Thanks AJ,
I do appreciate the advice.
1- I have been using 2257 to vaporize, but I did not consider the energy to
lift the water to boiling.
I am slowly beginning to understand.
Before I started trying to calculate the efficiency I thought, What is all
the fuss about all these details but now I understand how a little change
can effect the outcome.
2- A few points I am still trying to understand is how to calculate the pot
mass since metal has a different specific heat than water. I am using a 316
stainless pot so I am using .5 as a factor. Is that about right?
3- Also I still need the formula for calculating the loss from woods
moisture content. I an drying a batch of wood as I type. Before the moisture
here in NW Georgia USA has been but we have had a wet season and the last
batch was 14.5%. Wood sure does burn well at 14.5% in my stove.
4- Another confusing point is the difference between net and gross calorific
value.
Lanny
----- Original Message -----
From: <[email protected]>
To: "Discussion of biomass cooking stoves" <[email protected]>
Sent: Friday, October 04, 2013 4:33 PM
Subject: Re: [Stoves] Water heating fuel efficiency formula
[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
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