Paint covered cables - further derate or not?

Rockingit · Jun 6, 2022

So I spotted this earlier (venue in Holland) and it got me thinking, be interested to hear people’s thoughts. Painting over cables surely has to effect their thermal dissipation from manufacturers design, so do we apply a further derate, and if so based on what values?

[ElectriciansForums.net] Paint covered cables - further derate or not?

i=p/u · Jun 6, 2022

Depends on the paint type, but I'd say is fine and I'd leave alone. Just my opinion.

pc1966 · Jun 6, 2022

I would worry about some solvent-based paints potentially damaging the PVC, but I doubt that the thickness of paint would make that much of a difference.

Lucien Nunes · Jun 6, 2022

There are two opposing effects at work here - radial thickness and surface area, neither of them probably very significant. The additional thickness of the paint layer adds thermal insulation between the conductor and air; I don't have data but would not be surprised if the thermal resistance of the paint is relatively similar to the sheath material. Since it is applied as a liquid there should not be any significant voids between sheath and paint to create an extra boundary layer.

But the addition of the paint also increases the overall diameter and hence the surface area, the latter more so if the paint surface is less smooth than the sheath. Since a significant part of the thermal resistance from conductor to ambient comes from the interface to the air, it is quite possible that the two effects cancel partially or completely, or even that the additional surface area improves the heat dissipation.

Rockingit · Jun 7, 2022

Lucien Nunes said:
There are two opposing effects at work here - radial thickness and surface area, neither of them probably very significant. The additional thickness of the paint layer adds thermal insulation between the conductor and air; I don't have data but would not be surprised if the thermal resistance of the paint is relatively similar to the sheath material. Since it is applied as a liquid there should not be any significant voids between sheath and paint to create an extra boundary layer.

But the addition of the paint also increases the overall diameter and hence the surface area, the latter more so if the paint surface is less smooth than the sheath. Since a significant part of the thermal resistance from conductor to ambient comes from the interface to the air, it is quite possible that the two effects cancel partially or completely, or even that the additional surface area improves the heat dissipation.

Makes perfect sense to me.