[rescue] Cooling (Long Message, sorry)

[Jochen Kunz]

On Mon, Apr 15, 2002 at 10:21:49PM -0400, Dave McGuire wrote:

>   To be more exact about it, you can get a clamp-on ammeter and
> measure the current consumption of each device.  Multiply this by your
> (measured) line voltage to get power in watts.  Count on all of that
> being turned into heat.  
Aua! Dave you disappoint me. 'heat' != U * I You have AC here, 
SMPSUs with capacitors at the input that cause 'blind load'. 
'heat' == U * I is only valid if you have pure R, no C or L 
in the load impedance. 
So you have here: 'heat' == U * I * cos(PHI) 
cos(PHI), the phase angle between U and I, depends on the load 
impedance and can be measured e.g. with a oscilloscope.

Not to talk about the fact that I of a SMPSU has no sin()
wave form. So you get a fair amount of harmonics. That makes
the whole thing even more complex.
-- 



tsch|_,
         Jochen

Homepage: http://www.unixag-kl.fh-kl.de/~jkunz/

p.s. I am an electrician and hope to get my master degree in EE and CS
this year. So I know everything and I know it better than you. ;-)