 So, for an adiabatic flame we have the unburnt enthalpy should be equal to the burnt enthalpy which means we could now say sigma k equals 1 to n yk,u times hk,u equal to sigma k equals 1 to n ykb hkb where hkb is now the enthalpy of the burnt products ykb is something that we have seen so far. So what we then do is the normal thing which is we now say enthalpy is formation enthalpy plus sensible enthalpy right so using heats of formation and sensible enthalpies we can write sigma k equals 1 to n yk,u minus ykb times delta hfk0 which is the reference formation enthalpy for each species equal to now we look at the sensible enthalpy which is t ref time to t or tb that is cpb dt minus integral t ref to tu cpu dt right our job is to actually find out tb given to you okay that is what you are trying to do here. So where of course cpb is equal to sigma k equals 1 to n ykb cp cpk of temperature and pu is sigma k equals 1 to n yk,u cpk of t now so what we want to do is we want to try to find out what this is right and therefore and then we have already looked at how to do the products therefore we can do the same thing. So we had dyk divided by nu k double prime minus nu k single prime times capital Wk equals over many species to let us say one particular species dy1 divided by nu 1 double prime minus nu 1 single prime times capital W1 so integrating this integrating this we get yku minus ykb equals so let us suppose that you do this for the kth species in relation to the fuel so that means you take any kth species and then you take fuel the left hand side and right hand side respectively then you could you could write this as yfu minus yfb times nu k double prime minus nu k single prime Wk divided by nu f Wf now keep in mind your nu f here is actually the stoichiometric coefficient for the fuel and you do not have any fuel on the right hand side therefore your nu k double prime minus nu k single prime for the fuel is simply nu f right so you do not have to worry about that anymore because you know how the reaction looks like for this so from here if you know this difference now for the kth species multiplying by delta hfk and summing over all k sigma k equals 1 to n yku minus ykb times delta hfk not is equal to then you see the only thing that is really getting summed over on the right hand side is this part all right and this is all this is independent of k it is only for the fuel therefore why yfu minus yfb divided by nu f Wf times sigma k equals 1 to n u k Wk delta hfk not where we now use a symbol where of course it is going to get a little bit bigger so we tried to compress it by seeing nu k is nu k double prime minus nu k single prime just to just to kind of have our bearings on when you see nu k is equal to nu k double prime minus nu k single prime that is like a set that you are trying to allocate a particular symbol but we have also looked at the nu without a subscript which is a stoichiometric stoichiometric ratio mass ratio and nu f is actually the stoichiometric coefficient for fuel that is already been identified which is basically this one evaluated already right so these are these are the different news that we have we are looking at so you just do not get confused about these things all right so if you now look at this then the left hand side here is what this is effectively what it means is this essentially translates to just looking at this summation all right and therefore if you now say that summation if it is Q let that so we can we can now define heat of combustion heat of combustion Q equal to negative k equals 1 to n nu k wk delta Hfk not all right let us just see heat or the other way you can do is you can also say negative sigma k equals 1 to n nu k delta Hfk not where this is on a molar basis this is on a mass basis so mass basis multiplied by the molecular way is what you look in that so for simplicity let us now say t u equals t ref now we have already gone through this before in the context of premixed flames but it does not really lose much generality here by doing this and also zoom constant Cp constant Cp therefore now for a lean mixture for a lean mixture when yfb equal to 0 right means you are running out of the burnt fuel because you do not have any of it as an excess as for a lean mixture then all you do is you now say here why just plug in yfb equal to 0 here right that is what you would actually do for the left hand side on the right hand side there is a simply Cp times Cp times tb minus to you effectively right so that means a Cp can come to the other side in the denominator and then you will get tb minus to you equal to q yfu divided by Cp nu f wf okay now it is not for nothing that we chose to actually look at fuel as one of the sides of this equation so in this equation we chose one of them to be fuel with the other one being any kth species on the left hand side because we are looking at a lean mixture where you are going to substitute yfb equal to 0 but for a rich mixture you would want to actually substitute yo2b equal to 0 right which means we need to have a place in which we can substitute yo2b equal to 0 for which we should have started out having oxidizer integrated rather than fuel right so that is what we would do for a rich mixture we go back and integrate again as yku minus ykb trying to redo this again with oxidizer on the right hand side is to the fuel so ykb minus yku equal to yo2u minus y so this must be yo2u minus yo2b divided by nu o nu o2w o2 times times nu kwk divided by yeah that is it I will re-wrote this so nu k of course is again nu k double prime minus nu k single prime which I have not done before so if you do this and of course you now can take this step where you multiply by delta hfk and sum over and then you now recognize this is q and plug this back in here and write your tb minus tu I just say you now plug your yo2b equal to 0 right for the rich mixture and plug yo2b equal to 0 for rich mixture right tb minus tu is equal to q yo2u divided by cp nu o2w o2 now what we are really looking for is the burnt temperature in terms of the mixture fractions up because mixture fraction is kind of like our basic function now so we want to now represent this in terms of sorry yfu in terms of Z and yo2u in terms of Z which is not very difficult okay so we can say that this is basically Z times yf,1 and this is 1-Z times yo2,2 right so we know how to do that so so we we now write yfu and yo2u in the above expressions in terms of Z the mixture fraction then right also the unburned temperature temperature in terms of mixture fraction how does that work out you have tu of Z equal to t2-Z times t2-t1 right how do I know that this is right t2 is the oxidizer stream temperature and t1 is the fuel stream temperature right so if I now plug Z equal to 0 then it is all oxidizer then I get the unburned temperature for the oxidizer stream to be t2 that is what this is if Z is equal to 1 then it is the fuel stream right then the t2 gets cancelled and you get t1 for the unburned temperature so that is the temperature of the fuel stream so this is basically like a linear relationship between the two temperatures for the unburned temperature of a mixture of any given mixture fraction in between right so if with that being the case you now go back and plug that and in terms of Z and then plug your yfu in terms of Z and yo2 in terms of Z right so all of them put in put in terms of Z will be then tb of Z equal to tu of Z plus qyf, 1 divided by cp nuf wf times Z with Z much mean less than or equal to Z stoichiometric or tb of Z equals tu of Z plus q yo2, 2 divided by cp nuo2 wo2 times 1- Z for Z greater than or equal to Z st right now the maximum temperature you can calculate the maximum temperature so Z keeps on increasing you see tb will keep on increasing up to Z st and when you now cross Z st you now have to use this expression where as the Z keeps on increasing tb keeps on decreasing right and at Z equal to Z st you should actually get the same expression from both that means you should have like one temperature at which the stoichiometric mixture has a burn temperature so the maximum temperature and that will be the maximum temperature the max temperature T st is calculated for Z st from either expression from either expression above so how does this look like now we have been drawing these pictures at every now and then as in as in when we get a bunch of expressions are in terms of Z we try to plot them so here you see if Z is plotted along the horizontal axis and let us say go from 0 to 1 all oxidizer stream here and all fuel stream there and let us say the oxidizer temperature is T2 and the fuel temperature let us say is T1 the unburned temperature is not going to depend on whether you had a stoichiometric surface and so on okay what about it may be you just get a straight line that connects the two which is basically following this expression that is a linear expression in Z as well okay and you just get a straight line that connects these two this is TU as far as burnt temperature is concerned you have to now worry about whether your fuel rich or fuel lean right and that means you have to look at where your Z st will be so if your Z st is going to be here then the T st is going to be the peak and on the one side when you are now having only oxidizer you you do not have any reactions you are simply not having any fuel at all so you go back to T2 right and similarly on the other side you just have to look at a linear variation from T st to T1 so it just becomes that that simple all right so in some sense what this really means is we can now put the whole picture together in what we are doing in a bigger bigger graph let us just try to do that now need more space so let us now say we had a unburned situation and a mixing field in which you have the Z varying along the horizontal axis going from 0 to 1 all oxidizer to all fuel right and we now want to look at how this proceeds in a third axis where we are interested in constructing the same kind of picture for the burnt process so let us suppose that that is what it is right and along along the this third axis we now have a variable called reactiveness what is the level of reaction that is going on how do you monitor this by looking at yf minus yfu divided by yfb minus yfu so when you started out all right when you started out you have yf is equal to yfu so r is equal to 0 okay when you end yf becomes yfb the 4r becomes 1 all right so we can we can plot this along this axis maybe let us say along this axis maybe we just plot along this right that is a way the reaction is going all right and yf is turning into turning on turning from yfb yfu to yfb initially we are having your yfu actually starting starting from let us say yf1 here to 0 on this side and therefore you are going to have a straight line that looks like this so this is your yfu and somewhere in here you are going to have your zsd right and your yfb is going to be so maybe we can look at you do not have to worry about zsd at all here you just to just to get a perspective on how this kind of a linearly increasing yfu actually turns out to start increasing up to yf1 over here right and similarly you could think about a yo2u which starts from yo2 and goes all the way to 0 when z goes to 1 that is the unburned oxidizer right and that now is going to go all the way to so this is also yo2u and of course I should have marked yf1 there and then we could do a couple of things one let us assume that you have diluent only in the oxidizer like an air right so more the air more the oxygen and correspondingly more diluent if you do not have any air you do not have any oxygen you do not have any diluent either right so you could now think about a diluent concentration that is going to grow with more and more o2 as we go along backwards in this axis starting from 0 so so this is like ydu where d is for diluent right and this picture is not going to change at the end of the reaction it is about the same so you are now going to get this picture to look like ydu right and of course there is a value here that corresponds to how much diluent you have for how much oxygen so this could be for example 0.23 that could be like 0.77 and this could be now 0.77 alright okay and so on then we talked about temperatures oh no I think we talked about products so we talked about products we do not have any products here to talk about alright we are talking about products only here and the way we talked about it is we now said you are going to have your yco2b go like that and yh2ob go like that alright and finally we talked about temperature so the temperature we can plot two things the unburned temperature and the burn temperature right so the unburned temperature that supposed would be we now pick a scale like that this is your t2 and this is your t1 it is as if like the fuel stream is colder than the oxidizer stream right it could be and then you can mark these two points t2 here and t1 there maybe t1 is somewhere there according to this and now going to go all the way to TST and go back here so if you think about it this is like what we were saying for the vicinity of the Burke-Schuman flame so in the Burke-Schuman flame if you have a diffusion flame sheet in the Z axis it is going to be located at Z equal to ZST right and from Z equal to ZST the temperature is going to fall on either side on the fuel rich side one way and the oxidizer rich side the other way but in the flame we are looking at how it is behaving in space rather than in the space as in the physical space above the burner rather than looking at it in what is called as the mixture fraction space so in this picture everything looks very nice straight lines okay all linear dependences in the mixture fraction space but that is not how the curves were looking looking like in the physical space and that is one of the reasons why the mixture fraction space is quite valuable so question basically now is how do we transform this into or how do you utilize this in the physical space system that is what we are worried about like for example if somebody is giving you a burner and asking you to find out how the flame is going to look like and how the species concentrations in the temperatures are going to vary and all those things you cannot give them a picture back like this which is in the mixture fraction space you have to now translate this into the physical space so you now look at how the mixture fraction for example can be utilized in looking at the spatial variation that means we need to have a governing equation for how the mixture fraction varies in space that means we have to have an evolution equation for this what I am going to now write is basically the steady state counterpart but there is nothing that really stops you from doing an unsteady counterpart just like how you would do an unsteady Schwab-Zehler which formulation which we did not go through but we said it is possible to do so so similarly we can now write for example a evolution equation for the mixture fraction right so the mixture fraction essentially is like a conserved scalar and it is now being evolved by convective and diffusive processes okay now that is the physics behind how this goes you can actually derive this mathematically in two different ways one go back to the definition of mixture fraction mixture fraction is nothing but a fuel mass fraction kind of notion right so you can now find that you have like a new yf-y02 divided by something plus some plus new y02 2 divided by something so the all the things that are getting added up and divided by our constants which depend on the inlet concentrations but what is really happening is you are now forming this coupling function like new f-y0 new yf-y02 right if you now use the evolution equations for species concentration yf-y02 and you now multiplied by new subtracted one from the other added something else divided by something some other constant and so on you should now be able to get this and while you do that you can also find out that the right hand side being the reaction rate term is now going to go to is not going to get subtracted out pretty much like the way we did in the Berkshuman problem in the Berkshuman problem we started looking at the Schwab-Zeldovich species equation with the reaction rate on the right hand side and we now try to normalize things and subtract and form the coupling function such that you get the same expression on the right hand side you saw that when you subtract you get to get to 0 so you get you get 0 here similarly the other way of looking at it is while we were looking at the mixture fraction we also noticed that we could form element mass fractions like for example Zc Zh Zo and so on and we could form our mixture fraction based on the element mass fractions. So, so long as the mixture fraction is essentially defined based on the element mass fractions and in a combustion process the chemical reaction that we are going through conserves elements it does not really destroy the elements. So, no matter whether carbon is present in in the hydrocarbon or carbon dioxide no matter whether hydrogen is present in a hydrocarbon or water no matter whether oxygen is present as itself or as part of carbon dioxide or water right the mixture fraction gets conserved because it is based on element mass fractions and therefore the reaction rate term will not show up when you now try to write a evolution equation for this. So, we just write this out as it is and then the other thing that we need to do in order to actually look at the structure of the So, this is okay as far as the infinite chemistry kind of approach is concerned. So, what we talked about in the Berkshuman problem was that this is like a sheet that corresponds to infinite chemistry and on either side you have these things dropping and varying correspondingly. But if you want to now think about finite rate chemistry okay. So, if you are to think about finite rate chemistry you have to have the reaction rate show up somewhere and typically the problem with finite rate chemistry is the finite can the finite chemical reaction depends on species concentrations as well as temperature right. And the Schwab-Zeldovich formulation what we have noticed is we have to keep the reaction rate at least in one equation even while you are subtracting a lot of other things. Therefore if you now say let us now we have to consider the energy equation where the reaction rate depends on the temperature as well as react concentrations and the concentration is governed by reactions the sort of equations that have been subtracted from each third to get this okay. So, this represents a pair of species reactions the species equations and then we have the energy equation. So, if you now write this as so this is of course before we do that we can we can say this for a simple case of let us say the Buxchermann kind of problem if you now say rho u dou z divided by dou x minus right or if you want to do it in cylindrical polar this would be like this and 1 over r and so on that is okay equal to 0 now you could go ahead and solve this as it is and then finally look at the solution and find out wherever z is equal to z st is where the flame shape is in terms of so you are now going to get finally z of r, z equal to z st is the flame shape right. So, you could go ahead and do do your solution that way and once you do this on either side in the in the in the burner let us say you have your flame like this this is your z equals z st you can now get the solution to give other z values and so on that picture looks very good isn't it right that is how it actually turns out to be you can solve the show up Zondavich formulation and get this kind of picture and so these will correspond to different values of z with z equals z st being where it is to correspond to where the flame is and then you can now look at for different values of z what should be your yfb what should be your yf yo2b what should be your tb and so on all right so you could you could do this but that is actually the infinite rate chemistry approach now for the finite rate chemistry approach as I said you need to have at least one equation which contains the reaction rate that means you have to have the energy equation let us say minus 1 over r dou by dou r r rho d dou t by dou r right equal to 1 over cp sigma k equals 1 to n delta hfk0 wk this is directly coming from our Schwab-Zondavich energy equation right now what we have to do next and I am not going to do this in great detail I am just going to point to directions that that will take you there because this is just getting a bit too involved and I think the idea is essentially made and we could leave it there so what you now do is transform the equations with respect to Z that means we could say let one of the variables let us say Z here equal to Z which is the mixture fraction and let us say r is equal to Z1 so no big change there just changing symbols but here we are actually changing meanings okay and this is kind of like what is called as a croco transformation where you are replacing one of the independent variables by a dependent variable upon which you want other dependent variables to depend so involved that we have seen so far all the dependent variables that you are talking about are depending upon Z and Z itself is governed by this given that it is depending on X and sorry Z and R right so once you know how Z depends on the spatial variables we now transform our equation in terms of Z capital Z being the independent variable and now try to obtain the energy equation in terms of capital Z get your solution in terms of capital Z if you know how Z varies you know how your temperature varies right so this is not this is not so trivial you will actually end up with lots of so essentially what you have to go through is like if you now want to say dou by dou Z then let us say that that is actually dou by dou Z plus dou by capital dou capital Z by dou small Z times dou small Z and so on so you have to apply chain rule across and you have to bring in the dependence on capital Z everywhere by chain rule and then you plug this in these equations and this actually becomes a fairly huge term because of this and you will have lots of cross derivatives and so on and then you could make now an approximation you could make an approximation whereby you say gradients that are perpendicular to the flame are important gradients that are tangential to the flame or unimportant this is typically valid whenever you have a flame sheet kind of situation right and you can now throw away a lot of terms and to leading order you can even throw away terms that are there and other derivatives except these and finally what you will get is so neglect tangential gradients gradients and you will now get something like rho chi divided by 2 dou square T by dou capital Z square equal to minus 1 over Cp sigma k equals 1 to n delta hf not k wk where where chi is 2d dou Z by dou dou dou Z the whole square okay in general in 3d you could actually have this as a gradient square gradient Z squared and this would basically call be called whatever this is basically what is called as a scalar dissipation rate and this is essentially looking like a diffusion term in fact the scalar dissipation rate has dimensions of 1 over time and essentially it gives a it gives a characteristic diffusion timescale right and basically it tells us how the diffusion happens Z mixed Z is mixed a fraction keep that in mind and you now have the gradient coming in picture that means is essentially represents the transport phenomenon of mixing how the how the mixture fraction variable is now going to decay because of diffusion right while the mixing happens between fear and oxidizer and this is actually quite important in turbulent diffusion wherever you are looking at for example scalar transport by turbulence then you have the turbulent diffusion mass diffusion coming into picture and the scalar dissipation rate in and the turbulent conditions would be would be quite important quantity there where the moment let us not worry about all those things and the point I wanted to make is it looks like you have a diffusion term and reactive term and the effect of this would be to try to find out what goes on in between in a very narrow region in this in the Z space where things are varying due to finite rate chemistry and that is the effect of this okay and what you will find is in reality you do not really have the oxidizer concentration come all the way down to 0 it could leak across the reaction zone if you now take finite rate chemistry into account because it is not able to consume all the oxygen right it is not really infinite rate reaction now and then it will look like this is like a essentially a diffusive reactive balance that we saw for the reaction zone in the preheat premix flame and I also pointed out that this region is going to be like the reaction zone in the premix flame and whatever is outside of this in the near vicinity is going to be like the preheat zone in terms of transport processes dominating except it is going to be radial transport versus axial convection and so on in the Berkshuman problem or any co-flow diffusion flame problem like get diffusion flames and so on except this is not really a diffusion term alone you have it actually as a function of as a second derivative of capital Z which is the mixture fraction so this is a derivative in the mixture fraction space so it is a transformation that contains within it convection as well as diffusion but by making this transformation in terms of the mixture fraction we have essentially enabled it to look like a reactive diffusive balance all right but but the transformation contains in it the convective and the diffusive effects so we would like to stop here and then we can pick up on let us say droplet combustion from next class.