Ok, ok, to reply to all the previous messages. I apologize for my grievous error. When I said "units don't matter" I was referring to whether you used degrees Celsius or Kelvin. I forget that people still use imperial units, but I suppose that will change once I get into the working world. =20 As for this message, which was actually somewhat constructive. I know my "solution" may assume too much, but I only intended to give an idea of what the upper bound of the heat sink area might be, since the original request did not mention FEA or transient analysis. ;)=20 I guess I'll be triple-checking any further posts that I make! ----------------------------------------------------------------- Ryan Etwell -----Original Message----- From: pic microcontroller discussion list [mailto:PICLIST@MITVMA.MIT.EDU] On Behalf Of Spehro Pefhany Sent: December 23, 2003 9:00 AM To: PICLIST@MITVMA.MIT.EDU Subject: Re: [ee]: Are solid state relays TTL devices? 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, 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 -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads