 what we did last time I had given some example values for these heats of formation made a mistake there is not supposed to be a decimal place that is supposed to be a comma that means you are off by about three orders magnitude and what we said is last time okay. So I want we were talking about this equation that is used to find the adiabatic flame temperature and the point I was trying to make there is if you want to have a fairly high adiabatic flame temperature for the for the products which is given by having this ni double prime over here then what is given to us is ni single prime and T1 and maybe the pressure that is given to us which will determine ni double prime in a way that we will see okay but effectively what is happening is if you now have a T1 let us suppose there is significantly low could be 298 Kelvin or maybe 500 Kelvin or 700 Kelvin but it is not supposed to be it is not expected normally to be something like the adiabatic flame temperature itself. So effectively the sensible enthalpy of the fuel is going to be low because the temperature is the initial temperature is low okay if you want to now have a high value for this that means you want to have a high sensible enthalpy for the products that means we want to have a high negative value of the heats of formation standard heats of formation for the products okay. So those are the ones that actually give you stable products you get stable products mainly when the products are having high negative heats of formation yeah and then this total will be such that you get a fairly high sensible enthalpy for the products and therefore the adiabatic flame temperature but if the total itself has to be high then given that the sensible enthalpy of the reactants is low and let us say that one of the reactants is oxygen which is like a reference element therefore the heats of heat of formation standard heat of formation for that is 0 we are looking for this total to be high only by looking at the heat of formation the standard heat of formation of the fuel okay that means we want to have the standard heat of formation of the fuel to be high okay or in other words you look for a substance which has a high standard heat of formation then you can identify that as a fuel that is how we basically look at it right. So if you look at typical values of the heats of formation standard heats of formation for typical fuels let us say I have listed here taking the taking the source from S.R. Turn's book it is let us say we have methane acetylene ethene and ethane and propene and propene right so again keep on going to higher and higher hydrocarbons we see that some of them are negative some of them are positive okay but what we are looking for is essentially that even if they are negative they should not be high negative values okay so it could be somewhere around 0 negative or positive positive is better but we do not want them to be as high negative values as the product standard heats of formation so that we will get some high temperatures there okay so this is how you are essentially looking for identifying the fuel this is one point that I wanted to make the second point I wanted to make today is by actually looking at it this way we are constructing a hypothetical set of reactions that do not exist in reality but it is actually somewhat in our minds for a thermodynamic convenience because thermodynamics we are dealing with enthalpies here we recognize that these heats are all enthalpies and these are state variables right so it is efficient for us to actually go from one state to the other okay so the way we are due doing this in decomposing the enthalpy of the reactants into the standard heat of formation and the sensible enthalpy and we are also decomposing the enthalpy of the products into their standard heats of formation and their sensible enthalpies that is what we have done right so what is essentially going on is you are starting with a set of reactants which are at some temperature and pressure okay so be virtue of being at some temperature they are having a certain sensible enthalpy which is non-zero okay by looking at this kind of thing what we are doing is we are now bringing the reactants from their given initial temperature to the standard condition by which let us say we now take away some heat okay and cool this and we now bring it to the standard state so we now supply the heat of formation the standard heat of formation to these reactants at the standard condition all right and go through reverse formation reactions to make the reactants unwrap their bonds and give rise to reference elements so you now have reference elements at standard state okay and you now have the reaction go happen that is you now have reference elements these reference elements now regroup themselves to form the products at standard state that thereby releasing the standard heat of formation for the products okay and finally we still have in this budget okay we still have some heat that is available to us which will be used to raise the sensible enthalpy of the products from the standard state to wherever it can go so that you have heat used up all the heat that is available okay so most of the heat was actually available to us in this process when you now try to actually form the products you got a lot of heat okay that heat has to simply and then you have supplied some heat to actually unwrap the bonds of the reactants to the standard state so you have to you have negate that okay you got some heat by getting the reactants from their initial temperature down to the standard condition let us suppose that the reactants were heated up to 500k okay that is already some sensible enthalpy available with them which was been which could be taken into account when trying to bring them down hypothetically to the standard condition right keep that heat with you supply some heat for unwrapping the bonds to get the reference elements and then get the reference elements to form the products you get a lot of heat over here little bit of heat over here you have supplied some heat over there all that stuff now will add up to some net heat that is available to you which you now dump on the products to raise the temperature all the way to the airy barric flame temperature whichever temperature they can go to okay. So it is like a circuitous path if the thermodynamically speaking that we are taking for the reactants from their state to their standard state and then you form the standard the reference elements from the reactants and use the reference elements to form the products and then raise the temperature of the products to the to whatever okay so this is the circuitous path that we are having in our minds while we are doing this okay yes this is got to do with the way the bonds are in there okay so you have like some of them are saturated bonds some of them are unsaturated and so on so that is something that we would not get into in this in this class okay this is like that there is a thin line that is separating us from chemistry itself okay so we do not we would not get into that but the point I am trying to make is the numbers okay these are not supposed to be as high negative values okay so it is okay if they are negative values but they should not be very high negative values okay and if they are 0 or positive better okay as high as possible that is algebraically high is what we are looking for and the other point I want to make is so in all these things we have actually said there is like a adiabatic process that is going on that means you are not simply getting any heat at all right does it make sense when I burn stuff I get heat what are we talking about adiabatic so it is not as if I am trying to say that this is an ideal thing and in reality there is some heat loss so it is not adiabatic and all that stuff no in reality you are not very far away from the situation for most applications okay in the core of the combustion process in let us say in a rocket engine or a gas turbine engine or whatever it is you get temperatures that are pretty close to the adiabatic flame temperature within let us say 100 degrees or something all right which is which is not a lot of deviation when you are looking at something like 2500 right so we are not talking about a heat loss as a non adiabatic situation the point we are trying to look for is most of the time when you are looking at the heat that is generated in the combustion right we are primarily looking at the hot products that could be used the heat in the the high temperature of the hot products that could be used okay so when you are now looking at something like a steam turbine where you are actually essentially having a high temperature steam that turns turns the wheels right you are counting on the high temperature of the steam to the thermal energy that is associated with the high temperature of the steam to be imported into the turbine blades similarly for the hot combustion products by virtue of the fact that they have high temperature is what we are trying to utilize so if you just put a kettle on a stove okay what is actually heating the kettle is the hot products it is not the chemical reactions itself the chemical reactions are over many millimeters below below below the kettle okay the flame itself is not really near the kettle at all okay the flame is if the flame were to happen near the kettle the flame will get crunched and it will become non adiabatic the flame is actually on the stove okay on the burner and you get these hot products and the hot products are the ones that are actually flowing around the kettle and heating up the kettle right so the question basically then is what is the heat that you are going to get out of a chemical reaction or in other words in a now looking at these fuels with heats of formation and so on maybe in lots of practical applications as well as in your high school or wherever you have learnt any combustion you would have come across things like heating value for the fuel okay or calorific value to fuel and so on so when I now have a fuel in hand can I can you tell me how much it is going to how much heat it is going to give okay do not tell me all these things like the circuitous route over which it is going through and then heats of formation transaction reference elements and all these things that is that is what too confusing okay is it possible for me to see the calorific value or the heating value of the fuel from this can we can we you can it is all embedded in here right so what we are looking for is if you now did not really bother too much just for the sake of approximate argument okay if you did not bother about the fact that the the reactants and the products had different CP's or molecular weights and so on right so you just simply looking at sensible enthalpies you are simply saying I had a bunch of reactants at a let us say room temperature or initial temperature T1 okay and so that means the sensible enthalpy of that was pretty low and I had to now have this fuel release heat that will increase the sensible enthalpy of the of what were reactants to start with to a very high sensible enthalpy value for the products okay so the heating of the heating value of the fuel is primarily going into changing the sensible enthalpy of what were reactants to what was what will be products so essentially that is that is what is going on so if you now look at then this part minus this part that is a sensible enthalpy change that is being affected okay if you now take that to one side okay and keep this part and this part to the other side then it becomes automatically that the difference in the standard heats of formation weighted by the individual stoichiometric coefficients of the products and the reactants is effectively the heating value okay or in other words the heat that is actually released during chemical reactions in the combustion primarily comes from a difference of the heats of formation weighted by the amounts by which they are this is this is basically what is going on okay and it comes directly from here okay so finally the last point that I want to make here is this essentially says that all of combustion is all about a transaction between heats of formation and sensible enthalpy we started out with substances reactants which were low in sensible enthalpy high in heat of formation okay we now finally get to products which are high in sensible enthalpy and low in heat of formation so essentially it is like what you started out with low sensible enthalpy high heat of formation gets transformed into the other combination okay this is exactly what is going on in a in determining this adiabatic flame temperature okay. So now let us look at the question that I posed towards the end of last class which is where is the pressure showing up here right supposed to be a constant pressure process where is the pressure showing up here is really the question right so the answer to that is sitting in this ni double prime or the pressure is possibly the one that is going to dictate what should be the composition of the products is what we expect the question is how so the next point that we have to look at is composition we will start with a very simple example which is going to be puzzling to us in a minute right so example let us suppose that we consider a reaction where you have CH4 plus O2 gives CO2 plus H2O okay can we figure out first of all how many species are there in this system 4 species the capital N care is 4 right how many well I should not ask the next question now so can we now figure out let us suppose that we had one mole of methane okay and then we will do what is called as balancing the reaction right so we now have one carbon atom here one carbon atom here that is fine and then we have like 4 hydrogen atoms here we gave you have only 2 so we get 2 moles of water there and then we have like 2 oxygen atoms here and then 2 there we have only 2 so we have this is supposed to be a stoichiometric reaction yeah so this is what we this is what we expect now here it is pretty clear the composition of the products is just as clear as the composition of the reactants okay this is this is the stoichiometric stoichiometric right so CH4 plus O2 will be fuel rich because we are starving the system of one mole of oxygen and CH4 plus let us say 3 O2 is fuel lean right and correspondingly we should now be able to put in let us suppose that some excess excess in the case of fuel rich we now suspect that there must be some excess amount of methane that is remaining as products and in the case of fuel lean there is like some excess amount of oxygen that is remaining the products we should be able to find out yeah so for example we now have CH4 plus 3 O2 whereas one mole of CH4 is going to consume only 2 moles of oxygen there will be one mole there is excess okay and if there is one mole there is excess what is going to happen to the adiabatic flame temperature why would the adiabatic flame temperature decrease if you now had a excess amount of reactant that is of stoichiometric condition the answer is you are now actually soaking up so much amount of sensible enthalpy for the excess reactant that is remaining for it to be also taken to the adiabatic flame temperature you see therefore you are now going to reach a very high adiabatic flame temperature it is sort of like you now have this excess reactant that is remaining there which wants to share the pie on what is available for the sensible enthalpy rise okay so you are not going to rise a lot whereas if you did not have any excess okay it is only the these products the stoichiometric products that are remaining then they are the only ones that are going to actually go high up in the in the sensible enthalpy therefore they can actually reach fairly high temperature. So what it means basically is near the stoichiometric condition I am not saying at okay and let me qualify that a little bit near the stoichiometric condition is where we can expect to have the highest adiabatic flame temperature okay of stoichiometric conditions the adiabatic flame temperature drops because you are having excess of one of those reactants that also needs to be heated up to the adiabatic flame temperature therefore you do not get up to that high an adiabatic flame temperature as you would without the excess reactant right okay so why near instead of at it is it has got to do with how you look at the sensible enthalpy recall that we could look at the sensible enthalpy as an integral CPDT or if you are thinking calorifically perfectly perfect gas which is which is simpler it is like simply CP times T adiabatic minus TT difference right. So it now matters on what is the CP of your reactant okay so typically you will have and the CP will depend on the molecular weight okay and typically we expect that the fuels will have lighter molecular weight when compared to oxidizer and therefore a higher CP and so if it can actually hold more heat it is going to decrease the temperature because the CP is now holding the heat and not allowing for the temperature to rise okay so that is going to send and if you are now looking at how the adiabatic flame temperature is going to vary as a function of fuel air ratio okay you are going to actually have a peak that is slightly away from the stoichiometric value because the CP's of fuel and oxidizer and the reactants they will be different so you are you have to work that out okay. The other thing that you can also think about is in this reaction we are only looking at O2 but in reality it could be air which means you have a lot of nitrogen involved right. So you now have nitrogen which does not participate in the reactions that means it shows up on either side of this reaction okay what is the consequence of the presence of nitrogen for the adiabatic flame temperature relative to if you did not have the nitrogen obviously you are going to have much less adiabatic flame temperature because the nitrogen is now sitting there soaking up the heat okay so it is now going to contribute to the sensible enthalpy component as well. So when you now have a total amount of sensible enthalpy that is available for you to increase the temperature the nitrogen also comes in for its share therefore you do not get to a very high high sensible enthalpy keep in mind the nitrogen has a CP which corresponds to a diatomic molecule okay. So instead of nitrogen if you had let us say organ all right which is a monatomic molecule its CP is different because CP typically depends on for gases particularly on the atomic city of the molecule okay in monatomic diatomic polyatomic and so on okay when it is polyatomic whether it is linear or is it nonlinear and so on right. So it depends on whether you have rotational degrees of freedom that needs to be considered all right. So if you now replace the nitrogen by organ then you will now find that the way the diluent say effectively nitrogen as well as organ was acting like a diluent and the diluent is now going to the monatomic nature or the diaromic nature of the diluent is going to change the way that the adiabatic flame temperature comes down and even within a monatomic if you now try to replace organ by helium yeah. So organ has a higher molecular weight helium has a lower molecular weight correspondingly the CP changes right so on a per unit mass basis and then if you now calculate things on a per unit mass basis then you will again see differences. So I would like you to work out those kinds of details like for example in an exam we can directly ask you if I now replace so much amount of nitrogen by so much amount of organ what is going to happen to the adiabatic flame temperature increase or decrease okay so you should be able to figure that out okay. So those are things that we can look at what we were talking about is the composition of the products and as I was pointing out if you have a stoichiometric mixture where you have considered these two as the stable products finally you now know exactly what the composition is if you are now looking at off stoichiometric situation you could still expect one of the reactants to be in excess and find out what it is or how much it is excess right in which case you still know what the composition of the final products are including the excess reactant that was on the product side okay. So where is the problem we could find out the composition of the products we know what the ni double prime is the pressure does not then come into picture at all is there a problem or does not the adiabatic flame temperature does not not depend on the pressure at all right because you might expect because it is based in enthalpies and for ideal gases we showed that enthalpies only a function of temperature it is just a balance of enthalpies okay constant pressure yeah we are balancing enthalpies therefore constant pressure otherwise it does not matter so pressure does not come into picture at all that is not quite true okay in reality you know now we have to look at reality okay in reality these are not the only products that happen these are the products that we would like to keep okay these are the products that we would be expecting as the stable products but imagine when you now have these are the these as the final stable products and keep in mind that these are the final stable products that means they have the highest negative heats of formation which is now going to give rise to an estimate of a very high adiabatic flame temperature that is to say if in case they were the only ones to exist you are now going to have a high enough adiabatic flame temperature that is now bigging that is now going to cause a decomposition of these right they just cannot exist at those temperatures the way they are okay. So H2O could now get decomposed into OH and H for example right and that decomposition is going to soak up some energy and decrease adiabatic flame temperature that is one problem as far as the T adiabatic is concerned but the other big problem that we are talking about is how much H how much OH if I now have to factor in that you need to have some H and OH also thrown in there okay how much H how much OH because I need to know that otherwise I would not know first of all it increases the number of species right total number of species the second thing is what is the ni double prime for those and if you are not going to bring in those this is going to change so pretty much everything changes right. So how do we bring the existence of those kinds of ions or radicals in there okay as decomposition products of the stable stable products what we then assume next okay is that let us suppose that you started out with reactants and they give so that you could go through this fictitious path and all that stuff that we talked about you are now looking at the final state the final state of adiabatic flame temperature is such that these products are in equilibrium with their dissociation products okay so if you now have H2O okay it could be in equilibrium with OH and H right so if CO2 were to decompose it could be in equilibrium with its decomposition products again we do not know exactly what would be the nature of those decomposition reactions right there could be multiple ways by which these decomposition reactions could happen. So the simplest way again is for us to hypothetically assume equilibria that means if you now look at many more species of these unstable many more unstable species to be considered right we now have to actually consider their formation reactions as the equilibrium reactions okay so let us just go through that a little bit more so in reality we just do not have only the most stable products but many unstable products of dissociation right as well okay so in general we could assume say for the H2O2 reaction right H2O2 reaction if you now say H2 plus half O2 we have strictly speaking if I were to look at only this stable most stable product here so it has to be H2O okay so what I would actually have to say is I have only and I would like to say it is only one mole so that means I know A is equal to 1 yeah but in reality we have to now consider BOH plus CO2 plus DH2 plus EO plus FH alright. Now what happens here is because this is stoichiometric okay we were not expecting either O2 or H2 okay if it were off stoichiometric let us say if it is fuel lean right you could expect some O2 if it is fuel rich we could expect some H2 that is what we were thinking before but now what is happening is you could have H2O decompose into OH and H but H could combine to form H2 okay and you could also have O atoms come up okay during this decomposition which can recombine to form O2. So you now have a soup bowl of lots of things that we did not really bargain for alright and all these things are there so how do you handle them okay so effectively we now need to have 1 2 3 4 5 6 unknowns that we need to work out okay so previously how many unknowns that we actually work out by balancing the reactions we worked out about 2 or 3 okay 2 here if one of them were excess we would just be able to handle that as well how did we do that how did we actually end up finding out the composition of the products by what is called as balancing okay what were we doing when we were balancing we were actually looking at the counting at the number of atoms of a particular type okay so what did I do I said I have 1 carbon atom here I have 1 carbon atom here through 4 carbon hydrogen atoms here but only 2 therefore I put a 2 there and so on right so I was essentially looking at a particular atom type and I was balancing the number of atoms this is the hallmark of chemical reactions because chemical reactions are all about exchanges only at the electron level and not at the nuclei level okay and what we are looking for is a mass balance in a chemical reaction masses conserve it is the nuclei mass that we are looking at okay so we are actually mentally writing what is called as atom conservation equations for each atom okay and you can only write as many atom conservation equations as the number of atom types that are there in the system okay so in the case of so to give you an example so in the case of CH4 plus O2 how many atom types are there 3 okay we have carbon hydrogen oxygen so you know we can call this a C-H-O system okay CHO system okay in the case of the H2O2 system how many atom types are there only 2 okay so if you now were to write atom conservation equations for these okay in an attempt to balance the reaction we would we would get 2 equations but we still are actually having 6 unknowns that means there are 4 more unknowns for which we need to develop equations right so let us first do what we can which is atom conservation atom conservation equations which is on the left hand side for 4 H atoms we can write 2 that is given to us okay so 2 is equal to we have 2A plus B plus 2D plus F basically looking at wherever hydrogen shows up and for O atoms we have 1 equal to A plus B plus 2C plus E I want to note here that we could have also considered O3 ozone yeah H2O2 and there is this notorious unstable species called HO2 okay but they are too unstable okay if we were to consider them that means we will have 9 species okay and then we are still stuck with only 2 atom conservation equations that means we now have to develop 7 more equations right and what we expect to find is these will be in very trace quantities that means if you now had after H if you had GH and I for these these GH and I values will be very small is what we are expecting okay therefore we are we are neglecting this so what is happening now is the final answer for the adiabatic flame temperature actually depends on how many products you are willing to consider if you were to consider only 2 products you will have a fairly high estimate of the adiabatic okay if you want to have realistic estimates that means you have to consider more products how many more depends on how cumbersome you want to get yourself into right so we want to be reasonable that means we neglect some most unstable species but we have to consider some less unstable species also in the picture okay so and then we have to now look at equations for these so what did I say we now consider hypothetical formation equilibria for the other ones right so effectively what basically happens is you now look at OH O and H for example and H2 O okay these are ones that are not reference elements H2 and O2 are reference elements so in your mind you should be essentially like looking at these two equations actually serving the cost of O2 and H2 which are the reference elements for the two atoms atom types that we are looking at that means for every reference element that is there as part of the products okay it corresponds to a certain atom and the atom conservation equation for it is essentially tacked to its coefficient is like mental note it is all of them are coupled together it is like it is difficult for you to say this equation belongs to that and so on but for us to number or what do you call enumerate the number of equations that we have effectively any product that is not a reference element is what we will be writing the hypothetical formation equilibria based on the reference elements okay so then you will get the additional equations so that is what we are going to do now which is equilibrium formation equilibria reactions right so we now say H2 plus half O2 gives and takes H2 O half O2 gives and takes O half H2 gives and takes H and half O2 plus half H2 gives and takes OH so obviously when you write formation equilibria for H2 and O2 it will simply be H2 gives and takes H2 so there is that is that is no news okay or there is that is no news and therefore there is no use okay so here there is some use to actually considering hypothetical formation equilibria for non-reference element products okay that is exactly what you are trying to do here so with this what can we write okay we are assuming equilibria therefore we can write equilibrium reactions equilibrium equations okay so this means for these we can write KPF at T adiabatic we do not know what is T adiabatic but we suppose that at this adiabatic flame temperature is where the equilibrium is existing between the hydrogen oxygen and water high oxygen and its atomic form hydrogen atomic form and oxygen hydrogen and OH right so we have to write these at T adiabatic and then try to find out what it is so pH to O divided by pH to P O2 to the half KP, KPF T adiabatic we have to have the subscript here for let us say O then it is P O divided by P O2 to the half KPH KPF H T adiabatic equals pH divided by pH to the half and KP, KPF H to H I am sorry OH equal to P OH divided by P O2 to the half pH to the half right now we did not bargain for the pressure to come in the partial pressure these are partial pressures right we did not bargain for the partial pressures to come in there so we wanted to now we want to now change these partial pressures to the actual pressure the total pressure when I say total pressure it is not like the stagnation pressure in gas dynamics this is the total sum of all the partial pressures this is the thermodynamic pressure so now we are beginning to talk about the effective pressure you see the pressure at which this eclipse this the entire reaction from the reactants on to the products is happening is basically going to influence the proportion of the stable products versus the unstable ones okay so depending upon the pressure and depending upon the KP equations the proportions of these are going to get altered okay therefore the pressure is going to influence the the amount the existence and the amount of the unstable reactants that will soak up some heat and alter the adiabatic flame temperature of the products you see that is how the pressure is essentially coming to picture so here we can write pk equals xk times p where xk equal to nk double prime divided by n where n equal to sigma over all k nk double prime that means you add up all the number of moles of all the products together which we do not know yet okay this is what you are trying to find out you are trying to find out the nk double prime yeah but whatever it is it is going to have a total and the total will be n and then if you now take the value of n the kth species double prime divided by the total you are going to xk and xk times p so the p is the pressure at which the reaction is happening right there is a pressure at which the reaction is happening then we are beginning to get the p in the picture and therefore you can now plug this in here and all these things see effectively you are now going to get well in our case or for example if you now assigned numbers for k going from 1 to n let us suppose that one is a one corresponds to hydrogen 2 corresponds to oxygen 3 corresponds to water and so on then a would be n3 double prime okay so in this thing we would now plug in n3 double prime n4 double prime and so on okay so effectively everything is going to be in terms of some nk double prime all these things you can now rearrange them okay so how do you do this so in general in general we could have any stoichiometry right given by given by let us say half n h h2 plus half n o o2 gives we now have to write n h2 o h2 o plus n h2 h2 plus n o2 o2 plus n h h plus n o o plus n o h o h okay so I just do not want to use abc etc anymore because I might just cross 26 okay there is that is so many alphabet so many letters of the English alphabet so I will just keep n subscript whatever it is and then we get 2 n h2 o plus n o h plus 2 n h2 plus n h equals capital n o capital n o essentially refers to the number of oxygen atoms in the entire system okay we took a half n o because we have a oxygen molecule okay and n h2 o plus n o h plus 2 n o2 plus n o equals n h which is the I am sorry I made a mistake this is n h over there and n o over here these are the atom conservation equations okay n equals n h2 o plus n h2 plus n o2 plus n h plus n o plus n o h okay and kp let us say kp1 equal to n o divided by n o2 to the half p divided by n to the half you now plug in these mole fraction information okay in terms of pressure and n so strictly speaking we actually we can get rid of this and then say this is h2 o and no no I am sorry this is this is kp o or specifically we should say kp f o this is kp f h equal to n h divided by n h2 to the half p divided by n to the half and kp o kp f o h equal to n o h divided by n o2 to the half n h2 to the half and the p gets cancelled here so we do it does not show up and then kp f h2 o equal to n h2 o divided by n o2 to the half n h2 and here pressure shows up with a negative exponent minus half alright from this we will we will now do this rewriting in the next class so effectively what we want to do is write each of these n o n h n o h and n h2 o in terms of n o n o2 n h2 because the denominators are all in terms of n o2 and n h2 okay and then we can plug that back in here and you will now get two equations and two unknowns okay in n h2 and n o2 okay but those equations are going to look very ugly and then you have to solve that iteratively and then get the compositions once you get the composition you can go back and solve for the ad barric plane temperature using this equation okay so we will continue from here tomorrow.