 We want to look at this problem now that is what is particle size distribution in a fluid bed combustor particle size distribution fluid bed bed combustor. The context the context is the following you have coal going through ash coming you have gases then you have all this heat transfer tubes. Now, this is a very crude a crude way of trying to understand coal combustion. What we want to find out is what is the size distribution of solids that come out of the fluid bed? How do we do this? The example I have taken here is reaction control. Now just put the context of coal combustion in particularly in India coal combustions are not in large scale fluid bed combustions are not common yet, but BHEL has has good number of fluid bed combustors now in working. It has many advantages as I pointed out to you heat transfer coefficients are very large. It has disadvantages that also we have seen the power consumptions are very large. So, there are advantages there are disadvantages. Now our population balance looks something like this d by d s of f 1 r 1 t 1 bar this how we have written our population balance correct. What is f naught? f naught is what goes in f 1 it was comes out which is the same as what is in the equipment. This is what we have been writing for quite some time, but the context here is that we said whenever we write the population balance we like to see whether these distribution functions whether they are differentiable. If it is not a differentiable function we said we should write it slightly differently. Now let us try to understand what is f naught? What is f naught? It is the size distribution of inlet solids. If that size distribution is uniform of radius r then it is sum s minus 1 where s is r by r equal to 1 correct. If all the particles are uniformly distributed at one particle size then it is it is a delta function at s equal to 1 where all the particles are of radius r. Now second thing is what is our understanding of f 1? The context is we know that in reaction control our this is the reaction control we know this correct. We know that our particle sizes are determined by where this is the time for complete consumption of the particle. So, what is our understanding of f 1? We would expect that there is since our fluid beds are like stirred tanks stirred tank has got exponential RTD. Therefore, if this is the time for complete consumption of the particle this material would be completely consumed. Completely consumed means what? Its size is 0, which means your f 1 should have a continuous part and then there is a discontinuous part which is at 0. Is this clear to all of us? Therefore, our the size distribution of particles that emerged on the fluid bed this f 1 will have a continuous part which is g 1. It will have a discontinuous part corresponding to particle size corresponding to s equal to 0. So, keeping this in mind keeping this in mind we can say that therefore, integral f d s is 1. Therefore, that is equal to integral g d s plus k 0 therefore, 1 minus of k 0 is integral 0 to 1 g 1 d s. Is it alright? What is this area? Integral 0 to 1 1 by t bar e to the power of minus t by t bar sorry tau 2 tau 2 1. How much is this? We said what is this equal to? We have integrated many times is e to the power of minus of alpha. So, this area is e to the minus of alpha where alpha is tau r by t bar. We have done this. You can integrate and check for yourself, which means what? This fraction is completely consumed. Is that clear? Therefore, what is k 0? tau r 2. Thank you very much tau r to infinity. I hope you understand this because the time goes from up to infinity that is why alright. So, what we are saying is that this value of this k 0 is such that this actually this value of this k 0 is known because the fraction that is completely consumed is e to the power of minus of alpha. We know this or in other words even before you solve the population balance you have some understanding of what this solutions are going to look like. You are not totally in the dark alright. Let us now look at the population balance. We said whenever we have all these kinds of problems f naught minus of f 1 minus of d by d s of f 1 r 1 t 1 bar equal to 0. What we have said is that we will only look at the continuous part and then our f we will not look at f. We will look at the continuous part of the distribution. Therefore, we will only look at the g function g 1 plus d by d s of g 1 r 1 bar t 1 bar equal to 0. So, we will only solve this equation and not this equation. Why did we say that? We said whenever we have this kind of discontinuities then simply look at the problem minus all these problems all these difficulties and create an appropriate boundary conditions to account for wherever this kind of unboundedness discontinuities exist. Is this clear? So, let us integrate this. So, please integrate and tell me I am writing the final answers. I will integrate please tell me whether what I have run is correct. This is what I get tell me whether this is right. What is r 1? What is r 1 here? 1 minus please note 1 minus of s equal to t by tau r. Therefore, d s d t which is r 1 equal to minus of 1 by tau r. Is this clear? For reaction control 1 minus r by r is t by tau r. We have derived this r 1 minus of s is t by tau r and we are looking at the property. How do you determine the rate function? You simply look at the variation of that property with respect to time. So, we have done that you get 1 by t r. So, r 1 here is minus of 1 by tau r recognize that. So, here the r 1 here is r 1 equal to minus of 1 by tau r. Is this clear? What we are saying? For reaction control we have derived this. This we have derived in our class r by r is s. Therefore, 1 minus of s is t by tau r d by d t of gives you minus of 1 by tau r. Is it clear? For different rate controlling regimes you will get different kinds of rate functions. So, this is tau r. So, now tell me whether this solution is correct r 1 is minus of 1 by tau r. Put it here integrate and tell me if it is ok. Being quiet means what? It is not ok. Is it ok or not ok? It should be an emphatic s. Therefore, g 1 r 1 t 1 bar equal to q times exponential I will call it as alpha s, where alpha is tau r by t bar. It is only a single fluid bed. See our system is this. So, this is only one system here. Therefore, I did not bother about I should put t 1. I am putting t 1 everywhere. So, I may have put t 1 bar here ok. Thanks all right. How do you find q? So, to find q of course, there are many ways in which it can be done, but let us just formulate the boundary condition just to get a boundary all the b c s. Let me do a balance. I will do a balance between s minus of d s and s ok. Limit as s tends to 0 d s tends to 0 ok. I will write all the inputs. I will write the outputs and the generations equal to 0. So, what are the inputs? What are the inputs? V naught delta of s minus 1 this is input ok. What are the output? V naught what is going out? G 1 plus k naught delta of s minus 0 times d s. Is it all right? Inputs and outputs are correctly written. Now our rate function w r 1 at s minus d s g 1 at s minus d s minus w 1 ok r 1 at s g 1 at s equal to 0. This limit s tends to 0 d s tends to 0. I put d s d s is here all right. So s tends to 0. So this term goes away. This goes away. What else? What else goes away? This goes away. Is it all right? Is the argument clear to all of you? Why these terms disappear? This term disappears because it is at s equal to 1. We are talking about s tending to 0. So, this disappears. This is a continuous function d s tends to 0. Therefore, this goes away. But this is k naught. So this becomes minus V naught k naught minus w 1 r r 1 sorry r 1 g 1 at 0 equal to 0 or r 1 t 1 bar g 1 at 0 equal to minus k 0. Is it all right? And we can learn to do this at the other boundary also, but it is not necessary. Now tell me what is g 1 0 now? From this, what do we get g 1 0? What is r 1 minus 1 by tau r t 1 bar minus 1 by tau r t 1 bar equal to minus of k 0. Yes or no? So g 1 at 0 equal to tau r by t bar k 0. So this is alpha. This is alpha. k 0 we already said that it is e to the minus of alpha. So this is, is it okay? Yes or no? Now where is our solution? Let us look at our solution now. So we said g 1 0 from here. Can we find q? This is our solution here. I want q now. Let us find the value of q and tell me what is the value of q. This is our solution g 1 t 1 bar equal to q times e to the power of alpha s. We know that this is equal to alpha times e to the power of minus of alpha. So what is our solution? What is q? I get q as e to the power of minus of alpha and solution as this. This is my solution. All of you please see if you get this result. Alpha times s minus 1. See you look at this equation. From here you will be able to find q. Is q equal to e to the minus of alpha is right? G 1 s equal to this is okay. This solution. Find some more s s please. Q equal to minus. Thank you very much. Now we have got our solutions. Let us just look back at our solutions. Our solution is f 1 equal to alpha e to the power of alpha s minus of 1 plus k naught delta of s minus of 0. Once again I write this as alpha times s minus of 1 plus e to the power of minus of alpha delta of s minus 0. So this is our solution. So what is the distribution of sizes at the exit from the fluid bed? It is given by this function. Now we want to know what is the mean conversion at the exit? How do you find mean conversion? Let me write mean conversion like this. Is it okay? This is for each particle. Now we have to know there is all these problems now f. So the mean value is given by integral 0 to r 4 by 3 pi r cubed f of r dr times density divided by 4 by 3 pi r cubed. Do you all agree with this? What I have written? This is 3. What is the numerator? This is f of r dr. F r, what is F r? This is the distribution we have found out. This is what we have found out. Is this clear? We are going to put it in terms of s shortly but this is easier to understand that particle that enters the fluid bed because of the exponential RTD it has all these distributions. As a result we have to find the mean means we will have to find this integral to find out the mean value. Is this clear to all of us? Yes or no? The mean conversion is given by this integral. Is this okay with everybody? Why is this given by this? Because the particles that enter because of this exponential distribution we have all these features therefore we are doing this. Is this clear? Why are we having size distribution in a fluid bed? Now what is F of r? Suppose I want to put this in terms of s. Can I write this integral s cubed f of s ds 0 to 1? What I am saying is that this same I am writing it like this. Is it okay? Yes or no? All right. Help me now s cubed g 1 s ds sorry ds k naught delta s minus 0 ds. Is this correct? So this is the mean. Yes or no? It is fine with everybody. Now what happens this term? Why does it disappear? s cubed times k naught delta s minus of 0. That is why it is identically 0. Is this clear? s cubed times k naught delta s minus 0 ds is 0 because s is 0. Is this clear? The point is that a delta function the area it takes the value of the function what is the function is s cubed. Therefore the area takes the value of the function. That is something you would have learnt in your control theory anyway. So our answer is simply this. So the mean conversion is actually this. Is it okay? What is g 1 s? g 1 s this is integral 0 to 1 s cubed g 1 s we have found out which is alpha times e to the power of alpha s minus of 1 ds. This integral you have to do. Is it all right? Now I have done this integration. Therefore we will not waste time. So I have I will just put down the final form you can check at home. It comes all right no problem. It comes nicely. So this is the result which is there in books also. It comes nicely. There is no it is a little messy integration but it comes very nicely. Now to put it in the context the context is that in a fluid bed in a fluid bed you have size distributions and what you get is a mean number that is only message that you have to get rid of. Let us solve one of these problems mixed orders. Now this one is you have r 1 is minus of k 1 s, r 2 is k 2. Find f 1 and f 2. This is the problem. What are the problems you have solved so far? You have solved for the case both are first order. You have solved the case where both are 0 order that also we have done. When it is mixed how do we handle this? So what I have done is the following. I have written the population balance like this. Reactor f 1 tell me whether this is right. So this is our understanding of the distribution functions. That means here r 1 is minus of k 1 s, r 2 is plus k 2. For this case of reactor regenerator system we will find that f 1 is continuous, f 2 is continuous and a discontinuous part. Is this clear to all of us? Why is it that in f 2 there is a discontinuity at s equal to 1? Because it is 0 order. That means this is 2, this is 1. Therefore, whatever emerges from here will have a discontinuity at s equal to 1. But since this is first order whatever goes out of here there will be no discontinuity. That is why we have written they removed this term. Is it all right? Let us now rest of it is very straightforward. Let us write the population balance and we said we will write only the continuous part. So it is reactor g 2 minus of g 1 minus of d by d s of g 1 r 1 t 1 bar equal to 0. Regenerator g 1 minus of g 2 minus of d by d s of g 2 r 2 t 2 bar equal to 0. Now to solve this what we said we will add these two. Let us add these two. So when we add these two we get d by d s of g 1 r 1 t 1 bar plus g 2 r 2 t 2 bar equal to 0. When we integrate we get g 1 r 1 t 1 bar plus g 2 r 2 t 2 bar equal to constant. We have to find the constant of integration for which we will generate a boundary condition. Generate a boundary condition at let us say between s minus d s and s in the reactor. We will do it in the reactor now. What do we do? We write input minus of output plus generation equal to accumulation. What are the inputs? Let me write the inputs and you tell me whether this is right g 2 d s plus l 1 delta of s minus of 1. This is d s minus v naught g 1 d s input output. Generation is w g 1 s minus of d s r 1 s minus of d s minus of w g 1 s r 1 s equal to 0. Thank you. How does it look? Let us get rid of all the terms which we think it is not good. So limit as s tends to 0 d s tends to 0. It becomes this goes off this goes off this goes off this goes off. So what do we get g 1 r 1 t 1 bar equal to 0 at s equal to 0. So this I will just write here g 1 r 1 t 1 bar 0 equal to 0. So that is our boundary condition at 0. So let us do at repeat limit s tending to 1 d s tending to 0. We take limits between s and s plus d s. Once again input output generation equal to equal to 0. So let me write all the inputs v naught g 2 d s plus l 1 delta s minus 1. I will put d s here itself minus of v naught g 1 d s input output generation plus w r 1 s g 1 s in the reactor. I am writing the reactor only. Is it alright minus w r 1 r 1 sorry s plus d s g 1 s s plus d s equal to 0. Now limit as s tends to 1 d s tends to 0 s tends to 1 this goes off this goes off this does not exist. And only these 2 terms. So this gives you so I will write here. So g 1 r 1 t 1 bar at 1 equal to minus of l 1. Now we have 2 conditions therefore we should be able to find the we have done some integration we have constant of integration. Now tell me what is the value of the constant of integration g 1 r 1 g 2 r 2 t 2 r I see we are not done for the regenerator. Let us do for the regenerator then only we can solve the problem regenerator. Now I will take the balance between s minus of d s and s limit as s tends to 0 d s tends to 0 v naught input output g 2 plus l 1 of delta s minus of 1 times d s plus generation g 2 s minus of d s r 2 s minus of d s minus of w w 2 w 2 g 2 s s minus of d s sorry s sorry s r 2 at s equal to 0 alright. So I have got this off s tends to go away there is no material here. Therefore I gives you g 2 r 2 t 2 bar equal to 0 and you can do the same thing and show that g 2 r 2 t 2 bar at 1 equal to plus l 1. We will not do it but you know it is quite straight forward we have done this so many times. So what we are saying now is so if you want to find out the constant of integration if you want to find the constant of integration you find the g 1 r 1 t 1 bar at 0 this is also 0 at g 2 r 2. So you can find constant of integration 0 even if you take the other end that is at g 1 r 1 t 1 at 1 it is plus minus l 1 plus l 1. So it is consistent the important thing is that always see whether there is consistency in the way you formulate moment you make a mistake it will tell you that there is inconsistency. So that gives us the constant of integration to be 0. So from so g 1 r 1 t 1 bar equal to 0 g 2 r 2 t 2 bar equal to 0 so constant of integration equal to 0. So you get g 1 r 1 t 1 bar plus g 2 r 2 t 2 bar equal to 0 so g 2 equal to g 1 r 1 t 1 bar divided by r 2 t 2 bar with a minus sign. We have been having getting this kind of relationship for a long time so you know how to handle all this. So now we can substitute for g 2 and then integrate to get our answers. So it is slightly messy so make sure we do not make any mistakes. So in the population balance where is the population balance you know what it is let me write down minus of g 1 I am for the reactor I am writing g 1 r 1 t 1 bar divided by r 2 t 2 bar that is g 2 minus of g 1 minus of d by d s of g 1 r 1 t 1 bar equal to 0. So this is the population balance for the reactor is it alright. So that gives us d by d s of g 1 r 1 t 1 bar equal to minus g 1 t 1 bar 1 by r 1 t 1 bar plus 1 by r 2 t 2 bar is it okay. Now please we know r 1 equal to minus of k 1 s r 2 equal to plus k 2 okay. I am writing the solution please I am going to write the solution and you have to integrate and tell me whether this is right. I am simply going to substitute for r 1 and r 2 and integrate I am writing the final result please. So final solution is what I am writing please make sure that I am not wrong this is the final result is what I get this is my answer please see if all of you get this answer please do it yourself please do it yourself spend that few minutes and make sure that you all get this result it is a fairly simple integration it is not a difficult integration anyway okay good some more okay shall we go forward anyway I write the g 2 these are not difficult things minus of q beta s the power of alpha exponential of minus of beta s okay. So this is the g 1 and g 2 alright. So we have looked at various examples now just looking back there is another problem which is not so difficult to do you can do it yourself mixed orders where I have just changed the so you can do it yourself. So cutting the long story short the whole idea of this of looking at this at this dual bed circulating systems is really to sort of expose you to a very interesting developmental thing that will happen in the next number of years particularly for gasification of coal and biomass because of energy problems biomass has become very important around the world and if you do it in down draft grassy fires the nitrogen dilutes the gas therefore the fuel value becomes very low when you do it in the dual bed systems you are able to keep the nitrogen away and you get much higher thermal value. So that is why the two bed systems have great value in the years to come it is not yet commercial due to number of reasons but may be in your life time you might see all these okay. So that is the object of looking at and giving you a way to handle this kind of problems okay. Now we will take some simple problems in gas solid reactions I just put it down here I should have brought it and then circulated as solid react there is a very simple problems just to put all those things in context okay I will write down the data here DP 4 mm 12 mm temperature is 550 and 590 time for 50 percent conversion 15 minutes and 2 hours. Now what you have to find out is time for this is the question time for 98 percent conversion okay of 98 percent conversion of 3 mm particle okay. So let me state the problem once again data given is this is the experimental data particle size 4 mm 12 mm at this temperature this is the kind of results for 50 percent conversion and what is the time required for 98 percent conversion of 3 mm particle okay. Now it also says somewhere this moment says that neglect it says neglect neglect film diffusion control so this is what it says okay. Now put to put it in the context let me just write down here we have done it in class we know this what we know from our we know this are deleted film diffusion here that means if you have a particle which is undergoing chemical reaction under this and if film diffusion is not important the time required for complete consumption is given by this expression which you have derived in class now based on this data based on this data what are the unknown quantities here tau r and tau d okay but there are two particle sizes okay one is 4 mm one is 12 mm this is clear how do we find out so let me just put it in the context tau r1 that means tau r1 means for 4 mm particle and tau r2 which is for 12 mm particle now this is under reaction control how are they related equal to what is this ratio what is this ratio for reaction control time for complete consumption 4 by 12 exactly 4 by 12 1 by 3 isn't it if it is under reaction control the time for complete consumption of 4 mm particle divided by time for complete consumption 12 particle is simply 4 by 12 it is 1 by 3 okay similarly tau d1 by tau d2 which is 4 mm by this is 12 mm 12 mm sorry equal to what 4 square by is it correct yes or no 16 divided by 144 it is 1 by 9 okay is it alright okay now with these two results can we look at the data and tell me what you feel about this data now what is meant by what is meant by xb equal to 0.5 in terms of 1 minus of xb is what rc by r whole cube correct so if it is given as 0.5 what is rc by r value 1 minus of 0.5 to the power of 1 by 3 is how much yes or no if xb is given as 0.5 data says please look at the data where are we where are we where are we the data says 50% conversion data is given corresponding to xb equal to 0.5 what is the value for rc by r because everywhere rc by r is occurring so what is rc by r equal to 0.79 is it so point I have got 0.8 it is okay 0.79 okay is it alright now let us now look at this equation now we have data at two temperatures we have data for two particle size now can you tell me what is tau r1 tau r2 tau d1 tau d2 this is the question based on this data can you tell me what is the value of tau r1 tau r2 tau d1 tau d2 is it alright data is given please tell me yeah same rc no no what is important is what is important is this equality 1 minus of xb is this this is the conversion is xb but if you want rc by r you have to calculate this way is it okay yes or no so can you please look at the data look at the model and tell me what is the values of I am taking the first data first data is what where is the first data 15 minutes 15 minutes is means what 0.25 hours correct tau r1 1 minus of 0.8 I have taken not 0.79 plus tau d1 1 minus of 3 times 0.8 squared plus 2 times 0.8 cube is this correct is it alright no this is one data the second data is what this is for 4 mm particle okay for 12 mm particle I have got here what is the data is 2 hours equal to tau r2 times 1 minus of 0.8 plus tau d2 tau d2 is what where is that equation 1 minus 3.8 squared plus 2.8 cube is it okay this is for now can we solve and tell me what are the values of TR1 tau r1 tau r2 and all that please solve and give me all the answers the second data is at 4 590 C this is at 590 C correct first one is at 550 C now listen to me now see we are going from 550 to 590 correct so what do we expect tau r1 tau r1 we are going for 59 550 it is increasing the temperature what will happen to tau r1 what is our expectation it would decrease now if you look at the data data says that when we go from 4 mm to 12 mm which means it is the time for consumption increases by 15 to 2 hours from here are you able to make any guess about the role of temperature on tau r1 because activation energy is not given data it is so data expects you to sort of look at the data and then make some guess work as to what might be appropriate is that clear the data activation energy is not given and it could happen in any real situation all these things happen so you have to look at the data and make your judgment what is your judgment regarding the effect of temperature on tau r1 given the data you find that you see the time for 50% conversion is going from 15 minutes to 2 hours it is roughly how many times if it is 9 times if it is 9 times it should have been 135 minutes which is slightly more than 2 hours in which case we could have said there it is all diffusion control so we do not worry about reaction much so it is something close to 9 times see the second data seems to suggest that the temperature dependence on tau r1 is not all that significant this is the judgment you have to make because data is not given you understand what I am saying data is not given what is the temperature dependence of tau r1 so how do you make that judgment what I am saying is that by looking at this data it appears that tau r1 is not a very strong function of temperature that means the activation energies are not very large that is all it means okay see activation energy if it is large then its dependence is very strong okay so which means what you should be able to solve this quickly tau d1 tau d2 tau r1 tau r2 we need all these that kind of answers I have got what are the numbers you got tau d1 is how much point tau d1 tau d just a minute tau d2 tell me all the numbers in hours please I will write down what I have got tau d1 tau d2 is 2.25 hours is it correct and I this is correct sorry sir tau r1 this is what I get what do you get tau d1 2.25 what do you get how much are you getting I get tau d1 as 2.25 hours so tau d tau d1 is 2. d1 t2 na d2 is 2.25 what are you getting this is 2 is it and this one 18 is it okay okay now I have written tau d1 is 2.25 you are getting 2 is it okay okay tau d2 is 18 hours okay tau r1 0.208 you are getting is it okay I am getting a slightly higher value okay alright and how are the others which one tau r2 is how much hours now you have all the answers to solve this time for complete consumption of 3 mm particle how is this to be done if I give you tau d for 12 mm you can find for 3 mm no problem it is required only at 5 if similarly tau r if I give you 4 mm you can find for 3 mm so you know how to do this problem correct because that data is not given see when data is not given you have to make some assumptions and to that extent our answers would be unsatisfactory so we accept that till we get more data okay if it had come then we could have calculated this the value of tau r at 2 because at one temperature to another temperature you can calculate because we can use the Arrhenius factors to calculate okay alright stop there today