[Stoves] Heat Content of Ethanol Gel

[Tom Miles]

See http://www.bioenergylists.org/en/crispingelfuel

Tom


-----Original Message-----
From: Crispin Pemberton-Pigott
Sent: Sunday, August 26, 2007 6:13 PM
To: 'Discussion of biomass cooking stoves'
Subject: [Stoves] Heat Content of Ethanol Gel

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

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