Heat loss due to wet wood

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Wood Duck said:
Adios Pantalones said:
I am a scientist, and the question of taking water at one temp, and turning it to steam is actually one of those things you are taught in basic chem. I haven't gone through anyone's calculatuions though- mostly because it looks more like work :)

I am an engineer, and when we do calcs we leave out anything we don't think matters. We don't always choose the right things to leave out, but it makes the work a lot easier.

We do too for most scenarios, but someone else postulated that the starting temp was different. Fact is- water has a high heat capacity, and it's not a bad idea to see if it's on the order of magnitude to cause a difference.

Slow1 said:
That is the basic calculation that started this whole thread. I believe the key question (variable) that may be throwing the raw calculation off is exactly how much excess water may be in a given split. The second question (and perhaps more interesting) is how to put an objective measure on all the other factors related to the effects of burning the wetter wood. I'm not sure of how material some parts of the argument are (such as heating room air), but certainly the way that the boil-off affects the burn is material beyond the energy consumed converting water to steam.

That was my first post- that the wood isn't burning the same, and I think (my educated guess, and no more) that this is a really big effect in overall efficiency. There's still too many variables. If I know that my wood is wet (sideways rain or slush, etc) in my woodstove- I mix it in with dry or leave the air open more to compensate. I get a clean burn that way, but do sacrifice efficiency.
 
Battenkiller said:
skyline said:
What may be more useful to your friend that I have reported here before in my "wood drying experiments" is that a small house fan on splits even in a unheated garage and high humidities (85-95%) can remove lots of moisture in just a few days.

Sky, I've been waiting all season for the results of your "experiment". Surely this deserves its own thread, eh? I have many questions, but don't want to hijack this thread about heat loss. What say you?

Battenkiller, you have guilted me into it. My wood splits are in the oven so I can get a final dry weight to calculate their real moisture contents and not just how much water they lost.
I'll start a new thread when I finish the results.

As Wood Duck pointed out, the ~10.5% difference from just the btu calculation is not that great and many of us suspect, greater differences may be due to efficiencies in how our stoves burn dry vs. wet wood. But that is partly why the early fan results were exciting (this is a relative term as only us wood burners know!). The fact that a fan in an unheated garage with high relative humidities can lower the moisture content of really wet wood in a couple of weeks by a huge % means that the btu calculation between this "blown" wood and "properly seasoned" and may turn out to be just a few 1-2%. Mostly I thought it might be helpful to all those folks (not me, of course) that aren't quite as far ahead on their wood seasoning as they should be. The other plus is that I think a fan in 24 hours will remove most of the "surface moisture" from wood that has been well seasoned but suffered from more recent wind blown rain/snow wetting.

Now back to the aroma of baking Fir!
 

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Backwoods Savage said:
Billy, I fear you would baffle your friend with the numbers and reasoning. So why not try to keep it a bit simpler? Find out what type of wood he burns. Then find some green wood of that type or take some of his. Then find some of that type of wood that is dry and well seasoned. Now you and your friend should build two fires. One with the green wood and one with the dry wood. Note the difference in how it was getting the fire started and how the fire burned. Also take note of smoke from both fires. You might even try standing by both fires to see if there is any difference in the amount of heat both give. How about how long they burn? One could have lots of fun this way without trying to reason at all.

Hi Dennis

He's a retired H.S. auto shop teacher And is used to telling everyone what to do but doesn't listen much. And feels everything he is told is a exaggeration. For him a hard number will work. Actually he's my Uncle but he still doesn't listen well. And I am sorry to say he would be to busy talking to pay any attention to a wet/dry wood comparison.

Billy
 
Wood Duck said:
Here are my calcs: If you have 1 pound of dry wood, you have around 7000 BTUs of heat. This would be a small split of firewood. If you were to burn that in a 75% efficient stove, you'd get 5,250 BTUs into the house. If the same wood were at 20% moisture content, you'd be adding 0.20 pounds of water, which would require 0.20 x 1100 BTUs to evaporate, or 220 BTUs. That would mean you'd only get 5,250 - 220 = 5030 BTUs into the house, a loss of 220/5250 = 4.2%. If the same wood was at 50% moisture content, the 0.5 lbs of water would take 0.50 x 1100 BTUs to evaporate, or 550 BTUs, which is 550/5250 = 10.5% of the total heat that could have gotten into the house.

I agree with Cowboy's earlier comment- a 10% difference is much smaller than I would have expected. Perhaps the bigger change is due to the way you have to burn wet wood. If the wood is wet, additional air is needed, which would reduce the stove efficiency. Old 'smoke dragon' stoves are supposed to be around 40% efficient, so if burning wet wood caused you to operate the stove differently and get only smoke dragon efficiency, you get only 40% x 7000 BTUs = 2,800 BTUs into the house. This would mean a loss of (5,250 - 2,800)/5,250 = 46.7% loss of heat value. There is your answer! I am not sure it is the right answer, but a 47% loss is a lot more dramatic than a 10% loss of heat.


Ah Ha!! efficiency of the stove was something I did not think about at all!!! And that makes a lot of sense Wood Duck!! Like me he is running a Johnsons Energy Systems wood burning furnace.

I knew if I got this out and all of us kicking it around we would get something figured out! As my friend Rudi says "Confusion breeds Discussion which breeds Knowledge which breeds Confidence which breeds Friendship"

Billy
 
Sorry, I didn't read the whole thread yet, but the most simple way to figure this is to find a wood chart that has weight/cord both green and seasoned. I like this one: http://firewoodresource.com/firewood-btu-ratings/

So lets use cottonwood for example:

You have 2225 lbs of seasoned wood, or 3475 lbs of green wood. That means there is 1250 lbs of extra water that needs to be evaporated out of the wood. Because we're actually superheating the steam (heating it above boiling) and we have to heat the water before it evaporates as well, I'm going to guess that it takes 1200 BTU/lb (I'm too lazy to dig out my referrence books and look). So it's going to take 1.5 MBtu. So the 16.8 MBtu drops to 15.3MBtu.

Just do that calculation for whatever wood you want. As mentioned already though, that doesn't take into account the inefficient burning as well.
 
Wood Duck said:
Adios Pantalones said:
That wood still needs to be heated up- whether the fire does it before it gets in the stove or after- that's energy loss- so just use the outside temp.

For the same reason, an Outside Air Kit doesn't change overall efficiency. One way or another, cold air from outside your house is heated to combustion temperature before it takes part in combustion. That can happen in the stove or in the living room, but it still happens and absorbs some energy. It isn't a lot of energy, neither is warming up the wood. There are far bigger uncertainties in the calculation - such as moisture content of the wood, type of wood and stove efficiency - to make warming up the wood or the air in the stove a critical part of the calculation unless you are a scientist.
Actually your logic on that is flawed, you have to take into account the heat has to transfer out of the stove.

Let me ask you this, does a 40 degree temperature decrease make more difference in your room temperature or in the stove temperature? IE, would you prefer your stove temp go from 550 to 510, or your room temp go from 75 to 35?

That is putting it in really simple terms, but that demonstrates that temperature difference is not equally comparable inside and outside of the systems.
 
Cowboy Billy said:
So I'll start with poplar since I know it hold a lot of water. I believe I hear it was 60% water while growing.

You have to be very careful with those moisture content figures. Most all of the available data on moisture content in green wood was gathered by the U.S. Forest Service, and so is expressed as a percentage of the oven-dry weight of a given piece of wood. Wood heat values OTOH are computed using another formula, based on the actual weight of the water in the wood as a percentage of a green piece. Check out the table at the bottom to convert between the two methods. This table was taken from a book mentioned in a post above, "The Woodburner's Encyclopedia" by Dr. Jay Shelton. From the table, you can clearly see that wood that is quoted by the industry as being at 60% MC actually only has 37.5% water by weight when green.


So poplar has 6861 but's per pound (if I am doing that right) and say its 50% water thats .5 lbs of water and we need 575 btu's we have to take out of the poplars' total.

All hardwood has a heat value of about 8600 BTU per oven-dry pound. All computations must begin with this figure, with heat losses computed and subtracted from this number. Doing anything else will give wildly inaccurate results.

Oops if it wet 50% I only have a half pound of wood so 3430.5 btu's minus 575 leaves me 2855.5 btu's net.

Even if that split was really 50% water by weight, you are getting twice as much weight in a green cord, so it's a moot point. You will still have the same amount of wood fiber in that cord as in that same cord that has been oven-dried to 0% MC (and now weighs half what the green cord weighs), and the same amount of potential chemical energy will be locked into that wood.

Because of the incredible number of variables, it is pretty near impossible to put a figure on the heat loss. You can get a very close estimate of the theoretical loss due to heating up and evaporating the contained water, but you can't extrapolate this onto heating efficiencies in general. All you can be sure of is that you will lose at least all of the heat required to raise the water to 212º, plus the heat it takes to evaporate that same amount of water, plus all of the latent heat of condensation from the water that is formed as a product of wood combustion (this is a lot of heat loss that cannot be avoided no matter how dry your wood is).

Efficiencies of 75% are based on total combustion efficiency ("high heat value"), not heat transfer efficiency. All of the water that leaves as steam will carry away about 1000 BTU/pound, whether it was in the wood in the first place or if it was formed by combustion itself. This lowers the burn efficiency closer to the low 60% range (63% is the EPA estimate for non-cat stoves), with even further loses due to heat going up the stack.

These is absolutely no reason to use computations for this info, there are already industry-established tables that will give this info (called "low heat value") for any water content you will encounter. Below is a nice chart (from the same book mentioned above) that presents this info in an easy to see graphical format. As you can see, there is only a very small amount of additional heat loss due to burning wood at 60% MC (37.5% water by weight) compare to wood burned at 20% MC (16.7% water by weight).
 

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