Enclosure natural cooling?

I need to evaluate how much heat can be dissipated from an electrical enclosure. Inside I have 500W of losses and I cannot have no cooling devices. Only the enclosure itself.

Inside Tmax is 35°C (with circulating fan).
Outside Ta is 20 (not much airflow).

What should be the size of the enclosure?

What is the formula?

Thank you very much for your help.

Most enclosure suppliers that have fans and so on in their product line include thermal calculation charts or formulas in their catalogs. I am thinking spceifically of Rittal and Hoffman, but others do as well. I would look at them first.

A rough estimate.

Unregistered said:

Inside Tmax is 35°C (with circulating fan).
Outside Ta is 20 (not much airflow).

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Use the heating formula!

You will have losses of 5.5 W/m2*k (painted steel), that is 5.5 Watts per square meters per Kelvin.

If you want to loose 500W between delta of 35 and 20, you will need:

500
--------- = 6 square meters of surface free to the ambiant air
(15*5.5)

It will be a darn big cabinet.

Ph = ( A X T X k ) – Pv

Ph = Required heating power for your application in Watts (W)

Pv = Heating power generated by existing components in Watts

A = Exposed enclosure surface area in square meters (m2)

/\T = Temperature differential between the desired minimum interior temperature and the lowest possible external temperature of the enclosure in Kelvin (K). 1.8°F = 1°C = 1K

k = Heat transmission coefficient of the enclosure material used :

Painted steel : 5.5W/m2K
Stainless steel : 3.7W/m2K
Aluminum : 12W/m2K
Plastic : 3.5W/m2K

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