 in the last class we have been deriving generalized non-isothermal effectiveness factor for external mass transfer control step alone right external mass transfer control step alone. What was the last equation we have written was this Eta bar equal to exponential this multiplied by 1 minus Eta bar dA to the power of L okay Eta bar dA is a one group of course Eta bar what was the 12 equation 12 then we thought we have to still develop or write this expression in terms of observables because this T is T s by yeah T b where T s is still not known to me so that also we have to write and in terms of observables and for that what we have to write is the energy balance around the particle. So the energy balance around particle I think let me give this continuation otherwise you may not know what is happening in between yeah equation 12 cannot be used to calculate effectiveness factor since small t equal to T s by T b is not known right yeah to get surface temperature T s we can write the energy balance at steady state as heat generated due to surface reaction this is equal to heat transferred by convection sometimes radiation I think effects of we are neglecting okay only by convection we are taking. So now if I write heat generated by reaction is we know delta h r into rate of reaction okay r b this is equal to r r b r r b observed not b b means bulk we will say yeah observed r b yeah this is equal to heat transfer by convection that is h A T s minus T b if it is endothermic reaction that will be the reverse okay and that is also taken care of by this side minus delta h r and all okay so that is okay no problem there so this is equation 13 yeah equation 13 but now you have to be careful about this units this A is again meter square per meter cubed of the catalyst right so that is why heat transfer coefficient will not have straight forward you know the units you have to be more careful there in the examination I may give I may give constituent units but the constituent units will be different because of A and also delta h r can be expressed as calories per gram sometimes calories per mole also sometime so unless you know again that conversion all that you will make mistakes so that is why you have to be more careful good. So now we can use this equation and then start writing that if I have okay if I divide this one by k g b I will write here h A T s T b divided by I am just bringing this side k g A C b T b I divide this side this so then this side also we have to divide minus delta h r r o b divided by k g A C b T b right so now we can also write this one as minus delta h T b is also minus delta h r T b as one term the other term is r o b by k g A C b you know what this term is this is observable this is eta bar d A okay this equation okay let me also write here h A T s T b by k g A C b T b okay good so if I write if you just take only I think I will go here T s minus T b by T b equal to this I am just leaving so those things I am taking that side so then you will have here minus delta h r k g A C b divided by h A h A T b into r o b divided by k g A C b so this is another term for us yeah so some more also can be I mean this is enough for us but still we can use some more simplification taking k g and h A together there is a number also which relates Prandtl number and Schmidt number Lewis number okay good yeah so we will also try to write that because it should be as simple as possible for us for the use so that is why I really appreciate Carberry is the way he has done it is really wonderful so now you understood the idea no this is equation number 14 this also I will put 15 yeah okay so then that will be 16 so our idea now is whether we can use some other thing for k g A or h A A will get cancelled there so you will have simply k g by h okay so that is the term you will have and now for that we have to use the relationship J D too much to ask you for the definition of J D J D represents the mass transfer J D factor okay and J H represents heat and heat and J D equal to J H of course for the analogy wise J D equal to J H equal to f by 2 that analogy so but J D is k g by U U is the velocity and Schmidt to the power of 2 by 3 which is also equal to J H now J H definition is H by rho U C P into Prandtl to the power of 2 by 3 J D equal to J H good so now this can be also written now as k g we are taking only these two this and then this so if I take those two then k g by h h equal to we have Prandtl to the power by Schmidt whole to the power of 2 by 3 into 1 by rho C P this is another term okay yeah so this equation we will say 17 now we can substitute this in equation 16 because k g by h k g by h A A will get cancelled right k g by h then what do you get T S by T B by T B T S minus T B by T B you will get as minus delta h R C B rho C P T B Schmidt by Prandtl to the power of minus 2 by 3 yeah into into R O B divided by k g A C B correct or something missing you will get it now this is nice we also have now this okay I will also write this one as simply T S by T B equal to okay before that T equal to small t equal to T S by T B this is 1 plus minus delta h R C B divided by rho C P T B that is one group and this is 1 by Luise to the power of L E is Luise number into Eta bar D A so this is the equation this is equation 19 that is what is T S so we will write this entire thing as simply T equal to 1 plus very beautiful form okay beta bar that is supposed to be beta yeah beta bar Eta bar D A very simplified so this is equation number 20 where beta bar equal to minus delta h R C B divided by rho C P T B into 1 by Luise to the power of 2 by 3 yeah where of course Luise number Luise equal to Schmidt by Prandtl number this is the number which relates heat and mass transfer right that you know good so this is the one so now we have to substitute equation 20 in yeah here in equation 12 okay beta bar is it observable all those parameters we know okay yes so that is why now if this T is replaced by 1 plus beta bar Eta bar D A okay Eta bar D A then we have equation 12 everything is observable beautiful good so substituting equation 20 in 12 equation 20 in 12 what you get is Eta bar equal to 1 minus Eta bar D A okay whole to the power of n exponential minus epsilon naught yeah epsilon naught 1 by 1 plus beta bar Eta bar D A minus 1 this again yeah so this is the equation this is equation 21 beautiful derivation I like the derivation the reason is that you know whatever we studied in heat and mass transfer everything is used here for K G K H Luise number all these things you know independently you would have used here and there but here in this derivation so beautifully done so that you know I have this measurements R O B divided by K G A C B then we have beta bar again I know all these parameters and that is all what else we have epsilon naught which is nothing but your Arrhenius number where it is epsilon by R T B T B is the bulk temperature that also I know so then I can calculate what is Eta whether it is first order reaction second order reaction whatever order reaction good so I think we will plot those things later and then for discussion because just writing equation is not enough we have to plot and try to understand by the by how many parameters we have here Eta bar and D A will be Y axis and X axis Eta bar and this is observable right then I have epsilon naught as one parameter Eta bar is another parameter two parameters what Karthik looking clueless N also is a parameter okay thank you three parameters N also is a parameter so now we have to plot you know these what is simulations I say when you have the when you have derived the equation then you have to see what are the parameters and then try to of course dependent independent these two we have to first take then all other parameters you have to vary and then see how the effectiveness factor is changing with other parameters so that means first we have to vary Eta keeping beta beta bar and epsilon zero constant then you have to keep Eta constant N constant and then vary of course N and epsilon zero constant may be beta bar varied okay and of course again you keep Eta sorry N and beta bar constant and epsilon varied okay so these are the graphs which we will do and before that we can also very nicely use this information particularly equation 20 yeah equation 20 can it be the beta can it be negative or positive can it be negative yes it can be negative it can be positive positive means exothermic negative means endothermic please remember that so that also has come into picture good yeah and equation 20 will give me the relationship between concentration gradient and temperature gradient can you just look into that and tell is it really possible equation 20 t small t is t s by t b okay so if I bring one this side I will have t s minus t b by t b good yeah so that is the temperature gradient so the other side must be I should have concentration gradient find out where do you get that if you look into your notes you will get at all difficult yeah so here we have t minus one which is nothing but t s minus t b by t b equal to Eta bar sorry beta bar Eta bar d a where Eta bar d a is nothing but 1 minus yes beta bar 1 minus C s by C b beta bar right so for if I know this parameter beta bar then I can also calculate what will be the increase in temperature right and why we are doing that I think there is a reason for that okay so beta bar values we know and for okay we have to also find out what is the maximum temperature that is possible t t t s minus t b max maximum temperature difference or temperature gradient possible when do you get that from that equation okay so you will get the maximum temperature when t s equal to zero maximum temperature gradient when not here C s equal to zero then what we have here is this equation if I take this equation yeah then once only you have t s minus t b max equal to this t b I will take this side t b into beta bar right okay so beta bar we can substitute now beta bar equal to t b beta bar is minus minus delta hr C b divided by rho C p t b Lewis to the power of 2 by 3 so t b t b we can cancel out and for many gas phase for many gases okay gases Lewis number is approximately 1 for many gases that means Prandtl number and Schmidt number is same almost same okay so now you will have equation t s minus t b max which we call delta t max equal to what do you get yeah minus delta hr C b by rho C p so this is equation yeah this equation number I have to give 22 this is 23 yeah so now I can write this equation equation 22 as t s minus t b equal to delta t max of C b C s by C b so this is equation 24 okay simply I substitute it here that is all so what is the use of that equation 24 do you have any use or equation 23 apart from nicely writing equations is there any use of that or turn around equation 23 is there any use of that I will give you control the reaction sorry that is the equation but what is the use of that how can you control the bulk concentration okay control the bulk concentration what do you do what is required temperature how much I have committed what are the operating conditions yeah that is what finally you are slowly going okay good you know there are what are called hot spots what is the difficulty in hot spots why should we know why should we know hot spots yeah not only explosion there are many other things you know catalyst may get quiet okay so that is why we have to operate itself we can try to calculate whether this temperature is dangerous to either catalyst or deactivation may take place or explosion may take place so that is why in the beginning itself you can have an idea how to calculate I mean what is the temperature maximum you may get so then you should definitely you should not cross if there is a dangerous line like okay let us say 500 the difference and may be around 500 your explosion may start the reactants may explode or the catalyst may get damaged because of thermal sintering all these things we will know and when you get this estimation then we will know that okay now we should be safely around 450 you can operate otherwise always more the temperature more the rate of reaction you will be very happy okay but this will give you an idea how to calculate the thermal sintering or explosions of the other reactants all I mean products all that that is why every step what you have done is not waste of time okay so most of the time we will be happy to derive the equations but afterwards we do not know what to do where do you actually use so that is the reason why I am trying to explain all that so this is the overall picture of this derivation of non-isothermal effectiveness factors for external mass transfer control alone we have taken only taken one step right that one step alone is the mass transfer controlling step the rate of reaction as well as the if there is a diffusion there is no diffusion here because mostly non-porous even if there is diffusion diffusion also very very fast that means all surface area is freely available for you okay good now let us plot these graphs and then try to interpret any extra information you get from all these equation 21 we are now trying to plot various ways different parameters okay hopefully I do not need the other side graphical representation of equation 21 yeah so the first graph what we plot yeah is Eta bar versus Eta bar dA Eta bar dA is R O B K G A C B just to remind you once more so here we will start with log scale 0.1110 this side you cannot have more than one so this starts with 001 then I have 0.01 0.1 and 1 yeah so this is the coordinates so then we have yeah first isothermal we will take so isothermal means this will not be there no okay isothermal so if you plot isothermal okay it is just qualitative graph only so then I also have something like this this is n equal to minus 1 this is half this 1 equal to 1 this 1 equal to 2 okay very easy to plot this on XL right so only thing that changes is n and this is isothermal isothermal okay I will write Eta bar effectiveness factor good so let me also plot the other one so this is again Eta bar then we will have this is Eta bar dA I do not have to repeat every time R O B so this is same again 0.001 0.01 0.1 and 1 this side I have 0.001 8 I should get here okay 7 this is not not 1 0.01 0.1 this is 1 10 100000 okay yeah okay it can also go to 10,000 that is Eta bar so the parameters which we fix here is that we have epsilon not equal to Arrhenius number equal to 20 and Eta bar equal to 1 plus 1 that means exothermic reaction exo let us see how do you plot so this goes this is very nice to plot I say in the room also if you can plot this goes like this goes like this it goes like this so here the parameter is n equal to minus 1 n equal to half and n equal to 1 may also have something like this for n equal to n equal to 2 as usual I am dangerous so I can also ask you to plot this in the examination hall really why not okay and then interpret what is happening good so this is n equal to Eta bar equal to 1 so that means what you can say about this non-isothermal non-isothermal Eta bar okay what is the effect effect of this is what what you do in your research also effect of what what is that we are showing effect of order of reaction effect of n order of reaction good so now we have many more graphs like this we have now shown for that non-isothermal equation 21 the effect of n then we have to show effect of beta and then we have to show effect of epsilon okay that we will do so then all these are non-isothermal graphs this is Eta bar Eta bar dA so this side it is same 0.001 somewhere here 0.01 0.1 1 so here this is 0.1 1 10 and that right so the parameters what we fix here is epsilon not equal to 10 epsilon not equal to 10 n equal to 1 first order so then we have other I mean beta as the parameter so when you have the beta as parameter so we will give this number later I mean this parameter values later then we have so this is beta bar equal to minus 0.5 so this one equal to minus 0.2 this one minus 0.1 okay good so the so I think out of the graph we have to go okay this is beta bar equal to 0 this is beta bar equal to 0.1 this is beta bar equal to 0.2 this is beta bar equal to 0.5 this should be most horizontal line single line okay anyway you are going to plot all this in the computer okay using the computer okay so this one is effect of beta bar we will again have epsilon equal to 20 I have not written to the scale here I think we should not go touching there okay so exactly similar graph we will have eta bar dA 001 not 1 0.1 1 this is eta bar 0.1 1 10 100 so if I plot here also you will have so yeah so this is beta bar equal to 0.1 0.2 yeah so similarly this side I have beta bar equal to minus 0.5 why something wrong 0.3 yeah parameters I will write here epsilon not equal to 20 right and n equal to 1 so what is the difference between this graph and this graph this is y epsilon so that is why here approximately qualitatively this is going to almost may be around 70 80 and here it is crossing even 100 actual values when you plot you will know so this will tell us that always the effectiveness factors when I have epsilon 0 equal to 20 you will have higher effectiveness effectiveness factors the effectiveness factors values will be more if you take epsilon not equal to 20 when compared to this these are the same parameters and beta negative means endothermic and positive means exothermic okay now let us just see from this picture onwards what do you predict from this we know that as eta bar is increasing eta bar dA is increasing eta bar decreases eta bar dA or otherwise we will call that one as carburet number as carburet number is increasing we will have here eta bar falling why what explanation you tell me the physics of this one is it correct when a eta bar is increase I mean this observable carburet number is increasing this is effectiveness factor decreasing which one is controlling K gA is increasing means I mean yeah it is increasing means K gA is decreasing that means mass transfer decreases so that is why effectiveness factor that means there is no sufficient amount of mass going to the particle for reaction that means okay here are almost everything same and one that too which one is controlling here reaction is controlling there is sufficient amount of mass transfer but reaction is controlling and because amount of mass transfer is same throughout the particle okay so that is why all the surface area is occupied so then I have effectiveness factor equal to almost one so then as you increase this then K gA is decreasing right so that means all the surface area is not available effectiveness factor is falling that is another observation these are what you have to write in your research also after simulation or after doing some experiments when you plot the against the parameter this is what what you have to discuss okay thoroughly and then when you take n equal to 1 2 1 minus off and all that what can you say first of all n is negative always gives effectiveness factors more than one correct one or more why this why is very difficult I say every time you have to say why yeah that is the beauty in negative orders no so as the reaction as the concentration is decreasing rate of reaction is increasing that is why always it is very nice funny reaction okay so all positive orders as order is increasing you will have less and less effectiveness factors can you tell what is the reason concentration how do you say concentration is needed yeah that is true because the concentration is decreasing correct no when you come this side concentration is decreasing okay on the surface so when you have squared and cubed reactions like that definitely you will have that effect more for the rate of reaction so that is why it is decreasing with you know number all the observations here also same thing but here what is the striking thing here all the effectiveness factors are greater than one why n are greater so that means the temperature in the particle or on the particle is higher so rate of reaction is definitely higher and then you will have more and more rate there because the temperature is higher so that is why you are getting that all the things but now with order again same thing second order you will have less effectiveness factor first order half order and negative order that is why it can even go to calculations wise even 10,000 effectiveness factor when you actually calculate you will know that good but these are the parameters and also we have to tell here that here epsilon not what we have taken is high value see the parameters also is very funny always we think that we have to change from 0 to infinity or parameters it is not so like you see here one peculiar thing what is the maximum value of beta what we have taken 0.5 it cannot be 100 because physical parameters will not be there in the definition of beta bar here for many systems you will never get that kind of very large values like 100 200 like that it will be only around 1.5 0.7 0.8 like that these are the real physical parameters for many reactions that were taken and then plotted otherwise if you do not understand those physics then what you vary is from 0 to infinity so 0 to infinity you will get very funny funny readings so that is why one has to be careful when you are choosing these parameters and effectiveness factors also it is not thousands okay maximum most of the time it is less than 100 epsilon 0 100 means very very very very high reaction that means energy activation energy is very high a small change in temperature will be very very sensitive okay a small change in temperature you will have the reaction is very very temperature sensitive so all these things which you learnt already also we can use that for discussion here right and only thing when you compare this and this always another thing beta bar when it is negative always you will have reaction I mean the effectiveness factors less than what that is also there right why why anandukumar why for endothermic reactions always you have effectiveness factor less than 1 whatever always delta T between always delta T what is the temperature on the you have endothermic reaction what is the temperature on the surface is it less than T b or more than T b so naturally that will be having low rate of reaction so you will have that okay good so this is what this is what is that we can discuss and then of course here and here the difference is that this will give you a little bit low effectiveness factors and this will give you a little bit high effectiveness factors because here epsilon not equal to 20 so tomorrow if I give a separate test and ask you to draw these graphs qualitatively and then give the explanation you have to give all this but I can give you know what is wrong in that okay it is a qualitative you have to just see and then try to plot these graphs and explain each and every graph okay on these graphs on 45 points on this graph from 45 points okay this is in fact is very good training for research also this is what you do in your project whether you simulate or whether you do experiment or whether you do both you have to write only on the graphs like the discussion what we have done till now okay good with this I think this effectiveness factors are external effectiveness factor what we call interface effectiveness factor so the discussion on that is over but only thing is there are one or two points which you have to think what are the practical applications of this is there any practical application for this equation that means where I can use and calculate is there any practical application what you know what kind of particles we are talking here most of the time we are talking about non porous particles okay so is there any reaction where non porous particles are used as catalyst in all those places it is useful yeah so example which I have been telling already yeah ammonia is oxidized to NO for nitric acid production okay one example practical example you can take NH3 is oxidized to NO NO is nitric oxide nitrous nitric oxide over platinum catalyst wire meshes that is one and we also have another reaction synthesis of very dangerous hydrogen cyanide any cyanide is dangerous good not explosive poisonous HCN hydrogen cyanide synthesis of hydrogen cyanide from NH3 O2 oxygen and CH4 over platinum wire again over platinum wires wire mesh just repeat synthesis of hydrogen cyanide from NH3 that means reactants are NH3 oxygen and methane CH4 okay and catalyst is platinum wire mesh okay next one is third one is methanol oxidation to formaldehyde over non porous silver this is the reason why we teach you chemical technology chemical technology all these equations are supposed to come okay yeah so then the fourth one is very famous one where all of you know but you may not get the point now for cars you have that catalytic converters in catalytic converters also same equations are used why wire mesh is not put in catalytic no it is a monolith I have also drawn earlier I say it is a monolith where you will have it is a cylinder like this okay then you will have those are again platinum coated and you will have some other it is actually mixture catalyst so there will be many because you do not have only CO CO2 to CO2 it has to go then all other hydrocarbons also they should be decomposed to hydrogen and all that so that is the reason why it is a mixture catalyst and now see if I look at one box here okay so then you have the reaction taking place on the surface it is just a sheet one sheet another sheet another sheet another sheet like that so all the reaction is happening only on the surface it is not porous so then you have only film control only film is there so now under film control you have again same thing same equations valid so that is why fourth one is in automotive exhaust catalytic muffler in the automotive exhaust to catalytic muffler MUFF LERA muffler the oxidation of CO and hydrocarbons oxidation of CO carbon monoxide and hydrocarbons at high temperature occurs on the external surface of catalyst on the external surface of catalyst these are the examples that means what we have done is not waste but I think you know still it is useful for us just to tell you that you know all this analysis is a wonderful analysis because wherever you have this kind of mass transfer controlling alone film controlling this is the analysis what we have to do to find out the effectiveness of catalyst as I told you what is the use of this effectiveness in the beginning I also mentioned that first of all you will have an idea whether I have an effective catalyst or second is if I want to use that R O B R O B yeah and if someone gives me R intrinsic R based on bulk that means there is no mass transfer effect so then I can simply write R O B ETA bar R B and if you are using this information to design a packed bed catalytic reactor then we have W by F A not equal to 0 to X A D X A by minus R A my minus R O B which is nothing but ETA bar ETA bar R B which is simply nothing but ETA bar R B so now ETA bar equation you have to substitute here if it is isothermal if it is non isothermal you have to substitute this equation and you need here this is the mass balance equation you will have the temperature balance also that will tell you because finally this has to be converted everything in terms of X A for the integration right even though I have non isothermal it is temperature change okay so that is why to convert that we have all the relationships like including concentration yeah this equation 22 which relates concentration and temperature all that information you have to use here to integrate this expression and ETA bar if it is non isothermal you will have L because you have this equation okay so that is the reason why we have not done anything useless everything is practically useful if you want to design if you do not want to design and go to I T no problem okay yeah so good we will stop here and then tomorrow we will start the normal you know diffusion second step if diffusion is controlling through the pores how do you now again develop effectiveness factors and then both combined how do you develop okay