%CO versus AFR
Craig Dotson
crdotson at vt.edu
Wed Oct 31 14:04:08 GMT 2001
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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 -----=20
From: Rich M=20
To: Diy_Efi at Diy-Efi. Org=20
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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<DIV><FONT size=3D2>Stanford University has created a useful piece of =
freeware=20
that does combustion equilibrium calculations. There are a lot of=20
simplifications and assumptions made for the program to work, but it =
does give a=20
good representation of the trends you will see in the real =
world.</FONT></DIV>
<DIV><FONT size=3D2></FONT> </DIV>
<DIV><FONT size=3D2>I have used this program in my IC Engines class to =
calculate=20
exactly what you ask for. I didn't take a lot of data points, and =
the=20
combustion products are assumed to be at 1atm pressure and 2500 Kelvin =
(may or=20
may not be appropriate). For an equivalence ratio of 1.15 (12.6:1 =
is=20
approximately 1.17 equivalence) CO represents approximately 4.5% of the =
moles of=20
the exhaust mixture. I'm not quite sure how you convert from mole =
fraction=20
to ppm as I'm not a chemistry/chem eng type guy, but that's your mole=20
fraction.</FONT></DIV>
<DIV><FONT size=3D2></FONT> </DIV>
<DIV><FONT size=3D2>If you're interested in trends, CO concentration =
increases at=20
a very low slope (slight increase) as equivalence ratio increases to =
1. =20
Around 1, the slope increases and proceeds to increase almost linearly =
at a=20
higher slope as equivalence ratio goes further above 1. If you're=20
interested in knocking down your CO, bring it closer to stochiometric or =
put a=20
catalytic converter in the exhaust stream.</FONT></DIV>
<DIV><FONT size=3D2></FONT> </DIV>
<DIV><FONT size=3D2>Craig Dotson<BR><A=20
href=3D"mailto:crdotson at vt.edu">crdotson at vt.edu</A><BR>2002 VT=20
FormulaSAE</FONT></DIV>
<BLOCKQUOTE dir=3Dltr=20
style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV=20
style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
<A title=3Drsrich at cwcom.net href=3D"mailto:rsrich at cwcom.net">Rich =
M</A> </DIV>
<DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=3Ddiy_efi at diy-efi.org=20
href=3D"mailto:Diy_Efi at Diy-Efi. Org">Diy_Efi at Diy-Efi. Org</A> </DIV>
<DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Monday, October 29, 2001 =
11:57=20
AM</DIV>
<DIV style=3D"FONT: 10pt arial"><B>Subject:</B> %CO versus AFR</DIV>
<DIV><BR></DIV>
<DIV><FONT face=3DTahoma size=3D2><SPAN =
class=3D060275116-29102001>Knowledgeable=20
listers,</SPAN></FONT></DIV>
<DIV><FONT face=3DTahoma size=3D2><SPAN =
class=3D060275116-29102001>Does anyone know=20
the relationship in approximate terms for %CO in exhaust gas against=20
AFR?</SPAN></FONT></DIV>
<DIV><FONT face=3DTahoma size=3D2><SPAN class=3D060275116-29102001>eg, =
what is a=20
reasonable %CO to see at 12.6:1</SPAN></FONT></DIV>
<DIV><FONT face=3DTahoma size=3D2><SPAN class=3D060275116-29102001>A =
relationship,=20
graph or table for a range would be most useful...</SPAN></FONT></DIV>
<DIV><FONT face=3DTahoma size=3D2><SPAN class=3D060275116-29102001>I =
can imagine=20
that it will vary a bit depending on conditions, but an approximate =
idea at=20
the moment will be fine.</SPAN></FONT></DIV>
<DIV><FONT face=3DTahoma size=3D2><SPAN=20
class=3D060275116-29102001></SPAN></FONT> </DIV>
<DIV><FONT face=3DTahoma size=3D2><SPAN=20
class=3D060275116-29102001>Regards</SPAN></FONT></DIV>
<DIV><FONT face=3DTahoma size=3D2><SPAN=20
=
class=3D060275116-29102001>Rich</SPAN></FONT></DIV></BLOCKQUOTE></BODY></=
HTML>
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