My config:
400w power supply. (3 fans total in the case)
P4 3.0GHz 1G RAM PC at home.
Radeon 9800 Pro graphics card
2 x HHDs (80G & 200G).

I have always wondered what it costs (approx) to leave the home PC on most
of the evening and nights. Usually, on the weekends the PC would be on 12
hours Sat and again Sun.

Anyone know roughly?

Bennos wrote:
> My config:
> 400w power supply. (3 fans total in the case)
> P4 3.0GHz 1G RAM PC at home.
> Radeon 9800 Pro graphics card
> 2 x HHDs (80G & 200G).
>
> I have always wondered what it costs (approx) to leave the home PC on most
> of the evening and nights. Usually, on the weekends the PC would be on 12
> hours Sat and again Sun.
>
> Anyone know roughly?

Roughly? You can certainly get a top limit, per hour, by taking 40% of
what it costs for you to purchase a kilowatt-hour from your power
company. Realistically you will not be using the top rating of your
power-supply, especially when it's in an idle state, but it at least
sets a boundary on your expense.

In message <Fguef.576498$xm3.98314@attbi_s21> Grinder
<(E-Mail Removed)> wrote:

>Bennos wrote:
>> My config:
>> 400w power supply. (3 fans total in the case)
>> P4 3.0GHz 1G RAM PC at home.
>> Radeon 9800 Pro graphics card
>> 2 x HHDs (80G & 200G).
>>
>> I have always wondered what it costs (approx) to leave the home PC on most
>> of the evening and nights. Usually, on the weekends the PC would be on 12
>> hours Sat and again Sun.
>>
>> Anyone know roughly?
>
>Roughly? You can certainly get a top limit, per hour, by taking 40% of
>what it costs for you to purchase a kilowatt-hour from your power
>company. Realistically you will not be using the top rating of your
>power-supply, especially when it's in an idle state, but it at least
>sets a boundary on your expense.

Realistically it will be substantially less then that, most cheapo power
supplies can't put out more then 80% of their rated capacity for an
ongoing period of time.

If you have your power rates handy, you can work it out fairly easily,
Google will do the "tough" math for you:

http://www.google.ca/search?hl=en&q=...atts+*+1+month

So a 400w supply at full capacity would run you $17.28/month if the PC
was left on the entire time. Lets stick with those numbers for the
moment.

One other consideration is the climate. All 400w of power is being
converted to heat, so that $17.28 of heat isn't lost, it's released into
your house.

I'm currently $3.72/GJ to heat my house, so if my math is right, that's
about $0.014/kWh.

In other words, even though my electric company is charging me
$0.06/kWh, every $0.06/kWh I spend on electricity which is converted to
heat saves me $0.014/kWh on my gas bill, so the effective cost for
running my computer 24/7 is ~$14/month rather then $17.28/month.

However, if you're in a hot climate and are paying to cool your house
rather then paying to heat it, the reverse applies, every kW of energy
that your system uses results in another expenditure (of energy, and
cash) to remove that heat from your house.

--
They say you shouldn't say anything about the dead unless it's good.

"He's dead. Good."

In Australia, I just found some information - one of the major electricity
provide has a cost average for home appliances and electrical equipment.

The estimate for the home PC is AUD 0.0394¢/hr.



"Bennos" <(E-Mail Removed)> wrote in message
news:mDtef.18608$(E-Mail Removed)...
> My config:
> 400w power supply. (3 fans total in the case)
> P4 3.0GHz 1G RAM PC at home.
> Radeon 9800 Pro graphics card
> 2 x HHDs (80G & 200G).
>
> I have always wondered what it costs (approx) to leave the home PC on most
> of the evening and nights. Usually, on the weekends the PC would be on 12
> hours Sat and again Sun.
>
> Anyone know roughly?
>
>

By my calculations:

Computer probably draws 200 watts max most of the time. That's five hours of
operation for one KWH., which probably runs about 7 cents or about 35 cents
per day. Ours is 8.3 cents/KWH.

Couple of beers a month at your local bar. :-)

Never tried this but shut off everything in your house, except the computer.
Count the revolutions on the disc of your meter in a given time interval and
calculate that back to actual use. Check with your utility they can help and
also give KWH rates or look at your bill. The rate is probably given there.

An ammeter in the circuit will not give you true readings because of the
power factor but it would give you something to go by too. (I'd take a guess
and go with 80% of that reading because it will read slightly high due to
the power factor).


"Bennos" <(E-Mail Removed)> wrote in message
news:mDtef.18608$(E-Mail Removed)...
> My config:
> 400w power supply. (3 fans total in the case)
> P4 3.0GHz 1G RAM PC at home.
> Radeon 9800 Pro graphics card
> 2 x HHDs (80G & 200G).
>
> I have always wondered what it costs (approx) to leave the home PC on most
> of the evening and nights. Usually, on the weekends the PC would be on 12
> hours Sat and again Sun.
>
> Anyone know roughly?
>
>

In article <ZHxef.18791$(E-Mail Removed)>, "Bennos"
<(E-Mail Removed)> wrote:

> In Australia, I just found some information - one of the major electricity
> provide has a cost average for home appliances and electrical equipment.
>
> The estimate for the home PC is AUD 0.0394¢/hr.
>

I just tried my clamp-on ammeter on my P4 2.8C with FX5200.

DOS 1.9A @ 115V (I was running Seatools disk diagnostic...)
Linux 1.275A @ 115V (Xwindows desktop)
Win2K 1.353A @ 115V (Actually it is the same as Linux, as I had
to add my Win2K disk to do the test, and the
difference, is the power to run the disk.)

This does not factor in the phase angle, as mentioned by
"beach bum". Multiplying the above numbers together will be
over-estimating the cost, but it will give a rough idea.
The real figure might be 80% of this, but I don't have anything
to measure phase with.

1.275 * 115 = 0.147KwH x 0.057 CDN$/KwH x 24 x 365 = $73.21 CDN/yr or less.

If someone tries the above experiment with an "active PFC" supply,
then the measurements will be true without any correction being
needed. Unfortunately, PFC supplies aren't generally for sale
over here.

If the computer is in "Suspend to RAM" standby state, power
consumed will be about 10 watts - 0.010 x 0.057 x 24 x 365 = $5 per year.
My ammeter cannot measure this power, because the current waveform
is too distorted for accurate measurement (it is not even close
to a sinusoid).

Paul

>
>
> "Bennos" <(E-Mail Removed)> wrote in message
> news:mDtef.18608$(E-Mail Removed)...
> > My config:
> > 400w power supply. (3 fans total in the case)
> > P4 3.0GHz 1G RAM PC at home.
> > Radeon 9800 Pro graphics card
> > 2 x HHDs (80G & 200G).
> >
> > I have always wondered what it costs (approx) to leave the home PC on most
> > of the evening and nights. Usually, on the weekends the PC would be on 12
> > hours Sat and again Sun.
> >
> > Anyone know roughly?
> >
> >

Bennos wrote:
> My config:
> 400w power supply. (3 fans total in the case)
> P4 3.0GHz 1G RAM PC at home.
> Radeon 9800 Pro graphics card
> 2 x HHDs (80G & 200G).
>
> I have always wondered what it costs (approx) to leave the home PC on most
> of the evening and nights. Usually, on the weekends the PC would be on 12
> hours Sat and again Sun.
>
> Anyone know roughly?

Your monitor is probably the one single component that uses the most
electricity. Mine uses 75 watts.

"Paul" <(E-Mail Removed)> wrote in message
news:nospam-1611050358520001@192.168.1.178...
> In article <ZHxef.18791$(E-Mail Removed)>, "Bennos"
> <(E-Mail Removed)> wrote:
>
>> In Australia, I just found some information - one of the major
>> electricity
>> provide has a cost average for home appliances and electrical equipment.
>>
>> The estimate for the home PC is AUD 0.0394¢/hr.
>>
>
> I just tried my clamp-on ammeter on my P4 2.8C with FX5200.
>
> DOS 1.9A @ 115V (I was running Seatools disk diagnostic...)
> Linux 1.275A @ 115V (Xwindows desktop)
> Win2K 1.353A @ 115V (Actually it is the same as Linux, as I had
> to add my Win2K disk to do the test, and the
> difference, is the power to run the disk.)

That's interesting. Who would have thought that? DOS is a hog. :-)

> If someone tries the above experiment with an "active PFC" supply,
> then the measurements will be true without any correction being
> needed. Unfortunately, PFC supplies aren't generally for sale
> over here.

PFC. Power factor corrected? I would expect any PS to give a lagging PF due
to the transformers in it but many of the new supplies, smaller ones anyway,
like I have in an external HD enclosure don't use much for transformers
anymore. They use new technology that gets around the voltage reduction
necessity. These could very well operate close to unity PF. We have test
equipment at work and I should take a few minutes to test both.

With electric motors you can add capacitors to bring PF closer to unity. In
fact utilities will penalize large customers for poor power factor. It pays
those customers to install banks of capaitors to correct the problem. The
problem for the utility is that meters are designed to measure wattage (the
part of AC current the customer is billed for) but the electric company
facilities are providing volt amps, a higher (vector sum) quantity and they
don't get paid for that difference.

I would think one could add capacitors to the PS to correct PF but then why
bother? It would not make significant difference to the power company and
we're not billed for the extra current anyway.

>
> If the computer is in "Suspend to RAM" standby state, power
> consumed will be about 10 watts - 0.010 x 0.057 x 24 x 365 = $5 per year.
> My ammeter cannot measure this power, because the current waveform
> is too distorted for accurate measurement (it is not even close
> to a sinusoid).

Hmmm. That may throw off the phase angle instruments too.

>
> Paul
>
>>
>>
>> "Bennos" <(E-Mail Removed)> wrote in message
>> news:mDtef.18608$(E-Mail Removed)...
>> > My config:
>> > 400w power supply. (3 fans total in the case)
>> > P4 3.0GHz 1G RAM PC at home.
>> > Radeon 9800 Pro graphics card
>> > 2 x HHDs (80G & 200G).
>> >
>> > I have always wondered what it costs (approx) to leave the home PC on
>> > most
>> > of the evening and nights. Usually, on the weekends the PC would be on
>> > 12
>> > hours Sat and again Sun.
>> >
>> > Anyone know roughly?
>> >
>> >

In article <(E-Mail Removed)>, "beach bum"
<sunshine&(E-Mail Removed)> wrote:

> "Paul" <(E-Mail Removed)> wrote in message
> news:nospam-1611050358520001@192.168.1.178...
> > In article <ZHxef.18791$(E-Mail Removed)>, "Bennos"
> > <(E-Mail Removed)> wrote:
> >
> >> In Australia, I just found some information - one of the major
> >> electricity
> >> provide has a cost average for home appliances and electrical equipment.
> >>
> >> The estimate for the home PC is AUD 0.0394¢/hr.
> >>
> >
> > I just tried my clamp-on ammeter on my P4 2.8C with FX5200.
> >
> > DOS 1.9A @ 115V (I was running Seatools disk diagnostic...)
> > Linux 1.275A @ 115V (Xwindows desktop)
> > Win2K 1.353A @ 115V (Actually it is the same as Linux, as I had
> > to add my Win2K disk to do the test, and the
> > difference, is the power to run the disk.)
>
> That's interesting. Who would have thought that? DOS is a hog. :-)
>
> > If someone tries the above experiment with an "active PFC" supply,
> > then the measurements will be true without any correction being
> > needed. Unfortunately, PFC supplies aren't generally for sale
> > over here.
>
> PFC. Power factor corrected? I would expect any PS to give a lagging PF due
> to the transformers in it but many of the new supplies, smaller ones anyway,
> like I have in an external HD enclosure don't use much for transformers
> anymore. They use new technology that gets around the voltage reduction
> necessity. These could very well operate close to unity PF. We have test
> equipment at work and I should take a few minutes to test both.
>
> With electric motors you can add capacitors to bring PF closer to unity. In
> fact utilities will penalize large customers for poor power factor. It pays
> those customers to install banks of capaitors to correct the problem. The
> problem for the utility is that meters are designed to measure wattage (the
> part of AC current the customer is billed for) but the electric company
> facilities are providing volt amps, a higher (vector sum) quantity and they
> don't get paid for that difference.
>
> I would think one could add capacitors to the PS to correct PF but then why
> bother? It would not make significant difference to the power company and
> we're not billed for the extra current anyway.
>
> >
> > If the computer is in "Suspend to RAM" standby state, power
> > consumed will be about 10 watts - 0.010 x 0.057 x 24 x 365 = $5 per year.
> > My ammeter cannot measure this power, because the current waveform
> > is too distorted for accurate measurement (it is not even close
> > to a sinusoid).
>
> Hmmm. That may throw off the phase angle instruments too.
>
> >
> > Paul
> >

Xbitlabs has some info on the power factor:

http://www.xbitlabs.com/articles/oth...dology_10.html

This page and the one after it, show waveforms for no PFC,
passive PFC, and active PFC.

http://www.xbitlabs.com/articles/oth...odology_8.html

Paul

> >>
> >>
> >> "Bennos" <(E-Mail Removed)> wrote in message
> >> news:mDtef.18608$(E-Mail Removed)...
> >> > My config:
> >> > 400w power supply. (3 fans total in the case)
> >> > P4 3.0GHz 1G RAM PC at home.
> >> > Radeon 9800 Pro graphics card
> >> > 2 x HHDs (80G & 200G).
> >> >
> >> > I have always wondered what it costs (approx) to leave the home PC on
> >> > most
> >> > of the evening and nights. Usually, on the weekends the PC would be on
> >> > 12
> >> > hours Sat and again Sun.
> >> >
> >> > Anyone know roughly?
> >> >
> >> >

On Wed, 16 Nov 2005 17:39:45 -0600, "beach bum"
<sunshine&(E-Mail Removed)> wrote:


>> I just tried my clamp-on ammeter on my P4 2.8C with FX5200.
>>
>> DOS 1.9A @ 115V (I was running Seatools disk diagnostic...)
>> Linux 1.275A @ 115V (Xwindows desktop)
>> Win2K 1.353A @ 115V (Actually it is the same as Linux, as I had
>> to add my Win2K disk to do the test, and the
>> difference, is the power to run the disk.)
>
>That's interesting. Who would have thought that? DOS is a hog. :-)

It is to be expected on any modern system with bios and
operating system supporting ACPI, that is, at least when the
system is mostly idle inside the OS environment.


>
>> If someone tries the above experiment with an "active PFC" supply,
>> then the measurements will be true without any correction being
>> needed. Unfortunately, PFC supplies aren't generally for sale
>> over here.
>
>PFC. Power factor corrected? I would expect any PS to give a lagging PF due
>to the transformers in it but many of the new supplies, smaller ones anyway,
>like I have in an external HD enclosure don't use much for transformers
>anymore. They use new technology that gets around the voltage reduction
>necessity. These could very well operate close to unity PF. We have test
>equipment at work and I should take a few minutes to test both.
>
>With electric motors you can add capacitors to bring PF closer to unity. In
>fact utilities will penalize large customers for poor power factor. It pays
>those customers to install banks of capaitors to correct the problem. The
>problem for the utility is that meters are designed to measure wattage (the
>part of AC current the customer is billed for) but the electric company
>facilities are providing volt amps, a higher (vector sum) quantity and they
>don't get paid for that difference.
>
>I would think one could add capacitors to the PS to correct PF but then why
>bother? It would not make significant difference to the power company and
>we're not billed for the extra current anyway.

Here are some examples,
"Circuits for PFC with regards to mains filtering", ST app
note,
http://www.st.com/stonline/books/pdf/docs/3727.pdf

Taken on a system-by-system (personal PC) context, you're
right that it is not significant.