The efficiency of a valid wood stove at full power

[Philippe Schutt]

when we look at how performance is measured we come up against 2 problems

1) non-public standards, you would have to pay to have them: this principle is absurd! manufacturers boast that their device meets this or that standard but the customer cannot know what it contains

Yes, and this in all fields / professions! You have to respect the standards, but you have to pay to know them. Insofar as "no one is supposed to ignore the law", I wonder if the state does not have the obligation to make it freely available to all ??? If someone, having a copy of a standard, DTU, technical opinion etc ..., distributes such a document, he falls under the ax of copyright laws. Apparently they are well applied, I have not found them on the net.

That said, the standard sets manufacturing rules such as the minimum thicknesses of the hearth sheets, defines the measurement methods and the minimum performance to be achieved. For example, the EN14785 standard sets the quality of the pellets to be used, and for this requires a humidity of <12%.
So nothing prevents the manufacturer from providing pellets at 3% humidity, which will necessarily improve the result considerably!
I have a smoke analyzer which calculates the yield, it has a "wood" position. I have never succeeded in having the outputs announced by the manufacturers, at best I arrive a little above 80%, around 50% of the maximum power.

[Obamot]

Ok for standards, it's clear.

I have a smoke analyzer which calculates the yield, it has a "wood" position. I never managed to have the yields announced by the manufacturers, at best I get a little above 80%, around 50% of the maximum power.

Note: it seems to me that I understood that your measurements / controls would have demonstrated that the theoretical values ​​given by the manufacturers would be regularly higher than the actual yields obtained (subordinate relative to the quality of the burnt pellets, such as their rate of hygrometry).

But it's me where I didn't understand the end of your sentence?

If you got a little above 80% then declare that this efficiency would be 50% of the maximum power. Read like that, it could mean that the maximum power would then be 160%?

[dirk pitt]

If you got a little above 80% then declare that this efficiency would be 50% of the maximum power. Read like that, it could mean that the maximum power would then be 160%?

efficiency and power are two different quantities even if in this sentence both expressed in%.
the efficiency can be from 80% to 50% of the maximum power then be from 85% to 100% of the maximum power (or any other value)

[Obamot]

I was also thinking, thank you Dirk. As soon as we speak of calorific power which can vary, that the hygrometry rate plays its role there as well as the ambient temperature and the altitude, to express a power in "%" whereas until now we spoke in watts, didn't speak to me more than that. As in a stove, what part represents the unburnt compared to a mass stove, where do we say, they are better recovered so almost all participate in the combustion?

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The efficiency of a wood stove is only valid at full power

A cooker has a power of 2kw.
To heat a main room, 500w is enough

at 500W, 'there is 4 times less gas to produce the draft in the same volume.
which means that the print is divided by 4,
and yield by 2,

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We go from a yield of 70% to 35% at worst and
85% to 43% at best

We consume twice as much wood,
and the half that has not burned properly turns into bistre
- with risk of fire and pollution
and risk of poisoning by gas backflow,
the fire being too weak!

An outrageous mess!

From single to double, as Hic suggests, it's not nothing, so express such variables in "%"

[Philippe Schutt]

Expressing the power in% of the maximum power of the stove is justified because this maximum power is different from one stove to another.
More power means a larger burner, and therefore a greater minimum power too. As a result, it is always + - at half the maximum power that the stove is at its optimum.
By using very dry granules, the fumes are warmer at the inlet of the exchanger but less hot at the outlet. all good!

[Christophe]

From single to double, as Hic suggests, it's not nothing, so express such variables in "%"

The sentence was completely understandable ... I think ...

[Philippe Schutt]

It could be interesting to quantify the effect of the humidity of the granules on the T ° of the smoke entering and leaving the exchanger and therefore on the yield. Determine the decrease in the T ° of the fumes (of combustion?) Due to their greater caloric mass, then their least cooling in the exchanger always for the same cause. This would make it possible to estimate the lack of heat return and consequently the drop in yield per% more humidity.
The standard is very precise on the modalities of the tests, I hardly see any other way to improve the results than the fuel. The difference between the displayed performances and my measurements being of the order of 10%, I would be interested to know if the fuel can influence the results as much.