
S1: exciting isn't it? it is so exciting. <SS LAUGH> coming from the back forward, are, the blue sheet for today, and, the handout, which i promised you from King. the reference, not king as in, king of some country. on... the height equivalent to a theoretical transfer unit, quiet please. height equivalent to a theoretical transfer unit and, the height equivalent to a theoretical plate. so that's just, some fun reading for you. <P :05> <WALKS TO MICROPHONE> i really wanna go over here and, <SS LAUGH> oh sorry. <SS LAUGH> <P :04> but i'm gonna restrain myself. twenty-two, we had done all that example we had gotten the height of the column, calculated it out, and then i had gone through solute three concentrations, down at the bottom. <P :05> your new mass balance i did not do, but that is carried over, onto this sheet twenty-three. <P :05> there is, this one, point that i wanna make, sort of backing up, and i guess i'll do that first which is the log mean, driving force. okay, so let's start with this. we're doing absorption, or stripping columns packed columns, we have fluid flowing no equilibrium stages we want transfer between the two, phases liquid and gas. and we had those assumptions, C-E-L, right? so this was, constant flow rates we don't have any more equilibrium stages, and this one it can be linear... equilibrium or not. for this particular case for the simplest case i'm gonna just give you a quick equation you can use, instead of having to do the integration. okay and that's what the log mean driving force_ have you done heat exchangers yet? 
SS: (xx) 
S1: no?
SU-M: we're doing it now.
S1: oh you're doing it now. so have you done the log mean driving force for a heat exchanger?
SS: no 
SU-M: not yet 
S1: okay. well, you will, and you'll enjoy it. <WRITING ON BOARD THROUHGOUT UTTERANCE> <SS LAUGH> <P :04> so the, log mean driving force <P :05> it's just_ remember that integral that we had for the driving force one over Y minus Y-star? that's the driving force for absorp- absorption? remember that? so this is just, if you're if they're two straight lines you can r- you can integrate, that area very, simply. i'll show you how. <P :06> so the end result of this will be an equation and i'll just take you through how we'd come up with that. we have a certain Y-in, and a Y-out, and a linear, equilibrium line <P :05> you then have a particular, X-in, and, X-out... and if we're running a countercurrent system the two points on the operating line will be here and here, and it will be straight, if we have constant flow rates. so this is linear equilibrium and constant flow rates, no equilibrium stage, cuz it's a packed column. so with these two systems, sorry those two assumptions, you come out with a system that looks like this... now what you normally have to do is then get these Ys... and these Y-stars, and that's your driving force right? and do that equation, what equation is that? on twenty_ top of twenty-two that integral, one over Y minus Y-star D-Y, okay so i'm gonna eliminate that now, and just say let's call this, delta. okay, and let's just call_ so that's just Y minus Y-star it's just delta. and we'll call this one over here this Y minus Y-star, delta-in, and this one down here, delta-out. just by the Y-in we'll call that delta-in, and the Y-out we'll call that delta-out. <P :06> so if you plot delta... from let's say, delta-out, delta-in, from Y-out which is the low concentration to Y-in <P :08> looks something like that right? for what i've drawn there, approximately. okay? it's a straight line. <P :09> so this slope, is equal, to the change in delta with Y, and that is constant. okay <P :06> so now if i... if i have my, L-C equation where i had G over, let's say K-G-A-P, integral from Y-out, to Y-in i can change this instead of one over Y minus Y-star i'll just put one over delta <P :05> D-Y. and now i'll just convert variables so everything's in terms of delta <P :16> so this is from delta-out, delta-in, one over delta <P :05> and now i have to, put in a conversion <P :05> this constant that we're gonna multiply by is delta-in, minus delta-out, for Y-in minus Y-out. <P :07> so i've gotta convert from D-Y i solve for D-Y right that's, D-delta over D-Y is equal to that, quotient right. so i just solve for D-Y and stick it in for D-Y, so i get out, this Y-in minus Y-out, over delta-in minus delta-out, D-delta... and now i, integrate that, we still have this G over K-G-A-P out here. i integrate that i get, pull out this Y-in, minus Y-out, over delta-in, minus delta-out cuz that's a constant. that integral, is just the natural log, of delt-in over delt-out <P :07> and now if i rearrange this... this is all getting to the equation this is merely to help you understand where this equation came from i do not expect you to re-derive this. you get down to L-C is equal to G over, K-G-A-P or K-Y-A, times, Y-in, minus Y-out, over <P :04> Y minus Y-star, log mean. or, del-log-mean. where, Y minus Y-star, log mean, is equal to, del-in minus del-out, over the natural log, of del-in, over del-out. this log mean thing, is just that. it always has that form of, you take the difference of the two in the numerator, and then it's the log of the quotient of the two in the denominator, natural log, the quotient of the two yes?
S2: um on your graph i seen where you got delta from how that was ya- Y minus Y-star. but is del- delta-in, is that Y-in, just like the distance of Y-in to the equilibrium line?
S1: right delta-in is Y-in minus Y-in-star. it's just that, [S2: oh okay ] Y minus Y-star at that one end and Y minus Y-star at the other end. <P :06> is this okay? <P :04> the whole point is that, if you know Y-in and Y-out, and then you could find Y-in and Y-out-star, that means you'll have delta-in and delta-out and then you can just calculate the length. there is no, you know graphing or integrating or anything like that it's just a straightforward equation... the only thing weird, that you have to worry about, is if the two deltas are equal. then you get a zero up here, and like a zero down here and the thing blows up. so if they're precisely equal, which is really an easy, integral to do, you just have to realize that so if you ever get it that it's zero, just realize it's a straight horizontal line and you just have to take either of the dels and multiply it by the Y okay? <P :04> so that's the only caveat for this. caveat that's a good word. <SS LAUGH> that's an excellent word. any other, words you want to throw up here i can use. caveat wow. there's no quid pro quo that's for sure. <SS LAUGH> i never knew what that meant anyway. <WRITING ON BOARD THROUGHOUT UTTERANCE> um <P :04> so <P :04> so, let's... remember our example that we did...? right? and we, remember this? we got uh L-C is equal to twelve-point-something feet twelve-point, five? yeah twelve-point-five feet. let's do it, and pretend that that equilibrium is linear, when it really isn't we'll just take the del Y-star one end minus the other end so again you can go back and redo this i'll just put up the numbers here so you can check yourself, when you have time to go over that, so it's just L-C, then it's (the) forty which is G pound-moles, per hour, over the one-point-five feet squared, is a cross-sectional area so that's now the G. and then we had the K-Y-A down here, pound-moles per hour per feet squared, and now we have to multiply by, this thing which is Y-in minus Y-out so it's zero-point-zero-seven minus zero-point-zero-one, over that, Y minus Y-star log mean, which is, and i'll write it down, zero-point-zero-seven minus zero-point-zero-four-two, minus zero-point-zero-one minus zero... over, the natural log of, zero-point-zero-seven minus zero-point-zero-four-two, over zero-point-zero one minus zero. and that gives twelve-point-seven feet. so we went through all of that graphing that, stuff <P :07> hard to do here we just assume it was line- even though the equilibrium line curves i do it sp- specifically curved in there, so it should be off a little bit, that's how much it's off. okay? questions about that...? seem okay? 
<P :10> 
S3: excuse me? <P :04> how do you get the Y-in Y-star in this example?
S1: uh you just go over here and, Y-in is like point-zero-one and the Y-star is zero and then for the input it's point-seven, and you go down and it's point, zero-four-two. you just go, go over and go down, alright <P :07> other questions? <P :06> did i answer your question? is that okay? okay. everyone clear on that? <P :04> yes?
S4: what is the units on the uh seven-point-two?
S1: <WRITING ON BOARD THROUGHOUT UTTERANCE> uh pound-mole per hour per feet squared, feet cubed isn't it? or is it feet squared? [SS: cubed ] feet cubed right? yeah, thank you. <P :19> okay, that was, the assumption, of... linear equilibrium, and constant flow rates. <P :04> if we no longer have linear equilibrium, then we have to go back and graph that, one over Y minus Y-star, do the integral of that, to get that, nog, that number, of theoretical transfer units based on overall gas phase resistance. and now what we're moving to is what a- if we don't have the constant flow rates. okay? <P :05> so what we did is... define, the solute free flow rates <P :14> and i apologize i shoulda started with that simple log mean driving force equation. but i kinda like to do it after you do the f- the first equation so you know where it's coming from and just plug in for the simplification. now i'm gonna do the more complicated one <P :16> so we have, G is equal to G-S, over one minus Y, and L is equal to L-S, over one minus X, so our flow rates are now in terms are solute-free, and i think we went through and and derived, the concentrations <P :06> capital Y was, solute-free and one minus, sorry Y over, one minus Y, and X, was X over one minus X. <P :09> okay? <P :15> and we came down with the fact that if you had, a curve, now in units of X, capital X and capital Y... with the equilibrium data <P :07> plotted as capital X, and capital Y you can't do, little X versus little Y, on that plot, you have to convert, to capital X and capital Y. okay? so if you do capital X and capital Y plot the equilibrium data, now if you do your operating line... the slope of that is gonna equal L-S, over G-S. okay? yes? 
S5: is our equilibrium line straight or curved?
S1: doesn't matter.
S5: okay.
S1: okay. i do a curve purposefully to say that it can be linear or nonlinear. cuz yo- what you're gonna do is get those, Y_ now it's gonna be, capital Y and, capital Y-stars and then you convert them and plug them into the equation and that's what i'm gonna show you. okay...? okay. 
<P :07> 
S6: Mr Burns?
S1: yes?
S6: wh- what again are L-S and G-S?
S1: <POINTS TO BOARD> L-S and G-S are the solute-free streams. they're the streams of inert, gas and inert liquid that are flowing in, without the solute. okay <P :08> so now we can just, plot, the op line, the operating line, and the equilibrium, data on, our capital X, capital Y, plot. find, capital Y-star and capital Ys... you then actually need to convert, to the small Y-star and the small Y. <P :04> you plug into... a new equation, which i will now give you. okay? <P :09> and don't forget this conversion, you can just rearrange that equation and small Y is equal to capital Y, over one plus, capital Y, and X is equal to capital X over one plus capital X. <P :09> and what i handed out, what did i hand out? what'd i just hand out? [SU-M: (xx) ] i'll get it i got it... i, handed out, now that problem redone, with X, capital X and capital Y. okay? the op line's a little different, i used the other equilibrium line, the solid line now not the dashed line, and i plot the op line for both A and B. okay? <P :16> i have to do a new mass balance, to cut to the chase, this is what i'm gonna give you, for variable L over G, the new line is just now G-S, outside instead of G, G-S instead of G, and you have an extra one minus Y-squared, in the denominator. okay? you would think it would just be a one minus Y, because that would just be replacing G str- straightaway, but it's not, you have a squared in there and i'll show you why. okay? and you can see you just get slightly different results. <P :07> new mass balance, we have a particular cut, at Z, and Z plus delta-Z, coming out as G-at-Z plus delta-Z, going in is G-at-Z, flow rate changes now, so you have to keep track of the different flow rate going in and going out of the stage it's going to change. concentration going in is Y-at-Z, this is Y-at-Z plus delta-Z, the liquid side is going i don't care, and this is the flux. <P :10> mass balance now, Y-at-Z, G-at-Z minus Y-at-Z plus delta-Z, G-at-Z plus delta-Z, is equal to, the flux of A times little A, times, delta-Z that's the same thing as before. and now you take the limit, and rearrange things and the only difference you have, and here we're gonna plug in, just to be the same, N-A is equal to K-G-P, times Y minus Y-star, the only difference now is when we, do this <SS LAUGH> difference, Y-at-Z plus delta-Z times G-at-Z plus delta-Z minus, Y-at-Z plus G-at-Z divided by delta-Z. before we had just D-Y, D-Z, now we have D-Y-G, D-Z. because G is changing t- you cannot pull it out, of that derivative. this is exciting isn't it? <SS LAUGH> you guys just d- there's nothing like, a good derivation on a Friday, right? <SS LAUGH> am i wrong?
S2: yes.
S1: i'm wrong? 
<SS LAUGH> 
S2: i didn't mean to say that out loud
<SS LAUGH> 
S1: you just thought it really loud, and then heard it. <SS LAUGH> that happens to me a lot. there's a lot of things i don't mean to say out loud... K-G-A-P, Y minus Y-star. <P :08> okay? <P :09> so, let me just put in what that derivative is... cuz this is where that one minus Y-squared comes from, D-D-Z, of Y-G, is D-D-Z, of Y, G-S over one minus Y. which, if you do the_ what is it? the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator all over the denominator squared. and i'll save you the time. G-S over one minus Y, squared, D-Y, D-Z. <P :17> so that's what you gotta plug in, due to your, extremely, short, attention span i will now skip right to the end, unle- unless there's popular demand otherwise, which i fear, there is not going to be. that gives you just G-S, and you have that K-G-A-P down here, you're still integrating from, Y-out, to Y-in. it's one over you've got that Y minus Y-star is still there, D-Y but you have this correction term, mon- one minus Y, squared. it is not just replacing this with G-S, or its definition, it comes from, the Y-G varying, and then taking the derivative of that. <P :07> questions on that? 
S7: (i have a question.) will you repeat what you just said about, the varying? i don't know the very last thing (that you)
S1: oh yeah it's this, this term we're taking the derivative of it that's where we get that one minus Y, squared, [S7: oh okay ] in the denominator. it's not just_ the equation we had before was this, and this, so it's not just saying oh G is equal to G-S over one minus Y, let me just stick that in there. you you can't do that. okay? <P :04> any other questions? <P :15> what am i erasing here? so, if i go back to the example and i do the conversions now X-in was zero, so that's zero over one minus zero which is zero. Y-in, was zero-point-zero-seven now i convert it that's, one over_ sorry point-zero-seven over one minus point-zero-seven. what a lecture to translate it's going to be all, letters and numbers. sorry about that... <SS LAUGH> zero-point-zero-seven-five <P :08> which reminds me of that time i was out on route seventy-five and i was driving alon- oh sorry. <SS LAUGH> tr- trying to get in there non, technical stuff. one minus zero-point-one <P :07> what do you notice about low concentrations, of Y? 
S2: Y is (approximately equal to the) capital Y? 
S1: how does, Y compare to capital Y when the concentration is low. i'm sorry what did you say? 
S2: i said Y is approximately equal to the capital Y. 
S1: yes. they're about the same. once you g- start getting up above point-zero-five it starts changing and up above point-one, is really concentrated. okay? so typically the dilute, quote unquote, is less than about point-oh-five to point-one. why don't i give you an exact number? i can't. if it's below point-oh-five almost always you get the same answer whether you assume dilute or concentrated. between point-oh-five and point-one it's usually, off a little bit above point-one it's usually, you know significantly off five or ten percent. <P :04> so now you can see on this new graph instead of plotting Y-in at point-zero-seven i plot it at point-zero-seven-five. <P :04> exciting huh? <SS LAUGH> it's fascinating. the minimum operating, sorry the minimum flow rate, which is part A <P :06> you can calculate out the slope. remember it's L-S min, over G-S now. it is not L over G, these will errors that, if you're paying attention... <SS LAUGH> you will remember this on the exam... sorry. didn't mean to peg you there. if you don't convert back it will be the incorrect answer... this slope if you draw it out is zero-point, seven-eight... you then have to get G-S, which you don't have, which is one minus Y-in, G-in. which is twenty-four-point-eight, pound-moles, per hour per feet squared. and then you know that L-S min, is equal to, zero-point-seven-eight, times G-S. which is, nineteen-point-three... pound-moles per hour per feet squared. <P :08> which, did i convert back? i did not and i should've and i don't have a calculator. but you have to convert back that's, L is equal to L-S over one minus X. <P :06> so you have to convert back, the nineteen-point-three, divided by one minus X what is the X-in? oh it's zero. <P :05> right, so we're okay. <P :04> questions?
S4: is L-S
S1: sorry someone yeah sorry.
S4: is uh, L-S supposed to be, pound-moles per hour feet cubed?
S1: uh, no.
S4: squared
S1: per cross-sectional area so it's like a velocity. right this is the weird unit, that i always forget pound-moles per hour per feet squared. it was that K-Y-A that had the cube. [S4: okay ] other questions? <P :06> other questions? <P :04> 
S2: um, just like the Ys are similar for dilute solution is i- is it the same for L S and M? they're the same? 
S1: yes, Ys or Xs are similar, for dilute solution same, thing. this was zero so it doesn't matter, it was all solute-free coming in. right? so it's pure water coming in so that's solute-free.
S2: okay.
<P :09> 
S1: quiet please. if you go through and then do, part B <P :04> if you plug in all you have to find your Y_ capital Ys and capital Y-stars, convert them to small Ys and small Y-stars. you cannot not convert you will get the wrong answer. you convert them, and then you go forward and sum the integral you should get, L-C equals twelve-point-eight, feet. i will spare you the details but if you have questions you can always ask, your, humble and, proficient and uh, um, uh, good words, uh... <SS LAUGH> people. <SS LAUGH> to go over the details... now you also have uh HOG was, now different, i define HOG as G-S, over K-Y-A, which is that constant out front, and i had three-point-four-four, feet. <P :09> questions? yes? 
S2: um, is this equation L equals L-S over one minus X (i mean it's) or is it capital, capital X-in or small X? i know it doesn't matter for this case but 
S1: well if i want L-in then i use X-in. which is [S2: okay ] what i was looking for, okay. so if that helps yes. i was just writing it down in general terms.
S2: but does it matter if that's the small X or the capital X? i mean i know it doesn't matter in this case 
S1: that has got to be the small X. [S2: okay ] <P :06> there is so much talking going on. <SS LAUGH> but i will remain calm, <SS LAUGH> i will not scream... <SS LAUGH> i will not pick on anybody unless i really get upset and then i will. um, any questions on that? we've done now a packed column. we had assu- these assumptions constant flow rates equilibrium stages linear equilibrium. we no longer have equilibrium stages. we do the simplification for linear equilibrium. we then did the simplest case, for, nonlinear equilibrium. then we went forward what happens if the flow rates aren't constants? these flow rates are changing, because the solute is absorbing ten percent almost, of the solute might absorb through the length of that column that's gonna be a big change in the flow rates. turns out it doesn't affect our design very much, because we come out with, twelve-point-eight-feet, versus twelve-point-five. and actually, that is wrong. this is an example, of, and it's in books it's an example of, a little bit of knowledge is dangerous. cuz what you should also do, is, this, concentrated stream if we're taking into account, these high concentrations of solute, will actually do something which is called, diffusion induced convection... [SU-M: mm ] [SU-M: whoa ] you did not cover this probably yet, in mass transfer. so, we will cover it here. it will be so much fun, that you won't want me to stop, in like six or seven minutes you'll want me to go on, because it will be just riveting. <LAUGH> right? 
SS: yes
S1: and you won't want to go to heat and mass transfer because we're doing mass transfer here. so what's the point? alright so there's that heat thing so i guess you can go on there. <P :15> oh what to teach you in this section...? what to do? what to do? lemme do it this way. you will never stand for this derivation i know it. <WRITING ON BOARD THROUGHOUT UTTERANCE> not today. the weather's too nice, so i'm gonna cut to the chase and i might come back and fill in some details on Monday, Monday which is gonna be like probably dreary and rainy and, and you'll all be depressed that the weekend is over so, you'll have no other thoughts on your mind but, you know, depression and this class, so i, naturally make it kind of, low-key. <SS LAUGH> so, we have L-C, i gave you the first equation which was K-G oh sorry G over K-G-A-P, integral from Y-out, to Y-in, one over, Y minus Y-star, D-Y, correct? this was for, dilute solutions. <P :08> if the flow rates... are not, constant that's what we just did. so that's L-C is equal to it now we have to put a G-S up there <P :04> K-G-A-P, integral from Y-out, to Y-in, one over_ now we have that one minus Y-squared, term, Y minus Y-star, D-Y, right? <P :13> that's only part right cuz you have to include diffusion, induced, diffu- when my voice cracks will you write that down as sort of like an accent on the syllables? <SS LAUGH> the, diffusion, induced convection. this derivation usually takes a couple days, complete explanation even when i teach graduate heat and mass transfer, it takes a while for people to pick up so i'm gonna do it in like one and a half minutes right now, verbally. <GESTURE> you ready?
SS: no
SU-M: go 
S1: kay. so heat, if it's really hot and really cold over here heat goes that way right? [SU-F: yeah ] more heat goes it doesn't push anything right it's not a mass right? so if i have, like a mass here a concentration high concentration low concentration. it gets dilute, those molecules just kinda diffuse over. if i have a huge concentrated, stuff here, low concentration here now there's gonna be a lot of stuff moving. if there's mass moving is that diffusion or convection? 
SS: (xx) 
S1: i i don't know. right? it's kind of, diffusion cuz it's moving by diffusion but the diffusion itself causes some, convection of mass and if you were standing here if there were molecules here they keep getting <SS LAUGH> hit on this side by that concentrated stream. so they might push or sweep along, some other molecules. that is diffusion induced, convection. only happens in very concentrated systems. what's my time? anybody get my time? is that under a minute? it's about a minute and a half. okay. so what we accounted for was in the column these changing flow rates but we never accounted for this sweeping, of those molecules along. if we do that, what you have to do is go back in, and correct, the mass transfer coefficient, to include not only diffusion, through that film layer but some convection, through that film layer which, you can calculate, based on the diffusion. if you do that, the only thing that changes in this equation is, you have L-C we're using G-S now cuz we're in concentrated regime, K-G-A-P integral Y-out, to Y-in... on the deno- on the bottom we have Y, minus Y-star, D-Y, we have a correction one minus Y-squared for the nonconstant flow rate, and on top, we have an annoying term, one minus Y, star-M. which accounts for that diffusion induced convection. and that, what'd i call it? one minus Y-star-M is... it's like a, log mean driving force kinda thing it's, one minus Y-star... minus one minus Y, over, L-N, of, one minus Y-star, over one minus Y. alright, a- and i like to write it like this, you can also just, simplify the top, which turns out to be, Y minus Y-star, and then put that over L-N of one minus Y-star, over one minus Y. but to me, i can't remember where it comes from, whereas if i just remember, it's a it's a log mean thing, with a star in there then i can remember how to write that down. <P :07> so what i'll do is depending on how i feel on Monday, go into a little bit more detail on that term, this is actually what you'll find in your book but it won't look like this. i'm trying to get to where the book is, and i'm trying to make that comparison, okay? is that a question or a, thank goodness we're done, back there? <LAUGH> yes? 
S8: since um, the top
S1: quiet please.
S8: since the top of (xx) dilute solution, [S1: yes ] is the um, three, the new equation is also gonna have (D-X?)
S1: yes.
S8: okay.
S1: yeah the, we have the dilute solution equation we have this equation which you probably shouldn't use, unless there's a reason like it's too complicated to do the full equation. and we have the full equation. mkay? this is the one, that we're gonna focus on... have a good weekend. have a tremendous, superior, uh, jejune, prosaic, wait i don't know what i'm talking about weekend. 
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