<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>Jaakko cc list</div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>Thanks again for nice helpful response.</div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>The numerical computation at the end below is very helpful - and very rare on this list.</div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>You had this:</div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span><b style="font-family: 'Times New Roman', serif; font-size: 12pt;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"><i>Example, ignition front velocity is w=0.4mm/s and dry bed density is roob=150 kg/m3 , we get critical mass for rate of air/bed area Ma < 150 *0.0004* (1-0.2) */(1.1*0.23) =0.19 kg/(m2s)</i></span></b></div><div><br></div><div>[RWL: Trying to calculate a little further: w = .4 mm/s (0.0004 m/s) would mean a 6” = 150 mm fuel bed height would have to last t = 150/.4 = 375 secs = 6.25 minutes. (w = .4 mm/s = 2.4 cm/minute; a little less than 1 inch per minute)</div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>This is a very high rate I think for any practical TLUD discussed on this list. I would expect about 10 x longer, giving w = .00004 m/s</div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>Also think we might average a little higher dry bed density. So overall your key value of 0.19 kg/m2s might be as little as .04 or .05 for typical natural draft TLUDs. Sound true?</div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>You use a stoichiometric oxygen value of 1.1. Is there a similar rule of thumb number for the ratio of primary to secondary air? That is - how much of the 1.1 is used up right at/near the pyrolysis front, where we have only CO and no CO2?</div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>I find it interesting that you are doing everything by weight and not volume. I am only used to thinking of oxygen as 21%. I was surprised (being the wrong type of engineer) to learn that air at our pyrolysis temperatures is very nearly one-thousandth the density of water (i.e 1 kg/m3). And perhaps in the char zone about half that. And in the flame - again about another factor of 2 less dense. Do scientists like yourself have any such similar rules of thumb (for other variables) that can help us with better understanding TLUDs?. Thinking about viscosity especially.</div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>I think the problem for most of us (with almost no lab equipment) will be measuring air flow (primary secondary or both) in either weight or volume terms. If you or anyone can suggest a way to do this (a balloon with a weight forcing air out?), that could be a big help to carry these ideas further.</div><div><span class="Apple-tab-span" style="white-space:pre"> </span></div><div> <span class="Apple-tab-span" style="white-space:pre"> </span>Some TLUD possible numbers: Keeping your numbers for a 6”=.15 m TLUD diameter, with area pi r^2 = pi (.075)^2 = .0176 sam, and height = 6 inches = .15 m, so volume = .00265 m3 and the assumed density of fuel gives 150 times larger or nearly 0.4 kg for the fuel weight in your example.</div><div><br></div><div><span class="Apple-tab-span" style="white-space: pre;"> </span>Have I drawn the correct conclusions about the last part of your response?</div><div><br></div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>RWL2 I don’t yet understand your response to my first question on partial pressures (changing the number of particles when a secondary O2 molecule meets a pyrolysis gas molecule). I incorrectly used the term “draft” - but I feel sure there is a pressure difference that helps TLUD users to move both primary and secondary air in addition to the normal draft that you have emphasized. I am trying to understand the magnitude of this “chemistry” change in pressures.</div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>One way to help this list would be to give us some insights on pressure vs height inside a TLUD. At the pyrolysis front, the gas temperature jumps enormously, but also does the volume of gases. How large is the pressure change at the pyrolysis front? </div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>There is a similar large pressure change somewhere in the flame (presumably the brightest region?) as a chemical reaction occurs when the pyrolysis gases meet the oxygen in the secondary air. (I’m pretty sure we have to think about a diffusion flame - not a premixed flame). Because real numbers help me/most, let's issome a total height of the TLUD stove at 30, with the fuel bed and secondary air entry at 15 cm and a time when the pyroylsis front is again a half or 7.5 cm. Assume a primary air flow sufficient to have the flame disappear just at the top of the stove. Using torr or pascals, is it possible to say anything based only on analysis about the vertical distribution of pressures inside vs outside? Do we need to talk radial distributions as well? (That is - we observe conical shaped flames). Can these questions be answered without using the Dalton law of partial pressures?</div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>[RWL3. Near the top of your response you said: ‘… <span style="color: rgb(31, 73, 125); font-family: Calibri, sans-serif; font-size: 11pt;"><i>then the air rate is not given and it does not remain constant but depends on the balance equation.</i></span><font color="#1f497d" face="Calibri, sans-serif"><span style="font-size: 15px;"><i>”</i></span></font></div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>I agree that it need not stay constant, but experimentally we find that it (and the weight loss and power level) doesn’t change very much if primary air flow is not otherwise varied via control of a slot or set of holes). I am still hoping to get a handle on the amount of change for a well designed TLUD - because we can then control both primary and secondary air together.</div><div><br></div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>[RWL4: I believe you may be on vacation so I apologize for continuing (and I/we thank you as well). Sending us to a suitable text dealing with both pyrolysis and combustion may be the best approach, if you know of one that handles a TLUD geometry.</div><div><br></div><div><span class="Apple-tab-span" style="white-space:pre"> </span>Again, thanks for staying with me/us and, in advance, thanks for any further leads.</div><div><br></div><div>Ron </div><div><br></div><div><br></div><div>On Aug 22, 2014, at 5:24 AM, Saastamoinen Jaakko <<a href="mailto:Jaakko.Saastamoinen@vtt.fi">Jaakko.Saastamoinen@vtt.fi</a>> wrote:</div><div><br class="Apple-interchange-newline"><blockquote type="cite"><div lang="FI" link="blue" vlink="purple" style="font-family: Helvetica; font-size: 18px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: auto; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: auto; word-spacing: 0px; -webkit-text-stroke-width: 0px;"><div class="WordSection1" style="page: WordSection1;"><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">Dear Ron, see replies below.<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">Jaakko<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span></div><div><div style="border-style: solid none none; border-top-color: rgb(181, 196, 223); border-top-width: 1pt; padding: 3pt 0cm 0cm;"><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-US" style="font-size: 10pt; font-family: Tahoma, sans-serif;">From:</span></b><span lang="EN-US" style="font-size: 10pt; font-family: Tahoma, sans-serif;"><span class="Apple-converted-space"> </span>Ronal W. Larson [</span><a href="mailto:rongretlarson@comcast.net" style="color: purple; text-decoration: underline;"><span lang="EN-US" style="font-size: 10pt; font-family: Tahoma, sans-serif;">mailto:rongretlarson@comcast.net</span></a><span lang="EN-US" style="font-size: 10pt; font-family: Tahoma, sans-serif;">]<span class="Apple-converted-space"> </span><br><b>Sent:</b><span class="Apple-converted-space"> </span>19. elokuuta 2014 0:32<br><b>To:</b><span class="Apple-converted-space"> </span>Discussion of biomass; Saastamoinen Jaakko<br><b>Cc:</b><span class="Apple-converted-space"> </span></span><a href="mailto:mika.horttanainen@lut.fi" style="color: purple; text-decoration: underline;"><span lang="EN-US" style="font-size: 10pt; font-family: Tahoma, sans-serif;">mika.horttanainen@lut.fi</span></a><span lang="EN-US" style="font-size: 10pt; font-family: Tahoma, sans-serif;"><br><b>Subject:</b><span class="Apple-converted-space"> </span>Re: [Stoves] Request for help on TLUD operating data<o:p></o:p></span></div></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB">Jaakko cc List<o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><span lang="EN-GB">This is hugely helpful. Thanks.<o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><span lang="EN-GB">Yours below here is the third from you today - but the other two are easier to respond to, so I am starting here and do those next.<o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><span lang="EN-GB">See inserts below.<o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div><div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB">On Aug 18, 2014, at 10:06 AM, Saastamoinen Jaakko <</span><a href="mailto:Jaakko.Saastamoinen@vtt.fi" style="color: purple; text-decoration: underline;"><span lang="EN-GB">Jaakko.Saastamoinen@vtt.fi</span></a><span lang="EN-GB">> wrote:<o:p></o:p></span></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"><br><br><o:p></o:p></span></div><div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">Dear Ron,</span><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">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</span><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">draught = pressure losses (including inlet to the device, fuel bed and outlet)</span><span lang="EN-GB"><o:p></o:p></span></div></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><span lang="EN-GB">[RWL1: I believe there is another reason for draught, that I hope you will comment on. When the main pyrolysis gases CO and H2 react with secondary air just above the hot char bed, there is a reduction of the number of particles since<o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><span lang="EN-GB">2 CO + O2 = 2 CO2<o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><span lang="EN-GB">2 H2 + O2 = 2 H2O <o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> </span></span><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><span lang="EN-GB">In both cases 3 particles before the combustion (left side) turn into 2 particles after (right side) - with a consequent lowering of pressure, encouraging both primary and secondary air flows. Of course, reduced by the presence of Nitrogen in both the primary and secondary air streams, but bolstered by significant non-nitrogen gas release at the pyrolysis front. Is there a well-known “Law” to help put this additional helpful pressure drop into perspective? Note this doesn’t happen with a methane flame where we have:<span class="apple-tab-span"> </span><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><span lang="EN-GB">CH4 +2 O2 = CO2 + 2 H2O (3 particles on each side of the equation)<o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="color: rgb(31, 73, 125);">Jaakko: The equation above states that draught is consumed in pressure losses. The draught is due to weight differences of gas columns due to density differences explained (below for the chimney), which is the partly the same things as you explain above. The density difference is mainly caused by temperature difference. Gas density is proportional to 1/T, where T is in degrees of Kelvin, so the high temperature gas above the ignition from is much lighter than the gas outside the cooker and causes the draught.</span></b><b><span lang="EN-GB"><o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span></div><div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">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.</span><span lang="EN-GB"><o:p></o:p></span></div></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><b><span lang="EN-GB">[RWL2: Understood. I believe this happens quite quickly in a small cookstove of 30-40 cm height. I do not believe TLUD operators have complained about insufficient draft.</span></b><span lang="EN-GB"><br><br><o:p></o:p></span></div><div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">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.</span><span lang="EN-GB"><o:p></o:p></span></div></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><b><span lang="EN-GB">[RWL3: My question on the linearity of the stove operation (after setting the primary air for a final time) involved the parameters A, B, and C - seemed to be saying that C is generally small - and the “air rate” doesn’t change very much. That is, perhaps it is increasing viscosity effects (greater depth of char) offsetting the impact of decreasing fuel volume/depth.</span></b><span lang="EN-GB"><o:p></o:p></span></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="color: rgb(31, 73, 125);"> </span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">Jaakko: The main reason for the natural draught is the difference between gas densities. The pressure losses for flow through a bed of particles can be calculated by Ergun’s equation. Viscosity has some effect but the main factors are: gas velocity, bed height particle size, porosity of bed (void fraction) and particle shape (sphericity). See Wikipedia for this equation. So with larger particles one gets lower pressure losses (if gas velocity is constant) and higher gas velocity (if draught is constant). <o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"><br><br><o:p></o:p></span></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">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).</span><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><b><span lang="EN-GB">[RWL4: This was easy to find via Wiki - and then other similar equations - including Darcy’s equation. But none of these seem to handle the TLUD geometry and my issue of “C” normally being a small quantity (as experimentally observed).</span></b><span lang="EN-GB"><o:p></o:p></span></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="color: rgb(31, 73, 125);"> </span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">Jaakko: Look atr Ergun’s equation.<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"><br><br><o:p></o:p></span></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">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). </span><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><b><span lang="EN-GB">[RWL5: Except it seems that the draught is pretty constant - perhaps for the “viscosity” reason given above. I am not at all expecting a full balancing - but rather asking what conditions can we have (fuel, etc) that make the constant drought result possible (again for my reason of wanting to combine primary and secondary air control).</span></b><span lang="EN-GB"><o:p></o:p></span></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><span lang="EN-GB"><br><br><o:p></o:p></span></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">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.</span><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><b><span lang="EN-GB">[RWL6: Again my own personal testing experience is that the power level (weight loss per unit time) stays quite constant. I am looking for cases where this is true. Of course, stopping stove operation when the maximum amount of char has been achieved.</span></b><span lang="EN-GB"><o:p></o:p></span></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"><br><br><o:p></o:p></span></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">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. </span><span lang="EN-GB"><o:p></o:p></span></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><b><span lang="EN-GB">[RWL7: Not quite sure what you are saying here - but the last sentence “ balancing” is what I am after - how much can there be? How constant can the air flow be in a properly operating TLUD? I am asking the whole list for experimental data on linear decrease of stove weight.</span></b><span lang="EN-GB"><br><br><o:p></o:p></span></div><div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">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.</span><span lang="EN-GB"><o:p></o:p></span></div></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><b><span lang="EN-GB">[RWL8: Yes, Tom’s fan-powered TLUD was never intended to make char. In his fan stove, char combustion is occurring simultaneously with char production because of the vigorous air flow - no separation into primary and secondary components, I believe. When you say you have been cooking with a TLUD, is Tom’s the unit?</span></b><span lang="EN-GB"><o:p></o:p></span></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="color: rgb(31, 73, 125);"> </span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">Jaakko: There are two levels of air rate in this Tom Reed-type cooker. It is also possible to add adjustable flow resistance in the inlet to decrease the flow even further. One can get much char especially with the lower rate provided that:<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt 36pt; font-size: 12pt; font-family: 'Times New Roman', serif; text-indent: -18pt;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"><span>-<span style="font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-family: 'Times New Roman';"> <span class="Apple-converted-space"> </span></span></span></span><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">particles are dry enough so that energy from volatiles is enough to preheat the material to ignition<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt 36pt; font-size: 12pt; font-family: 'Times New Roman', serif; text-indent: -18pt;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"><span>-<span style="font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-family: 'Times New Roman';"> <span class="Apple-converted-space"> </span></span></span></span><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">particles are large enough. For small particles pyrolysis and char combustion can become overlapping (see Saastamoinen, J., Aho, M., and Linna, V., Simultaneous pyrolysis and char combustion. Fuel 72, 1993, 599 609). Small particles have larger surface area (which is inversely proportional to particle diameter) meaning higher gasification rate. I get low amount of char with pine needles, but with wood blocks the yield is high<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt 36pt; font-size: 12pt; font-family: 'Times New Roman', serif; text-indent: -18pt;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"><span>-<span style="font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-family: 'Times New Roman';"> <span class="Apple-converted-space"> </span></span></span></span><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">particles should be produced or sieved to the same size. It the bed consists of a mixture of sizes, small particles are pyrolysed fast and their char may partly be combusted in the ignition/pyrolysis front while the large particles are still in pyrolysis stage. If there is a mixture of different sizes, it is best to place particles of approximately same size at the same level in the bed, but one can have a different sizes in the vertical direction and still have high yield of char. The same applies for a mixture of different fuels (with different reactivity).<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">I have used this cooker to boil food with largish wood cylinders and blocks and when the flame reaches the bottom poured the char to a grill and used the char in grilling sausage.<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">In the following I present a rough calculation for primary air rate giving char. Higher air rates (excess air) consumed. The limit case all volatiles are assumed to be burned in the ignition front. Then equation is<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">Mass flux of volatiles = dry bed density * velocity of ignition front * (1 –mass fraction of char) = stoichiometric mass flow of oxygen = stoichiometric mass flux of air<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">or by using symbols: Mv=roob * w * (1-Fc) = Fo* Mo = Fo * Fa * Ma<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">My= mass flux of volatiles (kg/sm2),<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">roob = dry bed density = dry mass of fuel/volume of fuel bed (kg/m3)<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">w = velocity of ignition front (m/s)<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">Fc = mass faction of char = 0.2 (if 20% char is obtained)<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">Fo =stoichiometric mass ratio (mass of volatiles/mass of oxygen) in complete combustion =1.1 (calculated assuming char to consist only of carbon and using typical elemental analysis of wood <o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">Fa = mass faction of oxygen in air =0.23 (corresponds 21 % volumetric fraction)<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> Ma = mass flux of air (mass flow rate/cross section area of fuel bed)<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">This equation shows that the air rate must be below a critical value to avoid excess primary air for char combustion<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">Ma < roob * w * (1-Fc)/(Fo*Fa)<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">Example, ignition front velocity is w=0.4mm/s and dry bed density is roob=150 kg/m3 , we get critical mass for rate of air/bed area Ma < 150 *0.0004* (1-0.2) */(1.1*0.23) =0.19 kg/(m2s). With higher air rate, char is also burned. Of course, some char may also be burned at lower air rate, but the flux of volatiles hinders much of the oxygen to react the particle surface and react.<o:p></o:p></span></b></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"><br><br></span><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"><o:p></o:p></span></b></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">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.</span><span lang="EN-GB"><o:p></o:p></span></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><b><span lang="EN-GB">[RWL9: I agree. But we need more theory than we have had and you seem most qualified to help in this regard. Possibly also Prof. Horttanainen (cc’d) can chime in. My main interest now is in how we can equate the loss of flow resistance below the pyrolysis from with the gain above it - so that total flow resistance stays relatively constant. Seems that might be a suitable theoretical topic - that assumes relatively constant draft from other effects.</span></b><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><b><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></b></span><b><span lang="EN-GB">If there are any simple measurements you know of to help establish a best excess (secondary) air quantity, that would be helpful.</span></b><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><b><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></b></span><b><span lang="EN-GB">I am unaware of any natural draft stove where it seemed excess primary air was present. (Draft was never too much.) Anyone able to cite experimental TLUD work with that character?</span></b><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB"><br><br></span></b><span lang="EN-GB"><o:p></o:p></span></div><div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);">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.</span><span lang="EN-GB"><o:p></o:p></span></div></div></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><b><span lang="EN-GB">[RWL10: Paul Olivier (cc’d) is not a regular contributor on this list, but he has been using a small fan costing only a few $, with a nice speed controller. Perhaps he can comment for us on fan use. His regular fuel is rice husks - with a low energy density, so his units tend to be tall - not conducive to natural draft. His is one of the few TLUDs with a cap and external supply of (non-fan-driven) secondary air hitting the exhaust gases after they exit through a hundred or more small holes. His work builds on that of Prof. Alexis Belonio. One Olivier presentation is at </span></b><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><a href="http://www.slideshare.net/Jupiter276/paul-oliver-presentation-in-dalat" style="color: purple; text-decoration: underline;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><span lang="EN-GB">http://www.slideshare.net/Jupiter276/paul-oliver-presentation-in-dalat</span></a><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></span><b><span lang="EN-GB">In sum on fans - I am certainly in favor of exploring them, but I also feel we might sell more char-making stoves if we can get acceptable operation without fans.</span></b><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span class="apple-tab-span"><b><span lang="EN-GB"> <span class="Apple-converted-space"> </span></span></b></span><b><span lang="EN-GB">Again, many thanks for your many helpful comments. A bit more coming on an earlier message today.</span></b><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB">Ron</span></b><span lang="EN-GB"><br><br><o:p></o:p></span></div><div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><b><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span></b><span lang="EN-GB"><o:p></o:p></span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB"> </span></div></div><div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;"><span lang="EN-GB" style="font-size: 11pt; font-family: Calibri, sans-serif; color: rgb(31, 73, 125);"> </span><span lang="EN-GB"><o:p></o:p></span></div></div></div></div></div>_______________________________________________<br>Stoves mailing list<br><br>to Send a Message to the list, use the email address<br><a href="mailto:stoves@lists.bioenergylists.org" style="color: purple; text-decoration: underline;">stoves@lists.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/stoves_lists.bioenergylists.org" style="color: purple; text-decoration: underline;">http://lists.bioenergylists.org/mailman/listinfo/stoves_lists.bioenergylists.org</a><br><br>for more Biomass Cooking Stoves, News and Information see our web site:<br><a href="http://stoves.bioenergylists.org/" style="color: purple; text-decoration: underline;">http://stoves.bioenergylists.org/</a><br><br></div></blockquote></div><br></body></html>