 I think we have done the effectiveness factors for the external step that is the film control now we will take the internal diffusion, pore diffusion step and then we will try to develop effectiveness factors for this. So we first anyway try isothermal as usual, isothermal interface effectiveness factor I hope you understand interface means inside the particle, okay. So we start with the diagram ya of course there is a film outside this is isothermal so only one side is enough for me to draw, okay I will use this side this is C B, okay also equal to C S because there is no film resistance from then onwards it goes into the particle decreasing, okay. So this C is a function of R where R is increasing like this, this is 0 capital R small r, good. So what we have to now do is the effectiveness factor definition we know ya Swami what is 1 minus definition in words actual rate of reaction divided by ya actual rate of reaction divided by rate at bulk conditions, bulk conditions C B, right good. So here also we have to find out now what is the actual rate that is going on throughout the particle that divided by you have the rate based on bulk conditions like for example if you are talking about first order reaction then you have here K into C B, okay here the actual rate that because C will be very continuously right it is less than C B isothermal, okay this is less than C B so of course temperature is constant throughout C B is decreasing or C inside the particle is decreasing the procedure is we should be able to first develop this profile use this profile to find out how much A has entered into the particle, okay let me tell a like a story first this we have to use to find out effectiveness factor, okay. C B I know at bulk conditions so if it is first order reaction or second order reaction K C B square K C B or 0 third order reaction only K so all these conditions are known to me here and here I do not know what is the actual rate so to find out that actual rate I have to first develop the concentration profile, okay it is exactly like your mass transfer only, right if I give you a concentration profile in mass transfer through any either liquid or solid can you find out what is the flux I will simply give you concentration profile equation C as a function of some constant into R square or z square or z means you know that distance can you find out how do you find out d C A by d C A by d R alone will not give you any into division, okay that will give you the flux that is the same concept what we are trying to use here. So first thing is to develop the profile inside the particle and afterwards you differentiate that profile evaluate at any place may be at R equal to capital R, right that will tell you what is the flux that has entered into the particle that will tell me that should have reacted under steady state conditions we are talking about steady state conditions so that will be the actual rate that I substitute there that equation divided by rated bulk condition if it is first order simply k into C B cancel out and you will get what is phi I mean what is eta so when you are deriving an equation for concentration of reactant into the particle then you will get some other parameter what is called phi, Thiele modulus, okay when you are solving the equation as a non dimensional parameter so the first thing is to find out the concentration gradient how do you find out concentration gradient by the way where the mass balance we have to take a thin shell inside this is R this is R plus delta R delta R is the thickness, okay delta R is the thickness that is delta R so now we will write what is entering into the shell what is leaving in the shell and what is reacting in the shell it is steady state conditions so if you write that material balance we have okay let me also write this one input equal to output we are writing for A if the reaction is for example A going to be right, okay output reaction plus accumulation all abbreviations R X N A C C N and all that accumulation because it is a steady state reaction so this one will be 0 no accumulation then here input is through the shell outside and we are taking the coordinate in the opposite direction so plus we have 4 pi R square D E D C by D R at R plus delta R, okay R plus delta R that is outside this equal to again 4 pi R square D E D C sorry always A comes yeah at R at R plus we have the reaction step where we have pi R square delta R into R rate rate in the shell, okay if it is first order it is K C so in general nth order also we can write but actually this is nothing but R, okay rate of reaction in the shell, right so that can be simply K into C A if it is first order if it is nth order that is equal to K into where R equal to K C to the power of N, okay so this is equation 1 I crossed already boundary so R this is not K to the power of R yeah good so now we will take the limit as delta R tending to 0 what you get the equation which is familiar to you in many subjects not only this subject transport phenomena also is like this D E D square C by D R square plus 2 by R D C by D R minus actually you will get minus R then that I will write as minus K C to the power of N this is equal to 0 so this equation is 2 so this is the differential equation which you have to solve for nth order and here that is why I was telling you know chemical engineers you have to use lot of mathematics this R need not be only K C A to the power of N it can be also K C A to the power of N divided by another K plus C A to the power of M whatever then you have to only go for solving this as numerical yeah some numerical analysis numerical technique you have to use to find out concentration profile using the concentration profile you have to find out actual rate this rate anyway you will be knowing based on the bulk conditions, okay so the procedure is universal but we are taking the simplest one to discuss many things during the during the class because we will understand what is going on in the pellet etc good so this is a secondary differential equation I think all of you will agree that yeah, okay good so then you need two boundary conditions so what are the boundary conditions you can take there R equal to okay V C is R equal to R okay at R equal to R C equal to C S which is also equal to C B, okay yeah Saitha getting crash course the whole particle is R small r is the coordinate small r is the coordinate what you do in any problem where we have distributed parameter system may be I am using slight type under words in any system there is a change with space only beautiful system is the mixed flow reactor where there is no difference in concentration or temperature spatially so that is called lumped parameter all space everywhere you see the same concentration same temperature so there you write normally the balance in the input that is inlet and outlet because you can see the change only in the inlet and then outlet, okay whereas in a distributed parameter system like like a pipe where the reaction is going on where there is no mixing so that means at every point in the pipe along the length you will have change in temperature as well as change in concentration for that what we do is we take a small element like this and then we write first what is really happening inside that in this element then you will end up all the time with a differential equation, okay so why differential equation because you are taking a differential volume and then trying to find out what will be the change in that and by the using boundary conditions this boundary condition will give you one is here on the surface what is happening the other boundary condition you need at the centre what is happening, okay that means the entire particle is covered because of the symmetry entire thing is covered, right so that is the procedure I mean she may not be knowing this procedure so that is why I am just trying to repeat others also may not know but again you know just once more I am just repeating that, okay so what is the second boundary condition at r equal to 0 dc by dr equal to y fluxes what is happening, at the centre there is no flux, yeah why in the centre there is no flux, at the centre why it is 0, at the centre always it is 0, concentration gradient is 0 concentration is 0, yeah good concentration gradient is 0 but why even is there any reasoning for that, you know the answer but I think the correct point also is there where it is difficult for you may be to imagine because we are assuming that we have symmetric diffusion from all sides, okay that is why it is called also symmetric boundary condition if there is no symmetry in the particle you cannot say dc by dr equal to 0 at centre because it is if it is something offset so because of the symmetric diffusion, okay then dc by dr dc by dr at r equal to 0 that is another boundary condition I am not going to solve this I just finally give you what is the concentration profile but now you have to use all your mathematics knowledge what you have taken in mtech and btech together mtech mathematical methods in chemical engineering also you have taken no, yeah idea of giving that is to solve these problems, okay if I solve if they solve if everyone solves I think you do not solve so that is the reason why we leave some of these things to you and examination you cannot write that you know solving this you will get the final equation you will not get any more so I will tell you if you write solving that you will get the concentration profile but I will write there solving using the boundary conditions equation 3 and solving this there are many other beautiful techniques to solve this another point I want to tell you here when I write this equation, okay here you can say that let us say for simplicity it is k into c for first order what should be the units of k you should be very careful it is heterogeneous system heterogeneous reactions always you should be careful with the units, okay under these conditions where I have written the equation what should be value of k I mean dimensions of k, k, okay you substitute and then tell me whether you get the same units what are the units by the by here it is the flux multiplied by area so what do you get mass flow rate or molar flow rate, okay yeah that is fine, okay moles per second only you will get now put everywhere here here here and then tell me what should be the units of k then why do you say meters per second, okay now I will ask further funda, okay why do you say it is per second what is the assumption there is an assumption mole per second intensive rate not extensive what are the units of r yeah you know substitute moles per meter square second and then find out what is k value, this is what is wrong first order reaction k is time inverse is only strictly valid for homogeneous or when you express the rate based on volume of the particle, volume of the particle you see how much discussion is unnecessarily required here but I feel it is not unnecessarily required because I know your attention is not here not today very day and not you know if I do not mention that you will not even bother to think you will not even bother to think and examination you will have definitely doubt why this fellow k is given as meters per second I do not know this has got many many units sometimes I can give even in you know if the rate is based on moles per second per weight of the catalyst then density of the catalyst also will come into k density units those things also so if you look into the literature of the various books so different books will give you different values for k depending on how they have defined the rate because it is a catalytic rate you can express based on surface area you can express based on the catalyst of the particle you can express based on the whole bed volume bed volume also okay so that is the reason why you may get different units for k value I think that is necessary that definitely I have to tell you that okay. So now solving this equation 2 with B C is equation 3 okay B C is equation 3 you get I mean what we expect here is concentration as a function of you should expect no Anand Kumar what do you expect from this equation what are the variables here by the by and really asking you like it all jailed I say yeah only one distance so what do you get when you solve this concentration as a function of R with the parameter k and DE okay now you could have got some idea this k by DE is the Thiele modulus for first order reaction okay because I have to divide the whole thing by DE then you will have like in your normal format where all of us learnt DI by DX correct no I do not know who started this DI by DX all mathematics no first we learnt the differential equations so that DI by DX form if you want to change so this one will be D square by DX square X is R here Y is C there because I think many people are comfortable when you say DI by DX the moment we say DC by DR mind of okay so that is why you have to convert this into the solvable differential equation by using even dimensionless quantities okay even this equation can be converted into dimensionless differential equation like taking for example DC by what will be the logical parameter to non-dimensional C B always C not because I think you know transport phenomena effect yeah here C not is C B that is one okay R capital R size of the particle so when you use all that you will get automatically Thiele modulus so please do that please solve that it is not only this I will give you later for slab if you have a slab like this where the reaction is taking place this side reaction is taking place this side and all these things are sealed for reaction so again symmetric diffusion from this side from this side so you take half of the particle and then write the balance within that shell like again same small shell write the differential equation right and then take the boundary conditions solve to get concentration as a function of R that is one cylindrical particle okay this side this side both are closed not available for reaction now the diffusion throughout symmetric diffusion now write the balance here if I look at this point you will have a circle within the circle you have the small volume what you take element there again you have to write symmetric diffusion take the corresponding boundary conditions again you have to get a differential equation take the boundary conditions and then finally you have to solve these three are important anything can be asked in a separate test or in the final examination always I have to blackmail you with examination otherwise simply forgot good yeah so anyway so when I solve this what you get here is C by CB CB equal to CS okay sin H now I think I will use another term here phi 2 yeah phi 2 and R by R this entire thing is in sin H oh sorry sin H is there hyperbolic function sin H yeah this divided by R by R sin H phi 2 this is equation 4 where I will write generally phi 2 will have some meaning for you know 2 we will give when it is a spherical particle and 1 phi 1 will be for cylinder phi 0 will be for flat plate okay so that is why we will just give that notation but in general where phi is Thiele modulus Thiele modulus for nth order reaction it is expressed as yeah expressed as R square root of K CB to the power of n minus 1 by D E this is phi okay again I will write phi value this is Thiele modulus Thiele is the name of the person we have proposed that okay and in fact it is not fair you know to give only credit to Thiele Thiele is from US there was a person from again Germany Damkohler and another person in Russia Joldovish all these 3 people simultaneously solved this problem in their own places because there was no internet nobody knows what they were doing at that time this all happened in 1936, 1938, 37 during that time okay all 3 of them and but anyway Americans will have always the loud mouth so then they will get this name is given as Thiele modulus but other people also have worked on the same problem diffusion reaction who are the 3 people Thiele it is one Damkohler yeah and Joldovish all these 3 people okay good is also some kind of history also is required you know okay good so this is the one and yeah you can also beautifully plot this C by CB verses R by R as a function of phi because phi is the only function there no yeah function is the Thiele modulus but anyway this is when you say it is Thiele modulus is there any physical significance of this Thiele modulus let us say I will say Thiele modulus I have 200 what is the meaning so which is controlling diffusion control that means diffusion is very very small in the denominator so reaction rate must be high so phi 2 is large value phi is large value so that is why large values of phi means we have mass transfer through the force controlling right similarly it is 0.1 or less than 1 or may be 0.01 very small values 5 values so then diffusion is very very large but the reaction that is k value that is small so that is controlling okay so these things like Reynolds number it will always give you some idea okay diffusion I mean Thiele modulus is very large usually it is 10 beyond 10 100 definitely you are in very very strong diffusional regime very very strong diffusional regime okay the effectiveness factors will fall down to even 0.1 0.5 0.01 okay that means only 1% of the surface area is used for yeah it is really funny also when only 1% of the surface is used is when the reaction is very fast correct no when the reaction is so fast that the moment all the particles all the reactant molecules will come to the surface it is converted so it is not even allowing that to go inside the force but on the other hand if you are able to push more and more inside then you will have very very high effectiveness factors because rate is very high okay if you are able to push but that will not happen most of the time that is why very active catalysts very active catalysts like platinum or noble metals okay these are only distributed on a support but not used as single particle you know the reason because all these catalysts are very very fast so that is why you and most of the time even if you take a particle only outside surface area is used the remaining internal surface area internal material is totally waste platinum is very costly okay so that is the reason why you now take this platinum try to deposit on a support like silica activated carbon uniformly throughout the particle in a porous particle so then depending on the pore size you will have the reactant gases you have to find out that the design of the particle so then you will allow more and more gas inside so then entire particle coated with this noble metal will be used okay good so that is why I think you know we can also nicely plot and show how the concentration profiles look like and I think I will do that in a small graph this is C by CB versus R by R this is the advantage of having dimensionless variables then I will have here 0.2, 0.4, 0.6, 0.8, 0.2, 0.4, 0.6, 0.8 okay good so now the parameter is phi okay if phi is very small how the curve look like you should have some expectation no come on come on I think I thought I have do I do many things but Karthik any idea Prabhu like this sir like what where it starts where it ends is telling like this Ribbaka or Nike or tic Nike no Nike or Ribbaka? Nike yeah tic yeah no thinking where are you? where are you? where to start where to end are you? where is 0? 0 is here starting from 1 okay that is 1 how much difficulty you know Kavya Kavya will solve most of the problems yeah tell me think from way to way you have to this starting and then you have to show the end point no at low R it is a dimensionless coordinate what you are drawing is nothing but this concentration profile right yes sir okay where it has to start from 1 okay yeah because R by R equal to 1 is that okay so that means it has to start from here now tell me how does that go if it is very very low then you may get like this okay phi and then if it is slightly larger you may get like this much more larger like this like this so for example this may be phi 2 equal to may be 0.5 1 5 10 or this may be or okay I think I will also may be this is correct 5 10 100 means like that so what is the meaning you tell me meaning for 10 yeah before it is going into the particle may be around 50 percent of the particle only is occupied by the gas phi 2 y diffusion limitation reaction is fast if it is 100 almost near this much so this is the one why mass transfer limitation reaction is very fast beautifully one can predict I mean develop that kind of interest I say so many times I have telling you know instead of seeing google you can take the pen and paper and start drawing this kind of things and just imagine what is happening in the process and then try to write then automatically mathematics you can write and then check whether you are right or wrong anyway in the examination I can ask you if you are not doing in the assignment and all that so draw the profiles do not scold me at that time okay I will give you graph paper pencil I do not give you okay so then you have to calculate and you have to draw the profiles and explain what is the meaning of 10 5 2 and 0.5 if it is 0.1 you may get almost yeah what is the meaning reaction control the entire particle now is occupied by the gas so entire surface area is usual for the rate of reaction for the reaction to go on wonderful points good okay so now once we understand this concentration profiles now we have to take one concentration profile if this is the concentration profile at particular conditions given here in the in this when you are writing the balance and now you have to differentiate this at r equal to capital R this equation dc by dr at r equal to capital R and then multiply that one with de that will give you the flux like here okay but you are evaluating at r equal to capital R multiplied by de and 4 pi r square also and then here rate at bulk then you will get effectiveness factor because we are using simply this definition definition of effectiveness factor is actual rate divided by rate at bulk conditions now actual rate is what because the concentration is changing at each and every point you have to either integrate or differentiate one of the things so that is why I can give you two methods eta can be now found from actual reaction rate as 1 by r 0 to capital R yeah 0 to capital R kc dr divided by kcd if I cancel because isothermal if I cancel this and this k out what is the equation you get yeah also write here 1 by r integral 0 to r c dr by c b it is just nothing but concentration average integration throughout okay some kind of averaging this right yeah so that is the actual rate because here also you know kc is rate kc is the rate kcb is the rate okay so this is one method the other method is this also can be written as 4 pi r square de dc by dr at capital R that is the flux multiplied by 4 pi r square will give you the rate okay this divided by 4 third pi r cubed kcb either this equation that equation this is 5 this is 6 that is equation 7 any one method can be used both are same right yeah here why did you use 4 third pi r cubed and why 4 pi r square because that k is expressed based on volume of the part you have to be careful okay good so please do that this I am not doing again the differentiation or integration any one method you can use and then what you get if I substitute let us say from equation 7 that is easy for me normally Eta will be 3 pi 2 1 by tan h pi 2 1 by pi so this is equation 8 this is for spherical particle this is what we have started if you do the same exercise for cylindrical particle for cylinder Eta will be 2 by 5 1 so this is equation 9 for slab Eta is tan h 5 not by 5 not so this is equation 10 tan h yeah and any one remember what is that I 1 I 0 who said who said Bessel function you said how do you know we teaching you know multiphase push one okay yeah this is Bessel function why did you get that in effectiveness practice also you are doing there something else yeah so in this equation you know d square C by dr square plus 1 by r you will get for 2 by r for spherical particle right 1 by r for cylinder so you do not have a solution except you know you have to go for this Bessel function and I 1 is the modified Bessel function of first order I 0 is modified Bessel function of 0th order okay I have to write here all other things you know where I 1 is modified Bessel function of first order okay and I 0 is modified Bessel function of 0 order what you have to do is when this kind of problem is given if I ask you to calculate what is effectiveness factor you have to go to mathematical tables okay so in the mathematical tables you have to calculate 5 or 5 1 here and then go there correspondingly substitute in that constant and then get the corresponding values okay good so this is the one for 3 geometries oh I think I have to stop here okay I think we have to draw the figures and all that later