re: how much power to heat up a cabinet?
23 feb 2004
>>> >i have to build a cabinet for testing electrical devices at an
>>> >elevated temperature of 85 deg c. the ambient is around 25 deg c...
>>> (85-25) = 60 c is 108 f...
>>> >1. how many watts of power do we have to dissipate in the heaters to
>>> >achieve the intended temperature of 85 deg c?...
> or the 1.5 kw solution i recommended.
i doubt it, with the electrical devices off. you wrote:
>a 1 kw strip heater would do fine. you'll need a small fan to
>take heat off the element into the box. you could probably adapt a
>little residential 1.5 kw space heater...
with l^3dt = 6.5^3x108 = 30k > 63, it looks like the outside walls would
have turbulent airflow, with hcon = 0.19x108^0.33 = 0.9 btu/h-f-ft^2, from
equation 2.19us in kreider and rabl's 1994 heating and cooling of buildings
book. your fan would help ensure high airfilm conductance inside the walls
(if the motor doesn't burn up :-), and 85 c (185 f) box walls would lose
about 0.1714x10^-8((460+185)^4-(460+77)^4) = 154 btu/h-ft^2 by radiation.
>>> >2. the walls of this cabinet are made of sheet metal (mild steel). any
>>> >ideas on how to insulate the walls effectively...
>>> bolting 1" foil-faced "r7.2" polyisocyanurate board to the outside of
>>> the walls might reduce the power required by a factor of 10...
> so, he should just stick a 100 watt lightbulb or two in there ?
maybe two, with wall insulation and a thermostat near the top.
>how about 17 ashrae bunnies instead ?
i'm afraid the bunnies would suffer.
>>> insulated cabinet with a heater at the bottom may not need a fan.
> if you don't want stratification, it does. if you want
>anything resembling control or consistency, it does.
this might be a house with a heater downstairs and no heater upstairs, just
some vent openings to allow warm air to rise in the middle and fall near
the walls. on a cold day, the second floor is cooler than the first. with
r-value r walls, this box needs about 108fx60ft^2/r = 6840/r btu/h. we might
imagine warm air rising through a 2 ft^2 vent and falling through 2 ft^2,
so 6840 = 16.6(2)sqrt(6.5')dt^1.5, and dt = (76/r)^0.66. with r1 walls,
the top of the box would be 17.9 f cooler than the bottom; with r10 walls,
it would be 3.9 f cooler; with r20 walls, dt = 2.4 f, and so on.