 The first one, that we discussed is the gas solid reaction, right, so gas solid non catalytic reaction, right. So the example which we have taken was Carbon solid plus O2 gas giving us CO2 gas and the model imagination is something like this we have solid we have this is carbon then we have film and all around we have O2 and O2 has to diffuse to the surface and CO2 has to come out and we have also written the equations like yeah sorry this also can be modeled like this as solid this is the film gas film and then if I draw the profiles this is CAG that is the concentration of oxygen in the gas phase then I may have something like this this is CAS right so at steady state conditions the amount of oxygen which has diffused through the film must be equal to the reaction that is taking place on the surface so we have written that and then the equations what we have written was that minus r A equal to KG CAG minus CAS and if I assume I have first order reaction this is KS CAS okay this also we have written so now the global rate of reaction or observed rate of reaction this also okay is defined as the rate in which only the measurable variables are present right here the measurable variables is only CAG and I cannot measure the concentration on the surface which is CAS we have to why means because here I have to put some probes and then try to find out how many moles of oxygen staying on the surface right so that is not that easy and that too it is very very hot surface so any technique which can be used on the surface is normally very very costly technique only for that this Airtel from Berlin University he got noble price to show that how to measure concentration on the surface and he used the examples like on the platinum surface CO is getting oxidized as CO2 that example that is one example another example also ammonia how hydrogen molecules are at least getting adsorbed on the surface and how they are reacting the actual reactions he could show by constructing his own equipment that is why he got the noble price exactly that is surface phenomena what you could show that is available on his website if you want to go and then see that Airtel ERTL that is the noble laureate name so if you go you can also see that it seems during the reaction on the surface there are waves also generated that waves also he has picturized so that is why of course he got noble price okay good so that is why this is very difficult technique to measure on the surface and as I told you as engineers we should see that as simple as possible we should use the technique so that is why best thing for us is elimination of this and if we eliminate this and then try to write the equation that means I can take these two CAS is eliminated so if I eliminate these let me write the other steps also K G C A S K S C A S so then I have C A S equal to I am jumping the steps equal to yeah K G C A G by K G plus K S right okay so now this if I call this one as 1 2 equation 3 then equation 3 should be substituted in equation 1 because either this or this any one of the terms then I have minus R A O B observed rate equal to if I substitute here in this K S I have K S K G C A G divided by K G plus K S this also can be written as C A G divided by 1 by K G plus 1 by K S so this is the rate expression that is the rate expression for coal combustion and this rate expression expression is peculiar here because we have a mass transfer coefficient K G and also we have a reaction rate constant so in in general when you are talking about chemical reactions we never take mass transfer coefficient to account but still here in all heterogeneous systems mass transfer coefficients will be automatically coming into picture and if you remember my picture the famous picture which I draw here this is reactor input output kinetics contacting here we have chemical physical here we have batch continuous and in batch again we have P F and M F so this part we have understood what is batch what is continuous in continuous when do you choose P F when do you choose you know M F all that we have discussed now when you come to this there are two terms chemical and physical chemical is the one which represents K S physical is one which represents K G so that is why here you have both know chemical as well as physical that is the meaning of this diagram right so that means most of the time I have already told you sometime back that most of the combustion process this is coal combustion most of the combustion processes are mass transfer control yeah the reason is that the temperatures are very high so this K S is very very large because Arrhenius equation okay so when Arrhenius equation we have small change in temperature will be creating large coefficient and now at very very high temperatures this is very very large so at high temperatures high temperatures K S is large tending to infinity just as an example then we have minus R A O B equal to C A G by 1 by K G R K G C A G which we can write as mass transfer equation which is equal to 0 so this is 3 4 5 and then 6 now you see if you assume that for this gas phase you have plug flow we have to choose the contacting so if I am now thinking that my gas O 2 this is minus A is O 2 right yeah so this is moving in the form of plug flow so what is the equation we use here for the contacting V by F A not equal to 0 to X A D X A by minus R A minus R O B so now I have to substitute this equation that is 0 to X A D X A into K G C A G this is a reactor design equation but still we have only mass transfer coefficient coming into picture okay yeah so similarly if I have both the controls then I will have this equation C A G divided by 1 by K G plus 1 by K S that is what is here so on the other hand if I have at low temperatures or reaction rate is slow at low temperatures K S you know K G will go to infinity that means when compared to K S so this here also I will write this is mass transfer control and this is reaction control yeah so when K G is very large value from equation 5 from equation 5 minus R A O B equal to we have C A G this goes to yeah this goes to 0 1 by K G and then I have K S 1 by K S which is nothing but K S C A G so this is equation 7 this is equation 8 yeah at low temperatures the reaction may not be that fast K S is less reaction is not fast right at low temperatures okay so that is the reason why K G relatively is faster than compared to K S right so that is the reason why K G will be very large value so 1 by K G is neglected then it will be only K S which is reaction control I have been giving you this simple example like you know your hostel food okay I think assuming that you have conveyor belt for breakfast from the kitchen you send the I mean they will be continuously feeding let us say idlies okay so idlies will be put there on the conveyor belt you are there just around the conveyor belt and then as soon as idlies come on the conveyor belt if you are very hungry what will happen you will eat right so practically on the surface what will be the concentration 0 you do not have anything now which is controlling now which is the controlling is it the supply of idlies or is it your eating which is equivalent to reaction here supply mass transfer is controlling okay on the other hand after eating may be everyone 50 idlies you cannot eat anymore right so then you will have everywhere concentration that means your rate of reaction rate of eating is almost 0 now very very slow even if you eat because someone is putting a gun and then say that still you have to eat so then you know for the sake of eating you may take only piece of idly and then still you are eating but the rate is very very low that is what is reaction control it is relative both are relative whenever we talk in this case both are relative if both are equal then this is the equation which you have to use okay always in the controlling steps the lowest step will have the control slowest step okay in these steps these are simple examples which I give so in these steps the slowest step controls here the slowest step is mass transfer because reaction is very very fast and here the slowest step is yes reaction and mass transfer is very very fast and as you said these are all relative okay good so now this is the equation what we use and here we are comfortable it is K s C A g if I take it is like our first order reaction right so we say order of reaction with respect to concentration in this case it is first order with respect to oxygen if I want to use in the design expression again I go here and substitute this time I will have K s and C A g and C A g is measurable that is why we call this rate as observed rate observed or yeah another name global another name measurable very good is there any other name measurable or observed observed I have given observed global measured overall overall rate of reaction okay so rate of reaction yeah so all this will come there because when you have heterogeneous system you cannot avoid this observed rate or global rate or measured rate why we are telling observed rate is we are able to express this equation 5 in terms of measurable quantities C A g is measurable and 1 by this K g I should have some correlations any one of you know what are the correlations we use if I have single particle and then mass transfer is going through this particle or to the surface or inside the surface through the film I have come across that equations single particle we have just only one particle and around the particle oxygen is going and this oxygen is going to the surface of the particle through the film it is external film mass transfer not able to recall there is one equation called range and marshal equation it is in the third chapter of trible okay and of course Levenspiel non catalytic reaction also he has given that he calls that one range and marshal but trible calls that one as Frosling equation okay so that vacation tells us that we have Sherwood number equal to 2 plus 0.6 Reynolds to the power of 0.5 and Schmidt to the power of 1 by 3 or 0.33 that is the equation okay and I think I have to tell you here because most of the time now we will be talking about mass transfer coefficients this is one of the simplest equations in chemical engineering for because it is a single particle and in reality you will never have single particle right and in multi particle systems developing this equation is not that easy mass transfer equations that is why lot of researchers have spent lot of time to develop the correlations even now people are developing if it is a new system and surprisingly in chemical engineering if this is for single particle and I cannot use the same thing if I have packet bed the mass transfer through the packet bed is slightly different and if I take fluidized bed that is different and all the equations are in terms of only Sherwood number Schmidt number Reynolds number you take any equipment like packet bed fluidized bed moving bed rotating drums all these things you know the mass transfer equation contains only Sherwood number equal to the function of Reynolds number and Schmidt number that functionality we have to find out only through experiments right so that is why if you have the same process what we have discussed here if it is in fluidized bed then you have a different mass transfer correlation KG I am talking this KG because KG is here in Sherwood number I hope you remember Sherwood number and Schmidt number Reynolds number you know remembers that is only one number most of the time chemical engineers remember that is why in my interview whenever I sit for interviews and all that I ask them other than Reynolds number please tell me another number and it is really funny you know particularly there are also numbers in fluid mechanics so other than Reynolds number we cannot list out any other number in fluid mechanics other names of course very famous numbers next to Reynolds number is Schmidt number and Prandtl number yeah and why all these correlations should have Reynolds number and Schmidt number and here we have even heaters coefficient is also same exactly same thing here we have Nusselt number equal to same 2 plus 0.6 Reynolds number will be there and this will be Prandtl number that is the reason why you study transport phenomena because transfer phenomena tells you that all these transfers have some kind of similarities some kind of same phenomena so that is the reason why in transport phenomena you know you have the famous correlation J D equal to J H equal to F by 2 yeah why do you need that analogy we know I mean we know that equation but why do you need it for heat transfer mass transfer and heat transfer relation fluid mechanics also is related F by 2 J D equal to J H equal to F by 2 F is friction factor by 2 and we need that kind of thing as engineers if I do not have sufficient time to to find out mass transfer coefficient I can use friction factor but there are many conditions for you know that analogy right under those conditions I do not have to conduct separate experiments for heat transfer or mass transfer if I know any one if I know J D I can calculate again friction factor if I know J D I can calculate heat transfer coefficient J D is for mass transfer J H is for heat transfer and F is friction factor that is why that analogy comes automatically for us that is why I told you know chemical engineering is great the analysis the principles wonderfully demonstrated all the time through equations but only thing we do not know is that where they are applicable because I think in 50s and 60s that was excellent period for chemical engineering the reason was that the chemical engineering and industry work together beautiful love marriage yeah but after sometime in 80s divorce and the catalyst for divorce is you know what can you guess transport phenomena course that is the one because when transport phenomena that book was first published in 1961 or 1962 by birds to what light foot it really changed the thinking of many engineers not only chemical engineers mechanical engineers use that civil engineers use that and yeah metallurgical engineers use that many people use that phenomena I know this transport phenomena then people thought that particularly in academic institutions people thought that excellent now any system in engineering now I can model till then it is only empiricism do the experiment develop the correlations that is what because of that only this correlation has also has come later these coefficients have been justified by transport phenomena through boundary layer theory has come there theory also explained there but it was wonderful in the beginning that empiricism means you have a feeling for what you have done and theoretical highly complicated equations you solve and then finally you find out those coefficients which may not be exactly like you measure and of course any theory should be validated by experiment at the end so finally what has happened was many academicians thought that they can happily publish and get away without worrying about what is happening in the and when an industry person comes to an academic institution then the industry person was asking some questions and the moment he asks a question the professor always replies in terms of differential equations and differential equations in an industry person has forgotten long time back so that is why he thought that my god all the professor started talking a different language it is not industry language so that is why the distance between those two increased and that is when in some industries they started industrial research that means their idea is to solve immediate to the problems not like academicians academicians they take a problem and then they have their own pace because they don't worry about industry and time is not money in academic institution whereas in industry time is tremendous money one hour gone means maybe one million dollars also you may lose so that was the time when these people have become you know the divorce they are divorced and now of course with nanotechnology and all that I think no industry can understand what we are talking so that is also I mean that is the greatest contributions what we have with these new technologies so that was the reason why we are not able to get you know lot of consultancy and all that particularly in chemical engineering but in other engineering is like civil civil still lives with a lot of empiricism they have called I told you know one god's boon called factor of safety their factor of safety may be 5 times 10 times also do you know the reason why their factor of safety is high our factor of safety may be only 10 percent or you know may be 50 percent right I think in macabre theory method my efficiency normally how much you take you calculate it theoretically 10 plates right yeah divide by 0.8 that means 20 percent yeah otherwise sometimes 50 percent the reason is that you are not confident about your mathematical analysis because there are many assumptions like that in civil engineering there are many things first of all brick they have to assume some and every person produces brick in a different way so how can you generalize that then you know the concrete what they make you know the gravel so theoretically it may be only half an inch particle but I think it will vary from 1 nanometer to 1 kilometer maybe I am just exaggerating yeah so then where is the theory coming there on the top of it cement cement they specify something but you know particularly in we are great in mixing things so ash will be mixed with cement so then that also gone so that is why the factor of safety in a finally building has to live at least for some time no or bridge or road whatever so that is the reason why factor of safety is very very high in civil engineering but whereas factor of safety is minimum in aerospace engineering if they calculate 100 tons of weight finally for the flight and okay factor of safety equal to 0.5 50 percent okay yeah then weight will be 200 tons it will never fly it will only go as bullock cut again on the road so so that is the reason why that factor of safety is less for sophisticated things and electronics also I do not think electronics they do not have any factor of safety in electronic industry okay because the problem there advantage there with electronic industry is gravity will never bother them correct no it is only electronics flowing in this way current will come in this way but where is the gravity coming there gravity will never affect but in chemical engineering in civil engineering mechanical engineering metallurgy gravity affects and systems are generally very difficult because gravitational force you have to take and because of this gravitational force also we talk about again buoyancy and all that right and buoyancy we know what is buoyancy but we do not know how to really apply for our calculations so because of all these problems that factor of safety and all that will come there okay anyway so that is why this equation is one just I have written but I will write many more like that why I wrote that equation is from this you calculate what is the equation for Sherwood number Kg dp by capital D which is diffusivity dp is diameter of the particle and Kg is the mass transfer coefficient from here you can calculate that Kg substitute in equation 7 and then we can integrate because this is the simplest integration then you will get your volume of the system I mean in a simple design I am not saying that all combustors are designed like this but I want to connect that diagram where kinetics and contacting both we can take so that is the reason why I am trying to tell the simple example they are not real examples in the sense that real examples I will tell you when you come to non catalytic reactions okay good so this is how now let me also tell you here in this profile if I have reaction control maybe I will draw here again the profiles we will redraw this is solid that is carbon then this is the film and I have here CIG this is what I know this is bulk in bulk there is no concentration gradient okay so there is no resistance for the assumption is there is no resistance for the molecules to go through that film so that is why it is constant from there if I have reaction control how do I draw that profile that means you have to come here if it is reaction control the equation is this one Ks CIG equation 8 how do I draw the profile on the because the profiles are very important in heterogeneous systems if you draw a clear profiles it is very easy for you to develop the model kinetic model in fact this is called the kinetic model for gas solid reactions okay good yeah tell me now how do I draw that straight line why straight line this is correct yeah yeah you know just remember the idly example if your rate of heating is very very less you cannot heat anything so then entire conveyor belt will have the same concentration of idlies so that is why same everywhere you have the same concentration of CIG right okay on the other hand if you have the mass transfer control reaction is very very fast that means you are able to eat very quickly and practically you will see on the surface zero concentration of that idlies okay so you have to draw something like this may be straight line depends on the geometry so here CAS is approximately zero that is why I have drawn here I mean I have written here in equation 6 when you have much this is mass transfer control we will write here MT control yeah that is MT control and this is what I have given here you know always for mass transfer you will be very comfortable if you write CIG minus CAS but here CAS equal to zero and for reactor design expression I have to use now mass transfer coefficient there is no more you know reaction rate constant and what are the problems we may face with this if you do not know that there is a global rate of reaction if you know that mass transfer is not is playing a role in heterogeneous systems and all this we are talking about only at one temperature okay again heat transfer also comes into picture okay so if I take that only for isothermal system I am talking about only mass transfer in heterogeneous system and you do not know that you have the heterogeneous systems what do you imagine you imagine that the rate is proportional to some constant correct no I mean that is what all chemical reactions the first assumption is minus RA is proportional to CA to the power of N and CB to the power of M okay and those two coefficients you have to evaluate if N equal to 1 and M equal to 1 then you will have first order with respect to each overall second so that is why first assumption is directly proportional to concentrations right so that is why even here when you assume the same thing you have from equation 5 from equation 5 you have minus RAO B equal to K 0 CAG where K 0 equal to what is the number this is 9 this is 10 okay where K 0 equal to or 1 by K 0 equal to 1 by K G plus 1 by K S right this one I have written as this equal to 1 by K not which will go to the top and then this is K not okay good anyway so this is the equation what we have actually and you think that this is simply K not which is some constant but this K not has mass transfer coefficient and reaction rate constant also and suppose I now repeated this is for one temperature what we have done all this now I conduct the same experiment with another temperature like another temperature because I would like to find out the Arrhenius relationship but now if I and I am imagining that I have only K not which is a reaction rate constant because I have not taken heterogeneous terms I do not know whether chemical and physical kinetics both will really affect the process right so under those conditions I am imagining that I have K not and actually that is a combination of these two right so with temperature if I plot okay here this Arrhenius plot this is a ln K or K not equal to sorry not versus T 1 by T then you may get something like this or or instead of that I will just plot and then normally how we behave we will try to see okay so this one like this you may get the points right because I do not know anything about the presence of chemical and then physical factors affecting the rate of reaction normally what we will try to do we are all experts now going to excel and now choose that option linear because you know it must be straight line so you simply draw this which is totally wrong because there is a beautiful phenomena that is coming here right but on the other hand I will be carefully seeing that why another time repeating the experiments and then finally seeing that there is some problem with this why there is slight curvature so if I analyse that I have to plot like this so that means if I divide this as this zone or also yeah this zone right so here it is almost constant within the experimental error why mass transfer and the effect of temperature on mass transfer is not that much okay it comes as 2 to the power of 2 to the power of 2 by 3 like that because of the diffusion coefficient but whereas in this region when I have the reaction control now this goes as exponential so that is the reason why if you extend this this will be your normal minus E by RT line so you see now wonderful phenomena like empty control here and this is reaction control that means kinetics we are definitely evaluating wrongly if you do not understand the you know the chemical factors and physical factors that are coming into the rate equation that is the lesson what we want to I want to tell you here that is the only point I want to tell you that means unless you understand the difference between chemical factors physical factors that are affecting the rate of reaction by simply conducting experiment and then trying to find out the kinetics and using them blindly may make us lot of error may give us lot of errors this k value this k not which is a actually combination of mass transfer coefficient and reaction rate constant okay if I do not know this then this I am imagining that as a first order rate equation yeah like this is the one this k 0 I am simply because this is order 1 and here this now I have simply take I have the that is equivalent to first order rate constant but actually that first order rate contains two factors okay so that is the lesson which we have to learn that means in heterogeneous systems we have to be careful which is dominating like mass transfer or heat transfer or reaction or all three or all two and here if I have all two this entire equation I have to use right and most of the time for us likely we have the mass transfer coefficients available in the form of correlations that we do not have to do much work now because any system you take you have some information or other unless you want to design totally a new system okay and for new system anyway the whole research the whole work you have to do on your own and then confirm that what you have done logically right good so now you can see that what I have assumed here is yeah here equation 1 it is k s and c as okay it need not be that means here I have taken on the surface the reaction with as a first order reaction the moment I just change here I have second order reaction everything will change okay so everything changes so the moment I have the second order reaction assume second order reaction then that equation 1 is written as minus r a o b equal to k g c a g minus c as equal to k s c a square what is the procedure what is the procedure we have been discussing to find out the overall rate or global rate or you know other names measured rate yeah eliminate c as now if I want to eliminate c as this equation number 12 I will write 12 if I want to eliminate c as then I have the quadratic equation correct no I have quadratic equation so I think at this I will give you as an exercise and if you solve this and then get as a first you have to get c as and that c as you can substitute either here or here okay in any one so if you are able to do that then for second order reaction the rate expression what you get is r o b equal to k g by 2 k s this is 2 k s c a g plus k g minus square root of k g square plus 4 k s k g c a g this is the equation this is equation number 13 you see now how simple equation yeah and this is the one which you have to use here again in this design expression yeah this here minus r o b right so that is what what you have to integrate and cubic equation much worse because I think we do not know you may get various forms like negative positive or you know realistic route unrealistic routes imagine routes all kinds of things so that is why you have to really see the effect of physical and chemical factors on the reactor design I hope now you understood this entire diagram every term has a meaning now for you at this at this point of time I hope so okay why because input I know how do you get input market survey MBA guys okay they happily give us so we will take it then kinetics now we have to determine contacting you have to choose choose in the sense that you have to choose whether you have batch reactor or continuous reactor and we have already thumb rules for choosing gas I mean batch system or continuous system Kavya remember when do you choose batch when do you choose continuous that is why number of times I repeat do not worry I think you know do you remember no general thumb rules when do you choose a batch system when do you choose a continuous system I know because you know you may not remember when I told you that is why number of times I try to repeat that is what I am asking when do you need a continuous system we need okay yeah that is general one thumb rule small scale or large scale and you also have another thumb rule particularly for batch systems when you need flexibility in the product so today you may design you may produce particularly pharmaceutical one drug next day another drug next day another drug like that or pesticides insecticides and also you have the dice you know suddenly people all people may be interested in red colour if the movie red comes you know there is some movie called red okay so then afterwards if the movies are something about yellow and all the green comes everyone may be using that so this industry has to cater to the market for them it is flexible if you have batch systems whereas continuous system once you design you cannot use it for any other system unless hundred percent overlap of the parameters are there for these two processes process A process B right so that is the reason why once you design continuous system it is dedicated only to that unless otherwise you are very very lucky to have exactly the same parameters same temperatures same controls then only that is possible right so that already we have that information I know I am repeating this because it is not your interest to remember it is only my interest to remember and that kind of silly thing I may not give in the examination okay so that is why you may not remember that is why I repeat many times okay so once you have chosen between batch and continuous in continuous you have again two possibilities right plug flow and mixed flow so when do you choose plug flow and when do you choose mixed flow I mean do you remember or yeah see all of us want only always higher conversion okay yeah I have given some thumb rules again from the okay good why yeah the gas phase reactions have very small residence times in fact in the in the form of only seconds okay so and when I have in the form of seconds a residence time it is very easy for me to provide residence time in plug flow plug flow we know that it is definitely more efficient than mixed flow because of their residence time distributions that we have to bring residence time distribution definitely that is efficient that is why all of us would like to happily go only for plug flow but still we do not go many times the reason is if I want to provide 8 hours 10 hours 12 hours mean residence time in a plug flow reactor I told you the length may be 100 kilometers 200 kilometers because theoretically speaking what is the velocity which will give you plug flow again I am repeating yeah Swami theoretically what is the velocity which gives you plug flow theoretically okay why infinite and why turbulent tell me there must be some reason no just imagine plug flow flat velocity profile each and every particle spending exactly same time for each and every particle to spend exactly same time I should have a flat velocity profile flat velocity profile comes only theoretically if I have infinite velocity otherwise still near the wall you have small droplet I mean small drop there it is not straight straight line so that is why okay so when I use very very large velocities if not infinite velocities then the residence time will be length should be very very large for providing 8 hours 10 hours length of the reactor and always plug flow reactor the diameters are anywhere you go it will not be more than 8 inches 8 inches is maximum yeah 8 inches is really maximum so generally we will go in industry either 4 inches 5 inches 6 inches and that is why the reason also there is that you have to maintain that flat velocity profile so that you are very close to plug flow that is the reason why we do all that so this is the reason why the residence times must be smaller for plug flow reactor if I have of course liquid phase reaction still the residence time is very very small that means rate of the reaction time is very very small still I can go for plug flow that is again under thumb rule we use we have plug flow and mixed flow whenever we say that we have gas phase reactions where we have very small residence times required for this reaction to take place then you go for plug flow and very large residence time mixed flow is easy for me because it is a tank so any amount of residence time I can really give if I have a tank reactor because only length by diameter L by D we will just increase that is how I told you 3 weeks is the mean residence time for wastewater treatment plant and did you see any time wastewater treatment plant no never who has seen wastewater treatment plant you have seen yeah what is the size when you see when you saw that big parts yeah if it is normal water we can even swim there all of us can swim not only one okay so it is not a you know tub bath or something it you can really go one or two laps this side that side really that big that big is very easy for me because 3 weeks residence time easily I can give there so that is why that is why in industry for large productions even tank reactors will be 1 meter cube one is not smaller 1 meter cube is very very large thousand liters okay or 2000 liters you have to put the diameter the size so that is the reason why tank reactors I can provide any any kind of residence time that is one thumb rule the second thumb rule also I told you temperature control highly exothermic reactions even if it is gas phase but very very sensitive okay that means one temperature one degree that way this way may catch fire okay so I mean that fire definitely we have to avoid otherwise you know you will not be there or the persons working in the industry will not be there totally because explosions right so that is why under those conditions we can again go for a mixed system one of the beautiful examples are you know thalic anhydride I think to produce nylon or so it is a catalytic reaction but highly temperature sensitive it is a catalytic reaction so first they tried packet beds but when they tried packet beds it is very difficult in the sense that temperature control they are really worried every time when they go to the plant for operation the operators must go to 10 temples at least and then pray the God and then go to the industry otherwise there is no guarantee that they will come back safely even then there is no guarantee but it is only okay but only for our safe for our confidence we go there and then pray and then come back okay so that is why there they have gone to fluidized bed where fluidized system gives some kind of mixing temperature control is beautiful there in fact the same story what I have told you is also given in Levenspiel gas solid systems catalytic catalytic reactions okay so that is why so now we know that when do you choose batch system and when do you choose the continuous system and of course when do you choose plug flow when do you choose mixed flow please don't forget this okay please don't forget I think you know in a number of times I can repeat I will be repeating but next time when I ask you please answer that right but I think by repeating of course I don't blame you that is why I am repeating most of the time and at least this diagram we should thoroughly understand I say all other things are details calculation of volume if I am using mixed flow or plug flow or batch reactor that is the detail later but right now the concept of that what are the things that are involved in the diagram right I hope input we know contacting when do you choose between batch and continuous and in continuous again when do you choose mixed flow and plug flow that we know and this one is kinetics next one kinetics and chemical and physical kinetics the meaning is chemical and chemical and physical kinetics meaning is the steps that are involved the steps that are involved here I have actually that is in the last class I have given those three steps and all that that is what I have not written here so in the last class we have written there mass transfer from bulk to the surface then the rate of reaction then diffusion of oxygen sorry CO2 back but because it is irreversible reaction oxygen coming through the film will not affect my rate of reaction if it is reversible reaction the reversible reaction depends on again you know the yeah product concentration also they are going to be so I know here it is minus r A equal to K1 so that you can remain K2 C B right reversible reaction this is K1 and K2 yeah so if I have a reversible reaction the product also affects C B is the product B is the product so but here in this example I have CO2 coming but CO2 is not participating in the reaction that is the reason why I have to take only two steps mass transfer to the surface rate of reaction on the surface and under steady state conditions both must be same and those two we have written here this is the procedure for all heterogeneous systems I have given one but I think in the tomorrow's class I will also just draw the profiles for another two three systems so that you are familiar with simple drawing you know profiles and this imagination is a must I tell you please open up your mind and then have this imagination like I told you here how the carbon is burning you take one particle and there is a flame and anyway fluid mechanics will tell me that there is a film which you can't see around this but oxygen has to diffuse because oxygen is not available on the surface it is available only in the bulk it has to go through the film and then reach the surface then on the surface you should have sufficient temperatures and conditions for reaction then rate of reaction takes place then the product gases will come out if there is no product gas no problem that is fine okay and so many steps are involved out of that what are my important steps and how do I show them as pictures all this is imagination what I have done and now this is the picture and this picture leads to me just to draw the profile graph right this is the profiles like that I may have some three four steps and everywhere I have to draw the profiles and list out what is step one step two step three step four and you will see here most of the time reaction is only one step and mass transfer steps can be four or five that is why mass transfer is so important in heterogeneous systems particularly if you take our dhalda reactor salary reactor there are four five steps involved and you take normal catalytic reaction simple catalytic reaction if I have this is the particle and similarly the reactant is going that A going to B and it is not reversible reaction we assume that it is irreversible reaction catalyst particle A is going around so now this A has to diffuse through the film first because the reaction is taking place only inside the particle because this is porous particle here at least we have assumed it is non porous particle right so I have the particle and then going around this and you have now first step is mass transfer through the film second step is it is porous particle mass transfer is through the pores and when it is going through the pores it will also get adsorbed and if I look at one molecule this one molecule has to go from bulk to surface through the film step one step two it has to diffuse through the pores and after diffusing some distance it has to get adsorbed after adsorption it will form a complex and all that people say and then afterwards it will get dissociate and then product is formed that is reaction adsorption and then reaction after reaction the product which is formed there has to deserve so that is five and after deserving it has to come out through the pore that is six and then it has to again come out through the film to the bulk seven steps which one is controlling we do not know that is why in heterogeneous systems the kinetics is really hell that is why happily we ask chemist to do that okay you develop Langmuir initial wood kinetics you know Langmuir initial wood kinetics will give you all these seven steps and we are also going to derive some of them and I will also show you at the end one all seven controlling what kind of equation you get all seven steps controlling and unfortunately in all this is that you have these constants you know K g K s in this catalytic reaction you have two mass transfer coefficient steps okay and then adsorption desorption all these constants surface reaction all these constants are involved there so you have to determine from the experimental data these constants very very accurately that is why chemical engineers are also should be very good you know mathematicians I am not saying mathematicians to develop new theorems and all that but at least for using the mathematics that are available at least with us so in the in analyzing our data and you know how difficult it is to evaluate this constant K K 1 K 2 K 3 K 4 more than 2 we have always difficulty in evaluating the constant I am telling so how do you find out normally I have the rate equation I have the experimental data now rate equation has K 1 K 2 K 3 now I have to fit this entire data to give this K 1 K 2 K 3 right matlab is one which will help us right and then matlab we will go and then we will give some guess values and it evaluates K 1 K 2 K 3 and then gives you you bring those constants then I also done the same experiment experiment and same data in fact I also want to analyze using this K 1 K 2 K 3 to evaluate then I will give a different guess value then my values and your values may not be unique that is a physical system it must have only one unique set of values this is what is the biggest problem if I have more than two parameters that is why Levenspiel beautifully tells that you may produce any number of wonderful models for publications but for industry you should give at least the minimum you know one is the best one or two two parameter models he never complicates things because the moment you go more and more 4 5 6 and all that I do not have any physical meaning for that no physical meaning for these values okay I think I have already told you sometime back may be residence time distribution I do not know undock did you do C S T R with dead space and bypass a model for C S T R or mixed flow with dead space and bypass you are not doing okay there you will have three parameters which is called a bypass fraction a dead space fraction and dead space is not really dead definitely there will be you know mixing some kind of exchange between dead space and act to zone what we call exchange coefficient all these three bypass has a definite meaning 20 percent bypass 30 percent bypass okay and what is the other one dead space has a definite meaning may be 10 percent dead space or 20 percent dead space 100 percent cannot be dead space okay that is very bad design okay yeah so then this exchange coefficient depends on what kind of mixing is going on three parameters so if I evaluate data if I say if I get a set and if you evaluate if you are getting a different set what is the physical meaning for this I may be telling that I have 15 percent dead space you may come up with 30 percent dead space but that system at that particular conditions it must have only one set of values that is why that uniqueness in evaluating these constants is one of the biggest challenges in chemical engineering even if you do research at high funder level when you have more and more I do not know how really I will appreciate this bio technology people where they will have 25 parameters and mechanical engineers for combustion 400 parameters for combustion okay really if you see that there are so many models with a lot of constants I used to wonder how are you going to evaluate all these 400 parameters 400 constants so that is why as engineers we should present our system the experimental data modeling should be as simple as possible and beautifully Einstein told that model can be as simple as possible but not simpler really it is a wonderful statement what he has made you can make your model as simple as possible but not simpler so that we have to remember and then I think I will give you some examples in the next class and after the next class then we will go to non catalytic reactions first I think I have not told you how do I evaluate you till now because I thought I will give you this introduction for that particular figure and then tell you that