[rescue] Cooling (Long Message, sorry)

[Dave McGuire]

On April 16, Jochen Kunz 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.

  Disclaimer: I'm a computers-and-electronics-head, not a math-head...I
wouldn't know a cosine if it walked up and shook my hand.

  Nothing with a switching power supply is a purely resistive load, we
all know that.  I'm talking about rough calculations here.  Thanks for
clarifying the finer points.

    -Dave

-- 
Dave McGuire                                 "Mmmm.  Big."
St. Petersburg, FL                                -Den