 I think we will start now. This is the diagram I have also given in the last class. We have seen in mixed flow reactor how easily one can find out the relationship between temperature and conversion, okay. There is no trial and error involved. There is no trial and error involved. Why there is no trial and error involved in mixed flow reactor? What happened to third variable? Because of the space, you know, it is a lumped parameter system, right? Because of it is a lumping parameter system, that is why that changes in the space is they have disappeared. So that is why it became only two parameter problem, two variables, that is temperature and conversion. So normally you know conversion, you can calculate temperature. From the temperature and conversion, you can calculate volume of the reactor. But there is some other problem there when you just observed the equations what we have written earlier for, for q g and q r. What are the equations we have written for q g? Tell me minus delta h r, not that, final expression, minus delta h r, C A naught, V, yeah, whole thing divided by 1 plus, I think in terms of E power and all that we have written. E power E by RT by K naught plus, that is the equation, right? Volume must be there, volume is here, right, right. So volumetric flow rate has come into V, so tau, so that will not be there. That equation I do not have, that is why I am asking you, because I have slightly complicated equation. This is the simple equation. Okay, this is capital V, right? Of course, here again you have volume by volumetric flow rate, so both are there. So now, this is the one q g, q g, what was the equation number? 14, 14, 14, 14, okay. So heat removal line was, this is same what we have, yeah, F A naught, C P or otherwise, yeah, you tell me what is the equation I have given there? That also I have slightly different. Heat removal line, ah, summation of F I C P I plus U A, that is bracket there, into T, temperature minus, ah, yeah, sigma F I C P I T naught plus U A T C, so this is the equation, bracket, right, yeah, so this is equation 15. So under operation these two q g must be equal to q r, because for steady state, right, we are talking about only steady state mixture flow reactor. So this is what we have plotted. We separately plotted equation 14, here this is q g and you have q r lines because this is a straight line, this is y equal to m x plus type, this is y, this is m, all this is x, okay. So that is why you will have also this negative, you know this intercept, okay, minus and you have U A F I C P I and your slope depends on this, okay, good. Now let us look at these points, like where you have these three for this and one here and one here. Let us start with this is lowest one, where we have, yeah, at this point. At this point if I just look at, right and by chance let us say this is operating at 20 degree centigrade, it is operating at low temperatures because temperature is increasing in this. Let us say we have 20 degree centigrade and let us say because of the flow fluctuation because, you know, you could have seen either volumetric flow rate when you are measuring, if you are using, okay, what are the instruments you use for measuring flow rates? Rotometer. Rotometer. Rotometer. Venturometer. Venturometer. Surface. Surface meter. Yeah, what do you measure in venturometer and rotometer? Pressure. Pressure. Pressure. I do not know whether you looked at those pressures, you know, the manometer particularly when you are using, always that will be fluctuating a little bit, okay, slightly like this, like this, yeah. If that fluctuation causes may be one degree in this direction positive, that means 20 it is operating but it is going to 21. Our question is when you have the 21, that means there is a disturbance from the temperature, okay. This is the point what you get if I solve those two equations, okay, under this conditions, right. So under these conditions if I carefully look at that if there is one degree temperature this way, then the temperature is increasing, right. So naturally what should happen to rate of reaction? It should increase. When rate of reaction is increasing what should happen to heat release? It is exothermic, yeah. Also it is exothermic, okay. So temperature should increase. So once by the by if you do not have exothermic you will never get this shape, right, yeah. And once the temperature is increasing then is the temperature goes on increasing or will it subside? Again come back to this. Why it comes back? Temperature is increasing that means q g line is increasing, okay. Point in the q g line is increasing. So you have at 20 degrees, 21 degrees, temperature increase, rate of reaction increases, heat release increases and it will try to go on this line. But what is happening? This heat removal, the moment it crosses a little bit what is heat removal? Is it is less than q g or less than q r? Or greater than q g? Yeah. So then what should happen now when it is heat removal is more than heat generated? It should come down to because there is only one state. So at that point, so you should come back to that particular point e. Similarly even here it is falling a little bit the temperature by some reason because if volumetric flow rate is suddenly increased also it may fall, right? That means more heat is now taken by the convection, yeah. No, T C also suddenly increases, it may fall. So but even then heat removal line will be, yeah, here heat generation will be less and heat removal is, you know, at this point when you come back heat removal will be less and heat generation will be more. If I draw a line, I mean I am not able to draw that line, I will explain that to you, okay, there also. So then again it starts heating and then goes only to this point. So whether I move in this line, direction or here it is small, in this direction the stable point is only this one, right? So when it is reducing, like for example when the temperature is falling, falling down, okay, the temperature should, the rate of reaction must be less. So heat generation must be less, okay. So then, but when I move this way here, you see this point I have less than, this q g is less, q g at this point below, okay, q g is less. So when q g, sorry, this is q r, q r is less. When q r is less then what should happen again? Okay, it came to 19, right, but heat removal is less. So then what should happen? Temperature should increase. So that means again it will move to this direction, yeah. So that is why even if there are slight disturbances around this point, absolutely no problem, okay. Now let us do, let us talk about this particular case. I think here you can easily see that, yeah, same argument here, so when I am moving, let us say it is 100 degree centigrade and it move to 105, okay, more change. So when 105 it is going, then the rate of reaction should increase, right, heat release must increase, temperature must increase, so it will definitely move on this direction. But now, if I draw a line here, then heat removal is more than heat generated. So what will happen again? It will come back to this state. Similarly here, it move till this point, 95, right, heat generated is more, heat removal is less. So temperature should increase. So again it will move to this point. So that is why even this point also will be steady. It may be fluctuating around this but it will be only on that point. I do not know whether you have discussed this in your process control, okay, huh? Yeah, mathematical methods may be, yeah, you would have, for steady states. This is one of the very good examples for, you know, in chemical engineering steady states, okay, the reactor steady states and, okay, so now this we understood it is steady state and this we understood it is steady state. Now when I have this kind of situation, so we have three points here. Now just try to find out what is happening at C. See also same argument because when the temperature is moving this way, okay, yeah, slightly when it is, temperature is moving this way, so heat generation will be more but heat removal is more than heat generated, much more than heat generated. So again it will come back. That means there is no difference between A and C. Similarly you can also discuss here, when I am moving here, okay, when I draw this one slightly further, yeah, when I am moving this side heat removal is more and heat generated is more, even though there is a change in temperature and the temperature is trying to increase and heat generation will increase and, you know, the temperature anyway try to go on this way Q G will increase but at this point again Q R is greater than Q G. So that means again it has to come back. Same thing here again, when it comes back here then heat removal is less and heat generated is more, so it will move on this line and then stabilize there. So A C E B or A B and C E there is no problem when the reactor is operating at those temperatures, okay. Now without knowing you are at this point. Now the argument must be very easy, right. So let us say this is around 75 degree centigrade. It went to 76, this way, right, this way. So now the moment it went this way, so that means I am somewhere here, okay, I am somewhere here. So then what is the heat generated line this one? What is heat removed this one? So now what should happen to the temperature? It will again increase more, that means more rate of reaction, more heat release, more temperature increase. So you will go here because always here Q R is less, Q G is more, till it reaches this particular point. Similarly here it is same thing, the moment you move this side heat removal is more, heat generated is less. So more and more heat is removed, heat generated is not that much because you know this is the point and this is the point, this is the point and this is the point. So that means it becomes more and more cooling, temperature will fall again but there is a next steady state that is here, let us see, right. So that is why this particular point D is called unstable point or metastable point. So D is metastable or, okay, A, B, C, E stable point and D is unstable. In fact the first chapter of, you know there is a book process control by Stephanopoulos, yeah, to justify why chemical engineers must read process control, he gives this example, okay. Why should we read process control, right? So in the normal way when, even if you do not have control here, no problem for me. Why? Because if there is a disturbance it goes away but again comes back because of the conditions here. So why do I need a control? I do not need a control, right? So but here definitely I need a control, right? So why I need a control? Let us say my, the reaction temperature what I calculated was 75 and 75 is this one but I do not know because of this 14, 15 non-linearity you will get, when I solve this I will get this three points here that means three solutions you get for that mathematically speaking. So that is what is shown here whereas under these conditions that depends on what kind of slope you have, what kind of flow rates you have, what kind of CA naught and you know, yeah, what else, yeah, delta H R all these things will come into picture. So that is why theoretically speaking I can draw Q G by varying all those variables like for example tau changing, Q G alone, forget about the other one we will come back because just as a mathematical exercise how many parameters I have there, temperature is anyway, I am plotting here, Q G is here, the other parameter K naught is constant, E is constant and for a given system delta H R also is a constant then it will change with volumetric flow rate, right? Of course normally the stability and all that will come once you have the reactor, right? But even when you are designing a new reactor the procedure is this, you have calculated using the method what we have discussed in the last class, what was the method, write the energy balance, write the energy balance and energy balance is very simple because there is no trial and error involved for mixed flow reactor. So then calculate temperature for given conversion, take this temperature and conversion to design equation, okay. Design equation is V by F naught equal to X A by minus r A, minus r A has this T and X, right? So then calculate V. So now you know, you also know the, for that particular temperature what is T, that is the first step what you have done, right? From energy balance, T you have calculated for a given X. Now the question comes whether that temperature is stable or not, that is why this analysis is required, right? So how we have to do is, people who really like mathematics they simply go for these equations and try to find out by trial and error, okay or using some method to find out whether there is only one solution or more number of solutions, okay and this is, okay, I also have to write it is simple reversible, sorry, irreversible exothermic reaction, okay for this case. Again the moment I go to irreversible, sorry this is the irreversible, reversible and also if I go to A going to R, R going to S, right? So exothermic, all the cases will be different, good. So now the criteria given for necessary condition, okay by the by, I think you know this point A will happen point A and point A no one would like to operate, why? Conversion is very very small, okay corresponding to this, because temperature is very very small, right? So conversion also will be very very small. So that is the reason why normally we do not want to operate but here we would like to operate. So for that reason sometimes you have to move the T not, correct no T not values. So that is why by taking this equation T not is here, for different T nots and for fixing you know normally we do not change T C, once you fix T C and once you fix the heat exchanger this is constant, right and for a given flow rate and all that this is constant only T you can vary and then draw various lines, all will be parallel, why parallel? Because slope is not changing but T not depending on that, that is also important for us because all this is happening because of T not also, right? So I may have, if I fix T not like this one line, another line like this, another line like this, okay this also possible, another line like that, all that I have to do. So now when I am doing all this, the slope is falling, correct no, yeah and also yeah the operating temperature will be increasing because when I put here and then draw this line, I am talking about if I fix this, right and also it is definitely better to avoid this point because if I fix this point I have to draw a line like this otherwise it will touch here, correct no, it should not touch here, right, this line, Q G line, the moment it touches then you have a solution there. So to avoid that you draw the number of lines with T zero as parameter and then you will have an idea of where you have to fix that temperature, T not, okay, good, fine. So now, yeah the conditions for A, conditions for A are 1, low K value, K means reaction rate constant or 2, it can be low tau, low tau and it also can be small delta H R and also it can be small V into C A not, V into C A not is nothing but F A not, okay. So it also can happen with low T C, okay or it can also happen large, large U, large U A or separately U or A also, correct no, all these are common sense now once you understand that, why? Because low K will give me low conversion, low tau also will give me, okay, why? Why low T gives you low conversion? Because time is not provided sufficient time, then small delta H R because heat release itself is not much, so the temperature is very, very low because heat release is not that much temperature so again low conversions and small V into C A not, F A not, okay, F A not is small again, again that is low, then you will have low T C that means I am removing too much heat, T C low means, so naturally you will have low temperatures, so low conversions and large U A, large U A again will remove lot of heat, okay and exactly opposite points are opposite of A is what? Point E, exactly opposite, then only you will get that kind of thing, that means if you know that your values, delta H R values are large, your K value is large and you are not using that much U A, so then you may expect this condition, okay, good. So in between, yeah, unfortunately sometimes you have to fix only this condition because even though this gives me more conversion, if I draw a line here, temperature may be let us say 175 and that 175 will spoil my product, okay, so that means there may be side reactions or that may be, it may be charred because it depends on again you know what kind of products you are dealing with, so the physical properties of the reactants and products, so that will also once, yeah, ignition point, you know, there is a flash point where it may catch fire beyond certain point, so all this will happen here and I do not know whether you have observed this, how many of you used Bunsen burner, only used burner, not for cooking or doing experiments I am dealing with, yeah, Bunsen burner, only 3, yeah, what raised me you have never used Bunsen burner, used it, but the question is so stupid, so that is why you do not want to say anything, yeah, because you have used but did you observe something peculiar there, can you stabilize the way, when do you stabilize the way, low flow rate is this, you cannot, because there must be a balance between the amount of, you know, the gas coming and also the temperature that is able to sustain, there must be a balance. If you also use too much gas, you know, that also not sustainable, if you use too less gas then that is also not sustainable, okay, that means you are somewhere here, right, but if you correctly use the amount of gas and also, you know, there must be oxygen supply also, that is why you have, in the bottom there are 2 holes, 2 or 3 holes I think, okay, yeah, did you observe, you know, Bunsen at the bottom, that is only for oxygen to enter, so that sucking if it is not that much, then it cannot stabilize, so that means there is, there must be a balance between amount of oxygen, amount of fuel and then it will sustain for long time, otherwise you send too much oxygen, too low fuel, it won't, twice versa, too much oxygen but very, I am sorry, too much fuel and less oxygen again it won't sustain, right, and how many of you, it is a bad question, smoke, none of you, all good boys and girls are, okay, good, yeah, now a days girls also happy smoking, you have seen TV's and all that, and I think in Hyderabad there are many pubs it seems where you go and ask why are you smoking means what is wrong if I smoke, nothing is wrong, nothing will happen, only your lung will disappear after some time, that is all, I think, okay, yeah, anyway, any smoke of any kind for lung is very bad, lung does not like smoke, it only likes oxygen, okay, good, anyway, so when you are using matchstick, matchstick, right, when you hit like this, particularly rainy season and all that, will it catch fire, okay, when it is not catching fire you are under what condition here on this line, the lowest condition, that means here practically you don't have any reaction, okay, and then you may again strike and then wait for some time without even shaking and all that, then it catches fire, and also that phosphorus which is at the tip of this matchstick, it will also make some peculiar sounds like that, okay, so when you have that sound can take you either to this point or this point, right, I think, you know, you don't know, this is an experiment what you can do, you may not get in your house, okay, so you can buy a matchstick and then go on experimenting on this, whether it is really, you know, this point, right, sometimes when you strike you will get spark, but it will not catch fire, that means temporarily you are at this point and nowadays, I think not nowadays, in this season all of you are going to destroy your environment because of diwali, correct, no, diwali, you burn lot of crackers, I feel, I think practically you are burning money, that's all, nothing else, I think it is waste, as chemical engineers no one should do that, why? Pollution, bain karam pollution, I think in fact I think you should not, yeah, and you also find out, you know, the moment you, if you are standing there, you know, so many people may get wheezing and all that, right, so there also, you know, normally we have that, yeah, that crackers, red ones and Lakshmi is very, very Lakshmi cracker is very, very sound, lot of sound it gives and there are other things also, they tie with something and then also, what do you call that, that is called atom bomb, atom bomb without atoms there, okay, okay, so that one also will give you, so then what we do, we will put that here and then go to, like this, you take that very big stick, you know, with that fire right there and then slowly put, but I think, yeah, I mean you are afraid that suddenly it may blow up, so that is why you just touch and then come, touch and then come, so that is why it may, stop, stop, stop, I like that it may go, because that fuse is there outside, right, till, that means there must be a sustainable fire in the fuse itself, then the fuse goes and then opens up, okay and why you should give lot of sound, yeah, what is, what is the simple formula they use there in making crackers, there is a equation, beautiful equation that is involved here, huh, yeah, what, I am just asking only equation, no explanation, yeah, what equation? E equal to mc, yeah, he said only E equal to mc square, you are telling E equal to mc square plus xA plus xB plus xC, yeah, see how sad it is, you know, that means we are studying in the, this is high school question, okay, simply it is P v equal to R T, that is all, P v equal to R T, nothing else and we, I think till now 25 years we have burnt so many crackers without knowing what is the formula behind that, okay, anyway at the age of 2, 3, 4, 5 you may not know, but once you go to school you should know that equation P v equal to R T, why it is P v equal to R T, volume is kept constant, okay, that is not moving, yeah and temperature is increased, right, so P v equal to R T, volume constant, R is constant, temperature is increased because of explosion, then what will happen to pressure, pressure should increase, but you are having that, you know, papers or some kind of threads and all that around that, but you know if you have too much threading and where it is not at all able to crack that, then you do not get any sound, really you do not get any sound, so that means that pressure it is able to contain, like you know ammonia reactor, ammonia reactor thickness of the reactor is, I think you know it is 4 inches, 5 inches still, right, so it will not explode, if it explodes that is I think 1,000 times Lakshmi cracker, okay because 500 atmospheres, correct, 350 atmospheres or 250 atmospheres, okay, it will be very high, so that is the reason why that P v equal to R T, that is why in Lakshmi cracker you will have slightly tougher more number of papers, that smallest one is the red one, okay, so that would not give that much sound, so some people I think people like merit and all that they will have hold it in hand and then crack, not afraid at all, he is bold guy, so that is the reason why you know no problem at all, there are some more things you know that red small things, caps are something they call, that red very small one I think maybe 5 mm, 6 mm diameter and inside you have small thing and you take one stone and then heat it, okay, so I am training, I am giving training, I am giving training for you for Devali, okay, what are the fundas, so you put that and then take a stone and then heat it as if you know it is a great thing and no one will hear except you, so sound is so small because explosion is so small but still it gives sound because of the pressure you know which you are creating, what is Swami, cracking something there, cracking jokes on it, gun, gun, gun, gun, gun, yeah yeah, but gun is also mechanical, it is not purely chemical, gun is mechanical right, yeah, when bullet is leaving, what is that, that is okay, yeah, it is not the right question also because as you are applying the i thought he is talking about okay good so all these things are happening so that is why you are getting more and more sound now someone ask why let me use more sound than some other small one small one is called what not the round one that same cylinder type and then small one cracker only so Benzels don't know sound see all our diwali thing see either sound or light you will have pencils and you will have sudarshan chakras so all kinds of but poor fellows they do not make nice they will only make light so that is what i think you know even this particular curve is also responsible for that m-i-c disaster no m-i-c disaster in Bhopal civil engineers can be really dangerous they can kill many people civil engineer if he kills i think only building falls so all of us how many 60 people may not be there in the world if it falls or may be some more people down so now it deans room and all that just below this only deans room is there but if gas leaks and then you cannot stop bridge wherever it goes it kills you know what has happened there at that time there was that m-i-c stored and it is a catalytic reaction it seems in the presence of water the reaction is much faster water by mistake or so water entered so reaction started and it is highly exothermic reaction so the moment water small amount water has gone and reaction started then temperature is increased so that temperature again heat released exothermic reaction temperature increased that temperature increased again rate of reaction so that rate of reaction again created more temperature it goes on building it because there was lot of m-i-c stored but unfortunately if that kind of things happen they should have a safety wall that safety wall did not open but it is a concrete tank but the pressure inside is so much like this crackers okay if that tank is very very strong it could have not happened you know if that concrete tank if it is very very strong it could have not happened that is why if you go to this nuclear power plants that thickness of the walls where 3 meters, 4 meters, 5 meters depending on the amount of fuel you put there thick okay and that much is required because I think those everyday activity should not diffuse again through concrete I think that is why civil engineers also study diffusion of concrete through porous media like you know this concrete and those people who are doing this construction or that structural engineering concrete and all that they also study that diffusion because that should not come out right so all these things are just because of this you know exothermic reaction and temperature going on increasing so that is why endothermic people are very happy that is why most of us I think you know if there is an active fellow only problem if inactive fellow there is no problem okay he is anyway inactive he may not even laugh also because he may not have even patience to laugh that kind of endothermic reaction so that is why I think anyway this is also very good so all these things may happen that is why we are getting this particular state either go here and then go here and when I told you that when this unstable state is trying to go here then if this is explosion point if this point is explosion point then the entire reactor will explode okay so good so this is fine so then mathematically speaking the condition here why it is stable or here or here okay or here why it is stable that is given is as a q slope of du d t must be greater than this is the slope criteria this is sufficient condition no sorry necessary condition necessary condition for stability okay so that means if you take these two equations and then you can differentiate okay you can differentiate this and once you differentiate if you know all these values you will be knowing definitely you can also differentiate with respect to t and delta h are all these values you can substitute and then calculate which slope is more right so depending on that value you can find out whether the system is stable or not but that gives you only one condition that is necessary condition but the sufficient condition will be the sufficient condition is doing in terms of process control what do they do for process control they write unsteady state equations because i told you know all process control people are unsteady state they do not have any business in steady state okay only unsteady state so once you have that unsteady state equations and then you have to expand those terms you know the tile spirit and take like the way you do and take the first few terms and then write the equation you get one condition where change in temperature delta t will be some constant into m 1 t plus b into this you should have got many times i am not doing all that m 2 t when you say that the condition will be you know stable the point will be stable when m 1 and m 2 must be real values are negative right so that is the condition so now you have to do this and you have to also write the material and energy balances and then write the equation you know in terms of the tile is expansion and all that and then bring it to this level and then try to examine what is m 1 value what is m 2 value if you do that this m 1 and m 2 contains all these parameters this m 1 m 2 contains all these parameters like you know delta h r all these things will be there tau v okay all these parameters will be there so then m 1 and m 2 are negative real parts or negative real parts if then system is stable this is very good condition because you have to do this this is one condition necessary condition this is sufficient condition both one has to check whether it is really yeah stable or not okay good excellent i think this is nice this is very nice interesting thing but i tell you give the same thing in the examination you will have help because you do not even know how to plot and if you are going by trial and error okay thinking that okay you know i like mathematics and then i will do it but in the examination call you cannot do because it is exponential you have to find out three values and you do not know first of all there are three values there may be two sometimes depending on the shape of the curve right so that is why by trial and error if you do do not blame me because you may take three hours for one problem okay yeah for one problem you may take three hours by trial and error but if you draw the graph correctly you may get quickly you may get provided you also know how to calculate correctly q g q g is the main problem q r is not that problem because straight line fellow all curves are dangerous only straight lines very happy no problem at all right so that is what as far as examination is concerned so this is what is the condition in fact there is lot of information available on this you know you should have heard of limit cycles okay and also unsteady state you know when process control people analyze this system they start this is unsteady state at time t equal to 0 they start and till what time it comes to this stable point if that is the point that is there so if this is the point wherever you start so there are lines like this goes goes goes finally goes to this point and you may start this side this side means low temperatures high temperatures different flow rates all that okay so that phase diagrams are beautifully given wherever you start how this comes to finally to your stable point and there are also you know things called limit cycles I have not gone through all that because I think my interest is just to expose you to this and then move to residence time distribution because we are losing time now right yeah so that is why I just want to point out all these things to you those who are more interested because that goes into process control I mean I told you if you would have started with definitions already known very well that means 1st 2nd 3rd 4th 5th chapters of Lounspale is completed I should have started with 6th or 7th chapter then we should have done all this I also should have enjoyed that but I think now it is more fundage and also exposure to as much as possible okay that real things that are necessary so that is the reason why I am not able to go to that kind of analysis in terms of unsteady state otherwise it is not only the business of process control people even though I am telling many times but it is also a reactor design engineer duty to find out whether the system is stable or not and if you use this only one condition that may not be sufficient okay so that may be necessary but I think that condition again may not guarantee stability okay yeah so that is the reason why what is that yeah I mean see when you are starting this in the beginning itself no problem but as a chemical engineer I mean sorry as a reactor engineer what do I do I design this steady state reactor and then I go for unsteady state okay to check whether only for in this case right so that is the reason why I am telling this one first because my idea is first to calculate what is steady state temperature now go to the analysis so at that time if I say that yes I have one more point yes this is necessary I mean necessary condition this is stable right and this is problem for me because I mean I can do the second step this one to find out whether guaranteed stability or not okay good and for endothermic reaction this kind of things will not happen why endothermic reaction this kind of thing will not come into picture at all mathematically also you can tell what will be the value of dqg by dt for endothermic system can you say zero negative okay and what will be the slope dqg by dt is nothing but slope no will it be any time negative extend the brain a little bit I say expand the brain why ua cannot be negative and fi cpi cannot be negative I think you know that kind of expansion must be there in the brain I say only simply nodding hands and I mean heads and all that I think you know that expansion when does that come fire okay immediately before the teacher tells that you should have told the answer at this point of time you know it is very easy I told you know can it be negative and you have the straight there and that is why the physics unfortunately unfortunately is missing whenever in all the subjects and you are interested only in of course in examinations that is all may have you may be thinking that whether this fellow gives problem in the examination or not if it is not in the examination I will switch off the brain if it is as if there is lot of things in the brain so switching off because I think too much loading and all that is a problem right I think this can take you know this is only we have now 1 terabyte GB right as the hard disk only 1 terabyte increased more than 1 terabyte also available but this one has infinite terabyte but if you have willingness to accept willingness to learn then only it has terabyte unlimited terabyte but if you are restricting to only examinations all the time that will be only 256 MB that is why it will never grow more than that 256 is half of that also came in the beginning I think only 256 I saw the minimum one 64 umbers also was there so that is all so that is why I think may be at PHD and MS MTech may be that goes to 256 a little bit more so that is why don't imagine all these examinations and all that okay good so this is the one what we have about this one and so that is why endothermic reaction is not a problem another question I ask is why I have not talked anything about plug flow but why I am talking about only mixed flow is it possible to have multiple study states even in plug flow no expansion only 256 maybe 64 mb or 32 mb or 16 mb so what will happen why not so what happens okay it is varying continuously heat removal is not there heat removal is there you put outside jacket anybody interested in control here or all may be interested in control those who are taking I think PHD why the fact that we have not discussed that you know there must not be any problem I am asking why what area is yours real logical so you are not only reality she always flows rest me control yeah rest me control I saw only PHD control or something but now this has been rest me control shall I give one word it depends on the feedback now I can tell rest me measure temperature I mean any number of thermocouples I can put beautifully I can measure because here whatever temperature fluctuations are there that is going as feedback to the reactor and then slowly it is increasing but what will happen in a continuous flow along the length of the reactor if there is a temperature fluctuation what will happen to that temperature fluctuation you told like this so I thought you understood just wash it out so in it that means there is no feedback so there is no point for the accumulation of the heat there but here the heat that is which is about to go out will be brought back and then mix with the feed that is why it is called back mixing reactor so that is why this happens inside and temperature multiple study states is not a problem with normal plug flow but we have another problem there what is called parametric sensitivity in terms of plug flow and also batch reactor batch reactor also same problem in time forward it moves okay but I think that is what Ramakrishna has done I think none of you solved that problem did you solve the batch reactor problem which I gave you no no not simple adiabatic batch non-isothermal adiabatic batch nena reactor batch reactor none of you solved I am not asking about quiz I am talking about batch reactor which I gave you in the class for adiabatic system but did you do this by trial and error no no UA that is what I am telling adiabatic only you solved some of you but you do not even know that there is a problem for non adiabatic which I gave you I think it is 0 mbis so not even 64 absolutely no storage space there so then he has done it and he found that at one point the temperature will shoot up all the time you know normally what we expect when you have non-isothermal it has to go to maximum and then fall but that will not happen in mixed flow in batch reactor and mixed flow because you will get that point only under one UA condition UA heat removal because that is the parameter I have given you to change right but I also gave you in fact how much is 2 4 or something like that yeah 111 UA so then only you could get increasing and decreasing that will be stable points I mean not stable so that will be maximum and then fall but here a small change in UA will be shooting up you know that is called parametric sensitivity that means by chance there is a small temperature change okay along the line of parametric sensitivity it goes up and shoots up okay so I am not able to do that because again you know my time in other portions I have to do that I just want to mention first of all in plug flow reactor you will not get multiple steady states under normal conditions as you are going to r and all that okay yeah so then I think again I do not know about what happens in autocatalytic in plug flow okay yeah so but this one for normal reactions plug flow will not have multiple steady states but you have another problem called parametric sensitivity that is the problem what you will have there so that means there will be uncontrollable parameters after some conditions okay at some conditions that also one has to find out okay that is only the problem there and very beautiful information is available by Amundsen you know Amundsen is another grandfather of ours mathematical methods in chemical engineering there are books you know three books by Aries and Amundsen okay so I think Amundsen was the actually this multiple steady states first found out by Fan Hadin I think I do not know I will just put that Fan Hadin he is from Netherlands Netherlands is also a beautiful place for for chemical engineering in fact chemical reaction engineering name came from Netherlands I told you know you would have forgotten again 0 MB okay in the first few classes I told that and who was the person also I gave name the person who told chemical reaction engineering how beautifully he coined that name okay yeah I think I have to stop here but I think still I am not able to complete this also so I have to tell one more one or two things about this like for example adiabatic case what will happen UA is 0 this UA will be 0 then you will have only these two so then this must be equal to that equation equal to QR adiabatic case this is 0 this is 0 must be equal to this multiple steady states why no yes why yes because the line will still be there not only line will be there but QG will be there as usual QG a shaped curve will be anyway will be there okay so that is why only this slope will be different now because UA term is not there that intercept also TC and all that is not there intercept will be there but is only this term only this term T0 okay FICPA T0 so that is why even for adiabatic case also you will get multiple steady states okay good I think we will stop here because you have to go to class now