 I think we will start now in the last class what I have done was I just gave only discussion and analysis of this plug flow reactors in series but I thought now we have to also give that expression right that equation for number of PFRs in series and when do you use this PFRs in series that also we discussed merit do you remember yeah when you have to increase conversion you can add one more PFR the reason is that you are increasing the residence time so then naturally conversion will increase okay so on the other hand if you have parallel equation I mean parallel PFRs then you add another PFR if you want to increase the production rates okay so that is all I mean same thing with even mixed flow reactors if you put one more reactor in series with may be 2 or 3 already there then your conversion will increase but parallel you can put another MFR then you will get production rate increasing but condition there is maintain conversion same when you have either plug flow or mixed flow in parallel in parallel okay what are the equations you have to use you have to use the same equations of what you have already derived mixed flow we have derived series also right yeah but mixed flow series only first order we have done the other order I left to you which are more difficult algebraically more difficult otherwise it is not that difficult at all just algebraically so that is why again I am repeating many many many times I am repeating in this class that unless you work you cannot remember you cannot do anything in the examination at that time do not blame me and do not blame CRU okay it is now just depending on you I think you have got what is the basic concept in reaction engineering and what is plug flow and all that you know those concepts are with you but only thing now you have to see is your work from your side you have to work hard otherwise you cannot you cannot score marks also that anyway you will see this Friday okay 21st I am not threatening you but you know if you are able to do that will be easy for you if you are able to do able to spend sufficient time in solving the problems but if you think that mental solving is okay so that means reading the problem okay A is given B is given and okay volume is given conversion I will calculate because all the data given there you do not actually put on the paper so then you are going to lose at that time the first person to be blamed is me oh my god this guy gives very difficult questions they are not difficult questions only thing is you have not done your duty properly right okay so that is why I thought again I will start this PFRS in series to actually derive the equations it is not difficult it is same thing but only thing you may have some doubts is something like this I have one piece of PFR then I have another piece and then we have another piece let us say only three are there so here I have F A not and V not X A not equal to 0 etcetera yeah and C A not and at this end I have if I say that I have this is 1 2 3 here I have F A 3 V 3 for constant density systems it may be same okay and also X A 3 and C A 3 now if I in between if I write this this will be F A 1 X A 1 F A 2 X A 2 these are the inputs that is all yeah so how do I derive the equation for this I will take a small you know normal way of deriving an equation for PFRS because it is a distributed parameter system you take a small slice write what is entering what is leaving and all that so that we have done many times so that is why I will take a small strip here entering leaving and all that so this equation already we know that gives me F A not D X A equal to minus R A D V this is the equation where you get confused is why this F A not but I have here F A 1 okay you do not get you do not get that doubt why how do you define conversion yeah but with respect to input which input you are now talking is that this input or this input F A not because you are writing the conversion based on F A not that is why this F A 1 X A 1 will not come into picture unless otherwise it appears in terms of only boundary conditions okay so using this if I extend that logic that means this F A not D X A you got by defining F A equal to F A not into 1 minus X A at least these things you have to remember F A equal to F A not into 1 minus X A that is general 1 so that this definition is based on F A not not somewhere else say somewhere else inside okay so if I have that then I will have total volume V by F A not is sigma of V i F A not which is also equal to V 1 F A not plus V 2 F A not plus V 3 F A not okay good so now if I write this this will be V by F A not equal to integral 0 to X A I am writing for this X A 1 D X A by minus R A plus yeah exactly X A 1 to X A 2 D X A by minus R A plus X A 2 X A 3 D X A by minus R A this is the equation if I take that is starting equation yeah now in fact your condition is automatically met and those people who still bother that no no no I think he is wrong this F A 1 must come F A 1 automatically comes there because you have lower limit as X A 1 similarly here X A 2 F A 2 will automatically come because you have X A 2 entrance this is one of the problems you have when you are actually trying to derive in the 6th chapter beginning okay many students ask me so that is the reason why again I am you know trying to tell where you can have this kind of confusion and mathematically speaking this gap in between is imaginary gap for us so when I have 3 reactors that will also be equivalent to as 1 reactor where that is entering with F A not and coming out as F A 3 how do I write that equation assuming that I have only 1 you know this plus this plus this all 3 is just only 1 tube and even if you have the connections connections is directly given through flanges you do not have but to show as diagram we are writing this this does not mean that there is an arrow between these 2 pipes okay and through arrow it is going right through arrow it nothing can go only it can show that the direction where it is going so when I have all these 3 together what is the equation you have total if I take V by F A not equal to 0 to what is the upper limit X A 3 D X A by minus R A and if you are able to imagine this one as a just simple calculus you know integral expression cannot I split this as X A 1 D X A by minus R A plus X A 1 to X A 2 D X A by minus R A plus X A 2 X A 3 D X A by minus R A this is pure mathematical thing right because this integral I can split into 3 parts I mean even 10 parts also even 100 parts right okay because this one till X 1 then X A 1 X A 2 X A 3 like this that is why this equation also is exactly this equation that is why you should not worry that is why general thumb rule is that even if I have any number of mixed flow reactors in series so that is equivalent to as a single plug flow reactor expression that is what is this that is what is this and automatically that also comes as this yeah that is the one so I think there is no problem there good so yesterday after the class some students were coming and asking me how do you get that for M F R 1 by minus tau we plotted versus you know that slope on the curve on the graph where I have minus R A versus C A okay so this may be bothering some more people this is simple calculus again because we are excellent in forgetting whatever we taught earlier that is why you are not able to recall that but I think I do not blame that student there are many I told you know all brother or Indians or sisters brothers and sisters so some more people will have that doubt but E is bold enough to come and ask other people would have forgotten thinking that okay does not matter whether it is 1 by minus tau or 2 by minus tau we should okay so with that attitude you again you know you your cobs will not be removed so that is why I thought I will also tell that one in a more systematic way so mixed flow reactors in series okay they can be of different size or they can be of the same size does not matter okay for as it does not matter may be different size also we will take you know different size means you know the slopes will change okay good so now when I have this let me say this is one then I have another big one here not looking big one another small one here yeah and here the condition is that I have only C A versus minus R A data I do not have a rate expression okay normally what we do the moment I have minus R A versus C A data then I will try to find out what is the order of reaction and then substitute that in the equation and then do it so in industry you may not have that kind of time where you would search for a kind of rate expression and then use that rate expression and all that so that is why most of the time in industry particularly they say that exploratory research where they try different things for them the easiest one is once they have this minus R A versus C A or X A versus minus R A simply plot that and then try to find out which reactor is better you can also get very good idea which reactor is better even if I plot 1 by minus R A versus X A or 1 by minus R A versus C A C A also one can write only thing is coordinates reversed okay so but I think our usual thing is I think for any general case X A is better 1 by minus R A versus X A then depending on the shape of the car you can choose which reactor is the best okay so that means when you are using mixed flow you know what is the area you have to take when you are using plug flow you know what is the area you have to take wherever the area is coming in larger and larger that is bad for industry that means volume is larger so that is why you also have some increasing decreasing like autocatalytic reactions okay so that is why clearly the moment you plot 1 by minus R A versus X A or C A then you will very clearly know what reactor is the best reactor or what combination of reactors are the best reactors okay so that is why that remember always you know these are the general things I am repeating so many times right okay good so in this case I know that I have only 1 by minus R A versus C A data but here I do not have to plot as 1 by minus R A versus C A okay why I have a general expression for okay this is C A naught C A 1 C A 2 C A 3 okay different reactors so then if I write for this one this is C A naught minus C A 1 by minus R A this much you know definitely we have written in terms of V by F A naught but that can be easily converted I also told you how easily one can write the mass balance equation in terms of concentration and volumetric flow rate V C A so V into C A naught entering and all this is V because there is no volume change okay yeah volume change also can be easily dealt with that is not a problem here because volume change is already there in minus R A so that is why even if you have volume change the equation is same this will not change okay this is the one so why we have to plot in that way is now I will write try to write that equation as 1 by tau equal to minus R A by C A naught minus C A 1 okay and that is why minus R A versus C A if I plot this is minus R A versus C A when I plot so I may get normally as concentration increases for many reactions what will happen to rate rate should increase so that is why you may get something like this it need not be beautiful curve where you have only increasing you know linearly R with quadratic equation so that kind of things need not be there okay so simple this one now my C A naught is here because I am moving in this direction this is 0 C A naught is high right so now this will tell me that if I draw a line here with 1 by tau I think Ramakrishna was also pointing out sorry it must be minus and that minus comes because I told him I think that negative itself shows minus but anyway still equation also we can change so that equation now instead of writing this this also okay yeah so minus 1 by tau as minus R A into C A 1 sorry but divided by C A 1 C A naught okay why we should do that I will first draw this one because C A versus minus R A I know I have to also know the size of the reactor okay otherwise how do I know tau tau is volume by volumetric flow rate that I should know otherwise I cannot use this method but I can also use that method but it will be a trial and error method okay that I will tell you later so this one is this is minus 1 by tau which is nothing but or may be tau 1 because I have 3 different tanks okay this is tau 1 tau 2 tau 3 because volumes are changing volumetric flow rate is same so tau will change okay that is why tau 1 okay so this is tau 1 and I told that we have to draw a line here okay and this equation tells me this is minus R A what is this d y and this one is d x this slope is nothing but d y by d x okay d y here I am starting from 0 that is why this entire thing is d y what is the value of this minus R A 1 okay I mean this is some minus R A 1 here right and correspondingly what is it must be because yeah C A 1 is the one which is leaving and throughout also I have C A 1 so this is where some people are asking doubts are why we have to drop it down we have to drop it down because that is only possibility because you know from here to here that change inside the reactor will occur so this is C A 1 now you see this equation is just nothing but d y by d x because our mind is always in terms of d y by d x for slope and that is why you have this one as minus 1 by tau because negative thing will show me it is I mean negative slope this is nothing but minus R A 1 divided by C A 1 minus C A 0 that is why so similarly you can also go to yeah now as I told you here I have tau is different here tau is large so I may have 1 by tau is small right so I may have not same thing small means yeah like this it may go like that so this slope they look parallel I think they are not parallel anyway yeah so this is nothing but for me this is minus 1 by tau 2 now I have to drop again this is C A 2 now next one is smaller one so it may go like this right yeah so now you have okay this is the final one C A 3 so for a known this is for known tau this is for known tau good okay if the tau is not known to me what do I do by the by in industry I have to tell you this any industries won't use different volumes for C S T R okay so they go for almost all the time they go for equal sized tanks you can tell me the engineering reason why engineering reason why should we use almost same tank equation is not a problem no because equation is same thing instead of C A 1 I put C A 2 or instead of tau I put tau 1 I mean that's equation is not a problem equation is not engineering problem yeah that's okay that I am not also changing that is same everywhere same pop can be there I think that is only you know so what I mean it's not a problem for me conversion depends you know you already calculated conversion based on you have large tank one and small tank one that is not a thing but he he doesn't want to use he will hesitate to use no no no I think two different tanks I don't want to use can you say that only two equal tanks I would like to use yeah excellent that is only reason because as far as fabrication is concerned I can only design for one system and then say that okay this is the height this is the diameter and these are the fittings all that is fixed but the moment it is changing 1 meter cube 1.5 meter cube 1.7 meter cube many things will change including my cutting the plate to bend it and then again weld it to have the tank you know the cylindrical tanks most of the time that is why most of the time they don't want to use different reactors and fortunately for us when they you know there are some conditions where you have to use larger tank and smaller tank okay there are conditions I will just tell you that okay so now this is the one this is the method now if I don't know kinetic I mean tows I want to I know only conversion so that means I have another curve so this is minus r A versus C A so again I may have the rate you know minus r A versus C A I think there I know C A not here I don't know only X A is known so that means I know only C A final or X A 3 if I am taking 3 okay now how do I find out now this best thing is to assume that we have equal so that slopes are equal now try to draw those slopes okay here one here okay like this here one like this here one like this here one like this if you are able to match this fortunately then you will have I think I don't have to put here X A F I will put here then it is a same slope same slope same slope same slope then 4 tanks I have if I am very lucky then the first slope which you have taken may be giving me this is C A 1 C A 2 C A 3 etc so finally when it match because this line is fixed because that is your final conversion how do I draw and if it is not coinciding let us say it is going here or let us say it stopped here then again I will change my slope so once I can find out that kind of parallel slopes that is nothing but 1 by 2 so then you can find out yeah volumetric flow rate you know so volume you can find out and then how many number of tanks also you can find this is by trial and error method good excellent so I think these things are now very clear so this is over now so the next one is you do not know I mean you may not know that one earlier but if you know number of tanks then you have to draw the slopes again such that yeah only that 3 will come that means again you have to adjust the these things such that you will there you will get there yeah I think that can be done I think you know number of tanks now that is what is the problem there because it is by trial and error when you are doing it there are many possibilities okay so that is why if you are able to fix that that is easy for you if you are not able to fix that then as he said any number can be drawn but only thing is final one should match here okay or otherwise you can also start from here draw the line and move parallel lines again it will go through this point right that is by trial and error good okay so that is one thing now we have another problem this is also nice problem I have two tanks okay right and I gave you this one as an exercise I have two tanks in series I want to find out the optimal size of these tanks right optimal size means given conversion what will be the minimum volumes of these two right and what will be the the conversion is fixed there right but similarly if I have the the volumes fixed what will be the maximum conversion you get maximum conversion you get these are the two problems optimization problems what we have but the first one we will take two tanks two MFRs in series and I have in fact I do not know whether they are equal in volume also I do not know whether they are equal in volume also right so there is a nice again graphical procedure here for this kind of work for going to optimization I have to tell you another thing it is equivalent method to Machiavelli method okay where the stages and 45 degrees line and all that that is one of the oldest methods many people are not using but there what you know is you do not know the kinetics are known to me for example order of reaction is known to me in the in that case okay and now I would like to find out what is the conversion if I know the size so I mean kinetics are known to me and also the size is known to means tau also is known to me but that is very good for particularly complicated rate equations like second order if you put second order in this equation v by f n in this equation right you have here k c a square or k c a 1 square right then you will get a quadratic equation if you see Levenspiel book for tanks in series second order he will put many roots many square roots so under those conditions instead of using you know how many square roots you have to find you know the graphical procedure is one of the easiest and beautiful procedures very nice procedures okay so that procedure we will just discuss generally order one also one can put but it will be very useful for orders greater than one okay or of course less than one where you get algebraic equations are complicated equation so that I think this we will postpone and then we will take this one first this is two tanks in fact this so this one let us say for second order and above M F R in series second order and above okay so now first I would like to write the equation for a particular reactor so then we will have yeah so this is C A naught let me write for one first okay and F A naught V naught and this will be C A F A V and X A good so what do you write here V C A naught equal to V C A plus minus r A into V correct minus r A into V now this I know that second order and above let me take second order so that second order will be okay before second order I can also write this equation as C A naught equal to C A plus tau into minus r A where this one is written as C A naught C A it is second order so k tau C A square okay this is also 3 this is 4 this is 5 this is 6 okay all this okay this is 7 8 9 okay so this is for one reactor I have written but when I have n number of tanks like this and then I have okay okay this is nth tank this is 8th tank when I have this kind of situation I can simply write this one for 8th tank okay what what is that i minus 1 C A i k tau C A i square for 8th tank okay for 8th tank so now this is fine but now I do not want to solve this quadratic equation and then sometimes we also do not know which is positive you may get one positive root and another negative root and then you know so it is really difficult so that is why one of the easiest engineering way to solve is again using the graphs and this is very easy also to plot because for me I will simply try to find out what is C A i minus 1 versus C A i but for given tau it is second order I told that means kinetics I know I want to only find out what is the conversion right that means k I know tau I know because I know volume of the reactor so that I can find out what is the conversion okay yeah that is what is the normal may not be exactly normal all the time so here k tau is known so if k tau is known let me say that k tau equal to 1 just for simplicity so then what is the equation I have C A i minus 1 C A i C A i square so what I do is I assume C A i because that is easy otherwise every time if I assume C A not and then try to solve C A 1 again I am not solving my problem correct no again I have to do quadratic equation so I do not have to draw graph again so the simplest one is I think maybe I will write reverse because it is very easy I have also done using computer it is very easy but only thing you have to be careful is that when I have 3 reactors 4 reactors initial concentration you know but final concentration how much it will come so that is why your range of these values must be very very large where this entire all the tanks that must be within this range I am talking about numerical values if you have C A not equal to 100 that is what is entering here and you do not know what is outlet you have to find out some conversion right so under those conditions you should know what is the range because you have to draw the line yeah some line it is not straight line okay so like that you will have if I start with yeah some values I think I cannot write but numerical values because the moment I write C A 1 C A 2 again you will confuse with okay those things I do not want to write these are the numerical values based on this that means I will start with C A i C A i equal to 1 calculate this C A i equal to 2 4 6 8 12 like that you know you can simply dragging you go to excel and then simply drag you will get those values good so now what we have to plot is C A i minus 1 versus C A i C A i minus 1 versus C A i right and first I will have 45 degrees line this is 45 degrees line good okay 45 degrees line is easy for me to draw that is no problem good so the next one is I have to draw is this data this data I have again C A i and C A i minus 1 values okay so depending on what is the k tau you have given that means you know the reactor so you can find out k tau may be 2 may be 5 may be 10 may be 1.2 may be 3.5 like that then you will get a line something like this but only thing is this line should be beyond the input and output values of this entire system okay so now what is the starting point for I equal to 1 I equal to 1 means C A 0 so C A 0 I know that is what is entering so then I will draw this is C A 0 then I draw a straight line that is C A now that will touch this this particular line right so I cannot go further that means there is no change if I straight go here till 45 there is no change at all but there must be change because reaction is occurring so that is why only possible way for me is this and this distance that means this distance is nothing but you have here C A 1 that is what is coming out of reactor but if this curve is you know k tau equal to large for example then it may go like this so then automatically that value will change the distance between these two that is what know when you make a by till a diagram also the distance between equilibrium line and operating line if it is large then you will have minimum number or maximum number minimum same thing here also if k tau is large means what large size reactor when you have large reactor what will be the conversion definitely more more when compared to the small size reactor you know k tau large volume large so X will be more so for X will be more means C A will be small so naturally then you will have if it goes like this then it goes like this here only cutting so C A 1 will be less so conversion is more okay good so now what you do is this is 45 degrees line this same line is going to the next reactor why this C A 1 is also entering C A 2 that is not changing so that is why this line gives me this is again horizontal line there is no concentration change because this is the input we started with C A 0 next one input is C A 1 next one input is C A 2 so that is all like that you go and then if you if you get the you know conversion as let us say 90 percent if you are able to reach a 90 percent conversion equal to that corresponding concentrations then that is the final and you can also have so many tanks what is the assumption here because tau we have not changed tau is same so that means they are equal number of tanks I mean equal size the tanks so then you can easily find out this method as anyone seen this method you have done seen that which book you have seen that is the 1960s 70s same same I think old books only afterwards they stopped it okay but I thought you know that is really thrilling for me because it is the generalization of your yeah mecca bay and all that you know what are the equilibrium stage processes what we have used so that is the reason why I thought I have to just definitely mention this which is very easy and for second order I think I do not want to remember all that square roots square roots square roots okay in Levenspiel book and not only Levenspiel book any book so and we are capable of making mistakes there whereas here simple graph calculate and then plot and then you can beatfully solve okay I am a dangerous fellow so that is why in the examination I can give these kind of questions because more I like graphs I like pictures yeah anything change you cannot use that because you know each plate is equally same for us in in make up a feeling method no practical method between that thought between R A and C A from the first method which one but the total change means that you have to use this method because slope will change automatically but there you cannot okay yeah I mean we think there will be some drawbacks or some advantages so we have to take very good so this is the one which I want to tell you this is also very good so the next one is this M F R Sin series okay so M F R Sin series that two tanks so here I have two tanks my idea is to find out optimal volumes if X A is fixed okay good so that means I have here C A not this is C A 1 this is C A 2 my X A 2 is fixed okay good and analytical way also one can solve that but analytical you know differentiation for first order it is very simple that is what I have given to you okay okay let me tell analytical so I have to write the volume should be minimum so that means my function is V as a function of V as a function of which one is the variable here because C A not is fixed C A 2 is fixed so only C A 1 right and also the tau values yeah volumetric flow rate also is fixed so only thing I can manipulate is C A 1 so that my volume will be minimum so that is why I have here V 1 plus V 2 D by D C A 1 should be equal to 0 this is simple again calculus you know maxima minima problem you know optimization problem right okay so this is what is mathematical analytically we can solve that I was it getting confused okay I mean simply first derivative you take and second derivative you take to find out again whether you have maxima or minima that is all simple calculus you know that is what you have learnt may be in third year second year or first year of your degree but I think maybe you do not know how to use that here okay that is one method but using graphical method also Levenspiel does you know that wonderful work as far as that is concerned it is very nice problem you know that means can I now minimize the rectangles or not rectangles actually the area okay so what he does is you will take this kind of you know two reactors where I have let us say this is C A 1 this is tau 1 this is tau 2 yeah so then I have here we can use our normal method X A 1 sorry versus X A then you will have this kind of curve and X A 2 is fixed for me so this is the one right so if I do not know X A 1 so that means I mean I have to say simply somewhere here I have X 1 what are the areas I have to take I have to take this area plus I have to also take this area but this one can move this side that side so that you know if this is the total area if this is the total area how do you maximize this maximize this area why if I maximize this area then these things will become minimum okay so that is the one he has nicely given calculus method where I think that I do not want to give okay yeah I know the condition comes as equating slopes that means if this slope at this point and if this diagonal if they are parallel slope at this point because X A 1 is moving this way that way so if I draw the slope there and if I also draw the diagonal in this this is rectangle if they are parallel that means this will be the maximum and you will prove that with the simple you know X Y area equal to X into Y you differentiate this and then equate it to d Y by d X from this d Y by d X equal to water okay area should be maximum I think that is the condition what you get there yeah to maximize the d A okay anyway I think this is the one I think just go through that this part okay so now based on this that means what you have to remember is the slope at that point whatever you are able to take that point the tangent okay that is a tangent and this diagonal must be parallel if you have that then you have the minimum volume for the both the reactors so if I have so like this if I have maybe I will try this so this will be yeah so this is another area so I have now this triangle sorry this rectangle so now this diagonal and I will try to manage if these two are parallel okay so how do I write okay some of the volumes will be minimum total value that is why v 1 plus v 2 together is the minimum okay good so what I want to tell here is and what again I do not want to repeat that in the next class that is why I am just not given this quickly going is that if I have n equal to 1 first order reaction then v 1 equal to v 2 that means both the reactors will have same if n is greater than 1 v 1 will be greater than v 2 the reverse v 1 will be smaller and v 2 will be larger if n is less than 1 v 1 is greater than v 2 I think to give the simple idea this one is v exactly same v like this and in this case we have like this you have like this and in this case you have like this yeah that is the setup what you have why is very important there than one if r a is greater volume will be less for a given conversion that is why okay yeah so that is why whenever you have higher orders okay so then try to put smaller volume when you have two CSTRs so that you will get you know that means when you have a larger volume what is happening to the concentration decrease so when it is decreasing rate will not be very high okay because 1 by minus r a so that is why you have to maintain as high as concentrations possible for second order reaction orders greater than 1 because when you have C A square and if you are maintaining larger concentration what will be r r will be high so 1 by r will be less so volume will be less because all r's will be only in the denominator whether it is batch reactor or plug flow reactor or mixed flow reactor all these things are same only okay all are in the denominator so that is the reason but the reverse that means maintain high concentration okay now I will tell you if I have a plug flow reactor and then mixed flow reactor I have second order reaction how do I connect this first one you have understood why I have to keep a smaller volume for n greater than 2 sorry okay n equal to 2 for example first I have to put a small CSTR and then later larger you know CSTR why because here I want to use my concentrations as optimally as possible that means to maintain concentration as high as possible when I have a smaller CSTR then definitely concentration will be high when concentration is high I have minus r a equal to k C A square okay squared concentration large concentration into whole squared will give me more rate and 1 by rate will be small so that I can give for given conversion smaller volume or for given volume higher conversion I mean just opposite problem okay now you understood that no still not understood okay now if I have a plug flow and mixed flow in which reactor you can keep concentrations high because maximum possible that is all I think there is no other reactor which can beat that so but my order is more than 1 that means second order so I have to keep the concentrations as high as possible that is why keep first P f r and then later CSTR but the same thing is reverse if I have n less than 1 but I am not able to find out any physical explanation for that unless you calculate numerically you will see that this is easy to explain to me because keep concentrations as high as possible and all that last 30 years I am not able to get good explanation but there the explanation is through mathematics you solve for half order first order and second order you will beautifully find what is the difference whatever we told here exactly will happen there but then why for first order it is same even if I have 2 tanks smaller tank and larger tank okay first 2 tanks 2 tanks 2 CSTR for first order whatever way you keep does not matter and this exercise will give me that when I do this exercise it will give me if I take second order reaction analytically you can do if I take second order reaction here and then differentiate with respect to CA1 you will see that the first one will be smaller second one will be larger but if you do the same thing for n equal to 1 first one and second one will have the exactly same volume and you again repeat this same exercise for n equal to 0.5 then the first reactor will be larger and second reactor will be smaller okay so that is why whenever you have second order reaction keep the concentrations as high as possible and only for first order reaction I ask you to think why you get same conversion whether you put larger reactor or smaller reactor that means concentration really does not matter there right and not only that I gave you one problem I do not know how many of you have done it I think may be you are waiting for that day I think 28th is the last submission day means 27th night only you will do it okay yeah P f r because you have done you are able to tell that I think you know they have not done so they do not know that okay so I have given the exercise as plug flow reactor followed by C S T R okay does not matter volumes may be different okay or volumes same volumes also you can take I think it does not matter same thing is reversed first you put plug flow mixed flow then same thing instead of reversing you pump from this side now you will have first is mixed flow second is plug flow which one will give you more conversion with respect to first order reaction same now think about the reason why it is same so I will stop here