 So, just to reiterate what we had was a polynomial fit for C p and what you have been given are constants a 1, a 2, a 3, a 4, a 5 and in two ranges. So, this is just a pure polynomial you will get C p. Integrating C p d t you will get h you will need a constant of integration which depends on the enthalpy of formation at standard state which is 25 degrees. So, you will get a 1, a 2, a 3, a 4, a 5 and a 6 which comes from the constant of integration and again this is for h. Similarly, for s you will have a constant of integration a 7 which will tell you what the entropy is at 25 degrees C. So, if I put T is 298 Kelvin I should get my value for the entropy at 29, 25 degrees Celsius. So, we have this whole set of values for us and these are well tabulated values and we will use them very often for all our calculations. So, this is something that is a useful set of information to have. So, once this is done all I do is that let us say that I want the heat of a reaction I will say let us say H 4 plus twice O 2 C O 2 plus twice H 2 O. Then I find H F 0 for this at 25 degrees C this I do not need to look up because at 25 degrees C this is 0 this is assigned 0 I will get H F 0 at 25 degrees C I will get H F 0 at 25 degrees C I will just subtract this and this and I will get delta H R. So, this is how I get the heat of reaction. Now, once you have a fuel there is something which is called as a calorific value and what is the calorific value it is just going to be if you see it is just going to be the negative of the heat of reaction and expressed as per kg or per mole. So, for example, if I use this reaction and I get delta H R the delta H R would be for one mole for one mole of CH 4 burning this is the amount of energy released delta H R is going to be a negative quantity because it is an exothermic reaction, but we will say that the calorific value is the negative of that. So, we want to express the calorific value as a positive quantity. So, we will say the negative value of delta H R. So, minus delta H R is the C V for one mole of CH 4. Now, you can say that one mole of CH 4 is really 12 grams plus 4 grams this is carbon this is hydrogen. So, it is totally 16 grams. So, this much energy was released for 16 grams how much energy would be released for 1 kilogram and I can get the calorific value on a per kg basis. Now, it happens that if I have gaseous species species normally C V is expressed in molar or volume basis. Whereas, if it is a solid fuel or even a liquid fuel. So, you express C V on a per kg basis. So, for example, we say what is C V for 1 kg of carbon or something like this. So, we will just come to that and what I can write now is that if I use 1 kg of carbon and I convert it to CO 2 using 1 let us say 1 kg of carbon I should say that 12 kg of carbon really corresponding to 1 mole of 1 kilo mole of carbon sorry. So, 12 kg of carbon corresponds to 1 kilo mole of carbon 12 grams corresponds to 1 mole of carbon, but you realize that if I wrote it as a chemical reaction it would have been C plus O 2 giving CO 2. So, I would have written 1 mole of carbon plus 1 mole of O 2 gives 1 mole of CO 2, but if I write it in kilogram 1 mole sorry let me write it as all in kilo mole. So, that I can work in kilograms because I would rather work in kilograms here if I write it as 1 kilo mole this would have been 12 kilogram of carbon this would be 32 kilogram of oxygen and this would be 44 kilogram of CO 2. So, for 12 kilogram of carbon I require 32 kilogram of oxygen and this is to ensure complete combustion. So, what do we understand by complete combustion? So, if I have any species which is made up of a hydrocarbon C N H M then all C is converted to CO 2. So, that means if there is C N I should form N times CO 2 and all H should form H 2 O since there is M H in the fuel there will be M by 2 H 2 O in the product. So, once I have complete combustion only then I must have calculate the delta H R and that is called as a negative of this is the calorific value. So, we do not assign a calorific value if carbon is going to CO or something like that calorific value is defined if there is complete combustion. So, if I have one as getting back to my reaction if I have 1 kilogram of C then I will require. So, if you look here for 12 kilogram of carbon I require 32 kilogram of oxygen. So, for 1 kilogram I will require 32 by 12, 32 by 12 kilogram of oxygen and I will form 44 by 12 kilogram of CO 2. So, this is writing it in kilogram basis of course, you will see that it is all balanced here there is no problem all the mass is balanced on either side. I take 1 kilogram 32 by 12 kilogram if I add it up I get 44 by 12 kilogram. So, things are balanced because I use the balanced reaction and I just substituted the molecular weight for every species. So, that ensures that the mass is still balanced. Now, if I do this reaction it turns out that the energy that you release is 32.78 mega joules and hence the calorific value. So, this is delta H R would be minus 32.78 mega joules for 1 kilogram of carbon. So, the calorific value for pure carbon would be 32.78 mega joules per kg. So, you realize that you know this is a pretty high number and this is something that is you probably should remember as an order of magnitude I mean roughly somewhere in the thirties you will get the calorific value for carbon and you realize that this is expressed on a per kilogram basis. Now, what I will do is I will still use the same thing for hydrogen though it is a gas I will get it in terms of per kilogram basis. So, if I have 1 kilogram of H 2 again if I write the balanced reaction you see that H 2 plus sorry plus half O 2 gives me H 2 O. This means if I work in kilo moles this is 2 kilograms since this is half O 2 1 mole of O 2 would have been that 1 kilo mole of O 2 would have been 32 kilograms this is 16 kilograms and this is 18 kilograms. So, 1 kg of H 2 would have required 8 kgs of O 2 and they would have given in 9 kg of H 2 O. But if I use 1 kilogram of H 2 O and you know add 8 kilograms of O 2 and get 18 kilogram of H sorry not 18 9 kilograms of H 2 O the energy released is we normally express it in terms of 2 values it is either 141.96 mega joules or it is 120.0 mega joules and the 2 values are the values depending on when H 2 O is either in the liquid state or in the gaseous state. So, you realize that if it is in the liquid state then it would have given out even the H F G that is this so called latent heat also would have been released whereas in gaseous form that latent heat would still be there with H 2 O. So, if I go to the liquid form I will release far more energy and this is what is released if H 2 O is in liquid form and this is what is released if H 2 O is in the gaseous form. So, this is called as the higher heating value or the higher calorific value and this is what is called as a lower calorific value for hydrogen. Now, if I look on a per kilogram basis you let us see this value 141.96 mega joules are released for 1 kilogram of H 2 whereas if you look for carbon it was only 32.78 mega joules for 1 kilogram of carbon. So, you realize that for hydrogen the amount of energy released per kilogram basis is very very high more than 4 times that for carbon. Not only that in this case the by product was CO 2 here the by product is H 2 O which is considered as benign and this is one of the reasons that people always keep talking of this hydrogen economy. They say that if you have hydrogen as a fuel not only are we going to release lot of energy, but you know the pollution will be less, but you realize that you know this is going to be feasible only if we had hydrogen at our disposal. We do not have hydrogen at our disposal. We need to create hydrogen from water for example in which case we will actually provide this energy to water to get hydrogen. So, that means that you know there is no feasibility for using hydrogen as a fuel overall. What we can use it is we can use generate hydrogen at places where there is lot of energy and specifically at places where we require very high energy output using very small amount of fuel we can use hydrogen, but the overall economy just cannot run on hydrogen because we do not have hydrogen supply with us. So, but you should still see that hydrogen releases a lot of energy compared to carbon and this is something that you must keep in mind. Now, you will realize that most hydrocarbons are formed of sorry most hydrocarbons are formed of carbon and hydrogen. So, they will have a C v which is higher than 32 and power lower than you know 140. In fact, for most of the fuels that we look at that is gasoline or diesel you will see that the range is anywhere between 42 to 48 mega joule per kg. So, this is something that you will notice often that this is the range in which most of the calorific values will lie for the fuels that are of interest to us. So, this is again some kind of figure that people should be at least aware of that this is what is happening. So, once this is done we will come to other concepts which is stoichiometric amount of air and this is something which most students are aware of just something which needs to be reiterated. So, these are other concepts in combustion and used very often in IC engines etcetera. So, if I have a fuel C n H m then the for 1 mole of this fuel I will have 12 multiplied by n plus 1 multiplied by m or 12 n plus m sorry this is 1 kilo mole let us say I have 12 n plus m kg of the fuel I will require for each n I will require 1 O 2. So, I will require n O 2. Now, if I have m H then I will form m by 2 H 2 O which means I will require m by 4 oxygen. So, I will require m by 4 oxygen which means by default I will require m by 4 multiplied by 3.76 nitrogen. Now, if I want to find out what is the mass of this you will realize that I will have n plus m by 4 multiplied by 32 plus n plus m by 4 multiplied by 3.76 multiplied by 28 which is the molecular weight of nitrogen so many kilograms of air. So, this is assuming that I have formed n CO 2 and m by 2 H 2 O. So, you realize that 1 kilo mole of c n H m fuel means 12 n plus m kg of the fuel and if I achieve complete combustion which means I transform all my carbon into CO 2 which means I form n CO 2 and I form m by 2 H 2 O I will require n plus m by 4 kilo moles of oxygen which means n plus m by 4 multiplied by 32 kilograms of oxygen and correspondingly those many moles multiplied by 3.76 moles for nitrogen and I will just multiplied by 28 to get the amount of nitrogen. So, thus this is the mass of air required and this is the mass of fuel. So, what we have is what is standardly called as air to fuel stoichiometric ratio which is basically mass of air upon mass of fuel for stoichiometric combustion and stoichiometric combustion is nothing but when the reaction is exactly balanced and you form complete combustion that is you form CO 2 and H 2 O and not incomplete where you form CO or something like this. So, this is what is called as stoichiometric air fuel ratio and you will realize that that would be nothing but n plus m by 4 32 plus 3.76 multiplied by 28 upon 12 n plus m. So, this would be my stoichiometric air fuel ratio if I am using a hydro carbon fuel you will realize that if we consider most of our fuels that is liquid or gaseous fuels in many cases they will be alkenes I mean it is not that they are purely alkenes they are a mixture of alkenes, alkenes, alkenes etcetera. But if I consider them as alkenes you know the regular formula for alkenes is C n H 2 n plus 2, but when n increases to a large value this 2 will be a very very small number for example, C 20 H 42. So, you know tending if n starts going larger and larger the 2 will pale in significance to this n and roughly you know you can say that these formula are of the type C n H 2 n that is the alkene formula roughly this is only for calculation purposes. So, you will realize that irrespective of the number of carbon and hydrogen if the formula is roughly C n H 2 n then for every n a fixed amount of oxygen is required for every 2 H a fixed amount of oxygen is required. So, irrespective of the weight of the alkene I will require a fixed amount of mass for a fixed amount of mass of the fuel. So, the air fuel ratio here if I look at this this is n this is 2 n by 4 this is fixed this is if I write only in terms of n I will just write it as n plus n by 2 times 32 plus 3.76 multiplied by 28 and 12 n plus m is 2 n. So, 14 n I will just n I will take out and you will get roughly 1.5 multiplied by 32 plus 3.78. So, you will see that air fuel ratio for most alkanes is nearly constant and this will tend to roughly 14.7 you can check it out only for the lower alkane. So, for example, if I write C H 4 you realize that immediately H is actually 4 times the carbon if you go to C 2 H 6 by 2 which is 3 times. So, for the lower alkanes the air fuel ratio will be higher and you will realize it should be the highest for methane and it is around 17.16. So, if I plot air fuel ratio versus n I will see 17.16 and it will go to 14.7 for most of the higher alkane and you will realize that this is if you take any fuel like petrol and diesel it will be slightly greater than 14.7 it is n is still not that high, but it is tending to a higher value and you will see may be roughly around 15 or something as the air fuel ratio for petrol and diesel this is the stoichiometric air fuel ratio. Now, in most combustion processes whether in power plants or IC engines you do not provide stoichiometric amount of air in fact a lot of times you provide more air than required. So, for doing this you have another factor which is called as the equivalence ratio and you call it as phi. So, phi is defined either as mass of fuel upon mass of air actual upon mass of fuel upon mass of air stoichiometric or you can write it in terms of air fuel ratio a by f stoichiometric upon a by f actual. So, for example, if I consider the same amount of mass of air if my mass of fuel is more than the stoichiometric mass of fuel you will realize that phi will be greater than 1, but this means that I have provided more fuel than required which means the air is less than required for that fuel and since it is more fuel than required such a mixture is called as a rich mixture. So, basically it is rich in fuel. Now, mostly combustion is rarely done with rich mixture though you will see that in the beginning when you want to start the car usually some amount of rich mixture is required. So, only at that point you will slightly run with richer mixture and phi will be greater than 1, but in most diesel vehicles or in power plants you will not run with rich mixture. In fact your phi will be much less than 1 that means you would have provided far more air than required and such mixtures are called as lean mixture. So, these are some things that you must remember there is something else which is called as the excess air factor which is the same thing again you know told in a different manner. So, excess air factor is just air by fuel actual minus air by fuel stoichiometric upon air by fuel stoichiometric multiplied by 100. So, it is expressed as a percentage. So, if the air by fuel actual is the same as air by fuel stoichiometric this is 0. So, excess air factor is 0 which means you have provided exactly the amount of air that is required. If you provide let us say twice the amount of air then this and this would be exactly this will be twice of this. So, I would get air by fuel stoichiometric upon air by fuel stoichiometric into 100. So, it is 100. So, excess air factor of 100 means you have provided exactly twice the amount of fuel. Now, if you provide less than stoichiometric obviously, a by f actually would be lesser than a by f stoichiometric and you would have provided a lesser amount of air you will see that the numerator would be negative. So, whenever the excess air factor is negative it means you are provided less air than required and it is a rich mixture that is because it is rich in fuel or it is you know less of air required. So, these are some things. So, negative values here imply rich mixture. So, these are just some kind of terminologies or you know regular jargon which is associated with this place or with this field. What you must really remember is that what is stoichiometric air fuel ratio? Roughly what are the values for regular hydrocarbon and that you mostly will provide slightly more air than required and this is something that is to be kept in mind. Now, this is to do most of the discussion here we did with fuels which we could write as C n H 2 m. Now, what happens in solid fuels is that if I am using let us say coal. In this case usually we will talk of the composition of coal in mass basis. So, composition is given normally in mass basis and whenever composition is given in mass basis it is called a gravimetric basis. So, when I say gravimetric composition it means that you have given something in a mass basis. So, for example, I can say that some sample of coal of coal has 60 percent carbon, 4 percent carbon and hydrogen, 2 percent nitrogen, 8 percent oxygen remaining. So, coal can have various compounds in built it can even have nitrogen and oxygen. So, the normal thing that we say is that nitrogen and ash will not participate in combustion whereas, any oxygen in fuel will be used up first. So, what does this mean? This means that if I want to do an analysis typically what I say is that you have this amount of coal I want to find out air fuel ratio stoichiometric what should I do now. So, you say take 100 grams of that coal which means I will have 60 gram sorry 100 kilograms of coal which means I will have 60 kilograms of carbon. Now, we have already done this analysis before if you have 60 kilograms of carbon it means I have 60 by 12 kilo mole of carbon which means I will require 60 by 12 kilo mole of oxygen because for 1 kilo mole of carbon I require 1 kilo mole of oxygen which means I will require 60 by 12 multiplied by 32 kilogram of oxygen. So, similarly for 4 kilogram H 2 I will require 4 by 4 multiplied by 32 kilogram of or it means I will require 32 kilogram H 2. So, you realize that the net kilograms of oxygen would be 60 by 12 multiplied by 32 plus 32 which means 60 by 12 plus 1 which is 72 by 12 multiplied by 32 kilogram of oxygen. And you realize that 1 kilo mole oxygen corresponds to 3.76 kilo mole N 2 or 32 kilogram oxygen corresponds to 3.76 multiplied by 28 kilogram N 2. So, here you had 72 by 12 which is 6 times 32 kilograms of oxygen would mean 6 times 3.76 multiplied by 28 kilograms of N 2. But before this I realize that there was some oxygen in the fuel already. So, I should not be doing this step rather I should first subtract available oxygen which means I will say external oxygen required is actually 6 multiplied by 32 minus whatever is already there which is 8 kilograms of oxygen. So, these many kilograms of oxygen is required externally. So, then I will substitute that value here and accordingly calculate my N 2. So, the N 2 will not be this rather I will first calculate how many kilograms of oxygen are required from outside and hence I will calculate my N 2 and the external air required is based on this value is based on this value. After I do this I will just then I would have said for 100 kilogram of fuel I will require some whatever I get from here that is this is 6 let me just do it 6 2's are 12 6 3's are 18 192 kilogram minus 8 would be 184 kilogram of oxygen which means 184 multiplied by 3.76 multiplied by 28 upon 32 kilogram of N 2. So, I will add up these two numbers to get my external air and I will divide it by 100. So, let me say that if I add up these two numbers which is 184 plus 184 multiplied by 3.76 kilogram of air and then air to fuel ratio would be just this let me call as x. So, it will be just x by 100 and this is my stoichiometric air fuel ratio because I have converted all my carbon to CO 2 and H 2 to H 2 O and this is what I will get. So, this is what is called as then if I provide excess air the air will be more than this. So, this is the way to proceed now you realize that you know all of this would have given you some amount of exhaust gases made up of CO 2 and H 2 O and if you have incomplete combustion you would get CO. In fact, even if you provide good amount of air if complete combustion does not occur you will get CO and even O 2 in the products and normally what is done to test out incomplete combustion is what is called as an exhaust gas analysis and you will realize that you know I can always write equation C H 4 plus oxygen. Let us say this is A C H 4 plus B oxygen plus B multiplied by 3.76 nitrogen will give me A C H 4 plus B oxygen something like x CO 2 plus y CO plus z O 2 plus B multiplied by 3.76 N 2. So, I have undergone some kind of reaction I this must be balanced that means the amount of carbon here x plus y should be equal to A this is carbon balance the amount of oxygen which means x plus sorry this is y CO y CO. So, it is y by 2 plus z should be equal to B sorry there should be let us say u times H 2 O here. So, this is not only missed out. So, x plus y these are the carbon atoms on the right side they should be equal to the carbon atoms on the left side. Now, if I consider hydrogen you will see that there are 4 A hydrogens on the left side and I should have 2 u these are the amount of hydrogen atoms on the right side they should be equal to 4 A and then if I calculate the oxygen on the left side it is B, but on the right side I have x plus y by 2 plus u by 4 plus z which is unburned oxygen. So, the net oxygen if I add up in all these 4 these are the 4 things that I have put in they should add up to B and this is would be a balanced reaction. Now, what I should know is how much of each of these is there on the right side this gives me an idea of what kind of combustion is occurring and whether incomplete combustion is occurring whether good mixing is not taking place. So, one of the standard apparatus is that we normally talk of is what is called as the ORSAT apparatus. So, this is nothing, but a simple 3 bottle apparatus to analyze exhaust gases. So, bottle 1 is normally made up of NaOH and this absorbs CO2. Bottle 2 is usually made up of something called as pyrogallic acid in NaOH, AOH so solution and this absorbs CO and bottle 3 is made up of cuprous chloride in a solution of acid HCl and NH3 and this absorbs O2. So, what you do is you pass or you collect the exhaust gas above water which means the moment you collect it above water any water in the exhaust gases will dissolve in the water and only the gases will come out and you see that a typical in this reaction you would have had CO2, COO2, N2 and water. The water would get absorbed in the water that you are using to collect the gas and you remain with 4 gases which O2 and N2 mostly assuming the fuel is hydrocarbon and you have used only air. So, and again the order is important here because bottle 1 apparently can also absorb COO. So, you need to ensure that you pass once you collect the exhaust gas you pass it through these 3 bottles. So, the first bottle will absorb CO2, the second bottle will absorb COO, the third bottle will absorb O2 and you can figure out how much the absorption is and then whatever remains will be only N2. So, once we have this that means on the right side we will have information for x, y, z and this. So, all this will be known to us these numbers once we calculate this will be unknown because it has just mixed with water, but using these equations there are 3 unknowns A, B and U and we can easily find whatever we want what has happened or what kind of reaction we have occurred. So, this is something that we can find out whether we use more air than required whether we use less air than required whether in spite of using more air than required we have had incomplete combustion. Such information is then obtained using this exhaust gas analysis and a few of the problems that we have in the exercise sheet are related to this. So, this is as far as exhaust gas analysis is concerned. So, the last topic that we normally would cover in combustion is what is called as adiabatic flame temperature. So, let me just write that adiabatic flame temperature. So, what is this adiabatic flame temperature? So, all this time we were talking of the enthalpy of a reaction where all the species were brought back to the standard state and then we calculated the heat of reaction or the enthalpy of reaction, but let us say we do not do that and let the energy release increase the temperature of the species. So, what this means is that if I consider a control volume I have put in some fuel I have put in some air and this is completely adiabatic that is no q transfer here and here are 0. So, you can imagine let us say a regular boiler of a power plant or a combustion chamber in a gas turbine engine or a combustion chamber in an aeroplane. So, these are well insulated combustion chambers I mean and we will assume that such chambers are adiabatic and if I put in a certain amount of fuel and air the exhaust gases at that time will have a certain temperature and since this entire thing was done adiabatically whatever temperature the exhaust gases reach is called as the adiabatic flame temperature and what will the temperature be now since the entire thing is adiabatic you will realize that the enthalpy coming in should be the enthalpy going out because there has been no net q in and out and anyway work was assumed to be 0 net work in and out is assumed to be 0 and things go in and come out at the same pressure. So, what do we do then the simple way to look at this is that I just plot my H versus T here. So, I told you that for any species you know I could have just plotted a line like this of how H varies. So, this is 298 Kelvin and this is how H will vary as a function of temperature. Now, let us say I add up. So, let us say I have a set of reactants which is C H 4 O 2 and N 2. Now, if the reactants begin at room temperature then I know the enthalpy of reactants which means I add up the enthalpy of C H 4 O 2 and N 2. N 2 and O 2 at 298 Kelvin would be 0 and C H 4 would have some value you will see that it is negative. So, I will be here if I heated the air even oxygen and nitrogen would have some enthalpy and at some other temperature let us say in many combustion chambers the air is preheated in which case the enthalpy of the reactants would be different. So, I would have to calculate the enthalpy of C H 4 O 2 N 2 at the temperature in which they have entered the combustion chamber. So, for example, if the temperature of the combustion chamber is 200 degree Celsius which means it is 273 plus 200 or 473 Kelvin then I will have to calculate the enthalpy of C H 4 O 2 N 4 at 473 Kelvin, O 2 at 473 Kelvin, N 2 at 473 Kelvin and what I will do is I will just use the tables that have been given to you the curve fit values. So, for example, curve fit values have been given to you for most fuels as well as CO CO 2 H 2 H 2 O N 2. So, what you will see is that I just have to put the values of or the value of 473 as the temperature and I will get the enthalpy for all these species. So, I will get the enthalpy of the reactants. So, let us say this is the enthalpy of the reactants. Let me say now we are at 298 Kelvin. Now, if I look at the enthalpy of the products it better be lower than the enthalpy of the reactants at the standard state because the enthalpy of reaction has to be negative. So, the enthalpy of the products at 298 Kelvin is lower than the enthalpy of reactants at 298 Kelvin. So, this is where you are delta H R which is negative. So, this is the product. So, at 298 Kelvin the enthalpy of products will be lower than the enthalpy of reactants. So, let us say I have CO 2 H 2 O and N 2 at 298 Kelvin N 2 would anyway be 0. If I add up these two and I put the value here it will be here. Now, obviously as temperature changes the enthalpy of these species also increase and I will get some curve like this. Now, of course, I have drawn a certain curve smooth curve. The curves need not be smooth depending on how CT varies the curves will be in some wavy fashion, but you realize that I have plotted the enthalpy like this. This is the enthalpy of reaction at the standard state. Now, if I have the whole control volume as adiabatic then the enthalpy of product is the same as the enthalpy of the reactants which means that I will draw a horizontal line passing through the enthalpy of the reactants and see where it will cut the enthalpy line for the products. So, if I have some kind of an adaptive process then along this line the enthalpy is constant because this is a constant enthalpy line. I have made it a horizontal line here and what is happening is that whatever is the enthalpy of the reactant that is the same as the enthalpy of the products along this line and wherever it cuts here I see what temperature this is and this is the temperature that the species that the product species will reach if the situation is entirely adiabatic. This temperature is then called as the adiabatic flame temperature. So, this is as straight forward as it is if I just draw these two curves you realize that what is the adiabatic flame temperature. So, H products is the same as H reactants. For example, to find the H of the reactant is very straight forward because the temperature of the reactant is given to you. So, this is easy, but to find this temperature is not easy. All you know is that step one calculate H of reactants which is straight forward, step two find T such that. So, you have to ensure that the product H is the same as reactant H. Now, you realize that the product H can be or the temperature can be obtained. If I just put in some temperature in the curfit values I will get some product H. I will have to keep on varying and do trial and error to see at which temperature I will get the enthalpy of products to be the same as the enthalpy of reactant. So, you will realize that this can be a tedious trial and error process. Actually, we can simplify this process slightly by assuming a beginning value and carry out the iterations in two or three steps. We will do it in the afternoon session when we will do one problem, but just for the sake of knowledge what has been given to you is that if you look at the table for a certain fuels you will realize that some kind of rough value of the adiabatic flame temperatures have been given. T adiabatic for methane, acetylene, ethane, ethane, etcetera have been given and these adiabatic temperatures you realize are somewhere around 2200 Kelvin. They may be plus minus some value and one thing you must realize is that if I have any species let us say I have methane, CH 4 plus twice O 2 plus 7.52 N 2 and I think it should give you CO 2 plus twice H 2 O plus 7.52 N 2. If I assume this is the reaction at 298 Kelvin I know this is H is 0, this is H is 0. So, H reactant would only be the temperature of this. Now, I need to guess the temperature for this entire set such that if I add up the three H's I should get the same H here. Now, it turns out that at such high temperatures N 2 will react with oxygen form NO, NO 2, etcetera and whenever such species are formed they will eat up some energy and usually your adiabatic flame temperature that you calculate may not match what is given in the table and that is because in the table they have assumed certain other reactions to form. Whereas, we will keep it very simple we will say that N 2 will never take part in reaction and we will just say that all our reactions are such that we will not bother about any formation of N 2 O, NO, etcetera we will keep it very simple this much that fuel plus oxygen and nitrogen gives you CO H 2 O and N 2 and you just have to guess the value of T such that if you add up the three enthalpy here they should give you the enthalpy of the reactant. So, that is something that we will just do. So, we will keep it very simple the values that you will get will be only slightly off from the T adiabatic that is given in this table that you have been provided. So, this is the enthalpy of reaction for what is you know I have just depicted it graphically and something else that you must know is what is called as the constant volume T adiabatic. So, for example, in an IC engine let us say at a certain point the volume does not change for example, in a petrol engine you know it is considered as instantaneous combustion where before the volume changes the entire fuel is supposed to have combusted in a petrol engine. So, you realize at this point you are not letting the expansion work happen at all and it is a constant volume process and there is no expansion work and hence if the whole process is adiabatic rather than H being the same it is actually u reactant is equal to u products. So, the internal energy is the same not the enthalpy that is because there is no expansion work involved and it is just a pure closed system q is 0, w is 0. So, finally, whatever u is there that will be before the combustion and after the combustion the u should be the same. Now, what is normally done is that only H is tabulated. So, if H is tabulated and you want to use this equation then you will just say that H reactant minus p reactant v reactant is equal to H products minus p products v products. Now, obviously it is a constant volume process. So, this and this are same you will realize that because the temperature increases and the moles also may not be conserved these will not be the same, but the left hand side totally is the same as the right hand side. So, for the reactants since you know the initial conditions H r is known and instead of writing p v you will write n r t where this is n of reactants r is r universal and t is the initial t and here you will write n products into r u into t adiabatic where this is the constant volume t adiabatic not the regular definition of adiabatic flame temperature. So, now what you have reduced this is that the left hand side is now straight forward it is known you know the H you know n you know r you know t. So, LHS is known on the right hand side everything is now a function of temperature that is because I know how many products should form due to this set of reactants this is the universal gas constant and t adiabatic is the only unknown H is also a function of t adiabatic. So, I will now do trial and error and get my t adiabatic. So, it is only unknown I have you know guess a value of t adiabatic put it in the polynomial expression find my u adi you see if it matches the left hand side if not then I will choose another guess and continue this process till I get what I want and we will see a simpler process when we continue. But overall you know I just want you to remember that you know this is just H reactants is equal to summation of H of all the reactants and as a function of temperature this is H products this is delta H r this is H t if I am beginning here this is adiabatic plane temperature the H t is same as H r if you do this and remember what the adiabatic plane temperature is it is very straight forward using this polynomial fit and we will continue I mean I have finished whatever I wanted to t or del regarding combustion. So, we have talked about fuels we have talked about stoichiometric air fuel ratio we have talked about balancing we have talked about polynomial fits for C p. So, that you can get H and S and we have talked about excess air factors and finally, we have talked about exhaust gas analysis and adiabatic plane temperature. So, these are some of the basics that you will require regarding fuels and combustion and though you will realize that as far as adiabatic flame temperatures and enthalpy concerns these are standard thermodynamic things that probably you ought to teach in such a course, but most of the things regarding exhaust gas analysis and how much fuel you how much air to fuel ratio you require etcetera. Such concepts are normally kept to the IC engine course or the power plant engineering course, but in general I have covered most of this year and you will see that most of these are based on simple chemical reaction concepts as well as concept regarding enthalpy being the same etcetera. So, I want to end the topic of combustion here.