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<DIV>Greetings and a good day to you. </DIV>
<DIV> </DIV>
<DIV>Please do not send attachments. This one seems infected. </DIV>
<DIV> </DIV>
<DIV>I was led to understan the List policy was not to permit attachments.
</DIV>
<DIV> </DIV>
<DIV>Thank you, </DIV>
<DIV> </DIV>
<DIV>Bill Klein</DIV>
<DIV>3i</DIV>
<DIV><A
href="http://www.3iAlternativePower.com">http://www.3iAlternativePower.com</A>
</DIV>
<DIV> </DIV>
<DIV> </DIV>
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<DIV style="FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV
style="FONT: 10pt arial; BACKGROUND: #e4e4e4; font-color: black"><B>From:</B>
<A title=gfwhell@aol.com href="mailto:gfwhell@aol.com">GF</A> </DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A title=mark@ludlow.com
href="mailto:mark@ludlow.com">mark@ludlow.com</A> ; <A
title=gasification@lists.bioenergylists.org
href="mailto:gasification@lists.bioenergylists.org">gasification@lists.bioenergylists.org</A>
</DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Sunday, March 06, 2011 8:10
PM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Re: [Gasification] Benefits of
boosting compression ratio</DIV>
<DIV><BR></DIV><FONT color=black size=2 face=arial>
<DIV><FONT face="Arial, Helvetica, sans-serif"></FONT>Mark,</DIV>
<DIV>I always thought that it was possible to have a "slow explosion" as it
all depended on "flame speed" I</DIV>
<DIV>Which increases with compression. (a slow explosion can have certain
benefits) So run lean</DIV>
<DIV> </DIV>
<DIV><EM> <SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">It
would seem that if a greater quantity of fuel and air gas mixture is induced,
compressed and "exploded during" the power stroke of the
engine</SPAN></EM></DIV><BR>The point I am making is that the separation of
gasses has been successfully achieved with HRVT<BR>
<DIV style="CLEAR: both"></DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV>GF<BR><BR></DIV>
<DIV
style="FONT-FAMILY: arial,helvetica; COLOR: black; FONT-SIZE: 10pt">-----Original
Message-----<BR>From: Mark Ludlow <mark@ludlow.com><BR>To: 'Discussion
of biomass pyrolysis and gasification'
<gasification@lists.bioenergylists.org><BR>Sent: Sun, Mar 6, 2011 12:30
pm<BR>Subject: Re: [Gasification] Benefits of boosting compression
ratio<BR><BR>
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<DIV class=WordSection1>
<DIV class=MsoNormal><SPAN
style="FONT-FAMILY: 'Calibri','sans-serif'; COLOR: #1f497d">Just as a small
point, Geoff, the fuel does not “explode” but burns at a progressive rate.
Explosions do happen, inadvertently, but these detonations severely shorten
engine life.</SPAN></DIV>
<DIV class=MsoNormal><SPAN
style="FONT-FAMILY: 'Calibri','sans-serif'; COLOR: #1f497d">Mark</SPAN></DIV>
<DIV class=MsoNormal><SPAN
style="FONT-FAMILY: 'Calibri','sans-serif'; COLOR: #1f497d"></SPAN> </DIV>
<DIV
style="BORDER-BOTTOM: medium none; BORDER-LEFT: medium none; PADDING-BOTTOM: 0in; PADDING-LEFT: 0in; PADDING-RIGHT: 0in; BORDER-TOP: #b5c4df 1pt solid; BORDER-RIGHT: medium none; PADDING-TOP: 3pt">
<DIV style="MARGIN-LEFT: 0.5in" class=MsoNormal><B><SPAN
style="FONT-FAMILY: 'Tahoma','sans-serif'; FONT-SIZE: 10pt">From:</SPAN></B><SPAN
style="FONT-FAMILY: 'Tahoma','sans-serif'; FONT-SIZE: 10pt"> <A
href="mailto:gasification-bounces@lists.bioenergylists.org">gasification-bounces@lists.bioenergylists.org</A>
[<A
href="mailto:gasification-bounces@lists.bioenergylists.org?">mailto:gasification-bounces@lists.bioenergylists.org</A>]
<B>On Behalf Of </B>GF<BR><B>Sent:</B> Sunday, March 06, 2011 8:53
AM<BR><B>To:</B> <A
href="mailto:gasification@lists.bioenergylists.org">gasification@lists.bioenergylists.org</A><BR><B>Subject:</B>
Re: [Gasification] Benefits of boosting compression ratio</SPAN></DIV></DIV>
<DIV style="MARGIN-LEFT: 0.5in" class=MsoNormal> </DIV>
<DIV>
<DIV id=AOLMsgPart_2_4acca7ce-fe21-4489-9192-1726c94f505b>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">It
would seem that if a greater quantity of fuel and air gas mixture is induced,
compressed and exploded during the power stroke of the engine, a greater horse
power output can be expected. However, there is a considerable amount of
"inert gas" induced with this explosive mixture. Although this inert gas
plays a very small part in pushing the piston other than by thermal
expansion.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">What
effect could be expected, if we could somehow reduce some of this
"inert gas" from the induced explosive mixture?</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">There
are various methods of gas separation that could be considered.The most
promising would be HRVT Which has hardly any moving parts which imposes
high centrifugal forces upon gasses separating them by weight
differential.Here there might be a bonus as one of the separated gasses will
be cooled.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">GF</SPAN></DIV></DIV>
<DIV
style="MARGIN-BOTTOM: 12pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 12pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">-----Original
Message-----<BR>From: Daniel Chisholm <<A
href="mailto:dmc@danielchisholm.com">dmc@danielchisholm.com</A>><BR>To:
Discussion of biomass pyrolysis and gasification <<A
href="mailto:gasification@lists.bioenergylists.org">gasification@lists.bioenergylists.org</A>><BR>Sent:
Thu, Mar 3, 2011 7:33 am<BR>Subject: Re: [Gasification] Benefits of boosting
compression ratio</SPAN></DIV>
<DIV id=AOLMsgPart_3_61045ce8-4136-4a30-951e-e60ae2083283>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">First
the answer, then a long-winded explanation if you care for
it...</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<BLOCKQUOTE
style="BORDER-BOTTOM: medium none; BORDER-LEFT: #cccccc 1pt solid; PADDING-BOTTOM: 0in; PADDING-LEFT: 6pt; PADDING-RIGHT: 0in; MARGIN-LEFT: 4.8pt; BORDER-TOP: medium none; MARGIN-RIGHT: 0in; BORDER-RIGHT: medium none; PADDING-TOP: 0in">
<TABLE style="WIDTH: 463.2pt; MARGIN-LEFT: 0.5in" class=MsoNormalTable
border=0 cellSpacing=0 cellPadding=0 width=618>
<TBODY>
<TR>
<TD
style="PADDING-BOTTOM: 0in; PADDING-LEFT: 0in; PADDING-RIGHT: 0in; PADDING-TOP: 0in"
vAlign=top>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal>Then change the pressure
to 5 psig (say ~19 psia) with the same parameters and calculate the
mass of oxygen. Do the masses come out with the ratio of 9.8/19
or about 1/2 for my comparison between natural aspirated and (5
psig) low pressure?</DIV></DIV></TD></TR></TBODY></TABLE></BLOCKQUOTE>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">Yes.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<BLOCKQUOTE
style="BORDER-BOTTOM: medium none; BORDER-LEFT: #cccccc 1pt solid; PADDING-BOTTOM: 0in; PADDING-LEFT: 6pt; PADDING-RIGHT: 0in; MARGIN-LEFT: 4.8pt; BORDER-TOP: medium none; MARGIN-RIGHT: 0in; BORDER-RIGHT: medium none; PADDING-TOP: 0in">
<TABLE style="WIDTH: 463.2pt; MARGIN-LEFT: 0.5in" class=MsoNormalTable
border=0 cellSpacing=0 cellPadding=0 width=618>
<TBODY>
<TR>
<TD
style="PADDING-BOTTOM: 0in; PADDING-LEFT: 0in; PADDING-RIGHT: 0in; PADDING-TOP: 0in"
vAlign=top>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal> If so, then there
is twice the oxygen mass in a 5 psig pressure going into the engine,
as compared to 20 in hg (9.8
psia).</DIV></DIV></TD></TR></TBODY></TABLE></BLOCKQUOTE>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">Yes.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">And
you could therefore burn twice as much fuel, and therefore produce twice as
much gross horsepower. ("gross horsepower" means how much power your
pistons are extracting from the gas; most of this will make it to the
crankshaft output but some will be lost to sliding friction and pumping
losses)</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">Although
your net horsepower output at the crank won't be exactly twice as much (it'll
probably be a little bit better than that), it's a very good starting point to
make the approximation that it is.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><B><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">Bottom
line: power output is proportional to (absolute) manifold
pressure.</SPAN></B><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<BLOCKQUOTE
style="BORDER-BOTTOM: medium none; BORDER-LEFT: #cccccc 1pt solid; PADDING-BOTTOM: 0in; PADDING-LEFT: 6pt; PADDING-RIGHT: 0in; MARGIN-LEFT: 4.8pt; BORDER-TOP: medium none; MARGIN-RIGHT: 0in; BORDER-RIGHT: medium none; PADDING-TOP: 0in">
<TABLE style="MARGIN-LEFT: 0.5in" class=MsoNormalTable border=0
cellSpacing=0 cellPadding=0>
<TBODY>
<TR>
<TD
style="PADDING-BOTTOM: 0in; PADDING-LEFT: 0in; PADDING-RIGHT: 0in; PADDING-TOP: 0in"
vAlign=top>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal> </DIV></DIV>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal>Change the temperature
to 95C for both. Is it the
same?</DIV></DIV></TD></TR></TBODY></TABLE></BLOCKQUOTE>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">Yes.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">You
get twice as much air mass flow at 19.7psia as you do at
9.8psia.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">You
didn't ask this but since you probably know that your engine will make less
horsepower with 95C inlet air temperature than with 15C air temperature I'll
ask and answer a question here:</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">How
much power does an engine make at 95C inlet air temperature versus at 15C
inlet air temperature?</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">The
first thing to do is to convert these into absolute temperatures, 15C is
288 Kelvin (15+273), and 95C is 368 Kelvin.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">Your
engine's displacement is unchanged by the inlet air temperature. But the
density of the air that it moves is changed, in inverse proportion to the
(absolute) temperature ratio:</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">288/368
= 0.78</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">So
when your inlet air temp is 95C (368K), your engine will make 78% of the power
it makes when your inlet air temp is 15C (288K).</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">This
is because it is only flowing 78% as much air (mass-wise), therefore it is
only able to burn 78% as much fuel, and therefore there is only 78% as much
energy available to be extracted.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<BLOCKQUOTE
style="BORDER-BOTTOM: medium none; BORDER-LEFT: #cccccc 1pt solid; PADDING-BOTTOM: 0in; PADDING-LEFT: 6pt; PADDING-RIGHT: 0in; MARGIN-LEFT: 4.8pt; BORDER-TOP: medium none; MARGIN-RIGHT: 0in; BORDER-RIGHT: medium none; PADDING-TOP: 0in">
<TABLE style="WIDTH: 463.2pt; MARGIN-LEFT: 0.5in" class=MsoNormalTable
border=0 cellSpacing=0 cellPadding=0 width=618>
<TBODY>
<TR>
<TD
style="PADDING-BOTTOM: 0in; PADDING-LEFT: 0in; PADDING-RIGHT: 0in; PADDING-TOP: 0in"
vAlign=top>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal> </DIV></DIV>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal>I'm sorry to ask,
but as I read the ideal gas law equasions, my eyes
glass over and confusion sets
in.</DIV></DIV></TD></TR></TBODY></TABLE></BLOCKQUOTE>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">You'll
notice that I never did plug any numbers into "PV=nRT" (or P = rho*R*T) in any
of the above.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">A
great deal of useful insight can be had from looking at the form of the
equations, from this you can get useful relationships like</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">- If
temperature is constant, density is proportional to pressure (I used that in
the first part)</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">- For
any particular pressure, density changes inversely proportional to temperature
(that's what I used in the second part)</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">Now
to the first part of your email and longer-winded bit..</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">On
Wed, Mar 2, 2011 at 14:11, Toby Seiler <<A
href="mailto:seilertechco@yahoo.com">seilertechco@yahoo.com</A>>
wrote:</SPAN></DIV>
<TABLE style="WIDTH: 6.5in; MARGIN-LEFT: 0.5in" class=MsoNormalTable border=0
cellSpacing=0 cellPadding=0 width=624>
<TBODY>
<TR>
<TD
style="PADDING-BOTTOM: 0in; PADDING-LEFT: 0in; PADDING-RIGHT: 0in; PADDING-TOP: 0in"
vAlign=top>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal>Mark, Daniel C.,
list,</DIV></DIV>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal> </DIV></DIV>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal>OK I admit that I'm lost
when I get to "mole" in the ideal gas law. (A mole is a furry
creature my dog brings up on the porch to chew up.)</DIV></DIV>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal> </DIV></DIV>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal>So please help me in
another way if you understand the physics. First assume I use a
460 cubic inch engine and have a 10/1 compression ratio. In total
then all cylinders compress the gas/air to an area of 46 cubic
inches...easy enough. At 12/1 it is compressed to
38.3.</DIV></DIV></TD></TR></TBODY></TABLE>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">There's
a small error there which I'll point out for completeness.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">(firstly
good for you for choosing a big-block in your example, even if I would have
used 440 or 426 cubic inches in my 'fer-instance! :-)</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">Your
pistons *displace* 460 cubic inches, that is when they move from BDC to TDC
they have pushed through 460 cubic inches of volume.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">At
TDC the gas will be in the combustion chamber volume. At BDC the gas
will occupy this volume *PLUS* 460 cubic inches.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">So
we're naturally into algebra (sorry! ;-)</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">CR =
Vol_BDC / Volume_TDC</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">CR=
(displacement + CombChmbr_volume) / (CombChmbr_volume)</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">After
a bit of algebra we get:</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">CombChmbrvolume
= displacement/(CR-1)</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">For
displacement=460 in^3 and and CR=10 we get CombChmbr_volume = 460/9 in^3 or
51.1 in3</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">What
is happening is that we start with 511.1 in^3 of gas, which we then compress
by 460 in^3 down to 51.1 in^3 (we change the volume by 10:1 ---> our
compression ratio!)</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">For
displacement=460 in^3 and and CR=12 we get CombChmbr_volume = 460/11 in^3 or
41.8 in3 (we compress 501.8 in^3 of gas down to 41.8 in^3, i.e. by
12:1)</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<BLOCKQUOTE
style="BORDER-BOTTOM: medium none; BORDER-LEFT: #cccccc 1pt solid; PADDING-BOTTOM: 0in; PADDING-LEFT: 6pt; PADDING-RIGHT: 0in; MARGIN-LEFT: 4.8pt; BORDER-TOP: medium none; MARGIN-RIGHT: 0in; BORDER-RIGHT: medium none; PADDING-TOP: 0in">
<TABLE style="WIDTH: 463.2pt; MARGIN-LEFT: 0.5in" class=MsoNormalTable
border=0 cellSpacing=0 cellPadding=0 width=618>
<TBODY>
<TR>
<TD
style="PADDING-BOTTOM: 0in; PADDING-LEFT: 0in; PADDING-RIGHT: 0in; PADDING-TOP: 0in"
vAlign=top>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal> </DIV></DIV>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal>Now help me find the mass
of the oxygen (we will disregard the nitrogen, argon, etc. for
now),</DIV></DIV>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal>if my manifold is 20
hg (about 9.8 psi absolute) and I know oxygen weighs 32 grams
per mole and I assume 23C (of course it will be higher in a hot
engine)...how much oxygen mass is going into the cylinders (with
oxygen at 21% of air and for now disregarding the
fuel). </DIV></DIV></TD></TR></TBODY></TABLE></BLOCKQUOTE>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">It's
easier if you consider the air, and then when you're done that figure out the
oxygen.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">You
can calculate how much air flows in one cycle (which in a four stroke engine
is two crank rotations, or you can figure how much air flows when the engine
is turning at a certain RPM. I'll do the latter since it helps to have
in hand some real mass flow numbers for your air, which closely relate to fuel
flow rates, which also relate to power outputs - and fuel flow rates and power
outputs are something that we probably have a certain intuitive grasp
of.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">Let's
say you are turning your engine at 1800 RPM, a nice respectable synchronous
genset speed. So your crank makes 1800 rotations in a minute, which is
900 engine cycles. So your pistons will have displaced 460 cubic inches
times 900 cycles which is 414,000 cubic inches. Which is also a number
that means very little to me intuitively. It is 414000/1728 = 239.6
cubic feet, which might be a bit easier to grasp physically. Also for
anybody who has race engine experience, many carburetors are rated in cfm and
here we see that we need a 240cfm carbureter. A typical bigblock might
be fitted with a racing carburetor rated for 850cfm or 1050cfm (which is how
much air your engine will need at 6120rpm or 7875rpm respectively, which are
the sort of RPMs a race engine might be run at)</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">So
<B>your 460 c.i.d. engine is displacing 239 cubic feet of air per minute when
it turns 1800rpm</B>. It displaces, in its cylinders, 239 cubic feet per
minute no matter what the throttle setting is and no matter what the intake
manifold pressure is.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">In SI
units, 239 ft^3/minute is 6.77 cubic metres per minute. What Canadians
call a "45 gallon drum" and what Americans call a "55 gallon drum" is about
200 litres which is 0.2 cubic metres. This airflow is about 34 drums per
minute, or about half a drum of air per second.</SPAN></DIV></DIV>
<DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"><BR>In
US Customary units the density of air is .076 pounds per cubic foot (at 59F
and atmospheric pressure).</SPAN></DIV></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">So
let's say you set your throttle so that the manifold pressure is 20"Hg
absolute (on most engine vacuum gauges which go from 0"Hg at atmospheric
pressure to 29.92"Hg of vacuum, this will read as 10"Hg of
vacuum).</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">At
this throttle setting your manifold pressure will be about 9.8psia as you
said. This is 20/29.92 = 0.67 times atmospheric
pressure.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">When
you throttle a gas flow, its temperature remains unchanged, its pressure
drops, and its density drops by the same proportion. So if you are
throttling your air to 0.67 times atmospheric pressure, its density will be
0.67 times as big.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">So
your air mass flow rate is 239 ft^3/minute times .076 pounds/ft^3 times 0.67 =
12.17 pounds of air per minute.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">I did
promise to tell you the oxygen flow rate so this is where we do it: 0.21
(pounds of oxygen per pound of air) times 12.17 lbs/minute air = 2.56 pounds
per minute of oxygen. </SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">It
might be more tangible though to think in terms of air mass flow rates.
If we are burning gasoline we'd likely be using an air:fuel mixture
ratio somewhere in the range of 12:1 to 15:1. These mixture ratios are
"pounds of air per pound of fuel".</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">Since
we are flowing 12.17 pounds per minute of air, we'll need to flow about 1
pound per minute of gasoline.</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt">One
pound per minute is sixty pounds per hour, which is 10 US gallons per hour.
Does this seem "about right" for a 460 c.i.d engine turning 1800rpm and
producing about 65% of its wide-open power output at that speed? (It
passes my smell test, because it seems "close enough" to me to a 470 in^3
aircraft engine turning 2000-2300RPM, at 65% power setting, burning 12-15
USgph)</SPAN></DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<DIV>
<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"></SPAN> </DIV></DIV>
<BLOCKQUOTE
style="BORDER-BOTTOM: medium none; BORDER-LEFT: #cccccc 1pt solid; PADDING-BOTTOM: 0in; PADDING-LEFT: 6pt; PADDING-RIGHT: 0in; MARGIN-LEFT: 4.8pt; BORDER-TOP: medium none; MARGIN-RIGHT: 0in; BORDER-RIGHT: medium none; PADDING-TOP: 0in">
<TABLE style="MARGIN-LEFT: 0.5in" class=MsoNormalTable border=0
cellSpacing=0 cellPadding=0>
<TBODY>
<TR>
<TD
style="PADDING-BOTTOM: 0in; PADDING-LEFT: 0in; PADDING-RIGHT: 0in; PADDING-TOP: 0in"
vAlign=top>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal>Best regards,
</DIV></DIV>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal> </DIV></DIV>
<DIV>
<DIV style="MARGIN-TOP: 4pt" class=MsoNormal>Toby
</DIV></DIV></TD></TR></TBODY></TABLE>
<DIV
style="MARGIN-BOTTOM: 12pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"><BR><BR>_______________________________________________<BR>Gasification
mailing list<BR><BR>to Send a Message to the list, use the email
address<BR><A
href="mailto:Gasification@bioenergylists.org">Gasification@bioenergylists.org</A><BR><BR>to
UNSUBSCRIBE or Change your List Settings use the web page<BR><A
href="http://lists.bioenergylists.org/mailman/listinfo/gasification_lists.bioenergylists.org"
target=_blank>http://lists.bioenergylists.org/mailman/listinfo/gasification_lists.bioenergylists.org</A><BR><BR>for
more Gasifiers, News and Information see our web site:<BR><A
href="http://gasifiers.bioenergylists.org/"
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<DIV
style="MARGIN-BOTTOM: 0pt; MARGIN-LEFT: 0.5in; MARGIN-RIGHT: 0in; mso-margin-top-alt: 4.0pt"
class=MsoNormal><SPAN
style="FONT-FAMILY: 'Arial','sans-serif'; COLOR: black; FONT-SIZE: 10pt"><BR><BR
clear=all><BR>-- <BR>- Daniel<BR>Fredericton, NB
Canada</SPAN></DIV></DIV>
<DIV id=AOLMsgPart_4_61045ce8-4136-4a30-951e-e60ae2083283><PRE style="BACKGROUND: white; MARGIN-LEFT: 0.5in"><TT><SPAN style="COLOR: black">_______________________________________________</SPAN></TT></PRE><PRE style="BACKGROUND: white; MARGIN-LEFT: 0.5in"><TT><SPAN style="COLOR: black">Gasification mailing list</SPAN></TT></PRE><PRE style="BACKGROUND: white; MARGIN-LEFT: 0.5in"><TT><SPAN style="COLOR: black"> </SPAN></TT></PRE><PRE style="BACKGROUND: white; MARGIN-LEFT: 0.5in"><TT><SPAN style="COLOR: black">to Send a Message to the list, use the email address</SPAN></TT></PRE><PRE style="BACKGROUND: white; MARGIN-LEFT: 0.5in"><TT><SPAN style="COLOR: black"><A href="mailto:Gasification@bioenergylists.org">Gasification@bioenergylists.org</A></SPAN></TT></PRE><PRE style="BACKGROUND: white; MARGIN-LEFT: 0.5in"><TT><SPAN style="COLOR: black"> </SPAN></TT></PRE><PRE style="BACKGROUND: white; MARGIN-LEFT: 0.5in"><TT><SPAN style="COLOR: black">to UNSUBSCRIBE or Change your List Settings use the web page</SPAN></TT></PRE><PRE style="BACKGROUND: white; MARGIN-LEFT: 0.5in"><TT><SPAN style="COLOR: black"><A href="http://lists.bioenergylists.org/mailman/listinfo/gasification_lists.bioenergylists.org" target=_blank>http://lists.bioenergylists.org/mailman/listinfo/gasification_lists.bioenergylists.org</A></SPAN></TT></PRE><PRE style="BACKGROUND: white; MARGIN-LEFT: 0.5in"><TT><SPAN style="COLOR: black"> </SPAN></TT></PRE><PRE style="BACKGROUND: white; MARGIN-LEFT: 0.5in"><TT><SPAN style="COLOR: black">for more Gasifiers, News and Information see our web site:</SPAN></TT></PRE><PRE style="BACKGROUND: white; MARGIN-LEFT: 0.5in"><TT><SPAN style="COLOR: black"><A href="http://gasifiers.bioenergylists.org/" target=_blank>http://gasifiers.bioenergylists.org/</A></SPAN></TT></PRE></DIV></DIV></DIV></DIV></DIV></DIV><!-- end of AOLMsgPart_3_f293712e-96ae-4bf2-85e9-afa9f344c65f -->
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