[Stoves] Water heating fuel efficiency formula

Lanny Henson lannych at bellsouth.net
Fri Oct 4 16:14:31 MDT 2013

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 

----- Original Message ----- 
From: <ajheggie at gmail.com>
To: "Discussion of biomass cooking stoves" <stoves at lists.bioenergylists.org>
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"
> <rongretlarson at comcast.net> 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.
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