 Look at advanced reaction engineering, today we look at practice problems, look at tubular reactors, this is the exercise for today. We all know that tubular reactors are very very common in the process industry, fine examples of that would be sulphur dioxide plus oxygen giving a sulphur trioxide, then you have nitrogen plus hydrogen giving ammonia. These are finest examples of exothermic reversible, these are all exothermic reversible and catalytic, they all require catalyst. So, examples that our process industry use extensively, we want to look at some of these to understand some of the intricate details that go into it and some strategies that helps us to understand how to manage these reactions in the process industry. Now, to put it in the context, let us take an example, let us take an example a going to be 1 to a reversible reaction, let us say that r b equal to k 1 c a minus of k 2 c b. As an example, this is not illustrate how things would happen. So, c 1 c a 0 1 minus of x minus of k 2 c a 0 times x of course, assuming that c b 0 equal to 0, it is frequently this would be a case that applies. Now, if you look carefully at this reaction, we find that this term del r b with del t at constant x, what is del r b with del t at constant x, when you differentiate this 1, you will get k 1 e 1 by r t square c a 0 times 1 minus of x minus k 2 e 2 by r t square c a 0 times x. Now, it is known that for each x, this r b goes to a maximum. On other words, if you set it equal to 0, we get an equation, which is x m by 1 minus of x m equal to k 1 by k 2 k 1 by k 2 e 1 by e 2. What is k 1 by k 2? It is equilibrium constant times e 1 by e 2. So, what we are saying is that del r b by del t at constant x goes to 0. This implies x m by 1 minus of x m equal to k equal to constant times e 1 by e 2. Now, if you look at this reaction, this reaction at equilibrium r b equal to 0. So, at equilibrium r b equal to 0, which means that k 1 c a 1 minus of x minus of k 2 c b c a times x equal to 0, which means that x by 1 minus of x equal to k 1 by k 2 equal to k. Notice that x m by 1 minus of x m is k times e 1 by e 2 and at equilibrium, I will call it, these are at equilibrium. I should put x e here, so x e by 1 minus of x e equal to k. So, the forms are very similar forms are very similar and we have already shown this. So, there is no need to go through this once again. We have shown that plots for an exothermic reversible reaction. If you make a plot of of t versus x, if you make plot the equilibrium, this is the equilibrium curve, which means this curve is described by this equation. Now, if you make a plot of this equation x m by 1 minus of x m k e 1 by e 2, this curve will look something like this. This is the, will I call this del R b by del t at constant x equal to 0 or we can call it as locus of max rates. Now, why are we doing this? We are doing this to understand how to conduct this reaction in a tubular reactor. So, that is the, so essentially what we have done, we have found out the locus of maximum reaction rates that is appropriate to this reaction. So, clearly if you want to have a good design, you would like to see that our process runs along this line, so that the size of the equipment would be as small as possible. Let us say, let us say we have a tubular reactor. That means, look at a tubular reactor. What is our tubular reactor design equations? Our energy material balance, our material balance will look like F A 0 times d x d v equal to R 1 minus of R 2. This is a material balance. Our energy balance will look like this d t d v equal to R 1 minus of R 2 times minus of delta H 1 star, which is the heat of reaction plus q minus of W s. We say we do not look for any work from the system and if it is adiabatic, if adiabatic. So, let us look at the moment, the situation of a tubular reactor, material balance and energy balance under adiabatic situation. That means, we do not allow heat to go out of the system. Now, if I call this as equation 2 and if I call this as equation 1, if I do 1 divided by 2, we will get V C p d t d v divided by d x d v. So, I will call this as d t d x, d t d x. This is F A 0. R 1 minus of R 2 cancels off. So, we are left with minus of delta H 1 star. So, just set it up once again. So, I am just rewriting this equation once again. So, we get V C p. So, this is once again tubular reactor. We have V times C p d t times d x. There is a left hand side and the right hand side has got minus delta H 1 star. So, it is equal to minus of delta H 1 star. Now, V volumetric flow, if for a given reaction volumetric flow actually depends on so many factors particularly the temperature and so on. If there is change in moles, also volumetric flow will change and so on. So, suppose we consider situation as in this particular case, A going to B and B going to A, V equal to V naught. So, that in this case F A 0. So, that if I put V equal to V naught, this becomes V naught C p. F A 0 is V naught C A naught d t d x equal to minus of delta H 1 star or d t d x equal to cancels off. So, it becomes C A 0 times minus of delta H 1 star divided by C p. Now, the point of putting it in this form is that what we are trying to say here is that in a tubular reactor, variation of temperature with conversion depends equal to C A 0, which is a constant delta H 1 star by C p. This is typically is not a very strong function of a temperature or any other compositions. Similarly, C p the volumetric specific heat we have said it is not a very strong function of what happens during the chemical reaction. So, in other words what we are trying to say is that d t d x d t d x equal to C A 0, which is a constant minus delta H 1 star, which does not change very much. C p also does not change very much. Therefore, the right hand side is a constant. This is the point we are trying to put across. So, right hand side is a constant. You can see. So, as a result when you make a plot, you make a plot of x versus t and you have this equilibrium curve, you have this maxima and rate curve. This is equilibria. This is locus of max rates and if our temperature is t 0, we are starting here. As per this equation here, we should expect that we will move along a line like this. Or if your t 0 is here, then we will move along a line like this and it will go and touch the equilibrium curve. At this point, the rate reaction rates are 0, rates equal to 0. At this point, the rates are highest. So, you notice here that as you move along this line, this line is the adiabatic line. As you said, this is the adiabatic line. Why is it the adiabatic line? We have knocked out. We have knocked out the term Q from our equations because we are writing the equations for the case of an adiabatic process. So, as you move along the adiabatic line, from this point to this point, the reaction rate keeps on increasing. Here, it becomes maximum. As it moves from here towards the equilibrium curve, the rate keeps on decreasing and becomes 0 as it touches the equilibrium curve. Now, the question that you and I would like to know is, in our design, what do we do? How do we achieve a design which ensures that the reaction goes to see, this is x. So, we want to go as high a level of x as possible. That means, we want this to reach to the highest value. Now, if I say this is 1, this is 0.99, this point is 0.98, let us say and we want to go to as high a conversion as possible. How do we achieve this? Now, what our equations are telling us is that, an adiabatic process, temperature and conversion will move along this line and then, it will go and hit the point of equilibrium and stop. So, clearly if this point, the conversion corresponding to this point, this value of x here, let us say is 0.3, then clearly we are not able to reach as high a conversion as you and I would like to reach. Our point in this practice exercise is, how do we go to higher and higher levels of reaction, levels of extensive reaction. That is what we would like to do. Let us see how to achieve this? We achieve this by recognizing the following. You have equilibrium curve, you have your maximum reaction rate curve, this is equilibria, this is locus max rates and let us say our feed is available at t naught. Now, if feed is available at t naught and if you are going to do an adiabatic process, you would probably travel along this line and then, go and stop there. If instead, if we had heated up this t naught and gone up to this point t naught here, then once again this line is able to give you the average reaction rates here, the average reaction rates here, these reaction rates are much larger. Why are they much larger? Because the temperatures are much larger from t 1 0 to t 2 0. What we have achieved is that, we have achieved much higher reaction rates and we are able to move along this curve. Once again, we face the same problem that we have to stop somewhere because the reaction rates from this point, it starts decreasing. So, we will start somewhere here and then, we will start wondering what is it that we must do. So, what are we trying to say is the following that our tubular reactor equation says that d t d x on the right hand side is a constant and therefore, our the d t d x being a constant it is a straight line and this straight line, if you move along from t 0, if you move along you can keep going along or you can give going along up to the equilibria. And moment you reach equilibria, the reaction rates are nil at the point of corresponding to this maximum reaction, the rates are the highest. So, as you go from this point to this point rates keep increasing and from this point onwards, the rate starts to keep decreasing. And when you reach the end point, the reaction stops. The question is how is it that you and I can conduct this process. So, that we go to the highest level of conversion possible. Now, we notice that if the feed temperature is low, the reaction rates are low. So, that we can preheat it and take it up to t 2 0 as the initial temperature entering the equipment. Then, we run an adiabatic reactor. Once again, we will face the same problem that as it approaches the equilibria, the rate becomes very low. So, you have to stop here. Therefore, how do we go forward from here to much higher level of conversion for which what we decide. So, we run interstage cooling. So, if you can interstage cool this and then once again from here, we can run one adiabatic reactor. Once again, we can cool interstage and then run another adiabatic reactor. We can cool interstage and then run another adiabatic reactor and cool interstage. In other words, what we are trying to say is that interstage cooling interstage cooling interstage cooling interstage cooling interstage cooling is essential to be able to reach high levels of conversion for reversible reversible exothermic reactions. So, this is the. So, if you look at sulphur dioxide or if you look at ammonia, you will find that this is reactor 1, reactor 2, reactor 3, reactor 4. So, this is the instance of a 4 stage process, where we go from 0 conversion to a much higher level of conversion. So, this is the exercise that we want to illustrate. This is the exercise we want to illustrate and then the numbers are given below. So, we have a chemical reaction. We have an instance that data is given here C A 0 is given 1.6, specific heat is given as 1.0, mole per litre and then is calories per kg per C, T 0 is given as 21, V 0 is given as 5 cubic meters per hour, density is given as 0.91 grams per ml and then you have E 1 is given as 25,000 calories per mole, delta H is given as minus 20,000 calories per mole. On other words, E 1 is known, how is E 2? We know that E 1 and E 2 are related by our understanding of basics. So, if you have a chemical reaction which is reversible, so this is E 2 and this is E 1 and by definition E 1 minus of E 2 equal to delta H. So, for an exothermic reaction E 2 is greater than E 1 and therefore, delta H is negative. So, if all the data is given below and in the rate constants, rate constant k is given as 12 per hour at 21 C, equilibrium constant is given as 12.2 at 25 C. So, you can see here rate constants are given, equilibrium constant is given. Therefore, we can plot the equilibrium curve. So, we know, so we can do the whole exercise quickly, so that you know we have a good understanding of how this whole thing is to be done for a multistage tubular reactor design. Let us quickly run through this for the numbers. So, how do we do this design? We x versus t. So, what is equilibrium curve? We know that it is k by k plus 1. So, we can calculate the value of k, since k value is given at 12.2 at 25, the heat of reaction is given. So, we using the van toff's equation, we can calculate the value of k at various temperatures and using that, we can plot this curve. Then, we also recognize that the locus of maximum reaction rates, locus of maximum reaction rates is given by this x m by 1 minus of x m equal to k divided k 1 by e 2. So, using this relationship, we can plot the maximum reaction rate curve. This is x m, this is the x m curve, this is the x e curve. Then, if our feed temperature is given, feed temperature is given as 21. So, feed temperature is given as 21, but if you want to really go to very high conversions, if you start with this as 21 and if you run a tubular reactor, which is our dT dx, we know that dT dx is given by C A 0 minus of delta H 1 star divided by C p. C p is given, C p is volumetric specific heat, man. So, we can now run a tubular reactor corresponding to this slope. We can run a tubular reactor like this and we can stop somewhere here, but that does not take us to sufficiently high conversion. Then, we run a interstage cooling. Then, again another tubular reactor we run. Then, once again we run. Then, again we do this. So, this is how we keep doing this till we achieve the points of our interest. This is stage 1, stage 2 and stage 3. So, the exercise that we are looking at here is that. What is the rationale for multi-stage tubular reactor design? What is the rationale? The rationale is that given a feed temperature and an adiabatic reactor at that feed temperature, if you run, at best you can reach a conversion which is much lower than what is desired in a process. If you want conversions of 95 percent and so on, the process does not allow you to go to such high conversions unless you have multi-stage designs. So, the design feature that you must take into account is how much interstage cooling you would like to do between stage. You can cool up to here or you can cool up to here. Each decision implies a certain volume of the interstage cooling device and certain volume of the reaction. On other words, in multi-stage tubular reactor design, the volume of the reactor or size of the reactor and size of the cooling device are both important. Let us say for example, you look at ammonia which is a high pressure process or sulphur dioxide which is a low pressure process, but it is gas. So, if this cooling, it will be gas to gas cooling would mean a very large heat exchange surface. So, you have a large heat exchange surface and therefore, a large investment here and this is the reactor depending upon the cost of the catalyst. You will find this tubular reactor would have a certain size. So, you have to take into account the size of the reactor as well as the size of the exchanger. Both are important from the point of view of design. This is the point that we must bear in mind when we look at tubular reactor design. Both the size of this equipment and which is reaction equipment, this is the heat exchange equipment. We have to look at both to be able to decide what might be appropriate. So, this is the point I was trying to get across to you and there are several good examples in the literature shift conversion, steam reformation, ammonia synthesis, sulphur dioxide conversion and so on are fine examples of reactions, where the reaction is reversible exothermic and a catalytic. Therefore, we need to do a multi stage design to be able to give you an optimum size of the equipment at the same time achieve very high levels of conversion. Having said this, there are some features that we must appreciate. Let me sort of come back to this to illustrate what I want to say. What we want to recognize is that suppose we have a tubular reactor, let us say feed coming in going out and because it is exothermic A going to B as an example B going to A. So, this is exothermic. So, what happens as a result is that as it goes through, as it goes through you will find if you make a plot of temperature versus equipment and if there is wall cooling, let us say there is some cooling here. So, coming in and going out you will find the temperature goes through maximum and comes down. On other words there is a point called the hotspot. What is the hotspot? Hotspot is the position where the temperature of the catalyst becomes the highest. Generally, the reactor designs should take into account the maximum permissible, maximum permissible, permissible hotspot temperature. Now, there are several reasons for this. One could be that the catalyst is not meant to be used at a temperature higher than a certain specific temperature. And secondly, hotspot could be quite high and then there could be some bad effects on the material construction of the equipment itself. So, in both the cases we have to have a good idea of what the hotspot temperature is and how we can regulate this. Let us try and look at how to understand this by writing our equation once again. So, let me write the tubular reactor once again. We have our material balance which is F A naught times d x d v equal to r 1 minus of r 2. Then we have our energy balance which is d t d v equal to r 1 minus of r 2 in minus of delta h 1 star plus q minus of W s sorry plus q minus of W s. And q we have shown it can be replaced by this where it is T c minus of T where h is heat transfer coefficient T c is the temperature of the heating medium. And if it is a cooling medium this T c minus of T would be negative source it does not matter. So, if you look at this equation here and I ask you what happens at the hotspot temperature you will tell me that hotspot temperature this term is 0. On other words at the hotspot the rated heat generation is balance by the heat of heat removal. And consequently you find that you will see a temperature profile like this. On other words your tubular reactor temperature rises from T naught here T naught here to various temperatures. Now it might be useful to us if instead of allowing a temperature variation in the equipment if you can keep that temperature nearly constant. So, that we are able to ensure that hotspot problems I mean high excessively high temperatures are not encountered. At the same time we are able to ensure that the reaction occurs at the temperature of our choosing you know that is this is an advantage. The question that I would like to sort of put across to use that suppose we want to run our process at a constant temperature. This is a condition that we want to impose on our catalyst for the reasons being that we want to be absolutely sure that our catalyst is only exposed to temperature that we specify and nothing more than that we will we are willing to accept. So, if there are situations like this which you want to handle how do we handle this. So, this is an example I want to illustrate as an example. So, the example that I want to illustrate now is constant temperature operation of a tubular reactor. So, how do you operate a tubular reactor at constant temperature how can we do this and what does it tell us. So, tubular reactor catalytic I will say catalytic that is important catalytic tubular reactor. Of course, what is in my mind is reaction such as you know steam reformation shift conversion ammonia synthesis sulphur dioxide you know great variety of reactions in the process industry. Where you find that the temperature I mean that the catalyst is essential for the success of the process. So, we have here a reactor into which there is a catalyst and feed comes in and there is what is called as a jacket there is a jacket this is a jacket and through which the cooling fluid goes in and comes out. Now, because of the fact that there is a wall cooling we are seeing this temperature this kind of temperature I have pointed out to you T versus volume and we do not want this. How do we design and operate the process. So, that this variation in temperature is not an issue in our process this is what we want to do how do we do this. To be able to do this I just want to set up our equations once again and recall our interest in. So, we have a going to b as our reaction. So, we have this our material balance given by given by this and then this I call this r b as some k 1 c a 0 c a 0 1 minus of x. I will take for the moment a irreversible reaction we can relax it a little later. Now, let us say active to the catalyst alpha. So, this is our reaction is taking place. So, I will put it in the form of d x d tau equal to k 1 alpha c a 0 times 1 minus of x where alpha is catalyst activity. Now, this is the material balance now how does the energy balance look like our energy balance looks like this v c p d t d v equal to k alpha c a 0 1 minus of x minus of delta h 1 star plus q. I have just I mean now in our energy balance we have written a little earlier we have written earlier I have just replace the reaction rate with the rate function here. So, this is k 1 now what we said what we said let me say it once again or energy balance I just want to write this once again just to emphasize to emphasize d t d v equal to k 1 alpha c a 0 times 1 minus of x minus of delta h 1 star plus q of course, minus w s we are not going to derive work out of this and then d x d tau d x d tau equal to k 1 alpha c 0 times 1 minus of x. We want to run our process at constant temperature. On other words we do not want we do not want any variation of temperature in this equipment we do not want that which means what which means that we want this to be 0 correct. How do we get the left hand side to be 0 which means this term k 1 alpha c a 0 1 minus of x and this q let me write this in this form k 1 alpha 0 equal to c 1 minus of x times minus of delta h 1 star plus 4 h by d this we have done before I am just writing it once again this. So, this is our say t c is going in coming out this is t 0 and then temperature coming out is t. So, we do not want any variation between t 0 and t and the t 0 is chosen as per our understanding may be 200, 250, 300 whatever is the temperature at which we want the equipment to perform. So, t 0 whatever is the t 0 is the temperature at which it emerges and this temperature there is no change in temperature here only reaction occurs which means what this temperature this k 1 this temperature at which the reaction occurs this is the rate of heat release and the rate of heat being picked up by the cooling device. So, this rate of heat generation must be equal to rate of heat removal and this must be. So, at every point the point we are saying is that k 1 0 equal to k 1 temperature at any position alpha at any position c a 0 1 minus of x at any position minus of delta h 1 star plus 4 h by d t c minus of t this must hold at every point in the equipment this relationship must hold at every point in the equipment how do we do this this is the point that I am putting across to you because if we can do this the operation of tubular reactors that we will face in process industry constant temperature would have great advantage from the point of view of you know looking after the the the activity the catalyst and even running the process you know the the kind of benefits you will get in having constant temperature is a great advantage. So, we want to see by looking at this equation what is it that you and I can do to ensure that this equality is maintained at every point in the reaction equipment let us see how to do this. So, what we said we want we said we want this left hand side to be 0 what it means it means this is this is equal to q that means q must be equal to this term. So, let me write which means q equal to equal to k 1 alpha c h 0 1 minus of x times minus delta h 1 star. So, minus q equal to because q by definition is 4 h by d t c minus of t. So, minus q means t is greater than t c the solid means what is q now we also know we we also know that q is equal to which is heat generation heat removal per unit volume multiplied with the volume of the equipment must be equal to f a 0 minus delta h 1 star times x do we agree with this. What we are saying is that the total amount of heat release that heat is taken up by the cooling fluid the reason is why will this heat go anyway whatever is the heat that is released by this process where can it go because it is not increasing the temperature of the fluids because the fluids are at the temperature at which it is entering. So, what we are saying is that the reactor whatever is the temperature at which is entering it is leaving at that temperature only therefore, the only way this heat can escape is into the cooling fluid. So, this is the statement of that energy balance q times I put a minus sign here q times v equal to f a 0 times delta h 1 star times x. Let us understand whether this is positive minus delta h 1 star is positive for a exothermic reaction. So, right hand side is positive what is left hand side for x left hand side t is greater than t c for for a for the system where there is cooling therefore, this is also positive. So, it is quite consistent. So, what we what we get out of this is x equal to x equal to q v by f a 0 minus of delta h 1 star with a minus or it is also equal to q tau divided by c a 0 times minus delta h 1 star is that clear what we are saying. We know from our understanding of the physics of the process that whatever is the heat that is generated by the reaction is taken up by the cooling fluid. I put a negative sign here because it is first law convention q is formulated as heat put into the system here heat is removed from the system that is I put a negative sign. So, it says that the extent of reaction as by this definition is given by q v by f a 0 delta h r in this form. Now, we notice here we are going to substitute for x from this equation into this equation. So, I will call this equation as I will put a number just to I will put this as equation I will put this as equation 2 and therefore, this as equation equation 4 this is what I have in my notebook. So, you will be bear with me. So, I want to substitute for x from equation 4 into equation 2 that is all I am going to do now. When you do this you get q equal to k 1 alpha c a 0 1 minus x is what q tau divided by c a 0 times minus of delta h 1 star is this clear what are we doing it is minus q. I am substituting for x from 4 into 2. So, minus q equal to k 1 alpha I have just written exactly what I have written and minus of x since x is got minus sign I put a plus. So, this equation gives us alpha equal to minus of q by k 1 divided by alpha equal to q by k 1 divided by c a 0 within bracket. So, 1 plus q tau divided by c a 0 times minus of delta h 1 star. Let me just make small simplification just to make it look a little nice. So, I will just simplify it slightly. So, alpha equal to q by k 1 with a minus sign divided by I will take c a 0 common. So, it becomes c a 0 common. So, c a 0 plus q tau divided by q tau delta h 1 star with a minus sign. I have got it right and then there will be a minus delta h 1 star in the numerator is that correct. Please tell me I have got it right I have got it right. So, I have taken common. So, c a 0 cancels of minus delta h 1 star goes up. So, it is fine. So, what are we saying now? What we are saying is that what is alpha? Alpha is catalyst activity. Now, what can we do which means that we can change the activity of the catalyst by putting active catalyst along with some amount of inert. On other words by mixing the catalyst with some inert we can actually change the activity of the catalyst that we are going to put inside the reaction equipment. Now, what this equation is saying is that if you want to maintain the temperature of the reactor constant across the reactor if this temperature has to be constant everywhere then the catalyst activity that means the amount of catalyst that you will put here put here and put here this should be changed and that the program of change is going to be described by this equation. That means you must change the amount of catalyst that you are going to put into unit volume of the reaction equipment here and here and here and so on it should be described by this equation. On other words you have to exercise lot of care in filling your reactor with catalyst and if you fill that reactor with catalyst as per this equation then your reactor will perform at the temperature that you have chosen. Therefore, you have gotten rid of one of the most difficult problems of trying to operate a chemical reactor because this chemical reactor will run exactly at the temperature at which you have designed. Now, let us just put it in another slightly different form so that we can appreciate the usefulness of this equation alpha equal to minus of q by k 1 minus of delta h 1 star with a minus sign divided by minus of delta h 1 star times C a 0 plus q we can tau as q divided by v 0. So, I can put a v 0 here I can put a v 0 here. So, on other words what we are saying is that v 0 times minus of q by k 1 minus of delta h 1 star divided by v 0 minus of delta h 1 star C a 0 plus q q v equal to alpha. So, what we are saying now is that if you make a plot of alpha versus v as v increases as v increases alpha decreases please note the q is negative. Therefore, as v increases the denominator keeps on decreasing and therefore, you will find that you start here and then therefore, alpha keeps on increasing. Therefore, you are actually putting a higher and higher amount of catalyst at later positions in the equipment and this is because of this program of catalyst laying inside the reactor you are able to maintain constant here temperature. So, it is it is this profile or of catalyst loading inside the reaction equipment that ensures that you will get uniform temperature throughout the equipment. Let us ask one more question we have done our design for at a given temperature T equal to 220 as C a 0 equal to some value. Now, for some reason or v 0 equal to some value now in a process C a 0 might change v 0 might change. So, many things may happen in a process how do we manage to run a process at constant temperature despite variations in all these kinds of process. Now, we can understand this and we can also address these things quite easily now by looking at the equations we have written. Let us say let us take an example to illustrate what I am saying what we are saying now we are saying that C a 0 has changed C a 0 has increased or decreased let us say it is C a 0 has changed. That means, if you want to maintain if you want to maintain that this heat release this heat release this must be equal to Q. How do we ensure this equality how do we ensure this equality this Q is given by this quantity 4 H by D T C. So, on other words we can adjust T C to get our temperature T or alternatively if you do not adjust T your temperature will automatically will be reset instead of operating at 220 you might operate at 215 or 225. In both cases you will still be operating at a constant temperature that is the point I am trying to put across to you that this design gives you an opportunity to operate your process at a constant temperature. But you do not want to run at a temperature which is more than the temperature that you prefer. On other words if your design is only 220 C and you do not want it to be more than 220 clearly you do not want a higher value a different value of C a 0 if it is going to give you a higher temperature you do not want it in which case you will have to reset your T C. So, that this does not take place. On other words here is an instance of a design which gives you a lot of control over managing the quality of the catalyst. So, that the catalyst life is preserved catalyst performance remains as you have designed. And your process control becomes much much easier than what you would anticipate. So, just to cut this long story short what I am trying to say in tubular reactor design is that this approach of trying to maintain constant might be of great value particularly with expensive catalyst that we often encounter in the process industry. Thank you.