[Stoves] Fuel Testing

[AJH]

On Tue, 03 Oct 2006 23:33:36 -0500, Paul S. Anderson wrote:

>2.  If a kilo of wood has 10% moisture, then 0.27 MJ must be subtracted.
>0.27/18.6 = 0.0145 = 1.45 %.  Seems like a mighty small amount.  Did I miss
>something?

It's almost insignificant at these low moisture contents but you did
miss a bit. As the water forms 10% of the mass (given our mc was
calculated on a wet weight basis) the calorific value of the wood is
only 18.6 times the remainder after the water is removed so:

(18.6*.9)-(2.7*.1)=16.47 and the percentage of the wood needed to boil
off and discard the water is 0.27/16.47=1.61% of the net heat
available.

Of course my rule of thumb figures are a compromise and in most open
vented stoves the loss will be greater than 2.7MJ/Kg of unnecessary
water in the wood, bearing in mind in UK it's highly unlikely the wood
will dry much more than 20% mc (wwb).

The calculation becomes a bit complicated because the specific heat of
steam changes as its temperature rises  but appears to be about half
that of water. The calculation should be comprised of the rise in
temperature from ambient (say 10C) to boiling times 4.1kJ/kg plus the
latent heat of vapourisation at atmospheric pressure and 100C
*2258kJ/kg plus the specific heat of steam times the difference
between 100C and the rejection temperature (say 150C but like as not
far higher with most stoves)

(4.1*90)+(2258)+(2*100)=2727kJ so my 2.7MJ errs on the low side for
most cases as it assumes the steam is lost at 150C.

In the greater scheme of things this is no more than an interesting
calculation and small variations are lost in the noise  but the simple
calculation to indicate how much water in the wood displaces wood
available for energy is very significant for wood bought on weight (or
carried to the stove).

AJH