I started talking about this in another recent thread and realized I was probably hijacking the thread and this probably deserves its own thread. We were talking about possible efficiency loss by using double-wall connector pipe, so I wanted to try to calculate what this would be. That is, how much heat does a stovepipe at various temps radiate into a room ...
It might be interesting, if we tried to get an idea of how much heat is lost with double-wall. I guess we'd need folks with IR thermometers to try to report the outer-surface pipe temps they see with both types of pipe under similar conditions, then use some physics to figure the difference in BTUs/hr radiated.
The Stefan-Boltzmann law describes energy radiation by black bodies:
watts = 5.67x10^-8 * e * area * (Tr^4 -Tc^4)
... where ^ means exponent, area is in square meters, T’s are in temperature Kelvin (Centigrade + 273), Tr and Tc are temperatures of radiator and surroundings, respectively, and e is emissivity (probably pretty close to 1 for a black stovepipe).
So let’s say Tc = 27 degrees C, or 300 K and stovepipe is 100 degrees C hotter. Area of a 6†diameter pipe 6ft high is roughly 1 square meter. So the radiation is about 1 kilowatt which is about 3500 btu/hr. If the stovepipe is 200 degrees C hotter than the room this figure triples (about 10,000 btu/hour). If it is only 50 degrees hotter, it’s only radiating about 1000 btu/hr. It’s pretty non-linear, not surprising with exponents of 4 in the equation. I’ve never measured my stovepipe temps, so I’m not quite sure where on the curve we are operating when talking about typical outer temperatures of single- versus double-wall pipe.
Of course, this ignores convective effects - air being heated by flowing over the hot metal.
Anyone wants to check my math, PLEASE feel free !
It might be interesting, if we tried to get an idea of how much heat is lost with double-wall. I guess we'd need folks with IR thermometers to try to report the outer-surface pipe temps they see with both types of pipe under similar conditions, then use some physics to figure the difference in BTUs/hr radiated.
The Stefan-Boltzmann law describes energy radiation by black bodies:
watts = 5.67x10^-8 * e * area * (Tr^4 -Tc^4)
... where ^ means exponent, area is in square meters, T’s are in temperature Kelvin (Centigrade + 273), Tr and Tc are temperatures of radiator and surroundings, respectively, and e is emissivity (probably pretty close to 1 for a black stovepipe).
So let’s say Tc = 27 degrees C, or 300 K and stovepipe is 100 degrees C hotter. Area of a 6†diameter pipe 6ft high is roughly 1 square meter. So the radiation is about 1 kilowatt which is about 3500 btu/hr. If the stovepipe is 200 degrees C hotter than the room this figure triples (about 10,000 btu/hour). If it is only 50 degrees hotter, it’s only radiating about 1000 btu/hr. It’s pretty non-linear, not surprising with exponents of 4 in the equation. I’ve never measured my stovepipe temps, so I’m not quite sure where on the curve we are operating when talking about typical outer temperatures of single- versus double-wall pipe.
Of course, this ignores convective effects - air being heated by flowing over the hot metal.
Anyone wants to check my math, PLEASE feel free !