I have a single electrical distribution panel in an insulated metal panel building (a standalone shed) which has no active cooling or air vents. I need to determine whether the latter are required in a climate that has an average high and low temperatures in July of 22.6 and 7.0 degrees Celsius, respectively. A temperature of 28.5 has occurred in July in the recent past.

  • The total calculated capacity of the steel distribution panel is 414 amps and the MCB rating is 100A (if this matters, I don't know). The Mains Breaker is listed as 160A/230/400V.

  • The surface area (A) of the metal panel box is 0.855 m2.

I need to understand how to calculate the maximum heat that will be emitted from this panel.

By my calculations, with a 1 degree C temperature difference between the outside and the inside of the room (not the panel), the heat transfer is 41.9 Btu/hour (the total surface area = 129 m2).

My question is, in July will the room temperature exceed 40 degrees C with the heat emitted from this electrical panel?

  • $\begingroup$ you need to know the heat load in the box. How many amps is not it. And the room temperature is a completely different question. $\endgroup$
    – Tiger Guy
    Nov 7, 2023 at 1:49
  • $\begingroup$ "My question is, in July will the room temperature exceed 40 degrees C WITHOUT the heat emitted from this electrical panel?" Do you record the temperatures? should be easy enough. $\endgroup$
    – Solar Mike
    Nov 7, 2023 at 8:16
  • $\begingroup$ The issue really is how much heat can I expect this panel to emit? The total Watts calculated = 105,760. This is based on each circuit at its rated amperage and voltage. In reality it will never be maxed out. Is there a safe rule of thumb I can use to determine how many watts (which can then be considered heat energy) will be emitted? I can then add it to my building envelope heat transfer calculation to determine if the interior temperature will exceed 40C. Thanks for your help! $\endgroup$ Nov 7, 2023 at 14:33
  • $\begingroup$ 105,000 watts is the current flowing through the wires, you needs the i-squared-R losses in the panel. 105 kW would be about 35 kitchen ovens running flat out, you need a sense of scale here. $\endgroup$
    – Tiger Guy
    Nov 7, 2023 at 16:39
  • $\begingroup$ There must be industry guidelines for this. $\endgroup$
    – Drew
    Nov 8, 2023 at 14:38

1 Answer 1


Thanks all. I was able to find a couple of "rules of thumb" responses and averaged the outcomes from them, choosing a conservative number. I then made an hourly heat flow table to show how the constant heat gain from the panel impacted the heat flow between the building interior and exterior. Based on the worst historical temperature day, I was able to confirm that adding louvers wasn't critically needed. I also recommended ongoing temperature monitoring.


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