Emissivity (basic facts about heat radiation)

[precaud]

It's nice to be reminded from time to time of some of the basic facts that we are dealing with. This chart from The Woodburner’s Encyclopedia (1976) shows radiated heat per hour per square foot as a function of surface temperature. It explains to me again why I was so fond of the little Jotul 602 for so many years, which I normally ran at surface temps in the 800º-900º range...
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The amount of heat emitted per square foot is dependent on the temperature of the radiating body. Because surface temperatures are lower, massive heat exchangers need a lot of surface area to radiate useable heat into a room.

TEMPERATURE.....TOTAL ENERGY
......OF................TRANSFERRED
SURFACE (º F).....BTU / Hour - ft 2
----------------- ... ------------------
80º ......................... 15
100º.........................51
150º.......................168
200º.......................315
400º.....................1230
600º.....................2850
800º.....................5430
1200º...................9370
---------------------------------------
(sorry, it's hard to format columns here)

So... if you operate your stove in the 600º range and want to double your heat output, you can either a) run it at 800º or b) buy a stove with twice the radiating surface area.

 

[precaud]

In a roundabout way, I think it also gives support to the notion of buying a stove based on firebox dimensions rather than manufacturer's specs.

 

[EatenByLimestone]

On this same topic, is it possible to shape which direction the heat radiates? Is this what firebricks do? If you can reflect heat out the front, instead of letting it go out the back you should be able to make a small stove appear much larger.

 

[precaud]

On this same topic, is it possible to shape which direction the heat radiates?

One way is by geometry, yes? A shallower stove will radiate more heat forward.

 

[Rhone]

You can direct the way heat radiates, radiant energy is a form of light like sunlight. Think how much better it is in the sun when you wear white clothing vs. black. So, put a white or silver shield where you don't want it. After re-reading the post the above doesn't really apply. Firebricks insulate so the firebox stays hotter than the outside metal. Air turbulance makes a HUGE difference on the amount of heat transfer, think how quickly blowing on a spoonful of hot soup cools down vs. not blowing on it. Same with a stove, having air blowing across it transfers a lot more energy than not.

 

[Corey]

Good info, although I'm curious here...the text says 'radiated' energy, but the column heading is 'total energy'? As I understand it, there are three main modes of heat transfer...radiation, convection and conduction. So is this all inclusive or just based on pure radiation?

Corey

 

[Roospike]

Good information on heat transfer , its kinda misleading for wood stoves tho .

I'd like to add this information ,
Wile it may look to be the best option to have a 2/16 thick steel stove (for example) with a 4 cf fire box for the most heat and best heat transfer it just dont work that way for wood stoves and best options. There is a line from how thick a stove could be to should be and what works best for its performance.

With XXX amount of BTU a load of wood puts off through 2/16" steel plate is going to be much higher per sf per hour than 1/2" steel per sf per hour.

Take the 400° surface temperature for example , its going to take a lot more heat to make 1/2" steel 400° @ 1 sf then it would take the 2/16" to be 400° @ 1 sf . More heat transfer per 1 hour on the 2/16" steel @ 1 sf ? Yes. More heat transfer per 10 hours on the 2/16" steel over 1/2" ? NO.

Is more the heat transfer per hour better on thinner steel ? yes. is it better/best to have on a wood stove? No , not necessarily , not for long even heat.

 

[wg_bent]

Good information on heat transfer , its kinda misleading for wood stoves tho . Wile it may look to be the best option to have a 2/16 thick steel stove with a 4 cf fire box for the most heat and best heat transfer it just dont work that way for wood stoves and best options. There is a line from how thick a stove could be to should be and what works best for its performance.

I agree with you that there is an optimal, but that's also material dependent and heat storage dependent. Also, in a secondary effect, what other technologies are used in the stove such as CAT or secondary combustion or Masonry mass.

According to evidence here, a massive Tulikivi is a very good heater, but actually has very poor thermal transfer properties, but a very large surface area to radiate. I'd like to go on and on, but need to call someone on the otherside of the pond. Work does get in the way of my hearth posts.

 

[wahoowad]

precaud,

How does this apply to the "glass" on my stove? I am thinking heat is radiating thru it, and not necessarily off of it. I defiantely feel a lot more radiant heat a foot away from the "glass" than a foot away from the side.

 

[Andre B.]

precaud,

How does this apply to the "glass" on my stove? I am thinking heat is radiating thru it, and not necessarily off of it. I defiantely feel a lot more radiant heat a foot away from the "glass" than a foot away from the side.

Yep

Like on that Australian cook stove in a recent thread. Used the same door for the firebox and the oven, but I bet the glass in to oven door was different then that in the fire door. Not just for cost but the glass in the oven door is likely a lot less transparent in the inferred range, maybe even a double pane.

 

[Rhone]

precaud,

How does this apply to the "glass" on my stove? I am thinking heat is radiating thru it, and not necessarily off of it. I defiantely feel a lot more radiant heat a foot away from the "glass" than a foot away from the side.

Radiated energy's power is based on the temperature, size, and shape. For example, if your stove is biggest in the front and smaller on the sides, the front is going to have more power than the sides, if your sides are bigger, they'll have more power (if it's a unit without firebrick). As an experiment to show how radiant energy comes out from flat surfaces, approach your stove diagonally. Since there's almost no flat surfaces facing you, you should feel very little radiant energy and can get really close (and why stoves can be placed diagonally close to walls and have tighter clearances than square to a wall). Then, stand directly in the middle of the biggest flat side, same distance and see how long you can take it. Because radiant energy comes out most from flat surfaces, round/barrel stoves are better designs because it spreads the heat out more evenly around the room instead of creating hot spots. With flat surfaces though, you can strategically place people. For example, I know the seat directly in front of my glass is the hottest seat in my house and the wife who likes heat... that's her spot. I don't like it, so I sit diagonally to the front.

So, my guess is the glass in front is bigger than the sides of your stove, and your stove sides probably have firebrick insulating and protecting them from reaching the same temps as the glass in front, so your glass is emitting more energy. Also, radiant energy is a form of light, and follows the inverse square law which means every time you double your distance from the source, the intensity is reduced by 1/4th. It's still the same amount of energy in the end but it's being spread over 4x more area everytime you double your distance so you notice a pretty quick drop off, or gain depending on which direction from the source you're going in.

 

[Corey]

Good info, although I'm curious here...the text says 'radiated' energy, but the column heading is 'total energy'? As I understand it, there are three main modes of heat transfer...radiation, convection and conduction. So is this all inclusive or just based on pure radiation?

Unless thee is planning to warm thyself by touching said heated surface, radiation pretty much covers it.

I don't plan to warm myself by touching the stove, but I do need to warm the rest of the house besides the room the stove is in. For that I have to rely on convection and conduction to the surrounding air.

Corey

 

[benignvanilla]

So would it make sense then, that you can be more efficient in deriving heat from your pellet stove by running it at a higher temp for less time, instead of running it on low for a long time?

 

[Corey]

Good information on heat transfer , its kinda misleading for wood stoves tho .

I'd like to add this information ,
Wile it may look to be the best option to have a 2/16 thick steel stove (for example) with a 4 cf fire box for the most heat and best heat transfer it just dont work that way for wood stoves and best options. There is a line from how thick a stove could be to should be and what works best for its performance.

With XXX amount of BTU a load of wood puts off through 2/16" steel plate is going to be much higher per sf per hour than 1/2" steel per sf per hour.

Take the 400° surface temperature for example , its going to take a lot more heat to make 1/2" steel 400° @ 1 sf then it would take the 2/16" to be 400° @ 1 sf . More heat transfer per 1 hour on the 2/16" steel @ 1 sf ? Yes. More heat transfer per 10 hours on the 2/16" steel over 1/2" ? NO.

Is more the heat transfer per hour better on thinner steel ? yes. is it better/best to have on a wood stove? No , not necessarily , not for long even heat.

I think I see what you are trying to say...for short fires, you might prefer the thinner steel? Although, overall, btu into the steel has to equal btu out. A thick stove may take longer to heat up, but will stay warm longer after the fire goes out. Under steady state burning, the thickness of the steek should be of little relivance.

Corey

 

[precaud]

Good info, although I'm curious here...the text says 'radiated' energy, but the column heading is 'total energy'? As I understand it, there are three main modes of heat transfer...radiation, convection and conduction. So is this all inclusive or just based on pure radiation?

Corey

I don't think it makes any difference. Any energy taken by the other two methods will reduce the amount radiated.

 

[begreen]

Great thread. Thanks for the post Precaud. It also made me think a lot about the heat radiated out of our house, especially from the windows on a 24 degree morning.

 

[precaud]

Good information on heat transfer , its kinda misleading for wood stoves tho .

Why? The emissivity of steel (or almost any material) is the same, regardless of thickness. You're adding new elements to the equation; energy input, and mass. But this isn't about efficiency, or smooth btu output curves.

The point is, the radiating surface area and maximum acceptable surface temps will determine the maximum btus per hour you can ever hope to get from any given stove, regardless of design or materials. For instance, from a stove with 8 sq ft of area, you can't hope to get more than 14,250btu's per hour out of it IF you want to operate it at 600° max. Or... if a mfr claims 42,000 btu/hr heat output from a stove and you know it's dimensions, you can get a ballpark idea of how hot they were cranking that thing to get that rate of output from it!

I don't disagree with what you're saying. But the fundamental principles still apply.

 

[precaud]

It also made me think a lot about the heat radiated out of our house, especially from the windows on a 24 degree morning.

Yeah, that, unfortunately, is the other side of it... The heat will always flow from hot to cold... and big cold surface areas can suck alot of it!

 

[Andre B.]

Good information on heat transfer , its kinda misleading for wood stoves tho .

Why? The emissivity of steel (or almost any material) is the same, regardless of thickness. You're adding new elements to the equation; energy input, and mass. But this isn't about efficiency, or smooth btu output curves.

The point is, the radiating surface area and maximum acceptable surface temps will determine the maximum btus per hour you can ever hope to get from any given stove, regardless of design or materials. For instance, from a stove with 8 sq ft of area, you can't hope to get more than 14,250btu's per hour out of it IF you want to operate it at 600° max. Or... if a mfr claims 42,000 btu/hr heat output from a stove and you know it's dimensions, you can get a ballpark idea of how hot they were cranking that thing to get that rate of output from it!

I don't disagree with what you're saying. But the fundamental principles still apply.

But my stove has some extra tin around the back, sides, and the step on the rear of the top, that makes a very effective convection driven heat exchanger. Add the optional fan to force the convection and blow the radiation heat flow values right out of the water.

 

[precaud]

Here's an example. The outer dimensions of the Quad 2100 (not including the convection jacket) are roughly 20x19x16.5 . I think it's fair to throw the bottom out of the total area as a reasonable fudge factor. So it then has 11.57 sq ft of area. If the entire surface area was at 600º you would get 32,990 btu/hr from it.

By contrast, the non-rectangular little Jotul F602 is roughly 19x15x12, for approx. 8.04 sq ft. So 600º operation would give 22,920 btu/hr. The EPA tests gave an upper range of 42,000 btu/hr for this little thing, which means that the average surface temp of that thing was cranking at just shy of 800º to give that amount of output!

 

[precaud]

But my stove has some extra tin around the back, sides, and the step on the rear of the top, that makes a very effective convection driven heat exchanger. Add the optional fan to force the convection and blow the radiation heat flow values right out of the water.

Yes, but trapping part of the radiation to set up a convection only stops it from being radiated directly into the room and uses it to heat air in a cavity instead. There is no free lunch... or btus... There's no difference in heat output - just a difference in how it's getting into the room. Those things DO add some to the radiated area, as they're part of the stove structure. But they are cooler, hence radiate at lower temps.

 

[Rhone]

Andre B. that's how I understand it as well. With convection shrouds, the proportion of radiant energy reduced is directly proportional to the increase in convection energy for a net difference of 0.

What's always puzzled me is the effect of air blowing over vs. still air. My math may be wrong, maybe someone corrected me previously but I found the formula for heat transfer and found a 450F 1' square piece of metal will transfer 2,640 btu's/hr. Have that same piece maintain 450F with a fan blowing on it, and you increase your btu output to 6,471 btu's/hr.

 

[Andre B.]

Andre B. that's how I understand it as well. With convection shrouds, the proportion of radiant energy reduced is directly proportional to the increase in convection energy for a net difference of 0.

Not if you maintain the surface temp constant. If you hold the power level of the fire constant then yes, you only have X amount of heat and the only thing you can change is the ratio of into the house verses up the stack.

What's always puzzled me is the effect of air blowing over vs. still air. My math may be wrong, maybe someone corrected me previously but I found the formula for heat transfer and found a 450F 1' square piece of metal will transfer 2,640 btu's/hr. Have that same piece maintain 450F with a fan blowing on it, and you increase your btu output to 6,471 btu's/hr.

Why?
If the air is moving then colder air is always coming in closer contact with the surface, with still air the air near the surface heats up and heat flow slows down. Rate of heat flow is driven by temperature difference, be it thru a solid or across a solid fluid boundary. The boundary calculations are a lot more complicated then thru the solid but the concept is not all that difficult.

 

[Andre B.]

that unless you wanna maintain (And I will agree.) that a micro-amount of air is actually in CONTACT with the heated surface and thus acquires its heat via CONDUCTION (and then CONVECTS away), the VAST amount of heat is being RADIATED from said surface prior to its energizing of air, etc.....even those air 'molecules' that are but nano-meters away from said surface.

Yes in the absence of convection air's ability to conduct heat is not all that good which makes practical insulation possible. But in free air and a gravity field you cannot stop convection so new cold are moves in to be heated by conduction.


Related is water which also is not a very good conductor of heat. I have a 30HP modified scotch marine boiler, similar to this one but different brand.
(broken link removed to http://www.hurstboiler.com/400series.htm)

It is on wheels and was used by the township to thaw out culverts, the thing will burn 9 gallons of fuel oil an hour at full blast.
I removed the outer insulation on the barrel so I could measure its condition, rust and such.

Fill it with water.
Fire it up and burn until you have say 50psi on the gage, takes 1 to 1 1/2 hours (not burning all out).
You can still put your hand on the bottom of outer barrel under the main flue, and it is not much warmer then the water temp you started out with. There is only about 3 inches of water between the main flue and the outer barrel. Up the sides you get to a point where the temperature takes a big jump.
Now crack open the steam valve for a second and the water mixes and you burn yourself if you touch the same place on the bottom.

 

[precaud]

So, you're saying that it's not 'anti-heat' that does the flowing??

lol... you nailed me... and it!