# McCabe-Thiele Part I Binary Distillation

my topic today is the mccabe thiele system for analyzing steady state distillation columns so binary distillation is the topic today and this is a very classic subject in chemical engineering um there uh is a lot in the literature about this method and i wanted to just walk through so that we would understand uh what we were doing when we could do this calculation for the number of trades in a distillation column so here is my schematic of binary distillation and the mccabe thiele method uses the assumption of constant molar overflow which indicates uh that the liquid and gas flow rates in the upper section or the enriching section of the distillation column will be represented by a single liquid flow rate in a single vapor flow rate that's independent of which tray you're in as long as you're above the feed and also that below the feed there is a single liquid flow rate and a single vapor flow rate that is characteristic of any equilibrium stage below the feed and that greatly simplifies the mathematical analysis of a steady state distillation column it's of course not always true here's a list of the requirements for the chemical system binary system of a and b that the system would have to meet for this to be true and i i send you to the literature to understand better um where those constraints come from the reference that i used here is separation process engineering all right so i wanna have a guide post to take us through this calculation our goal is to determine the number of equilibrium stages that uh are needed in a distillation column to make the actual separation take place that we that we assert so um that is our goal and these will be the different constraints from the physical world from the real world that we have to match in order to have that be the case so we have to have masses conserved because we have no reaction those will be mole balances we're going to assume that all the stages are equilibrium stages and so equilibrium stages mean that we can go to the literature and get the physical property data for um for the material that we're dealing with and find out what kind of liquid vapor equilibrium that material actually exhibits and so for instance i'm going to draw the equilibrium stages like this and say we have a liquid and we have a vapor and in the liquid phase we have species a in mole fraction xa and 1 minus xa is the mole fraction b and in the vapor phase also mole fraction i'm going to call that y y a is the mole fraction of a and 1 minus y a is the mole fraction of b when i'm actually talking about uh two phases in equilibrium these are sometimes designated with a star that they represent the saturated case and this equilibrium curve is data from the literature of y star versus x star that indicates that there is an equilibrium so this is y star as a function of x star we're going to use y and x in this plot for a bunch of other things as well and so that'll be part of what i want to convey as we speak so this is quite crowded and we've got a lot to deal with in these calculations so we're actually just going to write x a here and y a here and we just have to remember that we're talking about a binary mixture actually even that's a little bit complicated we're actually going to just call it x and y for the particular stage we're on so if this is stage 9 this would be x9 and y9 those are the the two um the two compositions that are in equilibrium with each other so we're simplifying the nomenclature a little bit just to keep us on top of the calculations that we're doing so equilibrium stages all stages above the feed satisfy the same mole balances driven um written through the um the upper distillate stream so that means that if i if i draw um some kind of a envelope here to indicate where i'm doing my mole balance above the feed as long as i'm above the feed they all no matter which stage i pick pick above the feed they all have the same mole balance and then below the feed where they have the l bar and the v bar we can do a single um balance below the feet so that is a reflection of this constant molar overflow assumption that's part of the mccabe feely diagram that says in the enriching section the upper section there'll be this single value of the liquid flow rate at a single value of the vapor flow rate and in the stripping section below the feed there'll be a single value of the volt flow rate and the single value of the vapor so these uh we'll be applying those mathematically and then those two balances have to meet somewhere they meet at the feed stage so at the feet stage we have a special stage and we have to make the mass balances above and below match at that point and then there's the quality of the feed so the final piece of physics that that cinches this calculation is to know uh for the for the saturated mixture of saturated liquid and vapor that comes in in the feed how much of it is liquid and how much is it of this vapor so these are going to be the constraints that we use and i'll point out when we when we do each one so let's start with the overall mass conservation so let's start with this binary distillation column and the overall mass conservation would be us uh picking the overall balance so it's a classic first first coursing chemical engineering calculation so we have the flow in equals to flow out at steady state so f is equal to d plus b this is the total moles balancing but we also will balance the moles of a so the moles of a in is x f f in and x d flow d out xb moles b out and we've written down the constraints for the overall mass conservation so the next constraint is that we have equilibrium stages and so the equilibrium stages means that when i draw as i did a moment ago a particular stage uh the two the liquid and the gas are going liquid and the vapor are going to be at equilibrium in that stage and they therefore i should be able to um relate if i'm given the x value i should be able to get the y or if i'm given the y value at the equilibrium stage i should be able to get the x so we've got the um y star as a function of x star constraint so now we go to the third one all stages above the feed must satisfy the mole balances so here's our total distillation column but i'm now i'm going to do a balance just above the feed so that would look like this drawing it separately so now we have we've cut through the column and so there are in fact um liquid coming down and vapor coming up right and so i'm just going to um pick an arbitrary number for this stage i'm going to say that this is stage two it could be any number i could put a j a lot of books will say oh stage j but that's maybe a little confusing i'm gonna pick a number stage two okay so i'm gonna do i'm gonna do a balance at stage two so this stage has stage two it has liquid here and this stick liquid has composition x2 and it has vapor composition y2 but that's not where i'm drawing my balance i'm doing my balance all the way around the outside but this stream is the stream with molar flow rate v vapor coming up from below and it comes from the next stage so that's stage three so this is y3 so as if there were another stage outside of the view and it's its material comes up here as y3 so up here at the top it's not quite finished how it's drawn either we're going to consider the usual case of a total condenser so a total condenser means that this stream which came out of tray 1 which means that it's y1 every tray is an equilibrium tray and the compositions of the of the equilibrium phases are named for the tray number so it comes up as uh as y1 it goes through the total condenser and comes out as xd well that just must mean that y1 equals xd right because if i totally condense the stream it has the same composition when it was a vapor phase as it does when it's condensed and i know what that value is so y1 equals xd and then usually we reflux so we have to reflux a stream here and we're going to reflux by the amount l uh which is the downwardly flowing l uh that goes through the column i'm drawing those details so that we can pull out this y1 equal x2 but the but the actual balance i want to do is on the outside here it's good for any of the tray of the tray numbers that are above the feed okay so that's the balance i'm doing and so it's always in a complex problem it's always nice to draw the envelope so that we can get it right so we can draw the the total of moles balance so in there's v uh and and there's no that's the only one there is and out there's l and d then we can do moles of a and so n is y 3 v plus nothing equals x 2 l plus uh x d d okay so that's that's um the balance so we can work on that a little bit we'll combine we'll take that first one and divide through by v okay so we have y 3 equals l over v times x 2 plus d over v times x d d over v let's see we can go back up to we can go back up to this expression to get rid of the the d okay so i'm going to put in the d equals v minus l from the total mole balance up here at the top okay d is equal to v minus l and so i get l over v x 2 plus v minus l over v x d for equals y 3. so this is the balance across the upper part of the column and these two streams here at the bottom these two streams are passing are called passing streams so these two streams pass by each other so whether i drew this red loop anywhere in the upper section i would there would always be two passing streams and they would always be labeled with the liquid phase composition being um one less than the vapor composition so it would always be y3 x2 y4 x3 y5 x4 so those are the passing streams so i can write the balance i just did for these passing streams as generally true for y um and x in this upper section in the enric enriching section of the column so y uh l over v x plus um well i can say 1 minus l over v x d so this is a line y versus x it's a line that represents that we can draw here on the x y graph just because we want to and we could draw some kind of a line and it would tell us as a function of x how the x and y have to be related to each other in order for the um enriching section mass balance to be satisfied this line is called the operating line for the enriching section now let's notice that this passes through the point uh y equals x d i mean x y y one equals x d so let's try that so we talked about the fact that here at the top with the total condenser that uh because we totally condensed this stream which was a vapor into this liquid stream which we labeled as having a liquid phase composition of x d but they're actually equal to each other because of this total condenser situation so let's just uh let's say plug in that point um so there's our operating line and so let uh x equal xd let's just you know it's a function we can plug that in and we get that the y uh that that goes with it that's a passing stream in the upper in the enriching section is l over v x d plus one minus l over v x d and so we've got l over v x d plus x d minus l over v x d and so these cancel and we get y equals x d so the point y equals x t is on the line and that uh would be uh whatever stage we pick okay any stage we pick this is the this is the um operating line for the um upper section and the enriching section and we found that y equals x d is on that line so that inspires us to go over here to our graph and i'm going to switch to some smaller pens just because it's going to get crowded and i'm just going to say well i'm just going to say that xd is this number which is 90 something or rather i'm i'm just using an example um bit to show you how we use the graphical method okay so if this is xd then y equal xd means that on this 45 degree line that's sketched in um i would expect i you know this is one of the points y equal xd that's on that's on the operating line so this point is on the operating line and the operating line is a straight line y y equals slope okay so this is the slope y equals slope x plus intercept and so l over v is the slope and so i don't know what the slope is right now but there's some line that has a slope that gives y as a function of x for the upper region of the column so that's that constraint here is satisfied all stages above the feed satisfy the same mole balances drawn through the upper half so then we now go to the next uh step which is to talk about the the bottom of the column so the whole whole column looks like this and we can draw a uh a loop to indicate where we're going to do the balance on the stripping part of the column and as long as we're below the feed that will be the stripping section and it'll have to go through this point b this uh stream b that's here at the bottom and so here we have it so this is below the feed so the feed stream doesn't show up again the liquid streams have to come down and the vapor streams have to go up but as we said here with the constant molal overflow assumption these will be given by the barred flow rates that are not the same as the unbarred flow rates okay so the liquid stream of cutting through any tray that's below the feed comes down as l bar v bar and then there'll be some kind of uh whatever tray we're on and um so we we can i just need to use some kind of an example and so i think i'm going to pick uh that this is at uh this came down from tray five and uh this is coming up from so this tray here will be tray six so so this this y6 is in equilibrium with this x6 but this liquid stream comes from the one above so this has composition x5 so now we're just going to carry out that balance that uh that we want to do so we'll just draw ourselves a little envelope you don't have to worry about that total condenser business because we just have this uh stream at the bottom and then we do the the balance so we can do the total balance or the yeah total moles and so n is l bar and out is v bar and b and we can do the moles a balance so n is x 5 l bar and out is y 6 v bar plus x b b so we're going to follow the same ideas that we did the last time we'll just simplify this equation a little bit so um we can get the operating line for the bottom half of the column so i'm going to look i'm looking to have y as a function x so i'll say y 6 v bar equals equals x five l bar minus x b b and now divide by b bar y six equals l bar over v bar similarly to before only now it's the barred ones uh minus b over v and from the balance above here b is equal to l bar minus v bar right here l bar minus v bar and so we get uh y six equals l bar over v bar x five minus l bar minus b bar over b bar times x b i missed the x b here x b so as we did previously for the um the upper section we'll recognize that this could have been done for any for any stage below the feed and uh these are two passing streams okay these are also two passing streams and so uh we can just call them y and x passing streams below um below the mean and just to remind us that i'll call this one y bar and this one x bar and we end up with again a general equation which is the operating line for the stripping section so um so this this balance is true for all possible balances i might do through any equilibrium stage that's below the feed and so uh we don't know what this slope is just like we didn't know what the slope was for the upper section but there's that's we're heading towards that um but we do know one point again we can ask about the bottoms okay so let's see what happens just like we did last time let's let's try uh you know let x equal x b and let's plug it into the operating line okay so y equals l over v uh xb minus l bar over v bar minus 1 xb equals l bar over v bar x b minus l bar over v bar x b plus x b so these two cancel and we get the y bar equals x xb so again we've identified a point on our um xy graph i have to dig it out of my papers here somewhere i have there it is y bar equals xb so this is again a point on the on the 45 degree line so there's a y that's a function x there's some sort of a line down here in the stripping section that has some kind of a slope but we do know one of the points and that point is the point uh that's at xb so i'm going to use again just just for illustration a value of xb that's uh somewhere below 10 so i can i can say well let's let this be xb so if that's xb then this must be the y that goes with it so this is x b and this is the y of the last tray okay the absolute last tray so i'll just put that dot on there that's in the operating line for um for the stripping section so the stripping section has some sort of a line that has to go through that point and the enriching section has a line that has to go through this uh distillate point okay and then we have this other curve which represents all the equilibrium stages in between so we've now done the balances on the stages below the feed so the next step is that all the upper and lower lines um intersect they those two lines must intersect um at the feed okay so let's start by um drawing the feed tray just to keep track of what we're doing and i have i forgot that i had i have a little feed tray here all right so the feed tray is a very special tray that has again liquid coming down vapor going up from the upper section so those are the unbarred flow rates and then liquid going down and vapor going up in the bard section in the stripping section so that's at the feed tray okay at the feed tray there are uh different vapor and liquid phases because there's um there's a feed stream coming in that's has to be taken into account and doing the actual balance so i'm going to call so this is again a this is some kind of equilibrium stage i'm going to call this stage four let's say the stages of four again i'm using specific numbers just temporarily so that you don't have to put up with arbitrary js and fs and things like that but you know it's gonna be some it's gonna be some stage i don't know what stage it is but i'll i'll take the four out um maybe almost immediately so this would be uh this is an equilibrium stage so this would be x4 and this would be y4 and they're in equilibrium equilibrium with one another and so they uh they would be on the equilibrium curve i'm going to call this x tilde the general feed tray and this is y star of x tilde that's how i'm expressing that the two compositions in the feed tray are also an equilibrium stage okay so notice okay this x is liquid coming down here so this is x tilde or x4 and this is the vapor leaving the feed tray so this must be y4 which is the same as this equilibrium phase so i i don't want to clutter my diagram i'll get confused so i'll just call it y4 here for a second okay and now if this this is y4 then this stream must have come down from one stream above so this must be x3 coming down from the stream above and if this is x4 uh this must be um going up from the stream from the uh the stage below so this must be y five so you see i could have done n plus one and minus one all that stuff but let's just use numbers so that we don't we don't get confused all right so we want to apply the constraint um that the upper and lower operating lines intersect at the feed tray so i have the upper and lower operating lines uh here's here's the lower one with the bars and in my pile here here's the upper one without the bars and so i should be able to plug in the passing streams the passing streams uh for this um and so the passing streams are x tilde here and i'm going to call this one the other passing stream y tilde okay so x tilde and y tilde are passing streams so that what's important about that is we have an expression for the passing streams in the enriching section so i could just plug in x tilde y tilde there and then we also have the operating line for the stripping section so i can just plug in the x tilde y tilde there and then i'll have two equations all right so let's do that at the feed stage we have i'm i'm starting with um the enriching section i'm just going to copy the enriching equation with the y tilde in there so y tilde equals l over v x tilde plus one minus l over v x d and then i have the stripping section okay y tilde also has to be satisfied at the feed plate y tilde equals l bar over v bar x tilde minus l bar over v bar minus one x b so i'm going to um combine these two strings together and get an expression for a constraint that has to be held at the um at the feed tray so i'm going to subtract so first let me uh i actually before i subtract i i want to put these into a slightly different form i'm going to multiply that v through so it's y tilde v equals l x tilde plus v minus l xd well actually you know i'm actually just going to go back to the um operating line expression and rather than reverse all this algebra i'm going to use this version of the operating line so this version right here so i'm just going to stop myself for a second here and redo this page 8 and say at the feed i'm going to use so this is the operating line for the stripping section before a little bit of algebra this version so this is going to be y at the feed v bar equals x at the feed l bar minus x b b so that's uh for the bottoms and then if i go back into my algebra um if you'll just hang on for a second hopefully i can find it well i haven't found it so this is this is in the stripping section and a few steps above this maybe maybe i can find it here here we go no can't find yeah so a few steps above this no that's not the right one um there's the version for the enriching section which uh when you plug it in for the feed station is y b bar equals um excuse me not b bar y v for the stripping section uh equals x tilde l plus x d b so these are the versions of the equations that have the distillate phase in it and the bottom phase and then i apologize for not being able to find them in my notes but they're just before they're just before the derivations of the operating uprighting lines that i did and this is the the version that's going to lead to the mccabe feely i'm going to subtract them so there's no magic even though i i kind of had a little skip up there uh with the algebra but these are just the mass balances for the upper uh the upper section and the lower section so now back to uh the subtraction so we get y tilde equals we're going to subtract um the bottom from the top so this one well i shouldn't say bottom because i have a uh i that means something in this i'm gonna subtract two from one all right so y tilde times v minus v bar equals x tilde times l minus l bar uh plus x d b plus x b b so okay so i this is a typo this is a d the d's have to match and then therefore this one is a ds1 from the overall mass balance we know that these two streams must equal x f f so these are the two streams that come out of the distillation column and so we get the final expression for passing streams at the feed tray which is y tilde equals i'll divide through by this one l minus l bar over v minus v bar x tilde plus x f f divided by v minus v bar the upper and lower operating lines intersect at the feed tray which means at the feed tray uh the conditions the x and the y that show up in the feed tray have to follow this expression for um what their what the passing streams um have to be related between the two so we've now used that applied that constraint and we have just one more so we're almost done uh this uh this is also a line okay so this is a line of y versus x that expresses passing streams on the feed tray and so at the feed point which i'm going to put it you know approximately halfway here just for fun again i'm just making these numbers up for illustration so here at the feed so this is going to be x feed there's going to be some relationship that has to be held and it's it's that of that expression we just derived for the passing streams and let's put that x f in there just like we did the other two times okay so let's try let's let uh x tilde equal x f let's try that one okay so here's the here's the mass balance at the feed tray so that means that if x tilde equals um x f y tilde the passing stream at the feed tray has to be equal to l minus l um v minus v bar x f plus x f f over v minus v bar all right so um let's that that gets us only so far um we're going to look on the last the last constraint so the last constraint is about something called the feed quality and that's related to the issue of whether or not the feed comes in as a saturated liquid or a saturated vapor okay so we're going to define quality i need to take a quick break here so uh we'll move on to part two in the next video you

*2021-02-07 18:57*