 Okay, so this adiabatic analytical expression for, analytical expression for adiabatic PFR. Okay, so the equations which we know, V by F A naught equal to 0 to X A d X A by minus r A which can also be written 0 to X A d X A here I have to write K naught C A naught this may be already with you, so this is E 4 minus E by R T 1 minus X A, yeah this is equation 1, the other one you can remove sometime, I think why this equation I do not have tried this already you know, that already you know, so V by yeah, so the other energy balance equation what I have is sigma of F I C P I T minus T naught equal to minus delta H R into F A naught X A straight away after substituting you know that F naught X A I have written actually it is delta X A because X A naught equal to 0 that is why I am simply writing X A there so that you will come in our format, so T minus T naught is nothing but beta X A where beta equal to minus delta H R F A naught by sigma of F of I C P I yeah, so yeah I have to now substitute this T okay this is equation 2 equation 3 equation 4 okay, so now substituting this T there because I have to convert this one in terms of yeah here we are trying to convert in terms of T right yeah for this analytical expression otherwise we have to normally write this T in terms of X okay then integrate then you will have now all this we have now I mean this entire integral we are now trying to convert in terms of T and of course limits also will be automatically in terms of T, so yeah V by F A naught I think here let me write this equation yeah X A equal to T minus T naught T minus T naught by beta which we also write this one as lambda T minus T naught, so this is equation 5 where lambda equal to 1 by beta where lambda is 1 by beta yeah okay good, so now if I differentiate this equation then this will be D X A equal to lambda D T that is one expression then I have to substitute this D T D T will be in terms of okay D X A by lambda correct me here okay good, so then of course this from the same expression I can substitute for X A lambda T minus T naught and I am taking X naught equal to 0 I mean X A naught equal to 0, so in that derivation earlier you know which we have taken X A naught also was there I think I have not given that where lambda we have given an equation no alpha alpha equal to what yeah lambda E divided by 1 minus X A naught 1 minus X A naught that X A naught can be 0 that won't change anything okay, so yeah so after substituting what I have to say is substituting equation 5 and 6 in one substituting equations 5 and 6 in one one is this one okay V by F A naught equal to yeah I am giving the final expression should have some gap here you have to quickly someone has to check this lambda by K naught C A naught integral E power U okay I think in between I have to give I think another step so that you remember for continuity okay, so this will be E power E by RT D T whole thing divided by 1 plus lambda T naught minus 1 minus lambda T please check this correct already so fast correct no yeah, so now if I go here X equal to 0 fine X equal to 0 then I will have T equal to T naught because beta multiplied by 0 nothing T equal to T naught that is okay but now this final one in terms of T X equal to X means this equation only that equation is nothing but T equal to T naught plus beta X A, so instead of writing all that we have to only write this one here T naught okay good, so now this yeah now there is another substitution which we have to do to convert this into that exponential integral because it is not easy to integrate okay and analytical expression we are saying it is not correctly analytical expression it is still approximation because exponential integrate is an approximation they have tables where if you have that exponential of 1 may be it may go to may be 0.22227 something that is some value I am telling okay so that is why it is still an approximation it is not the truly analytical expression the way we are imagining okay yeah like for example our conversion X equal to for first order plug flow what is the equation for plug flow X A equal to remember that first order plug flow first order plug flow yeah X A equal to 1 minus E 4 minus K tau P first order okay so that you have to remember correspondingly what is for batch and batch is almost same and for mixed flow these things minimum you have to remember okay yeah any time so that is why that is an exact analytical expression right X A equal to 1 minus E 4 minus K tau P it would not be like this okay like that but still it is an approximation but in terms of exponential integrals okay so now if I want to convert this to in terms of exponential integrals I have to now put another substitution U as E by RT because this E by RT as it is so this has to why you have to put U I mean if you do not like U you can put some other variable that is not the one okay yeah why you have to write this one is that this one our exponential integral definition if you remember I think I would have given you that definition already right let me write exponential integral definition I gave X or Y E of Y is minus infinity to Y okay that is a dummy variable I can write U power T by T into T so that is the definition okay so that is why I have to this is important for me E power minus T by T form that is how you have to get and also the limits okay then only I can write this E I and this E I is listed in that abrumovic and stegan I think stegan stegan STI GEN handbook of mathematics or mathematical handbook okay yeah so that is the book that is very famous book probably we have not used you know till now I saw that book only with Professor Anand because I think his PhD was highly mathematical thermodynamics you would have not known him because he was the director last year you know you know because he is from your place AC tech he studied in AC tech came to AC tech many times yeah so only with him I have seen that book I asked him sir where did you get this book we have not seen this at all he said my and also sometimes he has to use that you know exponential integrals all kinds of or error functions error functions also are listed there the values so for that calculations only I bought that one and when he was doing PhD in Florida long time back okay so this is the substitution because idea is to convert this into in that format that is the reason why we are putting that so now we can yeah we will write this equation this equation 5 this is 7 yeah so then you are T from this I can write as E by R U yeah so why I have to write in terms of T is because this delta DT also I have to convert into U yeah now tell me what is DT yeah DT minus minus E E by R yeah I write E du by R U square yeah so that is the one what we have to substitute there that is equation number okay this is 8 this is 9 that is 1 and yeah so correspondingly for T and all that I have to substitute this E by R U here only we have okay so I think that is all with this substitution that means substituting equation 8 and 9 equation 8 because this T means here I have to substitute this one E by R U right yeah this is for this T and in 9 here I have to substitute 1 U will get cancelled right yeah here U take L C M and all that so here you have U square and here you have U in the denominator so that will get cancelled and the final expression what you get is yeah V by F A not equal to oh again I will write substituting equations 8 and 9 in 7 in 7 V by F A not equal to minus 1 by K not C A not integral E by U du 1 plus lambda not lambda T not lambda E by R that is correct only yeah minus 1 this again whole thing multiplied by U yeah now the limits here again you have to go to this one okay yeah so this will be when T equal to T not this will be U not so that is why what we write here is U not which is nothing but E by R T not similarly in the here U equal to that is all E by R T okay yeah so there is another again substitution that will come to convert exactly into this kind of exponential integrals okay I mean because I am taking a lot of time I said yeah so another one I think for easy remembrance this this entire thing let my god my time has gone whole 1 plus lambda T not by lambda E by R what I substitute was nu so now this equation will be simple equation like V by F A not equal to minus that minus is there 1 by K not C A not integral U not to U plus lambda U not lambda this is nu nu U minus 1 correct no check yeah so now how do I integrate that 1 so you have to that split into partial fractions okay so when you write that partial fractions I am solving the whole thing now almost okay let me write that partial fractions so now you know using partial fractions we have A by U plus B by nu U minus 1 this is how you write now yeah now tell me what is A and B instantaneous answer sir A is minus 1 B is nu correct what you said okay so now I have to now split this into 2 what saw me last time this is why this is the reason you know now it is a mathematics class we have forgotten about what we have done beyond this beyond this after writing energy and material balance okay now it is only solution solution solution there substitution to get another thing another substitution to get another thing another substitution like that it has to go okay but anyway these things also are required for definitely at a higher level when you go to slowly deeper and deeper into subject you cannot avoid mathematics and you should not avoid mathematics but all of us have this phobia for mathematics because we are not in touch with mathematics that is why if you are in touch with mathematics you are not afraid right even I could have not made that mistake of you know writing that limit only thing is we are not in touch with that so we will forget sometimes not sometimes many times we forget okay good so once I do this substitution the expression what I get here is V by F A not equal to 1 by K not C A not this is U not to U then E by DU that is one fraction then I have another one minus 1 by K not C A not I think this chap is are spoiling my handwriting otherwise it is not that bad okay 1 by K not C A not again U not to U nu E by U into DU so these two okay so here what I can do is again divide oh my god whole thing I have to write this I will divide by nu so this becomes U minus 1 by nu okay so I think I will write that one directly hopefully you do not mind no so this is U minus 1 by nu okay you have the equation for nu nu is this I lost naming this okay so this is 10 11 12 13 14 okay so here nu values I know nu is nothing but this one 1 by lambda 1 plus lambda T not by this one okay so there is another substitution where you have to make that substitution that is Z capital Z equal to U minus alpha where alpha equal to 1 by nu okay this one where alpha equal to 1 by nu also equal to lambda E by R 1 plus lambda T not correct no this is what what we have also written lambda alpha value okay good so this is the one so now again using this I have to convert these limits as well as I know into exactly this is no problem for me but this I have to again convert that so by putting Z equal to U minus alpha what is that I am trying to do now U equal to Z plus alpha right so here I will have E power Z plus alpha so that means E power alpha is a constant correct no that I do not have to integrate this all these things I know alpha is a constant that will come out right and then you will have this entire thing anyway Z right so E power Z by Z into DZ you know this okay what is DZ here DU so there is nothing so that is why this entire thing will become now E power Z by Z and you have here E power alpha which is constant which I can take out okay and limits also there is no problem that directly this will become Z 0 and this will become Z this will become Z 0 and this will become Z so then I will have the exponential integrals that is all I think anyway let me you have done let me do lot of only giving few steps but ended up giving the entire thing one over gone okay this is equation numbers I have to put this is equation 15 this is 16 and also we have another equation DZ equal to DU this is from equation okay from equation 15 okay so now substituting equations 15 and 15 16 and 17 in 14 substituting equations 15 16 17 in 15 oh in 14 yeah cannot write that okay so what you get is this expression V by F A naught equal to oh yeah anyway this tail is there always A naught C A naught so this will be U naught this will be U E power U DU minus again 1 by K naught C A naught yeah so this will be Z 0 Z oh yeah E power alpha I will take out here E power Z by I think I am writing chapters right but still the problem is that we do not have here that exponential form this is minus which minus you are talking limits okay that minus infinity how do I do that here I can split this this is U 0 to minus infinity and minus infinity to U you can split no in between right so that is all if you write this one again as splitting yeah minus infinity to U naught sorry U naught to minus infinity and plus minus infinity to U and this one you can write as minus minus infinity to U naught so everything is in place similarly there also that is how you got that final expression okay V by F A naught equal to after splitting that okay and substituting the corresponding things K naught C A naught exponential of E by RT minus exponential of E by RT naught okay this is 1 minus okay minus E power I will write here minus okay E power alpha by K naught C A naught and this bracket I have been here yeah this is E I of Z minus E I of Z A so that is the one is the final one okay so this is here and the exponential integral I know some of you would have been definitely switched off your brain okay this is what happens when you talk only about mathematics in the class after sometime 50 percent 60 percent then 70 percent finally 100 percent after sometime okay like this you know every class is there with a lot of mathematics because somehow we know we do not enjoy doing mathematics that is our problem if you are able to enjoy mathematics and you cannot avoid as engineers you have to definitely enjoy mathematics somehow you do not have that practice I mean frankly you just imagine I think all of us would have taken particularly we I think 5 year course we have taken math 6 M 1 M 2 M 3 M 4 M 5 M 6 how many you have taken M 6 so you also M 6 so mathematics portions are not reduced for M 4 for you M 4 or 3 4 years 4 years where it was okay so that is why even though we have done so many courses in mathematics but still we do not enjoy why rest me you enjoy I am not talking about you in general I am asking the answer you do not know yeah anyway that is the problem okay yeah may be yeah that is also required particularly when you know the real mathematics course they do not have to connect the real M 1 M 2 M 3 M 4 M 5 they do not have to connect but later somewhere that is why we have you know what is called chemical engineering mathematics they have that connection like that everyone I think I do not know whether civil engineers have a mathematics area is absent okay so may be last class I think everyone is absent after that one mathematics discussion okay so like that I think after sometime I think only 2 3 people only raise me will be there after sometime so this is the problem you know that is why always I try to talk more about physics more about how do you get the equation rather than the solution okay so this is minus infinity to y e power t by t d t this is also you know approximation after telling is this is 0.57722 plus ln y plus y square by 2 and 2 factorial plus y cube there is nice beautiful symmetry in all these things 3 3 factorial plus goes on series solution so if I say that y equal to 1 what is the value you can calculate right if I have 2 1.5 or 1.1 whatever even 1.1 1.2 whatever right yeah but in this case our value is e by r t right our value is e by r t so e by r t I mean e you know activation energy but be careful about the units particularly you cannot use calories here and here you cannot use joules or okay so that is why you and t is always in Kelvin all those problems will be there so for example if this value is 2.2 then you can go there and then calculate what is there so similarly this similarly this similarly this then you will get that yeah there is another also definition of this is noted as small i small e i this is also called exponential integral right but the definition of this is definition of this is x 2 infinity x 2 infinity this is e power minus u by u du this also has this expansion this is same thing but minus 7722 minus ln y plus y minus y square by 2 factorial now plus y cubed 3 3 factorial minus this is y okay right right so that is what you know actually these things in the latest book of Levenspiel he has included this I do not know whether you have seen that or not latest book means I am a third edition what you have you are not born when second edition came but third edition is the one which most of you are using and I do not know how many of you have read RTD chapter it is an RTD chapter it also comes in RTD there is one equation again which you cannot solve you cannot have analytical solution so you have to go for exponential integral okay in some places you have to go to error function right so that is why I told you know please remember all my words there are all beautiful words not only in the class I think even after in life also okay beyond certain point C R E is nothing but mathematics but unless you reach that point you do not enjoy mathematics okay yeah those who enjoy mathematics there are two books I think Aries is one the other one also who went very deep into mathematics is fromentent bishop I have written sometime back I think in the last class fromentent bishop chemical reactor analysis and design I think 79 or so 1979 but put a question mark I do not know exactly right but there are two 79 the second edition also has come first edition I think is 1979 second edition may be 1992-1993 okay but I think you also should know know the idea of any course is you have to be exposed more beautiful things and also you have to learn basic things that is the idea of any course right it is not that always only basics but basics plus exposure to complicated things this is one of the exposure to complicated things right those two books are very very deep I told you know we are losing you because unfortunately on our universities and colleges I think it is not that much even though there are two courses I think that much rigorous training in CRE is not there sorry to say this because the last 10-15 years I am conducting that zero examination okay that zero the test and I have shown you know most of the time it is only empty yeah out of 30 all those 30 questions are wonderful questions I can tell you I have not asked mathematics number one I just asked only simple questions like of course everyone knows only what is heterogeneous system homogenous system and other one is what is the other one I told also elemental law of mental also another thing is only differential and only two actually there are another two another two three okay yeah half life method is there and linear regression is there ideal reactors many people do not know differential analysis and that is what only they have done they have given only these two questions almost everyone answer but again you know even the other reactors they write MFR, CSTR, BMR and plus batch reactor plug they have forgotten all three are same only correct know some people call mixed reactor some people call continuous stirred tank reactor some people old people call be a back mixed reactor if one of the oldest professors are still teaching in your university then he may call it is only back mixed reactor okay so all those three names that is why that answer also not many by many people may be percentage wise slightly better than other places so if that basis is there and if you are excellent before I ask a question you are answering in basics so then I think we could have gone to this and this is highly mathematical okay so that is what is unfortunate thing that is happening last so many years that is why I thought after my retirement I have to go to I have to train the teachers that is one of the aim I thought in my mind anyway they have to call me I cannot go on my own because so I do not have that much money to travel and go okay I will now teach you no one will come also okay so that is why I thought you know teachers should be trained first in CRE then something may happen okay I think you know the overall picture I told you there is one beautiful quotation that the teacher not only teachers he gives an attitude towards the subject and there is another also I think sometime back also I told in my room it is there always right I think that one that is more important than simply solving problems like this approach to it okay okay and approach to it and also an attitude towards the subject so that means after taking that subject automatically your attitude towards that subject should change your approach should change okay approach I have already mentioned in CRE the beautiful approach is you should know only kinetics and you should know only contacting not grubs contacting that is all what you have and you know approach is the subject only not only you know you not day to day solving of the problem so that is why I have been telling you once you learnt that three contacting patterns batch and continuous two that is what I have been telling you you do not have to anything more to learn if you are trying to address only ideal contacting and I tell you 99 percent of the time you go only for ideal contacting then what is left now what is left is only kinetics and kinetics are very very troublesome very very complicated I can tell you right because first of all you do not know if I write okay Abdul what is the order of this reaction you do not know sir ya that is the correct answer really Abdul do not know or order do not know we do not know order you do not know ya so we cannot simply take get any information as a going to r okay absolutely we do not have any idea but on the other hand I write this equation ya what is this represent I know kinetics this is also very simple equation I know the kinetics similarly I can write for batch and I can also write for plug flow so the moment I look I know what is happening there yes that equation I get only when I have mixed flow no equation no other contacting pattern will give you like this that is why the here contacting is much easier because the moment I look at the equation then or even the equipment you know right most of the time it is a tank where you put the strutter and all that okay this must be CSTR if there is input and output if there is no input and output only continuous turning and all that you should have batch the other one always two lines for plug flow you know pipe so that is why you know much more information in contacting than kinetics absolutely kinetics you do not know anyway next time particularly when you come to catalytic reactions and also non catalytic reactions in heterogeneous systems you do not have any idea about what kind of reaction you get that is why all our approach should be how to solve this kinetics okay that attitude is liking that you know once I have this kind of clarity automatically my attitude changes ya this subject is very easy why it is very easy I should know only two things one is contacting other one is kinetics that is why my attitude is very simple now that is all what I have no complication but when you go to kinetics then go to deeper and deeper you will have more and more problems in the in getting the kinetic information okay good we will stop now because I do not want to start now that my mixed flow reactor so that is why I spent this time