computer loads?

[steveng]

hey guys, we have a project upcoming, computer lab, i am unclear on the calculated load assesment, can someone give some input on this.

33 computers 3.0 amp ea = 99a
1 server ? =05a
2 laser printers 10ea =20a

#1. 124a, would you agree this is the calculated load?




#2. i put an amprobe on a circuit supplying 6 computers that were rated a 5.0 ea the reading was approx 8 amps, the computers were not being used , but were on ,why would it be so low if the 6 are rated at 5.0 amp each? anyone

 

[LarryFine]

hey guys, we have a project upcoming, computer lab, i am unclear on the calculated load assesment, can someone give some input on this.

33 computers 3.0 amp ea = 99a
1 server ? =05a
2 laser printers 10ea =20a

#1. 124a, would you agree this is the calculated load?




#2. i put an amprobe on a circuit supplying 6 computers that were rated a 5.0 ea the reading was approx 8 amps, the computers were not being used , but were on ,why would it be so low if the 6 are rated at 5.0 amp each? anyone

#2. In your own home, what's your range rated, and what's the load when it's 'on, but not being used'? The load is determined by maximum usage. In your computer, and you burning a DVD while doing a hard-drive backup, or just posting in forums?

#1. At 120v, all on one line, yes. Does that mean that's the feeder size? It depends. What's the feeder and panel specs? 1ph or 3ph? 120v, 120/208v, or 120/240v?

"More input, Stephanie!" ~ Johnny Five in Short Circuit

 

[ultramegabob]

"More input, Stephanie!" ~ Johnny Five in Short Circuit[/QUOTE]


Boy havent thought of that in a long time..... No disassemble!

 

[dereckbc]

i put an amprobe on a circuit supplying 6 computers that were rated a 5.0 ea the reading was approx 8 amps, the computers were not being used , but were on ,why would it be so low if the 6 are rated at 5.0 amp each? anyone

I will take a shot at this. 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.

 

[ELA]

Here is a graphic example. A laser printer current waveform.

The real small peaks ( ~1A) are what the unit draws most if the time when on -but idle. The large peaks (~8A) are what it draws when actually printing.
Ever notice the lights blink when your laser jet prints? Don't put too many of those on one circuit .
Scale is 5 amps per division.

 

[bhsrnd]

hey guys, we have a project upcoming, computer lab, i am unclear on the calculated load assesment, can someone give some input on this.

33 computers 3.0 amp ea = 99a
1 server ? =05a
2 laser printers 10ea =20a

#1. 124a, would you agree this is the calculated load?




#2. i put an amprobe on a circuit supplying 6 computers that were rated a 5.0 ea the reading was approx 8 amps, the computers were not being used , but were on ,why would it be so low if the 6 are rated at 5.0 amp each? anyone

I assume that your calculations include monitors for said computers. This will add to the load (CRTs moreso than LCDs). A rule of thumb I have heard before is you can take the nameplate rating from the PC/server and derate it by about 70% to get the true load of the equipment under nominal conditions. I have used that myself and all worked out fine. That, of course, is for computers or servers which don't have a 100% workload.

As others have already stated, it will depend largely on what peripherals are attached to the computer and what the computer itself is actually doing though highly unlikely you would ever reach the power supply's full load potential.

 

[gar]

080417-0921 EST USA

Some data at 120 to 122 VAC:

My 20" CRT monitor is initially about 0.8 A (screen is black), somewhat later all blue is 1.0 A, and all white 1.2 A.

My 20" LCD is about 0.3 A and it does not matter what screen color is because there is a constant backlight.

A computer --- during initiallization ranges from 0.5 to 1.5 A. After Microsoft settles down it jumps around with a mean of 1.25 A. Fluctuation is maybe +/-0.2 A. This has 2 hard disks.

Connect several computers to your current meter and have students do typical work. See what the peak loading is.

The laser printers are a different story. If there is just one I would use its peak current for load. If communication is to a laser printer via Ethernet, then I would put the printer on its own circuit. If directly coupled, parallel port or RS232, then I would put it and the computer it was connected to on the same circuit.

If all the computers are interconnected by Ethernet, then there is usually electrical isolation between the computers. This is because the Ethernet connection is isolated with transformer coupled to the computer. In this case being on separate circuits presents no normal problems with ground path voltages.

For some discussion on Noise and Grounding see my web site www.beta-a2.com .

ELA:

Your waveform is interesting. I have never look at this and would not have realized that the heater would be modulated on a 1/2 cycle basis. I might have thought phase shifted modulation.

With whatever sequence of cords I use to power my HP 5SiMX I see about a 2 V change when the heater turns on.

.

 

[Lxnxjxhx]

If the computers don't have soft start the initial current pulse is very high because the line at peak voltage sees a diode bridge connected to large capacitors; theoretically a temporary dead short.
(120 x 1.41)/0.5 ohm =~300 A.
I guess it's something like a tungsten load, but instead of a few dozen milliseconds the caps probably build up in only one or two cycles.

You need a scope and a current sense resistor. Using an isolation transformer will raise the source impedance above 0.5 ohm and probably soften the startup current and so mess up your measurement.
If you put the resistor on the high side and you don't have a battery operated scope you should use two scope channels in a differential mode (Channel 1 minus Channel 2).

 

[ELA]

080417-0921 EST USA

ELA:

Your waveform is interesting. I have never look at this and would not have realized that the heater would be modulated on a 1/2 cycle basis. I might have thought phase shifted modulation.

With whatever sequence of cords I use to power my HP 5SiMX I see about a 2 V change when the heater turns on.

.

Gar,
That surprised me also. I find this a very interesting waveform to demonstrate
a distorted sinewave and crest factor.

I noticed my lights dim (oscillate) every time I printed for several years. I always wanted to capture the current waveform and finally got around to it.

 

[zog]

"More input, Stephanie!" ~ Johnny Five in Short Circuit

Boy havent thought of that in a long time..... No disassemble![/QUOTE]

LOL! Jonny 5 was so cool in 1984 or whatever that was, so corny now.

 

[Lxnxjxhx]

Your waveform is interesting

Your waveform is interesting

You can get this by monitoring the current through a diode that is charging a capacitor. The cap only pulls current when the peak line voltage exceeds the (capacitor voltage + diode voltage drop) so you get sort-of-a-halfsine pulse.
This should be the current waveform for all switchmode power supplies.

 

[gar]

080417-1512 EST USA

Lxnxjxhx:

The current pulses are clearly 8.3 MS, 1/2 cycle. This is not the shape or duration of a current pulse charging a capacitor from a 60 Hz line.

Clearly these are gated full half cycles. In the middle of the picture is one full cycle. Followed by off, on, off, on, off, off.


ELA:

What probe or other connection did you use to measure current? I use a Fluke Hall device which automatically provides isolation, is good to DC, but not good at frequencies above maybe 10 kHz (I do not remember my tests so this is a rough value). I also have a 3 phase wattmeter using Hall devices and it is worthless on vector drive motor lead measurements, but is good on a normal three phase motor to estimate torque. For short time pulses I built my own shunt and had to solve the problem of the induced 1 turn error voltage that exists from the magnetic field of the current flowing thru the shunt.

The little short pulses at the +/-90 deg points (the peak of each half cycle) are probably the current charging the DC supply in the printer. While the various staggered half cycle pulses are the current to the fusing heater. I believe if the small short pulses were removed that the half cycle current pulses would look like very good sinewaves.

The short pulses are causing the sinewave distortion. Also I would expect the short pulses to remain when the heater is off.

.

 

[dbuckley]

As I've noted before, last time in relation to a coffee machine making lights blink, burst mode power modulation is the preferred technique of controlling AC power where possible. There are very good reasons for this, for example, the controlling element gets less hot, and there is much less Radio Frequency Interference generated, so you dont need interference supression components, a saving of weight, real estate and money. And, of course, you don't get harmonic currents generated, as you have whole cycles, so no triplen current addition.

Unfortunately our eyes dont like burst mode controlled lighting, so for that we have to use phase shift dimming, or some other continuous technique.

Nice scope shot.

 

[steveng]

thanks

thanks

I will take a shot at this. 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.

thanks guys for the replies, it is a lot more clear now, funny how, experience and knowledge trump ignorance. im amazed at the wealth of knowledge i have found here on this site.

thanks
steve

 

[steveng]

#2. In your own home, what's your range rated, and what's the load when it's 'on, but not being used'? The load is determined by maximum usage. In your computer, and you burning a DVD while doing a hard-drive backup, or just posting in forums?

#1. At 120v, all on one line, yes. Does that mean that's the feeder size? It depends. What's the feeder and panel specs? 1ph or 3ph? 120v, 120/208v, or 120/240v?

"More input, Stephanie!" ~ Johnny Five in Short Circuit

larry, sorry for the lack of details,
the panel is 120/208 3 ph, 3 phase loadcenter new install, fed from 30kva 480/120/208 3ph

you made a good point about the range, thanks .

 

[ELA]

You can get this by monitoring the current through a diode that is charging a capacitor. The cap only pulls current when the peak line voltage exceeds the (capacitor voltage + diode voltage drop) so you get sort-of-a-halfsine pulse.
This should be the current waveform for all switchmode power supplies.

Yes I am well aware of that.
Here is an better trace of the idle waveform showing the cap. charging and the resultant distortion caused.

What probe or other connection did you use to measure current?

Gar,
I used a hall effect probe good from DC to 100Kz.
I included the expanded "idle" waveform for clarity. As you said the larger fuser waveforms are full half cycles (in the previous trace).
What surprised me was the 1/2 cycle bursts rather than full cycle bursts.

 

[ITO]

hey guys, we have a project upcoming, computer lab, i am unclear on the calculated load assesment, can someone give some input on this.

33 computers 3.0 amp ea = 99a
1 server ? =05a
2 laser printers 10ea =20a

#1. 124a, would you agree this is the calculated load?

#2. i put an amprobe on a circuit supplying 6 computers that were rated a 5.0 ea the reading was approx 8 amps, the computers were not being used , but were on ,why would it be so low if the 6 are rated at 5.0 amp each? anyone

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.

 

[gar]

080418-1132 EST USA

ELA:

Using the half wave pulses I would assume that they monitor the number and balance these to avoid an average DC component.

.

 

[dereckbc]

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.

Ito while I like your approach, it is overkill and does not account for the nameplate raings are maximum peak values. What I am saying is you can load factor because not everyone will turn on the PC or use them at the same time. IMO derate to 50%

 

[ITO]

Ito while I like your approach, it is overkill and does not account for the nameplate raings are maximum peak values. What I am saying is you can load factor because not everyone will turn on the PC or use them at the same time. IMO derate to 50%

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.

I did a community college building about 3 years ago and it had a computer lab with all the computers running full time, maybe it was a fluke, I dunno... but if this were a design build I would not risk de-rating them.

Just my 2 cents...