computer loads?

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ITO said:
Take that idea a step further and they are all continuous loads too.
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I have to disagree with the statement in bold. A computer is a nonlinear load and therefore its load fluxuates. A motor is a continuous load, not a computer.

For example, you have 100 computers in a room and decide to put a CD in the CDROM drive of only one of them. For the purposes of this example the load of that one PC will be greater than the other 99 computers because it's now reading that CD and requires additional power to use the CDROM.

I agree that using the nameplate rating is overkill. Even if all 100 computers in the latter example are turned on they will never use 100% of their nameplate rating.
 
thanks ITO!

thanks ITO!

ITO said:
Wow ask an engineer a simple question and there is no telling what you will get…

What you are looking for is a load calc, and when doing a simple load calculation for NEC leave the amprobe in your tool bag, and use the VA on the name plates, if it does not list VA or W, and only gives A, then convert it over for your calculation. Measured current of any device should only be used in a load calc IF there is no other way to determine the value.

33 Computers- 400va each (note this is taken from the 400va power supply which is middle of the road, sometimes I see 500va and generally use 500va for all my load calcs for a safety factor)

33 Monitors – 240va each (based on name plate of my monitor)

1 Server- 600va each (continuous load and may have duel hot swap power supplies)

2 laser printers 1100va each (this is pretty big laser printer)

33 x 400 = 13,200 va
33 x 250 = 8,250 va
(1 x 600)x 1.25 = 750 va
2 x 1100 = 2,200va

Total = 24,400va / (service voltage)

Here is your simple answer- 120V = 203.33a or 208V = 68.2a

Now what about heat load? If you add all that to a computer room will the A/C be able to handed the added heat load or is it already marginal? Typically when adding that many computers we forget how much heat they add, and any added cooling equipment should be part of your load calc.
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You rescued my thread, saved the day, if i could, id buy you lunch,
you are ok in my book, thanks again!:D
do you have a formula to calculate the heat load generated by 33 computers on at same time, that is something we deal wiith, but never do anything about, we do get lots of complaints from some areas with many computers.

would you put 4 per circuit?
 
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080418-2014 EST USA

ITO:

In my post #7 of 080417-0921 I measured a running average current of about 1.25 A at 120 V. This was done with a Fluke that measures the average value of the full-wave rectified sine wave, assumes it is a sine wave, then uses a fudge factor of 0.707/0.637 to convert the average value to display the RMS value of the sine wave.

This computer has two hard drives and relatively normal other items. This is in the 1.5 gig range for the processor. The power supply nameplate lists 100-127 V approx 4.0 A. No way am I going to draw 4 A.

An average per machine of the RMS current measured while a typical number of computers are being used should be a much better rating to use than what is on the power supply label. Then you add a reasonable safety margin to this. This is only good statistical system engineering.

Assuming the load can be considered resitive 4 A at 120 V is 480 W. Nothing like that is being dissipated in my computer example. Since the input to the power supply consists of a full wave rectifier connected to a capacitor, there are short input current pulses in phase with the input voltage. The average of this full wave rectified input current times the peak of the voltage is probably a reasonable estimate of input power.

Sizing your supply for these computers should be based on the RMS current to the computers using a reasonable estimate of their average requirment.

An important consideration is what will happen in the future if new computers with higher speed processors replace the current computers. These future computers can be expected consume more power. The central processor is getting to be a major power hog as processing speed increases.

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Power factor issue in the case of switching supplies is really more one of harmonic content than current and voltage being out of phase like motor loads. If the computers' power supplies are the older non PFC corrected ones, watch out for your feeder neutral current. Because they are higher order harmonics they don't cancel like resistive loads. There are stories about melted neutrals and even fires due to this. Since Europe cracked down on the nasty switchers a while back many computer power supplies (even those sold here) are a lot cleaner now. I've been working with AC/DC power factor correction recently (after protesting about not being given three phase). It consists of a high frequency boost converter placed between the rectifier bridge and the filter capacitors. They modulate the pulse width to force the current to stay in phase with the input voltage. More importantly the boost allows it to continue conduction through most of the cycle. When it's all working right the power supply should appear more like a resistive load.
 
I still catch grief for using averaging clamps compared to RMS clamps. I have used the method of using these two different clamps for harmonic issues. I did not have the advantage of using a PQA. You can use an averaging clamp compared to an RMS clamp for this purpose. Just need a pencil and paper to figure this out. This method won't show you what sequence or level of harmonic trouble you find, but it will show you if you have harmonic troubles at hand.

Anyone, not you electricalperson, know of this method? Don't answer this method I have told you of Chris, let others answer. You don't need a $2K PQA to show you this. An averaging clamp can determine this compared to an RMS clamp.

Anyone want to bite?????
 
ITO said:
Hmmm, I assumed since it was a computer lab, all the computers would be on at the same time. Take that idea a step further and they are all continuous loads too.
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They will likely be all on at the same time but they will never draw nameplate current.


dereckbc said:
First the nameplate rating is the absolute max under abnormal conditions. The only time it could ever reach that limit is when the PC has every optional bay filled up with drives (CD, DVD, Floppy, Tape, etc.. plus the internal hard drive) and everything is running at the same time which is a rare occurance if ever. The chance of all the PC doing this at once is impossible.
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I agree, how could I not. :)

The nameplate on a PC is the rating of it's power supply under full load, many PCs do not have enough equipment in them to reach the power supplies full load and even if they do have all the bells and whistles they will not all run at the same time.

Do some amprobe readings on real circuits and compare that with the name plate, they will not even be close.
 
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080419-0646EST USA

To visually see why you have a high neutral current in a Y system with an equal distribution of loads look at the scope picture ELA provided at post #16.

Copy this image and displace it 120 degrees and superimpose it with the orignal. Do the same for 240 degrees. There is no cancelation of these pulses. Because of the shortness of the pulses none overlap. Thus, the RMS current in the neutral is sq-root of 3 times the current in any one leg. (edit) I added the missing sq-root prefix on 3. The average power in the neutral increases by 3, and thus the average RMS by the sq-root. (end edit)

Do the same superposition of three sinewaves and what is the result?

(edit) Note: In a single phase center tapped transformer with equal loads the short pulses will will essentially cancel each other in the neutral. (end edit)

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080419-1158 EST USA

Lxnxjxhx:

Yes.

The instantaneous power is p = R*i squared for a time and current invariant resistor. To get average power of a steady state periodic waveform you integrate instantaneous power over one period of the waveform, and divide by the time period. This applies for any waveform.

For a non-linear load use p = e*i for the instantaneous power.

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http://en.wikipedia.org/wiki/Confidence_interval

http://en.wikipedia.org/wiki/Confidence_interval

IMHO, this is a probability problem. If the breaker is sized so that it is desired that it trips once/week (or once/month or once/year) then it is optimally sized. If you want to never trip you go with the statistically-unlikely max values and the high cost.

Probably POCO has the likelihood values that you need to do this for a roomful of computers.
The NEC has this implied in their loading tables. If you follow their tables I guess only one home or five homes out of 100 will occasionally trip breakers. But, before the table can be generated the level of confidence [50%, 95%, 99.9%] has to be decided by some knowledgeable VIP.

E.g., levees, downspouts, sewage systems and landscapes are designed to withstand a 1, 2, 5, 10, 20, 50, or 100 year floods or "events".
You only need 1 year to 50% confidence level? It's Home Depot.
You need 100 year to 99.9% confidence level? It's time to raise taxes or issue bonds.
 
080419-1523 EST USA

Lxnxjxhx:

It is not too hard to come up with a reasonable maximum input current to a computer. Start one program to write to a CD drive, another to do a useless continuous access to floating point, and another to copy from one hard disk to another. This will probably saturate the CPU, somewhat exercise 3 drives at once. In most case this is probably substantially below power supply specified current rating.

Using a value such as this with a 25% add on factor will provide reliable operation.

Note: designing for CRTs is not likely what you do today, but obviously one designs for the type of display that will be used. As I indicated earlier the current load of an LCD is virtually constant relative to what is displayed.

On your question about a vector voltmeter. It will need to be frequency specific or its readings may not mean much. With two correlation detectors, in other words a true multipler, and two reference signals 90 degrees apart you can extract the resistive and reactive components, and identify if the net reactive is inductive or capacitive. If you used this to measure the input current to a capacitor input filter it would be largely resistive and only somewhat reactive at base frequency.

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For discussions sake only…

I stand my preliminary calc, once full details of the scope are known a full calc could be done and then appropriate demand factor applied, but to be applying any kind of demand factor based on meter readings of similar utilization equipment in your office or home, is kinda putting the cart before the horse.

Article 100 pretty clearly defines “continuous load” and nowhere in the definition does it say that it only applies to motor loads. The argument could be made that a computer would not likely be operated at maximum for 3 hours or more, and I did not think it was in this application, however the monitors are likely to be operated at maximum current for 3 hours or more, so technically they are continuous loads.
 
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demand factor

demand factor

That's the word I was looking for, except that here I think you have the average steady state load, an occasionally high load when many computers happen to be on at the same time, and finally a high surge if several computers are turned on at the same instant. I don't think it would be an issue if non-soft-start computers didn't have such a high initial current.
 
ITO said:
Article 100 pretty clearly defines “continuous load” and nowhere in the definition does it say that it only applies to motor loads.
Click to expand...

If you really want to get into the definitions the load has to be at the maximum value for more then 3 hours, a computers load is always changing meaning it does not qualify as a continuous load.
 
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