[rescue] Cooling (Long Message, sorry)

[Joshua D Boyd]

On Tue, Apr 16, 2002 at 06:47:16PM +0200, Jochen Kunz wrote:
> 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.

> 
> 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. ;-)

It still works as a rule of thumb since |cos(PHI)|<1.  Sure, you buy a bigger
AC unit than needed, but it is better not to be running them at full load 
anyway, T.M.U.

-- 
Joshua D. Boyd