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<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Dear Ron,
<o:p></o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">my previous postings considered forced primary air condition. The air rate was kept constant with a fan. The velocity of the ignition front depends
on the primary air rate also in the case of natural draught, but then the air rate is not given and it does not remain constant but depends on the balance equation<o:p></o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">draught = pressure losses (including inlet to the device, fuel bed and outlet)<o:p></o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">A similar situation is when one heats a home with a stove. I have district heating but use sometimes also a masonry stove. The draught = gravity
* chimney height * density difference. The density difference =density of air outdoors minus density of air (averagely )in the chimney. When one ignites the wood logs in a stove, the draught is initially quite low, because the chimney is cool. One can increase
the draught by burning some paper in the chimney. One can decrease the pressure loss by opening the damper of the stove and the windows of the building. Then after some time the chimney gets hotter and draught becomes good and one can adjust proper air rate
by the damper.<o:p></o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">In TLUD (with natural draught) the draught and the pressure losses (thickness of the non-ignited fuel bed and the thickness of the char layer)
change with time (but they are equal according to the above equation). So the air rate changes with time. In principle it is possible to solve the air rate (or air velocity through the bed) from the above balance equation in dynamic conditions (as function
of time), because the pressure loss is proportional ~ air velocity ^2 or more precisely from a developed equation for a bed of particles ( Ergun’s equation, 1952). In practice it is difficult, because the situation is transient heating and the draught in
TLUD is also changing (in analogy with initially cool chimney). It has been noticed that the pressure losses are higher for moist fuel bed due to drying so drying of fuel in sunshine before combustion is beneficial to get lower pressure losses. In the case
of TLUD, the hot char layer gives also some draught because the gas is hotter in this layer than ambient air and gas density is lower (than ambient) but it causes also some additional pressure loss. Then, if no damper is applied along the burning, the air
rate (and the burning rate) will evidently increase during the burning (thickness of cool bed decreases and draught increases) and the ignition front velocity changes. The ignition front velocity may be low at first, then reach a maximum and then again could
become lower if the air rate becomes high over the maximum situation (discussed in previous post). However, in this case quenching with too high air rate is not likely to take place, because if the temperature begins to drop, then also air velocity decreases.
This balances the air rate to a certain level. <o:p></o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">So in the case of natural draught, the construction of the device and especially how the draught is obtained, determines the operation. I use wood
chips, pellets, small branches, wood blocks and pine needles in the TLUD (developed by Tom Reed) which has a small blower. All these fuels burn well, but is difficult to use a broad range of fuel types in a natural draught device. The operation of TLUD with
natural draught depends much on its construction and fuel type and it is difficult to develop any general theory or model unlike to a forced flow TLUD. It could be possible to simulate the operation numerically with theories of combustion and heat transfer
(including combustion of fuel bed, transient heating of different parts of the device etc.), but the calculations would be device-dependent, quite cumbersome and probably not very accurate and reliable so that experiments would be required for a specific stove
construction. <o:p></o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Some cheat device (fan) to replace the natural draught with a forced flow would be a major step to develop efficient cooking devices. Somebody
should invent a cheap fan that uses for example gravity to rotate the fan like in a clock using heavy weights. If high enough air rate is reached, it is then easy to use a damper to control the air rate suitably and get efficient burning.<o:p></o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Jaakko<o:p></o:p></span></p>
<p class="MsoNormal"><span lang="EN-GB" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
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<p class="MsoNormal"><b><span lang="EN-US" style="font-size:10.0pt;font-family:"Tahoma","sans-serif"">From:</span></b><span lang="EN-US" style="font-size:10.0pt;font-family:"Tahoma","sans-serif""> Stoves [</span><a href="mailto:stoves-bounces@lists.bioenergylists.org"><span lang="EN-US" style="font-size:10.0pt;font-family:"Tahoma","sans-serif"">mailto:stoves-bounces@lists.bioenergylists.org</span></a><span lang="EN-US" style="font-size:10.0pt;font-family:"Tahoma","sans-serif"">]
<b>On Behalf Of </b>Ronal W. Larson<br>
<b>Sent:</b> 16. elokuuta 2014 8:16<br>
<b>To:</b> Discussion of biomass<br>
<b>Subject:</b> Re: [Stoves] Request for help on TLUD operating data<o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB"><o:p> </o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB">Crispin, list<o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB"><o:p> </o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-GB">
</span></span><span lang="EN-GB">I am part way through understanding the paper under discussion (and two other Finnish similar papers). I now fear that the Figure 2 that I referenced was for a forced air situation. No mention of viscosity - since their interest
was mainly or only on the space below the pyrolysis front (termed a flame front). So I am back to hoping for more weight-time history natural draft data from this list. (on “B” and “C”).<o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB"><o:p> </o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-GB">
</span></span><span lang="EN-GB">There probably is considerable value still in understanding this work’s emphasis on theory. Note that the radiative forcing for the movement (speed) of the pyrolysis front varies (at least approximately) as the cube (!) of
the front’s temperature. (and of course many other variables - especially moisture content.)<o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB"><o:p> </o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-GB">
</span></span><span lang="EN-GB">The viscosity data will require a temperature difference much larger than the 200 oC shown below.<o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB"><o:p> </o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-GB">
</span></span><span lang="EN-GB">I found a typo in the next to the last sentence Crispin quoted<o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB"><o:p> </o:p></span></p>
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<blockquote style="margin-top:5.0pt;margin-bottom:5.0pt">
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<div>
<p class="MsoNormal"><b><span lang="EN-CA">And (a question) stage 3 is the 95% of the time period…</span></b><span lang="EN-CA"><o:p></o:p></span></p>
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</blockquote>
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<p class="MsoNormal"><span lang="EN-GB">should read <o:p></o:p></span></p>
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<blockquote style="margin-top:5.0pt;margin-bottom:5.0pt">
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<p class="MsoNormal"><b><span lang="EN-CA">And (a question) stage 2 is the 95% of the time period….</span></b><span lang="EN-CA"><o:p></o:p></span></p>
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</blockquote>
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<p class="MsoNormal"><span lang="EN-GB"><o:p> </o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-GB">
</span></span><span lang="EN-GB">More coming.<o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB"><o:p> </o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB">Ron<o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB"><o:p> </o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB"><o:p> </o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB">On Aug 14, 2014, at 7:56 PM, Crispin Pemberton-Pigott <</span><a href="mailto:crispinpigott@outlook.com"><span lang="EN-GB">crispinpigott@outlook.com</span></a><span lang="EN-GB">> wrote:<o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-GB"><br>
<br>
<o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Dear Ronal’n’All</span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="color:#1F497D">Good find!</span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<div>
<p class="MsoNormal"><a href="http://gekgasifier.pbworks.com/f/ignition+front+saasta.pdf"><span lang="EN-CA" style="color:purple">http://gekgasifier.pbworks.com/f/ignition+front+saasta.pdf</span></a><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA"> <o:p></o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-CA"> </span></span><span class="apple-converted-space"><span lang="EN-CA"> </span></span><span lang="EN-CA">b. Because of my background, I think of the upward flow of primary air and
pyrolysis gases as three resistances and a “current” (gas) generator. The lowest chip/pellet region has a resistance RL that continually gets smaller as its volume decreases. The upper (char) region keeps gaining in height, but is losing weight rapidly
as well; for height (and other) reasons it (RU) presumably is increasing (at the end of a run, RL is zero). The middle pyrolysis zone resistance would not seem to change much during a run. And the “current” source also would not seem to change much during
a run (But maybe it does.)<o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">I think that is a good description for the start of the conversation. The only think I would add is that the overall height changes with time because
of shrinkage. It means the RU (region in the upper section that contains char) will not only be smaller than the original fuel it will also be evolving some CO2 and CO (by cracking CO2 from below).</span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-CA"> </span></span><span class="apple-converted-space"><span lang="EN-CA"> </span></span><span lang="EN-CA">So my first (electrical analog) observation is that the only way that we can
have a straight line is if the sum of RL and RU is a constant (call RC)<o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">There is a reason why it should not be and that is because (assuming it is a natural draft device) the temperature of the vertical column varies
with time giving an increase in draft as the colder RL (Region Lower) develops into a hot RU. The draft increase can be calculated using the Draft calculator available on this site using the section on the right. All things being equal, the draft will increase
time, resulting in an slow increase in power which is what is observed using high frequency mass measurements (high in this case meaning per 10 seconds).</span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-CA"> </span></span><span class="apple-converted-space"><span lang="EN-CA"> </span></span><span lang="EN-CA">Second - If I had to guess that the change in either would “exactly” balance
the other, I would have said no way. But for us, it is decidedly serendipitous/fortuitous.<o:p></o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-CA"> </span></span><span class="apple-converted-space"><span lang="EN-CA"> </span></span><span lang="EN-CA">The reason for RU increasing must include viscosity changes. <o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">I have not investigated the influence of the temperature of the gases on the viscosity in an increasing temperature column. If the temperature
in RL is 293°K and 493°K in RU then perhaps the increase in viscosity of the (different) gases will approximately balance. It seems you should put those two variables into your formula so you can investigate the overall effect of holding the char at a different
temperature. Remember the height of RU as a % of total will change (as you describe) but also the ‘reference height’ will drop with time.</span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">This doesn’t undermine your initial conceptual description. </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<blockquote style="margin-top:5.0pt;margin-bottom:5.0pt">
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<p class="MsoNormal"><span lang="EN-CA">1. Ignition at the top. It takes some time that the ignition front propagation reaches a steady velocity.<o:p></o:p></span></p>
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</blockquote>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-CA"> </span></span><span class="apple-converted-space"><span lang="EN-CA"> </span></span><b><span lang="EN-CA">[RWL: Yes, but with controllable primary air, which most TLUDS allow,
there can be large early primary and then a cut back. And usually operated exactly that way.]</span></b><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">[CPP] Just a comment on the several failures to ignite TLUD’s in Ulaanbaatar this summer: there are some basic precautions to take. One is that
the secondary air ports have to be closed to get a rapid ignition. We had several cases of people (a) having secondary air only (!) as the recommended lighting method, (b) leaving it open or not fully closed during ignition, (c) not providing a large dominance
of primary air. A further refinement for ignition is that the kindling materials have to heat as much of the surface as possible as soon as possible. This is not readily accomplished by lighting a flat surface of fuel. In all cases the results of lighting
at the bottom of a conical depression or a ‘vee’ pushed into the surface results in more rapid ignition of a larger surface per minute. Radiant heat from the side of the flames is a booster compared with trying to radiate heat downwards.</span><span lang="EN-CA"><o:p></o:p></span></p>
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<blockquote style="margin-top:5.0pt;margin-bottom:5.0pt">
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<p class="MsoNormal"><span lang="EN-CA"><br>
Also, if the air rate is too high, the burning is quenched due to cooling by the air so that the flame temperature goes down giving less radiation to preheat the non-ignited fuel and also keeping the non-ignited fuel cool.<o:p></o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-CA"> </span></span><span class="apple-converted-space"><span lang="EN-CA"> </span></span><b><span lang="EN-CA">[RWL: This is the first time I have seen this. I can see a problem with
a fan/blower, but also natural draft?</span></b><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><b><span lang="EN-CA"><br>
</span></b><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">[CPP] We frequently see this when it is combined with the presence of secondary air. This season several stoves were ignited and ‘failed to thrive’ for more
than an hour because of the combination. Fuel moisture (which can be as high as 33%) is a major issue in the early phase.</span><span lang="EN-CA"><o:p></o:p></span></p>
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<blockquote style="margin-top:5.0pt;margin-bottom:5.0pt">
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<p class="MsoNormal"><span lang="EN-CA">The velocity of the ignition front has a maxim at certain air rate. There is accumulation of char above the pyrolysis front. The maximum amount char is obtained with low primary air rate (but high enough to keep the front
moving). If the air rate is high, also some char is burned above the pyrolysis front due to excess air especially if the fuel is moist.<o:p></o:p></span></p>
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</blockquote>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-CA"> </span></span><span class="apple-converted-space"><span lang="EN-CA"> </span></span><b><span lang="EN-CA">[RWL: I am sure you can help us with fuel moisture issues. Should we
be “curing/drying” all fuel (maybe above the cookstove)?</span></b><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA"><br>
</span><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">[CPP] This is not going to fly In many cases, many places. It is better to have a good ignition method so once started, it dries the fuel continuously. There
is already strong resistance at village level to trying to prepare fuel.</span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA">The excess oxygen that is not consumed in the pyrolysis front reacts with char giving less char. Even the ignition velocity is quite constant, the burning rate of the whole batch including the char may increase during
the burning with high primary air rate, since the amount and thickness of char layer accumulating above the pyrolysis front is increasing and can react with excess oxygen. Then you would have a positive value for C in your formula (considering the whole weight
loss of the batch), if there is much excess air.<o:p></o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-CA"> </span></span><span class="apple-converted-space"><span lang="EN-CA"> </span></span><b><span lang="EN-CA">[RWL: The design mod I have in mind requires small C, but keeping below
a certain primary air flow rate should not be a major constraint. By “excess air” in the last sentence, I presume you mean excess primary air? (we have been using “excess” with secondary air)</span></b><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">[CPP] I will add for Jaakko’s entertainment that we have been using the SeTAR HTP calculation of ‘excess air’ which is a chemically balanced calculation,
not the usual</span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">(O2-CO/2)/(21-(O2-CO/2)) [1]</span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">We are using all the available molecules as measured, to consider the influence on the EA value of O2 present in the fuel. Performing a chemically
balanced calculation provides a more realistic ‘EA equivalent’ telling us what is actually going on in the combustion and reaction zones. I can provide more details here if that is needed. It is in the lab manual which is on line at the WB site.</span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA"><br>
3. In the end, the pyrolysis front reaches the bottom and this is may also be accounted by the term C*t^2 in your model. In this stage the amount of pyrolysing particles at the bottom decrease leaving more excess air to react with the char. It seems that
sign of C depends on the air rate. With high air rate, the sign goes to more positive direction, since the rate of flame propagation is low in the stage 2, but the rate of char combustion becomes high at the stage 3.<o:p></o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><span lang="EN-CA"> </span></span><span class="apple-converted-space"><span lang="EN-CA"> </span></span><b><span lang="EN-CA">[RWL: I need help with the terms “stage 2” and “stage 3”. I think that
with controllable primary air, that we can avoid the "high rate of char combustion” at the end of a run - assuming (as I do) that we want to maximize char production. I presume stage 3 is this final stage as the pyrolysis front reaches the bottom. And
(a question) stage 3 is the 95% of the time period with a “constant” power level (constant rate of fuel conversion)? So “stage 1” is the short start up period as the pyrolysis from moves.</span></b><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">[CPP] Sort of related to the above: if the standard Excess Air is ‘x’ then a recalculated version including water vapour might provide some hints
as to where to set the airflow (assuming you are not only finding it by trial and error). You can maximise the char yield by dropping the system temperature. If you need more heat (Watts) just make it larger.</span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Regards</span><span lang="EN-CA"><o:p></o:p></span></p>
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<p class="MsoNormal"><span lang="EN-CA" style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Crispin</span><span lang="EN-CA"><o:p></o:p></span></p>
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