 Welcome, this advanced reaction engineering. Today, we look at further considerations in energy balance. Now, what we would like to do is look at what we have already learnt and then and apply them to situations that we might encounter in real life. Situations which I mean we have selected situations which are a little bit more involved. So, that we can really see how our equations apply. So, first example we would like to take is a multiple reactions A goes to B, B goes to A, then goes to A, C goes to D. So, this is the reaction we would like to look at. And all reactions for the moment we consider that these reactions occur in a stirred tank. These are occurring in a stirred tank, you have component A coming in and going out. And we will assume all reactions are instantaneous, which means that they are they are all in equilibrium, implying that C B by C A is K 1, C C by C B is K 2 and C D by C C is K 3. So, it is instantaneous therefore, the compositions are as per described by the equilibrium. And K 1, K 2, K 3 for example, take a situation where K 1, K 2, K 3 situation is something like this K 1 equal to 1, K 1 equal to 1, K 2 equal to 2, K 3 equal to 3 and delta H 1 equal to 10, delta H 2 equal to 20, delta H 3 equal to 30. These are all in units of K cal per mole, K cal per mole, K cal per mole. What do we have? We have a stirred tank, we have a stirred tank into which A is coming in and going out. The stirred tank is maintained at 25 C, this is the temperature 25 C is the temperature maintained. So, all these reactions as you can see from the delta H 1 values here, they are all endothermic therefore, you would expect that this reaction will proceed only if heat is added. So, we expect some amount of heat to be put in for this reaction to take place. Therefore, the question of interest to us is that what is the amount of heat that is to be put in number 1. And then what is the composition of the leaving stream assuming that the temperature of CSTR is maintained at 25. So, it is obvious that to be able to maintain this temperature at 25, you have to supply heat because all the reactions are endothermic. So, this is what we would like to do. So, A goes to B, B goes to C, C goes to A. If there are 3 reactions, if I call x 1, x 2 and x 3 are the extends of reactions in each of these, we can write the stoichiometry. So, let us write this stoichiometry now. So, we can write this stoichiometric table, stoichiometric table where this F A is F A 0 into 1 minus of x 1, the first reaction. And then F B is F A 0 times x 1 minus of x 2, of course F B 0 is nothing is stated. So, we think it is 0, F C is F A 0 x 2 minus of x 3 plus F C 0 which is 0, F D is F A 0 times x 3 plus F D 0 which is 0. This is stoichiometry. And we have said little earlier that all these reactions are instantaneous, reactions are all instantaneous. And we said that since they are all instantaneous that C B by C A is K 1 and all that, all that we have said already. So, we can now use these relationships to understand what has happened in the equipment. So, let us do that quickly. So, we have C B C B by C A which is C B is F A 0 times x 1 minus of x 2 divided by F A 0 times 1 minus of x 1. So, this is K 1 which is 1 equal to 1. Similarly, you have C C by C D is F A 0 times what is C C which is x 2 minus of x 3 divided by F A 0 C B by C A B sorry into x 1 minus of x 2 that is equal to K 2 that is equal to 2 C D by C C. C D is what F A 0 times x 3 and then C C is F A 0 times x 2 minus of x 3 equal to K 3 equal to 3. So, you have to solve these three equations. Let us do that quickly. So, the equations in front of us. So, the equations are the following. I am just putting the same thing again just to make it easier to solve. So, x 1 minus of x 2 equal to 1 minus of x 1 is one equation. Then you have x 2 minus of x 3 equal to 2 times x 1 minus of x 2 that is another equation. Third equations x 3 equal to 3 times x 2 minus of x 3 this is third equation. So, you have to solve these three equations and fairly elementary to solve you will find x 1 equal to 0.9 solving x 2 is 0.8 x 3 is 0.6. This is fairly elementary. So, nothing much needs to be done. So, you can solve this and find what is x 1 x 2 and x 3. Now, what we need to do is to find out how much heat is to be added or removed because that is what will ensure that our CSTR. So, what you have got we have got a CSTR in which you have this instantaneous reactions running. This instantaneous reaction is A goes to B, B goes to C, C goes to D, D goes to A. This is the this is running here at 25. So, we have to put in heat to be able to ensure that this happens actually correct. So, we need to write the energy balance to find out how much heat is to be put in etcetera. So, what is the energy balance equation V C p d t d t equal to V naught C p t naught minus of t plus sigma i equal to 1 to all the p reactions are I will put k here r k delta minus delta h k times V plus q minus of W s which is not there. So, this is at steady state therefore, this is not there this is V naught C p is volumetric specific heat volumetric we know all that inlet temperature, outlet temperature all the data is given. If the inlet temperature equal to outlet temperature as is the case here this is t naught here this is also t naught. So, this term is also not there. So, essentially what you have sigma r k times minus of delta h k V. I equal to 1 to p k equal to 1 to p plus q equal to 0. So, we can expand this essentially you know you have to substitute put all these find out how we can relate r k times V to x 1 x 2 and x 3 which are the extent of reactions in the different reactions. Once we do that our numbers are straight forward. So, let us recognize here to we want to find out sigma r k times minus of delta h k times V plus q equal to 0 I equal to sorry k equal to 1 to p reactions. Find q to be added this is what you have to do r removed. Now, we have been looking at instantaneous reactions and we want to calculate the value of q which is coming from the energy balance. And based on the energy balance we said that q plus sigma of minus delta h k r k V equal to 0. So, we have to calculate what is q and for that we must know what is r k. So, that we can substitute and find out the value of q. Now, we know that r k being an instantaneous reaction we do not know those numbers. And therefore, we must find that other ways of finding out r k. So, recognizing that we have a going to b going to c going to d going to a it is a rectangular reaction all of them are reversible. There is a procedure by which we can eliminate r k using our material balance. Let us see how to do this. Now, let us recognize that a going to b b going to c c going to d d going to a and all that. And we denoted this is reaction 1 reaction 2 reaction 3 reaction 4 reaction 5 reaction 6 and reaction 7 reaction 8. Now, we have denoted for our convenience that delta h 1 as the reaction going a going to b as the heat of reaction for a going to b. And delta h 2 as the heat of reaction for a going b going to c and delta h 3 for the heat of reaction c going to d and delta h 4 heat of reaction for d going to a. Now, having said this let us look at how we can determine the value of r k sigma delta h k v from our material balance. So, let us just expand r k minus of delta h k v what do we get v times r k is r 1. If you just expand this you will find that it is r 1 delta h 1 r 2 delta h 2 r 3 delta h 3 r 4 delta h 4 and so on up to r 8 delta h 8. But we know from our understanding that is coming back this the heat of reaction for a going to b and heat of reaction for b going to a just the reverse of each other keeping that in mind. We have simplified this summation we have simplified this summation as r 1 minus of r 2 delta h 1 r 3 minus of r 4 delta h 2 r 5 minus of r 6 delta h 3 r 7 minus of r 8 delta h 4. This is something that we all know that for the reaction a goes to b the heat of reaction for b going to a is simply the negative of heat of reaction for a going to b something that we all know. Now let us look at this representation of sigma r k delta h minus delta h k v given on the right hand side the equation 1. We notice that because being a rectangular reaction as we have said here being the rectangular reaction as we have said here delta h 1 plus delta h 2 plus delta h 3 plus delta h 4 must equal 0 that comes from our basic understanding thermodynamics. Therefore, we can say that since delta h 1 plus delta h 2 plus delta h 3 plus delta h 4 is equal to 0. We can write delta h 4 as minus of delta h 1 plus delta h 2 plus delta h 3. On other words we are able to eliminate delta h 4 from equation 1 that make a certain advantages in terms of trying to eliminate r k from this equation. We will tell shortly how it can be done. Now substituting just recalling recalling what we have done we have just replacing delta h 4 in from equation 2 and then simplifying that is what I am going to do now. So, you can see here that just the summation being written once again r 1 minus of r 2 delta h 1 star r 3 minus of r 4 delta h 2 star r 5 minus of r 6 minus of delta h 3 star and then r 7 minus of r 8 the delta h 4 is replaced as delta h 1 plus delta h 2 plus delta h 3. Please recognize that delta h 4 is minus of delta h 1 plus delta h 2 plus delta h 3. So, this delta h 4 is replaced from equation 2. Now having done this notice that delta h 1 and delta h 1 here delta h 2 and delta h 2 here delta h 3 and delta h 3. So, we can combine all that and make the whole thing look a little simpler. So, I have done that here minus of delta h 1 multiplied by r 1 minus of r 2 r 7 minus of r 8 with a negative sign because it is minus and this is no minus here. So, it becomes minus r 7 plus r 8 multiplied by minus of delta h 1 then delta h 2 minus delta h 2 multiplied by r 3 minus of r 4 you can see here and then minus of r 7 plus r 8. Similarly, delta h 3 here minus r 5 minus of r r v r 5 minus of r 6 and then minus of r 7 plus r 8. So, nothing new has been done simply you know assembling them together in forms that we can deal with them conveniently. Now what we recognize now is that each of these terms multiplying delta h 1 multiplying delta h 2 and multiplying delta h 3 we can find out what those numbers are using our material balance. Let us just do a material balance on component a to recognize how simple it becomes. For example, if you write a material balance please let us recall our reactions let us recall our reactions because it is important. Let us recall our reactions I am just putting it here just for the sake of understanding you have a going to b. So, I am writing the material balance for component a what is input output generation equal to 0 because there is no accumulation. So, it is in a stirred tank that is our problem. So, input minus of output based on our stoichiometry we have said that f a is f a 0 1 minus of x that is what is said here r 2 look at all the generations. Now a is generated from reaction to a is generated from reaction 7 that is why I have written r 2 plus r 7 they are both positive and a is consumed in reaction 1. So, minus of r 1 a is consumed in reaction 8. So, it is in minus sign. So, the signs are as we expect. So, we have r 2 plus r 7 minus of r 1 minus of r 8 multiplied b equal to 0. So, you get from here from this what you get is the following that f a 0 x left hand side plus I have just taken into the other side therefore, r 1 minus of r 2 plus r 8 minus of r 7 multiplied by b. So, essentially material balance on a gives us f a 0 x 1 on the other side it is r 1 r 2 r 8 r 7. So, you notice here that even though we do not know what is r 1 r 2 r 8 and r 7 because the reactions are instantaneous. Now the right hand side becomes equal to f a 0 x 1, but f a 0 x 1 is measurable. So, this is the advantage of the formulation that even though you do not know the right hand side because these numbers are very large. I mean numbers r 1 r 2 r 8 r 7 may be large, but the totality of it may be something that it is not very large that is why we are able to measure it. So, essentially what we have done is that since our numbers are very large we are able to replace in our equation by numbers that we can measure that is the advantage of writing the material balance. Now, just let us recognize what we have done. We have said that in a minus of delta h 1 multiplies r 1 minus of r 2 minus of r 7 plus r 8. You notice here that this term this term r 1 minus of r 2 plus r 8 minus of r 7 actually appears in the in the previous equation that we have done. So, we will come back to that in a minute. Now similarly, let us write a material balance for d input output and what is the generation of component d. Let us see how what is the generation of component d. So, what we are saying now is what we are saying now is that component d component d is formed in reaction 5. So, it is plus r 5 and consumed in reaction 6 it is minus of r 6. Now, component d is formed in reaction 8 plus r 8 it is consumed in reaction 7 is minus r 7. In other words, all the generations are positive all the consumptions are negative and that is the material balance of component d. So, that you get f a 0 x 3 I have taken it to the right hand side. So, that it becomes f a 0 x 3 becomes r 5 plus r 8 I can see here minus of r 6 minus of r 7. So, we have written material balance for a material balance for a and found out f a 0 x 1 in terms of all the reaction rate functions. Similarly, f a 0 x 3 in terms of reaction rate functions once again we recognize that r 5 r 8 r 6 r 7 are large quantities. So, by themselves are not easy to measure, but the left hand side is measurable that is advantage of the procedure. So, we have done material balance for a material balance for d. Let us now see whether we can do something more that is material balance for c. So, what I have done here what I have done here is that I have written a material balance for c. Please recognize that a going to b going to c going to d going to a component c component c is formed in reaction 2 and consumed in reaction 3 that is why it is written x 2 minus of x 3 input minus of output plus generation. What are the generations of component let us just put here. Notice that component c is formed in reaction 3 r 3 is plus it is consumed in reaction 4 r 4 is minus and then component c is formed in reaction 6 r 6 is plus and it is consumed in reaction 5 r 5 is negative. So, we have taken the generations is positive consumption is negative. Therefore, you have a material balance for component c which since. So, we have here the material balance. So, this since is 0 I have just simplified it and then it is written as f a 0 times x 2 minus of x 3 and it is taken and this how it appears. So, since negative sign therefore, I have written it in this form showing that f a 0 x 2 minus of x 3 is given by the right hand side that I have written the equation 5. Now, we have just done in the little bit earlier what is f a 0 x 3. So, I have just put equation 4 once again I have put the equation 4 once again which you already derived. What does it say f a 0 x 3 is r 5 plus r 8 minus of r 6 minus of r 7 that is equation 4 which you have derived a little earlier. Now, if you add equation 5 and equation 4 you notice here we add a equation 5 and equation 4 x 3 gets cancelled. So, you get f a 0 x 2. Now, let us put all the numbers here by r 3 r 4 r 6 r 5 I have written down r 5 plus r 5 plus r 8 minus of r 6 minus of r 7. Now, you notice carefully here r 6 cancels of r 5 cancels of. So, that f a 0 x 2 f a 0 x 2 becomes r 3 minus of r 4 plus r 8 plus r 8 minus of r 7 b. So, what is it we have done what we have done is that since our what we are saying here is that when we expand in we expand this function this sigma minus of delta h k r k i going from k going from 1 to p we find that right hand side r k's are very large. Therefore, we are able to eliminate them using the material balance and that is what we have tried to do. So, what we have done is that we have found out f a 0 x 1 f a 0 x 2 f a 0 x 3 in terms of rate functions. Now that we know them in terms of rate functions we can substitute in our equation and find out how they simplify. So, what I have done here is notice here notice here that this is how we have expanded delta h 1 I mean r k and put in terms of all the rate functions. Now, what we have done just now we have done just now is that all these rate functions all these r 1 minus of r 2 minus of r 7 minus of r 8 we have just now shown that this term we see here we have just now shown that r 1 minus of r 2 plus r 8 minus of r 7 is f a 0 x 1 we have shown using our material balance this is what we have done. So, I am able to replace this r 1 minus of r 2 minus of r 7 plus r 8 as f a 0 x 1. So, this is f a 0 x 1 that is what I have done. Now similarly, we have shown we have shown material balance for this is delta h 3 r 5 minus plus r 8 minus of r 7 minus of r 6 this is f a 0 x 3 we have shown that you see similarly, we have shown from our material balance on component c we have shown the following that f a 0 x 2 we have shown this similarly, we have shown that f r 3 minus of r 4 minus of r 7 plus r 8 we have shown as f a 0 x 2. In other words what we are trying to say here this is the point that we must recognize that if you recognize the number of independent reactions and then do all your stoichiometric table on the basis of these independent reactions. Then the term that multiplies the rate function that multiplies delta h 1 will be f a 0 x 1 the term that multiplies delta h 2 will be f a 0 x 2 the term that multiplies delta h 3 will be f a 0 x 3. So, this is how it is formulated and therefore, what happens is that this this whole term this whole term delta h k r k v now can be written as minus of delta h 1 f a 0 x 1 minus of delta h 2 f a 0 x 2 minus of delta h 3 f a 0 x 3. So, the whole idea of of using this approach of multiple reactions is that you can conveniently replace r k v as f a 0 x 1 you see that is the advantage and that is what makes it much simpler to manage multiple reactions. Now, our problem is very simple or the energy balance we have to calculate we we started if you recall we started our idea was to find out what is basically when we started we said the following that we want to calculate what is q and to calculate q we wanted to get rid of this term r k v then only we can calculate q that is what we have done what we have done is that we have we have found out what is sigma of r k v and all that and we found out that this whole term is delta h 1 f a 0 x 1 delta h 2 minus delta h 2 f a 0 x 2 minus delta h f a 0 x 3 which we can substitute in our energy balance that is what we have done here we have said q plus minus of delta h 1 f a 0 x 1 plus minus of delta h 2 f a 0 x 2 plus minus of delta h 3 f a 0 x 3. Now, we have already done a little earlier that x 1 is 0.9 x 2 is 0.8 and x 3 is 0.6 we already calculated that based on our understanding of equilibrium constants and so on. Now, all you have to do is that you substitute for delta h 1 substitute for delta h 2 delta h 3 what is delta h 1 what is given it is given as 10. So, it is minus 10 this is given as 20 therefore, it is minus 20 this is given as 30. So, it is minus 30. So, you can simplify this and find that the heat that you must supply or remove is 43 kilo calories per minute it is it is plus here it implies when it is plus it means that you must supply heat if it is negative it means you must remove heat. In other words what we have done what we have done is the following. Let me just explain this once again because this is something that we may have to do again and again that what we started with we started with a stirred tank we started with the stirred tank and then in this stirred tank we have this reaction a going to b b going to c c going to d d going to a. So, we call this 1 this is 2 this is 3 this is 4 this is 5 this is 6 this is 7 this is 8. So, because all these reactions this are instantaneous we said that the amount of and there are endothermic and so on. Therefore, we have a cooling coil cooling coil. So, heating coil in this case heating coil which supplies the heat that is required to run the process at the temperature at which we want to run the process. In other words what we trying to say is that when you have instantaneous reaction then the temperature at which you run the process it will determine what is the extent to which you will be able to drive the reaction whether it is endothermic or exothermic it is depending upon the situation that you will handle. But either case essentially the heat transfer will determine the extent to which you can push your reaction and this is the point that is trying to get across to you that if you have instantaneous reaction then you only have to manage the heat loads. So, that you are able to push the reaction in the direction of your interest. So, let us go to another exercise in this exercise it is also an illustration of energy balance this energy balance this is what we have here you have a fluid and there is a very very exothermic reaction is taking place. There is no outflow no outflow. So, the data given is the following reactor temperature is 38 C rate constant K is 0.36 per hour coolant temperature that means there is a there is a coolant that is 30 coolant temperature is 26 C is 26 C C P volumetric is 0.5 multiplied by density is 800. So, many kilo calories per cubic meter per C C P is given C A I A goes to B is the reaction C I is given as 8 kilo mole per cubic meter and initial volume in the reactor is given as 1.5 cubic meters. So, I mean this is an instance where we are trying to control a reaction simply by adding an inert coolant. So, very exothermic reaction therefore, we need to control the temperature and one way to control the temperature is to add your coolant I mean add your inert medium to keep the concentrations low. So, it does not heat up. So, we want to solve this problem and find out how we must vary the flow to take care of constancy of reaction temperature that means we must vary the flow in such a way that the reaction temperature does not change. So, that is the exercise in front of us. Let us recognize once more that we have we are adding and this temperature must be kept constant. So, let us write material by A goes to product is the reaction. So, material balance or material balance what does it give us input of material output of material plus reaction equal to d by T of material. So, problem statement says there is no accumulation of A in the equipment number 1 there is no out flow there is no out flow therefore, whatever is coming in is getting consumed is that right. Now, is there anything coming in this is inert this is inert. So, this also goes. So, you have R A equal to R A is what by definition R A is given as minus of k times C A. So, it is minus of k C A times V equal to d by d T of N A I am sorry is that clear what I am saying I am sorry I deleted this term is not the case there is d N A by d T that means initially you are putting same. So, much of C A I here that is getting reactor I deleted this term by mistake. Now, what we have is that by virtue of the fact that there is no out flow there is no component A in the inlet therefore, the rate at which chemical reaction occurs must account for the variation of component A in the equipment minus of k times C A times V equal to d by d T of N A. Now, we are V changes we know that V initial plus some V naught times T V actually changes this is well known but the point that is important to recognize here is that C A times V by definition is N A. Therefore, the left hand side is actually N A times k k N A is d by d T of N A. So, that N A equal to N A 0 or N A I into e raise to the power of minus of k T. See the important point to recognize here is that here is an equipment where component A component A is all in the equipment initially and when you add some inner coolant component total component A does not change. So, that is an important feature that you must recognize when you solve this problem. So, it is frequently what is observed is that you know we people write this N A as V times C A and then they substitute for this V and somehow get lost in the in the algebra and lose the lose the content of the problem content of the problem is this that N A equal to N A 0 this is the content we should realize that because A does not go out anywhere. So, A will decay as per a first order law that is what this means once you recognize that A will decay as per a first order law then the question of you know ensuring that the temperature of the reactor does not change is one of trying to understand what happens in the energy balance. So, let me write the energy balance this is the energy balance which says plus reaction and then some heat added and then you have this work and all that. Now, we want to maintain this at constant temperature. So, this is left answer is 0. So, we should expect the rated which we add it will be able to take care of the reaction. So, V naught is a function of time C p t naught minus of t. So, this term there is only one reaction therefore, I write r 1 times minus of delta h 1 star times V there is no other plus q. What is r 1 by definition k times C A. So, if you substitute this here we get V naught which is a function of time C p times t naught minus of t at the first term and this term is k times C A times minus of delta h 1 star multiplied by delta V plus q. So, C A times V is N A correct. So, we can simplify this further and write this as. So, our equation energy balance equation is 0 equal to V naught which is function of time C p t naught minus of t that is the heat transfer term plus k times N A k N A which is what we are saying is that C A times V is N A that is what we are saying times minus of delta h 1 star plus q. So, we do not know V naught is what we want to find out C p is given as 400 and t naught minus of t, t naught is 26, t is 30. So, it is minus 30 so minus 26, 38. So, it is minus 12 and then what is k N A N A is N A 0 times exponential of minus of k t times delta h 1 star with a minus sign plus q. So, 0 equal to V naught t is about 4800 with a negative sign. The next term is k N A is what is N A is 1.28 moles per liter it is 12, N A is 12 exponential of minus of k t delta h 1 star delta h 1 star what is delta h given as 20000 kilo calories 20000 k cal per mole. Essentially what we have done we know k, k is known k is known everything is known here. So, we have to find out q. So, when you do all that q turns out to be we do this calculations q is plus q. Now we notice here that in this problem in this problem the there is no there is no external transfer of heat therefore, this q is 0. So, all the heat is taken up by the fluid that we have added here itself. So, this is 0. So, you can calculate and find out V naught as a function of time. So, it shows that how V naught as a function of time can be handled in a situation like this. So, what we have tried to do here in this exercise is that how do you control temperature by adding a inert fluid that is the exercise. Third exercise I would like to look at. So, if here you have a reactor then it goes to a separator then you have pure product here and then whatever is not reacted is returned this is reactor separator. And data is given just put down the data for the sake of all of you data is F A 0 this is F A 0 it is pure pure a desired production 60 60 mole per minute feed temperature 49 C feed temperature molar density 20 mole per mole 20 moles molar density molar density is given as 20 mole per litre specific heat is given as 20 mole per litre 1 kilo cal per litre C. Heat transfer coefficient is given as 5 kilo cal per square meter per minute per centigrade coolant temperature T C is given as 5 C large quantities available. So, reaction E 1 is given as 20 kilo cal per mole per litre C. So, this is given as 20 rate constant is given as 0.4 per minute at 20 C K 2 rate constant is given as 0.1 per minute at 20 C gas constants are also ok. So, we have this data in front of us we have this data in front of us we have this data in front of us and then we have to see how best we can come up with a design which sort of is appropriate for our application. So, we have looked at similar problem at an earlier stage if you recall. So, this we do not have to do the whole thing again. So, you have a reactor you have a separator. So, it goes to the separator you can recycle. So, unreact this is separator this is reactor. So, the number of questions the questions here are that you know what is the best choice of conditions for running this process. Now, we have looked at problem similar to this where we have taken some objective function and try to optimize. There is something that we should do anyway, but when you look at exothermic reactions is reversible reactions. So, on then we will have to look at the fact that you know there are more constraints that we can place on the system and get some more optimal numbers for our design. So, a question that is frequently that is frequently asked is that what is the best temperature? What is the best temperature of operation? So, this is something that people will ask us. Now, for exothermic this instance of an exothermic reversible reactions we have done all these things. We know that if R B if our reaction is A goes to B and B goes to A 1 and 2 then R B is equal to K 2 C B minus of K 1 C A we can maximize R B with respect to conversion. So, this is same as K 2 times C A 0 times 1 minus of X minus of K 1 C A 0 X. So, what is this with respect to sorry temperature at constant X. So, when you do that you can differentiate the constant X and put it equal to 0 you will find that X m by 1 minus of X m where is equal to K 1 divided by K 2 multiplied by E 1. This is what we have shown at an earlier stage that if you have an exothermic reversible reaction the best choice of temperature for your reactor is this is the best choice. We can put all the numbers here for example, X m divided by 1 minus of X m equal to K 1 value of K 1 is 0.4 K 2 is 0.1 and what is activation energy is 20,000. So, this comes to 80 X m by 1 minus of X m comes out to be 80 X m given by K 2 E 1 by E 2 sorry E 1 by E 2 I am sorry we have shown this. What is E 1 E 1 is 20 and what is E 2 I am sorry let us do it properly I just E 2 E 2 equal to how do you find out let us find out something we have done anyway we may have forgotten we can do it again. So, what we have shown we have shown already so I am just doing it once again del R by del t at constant X equal to 0 gives us X m divided by 1 minus of X m equal to K equilibrium constant E 1 by E 2 is it clear. So, in this particular case equilibrium constant is 4 E 1 is 20 and E 1 minus of E 2 equal to delta H therefore, E 2 equal to E 1 minus of delta H. So, in this case delta H is what is the delta H here delta H is given as heat of reaction is 20,000 X atomic. So, delta H is 20 therefore, 25 plus 20 is 45 this is 25 this is 45. So, it comes out to be so it is 100 by 45 100 by 45 100 divided by 45 equal to 2.2. So, X m divided by 1 minus of X m equal to 2.2. Now, this K this K is at 20 degrees 20 C. So, on other words what we are trying to say is that if you have a stirred tank and if you have a choice at of temperature at which you want to operate then immediately this fixes the value of X m at which you must operate. So, X m equal to take it here. So, 3.2 X m equal to 2.2. So, X m equal to 2.2 divided by 3.2 that is equal to 2.2 divided by 3.2 equal to 0.67 0.687. So, X m equal to 2.2 So, what we are trying to say what we are trying to say here is that in an exothermic reaction you can choose the temperature which will give you the highest reaction rate in that environment. In this particular case our environment is this is the CSTR and what is the best temperature best what is the best way to operate this the best way to operate this moment you choose this temperature moment you choose this temperature it will tell you what is the conversion that you should run at moment you know the conversion you can specify the residence time. Basically you specify temperature that way you determine conversion and once you know conversion from your material balance you can determine what is the residence time. Let us do that. So, we have here our material balance which is F A 0 minus of F A plus R A V equal to 0. So, you know that best temperature of operation is 20 C we have taken and corresponding to that X m should be 0.687. What is this X m? This X m is with respect to inlet flow this X m let me just draw your attention here draw your attention here if you call this 0 1 2 3 and 4. So, what we have done is that the way we have defined X m is that the X m is defined with respect to this point position 1. So, F A 1 minus of F A 2 times R A 2 must be equal to 0. So, F A 1 minus of F A 2 by definition is F A 1 times 1 minus of 1 minus of X plus what is R A 2 by definition it is K times C A 1 times 1 minus of 1 minus of X at position 2 this is at position 2 this is at position 2 this is within brackets R A is minus R A. So, this is minus sign here plus K 2 sorry this is K 1 this is K 2 times C A 1 times X X 2 times V equal to 0. On other words F A 1 times X 2 then we have minus of K 1 C A 1 times 1 minus of X 2 minus plus K 2 C A 1 X 2 the whole thing multiplied by V equal to 0. Let me go through this properly now. So, what we have F A 1 X 2 minus of K 1 C A 1 1 minus of X 2 plus K 2 C A 1 X 2 the whole thing I will put V here V is to be here equal to 0. So, we should be able to find out the value of X 2 from here because we know K 1 and K 2 because temperature is fixed as 20 the best temperature. We have just taken the temperature as 20 we can choose any other temperature having chosen the temperature it tells you what is the value of X 2 with respect to X F A 1. Now, we also said at an earlier stage if you recall if you recall how F A 1 and F A 0 are related we have done all that. Let me do it once again we have F A F A 1 equal to F A 0 times F A 4 we have said that before because B is equal to F A 1 fully recovered therefore, this is pure A is that clear because this pure A the C A here C A 4 equal to C A 0 something that we have said. So, F A now that is F A naught plus F A 2 notice here that F A 4 is F A 2 only because B is fully recovered. So, that is equal to F A naught times F A 1 into 1 minus of X 2 where X is defined with respect to with respect to position 1 position 1. So, we have F A 1 times X 2 equal to F A 0 therefore, F A 0 F A 1 equal to F A 0 divided by X 2 is that clear. Now, please just look back this just look back for a minute and realize recognize this X M is defined this is defined with respect to position 1 we already said that. On other words this is the value of X 2 is 0.687 defined with respect to position 1. So, this is known as 0.687 is that clear. So, if you substitute value of X 2 as 0.687 F A 0 is given in our data our data F A 0 is already specified production is 60 mole per minute 60 mole per minute that is given. So, F A 0 is given which is 60 mole per minute. So, this is 60 mole per minute and X 2 is 0.687. So, F A 1 is so 60 divided by 0.687 that comes out to be 87 mole per minute is that clear 87 mole per minute. Now, with this we can put all our numbers here now we can put this is 60 this is 60 this is 87 and then and 0.687 87 87 multiplied by 0.687 that is equal to 59. So, this must be equal to 60 because whatever is coming must go out correct. So, what is returning here it is 27 is that right. So, this is 87 so 60 goes out. Therefore, 27 will come back here. So, if you ask you what is recycle ratio what will you tell us recycle ratio is defined as what is going out divided by what is coming in correct. So, recycle ratio is 27 divided by 60 27 divided by 60 that comes to 0.45 is that all right what we say. So, what is it that we have done what we have done is component A coming in and since it is continuously getting reacted unreacted recycle. So, whatever comes in must go out and therefore, what is recycle is unreacted alone is recycle and therefore, this becomes 87. Now, the question that we have to now ask is we have chosen this reactor temperature T as 20 20 C. Now, what is the best choice that we can make this is a question that we can ask and then of course, this many of these decisions are dependent on you know cost factors and so on. Therefore, we will have to look at different temperatures of operation and see what is the cost of doing this at different conditions. But before we do that we have to recognize that we also have to ensure that the temperature of our choice is the temperature at which we want to work. So, we want to make sure that the heat transfer that is needed is available. In other words we have this reactor we have this separator this is going out this is coming back like this and this is the CSTR. So, we want to operate this at this conversion everything is given to us. So, what is the amount of heat to be added or removed this question we need to answer plus sigma all the reactions r k times minus to delta h k times v i equal to 1 to p reactions we have done all these things plus q minus of w s. Now, this is steady state therefore, the amount of the heat generation this you have to balance by this T naught is given T is given everything is given. So, we have to calculate what is the heat to be added or removed let us do that now. So, how do we ensure that the temperature that we have chosen which is 20 that it operate at the temperature we ensure this by an appropriate choice of the heating or cooling mechanism. So, you have to substitute for all the numbers. So, that we can get. So, what is let us just quickly do all the numbers. So, let me 0 equal to molar density molar density is 20. So, 60 divided by. So, it is flow is 3 liters per minute T naught feed temperature is 40 reactor temperature is 20 C p what is the C p specific heat of reaction mixture is 1. So, 1 kilo calories per this 1 plus there are reactions r 1 minus of r 2 see r 1. So, I will write here r 1 minus of r 2 and what is minus delta h is 20 into v plus q. Now, what is r 1 minus of r 2 if you write your material balance what is your material balance f a 1 coming in minus of f a 1 times 1 minus of x 2 is going out. And then the reaction which is forward reaction is r a goes to b therefore, r 2 minus of r 1 multiplied by b equal to 0 this is clear. So, this becomes minus of f a 1 times x 2 plus x 1 plus r 2 minus of r 1 is it correct input minus of output f a 1 f a 1 f a 1 equal to 0. Therefore, f a 1 is equal to 0. Therefore, 1 times x 2 equal to r 1 minus of r 2 times v. And therefore, now that we know r 1 what is what we are trying to say here is that please understand that r 1 minus of r 2 times v is f a 1 x 2. And what is f a 1 x 2 equal to f a 0 that we already said in this particular case because of this recycle and so on. Therefore, we get 0 equal to 3 multiplied by 40 minus of 20 times specific heat plus f a 1 this is basically it is f a 0 multiplied by 20 plus q. Therefore, we calculate q equal to 20 60 and then f a is 60 moles per this is 60 mole per minute. So, it is 1200 is roughly the answer is about 11 40 with the minus sign kilo cal per minute is that minus sign means what minus sign means you have to remove heat is this clear. Now, the there are some more issues to be handled here which says that what is the heat transfer area. Now, what is the heat transfer area that is required for carrying out this reaction that means q is given as minus of 11 40 kilo cal per minute what is the heat transfer area you know that q is equal to u times a by delta t. Therefore, u is given as what is the heat transfer coefficient it is given as 5 kilo cal per square meter per minute degree c this is given a is not known. And what is the delta t reactor temperature t is 20 c and coolant temperature is 5 t c is given as 5 c. So, delta t is 20 minus of 5 equal to equal to heat this is heat to be removed. Therefore, it is an 11 40 output here therefore, a equal to 11 40 divided by 5 into 15. So, if you divide by 5 into 15. So, that comes out to be 11 40 divided by 75 it is about 15.2 square meters. So, in a sense what what we are trying to put across to put across to use that I mean this is something that we all will encounter in a process that we are feeding the material. So, undergoes reaction not does not get converted fully you will have to separate the desired product and then unconverted must go back because it is an expensive material. So, in a situation like this the you need to see that this particular the heat that is needed must be taken out. So, you will have to put a heating coil and take out the heat and that particular case in this particular case it turns out to be 15 square meters is to be put it. Now, the point is that is important is that to be able to put in 15 square meters of heat transfer area you can do it by putting a coil or you can also do it by a jacket. So, this choice is yours and then you might find in many cases the jacket may be better if the heat transfer area is not very large. If the heat transfer area requirements become very large you cannot do it by jacket you cannot do it by a cooling coil inside it has to be a separate external heat exchanger equipment. So, this is a matter of design that you will have to look at it could be external it could be internal or it could be jacket and these are issues that you will take or a design I will stop there. Thank you very much.