Stoich
Clare Snyder
claresnyder at home.com
Fri Dec 10 03:02:29 GMT 1999
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believe what you want to believe - I asked
see:
http://www.chemical-stoichiometry.net/Resource_Titlepage.htm
http://www.chem.ualberta.ca/~plambeck/che/course/p0304x.htm
http://www.compusmart.ab.ca/plambeck/che/p101/p01172.htm
http://www.chem.ualberta.ca/~plambeck/che/p101/p01035.htm
http://www.compusmart.ab.ca/plambeck/che/p101/p01033.htm
http://www.chemical-stoichiometry.net/Tutorial_Titlepage.htm
http://www.chemical-stoichiometry.net/
http://www.chemical-stoichiometry.net/tutorial.html
http://www.collingwood.org/chemistry/Curriculum%20Docs/ch1111.htm
http://www.chem.ualberta.ca/~plambeck/che/struct/s0201.htm
http://www.chemical-stoichiometry.net/begin_student.htm
http://www.compusmart.ab.ca/plambeck/che/course/vn41505x.htm
http://www.chem.ualberta.ca/~plambeck/che/p102/p0511x.htm
and
http://www.chem.ualberta.ca/courses/plambeck/p102/p02085.htm
to start with.
Says if anybody is full of BS it's not me. Stoiciometry in many points =
of reference - other than combustion.
An excerpt from the last reference:
Stoichiometric calculations involving redox reactions always begin with =
the balanced redox reaction. Once the balanced redox reaction has been =
obtained, the mole ratios will give the desired stoichiometric =
information. .............
Combustion is a Redox reaction (oxidation/reduction)
I've been told to make sure I've got my facts straight before posting =
"opinions" - the definition I gave may be a gross simplification - but =
the facts are the facts.
Now, as for stoich being a moving target - A theoretically stoich =
mixture may not behave as stoich under certain conditions - something to =
do with elemental vs radical oxygen - elemental oxygen is not as =
reactive as radical oxygen, so if the O2 can be dissassociated into 2 X =
O, and the Hydrogen and Oxygen of the hydrocarbon fuel dissaciated into =
hydrogen and carbon, the reactions of oxygen + carbon =3D carbon =
dioxide, and hydrogen + oxygen=3Dwater can become stoichiometrically =
correct, Changing the reaction of nitrogen + oxygen=3Dnitous oxides (or =
oxides of nitogen - NOX) in the cyl can and does change the =
stoichiometry of the reaction in the cyl. This is, in effect, what EGR =
does to an engine. By inhibiting the formation of NOX the actual mixture =
ratio of gasoline and air required can be can be changed. Decrease NOX =
production and the mixture required for a stoichiometrically correct =
combustion of the fuel can be richened because more oxygen is available =
for the desired reaction.
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----- Original Message -----=20
From: Phil Lamovie=20
To: Clare Snyder=20
Sent: Thursday, December 09, 1999 10:02 AM
Subject: Stoich
=20
=20
Hi Clare,=20
Stoich is a point that moves with conditions ?=20
ABSOLUTELY IMPOSSIBLE !=20
is the mixture required to convert ALL of 2 elements to a compound=20
TRUE !=20
I used the chemist's definition of stoichiometricly correct (sic) - =
which is when all of the (active) elements are combined (my italics) to =
produce a compound=20
In a word BULLSHIT ! There is no single compound produced and stoich=20
does not apply to any branch of chemistry except combustion.=20
=20
mixture is the key word.=20
If I had a pound of fuel and 14.7 pounds of air at STP I would have a=20
stoichiometric mixture even if it was held in a balloon. Absolutely no =
combustion is required for this condition it is a MOL. What happens =
when you try to ignite it is a completely different field of study. =
Combustion chemistry which I am sure you are familiar with has a =
multiplicity of factors that affect the results only one of which is the =
MIXTURE that is supplied to the combustion chamber.=20
Baffle them with accuracy and facts; that ought to do the job.=20
Phil=20
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<DIV><FONT face=3DArial size=3D2>believe what you want to believe - I=20
asked</FONT></DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV><FONT face=3DArial size=3D2>see:</FONT></DIV>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.chemical-stoichiometry.net/Resource_Titlepage.htm">htt=
p://www.chemical-stoichiometry.net/Resource_Titlepage.htm</A></FONT></DIV=
>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.chem.ualberta.ca/~plambeck/che/course/p0304x.htm">http=
://www.chem.ualberta.ca/~plambeck/che/course/p0304x.htm</A></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.compusmart.ab.ca/plambeck/che/p101/p01172.htm">http://=
www.compusmart.ab.ca/plambeck/che/p101/p01172.htm</A></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.chem.ualberta.ca/~plambeck/che/p101/p01035.htm">http:/=
/www.chem.ualberta.ca/~plambeck/che/p101/p01035.htm</A></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.compusmart.ab.ca/plambeck/che/p101/p01033.htm">http://=
www.compusmart.ab.ca/plambeck/che/p101/p01033.htm</A></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.chemical-stoichiometry.net/Tutorial_Titlepage.htm">htt=
p://www.chemical-stoichiometry.net/Tutorial_Titlepage.htm</A></FONT></DIV=
>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.chemical-stoichiometry.net/">http://www.chemical-stoic=
hiometry.net/</A></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.chemical-stoichiometry.net/tutorial.html">http://www.c=
hemical-stoichiometry.net/tutorial.html</A></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.collingwood.org/chemistry/Curriculum%20Docs/ch1111.htm=
">http://www.collingwood.org/chemistry/Curriculum%20Docs/ch1111.htm</A></=
FONT></DIV>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.chem.ualberta.ca/~plambeck/che/struct/s0201.htm">http:=
//www.chem.ualberta.ca/~plambeck/che/struct/s0201.htm</A></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.chemical-stoichiometry.net/begin_student.htm">http://w=
ww.chemical-stoichiometry.net/begin_student.htm</A></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.compusmart.ab.ca/plambeck/che/course/vn41505x.htm">htt=
p://www.compusmart.ab.ca/plambeck/che/course/vn41505x.htm</A></FONT></DIV=
>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.chem.ualberta.ca/~plambeck/che/p102/p0511x.htm">http:/=
/www.chem.ualberta.ca/~plambeck/che/p102/p0511x.htm</A></FONT></DIV>
<DIV><FONT face=3DArial size=3D2>and</FONT></DIV>
<DIV><FONT face=3DArial size=3D2><A=20
href=3D"http://www.chem.ualberta.ca/courses/plambeck/p102/p02085.htm">htt=
p://www.chem.ualberta.ca/courses/plambeck/p102/p02085.htm</A></FONT></DIV=
>
<DIV> </DIV>
<DIV><FONT face=3DArial size=3D2>to start with.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>Says if anybody is full of BS it's not =
me.=20
Stoiciometry in many points of reference - other than =
combustion.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>An excerpt from the last =
reference:</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=3DArial size=3D2>Stoichiometric calculations involving =
redox=20
reactions always begin with the balanced redox reaction. Once the =
balanced redox=20
reaction has been obtained, the mole ratios will give the desired =
stoichiometric=20
information. .............</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=3DArial size=3D2>Combustion is a Redox reaction =20
(oxidation/reduction)</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>I've been told to make sure I've got my =
facts=20
straight before posting "opinions" - the definition I gave may be a =
gross=20
simplification - but the facts are the facts.</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=3DArial size=3D2>Now, as for stoich being a moving =
target - A=20
theoretically stoich mixture may not behave as stoich under certain =
conditions -=20
something to do with elemental vs radical oxygen - elemental oxygen is =
not as=20
reactive as radical oxygen, so if the O2 can be dissassociated into 2 X =
O, and=20
the Hydrogen and Oxygen of the hydrocarbon fuel dissaciated into =
hydrogen and=20
carbon, the reactions of oxygen + carbon =3D carbon dioxide, and =
hydrogen +=20
oxygen=3Dwater can become stoichiometrically correct, Changing the =
reaction of=20
nitrogen + oxygen=3Dnitous oxides (or oxides of nitogen - NOX) in the =
cyl can and=20
does change the stoichiometry of the reaction in the cyl. This is, in =
effect,=20
what EGR does to an engine. By inhibiting the formation of NOX the =
actual=20
mixture ratio of gasoline and air required can be can be changed. =
Decrease NOX=20
production and the mixture required for a stoichiometrically correct =
combustion=20
of the fuel can be richened because more oxygen is available for the =
desired=20
reaction.</FONT><FONT face=3DArial size=3D2></FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2> </DIV>
<HR>
</FONT>
<BLOCKQUOTE=20
style=3D"BORDER-LEFT: #000000 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: =
0px; PADDING-LEFT: 5px; PADDING-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 href=3D"mailto:phil at injec.com" title=3Dphil at injec.com>Phil =
Lamovie</A> </DIV>
<DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
href=3D"mailto:claresnyder at home.com"=20
title=3Dclaresnyder at home.com>Clare Snyder</A> </DIV>
<DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Thursday, December 09, =
1999 10:02=20
AM</DIV>
<DIV style=3D"FONT: 10pt arial"><B>Subject:</B> Stoich</DIV>
<DIV><BR></DIV> <BR> =20
<P>Hi Clare,=20
<P>Stoich is a point that moves with conditions ?=20
<P>ABSOLUTELY IMPOSSIBLE !=20
<P>is the mixture required to convert ALL of 2 elements to a compound=20
<P>TRUE !=20
<P>I used the chemist's definition of stoichiometricly correct (sic) - =
which=20
is when all of the (active) elements are <I><U>combined</U></I> =
(my=20
italics) to produce a compound=20
<P>In a word BULLSHIT ! There is no single compound produced and =
stoich=20
<BR>does not apply to any branch of chemistry except combustion. =
<BR> =20
<P>mixture is the key word.=20
<P>If I had a pound of fuel and 14.7 pounds of air at STP I would have =
a=20
<BR>stoichiometric mixture even if it was held in a balloon. =
Absolutely no=20
<BR>combustion is required for this condition it is a MOL. What =
happens when=20
you try to ignite it is a completely different field of study. =
Combustion=20
chemistry which I am sure you are familiar with has a multiplicity of =
factors=20
that affect the results only one of which is the MIXTURE that is =
supplied to=20
the combustion chamber.=20
<P>Baffle them with accuracy and facts; that ought to do the job.=20
<P>Phil </P></BLOCKQUOTE></BODY></HTML>
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