 Now we will take the series reaction so our famous reactions are going to R R going to S K 1 K 2 just to prove that what kind of contacting is better for series reactions like exactly what we have done for the parallel reaction. what he takes is he takes a b car okay and then he puts in this all the I mean here it is only a I mean it can be also a plus b also okay let us take to understand easily only a is there initial at time t equal to 0 and then he imagines that this reaction is happening in the presence of light okay so with this solution is irradiated in the presence of you know by light so he puts a bulb here yeah that is the bulb where of course yeah this is the light here you have to put filament also okay good and of course there is also a stirring stirrer is moving there yeah so this is one setup and another setup also he just imagines that we have same size baker stirring but he would not put light here but he will take a small amount of solution this is the pump then it goes through a small pipe and here there is a glass very small one so here you put the bulb yeah this is the bulb where again you have light yeah okay so the difference is that here we have a small amount of solution coming and the same intensity here also used right so then the solution is exposed to that light whereas here the solution is exposed to this light entire solution is exposed but here it is only small part okay so now let us see and here also it is a here also a at time t equal to 0 right yeah you have discussed this in your b tech some of you would have discussed okay I mean most of you would have not discussed that so then what happens here let us see the reaction is happening in the presence of light right so when you switch on the light at time t equal to 0 after filling up and after stirring then what will happen light will go and attack the entire contents okay then what is there maximum now at time t equal to 0 which component A is there so then A is exposed that means A is exposed to the light okay and because of this presence of light the reaction proceeds right so let imagine afterwards you know may be after 30 seconds so that means some R would have formed because reaction is going on the moment you switch on the light so after that I have some amount of R and some amount of A but because we are talking about immediately after 30 seconds or 45 seconds which one will be more and which one will be less A will be more because I think not 100 percent condition has not occurred there A is more and then R is less so that is why still A gets more light because simply it is more there right so like that you will have again next time period if you discuss then imagine then what happens again is that you will have again slightly more A and less R but R composition is slowly increasing after sometime may be I think just values only after 10 minutes probably you will have more R and less A now the light R gets more light right and then A gets less light this reaction is still going on but this is started reacting to S so like that with more and more time when you analyze finally A will be everything converted then R also will be converted to S so you will be left with only S in the system okay so that means when I plot this C A okay C versus time with time only what you are plotting okay then you have A this is C A not decreasing like this correct no and this is C A and R we know initially it is 0 right then slowly some amount form then increasing then increasing and it goes like this yeah so afterwards when more light of more light is going to only R and less light to A then this will start decomposing more and more so then you will it will reach a peak point and then it goes down right but S will start this is C R and S will start slowly and then finally it may reach this is C S if all A is converted to S then you will not of course again you know only you will see this line all other things will become 0 that means at the end you know finally when you allow it for completion because it is not reversible reaction it is only irreversible so finally you may reach for long time if you wait only C S so that means this is the one and our interest here is R R is the desired product so now we have to find out what is the time that is maximum C R you get and beyond that we should stop the reaction okay so this is one way of operating the system and you get this kind of concentration profile yeah In the case of C R R will be decomposing right R will be decomposing Why it is more than C A T fall down is left and right No no no all three at any point of time all three must be same all three should satisfy this relation if I take pure A this is valid at any point of time if I take at this point all three concentrations must be this concentration okay so that is there I mean it may not be exactly to scale so that is why somewhere you may not get that kind of thing so but all three must be satisfying this condition C R not equal to C R and all that okay good so now if you go to this setup now what will happen let us see it started now time T equal to 0 you switched on this and then pump is running pump is running and switched on this light and when you switched on this light the intensity is same but here amount of liquid is more and here amount of liquid is less so that means in the presence of highly intensity light then this will be practically going to completion of the reaction what is the final completion of the reaction Yes only okay yeah so that means there may be small amount of or may be there but essentially it go to yes then that will come back here and again you are taking out and when it is coming here only S is coming inside right and then you have again practically more A only again A will come here when you are recirculating and okay same amount of A small amount of A will come goes here getting reacted and again S coming there okay right so under these conditions how do I plot this concentrations C versus time again right what will happen to C A this is C A not that is decreasing anyway this is C A and what will be A sorry R R will be may be very very small like this it may be this may be C R that means practically you do not see much concentration of R inside the here here inside this reactor okay yeah inside this time and S will be going on finally at this point it has to go may be beyond this it will go to C A not where again all 3 must be equal to 0 right okay so what is that lesson what you learn here this is this is C S what is the one what is that we have done in this case and what is done we have done in this case yeah here what we are doing is we are now just leaving the concentration on its own and just waiting for the reaction to occur in the presence of light anyway okay right this is one kind of contacting but here what are you trying to do you are now trying to do here a stream you are mixing with a different composition of A correct no I mean here practically there is no A everything is converted to either R or S but mostly S because intensity is very very high that means here two different concentrations are missing streams of two different concentrations are mixing that means this is not a stream here this is already there and then you are sending this stream into this with a different concentration whereas here you simply allow the system for the reaction to take place okay so if you want your R as the product which system is better for the first one so that means what you have to do to maintain this R high what is the kind of system what you choose reactor now you come to our reactors which reactor will maintain this kind of condition which reactor will maintain this kind of condition this is P F R or another alternative batch either batch or P F R will give you A going to R R going to S S R is maximum and compared to a mixed flow reactor where you have some reaction going on already you have some concentration which is different than the fresh liquid which is entering correct no the fresh liquid which is entering that is CA naught and it is mixing with the liquid which is inside the tank whose concentration is at the outlet concentration of the all components it may be S it may be R it may be A so that is what exactly we do here that is why here you do not get much R so do not use mixed flow reactor if you want to get more R okay so this is a wonderful thought experiment that means this is true everywhere so whenever you have an intermediate in fact is not only here I can also put T if you want intermediate this and this you go only for this one not this one okay and chlorination of benzene is one of the very good examples for this you know benzene chlorination where they have series parallel reactions okay or only series reaction benzene benzene chlorination monochloric benzene dichloric benzene dichloric benzene they are not parallel I think they are series yeah series that is one of famous example where what information also is there industrial information is there in that okay so this is the rule now let us let us record that rule so that you remember that you have drawn these figures no both the figures very good yeah you can write now just below that you know in the first beaker you can write this one as first this this is second beaker in the first beaker the contents remain homogeneous throughout all changing slowly with time whereas in the second beaker the reacted fluid is continuously being mixed with fresh fluid okay that is correct no this is fresh this is always at CA naught right and then you are now bringing another stream with a different concentration may be a little bit of R and mostly yes okay good yeah so pull stop there in other words we are mixing two streams of different compositions okay now the general rule you write just below next para for reactions in series the mixing of fluid of different compositions is the key to the formation of intermediate pull stop the maximum possible amount of any and all intermediates is obtained if fluid of different compositions and at different stages of conversion fluid of different compositions and at different stages of conversion are not allowed to mix that is the rule law right I do not know may be without understanding would have just taken down right you just go through once more for reactions in series the mixing of fluid of different composition is the key to the formation of intermediate okay how do you mix is important there the maximum possible amount of any and all intermediates it is need not be only one intermediate it can be even you know one more but all these intermediates when you want okay the maximum possible amount of any and all intermediates is obtained if fluid of different compositions and yet different stages of conversion are not allowed to mix so just leave it on its own okay good so next para please write based on this we can say that batch operation and P F batch and P F no operation not required batch and P F should give a maximum R because there is no mixing of fluid streams of different compositions on the other hand M F R should not give a high yield of R because a fresh stream of pure A is being mixed continuously with an already reacted fluid in the reactor I think this is clear no this I do not think you will forget also because you can just remember those two things and then always remember that for series reactions if the intermediate is one of the desired components right then do not mix the streams just allow either like in batch reactor or like in plug flow plug flow you are just making them enter and then just you are collecting at the end you are not interfering anywhere whereas in mixed flow you have reaction going on and continuously you are putting fresh okay that is stirring at very high speed and that means you are mixing two different compositions which will not give you very high intermediate conversion intermediate concentrations for intermediates okay good so now let us check that you know what we have learnt through small diagrams like this which contacting pattern will be better okay so I think shall I write here yeah I think you take this and then afterwards I draw which contacting pattern of of the following figures next paragraph you write example otherwise example which contacting pattern of the following figures can give you a higher concentration of any intermediate I mean that is you write left or right because I am going to draw two figures now left and right so that is why left you will have here right so one contacting pattern is this that is simple P F our reactions are this A going to R R going to S may be T also okay so right is so that is the kind of contact pattern that means either left or right which one is better so B is P F with recycle where R equal to 1 and here I have two tanks in series okay yeah another one is C where I have two tanks in series yeah but the feed is distributed like this and in this case yeah in this case this simply like this good that means between this and this which is better so D is this is a simple pipe but laminar flow and here also I have left side a pipe that is turbulent fluid inside moving okay so now we have to just discuss very simple things only which contact and pattern is better to get the intermediate like let us say R is our product what we want yeah so the first one is left is right okay like a symbol or tick yeah okay yeah this is one left is right so then the second one left is right you say right is right why how many tanks you require for almost plug flow yeah so how many you have there yeah and of course you can also ask me sir what is the recycle ratio okay yeah that is 0 to infinity right but what we feel is R is R equal to 1 is near to plug flow then this one near to yeah when compared to this one near plug flow so this one again left is right okay now what about this one right is right correct now tell me why right is right left is right oh that's why why left is right so because we are having a CA notice we get in two different what is our room to maximize R so that is the rule intermediate so maximize R is not the rule what is the rule we should not allow the intermediates to make sir that is the rule yeah but is it happening here or is it happening here sir that intermediates we should not allow them to not intermediates I think different streams of different concentration should not get mixed yeah here you know here we are again mixing the fresh one with some other concentration right and here the fresh one with some other concentration okay so that is why this is not right but right is right okay good what see me you are see me know I think between you again I am more confusing I think she is nithu right you are nithu okay you are laughing what tell me tikka okay anyway so next one right is right why right is right left is right Dinesha what is your name Adinarayana okay Adinarayana is Adinarayana is Adi means beginning so I think beginning in Adinarayana is telling that you know yeah you say left is right okay so you say right is right or left is right who is right centry is right why unnecessarily all the problems okay somewhere in the middle okay that is why I think this in Buddhism Gautam Buddha always told that you know always better take a center path middle path middle path is always safer so the puja right is right I am going to ask both of you why Adinarayana why you say left is right that confusion is still not done the turbulence in okay we are talking about velocity proof it is not the turbulence you imagine in the mixed flow reactor that is what do you know now I am afraid to say chaos okay so this is the problem you know I do not know whether I told you this also this also you know once you are aware of something it is really problem this turbulent I think Adinarayana that is the one I think sometime back also when we are discussing PFR some people came and asked I think merit he came and asked me okay so when you are talking about the pipe flow and then turbulence what kind of turbulence you are imagining there it is not the turbulence where this molecule is coming and mixing here right no in the actual pipe flow when you say turbulent pipe turbulent flow what are the renounce numbers you are talking first of all yeah about 10 to the power of 5 and all that okay very large values and what kind of you know the diagram you draw there to show the turbulence you also draw that many times turbulence means what you draw you draw only like that correct no it is not you do not draw like this like this like this like this right so this will give me this turbulence will give me almost to actual velocity profile okay I mean this is very good actually in fact I do not know you are not appreciating me I have to only appreciate myself only for this reason I have brought this example it is there in the Levenspiel book otherwise this very simple one I do not have to discuss that much I expected that thanks for you know fulfilling my expectation okay that I thought that you know this will happen some people still would may be thinking that laminar it may be better because there is no mixing here okay but there is residence time distribution when I have flat velocity profile here what is the residence time distribution 0 whereas here I have the laminar profile where the molecule which is sitting on the top of it it will come much faster and molecules which are coming near the walls they come late we also have that you know relationship average velocity and then maximum velocity maximum velocity is yeah twice the average velocity or otherwise half of the average velocity is half of the maximum velocity right so that means if I take 10 minutes is the mean residence time right so I will have molecules coming out 5 minutes afterwards till then you do not get anything right so that is why this one turbulent means this is the one watch what imagining in a pipe flow where the turbulence is not from that end to this end okay those molecules are only going like this and then moving forward and if I draw the velocity profile almost it is like a flat velocity profile but depending on the velocity you will see some slight bending near the walls and in our imagination of plug flow we are taking that also straight okay I do not know that is why this is more near to plug flow this one is more near to plug flow that is why again right is right okay yeah so we have two left right and two right right two two okay so this is very simple scheme what we have and then I think this is very nice simple things what we have now I think it is your problem now that means I am just going to write the equations you have to derive some of them you would have already done in your btech because for simple scheme like a going to r no this one a going to s you have already derived did you not derive what is cr max corresponding to that what is t max these are the favorite problems for gate examination and even other c re examination okay so that is what again I am going to repeat but later I think you know unless you work again once more at least you cannot solve in the examination okay good that is what what we are going to write now so now quantitative this is qualitative we have discussed what kind of reactor will give the best you know product right okay now quantitatively quantitative quantitative analysis okay so we have again same reaction K1 K2 yeah first we have to write the rate expressions so r a how do I write here I can also write minus r a okay otherwise r a equal to to maintain the uniformity some books write K1 C a this is equation one okay r r yeah who is your K1 K2 CR that is equation 2 and r s K2 CR K2 CR okay this is equation 3 so now either batch or plug flow because it is constant density system batch or plug flow will give me the same equations but you have to replace time by tau that is all okay please write there time by tau t or tau good so now I think this is very simple to solve because this is nothing but dc abide there is no minus plus only because minus I put there so the first one will be CA by CA not equal to e power minus K1 t or K1 tau if it is batch if it is a plug flow okay good so then CR by CA not but this is difficult for you you have to do that I do not know whether you have solved any problem using integrating factors and all that how many math scores you had in btech differential equation 4 differential equation is 1 you also had partial differential equations yeah okay I think that knowledge I think is required here so here you will have K1 K2 minus K1 e power minus K1 t minus e power minus K2 t I think this bracket is there right yeah so this is 4 this is 5 and of course CS will be yeah or CA not minus yeah CA not otherwise I have to write yeah either that I think someone told one that one you will get if I divide by CA not this is 1 minus CA by CA not minus CR by CA not okay simply you can substitute these two and then you get this is the sixth equation because now our interest is to find out the CR and CR has a maximum problem but whereas CA does not have that problem it will go to highest if you just leave it there is no optimization there optimization comes only for this where I have to find out which t and which CR maximum okay so that is why we have to now differentiate that equation 5 I think again I am requesting you you have to practice CR by CT that is right equation is not valid for K1 equal to K2 that is very good yeah this is only for K1 not equal to K2 right but you know this equation is also valid if you take the limits K1 K2 you know the ratio you take and then there is a mathematical derivation also for that but safest is K1 equal to K2 equal to some K and then solve the differential equation you will get the simplest one okay by the way that is one of my favorite problems in the examinations already gave an assignment already gave an assignment or examination so I think you know that is why only mathematics thank you for reminding so now this we have to differentiate so once we differentiate this what you get CR by CA okay I will write here CR by CA not max max I think CR max I will write equal to yeah some of you may be remembering this K1 by K2 yeah K2 by to the power of K2 by K2 minus K1 to the this is to the power please remember so this is again not valid for yeah so this is equation number 7 and corresponding to this Tmax corresponding to this Tmax is yeah very good that is a ln K1 or K2 by K1 divided by K2 minus K1 so this is equation number 8 yeah if you want you can arrange in the other way also okay so these are the things there and the concentration profiles and all that you know it will be similar to this only those concentration profiles so that is why whenever you have that maximum problem then we have to find out through differentiation what will be the corresponding value okay good now we will write the same thing for mixed flow I think mixed flow is a very nice guy we do not trouble you that much MFR yeah okay MFR we can write you know like the material balance like we have written earlier for the other one we have written here also we can write the material balance so here again we take normally CA not only CR not equal to 0 CS not equal to 0 here I have CA CR CS and volumetric flow rate V volumetric flow rate V this is V and we can also write of course in terms of FA not and all that okay so for a material balance yeah again I will write so that you know it will be practice for you V CA not equal to V CA plus minus RA into V where this minus RA can be written as in this case K1 C K1 C into V so from this equation 8 9 10 so then we can show this as CA by CA not equal to 1 by 1 plus K tau where tau equal to volumetric flow rate so this is 1 similarly I think okay R let me write for R MB for R so we know this is 0 entry V CR not which is 0 equal to V CR minus RR into V so this is equation number 12 and if you substitute for this RR this one okay and see also you will get there yeah this is V CR equal to RR into V okay so RR also equal to RR if I substitute this equation this is K1 CA minus K2 CR into V or CR equal to K1 CA minus K2 CR tau so this is equation 13 yeah so I can write of course I am not giving all the steps there the final steps CR by CA not yeah K1 tau by 1 plus K1 tau into 1 plus K2 tau yeah this is the equation this is 13 or 14 again CS by CA not also we can write let me write this time equation K1 K2 tau square whole thing divided by 1 plus K1 tau 1 plus K2 tau yeah so this is 15 we have now other problem like this one maximization right these are the standard nice beautiful problems so this I have to get now CR max maybe remove this some of you may be remembering also this one yeah nice equation CR max by CA not equal to 1 by K1 by K2 this is square this is equation number 16 yeah corresponding tau max no tau optimum will say not max there this is max this is optimum K2 K1 square so this is equation 17 okay so I think you have come with calculators no can you tell me what is the PFR maximum and CSTR maximum that CR max in PFR if I have K2 equal to 1 and K1 equal to 2 find CR max or CR max by CA not if K1 or K1 K2 okay for PFR and MFR and the corresponding times corresponding tau corresponding tau okay good so this is the one yeah what will be the value this one is same for one know this is square root of 2 1 by square root of 2 okay so first I write MFR which is easy yeah what is CR max by CA not point 171 okay and tau max optimal is not max point point 171 point 177 okay yeah what are the units units I am not told so when I have to give you here this is minutes and what's on second and also minute so this will let me say two seconds and 0.693s, 2 seconds I think very fastness, so minute signals is also minute signals, so this is minutes sir. so what is for P of R? 0.5 sir, CR Max P of R CR Max by CNR equal right? Ley. ఫివాచియా కివిఇవ కూతిలరు చిరిలు కూవా సిర్మా పారిటా. more mathematical but only thing is you have to go through that equations, Ok. Yeah, that is the one and afterwards we will discuss a little bit about DINB and Frandy Fussier. I will give the actual references. If someone is interested, started liking reaction engineering, they have to go to original papers and then see, Ok. But I can give derivations in the examination also about this. Simple derivations, like find out CR by CA not in even in DINB in some simple cases, Ok. Good. So this is the one, I think again this will prove our rules where we should not mix streams and all that. Now when I told you know the concepts part is over, right. So now it is only hard work part, sweating. The thought experiments have gone. So now you have to go to laboratory and then work. So that is what, you have to go home and then start deriving all these equations. So that is the hard work part, Ok. By the way, how many hours you are spending in CRE every day? Zero. Every day two hours? Every day two hours is excellent if you are able to do. Sit with us spending I think every day 27 hours. Ok. I think this stretches the clock and then goes to 24 to 27. Ok. See, discipline wise, every day two hours, two hours, two hours if you do, I think you are doing wonderfully well. You will do also in the examinations So now we have series parallel reactions and normal scheme. What we have is A plus B going to R, R plus B going to S, Ok. And this also can be represented in terms of series and parallel. How do you write as series? A going to R, R going to S. Here I have plus B plus B. In parallel scheme, Ok. See here surprising thing is I think there is another thought experiment. I think we do not discuss that. But only we believe that the experiments are right and then see whatever rules we have applied for parallel and series, they equally applied to even series parallel reactions. What is the rule for parallel? If I have a desired reaction with higher order, you should have higher concentrations, maintain high concentration. And in the case of this series, we have to now say that do not mix the reactants but just allow them to react on their own, Ok. So the two rules are again applied for these two. But it is not that easy, you know, our mind, I told you know someday, someday I think I have thrown some 10 chag pieces and asked you how many. So beyond that, you know, our mind may not accept. So that is why what you have to do here is every time a problem is given, you have to solve the problem. And that means you have to mathematically, you know, write all the equations and then get the solution if it is C R by C A naught or if it is C S by C A naught, Ok. Normally this is the desired product, normally that is what our convention, Ok. Why, you have headache? Under tension? Ok, yeah. So this is the one and we will try to write the equations first for this one. That means again, you know, some students may not be knowing also how to write R A, R R, R B and all that, Ok. So that is why we list out all the reactions. I think when you next do denby reaction, there definitely you have lot of possibilities. All the reactions, at least list out all the reactions and then we will take whatever is possible to write the equations. But I think again, you know, the plug flow and batch, you always go to integral expressions. You have to go to, you know, you have to integrate them, Ok. But if you go to mixed flow reactors, then there is no integration. Correspondingly you have to substitute what is minus R R, R R or R S or minus R A minus R B and then try to get the corresponding concentration, right. So the overall picture is that, Ok. Depending on which reactor you are talking, you have the equations, that is why I told you do not have to learn anything new because the same design expression is only valid here. You have also seen for parallel reactions what we did, right. And even just now, even these things also, we removed that mixed flow when you are writing equations for A going to R, R going to S also, the same design expression. So as far as design is concerned, that is why I told you, it is very very easy to remember contacting in the diagram. It is a long time I have drawn the diagram. I think sometime I have to draw again, otherwise you may forget, Ok. So that is why actual part is, difficult part is only reaction part. You can also see now, particularly when you have multiple reactions, it is a challenge. How do you get all, you know, all these rates, like, Ok, let me draw that and then you will see. How do you get R R, R S, R A, R B and all this? It is not that easy to get all the constraints. That is the real challenge, Ok. So now when you are writing the rate equations R A, I am writing only everything R A, R B like that, so that minus also will go this side. DC A by D T, if it is a batch reactor, equal to K 1 minus, I think, may be I will remove this. We will write general equation. K 1, yeah, can you tell me? Minus R A, K 1 C A C B and fortunately we are not taking any reversible reaction. If you take reversible reaction, Margea, Ok. Again I think, you know, the other term will come and then your integration will be held, all that. Ok, good. So continuation, where are the 17, 18, Ok. So R, R, no, no, sorry, R B, R B equal to, first it is participating here, yeah, minus K 1 C A C B minus K 2 C R C B, this is equation 19, excellent. Now R R, yeah, K 1, it is forming, K 1 C A C B and again it is excellent, C R C B. So this is 20. Then R S, R S is a nice guy at the end. It is like, you know, lost brother or lost sister, least affected for many things, correct, no? The smallest guy in the house, whether sister or brother, we leave them there on their own most of the time, Ok. So R S is like that, very nice. Yeah, what is R S? K 2 C R C B A C B, Ok. So this is equation 21, K 2, Ok. So this is definitely complicated when compared to our series alone and parallel alone. From here onwards you have to really think how to solve the problem. There are no simple rules of integration or simple rules of differentiation, Ok. Why I am saying that is, then, you know, sometimes, yeah, you have to get the relationship between, Ok, like one example I will tell you, I have now here C A C B C R, 3 R there here, right. But I have to convert into this one into only one, one component, either C B, Ok, mostly C B only. So then only I can solve that. That means you have to now find out what is the relationship between C A and C B and what is the relationship between C R and C B. That part is difficult part there. So to get that part what you have to do sometimes is that you have to manipulate these reactions to get, I think I will remove this, yeah. So we have to manipulate something like this like only remove the time, Ok, remove the time variable for example, that means, Ok, this I also told you, if it is a batch system when you are taking, this is nothing but minus d C A by d C A by d T. This is d C B by d T, this is d C R by d T, this is d C S by d T, Ok, good. So our relation, our idea is to get somehow the relationship between these, you know few expressions, like to eliminate one of them, like for example C A C B and if I want to express all this in terms of only C B alone, like that here it is only in terms of C R alone. So for that kind of thing you have to like R R by R A. What is the relationship? That means I have here d C R by d C A, negatives we have brought this, yeah. Now d C equal to tell me, Pooja is very good I say, I think without writing all thought experiments. Now good good Pooja, like that only you have to do, if you are able to imagine everything in the mind, nothing better than that, minus 1 plus C R by C A, Ok, yeah. So now to get a relationship between C A and C R, I have to integrate this, correct no? I have to somehow find out C R term separately, C A term separately and then I have to integrate. This is a differential equation, as far as possible it need not be always d C A by d T, right? So that is why here d C R by d C A, this is also a variable for me, this is a variable in terms of C R I have to find out. So that will give me the relationship between C A and C R, right? So now one relationship I will give, that means this differential equation has to be solved. Why I am telling all this? Probably now you know you stop, you stop listening to me, out sometime, Ok. Now again that is why, please, I think another may be 15, 20 minutes I think, then the reactions also will finish, Ok, yeah. So we are doing all that because finally we have to get the relationship between these concentrations like in this, in this also I have C R and C B and C A. So I have to convert this entire equation into one concentration possible, Ok and this also similarly. So that is why you have to look at all these equations and then see which one will give you that kind of relationship. Like if you solve this, if you solve this, what you get is C R by C A not equal to integrating, what is the number? 22, Ok. Let me also write here, integrating equation 22, yeah, what you get is C R by C A not equal to 1 by 1 minus K 2 by K 1 C A by C A not K 2 by K 1 minus C A by C A not which is not valid, Ok, which is only valid for K 2 by K 1 not equal to 1, Ok. So this is equation 23 and C R by C A not equal to C A by C A not, yeah, some mistake, C A not by C A log, yeah. So this is valid only for K 2 by K 1 equal to 1. So this is equation 24. You see this gives me the relationship between C R and C A, if K 2 is not equal to K 2, K 2, this one is K 2 not equal to K 1 and this one is K 2 equal to K 1. See now it is too much mathematics now, that is all. But again I tell you too much mathematics, what too much mathematics, they are not really too much mathematics. It is only mostly first order differential equations. So you have various techniques to solve this. You have integrating what is that, integrating factor. So what are the other methods? That is all we know, nothing else. Not partial, these are all straight forward order differential equations. Variable separable, yeah, variable separable then, yeah. So all that techniques only you have to use that here, Ok. Again I am repeating that it is only to get the relationship between two components, right? So that is what. So like that we have to solve this and then try to get, in fact it has been solved and this kind of, this kind of reactions, this is K 2, this kind of reactions and then the lot of information is available in any textbook. And Levenspiel is again, actually it is not Levenspiel, even before Levenspiel I think there are some German people who have done this you know, actually that graphs. Lot of graphs are available for the relationship between C R, C S, C B, C A in one graph, Ok. Given if it is a continuous reactor, I think it is expressed in terms of K 1 tau or so. K 1 tau is the Damkohler number for first order, yeah. That is one. In terms of reactor volumes is one. The other one is plotting C R by C A naught versus C A by C A naught, Ok. C R by C A naught versus C A by C A naught. C A by C A naught will give us what is the conversion, correct, no? And for that corresponding conversion what will be C R by C A naught which depends on K 1, K 2 variables. Of course it also depends on C B values, C B naught and all that. So these are the variables. So you are now increasing more and more variables because you have more complicated equations. So that is why you will go to multiple reactions more and more but it will be more and more mathematical techniques. Good, Ok. So that one I think kindly go through that and I will give only as far as possible in this section is concerned very simple things because I think I do not want to pain you also too much with this but I think again I can tell you actually this is what is more interesting in industry, Ok. But there is no simple reaction A going to R anywhere but in university always we start with A going to R. Luckily we are not talking about zero order because there is nothing to talk about zero order. So that is why because independent of concentration any reactor is Ok. So even if you put all those things in your pocket also they react, Ok. So that is why we are not talking anything there. So that is why this multiple reaction this is where I see the awkward face of C R E, awkward face. Why it is awkward? Because our brain has to work. So that is why we feel it is troublesome oh my god too much mathematics. But those people who love mathematics tremendous amount of information your passion can be put in this solutions. That is why that Fanda Fussier was the first person to find out for various conditions what could be the C R Max, that yield Max. Denby is another person, Ok, these two. And there are again few more reactions which I mean these are two are very famous most of the time, Ok. Now quickly shall we go to Denby reaction? I think it is not, it would not take much time. Correct? No, I think you are able to get these points now why we are in multiple reactions beyond simply A going to R going to S and also A going to R and A going to S. Those are the simplest schemes. Beyond that it is only the mathematics that is required rules are same only you have to carefully integrate and differentiate for maximum and all that but techniques are required for it, right. And slightly complicated like this if you go then you should be able to get you know to solve finally you should have also relationship between other components like C R, C B, C A, I mean C S and C A. Good? Yes. So now let us see Denby reaction. Oh, in fact I have not written here this is only for P F R what you have written, I mean these things P F R, Ok. Yes, so the same equation Ok this is P F R, Ok. I think that somewhere I have to mention this one integrating equation to far P F R, please write that otherwise now onwards you may accept anything whatever I say, P F R you will say S, M F R you will say S, Ok. Good. Yes, that is the one. If it is mixture flow what do you do with that equation? Pooja, mixture flow is the simplest flow. Then what are you thinking? What are you thinking? Yes, tell me then, that is the simplest one I think mixture flow, Raohal top, thinking is there integration in mixture flow? Yes, yes. Then what is that D C A and D C R and D C A? C R, C A, C A, C A, C A, C A, C A, That is all, that is delta also now. That is delta C R and delta C A and what is delta C A? C R F minus C R not, C R not equal to 0 here, Ok. And delta C A? C R F minus C A C R F minus C R not, C R not equal to 0 here, Ok. And delta C A? C R F minus C A. It is not C R, it is again, the definition is only C R not, you know, the definition is always C A, out and then in, Ok. So that is why it is the simplest one, that is why you do not have to really worry, that will give you very quickly the relationship between C A and C R, Ok. So that is why you can imagine now that, if at all I give this, mixture flow I may give, because there is no integration, very simple straight forward, Ok. So that is why, I think, you know, you can also write, for example, D C B by, or D C R by D C B, correct, no? Then you will still get the relationship, right? Yeah, so that is why, Ok, good. Now Denby reaction, let us take. So that is why M F R please make a note of that, it will be deltas, because there is no integration there. Denby reaction, Ok. Yeah, this is again simple scheme where you have A going to R, R going to S, this is K 1, K 2, oh, no, no, I think yes, I am using Levenspiel notation, this is T, this is K 2, this is 4, K 4 and this is U, yeah, so this is 1, 2, 3, yeah, 3, 4, this is the one, Ok. And here most of the time it is taken as elementary reactions, that means this, this, this, this, all these are first order reactions, Ok. So what kind of concentrations I have to maintain? Sir, independent of concentration. Independent of concentration. Then what is the very good, next very good for us? Temperature. Temperature. So when we have the temperature, then activation energy E 1, E 2, E 3, E 4 will come. Denby reaction is very famous for that arrangement, Ok. Depending on E 1, E 2, E 3, E 4 values, E means activation energy values, what kind of temperature scheme now you put? There is no one temperature scheme, like, you know, you cannot always use high temperatures. You have to use till some point high temperatures and beyond some point low temperature. That means you have to now reduce the temperatures. So somewhere you have heating, somewhere you have cooling, Ok. So that is why I think when time comes I will also just indicate, you know, and rule is same. What is the rule? If the desired product has, if the desired reaction has more activation energy, then maintain the temperature as high as possible. That same rule only. But how do you put that? Depending on you, because you have 4 parameters here, E 1, E 2, E 3, E 4. So whether E 1 is greater than E 2, like one simple thing is E 1 is greater than E 2, greater than E 3, greater than E 4. What kind of temperature scheme you put? How many she gave the answer? Ok, so then the other one is the reverse. Ok, E 1 is less than E 2, less than E 3, less than E 4. Then what kind of temperature progress? I think just think, I think there are possibilities, beautiful and also you have one of that, that is simple straightforward thing, E 1 greater than E 2, E 2 greater than E 3, E 3 greater than E 4. But there may be another combination like, Ok, E 1 is greater than E 2, but K 4 is greater than K 3. Ok, I mean not K 3, E 3 and E 4. So all combinations also are possible. Because it is not in our hands, you know, we cannot say that you reaction, you have only this kind of activation energy. If you are able to tell that it is fantastic, but we are not able to tell that really. So that is why rules are same. You have to believe me when I say that concepts you have already learnt. Because now temperature you should not forget, concentrations you should not forget. But in all these reactions where temperature, I mean concentration is not coming, it is independent because all the same, same order of reaction. But Fande Fussier reaction is not same order again, it is different order. One reaction is different order, another reaction is different order. There you have again concentration problem as well as temperature problem. So that is why it is more complicated. In 1960s and all that, you know, it is a beautiful playground for chemical engineers. Football, I think, you know, wonderfully you can play cricket, football, anything you can play there. Because there are so many wonderful problems, even now there are. But I think, you know, those people had really excellent way of solving the things because the first time they were doing all these problems. But now with our more knowledge, more technology, more cell phones, more, you know, pushing like this, pulling like this, tap like this, everything will work for us technology wise, but brain stop working. All thought experiments have gone. That is why it happens. Whenever technology increases, brain takes rest. That is the problem. That is why whether technology is really good or bad, highly philosophical people will discuss that. Till what extent technology is really good? Entire world is obese except Ethiopia and one more country is there. Somalia, Somalia, only skin. Ok, and only skeleton is there. Ok, very sad. Except those parts, most of the population, now we have 7.5 billions or something. Ok, I can tell you 7.3 billions are obese. Why? No physical work, no mental work. Correct, no? If you want to go, go and sit down in flight, that takes you. You do not have to work. Ok, from here to again airport, go by car, no walking. Right? And our body designed only for walking and working, I say. Not for travelling by flights. Why, really, it is not. It is not designed for that. It is only for working, hard work, hard work, hard work. When you do that, you will not have any, what is that? No, really, I think, you know, you have to work hard and hard and hard. So then only I think your health also is very good. Ok, otherwise what do you do except eating what else you are doing, you tell me? Only eating, you know, that is all. You come here and then you enjoy my drama and then go and eat. And go and eat in the tipa, what is that? Your mess. That is all. And then sleep. And tomorrow morning, Monday morning when I ask you, sorry, overslept. Naturally, if you eat more, you will oversleep. Ok. That is all what we are doing. That is why, you know, so many hospitals, so many doctors. If you want to avoid them, work all the time. Ok, so because as if we are saving this planet, I think we have designed cars, we have designed flights, we have designed cycles, we have designed many, many things. Destroying the planet indirectly, you know. All these are only indirectly destroying the planet. If our brain is not there, then happily like monkeys jumping here and there. And you know, you may have, we do not know whether they also may have IITs. We do not know. All monkeys. Because we do not know their language. So there may be a love spiel there in monkeys. There may be an Einstein who has really thought, you know, that is why there is a wonderful movie called Planets of the Apes. It is really a beautiful satire, you know, the other way. It is always, we are painting them. Now, at least once in the movie, at least let them paint us. At least in the movies. That is a real beautiful concept again. That is greatness of mind again, I can tell you. Ok. So that is why, we will just write now, actually concentration scheme is not that great here. But it is only the temperature scheme which is really good one. So we have to write the rate of reaction equations. And when I write R A, only once. Ok, of course this K 1 and K 2. Then I have here minus K 1 plus K 2 into C A, that is all, no? Yeah. And love spiel nicely, his brain is excellent. He will write. K 1 because he knows how to simplify things, you know. He writes K 1 to there. Ok. Because simply minus is there, minus is there. Ok. So now shall we continue or tired of so many things? I think. No, no, I am talking about equations. Not leaving. Equations only. Ok. So R, what is next one? R. Ok, R R, let me put. Ok. So R R is the desired product. Then it will have K 1 C A minus K 3 K 4. Yeah, what is that? C R. So love spiel language again if I write K 1 C A minus K 3 4 C R. So this is equation 2. Correct, no? I think this all of you got it, no? Because it is only participating in K 3 K 4 and forming here. Pooja. R is formed in this reaction, that is K 1 C A. This is reacting, this is reacting so minus C R. Ok, good. So then R S equal to? K 3 C R. See, I told you, you know, engaged brother, engaged sister. K 3 C A. See, C R, C R. K 3 C R. This is equation 3. Ok, then C, sorry R T. K 3 C A. Very good. K 2 C A. This is 4. And R U. Very nice, you have earned your lunch. Nicely told. Ok, I think few integrated expressions I will write for batch and Ok, for batch and P F R. Ok, so anyway beyond this we also have this material balance which is required for us, that is C A naught plus C R naught plus C S naught C R T naught plus C U naught equal to all that again, C A C R C S C T C U. Ok, and of course if you do not have R 0 S and all, that will be 0, then it will be only C A naught. So that relationship is there. And for batch C A by C A naught, when I integrate, this you should guess it know. What is the equation for C A by C A naught? By this time you should guess. Yeah, something happened to Pooja, I say. Yeah, K 1 2 R T R tau, I think T R tau, Ok. So this is what, I think this equation this is 6, this is 7. Yeah, so now C R by, I think here things are slightly complicated. Ok, so here let us take yeah, all these things are 0. 0 0 0 0 Ok, only we have pure component A Ok, so now C R K 1, K 3 4 K 1 2 E power minus K 1 2 T minus E power oh no, yeah, K 3 4, yeah K 3 4 T that is all. Yeah, do you, can you have some comparison between this and then what you have done earlier? This is K 1 by K 3 4 K 1 2 exponential K 1 2 T minus exponential K 3 4 T R right only. That is, it is in series. Yeah, Ok. So then C S by C A naught, this is slightly complicated because this is coming at the end. So we have K 1 K 3 by K 3 4 K 1 2 E power minus K 3 4 T by K 3 4 minus E power minus K 1 T by K 1 2 right, plus you have one more term here K 1 K 3 by K 1 2 K 3 4 so this is 9 Ok, so normally our desired product is this, we can leave it there but just to, I am just writing the other things how you get this. So then C T by C A naught equal to K 2 by K 1 2 1 minus exponential E power only I am writing, E power minus K 1 2 T. So this is equation 10 and of course C U you can get it C T R as minus of 1 minus of all those equations correct no? Yeah. So that you can write, I think that is C U by C A naught equal to 1 minus all this C R by C A naught C T by C A naught C R C T C S by C A naught and minus C A by C A naught, yeah. So now one more thing here is that you know our idea is to integrate to get the maximum of this C R Ok, so to get that maximum we have the equation C R max by C A naught equal to K 1 2 K 3 4 to the power of, divided by this is equation 12 and corresponding T max not a max actually optimal or tau optimal for P F R, so that is the one because that is again similar to your series reaction correct no? Exactly series reaction but only thing is you have more constant that are involved there Yeah, so quickly now writing for M F R this is very nice simple C A by C A naught equal to 1 by 1 plus K 1 2 tau this is 14 C R by C A naught equal to K 1 tau 1 plus K 1 tau 1 plus, no not K 1 tau this is K 1 2 tau this is K 3 4 so in fact you do not have to really worry about this because these equations are similar to only A going to R or going to S, Ok so the other reactions I do not want to write, no there are but you do not want to write but you can simply say C R max by C A naught equal to this one K 1 divided by K 1 2 yeah, now this is K 3 4 by K 1 2 this is half plus 1 this whole thing is squared, so this is 16 so corresponding tau optimal equal to 1 by K 1 2 by K 3 4 to the power of half, so this is 17, very good, right, so I think you know you do not have to worry that much even for examination because the same similar things you know A going to R or going to S only in denby reaction if R is the desired product, I can guarantee that I do not go to any other desired product or only I will ask, if I ask, Ok, yeah so the last one only just mention, front of this reaction denby reaction I have not given the reference so we note down I think when first title denby reaction you write C E S, you know what is C E S chemical engineering sin 9, Ok, Ok, I will write here chem inch science, Ok, normally we write like this volume 19, I think may be first time I have to tell you, Ok, volume 19 and page sorry, not 19, volume 8 page 125 here is 1958, Ok, I was 10 years old at that time no, that is normally we give starting page Ok, so normally we give starting page, I think some journals will give complete pages so then when you go then only you will know how many pages it is, Ok, yeah so this front of this reaction that is chemical engineering science this is volume 19, volume 19 page 994, this is 1964, actually see again that is beauty there, 60s and till 70s, whatever problem they did industry, afterwards only we lost touch with industry there is a beautiful story, I think you know to lose that connection with industry only one book is responsible can you guess that book transport that is the book which was responsible for chemical engineers to move away from the industry, particularly chemical academic because in that book what you have shown was, even earlier it was thought that all engineering problems can be solved only by either experiment or empiricism that is why you have so many empirical equations for heat transfer coefficient, mass transfer coefficients all these things, even friction factors, all these things only empirical correlations because they do experiment and then try to extend, scale up a little bit and then try to find out the correlation is valid, if it is not valid they add some more terms when they go bigger and bigger when 1962 when birds wrote light foot came and then wrote this book then all academicians thought that my God what is a beautiful idea because I think all the problems can be solved in terms of differential equations integration, it is wonderful idea is wonderful but we do not know how to use it so that is why all academicians now started working only in problem solving but not the problem formulation but in industry there was an existing problem they have to solve that but here we will simulate our problems so that is why we will say that one differential equation 10 boundary conditions each boundary condition will give me one paper paper publication only it became, it is not the solutions which we are providing to industry it finally became paper publication number increasing no challenging problems so that is why then industry people when they have a problem they used to come to academicians at that time in before 60s and all that perfect coordination between industry and university when this book came and when we started solving problems on the paper and also using different techniques for different techniques to solve the problems when the industry person comes and asks the professor then he will talk his own language what is his language matrices and vectors tensors that is our language in thermodynamics in transport phenomena you will see this in next semester only vectors and tensors then industry person ran away because what is this vector tensors I have never seen in industry he has not really seen in industry then they tried sincerely may be 10 years, 15 years, 20 years, 30 years and then they thought that all professors are a separate breed we are separated breed so you know gap widen so they stopped coming to universities and then they started their own industrial research because they need there are some problems where in industry it may take time it is not instantaneous solution you get it so that is why let us have our own R&D in industry so we will look into some long term problems and much more long term even then there was slight contact between industry and university much more complicated problems which you can they know that you cannot solve that problem that problem only will be given to academic institution they do not expect anything okay if something no really they do not expect anything but if they give that problem if a solution comes very happy if the solution does not come equally happy so that is how it has gone that one book which changed the minds of all chemical engineering academia okay now only again we are thinking whether how far how near we have to go to industry it is really destroyed because it is not the problem of the book it is the problem of our mindset in universities because we would be happy in solving the problems you know on the paper differential equations using differential equations all other techniques rather than actually going to industry and looking at the problem and solving it actually those problems also require these solutions solid when you have you know I do not know whether you heard what is called SMB simulated moving bed reactor highly mathematical industry people are using that particularly Germany is very good in that simulated moving bed highly programmed and highly controlled you know the simulated moving bed means you know normal adsorption column at one point they have four adsorption columns one adsorption column will be you know may be off filled up you know you have the adsorbent it is a batch process because adsorbent is in packet bed this packet bed moves now it is not literally moving but I think in the original laboratory it was literally moving also right so in one packet bed you will have adsorption going on in another packet bed you have adsorption almost about to complete it then in another packet bed you have the regeneration taking place because you have to use the same packet bed again for regeneration right so and then in other packet bed you have almost regeneration completed then that becomes again fresh adsorbent bed so all this is programmed like you know how the beds move or instead of beds moving people thought that let us move the solutions where they are getting adsorbed and this is widely used in bio chemical reactions where they have the enzymes are formed already in the liquid in the broth they call no reaction broth and then that reaction broth is pumped through these adsorbents and adsorbents will have different adsorbents not one single adsorbent so that means some adsorbents will take some enzymes some other adsorbents will take some other enzymes and then finally of course that illusion will come depending on what part you take out and then that is very very complicated you know this is really high tech processes in bio technology really high tech right that is after fermentation after the actual bio chemical reaction how do you separate the products so that is why even mathematics are really required in this in industry and I think another example also which I can I think I sent that paper Pooja was asking me that paper where you have haber reactor and all that chemical engineering science not free the word there was another paper which I said chemical engineering science paper no where is chemical engineering paper by that winter mantel one person road he was working in I think bio company also or hoaxed company I think so he has written a wonderful paper I think if all of you are interested I can send that paper it is a wonderful paper where how the chemical engineering started in Germany so at one point of time he also involved in that he first computer simulation of crystallization crystallization is another very very difficult problem it is very difficult problem if you want to get the type of crystals you want any kind of crystal no problem crystallization is the simplest I do not know I think a long time back there was a Tamil movie called Kamalasana Sri Devi okay these two people I do not know Tamil name is complicated but it is Balachandra movie where they go to Dilli and then he does not know how to live the Sivapala something will come that is correct I think other people may not understand that anyway so in that movie I think four fronts are there in Telugu it is Akalraj in Tamil is the original it is only I think dubbed there only but you know there is a wonderful dialogue where all the four fronts are there they cannot get jobs and they cannot live properly and all that so then the remaining three people will try to have whatever life they get but this fellow will not allow that I think he will always say that okay I want to have life only in this way that is why he says beautifully if you say idealism if I want to live like this only I will suffer but if I want to live like any other way that means by stealing by robbing by murdering all this also there is way of living okay so when I want to do that it is easiest because there are so many opportunities for that kind of thing but if you want to really live like an idealist there is only one way that is why I say you know truth is only one truth is only one whether you talk or I talk or he talks truth is only one but for all other people truth is I think you know when you want to tell lies so then you can tell you know for the same thing you can tell thousands of lies correct you are not come to the class okay how many lies you can tell you can first kill your father sorry I mean father I mean grandfather okay that is one lie you can kill your grandmother another lie you can kill someone also you know where some relative that is another lie and also you can say that you can and you may say that my leg was paining sir I cannot move another lie thousands of lies for the same one thing truth is only one sir correct reason may be only one you are sleeping but that is the problem okay yeah because it is over so then what happens to my Fanda Fusti reaction okay next class okay we will close it only Fanda Fusti reaction will take it thank you