>> I'm going to explain what adiabatic flame temperature is and how we calculate it
and some issues that can affect how accurate that calculation is.
Basic idea is we take a fuel, like in this case looking at methane, and we burn it.
This case we're looking at air.
And we burn it so that there's no heat transfer.
So Q is zero, making the system adiabatic.
And the temperature of the products leaving, this is the adiabatic flame temp.
And we calculate it from an energy balance.
And conceptually the idea is that, if we start -- well, let's look at this this way.
We, the heat of reaction, where we go from methane and oxygen to form CO2 and water,
heat reaction is typically viable at 298 because we calculate from heat
to formation, and tables are present at 298.
So if we're going to do an energy balance, so if we think of temperature as being on this axis
and reaction extent on this axis, then the fuel would enter
at some temperature above 298 typically.
And, likewise, the oxidizer will be at some temperature above 298.
And what we're going to have to do, what we're going to visualize
in our calculation is we're going to cool the oxidizer to 298.
Cool the fuel to 298.
Carry out reaction at 298.
And then take and heat the products up to the flame temperature, temperature out such that --
and the actual process being something that we know, delta H is zero, because Q is zero.
There's no work in this type system.
And so this calculation, this delta H plus this one plus this one should add up to zero.
And let's add a couple other things, right,
the oxidizer in this case is oxygen, but we also have nitrogen.
Because, looking at the calculation of airs, we have nitrogen here.
We have nitrogen here.
And if we have excess air, we also have un-reacted oxygen.
But the thing that we're considering here is that combustion is complete.
So complete combustion, meaning no methane is left.
So let's look at what the first law looks like.
So here's the first law for delta H for the process.
And we indicated this overall is zero.
Heat reaction term here times the moles of fuel, so [inaudible] moles of fuel.
And then the terms for each of the gases leaving integrate from 298
up to the flame temperature, the temperature out.
This is the number of moles of each of those components.
And the heat capacity is going to be a function of temperature.
This term accounts for the nitrogen, oxygen,
and in our case methane coming in at temperature above 298.
So the first law then is going to need heat capacity as a function of temperature.
And we're going to use the equations from the NIST WebBook.
Let me pause and show you what they look like.
So for each species, we're going to have a heat capacity that looks like this, where A, B, C, D,
and E is a constant over a certain temperature range.
And this one may have several heat capacities, each appropriate for a temperature range.
And then we have to integrate this equation.
And let me show you that, how the heat capacity changes for CO2
to understand why it's important to fit this way.
So this is heat capacity for CO2 as a function of temperature.
Notice particularly in the range below
about 1,500 how significantly it changes with temperature.
And so the NIST WebBook fits sections of this with this equation.
And what I'm showing you is an example, in this case the heat capacity for water vapor.
They have one equation from 500 to 1,700 Kelvin and another from 1,700 to 6,000.
If I calculate heat capacity at 600, this is the appropriate correct equation,
35 joules per mole per Kelvin, notice how much air there is if we were
to use the high-temperature model.
So it really matters that we use the appropriate temperature range.
So if we do this calculation for air,
to a pretty good approximation we can ignore other reactions that might occur.
So if I calculate the temperature out, the flame temperature, I get 2,326 Kelvin.
If I correct for the fact that other reactions start to take place at the high temperature,
then the correct value, a more correct value is 2,244 Kelvin.
So you can see it's not a large error if we ignore those disassociation reactions.
And so when air, this is air as the oxidizer.
When air is the oxidizer, it's not a bad approximation to ignore them.
This is for a stoichiometric feed that I've shown the calculation.
That's where the biggest difference is.
If we have excess air, than the difference becomes significantly smaller.
However, if we use oxygen as the feed and we do that same comparison,
we calculate using the heat capacity data, we get a very high temperature.
But if we correct for the fact that the other reactions, disassociation reactions,
we get a temperature that's almost 2,000 Kelvin lower.
So clearly if oxygen is the oxidizer, we can't ignore these reactions.
And the kinds of reactions that we're talking about, if we had a methane-oxygen feed,
and let's say it was stoichiometric,
then 73 percent of the CO2 that form actually dissociates.
Meaning that CO2 at that high temperature goes to an equilibrium mixture, CO and oxygen.
And 26 percent of water dissociates.
So water dissociates in the hydrogen and oxygen.
And we also have the potential to get a small amount
of additional dissociation where H2 can dissociate.
We can get hydrogen and oxygen [inaudible] disassociation and OH.
These are going to be much smaller.
But these disassociation reactions --
so important thing to remember is
that these reactions are endothermic, the disassociation reactions.
That's why the temperature is significantly lower here
when we take disassociation reactions into account.
And then I'll list the reference here that has more details on this type of behavior.
But it's something -- two important things to remember.
Heat capacity is a function of temperature, have to be for appropriate temperature range.
And if we're using oxygen as the oxidizer,
we have to take into account these disassociation reactions.
And here's the reference that discussions this in more detail.