Adiabatic Flame Temperature

[Music] all right so among all the different flavors or types of uh enthalpies reaction enthalpies that we've seen uh one of the most important is the enthalpy of combustion so combustion of course is a process that's important for a lot of different processes we use combustion uh to cook with we use it to generate a lot of our energy we use it in many cases to drive uh our cars so combustion as a process is very important and we can use thermodynamics to understand the details of combustion at a slightly deeper level in particular we can learn using what we know about thermodynamics uh to understand the temperature at which Flames burn when uh combustion takes place so to illustrate what I mean by that and why that's an interesting question you're familiar most likely with uh burning a number of different types of fuels you might burn um you know if you if you cook on a gas stove the natural gas is is mostly methane if you use a bunson burner in a chemistry lab those most likely are burning propane uh if you uh weld with an acetylene torch then you're burning acetylene if you drive a gasoline powered car you're using a mix of hydrocarbons that might be uh include some octane for example but a range of different hydrocarbons we can burn all of those those combustion in every one of those cases we combine those with oxygen the carbon turns into CO2 the hydrogen turns into steam or water so combustion is the same chemical process in each of those cases just the stochiometry of the reaction works out to be different and yet the Flames the temperature at which these burn the temperature of the Flames that are generated when uh these different hydrocarbons uh combust in a flame turn out to be very different so so uh to understand why that would be true let's take the the simplest of these as an example just to keep the stochiometry relatively easy for the first example if we burn methane so again maybe we're cooking on a gas stove and burning natural gas or or uh primarily methane if we burn methane and oxygen this will generate CO2 and H2O uh if I burn one mole of methane one carbon becomes one carbon so I've got one mole of CO2 uh four hydrogens need two molecules of water or two moles of water so I've got a total of 2 plus two is four oxygen so that means two o2 molecules on the reactant side of the reaction so there's my balanced reaction that's the combustion reaction for methane the enthalpy of combustion for methane so the enthalpy change associated with that reaction eny of combustion for methane isus 89.3 K per mole a number we can get by looking it up in a table or by calculating it from the Heats of formation of products minus reactants or by doing in the lab and measuring the the enthalpy In a calorimetry experiment we can get that number a number of of different ways that's the enthalpy change for this combustion reaction that's enough information that plus one additional assumption I'll mention in a minute is enough to let us calculate the temperature at which methane will burn uh so that assumption let's say let me draw a picture of what's going on in so I'll draw a flame which I've drawn as a as a candle but any in any of these examples if I'm burning methane I've got some source of methane my natural gas line on my stove or whatever the oxygen that's required to burn the methane that's just in the air so methane and oxygen both enter this flame and then this chemical reaction takes place the thing I need to um understand about the the flame what I eventually use this flame for if I'm burning methane on my stove I'm going to use it to heat a pot of water or whatever it is I'm cooking so there's some heat that's going to get transferred to something else at some point that process is relatively slow the the process of getting the heat from the flame to something else takes a significant amount of of time what's going to happen very quickly when this reaction liberates some heat so this enthalpy changes negative which means the enthalpy of the products is less than the enthalpy of the reactants so if I just do this reaction uh it has a negative enthalpy change so that would normally generate heat that gets given off to the environment and that's what heats my pot of water that process like I've said is relatively slow the first place that heat goes that energy in the form of enthalpy goes is to heating up the products of the reaction themselves so the first thing that gets heated up is the CO2 and the H2O so uh in fact those atoms that make up the molecules of CO2 and H2O those are the same atoms that started in my methane and O2 so it's the same um atoms that were there originally so in that sense the heat has not left the system and we can assume at least under some conditions if the reaction is fast enough that the reaction is adiabatic meaning that the heat in the reaction uh no heat was transferred to the surroundings in the reaction all of the energy stayed within all of the enthalpy stayed within the system which means that the enthalpy change of the products that energy was used to heat up the products and at least initially didn't transfer to anything outside of those individual uh atoms or molecules so if we do this process at constant pressure if it's open to the environment we know that heat and enthalpy change are the same thing at constant pressure that's why we Define the enthalpy is to be able to make statements like this one so if we do this at constant pressure enthalpy and heat are both equal to zero for this tic process so the thing that's zero the enthalpy that's equal to zero is the full enthalpy change for the reaction not just the enthalpy of combustion not that just the enthalpy change when the mo when the reactants get converted to products but also the enthalpy change that occurs when these products are heated up so the way physically to understand what's going on is I generate 890 kles every time I burn a mole of methane and then I spend all 890 KJ of that energy in heating up the CO2 and H2O to a much higher temperature than where they started that's why a flame Burns hot is because that energy is used to heat up the products so in this thermodynamic equation the total enthalpy is the enthalpy change of the reaction heat of combustion plus since I've got a mole of CO2 the if my ANL be if I'm sorry if my heat capacity is constant which is a big assumption we'll come back to that statement in just a second second but as a um an initial guess if I say CO2 has some heat capacity the total enthalpy change of the CO2 is its heat capacity times uh the change in temperature so for each mole of CO2 I have I've got one mole participating in this reaction I've got heat capacity times delta T if I've raised the temperature by some amount delta T I've also got two moles of H2O so I also need to include twice the heat capacity of water vapor times the same increase in temperature the whole system all of the products are going to be heated to the same temperature of this flame so this delta T and this delta T are the same number I've I've raised the temperature of the CO2 and the H2O by the same amount so if I rearrange this equation what I'm interested in I know that en of combustion I know I can look up the uh heat capacity of CO2 and H2O what I don't know is is the amount by which this reaction will have increased the temperature of the products so I want to solve for delta T so let's rearrange this equation first and say move the um enthalpy of combustion over the other side negative enthalpy of combustion is equal to these heat capacities so and heat capacity of CO2 and heat capacity of H2O twice and all times delta T then if I want to solve for delta T that's going to be negative eny of combustion divided by my heat capacities so these are numbers that I can look up and plug in to calculate the temperature change we're going to do that in just a second and this would be exactly how to proceed with the calculation if the chemical reaction occurred as I've drawn it if we're burning methane in oxygen and of course we are there's oxygen in the air that's the uh oxygen that's used in this combustion reaction but of course air is composed not just of oxygen but also of nitrogen nitrogen doesn't participate in the chemical reaction so normally we would avoid it when avoid thinking about it when talking about the chemistry of this reaction but once we start talking about the fact that the enthalpy gets transferred to the products uh and and heats them up it's unavoidable that if oxygen diffuses into this flame nitrogen is diffusing Into The Flame as well so uh in the immediate environment of the flame uh which is at several thousand degrees I've not only heating up the CO2 and H2O that I generate by burning the methane but I'm also going to need to heat up the nitrogen as well so for a more accurate um estimate of the flame temperature in air we're going to need to include the the effects of the nitrogen as well so remembering that air is 21 Parts oxygen so the O2 to N2 ratio is 21 Parts oxygen for every 78 Parts nitrogen and the other 1% is is argon and other Trace uh components so I need to know that ratio because for every two moles of oxygen that I need to diffuse into the flame and burn uh with the methane or help burn the methane I'm going to have so for every two moles of o2 there's 21 parts of o2 in air that ratio needs to be the same as 78 parts of N2 in the air so some number of moles of of N2 so roughly not exactly roughly four times as much N2 as O2 in the air if I work this ratio out it turns out to be that in instead of two moles of oxygen this will work out to be 7.4 moles of nitrogen are the stochiometric amount uh to the two moles of oxygen so if I go back and say the things I need to heat up in this flame include not only one mole of CO2 and two moles of H2O every time I burn a mole of methane but I'm also going to need to heat up 7 7.4 moles of N2 so again in this parentheses I've got 7.4 times the heat capacity of N2 and also in the denominator of this fraction I've got 7.4 times the heat capacity of N2 so that's accounting for the fact that I need to heat up the N2 that's a passive participant in this reaction just because it's in the same vicinity as the oxygen so now we're in a position where we'll get a reasonable answer if we plug the numbers into this expression we know that enthalpy of combustion 89.3 kles per mole let's pay attention to the units so that's 890 with a negative sign becomes positive in the numerator heat capacities of CO2 and H2O and N2 the heat capacities I'm going to need to use there are not the heat capacities at room temperature but the average heat capacity over the range of temperatures so this temperature increase is going to be on the order of thousands of degrees Flames are are relatively hot so we need the average heat capacity over the range from room temperature up to the temperature of the flame so I'll just let you know what those numbers are a reasonable number to use for the average heat capacity of of CO2 between room temperature and Flame temperature is about 55 Jew per mole Kelvin for water it's about 42 and for nitrogen it's about 33 Jew per mole Kelvin so a few things to notice about that as we've seen heat capacities are larger for these triatomic molecules than for diatomic molecules for reasons um related to the equip partition theorem um also um yeah that's enough information about the heat capacity so if we plug these numbers in notice that the units Jews per mole Kelvin in the denominator I've got kles per mole in the numerator so to keep the unit straight let's multiply the enthalpy of combustion by a th000 so that we've got units of Jews per mole that'll cancel the units of Jewel per mole in the denominator and what we get when we do that math 890,000 300 divided by the quantity 55 + 2 * 42 + 33 when I do that math with a couple of sigfigs oops that's not correct one step ahead of myself that number works out to be 2300 Kelvin is the increase in temperature so what that tells us is when I burn methane whatever the initial temperature was the temperature is going to increase by 2300 Kelvin so the actual flame temperature if I start out near room temperature that's where my 2600 kvin comes from so start out near room temperature increase the temperature by 2300 and I find out that the the temperature of this flame is going to be around 2600 Kelvin so or 2300 Celsius roughly if you prefer so a few comments about that number that that's a reasonable estimate in purely ideal conditions so the notes I'll make will be this calculation is for having made the assumption that this flame burns adiabatically the heat generated in this reaction immediately goes to raise the temperature of the products and the nitrogen in the air before any of it gets transferred to other things like um the the pot that's sitting on the flame and things like that so it's under relatively ideal tic conditions if the the reaction is not completely adiabetic then uh the temperature increase will be less so uh so if it's nonadiabatic a diabetic the flame temperature will will decrease another thing to notice is that we've done this calculation for Burning uh methane in air a 21 to 78 mixture of oxygen and nitrogen uh the calculation would be different if I were burning in pure oxygen if you take the effort to burn a flame in pure oxygen rather than oxygen with nitrogen then the flame temperature will be different and you can just leave out the nitrogen to do that so in oxygen I'll point out that the flame temperature will be considerably higher in oxygen because we don't have to go to the effort of the thermodynamic effort of heating up the N2 I'll also point out that with the same type of calculation we've done the ex Le for methane we could do it for propane we could do it for acetylene the only difference in those cases is the stochiometry of this reaction works out differently we combine it with a different number of moles of oxygen which bring with them a different number of moles of N2 and we generate a stochiometrically different number of moles of CO2 and H2O so every molecule every fuel Burns at a different flame temperature because the ratio of carbons to hydrogens is different in that particular fuel so the stochiometry of the reaction is a little bit different and then lastly I'll remind us that we used a constant just a single number for the heat capacity assuming that the heat capacity was a relatively constant number between room temperature and this flame temperature which is certainly not true or assuming that this average heat capacity is a reasonably good estimate in fact we know that the heat capacity depends on the temperature sometimes relatively strongly at these high temperatures so the more accurate way of doing the calculation if we want just a couple more than just a couple of sigfigs worth of uh accuracy out of this flame temperature would be to treat the the dependence of the heat capacity on the temperature so in other words instead of saying heat capacity times delt T we would replace this with something like an integral of the heat capacity over the temperature from some initial to some final temperature that makes the math of doing the calculation quite a bit more calc complicated but of course it gives you a much better uh more accurate estimate for the flame temperature but this uh overall procedure depending on whether you want to burn uh your fuel in oxygen or in air what your fuel is how accurately you want to treat the heat capacities can give you uh a pretty good estimate of the temperature at which different fuels will burn