5
$\begingroup$

Suppose I have a sealed enclosure, say this one from Hoffman:

The box is 16"x12"x6", made of galvanized steel with a wall thickness of .0635" or .0785" (gauge 16 or 14). I'm planning to put some electronics in this box, which will generate an arbitrary amount of heat. How can I use this information to estimate how much the air inside the box will heat up above ambient outside the box?

$\endgroup$
5
  • $\begingroup$ You can use the integral form of Fourier's Law to determine the rate of heat transfer. $$\frac{\delta Q}{\delta t} = -k_sA\frac{\delta T}{\delta x}$$ Where $Q$ is heat flux, $t$ is time, $-k_s$ is the thermal conductivity of your material, $A$ is the cross sectional area (in your case the surface area of each of your panels), $T$ is your temperature change (inside vs. outside) and $x$ is the thickness of the material. This will help you determine if your box will dissipate the heat when you know how much heat is generated from the electronics. $\endgroup$
    – William B.
    Feb 16 '15 at 21:44
  • $\begingroup$ Steel by itself is not a very good insulator, so the heat will transfer fairly quickly. That said, have you thought about venting the enclosure on the bottom (with or without a fan)? Is there a reason it must be 100% sealed? $\endgroup$
    – William B.
    Feb 16 '15 at 21:47
  • $\begingroup$ Well, it's a NEMA 3R enclosure, so I'm wary of punching extra holes in it and losing the environmental rating. $\endgroup$ Feb 17 '15 at 1:06
  • $\begingroup$ not knowing your specific application... a quick google turned up some NEMA 3R rated vents. I just had someone build an enclosure for me and they decided to vent it with something similar (my application is indoors but the air is dirty). $\endgroup$
    – William B.
    Feb 17 '15 at 2:53
  • 1
    $\begingroup$ A hard problem that depends on a lot of details you haven't given. (Are there holes in the box, amount of heat, orientation, heat fins...) In the past I've found it easier to measure the temperature rise at "worst case". $\endgroup$ Feb 17 '15 at 15:16
6
$\begingroup$

This will depend very much on the heat flow around the enclosure. There is a thing called the H factor which describes the heat transport properties between a surface and a fluid. The value of H varies with surface properties and flow.

So to do your calculation you need a few different assumptions. Let's simplify.

1) the air in the box is uniformly heated 2) air flows around five sides of the box the same way; no heat transfer takes place through the bottom

Now we can consider the box a "thermal resistance" (the walls) in series with a heat sink (the air flowing).

The thermal resistance (K/W) of the walls is given by

$$R=\frac{t}{kA}$$

For your box, t = 1.6 mm, A = 3.4E5 mm$^2$, k = 17 K/W, so R = 2.8E-4 K/W : the walls themselves won't present much thermal resistance. So we need to look at the air around it.

The heat transfer coefficient is a strong function of air flow rate (this is why you have fans for cooling!). Let's make a simple assumption that there is some natural air flow, but no forced convection. The heat transfer coefficient will then be at the lower end of the scale, or about 0.5 W/m^2/K. (See for example this reference). With the "used" area of the box of 0.34 m$^2$ (12x16+(12+16+12+16)x6 in$^2$), you get about 1.5 K temperature rise inside the box for every Watt of power dissipated. A little fan around the outside could easily cut that by a factor 10...

So just to drive the point home explicitly: this calculation is conservative and VERY APPROXIMATE; the number calculated can be easily an order of magnitude off from the number you get when you build your box, because there are SO many factors that will affect this. Controlling the air flow explicitly (versus relying on "whatever air flow lies around") is really essential to get any kind of believable limits on your answer. This is reflected in the comments above...

$\endgroup$
1
  • $\begingroup$ How did you arrive at the calculation "1.5 K temperature rise inside the box for every Watt of power dissipated" ? For conduction, you gave the equation R = t/kA. For convection, I believe the corresponding equation is R = 1/hA. Subsituting h = 0.5 and A = 0.34, I get 5.88 degrees per Watt. $\endgroup$
    – banjaxed
    Jan 28 at 18:14
1
$\begingroup$

http://www.automationdirect.com/static/specs/encstratusheatexchangers.pdf

According to this document, heat load transfer through the walls of a metal enclosure is estimated in BTU/H as 1.25 x surface area (sq ft) x temperature differential (degrees F).

Converting to W/degC, you would get about .66 times the surface area of the box in square feet, or 7.1 times the surface area in square meters.

Or in degC/W, that's 1.52 ft^2, 218.9 in^2, or .14 m^2. Divide by the surface area to get the thermal resistance of the box.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.