 So, in this lecture, we are going to look at the application of first law to combustion systems, basically first law and later on we will look at application of entropy balance also to combustion systems. In both cases, we actually will look at steady flow situations, but as I said before, application of first law to combustion systems which are non-flow systems is really straightforward. Many examples or at least one example of that is given in the textbook. So, I urge you to look at the textbook for that example, because all the examples that we are going to work out here will be steady flow examples. Now, before we can actually start applying steady flow energy equation to combustion systems, we need to revisit the concept of enthalpy as we have done it so far in the previous course as well as up to now in this course. So, you may recall that specific enthalpy h is defined as u plus pv where u is the specific internal energy and pv as you know is equal to RT as long as we take the gas to be perfect gas which obeys the ideal gas equation of state. Now, the specific internal energy for ideal gases is related to the modes of energy storage in molecules. This is something that we that I discussed in detail in the first level thermodynamics course. So, if you are not familiar with this idea, I suggest that you look at the video on pure substances modes of energy storage for ideal gases. So, in compact terms, the modes of energy storage in molecules are translation, rotation and vibration. These are the modes of energy storage that you will see in general. For a monatomic gas for example, there is only translation rotation and vibration are not available for diatomic and polyatomic gases, definitely rotation and vibration or modes of energy storage. Most importantly, the energy stored in these modes is dependent only on the temperature of the gas. And such a gas is typically called thermally perfect because its internal energy is dependent only on the temperature of the gas. So, for our purposes, we have actually simplified this in some instances some more and said that the specific heat capacity Cv or Cp is also a constant and which does not change its temperature that led to the so-called calorically perfect gas model, but that is not necessary. We have also done calculations in this course earlier using variable Cp for air and so on. So, that is a simplification, not necessarily a limitation. So, thermally perfect is adequate, thermally perfect plus ideal gas equation of state is what we prefer to use in this course. Now, we can fully account for energy that is stored in these modes because it is only temperature dependent and in the case of pure substances which are not ideal gases such as water for instance and refrigerant, we resorted to the tabulated form of internal energy because there the specific internal energy is a function of both temperature and pressure. It is a function of temperature only for thermally perfect gases. Now, if you look at the internal energy, it is made of generally we look only at the translation energy stored in translation, rotation and vibrational modes, even vibrational modes are not excited until the temperature, until and unless the temperature is very high, greater than let us say 1500, 1600 Kelvin or so. Now, normally the energy that is stored in the bonds, chemical bonds that make up the molecules is not accounted for. It is somewhat like dead storage in a reservoir or dam and we need not really account for it as long as there are no chemical reactions. For example, we may have O2, N2 and CO2 forming an ideal gas mixture or we may have O2 and N2, it is then mixed with CO2 to form an ideal gas mixture. In all these cases, CO2 remains as CO2, O2 remains as O2 and N2 remains as N2. Or if you consider psychrometry for example, we had moisture which was treated as a mixture of two ideal gases, air and water vapor. But the composition of the water vapor and the air or the individual components in the mixture remain the same, which means that the energy stored in the bonds remains intact and we need not really account for that. Now once chemical reactions begin to take place, we could have a new product species which are formed from the reactant species, which means that bonds in the reactant species are broken and new bonds are formed to create new product species, which means the energy that was stored in the bonds in the reactant species and the energy that goes into the bonds in the product species all has to be accounted for, all have to be accounted for and that is what we are going to do first. So this means that the specific enthalpy, the expressions that we use have to be modified. The definition itself does not change, the definition for specific enthalpy is always h equal to u plus pv. So we need to carefully look at the expression that we have been using and modify it so that we can take into account the energy that is contained in the bonds. Let us see how we do that. So for any chemical species, the enthalpy at a given temperature T is defined to be the sum of the enthalpy of formation, notice that we are encountering this term for the first time. So the enthalpy of formation is denoted hf0 of T rough, f refers to formation and the superscript refers to standard state. Standard state is 298 Kelvin and 1 atmosphere pressure. So the enthalpy at a given temperature is defined as a sum of the enthalpy of formation and a sensible enthalpy denoted delta h bar and again you may recall from the previous course that an over bar is used to denote the fact that this particular quantity is measured on a per k mole basis not per kg basis but per k mole basis so the over bar is used here also. So h bar of T which is a specific enthalpy on a molar basis so this means that this has units of for instance kilo joule per k mole and not the kilo joule per kg that we are familiar with. So h bar of T is a sum of enthalpy of formation plus sensible enthalpy. Notice that sensible enthalpy is a function only of temperature. So what we have been doing so far, what we have been calling enthalpy so far is actually the sensible enthalpy. So this was what we have been calling enthalpy so far but enthalpy is actually the sum of sensible enthalpy plus enthalpy of formation and as long as there are no chemical reactions we can pretty much ignore this and work only with the sensible enthalpy. Now sensible enthalpy is a function of temperature and for combustion calculations what we will do is use table for looking up values of sensible enthalpy. Of course curved field polynomial for Cp bar may also be used and we can then evaluate the sensible enthalpy h bar as Cp bar times T that is possible but we will use tabulated values of the enthalpy itself in this particular course. So that is the most important thing we need to account for the energy that is contained in the bonds of the molecules or the chemical species when we are doing combustion calculation because reactant species are being dissociated or bonds in the reactant species are being broken and bonds are being formed new bonds are being formed to create product species. So basically the enthalpy of formation is nothing but the energy that is is more or less the energy that is contained in the bonds or it is actually the energy that is required to form the species from its constituent elements and it is more or less very close to being the energy that is contained in the bonds as we will show in a minute but strictly speaking it is defined as the energy that is required to form the species from its constituent elements. Let us look at how the energy contained in the bonds how well this approximates the enthalpy of formation. So here is a table containing bond energies of different types of bonds it is a very short list and it contains bonds that we encounter customarily. So you see single bond H H single bond C C and double bond triple bond C C and so on and double bond C O and so on and so forth. So these are bond energies in the gas phase. So let us note a few things about the bond energy bond energy is defined as the energy that is required to break the bond. In other words if I have let us say a vessel which contains H 2. H 2 molecule is nothing but H H the single bond H H. So if I have a container with H 2 then if I supply energy or the energy that I must supply to break the bond or dissociate H 2 into two atoms of H is called the bond energy. So that is the energy that is used to that is required to break the bond and dissociate the molecule and convert it into its constituent atoms and our sign convention for heat given to a system is positive which is why this has a positive sign consistent with the sign convention that we have been following in thermodynamics. If for instance we have a container with let us say atoms of H and we want to form let us say molecules of H then energy must be removed from the system and we must form the bond. So that means energy is released when two atoms of H combine to form a molecule of H 2 that then in that case energy would be negative in keeping with our sign convention. Bond energy for the double bond are then those of a single bond for instance you see here single bond CO has a bond energy of 0.36 times 10 raise to 6 whereas a double bond CO has a bond energy of 0.804 times 10 raise to 6. A single bond CC has an energy of 0.348 times 10 raise to 6 whereas a triple bond CC has an energy of 0.814 times 10 raise to 6. So bond energy constitutes a major part of the energy contained in the molecule you will not go into the details or nuances of this but it constitutes a major part of the energy that is contained in the molecule. And which is why the bond energy is very well approximated or very well approximates the enthalpy of formation. Let us work out a couple of examples to indicate this. Determine the enthalpy of formation of gaseous methane and acetylene that is I will leave it to you as an exercise to do ethylene. Let us do gaseous methane and acetylene. Assume the enthalpy of formation of carbon in the gas phase from the element carbon in the solid phase to be 0.7184 times 10 raise to 6 kj per k bond. So gaseous methane contains four CH bonds. So you may recall that gaseous methane CH4 looks like this. So it contains four CH bonds. So the four hydrogen atoms have to come from breaking down the bonds in a hydrogen molecule. In other words we have to make this reaction happen. So to get the four hydrogen atoms. So two hydrogen molecules have to be broken to get four hydrogen atoms. So energies needs to be supplied for that. So now and the carbon here has to go from the solid phase to the gaseous phase. The energy has to be supplied to take the carbon in the solid phase to the gaseous phase. Now we have a soup of atoms in the gas phase carbon atoms and hydrogen atoms. So four hydrogen atoms will combine with the carbon atom through a CH, each through a CH bond. So that will cause energy to be released. So now we put together all these energies and then calculate the enthalpy of formation approximately of CH4. So here is the energy that needs to be supplied to take carbon in the solid phase to carbon in the gas phase. Notice that this comes with the plus sign because we have to supply energy to do this. This term represents the energy that is required to take to break down two molecules of hydrogen into four hydrogen atoms. Again it is positive because we have to supply energy to do this. So if you look at the previous table, HH has a bond energy of 0.436. So that is the energy that is required to break the bond, HH bond in a hydrogen atom. Now the formation of four CH bonds requires this much energy with a negative sign. Again notice that CH bond requires 0.413 times 10 raised to 6 or CH bond has a bond energy of 0.413 times 10 raised to 6 which means that I need to supply 0.413 times 10 raised to 6 kilo joule per kilo mole to break a CH bond. To form a CH bond I have to remove this much energy that is 0.413 times 10 raised to 6 kj per kilo mole. So or alternatively, stated alternatively when a CH bond is formed from its constituent elements 0.413 times 10 raised to 6 kj per kilo mole of energy is released which is why this goes in with a negative sign. So if you sum it all up we get the bond energy contained in CH4 to be minus 61600 kilo joule per kilo mole. The actual value is minus 74000 kilo joule per kilo mole it is reasonably close. Let us next look at acetylene. So acetylene contains so if you recall acetylene is C2H2. So it contains a carbon-carbon triple bond and two CH bonds single bond. So the two H atoms are or produced by breaking down one hydrogen molecule and the carbon two carbon atoms have to be taken from the solid phase to the gaseous phase. So from this soup of atoms one carbon-carbon triple bond and two carbon hydrogen single bonds have to be formed. So let us calculate the energy associated with each of this bond breaking and bond forming. So this is the energy that we need to supply to break the bond or to dissociate a hydrogen molecule. This is the energy that we need to supply to take two carbon atoms from the solid phase to the gaseous phase and both these terms go with the plus sign. This is the energy that is released when CC triple bond is formed from two carbon atoms and this is the energy that is released when two CH single bonds are formed from the constituent atoms. So the final value comes out to be 232800 kilo joule per kilo mole. This is the energy that is contained in the bonds and you can see that this is reasonably close to the enthalpy of formation of acetylene which is listed at 226736 kilo joule per k mole. So you can see that energy contained in the bonds accounts for most of the energy that is required in forming the species from its constituent elements. Accurate measurements and values are available but this procedure although it is slightly less accurate than the actual values indicates the nature of H of 0. So this shows that H of 0 is largely or the contribution to H of 0 comes largely from the bonds that make up the species. That is the important takeaway from this example. I will I will let you work out the bond energy contained in ethylene C2H4 and compare with the book value or literature value. So now we are in a position to carry out first law analysis of combustion systems. So the most important change that we are seeing here is the definition of is our understanding of the specific enthalpy of species. So far we had basically so far we had neglected H of so basically we had neglected H of so far and we had said that the specific enthalpy of species is simply its sensible enthalpy. So the modification that we are making now is that the specific is that the enthalpy of formation has to be accounted for and we will account for that in our in our expressions and calculations. But the modification to the energy equation or SFE that we derive is nil. So there is no modification to the energy equation that we derived. So whatever we had called enthalpy in our steady flow energy equation is still enthalpy. The only difference is the expression that we use for calculating the enthalpy. In the case of gases we said that enthalpy was equal to the so we said enthalpy was equal to the sensible enthalpy. Now we are saying enthalpy is enthalpy of formation plus sensible enthalpy. So only the expression for calculating enthalpy has changed not the terms that were you know that had enthalpy appearing in them. That is why I said the definition of enthalpy is still h equal to u plus pv nothing has changed that still holds.