noc19-ch17-lec50

[Music] [Music] welcome back we've been talking about vapor liquid equilibrium phase diagram calculations and particularly we have focused on the binary mixtures so today we'll start and again with the basic construct of the phase diagram or for the binary mixtures and we'll discuss about something called lever rule which many of you must be aware of it but for the sake of completion I will do that and subsequently we'll talk about non-ideal mixtures and how do we generate the phase diagram for that so as you already know that if it is binary mixtures okay which follows let's say cross flow okay in such case we can clearly draw pxy diagram okay where the liquid the line which we call as this is P and this is XY so this is it could be a straight line okay this is P XY hey let's say a px a particularly x and y here we are referring to a a which is let's say the one which could be more volatile so this is something which we have looked at earlier and this sometimes this line which is a liquid line also called bubble point curve okay and then you have this other line which is a curve which is for the gas phase this is a saturation kind of vapor line okay and this is P by a line okay you can clearly see that in if this is a bubble point curve that is the first bubble of vapor one can obtain from here for a given pressure right so if pressure is fixed P system is let us say fixed then we can have an equilibrium between the liquid phase here and a vapor phase here and this would be called dew point curve okay this is your P system now I mentioned about this is a liquid but this is something liquid and this is a vapor but so this is a saturation liquid line this is the saturation vapor line but this would be also one can say is the sub-cooled liquid okay and a similar line which we have done in in engineering thermodynamics so this would be your superheated vapour okay okay so this was a P XY curve we can do the similar analysis for t XY okay but note that we have clearly considered one of the component because he thing is here your the pure a the piece at for pure a is more than that of pure B okay that's something for a given temperature so this is when it's a P X Y means that T is constant right so so that's that's important that we have just considered that that X P of pure a okay saturation point of P saturation vapor for pure is more than that of a pure B okay okay now we can extend this thing for liquid for sorry t XY okay in this case we'll be having this kind of curve okay and the tie line this is also called tie line okay will be somewhere like this right so let's consider that for a given system we have something like tesis right for the particular case here all right so now if you look at it here this is x and y and the composition of Y has to be more if it is a volatile so that way this would be your T Y curve this would be your TX curve and this will be liquid this will be vapor again the liquid of the you sub cooled liquid and this will be a super heated vapor okay and this is that constant pressure P okay now make a note of this fact that for a given T C's you have your X equilibrium Y composition for considering a tie line this alright so this is the the points which represent the phases at equilibrium with this composition okay it's a vapor phase is a liquid phase and the component let's say a will be like this now if it is a mixture of the would somewhere let's say here okay then it's going to be the equilibrium but of course this will have so this is like an effective composition but this is something which is combining the vapor and liquid together but there is a phase separation indeed right so how do you find out that equilibrium or in general if you have this how do you connect this thing to diagrammatically if you've been asked that a composition of Zadok with is that equilibrium is given to you and it phase separate and the diagrams are given to you then how do you use this diagram to obtain the information such as a day of IA so one can use a lever rule liber rule essentially kind of uses these diagrams and let's say if this is at equilibrium and if I'm going to say that this is a length L and this is a length let us say V this is a branch or segment V and this is a segment L then I can use this information to obtain the informations of the compositions so how do I do that let me start with this simple material balance so if you considering the mass balance here okay the total mass would be something like you know mass in the vapor phase plus mass on the liquid phase and then you if you look just looking at the mass balance for component a so then it should be something like Z which is a total composition multiplied by mass is equal to Y a and V Plus X a and L so this is let us say one equation one and then the question two so if you multiply Zed here in the first one then I'm going to get Z multiplied by equation 1 we said n is equal to Z a and the law said a and L right so I can now subtract this to equation so or in other word I can do is in the substitute equation three in two right and I can get Z a and v+ z EI and L this is why a + V + X ok alright so this is now the expression I have I get za minus XA and this I'm now subtracting this with the common terms here so NL is common here and this coven this is going to be Y a - Z a and V therefore and V minus n L can be written as Z minus X a divided by Y a - that hey now if you look at it this is the mass information if it is available to you in the vapor phase or liquid phase and if you are aware of such a or one of these two composition then you can use this equation to find out so this is nothing but the segment if you look at this this there is nothing but the segment line segment V so this is nothing but this right because this is that a and this is your X a so this is a minus X a is this segment and why a - za is this segment so this is line segment V divided by line segment yeah okay so that's how we can use this approach to find out some of the unknowns so you can use graphs also if it is available to you so this is our famous liberal rule okay okay so now I can extend this understanding and we've been just talking about only the more or less idle case now let me just consider the case where gamma is not equal to 1 that means activity coefficient is not equal to 1 and that would be the case of non-ideal liquid okay so this is not very unusual because real system hardly follows Ross law since most of the cases you will find a not equal to B B so a is hardly unlikely related to or equal or identical to B B and that means that there's a non ideality is a usually a common in the real system so we'll be considering couple of cases to understand it so let's consider a a and B be interaction interactions are stronger then a B okay so a B interaction is an unlike interaction a and B B are like like interactions now if you have something like this your gamma of I should be greater than one okay that is something which suggests that these interactions are stronger hence the gamma is equal to one we'll talk about it this expect of that but usually or rather for such cases your gamma will deviate from one that's true but gamma will be more than one okay when like interaction okay are weaker that means a and B we are weaker than unlike in such case gamma is less than 1 okay so these are the two important cases the first one we often say is a positive deviation from ideality the second one we often say is a negative deviation from ideality we'll talk about this more once we understand is molecule in nature so let's consider the case of a binary mixture okay so we will consider a binary mixture to illustrate this understanding consider Prioleau pressure okay which essentially means I can consider a vapor phase as ideal right the second thing which we can consider is that you know we'll be using the reference of Lewis rondelle okay and we'll be considering that chemically dissimilar species okay so they are not similar species it's a dissimilar species okay so now given this that you're you have a Randall Lewis Randall reference what we can consider now is the ideal gas mixture that's a right hand side what about the liquid it's a gamma I X I P is sad since it's low pressure we are going to consider Phi ice at 1 and the pointing correction to be 1 okay so you have this now relation which we are going to use for such a case now if it is a binary mixture it means you have gamma a XA PA s is equal to Y a P gamma b XB PB s is equal to gamma Y B and P so these are the two equations you can add this up so adding would mean that you will get the following gamma a X a Y s plus gamma B 1 minus X a PB s and this is going to be so this is the relation we got okay now this is rather interesting because in our ideal cases in our ideal case scenario when we had at all store completely the same equation was used but with a gamma equal to 1 right but here and of course you could see that in the 4 discs such a case you had a linear relation of X with pressure but now with this the linear relation is not anymore true okay now how do you find out why a if you have this relation so why a is going to be from here is going to be gamma a X a PS divided by P right and this P now K is something which we can use this expression so you have gamma a X a pas divided gamma-x a PS plus gamma b1 my sexy PBS alright so this is the expression we have now okay so uh we can't a couple of cases and solve this for some such case where if we can use gamma relation with the composition so there we can use some of the models such as your two suffix models can be used or three suffix models can be used and we can solve this problem so we'll do that one of the cases in the later part of this particular topic okay so uh now we have two specific cases one is the gamma is less than one okay and the other one is gamma is less than so greater than one so let us consider the case gamma 1 is greater than one right now this essentially means that alike like interactions are stronger than unlike interactions now one can show that if gamma a is equal to greater than one then this also implies through gifts to him equation that gamma B should also be greater than one okay now if that is the case that means said if gamma 1 is greater than 1 this is also greater than 1 that means the pressure here okay would be more than the peer pressure ideal okay that essentially means that a pressure in the vapor phase of course is going to be more than ideal okay and this is why we call it as a positive deviation from raoult's law okay now what is a molecular description or understanding of this why does it happen if it is a pressure more now I the pressure is more that means the more particles have gone up in the air or rather vaporized it prefers the vapor phase of an answer is enhanced the pressure or a teenies increases the pressure now which essentially means if the particles are going up that means their interaction there in the liquid phase have kind of less compared to the case of the pure phase right now compared to the case of the ideal phase so this is this happens because of the fact that due to the weaker unlike interaction upon mixing a B is not held strongly right so you're mixing it and a and a of course is strong a BM B is stronger but when you mix you have a lot of coordination number between a and B and that means that there's a lot of particle between a a around beyond be around a however because of the weaker in a like interaction it is not held and does there is a preference to go in the vapor phase so it means that does the particles or molecules escape to the vapor phase which essentially means that a increase in pressure okay now similar to the understanding of what we did here for gamma greater than one one can also extend this understanding for gamma less than one where we call it this as negative deviation from raus now we have done this exercise for binary mixtures we can also try to extend this exercise for many many many components so it need not be just one because for each of them you will have this equality for the case of a Lewis trend of other case of a vapour phase so this expression will be extended for all different components and then you can sum it up for all different components so that means you can have pressure and this term here so here there are two terms these terms will increase if you increase the number of components so if we extend to M component we are going to get the following so this means simply summation X I now my PA said therefore what is y y is the same if you look at it why Y is always this term divided by total pressure so for y let's say I is going to be X I gamma I P I said a summation X I am I PS it okay so this is something which clearly make use of the assumptions and is a straightforward exercise for the case of the binary mixtures let us try to do a little bit of an example okay which is which will be same still simpler and then we will try to do more complicated where the iterative and other aspects of the numerical analysis will come into the picture so this is the first question so again we are talking about a non-ideal vapour liquid equilibrium and the first question is on the dew point calculation of a non-ideal liquid with t unknown so so here essentially what we need to find out is D point is that means D point calculation of a non-ideal liquid with T and T non sorry so so if you look at the Kawai question you have a binary vapor mixture which contains four air per 48 percent ethanol and in water 48 percent ethanol in water and the temperature is given determine the pressure at which the vapor develops the first drop of the liquid that's why it's a dew point calculation and what is the liquid composition so wise are given to you we don't know exhale right and what is being asked us to do is to use the three suffix modulus equation for the axis gives energy that means essentially we just have to use it otherwise you can solve the problem for different models also and the parameter of three suffix model is given here okay first so we'll start again with the fact that since it's a ideal vapour phase essentially so you have X I gamma I P is is why I okay so we assuming this to be a low-pressure that's why we ignore the the the fugacity saturation velocity of I at saturation and the pointing correction so now for the case of binary you will add it up this right hand side will become total pressure and the left hand side is basically it's given here so this is the first part and this is the second power second term this is the first term this is the second term and this is the total pressure so this comes from the first equation and they can also add here one and this will be X 2 gamma 2 P 2 s is equal to y2 okay now here the gamma information is given in the form of a bagra's equation the test book clearly has its expression so Ln gamma expression based on the three surface modulus equation is this similarly Allen gamma 2 is given to you like this so this you can derive it or you can take it from the test book so this is widely available the axis gives energy model for is also known it's also summarized or table eyes day in the test book so I've just directly taken it from there so this is the expression here okay I know a and B so essentially what I can do is I can plug in this expression here however these are depends on X 2 and X 1 okay okay so I that's what I did I brought represent gamma in terms of exponential term here and similarly here also exponential right and then I plug-in in this expression okay so this is how it looks like x1 exponential this multiplied by psi so this is nothing but gamma 1 and this is nothing but gamma 2 okay now as we have done for y y is again can be written as X 1 gamma 1 P 1 side divided by total pressure now total pressure is this so this is where it comes here gamma 1 is basically the exponential terms so this is now the expression we have so what we need to find out is clearly the first thing is that we have to determine the pressure to which this vapour develops the first drop at that particular temperature and the liquid composition so now what is remaining to be solved is the p1sat and p2 set so here the easiest way to do that is take the help of Antoine equation so one of them is it's an all okay without the ethanol so for that the internal equation is clearly available in the NASD wave data or of course you can also look at the appendix s of the test book which we are following the other thing which one can do is particularly for water one can take the P one set from the stream table okay so calculate the saturation pressure from n to n equation of for ethanol obtain the saturation pressure from the steam table for water since the sum of Y you would be one that means you can clearly take this and as well as you can also write y2 and then you can sum it up then you have this one one is equal to the first term which is this plus the second term and since X 1 can be written in terms of X 2 can be written in terms of X 1 so you will have this right-hand side only in one particular variable considering that you have calculated p1sat from the Antoine equation or pin that from the steam table for the case of water so you have a simple equation here which is a dependent on one variable which you can plug in you can find out you can you can plug those equations in the let's say excel sheet and you can use the solver to obtain X 1 okay so once you obtain X 1 you can plug back this expression to the same expression here pressure now X 1 is available that means X 2 is also available a and B we know T we know P one side and P 2 sat is known so that from there you can obtain the pressure so this is a simple from the mathematical part of is very simple this part of this analysis is Mary bit numerical and that's why you have to use some kind of tools such as Excel tool okay and that is very handy for that now if you look at the data which we obtained using the three suffix modulus equation and and compare that with the experiment it's very close by that tells you how well this analysis can be done using simple approximation which we have made and that is the one where we consider it's a low pressure we consider the ideal phase to be how the vapor phase to be I don't guess so I hope that you've got this understanding a bit this is a first set of examples which we solved we'll be using another one where you need to use iterative ways to obtain the answers okay so I will continue this exercise and I will stop here and we'll see you next time and with a more complicated and interesting the problems at hand okay so see you soon [Music] [Music] [Music]