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<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;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;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
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<p class="MsoNormal"><b><span lang="EN-US" style="font-size:11.0pt;font-family:"Calibri",sans-serif">“…</span></b>I went a lot further reporting on the test experts I talked to at ETHOS - NONE of whom agreed with Crispin on the topic of this reply - the way
to handle char in reporting efficiencies. <o:p></o:p></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">That provides a comment on the poor understanding of the principles of engineering and performance rating amount those who you contacted at ETHOS. It is sad that those how
claim to lead are so at sea when it comes to making such simple determinations. Perhaps you are not aware that the ‘char-deducted formula’ is unique in the world when reporting the % of fuel energy delivered as ‘work’.
<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Char making is not ‘work’ when it comes to cooking energy delivered.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal">>John Mitchell supports the existing “denominator equation” - as does EVERYBODY I talked to at the ETHOS conference. <o:p></o:p></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Ditto – if true, it is significant that the ETHOS participants you discussed this with do not follow this list, read about the problems and understand the implications, or
consider that scientific norms should apply to stove testing.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
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<p class="MsoNormal">><b>[RWL1: I am agreeing with decades of usage to make the “<i>cooking efficiency metric number"</i> more accurate. <o:p></o:p></b></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">The cooking efficiency is the ratio of cooking energy to the energy in the fuel fed into the stove. Comparing the rating of two stoves calculated on that basis gives a direct
comparison of the fuel consumption to accomplish a task. More on that below.<o:p></o:p></span></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b>>The ONLY accurate denominator is one with the char being subtracted - as the ratio is the heat into the cook pot divided by the total energy that COULD BE available for that measure of energy into the cookpot. <o:p></o:p></b></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">That is a description of the heat transfer efficiency. It is unfortunate you are not learning from this interchange. I have explained in detail how to calculate the heat transfer
efficiency and it starts by deducting the char energy (all of it, not just some recoverable portion). You are calling the heat transfer efficiency the cooking efficiency. That is the root of the problem. They are different when the stove produces a solid
residue containing unreleased energy.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><b>>Energy that is in the char was NOT available to go into the cookpot. <o:p></o:p></b></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">That is why it is deducted when calculating the heat transfer efficiency.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><b>>It HAS to be subtracted to get a valid efficiency when you are running different tests with relatively arbitrary and unintentional amounts of char being produced.<o:p></o:p></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">That is correct, if you want to calculate the heat transfer efficiency.<o:p></o:p></span></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal"><b>>You are the only person I have heard say this is improper. Your view has been dismissed by dozens of others - especially in “official”polling.</b><o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Obviously you are not in all conversations on the matter. You are in no position to conduct an ‘official poll’. Science does not operate on ‘official polls’ of people whether
or not they are informed on the subject. In fact, opinion counts for little in a mathematical calculation.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"><b>> </b></span><b>I think the “denominator equation” formula undervalues (not overvalues) the energy in the char. <o:p></o:p></b></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">The equation does not report the cooking efficiency, nor the energy in the char nor its fraction of the original energy. It is an error to use it for anything. It has no standard
name because it is not accepted as a standard calculation save as a rough guide to the heat transfer efficiency, which I remind you was the original intention of the authors of the VITA test. The approach was used in a much more refined form in the BUCT paper
of 1 year ago. When it was pointed out on this list that the calculation of the relative fuel consumption was in error because of this, one of the authors, Kirk Smith, made a comment on this group that the error would be corrected ‘if the paper was published’.
In fact the paper was already published. The error is to think that the relative heat transfer efficiencies of two stoves is the same as the relative fuel consumptions. This error is common to the WBT (all versions) the CCT and the KPT. (With the KPT it is
only considered in certain circumstances so there is a caveat there – the KPT sometimes gives the correct answer.)<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><b>>It says the inefficiency is larger than it is. <o:p></o:p></b></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">The formula doesn’t calculate the inefficiency of anything.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><b>> I accept the formula only because the tier structure is based on its use. <o:p></o:p></b></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">You are correct that the tier structure (which has its own additional defects) is based on the WBT 4.1.2. which miscalculates the cooking efficiency. I invite you and everyone
else on this list to obtain one of the (at least) three versions of the WBT 4.1.2 and run a set of measurements through it to get the ‘thermal efficiency’. Then run the same set of measurements through v 4.2.3 and see what the answer is. Be amazed. To get
the real answer, delete the contents of the ‘char’ cell and set the char catching container to zero. Compare that with the other results. Be shocked.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><b>>I would have been happier with a tier structure based on overall efficiency, but I know that is impractical - especially at this late date.<o:p></o:p></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Interesting. So getting the correct answer is not important, expediency is? How long would you be willing to wait under normal circumstances?<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">>></span><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">This deduction raises the reported fuel efficiency.</span><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p></o:p></span></p>
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<p class="MsoNormal"><b>>[RWL: For small amounts of char it makes the reported efficiency more accurate. <o:p></o:p></b></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">No, it mis-reports the metric that claims to represent the fuel consumption. The more char you make, the greater the misrepresentation of the fuel economy.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><b>>It undervalues the overall (more than heat transfer) efficiency of stoves that are trying to make char.<o:p></o:p></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">You are correct on this point insofar as the formula does indeed calculate the heat transfer efficiency (or a reasonable proxy of it). I am not cure why you have contradicted
your earlier points above.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">>></span><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Are you OK with that as the result?</span><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p></o:p></span></p>
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<p class="MsoNormal"><b>[RWL: Marginally. Only in the tier heat rating sense.<o:p></o:p></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">So having the wrong answer is not an issue as long as the heat transfer efficiency tiers are not changed? Did you ever buy a product based on the heat transfer efficiency?<o:p></o:p></span></p>
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<p class="MsoNormal"><span class="apple-tab-span"> </span><o:p></o:p></p>
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<p class="MsoNormal"><span class="apple-tab-span">></span><b>Example: If energy into the boiling water and charcoal each are one-third, then the sum of all inefficiencies MUST also be one-third. <o:p></o:p></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">That is incorrect. Please read on.<o:p></o:p></span></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b>>You argue for a heat transfer efficiency of 1/3 (dropping all consideration of the char). <o:p></o:p></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">No, I argue that the cooking efficiency is 1/3. Please correct your misunderstanding.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><b>>The “denominator equation” (used by everyone but yourself as near as I can determine) says the “heat transfer efficiency” is (1/3)/(1-1/3) = (1/3) / (2/3) = 1/2. <o:p></o:p></b></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">First, virtually no one outside the USA uses this formula in any official capacity (CDM/Gold Standard excepted) , and within the USA it is shunned for regulatory purposes by
the EPA.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Second, that is how to calculate the heat transfer efficiency, not the cooking efficiency.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><b>>I can live with this, but I also think it important to say that the inefficiency is NOT also 1/2. <o:p></o:p></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">The formula does not calculate an inefficiency, it calculates the heat transfer efficiency. The heat transfer efficiency is not useful for calculating fuel savings.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><b>>The overall efficiency, when one is trying to produce char, is 1/3 +1/3 = 2/3. This last is clearly NOT the “heat transfer efficiency”<o:p></o:p></b></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Correct. It is not the heat transfer efficiency. It does not have a name as you are adding the cooking efficiency to the % of energy in the original fuel that was not burned.
As I pointed out to you and Paul, this metric is non-standard and does not have a name.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><b>>but the overall efficiency should be reported as well if we are trying to promote more valuable stoves, and it is not being reported.<o:p></o:p></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">What do you consider to be the ‘overall efficiency’?<o:p></o:p></span></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
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<p class="MsoNormal">><b>Now the reverse question - WHY are YOU so unhappy with the subtraction in the denominator? <o:p></o:p></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Because as applied in the WBT (all versions) it gives a misleading number which is used to calculate the relative fuel savings of stoves, comparing them with a baseline product.
It gives the wrong answer. You indicated above that while the cooking efficiency is 1/3 and the WBT reports it to be ½, you are OK with that misrepresentation. I am not.<o:p></o:p></span></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b>>Is it your opinion that this char production was an inefficiency? <o:p></o:p></b></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">In standard terms char production is reported as a % of the dry fuel input. Where it is not a desired product, it is a mechanical loss.
<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><b>>You have expressed great unhappiness with the “denominator equation”, but I don’t recall ever seeing a reason. <o:p></o:p></b></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Then you have not been reading my posts.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><b>>The purpose of including a char term in the denominator is NOT to say anything about char - it is to get at the POTENTIAL heat transfer efficiency. <o:p></o:p></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">There is no such metric as the ‘potential heat transfer efficiency’ except to say that it is always 100%, until the stove is tested.<o:p></o:p></span></p>
<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b>>To repeat - too many will think that char-making stoves are much less efficient than they really are.<o:p></o:p></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">When it comes to fuel efficiency, char making stoves are usually much less fuel-efficient than stoves that burn all the char. As usual, you are looking for some way to over-report
the fuel efficiency by pretending the char is unburned ‘fuel’. <o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Then, you plan to bury the char in the ground (proving it was not ‘fuel’ after all) to accomplish what others call ‘sequestering carbon’.</span><b><o:p></o:p></b></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Using your example above:<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Cooking efficiency: 1/3<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Char energy retention, based on the recoverable mass of the solid residue: 1/3<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Heat transfer efficiency: 2/3<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Two stoves both have a cooking efficiency of 1/3. One of them makes some measurable amount of char. Applying the WBT formula raises the
<i>reported</i> cooking efficiency (not the actual cooking efficiency). The actual cooking efficiency shows that both stoves require exactly same amount of raw fuel to cook. Applying the WBT formula one finds that the char-making stove is
<i>claimed</i> to require less fuel to complete the cooking task. <o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">That claim is false. People are being induced to pay for a reduction in fuel use on the basis of the WBT calculation. They are being defrauded with false claims of fuel saving.
I don’t believe you are ‘OK’ with this situation.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Regards<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">Crispin<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"><o:p> </o:p></span></p>
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