 So, we will be looking at plug flow recycle reactors today is something that we have spoken about the earlier recycle. So, here is an instance of a plug flow reactor in which this material is taken out and bring put back. Now, why do we do recycle? Of course, recycles are common in process industry because the reactions do not go to completion. So, you separate and recycle. So, the separation part is not taken into account yet, but the fundamentals of recycle is what we want to understand and set up the basic approach to looking at recycle reactors. Now, just to make it a little easier what I have done here is I have taken position 0, 1, 2, 3 and 4 to indicate different positions in the equipment. And I have taken two instances where there is no catalyst in the reactor and second instance where there is a catalyst in the reactor. Now, case one when there is no catalyst in the reactor for example, if you take recycle ratio as 4 I will just put down what are the flows going through at different positions. Let us just go through it for instance I have taken F A 0 as 1 and therefore, F A 3 should also be 1 because in whatever comes in must go out. I have taken R as 4 therefore, F A 4 is 4 times F A 3 therefore, R is F A 4 is 4 and there is no reaction therefore, F A 1 should be F A 0 plus F A 4 which is 5. So, F A 0, F A 1, F A 2, F A 3, F A 4 this shows the number of units that flows through the equipment when there is no catalyst that means there is no reaction. Now, let us say there is an extent of reaction of something like x equal to 0.4 which means out of one unit that is coming in at position 0, 0.4 reacts therefore, the balance is 0.6. So, if balance is 0.6 therefore, at position 4 it should be 4 times 0.6 therefore, it is 2.4 if it is 0.6 4 times 0.6 is 2.4. Now, at position 2 it should be F A 4 plus F A 3 is F A 2 therefore, it is 2.4 plus 0.6 which is 3 and F A 1 it should be F A 0 plus F A 4 which is 3.4. So, this tells us based on material balance what are the flows at different positions. Now, what we would like to see is since there is no reactor between position 2 and position 3 common sense tells us that the value of conversion that we calculate must be the same at 3 and 2 this common sense. So, how do we make it this is the value of conversion to be the same we recognize that conversions are defined on the basis of a reference every time we have a reference basis of which we define conversion. Therefore, here if you take our reference as what flows into the equipment in the absence of reaction which is 5 units then we notice that conversion at position 2 becomes x A equal to 0.4 3 divided by 5 is 0.6 1 minus 0.6 is 0.4. So, common sense tells us that the reference on which we should define conversion inside the loop must be the units and the material that flows into the reactor in the absence of reaction or putting in terms of recycle ratio F A 0 is when there is no recycle there is R is 0 and when there is recycle then it should be R plus 1 F A. So, the basis for defining conversion inside the loop should be R plus 1 F A 0 if there is a different definition then our formulations will change, but the important thing is that as we will see as you go along our formulations do not change much. If you use this definition you find that when you put R equal to 0 you get what we would have gotten when we did not have any recycle. So, that is the advantage of this formulation, but it does not matter as long as you are very clear about the definition it does not matter. Let us just take it as in the absence if within the recycle loop we should have R plus 1 F A 0 as the basis. So, that when we go outside the loop R is 0 our basis becomes F A 0. So, it is consistency between inside and outside in the absence of which means what enters the reactor in the absence of reaction just to make it clear I said in the absence of catalyst there is no reaction therefore, what enters at 5 at 1 is 5 units that is how I explained it. So, I have calculated what is the conversion at 1 conversion at 1 should be 3.4 divided by 5. So, what is 3.4 divided by 5 1 minus 0.32. So, it shows that conversion here is 0.32 and here it becomes 0.4. So, important thing to recognize in recycle plug flow recycle is that per pass conversion going from 0.32 to 0.4, but overall conversion is 0.4 overall conversion 0.4 per pass conversion is 0.32 to 0.4 only 0.08. So, per pass conversion is low, but overall conversion is high and there are certain advantages for this we will look at this as we go along is there. So, I have just summarizing what we have said already that flow entering the reaction the absence of reaction with R plus 1 F A 0 therefore, the basis for defining conversion within the recycle should be R plus 1 F A 0. Therefore, value of x 1 is 0.32 x 2 is 0.4 and x 3 is 0.4 there is consistency between inside and outside the loop. So, we will go further now then I said we can calculate per pass conversion in this particular case for example, what is per pass conversion F A 2 it is a reverse F A 1 minus F A 2 I should have it is F A 1 minus F A 2 this per pass conversion. So, it is 3.4 minus of 3 this is how it will come out to be 0.11. So, now that we have said that the basis for defining conversion is R plus 1 F A 0 we can write this stoichiometry just like we have written stoichiometry before we can write stoichiometry again. That means, what comes in to the process is R plus 1 F A 0 R plus 1 F A 0 R plus 1 F C 0 R plus 1 F D 0 R plus 1 F I 0 therefore, the total is R plus 1 F D 0 when R is 0 it is same as what we had before. Now, what goes out since A is our reference R plus 1 F A 0 times 1 minus of x A is the flow rate of component A at any position inside the loop inside the loop it is at any position. Similarly, for component B it is R plus 1 time the F B 0 is coming in and this is the amount that is reacting for component C or R plus 1 time the F C 0 entering and this is the amount that is reacting. Similarly, component D similarly, component I so that when you add up all this the total moles of components at any position is R plus 1 F D 0 plus R plus 1 F A 0 delta or in other words you find that F T by F T 0 notice here what is this F T 0? F T 0 is actually what is entering outside the loop F T 0 is what is entering outside the loop you can see the way I have formulated it F T 0 is what is entering outside the loop F T is what is inside the loop and therefore, this term is exactly the term that we had earlier only thing it is multiplied by R plus 1. So, when you put R equal to 0 it is exactly what we had before the consistency is what is important outside the loop and inside the loop the consistency is simply put R equal to 0 you get what you got before F T is any position inside the react inside the recycle loop and we are talking about inside the recycle loop. So, that when you put R equal to 0 you find you get the same result that we got before at this point in the absence of reaction this is what is entering what we are trying to say is that when we are trying to look at the recycle system our basis should be R plus 1 times what enters the reactor in the absence of reaction I mean this is the fundamental difference that we must recognize towards recognizing the basis because we need a basis to define conversion and that is why we said this is the basis on which the reaction takes place. So, basis is R plus 1 F A 0 therefore, all our conversions are assuming that R plus 1 F A 0 is going through the react and how did we say come to this R plus 1 F A 0 we took the example to understand this that when there was no reaction 5 units is entering and therefore, our basis should be 5 units and we also found out that you know only when we do that there is consistency between position 2 and position 3 same value of conversion. So, that consistency will not be there if you do not define it this way. So, the important thing is only the choice of the references important in recycle let us look at the example the example that we have let us go back to the same example and ask all these questions let us ask all these questions when there was no catalyst we said 5 units is going through at 0.1 when there is a catalyst we find that at 0.2 it is 3. So, the conversion at 0.3 is 0.4 conversion at 0.3 is 0.4 therefore, between 2 and 3 you see consistency this consistency is fundamentally required for us to you know choose the reference the reference is to be chosen to get this consistency between 2 and 3 and that we have got by recognizing that 5 units was flowing at 0.1 when there was no reaction and F A 0 is what in the absence of reaction correct the meaning of F A 0 is what enters the process that means when there is no reaction at that point correct. So, in the same way what enters at position 1 in the absence of reaction is 5 units and in our normal class it is R plus 1 F A 0 is that clear answer to your question is we said between 0.2 and 0.3 we need consistency that means the value of conversion at 3 and 4 should be 2 and 3 should be the same because only then as you move from 2 to 3 the value of conversion does not change that is the consistency we require in our definition or in our choice of reference and we said that reference choice is R plus 1 F A 0 only then we get this consistency is that clear. So, consistency comes only when we choose our reference properly and reference is what enters the equipment in the absence of reaction and that makes sense also what enters the equipment in the absence of reaction should be our basis anyway correct. See if so much of sulphur dioxide is entering that is the basis so that we can find out how much of sulphur dioxide is coming correct. So, in the same way between 0.2 and 0.3 we must have consistency in the value of x that is why 5 units is chosen as the reference that is why R plus 1 F A 0 is chosen as reference when you put R equal to 0 it becomes our conventional definition that we have been using all the time. So, having said this let us go further now. So, what are we saying now the total moles that is coming out of our let me draw the recycle once again this is our recycle system. So, total moles at any position so F T is at this point F T is R plus 1 multiplied by 1 plus Y A 0 this is the position 0 Y A 0 refers to this position this is 1 this is 2 this is 3 this is 4. So, this Y A 0 refers to position 0. So, delta A refers to our stoichiometry X A is conversion at this point which is consistent with between 3 and 2. Now, what we are now saying is that if you want what is the conversion at position 1 if people ask our answer is at position 1 as per our definition it must be R plus 1 F A 0 1 minus of x 1 that is our definition our material balance tells us it must be F A 0 plus F A 4 what is F A 4 it is R times F A 0 times 1 minus correct because we are recycling R times 3. So, F A 0 plus R times F A 0 1 minus this is this is F A 3 we know that this is F A 3 correct. So, this is the statement of material balance therefore it gives you this relationship that x 1 is R by R plus 1 x 3. So, what does it mean it means that per pass conversion is quite low when overall conversion can be quite high depending on the choice of R depending on the choice of R. If R is very large what does it mean x 1 becomes very close to x 3 which means what the per pass conversion in the equipment is very low. So, you have very little conversion here between point 1 and point 2 per pass is low but the overall can be quite high let us go forward. So, what is per pass conversion if I say our position 2 this is our position 2 0 1 2 3 and 4. So, what is per pass conversion at 2 why F A 1 minus F A 2 divided by F A 1. So, it comes in terms of x 2. So, when R is large you can see that per pass conversion is quite low. So, if you ask yourself when would you be interested in as low per pass conversion when you do not want a very high temperature rise. So, on other words if you want to operate an equipment isothermally because you want to do chemical kinetic measurements. That means you want isothermal data you can get isothermal data in this kind of environment. So, if you have a catalyst which you have prepared you want to evaluate that catalyst and determine the chemical kinetics of that catalyst you can do measurements at isothermal conditions in a recycle kind of device. And it is relatively easy because it is catalyst does not get spoiled you know because it is it is not moving and so on. There is lot of advantages. So, let us look at the design equations. So, what is our design equation now you have to take a material balance in a volume delta v where the flow rate at that point is r plus 1 f a 0 1 minus of x a. This is our psychometric table this how we have defined and our material balance tells you that d f d v equal to r a this is something that we have done this is for a p of r. So, only thing that has changed now is that our flow at this point is r plus 1 f a 0 times 1 minus of x a correct. This equation d f d v equal to r there is no change, but value of f a at that point is r plus 1 f a 0 1 minus of x a because our basis for defining conversion is r plus 1 f a 0. Therefore, the design equation for a recycle reactor becomes what is mentioned here v equal to r plus 1 f a 0 integral x 1 to x 2 we are going for 1 to 2. So, let us look at this at some detail. So, design equation is r plus 1 f a 0 going for 1 to 2 where value of x 1 is r by r plus 1 x 2 and x 2 equal to x 3 as per our definition. Now, let us ask what happens to this when value of r is large what is the value of this integral what I have done limit as r tends to infinity I have just tried to find the value of this integral when r is very large I have just said d x a is simply x 3 minus r x 3 by r plus 1. So, this integral and I evaluated the denominator minus of r a at mean value of x what is the mean we have to take between x 1 x 1 plus x 2 divided by 2. So, this x mean refers to x 1 plus x 1 plus x 2 divided by 2 when r tends to infinity let me repeat we are looking we are trying to find out what happens to this when r is large this design equation what happens when r is large. So, I am taking limit r tends to infinity of r plus 1 multiplied by this integral this integral I have written as d x a I have written as x 3 minus r x 3 by r plus 1 denominator written as r x at x mean which is x mean means x 1 plus x 2 divided by 2 is this clear what we are saying. So, let us look at this what so x mean I have just calculated the value of x mean you can say here the limit as r tends to infinity of this it goes to x 3 you can do it yourself it is fairly elementary. So, the limit of this as x tends to infinity x 3 so that this whole integral now becomes x 3 divided by value of r a at x 3 is that clear what we are saying is this mean is at x 3 and you can see here this whole thing becomes x 3 only. So, numerator becomes x 3 r plus 1 gets cancelled therefore, then this whole thing becomes value of r a at mean value which is x 3 is that clear what we are saying this limit as r tends to infinity this whole thing becomes the numerator becomes x 3 and the denominator becomes value of r a at x 3 because the mean at r tends to infinity becomes x 3. So, what are we saying what we are saying is that when r tends to infinity this whole integral the whole thing actually becomes the equation becomes same as a stirred tank this integral this integral here when r tends to infinity by going through this procedure we have shown that it becomes the equation becomes same as a stirred tank. When r is large the recycled reactor equation becomes that of a stirred tank. So, what it means is that stirred tank is an instance of a plug flow recycle reactor with infinite recycle with infinite recycle that is when it becomes plug flow is that clear that is the point that is being made. Now, let us look at the other instance what happens when r is 0 when r is 0 x 1 becomes what is x 1 r is 0 x 1 becomes 0 and therefore, this equation becomes that of a PFR which we have talked about. So, you can see the consistency when r is 0 it becomes same as what we have done already when r is infinity it becomes another limit of what we have already done. So, in both cases it tells you that recycle is between a plug flow recycle plug flow reactor and a stirred tank. So, it tells you that recycle is a device which makes you achieve any property between plug flow and mixed flow that is what it says that our definition makes it possible to do this. So, let us plot this and see what it means. So, when you make a plot of 1 by r a versus x see 1 by r a versus x it means what it is the property of a reaction see 1 by r a comes from our experiment. So, we this data comes from our experiment. So, this curve comes from our experiment when recycle is small means what x 1 is what x 1 is r by r plus 1. So, x 1 is when r is small then you have this large area when r is large the very small area you can see here. So, r is very large it becomes a point it becomes a point. So, you can see when you increase r when r is 0 it becomes PFR you can see here when r is infinity it becomes a point which is CSTR you can see understand. So, going from 0 to infinity you are able to get various properties between mixed flow and plug flow. Now, let us just see quickly how does a first orders let us say you conduct a first order reaction in a recycle plug flow reactor let us say. So, what is our design equation. So, this minus of r a I have replaced it by a first order representation which is k times C a 0 times 1 minus of x a correct. So, x 1 is r by r plus 1 x 3 correct. So, I have just integrated and put them down. Therefore, our plug flow recycle for a first order reaction with recycle ratio r this is our integrated form of the design equation when it is a first order reaction. Beautiful question, but that answer I thought we have already done, but let me do it again now that you ask me I want to do this again I will do this here. So, your question is what is C a basically what is C a C a is what F a divided by volumetric flow what is F a r plus 1 F a 0 into 1 minus of x a correct. What is v let us say it is a gas phase reaction v by v 0 is F t by F t 0 t by t 0 z by z 0 p 0 by p and we now showed just now all this it is 1 this is equal to r plus 1 into what did we show here r plus 1 into 1 plus y a 0 x a 0 into 1 plus y a we have shown just now if you recall that means in the denominator you will have this term r plus 1 times 1 plus y a 0 x a times delta a. Now, I have taken the instance of a going to products therefore, all these are not there this cancels off therefore, it becomes C a 0 times 1 minus of x a v becomes v 0 times this r plus 1 cancels off this becomes 1 therefore, it becomes C a 0 times 1 minus of x a on other words inside the recycle loop the concentration dependence of x does not change because the formulation we have taken is like that that is the importance of the formulation that formulation in this formulation the form of the C a remains the same that is the advantage. If you take any other formulation then it is little bit it is little messy it is still give the same correct result only, but this formulation makes it very elegant is it shall we go forward. So, what are we saying what we saying is that if you have a recycle reactor then the integrated representation for a first order reaction is given by this is that clear. So, which means what if you are given a recycle ratio see here what are the unknowns k is unknown r is unknown x 3 is unknown tau r is unknown. Now, out of 4 if you are given 3 you can calculate the 4th that is the kind of we can have a design problem the operation problem whatever problem apparently appropriately you will have to put your numbers here to find out what the numbers that you are desire. Now, in order to understand the importance of recycle reactor in a larger context let us look at a small example. The example I have taken is an n tank sequence. So, what is an n tank sequence you have stirred tanks 1 to 1 1 to n in which I have taken a reaction which is first order I have taken a first order reaction which is r a is taken as minus k times c a a goes to products. Therefore, there is no volume change for which I have written the material balance input output generation equal to 0 is that clear. So, I have put the generation term here. So, c a 1 becomes c a 0 divided by 1 plus k tau 1 similarly, c a 2 equal to c a 0 divided by this. Let us just take this analysis forward a little because it is important what we saying is that c a n becomes c a 0 divided by k tau by n to the power of n. How does it come? It comes from here c a n from here c a n becomes like this number of terms tau 1 tau 2 they are all considered equal to tau divided by n that is the assumption. Therefore, you get this kind of that means the value of concentration at the end of the n th tank is c a 0 divided by this term. Now, we can do this manipulation I have just elementary manipulation I have just put this in this form in this form where k times tau n becomes n times 1 minus x to the power of x n to the power of 1 by n minus 1. It is a very elementary manipulation I have done because c a 0 by c a n is 1 minus x to the power of n common sense correct. Therefore, what it gives you is that a n tank sequence an n tank sequence the conversion at the end of the n th tank is related to system parameters like this. Now, let me just put this in the context of our interest what is our interest? Our interest is to understand the three types of equipment we have learnt so far. We have learnt this is a P F R we just now set up the equation for a recycle we set up this is a recycle reactor. Now, just now we talked about an n tank sequence where we got this relationship which I have put down here. So, you have a case of P F R where there is no mixing you have the case of a recycle reactor with R as the recycle ratio. Here the case of an n tank sequence where you have k tau n is the parameter k tau R is the parameter here k tau P as the parameter here. Now, what has changed from here to here from here to here what has changed is that it is a recycle we have put in a recycle. What has changed from here to here instead of a no mixing case it is infinite mixing case is an instance of n tank n tending to infinity is a huge amount of mixing or when the number of tanks is small it is intermediate mixing see when you go to n equal to infinity we have shown it is same as a P F R we have shown that when R equal to 0 we said it is same as a P F R. So, R equal to 0 is equivalent to n equal to infinity or in both cases R is a measure of mixing n is a measure of mixing because n is able to give you property between plug flow to mixed flow when n is infinity it is plug flow when n is 0 it is a single state tank infinite mixing. So, what means is that when you go from a P F R to a recycle reactor to an n tank sequence you are able to take into account various grades of mixing that can occur in your equipment which is what happens in the real life in real life you have to deal with mixing it may not be perfect it may be imperfect. So, well recycle reactor or value of R R as a parameter and similarly n as a parameter is a way of accounting for non idealities due to mixing in your equipment. So, to summarize this let us just look at an example let us say you have you have an arbitrary vessel what is this arbitrary vessel we do not know what it is. So, we are doing an experiment in which you are putting F A 0 you get a conversion of X A I ask you now what is this is the plug flow reactor what is it we do not know what we do now we simply make a plot now we can make a plot of this function we can make a plot of this function. So, I have done this here I make a plot of this function recycle reactor function here and I made a plot of the n tank sequence here. So, this is function relating n tank sequence this is for a recycle reactor as X versus R as R increases which means conversion decreases as R increases that means R is 0 is P F R R infinity C S T R conversion will keep decreasing n tank sequence X n infinity is a P F R. So, X will increase is that clear. So, value of R and value of n are measures of mixing suppose you get a conversion of X A in your experiment. So, you will read out a value of R from here you will read out a value of n from here that means this arbitrary vessel can be described by a value of R or by a value of n and in both in that case if it is value of n it will be described by this function if it is value of R it will be described by this function. So, which means we have a way of describing real life problems it is no longer idealized it is a real life problem whatever be the arbitrary system that you have you are able to relate it to some fundamental way of understanding reality. So, that is the advantage of these two equations an arbitrary system can be understood in terms of recycle ratio can be understood in terms of number of tanks of course, we should know the system parameters which is k tau r and k tau n which comes from independent measurements. So, having said this let us see what is the kind of use to which we can put recycle reactor to. So, now if you look at recycle reactor what it says is recycle reactor feeds process products into the process correct see what does it do it puts our products back into the process. Where is the advantage the advantage is in reactions where you require to put products into the feed otherwise reaction does not go forward there are auto catalytic reaction the good examples all biological reactions require products to be put into the feed and therefore, therefore it is very very useful in biological reactions. Now, there is another important is it allows adjustments of velocities through the reactor when you doing an experiment in the or laboratory development you can do your experiment in various velocities. Therefore, you can see how good is your catalyst to be able to stand the velocities of your industrial use. Therefore, you can appropriately design it for the kind of industrial operation that you are going to have of which I mentioned a little earlier is that see it allows you to do a very very low purpose conversion. When you have very large recycle purpose conversion is quite low which means essentially isothermal operation becomes possible and therefore, you can do kinetic measurements in a recycle reactor. And therefore, you will find in catalyst development for example, recycle reactor become particularly useful because you are able to get very good catalytic measurements. And essentially isothermal operation it may be very highly exothermic reaction, but because of a very large recycle purpose conversions are so low that you are able to get isothermal data. So, these are the three important advantages that you will have with recycle reactors. Having said this the real applications that you will see I mean we all have seen is that in auto catalytic reactions of recycle reaction reactor. So, when you make a plot reaction A goes to B where the rate function looks like this. When you plot R A versus x or concentration if it is plotted like this the progress reaction looks like this. If you plot it like this progress of reaction looks like this the important point is auto catalytic reactions show a maxima in reaction rate. This is the most important thing they show a maxima in reaction rate why do they show a maxima because as the products accumulate you will find that the reaction goes up and then comes down. So, this feature of maxima in reaction rate or minima in this kind of plot is useful in design. Because if you are going if you can design your system at this point your reaction equipment volume is very low highest because the reaction rates are very high. So, there is an advantage, but at the same time the disadvantages that the conversion may not be high enough you see where is issues like this which becomes important in design. Having said this let us let us just look at this in some detail this is 1 by R A versus x 1 by R A versus x. Now, this is the recycle reactor this is the point of maxima in reaction rate. Now, when you have a small recycle small recycle large recycle small recycle large recycle. Now, if I ask you what is the best value of recycle ratio you would employ what is the best value of recycle ratio you would employ our answer would be you say if you look at this function this function R plus 1 increases as R increases. But this function here d x by minus of R A what happens to this function d x by minus of R A this you can see this function here you can see it decreases and then increases. So, on other words this is an increasing function, but it has a decreasing and increasing function therefore, there is an optima. There is an optima that means recycle reactor employed in autocatalytic reactions there is an optimum value of recycle ratio which gives you the minimum volume. Now, if you ask what is that minimum volume of course, we can set up this equation put the appropriate functions here you will be able to you know find out the minimum where it occurs. But there are mathematical techniques by which you can show I have not shown it here it is an exercise in your problem sheet you can show this result what result is that optimum recycle that optimum exist it come by looking at the shape of this curve you can tell optimum exist. What is the optimum you will be able to show, but you have not done it here it is an exercise in that the optimum recycle puts into the reaction equipment that means optimum recycle the optimum recycle puts into position 1 this position 1 a feed you puts a feed into position 1 which is actually satisfies this relationship. What is this relationship the optimum recycle that means the reaction rate at this point which is 1 by r a at x 1 1 by r a at x 1 is the average of this whole thing what is this term d x by minus of r a x 1 to x 2 divided by x 2 minus of x 1 this is the average value of this rate over that interval is it clear the right hand side integral x 1 to x 2 d x by minus r a divided by x 2 minus of x 1 is the is the average value of this function in that interval x 2 to x 1 this is x 2 to x 1 is this which means what the optimum recycle puts into the reactor a feed whose 1 by r a at x 1 1 by r a x 1 is the mean in the range x 1 to x 2. We will show this mathematically also, but since it takes little while I put it as an exercise I have not shown it here, but we will do it as an exercise as we go along, but it is understood in physically what is important to understand is optimum means what you are choosing a value of r such that the reaction rate measured as 1 by r a at x 1 is the average that you would have gotten in the whole equipment. So, now that optimum recycle exists and the optimum recycle either can be found mathematically or graphically as possible. Now, there are some interesting issues that happens in autocatalytic reactions something that some of us may have appreciated, but I thought it is important to draw attention to this. See when you write the design equation for a CSTR in which an autocatalytic reaction is taking place or design equation looks like this v by f a 0 is equal to x by minus of r a r a have taken our r a value is minus of k times C a C b is what I have substituted with the minus therefore, it is become plus sign. Now, I replace C a and C b using our conventional stoichiometry. Now, if you look at this equation carefully what it says is if I specify a conversion x that I want straight away you will be able to find the size of the equipment. Now, in the way it is formulated here it does not say that the feed this feed should contain C b 0 it says C b 0 can be 0. Because this result tells us that it does not matter C b 0 need not be present in the feed is that clear what we are saying is that if you have an autocatalytic reaction being conducted in a CSTR as per the design equation that we have written C b 0 can be 0. That means, it is not necessary for you to put the product in the feed even then the reaction would occur and there the finite value of x you will calculate and for which the volume can be specified or in other words it seems a little difficult to understand why is it that autocatalytic reaction which should not take place if there is no product in the feed our design equation seems to suggest the reaction will occur. So, there is appears to be some inconsistency in the way we have formulated our problem. So, we needs to sort out this or at least to understand why this kind of inconsistency it appears to be inconsistent. So, we want to understand this how do we understand this to understand this what I have done is I have written the unsteady state equations. So, what happens in the unsteady state once we understand the unsteady state we can understand the steady state after all what is steady state after a very long time of the unsteady state is steady state is attained. So, what is this equation it says input output generation equal to accumulation correct. So, for component A component B I have written for component A this for component A is for component B this for component B look carefully at the two equations. So, I am saying t equal to 0 C A equal to C A i C B equal to C B i let us let us look at the situations when C A C B i is 0 what happens when C B i is finite what happens what is C B i and C A i C B i and C A i is what is in this equipment when we started this reactor some a million years ago as an example long time back, but today we are running it with C A 0 equal to some value, but C B 0 is 0, but when we started this reactor sometime back in the in the past there was C B i there was C A i we do not know what it was let us see what might have happened during the start up let us say if case 1 C A i is 0. So, C and C B i is not 0 C i is 0, but C B i is not 0 now what happens to this function C A at time t equal to 0 C A is 0 therefore, reaction rate is 0, but C A is coming in therefore, after sometime C A would become non 0 that means, if C B i was finite if C B i was finite the reaction would started to take place once C A starts coming into the process as per equation 1 you can see here input output generation accumulation when you started the reactor C B i is finite C A is 0, but C B i is not 0 therefore, this term it was 0 when you at 0 time, but little while later C A has started coming in therefore, there is C A in the reactor therefore, this term is finite therefore, this reaction would go this reaction would take place is that clear what we are saying this reaction would take place because after time t 0 plus this term is finite and therefore, this reaction would take place therefore, this equation can be integrated this equation also can be integrated and therefore, it will tell you how long it will take to reach steady state steady see here what we are saying is that continuous input of C B is 0 see what happens please is an important question please is a very important question let me answer this question see whenever we have an equipment we would start up the equipment sometime in the past after all we have to start the process some day at the past at the day when you first started that process if you had put C B i into your tank what it says is that that C B i would be active however, small it may be so that this C A time C B becomes finite after time 0 plus therefore, this reaction takes off therefore, this reaction takes off therefore, these 2 equations 1 and 2 can be integrated and it will tell you how long it will take to reach steady state at that steady state it may not recognize C A i and C B i it will get steady state values as described by our steady state equations where are we these are the steady state equations. So, the steady state equations will describe our steady state problem, but this equation here will tell you how long you would have taken to reach steady state on other words what we are saying is that if you have an autocatalytic reaction as long as you have C B i is non zero when you start the process sometime in the history of that it is enough as far as the reactor to reach a steady state value in which your products are positive. So, the reaction will start only when something like C A will C B C B i is not 0, but C A is there no C A is coming no C A is coming yes when a little C A come C A naught is from reaction will take reaction will take place that is what it says all it says is the time that is required to reach steady state will be determined by the value of C B i and C A i yes in during the history during the history. So, that means the day we start up the due reactor you have to put some C B i, but you do not have to put it every day once the reactor has started it has to be it is not required, but having said this I want to remind you say after all we do not want to wait for a very long time to reach steady state after all a process has to run it has to reach steady state quickly. So, for that to happen you may add C B i every day. So, that is a decision that you will take, but what our I mean our formulation tell you is that our steady state will not depend upon for depend upon the quantity of C B i and C A i steady state is independent of the choice of the initial state is that clear. So, what we are trying to point out here is that I have to stop here that steady states that you will reach for auto calculating reactions you do not require C B i and C A i in the process it is not required it is required only during start up I will stop there.