[Stoves] Air/fuel ratio for paraffin - correction

[Crispin Pemberton-Pigott]

Dear Perfectly Aspirated Friends

I have been struggling over a hot spreadsheet trying to get the air/fuel
ratio for paraffin right and I have the following conclusions regarding two
posts earlier this month from Dr Peter Verhaart and Prof Philip Lloyd in
answer to my question, "How much air is required for the stoichiometric
combustion of paraffin?"

Peter wrote:
>Stoichiometrically C16H34 needs 16 O2 for the C and 34 O (=17 O2) for the
hydrogen, makes 33 O2.

I am pretty sure it is:
C16H34 needs 16 O2 for the C and 34/2 O (=8.5 O2) for the hydrogen, makes
24.5 O2 to yield 16 CO2 and 17 H2O.

Peter again:
>One kmol equals 12*16 (for the C) and 34*1 (for the H) makes 226 kg.

Agreed.  

In response, the following was later added by Philip:

>That doesn't get you the air fuel ratio directly, however.  
>
>Air is 20.946 vol % oxygen. 33kmoles of oxygen will therefore require
>33/20.946% = 157.5kmoles air.  

If I correct the 33 to 24.5 that still leaves another problem: the air is
being measured by volume and the oxygen by molar mass.  Air is 23.1% oxygen
by mass http://scienceworld.wolfram.com/physics/Air.html which makes the
calculation:

24.5/23.1% = 106.06 kmoles of air

Philip again:
>A kmole of any ideal gas 22.414m^3 at NTP...

Taking this approach, the calculation is: 

In order to burn 1kmole paraffin I will need 
106.06 x 22.414 = 2377 M^3 of air

Or to burn 1 mole of paraffin (226 gm) I will need 2.377 cubic metres of
air.

There are 1000/226*0.87 = 3.850 moles of paraffin in a litre (density of
0.87 http://www.cmu.edu/gipse/materials/pdf-2003/8-9/float-M_Friedman.pdf
p.4)

So the volume of air needed to burn one litre of paraffin stoichiometrically
(average C16H34) is 
2.377 * 3.850 = 9.15 cubic metres.

Or 9.15 litres per cc
Or 10.52 litres per gm


Another way to do this without using the ideal gas volume is:

1 mole of paraffin = 226 gm (12 * 16 + 1 * 34 = 226)
And needs 784 gm of Oxygen to burn completely (49 * 16 = 784)

Density of paraffin  = 0.87
1 litre of paraffin = 3.85 moles (1000 / 226 * 0.87)
And needs 3,018 gm of Oxygen to burn completely (784 * 3.85)

As referenced above, air is 23.10% Oxygen by mass
Thus 13,065 gm of air is needed per litre (870 gm) of paraffin (3,018 /
0.231 = 13,065)

Air is 1.2 gm per litre http://en.wikipedia.org/wiki/Density_of_air
So the air requirement	 is 10,888 litres of air per litre of paraffin
(13,065 / 1.2 = 10,888)

Or 10.9 litres per cc
Or 12.5 litres per gm.

The two results are not as similar as I expected.  I prefer the second
method because I can see the chemistry based on mass all the way through to
the last formula where it is turned into volume. In order for the two final
figures to agree, the air in the second example would have to be about 50 C
so there is still a problem somewhere.

Any takers?

Regards
Crispin