 Today, we will look at some further considerations in energy balance, further considerations in energy balance. Now, the context to this is the following. Now, when you are running a chemical reaction equipment, we find that we need to control the process and then for controlling a process as you all recognize that we need good measurements and not just good measurements, measurements that will help us to regulate the process the way we are looking at. In other words, we really must be able to make such measurements on such variables which actually provides a good control over the process. So, in order to illustrate what we want to do now is to look at this problem in some context and understand how this can be done. So, to do this what I am going to do is let us consider a reaction equipment. Let us say it has a catalyst, I mean catalyst. I just put down some numbers just to say let us say the temperature k at 513 Kelvin, some data I have taken is 1800 per hour. Let us say the activation energy for this reaction is 20,000 calories per mole. Let us say the delta H for this reaction is minus 152 kilo cal per mole and heat transfer coefficient, the gas phase reaction, the heat transfer coefficient is 20 kilo cal per square meter hour C and the rate function R is some k times C H. So, some reaction, we look at this reaction what it is etcetera right now it may not be so crucial. Now, feed comes in and it goes out the reaction occurs and since it is an exothermic reaction very highly exothermic reaction we need to cool. So, there is a coolant which is flowing let us say it is a T C. Now, as this reaction takes place lot of heat gets generated and therefore, this gets is you know passes on to the coolant. And therefore, the reactor is maintained at temperature that is appropriate for the process go forward little and then let us say our energy balance or energy balance. Balance is I just write it down C p d T d v equal to some R 1 times minus of delta H 1 star plus 4 H by d T C minus of T. Let us just recall what is V is volumetric flow this is volumetric flow at any position C p is volumetric specific heat specific heat. And then R 1 is rate function rate function in this case we have taken the rate function as C H times k. And H is heat transfer coefficient and D is the pipe diameter pipe diameter and T C is the coolant temperature coolant temperature C is meters k cal k cal per square meter hour degree C rate function is moles per litre per hour. Let us say volumetric specific heat let us say it is k cal per litre per c volumetric flow some say cubic meter per hour whatever we do not know some number it is all the units. Now, when we are running this process let us say you have this is the tubular reactor this is the coolant going through coolant T C. Now, this temperature inside what happens to the temperature inside this reactor if it is an exothermic reaction if it is an exothermic reaction we should expect that when we make a plot make a plot of say volume of the or distance whichever is appropriate distance versus temperature. So, if it is a wall cooled here is an instance of a wall cooled reactor. That means, this reactor is cooled from the wall. So, we will expect the reactor to go through a temperature like this it goes through a this is called as the hot spot is a hot spot temperature. Now, we have said earlier and say it once again hot spot is the highest temperature that is exhibited by the reactor. And this is because of the fact that the process heat generation and heat accumulations are such that sorry heat accumulation heat removal is such. Now, this temperature at which this occurs this temperature may have an important bearing on the kind of catalyst that you are using. In many systems the there is a specification on the catalyst that is that you are not supposed to exceed a certain temperature. On other words hot spot temperature is a fairly important temperature from the point of view of controlling the reactor. So, if this is specified saying that this is the maximum temperature that we are willing to permit therefore, and if you can measure temperature along the length of the reactor which is not so difficult because we can put thermocouples and measure temperature along the length of the reactor. And this temperature read out and might be continuously available as the process is running. On other words as you are running the process at every point in the reaction equipment the temperature would be continuously measured and then it is available to you from your data. Now, how is it that we can make you good use of this temperature read to understand the process to understand how best to control this equipment that is what we would like to do by looking at this energy balance. This is the energy balance we want to look at this energy balance and see how best we can do this control. And to illustrate this we take an example that is what we are trying to do now. Now, what we have said just now is that if there is a wall cooled reaction if there is a wall cooled reaction there is hot spot which you have pointed out just now there is a hot spot and this hot spot. And if you look back at our equation V C P D T D T d V equal to R 1 times minus of delta H 1 star this is the heat generation due to reaction. And this is T C minus of T by D this is the heat removal the heat generation this is heat generation this is heat removal. Now, at the hot spot this must be 0 we know this at the hot spot at hot spot at hot spot we should have the left hand side as 0 this is well known on other words on other words what we are saying is that at hot spot at hot spot we have 0 equal to R 1 times minus of delta H 1 star plus 4 H by D T C minus of T or I can write this as 4 H by D T minus of T C equal to R 1 times minus of delta H 1 star at hot spot. So, what are we saying this is the heat generation heat generation and what is this this is heat removal. Now, please notice please notice that this term I should have put a negative sign here because in first law convention the heat addition is positive. So, I should have put a negative sign here. So, what we are saying is that heat removal equal to heat generation at the hot spot this at hot spot. Now, I have some data this is some data let me just put down this data what does this say this data says the following you have an exothermic reactor which is cooled by the wall it is cooling from the wall cooling T C and this is T. So, wall cooling this is wall cooling the data says D P T at hot spot and conversion at hot spot this numbers look like this 0.05 0.075 0.025 and then you have 240 you have 240 you have 300 0.43 0.79 and 0.61. What are you saying this is some data that is available from the process what is it that what are we expected to do with this data we are expected to review this data to see how best we can use this information to control the process. So, that is what we are trying to do. So, we will try to look at this data. So, and also some more information is given that is feed temperature is this is T T naught is equal to 140 C. And then we are expected to review this data to see how best we can use this information to control the process. So, that is what we are trying to do. So, we will try to look at this data. So, and also T T naught is equal to 140 C. And then feed y a 0 is coming in at 0.1 and pressure is 1 atmosphere. On other words we have a gas entering a reactor containing a catalyst at 140 C at 1 atmosphere pressure with a at 0.1 the rest. So, as an example we can look at this reaction acetylene plus hydrochloric acid this is a gas both are gas giving you vinyl chloride monomer C 2 H 3 C L this is V C L. So, the rest why this is H C L let us say and the balance is acetylene the reaction is taking place. And in this one is 5 centimeter diameter pipe it is 7.5 centimeter pipe and this 2.5 centimeter pipe link it 1 inch 2 inch and this is the observed data at this temperatures. Now, that we know that R R 1 times minus of delta H 1 star equal to 4 H by D T minus of T C. What is R we can say it is K times C H we said. So, it is K times C A 0 times 1 minus of X. So, I am doing a small simplifications because it is only 10 percent H C L I am not taking into account the volume change due to reactions. So, to that extent it is not such an accurate calculation, but that is not so important for the moment. What we trying to bring out is how we can understand the data from the point of view of controlling the process. That is why I am not doing the calculation to such data. Let us do this calculations. So, R 1 delta H. So, you can substitute for R 1. So, K C A 0 times 1 minus of X times minus of delta H 1 star equal to 4 H by D T minus of T C. Therefore, 1 minus of X equal to 4 H T minus of T C divided by K C A 0 D minus of delta H 1 star. So, at the this is at the hot spot at the hot spot. So, what is this equation give us? It tells us what is the conversion at the hot spot 1 minus of X is conversion. How it is related to the heat? This is the heat I mean cooling the reaction and the heat transfer you know how it is related H is heat transfer coefficient and so on. Now, we can put all our numbers here and find out what is the value of X. Let us just do that quickly. Let us do that quickly for the 3 sets of data. For the 3 sets of data let us quickly do that. So, set 1 what is set 1? Set 1 is let me just D is 0.05, temperature is 240, X is 0.43. So, 1 minus of X equal to 4 H value is given. So, it is 20 and then our equation is here 4 H T minus of T C. T is what is that this particular case what is the 240? 240 is the 240 minus what is the cooling temperature? Coolant temperature T C coolant temperature T C I have to write it down here T C is 25 C. So, 25 divided by diameter of the pipe is 0.05 rate constant at 513, 51240 is 513 is that right? This is 51240. So, it is 1800 this is 1800 per hour. See data is given at please note that data is given at 513 that is 240 C. So, it is 1800. So, that is what I am substituting here. So, K is 1800 and what is CA 0? We should calculate CA 0. It is not very complicated CA 0. Let me just do it here for you CA 0 equal to P by R T and then we have to multiply this by the mole fractions 0.1. So, that is turns out to be 2.34 10 minus of 3 moles per liter. Now, let us put all these numbers here that is 2.34 into 10 minus of 3 and heat of reaction heat of reaction please note here is fairly exothermic is 152 kilo calories per mole. So, I will put them in 152. I have to make it 152,000 to get all the units right. So, this is 20,000 I have put 20, 20 kilo calories per hour which can be Fahrenheit and so on and this is 152,000 and so on. So, if you put all these numbers and then we get x equal to 0.537. When you do this calculation we find that x equal to 0.537 1 minus x equal to 0.537 implies x equal to 0.46. Now, please notice from our data here x is given as 0.43 in the experiment these are the experimental and from our model we get a value of x of 0.46. So, I just want you to remember this we will come back to it as you go along. Let us do one more calculations this is for 240. Similarly, we can do for one more calculations let us do it quickly. This next data the next data once again let me just remember 1 minus of x equal to 4 h t minus of t c divided by d k c h 0 into minus of delta h 1 star. So, let us once again put these numbers here we have the second set of data 0.075 to 40 and 0.79 is the experimental value. So, we have to calculate notice here that the temperature here is 0.075 0.75. So, we can use this once again let us do this I will put all the numbers. So, numbers look like this equal to 4 times 20 times 240 minus of 25 divided by our pipe size 0.75 temperature is 240. So, it is 0.075 and what is c a 0 we have already calculated c a 0 where is c a 0 at 240 there is one more data which is missing y a 0 this is 0.1 this is 0.1 this is 0.2. So, that is what is missing. So, c a 0 corresponding to y a 0 equal to 0.2 this is 0.2 y a 0 that turns out to be point this turns out to be 0.0047. This is actually double the number that we have calculated see here it was 2.34. So, it is double that because the mole fraction is instead of 0.1 it is 0.2 that is why I have used this term 0.07 moles per liter. So, it is k value is 1800 and 0.0047 and then 152000. So, the whole turns out to be 1 minus of x equal to 0.178 implying x equal to 0.82. Notice here that the experimental value is 0.79 and then our model value is 0.82. Let us do one more calculation for this third set of data just bear with me for a little while because it is an important point to be made here. So, I am sort of going through this calculations in front of you. So, we have the third set of data once again 1 minus of x equal to 4 h t minus of t c this is at hot spotter d times k times c a 0 times minus of delta h. So, I am putting this as 4 times 20 and here the data says that it is 300 minus of 25 divided by 0.025 the size of this is 0.25 particle size of the pipe. So, that is why 0.025 k value corresponding to 300 we have to calculate k at 300 of course, since the activation energies are known I just write down the number k at 300 is 7.76 times k at 240 this is c 240 c. I would not calculate this because quite obvious. So, I am just putting down the numbers since activation energies are known you can do all these things yourself. So, we have 0.025 and then this is 7.76 times 1800 that is k value of 0.025 and then k value. So, you got d you got k then you got c a 0 corresponding to 300 c a 0 at 300 Kelvin sorry 300 c not Kelvin 300 c that also have calculated and found it to be 1.05 10 to the minus 3 mole per liter this is at 300 c what is the c a 0 corresponding to mole fraction y equal to 0.1 which is given. So, all this then I have to put this as 1.05 10 raise to minus 3. So, all this if you put our number comes out to be 0.394 therefore, x equal to 0.61. So, let me summarize by saying that we have we have here x x experimental x experimental model. So, we have d p d p I will write d p here. So, that you know 0.5 0.75 and then 0.25 solve meters d p in meters for the 3 experiments the last one the model experimental value this is the model values 0.61 and the previous case let put put down all the previous case the model value let me just put down all the model values. So, it is x experimental 0.61 0.79 0.43 and then the model values are model values are this is 0.60 not 0.60 model values of 0.60 and the other one is model value is 0.82 we have got that also and then model values 0.46 what in essence we have done is the following. We have let me just review this once again what we have got we started with the tubular reactor which contains a catalyst. We said this tubular reactor is cool from the wall and the coolant temperature is at 25 C. This is the catalyst and we have done experiments with different sizes of the pipe just to understand how we can regulate or control such a process. So, you have done 3 experiments one experiment with 5 centimeter pipe one with 7.5 centimeter pipe one experiment with 2.5 centimeter pipe in each case we have found out what is the hot spot temperature and what is the conversion at hot spot this is an experimental data. Now, based on our model what is our model our model said that at the hot spot the heat generation must be equal to heat removal heat generation equal to heat removal. And therefore, we have from this equality we are able to find out what is the value of conversion at the hot spot. See conversion measurements is more complicated while temperature measurements are relatively easy. So, if you can relate conversion to temperature and use this as a way of controlling your process it becomes a very valuable tool for the point of view of executing a process control. That is what I am trying to illustrate to you through this example. So, what we have done we have done 3 experiments and those 3 experiments they tell us I will write the hot spot temperature here is 240 and 300 this is hot spot temperature hot spot. Now, based on these 3 experiments what we find is that the experimental value of conversion and the model predicted value of conversions are nearly pretty good it is not bad at all. Showing that our way of predicting or in other words we can now look at this in a slightly different way. Now, we are now able to tell that if you know the hot spot temperature then we can tell what is the conversion. So, alternatively suppose we have a process in which we want to run the process so that we produce so many tons of a given product. So, that means the experimental conversion that we are getting it is actually a design that we have to ensure that we must produce so much. Therefore, this 0.79 or 0.61 or 0.43 is something that is given as requirement of the process. What is it that we need to do from our side is to see whether ensure that this is actually achieved. Now, what is being said from our model is that that if indeed our model is right and we are able to predict the experimental conversion simply by looking at the hot spot temperature. Now, we can say alternatively that in a process simply by controlling the hot spot temperature we can actually get conversions of our interest. So, this is the point of this whole exercise what we are trying to put across to you is that now since temperature measurements are relatively easy and very accurate. Therefore, we can continuously monitor hot spot temperature and once that monitoring of hot spot temperature and regulate we can regulate hot spot temperature. How do you regulate hot spot temperature? You look at this equality if you want to regulate this you have to control the coolant temperature which is also relatively easy in the sense that the coolant flow can be controlled coolant temperature at which is entering can be controlled. Therefore, T c is something that is in your hand moment you manage T c you are able to get hot spot temperature of your choice and therefore, productivity of your choice. So, this is the important point of this exercise which is a part of our further considerations in energy balance. So, this is the exercise that I wanted to explain to you in this. So, we want to consider the second related issue of what I call as further considerations in energy balance. So, let us look at that exercise. Now, let me before we get into this there are few points I want to draw your attention. Let us say you have a chemical reaction k whose rate constant is let us say 1800 per hour. Let us say you have an endothermic reaction it is an example. So, you are processing at let us say 1 mole per hour and your heat of reaction reaction 20 mole per hour 20 mole per hour heat of reaction is 20 kilo cal per mole. Therefore, if you wanted complete conversion you will say that heat release equal to heat to be delivered. So, 20 into 20 which is 400 kilo cal is 400 kilo cal per hour is what you have to supply is that clear. Now, let us say you have your heat transfer design heat transfer design what is your heat transfer design? Your heat transfer design is that you are able to supply 20 kilo cals per hour as an example 20 kilo cals per hour. And then what is heat of reaction what we are saying is that we want you want to process 20 kilo moles per hour. That means, you must be able to supply heat supply. So, heat supply you must be able to supply 400 kilo cals per hour to be able to process correct is this clear. So, what we are saying is that the amount of heat that you supply is what is going to determine the rate at which chemical reaction occurs. Because, your chemical reaction itself your k is something like 1800 per hour and then this reaction is relatively fast compared to what you are able to supply. What we are saying is that the rate at which you are able to supply is quite low compared to the rate at which the reaction occurs. On other words reaction only moves at the rate at which you supply heat. So, we are looking at an instance of a reaction reaction proceeds proceeds at the rate at which heat is supplied. So, if heat is not supplied reaction does not move. So, we are looking at situations where the rate of heat transfer essentially determines the rate of chemical reaction. Now, what are the situations that would be of such situations. Let us take an example an endothermic endothermic reaction which is instantaneous which is instantaneous. What do we understand by an instantaneous reaction? By understanding instantaneous reaction is that if the rate constant k 1 and rate constant k 2 are such they are very large. This is one way of understanding instantaneous reaction. That means therefore, the reaction is essentially at equilibrium. Because, both k 1 and k 2 are very large say a going to b a going to b and b going to a. This is a reversible reaction k 1 and k 2 both are very large and therefore, essentially at equilibrium. So, if a reaction is at equilibrium what is the conversion at equilibrium at equilibrium. We have done all these things. So, I will write down the answers if k if equilibrium constant is k then conversion is k by k plus 1. So, and k is a function of temperature and this dependence on temperature comes from our van toff's equation we know all that. We have done all that thing we know how k depends on temperature and so on. Because, it comes a matter of the equation. Therefore, let us say if you have a reaction let us say this is the jacket through which some fluid is circulating this is the reactor there is some catalyst may be may not be. So, this reaction is able to move forward only because this heat is getting supplied. If this supply is not there the reaction does not move. At every point in the equipment the conversion is as described by the equilibrium. Because, the reactions are very fast in relation to the rate of heat supply. So, that is the example we are considering. Now, what we would like to do is that let us say V C P I just want you to draw attention to our energy balance and sort of draw attention to your delta H 1 star plus R 1 plus R 2 minus of delta H 2 star plus 4 H by D T C minus of T. So, this is our energy balance equation which is also written as R 1 minus of R 2 times minus of delta H 1 star plus 4 H by D into T C minus of T. And we also know from our of our material balance that for a plug flow reactor it is. So, this we know this is the reaction is A going to be B going to A or this is also written as D X D V equal to R 1 minus of R 2. Now, the question that is in front of us is this reaction R 1 and R 2 are given as instantaneous reactions. Now, if R 1 and R 2 are instantaneous which means what R 1 is very large R 2 is very large. So, what is R 1 minus of R 2? There are two quantities both are very large. So, what is the difference between two large quantities? It is not easy to tell how large is large we do not know. So, clearly therefore, we are not able to make good use of the energy balance equation to solve a problem like this. One of the reasons why I am looking at this further considerations in energy balance is to be able to deal with variety of situations that we might encounter in daily life. Here is one of one such instance where we have an instantaneous reaction. Now, when you try to apply this into our energy balance we find that R 1 minus of R 2 which is 1 and 2. Now, R 1 is very large R 2 is very large and therefore, we do not know what is R 1 minus of R 2 and the rate functions themselves are not known. Therefore, we do not know what is R 1 minus of R 2 all that we know is that because R 1 and R 2 are very large. Therefore, it is essentially at equilibrium at every point in the process. So, we want to use that information to get around the difficulties that we face in using our energy balance and that is the point of interest to us in this exercise. Let us see how to do this. To do this we recognize the following. We know that d x d v equal to d x d t multiplied by d t d v. Now, we know that x equal to k by k plus 1. This since reactions are instantaneous. So, are at equilibrium. Now, so what is d x d t now? What is d x d t? d x d t equal to we have to differentiate this. I have done that. So, let me write down this k plus 1 minus k by k plus 1 whole squared into d k d t multiplied by 1 by k plus 1 whole squared delta h by r t squared. Now, this is I mean you might wonder you know how I have written this without giving you the details. So, we can understand this. So, when you differentiate this you get let us go through this 1 by 1. So, you get 1 by k plus 1 k by k plus 1 whole squared minus and d k d t. And what is d k d t? Let me write van toff's equation here van toff's equation. What does it say? d l n k by d t is delta h by r t squared or d k d t equal to k times delta h by r t squared. So, what we have tried to do is that we have substituted for all these and got that result. So, let us go through this now substitute and then simplify. We have got d x d t. Now, all we have to do is all we have to do is the following. So, let us write the energy balance now. The energy balance is v c p d t d v equal to minus of delta h 1 star f h 0 d x d v e plus 4 h by d t c minus of t. I hope you understand what we are saying. We just look back and recognize all these things. Please recognize that r 1 minus of r 2 is f a 0 d x d v. Now, since we have difficulty in managing r 1 minus of r 2, what I have done is that I have replaced this r 1 minus of r 2 from material balance and by f a 0 d x d v. On other words, our energy balance equation is this. So, first term this the r 1 minus of r 2 term I have replaced it at d x. Is that clear? So, our reaction of interest is a going to b and b going to a as an example. Now, we have written material balance which is d f a d v equal to r something that we have written for a long time. Now, if you substitute put f a and then recognizing that r a is what? r a is basically component a is consumed in reaction 1 and it is formed in reaction 2. They have put a negative sign here. Therefore, if you call f a as f a 0 times 1 minus of x something that we know if we call f a equal to f a 0 times 1 minus of x based on our understanding of stoichiometry and so on. Therefore, we can write the left hand side as f a 0 d x d v equal to r 1 minus of r 2. Something that we know means not very difficult to understand this. So, you have this is the material balance. I call this say let us say as equation 1. Now, what is our energy balance? Our energy balance is v c p d t d v. We are talking about a plug flow. Please understand the plug flow reactor. That means our reactor looks like this. Our reactor looks like this. May be there is a catalyst here and so on. Flow is coming in and flow is going out. So, v c p d t d v is r 1 minus of r 2 delta h 1 plus 4 h by d t c minus of t. This we have written for a long time. This heat is getting transferred from here. v c is the fluid that is flowing in the jacket through which there is a heat transfer fluid flowing. So, this is the statement of energy balance of for a plug flow reactor, where v c p d t d v is r 1 minus of r 2 minus of delta h 1 plus 4 h this is the heat transfer. Now, what we have done since r 1 minus of r 2 this is very large because this is very large. This quantity is very large. What do we do? We try and replace this from material balances. So, r 1 minus of r 2 is what? F A 0 d x d v. So, what is it that we have done? v c p d t d v is on the left hand side. We have knocked out this term and replacing it by from reaction 1. If I call this as reaction 2, what we have done is that we have replaced r 1 minus of r 2 from reaction 1 in equation 1 in equation 2. So, that this term r 1 minus of r 2 can be now written as F A 0 d x d v. Therefore, our energy balance now looks like v c p d t d v which is F A 0 d x d v minus of delta h 1 plus 4 h T c minus of t. Now, in this formulation in this formulation, what is the advantage that we see? We have got rid of the rate functions. We have gotten rid of the rate functions and they are given in terms of d x d v. In principle, x is measurable. See, the advantage of equation 3 is that now your quantities are all measurable. Therefore, even though your reactions may be very very rapid and therefore, the rate functions are difficult to handle, but x conversions are measurable, positions are measurable. So, you have replaced a difficult situation into something that you can handle. Now, we want to go further to see what more we can do. So, what we have done here is that this d x d v we have written as d x d t and d t d v d x d t and d t d v which is something that we can write. So, if I call this as equation 4, now in this form of equation 4 d x d t and d t d v, let us see what benefits that we get. Now, let us let us see what what what I have got here is, let us our reaction is a going to b a going to b it is instantaneous. Because it is instantaneous therefore, it is at its equilibrium and because it is at equilibrium, we can say that extent of reaction x is given by k by k plus 1 where k is the equilibrium constant k is the equilibrium constant. So, k is equilibrium constant and therefore, we can say that x is k by k plus 1. Now, what we are saying now what we are saying now is that in our previous representation in your previous representation our equation left hand side is rated which temperature changes its position or volume and the right hand side involves a d x d t and a heat transfer term. This d x d t is a term that we can understand more for the case of an instantaneous reaction. For the case of an instantaneous reaction, we said k can be given by k by k plus 1 where k is equilibrium constant and now d x d t d x d t which appears in our equation can be obtained by differentiating x that is what has been done here. So, you notice here the d x d t from our Vantos equation which is well known to you d x d t is simply k delta h divided by k plus 1 whole square r t square. Let us just understand how it comes now Vantos equation says that d l n k by d t is delta h by r t square that I have written as d k d t equal to k times delta h by r t square this is nothing very new to you. Now, we know that d k d t is k delta h by r t square now what is d x d t you have to differentiate k by k plus 1 this is what I have done I have taken the first term 1 by k plus 1 then I have taken the next term differentiation of this is k plus 1 whole square with the minus sign and the whole thing multiplied by d k d t is that clear what we are saying. So, you can simplify this common denominator is k plus 1 therefore, you get k delta h by k plus 1 whole square. So, on other words what we are saying here is that d k d t d k d t of our material balance or of our energy balance can now this term d x d t can be represented from this equation here d x d t is given by k delta h by k plus 1 whole square r t square. Therefore, now I am able to replace the d x d t term the d x d t term of our energy balance equation I am able to replace this d x d t term and simplify this what I have done here. So, I have replaced the d x d t term here and then put it as minus. So, anyway you know all that. So, I have replaced it I have found delta h f a 0 d t d v and then d x d t plus is that all right what I have done f a 0 f a 0 delta h f a 0 delta h d x d t d t d v the d x d t I want to replace d x d t is what I want to replace from here which I have replaced here. So, essentially what we have got what we have got is that because we are able to replace d x d t in terms of d t d v in terms of equilibrium constant and so on. So, the equation now looks like this v c p d t d v on the left hand side and on the right hand side also you have a term d t d v and then heat transfer term. Now, you can combine this d t d v terms that is what I have done now what it gives you is that the tubular reactor that tubular reactor that we have now that we have this is the catalyst may be now what we have now is that d t d v that rate at which temperature changes with position d t d v can be given as this heat transfer quantity divided by v c p and delta h squared f a 0 k by k plus 1 whole square the whole thing is in the denominator I have written it again I have just written it again here I have written it again here. So, that our tubular reactor let me just show it once again because this seems to be very important to understand this is the reagents are going this is where the heating or cooling medium is going these are the reactants reactants and products. So, what we are saying now is that the rate at which temperature changes with position inside the reaction equipment it depends upon the heat transfer that we that means the amount of heat that you are able to transfer or remove depending upon the in the direction you know whether it is t is greater than t c or less whatever it can be this way or the other way. So, heat transfer divided by v c p delta h 1 square f a 0 k by k plus 1 whole square and r t square. So, what we are now able to do is that if you have an instantaneous reaction if you have an instantaneous reaction then this if you want to trace what happens to our temperature what happens to our temperature with position now you can notice here on the right hand side at position 0 at v equal to 0 at v equal to 0 h is known t c is known t is known because t is t 0 v is known which is v 0 c p is known delta h 1 comes from thermodynamics f a 0 is known to you equilibrium constant is known to you at that temperature. So, on other words what we are saying is that d t d v at v equal to 0 is a known quantity. So, you have a differential equation where the initial state is known you can do a forward march lots of routines are available for it do the integration and therefore, you are able to tell if it is an exothermic reaction if wall cooling you might get a profile like this you can actually do the profiles simply by integration. So, what we are trying to put across to you here is that even when you have an exothermic reaction which is instantaneous you are able to handle that reaction because of the formulation that we have provided. Let us just take this idea little forward let us let us take I mean this is an example that you would like to see whether this applies to you in your practical situation rate constant Laplace is the first order reaction the rate constant k has a units of inverse of time or therefore, time constant for the reaction can be take 1 by k can be sort of as a time constant for reaction. Now, similarly you can look at time constant for heat transfer. So, what is time constant heat transfer? So, if I know the thermal diffusion coefficient and if you know the characteristic dimension r square by d r square by alpha alpha is the thermal diffusion coefficient. So, r square by alpha is a measure of the time constant for heat transfer this is time constant for heat transfer this is time constant for reaction. Now, if the time constant is time constant or inverse of time constant the reaction. So, if time constant is very large which means that that particular step is rate controlling therefore, if in a situation you find that heat transfer rates are very low it is it is it is you know heat transfer coefficients are very low and therefore, essentially the reaction moves only to the extent that you supply or remove heat. Therefore, you can actually design the system simply as a heat transfer equipment because the reaction rates are so fast that everything depends upon the rate at which heat is supplied or removed. This is the point I am trying to put across to you that every situation you will be able to determine whether you know what is it that controls your process if heat transfer is controlling you just look at the heat transfer problem and appropriately you can design your equipment this is the object of this exercise.