 So in diffusion flames we could actually look at two situations that are slightly different from each other, one is a co-flow diffusion flame or the other one is a counter-flow diffusion flame. We will talk a little bit about each of these, the reason why we need to talk about a co-flow diffusion flame is because essentially we are now talking about two different reactants that are approaching each other the question is when you now say flow you are now talking about how do they convict, previously earlier we were talking about how they mix with each other mixing is essentially depending upon primarily the concentration gradient that is prevailing for each species to mix into each other but the question about the configuration of the flame is how do the reactants flow to each other. In the case of premixed flames when we are saying premixed the means the reactants were mixed with each other so that means they flow together there is no way you are going to have the reactants flow from different sides or something of that sort right but the moment you are now talking about diffusion flames where the reactants are coming from, coming differently and then they have to mix at the flame then you could now begin to talk about these things right. So a co-flow diffusion flame would a typical configuration would be where you have for example in a duct you have a splitter plate and you have a fuel on one side and oxidizer on the other side and the splitter plate terminates at its lip and then you now have a mixing region that means as the flow is going up like this the fuel mixes like this and oxidizer mixes like that right and somewhere in there you are expecting a flame to happen because that is the flame is where they mix with each other right. So this is what is called as a co-flow configuration there are several situations which actually mimic this for example if you now think about like a Bunsen burner right. So you have a fuel that is coming in the oxidizer is actually ambient it is not exactly flowing but effectively you have a flow that is being set up by the fuel and of course the products are also actually flowing outward because they are hot and in a buoyancy driven flow they basically go outward and therefore you have an entrainment of the air. So the entrainment of the air is also causing a locally co-flow situation as far as the flame is concerned and you could also have like a burner with an adduct where you have a fuel in the middle and oxidizer in the outer annulus and so on so all these things are basically like a co-flow situation a counter flow situation is where you are essentially trying to have fuel come in this way and oxidizer coming this way and you now see if you do not want to have any shear layer effects in the co-flow case you would now say equal velocities for both the fuel fuel and oxidizer that means this is coming in at a velocity you this is coming in at a velocity you the same same velocity. So you do not necessarily have like a shear layer between the two and you are focusing mainly on the species mixing rather than any momentum mixing and all that stuff similarly you could also think about in the counter flow situation a counter flow at the same velocity for example. So if you now think about this in typical examples would be like what is called as in a post jet burner so if you now have like a jet here the jet here this is like the nozzle for the jet or the jet mouth if you will and then essentially they are now apart by about a distance h let us say right and then they are coming in with a certain velocity. So what this is going to do is you are going to have a flow that is going to go like that right and as the fuel is flowing like this and the oxidizer is flowing like this they are mixing with each other and when they mix with each there somewhere in there not necessarily in the middle of the flow but somewhere in there you are going to have a flame that is that is a diffusion flame right and this is a flame where you can actually hope to establish something that is very flat the flame would be quite flat because the mixing is going to happen equally on both sides except now the problem is you do not have a uniform velocity field for the flame in fact you are not going to have what is called as a high strain rate here and the flame is actually being stretched because of the flow wanting to actually go this way on either side as a matter of fact you could also have a frame-to-frame configuration here in the in the opposed jet a burner how would you do that you could now have a fuel and oxidizer coming from the same nozzle let us say from the bottom and then you now send a opposed jet of inert gas let us say nitrogen right and what would this do it would now create like a stagnation point flow. So this is essentially what is called as a stagnation point flow because by symmetry this would this would be as if like you had one jet that is impinging on a wall so if you now have a jet that is impinging on a wall it just spreads out and you have a stagnation point right at the middle right. So this creates like a stagnation point flow in the in the middle and essentially that both the flows together now are like mirror images of each other if the velocities are the same and so you have a stagnation point flow and you could have a premixed flame which is flat which is subjected to a strain right that that is an experiment to actually investigate the effect of the strain rate on the premixed flame but the case of diffusion flame you can actually hope to get a flat diffusion flame as opposed to in this case what is going to happen as the flame could actually be curved depending upon whether you have a we will talk about this in greater detail pretty soon depending upon whether depending upon the stoichiometry of the mixture right. So you could actually have a dilute mixture for the fuel or dilute mixture for the oxidizer for example if you use air then it is already diluted with lot of nitrogen for the oxygen and then it depends on the stoichiometric coefficients in the reaction like the stoichiometric coefficients means if you now we need more oxidizer for a given amount of fuel for stoichiometric reaction and so on so these things will actually dictate what the shape of the flame is but you cannot really expect to have a have a flat flame okay so if you want to have and the reason why we would like to have a flat flame for example is if you want to treat it in a one-dimensional manner right but there are ways of one-dimensionalizing these flames and so on we will discuss those things in greater detail but effectively you could think about a two configurations like a co-flow and a counter-flow geometry which makes more sense in the context of diffusion flames now what we will do now is actually pick the co-flow geometry and then look at what is called as the Berkschumann problem now the Berkschumann problem is a very fundamental problem in fact it was presented in the first symposium on combustion in 1928 by Berkschumann and it was subsequently published in the Journal of Industrial Chemistry so it is a very celebrated piece of work and it has had numerous citations subsequently and lots of different extensions to this problem and the name the phrase Berkschumann flame connotes a certain set of assumptions that it embodies and it refers to certain limits of like very high activation energy, infinite rate chemistry and so on so the Berkschumann problem is as follows I will describe the problem first and then list out the assumptions that are involved in this the Berkschumann problem is essentially that of two concentric ducts you could do this problem in a Cartesian coordinate system where you have like two channels right so this is like infinite plates and this is also infinite plates so these are essentially two coaxial channels or you can also do this in the in an axisymmetric manner that means you are actually having two circular pipes that are concentric to each other and the original Berkschumann problem was done as a pipe problem so it is an axisymmetric problem and the inner pipe terminates at a particular point the outer pipe extends to infinity so the inner pipe is a semi-infinite pipe it is coming from infinity below and up to this point the outer pipe is infinite all the way right only on either side and you have a fuel coming in at a certain mass fraction yf not inside on the inner duct and you have oxidizer coming in in the outer duct at a oxidizer mass fraction of yo not so these are the inlet mass fractions all right and they are also coming in at all all the flow both the flows are actually at the same velocity u so it is actually uniform velocity and we are basically supposing that you have a uniform flow everywhere and at this velocity all right we are going to adopt the Schwab-Zelovich formulation where we are going to solve only the species equation and the energy equation perhaps we will actually find that we will also make other assumptions that requires us to solve only the species equation and not the energy equation and we can decouple them therefore we are not going to solve the momentum equation we are going to prescribe the velocity field and the velocity field that you will prescribe here is just uniform velocity everywhere for every any any point in the flow for any mixture whether you have only reactant only product a mixture of reactant and product a only product or only fuel only oxidizer among the reactants whichever way right so everywhere you are basically going to say the same velocity exists now of course this is not the this is not a good assumption okay for several reasons you could first of all think wait a minute I am looking at pipe flow shouldn't I actually have something like a parabolic velocity profile of its laminar right and then parabolic velocity profiles for the outer outer flow and so on then what about the no slip boundary condition of these are these walls for this velocity profile that is not worry about all those things that solve a fluid mechanics persons problem okay I am not going to worry about all these things I am going to worry only about the combustion problem the second thing is we are we are not going to bother about the density variation with temperature so the density varies with the temperature then as if there is a flame and the flow goes through the flame it expands because of thermal expansion and therefore when the density changes the velocity will change as well right and then you the velocity field will now depend on the flame but that that is now like a convoluted problem I need to prescribe a flow field in order to actually work out the flame and the flame is now going to change the flow field so what do I prescribe so that that that is a problem that I that I will have to reckon with if I have to couple the flow and the combustion problems together right if I want to keep them separate I will simply have to prescribe a velocity field in spite of the flame that means I have to I shouldn't worry about the effect of the flame on the flow field this is reasonable because up to pretty much up to the point where the flow is heating the flame it is not expanding so and it is a convective situation of all the all the flow that is going out so all the expansion is happening downstream of the flame it is a it is a it is an approximation alright but still it is a reasonable one okay so this is typically the problem the question that we are asking this problem is what is the flame shape and this is very very important when compared to what is the flame structure okay in the context of premixed flames I have clarified the distinction between a flame shape and a flame structure so when we are looking at one-dimensional premixed flames right we were we were saying let us look at the preheat zone let us look at the reaction zone like let us look at how the temperature increases and then tapers off and how does the the reactant concentration falls in the preheat zone and then tapers off at a low value all these things and how does the reaction rate grow and all those things right now that is looking at the structure when you now look at the structure of the flame that means you are looking at the spatial variation in temperature profiles concentration profiles and reaction rate profiles and so on right shape of the flame in the context of premixed flames was something to do with like a conical flame above a Boonson burner or a wedge flame in a in the case of a ramjet and so on so that is like a gross thing and it is almost like we are looking at a sheet of a flame where in you the structure really happens right and then the way we were actually handling it is by looking at the G equation and how the flame propagates with this certain flow speed and balances the normal component of the flow those things that is what we were doing here again we are some it is sufficient for us the wave work and Schumann post this problem to work out the shape of the flame okay and so what we are expecting is the shape of the flame to look somewhat like this or somewhat like that right or somewhat like this I do not know that is something that I am going to work out so if you can imagine like a like a candle flame is something that is like a teardrop kind of shape right you could also have like a tent shape flame or if you had a fuel if you have a lot of fuel that is trying to actually mix with oxidizer the flame could open up right many times when you keep in mind when you when you are when you are looking at flames that are opening up it could be like a jet flame so a jet flame is where you could have like a very high velocity fuel and very low velocity oxidizer then you are mixing up in your mind the shear layer effects like momentum transport because of velocity gradients and you have a jet that is coming up in training and all those things okay it is possible to think about all these things that we are talking about in that context also but the fact that we are having having a uniform velocity is to remove any effects of shear layers from this and we are looking at only species mixing and no momentum of mixing effects no thermal mixing effects none of these things is brought into picture right so it is a very nicely post problem in the sense it clearly focuses on primarily two things one is that there are there are only two things that are happening here primarily one you have a axial convection of species and when I say species we are talking about reactants that are separate to begin with and then mixture of reactants somewhere in the middle and products also coming up as we go and then fully products are later on so you see this is not as nicely done as a one-dimensional premix flame previously so we had reactants here you had a flame you had products there right that was more and more of a black and white thing given in with very little fuzziness if you now go from here to there you have and then keep scanning across you will find all kinds of things that are happening up from the point where you had separate reactants and complete mixture products at the top right so we have to now evolve how this flame shape is happening so one of the things that is happening regardless of whether it is reactants that are separate or products that are combined is axial convection of species and it is balanced by radial diffusion of species you also have some axial diffusion alright I will explain this a lot more carefully some little later but the two primary things that are of most importance in a works human diffusion flame or axial convection and radial diffusion so this is the balance that we are basically talking about which is how well do the species diffuse while they are busy convecting right this is what is going to dictate how the flame shapes up right so frame shape is essentially dictated by the balance of axial convection and radial diffusion so once you get the flame shape the question of industrial interest for you is what is the flame height right so for example if you now want to design a furnace or a burner or a combustor and so on and you want to light up a flame at the burner you want to know how long you want to have the combustor walls right so you want to be able to cover up until the tip of the flame at least so that you can ensure complete combustion within the combustor so the type of the height of the flame is a practical interest right so the way Burke Schumann Burke and Schumann actually tried to get the flame height was to actually solve for the flame shape and once you get the flame shape you now find out where it actually stops at the central line or at the walls what is the vertical distance over which it spans and that is the height of the flame right so you can you can you can obtain this so this is basically the way the problem is posed and the solution procedure is to go through the shape of the flame so the adopt the adopt the Schwab-Zelowich formulation so right and the Schwab-Zelowich formulation recall comes with about 11 assumptions right and we did this adoption even when we started analyzing the premix flame structure and we may went through some additional assumptions maybe about two or three right so here again we will have to go through additional assumptions some of which I have already stated so the first assumption so further assumptions further assumptions number one if you now set up a coordinate system that is centered at the mouth of the fueled up in the middle and then say call this R and call this Z then let us say the VZ component of velocity is U which is a constant right given unlike in the case of premix flames where this is not given this is actually an eigenvalue you need to find out what is the flow velocity at which you need you have a flame that is stabilized we do not have to worry about the flame stabilization problem in the case of in the case of the diffusion flame as it is posed by Berkenshuman and that is very important for you to think about we will have to see later on what the stabilization mechanism is and we will bring in some elements of premix edness that is coming up that is the reason why we called it it is not exactly entirely diffusion flame we would simply call it a non-premix flame but so we will not worry about stabilization issues this is given and correspondingly we say VR is equal to 0 that means you do not have any radial velocities strictly speaking this is not true if you were to allow for the density to vary with temperature you will now have the flow diverge similar to the way the V we saw in the conical premix flame and that will acquire a radial velocity component but we are with disregarding this right the second assumption is the mass fluxes or equal that is rho V fuel equals rho V oxidizer right now putting things together what this basically means is that if the velocities of the two flows are the same right and this is the velocity field that is given there that means the fuel and oxidizer all of them have both of them have the same velocity if you say if the velocities are the same and the mass fluxes are the same that means that is simply means that the densities are the same and density is being the same is essentially what it amounts to saying the density does not vary right so the density does not vary not only because of temperature but also because originally the densities were the same so you do not really have to worry about any change in density in the first place and how does that the is it is it possible is it possible to have a fuel and oxidizer of the same density right the answer is in the laboratory it is possible to mix the fuel and oxidizer with different diluents in such a way that you achieve the same density for these diluted mixtures of fuel and oxidizer right so we say yf0 and yo0 so in this you have basically fuel with the diluent because 1-y yf0 should be the diluent in the outer at annular to a deduct you are having 1- yo0 is a diluent you could choose these diluents in such a way that the densities are the same right the next assumption is certainly a simplifying assumption which is rho D is a constant and this simplification is required for a from a mathematical standpoint and of course what this means is that if the density is constant D must be the same everywhere okay that is that is part of the Schwab-Zeldovich formulation anyway right so we are not looking at variation in D across the fourth assumption is diffusion in the axial direction is negligible diffusion in the axial direction is negligible that means we are saying it is negligible relative to the radial diffusion okay so that means we are saying wherever we have a du du z squared is much less when compared to partial squared yi over partial r squared right now this is to basically say that we are interested only in retaining the radial diffusion the convection is axial because we have disregarded any radial convection by this assumption right so we are prescribing a velocity field that is only axially convective and we now permit only radial diffusion so that we can clearly see the balance between axial convection and radial diffusion I want to point out that this assumption is not necessary you could keep the radial diffusion and sorry we could keep the axial diffusion and show that the axial diffusion is important when you have a small peclet number or it is negligible truly when compared to the radial diffusion at for large peclet numbers and large could be for laminar flows greater than 10 is large enough to 10 10 10 20 or 50 maybe is large enough right but less than 5 is something that you have to keep and what I will do is I will keep talking about the consequences of retaining the the axial diffusion but I would not solve that problem alright and interestingly I think at 19 this was actually published 1928 as I said I think it is 1982 or 84 there was a combustion science and technology paper by Chung and law that actually publishes the solution with axial diffusion retain okay so the interesting thing about these things is as and when people make more and more assumptions long later there could be somebody who relaxes an assumption and then you know you now have a expanded version of this or a more general version of this solution that is possible as a matter of fact we were talking about a Berkshuman problem adopting the Schwab-Zelda which formulation which essentially is a steady state problem right so you are looking at a steady state situation that means this this flame is not going to change shape with time and very recently in about three four years ago we relaxed the steady state assumption and actually did an oscillatory diffusion Berkshuman flame alright so what I mean to say is you could relax many of these assumptions and solve the Berkshuman problem and these things over a period of time begin to come into your exams and homework problems and assignments and all those things right the fifth assumption is perhaps the most important and insightful assumption that needs to be considered what we are thinking here is we say over the right words yes it is called the flame sheet assumption right flame sheet assumption means similar to how we adopted the G equation where we said the flame is essentially going to be a sheet it is a surface of discontinuity between reactants and products in the case of the premix flame as I said in this problem we are interested only in the shape we are not interested in the structure so it is sufficient for us to think about like a sheet right therefore we want to actually make the assumption that the flame is going to be a sheet the question is what is it that we are going to do in the equations that implements this assumption right what is the consequence of this assumption in terms of for analysis what does it really mean okay physically what this really means is this directly corresponds to what is called as infinite rate chemistry infinite rate chemistry or infinite kinetics infinite kinetics the other name for this is what is called as the mixed is burnt mixed is burnt approach so this requires some explanation what we are basically saying is this is very very important this assumption means means that the reactants react instantaneously when the mix in stoichiometric proportions okay and again that that needs to be further explained what we mean by stoichiometric proportions is you now have the fuel and oxidizer that are mixing right if you now think about even like a splitter plate problem let us suppose you do not even have these walls right you simply had only a splitter plate and this is a semi-infinite domain that is a semi-infinite domain and you had a plate that just stopped here and you had all fuel coming in this side all oxidizer coming in that side this is to say we are focusing on one lip here right let us not worry about what is what is on the other side then slowly what is happening is as the fuel and oxidizer are actually flowing out they are beginning to mix right. So as they begin to mix if you are somewhere very far away you are going to know that they are mixing that means it is all fuel if you are on this side very far away you are going to know that they are mixing it is all oxidizer but as you keep going from here to there all oxidizer becomes all fuel through a state where you have a mixture right that means there is going to be some particular contour where the fuel and oxidizer are actually finding themselves mixed in stoichiometric proportions right. And what we are saying in this assumption is the flame is going to be a sheet that is coincident with the stoichiometric contour in the mixing field right so let us just write that the flame is a sheet coinciding with the stoichiometric contour in the mixing field we will elaborate this a lot more after we go through the Berkschumman solution but at the stage I think I will just give you a peak of what is going to happen let us suppose that we do not have this assumption and we have what is called as a finite rate chemistry or finite kinetics okay and that is what is reality in reality if you now work out the Arrhenius law the law of mass action for the reaction rates you are going to get some number right then that number is going to be large all right for most combustion things but it is still finite let us say it is finite as the fuel and oxidizer are getting together right you are going to have more and more reactions happening if you find sufficient amount of fuel and oxidizer at a place because the law of mass action requires for the reaction rates to have concentrations of both fuel and oxidizer if you went a little bit further away and you found that you had only fuel and not know not much oxidizer that is mixed in you are going to have the reaction rates drop substantially if you go on this side and you find that you have more of oxidizer and you did not have hardly any fuel mixed to all the side you are not going to have concentration of fuel the reaction rates are going to drop substantially as a matter of fact if you think about the premixed flame you had a certain reaction zone thickness which was restricted because you are running out of the deficient fuel that means the reaction constant reactant concentration comes almost to zero right here the reactant concentration comes to zero from opposite sides for the two reactants there both the reactants were actually coming from the same side of the flame here the two reactants are coming or diffusing from opposite sides of the flame so both reactants will have to actually go to zero concentration on either side and then that is going to restrict the region of your reaction zone right where is it going to be like highly reactive so if you now think about the reaction rates they are kind of hitting zero at these ends where you are running out of fuel on one side and running out of oxidizer on the other side and you now have a peak in the reaction rate along the stoichiometric surface right that is where they are available in equal abundance so to speak right then we now say well let us want to have a flame sheet I do not want to have thick flame right so if you want to have a flame sheet I now have to collapse this to a particular due to a sheet right why would I collapse it obviously at the stoichiometric surface therefore we can imagine that this is a remarkable inside that Burkins-Schumann had back in 1928 when you know there wasn't too much combustion theory that was going on at the time right so the flame sheet obviously now can be convincingly thought of as coinciding with the stoichiometric surface in the mixing field if you now assume this we have a extremely simplified situation which is we do not have to solve the combustion problem at all it is sufficient for us to solve the mixing problem it could be as if we are doing a cold flow mixing without igniting so you just have to send the fuel in we just done to send the oxidizer there they mix and then they will they will mix at a stoichiometric proportion wherever they are mixing in stoichiometric proportion is where the flame is supposed to be right whether you are not okay I just want to find that therefore I do not have to actually worry about the energy equation along with the species equation in fact it is sufficient for me to actually worry only about the species equation of the fuel and oxidizer I do not here even have to worry about other species right and see how these species dynamics is going to work out for fuel and oxidizer alone I do not worry about energy so in the Schwab-Zelda which equation set of n plus 1 equations I will just focus only on two equations which correspond to the species conservation of fuel and oxidizer alone and then work out the mixing problem right so this is how this is going to happen so the Schwab-Zelda which set we just consider the species conservation equations of fuel and oxidizer species only and first of all for any species your Schwab-Zelda which formulation is divergence of rho v yi-rho d grad yi equals omega omega i or I think we have here the symbol wi that is wi right and what you are going to consider here is i equals o and f this is a step before we form the alphas and the betas if you go back to the Schwab-Zelda which formulation right so for the sake of refreshing your minds this is the species equation the Schwab-Zelda which species equation okay and then we formed an alpha such that you will now divide these by something and get only a omega here and that that is common for all the right hand sides and we would also do a particular alpha t for the energy equation should not going to consider here so that you will still get a omega there and then you can start subtracting one from the other two and by forming betas which are like alpha-alpha one or something of that sort we will try to do something like that here right now but at the moment we are going to consider only i equals o and f you know we don't worry we don't worry about anything else at the moment so we now form the alphas so let's say we have a alpha i which is why I divided by wi nu i single prime now of course the Schwab-Zelda which formulation is for a single step reaction so single step reaction we are looking at is nu f f plus nu o gives products we are interested only along this only the only the stoichiometric surface and the stoichiometric surface this is the stoichiometric reaction right so nu f and nu o or stoichiometric coefficients of f and o okay such that you get only products you don't get anything more right and therefore in this case if you now compare this with the template reaction reaction that we have which is like nu i single primes sigma i equals 1 to N nu i single prime script mi gives sigma y i equals 1 to N nu i double prime script mi and that's where you will get your nu i double prime minus nu i single prime for the alpha definition here nu i equal to nu i double primes are equal to 0 and nu i single prime or essentially nu f and nu o okay which means alpha f equals minus yf divided by wf nu f and alpha o equals minus y o divided by w o nu o all right and and wi divided by wi capital wi nu i double prime minus nu i single prime is equal to omega right therefore we get divergence of rho v yf sorry you can we can directly write alpha f alpha f minus rho d gradient alpha f equals omega and divergence of rho v vector alpha o minus rho d gradient alpha o equals omega so now form beta this is the last step in the Schwab-Zelovich formulation in this case we now say that beta equals alpha f minus alpha o if you do this then we get divergence of rho v beta minus rho d gradient beta equal to 0 this is amazing because this is a combustion problem where we don't have to solve we don't have to consider chemical reactions okay and look at what we have been doing so far the last time we solved the Schwab-Zelovich formulation for a premixed flame we looked at the preheat zone separately from the reaction zone and in the preheat zone we had a convective diffusive balance which means we had to consider convection and diffusion only and the reaction zone we had a reactive diffusive balance which means we had to consider only the diffusion and reaction only the full combustion problem in general always has a minimum of convection diffusion and reaction these are the three elements that constitute combustion primarily right we escaped try trying to solve all three of them together at the time by dividing it into two parts where we consider two of them to get at a time and here on the whole we are evading the consideration of reaction rates completely and we are looking at it as only a convective diffusive balance except here we are going to specifically say it is a axial convection versus a radial diffusion right so it is a two-dimensional problem and therefore that allows us to actually make these distinctions between axial convection and radial diffusion but there is a one-dimensional problem so we did not have that liberty okay so this is this is the this is a hallmark of classical combustion solutions we try to avoid situations where we have to deal with all three of them together that is what is called as a mixed problem okay so we try to avoid these mixed problems you always consider two processes at a time balancing each other wherever that is possible and you now get in this context a homogeneous linear equation right and we now adopt cylindrical polar coordinates so far we have not done that in the governing equations adopt cylindrical polar coordinates use v vector is equal to u e z cap and the dou square beta over dou z square much less when compared to dou square beta dou r squared that is going to happen here in the second term right so with all these assumptions we are simply going to have u divided by d partial beta over partial z equal to 1 over r partial derivative with respect to r of r partial beta over partial r this is the r component of the Laplacian in cylindrical polar coordinates and so it is a bit complicated there because of the cylindrical polar coordinates now for the first time we need to worry about boundary conditions we got the simplified governing equation which is this and let us now look at the boundary conditions okay the boundary condition that we require is in beta okay so we need boundary condition for beta beta is now our single variable and it is a single equation that is governing it and we need boundary condition for beta in r and z okay and since it is governing governed to second order and z sorry r you need to boundary conditions and since it is going to first order in z you need to bound you need only one boundary condition in z right so if you now look at the boundary conditions there are supposed to be in r for the two boundary conditions now we have to look at what is the domain the domain for us is going from r equal to 0 to r equals let us say you want to now distinguish between the inner duct radius so let us call this a and the outer duct radius which is b so our domain is going from r equals 0 to r equals b so we have to supply a boundary condition at r equal to 0 right and that is a symmetry boundary condition so b bc at for all for all z right at r equal to 0 partial beta over partial r is equal to 0 this is the symmetry boundary condition all right now what is partial what is the first derivative of beta actually imply physically if you now go back beta is nothing but alpha f minus alpha o and what is alpha alpha is yf divided by wf nuf and similarly alpha o so this ultimately goes back to taking partial derivative of r of yf and yo with respect to r right so partial derivatives of mass fractions amount to diffusion fluxes right so what we are saying by symmetry that partial beta over partial r should be equal to 0 means that you do not have diffusion across the center line because of symmetry right so whatever is happening here should actually have be the same thing that happens there and if you now have a diffusion over here that means that also means it you have to allow for diffusion from that side to this side that indicates in accumulation which is not possible right so you need to have that is what symmetry boundary condition is all about it is physical okay and that r equals b you have a wall I mean you have a wall you cannot have species diffuse species is not going to convert to the across because you have only you that is like kind of like the no slip I should say no slip this is like a inviscid flow that is slipping past the wall but it does not penetrate the wall by convection but it cannot diffuse this is a diffusion problem therefore we still have to say that partial beta over partial r is equal to 0 this is wall BC okay no diffusion across the wall now what we have seen here is we have two Neumann boundary conditions right we have to count on the third boundary where we need to have the boundary condition which is z equal to 0 to provide something dirichlet so that we will be able to pick the value of the problem if you are Norman BC here as well then you your your your solution will not be unique right now to get a unique solution we should look for some some Dirichlet boundary condition on the burner lip right here at z equal to 0 which we will consider tomorrow.