Heat Content of Ethanol Gel

Crispin Pemberton-Pigott, August 26, 2007

Dear Friends

I have been working on a formula to calculate the heat content (LHV) of ethanol gel. I would appreciate any comments on the text below drafted for the SABS Gel Fuel stoves Technical Committee.

Regards

Crispin

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Anyone who has tried the formula in the Draft SAND666EDI CD1 (2) PDF paragraph 6.4.7 will have seen that it is far from giving a true power rating of an ethanol gel stove.

There are large differences in the claimed heat content of ethanol. The Lower Heating Value (LHV) is the heat which can be obtained without condensing the combustion products and most closely represents a pot on a stove.

The Biomass Information Network http://cta.ornl.gov/bedb/pdf/Appendicies.pdf

reports the Ethanol energy content = 26.7 GJ/t = 21.1 MJ/litre The site http://www.members.aol.com/optjournal/ep3.doc says it is 26.9 MJ per Kg which is 21.22 MJ/Litre which is in general agreement.

The draft proposal uses the following variables:

PO = rated Power Output

Initial mass of stove before testing starts = Mi (i.e. 2145.9

gm)

Final mass of stove after 30 minutes at full power = Mf (i.e. 2070.0 gm)

PO = (Mi-Mf)*16/18,000 = Kw

It is out by a factor of more than 15 and does not take the water content into consideration.

Making the obvious correction:

PO = (Mi-Mf)*16/1,157 to get the correct answer for 5% water, gives an answer that is still out by 6% with 0 and 10% water content.

Using a divisor of 1092 which gives the correct answer for 0% water only creates other problems.

Here is proposed a formula that takes into account the water content of ethanol gel fuels (typically 5% or more) and other additives that displace ethanol.

Please try this and see if you get a realistic calculated power level:

W = water content (i.e. 5% = 0.05)

A = additives that are not water or ethanol (i.e. 2% = 0.02)

E = Ethanol content 1-(W+A) (i.e. 1-(0.05+0.02) = 0.93)

Proposed Formula:

PO = (Mi-Mf)/[(67/E)+(W*6)] = KW

Worked Example:

PO = (2145.9-2070.0)/[(67/0.93)+(0.05*6)]

PO = 75.9/[72.04+0.30]

PO = 75.9/72.34 = 1.05 Kw