Electronic – heat dissipation in watts of equipment in a box


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Here is the situation, some electrical equipment will be in low ambient temperatures and I need to make it warmer so that there is no damage to it.
I want to calculate how much power in watts is needed to heat up equipment inside a box made of aluminum.

The box is 5 inches X 5 inches X 11.5 inches (surface area is then 1.727 ft^2) it is made of aluminum and it is .1 inches thick and insulated.

from previous tests I recorded that when the ambient temperature was at -20 Celsius for a long time (at least 1hr) the surface temperature of equipment was steady at -10 Celsius and the air between the box and equipment was -11 Celsius.

Two questions:
1) How much power in watts is being dissipated by the electronics?
2) How much power in watts is needed to bring the temperature of the surface of equipment to 0 degrees Celsius? to 10 C?

I did some research and found this: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatcond.html

heat conduction formula

Q/t = kA(Thot – Tcold)/d

where k = thermal conductivity,
A = surface area,
Thot – Tcold = 10,
d = thickness,

The issue with this is that when I plug in k for aluminum = 205 or .5 by adjusting the units in the formula I get a large value for Watts either way…
am I doing this completely wrong is there another formula that would better model this problem?

Best Answer

Your problem here is that you're looking at the wrong part of the thermal equation. The heat dissipation of a heated metal box is dominated by the thermal resistance of the metal/air interface, not by the thermal conductivity of the box itself. Characterizing the thermal resistance of that interface will be difficult without taking a lot more measurements.

From a practical perspective, the simplest solution will be to overdimension the heater and use a thermostat to keep the enclosure at the target temperature -- this will make the exact size of the heater unimportant, and will also mean that the temperature will remain stable even when the ambient temperature changes.