DexterDay said:
olddawgsrule said:
Turbo-Quad said:
my sister just bought a Cumberland. I installed it a week ago. We have had a cold spell 40 at day and 20 at night. It has been burning an average of 1.8 lbs per hour to keep her well insulateghted pretty tight house at 73 degrees. I'm impressed, mostly with how well her house is insulated. The stove is ok too. Priced reasonbly also. The manual is confusing, It says the stove is 45000 btu but a burn test revealed it only burns 32000 btu's per hour. Tech's are telling me the manual is a mispprint. Who knows. She llikes it so far so I reckon that's all that matters.
my quad is a pellet pig it uses a bag a day on its lowest setting and keeps my uninsulated loose house at about 67.
Wow! 1.8 lbs an hour!
That's like 22 hours on a 40lb load!!
Impressed.
Almost half of what I'm burning...
Curious, how'd you measure the BTU output of the stove?
BTU's are measured by ---- Pounds per hour x BTU's of Pellets.
Example ---- 1.5 lbs per hr x 8,000 BTU pellet = 12,000 BTU's.
Most if not All stoves can go down to around 8,000-12,000 BTU's at there minimum setting.
Have to say, that's a bit easier than:
Technically, since 1997 the Standard air calculation has been revised to accommodate for humidy. The original factor of 1.08 is for dry air.
BTUH = 60 x 0.075 (0.24 + 0.45W) x CFM x Delta T
BTUH = 4.5 x .2445 x 3 x 10
BTUH = 33
Where:
60 = min/hr
0.075 = lb(dry air)/ft3
0.24 = specific heat of dry air, btu/(lb* degrees f )
W = Humidity ratio, lb(water)/lb(dry air)
0.45 = specific heat of water vapor, btu/(lb* degrees f )
When W = 0.01 the value = 1.10
When W = 0.02 the value = 1.12
When W = 0.03 the value = 1.14
Thus, because a value of W = 0.01 approximates conditions found in many air-conditioning problems, the Sensible heat change (in btu/h) can be found as;
BTUH = 1.10 x CFM x Delta T
The preceeding was extracted from 1997 ASHRAE Handbook "Fundamentals", page 28.15.
You're numbers have me blowing away my stove on high....