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/_/ / / / / __/ (__ ) / __/ /
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/_/ \___/ /____/ \___/
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.
Let's take a look at your figures.
With a diesel fuel oil consisting of 86.5% C, 13.2% H, 0.3% S and 0%O
86.5/12=7.21 where 12 is the molecular weight of carbon.
So ignoring the 0.3% sulphur, we can pretend your hydrocarbon is C7.21H13.2.
(I'm not saying the molecules are that short, I'm just thinking about the
carbon hydrogen ratio.)
CxHy + a(O2+0.79/0.21N2) => xCO2 + y/2H2O + a*0.79/0.21N2
where a=x+y/4
in this case, a= 7.21 + 13.2/4 = 10.51
AFR = a(MWO2+0.79/0.21*MWN2)/MWfuel
AFR = 10.51 (32 + 28*0.79/0.21) / (86.5+13.2)
= 10.51 * 137.33 / 99.7
= 14.42
14.55 Kg air to Kg fuel
So a rough calculation for your fuel (ignoring the sulpher in the fuel, argon
in the air, etc) gives 14.42 which is pretty close to the figure you gave - not
that I ever doubted you, ..., more a question of trying to understand why
things are the way they are.
So why the difference to the earlier estimate?
If hydrogens were about 2:1 with the carbons, we could expect an AFR a bit
below 15 as indicated earlier.
7.21*2= 14.42.
The actual percentage hydrogen you give is 13.2, ie less that 2 hydrogens per
carbon.
