# heat dissipation in watts of equipment in a box

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?

• Outdoor security cameras use resistors to warm the inside of the enclosure. A typical setup puts 12VDC into 20 ohms, for a heating power of 7.2 watts. For larger wattage needs, a thermoswitch put into the circuit. Use 12 or 24 volts and some 5 watt resistors in parallel to get get some preliminary results. If your aluminum box is insulated then conduction through the box is reduced, and the formula does not apply. Try experimenting instead. – John Canon Feb 18 '19 at 4:40

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.

You are using the wrong temperatures in your equation, and the wrong equation altogether. For that equation to work you have to measure the temperature of the internal and external aluminum surfaces, not the air temperature. But given the very low differential there, that exercise would be pointless.

As a first-order approximation you can simply calculate the dissipation of a block of material with a 10 degree differential and the same surface area of the box. The equation is very similar, but the important factor is not the heat conductivity of the box but the coefficient of heat convection for air and the total surface area.

$$Q = h * A * ( T_s - T_a)$$

$$\ T_s \$$ is the surface temperature, $$\ T_a\$$ the external air temperature, and A the total surface area. For a small temperature differential and still air h can be as low as $$\ 10 W/(m^2 K) \$$

But if you use the air temperature this is a gross estimate, as it ignores that the internal temperature also has to be coupled from the air to the box, and that some of the heat is being radiated as well.

You would do better if you directly measure the box surface temperature.

But in your application I would do two things:

1. Add some non-flammable foam insulation to the inside of the box. Even a few millimeters covering the internal surfaces would go a long way.
2. If that is not enough, add a resistor to dissipate a few extra watts.

You could save yourself some time if you simply experiment with a couple known power levels (a couple resistors and a power supply) and directly calculate the heat transfer coefficient air to air for your specific box.

If you want to regulate the temperature and you need relatively little power you can directly use a positive temperature coefficient thermistor. Or you can couple a thermistor with a transistor to increase dissipation.

Thermal conductivity of aluminum walls has very little to do with the problem (thermal impedance of walls is likely negligible). This is a question about thermal properties of a rectangular box under free convection in the field of gravity. As formulated, the problem has no easy solution and needs an involvement of CFD - computational fluid dynamics. The result will depend on box orientation and whether the ambient air is still or moving. There is a vast layer of engineering articles to assess thermal regimes of enclosures, like this one, although it is pretty clueless (fin spacing is too small for good natural convection to develop).

Here is a better article on how to access heat transfer across hermetically sealed electronics enclosure, here is the general sketch of the problem components,

On external side, you need to account for natural convection around vertical surfaces of the enclosure, from top and bottom horizontal surfaces (the transfer properties are different in all three cases), and radiative exchange. For inner convection the same factors are working and need calculations/estimation too. So, there are a lot of factors involved, and only crude estimations are possible.

• The article from heatsinkcalculator.com ends up with a spreadsheet. But if you want to get an estimated power based on your measurements, you will need to pay them \$49. – Ale..chenski Feb 18 '19 at 6:37
• Here is one more inspiring article that might help, electronics-cooling.com/2018/05/… – Ale..chenski Feb 18 '19 at 6:49