%CO versus AFR

[Programmer]

Do you have a copy ??

Lyndon.

----- Original Message -----
From: "Craig Dotson" <crdotson at vt.edu>
To: <diy_efi at diy-efi.org>
Sent: Wednesday, October 31, 2001 7:03 AM
Subject: Re: %CO versus AFR


Stanford University has created a useful piece of freeware that does
combustion equilibrium calculations.  There are a lot of simplifications and
assumptions made for the program to work, but it does give a good
representation of the trends you will see in the real world.

I have used this program in my IC Engines class to calculate exactly what
you ask for.  I didn't take a lot of data points, and the combustion
products are assumed to be at 1atm pressure and 2500 Kelvin (may or may not
be appropriate).  For an equivalence ratio of 1.15 (12.6:1 is approximately
1.17 equivalence) CO represents approximately 4.5% of the moles of the
exhaust mixture.  I'm not quite sure how you convert from mole fraction to
ppm as I'm not a chemistry/chem eng type guy, but that's your mole fraction.

If you're interested in trends, CO concentration increases at a very low
slope (slight increase) as equivalence ratio increases to 1.  Around 1, the
slope increases and proceeds to increase almost linearly at a higher slope
as equivalence ratio goes further above 1.  If you're interested in knocking
down your CO, bring it closer to stochiometric or put a catalytic converter
in the exhaust stream.

Craig Dotson
crdotson at vt.edu
2002 VT FormulaSAE
  ----- Original Message -----
  From: Rich M
  To: Diy_Efi at Diy-Efi. Org
  Sent: Monday, October 29, 2001 11:57 AM
  Subject: %CO versus AFR


  Knowledgeable listers,
  Does anyone know the relationship in approximate terms for %CO in exhaust
gas against AFR?
  eg, what is a reasonable %CO to see at 12.6:1
  A relationship, graph or table for a range would be most useful...
  I can imagine that it will vary a bit depending on conditions, but an
approximate idea at the moment will be fine.

  Regards
  Rich


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