 So, we are looking at this equation reaction where we are having a system of hydrogen and oxygen atoms where we have NH capital NH number of atoms of hydrogen capital NO atoms of oxygen and they react and then we are considering 6 products okay. So, we could consider a few more products and then the system is going to get little bit more complicated correspondingly and the temperature is going to change with the final answer is going to change correspondingly. So the accuracy to which you want the temperature depends on how much you want to go through the calculations and therefore you reside on the number of products you want to take right. So, what we already saw is these are the atom conservation equations right and then these are the hypothetical form a hypothetical formation equilibria equation or they are coming from there. So, in the last class we wrote actually these expressions for this Kpf, O, Kpf, H etc. We have now rearranged in such a way that we will now get expressions for NO, NH, NOH and NH2O each of them in terms of NO2 and NH2 okay. So, effectively we are now recasting everything that is there in the products in terms of only the reference elements okay. So, reference elements really is a key. So, once you do this you now plug these back in the atom conservation equations right and what are you going to get? So, you get 2 Kpf H2O NO2 to the half NH2 NO or P to the minus half that is for here plus KpNH2O to F OH NO2 to the half NH2 to the half plus KpF H that is for this sorry yeah NH2 is already there I am going to write this towards the end. So KpF HNH2 to the half NO or P to the half plus 2NH2 equal to capital NH right running out of space there. So, I am going to write the next equation here which is not very different now so or difficult now. So, KpF H2O NO to the half NH2N or P to the minus half plus for the OH we have KpF OH NO to the half NH2 to the half we do not have a NO or P for this yes plus then we write for NO and then we put the NO to later on. So, we can now write this is KpF OH it is here so NO2 to the half NO or P to the half plus 2N O2 equals capital NO right. So, what we have now is we actually have two equations okay two equations and explicitly speaking to unknowns the two unknowns being NH2 and NO2 but implicitly we have a N sitting here in a couple of places and that is a villain it contains all the unknowns right. So, it is like you really not you have really not eliminated more unknowns than we want therefore all the unknowns are there so we have to do this iteratively okay. So these are two equations and two explicit unknowns NH2 and NO2 but also contains contain N which has all the unknowns so which means we adopt a N iterative solution procedure right. So this is a bit of a cumbersome process if you are sitting in an exam you hope that you do not get a question on even one iteration to be manually calculated because it is a little bit of a nightmare so the first thing you have to do is do a initial guess for the N okay. So, assume an initial guess for N based on stable products okay stable products. So, we now have NHH2 plus half NOO2 gives a half NHH2O plus half I am sorry half NO minus one fourth NH times O2 if O2 is in excess right. So, if how much excess we can actually quantify this if half NO is greater than one fourth NH then initial guess for N equal to simply we are essentially looking at this clusters okay. So, these are the final stable products to expect that means we disregard all the unstable products right and therefore we can expect this to be half NH plus half NO minus one fourth NH and that would be one fourth N sub H plus one half N sub O yeah for this case this is so this is like a fulene situation. And or half NH plus I am sorry half NHH2 plus half NOO2 gives NO amount of H2O this is actually doing an algebraic balancing okay. So, this sort of like interesting instead of a high school way of numerically balancing you do an algebraic balancing right. So, so half NH minus NO times H2 where H2O is now the excess species if half NH is greater than NO this is the same as saying if one fourth NH was greater than one half NO okay we were comparing halves and quarters there. So, if you want to do the same thing you can do that this is fuleridge this implies initial guess N just add up these numbers. So, NO plus half NH minus NO is simply one one half NH right. So, you can start with this initial guess for N and then you have to proceed the other problem we see is these are actually functions of temperature here okay these these and so on right and that is the temperature we are trying to find out okay. So, we do not know what the temperature is but we need to take values for these so that is like a big problem that is a huge loop okay. So, what we have to do here is there are like about 2 or 3 loops that we have to think of we also are better off with actually so doing this. So, we do not know the KPFs because T adiabatic is unknown these are all evaluated at T adiabatic in fact we should specifically say over here right. So, therefore outer loop outer iteration loop would be to start assuming assume T ad okay. So, you have to make an assumption you start with some some some guess value for the T ad and so and then so the algo algorithm is now guess T ad to guess N3 calculate Ni double prime for calculate T ad okay by comparing with RHS of adiabatic flame or temperature equation which is essentially the enthalpy balance equation and then you go go around right of course you can you can actually check here check I am sorry between here and here you can check satisfaction of satisfaction of state equation or you can say equation of state right. So equation of state is essentially if you now have something like PV equals N or T ad so for the assumed T adiabatic okay if you have now if you have now converged on Ni double prime okay then you should now get an N that is different from the guess value okay and that final N that you get should satisfy this okay. So, there is something that we have not quite used up and therefore you can use this to check then you go through this loop right. Now it is easier said than done because solving this set of equations here is actually pretty like marriage we are looking at a highly nonlinear equation so you are looking at products of unknowns and powers of unknowns that are non unity so this is not very easy to solve so this is pretty difficult to do to begin with and the more the number of products that you assume the more difficult this equations are going to get the more the number of atom types okay you will now have more atom conservation equations and finally you will have in fact if you have more atom types you will have more products to consider all right the first place. So, the hypothetical formation equilibria equations will flare up to as many products you have assumed and you have to now rewrite them and plug them back to as many atom conservation equations as you have the number of atom types so all these things will become much bigger and so it is very difficult to do this. So, for example for a carbon hydrogen oxygen nitrogen system right we could assume the products as CO CO2 carbon in gaseous form at these temperatures okay H2O2 OH N OH N2 NO H2O etc right and so then you have to so there are some initial guesses that we can also think about so for C NC H NH O NO N NH system right so that means you have a system by which you have NC number of carbon atoms NH number of hydrogen atoms and so on so that means both reactants are included like if you have oxygen there so oxygen could be figuring out in the hydrocarbon or oxygen separately or both okay everything put together. So, this could be like you know if you now consider only the stable products among these look at how many products we have considered like 1 2 3 4 5 6 7 8 9 10 11 12 already yeah did I miss anything H2O2 OH N OH N2 NO H2O yeah of course I knew I was missing this we have a we have to assume carbon solid as well because that is actually the reference element for carbon gas without that you are getting it stuck right so that is about 13 species let me tell you that 13 species is actually not a very high number in reality okay so this is this is this is an easy problem okay so okay we are looking at what could be a good initial guess right so if you now assume that you are stable products or let us say carbon dioxide water and let us say nitrogen and of course if it is a fuel lean situation you should get some oxygen out if it is a fuel rich situation you might want to get some hydrogen out so I just put a reverse stick to indicate that you will have one of these right so if you now do this algebraic balancing you can get something like H2O plus half N N N2 plus half NO minus NC minus 1 4th NH O2 okay if half NO is greater than NC plus 1 4NH so it is kind of like this is the combination of the hydrocarbon for which you need to have so much oxygen in excess for a fuel lean combination to assume O2 as a final stable product right so this is a fuel lean situation for or okay I will just start with this right there and then say this could be NC CO2 plus NO minus 2 NC H2O plus half N N N2 that does not change and plus half NH minus NO minus 2 NC from H2 right this is if half NH is greater than NO minus 2 NC which is supposed to be but NO is much greater than so not greater than or equal to 2 NC right so we want to have this combination and then this should be less than the half NH this is the fuel rich situation so if you do this then you can you can now get your initial guess for the N because all you have to do the fuel lean case is to add NH plus half NH NC plus half NH plus half NN plus this thing and maybe cancel something and you now get the total N as the initial guess okay the number of moles is not the same okay so we are now basically looking at a initial guess for the total number of moles of the products based on the stable products effectively what you are saying is we should expect these things that is sorry the NH NO NO H etc to be in small quantities we expect largely the stable products but the other ones are going to be in small quantities higher the temperature more these will be okay and then they will try to bring down the temperature okay so there is an equilibrium that that gets attained by having more of these and getting the temperature down okay so there is a reason why we are actually starting with an initial guess based on the stable products composition okay so if it is a fuel rich situation you can add up these not to get your initial guess for the total number of N all right now the point is do we actually have to go through this Herculean task it is it is you try to do a computer code on this it would be a fairly good assignment for a course okay so and many times many students whoever not spending enough time on this do not succeed so how in the world that people can find the antibiotic claim temperature realistically when it is going to be a Herculean task to write a code to do these things so the answer is there are codes available okay so most of the time we do not we do not have to sit and do these calculations all the time we use some standard codes that are available so one of the most popular codes standard computer codes are available as open source okay well when I say open source typically bother only about getting the executable version do not worry about the source code itself so the most popular one to the extent I know is the CEA2 that is a current version of what is called the CEA that is that the CEA came out in 1996 the precursor to that was the CEC71 which came up in 1971 and it was published in this the CEC71 was published in this special publication called NASA SP273 so many times people refer to NASA SP273 calculations essentially they are talking about a latest derivative of this which is the CEA2 okay so it is it is actually available in the NASA Lewis Research Center code so you can look for these and download and so on we can I can I can give you more information about this there is also another one you can look at called Stan Jan so on so there are actually many more code than this okay available you can you can look up look up but effectively in fact I want to point out one thing about CEA2 it not only gives you things like the adiabatic flame temperature it also gives you other things like for example transport properties of the product mixture so the if you now have a product mixture it can also calculate of course typically you are expected to obtain molecular weight of the mixture for example the ratio of specific heats for the mixture and but the but what more this can do is to also get you thermal conductivity and viscosity those kinds of information also so it is extremely useful to have this code up and running for most applications that you want to do you get into doing some engine calculations or rocket calculations this is almost like the starting point okay you have to have this with you for you to be able to do anything more so what I will do later on is to put out some sample calculations in the website into which this video will be uploaded so that we can go through numbers on like let us say you now throw in some CP data like polynomial fits or something like that or make an assumption on calorically perfect gas and say constant CP but you need to look at typical numbers for these and so on so again we can do this and in fact I can also show you if you assume only the standard products let us say if you now take methane plus oxygen and give you carbon dioxide and water we can we can go through the calculation and you will find that the temperature is actually shooting up beyond about 3000 Kelvin but if you now add some more and then let us say you from a CEA2 kind of thing you now have a huge list of species that you assume for the products you will find what the temperature is for the same composition of methane and air that you consider and so on well so we can we can I will put up put out those kinds of exercises in the website as we as we go along which you can look at but we have spent a little bit more time than intended on the airbaric flame temperature calculation finally before I close this module I think I should mention something very very basic okay so one is this is the definition of what is called as equivalence ratio we did not quite get an opportunity to talk about this because we first talked about what is called as a stoichiometric ratio but then we were diverted at the time by what is meant by heats of formation of the products okay because the definition of the stoichiometric ratio involved that we have to actually get the most stable products which have the most negative heats of formation standard heats of formation then we were diverted into that and then that was right that is that is quite correct so to have that happen so we have to now get in get the jargon right so we now talk to the industry people they keep talking about something called the equivalence ratio okay so the equivalence ratio is typically given the symbol Phi and this is defined as the fuel fuel air ratio for a given condition so this is given given by fuel air ratio at stoichiometric stoichiometric proportions so in other words you do not really worry about exactly what the stoichiometric fuel air ratio is okay and of course you can also think about what if air is not the oxidizer okay and not all of air is oxidizer only the oxygen part of it is typically the oxidizer the nitrogen is like an inert does not matter you can look at a fuel oxidizer ratio or fuel air ratio okay and you now form a equivalence ratio that is like this which means Phi equals 1 is stoichiometric stoichiometric and Phi less than 1 is fuel fuel lean and Phi greater than 1 is fuel rich okay now the rocket people like to operate fuel rich whereas the breathing engines people like to operate fuel lean okay so the rocket people do not really talk about a fuel air ratio they talk about an oxidizer fuel ratio alright so when you talk about an oxidizer fuel ratio you could write this as oxidizer fuel stoichiometric divided by oxidizer fuel given so you now invert this because you have inverted your numerator and denominator so that this holds you always have this is something that I used to get confused to say as a student okay do I define it this way or do I define it that way or which way it should be it how do I invert and all that stuff do not think about all those things what matters finally is this if your Phi is less than 1 it is fuel lean you have to define your Phi like that okay that is how it works and Phi greater than 1 should be fuel rich this is like the mantra all over the world on this okay so you do not have to think about this at all then you work out all the other details okay then I also want to emphasize these are mass ratios they are not the number of moles or molecular weight nothing again nothing like that it is actually the mass okay and many times in flow situations you could also use mass flow rates m dots so you can say m dot m dot f divided by m dot a divided by m dot f stuck divided by m dot a stoichiometric alright all those things are permissible but it is all mass ratios so with this I should stop this particular module and let me now get into the next topic which is chemical kinetics so let us now define some so keep in mind that we are still in the preliminaries okay we have not got into combustion yet okay so we have to learn I told you we needed to learn some thermodynamics we needed to learn some some chemical kinetics some amount of mass transfer leading up to fixed law and then we will get into some conservation equations derivation that is when we are actually getting into combustion at the moment not yet but we need these preliminaries so some definitions first some definitions first is one a mass concentration of species I okay what is mass concentration what is meant by mass concentration mass concentration what does it mean in what way we are looking at mass right so mass of the species in the total mixture in in what of the total mixture is it the mass of the total mixture sorry that is the next okay so you are looking at mass concentration so what would concentration really mean is in volume so you want to say you now pick a unit volume of your species of your mixture okay and then you now look at what is the mass that is contained of a particular species in the mixture alright so what is this mass per unit volume density okay so this is something that we actually know right so this is this is basically this is given a symbol rho i which is the density okay this is equal to mass of species I in volume of the mixture right I am writing this in a congested manner because we are supposed to know this okay so you should not have to worry about this or maybe the equal to science here and then B because most of chemistry is done in moles okay so we also have now molar concentration concentration of species I which is given a symbol C I C sub C sub I right now this is pretty easy game okay so what do we what do we what do we know for C I would how do you how do you decide to define this not yet so this is this is basically number of moles number of moles of species I per unit volume so number of pi I per unit volume of the mixture okay now I want to actually get the mixture density what do I do do I take an average of all the all the species densities rho I can we think about this not yet yeah we want to know the mixture density so what is the mixture density okay you take a you take a unit volume of the mixture as we speak there is error round which is a mixture okay so I will take a unit volume of the mixture and then find out what does its mass okay but we know that it is actually composed of many species and each of those species has a mass rho I per in that volume because it we were now saying unit volume right so if I want to know the total mass of all the species in that unit volume I simply have to add all the masses of the individual species with the denominator being the same that is the volume okay so if I now split up the denominator for each of those I simply get back the sum of all the densities of individual species as the density of the mixture right so density of the mixture rho right is equal to sigma I equals 1 to n rho I just have to add do not do any averaging or anything like that all right so just add all the densities to get the density of the mixture and similarly molar concentration of the mixture C is just have to add the total number of moles there is there per unit volume right so total number of moles is like number of moles of this species plus number of moles of that species and so on in this volume therefore this is simply going to work out to I equals 1 to n C I that is it and what is the relationship between C I and rho I one of them is number of moles okay per unit volume and the other one is mass by the per the same unit volume so the volume is pretty much the same for both so the only difference between them is on the one hand you are counting the number of moles the other one you are counting the mass so what is the difference between the number of moles and the mass what is the factor that differs sorry the molecular weight yeah so the molecular weight basically tells you what is the mass per one mole right therefore if you have so many number of moles you need to have so much more mass as the molecular weight therefore the relationship is C I equals rho I divided by W I okay now these are things that are kind of confusing to begin with if you are not used to the reason why I am going through these things rather slowly okay so this where we now say W capital W I is the molecular weight of species I right the second definition that we want to have and I will try to use the next panel to a is mass fraction mass fraction and we use a symbol mass fraction of species I species I we use a symbol Y I what do you think is mass fraction now this this this is pretty indicative okay you can guess these things right so mass fraction is mass of species I divided by the total mass of the mixture right so of course you are not going to take like the mass of species I in this entire room and then divide by the total mass of the species all the species in this entire room you could now basically think I will take a unit volume of the mixture look at how much mass I have of species I look at how much mass I have for the mixture in this unit volume so per unit volume basis which should work just as well okay therefore you could we could say this is rho I divided by rho all right now if you now are looking at the contribution of the ith species mass per unit volume to the total mass per unit volume of the mixture then if I add up all of these mass fractions what should I get one good so this implies that sigma I equals 1 to n Y I equal to 1 and this is going to actually come in pretty handy to us later on because pretty soon a coming next week you will have lots of very dirty looking partial differential equations to stare at and one of the one one set of fun and then you will have sets of unknowns it is not just unknowns it is like sets of unknowns okay so we will now look at one equation and say that is actually going from I equals 1 to n therefore there are n equations and there are 3n unknowns and so on okay so let us be looking at n and n is actually not too small okay so and we will talk about this okay so n could be of the order of like let us say 10 or 100 okay let us about let us say 10 to 100 okay to be very modest so you know beginning to look at something somewhere between about 50 equations 500 equations and all those things like just write them so why why I is going to be one set of unknowns right so in all these things sigma I equals 1 to n Y I equal to 1 is going to now come come to us like a big relief because it is a simple nice algebraic equation good old middle school days not even high school days okay so just deal with algebraic equations it is great yeah so this is going to be one of the equations that we will actively consider among this dirty set of equations later on so similarly for B we could now say mole fraction of species I okay so the only thing that we are doing now is we know the trick what we want to do right so the only thing that we are doing now is to introduce the standard notation okay so it is given as Xi and as a matter of fact we introduced X as a most fraction last last class right when we are looking at the how to do how to relate the partial pressures to the total pressure so Xi Xi is what do you guess for this C I divided by C okay no no big prizes for this guess getting there so therefore therefore here again we can say sigma I equals 1 to n capital Xi equal to 1 right one more and then what is the relationship relationship is the Xi equal to Y I divided by W I divided by sigma I equals 1 to n Y I divided by W I alright so this is the relationship between X and Y and this is also pretty important as we will see later on there are these mass transfer equations and chemistry chemical kinetic equations that are based on molar concentrations and mole fractions whereas when we try to do mass conservation we are more interested in mass concentration namely density and mass fractions alright so on the one hand for the fluid dynamic kind of equations conservation equations of mass momentum energy we will be dealing with mass or a mass basis whereas for the chemistry and so on we will be dealing with the mole basis molar basis and there this set of equations that is relating the mole fractions to the mass fractions is going to come in handy okay we need to have these set of equations in fact what will happen is and I tell you this you have to recollect this and I again come back to this we will have to treat both Xi and Y is as unknowns okay so one set of equations that will link the two is this equation this set of equation this is actually valid for every species in the mixture right so we will have this relationship also come up in a big way alright so with these definitions we can now get into some kinetics and right now what I am going to do is to only bring in some notation okay. So we so we now talk about what is called as the law of law of mass action right and the axiom that you are working with is atoms or atoms and molecules cannot react unless they collide okay there are exceptions to this okay there are exceptions like for example you can heat a molecule and then it starts vibrating and dissociating and so on so it does not really collide with any other molecule it can also be photosensitive that means like light radiation or electromagnetic radiation can create these kinds of vibrations and stuff so there are exceptions but at the moment let us not worry about this and we have to start with this axiom now strictly speaking we need to actually get into the details of this axiom as in how the collisions happen and so on only when you are getting down to something like a kinetic theory level or kinetic theory of gases level or even further if you want to get into something like quantum statistical mechanics so if you have to now work out the collisions how they happen and what is the statistics of these collisions and how the energy exchange happens and all those things we would not do any of those things here thankfully right I just try to scare you there okay what we are going to do is to adopt a continuum approach so in a continuum approach what you essentially mean is at a particular point in the flow field you have a nicely defined density concentration mole fraction all these things how you got that is not your problem okay it is there again you can assume that it is there so with this we will actually be looking for laws that are just taken for granted at the continuum level rather than derived from molecular level so the thing that we were basically looking at is let us start with an example example is let us suppose that you now have H2 plus half O2 gives H2O here with this we can write a change in the number of moles of H2 okay and then we divide this by negative 1 to indicate that this change is in the manner of depleting this number and similarly we now write this as change in the number of moles of O2 divided by minus half to indicate that you have a half a mole that goes for every 1 mole of hydrogen and it goes therefore you have negative okay so this negative indicates depletion and the half of the 1 indicates the coefficient and that is equal to 1 for every mole of hydrogen that is produced you have plus 1 mole of hydrogen water that is produced right so this is actually three equations yeah so you can now look at this is one equation this is another equation and these two together is another equation each of these three equations effectively relates what should be the change in the number of moles of hydrogen for a corresponding change in the number of moles of oxygen and so on okay and the signs effectively tell you whether they are getting depleted versus getting produced okay so if you can write this like this then in general we can write dn1 divided by nu i double prime minus nu i single prime equal to dn2 divided by sorry nu 1 nu 2 double prime minus nu nu 2 single prime etc that is like you can write dn i nu i double prime minus nu i single prime where we had this notation or we can say sigma i equals 1 to n nu i single prime mi gives sigma i equals 1 to n nu i double prime mi so this is like a generalization of what we just went through and let us now call this dxi where xi is what is called as a degree of advancement of the reaction let me stop here for the day.