Yield wood stove and flue gas temperature

[Philippe Schutt]

On a pellet stove, in the heart of the vein I measure for example 140 ° while the stove indicates 108 ° (measured on the surface by a thermistor)
Another info that corroborates this difference is that I can hold a pipe carrying smoke at 85 ° measured in the flow.

at nominal power, some stoves announce an efficiency of 85% and a T ° of smoke of 300 °. So it is the quality of combustion that must be favored over the temperature of the fumes.

[dedeleco]

some stoves announce an efficiency of 85% and a T ° of smoke of 300 °. So it is the quality of combustion that must be favored over the temperature of the fumes.

85% cannot be obtained without good combustion quality, by not losing part of the wood in ash and smoke, safe, but with smoke at 300 ° C, and this 85% yield for stoves requires a combustion with little excess oxygen giving a lot of CO2 in proportion: 12% for dry wood !!
So if we introduce only a little air, just the mini and we have a stove using this little air to the maximum without sending unused in the chimney, we can have 85% efficiency with more fumes 300 ° C.
But I doubt that the combustion is so good without a lot of excess air under normal conditions because the air is sucked in by the chimney which draws and therefore cannot pass entirely over all the portions which burn to be used there. .
Rigorous method of performance measurement
see the curve with the coefficient f = 0,65 for wood, replacing that of 0,43 or 0,37 for gas! :
efficiency% == 100 - fx (Tfumes - Tamb) /% CO2

https://www.econologie.com/forums/post156304.html#156304
http://www.ecoconso.be/forum/showthread.php?t=947
in particular this very complete site giving the values
http://www.nano-sense.com/articles/anal ... alculs.htm
http://www.nano-sense.com/articles/anal ... .htm#chap1
http://www.nano-sense.com/articles/anal ... ements.htm

In practice, the combustion efficiency is often expressed by the Siegert formula:
Combination efficiency = 100 - fx (Tfumes - Tamb) /% CO2
or :
Tfumes = the temperature of the fumes leaving the boiler [° C]
Tamb = boiler room temperature [° C]
% CO2 = the CO2 content of the smoke [%]
f = factor mainly depending on the type of fuel (fuel oil: f = .. 0,57 ..; natural gas: f = .. 0,47 dry wood f = 0,65 ..)

Dry wood is in the same range of values ​​!!

In fact, the benchmark is the flame temperature for zero efficiency, complete combustion and optimal oxygen stoichiometry, the maximum possible CO2 12% for gas and 19,4% for wood (the rest compared to 21 % oxygen is hydrogen which also burns)
or Tmax flame = 100xCO2max / f which we find around 2800 for wood and 3000 ° C for gas.
This temperature seems very high for air with 79,1% nitrogen ???
But I doubt that the stoves can completely burn the wood without excess air, with a part passing away from the wood in turbulence, because the arrangement of the wood is not always the same tidy to see all the air passing enters against the wood which burns with all the oxygen without having other portions with the turbulence which lack oxygen.

Finally, he is important to measure the proportion of CO2 to assess the efficiency, because if there is an excess of air twice the necessary, we fall to 9% of CO2 for the wood and the temperature of the fumes drops to half 180 to 200 ° C!
We can have triple the air 6% CO and 150 ° C !!

A method without measuring CO2, if possible, is to reduce the air intake gradually and measure the T max reached with beautiful flames so CO2 must be close to CO2max. Below this oxygen optimum, combustion becomes incomplete with CO and has no meaning.

Cheap course method of measuring CO2 in flue gases:
http://sciencesphysiques.ifrance.com/mo ... le3_1.html
bridge 58-5

During the lime water test the reaction that occurs corresponds to the following unbalanced equation
"Ca (OH) 2 + CO2 CaCO3 + H2O".
A technician bubbled a quantity of air with CO2 in lime water. The lime water becomes cloudy because chalk (CaCO3) insoluble in water has formed. She decides to filter the solution to collect the chalk. After drying the chalk, she weighs it and notes that she has collected 125 g of chalk. How much CO2 had reacted with the lime water?

[bernardd]

But I doubt that the stoves will burn the wood completely without excess air, with a part passing away from the wood in the turbulence,

This is one of the very basic reasons that a pellet stove easily has better combustion.

[chatelot16]

But I doubt that the combustion is so good without a lot of excess air under normal conditions because the air is sucked in by the chimney which draws and therefore cannot pass entirely over all the portions which burn to be used there. .

that's the problem the conditions are not the same throughout the fire: if there is too much air it causes more heat in the smoke: it makes a small loss, calculable by the smoke temperature by measuring CO2

where there is not enough air the combustion is incomplete ca sending of CO or other imbrule in the smoke, it makes a huge loss of energy, and not measurable by the temperature of smoke and CO2

so always have a good amount of air to be sure you never have a shortage

it is only with the gas which mixes very well with the air that one can aim for the exact proportion of air (stoichiometric)

the lack of air not only decreases the yield worse than the excess, it may also cause pollution

it would be interesting to have a permanent indicator of the quality of the fire, by analysis of the smoke, but the CO and the unburnt ones are more difficult to measure than CO2: we can therefore be content to measure the CO2 rate: when the CO2 rate approaching the theoretical rate, it is that the lack of air is close and that it is already unburned: it is prudent to increase the air for a lower CO2 rate

the preferable level of CO2 should be indicated in the documentation of a stove, since the excess air required depends on the design of the stove