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I have a non functional PC power supply unit, and I would like to keep its heat sinks, but they have no model or part number in them. Is there a way to estimate, calculate or measure their thermal resitance? The PSU is a Sentey BCP 450-OC, here is a picture of the heat sinks

enter image description here

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    \$\begingroup\$ A look in heatsink catalogs will give you a reasonable idea. \$\endgroup\$ – Russell McMahon Feb 15 '20 at 1:04
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Measuring the effective heat resistance should be pretty straightforward: you'd affix load resistors to where you'd later put the components that should stay cool. You work in a controlled temperature environment (i.e. indoors at 20°C).

You put a measurable amount of power through the resistors (e.g. by using a lab voltage supply and measuring the voltage over and current through the resistors). You measure the temperature and wait until you've reached a stable state.

The difference of resistor temperature to ambient divided by the power is your thermal resistance.

But: honestly, I don't think this is worth it. Both the shape and the fact they're fixed right in front of a fan means that they are to be used with forced cooling, i.e. with a fan in a specially shaped enclosure. You'd have to measure in the enclosure and with the fans you're planning to use.

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Is there a way to estimate, calculate or measure their thermal resistance?

Yes. I am assuming that you want to know the heatsink to ambient thermal resistance.

The calculations are complicated, somewhat easier for forced air (fan), than natural convection. I wouldn't trust any calculation that I would perform without verifying it with a test. So, ...

It is easiest to test. This heatsink is designed for forced air, it won't be very good without a fan.

Attach something to the heatsink that you can be made to dissipate a known amount of power. A chassis mount resistor will be a good choice. Measure the temperature rise of the heatsink, subtract the ambient temperature, divide by the power.

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The fluid aerodynamics of forced air velocity overwing surface determines the thermal resistance coefficient in °C/Watt for case temp. rise.

The air velocity over the surface, not the fan exit, significantly lowers the resistance by some constant, k (< 1) times the Rca, Resistance case-ambient. This is more of an aerodynamic design requirement. One could come up with a more accurate answer, but based on my exposure to heatsink design and testing I would estimate 8 °C/W and with that airflow 1~2 °C/W.

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