 In this module on combustion thermodynamics, we will discuss how concepts that you learnt in the previous course, concepts such as steady flow energy equation as well as entropy balance of a control volume can be applied to combustion in particular combustors. The interesting aspect of this application is that we will take into account chemical reactions also. So, normally when we say for instance in the case of a Brayden cycle, we assume where we assume that there is a reservoir at a high temperature, but we really did not worry about how the heat was being supplied from the reservoir or how the reservoir itself was being maintained at a high temperature. In this case, we will take a closer look at such reservoirs where usually the high temperature is a result of burning of fuel and heat is released as a result of that. So, we are going to take a closer look at combustion of fuels and apply energy balance and also entropy balance to such devices. So, let us start with the first topic which is combustion stoichiometry. Combustion is an oxidation reaction that is accompanied by heat release. So, basically all combustion reactions are exothermally. So, for instance if you look at this reaction H2 plus half O2 giving H2O. This is a combustion reaction in which 1 kilo mole of hydrogen combines with half a kilo mole of O2 to form 1 kilo mole of water vapor. Now, since we are talking about combustion thermodynamics, we will not worry too much about the details of how the combustion takes place. For instance, whether the fuel and the oxygen or air, it has been mixed well, whether it has been ignited properly, whether the combustion can indeed propagate. Those are issues which are left to combustion text. So, combustion text typically would deal with these sort of aspects, whether combustion can be initiated in a particular scenario, whether it can be sustained in a particular scenario and so on. So, in this course we will not look at those types of details. This is a thermodynamics course, applied thermodynamics course. So, basically what we would say is when we write this we assume that combustion will take place and heat will be released and products will be formed. Now, the required oxygen may be provided in one of two different ways. The oxygen that is required for combustion, remember combustion is an oxidation reaction. So, the oxygen that is required for combustion may be provided either as a pure oxygen stream or as is commonly done the oxygen could be taken from atmospheric air. Atmospheric air as you know in the absence of any humidity, contains 21 percent O2 by volume and 79 percent N2 by volume. So, every kilo mole of O2 that is supplied is accompanied by 3.76 kilo moles of N2. So, if you are supplying air then for every kilo mole of O2, 3.76 kilo moles of N2 accompanies the oxygen. So, if we if we accomplish the combustion in the previous reaction by using oxygen from the air instead of a pure oxygen stream, the reaction would look something like this. So, we replace the O2 with O2 plus 3.76 N2 because every kilo mole of O2 is accompanied by 3.76 kilo moles of N2. What is that? N2 appears now in the product side also. Again in this course because it is a thermodynamic course, we will assume nitrogen to be inert that nitrogen does not participate in the chemical reaction. However, nitrogen will absorb the heat that is released from the reaction and so it will its temperature will go up. So, we have a mixture of ideal gases on the product side and nitrogen will participate there. In other words, the enthalpy changes in nitrogen will be accounted for as we have always done for a mixture of ideal gases. However, chemically nitrogen will be assumed to be inert. So, it will not combine with oxygen for the kind of situations that we are looking at. In reality, however, you know that as the temperature becomes higher, the nitrogen combines with oxygen to form oxides of nitrogen which are pollutants as you know since they can be harmful to the environment. But in this course, we will not deal with such complications. Now, let us consider the composition of liquid N octane which is something like petrol or gasoline not exactly, but close enough. So, liquid N octane would be C8H18 and subscript L denotes that it is in the liquid form. It is burnt in air. So, we have O2 plus 3.76 N2 and it is completely burnt in air. So, that is complete combustion of N octane. Complete combustion here means that the combustion products are fully oxidized, meaning that all carbon in the fuel is converted to CO2 and all hydrogen in the fuel is converted to H2O and there is no excess oxygen in the product stream. These are very important. So, all combustion products are fully oxidized and there is no excess oxygen in the product stream. So, the balanced combustion equation for this may be written like this. So, each kilo mole of fuel here requires 12.5 kilo moles of O2 or 59.5 that is 12.5 times 4.76, 59.5 kilo moles of air for complete combustion. So, this much amount of air is usually called theoretical or stoichiometric air. Stoichiometric what we mean is that complete combustion. All species are oxidized in the fuel oxidized and no excess oxygen in the product stream. So, this is called theoretical or stoichiometric air. So, the stoichiometric air fuel ratio on a molar basis is 59.5 kilo moles of air per kilo mole of fuel. On a mass basis, we may write it like this, 59.5 kilo mole of air per kilo mole of fuel and molecular weight of air is 28.97. So, 28.97 kg per kilo mole of air and molecular weight of N octane is 114. So, 114 kg per kilo mole of fuel. So, this allows us to convert this into a quantity on a mass basis. So, air-fuel ratio on a mass basis is 15 kilograms of air per kilo mole of fuel. Now, it is somewhat striking that the air-fuel ratio on a mass basis for almost all hydrocarbon fuels is very close to this value of 15. Even for fuels which are as light as methane to fuels which are as heavy as kerosene which would be dodecaine. So, from light fuels like methane to heavy fuels like diesel, the stoichiometric air-fuel ratio on a mass basis is around 15. I will not go through the derivation. Interested students may look up this derivation which is given in my textbook. It is very close to 50. So, that number is worth remembering that air-fuel ratio, theoretical or stoichiometric air-fuel ratio on a mass basis for hydrocarbon fuels is around 50. In many practical applications, the amount of air used is always in excess of the stoichiometric amount of air. You cannot just use the exact stoichiometric amount of air in practical situations. There are several reasons for this. One important reason is that when we use excess air, we have additional amounts of O 2 and additional amounts of N 2 on the product side. So, when we use air which is more than what is what is required as you can see here. So, this is the balanced chemical reaction for stoichiometric amount of air. So, when we supply excess air that means excess O 2 and excess N 2, we are going to have some amount of O 2 in the product stream and some amount of N 2 in the product stream. So, the presence of these additional amounts of oxygen and nitrogen serve to dilute the combustion products and also result in reduction in the peak temperature. So, the peak temperature can now be reduced and it can be reduced to manageable levels. For instance, in the case of a Brayton cycle or aircraft engine which uses the open form of the Brayton cycle, the air-fuel ratio is actually around 60 to 1 or so, which as you can see is 4 times the theoretical value. But that much amount of air is required to actually dilute the combustion products and keep the operating temperature at a reasonable level. Remember, in the combustion in a Brayton cycle, experiences the peak temperature continuously unlike an auto cycle or a diesel cycle. We mentioned this aspect earlier. So, the peak temperature in the case of an auto cycle or diesel cycle appeared to be more than that of Brayton cycle. But in the case of the auto cycle and diesel cycle or their real life counterparts or the counterparts which use these cycles in real life namely the SI engine and the CI engine, the peak temperatures are experienced only momentarily at the beginning of combustion and then the temperature comes down very rapidly. Whereas in the case of the gas turbine combustor, these temperatures are experienced continuously because that is operating at steady state. So, it is much more challenging. So, which is why much higher air-fuel ratios are used in the case of gas turbine engines to keep the temperatures at manageable levels. Now, in the case of SI and CI engines, the air-fuel ratios are closer to the stoichiometric value of 15 in order to ensure that combustion takes place and is steady. So, different environments, working environments on different engines require different types of different amounts of air or air-fuel ratio. Now, the other advantage of supplying excess air is that the presence of excess oxygen tends to stabilize the combustion. Sometimes, what may happen is that when you supply the exact amount of oxygen in a practical combustor, the oxygen may not spread uniformly to all parts of the combustor. So, that in some parts of the combustor, there may not be sufficient oxygen and the fuel may be starved for oxygen, whereas in other parts, there may be excess oxygen. So, this means that the combustion is going to be sort of non-uniform across the combustor and the performance will also not be very stable. In order to prevent this kind of a situation, if you supply excess air to begin with, then you are at least making sure that there is a reasonable chance that the fuel will always be burning under stoichiometric or with excess oxygen. So, the combustion is going to be much more stable for that reason. The downside or the flip side of supplying excess air is that because of the presence availability or free availability of oxygen in the reactant side, much more oxides of nitrogen are likely to be formed because there is plenty of oxygen in the product side. More of the nitrogen will combine with oxygen to form nitrous oxides, which is not very desirable. So, practical combustor designs as you can see are very challenging because of this conflicting considerations. So, the idea of excess air naturally leads to the definition of the equivalence ratio denoted by the letter uppercase phi, Greek letter uppercase phi. So, phi is defined as actual air fuel ratio divided by, I am sorry, actual fuel air ratio divided by stoichiometric fuel air ratio. So, fuel air ratio is the reciprocal of the air fuel ratio that we have been discussing so far. So, if phi is equal to 1, then of course, we have supplied the exact amount of air or theoretical or stoichiometric amount of air. If phi is less than 1, that means the fuel that we have supplied is actually less than what is required for theoretical combustion for the given amount of air. So, it is said to be fuel lean. And if phi is greater than 1, then we have supplied more fuel than what is required for stoichiometric combustion and so it is said to be fuel rich. So, if you look at the two engines that I mentioned so far, gas turbine engines typically tend to operate at equivalence ratios near 0.4 or so, which is very close to being on what is called the lean limit of combustion. As I said, we do not worry about such intricate aspects in this course because it is a course on thermodynamics, but if you do a course on combustion, there you would learn that phi equal to 0.4 is very close to the lean limit of combustion for most hydrocarbon fuels. What do we mean by lean limit? Which means that if you make the reactant seam or if you increase the amount of air anymore, the fuel will simply not burn at all. The fuel will burn only within certain flammability limits. So, below the value on the lean side, the fuel will not burn. Above the value on the rich side also, above a certain value on the rich side also, the fuel will not burn. There is too much fuel and too little oxygen on the rich side. On the lean side, there is too much oxygen and too little fuel. On the rich side, too much fuel and too little oxygen. So, these are the flammability limits. So, when you do a course on combustion, you will be taught all these sorts of aspects. It is not simply enough. If you just write down the combustion equation, then say that combustion will take place, not necessarily. The temperature has to be favorable. Mixing has to be favorable and flammability limit has to be favorable. All these conditions have to be satisfied. There must be sufficient time for the fuel to react and burn. So, there are many constraints, practical constraints that come when you actually design a combustor. Let us work out a couple of examples. First one reads like this. A dry analysis of the products from the combustion of methane with air is as follows. CO2, 9 percent, CO2 percent, O2, 2.11 percent and N2, 86.89 percent by volume. Calculate the excess air equivalent ratio and the dew point temperature of the products assuming the pressure to be 100 kilopascals. So, the product analysis is given on a volumetric basis. Since we are using the Dalton's model, it is the same as a molar basis. Now, combustion of hydrocarbon fuel such as methane would have produced water vapor, but this analysis is done on a dry basis. So, that aspect also needs to be taken into account. So, if you assume 100 mole of 100 k mole of dry products, then we may write the actual combustion reaction like this. A moles of CH4 combined with B moles of air to form this product stream. Remember, H2O is also present in this product stream, but this let me just use a slightly different here. So, this plus this plus this plus this add up to 100. And so, if we balance the equation, we are able to get A and B. So, the complete combustion reaction reads like this. So, 11 kilo moles of fuel is combined with 23.11 kilo moles of air to give rise to this product stream. So, the actual air fuel ratio, I am sorry not 23.11, but 23.11 times 4.76. I am sorry, hence the actual air fuel ratio is 23.11 times 4.76 divided by 11, which works out to 10.06 kilo moles of air per kilo mole of fuel. Now, for complete combustion, all the carbon in the fuel would have been oxidized, all the hydrogen in the fuel would also have been oxidized. So, the complete or stoichiometric combustion of methane with air reads like this. So, the stoichiometric air fuel ratio is thus 2 times 4.76, which is equal to 9.52 kilo mole of air per kilo mole of fuel. So, the excess amount of air that has been supplied is 10.06 minus 9.52 and on a percentage basis, it works out to 5.67 percent. And the equivalence ratio is 9.52 divided by 10.06. So, 9.946, which is less than 1. That means, mixture is lean, which is, which makes sense because excess air has been supplied. Now, you may recall from the module on psychrometry that the mole fraction of species, let us say H2O is nothing but partial pressure of the water vapor, which is denoted PV divided by mixture pressure. You may recall that this was what we had written earlier. So, we may actually calculate the partial pressure of water vapor. So, this has been given to be equal to 100 kilo Pascal in this problem. Now, the mole fraction of water vapor in the products may be evaluated simply by looking at this equation. So, if you look at the product stream, we have 9 plus 22 plus 86.89 plus 2 plus 2.11 as the total number of moles in the product stream. So, the number of moles of H2O divided by the total number of moles gives us the mole fraction of H2O in the product stream. So, 22 divided by the total number of moles, which is this. And from this, we may then evaluate the partial pressure of water vapor, which works out to 18.03 kilo Pascal. From the steam table, the saturation temperature corresponding to 18.03 kilo Pascal is 57.8 degree Celsius. Now, why are we calculating a dew point temperature in this problem on combustion of a hydrocarbon fuel? Now, the dew point temperature of the product is important because this determines whether the water vapor in the product stream. So, whenever you burn hydrocarbon fuel, there will always be water vapor in the product stream. So, the dew point temperature determines whether this water vapor will condense after it leaves the engine or whether it is going to condense in the exhaust tailpipe or even in the muffler or ahead of that. So, this is very important in the design of such equipment. So, if it condenses in the tailpipe or in any of the other components, this can cause corrosion damage, which is why dew point temperature is of particular interest in combustion applications, especially when you are burning a hydrocarbon fuel and when water vapor is likely to be present in the product stream. So, you want to keep the temperature of the exhaust gases well above this dew point temperature, so that there is no condensation of the water vapor in the components or before it leaves the exhaust. The next example reads like this Indian rice husk biomass fuel has the following percentage composition on a mass basis C 35 H 5.5 N 1.53 over 36 and sulphur 0.08. It is burnt with 185 percent theoretical air with air at 298 Kelvin, 100 kilopascal and 75 percent relative humidity. Remember, whenever we say that fuel is burnt with atmospheric air, we cannot always assume the air to be completely dry as we know from the module on psychrometry, there is always relative humidity, there is always water vapor in the air. So, this example illustrates how to do combustion calculations when there is water vapor in the incoming air. So, this is burnt with 185 percent theoretical air with air at 298 Kelvin, 100 kilopascal and 75 percent relative humidity. Determine the required mass flow rate of air per kg of fuel and the change in the dew point temperature of the products as a result of the humidity in the supplied air, assume the sulphur in the fuel to be inert. So, this is the information that is given in the problem statement, the composition of the fuel on a mass basis, percentage mass basis. So, if we convert this to mole, so we get the component distribution on a molar basis in the fuel like this. So, now we are in a position to write the balanced chemical reaction. So, the balanced chemical reaction for complete combustion of fuel, this fuel with dry air may be written like this with dry air. Now, with 185 percent theoretical air, notice that this for 185 percent theoretical air, this has to be multiplied by 1.85. So, times 1.85 times 3.1667 times 1.85. So, if you balance chemical reaction, this is what you get, notice that the nitrogen is inert, the sulphur is also inert. So, the partial pressure of water vapor in the product stream. So, now we have the product stream, we can calculate the mole fraction of water vapor from this. So, this is number of moles of water vapor. So, we may evaluate the mole fraction of water vapor to be 2.75 divided by 30.44275. So, this multiplied by the mixture pressure, which is 100 kilopascal, gives us the partial pressure of water vapor to be 9.033 kilopascal. And the dew point temperature of the product is nothing but T sat of 9.033 and that comes out to be 44 degree Celsius. So, this is the dew point of the product stream for combustion with 185 percent theoretical air, 185 percent theoretical air that to dry air for 100 kg of fuel. And that is what we did here, if you remember. So, 100 kg of fuel we converted to composition like this. So, this is the amount of air that is required, 185 percent excess air that is required for 100 kg of fuel. So, we can convert this into kg of air simply by multiplying by its molecular weight times 28.97 kg per k mole of air per 100 kg of fuel. Now, based on the given data, the partial pressure of water vapor in the air remembers the air has 75 percent relative humidity. So, the partial pressure of water vapor in the air is 0.75 times P sat of 298 Kelvin, it works out to 2.37675. And so, the mole fraction may be evaluated as the partial pressure divided by the mixture pressure, which is 100 k power. So, the mole fraction of water vapor is 0.0237675. So, basically the income of water vapor incoming air has O2, N2 and water vapor. Now, let us assume that the ratio of the mole fractions of N2 and O2 to be 3.76, which means that the mole fraction of O2 plus N2 plus H2O should be equal to 1. That is all that is in the incoming air. And if you use the fact that the ratio of mole fraction of N2 and O2 is 3.76, we can rewrite this expression like this. And we may finally get the mole fraction of O2 to be 0.2051. So, each k mole of O2 in the humid air is accompanied by 3.76 k moles of N2 and 0.118 kilo moles of water vapor. So, basically what this says is that the composition in the incoming air in mole fraction is YO2 equal to 0.2051, YN2 equal to 3.76 times 0.2051, YH2O equal to 0.0237675. So, that is the composition of the incoming air. So, every k mole of O2 is accompanied by 3.76 k moles of N2 and by 0.11588 k moles of water vapor. So, the reaction for complete combustion with 185 percent theoretical air may now be written like this. What is that? We have simply added this to the incoming air stream and the amount of water vapor has to be adjusted like this. So, now that we have the balanced chemical reaction, we may evaluate the mole fraction of water vapor in the product stream and from that we can work out the partial pressure of water vapor in the product stream and from that we can evaluate the dew point temperature of the products to be 47.7 degree Celsius. So, the humidity in the incoming air elevates the dew point by 3.7 degree Celsius. That is why we worked out the humidity of the product stream with dry air and humidity with humid air because the air that we are taking in is almost always humid. It would have the effect of elevating the dew point temperature of the exhaust stream or product stream which is actually not desirable because then the products of the water vapor in the products is much more likely to condense before it leaves the reactor. So, that is why we actually did this calculation. So, with 75 percent humidity we can see that for this particular case the dew point temperature is elevated by 3.7 degree Celsius. So, what we will do in the next lecture is to prepare ourselves to apply the steady flow energy equation to combustors in which combustion reactions are taking place. So, that would be application of first law to combustors. Although we will deal with steady flow reactors extensively it is not very difficult to apply whatever theory we are developing or whatever procedure we are developing in cases where the combustion reactions take place in a non-flow process. For example, the combustion reactions take place in a steady flow reactor in the case of a cast-turbine engine whereas in the case of an SI engine or CI engine combustion reaction takes place in a non-flow process. But the theory that we are developing applies without any difficulty or additional difficulties to non-flow cases as well.