I get some noise and less than perfect burn from 7 year old splits (species doesn't matter) if the wood has been exposed to rain water (even a little bit) and hasn't fully dried out before I use it. I try to avoid that but sometimes it happens.
Came home and started evening fire the other night. Have red oak that has been CSS for 3 years in a single row outside then in my wood shed since the end of Sept. Little triangular split with base (largest edge) only 1.5", kindling sized really. Yup, began steaming and boiling water like crazy. Go figure.
Yeah steady diet of red oak here too though I have some other stuff ready to play with.Red oak is most of what I burn. Mine is seasoned 2.5 years, and I haven't noticed many sizzlers. It burns like a dream, with very long active-cat coaling phase.
I'm stuck at home with a bad cold so I figured I'd finally work on this problem some more. What I did was put together an Excel file that calculated what percent of red oak's total heat value is expended in getting the wood up to the ignition point (and boiling off any water in the process). I looked at MCs from 0% to 25% and starting temps of -20C (-4F), 0C (32F), and 20C (68F). Not too many surprises in the numbers themselves but here's what I came up with...starting temperature has a small effect but not as much as the MC:I hear you. When I have more unallocated free time I would like to do just that. The heat of fusion (ice to water) is an eye-opener but pales in comparison to the heat of vaporization. And the energy loss as the water in your wood vaporizes hurts more than the lost energy would indicate. By consuming heat, the evaporation of water cools the combustion process and causes a less complete burn. So more unburnt gases escape up your chimney without contributing to the heating of your home. And some might say, that's ok, I'll burn the less dry stuff when the weather is mild and I don't need the extra heat. But that's the worst time to burn damp wood because that's when you want to be able to burn low and slow which is precisely when you don't want your combustion cooled further. Creosote city.
I'm stuck at home with a bad cold so I figured I'd finally work on this problem some more.
What I did was put together an Excel file that calculated what percent of red oak's total heat value is expended in getting the wood up to the ignition point (and boiling off any water in the process). I looked at MCs from 0% to 25% and starting temps of -20C (-4F), 0C (32F), and 20C (68F). Not too many surprises in the numbers themselves but here's what I came up with...starting temperature has a small effect but not as much as the MC:
View attachment 194473
What surprised me is that there's no solid-liquid phase change to worry about if the wood is below the fiber saturation point, according to the document. So that simplified the calculations a bit.
To break this down a little further (and allow others to check my work, because cold medicine makes me loopy), 1.2 kg of 20% MC red oak (1 kg dry equivalent) at 0C/32F would need about 66 kcal to heat it to 100C, 108 kcal to boil off the 0.2 kg of water, and 71 kcal to heat the 1 kg of dry wood to 250F. So 27% of the "wasted" energy goes into heating the damp wood and a whopping 44% goes into boiling off the water. The remaining 29% is "overhead" that's pretty much unaffected by the starting conditions. Add them all up and divide by 3574 and you get the 6.9% figure that's on the chart.
I hope I did this right--give me a shout if any of it doesn't make sense!
When you can't go outside and play with firewood, what else is there to do?Ha! I did my calculations when I was stuck at home with a bad cold. Great minds think alike! ;-)
Very nice. I think it would be even better if it went up to 35-40% MC because it would illustrate the effect of the drying curve better. I know you know this but I wanted to mention it for those who may not. Your table doesn't represent the true loss of heating value caused by the evaporation of water during the burn because it doesn't attempt to account for the fact that the cooling allows more volatile gases to escape unburnt that would otherwise have burnt without the evaporative cooling effect illustrated in your table. And this is a huge factor why well seasoned wood produces so much more heat than less well seasoned wood.
That's a great document and thanks for posting it! My interpretation of it is that various researchers results vary considerably on that point and the likely reality is that some of the frozen water bound in wood does go through solid-liquid phase but that as the MC content of the wood decreases the percentage of water molecules that are bound (frozen) to other water molecules declines and thus the effect of solid-liquid phase change declines as well. Oddly, it also appears that as the moisture content decreases, the temperature at which the phase change happens also decreases. Interesting stuff.
I haven't checked your numbers (I got sidetracked by unit conversions and then other related topics) but the figures seem very much in line with what I can believe intuitively. Keeping in mind the single biggest loss of wet wood is the way evaporative cooling that prevents all the volatile gases from burning! Which is also the source of most wood smoke pollution. Unfortunately, there is no good way to mathematically model that loss due to all the variables involved.
I guess if I were to add higher MCs I could start by running them at just 0C and above, which I think would still illustrate your point.
I think what is happening is that there is a fixed amount of sugars and other things dissolved in the unbound water. The water starts off fairly dilute but as the wood dries the concentration of dissolved materials increases, so you get freezing point depression.
Strangely enough, wood can have both freezing and unfreezing bound water. See http://forest.uef.fi/~karenlam/petri/measur/4361Naka.pdf for some fascinating reading!I'm not a chemist but I believe the water content of the wood would need to be at or above the fiber saturation point to consider that anything is dissolved in the water. Otherwise the sugars, etc. are not dissolved in the "water" but merely present in the wood. But I admit I don't know whether the presence of those sugars would affect the freezing point of the remaining moisture (or for that matter whether an individual water molecule can actually freeze). Seems to me, water is only "frozen" or solid when it is surround by other water molecules and bound into "ice" and thus the diminishing heat of fusion as the moisture content decreases is simply due to a lack of adjacent water molecules with which to adhere to.
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