My head is hurting trying to figure this out. I want to see if you guys get the same answer I did.

I upgraded my power supply. Using the "Kill-A-Watt" power meter, I measured the following statistics for each power supply under full load (old and new):

Old: 2.49A, 310VA, 0.65PF
New: 1.54A, 189VA, 0.98PF
Assume 124.0V for all calculations.
The cost of electric where I live is $0.09738/kWh.

Using a bunch of formulas, I want to calculate how much money on the power bill that using the new power supply will save me per month (30 days). This is theoretical because my PC is not always using the above amps, but this would be a "maximum benefit" kind of calculation. I got:
0.202 kW old
0.185 kW new
Savings? $12.24

What did you calculate?

My head is hurting trying to figure this out. I want to see if you guys get the same answer I did.

I upgraded my power supply. Using the "Kill-A-Watt" power meter, I measured the following statistics for each power supply under full load (old and new):

Old: 2.49A, 310VA, 0.65PF
New: 1.54A, 189VA, 0.98PF
Assume 124.0V for all calculations.
The cost of electric where I live is $0.09738/kWh.

Using a bunch of formulas, I want to calculate how much money on the power bill that using the new power supply will save me per month (30 days). This is theoretical because my PC is not always using the above amps, but this would be a "maximum benefit" kind of calculation. I got:
0.202 kW old
0.185 kW new
Savings? $12.24

What did you calculate?

Not quite...

You'll save 17 Watts * 24 hours * 30 days * $0.09738 = $1.91 per month.

Remarks:

Your Power is extremely cheap.

Why are You making it more difficult than needed? The Kill-a-Watt Meter can read out the power consumption directly in Watts. Save Yourself some headache and let the gadget do it for You :)

ChrisD

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My head is hurting trying to figure this out. I want to see if you guys get the same answer I did.

I upgraded my power supply. Using the "Kill-A-Watt" power meter, I measured the following statistics for each power supply under full load (old and new):

What did you calculate?

I also own a Kill-A-Watt. Good Choice.

But instead of working your way through power factor, VA, and amps, I suggest you make your head hurt less by clicking the Watt/VA button to the Watt display. You won't need to assume 124V (which would surprise me), as the meter measures watts, amperes, and their time relation (which is what Power Factor is about) and does that part of the math for you. Hint: you considerably overestimated your savings, as it is nearly your new total cost.

Obviously your new supply is power-factor corrected, and the old not, which is a good contribution on your part to better system power stability, and also reduces current loading and thus loss in all resistive paths uphill from your PC.

[edit: ChrisD beat me to the "use the watt" suggestion]

---

My head is hurting trying to figure this out. I want to see if you guys get the same answer I did.

I upgraded my power supply. Using the "Kill-A-Watt" power meter, I measured the following statistics for each power supply under full load (old and new):

Old: 2.49A, 310VA, 0.65PF
New: 1.54A, 189VA, 0.98PF
Assume 124.0V for all calculations.
The cost of electric where I live is $0.09738/kWh.

Using a bunch of formulas, I want to calculate how much money on the power bill that using the new power supply will save me per month (30 days). This is theoretical because my PC is not always using the above amps, but this would be a "maximum benefit" kind of calculation. I got:
0.202 kW old
0.185 kW new
Savings? $12.24

What did you calculate?

I can't calculate it easily, you're using VA and not Watts.

If I'm doing math correctly, the old power supply was right at 201 WATTS and the new one is 185 WATTS.

... but that's really only accurate to two significant figures because we don't have enough digits in the power factor.

Why do we have both?

Some of the power comes in from the power company, loops through and goes back out the wires. You have to size the conductors and the circuit breakers based on that value, and your UPS has to provide that much power. That's measured in volt-amps.

The watthour meter the utility uses only counts power that enters your house and stays. That's measured in watts.

The best way to measure this is to use the Kill-A-Watt in the "watthour" mode, where it actually tracks the number of watts consumed. Run for a known amount of time (i.e. 90 minutes) then multiply by 60/90 to get watt-hours actually consumed averaged over that 90 minutes.

My best guess is $9, plus or minus 30 percent.

---

My head is hurting trying to figure this out. I want to see if you guys get the same answer I did.

I upgraded my power supply. Using the "Kill-A-Watt" power meter, I measured the following statistics for each power supply under full load (old and new):

Old: 2.49A, 310VA, 0.65PF
New: 1.54A, 189VA, 0.98PF
Assume 124.0V for all calculations.
The cost of electric where I live is $0.09738/kWh.

Using a bunch of formulas, I want to calculate how much money on the power bill that using the new power supply will save me per month (30 days). This is theoretical because my PC is not always using the above amps, but this would be a "maximum benefit" kind of calculation. I got:
0.202 kW old
0.185 kW new
Savings? $12.24

What did you calculate?

So far, you calculated savings in kWh.... 12.24 kWh per month savings.

You forgot to apply cost per kWh... 12.24kWh/M * $0.09738/kWh = $1.19/Month

Here is my calculation factoring 365 days / year with 17W difference....

17W * 24 Hour/Day * 365 Day/Year / (12 Month/Year * 1000 W/kW) = 12.410 kWh/Month Savings...


12.41 * $0.09738/kWh = $1.21 / Month Savings

HTH...

---

Thanks for the lesson in linear algebra, guys. :) I'll have to run it for a while with the kWh mode for a certain few hours or so and then have a much better estimate (without the extra math).

I guess my savings isn't much at all...but this new power supply is a ton more efficient and has very low flutter/voltage variance compared to the old one. Hopefully, that will mean more life out of my PC.

What really surprised me was how much different the Amps were between the two power supplies. At idle, my PC uses less than 1 amp (0.96A), but using the old power supply, it uses 1.62A. According to your model, JDWhale, this PC costs me at most $13.29/month in electric. That's not bad at all.

And one last point. Since you all are telling me power is so cheap, I think I'll bring my old dual AMD CPU machine out of the closet as a dedicated cruncher. I'm sure it has a little life left in it. That thing at IDLE uses more electricity than my main PC at full load.

And one last point. Since you all are telling me power is so cheap, I think I'll bring my old dual AMD CPU machine out of the closet as a dedicated cruncher. I'm sure it has a little life left in it. That thing at IDLE uses more electricity than my main PC at full load.

[edit] You've got nearly 8.5% savings just with the new power supply... That's great! But don't bring back the "dead" just because of what others call "cheap electricity"...[/edit]

I've got a q6600 running diskless with onboard-graphics ASUS P5B-MX that is currently plugged into a Kill-A-Watt... Draw is 143W or $10 per month (a tad bit cheaper with my provider ($.0903/kWh), than yours)..

[edit2] This host generates 5 times production (RAC) when compared to my Prescott 3.2GHz that draws 160W. Newer, higher efficiency processors do much more computation for less cost! Combine with high efficiency power supplies for maximum impact! Compute Green!!! [/edit2]


BOINC Green Please!

And one last point. Since you all are telling me power is so cheap, I think I'll bring my old dual AMD CPU machine out of the closet as a dedicated cruncher. I'm sure it has a little life left in it. That thing at IDLE uses more electricity than my main PC at full load.

[edit] You've got nearly 8.5% savings just with the new power supply... That's great! But don't bring back the "dead" just because of what others call "cheap electricity"...[/edit]

I've got a q6600 running diskless with onboard-graphics ASUS P5B-MX that is currently plugged into a Kill-A-Watt... Draw is 143W or $10 per month (a tad bit cheaper with my provider ($.0903/kWh), than yours)..

[edit2] This host generates 5 times production (RAC) when compared to my Prescott 3.2GHz that draws 160W. Newer, higher efficiency processors do much more computation for less cost! Combine with high efficiency power supplies for maximum impact! Compute Green!!! [/edit2]

BOINC Green Please!

*** Quite the same here, ***

electra costs here €0,22 @ KWh.

Have 3 PC's, 1x Q6600 @ 2.4GHz, (MultiMediaPC) uses 210 Watt max, the CPU drops to 1580MHz, if NOT under full load =160Watt
1x Q6600 @ 2.74GHz, uses 180 Watt
PENTIUM D820 (Smithfield), with computes only a forth off Q6600 and uses
also 180 Watt's !
Better use an optimized version for that PC too .
With all (12) adapters (transformers:240V /17V /12V /5V , witch draw 110 Watt !

A total off 210W+180W+180W+110W=680Watt .0.68 x0.22== €0.15 x24 x30= €108,00/month

But it isn't ONLY thé amount off Watt's beïng used, due to the way, wich a PSU operates, current drawn from the
mains, gets (very) 'dirty', means a lot off spikes, pulses etc. are found in the
main supply.
Light 'dimmers'or 'softeners', also produce unwanted pulses and spikes.

---

*** Quite the same here, ***

electra costs here €0,22 @ KWh.

Have 3 PC's, 1x Q6600 @ 2.4GHz, (MultiMediaPC) uses 210 Watt max, the CPU drops to 1580MHz, if NOT under full load =160Watt
1x Q6600 @ 2.74GHz, uses 180 Watt
PENTIUM D820 (Smithfield), with computes only a forth off Q6600 and uses
also 180 Watt's !
Better use an optimized version for that PC too .
With all (12) adapters (transformers:240V /17V /12V /5V , witch draw 110 Watt !

A total off 210W+180W+180W+110W=680Watt .0.68 x0.22== €0.15 x24 x30= €108,00/month

But it isn't ONLY thé amount off Watt's beïng used, due to the way, wich a PSU operates, current drawn from the
mains, gets (very) 'dirty', means a lot off spikes, pulses etc. are found in the
main supply.
Light 'dimmers' or 'softeners', also produce unwanted pulses and spikes.

That's very interesting...that your Q6600 is 210W vs 160W between idle vs full. BTW, why do your two Q6600 PC's use different mounts of power but the faster chip uses less?

My dual AMD system has a much greater difference: 118W idle and 187W full. With the old power supply, the difference was even more: 123W idle and 202W full. I think the large difference could be explained by the fact that I purchased Opteron 248 HE chips instead of regular ones....HE = highly efficient. Still, it means when I'm not crunching, my system uses less than 120W...which is like having two incandescent lights on. :)

But it isn't ONLY thé amount off Watt's beïng used, due to the way, wich a PSU operates, current drawn from the
mains, gets (very) 'dirty', means a lot off spikes, pulses etc. are found in the
main supply.
Light 'dimmers'or 'softeners', also produce unwanted pulses and spikes.

Actually, it pretty much is just the number of watts being used if we're talking about paying for power. The typical meter doesn't react to pulses or spikes.

We also do not care about power factor if we're talking about the electric bill. There are other issues (like sizing UPSes) but they don't affect the bill.

---

Actually, it pretty much is just the number of watts being used if we're talking about paying for power. The typical meter doesn't react to pulses or spikes.

We also do not care about power factor if we're talking about the electric bill. There are other issues (like sizing UPSes) but they don't affect the bill.

So the electricity is "returned" via the neutral line in AC and is not counted by the meter? Perhaps I don't understand AC current perfectly. I thought the difference between VA and W was the wasted energy of the device.

Actually, it pretty much is just the number of watts being used if we're talking about paying for power. The typical meter doesn't react to pulses or spikes.

We also do not care about power factor if we're talking about the electric bill. There are other issues (like sizing UPSes) but they don't affect the bill.

So the electricity is "returned" via the neutral line in AC and is not counted by the meter? Perhaps I don't understand AC current perfectly. I thought the difference between VA and W was the wasted energy of the device.

Here's a wiki entry that I found that may explain things about Grounds and neutral: http://en.wikipedia.org/wiki/Ground_and_neutral

---

The T1 Trust, PRR T1 Class 4-4-4-4 #5550, 1 of America's First HST's

Actually, it pretty much is just the number of watts being used if we're talking about paying for power. The typical meter doesn't react to pulses or spikes.

We also do not care about power factor if we're talking about the electric bill. There are other issues (like sizing UPSes) but they don't affect the bill.

So the electricity is "returned" via the neutral line in AC and is not counted by the meter? Perhaps I don't understand AC current perfectly. I thought the difference between VA and W was the wasted energy of the device.

If the load on an AC line is NOT purely resistive the current and voltage waveforms are out of sync due to the reactive load. The reactive load can be capacitive or inductive. VA is Vrms * Arms (rms Root Mean Squared) and watts = rms(V * A) at the same point in time.
If the current waveform leads or lags the Voltage waveform by 20 degrees, when the voltage is at 90degrees it is at max (sine 90 =1) and the current waveform will be at 70 or 110 degrees sine = 0.94.

Word to remember is CIVIL, if load is Capacitive current (I) leads Voltage (V) and if voltage (V) leads current (I) then load is Inductive (L).

Actually, it pretty much is just the number of watts being used if we're talking about paying for power. The typical meter doesn't react to pulses or spikes.

We also do not care about power factor if we're talking about the electric bill. There are other issues (like sizing UPSes) but they don't affect the bill.

So the electricity is "returned" via the neutral line in AC and is not counted by the meter? Perhaps I don't understand AC current perfectly. I thought the difference between VA and W was the wasted energy of the device.

Exactly.

The wattmeter measures what you consume, and the power returned on the neutral is not consumed.

Here is why power factor matters:

While the power isn't consumed, it does flow through the wiring, so the wiring has to be sized according to the VA load.

So, let's say you have an office building, and you have five employees on a single 15 amp branch circuit. That limits each employee to a 360va load (at 120v, international users adjust appropriately).

If the power factor is 1, then that means a 360 watt load. If the power factor is a fairly awful 0.5, they're limited to 180 watts.

If 180 watts each isn't enough, you can either improve the power factor, or rewire.

Efficiency is a different measurement.

-- Ned

---

Actually, it pretty much is just the number of watts being used if we're talking about paying for power. The typical meter doesn't react to pulses or spikes.

We also do not care about power factor if we're talking about the electric bill. There are other issues (like sizing UPSes) but they don't affect the bill.

So the electricity is "returned" via the neutral line in AC and is not counted by the meter? Perhaps I don't understand AC current perfectly. I thought the difference between VA and W was the wasted energy of the device.

P.S. I've posted about this on other threads, but everyone who is concerned about this should know how to time their utility meter.

You need a decent stopwatch.

Time one rotation. If the meter is spinning too fast (bad!), time several rotations and divide by the number of turns.

Don't have a stopwatch? Use any timepiece with a second hand and count more turns to make up for inaccurate timing.

The basic formula is:

Power(in watts) = 3600 * Kh / seconds_per_rotation

Kh is constant. It is printed on the face of your meter.

-- Ned

---

Actually, it pretty much is just the number of watts being used if we're talking about paying for power. The typical meter doesn't react to pulses or spikes.

We also do not care about power factor if we're talking about the electric bill. There are other issues (like sizing UPSes) but they don't affect the bill.

So the electricity is "returned" via the neutral line in AC and is not counted by the meter? Perhaps I don't understand AC current perfectly. I thought the difference between VA and W was the wasted energy of the device.

P.S. I've posted about this on other threads, but everyone who is concerned about this should know how to time their utility meter.

You need a decent stopwatch.

Time one rotation. If the meter is spinning too fast (bad!), time several rotations and divide by the number of turns.

Don't have a stopwatch? Use any timepiece with a second hand and count more turns to make up for inaccurate timing.

The basic formula is:

Power(in watts) = 3600 * Kh / seconds_per_rotation

Kh is constant. It is printed on the face of your meter.

-- Ned

Ned,
Thats fine for you and those living in the twentieth century, but my electricity meter is about the size of a packet of cigarettes and has flashing led, and numerical display.

Andy

Actually, it pretty much is just the number of watts being used if we're talking about paying for power. The typical meter doesn't react to pulses or spikes.

We also do not care about power factor if we're talking about the electric bill. There are other issues (like sizing UPSes) but they don't affect the bill.

So the electricity is "returned" via the neutral line in AC and is not counted by the meter? Perhaps I don't understand AC current perfectly. I thought the difference between VA and W was the wasted energy of the device.

P.S. I've posted about this on other threads, but everyone who is concerned about this should know how to time their utility meter.

You need a decent stopwatch.

Time one rotation. If the meter is spinning too fast (bad!), time several rotations and divide by the number of turns.

Don't have a stopwatch? Use any timepiece with a second hand and count more turns to make up for inaccurate timing.

The basic formula is:

Power(in watts) = 3600 * Kh / seconds_per_rotation

Kh is constant. It is printed on the face of your meter.

-- Ned

Ned,
Thats fine for you and those living in the twentieth century, but my electricity meter is about the size of a packet of cigarettes and has flashing led, and numerical display.

Andy

Most Utilities in California still use that type of Electric Meter and I've seen a lot of them installed(The type with a rotating wheel in them), Their pretty standard here. Kh? What's that? I know of KWh of course, But not Kh.

---

The T1 Trust, PRR T1 Class 4-4-4-4 #5550, 1 of America's First HST's

Actually, it pretty much is just the number of watts being used if we're talking about paying for power. The typical meter doesn't react to pulses or spikes.

We also do not care about power factor if we're talking about the electric bill. There are other issues (like sizing UPSes) but they don't affect the bill.

So the electricity is "returned" via the neutral line in AC and is not counted by the meter? Perhaps I don't understand AC current perfectly. I thought the difference between VA and W was the wasted energy of the device.

P.S. I've posted about this on other threads, but everyone who is concerned about this should know how to time their utility meter.

You need a decent stopwatch.

Time one rotation. If the meter is spinning too fast (bad!), time several rotations and divide by the number of turns.

Don't have a stopwatch? Use any timepiece with a second hand and count more turns to make up for inaccurate timing.

The basic formula is:

Power(in watts) = 3600 * Kh / seconds_per_rotation

Kh is constant. It is printed on the face of your meter.

-- Ned

Ned,
Thats fine for you and those living in the twentieth century, but my electricity meter is about the size of a packet of cigarettes and has flashing led, and numerical display.

Andy

Most Utilities in California still use that type of Electric Meter and I've seen a lot of them installed(The type with a rotating wheel in them), Their pretty standard here. Kh? What's that? I know of KWh of course, But not Kh.

You'r correct, about measuring, with an 'old'type off KWh meter.

Me too have a 'modern' electricmeter , but you ONLY pay for actual amount off
Watts per hour. That is V x A= Watt .

But you have to have a Watt or KWatt meter and measure each HOST/PC, or measure them all, or the current in Amps, in order to calculate the amount off Watt or KWatt. The 'Volts', is 'fixed', 110- 130 or 220 -240, in Europe.And a Voltmeter to check that

But with a lot off hosts its difficult to measure the current, because you have
to measure it between powerline and PC/host. Then multiply the Amps x Volt.

PC's are also quite different in Power Saving Settings and/or behavior.
Measuring AC current isn't as easy as DC current.

---

Actually, it pretty much is just the number of watts being used if we're talking about paying for power. The typical meter doesn't react to pulses or spikes.

We also do not care about power factor if we're talking about the electric bill. There are other issues (like sizing UPSes) but they don't affect the bill.

So the electricity is "returned" via the neutral line in AC and is not counted by the meter? Perhaps I don't understand AC current perfectly. I thought the difference between VA and W was the wasted energy of the device.

P.S. I've posted about this on other threads, but everyone who is concerned about this should know how to time their utility meter.

You need a decent stopwatch.

Time one rotation. If the meter is spinning too fast (bad!), time several rotations and divide by the number of turns.

Don't have a stopwatch? Use any timepiece with a second hand and count more turns to make up for inaccurate timing.

The basic formula is:

Power(in watts) = 3600 * Kh / seconds_per_rotation

Kh is constant. It is printed on the face of your meter.

-- Ned

Ned,
Thats fine for you and those living in the twentieth century, but my electricity meter is about the size of a packet of cigarettes and has flashing led, and numerical display.

Andy

Most Utilities in California still use that type of Electric Meter and I've seen a lot of them installed(The type with a rotating wheel in them), Their pretty standard here. Kh? What's that? I know of KWh of course, But not Kh.

Kh is a constant. It's basically the gear ratio between the hands on the meter and the disk -- that ratio is chosen of course so that the hands read in KWh.

---

Actually, it pretty much is just the number of watts being used if we're talking about paying for power. The typical meter doesn't react to pulses or spikes.

We also do not care about power factor if we're talking about the electric bill. There are other issues (like sizing UPSes) but they don't affect the bill.

So the electricity is "returned" via the neutral line in AC and is not counted by the meter? Perhaps I don't understand AC current perfectly. I thought the difference between VA and W was the wasted energy of the device.

P.S. I've posted about this on other threads, but everyone who is concerned about this should know how to time their utility meter.

You need a decent stopwatch.

Time one rotation. If the meter is spinning too fast (bad!), time several rotations and divide by the number of turns.

Don't have a stopwatch? Use any timepiece with a second hand and count more turns to make up for inaccurate timing.

The basic formula is:

Power(in watts) = 3600 * Kh / seconds_per_rotation

Kh is constant. It is printed on the face of your meter.

-- Ned

Ned,
Thats fine for you and those living in the twentieth century, but my electricity meter is about the size of a packet of cigarettes and has flashing led, and numerical display.

Andy

Most Utilities in California still use that type of Electric Meter and I've seen a lot of them installed(The type with a rotating wheel in them), Their pretty standard here. Kh? What's that? I know of KWh of course, But not Kh.

You'r correct, about measuring, with an 'old'type off KWh meter.

Me too have a 'modern' electricmeter , but you ONLY pay for actual amount off
Watts per hour. That is V x A= Watt .

But you have to have a Watt or KWatt meter and measure each HOST/PC, or measure them all, or the current in Amps, in order to calculate the amount off Watt or KWatt. The 'Volts', is 'fixed', 110- 130 or 220 -240, in Europe.And a Voltmeter to check that

But with a lot off hosts its difficult to measure the current, because you have
to measure it between powerline and PC/host. Then multiply the Amps x Volt.

PC's are also quite different in Power Saving Settings and/or behavior.
Measuring AC current isn't as easy as DC current.

Measuring AC current isn't that tough, any good DMM will do it, and because it's AC and not DC, you can even read it inductively.

The problem is that the typical household has a whole bunch of loads, and it can be really hard to isolate them.

The good news is: sometimes you can isolate loads, and you can verify that your Kill-A-Watt is accurate, then use it for the individual loads.

---