At 01:16 AM 12/23/2003 -0500, you wrote:
>----- Original Message -----
>From: "Ryan Etwell" > Q =3D h x A x (Tb - Tf)
> >
> > Where,
> > Q - Heat Dissipated
> > A - Area of body
> > Tb - Temp. of body
> > Tf - Temp of surrounding fluid (air in this case)
> > h - Convection Heat Transfer Coefficient
> >
> > In this case Q is the heat generated by your device (in Watts), Tb would
> > be its operation temperature (units don't really matter here since the
> > equation just uses the difference of two temps)
>
>Sorry Ryan, but this doesn't pass the "common sense" test. If the units
>don't matter then I should get the same results with any units. Let's just
>pick a couple numbers out of the air- boiling and freezing water seem good
>as any. In celsius units thats 100 and 0 degrees respectivly and the
>(Tb-Tf) term in your equation would be 100. However in Farenhieght units,
>the same temps would be 212 and 32, the (Tb-Tf) term in your equation would
>be 180, resulting in a very different answer.

Only if you (erroneously) assume that h won't be different (and have=
 different
units). Yes, they have to be consistent units, but that's kinda obvious,=20
n'est pas?

> > The magical number in
> > this case is h, which is not a thermodynamic property, but an empirical
> > parameter.  For free convection (no fan being used) h is listed as being
> > between 10 - 100 W/m^2 K (Watts per square meter Kelvin).
>
>Ok, now we're referencing the Kelvin scale, without a conversion in the
>equation you should be using kelvin for the rest of the terms too.

Or =B0C, obviously, since the degrees are the same size as Kelvins, and it's
a difference in temperature.

Perhaps more important than nitpicking over his comment, note the=
 simplifying
assumptions that Ryan made:

1)      The heatsink is isothermal (it WON'T be even close unless the=
 heatsink
         is of suboptimal design (maybe it's used for something else).=
 Normally
         there's a temperature gradient on the heatsink.

2)      He's assuming steady-state, and ignoring transient effects if the
         heat source is not steady.

3)      He ignores radiation. Radiation can work for you (usually) or if you
         have something silly like a vacuum tube or another hot part nearby
         it can work against you. It depends on the emissivity of the=
 heatsink
         material's surface and the physical configuration (fins "see"=
 mostly
         each other).

4)      He also ignores conduction to other objects. This can be substantial
         if the heatsink is mounted to a metal housing.

5)      He was thoughtful enough to post a real number range for h.
         The coefficient h has a 10:1 range, which makes it pretty much=20
unsuitable
         for design purposes. Air is a viscous fluid and the type of
         air flow (laminar, turbulent or transition) depends on the physical
         configuration of the heat sink and how it is mounted in relation to
         gravity. You can see generally that if you have a thick heatsink
         and double the width you'll get more somewhat more than half the
         temperature rise, all other things being equal.

With a bit of instrumentation (many fine-gauge thermocouples) and perhaps
some Chinese jaw sticks you can do some experimentation. There are sometimes
surprising differences from minor changes in configuration. I was able to
double the expected MTBF on a product by a minor change in natural
convective airflow, with NO difference in product cost or weight.

The FEA software to do this analysis is still outrageously expensive,
so cut and try is the way to go for the time being for most small
(<$10M) electronics companies. Then there's *transient* thermal analysis...

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the=
 reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

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