sonnyinbc - 26 August 2008 08:17 PM

I`m getting gas just reading this thread

methane or CO2?= which end?

... definitely peroxide - my chimney is blond.

You didn’t name your chimney Eileen did you?  OH, sorry, that was another thread...nevermind.    Rick

sawdustburners - 25 August 2008 07:52 AM

Nitrogen N2 50.9%,
Carbon monoxide CO 27.0%,
Hydrogen H2 14.0%,
Carbon dioxide CO2 4.5%,
Methane CH4 3.0%,
Oxygen O2 0.6%.
assuming this is 1 log’s gas yield. can you assign approximate energy ratio % to components listed above. the CH4 will combust thru CO,i think, so a further analysis at this point would be great but anything helps...thanx
so far i have
C-CO=27KCAL/MOLE
CO-CO2=68KCAL/MOLE

chimney’s name is Beulah Gertrude but everybody calls her Hot & she complains about the blonde, claims it damages her hairs.

The real question- how much energy is left here since the starting material did not react to completion?

Nitrogen N2 50.9%,
Carbon monoxide CO 27.0%,
Hydrogen H2 14.0%,
Carbon dioxide CO2 4.5%,
Methane CH4 3.0%,
Oxygen O2 0.6%.

So- if we treat % as molar % (could be mass)- then just find the energy of combustion per mole for each starting compound, and multiply by the # moles.

N2 and CO2 don’t play a part.  O2 does, but there is also other O2 added for complete reaction.  To answer our question- we assume that enough O2 is provided for complete combustion- so the relative amount of O2 in this gas is not of consequence.

Methane combusting- CH4 + 2O2 -->CO2 +2H2O is -890.4 kJ/mol (hey- it was an example in the book- no calc reqd)
Carbon monoxide CO + 1/2 O2 --> CO2
= dH(CO2) - dH(CO)= -393.5 - (-110.5) = -283 kJ/mol
Hydrogen --> water (g) = -241.8 kj/mol

Assuming those % are molar- all we have to do is multiply each by the % and add

-890.4(0.03) + -283(0.27) + -241.8(0.14) = -137 kJ.  137 kJ is left unused when one mole of combustion gas is created (including N2, CO2, etc. in Saw-a-duck-burning’s question) as compared to complete combustion.

In kCal- just divide by 4.184 J/Cal

And you can keep the avatar- LOL

Adios Pantalones - 28 August 2008 04:23 PM

The real question- how much energy is left here since the starting material did not react to completion?

Nitrogen N2 50.9%,
Carbon monoxide CO 27.0%,
Hydrogen H2 14.0%,
Carbon dioxide CO2 4.5%,
Methane CH4 3.0%,
Oxygen O2 0.6%.

So, N2 and CO2 don’t play a part.  O2 does, but there is also other O2 added for complete reaction.  To answer our question- we assume that enough O2 is provided for complete combustion- so the relative amount of O2 in this gas is not of consequence.

Methane combusting- CH4 + 2O2 -->CO2 +2H2O is -890.4 kJ/mol (hey- it was an example in the book- no calc reqd)
Carbon monoxide CO + 1/2 O2 --> CO2
= dH(CO2) - dH(CO)= -393.5 - (-110.5) = -283 kJ/mol
Hydrogen --> water (g) = -241.8 kj/mol

Assuming those % are molar- all we have to do is multiply each by the % and add

-890.4(0.03) + -283(0.27) + -241.8(0.14) = -137 kJ.  137 kJ is left unused when one mole of combustion gas is created (including N2, CO2, etc. in Saw-a-duck-burning’s question) as compared to complete combustion.

In kCal- just divide by 4.184 J/Cal

And you can keep the avatar- LOL

NO HABLO INGLES! but ill read it again, thanx

If the question is- “what % of the heat do you lose, compared to complete combustion?"- we cannot calculate that with the information given.  There must be simple calorimetric data out there on the enthalpy per gram of wood from complete combustion.  If we find the enthalpy per mole of gas created from complete combustion, then just diviude -137 kJ by that number and multiply by 100.

the q = the combustible gases listed will each contribute a certain % of the total heat output with adequate O2 present.CH4,H,&CO;.
&i;promise i wont use my new avatar for a plume...duster...god forgive a mop!

Nitrogen N2 50.9%,
Carbon monoxide CO 27.0%,
Hydrogen H2 14.0%,
Carbon dioxide CO2 4.5%,
Methane CH4 3.0%,
Oxygen O2 0.6%.
above figures are gas/volume so ?
CH4= 20%
CO=55%
H=25%
approximate heat yield % of 3 gases if woodgas is fully combusted with adequateO2?

Berraco - 26 August 2008 01:21 PM

Adios Pantalones - 26 August 2008 08:22 AM

Oh man- well I have the tables of enthalpy in a book here, but I have work to do today.  Remember that N2 and CO2 don’t contribute as they don’t combust- so they’re just diluents.

definitely a headbuster for me!
i’m into this beacuse of the SEDORE stove which is a simple downdraft stove which has no secondary air provision unlike others i have seen.if not enough O2 is available for full combustion of CO the stove loses fuel up the chimney & is that much inefficient in spite of the fact that the smoke may be clear.GOOGLE patents has sedore
http://www.google.com/patents?id=-lcxAAAAEBAJ&dq=sedore

If the Sedore is being compared to the above patent, then the information is incorrect.  Please make an effort to see one work before making claims based on a 40 year old patent, and a website that neither sells or manufacturers the Sedores.  Special blogs, voicing one persons undocumented opinions?  If I had the time and energy to make undocumented claims, about a product I’d never seen, not to mention a special blog, for one uninformed opinion, then maybe I could do testing on my product that every citizen in the US asked me for??  Maybe one should be seen in action, so an informed opinion can be made and you could test the CO yourself?  My Sedores, also have a secondary burn, but I choose not to mention the specifics.  Could documentation be provided of other stove manufacturers making special arrangements’ to test for CO, or anything else, upon you’re request?  There are Sedores in Maine and maybe arrangements’ can be made to see one in action?

I have a recollection that CO combusts at about 1200F and H at about 1400F. I suspect the output of each is CO2 and H20.  I have no knowledge of the energy released in these chemical recombinations.

I don’t know, but doubt, that ordinary secondary combustion in a wood stove reaches these combustion temperatures in any sustained manner. I believe the gasification boilers not only reach but are designed to sustain these combustion temperatures.

Anecdotally, during the second half of the burn in my Tarm there is sustained “roaring” combustion through the nozzle and in the tunnel with little or no visible flame. I understand H has no visible flame when it combusts. I don’t know about CO.