Compare Fuels Chemically - Page 8

 

Jim,

I'm studying the notion of a small continuous still (~.5 liter/h) to
be primarily solar heated, with heat augmented as needed by a
secondary source, to run 24/7 120-150 days per year. Using your
concern which I have considered, the solar would supply all the heat
when it can, and when it can't, it would supply as much as it can,
augmented by something dependable, maybe electric.

Curbie

 

Curbie wrote:

Not to throw too many variables into your calculations, but you have to
be careful of your percentages for 'standard air'.  A common number you
might run across is 21% of air as O2, but that is by volume, not by
weight/mass.

So if you have 78% N2, 21% O2 and 1% Ar by volume, that works out to
about 75.4% N2, 23.2% O2 and 1.4% Ar by weight.

And of course the chemical reactions are mass balances, not volume, so
you might check this part.

daestrom

 

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)

=((Hydrogen - (Oxygen * 2)) / 2) + (Carbon * 2)
was used to calculate oxidation required for combustion.

Air mass for oxidation:
oxidation required for combustion * MaAir

I know fuels are not chemically monolithic, but in order to try an
objective "apples to oranges" comparison, it seemed that ideal
chemicals may get me close.

Curbie

 

Curbie wrote:

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



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

 

daestrom,

Thanks daestrom!

I didn't forget oxygen and nitrogen were diatomic I never took
chemistry in high school so I'll have chalk this up to being dumb not
forgetful. I'll straighten out the sheet and try to push it forward; I
don't know if in the end the spread-sheet will be helpful, but you
never know until you try.

Thanks again,

Curbie