 We're now going to talk about a topic referred to as being excess air, and this is something that we do with Rankin cycles for the Brayton cycle, where we would have a combustion chamber, and you put in excess air in order to lower the temperature and reduce NOx formation. You would not do this within an auto cycle with the internal combustion engine, because there when you have too much air, you go above stoichiometric quickly, you do get NOx formation. But that's not the case when you're dealing with a traditional burner. So this idea of excess air will take a look at it in this segment. So we talk about this as being excess air, and with the Brayton cycle, within the burner if you recall, when we looked at the Brayton cycle, we said that this helps reduce the hot spots of the gases coming out of our combustor, which then eventually go on into the turbine. And with the Rankin as well, quite often the Rankin, you'll be burning coal, and by adding excess air, we're reducing the temperature and reducing the formation of NOx, which we saw earlier, leads to photochemical smog, and we like to reduce that. So let's take a look at this idea of excess air. So you can have different amounts of excess air. For example, you might have 50% excess air. And what that would translate into is that we're 1.5 times the theoretical air, which we saw you were able to calculate by doing the stoichiometric balance. You could have another example, 150% excess air. And for that, what we would be doing would be 2.5 times the theoretical amount of air, or the stoichiometric amount. So the result of this, the net result, is that we are going to have excess oxygen within the product stream. And so if we're told that we have excess air, what we need to do is go through and do the stoichiometric balance, and then adjust it for the fact that we do have excess air. So we're going to start with an example by looking at a very simple combustion reaction, and that is of ethane gas. And so what we're told, we have combustion of ethane gas C2H6. And we're told that it is taking place in 200% excess air. So the way that we would go about solving this, we would do the stoichiometric balance first. So that's the first thing that we will do. So writing out our equation. OK, so there we have our equation. Now what we need to do is do a balance, a mass balance of each of the individual components that we have within the reaction. So like before, we'll begin with carbon. And so looking at carbon, we can see we have carbon here, and we have carbon there. And those are the only two places where we have carbon in our reaction equation. So we can write 2 on the left, and that is equal to x on the right. So that identifies x, which is good. Next thing we'll look at is hydrogen. And so balancing hydrogen, we have hydrogen here. The other place we have hydrogen in the water vapor, those are the only two places that we have hydrogen. So we can write 6 on the left is equal to 2y on the right. And with that, we can directly solve for y, and we get 3. So that gives us 3 for hydrogen. O2, we'll do diatomic oxygen. And we have O2 there. And then on the right hand side, we have a little bit of O2 there. And we have monotonic oxygen here. Well, it's not monotonic because it's bonded with H2O. But anyways, we have oxygen in the water vapor. And so then what we have for the left hand side, we have a theoretical. And on the right hand side, we have x. And then we have y. And we need to divide that by 2 because notice we don't have O2 in the water vapor. We only have oxygen. And so with that, we can then solve because we know the value of x. We've solved it here, and we know the value of y. So we can plug those into this equation. And what we end up with for the theoretical error is 3.5. Now, the last thing we will do is we will do a balance with nitrogen. So let's take a look at diatomic nitrogen N2. And on the left hand side of the equation, it appears here. And on the right hand side, it appears there. And so with that, on the left hand side, we have 3.76 times a theoretical is equal to on the right hand side, z or z, depending which country you are in. And with that, we get 13.16. And so that becomes the stoichiometric balance. And so what we can do is rewrite our equation before introducing this xs error that we have in the problem statement. So let's rewrite it with the values that we've just solved for. So that's our equation. Now what we're going to do, we're going to introduce this xs error. And we were told that we have 200% xs error. And that means that we're dealing with 3 times the theoretical error. So dealing with 3 times a theoretical error, what I'll do is rewrite our reaction equation. And if we're dealing with 3 times a theoretical, that means that we're dealing with 3 times a 3.5. So that's going to be 9, 10.5. So we have 10.5. Multiply by the atmospheric error. And then on the right hand side of the equation, the carbon is not going to change. So we can keep the 2CO2. The water vapor will not change. So we keep H2O. But what will change is that we're going to have some unknown amount of free oxygen. Because that's what we said happens with xs error. As well, the amount of nitrogen is going to be adjusted. Because we've added more kilomoles of atmospheric error on the left hand side with our reactant. So what we need to do is go through into our balance again. And for this, what we're going to do is a balance of nitrogen to begin with. So on the right hand side, we see we have nitrogen there. And on the left hand side, it is unknown. And that's what we're going to try to determine. So we have 10.5 times 3.76. And with that, we end up with the value that we can then put into our reaction equation. And that value is 39.48. So that gives us the value for nitrogen. And next, we will look at oxygen, diatomic oxygen. Locations of diatomic oxygen, we have it there. And we also have diatomic oxygen there. We have to account for it with the water. And then we have xs oxygen on the product side. And so with that, on the left, we have 10.5. And now that needs to be balanced with CO2. We have two kilomoles. In water, we have three, but we have to divide by two. Because we do not have diatomic oxygen there. And plus some unknown, and I will call that unknown y, which is what we have prescribed there for the oxygen. And when we go through and we solve this, we find y is equal to 7. And so with that, we can take it and put it into our reaction equation and we get that. And so there is our final reaction equation with xs air. And the main thing to notice is that we have a new free oxygen, diatomic oxygen on the product side. And we've also adjusted the amount of nitrogen because we have xs air coming through and consequently nitrogen, which is usually inert and along for the ride, if all goes well, then we would have to increase the amount on the right hand side as well. So that is an example of how you can deal with problems involving xs air. And again, remember, this typically is not done if you have an auto cycle. You can lead to increased NOx formation with the auto or the internal combustion, but it's used for rank and cycles burners where you have a burner and you want to reduce hotspots in the burner. So that is xs air and that will conclude this segment.