Need advice re. power load

Hi,

I'm in a bit a of a bind, or I think that I am, and hoping that someone here can tell me if I am or not :(....

The situation is that we got a new Sun server (a V880) in for a fairly short evaluation period. This was suppose to go into a new area, and there was some mis-communication between Sun and another guy at work, and then he left, and I'm stuck with getting things going now :(.

What happened was that he had asked Sun what kind of power was needed awhile, and they told him that their rack/power distribution (PDS) required 2 240V circuits @ 30 Amps each, with L60-30 receptacles/outlets, so that's what he went and got provisioned in the room.

Unfortunately, it turns out that Sun isn't able to provide their rack or their PDS, so I'm now stuck trying to figure out how to power the server, at least temporarily.

We're trying, but having a bit of a problem getting a proper PDU quickly, and meanwhile the Sun hardware arrived today. We'll be unpacking and inventorying tomorrow, but there's already a bit of pressure on me to try to power them up.

It turns out that besides the L6-30 outlets that we have, which we can't use until we get a PDU, we have some standard 120VAC outlets around the room. According to our electrical contractor, all the 120VAC outlets in that one room are wired back to a single 20Amp breaker.

I've been reviewing the following article on Sun's website:

formatting link
According to the above page, the "Maximum Configuration" for an 880 with

8 900MHz CPUs, and 12 36G drives draws 1923 Watts, or ~16 Amps @ 120V.

We have only 6 drives, so it looks like the system power draw would be about 1755 Watts, or ~14.6 Amps @ 120V. This would be total, for the 2 power supplies in a single 880 (the 880 can have up to 3 power supplies, but needs at least 2 minimum to operate), so each power supply would be drawing ~8 Amps @ 120V from a single 120V outlet, I think.

So, if I'm calculating correctly, and if the breaker for the room is really 20 Amps, then will I be ok plugging the two power supplies from ONE 880 into two of the 120V outlets, and be able to power the 880?

I'm assuming that if the breaker is really 20Amps, that all the wires to the outlets will be rated for 20 Amps? Is this correct?

Also, what is the "normal" power/amp rating for a 120V outlet (this is in an office setting)?

I *KNOW* that this will be "cutting it close", and I've told people at work that I want to wait for the PDUs to come in, but I'd really like to get some feedback on this, as soon as possible. I don't want to do anything unsafe!!

My apologies for this, but I hope that someone who has more experience can respond!!

Thanks, Jim

Reply to
ohaya
Loading thread data ...
040802 2115 - ohaya posted:

formatting link
4

The outlets in the room are for general purpose, such as desk light, computer, printer, electric shaver, vacuum cleaner, and etc. The Sun equipment would be considered to be a fixed appliance and would require their own circuits properly installed. Realizing that you are just trying them out for a short period -- maybe a couple of days -- try them on the general outlets. If the breaker goes out, then you will have to consider a separate set of circuits for the equipment even for tryout. If the equipment is going to use all of the power to the general outlets, even temporarily, then you can't have a desk light, computer, printer, electric shaver, vacuum cleaner, and etc. plugged into these outlets.

Reply to
indago

Indago,

Thanks.

Yes, I'm only intending to do this, if at all, until the PDUs get in, and yes, if we do this, I'll have to make sure that no one plugs anything else into any of the outlets in that room. We'll also make sure that the equipment is powered down and unplugged from the outlets when not in use and attended.

What I was (and am) a bit worried about is that it seems like a pretty heavy load, and doesn't leave much leeway on the power/current. If it does trip the breaker, is there any danger in doing so?

Could it damage the breaker, any of the electrical wiring, or the outlet?

Also, is there any possibility that if the breaker trips, that might cause some kind of "kick" or surge back into the server itself because of (maybe) arcing at the breaker if it trips?

I haven't purposely (at least not knowingly :)) gotten into a situation where I was tripping a circuit breaker through over-loading a circuit before (not AC anyway), so I want to get an idea of what the risks might be (damage the breaker, cause an electrical fire, etc.).

Better safe than sorry...

Thanks again, Jim

Reply to
ohaya

On Mon, 02 Aug 2004 21:15:20 -0400, ohaya put forth the notion that...

Normal 120 volt receptacles are rated for 15 amps, but can be used on a

20 amp circuit as long as the circuit feeds more than one receptacle. The assumption is that you won't be drawing more than 15 amps from any one receptacle. A 20 amp circuit breaker will not provide a full 20 amps continuously without tripping. The maximum continuous load you can put on a 20 amp breaker is 80%.
Reply to
Checkmate

Checkmate,

Thanks! Can you point to any place that would document what you're indicating?

It sounds like we'd be ok on the "15 amps from any one receptacle", but the "continuous current" seems like it's even closer than I first thought?

80% of 20 Amps = 16 Amps, whereas accord "The measurements are based on actual configurations running SunVTS[TM] (considered to be representative application)."

This seems to imply that what is running on the system would affect the power consumption :(, which makes this even more "iffy", i.e., something like if we put a heavy load on the hard drives, the system might start drawing more current and then pop the breaker.

As I indicated in my original post, this is NOT an idea that I'm advocating, or want to do, but there're a couple of people who've been telling me that they want me to go ahead with it. In fact, they've gone so far as to suggest that if I'm worried about it, we could plug one of the power supplies in at a time, to avoid inrush current from both supplies when the system is powered on.

So if you, or anyone could point to any where (e.g., some code or specification) that would indicate this, it would be a great help.

Thanks again, Jim

Reply to
ohaya

Hi,

I've found a reference at UL (under "Use" section):

formatting link
I think that should cover what I was looking for!!

Thanks!

Jim

Reply to
ohaya

formatting link

not so long ago i had one of these 15A "wonders" smolder in the wall when a copier was plugged into it. i replaced with it a comercial grade 20A duplex outlet.

it seem to me that a lot of $ is on the line here. an electrician could make a power distro, likely from parts laying around the ship, which would plug in to the 220 and have 2 20A outlets (or more). and do this for far less then you are probably going to pay for the Sun PDU.

Reply to
Tim Perry

Tim,

That (the "smoldering wonder in the wall") is the picture that I kept getting in my mind :)...

Re. the electrician, funny that you should mention that.

As I mentioned, the guy who started all of this originally put some dedicated (off of a UPS even) 30A @ 240V outlets in the wall (L6-30). The problem is we don't have any PDUs in-house that have the L6-30 connectors.

We do have some PDUs that have L5-20Ps on them, but when I asked about the possibility of having the electricians replace a couple of the L6-30 outlets to L5-20s, I got a resounding "no". I suspect (guess?) that if we had them change the outlets, they'd also have to change the wiring in the wall to meet code (i.e., so that the wiring matched the breaker and the receptacle?

Jim

Reply to
ohaya

| We do have some PDUs that have L5-20Ps on them, but when I asked about | the possibility of having the electricians replace a couple of the L6-30 | outlets to L5-20s, I got a resounding "no". I suspect (guess?) that if | we had them change the outlets, they'd also have to change the wiring in | the wall to meet code (i.e., so that the wiring matched the breaker and | the receptacle?

The wiring would not have to be changed. But the overcurrent protection device (the breaker) would. It's OK to have larger than required wire. But the L5-20's have to be protected at 20 amps. Then there is the possibility that this will require an electrical permit to be issued, inspection, etc, just because it is that much of a change. Then do it all over again to change it back. You might be better off rolling your own temporary PDU or "junction box". But you better be sure it is overcurrent protected at the rating of the smallest component.

Exactly what kind of plug does the server come with? L5-20? L6-20?

Reply to
phil-news-nospam

Reply to
Phil Munro

Hi,

Each of the servers has 3 power supplies, and a minimum of 2 of these have to be powered for the servers to work. The receptacles on the server power supplies are the 3-prong male type, similar to a PC (I think they're "IEC13"?).

As it turned out, we ended using a spare Dell PDUs, as you alluded to. This originally had a L5-20P connector on it, but we replaced that with an L6-30, and we're only going to use this arrangement temporarily, so I've told everyone that the single PDU should only be used for a single

880, with nothing else plugged into the PDU.

Seems to have worked fine. The Dell PDU splits the incoming 240V into 2 banks of outputs, which are also at 240V. Near as I can tell (from calculations) we'll only be drawing less than ~4 Amps (1755 watts/240V/2 power supplies) from each of the banks, so I think that we have good margin. The L6-30 outlets are coming off a UPS, and have 30 Amp breakers on them, so that should be fine also, I think.

This temporary arrangement will allow us to bring up one of the servers at-a-time, since we only converted one PDU, but that'll be ok until the final PDUs come in, since we'll be spending a day or so with each machine setting them up.

Thanks to all of you who replied! It was a real help, and I think that I was able to get us through this relatively safely.

Jim

Reply to
ohaya

Phil,

Believe me, I've been working with Sun, but they're kind of stuck themselves because they couldn't find a rack or PDU/PDS. As far as the equipment belonging to them, that (not causing any damage to their equipment) was one of my big concerns, especially because I ended up having to sign for it (> $200K worth), and that's why I've been trying to be very conservative in what I'm willing to do and not willing to do.

Thanks, Jim

Reply to
ohaya

It sounds like your PDU can handle 20A (or maybe only 16A continuous, that being 80% of 20A) if it came with a 20 A plug. Now you are feeding it with a 30A circuit and a 30A breaker. A 30A breaker doesn't properly protect a

20A PDU. A 20A breaker would. Can someone change the 30A breaker to 20A temporarily? In this case you don't have to worry about how much power your gear is consuming, the breaker will take care of that for you. And, no, I don't think there is anything to worry about in terms of a tripping breaker causing damage to the equipment.

Let me also suggest that electronic power supplies can draw a lot more Amps than you would get by calculating Watts divided by voltage, due to the distortion of the current. Your current could be 1.5* or higher than what you calculate. Conversely, there are also some servers coming out with pretty good power supplies, so maybe you're in luck.

j
Reply to
operator jay

| Let me also suggest that electronic power supplies can draw a lot more Amps | than you would get by calculating Watts divided by voltage, due to the | distortion of the current. Your current could be 1.5* or higher than what | you calculate. Conversely, there are also some servers coming out with | pretty good power supplies, so maybe you're in luck.

They average that out when they figure the watts. So an 1800 watt load at

120 volts would normally draw an RMS of 15 amps. But a switching power supply could draw a peak of say 45 amps for 1/3 the time, and nothing for other parts of the cycle. It would still average out to 1800 watts.

But where this can have an impact is in two areas. First, during the peak current, the I^2*R losses will be greater due to the higher current. So you'll get more heating of wires coming in. At triple the current you get nine times the heating for 1/3 the time, for an average of 3 times the heating. Then there is the overloaded neutral due to triplens on three phase circuits. Both effects can combine to make things worse.

Reply to
phil-news-nospam

I don't follow. If it draws 1800W, then it is pulling a lot more than 15A RMS. Fundamental must be 15A RMS. Harmonics probably equal 15A RMS.

Note to the O.P.: I let you off the hook for I^2*R because any extra heating going on in the circuit is also going on in the breaker's thermal element so a 20A breaker still protects. And the summing triplens you don't have to worry about because it's not a 3 phase load.

j
Reply to
operator jay

|> | Let me also suggest that electronic power supplies can draw a lot more | Amps |> | than you would get by calculating Watts divided by voltage, due to the |> | distortion of the current. Your current could be 1.5* or higher than | what |> | you calculate. Conversely, there are also some servers coming out with |> | pretty good power supplies, so maybe you're in luck. |>

|> They average that out when they figure the watts. So an 1800 watt load at |> 120 volts would normally draw an RMS of 15 amps. But a switching power |> supply could draw a peak of say 45 amps for 1/3 the time, and nothing |> for other parts of the cycle. It would still average out to 1800 watts. | | I don't follow. If it draws 1800W, then it is pulling a lot more than 15A | RMS. Fundamental must be 15A RMS. Harmonics probably equal 15A RMS.

You can also think about it in terms of the summation of harmonics. I would probably have to draw graphics to explain it further.

Reply to
phil-news-nospam

Well, there's the rub. I don't think your harmonic current components are delivering net power. I know the voltage and current waveform are always the same sign and it does seem that all the energy delivered should actually be consumed as net power. However I think a lot of the energy is flowing in and out as reactive power.

To deliver net power I believe you need a voltage component and a current component that are not orthogonal. Every harmonic current component is orthogonal with the fundamental voltage because they are a different frequency. In this case every harmonic voltage component is, in practice, going to be approximately zero. All this leaves is the fundamental current and the fundamental voltage to deliver some net power. The in-phase component of fundamental current will have to be ~15A RMS if 1800W is delivered.

This is all in keeping with the 'distortion power factor' calculation people sometimes use: dPF = (RMS fundamental current) / (RMS current) which, you can see, imlpicitly discounts the ability of harmonic current to deliver net power. I suppose it's only really correct when your voltage is pure fundamental, but it does pretty good. The classic 'golden arches' waveform of current going into a 6-pulse rectifier is lined up nicely with the voltage and has a displacement power factor of pretty close to one. Overall, the power factor is often 0.7 - 0.8. This is because of the distortion power factor.

j
Reply to
operator jay

| Well, there's the rub. I don't think your harmonic current components are | delivering net power. I know the voltage and current waveform are always | the same sign and it does seem that all the energy delivered should actually | be consumed as net power. However I think a lot of the energy is flowing in | and out as reactive power.

If the waveform is simply a chopped form of the normal sine wave, then at each point where current flows, it corresponds with a voltage of the same polarity. That's not returning power; it's a different kind of "reactive" than a phase shift.

What I suspect is happening is that each individual harmonic frequency is itself reactive, but that the summation of these harmonics is what is not reactive. In other words, all the harmonics cancel out when considered as power (multiplied by the voltage); they "react" against each other so the "sloshing" is really between harmonic frequencies. One frequency has one power factor at a given instant and another frequency has a different power factor.

Reply to
phil-news-nospam

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.