So you still want me to do the math, eh?
I will if you will.
(Using a thermometer, a postal scales, and a stopwatch, observing a local woodstove. Measure the temperature at the exhaust (or creosote or lack thereof), and the amount of wood burned in an hour, and eyeball the exhaust rate in feet/second. Then do the same math. No looking at the mfr specs first; the point is to see whether armchair estimates have anything to do with lab ones.)
My
answer: 75%, +/- 85%. (That's a range of about 10%-100%). Work is explained below.
My steps in solving this problem:
1) Recognize that what you want is a theoretical, high-school-science answer.
A much more relevant answer is the one we already gave.
In a particular building, a rocket stove may use about 1/8 the wood, produce almost no smoke, for the same level of indoor comfort in the same building, as a small woodstove circa 1975.
You could take the wood you use in a year, divide by 5 for a safety margin (your woodstove stove might be more efficient than the test example), and calculate the rocket stove size needed. Or build a one-room heater and see if you need another one later. This would be the practical answer to "how big does my stove need to be."
For insurance purposes, your problem is not the heat efficiency, but the building regulations. Basically, if you can get a permit from the local code officials, the insurers feel safe.
We're in conversation with the City of Portland to develop a code-friendly writeup comparable to masonry heater or woodstove specs, so an inspector could permit the stove. Will cost $$. Masonry heaters are exempt from DEQ, and can be built/designed by a licensed mason or architect who can certify its safety. It might be worth a conversation with your insurer about whether they're willing to add a rider given that it's a non-statistically measurable risk (few existing examples, and I don't know of any reported fatalities outside of maybe Hansel and Gretel).
Back to the recreational math:
We are comparing data with very low reliability (the weight of a bag of gas for Pete's sake). There's a 10% range of error in the fuel value for wood alone; air moisture content adds a weight variability of 60%. What error percentage would make the results meaningful?
I'd add another 50-100% error range for actual usage factors, like the operator's standards for 'dry' wood, the frequency of operation and cleaning, heat loss through walls, the tendency to forget the fire or damp it down, etc.
So I don't trust the industry data either, and I prefer practical answers.
But I do enjoy math.
So I'm willing to chew this cud through, as long as you recognize that the results are ... (what normally results from recreational mental digestion of unreliable data).
Solving the Story Problem:
1) Guess the weight of exhaust gas based on observations, using memory of observing two 6" systems.
(My assumptions are likely to be just as reliable as any bag-weighing experiment I can do with equipment on hand.)
Air weight vs gas coming out:
Our foggy flue gas at 95-100F (310K) is about equally dense with air at 70F(295K). (Very hard to run the stove then, though we can do it if the stove was pre-warmed in cool air the night before).
Flue gas at 110F(315K) is less dense than air at 55F(286K), even with condensed
water; rolls upward at about walking speeds.
Weight of air at 70F(20C)(293K) = 1.2 oz / cu. ft; (saturated with water vapor e.g. in vacuum above a pool is 1.9g/cu ft; Wikipedia says "Humidity ranges from 0 gram per cubic metre in dry air to 30 grams per cubic metre (0.03 ounce per cubic foot) when the vapor is saturated at 30 °C;" nobody will tell me what fog weighs, but dry ice vapor CO2 is 1.5 times denser than air.
One patent does claim that water droplets in a cloud are a small weight compared to the (saturated) vapor... Water compared to air: 780 times denser. ... I glipmse on a Google search that a gram per cubic meter is about normal for "wild" fogs ... that would be another 0.001 oz/cu ft, but what is the volume of vapor which squeezes down into that 0.001 oz? How much steam does an ounce of water make? ...
http://www.elmhurst.edu/~chm/vchembook/123Adensitygas.html 1a) I ran only the first comparison; got 1.4 oz of flue gas per cubic foot at 95 F. This is almost twice the weight I found
online, which makes sense (maybe) due to fog dew and CO2? The weight compared to ambient air would be a negative number at high temps, but heavier at STP.
1b) Rate of flow observed in a typical 6" system is about 1 foot/second for moderate burn, up to 3 feet/second for "rocketing" (fast, rushing burn). Cross section is 30 in^2; cubic inches per second = 360 in^3; 0.2 cubic feet/second = 750 cu ft/hour.
1c) Weight of gas coming out: 1.4 oz/cf * 750cf /16 oz/lb:
I'm getting about 66 lbs of exhaust per hour. I guess a few lbs of that is combustion products, and the rest is air in, air out.
2) Weigh some wood.
Our stove burns about 4.5 lbs of wood/hour in relaxed mode (one load of typical mix: cherry, fir, and
kindling), this would go faster in rocketing mode so we might use 7-8 lbs per hour. Weight in kg is: 2 to 3.5 kg of wood/hour, error range 2.5 kg +/- 40%
(For practical consideration:
A 4 hour burn (20-25 lbs of wood) heats our home for 24 hours, to 65-73 degrees in the various rooms and corners, with outside temperatures in the 40-55 range at this time of year. A 1-2 hour burn keeps the bench alcove comfortable for warm seating, and leaves the back bedroom at 55-60 degrees.)
3) Look up fuel values for cherry, doug fir, and alder. (There are no fuel values for punky sticks with the bark on, so we'll assume the oily paper from our fish and chips brings those up to average. We burn what we got, as dry as we can get it, and adjust.)
http://www.consumerenergycenter.org/home/heating_cooling/firewood.html Cherry and alder both turn out to be 8,200 BTU per lb, despite their different densities. Doug fir is 9,100 BTU/lb. Average fuel value for our 1-hour, moderate load: about 40,000 BTU/hour.
What was the question again?
mekennedy1313 wrote:
... Delivered efficiency is a simple calculation
(1 - (mass exhaust x temperature)/(kg wood x energy content/kg)) x100 to give a percent.
... Basically you are measuring how much heat escapes vs how much stays in the house. ....
I was interested because of the difficulty in getting insurance if you build your own stove.
I gotta say that my mass exhaust number feels unreliable - I got nothing to compare it to. If you don't care about the integrity of my exhaust number, skip these italics:
4) Question the integrity of my work in #1), and re-do it a different way:
Going with the chemistry, the wood is say
20% water (unbound & resin-bound),
1% mineral ash, and
80% carbohydrates and hydrocarbons (resin, cellulose, etc).
By dry weight, paper pulp woods are about 45% cellulose by weight, 27% lignin, 8-9% pentosan, and 3% other (plus that 1% ash). (Sorry, Internet source uncited)
Cellulose like sugar is basically CH2O;
lignin varies, but it's generally richer than cellulose as a fuel; (http://www.jbc.org/content/114/2/557.full.pdf) reports it as C40, H42-48, O 15-16. (C2,H2-2.4,O)
and some resin or oil would boost this further.
By a rough combination of these molecular weights, I make wood by weight:
10-20% water (unbound)
33% carbon (bound)
34% water (bound)
2% hydrogen (bound)
1% ash
10-20% other (oils? resins? variability in other components?)
Using the 33% carbon, and 2% available hydrogen, I make out that wood requires just over its own weight in oxygen to burn completely. (104% of the wood's weight in oxygen to burn 100% of the wood.)
Air is 20-21% oxygen, 78-79% nitrogen; 1% CO2; and less than 1% of other stuff.
The resulting flue gasses (of a pure wood reaction with no excess air) would be:
75-90 parts water by weight, plus atmospheric water, plus "other" - this might be either fog or vapor, denser or lighter than air, and releasing additional heat as it condenses...
130 parts CO2 by weight (includes about 10 parts from air)
520 parts nitrogen (inert) by weight
10-20 parts other (if inert).
(If the "other" is rich fuel, it could combine with another 80-150 parts oxygen and bring along another 300 - 700 parts nitrogen and a pinch of CO2.)
Total predicted exhaust weight/hour of a perfect wood/air reactor (perfect mixing, no excess of any reagent, just keep it on the heat until it's all incinerated) would be 7.5-15 kg exhaust per kg of wood.
Total predicted exhaust for a real-world fuel burner, where excess air must be provided to ensure that fuel is completely burned: say 2x to 4x that, with reasonably good mixing at high temperatures; so 20 to 60 kg exhaust/kilo of wood.
So any clean stove must pull at least 6x its fuel weight in air to burn clean, and 20-60x is more reasonable to get complete combustion. For 2 kilos/hr of wood, 40-120 kilos/hr of exhaust.
My weight guess in #1) (66 lbs exhaust/hr for burning 4-5 lbs of wood) represents about a 15x factor, on the low side but possible.
5) So what the heck, let's run the formula.
66 lbs is 30 kilos. (convenient.)
(1 - (mass exhaust x temperature)/(kg wood x energy content/kg)) x100 to give a percent.
(1- (30 kilos x 310K)/(40,000 - 46,000 BTU)) x 100
Ok, can we convert kilo.K's to BTU?
9300 kilos*K;
A BTU is 1055 Joules...
a joule is 1 kg*m2/s2 ...
Is a m2/s2 really equivalent to a degree Kelvin?
That would open up a whole new line of scientific inquiry about the universe...
Specific heat capacity is measured in joules/kg*K; that's the closest related unit that I can find online.
So please check the units on your equation.
It may just be "engineer math," in which you leave out anything inconvenient for estimating purposes. (As a scientist, when the units don't agree, it makes me nervous.)
But putting on my marketing / engineer hat, and running the equation as given:
(1- (30 kilos x 310K)/(40,000 - 46,000 BTU)) x 100
(1- (9300 kilo*K/43000BTU) x 100
(1-(0.216) x 100
(0.78) x 100
78 %
Efficiency of 78%, plus or minus....
(1-(100% variability in exhaust weight * 3% variability in temp)/(20% variability between fuel weight and fuel value/weight))
Aggregate error of +/- 85% ?
(I no savvy statistics, is this a meaningful way to combine error estimates?)
So the real value could be anywhere between
10% efficiency (or -7% ... )
and
130% efficiency (.... 82%? 91%? Or something....).
(Taking 100 lbs for exhaust weight gives efficiency of 70%. Taking a theoretical value of the exhaust mass - 2x extra air for complete burn - is pretty close to the first value listed above: 79%.)
Trained smoke observers (firefighters, DEQ) confirm that there is no detectable smoke in the exhaust, and the ash weights pretty closely correspond to the theoretical ash weights for the wood, about 1%. So clean combustion is definitely occurring.)
For comparison, a good auto engine has about 38% efficiency, a fuel cell about 78% efficiency, using a somewhat different definition of efficiency (conversion of fuel energy into useful motion).
Pellet stoves claim 75-90% efficiency, discounting the work (and expense) of the fuel manufacture.
But as we read here,
http://www.consumerenergycenter.org/home/heating_cooling/fireplaces.html Robert McCrillis of the EPA says, "...In the field, it's the installation and how the stove is operated that has the largest effect on how it performs..."
Examples of how operators abuse a lab-designed stove are given - the commonly practiced errors add up to an efficiency profile kinda like using your refrigerator to heat your house.
Rocket stoves built in the earth-friendly original materials are designed specifically to account for these human factors, and environmental ones rarely considered:
- very little transportation of materials (local earth and rubble),
- very low energy cost in manufacture of materials & components (re-used scrap metal and masonry components)
- no need to overheat and "damp down" to maintain comfortable temperatures; this practice alone is responsible for most woodstove pollution and wastes enormous amounts of fuel compared to a clean burn.
- burn at maximal efficiency for 4 hours, yet
experience the resulting warmth gradually released over 24-36 hours.
- heating the people directly, so that the entire house does not need to be kept as hot, reducing heat loss through walls & roof.
- can be fueled on 100% local biofuels; we use mostly
yard debris from a less than 2 acre property, with a little shop scrap and free local utility trimmings.
- low-cost building and fuel leaves money in the budget for insulation, etc.
You may quote me on "75% ... plus or minus 85%"; but please don't let people mistake that for accuracy.
Sooner or later, I'm sure someone will cough up to get one of these things lab tested.
If you got $6000 to burn, we'll build you one here at OMNI-labs and tell you what they say.
I'm interested to see their math, and how they measure actual flow rates without impeding draft. These things are touchy on draft (part of their efficiency comes from letting the exhaust wander through until it's almost sluggish), so putting a bag over its chimney might just stop it drafting altogether.
And now I realize that I
should have been running my stove while I type.
Good night and good luck.
-Erica Wisner
http://www.ErnieAndErica.info