 Okay, in this lecture what we're going to be doing, this is a second segment and we're going to be doing an example problem looking at applying the first law to a system whereby we have chemical reaction, oxidation, combustion. And so what I'll do, I'll begin like every other time writing out the example problem and then we'll work through what we know, what we're looking for is schematics and then we'll continue to solve it. So there is the problem statement. What we have is octane gas. Oops, sorry, there's a typo there. Octane gas. And that's C8H18, 25 degrees C, burns steadily. So we will have a uniform flow process, steady flow. We're told that we have 30% excess air and 60% relative humidity. And so we're going to have to do the excess air calculation that we saw in an earlier lecture. Assuming combustion is complete and products leave the combustion chamber at 600 Kelvin, determine the heat transfer for this process per unit mass of octane. So that's the problem statement. What I'll begin by doing is writing out what we know, what we're looking for as well as a schematic. So there is what we know, what we're trying to find, we're trying to find Q and we have our reactants coming in and our products leaving, we don't know what the products are yet. We will get that by doing these stoichiometric analysis. So let's proceed solving the problem now. The first place to start with these is doing the chemical balance with the stoichiometry. Then we'll proceed to excess air. And then after that we'll look at the moisture and then we'll proceed to the first law formulation. So there we have our reaction. What we're going to do is a mass balance for each of the different sides. So beginning with carbon, we have 8 on the left and on the right hand side, we have X. So that defines X for us. Now for hydrogen, we have 18 on the left and on the right. Now we only have hydrogen in the water and consequently we have 2Y. That defines Y as 9. Now for diatomic oxygen that helps us get theoretical air and we can plug in the values of X and Y and find that to be 12.5. And then finally for nitrogen on the left it's 3.768 theoretical and that equals Z or Z on the right hand side, 45.12. So there we have our stoichiometric balance. The next thing we're going to do, we have to rewrite the equation now and do our excess air calculation. And if you're called with excess air, we said it was 30%. So that means that we're dealing with 1.3 times the theoretical air, giving us 16.25 will be in the equation. That will leave free oxygen on the product side of the equation. So what I'm going to do in the next segment or right now, I'm going to rewrite that equation. So you'll notice we do not know how much oxygen, free oxygen we have on the product side. On the left we know diatomic oxygen has 16.25. And then on the right it is in the carbon dioxide, we have 8. We also have it in the water vapor 9 over 2 plus unknown, unknown we can solve for 3.75 and that enables us to take that and plug it into the equation. So there we go. We have our stoichiometric balance for the reaction equation now. Now if you're called for the question, we said that there is water vapor coming along as well. So we need to be able to handle the water vapor. It won't react, but it does get carried along. So let's take a look at that. So what we'll do, we'll look at the partial pressure. We know the relative humidity. So we can look up the saturation pressure from steam tables for the water vapor. And with that we can figure out the partial pressure attributed to the water vapor in our air. We get about 1.9 kPa for the partial pressure. So what we want to do, we want to figure out how many moles of water vapor is in our reaction equation. So that's what we're going to do next with this value. So the question that we ask is how many kilomoles of water vapor are in 16.25 times 4.76 kilomoles of dry air. The 16.25 is coming in from here and we're assuming that this here is dry air. And so what we want to do is figure out how many kilomoles of water vapor we have. And so what we're going to do, we use the ideal gas equation and we can use this now to come up with a ratio. And what we want to do now is solve for the number of moles of water vapor, which is nv in this equation. So we find this for the number of kilomoles of water vapor. We're now going to take that and we're going to plug it back into our stoichiometric equation with excess air, this one here. So let's go ahead and do that. And that will give us the equation with which we will apply the first law to. So this is our final reaction equation and this is the one that we will be using in order to do our analysis, which we'll do in the next part. We'll start applying the first law to this equation moving us towards enabling us to determine how much heat release is coming as a result of this reaction.