 See, we talked about reactor regenerated systems, just quickly run through what we have already said. For this type of rate functions, we have formulated the population balance equations sometime back. And we have got those results, we will not do it again, I will just set out what we have already done, where we have shown that f 1 is minus of q times alpha s of alpha minus of 1, 1 minus of s to the power of beta f 2 is q beta s alpha 1 minus of s beta minus of 1, where alpha is 1 by k 1 t 1 bar, beta is 1 by k 2 t 2 bar, this is something we have done. Now, for some cases we have taken an example and showed what are the mean values and so on. What I want to do now is use this result to look at some work that was done in this department sometime ago, that is use of reactions for recovery of carbon dioxide. I will just set out the equations k 2 c o 3 plus h 2 o plus c o 2 gives you twice k h c o 3 and then twice k h c o 3 giving you k 2 c o 3 plus h 2 o plus c o 2. Now, if you this is a model reaction, what we are having here is carbon dioxide is picked up by potassium carbonate to give you potassium bicarbonate and potassium bicarbonate decomposes to give you potassium carbonate. So, that the 2 equipments if I call this as R 1 reactor, if I call this as R 2 the regenerator, if you can appropriately bring them together, you can continuously remove carbon dioxide for some purposes. So, what you have let me just put it out in the context, you have a reactor, you have another regenerator, solids are circulating between the 2, you have carbon dioxide plus moisture goes in and comes out, then you have air comes out. So, that carbonate circulates between the 2 environments, is it clear? Now, this is R 1, this is R 2. Now, if instead of using air, if you use c o 2 itself, then this carbon dioxide you can recover for some uses. So, if you really want to recover carbon dioxide, then you can use c o 2 for the fluidizing gas and then recover the carbon dioxide, is that clear? Now, what we have done in the previous is that, we have assumed certain form of the rate functions in R 1 and R 2 and we determine the populations, the activity distributions and so on. So, we want to use that result to understand how we can explain this particular data. The question is, there is some experimental data that has been found and so, the data is given here for value of alpha 1.86, value of beta 1.03. The mean value, this is S 1 and S 2, they are mean values here, I have not put a bar on the top, S 1 is given as 0.45, S 2 is given as 0.873. So, you have to verify whether this model that we have developed, this is the model. The model is this, model says F 1 is this, F 2 is this and given the values of alpha and beta as 1.86 and 1.03. What is the mean values that we should expect from the model and what are the mean values given in this experiment? Can we tell how good is the model? This is the problem that we want to solve. How do we do this? The procedure is that, we first find the value of q given. So, to find the value of q, we do integral F 1 dS. What is F 1? F 1 is minus of q, what is the value of alpha? Value of alpha is 1.86, just to make the integration easier, I am just putting it as 2. When you go home and do this, you can do 1.86. Integral S to the power of alpha minus of 1, so S 2 minus 1 is 1, 1 minus beta, I take it as 1 dS. Is it alright? So, this is equal to 1. This is how we calculate or estimate the value of q. So, what is the value of q? q equal to let us integrate 1 equal to minus of 2 times q within brackets S square by 2 minus of S cube by 3, 1 equal to minus of 2 q. So, it is 1 by 6. So, it is minus of or q equal to minus of 3. Now, let us see if there is consistency whether we get the same value of q as minus 3 from the other from F 2. So, integral 1 equal to 0 to 1 F 2, where is F 2? F 2 is minus of q beta, beta is 1 S alpha S alpha is 2, 1 minus of S beta minus of 1 is 0. So, that is minus of q S square dS 0 to 1 that gives you 1 equal to q times S cube by 3 0 to 1 or it also gives you q equal to implies q equal to minus 3. So, it is consistent F 1 and F 2 are consistent. So, we know what is F 1. So, F 1 from our this one gives you 3 alpha is 2 S to the power of 2 minus of 1 that is 1 1 minus of 1 minus of S beta is 1. So, it is 6 S into 1 minus of S that is F 1 F 2 equal to F 2 is 3 beta is 1 3 S square is it all right. So, what is S 1 bar mean value of S 1 integral S F 1 dS 0 to 1 is it ok. The mean value of S and what is S what is the meaning of S in this problem meaning of S is carbonate. The carbonate fraction in the reactor meaning of S in the regenerator also is the same carbonate fraction. So, you would expect the carbonate fraction in the reactor to be less because it is getting consumed carbonate fraction in the regenerator to be more because it is getting regenerated is that clear. So, 0. So, what is this equal to. So, let me substitute for F 1. So, it is let me do it in the next page S 1 bar equal to integral S 6 S times 1 minus of S dS that comes to 6 times S square 1 minus of S dS integral 0 to 1. So, that is 6 times S cube by 3 S power of 4 by 4 0 to 1 that comes to S 6 1 by 12 is about 0.5. So, S 1 bar turns out as 0.5 what is S 2 bar S 2 bar is integral S times F 2 dS 0 to 1 integral 0 to 1 3 S cubed dS that is equal to 3 S to the power of 4 by 4 0 to 1. So, that is about 0.75. Now, the result says it is 0.45. So, we just compare S 1 bar and S 2 bar. So, experimentally it is 0.45 and then 0.873 and then this is the experiment model gives you 0.5 and 0.75 0.5 and 0.75. Now, the context to this exercise is whenever we find your model does something very different from the data. So, you should look at the model is not model is not satisfactory you have to look at what is been the assumptions you have made and what you can do to improve the model that is the important thing. And in this particular case I want to appreciate this problem. Problem is let me just do it once again materials coming in and going out. Now, if this is the regenerator this is the reactor. So, what happens in the regenerator carbonate is getting generated. So, you would expect that the way the model has been formulated. If you look back at the rate function the rate function says k 1 time says that means it reacts throughout the particle as per the model the reaction takes place throughout the particle. But in the shrinking core as we have said with gas solid reaction there is a an interface between the reactants and the solid. There is an interface where the reaction occurs. On other words we have model the system using an assumption of uniform reaction throughout the particle. Well in shrinking core you find that the particle actually reacts at the gas solid interface. So, this is the weakness of the assumption. Now, when would this assumption be very good this would be very good if it is a highly porous particle. So, that the reactor is a reaction able to take place everywhere. The fact that this data does not fit so well implies that that assumption that this rate function may not be as good that is the meaning of this. So, the important point I want you to appreciate here is that every time you postulate a model the model does not quite describe the data. Then we look at the model and make refinements to the model. So, that is the point of this exercise. Sir, I am looking at the reaction as how do you say it is uniform. How do you say it is formulate because it is a k 1 times s which means that there is a when we say reaction rate is k 1 times c which means that everywhere in the equipment it is reacting at that rate. But in the shrinking core the particle and the gas wherever is the interface only it is reacting it is not reacting everywhere. So, that assumption that we have made in this formulation is not what is happening in the reaction. So, the two are not consistent that is why we are seeing a difference. Is this point laid all of you that is why we are not finding such a good agreement. Let us go to an equally important exercise which is of course those of you who have looked at the cat cracking what is in the cat cracking process. We have a catalyst and this catalyst is porous and as the reaction proceeds the catalyst has a deposition of coke. So, let me this is problem number 6. So, this is the cracking reactor where gas oil is coming in this is gas oil say and it is regenerated with air. So, as this reaction proceeds as this reaction proceeds the catalyst which is present in the reactor there is coke deposition and because of coke deposition which is uniform throughout the particle and this deactivates the catalyst. So, the first part of the exercise is that what is the extent to which this catalyst is getting deactivated that is the first part. The second part is how do we run this process second part. Third part is that what kind of I mean conversions we should expect in our process. There are three aspects to this let us do it one by one. First part says the reactor gets deactivated by this rate function k 1 time s 1 where s 1 now refers to mean values. What do we mean previously we talked about distributions now we are not talking about distribution we are saying r 1 is simply k 1 times s 1 s 1 bar with a minus sign indicating that the rate at which this is how we are formulating the problem because we already found out how to determine s 1 bar we have done that problem already correct. For the case we already we know what is the value of the mean value of the catalyst activity because we have done this problem just now. Now, what we are saying is that let us now look at this reactor deactivating by this function k 1 time s 1 bar s 1 bar is known. Similarly, r 2 which is the regenerator where it is getting regenerated by this kind of function k r 2 equal k 2 time s 2 bar this is given to you k 2 and k 1 values are given to you. Now, what is problem now says is if s 2 bar minus of s 1 bar the mean values s 1 and s 2 are mean values this is let us say it is 0.3 what is being said the mean value of activity s 1 bar and mean value x 2 s 2 bar is about 0.3 this how the process is run you can run the process at different values depending upon what you feel as appropriate to you in this particular problem we are saying it is s 2 bar minus of s 1 bar is 0. It is given the question is show that the optimum size ratio of w 1 that means optimum size ratio of the reactor and regenerator that is given by this square root of k 2 by k 1 this is the first part of the exercise. So, let us try to do that problem. So, I will do it once again we have reactor we have regenerator solids are circulating between this two. Now, we are writing a material balance for coke deposition this is what is responsible for the loss of activity correct or we can also write material balance for activity itself because we understand that activity is because of coke form loss of active because coke formation. So, v naught times a 2 bar or s 2 bar I mean I have written a 2 bar here v naught times a 1 bar minus k 1 w 1 a 1 bar equal to 0 is it all right what we are saying v naught times this is reactor 2 this is reactor 1 material is from reactor 2 material is coming at an activity of a 2 bar it is flowing at v 0. Similarly, it is going out at v 0 at a 1 bar and the rate at which the activity is lost is given by k 1 bar a 1 bar. So, this is equation 1 this further reactor similarly, we can do for v naught a 1 bar minus v naught a 2 plus k 2 w 1 a 2 bar equal to 0 is it all right but 1 minus 1 1 is it ok yes or no all right. Now, we want to solve these two let us quickly solve. So, what I have written please tell me it is t 1 bar equal to a 2 bar minus of a 1 bar divided by k 1 a 1 bar t 2 bar equal to a 2 bar minus of a 1 bar divided by k 2 times 1 minus of a 2 bar is it look at this. So, I just call this as equation 3 and equation 4 do you all agree with this yes or no now a 2 bar minus of a 1 bar is actually a process design decision you will decide how you will run your process. That means, a 2 bar gives you the average activity in the react and regenerator a 1 in that difference is a process decision you will find various values in this case it is given as 0.3. On other words t 1 bar plus t 2 bar if I add these 2 a 2 minus of a 1 bar within brackets 1 by k 1 a 1 bar plus 1 by k 2 times 1 minus of a 2 bar. The question we ask now is we have made a decision of what is a 2 bar minus a 1 bar. Let us say that is fixed for a process therefore, if you want to find the best choice or the minimum size of this equipment w 1 and w 2 you have to minimize this quantity with respect to a 1 bar or a 2 bar. But, the difference is fixed that means what we are saying is that a 2 bar minus of a 1 bar is fixed it is omega and then what is the best choice that you can make for a 1 and a 2. How do you solve that problem you solve this problem by recognizing that t 1 bar plus t 2 bar t 1 bar sorry. So, this is omega 1 divided by k 1 a 1 bar I can say plus 1 by k 2 times what is this a 2 is I will say omega plus a 1 is it all right. So, let me say a 2 minus of a 1 bar is omega. So, a 2 bar equal to you are right you are right you are right minus minus minus all right. Now, we can differentiate this and find the best values please do that differentiate this and bind the best values. So, you differentiate with respect to a 1 set it equal to 0 I have done this here. So, 0 equal to minus omega this is what I get please tell me whether I got it right a 1 square plus 1 by k 2 I have removed the a 2 bar I have removed the omega from here is it ok please tell me I have differentiated this and then replace omega in terms of a 1 a 2 s n o please differentiate with respect to a 1 and in the result replace omega in terms of a 1 a 2 this is the result I get s n o please 1 or 2 is not enough is it ok. So, what do we get now. So, we get k 1 divided by k 2 divided by k 2 divided by k 2 to the power of half equal to 1 minus of a 2 bar divided by a 1 bar is this correct s n o all right. So, shall we go forward let us look at our first equation t 1 bar is given t 2 bar is given by this. So, from here k 1 t 1 bar I can write. So, I will write I will write like this please tell me if it is ok k 1 t 1 bar divided by k 2 t 2 bar equal to 1 minus of a 2 bar divided by a 1 bar is this ok that comes in equation that we written earlier is it ok all right. Now, you also have this let me put this in here k 1 divided by k 2 to the power of half equal to 1 minus of a 2 bar divided by a 1 bar. So, with this what kind of simplifications can we make regarding t 1 by t 2. So, I mean it follows directly from here t 1 by t 2 bar is simply k 2 by k 1 this is what is asked if you look at here see w 1 by w 2 is simply t 1 by t 2 correct. So, k 2 by k 1 that is what we get. So, what do we get here. So, it simplifies as t 1 bar divided by t 2 bar equal to k 2 divided by k 1 to the power of half that is same as what you have to show you have to show is w 1 by w 2 is same as t 1 bar by t 2 bar. So, what we are saying is that the size ratio the point that is being put across here the size ratio for this recirculating systems really depends upon the rate constants. So, essentially your catalyst design see if you have a rate constants. So, essentially depends upon your catalyst how good or how bad it is on that basis only the size ratio will be determined I mean this is nothing unusual I mean this is something that we all know the rate constants are high then you can have smaller equipments and so on that is what it is saying is it. Now, the next thing is for the optimal size ratio you find out these numbers s 1 s 2 t 1 t 2 and so on for optimal size ratio. So, if the optimal size ratio is given what is a 1 bar and a 2 bar that is our next question that means the size ratio is optimal is given first you have to find out what is s 1 bar what is s 2 bar or a 1 bar a 2 bar what are the t 1 bar and t 2 bar this is what we have to the next let us do that now. What is given is a 2 bar minus of a 1 bar is 0.3 we know that k 1 t 1 bar by k 2 t 2 bar we have derived just now is 1 minus of a 2 bar by a 1 bar we have derived just now. Now, what is k 1 from our exercise what is k 1 value 0.0 far per second at 400 degree c for c a equal to 1.5 gram moles per liter that means see what happens please appreciate in the cat cracking system the rated which coke deposits really depends upon the concentration of gas in contact with the solid. So, if the gas concentration is high to that extend the coke deposition is high if the gas concentration is low all those issues are there with cat cracking system. So, here it says if c a is 1.5 it is 0.05 per second actually what we are given a little later in fact these two are related that we are operating at 5 grams per liter that is the concentration and we wants to operate at 60 percent conversion. So, what is the value of c a for this for the problem we want to solve see 5 grams per liter is the extent of reaction is 60 percent what is this a value in contact with the solid in the reactor c a corresponding to 60 percent conversion is what c a equal to c a 0 times 1 minus of x which is 5 multiplied by 0.6 which is 3 grams per liter correct is it ok what we are saying. So, thank you 2 grams per liter. So, if k 1 is 0 5 at c a equal to 1.5 what is k 1 value corresponding to 2 grams per liter this something common sense says that it will be proportionally higher. So, k 1 value k 1 at c a equal to 2 grams per liter equal to 0.05 multiplied by 2 divided by 1.5 is it ok you all agree is it all right how much is it 0.06 sorry 0.066 shall we say per second is it all right. So, k 1 so for us to proceed further the value of k 1 is 0.06 and k 2 is 0.04 is it ok k 1 is 0.06 that is what we calculated just now 1.5 our value is 2 just now we said what I am saying is see we have said a little later that it is actually going to be operating at 2 grams per liter it is said a little later it should have been said earlier unfortunately it is said later. So, we calculate value of c a corresponding and therefore, calculate k 1 at 2 grams per liter is it ok. So, that is 0.06 k 1 is 0.06 k 2 is 0.04. So, let us find out what is k 1 t 1 bar let me state once again k 1 t 1 bar divided by k 2 t 2 bar equal to 1 minus of a 2 bar divided by a 1 bar k 1 is 0.06 t 1 bar and what is k 2 what is k 2 0.04 t 2 bar what is t 1 bar by t 2 bar we also done that what is t 1 bar by t 2 bar. So, let us substitute that also 0.06 t 1 bar by t 2 bar is k 2 by k 2 by k 1 correct k 2 by k 1 to the power of half. So, that becomes 0.04 by 0.06 to the power of half equal to 1 minus of a 2 bar divided by a 1 bar is it all right yes or no. So, what is the value no please tell me all the numbers. So, we know that a 2 bar minus of a 1 bar is 0.3 is given. So, this gives us a 2 bar equal to a 1 bar equal to what tell me is this clear what we are saying a 2 bar this a 1 bar a 2 bar minus a 1 bar is given. Therefore, you can do all the manipulations and give me the results what is a 1 bar and what is a 2 bar what is the answer. I get a value of a 1 bar is 0.307 a 1 bar is 0.307 I got and a 2 bar is 0.607 anybody else anybody else please is it all right. So, that means we are regenerate c says just a minute wherever how do you say that 6.2 6.2 yeah k 2 yeah yeah see the temperatures of regenerate is always higher than the reactor because you have to burn off the coke that data is given data is given. So, that is why k 1 is 0.066. Why k 1 is point yeah yeah yeah good good see what I am saying is this 6.2 is related to 6.3 the should have been said earlier the 6.3 what they are saying is this gas oil cracking that goes on in the petroleum industry with the inlet gas are coming in the 5 grams per liter and 60 percent conversion is typical. Therefore, the gas solids in contact with the gases gas concentration is about 2 grams per liter because it is 2 grams per liter and that data k 1 is a 0 5 corresponding to c s 1.5 and our c a values are more like 2 that is why we have upgraded this number and said that it is this is of this 0 5 multiplied by 2 by 1.5. That means higher the concentration higher is the coke formation that is what is being said is that clear is it ok now. Now, we have a 1 bar is 0.607 a 2 bar is 0.307. Then the next we have to find out is what are the find s 1 s 2 we have done t 1 bar t 2 bar. How do you find t 1 bar t 2 bar we go to our equation number 1 k 1 t 1 bar equal to a 2 minus of a 1 bar divided by k 1 a 1 this is our equation 1 correct this is something that we have written earlier. Now, can we find out t 1 bar. So, t 1 bar equal to 0.3 k 1 is 0.06 a 1 bar is 0.307 is it correct 14.3 correct 14.8. Now, what is t 2 bar t 2 bar equal to can do the same thing a 2 bar minus of a 1 bar divided by k 2 times 1 minus of a 2 bar. So, what is t 2 bar equal to 0.3 divided by 0.04 multiplied by 1 minus of 0.607. How much is it 1 9 everybody 1 9 is it all right. Now, we have set up the equations for the dual bit circulating systems. And we have said that we have this reactor and regenerator and material going between the term at v 0 and in the reactor the reaction takes place and in the regenerator the coke deposited gets regenerated and so on. Now, our equation is r 1 is minus of k 1 s bar r 2 is k 2 times 1 minus of s 2 bar I sometimes use a here instead of s. So, you know both are interchangeably used. So, do recognize that our reaction rate expression we said is minus of r a k r c a times s 1 bar where k r c a 0 is 8.67 per minute. We would like to find out what is the size of w 1 what is the size of w 2 I think that is the point that we want to answer now, because we have done the modeling and then found out what is t 1 bar and what is t 2 bar that we already done. Let us see how to do this. Now, what is our material balance for component a in the reactor what is our material balance for component a in the reactor you have this reactor here you have this regenerator here. So, this is reactor this is regenerator this is reactor this is regenerator. Now, you have this products coming out products this is a phase 0. So, what is our material balance input output generation equal to 0 at steady state. Therefore, size of the equipment w 1 is f a 0 x a by minus of r a I will put r a 1 here r a 1 this r a 1. So, what is f a 0 we are saying f a 0 is 16 kg per second we have already said that 16 kg per second and the reaction extent is x a is 0.6 that is also given and what is k r c a s 1 bar what is k r c a we have given k r c a 0 is 8.67 per minute. So, what is c a then c a 0 times 1 minus of x therefore, c is that number is c a 0 is 5 and 1 minus of x is 0.4 because x is 0.6. Therefore, we find that k r c a is 8.67 multiplied by 0.4 which is k r c a and s 1 bar is already we have derived this 0.307 this is already done therefore, how do you find out r a 1 r a 1 is k r c a times s 1 bar k r c a is given here s 1 bar is given here. Therefore, we can substitute here and find out what is the size of w 1 let us do that now let us do that now when you do that what do we find f a 0 is 16 this is 0.6 now 8.67 by 0.4 points this multiplied by 60 to convert this is per minute that is what we have because our numbers are given in minute. Therefore, 8.67 where is 8.67 where is 8.67 8.67 per minute. So, 8.67 is per minute not per hour therefore, what I have done is that 8.67 per minute I have divided by 60 to get per hour because is that clear. So, that we get v 0 equal to w 1 by 2 1 bar this is 540 k g therefore, divided by 14 we get 38 k g per second let us go through this once again what are we done what we have done is w 1 is f a 0 x a by minus of r a minus of r a 1 sorry minus of r a 1 we only have to substitute all the numbers to get our answers this is 16 k g per second while this 8.67 is given per minute I just converted to second that is why I divided by 60 that is why we get size of this equipment w 1 is 540 k g and what is v 0 v 0 is circulation. So, v 0 is simply w 1 divided by t 1 bar we already found t 1 bar to be 14 seconds you already done this therefore, we get v 0 to be 38 k g per second which means we must circulate we must circulate the solids between the two reaction equipment let me just run through this once again what we are saying is that this is v 0 this is v 0 that means we are saying that this v 0 must be circulated at 38 k g per second that is what we are saying is that clear what we are saying. So, because t 1 bar is already calculated as 14 seconds we find v 0 is 38 k g per second now what is v 2 v 2 by definition is v 0 multiplied by t 2 bar we have also found out t 2 bar to be 19 seconds therefore, we find w 2 equal to 722 k g. So, what we have done now we have done this design for a reactor regenerated system where we have done an optimal choice of t 1 bar and t 2 bar based on that optimal choice of t 1 bar and t 2 bar we determine w 1 because we knew the reaction kinetics and based on that we found out v 0 and hence we found out w 2 therefore, what is the production rate of the products from reactor because that is what we want we want to convert gas oil to some useful products therefore, our production rate is 16 multiplied by 0.6 9.6 k g of products are coming out of reactor and therefore, production rate divided by v 0 what is this ratio 9 points p divided by v 0 how much is that v 0 this is the reverse v 0 by v 0 by p. So, the circulation divided by production 38 by 9.6 is 3.8. So, the important point that I want to put across to to you here is the following that is if you have if you have a reactor regenerate this is reactor I will call it R 1 this is regenerator. So, reactor is reactor and R 2 is regenerator. So, what we are saying is that in reactor regenerator system v 0 divided by f a 0 is typically supposed to be 4 to 6 it is supposed to be 4 to 6. In our case we have found v 0 divided by production rate is about 3.8 or v 0 divided by f a 0 is slightly less. What we are trying to say is that the industry norm is in the range of 4 to 6 which means we must circulate the solids which at about 4 to 6 times the rate at which you want to process the gases this is the industrial practice. This is for the case of k 1 is so much and k 2 is so much k 2 is k 1 is 0 5 per second and k 2 is 0 4 per second at 500 degree centigrade this is 400 degree c. Now of course there is a great interest in improving your catalysts and so on and we can do that and then moment you have better catalyst of course you will have to redesign a systems to give you the appropriate optimal choices. And another point to be borne in mind is that our reaction rate in this in the reactor where the gas oil is coming in is s 1 bar k r c a which is the rate at which the reaction takes place in this equipment where this k r c a 0 is experimentally found to be 8.67 per minute. On other words we can improve the catalyst to improve this reaction rates so that we get higher productions that is the important thing that means for a given circulation if you want higher production we have to improve the catalyst. Of course these are all the important features people understand this so it cuts this long story short what we are trying to say is that in reactive regenerator system we must circulate the solids at roughly 4 times the rated which we process the gases this is the industry practice and that is something that we would do in our design so that we get our results of our choice. Alright let us go to the next one our next we have a reactor please talk to me talk to me we have a reactor regenerator system in which our rate function r 1 and our rate function r 2 are k 1 minus k 1 plus k 2 we started this exercise sometime back where we said that whenever we have a rate function which are 0 order which are 0 order what are 0 order mean 0 order means that the time required for complete consumption of the particle is finite. And these are all these are all well mixed vessels that means we are looking at well mixed well mixed vessels well mixed vessels means what that the r t d the r t d is the exponential r t d correct we have done that the r t d is exponential r t d and then if tau is the time for complete consumption there is a finite fraction of material which is in the equipment which is completely consumed. And what is that fraction we said that fraction which is fine completely consumed is 0 to infinity sorry tau to infinity e raise to power of minus of t by t bar d t divided by t bar we have done that correct this is the fraction which is completely consumed. Because any material which is staying for a time more than tau will be completely consumed therefore this what is this fraction we have integrated this and if we call alpha equal to tau by t bar we said that fraction is e raise to power of minus of alpha. What we are saying is that in a stirred vessel there is finite fraction of material which has a time of residence greater than tau and therefore that material will be completely consumed. Because it is completely consumed it will you can be seen as either completely consumed or completely regenerated whichever may be the case depending upon what is the situation. Therefore the fraction that is completely consumed is e to the power of minus of alpha in a stirred tank or in other words if you have a single vessel let me state this once again because it is important. If you have a single equipment material coming in material going out e to the power of minus of alpha tau is time for complete consumption and then t bar is the residence time. Then this is the fraction which is completely consumed or completely regenerated is this clear. Therefore in the context of population balance if it is the instance of a reactor in which there is only deactivation that means your rate function of is of this form then distribution function will have a discontinuity. The discontinuity will be at s equal to 0 because it is completely consumed. Completely consumed means what the property value is s equal to 0 and what is that fraction we said that fraction is e to the power of minus of alpha. Is this clear to all of you if you have a fluid as bed for example if you have solids coming in and solids going out and the solids are reacting as per this rate function. Then there is a finite fraction of material which is completely consumed and that fraction is given by not no there is no s here sorry no s there that fraction is given by e to the power of minus of alpha. Therefore the distribution function f 1 will have a discontinuity at s equal to 0. Therefore we say that if that quantity is delta of sorry I am not writing let me write it in k 0 times delta of s minus of 0 plus a continuous part. Is this clear physically what it means that there is a finite fraction which belongs to delta s minus 0 that means its property value is s equal to 0. If this rate function is of the form r 2 equal to k 2 if this form then the fraction that is completely consumed will be let us say it is f 2 equal to I am denoting this at s l 1 I will give you a reason why I denote it as l 1. The fraction that is completely consumed if it is of this form it is regenerated. So, the discontinuity is at s equal to 1 is it clear is it okay now the context the context is this problem or context is this or context is this okay this is reactor this is regenerated r 1 is minus of k 1 r 2 is plus k 2. Therefore f 1 is some g 1 at s plus k 0 delta of s minus 0 f 2 is g 2 of s equal to 0. So, this is plus sorry plus l 1 delta of s minus 1. I draw your attention to this once again that I said please just give me a minute I said when there is a single fluid bed this fraction k 0 is e to the power of minus of alpha okay when there is a single when there are two of them then this fraction is determined by the interaction between both. Therefore it is not e to the power of minus k 0 alpha it is not something else we will derive that now what is important to recognize here is that function f 1 and f 2 have a discontinuous part and a continuous part okay and then we now we should be able to derive what is k 0 what is l 1 what is g 1 what is g 2 this is what we want to do now is this clear okay. Now the general approach here is that whenever we have a function which is unbounded we try to get rid of it you see we try to get rid of it and look at the homogenous equation generate boundary conditions and appropriately account for the discontinuity okay the same procedure. We have done part of this before but I will run through this once again we have already derived I will write it down but you know we have done this when we met last time so I thought I will save some time we did this last time sorry thank you okay is this something that we have how did we derive this by setting up what happens at the boundaries okay we can do it again but we will not do it now because we will lose some time Now what is our population balance let me just write it here just for g 1 sorry for this is for reactor g 2 minus of g 1 minus of d by I have written d sometimes I write s sometimes g 1 r 1 t 1 bar equals maybe I will write it again one minute I will just write it again for reactor is g 2 minus of g 1 minus of d by d I have written a and s I do not I will write I have I have written a here so let us continue with that g 1 r 1 t 1 bar equal to 0 similarly g 1 minus of g 2 minus of d by d a of g 2 r 2 t 2 bar equal to 0 okay is it alright now what is the procedure that we have adopted for solving this we simply add these two correct and then so let us add 1 and 2 what is the equation number I have given here I call this is 1 I call this is 2 suppose you add 1 and 2 what do we get so we get g 1 r 1 t 1 bar plus g 2 r 2 t 2 bar d by d a equal to 0 so that gives us g 1 r 1 t 1 bar plus g 2 r 2 t 2 bar equal to constant is it alright okay we agree now to find the constant of integration we have got the all the boundary conditions here we can see how it looks g 1 r 1 t 1 bar at 0 g 1 r 2 t 2 1 is k 0 1 is minus k 0 therefore constant of integration is 0 is it okay because we have seen g 1 r 1 t 1 bar at 0 and g 2 r this is minus k 0 this plus k 0 therefore constant of integration is 0 is it okay so since g 1 r 1 t 1 bar equal to minus k 0 g 2 r 2 t 2 bar is plus k 0 we have constant equal to 0 therefore this our equation becomes g 1 r 1 t 1 bar plus g 2 r 2 t 2 bar equal to 0 so what I have done I have said g 2 equal to g 1 r 1 t 1 bar divided by r 2 sorry r 2 t 2 bar okay with the minus thank you now I put alpha equal to 1 by k 1 t 1 bar and beta equal to 1 by k 2 t 2 bar with these I get g 2 equal to g 1 beta by alpha is it correct please tell me g 2 equal to g 1 beta by alpha seems alright okay good now we can substitute for g 2 in our equation 1 correct we can substitute for g 2 here okay and then simplify and so on I have done that this one next one sorry sorry sorry I have written it correctly here I did it wrongly here okay let us let us go further now please substitute for this in our first equation I get please see if it is alright g 1 by alpha beta minus of g 1 plus d g 1 by alpha d a so this how it looks our equation 1 please see if it is okay say substituting for g 2 this is all it is substituting for g 2 in our equation 1 our equation 1 is this this is equation 1 and it simplifies like this please tell me if it is okay so finally it simplifies as d g 1 by d a is g 1 times alpha minus of beta this is okay whether this representation is correct you please tell me listen listen you should you should not make any mistake here I have written g 2 as g 1 beta by alpha this is okay g 1 is alright minus d by d a of g 1 r 1 so r 1 is negative therefore this is become a plus so this statement is correct so this equation is correct okay now please help me to simplify this can we can we solve this now what is the solution so you get l n g 1 by q equal to alpha minus of beta times a is it alright can we write this okay now therefore g 1 equal to q times exponential of alpha minus of beta times a so this is the solution okay okay thank you what is g 2 I written as q times beta by alpha exponential of alpha minus of beta times a is this correct our solution is complete if you find the values of k naught and l 1 in terms of system parameters which is alpha and beta are the system parameters so let us try to do that for that what I have done integral 0 to 1 g 1 d a is 1 minus of k 0 integral 0 to 1 g 2 d a is 1 minus of l 1 is it alright it is by definition please our definitions are here this is our definition so integral f 1 d a is 1 therefore g 1 d a 1 minus of k 0 g 2 d a will be 1 minus of l 1 is it okay okay so let us go further now so what is plus it has to be plus now see please do not forget our physics is what the function f 1 and function f 2 has a continuous part and a discontinuous part so it is it is plus okay alright now what I have done is that integral g 1 d a can we integrate g 1 d integral 0 to 1 what is g 1 q times exponential of alpha minus of beta times a d a okay that is equal to 1 minus of k 0 I have integrated this and you please see whether we are done it right or not q minus of alpha minus of beta within brackets of this is what I get is it alright okay we will keep this some manipulations I have done the manipulations okay now let us look at g 1 where is g 1 little messy but anyway this is g 1 what is g 1 at 0 g 1 0 g 1 0 is q I will write okay g 1 0 is q okay now what else do we have I have got something else I have done I have written g 1 0 by alpha from 3 what is 3 g 1 t 1 r 0 equal to minus k is it correct so can I write this as minus of g 1 at 0 by alpha equal to minus k 0 can I write this okay so that tells me that g 1 0 is k 0 alpha and g 1 0 is also equal to q therefore our solution looks like this so what I am saying is that this q I will replace it as k 0 alpha is it alright so help me now so I am saying 1 minus of k 0 equal to k 0 alpha divided by alpha minus of beta e to the power of minus of 1 so this simplifies and gives you k 0 as 1 divided by 1 plus alpha minus of beta e to the power of alpha minus of beta minus of 1 okay now just to cut this long story short and we can do some more manipulations I will not do this I will simply write this you can do it yourself it is not so difficult okay I get this 1 plus alpha by alpha minus of beta e to the power of alpha minus of beta minus you can do this yourself okay so what we have tried to do is we have found out g 1 we have found out g 2 we have found out k 0 we have found out l 1 on other words you have completely specified the distributions in terms of system properties alpha and beta okay the important point is to recognize that k 0 and l 1 depends on choice of the operation alpha and beta okay that is the point that is that comes through this exercise the property distributions are done so okay the property distributions I will just all the numbers it will write down alpha is known k 0 is known l 1 is known g 1 is known g 2 is known correct let me just write here g 1 you have done this g 1 is k 0 alpha e to the power of alpha minus of beta times a and g 2 is l 1 beta e raise to the power of minus of alpha this is we know all this so we know the complete solution is this clear the complete solution is in front of you so you can see a g 1 g 2 k 0 l 1 okay this is alright the what it tells us is that if you want the mean values if you want the mean values now you have to take the first moment of the distribution okay so if you want mean value of in reactor so it will be s s times so I just want you to remind let me just do one of them s times you have k 0 alpha minus of beta times a plus this is k 0 times delta of s minus of 0 d s this is the integral which gives you s 1 is it alright what is this integral s times k 0 times delta s minus 0 d s what is the meaning of the delta function s equal to 0 is s multiplied by k 0 delta s minus 0 d s that integral is what it is 0 do you understand that do you understand s times k 0 times delta s minus 0 d s is 0 is that clear that is the definition of delta function the delta function takes the value see when you multiply s delta s times delta s minus 0 is 0 okay do not make that mistake when you do this integration this term actually disappears is this clear is this clear to all of you see what is important is that you must appropriately use the property of the delta function okay so that is the point that I point out okay so what it says is that we have done the property distribution and then boundary conditions and all that we have done the mean value I have not done you can do it yourself and show it to me so what are the things we have done we have done this we have we have done this zero order we have done okay carbon recovery we have done mixed orders see this this four and five they are exercises for you because we have set up we have set up for various cases so four and five are two exercises that you should do okay so that you know how to apply population balance how to write the balances properly take things appropriately into account so these are very good exercises you know unless you write it properly you do not get the results okay and all these are analytical solutions you will get analytical solutions so you get you know there is no so it will be a good exercise for both all of you I said it is plus k two which means that I mean plus I need not have written okay is it alright now r2 equal to k2 no no no this is wrongly written wrongly written it should be r1 is only correct is wrongly written you are right I will make that change it is only minus sorry now it is okay r1 is minus k1 s r2 is plus k2 okay please make that change please make that change question number five this is correct five is okay so this in this of course you have done all these things you can do it once again and show it to me but most importantly four and five you must submit because that I have not done here so whatever I have not done here you should definitely do and what I have not done fully also you should submit I have not done fully I think this this particular question I have not done fully this part I have not done okay we will stop there we will take it next time.