 So, this diagram I hope you remember, right, yesterday also we have been discussing this, this is reversible exothermic reaction where this line is the locus of maximum rates, right, where dou r by dou t equal to 0 and we also discussed that, I mean how I can use various sections for various stages for the minimum volume but I want absolutely minimum volume what I should do with the diagram, you cannot have less than that, yesterday we had 3 stages, okay. So, how many stages we should have if you want to have absolutely absolute minimum, it is infinite why? It cannot be isothermal here, it cannot be isothermal, so but where I should be, where is t allowed, here this is reversible exothermic reaction, where is t allowed there, I should be on this line, then you will get absolute, that means all these, you know the rates just touching all these points corresponding excess, okay that is the maximum rates that are possible, all maximum rates will be on this line, so when I have all maximum rates and then if I am able to operate my reactor with all that possible maximum rates along the length of the P f r, then I will have 1 by r, r is maximum, so 1 by r will be minimum, so you will get minimum reactor volume, okay. How do you get this equation, I will just give you for a reaction A going to r, this is K 1 and K 2, K 1 and K 2, you can show that, I am not giving you the final thing, how optimum equal to E 2 minus E 1, instead of changing all these always you write E 1, E 1 minus E 2, this minus will be there divided by r, so ln of K 2 not by K 1 not, that is you know meaning of this K 1 and E 2 by E 1 X A by 1 minus X A, this is the equation, very beautiful equation which connects tau optimum, tau optimum is corresponding to maximum rate or the rate corresponding to this tau optimum will be the maximum rate, right and this is now related with conversion, T optimum, not tau optimum, T T T optimum, yeah temperature optimum, temperature optimum is you know corresponding to that optimum temperature you will have the rate is maximum, so if I am able to draw this profile X versus T or other things are known, in fact E 1 minus E 2 is nothing but delta H r, this is minus delta H r, this entire thing is minus delta H r, so you can also write this one as minus delta H r by r, okay and then E 1, E 2 I know, K 2 0, K 1 0 I know, X A versus T, I can imagine I mean I can calculate for X A equal to 0.1, what is tau, T not tau, okay and again 0.2, 0.3, 0.4 that is what is this line, if I am able to stay exactly on that line then I will have absolute minimum that is possible and if you have mixed flow reactors it is very easy to operate, right. So I will have large number still but in each reactor I can show that okay this optimum is 1, next one here, next one here, next one here, next one here, right, yeah so like that it is very easy to operate that kind of thing in mixed flow reactors whereas for plug flow reactors that is not that easy to maintain at each point what is the how optimum. So that means you just imagine now, you have to put the heat exchangers such that at every cross section you should have that T optimum, correct no, you should have your scheme either removing heat or adding heat such that always you maintain in that small cross section, theoretically speaking it is 0 bit through each cross section, right, you should be able to maintain that T optimum where you can get the minimum absolute minimum reactor volume that is possible, okay because we cannot do that that is the reason why yesterday we have gone and then drawn that okay I will start from here, go here, start, start, start, start, if this is the final conversion that is required, this is X AF, okay this derivation is very beautiful derivation please do that, right it is just nothing but again dou R by dou T equal to 0, okay good very nice, yeah now I also told you that we should have some practice for drawing these lines, right so I will give this exercise please take that develop conversion develop this is the problem this is assignment to you develop conversion X A versus temperature T T plots develop conversion X versus temperature T plots with rate of reaction R as parameter as parameter using the following data using the following data I think I will write here, okay, A going to R then of course minus R A equal to this is K 1 K 2 C A this is K 2 C R of course C R not equal to 0 here that is the pure C A only what we are taking and equilibrium constant K capital K is E power 18000 18000 by R T minus 24.7 yeah everything is in exponential that is K rate constants K 1 is E power 17.2 minus 11600 by R T units are minutes inverse K 1 K 2 because both are first order I think better write minutes inverse K 2 is again E power 41.9 41.9 minus 29 by 6 second by R T this is also minutes inverse yeah this is now where this activation energy these values where activation energy energy is in calories per mole, okay. So I think normally in this when you are drawing these graphs I think your range of temperatures will give you lot of problem so that is why in this problem I am specifying the range of temperatures to draw that graph you can try between minus 10 to 110 degrees Celsius and of course that interval you can have interval 10 degrees that means minus 10 0 10 20 30 40 like that yeah, okay. So then the parameter values rates that is minus R A or R A they will be 0.001 0 0 2 0 0 3 point is there 0.002 0.003 then we can have 0.005 0.01 0.005 0.02 0.03 maybe I think we can shift to 0.05 yeah this is 0.1 0.2 0.5 then 1 2 3 5 how many lines you will have 4 5 6 7 8 9 10 11 12 13 14 15 lines you will have here in that graph draw that that is an excellent experience if you are able to draw that and minus R A is in moles later minute because this our concentration also in moles per later C A also is in moles per later, okay this one please draw and few of the graphs I may also ask in the examination because you have 3 hours no in the final examination right. So what do you do you have to draw this kind of correction all that 3 hours what do you do. So that is why I may give this examination there, okay K 1 by which K you are talking capital K yeah why run it is correct only check because that is given in a different way that is correct only 17 point 2 and 41 point 9 correct only sir and yeah the other one is 7 point 600 yeah it will it is a standard problem it is not a new problem created by me okay good so that is what you can do it and then yeah so with this this general discussion on graphical design is over right I hope you understood the essence of what we are trying to do in the last 3 4 classes this will give you the overall picture for the actual analytical work also now what we are going to do is for batch reactor separately plug flow reactor separately and also mix flow separately we are going to write the equations but this gives the framework for the coming classes but I think here if you are able to use even graphical one even if it is possible to draw in the examination hall it is excellent because do not think that graphs are going to take more time but I think if you are good artists quickly you can draw that and all engineers must be artists otherwise you do not even know how to draw a straight line right and if you have a B Tech degree or M Tech degree or not able to draw the straight line I think it is waste yeah because I think straight line is the simplest one what you can draw good yeah so that is the one now let us take batch reactors and if you have any questions I am sure you will not have any questions because most of the time I have to question I have to answer okay and you will be seeing there sitting okay now let us take batch reactor good so I think to summarize just Abhishek which will be really thrilling out of all this reversible irreversible reaction reversible endo reversible exo reversible exo reversible exo why why thrilling just thrilling dynamic programming dynamic programming yeah that condition comes through dynamic programming because here the problem is challenging right the challenging in the sense that there are two opposing things you know high temperature yes we like it okay high rate of reaction we like it but you do not have sufficient conversion finally at the end you cannot get if I start here at a very high temperature rate of reaction will be very high but maximum temperature is even if you go vertically up also it is only this much temperature if you go to adiabatic and all that only this much okay so that is why how do you make this one use high rates that means high temperatures plus still get maximum conversion so you start here and then go slowly and you find out what is the required temperature and how many stages you need that is what exactly the question I have asked in the first zero eighth examination where I asked you know why for SO 2 to SO 3 multi bed adiabatic reactor is used multi bed adiabatic reactor for the V 2 O 5 vanadium peroxide catalyst and is the catalyst catalytic reactor and they need generally 2 to 3 beds in series in between and they have heat exchanger by also asked another question why in ammonia reactor heat exchanger type is used it is not adiabatic reactor that also I asked have you thought any type that also comes under non isothermic question is clear first of all manikanta question also not clear what did I tell you just now why in ammonia so then why is a question is not clear that is to avoid answer first thing students says is that question is not clear sir even the simplest question we ask in the interview they say no sir I think as if drama acting as if he has not understood at all no sir not understood okay then again I have to tell the same question in a different way okay answer may not come but anyway question will be repeated many times so but why why should we use now can you expand your brain and then try to think now you know why we have to use for SO 2 to SO 3 this multi bed adiabatic reactor there we need high conversions in SO 2 to SO 3 usually around 85 90 percent conversion it is reversible exothermic right V 2 O 5 you know on the and that catalyst is reversible exothermic right so if I straight away start here because here I have 85 percent conversion if I just use only isothermal and then 85 what are the rates here very less than what will be the volume equivalent to IIT campus you know what is the area of IIT campus very good yeah 632 acres okay so one reactor you cannot have that area so that is the reason why we have to go for high use you know high means an high rate of reaction and then at least the first part will be less and next part will be slightly bigger second reactor third will be slightly bigger than the second one so like that it increases because you are moving towards lower region whereas in ammonia for example that is very very highly exothermic reaction so that is why moderately exothermic reactions where you have a reversible then only we use adiabatic reactors and when you use adiabatic reactors of course reversible exothermic then you will have definitely have to move from low conversions to high conversions so that is why you use multi bed stage 1 at this level at this range high ranges high rate range so then move to low range still further move to low range depending on your conversion okay so that is the reason I think now you appreciate these reasons I say so wonderful theory behind that and everything is you know thermodynamic and kinetic data wonderful okay good now let us take batch reactor so I think we will remove this where is this now let us write these material and energy batch material and energy balance equations for each reactor first let us take batch reactor batch reactor we have been telling that normally we will show like this okay then you may have the yeah may be coolant in T C T C then we will have this stirrer yeah here I may have temperature probe may be thermocouple somewhere here I will have pressure gauge okay if possible we have to also use internal coils again T C T C etc not only bottom I think you can also use side you can also use side right all this is possible okay good so now we need the material and energy balance for this and of course I do not have to draw again the profiles you know like with here with time yeah again depending on if I have adiabatic reactor what kind of temperature profiles you get conversion and temperature profiles if I have a adiabatic reaction that means all this is insulated I am simply so okay draw in the air air means like this yeah XA versus what is the other variable temperature I have asked me I have asked Abdul means Abdul represents the entire class everyone has to draw okay yeah I mean you are enjoying only we are talking and all of you are laughing happily yeah yeah like this what is this what is this first you have to draw like this like this so now you have to draw type temperature like this okay that is adiabatic now tell me for non adiabatic non isothermal non adiabatic non isothermal bala adiabatic also increases non adiabatic non isothermal also increases what do you mean by non isothermal non adiabatic what is the reaction okay dhanavi non isothermal can increase or decrease how under what conditions it will increase under what condition it decreases I am asking I have a system where it is not adiabatic it is non isothermal non adiabatic what is the meaning of nena good so what is the meaning what is the real meaning you have heat exit okay so when they have exothermic reaction now we are removing some heat now bala depending on the heat removal it may go maximum and then decrease okay yeah if you are removing only very very very small amount of heat then also it may slowly increase and then may be flattening out so there are many many possible curves isothermal that is very easy straight line okay good so that is what is the plant and conversion always increases so that is no problem so now we have to have the mass balance equation always we have a reaction something like this a going to r for the simplest cases okay so mass balance of a only normally we write only for one component right mass balance of a this is n a not d x a but this we have already done it so that is why I am not writing again elements and all that okay minus r a and this equation can also be written in terms of delta hos that is required delta x a equal to minus r a v delta t so this is equation 1 okay this is 1 a this is 1 b okay good so this minus r a can be also you know variable volume means it will have k into c a where variable volume means c a by c a not equal to 1 minus x a by 1 plus x a and all that are there okay good so energy balance this is input plus generation equal to generation will come if it is a big heat right yeah then output plus accumulation oh sorry this fellow is inherently unsteady fellow batch reactor plus accumulation and heat removed or added okay of course again here also have to write you know in general minus generation means you know endothermic right okay good so this is the general one so now let us write for only exothermic so that you know we will have some feeling for that so yeah and we know that it is a batch input 0 and output 0 we are writing for exothermic exothermic let us maintain minus delta hr as a positive quantity that is our convention please write there somewhere minus delta hr equal to some positive maybe 30,000 calories per mole or something like that okay good so generation when I write for this is minus r a v minus delta hr that is equal to yeah that is generation that is equal to then I have sigma of sigma of mi cpi dt by small dt that is the accumulation plus I have heat removed u a t minus t c so we have only three terms now no generation accumulation and heat removed because we are talking about exothermic reaction minus delta hr so that is why this is u a t minus t c good so this is equation and now let me go to other side so this equation this is maybe 1 and this is 2 this is equation number 2 okay so this equation I will just simply try to write like that sigma see if you have the average properties for the entire bulk reaction mixture okay so then that will be simply m c p that is otherwise if you want to write for individual components m a okay m is the mass and c p or m also can be moles but cpi will be in terms of moles okay yeah that calories per mole per degree so that is why that one please do not get confused so this is dt by this is temperature and this is time that is equal to minus r a v minus delta hr yeah minus u a t minus t c yeah so I think maybe for continuity sake or as a clue I can also write here mi cpi where m is mass cpi is the specific heat express it per mass this also equal to mi this is ni cpi dash di where cpi dash di is in terms of moles okay normally here you will make mistakes that is why I am trying to write all those simple things also okay good so you know most of the time we have been talking about energy balance in terms of energy balance that gives me the relationship between x a and but this equation 3 will it give me relationship between temperature and conversion it would not give me so then how do you make that now you use yeah you use this relationship okay and then substitute there so that you will have the relationship between conversion and temperature so that is why substituting equation 1 a in 3 substituting equation 1 a in 3 this is dt by small dt equal to minus r a v no minus r a v only have to change this is minus delta hr this is minus r a into v so this is n a not dx a by dt minus u a t minus t c so this is equation number 4 okay but now the problem here is that you know I have to use these 2 equations this is energy balance equation this is this is material balance equation this is energy balance equation to solve my you know conversion versus rate right finally the design is only 1 by minus r a versus x a only okay then area under the curve will give me here total time that is required for certain conversion okay but here I have only 2 equations and how many variables what are the 3 variables temperature and time 3 are there so that is why we have to go by trial and error we do not have third equation so that is why what we write here is in terms of deltas sigma m i c p i delta t equal to minus delta hr n a not delta x a minus u a t minus t c into delta t this is time small okay I just divided by small dt and then written this in terms of difference equation okay so now what is the procedure because I have to get the relationship between temperature conversion and rate right temperature conversion and rate and of course indirectly that you know this time rate has dx by dt you know okay so that comes there so that is why there is a procedure now by trial and error what we have to do is that procedure to solve equations 1 anyway this 1 b that is also difference equation and 5 okay so what is step 1 is we will just try to have you can start with anyone but I think I am starting with conversion I will assume that I have delta x a equal to 0.1 choose delta x a equal to 0.1 and Abhishek may say no sir I do not want to take delta x a equal to 0.1 I want to take only delta t temperature 5 degrees capital t and Abdul may say no no no sir I do not want to follow him so I want to take small delta t delta small t so then that is time then probably you may think that okay in 5 minutes what is now you have to whether if you assume delta x now you have to imagine that or guess that in this 0.1 increment of conversion what is the increase in temperature time also you can anyone okay then the other one you will check whether your guess is correct or not so now here guess a value for right delta t small t I am taking to calculate temperature small t delta t corresponding to corresponding to this delta x a 0.1 okay now so that means I know delta x from this equation 0.1 and I know delta t 0.5 and other things you should know right so other things means you know UA you should know heat transfer coefficient and area and here of course delta h are I know NA not I know and these properties I know here I have t 0 minus t 0 I know t c I know t 0 is the initial temperature so that is why I know everything except t here so you calculate calculate t t using equation equation 4 how do I know that my guess is right so this step I have temperature conversion x a x a initially equal to 0 right delta x x 0 equal to 0 so x equal to 0.1 corresponding to that I have guessed time 5 minutes right so then I know everything now I have to check this so I will go here and then check in this that means I will substitute whatever values I have there I will put here I know delta x and NA not I know and here in minus r a it can be very complicated need not be simple k c a it can be any complicated equation right so here this minus r a contains temperature and conversion if it is the simplest case if you take this is k c a k is nothing but k 0 e power minus e by r t x so I know x now I know t now and I can calculate minus r a and v you know in this problem you have to assume some v in the beginning and also you know that whatever volume you assume okay and you know you are finally calculating time right time will not change for a batch reactor time whether you take large reactor or small reactor time is same for given conversion okay even if you take 10 meter cube reactor for 90 percent what is the what is the time and even if you take 1 liter time is same but volume will come only when you are calculating productivity right that is what what we have also done know that thing how many batches per day and each batch volume and if you know the densities and all that you can calculate volume of the actual reactor right so that is why some volume definitely you have to assume there so that means I will now put all those values and then check whether l h s equal to r h s okay if not again okay good so that is the procedure so now yeah then step 4 is check equation 1 b 1 b with at this time with known x a small t and t okay so if 1 b is not satisfied equation 1 b you know if equation 1 b is not satisfied then go to yeah then you have to go to step 2 satisfied go to and you will easily get this things and of course next one is repeat 2 2 5 1 because this anyway every time you are choosing that so 2 2 5 yeah so that is the procedure very good this is actually the non-isothermal non adiabatic reactor because you have that you have the c terms for term here good so now let us see for adiabatic system what will happen right for adiabatic system what should be missing from equation 3 for adiabatic reactor u a equal to 0 then this term will not be there then we will have sigma of m i c p i d t by d t equal to not 3 4 4 is easy for me because I want relationship between conversion and temperature right so equation 4 gives me that so this is minus delta h r n a not d x a by d t so this this can get cancel so this is 6 yeah so now if I take this side this is integrate that t minus t not equal to minus delta h r n a not into x a minus x a not x a not usually we take 0 okay x a not so this is the equation so this is what what we have been telling also earlier that you get t minus t not equal to this is beta x a this is easy to solve for me because time is getting cancel here okay in this equation right okay so this is the one and you know how to use this this is the beta is the slope and you have to plot if you want to plot you know on the x versus time diagram so then you can plot that 1 by beta as the adiabatic line and then see what are the intersection points and now you take corresponding x corresponding r 1 by r plot 1 by minus x r a versus x a area under the curve will give you yeah t by c a not or the way how you write the that equation okay good so this is what I think we already discussed about adiabatic system long time so I think we do not have to discuss anything further good yeah so one problem quickly you can do it yeah I will just dictate and I will give the answers you can check okay this is adiabatic reactor batch problem on adiabatic okay good yeah reagent a undergoes an essentially irreversible first order reaction a going to r to stop both a and b are liquids at room temperature and both have extremely high boiling points determine the reactor volume necessary to produce okay I just want to give this one as it is 2 million pounds of r I hope you know the conversion between pounds and yeah kgs okay so that I think you know still this kind of units are required in chemical engineering because we have still lots of data in various units still lot of heat transfer data and all that available in BTUs okay no more joules and all that so that is why I think we should be familiar with all that that is why I just would do tell this 2 million pounds of r in 7000 hours of operation in 7000 hours of operation okay that means you have to determine the reactor volumes necessary to produce 2 million pounds of r in 7000 of operation one if reactor operates isothermally at 163 degrees centigrade if reactor operates isothermally at 163 degrees Celsius B 2 okay 1 I said 2 if the reactor operates adiabatically so that means 2 you have to do one is isothermal case other one is adiabatic case okay good use the following data use the following data just below right minus r A equal to k into C A first order reaction very simple okay and liquid phase also we told minus r A equal to k C A then next one is k at 163 k is small k at 163 degrees Celsius is 0.8 hours k at 163 is 0.8 hour inverse okay activation energy E equal to 228960 60 calories per gram more calories per mole mole means always gram more okay so then minus delta H R equal to 83 calories per gram okay molecular weight M W 250 okay the heat capacities of A and R may be assumed constant equal to C P 0.5 calories per gram for a degree centigrade okay that is C P both are same C P value it does not look like C P okay good the density of the reaction mixture may be okay the density of the reaction mixture is 0.9 grams per C C all kinds of units we are giving okay the density is the density of the reaction mixture is 0.9 grams per centimeter centimeter okay next line the times necessary to fill and drain the reactor may be assumed okay next sentence it may be assumed that it may be assumed that negligible reaction occurs during occurs during the 14 minutes during the 1414 14 minutes it takes to heat the feed from from the temperature at which it enters the reactor to 163 degrees Celsius okay in simple words it is heating time required is 14 minutes heating time okay tool to 163 after 97% of A has been converted after 97% of A has been converted the product mixture is discharged so the answer to this problem is that now I think you have to go now see for the first case isothermal the answer is T the time time is 4.383 hours okay T is 4.383 hours and the volume is 740 liters plus or minus 5 this way that way the volume required is 740 liters for non-isothermal case that is adiabatic reactor you expect lower volume or higher volume which one Rahul higher volume higher volume why anyone Dania okay Rajasri you said decree Savita why exothermic decreases mind mind mind because in exothermic reaction temperature increases when temperature increases then you have rate of reaction increasing when rate of reaction increasing in the design expression that fellow is there in the denominator so time will decrease so total volume also will decrease in fact time you will get only 0.12 hours 7.2 minutes okay for adiabatic case time required is for 97% conversion is only 7.2 minutes and what is the earlier number for isothermal yeah see okay good so that is one and volume required is may be around 105 or 110 liters volume required for adiabatic this is a wonderful problem I have chosen the reason is you will know the effect of this adiabatic reaction for exothermic reactions particularly because naturally it also increases the rate of reaction only reversible exothermic only is the problem as the temperature increases conversion falls but this is irreversible reaction so temperature increases rate of reaction increases your time decreases and volume decreases see those things I think you know that should be permanently there in your brain no should be but I think you know okay good we will stop here