[Stoves] Request for technology proposals - Clean Stove Initiative, Indonesia
Ronal W. Larson
rongretlarson at comcast.net
Sun Feb 23 01:35:25 CST 2014
Crispin and List:
1. Sorry, but I am not happy with the response. Let me repeat what I asked:
> 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.
[RWL: I still can’t find the answer to my question. I think below you are answering my question saying that the energy efficiency equation to be used with the water heaters is the same as for the stoves efficiency computations. I think there is no counting of the char value in any of your efficiency computations. I think this is unfair and unwise. Please advise if I have the correct interpretation of what you have written below. I hope I am wrong.
If I am I correct in this interpretation of what you have written, then I recommend that those interested in char-making stoves stay away from this Indonesian opportunity. Char-making stove folk will be universally unhappy with their reported efficiency results, is the only conclusion I can draw. I recommend they stick with the EPA/GACC approach instead. The following is to explain my reasoning.
2. Let’s start with a standard nomenclature (I have violated a bit maybe, due to lack of time. Apologies in advance for any typos.
a. Energies (MJ) : A=energy into pot, B= energy in input fuel (pellets, etc), C= energy in char, D = non useful energy
b. Energy densities (MJ/kg) ED: EDb = 18 MJ/kg, EDc = 30 MJ/kg (designed to give easy ratio of 5/3)
c. Energy efficiencies(%) E1= A/B (this is apparently the Indonesian standard); E2= C/B; E3=E1+E2 = (A+C)/B; E4 = A/(B-C) (the present GACC form)
d. Use subscripts r= rocket or similar stove, and t=TLUD - for any char-making stovealthough this might be BLDD, etc), so
another set f=fuel, c = char
e. Weights (kg) Wfr (weight of fuel in rocket), Wft, Wcr, Wct (and these depend on which efficiency formula we use: E1, E2, E3, or E4.
f. Energies Afr, Bfr, Cfr, Dfr; Aft, Bft, Cft, Dft
g. Efficiencies E1r, E1t; E2r, E2t; E3r, E3t; E4r, E4t (using row 2c)
h. Char weight conversions Wc/Wf = .2, .25, .3 for different “T” stoves
i. Char E2t= .333, .417, .5 for different “T” stoves (this is 4/12, 5/12, 6/12)
j. Cases to analyze Water boiling test at 1*, 2*, and 3* minimum levels (45%, 55%, 65% I think - find all 8 E’s) for Wfr = 1 kg, with Bfr = 18 MJ/kg;
In sum, I am heading to find the 27 numerical inter-relationships for three each of these variables:
1. the assumed defining efficiency equations, E1, E3, E4
2. the 1*, 2*, and 3* minimum efficiencies in your “water heater” option (I think this will end up calculating the needed heat transfer E1 to the pot.
3. different “t” stove char weight conversions: .2, .25, .3
4. I might be able to drop this by concentrating on the central value of E3, the .55 2-star scenario, and a stove providing 25% conversion by weight, and 41.7% conversion efficiency by energy
3. I don’t see in your computations that the input fuel weight has to change as you go from no char to char-making.
4. I have scrubbed a bunch of computations here - and will start them again, after your response to the following. The main point is at the bottom - non-physical results are being offered.
more below. Sorry, for so many new bold lines below - just a way of making sure I followed what you were calculating.
On Feb 21, 2014, at 12:16 PM, Crispin Pembert-Pigott <crispinpigott at outlook.com> wrote:
> Dear Ron
>
> 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.
RWL: I did not make that assumption. I only want to know what happens during computations of repaired efficiency if the stove user wants char - for whatever reason.
>
> 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%.
{RWL: All true, but I am asking only for what energy efficiency equations are to be used.
>
> 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.
[RWL: Still not answering my question.]
>
> 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.
[RWL: OK. I know some big units have efficiencies like that.
>
> 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.
>
> 1/3 x (65/17) = 1.3
[RWL: Not sure why this assumption - 17% sounds low for TLUDs, even without counting the char value in the efficiency formula. We are down now by a factor of almost 4. Or maybe your baseline is for a rocket?
>
> I am not saying there is a local baseline pellet stove, just using biomass fuel with the same energy per kg.
[RWL: OK. Obviously we have to do this.
>
> 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.
[RWL: I assume from this sentence that the “overall thermal efficiency of 40%” is calculated by giving zero credit for the “unburned charcoal”. This is not fair to a manufacturer trying to make char, or a user hoping to compare stoves on a char-making basis.
> Take a mass of raw fuel, calculate the energy content. Suppose it is 20 MJ from 1 kg.
[RWL: a little high in my experience, but OK. You assume substantially lower below.
>
> Transfer 40% of that to the water vessel, that is, 8 MJ to boil 22 litres of water. [RWL: Above we had 10 liters.]
> Actual heat transfer efficiency is 65%, i.e. 13 MJ available were all the fuel to be burned.
> The difference, 5 MJ, could instead be char. At a heat content of 172 g of char or 17.2% char yield.
[RWL: This could probably read (to the same accuracy as 20 MJ/kg) approximately “ x kg*30MJ/kg = 5MJ - so x = 5MJ/30MJ/kg = 1/6 kg= .167 kg, but 172 gm is OK, since you apparently assumed the char energy was 5MJ/.172 = 29.07 MJ/kg. This 17.2% is definitely on the low side for TLUDs; I always got near 25% and we hear 30% sometimes (and even more).
>
> 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.
[RWL: Answering my question of what efficiency equation should be used, Jim Jetter would report eff4 =E4 = 8/(20-5) = 8/15 = 53%.
How do we calculate the efficiency of charcoal making? I say it is not 53-40 = 13%. To me, it is 5/20 = 25%. The total efficiency would be promoted by many users as 40+25 = 65%. 35% is the lost energy
To summarize:
Eff1= 40% your assumption
Eff2= 25% your assumption (the weight conversion efficiency would be about 25% *3/5 = 15% - very low for a char-maker.
EFF3 = Eff1+Eff2 = 65% Jetters does not endorse this method - and of course you don’t either
Eff4= 53% I don’t see how this can be controversial; you just gave all three A, B, C numbers as facts
This set of numbers was easy to do - because everything was given in MJ terms. I can endorse these, as reasonable.
The point is that using your rule#1 for scoring, this stove fails to get 1 star. Depending on using the efficiency equation 3 or 4, you get three or one stars.
> 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.
[RWL: Of course, I do not consider either 40% or 25% as being the “overall thermal efficiency”. But even if your eff1 does go down, my eff.2 = 41.7% need not go up, if we are talking the same “T” stove as earlier. But assume it does, I see no reason to assume your “additional 138 g”. That is your number, not one I would conclude possible, since we clearly should need more input fuel as we go from an “r” stove to a “t” stove. But for completeness, then
Eff1= 25% (your assumption)
Eff2 = 41.7% (my assumption, “T” stoves can’t change their production efficiency in the arbitrary manner you have assumed.)
Eff3 = 25% + 41.7% = 66.7%
Eff4 = .25/(1-.417) = 42.8% This is the EPA-standard reported efficiency, and again fails to get one-star.
> 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.
[RWL: If I didn’t stick with my assumed stove with a weight ratio of 25%, I would get the same nonsense answers as given below. t think your example above shows yours is not the way to produce meaningful results. The right way is to measure the char weight and use the proper energy content of that char. Agree on 8.235. I am not agreeing that you can predict something about char production, in large part because no one is measuring with small stoves the efficiency numbers you are quoting as possible.
> 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.
[RWL: No disagreement. But we are talking a very large system to get these efficiencies - I would say much higher than 20 liters. Do you have a citation for the 80% or (below) 85% with wood/pellet fuels?
>
> To calculate that, let’s use 85% HT eff and 3.5 MJ/10 litres:
[RWL: If we use your previous 8MJ/22 liters, that would be 3.64 MJ for 10 liters, so we are close to above.
> 1 kg fuel @ 17MJ/kg AR (As received) total 17 MJ
[RWL: Previously assumed at 20 MJ/kg, but this is OK. I prefer 18 MJ/kg to get simple ratios. Below we assume less than 1 kg
> 2 Stars needs 55% overall thermal efficiency (Heat in the fuel v.s. heat in the pot)
[RWL: I’d prefer to say 55% = ratio of heat in the pot divided by heat in the fuel, but I understand.
> 20 litres needs 7 MJ in the pot. 7/0.85 = 8.235 MJ of fire heat
[RWL: OK 7 = 2*3.5. And yes to 7/.85 = 8.235
> 2 Stars means using up to 7/.55 = 12.727 MJ of heat
[RWL: OK
> Difference is 4.49 <J or 155 g of char
[RWL: 4.49 MJ=12.727 MJ - 8.235 MJ; .155 kg = 4.49/29.07
> Mass of fuel used including making the char = 749 g with 155 g left as char (21%)
[RWL: 155/749 = 20.6%, but this is apples weight (char) over oranges weight (wood) , and only the start of a proper efficiency calculation. Using your previous assumptions, the 155 gm of char has an energy content of .155*29.07 = 4.506 MJ (slightly higher than your number). The input wood has an energy content of 749*17 = approximately 12.73 MJ, which checks above.
Eff1 = 85% = [7 MJ/8.235 MJ]
Eff2 = 4.49 MJ/8.235 MJ = 54.5% (This number is OK by itself, but not when 85% means an E1=Eff1 computation. You did not make this Eff2 computation or the next two either. Jim Jeffers certainly would.
Eff3 = 85%+ 54.5% = 139.5% - Clearly something wrong - the problem was just over-specified I think.
Eff4 = 7/(8.235 - 4.49) = 7/3+ - >>100%; not even worth calculating the GACC number. D (losses) would be negative.
>
> This is technically possible.
[RWL: and I say not, based on these calculations. But this can be salvaged, if by 85% you mean either Eff 3 or Eff4. Did you? Could you? I’d be glad to try recalculating everything if that was the case.
>
> For reference: Baseline 17% eff = 1.211 kg fuel per 10 litres boiled
[RWL: Exactly 20% of the 85% - factor of five improvement - while making char, but not using it in any efficiency calculation?
>
> 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.
[RWL and may not. Everyone will be happy with any efficiency number > unity.
>
> 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.
[RWL: didn’t make sense anyway - seemed like a number pulled out of the air, and I think that even more strongly now. I see no validity in the 45,55, and 65% values for the three water heating tiers.
>
> I hope this clarifies everything.
[RWL: I think you can see I don’t think it has clarified anything. Unless these high efficiency numbers are associated with something other than the Eff1 = A/B value. Apologies again for typos. It is late. Ron
>
> Regards
> Crispn
>
> From: Ronal W. Larson [mailto:rongretlarson at comcast.net]
> Sent: Friday, February 21, 2014 1:22 PM
> To: Discussion of biomass; Crispin Pemberton-Pigott
> Subject: Re: [Stoves] Request for technology proposals - Clean Stove Initiative, Indonesia
>
> Crispin:
>
> 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.
>
> Ron
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