 Today, we will look at tubular reactors heated and cooled from the wall. Now, we know our tubular reactor equation looks like this. So, q is heat removal per unit volume and we represent this in this form r i. This goes from i equal to 1 to p there are p rate processes. So, delta H star is the enthalpy change for reaction for a pipe our equation looks like this. Where T c is the temperature of the cooling fluid this is the reactor this is the jacket. Jacket is a T c the reactor is a T. Therefore, this heat transfer is this is the heat transfer and if it is cooling it will be negative. So, T will be greater than T c. So, if you have to integrate this we need to replace this T c appropriately. So, that we can integrate this equation. So, how do we do this? So, this is the case we want to consider in this lecture. So, let us look at the situation in some detail. So, we have here a representation of a tubular reactor cooled from the jacket and you have the cooling fluid entering at T c 0 and going out at T c shall we say T c l and then you have the reactor in which fluids are entering at T naught and leaving at T l. The flow is co-current as you can see both are co-current and there is no phase change is that we are assuming situation where the phase changes are not there. So, how does our equation look like? Now, we have this material balance equation which you have written number of times in the past it looks like d by d b of x j where j is the extent of reaction in the different independent reaction. There are such reactions which you have to consider. So, the right hand side becomes f j which is the function that determines the variation of x s with respect to temperature in the equipment. This we know how to get we have done this before. Now, the energy balance equation for the case of no phase change looks like v c p d T d v this c p is volumetric specific heat and v is the volume of the equipment and sigma i equal to 1 to p there are p rate processes r i is the intensive rate of the different reactions 1 to p and 4 h by d times T minus T c. This is the equation which you want to handle if you want to integrate this r plus 1 equations we must hand know how to manage replace T c in terms of capital T and x 1 x 2 to x r. So, that is what we would like to do now. Now, to do this what we recognize is that since there is no phase change we can write what is called as an overall balance which says this is T naught is the entry this is the energy that is coming into the fluids that is entering. So, this is v naught all right. So, we call this as v c this volumetric flow at the inlet v naught and v c. So, we have v naught time c p multiplied by T naught minus of T r. So, this is the energy that is coming in with the reagents and v naught v c time c p c multiplied by T c naught minus of T this is the energy that is coming in with the cooling fluid. So, at any position suppose we take any position here any position here and then we can write the balance up to that position. So, what is the energy that is released by the chemical reaction i equal to 1 to r f a naught x i multiplied by delta h i star. So, what is this what we are doing is that we are looking at between position 0 and position say any position say how much is the energy that is coming in with the reacting reagents how much is the energy that is coming in with the cooling fluid and how much is the energy that is released by the chemical reaction in this equipment. So, that is what this term f a 0 x i minus delta h i. So, if this plus this is total amount of energy that is coming in. So, how much of energy is going out the going out stream as you can see here the going out stream both the streams are going out here this is the going out stream going out stream. So, the temperature here is T c temperature here is T therefore, the going out stream goes out with v naught c p t minus of t r assuming the c p is the same for the inlet fluid and at any position. So, this is the energy that is that is taken by the the reagents and v c c p c multiplied by t c minus of t r this is the energy that is taken up with the cooling fluid or heating fluid. So, this is the energy out. So, we have energy in and energy out. So, statement of energy balance is there is energy conservation therefore, what comes in must go out. Now, what what is being said here is that f a naught x i is in the terms of moles per second in the i th independent reaction and there are r independent reactions we are looking at i equal to 1 to r. Now, we also recognize that as per the first law convention we are looking at first law convention which means the energy is going from the cooling fluid into the heating fluid that is the first law convention. So, if T c is less than T this becomes negative. So, it does not change the form of the equations all these equations will stand whether delta h i is negative or positive it does not matter appropriately the temperatures will adjust themselves. Now, having said this if you look at this equation 3 from equation 3 you can easily find out what is temperature T c at any position simply by rearranging all these equations you can find temperature T c at any position which is simply T c 0 plus this term v naught c p t naught minus of T sigma what is the heat generation or release divided by g c p c or in other words what we are saying now is that by writing the overall balance we have been able to find out what is T c in terms of variables which is in our differential equation our differential equation involves x 1 to x j x 1 to x r and temperature T there are r plus 1 variables here and this T c what we have done by this overall balance we have replaced in terms of quantities we know and quantities we do not know which is T and x i going from 1 to r. In other words we are now able to replace T c in equation 2 in equation 2 in terms of capital T and capital x 1 to x r and therefore all the equations 1 to r equations and r plus 1 equations the left hand side which involves the derivative involves x 1 x 2 to x r and T and the right hand side involves functions with dependence on x 1 to x r and temperature the T c has now been removed because of this relationship. So, this makes it possible for us to do what is called as forward integration. Now we have virtue of this relationship of this relationship T c equal to what we have written down just now we are now able to integrate this forward because x 1 to x r is 0 at v equal to 0. That means at the entrance to the reactor all the x s are 0 and then T equal to T c that means we have fully specified the initial state of x 1 to x r and T and therefore the right hand side that means we can calculate the right hand side at T at the entrance therefore we can do forward march and then solve these equations using well known techniques of integration of these equations. So, this is the general procedure that is adopted to take care of the effect of the jacket fluid on the reactor performance. Let us look at the second case the second case of tubular reactor heated and cooled from the wall is you have feed entering the reactor this is the reactor. So, this is the jacket as you can see here fluid enters at v 0 and v 0 and goes out at v and T l and fluid is entering at T c l and leaving at T c 0. So, this is the familiar counter current case which we must now learn to deal with. Once again let us recognize that our equations are that you have r d x d x j by d v equal to f j x 1 to x r there are r such equations and then you have the energy balance v c p d t d v equal to sigma i equal to 1 to p r i minus delta h i this is the reaction term and 4 h by d t c minus t this is the heat transfer from the cooling fluid or heating fluid into the reactor equipment. This is by first law convention heating of the control volume is taken as positive. So, if that means t c minus of t if t c is less than t this will be appropriately negative sign showing that the heat flows from the reactor into the cooling fluid. So, depending upon the direction of transfer the signs take care of themselves therefore, there is no problem here. Now, how do we solve equation 1 and 2 now that t c is coming from this counter current relationships which we have to appropriate the account for. For this what we do is once again the same procedure we write the energy balance. So, we have done here overall balance between 0 and any position. So, you have position 0 this is position 0 and any position here. So, what are the inputs what are the outputs you can see here the inputs are coming from the reagents and coming from cooling fluid a heating fluid. So, you have 2 inputs coming here the 2 outputs reagents here and reagents here this is what we have to appropriately account for that is what I have done. So, v c p so you can v c p t naught minus of t I should put v naught here v naught c p. So, v naught c p t naught minus of t is the reagents entering once again no phase change has been assumed. Similarly, you have what is the energy that is coming in because of the cooling fluid or heating fluid at position any position. So, that is that is written as v c c p c t c minus of t r. So, you have this input coming from the reagents you have the input coming from the energy input coming from the heating or cooling fluid. And now between i equal to 1 to r you can see we are going from 0 to position any position. So, in that position we would have had reaction since 1 to r independent reactions which is taken as therefore, we sum over i equal to 1 to r f a 0 x j multiplied by delta x j with this shows the heat released or absorbed depending upon whether delta is positive or negative. So, the left hand side represents 3 terms term 1 is what happens to the reagents the energy picked up by the reagents term 2 is energy picked up by the heating or cooling fluid third is the energy released by the chemical reaction. Therefore, this must be equal to the leaving stream what are the leaving streams the leaving streams are t c 0. So, v c c p c multiplied by t c 0 minus of t r. So, this is the energy that is in the leaving stream and then we have the other leaving stream is the reagents which is v c p multiplied by t minus of t r. So, we have energy in equal to energy out same statement of energy balance. Now, let us see what this gives us. So, by appropriately rearranging we can get what is t c after all we need to replace this t c the whole idea here is that get rid of this t c. So, that we can integrate. So, that is what we are trying to do here also we replacing this t c from equation 3. So, by rearranging this you can get t c in terms of others which is what I have done here. You can see what I have done here is that I have put t c in terms of others t c 0 I mean you can look at this and then perhaps understand what I have done see you can see here. So, from this equation I get t c I replace and rearrange this and find t c. So, it is what I have done here when you do that manipulation you get t c in terms of others t c 0 t 0 temperature in any position and f a 0 i equal to i equal to 1 to r independent reaction these r is independent reactions. Now, that we know t c. So, it is now possible for us to replace in this equation 1 this this relationship this relationship can be replaced into equation 1 and once we do that. So, we find that you have this r plus 1 differential equation the right hand side involves terms that you know all the terms that you know x 1 to x r x 1 is 0 to x r is 0 at t equal to 0 at the inlet. Now, only thing that you do not know in this integration is t c 0 or in other words when we want to integrate these r plus 1 equations by replacing t c from equation 4 into equation 1 the integration now becomes possible because you know what is x 1 to x r initially they are all 0 you also know what is t equal to t 0 at the inlet. Therefore, the right hand side involves terms that you know fully accepting t c 0 because our t c now involves t c 0. So, in other words what we are now saying is that to be able to integrate this equations 1 to r plus 1 equations you have to assume a value of t c 0. Once you assume t c 0 now you can integrate and then complete the exercise. Now, the question is how do you know how whether this assumption that you have made is right or wrong. Now, to be able to take care of that what we do is that we do what is called as the overall balance. That means we look at overall balance between position 0 and position l we do a overall balance between position 0 and position l once again overall balances what is the energy coming in what is the energy going out, but it is between position 0 and position l. Now that means this is what the energy coming in. So, V C P C T C L this is the energy coming in I could 1 to R F A 0 X J is the energy coming in notice here this X J refers to X J at L. And then the output streams are V C P the reagent stream output and then you have the coolant stream, but the going out stream the temperature is T C naught. So, what have we done now we have now by assuming T C 0 we have integrated equations 1 and 2. Then at the end of this integration what do we get we will get the value of X 1 to X R at every position including at position L. Similarly, you will get temperature at every position including at position L. On other words you from this in this equation you know all the numbers T C L you know from your integration X L X 1 to X R at position L you know from your integration. Therefore, everything that is here is known from your integrations. Therefore, if the assumption of T C 0 was correct equation 5 would be satisfied. So, therefore, what we do is that we perform the integration determine the numbers X 1 to X R and T at every position including the end point. Then we come back to equation 5 to check whether our assumption of T C 0 is right or wrong. If it turns out that it is incorrect then this equality will not hold. Then we know that we have a bad assumption and therefore, we have to reset our assumptions. Now the question is how do you reset your assumptions correctly every time you make. So, this is something that comes by experience and you will have to move the assumption assumed value of T C 0 in the right direction. So, that your answers converge. So, experience will tell you how to move the T C 0 assumptions iteration into iteration. So, what we have tried to do in this exercise of tubular reactor heated or cooled from the wall is that we have been able to determine the positions and the temperatures the compositions the extensive reaction for all the independent reactions at every position including the end points. Whether it is co current whether it is counter current it does not matter because the procedure tells you how to handle this. Having said this there are few points that we might like to recognize that is why are we heating or cooling a fluid. We are heating or cooling because the reactor the catalyst perhaps which is inside the reactor requires that the temperatures must be appropriately regulated or the temperatures must be appropriate to the reaction that you are carrying out. Now frequently what may be of concern to all of us is that the temperature let me just make a small plot if this is volume versus temperature. So, if you had an adiabatic reactor the temperature keeps on increasing it is an exothermic reaction. If it is cooled from the wall you will find there is a cooling that means there is a point of maxima in reaction rate and maxima in temperature in the reaction equipment. Frequently this hot spot it is called hot spot temperature is a specification on the catalyst that we will engage. Now if the specification of the temperature maximum temperature that the catalyst is able to perform is given to you our design of the system must ensure that this temperature is never exceeded. And it is that interest that we line try and look at all the simulations to see how best we can adjust this term. So, that the maxima in this temperature is well within what is specified. So, our simulations by and large are towards ensuring that our system performs as expected. And towards to do that we adjust our process to ensure that this T c is so chosen that our temperatures are maximum temperatures are well within what is specified. So, trying to understand how to manage the hot spot temperature is one of the important issues that we try to do in our design. And more importantly we ensure that this temperatures are not exceeded in our process operation. So, having said this we will go on to the next item of our interest which is. So, our next topic of interest is we want to look at we want to look at this whole issue of transient behavior of exothermic stirred tanks. Now stirred tanks are very common devices particularly in small scales. And, but the interesting about stirred tanks that it brings out very interesting features of the reaction which were able to use effectively in our design of even other types of reactors. So, that is the object of trying to look at exothermic stirred tanks. Now, what is our system? Here we want to look at this system in a very simple system so that in all the issues are fairly understood. So, we have a tank which is well stirred there is a cooling coil through which a coolant circulates. And this circulation of this coolant let us say is very large. So, that the temperature of the coolant coming in and the temperature of the coolant going out is not very different so that T c in layer equal to T c at the exit. This need not be, but it is an assumption that we may exist to make our mathematics a little simpler to handle. So, you have a reaction a going to be we are only looking at case of a single independent reactions when there are more reactions of course, there are more features which we are not going to consider now. There is a single independent reaction which means that even if there is reaction which is reversible like this, this analysis is quite satisfactory, but we will by and large be looking at one reaction a going to be a going to be single independent in we just look at a simpler situation. So, our statement of material balance is what is coming in F a 0 what is going out, what is the rate of generation equal to the accumulation. So, this statement of energy of material balance we can write in various ways, but for this case of trying to understand the unsteady behavior of a stirred tank. We have written this in the form of a variable x which is defined as c a 0 minus of c a divided by c a 0. Now, this definition of x is commonly done and it is has the meaning of conversion whenever we operate the process at steady state. When the process is not at steady state then x does not take the meaning of conversion therefore, it should be treated as a variable whose definition is given by x equal to c a 0 minus of c a by c 0 beyond that no other meaning can be attached in the unsteady state. So, we have variable x which define this way and therefore, we can replace this equation one appropriately in terms of x is what I have done. So, replacing F a 0 as v 0 c a 0 F a as v 0 c v equal to v 0 is assumed assuming that there is no volume in volume change to the reaction this a to b there is no volume change anyway. So, this is a fairly satisfactory appreciation. Now, the right hand side right hand side is d by d t of n a or n a is simply v times c a therefore, since there is no volume change since our reactor volume does not change therefore, v can be taken out of the derivative therefore, we write the equation in this form. Now, you notice that the term r a the term r a is replaced as r 1 with a minus sign that is because our rate of formation of a this is the reaction a going to b our rate of formation of a rate of is equal to minus 1 times r 1 where r 1 is the intensive velocity of the reaction then minus sign is put because a is a reactant. So, we have r a is minus of r 1 volume of the system is constant v equal to v 0 is assumed with all these assumptions this equation becomes something like this you can very easy to notice this. So, you divide throughout by v 0 if you divide throughout by v 0 v equal to v 0 therefore, this goes away and becomes v by v 0 equal to tau v by v 0 which is equal to tau. So, this equation becomes minus of tau d x d t is x minus of r 1 tau by c 0 or equation 1 this is final representation of the material balance equation for a stirred tank which says d by d t of x multiplied by tau is minus of x plus r 1 by c a 0 times tau. And this something that we know is not new to us we have been doing this for a long time, but in this form there is certain advantages. So, we want to retain it in this form now what is our energy balance our energy balance we have written many times. So, I am writing it again which says this is a single reaction. So, the left hand side is v c p d t d t and the right hand side is first term is v 0 c p t 0 minus of t and then the reaction the heat generation term in this case is r 1 b minus delta h 1 star and q is the heat inputs and w is the work output which is taken as 0. Each of these terms I have not explained it again because we have done this so many times. So, this is the accumulation of energy and this is the energy that is coming in from the fluids and it is getting heated from t naught to t. So, that is the energy picked up by the fluids that is entering the system and this is the energy that is released because of chemical reaction this term is the is the intensive velocity is only one reaction we are taken therefore, it is only r 1 v minus of delta h 1 star q is the heat added to the system as per first law convention. Now, if there is if you had one independent reaction, but another two rate process suppose in this case instead of r 1 it will become r 1 minus of r 2 something we have done in our earlier classes. So, this is not something new to you, but in this case we are only looking at a single reaction a going to b and not not this. Now, if you divide throughout by v naught c p c p is volumetric specific heat something that we know. So, if you divide throughout by v naught c p what do we get v by v naught c p and then this disappears here r 1 v by v naught c p q by v naught c p. Now, what I have done here v by v naught I have taken it as tau the residence time. So, the left hand side is simply tau times d t d t. So, first term is t naught minus of t as can be seen here the second term is r 1 tau and I denote this j delta h 1 star by c p as j 1 is a convention that has a I mean using in the literature. So, I am retaining that minus delta h 1 c p is j 1 therefore, this whole term becomes r 1 tau j 1 this term q by v naught c p what is q is the heat transferred. So, I have written it as h times a heat transfer surface t c minus of t. So, this is the amount of heat that is transferred by the cooling or heating fluid to the reaction mixture h a by v naught c p multiplied by t c minus of t. So, it is sort of simplified it appropriately. So, that it is convenient for us to deal with it in a certain way. So, equation 2 let me write equation 2 once again, but only here what I have done is that I have replace this whole term h a by h a by v naught c p v naught c p equal to beta I have just called it as beta. So, this equation looks like this in the literature the term beta which is h a by v naught c p if you put all the units you notice that it is dimensionless. So, it is a dimensionless parameter beta. So, I have replaced it as beta here and now this is also a form of you know energy a picked up or moved out and this is the energy there. So, it is convenient to combine the first term and the second third term and literature does this in this form put this in the form of 1 plus beta t c star minus of t. That means, t c star minus of t is equal to t naught plus beta t c that is how it is written. So, that it looks like you know the whole term first on the third term looks in a more simple form 1 plus beta times t c star minus of t where t c star is defined by this equation 3 let us say 3 a. So, if you are given t naught beta and t c you can find out t c star because you know all this. On other words t c is some kind of a fictitious temperature which sort of enables us to combine equation the terms 1 and 3 in forms that is makes it convenient to handle mathematically. So, what are we having our equation now is tau d t d t is 1 plus beta t c star minus of t plus r 1 tau j 1. So, this is the heat generation and in a sense this is the amount of heat that is removed. So, you can say that heat generation this is heat it is not heat removal this is minus of this is heat removal. They will just recall what we have done what we have done is that we had our material balance equation which looks like this. So, we have the energy balance equation which looks like this. So, there are 2 equation which describes our unsteady state which is tau d x d t is minus of x r 1 tau by c a 0 that is this is important to write this down because it is. So, let me just write this down our differential equations are tau d x d t equal to minus of x plus r 1 tau by c a 0 this is the material balance equation. The energy balance equation looks like this tau d t d t equal to 1 plus beta t c star minus of t plus r 1 tau j 1. So, there are 2 equations which will govern what happens to our process. Now, if I ask you what happens to this process at steady state? Now, clearly steady state means what the left hand side is 0, left hand side 0 means what left hand side 0 means that 1 plus beta t c star now that is at steady state. Therefore, the temperature becomes t becomes t s r becomes r s. So, that at steady state our equations these 2 equations can be written as 1 plus beta t c minus of t c star minus of t s which is steady state temperature. And similarly, tau d x s d t equal to 0 because steady state means 0 x becomes x s. Now, in this form it has a meaning of conversion now r 1 s is the rate of generation of rate of consumption of component a or rate of generation of component b whatever it is r 1 s. Now, what meaning can we attach to this? The meaning we can attach to this is that it is rate of generation of heat. So, it is r 1 s j 1 tau we can call it as rate of generation of and then what is this term t c star minus of t s it is in a sense 1 plus beta if I put it as a reverse on other was I call this as minus u r s I call this as u g s then this becomes heat of heat removal. On other words what we are trying to say is that this equation 5 can be thought of this steady state can be thought of as 2 terms u g s equal to minus of u r s equal to 0 this is how we can look upon equation 5 u g s minus of u r s equal to 0 or steady state is described as u g s equal to u r s. So, this is how we can understand steady state. Now, let us take a small example to understand how we can deal with all these things say we have let us say you have a first order reaction first order reaction r 1 the rate of generation I mean component r 1 is given as k 1 s c s we know this r 1 s means k 1 s c s what is k 1 s c s c s by our definition is c s 0 times 1 minus of x s we know this. Now, we know we also know from our material balance say for example, we have done this just now this come back to you in a minute from this equation x s given by this we can substitute for r 1 s and we will find that x s is actually given by k 1 s tau divided by x s is given by this expression. So, we can solve our material balance equation and find that our x s can be given simply by k 1 s which is a function of temperature tau times 1 plus k 1 s tau. In other words the steady state conversion in the equipment is k 1 s tau divided by 1 plus k 1 s tau this comes from our material balance. Therefore, the rate of generation of heat or rate of generation of energy which is r 1 s tau j 1 which is once again we have said this we have said all these things this is u g s this is u r s what is u g s for a first order reaction we can now calculate this by substituting for r 1 s r 1 s comes from here this is r 1 s. So, you can substitute all this and find that the rate at which u g s is actually c a 0 x s j 1 or u g s is given by this expression which is k 1 s tau divided by 1 by k 1 s tau multiplied by c a 0 j 1 we know all these things. So, what we have done we are now able to express our rate of heat generation u g s in terms of k 1 s tau divided by 1 plus k 1 s tau and this k 1 s depends only on temperature on other words by our manipulations you are able to express the rate of heat generation in terms of one variable which is temperature. Now, if you look at the other term which is the other term the rate of heat generation see this is the heat removal which is 1 plus beta times t c star minus of t s heat removal is minus of u r s. So, u r s is 1 plus beta t s minus of t c star on other words if I ask you what is the dependence of u r and temperature depends on u r and temperature it is simply multiplied by t s minus of t c star t c star is known it only depends on t s on other words what we have done now is that u g which is the rate of heat generation rate of heat generation is got as a function of temperature. Similarly, you have rate of heat removal which is u r s which is also as a function of temperature. So, once you have both u g s and u r s as a function of temperature now you can plot u g s. So, this let us say this function looks like this say u r s u r s is what 1 plus beta. So, it has a let us say what is the slope of this the slope of this is 1 plus beta. So, what are we done. So, this is started at t 0 t c star. So, this is. So, it starts at t c star and then it and it seems to intersect at 3 points. Suppose, our t c star is here. So, it intersects at this point. Suppose, your t c star is here. So, this is equations like this. So, it intersects at only 1 point. So, what we are trying to say is that depending on choice of t c star you can have the intersection or you can have the u g s equal to u r s this equality is satisfied. This equality is satisfied when it is satisfied at only 1 point when you have choice of t c star at a very high temperature or it is at 1 point at this choice of t c star at a very low temperature or at 3 points when you make a choice of t c star appropriately. So, you can have a steady state whose temperature is this or a steady state where temperature is this. So, you can have case of a steady state temperature which is very high or a steady state temperature is very low or a steady state temperature which is intersecting at 3 points. So, what we want to conclude is that if you have an exothermic stirred tank, this is the point we would like to get across to you. If you have an exothermic stirred tank, you can have what is called as one steady state. We can call it lower steady state. We can have one steady state which is called as upper steady state or we can have 3 steady states which is intermediate. We have upper lower and intermediate that is what we have tried to show here. It is an upper steady state. It is a lower steady state or it can be 3 upper lower and an intermediate. Now, let us try to understand this in another way. Now, let us say I want to plot u g s versus t s. So, let us say you have this is one such curve. Now, if you change this is for let us say tau equal to some tau 1. Suppose, you increase tau. So, you will get curves like this. This is increasing tau. It stands to reason why it is increasing tau. If you increase tau, you will get curves like this. In other words, you can generate curves of rate of heat generation versus react to steady state temperature for different values of tau at which you will operate your process. Now, suppose your heat removal curve is here. Let us say this is your heat removal curve. Now, you will find that if you are operating here, you will get only once. If you are operating the reactor at this tau equal to tau 3, you will only get the upper steady state or tau equal to tau 2. You will get this upper steady state tau equal to tau 1 upper steady state. When will you get the lower steady state? The lower steady state you will get only when you choose this t c star sufficiently low. Then only you will get over steady state. In other words, the steady state that you will get in a process is something that you have to think deeply and find out which steady state is what you will reach in your process. That will come from an understanding of the transient which we will do shortly. Let us look at one more exercise. Let us say you are operating at this steady state. In other words, this is u g and this is u r and this is u g. Now, suppose you want to travel down to a lower steady state means by changing t c star, you can change t c star, keep on changing t c star and depending on the choice of t c star, if you choose here, you will get a steady state here. If you choose a lower steady state, you can come to lower point, lower point, lower point, lower point. That means depending upon the choice of t c star, you will start moving down from your upper steady state to lower and lower temperature. You will come to a point at which you will find that the next intersection is only here. What are we saying now? If you have an exothermic stirred tank, then as you are changing your t c star from point point point and you will finally come to a particular value of t c star after which a further small decrease in t c star, you will intersect at a much lower point. Which means this change from a temperature, a high temperature to a very low temperature is an instance of what is called as sudden cooling. There is a sudden cooling. In the same way, if you are changing temperature from here, from here, from here, from here, from here, from here and then changing t c star, a point will come suddenly it gets heated up. So, this is what is called as the ignition. What we are saying? That means what we are saying is that, let us say this is t c star. There is a particular value of t c star as you are operating from lower steady state and keeping on increasing t c star. There is a particular value of t c star when you reach, suddenly the reactor heats up and then goes to the upper steady state. You understand that, this behavior of exothermic tank tanks, sudden drop in reactor temperature and sudden increase in reactor temperature. These are all issues associated with an interplay between heat generation and heat removal. These are crucial issues which can lead to accidents, etcetera, if you are not careful. But, these issues are well understood nowadays and then you can take care of it by an appropriate design. These are the kinds of issues which we must take care when you try to understand what is called as a transient behavior of a stirred tank. Not only that, we want to understand the transient behavior stirred tank, we want to see how best we can ensure that the process operates in a stable way. So, that if there is a disturbance, if the disturbance does not grow, it starts to decrease rapidly. So, the steady state that you want to operate on, you are able to reach in that point without too much of difficulty. On other words, the whole issue of stability of steady states is a matter that is of great interest to us, stability of steady states. This is something that we want to understand and then take appropriate measures. Now, how do we understand stability? To understand stability, what is done is that we should look at the steady state behavior. We should look at the unsteady state behavior. We should look at the difference between steady state and unsteady state behavior and find out how best these differences can be kept as small as possible in a process. So, to do that, let me write the steady state equation and steady state. So, our steady state equation, let me write d x d t minus of x plus r 1 tau c a 0. This is our material balance tau d t d t equal to 1 plus beta t c star minus of t plus r 1 tau j 1. This is our energy balance. So, this is the values of x and t in the unsteady state. Now, at steady state, at steady state we said tau d x d t, this is d x s d t equal to 0 is equal to minus of x s plus r 1 s tau by c a 0 and then tau d t is equal to t by d t s equal to 0 equal to 1 plus beta t c star minus of t s plus r 1 s tau j 1. So, you have this is the unsteady state equations, unsteady state these are the steady states. So, our interest now is to actually find out what is the deviation between x and x s t and t s. Why are we interested in this deviation? We are interested in this deviation because our interest is to operate the process at steady state, but due to variety of reasons this steady state gets disturbed. Therefore, we are away from steady state that means we are at x and not at x s. Therefore, our interest is always what is x minus x s, what is t minus of t s and indeed how we can operate the process. So, that x s minus of x s t minus of t s remains as small as possible or within the limits that you and I will specify. So, let us try and understand deviation from steady state. So, to do this what I do is that I subtract 1 minus 2. So, I call this as 1, I call this as 2, call this as 3 and call this as 4. So, I do 1 minus 3 and 2 minus of 4. So, I will do 1 minus 3 and 2 minus of 4. So, let me see how it looks like. So, it becomes tau d by dt of x minus of x s equal to minus of x minus of x s plus r 1 minus of r 1 s tau by c 0. The other 1 is tau d t minus of t s by dt equal to minus 1 plus beta t minus of t s plus r 1 minus of r 1 s tau by c 0. The other 1 is tau d t minus of t s by dt equal to minus 1 plus beta t minus of t s plus r 1 r 1 s multiplied by tau and c 0. So, what we have done? What we have done is that we are now looking at the deviations from the 2 steady states looking at deviations. So, the various things we can do with this right now. So, x minus of x s and t minus of t s we like to see that x minus of x s and t minus of t s are within limits that you and I will specify. So, we want to now understand the solution to 5 and 6 and how these solutions can be understood as stable within our way of understanding the process. So, this is what we will do when we meet next. Thank you.