 in the last class last 10, 15 minutes we have derived one equation for interface, interface together, right. Yeah and in the graph I have drawn I have mentioned that that is 8 of i square that must be 8 of 0 of i square in the graph x axis swami did not draw the graph at all did you draw the graph x axis x axis is 8 of i square I have written there that must be 8 of 0 of i square, right. So I think again I will quickly derive this also can be derived in another way in terms of 8 of 0 starting yeah starting with 8 of 0 this is again interface effectiveness continued, okay what we have done in the last class last minute you know last 10, 15 minutes we have done that so in the hurry we have done but I think you know all the equations are right but only thing one step I have to prove that why that r, o, b, l square by d, e, c, b equal to 8 of 0, 5 square where 8 of 0 is overall effectiveness factor, okay we have used 8 of r for external 8 of r internal interface and overall is 8 of 0, okay so that is what we have to just yeah the first definition we had 8 of 0 equal to r, o, b by r, b this equation did I write last time no not, okay yeah good so this is the one which missing so 8 of 0 is defined in the normal fashion but this is overall from film to centre of the particle so the entire dimension is taken for defining this so then you have the equation already in terms of c s by c b I do not know c s by c b you have an equation already please tell me that k g a divided by k g a plus eta k okay now substitute this equation what was this number this one is 2 and this one what was the last number we will just continue so this we will put as 8, okay good so from this from 8 we can write r, o, b equal to eta 0 r, b which is if you take for example first order reaction and that reaction also it will be same no problem eta 0 k c, b for first order reaction correct no this is r, b r, b is rate based on yeah but isothermal only we are talking please remember that okay this overall interphase that is interphase plus interphase together we have been talking only overall I mean only isothermal condition so that is why so this is no temperature there it is only k into c, b now substitute yeah substitute now for this c s okay sorry this is eta 0 k c, b okay one more equation I think what you have there so that is why I have been searching that yeah r, o, b equal to r, o, b equal to what is that you have given written there not that in the beginning r, o, b equal to k g a yeah c, b is also equal to eta k c s eta k c s so r, o, b is also eta k c s good substitute now what is this equation number 1 no the first one okay you do not have to write that equation number substitute equation 2 here and tell me what is eta what is r, o, b sabita not able to follow still yeah this equation you have this equation you have what is that you have here c s right yeah like sabita many people also may not understand this c s c s substitute here and then what you get for r, o, b yeah r, o, b equal to this whole thing right eta k eta k that is this yeah for c s I have to write c, b k g a c, b divided by this one k g a plus eta k so this is the one now so substitute this in eta 0 yeah you do that later but I think substitute what is eta 0 eta 0 equal to r, o, b by r, b or in this equation also this is equation 3 oh no not 3 9 Karthik what was this equation number this is already there that is what I am 3 yeah this is 3 we are mixing that and this now so now substitute for eta 0 and then give me the equation what do you get that is equation 8 substitute equation 3 in 8 with r, b also r, b also you know r, b is k, c, b substitute for r, b and r, o, b what do you get 1 by eta plus 1 plus 1 you will get yeah everyone is giving his own so I will give my own so this is eta k g a k g a plus eta k excellent that is one thing now divide this is also no no not 1 by just to make some 1 plus eta k by k g there must be some meaning when writing that what is k by k g a yesterday we have done something else it will become phi square yes that becomes phi square by biotomba so eta 0 equal to eta by 1 plus eta phi square by biot number mass okay good so now here you can substitute for tan h tan h phi by phi for slab or otherwise for other one spherical particle you have 1 by 3 by phi into whole that that you have to remember examination is coming next week okay yeah so that also you can substitute for any shape now you can substitute because we have not taken any shape when we are developing this equation so only thing that is left you know what should be proved was that r, o, b c, b d, e by by d, e r, o, b l square right r, o, b l square by yeah that is equal to eta 0 phi square I said can you prove that eta 0 phi square what do you start with this is the equation to start with yeah in equation 9 you can substitute and then get equation 9 you have r, o, b multiplied by l square divided by Karthik searching here so what do you get so we get eta 0 is 1 by phi square by bi plus 1 by kaa all that is over by d e c b yeah so that is one way here also you can use the equation this equation divided by l square sorry multiplied by l square divided by d e and what is else the other one c, b both sides anyway c, b also we get cancel you will also get eta 0 phi square okay so anyway let me do that so from equation 9 we have r, o, b l square d e c b equal to this if I take eta 0 k c b l square same thing d e c b c b c b we will get cancel this k l square d e is nothing but phi square so that will be eta 0 phi square yeah because there are mistakes even in Carberry sometimes he writes this one as eta sometimes he writes mistakes as eta 0 eta 0 is the correct one for overall so in that graph you should have eta 0 okay eta 0 and please think carefully about that graph that is in a graph is looking good but I think it is not that easy to generate I may ask you how do you generate that graph you know you have eta 0 verses the plot was eta 0 verses eta 0 phi square for various biote numbers for various biote numbers how do you really generate that graph make a question mark there you know how do you generate that graph I want you to think good I think somewhere numbers have gone now 9 so this was not there right it was not there yesterday so this is 10 this is 11 this is 12 yeah okay because I realized that you know you also should have eta 0 and eta related that is the reason why I wanted to do that otherwise yesterday it was not obvious eta 0 I have not mentioned so that is the reason why I thought again I will do it okay so now next one what is left was is taking all the equations what we have done till now that means interface effectiveness factor the temperature profiles concentration profiles you know particularly temperature difference concentration difference and also interphase temperature difference concentration difference you have the relations right can you get some more information from this like for example I told you thumb rule is for many of the processes in catalytic reactions the controlling factor is mass transfer through the pores and heat transfer through the film okay these are the thumb rules but how do you know that what I am saying is right simply believe me so you say yes you have to believe me because you think that you know if I if you do not believe if you write wrong things in examination then I will not give marks I think at least for that reason you have to believe but we have to prove that we can also prove that without trying to use any new equation all the equations already with you till now in your notes whatever equations what we have so that is why what we do now is this is interplay of interphase okay inter and interphase resistances yeah see now resistances we are talking either mass transfer resistance when do you get or heat transfer resistance when do you get so that is the kind of information so please take this as an introduction we have seen till now that we have seen till now that the general problem of diffusion reaction is quite involved so that means you know too much information we have generated it is quite involved in the derivations and all that fortunately however the physical and chemical processes at work under realistic conditions favour isothermal pellets and negligible external mass transfer resistances let us now prove the same Reneeta what did you understand from this para what did you understand I think I gave already the answer very nice answer read the second sentence what you have written yeah why should we assume you have isothermal particle thumb rules are there nicely see I mean here safely after doing all these if you are interested in reaction engineering and if you are interested in catalytic reactions in future life unfortunately if you get a job somewhere in chemical industry then you can happily tell this thumb rule you know when you have a catalyst particle you can always assume that the particle is isothermal because heat transfer is not controlling or the other way to tell is mass transfer control is more important in the particle force whereas if you take the external film it is the heat transfer which is important but mass transfer is not a problem you will have sufficient amount of mass that is going through the film okay through the film to reach the surface then afterwards is the diffusion okay mathematically also we can prove that with the equations what you have already used one equation what you have already used is this that is T s for external that means external film what you have the equation is T s minus T b by T b if you have nodes you can check equal to beta bar beta bar theta bar dA which we call now as theta bar C A where C A is carberry number some people use this that is the spelling of carberry car plus berry okay yeah so this is one equation what you have similarly you have for external in fact theta bar dA is nothing but again theta bar dA in terms of concentrations 1 minus C s by C b okay that is also there good so then the next one is we have internal that means interphase you have T c minus T s T c is the centre yeah equal to this I think I have to just write maybe I think I can also write this okay T b equal to this is beta C s minus C c by C b yeah C b where beta where beta equal to you have the equation minus delta h minus delta h r d e C b by yeah K e T b similarly beta bar also you have okay yeah so now this is another equation beta this equation number is 2 yeah so we have also seen that this we will get T c minus T s max by T b equal to eta where is eta beta into beta into C s by C b okay so this is equation number 4 1 or 4 3 equation number 3 so then C s minus C c because now earlier you assumed C s equal to C b but now it is not Kavya we are now breaking the problem till T c I mean T b T s and T s to T c similarly concentration is C b C s C c and C c is 0 for the maximum temperature difference okay so that is why this C s will be there this C s by C b is also nothing but 1 minus eta bar d e that also we have this C s is so you have same thing I am writing here no T s by T b equal to delta T here I have to also write yeah here this one is just nothing but delta T I max I will write delta T I max I that means internal internal temperature by T b so okay so this also I am writing as max by T b equal to beta 1 minus Carberry number so this is 4 okay actually that is eta bar d e okay so which is as yeah I think here also shall I write Carberry number equal to eta bar d e otherwise you will get confusion later what is this good so now we will add these two equation that is 4 and then 1 yeah 4 and 1 so when you add that you will have T s minus T b by T b plus T c minus T s by T b equal to beta bar C a plus beta 1 minus C a this is equation number okay let me simplify that some more so this becomes T c minus T b by T b equal to beta bar C a plus beta 1 minus C a so yeah this is equation number 5 okay now there is another thing here yeah this one also please add here this one as delta T x by T b I do not want to write another equation there itself in the first place itself you can write x means external yeah good and this one this one can be written as this is delta T not max T c minus T b by T b that is overall no T c to T b that is why T 0 is overall m m m refers max okay so now if I take the ratio of 5 and 1 1 as delta T x by delta T not max equal to beta bar C a divided by beta bar C a plus beta 1 minus C a which also can be written as beta bar by beta into C a by beta bar by beta C a plus 1 minus C a yeah so one more step here that is delta T x by delta T max equal to we will write another number b r into carberry number by 1 minus C a carberry number plus b r C a so this is equation number 6 where b r is anyone heard b r is what number that is what it is brinkman number b r a and k brinkman number how do you know all these numbers what is the problem most of the time how do you know means you do not know how do you know transport phenomena okay good yeah b r is brinkman number brinkman number yeah this brinkman number can be beautifully proved I just leave it to you all the information with you all the equations are with you okay brinkman number using all these things you have mass biot number by feedback number which is nothing but here beta bar by you have all that you know what is beta bar all the things are there with you the definition of beta bar you know definition of beta you know okay using those two quantities you try to get this ratio as bi m and bi h and you know what is bi m okay bi m is just I am giving for easy thing this is k l by d and k g and the other number heat h l by k so you have to arrange this beta bar by b b beta bar by beta in this format then you will get biot number okay good so now I think it is nice you prove that it is very nice very nice because I think you have to use again relationships from external the treatment for external effectiveness factor and then do it yeah which ratio is bi r oh sorry correct correct correct see abhinav reminded you abhinav is today controlling okay thank you okay good this is the equations 6 may be this is 7 this is only definition brinkman number this is yeah these two definitions so anyway so now what we do after deriving the equation we have to plot that is our life all the time after plotting you have to try to find out what is happening in that plot sometime research also is boring all the time you find something plot the graph and explain the graph and publish okay so that is why unless you have some kicks in any problem I told you know simply increasing simply decreasing will not give you any thrill okay so now this is delta x by delta t i also delta t o max what is the other parameter I have to plot here that is that equation 6 what is the other parameter c a c a is nothing but data bar d a okay c a is nothing but data bar d a and data bar d a also you know which you can measure and all that but anyway that is not required here this is c a and you have this side 0 to 1 data bar d a cannot be more than 1 I do not know whether you have checked earlier data bar d a all the plots are between 0 and 1 now go back at least you will go back next week anyway you have the exam okay so this is 0 to 1 the may be 0.2 0.4 0.6 0.8 is there 2 4 6 approximately 0.2 0.4 0.6 0.8 okay 0.8 and this side this we can plot as a percentage then we have 0 and 100 then I have here may be 20, 40, 60, 80 so if I plot this the parameter will be Brinkman number which is the ratio of mass to heat biot numbers so then you will have something like this I will put the parameter later this is 4 Brinkman number 4 this is 10 this is 20, 40, 100, 400 good so what do you predict from this nothing there are some lines that is all what do you predict there what is d T x by d overall thing what does that tell you that coordinate how much is the external external heat transfer okay that is why we plotted like that okay so now if the Brinkman is 4 then we have sufficient gradients inside and also outside that means in the film as well as in the particle but as you move towards Brinkman number higher numbers 100 so at this point for example you have almost 70 to 80 percent of the temperature drop within the film okay so that is why most of the time these things will come to very close to this so all the resistance will lie only within the film okay so that is what is the message what you have to give and generally Brinkman number 4 is very rare okay particularly for gases this Brinkman numbers will be 400, 500, 600, 800 like that okay large numbers so that is the reason why we neglect the external mass transfer heat transfer is there definitely heat transfer is the controlling and inside we have mass transfer controlling okay that also mass transfer control also I will tell you but quickly we will distribute now these things yeah I think I have to give Rahul come because there are two sets one set I will give and one other set you give yeah you give this I will give this yeah send how many back yeah extra Kavya you give to me you will get confused you got this that grab what I have just plotted give it to one to him and one to yeah yes good you see for example the this graph this graph you take this side A okay another what is on side A actually these are the measured profiles within the particle and also in the bulk measured okay beautiful work by butt and coey yeah you see there A is low Brinkman number so that is why there is some temperature gradient within the film and also there is temperature gradient within the particle thus radial position I say R A R by A 0 means centre the other side 1 means surface if you want you can write there bulk is written there right so it is beautifully demonstrated here that low Brinkman numbers you will have the concentration gradient I mean temperature gradients within the pellet and also within the film okay understood appreciated no not appreciated only understood okay good the next side next side is Brinkman number equal to 300 now what is happening inside the particle so that means what isothermal the particle is isothermal all the gradient is only within the film and this is so beautiful I think they have actually measured wonderful is really wonderful for chemical engineers okay so that is one information and the other information is this that is next graph this is the graph what we have I have drawn there see anyway we have I have explained what is that you know most of the time you will get the Brinkman number 400 500 like that you know yeah now see the table that is very important range of parameters what is that you are discussing no tell me tell me A graph is there anything wrong yeah what is the logic you tell me film is still contributing more but inside the particle also you have the temperature more gradient there yeah so that is why most of the time it is only the film which is controlling for as far as temperature is concerned okay yeah Prabhu the table what do you have in the table yeah so far what are the you know in the parameters normally this is a very good table for those people who are really want to do chemical engineering later okay you see the range of parameters are given for example diffusivities how they vary in systems like gas gas gas liquid gas solid you see now diffusivity coefficients for gas gas it is 1 to 0.1 meter centimeter square per second they are large okay and meter square also you can convert that you know 10 to the power of minus 4 you get there okay so then you see liquid 10 to the power of minus 5 to 10 to the power of minus 6 but porous particle you just see 10 to the power of minus 1 to 10 to the power of minus 3 increase it then liquids you know why the porosity what we are developing we are making the catalyst pellet and we allow the gas molecules to enter into the pores so that is why we cannot have very very small pores where it can be more than liquid right but liquid there are no pores it has to anyway diffuse between the molecules in the liquid so that is the reason why you know these numbers and of course thermal conductivity is given lambda is thermal conductivity k e what we used okay thermal conductivity range also is given and we have the rho c p value that heat capacity then we have arhenous numbers how normally they change epsilon, epsilon is arhenous number okay and beta is internal beta what we have beta x is beta bar because this is some other book there is a book called Li H H Li heterogeneous chemical reactors so I have taken from that this graph so I have collected information for way from various books H H Li chemical heterogeneous chemical heterogeneous reactor design okay then Brinkman numbers also are given you see there for gas solid system how they vary 10 to 10 to the power of 4 so if you take in between value how much it will be 10,000 no 10 to 10,000 so for many systems it may be in 1000 so that is why that line will be almost on the y axis covering 100 percent delta T x only in the only in the film so I feel this is very beautiful information other things of course dimensionless groups and all that is given okay okay so the other one I think may be tomorrow I will try to prove that you know for mass transfer always this is the internal but not external mass transfer controlling that I think I will do it okay now run and Levenspiel gave many problems in different notations but the same thing what we have discussed in catalytic 2 3 chapters okay 2 3 chapters so for examination you have to really thoroughly go through that tomorrow what I do is I just try to prove that always it is internal effectiveness factor sorry internal mass transfer which is controlling internal external mass transfer does not play a role at all because you will have easily molecules diffusing through the film so that we will prove and afterwards I will just tell you the laboratory reactors which they use for finding out observed rates okay so with that catalysis will be over and then we will go to packed bed fluidized bed slurry reactor if there is time and you have to prepare well for the exam because exam will not be easy because you have no so much information it is not that only module 1 module 2 like that from the beginning again till now