 We will adopt what is called as the mixture fraction formulation in trying to describe diffusion flames this is actually a powerful idea for combustion in general and with particular context of diffusion flames and partially premixed flame. So the idea basically is we now try to define this variable called the mixture fraction mixture fraction extremely useful than in diffusion flames and partially premixed flames. So what we will do is let us consider a two-feet system that means we have two inlets that are allowing gases to come so we are looking at a homogeneous gaseous system and let us say that one represents fuel stream and two represents oxidizer stream in fact we will try to adopt this for a general hydrocarbon fuel hydrocarbon oxidation that means when we now say oxidizer we will mainly be thinking about O2 and for the fuel we might mainly be thinking about a Cm Hn kind of hydrocarbon right. So then the way the mixture fraction is defined is essentially like a mass fraction okay. So the mixture fraction mixture fraction represents the mass fraction mass fraction of the fuel in the mixture fuel in the mixture right that means let us say Z is the symbol given to mixture fraction then this is basically m dot 1 divided by m dot 1 plus m dot 2 where m dot 1 is the mass flux of the mass flow rate of the fuel stream m dot 1 is m dot 2 is the mass flow rate of the oxidizer stream and of course what this means is the mass fraction of the oxidizer then is oxidizer in the mixture is then 1-Z alright it would be defined as m2 dot divided by m1 dot m2 dot therefore it is 1-Z now each stream contains diluents it is not as if like the fuel stream is fully fuel you could have some nitrogen you could also have the oxidizer stream is like a let us say air and most of it is nitrogen so you have only part of it is oxygen so contains diluents different extents. And what you are looking at is the mass fraction of the unburned fuel in the mixture what we expect is it should be proportional to the inlet mass fraction of the fuel in the fuel stream times the mixture fraction so the mixture fraction is a fraction of fuel when compared to the total so the mass fraction of the unburned fuel in the mixture is proportional to the inlet fuel mass fraction so that is to say yf,u that is unburned fuel mass fraction anywhere in the mixture can then be written as Z times yf,1 so yf,1 is essentially the mass fraction fuel mass fraction at in at the inlet to stream 1 right now previously we were using the symbol yf,0 for this purpose the corresponding oxidizer would be yo,0 well this allows for us to actually think about a yf,2 which means you could also dope some fuel into the oxidizer stream and vice versa all right. So this notation some somewhat is a little bit more general of course it is not too general it is still confining ourselves to two streams all right but we could say yf,1, yf,2, yf,3 and so on okay so this is a little bit more general notation then and similarly the unburned the unburned oxidizer mass fraction the unburned oxidizer mass fraction is yo,2,u is equal to 1-z times yo,2 right so yo,2 is actually the oxidizer mass fraction at inlet to stream 2 so for example if it were air this number would be like 0.23 or so with this if you now think about it what happens is your unburned fuel mass fraction or the unburned oxidizer mass fraction both vary with the mixture fraction z linearly in other words if z were more your yf,u would be more at any place proportional to yf,1 right if z were more yo,2 unburned will be less correspondingly proportional to yo,2,2 so if this is represented in a graph that could go like this so if you now plot z along the horizontal axis it goes from 0 to 1 so if your mixture fraction is going all the way from 0 to 1,0 would correspond to no fuel and one would correspond to all fuel all right no oxidizer and therefore as z keeps on increasing then you have a straight line that keeps on increasing from 0 all the way to yf,1 this is to say in the mixing field at the inlet you would have only oxidizer in the oxidizer stream you will have no fuel at all in the you would have yf,1 in the fuel stream and you would have no fuel in the oxidizer stream and anywhere else in the mixing field you would now have a mixture fraction which is having different values and corresponding to that the fuel mass fraction would lie on this line essentially all right. If you want to think about how the so this is this is the plot of yf,u similarly if you want to look at the plot of yo,u that is another straight line that starts with the value of yo,u let us say at the oxidizer stream inlet why you do not have any fuel and goes all the way to 0 at the fuel inlet where you do not have any oxidizer anywhere in the mixing field your mass fraction of the oxidizer lies on this line corresponding to whatever is the mixture fraction locally right. So this is regarding how the unburnt fuel and unburnt oxidizer mass fractions work out what we want to now do is to further proceed and see if we can actually get mass fractions of burnt products right and where you have when where you are in non-starch emetric proportions you will have excess fuel and excess or excess oxidizer depending upon which is in excess whether you are in fuel rich or fuel lean conditions and we would like to also find out what is the burnt fuel concentration and the burnt oxidizer concentration depending upon whether it is excess or not. So for which we need now have to actually go through what happens when you have a reaction so in a reaction we had this is long ago we came up with this d y1 divided by w1 times nu1 double prime minus nu1 single prime etc you could also write of course one more d y2 divided by w2 times nu2 double prime minus nu2 single prime etc and the general ith species you could say d yi divided by w i times nu i double prime minus nu i single prime you recall this so let us suppose that you can call this set of equations a star we will pick at least two of these in order to integrate it at different times in the near future. So for a single single component general hydrocarbon that is to say you do not have different mixtures of hydrocarbon for example if you take liquid petroleum gas it is actually a mixture of hydrocarbons butane and propane so instead of that if let us suppose that we are considering only one hydrocarbon at a time hydrocarbon fuel and let us suppose that we write this equation as nu of single prime Cmhn for the generalized hydrocarbon that we considered plus nu O2 single prime O2 gives you CO2 double prime CO2 plus nu H2 O double prime H2 O right with nu F single prime equal to 1 fixed arbitrarily let us suppose that we just say we are interested in one mole of this hydrocarbon right we could get nu O2 single prime equal to m plus n divided by 4 nu CO2 double prime equal to m and nu H2 O double prime equal to n over 2 as stoichiometric coefficients that means these values correspond to the reaction happening stoichiometrically right that means you do not have any of the fuel or oxidizer left over as part of the products right. Now at the stoichiometric surface in the mixing field we then have the number of moles of oxygen unburnt divided by the number of moles of fuel unburnt at stoichiometry equal to nu O2 single prime divided by nu F single prime right as given above so in terms of mass fractions so this is this is moles fractions so in terms of mass fractions where the above is in terms of moles not most fractions I am sorry but of course you can divide by the total number of moles and get the most fractions as well so here we are going to be looking at the mass fractions why O2 unburnt divided by why F unburnt at stoichiometry is nu O2 single prime molecular way times molecular weight of oxygen divided by new fuel single prime times WF which let us we let us say we call us nu which is the stoichiometric mass ratio so what we want to do is we want to see if we can integrate this equation for a situation where you have a stoichiometric reaction so that we can we can make use of the fact that you are not going to have any of the reactants left over all right. So for a stoichiometric reaction for a stoichiometric reaction both fuel and oxidizer are completely consumed right therefore we use this to integrate the this equation here star at any mixture condition relative to the unburnt mixture because you are not going to have any of the reactants left over so you can you can you can now say we integrate star at any mixture condition that means you look at any intermediate mixture condition mixture condition relative to the unburnt condition right that means the limits of integration go from unburnt values to any value during the reaction as Y O2-Y O2, U divided by nu O2 prime W O2 right you should you can you can just pick let us say oxidizer and fuel then you would get YF divided by-YFU divided by nu of single prime WF or rearranging this all this means is nu YF-Y O2 U right. So what this tells us is this is sort of a preserved quantity right you started out with unburnt values YFU and Y O2U at a point in the mixture mixing field in a proportion with a nu here right where should I should I have yes nu that is right you call this nu. So and that is the same even during the reaction between the two that is essentially the idea okay so whatever is the unburnt values that difference is what is preserved so what you can then do is we are interested in writing everything in terms of the mixture fraction so we had the YFU in terms of mixture fraction and the inlet mass fraction so here Y O2U in terms of mixture fraction and the inlet mass fraction and the oxidizer so substituting the expressions for YFU and Y O2U in terms of Z in here we get and then of course rearrange so you say Z equals nu times YF-Y O2 plus Y O2, 2 divided by nu YF, 1 plus Y O2, 2 so you can see that Y O2, 2 YF, 1 YF, 1 these things are showing up from the inlet mass concentrations mass fractions of the oxidizer and fuel in the respective streams. So for stoichiometric mixture for stoichiometric mixture nu F nu YFM sorry should be equal to Y O2 in fact this is what we had also got for the Berkschumman solution where we said gamma is equal to 0 means beta is equal to 0 that means alpha F-alpha O is equal to 0 right and that amounted to exactly this we say YF divided by Y O2 is or rather Y O2 divided by YF should be equal to nu and that should be the same as YFU divided by YFU right the unburnt mass fractions should be in stoichiometric proportions just as well as whatever is in the reaction okay. So from this what we can do is this will actually now tell us what should be the stoichiometric mass fraction right. So the stoichiometric mass fraction stoichiometric mass fraction is ZST equal to 1 plus nu YF, 1 divided by Y O2, 2 the whole to the – 1 it should be able to show that right if you now say this is going to go away right then you rearrange what this tells us is there is a particular value of Z which is the stoichiometric mixture fraction naturally and it is related to how the inlet stream mass fractions of fuel and oxidizer work out right. So if Z is less than ZST right we are less than fuel so as Z keeps on increasing you have more and more fuel that is essentially the idea. So if Z is less than ZST we know clearly that it is fuel deficient right deficient or fuel lean whichever way you want to right then YF burnt is going to be 0 this is now the mass fraction of fuel in the burn products so you are not going to have any excess fuel in the products therefore you can now substitute YFB equal to 0 which terminates a combustion right. So the combustion essentially does not proceed simply because you run out of fuel you do not have any more fuel. So plug in plug this in Z expression and use definition of ZST you should now get if you now plug in the expression for Z YFB is equal to 0 then from there you now rewrite you should get YO2, B equal to YO2, 2 times 1- Z divided by ZST for Z less than or equal to ZST similarly for fuel rich mixture combustion terminates when YO2B equal to 0 so you now plug this in the expression for Z and rewrite for YFB you now get an expression YF, B equal to YF, 1 times Z- ZST divided by 1- ZST and this is valid for Z greater than or equal to ZST that means we can now draw a picture similar to this this is for the unburned fuel and unburned oxidizer in a mixing field but we can now anticipate how the unburned sorry the burnt fuel unburned oxidizer should look like if you now have Z going in the horizontal axis from 0 to 1 somewhere in there you should now be able to locate Z equal to ZST now that that become that becomes an important thing whether or we on this side of ZST or that side of ZST because this is going to correspond to a fuel lean situation that is going to correspond to an oxidizer lean situation therefore we should now get your YFB is valid only for Z greater than ZST right on this side your YFB is equal to 0 therefore you and this is also linear in Z okay so this goes so your YF, B is going to go all the way to from 0 to YF, 1 right as a matter of fact when you go to this corner when Z is equal to 1 your YF is equal to YF1 you know and so sorry Z goes to 1 so YFB goes to YF1 all right and this is what we were looking at the last class where at the flame you now have the fuel concentration go all the way to 0 and then stay flat on the other side okay so here the fuel is going to go all the way from the fuel inlet stream all the way down to 0 at the stoichiometric surface and then lie flat on the oxidizer which side right and similarly you now have a straight line that goes from 0 here backwards to Y0 to 2 in the mixing field so essentially if it is possible for us to now go back and redo the Berkshuman problem in terms of Z instead of beta which is not very different if you think about it all right and that is also possible then you can actually look at contours of Z several contours of Z and depending upon the contours of values of Z you now try to actually look at what the values of YFB and YO2B should be you now can map how the fuel concentration and oxidizer concentration should show up this is still for the excess fuel the side the products and excess oxidizer in the products for fuel rich side and fuel lean side still have not talked about the actual products of combustion right like CO2 and H2O so for getting the product concentrations we should do something similar to what we have done here in order to obtain your new CO2 double prime and new H2O double prime all we have done by saying we are balancing this reaction is to basically conserve atoms we compared carbon atoms hydrogen atoms oxygen atoms and then came up with these things so we conserve atoms across the chemical reaction to obtain product mass fractions as a function of Z still we are interested in what happens as a function of the mixture fraction Z that is a hero for the day mixture fraction so we keep on looking at everything as a function of Z so element mass fractions so you look at what happens in the unburnt mixture element mass fraction in the unburnt mixture so look at the fraction of carbon atoms so you had m carbon atoms in the hydrocarbon right so this is m times molecular weight of carbon atom divided by the molecular weight of the fuel right times yf, u so this is the mass fraction of the unburnt fuel times this fraction is what is going to give you the carbon fraction similarly if you now look at ZH that is going to be n times WH divided by W fuel times yf, u for the oxygen it is mainly coming from the oxidizer stream so you might as well just write ZO equal to YO to U whatever is the mass fraction of the oxygen in the oxidizer stream unburnt right is the ZO, ZO of course from here we should now be able to also write CC that Y of U is equal to WF times n ZC divided by WC plus m the dynamic of a stick ZH by WHH all right on the burnt side on the burnt side for the products the element mass fraction for the carbon we have ZC equal to m WC divided by WF yf, B that is actually for if you have excess burnt fuel right plus if carbon is also there in carbon dioxide that is what we are more interested in WCO2 times CO2 burnt and this is what we are trying to find out in terms of Z okay and hydrogen is going to be like n times WH divided by WF yf, B plus twice because water contains two hydrogen atoms so WH divided by WH2O times YH2O, B is also the next thing that we are trying to find out in terms of Z currently so and finally we have to conserve oxygen atoms so ZO from the products would be twice WO divided by WO2, YO2, B plus twice WO divided by WCO2 because there are two oxygen atoms in carbon dioxide YCO2, B plus 1WO divided by WH2O, YH2O, B right so now if you now equate these so use YFU and YO2U in terms of Z in ZC, ZH, ZO and rewrite okay so you then get YCO2B would be YCO2 stoichiometric times Z divided by ZST and YH2O, B equal to YH2O stoichiometric times Z divided by Z stoichiometric this is for Z less than or equal to ZST or YCO2, O, B equal to YCO2 stoichiometric times 1 minus Z divided by 1 minus Z stoichiometric and YH2O, B equal to YH2O stoichiometric times 1 minus Z divided by 1 minus ZST this is for Z greater than or equal to ZST where we have to say what is YCO2 stoichiometric and YH2O stoichiometric so YCO2 stoichiometric is YF, 1 ZST times MWCO2 divided by WF and YH2O, stoichiometric is YF, 1 ZST NWH2O divided by WF alright so what do you what do you expect we can still see that YCO2 is still going linearly as Z, YH2O is going linearly as Z so long as Z is less than ZST, YCO2 is still going linearly but decreasing with Z, YH2O is decreasing linearly with Z as Z is greater than right. So let us think about plotting this so you still have Z along the horizontal axis going from 0 to 1 and we now locate the stoichiometric surface at somewhere in between and we noticed earlier from that picture just to reproduce your YF, B goes from 0 at the Z equal to Z stoichiometric to YF, 1 when Z is equal to 1 this is the excess burnt fuel and you now have YO2, B going all the way up to YO2, 2 at Z equal to 0. But what we are looking for now is how this varies YCO2 goes to a peak value at Z equal to ZST and then comes all the way down to 0 so this is your YCO2, B and similarly of course depending upon the numbers you should now get a similar set of lines for YH2O, B. So it reaches a peak value you get the same expression same numbers I should say from both the expressions at Z equal to ZST you can plug that and then see what happens right. Now we are still interested in the temperature so for temperature all these things are essentially like looking at an equilibrium you see so for the temperature where we now say for an adiabatic flame right for an adiabatic flame we have the unburnt enthalpy equal to the burnt enthalpy and it is probably a good time to stop we will just stop here.