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<div class="moz-cite-prefix">Crispin,<br>
<br>
Thank you for that carefully expressed statement with examples.
More of us need to master these calculations, or find simple ways
(spreadsheets perhaps?) to make the calculations easier.<br>
<br>
Still 10 days before I get back home from Brazil. This will be
one of my priority tasks then.<br>
<br>
Paul<br>
<pre class="moz-signature" cols="72">Doc / Dr TLUD / Prof. Paul S. Anderson, PhD
Email: <a class="moz-txt-link-abbreviated" href="mailto:psanders@ilstu.edu">psanders@ilstu.edu</a>
Skype: paultlud Phone: +1-309-452-7072
Website: <a class="moz-txt-link-abbreviated" href="http://www.drtlud.com">www.drtlud.com</a></pre>
On 2/21/2014 1:16 PM, Crispin Pembert-Pigott wrote:<br>
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<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">Dear
Ron<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">I
think there is an assumption which is not always true: that
a TLUD makes char. It is true that stoves can be designed to
produce char from the fuel placed in them – many types – but
it is not the case that all TLUD’s have to make char. <o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">Where
there is a fuel consumption metric based on the total fuel
used per cycle, whether the stove makes char or not is up to
the producer. A water boiler could have an heat transfer
efficiency that was very high, make some char, and still
meet the minimum 1-Star target of an overall thermal
efficiency of 45%. <o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">You
will perhaps recall the discussions we have had in the past
where the calculations were laid out in such a way that the
energy accounting is plainly stated so that each portion of
the process is identified and the overall raw fuel needed to
achieve the result also given.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">If
you have a 10 litre water boiler, it will require about 3.5
MJ to be delivered to the water container. At a fire heat to
water container efficiency of 65% it means a fire generating
a bit more than 5.2 MJ. Suppose that is 1/3 of a kg of wood
pellets. <o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">The
baseline device has an efficiency (based on raw fuel
consumed) of about 17% which means it would take 1.3 kg of
those same pellets to do the same thing. <o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">1/3
x (65/17) = 1.3<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">I
am not saying there is a local baseline pellet stove, just
using biomass fuel with the same energy per kg.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">So
if you were to achieve an overall thermal efficiency of 40%,
while actually having a heat transfer efficiency of 65%, the
difference can be produced as unburned charcoal. In fact a
dedicated water boiler should be able to achieve a heat
transfer efficiency of 75% or better so this is not a
stretch.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">Take
a mass of raw fuel, calculate the energy content. Suppose it
is 20 MJ from 1 kg.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">Transfer
40% of that to the water vessel, that is, 8 MJ to boil 22
litres of water.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">Actual
heat transfer efficiency is 65%, i.e. 13 MJ available were
all the fuel to be burned.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">The
difference, 5 MJ, could instead be char. At a heat content
of 172 g of char or 17.2% char yield.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">In
this example, the water boiler would complete the boiling
task, it would receive 1 Star for overall thermal efficiency
and it would also produce 172 g of char.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">If
the overall thermal efficiency was lower (not for this
programme however) say, 25%, the char producible would quite
a bit higher: an additional 138 g for a total of 310 (31%).
I think at the moment no TLUD is produce 31% char yield.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">I
have encouraged everyone for several years to make this sort
of calculation because if you have a char production target,
say 18-20%, and you have a fuel efficiency target, there is
a heat transfer efficiency target that will produce all the
results simultaneously. <o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">I
am confident that with the larger boilers, 20 litres, the
heat transfer efficiency will be in the high 80’s. It is,
after all, a gas flame burning in constant, perfect
circumstances each time with no need for control.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">To
calculate that, let’s use 85% HT eff and 3.5 MJ/10 litres:<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">1
kg fuel @ 17MJ/kg AR (As received) total 17 MJ<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">2
Stars needs 55% overall thermal efficiency (Heat in the fuel
v.s. heat in the pot)<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">20
litres needs 7 MJ in the pot. 7/0.85 = 8.235 MJ of fire
heat<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">2
Stars means using up to 7/.55 = 12.727 MJ of heat<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">Difference
is 4.49 <J or 155 g of char<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">Mass
of fuel used including making the char = 749 g with 155 g
left as char (21%)<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">This
is technically possible.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">For
reference: Baseline 17% eff = 1.211 kg fuel per 10 litres
boiled<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">Someone
who built a boiler that burned the great majority of the
char would of course use far less raw fuel. Customers may
prefer that. <o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">You
were correct in your earlier comment that I proposed a 10%
efficiency gift to the char makers but that was not accepted
by the programme. Maybe next year, if there is judged to be
a benefit from having the char, and people are willing to
deal with handling it – something not yet proven.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">I
hope this clarifies everything.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US">Regards<br>
Crispn<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<div>
<div style="border:none;border-top:solid #E1E1E1
1.0pt;padding:3.0pt 0mm 0mm 0mm">
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif""
lang="EN-US">From:</span></b><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif""
lang="EN-US"> Ronal W. Larson
[<a class="moz-txt-link-freetext" href="mailto:rongretlarson@comcast.net">mailto:rongretlarson@comcast.net</a>] <br>
<b>Sent:</b> Friday, February 21, 2014 1:22 PM<br>
<b>To:</b> Discussion of biomass; Crispin
Pemberton-Pigott<br>
<b>Subject:</b> Re: [Stoves] Request for technology
proposals - Clean Stove Initiative, Indonesia<o:p></o:p></span></p>
</div>
</div>
<p class="MsoNormal"><o:p> </o:p></p>
<div>
<p class="MsoNormal">Crispin:<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"><span class="apple-tab-span">
</span>I think TLUD companies might find it interesting to
enter on the water-boiling side of this RFTP. But it is
not clear how charcoal will be counted. I can see at least
three formulas for calculating the efficiency; one ignoring
the char, by which no TLUD could ever win if they wanted to
save the char. The two formulae that count char as a useful
co-product could either (or both) place the char in the
numerator or the denominator. One of these formulas will
make the proscribed efficiencies quite achievable. I can’t
find the intended formula(e) in what you have sent.<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal">Ron<o:p></o:p></p>
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