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 From:  "James W. McKeand" <james at mckeand dot biz>
 To:  "'Peter Parnican'" <peter at procad dot sk>
 Cc:  "'M0n0wall'" <m0n0wall at lists dot m0n0 dot ch>
 Subject:  RE: [m0n0wall] Saving energy
 Date:  Thu, 21 Oct 2004 14:31:45 -0400
I think your calculation should be close if not right on. 

The formula I could find was explained at

I am assuming that fi in your formula is phase angle. If power is
single phase you can assume a phase angel of zero. Thus, P = 230V x
.15A x 1 = 34.5W. With either a phase angle of zero or what ever you
used, you should be consuming less than 40 Watts of power. The only
question I have on your calculation is the amperage - drawing .15A?

110 - 150 Watts is probably based on an active machine with monitor
and some hard disk activity. Monitor alone could draw 70W or more (my
CTX 15" monitor - 95W).  My Intel Pro100+ NIC's docs state that the
power requirement is 1.06W at 5 VDC. The docs for a Seagate ST320014A
hard disk state a max power draw of 6W during seek and 4.5 for an
idle. With my 15" monitor, 2 NIC's and an idle HD I should be drawing
over 100W. This would not include that little heater called a CPU. I
could not find specs on the Celeron 333, but the 1.4 GHz should draw
34.5 W (see:

If power consumption is an issue, Soekris boards only draw 10 to 15W. 
James W. McKeand

-----Original Message-----
From: Peter Parnican [mailto:peter at procad dot sk] 
Sent: Thursday, October 21, 2004 1:22 PM
To: M0n0wall
Subject: [m0n0wall] Saving energy

Hi everybody, this is not realy technical "problem" question. Im using
(celeron 333, 64MB, 20GB, 2xNIC) ...none monitor, keybord, mouse.

So, im measuring the current which use my PC (230V / 0.15A). ...so it
be P=U * I * cos (fi) = 230 * 0.15 * 0.76 = 26.2 W.
...I read that average consumption for PC is 110 - 150 W (max. 3 PCI

That is quite big differences 26W to 110W.  ..Is my calculation OK ?

(m0n0 will work on my friend roof, so i want pay him all expenses for


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