[Default] On Thu, 3 Oct 2013 07:14:05 -0600,"Ronal W. Larson" <[email protected]> wrote:
>Andrew: cc list and Lanny > > This is the first time I can recall seeing the number 2.3 MJ/kilo. This > must be associated with some initial moisture? It's associated with any water that leaves the stove as vapour, it's the amount of energy that was necessary to turn the liquid water in the wood chemistry or water associated as free water in the wood into a vapour. In fact this enthalpy of vaporisation varies with temperature but it's around 2260kJ/kg. plus of course you have to add the necessary amount of heat to lift it from local boiling point to the exhaust temperature, I use 2.7MJ (.75kW) as a rule of thumb. In fact one should probably express it as the heat necessary to vaporise a mole of water and relate that back to the heat energy in a mole of wood. If we assume wood chemistry to be in the ratio CH1.4O0.6 and we use whole numbers C5H703 then we have a mole of wood with a weight of 2*(12*5+1*7+16*3)=230grams which when completely oxidised yields 10CO2 and 7H2O. This 7H20 weighs 7*18=126grams and contains about 0.3MJ of heat in the vapour. As I proposed the original calorific value of red oak would be about 18.6 MJ/kg as a generally accepted figure but this is normally the lower heating value after this latent heat has been accounted for. so in fact the 230grams of wood burned probably yielded 4.3MJ of heat which was available to the pot plus the 0.3MJ of latent heat of vaporisation. Hence the LowerHeatingValue of the sample is the HigherHeatingValue of the sample minus the latent heat of vaporisation of the water released in combustion. Now of course we know that we never get to burn wood that is completely dry of free water. So if your sample is burned "Denver dry" with about 10% of the wet weight being water then our initial sample will now weigh 255.5grams with the same calorific value but having to now vaporise 126+25.5 grams of water and lose 0.35 off the HHV of the wood. > > I think Lanny is looking for ways to do testing with wood of different > moisture content. Doesn't he have to do some wood drying in an oven, > measuring weight loss? Yes the generally accepted way is to heat the wood to around 110C-120C (any higher and you start losing the mass of volatile compounds, any less and you don't boil off the water that is loosely bound to the cell walls) for long enough that there is no further weight loss. Subtract this final weight from the initial wet weight and that gives the original moisture content, divide this water weight by the initial wet weigh and you have the moisture content on a wet weight basis. AJH _______________________________________________ Stoves mailing list to Send a Message to the list, use the email address [email protected] to UNSUBSCRIBE or Change your List Settings use the web page http://lists.bioenergylists.org/mailman/listinfo/stoves_lists.bioenergylists.org for more Biomass Cooking Stoves, News and Information see our web site: http://stoves.bioenergylists.org/
