[Gasification] Greg's reply to my request for help on air ratio.

[John Bertl]

Greg, I believe I understand what you are saying.

The formula for gasification is given as:

CH1.4O0.6 + 0.4O2    =>

     0.7CO + 0.3CO2 + 0.6H2 + 0.1 H2O

This formula, of course, does not have
the Nitrogen from atmospheric air on
both sides of the equation which would
be in the same ratio as used in the
combustion equation.  (1.05 / 3.95)

1.05 / 3.95 is equal to 0.4 / X

X = 1.50

For Gasification:
Nitrogen in air = 1.50 * 14 * 2 = 42.1 atm wt
Oxygen in air   =   0.2 * 16 * 2 = 12.8 atm wt

         Total atomic weight of air = 54.9

>From previous work, atomic weight of wood = 23

  23 / 54.9 = 2.39

This was my earlier number

Now you say I should only apply the ratio to
the 30% wood that is combusted.

“... apply that air ratio to the 30% that is
combusted... make sense now ? “

Let see

30% of 23 is 6.6

54.9 / 6.6 = 8.32

I believe this ratio deviates even more???

More importantly, do I have your thoughts correct??

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