D
daestrom
- Jan 1, 1970
- 0
Curbie said:daestrom,
The components of air are by volume:
Nitrogen Component of Air by volume (Pn) 78.084%
Oxygen Component of Air by volume (Po) 20.946%
Argon Component of Air by volume (Pa) 0.934%
Carbon Dioxide Component of Air by volume (PCo2) 0.033%
... but I also have the mass of each component:
Atomic Mass of Carbon (MaC) 12.0107
Atomic Mass of Nitrogen, (MaN) 14.0067
Atomic Mass of Oxygen (MaO) 15.9994
Atomic Mass of Argon (MaA) 39.9480
Atomic Mass of Carbon Dioxide (MaCo2) 44.0095
... then I tried to convert volume to mass by:
Atomic Mass of Air (MaAir) 14.6759 =(MaN * Pn) + (MaO * Po) +
(MaA * Pa) + (MaCo2 * PCo2)
You seem to have forgotten that N2 and O2 are diatomic gases and have
two atoms per molecule (remember the '2' ;-).
MaAir = 28.0134*78.084% + 31.9988*20.946% + 39.948*.934%+44.0095*.033%
MaAir = 28.96409
=((Hydrogen - (Oxygen * 2)) / 2) + (Carbon * 2)
was used to calculate oxidation required for combustion.
Air mass for oxidation:
oxidation required for combustion * MaAir
As an example, For a mole of octane (c8H18) to burn stoically, you need
12.5 moles of O2 (moles O2 = 1/4Moles H + moles C). MaOctane is 114.23
(8*12.0107+18*1.00794).
So for 114.23 lbm of octane, you need 399.985 lbm O2 (12.5*31.9988).
Air is 23.14% O2 by mass ((20.946%*31.9988)/28.96409) so for 399.985 lbm
of O2 you need 1729.498 lbm of air (399.985/23.14%).
That ratio can be reduced down to 15.13 lbm of air for every 1 lbm of
Octane.
'Gasoline' is really a mixture of many different compounds, but this
gives you the idea. Longer-chain compounds (heavier oils) with a H-C
ratio very close to 2:1 would tend towards 14.7:1. (MaCH2 = 14.03,
needing 1.5 moles O2)
daestrom