I read somewhere that in even the very tightest of modern homes the air turn-over needs to be at the level of at least one complete change-out of the air every 3 hours at a minimum, and if it is less than that it is considered to be very unhealthy for the occupants.
If a home has 2,000 sq-ft of floor space and 8 ft. ceilings that is 16,000 cubic feet of minimum air leakage that is actually required by the home every 3 hours (not counting the basement). That comes to 5,333 cubic feet of infiltration by cold air rushing in and the same volume of heated air drafting out, stove or none. For an older home of comparable size the air infiltration (also stove or none) could be perhaps 50% more than the requisit minimum for modern tight homes, or about 8,000 cubic feet per hour.
Every pound of coal that you burn requires the draw of 123 cubic feet of air through the stove to combust it completely. If you are burning 50 lbs. of coal per day thats only about 256 cubic feet required per hour going through the stove, into the coal, and out the chimney as combustion byproducts vs. from 5,333 to 8,000 cubic feet or more going out of the home every hour due to all of the homes other leaks and natural drafts. Even if allowing for 15% more air than is required in a perfect world (as I've read that real world combustion is best when 15% excess air is present) that means that only about 300 cubic feet of air per hour are drafted through a stove that is burning 50 lbs. of coal per day.
Though this excercise does not measure how much a home loses to a barometric damper, it might be in the ballpark of the amount being consumed through the stove, and if so It seems negligable overall vs. the natural turn-overs of air for a home.
Last edited by lsayre
on Fri Dec 13, 2013 1:56 pm, edited 1 time in total.