 in the last class we have developed the equations for the design of mixed flow of solids but single size what we do in this class is that again we extend the same analysis for mixture of particles we have mixed flow of solids and most of the time mixed flow of solids the excellent example is fluidized bed where you have perfect mixing of solids and I hope you know what is fluidization right so that is why only based on that we are moving and all chemical engineers must know what is fluidization and I am sure most of you would have done the experiments even in fluid mechanics laboratory but your experiments at that time probably would have been liquid solid fluidization you have not done you have done yeah androck you have done fluidization liquid solid yeah ranita done it yeah gas solid is most widely used in industry so what we are talking here is gas solid where perfect mixing assumption is always very well valid okay so now what we are going to do is if you have 2 or 3 types of particles that means you know sizes like 1 mm particle 5 mm particle 10 mm particle so how do you get fluidization is a totally different matter but we assume that we have uniform fluidization for all these 3 sizes of the particles and then can you also develop design expressions for this mixture of particles right so that is the one but in general the realistic situation in fluidized bed is when you have 1 mm particle and then 10 mm particle I have just exaggerated there 10 mm particle will never fluidize and 1 mm particle may elutrate it may go out okay so that is why the realistic situation is that even if you have very narrow size particles and particularly when you have gas solid fluidization you will have lot of attrition because the solid particles are moving vigorously in the bed so 1 particle will go and hit the other particle so if there are edges for these particles they are not perfectly spherical particles you never get perfectly spherical particles except on the board when you draw or in the textbook okay in industry you never get that so always you will have some kind of edges rough edges for any particle and if you want to remove those edges then go to fluidized bed fluidized all the corners will be cleared then you will get almost spherical particle okay and during that time what is happening is the lot of powder is generated that powder comes out in the cyclone and sometimes if it is catalyst particles so this powder again they take make particles and again they may send it okay so that is why gas solid fluidized beds we have this attrition that means small particles may get generated but on the other hand to start with the process contains small particles and large particles small particles will go out of this terminal velocity that is why I have been asking you know tell me one use of terminal velocity in industry this is one of the uses you will calculate and then find out at what velocities these particles will go out right so elutration is a normal phenomena in fluidization now in the next after finishing the distribution of solids next one is extending the same thing to elutration if I have elutration can I design the fluidized bed reactor this is more complicated but you know as a teacher I have to take you from simple to complicated right this is complicated design the other one so that is what what we do in this class right so the problems that may we may face now is that the residence time of the particles may not be uniform in the bed right so the first case is single particles uniform gas composition and they are not changing with time that is assumption right so that is why you do not have attrition powder coming and all that is not there ideal situation so that we can easily understand what is going on in the bed okay so that one we have derived already so now the second situation is I have mixture of particles may be 1 mm, 2 mm, 3 mm particles right so now first we have to prove that this 1 mm, 2 mm, 3 mm particles will have the same residence time or if they do not have how do you take that into account that is the problem and next one if you go elutration definitely you will not have uniform residence time for all the particles why because when you are continuously feeding to the fluidized bed the light particles elutration we are taking into account that means some of the fine particles will definitely go may be if you have 200 micron particles 500 micron particles and may be 800 micron particles okay fluidizing so then 200 particles may go to the top may not be all some of them okay why because of this perfect mixing of all the solids some 200 micron particles may stay inside some 200 particles which may got caught into this outlet gas then they may go out so that means those particles are definitely not spending same residence time as other particles that we have to bring into picture when you are designing this that is the only complication if you are able to understand this story first then equations we will derive okay mixture of particles we have to prove that all the particles will stay same time or that means residence time is same but for elutration definitely there will be residence time distribution or residence time changes for small particles and large particles how do you take that into account and then try to design the reactor what do you mean by design again always given the conversion find out the volume or the total hold up here okay or given the reactor find out conversion that is all what do you mean by design equation is same for both either this will be given or that will be given so now let us take that mixture of particles mixture of particles of different sizes different sizes but unchanging unchanging size mixed flow of solids okay mixed flow of solids and other assumption is uniform gas composition okay good so we will imagine our bed something like this we have the fluidizer bed this is gas gas may come out there then we have the solids entering okay so these are the solid particles I am putting something slightly bigger something smaller yeah they are not uniform size then we also have what are called bubbles this bubble is breaking somewhere here we have bigger bubble may be this side another bubble and solids also come out so if I take this one as f f is coming out so then we have three particles you know three different size particles or may be four different size particles or may be two different size particles so if I just show the distribution of the particles so this is f of r i versus r i r i is the particle f of r i you know the flow rate okay so that distribution may be if it is something like this this is very nice distribution okay and this side also we can show the distribution this is again f of r i versus r i okay yeah so will this distribution really change because all the particles are coming in they are not changing in size and getting reacted coming out will the distribution this distribution will it change here just in case will it change you know normally this is what is the problem with all of us the moment we say distribution our mind will block we cannot imagine why are you thinking so much time to answer this question yeah it must be same but why are you keeping quite sleeping time okay so this must be the same okay because there is nothing no change is happening those are the assumptions very clearly right so this distribution will be exactly yeah same good this is one thing now I will ask another question now I have this distribution of solids inside this is distribution in the outlet this is distribution in the inlet and now this is distribution in the bed okay will the composition will be same there also that means this is in the bed means it is W W is the hold up hold up of 1 mm particles 2 mm particles 3 mm particles will that distribution will be different than this and this will it be different yes sir why yes here is the elutration is it we have taken that case are you chess player chess player think 10 steps ahead you are thinking about the next problem which I am going to tell you okay this is the present problem where we are only assuming that you know no elutration nothing right where do we say that particles are going out it is the same size uniform size nothing is happening so yeah why it is same yeah that is again beauty of assuming perfect mixing ideal condition okay what is the meaning of this perfect mixing the outlet conditions and also bed conditions are exactly same otherwise I told you know you cannot even write the material balance in the basic mixed flow reactor also I have been repeating these things many times but still it will not record because simply I think you come and sit and then enjoy the class and go that is all no yeah not for learning so that is why here also you will get exactly the same distribution right so this bed distribution so now this tells me that we do not have any change in the residence times okay but anyway mathematically also we can prove that please take this otherwise you know you do not understand I know that at least at least once in before the examination at least you read it yeah please take this since the system is assumed to be mixed flow of solids since the system is assumed to be mixed flow of solids and the size particle is unchanging the size of particle is unchanging the size of particle is unchanging the exit stream represents the bed conditions the exit stream represents the bed conditions the bed conditions full stop we can also say that we can also say that the size distribution of the solids are okay the size distribution of the bed we can also say that the size distribution of the bed feed bed, feed and exit streams are all alike okay are all alike are mathematically f of r i by f equal to w of r i by w okay so then which can be this is 1 a which can be written as w by f equal to f of r i good so our t bar m equal to t bar m of r i because w by f is nothing but your t bar right since these two are same we also have w by f so this is equation 2 this is hold up divided by mass flow rate f is mass flow rate good so now once we know that the thing is same okay the residence time for each and every particle is same we have an equation for calculating conversion 1 minus x bar b equal to what is that equation 0 to tau 1 minus x b r i into e power minus t by t bar m by t bar m dt right so now the procedure is same that means 1 minus x b r i for each particle I have to substitute and then add up all the particles I mean this is for a single size particle and because we have the now the distribution you will have this is equation number 3 since we have distribution you will have now 1 minus x bar 2 bars sigma r equal to 0 to r m then we have f of r i so this is equation number 4 good so now we already know what is the solution for this we have but again you know independent steps controlling film control reaction control diffusion control we can substitute here and correspondingly write the equations let me write that those equations and I think this is clear no that integration and all that you have done for single size particles individual rate controlling steps so now we have to just only take the weighted average but only thing you have to prove is that your residence times are same so that your t by t bar here which is coming later you get most of in all the equations you get t bar by tau t bar m by tau so that t bar you will know for each individual size okay so that is all what you have to prove that is why you are not learning anything extra here in terms of equations except that you are only extending your knowledge of understanding that is all okay other than that we do not have anything new here good so if I write again for practice you have to anyway note down this for film control we have an equation this is maximum r m 2 factorial tau of r i t bar m minus 3 factorial tau of r i by t bar m whole square plus etc oh sorry I have given only 2 terms please check if I am writing something wrong you have to tell me so this is equation 5 you have to remember all this so then 1 minus x bar double b oh sorry for reaction control number 2 I have to write reaction control for reaction control we have 1 by 4 tau of r by t bar m minus by 20 for a by f this is equation 6 and for diffusion control hash diffusion control we have again x bar b t bar m minus 19 by 420 whole square f of r i by f so this is equation 7 and definitely you will have much more difficulty if you have 2 control or 3 control from this equation okay but one has to do it is only mathematical technique that is all conceptually you do not have to understand anything more but only mathematics will be messy so that is why there is nothing new to learn conceptually that is fine good so this is the one and as usual I can tell you that if I know the reactor volume or hold up and then flow rate flow rate we should know anyway so then you will know t bar m and then is easy to calculate x bar for a new reactor x bar will be given normally that is what the problem also given that is for plug flow right then x bar b you will know then you have to solve this equation to get t bar m once you know t bar m this is the equation you have to use to calculate w right so f is known to you and w w let us say you got 2 tons of solids so how do you put them as a fluidized bed so general thumb rule is fluidized bed packet bed before packing you know fluidized bed when fluidization starts before that all the particles will be in packet condition when the drag force of this gas which you are sending from the bottom of the distributor bottom of the bed when it is equal to weight of the particles total weight of the bed then starts moving okay so that means you are floating you are making these particles float in fluid stream and at that point of time if you look at this mixture now the solids as well as gas all the liquid properties or fluid properties you will see for this mixture mixture is now gas and solids or liquid or liquid and solids that is why the name fluidization is given that means imparting fluid properties for normally immobilized solids otherwise solids cannot move right so by putting this fluid then you can transform these solids into a fluid like state okay good so for that of course even now if you go to research papers in chemical engineering and see journal will have atleast minimum 2-3 papers in fluidization so that means we have not yet understood thoroughly because we have put so many assumptions it is very easy for us to understand the problems but in reality I have shown you here these bubbles those bubbles are the biggest headaches in any fluidized bed but these bubbles are also good for mixing unless there are bubbles there is no good mixing so that is why where do you cut I mean how many bubbles you need and how many bubbles you do not need that is very important thing in fluidization okay because if you have very very large bubbles and all that you know there will be chaotic conditions you do not know what is happening in the bed and that is why actually these models will not be exactly suitable for calculating there is what is called bubbling bed models and also two phase models so these are the models which we will also discuss when we come to fluidized bed reactors okay but right now it is a simple decomposition I told you know I think I promised you any reactor you bring on this planet to me then we can definitely imagine that that one to be in mixed flow reactor mixed flow conditions are plug flow conditions that is why this fluidized bed has also has been imagined as a fluidized as a mixed flow reactor because the conditions look like mixed flow but the actual phenomena if you go and see inside the bed it is slightly different than complete mixing okay solids are complete mixing but how gas is transported to the solids and all that will be slightly different that we will read when you come to fluidized beds okay so many times I am repeating about the design design objective always you know either X is given calculate the volume or W here hold up or when hold up is given that means already you then calculate conversion sorry these are the things okay so now let us take with elutriation that is slightly more realistic when the bubbles are just going from the through the particles and when the bubbles break here some of the solids also will just jump up okay into the space so at that time if there is slight change in the particle size due to attrition and all that particles will be carried away and it is inevitable in any fluidized bed okay so that is why that is more realistic condition with elutriation and let us take now design of fluidized bed with elutriation same conditions but with elutriation good again size is not changing the way what we are imagining I told you that in reality you will have size change because of the collisions, attrition and all that but in our problem here when you are talking about elutriation particles are not changing size only thing is I have some small particles and large particles some of the small particles may get elutriated right so that means those particles are not going to spend more time in the bed so under those conditions how do we now design the reactor that is what what we do and I think I have to remove all this if I use this diagram I think you will also draw the same diagram I think there so you have to draw a new diagram so I will just I use this diagram but please draw this diagram again for you in your notes okay but only thing here is that you have that also I have to show gas is going up anyway but there are some solids also this is f0 f1 f2 solids are just going up the distribution because now we have little bit more number of different sizes so I may have distribution something like this this is f of r i what is this r i f0 r i f0 r i so then in the bed as well as outlet I have the same composition because again even here we have perfect mixing right so in the bed I may have like this that is in the outlet and in the bed we have again ya so this is w of r i versus r i this is f1 r i but now here also you will have the distribution in the outlet here also so that will be if I plot here this is f2 of r i will be narrow size like this this is r i because only fine particles are going small diameters so that is why this moves this side this moves slightly the other side the distributions qualitative distributions ya so now we will write the material balance for the solids the material balance for the entire streams f0 equal to f1 plus f2 this is that we know very easily then we also have the size distribution mass conservation that is f of 0 of r i equal to f1 of r i plus f2 of r i so this is equation 2 that means one among particles if they are there in the inlet how many coming outlet here the stream that balance ya please again take this since the mixed flow of solids is assumed the composition of the under stream this is called under stream f1 you can write there since the mixed flow of solids is assumed the composition of the under stream f1 still represents the composition within the bed so that we have f1 of r i by f1 equal to w of r i by w this is equation 3 ya please take this again next the mean residence time of material of different size may not be the same in fact small particles will be blown out or elutriated out of the bed and these particles spend less time in the bed than large particles next paragraph you write hence for particles of size r i for particles of size r i t bar of r i equal to w of r i divided by f0 of r i w is in the bed and f0 in the inlet f0 of r i so this is ya which can also be written we have f0 r i this equation 2 which also can be written this as w of r i f1 of r i plus f2 of r i this is simple substitution only so f0 is simply written like this so this also can be written just by adjustment of the terms you will have here this is 1 by f1 of r i by f1 of r i plus f2 r i w of r i ya this is equation 5 but do you have anything here connected in terms of easily measurable quantities this is individual solids r i 1 mm, 2 mm, 3 mm like that f1 of r i f1 of 1 mm particles divided by hold up of and again you can write somewhere substituting equation 3 in 5 substituting equation 3 in 5 okay somewhere it is okay I have to write here mm to be specific I have been maintaining that so let me write that so T bar this is not our T bar f2 1 by w by f1 plus f2 of r i by w of r i ya f1 by w f1 by w good so this is one equation what we have so that means if I am able to find out f1 and w and f2 f2 of r i and w r i then I have time for each size residence time so once I know this now I will go to 1 minus x bar equations and then try to find out for 1 mm what is the conversion 2 mm what is the conversion 3 mm what is the conversion that is all but now problem is w w r i at least they are not that difficult okay so that is why we have to try to find out those things but before that what we do is we will now assume that we know this T bar of r m T bar m r i and then write the design expression then question how do I use the design expression right question is do I know T bar m T bar m of r i for a particular size if I do not know what else I can do with this equation to find out T bar T bar m of r i okay so that is why design expressions first let us write again design expressions are not new same thing if it is film control reaction control as diffusion control those are the same equations same summation because you have 3 different sizes okay that will not so that is why the 1 minus okay that is called design equation design equations for m f of solids is 1 minus okay I think let me also write once more x bar b equal to again 0 to T bar of r i here now you see T bar of r i then 1 minus not x bar so this equation again T by T bar m now please remember this this will be r i divided by ya r i r i d T okay so if I know T bar m of r i for each particle right so now this equation everything I know is kinetics so that means whether as diffusion control or reaction control or whatever control we take and from this we know the extension for multi particles that means summation of that equation in terms of okay sigma of all so that is equal to sigma of 1 minus x bar b of r i into f not r i by f not so somewhere I have missed this one this is equation 7 this is equation 8 right I think we will write once more the equations that is for as diffusion control sorry first film control for film control number of times if you write you will remember so let me write that so 1 minus for film control ya can someone tell me ya 1 by 2 factorial tau of r i by T m of r i minus 1 by 3 factorial tau of r i by T bar m of r i whole square right ya so other terms will come into f not of r i by f not good so I think the other one other 2 also let us write so that I think at least you will remember once more reaction control number 2 chemical reaction control when you have then we have 1 minus x bar double b equal to sigma of all particles so 1 by 4 tau of r i by T bar m of r i minus 1 by 20 tau of r i by T bar m of r i plus we have other terms this is whole square right so we also tied up so this is f not of r i divided by f not so the other one is diffusion control hash diffusion control so here we have 1 minus sigma of all so then we have 1 by 5 tau of r i T bar minus 20 whole square plus f not of r i by f not so if I put number this is 9 this is 10 this is level okay so now of course in the next class we will try to find out the T bar of r i for these different sizes no definitely I do not know because I do not know what is f 2 okay and also w r i and w of course as total hold up I may know and f 1 as the under flow I may know but we have to use this equation and then try to find out this T bar and that is what what we do in the next class okay okay thank you