You are correct that this isn't an end all be all calculation. There is much more that can be taken into account. This is just a good method to use since most of us know down to the watt how much power we are using. And we mostly know our temps. Sure there are volumetric calculations that can be taken into account along with how many times you want the air changed out per minute/hour. But I am waaayyy to lazy to go through all of that when I'm at home! hehe... I get enough of that at work!
Thanks,
it's nothing short of a miracle that your butchered equation gives the correct CFM with every unit and constant wrong, but somehow it does... thats not the density of air, or the density of anything mind u. its either about 1.2 kg/m^3 or 0.0746 lb/cuft. (must be in lb/cuft for english conversion)
THIS is the exact equation, from a thermodynamics standpoint, no matter what u here:
Vrate (cuft/min) = [W (watts) x 0.0569 (BTU/min/W)]/[(0.0746 lb/cuft) x (0.24 BTU/lbR) x (T2 +460 - T1 - 460 R)]
Therefore:
Vrate = (3.178 x W)/(T2 - T1)
Vrate is in CFM
W is in watts
T's are in degree F
the density of air, assume atmospheric P and T are 14.65 psia and 70 F, is roughly 0.0746 lbm/cuft.
when using a 500 W source, 72 F intake temp, and 82 F exhaust temp, the volumetric flow rate of the air required to remove the heat input is 159 CFM. It's just by luck that socal70's equation somehow works out. thats of course with air running into and leaving the box in a steady flow; ur heat removal will be less because you are just sucking air from the top instead of completely exchanging it, hence a much larger fan will be needed to move all that air
U should seriously think about swapping out ur hood for either an inline one, or one with a 6" flange directly above the bulb, so u can hook ur exhuast fan and ducting to the top (venting and cooling at the same time), and still bring in fresh air with smaller fan