 So, we will quickly revise what we have learned in the last lecture. In the last lecture, we looked at dispersion model. Overall, we are just looking at models for non-ideal reactors. The first model was Tanksin series. The second one is dispersion model. So, in dispersion model, we assume a tubular reactor geometry. And then, of course, we say, we characterize this flow pattern through a parameter called Peclet number. So, we say that fine, we have a tracer that is injected at the inlet which gets into the form like this. So, it becomes flat when it comes out flat in the sense. Of course, it gets distributed, I would say. Now, we want to see the behavior. And this particular behavior depends on the extent of back mixing which in turn depends on the Peclet number for the given system. So, we derived equation for this or rather we formulated a differential equation for this which was 1 by P e del psi del 2 psi by del lambda square minus del psi by del lambda is equal to del psi by del theta, where theta is dimensionless number, lambda is dimensionless length. And Peclet number is something that characterizes the extent of mixing which is nothing but U L by dA, where dA is the dispersion coefficient, L is the length and U is the velocity. And this dispersion coefficient is nothing but something similar to a diffusion coefficient, but that the diffusion is by fixed law where you are talking about migration due to concentration driving force. Here, it has a slightly broader meaning in the sense not just diffusion, but it takes care of other effects also and somehow characterizes the extent of mixing that is occurring or the flow that is occurring outside the convective flow. So, the two types of flow now convective and dispersive. Now, so this is something that takes care of dispersion and convection of course, taken care of by the velocity term. So, this is that particular term where you have velocity coming in picture. So, now this Peclet number I told you that if it goes to infinity means you have a typical ideal plug flow reactor, if it goes to 0 means you have a CSTR. So, the two extremes of plug flow sorry Peclet number and in between these two 0 to infinity you will have number of Peclet. So, Peclet number giving an idea about the extent of back mixing occurring in the real reactor. Now, if you have this equation and of course, the corresponding boundary conditions you solve them and you get you get a behavior like this for different values of Peclet number. This is the exit concentration of course, and for Peclet number equal to 0 you have a CSTR then for Peclet number higher than that you have a behavior like this Peclet number say 5 then you start seeing some lag then Peclet number say 100 or so and then you have something like this going up and then at value of Peclet number close to infinity or very large you see a plug flow behavior. So, this is a quick revision of what we have learned in the last lecture. So, I told you that once Peclet number is known I can draw this profile at the way round if the profile is known I can draw I can get a value of Peclet number. Now, how is the profile known you do experiments in laboratory. So, you have a reactor you do tracer injection and you measure the exit concentration and get this profile. Once you have this profile from this profile you have go back and calculate Peclet number. So, you know what this reactor is all about or what kind of mixing is occurring here and then you can predict the conversion you applied for the reactive flow that is what we are going to see now. So, before that now Peclet number is related to deviation or variance. Now, what is that relation I told you last time sigma square divided by tau m square tau m is nothing, but tau sorry t m is nothing, but tau is a mean residence time 2 by Pe minus 2 by Pe square 1 minus e raise to minus Pe sorry. So, this is 1 minus e raise to minus Pe ok. So, this is the relationship. So, you know this is a residence time this is Pe and sigma. So, in the case of Tanksian series I had relationship between Pe and sorry sigma and n now instead of n I have Pe. So, in Tanksian series I have a parameter called n here it is Peclet number. So, once you know the Peclet number how to get a conversion. So, let us go back to the reactor now I have a tubular reactor I write differential balance for it and get a conversion. So, let me write differential balance for the reaction case now earlier it was only non-reactive tracer injection now I have a reaction case ok. Before that sorry before that this is something that I have obtained for obtained for close, close vessel you know what I mean by close close that means before the inlet and after the outlet dA is equal to 0 there is no dispersion happening and accordingly we wrote the equations at those boundaries and got a boundary condition. But then it can be open vessel as well it can be open close it can be close open. So, depending on what I assume I will get different boundary condition and if your boundary condition is different then your value of Peclet number is also going to change. So, the expression that I have got always remember the expression that I have got is for a closed, closed vessel very important for open upon vessel you can derive it, derive it. But then of course I am not going to spend some time on this it is all given in standard text books. So, what I get there is sigma square divided by tau m square or sorry tm square is equal to 2 by Peclet number plus 8 by Peclet number raise to 2 alright. So, I will get a relationship for open open vessel like this what you need to remember is there is a relationship is sigma and p it changes as per the boundary condition that I am using for an open open system it must be noted that tm is not equal to tau. So, once you get sigma from e curve you can calculate a value of p if you are not sure about what happens in inlet what kind of vessel it is open open open try and get all values of possible values of Pe depending on which boundary condition you are depending on the boundary conditions and for every Peclet number calculate a conversion. So, you will get some idea at least fine. So, let us go ahead for the reactive case reaction happening in a tubular reactor in fact our main aim is to design a reactor, but time that we have spent so far is to this character is a flow pattern alright. So, let us go ahead if I take a differential balance at steady state now it is very important earlier all those unsteady state now it is a continuous reactor it is steady state operation. So, it is a plug is similar to what we see in plug flow only thing is now the f a is different anyway. So, let us write it d a f a by d a z now it is not del f a it is d f a because it is steady state no time no change with respect to time alright why r a r a is the rate of reaction it comes from the rate equation r a is equal to minus k c a or whatever depending on what kind of reaction it is first order second order it can be anything alright. So, I have this differential equation now in this differential equation this f a is not u into c a it is not just u into c a, but it has the diffusion or dispersion term also present there. So, if I substitute for f a based on those two terms convective and dispersive or dispersion rather then what I see is this d a into u divided by u d c a by d z square plus minus d c a by d z plus r a by u is equal to 0 u will come here anyway so that is. So, these two terms is nothing but what is it d a f a by d z. So, this is 1 by u yeah 1 by u is equal to this alright. So, d f a by d z is equal to d a into this into u into d c a by d z. So, I have just substituted for this and then I have divided by u also for this term. So, I am dividing it by u dividing the entire LHS by u I hope this is clear. So, you just substitute for f a you know what is f a right f a is equal to you have you already seen this f a is equal to u into c a minus d sorry d a d c a by d z into a c you have seen this right fine. So, this is a this is the equation now for the reactor because I have a reaction term I have not expanded this it can be rate equation minus k c a or whatever depending on first order or second order. Now, let me write r a is equal to minus k c a first order reaction and go ahead. So, what do I see d a by u into d 2 c a by d z square this minus d c a by d z minus k c a divided by u is equal to 0 I have just substituted for r a I had non dimensionalized this further on I non dimensionalized this further what do I get if I just non dimensionalized it I would get I divided by L. So, 1 by p e d 2 c a sorry d 2 psi by d lambda square minus d psi. See look at this I am going first, but I think it is because you already done similar exercise before now the difference is that you have the reaction term appearing you have the reaction term appearing here. So, for c a it will be psi into into k sorry k psi into k into L divided by u why see what I am doing here is I am multiplying or rather since I am writing p e here that means I have to multiply it by L here. So, I am multiplying by L here. So, L L got cancelled right. So, I have L appearing here right is equal to 0. So, this is my dimensionless equation the dimensionless equation and I go ahead and solve this of course it needs boundary condition, but before that we will define one more parameter or one more dimensionless number which is called as of course it has many other definitions as well, but here psi into d a is equal to 0 what is this d a do not confuse this d a with dispersion coefficient. So, this is a dimolar number this is a dimolar number and you have already seen this number before probably in reaction engineering part 1. So, this becomes a non-dimensional less sorry non-dimensional or dimensionless equation dimolar number let us define it like what we did for Peclet number what is dimolar number is rate of consumption of a by reaction see in the numerator you had k remember let me write it k L by u. So, in the numerator you have k. So, this divided by rate of transport of a by convection. So, this is dimolar number. So, see is what is happening to a in the reactor is moving ahead because of convection and is reacting. So, d a tells you the relative important or relative potential of these two terms or magnitudes of these terms what is important in convection is important it comes in denominator reaction is less. So, d a is less higher the value of larger the value of d a it says that the reaction is more compared to convection. So, dimolar numbers I am not going to spend more time on this because you already discussed this or learned this in chemical engineering chemical reaction in part 1. So, this is a definition for d a for first order reaction for second order or nth order reaction d a is nothing but or dimolar number is nothing, but k into c a 0 n minus 1 in divided by u. So, this term has appeared here because that concentration will also come in picture and the initial concentration or inlet concentration rather, but raise to n minus 1 n is the order. So, where are we? So, we have this equation sorry I will get back to the main equation because that we should not forget. So, this is the main equation that I have got in that I have two dimensionless numbers Peclet number that characterizes the extent of back mixing or dispersion and dimolar number that characterizes the importance of reaction as against convection. And for in order to solve this equation and need boundary conditions again two boundary conditions 1 divided by p e remember this close, close vessel we have we have used this before right at lambda is equal to 0 that means inlet and at the outlet at the outlet we have d psi by d lambda is equal to 0 that is at lambda is equal to 1 outlet close, close vessel right. So, these two boundary conditions I do not need a boundary condition I do not need initial condition now this equation because the steady state equation, steady state equation two boundary conditions I solve this equation for psi that is dimensionless concentration of say a what I get is this psi at why see psi is changing with respect to length, but where at what position I do I want the value of psi, I want the value of psi at the exit, exit because that is going to decide a conversion psi is a conversion concentration. So, correspondingly so best so if I know the concentration I can calculate the conversion. Now this psi at the outlet that means at lambda is equal to 1 or z is equal to l. So, psi at the lambda equal to 1 is equal to after I solve this equation what I get of course, this is nothing but C a l that is outlet concentration divided by inlet concentration and in terms of conversion it is going to be 1 minus x and you get a very big expression analytical expression if I solve this equation and I am just writing it I am not spending time for this mathematical derivation here what is more important is I get a value of conversion based on what there will be two numbers here which are the two numbers Peclet number and Daimkohler number. So, this is nothing but look at this expression there is no need to buy had this expression, but then you should know that there is an equation like this possible what are we doing we are looking at a tubular reactor non-ideal reactor with axial mixing with axial mixing and this mixing is characterized by a number called Peclet number this is a close close boundary condition sorry it is very big expression. So, what do you see here on the right hand side Peclet number what else you expect here Daimkohler number this is not Daimkohler, but something called Q why because it is quite complicated now this Q is nothing but a function of Daimkohler number let me write it here. So, where Q is equal to root of 1 z is 1 plus Da divided by Pe into 4 sorry it should be 4 Da by Pe right. So, I get expression for psi which is nothing but 1 minus x right. So, x is nothing but 1 minus psi. So, 1 minus this quantity will give me the conversion shall I just write it again yeah it is x is equal to 1 minus 4 Q exponential Pe by 2 divided by 1 plus Q square exponential Pe by 2 divided by 2 minus 1 minus Q square exponential minus Pe Q by 2 this is a conversion. So, now compare this with what you got in the case of tanks in series model what was it x is equal to 1 minus 1 divided by 1 plus tau i k raise to n. So, this is dispersion model this is tanks in series models. So, obviously this is simpler compared to this, but does not matter these is know like you have everything fed in computer. So, you can calculate a convergence is not an issue what is more important how close are you to the reality and so in. So, look at the entire exercise what have we done so far or if I want to design a reactor how do I do it based on this one parameter model. So, let me write down in steps design of a non-ideal reactor using one parameter model what do I do first I do tracer experiment in the lab get C T sorry C T versus T curve. Then determine E T third determine where the variance sigma or sigma square and you know the definition of this shall I write it here once you get sigma determine if it is tanks in series n you know what you know this is related to this or Pe this is again related to this up to this point I am still looking at a flow pattern now using this parameters what I need to do is just get a conversion. Now, for first order reaction conversion x sorry first order reaction I have an analytical expression, but depending on the order I will have probably I have to do it numerically especially in the case of dispersion model. So, get conversion by solving reaction enabled balance reaction enabled balance for a real case. So, in this particular for a first order of course, I will repeat 1 minus 1 by 1 plus tau i k raise to n this is tanks in series model and x is equal to sorry I do not want to write it here the big expression for a dispersion model dispersion model this will be function of Pe and da. So, this is the overall procedure this is for a design now sorry I started I said that this is a design problem and I have written an algorithm to calculate a conversion. So, this is not really a design problem see it is like you are given the length or the volume reactor and you are calculating the conversion, but then it can be other way round somebody says I want 50 percent conversion design the reactor for me. So, what will I do? I will just do the entire exercise assume some flow rate calculate a conversion it is not matching play with the flow rate, but I will do an entire exercise again. So, this becomes a trial and error trial and error method and it is it is a big complicated, but there is no way out because the flow pattern or your flow rate that you are going to decide the residence time is likely to change your e curve also it is likely to change your mixing also. So, every time whatever flow rate that you want to operate it at is always better to do experiment and see whether you have the right kind of e curve or you have taken into consideration the non-ideality and accordingly the value of p e and p e or n. So, this procedure tells you for a given volume what is the conversion possible problem can be solved iteratively to get a particular conversion sorry to get volume for a desired conversion. So, this is how it works because you want to incorporate a flow patterns right if we had assumed the reactor to be ideal close to p f r or c s t r things would have been much easier like what we did before, but now I want to take into consideration the flow patterns and this is remember this is only for reactor which is giving me a behavior close to a tubular reactor. If you have a very irregular geometry then this is not going to fit your p e or n values are not going to they are not enough one parameter models are not enough or they are not it is not adequate to rather express or characterize your flow patterns you have to go for multi parameter models we will see that later. So, this works for particular e curve always remember that I told you in the beginning of this particular first lecture of this chapter that if you have very weird like e curve you do not think of using a one parameter model for it it should look like c s t r then going from c s t r to p f r all right. So, let us solve one problem of course, I am not going to do numerical calculations here, but again like whatever I have written words here let me try and explain more detail if one has to solve a problem. So, the first so who so what is given to you c t versus t. So, this is what you would expect in a problem the tracer concentration the tracer concentration time and concentration the unit can be grams per liter or moles per liter here it is can be second minute order. So, 1 2 3 4 3 4 5 order and here again you have some numbers right. So, at 0 time it will be 0 then probably 1 then 7 20 and it may come down again say 8 3 what. So, this is the typical behavior like this. So, this is given to you what else do you need to design a reactor the value of k it is rate constant and what order. So, that independently somebody would have determined it for you in a differential reactor or in laboratory and then they would have estimated values of k rate constant activation energy if it is exothermic endothermic reaction and all frequency factor and order. So, these are all parameters for the reaction kinetics fine. So, this is what is given to you and from this you need to determine the conversion using different models. Now, a problem may be calculate conversion for a CSTR PFR 1 parameter models it can be tank in series and can be dispersion. Now, just one question before we go ahead see it is not that always solve equations whatever we know and get a values, but then if it is a first order reaction or if it is a positive order reaction CSTR will give me one bound conversion will be minimum or maximum conversion will be for CSTR positive order reaction conversion will be minimum. PFR will give me another bound conversion will be maximum and in between you have this mixing happening because of which conversion will be in between PFR and CSTR right the tubular reactor. So, only thing is because of partial back mixing the conversion is going to get affected and since this is a positive order reaction back mixing is not desired for a conversion like higher conversion will get affected adversely with back mixing right. Of course, you need initial inlet concentration and all for all this exercise. So, how do I proceed I have T 1 2 3 4 and so on C T concentration measured concentration 1 0 1 7 20 and so on I need to get E T I need to get E T how do I get E T E T is equal to C T divided by integration 0 to infinity C T D T. Now, how so C T is known C T is known for a particular time I need to know this how much is the tracer that I have injected. So, the entire concentration I have a plot of C T versus time if I want to get this value what do I do I just take the area under the curve from 0 to infinity may meet the x axis I do not need to go further. So, this area is nothing but this denominator I am just explaining you the procedure. So, this gives me E T. So, I will have some values here I am not going to write those values I will have some values here right. So, I got E T from E T what do I find out first thing is I need to get average residence time I need to get average residence time I want the value of tau remember everywhere tau came in picture that tau is required how do I get average residence time or mean residence time mean residence time is this the expression for mean residence time tau T m is equal to 0 to infinity T E T D T again tau is equal to T m is valid only for closed closed system. So, what I need here is from this plot I will calculate T into E T and I will have a value. So, T into E T I will have some values oh sorry it looks like 0. So, let me write cross I am just simply multiplying these two numbers T T E T D T I am explaining though only a procedure because doing calculations in the lecture will be difficult. So, T E T D T and then from this once I know T E T I can plot see I can plot T E T versus T. So, what a plot I get I get area under the curve I do not know the nature of this it can be anything all I need to do is get area under the curve and that is nothing but tau. So, in this step I have determined the value of tau that is mean residence time mean residence time. So, this is a this particular row would give me the T E T by multiplying these two and then area under the curve when I plot this against this that is time this against this and I get mean residence time. What do I need next then what I do is I take T square now you know why I am doing this instead of T I have T square why will I need T square because I need to calculate variance and you know the expression for variance. What is the expression for variance the expression is this sigma square is equal to 0 to infinity T minus tau E T D T. If I expand this it is T square E T D T minus 2 tau sorry T E T D T plus tau square 0 infinity E T D T. You know what this is this is nothing but 1. So, this becomes tau square and what is this is tau. So, this is minus 2 tau square plus tau square that means it is minus tau square. So, these two terms I know what I need is only this right. So, this is to be found out by plotting T square E T D T sorry T square E T versus time and whatever plot I am going to see sorry I do not know the nature of this I take the area under the curve and this is nothing but this term. This term minus tau square this is nothing but tau square minus tau square. So, this term this area minus tau square nothing but the variance. So, I did I get variance I hope it is clear. So, I have these values I plot these values against these values this versus time get a curve and area under the curve will give me this particular term this term minus this all is tau square I have already determined tau that is mean residence time in my earlier exercise right here tau. So, from that I get sigma square that is variance why I want variance because variance tells me the extent of distribution or dispersion right. So, from variance I calculate get P E or N from sigma square this is required sorry these two are required because I want to get conversion from this. So, from this there are expressions for conversion and you know expression. So, here it is nothing but 1 minus 1 by 1 plus tau i k raise to N for first order reaction and similarly you have a big expression here you will get these two values. Now, you may ask me these values will be different or they are same see if they should be quite close to each other these two values if you exercise is correct these two values should be very close to each other. So, you make it say I am I. So, I will just write a value say CSTR you may get a conversion point 5 P F R you may get a conversion point 7 for a first order reaction or positive order reaction and it is quite possible that for a tank in series model you get point 6 3 and dispersion for a dispersion model you may get something close to say point 6 1 or 6 2. So, they are quite close to each other just to give an idea and they these two are going to be in between these two positive order reaction effect of mixing on the reaction conversion. So, this is how it going to be as far as effect of mixing on reactor is concern sorry on conversion is concerned when you have a tubular reactor and we are talking about first order reaction of course, it can be any other reaction, but the algorithm or the procedure will be quite similar be quite similar. So, there are many examples in assignment there are many example in text books like Levenspiel, Fogler and all like and you can solve these examples and if you are clear about the overall procedure then there should not be much of a problem. Now, there is some correlation between this N and Peclet number. So, approximately this N is nothing, but Peclet number by 2 plus 1 approximately this is approximate relationship. So, they are related. So, if you if you want to compare these two approaches where the question is like whether I should go for dispersion model or tanks in series model. So, that will depend on your choice whether you like differential equation or you like algebra equation because in the case of tanks in series model you need to solve algebra equation case of dispersion model you need to deal with differential equations. Typically the calculation wise or if you want to remember formulae and all that probably tank in series model is relatively easy compared to dispersion model fine. Now, let us go ahead. So, what we have seen here is that we can use one parameter model if you have a typical E curve. So, it is always it is very important like once you do experiment tracer experiment you get an E curve you should know whether you should apply one parameter model or some other model because sometimes one parameter model may not fit I have been repeating this again and again because it is very important. So, sometimes students are in impression that one parameter model can be used for any reactor. So, after you do entire exercise realize that you have got a value which is or you come with a result which are of no use because the E curve does not represent a tubular reactor which can fit in either dispersion model or tanks in series model fine. Now, we are going to go one step ahead and look at two parameter model. Now, why do we need extra parameter because one parameter is not able to explain the flow behavior properly as simple as that as I said like if I go on changing the value of Peclet number or n I get a typical behavior like this of course this peak is going to go increase, but it is not necessary that E curve falls in these categories only or you will have E curve of this nature only as I said before I may have E curve like this I may have E curve like this it all depends on the geometry that you have it all depends on what is happening inside. So, in that case you should be quick enough to know that my one parameter model is not going to have I have to go for a two parameter model or a multi parameter model. So, here we need to know what happens inside reactor you have should have some information about what is happening inside a reactor. So, that I have proper guess as to how to go ahead now I will give you an example here. So, you have you have a normal CSTR like look reactor all I would not call it a CSTR it is a mixing tank or mixer or reactor. Now, this is the this is the way I represent CSTR, but in reality I will have a vessel I will have a vessel in that I may have a reactor sorry agitator I will have an agitator. Now, I am looking at the geometry also. So, it is not a schematic, but it is somewhat real picture it will have a nozzle here I will have a nozzle here and I have field coming in and the product stream going out from here. Look at this nozzle is now close to inlet now this is a way. So, it may be a overflow. So, the whatever extra comes in goes out from here and there is intense agitation inside. So, you have a nice mixing happening so called nice because most of the times like you do not have proper mixing in each and every corner of the CSTR or in a mixing tank for that matter. There are always some pockets in the reactor which do not see the kind of turbulence intensity which is there near the impeller you can very well appreciate this fact. See if you want exactly same amount of turbulence intensity or mixing level everywhere in the tank then you need to put agitators as many agitators as possible at every position in the reactor is not it that should be the case, but that is not practical. So, normally you have only one agitator for some time you may have one shaft on which you have multiple agitators or impellers mounted. So, it is quite possible that if you are away from the impeller especially in the corners then there are stagnant pockets. So, what is likely to happen in this particular case see the feed is coming in and you have certain dead zones possible. So, at this point is away from agitator this point is away from agitator. So, somewhere in this part you will have sorry I will use another pane you will have dead zones. This is not a only effect another effect can be that some of the stream see there is a stream you can imagine now this stream gets divided in several sub streams. So, it is quite likely that some part of the inlet stream will directly go here and will not see the agitator at all what is this is bypassing these are dead zones and in between you have perfect back mixing in between or other in this particular region sorry. So, now I have three things happening bypass dead zones and mixed zone. I know how to model this this is like an normal CSTR how do I incorporate this how do I incorporate this in my model. So, I need to have a combined effect of all this happening here. So, what I am going to see at exit is a effect of what happens here what happens here and what happens here. So, all these three effects are going to tell me the conversion or go to decide a conversion. So, if I just assume a normal CSTR a mixed zone I may go wrong. I need to consider these two effects also which are probably not desired, but definitely I have to consider them to design the reactor bypass and dead zones and there is no provision to consider these effects in my one parameter model that is tanks in series or dispersion model there is no provision for these. So, I need to have additional parameters I need to complicate my model further. So, that I get these effects incorporated in my model and my prediction of the conversion will be more or it will be closer to the reality. Thank you will discuss this further in the next lecture.