 So, now we have a combustor operating at steady state. So, air at a temperature T A and fuel at a temperature T F enter the combustor with the steady mass flow rates and steady mass flow rates and properties and chemical reaction takes place inside the combustor and products. So, some heat is removed from the combustor so that products at a temperature T P leave the combustor. So, depending on the amount of heat you remove the product temperature will be different ok. So, we assume that so because the combustor operates at steady state we assume that sufficient heat is removed to maintain the product temperature at T P. Now, we apply steady flow energy equation to the combustor we get 0 equal to Q dot minus W X dot plus M dot F times H F plus M dot A H A minus M dot T H P no change so far W X dot is 0 no external work is being supplied or generated in the combustor so that is set to 0. Now, chemical reactions are taking place in this in this combustor and based on what we have discussed so far if you recall all the combustion reactions that we wrote down in terms of return on a molar basis ok. So, chemical reactions basically what we say in chemical reaction is the following that 1 mole of hydrogen combines with half mole of oxygen to form 1 mole of water vapor in a chemical reaction. So, it is always convenient to or it is always customary to write chemical reactions in terms of number of moles. So, the coefficients that appear in the chemical reactions are actually the number of moles of each one of the species ok. So, it would be convenient to write this expression also on a molar basis. So, for that purpose what we do is the following. So, we take this term and multiply and divide by its molecular weight for instance we write the first term as M F dot times molecular weight of fuel divided by molecular weight of fuel times H of fuel. Remember H here as written is without an over bar so that means it is in units of kilo joule per kilogram. So, if I combine remember M dot as units of kg per second. So, this quantity here as units of kg per k mole and so does this one. So, this also as units of kg per k mole. So, if I combine these two terms the resulting quantity would have units of k mole per second or that can that may be identified as the molar flow rate of the fuel. So, M dot F is the mass flow rate of the fuel and M dot F divided by the molecular rate of fuel may be identified as the molar flow rate of fuel and the remaining two terms may be combined and that may be identified. So, this has units of kilo joule per kilo mole. So, this has units of kilo joule per kilo mole and that may be identified as H F bar. So, each one of these terms may then be written as instead of M dot F H F we may write it as N dot F times H bar of F. So, we may write it as N dot F times H bar of F like this. Now, we divide through by N dot of F because you may recall that whenever we write combustion equations we always write on a per kilo mole of fuel basis we would always like to do that. So, you always write on a per kilo mole of fuel basis which is why I am dividing through by N dot of F. So, if you divide through by N dot of F you are left with this expression and the H bar in each one of this term this one this one and this one may itself be written based on the expression that we wrote down just a couple of slides ago. So, that H bar of F is nothing but enthalpy of formation of fuel plus delta H F evaluated at T F and so on. And for the reactant stream the term has to be evaluated for each one of the reactant species. So, I here denotes the reactant species. So, we have to denote it for each one of the reactant species and again for the products we need to write this for each one of the product species. Notice that the reactant stream consists of in this case air which means it consists of O 2 plus N 2 which means we have to write the we have to break this down into its individual species and then treat each individual species like this and similarly for the product stream. So, we can actually combine this term here or these two terms and identify this as the specific enthalpy of reactants H bar on a molar basis and write this as H bar of products. So, this is steady flow energy equation applied to the combustor. So, now we are in a position to calculate for instance Q dot which is what is usually desired. If you recall when we had a Brayton cycle or any of the other applications where we supplied heat basically fuel was burned to generate that heat and we said you know so many kilo joules are supplied from a reservoir at such and such a temperature. So, now we are in a position to actually calculate Q dot and also we get an idea about the reservoir temperature. Notice that reservoir temperature that we are talking about earlier would more or less be represented by T p. So, the temperature is the product stream. So, that is the temperature at which the reservoir is maintained so that we can we remove the certain amount of heat to maintain the products and that temperature. So, now we are in a position to calculate given say the temperature of the products we can actually apply steady flow energy equation and calculate Q dot or vice versa. Now, I mentioned earlier that combustion is an exothermic reaction. So, which means that H bar of reactants is actually greater than H bar of products. So, that heat has to be removed to maintain the products at a certain temperature. So, the reactants is greater than the enthalpy of the products. So, heat is released in the combustion reaction. So, this shows the variation of enthalpy of reactants and the variation of the enthalpy of products on a molar basis. Remember in the module on combustion everything is done on a molar basis because chemical reactions are taking place and chemical reactions are always written on a molar basis. So, property calculations, specific enthalpy, specific entropy, everything is on a molar basis. So, this is a departure from how we have done this so far. So far we have always used specific enthalpy in units of kilo joule per kilogram or specific entropy in units of kilo joule per kilogram Kelvin. The only exception is this module on combustion and the reason for that comes from chemical reactions. Now, this difference for example, the difference in enthalpy between the reactants and products at any temperature for example, I may evaluate it at this temperature. So, this then would be delta H bar of the reaction at this temperature which is this value. So, delta H R is actually a function of temperature. So, enthalpy of reaction or heat of reaction typically is measured at T ref which is 298 Kelvin. So, it is in general defined as H bar of products minus H bar of reactants and the important thing is the products, reactants and fuel all must be at the same temperature. So, that the heat that we are getting is only due to chemical reaction, not due to temperature changes in the product stream or reactant stream. Everything must be at the temperature at which we want the enthalpy of reaction. So, for example, in this reactor, if air comes in at let us say 700 Kelvin, fuel comes in at 298 Kelvin, products leave at some other temperature. See, there is a change in enthalpy of the air and fuel due to changes in temperature that is you know that is taking place in the combustor. So, then this would strictly speaking not be the enthalpy of the reaction because some of the energy goes into heating the air or the fuel has to be heated more than the air stream in this case and so on. So, if you want enthalpy of reaction at a particular temperature, let us say 1000 Kelvin, air must come in at 1000 Kelvin, fuel must come in at 1000 Kelvin, products must leave at 1000 Kelvin. If you want heat of reaction at 298 Kelvin, then air, fuel and products must all enter and leave at 298 Kelvin that is very important. So, delta HR as we have already mentioned is negative because H bar products is greater than H bar reactants for combustion reactions which are exothermic and as I mentioned delta HR is usually evaluated at the reference temperature. Now, enthalpy of combustion naturally leads us to the notion of a calorific value of the fuel. In combustion since reactants are fuel plus air, the calorific value of the fuel is numerically equal to delta HR without the negative sign. We always say calorific value of a fuel is such and such a number and that is always a positive number. So, calorific value is always equal to absolute value of delta HR bar. However, distinction needs to be made on the calorific value depending on whether the water in the product stream is in vapor form or liquid form. If the water leaves in vapor form, then we get what is called a lower calorific value. Now, if we remove more heat so that the water also condenses and leaves in liquid form, then we get the higher calorific value. So, if you look at this reactor, so if the water leaves as vapor and we remove certain amount of heat that is called the lower calorific value. If I remove more heat, then the water vapor in the product stream will also condense before it leaves which means the resulting calorific value will be higher. That is why we have two calorific values, lower calorific value and higher calorific value. The difference between the two is the latent heat of vaporization at that temperature. So, let us work out a few examples to illustrate the ideas that we have discussed so far. Determine the higher and lower calorific value of methane at 298 Kelvin. Enthalpy of formation of methane is given to be minus 74850 kilo joule per kilo mole. Unless otherwise stated, we assume stoichiometric amount of air. So, combustion of methane with stoichiometric amount of air is represented by this chemical reaction. Now, if I say vapor, then I get the lower calorific value. If I account for latent heat, then I get the higher calorific value. So, SFEA applied for this case with the Tp equal to Ta equal to Tf equal to 298 Kelvin reads like this. So, 0 equal to q dot over nf dot plus enthalpy of formation of fuel, enthalpy of formation of the air that is coming into the combustor broken down into its consumed elements O2 and N2. And this is the enthalpy of the product stream which consists of as you can see from here CO2, H2O and N2. All are leaving at 298, the products are leaving at 298 Kelvin, reactants are coming in at 298 Kelvin. Which means that the sensible enthalpy of CH4 is 0 because it is at 298 Kelvin. Now, enthalpy of formation of O2 is 0 because it occurs naturally. So, there is no enthalpy of formation for O2. The sensible enthalpy for O2 is 0 because it is coming at 298 Kelvin. And the same is true for nitrogen also. Enthalpy of formation of nitrogen is 0 because it occurs naturally and in this case delta H bar is 0 because it comes in at 298 Kelvin. Enthalpy of formation of CO2 is non-zero, but sensible enthalpy is 0 because it leaves at 298 Kelvin and again 0 for H2O also. And for the nitrogen in the product stream, both the enthalpy of formation and delta H bar are 0. Now, the enthalpy of formation of CO2 and H2O are given in the combustion table. Let us take a look at the combustion table. So, as you can see this table lists sensible enthalpies and enthalpy of formation of just limited number of gases. Of course, if you consult this website, this I consider to be the most authoritative, webbook.nist.gov contains all the, I mean values for practically any species that you can think of. This is an excellent source and I urge you to consult this source if required for any use in the future. So, enthalpy of formation of as you can see O2 is 0, N2 is 0, CO2 minus 393.52 mega joule per kilo mole and so on. So, this also gives delta H bar as a function of temperature. So, if you retrieve the value from the table, so you may retrieve the value for enthalpy of formation of CO2 and so on from the table and enthalpy of formation of methane is already given in the problem statement. So, if you substitute the numbers, you get the lower calorific value to be 802330 kilo joule per kilo mole. Remember, we have taken water to be in the vapor phase. So, the combustion table that we just looked at lists delta H bar for water vapor. So, on a mass basis, we get the calorific value of methane to be approximately 50 mega joule per kilogram of CH4. So, the calorific value of hydrocarbon fuels generally tend to be in this range between 40 to 50 mega joule per kilogram. Now, the higher calorific, we want to calculate the higher calorific value also and there is no change in any of these terms except the one for the water vapor. So, here this value here is Hg minus Hf from the steam table at 298 Kelvin and converted to kilo joule per kilo mole basis. What is that? This goes with the negative sign because we are removing heat. So, the higher calorific value comes out to be 890226 kilo joule per kilo mole of fuel or 55 mega joule per kilogram of CH4. That is the only difference. So, basically the enthalpy of formation of liquid water, there is nothing but enthalpy of formation of water vapor and then we need to remove the latent heat to get it to the liquid state. Just like what we did for the carbon when we took it from solid phase to gaseous phase, we needed to supply heat. Here we take the water vapor from the gaseous phase to the liquid phase. So, we need to remove heat. So, that is what we have done here. Next example involves liquid n-dough decaying. So, n-dough decaying basically is kerosene. So, liquid n-dough decaying at 298 Kelvin and air at 700 Kelvin steadily enter the combustor of a gas turbine engine. Determine the required mass flow rate of air and the corresponding equivalence ratio. If the temperature of the products is A, 1700 Kelvin, B, 2200 Kelvin. Heat loss from the combustor may be neglected. Enthalpy of formation of liquid-dough decaying may be taken as minus 352100 kilojoule per kilo mole. So, chemical reaction for the combustion of liquid n-dough decaying with air may be written like this. Now, we are asked to calculate the required mass flow rate of air. So, this is for stoichiometric combustion. Now, you notice that the temperature of the product varies and there is no heat loss or heat removal from the combustor. So, basically if you look at the combustor, so this is air at 700 Kelvin, this is fuel at 298 Kelvin. Now, the products are leaving either at 1700 Kelvin or 2200 Kelvin. So, the temperature of the product is being adjusted by adjusting the amount of air. So, by supplying excess air or by reducing the amount of air, whatever the case may be, the product temperature is being adjusted. So, we rewrite this equation like this. So, the 18.5, the coefficient 18.5 in the second term is now multiplied with the 1 plus x. So, if the excess air supplied is 0, then we revert to the stoichiometric combustion case. If it is more, if excess air is supplied, then we get it to be a positive number, otherwise we get it to be a negative number. The quantity x otherwise becomes a negative number. So, the balanced chemical reaction for this case looks like this. So, we apply SFE to the combustor. Fuel comes in at 298 Kelvin. So, sensible enthalpy is 0 for the fuel. And this corresponds to the enthalpy term for the reactants, for the air, I am sorry, for the air. Notice that the sensible enthalpy for O2 and N2 are not 0 because they both come in at 700 Kelvin. And this is the enthalpy term for the products which comprises of CO2, H2O, O2 and N2. CO2, H2O, O2 and N2 as you can see here, and the products leave at a temperature Tp. So, if you substitute the known values, which are basically the enthalpy of formation values and rearrange, we get an expression for x like this. So, here all the delta H values are to be evaluated at the product temperature. So, if the product temperature is 1700 Kelvin, then we get x to be 1.2695. And the required mass flow rate, remember this is in on a molar basis. So, 1 plus x times 18.5 times 4.76 kilo moles of air or supplied per kilo mole of fuel. So, we multiply that by the molecular rate of air to convert it to mass basis. And remember here we are supplying the molecular rate of fuel is 170. So, we get this to be on a mass basis like this, 23.905 kg per kg of fuel. So, this is basically kg of air per kg of fuel. And remember equivalence ratio is the fuel air actual fuel air ratio divided by stoichiometric fuel air ratio. So, actual fuel air ratio is 1 plus x stoichiometric is 1. So, this comes out to be 0.44 which is on the very much on the lean side probably being quite close to the lean limit of the fuel. So, I am going to leave it as an exercise for you to rework this problem for a product temperature of 2400 Kelvin. The calculation for product temperature of 2400 Kelvin or worked out in the book. So, you may compare your answer with what is given in the book.