Solving for CFM

We recently installed a wood stove in our living room. This room is very comfortable, but the hot air does not move into the adjacent room, despite there being a large opening between the two rooms.

This post documents my attempt to figure out how to move the air so both rooms are warmed.

Conclusion

After looking into the heat capacity ⊕ It takes 0.24 BTUs to raise a pound of air 1°F. Air is remarkably heavy, weighing in at 14.7 pounds per square inch. of air, the actual calculation turned out to be simple.

Calculate the cubic feet of air in the cold room, pull that volume of air out, and replace it with the warmer air from the fireplace room.

The total volume of air in the room adjacent to the wood stove plus the upstairs room next to the bathroom fan is 8,000 cubic-feet.

Panasonic sells a bathroom fan that can pull 390 ⊕ $\mathrm{20\; min.}\times {\mathrm{390\; ft}}^3\mathrm{/min.}={\mathrm{7,800\; ft}}^3$ cubic-feet per minute (the VN-40VQ4), so in an ideal world, running that fan for 20 minutes will do the trick.

I’ll also get a fan to push air from the wood stove room into the middle room to see how much that on it’s own will help.

Research Notes

• The rule of thumb for a whole-house fan is 2 - 3 cubic-feet per minute.
• Houses with higher ceilings should size up.
• I have a volume of air at 60°F, and another volume of air at 70°F. I want both rooms to be at 70°F.
• Assuming the air in the cold room mixes evenly, one part 60 and one part 70 should result in a single part of 65.
• A single, full air exchange that removes the 60° air and replaces it with 70° air should make both rooms the same temperature.
• The volume of air in the kitchen and living room is roughly 5,500 feet³. The volume of air in the upstairs room is 17 x 14 x 8 + 1/2(14 x 17 x 6) = 1904 + 294 = 2198 for a total air volume of roughly 8,000 cubic-feet.
• Panasonic sells a 390CFM bathroom ceiling fan (the FV-40VQ4), which would need to run for 20 minutes to pull out 8,000 cubit-feet of air, assuming no leakage.
• The heat output of the Jotul F 445 Holliday ranges from 13,700 BTU/hr to 35,700 BTU/hr.
• Note the stove is rated at .49 grams of particulate emissions / hour, which is about the same as a cigarette and less than a car. However, this rating must have been measured once the stove is heated up; the first 20 to 30 minutes there is a lot more smoke coming out the chimney. Source: Burning Issues, Chart: Comparing Particle Emissions from Traffic, Cigarettes and Heating.

Heat Loss in Dining Room

• l = heat loss, in BTU/hour.
• a = surface area, in feet²
• δT is difference in temperature between inside and outside.
• R = average R value of the wall.

Estimated R-value for walls in kitchen and dining room is 12. So if it is 25 outside and 70 inside, the BTUs lost per hour is $l=\mathrm{640}\frac{\mathrm{45}}{\mathrm{12}}=\mathrm{2,400}$

Adding thermal curtains can increase the windows R-value from 2.5 (estimated) up to six.

Our dining room/kitchen has 135 square-feet of class (windows + sliding doors) and 505 square-feet of R15 walls.

If we increase the R-value of the the sliding glass doors from 2.5 to five, the R-value goes up to

and the BTUs lost / hour is reduced to 2,215.

Tags: eco