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Ron, Crispin,<br>
Many moons ago, when I had access to an IR CO,CO2 monitor, I
filtered the gasses before the flame on a TLUD. If memory serves me
I got 9% CO and 15% CO2. But memory sometimes seems to be serving
others. If only we had access to the old archives, I reported it to
this list at the time, but alas the 'server' changed. It amazes me
that nobody has done this since.<br>
Alex<br>
<br>
<br>
On 24/10/2011 10:41 PM, Crispin Pemberton-Pigott wrote:
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<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Dear
Ron<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<div>
<div>
<div>
<p class="MsoNormal"><span
style="font-family:"Arial","sans-serif";color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">>></span><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">“For
those not having experience with TLUDs, Dean's
reference to "no primary air can make it up", means
that the oxygen is "entirely" used to produce carbon
monoxide.”</span><span style="color:black"><o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">…<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">>></span><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">It
is really rare to find a normalised CO emissions
factor (not concentration in the emerging gases) above
100,000 ppm. I have only see it once and I work with
some of the wildest devices the imagination has
produced. </span><span style="color:black"><o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">></span><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">
<b>[RWL: This part I don't understand. Neither Dean
or I were talking about anything other than primary
air. </b></span><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">I
see that you did not follow. I am sure is it my
method of describing it. Here is a short version:
You can’t get pure CO from biomass pyrolysis for
inherent chemical reasons. <o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Proof
that I offer: I have measured CO production across
a wide range of conditions and it is almost
impossible to get more than 10% CO even when it is
theoretically possible (from the elemental
composition) to get 40%.<o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="color:#1F497D">>></span><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">CO(ppm)
* (EA+100%) = CO(EF) at O2=0% (the O2 is factored
out). </span><span style="color:black"><o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">></span><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">
<b>[RWL3: The subject of excess air for testing the
completeness of combustion (after adding secondary
air and releasing the majority of the energy) is
extraneous to the sentence under discussion.]</b></span><span
style="color:black"><o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">That
is how to work out what the CO level is, in ppm.
100,000 ppm is 10%.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">>></span><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">I
mention this to support my conclusion that the O2
tends to create ‘fuel moisture’ very easily.</span><span
style="color:black"><o:p></o:p></span></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">></span></b><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">I
have personally measured the gases coming up through
the fuel bed in a TLUD (a borrowed high quality
tool) and the dominant gas was CO (many millions of
ppm). </span></b><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">That
is impossible. 1 million parts per million is 100% CO.<o:p></o:p></span></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">>></span></b><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">
RWL4: My main concern is with Crispin's above
next-to-last sentence: "</span></b><i><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">I
am expressing doubts that level could be created in
a TLUD that was not first run as a regular fire.</span></i><span
style="color:black"><o:p></o:p></span></p>
<p class="MsoNormal"><b><span style="color:black"> The
word "TLUD" should say to all that the test
operation was NOT run as a regular fire. They are
as near to polar opposites as the stove world can
get. So this is to ask Crispin what he is saying
here and what part of my response he is objecting
to?<o:p></o:p></span></b></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">I
am offering a method of how to get a CO content as
high as 10%. Start a regular fire, get it going well,
then enclose it in a vessel while hot and running.
This can produce 10% CO, but a TLUD cannot. That is my
contention.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Biomass
needs just a little more air (Oxygen) to completely
use up the H2 and then breathe in whatever additional
air would burn all the Carbon. In any real file, some
of the C becomes CO and CO2 (surface reactions
mentioned by Dr Tom Reed in a previous discussion).</span><span
style="color:black"><o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">>></span><span
style="color:black"> <b>[RWL: I hope we can get a
specific citation for/from Tom here. </b></span><b><span
style="color:#1F497D"><o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">He
addressed it directly previously on two occasions. <o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b><span style="color:#1F497D">></span><span
style="color:black">At the hot surface from which
(very complicated and numerous [1000's of
species??]) pyrolysis gases are emerging, my
understanding of the pyrolysis surface effect
literature is that "all" (given control of the
incoming oxygen flow) are turned into CO and water.
</span><span style="color:#1F497D"><o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">My
point is that if there is NO air entering, virtually
all the O2 in the fuel is turned into water. The
great proportion. Some of the H2 is left in the gas
but is it hard to find an H2(EF) of 15,000 (1.5%). I
have no problem with others contradicting this with
real measurements.<o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b><span style="color:#1F497D">>></span><span
style="color:black">The relatively small amount of
CO2 that is produced near the surface (not ON) is
converted back to CO as it interacts with the hot
char above it</span><span style="color:#1F497D"><o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">This
only happens under certain conditions and not when
it is cool. Dr Tom was mentioning C=> CO2
reactions taking place above 400 C on the surface.
CO can also be formed, and volatiles (which contain
carbon) can break down into CO as well. The CO in
the gas is not necessarily produced from CO2 and is
unlikely if the temperature is low because it has to
absorb a lot of heat to do so (24 MJ/kg).<o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b><span style="color:#1F497D">></span><span
style="color:black"> So I repeat - I am mystified by
this message and about what is at dispute. </span><span
style="color:#1F497D"><o:p></o:p></span></b></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">I
am not sure there is a dispute. Perhaps the
clarification will suffice.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Regards<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Crispin<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
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