 We are going to be looking at gas solid reactions effect of equilibria. Now, just give a background on this whole effect of equilibria. Let us just look at what we have learnt in our school on the blast furnace. Blast furnace technology is rather new perhaps last 50, 60 years, but steel making has been known for a very long time. For example, I mean if you look at the history of India for example, there are lot of reports that is available around the world when it says that India was the largest exporter of steel. Now, why is it that you know these people have been able to make so much of steel is because raw materials are available and so on. Now, to be able to get good quality steel just look at some of the reactions that are taking place here. You have ion oxide in the form of ion ore it is coming in. You have charcoal is a reducing agent and some fluxes to take care of some other reactions. And then as this material keeps moving down and more and more reactions take place. And then the reaction for example, ion oxide reacts with carbon monoxide to give you ferric oxide and then carbon dioxide as goes down and found you find finally, this ion oxide reacts with CO and finally, give you ion and then carbon dioxide and finally, molten metal comes out. But the traditional process of steel making in India was very very different. For example, what is seen around the world today is that you get a hot metal. And because the hot metal and which is protected by slag certain amount of dissolution of these materials also take place in the hot metal and which create certain problems in the point of view of using this ion. But the traditional process that was happening in India was that this ion Fe was actually harvested as a solid and not as a liquid. This is what was unique about the Indian process I mean very old perhaps going on for last 1000 years. And even more interesting perhaps some of you may not know is that beginning may be 4th or 5th century AD to as late as 1500s. All the wars particularly in Europe in West Asia were fought with what is called as the Damascus word and the steel that was used for the Damascus word was actually made in a some places around Andhra Pradesh and from where it was exported for a very very long time. In other words this technology of steel making that is traditional to India is interesting is that they were making solid ion rather than liquid ion. So, in other words what we are trying to say is that if you know how to handle the equilibria then we can drive the reaction in the appropriate direction. This is something that we all know and this and something that our you know people who have I mean created this process or understood this. We want to quantitate this by looking at some examples some more examples. Let me just quick put down the next example which is also very familiar to you which is calcium carbonate decomposition C O 2 sorry C A C O 3 giving you C O 2 plus C A O. I call this as A call this as B call this as C. Let me write down the K P for this reaction at different temperatures K P is in M M H G T is in Kelvin 0.073, 1.84, 22.0, 1 167, 1 793, 2942, 1 sorry 773, 873, 973, 1073, 1173, 1273. Let me write down the rate of formation as K 2 times C C minus K 2 dash C B times C C minus K 1 dash C A. So, this R A refers to the formation of calcium carbonate. So, that K 2 dash C C B minus K 1 dash C A. Now, since this is a solid this is a solid and we take solid as unit activity of course in actual practice C is solid and B is gas sorry. Now, if you take solid as unit activity we can simplify this let me just quickly write that down. So, with solid as unit activity our R A becomes K 2 C C minus of K 1 or I will put this as K 2 times C C minus of K 1 divided by K 2. Let me simplify it further which is K 2 times C C minus of K C and I will write it as K 2 by R T P C minus of K P. So, the rate at which the decomposition reaction C A C O 3 equal to C O 2 plus C A O takes place really depends. Therefore, the decomposition reaction takes place when the formation takes place when P C is greater than K P. Therefore, the decomposition takes place when K P is greater than P C is it clear. So, the most important thing is that when P C what do we conclude from here R A refers to rate of formation of calcium carbonate. So, when would calcium carbonate formation take place when P C is greater than K P when would the reverse reaction takes place. So, we say minus of R A equal to K 2 by R T K P minus of P C is it all right. So, the decomposition reaction takes place when the choice of K P is such that this term is positive is this point clear to all of you. So, let us look at this once again. So, if you look at the decomposition K P values for decomposition we find that around 1073 K P values are about 167 correct. Now, the partial pressures of carbon dioxide in combustion gases something like 9 percent 10 percent. So, it is about 65 to 70 mm is what we expect in the combustion gases. Therefore, if reaction has to take place at all then the temperature at which we must perform this must be greater than the K the about 65 mm which means you must choose temperatures around 1073 at least to give us the driving force for the decomposition. So, this is why you will find in all the lime kilns around the world the temperatures are well above 1073 well above. Now, if you look at this planet earth all of you know that carbon dioxide in this planet is primarily present in rock. That means all the carbon dioxide is present in the form of calcium carbonate rocks. If you go to the martian atmosphere carbon dioxide is in the atmosphere that is how its whole thing is organized. That means depending upon this K P values you will find the carbon dioxide exists in appropriate forms in different planet to the world of the universe effect of changing gas composition. Now, the context to this let me put it down once again the context to this is as follows. Let us say you are conducting a reaction like zinc sulfide zinc blend is a very important ore in this country it is available in Udaipur area in the Stansink it is used to be now it is worn by Vedanta group. So, they oxidize zinc sulfide zinc sulfide with oxygen make zinc oxide. So, you get sulphur dioxide what is it that they do they have a fluidized bed in which they contact this the zinc sulfide is fed like this and air is coming like this it gets oxidized and so on. So, what is important here is that the effect of the composition which means the effect of what is the oxygen concentration that is in contact with zinc sulfide is important for the rate of chemical reaction. We must be able to understand this. So, that we want to put those numbers in the appropriate perspective. So, let us say the reaction is A gas plus B B solid giving you C gas plus D solid. Let us assume that it is in contact with for the moment you have gas and you have solid that means gas and solid for the for the moment it is in co-current flow of course, we can look at various other features later on co-current flow. So, our stoichiometry let me put down stoichiometry F A F B F C and F D are usual nomenclature nothing new here 1 minus of X A F B 0 F L 0 times X A F C 0 plus F C 0 plus F C 0 X A F D 0 plus F A 0 X A. So, these are stoichiometry and we notice that if you want C A it is simply F A by V. So, this is F A 0 in our nomenclature divided by V we take V as V naught taking the gas law into account V naught it becomes T by T naught. This is the effect of the temperature on gas flow and so on. Similarly, for we can do for others let us do for others C A and C C C C C equal to sorry C C equal to F C 0 plus F A 0 X A times T naught divided by V naught times T. So, C C becomes C C 0 plus C A 0 X A times T naught by V naught. So, this is T. Now, we just want to see how the equilibria effects the process. So, we recognize that K P for this reaction for this reaction let us look at it once again for this reaction K P is P C by P A that is what I have written here P C by P A. So, K P is this now this is written as in this form C C 0 plus C A 0 X A divided by P A which is C A 0 times 1 minus of X A. So, at equilibria I am putting a star to denote its equilibria accordingly we get K P now depends on C C 0 plus C A 0 times X A divided by P A. So, C A 0 times 1 minus of X A it is at equilibria. So, this gives us X A star this fairly simple arithmetic. So, it divided by 1 plus K P where theta C is simply F C 0 divided by F A 0. Now, what we are trying to say here is that the value of X A star at the solid surface depends on the choice of theta C. On other words if the product what is our product please recognize here our product is product is C if product is present in the feed it has a bad effect on the process because it affects the extent to which we can drive the reaction at equilibrium. So, this is one important message that we all know from basic thermodynamics that is stating the whole thing once again. Now, let us look at the chemical reaction A gas plus B B solid equal to C gas plus D solid this is the reaction we are considering. So, the rate at which reaction occurs we want to do it in this co-current flow in a rotary kiln for example, it is R A dash times A s. What is R A dash? R A dash is the rate of chemical reaction per unit surface area and A s is the surface area per unit volume. So, that is how and if R A dash is due to a film diffusion control of which we have talked about earlier. So, it is taking a form of this nature therefore, you will get R A dash equal to minus of K G I forgot the minus sign here K G times C A 0 times 1 minus of X A minus of C A naught times 1 minus of X A star. So, this is the rate at which reaction occurs and I have to multiply it by this is R A. Therefore, D F A which is the rate at which chemical reaction takes place in our equipment is now K G C A naught within brackets of X A star minus of X A multiplied by the surface area per unit volume. Now, this X A star we have already said let me put this in this form D X A by D V with a minus sign equal to with the minus sign K G C A 0 times X A star minus of X A times A s. Now, what we say is that this X A star we have already shown little earlier if you recall here the X A star we have already shown is given by K P minus of theta C by 1 plus K P. Therefore, we are now in a position to tell what will be the rate at which our chemical reaction will occur in the rotary kiln in the co current flow of gas and solid. Now, also recognize that this X A star depends upon the choice of temperature it depends on the choice of the conditions under which we will run the process. If there is product in the feed to that extend you lose reaction rate. So, in the context is that combustion gases contains carbon dioxide and therefore, you have to deal with the fact that carbon dioxide will have a negative influence on the rate of reaction. Therefore, you have to choose the temperature at which you will run the process having high K P values. So, the driving forces are satisfactory that is why people run rotary kilns at up something like 1150-1200. Now, let us take this forward a little we have D X A by please notice that the left hand side can now be written as D X A by D where it is a gas residence time K G C A 0 cancels off. So, you get X A star minus of X A times A S. So, what is tau G tau G is simply gas residence time where tau G is given as sorry tau G is simply V divided by V 0. Now, let me restate this equal to K G times X A star minus of X A times A S what is A S? A S is surface area per unit volume for film diffusion control we are looking at film diffusion control here where the surface area which is relevant to the process is the external surface area. So, we said A S is given as 4 pi r square which is the surface area of the particle there are n particles per unit volume of the reaction equipment. So, this is the surface area of our interest experimentally we can determine this quantity what is called as 4 by 3 pi r cubed this is the volume number of particles divided. Generally for rotary kilns in fact, for any reacting equipment this epsilon r is an experimentally determinable number. What means how much solids are held per unit volume of the equipment? Suppose you stop the rotary kiln and then go and collect all the solids you will find there are so many cubic meters of solid per unit volume you can calculate the epsilon r is a well documented number. Therefore, depending upon the rotation speed this number can be obtained. So, accordingly A S can now be written in terms of the epsilon r. So, it comes out to be how much in terms of epsilon r is how much is it? 3 epsilon r by r is it all right. So, let us substitute for for this one here. So, you get k g x a star minus of x a A S is thrice epsilon r by r is it ok. If you have a rotary kiln in which there is gas and solids are in co-current flow it is a co-current flow. Then directly this equation will tell us what will be the extent to which we can drive the process. So, once you know gas residence time to find out solids residence time how do you find solids residence time? Solids residence tau S is simply volume of the equipment divided by volume of solids all the all the solids you have to find the volume. So, this is known the solid volume is known because you are putting in solids at a certain rate. So, you know the solids volume you know the reaction volume of the equipment what is v equal to v equal to v naught times tau g. So, you can calculate the solids residence time as well. So, given given gas residence time data you can calculate gas solids residence time. Therefore, you can specify the whole process on the basis of this equation. Let us say this is equation 1. If instead of gas film control suppose let us say you have a reaction control process what do we have? You have a gas plus b b solid giving you c gas plus d solid is it all right d solid correct. Now, if it is under reaction control once again our basic equation that describes our process is r a dash times a s. What is r a dash for reaction control r a dash for reaction control r a dash for reaction control looks something like this is k s times c a minus c s star. Once again the form is identical to what we have written for film diffusion control. And what is a s for reaction control we have said this before it is 4 by 3 by r cube n divided by v is this correct this is epsilon r this is epsilon r. So, a s is 4 pi r c square n divided by v this is a s and this is a epsilon r this is an experimentally measured experimental quantity all right this is an experimental quantity. So, this comes from experiment a s for reaction control is 4 pi r square n by v. So, we can simplify this and now write a s as thrice epsilon r divided by r within brackets of r c square by r square is it all right. So, this I will simplify this as thrice epsilon r divided by r 1 minus x b to the power of 2 by 3 is it all right. So, let me write a s equal to thrice epsilon r divided by r multiplied by 1 minus of x b to the power of 2 by 3. How is x b related to x a from our stoichiometry we have already done this. So, it is simply divided by theta b. So, this comes from stoichiometry. So, that our now our equation now looks like this d f a by d v equal to k s times c a star c a minus of c a star times a s which is thrice epsilon r divided by r 1 minus of x a by theta b to the power of 2 by 3 with a minus sign. So, r it is d x a by d v f a 0 with a minus sign here equal to minus k s c a minus of c a star thrice epsilon r by r 1 minus of x a by theta b to the power of 2 by 3. So, what is it that we have done identical to what we have done for reaction control I mean film diffusion control a similar equation is there for reaction control. Now, we can simplify this and write it in this form of d x a by d tau g please recognize that it is tau g is gas residence time. And c a can be written like this k s times x a star minus of x a multiplied by this effect which is by r 1 minus of x a by theta b to the power of 2 by 3 is it all right is it. So, what are we saying now if we have reaction control instead of film control the form of the equation the describes your process is given by the right hand side k s is the reaction velocity x a star is your equilibria which is already determined. And there are all the other things refers to the process of epsilon r comes from experiments capital R is the size of the equipment. So, this can be integrated if you know the initial condition how do you integrate this x a equal to 0 tau g equal to 0. Therefore, a forward march this can be integrated and you can get how x a changes with tau g. Now, third example is the case of what is called as ash diffusion control. What is the ash diffusion control we talked about it which said that it is the resistance to the supply of gas to the product layer. So, let us write our differential equation once again you have d f a d v equal to some r a dash times s which what is this r a dash times s it is the rate per unit per particle. So, I am just multiplying this by thrice epsilon r divided by 4 pi r cubed. Please understand what I am saying what is epsilon r what is epsilon r epsilon r is hold up of solids per unit volume. So, that means every unit volume has epsilon war cubic meters of solids what is the volume of each particle 4 by 3 pi r cubed. Therefore, this 3 epsilon r by 4 pi r cubed refers to the number of particles where r a dash s refers to reaction per particle. So, this is the rate at which reaction occurs to unit volume per unit time. Now, what is this r a dash s we have already done this we said this is minus of 4 pi d c a minus of c a star we have done this before divided by 1 by r c minus 1 by r we have done this also and then thrice epsilon r divided by 4 pi r cube. So, this r a dash s this form we already derived you only multiplied by the appropriate number to take care of the number of particles per unit volume. Now, we can integrate this. So, help me now. So, left hand side becomes f a 0 d x a by d tau g with a minus sign equal to right hand side is minus of 4 pi d this becomes c a 0 x a star minus of x a divided by 1 by r c minus of 1 by r and put this here thrice epsilon r 4 pi r cube. Thank you is it ok now is it all right. So, let me write this in this form d x a by d tau g and the right hand side signs go off c a 0 cancels off I get 4 pi d x a star minus of x a thrice epsilon r divided by 4 pi r square r by r c minus of 1. So, this d x a by d tau g equal to thrice epsilon r d divided by r c minus of 1. So, this r square within bracket I am just putting this to that x a star in this form minus of x a divided by within brackets 1 minus of x a by theta b minus 1 by 3 minus of 1. Do we all agree with this from here r by r c is written in this form s or no. So, this term this term is r by r c square is it r by r c r by r c is what what is the r by r c 1 minus of x b to the power of 1 by 3 and x b is x a divided by theta b comes from stoichiometry is it all right s and o. So, for the case of as diffusion control our final form looks like this this is as diffusion control. Now, we can integrate this how do you integrate this at tau g equal to 0 x a equal to 0. So, for a therefore, if you have a range kutta routine if you want to integrate forward you only require the right hand side at tau g equal to 0 right hand side is fully known. Therefore, you can forward march and then complete the integration. So, all the 3 cases case of reaction control case of film diffusion control case of as diffusion control we have forms by which we can integrate forward and determine the size of the equipment for a given process that you have chosen is that clear. Now, we can take this a little forward and by and look at combination of resistances it is it is fairly elementary we have done this in earlier. But, let me run through this once again for the case of as diffusion control for combination. Let me just put down all the things once again because that makes it a little easier equal to minus of k g c a 0 x a star minus of x a is this writing it again thrice epsilon r by r this is for film diffusion. So, we have done this before now I am writing it again for the case of reaction control c a 0 x a star minus of x a thrice k s epsilon r by r 1 minus of x a by theta b to the power of 2 by 3 this is reaction control this is alright reaction control. Now, for the case of what is this this is as diffusion control our numbers look like this thrice epsilon r d c a 0 x a star minus of x a divided by I am just writing it once again exactly what we have done before 1 minus x b it to the power of minus 1 by 3 minus 1 this is something that we have done just now. So, this is for the this is ash control therefore, if you have a rotary kill in which all the 3 are important how do we combine them we combine them by recognizing the following we write resistance equal to potential by flux. So, for the case of film diffusion our resistance this becomes 1 by thrice k g epsilon r by r please look at the form look at the form here. That means, thrice k g epsilon r by r is this clear I am just writing the resistance by looking at this form itself the resistance is potential divided by flux therefore, it becomes 1 by with the minus sign k g thrice epsilon r by r. Similarly, for the next case which is film and reaction let me write this is for film and for epsilon omega 2 I am writing it as 1 divided by thrice k s epsilon r by r within brackets 1 minus x a by theta b to the power of 2 by 3 this is reaction. Let us just check this once again c s divided by this it becomes 1 divided by thrice k s epsilon r by r exactly is what I have written is it all right it comes it comes from the previous form please I cannot show both at the same time. But, if you look at this here c a 0 x a star divided by this becomes just inverse of this and for the third case which is epsilon third that becomes minus 1 minus of x b to the power of minus 1 by 3 minus 1 divided by thrice epsilon r divided by r square. So, this is the form in which the resistance for ash control. So, if you want to combine all the 3 three our procedure is what is our procedure that flux equal to total potential divided by the by the total resistance. So, let me write it in that form which means for the case of combined resistances combined resistances resistance flux which is d f a by d f a d v should be equal to potential divided by summation of sigma s n o. So, once we put all the resistances together our numbers look something like this I will not write this in fact I will write the final form because it is fairly elementary. So, the final result looks like this it is not necessary to do the whole thing again and again because we already done that becomes x a star minus of x a within brackets of thrice k g epsilon r by r the first one. Then you have thrice k s epsilon r by r within brackets of 1 minus of x a by theta b to the power of 2 by 3. So, this is reaction third one is ash diffusion which is epsilon r and diffusion coefficient divided by r squared 1 minus x a by theta b is a little messy, but you know it is something that we have to get used to minus 1 by 3 minus 1. Is it all right? So, is exactly similar to what we have done for single particle. So, if you have a rotary kill gas co-current flow of gas and solids the gas conversion with respect to gas residence time is this is the potential divided by resistance. The right hand side once again x a equal to 0 at tau g equal to 0 which means for a forward march ranje kutta routine the right hand side is fully specified. Therefore, you can integrate forward and complete the process by whatever is specified the rest of it can be done through the appropriate integration. Notice here that k g is an experimentally known quantity k s is a known quantity diffusion coefficient is a known quantity epsilon r the hold up is a known quantity. So, right hand side everything is known and therefore, it will be able to determine the extent of reaction for a given residence time. Once gas residence time is known you know the volume of the equipment and if you know the volume of the equipment you know the solids residence time. So, your process is fully specified. So, for the case of gas solid reaction taking place in a rotary kill under co-current flow we have the process design completely specified. Now, if I ask you what is the way by which we can tell what resistance is controlling out of the three whether it is film diffusion is controlling whether it is reaction is controlling whether it is ash diffusion is controlling. We said one way of knowing this is to try and do an experiment where we change velocities. When you change velocities mass transfer coefficient changes generally to the power of 0.8 of Reynolds number. So, you will find velocity effects generally affecting the mass transfer coefficient. Therefore, if you do three experiments in three different velocities you will be able to tell whether the mass transfer is an important resistance or not. Similarly, if you have control due to chemical reaction chemical reactions are very strong functions of temperature. You do experiments at three different temperatures you will find that if it is important then temperature effects will show up because reaction rates will change rapidly because of the choice of temperatures. If it is ash diffusion control we know that it depends on square of particle size. Therefore, if we choose different particle sizes immediately that effect will come out. Therefore, to discern the importance of controlling regime is essentially do some experiments to find out what is important. Now, having said this let us look at an example from see we did at in a. So, this is conversions from R T D. See what we have done so far if you look here what is it that we have done so far. Here we said gas and solids are in coherent flow and the implicit assumption here is that both are in plug flow. Therefore, the residence times are the same for every particle. This is implicit in this formulation, but this may not be the case. Therefore, we will have to see what can we do in case there is an R T D. The R T D can be for gas it can be for solids. So, we have done this I just said it down for the case of film diffusion. Let us say there is a single particle and we have said this that if there is a single particle that particle behaves like this. We have derived this if there is a single particle and if it reacts and it will react in this form. Now, if you now put this inside a reaction equipment this is the reaction equipment where you know that it has R T D is the R T D of the reacting equipment is some E function. Let us say this is known to you. How does it come? It comes from an experiment. We have done these experiments. So, whatever be the equipment we can determine what is the residence time distribution for that particular. If it is for solids we do an experiment by putting a tracer on the solid state. If it is for gas we put a gas tracer. Both types of experiments we may have done in our undergraduate course or it can be done is not a very difficult experiment to do. On other words we know what is the residence time distribution. Now, what is it that we want? We want so we have 1 minus of x B equal to 1 minus of x B 0 to infinity. Can we say this? We said this in the context of we what we have said so far is I just recall what we have said. We said C A average equal to C A element multiplied by E T D T. So, exactly what I have written is same thing is being written. So, the average you will see is average each particle multiplied by the E T D T of that particle. Now, what happens in a gas solid reaction is that let us recognize that once again in the gas solid reaction we have 1 minus of x B equal to 0 to tau 1 minus of x B to the E T D T plus tau to infinity 1 minus of x B E T D T. Can I write this? Now, what happens to this integral tau to infinity 1 minus of x B E T D T? What happens to the integral? We know that at tau x B is 1. Therefore, 1 minus of x B is 0 for all particles with time of residence greater than tau. Therefore, the second integral goes to 0. Therefore, in gas solid reactions where the time for complete consumption is finite the integral has to go from 0 to tau and not 0 to infinity. Is this clear? So, recognize that the integral goes from 0 to tau and not 0 to infinity. What we have said is that this integral 0 to infinity have a broken up into 0 to tau and tau to infinity. Now, 1 minus of x B what is the value of 1 minus of x B? For time of residence greater than tau every particle is fully converted for time of residence greater than tau. Therefore, 1 minus of x B for that particle is 0. Therefore, the second integral is identically 0. Therefore, the second integral disappears. Is that clear? Yes or no? 1 minus of x B is 0 because the time of residence is greater than tau. When time of residence is greater than tau the particles fully consumed, fully reacted x B is 1. Therefore, 1 minus of x B is 0. Therefore, 1 minus of x E t d t is 0. Therefore, we delete that term. Therefore, 1 minus of x bar B which is the average extent of reaction you will find on the particle is 0 to tau of 1 minus of x B E t d t. What is 1 minus of x B for the case of film diffusion control? We have already written 1 minus of x B. We have said it is 1 minus T by tau f. It has come from our single particle analysis. Therefore, 1 minus of x bar B equal to integral 0 to tau 1 minus of T by tau f E t d t. This is all right. What is the first term? First term is how the particle behaves. What is the second term? Second term talks about how much time this particle is spending in the equipment and it is this product which gives you the average. Is this clear? First gives you the behavior of the particle. Second term gives you how much time the particle is spending in the equipment. Therefore, that product gives you the average integrated over 0 to tau. This is clear. So, that is what this whole thing about. Now, if you have done this for the case of film, this is for film diffusion control. And similarly, you can do for reaction control and so on. See, I just want to begin what is called as population balance modeling. See, if you look at a chemical reaction particularly with respect to particulates like gas solid reactions. Where solids are moving, gas is moving. And we want to understand how the solids, how much time it is spending in the equipment, how much time gas is spending in the equipment. So, basically we want to get more clarity on what happens to each element that is going into the equipment. So, this is a very nice technique and I want to present this in the simplest form. There is lot of material in the literature. In this simplest form, I have taken an example, a simple example to illustrate what we want to do. Now, we have been talking about stirred tanks for a long time. So, what do we have? In stirred tank, we have a fluid entering the equipment at some flow. And there is some concentration whatever that may be and comes out at some other concentration because of the reaction. Now, if I ask you, what is this S naught? What will you tell me? You say it is concentration of material that is entering the equipment. Suppose, I ask you how do you know it? What that number is? You will answer saying that I have measured this. This is the measurement that I have done by taking samples. Now, we write our material balance. We write like this. This is how we generally write our material balance. So, this is a material coming in. This is material going on. This is the material that is generated and this is what is accumulating. Now, what I want to say now? You do not have to agree with me, but the argument is like this. What we are measuring is not S naught. We are measuring some average. What we are measuring is some average. Would you agree with me? Any measurement we are doing is an average of the samples we have taken. What I now want to say is that this number that we are measuring is actually this. What is F naught? The F naught is the distribution of that property S of which we have taken samples. That means, there are many fluid elements in our sample. What you have measured is some average and that average is defined as the first moment of the distribution. Is this clear to what we are saying? Every measurement we do is an average and that average is obtained by integrating that distribution property first moment of the distribution property. Therefore, if F naught is the distribution of this property S at the feed. Therefore, the average we measure is actually integral S F naught of S d S. Similarly, what we measure on the other side is I will call this as F. I have put this as F 1 or norm of F 1. Is it okay? Yes or no? Is it okay? How do we understand this R bar? We understand R bar equal to integral R F 1 S d S. Is it okay? The meaning of R bar what is F 1 at the exit is also the F 1 in the equipment. That is the meaning of a stirred tank. We are looking at a stirred tank. We will relax all this as we go along. We will look at other situations where we can take care of all this. For the moment F 1 at the exit is same as F 1 in the equipment. That means, the distribution of the property that we are trying to understand it is the same inside the equipment as it is at the exit because that is the property of stirred tanks. Is it okay? So, what I want to do now is put this definitions. What I want to do now is that now that we know what is S naught, what is S bar and what is R bar. We can substitute in this equation. Yes or no? So, let us substitute and see how it looks like. So, let me replace V naught integral S F naught S d S minus V naught integral S F 1 S d S plus integral V R F 1 S d S equal to del by del t of V S F 1 F 1 d S. I will put it like this. Is it alright? What I have written? I have written it correctly. Please tell me. Now what I am saying is that I will call this. I want to integrate by parts. This integral I want to integrate by parts. So, this is dS is our first function. So, dS is our second function and this is the first function. So, integrating the parts, please help me. Let me just write down and you tell me whether what I have done is right. I am integrating by parts. So, I am writing the first two terms as such. No change. Now I am integrating by parts. So, first function into the integral of the second. So, I write this as first function into integral of the second. I have written it like this R V F 1 S times S minus integral of differential of the first. Differential of the first is S del by del S R F 1. Is it alright? What I have written? S please tell me. Is it ok? Is this integration correct? And then equal to right hand side del by del t because all the rest is very straight forward once we are clear about this then all the rest is very straight forward. Here, no? Alright. That is a mistake I made. So, first function into integral of the second. Now what I have done is the following. What I am saying now is that let us look at carefully what is this term? First term is S F naught S D S. Second term is S F 1 S D S. Third term is this term minus of S naught del by del S of this whole term. Now what I am saying is please tell me whether I do it correctly. I am just write this. Please observe what I am writing? F naught minus of V naught F 1 minus of del by del S of R 1 V F 1 equal to del by del t of F 1 V. What I am saying is this equation here is first moment of this equation. Do we agree? Suppose you take first moment of this equation. Do you get this? What is first moment? You multiply by S and integrate over the interval. First moment of this equation is what we have got above. Is it alright? We all agree with this. First moment of this equation I call this as equation star then I say that first moment of star is what I have written above. Do we all agree with this? Yes or no? See carefully. First moment of this is this excepting that this term is extra. From this first moment of this term is coming out as extra. Is this clear? I am going to delete that term for the moment. I will talk about it later. For the moment I am going to delete this term. I will give reasons for it a little later. So what I am saying is that equation star represents a more fundamental statement of conservation. So far we have always talked about conservation by talking about averages. Now we have an equation which talks about the distribution of the property which means now if you have a population if you know the distribution of the property of interest in that population then we can now understand how the distribution changes because of whatever happens inside that population. It can be birth, it can be death, it can be grow, whatever various kinds of things that happen in a population we can understand. So this is the fundamental statement of what we call as population balance modeling. You will find in population balance literature people will start with this equation. They will give you no proof. This equation is assumed. What I have tried to do is that how this comes from our basic understanding of material balance in a stirred tank. What I have tried to tell you is that there is one term that we have deleted that people delete. They delete this one term and then then write this star as the statement of conservation when we talk about populations. So henceforth we will write our material balance for population distributions in this form where f will refer to it can be activity of a catalyst or it can be size of a particle in a process where it undergoing combustion or it could be you know how if f refers to a share in a market how the share gets distributed among various people. So we can talk about dynamics in an economic system. We can various things you can talk about really once you understand what is this f then we can do lot of these things. Now I will prove this term why I have deleted. I will prove that shortly as we go along and look at some examples. For the moment let us assume that this is ok. Now having said this what people do in the literature I will tell you. People do not derive it in this form. This form is it is not derived instead they do it in a slightly different way what they do is the following. So I will call this a setting up population balance from basics. So what they will do is like this I will call this s I will call this s plus ds ok. So what are we doing we are trying to look at an interval between s and s plus ds and trying to find out what happens to this interval due to flow and due to reaction. Now what happens to this so I am writing convective flow. Convective flow is so much material enters this interval due to convective flow do we agree yes or no do we agree or not what I have written. See we have a stirred tank into which material is entering this is our stirred tank this is our stirred tank material is entering material is leaving this we say is f naught this say is f 1 correct. So if you take an interval between s and s plus ds we want to know what happens to this interval because of flow and reaction ok. Because of flow material so much is material once again it is input minus of output plus generation equal to accumulation ok. For the moment we can write for steady state may be. So how much material is entering interval between s and s plus ds I say it is v naught f naught s ds is this clear to all of you so much material is entering. How much is leaving v naught f 1 s ds ok. Now what happens to this interval because of reaction ok. Let us say our reaction is r 1 ok. So material the reaction is written like this so much of material at s is entering so much of material is leaving. Can we say that in the interval between s and s plus ds so much is whatever reacting at s will contribute to s plus ds correct r 1 this is that means whatever material is happening at s it will contribute to s plus ds. Therefore, whatever is accumulating between here is this difference is what will accumulate. So that is equal to this how our friends will write the population balance in the literature whether we understand this you have to understand this how they will write that means this what is this reaction rate function it increases the property the property increases ok. Therefore, s contributes to s plus ds it is this difference which will contribute to the interval between s and s plus ds. So this is how in population balance modeling they will write the effect of chemical reaction. Now if you take the limit as s tends to 0 our equation will look like this. So this is how they would write is this clear how do we derive population balance equation for a stirred tank. So the interesting thing is only this to understand how do we represent the reaction term contributing to the interval between s and s plus ds this how it is represented. Whatever reacts at s contributes to s plus ds whatever reacts at s plus ds goes out it is this difference which will accumulate in the interval. So this how it is represented we have derived the same thing from fundamentals already previously we derived the same thing from fundamentals is that clear. But we said we will knock out one term we will knock out I have not given you reasons for that we will come to that shortly. But in the literature they would do it like this and not recognize that there is one term which you have knocked out is that clear what we are saying. So basically I mean what they have left out we will prove that is not wrong it is ok. But what you will see in the literature is this. Therefore if you have a stirred tank where the distribution of property at entrance is f naught the distribution of property at the exit is f 1 and the rate at which the property changes because of this reaction is represented by r 1 and t 1 bar represents the residence time of that property in the equipment. So this is what describes how the distribution changes because of chemical reaction is that clear we have run out of time. So what we will do when we meet tomorrow is that we will go through this once again and then try and apply this to one or two situations of you know practical interest in our process industry. Try and understand how we can actually determine the distribution functions and how those distribution functions are affected by the process parameters and how they are useful in designing equipments and in you know understanding how the equipments in performance so on I will stop there.