 So we have been looking at the solution for the Berkschumann problem and what we can now figure out is we want to actually interest we are interested in the flame height surprisingly for tall laminar flames that means we still have the situation of Peckley number being quite large but still not into the turbulent regime the Berkschumann solution gives a very good prediction of the flame height and this was significant back in 1928 if you think about it so the way you actually can find out the flame height as I said ? equal to 0 is going to give you the flame shape and of course it is going to change switch from a under ventilated flame on the one side to a over ventilated flame on the other side so depending upon whether it is under ventilated or ventilated you need to evaluate this expression setting ? equal to 0 for the ? all right now plugging in ? equal to either 0 or 1 right so with this you can get the flame shape so let us do this for at least one of those so for the over ventilated case say ? equal to 0 it a max that is a height of the flame so you can now plug this in here of course what happens is you now have a ? equal to 0 and then you have this expression which is a series summation and we are trying to find out ? on top of this exponent sitting there with the negative sign in each and every term right that is quite difficult to do so one thing you can do is to recognize how the series is going to behave and then say as a first approximation let us not worry about the series let us consider only the first term in the series right as a leading term and then the remaining terms are corrections rate therefore approximate to the first term in the summation with ?1 equals 3.83 this is something that you can find out from the bezels function tables so there is like a table of special functions which will give you values of these zeros so we are looking at the bezel function of first kind the first zero of that you will find that in the table of ?1 is equal to 3.83 keep this in mind and then you get a eat a max all you have to do is get rid of the summation wherever you have n you just put equal to 1 because you look at only the first term and this thing goes to the left hand side this thing comes down and then you have this then you find that this is actually either the negative right so if you now want to flip everything and take a natural algorithm then you will get a ? and ? and squared ? ? equals ? max and therefore you now get the ? and squared also to the denominator on the other side you will get it as 1 over ?1 squared natural algorithm 2 plus sorry 2 twice of 1 plus ? times C j1 of C ?1 divided by ? minus ? minus 1 plus ? C squared when this goes to the left hand side you get a negative sign so this slips times ?1 j0 of ?1 right so this gives you a good idea of what the flame length should be but what do we learn from this okay that we know how the flame is going to how the flame should length is gone as a matter of fact it is difficult to find out how the flame shape is going to be with this expression and as I said this is how the flame looks like but if you now look at this expression for the flame length you cannot see the dependencies in the problem right so how the how do you set up this problem you now have these coaxial pipes and then you want to send fuel and oxidizer together at particular since they together I mean in each of these pipes at the same velocity that is what I meant by together right so the same velocity and the C is your variable as I said and your YF0 and YO0 order or your control variables these are the ones that you are trying to control depending upon YF0 and YO0 the new is going to get fixed right so you can you can do all that but how is it how does it vary or for that matter is the flow velocity showing up here no because the flow this parameter did this this equation was primarily did this solution was primarily obtained for neglecting the axial diffusion alright so we will do a couple of things now first let us think about what happens when you now try to keep axial diffusion alright and I am not going to solve the problem I let it let you let you figure that but I will tell you what are this what are the possible steps that that can get in there and the second thing is and then what are the consequences and the second thing is let us look at the dependencies okay so the first thing is if you consider axial diffusion, diffusion is included rather than neglected right we would have we would have u over d dou beta by dou z equal to dou square beta by dou z square plus 1 over r dou by dou r r dou beta by dou r previously we did not have this term we are not trying to keep this right so what is the consequence of this first of all you could say that r is still governed to second order and therefore it requires two boundary conditions the two boundaries in r or r equals 0 and r equals b and I would like to supply boundary conditions there and I expect to have symmetry boundary condition at r equal to 0 so dou beta by dou r is equal to 0 that is a Neumann condition and at the wall I have a rigid non porous wall so I cannot have diffusion through that wall so I still have no diffusion mass flux which amounts to dou beta by dou r is equal to 0 at r equals b at the outer wall and so I get Neumann boundary condition there as far as the z boundary is a concern this is now suddenly begin beginning to be governed by second order previously it was governed only to first order now you have actually a second order term that is governing this right so whatever we neglected was somewhat pretty important I mean it actually reduces the order of the equation by one which means it does not permit an additional boundary condition whereas this one demands an additional boundary condition that means you have to give two boundary conditions in Z okay the domain for Z is Z equal to 0 which is something that we considered earlier and Z equals infinity so you now can actually go all the way up to infinity here and that is where you have to supply the boundary condition what do I know about beta at Z equal to infinity or it is derivative and now it now you have a very important situation you see because you have this governed to second order not only it demands two boundary conditions but it can also permit a boundary condition in the derivative okay so you already had two boundary conditions and derivatives here and if you now allow if this admits boundary conditions in its in derivative that means you can you can specify dou beta by dou Z at R equal to sorry Z equal to 0 and or dou beta by dou Z at Z equals infinity then you do not have a unique solution because you are you are supplying Neumann boundary conditions everywhere therefore you need to supply Dirichlet boundary condition somewhere okay now previously we could supply only Dirichlet boundary condition because this is governed to first order and you could give only Dirichlet data now it is supply it is governed to second order that means you can have a not only Dirichlet but also a mixture of Dirichlet and Neumann what is that how could I get that so now you have a choice of different boundary conditions that you can give okay and they should mean something physically or in other words we should now interpret that physically okay so if you go back to your original problem so beta is now governed to second order in Z so need to be sees in Z for beta which implies Z equal to 0 and Z equal to infinity okay also can admit Neumann BC's in beta that is could specify dou beta by dou Z at Z equal to 0 or infinity but cannot specify all Neumann BC's that is keeping the beta by dou R equal to 0 at R equal to 0 and B as others okay because this will lead to a non-unique solution therefore retain Dirichlet BC or mixed which is Dirichlet plus Neumann BC all right all we can say in fact at Z equals infinity you do not have too much of a room to play with on what should be the boundary condition all we can say and hope for at Z equal to infinity is that beta should be bounded we cannot specify values we cannot specify derivatives simply because you are expecting to get a exponential solution and since you have a second derivative now you can actually admit a e to the plus some constant times let us say phi n squared e to the plus phi n squared eta plus some constant times e to the minus phi n squared eta and if you now admit the coefficient to e to the plus phi n squared eta as eta tends to infinity that solution is going to blow up so by specifying that Z equals infinity beta should be bounded we can get rid of the exponentially growing part of the solution and retain only the exponentially decaying part of the solution as before all right. So at Z equals infinity we merely specify that beta remains bounded which eliminates the exponentially growing solution in beta in e term sorry right and retains only the exponentially decaying solution eta as before a matter of fact that is what boundary conditions are supposed to do boundary conditions are supposed to evaluate constants of integration that are appearing as coefficients to solutions and by just merely saying that beta should remain bounded as a boundary condition at Z equal to 0 we evaluate the coefficient to the exponentially growing solution in eta as 0 so we have done the job as far as that particular boundary condition is concerned then comes this right here is where it is important for us to decide whether we want to have a Dirichlet boundary condition or a mixed boundary condition so this question of whether we want to have a Dirichlet boundary condition or mixed boundary condition arises primarily at Z equal to 0 at the lip of the burner right that means it is now possible when you when you admit or include axial diffusion that you have a choice of boundary conditions either it could be the Dirichlet boundary condition that we use before which we did not have a choice about earlier or we can use the mixed boundary condition now the question is what is the mixed boundary condition really mean right so let us now think about the flow that is coming through one of these ports and now we are admitting axial diffusion right so we now go back and look at the solution so or if you now try to map the solution and say you have a flame that is supposed to be here right what does that mean this really means that you have a field rich region over here this is the stoichiometric surface there is a fuel fuel lean region around okay and there is a progressive change in the mixture fraction from a pure fuel or more fuel in the middle progressively the stoichiometric and then fuel lean okay it is a varying region over there what that means is if you now look at it along the axial direction or the stream wise direction you have more fuel here than here right so it is sort of like as the as the fuel is coming out and flows up into the into the domain and you do not have the tube anymore it is now got into the domain it looks around and then sees wait a minute I am not there so let me go there right and then once it goes there it says hey wait a minute I am not there let me go there what is that that is axial diffusion okay now of course you might think wait a minute we do not we have all the flow kind of coming up yeah that is that is what we had in premixed flames also right so but at that time what happened we we now decided that you are going to have a flame and the and if and the reactants are coming in and then they all getting converted to products and when the products are formed they suddenly get formed and then look around and then find it is all products over here no products over here can I diffuse backwards right and it tries to diffuse against the current current meaning the convection right and it succeeds to some extent just as well as the heat gets conducted upstream even as the convection is actually carrying the enthalpy this way if the heat could penetrate why cannot mass right it is after all both of them are transport processes similarly here you could have a current that is that is setting up set up set up upwards but you could have a reverse diffusion that is for the fuel but think about the oxidizer the oxidizer does not even have to fight the current right the oxidizer is here and it finds wait a minute there is more fuel over there right can I now go go up right so along with actually convecting it begins to diffuse up right so these axial diffusion processes the question is how good are they is a question of how well are they competing with convection so when you do not have a large convective effect that means you are at fairly low velocities right in simple English huh then you can now expect the axial convection axial diffusion to be more predominant all right so this is a problem where the convection could predominate axial diffusion but balances radial diffusion all right so the balance between radial diffusion and convection is is the centerpiece of this but alongside at low convective effects you could have a significant contribution from axial diffusion as well if you now think about that so you now say fuel is coming out like this diffusing like this and then going backwards oxidizer comes like this diffuses like this and or rather even carry can even diffuse like this and then goes backwards if it did then the question is how good is a Dirichlet data correct we were imposing saying at the at the lip of the burner you have only fuel here and only oxidizer here that is what we did right but that need not be the case you could have the incoming fuel get contaminated by oxidizer that is diffusing from the other side right or products as a matter of fact huh and similarly the oxidizer so where can I expect to be sure that I do not have any contamination for upstream because you have a convection there is a certain length scale associated with the diffusion in competition with the convection right and beyond that length scale you can hope to have pure fuel and pure oxidizer but how do I know what that is unless I solve and for me to solve I need to know the boundary conditions so where do I go in search of the boundary do I want to now take like a minus infinity to plus infinity in Z for the domain when in fact I am interested in what is happening here that is where what is called as a flux BC comes into picture right so we could now think about a flux boundary condition which is essentially a species balance that is integrated over a control volume for each of the species from minus infinity for the Z to over here knowing that this is going to be all fuel all right so if you now take the governing equations and integrate within this control volume what you can now expect is we now know that you can say rho u yf minus rho d dou y right at Z equal to 0 should be rho u yf minus rho d yf by dou Z at Z equals minus infinity there is hardly any thing that is going on along the walls you neither have convection nor diffusion along these walls right so the control volume here that we are looking at is having exchange only at this surface and this surface at infinity for upstream for upstream we know that you do not have you have your fuel and therefore you do not have any fuel concentration gradients so you do not have any diffusion to talk about right so this goes away and you can directly now say this is equal to so you can say rho u yf at Z equal to 0 minus rho d dou yf by dou Z at Z equal to 0 equals rho u yf not right and then I can say yf at Z equal to 0 minus d over u dou yf by dou Z Z equal to 0 equals equals yf not so what has happened now previously we ignored this term we did not consider any diffusion that was happening axially across the inlet to your domain we simply said yf at Z equal to 0 is yf not but now you have to subtract that amount that is actually if you diffused you see and what is that going to be like in fact if you think about axial diffusion including axial diffusion then the problem becomes somewhat symmetric in in both Z and Z and eta that means you do not have to worry about sorry Z and R that means if you now thought that the length scale of your R dimension was B the burner width what you are essentially saying is it is diffusing along the radial direction just as well as it is diffusing along the axial direction at least to begin with the way that is a consideration we are to evaluate how much this is versus that okay and that will be done by the Peclet number so if you were to now say that Z is also going to be of the order of B and then say so for non-dimensionalization non-dimensionalize a of Z as sorry eta equals Z by B okay same length scale for axial as well as diffusion this was not the case before we did not have an axial diffusion length scale unless we actually started looking at what is the so flow versus the diffusion length scale so we had to do something like Z by B divided by U by sorry B squared by D and then we came up with a new non-dimensional number sorry coordinate eta last time but if you do this then YF at Z equal to 0 minus you now get a B out here right so we can have a D YF0 or YF minus 1 over Peclet dou Y I am sorry dou YF by dou psi at Z equal to 0 equals YF0 right now this is what is called as a flux boundary condition and it is a mixed boundary condition it now has a linear combination of value and derivative together right this is sort of like why a YF plus B dou YF by dou psi where a is equal to 1 B is equal to minus 1 over Peclet okay it is a linear combination that is what is a mixed boundary condition so the summary of what we are talking about is that axial consideration of axial diffusion permits permits diffusion at the inlet boundary right and leads to flux sorry a mixed boundary condition now mixed boundary condition is all right because it still involves some Dirichlet data right yes I am sorry that keep making that mistake psi and eta agree right as a matter of fact psi psi is equivalent of X and eta is actually equivalent of what it is it is not z it is actually H so that is why the confusion we should have been using zeta all right so all right so now question we can ask two questions one all right I have axial diffusion okay but let me insist on having only the Dirichlet boundary condition as before that is possible mathematically that is correct okay may not be physically correct in some situation but mathematically that is okay you can do that what would be the effect of considering axial diffusion okay with just a Dirichlet boundary condition as before the answer is if you that now begins to depend on a new parameter which is called Peclet number before we did not have Peclet number we had a non-dimensional governing equation without any parameters but now what will happen is you will have a 1 over Peclet square showing up and clearly what that means is as Peclet number is large the axial diffusion effect relative to axial convection is going to be small okay that means axial diffusion is important mainly for small Peclet numbers right so small Peclet number here is obviously ubu divided by d this is the mass diffusion counterpart of Reynolds number that I pointed out yesterday and what that means is it small Peclet numbers your convection is not as convection is not predominating over diffusion okay diffusion is quite important and therefore you will now end up with a flame that is a bit fatter and shorter right so this is for so this is the way you are going to get things to go as Peclet number increases and the Berkschumman solution that we saw so far in neglecting axial diffusion is in the limit of infinite Peclet number alright so the job of axial diffusion is to make the flame shorter because you have more mixing happening actually as well as radially and that therefore you get a shorter and a bit fatter flame correspondingly then the question the next question that we have to ask is well fine now I have axial diffusion and then I look at the problem and then decide that I want to have a mixed boundary condition which means I want to have a flux boundary condition like this what do I do how does how does the flame look like right the answer is if you now had a Berkschumman solution with axial diffusion and Dirichlet boundary conditions if you had a flame that look like that your flux BC's are going to actually make sure that you your course your domain starts here right your flame starts somewhere there that means the flame is no longer going to be attached to the rim of the burner right why was it attached to the rim of the burner before because we did not permit axial diffusion across the interface sorry across the across the inlet and therefore when the fuel and oxidizer met the first opportunity at which they could meet was right at the burner lip and they meet at stoichiometric stoichiometric proportions there and then on in a certain curved manner and therefore the stoichiometric surface starts from the lip of the burner like what we have shown here but when you have axial diffusion taken into account you can now permit mixing to happen this way across upstream of the inlet to the domain and therefore you could find the fuel and oxidizer being in stoichiometric proportions away from the lip of the burner and that is possible and we are not strictly speaking drawing what is happening inside because we are not solving for it our boundary condition is still applied only here and our solution starts only into this domain right so we do not know we do not know exactly what is happening but this is permitted okay so this is the consequence of having axial sorry flux boundary conditions along with axial diffusion taken into account in the case of the Berkschumann problem in general if you look at the literature the term Berkschumann flame stands for infinite chemistry right that means we have not really bothered about solving anything to do with finite rate chemical kinetics so you can have for example Berkschumann spray flame like like you have a spray of droplets and the droplets are burning and what it means is basically that as the droplets evaporate and and the vapor from the droplet is mixing into the oxidizer let us say you are looking at fuel fuel droplets fuel spray right wherever you find the fuel vapor in stoichiometric proportion with the oxidizer ambient gaseous oxidizer mixing into this you now form a flame that is a Berkschumann flame okay and that is what you would call as a Berkschumann spray flame right you can also have something called a Berkschumann jet diffusion flame or a jet Berkschumann flame so what that would mean is I could I do not have to have walls Berkschumann were very clever you see they made these walls so that they do not worry about entrainment from the surroundings right this is a purely species mixing and convection problem the isolated the most important aspects of this very very well okay. So you could now think about a jet flame in which you have only the fuel coming out you have quiescent oxidizer everywhere right and it now gets entrained as well as diffuse and in this flow field you could now think about wherever they feel an oxidizer are present in stoichiometric proportions and that would be your Berkschumann jet flame. So on the one hand the Berkschumann problem would actually mean this problem a Berkschumann flame in the literature now comes to mean any stoichiometric surface that is now coincident with a Berkschumann flame that is essentially what it means that means we are adopting the infinite rate chemistry assumption or the flame sheet assumption or the mixed disburnt approach or whatever it is all you can you can call it in any other way anyway this is what we are talking about the second thing that we decided to talk about when we looked at this expression was what about the dependence on the flame height right do we get these dependencies we do not get to see that right there okay and part of the reason was Peclet number did not show up because we had we had supposed infinite Peclet number there in this kind of a formulation you could hope to actually begin to see Peclet number show up in places and that would actually now denote the flow velocity relative to diffusion right and then you can begin to see what is the effect of flow velocity but still the expressions are going to be so complicated you do not get the physical feel right. So it is easy for us to actually look at what happens in a in a order of magnitude manner so order of magnitude all we are saying in this problem is a axial convection balance is radial diffusion right so what that means is you can now look at the axial diffusion axial convection timescale and axial convection timescale is essentially let us suppose that you now have the flame length as LF so axial convection time is essentially the residence time of reactance in the flame the residence time is actually a very very basic and important idea that you will find happen to think thought about in design of combustors and so on so whenever you want to actually design the length of a furnace right particularly for diffusion flames previously we saw what happens when you have ramjets and so on or after burners the flame is here the premixed flame is held at the flame holder and then it gets inclined of trying to burn into the reactant flow so that the normal component of the flow velocity balances the flame speed of course it could be turbulent flame speed so you get the shape of the flame and therefore you get the length of the combustor all those things right essentially the idea basically there is what is the residence time of the reactants within this region okay so there what happens is you are looking at LF over U so if this is the length of the flame then as the flow goes along the along this it is going to be there for so long while it is there it is now diffusing this way right so the diffusion timescale is as we saw yesterday we have B squared over D right therefore therefore if you now say LF over U is of the same order as B squared over D then LF which is the same as eta max here okay so long as we were doing a heavy-duty mathematics we were we were using Greek like eta eta max and all those things but this is so just order of magnitude just thinking about this so we just so you started using LF right this is the same thing so LF is UB squared divided by D right now this is an order of magnitude idea and it works reasonably well not just for Berkshuman problem or the Berkshuman problem. So if you have a jet that is coming out of a fuel jet that is coming out of a pipe of certain radius or diameter D you could still say do not worry about the fact that you have to use outer diameter outer duct diameter the order is not going to be significantly different if you use the inner duct diameter all right so in a jet diffusion flame you have only one duct you do not have the outer duct it is a it is just in training and diffusing with atmospheric quiescent air therefore you could simply say U U small D squared divided by capital D where small D is like the diameter of the jet right or the inner pipe that is fine now what that means is two things you can look at one the diameter squared is proportional to the cross sectional area right and the cross sectional area times velocity is essentially the volume flow rate like what you would measure with a flow meter let us say like liters per minute or meter cubes per minute or whatever it is right so this is basically volume flow rate divided by the diffusion coefficient so you can look at it in two ways one either I say velocity burner diameter split or volume flow rate if you look at the first way you learn two things one the flame height is going to be proportional to velocity and proportional to square of the duct diameter all right if you look at the second way all you get is the flame is going to be proportional to the volume flow rate all right so larger faster the flow that is coming taller the flame is going to be or larger the duct diameter much taller the flame is going to be or whatever is more the volume flow rate larger the flame longer the flame is going to be right that is what it means what about we have done we have been talking whenever we talk about flow velocity we have to bring in the Peclet number in our in our discussion and so on but all the when we do that and then I also said Peclet number is kind of like Reynolds number and so on but we never talked about turbulence right we are not really worried about that so where is the question of that that and that why are we talking about it the reason why we are talking about it is as we keep on increasing the velocity at some stage or keep on increasing the volume flow rate or your diameter duct diameter right at some stage you are now getting to turbulent flows and then what happens well if you now have turbulent flows the diffusion that is going to happen is going to be a turbulent diffusion right and there is not going to be molecular diffusion anymore so we have to now factor in the turbulent diffusion so the way you can do this is you can all look at the Schmidt number s e is essentially equal to nu over d or d goes as nu over s e right so now for a constant Schmidt number let us not worry about how the Schmidt number goes typically we do not have to worry about that and therefore Lf goes as let us say u b square divided by nu right so if nu changes d changes through Schmidt number therefore if you want to talk about d as a turbulent mass diffusivity you can think about it in terms of turbulent kinematic viscosity and a turbulent kinematic viscosity is a flow dependent parameter right and the way it goes is this goes is u b so sorry you right the greater the velocity more the turbulent viscosity greater the diameter because it it depends on Reynolds number right so if you now think about this then Lf simply becomes proportional only to b that means in fact I think I think I would feel a lot more comfortable you can say also Lf then is u d square divided by d where d is fuel duct diameter I think it makes a lot of sense to talk about it in terms of in the context of jet flames and keep it as d rather than b because I am not thinking more about the outer duct diameter anymore mainly talking about the fuel duct diameter so what this tells us is the flame the flame length is no longer going to be dependent on the flow velocity it simply is going to scale only with the duct diameter right so if you look at the data what you should find is if you now plot your Lf versus you what you will find here is Lf is proportional to u u b squared by d or u d squared by capital D so this is linearly increasing with d with you right so for small u you get a linear increase all right but then you get into a transition region where you transition to turbulent flows and you now have a transition that now makes it insensitive ultimately to why is that because you have more and more mixing that is happening and as more and more mixing is happening you are having the fuel and oxidizer burn in stoichiometric proportions much closer to the burner and that does become insensitive to the flow because more and more flow more and more burning can happen and all these things are happening within a very short within a pretty much constant distance right so that is essentially what is going on as far as turbulent mixing and burning is concerned and therefore you have a pretty insensitive flame length to the flow velocity so the interesting picture I have in my mind about these things is about our cartoons that we watch when we are kids like you have this dragon that spews fire right like the spitfire dragon then it is a fool and then you now get a fire out of it they can clearly see from there that the greater the velocity greater is the flame length right and of course the and then the dragon really wants to hurt you and then keeps increasing the velocity and if the flame length does not does not increase anymore because it is become turbulent and it starts blinking what the hell how am I not able to make any impact so it starts opening up its mouth more and more and then the flame length increases and then it attacks the enemy right so we can learn quite a bit of diffusion flames watching cartoons thank you very much.