 I think this equation already we have right so where t bar and all that also we have written there but you see as I told you the procedure we have drawn the graphs somewhere there and then for different u by l and the best method is I think plotting on the same graph so somewhere it will match with one of those curves so that will be d by u l but there are many other techniques that is not the only technique because it is only you know Gaussian distribution if you are aware of the Gaussian distribution curve 68 percent of the area middle area okay when I plot like this this is e theta versus theta so the 68 percent of this area 68 yeah we will have 2 sigma yeah right 68 percent we will have 2 sigma you know this width sigma is the variance right so now if you are able to connect variance and d by u l then definitely you can estimate d by l our idea is to find out what is the actual value of d by u l okay for that axial mixing which you are measuring experimentally right so what we did earlier was you plotted so many graphs and then one of the d by u l will match with the that simulated so many curves one of the d by l will match with that so that is the d by u l that is existing for that set of run okay that particular run then once you know the d by u l what is the next step next step is using this d by u l to calculate actually what is the ultimately it is conversion what we want to find out right Ramakrishna why not able to follow okay or no following okay good so this is like that there are many other techniques so that is why how do I now relate the variance and d by u l we have an equation sigma theta square equal to we have from that equation you know sigma theta square or sigma square what is the equation 0 to infinity anyone I think you know no one remember that theta square e theta d theta that is all something else is there minus 1 that theta bar will be 1 okay minus theta bar square actually minus theta bar square that will be 1 right so this is the 1 now you see I have this equation e theta here substitute this equation here integrate and it is 0 to infinity anyway minus 1 is there then you will get that beautiful equation if you do that 2 d by u l please remember in r 3d our idea is when you are modeling our idea is to estimate the non-idealities for axial dispersion model there is only one parameter called dispersion number which contains d u and l of course I know I should know right so d is the diffusivity if diffusivity is very very large you have large dispersions I mean I know spray tracer will spread more and if d by u l is or d is small you will have very narrow distribution that means you are towards plug flow yeah towards p f that is the idea right so that is why you should not forget the actual purpose of the model the model purpose here is to estimate actual quantity of d by u l non-idealities here it is d by u l okay in some other case it will be dead space in some other case it may be bypassing so every time when you have bypassing here we assumed only axial mixing that is why this continued only a parameter which describes which describes axial mixing in terms of d by u l indicating d by u l is small means you are towards plug flow d by u l large means you are towards mixing flow right in between you can also have 0 to infinity these values right that is the idea so that is why the best method is going to matlab giving this equation and the experimental data ask it for match but you have to give the guess guess value that is one of the excellent methods otherwise if you want to have a field simulate all the graphs you know earlier you used to have tracing papers I do not know whether you have seen any you know anyone now know about tracing papers so it is transparent paper you draw all those lines on that and then put on this your experimental curve but scale must be same your experimental curve and all that then one of that thing will definitely match or if it is not matching exactly it will be between two curves so that curve value approximately you take and then go to matlab give it and then you get the final d by u l value right good so and here this is very nice equation only for small dispersions dispersion less than d by u l less than 0.0 that is valid for that okay so here boundary conditions will not affect whatever boundary boundary conditions you get yeah whatever boundary conditions you used all those to solve that equation you will get this kind of equation this is Gaussian distribution it is logical you know this is where engineering you know if you have engineering intuition you will simplify mathematics what is engineering intuition here I am expecting axial mixing right so axial mixing very small mixing that means it is towards plug flow but a little bit of spread is there so what kind of spread you can expect I can expect symmetrical spread that means if I draw the line in between at mean so both sides you will have symmetric that is Gaussian curve okay and that can happen only when you have small dispersions that dispersions people have tried long time back and then said that must be d by u l less than 0.01 if you go beyond that you will have again problems so this is one of the methods and now you also have to how to calculate sigma theta square using straight concentration versus time data do you have that in fact this is nothing but sigma t square by t bar square how do I get sigma t square directly because you do not have to even calculate e theta and all that right so you have that sigma okay 0 to infinity t square c d t divided by again 0 to infinity c d t so minus of t bar square right that equation itself you can use to find out sigma t square and then divide by t bar square you will get sigma theta square which is 2 d by u l okay so that way also you can find out d by u l there are many ways how to find out d by u l so this is the one and this is about small dispersions and if I go to large dispersions that is d by u l okay I think I have tried large dispersions large dispersions d by u l greater than 0.01 here boundary conditions play a role b c is play a role and what kind of b c is we have not discussed what kind of b c is till now right but I said there the boundary conditions will not affect so that is why we have not discussed there but here what kind of even there also the same boundary conditions will come but any boundary condition use it does not matter but here what kind of boundary condition is I discussed this sometime back what is called open open vessel closed closed vessel so that means whether you have open open boundary or closed closed boundary is the discussion okay so when you have when you say closed closed boundary because already we did this so let me write that closed closed boundary closed closed boundary so you can imagine something like this you have a plug flow reactor and then okay so here material enters by P f and here material enters by leaves by P f this is closed closed boundary this is closed closed closed it does not mean English meaning closed closed means you know okay so you close and then nothing will happen there you know nothing will enter nothing will come out okay so not that closed closed boundary means P f material is entering by that means till here I have flat velocity profile oh now this this is closed okay okay and also from here when material comes out you have flat velocity profile so that means what is dispersion in that zero right that is the boundary condition so this is closed closed now you have open open open open is straight so material enters like this leaves like this but now whatever yeah okay whatever dispersion is there here okay same dispersion also you will have here and also same dispersion you have here that is what is open open okay so this can happen in packed beds for example packed bed diameter will be 1 meter then the entry may be 7 inches 8 inches okay so then uhh you know the velocity inside the packed bed will be less definitely because area is more and uhh the entry point is small velocity will be very high we can always assume that we have plug flow there so most of the time for packed beds you have closed closed boundary and most of the time for tubular reactors where you have you know homogeneous tubular reactors where I have given you that problem you know your design problem is homogeneous reaction there you do not use feed pipe small and also other pipe bigger you know actual reactor you do not use so whether you use 4 inches or 5 inches or 6 inches uniformly like this like this use okay so test section we are assuming we have axial mixing and even before test section this side also you have dispersion this side also you have dispersion okay so now I told you in the on this uhh oh the differential equation I have not written the differential equation which I have written earlier that differential equation I was telling you there are many papers on that okay so people try to I mean because academicians they do not have any much work know so in the sense that okay let us try what will happen see same dispersion number throughout here here here that is one paper right okay otherwise you use some different dispersion here number here different dispersion number here same dispersion number here and here another paper okay so now you use the different dispersion number different dispersion number different dispersion number 3 different dispersion number or D by U L okay or diffusivity then you have another paper so like that you know just to show that what are the conditions you may get what kind of responses you may get when you change diffusion coefficient that is the reason why I have so many combinations of I told you so many papers have come earlier and this is taking open open and the closed closed right see one example is this is the boundary right we have dispersion and plug flow what should be the equation equation should be U C minus or okay first minus I will write minus D D C by D X if I take this one as X direction plus U T C that is the flux that is entry okay this diffusivity I am talking this diffusivity is one here so because it is open open you also expect the same equation to be valid here again D C by D X plus U C okay now you can ask okay and inside also I have same dispersion some dispersion and velocity velocity normally taken throughout same okay so what I was telling was that this will be D 1 D 2 D 3 so in the outlet pipe there is some kind of dispersion may be size change and all that and here inlet pipe you will have some other dispersion and here you will have some other dispersion or otherwise even if it is a packed bed with same cross section okay or even different cross sections to you will put some packing here some other packing here some other packing here that will change the diffusivity coefficients right okay now extending the same thing here what should be closed closed vessel equation I mean yeah closed closed boundary equation this is plug flow okay yeah this must be simply U into C that must be equal to because here you have dispersion right the moment it enters this is actual this is dispersion because please remember please remember we are talking about only dispersion equation actual dispersion is there that is what is the model so I have dispersion here but here when it is entering the flux okay alone by U C and the moment it comes it will split into because actual mixing is due to so flux due to or okay dispersion flux and also convective flux here the moment it crosses there again U C only there is no dispersion there that is what is the different boundary conditions what we have and now you know our all intelligent people know why not open closed closed open all combinations okay so now if you say yeah open closed you will have the same pipe till here okay same pipe but here what happens plug flow so this is open yeah this is open this is closed reverse also all you know we have academicians are highly intelligent people in doing all this yeah so this is sorry sorry yeah so now till here this is closed this is open so those are the boundary conditions what we have been talking now when you have this kind of dispersion d by u l less than 0.01 all this does not matter whatever dispersion you know the boundary conditions what we have used you will get only Gaussian distribution which is described by this equation that is why in engineering you have to see how to use mathematics in a proper sense mathematics in need not care that is his job not to care for everything he may complicate but as engineer you should be able to tell that okay so now for small dispersions all these boundary conditions will not you know will give me only Gaussian distribution and what is the equation that will describe Gaussian distribution this is the equation that gives the Gaussian distribution nowadays you have the matlab or not matlab what is that laptop and all that you have XL okay just simply go d by u l assume 0.001 and then theta vary from 0 to may be 5 or 6 okay and then with a thing 0.1, 0.2, 0.3, 0.4, 0.5 intervals you just calculate E theta and then thought beautifully you can see I tell you then only you feel actually otherwise you do not see okay good now with this boundary conditions that means that yeah open-open boundary and closed-closed boundary and when I have large dispersions what are the equations we are going to get understood not understood okay so now first we will take closed-closed vessel for closed-closed vessel what is the equation we are going to get you are very happy at least for the examination there is no equation there okay but you know the problem is we cannot solve that you do not have an analytical expression then what do you do so numerically they solved and constructed the curve and took the variance and mean right so please write that then for closed-closed vessels yeah now that means we are talking about large dispersions large dispersions case 1 case 1 is closed-closed vessel closed-closed boundary under these conditions write that for closed vessel analytical expressions not available not expressions expression for closed vessel closed vessel means you know both sides closed-closed boundary for closed vessel analytical expression not available is not available however we can construct a curve by numerical methods and evaluate evaluate its mean and variance mean and variance exactly we can evaluate so the mean will be t bar also equal to t bar e the meaning of t bar e is there you know t bar is volume by volumetric flow rate t bar e is yeah from e curve how do you get e curve from the experiment sigma you know all that that same equation right okay so this is equal to yeah volume by volumetric flow rate both are same again here in closed-closed vessel so that means t bar e experimental and t bar theoretical that is I mean I do not say theoretical volume by volumetric flow rate and this is from experimental observation both are same okay so that is one thing and yeah naturally what will be theta bar e when t bar equal to t bar e okay so theta theta bar e equal to 1 okay good so now the other one variance sigma theta square which is nothing but sigma t square by t bar square equal to yeah you see now this equation 2 d by u l minus 2 d by u l square okay into 1 minus e power minus u l by d that is the equation okay I think if you have just compared that one with this equation what is that because here dispersions are small in this case yeah only first term is taken because because dispersion is small d by u l is small square of that is much smaller so that is why this term is neglected for you know small dispersions okay good next one also we are happy case 2 is yeah case 2 case 2 will be closed open or open closed both are same so that means this case yeah this case and this case right this is open open this case and this case both are clubbed together but only thing is one end is changing that is all here this side is closed and here this side is closed but again here no solution is available for that closed closed open or open closed or open closed right please take that here also there is no solution available just below that you are writing here also yeah but here I can tell you there are very nice things here t bar e o c o c means open closed equal to t bar e c o closed open right equal to volume by volumetric flow rate that is t bar 1 plus d by u l and sigma theta square equal to sigma t square yeah for both of course we are telling o c and also r c r c o this is sigma t o c c o by t bar square this must be equal to we have 2 d by u l plus 3 d by u l square this is the equation what you have to use okay and here see we tell always that volume by volumetric flow rate equal to t bar okay correct no and no one knows how it has come correct no you never question how it has come because we are happy because volume has meter cube volumetric flow rate as meter cube per second then you get seconds very happy because time is there so we are very happy that is all what we do but actually that is not right that is only strictly valid for what condition not steady state we have the answer on the board ideal what what kalpana where no dispersion it is not for plug flow it is also valid for even mixture flow yeah close to close to vessel why close to close to vessel again there is beautiful physics there because in open open vessel why it is not valid is that the material because of dispersion may go outside and inside the reactor but your volume by volumetric flow rate is only by for volume is for the reactor alone not for the entry section correct no swami last time where did you lose I think I am talking only English shall I tell you or what what is the problem here I think so nice diagrams we have given here okay if you are lost means you have not come with your brain go and check somewhere you left it okay that may be the problem or may be in the suitcase or something somewhere you have the bag row you do not have bags so then definitely there is somewhere somewhere you left okay this is so simple to understand no because always we say volume by volumetric flow rate as T bar which is not correct that is only correct for close to close to vessel because the volume what you are taking is between these two boundaries but actually when you have that open open vessel then the material is going like this that is what is the dispersion so that is why truly it is not valid and how much it is valid is given by this that is why I told you answer is there on the board itself this is valid only when d by u l equal to 0 right and that will be only for close to close to vessel here you see here T bar equal to T bar equal to volume by volumetric flow rate this is another beautiful thing what you learn okay and in fact you can also derive there is another equations I think in one of the old books I found when it is valid it is valid only for considering the entry level and also exit level but not in the feed pipes but open open vessel you will get this problem because material may exchange across the boundaries at least at this point of time you should be able to tell your juniors in who are there in the other colleges do not tell that volume by volumetric flow rate equal to T bar it is not correct it is only correct for close to close to vessel he will say that when you close what where is the vessel he will say or here see okay then you have to explain again what is you know close to close to vessel what is open open vessel so that is what is beauty in learning I say one point in a course where you should not forget I am sure your residence time is only 24 hours maximum that too I think before the exam if there is no exam zero residence time nothing will be there just listen and then go away forget so that is very very nice things you know you have to learn here okay so that is why you will not get this T bar as volume by volumetric flow rate but at least that point you please remember this blind definition of volume by volumetric flow rate in fact I told this in the first few classes who told us volume by volumetric flow rate equal to mean residence time unless otherwise because mean residence times you should take mean size correct no either integral or sigmas you have to take we never take that and simply someone told teacher and you are happy because as I told you again it is volume by volumetric flow rate so volume volume get cancelled seconds will come we are happy okay hours also we are happy time is coming we are happy that's all but there is a lot of things here you know behind that definition when it is valid when it is not valid okay then these two we covered even though there are four cases this case in case two we have the both combined closed open and open closed the other one is case three where we have open open okay Swami open open open open do not open both sides only one side open otherwise it will enter and then go away okay so in open open vessel analytical solution is possible again so that means they have they have used these boundary conditions okay for open open vessel and then solute that equation the analytical expression is E theta equal to 1 by 4 pi oh my God 4 pi D by U L into theta square root that has that is there as you shown so exponential then we have minus so minus 1 minus theta square here we have 4 D by U L D by U L and theta and yeah okay so that is the expression what is the difference between this and that actually that will stretch the curve stretching the curve means for open open the response what you get E theta versus theta will be it is not symmetrical but it will be like this it is not symmetrical symmetrical means you should have come to zero here itself okay and why that is happening why that is happening yeah because more dispersion so it is not nicely only is moving like this like this it is moving like this like this but one side you will have skewness okay so that is the reason why you will get that kind of response there when you have large dispersion if you have infinite dispersion how do you draw if you have an infinite dispersion yeah you time to leave zero straight line yeah straight line is vertical or horizontal horizontal any idea Swami vertical yeah why straight line see infinite dispersion E theta what is the equation I have infinite dispersion in the equipment that means it is equivalent to mixed flow mixed flow has beautifully decreasing like this exponential decay without touching it goes to infinity I cannot go and draw okay so I think okay so that is what is E power minus theta you see I think these small extensions is the one where I am always telling you you should be happy to learn those small things right yeah small dispersions large dispersions infinite dispersions and all that okay good so this is the curve now you see our idea is to estimate again D by U L we have discussed it already how to estimate go to math lab or any any lab and then try to match okay or I think Ranganathan lab that is what you have just done the experiments right okay yeah that is that is computer lab only right yeah I mean actually I will tell Ranganathan next time I think you have to include this exercise in that that is more relevant than telling stories okay no no I think you all the time when you are doing project if you have mathematical model as well as experimental data may not be your experimental data or may be your experimental data what you do is this and in in see M Tech we may not stress that much okay either you do modeling or experiment generally not both but for PhD it is a must PhD we able to see whether they are good in doing experiment and also analyzing the data and also using the mathematics to describe that experiments all these three we check then only degree is given okay theoretically speaking anything so we are not checking here that is different okay so that is why this is a must the matching the data with the experiments is a must and always when you do you know in a PhD also it is not your data in fact if it other data is falling on your equation that is fantastic people will really appreciate your data your model there is no thrill right but your data some other model or your model with some other data jump up and down thrill okay that will be very good because people really appreciate my god because someone in America or someone in Russia doing that experiment okay that data still your model is able to describe or vice versa your data which you have taken in India described by an American model or maybe Russian model or Japanese model okay so that is why all these things are very important for us okay good now the next one is to estimate this equation I have just forgotten to mention I am repeating many times so I have to find out sigma T square by T bar square because from my C verses T data right sigma C and all that then that will give me let us say point 2 for example then calculate D by U L this is a quadratic equation you have to take the reasonable value okay normally you will get one negative one positive so then you can find out which one is the correct value D by U L for that particular you know equipment okay good so here also same thing the procedure is same now so here also you will not get sigma theta square okay this is also sigma theta square open open O O okay this is equal to we have 1 plus okay I think I will also write this one this is T bar E by T bar equal to 1 plus 2 D by U L or otherwise T bar E same equation I am writing this is not a new equation equal to T bar that is volume by volumetric flow rate 1 plus 2 D by U L okay what is the difference between this and that you have twice the D by U L because large dispersion open open but sigma T square by O O that is the next equation this is theta 1 thank you so sigma theta square is the next one sigma theta square will be this is sigma T square by T bar square which is equal to 2 D by U L plus 8 D by U L square so that is what is the story of this dispersion model okay so we covered all the conditions and this is enough and there is a lot of literature on this and corresponding to this car there is an F car right Swami how do you do from E theta to F theta that is all F theta equal to 0 to theta E theta E theta so you have to integrate and then you will get F theta similarly here also you have get F theta right so either F theta or E theta whichever is convenient to you one has to use that right good again here D by U L you can calculate using sigma theta square which you get from your experimental data and then say that this is the D by U L now the story goes to the conversions okay so now I have a reactor I designed it and then I was not very sure whether I have plug flow or how much axial mixing right then we conducted the experiments we already designed please remember we already designed and may be assuming that you have plug flow you designed right and then you conducted experimental you know RTD experiments then you know what is the response from the response you will see you saw that there is a axial mixing there is axial mixing now that you quantify after quantifying what you have to do is okay I designed for plug flow but actual mixing actually axial mixing is there so now what is the actual conversion that is curiosity definitely we designed for 95 percent assuming plug flow when you have axial mixing so what will happen conversion reduces or increases so I would like to find out how much is that exactly 95 I designed right yeah you are not lost I think you are there very much so when you are yeah when you have 95 percent and then when you are able to find out what is the axial mixing now it is our interest to find out whether it is 90 percent or 92 percent or 93 percent that is the actual conversion so for that there is another model required correct no D by L you have now so using that D by L you should be able to calculate how do you calculate without an equation how do you calculate because it is calculation it is not estimation and of course you can do experiment and then find out what is the actual conversion right that will not be 95 percent if you have axial mixing but can I also calculate so for calculation we have axial dispersion model with reaction right that we will do very quickly now you can assure it will be till to our midnight you are able to understand the sequence no see RTD will give us the non-ideal parameters using these non-ideal parameters we should be able to calculate the actual conversion after all chemical reaction engineering only to calculate conversions for a given volume or for a given conversion calculate volume that is all the entire thing so now axial dispersion with reaction axial dispersion model dispersion model with reaction okay here I think I will just draw the figure okay this is plug flow what we have good so input okay this is volumetric now now I can write F A not C A not of course V and T not of course isothermal we are talking about isothermal and X A not equal to zero so this is F A C A X A and of course volumetric flow rate may be V or V F good so input output plus reaction plus accumulation all four will be there or which one I should take out now because our reactors are always steady state reactors so this will not be there in the earlier case reaction was not there but accumulation was there because tracer is by definition that experiment is unsteady state experiment good so this is the one and here also the same thing what we have written in the morning D C A by you see now A has come D C A by D X plus U into C A this is A but this is at Z at X this is X this is X plus delta X yeah so that is input and output is exactly same U C A A but this is X plus delta X plus reaction in that volume yeah minus R A A D X A delta X okay that is the volume so I think this we must be having so now once you simplify that the equation what you get is that D that is the diffusivity coefficient D C A by D X square minus U D C A by D X minus R A so now this minus R A I am taking as K C A to the power of n which is equal to zero so minus R A equal to K C A here nth order reaction but as usual only first order reaction zero order reaction will have the solution other orders will not have an analytical expression and you will have only you know for first order for n equal to 1 we will have solution right and with B C is right okay we need 2 B C C here because second order yeah so that means at the entry at the outlet it is closed closed vessel but now you see dank words comes into picture and then you tell some slightly different kind of argument for these boundary conditions and there were on that paper alone around 50 60 papers again on the dank words boundary conditions let us see so when this is closed closed vessel what is at the entry equation you yeah what you into so I see a knot this is the reaction you into C A knot that is the flux which has been dispersed into minus D D C A by D X plus you into C okay so that is what and at the outlet I expect exactly same thing minus D D C A by D X plus U C must be equal to U C but now if I take this as L this entire thing as L length that is what no D by U L when it comes this is at C A L just for example okay so this is C A L now there was lot of discussion on this boundary condition and dank words said that you know he is an excellent engineer his intuitions are fantastic he said that simplest manner what you can understand but anyway so this boundary condition is okay but this boundary condition he says that it is not physically correct right so why he says that he physically correct is not correct is that this one because when you have this C A U into C A that is the concentration where is that concentration when I look at this point you have this one L plus this is L minus let us say right so at that very very very very very thin boundary thin boundary and because till here only you have reaction and the afterwards there is no reaction what should be the concentration C A must be equal to C A L otherwise what will happen either reaction or there is concentration discontinuity concentration discontinuity should not be allowed because unless there is a reason you cannot say that there is concentration discontinuity so what should be the correct DC BC then excellent that is all DC by DC must DC A by DC A by DX must be 0 and these are the two boundary conditions what he has used this one and also this one afterwards many people try to you know prove that he was wrong no I think there are so many papers and there is a big article also review article where finally he collected all the papers and finally said Danquats intuition was right okay so these two are called Danquats Danquats Danquats intuitive intuitive boundary conditions BCs I think handwriting is very bad Danquats I think I will write clearly because I think Danquats is very important so these two are Danquats intuitive BCs intuitive okay INTU T8 VA INTU ITI V INTU to boundary conditions okay so I think you know I tell you when you go for an interview somehow you bring them to RTD tell Danquats boundary conditions your job is guaranteed because they don't understand much I think high fund a guy so that there must be something but you should explain very clearly without you confuse if you confuse yourself and then you know no job I think other job also they will ask them to go away from there so if you have already some job so that's why this is very clear and you know this and later many people try to prove that he is wrong but found that you know experimentally the conversions what they get was right is depending on is better than quotes boundary conditions now using these two what should be the equation so that we can use that for our okay yeah the analytical solution is for n equal to 1 is this this already I have written once okay I also told you the beauty in assuming plug flow that is the equation so now using those two boundary conditions you get CA minus CA not equal to 1 minus XA and I think today I don't know somehow I stopped writing the equation numbers exponential of ul by D 1 plus a square exponential a by 2 ul by D minus 1 minus a whole square exponential minus a by 2 where a equal to square root of 1 plus 4 k tau D by ul okay very simple equation know yeah very simple and happy equation that is what you have to use just by taking one non-ideality called axial mixing just one non-ideality right so instead of that I think you can happily assume plug flow and correspondingly what is the equation for plug flow first order 1 minus XA is equal to yeah for P f CA by CA not CA P I will write equal to 1 minus XA equal to T bar equal to tau you see how beautiful innocent looking expression and how complicated idiotic expression you see you choose what is that you need beauty or you need complication beauty yes very good yeah you need that so this is so a happy simple equation know yeah so this is why you know this is complicated anyway but anyway after sometime I think life goes to complications so then that is why you have to conduct RTD experiment and yeah normally know D by ul you have to estimate go to this particular equation and then calculate what is the exact conversion because for a given reactor you know tau right volume by volumetric flow rate you know you know volume because given given volume so that means you already have the reactor and now you want to find out what is the conversion in that that is what normally we do this comes only after designing assuming you have plug flow but you are not very sure again I am repeating so then you conduct an RTD experiment right so I think Arya getting tired okay yeah so then you conduct an experiment and then you see the response curve it may be very nice small nice Gaussian curve and then you may say that you have small amount of axial dispersion and that dispersion number is estimated D by ul and you go to go to this equation and calculate conversion just imagine for a new reactor what you do in the beginning itself you want to take this axial mixing into into account and then design a new reactor what are the procedures for me I am asking you because I think you know you said you are lost I want to bring you to the closet what is the procedure no question is clear or not clear smiling is not the answer you have to tell something yeah I think first of all you tell me what is the procedure that I will answer later yeah how do you assume because you are now considering that already you have D by ul okay so now there is lots of data available in terms of Reynolds numbers and D by ul if you go to Levenspiel book for packet bits there is separate graph for tubular reactors there is separate graph Reynolds number versus D by ul okay it is not exactly D by ul it is D by U DT DT is the diameter of the tube multiplied by DT by L DT DT will get cancelled you will get D by ul okay that is the geometry aspect ratio what we call this is the dispersion number expressly in a slightly different way okay good so that is what so what are the procedures for me you know the other procedure you know I already have the volume I want to calculate conversion straightforward yeah I am asking now I want to have the volume but conversion I assume 90 percent and also dispersion number I will assume because normally you know Reynolds number means sorry plug flow means what kind of Reynolds numbers you should have very high so you have to have definitely some Reynolds number when you are calculating based on your plug flow alone okay so you have to assume okay my plug flow will be only 40,000 Reynolds number then go to that graph and read that D by ul okay so then even if it is packet bit we have that data then what do you do now I know D by ul so what else you should know there only you have to calculate right so because conversion I know conversion I know okay and D by ul I know of course here A I do not know A means again I have tau okay K I will know because whether I have the kinetics so now you have to solve this equation for tau see how difficult it is that is why best thing is assume plug flow again same story okay same thing okay so now of course here you have A you have tau and this year you have tau this year you have tau all that right D by ul you are assuming D by ul you are assuming okay you do not know L at that time but D by ul has a number you are assuming right okay good so like that we have to do that is why again one of the other greatest engineers after Dan Quartz a logical engineer is Levenspiel okay and I relate to him also people have done he now said that this equation need not be used completely if you have again small dispersions we can now expand this this exponential expansion you remember that is why we read mathematics earlier okay M1 M2 M3 M4 M5 only for this okay that exponential expansion is what is that X X square that expansion is used and drop out all D by ul squares because dispersion is small we are assuming that using small dispersions that means dispersion square will be smaller D by ul square will be smaller and you take only D by ul values so when you do that for small dispersions for small dispersions the above equation will be CA by CA not equal to exponential minus k tau plus k tau square D by ul so this is the equation okay and this is very easy to get from this that is not difficult at all when you expand this this this and then finally you will get only this if you expand and then simplify okay no no I think his problem is you know you wrote there ul by D but below you write D by ul how do you get then what else yeah so if you don't like this I will write here ul by D below that is all this ul by D I can write there so that is not a problem but you do it is very simple it is not that complicated that equations are threatening but if you put on the paper and then write you know that expansions it is not that easy because all second order terms you are removing whenever you have D by ul square you are removing or ul by D square then you will end up with this okay so this is the equation what you have you see now straight away for even non ideal reactor I can now calculate what is the conversion right so D by ul I estimate it okay and if I want to find out conversion now you know tau I know I can calculate easy CA by CA not otherwise even reverse is also not difficult reverse is what I know CA by CA not I don't know tau D by ul I am taking you know like from that graph and you can calculate what is tau good okay so now using this information we can even find out what will what should be the logical length of or L by DP and also L by DT if it is empty normal homogeneous reactors what should be the length by diameter value for maintaining almost negligible dispersion that is again some simple exercise is there using this okay are you able to understand me yeah this is the straightforward equation simplification of this but using this and also this equation CAP this is ideal this is non ideal okay because D by ul is there for me right so now if I take the ratios of these two for a given conversion for a given conversion 1 minus XA equal to 1 minus XA this is also 1 minus XA okay yeah these two equations I will take and now for a given X at least I think I have to write this equation I think A I will say because there is no serial number yeah from A from A and B for same XA okay Pooja anything sir this one for small dispersion actually we are using the dampers boundary condition then only we are getting that equation and in that equation the upper one you are neglecting the higher terms of exponent till further simplification so it is not visible for small dispersion we are actually neglecting the higher components of dispersion no no D by ul when it is smaller square of that is much smaller it is mathematical simplification it is not we are neglecting dispersion valid for small dispersion this equation is valid for large dispersion only but we are neglecting the higher components of dispersion yes so you have written for small dispersion we are neglecting the small dispersion yeah for small dispersion I can further yeah for small dispersion means D by ul is value must be what very small small so then only I can expand and then D by ul square I can neglect no you will get only D by ul here D by ul here only there also you will get only yeah here this is an analytical expression that is fine if you expand that term then sir it will be coming one time no no I think you know this mathematical details you can find out you will get only this expression okay yeah so then there are other equations here Pooja there are other terms here D by ul squares those things are neglected because that is not contributing for the value let us say let us say I have k T square or k T equal to 10 okay and D by ul equal to 0.0000001 what do you do so you do not get any value there right in engineering sense so that is why we will neglect those terms right so that is how only we are simplifying this and I think thereby I am still not come out of ul by D and D by ul okay you expand this and you will get that I think you know yeah that is not a problem at all okay good so now the question is this is non ideal this is ideal for a given conversion can I find out length or volume which one will be more for a given conversion I have non ideal reactor okay I want 95 percent conversion and I have ideal reactor which length or which volume will be more non ideal that one now you can estimate using these two equations that length you can estimate how do you estimate that how do you estimate from this these two are equal right correct no so 1 minus x a is 90 percent 95 percent this is also 1 minus x a 95 percent conversion so now both are equated and then separated for tau p and tau tau p and tau nothing but L p by or okay tau by tau p is nothing but L by L p correct no same volume to flow rate so that is what is the estimation now so that will be V by V p equal to L by L p if you want you can also write here tau by tau p okay this is equal to from these two equations what you get is 1 plus k tau again D by ul that is the one okay this equation good and this is again very simply one can do that I think it is not that difficult to do that again here when you are doing this when you are equating you have to also expand this I am just giving you the clue you have to this is e power minus k tau plus k tau by D L no you expand this again in terms of that 1 plus x 1 minus x square a x then plus x square by 2 or 2 factorial okay all that expansion again neglect that D by ul squares and then you will get this expression you see now so the length of non ideal reactor will be 1 plus this is L p plus this is the extra tau correct no if I just divide multiply by L p L equal to L p into the whole thing so L equal to L p plus this term will be the extra length right see these are all just to give an idea you know how nicely how simply how beautifully one can simplify the mathematics for actual engineering purposes okay so that is the one so there is another question also that can rise now what is the other question here we have conversion same so it need not be that always conversion same it can be also yeah volume same then how do you calculate that means tau equal to tau p so for same volume that means v equal to v p right volume same but in one you have non ideal parameter like action mixing in other case it is ideal so now conversion must be different right so how do I find out that of course we can write this in terms of C A by C A p that means the concentration coming out of ideal reactor and the concentration coming out of the non ideal reactor so that is 1 plus K tau square D by U L that is the equation right okay good now let us take this equation and now try to estimate for tubular reactors for tubular reactor that means you know empty cross section you have right that is it is not a packed bed homogeneous tubular reactor then I may have action mixing in this and here we would like to find out what is the effect of this term K tau square by D by U L what is the real effect of that so that means what should be the ratio of okay or what percentage of conversion we will lose or concentrations we will have when you have depending on this values of D by U L D by U L need not be one value it can be different values which are connected with we are able to see not able to see then I sure yeah this is for tubular reactor and this one is for packed bed this one is for packed bed right otherwise I think if I keep here this is Versa I think you cannot project that yeah this is for packed bed this is Reynolds number and this side we have dispersion number similarly here we have dispersion number and this side we have Reynolds number so if I know my Reynolds number some value I can go and get that D by U L right and then we will do some kind of slight small calculations to get some important points here okay yeah please take this one let us estimate the rough order of magnitude the rough order of magnitude of effect of the term k tau square D by U L k tau square into D by U L that entire term for tubular reactors let us also assume Reynolds number equal to 2 into 10 to the power of 4 how many 20,000 yeah 20,000 Reynolds number so now go to that Levenspiel graph okay I will also write from graph okay from graph that is figure 13, 15 from Levenspiel third edition most of you have third edition okay so there it is not D by U L but it is D by U D T okay so D by U D T value D by U D T value is approximately 0.3 from graph good okay so I can tell third edition page number 310 okay good so now D by U L of this will be D by U D T into D T by L so this value I know so this will be 0.3 into D T by L simply substituted this value only we substituted now let us assume that we have X A equal to 0.95 X A equal to if X A equal to 0.95 what will be K tau P okay K tau P ideal ideal reactor X A is given what is K tau P I know you tell me the value 3 3 3 excellent so K tau P for X A equal to 0.95 K tau P equal to 3 okay now what is our assumption this equation is valid for V equal to V P then that is also must be equal to K tau still some minds are working yeah so okay I think this side totally dead I think they went to darkness but they went they are still in with the speculate worried about yeah so that is also K tau now substitute that K tau and also D by U L there Y L you will find out you find out now D T by you know D T by L to find out D T by L Sushmita Gana okay you understood now why we are doing that we would like to find out what should be the logical D T by L or L by D T so length by diameter of the 2 you are designing that new reactor okay yeah so there it comes nicely yeah you have the equation what equation what equation thereby what equation you lost it already what did I say there K tau is 3 yeah K tau is 3 but what do I do with that 3 idea is to find out L by D T I have this equation right you know for D by U L it is 0.3 D T by L and for K tau it is 3 and where is the equation this equation so now what is C A by C P now you tell me yeah how much is that 1 plus 1 plus 2.7 D T by L now you calculate if I need 1 percent deviation from ideal C A P C A P is the concentration that is coming out from ideal plug flow right I think see Pooja with time becoming more and more active okay good Pooja for you I think till 12 o clock class because you are you are becoming more and more active now it is very good I also become like that you teach more and more I think you know go on teaching that is all life long okay good so now the deviation is from C A P C A should be 1 percent deviation right so you tell me what is L by D T that is beautiful information this is I say C A P and C A C A must be 0.01 C A P yeah 1 percent how much L by D T 20 yeah 270 270 excellent L by D T equal to for 1 percent deviation 269.9999 0.9999 okay so D T by so for 5 percent because as engineers you can discuss you can decide whether you want 1 percent deviation or 5 percent deviation yeah for 5 percent what is that very easy I say divided by 5 you know 54 okay so what is the conclusion simply I wrote something you said 70 270 and then I wrote I mean is there any meaning in this what meaning you can make out of this if you want 1 percent deviation from ideality you maintain 270 times L by D T value then you do not have to worry about that complicated equation so is it logical 270 times what is the diameter of the normal tubular reactors normal yeah you take 6 inches 150 okay 15 centimeters or 150 is 150 mm okay 6 inches that is 0.15 meters now you multiply how many meters length you have to maintain 0.15 into 270 40 meters is it logical 40 meters I mean will you get this kind of values in industry Arshad I do not know is he seeing like this or like this what is that you said like this or like this will you get around this 40 meters length in plug flow reactors in the industry and what is the design problem ask him what is the length he got half kilometer yeah usually you will have kilometers length if you need you know that kind of big production rates what we are trying to do in industry okay so that is why that condition is easily met condition is very very easily met this is what what you have to really remember so okay I will maintain my L by DT as more than 300 approximate value then I do not have to worry about my non-ideal axial mixing at all wonderful no it is not there in any book except in one book called foment and not foment Himmelblow and Bischoff Himmelblow and Bischoff yeah or anyone has seen this book you would have not even known that that book exists but once upon a time this was an excellent book particularly for controlled people and also mathematical models same Bischoff Himmelblow is different why it is unpopular we never had this book because no one is teaching you from that that is the problem but Himmelblow and Bischoff Himmelblow separate you may have some book and you know that process calculation same guy only Himmelblow this is called deterministic deterministic systems and above that there is another title process analysis or something process analysis control is there RTD is there, transport phenomena is there wonderful book exercise it is not small book from there about reactors this particular thing when I saw I thought what a wonderful analysis I have to tell and I have been telling this last 20 years okay so that is the kind of wonderful information Himmelblow and Bischoff I think in our library there must be one copy see same thing let us extend for packed beds packed beds also you know plug flow reactors so the same calculations let me tell yeah DL by UDP or D by UDP UDP equal to for you know 0.5 from the other graph in the same page we have for packed beds also another graph yeah and K2 is same again let us say for 95 percent K2 equal to K2 P3 now tell me what is the equation for the point okay first we have to calculate for C A C A by C A P yeah so now this equation I can write now D by UDP no no D by U L equal to D by UDP into D P by L okay so this will be 0.5 D P by L for 1 percent deviation it is 450 I think anything thereby an accept no whatever you say you say minus 450 also you will say yes so you have to come to that situation now okay not over another thing also I have to do now okay so yeah I think the equation is you have C A C A by C A P equal to for others who are still sleeping yeah 1 plus D P by L so for 1 percent deviation okay L by D P equal to 450 for 5 percent okay is the value is logical what is the cactus diameter Swami what is normal cactus diameter we have our what you have done you dealt with only batch reactors sir 1 m to 5 mm is only half an inch also is higher side most of the time it is around 8 to 10 millimeters okay maximum half an inch is 12 maximum because larger and larger 12 mm okay 8 to 12 mm in the industry okay let us take average as 10 mm 1 centimeter how many centimeters length we will be there 4.5 meters 4.5 meters again is quite logical in industry height of the packet bed 4.5 feet may meters means how many feet that is all around 15 feet around 15 feet that is why axial mixing is not a serious problem in industry that is what is the conclusion from academic understanding so what you have to tell is if you get unfortunately a job in chemical industry you have to because most of you want to go for some other jobs so then what you have to tell your boss do not worry about axial mixing okay worry about only axial mixing radial mixing is very very complicated and also difficult to analyze as well as radial mixing is always present mixing means radial non-uniformity in mixing why all packet beds all tubular reactors whether you have endothermic reaction or that means you have to remove heat or add heat so that will definitely create at any cross section some kind of radial non-uniformity it is a wonderful conclusion at the end okay really wonderful conclusion that axial mixing is not a major problem again I will tell you this kind of explanations when you give when you go for even faculty positions and all that most of you I tell you again faculty is wonderful position okay yeah when you do phd and after phd you go for faculty meetings some how you have to bring them to RTD you know the entire interview then you tell what is ET what is FT and all that finally you end up axial mixing absolutely do not worry I will take care of it okay then I think jab is guaranteed good so that is what is the conclusion here right so the main conclusion is that axial mixing is not a serious problem because this kind of dimensions L by D equal to 450 in industry is easily yeah available for all reactors similarly the other one what is that 270 is also easily available most of the time they use either 3 inches or 2 inches 6 inches also is I am telling very large you know for plug flow for plug flow homogeneous 6 inches is definitely large normally 4 inches many many places 4 inches means 200 mm right yeah 20 100 mm 4 inches is 100 mm so 100 mm means 0.1 0.1 meters right yeah 0.1 meters into just yeah 27 meters 27 meters is change play in industry right so that is why it is very asking anything okay good so the next model it do not take much time is till 8 o clock we have time and also you came at 6.30 okay we started at 6 anyway the other one will not take it is now tanks in series model very simple not much complicated not much complicated okay actually I thought I will finish today even the other one also tanks in series model as well as CSTR with dead space and bypass these two I thought I will finish then I think happily Monday you do not have to come Friday also we can play here okay so in tanks in series model okay we have n number of tanks yeah all equal number equal volume so here we want to find out tanks in series model E t or E theta that is what we want to find out so this is 1 2 this may be ith tank this may be nth tank yeah we have volumetric flow rate v volumetric flow rate v there and we give a pulse input so this is the symbol for pulse input definitely you do not get pulse input out there so you will have of course n equal to infinity you will get there otherwise you won't get so now we have to write for the first tank what is concentration entering concentration leaving and there is no reaction because it is tracer right so I think I will ask you to derive that it is after sometime t right after sometime t only you have you have to write the one material balance and here you have pulse input right for first tank and also v1 equal to v2 equal to vi equal to vn and total v equal to n into vi total entire thing if you take good ok for first reactor what is input 0 ok let me write input equal to output plus accumulation plus reaction so this fellow is 0 then accumulation we have this is 0 for pulse input pulse ok so output ok c1 c1 c2 this is ci minus 1 ci cn minus 1 this is cn ok so what we have to do is we have to find out c1 and so accumulation is v into small v vi into vi is uniform for everything so vi into dc by not dc dc1 by dt dc1 we are writing for first so what is the solution for this Ramakrishna will tell you instantaneously c1 by c0 equal to s e power minus t by t bar i for individual tanks right yeah I think this you should have done now of course if I ask you to find out et for this then you have definition for et et equal to c divided by cdt right that you have to find out but we are not finding out et immediately so then for c2 similarly you write but that will be slightly complicated than this because there is input continuously input will be there continuously c2 by c1 equal to so again I think we also express it here in the other one it is not c1 by c0 I have but again I have c2 by c0 that means two tanks together ok two tanks together e power minus t by t bar i by t bar i ok c3 by c0 all these things can come in the examination also because I may give two tanks and then try to find out what will be the concentration coming out for tracing c3 by c0 equal to Abhishek will say not phase or you are blackmailing all the time no ok serious 2 t bar square i and e power minus t by t bar i so like that for n tanks one can do and what you get is after finding out finally cm by c0 and then converting that into e t and e t converted into e theta we will finally give only e theta you understood no first concentrations and converting to e t e t is the same definition for n number of tanks you have to write cn by c0 ok yeah so then you will have I think after deriving this you will have some logic it will be n here and all that you know this is n minus 1 for third tank this is n minus 1 this is also n minus 1 but this will be t by t bar i so it is logically it will come so once you get for cn by c0 convert into e theta then you will get this is nice expression some of you may be remembering this equal to n n theta to the power of n minus 1 by n minus 1 into e power minus n theta so where theta equal to t by n into t bar that is overall theta so t is volume by volumetric flow rate of individual tanks so this is the equation so this equation is it is again one can plot this equation for n equal to 1 e theta versus e theta so let me say this is theta equal to 1 that is t by t bar equal to i am sure for n equal to 1 n equal to 0 0 0 0 0 there is no reactors equal to 1 exponential so where did this start 1 1 at theta 1 to 1 1 to 1 you will have somewhere here then exponentially that is n equal to 1 n equal to 2 what is the value of e theta for n equal to 2 e theta is exit exit substitute for n equal to 2 theta when this starts what is theta equal to 0 e theta or theta equal to 0 e theta equal to 0 but what is that same thing for n equal to 1 that is 1 because that equation is simply that means I have only one tank right so that is e power minus theta for 2 tanks but theta comes almost in the equation not almost it is there so theta equal to 0 it will become 0 so it starts from here it goes like that this is n equal to n equal to 2 n equal to 3 again 0 it goes like this 3 okay n equal to 20 almost this is n equal to see even though I am telling many times you are not really enjoying what is happening in the subject because that enjoyment comes when you plot and plotting nowadays is not that difficult all these things I do not want you to plot with your hand give that equation in your laptop excel and then give some theta values and n equal to 1 what is the plot n equal to 2 then only you will have feelings otherwise you will never have any feelings you are all living without any feelings correct and you will have that feelings when you first fall in love then only atleast fall in love first and then fall in love with the subject later because the same thing can be extended there subject is very seriously thinking very seriously thinking okay so like this again you see I gave this RTD assignment and atleast particularly the first assignment can be solved on your computer after Rahul you left I plotted all that it comes almost exponentially that shape is that fourth problem in seventh fourth problem in seventh assignment okay it must be only ideal CSTR because the question is what is the ideal reactor there so I have not fit the model but when I plotted concentration versus time ET versus T and also FT versus T it did not take more than 6 or 7 minutes for me because excel you just add data and go and you have the corresponding equations and then just click you will get the answers and plot very fast so like that all of them when you calculate it will be very good of course calculating T bar and all that it may take time but plotting itself is very very simple so like that even dispersion numbers these values I mean this equation all these things are very easy to plot but I told you again no you are losing in your life something okay if you don't plot copy really you are losing something if you are not doing experiment okay really in project also if you are not doing experiment you are losing many things in fact simulations are good but when you do experiment first of all you have to deal with people in our workshop if it is not in our workshop you have to go to central workshop if that is also not possible then you have to go outside the workshops outside workshops all of our students are doing okay and we wanted to find out small particles those particles I think these people are I think entire world they are such okay world is of course through google and madras of course they know each and every corner who is making what kind of particle and that will teach you lot of lessons how to handle people otherwise if you are sitting before computer all the time doing only computer simulations you don't know how to deal with people that will be a drawback in the actual job unless you go to IT where you always talk to your own computer that is all nothing else then it is fine no problem you and your computer okay but otherwise if you want to go to management job or some industry job this is very important or even if you want to go for faculty positions this is important talking to people and you know managing them means don't we are not manipulating them right yeah manipulation is very dangerous word that is not the one what I am talking just looking at them and talking to them and then getting answers or also you are trying to explain that makes lot of difference I feel one of the reasons why most of us most of the young people are nowadays not able to get the jobs also one of the reasons is they don't know how to talk because where are you talking the moment you go to your room go to google that is all google who is there to talk maximum you can only sing songs that is all because you know same humming and then doing all google work what songs you use some kind of songs you know these are all natural reactions when you are seeing because you don't have any other thing so you click and then song and all that is all but you are not even going to your next room and talking to people so really when you want to do all this then only you will become perfect man or perfect man or perfect woman otherwise it is not possible I think technology is really killing this planet really because you don't know how to talk to people how to manage people how to move with people so what is technology connected that is all actually it also connects no kia connecting people that also advertisement I really like but there is no physical it is virtual connection matrix connection it is not the real connection yeah I think side by side both of them this fellow will be talking turning this way both are standing in the same place they don't they can switch off the cell phone and then talk they don't do that they are side by side that is also possible that is another thing group talk offer technology is really spoiling our planet anyway we will come back here this one this graph what we have again my question is how do you have to find it here what is n because I may have physically here n number of tanks I know but even all physical tanks are behaving as ideal tanks 10 means 10 tanks in series are they really behaving as 10 tanks in ideal tanks in series okay that is one thing otherwise the same model we can also imagine for packet byte instead of axial mixing we can use this model saying that the performance is equivalent to 10 tanks in series because if you plot this is nothing but your axial mixing curve very very low axial mixing also you get the same curve experimentally if you do for 10 tanks you get only this kind of curve and for packet byte also if you do you get this kind of curve that is why that equivalence in modeling same n number of tanks can also be used for axial mixing or axial mixing can also be used for n number of tanks but normally the other way we do that is for axial mixing using this what is the advantage advantage is first we use this graph again same techniques go to matlab and then find out what is correct here okay let me say that n equal to 7 now I have a parameter now called number of tanks in series this has to be used now for calculating ultimately what is that you have to calculate conversion so how do I now use this number for calculating conversion if it is first order tell me if it is first order this you should remember 1 by 1 plus 1 by 1 plus k tau i 40 bar i that is all that is why beauty here you do not have to go to that terrible equation where you have exponential terms and all that that is why many people prefer this model even for packet bats when compared to axial mixing model simple no for first order even for second order you will get that quadratic terms and all that but again 7 square roots will come there let us beautifully use that so that is all that is why this model is very simple than the other complicated model right so that is what I just want to tell and we have this equation sigma theta square how do you get that that you get only because of this theta theta square e theta no e theta d theta minus 1 compute this equation here and then integrate what you get is sigma theta square equal to 1 by n which is nothing but oh no no this is total that is total so it is much easier now that means sigma theta square how do I get sigma theta square I get from my c versus t data itself I do not have to go to this graph and then plot also simply go to c versus t graph calculate data calculate sigma t bar square sigma t square divided by the average residence time that will give you sigma theta square that is nothing but 1 by n this is another model again I beautifully love in chemical engineering sigma theta square equal to 1 by n simplest equation so now if n equal to sigma theta sigma theta square equal to 0.1 n will be 10 so calculate from the data let us say you got 0.1 then you will have number of tanks 10 so go to this equation if it is first order this is 10 to the power of 10 this is not 10 to the power of 10 1 plus k to the power of 10 calculate c A by c A naught what is x A okay so this is what I think the single parameter single parameter models okay thank you for your patience I hope you also enjoyed