 The next example that we are going to look at involves gasoline E10 which is actually a blend of 90 percent N octane and 10 percent ethanol as many of you may be aware this sort of blended fuel is becoming very popular nowadays. So, gasoline is blended with ethanol to not only reduce the dependence on oil but also you have to reduce emissions and so on ok. So, it is a very popular fuel nowadays and we are asked to determine its lower calorific value enthalpy of formation of liquid N octane and liquid ethanol are given and the specific gravity of the 2 fuels in the blend are also given ok. So, 1 liter of the fuel contains 0.9 liter of N octane and 0.1 liter of ethanol. So, if we multiply this by the density then we get that 1 liter of the blended fuel contains 0.6327 kg of ethanol I am sorry of N octane and 0.1 times 0.79 or 0.079 kg of ethanol of ethanol ok or on a molar basis this works out to 5.55 moles of N octane and 1.7174 moles of ethanol in 1 liter of the blended fuel or the mole fraction may be worked out to be 0.7637 and 0.2363. So, the so the chemical reaction for the complete combustion of the blended fuel with stoichiometric here may be written like this. So, we worked out the mole fraction to be 0.7637 and 0.2363. So, we are using that here and after balancing the chemical reaction you can see that the amount of the number of moles of air that is required is 10.25515 times 4.76 that is number of kilo moles of air I am sorry. So, the products are shown here and SFE applied to the combustor in this case reduces to something like this. Remember, you will see we calculated with reactants and products being at 298 Kelvin ok. So, this comes out to be 4167335 kilo joule per kilo mole of fuel or using the values in the combustion tables. Notice that these values are given in the problem statement. So, Hf of Hf naught of CO2 and Hf naught H2O are available in the combustion table. So, we may work out that the lower calorific value of the fuel comes out to be 42,553 kilo joule per kg of fuel puts it in the same range as other hydrocarbon fuels. The only difference being that the ethanol fuel brings in more I mean it brings in oxygen which was not present before in the fuel stream itself ok. So, the advantages and disadvantages of having fuel like this will be discussed is usually discussed in course on combustion because using a fuel like ethanol has implications on the emissions aspect. It lowers certain kinds of emissions but it also has certain disadvantages ok and these are usually discussed in great detail in courses on combustion. So, the next topic that we are going to discuss is the following. So, let us say we have a combustor like this which is fully insulated. So, we have let us say air and fuel which enter the combustor let us say at the same temperature at a temperature T1 let us say. So, air enters at temperature T1 fuel also enters at temperature T1 and products leave at a certain temperature which is to be determined. So, basically once a reactor is completely insulated all the enthalpy of the reaction goes into the products. So, the resulting temperature of the products is known as the adiabatic flame temperature. That is the maximum product temperature that we are going to see or we will see with that is the maximum temperature that we will see as a result of combustion of the particular fuel that we are looking at ok. So, if you apply SFE to this reactor you can see that in this case H reactance is equal to H products. So, the temperature of the product in this case is called the adiabatic flame temperature. It is an important concept since it represents the theoretical maximum product temperature that can be realized at the combustor exit. So, if you want to get an idea of what the maximum temperature possible is with the particular fuel we calculate the adiabatic flame temperature and that gives us an upper bound on the temperature that we are likely to see. So, this expression itself H reactance equal to H products is illustrated graphically in this diagram ok. So, as we said because combustion is an exothermic reaction H reactance is always greater than H products and the variation of H bar reactance and H bar products with temperature is shown in this ok. Since H bar products is equal to H bar reactance in this case let us say that the reactance coming at a temperature T1. So, this is H bar reactance right. So, this is H bar reactance. So, the products leave at a temperature which is such that H bar products is equal to H bar reactance. So, you can see that the reactance come in at much lower temperature, but the products on account of the fact that H bar products equal to H bar reactance leave at a much higher temperature. Let us work out an example illustrating this concept. A stoichiometric mixture of methane and air at 298 Kelvin enters an insulated combustor where it undergoes complete combustion determine the temperature of the products leaving the combustor ok. So, SFE applied to this combustor gives us the following and for stoichiometric combustion of methane you know that. So, this is the balance chemical reaction for the combustion of methane with air. So, SFE applied to the combustor gives us this. The air and fuel enter at 298 Kelvin. So, that means a sensible enthalpy is 0 on the reactant side which is what we have done here. And of course, as before enthalpy of formation of O2 and N2 are 0 because they occurred naturally. So, this is on the reactant side. On the product side we have this expression which is the some of the enthalpies of CO2, H2O and N2. So, if you substitute the known values notice that the enthalpies of formation are all known here. So, this is given in the problem statement. This is known, this is known and the sensible enthalpy on the product side is not known because the exit temperature is not known. We have to calculate the exit temperature. So, the delta H bar CO2, delta H bar H2O and delta H bar N2 are not known. So, these are not known since the temperature is not known. So, if you gather the unknown quantities on the left hand side this is the expression that we get. And the resulting temperature has to be determined iteratively. So, we start with the guest value for the temperature. Let us say 1700 Kelvin and if you evaluate the left hand side of this expression at 1700 Kelvin we get this number which is less than the target value that we are seeking. So, now we guess another value which is higher let us say 2400 Kelvin and the left hand side of this expression comes out to be higher than the target value. So, now we know that it is between 1700 and 2400. So, we take a slightly lower guess which is 2300 Kelvin and we get the left hand side to be just below or just less than the target value. So, we have one temperature at 2400 the left hand side is greater than the target value and at 2300 the value that the left hand side is less than the target value. So, we can actually interpolate for the target value and estimate the temperature to be approximately 2330 Kelvin, 2330 Kelvin the temperature of the product. And as you can see this is relatively high temperature. So, the adiabatic flame temperature can thus be seen to be the upper bound on the product temperature. So, this completes the discussion on the first law analysis of combustion systems. Let us take up second law analysis of combustion systems next. I discussed at the beginning of the first law analysis of combustion systems that energy balance for a combustion process must take into account enthalpy of formation of the participating species because the product composition is different from the reactant composition. So, species that are present in the reactant stream or absent in the product stream and vice versa. So, which means bonds are being broken in species and new bonds are being found in new species. So, the energy balance must take the energy in the bonds into account in this case and not just the sensible enthalpy alone. We have done all the calculations so far with sensible enthalpy alone which suffice because the composition of the mixture that we were considering did not really change CO2 remained as CO2, O2 remained as O2 and so on. So, sensible enthalpy is sufficient in that case, but now we have to take we have to take into account enthalpy of formation along with sensible enthalpy. In the same manner and for the same reason entropy balance for a combustion process requires absolute entropy for each species because we are no longer calculating just S and just delta S we are actually looking at S for each species because there are new species on the product side different species on the reactant side. So, we need the absolute entropy of species we are not merely calculating delta S. Now, third law of thermodynamics states that the absolute entropy of a pure crystalline substance 0 at 0 Kelvin which provides a datum against which the entropy absolute entropy of any species may be evaluated. Now, in the context of combustion thermodynamics the absolute entropy of substances at the reference temperature of 298 Kelvin and reference pressure of 1 atmosphere have been measured and or theoretically calculated for most species the absolute entropy at this temperature at the reference temperature and reference pressure are available. So, in the same manner as enthalpy of formation entropy of a substance at any other temperature and pressure is evaluated with respect to this reference state we have the TDS relationship. So, once I have the absolute entropy at the reference pressure temperature that may be used to evaluate the entropy at any other pressure and temperature. Let us see how that is done remember TDS is equal to DH minus V Dp. So, the absolute temperature of an ideal gas at any temperature and pressure may be written like this the absolute I am sorry the absolute entropy of any ideal gas at a temperature of T and P may be written as absolute entropy at T ref comma P ref. So, from T ref to P ref we go to so from a state or from the reference state of T ref to P ref we then move on to a state of T comma P ref where T is the temperature at which we want to evaluate the absolute entropy. So, from T to P ref we then move on to T to P. So, notice that here pressure is constant here temperature is constant. So, we use a two step process to go from T ref comma P ref to T comma P any temperature or pressure. So, if we apply a TDS relationship for the first step since the pressure is constant this term will be absolute. And in the second step since the temperature is constant the first term will be absolute. So, we may this evaluate the absolute entropy at a temperature T comma P as S bar T ref comma P ref plus integral T ref to T DH over T right. So, for the first step let me just write it with a different colour. So, the so if I apply TDS relationship for the first step right I get DS equal to DH over T which can then be integrator. For the second step I may write DS equal to minus V over T DP or since we are using ideal gas equation of state V over T on a molar basis this is on a mass basis. So, on a mass basis this is simply minus r times DP over P and on a molar basis they are here instead of being in the particular gas constant becomes the universal gas constant. So, that is applicable to this step the second step. Now, the term the first two term together is usually identified as S bar of T comma P ref or S bar 0 of T where 0 represents the reference state. So, this is actually tabulated in the absolute entropy table let us just quickly see what is available in the table. So, this table lists absolute entropy of some gases as a function of temperature. So, S 0 a different values of temperature are given in this table. So, we can just simply go into the table look up the value for S 0 of T up to a temperature of 4000 is what is given here values are available for I think slightly high temperatures in the source that is listed here. So, this is how we are going to do the calculations. So, we will use this table to do the calculations. Now, in case we have a mixture of gases the absolute entropy of each of the species has to be evaluated using this expression and then multiplied by its mole fraction. Remember the pressure that you use in the expression is the partial pressure of each species we are using Dalton's model. So, the pressure in this expression then used for individual species in a mixture must be replaced by the partial pressure of that particular mixture and we have replaced the partial pressure with the mole fraction here. Now, let us look at application of second law analysis of combustion system. So, we here we have the combustor the same one as what we considered when developing the theory for first law analysis of combustion systems. So, we have a combustor air comes in at a temperature T a fuel comes in at a temperature T f and products leave at a temperature T p. So, entropy balance for this control volume gives us this expression. Remember Q dot surrounding is equal to minus Q dot. So, we use the same step or idea as before we multiply and divide the this group of terms by the molecular weight of that particular stream to convert this into a molar basis. So, if you write this on a molar basis, we get the left hand side which is the rate of entropy generation divided by the fuel flow rate on a molar basis is equal to ni over n of ni dot over n of dot times Si of T p comma Pi. This is the absolute entropy of each species in the product stream which is why we are using the partial pressure of that particular species minus absolute entropy of the fuel stream minus ni dot over n of dot sigma absolute entropy of each of the species in the incoming air stream. In this case, we have only O 2 and N 2 plus this term. So, basically this expression simplifies to something like this very nicely. Sigma bar which is the entropy generator per kilo mole of fuel comes out to be something like this. Remember our interest is always in the entropy generation in the universe. So, and that is what this expression that is what this expression gives us because as we as we saw in the module on exergy, entropy generation in the universe is tantamount to exergy destruction. Let us go through an example. Determine the entropy generator per k mole of fuel in the earlier example, if the gas turbine combustor operates at 3 MPa and the products leave at 1700 Kelvin, the specific entropy of liquid N dodecane at entry may be taken as 490.660. This example involved a gas turbine combustor. So, air came in at 700 Kelvin and now additionally we are also told that the air enters at 3 MPa. So, fuel which is liquid N dodecane entered at 298 Kelvin and the products leave at 1700 Kelvin and there was no heat loss from the combustor. The percentage excess air was also determined earlier. So, that can be made use of now. We need the mole fractions of the individual species in the reactant and the product stream. So, when we calculate the mole fraction of the reactant stream in this case, since the fuel enters as a liquid, it is not used in the calculation of mole fractions. So, on the reactant side we have two species O2 and N2 and these numbers were obtained based on the calculations that we did earlier with the excess air and this gives mole fraction to be 0.21 and 0.79. So, on the product side from the balance chemical reaction we can retrieve these numbers and the mole fractions may be evaluated like this. So, S bar of reactants is composed of S bar of the fuel in liquid form plus the air stream. Air stream itself involves S bar of O2 times N O2. So, this is S bar of O2 and this is N O2. Similarly, this is S bar of N2 in the inlet side and this is N of N2 on the inlet side. This has to be evaluated at 3 MPa 700 Kelvin. So, we have assumed the pressure to remain constant in the combustor which is a good assumption anyway we are looking at a steady flow system. So, the product stream is at leaves at 3 MPa 1700 Kelvin. So, the product stream again consists of CO2, H2O, N2 and O2, mole fractions are known, partial pressures are known and the number of moles of each of the species are known. So, all these quantities are known since the temperature is known we basically retrieve Si OO at 1700 Kelvin for each of the species and so, this gives us the S bar of products to be 47,506.56 kilo joule per kilo mole Kelvin of fuel. So, entropy generation on a molar basis, so it is not rate of entropy generation, it is entropy generation on a molar basis comes out to be so many 7, 8, 5, 2 kilo joule per kilo mole of fuel Kelvin. Remember this itself is nothing but sigma dot rate of entropy generation divided by NF dot which is molar flow rate of fuel. So, it is clear that whatever ideas and theory that we developed earlier basically this equation which we just wrote down simply was developed in great detail in the beginning of this course. We started with entropy generation in a control volume and entropy balance for a control volume. So, you can see that the theory that is developed in the first level course has simply been extended to cover many more applications. Basically, we looked at air standard cycles, Rankine cycle, vapor compression cycle where for which we calculated the second law efficiency which is a very, very important performance metric. And the second law efficiency itself was calculated based on exergy which is a new concept that was introduced but the theory behind exergy was all taken from whatever was developed in the first level course. Then we looked at an interesting application namely psychrometry where again we dealt with the mixture of ideal gases, air moist here which is nothing but air plus water vapor that is a mixture of ideal gases. Then the concepts developed in the first level course are applied to combustion systems first law analysis as well as second law analysis where we were able to calculate the entropy generation in the universe due to chemical reactions or in a combustor study flow situation. So, these are very important concepts. We have not looked at a non-flow application or example in this module for the sake of brevity. The textbook does contain a non-flow example related to IC engines. So, I urge you to consult the textbook and go through that example. The only difference that you are going to see especially in the first law analysis is that you will encounter u instead of h as you are familiar. And again here also there will be some minor changes when you calculate the entropy change but apart from that you know there are no big changes. So, I urge you to consult the textbook for this example. So, what we will do in the next lecture is to begin our discussion of compressible flow through nozzles.