 Okay I think we will start now our non-isothermal reactor design graphical design only one part is left I hope you have understood okay or did you enjoy doing that it is so beautiful I say very simple very simple way of doing things but it is not easy for you to do on your own because I think I have drawn the lines just like that but you have to calculate those lines and then you have to draw and this activation energy and also you know this delta h r s will give you help particularly the units because you have activation energy coming there in k values and k by mistake if you not able to put even proper dimensions you will never get those lines so that is why you have to be very careful in drawing those lines anyway today I will give you also one exercise so that you know you will practice how to draw those lines and we will draw I will give the exercise only for irreversible reaction because that is slightly complicated no reversible reaction reversible reaction which is complicated but then I think you know when you do complicated things non uncomplicated things are very simple so that is the reason okay good so the last part that is left in graphical design is is there an optimal design for this you know using graphs okay analytically also we have the same question is there an optimal design okay what do you mean by optimal design what do you mean by optimal design ya okay that minimum volume that is possible for a given conversion or maximum conversion that is that is possible for a given volume okay and if it is multiple reaction what is the optimal problem there this is straight forward single reaction when I have multiple reaction ya it is not the volume there it is the yield there okay yield or selectivity I mean both are same the way you define okay ya so there the question is but you do not worry about whether the reactor is smaller or larger there okay so that is why you always question there that okay how much yield I can get because that particular product may be having very very high cost so that is why even if you forget about the total volume of the reactor by producing more and more of that particular compound in terms of yield okay then you have to I mean then you may your plant may get lot of profit so that is the reason why that question is different there right but still you have to also find out what is the volume of the reactor for that equivalent yield that equivalent yield given by what conversion and use that conversion to again calculate the volume using the same procedures what we have done so here now optimal design what we discuss is that if it is irreversible exo or endo what is the optimization here so I have here X A versus T graphs so now for irreversible you know what kind of graphs you get what kind of graphs you get like this you get good yeah now how do I use this information for getting minimum volume for the given conversion what is the strategy you have already some some information in your brain okay so when do you get the minimum volume excellent okay so when the rate is maximum where do you get rate maximum here at the high temperatures at the highest temperature that is possible because as the temperature is increasing here the rate is increasing this is R okay so that means I can go on drawing till that end you know the line and then draw draw draw draw that the temperature but is there a limit highest temperature you know the first question what I asked you was what do you mean by the minimum volume here when do you get so if you remember the design expressions and you remembered and you told that when you get the highest rate then you will get the minimum volume and the highest rate will be given by here highest temperature because temperature is involved here okay yeah so and what is that highest temperature that is possible another question here so the highest temperature means is it you told that highest rate do you get this rate at one temperature or at various temperatures in this case we are talking irreversible reaction so that means what for irreversible exothermic or endothermic what should be the now strategy very good I am very happy that you are able to answer all correctly exothermic or endothermic what is the strategy I mean you know the temperature you said corresponding to a particular temperature you will get the highest rate okay so now what is that temperature and what kind of temperature scheme I should have in the reactor is it one temperature so that means what do you call that if you throw out the reactor if you have one temperature that is one you should have isothermal temperature highest that is possible this is T max okay how do I decide now the T max yeah you come at the end but before that taking care how do I choose that T max you have ideas I say already you know I told you also just think conversion is fixed for a given conversion how do I choose that increase will be different that's why we told that T max after certain temperature after certain temperature you don't have a choice because after certain temperature the reactor may decompose rate won't be that much increasing oh no here as temperature again you are coming back to the other wrong answer because we already decided that as temperature increases rate will be the rate will be increasing continuously again you cannot say rate decrease after some time no so no telling rate is there rate of yeah how do you decide that operating temperature Abhishek already given the answer yeah highest temperature reactor should be stable that means what is the factor that comes materials of construction that is one other one products are the the reacting should be a stable yeah at high temperature they may be side reactions which you don't like okay even though there is a multiple reaction sorry single reaction so at one point of time you don't know you know there may be a separate another reaction coming that not possible you should not allow that then any other thing that is what he said I think you know that is the materials of construction beyond which the you know the material should withstand that kind of temperature okay and also another thing the product formation and also charring of the products you know some of the products may get burned and everything you get only carbon okay at the end so that is why we don't want carbon including reactor if you go beyond certain values okay so that is why T max is decided by now yeah materials of construction yeah that is one and also side reactions and number three charring of products yeah this is true even for heterogeneous reactions the similar things only what same rules can be extended but if it is catalytic reaction what is the fourth one catalyst should withstand I think most of the time catalyst will control material of construction and all that there is no problem okay catalyst will control because catalyst should not be thermally deactivated so catalyst deactivation okay so all these things will come for this irreversible so it is very easy now for us to remember that okay graphical design will tell me that I have to go for highest temperature because the rates are increasing as I increase the temperature and now the scheme should be isothermal that means throughout the reactor I should have the same temperature now the question is how do I maintain that that particular temperature because throughout I should maintain the same highest possible temperature how do I do that if it is exothermic reaction or endothermic reaction okay yeah I think before answering those questions please take these yeah I think you know the optimal design optimal design means you know minimum volume or maximum conversion depending on what is the parameter which we are going to use optimal design depends on the optimum temperature progression which minimizes V by F A naught for a given conversion the optimum may be an isothermal or it may be a change in temperature in time for batch reactor, along the length in P F R and stage to stage for a series of M F R okay next part you can write it is important to know what this progression is because it is the ideal which we try to approach with a real system it also allows us to estimate how far any real system departs from the ideal okay good next part please write the condition for optimum temperature progression in a given type of reactor is as follows whatever the composition always have the system where the rate is maximum very simple rule so I think inverted commas if you started you just close it okay so that means you know I mean whatever may be the composition you look at the temperature where you get the maximum for the temperature what is the maximum rate so when you are putting together all these rates for example in a P F R okay all the maximum rates then the reactor volume will be minimum because in the denominator average of all that rates only will be there if it is a P F R if it is batch same thing again batch and P F R simply replaced by you know space which is replaced by time okay yeah and if it is M F R we have only one rate and also you know one temperature and one conversion correspondingly so we know how to find out that maximum rate because we know what is the corresponding temperature and conversion and also corresponding rate so whether that is highest or not okay so now next one what you just write is after that the locus of maximum rates is found by examining X A versus T graphs okay that is what what we have done here this is X A versus T graph right and now we thought that rate is increasing as you increase the temperature so you can go and go further whatever temperature you want but we have restrictions these things so you have to choose the highest any one of the one okay the lowest of these things because even if catalyst is getting deactivated and materials of construction it may withstand may be 2000 degrees Celsius but catalyst is getting spined up to 600 you cannot go there okay so that is lowest of this we will decide which one will be the or side reaction may be taking place at 550 degree solo right so that is one what you have to choose and within that now you have to try to get this what is the minimum volume for a given conversion or vice versa the other also is true okay good so now for irreversible reactions please take this for irreversible reactions that we have already discussed for irreversible reactions the rate always increases with temperature at any composition so the highest rate occurs at the highest allowable temperature this temperature is set by materials of construction side reactions catalyst deactivation etc good excellent so now the second one this is the first one irreversible we have now endothermic endothermic reversible reaction endothermic reversible okay yeah before going to that endothermic reversible you see here yeah I think this practical thing I wanted to ask you how do you maintain this isothermality if I have a plug flow it is easy to maintain in mixed flow right yeah so whether I have to heat or cool to get that temperature if it is highly exothermic but you know sometimes if the rate is low beyond certain temperature I have to remove some heat and then maintain the temperature so that I will get the highest rate that is no problem but for plug flow and also batch reactor what do you do just I want you are thinking that is all I think you know how do you imagine this see by putting just blindly one heat exchanger like this this is plug flow yeah so may be this is out you can put in the other way also no problem yeah so this way this is of course f a not entering f a coming out so now how do I maintain that isotherm that means every cross section I should have the same temperature then only you will get the real optimum that is the reason why I told this is the ideal situation that is the most ideal case and now how do you maintain that kind of temperatures I think theoretically we have also told you there is no isothermal condition anywhere in plug flow unless you put infinite number of heat exchangers that means every small cross section I have to take and then I have to put an heat exchanger to effectively control the temperature in that cross section okay and cross sections you can put infinity along the length so right so that is the reason why you don't have that kind of ideality so that is why you have to know in reality you will just go slightly away from the ideality that is what no that is what I think all of us do okay when you are very you are all young when we are all very young one of what we imagine is that the most beautiful girl on this planet should be my wife okay true boys girls also same thing the most handsome guy on this planet should be but the reality is different okay reality is totally different okay because I think how many most beautiful girls you can find out in this on this planet so whoever you get is most beautiful for you okay so that is the that is the actual adjustment with the ideal world correct no minute yeah so that is what is the ideality you know deviation from ideality that is what you know what we can do here so one way of doing that is if I have exothermic reaction how do I maintain but this is very nice very beautiful if I have exothermic reaction how do I maintain isothermality so that means every number of the heat exchangers I cannot put so I have to go to some finite number may be 10 may be 5 how do I do that how we do is you cannot cross this because the moment you cross some one of these things will control and you know you get spoil the entire reaction so you start as much as possible yeah if it if you are using adiabatic reactor that is one of the simplest ones instead of putting jacket around you put adiabatic reactor no jacket or no heat transfer no no heat removal system around the reactor but you do all that heat transfer after the reaction that section is over so that means you use now multi stage small portion heat exchanger small portion heat exchanger how much is the small portion that depends on again your heat transfer and coolant and all that so you start that means adiabatic system you are using reactor is adiabatic but in between you have heat exchanger so if it is exothermic reaction I will start even this is also a parameter even T not also which one is the correct T not also is a parameter through optimization one can find out that so now because it is adiabatic exothermic how it moves it moves like this okay but then I have to that means I have to see my calculations that the temperature increases along with the along the length of the reactor with conversion and then it goes to maximum this value correct no beyond that I cannot go then what I have to do heat exchanger I have to use when I am using heat exchanger there is no reaction right so then what kind of line I can draw here how it can come down okay it can come down because I think conversion I told you there is no reaction so what should be conversion same it is same so now it has reached this conversion so that means again how far you have to go again it is a optimization problem and then you let us say that you found that and again you go here and again go here so what is the type of reactor now what we have here section yeah this is heat exchanger again another section heat exchanger another section like that you know how many you need so this is the reactor and it need not be same length sometimes you know it will be decreasing in progression sometimes it will be increasing in progression probably if it is endothermic reaction now endothermic reaction what do you do it is not true isothermal how it is isothermal because I cannot get that kind of isothermal okay yeah I think this question is good so good yeah to really become isothermal I have to move only yes that is why we are telling we have infinite number of times right no here I have put but that means I have to start here go here start here go here start here go here so that means infinite number I want what do I say yeah I know the assumption is that you know there is no the moment it comes out of the reactor okay out of the reactor we have to assume that there is no reaction which is not correct particularly for homogeneous reactions okay that is correct for catalytic reaction because there is no catalyst no reaction that is why you know I think that sulphate dioxide to sulphate trioxide also it is used I will tell you that one later so like this but now if I have endothermic reaction what do I do I start with this point that is highest possible right but now temperature decreases though this goes this side so this one so this goes this side right that means temperature falls and also conversion of course is increasing then you have to see where you have to again heat now because you have to be on this line as much as possible then again so like this till your conversion that is the line that is the final conversion okay so like that you have to do see how thrilling it is if you really do because you are not doing you don't really enjoy it okay I don't know whether it is fortunately not doing or doing you know unfortunately not doing because this is what is true chemical engineering I say there are so many things there are so many wonderful wonderful things that are happening right and people have been using this already but only thing is I think even at this point of time I know you will get excited provided you have interest in designing this chemical engineering equipment so that is how what we do for exothermic or endothermic and red one I will write is endo and this white one we will write as exo so that is how and as I told you infinite means you will practically lie on this that is isothermal throughout many can turn so inside you have definitely there is some small variation of temperature but if I take this kind of length from here to here there may be 15 20 degrees variation but if I take very close here and then try to do there may be only one degree variation so practically almost you may get isothermal conditions when you go to the overall length so that is how for exothermic endothermic it is done okay good now reversible this is endothermic reaction reversible what is the kind of x versus 2 you get here this is x a versus 2 okay then we have r equal to 0 something like this correct know this is r equal to 0 that is equilibrium it is 0 equilibrium conversion and corresponding temperatures good so now if I draw this will be like this again how the rate is it cannot touch there yeah so rate is increasing like this only r equal to 0 may be r equal to 1 here so like that it increases now what is the condition for here optimal progression that means even here we can clearly see that as temperature is increasing rate is increasing so limit is again tmax yeah straight tmax so this is tmax oh my god okay so now of course here again if you want to be exactly on the line so there will be infinite number of heat exchangers along the length okay but this is endothermic it is very easy now to draw because we have already done here so we start with the highest rate I mean highest yeah temperature corresponding to that highest rate right so then I will draw like this like this like this if I am going for yeah may be the conversion required is this much yeah then I need 1 2 3 reactors and 3 heat exchangers okay good so this is actually not that thrilling when compared to your reversible exothermic reversible exothermic you know the plots here xa versus t here yeah this is r equal to 0 and we know that as temperature is increasing you will have conversion falling equilibrium conversion itself is falling and you cannot cross equilibrium conversion so that is why conversion falls in a reactor where temperature is increasing and the rates if you calculate and then plot you may get something like this yeah rates are increasing like this this is 1 okay 10 100,000 10,000 okay this is 10 to the power of 4 I will just start okay in between you have the lines yeah so here this is exciting the reason is that it is not blindly as temperature is increasing you have you know rate also increasing right rate is increasing there is no problem at all as temperature is even compared to this rate this is definitely more right so but the achievable conversion decreasing that is thermodynamic limitation that is why this is one problem where kinetics and thermodynamics both play a role okay there also it is playing but this is quite dramatic in the sense that you cannot cross thermodynamics here right so that is why what we do here is that we start at high temperatures okay we start definitely at high temperatures yeah okay by the way before that when we start at high temperature will come later but what are the maximum rate curves here maximum rate line will be connecting this maximum this is locus of maximum rates maximum rates some of you may not be easily you know may not be understanding this one that easily why they are called actually this is a different rate this is different rate this is different rate and I am telling that that is the maximum rates okay if you plot x versus t it is not quite obvious but x versus t is the best way for design right so but actually if I plot rate versus temperature and this line is this line is nothing but dou r by dou t equal to zero that maximum locus line okay that is what you can actually derive you cannot actually differentiate that and try to find out the temperature and corresponding rates okay or corresponding conversion also okay good so yeah this is the one and just to tell you I mean so that you know your imagination will be easier it is not that easy to imagine that how these lines are coming this is coming but that comes if I just draw a line if I just draw a line yeah if I just draw a horizontal line what is the meaning conversion is the same that means I am now talking about at a particular conversion right let us say 60 percent conversion how the rates are changing okay how the rates are changing see this is t starting from normally we do not start from initial you know absolute temperature zero but imagine that you are starting from absolute temperature zero so what is the rate there zero okay right then there is slightly more rate then slightly more rate then slightly more rate at that conversion any line horizontally it will go through all this history right so this is slightly lower slightly higher more higher then what is happening after crossing this again it is decreasing how do I plot this that means I have to plot this for a given conversion rate versus temperature how do I plot that this is rate versus temperature so how do I draw that so yeah so it may go may not be parabola exactly right so this is what then what is this point how do I get this point how do I get mathematically you cannot say d it should be doh why all that is happening at one conversion x this is at x constant x a constant correct x a is the parameter so similarly if I draw another line may be somewhere above right what do I get above this I get or below this I get conversion below it below it yeah I mean if I draw this here below that below that because these rates are definitely smaller than these rates if I go here these rates are much higher so that means lower conversion if I go to lower conversion that is what is reversible reaction because at higher temperature you get high rates but conversion will be lower that same thing only is reflecting there so then if I draw another line here and another line here so how in which direction I have to show the conversion how it is increasing from top to the this is x a increase so this is what what we are telling you know these points are nothing but doh r by doh t equal to 0 and that is what is the line which you get now I think is understood otherwise you know even last year also some people are asking sir you are telling this but we are not able to get what is this doh r by doh t that is what is the line so there if I plot r versus temperature it is easy for me to tell I mean for you also it is obvious because doh r by doh t that is where what you are but here there is no doh r by doh t that means that this line is doh r by doh t but r is as a parameter that is coming there right but if you plot the same information in this x versus t now you have three parameters know now I take at a constant r and then plot all that information then you will get this maximum line okay so that is why good so this is the one now under these conditions what we normally do is our okay our information tells us that go to the highest temperature that is possible because even though conversion is is less but rate is high and in my design expression I have this rate in the denominator so I have to have definitely more but if you want to get more conversion what you do you have to come back now so that means maybe I am starting some somewhere here right and it is again let us say adiabatic reaction that is easy one so then you have going like this right but if I proceed like this this is the maximum conversion I get but my maximum conversion may be somewhere here this is the conversion which I want okay for my design may be 85% 90% is the conversion which I want so then what I have to do I will start as far as possible the highest even I can start here right but where do you start is also optimization problem t0 what is the first t0 otherwise the overall may not be optimal unless this is t0 is optimal right so in this case let us assume that this is optimal and then yeah it is one reactor this is one reactor PFR right so then I have reactor and then I have to now cool this till what point here it is very clear to what point we have to cool the reason is you will get this information from what is called dynamic programming aris is very famous for that you only gave these conditions but didn't we gave this condition what I am going to write a little bit later is the condition without mathematics by simple logic he has written that condition because we know for optimality dou r by dou t equal to 0 so similar condition he has is also wrote for this optimality which has been proved by later with people you know with actual mathematics okay that condition I will tell but because it is a there is no reaction in the heat exchanger this is cooling now it will go and stop only at this point because the rate is same this rate line correct only one rate I have right on this it has to stop only on this otherwise it is not economic okay but by logical common sense I think you know then we had that one even without doing mathematics also you could tell that that it has to stop only on this reactor line and they did lot of work on this particular exothermic reversible reactions particularly for catalytic reactions catalytic reactions we can definitely guarantee that the moment you come out of the reactor there is no reaction so that is why you have to have the same conversion same rate but only temperature is decreasing because you are no cooling the entire reaction mixture then you draw again parallel line where you have to stop again depends on the mathematical nothing then you have this line now because this line this line right so then if you want this conversion again you have to go further maybe you have to just stop there okay it may not be economical but I think correctly if you draw this then you will also go there and then that finally it has to be you know for optimality the general common sense is that we have to be both sides of this maximum this is this side this is this side if you are only one side you will never get optimum right so the condition for this to be optimal the condition for this to be optimal that means this casc crossing this side is integral of course you have to be if I say this is A this is B dou of 1 by minus r A by dou t into dx A equal to 0 this condition you get from yeah dou by dou t okay 1 by this is 1 by minus r A okay of course which also can be written if I just expand that this will be A to B 1 by minus r A whole square into dou t into dx A equal to 0 same thing okay so that is the condition so in fact this one will be a condition where if I plot this one for example dou r by dou t versus x A okay I am plotting this right how it should be because the area should be 0 so that means there somewhere it should be positive somewhere it should be negative so it will be like this where this is A this is B okay so positive area this area must be equal to this area so that is the condition so that means you do this try to find out the rates and differentiate and then try to find out this area going on crosses goes and at this point you have to stop here if you are doing graphically if you are doing by dynamic programming that will automatically fix the corresponding temperature okay good so this is the one so that means this is exactly what is happening in SO 22 SO 3 plus half O 2 SO 3 this is gas this is gas this is gas all gases and it is a catalytic reaction and they use adiabatic reactor I will tell you I think why adiabatic are used a little bit later when you are coming to individual reactors okay adiabatic reactors are generally used if you do not have very high exothermic heat so that means around 25 to 30 kilocalories per mole that is medium if you have 60 70 kilocalories per mole that is very high 60 70 kilocalories that is highly exothermic right so almost isothermally you will get if you have 5 6 or 10 kilocalories per mole delta HR I am talking about delta HR okay so those conditions yeah so this is what is this and even this line how to get is simply differentiate this like what I have minus r A equal to simple reversible reaction K 1 C A minus K 2 C R dough R by dough T because K 1 is nothing but K 1 0 E 4 minus E by RT you know by minus E by RT and K 2 also is simply K 2 0 E 4 minus E 2 by RT that you have to differentiate that I can give in the examination this dough or by okay yeah I think maybe you have class