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I would like to cool a small device, with a size of 45x45 mm² and a thermal load of ~40 W. Currently, I have a cooling system available, capable of cooling a thermal load of 52 W, and a cold plate with the dimensions of 60x80 mm². Due to the size difference I expect a small drop in the efficiency.

Is there a way to calculate that drop, and to estimate if I still can use that cooling device? As a naive approximation I would take the difference in the area size ($80\cdot60-45^2=2775$), which is approximately 60 % of the original cold plate. Can I therefore expect a loss of 60 % of the original cooling capacity, i.e. I only can cool a thermal load of 30 W for the given dimensions, or is that approach wrong?

arc_lupus
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1 Answers1

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Summary

Assume that you put the heating element directly on the cooling element. The cooling element can handle an input flux (mW/mm$^2$) of

$$ \frac{\dot{q_c}}{A_c} = 52/(80*60) = 10.8 $$

The area of the heating element is $45^2 = 2025$ mm$^2$. You will pull a maximum of $2025*10.8 = 21.9$ W power from the heater.

Assume that you can "pipe" the heat from the heating element directly to the cooling element through a tapered conducting path (that is otherwise insulated on all sides). At that point, you can pull 52 W into the cooling element. Your heating element is only giving 40 W maximum. The input flux (mW/mm$^2$) to the cooling element is

$$ \frac{\dot{q_h}}{A_C} = 40/(80*60) = 8.33 $$

Conclusion

To improve the efficiency of the system, pipe the heat from the heating element to the cooling element using a tapered conducting path that is insulated on all sides.

Jeffrey J Weimer
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