Computer Power Usage

Hello,

Today I plugged in a power meter I bought a while ago, and decided to check out how much power my old Pentium 233MMX was using, here are the results (UK Power):

Normal Operation

244 Volts, 0.21 Amps, 31 Watts, Power Factor = 0.60

But if I disconnect the hard drive, I get:

250 Volts, 0.17 Amps, 41 Watts, Power Factor = 1.00

How could it be using more power when the hard drive is disconnected?

41 Watts is how much power it is using, right? Which one (if any) is accurate?

Thanks!

paul

Reply to
paul
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If that is the case - that it can't handle switching power supplies, would the data be completely wrong? I'm trying to work out how long I could power it from a 12v battery on a full charge, but need to know how much power it is drawing. Is there a standard/cheap way of working out exactly how much power the pc is using? What equipment is needed?

Cheers,

paul

Reply to
paul

Just a thought, but I wonder if your power meter is able to handle switching power supplies (which are used by all computers). The power meter may be expecting simple resistive loads, or maybe motors where the current and voltage are a bit out of phase, but not be able to handle loads where the waveforms are chopped up by a switcher.

Best regards,

Bob Masta dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis

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Home of DaqGen, the FREEWARE signal generator

Reply to
Bob Masta

This is a very good question, but I unfortunately have no answer. Note that if you were actually measuring the current from the battery the task would be much simpler, since you could assume that the voltage was pretty much a constant during the current draw. Then all that would be needed would be to integrate the total current in amp-hours. But if you already had the battery, the inverter, and the computer you could skip all that and just time it directly!

I think the power meter has to integrate the product of instantaneous voltage times instantaneous current, averaging over one or more cycles of the mains frequency. I don't know what typical commercial power meters do. Anyone else have any info on that?

Best regards,

Bob Masta dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis

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Home of DaqGen, the FREEWARE signal generator

Reply to
Bob Masta

I wanted to work out the power usage, before obtaining the battery & inverter, so I would know exactly what size/types I'd need to get :)

Regards,

paul

Reply to
paul

Based on my limited knowledge (having just finished an AC circuit theory course at my local community college), I believe there are 3 ways to measure the power drawn by a load from an alternating voltage source:

  1. Apparent power. This is RMS AC Volts x RMS AC Amps. It is reported not in Watts, but in "VA" = Volt-Amperes.
  2. True power. This is the resistive component of Apparent Power, and is reported in Watts. If the load has no capacitive or inductive elements, then all the load is resistive. True power is the power that is dissipated in heat through your load.
  3. Reactive power. This is the inductive and capacitive (= reactive) component of Apparent Power, and is reported in "VAR" = Volt-Amperes Reactive. Reactive power isn't dissipated in heat, but cycles back and forth between you and the power company as alternating current flow.

Apparent power is the vector sum of True Power and Reactive Power: the square root of (True Power**2 + Reactive Power**2).

The ratio of True power to Apparent power is the Power Factor.

Motors (hard drives, fans, etc.) are typically inductive loads.

Now let's look at the figures your meter is reporting:

244 V x .21 A = 51.2 VA apparent power. Of that, the 0.60 Power Factor x 51.2 VA = the 31 Watts true power being consumed by the resistive loads in your computer and hard drive. At the same time, there is a reactive (inductive and capacitive) power draw of sin(arccos 0.60) x 51.2 VA = 41 VAR. The square root of (true power**2 + reactive power**2) is indeed 51.4 VA, which matches your figure of 51.4 VA.
250 V x .17 A = 42.5 VA apparent power. Since the Power Factor is 1.00, all of that apparent power is due to the resistive loads in your computer and hard drive; the true power is 42.5 Watts, or approximately the 41 Watts you reported. There is little or no inductive and capacitive load. That makes sense, because the hard drive motor was a major part of the inductive load when it was connected.

So, you can see that the apparent power draw does decrease, from 51.2 VA to

42.5 VA, when you disconnect the hard drive.

Can somebody else explain why the true power (resistive load) -increases- from 31 Watts to 41 Watts when the hard drive is disconnected?

Wayne

Reply to
Wayne Farmer

Back in the ancient days, and thus this information may or may not apply to the current day, computer power supplies had big heavy expensive iron transformers. When the slightly more modern, but still ancient, "pc was invented" transformers were replaced by lighter cheaper and often much poorer quality switching supplies. The widespread knowledge of those old days was that you NEVER turned on one of these supplies without some load, typically having at least a 5" full height hard drive attached to do this.

Not having a load on a cheap low quality switching supply can lead to wild and unhappy oscillation within the supply, often coming to an end in a few seconds when you go and get a replacement for your supply.

Perhaps the supply operating without a load is demonstrating its unhappy state by increasing dissapation by 35%+. Adding resistive load in small increments and seeing whether the true power decreased before increasing might help test this guess.

I once destroyed one of these supplies just by switching it off and then back on too quickly.

Reply to
Don Taylor From:

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