 We go on with advanced reaction engineering time dependent operations. We said last time in we met that there are several situations where we have to deal with time dependent operations. Of course, the classic example would be a batch reaction where we are concerned with how long it takes for the process to get completed to the level that you and I require. The other situation would be that you know you have a process which you want to start up. The continuous process will take some time for it to stabilize. So, start up of a process also is a time dependent operation. A third situation which we also looked at is a case where this process is running and a steady state, but it becomes time dependent because the catalyst tends to deactivate. And the deactivation brings in a time dependent element which we must appropriately take into account so that the process runs as per our requirements. We did speak about this briefly by just want to run through this once again just to emphasize a few issues. So, we are looking at deactivating catalyst. So, we are looking at deactivating catalyst and what you like to do is quickly get a view for how we handle deactivating catalyst for which we require some understanding of what we call as deactivation kinetics. How do you get some information on deactivation kinetics as an example there are very ways it can be done. Example is let us say we have a catalyst which is sitting on a spinning basket and to which let us say you have material coming in at some and material going out at some. So, it is undergoing some reaction. You want to understand what is the kinetics of this deactivation. So, to just to understand this I am just writing the material balance input output then you have generation equal to accumulation. This is something that you and I have been writing for a very long time. So, it is nothing very new suppose I say that this C A by C A naught and if I also say that time theta divided by tau d where tau d is the characteristic time characteristic time for deactivation. What is meant by characteristic time we can understand it like this that let us say catalyst has a life of let us say 1 year then its characteristic time is 1 year. So, essentially it is a time which is characteristic of the process and I have taken it as tau d. So, that we try to understand this whole thing with respect to the characteristic time of the process. Let us see what it gives us. So, let me simplify and write this as this is fairly elementary. So, I will not go through all the steps. So, you can say I am just write this as 1 minus and just simplify this and write this plus r a divided by v divided by v naught C A naught equal to v divided by tau d d psi by d theta times v naught. So, it can be written in this form plus r a times tau r divided by C A 0 equal to tau r divided by tau d d psi by d theta. So, a catalytic reaction in which deactivation is important this the kind of material balance which describes what happens during the process of deactivation. Now, if it so happens that tau r is less than tau d on other words what we are saying is that the reaction time typically for 1 or 2 seconds in a most of these catalysts the reaction times in the equipment is not more than few seconds. So, if the reaction time is few seconds while the deactivation time or the characteristic time deactivation of the order of months or years then this tau r by tau d is quite small. So, that we can neglect this on other words when we do experiments on trying to understand deactivation we might recognize that the right hand side is not important because it is 0. Therefore, we can understand the left hand side and then simply look at what happens to psi as a function of time. Let us just take this forward this whole idea forward. So, what it means is that 1 minus of psi plus r a by tau r divided by c a 0 equal to 0 this is what is called as quasi steady state approximation. So, that r a as a function of time is 1 minus of psi as a function of time divided by multiplied by c a 0 divided by tau r. On other words what we are saying is that simply by measuring psi t as a function of time we can get reaction rate as a function of time. On other words we are able to plot reaction rate as a function of time and we find that this reaction rate does this over a period of time. This is what we are trying to say and we want to know how activity of the catalyst changes as the reaction takes place. We pointed out yesterday that we can understand this simply by trying to look at activity of the catalyst which is reaction rate at any time divided by reaction rate at 0 time. We mentioned this yesterday that by 0 time what we mean is that some standard state this is some standard state the standard state. Therefore, we can understand the activity with respect to the standard state. Therefore, what we are saying is that reaction rate at any time is some activity multiplied by a standard state. Therefore, we can say if this standard state this is some function of concentration multiplied by an activity. So, essentially what we need from our experiments from our experiments is that we need reaction rate. We need reaction rate as a function of time I will put a minus sign here. Even here also I will put a minus sign because our reaction may be a going to products. Therefore, we expect that I will put a minus sign here. So, we find that this decreases and therefore, if you want to find activity at any point if you want to activate at any point. How do you find activity you simply find reaction rate at this point what is reaction rate is 0 time. Therefore, reaction rate at any time divided by reaction rate 0 time gives you the activity or in other words our experiments can give you activity as a function of time or in other words from this data we can generate what is called as A versus time which shows this kind of data. So, this is the kind of data we can generate from our experiments. We pointed out yesterday how to collect this data from the equipment like a CSTR and so on. But for the moment we know how to do it. So, all we have got now is that activity is a function of time. Now, what is required for us is to be able to find a function R D which tells us what is the rate at which the deactivation occurs. So, some a to the power of m some function of. So, I will say c to the power of n some concentration a to the power of n concentration b to the power of some p and so on. On other words just like we are talking about reaction rate function. Similarly, we can talk about deactivation rate function. Therefore, all these are parameters k d m and p are all parameters which we can determine by this kind of data which comes from the experiments. On other words we can generate this data A versus time of under different kinds of condition at different temperatures at different compositions and so on. Therefore, we can generate this function. So, this we have said already. So, what we did point out yesterday and this is something that is of interest to us when we trying to understand process is that if we have a rate function a rate function a I will put a minus sign here to say that it is some function of time and then some function of a and then some function of concentration where our R D is some function this is some function a to the power of m c a to the power of n and so on. On other words we have from our experiment the parameters of deactivation models are known. Therefore, we know we know how activity d a d t changes as time changes and therefore, we know from this data we know how a changes with time. Now, frequently our concern is that we want to run a process let us say we have a process and then we want to run this process feed coming in and going out this process should run. So, that the composition of the exit does not change much with time because that is useful because all the downstream equipment would have been designed for certain capacities and if this composition keeps changing with time then there are problems with downstream equipment. So, on other words that means you do not want this function f c to change much during your process which will run for several months several years sometimes. What I want to draw your tension is an interesting result that you and I will see when we go to a process plant. Let me just draw this and then perhaps you know explain to you something that you all know it is not that it is new to you, but I want draw tension to some results which would be very useful for all of us. So, let me write here catalyst temperature v naught composition percent a at reactor exit. So, you have fresh catalyst fresh temperature is 450 k this is kelvin 0.39 liters per second is liters per second composition is 0.25 percent and spent catalyst spent catalyst temperature is 477 0.38 and composition is 32.2 regenerated catalyst temperature is 450 0.28 and 19 percent. So, what I what I have got here is is the behavior of a reactor separator it is behavior of a reactor in which there is a deactivating catalyst and the composition at reactor exit that is a position 2 is given here. That when you run the reactor at 450 k v 0 v 0 coming in at 0.39 liters per second we were getting 25 percent at position 2 that is when the catalyst was fresh. Now, when the the reactor is operating for months and months we find that catalyst is deactivating. So, in order to annul the effect of deactivation what the operators have done is to increase the temperature to 477 they have increased the temperature to 477 and you can see that they have adjusted the v 0 what was 0.39 they reduced it to 0.38 and in spite of that the composition at the exit is 32.2 which what was 25 that means the conversion was 75 percent here that has become much less even though the temperature is increased the flow has come down and the composition is also come down. So, what do we conclude from here we what we can conclude from here is that, but in to arrest the effect of catalyst deactivation the the operator has increased the temperature. But in spite of that it is performance is not very satisfactory you know in the sense that the catalyst deactivation is quite large as you can see here. So, I mean there is an alternative process that the operator is looking at is that there is a regenerated catalyst whose performance is like this which says at 450 the the flow rate is 0.28 and then the composition at reactor is this is 19 19 means what it is 25 this is 25 let me write it properly here this is 25 percent 19 means what we are getting 81 percent conversion here. So, we might like to see I mean in spite our major interest would be in reactant engineering and evaluating performance of process and so on to see how the performance of a fresh catalyst performance of a spent catalyst performance of a regenerated catalyst compare on some common basis and what could be a common basis for a catalyst the activity of the catalyst. In fact therefore, when we look at problems like this we need to be able to look at what is the catalyst activity in something that we would like to calculate for which of course we will set out procedures. This is the kind of problem that we are trying to understand trying to solve and for which we want to set up procedures. So, what is the procedure I mean we have said this before, but you know. So, we have this reaction a going to let say b now we said under quasi steady state approximation our equations look like this if the experiments were done in a stirred tank they look like this equal to 0 under quasi steady state approximation. So, that our reaction rate will look like this and therefore, therefore we are able to you know conduct this kind of experiments we are able to conduct this kind of experiments and find out how the reaction rate changes with time that there is something we already said having said that having said that what we have we also said is the activity of the catalyst which is given like this at 0 time or standard standard state standard state can be found out. On other words what we are saying is that the rate at which the catalyst is reaction rate rate of rate of chemical reaction any time is r a at t r a at 0 or we can also say it as some function of temperature some function of concentration and a some function of activity. On other words the rate at which chemical reaction occurs depends upon temperature depends upon activity depends upon the compositions to which the catalyst is exposed. And we also said that we will be able to determine the deactivation function by doing the deactivation kind of experiments which we already talked about which we already talked about. So, that we know what are these parameters because this k d we said this k d equal to minus of d a d t equal to r d equal to k d times a to the power of m c a to the power of n c b to the power of p where k d is here deactivation velocity constant which comes from. So, we have we have the parameters of the deactivation velocity which is k d e d a m m n and p n so on. So, all these are known therefore r d is known therefore a at any time is known. And therefore therefore we know what is the reaction rate because reaction velocity constant this is something that is known from our reaction kinetic experiments at using standard state and so on. Therefore, k t is known a function is known a function is known because it comes from a reaction kinetic experiment. Therefore, our reaction rate at any time is known fully the point I was trying to put across to you point I was trying to put across to use that here is an instance here is an instance of a catalyst which is deactivated. And the operators have tried to anal anal this effect by changing this temperature and in spite of that it is not been satisfactory. On other words what we find is that the composition at position two the composition at position two it is changing because the composition at position two is changing it is not a not as per design. Therefore, what is coming out as product at position three is also not as per your design. And therefore this separator is not be used to the capacity that you would have like to use it to. On other words you are you are having a sub optimal kind of performance. Now what is it that you and I can do to see that this problem can be overcome. Now the our friends in catalyst deactivation have a nice suggestion what is that suggestion the suggestion is that if you look at this function if you look at this function r a at any time t that which is a as a function of time. And this k which is a function of temperature f c which is a function of composition. Now what is it that your this reactor this reactor why why does this change compositions here change the composition here changes. Because this activity keeps decreasing how does this activity decrease we said this activity decreases as per this deactivation rate function. And the parameters of this deactivation rate functions are e d k d 0 m n p where r d is equal to k d times a to the power of m c a to the power of n c b to the power of p n. So, all these are known now the question is that if I want to keep this concentration here the same at all times how do I achieve that which means I want this function f c here not to change. Now to ensure that what I do is that if I can keep this product constant. Then what I am seeing here is that whatever is a reaction rate at 0 time I will see that reaction rate at all times. So, this strategy for managing catalyst deactivation in a process is essentially to see that the product of this reaction velocity and this activity is held constant. Now if we of course something that we all know we have done in our reaction kinetic experiments how the reaction the rate constant k this rate constant k which is simply this is some k 0 exponential of minus of e by r t this is how the rate constant will change the temperature. Our k d which is changing like this exponential of minus of e d by r t where and this a this activity a itself is changing with time. Suppose I say that it is an exponential d k then I say that it is exponential of minus of k d f t because it is an exponential d k. Therefore, our rate function r a which is a function which is k t a t and some function of c you find that this changes the temperature dependence of this is given by this the temperature dependence of this is taken care through this expression. Therefore, as a result of this you are able to maintain r at any time same as r at 0 time by simply increasing the temperature t you increase t you increase t that is which time you increase temperature and the idea of increasing temperature is that the reaction velocity will go up and the deactivation velocity constant also goes up. But, experience says that the increase in deactivation velocity constant is not as large as increase in the reaction velocity that is the experience of reaction velocity constant in the in the process. And as a result it is possible to find temperature time trajectory that will give you constant value of this function this product. Therefore, in processing industry we try and maintain such the reaction velocity to remain reasonably constant by ensuring that this product is held constant this is the strategy. Now, under this strategy under this strategy if you look at this data if you look at this data what is this data say this data says that when we conducted the experimented 477 spent catalyst which is getting deactivated we find that even at 477 it is giving at a velocity at a volumetric flow 0.3 at only 0.32.2 showing that the spent catalyst has not been performing as well. And therefore, we should have gone to a higher temperature to a higher temperature. So, that you would have gotten results similar to what you expect with fresh catalyst. Now, the point is that this kind of variations of temperature on a catalyst may not always be possible in an industry. The reasons are many most important reason could be there could be some specification on the catalyst which will tell us that you know temperatures more than this is not permissible. And therefore, this particular catalyst is sort of done its work and done its life either it should be regenerated or discarded this kind of answers may be there in process industry. But, what we are trying to say here is that there is a strategy to manage this. And to the extent that you know your catalyst permits you to change temperature in that direction to that level and so on you will be able to handle this deactivation. If it is not possible you will have to stop and then use a fresh catalyst and discard the old and get a fresh catalyst or look at a regeneration process which might give you a regenerated catalyst of reasonably good activity. As you can see here that the regenerated catalyst at 450 is able to give you 81 percent conversion or 19 percent at the exit here. But, it is able to process less it is able to process less or in other words what we are trying to say here is that in when you have a deactivating catalyst you have to contend with the fact that you know the downstream equipment may have to be utilized at a capacity lower than what you may have designed for or alternatively you may have to factor this in your design. So, that some of these problems can be adequately taken care in your design and in your operation. So, we want to illustrate this through a example we started last time, but you know we did not get far enough I thought you know we should sort of look at this very effectively because it is an important example from process industry point of view. So, here is an example of a process this is the process this is the PFR this is the separator. And now we have this is position 0 position 1 2 let us say it is 3 and 4 and your feed coming in v naught and then this is pure product pure b. Let us say the reaction taking place a going to b and b going to a the data given is as follows rate constant k is exponential 8.85000 by T this is in Kelvin and then deactivation velocity constant is exponential 1.3 minus 2500 divided by T this function of k. And then you have the equilibrium constant k this is exponential minus 19.5 plus 10000 divided by T which is also in Kelvin this whole exercise the context this whole exercise is as follows. There are instances in which you will be running a process where this catalyst which is inside this PFR is used to produce this product b. Because this catalyst deactivates you would have to take care of this catalyst periodically to see that you get the product at the capacity rates that you have designed for. So, in this exercise in this exercise what is been said by the manufacturer is the following says that this reaction a goes to b let me just write down the data I come back to you in a minute catalyst temperature feed and a at reaction. This is the factor exit percent this is liters per second F A 0 this is T in Kelvin. So, it is fresh catalyst spent catalyst regenerated catalyst says 450.39 25 percent 477.5 percent 38 32.2 percent 450.28 19 percent. Now, what is being said is the following that here is a process which is been running when it runs with fresh catalyst when it runs with spent catalyst when it runs with regenerated catalyst data is given for 3 situations 3 situations when you run this is fresh catalyst process runs at 450 c and then F A 0 F A 0 means feed at position 0 turns out to be 0.39 liters per second and exit composition of a is 25 percent you understand what is it says when you run this process at we using PFR V naught is given as 0.39 liters per second and then the composition at reactor exit which is at position 2 is 0.2 for 25 percent. Now, when the catalyst gets spent in the sense after few months of few I do not know how long let us say few months then it is found the catalyst is sort of its activity has become low at that time what is observed is that at 477 c at 477 c the composition the composition here is 32.2 here and then flow is 0.38 Now what is being said is that when you run a process let us come back to this when you run a process with the deactivating catalyst we said just now that you have to increase temperature with respect to time. So, that this product remains constant and in fact if you look at this data here what they have said is that they started the process at 450, but continue to increase the temperature to 477 at that time when catalyst has become a little low in activity the composition at the exit has increased from 25 to 32. You can see that the catalyst is performing poorly as therefore, we have increase the temperature and so on despite this increase in temperature we are performing poorly because 25 has become 32. Problem statement says is that a spent this is a spent catalyst that means this catalyst should have be discarded. Now, the question is now whether you discard this catalyst and then buy a new catalyst or alternatively in this problem the suggestion is that there is a nice process is available in the industry by which you can regenerate this catalyst. And then what has been done here is that the regenerated catalyst has this kind of performance that is at 450 c, but flow is only 0.28 the composition at reactor exit is 19. On other words what is being said is that we have some data on a catalytic reactor. And they ask you to evaluate they ask you to evaluate whether this regeneration process this regeneration process is good enough. Now, how do you judge whether it is good enough or not they give a criteria the criteria is the activity of this regenerated catalyst the activity of the regenerated catalyst should be 90 percent of the activity of fresh catalyst or it should be 2.25 times the activity of the spent catalyst. Let me repeat what is being said in this problem statement is this we will accept the regeneration process. If the activity of the regenerated catalyst is 90 percent of the activity of fresh catalyst or alternatively it is 2.25 times the activity of spent catalyst. In that case only we will accept the regeneration process. So, what is being asked of you is we will have to evaluate the data given to see whether the regeneration process is satisfactory with respect to the 2 criteria that is specified. So, what in essence expected of us is to be able to set up the equations and find out what is the activity of the catalyst. So, we have to find the activity of spent catalyst activity of regenerated catalyst with respect to activity of fresh catalyst and then see how the regeneration process is performing with respect to the criteria that is specified. So, this is the exercise that we want to do and let us go through this exercise. This is not very difficult, but you know some of these things once some procedures are required. So, I am just trying to do that I did do that to some extent last time, but I was not satisfied I am just doing it again. So, it does not mean that what I have done last time is not satisfactory, but you know I am just sort of doing it again. So, that I am satisfied with the procedure I have done 0 1 sorry 0 1 2 3 and 4. The problem statement says that this PFR with the catalyst and then it goes to a separator and the separator is able to recover pure B. So, here what we have pure A pure A and then you request pure. Therefore, what comes here is pure A this is. So, let us sort of go through this whole exercise. So, what 1 by 1 what is F A 1 F A 1 equal to F A 0 plus F A 2 this is F A 1 equal to F A 0 plus F A 4 sorry F A 4 we know this. Now, let us say that F A 2 equal to F A 1 multiplied by 1 minus of y 2, where y 2 is conversion at position 2 with respect to with respect to 2 F A 1 at position 2. What are we saying see in order to define conversion we have said this before we have said this many times before that we need a reference. Conversions are always with respect to a reference what is the reference that we have taken we have taken position with respect to position 1 that is F A 1 at position 1 position 1 is our reference. So, we can define conversion with respect to position 1 that is what we have done. So, in that case now that we know this I can substitute it here therefore, F A 1 equal to F A 0 plus F A 1 times 1 minus of y 2 why is that whatever is A here will be the A here because all the B is recovered in this process. So, therefore, I have written F A 2 see this is F A 2 is F A 1 this is also equal to F A 4 we know this therefore, I have written this. So, this implies what this implies F A 1 is equal to F A 0 by y 2 is it correct it comes directly from here. Therefore, implies F A 1 is F A 0 by y 2 we know that C A 4 is C A 0 we agree with this y is C A 4 equal to C A 0 y is C A 4 equal to C A 0. The reason is that at 2 we have both A and B unreacted A and react and then you are recovering completely the B here. Therefore, here it should be pure A therefore, the composition C A 4 here should be equal to C A 0 that is what I am saying C A 4 should be equal to pure A equal to C A 0. So, C A 0 therefore, C A 1 is equal to C A 0 that is point I am trying to make. So, implies C A 1 equal to C A 0 I hope this is clear to all of you that because if C A 4 is C A 0 and then what is coming in is C A 0 pure A therefore, at position 1 the concentration is C A 0 that is common sense. So, let us see what else we can do what is V 4 by our definition is F A 4 divided by C A 4. F A 4 we said is F A 2 by C A 4 this we have said that book this is equal to F A 2 is what F A 1 times 1 minus of y 2 divided by C A 4 which is C A 0. So, that is equal to F A 1 is what F A 0 by y 2 C A 0 into 1 minus of y 2. So, that is equal to V 0 by y 2 into 1 minus of y 2 is it all right what we are saying what we are saying is that the flow at position 4 flow at position 4 what is the flow at position 4 by definition it is F A 4 by C A 4. Now, a C A 4 is C A I mean C A 4 is C A 0 we have already proved that saying because B is completely recovered therefore, this must be C A 0 then we said that F A 2 F A 2 is what F A 2 is component at position 2 by definition F A 2 is F A 1 times 1 minus of y 2 by definition. So, that is what has been put in here now what is V 1 V 1 is V 0 plus V 4 that is V 0 plus V 0 y 2 multiplied by 1 minus of y 2. Which is equal to V 0 by y 2. So, what we have done we have found out that F A 1 equal to F A 0 by y 2 already we have proved that similarly, V 1 is V 0 by y 2. So, we have got some information about what are the flow rates at position 1 what are the compositions in position 1. Therefore, we are now in a position to write the reactor design equation for P F R. So, that we can understand what is happening in this P F R and address all the questions that has been raised by the problem. Let us do one by one let us address this one by one. So, what we want to do now is we want to write the design equation for P F R and then understand and then use that data to be able to give a proper explanation for the data. So, this is what we are trying to do. So, what we are trying to do is that we will write the design equation for P F R. I am making use of this relationship that we already established and then look at the data to see how we can explain what is going on. Let us see how to do this. So, design equation for P F R. What is the design equation for P F R? We have done this many times. So, I say that it is D F A by D V E equal to R A. This is a well known design equation that we have written for a long time. Now we know that C A 1 equal to C A 0. What is C A 2? By definition is F A 2 by V 2. What is F A 2? It is F A 1 multiplied by 1 minus of Y 2 and what is V 2? We said this V 1 is V 2 is V 1 because there is no volume change. That is V 1 is equal to F A 1 into 1 minus of Y 2. V 1 is what? V 0 divided by Y 2. I put it on this. What is F A 1? F A 1 is F A 0 divided by Y 2 V 0 into 1 minus of Y 2 into Y 2. Therefore, if this becomes C A 0 times 1 minus of Y 2. That is clear. So, what we have is that C A 1, C A 2 equal to C A 0 1 minus of Y 2. Similarly, C B 2 equal to C A 0 into Y 2. Similarly, is that clear? What we are saying? This is coming from basic stoichiometry. So, we are not saying anything new here. So, we just calculated what is concentration at position 2, which is C A 0 1 minus of Y 2, concentration of B at position 2, C A 0 1 minus of Y 2. So, this is at the exit from the PFR. Therefore, if I ask you what is C A 2? What is C A at any position inside reactor? What is any position inside reactor? Then we will say it is C A 0 into 1 minus of Y. So, Y goes from 0 here to Y 2 here. Is that clear? Similarly, what is C B at any position? We say it is equal to C A 0 times Y, where Y is a variable that goes from 0 at this point to Y 2 at the end. And Y is defined as conversion with respect to position 1. So, if I want to substitute, so if I want to take R A, now the rest is fairly straight forward and we have done this many many times before. So, this is not new. Let us go through this D F A D V equal to R A. What is R A? R A by definition is K 2 times C B minus of K 1 times C A, where C A and C B are concentrations in the, concentrations in the equipment. Now, we can put, we know that F A by definition is F A 1 times 1 minus of Y. Y is the conversion at any position in the reaction equipment and that Y is defined with respect to position 1. So, Y is the conversion in our nomenclature, but this time conversion is defined with respect to position 1. That is the only difference. Now, let us put all these things here. The left hand side becomes F A 1 times D Y 1 by D V equal to the right hand side is K 2 times C A 0 times Y minus of K 1 times C A 0 times 1 minus of Y. Is this alright? Now, what is the F A 1 by definition? F A 0 by Y 2 D Y 1 D Y. So, not Y 1 D Y D Y D Y sorry D V equal to K 1 C A 0 1 minus of Y minus K 2 C A 0 Y. Now, we can cancel off the C A 0, which will give you V 0 D Y D V Y 2 equal to K 1 1 minus of Y K 2 Y. Now, what we have done? What we have done is in order to be able to explain this data. See, we have to explain this data and using the data calculate what is the activity of the catalyst here and here. So, that we can judge whether the appropriately the criteria is satisfied or not. So, to be able to do that, we will have to use this process data for which we need to do a process analysis. We have done the process analysis and based on that process analysis, we find that the differential equation that governs the performance of this reactor is given by this. This is the differential equation, which governs the performance of this reactor. Now, once you know this differential equation performs equations to an equipment, the rest is very straight forward. Let us integrate this and then. So, when we integrate this please notice here that this K 1, this K 1 is important here. This K 1 I will put a bracket here just to indicate that these are reactions 1 and 2. These are reactions A going to B is reaction 1, this is reaction 2. This is not to be confused with positions 1, 2, 3, etcetera in the equipment. That is important. That is why I put a bracket here to say that this is reaction 1, this is reaction 2. Now, can we integrate this? Answer is yes. Let us integrate this and then find out what is the integral. I have done this integration. It is fairly straight forward. So, I will not show you the details now. There is one more point that we must bear in mind right here. That is this representation of this equation we must take into account the fact that the catalyst that is now being used activity seems to change and it does change depending upon the operating conditions. Therefore, we recognize that there is catalyst activity that we must account for in our analysis. So, I am putting that alpha here so that we take into account the catalyst activity that is on other words what we are trying to say here is that if you operate the process is one condition and change the conditions then the catalyst activity will also change because of the conditions are changed. Therefore, I have put the catalyst activity alpha is activity. I have put that effect so that we are in a position to see how that activity is contributing to our process. So, let us quickly write down once again the differential equation that governs our process is this. This is y k 2 alpha y. So, we can integrate this. So, I have the integrated form let me integrated form looks like this. Now, you might ask how I have got this integrated form. Now, this is the first order differential equation is very simple to integrate. There is nothing new about this it is alpha multiplied by y there is no terms like this is not there. Now, beta is k e plus 1 by k e where k e is equilibrium constant equilibrium constant. Now, the most important thing is now try to understand how we can understand the data now that we know the performance equations. Now, what is performance equation tell us it tells us that k 1 alpha v what is k 1 reaction velocity constant what is alpha catalyst activity what is v reactor volume what is v naught it is the inlet flow rate is the inlet flow rate. And then what is y 2 y 2 is the mole fraction in percentage at position 2. So, on other words we have express the performance of the reactor to make good use of the data that we see our data is in this form only you can see here our data is given. So, nicely our data is temperatures are given flow rate see this is actually written f a 0, but it is given in terms of flow v 0. So, we have got v 0 here and similarly we have compositions of the reactor exit which is y 2 which is given what is beta is the equilibrium constant. And we have got data all the data is also here we can calculate equilibrium constant any temperature. On other words what we are saying is that we have using the fundamentals of the reactor design equation that we have we have generated a performance equation which is now in a position to tell us how to evaluate the data that is in front of us. Now, it is a question of quickly doing some calculations to find out what is the value of alpha for the different situations for which data is available in front of us. So, let us do that because that would give us a way of I mean evaluating whether the our regeneration process is good enough or not. So, let me let me do that calculations. So, let us do fresh catalyst what I am doing now I am trying to find the value of I am trying to find the value of k 1 alpha v for fresh catalyst fresh k 1 alpha v. What is now that is equal to we have said v naught l n 1 minus of beta y 2 divided by y 2 times beta. What is beta beta is k e plus 1 divided by k e and fresh catalyst data is given at 450 c. So, we have to do k e at 450 I have calculated k e at 450. So, it is 8.5 plus 1 divided by 8.5 the reason is k e at 450 k equal to 8.5 from data this comes from data given. So, beta is known. So, let us put all the numbers and then find out let me put all the numbers please help me. So, fresh catalyst continued fresh catalyst continued fresh catalyst catalyst continued. So, we have k 1 alpha v equal to I have written 0.39 help me please 0.75 l n 1 minus 1.117 multiplied by 0.75 that k 1 alpha v equal to 0.847 this was fresh catalyst what I have done. Of course, I have not done the calculations in front of you, but I am sure this is very elementary nothing to do with anybody can do this. So, by putting the numbers by putting the numbers for v naught which is given notice here y 2 I have taken y 2 as 0.75 look at the data look at the data y 2 is 0.75 is it correct I have taken y 2 as 0.75 why is it 0.75 why is it 0.75 it is the way to the way to handle this is it is a data says the composition here is 25 percent. So, that means the mole fraction here is 25 percent let us just say what is mole fraction let us just quickly calculate what is mole fraction then you can tell. So, mole fraction mole fraction at position 2 what is it f a 1 1 minus y 2 divided by f a 1 that is mole fraction of a position 2 of a this is the moles of a coming out total moles f a 1 1 minus y 2 plus f a 1 y 2 that is 1 minus y 2 is a mole minus 1 minus of 0.25 is 0.75 that is how we have substituted the value of y 2 what is given see please understand what is given is not done properly. So, this is this whole thing is given as 0.25 I hope you understand this what is given is that mole fraction at position 2 is given as 0.25. So, by definition mole fraction is 1 minus of y 2 therefore, 1 minus of y 2 is 0.25 therefore, y 2 is 0.75. So, what we are saying is that in our process in the data that is given in the data that is given 25 percent means at this position 2 at this position 2 mole fraction of a is 25 percent means conversion y 2 is 75 percent that is why y 2 is taken as 0.75. So, you put all the numbers here you get k 1 alpha v equal to 0.847 all right let us go forward how do you find out. So, the important thing is to find out activity means we need a reference we can take 450 data fresh catalyst as reference therefore, essentially we must compare with respect to this data. On other words since we do not know the volume of the equipment is not given to us therefore, and then therefore, the best would be to use this as the reference fresh catalyst and evaluate the other catalyst. Let us do that. So, let us do that for the case of let us look at our spend catalyst. So, some spend catalyst data is given if we write it down once again spend catalyst what is spend catalyst spend catalyst data spend catalyst here you spend catalyst data is here 477.38 and 32.2 let us do that. So, spend catalyst is spend catalyst data is given at 477 k and then equilibrium constant equilibrium constant at 477 k equilibrium constant see I showed you the this one just somewhere here it is he said equilibrium constant is described by this equation this equation. So, you can calculate at 477 what is the equilibrium constant I have done that I have done that. So, that number turns out to be so equilibrium constant 477 turns out to be is 6. Therefore, beta turns out to be 6 plus 1 divided by 6 that is equal to 1.16 and now k 1 is equal to alpha v at 477 we have to look at our design equation what is the design equation somewhere I wrote down some equation is v naught beta y 2 l n of 1 minus of beta y 2. Now, I have to substitute for these numbers from this data here from this. So, y 2 is what is the value it is 32.2 means y 2 is 1 minus of 32.2. So, it is 0.678 I will put all the numbers here right now in front of you. So, this notice here this v naught is 0.38. So, I will substitute these. So, it looks like minus sign is that minus 0.38 divided by 1.16 which is beta and then what else then you have y 2 0.678 l n 1 minus 1.16 and then multiplied by 0.678. So, then you have y 2 0.678 l n 1 minus 1.16 and then multiplied by 0.678. So, when you do all this calculations k 1 alpha v turns out to be 0.75 what are we saying if we go through this once again what we are saying is that the data given please let us look at the process once again. So, when you are employing fresh catalyst you are running the process at 450. As the catalyst deactivates you are increasing the temperature slowly over a period of time and what is this what is the program of increasing the temperature we talked about that also we said that we would increase temperature as per this program and as per this time temperature program is given by this equation. And when you do that and it has become 477 at 477 our data our data says our data says that this is the data. And for that data we have done the calculation using the numbers given and it turns out it turns out that our results are something like this the k 1 alpha v at 477 is 477 is 0.75. Now, if I ask you if I ask you now what is the activity of the spend catalyst with respect to the fresh catalyst what will you say you will say that we have done the calculation for fresh catalyst we have done this just now I hope I can find it here it is. So, we have said this is fresh catalyst correct this is fresh catalyst 0.847 suppose I ask you what is the activity of spend catalyst with respect to the fresh catalyst then fresh catalyst is at 477 for 450. So, the activity with respect to fresh catalyst because the standard state is 477. So, we should say that k 1 alpha v at 477 this is 0.75 we have done this calculation we have done this calculation here. Now, we wanted with respect to fresh catalyst what is fresh catalyst k 1 alpha v is 0.847. So, it is 0. divided by k 1 alpha v 450 this is 0.847. So, this ratio comes out to be 0.88 this is clear. So, what are we getting out of this we are getting that k 1 alpha at 477 divided by k 1 alpha at 450 equal to 0.88. But what is it that we want we want activity alpha at 477 in relation to. So, we need to find out what in other words we should active to spend catalyst at 477 you have to calculate that is important. So, we will come to that in a minute what is the activity of spend catalyst what is the activity of spend catalyst. So, I will write here fairly straight forward. So, we would not spend too much time activity of spent catalyst at 477 k what is it equal to 477 k what shall we say based on this what shall we say activity of spend catalyst at 477 divided by activity what we are asking for is the ratio of alpha at 450 477 to alpha at 450. So, what you have to do you simply have to multiply 0.488 by value of k 1 at 450 and divided by value of k 1 at 477. So, activity of spend catalyst at 477 is equal to 0.88 I will write here it is already calculated. So, it is 0.88 multiplied by k 1 at 450 k 450 k divided by k 1 at 477 k. Now, we can calculate all these things because they are not equal to the data is given. So, I have done all that. So, it turns out that this ratio becomes 0.88 divided by 2.6 I have done this calculation this ratio upon that turns out to be 0.33. So, what we are saying now is that the activity of spend catalyst at 477 in relation to activity. So, I will write this as alpha at 477 divided by alpha at 450. So, what we are saying what we have done is that the activity is calculated with respect to the standard condition which is 450 that is important. So, this ratio turns out to be 0.33. Now, let us quickly look at this quickly look at what is the regenerated catalyst. Now, what is our regenerated catalyst k 1 alpha v regenerated that is equal to once again v naught l n 1 minus 1 minus 1 minus beta y 2 divided by y 2 beta. And we can put all the numbers. So, please our data is in front of us. So, that we can put all the numbers I have done that here. So, it becomes minus of 0.28 and then l n of 1 minus of beta beta value is this is at 450. So, that beta value is known which is 1.117 1.117 and then y 2 y 2 value please note here it is 90. So, 1 minus 0.19 is 0.81 correct 0.81. So, it is 0.81 divided by divided by y 2 is 0.81 and then beta is 1.117. So, we calculate this it comes out to be 0.727. So, k 1 alpha v regenerated is 0.727. So, k 1 alpha v regenerated is 0.727 this is clear what we are saying. Now, what we want is what is the activity with respect to fresh catalyst. So, let us calculate what is the activity with respect to fresh catalyst. The activity with respect to fresh catalyst is simply how do you do that. So, you have k 1 alpha v regenerated. Now, this is at 450. So, this is actually at 450 data is given at 450 then we have k 1 alpha v fresh catalyst which is also at 450 fortunately. So, this is 0.727 and if you recall we had calculated for the case of fresh catalyst this turned out to be 0.847 sorry 0.847 we had calculated that that is equal to 0.858. So, what we are saying now is that there are two situations we have considered situation one in which we find that the regenerated catalyst this is regenerated catalyst the activity with respect to fresh catalyst is 0.858. The other situation we have found out is that the regenerated catalyst the spend catalyst this is spend catalyst spend catalyst. The spend catalyst activity with respect to fresh catalyst this is fresh that is 0.33 and the data says please let us look at the data it says as long as it is 2.25 times at is 2.25 2.25 times 0.33 should be 2.25 times a fresh catalyst. So, what we are saying regenerated catalyst is 0.727 if you just multiply by 2.25 you find that it is 2.25. So, let us say 2.25 multiplied by 0.33 this comes to approximately 0.7 0.71 or so. So, it is that on other words this is more 0.2727 is more than 2.25 times 0.33 showing that the regenerated catalyst has activity more than 2.25 times the activity of the spend catalyst and that criteria is satisfied. So, what we are saying is that the criteria let me just write down what we are saying is the criteria what is the criteria regenerated catalyst activity should be greater than 2.25 times spend catalyst. So, this activity is 0.727 this is 0.71 therefore, this is greater than 0.71. So, criteria criteria in this particular criteria 2 is satisfied. Now, the point of going through this exercise let us not forget the whole thought. The thought is important the details sometimes is unwieldy and so on. See in a process you will always find situations where the catalyst is deactivating and therefore, you will have to change the conditions and those conditions have to be changed as per a program which I pointed out to you the program is given by this. This is the program which you have done. Now, but more important in many situation is that you may have to change the catalyst and because there is cost implications here therefore, there are criteria that you might like to look at. So, what we have done in this exercise we have set up an equation which describes the process how it performs and we have related the process performance to measurable quantities. And therefore, by doing that we are able to evaluate how the decision that we have taken with respect to spend catalyst, regenerated catalyst and fresh catalyst how they all come together. Thank you.