aquarium cooling SOLAR.KnowledgePublications.com

aquarium cooling
8 jul 1997
i wrote:
 
>you might drill some holes in the lighting fixtures so most of the heat
>doesn't end up in the water... and let some of the water evaporate into
>room air. suppose the tank were 2' deep x 4' wide, with 8 ft^2 of water 
>surface exposed to 78 f 50% rh room air with a 60ish wet bulb temp, and
>you blew some air over the water with a tiny fan and a cooling thermostat...
 
how much heat needs to be added to an uncovered aquarium to keep it the same
temperature as the air around it? duffie and beckman's equation for the ratio
of evaporative to convective heat loss is qe/qc = 1000(pp-pa)/(tp-ta), where
pp and pa are vapor pressures of water near a water surface and in surrounding
air (in "hg), and tp and ta are pond and air temps (f). this is probably only
intended to be used when tp>ta... with pa = 50%x0.96733 "hg, this equation
predicts qe=516 qc with pp = 0.99970 at an aquarium temp of 79 f, and qe=-452
qc at 77 f, with pp = 0.93589, but what's the evaporative loss at 78 f?

the convective loss in still air is about (tp-ta)1.5 btu/h, according to
fig. 1 on page 22.1 of the 1993 ashrae hof, so the aquarium would lose
1.5 btu/h-ft^2 by convection at 79 f, and another 774 btu/h by evaporation,
a net loss of 775.5 btu/h. at 77 f, it would gain 1.5 btu/h by convection
and lose 678 by evaporation, a net loss of 676.5. the average loss is
726 btu/h. would that apply at 78 f? that's 200 watts per square foot of
water surface, 1600 watts for the aquarium. seems like a lot, doesn't it...

nick