Joseph Herbert on sun 20 nov 11
We should remember that for every molecule of methane burned, two molecules
of water are produced. The amount of water in the ware is inconsequential
compared to the MANY moles of water created as the fuel is burned. Of
course, things don't dissolve in gaseous water but could when that steam
condenses, if the material were present is some significant concentration a=
t
that time. (which I doubt)
I love these speculative arguments! Remember, my assertions are always
better than your assertions because, after all, they are MY assertions.
Also, it says so on my t-shirt.
Joe
Joseph Herbert
Training Developer
Eric Koenig on mon 21 nov 11
Well, There is probably more to it than that. How much water buildup are=
=3D
we=3D20
talking about? You figure you have to burn half a mole (the equivalent o=
=3D
f 22=3D20
grams or ~11 liters of pure uncompressed propane at STP or ~20 liters at =
=3D
450=3DB0
F); and it has to be within the flammability concentration limits, or 2-1=
=3D
0% to=3D20
burn, just to make 18 grams of water, assuming the burners are operating =
=3D
at=3D20
optimum efficiency; and how much of that just goes out with the hot=3D20
exhaust?=3D20=3D20
Also, dinnae forget all of that molecularly bound water in the clay that =
=3D
isn't=3D20
apparent when the ware feels bone dry. You are talking about at least 4-=
=3D
6%
percent by weight there, i am thinking; and what if you load your kiln to=
=3D
the=3D20
gills? Let's say we've loaded 52 lbs of unfired air-dry un-candled ware =
=3D
for=3D20
argument's sake, which might yield up 2 or 3 lbs of water, or about a 1-1=
=3D
.5=3D20
liters.=3D20
Or something.
Eric
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