[TheForge] Chip forge

[Bruce Freeman]

But Steve's point, if I understand it correctly, is quite valid.  

Get one "chip" up to temperature, and, isolated in space, it will radiate away its heat more rapidly if irregular than if a sphere (i.e., if it has more surface area than if it has less).  

HOWEVER, within the mass of the chips of the forge where the action is, there is no "radiating into space" going on.  One chip radiates straight onto another.  At least untill you insert cold steel to be heated.  

So surface area of the chips is not all that big a concern.  Rather, what matters is that a bed of uniform spheres will not pack and restrict burner flow as would a bed of irregular chips.

Bruce
NJ

>>> osan at netlabs.net 3/4/2006 7:21:21 PM >>>


Steve Smith wrote:
> I could be completely off base here, but why does the surface area 
> matter in radiative transfer? If the chips are all pretty much at the 
> same temperature, isn't the steel already getting its max radiation, 
> regardless of the chip surface area? Isn't the air (whatever) itself 
> between chips radiating at the same temp as the chips? I think the steel 
> is already getting the max IR since it is completely surrounded by a 
> constant temperature environment.

	Area has a direct bearing on the rate of thermal dissipation, all else 
equal.  Let's use an absurd example to illustrate.  suppose you have two 
equal masses of steel (same alloy and all that so that the volumes are 
precisely equal).  one is in the form of a cube (let's say 1 cu in.) and 
the other rolled into foil 1" wide x .001" thick x 1000 inches long. 
Surface area of the cube will be 6 sq in, and the foil will be 2002.002 
sq. in.

Forget factors such as scaling and all that other real world trouble 
that throws small wrenches into the realm of the ideal) and assume we 
have a furnace or forge that will precisely raise both pieces to, say, 
1000*F.  Assume perfectly still air.  At temperature, each mass will 
hold precisely the same heat volume, but on removal the foil will 
dissipate its thermal load far more rapidly than will the cube because 
for any given set of conditions, such as temperature gradient between 
metal and air, the specific rate of heat transfer per unit area will be 
basically constant, assuming an effectively infinite air volume and 
convection.  But if the area be increased, the same specific rate of 
heat transfer will now be applied over that larger area and heat will 
therefore dissipate at a proportionally higher rate.

	-Andy

PS: anyone have Ralph Douglas' email address?  I have it on the machine 
I left in AZ.  I'd just like to drop a line to say hi.
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