Thermodynamics: Combustion with excess air, dew point of combustion products (50 of 51)

>> OK. Well, good morning everybody. So once again anybody who didn't pick up homework for the last week they're up here at the front. Also a reminder that homework solutions right to the end of the quarter have been posted, in the case, outside my office. So please make sure you look at those. They'll be there, you know, right until your final exam is done, so no need to rush over there. Nonetheless and also a reminder that you're going to be turning in homework from last week to this week, certainly you're responsible for that material I explain this last week but, you know, you're not going to turn it in.

So the homework you've already turned in will represent your homework grade for this class. OK. So, today we're going to continue talking about combustion. We may get through it all. We may have to spend the few minutes in the beginning on Friday. Nonetheless, let me also just remind you that your final exam is a week from today at 9:10 in the morning, so not the usual class time, right? Where the morning final exam period, so please make sure you're here on time.

Also let me note that this Friday is the Engineering project symposium, it does not affect this class at all we're still going to meet here I assume in the same classroom. Certainly I would encourage you to look at some of your friends or classmates projects but you know hopefully they're going to give their presentations at 8:30 or 9 in the morning and then it's not going to interfere with this particular class. Nonetheless, class will still be held on Friday just so you're aware of that. OK. Also let me just note that Friday will be a general review day, question and answer day, teacher evaluation day. So there's a lot of things that we're going to be dealing on Friday and certainly we're going to talk more about the final exam and make sure you understand the format. So, anyway and of course I already mention many times that on the final you can use a three and a half by five note card with equations and diagrams only written on both sides if you choose to.

So certainly you can start putting those together. All right, so now back to combustion. So remember that when we're talking about a combustion process we're using a molar analysis. We're looking at the number of moles, if you will– the equivalent number of atoms or molecules of a particular substance. The mass of these substances doesn't really matter from the standpoint of our combustion equation. Certainly we're interested in the ratio of air to fuel and that's something we can calculate as we've seen last week. But the basic analysis always starts with a molar analysis. So we're going to have some fuel, right, or the form CaHb.

We're going to have a certain amount of oxygen that's required and that amount of oxygen is this a plus b over 4, we find this based on our balance equations that we've seen last week. We also recognize that in general we have air that we use in for combustion and not just pure oxygen. So we know that in the atmosphere the ratio of nitrogen to oxygen by volume which is the same as the molar fraction or molar percentage it is just 3.76.

So we've got 3.76 times more nitrogen than oxygen, right? The 79% versus 21%, so this would be the reactant side of the equation. And then on the product side, you know, all of the carbon is going to end up with carbon dioxide. All of the hydrogen is going to end as water. In this particular case because we're talking about chemically correct, chemically balance what we call the stoichiometric. There's not going to be any excess oxygen that's going to be showing up on the right-hand side as a combustion product.

But certainly all of the nitrogen that we start with is going to end up as a combustion product. So we have the same a plus b over 4 times 3.76 N2. So this is the general equation that we're going to be solving for. Again though, only if we are stoichiometric, so we will just call this the stoichiometric equation. I'll often just abbreviate stoichiometric with just stoic and that's the equation we've seen so far. Now we do understand that in the real world we are going to need some excess air to guarantee or at this mostly guarantee that we are going to be able to move to completion in a combustion process. So we'll get to that real shortly but let me just start or I should say continue with an example problem. In fact before that– I do want to also put down the finals we've got this hour. Since both the final exams that I'm giving this quarter on Wednesday, I'm going to have all my office hours on Monday and Tuesday.

So, Monday from 11 to 1 and then Tuesday 1:30 to 3:30 p.m., so hopefully if you want to see me next week before your final exam you'll be able to come see me then. OK. So, back to combustion. Let's write this stoichiometric equation for propane and let's find there the fuel ratio. So it's really pretty straightforward, I mean the first thing you want to do is go into your tables in the appendix of your book, table A1 or A2 has this data and just look at propane and you'll find that propane is C3H8. So from table A1 we can immediately find what our chemical formula is for propane. We can also find our a plus b over 4 term, please note that we can also call this the stoichiometric coefficient and obviously this is just 3 plus 8/4, so that's just 5. OK. So, we can go ahead and write the combustion equation now.

So, we know that we need 5 oxygens along with the nitrogen. Let me just rewrite the left-hand side of the equation. Again C3H8, 5 O2 plus 18.8 N2, so those are the reactants, right? And then the combustion products that result from this, well we're going to have 3 carbon dioxides, we're going to have 8 over 2 waters and then again we're going to have all of the nitrogen. So just rewriting this, we have 3 CO2s plus 4 waters plus 18.8 nitrogens and there's the stoichiometric equation that you have to generate.

OK. Now, to find the air to fuel ratio we need to use one of the equations that was presented last time. The air to fuel ratio is just the mass ratio, right? This is the mass of the air divided by the mass of the fuel and there's many different ways that we can find this. I mean probably the easiest way is just to note that the amount of air that we have is going to be the stoichiometric coefficient times 4.76, right? We have 1 oxygen and 3.76 nitrogens. So we have a plus b over 4 and then times 4.76.

So this just gives me the number of moles of air but then we have to multiply that by the molar mass of air and then the denominator will have the number of moles or the fuel times the molar mass of the fuel. By the way, this a plus b over 4 times 4.76, this is the number of moles of the air, right? So it's an air times molar mass of air, gives you the mass of air and the number of moles in the fuel times molar mass of the fuel, gives you mass of the fuel, right? So it's just a matter of plugging in numbers, 5 times 4.76.

I have 28.97 is the molar mass. In fact, I should really put units up here. So this is the number of moles, I'll just write this as kilomoles of air and then 28.97 kilograms per kilomole of air. So there's a numerator and then the number of moles of fuel. Well, we're always starting with just one single few molecules, so just one mole of the fuel. And then the molar mass is something that we'll actually have to get out of table A1, so when I was back here in table A1 and I pulled out the, you know, chemical formula from– for propane, I should have also pulled out the molar mass. So the molar mass is going to be 44.1 kilograms per kilomole.

So also from A1, it's also in table A2. So 44.1 kilograms per kilomole of the fuel, and again just a little bit of math we got 15.6. So this is kilograms of air per kilogram of fuel. This is not a dimensionless member, even though you have mass unit than both numerator and denominator. You should get into the habit of always writing out your mass units. I mean we've talked about this with regards to ideal gas mixtures and certainly we've talked about this with regards to air, water vapor mixtures which is you know, the primary ideal gas mixture we're talking about in this class.

But it certainly applies here too. So we've got air in the numerator, fuel in the denominator. They'll cancel units out. They're not the mass of the same thing. They're different masses of different things, right. So, are there any questions just on how or when to approach this type of problem? OK. So now we have to ask ourselves, is this really reasonable? Is this stoichiometric equation appropriate for our combustion process? And the answer is generally not. That doesn't mean we can't burn a stoichiometric mixture of air and fuel but in the real– in this real world of ours it's not likely that that combustion is going to go to completion. We're not going to end up with just carbon dioxide and water vapor, right? We're going to end up with other oxides of carbon and hydrogen, HO, CO and those are not representative of complete combustion.

We don't get as much energy out of a fuel unless it burns entirely and becomes carbon dioxide and water vapor. So we're going to add some excess air for our combustion processes. So now, let's look at actual combustion– — using excess air. Now when we talk about excess air, we're talking about something in addition to the air that's required for stoichiometric combustion. OK. And it's always going to be presented either as a fraction or as a percentage. So if we're talking about 20% excess air then we're just going to go here and to our stoichiometric coefficient and we're going to multiply that by in additional 20%.

In other words, there's going to be a term right in here which is going to be 120%, right? If we have 20% additional, we need 100% of the stoichiometric coefficient for the combustion to complete. But then we need that extra 20% to make sure that we have enough air molecules, really oxygen molecules in the vicinity of the fuel especially combustion actually moves forward. So as for combustion using excess air, we basically just multiply the stoichiometric coefficient by a fraction.

And this fraction would appear in our combustion equation. So let me first just write this in general terms. So we have CaHb and now we're going to have instead of just a plus b over 4 and the O2 plus 3.76 N2. Instead of just having that, this is where we're going to have an additional fraction. So this fraction of it is what I personally call the theoretical air fraction. OK or just TAF for short. The author talks about theoretical air but always refers to it as a percentage and doesn't actually use the fraction form. But you have to use a fraction form and you can't use percentages in calculations, you're using fractions.

So this theoretical air fraction is what's going to go right in here. OK. So, for instance, if the theoretical air fraction equals 120% or let's say 1.2 then how much excess air we use? Well, 20% excess air, right? We've got the 100% needed for combustion and then the extra 20% to make sure that the mixing takes place properly. So if this is the case, well then the excess air is going to equal 20%, right? Twenty percent or just 0.2, in fact, we call this 20%, we call this the excess air fraction. And personally I just call it EAF for excess air fraction and you should be able to see pretty clearly then that the theoretical air fraction is just 1 plus the excess air fraction, right? Now, you can certainly write this as a percentage if you're given percentage of theoretical air of a 120% then you know we would just say that the theoretical percent or the theoretical air percentage is 100% plus the excess air percentage. So read the problems carefully and make sure you know whether you're being given a fraction or whether you're given a percentage in your particular problem.

OK. So let's go back to the combustion equation and let's finish it up. So what I'd like to do is go to the same general derivation as I did last time but now considering excess air. So let's just write it this way, CaHb and then plus. Now here we're going to have some sort of a coefficient and why don't I just call this coefficient d. So, that's the amount of oxygen we need plus 3.76 nitrogen that goes along with it. Now, frankly we already know what that is. I mean logically we talked about it, I'm going to derive it but that's just going to be the theoretical air fraction times a plus b over 4 times the stoichiometric coefficient.

We know that's what it's going to be but let's actually derive that so there's no confusion. Anyway and then this is going to create a certain amount of carbon dioxide, certain amount of water. Now here, there is going to be excess oxygen, right? So we can't ignore like we did in the stoichiometric equation. In the stoichiometric equation there was no excess oxygen, all of it was used to turn carbon and hydrogen into carbon dioxide and water vapor. But now we have excess, so now we have to include the excess. And then lastly, there's going to be a certain amount we'll call it w of nitrogen.

And then what I want to do is the same thing that I've done before is I really just want to do my molar balance and like I did before when we talk about stoichiometric. I'm going to put the right-hand side first which is our product and then the left-hand side on the second column which is our reactants. And we'll just start with carbon.

So, well this we know, right? We know that we're going to end up with x carbon dioxides and to do so we need to have the same number of carbons in the fuels we have carbon dioxides so clearly we're just going to have x equals a. And that's my carbon balance. Next, I'm going to do the hydrogen balance. So here we're going to end up with y hydrogens and for this to balance properly this is going to have to equal v over 2, right? So if for instance if we have let's say 4 hydrogens but then 4 hydrogen atoms means that we have to have 2 H2Os. So we have exactly half the number of this, half the number of hydrogen reactants as hydrogen as H2O in the products.

All right, so that's the hydrogen balance and next, let's do our oxygen balance. So, on the right-hand side, we've got x O2 as part of the CO2. We have y over 2, O2 as part of the H2O again we're doing an O2 balance. So if we have y elemental oxygens, we'll have half that number of O2 molecules, right? Molecule is twice as big. We have had for the many. All right, so we have x plus y over 2 and then the excess oxygen z and then this all has to equal d. Now here, I just want to go through a little bit of mathematics. So first, I just want to rearrange. So, z is going to equal d minus a plus b over 4. OK. But isn't d just my stoichiometric coefficient times the theoretical air fraction. So that's the theoretical air fraction times the stoichiometric coefficient and again the stoichiometric coefficient we know it's a plus b over 4. OK. So this is going to actually be rewritten.

Now, the theoretical air fraction is this 1 plus the excess air fraction. So 1 plus EAF times the stoichiometric coefficient or since we know what this stoichiometric coefficient is, this is just a plus b over 4. All right, and just continuing. Now we can actually plug this in above. OK. So we're going to end up then with z equals and then 1 plus the excess air fraction times the stoichiometric coefficient and then minus a plus b over 4. Well but that's the stoichiometric coefficient again, right? So basically this 1 times the stoichiometric coefficient is going to cancel with the other stoichiometric coefficient. So we just end up with z and this is just going to equal the excess air fraction times the stoichiometric coefficient. OK. So we have the value of z and then last but not least, let's go to our nitrogen balance. We have w nitrogens and this is got to equal d times 3.76.

But we know that d is related, it's going to be this term up here, so this is a 3.76 times the theoretical air fraction. I'm sorry, 3.76 times the theoretical air fraction times the stoichiometric coefficient. So, 3.76 and then the theoretical air fraction and then the stoichiometric coefficient which is again just a plus b over 4, right? So, 3.76 times the theoretical air fraction times a plus b over 4.

OK. All right, so now we can go back to our general equation and just simply rewriting. So we have CaHb plus– and that we know that d is just our theoretical air fraction, times the stoichiometric coefficient which is just a plus b over 4. So that term n is multiplied by O2 plus 3.76 N2. OK. And then on the right-hand side, well we have x which is just a carbon dioxides. We have y which is b over 2 waters. We have z and then z is just going to be– well the x is their fraction times the stoichiometric coefficient. I'll just write this is a plus b over 4 and this is the oxygen term. And then lastly we have w which you've already seen, so that's just 3.76 and then multiply it by the theoretical air fraction and then multiply it by a plus b over 4. And then nitrogen, OK, all of the nitrogen and that's coming in as a reactant has to leave as a product.

So we're going to actually modify this equation just a little bit if we wanted to but you don't absolutely have to. It really just depends on you, why don't we just leave it in this form. So as long as we know what our fuel is and as long as you know how much excess air we have, we should be able to go through this complete combustion equation, right? We know a, we know b, we know the theoretical air or the excess air fraction one of the other, everything else is just based on those, right? So we should be able to go to this calculation pretty easily.

Now, once we've gone through our combustion equation and I would not call this a stoichiometric equation anymore. This is a combustion equation, it's not stoichiometric. So we don't use the word stoichiometric. This is our combustion equation. Once you've gone to our combustion equation our molar analysis, well now we can do the other things that we're interested in. One of them is to find the air to fuel ratio. Now the air to fuel ratio is still defined the same way, it's still just the mass of the air divided by the mass of the fuel. It's still just going to be the number of moles times the molar mass of the air divided by the number of moles times the molar mass of the fuel. So none of this really changes, we would still note that the number of moles of the fuel is still just going to equal 1.

I mean we're always talking about a single fuel molecule, so that's never going to change. But the number of air moles is different than it was previously when we talked about stoichiometric combustion and that's simply because we have excess air, right? So we're going to take the same numbers that we had before and we're still going to have 4.76 times the stoichiometric coefficient that's going to represent the number of moles of air we would have if it was stoichiometric combustion. But then we're just going to simply multiply by the theoretical air fraction to give us the extra number of moles that we require. So this is just going to be the theoretical air fraction then times 4.76 and then multiply it by the a plus b over 4. OK. Previously we just had 4.76 times a plus b over 4, now we have to consider the excess air. So, theoretical air fraction times that. So, when you're solving a problem like this, please make sure you use the right number of moles for your air. If you're just using the stoichiometric data, your numbers are going to be wrong.

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OK. So this is just something that you need to note. And then one other thing that we're interested in and this is the dew point. Now, I know what you think and wait a minute dew point has to do with air and water vapor mixtures, doesn't it? Well, it has to do with air and water vapor mixtures as we've discussed it to this point but that doesn't mean that you can't have a dew point for other mixtures that contain water vapor. Like for instance, I don't know how about combustion products? There's a reasonable amount of water in the combustion products, right? All of the hydrogen that was in the fuel is going to become water vapor. We certainly don't want that water vapor to have a temperature of below the dew point because if we do then the water is going to condense, right? The water vapor is going to condense into a liquid water. Now, that may not sound like a big deal.

It's like. OK, so what, right? But I think I mentioned this last week when you have a combustion process in the real world, some of this nitrogen is actually going to dissociate and then recombine with oxygen in the form oxides and nitrogen, NOx if you will which when interacting with water in the environment is going to turn into nitric acid or various acids of nitrogen. If there is, let's say a fuel like I don't know, fuel oil or coal, then you're going to have sulfur here as well. And the sulfur is going to turn into sulfur dioxide. Now unless you remove that sulfur dioxide which is not uncommon, I mean there's just a system that could cause a couple of hundred million dollars called sulfur scrubbers where you could actually remove the sulfur dioxide from a combustion gas stream. In fact, you really have to do that now by law but if you didn't then the sulfur dioxide is going to mix with water in the same way that the NOx is going to mix with water and this is going to form oxides– I'm sorry then these oxides and sulfur are going to form well sulfuric acid in the environment.

So we need to make sure that we don't drop below the dew point. If we do then these acids could actually form on the inside of your smokestack, they could form inside your furnace, they can collect on your metal tubes and of course acid eats away metal. So you don't want to damage your 2 billion dollar power plant just because you weren't paying enough attention to the dew point. It does happen, believe it or not. So with all these in mind, how do we calculate the dew point? So the dew point is important. We know how to calculate the dew point. I mean we've seen the general equations before just because we have a different mixture than in air and water vapor mixture doesn't mean that our equation changed.

We would know that the dew point is going to be the saturation temperature at the partial pressure of the water vapor, right? I mean that's how we define dew point and this would be the partial pressure of water vapor in the products. OK. These are combustion products where we're creating water, so we have to calculate the partial pressure of water vapor. Now, that's not hard to do. We know that the partial pressure is related to the total pressure and the mole fractions, right? And we know that if you just took the mole fraction of water vapor and multiply that by the pressure in the environment, let's just call this the mixture pressure or how about just the atmospheric pressure or the product pressure.

But you'll see it's just a pressure of the environment that you're, you know, participating with this combustion process. But it doesn't have to be. I mean if this is an internal combustion engine, the combustion process might take place at a pressure of a thousand or more pounds per square inch. So I mean it's certainly possible to have a high pressure of the combustion products.

But, nonetheless, we know that the mole fraction times the total pressure is going to equal the partial pressure. And we should also recognize that mole fractions are easy to find, right? Just look at the right-hand side of the equation in the numerator. Well this is a mole fraction of water vapor, so this is going to be the number of moles of water vapor.

And then in the denominator it's just going to be the sum of the number of moles of all the components in the products. OK. So, in this particular case, we got to multiply by the total pressure. OK. So you look at the equation there and it's easy to do, right? The number of moles of water vapor is just b over 2 and the number of moles of all the products and sum together. Well, you've got 8 carbon dioxides and b over 2 waters, you've got the excess air fraction times your stoichiometric coefficient times well nothing.

That's your oxygen and then you've got your theoretical air fraction times your stoichiometric coefficient times 3.76. And that's going to represent the number of moles of the nitrogen, right? So carbon dioxide, water, oxygen and nitrogen, so it's really just a matter of taking the ratio of these two numbers multiply it by the pressure, that gives you the partial pressure. Go into your table A5 at that partial pressure and look up the saturation temperature and you're done. Now again, we've seen these types of problems before, not with combustion products but certainly we should understand how to do it for combustion products as well.

OK. So let me go through some examples now. You know, first just a simple one. Let's just write for combustion equation, for methane gas– — burning with 20% excess air. OK. So, just a simple combustion equation. Well, as usual we have to go into table A1 and we'll find that methane which is CH4. Now in this problem I haven't asked you to find the air to fuel ratio, I've not asked you to find the dew point. So we don't need to look at any other data, no molar mass data is necessary. So it's really just a matter of going through the equation. So we have CaHb, the theoretical air fraction, well we're given an excess air fraction of 20% or a fraction of 0.2. So that means the theoretical air is 20% more than the stoichiometric coefficient. So again this is b, 1.20 for 20% excess air and then the a plus b over 4 gives us 1 plus 4 over 4 and O2, 3.76 N2.

And in fact, let me just rewrite the left-hand side of the equation. So we have CH4 plus and now we have 2 times 1.2, so 2.40 oxygens and then 1.2 times 2 times 3.76 is 9.024 nitrogens. And now let's look at the right-hand side of the equation. So, well we've got 1 carbon, so 1 carbon dioxide. We have– well for hydrogen, so 4 over 2. In other words, 2 H2Os and in fact let me just show this as 4 over 2. And then we have our excess oxygen, so we have 0.2, times 1 plus 4 over 4 O2. And then lastly we have 1.2 times the stoichiometric coefficient times 3.76 nitrogens. And then let me just rewrite. So we have 1 carbon dioxide, we have 2 waters and now we have 0.2 times 2, so 0.4 oxygens. And then of course all of the nitrogen that we start with, we end with. So we have the same 9.024 nitrogen. And this would be the combustion equation then for this particular process. So as long as you understand the concept of excess air and theoretical air, I don't think these combustion problems are really that difficult although these are new types of problems and make any new type of problem and it takes a little bit of time to get used to them.

Let's go through one– possibly two example problems, we'll see in this time we have. So again, any questions just in general about this type of process? All right. So let's do this one. So let's say we have hexane gas burned with 200% theoretical air. At 100 kilopascals and we want to find the air to fuel ratio and the dew point. OK. So I suppose the first thing we need to do is write the combustion equation, right? I mean there's nothing in this problem that says, write the combustion equation and then figure out everything else. But quite frankly if you don't write the combustion equation how would you begin a problem like this. So always start with a combustion equation. Note that we have 200% theoretical air, not 200% excess air. So that means we have 100% excess air, in other words we've got twice as much air as we need at least as we need for stoichiometric combustion. So as usual we'll go into the appropriate table.

We're going to A1 or A2, I don't recall which one. But we can look up the chemical formula and it's C6H14. So that's the chemical formula, why don't we also look up the molar mass since we're going to need that in just for a little while. So this is 86.179 mass per unit mole. So that's the data that we're going to have. We might as well find the stoichiometric coefficient. So, that's just 6 plus 14 over 4, so it's that 9 and a half. So now we can write the combustion equation. So C6H14 plus– all right, so first we would have all of our excess air, right, 200% theoretical air is the same as 100% excess air. Anyway, this is just 2.0 and then this is multiplied by the stoichiometric coefficient which is 9 and a half and then O2 plus 3.76 N2.

And so again I always like to rewrite the left-hand side. I'm comfortable and I have all the data. So C6H14 and then plus, and now here we have 19 oxygens and then 71.44 nitrogens. On the right-hand side, we would have 6 carbon dioxides. We would have 14 over 2 waters. We would have our excess oxygen, so this is going to be the theoretical air fraction minus 1, right? If you're given the excess air fraction like the previous example then you would just put the excess air fraction right in here. But here we're given a theoretical air fraction so we just subtract 1 from it, times O2. And then lastly, we have all the nitrogen, so theoretical air fraction and then the stoichiometric coefficient– I'm sorry, I forgot my stoichiometric coefficient in here, didn't I? Let me finish that. So we have that and then the stoichiometric coefficient. OK. And right here, for the nitrogen, the theoretical air fraction and again times the stoichiometric coefficient and then 3.76 N2. OK. So as I've done this. Let's just rewrite one more time.

So we have 6 carbon dioxides and 7 waters and then let's see this is just going to be 1 times 9 and a half, so 9 and a half oxygen and then all the nitrogen. OK. So, it's always good to start with your combustion equation– well it's not good, it's just necessary. Now that we have this number, let's calculate the air to fuel ratio, so again, nm over nm, air to fuel. All right, so what are the number of moles of air? Well, again now it's the theoretical air fraction times the stoichiometric coefficient times the 4.76 and there's still just 1 mole of fuel times the molar mass of the fuel. So this is what we're calculating. So we have 2 times the stoichiometric coefficient which is 9 and a half times 4.76. So this is the number of kilomoles of air and then the usual 28.97 kilograms per kilomole on air and then in the denominator 1 kilomole of fuel and then the 86.179 kilograms per kilomole of fuel. So really just go to a little bit of mathematics now. We end up with 30.4 kilograms of air per kilogram of fuel. So there's your air to fuel ratio.

Now, again, we understand that this is not just an academic exercise if we're burning something in a power plant, in an engine, whatever it happens to be, we need to know how much air we need for every unit of fuel. I mean the amount of fuel we have is going to determine the energy that's released by combustion that certainly going to be important we need to know how much steam we're going to make. So we can buy the right turbine and so it spins at the right speed and produces the right amount of power. But it's not just about the fuel, right? It's about the air. We need to know how much air we need to burn that right amount of fuel to give us that amount of heat energy to power my power plant or my engine or whatever. So this is important. And then last but not least, let's calculate the dew point. So why don't we just start by finding the mole fraction of the water vapor.

So we just go to the equation, we see that we have 7 kilomoles of water and then we divide it by the sum of everything in the numerator. We've got 6 plus 7 plus 9 and a half plus 71.44 and this is kilomoles of the entire mix. So this is kilomole of water vapor per kilomole of mix and with this a little math. We get 7 over 93.94 or 0.0745, so that's the partial pressure– I'm sorry that's the mole fraction, did I say partial pressure? This is the mole fraction which we multiply by the total pressure to get the partial pressure. So now the partial pressure of the water vapor is just mole fraction times the pressure of the mix, 100 kilopascals is the given pressure. So 0.0745 times 100 equals 7.45 kilopascals and now it's just a matter of going into our appropriate tables.

So at PV we're going to table A5. And we get that saturation temperature and it's 40.1 degrees Celsius. Yeah you have to do a little bit of an interpolation but this is the dew point. And that's the end of this particular problem. Now, that temperature might sound low but that's not unusual. One thing that might be noted, the temperature of the combustion products can drop below the standard temperature associated with the water phase change at atmospheric pressure, right? The atmospheric pressure, water changes phase at 100 degrees Celsius, you might think that if the combustion products in your smokestack drop below a 100 degrees Celsius then that means you're going to start to get some condensation.

But that's not true, right? Like any other gas, the water behaves as if it existed alone in the atmosphere. It exists as if it behaved at that particular partial pressure that's the pressure at which that the evaporation process or the reverse, the condensation process is going to take place. So we can drop all the way down to 40 degrees Celsius before any condensation occurs. That's not too bad was that 105 degrees more or less, so pretty good. By the way it's always in your best interest to try to get the temperature of the combustion products as low as possible because that means you can transfer the most heat as possible from those combustion products into whatever it is that requires that heat. Like let's say the boiler of your power plant. If the temperature is elevated then that means you haven't transferred as much heat as this possible and the efficiency of your entire cycle is going to suffer.

On the other hand, you can't let that temperature get to low, right? If the temperature gets too low then you're going to have all these maintenance issues associate with the acid forming in your furnace or in your smokestack and that could be a problem. So there's kind of a balancing act here. You want that temperature of the combustion products to be low but you don't want it so low that you're going to start to eat away the inside of your furnace because of the acids, anyway, another good example problem. So, any questions on this one? All right, how about one more example then? And this one is actually from your book. Well, this is problem 15-23 and it will be up shortly [inaudible]. Well I guess practically, it's 15-23E. I have assumed you've noticed this that E stands for English units.

But nonetheless, so here's the problem and this is a little bit different than the problems we've already looked at. Think about this as kind of a backwards problem compared to the ones we've looked at or maybe sideways would be a better analogy. So here we're given the actual mass, all right. We're given 1 pound mass of butane. It even gives us the chemical formula though we don't need that. We can look it up. And it's burned with 25 pounds mass of air and that air is that 90 degrees Fahrenheit and 14.7 psia, actually I think we're assuming the whole mixture is at that particular temperature of pressure. Now it says, assuming that combustion is complete and that the pressure of the products is 14.7 psia, find the percentage of theoretical air used and find the dew point. OK. So here it kind of backwards, right? On the previous problem we're given the ability to find the number of moles of this component having been given the theoretical air fraction or excess air fraction. Here it's the reverse, right? Here we are told what the air to fuel mass ratio is and we have to kind of work backwards and figure out the amount of theoretical air that must have been used for that particular combustion process.

But it's the same basic procedure. So, let's go to this example and I suppose the first thing we want to do is just go through our combustion equations. So, C4H10, now that's we're trying to find here. Well, the stoichiometric coefficient is 4 plus 10 over 4. So that's just going to be 6 and a half of this particular problem, so plus. And now here, we don't know what the theoretical air fraction is. So all we can really do is just plug in the number, I suppose. OK. Yeah, right. So theoretical air fraction, OK, before I go any further. Well, yeah OK, I guess I'm going to try that. I didn't actually do it the way I'm about to do it in my notes. So I keep my fingers crossed.

But we've got theoretical air fraction and our stoichiometric coefficient and our O2 plus 3.76 N2. OK. And this will equal 4 carbon dioxides and 5 waters. And then we have our excess air fraction times 6.5 O2 and then we have our theoretical air fraction 6.5, 3.76 N2. OK. So we really can't go any further in this particular problem. All right. So, now let's leave it this way but let's see if we can find a solution to this problem. So how are we going to find this? How are we going to find the theoretical air? Well, maybe we could use the air to fuel ratio that's been given.

We know that the air to fuel ratio is 25 pounds of air for every 1 pound of fuel. So that's just going to be 25. But we also know this air to fuel ratio is going to be the number of moles times the molar mass of air and the number of moles times the molar mass of the fuel. So what would be the number of moles of the air? Well, this is just going to be our theoretical air fraction times stoichiometric coefficient times 4.76 and then we multiply that by the molar mass of the air, so 28.97 and then in the denominator we have just 1 mole of the fuel and the molar mass of the fuel. Well they haven't exactly been given the molar mass. So what we'll do is we'll actually do this the other way. Now keep in mind that we can instead of looking at the number of moles of fuel times the molar mass of the fuel, we can look at this as the number of moles of carbon times molar mass of carbon in the fuel or plus the number of moles of hydrogen times the molar mass of hydrogen in the fuel.

So that would be a way that we can actually solve this. Now I suppose we can look in the textbook in the appendix and look up the data and hopefully this particular gas butane is there. I'm not positive. I'm going to do it the other way anyway. So, yeah? >> Is the molar mass of air still 28.97 in English units? >> Yeah. That's just mass per unit mole, so it doesn't matter what the units are. So we can also write this as nm air over the number of moles of carbon times molar mass of carbon plus moles hydrogen– molar mass of hydrogen. This is another way to do the same thing. So let's do that.

So we have theoretical air fraction times 6.5 times 4.76 times 28.97 that should be the mass of the air and what about the fuel? Let's see. So we have 4 carbons and the molar mass of carbon is 12.0 and then we have 10 hydrogens and 1.01 is the molar mass of hydrogen, so you know all this data you can certainly look up in your book. Well, nonetheless, we've got one equation with one unknown and the only unknown is the theoretical air fraction. So it worked.

And we solve for the theoretical air fraction, 1.61 or 161% if we want to write it as a percentage. OK. So now that we have this, now we can actually go ahead and complete our combustion equation that will then allow us to find the number of moles of the various components. Well components in the products. And then we can find the partial pressure and we can find the dew point. So we'll do that next. So just completing the combustion equation, C4H10 plus– now here we have 1.61 times 6.5 times O2 plus 3.76 N2, 4 CO2s, 5 H2Os plus 0.61 times 6.5 O2 and 1.61, 6.5, 3.76, yup [inaudible]. OK. Now you can go through all the mathematics on the left-hand side if you want to, for some reason I'm looking into my notes and I'm not sure why but I didn't complete the left-hand side.

I mean we could do it with the calculator. But it's really the right-hand side with this importance now because I need to calculate the dew point. So, anyway, we get 4 CO2, 5 H2O and this actually becomes 4 oxygens and then plus 39.5 nitrogens. OK. So now we can find our partial pressure of water vapor. So we got 5, 4 plus 5 plus 4 plus 39.4, so there is the mole fraction. We're given 14.7 psia as the pressure. So we find that the partial pressure is 1.4 psia and now we can find the dew point which is a saturation temperature at 1.4 psia. So, of course, we're going to go into the appropriate tables. Now, I do need to caution you. There are some issues and I think I mention this before. We only go down to so low of a pressure, I think it goes all the way down to 1 psia in table A5E, if not then you might have to use A4E and just interpolate within that table. But nonetheless, typically we would use table A5 or again A5E here and we go through the math.

And at this partial pressure we go to 113.2 degrees Fahrenheit and there's your dew point. So there's another example problem that, you know, just a little bit different from what you've seen previously but still good problem. So, any questions on this material at all? OK. Why all– We've actually gotten through everything. I do have about 5 minutes so I can get it, go into some new material but that wouldn't be any fun. Plus there's not enough time. So I think that would then do it for today. If there's no questions on anything, there will be no new material on Friday. So Friday will be a review, Q&A day so come prepared to ask questions. I think you've all seen my finals week office hours now. So with that I'll see you all next time..

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