 I hope you remember what is going on in this class so we have completed the elutration problem also in the last class elutration is the more realistic problem where in mixed flow if you have plug flow reactor sorry mixed flow mixed solids mixed of solids so then you will have some fine particles and some large particles so the fine particles may go out which is more realistic so that is the problem which we have solved okay so with that about mixed flow slight complications have been done there may there are many many problems still which we cannot discuss here okay so now we will also try to do for any RTD is it possible to have the design equations that means the general performance equation what we have general performance equation for mixed flow of solids for any flow actually I hope you remember this equation very general performance equation anyone can tell for specifically ask non-catalytic gas solid reactions what say sir design equation performance equation is the design equation yeah 1 minus x b equal to or 1 minus x bar b single particle then ET DT this is the one and this ET takes care of any kind of reactor this we have done already okay in the last few classes also what we have done is only this I think maybe you would have forgotten so this ET for plug flow we have substituted right equation is same okay in the sense that equation you will get one equation if you have ET for plug flow then you use the same equation for multi particles mixture of particles is only weighted average so then we have changed this equation to this ET to mixed flow right again in mixed flow we have done for mixture of particles and mixed flow normally good representation is the fluidized bed in fluidized bed you will have the elutriation okay so that also we have taken into account and then solved but equations is the equation is only one in fact only one that you take it and then take it weighted average for mixture of particles and also the same equation but only thing is you have to now calculate for each particle what is T bar in elutriation case okay so essentially you learnt only two equations one plug flow another one is mixed flow then various conditions we have applied so now we can also extend this ET for number of tanks in series if I have number of mixed flows in series right one situation can be the as I told you mixed flow good representation is fluidized bed if I keep these fluidized bed one above the other or side by side so then I can take a case where you have tanks in series model like if I have a fluidized bed you have that is the distributed bed this is the down comer down comer down comer okay so this is the outlet for solids gas solids coming out and this is inlet for solids and of course gas goes out how the solids move they go like this come here again they go like this come here they go like this come here good so each plate is a fluidized bed and on each plate you have perfect mixing of solids and this is considered as one one mixed flow okay one mixed flow then it goes to the second third this is as if we have yeah tanks in series that is one so the other model also what we can have is you have normal fluidized bed this is gas gas solid here this solids will go to another fluidized bed again gas and this will go to another fluidized bed this is solids coming yes yes and finally solids will be coming out so this is solids solids this is also another type of multi say system okay multiple tanks in series good yeah so in these two cases either this or this yeah what is the difference yeah so here you have fresh gas contacting every time whereas here you have the same gas that is going but what was our assumption whether the in deriving shrinking core model we have one concentration CAG whether it is here or here that concentration is not changing so the same analysis can be also applied if I know what is ET I think it is too much to ask you what is ET for tanks in series because I think single tank single tank what is the ET Kavya yeah that is for single so if I have n number of tanks we also have a equation for ET so that I will write first and then we will substitute I will ask you to integrate in the examination hall but right now I will give you the final okay good so ET let me say this is equation 1 I will write here RTD function for n number of tanks n tanks in series will be this is called RTD function 1 by n minus 1 factorial T by T bar to the factorial sorry not factor to the power of n minus 1 then we have E minus to the power of T by that is the equation so this is for ET this is slightly complicated equation in the sense that because we have n number of tanks so this will come in that recursive formula n number of times so then you will get a series solution and if I write equation that is 2 equation 1 in terms of 2 that will be 1 minus X bar B equal to 0 to tau 1 minus X B for single particle and all of these factorial n n minus 1 okay then E power minus T by into DT this is equation number 3 in reality you can see how complicated it is you know if I put here all 3 controlling for single particle okay or it may be 2 controlling or the simplest one which we take here is reaction control the same equation for reaction control what is the equation for 1 minus X B for reaction control yeah 1 minus X B single particle yeah single particle 1 minus whole cube would write yeah so this is equation number 4 so that you have to substitute there so that this entire function will be in terms of T then you have to integrate madam please make a note there that mathematical details I can ask you cannot also write that after integrating without integrating okay because I think I do not want to give marks for the memory you have to show the details good so for reaction control if I substitute equation 4 here okay let me also write that equation 0 to tau so 1 minus T by tau it is a single particle so only tau this one plus I mean into all this E power minus T by T bar M T by T bar of M by n minus 1 then I have E power minus T by correct this equation you have to integrate yeah so if I write the solution for this then 1 minus X bar B equal to sigma of M equal to 0 to M equal to n minus 1 M factorial T bar M I by tau whole to the power of 3 M then E power minus tau by T bar M I that is one term plus M equal to 0 to M equal to 3 and the top of it you have factorial there n minus 1 factorial then you have 3 factorial M factorial 3 minus M factorial minus of that is the equation so this equation number 6 this one yeah it will be T M I by tau this one yeah yeah this one is no to the power of M to the power of M that is minus T bar M I by tau only so we can simplify this for n equal to 2 for n equal to 2 you know for n equal to 1 you know for n equal to 2 for n equal to 2 what you get is X bar B I think now you cannot drop course now 6 by Y square 3 plus E power minus Y plus 24 Y cubed 1 minus E power minus Y yeah this is equation number 7 for n equal to 3 yeah I will write where Y equal to tau by T bar M I okay so for n equal to 3 X bar B equal to 3 Y 3 minus E power minus Y minus tau by Y square 3 plus 2 E power minus Y plus 60 by Y cubed 1 minus 4 so this is equation 8 okay like that of course n equal to 4 also equations are available how can I change fluid as a plug flow I cannot fluid as a mixture flow single stage if you go to 7 stages here okay then you can use plug flow equation so that is why you would like to have 7 stages we can write like we mix all fluid as and we can get overall equation for plug flow yeah that is what for n equal to see this one if I write n equal to 4 5 6 7 onwards if you try to simplify that that will tend to plug flow that equation will tend to plug flow okay yeah but you cannot avoid 2 3 4 and all that yeah beyond 6 7 only you will have plug flow good okay for comparison I will ask you okay 2 and 3 with plug flow how much deviation between these 2 plug flow and 2 tanks in series or 3 tanks in series okay good so that is the one and industry who told you it will be more than 7 where did you see first of all SR oil limit SR oil where they are using fluid as inputs first of all purification I remember you are not reactors purification sorry purification steps are not reactor steps reaction steps I do not exactly remember the purpose but I remember they were using a fluid as bed and they had some 10 to 15 stages in that so I do not exactly remember what is the process that I cannot remember exactly handling one fluid as a bed in industry is a hell if you want to have 17 hells yes not in series like the second arrangement the first one with 7 stages like that one I am telling but what for they are using fluidization is such a beautiful technique where this can be also used for adsorption it also can be used for drying but those are physical operations okay yes so drying operations I do not know which process you are talking in fact is it reaction you have only one drying one can use but drying these things will not apply because this is reaction in cracking SR oils in cracking they use first fluidized bed where they have a single tube where solids will go very fast like moving bed then there I have also discussed that actually the contacting pattern for that will be both are in plug flow solids will be in plug flow and gas also will be plug flow yes so come on you are asking only cracking only one reactor but you know depending on the capacity it will be only one single tube and okay because you asked me let me tell that so this is how cracking reactor looks like there are many variations of slight complication but simplest thing is here you have the gas entering then solids will move so there is a cyclone okay so then you have the regenerator and after regeneration solids will come here so solids line is this solid of course here here solid then here solids will come this is again solid back okay gas line is gas gas okay this is how but why they have to do this this is regenerator yeah catalyst catalyst is getting deactivated because yeah because of the coke formation so then you have to regenerate this that means the coke deposit again will be burnt out actually it is not burnt out it is gasification reaction C plus H2O what they send here is H2O okay steam so then you will get CO plus H2 there that can be used for some other energy management because that can be burnt and then that can be used as some other energy management and here again this can be a fluidized bed right that can be a fluidized bed and there are many reasons why a fluidized single stage fluidized bed cannot be used there also right packet bed is the best or moving bed is the best there because if I use a single fluidized bed here then the because of the fluidization some particles will get regenerated quickly some particles will not get regenerated at all that means those particles which are straight away entering into the is a mixture flow right so some particles will immediately come into this line outlet so under those condition those particles will not be dig I mean that carbon cannot be gas fried at all that reaction is time is very very small so reaction may not take place so that is why you will have particles with carbon coated again plus completely converted particles also that means all carbon deposit burnt so that is why there they can use generally moving beds where you have plug flow plug flow gives for each and every particle the same residence time so that you know exposure reaction conditions are same for each and every particle whereas single stage fluidized bed will give you distribution of solids okay good so that is why I do not know what you are talking about but I think 17 I have not heard of in any industry using 17 fluidized beds by the way what is the size do you have any idea what is the diameter of the circuits this is diameter 1 cm2 logical guess so I am just trying to check you whether you have that estimation of sizes okay so if you are able to estimate this approximately there also probably you can but beyond 15 10 you cannot differentiate but definitely 2 and 5 you can differentiate okay beyond 7 there is no use because you get almost plug flow with 6 or 7 this in react theory course we have done it you know we have actually calculated 2 stages together 3 stages together and all that I also would like to do now on demonstrative problem where we can see what will happen if I take one tank and two tanks okay so that I would like to tell you so this is how and these equations when I take equation 7 and 8 if y is small then we can also expand that in series like for example these things there okay so when we do that then my diagrams are nice diagrams I think I will remove this okay at high conversions normally we need high conversions y is small right what is y tau by t bar m so for large conversions t bar must be large so that is why y is small okay so y is small then the equations 8 7 and 8 can be written as equations 7 and 8 can be written as for n equal to 2 again we come back here 1 minus x bar b equal to y square by 20 minus y cubed by 60 plus to the power of 4 by 280 happily sleeping trying to sleep someone so this is 9 so for n equal to 3 1 minus x bar b equal to y cubed by 120 minus plus okay this is equation 10 for reaction control okay then I can also write for film control think I will go there for film control this I will remove this is all reaction control for film control what is the equation x b equal to or 1 minus x b equal to 1 minus t by tau that is the equation that you have to substitute in the general expression here yeah here and then for n number of tanks okay so e t is for n number of tanks then you will have for n equal to 2 these are slightly better equations small e power minus y so I lost my count this is 11 this is 12 then for n equal to 3 x bar b equal to 3 by y minus equation 13 let me also write for 4 stages 4 by y minus e plus 3 plus y plus y square by 6 into e power minus y of course one can also expand this e power minus y for small y and then also write okay I think you should appreciate it is not simply you know I am writing there you are writing there because how much mathematics we also use in chemical engineering because beyond certain concepts then everything is only mathematical okay but in a classroom I cannot make it as a mathematical course you know rather than conceptual course right so that is why in any academic institution the concepts are told and at a still if you go to higher level then you know the mathematics you have to use and then solve them okay yeah even now of course for examination you know to solve them and that is why you know we have only 6 courses per semester no maximum I think in chemical engineering we have reduced it to 4 or 5 I think for m take m take so that you can work more you have to work more okay good so this is the one and you know it is not that easy to write these equations you will not get in fact this kind of closed form solutions are analytical equations if you have the other devil what is the other devil this is film control the previous one was reaction control next one is as diffusion control so you do not have equations I can give you numerical problems there good okay so now let us illustrate the effect of multi staging okay the effect of multi staging very simple thing what we take yeah for example I have to solve this equation I can write this y square in terms of what is that tau by t tau by t bar right yeah so I can write this tau by t bar square cube to the porof and all that but we do not take all these terms to demonstrate the concept but we will take the first term this one first term this one and also you have similar term for single tank when you expand that in series okay when you expand that in series so we will take one tank two tanks and three tanks and we will try to find out what is the hold up flow rate is same for the same given flow rate and if I take and conversion if I take one tank so what will be the hold up if I take two tanks what will be the hold up because conversion we are assuming right and three tanks what will be the hold up so hold up means it is indirectly your volume of the reactor size of the reactor okay good let us do that so please write this let us illustrate the size reduction achieved by multi staging comma comparing the size requirements for single stage two stages and three stages for 99 percent conversion of solids where reaction at the sinking core controls where the reaction at sinking core controls good so I have to also write here for n equal to 1 you have in your notes can you kind tell me at least you can once more see what you have written there excellent y by 4 minus y square by 20 okay etc and y cubed by ya plus 1 ya 120 good okay but for our present discussion we take only these terms actually what you have to do is that you can you have to solve this equation either cubic equation or quadratic equation and then calculate what is t bar by or tau by t bar okay that we can do what is the conversion conversion equal to 99 percent I think nela is not in the class I have told okay for 99 percent conversion right okay so can you tell me what is t bar by tau for 99 percent conversion if I have only single stage take only one term what is w what we are taking is 1 minus x bar b equal to for n equal to 1 ya what is the equation here y by 4 y by 4 means tau by t bar m by 4 right ya so this is equal to 0.01 because x bar b equal to 0.99 okay so now tell me w that is t tau equal to 0.04 t bar t bar m t bar m t bar m equal to w by w by f not so w equal to w by f not correct no so w equal to 25 this is 50 okay very easy now for 2 tanks you can find out because we are taking only one term otherwise you know that quadratic equation you have to solve for 2 tanks for n equal to 2 1 minus x bar b equal to tau by t bar m i whole square by 120 correct no ya how much you got waiting for someone else to tell you so 2020 2020 2020 ya good so here w equal to 4.5 tau f not so this is equation number 16 okay this is w n equal to 2 to be more specific for 2 tanks together you calculated for 1 tank t bar i okay so 2 tanks you have to multiply then you will get that is where you make the mistake in the examination and you think that you will you have solved correctly and then you scold me i am not giving the mark okay ya now take the ratio between these 2 what will be the ratio of the w's w n 2 by w n 1 w n 2 w n 1 0.1 0.18 is correct ya 0.18 now what is the meaning w for 2 tanks equal to point wonderful result so that means the hold up which you have to use is 18 percent of single tank what Kavya able to get that what did i say no ya now the question is why sir is the hold up for the 2 tanks in series in total or for each one of those that is what what you know for each one that is half of that it is 2 point 2 point 2 4 6 for 1 tank okay w equal to so for 2 tanks multiplied by 2 okay so then this is the total for 2 tanks together that is why n equal to 2 i have put that means n equal to 1 n equal to 2 together in one setup you will have 4 point 5 tau f n 0 tau is constant because same particle size and same conditions and all that so f n 0 is the given flow rate conversion is 19 percent so then you see if you are just to tell you if you are using 100 tons in one single tank and you are going to use only 18 tons in for 2 tanks okay now n equal to 3 can you kindly repeat then we can discuss for n equal to 3 n equal to 3 what is the equation ya i think let me write 1 minus x bar b equal to tau by t bar of mi whole cubed t bar of mi whole cube divided by 120 120 okay this is equal to point not 1 now that you have to solve and what you get for w n 3 all 3 together what you get for single stage how much on rock 2 point 8 2 for 3 tanks how much okay this is how much you tell i do not have that calculation 2 point 8 2 so this is the equation now if i am writing this you know for n equal to this is 17 so now w n 3 n equal to 3 by w n equal to 1 point 1 1 3 point 1 1 3 point 1 1 3 this is another wonderful result now you can discuss a lot and for n equal to infinity single stage reaction control n equal to infinity so very nice you know n equal to infinity ya n equal to infinity plug flow correct for n equal to infinity plug flow of solids single size you have the equation already single pallet equation itself is useful reaction control ya what is that before i ask the question you should have told that first tell me the equation t bar p by tau equal to 1 minus x b should be there x b equal to 0.99 calculate t bar p t bar p equal to w p by t bar p equal to w p by f not w p means you know plug flow how much is this what is 0.184 t bar p you are telling or w p equal to 0.784 okay so now w p by w n 1 also equal to w n infinity w n equal to 1 equal to 0.03 ya this is another very beautiful equation so that equation is okay I am not giving this this is the one which I am giving numbers so this equation is 18 you see now when you use a single mixed flow and then if you make the same thing same conversion and same flow rate make that into 2 equal reactors you know size is equal for both because t bar i is same okay so then only 18 percent of single stage hold up you have to use that means indirectly the size reduces by so much 82 percent okay from there if you want to put the same thing into 3 reactors single stage you make now into 3 small reactors so then it will be 11.3 percent right and then you go to infinity it is ya 3.1 percent okay good but the maximum change that is from single tank to second tank that is very very drastic from there it would not change much right you can calculate for example for 4 it would not still further reduce this is n equal to 3 this is n equal to 1.18 and 0.11 then next one may be 0.09 okay then may be 0.07 may be 0.06 like that comes till 0.031 beyond this you cannot this is one extreme single stage is another extreme where is my single stage ya here equation number 15 okay why do you think this is really happening it is good equations are beautiful nice why do you really think that this is happening particularly from 1st to 2nd this I discussed also last semester again mathematics what is really happening not exponential decay and all that what is that really happening to the particles so that there is so much drop in the size just 2 tanks if you take instead of 18 tons I mean instead of 100 tons I am taking only 8 tremendous reduction I say all intelligent chemical engineers will do that whenever they have mixed flow it is really beautiful concept okay now how do you define mixing how do you define mixed flow uniform temperature and uniform concentration throughout and what is happening to the particles inside uniform concentration means uniform conversion uniform temperature okay good but still what is happening when it is coming to the 2nd reactor and do you know did you calculate at any time even now you can calculate why at any time okay what is ET DT or how much of the material will come out in the 1st 1 mean residence time 63 percent the moment you take 2 reactors that means what see the by definition mixed flow is a bypass type reactor bypass what is bypass not spending sufficient time in the reactor just coming out by definition of mixing itself 63 percent has to come out within 1 mean residence time and the same thing when you extend to plug flow what is the percentage that comes in 1 mean residence time nothing will come only after 1 mean residence time only all that batch will come so that means each and every particle exactly spends same amount of time the conversion in that is same then the average also is same but when you look into mixed flow when you are measuring the mean residence time 0 to 1 second you get some fraction we can calculate you can calculate in fact using that equation ET equal to E power minus T by T bar M by T bar M this is the only equation now the fraction will be ET delta T equal to E power minus T bar M delta T 1 can calculate from this so this is the fraction this delta T is between 0 to 1 second 1 second to 2 seconds 2 seconds to 3 seconds so like that whatever time interval you take so you can calculate how the concentration is coming out or the fraction is coming out if I plot ET versus T then you will have very steep drop and then slowly going this is what you are doing but that is related with bypass so inherently mixed flow reactors are the reactors where bypass is already there but that bypass is a benign bypass that means benign bypass in the sense that I know clearly what is the bypass okay if someone asks me okay your mean residence time is 10 minutes can you tell me how much material has come between 0 to 1 minute I can tell okay because of this equation because of that equation I can calculate that so clearly I will find out and the same time if I take 0 to 1 minute 0 to 1 minute I can calculate what is the fraction here the same 0 to 1 if I go to plug flow reactor and look at what will happen you will not see any that material that batch material coming out in the outlet so that is why mixed flow inherently gives the bypassing and when you have bypass solids will not have sufficient residence time to get converted so that is why you get some unconverted particles practically no conversion of particle also 0 to 1 second or 0 to 5 seconds 0 to 1 minute depending on mean residence time okay so what you see at the outlet is on the average the concentration of all these particles what you see at the outlet of the mixed flow okay so that is why the moment you put the second reactor that bypass is tremendously limited why the same particles are not going to come out the chances are very very very very less okay for the second the moment you put second reactor after first reactor the particle which has bypassed the first reactor need not bypass in the second reactor in perfect mixing conditions okay that is the reason why you put n equal to infinity we say you get plug flow what is the meaning of that meaning is all the particles will experience the same bypass the same you know time and at the end if you look at all the particles every particle would have spent exactly same time that is the funder there why you put n number of tanks equal to as n tends to infinity we have n number of tanks as n tends to infinity you will get plug flow the particle residence time the probability for the particles to stay more time is more when you put more and more tanks in series so wonderful concepts you know really wonderful concepts mathematically you can just try to enjoy but I think you know physically if you are able to imagine that is the thing what is happening so that is why whenever you have a mixed flow reactor if there is no other constraint definitely better it is better to go for multi-staging in fact plug flow never really you know you cannot maintain plug flow in reactors but whereas mixing you can fairly do that so that means if you have you can calculate and then show that when it is almost equal to this okay normally you know beyond 6 only it is almost may be 95 percent 98 percent you can touch this value okay and if you want to get pure plug flow you have to have reactors from 7 to infinity 7 to infinity but if you are able to say that okay around 5 percent also I can tolerate as an engineer then 6 tanks are used and where used in industry is in ammonia fertilizers you know ammonium fertilizers ammonium sulphate when they are making ammonium sulphate in ICI company they were the first people I think 6 tanks in series okay to make ammonium sulphate as a fertilizer okay so 6 tanks are just side by side not one below the other these are liquid phase reactions so one tank another tank another tank another tank by gravity it flows and you will get almost plug flow condition they cannot use directly plug flow there because some precipitation comes during the reaction so if you have plug flow and precipitate single pipe reactor as plug flow reactor the precipitation will damage the plug flow okay so you cannot have that so that is why this tanks many wonderful things are there in chemical engineering if you are able to think properly okay good so this is what I wanted to do in non catalytic reactions but what I have not done also I will tell you okay I just only try to tell you the beauty in non catalytic reactions by taking simple concepts all other things are the same type of analysis only we use okay but only thing is mathematics will be slightly complicated the moment you relax the assumptions one by one like you know Levenspiel in his book also gave that if it is not spherical particle if it is a cylindrical particle there are equations if it is flat plate there are equations so that takes care of geometry okay of course we are assuming that it is isothermal reactions all the time non-isothermal reactors analysis is more because energy balance material balance both is required but most of the time in non catalytic reactions the particle maintains isothermality because the conductivity of solids are very high so whatever temperature is generated there immediately it is distributed across the particle so that you can always think that the particle is isothermal particle same thing even with catalytic reaction because the thermal conductivity of the solids is high so that is why we can take isothermal particle that I will also prove when we come to catalytic reactions okay that is one and what we have not also done was that you know like for example calcium carbonate particle that will react in the presence of heat that is heat transfer control problem all the time we have been talking about mass transfer but analysis is again same but only thing is energy balance you have to write similarly gasification is again a kind of heat transfer problem it is endothermic reaction so you have to supply sufficient heat first of all to the coal particles if there is not sufficient amount of heat that is going to the particles gasification reaction cannot take place that is again mostly heat transfer control most of the time endothermic reactions okay then we also have some reactions where gas phase solids are getting formed this is again wonderful reaction for example your silica you know chips what you make pure silica is made only because of this particular reaction in the vapor phase reaction is taking place they send this gas towards a surface like a coating okay like jets and then their nucleation starts and then fine particles will form again wonderful reaction but mathematically it is also very complicated because what we have to use probably is population balance models and all that where different particles at what time what size at all that coming so that we have not done and what else type A type B type D we have done type C is your calcium carbonate particle okay then E F E F R you know I think E is that bombs yeah ammonium nitrate and all that explosives so there you do not need any reactor okay because the reaction is instantaneous there again it is mass transfer I am sorry heat transfer control that is why during Diwali time if your fuse is not sufficiently getting heated up you know fuse means that white thing will come out like tail yeah and then we stay half kilo meter and then try to burn it okay so if there is not sufficient amount of heat that is passing through it would not you think that it will catch the fire it has caught the fire and then you start running you start running but nothing happens to that so again you have to come and heat it okay so those reactions so these are the overall pictures of you know our non-catalytic reactions but most of the reactions are iron ores iron ore zinc ore copper ore all these reactions most of them behave as shrinking core models okay then the next one with the particle change is all coal combustion process coal gas good so in the next class we will happily do catalytic reactions and for the examination and all that I think you have to prepare a lot for non-catalytic reactions okay good thank you