 I think we have done the non catalytic reactions and we started with that performance equation diagram now again I am going to draw the same diagram this is reactor I think this diagram will come even in your dreams I think output kinetics interacting then we have here chemical physical batch continuous and here we have in continuous P of M of and the performance equation is given as output as a function of input kinetics contacting so this is the equation when you are discussing about non catalytic reactions this equation has been beautifully used I do not know whether you appreciated or not I appreciate myself because there is information that is required from kinetics that one we have seen for single particles and contacting mainly comes through RTD in fact okay yeah ET either ideal or non ideal so that means we have a broad framework for all reactor design right now let us see in catalytic reactions what is that that is required input again is definitely required because that determines the plant capacity and kinetics we need now a rate expression and we also had a rate expression for non catalytic reactions okay that is mainly in terms of X B and time here we will have in catalytic reactions the rate equation will be in terms of minus R A and as a function of you know the partial pressures or concentrations again you need a rate expression but what form it will be we will now see okay then contacting anyway it will come if it is a packed bed if you have a rate expression packed bed catalytic reactor where the solid particles are catalysts now and the difference between non catalytic and catalytic is in non catalytic reactions the solid will finally be converted into a product right so and it may even disappear like in coal combustion or coal gasification whereas in catalytic reactions the solid will not change the size is same and if you assume that activation deactivation is not there so then lifelong you can use this catalyst as a catalyst particle okay so under those conditions somewhat it is easier than non catalytic reactions because the solid here is a inert as far as reaction is concerned but only it is increasing the rate of reaction inside the particle when the molecules go but it is not participating directly in the reaction okay so it is easy for us that means when I put a packed bed let us say 1 centimeter particles I pack and then if there is no deactivation till I day I can use of course now I have only short span but even if you also start till your end you can always use packed bed okay so like that even fluidized bed moving bed again all the reactors that is why I have been appreciating chemical engineering so much because you have the contacting patterns any reactor you bring you can bring you can break into only 2 ideal reactors so that is why in one sense you do not have to learn much as far as the contacting pattern is concerned only you have to learn is mathematics analysis depending on the complications right so I mean even non catalytic reactions also I have done only simple things and one of the simplest thing what we have done is only single particle and change is only in the particle there is no gas change that means there is no concentration change in the gas phase okay normally we can maintain that in industry by sending large amounts of gas particularly when the gas is not very costly like air you just compress and then blow over the coal particles and you do not have to recover anything with that but whereas hydrogen and all that when you use definitely we have to take that also into account but concepts are same only mathematical analysis will be a little bit difficult that is all and that is the reason why you know we are giving you 6 or 7 courses mathematics you know in B Tech 4 5 you I think some of you will have done 6 and in M Tech also you have mathematical methods in chemical engineering and I think and every transport phenomena is nothing but it again mathematical course if you do not appreciate the physical phenomena of the processes so everywhere you see mathematics in chemical engineering because the process are complicated right so now let us see how we develop this kinetic expression as I have been telling you this we know only 2 reactors and batch that is all 2 continuous and 1 batch and this also I know the job we will give to MBA people they will go to market and then say that so much you have to produce okay so then we will know our production then this is where we have to concentrate most of the time depending on what kind of heterogeneous system you have we had gas solid non catalytic reactions liquid solid also same except that when you have that pseudo steady state assumption that is no more valid so again mathematically complicated that is all other than that you do not have to learn anything new there for example gas liquid systems same procedures all the time develop a rate expression go to contacting pattern and you know the input and substitute this kinetics in terms of rate in the contacting pattern equation and then you have to integrate good okay so in the kinetics of catalytic reactions most of the time we know that we have only solid catalyst we also have liquid catalyst but I think you know that is that is called homogeneous catalysis okay for example biodiesel is one of the examples they use liquid biodiesel from seeds and all that they are doing now jethropa and all that so there the catalyst also is in the form of liquid but the liquid catalyst should be dissolved in one of the liquid phases you cannot really identify whether there is a catalyst or not only through reactions only you can see because it is completely dissolving which you cannot see that okay so but we are talking about gas solid catalytic reactions because that kind of reactions are many many in industry can you give me examples gas solid FCC FCC what is the catalyst? zeolite zeolite okay one more example hydrogenation reaction is there hydrogenation reaction is there I am asking gas solid I think very simple example my favourite example all the time Kavya ammonia ammonia what is the catalyst? iron catalyst I think you know the simplest one what I have been telling all the time is sulfuric acid sulfuric acid okay in one step you have to use catalytic reaction what is that SO2 SO3 SO2 SO3 and what is the catalyst? so like that of course many many many catalyst are used in industry because just to increase the rate of reaction and it is heterogeneous system so what we imagine again here is one single particle right so I may have a particle okay something like this again beautiful spherical particle we are trying to draw and then we have here the film and because most of the time we use porous particles we have to also show that pores so this may be something like this or okay it may be whatever way you want to draw you can draw it okay so here also may have so this may be solid this may be solid so something like this you will have the solid particle and now if we the gas will be flowing like this this is gas reactant gas now as usual we can imagine what are the number of steps that are required for the reaction to take place step one will be okay so step one is yeah mass transfer of okay let me also say that we have a reaction something like for simplicity again A going to R this is catalyst A going to R that is catalyst so now this is actually gas A then step one is mass transfer of reactant to the surface correct yeah step two yeah where yeah so unless it goes it does that it cannot so this is adsorption or diffusion we will say not adsorption diffusion of A through the pores yeah and step three adsorption of adsorption of A through the surface step four yeah reaction on the surface that is step four and step five desorption of product from the surface step six diffusion of product gas through the pores and step seven is again mass transfer of products through the film through the film to bulk gas seven steps seven commandments yeah so all of them are they are you know series steps parallel steps or series parallel all of them are series okay so if I take one molecule the molecule has to first go through the film then diffuse through the pores then get adsorbed and our imagination is after getting adsorbed and an active site active site then that will dissociate into some kind of intermediate and then the reaction takes place okay and after the reaction the product is formed at the same site but now product has to leave the site why you can happily sit there catalyst yeah but why you should leave yeah concentration gradient because outside you have less concentration of product and inside you have more concentration of product because of the generation there because this is converted reactant converted to product so that is why it has to leave so because of the concentration gradient it comes out and then again the diffusion through the pores and diffusion through the film and then comes out okay good out of this how many are mass transfer steps and how many are reaction steps which is the one step four but what about desorption and adsorption what do you tell them is it mass transfer step or reaction step yeah so we call them 4 5 and 6 sorry 3 4 5 are the generally reaction steps which contains you know adsorption desorption and surface reaction okay adsorption surface reaction and desorption all 3 so that is why all 3 first we take because that is where the reaction taking place the remaining 4 are mass transfer steps right and through the film we already have the equation same frosting equation again same range and marshal correlation here also why I have the film surrounding that I have the I mean I have the particle surrounding that I have the film then reactant has to go through that and at steady state when the continuous process is going on we also have the products coming out so now you have again assuming that equivalent or counter diffusion right so products are coming out reactants are going in so a kind of dynamic equilibrium is established there steady state again okay steady state so that mass transfer coefficient I can find out through correlations and always mass transfer depends on hydrodynamics I do not know whether anytime you realize that first of all you should know what is hydrodynamics okay hydrodynamics means it is the fluid mechanics around the particle are inside the reactor okay so mass transfer coefficients and heat transfer coefficients always depend on hydrodynamics are in general fluid mechanics that is why first fluid mechanics we teach before going to heat transfer and then mass transfer okay why why it should depend because if the particle is stationary and the gas is moving around that or fluid is moving around that we will have a kind of relationship between the particle and then the fluid this is in the packing if I use the same particle in fluidized bed again the relationship between the gas how it moves and then you know the solid how it moves is totally different so here in a fluidized bed the particle vibrates particle hits other particles and all kinds of things will happen okay it may temporarily jump and then again fall down all kinds of things so the mass transfer process or heat transfer process surrounding this will be different when compared to a packed bed where the particle once you put there it is there all the time it is not changing but the same thing again if I put in a moving bed moving bed that means now packed bed is no more packing then the entire bed is slowly sliding down then again how the hydrodynamics what we call or fluid mechanics around the particle again different right same thing and now you can put the same particle in rotary kill again mass transfer will be heat and mass transfer will be different so that is why heat and mass transfer coefficients or the process depends on what kind of hydrodynamic conditions what you have that is what actually my research area for long you know so many years I am always trying to find out in a reactor before the reaction starts what kind of hydrodynamics are established and not only this particularly heterogeneous systems packed bed for example we say wide age of or wide fraction of solids and wide fraction of next phase liquid or fluid in general okay it can be liquid or it can be gas then how do I know that because it is packing here and is not moving I know because I can find out exactly what is the wide age but the moment I go to fluidized bed where particles are moving how do I know that what is the wide age the fractions again wide age means the fractions okay fraction of solids and fraction of gas how do I know and depending on my velocity in the fluidized bed the bed may expand because normally in a fluidized bed it is open okay there is no top distributor whereas in packed bed between two distributors it is tightly packed so that is why even though you use velocities much higher than terminal velocities of the particles still particles will not go still it is packed only because you are not allowing it to move at all it is tightly packed so that kind of situation will not be there in fluidized bed so the particles will expand the bed will expand so naturally the fractions of gas and also solids will be different now just imagine for three phase you have two continuous phases gas is continuously going liquid is continuously coming then you have the solid okay and again many varieties are there like you heard of trickle beds catalytic reactors where do they use I mean in a sense we do not know the oil hydrogen desulfurization desulfurization of you know oils sulfur is removed using zinc oxide catalyst or so so there you have hydrogen and also oil as well as the solid catalyst but you know trickle bed is a packed bed but the other two phase are moving gas and liquid but again your fraction of gas and fraction of solid liquid depends on again what kind of flow rates we have so that is why hydrodynamics are very very important and many people in industry never bother about that and this because the establishment of volume fractions depend on the flow rates at a particular flow rate automatically the phases will have certain fractions that is established due to steady state conditions at that particular flow rates if I change the flow rate again gas volume may change solid volume may change solid volume is constant generally when it is not when it is catalyst when it is not reacting in the reactor but most of the time catalyst but gas and liquid will change because even in a fluidized bed you have the solid particles and you are not allowing them to go because even if they are going out like you know FCC there is a lot of solid particles which are going out I mean old FCC process not recent one okay so then you collect them and again send back now the recent one is first fluidized bed anyway they are going out let me allow to go out they are going out and again collecting in a cyclone because of deactivation again regenerate and then we will send it back okay so but the old process only normal not first fluidized bed normal fluidized bed when they use that means the velocities are much lower then they will put a cyclone and then try to collect not only of course FCC in any other catalytic process you know catalyst should be as porous as possible the moment you have as porous as possible what will happen to the mechanical strength of the particle decreases so when mechanical strength decreases because of attrition you will have lot of fines going out so that you have to collect in the cyclone and if it is possible again you have to recycle back otherwise you have to take them away and then again make similar particles and then send it back so that is why any these mass transfer steps in any reactor will be different depending on its own hydrodynamic conditions that is why mass transfer even though we say that you know we have understood it is very difficult for a totally new reactor now I do not know whether you heard of platinum mesh type of reactors mesh you have seen mesh no where they see in your house also I think nowadays no housewife also is doing to separate that rice good rice and bad rice you may have very fine powder rice not very fine powder small small pieces okay and you will have normal rice so to separate them in houses villages they use to see that kind of sieves are used as packed bed one sieve above one sieve above one sieve it is the production of hydro production of nitric acid in one of the steps there because the hydrodynamic conditions are different mass transfer coefficients like you know range and marshal correlation will not be useful you have to develop on your own another correlation so and you know the of course we drive cars and then we use cars but we do not know there is a catalytic reaction okay in exhaust pipe and that is monolith entire catalyst is only one particle that particle is simply put into the exhaust pipe okay and it is not solid again you know it is a porous one it is a wonderful design where you will have this is the cylinder like this then you have different plates arranged like this and gases go through this and still good engineers will try to put this one as like this what is the advantage some more surface area always chemical is always how to get more surface area whatever you do okay that is why I think you know by the time you come to extraction you are tied in mass transfer and maximum amount of time we spend only on distillation everyone knows that afterwards we do not know many operations okay next maximum is only absorption but in extraction if you go you know what kind of equipment they use mixture settlers of course again packed beds cu plates in all that this hydrodynamics is very very very difficult in liquid-liquid extraction why do you think so I mean at least try to think in liquid-liquid extraction it is not that easy to maintain flows through these columns why do you think so in absorption where you have gas-liquid system it is easy to do that it is not that difficult as liquid-liquid just can you try to expand your brain why it can be solubility is fixed actually in liquid-liquid extraction what we will not solve in the other okay only one of the components in one phase will only try to go to the other phase these two are supposed to be immisible okay we are now right now facing also we have liquid-solid system but still you know it is like a moving bed type where we want to use that one for of course it can be used as a bio-chemical reactor or it can also be used as adsorber to remove colour and all that because it gives a beautiful efficiency as far as adsorption is concerned because it is a steady state process normally all adsorbers are unsteady state process because we put the adsorbent as batch here we are trying to make this one as continuous and when you have the continuous solid is continuously coming steady state liquid is continuously going so whatever the pollution control boat tells us that okay you should have 10 ppm in the outlet we can design for that whereas in the other system where the solids are in batch condition initially you will get zero because large amount of adsorbent and the same concentration right so you will get zero very pure water then after some time slowly the concentration in the outlet goes on increasing and then finally it will reach your original concentration in the inlet so this pollution control boat people will not allow you that so they will say that yes always fixed only 10 ppm example okay ya you are not answered me why do you think that why you have to expand I say you have to even if it is wrong answer tell me no problem try to think diffusion who told you such a diffusion of what gas into the how the two phase are flowing flowing in the reactor or in liquid-liquid extractor of course we have liquid-liquid reactions also they are also again hydrodynamics come what is the driving force we have two liquids what is the driving force if I am not pumping them how they flow ya it is only the density difference for liquid-liquid systems most of the time the density difference is not that much so that is why it is very difficult to make them flow the way you want and the separation also takes place lot of time because particularly when you have very close densities whereas if I have gas and solid solid will have normally you know 1000 kg per meter cube and gas will have only 1 kg per meter cube so then separation is so fast that is why gas solid fluidized beds are very easy to design whereas liquid solid fluidized beds are again not easy that is what the right now we are facing my research student you are not able to separate the solids from liquid when he is conducting the experiments on hydrodynamics okay so that is why mass transfer and hydrodynamics are very well related okay and one has to remember that for every system you have a different correlation it is not universal the way I was telling you always range and marshal correlation range and marshal correlation is only for one particular particle and around that you have ya gas going like that even in a packed bed you have a different correlation range and marshal correlation is not valid only place it may be valid slightly easier fluidized bed why because in fluidized bed particles are supposed to ya move around independently so the gas may be going around each and every particle so if no correlation is available for a actual fluidized bed then range and marshal correlation is generally used that is single particle correlation generally used okay good so now what we do is we take step 3, 4, 5 and combine them as reaction step and this is where what you get is that Langmuir Hinshelwood kinetics okay what we do is we take first these 3 steps and then we develop a rate expression you will get minus r a equal to in terms of you know k a k b k c you have I think I also told in the beginning of the course what kind of rate expressions you get for catalytic reactions okay minus r a equal to p a minus p b by k divided by you have adsorption desorption constants okay so that is the kind of equations what we are going to develop if you take steps 4 steps 3, 4, 5 and then once I have that rate expression that means it is totally not affected by these 2 steps these 2 steps what is the meaning mass transfer is not controlling the rate of the reaction okay so then once you have this understanding of these 3, 4, 5 steps then you can have how to combine this diffusion and may be you know diffusion is may be rate controlling most of the time depending on the pore size of your catalyst so that means the catalyst particle is able to react because of large surface area that is available in the particle or exothermic reaction the temperature may be high in the particle so rate of reaction may be fast but you do not have mass transfer coming through the sufficient mass transfer through the pores sufficient reactant coming through the pores for the reaction so then how do we combine that reaction as well as diffusion and we have a general thumb rule whenever you have this kind of porous particle the pores through the particle right and when I have film which will be controlling generally diffusion through the pores are controlling I mean these are the things which you have to remember there is no choice and if you have a good decent interview these are the questions that will be asked Lousy interview is asking how many mothers how many fathers how many sisters how many brothers so this is very good very well answered so take the job and then you end up in IT okay that is the kind of jobs beautifully they give you right but here whenever you have this kind of interviews in chemical engineering these are the things only asked I mean do you have any thumb rule like you know what step you can neglect and you know there are questions I mean there are interviews where what are the possible steps in any catalytic reaction I myself asked many times but I think even PHD interviews also we used to ask what is the difference between homogeneous reaction and heterogeneous reaction okay that they will say oh that answer straight forward comes what is that one phase are more than one phase that is excellent then we ask okay so what I think if there is more than one phase what will happen what is the extra information you require for the design okay so that is why all these things are very important for us also to remember and I am not saying that you know these are very difficult this is very very simple things to remember provided you imagine the process that is why always I say always in my teaching I try to ask you to imagine what is happening in the process automatically mathematics come easily the moment you imagine the process okay most of the time we start with mathematics and we forget about the actual process good so let us take this steps 3 4 5 and then develop the equations for that I think you have to take some continuity notes yeah we call them as surface reaction models we call them as surface reaction models which involve adsorption reaction and desorption all 3 good here also we will try you know all 7 all 7 may be rate controlling loziest situation possible but I think you know it is possible then we will derive one equation that will give you a wonderful idea okay please take this physical mass transfer steps are not rate controlling when surface reaction models are developed okay or in other words these steps are very fast compared to surface reaction okay I think slightly I found out words we will use here in the bracket you can write that is interface and interface transport processes are faster these are slightly I found out words people who want to impress they say always interface interface okay because again you know our subroutines have to work you know what is interface the moment I said interface interface nela like this so that means you know trying to get all of us do that nothing wrong okay so interface is what interface is what by the way yeah interface is this and this interface is yeah through the pores okay so those 2 okay good and isothermal condition is assumed good next one you write surface reaction rates can be expressed in 2 ways one langmuir hinshelwood formulation one is langmuir hinshelwood formulation please write here in this just below that in this the rate is expressed in terms of in this the rate is expressed in terms of surface coverage theta surface coverage theta, and then employing the langmuir isotherm to relate to relate theta to fluid concentrations second one is Watson approach Haugen Watson approach that is second one just write below that Haugen and Watson derived rate equations in terms of surface concentrations of absorbed species and free sites and then expressed these concentrations in terms of langmuir isotherm I think in your physical chemistry book that theta approach you could have seen when you are deriving langmuir isotherm langmuir isotherm okay but you have seen Haugen Watson book you have used I think 8 meters you would have never used I think you do not even know know there are 3 volumes volume 1, volume 2, volume 3 volume 1 is material balance okay energy material balance volume 2 is thermodynamics volume 3 is kinetics there is no heat transfer you are talking about Coles and Richardson only these 3 in volume 3 you have lots of catalytic reaction formulations kinetics and catalysis I think that is the title of volume 3 remember correctly okay volume 2 is thermodynamics and volume 1 is material and energy balances and those are almost the first books in chemical engineering and what a wonderful books you know all the books because directly the data from industry were used at that time most of the I think Shankar was asking me he wanted to give some reactor problems for I think that optimization course or so so he wanted to have good kinetic data then I told him just go and see catalysis that book by Haugen and Watson they have wonderful data given actual experiment on the industrial data afterwards anyway we went very far from industry and industry also ran away from us and now again we are trying to come together because now industry I do not know whether you have observed industry has now intellectual people not the I am talking about percentage one because of the computation they have taken the best people that are possible international computation so that is why their processes must be totally new and updated and always they have to be one step ahead when compared to their competitors so that is why even though 100 percent basic theory they have not understood wonderful information from chemical engineering has been used to develop these processes okay and also their techniques are academic institutions nowhere can come because of course they have lot of money so they can use those techniques and techniques means I am talking about analytical instruments for the analysis and all that because I told you know small 500 ml medicine also may be costing 1 lakh 2 lakh rupees so that is why they have to be very very that is why industry has now excellent talent now they want to come and then of course now the again you know always life is cyclized you know in the beginning they were together in between separated that is also most of the time happens you know first love you come very close together get married fight every day then go away then general will come again come together for the sake of children everywhere any process you take the cyclic things may happen okay so that is why industry nowadays you know we have how much money that is coming to IIT from either industry or other research laboratories private laboratories or government laboratories tremendous amount okay anyway so that is why I think this history also sometimes I think you require you know Hogan Watson or the people and what is the third name it is not only 2 that is why when I was HOD I used to push you know these B Tech boys to have quizzes chemical engineering quiz and some of us used to give all those questions now they will also run away the moment you say chemical engineering you tell me MBA quiz they will run yeah run towards that not chemical engineering yeah what is the third name no one remembers okay I think very nice you know I think you do not remember even our basic books no idea at all trying to recall all files are searching in the mind Rangan's R A G A T Z yeah now you remember no one remembers yeah these 3 people you know because of this and another 3 people also wrote wonderful book who are the 3 McCabe and Smith is only 2 I say Harriet is I think recently yesterday he was born the original one Harriet I think you know Smith and van is only 2 again 3 BSL Bud Stewart Lightfoot of course there are 4 others mass transfer book I do not know whether you remember that or not those are 3 these are 3 Faust Faust there is a book you know that is what you know we are even going away because of the computers we are going away from the text books also because all your information is only google no if you want mass transfer correlation you first google only right or e-transfer correlation only google I think the beauty is going but anyway again it will come back definitely please remember everything will come back to the books again yeah good okay so yeah take this one even though there is not much difference next paragraph you can write even though there is not much difference between these models Hogan and Watson models are easy to handle in which surface deactivation can be incorporated surface deactivation can be incorporated full stop to do justice to both these schools that means Hogan Watson school and Langmuir Hinshelwood school to do justice to both these schools let us refer them as Langmuir Hinshelwood Langmuir Hinshelwood Hogan Watson Hogan Watson formulations we call LHHW equations LHHW kinetics LHHW models all these things LHHW means Langmuir Hinshelwood Hogan Watson models okay this is what one of the questions I gave also in the zero test where what information you get from LHHW models what information you get you get a kinetic expression for catalytic reactions okay good let us first derive Langmuir Hinshelwood Hogan Watson what are the assumptions of Langmuir Hinshelwood Hogan Watson very good monolayer coverage okay uniform surface conditions that means I think all surface behaves in the exactly same way isothermal conditions these are all assumptions okay Sukumar last monolayer coverage uniform surface activity thermal conditions these are the assumptions in LHHW next one you can write the surface may be divided into Theta and 1 minus Theta where Theta is fraction of surface covered by Theta is fraction of surface covered by adsorbed molecules and 1 minus Theta is bare surface 1 minus Theta is bare surface that means the surface is free of any adsorption so rate of adsorption rate of adsorption we call minus rA equal to k into p the partial pressure of certain gas into 1 minus Theta okay rate of desorption equal to minus rD equal to minus also minus you do not have to put can I tell this one yeah some okay see Theta is the surface that is covered with the molecules now rate of desorption depends on how many molecules you have in the surface more coverage means more desorption and the other one is the rate of adsorption depends on what is the free surface because it is only monolayer only one molecule can sit there okay at one particular site so that is why this is proportional to the surface area that is available that is 1 minus Theta and that is also proportional to the concentration of that is I think PA by RT is concentration partial pressure okay the pressure on the surface of a particular gas okay so at study state so rA rA equal to rD you just equate those two and then k because this is easy so I thought I will cover this and then leave you so do you remember the equation Theta equal to capital K p by 1 plus K p where capital K equal to what yeah k by k dashed okay of course sometimes this is also imagined as small v by v where small v equal to volume of gas adsorbed divided by volume of gas adsorbed with a monolayer coverage with a monolayer coverage of surface that is always the fraction good anyway I think we will stop here this is what is Langmuir isotherm and even if you derive an equation for Haugen-Watson in terms of surface concentrations you will get a similar equation as Theta equal to only one equation K c into C g is the gas concentration and 1 plus K c C g this is Langmuir Hinshelwood this is Haugen-Watson it is same thing but that is expressed in terms of partial pressures and this K c is written because it is in terms of concentrations okay format everything is same good so these two the first equations and then we will actually derive the equations rate expressions for surface reaction where we have step 3 step 4 step 5 we will come into picture okay thank you