power cosumption for my pc ? UPS config needs...


R

raghav

Hi,

I have this motherboard installed with 2 GB RAM, a 250 GB Seagate
Baracuda hard disk and LG DVD RW drive.

My power-supply says its 450W. My UPS has a rating of 650 VA (should
be around 350 watts).

My display is also plugged into the UPS, which is a LG flatron 19"
TFT.

My UPS is not able to support my system during a power fluctuation.

my question is, how do I measure as to how much is my system using
from the mains ? That seems to be the only way I could decide on my
power usage and then only I can choose the right UPS configuration for
my system.

Please suggest...

regards
raghav..
 
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D

david

Hi,

I have this motherboard installed with 2 GB RAM, a 250 GB Seagate
Baracuda hard disk and LG DVD RW drive.

My power-supply says its 450W. My UPS has a rating of 650 VA (should
be around 350 watts).

My display is also plugged into the UPS, which is a LG flatron 19" TFT.

My UPS is not able to support my system during a power fluctuation.

my question is, how do I measure as to how much is my system using from
the mains ? That seems to be the only way I could decide on my power
usage and then only I can choose the right UPS configuration for my
system.

Please suggest...

regards
raghav..

To get an accurate power measurement, you'll need to measure both the AC
RMS voltage, and the AC RMS current. To measure the current, you'll need
to connect AC ammeter in series with the hot side of the line. Measure
the RMS current value with your system operating. Then, measure the AC
voltage between hot and neutral. Multiply those two readings, and that
is the power that your system is drawing (in VA, or volt-amps). Multiply
that number by the power factor (typically 70% for a switching power
supply) and that result is the real power in watts that your system is
using. The difference between the total power (in VA) and the real power,
is the reactive power, which is lost as heat in the power supply.


Or you could just go buy the next bigger sized UPS.
 
F

frischmoutt

david said:
[...]

the RMS current value with your system operating. Then, measure the AC
voltage between hot and neutral. Multiply those two readings, and that
is the power that your system is drawing (in VA, or volt-amps). Multiply
that number by the power factor (typically 70% for a switching power
supply) and that result is the real power in watts that your system is
using. The difference between the total power (in VA) and the real power,
is the reactive power, which is lost as heat in the power supply.

Not exactly. The form factor of an AC/DC power supply is not good. The
current if far away a sine wave, even distorted. Depending on the load, it
might be short peaks, approaching the sine for higher loads. The reason for
that is that the mains, after rectification is charging a capacitor. The
current may only flow on the peaks of the voltage, when it exceeds the value
stored in the filtering capacitors. An usual Am-meter won't be able to
measure such currents because it measures the mean value and it display the
RMS calculated value FOR A SINE wave.
Unless you're using a true RMS Am-meter (reasonnably expensive instrument),
you'll get important errors. You, also can't compute a cos Phi because of
the bad and changing wave form.
Your rationale, however, allows to provide an approximation that may be
sufficient for the use of the poster.
Regards
 
D

Dave

raghav said:
Hi,

I have this motherboard installed with 2 GB RAM, a 250 GB Seagate
Baracuda hard disk and LG DVD RW drive.

My power-supply says its 450W. My UPS has a rating of 650 VA (should
be around 350 watts).

My display is also plugged into the UPS, which is a LG flatron 19"
TFT.

My UPS is not able to support my system during a power fluctuation.

I wouldn't expect it to. Your monitor will use about 50W. At most, your
computer might draw about 200W. So that's ~250W maximum. But, that is
before you figure in the power loss due to the power supply not being 100%
efficient. If we assume 60% efficiency (hopefully it's better than that,
but it might not be), then your system is drawing a maximum of about 370W.

I wouldn't expect that to run off of a UPS with a 650VA rating, as this is
peak power of 650W and nominal power of about half that.

For a UPS, you should find one with a minimum 1000-1500VA rating. 1000VA
will be about the minimum that will allow you time to do a safe shutdown.
At best, your UPS will only run your system for several minutes, unless you
throw enough money at it to get an extremely high VA rating approaching
overkill. :) -Dave
 
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D

david

david said:
[...]

the RMS current value with your system operating. Then, measure the AC
voltage between hot and neutral. Multiply those two readings, and
that is the power that your system is drawing (in VA, or volt-amps).
Multiply that number by the power factor (typically 70% for a switching
power supply) and that result is the real power in watts that your
system is using. The difference between the total power (in VA) and the
real power, is the reactive power, which is lost as heat in the power
supply.
Not exactly. The form factor of an AC/DC power supply is not good. The
current if far away a sine wave, even distorted. Depending on the load,
it might be short peaks, approaching the sine for higher loads. The
reason for that is that the mains, after rectification is charging a
capacitor. The current may only flow on the peaks of the voltage, when
it exceeds the value stored in the filtering capacitors. An usual
Am-meter won't be able to measure such currents because it measures the
mean value and it display the RMS calculated value FOR A SINE wave.
Unless you're using a true RMS Am-meter (reasonnably expensive
instrument), you'll get important errors. You, also can't compute a cos
Phi because of the bad and changing wave form.
Your rationale, however, allows to provide an approximation that may be
sufficient for the use of the poster. Regards

Yes, exactly.

A decent true RMS multimeter, such as a Fluke 175, will measure a
distorted waveform caused by a switching power supply with acceptable
accuracy.
 

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