# Stove temperature vs thermal radiation

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#### Renaissance

##### Member
Hearth Supporter

Where P is energy radiated and T is temperature in Kelvin.

300F = 422K
600F = 588K

588^4/422^4=3.77

If I have this right, a stove at 600F is radiating 3.77 times the heat of one at 300F.

I've noticed my stove doesn't consume that much more fuel at 600F than at 300F. Lately, instead of trying to maintain a long, low overnight burn, I run the stove even hotter (650F) at bedtime when I leave the main living space. While I'm sure the room gets uncomfortably warm as I sleep, it's around 68-70F when I wake up. The stove is under 200F and the glass is just slightly hazy. I find that dumping heat into the room quickly gives me the same result in the morning with less wood and a cleaner stove. Any thoughts?

Yes, I have a thought; I thought if I was that smart, I wouldn't be burning wood.

Sorry man- I'm going through the threads this morning trying to find someone to help and even the ones who don't know, know more than I.
I'm going over to my other forum, The Pre-K Mentor.

I guess it would work in a well insulated house but you lose more heat at a higher temperature so for some it might not work as well.

I've been thinking a little about this myself. I've noticed that when my stovetop (soapstone) reaches upwards of 600*, the rate that the stovetop cools to say 400* is fairly quick - just a couple hours or so, but when the stove reaches ~ 350 or so, the rate of cooling of the stones drops off dramatically (the stove seems to hold at 300 or so for a long, long time). I was thinking that the rate of heat loss from the stone must be proportional to the temperature of the stone itself. The higher the stone temp, the faster the heat loss from stone to room. I spent some time looking into this, and I think that Newton's law of cooling might be a near, or at least partial explanation. Seems that the rate of heat exchange is directly proportional to the temperature of the object and ambient temperature. Assuming my house is at a constant 70* (not true, but let's assume this), a 600* stovetop has a much higher rate of heat loss to the room than the 300* stovetop, which is why it holds the 300* temp for so much longer. At least this is how I'm reading the info. At some point I think I'll plot the change in stovetop T over time and come see what the data are. I think it will show a steep rate of heat loss during the initial cooling phase of the stones, and then decrease in rate (a more gentle slope) as the stovetop T gets lower and lower. Cheers!

NH_Wood, you're right about that. I've noticed it too, even with hot beverages or engines. That holds true for the convection/conduction side of things, and it also backs up what oldspark said about heat loss through walls.

The formula in my first post only applies to radiant heat. The fourth power seems to make a dramatic difference in energy radiated vs. temperature.

The other piece of the puzzle is stove temperature vs fuel consumption.

I agree somewhat.. I also try to burn at a higher temp.. for the volume and layout I am trying to heat low burns don't really work ... during the day they are ok but if I don't get that puppy up to 700 my the far parts of the house would be to cold at same up for the lil one..
But I fill her up with rounds at night and slide skinny pieces of pine in between.. eventually it gets to 700!

I've noticed when it is cold I tend to burn the stove hot. When it is not so cold we don't burn that hot of a stove. This is a very simple formula that has worked well for us for a long, long time.

Your formula is only valid in a vacuum. In air, and at typical stove temps, only about 60-65% of the heat leaving a radiating surface actually leaves as radiant energy. The rest enters the room through convection transfer. You can feel this in many stoves (most notably steel or cast iron ones) when you close the bypass. That blast of heat you feel is convection at play. As the heat is further contained inside the firebox, it quickly gets conducted through the metal and to the surface, where is heats up the air surrounding the stove. Even without fans, the air will initially have enough of a cooling effect on the stove that surface temps will be stable for a short period before they begin to rise again.

The table below shows how much, and in what proportions, heat energy leaves a radiant heater. This is different for convection stoves, where the percentage of convective transfer is obviously much higher.

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• Stove Energy Transfer Table.jpg
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Although the radiative energy output of a stove at 600F should indeed be almost 4 times larger than the radiative emission at 300F, it should be pointed out that even at 600 F the radiative energy output is likely to remain many times smaller than the convective output.

Reliable quantitative estimates are hard to make because of the many size/shape/construction/design factors influencing both types of thermal energy output. However, for a large box stove without forced external convection or big windows and producing 100,000 btu/hr while having 4-5 square feet of black outer mantle and/or flue pipe reaching 600F I believe that the total radiative contribution is likely to remain under 10,000 btu/hr; i.e. 10%.

Considering the many potentially negative side effects of running large stoves that hot that long, it seems to me that there must be easier strategies to achieve the desired result.

Hmm, didn't see Battenkiller's table before I posted. Will have to do some more calculation to see where the values in the table and my own estimate part ways.

PyMS said:
Hmm, didn't see Battenkiller's table before I posted. Will have to do some more calculation to see where the values in the table and my own estimate part ways.

Ah... that ain't my table. I plagiarized it from some third-rate university physicist, whose wacky notions regarding wood heat have been thoroughly discredited on this forum. Feel free to improve on it.

Thanks for that clarification, Battenkiller

I did some checking on my math and I concluded that a 4-5 square feet area of blackened steel surface heated to a surface temperature of 600F at normal ambient temperature will produce no more than 6,000 - 7,500 btus worth of thermal radiation per hour.

Then I guesstimated as follows: to get 4-5 square feet of stove wall at a temperature of 600 F requires a pretty large stove, let's say with a total surface area around 10 square feet and a fire box of up to 2 cubic feet volume. Fill that 2/3 with wood that has a heating vallue of approx. 20 million btu per cord and you should be able to produce around 200,000 btu from that burn. Morever, with the stove at such a high temperature (600 F on much of the outside surface) that burn is probably not going to last more than 2-3 hours, or so, thus producing on average some 66 - 100 kbtu per hour.

How much of that is going out of the chimney depends very much on the type of stove used, of course. However, the point I am trying to make is that of that pretty massive total energy output, less than 10% is emitted into the room as radiative energy and can thus be used to achieve specific home heating goals.

Perhaps some of my assumptions were unrealistic. If so, I hope that the forum members -- many of whom have tons more practical experience with wood heating stoves than I do --will be willing to help educate me.

A black steel/iron stove at X degrees should produce the same balance of radiative/convective heat regardless of the heat source. The wood and flue add way too many variables.

I fully agree with that statement. There are far too many variables to try and use a mechanistic model. Thus, my empirical "black box" aproach where I make the (reasonable?) assumption that what goes into the box (energy-wise) also has to come back out, one way or the other.

What's going in is an amount of fuel with a heat of combustion value that can be estimated (I just used wood because we are talking about wood stoves). What comes out are various forms of matter and energy. Assuming negligible amounts of char and creosote plus no combustion heat left in the ash, the energy must have come out in the form of thermal convection, radiation and conduction. Since conduction losses in a well designed combustion system can often be neglected, we can probably write I = C + R, where I is the enrgy input, C is the convective energy produced and R is the radiative energy (at least as a first order estimate).

So, unless I made a bad mistake, all I need to do now is to estimate either the radiation energy or the convection energy produced and subtract that amount from the estimated total energy input in order to "know" all three. Since radiative energy yields (based on temperature, emissivity and surface area) are much easier to estimate than convective energy yields, I decided to estimate convection energy by difference; i.e. C = I - R.

As far as I can see, the main limitation of this approach is the need to accommodate different stove sizesl let alone shapes/designs, since the amount of fuel input will always be largely a function of stove volume whereas surface radiation will be primarily a function of surface area. In other words, for smaller but similarly shaped stoves the fuel energy input will fall faster than the radiation loss. As a consequence, radiation/convection yield ratios will tend to be higher for small stoves.

So, if your firebox is a lot smaller than the 2 cubic feet used in my posted example, you may well be getting more than 10% thermal radiation yields (provided your smaller stove can still reach and maintain 600F long enough.........)

There is at least one other important factor to consider in all this...

It is true that the rate of energy transfer across a barrier is directly proportional to the temperature difference across that barrier. As others have noticed with beverages and stoves, a hot object cools the most at first, then slower and slower the cooler it gets.

The additional factor is this applies to your house as well as your stove--the hotter it is the more energy you lose. All other things being equal, this argues for constant house temps, as heating it hotter than average results in faster heat loss, that will not be fully compensated by slower loss at below average temps. In other words, a house at a steady temp will lose less heat than one with the same average temperature, but with temperature swings.

This argues for a constant temperature stove--rock, or thermostatically regulated washing machine.

This is just one consideration, and I don't know what is most efficient overall.

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