I am contemplating whether or not I need chassis perforations for heat dissipation in a project. The box is roughly 440 mm wide, 280 mm deep and 85 mm high (2 rack height units aluminum enclosure.)

The continuous TDP is 40W originating from spread out sources. There is some direct contact between the chassis side profiles and a metal mounting plate, which carries the electronics. The contact is through several (~8) short bolts of about 4mm diameter. Otherwise, the thermal dissipation is mainly through the interior air.

I could estimate the internal temperature rise if I knew the thermal impedance of such a closed chassis but after searching a bit, I have found no clues. I am looking for a ballpark number such as 1 K/W (fine) or is it more like 10 K/W (insufficient).

I also don't have much experience with closed chassis. A similar power project in a similar perforated chassis (with no intentional cooling) didn't even get warm to the touch.

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    \$\begingroup\$ Put a heater inside it, run it continuously until you reach steady state, measure the temperature and then calculate the delta to ambient and divide by power? \$\endgroup\$
    – winny
    Commented Sep 29, 2021 at 9:39
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    \$\begingroup\$ I would estimate closer to 10 than 1 K/W for that geometry. \$\endgroup\$
    – winny
    Commented Sep 29, 2021 at 9:46
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    \$\begingroup\$ @jonk The ambient temperature is 5 - 40 °C (room temperature). The external dissipation will be dominated by passive convection and/or conduction. Radiation will be irrelevant at this temperature range. \$\endgroup\$
    – tobalt
    Commented Sep 29, 2021 at 9:56
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    \$\begingroup\$ @tobalt I believe radiation will be worth a look-see. But if you are sure about the irrelevance, then that's the end of the discussion. \$\endgroup\$
    – jonk
    Commented Sep 29, 2021 at 9:57
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    \$\begingroup\$ 40W in a 440x280x85mm box is a lot. In practice, the box will have another appliance sitting on top, or a stack of magazines, or a cat, so only the sides will be available for dissipation. Vents aren't just for airflow, they also dissuade the user from stacking too much stuff on top. \$\endgroup\$
    – bobflux
    Commented Sep 29, 2021 at 10:07

1 Answer 1


I like the following approach: You can calculate the thermal resistance of convection on a heated wall as

\$R_{th} = \frac{1}{A \cdot h}\$

where A is the wall surface and h is the heat transfer coefficient. In case of natural convection you set the value of h dependent on wall angle and wall temperature, typically in a range between 5 and 8 for electronic heat sink problems.

Now with two side walls with \$\frac{1}{A_Side1} = 26.7\$, \$\frac{1}{A_Side2} = 42.0\$ and the top wall with \$\frac{1}{A_top} = 8.1\$ and an approximated h = 5 (kind of conservative), you can calculate the wall's thermal resistances of convection: \$R_{th,wall,Side1} = 5.3\$, \$R_{th,wall,Side2} = 8.4\$, \$R_{th,wall,top} = 1.6\$

Looking at an individual wall, you have convection internally, conduction through the wall, and convection outside. Since the wall is aluminium, the conduction resistance can be neglected, and so so can in good approximation double the resistances to get the total resistance of the wall as

\$R_{th,Side1} = 10.6\$, \$R_{th,Side2} = 16.8\$, \$R_{th,top} = 3.2\$

Assuming that the bottom wall sits on ground and does not contribute to heat dissipation, the whole box resistance is

\$\frac{1}{R_{th,box}} = \frac{2}{R_{th,Side1}} + \frac{2}{R_{th,Side2}} + \frac{1}{R_{th,top}} = \frac{2}{10.6} + \frac{2}{16.8} + \frac{1}{3.2}\$

\$R_{th,box} = 1.6 \frac{K}{W}\$

If the internal power is 40 W, then the air internally is about 64 K higher than outside.

  • \$\begingroup\$ Thanks for the answer! I will have to do some detailed checks and repeat your calcs for myself (to fully comprehend them) but it looks a nice instruction.om how to estimate this kind of problem. In addition, internal radiation also contributes as suggested by jonk. I have moved on in the meantime and planned for active cooling anyway, but I might close the perforations for a test and see how hot it gets. \$\endgroup\$
    – tobalt
    Commented Oct 31, 2021 at 7:14
  • \$\begingroup\$ This procedure is usually confirmed by my CFD simulations. The outside surface temperature, which is relevant for radiation (which I did not take into account here), is increased by 32 K as compared to ambient. Internal radiation depends a lot on the usually complex internal geometry, and might be not be dominating here because all internal surfaces will be on not too different temperatures. If there are openings in the box, the model becomes a lot more complex and I prefer to employ CFD instead. \$\endgroup\$
    – UweD
    Commented Oct 31, 2021 at 7:51

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