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Convection Time-Constant in Air 1

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logbook

Electrical
Sep 8, 2003
764
My book on heat flow analysis does not help me with respect to the time-constants involved with natural or forced convection in air. My application is just an ordinary electronic circuit bolted to an ordinary finned aluminium heatsink. I could calculate the steady state temperature rise from equations in the book, but not the time constant. Obviously the mass of the heatsink needs to be heated up to the steady state value and additionally the heat is simultaneously being convected away.

I have previously experimented with this subject and it seemed to be the case that the more fan cooling I applied, the shorter the resulting time constant.

In my current application rapid temperature changes of the circuit are very bad, but slow changes are fine. The fan cooled heatsink I am using responds too rapidly to ambient changes and I want to get a feel for the problem.
 
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In transient calculations I don’t think you take into account the heat dissipating from the fins (convection). The time constant determines how fast the temperature rise will occur with respect to time and mass. The time constant is really the thermal resistance (of the heat sink) multiplied by the thermal capacitance (of the heat sink). By the time you reach steady state, heat will start to dissipate from the fins. IMO
 
Let me take that back, sorry it was an off the cuff remark. You can calculate the time constant with respect to convection. The thermal resistance will equal 1/(hcxA) the hc is the convection coefficient of the fin. If you can get your hands on Cooling Techniques for Electronic Equipment by Dave S. Steinberg, there is a method to calculate the time constant in respect to convection.

I apologizes about my first comment.

Good Luck
Twoballcane
 
Thermal time constant = RC

C = thermal capacitance = specific heat x mass
R = thermal resistance = 1/hA (assumes convection dominates)
h = heat transfer coefficient
A = surface area exposed to fluid

So your previous experiments were correct, logbook. For example, if you improve convection (reduce "h") by a factor of 2, you'll need about twice the mass to maintain the same time constant.

If you change from aluminum to copper, heat transfer will improve but typically only 10-20% for most heat sinks (depends on your geometry) and the mass will triple.

ko (
 
Thanks Twoballcane and ko99 for the fast response and for taking the time to respond. It's helpful to get my vague ideas clarified. Once there is a model to look at, one can really understand it better :)
 
You might want to downloaded It's a really slick tool that predicts heat sink performance. You can try different materials, shapes and airflow and it predicts thermal resistance, weight, etc
 
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