*>Thanks guys. I guessed there would be too many factors involved to*

*>come up with a strong estimate...*

Daestrom suggests a way:

*>You can get an estimate of one of these factors, the thermal resistance, by *

*>noting how much energy you use to keep the place warm for some period of *

*>time. If you can estimate how much wood you burn over a week (along with *

*>the average outdoor temperature), we can figure out what the capacitance is *

*>(once the time-constant itself is determined).*

If the average indoor and outdoor temps are (say) 65 and 40 F and you burn

210 lb/week of wood at 60% efficiency, that makes the house conductance G

= 0.6x10Kx210/(7dx24h)/(65F-40F) = 300 Btu/h-F. If the house cools from 72

to 61 in 18 hours on a 40 F night, RC = C/G = -18/ln((61-40)/(72-40))

= 42.7 hours, so C = 42.7hx300Btu/h-F = 12.8K Btu/F.

Cooling from 72 to 61 on a 30 F night makes RC = -18/(ln((61-30)/(72-30))

= 59.3 hours, so C = 59.3x300 = 17.8K Btu/F, an increase of 5K Btu/F, eg

5K pounds or 600 gallons of water, or less, if the water cools from (say)

100 vs 72 F, or still again less, if the house temp instantly drops from

72 to 61, where it stays for 18 hours, with controls to regulate the heat

release from the mass to maintain a constant vs drooping house temp.

Nick

>Thanks guys. I guessed there would be too many factors involved to>come up with a strong estimate...