Richtee Posted October 25, 2010 Share Posted October 25, 2010 OK all you chem guys... to a first approximation, how much air does a pound of charcoal need for combustion? How about a CFM number if that can be approximated. Yes, I know charcoals ain't the same, but assume they are. Link to comment Share on other sites More sharing options...
Mumbles Posted October 25, 2010 Share Posted October 25, 2010 There are a few assumptions that we're going to have to make to get a relative value. The thing to remember about charcoal is that it isn't pure carbon. I've often seen approximations around 70-90% or so, so lets call it 80% with the rest being ash, volatiles, and incompletely charred wood. That makes 1lb of charcoal contain 363.2g of pure carbon. Also, we'll have to pretend that the remaining voloatiles aren't going to contribute to the combustion. 363.2g C / 12.01 g/mol C = 30.24 moles (which is a conversion to number of atoms from weight. A mole is a specific number of atoms to make things easier to work with, and relate atomic mass units to grams.) C + O2 --> CO2 So, we'll need one mole of oxygen per mole of carbon even though there are two atoms, so 30.24 moles again. A mole of gas in theory takes up 22.4L in volume. That corresponds to 677.4L, or 23.92 cubic feet. Air is only 21% oxygen, so you really need 113.9 cubic feet of air. The rest of your question depends on combustion efficiency, which you may know a bit better about your system than anyone else would. All the values I am finding are about how much of the heat produced is transferred to a heating surface, not how efficiently it actually burns. Link to comment Share on other sites More sharing options...
Richtee Posted October 25, 2010 Author Share Posted October 25, 2010 (edited) There are a few assumptions that we're going to have to make to get a relative value. The thing to remember about charcoal is that it isn't pure carbon. I've often seen approximations around 70-90% or so, so lets call it 80% with the rest being ash, volatiles, and incompletely charred wood. That makes 1lb of charcoal contain 363.2g of pure carbon. Also, we'll have to pretend that the remaining voloatiles aren't going to contribute to the combustion. 363.2g C / 12.01 g/mol C = 30.24 moles (which is a conversion to number of atoms from weight. A mole is a specific number of atoms to make things easier to work with, and relate atomic mass units to grams.) C + O2 --> CO2 So, we'll need one mole of oxygen per mole of carbon even though there are two atoms, so 30.24 moles again. A mole of gas in theory takes up 22.4L in volume. That corresponds to 677.4L, or 23.92 cubic feet. Air is only 21% oxygen, so you really need 113.9 cubic feet of air. The rest of your question depends on combustion efficiency, which you may know a bit better about your system than anyone else would. All the values I am finding are about how much of the heat produced is transferred to a heating surface, not how efficiently it actually burns. I knew it was going molar at some point... heh! OK..that's plenty close enough. I am looking at controlling the combustion rate of a fairly sealed charcoal container for temp regulation purposes using a valve setup to meter air to the vessel. Just needed some outside values to have an idea of the intake size and valve body I'll need. It will be naturally aspirated depending on convection to create draw. Hmm... a few calcs to fudge, but I bet a 2" dia. short supply tube will do it Thank you Kind Sir! Edited October 25, 2010 by Richtee Link to comment Share on other sites More sharing options...
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