>> Here, we're going to calculate the adiabatic flame temperature
for burning butane, normal butane, 100% excess air.
And both the butane and the air enter at 25 degrees C.
So given the heat capacity parameters,
the heat capacity is form A plus B times T plus C times D squared plus D times T cubed.
And what's important and we want to keep the mind,
these heat capacity parameters are only good over this temperature range.
So if we were to end up calculating a temperature higher than 1,800,
it's not going to be a valid answer.
We'd have to go back and find different heat capacity values.
So the reaction that we're interested in, normal butane C4H10
and 13/2 oxygen gives us 4CO2 and 5 water.
Now, because we're going to use 100% excess air, that means if we have one mole of butane,
we're going to have 13/2 times 2, 13 moles of O2 fed to the system.
And then 13 times 79/21, which is 48.9 moles of nitrogen.
We're assuming air is 21% oxygen, 79% nitrogen, that's a reasonable approximation.
So the way we're going to do the calculation, we take the advantage of this idea
of state function, and we feed in 13 O2, 48.9 moles of nitrogen,
and 1 mole of C4H10, feeding that 298 Kelvin.
So we're going to visualize we carry out the reaction at 298 Kelvin.
And the reason we do that is because now we can get the heat reaction at 298 from heat
to formation because we have tables at that temperature.
we're going to end up with we're going to have 4 moles of CO2, 5 moles of water.
We're going to have unreacted oxygen, 6.5 moles.
We had 100% excess.
And, of course, all the nitrogen un-reacts, we're going to have that many moles of nitrogen.
We're going to take then all of those species and heat them up to our flame temperature.
And that means we're going to calculate this integral to go from 298
to the temperature we're trying to determine.
And what we're going to write is the sum of the number
of moles times the heat capacity of each of those species.
So if we add these two together, that's the same as carrying
out this reaction, where delta H is equal to 0.
So this is the adiabatic flame temperature, assume no heat loss.
So let's first calculate the heat reaction at 298.
And the heat reaction at 298 then is going to be 4 times the heat of formation of CO2,
since it's 4 moles, 5 times heat of formation of water, minus the heat of formation of butane,
minus 6.5 heat of formation of O2.
Of course, this is 0 because oxygen is an element.
Now I can look up the values.
Let me pause and write down these values I take from the table.
So here's heat of reaction of 298.
And the other term with the heat capacity calculation that's is integral,
that's 298 up to the flame temperature.
The sum number of moles heat capacity DT, which I can write as the sum of NI,
and then the heat capacity for each of the components.
This is BI, CI, and DI.
So if I integrate that, I'm going to end up with the sum of NI, AI, times T, plus BI,
T squared over 2, CI cubed over 3, and DIT to the 4th over 4.
And then we have to apply the limits of integration.
I won't write out all these terms, but let me write out one just
so we understand what we're talking about.
The first one would be 4 times A for CO2, plus 5 times A for water,
plus 6.5 times A for oxygen, plus 48.9 times A for nitrogen.
And I apply the limits, would be T minus 298.
I won't write the rest of the terms, but I've written B.
So this would be the C terms and then the D terms.
So then what we're doing is we're solving our equation, delta H for reaction at 298,
plus this integral of the sum heat capacity terms, from 298 to T equals 0.
And I'm going to use the spreadsheet because now we have a polynomial,
temperature's unknown to the 5th power.
It's easiest to use Solver in a spreadsheet.
When I do that, I end up with adiabatic flame temperature 1,510 Kelvin.
Let me just show you the spreadsheet.
So here we see I put in the heat capacity constants,
the sum of the NI/AI terms, then the integral terms.
Notice that the heat capacities are in joules per mole per Kelvin.
Heat reaction in kilojoules.
So this sum, I divide it by 1,000 to put in the correct units.
The spreadsheet's available at the LearnChemE.com site.