This subject has been touched on in many places and in bits and pieces throughout the forum. I have noticed some extraordinary claims of stove burn times, heat output and efficiency.
For this thread, I would like to apply some basic math and physics and remove some variables to prove a point.
In lab testing it is basically a well accepted rule of thumb that wood (fire wood) that has been dried to 0% moisture content in a 100% oxygenated environment will produce 8660 BTU per pound of wood.
Lets assume that these "perfect" conditions can be met with our stove.
Lets assume that we are trying to heat a large old farmhouse on the northern boarder of Minnesota, that requires 45,000 BTU per hour to maintain the inside temp during a nasty storm in Feb. (am I painting a picture yet?)
Lets assume we have a large stove that can take loads of fuel of 60 pounds. Thats a pretty large load even in a Blaze King, King. If you don't believe me, load your cold stove up and then remove the wood and weigh it.
8660 BTU x 60 pounds = 519600 BTU
519600 BTU / 45,000 BTU per hour = 11.5 hours.
This is in PERFECT conditions.
Lets add in some variables:
85% efficient stove (pretty darn good cat stove)
519600 BTU is reduced to 441660 BTU
441660 / 45,000 BTU per hour = 9.8 hours
Now wood - lets say 18% on the ol meter. Which will reduce heat output of said wood by roughly the same number. 18%
So we will start again:
519600 BTU output reduced by 18% = 426,072 BTU is available from the wood.
426,072 being burned in an 85% efficient stove = 362161 BTU output of the stove.
362161 BTU / 45,000 BTU per hour = 8 hrs.
This represents the best of the best that can be achieved by one of the most efficient stoves on the market and a very dedicated operator. That is VERY good burn times IMHO for trying to heat an old house on a brutal day.
Whats my point? You can't magically produce more heat than the load of fuel can deliver. It doesn't matter if you stretch the burn times out to 15 hours, you will HAVE to reduce hourly btu output to obtain that. Same holds true if you need more BTU per hour to maintain house temps, you WILL reduce burn time.
This post stems from my knee jerk reaction to some posts I have seen recently. "Yup, I can burn my stove for 20 hours", so what! I can burn a bic lighter for 20 minutes on two table spoons of fuel, but it ain't gonna heat my house.
It comes right on down to how much BTU is available in your fuel load, and how efficiently your stove can burn and deliver that to your living area. Some stove are bigger or smaller. Some stoves can boast a higher efficiency but NONE of them can generate MORE heat than the fuel load that it was feed with. PERIOD.
I know this seems very obvious to many, I simply wanted to clarify the term "Burn time" versus "Heat output"
Sorry for the rant I just felt a need to make this perfectly clear.
For this thread, I would like to apply some basic math and physics and remove some variables to prove a point.
In lab testing it is basically a well accepted rule of thumb that wood (fire wood) that has been dried to 0% moisture content in a 100% oxygenated environment will produce 8660 BTU per pound of wood.
Lets assume that these "perfect" conditions can be met with our stove.
Lets assume that we are trying to heat a large old farmhouse on the northern boarder of Minnesota, that requires 45,000 BTU per hour to maintain the inside temp during a nasty storm in Feb. (am I painting a picture yet?)
Lets assume we have a large stove that can take loads of fuel of 60 pounds. Thats a pretty large load even in a Blaze King, King. If you don't believe me, load your cold stove up and then remove the wood and weigh it.
8660 BTU x 60 pounds = 519600 BTU
519600 BTU / 45,000 BTU per hour = 11.5 hours.
This is in PERFECT conditions.
Lets add in some variables:
85% efficient stove (pretty darn good cat stove)
519600 BTU is reduced to 441660 BTU
441660 / 45,000 BTU per hour = 9.8 hours
Now wood - lets say 18% on the ol meter. Which will reduce heat output of said wood by roughly the same number. 18%
So we will start again:
519600 BTU output reduced by 18% = 426,072 BTU is available from the wood.
426,072 being burned in an 85% efficient stove = 362161 BTU output of the stove.
362161 BTU / 45,000 BTU per hour = 8 hrs.
This represents the best of the best that can be achieved by one of the most efficient stoves on the market and a very dedicated operator. That is VERY good burn times IMHO for trying to heat an old house on a brutal day.
Whats my point? You can't magically produce more heat than the load of fuel can deliver. It doesn't matter if you stretch the burn times out to 15 hours, you will HAVE to reduce hourly btu output to obtain that. Same holds true if you need more BTU per hour to maintain house temps, you WILL reduce burn time.
This post stems from my knee jerk reaction to some posts I have seen recently. "Yup, I can burn my stove for 20 hours", so what! I can burn a bic lighter for 20 minutes on two table spoons of fuel, but it ain't gonna heat my house.
It comes right on down to how much BTU is available in your fuel load, and how efficiently your stove can burn and deliver that to your living area. Some stove are bigger or smaller. Some stoves can boast a higher efficiency but NONE of them can generate MORE heat than the fuel load that it was feed with. PERIOD.
I know this seems very obvious to many, I simply wanted to clarify the term "Burn time" versus "Heat output"
Sorry for the rant I just felt a need to make this perfectly clear.