 Today, we will be looking at residence time distribution. Now, why are we looking at residence time distribution? We know from our understanding of real life that the longer system I mean fluid element spends in the equipment and goes greater amount of reaction. And therefore, if you know the residence time, you will be able to understand how the reaction system is performing. So, what we want to do here is to set up methods by which we can understand how long a fluid element spends in the equipment. Now, for this we use tracer. Tracers, what are the tracers that are used? Tracers used in residence time distribution technique are those that enter the equipment and behave exactly like the fluid elements that would be entering whether it is solid liquid or gas, but does not undergo any chemical reaction. Therefore, it gives you the time the fluid element spends in the equipment. So, examples are let us see you have a gas. If you have a gas, we can have tracers which you can measure. Any gas tracer that you can put into the equipment which you can measure is a useful gas tracer solid. So, various types of suppose you are like a fluid bed for example, if you want to understand the solid residence time distribution, you can put tracer. It can be a radio tracer. It can be simple things like sodium chloride or potassium carbonate. Anything that has similar density as that of the particles of the fluid bed and does not undergo chemical reaction is what is appropriate. If it is a liquid, it can be a colored dye which behaves just like the other fluid similar densities and so on. Therefore, it is able to tell you the distribution of residence time for the liquid. So, it can be a dye here. It can be a radio tracer or it can be let us say sodium chloride. The important thing is that the density must be. So, it can be any gas which can be measured accurately. So, this is. So, a tracer is appropriately chosen so that we are able to put the tracer into the system and measure the concentration of the tracer. Whether it can if it is a tracer gas or the tracer solid or tracer dye, it should be able to measure appropriately. Let us say you have an equipment. Fluids are coming in. Fluids are going out and our interest is to find out how this fluid moves, whether it moves like this or whether it moves like this, like this, like this and goes out. We do not know. So, what we do is that we put a pulse of tracer. What is meant by a pulse of tracer? A pulse of tracer is a tracer of small quantity which is very quickly or suddenly introduced into the fluid stream and this fluid stream it is able to mix quickly and therefore, follow the path of various fluids that is going through the system. Let us say we have some data. I am just putting down some data just to illustrate what we want to say. So, this is some experimental result which I am just putting down some measurement we have made. So, I am just writing down all the numbers. So, suppose we have an experiment in which for various times these minutes, these are the measured concentration of tracer that comes out. These are all done at the exit. These are all exit concentrations. So, as soon as you put this tracer, you start measuring at the exit. This is what you have observed. Suppose if I say that the flow is something like 10 liters per minute. If I give you the flow, you can calculate how much material in milligrams per minute that has come out by simply multiplying the concentration by V. So, you can get all the flows here. So, if you want to find out the grams of tracer that is exiting the system, you simply have to multiply the exit concentration by the flow that you have also measured. Now, let us just plot this data in this form. Let us say we have time versus V times C. What is V times C? V times C by is M which is the mass. So, this will look something like this showing that the amount of material. What are the units of this? It is milligrams per minute. So, if I ask you at any time t, suppose this is at any time t, what is this material? So, this refers to material that has exited the vessel. So, many this area refers to the grams of material that is left the vessel. Now, suppose I ask you what is this area? What will we tell as this area? What is the meaning of this area? This is the grams of material that has left the equipment between t and t plus dt. If this is the time t, in this so many grams has left the vessel. So, if I now ask you what is this term? M d t divided by M 0. What is M d t? M d t is V times C d t divided by what is M 0? M 0 is simply integral 0 to infinity of V times C t d t. So, the denominator is the total amount of material that we have injected. While the numerator is the amount of material that has left the vessel between time t and t plus dt or this is the material that has spent between time t and t plus dt in the vessel. This term is often termed as the E t, the residence time distribution function. So, let me go through this once again. What have we done? Let us just run through the experiment once again. We have an arbitrary vessel for which we want to determine the residence time distribution function. Now, what we do is that since there is a continuous flow input and continuous flow output, we put a tracer appropriately chosen that it behaves exactly like that of the fluid, but we are able to measure it at the exit. That means, at this exit point can be measured. So, we have measured the concentration of the tracer at the exit and this is the concentration that we have obtained. What we have done? We have plotted this data in this form where we say V times C, V is the mole of the flow rate which is known in this particular case is given as 10 liters per minute. In any experiment you know the flow rate anyway. So, you can plot V C which is the grams of tracer per minute versus time. So, when you plot this, so from there we can get what is m d t? What is m d t? m d t is the fraction of material that leaves between time t and d plus d t. What is m 0? By definition is the total amount of material is injected which is 0 to infinity time V C d t. So, both numerator and denominator are known quantities because it comes from an experiment. Therefore, this ratio V C d t by definition this is the grams of material that is left the vessel between t and t plus d t and denominator is the material that is the total amount of material is injected. Therefore, this ratio represents the fraction of material that leaves the vessel between t and t plus d t or the fraction of material that spends time between t and t plus d t which means it spends time t in the equipment. Therefore, this is the fraction which has the residence time which is t and t plus d t and that fraction is given as e t d t. The important message from this experiment is that residence time distribution function that is e function is an experimentally measurable quantity. Now, let us ask one more question here. What is this integral e t d t? What is this integral between 0 to t? By definition what is this meaning? This refers to all material that is entered between time t 0 t d and is left the vessel. So, this is often called as the f function. So, this the integral between 0 to t of the e function represents the total amount of material that is left the vessel between time in this time interval t. So, another property that is obvious from the definition of e function is that e t d t 0 to infinity is 1. It is very obvious from this definition that when we integrate e t d t by 0 to infinity it is 1 which because it refers to the total material that we have injected. Now, we have understood how we can obtain the residence time distribution function. So, let us look at some ideal situations this r t d of stirred tank. What is it that we want to do now is? We want to determine what is the r t d of stirred tank? So, we have a stirred tank to which we have material coming in and material going out. Let us say it is coming in at v naught and let us say it is also going out at v naught it is a well stirred vessel. We are putting a pulse of tracer a pulse of tracer let us say q 0 milligrams of tracer we have put into the equipment. So, we should expect that at the instant let us say volume of the equipment is v at the instant that we have put into the equipment. So, must by definition we should have the total amount of tracer that we have injected must be equal to the volume of the fluid in the equipment multiplied by c naught this is by definition. Therefore, by definition at any other instant of time we should have that any quantity of tracer in the equipment is where c is the concentration of tracer v is the volume. Therefore, any other instant of time we should have this the amount of tracer in the equipment is v time c where c is the concentration of tracer and therefore, total amount of tracer is v c equal to q. So, what is q 0? q 0 is the total amount of tracer that we have injected at 0 time and because of the flow you will find that the q 0 keeps decreasing as time proceeds. Therefore, any instant of time the amount of tracer in the equipment is v time c which is q and q will be less than q 0 by common sense. So, now let us write a material balance to find out what is the residence time distribution. So, let me material balance which is input minus the output plus generation equal to accumulation. So, input of amount of tracer that is input amount of tracer is going out amount of tracer that is generated equal to the amount of tracer that is accumulating. There is no generation of tracer because there is no chemical reaction. Therefore, input of tracer minus the output of tracer plus generation of tracer equal to 0. Now, are we putting continuously any tracer into the system? Our tracer is coming in as a pulse. Therefore, there is no continuous input of tracer. So, what we have is 0 and output. So, what is 0 minus what is the amount of tracer going out which is v time c equal to d by d t of v time c. So, this is the statement of material balance for the tracer. How did we get this? We said that if you put q 0 grams of tracer into the system by definition q 0 equal to v time c naught. At any other instant of time v c equal to q and material balance for tracer is input output generation equal to accumulation. There is no generation of tracer because there is no chemical reaction. There is no continuous input of tracer. Therefore, there is no input. Therefore, you have 0 minus the v c. v c is the output equal to d by d t of v c. So, this is the differential equation which governs the variation of tracer in the equipment. Now, we know that v time c is q. So, we have v and by definition c is given by q by v equal to d by d t of v time c is q. Now, we know that v by v equal to residence time. So, we have here q by tau equal to d q d t or q equal to e to the power of minus t by tau with some constant of integration. This is solution of this is this and then at time t equal to 0 q is q 0. Therefore, this solution is q equal to q 0 e to the minus of t by tau. So, what are we saying now? That if you have a continuous stirred tank ray, then if you have a pulse of tracer that is introduced at 0 time and then the variation of the hold up of tracer in the equipment which time is given by this equation q equal to q 0 times e to the power of minus t by tau. Now, our interest is what is the meaning of residence time distribution? What is the meaning of residence time distribution? By definition is it r t d by definition is what? By definition is v c t d t divided by q 0. This definition is important. What is the r t d? R t d by definition is what is the amount of material that is leaving the equipment between time t and t plus d t. So, between time t and t plus d t, this amount of material will leave the equipment. What is the total amount of tracer that we have introduced in the system which is q 0? Therefore, residence time distribution by definition is the amount of material that leaves during the time t and t plus d t which is this divided by the total amount of material which is injected. Therefore, this is by definition the e t function. So, we have v. What is c t? d t divided by q 0 is also equal to v. C is defined by q at time t divided by capital V times q 0. Therefore, this is 1 by tau q by q 0. q by q 0 we have just now shown it is q by q 0 e to the minus t by tau. Therefore, I am putting it as minus of t by tau. This is the e t d t. So, I have forgotten the d t here. So, e t d t is 1 by tau e to the power of means t by tau times d t. So, this is the R t d for a stirred tank. So, if you want to remove the d t here and d t here, you can say that the e function for a stirred tank is 1 by tau e to the power of minus t by tau. How does this function look like? We just quickly see how this function looks like. Suppose, I plot suppose I make a plot of t versus e t function. So, you can see here e function is 1 by tau e to the power of minus of t by tau. Therefore, at time t equal to 0, time t equal to 0 is here. This is 1 by tau and at infinite time. So, it is an exponentially decaying function like this. So, this is the e function. Let me ask you one more question. So, the next question we want to answer is, what is the mean value of e t? How do you define a mean? What is the meaning of mean? Mean is always defined as t times e to the power of minus of t by tau t t 0 to infinity. This is what is called as mean. First moment of the distribution function is called as the mean. It is a fairly simple thing. You can integrate and check for yourself. This mean will turn out to be tau. So, the mean first moment of the e function is very fairly simple integration. You can calculate and check for yourself. So, the mean of this can be found out to be tau. Now, this next question we would like to answer is, let us say we have a sequence of stirred tanks. We have a sequence of stirred tanks into which fluids are coming in, fluids are going out, fluids are going out, fluids tank 1, tank 2, say it is tank n. There is an n tank into which you have fluids are coming in and therefore, fluids are going out. If you put a pulse of tracer here, then what happens to the tracer at any instant of time in this sequence of stirred tanks? Now, this is something that we can do very quickly. Let us try to do that, because it has some value in terms of understanding real vessels. Let us for the moment assume that v 1 equal to v 2 equal to up to v n. We make this generalization because there is certain advantages, which we will see shortly after going through this exercise. We will see the advantage of these simplifications. So, you have tank 1. We have already done this tank 1, where we have said q 1 equal to q 0 e raised to the power of minus of t by tau. I am call this as tau i, because I am assuming that v 1, because they are all equal. Therefore, v 1 by v 0 is tau 1, which is also equal to tau 2 equal to tau i. In other words, the residence time in each of these tanks are the same. That is the assumption. The advantage is we will see shortly. So, based on this, we already derived this q 1 equal to q 0. It is already done from the previous exercise. So, we will not do for tank 1. We will try to do for tank 2. That means, we want to write a material balance for tracer entering tank 2. So, let me write input minus of output plus generation equal to accumulation. What is the amount of material entering this one? There is tank 2, which is v 0 times c 1. What is going out? v 0 times c 2. There is no generation equal to d by d t of v times, which is v 2. So, I am just calling all the v, same volume v. That means, v 1 by v 0 is equal to all have the same value. I have taken it as v, v times c 2. So, this is the volume. That means, I have taken v 1 equal to v 2 equal to v. So, that is the volume that is mentioned here. That is v 1 equal to v 2 up equal to v, all of the same volume. Or we can keep it as v 2 for the moment. There is nothing wrong with that. Now, if I divide throughout by v 0. So, what do I get? Let me write this as, how should I write this? I want to write this as q by v. So, v 0 times q 1 by v minus q 2 by v equal to d by d t of q 2. I put this c 1 as q 1 by v, c 2 is q 2 by v. I can write this as v 1. I can write this as v 2 and v 1 equal to v 2. So, there is no… Let us take this argument further and say that this becomes q 1 divided by tau i minus of q 2 divided by tau i equal to d by d t of q 2. q 1 is what? q 0 e to the power of minus of t by tau i by tau i minus q 2 by tau i equal to d q 2 by d t. Let me write it in this form, d q 2 by d t plus q 2 by tau i equal to q naught e to the minus t by tau i by tau i. Let me just see whether I have got all these things properly. So, you will d q 2 by d 2 q 2 by tau i equal to q 1 by tau i, which is known from the previous. So, this for you to solve this, you only need to know what is q 2 at time t equal to 0, which you know as 0, because at time t equal to 0, your tracer we have put into our initial condition is that the tracer was put into tank 1. Therefore, in tank 2, the amount of tracer at time t equal to 0 is 0. Therefore, q 2 equal to this is the initial condition for this problem. So, you can solve this, there is fairly elementary. So, it is nothing very difficult about solution of this equation. Solving this, we will get I will write the solution, because it is not very complicated. The solution to this is that q 2. So, let me write the equation once again just to remind you, this is equation that we have q 2 by tau i equal to q 0 by tau i e to the power of minus of t by tau i. So, solution is I will just write the solution. Solution looks like this. So, q 2 divided by q 0 equal to t by tau i e raise to the power of minus of t by tau i. So, this is the solution for the case of for the second tank. Now, let us just write the equation for the third tank. So, that will make it a little easier to understand what we are trying to prove. What we are trying to prove? We will quickly write down for third tank. Now, our third tank input minus of output plus generation equal to accumulation. So, what comes into our second third tank? This is for tank 3. We are writing for tank 3 input. This is the output. There is no generation equal to d by d t of v times c 3. So, this is v naught q 2 by v 2 and then v naught q 3 by v 3 equal to d by d t of v times c 3. So, we can write this as q 2 by tau 2 minus q 3 by tau 3 equal to d by d t of v times c 3. So, we can write this as q 2 by tau 2 minus q 3 by tau 3 equal to d by d t of q 3. Now, all these tau's tau 3 equal to tau i. Therefore, I can write this as q 2 by tau i minus of q 3 by tau i equal to d by d t of q 3. And what is the initial condition? Let me just set it up once again. This is tank 3 continued. We have q 2 by tau i q 3 by tau i equal to d by d t of v times c 3. So, this is what we are trying to prove of q 3. q 3 at time t equal to 0 equal to 0. As we have, if you please recall in our stirred tank, in our stirred tank, we have put our tracer only into the first tank. Please recognize that our tracer we put into the first tank only. Therefore, in tank 3 at time t equal to 0, the tracer is 0. Therefore, this is 0. So, solution of this once again is not very complicated. I will just write down the solution. So, the q 3 q 3 equal to q 0 t squared by tau i squared e raise to power of minus of t by tau i. So, what is e 3? e 3 by definition is v times c 3 d t divided by the total number of the tracer. So, the total amount of material that we have put in or this v which is q 3 divided by capital v d t times q 0 or e 3 equal to v by v star of q 3 by tau e d t. If you want to put, you can put q 3 d t e d t. So, this becomes q 0 q 3 d q 0. So, q 3 is q 0. We have just now mentioned q 3 is q 0. So, q 3 by q 0 we can put right away as straight away substitute from here as t squared by twice tau i squared multiplied by this is tau i. Therefore, another tau i is coming here e raise to the power of minus of t by tau i. Let me write this properly once again tank 3, tank 3 continued, tank 3 is e 3 d t equal to t squared by 2 tau i cubed e raise to the power of minus of t by tau i or e n, the n th tank I am just writing by induction t n minus of 1. This is actually a factorial 1. So, n minus 1 factorial tau i to the power of n e raise to the power of minus of t by tau i, where tau i by definition is by definition is tau divided by the number of tanks. What is tau? Tau is v total volume divided by v naught divided by number of tanks or tau equal to n tau i. So, this is another relationship that we keep in mind. So, what we have said, let me just run through this once again. So, that you know we have the context, the context is not lost. We have started by saying that there are n tanks, there are n tanks in series and volume and flow is coming into tank 1 and our tracer is put into first tank that and then we are measuring what is happening to the tracer in different tanks. And by writing the material balance, we found out what happens to all of them and then based on our definition of the e function, we have derived that the RTD function for the n th tank. That means, the n th tank that means, whatever the fluid that comes at the end of the n th tank, if you make the measurement of RTD here, this is what we will find that e n is t to the power of n minus of 1 factorial n minus 1 tau i to the power of n e to the power of t by tau i, where tau i is tau divided by n, where tau is simply volumetric flow to the whole, this is the volume of the entire n tank system. So, this v is n times v i, this is something we know. So, tau equal to n tau i is a result we know from first principles. Having said this, let us see what is why are we doing all this? How does it help us? To understand this, let us recognize some simple algebraic relationships. Our e function for n tank sequence, we have derived as factorial n minus of 1 tau i to the power of n e raise to the power of minus t by tau i. This we have derived, this small n and capital N are interchangeably used. Now, there is one interesting relationship that we know from our basic calculus, which I will write down without proof. See, this relationship is very useful for determining values of various integrals that we will experience in RTD. That means, integral 0 to infinity t n e to the minus t by tau is factorial n tau raise to the power of n plus 1. Now, using this relationship, let us now calculate what is the first moment or that means, what is called as first moment mu n. What is the meaning of mu n? Mu n is the first moment of the e function. First moment of the e function is called mu. That means, this essentially tells us what is the mean value of the function. Now, we can do this integration right away. So, we have integral 0 to infinity t multiplied by t raise to the power of n minus of 1 by factorial n minus of 1 e raise to the power of minus t by tau i and then tau i to the power of n d t or m d t. That is just do this integration because it is important. So, mu n of t is integral 0 to infinity t e t d t. I am just substituting them here 0 to infinity t t raise to n minus of 1 factorial n minus of 1 tau i to the power of n correct and then first moment just have got it right. t to the power of n e raise to the power of minus t by tau d t. You call it correct tau raise to the power of n and then e raise to the power of minus of t by tau i d t. Now, you can simplify this and then it is very easily seen. I am just writing down the answers here factorial n tau i to the power of n plus 1 factorial n minus of 1 tau i to the power of n. So, I have just integrated this and using this relationship that we have taken from basic calculus. This is a basic calculus relationship and then it comes out to be please help me to simplify this n tau i equal to tau. What we are saying is that if you have an n tank sequence where the e function is given by this. If you want to find out the first moment as defined by this equation then the first moment turns out to be n times tau i which is tau. So, essentially what it says is that first moment of the e function is tau. This is what we expect based on our understanding that v equal to this tau equal to v by v naught that is it comes from the basic understanding. Now, the second result which you would like to know is what is the variance. So, variance of variance of e n function. Now, sigma n squared by definition is 0 to infinity t minus of mu n squared e n d t. This is how variance is defined as t minus of mu n whole square e n d t. Now, this also not a very complicated integration. So, we just do some simple things. So, that I am just expanding this twice mu n t e n d d t. So, I put d n t afterwards twice mu n then my plus mu n squared e n. Now, therefore, first term becomes integral 0 to infinity t squared e n. Second one is t e n d t is twice mu n squared. So, plus mu n squared e n d t. So, this becomes equal to t squared e n d t integral 0 to infinity minus of mu n squared. So, sigma is simply t squared e n d t minus of mu n squared. Now, we have once again this integral this integral is available to us this integral is available to us. So, we can use the same integral to simplify and find out what is this variance of this function. Let us do that because it is important. So, sigma n squared equal to integral 0 to infinity t squared. What is e n? The e n function is here. Our e n function is here. I will write it down. So, t raise to power n minus of 1 by factorial n minus of 1 e raise to power of minus of t by tau i divided by tau i to the power of n d t. So, it is the first term I have written down minus of mu n squared. We can simplify this further which is t to the power of n plus 1 e raise to minus of t by tau i divided by factorial n minus of 1 tau i to the power of n integral 0 to infinity d t minus of mu n squared. Now, we can use this integral once again the same integral the t n e to the minus t by tau is factorial n n plus 1 tau. So, that same result we can use here when I do that the result turns out as. So, I will expand this and I get the sigma n squared equal to it becomes 1 by tau i to the power of n factorial n minus of 1 denominator stays numerator factorial n plus 1 tau i to the power of n plus 2 minus of mu n squared. So, t tau to the power of n plus tau is comes out. So, it is it is factorial n plus 1 and then tau i square simple. So, this can be simplified further let me simplify this further. So, sigma n squared just help me now. So, sigma n squared equal to it simplifies I have to simplify this. So, tau i n plus 2 tau i squared will stay on the numerator factorial n plus 1 factorial n minus of 1. So, n n plus 1 will stay on the numerator. So, it becomes n into n plus 1 tau i squared minus of mu n squared. Now, n into n plus 1 and then n into n plus 1 is it correct what is mu n squared mu n squared mu n is you have written mu n as n tau i. So, it is minus of n tau i whole squared mu n by definition is n tau i. So, that is what is written here. So, this is tau i squared. So, tau i squared within brackets of n squared plus n minus of n squared that is called n tau i squared or n times tau i is tau by n whole squared or tau squared by n is equal to sigma n squared or n equal to n equal to tau squared by sigma n squared. In fact, this is the result for which we have done all this homework. The result is the very important result that if you have a n tank sequence then the mean mu n mu n is n tau i this is one result and then sigma squared n is given by tau squared by n. So, let us just look back at what we have done. What we have done is we have taken an n tank sequence. We have put our tracer into first tank. We have measured the output at the end of n tanks and then we have plotted that function and found out the mean and variance of the response that we have got. But, in other words what we are saying is that the response at the end of these n tanks is something that you and I can measure. So, that response from that response we have determined what is the mean and variance of the response that we have got. This is the variance of the response this is the mean of the response both mean and that means mu n mu n and the variance sigma squared n both are experimentally measured quantities. Let me repeat what we are saying is that we have determined from for our n tank sequence we have number of tanks number of tanks. We have put our tracer here v naught and then we measured our response at the end of n tanks and then our response in the form of concentration at the end of verses time we have got this response. Based on that response some response we got we found out the mean we have found out the sigma squared or mean of the response the variance of the response we got and then we said that n equal to tau n squared by sigma n squared. This is what we have got that the number of tanks n equal to tau n squared by sigma n squared. So, this is the result that we have got from our tracer. We got this result we got this n is equal to tau squared by sigma n squared. So, which is mu n squared by sigma n squared. So, if you do a tracer experiment if you do a tracer experiment you are able to determine the number of tanks that are required to describe the performance of this equipment and that number of tanks that you determine from a tracer experiment is simply mu n squared by sigma n squared both are experimentally determined quantities this is the important result. Let us see how to make use of this result to understand performance of an equipment we will do that shortly. Let us say now we have a arbitrary vessel we have an arbitrary vessel into which fluids coming in fluids going out and there is a reaction taking place a going to b. Now, we want to know how this reaction will perform in this arbitrary equipment. If it is a first order reaction and let us say what we have done now let us say we divide this assume that you know this tank can be divided into n such tanks 1 to n. So, we have determined the RTD of this for which we know sigma squared n and then we squared mu n squared both are experimentally determined quantities. And on this basis we also know that mu n squared by sigma n squared equal to the number of tanks. That means this particular arbitrary vessel can be described using the tracer curve that number of tanks that will take care of this arbitrary flow is given by mu n squared by sigma n squared. So, we want to use the result to understand this let us do how to do this. Say c 1 we know is c naught divided by 1 plus k tau i this is for a first order reaction. Similarly, you know that c n equal to c naught divided by 1 plus k tau i to the power of n this also we know for a first order reaction. Now, that we know this as a first order reaction what can we do with this what is being said by this model of understanding this non idealities that this c n is c naught divided by 1 plus k tau i to the power of n. We also say this equal to c naught divided by 1 plus k what is tau i it is tau by n tau by n to the power of n. Now, we also write this as c naught divided by 1 plus k what is tau tau we say it is mu n what is number of tanks number of tanks we have just said from our understanding of R T D theory this can be approximated as mu n squared by sigma n squared. So, I will replace this capital N as mu n squared divided by sigma n squared and to the power of n to the power of n also to be determined is sigma n mu n squared by sigma n squared or in other words what we are saying is let me write it once again c n equal to c naught divided by 1 plus k mu n divided by mu n squared sigma n squared to the power of n which is mu n squared divided by sigma n squared. So, this simplifies as c naught divided by 1 plus k mu sigma n squared by mu n to the power of mu n. So, this simplifies as mu n squared by sigma n squared let me write it once again then write very neatly. So, we have c n equal to c naught divided by 1 plus k tau is sigma n squared by mu n to the power of n which is mu n squared divided by sigma n squared. So, this result has the advantage that mu n and sigma n are experimentally determined experimentally determined quantities which comes out over experiments using simply the RTD theory. On other words if you have an arbitrary lesser we can simply do a tracer test to find out the number of tanks that are required to describe the flow field. And that comes from the RTD theory that is described in terms of mu n and sigma n. So, once you know mu n and sigma n and you know the rate constants you can say what will be the extent of reaction or what will be the concentration at the end of therefore, conversion is simply 1 minus of by c naught. So, this is the conversion. So, we can tell based on RTD theory what is the extent to which the reaction will take place in this arbitrary vessel without knowing the details of the flow field. So, it is no longer needs to be an ideal reactor it can be an arbitrary reactor only thing is that we have treated it as a sequence of stirred tanks. And we determined the number of tanks that are required to equivalently describe this arbitrary vessel. So, that is the advantage of this treatment. I will stop there and we will take up other things in the afternoon. Thank you.