<html><head><meta http-equiv="Content-Type" content="text/html charset=windows-1252"></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; "><div>Paul: Your and my last were also to the stove list, so I re-insert the list </div><div><br></div><br><div><div>On Aug 17, 2013, at 8:06 PM, Paul Olivier <<a href="mailto:paul.olivier@esrla.com">paul.olivier@esrla.com</a>> wrote:</div><br class="Apple-interchange-newline"><blockquote type="cite"><div dir="ltr">See comments below.<br><div class="gmail_extra"><br><br><div class="gmail_quote">On Sun, Aug 18, 2013 at 8:03 AM, Ronal W. Larson <span dir="ltr"><<a href="mailto:rongretlarson@comcast.net" target="_blank">rongretlarson@comcast.net</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div style="word-wrap:break-word"><div>Paul and List:</div><div><br></div><div> Three comments/questions:</div><div>
<br></div><div> 1. The gas analysis from Belonio was apparently at 1000C in the hot char, but you believe you are closer to 500 C?</div></div></blockquote><div><br></div><div>This 1000 C is not the temperature of the gas that exits the reactor. It is the process temperature in the zone where C combines with O2 to form CO2. This reaction supplies the heat for the endothermic reactions that follow. These endothermic reactions cool down the gas as it exits the reactor. The temperature of the gas as it exits the reactor (and prior to combustion) reaches as high 500 C. This we were able to measure.<br></div></div></div></div></blockquote><div> <b>{RWL1: I am comfortable with the 500 C number, but I remain concerned that 1000 C may be too high for the (or most) pyrolysis from max temps. Numbers I recall seeing with thermocouple readings have never been that high.</b></div><br><blockquote type="cite"><div dir="ltr"><div class="gmail_extra"><div class="gmail_quote"><div>
</div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div style="word-wrap:break-word"><div><br></div><div> 2. Is there any way to know what the air equivalency ratio is as you are operating? even if you are above or below the optimum (of 0.3)? I guess this is determined by the CO measurements, but I haven't seen any data for either TLUDs or rockets on that.</div>
</div></blockquote><div><br></div><div>If too much oxygen is supplied to the process as is the case of an air equivalency ratio of 0.6, the amount of carbon monoxide produced can drop by over 50% and the amount of H2 produced can drop by almost 60%. The reactor will heat up to dangerously high levels. We can try to correlate the specific rate of solids consumption to the flow of primary air to determine the equivalency ratio. But I have never done this.<br></div></div></div></div></blockquote> <b>[RWL2: I hope that they will jump into this dialog if they have seen anything on this ratio in tests or literature. I am still unsure what "equivalency" means, does it include excess air concepts?</b><br><blockquote type="cite"><div dir="ltr"><div class="gmail_extra"><div class="gmail_quote"><div>
</div><div> </div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div style="word-wrap:break-word"><div><br></div><div> 3. Some reading this exchange may not realize that you light the pyrolysis gases before adding the burner assembly, </div>
</div></blockquote><div><br></div><div>Before adding the burner assembly, I do not light the gases. I am merely lighting the biomass. When the biomass lights, a large flame rises out of the reactor.</div></div></div></div></blockquote><div> <b>[RWL3: This is what most TLUD users live with (and same for rockets). A tall diffusion flame</b></div><br><blockquote type="cite"><div dir="ltr"><div class="gmail_extra"><div class="gmail_quote"><div> When the burner is placed on the reactor, this large flame within the reactor must go out. There should never be open flames within the reactor, otherwise I get burner holes that do not support a flame.<br>
</div><div> </div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div style="word-wrap:break-word"><div>then you drop the fan speed to extinguish the interior burning and can then relight the 80 flame lets. </div>
</div></blockquote><div><br></div><div>Yes. At this point, I do not "relight" but "light" the burner holes for the first time.<br></div></div></div></div></blockquote> <b>[RWL4. This was point I was making- because only you and Belonio are doing this (a top surface with lots of small holes - that seems to work very well), I think.</b><br><blockquote type="cite"><div dir="ltr"><div class="gmail_extra"><div class="gmail_quote"><div> </div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div style="word-wrap:break-word">Other than Belonio, I don't know anyone else doing this. </div></blockquote><div><br></div><div>I am totally confused. How then do they get the open flames within the reactor to go out?<br>
</div><div> <b>[RWL5. They are "happy" with the flame as is. For one thing, the "chimney" they are using provides draft that you (with a fan) don't need.</b><br> </div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div style="word-wrap:break-word"><div>In your final sentence, people may not realize that your flamelets are still diffusion type, not premixed. I know no-one getting premixed flames, either rockets or TLUDs.</div>
</div></blockquote><div><br></div><div>This morning I will test a 250 unit with a secondary air pipe that runs from the reactor grate at the bottom of the reactor into the burner at the top. This pipe is situated fully inside the reactor. This burner has 50% more burner holes than a normal burner to account for the added flow. If this works, this will be true premix burner.<br></div></div></div></div></blockquote> <b>[RWL: I look forward to hearing (actually I did hear - and not so good, for reasons we have discussed - which I will let you report). Ron</b><br><blockquote type="cite"><div dir="ltr"><div class="gmail_extra"><div class="gmail_quote"><div>
<br></div><div>Thanks.<br></div><div>Paul<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div style="word-wrap:break-word"><div class="h5"><div><br></div>
<br><div><div>On Aug 17, 2013, at 5:54 PM, Paul Olivier <<a href="mailto:paul.olivier@esrla.com" target="_blank">paul.olivier@esrla.com</a>> wrote:</div><br><blockquote type="cite"><div dir="ltr"><div>It is challenging to try to
understand what happens in a char-making TLUD. My exposure to stoves has
been entirely limited to the work of Belonio, both from a practical and
theoretical side. On the theoretical side, the following is what I have
gleaned from Belonio with the help of a young engineer from the
University of Delft. I throw this out to the list with great trepidation, since I have only been working on this reflection for about a week.<br>
<br>Temperature is very important, and it is generated as C reacts with O2 giving rise to CO2 (initial combustion that supplies heat to the process). The O2 is supplied from the primary air and from the H2O within the biomass. The temperature has to be high enough to
optimize the endothermic reactions that take place within the process.
The endothermic reactions are the water gas reaction (C combines with H2O to form CO and H2) and the Boudouard
reaction (C combines with CO2 and to form CO). If the temperature is high enough, C will not combine with H2 to form methane. If the temperature is high enough, there will be little tar and oil formation. The goal is to create a high
percentage of CO and H2. <br>
<br>Then there is the moisture content of the biomass. A moisture content of 10% is ideal. If there is too much water in the biomass, water is transformed from a liquid to a gas within the process, and the process temperature is lowered. Also if there is too much water, the water gas shift reaction is favored giving rise to CO2 and H2. So if the
moisture content increases beyond what is optimal, there is less CO, more CO2 and more H2O in the gas. <br><br>
Then there is the amount of oxygen being supplied to the process. If too
much oxygen is supplied, the amount of CO and H2 decreases, and the
amount of CO2 and H2O increases. Excess oxygen burns up CO and H2
within the reactor. This translates into a big inefficiency, since the
heat generated here is generally quite far away from the bottom of the
pot. Part of the oxygen comes from the water, and the rest from the
primary flow of air. An air equivalency ratio of 0.3 is ideal. <br>
<br>But air must be supplied uniformly up through through the biomass.
Channeling (too much air in some places and not enough in other places)
severely disrupts the entire process. In such a case, the concept of an ideal air equivalency ratio becomes somewhat meaningless. Some people design TLUD stoves
that handle all types of biomass. But I only know of about 4 or 5 types
of biomass that are sufficiently uniform to be run through a TLUD in
their raw state. Everything else has to be prepared (splitting, cutting,
chipping or pelletizing) to be rendered sufficiently uniform. Of all
forms of preparation, pelletizing appears to be the best.<br>
<div>
<br></div>If rice hulls are processed at 1000 C, at an equivalency ratio
of of 0.3 and at a moisture content of 10%, the gas content consists of
26.1% CO, 20.6% H2, 0% CH4, 6.6% CO2 and 8.6% H20 (numbers from
Belonio). This adds up to 61.9% of the total gas. The remainder is
mostly N2. <br>
<br>The presence of CO2 and H2O in the gas gives rise to a
dirty gas. In a stove test, it would be interesting to measure the CO2
and H2O content of the gas prior to combustion at the burner. If CO is
intimately mixed with CO2 and H2O, the combustion of CO at the burner is compromised.<br>
<div>
<br></div>When the gas is burned at the burner, heat is generated by the combustion of
CO and H2. Air is about 21% oxygen and 79% nitrogen, and it takes
considerably less oxygen to burn CO and H2 than other more complex forms
of gas such as methane, propane or butane. The molar ratio of air to
gas to burn the CO and H2 in the above proportions is roughly 1.11
mol/mol. The mixing ratio of air to gas by volume is roughly 0.42 m3/m3.
Also if the gas prior to combustion has a temperature in
excess of 500 C, this facilitates the combustion of CO and H2. If anyone would like to see these calculations, I will supply the spreadsheet off-list.<br>
<br>This might explain why the Belonio burner with the burner
housing I added to it functions reasonably well in spite of the fact
that the premixing of air and gas does not take place. So little
secondary air is required, the gas is hot, and the mixing takes place
all along the periphery of the two off-set rings of burner holes. As the gas exits the 80 burner holes, it does so under mild pressure and sucks in air from the burner housing.<br><a href="http://www.youtube.com/watch?v=84qDsbBO9p8" target="_blank">http://www.youtube.com/watch?v=84qDsbBO9p8</a><br>
<br>I have seen several rice hull gasifiers where gas exits through one large burner hole in the middle of the burner. This produces a single flame with a long diffusion tail, and the transfer of heat to the pot under such conditions cannot be optimal.<br>
<br></div><div>So in conclusion, the process temperature within the reactor should be higher than 700 C, the moisture content of the biomass should be less than 12%, the air equivalency ratio should be about 0.3, the biomass should be sufficiently uniform, the temperature of the gas prior to combustion should be in the range of about 500 C, the gas prior to combustion should contain little CO2 and H2O, and the mixing of secondary air with gas should as thorough as possible.<br>
</div><div><br></div><div>Thanks.<br></div><div>Paul Olivier<br></div><div>
<br><br></div></div><div class="gmail_extra"><br><br><div class="gmail_quote">On Sun, Aug 18, 2013 at 12:19 AM, Ronal W. Larson <span dir="ltr"><<a href="mailto:rongretlarson@comcast.net" target="_blank">rongretlarson@comcast.net</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div style="word-wrap:break-word"><div><br></div><div><a href="http://www.et.byu.edu/~tom/classes/733/ReadingMaterial/Jenkins-Baxter.pdf" target="_blank">http://www.et.byu.edu/~tom/classes/733/ReadingMaterial/Jenkins-Baxter.pdf</a></div>
<div><br></div><div><i>"Stoichiometric air fuel ratios …………..for biomass they are 4 to 7,"</i></div><div><br></div><div>I have seen "6" a lot, and the inverse (fuel to air weights) would be 17%</div>
<div><br></div><br><div><div><div>On Aug 17, 2013, at 5:49 AM, Alex English <<a href="mailto:english@kingston.net" target="_blank">english@kingston.net</a>> wrote:</div><div><br></div><br><blockquote type="cite">
<div text="#000000" bgcolor="#FFFFFF">
<div>
<div>
<div>Ron, Paul,<br>
Below; Paul refers to 'equivalency ratio'. This would be the
amount of primary (under fuel air) </div></div></div></div></blockquote></div><div> <b>[RWL: Alex, thanks _ I wasn't thinking this way. For your moving grate design, this term "under fuel air" makes sense. But for TLUDs, I believe the term "under" makes less sense, as all the O2 is used up at the pyrolysis front, regardless of its magnitude in volume per unit time. Since it would seem that CO needs about half the oxygen as CO2 (except some O2 is coming from the biomass and we have to account for H2 going to H2O), maybe a number near half (meaning the 30% and 60% numbers below) makes sense. Or, maybe Paul's definition of equivalency ratio includes excess air - not stoichiometric air. Paul - do you have a cite we can go to?</b></div>
<div><div><b><br></b></div><blockquote type="cite"><div text="#000000" bgcolor="#FFFFFF">divided by the theoretical
amount of air (stoichiometric) for complete combustion of that
fuel. Then he speaks of CO2, CO and H2 production and syngas
quality and variable fuel moisture contents. It would be nice
to see data that would correlate to his instance #2. I have
yet to see "Syn" gas composition measurements from a TLUD.
"process temperature might be below 500C" Where does this
number come from?<br>
<br></div></blockquote></div> <b>[RWL: I am going to stay away from this, due to press of other business. The above cite with Tom Miles as co-author might have some of this. I think the 500 C term means at the pyrolysis front. Would you go higher?</b></div>
<div><b><br></b><blockquote type="cite"><div text="#000000" bgcolor="#FFFFFF"><div><div>
"A lot of CO is emitted by the stove" <br>
Here he refers to CO that fails to be combusted in the burner
portion of a stove making it sound like it is a consequence of
conditions that occur in the fuel bed. "Syn"gas quality does
affect burner performance but burner parameters also affect
stack CO emissions.<br></div>
<b>[RWL: Maybe, but I think Paul is repeating what I heard often at the Stove Camp. All the stoves burning char (not done in TLUDs usually) suffer from very high CO production. (emphasis added below in Paul's comment).</b></div>
</div></blockquote><div><br></div><br><blockquote type="cite"><div text="#000000" bgcolor="#FFFFFF"><div>
Instance #3 seems plausible.<br>
<b>[RWL: Agreed. but there should be a paper to see the details and definitions.] </b>Whew - this is a good topic - but I need something more to read. Thanks to both Paul and Alex. Ron</div>
</div></blockquote><blockquote type="cite"><div><div text="#000000" bgcolor="#FFFFFF"><div><div><div>
<br>
<br>
Alex<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
Paul writes;<br>
<br>
Ron,<br>
<br>
</div>
One should look at a stove according to what it is designed to
use as fuel. Let us look, for example, at stoves that process
rice hulls. <br>
<br>
In a first instance, the stove might simply burn rice hulls.
Here we are talking about direct combustion where an air
equivalency ratio situates close to 1. Such a stove will produce
a lot of CO2 and H2O as well as relatively high levels of CO.
The fuel for such a stove is rice hulls.<br>
<br>
In a second instance, the air equivalency ratio might be 0.6,
the process temperature might be below 500 C, the moisture of
the biomass might be 20% or more, and too much secondary air
might be applied to the combustion of a dirty syngas containing
a lot of CO2 and H2O. Since the production of CO and H2 is
suboptimal, it might make sense in this instance to burn the
char in order to maximize the production of energy. <b><u>But
unfortunately burning the char has serious problems: a lot of CO
is emitted by the stove</u>,</b> and heat is generated far below the
pot. If the char is burned within this second stove, the fuel
for such a stove is rice hulls.<br>
<br>
</div>
In a third instance, the air equivalency ratio situates close to
0.3, the process temperature rises above 800 C, the moisture
content of the biomass situates at 10%, and the supply of
secondary air is kept low, but still adequate, to achieve total
combustion of the syngas. Here the production of CO and H2 is
optimized, the temperature of the syngas prior to combustion at
the burner reaches as high as 500 C, and not too much secondary
air is mixed in with the syngas. In this instance, up to 30% of
the weight of the rice hulls would still remain as biochar. But it
would make no sense to burn this biochar, since the production and
combustion of the syngas were optimized.<br>
<br>
</div>
<blockquote type="cite">
<br>
</blockquote>
<br>
</div></div><div>
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