Here's a potentially useful expression for the "rule of thumb" energy
dissipation capacity of a free air cooled heatsink.

Merciless (and possibly informed) criticism expected :-).

This expression is an entirely empirical one based on inspection of heatsink
profiles and degrees C / watt ratings in a catalogue. It is solely an
attempt to fit the observed figures to reality and makes no claim to actual
correctness in a given situation. That said, this is the sort of thing that
helps make an engineer an engineer in the long term With suitable experience
and noting where this falls down you should be able to pick up a heatsink
and guesstimate its capacity within a factor of two or so. For those who can
do this already by inspection (as can many who have much experience in the
field), please move along, these are not the ones you want .... :-)

The formula works ROUGHLY for heatsinks in the 0.2 to 40 C/W range with
increasing inaccuracy at either end.

        Degrees Celsius rise per watt = 1/(Volume in litres)^ 2/3 x G

G is a "goodness factor" of your estimate based on the experience you are
about to get :-). A heatsink with good webs for heat flow and ample finning
within the volume profile will probably work better than a very thing unit
with minimal fins. For typical commercial heatsinks G ~~ 1.


ie Measure outer dimensions of heatsink profile in decimetres (0.1m = 10cm =
100mm = 1 decimetre)
Multiply these together to get litres volume.
Take 0.666 power (or square then cube root).
Take inverse.

eg 100 x 150 x 300 mm
Typical design so estimate G = 1

1/(1 x 1.5 x 3.0)^0.666   x  1 = 0.22 C/W

A Largish flag plus fins unit say 30mm x 30mm x 10mm =

1/(0.3 x 0.3 x 0.1)^0.666 = 23 C/W
If this is not especially well endowed with fins or rather thin a value of
50% to 100% higher may be in order.



        Russell "all models are wrong, this model looks like it might be
useful" McMahon

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