 Welcome to this edition of chemical reaction engineering. In this edition, we will be looking at deactivating catalytic systems. And of course, I mean deactivating catalytic systems are not new to us. I mean chemical cross industry has made great strides in this area of catalysis and so on. But our concern here is how do we deal with designs for deactivating catalyst. And then how do we manage systems in which there is catalyst deactivation and so on. So, we will try to go through this in a simple way so that we get a grip on what happens and so on. So, we will begin and let us look at this problem of our interest is that we have a catalyst which deactivates. And we want to understand what is the kinetics of catalyst deactivation. Why is this required? It is required because that data would be that model the data whatever is useful for us to design systems in which catalyst undergoes deactivation. So, to be able to quantitate a situation like this, what we have got here is a stirred tank. There is a stirrer in the there is a stirrer and the stirrer has a small basket which contains the catalyst. Now, this stirrer is spinning the idea of making it spin so that the temperature and compositions inside the equipment remains reasonably the same everywhere. So, that or data can be meaningfully used for other purposes. We have feed coming in at a certain concentration and going out a certain concentration and that is the reaction that is taking place inside the equipment. And we want to quantitate the extent to which catalyst deactivates. We want to make these measurements so that we can make use of it in other systems design and so on. So, to be able to do this I have written down the material balance for material coming in input it is output V naught C A naught is input. So, many moles per time so many moles per time is going out and this is the rate at which chemical reaction is occurring where r a is the rate function V is the volume of the equipment. In this case this volume refers to the volume of the catalyst and this is the accumulation term which is V time d by d t of C A. I have taken V outside the derivative because V is constant that is why I have taken it outside. And I have written V naught same at the inlet and the outlet under the assumption that there is not much volume change because you are looking at the reaction which is a going to be. Now we want sort of take it forward by doing some simple manipulations so I have said this psi is C by C A 0 it is non-dimensional. And I have defined a what is called as a characteristic time tau d is the characteristic time for deactivation. What is meant by characteristic time? For example, in our commercial processes our catalyst perhaps last for a year sometimes for 2 years. While the reaction time that means the time that these gases spend inside the equipment and the time that these gases spend inside the equipment is generally quite small. But what is really of great interest to us is what is the time for which the catalyst would last. So, it makes sense for us to be able to look at what happens to this equation in the time scale of deactivation. That time scale is taken as tau d what is tau d tau d is the characteristic time for deactivation. For example, if the catalyst last for 1 year our characteristic time is 1 year that is the meaning of tau d. So, what you have done here is that we have done non-dimensionalization of chronological time t with respect to the characteristic time for deactivation that is theta. So, what we have done we have defined a variable psi with the c a by c a 0. We have defined a variable theta with this t by tau d therefore, we can now write this equation in in forms like this 1 minus of psi r a tau r c a 0 tau r by tau d d psi by d theta. What is tau r tau r is volume of by v 0 in this case this v refers to the volume of catalyst that we have put inside the equipment. Now, if you look at this equation here let me go to the next line because that. So, we have the unsteady state representation of our stirred time we have our stirred time here. Let us not forget our system we have stirred time our catalyst is sitting here material is coming in material is going out due to reaction a going to b. So, this represents 1 minus of psi r a tau r c a by c a 0 equal to tau r by tau d d psi by d theta is the statement of the unsteady state material balance for the material that is happening inside this reaction equipment. Now, if you look carefully if you look carefully what we find we find that the order of magnitude of psi is order of magnitude 1 why is it 1 psi c a by c a 0 at best psi can go to 1 I mean c a by c a 0 can be 1. Therefore, values are roughly of the order magnitude of 1 if you look at theta this ratio at best can be 1. So, order of magnitude is 1 once again we recognize that r a tau r by c a 0 what is tau r tau r is some kind of a reaction time we can all say the inverse of reaction time is rate constant. Therefore, what what this what this whole thing refers to is that is some kind of a ratio of reaction rate divided by the maximum reaction rate that is a kind of meaning this gives also has an order of magnitude of 1. So, what we are saying is that in this unsteady state representation of the material balance for this third tank all the terms are first term is magnitude 1 order of magnitude 1 order of magnitude 1 order of magnitude 1 order of magnitude 1 order of magnitude 1 the only thing that is not this tau r by tau d you see all terms being of the order of magnitude of 1 if any of these terms are very small we can knock them out. For example, if we find that this term tau r by tau d is very small then we can knock out this term that is that is the point that we are trying to get across what we are saying is that under the non dimensional representation we are able to look at the values of each term and since the order of magnitude is all made up to 1 any term which is very small we can knock out and therefore, we can understand what is happening to the system under those conditions under which this particular assumption holds or in other words in this particular case tau r by tau d less than less than 1 means that this right hand side can be knocked out. Therefore, we can look at the material balance as equal to 1 minus of psi r a tau r by c a 0 equal to 0 equal to 0 since the right hand side is near a very small we call it equal to 0 when this equality becomes 0 we call this as the quasi steady state approximation on other words what we are trying to say here is that under the condition that tau r by tau d is very very small the accumulation inside the reaction equipment can be neglected. Therefore, we might be able to quite satisfactorily say that 1 minus of psi plus r a times tau r by c a 0 equal to 0 or in other words we can say r a equal to 1 minus of psi multiplied by c a 0 divided by tau r. So, what we are trying to say here is the following now here is an instance of a reaction which undergoes a catalyst which undergoes deactivation which means what the reaction rates a keep on changing because catalyst is losing its activity. Now our interest in this whole problem is to be able to measure this catalyst deactivation and quantitate this through a model this why we are doing all this experiment. So, what we are saying now is that under conditions where tau r by tau d is very small tau r by tau d is very small the accumulation of material inside the equipment is very very small therefore we can neglected. Therefore, reaction rate r a can be measured simply by measuring the right hand side do we know the right hand side the answer is we know c a 0 we know psi because psi is c a by c a 0 c a is by being measured tau r what is tau r tau r is simply v by v 0 we have said all this v by v 0 is tau r and then psi is c a by c a 0 and both these are measured quantities and as a result we are able to measure the reaction rate at any time. Therefore, if the catalyst deactivates which means the reaction rate keeps changing with time now we are able to measure what is the reaction rate at any instant of time because all the points all the quantities in the right hand side is known to cut this long story short what we are trying to say here even for the case of a deactivating catalyst under assumption of quasi steady state which means that the accumulation inside the equipment is small we can measure reaction rate simply by measuring psi which is c a by c a 0 is measured c a 0 is measured tau r is measured. Therefore, all the quantities in the right hand side is known and therefore, reaction rate is known therefore, we are able to quantitate reaction rate as a function of time this is what we are trying to say having said this. So, what we are saying now now we have a catalyst we have a catalyst which is deactivating now our interest is to measure how this deactivation is taking place for doing that what are we doing we are measuring reaction rate how do you measure reaction rate by measuring all the quantities on the right hand side what are the quantities on the right hand side psi which is c a by c a 0 which is c a by c a 0 and then tau r which is v by v 0 v is the volume of the catalyst v is the volumetric flow. Therefore, by measuring on the quantities on the right hand side we are able to measure reaction rate as a function of time therefore, the data that is required for us to quantitate catalyst deactivation is in front of us that is the advantage of what we call as a stirred tank stirred tank as you can see here the stirred tank because we are able to put our catalyst inside this basket we are able to spin it around we can put it in an equipment we can put it in an equipment and put our material inside and then take out our material and by appropriately changing the slow rates etcetera we can measure reaction rates reaction rates for different conditions of the system of our interest. Having said this let us just look at some interesting features that might be of interest to us let us say let us say if this reaction of that we are studying a goes to b let us say this is the first order reaction. If this is the first order reaction we know from our basic understanding that the rate the conversion is given by k tau divided by 1 plus k tau or in other words if it is the first order reaction then our conversion does not depend upon concentration it only depends upon what is called as the residence time and the reaction rate constant that is for a first order reaction. Suppose you have a second order reaction you notice for a second order reaction we can do all these the extent of reaction x depends upon what is the composition at the inlet depends upon the composition of that for a PFR this is for a PFR this is for a CSTR this I have written for a CSTR. What I am trying to say here is we can devise experiments to understand what is the extent to which reaction is occurring we can measure those reactions depending upon the device if it is a CSTR we measure conversion like this if it is a PFR and the second order reaction we measure conversion like this. So depending upon the order of reaction the dependence of concentration will appear suitably in our expression for conversion in our expression for rate constant for the rate of reaction for example if you have a PFR and you have a rate function is k C A squared as is written here then the rate function will contain the concentration term in the second part. So these things we know the point of trying to draw this attention in this case is that if you have a first order reaction then there are certain benefits in terms of managing it if it is second order then it is little bit more complicated as you wanted to draw attention to those differences by pointing this out. Let us continue let us continue let us say we continue with our first order representation for the moment so we have a first order reaction in which the catalyst undergoes deactivation. Now if you say that this rate of reaction rA is given by kR times C A where kR is the rate constant then we know that because of this of the deactivation this the rate at which chemical reaction occurs changes it changes because this rate constant is changing with time. See that is why we are seeing the effect of the changes in rate constant because of that the reaction velocity change. Now I have just put it in slightly different form to make it easier to understand so I would put it in this form rA which is rate of chemical reaction is kR times C A and I have replaced C A in this form because we know all these from our first order understanding. Therefore the rate at which chemical reaction occurs for a first order process is given by this and then since the deactivation is occurring we can now see that the reaction rate at any time the reaction rate at any time depends on the rate constant kR which is the function of time multiplied by C A 0 divided by kR is the function of time multiplied by the residence time. Now sometimes it is difficult to understand this why is rate constant a function of time. The rate constant is function of time is because our rate constant actually consists of a kR 0 multiplied by a term which is called activity. Because of the fact that this activity is a function of time this product is a function of time. This is why we see because this kR is a function of time because kR 0 is multiplied by this activity which is changing because of the process we find that reaction kR is changing with time because A is changing with time and therefore reaction rate is changing with time. So what is it that we are saying now that if you have a catalyst which is deactivating then reaction rate will change with time. Our interest in this whole exercise is that can we quantitate and find a model which takes care of this deactivation phenomena. If we can do that then there are lot of things that we can do in reactor design operation evaluation and so on which makes the whole thing much easier and that is the object of this particular module. Stating it once again we are used to stating that reaction rate is kR CA 0 divided by 1 plus kR tau r for a first order reaction first order in a CSTR all these are in a CSTR. So this is all in a CSTR. Now if this reaction is undergoing deactivation then we recognize that this is going to change with time because kR is going to change with time. So these two are representations in which in here we have recognized that the deactivation is occurring here we have recognized that deactivation is not occurring. We have two statement of the same situation. What are we saying now? What we are saying now is we have said just now that reaction rate is changing with time. So we would like to know what is the reaction rate at any time divided by what is the reaction rate at 0 time. So what meaning can we attach to reaction rate at any time to reaction rate at 0 time? By definition it is activity of the catalyst. So we can say that this is the activity of the catalyst. The activity of the catalyst is reaction rate at any time divided by reaction rate at 0 time. Now just now we derived what is reaction rate at any time which is this function kR T CA 0 divided by 1 plus kR tau. And what is reaction rate at 0 time? It is 1 plus kR CA kR 0 which means that at 0 time we have taken this kR as kR 0 and therefore I have written it in this form. So if I ask you now this quantity rA what is the functionality on which this function this A depends. Now we can tell by looking at the right hand side, looking at the right hand side that the right hand side depends on number of factors. Very clearly it depends upon temperature because kR is involved. Very clearly we can expect that it depends upon concentration because CA 0 is involved. Now we can say since tau r is involved. So we can in general say that this function this quantity A depends upon the composition of the system in which the reaction is taking place which is temperature and composition. Now that composition is determined by residence time therefore it has an indirect dependence on the residence time. So what we are trying to say here is that we want to quantitate how activity of the catalyst changes and what is that function which determines this reactivation rate phenomena. So if you look at the right hand side, the right hand side if we can measure and quantitate then we will be able to get A as a function of time in data for which we can model. So this is what we want to do in this. Let us go further and see what we can do with this. So let us let us just look at what is the phenomena we are trying to understand. The phenomena we are trying to understand is what happens to rA as a function of time as CA increases. We know that just looking at I just write this function once again here so that we can remember this. We write it here rA as a function of time is for a first order kr CA0 divided by 1 plus kr tau r this is in a CSTR all these are in a CSTR. So if you look at this function and say what happens to rA as a function of time as CA increases you can see very clearly that the reaction rate will increase therefore we will get curves like this. Now suppose we keep temperature constant CA0 constant and change or increase the residence time then we will get curves like this. In other words for a given one residence time it will be this for a higher residence time it will be this another higher time it will be this. Is that clear? Because simply by looking at this expression we can tell what is the direction in which the rate functions will move. Similarly when time temperature is increased we will see a behavior like this that at high temperature it will be like this at lower temperature it will be like this. So what we are trying to say now is that if we have a first order reaction we can measure rA as a function of time and then we can plot these to understand how the reaction rate is changing with time and once we do that then we can measure what is called as activity. What is activity? It is reaction rate at any time divided by reaction rate at 0 time. Now moment you do it in this form you will find that at 0 time all the activity at 0 time should be 1. So we can actually normalize the point that we can normalize the activity as 1 so that we can plot all the data in one curve. So that is the advantage of this normalization. So what we have tried to do is that we have tried to conduct an experiment and this experiment we have done in a CSTR. What is the advantage of doing this in a CSTR? We are able to maintain temperature to that level that we want and then we are able to change the flow rates or residence time and therefore we can measure reaction rate at under different conditions. Now when we do reaction rate at different conditions we said that we can normalize the reaction rate at 0 time and we make plots like this. What we are saying here is that if you normalize at 0 time we can say that the activity at 0 time is 1 it is a nomenclature activity at 0 time is 1 then in that case we can plot all of them using the same plot or we can plot differently as well that choice is ours. So what we are trying to say is that we can plot activity as a function of time of our different conditions. Now having done this we have activity as a function of time for different conditions. What are the conditions under which we have done the experiment? We have done the experiment for different times. We have done the conditions for different compositions. We have done the experiment for let us say for these two for the moment. So our model we need a model let us say that model is this that the rate at which the activity changes minus of d by d t of d a by d t is this r d and the r d is given by this function k d times a to the power of m c a to the power of m. We do not know that function so we want to find out what is m what is n what is k d that is our objective doing this whole experiment. Now suppose we find that when we make a plot of a versus time we get a curve like this the points like this we get a curve. Let us say this corresponds to an experiment corresponding to c a equal to some c a 1. Now we repeat this for another set of experiments in which we put c a equal to some c a 2 then we get let us say all these triangles. Let us say c a equal to c a 3 we get let us say all these stars. Now when we do experiments under different conditions let us say this is for one temperature then we can do for another temperature. Let us say this is very sharp a versus t some data like this. So what in essence we are able to do through these experiments is that we are able to get a versus t and because of this experimental data we are able to determine what is the value of m what is the value of n what is the value of k d at different temperatures. So what we have got from our experiments we find k d m n this. So our deactivation experiments essentially what is the experiment we have done r a at time t divided by r a at time 0 and made a plot of a versus time and we got data we got data at one temperature maybe this may be a temperature t 1 and this may be a temperature t 2 is that clear. So from the data of a versus t we fitted the curve fitted this curve to this to this model and found out k d m and n and then k d values at one temperature that is k d 1 similarly another temperature we got k d 2 another temperature we got k d 3 and so on. So in other words we had value of k d at different temperatures and m and n is this clear. So the whole object of doing this CSTR kind of experiments is that CSTR allows you to operate at constant temperature comfortably. And therefore we are able to make composition measurements comfortably we are able to make changes in the external environment comfortably. Therefore we are able to get lot of data which spans a great variety of compositions and temperatures. So that we can fit to this model and find out k d at different temperatures find out m and n and we think I mean most of the data is suggest that m and n values as remain reasonably consistent across all these conditions. Therefore we have a value of m a value of n and a value of k d for different temperatures. Now what we normally do is that we look for a dependence of temperature on dependence of k d on temperature and this is something that we have learnt from our early chemical kinetics that rate constants against temperatures typically have an Arrhenius kind of dependence and it is not unusual to find this kind of dependence. And then when you fit it to a model like this we find what is called as the activation energy for deactivation. So what do we get from these experiments we are able to find out because we know the value of k d at different temperatures because these measurements have been done at different temperatures. And we have found out the k d from our fitting we have plotted k d versus 1 by t and we find this kind of linearity and then from the linearity slope we find out what is e d the activation energy for deactivation. So cut this long story short what we have done is the following what we have done is we have done an experiment in a CSTR. We have done measurements at different compositions in different temperatures from that reaction rates at different temperature at different conditions. We have found out normalized that reaction rate with respect to 0 time reaction rate and found out activity as a function of time. From the activity as a function of time data we have fitted to a model like this and found out the value of m n and k d this k d is at different temperatures. And then we have plotted this k d versus 1 by t and found out the activation energy for deactivation. So that our deactivation experiments gives us what is called as m n and e d and k d 0. So these four parameters come from our deactivation experiments. Now the question I mean the reason why we have done all these things is to be able to understand how deactivating catalytic reaction systems perform number one. Number two more importantly can we ensure that in spite of deactivation our process our process runs as per design. Because after all we have designed it for so many tons per day and if it does not give us that many tons per day clearly our economics do not work out for us. And our process equipments that we have put in place are not performing to the optimality that we have designed for. Therefore our prime interest in looking at catalyst deactivation is to be able to understand them number one of course number two to be able to make good use of that information so that our designs are as per our requirements. Now we want to address that we want to address that question by just looking back at what we have done. What we have done if you recall a little earlier we said our reaction rate we did this for the case of a CSTR we wrote this for the case of a CSTR. We said our reaction rates can be described by K R C A 0 divided by 1 plus K R T R. So I put K R 0 here multiplied by A. So notice here the K R 0 times A is our K R or in other words this is our K R that we have been talking about. Now what we are trying to say here is that if I ask you I want to keep the left hand side constant irrespective of what happens to the catalyst. So what is the kind of answer that we can think of if R A has to remain constant which means what then only our designs will give us the production that we have designed for. So left hand side must stay constant which means what on the right hand side on the right hand side the term K R 0 times A should not change. So we should have K R 0 times A should be constant only then the reaction rate as a function of time will not change only then our equipments will deliver productivity that we have designed for. On other words what we are trying to say here is that in the case of a deactivating catalytic systems we recognize that the catalyst deactivate therefore activity will change. But is it something that we can do to ensure that K R 0 times A does not change. In other words can we see what we can do to this product K R 0 times A so that K R 0 times A does not change. A might decrease but which means what we are trying to say here is that can we increase K R 0 so that it is able to compensate for the loss of activity. This is the point that we would like to investigate understand and see how best we can integrate these ideas into a design. Let us go forward what we are saying now what we are saying now is that our reaction rate R A it is changing with time and this is something that we have to accept because as the process is running due to variety of factors on the catalyst sintering may be may be coke deposition may be may be poisoning we do not know it could be different from different situations the activity is on the way down. So, in order to be able to anal effect of decrease in activity what is being suggested is that if you can keep this K R times A constant on other words if this product is kept constant then very clearly as you can see here when you can keep this K R 0 times A constant or we can keep K R constant then clearly the right hand side does not change and therefore left hand side does not change therefore our productions will be as per design. In other words what we are trying to say here is that the strategy for managing deactivation should be keeping this K R times A constant moment we do that we have a way of managing deactivation. Now how do we do this what is being suggested is if we assume that this A is decaying exponentially which means what we are saying that d A by d t d A by d t equal to K D times A therefore A is decaying exponentially if A decays exponentially then we can write K R times A equal to K R times E d t therefore this must be held constant how do you keep it constant we keep it constant by recognizing that K R which is rate constant which is an Arrhenius K R 0 times exponential of E by R t and K D is given by this expression therefore K R times A can be written as what I have written here A R 0 exponential E by R t and exponential K D t where this K D has an Arrhenius dependence as mentioned here. So if you look at the last line here what we are saying here is that well this term decreases with time because of K D this term increases the temperature because it is a very strong function of temperature. If this product has to stay constant it only means that the effect of temperature on increasing this quantity more than makes up for the effect of decrease in this term because of effect of temperature let me state this once again. We know that as temperature rises as t rises K D increases and also as t rises K R also increases but what seems to be happen in real life is that increase in K R increase in K R 0 K R 0 because of temperature is generally much much larger than the effect of K D increase in the activity of the catalyst. On other words what we are saying is that when we rise take temperature from x 1 to x 2 the increase in this first term you see this in first term increases second term also decreases you know because of rise in temperature but the increase in the first term compensates for the decrease in the second term so that the product is kept constant. On other words what we are trying to say here is that deactivation data that we gather from our experiments gives us deactivation parameters. Similarly reaction rate data in the absence of catalytic deactivation gives you reaction rate parameters so we have reaction rate parameters we have deactivation parameters. What is being said is that when you increase temperature the rate constant in the reaction rate system increases similarly when you increase temperature the rate constant in deactivation also increases but what is important to recognize is that the effect of increase in the rate constant because of increase in temperature compensates for the decrease in the activity because of the increase in deactivation rate constant so that the product can be held constant so that our designs ensures the reaction rate does not change its time. Moment reaction rates do not change its time then our designs gives us production that we have designed for and therefore our economics and our optimalities etcetera that we have taken into account none of them are affected. We will take an example to illustrate how this is actually achieved and some of the issues that arise from some of these features. So, let me illustrate this through an example what I want to do now is to illustrate interesting example of catalyst deactivation that we might have to encounter in process design process evaluation process optimizations and so on. So, here is an instance of an illustration where you have a PFR is a plug flow reactor I call it PFR it has a catalyst which deactivates. Now, here is a separator what do we have we have a plug flow reactor and a separator now we have this reaction A going to B. So, there are A going to B you know it is a reversible reaction therefore every reaction is reaction 1 and reaction 2 and the data basic data is given activation energy for reaction 1 is given equilibrium constants are given at 2 temperatures and so on. Now, the situation of our interest is the following catalyst deactivates. So, in a process when catalyst is deactivating that after some time the activity of the catalyst is quite and not satisfactory to be able to make good use of the downstream equipment therefore you might want to think about disposing it and buying a fresh batch of catalyst catalyst being expensive there are processes available in which catalyst can be regenerated. And you might want to see whether a regeneration process can be thought of. So, that you can regenerate and use that catalyst here is an instance of a problem in which problem statement says that this catalyst deactivates and then this the data on fresh catalyst and spent catalyst is available. And when you regenerate a spent catalyst it has certain types of activity. So, what is important for us in this particular problem is to find out whether the regeneration process that we have put in place is satisfactory for our economics to be satisfactory for which some data is given what is the data given the data says the following it says it says the following that here is a catalyst here is a catalyst fresh catalyst spent catalyst regenerate catalyst the performance is given. Now, we are asked to judge whether the regeneration process that we have adopted is good enough to be purchased. Now, whenever you want to evaluate we must say what is the criteria on which we will judge the regeneration process the kind of criteria that is suggested in this particular exercise is the following it says the following it says the catalyst that we are using in this reactor is proprietary the catalyst undergoes deactivation. And the regeneration process is available and it says that we can accept the regeneration process if the regenerated catalyst if the regenerated catalyst has 90 percent activity of fresh catalyst what are we saying what we saying is that we will accept the regeneration process if the activity of the regenerated catalyst is 90 percent of fresh catalyst. Alternatively if the activity is regenerated catalyst is 2.25 times the activity of the fresh catalyst of the spent catalyst also we will accept the process. So, what is being said is the data that is in front of us we have to evaluate the data on fresh catalyst spent catalyst regenerated catalyst with respect to the 2 criteria mentioned here which means what we must judge the activity of regenerated catalyst estimate the activity based on the data given and see whether it is 90 percent of the fresh catalyst or you can check the activity of regenerated catalyst and see whether it is 2.25 times or more than the activity of spent catalyst. If either of these two criteria are met we are willing to accept the regeneration process. In other words what we are trying to say here is that we have to deal with deactivating catalyst in our industry. So, we need a systematic procedure to evaluate process data. Now, here is an instance of process data in front of us all the data is given to us. Now, we have to set up the design equations to see how shall we evaluate the performance of the catalyst under different conditions that is given in front of us. So, this is what we want to do now this is what we want to do now that means the following that we have a catalyst which is fresh performs like this we have a catalyst which is spent which perform like this we have a regenerated catalyst which performs like this we have to see the 3 data and advise whether we should buy the regeneration process if so why if not why not this is the kind of questions that we want to answer. Let us get ahead with this problem let us get ahead with this problem how do we solve this problem how do we solve this problem. Now, to be able to solve the problem in reactor design reactor optimization reactor evaluation so on we will have to develop equations that will describe the performance of this equipment this is all once we have the equation that describes the performance then we can answer various kinds of questions. So, we will go through the algebra that is required to set up the design equation for understanding the performance of this equipment which is a P F R containing a deactivating catalyst let us see how we can do this. So, to do this so what I have done is of course I have done this drawn the same figure once again may be. So, we have this catalyst I have denoted this as 0 1 2 3 4. Now, there are few things that we can recognize very quickly without losing too much time that this is a separator separator is the reactor. Now, this is pure pure B now since this is pure B and this is C A 0 is coming in. So, we can say that since B is fully recovered this stream is only C H 0 this common sense because B is fully recovered our reaction is A going to B and B going to A therefore, this is C H 0 something very elementary nothing very complicated. Now, we can write material balances we can write material balances at various points. So, I have done just very elementary. So, this is not difficult to do. So, please notice here this is 0 this position 1 position 2. Now, we are defining conversion conversions are defining with respect to conversion conversion y defined with respect to position 1 what do you mean what we mean is that if I want conversion at position 2 and our reference is position 1 then I say F A 2 equal to F A 1 multiplied by 1 minus of y 2 that is what I am saying that is what I have written F A 2 is F A 1 multiplied by 1 minus of y 2. Why is why is that that is how we are defining conversion here is that clear. Similarly, if I ask you what is F B 2 that means what is the amount of B at position 2 we say the F B 2 is F A 1 times y 2 because whatever A is disappearing it becomes B. So, F A 1 and F A 2 and F B 2 are F A 1 1 minus of y 2 and F A 1 1 minus of y 2. If I ask you please look at this data it says composition at reactor exit is 25 percent. So, if composition is 25 percent means what this is a gas gas gas reaction therefore, if I find the mole fraction this is the number of moles that they exit. So, if this is this is mole fraction y 2. So, what what is this equal to say this is not y 2 this ratio is equal to this 1 minus F A 1 1 minus of y 2 divided by F A 1 you can see this 2 is F A 1. So, this is mole fraction. So, mole fraction at 2. So, mole fraction at 2 is 0.25 that is that is the data given please this data is given this is given here. Therefore, you can find out from this that y 2 value at reactor exit is 0.75 what I will be saying what we are saying is that if I give you mole fraction at position 2 I can calculate conversion simply from this relationships therefore, if I say mole fraction is 0.25 then conversion is 0.75 or in other words what we are saying is that in this data mole fraction at the reactor exit is 25 percent means our conversion is 75 here. Here it is 32.2 means our conversion is 67.8 in this is 19 percent means our conversion is 81 percent. So, this is the kind of data that is given that is what we are saying. So, what we have tried to say is we have found a value of y 2 because that is given to us in the from the data we have converted the data into value of y 2. So, that we can go ahead and do in the evaluation. Now, let us go forward what we is now saying is what we are now saying is all right now we know how to make good use of the data only we have to write various simple expressions for various flows. So, that we can understand what is going on in the equipment. So, please once again look at our look at our figures that we can understand what we are saying here is that F A 1 is F A 0 plus F A 4 please see here F A 1 this is F A 1 F A 1 is equal to what is coming at 0 and what is coming at 4. So, F A 1 is F A 0 plus F A 4 what is F A 4 we know that F A 4 is only F A 2 because all the 2 here will come to 4 what is F A 2 we just now said F A 2 equal to F A 1 times 1 minus of y 2 we said that just now. Therefore, and similarly what is what is coming here all the 2 that is coming here is also going to come here. Therefore, F A 2 is also equal to F A 4 this is also equal to F A 4 correct. Therefore, what is coming at position 1 is F A 4 which is F A 1 times 1 minus of y 2 plus F A 0. Therefore, simplifying you find F A 1 that means the molar flow at position 1 is simply F A 0 divided by y 2. What we are saying is that from elementary material balance we find out what is coming at position 1. Now, what is the concentration of A at position 4 we said all the B is fully recovered and therefore, the concentration at 4 is same as C A 0. So, what we are saying now is that C A 0 is C A 4 is C A 0 and therefore, we can calculate now what is the flow at position 4 what is flow at position 4 by definition is F A 4 divided by C A 4 molar flow divided by concentration C A 4 is C A 0. Therefore, it is F A 4 means F A 1 times 1 minus of y 2 we have just now said that C A 4 is C A 0. Therefore, we find the molar flow at position 4 is V 0 times 1 minus of y 2 divided by y 2. So, what we are saying is that molar flow at position 4 we have found out as V 0 times 1 minus of y 2 divided by y 2 that is V 4. So, what is the molar flow at position 1 V 0 plus V 4 that is what we have done V 1 equal to V 0 plus V 4 when you do that we say it is equal to V 0 by y 2. In a sense what we have done is that we have tried to put all the terms that are required for us in terms of quantities that we know from our experiments y 2 is known from our experiment V 0 is known from our experiment from our data V 0 is given y 2 is given. Therefore, we should express all numbers in terms of what we know that is what we are trying to do in this algebra. Having done that let us let us go forward what we are saying now what we are saying now is that we know from our material balance that concentration at 1 is simply C A 0 because this is also C A 0 we know that C A 1 is C A 0 we know and what is the concentration at any position see after all if you want to write the design equation for a for a equipment we should know concentration at any position. Therefore, we have what is C A by definition is F A by V what is F A F A 1 times 1 minus y at any position what is V since there is no change in volumetric flow V at any position is V 1 which is coming in or in other words what we are saying is that in this in this PFR the flow at any position is same as V 1 because there is no volume change in the reaction and what is the concentration at any position F A F A divided by V that is what I have written here. Therefore, concentration is F A by V which is C A 0 times 1 minus a y for component A and for B it is C A 0 times y. Therefore, what we have tried to do is that we are now able to express concentrations in terms of y now we have expressed all the other flows in terms of known quantities. Therefore, we are now in a position to be able to write the reactor design equation that is required to understand what is going on. So, this is what I have done here. So, our reactor design equation that D by D V of B is R B therefore, I have written F B we know that what is F B by definition is F A 1 times y I have written that here. So, F A 1 times D by D. So, I put all the terms and notice here notice here that I have put one inside brackets to indicate that this refers to reaction and not position. So, this is reaction one reaction two and this term is a catalyst activity. What we are saying here is that that the reaction rate function R B depends on k 1 alpha times C A minus k 2 alpha times C B. Notice that the catalyst accelerates both the forward reaction and the reverse reaction. Therefore, we find F A 1 D by D V is given by this expression. So, what we have now been able to do is that we are able to write the design equations for the reactor in terms of y see D by D V in terms of y and all other terms excepting y are known that is the interesting point that means we are able to write the design equation in terms of quantities that we know and the quantities we would be able to measure in our experiment. So, this differential equation in this form you can see here F A 0 by y 2 D y D V equal to the right hand side. Now, we are able to integrate and then get the results in the form in which we require. When we go further we recognize that we are able to simplify the equations in terms of numbers that we know B 0 we know y 2 we know. Now, D y D V we can put in terms of beta beta is simply beta is simply k e plus 1 by k where k e is the equilibrium constant. Therefore, the integrated form the integrated form of the equation which describes the performance of our equipment is given by this. Now, see what we have done what we have done is that based on our understanding of reaction engineering we are able to write a design equation which is able to describe our equipment. Notice here that k 1 is rate constant for forward reaction alpha is the activity the catalyst v is the volume of catalyst that we have used in our experiment v 0 is the volumetric flow at the inlet y 2 is the y 2 is the extent of reaction at the reactor exit beta is simply k e by k e plus. So, all the terms we know see all these are given to us in the experimental data. The experimental data you can see here very clearly all these are given to us temperature is given v 0 is given this is v 0 this is v 0 and what is this this is actually this is 1 minus of y 2 multiplied by 100. So, everything is given understand 1 minus of y 2 we will not write it like this that means if it is 25 percent this 0.25 1 minus of 0.25 is y 2 we know all that. So, what I trying to say here is that the data that is given can be now utilized to understand how the system is performing to understand what is the activity of the catalyst because our question is you are supposed to check the activity of the catalyst and compare with the criteria that is given to us. So, this equation helps us to find out the activity of the catalyst because all the rest of the quantities are known and that is the important point of being able to formulate your mathematics. So, that you are able to represent the physical system that is in front of you. Now, that we know all this we are now in a position to make an evaluation of the data that is given to us. So, let us evaluate the fresh catalyst fresh catalyst your y 2 is given as 0.75 y 2 is given as 0.75 it is given as 25 percent therefore, 0.25 1 minus 0.25 is 0.75 now beta is K e plus 1 by K e beta values are given you can see here our our equilibrium constant K e at 450 is given as 8.5 therefore, 8.5 plus 1 is 9.5 therefore, beta is 1.17 at 450. So, we are able to calculate the value of K 1 alpha v at 450 because all the terms on the right hand side all the terms on the right hand side is given v naught is given y 2 is given beta is given everything is given on the right hand side you can substitute and calculate the value of the right hand side which is comes out to be 0.847. So, fresh catalyst K 1 alpha v given as we have found out as 0.847 similarly, we can do for a spent catalyst spent catalyst data is also given. So, when I do when you do the same thing for spent catalyst we find the value for alpha K 1 v at 477 that turns out to be 0.750. Because y 2 is 0.322 is given and then this value of beta where is beta values beta value of calculated because of the from the data given. So, accordingly we find that the values are 0.678 you know I have not given the detail here 0.678 is y 2 and beta is this K e by K e plus 1 K e plus 1 by K e which is beta at 477 it turns out to be 7 by 6 I have done this calculation the details I have not given here you can check it for yourself. So, what in essence it means is that at 477 spent catalyst our value of alpha K 1 v turns out to be 0.75 for spent catalyst it is for spent catalyst 0.475 and for fresh catalyst we just now said it is 0.847 we said that just now just now we said that fresh catalyst 0.847. Now, we have to do the same thing for the regenerated catalyst correct now before doing that let us just go forward a little and find out what is the activity of spent catalyst because this notice here that this at this alpha is at 477. So, we always like to express everything in terms of fresh catalyst. So, we want to find out the value of alpha at 477, but we want to see what is alpha value with respect to our fresh catalyst to do that I have just found out K 1 alpha v divided by K 1 alpha v 450 this ratio is 0.75 or 0.4 this turns out to be 0.88. So, what we are saying is that the value of alpha at 477 divided by value of alpha at 450 is actually what K 1 K 1 at 450 I just multiplied K 1 at 450 divided by K 1 at 477. So, that I am able to find out what is the activity of catalyst at 477 in relation to activity at 450 and that turns out to be 0.33. What we are trying to say here is the following what we are trying to say here is that in this experiment in this catalyst evaluation we are given a spent catalyst we want to find out what is the activity of spent catalyst. Now, we find that the spent catalyst is working at 477. Therefore, we want to check the activity of spent catalyst at 477 and we want to compare this is the activity of the catalyst at 450 and to do that is when we have done this through this manipulation we have been able to find out what is the activity of spent catalyst at 477 in relation to the activity of the catalyst at 450 by multiplying by the appropriate reaction rates at the two temperatures that is what I have done. So, what we find what we find is that the activity of spent catalyst turns out to be 0.33. So, what we are saying here is that the spent catalyst with respect to front catalyst is quite inactive and the reason why we are looking at the way of regenerating the catalyst let us go forward and look at the activity of regenerated catalyst. Now, the regenerated catalyst data is given at 450 that is the right hand side all the data is given when you calculate we find that k 1 alpha v at 450 turns out to be 0.727. Therefore, we can compare the k 1 alpha at 450 with fresh catalyst that turns out to be 0.858. In other words what we are saying now is that the activity of regenerated catalyst in relation to activity of fresh catalyst both are at same temperature it is not 0.9 we said we can accept the regeneration process if the activity is 0.9 or higher now it is only 0.858 and therefore, criteria 1 is not satisfied. So, criteria criteria 1 is not satisfied. So, that is the first point that we recognize now let us see about criteria 2 what is criteria 2. So, criteria 2 says criteria 2 says the activity of regenerated catalyst is 0.858 and activity of spent catalyst is 0.33 and criteria 2 says if it is 2.25 times the activity of spent catalyst it is ok. If you multiply 0.33 by 2.25 you find that it is less than 0.858. Therefore, at criteria 2 is satisfied on other words what we are trying to say here is that out of the 2 criteria criteria 2 is satisfied therefore, we can accept the regeneration process. So, by going through this whole exercise by going through this whole exercise what we have tried to bring out to your attention is that in a deactivating catalyst catalyst will deactivate. Therefore, you are increasing the temperature of operation of the catalyst and therefore, the spent catalyst when we are discarding it the temperature is 477. What was 450 initially? We had to rise it up to 477 to be able to take care of deactivation. On other words what we are trying to put across to you here is that in deactivating catalytic systems increasing the temperature of operation. So, as to annul the effect of catalyst deactivation is the strategy generally employed, but this strategy may not be successful for the entire duration of operation. You may have to stop it after some time when the activities of the spent of the catalyst is unsatisfactory. You will have to take it out and discard it or replace it by an appropriate regeneration process. Thank you.