 Now we will start deriving the continuous I mean we will continue shrinking core model derivation. We have written three equations that again I will write here for connectivity R C equal to 4 pi R square C A S this is 1 R C square D at R equal to R C this is 2 again 4 pi R C square K S C A C this is equation 3. So now our general procedure is to eliminate the intermediate concentrations and then express this in terms of only measurable quantities like K S D E K G and this concentration that is the usual procedure right. But here we have a problem we have D C A by D R where I cannot easily eliminate using like C A C and C A S so we have to develop an equation for D C A by D R in terms of C A S and C A C right. So that means we have to have a concentration profile C A versus R R is the radial distance and then you differentiate that and evaluate at R equal to R C then you will get the concentration C A C and also C A S right. So using that you can eliminate the intermediate concentrations. So now the question is how do we write how do we derive an equation for C A the concentration profile of A through the ash layer right okay. So that is why we first draw this picture for ash layer so this is the particle then we have here B that is the sinking core actually and here we have the film so this is centre this is R equal to 0 I have capital R here yeah. So then we also have R C this is capital R yeah. So if I plot the profile which we have done many times this will be C A S and next one will be yeah this is C A as a function of R and here I have C A C this is the profile what we require if I have this profile then I can differentiate and evaluate either at R equal to R C or at R equal to capital R so that is okay I mean that is left to us whichever is convenient we can do that how do we do this that means when A is diffusing from surface to the core through the ash layer ash is inert there is no reaction. So it goes as if there is an inert porous body through which A diffuses right so we have to write the diffusion equation for A going through ash layer okay let me also write that this is ash layer ash layer and of course this is film okay so now through the ash layer when you are writing the equation again I have to identify a shell like for example this one yeah so what we do here is we have from this is R and this one is R plus delta R okay this is R plus delta R so that the thickness is delta R for me good yeah now how the diffusion is taking place diffusion is taking place all through like this okay maybe just we draw few that is the diffusion that is going through A right able to follow okay yeah so now I have to write what is the amount of A that is diffusing through this line that is the shell actually shell boundary through that that means through R plus delta R then it is coming through this then I have R and what is the rate of reaction if there is rate of reaction or what is accumulation right so mass balance for A I think we better take this one the concentration gradient in equation 2 the concentration gradient in equation 2 DCA by DR at R equal to RC can only be evaluated if the concentration profile is known through ash layer through ash layer the derivative of the concentration profile at R equal to RC derivative of concentration profile at R equal to RC will provide the equation in terms of CAC and CAS CAC and CAS yeah so then the material balance the material balance of A the shell of ash layer can be written as can be written as okay input input equal to output plus reaction plus accumulation I am using the short form so reaction equal to 0 for the reason that ash is inert and there may be accumulation through the shell and if I write this this is diffusion so it is only pixelar right diffusion through the shell so that is 4 pi R square DE DCA by DR at R equal to R plus delta R okay that is 1 okay I think R equal to I do not have to write at R plus delta R this is also equal to 4 pi R square DE DCA by DR at R that is output that is coming out okay input output and reaction equal to 0 and accumulation is yeah 4 pi R square delta R DCA by DT yeah I think actually I have to write DOE here okay yeah but anyway I am not writing DOE there because I will tell you that one of our assumptions is pseudo steady state okay so that means I have this one pseudo steady state means there is actually steady state but it may not be pseudo but why we call that pseudo steady state I will explain so this also we will take it as 0 based on assumption of pseudo steady state okay I will explain to you what is pseudo steady state later right okay so this is the one please see that in the equation this is equation 4 accumulation term is neglected based on the assumption of pseudo steady state theory right so what do you mean by this pseudo steady state theory I will just explain and mathematically also we can prove that later so this is the one what is our idea we have to evaluate what is the concentration here and also what is the concentration here so that I can now use those two concentrations in this equation and eliminate CAS CAS and here finally express in terms of CAG okay that is the idea so now when the diffusion is occurring reaction also taking place when reaction is taking place this core is moving that is going inside so when it is going inside then I do not have a fixed boundary correct now this is fixed anyway here diffusion is taking place and we have to integrate from here to here R to R C but R C is not constant R C is continuously moving so the pseudo steady state assumption state says that if I have the diffusion velocity through this ash much much much faster than may be thousand times faster than this shrinking core velocity core also is shrinking so if that rate of shrinking should be thousand times smaller than velocity of diffusion if that condition is satisfied then practically what happens at any time when the gas is coming this is practically constant that is what is pseudo steady state and this condition can be easily used for gas solid systems because the gas diffusion through porous layer is much much faster but this condition cannot be used for liquid solid systems like for example it can be liquid you know in over leaching over leaching also is done you know you take a over and then put sulphuric acid or hydrochloric acid now this hydrochloric acid has to go inside and then react now the diffusion of this hydrochloric acid through that is not that fast when compared to core so under those conditions we cannot assume pseudo steady state for liquid solid reactions why because the densities are comparable whereas the density of gas and then density of solids will be thousand times okay yeah under I have any idea what is the density of air at room temperature under all these only one okay this is normally the problem in all classes you know when I ask a question X X will not answer all other people will answer okay yeah and for solids normally it can vary a lot for sand what is the density because every day we see sand somewhere other probably you have any idea density of sand may be around thousand two thousand thousand or two thousand order of thousand means yeah coming up fourteen hundred fifteen hundred come closer yeah Renitha sixteen hundred it is like auction okay we is going higher and higher it is two thousand six hundred this what is unfortunate with most of us because as engineers we cannot really estimate you know we will not have anything in our mind about some values like for example what is the diameter of this check piece? five mm five mm five mm opening I am happy you are not told one kilometer anyway yeah so that is what is the meaning of the pseudo steady state I think you know in a very simple terms I am explaining that okay now our idea is to again correctly get what is CAC on the surface if CAC is okay if the score is moving CAC also continuously changes but under pseudo steady state conditions the velocity of diffusion of this gas is so fast at any time this profile is established at that instant of time so that means this is constant at that instant of time it is not moving but whereas if the velocity is diffusion velocity is smaller comparable with core receding velocity then you cannot assume that it is exactly like you know there are many beautiful this is called moving boundary problem okay there are many beautiful problems like one problem is your eyes you know when you are taking a drink you put eyes you know yeah how the eyes boundary is slowly moving there is heat diffusion right otherwise it would not melt you know yeah so that problem also you solved using not pseudo steady state theory but you know you have to take the actual this term also into account mathematically it will be complicated but particularly for gas solid reactions this will not come into picture we can prove that later because normally it happens if the density of that is why I was asking. So typically if you have either 1000 or may be 2000 or sometimes 4000 for solids like you know iron ore and all that iron ore density will be around 4.8 iron ore okay so we are using also iron ore for the catalytic non catalytic reaction right what is the reaction to make ore yeah what is the reaction so many times I have written there is iron ore yeah Fe2 yeah good balance is awesome it will do those numbers okay anyway I am happy atleast you could tell that this Fe2 O3 is it magnetite or hematite there are two ores you know why you care why you say hematite I remember always H comes first so Fe2 O3 what is the Fe3 O4 magnetite okay anyway so this that is also one of the reactions where the density is 4 or 4.8 or so so that divided by 1.2 will give me again you know 2000 times also so that is why most of the time it is very well valid but it is also not valid if there is very high pressure gas solid reaction if the pressure is very high you know again density P is yeah Prabhu what is the equation you said for density Pm by RT so the P is very large for high pressure reactions then rho become very very large if it is 100 then again it may not be you know 1000 times even with 2000 3000 if you take so that is the reason why it may not be valid mathematical also we will prove that after some time okay so now when I take this equation equation 4 and this is 0 and that can be written as 4 pi r square d e dca by dr at r plus delta r minus 4 pi r square d e dca by dr okay at r equal to 0 no one ask me why I have not put a minus there only I have to answer questions and I have to question myself and also give the answers this is equation 5 why I have not put minus there here no normally the diffuser equation will have minus no why Renitha any idea Renitha is only one you are not Renitha excellent j e solution all of them are minus you do not have to write no there is a wonderful physical significance for that by the by why do you write minus there why do you write j flux equal to minus d e dca by dr what is flowing negative side what is negative there I mean I know you are now searching for words but may be you understand tell me what is happening the concentration how the diffusion takes place from high concentration to low concentration or yeah but how do you define differential yeah because you know it is always we remember why 2 minus why 1 right so but I think this is a physical quantity it cannot be negative flux so that is the reason why we put minus there to take care of dca by dr which is negative the way we define mathematically but here you see my coordinate is increasing in this direction right and it is in the opposite direction what we write diffusion so that is why we do not write not simply everywhere minus and then I cancel it okay it is the coordinate system which I have chosen here it is in the opposite you know the diffusion is in this my coordinate is increasing in this side in this direction so that is the reason okay anyway so this equation if I write in terms of differential form r square d e equal to 0 so this is equation 6 so now this has to be solved to get the concentration profile so now I need to boundary conditions what can be the boundary conditions here ca equal to ca s where at r equal to capital R ca equal to ca c at r equal to rc small rc so this 2 is equation 7 I am not giving the solution sorry I am not showing the steps of the solution and if you are able to integrate this twice and then substitute the corresponding boundary conditions what you get is ca minus ca c ca minus ca c ca c equal to 1 minus rc by small r divided by 1 minus rc by capital R or for uniformity I will also write here divided by so this is the equation right actually the concentration profile which I draw here can now be plotted if I know these values ca versus r okay that gives me exactly the shape whether it will be like this or whether it will be like this okay that gives me the correct mathematical yeah from this mathematical equation you can also draw that provided I know what is r what is small rc and ca s and ca c ca versus r we can plot that okay good so now this has to be integrate sorry the equation 8 has to be differentiated and evaluated at r equal to rc okay so now equation 8 differentiated with respect to r with respect to r and evaluated at r equal to rc okay so this I will give you in my separate test but then what you get equation is okay so that equation is dca by dr at r equal to rc is ca s ca c divided by rc 1 minus rc by R this is equation number 9 now I have equation 9 we can substitute here equation 2 then you know this is ca s ca c I can now eliminate intermediate concentrations right so I can also write this substituting equation 9 substituting equation 9 in 2 what you get is minus dna by dt equal to 4 pi de yeah okay ca s ca c divided by 1 by rc minus 1 by R that is the one so this is equation 10 so now the equations what I have 1 by rc minus 1 by R that is correct only we can just check yeah so this is right okay so now the equations what we have to solve is equations 1 and 10 and 3 right so if I write that equation here then I will have 4 pi de and ca s minus ca c that is the one divided by 1 by rc minus 1 by R that is the equation so we have to now eliminate eliminate concentrations of you know concentrations ca c ca s and express in terms of ca g right this again I will leave it to you as mathematical exercise how do you eliminate that eliminate there are simple techniques and also very complicated techniques right so in the examination do not blame me you are supposed to know that so equation yeah using equation 1 10 3 are used to eliminate intermediate concentrations intermediate concentrations ca c and ca s good to get the equation that equation I will write here I think it may be written may not be sufficient here to get minus dna by dt equal to ca g whole thing divided by 1 by 4 pi R square kg plus R minus rc divided by 4 pi de R Rc plus 1 by 4 pi R square k s this is the equation so this is equation number 11 but still this is not you know in usual form in the sense that it is extensive equation or intensive equation by the way what is the difference between extensive and intensive I am talking as if you know if you do not know Relta you are telling something depends on the mass how do you express the rate of reaction for homogeneous reactions definition of rate I am asking yeah is it intensive or if I only express as dna by dt is it extensive and intensive no yeah that means dna by dt depends on if I take a large size particle or small size particle or if you take 10 mm particle I take 1 mm particle nil or may take 5 mm particle all 3 we will get different rate but how do you normalize that so that we should talk the same language by taking some yeah either volume mass or whatever okay per mass or volume okay so that is why here what is the logical one area so we can also express this as minus r a equal to minus 1 by 4 pi R square yeah dna by dt equal to after cancelling and all that a g 1 by k g plus r r minus r c by d e r c plus 1 by k s into r by r c whole square r c whole square so this is the equation what we have for based on a we have not talked anything till now about b but we would like to find out when the b is disappearing or when the disappearing in the sense that when the reaction is going to complete for b right this is not directly useful for me because this gives me only the rate of reaction of gaseous phase if you remember our stoichiometric equation but we are also interested this is useful if I am only worried about how the reaction you know gas phase reaction is taking place provided this r c at this particular r c at that point of time you can calculate what is the rate of reaction with respect to a we can calculate right but I also would like to find out how this r c is shrinking and when finally I will get 100 percent conversion or at any time what is the relationship between this r c that core and then the time this will not give me obviously now that information so that is the next one using this information now you have to relate these two phases right so when we relate that phases now we have to use the basic material balance that material balance is coming from this stoichiometric equation b b s giving me r r solid plus s s gas what is the relationship minus d n a by d t equal to 1 by b minus 1 by b d n b by d t equal to 1 by r d n r by d t 1 by s d n s by d t this is equation number 13 okay I think if I say this one as equation number 13 then this will be 14 okay so that is the relationship no good so now again I can take these two and then relate d n a by d t I know already and if I that is nothing but 1 by b d n by by d t but still this is not useful to me because this n b is not giving me much information so this I have to write in terms of r c in terms of r c right our ultimate aim is to finally express this r c in terms of time okay correspondingly what is time and what is r c r c gives me an idea about my conversion if r c equal to 0 conversion equal to 100 percent if r c equal to capital R 0 conversion so in between that definition also we will make for x b okay good so that is why we take this this relationship and we have n b what is n b by the by kavya excellent number of moles of b how do I find out number of moles of b how do I find out number of moles of b I have the particle and in this particle there are moles definitely how do I find out n b ya if I know there we know that we have molar density and also mass density if I know molar mass density then I have to divided by molecular weight and if I know molar density that is rho b we call rho b is the molar density this is rho b into v c c is the volume of core okay so of course this v c equal to v volume of the particle at the beginning then it slowly shrinks because we are only now trying to find out how the number of moles are getting slowly reacted right okay so now this equation please remember only one thing you have to important thing remember is rho b is not mass density rho b is moles per unit volume that you have to write make a note of that somewhere okay so now d n b is differential of I am writing very very small steps also rho b v c whether rho b is constant when the reaction is taking place rho b is it constant rho b is it is the intrinsic property or extensive property rho b per unit volume I say ya will it change the same so that is why it is constant we will take it out okay so this is rho b d of v c which is nothing but 4 third or 4 third no pi r c cube ya correct correct kavya still doubts rho b okay you have taken let us say molar density let us say 10 okay right that is moles per unit volume okay does that matter when you have this size this size this size does this matter or not assuming I think it is constant it will not change during the reaction if something happens it may change that is not I think you know that things we are not taking into account right so the number of moles per unit volume of b is constant whether I have 1 centimeter cubed particle or 10 centimeter cubed or 100 centimeter cubed it is constant no it is constant still not able to get ya so that is why it is constant throughout you know please do not think that you know it is like normally with reaction concentration should change concentration will change if you have everywhere you know for porous particle it changes this is a non porous particle very clearly we can identify this is the b may be logical doubt what she has if it is a porous particle b then I do not know where the how much b is available where inside the particle ya you are thinking it ya a is gas ya for gas ya that is what the profile no and at r equal to r c at r equal to r c okay whatever r c that comes there that will react with b but this molar density is constant because moles available in that unit volume is constant okay so this is nothing but ya 4 pie r square 4 pie rho b r c square d r c ya so this is equation number 16 good so now we will go to this side again okay what is that next thing we have to do d n by d t you can calculate d n by d t equal to we have 4 pie rho b r square d r c by d t right so 4 pie so this is equation 17 correct 17 ya that is right ya so now we have the relationship minus d n a by d t equal to minus 1 by b d n by by d t equal to minus 4 pie r c square rho b by b small b b is the stoichiometric coefficient d r c by d t this is equation 18 now we have some relationship between 18 and then that equation 12 right this one or this one equation 11 minus d n a by d t so now I will equate 11 and 18 with appropriate multiplications right of course here I have minus 1 by b d n a by d t I do not have to multiply anything so now equating 11 and 18 now I have the relationship between r c I have the relationship between r c ya so we have to integrate after separating the variables because now this one okay let me write this step 4 pie r c square rho b by b ya d r c by d t equal to I have here c a g 1 by whole thing divided by 4 pie r square ya plus capital R minus r c divided by 4 pie d e r r c plus I have another term 1 by 4 pie r c square k s this is the one these two we have equated now we know r c and all this contains r c so I can bring this side and other constant terms I can bring that side so d t will come this side d r c will be here then we can integrate and after integration that also I do not give integration I will give after integration and rearranging right I think this is equation number 19 okay ya integrating equation 9 with appropriate arrangement integrating appropriate modifications okay what you get is this equation that is may not be sufficient here what you get is c a g b by rho b into t equal to r by 3 k g plus r square 6 d e 1 minus 3 1 minus 3 r c by r whole square whole cubed ya plus r by k s 1 minus r c by capital R this is equation number 20 this is the equation this is the one so now I have a relationship between r c and then the t but all this mathematics you have to do do not blame me when I give this in the examination at least you have to do that okay this simple ya know high school integration it is not at all great difficult integration so now this also as engineers particularly chemical engineers we are interested happily in terms of conversions so how do I find out this equation in terms of conversions okay how do you define conversion conversion x equal to initially what are the moles minus moles left divided by initial moles even here when I have x b we are now interested r c has come in terms of x b okay so x b equal to n b not minus n b by n b not please give me the equation you know in terms of r c that is the one just write and then tell me in terms of r c 1 minus x l i 1 minus r c by that is equation okay so now I have r c by r cubed r c by r square r c by r square so this is r c by r so then correspondingly I can write this equation in terms of this is equation 21 so writing the equation 20 in terms of conversions what you get is c a g b rho b into t equal to r by 3 k g ya this one will be x b plus r square by 6 d 1 minus 3 1 minus x b 2 by 3 plus 2 1 minus x b ya that is all plus we have r k s 1 minus 1 minus x b to the power of 1 by 3 ya so this is the expression which gives me the relation between conversion and time so you can now calculate when x b if you want 100 percent you know x b equal to 1 you can substitute there and then it will be in terms of k g d e k s and capital R okay this is beautiful this one I have solved this equation using all 3 resistances same time but if you see Levenspiel which you would have studied definitely and you would have forgotten totally and you know if you see that every step we take separately diffusion you know first mass transfer of a through film controlling then diffusion controlling then reaction controlling the same thing even now we can do that here okay after taking the general expression like for example if k s is very very large which is controlling reaction is controlling mass transfer which mass transfer is controlling there are two one is diffusion other one is ya convection through the film so that is why every time you know we can take of course individual steps controlling means if it is mass transfer control this k g d e is very large k s is very large so like that simplifications we will discuss in the next class ya and ya just before that one more thing this equation can also be written in terms of dimensionless quantities which is very important for chemical engineers dimensionless so how we write this equation only one equation I just write is something like this write this in terms of dimensionless so this is theta time equal to 2 bio number into x b plus I have 1 3 1 minus x b to the power of 2 by 3 plus 2 1 minus x b plus I have 6 by d a 1 minus 1 minus x b of 1 by 3 this is equation number 23 where theta equal to 6 b c a g d e divided by rho b r square that is the definition of theta that is 1 and bio number bio number b a ya anyone knows about this bio number k g r by d e b i is bio number b i o t and d a is damkohler number which is k s r by d e okay so I think I can also write here b i equal to bio number and d a equal to damkohler number damkohler number right I think next class we will discuss the significance of this the meaning of bio number very large for example it should be controlling and most of the time in gas solid reactions you first calculate bio number and then if it is very large you neglect film step that means neglect in the sense that film is not controlling that means bio number large means k g must be very large or diffusion must be very small diffusion very small means diffusion control k g is anyway very large for large bio numbers so that is why one has to get this kind of estimation in the beginning which one is rate controlling step damkohler number means it is reaction and diffusion right if damkohler number is small that means d e should be very large and k s should be small that means it is reaction control so that is why you have this numbers which will give you some physics to understand okay I think we will stop here tomorrow we will discuss about the individual steps controlling and how do you find out the diffusivity how do you find out k g how do you find out k s all that is important otherwise we do not know how to use that equation 22 is a 22 and 23 are beautiful equations okay for gas solid reactions non catalytic okay thank you