sawdustburners said:
SteveD said:
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: (broken link removed)
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: (broken link removed)
Putting this all together, we need 7.81 x 1.19 x 4.29 x 387 = 15430 cubic feet / hour or 257 cfm.