 so design equations for mf for or mf of solids till there are some more condition uniform gas throughout uniform gas composition any other condition constant particle size you can also add here constant single particle size and scm right good the design expression already you have derived right we have given the equation what is the equation tomorrow examination 1 minus x bar b equal to 0 to tau 1 minus x b single particle sp is single particle e power minus t by t bar m by t bar m dt that is the design expression and the general procedure is if I have 3 steps controlling we have that equation t by tau equal to in terms of 3 resistances re-lengthy equation so that equation you have to take but this t by tau is a function of x b you will have 1 minus x b to the power of 2 by 3 and also 1 by 3 all kinds of combinations of 1 minus x b is there so we have to solve for 1 minus x b and then substitute here for single particle and then integrate that is the general procedure so to simplify to understand you know easily for example if it is film controlling or reaction controlling or diffusion all over controlling all 3 steps single steps controlling we will have the equations and complicated things I will give it to you so that you can do in the examination simple things I do in the class complicated things you have to do in the exam that is an examination is always like that you know you tell in the class simple things and then allow you to expand your brain thinking how to extend the situations to various complicated situations in fact true education is that and if you are not doing that what you are doing is coaching what is that give the problem and ask you to solve 1000 times the same problem so that your speed will go to 0 to infinity okay even before you see the question you will answer that is what is coaching centers do you know but I think in real education expansion of brain so that is why examination always complicated problems you have to extend what you have understood the knowledge good so now that is why simple things for me first film control if you have film control you know that the equation for 1 minus x b what is the equation for film control if I have only I mean in terms of 1 minus x b 1 minus x b equal to t about to funda is okay that is okay now you have to write in this fashion for single particle right so 1 minus t by tau that is the one so we have to substitute this equation 2 in equation 1 yeah probably finally you got your assignment we will do it today because tomorrow exam now but did you see you said that you know email you could not get it and all that now first of all you have to accept that you should be in the group you already accepted that then why did you not get all of you got it or not got so because all all other assignments also I will be sending only through this group otherwise Rahul I think we can do is simply taking their actual IDs and then make as a group I say we should we go to Google and take permission from them us or no and all that and then finally put it that is the best one you have you know you are not using smail right you use so then give a smale I think you have all the role numbers if you add all the number that is your automatically smail right yeah that is officially always better Google is difficult we don't know I think everyone has to accept first if they don't accept they don't accept yeah but I prefer smail instead of Google group what is the advantage of Google Google group this also I can make it as a group all these numbers together in one group made so I can how many everyone forwards right your smail whatever you are using for that mail smail is forward you can also do that okay good so substituting equation to here we have for single particle 0 2 tau and this one 1 minus t by tau e 4 minus t by t bar m by t bar m dt so this is equation 3 and I am going to write here after integration you are going to get the final expression but if you write the same thing in the examination I don't give you marks okay yeah so it's not memory you should also use that mathematics so that is why please don't write that after integrating you are getting the one now I am going to write after integrating okay so after after integration yeah integration by parts you know this is e so after doing that what you get is x bar b equal to you can also write in terms of 1 minus e 4 minus 2 by t bar m where tau is the single particle time required for complete conversion and t bar m is yeah you see this is the beautiful integration of kinetics and contacting okay yeah contracting comes through t bar m okay yeah and kinetics comes through tau okay good so this is 4 so practical engineers have also used this this e power x if you take e power minus x you have some expansion right e power minus x equal to 1 minus x and x square factorial by 2 and all that is okay x square by 2 factorial that expansion if you do and then write you will get again in terms of 1 minus x bar b equal to 2 factorial tau by t bar m 1 by 3 factorial tau by t bar m square again plus alternately 1 by 4 factorial factorial tau by t bar m whole cubed minus now etc so this is the equation this is equation 5 generally we can prefer this and this also sometimes can be used you know this is by trial and error only how to solve in this case of course it is easy you can bring this side here and because you know either you have to find out either x b or t bar m okay so given t bar m you find out x bar b otherwise for a new reactor x bar b you have to assume okay 90 percent conversion for example or 99 percent conversion and then solve for t bar m tau comes through kinetics total time you should know okay so that is why that is why I think it is may be trial and error problem so this is good guess in the beginning and then you can refine retune it for exact answer so that is what what we do so this is for film control next one is reaction control diffusion is slightly cumbersome so that is why reaction control we are going this is again you will get an analytic expression here for reaction control you have the equation for single particle x b s b yeah anyone remembers 1 minus t by tau yeah whole cubed okay yeah if you see the actual equation in terms of t by tau equal to 1 minus 1 minus x b to the power of 1 by 3 okay so that we are writing in this passion so now equation 6 is again substituted in equation 1 substituting equation 6 in 1 let me write the equation 1 minus x bar b equal to 0 to tau for this one it is 1 minus t by tau whole cubed t bar m by t bar m d t so this is equation 7 good and this mathematical technique you have to learn it comes through the trickers in formula because this is cubed and you have e power minus t here okay anyway this is constant so that mathematical technique you have to show when I ask derive that corresponding equation okay so after this is equation 7 yeah again after integrating equation 7 through the trickers in formula okay brush up you know you learnt mathematics only not to write the examination later to use okay so that is why I think you have to brush up all those things here yeah what you get is an unethical expression x bar b equal to yeah I do not know how do you remember but you have to remember tau minus 6 t bar by tau whole square plus 6 t bar m by tau whole cubed yeah and here we have 1 minus e power minus so this is equation this is equation number 8 you see now you have 3 t bar by tau and next one is minus 6 t bar m by tau square plus 6 t bar by tau so t bar cubed and into that exponential tau right this is truly difficult if I give you t bar it is very easy for you to calculate simply substitute t bar by tau because tau through kinetics you know and then you can easily calculate x bar b but for a new reactor it is hell you have to solve this equation okay to calculate what is t bar m and for that unless you have practice you are always going to complain in the examinations or time is not enough yeah if your velocity of writing is 0 you need an infinite time I say infinite time you cannot give for examination okay so that is why you practice practice practice that is why assignments are given good so again we can expand this and then write this in a more easy manner for solving the problem as engineers so we have in series 1 minus x bar b 1 by 4 tau by t bar m minus 21 by 20 tau by t bar m square plus 1 by 120 tau by t bar m whole cubed etc so this is equation number 9 so that is the another case okay let me underline this so that you will see difference yeah next one is diffusion control as diffusion control yeah what is the equation for as diffusion control yeah I will write tell me t by tau equal to 1 minus 1 minus 3 into yeah right 3 into 1 minus x b 2 by 3 plus yeah this is for single particle right what is the equation number 10 this I cannot write so easily the way we have written here okay yeah I am telling that many times so you have to solve for this 1 minus x b and then substitute equation 1 and then you have to get the final solution so that is not the easy one so that is why people have already expanded that in you know some common series solution then we have the solution for as diffusion control as 1 minus x bar b equal to 1 by 5 tau by t bar m minus 19 by 420 tau by t bar m whole square across that okay plus I will write here 41 by 4620 tau by t bar m whole cubed that is plus yeah then we also have another term minus 0.00149 tau by t bar m to the power of still some more able to see this is equation number 11 only thing you have to preach that you know you should never be diffusion control okay but unfortunately most of the time it is diffusion control why why most of the time it is diffusion control yeah there are thumb rules you know to remember very easily why because most of these gas solid non catalytic reactions takes place at very high temperature so reaction will be generally very very fast because idealistic equation then whenever you have the product as ash where you know structure is porous then outside film is always negligible so that is why most of the time this is equation you have to remember not for the examination I am talking I am talking about in general examination I can give any control okay so this is how you have to solve this problem and then you have to remember all this and again as usual I am giving you if I give you t bar m that is a for existing reactor where you know already t bar m t bar m is nothing but w by f where w is the hold up of solids inside the reactor and f is mass flow rate of the solids okay normally it is in homogenous system it is volume by volumetric flow rate but here we go for mass yeah weight by mass flow rate of solids okay good so if already have the reactor t bar m I know then I can calculate x bar b which is easy but to design a new reactor you have to assume whether you are going for 99 percent conversion 99.99 percent conversion and then solve this equation for t bar m good okay so because you have exam tomorrow let us do one or two problems so that you will have at least some experience of doing a problem I know this may be the first and last time you do the problems so that is why I think I will tell you some problems one plug flow reactor other one is mixed flow reactor we will take okay for plug flow reactor please take the problem statement I will just dictate yeah you remember plug flow for plug flow the single particle equations like equation 2 equation yeah 6 and equation 10 all of them are straight straight forward you will get it okay only thing is t is replaced by t bar p t bar p okay so that is mean residence time of solids within the plug flow reactor so that is all and depending on which one is controlled you have to substitute that and I also told you that we may have distribution of particles that we have done if you have simply distribution take the average weighted average right so that is why we will do one problem because the other problem is simple single size particles we will take for plug flow distribution of particles right that one is okay please take this problem a feed consisting of 30 percent of a feed consisting of consisting of 30 percent of 50 micron 50 micron radius particles comma 40 percent of 100 micron particles okay micron radius okay 100 micron radius particles and 30 percent of 200 micron radius particles is to be fed continuously is to be fed continuously in a thin layer in a thin layer onto a moving grate onto a moving grate here it is moving grate cross current okay onto a moving grate cross current to the flow of reactant gas cross current to the flow of reactant gas to stop okay for the planned operating conditions for the planned operating conditions the time required for complete conversion is 5 comma 10 and 20 minutes for the 3 sizes of the particles 5 comma 10 and 20 minutes for the 3 sizes of the particles that means here I have tau minutes 5 10 20 for 3 sizes of particles okay full stop find the conversion of solids for a residence time of 8 minutes in the reactor very nice problem find the conversion of solids for a residence time of 8 minutes in the reactor so T bar P is 8 minutes yeah cows are given tau for 50 tau for 100 tau for 200 please try you have the calculators know you have to bring calculators every day because separate test anytime I can give what have you know no calculator read the problem is there anything missing which is not specifically given yeah how do you find out which one is rate which one is controlling billa reaction control why why do you say reaction is control tau is directly proportional to R yeah but now you have crossed LKG because other one when you are talking about three controls tau was also proportional to R for film also but later you graduated when you are talking about changing size particle okay for if the particle is becoming smaller and smaller okay so that means if you take very small particle constant size then the exponent may be different so that is why for film control the exponent changes from 1.5 to okay so that is why the logical conclusion here is reaction control how how can you say that yeah feed is spent in a control manner this is moving rate in fact okay so this is moving rate where so here only you are feeding of course the depth I have not shown you so this is this is moving rate comes out so I told you know this is wonderful example for plug flow because every particle definitely will spend exactly same time here in normal flow when you have fluid that is not possible but solid this is one of the excellent examples okay yeah so this is what but you know how can you say that when it when I put on the moving rate how can you say that it is reaction control and yeah gas is moving cross current to this yeah so what even then how do you conclude that it is reaction control you cannot say okay thin means what you are putting thin so what it can be film controlling it can be diffusion controlling as diffusion controlling or it can be reaction control also that will not tell you anything unless I say that I have infinite velocity one will go which one will go infinite gas velocity film control will not be there because the the film thickness is so small that may not be contributing but you know that is a lousy assumption if I have infinite velocity solids will not be there they will be fluidized they will go to some other solar system okay so that is why that is also you cannot say only from data you have to see how is proportional to or yeah we have another one for film control I have given you some data I do not know whether you are there on that day or if you have film control the exponent falls it will change from 1.5 to 2 that is also one of my favorite problems I can ask that also prove that for film control if for small particles and large particles show that the exponent changes that you already you should have that knowledge because for film control we have shown that from mathematical equations derivations derivations we have derived for small particles large particles I did it for n you know exponent n large particle here we told that we have a particular transition and a particular numerical value which is whatever it is excellent expansion of this range good that is nice if you do that but we are not going to that complicated way in transition between this particle to that particle if you are able to do that you will get 100% marks number generally at least in order of size so like an mm would be considered one mm particle would be considered yeah but how do you find out that you tell me I have already told that I say that order also how do you find that is what I am asking what is the starting point what are the equations we have used for small particles and large particles film controlling you heard of frosting equation frosting equation or range and I have told that already yeah that is why you have to continuously come to the classes not alternatively you are sign waves okay yeah up class down no class up class okay and regular period one class attend next class don't attend another class attend by the by if you don't get that 85% attendance I am going to send this information to Ramboorthy Dean AC okay so I think definitely you have to follow I mean the continuity I say subject continuity must be there range and martial correlation you remember range and martial correlation what is that 2 plus something that something also you have to remember in the examination now we can say something okay from there you can find out whether 2 is dominant and compared to the other term you know for small particles Reynolds number can be neglected based on that you can really find out what is the transition range and all that but these are the simple problems for your exercise understanding straightforward r is proportional to tau or tau is proportional to r that is what I say film control if you have film control the exponent will change from 1.5 to 2 for small particles and large particles so that kind of situation is not here okay so that is why read my notes even if you don't come to the class takes notes from someone and then read large particles 1 1 to 1.5 correct yeah I am sorry I was telling 1.5 to 2 it will change from 0.5 to I have given also that one that notes I don't have done it no one has done it so much time 0.9352 you have done it 0.966 0.966 is not correct how do you do we should have 3 terms there for 3 contributions so I will write the general expression I think I will go that side 1 minus X bar B equal to we have 1 minus XB for 50 micron particles into F of 50 micron particles by F plus 100 micron particles again F of 100 microns by F I think I have to write here only so 1 minus XB of 200 micron particles yeah that is the equation okay yeah and 1 minus XB of each general expression is for plug flow 1 minus yeah tau full cubed right so you have to substitute this for each you know T bar P you know tau for each particle 3 terms and F by F naught is 30 percent 40 percent another 30 percent for yeah different 3 sizes yeah how much you got please remember there is a condition that if a particle has tau less than T bar P so that will not contribute to those terms okay and to check if you substitute that and calculate you will get some illogical answer conversion coming more than 1 yeah exactly conversion coming more than 1 is illogical you know you cannot have conversion 120 percent or 200 percent okay what say sir you will get correctly 93.2 see actually this term will not be there yeah that term will not be there so then I think you know for the 1 minus X bar B equal to 1 minus 8 by 10 whole cubed into 0.4 plus 1 minus 8 by 20 whole cubed into 0.3 so this this was yeah this how much it comes you tell me this one 0.068 so XB X bar B will be 0.932 or 93.2 percent yeah that is the answer good shall we take another problem quickly yeah we have to do this because in examination for you tomorrow okay this fine now promo able to get it yeah simple only essay for you simple if you concentrate it is not difficult at all okay let us do for mixed flow reactor another problem for mixed flow reactor okay please take this yeah a stream of solid particles R equal to 1 mm radius X equal to 1 mm passes through a bench scale fluidized bed reactor a bench scale fluidized bed reactor full stop the solids react with gas solids react with gas to give a solid product according to according to SCM shrinking core model slash reaction control so it is straight away given here it is shrinking core model reaction control as follows okay the solids react with gas to give a solid product according to shrinking core model SCM reaction control as follows so that as follows is F not equal to 100 grams per minute that is 1 W hold up 1 more 0 and X bar P equal to 0.65 okay this data given for the bench scale laboratory reactor so next one assume uniform gas composition throughout the reactor okay assume uniform gas composition throughout the reactor then one calculate the expected conversion calculate the expected conversion in a commercial sized fluidized bed reactor calculate the expected conversion in a commercial sized fluidized bed reactor of 4 tons that means W equal to 4 tons 4 tons treating treating 40 kgs F not equal to 40 kg per minute of 1 mm particles same thing 1 mm particles 1 mm particles yeah that is 1 okay you have to calculate the expected conversion of this 1 mm particle in the actual industrial reactor where I have 4 tons of hold up and 40 kg per minute we are processing okay that is the input to the reactor good number 2 number 2 find the size of a single size fluidized bed reactor find the size of find the size of a single fluidized bed reactor needed to treat 4 tons per hour that means here F not is given 4 tons per hour of R equal to 1 mm particle okay same particles 1 mm particles to 99 percent conversion to 99 percent conversion so X bar B this is a nice problem covering all that you know like you have a fluidized bed laboratory reactor this is what normally we do for any new process we develop a small reactor in the laboratory and then try to find out what is happening in that and then extend that data to large scale systems okay yeah so in the first data is given to you to find out TM is given TM is 1000 tons by sorry 1000 grams by 100 grams per minute so tau you will know and is only single size particle no problem reaction control everything is clear that means tau you have to calculate from this data and in the first this industrial problem it is asked to calculate expected conversion X B equal to X bar B equal to how much here W equal to how much yeah yeah T bar B is given there T bar M T bar M is given here because it is again no no T bar M is not given here only tau you know X B is given here X bar B is given here so you have to find out T bar M we will do that and you see from the laboratory reactor you have to use this equation or this equation to calculate T bar M by tau T bar M is given anyway here and then calculate tau okay so from equation 8 by trial and error or this you can start and then put it here by trial and error and then you have to find out right yeah so from laboratory data bench scale or lab data yeah lab data what is this T bar M T bar M equal to 10 minutes right X bar B is 0.65 yeah that depends on the magnitude of tau by T bar M because if tau by T bar M is very small so cube will not contribute that much yeah so that you have to decide for each problem when you have this idea what is the safe bet what do we start with safe bet is 8 equation 8 that is analytical formula closed form solution so you have to start that is why mathematical methods in chemical engineering you learn you know so all that is only to use here you know all the time you do not have to have a computer you can also use brain tau is 18.8 yeah approximately 20 okay 19.88 I got by trial and error depends on how many terms they have taken okay yeah you have to do that good yeah so from this first part what you get tau is 19.88 minutes or approximately 20 minutes engineering good by trial and error method there are many methods you know I think you can guess a value and then what is that next one whatever you get what is that what is that some method yeah correct so with the 2 or 3 or 3 times also if you then you will converge to one value like that okay good yeah so then the other one what this bit if you want to find out again you have to use the same equation but what is known there tau is known to me because same particles that is not going to change okay then T bar M is known to me because T bar M for the first case is 4000 divided by 40 that will give me T bar M tau is known T bar M is known so for first bit here this is easy I think everyone likes that T bar M yeah can someone tell the term sir by the way what is T bar by tau T bar by tau is how much or T bar M by tau 100 by 20 yeah 19.88 if you take almost 5 T bar by tau I am asking 5 yeah of course when you are substituting there you have to substitute reverse now here it is okay T bar by tau correct how much what is the answer there X bar be any other answer may be seeing me you may think that I will give only always difficult problem got it no calculator why if I have given you supply test what you would have done exact I also got 0.952 95% 95.22% okay so X bar equal to 0.9522 good so the other one because you do not have time now to check all this the other one is calculating W to calculate W first you should solve this equation for T bar M okay because tau B is known X bar B is given so X bar B is how much 0.99 so you have to solve again same formula only one formula that is all now you have to calculate T bar M if you calculate T bar M that will be that you check that think T bar M is approximately anyone did it no 25 tau yeah around 25 so now you know T bar okay what is W if you take that answer we calculate again 25 tau means 25 into 20 500 yes 33.1 ton okay so W equal to here 33.1 tons okay and now you can imagine how do you put this 33.1 ton in the reactor it is a fluidized bed so I told you know that example all our Indian movies end with a what is that marriage okay all the problems start later so here also exactly same thing after calculating W all the problems start how do I put that is it one particle one above the other where it can go to Mars or moon somewhere yeah or is it in this direction any infinite diameter and then you know you will have one layer or less than one layer how do you really choose it this no one talks generally most of us only say that find the volume or find the hold up then how do you how do you really put that in the reactor so that shall we wait till fluidized bed reactor design is done okay there are some thumb rules what is the diameter and what will be the height of the bed diameter generally L by D people take 1 to 2 why pressure drop is one criteria but I think you know not that one no what is the idea what is the assumption for fluidization perfect mixing so to maintain if you put 1 kilometer height and then you know perfect mixing I cannot expect so similarly I put 1 kilometer diameter and then again fluidize I may not expect again perfect mixing so that is why L by D equal to 2 even though industry they may go to 3 4 times also because they do not bother approximation because in the same area they are able to accommodate more number of area is important there no so that is why sometimes 4 5 also they go but 4 5 you will not get definitely good mixing because our assumption entire thing is based on perfect mixing so when you are designing in the reactor the final fluidized bed reactor in the industry that also should you have perfect mixing of solids when you do not have your assumptions are wrong you will not get exactly same conversions or you need more than this if the assumption of perfect mixing is wrong you may need 40 tons instead of 33 tons okay so those things I think we will discuss when fluidized bed is designed but this is how like for example even in this is one here if the problem is reverse I asked you know in the plug flow reactor design if I ask you the reverse question that means T bar P is given here but now I will ask you to calculate T bar P and you have to calculate conversion sorry given conversion given conversion calculate T bar P right so T bar P you got 8 minutes now I tell you that okay let us use rotary kill also is a plug flow reactor right what is the rotary kill you have a cylindrical tube moving slowly like this okay so it will rotate slowly it will be rotating feed here solids will come out of course we are talking about large amount of gas so either you can send this way or this way to maintain that uniform composition of gas throughout the reactor okay so then yeah I think listen to this is this is very important rotary kill design right so that 33 tons for example if I have that 33 tons and that 33 tons also I have to accommodate here how do you accommodate what is the length I have to take what is the diameter you have to take and you know hold up in a rotary kill how much free volume is given for the particles to move because it has to rotate like this go up and the particles will be taken to the top and then they fall okay so to give that that is a good guess less than 10 percent you will have hold up 90 percent is free volume 67 and you know the rotation I mean a rpm very high this I think one of my problems must be there in your plant design I know that problems you have 60 70 problems this rotary kill design also this should be there one of my problem at the time which I know Margea if you have go to critical speed and all the totally Margea how much below it is actually 5 rpm 4 rpm 6 rpm like that very very low slow very very slow particles should slowly go up if you are almost going to critical speed what will happen I am just example as an example so solids will go on stick to the walls and all gas will go through the centre so bypassing now you have to also provide the residence time so that is why length will come in the residence time and you know angle how much angle do you think it will be 30 is too too large it is again 3 4 degrees to provide you know maximum residence time 30 is very slant everything will come out so quickly yeah so you know all these things you know very very simple things what we think that is why I told you my example you know that marriage movies and with marriage but afterwards so many problems are there starting with you know we have to go and buy vegetables after marriage before that both will go happily okay so like that here also everything is a problem how do you fix them how do you rotate them even this also so what should be the diameter of this wheel what should be the diameter of this wheel okay and the distance length and width all these things are problems they should not be if you have mechanical vibrations particle may go forward particle may go yeah and also this way you cannot expect again ideal plug flow we have to be close to our ideal assumptions right so that is the challenging for engineer you assume something how do you maintain the same thing in industry with the same assumptions same conditions like ideal plug flow same conditions like ideal mixed flow beauty beauty chemical engineering is beautiful because only chemical engineers deal with so many types of equipment not others because every process different for us if you take sulphuric acid the same sulphuric acid things you cannot use for nitric acid you may say both are acids or I cannot use you can never use and even if you have the both the process nitric acid and sulphuric acid both process together also you can never use it for hydrochloric acid that is that independent that is independent these are independent so that is why every time for every flow chart you should have a different kind of equipment where beautifully everything is that is why chemical engineering curriculum I think a subject is so beautifully designed okay everything comes under mechanical operations are you know that other unit operations correct no any kind of equipment we can put that because the basic phenomena is same but you can use different kind of equipment so that is why simply heat transfer mass transfer momentum transfer and CRE if you read you know most of the things of course thermodynamics is God you cannot cross beyond that process control is the final control and process calculations will give you what is entering what is leaving you know energy balance material balance and all that really I love chemical engineering you know the subjects everything anything that is what you know only two ideal reactors any kind of reactor you bring I can divide I can tell you whether it is either plug flow or mixed flow what your wonderful assumption or assumptions so that is why chemical engineering that design of curriculum and it is really great I do not know you may not like it most of you when I say this because you say that management is great because 30,000 30 lakhs 40 lakhs after MBA what is your 1.2 crores that is what people see there but the beauty in subject I say that is very very beautiful chemical engineering is wonderful okay anyway is my passion but it may not be your passion at all okay thank you