore it is called as vj minus 1. So for any plate let us just revise for any plate lj and vj are the liquid and vapour leaving this plate. This plate receives liquid lj plus 1 and this plate receives vapour which is vj minus 1 alright. The vapour and the liquid on any plate lj and vj are in thermal equilibrium. We have seen this earlier that for any plate the liquid and the vapour are in thermal equilibrium that means their temperatures will be remaining the same that is lj and vj. Ok now let us take a control volume and as we have done we will do the mass balance and the energy balance etc. So consider a control volume enclosing the condenser and top section of the nth plate. So this is the nth plate and this is the top part of it. So this is the condenser on the top and let us have any plate nth plate over here and this is our control volume ok. As explained earlier for this nth plate the vapour leaving is vn. So this is the nth plate and therefore whatever vapour is leaving up is the vn and whatever liquid is coming down is ln plus 1. I think this is very clear now. So applying the mole balance across the control volume what we get is now what is coming in in the control volume is vn and what is leaving this control volume is d and ln plus 1. This is very important to understand again. What is leaving is d and ln plus 1 and if we do the mass balance now what we get is vn is equal to ln plus 1 plus d alright. So this is the most important whatever vapour goes up ultimately it is equal to whatever condensate you receive at the top and whatever liquid comes down very understandable. Multiplying the mole fractions a mole balance equation with mole fraction. Now this was a molar balance which comprises of both the components v moles are going up alright and d moles are received but all these moles will have both the fractions of let us say component a and component b. Now if I want to make a molar balance of only one component of let us say component a or let us say nitrogen for example we have to multiply by its mole fraction. So what is the mole fraction of nitrogen in a feed mixture? So multiply the mole balance equation with mole fraction of a particular component in a mixture we get mole balance for that component. Now we are talking about component balance component mole balance. Therefore yn vn is the mole number of moles of a component which is going up. So let us say nitrogen if I say so yn vn because I know yn is a fraction of nitrogen in the mixture. So yn vn is going the vapour fraction yn vn gives the vapour fraction of the one component of this mixture and xn plus 1 ln plus 1 is number of moles of that component which is coming down and at the same time xd into d is the moles of that particular component which is leaving this control volume. So now we are doing a component mole balance from the earlier complete mole balance. So where yn xn plus 1 and xd are mole fractions of particular component in vapour, liquid and top product respectively alright that means if I got a component xd which is leaving over here. So xd actually represents amount of nitrogen in d for example alright it is very important to understand that xd therefore if I say I want d is the top product if I say that d should have 98 percent nitrogen then my xd becomes 98.98 actually which is what it says it automatically means that xd which is the mole fraction of a particular component is a desired purity of the top product alright. So this is basically desired purity of that component in d. For control volume taking into account qd watts now we have got a condensation happening out here and therefore some refrigeration effect has to be supplied and this is qd because some heat has to be taken out from here. So qd has to be taken into account when I do energy balance. So qd watts as the heat rejected by the condenser the enthalpy balance is given by now I am doing energy balance alright first we did the molar mass balance now we do the energy balance and multiply all the component with respective enthalpies and include qd in that. So what is the energy of the molar moles Vn which is coming up hn Vn is equal to the liquid moles which are coming down and therefore corresponding enthalpy of small h we know about this we have seen the enthalpy concentration diagram enthalpy composition diagram. So small h denotes the enthalpy associated with liquid the capital H associates the shows the enthalpy associated with vapor. So hn Vn is equal to hn plus 1 ln plus 1 plus hdd which is the enthalpy associated with the top product plus qd. So this is entering and all these are leaving and therefore the energy balance has been done. Now do the mathematical jugglery and therefore we have some algebraic rearrangements over here dividing above equation by d what you get is this. So you have got now d at the bottom over here rearing the total mole balance equation we have already got a total mole balance which was this and if I rearrange that or divide that by d I will get expression. So what is important to see here is you have got Vn by d over here you have got Vn by d here you got ln by d over here you got ln by d over here. So if I replace this ln by d by this you will get a further expression just replace put this values over here and eliminating ln plus 1 by d from the earlier equation what you get is this. So now you got the two terms which has got Vn by d common and doing that I will get an expression like this taking capital H and small h difference that is the enthalpy difference between the vapor and the liquid Vn by d and you got all energy terms over here. So here I can rearrange the whole thing and get d by v as this expression what does it mean? What is d? It is the moles of a top product coming out and what is Vn? The moles of in vapor form which is entering the control volume. So d by v actually represents something Vn is entering the control volume d is living the control volume. Condense low boiling component which I am getting so many moles I am getting at the top and it depends on the enthalpy difference between the vapor and the liquid and you got all the energy terms coming over here Qd, Hd and Hn plus 1. So all the enthalpy terms plus the cooling effect that has been supplied coming over here and this gives a very important term called d by vn. Now let us see the enthalpy terms. You know the enthalpy composition diagram enthalpy and mole fraction diagram is shown like this. The enthalpy composition diagram for a mixture of nitrogen and oxygen let us say at one atmosphere is shown over here. So this is vapor and this is liquid and there are different temperatures knowing that we know the change of phase is not at constant temperature alright. If we neglect enthalpy variation with the mole fraction the bubble and dew lines can be taken as horizontal these are the bubble and the dew line you know understand from here you got some inclination over here and can they be approximated can they be approximated you can see the enthalpy changes with mole fraction. But can I say that you can understand that change in enthalpy is not so much over here it is not very it is not so much. So understanding from that you can see that this is this. If I approximate as a straight line I can come to a conclusion that now both the enthalpies are independent of mole fraction and if I do this matter become very simple for me. So now coming back to the expression you have got this and if I say that also one more expression and the argument which we just did that it leads to the fact that liquid enthalpy vapor enthalpies are constant they are independent of the mole fractions and hence we can say that because they are all enthalpy dependent and if we say that they do not change at all we can say that d by v and therefore l by v is constant that these terms are constant they will not change from plate to plate because the enthalpies will remain constant they will not change with molar fraction which is what happened from every plate right please understand this concept. So we come to the conclusion therefore that both d by v and l by v are the terms which are constant and if I rearrange my mass balance expression early I now get an expression which is y n is equal to l n plus 1 by v n x n plus 1 plus d by v n x t. So here we have got two components which is l by v and d by v and both we have just proved that they are constant what does it mean what does this equation tell you y is equal to m x plus some constant term so it is a basically equation of straight line and that is with this assumption we are leading to this conclusion that this represents an operating line for the top section of a column which will give me what is the vapor fraction at every corresponding to that vapor fraction I will get what is the liquid fraction. So I can now calculate y and x at every location in the top section and this is what we were aiming for. The above equation represents a straight line and is called as operating line for enriching section gives you a relationship between y and x vapor and liquid at every stage and this is what we wanted to do basically when I told about what do we basically aiming for in a design of a rectification column. So this is an expression which we have just derived also understand that at every at the top most plate I have got x n plus 1 is equal to x t at the top most plate this is x d which is coming at its condensed set alright because there is no other plate and at the top most plate the purity is going to go on increasing and this is equal to x t substituting this value x n plus 1 is equal to x d at the top plate what do we get I just replace x n plus 1 as x d I get if I put the value of x d taking common and if I put the relationship between this 2 this is nothing but equal to 1. So what do I get y n is equal to x t what does it mean for the top most plate y n is equal to x t not for any other plates only for the top most plate I get y is equal to x t and what is this intercept of this line what so we have located one point on the top which is at the top most plate and other point if I want to draw this line is the y intercept and this is nothing but the y intercept d by v n into x t and this will happen when x n plus 1 is equal to 0 and that is why it is called y intercept. So you have got a y n is equal to d by v n x t and therefore what I conclude from here I know 2 points to draw this operating line for a enriching section what are these 2 points y n is equal to x d at x n plus 1 is equal to x d which we just derived and also the other point is the y intercept which is y n is equal to d by v n x d at x n plus 1 is equal to 0. So if I know these 2 points I can draw the operating line for the enriching section if I want to do that now I am plotting y versus x for a component let us say a component. So a plot of vapor versus liquid mole fraction for a particular component says a is as shown in the figure. So what I do first I will just draw y is equal to x line because I know that y is equal to x for the top most plate. What I also know the desired purity of this component a in the top product is x t this is the problem definition for me that I want to design this column for getting a purity of x t for the top product or for the low boiling component alright. So I got I can locate this point which is x d over here on the x axis and if I want to locate that point on the on this scale I got this equation and as you know I got other point which is at x is equal to 0 I got a y intercept. So what are these 2 points the first point is this y is equal to x. So y d is equal to x y n is equal to x d I can locate on this I am sure because it lies on the y is equal to x line. So look at this point the other point is this and if I join these 2 points I will get the operating line for this n reaching section I will know the how much y is there for every x at every location in the n reaching section. So if I know my x is this much corresponding to that I can calculate y which is also given by this operating line equation for the n reaching section what I am doing ultimately with some reasonable assumption I am trying to get the liquid and the vapor at every plate in the n reaching section and now I will do the same exercise for stripping section also that means for the bottom part of the column also this is what we are aiming for basically. Similarly the slope of this operating line is L by V alright slope is basically how much liquid is coming down how much vapor is going up for every plate. We are saying that the slope of this line is same that means what we assume is L by V is common for the entire n reaching section this is what there are some reasonable assumption and that is why we say that with enthalpy remaining constant this will automatically come. Now what we are concentrating on is the bottom part. So if I take a control volume for now n reaching section similarly for the analysis of mth plate and boiler in the lower part we have the following equations. So what is the mole balance now L m plus 1 is entering. So I have got a mth plate over here and I am taking top of mth plate and boiler over here we have got what is entering is equal to what is leaving. So mole balance is L m plus 1 is coming down V m is leaving this plate and B is leaving this control volume. So if I do mole balance entire mole balance for both the components L m plus 1 is equal to V m plus B and then if I do component balance as we have done earlier for a given component A I will have X m plus 1 into L m plus 1 is equal to Y m V m plus X B B. Again X B is now denoting how much A is there in bottom. As you know X D we are assumed to be 98 percent but X B is now going to be very small because I do not want this nitrogen to be there in oxygen. So X B is going to be only 0.02, 0.03 etcetera right because the purity of B has to be 97 percent and I am talking about component of nitrogen in oxygen here and that is what X B will denote because we are doing mass balance for the same component over here. The third is energy balance multiplied by enthalpies and also I am giving some heat to this boiler that also would figure in this as it had figure Q D in the condensation in the earlier case. So if I do enthalpy balance H m L m plus 1 plus Q B which is entering the control volume is equal to H m V m which is leaving over here and H B B which is leaving the place alright where B and Q B are the mole flow rate out of the bottom and heat input to the boiler respectively. Now I will do the similar rearrangements as what we have done in the earlier enriching section rearranging the above equations we have the following I will get B by V m is equal to H m minus H m plus 1 divided by Q m upon B minus H B plus H m plus 1 and again we have got a similar assumption that enthalpy values are independent of the molar fractions we can again show that B by V is constant and therefore L m by V m also will be constant. Applying the assumption we have capital H m and small H m plus 1 as constant which implies that B by V m and L m plus 1 by V m are constant the operating line for stripping section therefore is Y m is equal to L m plus 1 by V m into X m plus 1 minus B by V m into X B. So similarly what we had done earlier I have got an operating line for the enriching section which will be giving me amount of vapor for every X amount of vapor for liquid on every plate as we march down and this L by V is going to be the slope of that line which is constant throughout and this is the Y intercept. So again if I want to draw this line graphically I have got two points which is what we will see down. So this is my expression for the equation of line for the stripping section again putting the same assumption for the bottom or the lower most plate near the boiler we got X m plus 1 is equal to X B it basically shows how much nitrogen is there in oxygen which is going to be very small amount alright it going to be 0.02 or 0.03 it is actually talks about impurity now for the down product and which is a purity for the top product substituting this value. So if I write this equation only for the bottom most plate I will get X m plus 1 is equal to X B in that case putting that I will again get an expression as 1 and I will get Y m is equal to X B that means this line again has Y m is equal to X B or Y is equal to X at the bottom most plate. So again this line will intersect Y is equal to X line at the bottom most plate as we saw earlier this is one point I will get another point again is a Y intercept which is minus B by V m into X B. So other point is a Y intercept which is definitely going to be at X m plus 1 is equal to 0 which is the Y intercept. So again I have got two points which are Y m is equal to X B at X m plus 1 is equal to X B basically the intersection point of the operating line at Y is equal to X and other point is the Y intercept Y m is equal to minus B by V m X B at X m plus 1 is equal to 0. So now I have got two line two operating lines and I will draw those two operating lines graphically. So this is my earlier line which is operating line enriching section the plot of vapour versus liquid mole fractions for a component A with operating line and 45 degree angle as shown over here. The purity of the component A in the bottom product is X B. So what is the purity now actually is going to be impurity for this product it going to be 0.02 how much of component A exist over here is going to be X B which is very small. So if I look at that and now look at this point on a Y is equal to X because at this bottom most plate it going to be the Y is going to be equal to X. So it is going to be on a Y is equal to X line alright. So that point is going to be the Y intercept which is minus B by V m into X B this is this. So one point is going to be here other point is going to be here and I am just going to draw this point. So this is my one point and other point is this. So if I draw a line between this two I will get operating line for the enriching section and this is my operating line for enriching. So this line gives me what is Y for every X in the stripping section and what does this line give me what is my Y for every X in the stripping section and what does this point indicate this point indicates where both of them beat and this is thing called feed point where the feed inlet happens. The slope of the operating line is given by L m by V m as shown in the above expression. So now after seeing that how this operating lines for stripping section as well as for enriching sections are drawn now we will come to the third point which is third line which is called a feed line which will denote the feed condition for the column and this is very important because this is the point at which both this line intersect. The mixture that is to be separated is called a feed it is introduced in the column through an opening called as feed inlet as shown in the figure. So you can see here F this F denotes this is the feed inlet through which the mixture enters and this feed point could be anywhere it need not be at the center of this column it can be at this point at this point depending on the feed condition this feed inlet can vary alright. Now let us consider a control volume across this till now we had a control volume including the condenser or the other control volume we had including the boiler and that was above the feed or below the feed. Now let us take a control volume which encloses some part of the some plates in the enriching section and some plates in the stripping section. So you got some 1 n plate and 1 m plate over there and let us do again the mass balance and the energy balance and the difference of this is going to come from the feed because inlet is feed what is leaving here is going to be n what is leaving here will be m. So consider a control volume including the nth and mth plates and feed inlet as shown over here let F be the number of moles in the feed. So the total number of moles that are coming are F let us see the control volume like this. So you got a nth plate and you got a mth plate and you got a control volume as shown over here comprising of let us say just 2 plates above near this feed. So what is coming in the control volume and what is leaving the control volume have a look at this. So the above control volume we have what is in is feed inlet which is F. What is entering from this mth plate is V m and also what is entering from the top here is ln plus 1 this is the nth plate 1 nth plate what is entering this nth plate is ln plus 1 liquid. So what is entering this control volume is F, V m and ln plus 1 F, V m and ln plus 1 are entering the control volume and what are leaving control volume what are what things are leaving the control volume is V n which is the vapor which is leaving the nth plate and the liquid ln plus 1 which is leaving or which is entering the mth plate. So this is ln plus 1 over here and this is V n which is leaving. So if we do the mass balance of the molar balance of all these components what you get applying the molar balance what you get is equal to F is equal to V n minus V m plus ln plus 1 minus ln plus 1 have a look is F is a number of moles that are coming in the feed different between the vapor component which is V n minus V m different between the liquid component what is ln plus 1 minus ln plus 1 all right very important to understand that this is the feed and we can have some the feed can have some vapor component as well as some liquid component. So what is the vapor component V n minus V m what is the liquid component ln plus 1 minus ln plus 1 sometimes vapor may not be there. So only liquid component there will could be there in the fluid F that time ln plus 1 minus ln plus 1 is to 0 sometimes only vapors can come in the feed and that time ln plus 1 minus ln plus 1 could be equal to 0 all right. So feed conditions will decide what is the form it has vapor plus liquid only vapor or only liquid and that will decide what we call as feed quality all right. So please have a look at these things. So if you come to the same thing again we define a parameter called q all right we define a parameter called q which can denote the quality of the feed also ok q as the ratio what is how do you define q it is the ratio of liquid moles in the feed to the total number of moles in the feed. So how many liquid moles are there in the feed is called as q or the quality what does it actually denote it actually denotes how many liquid moles are there in the feed. In short it called it actually shows the wetness fraction of the feed how much wetness is there how much liquid is coming with the feed. So it is the ratio of liquid moles in the feed to the total number of moles in the feed. So mathematically if I want to write q q is equal to ln plus 1 minus ln plus 1 upon F. If we just saw the definition earlier the liquid components are going to be given as ln plus 1 minus ln plus 1 and therefore q is equal to what is the amount of liquid moles to the feed number of moles coming in the feed. It also denotes something if I have q is equal to 0 that means ln plus 1 minus ln plus 1 is equal to 0 that means there are no liquid components in the feed what does it mean if q is equal to 0 the feed is completely vapor because your numerator is going to be 0 and no liquid moles are coming in the feed. So if q is equal to 0 your feed is completely vapor or if q is equal to 1 that means ln plus 1 minus ln plus 1 is equal to F that means feed is 100 percent now liquid F upon F and therefore you got q is equal to 1 it means therefore that the feed is completely liquid. So it has a meaning to the value of q if q is equal to 1 it is completely wet that means it is only liquid it is bringing only liquid and this liquid could be saturated liquid. If q is equal to 0 the feed is bringing in only vapor and if something in between then you got some v and some l also that means you got a two phase feed now got a vapor and liquid component also we will study that in detail in the subsequent slides but understand the concept of q because very important. So this is what we have already studied that we have got operating line for the stripping section we got operating line for the enriching section from the earlier slides we know these equations for both these sections. Now you have got an intersection point and the locus of intersection of this operating line denote the feed condition. So where do they meet at this point and they can meet anywhere depending on the slope of these lines and therefore slope of these lines also play a very crucial role the point of intersection however we denote the feed inlet condition alright. The condition of the feed is vital to determine the number of plates. So what is this locus this locus could be like this the feed line could be like this like this or anywhere in whatever direction and this is very important to understand how many plates are going to be there in the enriching section or in stripping section this plays a very vital role in this and how does it. So let us see based on the feed equation and q definition we have got a feed equation f is equal to something and we have got a q definition also which is l m plus 1 minus l m plus 1 divided by f what we have is this is my f definition this is my q definition and if I want to calculate from here v n minus v m is equal to and putting this value as l m plus 1 minus l m plus 1 upon f I get an expression like this which is v n minus v m the vapor difference vapor mole difference is equal to I am representing this vapor mole difference in terms of q and f 1 minus q into f. Again from the operating lines of upper and lower section we can rearrange to get v n and v m. So I can get v n and v m from the feed line or the feed expression for the feed and I can also get expression for v n and v m as this because I have already got a equation as y n is equal to l m by v and etcetera I am just rearing that to get the value of v n. So v n is equal to l by y x n plus 1 plus d by y x d and similarly I will get v m is equal to l m plus 1 by y m x m plus 1 minus b by y m x b. So I got two expressions for v n and v m and from here it is very important to note that v n minus v m is the vapor content in the feed we have just seen that thing. So v n minus v m is going to be the vapor content in the feed if we are writing this equation for the n th and m th plate what we have defined earlier. In the calculation of point of intersection now I am very keen to know my feed inlet condition which is denoted by the point of intersection of these two operating lines. So operating lines will have some coordinate x and y and they should be satisfied by both this equation of operating line for the stripping section and also the operating line for the enriching section. So we choose a common point and let us call it as x and y. So hence now we do not have to write x n plus 1 x m plus 1 y m and y n they will be replaced by x and y over there and now I get v n minus v m is equal to l n plus 1 minus l m plus 1 upon y into x plus x d d plus x b b by y is equal to 1 minus q into f. The locus of this point of intersection is the feed line or q line and is calculated as explained in the next slide. So here for a column as a whole using mass balance we can now have an entire molar balance for the column which is x f f is equal to x g d plus x b b right these are the most coming in these are most leaving the column. Again q is equal to l m plus 1 minus l m plus 1 upon f rearrange in this term you get this expression I will replace this by this and this by q I got this expression and rearrange in this expression I will get an expression as y is equal to q upon q minus 1 x plus x f upon 1 minus q this is also a constant this is also a constant and this line therefore is called as feed line actually which has the above equation represents a straight line with q upon q minus 1 and x f upon 1 minus q as slope and y intercept respectively and it will pass through the point of intersection of the two operating line. So we got three lines now where is operating line for stripping section operating line for enriching section and this is now what we call as feed line. More importantly it is the locus of the point of intersection of operating lines this is line is called as feed line or q line. So we got a third line now which represents the locus of the point of intersection of the operating line but it has got lot of physical significance which depends on the value of q or the quality or as I said weightness of the feed and let us see how it denotes that thing. So this is the column it is clear that the value of parameter q is yet to be determined applying the energy balance to the control volume here I got enthalpy balances for everything h f f is equal to v n h n minus v m h m. So I got an enthalpy associated with this rearranging this and we say that all enthalpy are same or capital H is equal to h all small h is equal to h putting them together I will get this expression I will get this expression as again rearranging the term and ultimately I know also v n minus v m is equal to 1 minus q into f I know the quality expression or q expression basically putting these values over here and putting this value over here what I get q is equal to h minus h f upon h minus h. So this is enthalpy difference h minus h is a complete enthalpy difference of the latent part of it difference of enthalpy constant due and a bubble point and this is the feed enthalpy h f is the feed enthalpy which will determine what is the q otherwise this h this h and this small h are constant. So this q will be determined what on the fact what is the enthalpy of the feed. So this is our enthalpy composition diagram and we know q is equal to h minus h f upon h minus h also we know that the slope of the line slp is equal to q upon q minus 1 this slope of the q line or the feed line depending on the feed condition q can take any value. So depending on the feed condition h f will be determined and depending on h f value q will be determined. So now we have got a feed at various possible conditions various possible condition will decide the value of h f and which will decide the value of q which will decide the value of slope of the q line. Let us see those conditions the first condition is the feed is coming at saturated vapour in this case h f is going to capital H and this is the point this any point on this line is going to define the saturated vapour condition. If I put h f is equal to capital H over here I get q is equal to 0 and if I put q is equal to in this line I get slope as equal to 0 slope of the q line slope of the feed line is going to be 0 which means it is parallel to x axis if I plot in a graphical way. Second point is if I got a saturated liquid which is coming with the feed. So everything is liquid and therefore h f is equal to small h and the line will be at this point the feed line will be the feed point will be somewhere on this line. If I put h f is equal to small h I will get q is equal to 1 and if I put q is equal to 1 in the slope I will get slope as infinity and infinity line therefore will be parallel to y axis. So if feed line if feed has saturated liquid it will be the line will be parallel to y axis. If I got 2 phase now my h f will be somewhere between h and h and therefore q will be somewhere between 0 and 1 which can be calculated from here and the slope will be negative if I get the value from over here. Similarly, if I got a subcooled liquid that means h f is going to be less than small h I got a q is equal to now more than 1 and therefore the slope will be now a positive quantity slope line direction will denote a positive slope. Similarly, if I got a super heated vapor I got point feed point somewhere over here above capital H enthalpy q is less than 0 and the slope according to this equation will be positive slope and if I want to now denote all these conditions graphically please understand this it will look like what I show in the next slide. The equation of feed line therefore is this this is what we know and q upon q minus 1 is the slope and this is my y intercept. So, if I condition saturated vapor condition we have already seen that q is equal to 0 slope is equal to 0 and therefore feed line is going to be horizontal to x axis alright. If I got a saturated liquid coming my slope is infinity and therefore feed line is going to be parallel to y axis. If I got a 2 phase flow I got a slope which is going to be negative the feed line is going to be passing through this point and in this direction and sub cooled liquid will be shown by this and super heated vapor is going to be shown again by a positive slope. And therefore this point or the q line makes a very very big impact in determined number of plates for the enriching section and for the stripping section which we will see in the next lecture also. So, this is the graphical representation of all the possible line. Enriching section enriching section and one of this line will constitute a q line or a feed line for your problem. The point of intersection also what we of interesting is where does this line hit y is equal to x. The point of intersection of the feed line or q line and y is equal to x it gives the liquid content of the component a in the feed x f how do we conclude this. So, let us find mathematically it is calculated by substituting y is equal to x in the feed line. So, if I put x is equal to y is equal to x in this what you get here is nothing but x is equal to x f by rearrangement I will get x is equal to x f. It means that when this q line wherever it hits y axis y is equal to x line you get corresponding to that is your x f that means amount of component a the liquid content of the component a in the feed is given by this line. So, I know now two points if I want to draw the feed line I know the liquid content of a component in given feed I can locate this point also if I develop equation for the two operating lines I know in point of intersection and joining this two will give me what is the q line what is the equation also for the q line or the feed line. So, I know all the equations I can plot them graphically and this is what I wanted to cover in this lecture. So, graphically it is clear to draw a line through this two given points rather than using a given slope and a point. So, I got two points basically over here this intersection point is used to draw the feed line as given in the figure. So, I got two points join this two point what I get is a q line or a feed line. So, summarizing what I have just talked about in this lecture the plate calculation procedures requires the data line data like number of components, phase diagrams, property data variations of mixtures, heat transfer correlations macapthil method is less general and is widely used for binary mixture at cryogenic temperature and this is what we are going to study in this particular topic. The major assumption in this method is that the liquid and vapor enthalpies are independent of mole fraction and that is what simplified various calculations. The equations of operating line for stripping and enriching sections are this for the enriching section and this for the stripping section. The locus of intersection of this operating line denote the feed condition or the q line or the feed line it is given as this which denotes basically the equation for the q line or the feed line and the point of intersection of the feed line or q line and y is equal to x gives the liquid content of a component in the feed except which is what just saw in the last two slides and this is what our graphical representation of all the three lines are and this is what the slope of the line and this is what we understood depending on the feed conditions you have got various value of h f which determines the value