# How to calculate the volume of air required to keep electrical box cool?

I have a steel electrical box that is 15 cm x 15 cm x 10 cm. The weight of the box is 1kg. Inside the box is a relay that loses 30 Watts of heat while in operation (it's attached to a heat sink that limits the temperature rise to 2 $$^\circ$$C/W, though I don't think that matters).

I'm trying to figure out the required volume of air surrounding the box in order to keep the temperature rise outside the box less than 10 degrees C (assume air temperature to start is 25 $$^\circ$$C).

How would I go about calculating this? I thought about using specific heat equation but I can't figure it out.

• Is this air outdoors in the wind? Indoors in a cupboard? Nov 11 '19 at 19:51
• Assume indoors in a big room where the ambient room temperate change will be negligible. Nov 11 '19 at 20:26
• If the relay and its heat sink are entirely inside the box, then the inside of the box will get hot. The heat sink needs to be on the outside of your box, or you need really good thermal coupling between the heat sink and the box. Nov 11 '19 at 21:17

This question can only be answered after many tests as to the convective properties of the box, it's geometry with respect to the geometry of the void, ignoring radiation. Because convection which is going to be the major heat transfer mechanism is basically a complex movement of air flowing up in a widening plume, and losing its temperature. Here is a link to imperical formulas for a single source convection from Engineering Toolbox site. Link

But basically comparing your box to a 60 watts recessed light bulb housing which is usually an 8 inch by 8 inch cylinderical can, one could assume a 4- 6 inchclearance should be ok.

If I get you right, there is a box having a small device that dissipates 30 Watts of heat continuously into the air. The air is blocked into a box, no way in or out. There are no fans/blowers trying to cool the device itself nor circulating the air inside the box. The size of the box is to be defined and the air temperature inside the box has to be figured out.

If the air volume were infinite - we could assume that its ambient temperature is fixed (25 deg), approximate a natural heat convection coefficient (something around 7-10 watts $$\frac{Watt}{m^2K}$$) and find out the temperature over the heatsink by using the well-known $$q=hA(\Delta T)$$ equation. When q is the dissipated heat, A is the total area of heat convection and $$\Delta T$$ is the difference between the ambient temp and the temp over the heatsink surface (assuming it is uniformly distributed).

However, in your case the temperature of the air inside the box will rise up since its volume is finite. In the long run, the steady state, the total amount of heat flowing out of the box must be exactly 30 watts and the air temperature will converged to a constant value. Choosing the desired walls are is a bit tricky procedure: Assuming the walls have a constant uniform temperature of 25 degrees, and guessing the air temperature is something between 25 deg to the heatsink temp - we can find the total area needed for allowing the 30 watt heat to flow outside the box under those assumptions. It is a kind of back and forth calculation for getting an order of magnitude. For whatever area you finally choose, you can find out the air temperature close to the walls by using $$q=hA(\Delta T)$$. This time A is the walls area. We can take an assumption that the air temperature inside the box is uniform and having that value.

Now you have a decent approximation of the air temperature and you can go on finding the heatsink temperature.

Of course, this is only an approximation. For more precise calculation you may use a CFD tool.

Using the specific heat and heat capacity are needed only if you want to analyze transient problem. I.e - checking the temperature behavior during time before reaching the steady state.

So sort of like a computer ; put a computer fan in it (preferably with an opening) . I have a few fans that I pulled out of junked electronic equipment, doesn't everyone ?

• No. Someone pointed out to me that the brand of industrial label printer we were using had a heatsink mounted on the back and no fan. This was a "case" of proper thermal design and one less electro-mechanical component to fail. There are many reasons why a fan may not be applicable - water-proofing, dust ingress and noise for starters. Jun 4 at 20:56