 Good morning. So, in the last lecture we looked at how to incorporate the effect of external mass transfer. We are talking about a particle, a catalyst particle around or there is a film around it, this surrounds a catalyst particle which is a hypothetical film and the concentration varies inside a film. Outside of film, the concentration is a bulk concentration. At the surface, the concentration is C A S and inside a particle, of course you have the reaction taking place. You may have effectiveness factor being significantly or being substantially lower than 1 or it can be equal to 1 depending on what is the pore structure and all that. But we are not worried much about that. We have, we are worried about the external mass transfer resistance. So, in that case, the overall rate can be given as if you have a reaction which is first order then R A is equal to minus K C K R C A divided by K C plus K R. Let me quickly tell you what we mean by K C. K C is the mass transfer coefficient, mass transfer coefficient and K R is the reaction rate constant, rate constant. And of course, we are talking about a bulk concentration which would be C A B there because everything we are expressing in terms of the bulk concentration and that is the main purpose or motivation behind deriving this equation. Now, again like I will repeat like if you write expression in this particular form, the units of K C and K R should be same. So, the mass transfer unit, you remember D A divided by delta, what is the unit of D A? D A is say meter square per second or centimeter square per second, delta is meter. So, unit of mass transfer coefficient is meters per second, it is meter per second. Now, the K R that is a rate constant of the first order reaction, it should be in this particular case it should have the same unit as that of the mass transfer coefficient. Now, the K R here is the rate per unit external surface area, rate per unit external surface area. So, if it is a volumetric rate constant, the unit is second inverse whereas, in the case of in this particular case, we have the unit which is equal to K C right fine. Now, let us go ahead and of course, we further try to interpret this by saying that R A is equal to C A B divided by 1 by K C by 1 by K R sorry K R and this is a negative sign here because, we are talking about the rate of reaction for A. Now, depending on the relative magnitudes of these constants reaction is either external mass transfer control or reaction control. So, if K R is very large then R A is equal to K C C A B where C A S is almost 0. If K R is very small or yeah if K R is very small then it is governed by the reaction, the overall rate is governed by the reaction it is going to be like this. In this case C A S is equal to C A B right the rate is expressed in units moles per second meter square very important and meter square is a is a is a cross sectional area external surface external surface and not the internal surface. So, always remember that we have been talking about different surface areas for internal mass transfer, internal diffusion sometimes we express our rate constant in terms of internal surface area which is significantly large substantially large compared to the external surface area. But, right now this is this has a unit of flux that is moles per meter square per second all right. So, let us go ahead now and try and understand is external mass transfer in more detail. Now, external mass transfer suppose external mass transfer is important now it is the rate given by this particular expression. Now, this is a very simple expression where K C is very important is a mass transfer coefficient how do we get it this expression looks quite similar to a first order rate equation when reaction is intrinsically kinetically controlled right adsorption is negligible desorption is negligible at all other adsorption coefficients are smaller. So, the denominator is unit. So, R is equal to minus K C. So, again the expression looks quite similar, but meaning is quite different ok. Now, this K C here in this expression is different from the K R from the K R. Now, K R we know K R is a rate constant which is of course, a frequency factor has a same unit as that of K R. So, if it is per unit external surface this will also be per unit external surface area then exponential minus E by R T. Now, it is very easy or it is it we know how to calculate this or we know how to determine this through the experiments in laboratory I do experiments at different temperatures I get plot for 1 by T versus L N K and this plot the slope will give me delta E or E rather and then you have the intercept giving A of course, this is semi log plot L N K versus 1 by T right. So, you know how to get K R you know how to get A delta E by performing experiments at different temperatures in laboratory and get a value of K R. So, for a normal reaction which is controlled by intrinsic kinetics I do not have any problem designing a reactor. Now, if it is external mass transfer which is playing an important role how do I design a reactor that is a question. So, first of all you need to know the external mass transfer coefficient the value of external mass transfer coefficient under the conditions that you are doing the reaction in fine. Now, let us talk let us spend some time on this external mass transfer coefficient because that seems to be the important parameter here mass transfer coefficient right. Now, it depends on which parameter as now always remember this relationship easy to know what K c is dependent on. As I said depends on the diffusion coefficient and it depends on delta that is film thickness and this film thickness is in turn the function of say turbulence or how well is the mixing what is the flow pattern. Now, who is going to decide all this this is this in turn decided by the reactor geometry flow patterns are decided by the reactor geometry actual flow rates. So, in term of dimensionless numbers which is the dimensionless number which is important in this case it is a renewal number n r e or r e and is another number which is something to do with the properties like viscosity, then density, then diffusion coefficient this is a dimensionless number in mass transfer which is called as smith number right. So, the mass transfer coefficient depends on these parameters or these particular factors the smith number, renewal number and the reactor geometry. So, there are correlations for mass transfer coefficients. So, what people do know normally that perform experiments under different conditions they know that mass transfer coefficient is going to be function of renewal number is going to be function of smith number. So, what I do is that perform experiments in the laboratory at different values of renewal number different values of smith number under otherwise similar conditions get a functionality get the dependence and that is a correlation for the given reaction geometry. See, I have a tubular reactor say a fixed bed reactor this is the correlation for the mass transfer coefficient. I have a fluidized bed reaction this is the correlation for the mass transfer coefficient. In every correlation you want to see these two numbers, but that correlation is specific to the reactor of interest fine. Now, let us make an analogy because you have probably learned heat transfer before mass transfer. So, we know there is something called as the Nusselt number in heat transfer this is something called as Nusselt number how do we define this? This is something to do the convection or heat transfer coefficient. So, it is nothing but H which is the heat transfer coefficient H then some D now D is a characteristic dimension it can be a diameter it can be a diameter of the tube by k. Now do not do not confuse this k with the rate constant we are talking about heat transfer here. So, this is nothing but a thermal conductivity and this D is not diffusion coefficient this is diameter or some dimension this is the conventional heat transfer coefficient right and this is the Nusselt number which is given by this particular ratio of product of H in D to k. So, Nusselt number tells you about the heat transfer coefficient now this Nusselt number is related to again two numbers a non number and something called as Prandtl number. Now of course, this is not just the product we will see that but there is something some raise to m some m and a some constant a. So, this is a correlation this is a correlation so, you will always see a correlation in this particular form now from reactor to reactor from geometries to geometries these constants would vary right. But the otherwise the terms are same terms are same now you may ask me a question that suppose this is stagnant medium there is no flow. So, Reynolds number which is functional velocity is 0. So, does it mean that there is no heat transfer at taking place does it mean that heat transfer coefficient is 0 that is not true because I have a stagnant medium say water I am heating it from one side it is going to get transferred right. So, for very low values of Reynolds number or for a stagnant medium you have a relationship given by 2 plus a into R e raise to n P raise to m. So, this is a number 2 it so happens that at very low value of Reynolds number Nusselt number becomes equal to 2. Now for mass transfer you have see this is an analogy heat momentum and mass. So, this dimensionless numbers see there is always a counterpart for one number in heat transfer to another number mass transfer. So, which is a counterpart for Nusselt number in mass transfer the counterpart is a number called Sherwood number Sherwood number very important what is the counterpart R e no there is no counterpart. So, R e is there in mass transfer also what is P R P R is C P mu by K where K is thermal conductivity C P is specific it mu is viscosity for P R there is a counterpart in mass transfer which is nothing, but we already mentioned it that is mid number which is given by mu by rho d where d is now the diffusivity d is diffusivity and mu is viscosity. So, mu by rho is kinematic viscosity sometimes you denote it as this mu divided by d I hope it is clear. So, there is an analogy for Nusselt number the analogous number in mass transfer is Sherwood number for P R this analogous number in mass transfer that is Schmidt number. So, like this correlation I will have a similar correlation in mass transfer given as Sherwood number is equal to 2 plus A into Reynolds number raise to M and Schmidt number raise to M. So, I got a relationship and this is the relationship that I want this is something that I am going to use to estimate or to calculate the value of K c because that is what is required for my reactor design. And these constants are known for the given geometry. So, what I will do for example I want to design a fixed bed reactor say fixed bed reactor which is packed with the catalyst and there is a gas or fluid or rather liquid going inside. I will perform experiments well planned well designed experiments to estimate or get a values of Sherwood number for different values of Reynolds and Schmidt number. And once I have this I can estimate the values of this like what I do for normal kinetics I determine rate constant I determine activation energy I determine frequency factor from the laboratory data. So, I am doing a similar thing here I am generating I am generating laboratory data by doing experiments under different conditions that is at different values of Reynolds number and Schmidt number and getting the value of Sherwood number I do not need to do experiments under reactive condition can be a non-reactive also. So, simple example say I will take naphthalene balls naphthalene has a tendency to sublime. So, I will just pass a gas at a certain temperature I will see what happens how much naphthalene gets vaporized what is the rate at which it gets vaporized that depends on the mass transfer camphor. So, the many possibilities you can get a value of mass transfer coefficient. So, need not be the reactive experiments conducted under reactive conditions it can be experiment for just mass transfer just mass transfer from solid to liquid or solid to gas. So, doing such experiments you determine the values of by doing some such experiments determine the values of this a n and m and get a final correlation for the fixed bed reactor and not just fixed bed it can be fluidized bed. So, I have a certain correlation. So, typically for a fixed bed reactor now this value if you are doing it at very large or relatively high velocities then this factor becomes very high compared to this. So, I can neglect this two. So, I may write this as say 0.6 and just typical value for a fixed bed reactor with cylindrical or rather with circular cross section 0.5 S3 raise to 0.33. So, this is a correlation this is a correlation for Sherwood number or in other words this is a correlation from which I determine the value of mass transfer coefficient. How do I determine? I can calculate the reward number for given velocity I can calculate smith number I know diffusion coefficient I know characteristic dimension. So, I know the mass transfer coefficient if I know the Sherwood number how do I define Sherwood number by the way I did not tell you that sorry. So, Sherwood number is something similar to Nusselt number. So, here instead of convective heat transfer coefficient I have a mass transfer coefficient k c into some characteristic dimension if it is a fixed bed reactor with spherical catalyst particles then it is d p divided by instead of conductivity now I have diffusivity I have diffusivity. Now, this is diffusivity here do not get confused with diameter I will use this for diameter while defining Nusselt number, but now it is diffusivity. So, many variables and not many alphabets that is why this is you have same alphabets coming again and again all right Sherwood number. So, once I have the value of Sherwood number from the correlation I know the diffusivity let me denote it as d a for component a I know the particle diameter I can get a value of k c all right now let us see k c depends on what. So, as I said before let it is the something due to the flow pattern flow patterns are decided by the reactor geometry and velocity. So, let say I have a cylindrical reactor a fixed bed reactor how do I determine. So, which which are the parameters or factors or variables rather that I can play with to manipulate value of k c. So, for rate constant for the reaction I change the temperature because I know Arrhenius law tells me that if I increase the temperature k increases drastically. Now, here which are the parameters that I can play with I have already told you Reynolds number and Schmidt number. Now, Schmidt number is what mu by rho d these are all properties of the components that I am using I cannot really change them of course, for the gas phase diffusivities and viscose it is are functions of temperature they vary, but of course, not significantly we will see that later. So, which is the main parameter that I can play with the Reynolds number and Reynolds number in this case the Reynolds number that we are going to define is u that is velocity meters per second d p is a particle Reynolds number. So, let me write it as d r e p particle Reynolds number d p right rho divided by mu or u d p divided by mu kinematic viscosity right. This is how I am defining Reynolds number you know definition on d v rho by mu instead of d now I have d p here because of particle Reynolds number because I am taking particle radius or diameter as a characteristic dimension here to define Reynolds number because that is the that is the most appropriate dimension here as far as the mass transfer coefficient is concerned because I am talking about mass transfer coefficient around the catalytic particle and not for the mass transfer of wall to the fluid inside fluid to the wall I am talking about particle. So, now it is a particle diameter that is the more important fine. So, this is Reynolds number. So, this u u is something that is going to decide a value of s h and hence k c why because s h is a strong function of Reynolds number you already seen that r e raise to 0.5 and what else in all this again there is another parameter that is coming again and again which I can play with is d p is the d p here is the d p here. So, let us put everything together and see how the mass transfer coefficient varies with d p and u together. So, s h is equal to k c right into what d p divided by d a is equal to 0.6. Reynolds number u d p nu raise to 0.5 right into smith number raise to 0.33. So, how is the dependency now this diameter is coming here also here also. So, s h is equal to sorry let me not write s h let me write k c is proportional to u raise to 0.5 and if you take d p here d p raise to 0.5. In other words u by d p raise to 0.5 here it is minus 0.5 because d p comes here. So, this is the relationship or proportionality that I see k c is proportional to u by d p raise to 0.5. Of course, remember this is for fixed bed reactor cylindrical geometry and circular cross section. It may slightly vary here and there, but then I am just showing you the importance of these parameters as far as external mass transfer is concerned. So, let us go ahead. Now, let me do experiments in laboratories. Now, I know which is a factor that is affecting or making more impact on mass transfer coefficient. For reaction intrinsic reaction is the temperature whereas, in this particular case it is u by d p it is a u by d p. So, I do experiments in laboratory for a fixed bed reactor at different values of u by d p I change u I change d p I do it right. And I plot this graph by getting the rate under otherwise similar conditions. Now, you may ask me like it may vary along the length and all that again same principle probably I will just take small length small length of the catalyst or small amount of the catalyst. So, for that I need to plan experiments very well. So, that I do not bring in other effects. So, under otherwise similar conditions if I perform experiments for different values of u by d p raise to half sorry raise to half and determine the rate say r a or minus r a rather. What am I going to see? Can we predict the nature of this graph? It will come down like this or we go like this remain constant or it will go up come down. What can we say? If I have very small value of u by d p what it means is the mass transfer coefficient is going to be very small. I have a reaction taking place now I am talking about reaction that is taking place on the catalyst inside or outside. Why? Because catalyst is responsible for the reaction I am doing the experiment here say cyclohexane dehydrogenation, cumene decomposition, ethyl benzene dehydrogenation whatever. There are so many reactions possible. I am doing those reactions and I am determining the rate. Now, this where I am going to value of u by d p is very small means if I want to compare intrinsic reaction rate and mass transfer which is going to govern the rate it is the external mass transfer that is going to govern the rate. Why? Because value of external mass transfer is very small remember r a is equal to minus c a for a first order reaction 1 by k c plus upon 1 by k r. So, if k c is very small this is going to govern because this is very large this is going to be 0 or negligible compared to this and the reaction will become mass transfer controlled. I go on increasing u by d p what will happen? The value of k c will increase the rate will increase because k c will come in the numerator as I go on increasing the value of u by d p k c will increase the rate will increase. So, it is going to be like this I go on increasing now look at the nature then look at the train what is happening here? It is going it is attaining saturation here it is not increasing it further after certain value of u by d p I do not see a rise in rate can you say why or can you explain why? It is all because of the fact that after some value of u by d p the k c becomes so large that now the reaction is controlled by k r. So, it is crossing the value because u by d p are not affecting k r remember that k r is affected by the k r is affected by the temperature. So, if I change u by d p only k c is changing and because of which I am getting increase in rate, but after some time this k c becomes so large that it will overtake k r and k r starts controlling. So, in this region I have external mass transfer control and in this region I have reaction control. So, this is how it is going to be like. So, you it is quite similar to what we have looked at before as well like for intraparticle diffusion remember that you have two regions if you plot d p versus r and you see that this in this range you have intraparticle and this say you have reaction is a reaction control it is in this case you have intraparticle diffusion control. So, something similar that you are seeing here here it is reaction control here is a mass transfer control. Now you may you may ask me the question where is the intraparticle it is here right now we are just clubbing all those other effects in this remember this graph only tells me that if I am here the reaction is not controlled by external mass transfer it is controlled by either reaction or internal diffusion or adsorption desorption whatever. So, I have already told you while doing this analysis I am clubbing all those effects I am putting together all those effects in this reaction term and external mass transfer is treated separately. Now, for a first order reaction tell me like what will happen if I have some effectiveness factor. So, it is very simple what will I do is this the rate of the reaction is C A 1 by k c plus 1 by k r. Now, if I want to incorporate effectiveness factor also what will I do I do not need to do anything much I need to just do this why because you already seen this is the definition of eta definition of eta. Now, this what I have done is I have isolated effects of now I may call this as k r dash because this is intrinsic when the molecules are near the catalytic site and this is taking care of the pore diffusion effects this means the eta value and why it appears C again look the way they are appearing in this equation. Why eta is appearing as a product and not like this separately 1 by eta or something like that why because diffusion and intrinsic reaction they go in parallel whereas, reaction and mass transfer they are in series. So, when they are in series then you see the effects as to separate terms then parallel you see their product. This is how you should read equations this is how you should read or interpret the final equation that you get see there is no point in biasing this equation is a much in this like if you understand it very well you would be able to write it properly. But of course, remember this is for the first order reaction. So, we have so we know what is mass transfer coefficient we know how it changes with velocity and particle diameter and how do we know that reaction is controlled by mass transfer resistance or by intrinsic reaction. So, now once we know that reaction is controlled by mass transfer coefficient or sorry external mass transfer then how do we design a reactor. So, let us look at that I have a fixed bed reactor which is packed with the catalyst I want to determine the length or height required or weight of the catalyst required to achieve certain conversion. This is our problem this is our problem finally it boils down to this every reactor engineer reaction engineer should think of a reactor design as his final objective all we are doing is to somehow to be able to design this particular reactor that we have in mind fine. So, if I am in this particular region u by d p raise to point 5 I am here then how do I design the reactor. What is the rate equation r a is equal to minus k c a divided by 1 plus k c plus 1 sorry 1 by k c plus 1 by k r and k r is very large compared to k c. So, it is going to be minus k c c a right now what is the unit by the way what is the unit of rate here the unit of rate here we have been using that right from the big thing the unit is moles per meter square second here. Now, I want to design the reactor. So, let me write a mass balance this particular term looks quite similar to the reaction term of course I have to take care of the units fine. So, let me take the differential balance it is like like my normal plug flow reactor what is the equation for the plug flow reactor it is going to be like if I take the balance coming in going out and delta z becoming almost 0. What I have is d f a by d z divided by ac that is cross sectional area plus r a into ac let me define all these terms is equal to 0. Now, it would be very clear for see in normal plug flow reaction plug flow is like d a f a by d v d a f a by d v plus r a sorry it should be minus here is equal to 0. See remember like for a normal plug flow reactor d a f a by d v is equal to r a this is what we have been using reaction engineering part 1 right see the difference here it is the same equation except the fact that instead of r a I have r a into ac why because the unit of r a here is per unit area and here it is per unit volume right. Now, look at this units here f a is moles per second this has unit meter cube right. So, it is same as r a into ac right. So, same as this. So, r a also has a unit moles per second meter cube. So, this fellow this particular term will also have the same unit moles per meter cube second, but you already say that r a has the unit moles per meter square second I would not I am I am multiplying it by ac to bring it to this unit. So, what is the unit of ac? Ac has a unit meter square per meter cube. So, how do I define ac? Ac is the mass transfer area per unit volume volume of what? Volume this volume volume of the contactor volume of the contactor volume of the reactor. So, this volume is volume of reactor. So, how much mass transfer area I am providing per unit volume of the reactor that becomes very very important right. So, how so ac now let us talk about ac, ac is interfacial area or mass transfer area empty area rather. So, in the reactor I have this particles which area am I talking about I am talking about this area which is at the external surface right. So, it is easy know I know how many particles I have I know what is the particle diameter I know void age for a fixed bed. So, from that I can calculate this. So, it is nothing but the property of the reactor and the packing that I am using the catalytic packing that I am using. So, it is something that is given to you or it needs to be found out depending on what is given and what is to be found out right. So, for a given the extent of conversion I can get a value of ac that is required or for ac given I can calculate a conversion. So, it is quite similar to what we did for reactor design again go back to your reaction engineering part 1 for the given volume I calculate conversion or for given conversion I calculate volume or later what we did was like when was late was expressed per unit weight of the catalyst I said fine if you have conversion given you can calculate a weight of the catalyst required or if the weight of the catalyst is given you calculate a conversion. Now, here in this problem I am calculating the area required for the mass transfer for the given conversion or for the given area I can calculate a conversion. Now, this area depends on what area depends on the amount of catalyst that I am taking it depends on the particle diameter and a wattage because it is per unit volume of the reactor volume the reactor right. So, this is the meaning of interfacial area per unit volume interfacial area sorry I should have interfacial area per unit volume nothing else is important here because the entire reaction the rate of the reaction is governed by the external mass transfer. So, it does not depend on the pore structure it does not depend on the activity of the catalyst I have already I have already concluded by doing experiments that it is governed by external mass. So, nothing to do with the temperature of course, like temperature will slightly make an impact on diffusivity and all, but catalytic activity pore structure pore volume tortuosity adsorption coefficients none of them are making impact on the rate it is external mass transfer. So, which are the parameters then remember the equation for the mass transfer coefficient Sherwood number ok. So, Reynolds number and Schmidt number for the given properties diffusivity viscosity and density it is the velocity and the particle diameter all right. So, A c is a very important parameter here and when if you look at the equation you have this which is similar to a normal plug flow equation only difference is I have this correction here term of A c because I am expressing R A per unit area. So, let us go ahead and derive this further F A can be converted to concentration this can be written as say F A F A will be multiplication of flow rate and concentration right. So, it is this can be written as D C A into E U divided by or D by D Z of C A into U U is the velocity it would be volumetric flow rate, but since area is there volumetric flow rate divided by area is velocity plus R A into A c is equal to 0. You can have a separate notation for R A because the way we are defining right now it is per unit area as long as you remember the units of R A that are being used you do not have to worry much about it. So, here I am multiplying it by A c means it is per unit area fine. So, I can take U out of the derivative if it is constant now you know when it is constant in there is no pressure drop velocity is not changing as we go from inlet to outlet temperature is not changing isothermal conditions. So, many assumptions are there, but if I am allowed to do that then I take U out D C A by D Z plus R A A c is equal to 0. Now, suppose I substitute for R A now what is R A? R A is equal to minus K c into C A minus C A S right remember C A C A B minus C A S sorry C A B here it is C A S linear difference between these two and if it is external mass transfer control you know approximately this is equal to 0 right. So, since C A B is very large compared to C A S this becomes 0. So, I can I can say this as C A B minus C A S sorry C A B here it is C A S linear difference between these two and if it is external mass transfer control you know approximately this is equal to 0 right. So, since C A B is very large compared to C A S this becomes 0. So, I can I can say this I can simplify this further rate fine. So, let us substitute for R A what I get here is minus U D C A by D Z is equal to K c K small k c that is the mass transfer coefficient C A B or C A C A B C A B here and then A C. Now, I am working in terms of bulk concentrations. So, earlier also I should have written C A B let me do that C A B bulk concentration C A B here everywhere you can make C A B is a bulk concentration that I am working with fine. So, this is a very familiar equation right. We know as Z increases C A B goes down. Now, this looks like our normal rate equation for a first order reaction only difference is there it is intrinsic rate constant. Now, I have a mass transfer coefficient multiplied by interfacial area per unit volume right. So, I can solve this equation analytically C A by C A 0 is equal to exponential minus K C A C divided by U into Z. This is how the concentration varies of course, you need to know the boundary condition what is that boundary condition at Z is equal to 0 C A is equal to C A 0. If I use this boundary condition for this differential equation is O D ordinary differential equation I get this profile of concentration along Z. Let us all are known once I start my experiment or reactor right. So, this is quite similar to what we have got before for a first order reaction in a plug flow reactor only difference that you should remember is that meaning of these constants are different mathematically they look similar fine. Now, so as the length increases from 0 to 1 what is L L is a length of the reactor C A by C A 0 goes like this a typical plug flow type behavior I can I can convert this equation I can convert this equation to conversion form and you know how to do that 1 upon 1 minus x is equal to K C A C divided by U into L that means for the given length what is the conversion. So, substitute L for Z in that equation in this equation substitute L for Z in this equation what is C A C A is equal to C A 0 minus sorry C A 0 to 1 minus x all right again I can write B here sorry I keep forgetting that, but it is like you know the meaning of it C A B is equal to C A 0 clear. So, I get a relationship between length and conversion length and conversion I need to know K C I need to know A C I need to know U velocity. Now, this K C will again I have spent enough time how will you get this K C K C will be obtained from the correlations Sherwood number Sherwood number fine. Now, so you have this correlations given may be we can solve some problem later, but before that let us quickly look at one nice example given in Fogler's book all right before that like let me talk about every time we have been talking about porous catalysts and all that, but then if you remember first or second lecture on catalytic introduction to catalysis I said that different types of catalysts apart from porous there are some non porous catalysts also where you can have a gauze or you can have a plate or you can have a monolith where pore diffusion is not important reaction takes reaction is so fast or you do not need many many catalytic species or atoms for the reaction to for them to catalyser reaction. So, I can just take a gauze a wire gauze or I can take a foil or which reaction will take say alcohol to aldehyde silver gauze you have ammonia to nitric acid reaction is instantaneous, but then there is a competition between intrinsic reaction and mass transfer because once the once the reactant reaches the catalytic site reaction takes place immediate. So, most of the times these reactions are governed by mass transfer they are governed by mass transfer. So, you have to identify the area for the mass transfer how much area you are providing and that is good enough or that tells you about a rate of mass transfer if you are increase the mass transfer rate or overall rate because rate is governed by mass transfer you can increase the area available or you can increase the overall number and you may get higher rate fine. So, there is one example or a case study that is given in Fogler's book which is quite good. So, I thought I will just discuss that now you have two reactors say fixed bed reactors I want to put them in series and I will I will realize certain conversion here right I will realize certain conversion here. Now, instead of using this reactors in series I may use them in parallel I may use them in parallel right what are you going to see now parallel means I am going to split my feed in two parts equally and these reactors are also of equal length let us assume that and I do this. So, the conversion here x 1 and conversion here x 2 are they going to be same are they going to be different if at all x 1 is higher than x 2 or x 1 is lower than x 2 we need to see that. Now, we already seen in reaction engineering part 1 that for normal reactions which are governed by intrinsic kinetics this does not matter for a first order reaction this does not matter irrespective of fact that you put it they put them in series of parallel you get a same conversion. But now the reaction is controlled by mass transfer external mass transfer I would say external mass transfer if that is the case then you will see a difference why will you see a difference look at this I have very large flow rate here I am using the same reactor of this dimension like the velocity here say u 1 and the velocities here say again the velocities are going to be same here say let me call them as u 2 u 2 right u 1 and u 2 which is greater u 1 is greater than u 2 right. And if the reaction is controlled by mass transfer the same volume now which is provided for the same feed it is likely that this combination is going to give me higher rate compared to this of higher conversion compared to this so x 1 is going to be greater than x 2 you can prove that. So, x 1 is likely to be greater than x 2 so do it as homework let us see what happens look at equations for the write down the equations the plug flow reactor or whatever like you are not plug flow your tubular reactor just now we have derived that you just derived this know this is the equation that we have derived now this equation you write it for every reactor reactor 1 reactor 2 sum them up and similar thing you can do it for these two simultaneous and see what happens you will realize that because the velocity is higher here than here for every reactor here individual reactor the mass transfer coefficient is large because the number is large the reaction rate here is higher compared to these two. So, this is going to give you better conversion than this all right. So, we will continue this discussion later and we will solve some examples thank you.