1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.