Can someone check my math please?

[Poindexter]

Imagine I got three cords of paper birch in the back yard. One measure 20% MC, one measures 16% MC and one measures 12% MC. Each of them is exactly 85cf of wood in a 128cf box, 1020 board feet in each cord.

I "know" the 20% MC is the heaviest one. I also "know" I should get more heat out of the driest cord, because less water will need to be boiled out of the drier wood before combustion can take place.

I finally found a online calculator that i think works, I had to scribble on an envelope to come with 1020bf = 85cfm but there you have it: http://www.woodweb.com/cgi-bin/calculators/calc.pl

So according to these guys @ 20% MC the cord of paper birch weighs 3235#, at 16%MC it is down to 3200#, at 12% MC, 3164#.

So dropping from 20% to 12% MC (I ass/u/me they mean dry basis), only drops 70# of water out of the entire cord. At 1000BTUs to evaporate 1# of water (I think it is 876BTUs or 906BTUs or so, a bit less than 1000), but even over estimating at the 1000BTUs per pound I can maybe get an extra 70,000 BTUs out of the 12% MC cord compared to the 20% MC cord.

It will be 70 fewer pounds to carry from the shed to the stove, that's a good thing.

On the other hand, with #2 oil at 3.82/ gallon (same as electric at 11.1 cents per kwh, or propane at $2.40 per gallon) I am saving a whopping $2.33 burning the entire cord of 12% instead of at 20%.

Did I make a computational error somewhere? It "seems" like 12% MC burns a LOT better than 20% MC.

Thanks

 

[TheRambler]

For starters a cord is 1536 board feet approximately

 

[Poindexter]

For starters a cord is 1536 board feet approximately

I was allowing for 85cf of actual wood in the cords. I agree that 128cf of wood = 1536 board feet.

 

[ewdudley]

So according to these guys @ 20% MC the cord of paper birch weighs 3235#, at 16%MC it is down to 3200#, at 12% MC, 3164#.

I think this is where it starts off track.

(All weights in pounds. )

3235 sounds about right for 20% MC DB paper birch, since it agrees with other sources, for instance:

http://mb-soft.com/juca/print/firewood.html

20% moisture dry-basis is 20 parts water and 100 parts wood, so the weight of the dry wood in 3235 pounds of wood at 20% DB is (100 / 120) * 3235, or 2696.

So the weight a 16% DB would be 1.16 * 2696, or 3127, an for 12% weight would be 3020.

So dropping from 20% to 12% MC (I ass/u/me they mean dry basis), only drops 70# of water out of the entire cord.

But I'm coming up with 215 pounds of water per cord difference between 20% and 12%.

So using your 1000 btu per pound for easy figuring you're saving 215,000 per cord (as opposed to the previously calculated 70,000 btu per cord), which still seems like squat when we're talking 20,000,000 btu per cord.

And it is pretty much squat.

Except depending on what type of wood burning equipment you're using, it's generally much more a matter of how well and efficiently you can burn with drier wood as opposed to how much less water vapor you're sending up the flue. For instance a big biomass plant can burn 70% MC DB fuel quite efficiently, but a wood stove can't.

 

[GeHmTS]

Using the information given in the problem, it breaks down as follows:


sample 1 sample 2 sample 3
moisture content C1=.20 C2=.16 C3=.12
total mass (Mt) 3235 pounds 3200 pounds 3164 pounds
mass of H2O Md1*C1 Md2*C2 Md3*C3 (Md is dry mass of sample)
Volumes V=V1 V=V2 V=V3
density of cord 3235/V 3200/V 3164/V

difference in mass of H2O: ΔH2O = Md1*C1-Md3*C3

Note that Mt1 is Md1*(C1+1); Mt3 is Md3(C3+1). This is from ((mass of H2O)/(Md))=C and Mt = (Md + mass H2O)

putting it all together you end up with ΔH2O=[Mt1*C1]/(C1+1) - [Mt3*C3]/(C3+1)

If you plug the values above into this equation, you get 200 lbs of H2O between 20% and 12%.

 

[brad wilton]

your giving me a headache

 

[Babaganoosh]

What is it you are trying to do exactly? What is the purposeof all this?

 

[Poindexter]

What is it you are trying to do exactly? What is the purposeof all this?

To measure is to know. Knowledge is power. Thus, borrowing from Francis Bacon and Lord Kelvin both, to measure is power.

What I want really is good data to take to my back fence neighbor so that I can explain to him nonconfrontationaly, gently, what is in it for him to burn drier wood. He doesn't, or many of the folks around me don't, care about my kids "catching" asthma, but if I can convince them to clean up their stack exhaust for their own benefit they are more likely to do it.

Dropping 200# off carrying a cord of wood to the stove is a start, both the respondents who did the math came up with 200# of missing water.

Not nearly as many theoretical BTUs as I had hoped. I do notice one BTU is the energy required to heat one pound of water from 62 to 63 dF. So I bet that 906 number of BTUs is the latent heat of vaporization for water. First all that water has to be heated from say -20dF ambient in my backyard to +55dF ambient in my garage - I suspect the latent heat of crystalization will come into play there - and then once in the stove heated from +55dF to +212dF before the latent heat of vaporization can be applied.

I think we are still leaving some BTUs on the table without allowing for those, but getting late for me.

 

[ewdudley]

To take a pound of water from 0 degF to 350 degF:

(32 degF * 0.5 btu_per_pound_per_degF_specific_heat) +
(144 btu_per_pound_heat_of_fusion) +
(180 degF * 1.0 btu_per_pound_per_degF_specific_heat) +
(970 btu_per_pound_heat_of_vaporization) +
(138 degF * 0.5 btu_per_pound_per_degF_specific_heat)

16 + 144 + 180 + 970 + 69 = 1380

Times 215 pounds of water per cord equals 300,000 btu per cord up the flue (20% MC DB vs 12% MC DB).

Again, not a big deal when we consider the cord of wood is bringing 20,000,000 btu to the table.

Again, the problem has much more to do with the incomplete combustion and the need for excess combustion air associated with trying to burn wet wood in a wood stove.

 

[rwhite]

I think the 300000 btus is pretty close to what would be required to bring 20% to 12%. But that is all based on lab setting with 100% effenciency and wood of a known surface area. There are way to many variables with a wood stove. This is purely conjecture but when I burn wood that is 20% vs 12% I find that my consumption rate increases by 1/2 to 2/3 ( 1-2 hour burn times vs 4-6) because I have to increase the air to get it to burn and all my heat is going out the chimney. Assuming I burn 1 cord a month and it cost $200. Not only will I need 1/2 again as much wood I am spending an extra 100 per month.

 

[brad wilton]

just tell them how much money would be saved in wood

 

[Poindexter]

Thanks for getting the heat of fusion in there rwhite. I am also looking at bringing wood straight from the shed (zero dF give or take) and putting it in the wood box right next to the stove upstairs.

Currently I am running a ready rack in the oil heated garage downstairs, I bring in about a face cord at a time, then the oil furnace pumps BTUs into the garage to heat the latest facecord from 0dF up to 55dF, then I have been feeding the stove out of that preheated/ pre-thawed rack.

I am thinking the man door out from the garage to the woodshed will be open the same amount of time whether I got thrugh it twice daily stocking the woodbox by the stove, or 21 times on Saturday refilling the rack in the garage.

I am essentially heating 1200sqft of downstairs with oil and1200 sqft of upstairs with wood. My cost per BTU for wood is about1/3 the price of oil, so it _might_ be economical for me to bring wood a little bit at a time straight from the shed to the woodbox by the stove. This calculation is complicated by my wife opening the overhead garage door twice daily. In the morning she leaves the garage in a car that was warmed to 55dF by the oil furnace, every night she brings in a car that was parked outdoors at her office all day.

On seven cords a year I wonder how much it costs me to pre heat it in the garage. Possibly not much, especially with her car going in and out daily anyway.

 

[Spletz]

IMHO it maybe easier to just have a frank conversation with your neighbor regarding the well documented relationship of wood smoke and health issues.

 

[tarzan]

Is your neighbor burning wood at 20% moisture? If so, would'nt the problem be the stove he is burning it in?

I can most definately burn 20% moisture wood cleanly. By cleanly I mean up to temp in a reasonable time and no smoke once fire is established.

 

[Thislilfishy]

What Tarzan says. The other side of the debate is how fast that wood creates the bulk of its BTU output. I'm finding with my stove that very dry wood burns very hot (somewhat good until it starts climbing over 700*!) but very fast(bad). Where if my wood is closer to 20% then I get slower burns and at lower more controlled temps. So although some of the btu's are lost to water, I get longer btu output. Seems better that way. There are so many variables it's impossible to say any one burning method is best for the hundreds or thousands of different stove, and millions of different chimneys, in the many different climates. Without even getting into wood types! If you're neighbor is smoking you out, then approach him about it. Maybe he's just got an old non epa stove.

Ian