Rather than edit it out, I’d like to nominate the phrase “regardless of whether or not” to the awkward phrase hall of fame.
Actually, Tom, I don’t see anything wrong with that phrase. Now, if you had said, ”irregardless of whether or not”, that might be a different story. I’m of a mind that use of the word “whether” pretty much demands the “or not”. Rick
Oh wait, I get it now...you’re saying it’s redundant because “regardless” means pretty much the same thing as “whether or not”. OK, I agree, into the Hall of Fame it goes. Rick
thats enough to hold a serious pyro’s attention for a few days anyway. i got through 4 pages before i thought… “wait a minute i dont need to know this much detail” and clicked out.
...thats enough to hold a serious pyro’s attention for a few days anyway. i got through 4 pages before i thought… “wait a minute i dont need to know this much detail” and clicked out.
You got farther into it than I did, Mike , but I’m gonna hold onto the link for some of those winter nights when I just want to sit by the stove and I’m having trouble dozing off. Rick
...thats enough to hold a serious pyro’s attention for a few days anyway. i got through 4 pages before i thought… “wait a minute i dont need to know this much detail” and clicked out.
You got farther into it than I did, Mike , but I’m gonna hold onto the link for some of those winter nights when I just want to sit by the stove and I’m having trouble dozing off. Rick
i hear ya rick, gawd, but burning a stick is so complicated. i say “load wood, remove ash, repeat.”
as for the CO released by a woodstove , i would bet that a mid 1980’s automobile releases more minute running for minute running. especially with a modern reburn stove , which burns away a large amount of CO in the secondary combustion function of the unit.
thats enough to hold a serious pyro’s attention for a few days anyway. i got through 4 pages before i thought… “wait a minute i dont need to know this much detail” and clicked out.
THANX!
wood [C] to CO = 27 kilocalories / mole
CO- CO2 = 68 kcal/mole
so it looks like most of the heat in the wood’s carbon can be lost due to inadequate O2 & temp. to combust the CO
couldnt copy the script from the tome but it was about 3/4 of the way down it.
Thanks for the replies. I am puzzled, however, by the piece on outside air recommended by Sawdustburners. I read it, but it seems to me that a wood stove running at high heat (50,000 Btu/hr) will require, by my calculation, a minimum of 250 cfm of make-up air to supply the necessary oxygen. For a 2000 sq ft house, that means a complete change of air in a little over an hour. If we try to supply that air from just the leaks in the house, it will certainly cool the house away from the wood stove, right?
50 cfm is what a cheap bathroom vent fan blows & i dont think a chimney blows/pulls that hard. wood is 1/2 air as opposed to coal which requires less space to store. by pulling air from other parts of the house, u get fresh air in those parts as it will cool those parts.1 cubic foot of air with a 50*F difference will lose .6 BTU as told to me by local university prof. another link from EPA disscusses indoor air pollution.
how’d u get that 250cfm ? woodheat link says woodstove consumes 10-25 cfm. so >.6 x 10 = 6btu/min or 360/hr. on low. for when its 20*F outside & 7o inside.
Here’s how I got the 250 cfm, it is based on the oxygen needed for combustion to supply a certain amount of heat, which I chose to be 50,000 Btu/hour for a wood stove burning at the higher part of its range.
The energy content of wood is about 6,400 Btu/lb for air dry wood, 20% moisture, according to this source: http://bioenergy.ornl.gov/papers/misc/energy_conv.html
That means we need 50,000 Btu/hour / 6,400 Btu/lb or about 7.81 lbs of wood per hour.
(note that the numbers following chemical symbols below are subscripts, they didn’t come out right.)
To get the amount of oxygen needed, I assume wood is largely cellulose, a polymer with repeated units of (C6H10O5), so we need 6O2 to convert (burn) each unit completely to CO2 and H2O. The molecular weight of each unit of cellulose is (6x12 + 10 + 5x16) = 162 and the molecular weight of 6O2 is (6x16x2) = 192 , so for each pound of wood, we need 192/162 = 1.19 pounds of oxygen to burn it completely (and even more if the burn is not 100% efficient and some oxygen escapes up the chimney).
Air is 21% oxygen and 78% nitrogen with the last 1% some trace gases. Again, using molecular weights and ignoring the trace gases, the relative weights of air’s components are 21%x16 = 3.36 for oxygen and 79%x14 = 11.06 for nitrogen, for a total of 14.42 in air. This means for every pound of oxygen, we need to supply 14.42 / 3.36 = 4.29 pounds of air. Now a pound of air at standard temperature and pressure (about 70 degrees F and 14.7 lb/sq. in.) is 387 cubic feet. See this source for example: http://www.newton.dep.anl.gov/askasci/gen99/gen99510.htm
Putting this all together, we need 7.81 x 1.19 x 4.29 x 387 = 15430 cubic feet / hour or 257 cfm.
Catching up on this thread I want to thank all contributers for a great discussion. And tip of the hat to SteveD for the great calculations that saved me from the head throb of finding the correct formulas and running the numbers. I’m going to OAK the PE T6 with a valve so that this winter I can start gathering some data on OAK vs non-OAK operation in this old farmhouse. Should be pretty interesting.
I welcome suggestions for monitor points to establish baseline metrics for this experiment. I don’t have a lab load of sophisticated equipment, but can do IR thermometer readings of door, window, recessed ceiling light penetrations to see if there is a temp difference. And I can do observations of airflow based on incense stick smoke. Any other suggestions for data points?
A real truthful answer to… Does it feel warmer in the outer areas?
I mean a real rise in temps would be nice but 50 degrees doesn’t always feel the same.
BeGreen, here are some random thoughts about your experiment:
One of the main objects of this experiment (at least for me) is to see if using outside air keeps the remote rooms from getting cold. Therefore, I would get temperature measurements in various places around your home, from near the stove to the rooms farthest away, and on different floors if it is a multistory structure. The temperature measurements need to be made with and without outside air, and should probably last for at least an hour for each situation to allow time for the airflows and temperatures inside the house to equilibrate. You probably want to turn your regular furnace off during the tests so you can see just the effect of outside vs inside air with no heat sources other than the wood stove. To make any differences easier to see, you probably want to run your stove at moderately high heat and keep the burn constant, say by monitoring the stove temperature.
During the measurements, with and without outside air, I would try to control and keep constant all the important variables you can think of:
1) See that the temperature outside is approximately constant during the tests.
2) See that the wind is very low and stays that way, at least during your baseline cases. Later, if you want, you can run the experiment again when the wind is blowing at a constant rate.
3) Keep all the doors in the house in the same position and keep outside doors closed.
4) etc.
If you don’t want to run around with a single thermometer to make the temperature measurements, you could just buy a number of inexpensive digital thermometers and put them in each place you want to measure. To avoid systematic errors, initially place all the thermometers together for awhile and see that they read the same temperature. If they don’t, you can simply label the outliers with their temperature error and just account for that error when you make your reading. Almost surely the thermometers are all reasonably linear, even if their set points are a little off.
You might also rig up a poor man’s manometer, a U-tube filled with colored water, and measure the pressure difference from inside to outside the house for each situation. Each mm of difference in water levels means a pressure difference of 9.8 Pascals. If your house is very leaky, like mine, I wouldn’t expect much difference, but if your house is tight, you should probably see an effect.
A really good thing to measure and know is the actual air flow in the pipe which brings outside air to your stove. I’m not sure what the best, inexpensive way to make that measurement (anemometer, hot wire, etc.), but it would be worth looking into to resolve the debate in this forum.
Rick
