 So, we stopped here, let me just go over just of what we have done so far, essentially we are making a flame sheet assumption in this problem of co-flowing a field of an oxidizer at the same velocity and so the flame sheet assumption basically is a infinite trade chemistry assumption or infinite kinetics and that basically boils down to the mixing problem and we adopt this notion called the mixed is burnt approach as far as this solution is concerned. So, if you now solve only the mixing problem all we need to do is to locate the stoichiometric surface in order to get the flame shape. So, we just take the fuel and oxidizer species conservation equations in the Schwab-Zeldovich formulation and form the alphas corresponding to them for a one step chemistry of new F F plus new O gives P and then of course you also manipulate the right hand side to look the same in these two equations in for alpha and then you now have the Schwab-Zeldovich coupling function beta that is formed as alpha F minus alpha O and then you get a homogeneous linear equation for beta and if you now unwrap this vector equation sorry vector calculus equation in terms of vector in this equation in terms of vector identities in terms of cylindrical polar coordinates and then we also have the fourth assumption that we listed which is to a neglect axial diffusion and preference to radial diffusion then we are left with these two terms. So, these two terms basically signify the balance between axial convection and radial diffusion. So, essentially the governing equation boils down to this particular balance that we have been discussing that dictates the flame shape we will talk about pretty soon we will talk about what is the consequence of retaining axial diffusion as well but let us just proceed this equation now is first order in Z and to second order in R that means it requires one boundary condition in Z and two boundary conditions in R and moreover since the boundary conditions that we can supply or at least to at most to one order less than the leading order. So, the leading order here is second order in R that means we can supply boundary conditions in the value of beta or its derivative its first derivative and the first derivative in beta signifies diffusion mass flux because if you now go back and see beta is alpha f minus alpha o alpha alphas are basically yi's with the normalization. Therefore, a derivative in the first derivative in beta will essentially boil down to indicating a first derivative in yi's which essentially means diffusion mass flux in terms of the fixed law to which we have reduced the from the which formulation and since the Z boundary condition is only to first order we can supply boundary conditions only in the value we cannot supply boundary conditions in the derivative this is going to be pretty important for us to think about when we start looking at retaining axial diffusion sometime later but at this stage we are recognizing that we need to have two boundary conditions either in the value or derivative for beta in R two boundaries of R which are R equal to 0 and R equal to b R equal to 0 is the center line R equal to b is the outer duct radius and in both the cases it turns out of course this is a Neumann boundary condition required for symmetry and this is a no mass flux penetrating the wall because the wall is a rigid wall it is a non porous wall and you are not looking at any penetration of mass by diffusion there. Therefore, these are the boundary conditions and we can look at boundary conditions the first derivative now if you now have these two boundaries subjected to Neumann boundary conditions we have to make sure that the other boundary that we are looking at which is Z equal to 0 for the Z boundary is provided with the Dirichlet condition and that is what is actually required you cannot provide a Neumann boundary condition but when you now think about a axial diffusion taken into account you will also have a second order in Z which will admit two boundary conditions up to first order in beta but at that stage we cannot give Neumann boundary conditions everywhere then you will have a non unique solution therefore we will think about the kind of boundary conditions admitted in that case next but at this stage we can only provide Dirichlet boundary conditions for this and in beta so need 1 BC in Z at Z equal to 0 so at Z equal to 0 which is a Dirichlet data which should be in the form of Dirichlet data so at Z equal to 0 we have to now say what should be the value of beta so we now have to go back and look at how to form our beta beta is alpha f-alpha o right and what we find is if you are now looking at Z equal to 0 that is at the same plane as the lip of the inner duct where the fuel is coming in and beginning to mix with the oxidizer from the outer duct so and we have to now look at how this the beta should span from 0 to B across r equal to a so what we find is between 0 r equals 0 and r equals a you have only fuel at a mass fraction of y of 0 and you do not have any oxidizer therefore your y o is 0 and your therefore alpha o is 0 so if alpha o is 0 you only are left with alpha f so beta would be yf0 divided by Wf nuf right with a negative sign because you have a nu i single prime is nu f and you have a denominator nu i double prime minus nu i single prime nu i double prime is 0 so you have a negative sign over here therefore we are going to get a negative sign there this is for 0 less than r less than a now if you now go back to what happens between r equal to a to r equals B you have all oxidizer that means you have a y o0 over here but you have yf0 is 0 right so if you now go back and see your alpha o will be alpha o0 and your alpha f would be 0 alright so if you now plug alpha f equal to 0 and alpha o is equal to alpha o0 and alpha o0 is nothing but y o0 divided by W o nu o with a negative sign and then you have a negative sign here as well right so therefore you will get a plus y o0 divided by W o nu o this is for r a less than r less than B so what this really means is you have now a jump in the value of beta at r equal to a alright it is discontinuous but that is exactly what the mixing problem is all about you have all fuel on one side you have all oxidizer on the other side and they are now going to begin to mix right so and the mixing is essentially by diffusion and diffusion is essentially a transport process and like any other transport phenomena the job of diffusion is to mix things and even out discontinuities right so it is okay to admit discontinuities at the boundary so what is going to happen is you have all fuel all oxidizer so the beta is discontinuous at the boundary but as you now get into the domain the job of this governing equation particularly this term is to even out or smoothen out this discontinuity alright so that is alright so considering that we have these boundary conditions the next step that we want to do is to non-dimensionalize non-dimensionalize this equation now many times as a student you feel that this is this kind of a burden like why would I want to do that can I can I just go ahead and solve this right away the answer is sometimes non-dimensional non-dimensionalization is done for elegance so your governing equation will look very very nice as if it has been to a hairdresser the other times most most most important technical reason for an engineering point of view is you can now actually come up with non-dimensional parameters and the non-dimensional parameters are essentially a group of quantities so which are now grouped together in such a way that it is okay to vary one of them and not necessarily the other or you can vary them in combination so that the non-dimensional number is remaining constant and so on so you can begin to see how the different parameters that are in the actual problem can group together in influencing the solution and the solution that you obtain could be just obtain for one value of the parameter as a whole non-dimensional parameters a whole you do not have to do this for many different parameters which are not non-dimensionalized so it is pretty good here to basically say that let us now say psi equals r divided by b now because you have a characteristic length scale in the r direction you can non-dimensionalize by that. Fortunately in this problem since you have neglected radio axial diffusion right a neglected axial diffusion it is a infinitely long duct so pretty much do not have a good length scale corresponding to the radial the axial direction it is almost infinite right so what you are basically looking for is what is the diffusion length scale because that is the process that we are basically looking at in fact what we should be looking at is what is the competition between diffusion and convection or how fast are we able to diffuse as we are convecting okay it is kind of like a race between the two the question is how much could I have diffused out how this way while I am convecting at a particular rate that is now going to give me a certain length scale associated with this right so that is going to be made by saying if I now have my eta equals z let us say z by u divided by b squared over d right. So z by u is essentially a convective length scale and b squared by d keep in mind d has the units of like meter squared per second right so b squared as meters square meter squared therefore this has a units of time right and so z by u also has units of time and therefore eta becomes non-dimensional but in this the only variable is z okay so the farther you go and as you convect you know how your the more time you are taking but in this time you are also diffusing this way that is because b is the length scale corresponding to diffusion because the diffusion is predominantly radial right so if you want to put these two together then this is the zd divided by u b squared as a matter of fact you could begin to think about this as if I were to say I do not know any of this let us say I am not interested in thinking about the actual problem the the professor asked me to non-dimensionalize I see b as my parameter the length scale here and that fitted in fairly well with xi why cannot you do the same thing with eta I would like to think of this is this z by b then you have all the stuff sticking out which is essentially d divided by u b now u b divided by d essentially is the counterpart of your Reynolds number that you would use in in momentum mixing but here this is species mixing so you would actually have to use the Peckley mass transfer number which is the corresponding part of this so this is 1 a 1 over Peckley where pe is u b divided by d so from here we can begin to think if Peckley is kind of small alright then eta becomes proportionately larger okay so Peckley small is basically meaning that you have less convection relative to diffusion so you would expect that you have more and more diffusion happening this way but less convection happening that way so you should actually be able to get your flame to be shorter so that is what you should expect for smaller Peckley numbers because you are not going too far along the convective path rather you are diffusing so your flame should all be confined to closer to the burner because you have diffused a lot therefore a smaller Peckley number will correspond to a a a shorter flame effectively and we are trying to cover the shorter flame by artificially blowing up the eta having this 1 over Peckley there essentially so effectively this takes into account that the effect but your governing equations do not take into account axial diffusion so that is something that you have to keep in mind that is what should really make the flame shorter strictly speaking how the governing equation turns out to be further you can form some more non dimensional parameters let C be equal to A by B and I will make a big deal about this and the next one which is nu equals alpha O0 divided by alpha F0 okay alpha O0 divided by alpha F0 if you want to now go back and see what those are this is nothing but alpha F0 would be YF0 divided by WI nu I with a negative sign alpha O0 would be YF0 YO0 divided by WO nu O with a negative sign so the negative signs cancel each other all you are going to be dealing with is Y O0 divided by WO nu O divided by YF0 divided by WF nu F alright and we will come back to this this and this quite soon we have non dimensionalized or we have non dimensionalized Z and then we have a couple of now new non dimensional parameters in the problem but we have not non dimensionalized beta okay so let ? be equal to beta divided by alpha F0 that is to say if you now were to basically go back and say as we just have to pick one of those two right we just pick alpha F0 so beta divided by alpha F0 would be alpha F divided by alpha F0 minus alpha O divided by alpha O0 is what we are going to so alpha O0 alpha F0 okay so you just have a alpha F0 in the denominator just to go with this you can also see that this goes with this here we have divided alpha F0 in the denominator therefore we are using this so that the way the nu is defined is how that way ? is also defined okay now the interesting thing if you now do all these things the governing equation then becomes then becomes ? ? by ? ? equals 1 over ? ? by ? of ? ? by ? so essentially we got it a few by D completely alright so that is because that information is actually buried in ? if we have scaled over ? corresponding to the competition between axial convection and radial diffusion that is what this signifies right so we now have a very nice looking equation without any parameters so this solution is now going to give you so this equation is now going to give you a solution without any parameters except the boundary conditions are now going to carry some parameters so what are they the boundary conditions or the non-dimensional boundary conditions are so now we have to write in terms of ? so ? equals well let us first write the Neumann BC's for ? that is very easy at ? equal to 0 we have do ? by do ? equal to 0 and at ? equals 1 we have do ? by do ? equal to 0 for all either greater than 0 right in fact we should have gone back and said right here for on Z greater than 0 that that is true for the entire domain for the route then so this is the boundary condition for ? in so for ? in ? but what happens to ? in ? right so and ? we now have to translate this boundary condition in non-dimensional form that simply turns out to be 1 recall we hand this is basically this is basically a f0 okay and then we just form a ? by dividing by a f0 so we need to get a do we have a negative sign with the negative sign is a f0 okay so with the negative sign is a f0 and we just divide by a f0 therefore you simply get a 1 and similarly if you now go back and plug in here this is actually minus a o0 okay with the positive sign this is minus a o0 and then for trying to find the ? you divide by a f0 so minus a o0 divided by a f0 is nothing but minus ? okay but when where are they we can now write this is actually a 0 less than style less than C C less than style less than 1 so even though your governing equation did not have any parameters there are a couple of parameters that have crept into the problem in terms of ? and C through the boundary conditions all right and this is going to be very very important in fact this is actually only one boundary that is set at ? equal to 0 right basically that is the boundary that is that is describing the problem for you if you did not have fuel and oxidizer initially unmixed entering the domain at ? equal to 0 you did not have a problem okay so this essentially tells you that the fuel and oxidizer are in as they enter the domain or unmixed and that is what the problem is so all the parameters in the problem are now reduced to just these two okay so with this I am not going to derive the solution for you and I am going to assume that you know how to how to derive this okay and then the way to do this is you seek a product solution that means you now say ? is a function of ? is now a product of two functions one function which is a function only of ? another function that is a function only of ? so you now have let us say seek a product solution of the form that is suppose ? equal to some function ? that is only a function of ? and another function let us say I do not know what is a good symbol just go back with angle of Saxon variable so you can simply say capital X of ? and capital Y of ? all right so if you did this plug it back in here go through the product solution approach and substitute these boundary conditions noting that you are expecting to have a periodic solution in ? because you have homogeneous boundary conditions there this is a inhomogeneous boundary condition here so you are not expecting periodic solution in ? and so on you do all those things and you can now find the solution to be solution is ? equal to 1 plus ? c2 minus ? plus twice of 1 plus ? times c summation n equal to 1 to infinity that means it takes integer values 1 over ? n j1 of c ? n divided by j0 of ? n squared times j0 ? n ? e to the minus ? n squared ? I do not even understand this right well let us let us just go step by step I mean of course we are not as intelligent as we would like us ourselves to be so all these things are something that we have seen okay these are part of the parameters of the problem the first time we encountered a problem now is what is ? n so ? n is what is called as the nth 0 of j1 of ? where j1 is a Bessel's function right so that is a Bessel's function of first order so obviously then j0 is the Bessel's function of 0th order right now what is Bessel's functions Bessel's functions are those that look like signs and cosines in fact j0 and j1 would look like signs and cosines except that they do not have a constant amplitude so you will now find that they kind of keep decaying but keep alternating about the horizontal axis right so that is the kind of solution that you are going to get and you now have a summation over a series of those so ? 0 sorry ? n being the nth 0 of j1 of ? is essentially means that this function keeps on alternating up and down about the horizontal axis at specific values of ? the argument right so when you say 0s those are the values so where are where all do you have the j is going to 0 is essentially where you get these values of ? okay so it is kind of like in sign signs and cosines if you take a sine wave you know that is the sine wave passes 0 at 0 ? 0 ? 2 ? 3 ? 4 ? and so on so those are the 0s of this the sign function so similarly we will have these what will happen is you will find that these are not actually equally spaced they have a certain pattern alright so that is how the Bessel's function behaves and typically we get into a Bessel's function mainly because you are having this kind of a derivative for cylindrical polar coordinates and that is mainly because we are using axi-symmetric pipes okay so if instead if you now had like vertical plates like channels right you will see you will not have a complicated looking derivative for the Laplacian you will simply have a partial squared beta divided by partial squared partial squared gamma divided by partial squared psi partial psi squared and you will get signs signs and cosines and typically for these kinds of boundary conditions you should get cosine a cosine function here and that would actually imply that we are fitting strictly speaking what what it really means is we are fitting this particular discontinuity by a Fourier expansion and then letting the Fourier expansion decay so essentially you now have a step that is represented by a Fourier series and then it is subjected to boundary conditions of Neumann B C's on either side and then as you keep on mixing more and more this this particular function now changes and that essentially the Fourier representation keeps changing as you go along until you get a more and more uniform variation in the mixed diffraction in the instead of a Fourier series you now have a series that is based on Bessel's functions and that is possible anytime you have you are looking at functions that are orthogonal to each other like in terms of signs and cosines you will have orthogonality in Bessel's function also you have orthogonality so based on these you can actually form say series that look like Fourier series but not necessarily in science and cosines but some other orthogonal basis functions that is what we have done here okay or that is how it comes out to be when you now pursue the product solution how do you know it is a product solution because you now see the psi dependence actually here and the eta dependence here it is a product okay so you do not have psi and eta mixing up with each other they are separate but they are multiplying the functions containing them are multiplied with each other and we also see that you have a e to the minus phi and square sorry e to the minus phi and square eta that is a exponentially decaying term in eta so what that basically tells us is as you now go to e to tending to infinity further and further from the burner lip this entire term is going to go away and you are now going to get a a gamma that is only this this term all right and gamma is nothing but beta normalized and beta is nothing but a difference in normalized mass fractions and normalized mass fractions are this so essentially it tells you that as you go further and further out the mass fractions are going to differ by a constant all right and the constant depends on two things new and see so now we have to think about what does what does new and see basically mean they mean something very interesting new is telling us why oh not divided by w o new over by yf not divided by w of new f this has got nothing to do with the geometry of the problem it is got to do with our inlet mass fractions of oxidizer and fuel and their respective molecular weights and the stoichiometric coefficients in the reaction in the single step stoichiometric reaction okay this is a stoichiometric reaction that means you do not have either fuel or oxidizer left over as part of the products so what are the stoichiometric coefficients required for them to react with each other completely right that is what is contained in this so this is a mixture quantity and it is a inlet quantity because it has yf not and yo not see on the other hand is purely a geometric parameter it tells you how small or large is the inner duct relative to the outer duct okay so now you have to think about many different combinations is it possible for me to have a very small inner duct relative to a very large outer duct but send a high concentration of fuel inside when compared to a very diluted concentration of oxidizer outside let us suppose that that is one possibility the other possibility I now have a fairly fat inner duct and just a little bit bigger outer duct that means most of the outer duct is contained by the inner duct and you are essentially trying to send a lot of fuel but let us suppose that the fuel is highly diluted but the oxidizer is highly concentrated what kind of flames will we get right on the one hand you are trying to send a very concentrated amount of fuel but in a very small region on the other hand you are sending in a lot of fuel alright but it is actually highly diluted is would you happen to have the same kind of flame right so that is what this this is going to dictate okay so what are we going to look for in this what we want to see is interestingly this is now a ? that is a function of ? and ? that means we should be able to plot ? in this domain in where the mixing is happening and all we are looking for is look for ? equal to 0 right as the stoichiometric surface which is essentially the flame shape so that means if you are able to plot this particular function in the domain and have a contour map of ? you now look for the ? equal to 0 contour the ISO contour of ? equal to 0 corresponding to ? equal to 0 so what is ? equal to 0 mean right ? equal to 0 means or ? equal to 0 if ? is equal to 0 then your a f equal to a O back here right so ? equal to 0 means ? equal to 0 ? equal to 0 means a f equal to a O and a f is nothing but minus y f by w f nu f which is equal to minus y o divided by w o nu o all right so what this basically means is we want our y f by y o along ? equal to 0 to be in the ratio of w f divided w f nu f divided by w o nu o and what does that mean if I were to simply say nu f by nu o that would mean the stoichiometric ratio in terms of moles right all I have done here is to multiply nu f by w f and multiply nu o by w o so I am essentially looking at the mass ratio so this is the mass ratio of stoichiometric proportions of fuel and oxidizer in a reaction for stock for the for complete reaction to happen and what you are basically saying is when nu f by nu o anywhere in the domain is in this ratio you have a stoichiometric surface and that is what the flame is that is what the flame sheet assumption is that is what infinite chemistry means that is what mixed is burnt approaches all about right so effectively it all translates to saying just look at the ? equal to 0 contour all right further let us look at go back and see what nu is in the light of what we have done right so this implies we can write or we can simply write this equal to nu o divided by nu f o right divided by w o nu o divided by w f nu f so what nu signifies is the incoming oxidizer fuel ratio by mass to the stoichiometric oxidizer fuel ratio by mass right so effectively nu tells us something about what is coming into the domain relative to this stoichiometric mixture ratio and C tells us how much of what you are getting all right so what you are then saying is we can now we begin to do a few things here one how does the height of the flame get determined right so how does the flame look like what what will happen if you now try to plot this contour for ? equal to 0 right so depending upon the values of your nu and C right so if you now say this is your ? equal to 1 ? equal to 0 and you now fix your C as ? equal to C that is where the inner tube ends and then the outer tube is there you could now get a flame that either looks like this or in fact we have a wall there you could have a flame that looks like that as you now change your new so for different values if you now fix your C at a particular value and change your new you can get the flame to close over the fuel or you can get the flame to open up and close over the oxidizer open up from the middle and close over the oxidizer all right so this is now called the over ventilator flame and this is what is called the under ventilator what does that mean it is essentially having a concept of ventilation as in opening up more air or so when you now say over ventilator that means you have a lot of air when compared to fuel so the the flame is actually closing over the fuel and consuming all the fuel and we now say under ventilator that means you do not have enough air so the fuel is spreading out and then mixing with whatever little air it can and then having and meeting it at stoichiometric proportions along this line that closes over the oxidizer and opens up over the fuel so by this what it means is if you now have a flame that looks like this you now have a fuel rich region over here or pretty much all fuel in the actual case and outside is going to be oxidizer because the flame is separating the fuel than oxidizer and getting them to meet at stoichiometric proportions in between all right so if you now have a under ventilated flame you are simply going to have a lot of fuel that is going to go out unreacted over here so when would you expect a fuel rich sorry when would you expect a under ventilated flame or a over ventilated flame so now think about those two limits that we were talking about if you now have a very short duct for the inner duct very very small duct for the inner duct when compacted out a duct and you have an infinite domain pretty much no matter how concentrated the fuel is relative to how dilute the oxidizer rich you are very likely to have a if a flame that is closing over the fuel that means it is going to be an over ventilated case if you now have a very very broad inner duct when compared to the outer duct being just a little bit bigger you are very likely to have a opened up flame that is under ventilated right no matter how much you are diluting things but there is an effect of that and that effect is going to come up when you have you are at a 50-50 kind of situation that is you need a too small not to large the message here is the slot width is going to dictate whether the flame is going to be over ventilated under ventilated more than new or in other words the far field behavior of the flame is more sensitive to see rather than new so what is the job of new in this then you have to ask yourselves look at this picture this flame if you are now somewhere here you are just beginning to look at how this mixing is happening and then the flame is shaping up both of them are beginning to look pretty much the same they are going like that over the oxidizer starting from the fuel right but then as we keep going that went that way this went this way and then we now call those over ventilated that one under ventilated but why did they start going like that in the first place right why couldn't you have a flame that what that was like this that could have been a possible solution as well so what is the difference between this and this this flame started going over the fuel and closed over the fuel whereas this flame started going over the oxidizer and then closed over the fuel this flame goes over the fuel and closes over the fuel so you now have multiple possibilities you could have a flame that starts going over one of the ducts and then closes over the other or it goes over the same duct and closes over the same duct right so what governs this is new when they are just beginning to mix and react they are just beginning to consume the reactants and it is the final large scale availability of reactants that is going to shape up how the flame is going to end up and as we go along and the large scale availability is dictated by the sea how much fuel you have on the whole versus how much oxidizer you have but locally as they just begin to mix and react they are not worried about that they are not worried about how they are going to end up and all the less deficient species is consumed reactant is consumed they are going to first be dictated by what is the new how how concentrated the oxidizer is relative to the fuel as it comes in relative to what their stoichiometric ratio demands so if you had a fairly concentrated fuel but coming in a very very small region it is going to actually go out in search of the oxidizer because oxidizer is dilute all right and ultimately your as you go along you are having a short fuel duct is a small fuel duct so you are running out of fuel and then it closes over so you can you can see that the new dictates the near field behavior of the flame and the sea dictates none both of them are together it is like it is difficult to separate them but you look at the sensitivity right you look at the sensitivity of this the law the far field behavior of the flame is more sensitive to see the near field behavior of the flame is more sensitive to new all right and then from this you can also find out what should be the shape of the flame so what should be the combination of new and see together that will give rise to a optimally ventilated flame right a flame that is kind of like that but I just went too fast I did not tell you whether it is going like that or like this right that depends on the new what I am talking about is for a given new what should be the sea such that the flame goes vertical and does it go to infinity we will talk about it later.