 we have completed the shinking core model I think except Renitha others were not there so you have to take notes from her and we also discussed the extreme conditions like which are rate controlling how do you find out the mechanism which one is rate controlling all that we have discussed so based on that now I would like to give you an assignment so I will give you the parameters you have to plot this assignment it is graphical but each graph at least you should able to you should able to write on that graph at least 10 to 15 sentences it is not that simply go to excel and then plot and then give it okay so that anyone can do whoever knows the computer programming but what your expertise comes there is that if you are able to predict why something is increasing something is decreasing something is not increasing decreasing all that physical phenomena right so the assignment is you take for example that equation number dimensionless equation number what is that number 23 or 24 in terms of biot number and damkohler number s equation number 23 okay take equation 23 and plot x b versus theta theta is the dimensionless time x b of course is the conversion okay so the parameter the range for theta will be 0 to 5 in the steps of 0.2 okay that means you know 0.2 0.4 0.6 0.8 1 1.2 like that right okay and x b is of course 0 to 1 anyway that I do not have to tell so what we have to do is a 1 biot number equal to infinity there are 2 numbers biot number and damkohler number so and damkohler number equal to 100 you have to keep biot number infinity and change damkohler number 0.1 and 0.01 so then come to biot number equal to damkohler number is same 0.01 then biot number equal to 10 damkohler number is very does same ranges 10 1.1 0.01 then we have biot sorry biot number equal to 1 damkohler number equal to 100 10 1.01 then we have the fifth one biot number equal to 0.1 damkohler number is same again 110 1.1 0.01 I think from this you will get almost all the information x b is varying with biot number in that equation x b is a function of theta biot number and damkohler number that is all nothing else is there in that all are in the form of dimensionless numbers so then you vary and now you have to say that for the comparison when you are plotting these graphs for example if there is a change between this and this why there is change the physical phenomena or if there is no change between this and this why there is no change what is actually happening and you know very large damkohler number means which one is controlling what is the damkohler number definition yeah large number means mass transfer diffusion control diffusion through the ash layer controlling okay so when diffusion through ash layer controls then what kind of phenomena what kind of things are going on within the particle so all that information you have to write you have to write at least you know 1 2 or 3 or 4 pages and then attach these graphs and then submit okay it is left to you okay if you do not submit then examination if you are not doing well please do not ask me right yeah for the people who want to definitely pass I think it is very difficult I do not consider at all if you are not doing all this this is only for your help I may give the same thing in the examination okay at that time this is a practice and the physics if you understand after plotting and after writing then definitely you will understand and do not think that you know you are heroes when you copy copying is not heroism actually solving and writing is heroism but most of the time we will keep quiet till last day and then last night we will just ask someone else's assignment copy and give as if you fooled me actually you are not fooled me you fooled yourself first without doing so that is why these assignments even though there are marks I think you cannot only measure everything in terms of marks you do not learn anything if you do really that okay so after this what we have to do is one other type reaction type D reaction if you have no notes can you tell me what is type D reaction we will derive an equation for type D reaction so yeah so this will be a gas plus b b solid going to gaseous products right yeah what is the example gaseous yeah combustion gasification there are very nice reactions where you are the end you will not see any particle everything will be disappearing so that is the one for that kind of reactions now we have to develop a sinking core model right what is the difference what could be the difference gaseous diffusion will not be there okay because every time the particle outer layer is first converted then slowly it moves and there is no because everything is becoming gas and coming out there is no solid product so that is why and how we derive that we will see and before that let us imagine our particle how it is moving this is particle at time t equal to 0 may be cool particle so then after sometime t then we will have this is the size then after sometime t this is the one and finally disappears into nothing okay yeah so now this is t sometime t yeah okay good so what is the difference is at time t equal to 0 just when you have the reaction started t0 plus for example that is the film then this is the film then we will have this is the gas film right it must be easy for us now to derive this because there is no one more resistance there there is no ash forming so there is no diffusion and if you are able to plot the profiles this is r equal to 0 that is the and this is at any time I will call it as rc and here we will call this one as capital R that is initially at time t equal to 0 then only the core is the particle is shrinking overall but I am trying to maintain the same notation rc is the diameter of the core which is shrinking but the overall particle itself is core for us not like in the other case where inside the particle you have your core unreacted core which is shrinking but here the entire particle is shrinking so as usual now if I draw the profile like that that is all this is c as when it touches the surface then reaction is taking place and particle is slowly disappearing because everything is becoming gas gas is product so you will have this kind of situation at up to sometime t good so what are the steps now we have step one is mass transfer m t of go to abbreviations now gaseous product sorry m t of gaseous mass transfer of gaseous reactant step two is reaction on the surface reaction on the surface is there step three why you know yeah see the gaseous products have to come through the film again right yeah they have to yeah but we do not take that one step three equal to m t of product gaseous product gaseous through film but it does not matter for us because it is not reversible reaction so that is why okay so that is why only these two steps you have to take and at steady state mass transfer of reactant a going through the film must be equal to rate of reaction on the surface so that we know we already are masters of that so then at steady state rate of mass transfer through the film equal to rate of reaction on the surface correct yeah so we know the equations now so this will be minus d n a by d t if I write in terms of equations the simple rate 4 pi r c square okay and that is the area and mass transfer equation this is k g c a g minus c a s equal to if I assume first order reaction then this will be 4 reaction is on the surface every time 4 pi r c square okay yeah so 4 pi r c square k s c a s okay so this is equation 1 so now of course we have to eliminate this I think I will write that in a slightly different way so that some of you can get what is going on so this is if I write like this this will be minus d n a by d t divided by all this 4 pi r square pi r c square k g this is equation 2 okay you have to remember that c a c also equal to c a s so I think let me write that at least once okay then you are comfortable no problem yeah it is on the core but now for us core itself on the outer surface okay good so let me write this once yeah so now c a s equal to again minus d n a by d t divided by 4 pi r c square k s correct yeah so from these two equations from equation 2 and 3 what we can do is minus d n a by d t equal to 4 pi r c square c a g divided by 1 by k g plus 1 by k s so this is equation 4 so this is our rate equation but of course if you want to make this one as a intrinsic rate so then I have to divide by some area or volume of the particle or weight of the particle whatever okay so 4 pi r square that means I have just added these two and then separated again d n a by d t that is all very simple okay so now there is a problem here I cannot directly integrate and of course you have another thing here our normal procedure is this equation just gives me what is the concentration profile or how the rate of reaction is taking place with respect to gaseous a it is d n a by d t right d n a by d t but what we would like to have is also d n b by d t how the particle is shrinking how the particle finally disappears so that is what what we have to relate so for that we already have that equation minus d n a by d t from this stoichiometric this comes from this stoichiometric equation equal to minus 1 by b d n b by d t which is nothing but minus 1 by b d of once more I will write rho b v c correct no d t which is again nothing but minus 4 pi r c square by b d r c by d t here also we have rho b yeah correct thank you rho b good so this is equation number 5 now if I take equation number 4 and 5 d r c by d t can be written as from equations 4 and 5 minus d r c by d t equal to yeah can someone quickly tell yeah b c a g by rho b into 1 by k g plus 1 by k s so this is the equation which I am supposed to integrate to get the relationship between conversion and time or shrinking core and then time this is equation number 6 yeah here only straight forward integration I cannot do okay why because now the procedure is take if there is any r c terms I have to take this side and time I have to take this side and then simply integrate is there anything which is changing with r c in the other side yeah why why it is changing with k g changing with r c film thickness is changing yeah so film thickness changes means what is directly changing what is it changing what is the quantity yeah k g k g is changing with r c do you have any relationship between k g and r c yeah we go again to Ranjana marshal correlation okay so then we will see for small particles for large particles what is the relationship between k g and r c and that relationship we have to substitute here in k g and finally we have to integrate because now this k g is varying with r c that means if you take small particle or large particle you will have and I do not know some of you would have done already something in Levenspiel using Levenspiel book in non catalytic reactions you have done this non catalytic reactions he gives some very nice conclusions there at the end of the kinetic chapter he has divided that into one is the development of kinetic equations other one is design two chapters are there are non catalytic reaction so at the end of the first you know kinetic model chapter he he comments that when you have film film control the power to the the power to the r to the power of you know t is directly or there t is proportional to r to the power of 1.522 it goes many people do not understand that I think now this is one of the basis is how you can derive that that I do not derive but I will just mention and then give it to you so that I can ask that in the examination okay so t is t time is proportional to r to the power of 1.522 that means as the particle size is decreasing the exponent falls okay yeah so that you can get from this derivation good okay now what we have to write here is k g is a function of r c in equation 6 and you cannot directly derive that I mean integrate that unless you know that functionality right. So that functionality can be this functionality can be obtained from from range and martial correlation what is that Sherwood number equal to 2 plus 0.6 Reynolds to the power of half and Schmidt to the power of 1 by 3 this is equation 7 the same equation I can also write for me for easy solving this is d p k g divided by d that is diffusivity equal to 2 plus 0.6 Reynolds number it is d p u rho by mu to the power of half and Schmidt number is mu by rho d whole to the power of 1 by 3 so this is equation 8 okay now we have to solve this okay let us see first for small particles for small particles the other term Reynolds number term is neglected small particles means Reynolds number may be very very small so this entire term may be smaller when compared to 2 the one extreme very very small particles right for small particles for small particles Sherwood number equal to 2 which is nothing but again d p k g by d equal to 2 so from this what you find out k g is proportional to 1 by d p correct or proportional to 1 by r c radius that is 1 and for large particles this is what is the number 9 this is equation 9 for large particles Sherwood number equal to d p k g by d 0.6 d p u rho by mu to the power of half mu by rho d to the power of 1 by 3 yeah now can you tell what is k g proportional to yeah this is 1 by d p to the power of half correct also proportional to 1 by r c to the power of half good so if initially our radius equal to capital R right initially then in both the cases initially k g is okay I want to convert this into based on initial particle yeah so for small particles I think I may write here for small particles I can write initially k g by k g not equal to k g not I will tell you k g not is the mass transfer coefficient initially right so you will have here R by r c to the power of 1 right initially and similarly here so this is of course also an equation 9 a 9 b I will write here yeah so here also I can write the same thing k g I think below also I can write k g by k g not equal to for large particles yeah can someone tell capital R by r c to the power of half okay so in general what I can write this may be 9 okay this I will say 10 so in general what I can write is what I can write so in general what I can write is k g by k g not equal to if I take the ratio R by r c to the power of n so this is equation 11 and we have the two cases if n equal to 1 the other way for small particles n equal to 1 correct for large particles n equal to half so I try to integrate this equation using substituting this and then you will have both the cases by substituting for n equal to half you will have large particles and vice versa there is another important thing also which I have not mentioned to you there are some books like carberry books carberry carberry is slightly I found a book so there he also takes u variation with dp or dp variation u particularly that is correct when you have you know moving bed reactors and also fluidized bed reactors it is reacting becoming smaller so when it is becoming smaller then the terminal velocity yeah I think is higher and higher because the particle can be easily thrown out so that variation of u with respect to u how the r c is changing the relationship between u and r c that also is taken care of but here we are assuming that u is constant for us and only d is changing the other things are properties rho and mu they will not change right here also diffusivity and all that they will not change only u can be changing and that probably I will give in the examination some other nice trick there and then I will ask you to do that okay yeah good because everything if I tell you know you may not appreciate that okay so now it is only integration that is left now so that means from equation 6 I will simply write from equation 6 minus integral I have 1 by k g plus 1 by k s d r c equal to integral I will take the other thing that is okay b c a g rho b integral d t this is 0 to t correspondingly 0 means capital R at any time t means r c that is what I have to integrate but I cannot integrate this straight forward because this k g is a function of r c so that is why now again minus r 2 r c here I will substitute for k g equation 11 so if I substitute that it becomes 1 by k g not into r c by capital N that is 1 term plus 1 by k s into d r c equal to b c a g rho b into t okay yeah we also have an equation we also have an equation for r c in terms of x b or x b in terms of r c correct no you know that so what is that equation yeah we have r c by capital R equal to 1 minus x b power 1 by 3 yeah this equation I will call it as 12 yeah so now this is 13 so substituting equation 13 in this and then integrating directly you will get in terms of time and then conversions okay so when you do that what you get is substituting equation 13 in 12 okay I will write here using equations 12 and 13 what you get is t equal to this mathematics you have to do r b b k g not c a g 1 minus 1 minus x b to the power of N plus 1 by 3 by N plus 1 plus 1 minus 1 minus x b k s by k g not so this is the equation this is equation number 14 that is the equation I think we will stop here you have to go no yeah we will stop here and then try to discuss this in the tomorrow tomorrow we will try to discuss about this particular equation and k g not you know k g not is based on or initially a time t equal to 0 right yeah so again of course you can also simplify this saying that okay when film alone controlling or reaction alone controlling that we will discuss tomorrow