 We are going to start a new chapter today, which deals with design of non-ideal reactors. So, we are going to look at reactor models for non-ideal reactors. So, let us take a review of what we have learned so far as far as non-ideality in reactors is concerned. As you know, there are ideal reactors like CSTR, PFR, in which the flow pattern is well defined. In these reactors, the flow pattern is well defined. Whereas, non-ideal reactors, the flow pattern can take any particular shape in terms like if you look at the E-curve. Now, this flow pattern tells you how the fluid is flowing inside a reactor or somehow it does not follow a very particular or specific pattern. Now, in order to get a conversion, final aim is to have a reactor design and this reactor design somehow we have to incorporate this non-ideality. So, I cannot assume any reactor to be plug flow reactor or a CSTR. So, there can be non-ideal flow patterns which will affect the reactor designs. So, in the last chapter, we looked at zero-dimensional models or zero-dimensional models for non-ideal reactors. So, in this, you have a typical E-curve and this E-curve of course, as you know E is the exit edge distribution. That means, I have a reactor, this inlet is going outlet and what you are seeing at outlet gives you this particular pattern. Now, there can be many different possibilities which of the flow patterns which would give rise to this E-curve. And what happens inside a reactor is not exactly reflected in the exit edge distribution. This is something that we already learned. So, there are different possibilities in terms of what happens inside a reactor, the mixing pattern inside a reactor. So, mixing is not well incorporated in the E-curve. So, then we looked at two different extremes. So, one extreme is complete segregation model and another extreme is the maximum mixedness model. Now, let me quickly revise so that it forms a nice platform for further discussion as far as this chapter is concerned. So, let me quickly revise what these two models are. So, maximum mixedness model means that mixing is maximum. So, mixing between what? So, if you look at this E-curve, there are different segments. There are different segments. Every segment, these segments will be flowing through the reactor. Every segment will have residence time and that is why you will have a fluid element which will have a specific residence time and there is a distribution of residence time. So, you will have different segments spending different residence times in the reactor and the extent of in terms of the mass or volume of this particular element spending this much residence time in the reactor that may vary and that is nothing but E-curve. So, you have these different segments or different elements spending different residence time. Now, we are talking about the mixing between these two elements. We are talking about the mixing between these elements or several elements. Now, there is an extreme where these elements they mix they are well mixed and there is another extreme where these elements they do not mix at all. So, they go in parallel, they do not talk to each other, they do not interact with each other. So, that is segregation, complete segregation and there is another extreme where these are completely mixed. Now, how it happens in all? We will not look at that in detail but that is the meaning of it. Why we look at these extremes? Because these two extremes would give us a bound on conversion. So, it gives us a range. So, given an E-curve, given an E-curve like this, I make an assumption there is a complete segregation of these elements and then calculate a conversion and then another extreme where they mix thoroughly mix before they come out inside a reactor and calculate a conversion and these two conversions are likely to be different and they are going to be different for most of the reactions but they are going to be same for just one case. Remember what is it? When the reaction is first order and I think it has been discussed well before. Why first order reaction? Internal mixing does not matter whereas any other order internal mixing matters a lot. So, if you go for a complete segregation model and maximum mixing this model, these two models are going to give you different convergence, different extent of reactions. Intensive current is same. Rest all the volume of the reactor everything is same. E-curve is same but internal mixing will matter. So, what happens in this case? I am not going to get an exact conversion. Why? Because it is zero dimension model I am just going to look at the extremes. So, I get bounds. I get a range. I may say that fine, for a given volume this is a possible range of conversion that I am likely to get if the E-curve is like this for a non-ideal reactor. So, it may vary from 0.45 to 0.55. So, that is the idea I get but I do not get exact conversion. That is a limitation of this model. It is a zero dimensional model. It talks about the extremes as far as the mixing inside a reactor is concerned. E-curve only gives you partial information of the flow pattern. It does not tell you about the internal mixing. And for the first order reaction of course, it does not matter. So, instead of getting a bound I will just get a single value. I do not have to worry about whether it is a complete segregation or whether it is maximum mixedness. So, this is a quick revision of what we have learned before and now let us go ahead. Now, this is a problem with zero dimensional model where it gives most of the times for a non first order reaction the range of conversion because we consider it two extremes. We are not really looking at what is happening inside. Now, in this chapter we are going to look at different models. One of them is one parameter model. So, I have an additional parameter that is going to give me some idea about the conversion. It will help me get the exact value of conversion for the given E-curve. Now, remember that this model that we are going to discuss or the two types of one parameter model we are going to discuss they are applicable to a particular geometry. I will elaborate this point later, but remember that. In one parameter model, we have two models that we are going to discuss that is tank in series and dispersion model. Now, let us consider a tube. Let us consider a tube. That is why I said it is applicable to a particular geometry. Again, I will tell you why. Later, we will just talk more on this. A tube is a flow that is taking place and you may have different extents of back mixing that is occurring. Now, if you have very flat profile, what it means is you have a plug flow reactor. What is the E-curve that I am going to get is a direct delta function. So, the E-curve for the plug flow reactor is direct delta function, but this tubular reactor that I am talking about is not a plug flow reactor. Not necessarily behaving close to a plug flow reactor. You know not a single real reactor is behaves like an ideal CSTR or PFR. It is close to those reactors in extreme situations, but they do not exactly follow the pattern that we assume. But anyway, for if you are very close to a plug flow reactor, if you are getting a E-curve which is matching with a direct delta function to some extent, you can make an assumption that behave like a plug flow reactor and I can design the reactor accordingly considering some possibility of conversion being plus minus. But that is because by making the assumption of plug flow reactor, it reduces the complexity or regaling calculation. It makes my calculations simpler. It helps me to get some quick estimates of conversion for a given volume or for given conversion the estimate for the volume. So, let me get back to this. You have a tubular reactor. You have a tubular reactor in which there is a possibility of back mixing. That means, I am not going to get an E-curve which is similar to a direct delta function or a plug flow reactor. So, what is likely to happen? So, you have E-curves E t versus t. Now for a plug flow reactor, I am going to get something like this, infinite. You know the meaning of it and this is nothing but tau which is volume divided by volumetric flow rate. This is an ideal situation. Now, what I am going to see in reality is somewhat like this possible. I may see something like this. I may see something like this. Why does this happen? It happens because there is mixing in axial direction. If there is no mixing, I am going to see something like this. But because of mixing, some fluid elements may spend more time because they go back and then forth. So, it is possible that they spend more time in the reactor and they come later. Some fluid elements may spend less time just to compensate for those who have gone ahead or those who are lagging behind. So, it is possible that you get a distribution. So, I am talking about a tubular reactor where there is a possibility of back mixing. You may have packing. It may provide some because of tortuosity and all. It is quite possible that you do not have the exact plug flow type behavior. Now, these are the different e-curves. How do I interpret these e-curves? This is happened only because of back mixing. So, I need to incorporate the effect of back mixing in this particular behavior that I have observed. So, let us consider any e-curve. Once this e-curve is obtained, I have a tubular reactor. I do a pulse injection experiment and I look at exit as distribution. I am going to see this. Now, this is something given to you from this you are going to come up with a model. I am going to determine a parameter. Now, mixing is characterized by C s like you have a say let us have a hypothetical mixer or agitator in the reactor. So, in the tube itself, I can say that tube is consisting of various compartments. If you have infinite such compartments in a given volume, total volume, small compartments, what does it mean? Infinite means that you have a flow similar to a plug flow reactor. Now, you reduce the number of compartments. What does it mean? That means there is some back mixing happening. You go on reducing, go on reducing. Mixing, extent of mixing would increase. Consider an extreme where you have just one compartment. There has to be a compartment. So, one compartment. What does it mean? A compartment with agitator in it, good mixing, complete back mixing in that compartment. What does it mean? It is a CSTR. This is another extreme, another extreme. So, what does it mean? So, on one side you have a CSTR, on the other side you have PFR. CSTR means one compartment, PFR means infinite compartments in between you have a reactor which has partial back mixing. So, now you would have guessed what is the parameter that I am talking about? It is a number of compartments. Every compartment is a CSTR, perfect back mixing. So, it is a number of CSTRs or number of tanks which are in series that are going to, that number is going to be a parameter for this model, fine. So, why I am saying it is a tubular reactor, it is a tubular reactor. Let us consider a reactor which is a very irregular geometry. Say I have a reactor like this. This is an inlet, this is an outlet. Now, there are flow patterns. Then there are, there is a possibility of some isolated zones and you have the output here. What you are going to see is the behavior like this, any general E curve. Now, this particular E curve does not look like the E curve of a tubular reactor. Why? Because for a tubular reactor, let us look at all the possibilities. Now, you have a plug flow reactor, you have a plug flow reactor. CSTR, E curve, right, CSTR. One CSTR is this. What will happen to two CSTRs? Like in the case of two CSTRs, you are going to get something like this. Then three CSTRs, four CSTRs. See, what is happening now? As you go on increasing number of CSTRs, you are approaching the plug flow. This is of course going to infinity. This is becoming narrower. You have started seeing a delay here. So, what is happening? As we go on increasing the value of n, that is number of compartment, number of CSTRs in series, the behavior, it is increasing it. It goes, it follows this particular train, goes from CSTR to PFR. But look at these curves, the nature of these curves. These curves and compare it, compare these curves with this particular curve. I am getting a very irregular shape here or rather very unusual. There are many fluctuations, ups, downs and all that. Why does that happen? It happens because of the irregular geometry here. I have not defined the geometry. Flow pattern is quite complex. So, this flow pattern and this flow pattern, they are not matching. So, can I apply a tank in series model for this reactor? I cannot do that. Why? Because there is no curve that fits well in this particular shape. So, again, the one parameter model has its own problems. It is applicable to most of the times, a tubular reactor where the e curve will be like this or one of these rather. It is a continuous thing. It is a continuous thing. There are no ups and downs, no recirculations or recycled type of behavior. So, I will go back to my statement that one parameter models are good for tubular reactors where you get e curve with a nature like this. So, let us go ahead and do some mathematical derivations. How do I find out number of tanks that I have in series for a given tubular reactor? Say I have an e curve. How do I find it? So, let us derive an expression for the e curve for a tank in series. Suppose I have n number of tanks in series. Let us try with three tanks in series first and then extend this concept to n tanks. So, let us have three tanks in series. So, let me write it like the expression for the e curve. That means, I am going to give a pulse here. I am going to give a pulse here and I am going to see what happens here. So, this is my e curve and this is what I am going to get, likely to get. Now, can I get an expression for this? I know how CSTR behaves. So, let us write e t into delta t. This is a fraction. You know the meaning of this. I am not going to repeat. So, e t delta t is a fraction that comes in between t and t plus delta t. That is equal to the volumetric flow rate. I am going to assume that volumetric flow rate remains constant throughout into C3 t. That means, C3 is a concentration at the outlet of tank number 3. So, 1, 2 and 3. C3 t into delta t divided by n 0. What is n 0? n 0 is a total number. So, I am going to get a total amount of pulse that I have injected. How do I calculate n 0? n 0 is nothing but based on C3 t because now I am going to get rid of n 0. I am just looking at concentration at the outlet. See, try and visualize like realistic experiment that I am doing or real experiment that I am doing. I am looking at the concentration at the outlet. So, try and express everything in terms of concentration at the outlet. Now, n 0 is a total amount of pulse or the tracer that I have injected. n 0 is equal to 0 to infinity C3 t dt. Total amount that is come out into of course, the volumetric flow rate. So, concentration at the outlet of tank number 3. Concentration into volumetric flow rate into time. So, I just integrate. I just integrate from 0 to infinity. So, that much amount I have injected as tracer. So, that becomes n 0. So, let us go ahead. Let me substitute for n 0. So, what I get is E t is equal to V C3 t. It says, sorry, this is delta t. So, E t is equal to C3 t divided by. So, I have got a expression for E t based on the concentration at the outlet. Now, this C3, I need to get that in terms of the residence time, initial concentration or inlet concentration. And I know all these are CSTRs. Let me write n steady state balance for a CSTR. CSTR 1, V1 is the volume of CSTR. DC1 by dt, you have learned this. Unsteady state balance for a CSTR is equal to minus V C1. Shall I write here? This is what? This is going out. This is going out. This is accumulation. Coming in, shall I write this? No, because what I am looking at, see it is a tracer experiment, the pulse experiment. I am injecting a tracer and then after that, at 0 time, I am injecting a tracer and after that, I am observing the response. So, after that, is there anything that is coming in as far as the tracer is concerned? No. So, I have injected and stopped it. Now, it is only the inner that is flowing, the solvent that is flowing. The tracer is not coming in after 0 plus. And this equation I have written is for time 0 plus onwards, let me 0 onwards. So, this term is not there. This term is not there. So, I have only this equation for the concentration in the first time. I am writing it for the first time now. Later on, I will do it for second, third and then finally, I will get an expression for C 3. That is my objective. So, here from this, I get a concentration at the outlet of tank 1, where I have the boundary condition at time is equal to 0, sorry, time is equal to 0, C 1 is equal to C 0. If I solve this equation, it is very simple and that is nothing but C 1 is equal to C 0 e raise to minus T by tau 1. What is tau 1? Tau 1 is equal to V 1 divided by small v. This is the volumetric flow rate. This is the volume of first tank. This is the expression for C 1. That is the concentration at the outlet of first tank. So, let us continue now. So, this acts as inlet for the second tank. So, let us write an expression for the second tank. Tank number 2. What is it? V 1, sorry, V 2 d C 2 by d T is equal to small v C 1 coming in minus small v C 2. I cannot neglect this. This is changing with respect to time. At 0 plus C 1, we will have some value. So, there will be two terms as far as tank 2 is concerned. For tank 1, inlet was 0 because tracer was 0 at the outlet of tank at time 0 plus. Before we go ahead, let me make an assumption that v 1 is equal to v 2 is equal to v 3. All v's are equal, say v. This was simplicity or let me say v i because I am going to use v for something else later. I am going to use v for the total volume. Let me call this as v i. So, it becomes v i here. So, let me simplify or rather solve this further. For C 1, I have the expression. For C 1, I have the expression. What is that? This one. So, this expression I substitute for C 1. What I get is this d C T 2 by d T is equal, sorry, plus C 2 by, now I will say tau i. Why tau i? So, what is tau i? Tau i is equal to v i divided by small v which is constant. That is volumetric flow rate given to you. So, C 2 divided by tau i is equal to C 0 divided by tau i into e raise to minus T by tau i. I am just substituting for C 1 and expressing everything in terms of tau i now. So, you know what is tau i? So, I get an expression for a differential equation for C 2. Now, again for this, I need a boundary condition. At T is equal to 0, C 2 is equal to 0. So, this equation, I solve this using this boundary condition. This is a famous or popular method of integration factor e raise to T by tau i. So, this expression can be solved using integration factor or integrating factor e raise to tau i. So, T by tau i and I get a solution for it. I get a solution for it which is given by C 2 is equal to C 0 T divided by tau i into e raise to minus T by tau i. See, concentration and outlet of the second tank is expressed in terms of initial concentration time. Anyway, I want to know how it changes with time and of course, the tau i. I have eliminated C 1. What is my aim? My aim is to get C 3 in terms of C 0 T and tau i. So, let us do it now. I follow the same methodology for tank 3 substitute for C 2 because now in tank 3, the inlet is C 2 and I solve this equation. So, what I get is C 3 is equal to. So, you can try it out. It is again the same methodology. I get, what I get is this, oh sorry, e raise to minus T by tau i. So, you have two appearing here. See the difference, not much difference, but of course, you have the exponential term there, but then instead of T, now it becomes T square. I have two appearing here, tau instead of tau i, I have tau i square. So, let us go ahead. Now, I have the expression for the e curve or whether the e for the entire reactor, the tanks in series. What is that? This one. Now, I have the expression for e. I have the expression for C 3. Substitute the C 3 in e curve or rather the e expression for e. So, what I get is, see the objective is to get expression for e. So, I substitute for C 3. Let me write this first. So, that C 3. Now, what is this? This is the total amount of total amount of the tracer that I have injected. The amount of tracer that I have injected. Can I get it in terms of initial concentration C 0? It is possible. What is it? If you look at what happens at time 0, you have C 0 is equal to N 0. That is the total number of moles or whatever unit can be grams or moles or whatever. You have that much amount of tracer divided by V 1. So, this is something present at time is equal to 0 sharp in the reactor. V 1 is a volume. It can be V i also in our case. So, let me call this as V i, which is nothing but the total amount is V 0, which is nothing but V rather into C 3 t dt divided by V i. So, from this, what is this? V by V i tau sorry 1 by tau. So, this tells me that this integration in the thing, but tau i into C 0. I am just expressing it in terms of initial concentration, because this is appearing in the final equation. So, let me get back to this equation. This is for the e curve. This is for the e curve. I am substituting for C 3 that I have just derived. C 3 that I have derived here. I will substitute it here. And then I will have a very simple equation. E t is equal to C 0, C 0 will get cancelled and I will have t square divided by 2 tau i. Now, it is cube, because it is 1 tau i which is come from the denominator integration term e raise to minus t by tau i. Now, this is for 3 tanks. If you apply the same logic, do it for 4 tanks, 5 tanks, 6 tanks, n tanks, then what I am going to get is for n tanks is t raise to see 3 tanks 2 here n minus 1 divided by see n minus 1 factorial into tau i raise to n. See the logic then e raise to minus t by tau i. This is for n tanks in series. So, I have got an expression for e curve for n tanks in series. A general expression for e curve for n tanks in series and that is what I want. So, when I get a e curve, I have e curve. So, in this, this is only one parameter. That is n that I can calculate from this. So, this will obtain from experiment in the laboratory for a reactor, tubular reactor under the flow conditions that are desired for the reaction. But, of course, I do the experiment in non-reactive condition. You know that just to get a residence time distribution. So, from this experiment, I get e curve. From e curve, I get n because I have an expression for tanks in series e curve. And once I get n, then I can calculate a conversion. What is n? n tells you the extent of back mixing. n infinity means is P F R and 1 means is C S T R. And in between 1 and infinity, you have the extent of back mixing characterized by the number n. Fine. So, let us go ahead and simplify it further because this equation looks a bit complicated. So, from this, we are going to define a term called variance. Of course, you know what is variance when you have distribution. The variance that means how much at a particular time the value is away from the average. So, there is something called as a variance. Before that, we will define a term, a non-dimensional term tau, sorry theta equal to T by tau, where tau is the total residence time. What does mean? That means tau is n into tau i. It is the individual residence time of every tank. n into tau i is the total residence time. So, if tau i is equal to V i by small v, then tau is equal to capital V, which is nothing but V 1 plus V 2 and so on divided by small v. That is why I said I will be using V for something else, the total volume. Why? Because I do not have tanks. I am just assuming a tubular reactor to be set of tanks. So, all I know is the total volume. So, I need to express everything in terms of total volume finally. This total volume is V 1 plus V 2 plus V 3 plus V 4 and so on. So, the residence time based on this total volume is tau. For individual reactor, it is what is tau i. So, this is tau i. Now, it is tau and then I have a dimensionless number theta expressed in terms of tau that is T by tau. So, I have the expression for E t. I will write it again T raise to n minus 1 divided by n minus 1 factorial tau i n e raise to minus T by tau i. Now, this will get reduced to E theta. Now, T will be expressed in terms of dimensionless time theta. If I do that, then all this will get converted to n into n theta raise to n minus 1. See, T will become n theta because see T then e raise to minus n theta divided by n minus 1 factorial. So, this is the expression for E in terms of dimensionless time instead of actual time. You are going to use this further that is why you expressed it like that. Now, there is a variance that we define. Now, how do we define variance? If you have a distribution, the expression for the variance is in terms of dimensionless time variance square is equal to actual variance in time divided by tau square. Now, this is nothing but 0 to infinity theta minus 1 square E theta d theta. I hope it is clear. If you write just variance in time, you will have T minus tau. You have to just divide it by tau square and then you get this. It is a variance. I will say how much you go away from tau mean residence time. So, if you expand this further, what you get is 0 to infinity theta square E theta d theta minus 2. I am just expanding this. So, theta minus 2 theta plus 1. So, you get this. Sorry. So, what is the value of this? This is equal to 1, this particular term. What about this? This particular term. This is dimensionless average residence time dimensionless. So, it is going to be 1. So, because it is the reference is tau for the residence time. If it was T E t d t, then it was tau. Now, it is theta E theta d theta. That means, it is tau divided by tau. That means, 1. So, this is going to be again this integration is going to be 1. This is 1. This is 1. So, minus 2 plus 1 is minus 1. So, 0 theta minus 1. So, this is the expression. I am going to simplify this further. So, let us solve this integration. What I get is, before that let me just write it again. Now, I will substitute for E theta. See, we have derived the equation for E theta. So, this is E theta. I will just substitute for it n n theta n minus 1 divided by n minus 1 factorial e raise to minus n theta d theta minus 1. Now, if you do all this, what you get is, of course, take n. So, you will have n raise to n divided by n minus 1 factorial. You come out and you have 0 infinity theta raise to n plus 1 e raise to minus n theta d theta. So, if you solve this further, what you get? You can do it on your own. So, it gets simplified to very simple term 1 by n. And this is what I want. Sorry, variance e raise to minus is 1 by n. Variance will be obtained from the e curve. So, once you get variance, you get value of n, number of tanks in series. So, n is equal to 1 by sigma theta square, which is nothing but tau square divided by sigma square. So, I get a value of n. This is what I want. So, given the e curve, now we will see how to calculate sigma from e curve. I have already told you the expression for it, but we will solve or I will tell you the procedure to get sigma from that or this sigma, from that I get n. So, e curve will give you sigma and sigma will give you the value of n. How to calculate conversion for the given n? Now, we are talking about a reactor. Tanks in series is very simple. I have n number of tanks. How do I calculate conversion for a CSTR? And I have n number of CSTRs in series. The conversion, say for a first order reaction, the conversion is 1 minus 1 by 1 plus tau i k raise to m. Tau i is the, again, see you do not get confused between these two tau. This tau is the residence time for a single tank. What is k? k is the rate constant. And this is something that you have done before. So, let me summarize. I have a tubular reactor. I have a tubular reactor. I want to get the conversion. The problem can be other way around. For a given conversion, find out a length or sorry, the volume of the tubular reactor. So, when I have a tubular reactor, let us talk about given reactor and calculating conversion. So, I have a tubular reactor. What will I do? I will not assume it as a PFR. Now, we are talking about a non-ideality. There can be possibility of backmixing. What I will do is, I will just pass a fluid. I will do residence time distribution experiment. I will inject a tracer. Look at its response and outlet. I will get a e-curve from it. Once I have a e-curve, then I can get a variance from the e-curve. From the variance, I will get a number of tanks in series for that particular tubular reactor. If it is close to PFR, the value of n will be very large. If there is so much backmixing happening for some reasons, the value of n will be close to 1, 1, 2, 3, whatever. So, I get a value of n. Once I know the value of n, I have the expression for conversion. I have the expression for conversion. That tells me how much is the conversion based on n. Rest all you know, what is this rate constant? Tau i is the residence time for the individual reactor. How do I calculate tau i? I know the total volume for the tube. I know the total residence time from total residence time. I calculate the individual residence time. The individual reactor by dividing it by n. The n is known. So, I get an expression for the conversion. That is for the first order. But of course, for second order, third order, I will have different expression. It is just solving a problem for CSTR. So, I have converted a non-ideal reactor to an ideal reactor problem. Tanks in series, ideal reactor in series. n varies. n is the parameter. Thank you. We will continue this discussion. We will solve a small problem. Of course, not numerically, but I will tell you the procedure so that the things will be clear to you. Thank you.