> -----Original Message----- > From: Wan Juang Foo [SMTP:[email protected]] > Sent: Tuesday, May 13, 2003 3:09 AM > To: [email protected] > Subject: RE: Heat Sink Colour. (IR scanners a little O.T.) > > > > I was thinking about those far end IR scanners used to monitor travelers > for 'fever'. So this may be a little O.T. > > A prctical 'blackbody' radiates more heat. A good reflector is a poor > emitter. > I just read more about it from a Physics text (Sears, Zemansky and Young). > :-) > Tim et al *Commercial warning*- my employer is a leading player in the SARS ( "fever") imager market. The favoured technique is to monitor the temperature of the eyes. A typical thermal image of a face will show these to be the "hotspots". Hence skin colour, facial hair, surface chilling etc are not an issue. Assuming no transmission then Emissivity + Reflectivity = 1 so as you said, the key to a good emitter is to think of a bad reflector. To increase Emissivity, the surface should be as black as possible -at the wavelength of interest. As others have pointed out, there is not necessarily any correlation between being optically black at visible wavelengths and being "black" in the infrared. In our products we use different grades of black anodized finish. The finest for cosmetic finishes is smooth and quite shiny. The coarsest grade used on internal components is sand blasted before anodizing and gives a very matt finish, almost like a blackboard (chalk board). It gives the same sensation if you run your finger nails along it! When a similar thread arose previously, there seemed to be some confusion as to how airflow affects the radiated energy. W= SB x A x (T^4 - Tamb^4) where W =power radiated in Watts, SB=Stefan Boltzmans constant 5.67E-8, A=area in m^2, T is temp of body in K, Tamb is temp of ambient in K. If I take an example of a surface of 0.0058m^2 ( approx 3inch x 3 inch) at a temperature of 95C (368K) in an ambient of 25C (298K) Then a black body surface would radiate 3.5W. It doesn't matter whether this surface is in a vacuum or a blowing hurricane. At 368K it radiates 3.5W. For black anodised aluminium assume Emissivity is 0.95, so it radiates 3.3W For oxidised aluminium, assume Emissivity is 0.35. It then radiates 1.2W For shiny unoxidised aluminium, assume Emissivity is 0.025. It then radiates 0.09W. If this surface was a simple heatsink, it might be dissipating 10W for this 70C rise above ambient. In this case 35% of the energy would be lost via radiation for a black surface. If by good design and increased conduction and forced air cooling etc, this heatsink was dissipating 1kW for the 70C rise: the surface would still be radiating 3.5W, but this would now represent only 0.35% of the total. Hence, the designer will probably decide that the additional cost of anodising etc is not worth while. (Note that the total physical surface area is not same as the effective area for radiation loss. eg for two parallel sheets of metal there are four surfaces available for heat loss via convection. For the two inner surfaces there would be no net loss of radiated energy to the environment, they would simply exchange energy with each other. Therefore complicated profiles with lots of fins etc help greatly with convection loss but make little difference to radiated loss) Also for a large high power heatsink, the radiated energy (although small as a percentage of the total dissipation) may be large enough to interfere with nearby components. In this case it would be better to have low radiation so that the energy can be taken away to the "safe" area and not "leak" to nearby components. I am not a heatsink designer, but from my experience of measuring infrared radiation, it makes sense that most small heatsinks I have seen are black while most large high power heatsinks are not. Regards Andy Wood Land Infrared division of Land Instruments International Dronfield England www.Landinst.com This e-mail and its contents may be confidential, privileged and protected by law. Access is only authorised by the intended recipient. The contents of this e-mail may not be disclosed to, or used by, anyone other than the intended recipient, or stored or copied in any medium. If you are not the intended recipient, please advise the sender immediately. This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. Visit our web site at: http://www.ewh.ieee.org/soc/emcs/pstc/ To cancel your subscription, send mail to: [email protected] with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: [email protected] Dave Heald: [email protected] For policy questions, send mail to: Richard Nute: [email protected] Jim Bacher: [email protected] Archive is being moved, we will announce when it is back on-line. All emc-pstc postings are archived and searchable on the web at: http://www.ieeecommunities.org/emc-pstc
